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Stick to \[ ... \], equation, equation*, align, or align* environments in math environments. +Failed to remove ans_at_end environment +Failed to remove sol_at_end environment +output in chapter11.ipynb diff --git a/doc/BookChapters/chapter11.do.txt b/doc/BookChapters/chapter11.do.txt new file mode 100644 index 000000000..487acc23b --- /dev/null +++ b/doc/BookChapters/chapter11.do.txt @@ -0,0 +1,2319 @@ +======= Solving Differential Equations with Deep Learning ======= + +The Universal Approximation Theorem states that a neural network can +approximate any function at a single hidden layer along with one input +and output layer to any given precision. + + +An ordinary differential equation (ODE) is an equation involving functions having one variable. + +In general, an ordinary differential equation looks like + +!bt +\begin{equation} \label{ode} +f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 +\end{equation} +!et + +where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$. + +The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in (ref{ode}). +The highest order of derivative, that is the value of $n$, determines to the order of the equation. +The equation is referred to as a $n$-th order ODE. +Along with (ref{ode}), some additional conditions of the function $g(x)$ are typically given +for the solution to be unique. + + + +Let the trial solution $g_t(x)$ be + +!bt +\begin{equation} + g_t(x) = h_1(x) + h_2(x,N(x,P)) +\end{equation} +!et + + +where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set +of conditions, $N(x,P)$ a neural network with weights and biases +described by $P$ and $h_2(x, N(x,P))$ some expression involving the +neural network. The role of the function $h_2(x, N(x,P))$, is to +ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is +evaluated at the values of $x$ where the given conditions must be +satisfied. The function $h_1(x)$ should alone make $g_t(x)$ satisfy +the conditions. + +But what about the network $N(x,P)$? + + +As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation. + + + +For the minimization to be defined, we need to have a cost function at hand to minimize. + +It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in (ref{ode}). +We can choose to consider the mean squared error as the cost function for an input $x$. +Since we are looking at one input, the cost function is just $f$ squared. +The cost function $c\left(x, P \right)$ can therefore be expressed as + +!bt +C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 +!et + +If $N$ inputs are given as a vector $\bm{x}$ with elements $x_i$ for $i = 1,\dots,N$, +the cost function becomes + +!bt +\begin{equation} \label{cost} + C\left(\bm{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 +\end{equation} +!et + +The neural net should then find the parameters $P$ that minimizes the cost function in +(ref{cost}) for a set of $N$ training samples $x_i$. + + + +To perform the minimization using gradient descent, the gradient of $C\left(\bm{x}, P\right)$ is needed. +It might happen so that finding an analytical expression of the gradient of $C(\bm{x}, P)$ from (ref{cost}) gets too messy, depending on which cost function one desires to use. + +Luckily, there exists libraries that makes the job for us through automatic differentiation. +Automatic differentiation is a method of finding the derivatives numerically with very high precision. + + +=== Example: Exponential decay === + +An exponential decay of a quantity $g(x)$ is described by the equation + +!bt +\begin{equation} \label{solve_expdec} + g'(x) = -\gamma g(x) +\end{equation} +!et + +with $g(0) = g_0$ for some chosen initial value $g_0$. + +The analytical solution of (ref{solve_expdec}) is + +!bt +\begin{equation} + g(x) = g_0 \exp\left(-\gamma x\right) +\end{equation} +!et + +Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (ref{solve_expdec}). + + + +The program will use a neural network to solve + +!bt +\begin{equation} \label{solveode} +g'(x) = -\gamma g(x) +\end{equation} +!et + +where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values. + +In this example, $\gamma = 2$ and $g_0 = 10$. + + +To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be + +!bt +g_t(x, P) = h_1(x) + h_2(x, N(x, P)) +!et + +with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer. + + + +In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}}, P_{\text{output}} \}$. +If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer. + +The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. +If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix. + +Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron. + +It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of (ref{solveode}). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution: + +!bt +\begin{equation} \label{trial} +g_t(x, P) = g_0 + x \cdot N(x, P) +\end{equation} +!et + +=== Reformulating the problem === + +We wish that our neural network manages to minimize a given cost function. + +A reformulation of out equation, (ref{solveode}), must therefore be done, +such that it describes the problem a neural network can solve for. + +The neural network must find the set of weights and biases $P$ such that the trial solution in (ref{trial}) satisfies (ref{solveode}). + +The trial solution + +!bt +g_t(x, P) = g_0 + x \cdot N(x, P) +!et + +has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that + +!bt +\begin{equation} \label{nnmin} +g_t'(x, P) = - \gamma g_t(x, P) +\end{equation} +!et + +is fulfilled as *best as possible*. + + +The left hand side and right hand side of (ref{nnmin}) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible. +This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. +In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network. + +This gives the following cost function our neural network must solve for: + +!bt +\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} +!et + +(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$) + +or, in terms of weights and biases for the hidden and output layer in our network: + +!bt +\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} +!et + +for an input value $x$. + + + +If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes + +!bt +\begin{equation} \label{min} +\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} +\end{equation} +!et + +Letting $\bm{x}$ be a vector with elements $x_i$ and $C(\bm{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes + +!bt +\min_{P} C(\bm{x}, P) +!et + +In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as + +$$ +\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\bm{x}, \{P_{\text{hidden} }, P_{\text{output} }\}) +$$ + + +For simplicity, it is assumed that the input is an array $\bm{x} = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills (ref{min}). + +First, the neural network must feed forward the inputs. +This means that $\bm{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. +The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\text{hidden} }$. + + +For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is: + +!bt +\begin{aligned} +z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ +&= +\begin{pmatrix} +b_i^{\text{hidden}} & w_i^{\text{hidden}} +\end{pmatrix} +\begin{pmatrix} +1 \\ +x_j +\end{pmatrix} +\end{aligned} +!et + + + +The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector: + +!bt +\begin{aligned} +\bm{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ +&= +\begin{pmatrix} + b_i^{\text{hidden}} & w_i^{\text{hidden}} +\end{pmatrix} +\begin{pmatrix} +1 & 1 & \dots & 1 \\ +x_1 & x_2 & \dots & x_N +\end{pmatrix} \\ +&= \bm{p}_{i, \text{hidden}}^T X +\end{aligned} +!et + + +The vector $\bm{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to (ref{min}). + +After having found $\bm{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\bm{z})$. + +In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: + +!bt +f(z) = \frac{1}{1 + \exp{(-z)}} +!et + +It is possible to use other activations functions for the hidden layer also. + +The output $\bm{x}_i^{\text{hidden}}$ from each $i$-th hidden neuron is: + +$$ +\bm{x}_i^{\text{hidden} } = f\big( \bm{z}_{i}^{\text{hidden}} \big) +$$ + +The outputs $\bm{x}_i^{\text{hidden} } $ are then sent to the output layer. + +The output layer consists of one neuron in this case, and combines the +output from each of the neurons in the hidden layers. The output layer +combines the results from the hidden layer using some weights $w_i^{\text{output}}$ +and biases $b_i^{\text{output}}$. In this case, +it is assumes that the number of neurons in the output layer is one. + + + +The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously. + +!bt +\begin{aligned} +z_{1,j}^{\text{output}} & = +\begin{pmatrix} +b_1^{\text{output}} & \bm{w}_1^{\text{output}} +\end{pmatrix} +\begin{pmatrix} +1 \\ +\bm{x}_j^{\text{hidden}} +\end{pmatrix} +\end{aligned} +!et + + +Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer: + +!bt +\bm{z}_{1}^{\text{output}} = +\begin{pmatrix} +b_1^{\text{output}} & \bm{w}_1^{\text{output}} +\end{pmatrix} +\begin{pmatrix} +1 & 1 & \dots & 1 \\ +\bm{x}_1^{\text{hidden}} & \bm{x}_2^{\text{hidden}} & \dots & \bm{x}_N^{\text{hidden}} +\end{pmatrix} +!et + +In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\bm{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\bm{z}_{1}^{\text{output}}$ is the final output of the network. + + +The next step is to decide how the parameters should be changed such that they minimize the cost function. + +The chosen cost function for this problem is + +!bt +C(\bm{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 +!et + +In order to minimize the cost function, an optimization method must be chosen. + +Here, gradient descent with a constant step size has been chosen. + +=== Gradient descent === + +The idea of the gradient descent algorithm is to update parameters in +a direction where the cost function decreases goes to a minimum. + +In general, the update of some parameters $\bm{\omega}$ given a cost +function defined by some weights $\bm{\omega}$, $C(\bm{x}, +\bm{\omega})$, goes as follows: + +!bt +\bm{\omega}_{\text{new} } = \bm{\omega} - \lambda \nabla_{\bm{\omega}} C(\bm{x}, \bm{\omega}) +!et + +for a number of iterations or until $ \big|\big| \bm{\omega}_{\text{new} } - \bm{\omega} \big|\big|$ becomes smaller than some given tolerance. + +The value of $\lambda$ decides how large steps the algorithm must take +in the direction of $ \nabla_{\bm{\omega}} C(\bm{x}, \bm{\omega})$. +The notation $\nabla_{\bm{\omega}}$ express the gradient with respect +to the elements in $\bm{\omega}$. + +In our case, we have to minimize the cost function $C(\bm{x}, P)$ with +respect to the two sets of weights and biases, that is for the hidden +layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output} +}$ . + +This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by + +!bt +\begin{aligned} +P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\bm{x}, P) \\ +P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\bm{x}, P) +\end{aligned} +!et + +=== The code for solving the ODE === + +!bc pycod +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +# Assuming one input, hidden, and output layer +def neural_network(params, x): + + # Find the weights (including and biases) for the hidden and output layer. + # Assume that params is a list of parameters for each layer. + # The biases are the first element for each array in params, + # and the weights are the remaning elements in each array in params. + + w_hidden = params[0] + w_output = params[1] + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + ## Hidden layer: + + # Add a row of ones to include bias + x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_input) + x_hidden = sigmoid(z_hidden) + + ## Output layer: + + # Include bias: + x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) + + z_output = np.matmul(w_output, x_hidden) + x_output = z_output + + return x_output + +# The trial solution using the deep neural network: +def g_trial(x,params, g0 = 10): + return g0 + x*neural_network(params,x) + +# The right side of the ODE: +def g(x, g_trial, gamma = 2): + return -gamma*g_trial + +# The cost function: +def cost_function(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial(x,P) + + # Find the derivative w.r.t x of the neural network + d_net_out = elementwise_grad(neural_network,1)(P,x) + + # Find the derivative w.r.t x of the trial function + d_g_t = elementwise_grad(g_trial,0)(x,P) + + # The right side of the ODE + func = g(x, g_t) + + err_sqr = (d_g_t - func)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum / np.size(err_sqr) + +# Solve the exponential decay ODE using neural network with one input, hidden, and output layer +def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): + ## Set up initial weights and biases + + # For the hidden layer + p0 = npr.randn(num_neurons_hidden, 2 ) + + # For the output layer + p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included + + P = [p0, p1] + + print('Initial cost: %g'%cost_function(P, x)) + + ## Start finding the optimal weights using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_grad = grad(cost_function,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of two arrays; + # one for the gradient w.r.t P_hidden and + # one for the gradient w.r.t P_output + cost_grad = cost_function_grad(P, x) + + P[0] = P[0] - lmb * cost_grad[0] + P[1] = P[1] - lmb * cost_grad[1] + + print('Final cost: %g'%cost_function(P, x)) + + return P + +def g_analytic(x, gamma = 2, g0 = 10): + return g0*np.exp(-gamma*x) + +# Solve the given problem +if __name__ == '__main__': + # Set seed such that the weight are initialized + # with same weights and biases for every run. + npr.seed(15) + + ## Decide the vales of arguments to the function to solve + N = 10 + x = np.linspace(0, 1, N) + + ## Set up the initial parameters + num_hidden_neurons = 10 + num_iter = 10000 + lmb = 0.001 + + # Use the network + P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) + + # Print the deviation from the trial solution and true solution + res = g_trial(x,P) + res_analytical = g_analytic(x) + + print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) + + # Plot the results + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(x, res_analytical) + plt.plot(x, res[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('x') + plt.ylabel('g(x)') + plt.show() +!ec + + +===== The network with one input layer, specified number of hidden layers, and one output layer ===== + +It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers. + +The number of neurons within each hidden layer are given as a list of integers in the program below. + +!bc pycod +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +# The neural network with one input layer and one output layer, +# but with number of hidden layers specified by the user. +def deep_neural_network(deep_params, x): + # N_hidden is the number of hidden layers + + N_hidden = np.size(deep_params) - 1 # -1 since params consists of + # parameters to all the hidden + # layers AND the output layer. + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + # Due to multiple hidden layers, define a variable referencing to the + # output of the previous layer: + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output + +# The trial solution using the deep neural network: +def g_trial_deep(x,params, g0 = 10): + return g0 + x*deep_neural_network(params, x) + +# The right side of the ODE: +def g(x, g_trial, gamma = 2): + return -gamma*g_trial + +# The same cost function as before, but calls deep_neural_network instead. +def cost_function_deep(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial_deep(x,P) + + # Find the derivative w.r.t x of the neural network + d_net_out = elementwise_grad(deep_neural_network,1)(P,x) + + # Find the derivative w.r.t x of the trial function + d_g_t = elementwise_grad(g_trial_deep,0)(x,P) + + # The right side of the ODE + func = g(x, g_t) + + err_sqr = (d_g_t - func)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum / np.size(err_sqr) + +# Solve the exponential decay ODE using neural network with one input and one output layer, +# but with specified number of hidden layers from the user. +def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): + # num_hidden_neurons is now a list of number of neurons within each hidden layer + + # The number of elements in the list num_hidden_neurons thus represents + # the number of hidden layers. + + # Find the number of hidden layers: + N_hidden = np.size(num_neurons) + + ## Set up initial weights and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 ) + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: %g'%cost_function_deep(P, x)) + + ## Start finding the optimal weights using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_deep_grad = grad(cost_function_deep,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases + # in the hidden layers and output layers evaluated at x. + cost_deep_grad = cost_function_deep_grad(P, x) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_deep_grad[l] + + print('Final cost: %g'%cost_function_deep(P, x)) + + return P + +def g_analytic(x, gamma = 2, g0 = 10): + return g0*np.exp(-gamma*x) + +# Solve the given problem +if __name__ == '__main__': + npr.seed(15) + + ## Decide the vales of arguments to the function to solve + N = 10 + x = np.linspace(0, 1, N) + + ## Set up the initial parameters + num_hidden_neurons = np.array([10,10]) + num_iter = 10000 + lmb = 0.001 + + P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) + + res = g_trial_deep(x,P) + res_analytical = g_analytic(x) + + plt.figure(figsize=(10,10)) + + plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') + plt.plot(x, res_analytical) + plt.plot(x, res[0,:]) + plt.legend(['analytical','dnn']) + plt.ylabel('g(x)') + plt.show() +!ec + + + +=== Example: Population growth === + +A logistic model of population growth assumes that a population converges toward an equilibrium. +The population growth can be modeled by + +!bt +\begin{equation} \label{log} + g'(t) = \alpha g(t)(A - g(t)) +\end{equation} +!et + +where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment. +Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant. + +In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability +and high execution time (this might be more apparent in the examples solving PDEs), +using a library like TensorFlow is recommended. +Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method. + + + +Here, we will model a population $g(t)$ in an environment having carrying capacity $A$. +The population follows the model + +!bt +\begin{equation} \label{solveode_population} +g'(t) = \alpha g(t)(A - g(t)) +\end{equation} +!et + +where $g(0) = g_0$. + +In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$. + + +We will get a slightly different trial solution, as the boundary conditions are different +compared to the case for exponential decay. + +A possible trial solution satisfying the condition $g(0) = g_0$ could be + +$$ +h_1(t) = g_0 + t \cdot N(t,P) +$$ + +with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$. + +The analytical solution is + +$$ +g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} +$$ + + + +The network will be the similar as for the exponential decay example, but with some small modifications for our problem. + +!bc pycod +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +# Function to get the parameters. +# Done such that one can easily change the paramaters after one's liking. +def get_parameters(): + alpha = 2 + A = 1 + g0 = 1.2 + return alpha, A, g0 + +def deep_neural_network(P, x): + # N_hidden is the number of hidden layers + N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + # Due to multiple hidden layers, define a variable referencing to the + # output of the previous layer: + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = P[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = P[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output + + +def cost_function_deep(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial_deep(x,P) + + # Find the derivative w.r.t x of the trial function + d_g_t = elementwise_grad(g_trial_deep,0)(x,P) + + # The right side of the ODE + func = f(x, g_t) + + err_sqr = (d_g_t - func)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum / np.size(err_sqr) + +# The right side of the ODE: +def f(x, g_trial): + alpha,A, g0 = get_parameters() + return alpha*g_trial*(A - g_trial) + +# The trial solution using the deep neural network: +def g_trial_deep(x, params): + alpha,A, g0 = get_parameters() + return g0 + x*deep_neural_network(params,x) + +# The analytical solution: +def g_analytic(t): + alpha,A, g0 = get_parameters() + return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) + +def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): + # num_hidden_neurons is now a list of number of neurons within each hidden layer + + # Find the number of hidden layers: + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 ) + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: %g'%cost_function_deep(P, x)) + + ## Start finding the optimal weigths using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_deep_grad = grad(cost_function_deep,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases + # in the hidden layers and output layers evaluated at x. + cost_deep_grad = cost_function_deep_grad(P, x) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_deep_grad[l] + + print('Final cost: %g'%cost_function_deep(P, x)) + + return P + +if __name__ == '__main__': + npr.seed(4155) + + ## Decide the vales of arguments to the function to solve + Nt = 10 + T = 1 + t = np.linspace(0,T, Nt) + + ## Set up the initial parameters + num_hidden_neurons = [100, 50, 25] + num_iter = 1000 + lmb = 1e-3 + + P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) + + g_dnn_ag = g_trial_deep(t,P) + g_analytical = g_analytic(t) + + # Find the maximum absolute difference between the solutons: + diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) + print("The max absolute difference between the solutions is: %g"%diff_ag) + + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(t, g_analytical) + plt.plot(t, g_dnn_ag[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('t') + plt.ylabel('g(t)') + + plt.show() +!ec + +===== Using forward Euler to solve the ODE ===== + +A straightforward way of solving an ODE numerically, is to use Euler's method. + +Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$: + +$$ +f(x + \Delta x) \approx f(x) + \Delta x f'(x) +$$ + +In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields + +!bt +\begin{aligned} + g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ + &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) +\end{aligned} +!et +along with the condition that $g(0) = g_0$. + +Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$. + +For $i \geq 1$, we have that +!bt +\begin{aligned} +t_i &= i\Delta t \\ +&= (i - 1)\Delta t + \Delta t \\ +&= t_{i-1} + \Delta t +\end{aligned} +!et + +Now, if $g_i = g(t_i)$ then + +!bt +\begin{equation} + \begin{aligned} + g_i &= g(t_i) \\ + &= g(t_{i-1} + \Delta t) \\ + &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ + &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) + \end{aligned} +\end{equation} \label{odenum} +!et +for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$. + +Equation (ref{odenum}) could be implemented in the following way, +extending the program that uses the network using Autograd: + +!bc pycod +# Assume that all function definitions from the example program using Autograd +# are located here. + +if __name__ == '__main__': + npr.seed(4155) + + ## Decide the vales of arguments to the function to solve + Nt = 10 + T = 1 + t = np.linspace(0,T, Nt) + + ## Set up the initial parameters + num_hidden_neurons = [100,50,25] + num_iter = 1000 + lmb = 1e-3 + + P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) + + g_dnn_ag = g_trial_deep(t,P) + g_analytical = g_analytic(t) + + # Find the maximum absolute difference between the solutons: + diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) + print("The max absolute difference between the solutions is: %g"%diff_ag) + + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(t, g_analytical) + plt.plot(t, g_dnn_ag[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('t') + plt.ylabel('g(t)') + + ## Find an approximation to the funtion using forward Euler + + alpha, A, g0 = get_parameters() + dt = T/(Nt - 1) + + # Perform forward Euler to solve the ODE + g_euler = np.zeros(Nt) + g_euler[0] = g0 + + for i in range(1,Nt): + g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1])) + + # Print the errors done by each method + diff1 = np.max(np.abs(g_euler - g_analytical)) + diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical)) + + print('Max absolute difference between Euler method and analytical: %g'%diff1) + print('Max absolute difference between deep neural network and analytical: %g'%diff2) + + # Plot results + plt.figure(figsize=(10,10)) + + plt.plot(t,g_euler) + plt.plot(t,g_analytical) + plt.plot(t,g_dnn_ag[0,:]) + + plt.legend(['euler','analytical','dnn']) + plt.xlabel('Time t') + plt.ylabel('g(t)') + + plt.show() +!ec + + + + +===== Solving the one dimensional Poisson equation ===== + +The Poisson equation for $g(x)$ in one dimension is + +!bt +\begin{equation} \label{poisson} + -g''(x) = f(x) +\end{equation} +!et + +where $f(x)$ is a given function for $x \in (0,1)$. + +The conditions that $g(x)$ is chosen to fulfill, are +!bt +\begin{align*} + g(0) &= 0 \\ + g(1) &= 0 +\end{align*} +!et + +This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. +The results from the networks can then be compared to the analytical solution. +In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks. + + +Here, the function $g(x)$ to solve for follows the equation + +!bt +-g''(x) = f(x),\qquad x \in (0,1) +!et + +where $f(x)$ is a given function, along with the chosen conditions + +!bt +\begin{aligned} +g(0) = g(1) = 0 +\end{aligned}\label{cond} +!et + +In this example, we consider the case when $f(x) = (3x + x^2)\exp(x)$. + +For this case, a possible trial solution satisfying the conditions could be + +!bt +g_t(x) = x \cdot (1-x) \cdot N(P,x) +!et + +The analytical solution for this problem is + +!bt +g(x) = x(1 - x)\exp(x) +!et + + +!bc pycod +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + # Due to multiple hidden layers, define a variable referencing to the + # output of the previous layer: + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output + +def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): + # num_hidden_neurons is now a list of number of neurons within each hidden layer + + # Find the number of hidden layers: + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 ) + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: %g'%cost_function_deep(P, x)) + + ## Start finding the optimal weigths using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_deep_grad = grad(cost_function_deep,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases + # in the hidden layers and output layers evaluated at x. + cost_deep_grad = cost_function_deep_grad(P, x) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_deep_grad[l] + + print('Final cost: %g'%cost_function_deep(P, x)) + + return P + +## Set up the cost function specified for this Poisson equation: + +# The right side of the ODE +def f(x): + return (3*x + x**2)*np.exp(x) + +def cost_function_deep(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial_deep(x,P) + + # Find the derivative w.r.t x of the trial function + d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) + + right_side = f(x) + + err_sqr = (-d2_g_t - right_side)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum/np.size(err_sqr) + +# The trial solution: +def g_trial_deep(x,P): + return x*(1-x)*deep_neural_network(P,x) + +# The analytic solution; +def g_analytic(x): + return x*(1-x)*np.exp(x) + +if __name__ == '__main__': + npr.seed(4155) + + ## Decide the vales of arguments to the function to solve + Nx = 10 + x = np.linspace(0,1, Nx) + + ## Set up the initial parameters + num_hidden_neurons = [200,100] + num_iter = 1000 + lmb = 1e-3 + + P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) + + g_dnn_ag = g_trial_deep(x,P) + g_analytical = g_analytic(x) + + # Find the maximum absolute difference between the solutons: + max_diff = np.max(np.abs(g_dnn_ag - g_analytical)) + print("The max absolute difference between the solutions is: %g"%max_diff) + + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(x, g_analytical) + plt.plot(x, g_dnn_ag[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('x') + plt.ylabel('g(x)') + plt.show() +!ec + +=== Comparing with a numerical scheme === + +The Poisson equation is possible to solve using Taylor series to approximate the second derivative. + +Using Taylor series, the second derivative can be expressed as + +$$ +g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) +$$ + +where $\Delta x$ is a small step size and $E_{\Delta x}(x)$ being the error term. + +Looking away from the error terms gives an approximation to the second derivative: + +!bt +\begin{equation} \label{approx} +g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} +\end{equation} +!et + +If $x_i = i \Delta x = x_{i-1} + \Delta x$ and $g_i = g(x_i)$ for $i = 1,\dots N_x - 2$ with $N_x$ being the number of values for $x$, (ref{approx}) becomes + +!bt +\begin{aligned} +g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ +&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} +\end{aligned} +!et + +Since we know from our problem that + +!bt +\begin{aligned} +-g''(x) &= f(x) \\ +&= (3x + x^2)\exp(x) +\end{aligned} +!et + +along with the conditions $g(0) = g(1) = 0$, +the following scheme can be used to find an approximate solution for $g(x)$ numerically: + +!bt +\begin{equation} + \begin{aligned} + -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ + -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) + \end{aligned} +\end{equation} \label{odesys} +!et + +for $i = 1, \dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\exp(x_i)$, which is given for our specific problem. + +The equation can be rewritten into a matrix equation: + +!bt +\begin{aligned} +\begin{pmatrix} +2 & -1 & 0 & \dots & 0 \\ +-1 & 2 & -1 & \dots & 0 \\ +\vdots & & \ddots & & \vdots \\ +0 & \dots & -1 & 2 & -1 \\ +0 & \dots & 0 & -1 & 2\\ +\end{pmatrix} +\begin{pmatrix} +g_1 \\ +g_2 \\ +\vdots \\ +g_{N_x - 3} \\ +g_{N_x - 2} +\end{pmatrix} +&= +\Delta x^2 +\begin{pmatrix} +f(x_1) \\ +f(x_2) \\ +\vdots \\ +f(x_{N_x - 3}) \\ +f(x_{N_x - 2}) +\end{pmatrix} \\ +\bm{A}\bm{g} &= \bm{f}, +\end{aligned} +!et + +which makes it possible to solve for the vector $\bm{g}$. + + +We can then compare the result from this numerical scheme with the output from our network using Autograd: + +!bc pycod +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + # Due to multiple hidden layers, define a variable referencing to the + # output of the previous layer: + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output + +def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): + # num_hidden_neurons is now a list of number of neurons within each hidden layer + + # Find the number of hidden layers: + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 ) + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: %g'%cost_function_deep(P, x)) + + ## Start finding the optimal weigths using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_deep_grad = grad(cost_function_deep,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases + # in the hidden layers and output layers evaluated at x. + cost_deep_grad = cost_function_deep_grad(P, x) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_deep_grad[l] + + print('Final cost: %g'%cost_function_deep(P, x)) + + return P + +## Set up the cost function specified for this Poisson equation: + +# The right side of the ODE +def f(x): + return (3*x + x**2)*np.exp(x) + +def cost_function_deep(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial_deep(x,P) + + # Find the derivative w.r.t x of the trial function + d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) + + right_side = f(x) + + err_sqr = (-d2_g_t - right_side)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum/np.size(err_sqr) + +# The trial solution: +def g_trial_deep(x,P): + return x*(1-x)*deep_neural_network(P,x) + +# The analytic solution; +def g_analytic(x): + return x*(1-x)*np.exp(x) + +if __name__ == '__main__': + npr.seed(4155) + + ## Decide the vales of arguments to the function to solve + Nx = 10 + x = np.linspace(0,1, Nx) + + ## Set up the initial parameters + num_hidden_neurons = [200,100] + num_iter = 1000 + lmb = 1e-3 + + P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) + + g_dnn_ag = g_trial_deep(x,P) + g_analytical = g_analytic(x) + + # Find the maximum absolute difference between the solutons: + + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(x, g_analytical) + plt.plot(x, g_dnn_ag[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('x') + plt.ylabel('g(x)') + + ## Perform the computation using the numerical scheme + + dx = 1/(Nx - 1) + + # Set up the matrix A + A = np.zeros((Nx-2,Nx-2)) + + A[0,0] = 2 + A[0,1] = -1 + + for i in range(1,Nx-3): + A[i,i-1] = -1 + A[i,i] = 2 + A[i,i+1] = -1 + + A[Nx - 3, Nx - 4] = -1 + A[Nx - 3, Nx - 3] = 2 + + # Set up the vector f + f_vec = dx**2 * f(x[1:-1]) + + # Solve the equation + g_res = np.linalg.solve(A,f_vec) + + g_vec = np.zeros(Nx) + g_vec[1:-1] = g_res + + # Print the differences between each method + max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical)) + max_diff2 = np.max(np.abs(g_vec - g_analytical)) + print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1) + print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2) + + # Plot the results + plt.figure(figsize=(10,10)) + + plt.plot(x,g_vec) + plt.plot(x,g_analytical) + plt.plot(x,g_dnn_ag[0,:]) + + plt.legend(['numerical scheme','analytical','dnn']) + plt.show() + +!ec + + +===== Partial Differential Equations ===== + +A partial differential equation (PDE) has a solution here the function +is defined by multiple variables. The equation may involve all kinds +of combinations of which variables the function is differentiated with +respect to. + +In general, a partial differential equation for a function $g(x_1,\dots,x_N)$ with $N$ variables may be expressed as + +!bt +\begin{equation} \label{PDE} + f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 +\end{equation} +!et + +where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given. + +=== Type of problem === + +The problem our network must solve for, is similar to the ODE case. +We must have a trial solution $g_t$ at hand. + +For instance, the trial solution could be expressed as +!bt +\begin{align*} + g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) +\end{align*} +!et +where $h_1(x_1,\dots,x_N)$ is a function that ensures $g_t(x_1,\dots,x_N)$ satisfies some given conditions. +The neural network $N(x_1,\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$ is an expression using the output from the neural network in some way. + +The role of the function $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$, is to ensure that the output of $N(x_1,\dots,x_N,P)$ is zero when $g_t(x_1,\dots,x_N)$ is evaluated at the values of $x_1,\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\dots,x_N)$ should alone make $g_t(x_1,\dots,x_N)$ satisfy the conditions. + + + +=== Network requirements === + +The network tries then the minimize the cost function following the +same ideas as described for the ODE case, but now with more than one +variables to consider. The concept still remains the same; find a set +of parameters $P$ such that the expression $f$ in (ref{PDE}) is as +close to zero as possible. + +As for the ODE case, the cost function is the mean squared error that +the network must try to minimize. The cost function for the network to +minimize is + +!bt +\begin{equation*} +C\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 +\end{equation*} +!et + + +If we let $\bm{x} = \big( x_1, \dots, x_N \big)$ be an array containing the values for $x_1, \dots, x_N$ respectively, the cost function can be reformulated into the following: +!bt +\[ + C\left(\bm{x}, P\right) = f\left( \left( \bm{x}, \frac{\partial g(\bm{x}) }{\partial x_1}, \dots , \frac{\partial g(\bm{x}) }{\partial x_N}, \frac{\partial g(\bm{x}) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\bm{x}) }{\partial x_N^n} \right) \right)^2 +\] +!et + +If we also have $M$ different sets of values for $x_1, \dots, x_N$, that is $\bm{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)$ for $i = 1,\dots,M$ being the rows in matrix $X$, the cost function can be generalized into +!bt +\begin{equation*} +C\left(X, P \right) = \sum_{i=1}^M f\left( \left( \bm{x}_i, \frac{\partial g(\bm{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\bm{x}_i) }{\partial x_N}, \frac{\partial g(\bm{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\bm{x}_i) }{\partial x_N^n} \right) \right)^2. +\end{equation*} +!et + + +===== Example: The diffusion equation ===== + +In one spatial dimension, the equation reads +!bt +\begin{equation*} + \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation*} +!et + +where a possible choice of conditions are +!bt +\begin{align*} +g(0,t) &= 0 ,\qquad t \geq 0 \\ +g(1,t) &= 0, \qquad t \geq 0 \\ +g(x,0) &= u(x),\qquad x\in [0,1] +\end{align*} +!et +with $u(x)$ being some given function. + + + +For this case, we want to find $g(x,t)$ such that + +!bt +\begin{equation} + \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation} \label{diffonedim} +!et + +and + +!bt +\begin{align*} +g(0,t) &= 0 ,\qquad t \geq 0 \\ +g(1,t) &= 0, \qquad t \geq 0 \\ +g(x,0) &= u(x),\qquad x\in [0,1] +\end{align*} +!et +with $u(x) = \sin(\pi x)$. + +First, let us set up the deep neural network. +The deep neural network will follow the same structure as discussed in the examples solving the ODEs. +First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions. + + + + +The only change to do here, is to extend our network such that +functions of multiple parameters are correctly handled. In this case +we have two variables in our function to solve for, that is time $t$ +and position $x$. The variables will be represented by a +one-dimensional array in the program. The program will evaluate the +network at each possible pair $(x,t)$, given an array for the desired +$x$-values and $t$-values to approximate the solution at. + +!bc pycod +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # x is now a point and a 1D numpy array; make it a column vector + num_coordinates = np.size(x,0) + x = x.reshape(num_coordinates,-1) + + num_points = np.size(x,1) + + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assume that the input layer does nothing to the input x + x_input = x + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output[0][0] +!ec + + +The cost function must then iterate through the given arrays +containing values for $x$ and $t$, defines a point $(x,t)$ the deep +neural network and the trial solution is evaluated at, and then finds +the Jacobian of the trial solution. + +A possible trial solution for this PDE is + +$$ +g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) +$$ + +with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$. + +To fulfill the conditions, $A(x,t)$ could be: + +$$ +h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) +$$ +since $(0) = u(1) = 0$ and $u(x) = \sin(\pi x)$. + + + +The Jacobian is used because the program must find the derivative of +the trial solution with respect to $x$ and $t$. + +This gives the necessity of computing the Jacobian matrix, as we want +to evaluate the gradient with respect to $x$ and $t$ (note that the +Jacobian of a scalar-valued multivariate function is simply its +gradient). + +In Autograd, the differentiation is by default done with respect to +the first input argument of your Python function. Since the points is +an array representing $x$ and $t$, the Jacobian is calculated using +the values of $x$ and $t$. + +To find the second derivative with respect to $x$ and $t$, the +Jacobian can be found for the second time. The result is a Hessian +matrix, which is the matrix containing all the possible second order +mixed derivatives of $g(x,t)$. + +!bc pycod +# Set up the trial function: +def u(x): + return np.sin(np.pi*x) + +def g_trial(point,P): + x,t = point + return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) + +# The right side of the ODE: +def f(point): + return 0. + +# The cost function: +def cost_function(P, x, t): + cost_sum = 0 + + g_t_jacobian_func = jacobian(g_trial) + g_t_hessian_func = hessian(g_trial) + + for x_ in x: + for t_ in t: + point = np.array([x_,t_]) + + g_t = g_trial(point,P) + g_t_jacobian = g_t_jacobian_func(point,P) + g_t_hessian = g_t_hessian_func(point,P) + + g_t_dt = g_t_jacobian[1] + g_t_d2x = g_t_hessian[0][0] + + func = f(point) + + err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 + cost_sum += err_sqr + + return cost_sum +!ec + + +=== Setting up the network using Autograd; The full program === + +Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution. + +The analytical solution of our problem is + +$$ +g(x,t) = \exp(-\pi^2 t)\sin(\pi x) +$$ + +A possible way to implement a neural network solving the PDE, is given below. +Be aware, though, that it is fairly slow for the parameters used. +A better result is possible, but requires more iterations, and thus longer time to complete. + + +Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. +Using TensorFlow results in a much better execution time. Try it! + +!bc pycod +import autograd.numpy as np +from autograd import jacobian,hessian,grad +import autograd.numpy.random as npr +from matplotlib import cm +from matplotlib import pyplot as plt +from mpl_toolkits.mplot3d import axes3d + +## Set up the network + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # x is now a point and a 1D numpy array; make it a column vector + num_coordinates = np.size(x,0) + x = x.reshape(num_coordinates,-1) + + num_points = np.size(x,1) + + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assume that the input layer does nothing to the input x + x_input = x + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output[0][0] + +## Define the trial solution and cost function +def u(x): + return np.sin(np.pi*x) + +def g_trial(point,P): + x,t = point + return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) + +# The right side of the ODE: +def f(point): + return 0. + +# The cost function: +def cost_function(P, x, t): + cost_sum = 0 + + g_t_jacobian_func = jacobian(g_trial) + g_t_hessian_func = hessian(g_trial) + + for x_ in x: + for t_ in t: + point = np.array([x_,t_]) + + g_t = g_trial(point,P) + g_t_jacobian = g_t_jacobian_func(point,P) + g_t_hessian = g_t_hessian_func(point,P) + + g_t_dt = g_t_jacobian[1] + g_t_d2x = g_t_hessian[0][0] + + func = f(point) + + err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 + cost_sum += err_sqr + + return cost_sum /( np.size(x)*np.size(t) ) + +## For comparison, define the analytical solution +def g_analytic(point): + x,t = point + return np.exp(-np.pi**2*t)*np.sin(np.pi*x) + +## Set up a function for training the network to solve for the equation +def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): + ## Set up initial weigths and biases + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: ',cost_function(P, x, t)) + + cost_function_grad = grad(cost_function,0) + + # Let the update be done num_iter times + for i in range(num_iter): + cost_grad = cost_function_grad(P, x , t) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_grad[l] + + print('Final cost: ',cost_function(P, x, t)) + + return P + +if __name__ == '__main__': + ### Use the neural network: + npr.seed(15) + + ## Decide the vales of arguments to the function to solve + Nx = 10; Nt = 10 + x = np.linspace(0, 1, Nx) + t = np.linspace(0,1,Nt) + + ## Set up the parameters for the network + num_hidden_neurons = [100, 25] + num_iter = 250 + lmb = 0.01 + + P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) + + ## Store the results + g_dnn_ag = np.zeros((Nx, Nt)) + G_analytical = np.zeros((Nx, Nt)) + for i,x_ in enumerate(x): + for j, t_ in enumerate(t): + point = np.array([x_, t_]) + g_dnn_ag[i,j] = g_trial(point,P) + + G_analytical[i,j] = g_analytic(point) + + # Find the map difference between the analytical and the computed solution + diff_ag = np.abs(g_dnn_ag - G_analytical) + print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag)) + + ## Plot the solutions in two dimensions, that being in position and time + + T,X = np.meshgrid(t,x) + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) + s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Analytical solution') + s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Difference') + s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + ## Take some slices of the 3D plots just to see the solutions at particular times + indx1 = 0 + indx2 = int(Nt/2) + indx3 = Nt-1 + + t1 = t[indx1] + t2 = t[indx2] + t3 = t[indx3] + + # Slice the results from the DNN + res1 = g_dnn_ag[:,indx1] + res2 = g_dnn_ag[:,indx2] + res3 = g_dnn_ag[:,indx3] + + # Slice the analytical results + res_analytical1 = G_analytical[:,indx1] + res_analytical2 = G_analytical[:,indx2] + res_analytical3 = G_analytical[:,indx3] + + # Plot the slices + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t1) + plt.plot(x, res1) + plt.plot(x,res_analytical1) + plt.legend(['dnn','analytical']) + + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t2) + plt.plot(x, res2) + plt.plot(x,res_analytical2) + plt.legend(['dnn','analytical']) + + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t3) + plt.plot(x, res3) + plt.plot(x,res_analytical3) + plt.legend(['dnn','analytical']) + + plt.show() +!ec + + +===== Solving the wave equation with Neural Networks ===== + +The wave equation is +!bt +\begin{equation*} + \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation*} +!et + +with $c$ being the specified wave speed. + +Here, the chosen conditions are +!bt +\begin{align*} + g(0,t) &= 0 \\ + g(1,t) &= 0 \\ + g(x,0) &= u(x) \\ + \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) +\end{align*} +!et +where $\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions. + + +The wave equation to solve for, is + +!bt +\begin{equation} \label{wave} +\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation} +!et + +where $c$ is the given wave speed. +The chosen conditions for this equation are + +!bt +\begin{aligned} +g(0,t) &= 0, &t \geq 0 \\ +g(1,t) &= 0, &t \geq 0 \\ +g(x,0) &= u(x), &x\in[0,1] \\ +\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] +\end{aligned} \label{condwave} +!et + +In this example, let $c = 1$ and $u(x) = \sin(\pi x)$ and $v(x) = -\pi\sin(\pi x)$. + + + +Setting up the network is done in similar matter as for the example of solving the diffusion equation. +The only things we have to change, is the trial solution such that it satisfies the conditions from (ref{condwave}) and the cost function. + +The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is + +$$ +g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) +$$ + +where + +$$ +h_1(x,t) = (1-t^2)u(x) + tv(x) +$$ + +Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example. + + +The analytical solution for our specific problem, is + +$$ +g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) +$$ + + + +!bc pycod +import autograd.numpy as np +from autograd import hessian,grad +import autograd.numpy.random as npr +from matplotlib import cm +from matplotlib import pyplot as plt +from mpl_toolkits.mplot3d import axes3d + +## Set up the trial function: +def u(x): + return np.sin(np.pi*x) + +def v(x): + return -np.pi*np.sin(np.pi*x) + +def h1(point): + x,t = point + return (1 - t**2)*u(x) + t*v(x) + +def g_trial(point,P): + x,t = point + return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point) + +## Define the cost function +def cost_function(P, x, t): + cost_sum = 0 + + g_t_hessian_func = hessian(g_trial) + + for x_ in x: + for t_ in t: + point = np.array([x_,t_]) + + g_t_hessian = g_t_hessian_func(point,P) + + g_t_d2x = g_t_hessian[0][0] + g_t_d2t = g_t_hessian[1][1] + + err_sqr = ( (g_t_d2t - g_t_d2x) )**2 + cost_sum += err_sqr + + return cost_sum / (np.size(t) * np.size(x)) + +## The neural network +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # x is now a point and a 1D numpy array; make it a column vector + num_coordinates = np.size(x,0) + x = x.reshape(num_coordinates,-1) + + num_points = np.size(x,1) + + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assume that the input layer does nothing to the input x + x_input = x + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output[0][0] + +## The analytical solution +def g_analytic(point): + x,t = point + return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t) + +def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): + ## Set up initial weigths and biases + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: ',cost_function(P, x, t)) + + cost_function_grad = grad(cost_function,0) + + # Let the update be done num_iter times + for i in range(num_iter): + cost_grad = cost_function_grad(P, x , t) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_grad[l] + + + print('Final cost: ',cost_function(P, x, t)) + + return P + +if __name__ == '__main__': + ### Use the neural network: + npr.seed(15) + + ## Decide the vales of arguments to the function to solve + Nx = 10; Nt = 10 + x = np.linspace(0, 1, Nx) + t = np.linspace(0,1,Nt) + + ## Set up the parameters for the network + num_hidden_neurons = [50,20] + num_iter = 1000 + lmb = 0.01 + + P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) + + ## Store the results + res = np.zeros((Nx, Nt)) + res_analytical = np.zeros((Nx, Nt)) + for i,x_ in enumerate(x): + for j, t_ in enumerate(t): + point = np.array([x_, t_]) + res[i,j] = g_trial(point,P) + + res_analytical[i,j] = g_analytic(point) + + diff = np.abs(res - res_analytical) + print("Max difference between analytical and solution from nn: %g"%np.max(diff)) + + ## Plot the solutions in two dimensions, that being in position and time + + T,X = np.meshgrid(t,x) + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) + s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Analytical solution') + s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Difference') + s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + ## Take some slices of the 3D plots just to see the solutions at particular times + indx1 = 0 + indx2 = int(Nt/2) + indx3 = Nt-1 + + t1 = t[indx1] + t2 = t[indx2] + t3 = t[indx3] + + # Slice the results from the DNN + res1 = res[:,indx1] + res2 = res[:,indx2] + res3 = res[:,indx3] + + # Slice the analytical results + res_analytical1 = res_analytical[:,indx1] + res_analytical2 = res_analytical[:,indx2] + res_analytical3 = res_analytical[:,indx3] + + # Plot the slices + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t1) + plt.plot(x, res1) + plt.plot(x,res_analytical1) + plt.legend(['dnn','analytical']) + + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t2) + plt.plot(x, res2) + plt.plot(x,res_analytical2) + plt.legend(['dnn','analytical']) + + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t3) + plt.plot(x, res3) + plt.plot(x,res_analytical3) + plt.legend(['dnn','analytical']) + + plt.show() +!ec + + +===== Resources on differential equations and deep learning ===== + +o "Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al":"https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf" +o "Neural networks for solving differential equations by A. Honchar":"https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c" +o "Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener":"http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf" +o "Introduction to Partial Differential Equations by A. Tveito, R. Winther":"https://www.springer.com/us/book/9783540225515" + + + + diff --git a/doc/BookChapters/chapter12.dlog b/doc/BookChapters/chapter12.dlog new file mode 100644 index 000000000..06b7d3cc2 --- /dev/null +++ b/doc/BookChapters/chapter12.dlog @@ -0,0 +1,6 @@ +Translating doconce text in chapter12.do.txt to ipynb +*** error: figure file "figslides/nn.jpeg" does not exist! +Translating doconce text in chapter12.do.txt to ipynb +Failed to remove ans_at_end environment +Failed to remove sol_at_end environment +output in chapter12.ipynb diff --git a/doc/BookChapters/chapter12.do.txt b/doc/BookChapters/chapter12.do.txt new file mode 100644 index 000000000..b3815444f --- /dev/null +++ b/doc/BookChapters/chapter12.do.txt @@ -0,0 +1,578 @@ +======= Convolutional Neural Networks (recognizing images) ======= + + +Convolutional neural networks (CNNs) were developed during the last +decade of the previous century, with a focus on character recognition +tasks. Nowadays, CNNs are a central element in the spectacular success +of deep learning methods. The success in for example image +classifications have made them a central tool for most machine +learning practitioners. + +CNNs are very similar to ordinary Neural Networks. +They are made up of neurons that have learnable weights and +biases. Each neuron receives some inputs, performs a dot product and +optionally follows it with a non-linearity. The whole network still +expresses a single differentiable score function: from the raw image +pixels on one end to class scores at the other. And they still have a +loss function (for example Softmax) on the last (fully-connected) layer +and all the tips/tricks we developed for learning regular Neural +Networks still apply (back propagation, gradient descent etc etc). + +What is the difference? _CNN architectures make the explicit assumption that +the inputs are images, which allows us to encode certain properties +into the architecture. These then make the forward function more +efficient to implement and vastly reduce the amount of parameters in +the network._ + +Here we provide only a superficial overview, for the more interested, we recommend highly the course +"IN5400 – Machine Learning for Image Analysis":"https://www.uio.no/studier/emner/matnat/ifi/IN5400/index-eng.html" +and the slides of "CS231":"http://cs231n.github.io/convolutional-networks/". + +Another good read is the article here URL:"https://arxiv.org/pdf/1603.07285.pdf". + + +As an example, consider +an image of size $32\times 32\times 3$ (32 wide, 32 high, 3 color channels), so a +single fully-connected neuron in a first hidden layer of a regular +Neural Network would have $32\times 32\times 3 = 3072$ weights. This amount still +seems manageable, but clearly this fully-connected structure does not +scale to larger images. For example, an image of more respectable +size, say $200\times 200\times 3$, would lead to neurons that have +$200\times 200\times 3 = 120,000$ weights. + +We could have +several such neurons, and the parameters would add up quickly! Clearly, +this full connectivity is wasteful and the huge number of parameters +would quickly lead to possible overfitting. + +#FIGURE: [figslides/nn.jpeg, width=500 frac=0.6] A regular 3-layer Neural Network. + + + +Convolutional Neural Networks take advantage of the fact that the +input consists of images and they constrain the architecture in a more +sensible way. + +In particular, unlike a regular Neural Network, the +layers of a CNN have neurons arranged in 3 dimensions: width, +height, depth. (Note that the word depth here refers to the third +dimension of an activation volume, not to the depth of a full Neural +Network, which can refer to the total number of layers in a network.) + +To understand it better, the above example of an image +with an input volume of +activations has dimensions $32\times 32\times 3$ (width, height, +depth respectively). + +The neurons in a layer will +only be connected to a small region of the layer before it, instead of +all of the neurons in a fully-connected manner. Moreover, the final +output layer could for this specific image have dimensions $1\times 1 \times 10$, +because by the +end of the CNN architecture we will reduce the full image into a +single vector of class scores, arranged along the depth +dimension. + +#FIGURE: [figslides/cnn.jpeg, width=500 frac=0.6] A CNN arranges its neurons in three dimensions (width, heigh#t, depth), as visualized in one of the layers. Every layer of a CNN transforms the 3D input volume to a 3D out#put volume of neuron activations. In this example, the red input layer holds the image, so its width and heigh#t would be the dimensions of the image, and the depth would be 3 (Red, Green, Blue channels). + + + + +A simple CNN is a sequence of layers, and every layer of a CNN +transforms one volume of activations to another through a +differentiable function. We use three main types of layers to build +CNN architectures: Convolutional Layer, Pooling Layer, and +Fully-Connected Layer (exactly as seen in regular Neural Networks). We +will stack these layers to form a full CNN architecture. + +A simple CNN for image classification could have the architecture: + +* _INPUT_ ($32\times 32 \times 3$) will hold the raw pixel values of the image, in this case an image of width 32, height 32, and with three color channels R,G,B. +* _CONV_ (convolutional )layer will compute the output of neurons that are connected to local regions in the input, each computing a dot product between their weights and a small region they are connected to in the input volume. This may result in volume such as $[32\times 32\times 12]$ if we decided to use 12 filters. +* _RELU_ layer will apply an elementwise activation function, such as the $max(0,x)$ thresholding at zero. This leaves the size of the volume unchanged ($[32\times 32\times 12]$). +* _POOL_ (pooling) layer will perform a downsampling operation along the spatial dimensions (width, height), resulting in volume such as $[16\times 16\times 12]$. +* _FC_ (i.e. fully-connected) layer will compute the class scores, resulting in volume of size $[1\times 1\times 10]$, where each of the 10 numbers correspond to a class score, such as among the 10 categories of the MNIST images we considered above . As with ordinary Neural Networks and as the name implies, each neuron in this layer will be connected to all the numbers in the previous volume. + + + +CNNs transform the original image layer by layer from the original +pixel values to the final class scores. + +Observe that some layers contain +parameters and other don’t. In particular, the CNN layers perform +transformations that are a function of not only the activations in the +input volume, but also of the parameters (the weights and biases of +the neurons). On the other hand, the RELU/POOL layers will implement a +fixed function. The parameters in the CONV/FC layers will be trained +with gradient descent so that the class scores that the CNN computes +are consistent with the labels in the training set for each image. + + + +=== CNNs in brief === + +In summary: + +* A CNN architecture is in the simplest case a list of Layers that transform the image volume into an output volume (e.g. holding the class scores) +* There are a few distinct types of Layers (e.g. CONV/FC/RELU/POOL are by far the most popular) +* Each Layer accepts an input 3D volume and transforms it to an output 3D volume through a differentiable function +* Each Layer may or may not have parameters (e.g. CONV/FC do, RELU/POOL don’t) +* Each Layer may or may not have additional hyperparameters (e.g. CONV/FC/POOL do, RELU doesn’t) + +For more material on convolutional networks, we strongly recommend +the course +"IN5400 – Machine Learning for Image Analysis":"https://www.uio.no/studier/emner/matnat/ifi/IN5400/index-eng.html" +and the slides of "CS231":"http://cs231n.github.io/convolutional-networks/" which is taught at Stanford University (consistently ranked as one of the top computer science programs in the world). "Michael Nielsen's book is a must read, in particular chapter 6 which deals with CNNs":"http://neuralnetworksanddeeplearning.com/chap6.html". + + + + +===== CNNs in more detail, building convolutional neural networks in Tensorflow and Keras ===== + + +As discussed above, CNNs are neural networks built from the assumption that the inputs +to the network are 2D images. This is important because the number of features or pixels in images +grows very fast with the image size, and an enormous number of weights and biases are needed in order to build an accurate network. + +As before, we still have our input, a hidden layer and an output. What's novel about convolutional networks +are the _convolutional_ and _pooling_ layers stacked in pairs between the input and the hidden layer. +In addition, the data is no longer represented as a 2D feature matrix, instead each input is a number of 2D +matrices, typically 1 for each color dimension (Red, Green, Blue). + + + +It means that to represent the entire +dataset of images, we require a 4D matrix or _tensor_. This tensor has the dimensions: +!bt +\[ +(n_{inputs},\, n_{pixels, width},\, n_{pixels, height},\, depth) . +\] +!et + + + +The MNIST dataset consists of grayscale images with a pixel size of +$28\times 28$, meaning we require $28 \times 28 = 724$ weights to each +neuron in the first hidden layer. + +If we were to analyze images of size $128\times 128$ we would require +$128 \times 128 = 16384$ weights to each neuron. Even worse if we were +dealing with color images, as most images are, we have an image matrix +of size $128\times 128$ for each color dimension (Red, Green, Blue), +meaning 3 times the number of weights $= 49152$ are required for every +single neuron in the first hidden layer. + + + +Images typically have strong local correlations, meaning that a small +part of the image varies little from its neighboring regions. If for +example we have an image of a blue car, we can roughly assume that a +small blue part of the image is surrounded by other blue regions. + +Therefore, instead of connecting every single pixel to a neuron in the +first hidden layer, as we have previously done with deep neural +networks, we can instead connect each neuron to a small part of the +image (in all 3 RGB depth dimensions). The size of each small area is +fixed, and known as a "receptive":"https://en.wikipedia.org/wiki/Receptive_field". + + + +The layers of a convolutional neural network arrange neurons in 3D: width, height and depth. +The input image is typically a square matrix of depth 3. + +A _convolution_ is performed on the image which outputs +a 3D volume of neurons. The weights to the input are arranged in a number of 2D matrices, known as _filters_. + + +Each filter slides along the input image, taking the dot product +between each small part of the image and the filter, in all depth +dimensions. This is then passed through a non-linear function, +typically the _Rectified Linear (ReLu)_ function, which serves as the +activation of the neurons in the first convolutional layer. This is +further passed through a _pooling layer_, which reduces the size of the +convolutional layer, e.g. by taking the maximum or average across some +small regions, and this serves as input to the next convolutional +layer. + + + +By systematically reducing the size of the input volume, through +convolution and pooling, the network should create representations of +small parts of the input, and then from them assemble representations +of larger areas. The final pooling layer is flattened to serve as +input to a hidden layer, such that each neuron in the final pooling +layer is connected to every single neuron in the hidden layer. This +then serves as input to the output layer, e.g. a softmax output for +classification. + + + +=== Prerequisites: Collect and pre-process data === +!bc pycod +# import necessary packages +import numpy as np +import matplotlib.pyplot as plt +from sklearn import datasets + + +# ensure the same random numbers appear every time +np.random.seed(0) + +# display images in notebook +%matplotlib inline +plt.rcParams['figure.figsize'] = (12,12) + + +# download MNIST dataset +digits = datasets.load_digits() + +# define inputs and labels +inputs = digits.images +labels = digits.target + +# RGB images have a depth of 3 +# our images are grayscale so they should have a depth of 1 +inputs = inputs[:,:,:,np.newaxis] + +print("inputs = (n_inputs, pixel_width, pixel_height, depth) = " + str(inputs.shape)) +print("labels = (n_inputs) = " + str(labels.shape)) + + +# choose some random images to display +n_inputs = len(inputs) +indices = np.arange(n_inputs) +random_indices = np.random.choice(indices, size=5) + +for i, image in enumerate(digits.images[random_indices]): + plt.subplot(1, 5, i+1) + plt.axis('off') + plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest') + plt.title("Label: %d" % digits.target[random_indices[i]]) +plt.show() +!ec + + +=== Importing Keras and Tensorflow === +!bc pycod +from tensorflow.keras import datasets, layers, models +from tensorflow.keras.layers import Input +from tensorflow.keras.models import Sequential #This allows appending layers to existing models +from tensorflow.keras.layers import Dense #This allows defining the characteristics of a particular layer +from tensorflow.keras import optimizers #This allows using whichever optimiser we want (sgd,adam,RMSprop) +from tensorflow.keras import regularizers #This allows using whichever regularizer we want (l1,l2,l1_l2) +from tensorflow.keras.utils import to_categorical #This allows using categorical cross entropy as the cost function +#from tensorflow.keras import Conv2D +#from tensorflow.keras import MaxPooling2D +#from tensorflow.keras import Flatten + +from sklearn.model_selection import train_test_split + +# representation of labels +labels = to_categorical(labels) + +# split into train and test data +# one-liner from scikit-learn library +train_size = 0.8 +test_size = 1 - train_size +X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size, + test_size=test_size) +!ec + + +!bc pycod +def create_convolutional_neural_network_keras(input_shape, receptive_field, + n_filters, n_neurons_connected, n_categories, + eta, lmbd): + model = Sequential() + model.add(layers.Conv2D(n_filters, (receptive_field, receptive_field), input_shape=input_shape, padding='same', + activation='relu', kernel_regularizer=regularizers.l2(lmbd))) + model.add(layers.MaxPooling2D(pool_size=(2, 2))) + model.add(layers.Flatten()) + model.add(layers.Dense(n_neurons_connected, activation='relu', kernel_regularizer=regularizers.l2(lmbd))) + model.add(layers.Dense(n_categories, activation='softmax', kernel_regularizer=regularizers.l2(lmbd))) + + sgd = optimizers.SGD(lr=eta) + model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy']) + + return model + +epochs = 100 +batch_size = 100 +input_shape = X_train.shape[1:4] +receptive_field = 3 +n_filters = 10 +n_neurons_connected = 50 +n_categories = 10 + +eta_vals = np.logspace(-5, 1, 7) +lmbd_vals = np.logspace(-5, 1, 7) +!ec + + +!bc pycod +CNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object) + +for i, eta in enumerate(eta_vals): + for j, lmbd in enumerate(lmbd_vals): + CNN = create_convolutional_neural_network_keras(input_shape, receptive_field, + n_filters, n_neurons_connected, n_categories, + eta, lmbd) + CNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0) + scores = CNN.evaluate(X_test, Y_test) + + CNN_keras[i][j] = CNN + + print("Learning rate = ", eta) + print("Lambda = ", lmbd) + print("Test accuracy: %.3f" % scores[1]) + print() +!ec + + +=== Final visualization === + +!bc pycod +# visual representation of grid search +# uses seaborn heatmap, could probably do this in matplotlib +import seaborn as sns + +sns.set() + +train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals))) +test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals))) + +for i in range(len(eta_vals)): + for j in range(len(lmbd_vals)): + CNN = CNN_keras[i][j] + + train_accuracy[i][j] = CNN.evaluate(X_train, Y_train)[1] + test_accuracy[i][j] = CNN.evaluate(X_test, Y_test)[1] + + +fig, ax = plt.subplots(figsize = (10, 10)) +sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis") +ax.set_title("Training Accuracy") +ax.set_ylabel("$\eta$") +ax.set_xlabel("$\lambda$") +plt.show() + +fig, ax = plt.subplots(figsize = (10, 10)) +sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis") +ax.set_title("Test Accuracy") +ax.set_ylabel("$\eta$") +ax.set_xlabel("$\lambda$") +plt.show() +!ec + + + + +===== The CIFAR01 data set ===== + +The CIFAR10 dataset contains 60,000 color images in 10 classes, with +6,000 images in each class. The dataset is divided into 50,000 +training images and 10,000 testing images. The classes are mutually +exclusive and there is no overlap between them. + +!bc pycod +import tensorflow as tf + +from tensorflow.keras import datasets, layers, models +import matplotlib.pyplot as plt + +# We import the data set +(train_images, train_labels), (test_images, test_labels) = datasets.cifar10.load_data() + +# Normalize pixel values to be between 0 and 1 by dividing by 255. +train_images, test_images = train_images / 255.0, test_images / 255.0 + +!ec + + + + +To verify that the dataset looks correct, let's plot the first 25 images from the training set and display the class name below each image. + +!bc pycod +class_names = ['airplane', 'automobile', 'bird', 'cat', 'deer', + 'dog', 'frog', 'horse', 'ship', 'truck'] +​ +plt.figure(figsize=(10,10)) +for i in range(25): + plt.subplot(5,5,i+1) + plt.xticks([]) + plt.yticks([]) + plt.grid(False) + plt.imshow(train_images[i], cmap=plt.cm.binary) + # The CIFAR labels happen to be arrays, + # which is why you need the extra index + plt.xlabel(class_names[train_labels[i][0]]) +plt.show() +!ec + + +The 6 lines of code below define the convolutional base using a common pattern: a stack of Conv2D and MaxPooling2D layers. + +As input, a CNN takes tensors of shape (image_height, image_width, color_channels), ignoring the batch size. If you are new to these dimensions, color_channels refers to (R,G,B). In this example, you will configure our CNN to process inputs of shape (32, 32, 3), which is the format of CIFAR images. You can do this by passing the argument input_shape to our first layer. + +!bc pycod +model = models.Sequential() +model.add(layers.Conv2D(32, (3, 3), activation='relu', input_shape=(32, 32, 3))) +model.add(layers.MaxPooling2D((2, 2))) +model.add(layers.Conv2D(64, (3, 3), activation='relu')) +model.add(layers.MaxPooling2D((2, 2))) +model.add(layers.Conv2D(64, (3, 3), activation='relu')) + +# Let's display the architecture of our model so far. + +model.summary() +!ec + +You can see that the output of every Conv2D and MaxPooling2D layer is a 3D tensor of shape (height, width, channels). The width and height dimensions tend to shrink as you go deeper in the network. The number of output channels for each Conv2D layer is controlled by the first argument (e.g., 32 or 64). Typically, as the width and height shrink, you can afford (computationally) to add more output channels in each Conv2D layer. + + + +To complete our model, you will feed the last output tensor from the +convolutional base (of shape (4, 4, 64)) into one or more Dense layers +to perform classification. Dense layers take vectors as input (which +are 1D), while the current output is a 3D tensor. First, you will +flatten (or unroll) the 3D output to 1D, then add one or more Dense +layers on top. CIFAR has 10 output classes, so you use a final Dense +layer with 10 outputs and a softmax activation. + +!bc pycod +model.add(layers.Flatten()) +model.add(layers.Dense(64, activation='relu')) +model.add(layers.Dense(10)) +Here's the complete architecture of our model. + +model.summary() +!ec +As you can see, our (4, 4, 64) outputs were flattened into vectors of shape (1024) before going through two Dense layers. + +Compile and train the model. + +!bc pycod +model.compile(optimizer='adam', + loss=tf.keras.losses.SparseCategoricalCrossentropy(from_logits=True), + metrics=['accuracy']) +​ +history = model.fit(train_images, train_labels, epochs=10, + validation_data=(test_images, test_labels)) + +!ec + +Finally, we evaluate the model. + +!bc pycod +plt.plot(history.history['accuracy'], label='accuracy') +plt.plot(history.history['val_accuracy'], label = 'val_accuracy') +plt.xlabel('Epoch') +plt.ylabel('Accuracy') +plt.ylim([0.5, 1]) +plt.legend(loc='lower right') + +test_loss, test_acc = model.evaluate(test_images, test_labels, verbose=2) + +print(test_acc) + +!ec + + + +===== Recurrent neural networks: Overarching view ===== + +Till now our focus has been, including convolutional neural networks +as well, on feedforward neural networks. The output or the activations +flow only in one direction, from the input layer to the output layer. + +A recurrent neural network (RNN) looks very much like a feedforward +neural network, except that it also has connections pointing +backward. + +RNNs are used to analyze time series data such as stock prices, and +tell you when to buy or sell. In autonomous driving systems, they can +anticipate car trajectories and help avoid accidents. More generally, +they can work on sequences of arbitrary lengths, rather than on +fixed-sized inputs like all the nets we have discussed so far. For +example, they can take sentences, documents, or audio samples as +input, making them extremely useful for natural language processing +systems such as automatic translation and speech-to-text. + + + + +=== A simple example === + +!bc pycod +# Start importing packages +import pandas as pd +import numpy as np +import matplotlib.pyplot as plt +import tensorflow as tf +from tensorflow.keras import datasets, layers, models +from tensorflow.keras.layers import Input +from tensorflow.keras.models import Model, Sequential +from tensorflow.keras.layers import Dense, SimpleRNN, LSTM, GRU +from tensorflow.keras import optimizers +from tensorflow.keras import regularizers +from tensorflow.keras.utils import to_categorical + + + +# convert into dataset matrix +def convertToMatrix(data, step): + X, Y =[], [] + for i in range(len(data)-step): + d=i+step + X.append(data[i:d,]) + Y.append(data[d,]) + return np.array(X), np.array(Y) + +step = 4 +N = 1000 +Tp = 800 + +t=np.arange(0,N) +x=np.sin(0.02*t)+2*np.random.rand(N) +df = pd.DataFrame(x) +df.head() + +plt.plot(df) +plt.show() + +values=df.values +train,test = values[0:Tp,:], values[Tp:N,:] + +# add step elements into train and test +test = np.append(test,np.repeat(test[-1,],step)) +train = np.append(train,np.repeat(train[-1,],step)) + +trainX,trainY =convertToMatrix(train,step) +testX,testY =convertToMatrix(test,step) +trainX = np.reshape(trainX, (trainX.shape[0], 1, trainX.shape[1])) +testX = np.reshape(testX, (testX.shape[0], 1, testX.shape[1])) + +model = Sequential() +model.add(SimpleRNN(units=32, input_shape=(1,step), activation="relu")) +model.add(Dense(8, activation="relu")) +model.add(Dense(1)) +model.compile(loss='mean_squared_error', optimizer='rmsprop') +model.summary() + +model.fit(trainX,trainY, epochs=100, batch_size=16, verbose=2) +trainPredict = model.predict(trainX) +testPredict= model.predict(testX) +predicted=np.concatenate((trainPredict,testPredict),axis=0) + +trainScore = model.evaluate(trainX, trainY, verbose=0) +print(trainScore) + +index = df.index.values +plt.plot(index,df) +plt.plot(index,predicted) +plt.axvline(df.index[Tp], c="r") +plt.show() +!ec + + diff --git a/doc/BookChapters/chapter12.do.txt~ b/doc/BookChapters/chapter12.do.txt~ new file mode 100644 index 000000000..6a816eda7 --- /dev/null +++ b/doc/BookChapters/chapter12.do.txt~ @@ -0,0 +1,578 @@ +======= Convolutional Neural Networks (recognizing images) ======= + + +Convolutional neural networks (CNNs) were developed during the last +decade of the previous century, with a focus on character recognition +tasks. Nowadays, CNNs are a central element in the spectacular success +of deep learning methods. The success in for example image +classifications have made them a central tool for most machine +learning practitioners. + +CNNs are very similar to ordinary Neural Networks. +They are made up of neurons that have learnable weights and +biases. Each neuron receives some inputs, performs a dot product and +optionally follows it with a non-linearity. The whole network still +expresses a single differentiable score function: from the raw image +pixels on one end to class scores at the other. And they still have a +loss function (for example Softmax) on the last (fully-connected) layer +and all the tips/tricks we developed for learning regular Neural +Networks still apply (back propagation, gradient descent etc etc). + +What is the difference? _CNN architectures make the explicit assumption that +the inputs are images, which allows us to encode certain properties +into the architecture. These then make the forward function more +efficient to implement and vastly reduce the amount of parameters in +the network._ + +Here we provide only a superficial overview, for the more interested, we recommend highly the course +"IN5400 – Machine Learning for Image Analysis":"https://www.uio.no/studier/emner/matnat/ifi/IN5400/index-eng.html" +and the slides of "CS231":"http://cs231n.github.io/convolutional-networks/". + +Another good read is the article here URL:"https://arxiv.org/pdf/1603.07285.pdf". + + +As an example, consider +an image of size $32\times 32\times 3$ (32 wide, 32 high, 3 color channels), so a +single fully-connected neuron in a first hidden layer of a regular +Neural Network would have $32\times 32\times 3 = 3072$ weights. This amount still +seems manageable, but clearly this fully-connected structure does not +scale to larger images. For example, an image of more respectable +size, say $200\times 200\times 3$, would lead to neurons that have +$200\times 200\times 3 = 120,000$ weights. + +We could have +several such neurons, and the parameters would add up quickly! Clearly, +this full connectivity is wasteful and the huge number of parameters +would quickly lead to possible overfitting. + +FIGURE: [figslides/nn.jpeg, width=500 frac=0.6] A regular 3-layer Neural Network. + + + +Convolutional Neural Networks take advantage of the fact that the +input consists of images and they constrain the architecture in a more +sensible way. + +In particular, unlike a regular Neural Network, the +layers of a CNN have neurons arranged in 3 dimensions: width, +height, depth. (Note that the word depth here refers to the third +dimension of an activation volume, not to the depth of a full Neural +Network, which can refer to the total number of layers in a network.) + +To understand it better, the above example of an image +with an input volume of +activations has dimensions $32\times 32\times 3$ (width, height, +depth respectively). + +The neurons in a layer will +only be connected to a small region of the layer before it, instead of +all of the neurons in a fully-connected manner. Moreover, the final +output layer could for this specific image have dimensions $1\times 1 \times 10$, +because by the +end of the CNN architecture we will reduce the full image into a +single vector of class scores, arranged along the depth +dimension. + +FIGURE: [figslides/cnn.jpeg, width=500 frac=0.6] A CNN arranges its neurons in three dimensions (width, height, depth), as visualized in one of the layers. Every layer of a CNN transforms the 3D input volume to a 3D output volume of neuron activations. In this example, the red input layer holds the image, so its width and height would be the dimensions of the image, and the depth would be 3 (Red, Green, Blue channels). + + + + +A simple CNN is a sequence of layers, and every layer of a CNN +transforms one volume of activations to another through a +differentiable function. We use three main types of layers to build +CNN architectures: Convolutional Layer, Pooling Layer, and +Fully-Connected Layer (exactly as seen in regular Neural Networks). We +will stack these layers to form a full CNN architecture. + +A simple CNN for image classification could have the architecture: + +* _INPUT_ ($32\times 32 \times 3$) will hold the raw pixel values of the image, in this case an image of width 32, height 32, and with three color channels R,G,B. +* _CONV_ (convolutional )layer will compute the output of neurons that are connected to local regions in the input, each computing a dot product between their weights and a small region they are connected to in the input volume. This may result in volume such as $[32\times 32\times 12]$ if we decided to use 12 filters. +* _RELU_ layer will apply an elementwise activation function, such as the $max(0,x)$ thresholding at zero. This leaves the size of the volume unchanged ($[32\times 32\times 12]$). +* _POOL_ (pooling) layer will perform a downsampling operation along the spatial dimensions (width, height), resulting in volume such as $[16\times 16\times 12]$. +* _FC_ (i.e. fully-connected) layer will compute the class scores, resulting in volume of size $[1\times 1\times 10]$, where each of the 10 numbers correspond to a class score, such as among the 10 categories of the MNIST images we considered above . As with ordinary Neural Networks and as the name implies, each neuron in this layer will be connected to all the numbers in the previous volume. + + + +CNNs transform the original image layer by layer from the original +pixel values to the final class scores. + +Observe that some layers contain +parameters and other don’t. In particular, the CNN layers perform +transformations that are a function of not only the activations in the +input volume, but also of the parameters (the weights and biases of +the neurons). On the other hand, the RELU/POOL layers will implement a +fixed function. The parameters in the CONV/FC layers will be trained +with gradient descent so that the class scores that the CNN computes +are consistent with the labels in the training set for each image. + + + +=== CNNs in brief === + +In summary: + +* A CNN architecture is in the simplest case a list of Layers that transform the image volume into an output volume (e.g. holding the class scores) +* There are a few distinct types of Layers (e.g. CONV/FC/RELU/POOL are by far the most popular) +* Each Layer accepts an input 3D volume and transforms it to an output 3D volume through a differentiable function +* Each Layer may or may not have parameters (e.g. CONV/FC do, RELU/POOL don’t) +* Each Layer may or may not have additional hyperparameters (e.g. CONV/FC/POOL do, RELU doesn’t) + +For more material on convolutional networks, we strongly recommend +the course +"IN5400 – Machine Learning for Image Analysis":"https://www.uio.no/studier/emner/matnat/ifi/IN5400/index-eng.html" +and the slides of "CS231":"http://cs231n.github.io/convolutional-networks/" which is taught at Stanford University (consistently ranked as one of the top computer science programs in the world). "Michael Nielsen's book is a must read, in particular chapter 6 which deals with CNNs":"http://neuralnetworksanddeeplearning.com/chap6.html". + + + + +===== CNNs in more detail, building convolutional neural networks in Tensorflow and Keras ===== + + +As discussed above, CNNs are neural networks built from the assumption that the inputs +to the network are 2D images. This is important because the number of features or pixels in images +grows very fast with the image size, and an enormous number of weights and biases are needed in order to build an accurate network. + +As before, we still have our input, a hidden layer and an output. What's novel about convolutional networks +are the _convolutional_ and _pooling_ layers stacked in pairs between the input and the hidden layer. +In addition, the data is no longer represented as a 2D feature matrix, instead each input is a number of 2D +matrices, typically 1 for each color dimension (Red, Green, Blue). + + + +It means that to represent the entire +dataset of images, we require a 4D matrix or _tensor_. This tensor has the dimensions: +!bt +\[ +(n_{inputs},\, n_{pixels, width},\, n_{pixels, height},\, depth) . +\] +!et + + + +The MNIST dataset consists of grayscale images with a pixel size of +$28\times 28$, meaning we require $28 \times 28 = 724$ weights to each +neuron in the first hidden layer. + +If we were to analyze images of size $128\times 128$ we would require +$128 \times 128 = 16384$ weights to each neuron. Even worse if we were +dealing with color images, as most images are, we have an image matrix +of size $128\times 128$ for each color dimension (Red, Green, Blue), +meaning 3 times the number of weights $= 49152$ are required for every +single neuron in the first hidden layer. + + + +Images typically have strong local correlations, meaning that a small +part of the image varies little from its neighboring regions. If for +example we have an image of a blue car, we can roughly assume that a +small blue part of the image is surrounded by other blue regions. + +Therefore, instead of connecting every single pixel to a neuron in the +first hidden layer, as we have previously done with deep neural +networks, we can instead connect each neuron to a small part of the +image (in all 3 RGB depth dimensions). The size of each small area is +fixed, and known as a "receptive":"https://en.wikipedia.org/wiki/Receptive_field". + + + +The layers of a convolutional neural network arrange neurons in 3D: width, height and depth. +The input image is typically a square matrix of depth 3. + +A _convolution_ is performed on the image which outputs +a 3D volume of neurons. The weights to the input are arranged in a number of 2D matrices, known as _filters_. + + +Each filter slides along the input image, taking the dot product +between each small part of the image and the filter, in all depth +dimensions. This is then passed through a non-linear function, +typically the _Rectified Linear (ReLu)_ function, which serves as the +activation of the neurons in the first convolutional layer. This is +further passed through a _pooling layer_, which reduces the size of the +convolutional layer, e.g. by taking the maximum or average across some +small regions, and this serves as input to the next convolutional +layer. + + + +By systematically reducing the size of the input volume, through +convolution and pooling, the network should create representations of +small parts of the input, and then from them assemble representations +of larger areas. The final pooling layer is flattened to serve as +input to a hidden layer, such that each neuron in the final pooling +layer is connected to every single neuron in the hidden layer. This +then serves as input to the output layer, e.g. a softmax output for +classification. + + + +=== Prerequisites: Collect and pre-process data === +!bc pycod +# import necessary packages +import numpy as np +import matplotlib.pyplot as plt +from sklearn import datasets + + +# ensure the same random numbers appear every time +np.random.seed(0) + +# display images in notebook +%matplotlib inline +plt.rcParams['figure.figsize'] = (12,12) + + +# download MNIST dataset +digits = datasets.load_digits() + +# define inputs and labels +inputs = digits.images +labels = digits.target + +# RGB images have a depth of 3 +# our images are grayscale so they should have a depth of 1 +inputs = inputs[:,:,:,np.newaxis] + +print("inputs = (n_inputs, pixel_width, pixel_height, depth) = " + str(inputs.shape)) +print("labels = (n_inputs) = " + str(labels.shape)) + + +# choose some random images to display +n_inputs = len(inputs) +indices = np.arange(n_inputs) +random_indices = np.random.choice(indices, size=5) + +for i, image in enumerate(digits.images[random_indices]): + plt.subplot(1, 5, i+1) + plt.axis('off') + plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest') + plt.title("Label: %d" % digits.target[random_indices[i]]) +plt.show() +!ec + + +=== Importing Keras and Tensorflow === +!bc pycod +from tensorflow.keras import datasets, layers, models +from tensorflow.keras.layers import Input +from tensorflow.keras.models import Sequential #This allows appending layers to existing models +from tensorflow.keras.layers import Dense #This allows defining the characteristics of a particular layer +from tensorflow.keras import optimizers #This allows using whichever optimiser we want (sgd,adam,RMSprop) +from tensorflow.keras import regularizers #This allows using whichever regularizer we want (l1,l2,l1_l2) +from tensorflow.keras.utils import to_categorical #This allows using categorical cross entropy as the cost function +#from tensorflow.keras import Conv2D +#from tensorflow.keras import MaxPooling2D +#from tensorflow.keras import Flatten + +from sklearn.model_selection import train_test_split + +# representation of labels +labels = to_categorical(labels) + +# split into train and test data +# one-liner from scikit-learn library +train_size = 0.8 +test_size = 1 - train_size +X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size, + test_size=test_size) +!ec + + +!bc pycod +def create_convolutional_neural_network_keras(input_shape, receptive_field, + n_filters, n_neurons_connected, n_categories, + eta, lmbd): + model = Sequential() + model.add(layers.Conv2D(n_filters, (receptive_field, receptive_field), input_shape=input_shape, padding='same', + activation='relu', kernel_regularizer=regularizers.l2(lmbd))) + model.add(layers.MaxPooling2D(pool_size=(2, 2))) + model.add(layers.Flatten()) + model.add(layers.Dense(n_neurons_connected, activation='relu', kernel_regularizer=regularizers.l2(lmbd))) + model.add(layers.Dense(n_categories, activation='softmax', kernel_regularizer=regularizers.l2(lmbd))) + + sgd = optimizers.SGD(lr=eta) + model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy']) + + return model + +epochs = 100 +batch_size = 100 +input_shape = X_train.shape[1:4] +receptive_field = 3 +n_filters = 10 +n_neurons_connected = 50 +n_categories = 10 + +eta_vals = np.logspace(-5, 1, 7) +lmbd_vals = np.logspace(-5, 1, 7) +!ec + + +!bc pycod +CNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object) + +for i, eta in enumerate(eta_vals): + for j, lmbd in enumerate(lmbd_vals): + CNN = create_convolutional_neural_network_keras(input_shape, receptive_field, + n_filters, n_neurons_connected, n_categories, + eta, lmbd) + CNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0) + scores = CNN.evaluate(X_test, Y_test) + + CNN_keras[i][j] = CNN + + print("Learning rate = ", eta) + print("Lambda = ", lmbd) + print("Test accuracy: %.3f" % scores[1]) + print() +!ec + + +=== Final visualization === + +!bc pycod +# visual representation of grid search +# uses seaborn heatmap, could probably do this in matplotlib +import seaborn as sns + +sns.set() + +train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals))) +test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals))) + +for i in range(len(eta_vals)): + for j in range(len(lmbd_vals)): + CNN = CNN_keras[i][j] + + train_accuracy[i][j] = CNN.evaluate(X_train, Y_train)[1] + test_accuracy[i][j] = CNN.evaluate(X_test, Y_test)[1] + + +fig, ax = plt.subplots(figsize = (10, 10)) +sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis") +ax.set_title("Training Accuracy") +ax.set_ylabel("$\eta$") +ax.set_xlabel("$\lambda$") +plt.show() + +fig, ax = plt.subplots(figsize = (10, 10)) +sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis") +ax.set_title("Test Accuracy") +ax.set_ylabel("$\eta$") +ax.set_xlabel("$\lambda$") +plt.show() +!ec + + + + +===== The CIFAR01 data set ===== + +The CIFAR10 dataset contains 60,000 color images in 10 classes, with +6,000 images in each class. The dataset is divided into 50,000 +training images and 10,000 testing images. The classes are mutually +exclusive and there is no overlap between them. + +!bc pycod +import tensorflow as tf + +from tensorflow.keras import datasets, layers, models +import matplotlib.pyplot as plt + +# We import the data set +(train_images, train_labels), (test_images, test_labels) = datasets.cifar10.load_data() + +# Normalize pixel values to be between 0 and 1 by dividing by 255. +train_images, test_images = train_images / 255.0, test_images / 255.0 + +!ec + + + + +To verify that the dataset looks correct, let's plot the first 25 images from the training set and display the class name below each image. + +!bc pycod +class_names = ['airplane', 'automobile', 'bird', 'cat', 'deer', + 'dog', 'frog', 'horse', 'ship', 'truck'] +​ +plt.figure(figsize=(10,10)) +for i in range(25): + plt.subplot(5,5,i+1) + plt.xticks([]) + plt.yticks([]) + plt.grid(False) + plt.imshow(train_images[i], cmap=plt.cm.binary) + # The CIFAR labels happen to be arrays, + # which is why you need the extra index + plt.xlabel(class_names[train_labels[i][0]]) +plt.show() +!ec + + +The 6 lines of code below define the convolutional base using a common pattern: a stack of Conv2D and MaxPooling2D layers. + +As input, a CNN takes tensors of shape (image_height, image_width, color_channels), ignoring the batch size. If you are new to these dimensions, color_channels refers to (R,G,B). In this example, you will configure our CNN to process inputs of shape (32, 32, 3), which is the format of CIFAR images. You can do this by passing the argument input_shape to our first layer. + +!bc pycod +model = models.Sequential() +model.add(layers.Conv2D(32, (3, 3), activation='relu', input_shape=(32, 32, 3))) +model.add(layers.MaxPooling2D((2, 2))) +model.add(layers.Conv2D(64, (3, 3), activation='relu')) +model.add(layers.MaxPooling2D((2, 2))) +model.add(layers.Conv2D(64, (3, 3), activation='relu')) + +# Let's display the architecture of our model so far. + +model.summary() +!ec + +You can see that the output of every Conv2D and MaxPooling2D layer is a 3D tensor of shape (height, width, channels). The width and height dimensions tend to shrink as you go deeper in the network. The number of output channels for each Conv2D layer is controlled by the first argument (e.g., 32 or 64). Typically, as the width and height shrink, you can afford (computationally) to add more output channels in each Conv2D layer. + + + +To complete our model, you will feed the last output tensor from the +convolutional base (of shape (4, 4, 64)) into one or more Dense layers +to perform classification. Dense layers take vectors as input (which +are 1D), while the current output is a 3D tensor. First, you will +flatten (or unroll) the 3D output to 1D, then add one or more Dense +layers on top. CIFAR has 10 output classes, so you use a final Dense +layer with 10 outputs and a softmax activation. + +!bc pycod +model.add(layers.Flatten()) +model.add(layers.Dense(64, activation='relu')) +model.add(layers.Dense(10)) +Here's the complete architecture of our model. + +model.summary() +!ec +As you can see, our (4, 4, 64) outputs were flattened into vectors of shape (1024) before going through two Dense layers. + +Compile and train the model. + +!bc pycod +model.compile(optimizer='adam', + loss=tf.keras.losses.SparseCategoricalCrossentropy(from_logits=True), + metrics=['accuracy']) +​ +history = model.fit(train_images, train_labels, epochs=10, + validation_data=(test_images, test_labels)) + +!ec + +Finally, we evaluate the model. + +!bc pycod +plt.plot(history.history['accuracy'], label='accuracy') +plt.plot(history.history['val_accuracy'], label = 'val_accuracy') +plt.xlabel('Epoch') +plt.ylabel('Accuracy') +plt.ylim([0.5, 1]) +plt.legend(loc='lower right') + +test_loss, test_acc = model.evaluate(test_images, test_labels, verbose=2) + +print(test_acc) + +!ec + + + +===== Recurrent neural networks: Overarching view ===== + +Till now our focus has been, including convolutional neural networks +as well, on feedforward neural networks. The output or the activations +flow only in one direction, from the input layer to the output layer. + +A recurrent neural network (RNN) looks very much like a feedforward +neural network, except that it also has connections pointing +backward. + +RNNs are used to analyze time series data such as stock prices, and +tell you when to buy or sell. In autonomous driving systems, they can +anticipate car trajectories and help avoid accidents. More generally, +they can work on sequences of arbitrary lengths, rather than on +fixed-sized inputs like all the nets we have discussed so far. For +example, they can take sentences, documents, or audio samples as +input, making them extremely useful for natural language processing +systems such as automatic translation and speech-to-text. + + + + +=== A simple example === + +!bc pycod +# Start importing packages +import pandas as pd +import numpy as np +import matplotlib.pyplot as plt +import tensorflow as tf +from tensorflow.keras import datasets, layers, models +from tensorflow.keras.layers import Input +from tensorflow.keras.models import Model, Sequential +from tensorflow.keras.layers import Dense, SimpleRNN, LSTM, GRU +from tensorflow.keras import optimizers +from tensorflow.keras import regularizers +from tensorflow.keras.utils import to_categorical + + + +# convert into dataset matrix +def convertToMatrix(data, step): + X, Y =[], [] + for i in range(len(data)-step): + d=i+step + X.append(data[i:d,]) + Y.append(data[d,]) + return np.array(X), np.array(Y) + +step = 4 +N = 1000 +Tp = 800 + +t=np.arange(0,N) +x=np.sin(0.02*t)+2*np.random.rand(N) +df = pd.DataFrame(x) +df.head() + +plt.plot(df) +plt.show() + +values=df.values +train,test = values[0:Tp,:], values[Tp:N,:] + +# add step elements into train and test +test = np.append(test,np.repeat(test[-1,],step)) +train = np.append(train,np.repeat(train[-1,],step)) + +trainX,trainY =convertToMatrix(train,step) +testX,testY =convertToMatrix(test,step) +trainX = np.reshape(trainX, (trainX.shape[0], 1, trainX.shape[1])) +testX = np.reshape(testX, (testX.shape[0], 1, testX.shape[1])) + +model = Sequential() +model.add(SimpleRNN(units=32, input_shape=(1,step), activation="relu")) +model.add(Dense(8, activation="relu")) +model.add(Dense(1)) +model.compile(loss='mean_squared_error', optimizer='rmsprop') +model.summary() + +model.fit(trainX,trainY, epochs=100, batch_size=16, verbose=2) +trainPredict = model.predict(trainX) +testPredict= model.predict(testX) +predicted=np.concatenate((trainPredict,testPredict),axis=0) + +trainScore = model.evaluate(trainX, trainY, verbose=0) +print(trainScore) + +index = df.index.values +plt.plot(index,df) +plt.plot(index,predicted) +plt.axvline(df.index[Tp], c="r") +plt.show() +!ec + + diff --git a/doc/LectureNotes/_build/.doctrees/chapter1.doctree b/doc/LectureNotes/_build/.doctrees/chapter1.doctree index 9cc4eb4c8..91b118b94 100644 Binary files a/doc/LectureNotes/_build/.doctrees/chapter1.doctree and b/doc/LectureNotes/_build/.doctrees/chapter1.doctree differ diff --git a/doc/LectureNotes/_build/.doctrees/chapter10.doctree b/doc/LectureNotes/_build/.doctrees/chapter10.doctree index 75555916a..51ed9de1d 100644 Binary files a/doc/LectureNotes/_build/.doctrees/chapter10.doctree and b/doc/LectureNotes/_build/.doctrees/chapter10.doctree differ diff --git a/doc/LectureNotes/_build/.doctrees/chapter11.doctree b/doc/LectureNotes/_build/.doctrees/chapter11.doctree new file mode 100644 index 000000000..df4f51c9c Binary files /dev/null and b/doc/LectureNotes/_build/.doctrees/chapter11.doctree differ diff --git a/doc/LectureNotes/_build/.doctrees/environment.pickle b/doc/LectureNotes/_build/.doctrees/environment.pickle deleted file mode 100644 index 8974a9d84..000000000 Binary files a/doc/LectureNotes/_build/.doctrees/environment.pickle and /dev/null differ diff --git a/doc/LectureNotes/_build/html/reports/chapter10.log b/doc/LectureNotes/_build/html/reports/chapter10.log new file mode 100644 index 000000000..5c0d14ab9 --- /dev/null +++ b/doc/LectureNotes/_build/html/reports/chapter10.log @@ -0,0 +1,29 @@ +Traceback (most recent call last): + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/jupyter_cache/executors/utils.py", line 51, in single_nb_execution + executenb( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 1087, in execute + return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 74, in wrapped + return just_run(coro(*args, **kwargs)) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 53, in just_run + return loop.run_until_complete(coro) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/asyncio/base_events.py", line 616, in run_until_complete + return future.result() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 540, in async_execute + await self.async_execute_cell( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 832, in async_execute_cell + self._check_raise_for_error(cell, exec_reply) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 740, in _check_raise_for_error + raise CellExecutionError.from_cell_and_msg(cell, exec_reply['content']) +nbclient.exceptions.CellExecutionError: An error occurred while executing the following cell: +------------------ +pip3 install tensorflow +------------------ + + File "", line 1 + pip3 install tensorflow + ^ +SyntaxError: invalid syntax + +SyntaxError: invalid syntax (, line 1) + diff --git a/doc/LectureNotes/_build/html/reports/chapter5.log b/doc/LectureNotes/_build/html/reports/chapter5.log new file mode 100644 index 000000000..828ec304d --- /dev/null +++ b/doc/LectureNotes/_build/html/reports/chapter5.log @@ -0,0 +1,41 @@ +Traceback (most recent call last): + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/jupyter_cache/executors/utils.py", line 51, in single_nb_execution + executenb( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 1087, in execute + return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 74, in wrapped + return just_run(coro(*args, **kwargs)) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 53, in just_run + return loop.run_until_complete(coro) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/asyncio/base_events.py", line 616, in run_until_complete + return future.result() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 540, in async_execute + await self.async_execute_cell( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 832, in async_execute_cell + self._check_raise_for_error(cell, exec_reply) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 740, in _check_raise_for_error + raise CellExecutionError.from_cell_and_msg(cell, exec_reply['content']) +nbclient.exceptions.CellExecutionError: An error occurred while executing the following cell: +------------------ +# Import the necessary packages +import numpy +from cvxopt import matrix +from cvxopt import solvers +P = matrix(numpy.diag([1,0]), tc=’d’) +q = matrix(numpy.array([3,4]), tc=’d’) +G = matrix(numpy.array([[-1,0],[0,-1],[-1,-3],[2,5],[3,4]]), tc=’d’) +h = matrix(numpy.array([0,0,-15,100,80]), tc=’d’) +# Construct the QP, invoke solver +sol = solvers.qp(P,q,G,h) +# Extract optimal value and solution +sol[’x’] +sol[’primal objective’] +------------------ + + File "", line 5 + P = matrix(numpy.diag([1,0]), tc=’d’) + ^ +SyntaxError: invalid character in identifier + +SyntaxError: invalid character in identifier (, line 5) + diff --git a/doc/LectureNotes/_build/html/reports/chapter6.log b/doc/LectureNotes/_build/html/reports/chapter6.log new file mode 100644 index 000000000..b9dbedf5f --- /dev/null +++ b/doc/LectureNotes/_build/html/reports/chapter6.log @@ -0,0 +1,66 @@ +Traceback (most recent call last): + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/jupyter_cache/executors/utils.py", line 51, in single_nb_execution + executenb( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 1087, in execute + return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 74, in wrapped + return just_run(coro(*args, **kwargs)) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 53, in just_run + return loop.run_until_complete(coro) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/asyncio/base_events.py", line 616, in run_until_complete + return future.result() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 540, in async_execute + await self.async_execute_cell( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 832, in async_execute_cell + self._check_raise_for_error(cell, exec_reply) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 740, in _check_raise_for_error + raise CellExecutionError.from_cell_and_msg(cell, exec_reply['content']) +nbclient.exceptions.CellExecutionError: An error occurred while executing the following cell: +------------------ +import os +from sklearn.datasets import load_breast_cancer +from sklearn.tree import DecisionTreeClassifier +from sklearn.model_selection import train_test_split +from sklearn.metrics import confusion_matrix +from sklearn.tree import export_graphviz + +from IPython.display import Image +from pydot import graph_from_dot_data +import pandas as pd +import numpy as np + + +cancer = load_breast_cancer() +X = pd.DataFrame(cancer.data, columns=cancer.feature_names) +print(X) +y = pd.Categorical.from_codes(cancer.target, cancer.target_names) +y = pd.get_dummies(y) +print(y) +X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=1) +tree_clf = DecisionTreeClassifier(max_depth=5) +tree_clf.fit(X_train, y_train) + +export_graphviz( + tree_clf, + out_file="DataFiles/cancer.dot", + feature_names=cancer.feature_names, + class_names=cancer.target_names, + rounded=True, + filled=True +) +cmd = 'dot -Tpng DataFiles/cancer.dot -o DataFiles/cancer.png' +os.system(cmd) +------------------ + +--------------------------------------------------------------------------- +ModuleNotFoundError Traceback (most recent call last) + in  + 7  + 8 from IPython.display import Image +----> 9 from pydot import graph_from_dot_data + 10 import pandas as pd + 11 import numpy as np + +ModuleNotFoundError: No module named 'pydot' +ModuleNotFoundError: No module named 'pydot' + diff --git a/doc/LectureNotes/_build/html/reports/chapter7.log b/doc/LectureNotes/_build/html/reports/chapter7.log new file mode 100644 index 000000000..727b82aa1 --- /dev/null +++ b/doc/LectureNotes/_build/html/reports/chapter7.log @@ -0,0 +1,46 @@ +Traceback (most recent call last): + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/jupyter_cache/executors/utils.py", line 51, in single_nb_execution + executenb( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 1087, in execute + return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 74, in wrapped + return just_run(coro(*args, **kwargs)) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 53, in just_run + return loop.run_until_complete(coro) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/asyncio/base_events.py", line 616, in run_until_complete + return future.result() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 540, in async_execute + await self.async_execute_cell( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 832, in async_execute_cell + self._check_raise_for_error(cell, exec_reply) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 740, in _check_raise_for_error + raise CellExecutionError.from_cell_and_msg(cell, exec_reply['content']) +nbclient.exceptions.CellExecutionError: An error occurred while executing the following cell: +------------------ +heads_proba = 0.51 +coin_tosses = (np.random.rand(10000, 10) < heads_proba).astype(np.int32) +cumulative_heads_ratio = np.cumsum(coin_tosses, axis=0) / np.arange(1, 10001).reshape(-1, 1) +plt.figure(figsize=(8,3.5)) +plt.plot(cumulative_heads_ratio) +plt.plot([0, 10000], [0.51, 0.51], "k--", linewidth=2, label="51%") +plt.plot([0, 10000], [0.5, 0.5], "k-", label="50%") +plt.xlabel("Number of coin tosses") +plt.ylabel("Heads ratio") +plt.legend(loc="lower right") +plt.axis([0, 10000, 0.42, 0.58]) +save_fig("votingsimple") +plt.show() +------------------ + +--------------------------------------------------------------------------- +NameError Traceback (most recent call last) + in  + 1 heads_proba = 0.51 +----> 2 coin_tosses = (np.random.rand(10000, 10) < heads_proba).astype(np.int32) + 3 cumulative_heads_ratio = np.cumsum(coin_tosses, axis=0) / np.arange(1, 10001).reshape(-1, 1) + 4 plt.figure(figsize=(8,3.5)) + 5 plt.plot(cumulative_heads_ratio) + +NameError: name 'np' is not defined +NameError: name 'np' is not defined + diff --git a/doc/LectureNotes/_build/html/reports/linalg.log b/doc/LectureNotes/_build/html/reports/linalg.log new file mode 100644 index 000000000..4cb123d39 --- /dev/null +++ b/doc/LectureNotes/_build/html/reports/linalg.log @@ -0,0 +1,31 @@ +Traceback (most recent call last): + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/jupyter_cache/executors/utils.py", line 51, in single_nb_execution + executenb( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 1087, in execute + return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 74, in wrapped + return just_run(coro(*args, **kwargs)) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 53, in just_run + return loop.run_until_complete(coro) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/asyncio/base_events.py", line 616, in run_until_complete + return future.result() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 540, in async_execute + await self.async_execute_cell( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 832, in async_execute_cell + self._check_raise_for_error(cell, exec_reply) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 740, in _check_raise_for_error + raise CellExecutionError.from_cell_and_msg(cell, exec_reply['content']) +nbclient.exceptions.CellExecutionError: An error occurred while executing the following cell: +------------------ +import numpy as np +x = np.log(np.array([4.0, 7.0, 8.0]) +print(x) +------------------ + + File "", line 3 + print(x) + ^ +SyntaxError: invalid syntax + +SyntaxError: invalid syntax (, line 3) + diff --git a/doc/LectureNotes/_build/html/reports/statistics.log b/doc/LectureNotes/_build/html/reports/statistics.log new file mode 100644 index 000000000..279cbb99d --- /dev/null +++ b/doc/LectureNotes/_build/html/reports/statistics.log @@ -0,0 +1,58 @@ +Traceback (most recent call last): + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/jupyter_cache/executors/utils.py", line 51, in single_nb_execution + executenb( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 1087, in execute + return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 74, in wrapped + return just_run(coro(*args, **kwargs)) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/util.py", line 53, in just_run + return loop.run_until_complete(coro) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/asyncio/base_events.py", line 616, in run_until_complete + return future.result() + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 540, in async_execute + await self.async_execute_cell( + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 832, in async_execute_cell + self._check_raise_for_error(cell, exec_reply) + File "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/nbclient/client.py", line 740, in _check_raise_for_error + raise CellExecutionError.from_cell_and_msg(cell, exec_reply['content']) +nbclient.exceptions.CellExecutionError: An error occurred while executing the following cell: +------------------ +# Blocking + @timeFunction + def blocking(self, blockSizeMax = 500): + blockSizeMin = 1 + + self.blockSizes = [] + self.meanVec = [] + self.varVec = [] + + for i in range(blockSizeMin, blockSizeMax): + if(len(self.data) % i != 0): + pass#continue + blockSize = i + meanTempVec = [] + varTempVec = [] + startPoint = 0 + endPoint = blockSize + + while endPoint <= len(self.data): + meanTempVec.append(np.average(self.data[startPoint:endPoint])) + startPoint = endPoint + endPoint += blockSize + mean, var = np.average(meanTempVec), np.var(meanTempVec)/len(meanTempVec) + self.meanVec.append(mean) + self.varVec.append(var) + self.blockSizes.append(blockSize) + + self.blockingAvg = np.average(self.meanVec[-200:]) + self.blockingVar = (np.average(self.varVec[-200:])) + self.blockingStd = np.sqrt(self.blockingVar) +------------------ + + File "", line 2 + @timeFunction + ^ +IndentationError: unexpected indent + +IndentationError: unexpected indent (, line 2) + diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter1.ipynb b/doc/LectureNotes/_build/jupyter_execute/chapter1.ipynb index cb673cd61..ebd1120c1 100644 --- a/doc/LectureNotes/_build/jupyter_execute/chapter1.ipynb +++ b/doc/LectureNotes/_build/jupyter_execute/chapter1.ipynb @@ -441,8 +441,8 @@ "name": "stdout", "output_type": "stream", "text": [ - "[ 0.82672272 0.45257846 -3.50535083 -0.60868993 -0.32256204 0.35390727\n", - " -0.8063309 -0.22751293 -0.26205577 -0.65826338]\n" + "[ 1.3280646 -0.56501263 0.21077998 -0.55360402 0.06730225 1.01324127\n", + " -0.21780848 -0.34317041 0.45298301 0.08215028]\n" ] } ], diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter10.ipynb b/doc/LectureNotes/_build/jupyter_execute/chapter10.ipynb index e0377854a..85d81c6c5 100644 --- a/doc/LectureNotes/_build/jupyter_execute/chapter10.ipynb +++ b/doc/LectureNotes/_build/jupyter_execute/chapter10.ipynb @@ -2378,10 +2378,6 @@ "Learning rate = 1.0\n", "Lambda = 10.0\n", "Accuracy score on test set: 0.05\n", - "\n", - "Learning rate = 10.0\n", - "Lambda = 1e-05\n", - "Accuracy score on test set: 0.08888888888888889\n", "\n" ] }, @@ -2389,6 +2385,10 @@ "name": "stdout", "output_type": "stream", "text": [ + "Learning rate = 10.0\n", + "Lambda = 1e-05\n", + "Accuracy score on test set: 0.08888888888888889\n", + "\n", "Learning rate = 10.0\n", "Lambda = 0.0001\n", "Accuracy score on test set: 0.08611111111111111\n", @@ -2396,7 +2396,13 @@ "Learning rate = 10.0\n", "Lambda = 0.001\n", "Accuracy score on test set: 0.08888888888888889\n", - "\n", + "\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ "Learning rate = 10.0\n", "Lambda = 0.01\n", "Accuracy score on test set: 0.08888888888888889\n", diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11.ipynb b/doc/LectureNotes/_build/jupyter_execute/chapter11.ipynb new file mode 100644 index 000000000..344f752d3 --- /dev/null +++ b/doc/LectureNotes/_build/jupyter_execute/chapter11.ipynb @@ -0,0 +1,3310 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Solving Differential Equations with Deep Learning\n", + "\n", + "The Universal Approximation Theorem states that a neural network can\n", + "approximate any function at a single hidden layer along with one input\n", + "and output layer to any given precision. \n", + "\n", + "\n", + "An ordinary differential equation (ODE) is an equation involving functions having one variable.\n", + "\n", + "In general, an ordinary differential equation looks like" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{ode} \\tag{1}\n", + "f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right) = 0\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$.\n", + "\n", + "The $f\\left(x, g(x), g'(x), g''(x), \\, \\dots \\, , g^{(n)}(x)\\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \\ g'(x), \\ g''(x), \\, \\dots \\, , \\text{ and } g^{(n)}(x)$ on the left side of the equality sign in ([1](#ode)).\n", + "The highest order of derivative, that is the value of $n$, determines to the order of the equation.\n", + "The equation is referred to as a $n$-th order ODE.\n", + "Along with ([1](#ode)), some additional conditions of the function $g(x)$ are typically given\n", + "for the solution to be unique.\n", + "\n", + "\n", + "\n", + "Let the trial solution $g_t(x)$ be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + "\tg_t(x) = h_1(x) + h_2(x,N(x,P))\n", + "\\label{_auto1} \\tag{2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set\n", + "of conditions, $N(x,P)$ a neural network with weights and biases\n", + "described by $P$ and $h_2(x, N(x,P))$ some expression involving the\n", + "neural network. The role of the function $h_2(x, N(x,P))$, is to\n", + "ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is\n", + "evaluated at the values of $x$ where the given conditions must be\n", + "satisfied. The function $h_1(x)$ should alone make $g_t(x)$ satisfy\n", + "the conditions.\n", + "\n", + "But what about the network $N(x,P)$?\n", + "\n", + "\n", + "As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.\n", + "\n", + "\n", + "\n", + "For the minimization to be defined, we need to have a cost function at hand to minimize.\n", + "\n", + "It is given that $f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right)$ should be equal to zero in ([1](#ode)).\n", + "We can choose to consider the mean squared error as the cost function for an input $x$.\n", + "Since we are looking at one input, the cost function is just $f$ squared.\n", + "The cost function $c\\left(x, P \\right)$ can therefore be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(x, P\\right) = \\big(f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right)\\big)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If $N$ inputs are given as a vector $\\boldsymbol{x}$ with elements $x_i$ for $i = 1,\\dots,N$,\n", + "the cost function becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{cost} \\tag{3}\n", + "\tC\\left(\\boldsymbol{x}, P\\right) = \\frac{1}{N} \\sum_{i=1}^N \\big(f\\left(x_i, \\, g(x_i), \\, g'(x_i), \\, g''(x_i), \\, \\dots \\, , \\, g^{(n)}(x_i)\\right)\\big)^2\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The neural net should then find the parameters $P$ that minimizes the cost function in\n", + "([3](#cost)) for a set of $N$ training samples $x_i$.\n", + "\n", + "\n", + "\n", + "To perform the minimization using gradient descent, the gradient of $C\\left(\\boldsymbol{x}, P\\right)$ is needed.\n", + "It might happen so that finding an analytical expression of the gradient of $C(\\boldsymbol{x}, P)$ from ([3](#cost)) gets too messy, depending on which cost function one desires to use.\n", + "\n", + "Luckily, there exists libraries that makes the job for us through automatic differentiation.\n", + "Automatic differentiation is a method of finding the derivatives numerically with very high precision.\n", + "\n", + "\n", + "### Example: Exponential decay\n", + "\n", + "An exponential decay of a quantity $g(x)$ is described by the equation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solve_expdec} \\tag{4}\n", + " g'(x) = -\\gamma g(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $g(0) = g_0$ for some chosen initial value $g_0$.\n", + "\n", + "The analytical solution of ([4](#solve_expdec)) is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " g(x) = g_0 \\exp\\left(-\\gamma x\\right)\n", + "\\label{_auto2} \\tag{5}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of ([4](#solve_expdec)).\n", + "\n", + "\n", + "\n", + "The program will use a neural network to solve" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solveode} \\tag{6}\n", + "g'(x) = -\\gamma g(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(0) = g_0$ with $\\gamma$ and $g_0$ being some chosen values.\n", + "\n", + "In this example, $\\gamma = 2$ and $g_0 = 10$.\n", + "\n", + "\n", + "To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x, P) = h_1(x) + h_2(x, N(x, P))\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.\n", + "\n", + "\n", + "\n", + "In this network, there are no weights and bias at the input layer, so $P = \\{ P_{\\text{hidden}}, P_{\\text{output}} \\}$.\n", + "If there are $N_{\\text{hidden} }$ neurons in the hidden layer, then $P_{\\text{hidden}}$ is a $N_{\\text{hidden} } \\times (1 + N_{\\text{input}})$ matrix, given that there are $N_{\\text{input}}$ neurons in the input layer.\n", + "\n", + "The first column in $P_{\\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer.\n", + "If there are $N_{\\text{output} }$ neurons in the output layer, then $P_{\\text{output}} $ is a $N_{\\text{output} } \\times (1 + N_{\\text{hidden} })$ matrix.\n", + "\n", + "Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.\n", + "\n", + "It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of ([6](#solveode)). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \\cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{trial} \\tag{7}\n", + "g_t(x, P) = g_0 + x \\cdot N(x, P)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Reformulating the problem\n", + "\n", + "We wish that our neural network manages to minimize a given cost function.\n", + "\n", + "A reformulation of out equation, ([6](#solveode)), must therefore be done,\n", + "such that it describes the problem a neural network can solve for.\n", + "\n", + "The neural network must find the set of weights and biases $P$ such that the trial solution in ([7](#trial)) satisfies ([6](#solveode)).\n", + "\n", + "The trial solution" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x, P) = g_0 + x \\cdot N(x, P)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{nnmin} \\tag{8}\n", + "g_t'(x, P) = - \\gamma g_t(x, P)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "is fulfilled as *best as possible*.\n", + "\n", + "\n", + "The left hand side and right hand side of ([8](#nnmin)) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible.\n", + "This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero.\n", + "In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network.\n", + "\n", + "This gives the following cost function our neural network must solve for:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P}\\Big\\{ \\big(g_t'(x, P) - ( -\\gamma g_t(x, P) \\big)^2 \\Big\\}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "(the notation $\\min_{P}\\{ f(x, P) \\}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$)\n", + "\n", + "or, in terms of weights and biases for the hidden and output layer in our network:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P_{\\text{hidden} }, \\ P_{\\text{output} }}\\Big\\{ \\big(g_t'(x, \\{ P_{\\text{hidden} }, P_{\\text{output} }\\}) - ( -\\gamma g_t(x, \\{ P_{\\text{hidden} }, P_{\\text{output} }\\}) \\big)^2 \\Big\\}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for an input value $x$.\n", + "\n", + "\n", + "\n", + "If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \\dots, N$, then the *total* error to minimize becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{min} \\tag{9}\n", + "\\min_{P}\\Big\\{\\frac{1}{N} \\sum_{i=1}^N \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2 \\Big\\}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Letting $\\boldsymbol{x}$ be a vector with elements $x_i$ and $C(\\boldsymbol{x}, P) = \\frac{1}{N} \\sum_i \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2$ denote the cost function, the minimization problem that our network must solve, becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P} C(\\boldsymbol{x}, P)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In terms of $P_{\\text{hidden} }$ and $P_{\\text{output} }$, this could also be expressed as\n", + "\n", + "$$\n", + "\\min_{P_{\\text{hidden} }, \\ P_{\\text{output} }} C(\\boldsymbol{x}, \\{P_{\\text{hidden} }, P_{\\text{output} }\\})\n", + "$$\n", + "\n", + "\n", + "For simplicity, it is assumed that the input is an array $\\boldsymbol{x} = (x_1, \\dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills ([9](#min)).\n", + "\n", + "First, the neural network must feed forward the inputs.\n", + "This means that $\\boldsymbol{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further.\n", + "The input layer will consist of $N_{\\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\\text{hidden} }$.\n", + "\n", + "\n", + "For the $i$-th in the hidden layer with weight $w_i^{\\text{hidden} }$ and bias $b_i^{\\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "z_{i,j}^{\\text{hidden}} &= b_i^{\\text{hidden}} + w_i^{\\text{hidden}}x_j \\\\\n", + "&=\n", + "\\begin{pmatrix}\n", + "b_i^{\\text{hidden}} & w_i^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 \\\\\n", + "x_j\n", + "\\end{pmatrix}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "\\boldsymbol{z}_{i}^{\\text{hidden}} &= \\Big( b_i^{\\text{hidden}} + w_i^{\\text{hidden}}x_1 , \\ b_i^{\\text{hidden}} + w_i^{\\text{hidden}} x_2, \\ \\dots \\, , \\ b_i^{\\text{hidden}} + w_i^{\\text{hidden}} x_N\\Big) \\\\\n", + "&=\n", + "\\begin{pmatrix}\n", + " b_i^{\\text{hidden}} & w_i^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 & 1 & \\dots & 1 \\\\\n", + "x_1 & x_2 & \\dots & x_N\n", + "\\end{pmatrix} \\\\\n", + "&= \\boldsymbol{p}_{i, \\text{hidden}}^T X\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The vector $\\boldsymbol{p}_{i, \\text{hidden}}^T$ constitutes each row in $P_{\\text{hidden} }$, which contains the weights for the neural network to minimize according to ([9](#min)).\n", + "\n", + "After having found $\\boldsymbol{z}_{i}^{\\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\\boldsymbol{z})$.\n", + "\n", + "In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "f(z) = \\frac{1}{1 + \\exp{(-z)}}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "It is possible to use other activations functions for the hidden layer also.\n", + "\n", + "The output $\\boldsymbol{x}_i^{\\text{hidden}}$ from each $i$-th hidden neuron is:\n", + "\n", + "$$\n", + "\\boldsymbol{x}_i^{\\text{hidden} } = f\\big( \\boldsymbol{z}_{i}^{\\text{hidden}} \\big)\n", + "$$\n", + "\n", + "The outputs $\\boldsymbol{x}_i^{\\text{hidden} } $ are then sent to the output layer.\n", + "\n", + "The output layer consists of one neuron in this case, and combines the\n", + "output from each of the neurons in the hidden layers. The output layer\n", + "combines the results from the hidden layer using some weights $w_i^{\\text{output}}$\n", + "and biases $b_i^{\\text{output}}$. In this case,\n", + "it is assumes that the number of neurons in the output layer is one.\n", + "\n", + "\n", + "\n", + "The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "z_{1,j}^{\\text{output}} & =\n", + "\\begin{pmatrix}\n", + "b_1^{\\text{output}} & \\boldsymbol{w}_1^{\\text{output}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 \\\\\n", + "\\boldsymbol{x}_j^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Expressing $z_{1,j}^{\\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\boldsymbol{z}_{1}^{\\text{output}} =\n", + "\\begin{pmatrix}\n", + "b_1^{\\text{output}} & \\boldsymbol{w}_1^{\\text{output}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 & 1 & \\dots & 1 \\\\\n", + "\\boldsymbol{x}_1^{\\text{hidden}} & \\boldsymbol{x}_2^{\\text{hidden}} & \\dots & \\boldsymbol{x}_N^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\\boldsymbol{z}_{1}^{\\text{output}}$ the neural network has finished its feed forward step, and $\\boldsymbol{z}_{1}^{\\text{output}}$ is the final output of the network.\n", + "\n", + "\n", + "The next step is to decide how the parameters should be changed such that they minimize the cost function.\n", + "\n", + "The chosen cost function for this problem is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C(\\boldsymbol{x}, P) = \\frac{1}{N} \\sum_i \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In order to minimize the cost function, an optimization method must be chosen.\n", + "\n", + "Here, gradient descent with a constant step size has been chosen.\n", + "\n", + "### Gradient descent\n", + "\n", + "The idea of the gradient descent algorithm is to update parameters in\n", + "a direction where the cost function decreases goes to a minimum.\n", + "\n", + "In general, the update of some parameters $\\boldsymbol{\\omega}$ given a cost\n", + "function defined by some weights $\\boldsymbol{\\omega}$, $C(\\boldsymbol{x},\n", + "\\boldsymbol{\\omega})$, goes as follows:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\boldsymbol{\\omega}_{\\text{new} } = \\boldsymbol{\\omega} - \\lambda \\nabla_{\\boldsymbol{\\omega}} C(\\boldsymbol{x}, \\boldsymbol{\\omega})\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for a number of iterations or until $ \\big|\\big| \\boldsymbol{\\omega}_{\\text{new} } - \\boldsymbol{\\omega} \\big|\\big|$ becomes smaller than some given tolerance.\n", + "\n", + "The value of $\\lambda$ decides how large steps the algorithm must take\n", + "in the direction of $ \\nabla_{\\boldsymbol{\\omega}} C(\\boldsymbol{x}, \\boldsymbol{\\omega})$.\n", + "The notation $\\nabla_{\\boldsymbol{\\omega}}$ express the gradient with respect\n", + "to the elements in $\\boldsymbol{\\omega}$.\n", + "\n", + "In our case, we have to minimize the cost function $C(\\boldsymbol{x}, P)$ with\n", + "respect to the two sets of weights and biases, that is for the hidden\n", + "layer $P_{\\text{hidden} }$ and for the output layer $P_{\\text{output}\n", + "}$ .\n", + "\n", + "This means that $P_{\\text{hidden} }$ and $P_{\\text{output} }$ is updated by" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "P_{\\text{hidden},\\text{new}} &= P_{\\text{hidden}} - \\lambda \\nabla_{P_{\\text{hidden}}} C(\\boldsymbol{x}, P) \\\\\n", + "P_{\\text{output},\\text{new}} &= P_{\\text{output}} - \\lambda \\nabla_{P_{\\text{output}}} C(\\boldsymbol{x}, P)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### The code for solving the ODE" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Initial cost: 367.01\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Final cost: 0.0666807\n", + "Max absolute difference: 0.0437499\n" + ] + }, + { + "data": { + "image/png": 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zMann6MEhhL1DCPmkyt0SUsustK9Y/bZ5dfO1tutB1rJ4rYP0CnAYqd1b00m9w7iYtXs8/0XqEyLDSH1C8cv03zZEvnmkNm7Pk3rh7Enq3WtZXZTONIjULoMbSB2rVeb5jjEuJVW0DiQ1inY3qWN8vlu3ufrNbReRepHoTOoDBr+QekGsm77IE6Q+cTkGeI/UQaaru72RpI5/+YLUE38rUsckbQi3kXrs3wshzCN1QPyO6ftdSPrTr+ldD93S1/mE1Aay7yp+h/VYf2KM75M6+PuNEMJ2K7nI9sCAEML8dPbzYow/xRhnkHrcLyS1i+sS4JAY40p3u8UYB5B6h7sRqReCtRZj7A3cTmrX5WhSjx+kNuylL7u+633p23uO1HFtF5Ca35GkjvHcJf1YrLBT+rGaS6ok1wG2jzEOL3WTs8Nvz+P1l1Xc9XGkjt36jtQxaOen83xAqni+RGoUqS3wx3WdP1IjqbNIjSQ8Repg8xXPzzOBa9Lr7D9IPaarFGN8hdR24tmQ2oX1DannPun1owdwPanHsT2rf36tKH8zQghfpn8+htTxPpNI7bK+Mv14rMxDQMf0c+rV1eVO5xtM6sNJd5J6PEaTOt5qZd4F3gG+J7WbazFl3G26jtuZx9KXK72b8TpSbwpmhxAuWsn1VroOkTpWrzqpbWb/9LyU1QHAiPS6fhup41gXxRhHkdoVekf6dn9P6jRMS1dyG2vaNq9pvtZmPchaIfXGTpKyXwhhC1Iv6tViwueAq8jSI9BPxhjXZpReGZYeqXqS1Kk5fLGuJBzxkpTVQgiHhxCqhdRJS28gdW4mS5cqtfRuu/OABy1dlYvFS1K2O53U7pIfSX0i9M/JxpHKV3pkdzapT9DemmgYbXDuapQkScoQR7wkSZIyJOtOLLYyjRo1iq1bt046hiRJ0hoNGTLklxhj45VNqxDFq3Xr1gwePDjpGJIkSWsUQij9LR+/clejJElShli8JEmSMsTiJUmSlCEV4hgvSZK0/pYtW8aECRNYvHhx0lEqhcLCQlq2bEl+fn6Zr2PxkiSpipgwYQK1a9emdevWhBCSjlOhxRiZMWMGEyZMoE2bNmW+nrsaJUmqIhYvXkzDhg0tXRtACIGGDRuu9eihxUuSpCrE0rXhrMtjafGSJEnKEIuXJEmqcB599FHOPvvsNV5m0qRJv/5+yimnMHLkyLW+rz59+nDIIYes9fVWxuIlSZIqpdLF68EHH6Rjx44JJrJ4SZKkDOvevTtdunShU6dO3H///QDUqlWLK664gm222YZu3boxdepUAN544w123HFHtt12W/bZZ59f/77CvHnzaNOmDcuWLQNg7ty5tGnThhdeeIHBgwfTq1cvOnfuzKJFi9hjjz1+/QrCd955h+22245tttmGvffeG4CBAwey0047se2227LzzjszatSoDT7vnk5CkqQq6Oo3RjBy0twNepsdN6rDlb/vtMbLPfzwwzRo0IBFixax/fbbc+SRR7JgwQK6devGv//9by655BIeeOAB/va3v7HrrrvSv39/Qgg8+OCD3Hjjjdxyyy2/3lbt2rXZY489eOutt+jevTvPPvssRxxxBD169OCuu+7i5ptvpmvXrr+5/+nTp3PqqafSt29f2rRpw8yZMwHYfPPN+fTTT8nLy+ODDz7g8ssv56WXXtqgj5HFS5IkZdTtt9/OK6+8AsD48eP54YcfKCgo+PU4qi5duvD+++8DqXOP/eEPf2Dy5MksXbp0pefMOuWUU7jxxhvp3r07jzzyCA888MBq779///787ne/+/W2GjRoAMCcOXM44YQT+OGHHwgh/DqKtiFZvCRJqoLKMjJVHvr06cMHH3zAF198QY0aNdhjjz1YvHgx+fn5v56eITc3l+XLlwNwzjnn8Je//IVDDz2UPn36cNVVV/3Pbe6yyy6MGTOGPn36UFRUxJZbbrlO2f7+97+z55578sorrzBmzBj22GOPdZ3NVfIYL0mSlDFz5syhfv361KhRg++++47+/fuv8fItWrQA4LHHHlvl5Y4//nh69uzJSSed9Ovfateuzbx58/7nst26daNv3778/PPPAL/uaix5X48++uhazVdZWbwkSVLGHHDAASxfvpwtttiCv/71r3Tr1m21l7/qqqvo0aMHXbp0oVGjRqu8XK9evZg1axbHHHPMr3878cQTOeOMM349uH6Fxo0bc//993PEEUewzTbb8Ic//AGASy65hMsuu4xtt9321xG3DS3EGMvlhjekrl27xhWfQpAkSevm22+/ZYsttkg6Rrl48cUXee2113jiiScyer8re0xDCENijF1XdnmP8ZIkSRXaOeecQ+/evXn77beTjrJGFi9JklSh3XHHHUlHKDOP8ZIkScqQciteIYSHQwjTQgjflPhbgxDC+yGEH9L/1y+v+5ckSco25Tni9ShwQKm//RX4MMbYHvgw/XtWWL5sadIRJElSJVduxSvG2BeYWerPhwErTsLxGNC9vO5/bfR/9HJ+vmFniouKko4iSZIqsUwf49U0xjg5/fMUoOmqLhhCOC2EMDiEMHj69OnlGiq3wca0X/4DX3/wVLnejyRJqtoSO7g+pk4gtsqTiMUY748xdo0xdm3cuHG5Ztn2wJMZFzai7sBbiMWOekmSpPKR6eI1NYTQHCD9/7QM3/9K5eXnM6XzuWxaNIavP3g66TiSJFVaY8aMYYsttuDUU0+lU6dO7LfffixatIg99tiDSy+9lB122IEOHTrw6aefJh21XGT6PF6vAycA16f/fy3D979K2x10MuO/up06A24h7tOTkJObdCRJkspP77/ClOEb9jabbQUHXr/Gi/3www8888wzPPDAAxx99NG89NJLACxfvpyBAwfy9ttvc/XVV/PBBx9s2HxZoDxPJ/EM8AWwWQhhQgjhZFKFa98Qwg/APunfs0JefgGTtjmLTYt+ZtiHzyQdR5KkSqtNmzZ07twZgC5dujBmzBgAjjjiiP/5W2VTbiNeMcZjVjFp7/K6z/W13cGnMf7rO6nd/xbi3j0JOZ5fVpJUSZVhZKq8VKtW7defc3Nzf/0C6xV/z83NLbcvqU6azaKE/PwCJm19FpsW/cQ3Hz2bdBxJklTJWLxK2fbg05kQmlGj/83E4uKk40iSpErEL8kupaCggPFbnsVOw//O8I+fY6u9V7XHVJIkra3WrVvzzTe/fpsgF1100f9cplGjRpX2GC9HvFaiy+/PSI16feGolyRJ2nAsXitRUFDAuE5n0nb5aEZ+8nzScSRJUiVh8VqFLr8/nQk0pbCfo16SpMoj9cUx2hDW5bG0eK1CtWqFjOv0Z9ou/4Fv+76YdBxJktZbYWEhM2bMsHxtADFGZsyYQWFh4Vpdz4PrV2O73/+ZiSPupuCzG4m/O8rzekmSKrSWLVsyYcIEpk+fnnSUSqGwsJCWLVuu1XUsXqtRWFjImI5/ZpeRV/Ptpy+xxe49ko4kSdI6y8/Pp02bNknHqNIcwlmDLoeeySSakP/ZjeDQrCRJWg8WrzUoLCzkpy3OoN2y7/nu05eSjiNJkiowi1cZdD3sTCbRmLxPHfWSJEnrzuJVBoWF1flx8zNot2wU3/d7Jek4kiSpgrJ4lVHXw85iMo3I6XuDo16SJGmdWLzKqHr16vyw2em0W/odP3zxWtJxJElSBWTxWgtdu5/NZBoR+lzvqJckSVprFq+1UKN6Db7vcBrtln7L6P6vJx1HkiRVMBavtdS1+zlMpiHRUS9JkrSWLF5rqWaNGoxqfyrtl4zkxwFvJh1HkiRVIBavddC1+7lMoSHFH1/nqJckSSozi9c6qFWzJt+1PYX2S0bw08C3ko4jSZIqCIvXOtru8HOZSgOKPnLUS5IklY3Fax3VqVWLkZueQvsl3/DzoLeTjiNJkioAi9d6WDHqtdxRL0mSVAYWr/VQt3ZtRmx6Mu0XD2fskHeSjiNJkrKcxWs9del+HlNjfZZ8cK2jXpIkabUsXuupbp3ajNj0T3RYPIxxX76bdBxJkpTFLF4bwHaHn8+0WJ/FH1ybdBRJkpTFLF4bQL06dRje5iQ6LPqaCY56SZKkVbB4bSDbdj+f6bEeC9931EuSJK2cxWsDaVCvLl+3PokOi75iwtD3ko4jSZKykMVrA3LUS5IkrY7FawNqWL8eX29yIh0WDmXi1x8kHUeSJGUZi9cG1vnwC5ge67Lg3X8nHUWSJGUZi9cG1qh+PYa2OoEOC79k0rCPko4jSZKyiMWrHHQ+IjXqNf/dfyUdRZIkZRGLVzlo0qABQ1sdT4cFQ5gy3FEvSZKUYvEqJ50Pv4BfYl3mvuOxXpIkKcXiVU6aNGzIly2Po8OCwUz9pk/ScSRJUhaweJWjbQ7/CzNiHea847FekiTJ4lWumjZqyOAWx9Jh/iCmjvgk6TiSJClhFq9yts0RFzIz1mZOb0e9JEmq6ixe5axZo0YM3OhYOswfyLRvP0s6jiRJSpDFKwO2PvwiZsbazH77mqSjSJKkBFm8MmCjJo0Y0LwXHeYN4Jfv+iUdR5IkJcTilSFbdb+ImbEWMx31kiSpyrJ4ZUjLZo0Z0KwXHeb2Z8aoz5OOI0mSEmDxyqCtDr+IWbEWM95y1EuSpKrI4pVBLZs14YumPekw9wtmOuolSVKVY/HKsC3To16/vP3PpKNIkqQMs3hl2MbNm/JFk2PoMOdzZv7wRdJxJElSBlm8EtDx8IuYHWsy401HvSRJqkosXglovVEzPm9yDO3n9GP26AFJx5EkSRli8UrI5t1To17T3/QTjpIkVRUWr4Rs2qI5/Rr/gfazP2P2j4OSjiNJkjLA4pWgLQ67iDmxBtPfvDrpKJIkKQMsXgnatFULPmv0B9rP+pS5Pw1OOo4kSSpnFq+EbZYe9Zr6hqNekiRVdhavhLXbuCWfNTya9rP6Mu9nR70kSarMLF5ZoP1hFzE31mDKG37CUZKkyszilQU6bNKKvg2Oov3MT5j385dJx5EkSeXE4pUl2h16SXrUy2O9JEmqrCxeWWLzNq34pP6RtJ/Zh3ljhyYdR5IklQOLVxZpe+jFzI3Vmfq6o16SJFVGFq8s0nHTTfik/pG0m/ExC8Z9lXQcSZK0gVm8skzbQy5hXqzO5Nf9hKMkSZWNxSvLdGy3CX3qHUG7Xz5k4fhhSceRJEkbkMUrC7X5fWrUa5LHekmSVKlYvLLQlu1a83Hdw2k3/QMWThiedBxJkrSBWLyyVOtDLmZ+LHTUS5KkSsTilaW27rApH9U9nE2nfcCiid8kHUeSJG0AFq8stvHBF7MwVmPSa456SZJUGVi8sljnzdryUZ3utJn2Posnjkg6jiRJWk8WryzX6uCLWRQLmOixXpIkVXgWryy37ebt+LD2YbSZ+h5LJo9MOo4kSVoPFq8KoMVBqVGvCa866iVJUkVm8aoAunTswAe1DqXN1HdZMvnbpONIkqR1ZPGqIJofdDGLYwET/ISjJEkVlsWrgti+Ywfer/V7Wk9x1EuSpIrK4lVBhBBofuDFLIn5THz9mqTjSJKkdWDxqkC277QZ79c8hE0mv8PSqd8lHUeSJK0li1cFEkKgyQGXsDTmMeG1fyYdR5IkrSWLVwXTbavNeLfGIWwy6W2WTv0+6TiSJGktWLwqmBACjfe/mKUxz2O9JEmqYCxeFdDO22zBuzUOZuOJb7Fs2g9Jx5EkSWVk8aqAQgg02u8ilsVcJjjqJUlShWHxqqB26dyJd6ofTKsJb7Js+uik40iSpDKweFVQIQQa7Hsxy2Oun3CUJKmCsHhVYLtt14l3qh9Eqwmvs3z6j0nHkSRJa2DxqsBCCNTb56LUqNfrjnpJkpTtLF4V3O5dtuKdwgNpNf41R70kScpyFq8KLoRA3X0uTI16vfGvpONIkqTVsHhVArt32Ybe1Q6g1bhXKZrxc9JxJEnSKli8KoGcnEBtj/WSJCnrWbwqiT27bkPvavvRYqyjXpIkZSuLVyWRkxOoudfFFMXAhNc91kuSpGxk8apE9tqhM70L9qPF2Fconjkm6TiSJKkUi1clkpsTqL7XRY56SZKUpSxelcw+O25H74J92WjMyxTPHJt0HEmSVILFq5LJzQkU7nkhRTEw0fN6SZKUVSxeldA+O3bhnfx9afazo16SJGUTi1cllJebQ/4eFxJjZOKb/046jiRJSrN4VVL77dSF3vn70uynl4izxyUdR5IkYfGqtPJyc8jdPT3q9ca1SceRJElYvCq1/Xfuytv5e9P0xxeIs8cnHUeSpCovkeIVQrgghDAihPBNCOGZEEJhEjkqu/zcHHJ3W3Gsl6NekiQlLePFK4TQAjgX6Bpj3BLIBf6Y6RxVxQG77kDvvL1pOvp5R70kSUpYUrsa84DqIYQ8oAYwKaEclV5+bg787kKKY2DqixcnHUeSpCot48UrxjgRuBkYB0wG5sQY3yt9uRDCaSGEwSGEwdOnT890zErloF134LnCHjSb0JvFoz5IOo4kSVVWErsa6wOHAW2AjYCaIYRjS18uxnh/jLFrjLFr48aNMx2zUsnPzaFjj38wprgpC165AJYvSTqSJElVUhK7GvcBfo4xTo8xLgNeBnZOIEeV0rVdcz5qcxENF49j2nu3JB1HkqQqKYniNQ7oFkKoEUIIwN7AtwnkqHIOP/pEPmIH6g681a8SkiQpAUkc4zUAeBH4EhieznB/pnNURfVrFrBo739THCMTnzs/6TiSJFU5iXyqMcZ4ZYxx8xjjljHG42KMHnSUIQftuj2v1ulJq6kfMWfYW0nHkSSpSvHM9VVMCIHte/6DH+NGLHvjIli2OOlIkiRVGRavKqhd84YM6XQ5jZZNYpzf4yhJUsZYvKqoQw/vyYe5u9Js2N0smTY66TiSJFUJFq8qqjA/l+qHXMfSmMvkZ86FGJOOJElSpWfxqsJ23nZr3mt8Eq1n9WPaoJeSjiNJUqVn8aridul1BT/EVuS8exlxyfyk40iSVKlZvKq4pvVr8+P2V9OoaBqjX7om6TiSJFVqFi+x70FH8FG1vWj9/UPMmzAy6TiSJFVaFi+RmxNofuRNLIoFTHvuHA+0lySpnFi8BMAWHdrxWaszaDtvMGP6PpV0HEmSKiWLl361W89LGBXaUKvP31m+cE7ScSRJqnQsXvpV7RrVmbH7tTSKM/n2ub8lHUeSpErH4qXf2Gn3A/mk1gFsPuYppv84NOk4kiRVKhYv/UYIgbZ/vJn5VGfWC57RXpKkDcnipf/RsmUrvupwHh0WD2PEuw8kHUeSpErD4qWV2qXHBXyX055m/f/Norkzk44jSVKlYPHSShUU5LP8gJuoH+cw4ulLk44jSVKlYPHSKm25w570b3AY205+gbEj+icdR5KkCs/ipdXa4tibmBNqs/jV8ykuKko6jiRJFZrFS6tVv2ETRm99CZst+5Yhr92ZdBxJkio0i5fWqOthZ/JdfkfaDruJWb9MTTqOJEkVlsVLa5STm0th91upG+fz7VMXJx1HkqQKy+KlMmndaUeGNj+abjNfZ/jAj5KOI0lShWTxUpl16nkDM0M9Ct65mKVLlyUdR5KkCsfipTKrXqc+U7v9jc2KR/P5C7ckHUeSpArH4qW10mn/kxlVvTOdv7+dCRPGJR1HkqQKxeKltRMCDXrcTk0W8+MzFxP9Em1JksrM4qW11njTbfi29bHsvuAdvujzdtJxJEmqMCxeWicd//gvpuc0otEnlzNv4aKk40iSVCFYvLRO8qrXYcEe19CBMXz29PVJx5EkqUKweGmdtd6tJ6Nr78Cu4+9j5Pejko4jSVLWs3hp3YVAs2PuoFpYxtQXL6Go2APtJUlaHYuX1kutjTZnzGYns+fSPrz31otJx5EkKatZvLTe2h95JdNzm9Ju8NVMnTUv6TiSJGUti5fWWyioSfEB19M+jKffU/9KOo4kSVnL4qUNoun2R/Bzg93Yf/ojfDH066TjSJKUlSxe2mA2OuZ28kJk0Zt/ZfGyoqTjSJKUdSxe2mCqNd6UqducyV5Fn/PGy08lHUeSpKxj8dIGtfEhl/FLfgu6jLyW0ZN/STqOJElZxeKlDSu/kPxDb2HTMJlBT//TL9GWJKkEi5c2uLpbHcj4pvvQfe7T9P5sYNJxJEnKGhYvlYsWf7yVEALVP7yCWQuWJh1HkqSsYPFSucip34q5O/6FPRnEq88/nHQcSZKygsVL5abJvn/hl8JN2OvnWxg8elLScSRJSpzFS+Unr4BaR9zKJjnT+PaFq1m6vDjpRJIkJcripXJV2GEvprQ6mKMXv8QL7/dNOo4kSYmyeKncNetxC8U5+bTsfyXjZyxIOo4kSYmxeKn81WnO0t0uZffwFa8++4Dn9pIkVVkWL2VE3d3PYmbNdhw+7Q4++PqnpONIkpQIi5cyIzefOkfdRsvwCxNf/xfzlyxPOpEkSRln8VLG5LXZlRntjqBn0Ws89vp7SceRJCnjLF7KqIbdb6Aotzqdh/+LbybMTjqOJEkZZfFSZtVqAnv9nV1yRtD7ubspKvZAe0lS1WHxUsZV3/lUZtfdguPm3s/z/UYmHUeSpIyxeCnzcnKpe9TtNAuzWPLBtUybtzjpRJIkZYTFS4kIrXZgbseeHMvbPPjSW0nHkSQpIyxeSkydg//Fsvza7P3TjfQdNS3pOJIklTuLl5JTsyF5+13Fjjnf8elLd7J4WVHSiSRJKlcWLyUqv+uJzGu4DacteYSH3h+adBxJksqVxUvJysmh9pG30TDMo/YXNzJ62vykE0mSVG4sXkreRtuyZJsT6ZX7Hg+98Kpfoi1JqrQsXsoK1Q+4kmUF9egx9b+88uX4pONIklQuLF7KDtXrU3Dgv9kuZzTfvHk3sxcuTTqRJEkbnMVLWSOnc08WNtues4uf4PY3ByUdR5KkDc7ipewRAjW630q9sIi2w25hyNiZSSeSJGmDsngpuzTbkqLtT+WYvI947IWXWVZUnHQiSZI2GIuXsk7+3lewtLARp8y9k0c+HZ10HEmSNhiLl7JPYR2qHXQdW+f8zOSP7mXCrIVJJ5IkaYOweCkrha2OYnHLXTg/PMvNr3yedBxJkjYIi5eyUwgUHvZfauUsYeefbufdEVOSTiRJ0nqzeCl7Nd4Mup3J0Xmf8PKrL7JgyfKkE0mStF4sXspquXtcytIazTl/yX3c9v7IpONIkrReLF7KbtVqUXDIDWyRM46i/g8wctLcpBNJkrTOLF7KflscyrI2e3JB3gvc9OInFBf7JdqSpIrJ4qXsFwL5h9xC9ZzlHDr9Hp4ZNC7pRJIkrROLlyqGhm3J2eU8Ds/tx4e9X2L6vCVJJ5Ikaa1ZvFRhhN0uZFntVlwWH+S6N4clHUeSpLVm8VLFUVCD/ENuon2YSKNvHqLf6F+STiRJ0lqxeKli2exAitofwPn5r3Dry31YvKwo6USSJJWZxUsVTu5BN1AtN3LCvPu595Mfk44jSVKZWbxU8dRvTe7vLuaQ3AF83edlfv5lQdKJJEkqE4uXKqadz2F5vTZcmfsoV70yhBg9t5ckKftZvFQx5ReSd8jNtA6T2XLME7z+9aSkE0mStEYWL1Vc7fYhbnEo5+a/xoNv9GHOwmVJJ5IkabUsXqrQwgHXkZeXyzlLH+Jfb/kl2pKk7GbxUsVWtyW5e1zKfrlDmDH0dZ4fPD7pRJIkrZLFSxVftzOJjTfnlsKHue3VzxgxaU7SiSRJWimLlyq+vALCUY9QL2cRt+XfyVlPDGLOIo/3kiRlH4uXKoemHQmH/Ieu8Rt6zH+CC5//muJiTzEhScouFi9VHp17wrbHcVbuqywf9S739f0p6USSJP2GxUuVy0E3EZtuyZ2F9/LUu5/x+Y9+kbYkKXtYvFS55FcnHP04NfOKub/6nfzl6YFMmbM46VSSJAEWL1VGDdsSut9Nx+LvOXP545z99JcsKypOOpUkSRYvVVIdD4Md/8zxoTeNx7/D9b2/SzqRJEkWL1Vi+14DLbfn1sIH+Kjf57w1bHLSiSRJVZzFS5VXXgEc9QgF1arxcM07+fuLg/hx+vykU0mSqjCLlyq3eq0IRzxAm+U/84/cRzjjiSEsWLI86VSSpCrK4qXKr/2+sNtFdI8fsc2Mt7js5eHE6MlVJUmZZ/FS1bDn5dB6N66r9iijhvXnif5jk04kSaqCLF6qGnJy4ciHyKtRj8dq3sl/3xzM0HGzkk4lSapiLF6qOmo3JRz1ME2LJnFz4cOc+eQQZsxfknQqSVIVYvFS1dJ6V8Jef2fvon4cuOgNzn/uK4r8Mm1JUoZYvFT17HI+dDiAK/KeZO7o/tz2wfdJJ5IkVREWL1U9OTnQ/R5y6jTnkVp389hHX/Hxd9OSTiVJqgIsXqqaajQg9HiM+kUzuL/2A1zw7JeMn7kw6VSSpEoukeIVQqgXQngxhPBdCOHbEMJOSeRQFdeyC2H/a9lx2SBO5DXOfOpLFi8rSjqVJKkSS2rE6zbgnRjj5sA2wLcJ5VBVt8Op0OlwzuNZakzqz9VvjEw6kSSpEst48Qoh1AV+BzwEEGNcGmOcnekcEgAhwKF3EBq25aFad/PBwGG8OGRC0qkkSZVUEiNebYDpwCMhhKEhhAdDCDVLXyiEcFoIYXAIYfD06dMzn1JVR7XacPTj1IwLebTOffz9la8ZOWlu0qkkSZVQEsUrD9gOuCfGuC2wAPhr6QvFGO+PMXaNMXZt3LhxpjOqqmnakXDIf+i09GsuLniJM58awtzFy5JOJUmqZJIoXhOACTHGAenfXyRVxKRkde4J2x7Hn4pfYtPZX3DR81/7ZdqSpA0q48UrxjgFGB9C2Cz9p70Bj2hWdjjoJmi6FXfXuJdvRo7g/r4/JZ1IklSJJPWpxnOAp0IIw4DOwLUJ5ZB+K786HP0Y1UIxT9W7h/+88w1f/Dgj6VSSpEoikeIVY/wqffzW1jHG7jHGWUnkkFaqYVtC97tos/hbrq31Auc8M5RpcxcnnUqSVAl45nppZToeBjv+mSOXvcEuSz7jrKe/ZFlRcdKpJEkVnMVLWpV9r4GW23NztfuZPnYkN77zXdKJJEkVnMVLWpW8AjjqEfLzC3i27j08/uko3h4+OelUkqQKzOIlrU69VnDEAzRbPJo76z3NJS8O46fp85NOJUmqoCxe0pq03xd2u4h9F7/HETmf8Ocnv2Th0uVJp5IkVUAWL6ks9rwcWu/GlTkPEaaP4IpXvvHkqpKktVbm4hVCqB9C6BRC2DSEYGFT1ZKTC0c+RG71ujxT9x7eGzqaJweMSzqVJKmCWW2BCiHUDSFcHkIYDvQH7gOeB8aGEF4IIeyZiZBSVqjdFI56mHqLx/Nwgye45o1v+Gr87KRTSZIqkDWNXL0IjAd2izFuFmPcNX3i01bA9cBhIYSTyz2llC1a70rY6+/suLAPZ1T/mLOe+pKZC5YmnUqSVEHkrW5ijHHf1UwbAgzZ4ImkbLfL+TB+ABeMfpTP57fmvGdr8uhJO5CbE5JOJknKcmU6Vqv0qFYIITeEcGX5RJKyXE4OdL+HnNrNeaz2XQz7YQy3f/hD0qkkSRVAWQ+S3zuE8HYIoXkIoROp471ql2MuKbvVaAA9HqXmkuk83ehR7vhoFH1GTUs6lSQpy5WpeMUYewKPAcOBt4HzY4wXlWcwKeu17ELY/1o6zf+cK+q+z/nPfcWEWQuTTiVJymJl3dXYHjgPeAkYCxwXQqhRnsGkCmGHU6HT4fxpyZN0LhrBmU99yZLlRUmnkiRlqbLuanwD+HuM8XRgd+AHYFC5pZIqihDg0DsIDdpwb/W7mDRhHNe8MTLpVJKkLFXW4rVDjPFDgJhyC3B4+cWSKpBqteHoxylcPp8XmzzEMwPG8PKXE5JOJUnKQms6gequADHGuaWnxRi/DyHUCSFsWV7hpAqjaSc4+BZazx3MTY3e4vJXhvPdlP952kiSqrg1jXgdGUL4PITwjxDCwSGEHUIIvwsh/CmE8ATwJlA9Azml7LdtL9j2WI6c/wz7Fwznz09+ydzFy5JOJUnKIqstXjHGC4BDgMlAD+Aa4AKgHXBfjPF3MUaP9ZJWOOhmaLolt+TdxbKZ47jkhWF+mbYk6VdrPMYrxjgTqAMMA94HPgN+ATYLIXQu13RSRZNfHY5+nLxYxMuN7ufDERN48NOfk04lScoSZT24vgtwBtAc2Ag4HTgAeCCEcEk5ZZMqpoZt4bA7aTJ3OPc1fZXr3/mOAT/NSDqVJCkLlLV4tQS2izFeFGO8kFQRawL8DjixnLJJFVen7rDjGew152WOqzOUs58ZyrS5i5NOJUlKWFmLVxNgSYnflwFNY4yLSv1d0gr7/hNadOXvRXfTcPF4zn5mKMuLipNOJUlKUFmL11PAgBDClekvx+4HPB1CqAl4tkhpZfIKoMej5Obl81z9e/j65ync9O6opFNJkhJU1u9q/CdwGjA7/e+MGOM1McYFMcZe5RdPquDqtYIjHqDu3FE8tdGL3Nf3J975ZnLSqSRJCckr6wVjjIOBweWYRaqc2u8Lu11I109v4fzGHbj4hTw2a1aHNo1qJp1MkpRhZd3VKGl97HE5tN6Ncxfdw+Y54/jzk0NYtNQv05akqsbiJWVCbh4c+RA5hXV4rPbdTJg6jSteGe7JVSWpirF4SZlSuykc+RA15o3hpRbP8vLQCTw9cFzSqSRJGWTxkjKpzW6w19/Z7Jf3+edGX3D16yMZNmF20qkkSRli8ZIybZfzocMBHDv7PnarOY4/P/klsxYsTTqVJCkDLF5SpuXkQPd7CLWbc0/B7SydN4Pzn/uK4mKP95Kkys7iJSWhRgPo8SgFC6fyykaP0/f7qdzx0eikU0mSypnFS0pKyy6w/7W0nN6XOzb+lFs//J5Pvp+edCpJUjmyeElJ2uFU6HQ4B09/kCMbjuX8Z4cycfaipFNJksqJxUtKUghw6B2EBm24vvhW6hbN4synvmTJck+uKkmVkcVLSlq12nD04+QtnctLTR5m+PiZ/OvNb5NOJUkqBxYvKRs07QQH30LD6f15YtMPeaL/WF4dOjHpVJKkDcziJWWLbXvBtseyy6RHOLX5T1z28nBGTZmXdCpJ0gZk8ZKyyUE3Q9MtuWzRf2hbbTZ/fnII8xYvSzqVJGkDsXhJ2SS/Ohz9ODnFy3im3r1MmjmXS14c5pdpS1IlYfGSsk3DtnDYndT+ZSgvtO1N72+m8NBnPyedSpK0AVi8pGzUqTvseAZbjX+ayzb5nut6f8fAn2cmnUqStJ4sXlK22vef0KIrp82+hW715nDW018ydsaCpFNJktaDxUvKVnkF0ONRQk4eD1e/nZzli+j5wADPbC9JFZjFS8pm9VrBEQ9QbcZIerd/k7mLl9Lrgf5Mnbs46WSSpHVg8ZKyXft9YbeLaPD9s7y3zWdMm7eEXg8O4Jf5S5JOJklaSxYvqSLY8wrY7gSaf30H727XnwmzFnLsgwOYvXBp0skkSWvB4iVVBDk5cMitsE1PWn31X97u8iU/TV/A8Q8PZK4nWJWkCsPiJVUUOTlw2J2w5VFs+tWNvLH9MEZOmstJjwxiwZLlSaeTJJWBxUuqSHJy4fD7YItD2eyra3l1h28ZOm4Wpzw2mMXLipJOJ0laA4uXVNHk5sFRD8NmB7HlV9fw0o6j6f/zDE57YghLllu+JCmbWbykiig3H3o8Cu32ZduvruTZHX6i7/fTOfvpoSwrKk46nSRpFSxeUkWVVw3+8CRsujs7DvsHj+8wjvdHTuX8575iueVLkrKSxUuqyPIL4Y/PwMY787vhV/BA14m8NWwyl7w4jOLimHQ6SVIpFi+poiuoAT2fg5bbs+/Iy7hju8m8PHQiV7z6DTFaviQpm1i8pMqgWi3o9QI034ZDRl3GTZ2n8czAcVz9xkjLlyRlEYuXVFkU1oFjXyI03pyjfriUf241nUc/H8ON746yfElSlrB4SZVJ9fpw/GuEhu049ue/ckWnGdzT50du/3B00skkSVi8pMqnRoNU+aq/CaeM+ysXbTGL/37wPfd98mPSySSpyrN4SZVRrcap8lW7GWdN/Ctnd5jDdb2/49F+PyedTJKqNIuXVFnVbgYnvEGo2ZALp/2VU9rN4ao3RvLMwHFJJ5OkKsviJVVmdVukyle1Olwx43KObTOPy18ZzitDJySdTJKqJIuXVNnV2zhVvvKq8885V3BUq/lc+PzXvDVsctLJJKnKsXhJVUGDNnDim4ScXG5c8HcO3mgB5z07lPdHTk06mSRVKRYvqapo2BaOf50Qi7ht6T/Yu+kCznrqSz75fnrSySSpyrB4SVVJk83hhNfJWb6Ee4quYqeGCznt8cF88eOMpJNJUpVg8ZKqmqad4PhXyVk6j4fD1WxXbyEnPzaIIWNnJp1Mkio9i5dUFTXfBo59hdxFM3ki/99sXnMBJz48iGETZiedTJIqNYuXVFW17ALHvkTe/Ck8V/06Nqm+gOMeGsi3k+cmnUySKi2Ll1SVbbwj9HqB/LkTeLnmjTTNW8ixDw5g9LR5SSeTpErJ4iVVda13gWOeoWD2T7xR9ybqsICeDwxgzC8Lkk4mSZWOxUsStN0T/vg01WZ9z9sN/0vB8vn0enAAE2YtTDqZJFUqFi9JKe33gaMfp/ov3/Buk9tZvngevR4cwJQ5i5NOJkmVhsVL0v/b7EA46mFqTvuKD5rdxYL58+j1YH9+mb8k6WSSVClYvCT9VsfD4Ij7qT11EB9udC+/zJ7LsQ8OYNaCpUknk6QKz+Il6X9tdRQcdjd1J3/Ox60eZPwvszn+4YHMWbQs6WSSVKFZvCStXOdj4Pe30WDSJ/TZ5FF+nDKTkx4ZyPwly5NOJkkVlsVL0qp1OQEOupnGEz/ko9ZP8s2EmZz86CAWLS1KOpkkVUgWL0mrt8OpsP91NJv4Lh+1eZrBY37htCcGs3iZ5UuS1pbFS9Ka7XQm7HM1LSe+zQdtX+CzH6Zx9tNfsnR5cdLJJKlCsXhJKptdz4c9r6DNhNd4r+3LfPjtFM5/bijLiyxfklRWeUkHkFSB7H4JLF9C+09v5q22eRw0/DCq5eVyc49tyM0JSaeTpKxn8ZK0dvb6GxQtpePnt/Nq2zy6Dz2Yank5XHv4VuRYviRptSxektZOCLDvNVC0jM4D7uG5TfP4w6D9qZaXw1WHdiIEy5ckrYrFS9LaCwEOuA6KlrLj4Id4vE0ex3+xD4X5ufz1wM0tX5K0ChYvSesmBDjoZihayu+GPsyDrfM4pS9Uy8/lL/t2SDqdJGUli5ekdZeTA7+/DYqWsc+w+7mzdS5nfwiF+TmcuUe7pNNJUtaxeElaPzm5cNhdULSUQ0bcw5KN87jwHSjMy+VPu7ZJOp0kZRWLl6T1l5sHR9wPRUs58rs7WNwylyvehML8XHruuHHS6SQpa3gCVUkbRm4+HPUIdDiAXr/cyj9aDOGKV4fz0pAJSSeTpKxh8ZK04eQVQI/HoO3enDTjP1zS7CsufvFr3vh6UtLJJCkrWLwkbVj5hfDHpwhtduOM2TdzTpNvOP+5r3hvxJSkk0lS4ixekja8/OpwzLOEVt04f+4NnNJoBGc/PZQ+o6YlnUySEmXxklQ+CmpCr+cJLbrw1/k30LP+SE5/Ygifj/4l6WSSlBiLl6TyU602HPsiodmWXLnoeo6sM4qTHxvM4DEzk04mSYmweEkqX4V14diXCY03499Lr+Ogmt9z4iOD+Hr87KSTSVLGWbwklb8aDeC41wgNNuWm5deye+EPHP/wQEZOmpt0MknKKIuXpMyo2RCOf42ceq24o/hadswbzbEPDeCHqfOSTiZJGWPxkpQ5tZrA8a+TU7sp94Zr2ZIf6fngAH7+ZUHSySQpIyxekjKrTnM44Q1yajTgkbxraVv0E70e6M/4mQuTTiZJ5c7iJSnz6raEE94gt1ptnsy/lmZLfqLng/2ZMMvyJalys3hJSkb9TeCE18krKOS5wuupv3AM3e/qx5Cxs5JOJknlxuIlKTkN28IJb5Cfm8PLNa6nU/5kjrm/v1+sLanSsnhJSlaj9qmRrxB5tOgyTms6igtf+Jrren9LUXFMOp0kbVAWL0nJa7IFnNaH0Kg9F868igdbf8T9n4zm9CcGM3/J8qTTSdIGk1jxCiHkhhCGhhDeTCqDpCxStwWc1Juw9R/YZ8qD9NnkUQaMGs+Rd3/uJx4lVRpJjnidB3yb4P1Lyjb51eHwe2H/a9lk2kf0b3I9uXPGcthd/Rj4s9/vKKniS6R4hRBaAgcDDyZx/5KyWAiw01lw7MvUXDyVN6r9jb0KRtDrwf48N2hc0ukkab0kNeJ1K3AJULyqC4QQTgshDA4hDJ4+fXrGgknKEm33hNM+JrfuRty0+GquatyHS18axj/fHOlB95IqrIwXrxDCIcC0GOOQ1V0uxnh/jLFrjLFr48aNM5ROUlZpsCmc/D5h84PpNfs+XtvoCZ78bBR/enQQcxcvSzqdJK21JEa8dgEODSGMAZ4F9gohPJlADkkVQbVa0ONx2PMKtpn5Dp83vYkfR4/i8Lv6McbveJRUwWS8eMUYL4sxtowxtgb+CHwUYzw20zkkVSA5ObD7JfDHZ2i4aBwf1bmKVvOH0f3ufnz+4y9Jp5OkMvM8XpIqjs0PglM/pKBGXR4J13BCwccc/9BAnuw/NulkklQmiRavGGOfGOMhSWaQVME03gxO/Yiw6e5csPhu7m/wFFe/+hVXvvYNy4tW+XkdScoKjnhJqniq14Oez8Mu57PX/Df5sNEtvPXFME58ZBBzFnrQvaTsZfGSVDHl5MK+V8ORD7Hx4u/pW+8q5v88iO539+PH6fOTTidJK2XxklSxbXUUnPwuNaoV8HLhNeyy4EMOv6sfn/7g+f8kZR+Ll6SKr/k2cFofclpuz7/i7fyj2tOc/Eh/Hu33MzF6slVJ2cPiJalyqNkIjn8VdjiNo5a8yit1/st/3xjIFa9+wzIPupeUJSxekiqP3Hw46CY49A46LhtOn7rXMHhgP457aACzFixNOp0kWbwkVULbHU848S3q5y/j7RpX0XD8exx2Vz9+mDov6WSSqjiLl6TKqdUOcFof8pp15K7c/3Dsoqc58u7P+HjUtKSTSarCLF6SKq86G8GJb0PnXpwWn+e+gls599G+PPjpTx50LykRFi9JlVt+IRx2FxxwA92WD+Ld2tfwxNsfc+lLw1iyvCjpdJKqGIuXpMovBOh2BuG4V2ieO493a1zJlC/f4tgHBzBj/pKk00mqQixekqqOTXcnnPYxhY024bGCm9h+4pMcesdnfDdlbtLJJFURFi9JVUv91nDye4SOh3JJ7lP8Y9l/6HX3x7w/cmrSySRVARYvSVVPQU3o8Sjs/Q/2K+7H8/lXc9UT73BPnx896F5SubJ4SaqaQoDdLiT0fI5N86bxTo1/8PG7r3Dh81+zeJkH3UsqHxYvSVVbh/0Jp35MrXqNeabwWmoMe5Rj7v+CafMWJ51MUiVk8ZKkRu0Jp35Ibrt9+Ff+I/Scegs97ujDNxPnJJ1MUiVj8ZIkgMK6cMwzsNtF9Mj5iDuX/YMz732bd76ZnHQySZWIxUuSVsjJhb3/Dj0eZcvccbySfwX3PPU8d3z4gwfdS9ogLF6SVFqnwwmnvE/92jV5sfBfjPnwQc599isPupe03ixekrQyzbYi57Q+5LXuxi0F97LtiOs55t7PmDrXg+4lrTuLlyStSs2GhGNfgR3/zJ/y3uHSXy7n2Dt6M2zC7KSTSaqgLF6StDq5eXDg9XDY3eyQ9wOPLb+Ev937LG98PSnpZJIqIIuXJJXFtr3IOak3TWvm8HzeP+j93D385/3vKS72oHtJZWfxkqSyatmF3NM/oaDl1txdcDsFn/yLs58azMKly5NOJqmCsHhJ0tqo3YycE98ibnscZ+e9xpHfX8wJd3/A5DmLkk4mqQKweEnS2sqrRjj0DjjoZvbMG84Nsy7gnDueZ+i4WUknk5TlLF6StC5CgB1OJeeE19i4+mIeXf5X7n7gHl4dOjHpZJKymMVLktZH613JO+MTCpu05b7cG/n+xau5sfe3HnQvaaUsXpK0vuptTN4p7xE7Hs4l+c/R8fPzOefxz1iwxIPuJf2WxUuSNoSCGuT2eJi4z9UcnDuAs346izPvfIUJsxYmnUxSFrF4SdKGEgJh1/MJvV6kQ7WZ3DrvAv55x30MHjMz6WSSsoTFS5I2tPb7kHfGJ9Ss34y7i6+h90NX8cKgcUmnkpQFLF6SVB4atqXg9I8obrcff899DF47i3+9+iWLlhYlnUxSgixeklReCuuQ3/MZina7hB55fTn8yxM597+PuetRqsIsXpJUnnJyyN37CjjmWdrXWMg9iy5kwIPnc+3rQx39kqogi5ckZcJmB1Jw3iDi1n/krLzX+MPgY7j4P/cx8GdHv6SqxOIlSZlSvT75R9wDx75Mi9o53L74cr59+HSue3WgX7QtVREWL0nKtHZ7U3juQIq6nspxue9z3NBj+McttzPgpxlJJ5NUzixekpSEarXIP+Qmcv70Dg3q1uXmJVcz/pETueHlLxz9kioxi5ckJWnjbtQ453OW7fwXDs/tx5++/gPX3XwD/R39kioli5ckJS2/kPz9riT39D5Ub9iKfy69kVmP/IGbXuzj9z1KlYzFS5KyRfOtqXXWJyzd8x/sm/c1pw0/httvuZLPR09POpmkDcTiJUnZJDePgt0vJO+sLwhNO3HZ0jtZ/tjh3PLc+45+SZWAxUuSslGjdtQ54z2W7n8TO+b/yBkje/HAzZfw+fdTk04maT1YvCQpW+XkULDTaVQ7dyDLWu7E+cseouDJQ7jt2TeY7+iXVCFZvCQp29VrRb1TXmXpoffSMX8KZ3x7Ik/ddA6ffz856WSS1pLFS5IqghAo2O4YalzwJQva7M/py5+m/pP7cedTLzj6JVUgFi9JqkhqNabBiU+z9KgnaFltIWd8fxqv3HgKn387IelkksrA4iVJFVDBlodS+y9DmN2hB8cVvUKzZ/bmvscfZ97iZUlHk7QaFi9Jqqiq16NRr/tZ2vMV6lfP4fSfzuG9G4+l34ifkk4maRUsXpJUwRV02Iv6Fw5maqeTObz4XVo/vw+PPHIfcx39krKOxUuSKoOCmjTt8R+WnfgOBTXqcNLYS+h345H0GzYq6WSSSrB4SVIlUq11NxpfOIAp257HvsWfsdlL+/DUQ7cyd9HSpKNJwuIlSZVPXjWaHXYNRaf0YVmtFvQafyVDbzqYfkOHJ51MqvIsXpJUSVVruTXN//IZk3a4nG7FX7HVq/vx/P3/Zs5CR7+kpFi8JKkyy81jo4MuhTM/Z3bdzTl60o18f9NefDF4cNLJpCrJ4iVJVUC1Ju3Z+PwPmbDLtXSMo+n8xoG8ds8VzJm/OOloUpVi8ZKkqiInh5b7nkXeuQOZ1GB7Dpt6J+Nv2ZUBAz5LOplUZVi8JKmKqdZgY9qe+xbj9ryDlkxl27cP5Z27zmfOvAVJR5MqPYuXJFVFIbDx7sdT4/wh/NR4bw6Y/gjTbunGwM/eTzqZVKlZvCSpCiuo24TNz36Bsfs9RL2wgC7v9+Cj209n9pzZSUeTKiWLlySJTXY+iroXDmFEs8PZa+azzPvvjgzu83rSsaRKx+IlSQKgoFZ9tv7zI/x8yLPk5gS69jmOz289llkzf0k6mlRpWLwkSb/RpuuBNLp4MINbHMuOs95k6e3b8+X7zyQdS6oULF6SpP9RUL0WXU+9i7GHv8ainNps1+8MBt9yBLOnT0o6mlShWbwkSau0aefdafHXgfTf+HS2ntuHeNeOfP32AxBj0tGkCsniJUlarfyCQrr96UYmHP0O0/Kasc3Aixh+84HMmjIm6WhShWPxkiSVyaaddmDTSz/n000voN38IeTf241vXvsvFBcnHU2qMCxekqQyy8/PZ7fjr2JSr4/4Ma89Ww69iu9v2pNZ479NOppUIVi8JElrrW2Hrej41z581OFvNF84iuoP/Y6RL/6TuHxJ0tGkrGbxkiStk/y8XPbqeTHTjuvL1wXb0vGbm5l27Vb89OGDUFyUdDwpK1m8JEnrpW27DnS5pDcfdr2XmcU12fTTC5l43XZM+OIFP/0olWLxkiStt7y8XPY+5BhaXzaQdztez9JlS2j57imMuXFnpg37IOl4UtaweEmSNpjq1fLZ/+g/U/+iL3m7zeUULpxCk5ePZPR/9mPW6EFJx5MSZ/GSJG1w9WrV4KATLoVzv+St5mfRcM4I6j+5D6PuOJL5k/wEpKoui5ckqdw0a1ifg0+/ltmnDuLtBsfR8pdPKbx/Z767/yQWzxiXdDwp4yxekqRy16blRhx07p2M7dWPD2sdyqYTXyPc0YVRj59H0fwZSceTMsbiJUnKmI4d2rP/RY8x7PCP+KzabrT/8TEW37wlP7zwD+KSeUnHk8qdxUuSlHFdO3dmr7++RL/9Xmdo3la0H3Ebc67fkp/f+g94ElZVYhYvSVIiQgjstsvv6HbZO3yw85P8SEvaDLqa6ddtzYSPH/IkrKqULF6SpETl5eawz36/p9Nln/DWNncxvag6LT/5C5Ou346pA1/yJKyqVCxekqSsUFiQx8GHH0uLSwbweodrWbJkCU3f/hPjbtqFWSM+TDqetEFYvCRJWaVujWoc2vMsav5lMK+2upSCBZOp/8IR/PTf/Zn38+Ck40nrxeIlScpKTerWovvJl7P0zEG82vjP1J/9DbUf25sf7z6KJVNGJR1PWicWL0lSVtu4aSO6n3U9U07qz2t1e9Fsal9y7+3G6If+xPJZ45OOJ60Vi5ckqULYonUrDrvgbr47ui/vVD+Ejce9SvFt2/LTUxcQF3gSVlUMFi9JUoXSpdPmHHzJ4ww45H365O/GJt8/wsKbt2Tsy1fBkvlJx5NWy+IlSapwQgjstn0X9r7sZT7c81UGh63YZNh/mXNDJya++19PwqqsZfGSJFVYuTmB/fbYg26X9+aN7R/n++IWtPjiKmZcvxVTP33Ek7Aq61i8JEkVXrW8XH5/8GFsfmkfXup4B1OW1aDph+cz5YYuzPzyFU/Cqqxh8ZIkVRq1qxdw5NHH0/SiL3hh03+xcPFiGrx+IhNv3pX5332cdDzJ4iVJqnwa1a5Oj+PPIf+cgTzX/GLC/EnUerY7427bn0VjPQmrkmPxkiRVWq0a1eEPp/+N+acO4PkGp1N75jdUf2Rvxtzbg2VTPQmrMs/iJUmq9Dq0bMLR597ImF79eLHmMTSe/Anhnm6MfeRkimdPSDqeqhCLlySpyti2Q2uOvOgehh7+CW8WHEzzMa+w/NbOjH/2L56EVRkRYgX4pEfXrl3j4MHuk5ckbTjFxZEPvhhE0UfXst/yPizJqc7sbc9go/0vhGq1ko6nCiyEMCTG2HVl0xzxkiRVSTk5gf122YG9L3uZN3d9kQFsyUZf/oe5N3Zi2vu3ehJWlQuLlySpSivIy+Gwffdhh7/25vltHuG75S1o0u9KZt6wNTM/f8yTsGqDsnhJkgTUrJbH0YcfQbuLP+bpDrcycWkNGrx3LtNu7Mq8r171JKzaICxekiSV0KBWNXr2PImG5/fjiVbXMH/hImq/egKT/7MrC795E4qLk46oCsziJUnSSmxUvwbHnXwexWd+wRONLyTOnUSNF3sx4/otmf7+rbB4btIRVQH5qUZJksrgm/G/MPSdx+k0/mm2y/mBRaEG09sdRYv9ziO3cbuk4ymLrO5TjRYvSZLWwoz5S/jow3eo9fVD7F30GXmhmAkNd6XhXudSs+O+EELSEZUwi5ckSRvY8qJiPhkynJmf3Mce89+gcZjL1GqtiTucTrPdToCCmklHVEKyqniFEFoBjwNNgQjcH2O8bXXXsXhJkrLZt+OnM+ydR+g04Wm2DD8zP9RievujabX/eeQ1bJ10PGVYthWv5kDzGOOXIYTawBCge4xx5KquY/GSJFUEs+Yv4ZMP36T21w+xe9EXhABjG+1O433Oo/Zme7gbsorIquL1PwFCeA24M8b4/qouY/GSJFUkRcWRz4Z8xexP7uF3896ifpjPpMJ2xB1Op8Vux0N+YdIRVY6ytniFEFoDfYEtY4xzS007DTgNYOONN+4yduzYzAeUJGk9fT9hGt+88yBbjn+aDmE8c0MdpnY4htYHnEt+/ZZJx1M5yMriFUKoBXwC/DvG+PLqLuuIlySpopuzYCmfffAKtYc9xK7LB1Iccvip0V402e986rXfxd2QlUjWFa8QQj7wJvBujPE/a7q8xUuSVFkUFUf6DxnCnE/uZtd5vakTFjKucDPijmewyW7HQl5B0hG1nrKqeIUQAvAYMDPGeH5ZrmPxkiRVRj9OnMrI3vfRafwzbBomMTOnPtM69GTTA86loF6zpONpHWVb8doV+BQYDqz4wqvLY4xvr+o6Fi9JUmU2b9ESPn/vReoOe4huRUNYSh6jG+9Hs/3Op0H7HZOOp7WUVcVrXVi8JElVQXFxZNCQgczrexfd5r5LrbCYH6tvSdjxDNrs+geCuyErBIuXJEkVzNiJk/mu9z10HP8srcJUpuc0YkqHY2l/4NkU1m2cdDythsVLkqQKasGiJQx87xnqDnuY7Yq+ZjEFfN/kQJrvfz6N226XdDythMVLkqQKLsbI0MGfM6/vXeww932qh6WMqr4todsZtN/1KEJuXtIRlWbxkiSpEpk4aSKjet/FFuOfpTkzmJLTlCmbHcfmB51FYe0GScer8ixekiRVQosWL2HIu09Qd9jDbFU0goVU49smh9Bi//No1nabpONVWRYvSZIqsRgjwwd/yvy+d9Jl7odUC8sZUX17wk5nsMWuhxNycpOOWKVYvCRJqiImTxrH6LfvYPMJL9CYWYzPacGUzY+n04GnU6N2/aTjVQkWL0mSqpjFixfx1buPUW/Yw2xeNIp5VGdkk0NpdcD5bLRpx6TjVWoWL0mSqqgYI98N/oj5fe+i89w+5FLMsBrdyNnpz2y16+8JOTlJR6x0LF6SJIlpE8fwY+/b2GzCizRgLj/nbMyUzU9gq4NOo1atOknHqzQsXpIk6VdLFi9g+DsP02D4w2xa9BNzYk2GN+tOy33PpnU7d0OuL4uXJEn6XzHy/aD3WPjpnWw191NyQ2RE7hbMaPN72u5xLC1abpJ0wgrJ4iVJklbrl4mjGfPRIzQe8wabFI2lKAaGFXRmbrvD2GyPY2jWtFnSESsMi5ckSSqzKT98ycTPnmSj8W/RvHgKS2IeXxduz6LNurPFHkfTpIFnx18di5ckSVp7MTJxZD+m9XuSjSe/S8M4kwWxGl/V2JnlWxzBVnscQYM6tZJOmXUsXpIkaf0UFzH+qw+YNeAZNpn6AXWZx+xYk69q/Q62OoptdzuEujULk06ZFSxekiRpg4nLlzB20FvMH/wcm87oQw0WMy3WY1jdPSnofDRddt6XmoX5ScdMjMVLkiSVi7h0AWO+eIUlQ5+nzezPqcYyxscmjGiwDzW7/IHtd9yNwvyq9V2RFi9JklTuihfOZky/5yka9iJt5g0ij2J+iC35vvF+1N2hJ9tvtx3V8ip/CbN4SZKkjFo+dypjP32GnBEv0WbhMACG05afmh5Ik53+SNetOpGfWzm/rsjiJUmSErNs5jjG9n2Sat+9QqvF31McA1+GLRjf4iBa7PJHumzejtyckHTMDcbiJUmSssKSKaMY3/cJao9+laZLx7Ms5jIwZ2umbHwIbXY9ms5tW5FTwUuYxUuSJGWXGFk84WsmfvoE9X56nYbLp7E45vN5bldmtDmUzXY9gq1aNyWEilfCLF6SJCl7FRez8OcvmPLZkzQa+zZ1imczL1anX1435rY/jC13PZQtWjSoMCXM4iVJkiqGouUsGPUx0754kiYT3qdmXMCMWJvPCnZlyWbd2W63A2nXtG7SKVfL4iVJkiqe5UuY+01vZg54huZTPqZaXMKk2IDPC39Hcaej2HHnPdmkUfZ9ZZHFS5IkVWxL5jPnq9eYO+hZmv3yOfks56fiZgysuQc5Wx/FLjvtSot61ZNOCVi8JElSZbJwJrO+fJmFQ56j+axB5BAZWbwJQ+rsRbXOPdhjhy40qZPc90ZavCRJUuU0bwozBz3H0qEv0GzecACGFLdnWL19qN3laPbs0omGtaplNJLFS5IkVX6zxjCj/zMUD3+Jxgt/oCgGvoid+LbhfjTa/kj26rwZdWuU/5d3W7wkSVKVEqd9y4z+T5M74mXqL5nAkpjHp3EbGne/jm2227Fc73t1xSuvXO9ZkiQpAaHJFjQ69J/w+2uIE79kbv+n2eH718lv1STRXBYvSZJUeYVAaNmFxkd1gXgzJHwS1sr5teCSJEmlZcGZ7y1ekiRJGWLxkiRJyhCLlyRJUoZYvCRJkjLE4iVJkpQhFi9JkqQMsXhJkiRliMVLkiQpQyxekiRJGWLxkiRJyhCLlyRJUoZYvCRJkjLE4iVJkpQhFi9JkqQMsXhJkiRliMVLkiQpQyxekiRJGWLxkiRJyhCLlyRJUoZYvCRJkjLE4iVJkpQhFi9JkqQMCTHGpDOsUQhhOjC2nO+mEfBLOd+H1p7LJfu4TLKTyyX7uEyyT6aWySYxxsYrm1AhilcmhBAGxxi7Jp1Dv+VyyT4uk+zkcsk+LpPskw3LxF2NkiRJGWLxkiRJyhCL1/+7P+kAWimXS/ZxmWQnl0v2cZlkn8SXicd4SZIkZYgjXpIkSRli8ZIkScqQKle8QggHhBBGhRBGhxD+upLp1UIIz6WnDwghtE4gZpVShmXylxDCyBDCsBDChyGETZLIWdWsabmUuNyRIYQYQvBj8+WsLMskhHB0+vkyIoTwdKYzVkVl2IZtHEL4OIQwNL0dOyiJnFVJCOHhEMK0EMI3q5geQgi3p5fZsBDCdpnKVqWKVwghF7gLOBDoCBwTQuhY6mInA7NijO2A/wI3ZDZl1VLGZTIU6Bpj3Bp4EbgxsymrnjIuF0IItYHzgAGZTVj1lGWZhBDaA5cBu8QYOwHnZzpnVVPG58rfgOdjjNsCfwTuzmzKKulR4IDVTD8QaJ/+dxpwTwYyAVWseAE7AKNjjD/FGJcCzwKHlbrMYcBj6Z9fBPYOIYQMZqxq1rh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" + ] + }, + "metadata": { + "filenames": { + "image/png": "/Users/mhjensen/Teaching/MachineLearning/doc/LectureNotes/_build/jupyter_execute/chapter11_47_2.png" + }, + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "\n", + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# Assuming one input, hidden, and output layer\n", + "def neural_network(params, x):\n", + "\n", + " # Find the weights (including and biases) for the hidden and output layer.\n", + " # Assume that params is a list of parameters for each layer.\n", + " # The biases are the first element for each array in params,\n", + " # and the weights are the remaning elements in each array in params.\n", + "\n", + " w_hidden = params[0]\n", + " w_output = params[1]\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " ## Hidden layer:\n", + "\n", + " # Add a row of ones to include bias\n", + " x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_input)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " ## Output layer:\n", + "\n", + " # Include bias:\n", + " x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_hidden)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial(x,params, g0 = 10):\n", + " return g0 + x*neural_network(params,x)\n", + "\n", + "# The right side of the ODE:\n", + "def g(x, g_trial, gamma = 2):\n", + " return -gamma*g_trial\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the neural network\n", + " d_net_out = elementwise_grad(neural_network,1)(P,x)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = g(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# Solve the exponential decay ODE using neural network with one input, hidden, and output layer\n", + "def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):\n", + " ## Set up initial weights and biases\n", + "\n", + " # For the hidden layer\n", + " p0 = npr.randn(num_neurons_hidden, 2 )\n", + "\n", + " # For the output layer\n", + " p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included\n", + "\n", + " P = [p0, p1]\n", + "\n", + " print('Initial cost: %g'%cost_function(P, x))\n", + "\n", + " ## Start finding the optimal weights using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of two arrays;\n", + " # one for the gradient w.r.t P_hidden and\n", + " # one for the gradient w.r.t P_output\n", + " cost_grad = cost_function_grad(P, x)\n", + "\n", + " P[0] = P[0] - lmb * cost_grad[0]\n", + " P[1] = P[1] - lmb * cost_grad[1]\n", + "\n", + " print('Final cost: %g'%cost_function(P, x))\n", + "\n", + " return P\n", + "\n", + "def g_analytic(x, gamma = 2, g0 = 10):\n", + " return g0*np.exp(-gamma*x)\n", + "\n", + "# Solve the given problem\n", + "if __name__ == '__main__':\n", + " # Set seed such that the weight are initialized\n", + " # with same weights and biases for every run.\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " N = 10\n", + " x = np.linspace(0, 1, N)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = 10\n", + " num_iter = 10000\n", + " lmb = 0.001\n", + "\n", + " # Use the network\n", + " P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " # Print the deviation from the trial solution and true solution\n", + " res = g_trial(x,P)\n", + " res_analytical = g_analytic(x)\n", + "\n", + " print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))\n", + "\n", + " # Plot the results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, res_analytical)\n", + " plt.plot(x, res[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## The network with one input layer, specified number of hidden layers, and one output layer\n", + "\n", + "It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.\n", + "\n", + "The number of neurons within each hidden layer are given as a list of integers in the program below." + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Initial cost: 324.246\n" + ] + }, + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/numpy/core/_asarray.py:83: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray\n", + " return array(a, dtype, copy=False, order=order)\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Final cost: 0.119936\n" + ] + }, + { + "data": { + "image/png": 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" + ] + }, + "metadata": { + "filenames": { + "image/png": "/Users/mhjensen/Teaching/MachineLearning/doc/LectureNotes/_build/jupyter_execute/chapter11_49_3.png" + }, + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# The neural network with one input layer and one output layer,\n", + "# but with number of hidden layers specified by the user.\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + "\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consists of\n", + " # parameters to all the hidden\n", + " # layers AND the output layer.\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial_deep(x,params, g0 = 10):\n", + " return g0 + x*deep_neural_network(params, x)\n", + "\n", + "# The right side of the ODE:\n", + "def g(x, g_trial, gamma = 2):\n", + " return -gamma*g_trial\n", + "\n", + "# The same cost function as before, but calls deep_neural_network instead.\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the neural network\n", + " d_net_out = elementwise_grad(deep_neural_network,1)(P,x)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial_deep,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = g(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# Solve the exponential decay ODE using neural network with one input and one output layer,\n", + "# but with specified number of hidden layers from the user.\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # The number of elements in the list num_hidden_neurons thus represents\n", + " # the number of hidden layers.\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weights and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weights using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "def g_analytic(x, gamma = 2, g0 = 10):\n", + " return g0*np.exp(-gamma*x)\n", + "\n", + "# Solve the given problem\n", + "if __name__ == '__main__':\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " N = 10\n", + " x = np.linspace(0, 1, N)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = np.array([10,10])\n", + " num_iter = 10000\n", + " lmb = 0.001\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " res = g_trial_deep(x,P)\n", + " res_analytical = g_analytic(x)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, res_analytical)\n", + " plt.plot(x, res[0,:])\n", + " plt.legend(['analytical','dnn'])\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example: Population growth\n", + "\n", + "A logistic model of population growth assumes that a population converges toward an equilibrium.\n", + "The population growth can be modeled by" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{log} \\tag{10}\n", + "\tg'(t) = \\alpha g(t)(A - g(t))\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(t)$ is the population density at time $t$, $\\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment.\n", + "Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant.\n", + "\n", + "In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability\n", + "and high execution time (this might be more apparent in the examples solving PDEs),\n", + "using a library like TensorFlow is recommended.\n", + "Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method.\n", + "\n", + "\n", + "\n", + "Here, we will model a population $g(t)$ in an environment having carrying capacity $A$.\n", + "The population follows the model" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solveode_population} \\tag{11}\n", + "g'(t) = \\alpha g(t)(A - g(t))\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(0) = g_0$.\n", + "\n", + "In this example, we let $\\alpha = 2$, $A = 1$, and $g_0 = 1.2$.\n", + "\n", + "\n", + "We will get a slightly different trial solution, as the boundary conditions are different\n", + "compared to the case for exponential decay.\n", + "\n", + "A possible trial solution satisfying the condition $g(0) = g_0$ could be\n", + "\n", + "$$\n", + "h_1(t) = g_0 + t \\cdot N(t,P)\n", + "$$\n", + "\n", + "with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$.\n", + "\n", + "The analytical solution is\n", + "\n", + "$$\n", + "g(t) = \\frac{Ag_0}{g_0 + (A - g_0)\\exp(-\\alpha A t)}\n", + "$$\n", + "\n", + "\n", + "\n", + "The network will be the similar as for the exponential decay example, but with some small modifications for our problem." + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Initial cost: 0.221805\n" + ] + }, + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/numpy/core/_asarray.py:83: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray\n", + " return array(a, dtype, copy=False, order=order)\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Final cost: 0.000417932\n", + "The max absolute difference between the solutions is: 0.00424909\n" + ] + }, + { + "data": { + "image/png": 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\n", + "text/plain": [ + "
" + ] + }, + "metadata": { + "filenames": { + "image/png": "/Users/mhjensen/Teaching/MachineLearning/doc/LectureNotes/_build/jupyter_execute/chapter11_55_3.png" + }, + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# Function to get the parameters.\n", + "# Done such that one can easily change the paramaters after one's liking.\n", + "def get_parameters():\n", + " alpha = 2\n", + " A = 1\n", + " g0 = 1.2\n", + " return alpha, A, g0\n", + "\n", + "def deep_neural_network(P, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = P[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = P[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial_deep,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = f(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# The right side of the ODE:\n", + "def f(x, g_trial):\n", + " alpha,A, g0 = get_parameters()\n", + " return alpha*g_trial*(A - g_trial)\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial_deep(x, params):\n", + " alpha,A, g0 = get_parameters()\n", + " return g0 + x*deep_neural_network(params,x)\n", + "\n", + "# The analytical solution:\n", + "def g_analytic(t):\n", + " alpha,A, g0 = get_parameters()\n", + " return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nt = 10\n", + " T = 1\n", + " t = np.linspace(0,T, Nt)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [100, 50, 25]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(t,P)\n", + " g_analytical = g_analytic(t)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%diff_ag)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(t, g_analytical)\n", + " plt.plot(t, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Using forward Euler to solve the ODE\n", + "\n", + "A straightforward way of solving an ODE numerically, is to use Euler's method.\n", + "\n", + "Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\\Delta x$ from $x$:\n", + "\n", + "$$\n", + "f(x + \\Delta x) \\approx f(x) + \\Delta x f'(x)\n", + "$$\n", + "\n", + "In our case, using Euler's method to approximate the value of $g$ at a step $\\Delta t$ from $t$ yields" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + " g(t + \\Delta t) &\\approx g(t) + \\Delta t g'(t) \\\\\n", + " &= g(t) + \\Delta t \\big(\\alpha g(t)(A - g(t))\\big)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "along with the condition that $g(0) = g_0$.\n", + "\n", + "Let $t_i = i \\cdot \\Delta t$ where $\\Delta t = \\frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \\in [0, T]$ for $i = 0, \\dots, N_t-1$.\n", + "\n", + "For $i \\geq 1$, we have that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "t_i &= i\\Delta t \\\\\n", + "&= (i - 1)\\Delta t + \\Delta t \\\\\n", + "&= t_{i-1} + \\Delta t\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Now, if $g_i = g(t_i)$ then" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\begin{aligned}\n", + " g_i &= g(t_i) \\\\\n", + " &= g(t_{i-1} + \\Delta t) \\\\\n", + " &\\approx g(t_{i-1}) + \\Delta t \\big(\\alpha g(t_{i-1})(A - g(t_{i-1}))\\big) \\\\\n", + " &= g_{i-1} + \\Delta t \\big(\\alpha g_{i-1}(A - g_{i-1})\\big)\n", + " \\end{aligned}\n", + "\\end{equation} \\label{odenum} \\tag{12}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for $i \\geq 1$ and $g_0 = g(t_0) = g(0) = g_0$.\n", + "\n", + "Equation ([12](#odenum)) could be implemented in the following way,\n", + "extending the program that uses the network using Autograd:" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Initial cost: 0.221805\n" + ] + }, + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/numpy/core/_asarray.py:83: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray\n", + " return array(a, dtype, copy=False, order=order)\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Final cost: 0.000417932\n", + "The max absolute difference between the solutions is: 0.00424909\n", + "Max absolute difference between Euler method and analytical: 0.011225\n", + "Max absolute difference between deep neural network and analytical: 0.00424909\n" + ] + }, + { + "data": { + "image/png": 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\n", 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" + ] + }, + "metadata": { + "filenames": { + "image/png": "/Users/mhjensen/Teaching/MachineLearning/doc/LectureNotes/_build/jupyter_execute/chapter11_63_4.png" + }, + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "# Assume that all function definitions from the example program using Autograd\n", + "# are located here.\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nt = 10\n", + " T = 1\n", + " t = np.linspace(0,T, Nt)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [100,50,25]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(t,P)\n", + " g_analytical = g_analytic(t)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%diff_ag)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(t, g_analytical)\n", + " plt.plot(t, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " ## Find an approximation to the funtion using forward Euler\n", + "\n", + " alpha, A, g0 = get_parameters()\n", + " dt = T/(Nt - 1)\n", + "\n", + " # Perform forward Euler to solve the ODE\n", + " g_euler = np.zeros(Nt)\n", + " g_euler[0] = g0\n", + "\n", + " for i in range(1,Nt):\n", + " g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))\n", + "\n", + " # Print the errors done by each method\n", + " diff1 = np.max(np.abs(g_euler - g_analytical))\n", + " diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))\n", + "\n", + " print('Max absolute difference between Euler method and analytical: %g'%diff1)\n", + " print('Max absolute difference between deep neural network and analytical: %g'%diff2)\n", + "\n", + " # Plot results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.plot(t,g_euler)\n", + " plt.plot(t,g_analytical)\n", + " plt.plot(t,g_dnn_ag[0,:])\n", + "\n", + " plt.legend(['euler','analytical','dnn'])\n", + " plt.xlabel('Time t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Solving the one dimensional Poisson equation\n", + "\n", + "The Poisson equation for $g(x)$ in one dimension is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{poisson} \\tag{13}\n", + " -g''(x) = f(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f(x)$ is a given function for $x \\in (0,1)$.\n", + "\n", + "The conditions that $g(x)$ is chosen to fulfill, are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + " g(0) &= 0 \\\\\n", + " g(1) &= 0\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used.\n", + "The results from the networks can then be compared to the analytical solution.\n", + "In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks.\n", + "\n", + "\n", + "Here, the function $g(x)$ to solve for follows the equation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "-g''(x) = f(x),\\qquad x \\in (0,1)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f(x)$ is a given function, along with the chosen conditions" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "g(0) = g(1) = 0\n", + "\\end{aligned}\\label{cond} \\tag{14}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this example, we consider the case when $f(x) = (3x + x^2)\\exp(x)$.\n", + "\n", + "For this case, a possible trial solution satisfying the conditions could be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x) = x \\cdot (1-x) \\cdot N(P,x)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The analytical solution for this problem is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g(x) = x(1 - x)\\exp(x)\n", + "$$" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [ + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/numpy/core/_asarray.py:83: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray\n", + " return array(a, dtype, copy=False, order=order)\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Initial cost: 457.256\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Final cost: 0.00310113\n", + "The max absolute difference between the solutions is: 0.000464088\n" + ] + }, + { + "data": { + "image/png": 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\n", + "text/plain": [ + "
" + ] + }, + "metadata": { + "filenames": { + "image/png": "/Users/mhjensen/Teaching/MachineLearning/doc/LectureNotes/_build/jupyter_execute/chapter11_76_3.png" + }, + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "## Set up the cost function specified for this Poisson equation:\n", + "\n", + "# The right side of the ODE\n", + "def f(x):\n", + " return (3*x + x**2)*np.exp(x)\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)\n", + "\n", + " right_side = f(x)\n", + "\n", + " err_sqr = (-d2_g_t - right_side)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum/np.size(err_sqr)\n", + "\n", + "# The trial solution:\n", + "def g_trial_deep(x,P):\n", + " return x*(1-x)*deep_neural_network(P,x)\n", + "\n", + "# The analytic solution;\n", + "def g_analytic(x):\n", + " return x*(1-x)*np.exp(x)\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10\n", + " x = np.linspace(0,1, Nx)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [200,100]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(x,P)\n", + " g_analytical = g_analytic(x)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " max_diff = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%max_diff)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, g_analytical)\n", + " plt.plot(x, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Comparing with a numerical scheme\n", + "\n", + "The Poisson equation is possible to solve using Taylor series to approximate the second derivative.\n", + "\n", + "Using Taylor series, the second derivative can be expressed as\n", + "\n", + "$$\n", + "g''(x) = \\frac{g(x + \\Delta x) - 2g(x) + g(x-\\Delta x)}{\\Delta x^2} + E_{\\Delta x}(x)\n", + "$$\n", + "\n", + "where $\\Delta x$ is a small step size and $E_{\\Delta x}(x)$ being the error term.\n", + "\n", + "Looking away from the error terms gives an approximation to the second derivative:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{approx} \\tag{15}\n", + "g''(x) \\approx \\frac{g(x + \\Delta x) - 2g(x) + g(x-\\Delta x)}{\\Delta x^2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If $x_i = i \\Delta x = x_{i-1} + \\Delta x$ and $g_i = g(x_i)$ for $i = 1,\\dots N_x - 2$ with $N_x$ being the number of values for $x$, ([15](#approx)) becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "g''(x_i) &\\approx \\frac{g(x_i + \\Delta x) - 2g(x_i) + g(x_i -\\Delta x)}{\\Delta x^2} \\\\\n", + "&= \\frac{g_{i+1} - 2g_i + g_{i-1}}{\\Delta x^2}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Since we know from our problem that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "-g''(x) &= f(x) \\\\\n", + "&= (3x + x^2)\\exp(x)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "along with the conditions $g(0) = g(1) = 0$,\n", + "the following scheme can be used to find an approximate solution for $g(x)$ numerically:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\begin{aligned}\n", + " -\\Big( \\frac{g_{i+1} - 2g_i + g_{i-1}}{\\Delta x^2} \\Big) &= f(x_i) \\\\\n", + " -g_{i+1} + 2g_i - g_{i-1} &= \\Delta x^2 f(x_i)\n", + " \\end{aligned}\n", + "\\end{equation} \\label{odesys} \\tag{16}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for $i = 1, \\dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\\exp(x_i)$, which is given for our specific problem.\n", + "\n", + "The equation can be rewritten into a matrix equation:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "\\begin{pmatrix}\n", + "2 & -1 & 0 & \\dots & 0 \\\\\n", + "-1 & 2 & -1 & \\dots & 0 \\\\\n", + "\\vdots & & \\ddots & & \\vdots \\\\\n", + "0 & \\dots & -1 & 2 & -1 \\\\\n", + "0 & \\dots & 0 & -1 & 2\\\\\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "g_1 \\\\\n", + "g_2 \\\\\n", + "\\vdots \\\\\n", + "g_{N_x - 3} \\\\\n", + "g_{N_x - 2}\n", + "\\end{pmatrix}\n", + "&=\n", + "\\Delta x^2\n", + "\\begin{pmatrix}\n", + "f(x_1) \\\\\n", + "f(x_2) \\\\\n", + "\\vdots \\\\\n", + "f(x_{N_x - 3}) \\\\\n", + "f(x_{N_x - 2})\n", + "\\end{pmatrix} \\\\\n", + "\\boldsymbol{A}\\boldsymbol{g} &= \\boldsymbol{f},\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "which makes it possible to solve for the vector $\\boldsymbol{g}$.\n", + "\n", + "\n", + "We can then compare the result from this numerical scheme with the output from our network using Autograd:" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Initial cost: 457.256\n" + ] + }, + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/numpy/core/_asarray.py:83: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray\n", + " return array(a, dtype, copy=False, order=order)\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Final cost: 0.00310113\n", + "The max absolute difference between the analytical solution and DNN Autograd: 0.000464088\n", + "The max absolute difference between the analytical solution and numerical scheme: 0.00266858\n" + ] + }, + { + "data": { + "image/png": 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\n", + "text/plain": [ + "
" + ] + }, + "metadata": { + "filenames": { + "image/png": "/Users/mhjensen/Teaching/MachineLearning/doc/LectureNotes/_build/jupyter_execute/chapter11_88_4.png" + }, + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "## Set up the cost function specified for this Poisson equation:\n", + "\n", + "# The right side of the ODE\n", + "def f(x):\n", + " return (3*x + x**2)*np.exp(x)\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)\n", + "\n", + " right_side = f(x)\n", + "\n", + " err_sqr = (-d2_g_t - right_side)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum/np.size(err_sqr)\n", + "\n", + "# The trial solution:\n", + "def g_trial_deep(x,P):\n", + " return x*(1-x)*deep_neural_network(P,x)\n", + "\n", + "# The analytic solution;\n", + "def g_analytic(x):\n", + " return x*(1-x)*np.exp(x)\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10\n", + " x = np.linspace(0,1, Nx)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [200,100]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(x,P)\n", + " g_analytical = g_analytic(x)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, g_analytical)\n", + " plt.plot(x, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + "\n", + " ## Perform the computation using the numerical scheme\n", + "\n", + " dx = 1/(Nx - 1)\n", + "\n", + " # Set up the matrix A\n", + " A = np.zeros((Nx-2,Nx-2))\n", + "\n", + " A[0,0] = 2\n", + " A[0,1] = -1\n", + "\n", + " for i in range(1,Nx-3):\n", + " A[i,i-1] = -1\n", + " A[i,i] = 2\n", + " A[i,i+1] = -1\n", + "\n", + " A[Nx - 3, Nx - 4] = -1\n", + " A[Nx - 3, Nx - 3] = 2\n", + "\n", + " # Set up the vector f\n", + " f_vec = dx**2 * f(x[1:-1])\n", + "\n", + " # Solve the equation\n", + " g_res = np.linalg.solve(A,f_vec)\n", + "\n", + " g_vec = np.zeros(Nx)\n", + " g_vec[1:-1] = g_res\n", + "\n", + " # Print the differences between each method\n", + " max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " max_diff2 = np.max(np.abs(g_vec - g_analytical))\n", + " print(\"The max absolute difference between the analytical solution and DNN Autograd: %g\"%max_diff1)\n", + " print(\"The max absolute difference between the analytical solution and numerical scheme: %g\"%max_diff2)\n", + "\n", + " # Plot the results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.plot(x,g_vec)\n", + " plt.plot(x,g_analytical)\n", + " plt.plot(x,g_dnn_ag[0,:])\n", + "\n", + " plt.legend(['numerical scheme','analytical','dnn'])\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Partial Differential Equations\n", + "\n", + "A partial differential equation (PDE) has a solution here the function\n", + "is defined by multiple variables. The equation may involve all kinds\n", + "of combinations of which variables the function is differentiated with\n", + "respect to.\n", + "\n", + "In general, a partial differential equation for a function $g(x_1,\\dots,x_N)$ with $N$ variables may be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{PDE} \\tag{17}\n", + " f\\left(x_1, \\, \\dots \\, , x_N, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1}, \\dots , \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_N}, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(x_1,\\dots,x_N) }{\\partial x_N^n} \\right) = 0\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given.\n", + "\n", + "### Type of problem\n", + "\n", + "The problem our network must solve for, is similar to the ODE case.\n", + "We must have a trial solution $g_t$ at hand.\n", + "\n", + "For instance, the trial solution could be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + " g_t(x_1,\\dots,x_N) = h_1(x_1,\\dots,x_N) + h_2(x_1,\\dots,x_N,N(x_1,\\dots,x_N,P))\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $h_1(x_1,\\dots,x_N)$ is a function that ensures $g_t(x_1,\\dots,x_N)$ satisfies some given conditions.\n", + "The neural network $N(x_1,\\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\\dots,x_N,N(x_1,\\dots,x_N,P))$ is an expression using the output from the neural network in some way.\n", + "\n", + "The role of the function $h_2(x_1,\\dots,x_N,N(x_1,\\dots,x_N,P))$, is to ensure that the output of $N(x_1,\\dots,x_N,P)$ is zero when $g_t(x_1,\\dots,x_N)$ is evaluated at the values of $x_1,\\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\\dots,x_N)$ should alone make $g_t(x_1,\\dots,x_N)$ satisfy the conditions.\n", + "\n", + "\n", + "\n", + "### Network requirements\n", + "\n", + "The network tries then the minimize the cost function following the\n", + "same ideas as described for the ODE case, but now with more than one\n", + "variables to consider. The concept still remains the same; find a set\n", + "of parameters $P$ such that the expression $f$ in ([17](#PDE)) is as\n", + "close to zero as possible.\n", + "\n", + "As for the ODE case, the cost function is the mean squared error that\n", + "the network must try to minimize. The cost function for the network to\n", + "minimize is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(x_1, \\dots, x_N, P\\right) = \\left( f\\left(x_1, \\, \\dots \\, , x_N, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1}, \\dots , \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_N}, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(x_1,\\dots,x_N) }{\\partial x_N^n} \\right) \\right)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If we let $\\boldsymbol{x} = \\big( x_1, \\dots, x_N \\big)$ be an array containing the values for $x_1, \\dots, x_N$ respectively, the cost function can be reformulated into the following:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(\\boldsymbol{x}, P\\right) = f\\left( \\left( \\boldsymbol{x}, \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_1}, \\dots , \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_N}, \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(\\boldsymbol{x}) }{\\partial x_N^n} \\right) \\right)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If we also have $M$ different sets of values for $x_1, \\dots, x_N$, that is $\\boldsymbol{x}_i = \\big(x_1^{(i)}, \\dots, x_N^{(i)}\\big)$ for $i = 1,\\dots,M$ being the rows in matrix $X$, the cost function can be generalized into" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(X, P \\right) = \\sum_{i=1}^M f\\left( \\left( \\boldsymbol{x}_i, \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_1}, \\dots , \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_N}, \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(\\boldsymbol{x}_i) }{\\partial x_N^n} \\right) \\right)^2.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: The diffusion equation\n", + "\n", + "In one spatial dimension, the equation reads" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial g(x,t)}{\\partial t} = \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where a possible choice of conditions are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "g(0,t) &= 0 ,\\qquad t \\geq 0 \\\\\n", + "g(1,t) &= 0, \\qquad t \\geq 0 \\\\\n", + "g(x,0) &= u(x),\\qquad x\\in [0,1]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $u(x)$ being some given function.\n", + "\n", + "\n", + "\n", + "For this case, we want to find $g(x,t)$ such that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\frac{\\partial g(x,t)}{\\partial t} = \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "\\end{equation} \\label{diffonedim} \\tag{18}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "and" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "g(0,t) &= 0 ,\\qquad t \\geq 0 \\\\\n", + "g(1,t) &= 0, \\qquad t \\geq 0 \\\\\n", + "g(x,0) &= u(x),\\qquad x\\in [0,1]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $u(x) = \\sin(\\pi x)$.\n", + "\n", + "First, let us set up the deep neural network.\n", + "The deep neural network will follow the same structure as discussed in the examples solving the ODEs.\n", + "First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions.\n", + "\n", + "\n", + "\n", + "\n", + "The only change to do here, is to extend our network such that\n", + "functions of multiple parameters are correctly handled. In this case\n", + "we have two variables in our function to solve for, that is time $t$\n", + "and position $x$. The variables will be represented by a\n", + "one-dimensional array in the program. The program will evaluate the\n", + "network at each possible pair $(x,t)$, given an array for the desired\n", + "$x$-values and $t$-values to approximate the solution at." + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The cost function must then iterate through the given arrays\n", + "containing values for $x$ and $t$, defines a point $(x,t)$ the deep\n", + "neural network and the trial solution is evaluated at, and then finds\n", + "the Jacobian of the trial solution.\n", + "\n", + "A possible trial solution for this PDE is\n", + "\n", + "$$\n", + "g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P)\n", + "$$\n", + "\n", + "with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$.\n", + "\n", + "To fulfill the conditions, $A(x,t)$ could be:\n", + "\n", + "$$\n", + "h_1(x,t) = (1-t)\\Big(u(x) - \\big((1-x)u(0) + x u(1)\\big)\\Big) = (1-t)u(x) = (1-t)\\sin(\\pi x)\n", + "$$\n", + "since $(0) = u(1) = 0$ and $u(x) = \\sin(\\pi x)$.\n", + "\n", + "\n", + "\n", + "The Jacobian is used because the program must find the derivative of\n", + "the trial solution with respect to $x$ and $t$.\n", + "\n", + "This gives the necessity of computing the Jacobian matrix, as we want\n", + "to evaluate the gradient with respect to $x$ and $t$ (note that the\n", + "Jacobian of a scalar-valued multivariate function is simply its\n", + "gradient).\n", + "\n", + "In Autograd, the differentiation is by default done with respect to\n", + "the first input argument of your Python function. Since the points is\n", + "an array representing $x$ and $t$, the Jacobian is calculated using\n", + "the values of $x$ and $t$.\n", + "\n", + "To find the second derivative with respect to $x$ and $t$, the\n", + "Jacobian can be found for the second time. The result is a Hessian\n", + "matrix, which is the matrix containing all the possible second order\n", + "mixed derivatives of $g(x,t)$." + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Set up the trial function:\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)\n", + "\n", + "# The right side of the ODE:\n", + "def f(point):\n", + " return 0.\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_jacobian_func = jacobian(g_trial)\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t = g_trial(point,P)\n", + " g_t_jacobian = g_t_jacobian_func(point,P)\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_dt = g_t_jacobian[1]\n", + " g_t_d2x = g_t_hessian[0][0]\n", + "\n", + " func = f(point)\n", + "\n", + " err_sqr = ( (g_t_dt - g_t_d2x) - func)**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Setting up the network using Autograd; The full program\n", + "\n", + "Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution.\n", + "\n", + "The analytical solution of our problem is\n", + "\n", + "$$\n", + "g(x,t) = \\exp(-\\pi^2 t)\\sin(\\pi x)\n", + "$$\n", + "\n", + "A possible way to implement a neural network solving the PDE, is given below.\n", + "Be aware, though, that it is fairly slow for the parameters used.\n", + "A better result is possible, but requires more iterations, and thus longer time to complete.\n", + "\n", + "\n", + "Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE.\n", + "Using TensorFlow results in a much better execution time. Try it!" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [ + { + "name": "stderr", + "output_type": "stream", + "text": [ + "/Users/mhjensen/opt/anaconda3/lib/python3.8/site-packages/numpy/core/_asarray.py:83: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray\n", + " return array(a, dtype, copy=False, order=order)\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Initial cost: 41.05505310046363\n" + ] + } + ], + "source": [ + "import autograd.numpy as np\n", + "from autograd import jacobian,hessian,grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import cm\n", + "from matplotlib import pyplot as plt\n", + "from mpl_toolkits.mplot3d import axes3d\n", + "\n", + "## Set up the network\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]\n", + "\n", + "## Define the trial solution and cost function\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)\n", + "\n", + "# The right side of the ODE:\n", + "def f(point):\n", + " return 0.\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_jacobian_func = jacobian(g_trial)\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t = g_trial(point,P)\n", + " g_t_jacobian = g_t_jacobian_func(point,P)\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_dt = g_t_jacobian[1]\n", + " g_t_d2x = g_t_hessian[0][0]\n", + "\n", + " func = f(point)\n", + "\n", + " err_sqr = ( (g_t_dt - g_t_d2x) - func)**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum /( np.size(x)*np.size(t) )\n", + "\n", + "## For comparison, define the analytical solution\n", + "def g_analytic(point):\n", + " x,t = point\n", + " return np.exp(-np.pi**2*t)*np.sin(np.pi*x)\n", + "\n", + "## Set up a function for training the network to solve for the equation\n", + "def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):\n", + " ## Set up initial weigths and biases\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: ',cost_function(P, x, t))\n", + "\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " cost_grad = cost_function_grad(P, x , t)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_grad[l]\n", + "\n", + " print('Final cost: ',cost_function(P, x, t))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " ### Use the neural network:\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10; Nt = 10\n", + " x = np.linspace(0, 1, Nx)\n", + " t = np.linspace(0,1,Nt)\n", + "\n", + " ## Set up the parameters for the network\n", + " num_hidden_neurons = [100, 25]\n", + " num_iter = 250\n", + " lmb = 0.01\n", + "\n", + " P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " ## Store the results\n", + " g_dnn_ag = np.zeros((Nx, Nt))\n", + " G_analytical = np.zeros((Nx, Nt))\n", + " for i,x_ in enumerate(x):\n", + " for j, t_ in enumerate(t):\n", + " point = np.array([x_, t_])\n", + " g_dnn_ag[i,j] = g_trial(point,P)\n", + "\n", + " G_analytical[i,j] = g_analytic(point)\n", + "\n", + " # Find the map difference between the analytical and the computed solution\n", + " diff_ag = np.abs(g_dnn_ag - G_analytical)\n", + " print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))\n", + "\n", + " ## Plot the solutions in two dimensions, that being in position and time\n", + "\n", + " T,X = np.meshgrid(t,x)\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))\n", + " s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Analytical solution')\n", + " s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Difference')\n", + " s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " ## Take some slices of the 3D plots just to see the solutions at particular times\n", + " indx1 = 0\n", + " indx2 = int(Nt/2)\n", + " indx3 = Nt-1\n", + "\n", + " t1 = t[indx1]\n", + " t2 = t[indx2]\n", + " t3 = t[indx3]\n", + "\n", + " # Slice the results from the DNN\n", + " res1 = g_dnn_ag[:,indx1]\n", + " res2 = g_dnn_ag[:,indx2]\n", + " res3 = g_dnn_ag[:,indx3]\n", + "\n", + " # Slice the analytical results\n", + " res_analytical1 = G_analytical[:,indx1]\n", + " res_analytical2 = G_analytical[:,indx2]\n", + " res_analytical3 = G_analytical[:,indx3]\n", + "\n", + " # Plot the slices\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t1)\n", + " plt.plot(x, res1)\n", + " plt.plot(x,res_analytical1)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t2)\n", + " plt.plot(x, res2)\n", + " plt.plot(x,res_analytical2)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t3)\n", + " plt.plot(x, res3)\n", + " plt.plot(x,res_analytical3)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Solving the wave equation with Neural Networks\n", + "\n", + "The wave equation is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial^2 g(x,t)}{\\partial t^2} = c^2\\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $c$ being the specified wave speed.\n", + "\n", + "Here, the chosen conditions are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "\tg(0,t) &= 0 \\\\\n", + "\tg(1,t) &= 0 \\\\\n", + "\tg(x,0) &= u(x) \\\\\n", + "\t\\frac{\\partial g(x,t)}{\\partial t} \\Big |_{t = 0} &= v(x)\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $\\frac{\\partial g(x,t)}{\\partial t} \\Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions.\n", + "\n", + "\n", + "The wave equation to solve for, is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{wave} \\tag{19}\n", + "\\frac{\\partial^2 g(x,t)}{\\partial t^2} = c^2 \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $c$ is the given wave speed.\n", + "The chosen conditions for this equation are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "g(0,t) &= 0, &t \\geq 0 \\\\\n", + "g(1,t) &= 0, &t \\geq 0 \\\\\n", + "g(x,0) &= u(x), &x\\in[0,1] \\\\\n", + "\\frac{\\partial g(x,t)}{\\partial t}\\Big |_{t = 0} &= v(x), &x \\in [0,1]\n", + "\\end{aligned} \\label{condwave} \\tag{20}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this example, let $c = 1$ and $u(x) = \\sin(\\pi x)$ and $v(x) = -\\pi\\sin(\\pi x)$.\n", + "\n", + "\n", + "\n", + "Setting up the network is done in similar matter as for the example of solving the diffusion equation.\n", + "The only things we have to change, is the trial solution such that it satisfies the conditions from ([20](#condwave)) and the cost function.\n", + "\n", + "The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is\n", + "\n", + "$$\n", + "g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P)\n", + "$$\n", + "\n", + "where\n", + "\n", + "$$\n", + "h_1(x,t) = (1-t^2)u(x) + tv(x)\n", + "$$\n", + "\n", + "Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example.\n", + "\n", + "\n", + "The analytical solution for our specific problem, is\n", + "\n", + "$$\n", + "g(x,t) = \\sin(\\pi x)\\cos(\\pi t) - \\sin(\\pi x)\\sin(\\pi t)\n", + "$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import hessian,grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import cm\n", + "from matplotlib import pyplot as plt\n", + "from mpl_toolkits.mplot3d import axes3d\n", + "\n", + "## Set up the trial function:\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def v(x):\n", + " return -np.pi*np.sin(np.pi*x)\n", + "\n", + "def h1(point):\n", + " x,t = point\n", + " return (1 - t**2)*u(x) + t*v(x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)\n", + "\n", + "## Define the cost function\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_d2x = g_t_hessian[0][0]\n", + " g_t_d2t = g_t_hessian[1][1]\n", + "\n", + " err_sqr = ( (g_t_d2t - g_t_d2x) )**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum / (np.size(t) * np.size(x))\n", + "\n", + "## The neural network\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]\n", + "\n", + "## The analytical solution\n", + "def g_analytic(point):\n", + " x,t = point\n", + " return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)\n", + "\n", + "def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):\n", + " ## Set up initial weigths and biases\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: ',cost_function(P, x, t))\n", + "\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " cost_grad = cost_function_grad(P, x , t)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_grad[l]\n", + "\n", + "\n", + " print('Final cost: ',cost_function(P, x, t))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " ### Use the neural network:\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10; Nt = 10\n", + " x = np.linspace(0, 1, Nx)\n", + " t = np.linspace(0,1,Nt)\n", + "\n", + " ## Set up the parameters for the network\n", + " num_hidden_neurons = [50,20]\n", + " num_iter = 1000\n", + " lmb = 0.01\n", + "\n", + " P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " ## Store the results\n", + " res = np.zeros((Nx, Nt))\n", + " res_analytical = np.zeros((Nx, Nt))\n", + " for i,x_ in enumerate(x):\n", + " for j, t_ in enumerate(t):\n", + " point = np.array([x_, t_])\n", + " res[i,j] = g_trial(point,P)\n", + "\n", + " res_analytical[i,j] = g_analytic(point)\n", + "\n", + " diff = np.abs(res - res_analytical)\n", + " print(\"Max difference between analytical and solution from nn: %g\"%np.max(diff))\n", + "\n", + " ## Plot the solutions in two dimensions, that being in position and time\n", + "\n", + " T,X = np.meshgrid(t,x)\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))\n", + " s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Analytical solution')\n", + " s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Difference')\n", + " s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " ## Take some slices of the 3D plots just to see the solutions at particular times\n", + " indx1 = 0\n", + " indx2 = int(Nt/2)\n", + " indx3 = Nt-1\n", + "\n", + " t1 = t[indx1]\n", + " t2 = t[indx2]\n", + " t3 = t[indx3]\n", + "\n", + " # Slice the results from the DNN\n", + " res1 = res[:,indx1]\n", + " res2 = res[:,indx2]\n", + " res3 = res[:,indx3]\n", + "\n", + " # Slice the analytical results\n", + " res_analytical1 = res_analytical[:,indx1]\n", + " res_analytical2 = res_analytical[:,indx2]\n", + " res_analytical3 = res_analytical[:,indx3]\n", + "\n", + " # Plot the slices\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t1)\n", + " plt.plot(x, res1)\n", + " plt.plot(x,res_analytical1)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t2)\n", + " plt.plot(x, res2)\n", + " plt.plot(x,res_analytical2)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t3)\n", + " plt.plot(x, res3)\n", + " plt.plot(x,res_analytical3)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Resources on differential equations and deep learning\n", + "\n", + "1. [Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al](https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf)\n", + "\n", + "2. [Neural networks for solving differential equations by A. Honchar](https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c)\n", + "\n", + "3. [Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener](http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf)\n", + "\n", + "4. [Introduction to Partial Differential Equations by A. Tveito, R. Winther](https://www.springer.com/us/book/9783540225515)" + ] + } + ], + "metadata": { + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.8.5" + } + }, + "nbformat": 4, + "nbformat_minor": 4 +} \ No newline at end of file diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11.py b/doc/LectureNotes/_build/jupyter_execute/chapter11.py new file mode 100644 index 000000000..17470c2a0 --- /dev/null +++ b/doc/LectureNotes/_build/jupyter_execute/chapter11.py @@ -0,0 +1,2345 @@ +# Solving Differential Equations with Deep Learning + +The Universal Approximation Theorem states that a neural network can +approximate any function at a single hidden layer along with one input +and output layer to any given precision. + + +An ordinary differential equation (ODE) is an equation involving functions having one variable. + +In general, an ordinary differential equation looks like + + +
+ +$$ +\begin{equation} \label{ode} \tag{1} +f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 +\end{equation} +$$ + +where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$. + +The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in ([1](#ode)). +The highest order of derivative, that is the value of $n$, determines to the order of the equation. +The equation is referred to as a $n$-th order ODE. +Along with ([1](#ode)), some additional conditions of the function $g(x)$ are typically given +for the solution to be unique. + + + +Let the trial solution $g_t(x)$ be + + +
+ +$$ +\begin{equation} + g_t(x) = h_1(x) + h_2(x,N(x,P)) +\label{_auto1} \tag{2} +\end{equation} +$$ + +where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set +of conditions, $N(x,P)$ a neural network with weights and biases +described by $P$ and $h_2(x, N(x,P))$ some expression involving the +neural network. The role of the function $h_2(x, N(x,P))$, is to +ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is +evaluated at the values of $x$ where the given conditions must be +satisfied. The function $h_1(x)$ should alone make $g_t(x)$ satisfy +the conditions. + +But what about the network $N(x,P)$? + + +As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation. + + + +For the minimization to be defined, we need to have a cost function at hand to minimize. + +It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in ([1](#ode)). +We can choose to consider the mean squared error as the cost function for an input $x$. +Since we are looking at one input, the cost function is just $f$ squared. +The cost function $c\left(x, P \right)$ can therefore be expressed as + +$$ +C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 +$$ + +If $N$ inputs are given as a vector $\boldsymbol{x}$ with elements $x_i$ for $i = 1,\dots,N$, +the cost function becomes + + +
+ +$$ +\begin{equation} \label{cost} \tag{3} + C\left(\boldsymbol{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 +\end{equation} +$$ + +The neural net should then find the parameters $P$ that minimizes the cost function in +([3](#cost)) for a set of $N$ training samples $x_i$. + + + +To perform the minimization using gradient descent, the gradient of $C\left(\boldsymbol{x}, P\right)$ is needed. +It might happen so that finding an analytical expression of the gradient of $C(\boldsymbol{x}, P)$ from ([3](#cost)) gets too messy, depending on which cost function one desires to use. + +Luckily, there exists libraries that makes the job for us through automatic differentiation. +Automatic differentiation is a method of finding the derivatives numerically with very high precision. + + +### Example: Exponential decay + +An exponential decay of a quantity $g(x)$ is described by the equation + + +
+ +$$ +\begin{equation} \label{solve_expdec} \tag{4} + g'(x) = -\gamma g(x) +\end{equation} +$$ + +with $g(0) = g_0$ for some chosen initial value $g_0$. + +The analytical solution of ([4](#solve_expdec)) is + + +
+ +$$ +\begin{equation} + g(x) = g_0 \exp\left(-\gamma x\right) +\label{_auto2} \tag{5} +\end{equation} +$$ + +Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of ([4](#solve_expdec)). + + + +The program will use a neural network to solve + + +
+ +$$ +\begin{equation} \label{solveode} \tag{6} +g'(x) = -\gamma g(x) +\end{equation} +$$ + +where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values. + +In this example, $\gamma = 2$ and $g_0 = 10$. + + +To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be + +$$ +g_t(x, P) = h_1(x) + h_2(x, N(x, P)) +$$ + +with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer. + + + +In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}}, P_{\text{output}} \}$. +If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer. + +The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. +If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix. + +Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron. + +It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of ([6](#solveode)). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution: + + +
+ +$$ +\begin{equation} \label{trial} \tag{7} +g_t(x, P) = g_0 + x \cdot N(x, P) +\end{equation} +$$ + +### Reformulating the problem + +We wish that our neural network manages to minimize a given cost function. + +A reformulation of out equation, ([6](#solveode)), must therefore be done, +such that it describes the problem a neural network can solve for. + +The neural network must find the set of weights and biases $P$ such that the trial solution in ([7](#trial)) satisfies ([6](#solveode)). + +The trial solution + +$$ +g_t(x, P) = g_0 + x \cdot N(x, P) +$$ + +has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that + + +
+ +$$ +\begin{equation} \label{nnmin} \tag{8} +g_t'(x, P) = - \gamma g_t(x, P) +\end{equation} +$$ + +is fulfilled as *best as possible*. + + +The left hand side and right hand side of ([8](#nnmin)) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible. +This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. +In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network. + +This gives the following cost function our neural network must solve for: + +$$ +\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} +$$ + +(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$) + +or, in terms of weights and biases for the hidden and output layer in our network: + +$$ +\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} +$$ + +for an input value $x$. + + + +If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes + + +
+ +$$ +\begin{equation} \label{min} \tag{9} +\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} +\end{equation} +$$ + +Letting $\boldsymbol{x}$ be a vector with elements $x_i$ and $C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes + +$$ +\min_{P} C(\boldsymbol{x}, P) +$$ + +In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as + +$$ +\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\boldsymbol{x}, \{P_{\text{hidden} }, P_{\text{output} }\}) +$$ + + +For simplicity, it is assumed that the input is an array $\boldsymbol{x} = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills ([9](#min)). + +First, the neural network must feed forward the inputs. +This means that $\boldsymbol{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. +The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\text{hidden} }$. + + +For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is: + +$$ +\begin{aligned} +z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ +&= +\begin{pmatrix} +b_i^{\text{hidden}} & w_i^{\text{hidden}} +\end{pmatrix} +\begin{pmatrix} +1 \\ +x_j +\end{pmatrix} +\end{aligned} +$$ + +The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector: + +$$ +\begin{aligned} +\boldsymbol{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ +&= +\begin{pmatrix} + b_i^{\text{hidden}} & w_i^{\text{hidden}} +\end{pmatrix} +\begin{pmatrix} +1 & 1 & \dots & 1 \\ +x_1 & x_2 & \dots & x_N +\end{pmatrix} \\ +&= \boldsymbol{p}_{i, \text{hidden}}^T X +\end{aligned} +$$ + +The vector $\boldsymbol{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to ([9](#min)). + +After having found $\boldsymbol{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\boldsymbol{z})$. + +In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: + +$$ +f(z) = \frac{1}{1 + \exp{(-z)}} +$$ + +It is possible to use other activations functions for the hidden layer also. + +The output $\boldsymbol{x}_i^{\text{hidden}}$ from each $i$-th hidden neuron is: + +$$ +\boldsymbol{x}_i^{\text{hidden} } = f\big( \boldsymbol{z}_{i}^{\text{hidden}} \big) +$$ + +The outputs $\boldsymbol{x}_i^{\text{hidden} } $ are then sent to the output layer. + +The output layer consists of one neuron in this case, and combines the +output from each of the neurons in the hidden layers. The output layer +combines the results from the hidden layer using some weights $w_i^{\text{output}}$ +and biases $b_i^{\text{output}}$. In this case, +it is assumes that the number of neurons in the output layer is one. + + + +The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously. + +$$ +\begin{aligned} +z_{1,j}^{\text{output}} & = +\begin{pmatrix} +b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} +\end{pmatrix} +\begin{pmatrix} +1 \\ +\boldsymbol{x}_j^{\text{hidden}} +\end{pmatrix} +\end{aligned} +$$ + +Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer: + +$$ +\boldsymbol{z}_{1}^{\text{output}} = +\begin{pmatrix} +b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} +\end{pmatrix} +\begin{pmatrix} +1 & 1 & \dots & 1 \\ +\boldsymbol{x}_1^{\text{hidden}} & \boldsymbol{x}_2^{\text{hidden}} & \dots & \boldsymbol{x}_N^{\text{hidden}} +\end{pmatrix} +$$ + +In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\boldsymbol{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\boldsymbol{z}_{1}^{\text{output}}$ is the final output of the network. + + +The next step is to decide how the parameters should be changed such that they minimize the cost function. + +The chosen cost function for this problem is + +$$ +C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 +$$ + +In order to minimize the cost function, an optimization method must be chosen. + +Here, gradient descent with a constant step size has been chosen. + +### Gradient descent + +The idea of the gradient descent algorithm is to update parameters in +a direction where the cost function decreases goes to a minimum. + +In general, the update of some parameters $\boldsymbol{\omega}$ given a cost +function defined by some weights $\boldsymbol{\omega}$, $C(\boldsymbol{x}, +\boldsymbol{\omega})$, goes as follows: + +$$ +\boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega}) +$$ + +for a number of iterations or until $ \big|\big| \boldsymbol{\omega}_{\text{new} } - \boldsymbol{\omega} \big|\big|$ becomes smaller than some given tolerance. + +The value of $\lambda$ decides how large steps the algorithm must take +in the direction of $ \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})$. +The notation $\nabla_{\boldsymbol{\omega}}$ express the gradient with respect +to the elements in $\boldsymbol{\omega}$. + +In our case, we have to minimize the cost function $C(\boldsymbol{x}, P)$ with +respect to the two sets of weights and biases, that is for the hidden +layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output} +}$ . + +This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by + +$$ +\begin{aligned} +P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\boldsymbol{x}, P) \\ +P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\boldsymbol{x}, P) +\end{aligned} +$$ + +### The code for solving the ODE + +%matplotlib inline + +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +# Assuming one input, hidden, and output layer +def neural_network(params, x): + + # Find the weights (including and biases) for the hidden and output layer. + # Assume that params is a list of parameters for each layer. + # The biases are the first element for each array in params, + # and the weights are the remaning elements in each array in params. + + w_hidden = params[0] + w_output = params[1] + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + ## Hidden layer: + + # Add a row of ones to include bias + x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_input) + x_hidden = sigmoid(z_hidden) + + ## Output layer: + + # Include bias: + x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) + + z_output = np.matmul(w_output, x_hidden) + x_output = z_output + + return x_output + +# The trial solution using the deep neural network: +def g_trial(x,params, g0 = 10): + return g0 + x*neural_network(params,x) + +# The right side of the ODE: +def g(x, g_trial, gamma = 2): + return -gamma*g_trial + +# The cost function: +def cost_function(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial(x,P) + + # Find the derivative w.r.t x of the neural network + d_net_out = elementwise_grad(neural_network,1)(P,x) + + # Find the derivative w.r.t x of the trial function + d_g_t = elementwise_grad(g_trial,0)(x,P) + + # The right side of the ODE + func = g(x, g_t) + + err_sqr = (d_g_t - func)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum / np.size(err_sqr) + +# Solve the exponential decay ODE using neural network with one input, hidden, and output layer +def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): + ## Set up initial weights and biases + + # For the hidden layer + p0 = npr.randn(num_neurons_hidden, 2 ) + + # For the output layer + p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included + + P = [p0, p1] + + print('Initial cost: %g'%cost_function(P, x)) + + ## Start finding the optimal weights using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_grad = grad(cost_function,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of two arrays; + # one for the gradient w.r.t P_hidden and + # one for the gradient w.r.t P_output + cost_grad = cost_function_grad(P, x) + + P[0] = P[0] - lmb * cost_grad[0] + P[1] = P[1] - lmb * cost_grad[1] + + print('Final cost: %g'%cost_function(P, x)) + + return P + +def g_analytic(x, gamma = 2, g0 = 10): + return g0*np.exp(-gamma*x) + +# Solve the given problem +if __name__ == '__main__': + # Set seed such that the weight are initialized + # with same weights and biases for every run. + npr.seed(15) + + ## Decide the vales of arguments to the function to solve + N = 10 + x = np.linspace(0, 1, N) + + ## Set up the initial parameters + num_hidden_neurons = 10 + num_iter = 10000 + lmb = 0.001 + + # Use the network + P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) + + # Print the deviation from the trial solution and true solution + res = g_trial(x,P) + res_analytical = g_analytic(x) + + print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) + + # Plot the results + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(x, res_analytical) + plt.plot(x, res[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('x') + plt.ylabel('g(x)') + plt.show() + +## The network with one input layer, specified number of hidden layers, and one output layer + +It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers. + +The number of neurons within each hidden layer are given as a list of integers in the program below. + +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +# The neural network with one input layer and one output layer, +# but with number of hidden layers specified by the user. +def deep_neural_network(deep_params, x): + # N_hidden is the number of hidden layers + + N_hidden = np.size(deep_params) - 1 # -1 since params consists of + # parameters to all the hidden + # layers AND the output layer. + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + # Due to multiple hidden layers, define a variable referencing to the + # output of the previous layer: + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output + +# The trial solution using the deep neural network: +def g_trial_deep(x,params, g0 = 10): + return g0 + x*deep_neural_network(params, x) + +# The right side of the ODE: +def g(x, g_trial, gamma = 2): + return -gamma*g_trial + +# The same cost function as before, but calls deep_neural_network instead. +def cost_function_deep(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial_deep(x,P) + + # Find the derivative w.r.t x of the neural network + d_net_out = elementwise_grad(deep_neural_network,1)(P,x) + + # Find the derivative w.r.t x of the trial function + d_g_t = elementwise_grad(g_trial_deep,0)(x,P) + + # The right side of the ODE + func = g(x, g_t) + + err_sqr = (d_g_t - func)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum / np.size(err_sqr) + +# Solve the exponential decay ODE using neural network with one input and one output layer, +# but with specified number of hidden layers from the user. +def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): + # num_hidden_neurons is now a list of number of neurons within each hidden layer + + # The number of elements in the list num_hidden_neurons thus represents + # the number of hidden layers. + + # Find the number of hidden layers: + N_hidden = np.size(num_neurons) + + ## Set up initial weights and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 ) + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: %g'%cost_function_deep(P, x)) + + ## Start finding the optimal weights using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_deep_grad = grad(cost_function_deep,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases + # in the hidden layers and output layers evaluated at x. + cost_deep_grad = cost_function_deep_grad(P, x) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_deep_grad[l] + + print('Final cost: %g'%cost_function_deep(P, x)) + + return P + +def g_analytic(x, gamma = 2, g0 = 10): + return g0*np.exp(-gamma*x) + +# Solve the given problem +if __name__ == '__main__': + npr.seed(15) + + ## Decide the vales of arguments to the function to solve + N = 10 + x = np.linspace(0, 1, N) + + ## Set up the initial parameters + num_hidden_neurons = np.array([10,10]) + num_iter = 10000 + lmb = 0.001 + + P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) + + res = g_trial_deep(x,P) + res_analytical = g_analytic(x) + + plt.figure(figsize=(10,10)) + + plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') + plt.plot(x, res_analytical) + plt.plot(x, res[0,:]) + plt.legend(['analytical','dnn']) + plt.ylabel('g(x)') + plt.show() + +### Example: Population growth + +A logistic model of population growth assumes that a population converges toward an equilibrium. +The population growth can be modeled by + + +
+ +$$ +\begin{equation} \label{log} \tag{10} + g'(t) = \alpha g(t)(A - g(t)) +\end{equation} +$$ + +where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment. +Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant. + +In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability +and high execution time (this might be more apparent in the examples solving PDEs), +using a library like TensorFlow is recommended. +Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method. + + + +Here, we will model a population $g(t)$ in an environment having carrying capacity $A$. +The population follows the model + + +
+ +$$ +\begin{equation} \label{solveode_population} \tag{11} +g'(t) = \alpha g(t)(A - g(t)) +\end{equation} +$$ + +where $g(0) = g_0$. + +In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$. + + +We will get a slightly different trial solution, as the boundary conditions are different +compared to the case for exponential decay. + +A possible trial solution satisfying the condition $g(0) = g_0$ could be + +$$ +h_1(t) = g_0 + t \cdot N(t,P) +$$ + +with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$. + +The analytical solution is + +$$ +g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} +$$ + + + +The network will be the similar as for the exponential decay example, but with some small modifications for our problem. + +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +# Function to get the parameters. +# Done such that one can easily change the paramaters after one's liking. +def get_parameters(): + alpha = 2 + A = 1 + g0 = 1.2 + return alpha, A, g0 + +def deep_neural_network(P, x): + # N_hidden is the number of hidden layers + N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + # Due to multiple hidden layers, define a variable referencing to the + # output of the previous layer: + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = P[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = P[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output + + +def cost_function_deep(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial_deep(x,P) + + # Find the derivative w.r.t x of the trial function + d_g_t = elementwise_grad(g_trial_deep,0)(x,P) + + # The right side of the ODE + func = f(x, g_t) + + err_sqr = (d_g_t - func)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum / np.size(err_sqr) + +# The right side of the ODE: +def f(x, g_trial): + alpha,A, g0 = get_parameters() + return alpha*g_trial*(A - g_trial) + +# The trial solution using the deep neural network: +def g_trial_deep(x, params): + alpha,A, g0 = get_parameters() + return g0 + x*deep_neural_network(params,x) + +# The analytical solution: +def g_analytic(t): + alpha,A, g0 = get_parameters() + return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) + +def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): + # num_hidden_neurons is now a list of number of neurons within each hidden layer + + # Find the number of hidden layers: + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 ) + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: %g'%cost_function_deep(P, x)) + + ## Start finding the optimal weigths using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_deep_grad = grad(cost_function_deep,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases + # in the hidden layers and output layers evaluated at x. + cost_deep_grad = cost_function_deep_grad(P, x) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_deep_grad[l] + + print('Final cost: %g'%cost_function_deep(P, x)) + + return P + +if __name__ == '__main__': + npr.seed(4155) + + ## Decide the vales of arguments to the function to solve + Nt = 10 + T = 1 + t = np.linspace(0,T, Nt) + + ## Set up the initial parameters + num_hidden_neurons = [100, 50, 25] + num_iter = 1000 + lmb = 1e-3 + + P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) + + g_dnn_ag = g_trial_deep(t,P) + g_analytical = g_analytic(t) + + # Find the maximum absolute difference between the solutons: + diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) + print("The max absolute difference between the solutions is: %g"%diff_ag) + + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(t, g_analytical) + plt.plot(t, g_dnn_ag[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('t') + plt.ylabel('g(t)') + + plt.show() + +## Using forward Euler to solve the ODE + +A straightforward way of solving an ODE numerically, is to use Euler's method. + +Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$: + +$$ +f(x + \Delta x) \approx f(x) + \Delta x f'(x) +$$ + +In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields + +$$ +\begin{aligned} + g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ + &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) +\end{aligned} +$$ + +along with the condition that $g(0) = g_0$. + +Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$. + +For $i \geq 1$, we have that + +$$ +\begin{aligned} +t_i &= i\Delta t \\ +&= (i - 1)\Delta t + \Delta t \\ +&= t_{i-1} + \Delta t +\end{aligned} +$$ + +Now, if $g_i = g(t_i)$ then + + +
+ +$$ +\begin{equation} + \begin{aligned} + g_i &= g(t_i) \\ + &= g(t_{i-1} + \Delta t) \\ + &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ + &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) + \end{aligned} +\end{equation} \label{odenum} \tag{12} +$$ + +for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$. + +Equation ([12](#odenum)) could be implemented in the following way, +extending the program that uses the network using Autograd: + +# Assume that all function definitions from the example program using Autograd +# are located here. + +if __name__ == '__main__': + npr.seed(4155) + + ## Decide the vales of arguments to the function to solve + Nt = 10 + T = 1 + t = np.linspace(0,T, Nt) + + ## Set up the initial parameters + num_hidden_neurons = [100,50,25] + num_iter = 1000 + lmb = 1e-3 + + P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) + + g_dnn_ag = g_trial_deep(t,P) + g_analytical = g_analytic(t) + + # Find the maximum absolute difference between the solutons: + diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) + print("The max absolute difference between the solutions is: %g"%diff_ag) + + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(t, g_analytical) + plt.plot(t, g_dnn_ag[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('t') + plt.ylabel('g(t)') + + ## Find an approximation to the funtion using forward Euler + + alpha, A, g0 = get_parameters() + dt = T/(Nt - 1) + + # Perform forward Euler to solve the ODE + g_euler = np.zeros(Nt) + g_euler[0] = g0 + + for i in range(1,Nt): + g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1])) + + # Print the errors done by each method + diff1 = np.max(np.abs(g_euler - g_analytical)) + diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical)) + + print('Max absolute difference between Euler method and analytical: %g'%diff1) + print('Max absolute difference between deep neural network and analytical: %g'%diff2) + + # Plot results + plt.figure(figsize=(10,10)) + + plt.plot(t,g_euler) + plt.plot(t,g_analytical) + plt.plot(t,g_dnn_ag[0,:]) + + plt.legend(['euler','analytical','dnn']) + plt.xlabel('Time t') + plt.ylabel('g(t)') + + plt.show() + +## Solving the one dimensional Poisson equation + +The Poisson equation for $g(x)$ in one dimension is + + +
+ +$$ +\begin{equation} \label{poisson} \tag{13} + -g''(x) = f(x) +\end{equation} +$$ + +where $f(x)$ is a given function for $x \in (0,1)$. + +The conditions that $g(x)$ is chosen to fulfill, are + +$$ +\begin{align*} + g(0) &= 0 \\ + g(1) &= 0 +\end{align*} +$$ + +This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. +The results from the networks can then be compared to the analytical solution. +In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks. + + +Here, the function $g(x)$ to solve for follows the equation + +$$ +-g''(x) = f(x),\qquad x \in (0,1) +$$ + +where $f(x)$ is a given function, along with the chosen conditions + + +
+ +$$ +\begin{aligned} +g(0) = g(1) = 0 +\end{aligned}\label{cond} \tag{14} +$$ + +In this example, we consider the case when $f(x) = (3x + x^2)\exp(x)$. + +For this case, a possible trial solution satisfying the conditions could be + +$$ +g_t(x) = x \cdot (1-x) \cdot N(P,x) +$$ + +The analytical solution for this problem is + +$$ +g(x) = x(1 - x)\exp(x) +$$ + +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + # Due to multiple hidden layers, define a variable referencing to the + # output of the previous layer: + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output + +def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): + # num_hidden_neurons is now a list of number of neurons within each hidden layer + + # Find the number of hidden layers: + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 ) + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: %g'%cost_function_deep(P, x)) + + ## Start finding the optimal weigths using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_deep_grad = grad(cost_function_deep,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases + # in the hidden layers and output layers evaluated at x. + cost_deep_grad = cost_function_deep_grad(P, x) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_deep_grad[l] + + print('Final cost: %g'%cost_function_deep(P, x)) + + return P + +## Set up the cost function specified for this Poisson equation: + +# The right side of the ODE +def f(x): + return (3*x + x**2)*np.exp(x) + +def cost_function_deep(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial_deep(x,P) + + # Find the derivative w.r.t x of the trial function + d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) + + right_side = f(x) + + err_sqr = (-d2_g_t - right_side)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum/np.size(err_sqr) + +# The trial solution: +def g_trial_deep(x,P): + return x*(1-x)*deep_neural_network(P,x) + +# The analytic solution; +def g_analytic(x): + return x*(1-x)*np.exp(x) + +if __name__ == '__main__': + npr.seed(4155) + + ## Decide the vales of arguments to the function to solve + Nx = 10 + x = np.linspace(0,1, Nx) + + ## Set up the initial parameters + num_hidden_neurons = [200,100] + num_iter = 1000 + lmb = 1e-3 + + P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) + + g_dnn_ag = g_trial_deep(x,P) + g_analytical = g_analytic(x) + + # Find the maximum absolute difference between the solutons: + max_diff = np.max(np.abs(g_dnn_ag - g_analytical)) + print("The max absolute difference between the solutions is: %g"%max_diff) + + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(x, g_analytical) + plt.plot(x, g_dnn_ag[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('x') + plt.ylabel('g(x)') + plt.show() + +### Comparing with a numerical scheme + +The Poisson equation is possible to solve using Taylor series to approximate the second derivative. + +Using Taylor series, the second derivative can be expressed as + +$$ +g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) +$$ + +where $\Delta x$ is a small step size and $E_{\Delta x}(x)$ being the error term. + +Looking away from the error terms gives an approximation to the second derivative: + + +
+ +$$ +\begin{equation} \label{approx} \tag{15} +g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} +\end{equation} +$$ + +If $x_i = i \Delta x = x_{i-1} + \Delta x$ and $g_i = g(x_i)$ for $i = 1,\dots N_x - 2$ with $N_x$ being the number of values for $x$, ([15](#approx)) becomes + +$$ +\begin{aligned} +g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ +&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} +\end{aligned} +$$ + +Since we know from our problem that + +$$ +\begin{aligned} +-g''(x) &= f(x) \\ +&= (3x + x^2)\exp(x) +\end{aligned} +$$ + +along with the conditions $g(0) = g(1) = 0$, +the following scheme can be used to find an approximate solution for $g(x)$ numerically: + + +
+ +$$ +\begin{equation} + \begin{aligned} + -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ + -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) + \end{aligned} +\end{equation} \label{odesys} \tag{16} +$$ + +for $i = 1, \dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\exp(x_i)$, which is given for our specific problem. + +The equation can be rewritten into a matrix equation: + +$$ +\begin{aligned} +\begin{pmatrix} +2 & -1 & 0 & \dots & 0 \\ +-1 & 2 & -1 & \dots & 0 \\ +\vdots & & \ddots & & \vdots \\ +0 & \dots & -1 & 2 & -1 \\ +0 & \dots & 0 & -1 & 2\\ +\end{pmatrix} +\begin{pmatrix} +g_1 \\ +g_2 \\ +\vdots \\ +g_{N_x - 3} \\ +g_{N_x - 2} +\end{pmatrix} +&= +\Delta x^2 +\begin{pmatrix} +f(x_1) \\ +f(x_2) \\ +\vdots \\ +f(x_{N_x - 3}) \\ +f(x_{N_x - 2}) +\end{pmatrix} \\ +\boldsymbol{A}\boldsymbol{g} &= \boldsymbol{f}, +\end{aligned} +$$ + +which makes it possible to solve for the vector $\boldsymbol{g}$. + + +We can then compare the result from this numerical scheme with the output from our network using Autograd: + +import autograd.numpy as np +from autograd import grad, elementwise_grad +import autograd.numpy.random as npr +from matplotlib import pyplot as plt + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assumes input x being an one-dimensional array + num_values = np.size(x) + x = x.reshape(-1, num_values) + + # Assume that the input layer does nothing to the input x + x_input = x + + # Due to multiple hidden layers, define a variable referencing to the + # output of the previous layer: + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output + +def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): + # num_hidden_neurons is now a list of number of neurons within each hidden layer + + # Find the number of hidden layers: + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 ) + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: %g'%cost_function_deep(P, x)) + + ## Start finding the optimal weigths using gradient descent + + # Find the Python function that represents the gradient of the cost function + # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer + cost_function_deep_grad = grad(cost_function_deep,0) + + # Let the update be done num_iter times + for i in range(num_iter): + # Evaluate the gradient at the current weights and biases in P. + # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases + # in the hidden layers and output layers evaluated at x. + cost_deep_grad = cost_function_deep_grad(P, x) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_deep_grad[l] + + print('Final cost: %g'%cost_function_deep(P, x)) + + return P + +## Set up the cost function specified for this Poisson equation: + +# The right side of the ODE +def f(x): + return (3*x + x**2)*np.exp(x) + +def cost_function_deep(P, x): + + # Evaluate the trial function with the current parameters P + g_t = g_trial_deep(x,P) + + # Find the derivative w.r.t x of the trial function + d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) + + right_side = f(x) + + err_sqr = (-d2_g_t - right_side)**2 + cost_sum = np.sum(err_sqr) + + return cost_sum/np.size(err_sqr) + +# The trial solution: +def g_trial_deep(x,P): + return x*(1-x)*deep_neural_network(P,x) + +# The analytic solution; +def g_analytic(x): + return x*(1-x)*np.exp(x) + +if __name__ == '__main__': + npr.seed(4155) + + ## Decide the vales of arguments to the function to solve + Nx = 10 + x = np.linspace(0,1, Nx) + + ## Set up the initial parameters + num_hidden_neurons = [200,100] + num_iter = 1000 + lmb = 1e-3 + + P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) + + g_dnn_ag = g_trial_deep(x,P) + g_analytical = g_analytic(x) + + # Find the maximum absolute difference between the solutons: + + plt.figure(figsize=(10,10)) + + plt.title('Performance of neural network solving an ODE compared to the analytical solution') + plt.plot(x, g_analytical) + plt.plot(x, g_dnn_ag[0,:]) + plt.legend(['analytical','nn']) + plt.xlabel('x') + plt.ylabel('g(x)') + + ## Perform the computation using the numerical scheme + + dx = 1/(Nx - 1) + + # Set up the matrix A + A = np.zeros((Nx-2,Nx-2)) + + A[0,0] = 2 + A[0,1] = -1 + + for i in range(1,Nx-3): + A[i,i-1] = -1 + A[i,i] = 2 + A[i,i+1] = -1 + + A[Nx - 3, Nx - 4] = -1 + A[Nx - 3, Nx - 3] = 2 + + # Set up the vector f + f_vec = dx**2 * f(x[1:-1]) + + # Solve the equation + g_res = np.linalg.solve(A,f_vec) + + g_vec = np.zeros(Nx) + g_vec[1:-1] = g_res + + # Print the differences between each method + max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical)) + max_diff2 = np.max(np.abs(g_vec - g_analytical)) + print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1) + print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2) + + # Plot the results + plt.figure(figsize=(10,10)) + + plt.plot(x,g_vec) + plt.plot(x,g_analytical) + plt.plot(x,g_dnn_ag[0,:]) + + plt.legend(['numerical scheme','analytical','dnn']) + plt.show() + +## Partial Differential Equations + +A partial differential equation (PDE) has a solution here the function +is defined by multiple variables. The equation may involve all kinds +of combinations of which variables the function is differentiated with +respect to. + +In general, a partial differential equation for a function $g(x_1,\dots,x_N)$ with $N$ variables may be expressed as + + +
+ +$$ +\begin{equation} \label{PDE} \tag{17} + f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 +\end{equation} +$$ + +where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given. + +### Type of problem + +The problem our network must solve for, is similar to the ODE case. +We must have a trial solution $g_t$ at hand. + +For instance, the trial solution could be expressed as + +$$ +\begin{align*} + g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) +\end{align*} +$$ + +where $h_1(x_1,\dots,x_N)$ is a function that ensures $g_t(x_1,\dots,x_N)$ satisfies some given conditions. +The neural network $N(x_1,\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$ is an expression using the output from the neural network in some way. + +The role of the function $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$, is to ensure that the output of $N(x_1,\dots,x_N,P)$ is zero when $g_t(x_1,\dots,x_N)$ is evaluated at the values of $x_1,\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\dots,x_N)$ should alone make $g_t(x_1,\dots,x_N)$ satisfy the conditions. + + + +### Network requirements + +The network tries then the minimize the cost function following the +same ideas as described for the ODE case, but now with more than one +variables to consider. The concept still remains the same; find a set +of parameters $P$ such that the expression $f$ in ([17](#PDE)) is as +close to zero as possible. + +As for the ODE case, the cost function is the mean squared error that +the network must try to minimize. The cost function for the network to +minimize is + +$$ +C\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 +$$ + +If we let $\boldsymbol{x} = \big( x_1, \dots, x_N \big)$ be an array containing the values for $x_1, \dots, x_N$ respectively, the cost function can be reformulated into the following: + +$$ +C\left(\boldsymbol{x}, P\right) = f\left( \left( \boldsymbol{x}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}) }{\partial x_N^n} \right) \right)^2 +$$ + +If we also have $M$ different sets of values for $x_1, \dots, x_N$, that is $\boldsymbol{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)$ for $i = 1,\dots,M$ being the rows in matrix $X$, the cost function can be generalized into + +$$ +C\left(X, P \right) = \sum_{i=1}^M f\left( \left( \boldsymbol{x}_i, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}_i) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}_i) }{\partial x_N^n} \right) \right)^2. +$$ + +## Example: The diffusion equation + +In one spatial dimension, the equation reads + +$$ +\frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} +$$ + +where a possible choice of conditions are + +$$ +\begin{align*} +g(0,t) &= 0 ,\qquad t \geq 0 \\ +g(1,t) &= 0, \qquad t \geq 0 \\ +g(x,0) &= u(x),\qquad x\in [0,1] +\end{align*} +$$ + +with $u(x)$ being some given function. + + + +For this case, we want to find $g(x,t)$ such that + + +
+ +$$ +\begin{equation} + \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation} \label{diffonedim} \tag{18} +$$ + +and + +$$ +\begin{align*} +g(0,t) &= 0 ,\qquad t \geq 0 \\ +g(1,t) &= 0, \qquad t \geq 0 \\ +g(x,0) &= u(x),\qquad x\in [0,1] +\end{align*} +$$ + +with $u(x) = \sin(\pi x)$. + +First, let us set up the deep neural network. +The deep neural network will follow the same structure as discussed in the examples solving the ODEs. +First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions. + + + + +The only change to do here, is to extend our network such that +functions of multiple parameters are correctly handled. In this case +we have two variables in our function to solve for, that is time $t$ +and position $x$. The variables will be represented by a +one-dimensional array in the program. The program will evaluate the +network at each possible pair $(x,t)$, given an array for the desired +$x$-values and $t$-values to approximate the solution at. + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # x is now a point and a 1D numpy array; make it a column vector + num_coordinates = np.size(x,0) + x = x.reshape(num_coordinates,-1) + + num_points = np.size(x,1) + + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assume that the input layer does nothing to the input x + x_input = x + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output[0][0] + +The cost function must then iterate through the given arrays +containing values for $x$ and $t$, defines a point $(x,t)$ the deep +neural network and the trial solution is evaluated at, and then finds +the Jacobian of the trial solution. + +A possible trial solution for this PDE is + +$$ +g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) +$$ + +with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$. + +To fulfill the conditions, $A(x,t)$ could be: + +$$ +h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) +$$ +since $(0) = u(1) = 0$ and $u(x) = \sin(\pi x)$. + + + +The Jacobian is used because the program must find the derivative of +the trial solution with respect to $x$ and $t$. + +This gives the necessity of computing the Jacobian matrix, as we want +to evaluate the gradient with respect to $x$ and $t$ (note that the +Jacobian of a scalar-valued multivariate function is simply its +gradient). + +In Autograd, the differentiation is by default done with respect to +the first input argument of your Python function. Since the points is +an array representing $x$ and $t$, the Jacobian is calculated using +the values of $x$ and $t$. + +To find the second derivative with respect to $x$ and $t$, the +Jacobian can be found for the second time. The result is a Hessian +matrix, which is the matrix containing all the possible second order +mixed derivatives of $g(x,t)$. + +# Set up the trial function: +def u(x): + return np.sin(np.pi*x) + +def g_trial(point,P): + x,t = point + return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) + +# The right side of the ODE: +def f(point): + return 0. + +# The cost function: +def cost_function(P, x, t): + cost_sum = 0 + + g_t_jacobian_func = jacobian(g_trial) + g_t_hessian_func = hessian(g_trial) + + for x_ in x: + for t_ in t: + point = np.array([x_,t_]) + + g_t = g_trial(point,P) + g_t_jacobian = g_t_jacobian_func(point,P) + g_t_hessian = g_t_hessian_func(point,P) + + g_t_dt = g_t_jacobian[1] + g_t_d2x = g_t_hessian[0][0] + + func = f(point) + + err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 + cost_sum += err_sqr + + return cost_sum + +### Setting up the network using Autograd; The full program + +Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution. + +The analytical solution of our problem is + +$$ +g(x,t) = \exp(-\pi^2 t)\sin(\pi x) +$$ + +A possible way to implement a neural network solving the PDE, is given below. +Be aware, though, that it is fairly slow for the parameters used. +A better result is possible, but requires more iterations, and thus longer time to complete. + + +Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. +Using TensorFlow results in a much better execution time. Try it! + +import autograd.numpy as np +from autograd import jacobian,hessian,grad +import autograd.numpy.random as npr +from matplotlib import cm +from matplotlib import pyplot as plt +from mpl_toolkits.mplot3d import axes3d + +## Set up the network + +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # x is now a point and a 1D numpy array; make it a column vector + num_coordinates = np.size(x,0) + x = x.reshape(num_coordinates,-1) + + num_points = np.size(x,1) + + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assume that the input layer does nothing to the input x + x_input = x + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output[0][0] + +## Define the trial solution and cost function +def u(x): + return np.sin(np.pi*x) + +def g_trial(point,P): + x,t = point + return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) + +# The right side of the ODE: +def f(point): + return 0. + +# The cost function: +def cost_function(P, x, t): + cost_sum = 0 + + g_t_jacobian_func = jacobian(g_trial) + g_t_hessian_func = hessian(g_trial) + + for x_ in x: + for t_ in t: + point = np.array([x_,t_]) + + g_t = g_trial(point,P) + g_t_jacobian = g_t_jacobian_func(point,P) + g_t_hessian = g_t_hessian_func(point,P) + + g_t_dt = g_t_jacobian[1] + g_t_d2x = g_t_hessian[0][0] + + func = f(point) + + err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 + cost_sum += err_sqr + + return cost_sum /( np.size(x)*np.size(t) ) + +## For comparison, define the analytical solution +def g_analytic(point): + x,t = point + return np.exp(-np.pi**2*t)*np.sin(np.pi*x) + +## Set up a function for training the network to solve for the equation +def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): + ## Set up initial weigths and biases + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: ',cost_function(P, x, t)) + + cost_function_grad = grad(cost_function,0) + + # Let the update be done num_iter times + for i in range(num_iter): + cost_grad = cost_function_grad(P, x , t) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_grad[l] + + print('Final cost: ',cost_function(P, x, t)) + + return P + +if __name__ == '__main__': + ### Use the neural network: + npr.seed(15) + + ## Decide the vales of arguments to the function to solve + Nx = 10; Nt = 10 + x = np.linspace(0, 1, Nx) + t = np.linspace(0,1,Nt) + + ## Set up the parameters for the network + num_hidden_neurons = [100, 25] + num_iter = 250 + lmb = 0.01 + + P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) + + ## Store the results + g_dnn_ag = np.zeros((Nx, Nt)) + G_analytical = np.zeros((Nx, Nt)) + for i,x_ in enumerate(x): + for j, t_ in enumerate(t): + point = np.array([x_, t_]) + g_dnn_ag[i,j] = g_trial(point,P) + + G_analytical[i,j] = g_analytic(point) + + # Find the map difference between the analytical and the computed solution + diff_ag = np.abs(g_dnn_ag - G_analytical) + print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag)) + + ## Plot the solutions in two dimensions, that being in position and time + + T,X = np.meshgrid(t,x) + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) + s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Analytical solution') + s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Difference') + s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + ## Take some slices of the 3D plots just to see the solutions at particular times + indx1 = 0 + indx2 = int(Nt/2) + indx3 = Nt-1 + + t1 = t[indx1] + t2 = t[indx2] + t3 = t[indx3] + + # Slice the results from the DNN + res1 = g_dnn_ag[:,indx1] + res2 = g_dnn_ag[:,indx2] + res3 = g_dnn_ag[:,indx3] + + # Slice the analytical results + res_analytical1 = G_analytical[:,indx1] + res_analytical2 = G_analytical[:,indx2] + res_analytical3 = G_analytical[:,indx3] + + # Plot the slices + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t1) + plt.plot(x, res1) + plt.plot(x,res_analytical1) + plt.legend(['dnn','analytical']) + + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t2) + plt.plot(x, res2) + plt.plot(x,res_analytical2) + plt.legend(['dnn','analytical']) + + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t3) + plt.plot(x, res3) + plt.plot(x,res_analytical3) + plt.legend(['dnn','analytical']) + + plt.show() + +## Solving the wave equation with Neural Networks + +The wave equation is + +$$ +\frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} +$$ + +with $c$ being the specified wave speed. + +Here, the chosen conditions are + +$$ +\begin{align*} + g(0,t) &= 0 \\ + g(1,t) &= 0 \\ + g(x,0) &= u(x) \\ + \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) +\end{align*} +$$ + +where $\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions. + + +The wave equation to solve for, is + + +
+ +$$ +\begin{equation} \label{wave} \tag{19} +\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} +\end{equation} +$$ + +where $c$ is the given wave speed. +The chosen conditions for this equation are + + +
+ +$$ +\begin{aligned} +g(0,t) &= 0, &t \geq 0 \\ +g(1,t) &= 0, &t \geq 0 \\ +g(x,0) &= u(x), &x\in[0,1] \\ +\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] +\end{aligned} \label{condwave} \tag{20} +$$ + +In this example, let $c = 1$ and $u(x) = \sin(\pi x)$ and $v(x) = -\pi\sin(\pi x)$. + + + +Setting up the network is done in similar matter as for the example of solving the diffusion equation. +The only things we have to change, is the trial solution such that it satisfies the conditions from ([20](#condwave)) and the cost function. + +The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is + +$$ +g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) +$$ + +where + +$$ +h_1(x,t) = (1-t^2)u(x) + tv(x) +$$ + +Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example. + + +The analytical solution for our specific problem, is + +$$ +g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) +$$ + +import autograd.numpy as np +from autograd import hessian,grad +import autograd.numpy.random as npr +from matplotlib import cm +from matplotlib import pyplot as plt +from mpl_toolkits.mplot3d import axes3d + +## Set up the trial function: +def u(x): + return np.sin(np.pi*x) + +def v(x): + return -np.pi*np.sin(np.pi*x) + +def h1(point): + x,t = point + return (1 - t**2)*u(x) + t*v(x) + +def g_trial(point,P): + x,t = point + return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point) + +## Define the cost function +def cost_function(P, x, t): + cost_sum = 0 + + g_t_hessian_func = hessian(g_trial) + + for x_ in x: + for t_ in t: + point = np.array([x_,t_]) + + g_t_hessian = g_t_hessian_func(point,P) + + g_t_d2x = g_t_hessian[0][0] + g_t_d2t = g_t_hessian[1][1] + + err_sqr = ( (g_t_d2t - g_t_d2x) )**2 + cost_sum += err_sqr + + return cost_sum / (np.size(t) * np.size(x)) + +## The neural network +def sigmoid(z): + return 1/(1 + np.exp(-z)) + +def deep_neural_network(deep_params, x): + # x is now a point and a 1D numpy array; make it a column vector + num_coordinates = np.size(x,0) + x = x.reshape(num_coordinates,-1) + + num_points = np.size(x,1) + + # N_hidden is the number of hidden layers + N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer + + # Assume that the input layer does nothing to the input x + x_input = x + x_prev = x_input + + ## Hidden layers: + + for l in range(N_hidden): + # From the list of parameters P; find the correct weigths and bias for this layer + w_hidden = deep_params[l] + + # Add a row of ones to include bias + x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) + + z_hidden = np.matmul(w_hidden, x_prev) + x_hidden = sigmoid(z_hidden) + + # Update x_prev such that next layer can use the output from this layer + x_prev = x_hidden + + ## Output layer: + + # Get the weights and bias for this layer + w_output = deep_params[-1] + + # Include bias: + x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) + + z_output = np.matmul(w_output, x_prev) + x_output = z_output + + return x_output[0][0] + +## The analytical solution +def g_analytic(point): + x,t = point + return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t) + +def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): + ## Set up initial weigths and biases + N_hidden = np.size(num_neurons) + + ## Set up initial weigths and biases + + # Initialize the list of parameters: + P = [None]*(N_hidden + 1) # + 1 to include the output layer + + P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias + for l in range(1,N_hidden): + P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias + + # For the output layer + P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included + + print('Initial cost: ',cost_function(P, x, t)) + + cost_function_grad = grad(cost_function,0) + + # Let the update be done num_iter times + for i in range(num_iter): + cost_grad = cost_function_grad(P, x , t) + + for l in range(N_hidden+1): + P[l] = P[l] - lmb * cost_grad[l] + + + print('Final cost: ',cost_function(P, x, t)) + + return P + +if __name__ == '__main__': + ### Use the neural network: + npr.seed(15) + + ## Decide the vales of arguments to the function to solve + Nx = 10; Nt = 10 + x = np.linspace(0, 1, Nx) + t = np.linspace(0,1,Nt) + + ## Set up the parameters for the network + num_hidden_neurons = [50,20] + num_iter = 1000 + lmb = 0.01 + + P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) + + ## Store the results + res = np.zeros((Nx, Nt)) + res_analytical = np.zeros((Nx, Nt)) + for i,x_ in enumerate(x): + for j, t_ in enumerate(t): + point = np.array([x_, t_]) + res[i,j] = g_trial(point,P) + + res_analytical[i,j] = g_analytic(point) + + diff = np.abs(res - res_analytical) + print("Max difference between analytical and solution from nn: %g"%np.max(diff)) + + ## Plot the solutions in two dimensions, that being in position and time + + T,X = np.meshgrid(t,x) + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) + s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Analytical solution') + s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + + fig = plt.figure(figsize=(10,10)) + ax = fig.gca(projection='3d') + ax.set_title('Difference') + s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis) + ax.set_xlabel('Time $t$') + ax.set_ylabel('Position $x$'); + + ## Take some slices of the 3D plots just to see the solutions at particular times + indx1 = 0 + indx2 = int(Nt/2) + indx3 = Nt-1 + + t1 = t[indx1] + t2 = t[indx2] + t3 = t[indx3] + + # Slice the results from the DNN + res1 = res[:,indx1] + res2 = res[:,indx2] + res3 = res[:,indx3] + + # Slice the analytical results + res_analytical1 = res_analytical[:,indx1] + res_analytical2 = res_analytical[:,indx2] + res_analytical3 = res_analytical[:,indx3] + + # Plot the slices + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t1) + plt.plot(x, res1) + plt.plot(x,res_analytical1) + plt.legend(['dnn','analytical']) + + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t2) + plt.plot(x, res2) + plt.plot(x,res_analytical2) + plt.legend(['dnn','analytical']) + + plt.figure(figsize=(10,10)) + plt.title("Computed solutions at time = %g"%t3) + plt.plot(x, res3) + plt.plot(x,res_analytical3) + plt.legend(['dnn','analytical']) + + plt.show() + +## Resources on differential equations and deep learning + +1. [Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al](https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf) + +2. [Neural networks for solving differential equations by A. Honchar](https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c) + +3. [Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener](http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf) + +4. [Introduction to Partial Differential Equations by A. Tveito, R. Winther](https://www.springer.com/us/book/9783540225515) \ No newline at end of file diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11_47_2.png b/doc/LectureNotes/_build/jupyter_execute/chapter11_47_2.png new file mode 100644 index 000000000..5667f3351 Binary files /dev/null and b/doc/LectureNotes/_build/jupyter_execute/chapter11_47_2.png differ diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11_49_3.png b/doc/LectureNotes/_build/jupyter_execute/chapter11_49_3.png new file mode 100644 index 000000000..d8fe671bb Binary files /dev/null and b/doc/LectureNotes/_build/jupyter_execute/chapter11_49_3.png differ diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11_55_3.png b/doc/LectureNotes/_build/jupyter_execute/chapter11_55_3.png new file mode 100644 index 000000000..25118b436 Binary files /dev/null and b/doc/LectureNotes/_build/jupyter_execute/chapter11_55_3.png differ diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11_63_3.png b/doc/LectureNotes/_build/jupyter_execute/chapter11_63_3.png new file mode 100644 index 000000000..25118b436 Binary files /dev/null and b/doc/LectureNotes/_build/jupyter_execute/chapter11_63_3.png differ diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11_63_4.png b/doc/LectureNotes/_build/jupyter_execute/chapter11_63_4.png new file mode 100644 index 000000000..7b1474c1b Binary files /dev/null and b/doc/LectureNotes/_build/jupyter_execute/chapter11_63_4.png differ diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11_76_3.png b/doc/LectureNotes/_build/jupyter_execute/chapter11_76_3.png new file mode 100644 index 000000000..40d7e8815 Binary files /dev/null and b/doc/LectureNotes/_build/jupyter_execute/chapter11_76_3.png differ diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11_88_3.png b/doc/LectureNotes/_build/jupyter_execute/chapter11_88_3.png new file mode 100644 index 000000000..40d7e8815 Binary files /dev/null and b/doc/LectureNotes/_build/jupyter_execute/chapter11_88_3.png differ diff --git a/doc/LectureNotes/_build/jupyter_execute/chapter11_88_4.png b/doc/LectureNotes/_build/jupyter_execute/chapter11_88_4.png new file mode 100644 index 000000000..0af57c0aa Binary files /dev/null and b/doc/LectureNotes/_build/jupyter_execute/chapter11_88_4.png differ diff --git a/doc/LectureNotes/_toc.yml b/doc/LectureNotes/_toc.yml index 2754ecaff..52a203dfd 100644 --- a/doc/LectureNotes/_toc.yml +++ b/doc/LectureNotes/_toc.yml @@ -31,3 +31,5 @@ chapters: - file: chapter9.ipynb - file: chapter10.ipynb + - file: chapter11.ipynb + - file: chapter12.ipynb diff --git a/doc/LectureNotes/chapter11.ipynb b/doc/LectureNotes/chapter11.ipynb new file mode 100644 index 000000000..c03e4a368 --- /dev/null +++ b/doc/LectureNotes/chapter11.ipynb @@ -0,0 +1,3023 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Solving Differential Equations with Deep Learning\n", + "\n", + "The Universal Approximation Theorem states that a neural network can\n", + "approximate any function at a single hidden layer along with one input\n", + "and output layer to any given precision. \n", + "\n", + "\n", + "An ordinary differential equation (ODE) is an equation involving functions having one variable.\n", + "\n", + "In general, an ordinary differential equation looks like" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{ode} \\tag{1}\n", + "f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right) = 0\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$.\n", + "\n", + "The $f\\left(x, g(x), g'(x), g''(x), \\, \\dots \\, , g^{(n)}(x)\\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \\ g'(x), \\ g''(x), \\, \\dots \\, , \\text{ and } g^{(n)}(x)$ on the left side of the equality sign in ([1](#ode)).\n", + "The highest order of derivative, that is the value of $n$, determines to the order of the equation.\n", + "The equation is referred to as a $n$-th order ODE.\n", + "Along with ([1](#ode)), some additional conditions of the function $g(x)$ are typically given\n", + "for the solution to be unique.\n", + "\n", + "\n", + "\n", + "Let the trial solution $g_t(x)$ be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + "\tg_t(x) = h_1(x) + h_2(x,N(x,P))\n", + "\\label{_auto1} \\tag{2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set\n", + "of conditions, $N(x,P)$ a neural network with weights and biases\n", + "described by $P$ and $h_2(x, N(x,P))$ some expression involving the\n", + "neural network. The role of the function $h_2(x, N(x,P))$, is to\n", + "ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is\n", + "evaluated at the values of $x$ where the given conditions must be\n", + "satisfied. The function $h_1(x)$ should alone make $g_t(x)$ satisfy\n", + "the conditions.\n", + "\n", + "But what about the network $N(x,P)$?\n", + "\n", + "\n", + "As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.\n", + "\n", + "\n", + "\n", + "For the minimization to be defined, we need to have a cost function at hand to minimize.\n", + "\n", + "It is given that $f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right)$ should be equal to zero in ([1](#ode)).\n", + "We can choose to consider the mean squared error as the cost function for an input $x$.\n", + "Since we are looking at one input, the cost function is just $f$ squared.\n", + "The cost function $c\\left(x, P \\right)$ can therefore be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(x, P\\right) = \\big(f\\left(x, \\, g(x), \\, g'(x), \\, g''(x), \\, \\dots \\, , \\, g^{(n)}(x)\\right)\\big)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If $N$ inputs are given as a vector $\\boldsymbol{x}$ with elements $x_i$ for $i = 1,\\dots,N$,\n", + "the cost function becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{cost} \\tag{3}\n", + "\tC\\left(\\boldsymbol{x}, P\\right) = \\frac{1}{N} \\sum_{i=1}^N \\big(f\\left(x_i, \\, g(x_i), \\, g'(x_i), \\, g''(x_i), \\, \\dots \\, , \\, g^{(n)}(x_i)\\right)\\big)^2\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The neural net should then find the parameters $P$ that minimizes the cost function in\n", + "([3](#cost)) for a set of $N$ training samples $x_i$.\n", + "\n", + "\n", + "\n", + "To perform the minimization using gradient descent, the gradient of $C\\left(\\boldsymbol{x}, P\\right)$ is needed.\n", + "It might happen so that finding an analytical expression of the gradient of $C(\\boldsymbol{x}, P)$ from ([3](#cost)) gets too messy, depending on which cost function one desires to use.\n", + "\n", + "Luckily, there exists libraries that makes the job for us through automatic differentiation.\n", + "Automatic differentiation is a method of finding the derivatives numerically with very high precision.\n", + "\n", + "\n", + "### Example: Exponential decay\n", + "\n", + "An exponential decay of a quantity $g(x)$ is described by the equation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solve_expdec} \\tag{4}\n", + " g'(x) = -\\gamma g(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $g(0) = g_0$ for some chosen initial value $g_0$.\n", + "\n", + "The analytical solution of ([4](#solve_expdec)) is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " g(x) = g_0 \\exp\\left(-\\gamma x\\right)\n", + "\\label{_auto2} \\tag{5}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of ([4](#solve_expdec)).\n", + "\n", + "\n", + "\n", + "The program will use a neural network to solve" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solveode} \\tag{6}\n", + "g'(x) = -\\gamma g(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(0) = g_0$ with $\\gamma$ and $g_0$ being some chosen values.\n", + "\n", + "In this example, $\\gamma = 2$ and $g_0 = 10$.\n", + "\n", + "\n", + "To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x, P) = h_1(x) + h_2(x, N(x, P))\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.\n", + "\n", + "\n", + "\n", + "In this network, there are no weights and bias at the input layer, so $P = \\{ P_{\\text{hidden}}, P_{\\text{output}} \\}$.\n", + "If there are $N_{\\text{hidden} }$ neurons in the hidden layer, then $P_{\\text{hidden}}$ is a $N_{\\text{hidden} } \\times (1 + N_{\\text{input}})$ matrix, given that there are $N_{\\text{input}}$ neurons in the input layer.\n", + "\n", + "The first column in $P_{\\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer.\n", + "If there are $N_{\\text{output} }$ neurons in the output layer, then $P_{\\text{output}} $ is a $N_{\\text{output} } \\times (1 + N_{\\text{hidden} })$ matrix.\n", + "\n", + "Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.\n", + "\n", + "It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of ([6](#solveode)). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \\cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{trial} \\tag{7}\n", + "g_t(x, P) = g_0 + x \\cdot N(x, P)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Reformulating the problem\n", + "\n", + "We wish that our neural network manages to minimize a given cost function.\n", + "\n", + "A reformulation of out equation, ([6](#solveode)), must therefore be done,\n", + "such that it describes the problem a neural network can solve for.\n", + "\n", + "The neural network must find the set of weights and biases $P$ such that the trial solution in ([7](#trial)) satisfies ([6](#solveode)).\n", + "\n", + "The trial solution" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x, P) = g_0 + x \\cdot N(x, P)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{nnmin} \\tag{8}\n", + "g_t'(x, P) = - \\gamma g_t(x, P)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "is fulfilled as *best as possible*.\n", + "\n", + "\n", + "The left hand side and right hand side of ([8](#nnmin)) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible.\n", + "This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero.\n", + "In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network.\n", + "\n", + "This gives the following cost function our neural network must solve for:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P}\\Big\\{ \\big(g_t'(x, P) - ( -\\gamma g_t(x, P) \\big)^2 \\Big\\}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "(the notation $\\min_{P}\\{ f(x, P) \\}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$)\n", + "\n", + "or, in terms of weights and biases for the hidden and output layer in our network:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P_{\\text{hidden} }, \\ P_{\\text{output} }}\\Big\\{ \\big(g_t'(x, \\{ P_{\\text{hidden} }, P_{\\text{output} }\\}) - ( -\\gamma g_t(x, \\{ P_{\\text{hidden} }, P_{\\text{output} }\\}) \\big)^2 \\Big\\}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for an input value $x$.\n", + "\n", + "\n", + "\n", + "If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \\dots, N$, then the *total* error to minimize becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{min} \\tag{9}\n", + "\\min_{P}\\Big\\{\\frac{1}{N} \\sum_{i=1}^N \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2 \\Big\\}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Letting $\\boldsymbol{x}$ be a vector with elements $x_i$ and $C(\\boldsymbol{x}, P) = \\frac{1}{N} \\sum_i \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2$ denote the cost function, the minimization problem that our network must solve, becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\min_{P} C(\\boldsymbol{x}, P)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In terms of $P_{\\text{hidden} }$ and $P_{\\text{output} }$, this could also be expressed as\n", + "\n", + "$$\n", + "\\min_{P_{\\text{hidden} }, \\ P_{\\text{output} }} C(\\boldsymbol{x}, \\{P_{\\text{hidden} }, P_{\\text{output} }\\})\n", + "$$\n", + "\n", + "\n", + "For simplicity, it is assumed that the input is an array $\\boldsymbol{x} = (x_1, \\dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills ([9](#min)).\n", + "\n", + "First, the neural network must feed forward the inputs.\n", + "This means that $\\boldsymbol{x}s$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further.\n", + "The input layer will consist of $N_{\\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\\text{hidden} }$.\n", + "\n", + "\n", + "For the $i$-th in the hidden layer with weight $w_i^{\\text{hidden} }$ and bias $b_i^{\\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "z_{i,j}^{\\text{hidden}} &= b_i^{\\text{hidden}} + w_i^{\\text{hidden}}x_j \\\\\n", + "&=\n", + "\\begin{pmatrix}\n", + "b_i^{\\text{hidden}} & w_i^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 \\\\\n", + "x_j\n", + "\\end{pmatrix}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "\\boldsymbol{z}_{i}^{\\text{hidden}} &= \\Big( b_i^{\\text{hidden}} + w_i^{\\text{hidden}}x_1 , \\ b_i^{\\text{hidden}} + w_i^{\\text{hidden}} x_2, \\ \\dots \\, , \\ b_i^{\\text{hidden}} + w_i^{\\text{hidden}} x_N\\Big) \\\\\n", + "&=\n", + "\\begin{pmatrix}\n", + " b_i^{\\text{hidden}} & w_i^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 & 1 & \\dots & 1 \\\\\n", + "x_1 & x_2 & \\dots & x_N\n", + "\\end{pmatrix} \\\\\n", + "&= \\boldsymbol{p}_{i, \\text{hidden}}^T X\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The vector $\\boldsymbol{p}_{i, \\text{hidden}}^T$ constitutes each row in $P_{\\text{hidden} }$, which contains the weights for the neural network to minimize according to ([9](#min)).\n", + "\n", + "After having found $\\boldsymbol{z}_{i}^{\\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\\boldsymbol{z})$.\n", + "\n", + "In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "f(z) = \\frac{1}{1 + \\exp{(-z)}}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "It is possible to use other activations functions for the hidden layer also.\n", + "\n", + "The output $\\boldsymbol{x}_i^{\\text{hidden}}$ from each $i$-th hidden neuron is:\n", + "\n", + "$$\n", + "\\boldsymbol{x}_i^{\\text{hidden} } = f\\big( \\boldsymbol{z}_{i}^{\\text{hidden}} \\big)\n", + "$$\n", + "\n", + "The outputs $\\boldsymbol{x}_i^{\\text{hidden} } $ are then sent to the output layer.\n", + "\n", + "The output layer consists of one neuron in this case, and combines the\n", + "output from each of the neurons in the hidden layers. The output layer\n", + "combines the results from the hidden layer using some weights $w_i^{\\text{output}}$\n", + "and biases $b_i^{\\text{output}}$. In this case,\n", + "it is assumes that the number of neurons in the output layer is one.\n", + "\n", + "\n", + "\n", + "The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "z_{1,j}^{\\text{output}} & =\n", + "\\begin{pmatrix}\n", + "b_1^{\\text{output}} & \\boldsymbol{w}_1^{\\text{output}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 \\\\\n", + "\\boldsymbol{x}_j^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Expressing $z_{1,j}^{\\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\boldsymbol{z}_{1}^{\\text{output}} =\n", + "\\begin{pmatrix}\n", + "b_1^{\\text{output}} & \\boldsymbol{w}_1^{\\text{output}}\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "1 & 1 & \\dots & 1 \\\\\n", + "\\boldsymbol{x}_1^{\\text{hidden}} & \\boldsymbol{x}_2^{\\text{hidden}} & \\dots & \\boldsymbol{x}_N^{\\text{hidden}}\n", + "\\end{pmatrix}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\\boldsymbol{z}_{1}^{\\text{output}}$ the neural network has finished its feed forward step, and $\\boldsymbol{z}_{1}^{\\text{output}}$ is the final output of the network.\n", + "\n", + "\n", + "The next step is to decide how the parameters should be changed such that they minimize the cost function.\n", + "\n", + "The chosen cost function for this problem is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C(\\boldsymbol{x}, P) = \\frac{1}{N} \\sum_i \\big(g_t'(x_i, P) - ( -\\gamma g_t(x_i, P) \\big)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In order to minimize the cost function, an optimization method must be chosen.\n", + "\n", + "Here, gradient descent with a constant step size has been chosen.\n", + "\n", + "### Gradient descent\n", + "\n", + "The idea of the gradient descent algorithm is to update parameters in\n", + "a direction where the cost function decreases goes to a minimum.\n", + "\n", + "In general, the update of some parameters $\\boldsymbol{\\omega}$ given a cost\n", + "function defined by some weights $\\boldsymbol{\\omega}$, $C(\\boldsymbol{x},\n", + "\\boldsymbol{\\omega})$, goes as follows:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\boldsymbol{\\omega}_{\\text{new} } = \\boldsymbol{\\omega} - \\lambda \\nabla_{\\boldsymbol{\\omega}} C(\\boldsymbol{x}, \\boldsymbol{\\omega})\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for a number of iterations or until $ \\big|\\big| \\boldsymbol{\\omega}_{\\text{new} } - \\boldsymbol{\\omega} \\big|\\big|$ becomes smaller than some given tolerance.\n", + "\n", + "The value of $\\lambda$ decides how large steps the algorithm must take\n", + "in the direction of $ \\nabla_{\\boldsymbol{\\omega}} C(\\boldsymbol{x}, \\boldsymbol{\\omega})$.\n", + "The notation $\\nabla_{\\boldsymbol{\\omega}}$ express the gradient with respect\n", + "to the elements in $\\boldsymbol{\\omega}$.\n", + "\n", + "In our case, we have to minimize the cost function $C(\\boldsymbol{x}, P)$ with\n", + "respect to the two sets of weights and biases, that is for the hidden\n", + "layer $P_{\\text{hidden} }$ and for the output layer $P_{\\text{output}\n", + "}$ .\n", + "\n", + "This means that $P_{\\text{hidden} }$ and $P_{\\text{output} }$ is updated by" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "P_{\\text{hidden},\\text{new}} &= P_{\\text{hidden}} - \\lambda \\nabla_{P_{\\text{hidden}}} C(\\boldsymbol{x}, P) \\\\\n", + "P_{\\text{output},\\text{new}} &= P_{\\text{output}} - \\lambda \\nabla_{P_{\\text{output}}} C(\\boldsymbol{x}, P)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### The code for solving the ODE" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "%matplotlib inline\n", + "\n", + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# Assuming one input, hidden, and output layer\n", + "def neural_network(params, x):\n", + "\n", + " # Find the weights (including and biases) for the hidden and output layer.\n", + " # Assume that params is a list of parameters for each layer.\n", + " # The biases are the first element for each array in params,\n", + " # and the weights are the remaning elements in each array in params.\n", + "\n", + " w_hidden = params[0]\n", + " w_output = params[1]\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " ## Hidden layer:\n", + "\n", + " # Add a row of ones to include bias\n", + " x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_input)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " ## Output layer:\n", + "\n", + " # Include bias:\n", + " x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_hidden)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial(x,params, g0 = 10):\n", + " return g0 + x*neural_network(params,x)\n", + "\n", + "# The right side of the ODE:\n", + "def g(x, g_trial, gamma = 2):\n", + " return -gamma*g_trial\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the neural network\n", + " d_net_out = elementwise_grad(neural_network,1)(P,x)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = g(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# Solve the exponential decay ODE using neural network with one input, hidden, and output layer\n", + "def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):\n", + " ## Set up initial weights and biases\n", + "\n", + " # For the hidden layer\n", + " p0 = npr.randn(num_neurons_hidden, 2 )\n", + "\n", + " # For the output layer\n", + " p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included\n", + "\n", + " P = [p0, p1]\n", + "\n", + " print('Initial cost: %g'%cost_function(P, x))\n", + "\n", + " ## Start finding the optimal weights using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of two arrays;\n", + " # one for the gradient w.r.t P_hidden and\n", + " # one for the gradient w.r.t P_output\n", + " cost_grad = cost_function_grad(P, x)\n", + "\n", + " P[0] = P[0] - lmb * cost_grad[0]\n", + " P[1] = P[1] - lmb * cost_grad[1]\n", + "\n", + " print('Final cost: %g'%cost_function(P, x))\n", + "\n", + " return P\n", + "\n", + "def g_analytic(x, gamma = 2, g0 = 10):\n", + " return g0*np.exp(-gamma*x)\n", + "\n", + "# Solve the given problem\n", + "if __name__ == '__main__':\n", + " # Set seed such that the weight are initialized\n", + " # with same weights and biases for every run.\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " N = 10\n", + " x = np.linspace(0, 1, N)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = 10\n", + " num_iter = 10000\n", + " lmb = 0.001\n", + "\n", + " # Use the network\n", + " P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " # Print the deviation from the trial solution and true solution\n", + " res = g_trial(x,P)\n", + " res_analytical = g_analytic(x)\n", + "\n", + " print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))\n", + "\n", + " # Plot the results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, res_analytical)\n", + " plt.plot(x, res[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## The network with one input layer, specified number of hidden layers, and one output layer\n", + "\n", + "It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.\n", + "\n", + "The number of neurons within each hidden layer are given as a list of integers in the program below." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# The neural network with one input layer and one output layer,\n", + "# but with number of hidden layers specified by the user.\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + "\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consists of\n", + " # parameters to all the hidden\n", + " # layers AND the output layer.\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial_deep(x,params, g0 = 10):\n", + " return g0 + x*deep_neural_network(params, x)\n", + "\n", + "# The right side of the ODE:\n", + "def g(x, g_trial, gamma = 2):\n", + " return -gamma*g_trial\n", + "\n", + "# The same cost function as before, but calls deep_neural_network instead.\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the neural network\n", + " d_net_out = elementwise_grad(deep_neural_network,1)(P,x)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial_deep,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = g(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# Solve the exponential decay ODE using neural network with one input and one output layer,\n", + "# but with specified number of hidden layers from the user.\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # The number of elements in the list num_hidden_neurons thus represents\n", + " # the number of hidden layers.\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weights and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weights using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "def g_analytic(x, gamma = 2, g0 = 10):\n", + " return g0*np.exp(-gamma*x)\n", + "\n", + "# Solve the given problem\n", + "if __name__ == '__main__':\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " N = 10\n", + " x = np.linspace(0, 1, N)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = np.array([10,10])\n", + " num_iter = 10000\n", + " lmb = 0.001\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " res = g_trial_deep(x,P)\n", + " res_analytical = g_analytic(x)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, res_analytical)\n", + " plt.plot(x, res[0,:])\n", + " plt.legend(['analytical','dnn'])\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example: Population growth\n", + "\n", + "A logistic model of population growth assumes that a population converges toward an equilibrium.\n", + "The population growth can be modeled by" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{log} \\tag{10}\n", + "\tg'(t) = \\alpha g(t)(A - g(t))\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(t)$ is the population density at time $t$, $\\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment.\n", + "Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant.\n", + "\n", + "In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability\n", + "and high execution time (this might be more apparent in the examples solving PDEs),\n", + "using a library like TensorFlow is recommended.\n", + "Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method.\n", + "\n", + "\n", + "\n", + "Here, we will model a population $g(t)$ in an environment having carrying capacity $A$.\n", + "The population follows the model" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{solveode_population} \\tag{11}\n", + "g'(t) = \\alpha g(t)(A - g(t))\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $g(0) = g_0$.\n", + "\n", + "In this example, we let $\\alpha = 2$, $A = 1$, and $g_0 = 1.2$.\n", + "\n", + "\n", + "We will get a slightly different trial solution, as the boundary conditions are different\n", + "compared to the case for exponential decay.\n", + "\n", + "A possible trial solution satisfying the condition $g(0) = g_0$ could be\n", + "\n", + "$$\n", + "h_1(t) = g_0 + t \\cdot N(t,P)\n", + "$$\n", + "\n", + "with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$.\n", + "\n", + "The analytical solution is\n", + "\n", + "$$\n", + "g(t) = \\frac{Ag_0}{g_0 + (A - g_0)\\exp(-\\alpha A t)}\n", + "$$\n", + "\n", + "\n", + "\n", + "The network will be the similar as for the exponential decay example, but with some small modifications for our problem." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "# Function to get the parameters.\n", + "# Done such that one can easily change the paramaters after one's liking.\n", + "def get_parameters():\n", + " alpha = 2\n", + " A = 1\n", + " g0 = 1.2\n", + " return alpha, A, g0\n", + "\n", + "def deep_neural_network(P, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = P[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = P[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d_g_t = elementwise_grad(g_trial_deep,0)(x,P)\n", + "\n", + " # The right side of the ODE\n", + " func = f(x, g_t)\n", + "\n", + " err_sqr = (d_g_t - func)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum / np.size(err_sqr)\n", + "\n", + "# The right side of the ODE:\n", + "def f(x, g_trial):\n", + " alpha,A, g0 = get_parameters()\n", + " return alpha*g_trial*(A - g_trial)\n", + "\n", + "# The trial solution using the deep neural network:\n", + "def g_trial_deep(x, params):\n", + " alpha,A, g0 = get_parameters()\n", + " return g0 + x*deep_neural_network(params,x)\n", + "\n", + "# The analytical solution:\n", + "def g_analytic(t):\n", + " alpha,A, g0 = get_parameters()\n", + " return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nt = 10\n", + " T = 1\n", + " t = np.linspace(0,T, Nt)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [100, 50, 25]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(t,P)\n", + " g_analytical = g_analytic(t)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%diff_ag)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(t, g_analytical)\n", + " plt.plot(t, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Using forward Euler to solve the ODE\n", + "\n", + "A straightforward way of solving an ODE numerically, is to use Euler's method.\n", + "\n", + "Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\\Delta x$ from $x$:\n", + "\n", + "$$\n", + "f(x + \\Delta x) \\approx f(x) + \\Delta x f'(x)\n", + "$$\n", + "\n", + "In our case, using Euler's method to approximate the value of $g$ at a step $\\Delta t$ from $t$ yields" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + " g(t + \\Delta t) &\\approx g(t) + \\Delta t g'(t) \\\\\n", + " &= g(t) + \\Delta t \\big(\\alpha g(t)(A - g(t))\\big)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "along with the condition that $g(0) = g_0$.\n", + "\n", + "Let $t_i = i \\cdot \\Delta t$ where $\\Delta t = \\frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \\in [0, T]$ for $i = 0, \\dots, N_t-1$.\n", + "\n", + "For $i \\geq 1$, we have that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "t_i &= i\\Delta t \\\\\n", + "&= (i - 1)\\Delta t + \\Delta t \\\\\n", + "&= t_{i-1} + \\Delta t\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Now, if $g_i = g(t_i)$ then" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\begin{aligned}\n", + " g_i &= g(t_i) \\\\\n", + " &= g(t_{i-1} + \\Delta t) \\\\\n", + " &\\approx g(t_{i-1}) + \\Delta t \\big(\\alpha g(t_{i-1})(A - g(t_{i-1}))\\big) \\\\\n", + " &= g_{i-1} + \\Delta t \\big(\\alpha g_{i-1}(A - g_{i-1})\\big)\n", + " \\end{aligned}\n", + "\\end{equation} \\label{odenum} \\tag{12}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for $i \\geq 1$ and $g_0 = g(t_0) = g(0) = g_0$.\n", + "\n", + "Equation ([12](#odenum)) could be implemented in the following way,\n", + "extending the program that uses the network using Autograd:" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Assume that all function definitions from the example program using Autograd\n", + "# are located here.\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nt = 10\n", + " T = 1\n", + " t = np.linspace(0,T, Nt)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [100,50,25]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(t,P)\n", + " g_analytical = g_analytic(t)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%diff_ag)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(t, g_analytical)\n", + " plt.plot(t, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " ## Find an approximation to the funtion using forward Euler\n", + "\n", + " alpha, A, g0 = get_parameters()\n", + " dt = T/(Nt - 1)\n", + "\n", + " # Perform forward Euler to solve the ODE\n", + " g_euler = np.zeros(Nt)\n", + " g_euler[0] = g0\n", + "\n", + " for i in range(1,Nt):\n", + " g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))\n", + "\n", + " # Print the errors done by each method\n", + " diff1 = np.max(np.abs(g_euler - g_analytical))\n", + " diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))\n", + "\n", + " print('Max absolute difference between Euler method and analytical: %g'%diff1)\n", + " print('Max absolute difference between deep neural network and analytical: %g'%diff2)\n", + "\n", + " # Plot results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.plot(t,g_euler)\n", + " plt.plot(t,g_analytical)\n", + " plt.plot(t,g_dnn_ag[0,:])\n", + "\n", + " plt.legend(['euler','analytical','dnn'])\n", + " plt.xlabel('Time t')\n", + " plt.ylabel('g(t)')\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Solving the one dimensional Poisson equation\n", + "\n", + "The Poisson equation for $g(x)$ in one dimension is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{poisson} \\tag{13}\n", + " -g''(x) = f(x)\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f(x)$ is a given function for $x \\in (0,1)$.\n", + "\n", + "The conditions that $g(x)$ is chosen to fulfill, are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + " g(0) &= 0 \\\\\n", + " g(1) &= 0\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used.\n", + "The results from the networks can then be compared to the analytical solution.\n", + "In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks.\n", + "\n", + "\n", + "Here, the function $g(x)$ to solve for follows the equation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "-g''(x) = f(x),\\qquad x \\in (0,1)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f(x)$ is a given function, along with the chosen conditions" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "g(0) = g(1) = 0\n", + "\\end{aligned}\\label{cond} \\tag{14}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this example, we consider the case when $f(x) = (3x + x^2)\\exp(x)$.\n", + "\n", + "For this case, a possible trial solution satisfying the conditions could be" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g_t(x) = x \\cdot (1-x) \\cdot N(P,x)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The analytical solution for this problem is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "g(x) = x(1 - x)\\exp(x)\n", + "$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "## Set up the cost function specified for this Poisson equation:\n", + "\n", + "# The right side of the ODE\n", + "def f(x):\n", + " return (3*x + x**2)*np.exp(x)\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)\n", + "\n", + " right_side = f(x)\n", + "\n", + " err_sqr = (-d2_g_t - right_side)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum/np.size(err_sqr)\n", + "\n", + "# The trial solution:\n", + "def g_trial_deep(x,P):\n", + " return x*(1-x)*deep_neural_network(P,x)\n", + "\n", + "# The analytic solution;\n", + "def g_analytic(x):\n", + " return x*(1-x)*np.exp(x)\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10\n", + " x = np.linspace(0,1, Nx)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [200,100]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(x,P)\n", + " g_analytical = g_analytic(x)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + " max_diff = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " print(\"The max absolute difference between the solutions is: %g\"%max_diff)\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, g_analytical)\n", + " plt.plot(x, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Comparing with a numerical scheme\n", + "\n", + "The Poisson equation is possible to solve using Taylor series to approximate the second derivative.\n", + "\n", + "Using Taylor series, the second derivative can be expressed as\n", + "\n", + "$$\n", + "g''(x) = \\frac{g(x + \\Delta x) - 2g(x) + g(x-\\Delta x)}{\\Delta x^2} + E_{\\Delta x}(x)\n", + "$$\n", + "\n", + "where $\\Delta x$ is a small step size and $E_{\\Delta x}(x)$ being the error term.\n", + "\n", + "Looking away from the error terms gives an approximation to the second derivative:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{approx} \\tag{15}\n", + "g''(x) \\approx \\frac{g(x + \\Delta x) - 2g(x) + g(x-\\Delta x)}{\\Delta x^2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If $x_i = i \\Delta x = x_{i-1} + \\Delta x$ and $g_i = g(x_i)$ for $i = 1,\\dots N_x - 2$ with $N_x$ being the number of values for $x$, ([15](#approx)) becomes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "g''(x_i) &\\approx \\frac{g(x_i + \\Delta x) - 2g(x_i) + g(x_i -\\Delta x)}{\\Delta x^2} \\\\\n", + "&= \\frac{g_{i+1} - 2g_i + g_{i-1}}{\\Delta x^2}\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Since we know from our problem that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "-g''(x) &= f(x) \\\\\n", + "&= (3x + x^2)\\exp(x)\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "along with the conditions $g(0) = g(1) = 0$,\n", + "the following scheme can be used to find an approximate solution for $g(x)$ numerically:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\begin{aligned}\n", + " -\\Big( \\frac{g_{i+1} - 2g_i + g_{i-1}}{\\Delta x^2} \\Big) &= f(x_i) \\\\\n", + " -g_{i+1} + 2g_i - g_{i-1} &= \\Delta x^2 f(x_i)\n", + " \\end{aligned}\n", + "\\end{equation} \\label{odesys} \\tag{16}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "for $i = 1, \\dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\\exp(x_i)$, which is given for our specific problem.\n", + "\n", + "The equation can be rewritten into a matrix equation:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{aligned}\n", + "\\begin{pmatrix}\n", + "2 & -1 & 0 & \\dots & 0 \\\\\n", + "-1 & 2 & -1 & \\dots & 0 \\\\\n", + "\\vdots & & \\ddots & & \\vdots \\\\\n", + "0 & \\dots & -1 & 2 & -1 \\\\\n", + "0 & \\dots & 0 & -1 & 2\\\\\n", + "\\end{pmatrix}\n", + "\\begin{pmatrix}\n", + "g_1 \\\\\n", + "g_2 \\\\\n", + "\\vdots \\\\\n", + "g_{N_x - 3} \\\\\n", + "g_{N_x - 2}\n", + "\\end{pmatrix}\n", + "&=\n", + "\\Delta x^2\n", + "\\begin{pmatrix}\n", + "f(x_1) \\\\\n", + "f(x_2) \\\\\n", + "\\vdots \\\\\n", + "f(x_{N_x - 3}) \\\\\n", + "f(x_{N_x - 2})\n", + "\\end{pmatrix} \\\\\n", + "\\boldsymbol{A}\\boldsymbol{g} &= \\boldsymbol{f},\n", + "\\end{aligned}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "which makes it possible to solve for the vector $\\boldsymbol{g}$.\n", + "\n", + "\n", + "We can then compare the result from this numerical scheme with the output from our network using Autograd:" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import grad, elementwise_grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import pyplot as plt\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assumes input x being an one-dimensional array\n", + " num_values = np.size(x)\n", + " x = x.reshape(-1, num_values)\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + "\n", + " # Due to multiple hidden layers, define a variable referencing to the\n", + " # output of the previous layer:\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output\n", + "\n", + "def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):\n", + " # num_hidden_neurons is now a list of number of neurons within each hidden layer\n", + "\n", + " # Find the number of hidden layers:\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 )\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: %g'%cost_function_deep(P, x))\n", + "\n", + " ## Start finding the optimal weigths using gradient descent\n", + "\n", + " # Find the Python function that represents the gradient of the cost function\n", + " # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer\n", + " cost_function_deep_grad = grad(cost_function_deep,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " # Evaluate the gradient at the current weights and biases in P.\n", + " # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases\n", + " # in the hidden layers and output layers evaluated at x.\n", + " cost_deep_grad = cost_function_deep_grad(P, x)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_deep_grad[l]\n", + "\n", + " print('Final cost: %g'%cost_function_deep(P, x))\n", + "\n", + " return P\n", + "\n", + "## Set up the cost function specified for this Poisson equation:\n", + "\n", + "# The right side of the ODE\n", + "def f(x):\n", + " return (3*x + x**2)*np.exp(x)\n", + "\n", + "def cost_function_deep(P, x):\n", + "\n", + " # Evaluate the trial function with the current parameters P\n", + " g_t = g_trial_deep(x,P)\n", + "\n", + " # Find the derivative w.r.t x of the trial function\n", + " d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)\n", + "\n", + " right_side = f(x)\n", + "\n", + " err_sqr = (-d2_g_t - right_side)**2\n", + " cost_sum = np.sum(err_sqr)\n", + "\n", + " return cost_sum/np.size(err_sqr)\n", + "\n", + "# The trial solution:\n", + "def g_trial_deep(x,P):\n", + " return x*(1-x)*deep_neural_network(P,x)\n", + "\n", + "# The analytic solution;\n", + "def g_analytic(x):\n", + " return x*(1-x)*np.exp(x)\n", + "\n", + "if __name__ == '__main__':\n", + " npr.seed(4155)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10\n", + " x = np.linspace(0,1, Nx)\n", + "\n", + " ## Set up the initial parameters\n", + " num_hidden_neurons = [200,100]\n", + " num_iter = 1000\n", + " lmb = 1e-3\n", + "\n", + " P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " g_dnn_ag = g_trial_deep(x,P)\n", + " g_analytical = g_analytic(x)\n", + "\n", + " # Find the maximum absolute difference between the solutons:\n", + "\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.title('Performance of neural network solving an ODE compared to the analytical solution')\n", + " plt.plot(x, g_analytical)\n", + " plt.plot(x, g_dnn_ag[0,:])\n", + " plt.legend(['analytical','nn'])\n", + " plt.xlabel('x')\n", + " plt.ylabel('g(x)')\n", + "\n", + " ## Perform the computation using the numerical scheme\n", + "\n", + " dx = 1/(Nx - 1)\n", + "\n", + " # Set up the matrix A\n", + " A = np.zeros((Nx-2,Nx-2))\n", + "\n", + " A[0,0] = 2\n", + " A[0,1] = -1\n", + "\n", + " for i in range(1,Nx-3):\n", + " A[i,i-1] = -1\n", + " A[i,i] = 2\n", + " A[i,i+1] = -1\n", + "\n", + " A[Nx - 3, Nx - 4] = -1\n", + " A[Nx - 3, Nx - 3] = 2\n", + "\n", + " # Set up the vector f\n", + " f_vec = dx**2 * f(x[1:-1])\n", + "\n", + " # Solve the equation\n", + " g_res = np.linalg.solve(A,f_vec)\n", + "\n", + " g_vec = np.zeros(Nx)\n", + " g_vec[1:-1] = g_res\n", + "\n", + " # Print the differences between each method\n", + " max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))\n", + " max_diff2 = np.max(np.abs(g_vec - g_analytical))\n", + " print(\"The max absolute difference between the analytical solution and DNN Autograd: %g\"%max_diff1)\n", + " print(\"The max absolute difference between the analytical solution and numerical scheme: %g\"%max_diff2)\n", + "\n", + " # Plot the results\n", + " plt.figure(figsize=(10,10))\n", + "\n", + " plt.plot(x,g_vec)\n", + " plt.plot(x,g_analytical)\n", + " plt.plot(x,g_dnn_ag[0,:])\n", + "\n", + " plt.legend(['numerical scheme','analytical','dnn'])\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Partial Differential Equations\n", + "\n", + "A partial differential equation (PDE) has a solution here the function\n", + "is defined by multiple variables. The equation may involve all kinds\n", + "of combinations of which variables the function is differentiated with\n", + "respect to.\n", + "\n", + "In general, a partial differential equation for a function $g(x_1,\\dots,x_N)$ with $N$ variables may be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{PDE} \\tag{17}\n", + " f\\left(x_1, \\, \\dots \\, , x_N, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1}, \\dots , \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_N}, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(x_1,\\dots,x_N) }{\\partial x_N^n} \\right) = 0\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given.\n", + "\n", + "### Type of problem\n", + "\n", + "The problem our network must solve for, is similar to the ODE case.\n", + "We must have a trial solution $g_t$ at hand.\n", + "\n", + "For instance, the trial solution could be expressed as" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + " g_t(x_1,\\dots,x_N) = h_1(x_1,\\dots,x_N) + h_2(x_1,\\dots,x_N,N(x_1,\\dots,x_N,P))\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $h_1(x_1,\\dots,x_N)$ is a function that ensures $g_t(x_1,\\dots,x_N)$ satisfies some given conditions.\n", + "The neural network $N(x_1,\\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\\dots,x_N,N(x_1,\\dots,x_N,P))$ is an expression using the output from the neural network in some way.\n", + "\n", + "The role of the function $h_2(x_1,\\dots,x_N,N(x_1,\\dots,x_N,P))$, is to ensure that the output of $N(x_1,\\dots,x_N,P)$ is zero when $g_t(x_1,\\dots,x_N)$ is evaluated at the values of $x_1,\\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\\dots,x_N)$ should alone make $g_t(x_1,\\dots,x_N)$ satisfy the conditions.\n", + "\n", + "\n", + "\n", + "### Network requirements\n", + "\n", + "The network tries then the minimize the cost function following the\n", + "same ideas as described for the ODE case, but now with more than one\n", + "variables to consider. The concept still remains the same; find a set\n", + "of parameters $P$ such that the expression $f$ in ([17](#PDE)) is as\n", + "close to zero as possible.\n", + "\n", + "As for the ODE case, the cost function is the mean squared error that\n", + "the network must try to minimize. The cost function for the network to\n", + "minimize is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(x_1, \\dots, x_N, P\\right) = \\left( f\\left(x_1, \\, \\dots \\, , x_N, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1}, \\dots , \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_N}, \\frac{\\partial g(x_1,\\dots,x_N) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(x_1,\\dots,x_N) }{\\partial x_N^n} \\right) \\right)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If we let $\\boldsymbol{x} = \\big( x_1, \\dots, x_N \\big)$ be an array containing the values for $x_1, \\dots, x_N$ respectively, the cost function can be reformulated into the following:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(\\boldsymbol{x}, P\\right) = f\\left( \\left( \\boldsymbol{x}, \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_1}, \\dots , \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_N}, \\frac{\\partial g(\\boldsymbol{x}) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(\\boldsymbol{x}) }{\\partial x_N^n} \\right) \\right)^2\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "If we also have $M$ different sets of values for $x_1, \\dots, x_N$, that is $\\boldsymbol{x}_i = \\big(x_1^{(i)}, \\dots, x_N^{(i)}\\big)$ for $i = 1,\\dots,M$ being the rows in matrix $X$, the cost function can be generalized into" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "C\\left(X, P \\right) = \\sum_{i=1}^M f\\left( \\left( \\boldsymbol{x}_i, \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_1}, \\dots , \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_N}, \\frac{\\partial g(\\boldsymbol{x}_i) }{\\partial x_1\\partial x_2}, \\, \\dots \\, , \\frac{\\partial^n g(\\boldsymbol{x}_i) }{\\partial x_N^n} \\right) \\right)^2.\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: The diffusion equation\n", + "\n", + "In one spatial dimension, the equation reads" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial g(x,t)}{\\partial t} = \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where a possible choice of conditions are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "g(0,t) &= 0 ,\\qquad t \\geq 0 \\\\\n", + "g(1,t) &= 0, \\qquad t \\geq 0 \\\\\n", + "g(x,0) &= u(x),\\qquad x\\in [0,1]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $u(x)$ being some given function.\n", + "\n", + "\n", + "\n", + "For this case, we want to find $g(x,t)$ such that" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation}\n", + " \\frac{\\partial g(x,t)}{\\partial t} = \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "\\end{equation} \\label{diffonedim} \\tag{18}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "and" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "g(0,t) &= 0 ,\\qquad t \\geq 0 \\\\\n", + "g(1,t) &= 0, \\qquad t \\geq 0 \\\\\n", + "g(x,0) &= u(x),\\qquad x\\in [0,1]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $u(x) = \\sin(\\pi x)$.\n", + "\n", + "First, let us set up the deep neural network.\n", + "The deep neural network will follow the same structure as discussed in the examples solving the ODEs.\n", + "First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions.\n", + "\n", + "\n", + "\n", + "\n", + "The only change to do here, is to extend our network such that\n", + "functions of multiple parameters are correctly handled. In this case\n", + "we have two variables in our function to solve for, that is time $t$\n", + "and position $x$. The variables will be represented by a\n", + "one-dimensional array in the program. The program will evaluate the\n", + "network at each possible pair $(x,t)$, given an array for the desired\n", + "$x$-values and $t$-values to approximate the solution at." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The cost function must then iterate through the given arrays\n", + "containing values for $x$ and $t$, defines a point $(x,t)$ the deep\n", + "neural network and the trial solution is evaluated at, and then finds\n", + "the Jacobian of the trial solution.\n", + "\n", + "A possible trial solution for this PDE is\n", + "\n", + "$$\n", + "g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P)\n", + "$$\n", + "\n", + "with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$.\n", + "\n", + "To fulfill the conditions, $A(x,t)$ could be:\n", + "\n", + "$$\n", + "h_1(x,t) = (1-t)\\Big(u(x) - \\big((1-x)u(0) + x u(1)\\big)\\Big) = (1-t)u(x) = (1-t)\\sin(\\pi x)\n", + "$$\n", + "since $(0) = u(1) = 0$ and $u(x) = \\sin(\\pi x)$.\n", + "\n", + "\n", + "\n", + "The Jacobian is used because the program must find the derivative of\n", + "the trial solution with respect to $x$ and $t$.\n", + "\n", + "This gives the necessity of computing the Jacobian matrix, as we want\n", + "to evaluate the gradient with respect to $x$ and $t$ (note that the\n", + "Jacobian of a scalar-valued multivariate function is simply its\n", + "gradient).\n", + "\n", + "In Autograd, the differentiation is by default done with respect to\n", + "the first input argument of your Python function. Since the points is\n", + "an array representing $x$ and $t$, the Jacobian is calculated using\n", + "the values of $x$ and $t$.\n", + "\n", + "To find the second derivative with respect to $x$ and $t$, the\n", + "Jacobian can be found for the second time. The result is a Hessian\n", + "matrix, which is the matrix containing all the possible second order\n", + "mixed derivatives of $g(x,t)$." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Set up the trial function:\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)\n", + "\n", + "# The right side of the ODE:\n", + "def f(point):\n", + " return 0.\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_jacobian_func = jacobian(g_trial)\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t = g_trial(point,P)\n", + " g_t_jacobian = g_t_jacobian_func(point,P)\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_dt = g_t_jacobian[1]\n", + " g_t_d2x = g_t_hessian[0][0]\n", + "\n", + " func = f(point)\n", + "\n", + " err_sqr = ( (g_t_dt - g_t_d2x) - func)**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Setting up the network using Autograd; The full program\n", + "\n", + "Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution.\n", + "\n", + "The analytical solution of our problem is\n", + "\n", + "$$\n", + "g(x,t) = \\exp(-\\pi^2 t)\\sin(\\pi x)\n", + "$$\n", + "\n", + "A possible way to implement a neural network solving the PDE, is given below.\n", + "Be aware, though, that it is fairly slow for the parameters used.\n", + "A better result is possible, but requires more iterations, and thus longer time to complete.\n", + "\n", + "\n", + "Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE.\n", + "Using TensorFlow results in a much better execution time. Try it!" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import jacobian,hessian,grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import cm\n", + "from matplotlib import pyplot as plt\n", + "from mpl_toolkits.mplot3d import axes3d\n", + "\n", + "## Set up the network\n", + "\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]\n", + "\n", + "## Define the trial solution and cost function\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)\n", + "\n", + "# The right side of the ODE:\n", + "def f(point):\n", + " return 0.\n", + "\n", + "# The cost function:\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_jacobian_func = jacobian(g_trial)\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t = g_trial(point,P)\n", + " g_t_jacobian = g_t_jacobian_func(point,P)\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_dt = g_t_jacobian[1]\n", + " g_t_d2x = g_t_hessian[0][0]\n", + "\n", + " func = f(point)\n", + "\n", + " err_sqr = ( (g_t_dt - g_t_d2x) - func)**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum /( np.size(x)*np.size(t) )\n", + "\n", + "## For comparison, define the analytical solution\n", + "def g_analytic(point):\n", + " x,t = point\n", + " return np.exp(-np.pi**2*t)*np.sin(np.pi*x)\n", + "\n", + "## Set up a function for training the network to solve for the equation\n", + "def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):\n", + " ## Set up initial weigths and biases\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: ',cost_function(P, x, t))\n", + "\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " cost_grad = cost_function_grad(P, x , t)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_grad[l]\n", + "\n", + " print('Final cost: ',cost_function(P, x, t))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " ### Use the neural network:\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10; Nt = 10\n", + " x = np.linspace(0, 1, Nx)\n", + " t = np.linspace(0,1,Nt)\n", + "\n", + " ## Set up the parameters for the network\n", + " num_hidden_neurons = [100, 25]\n", + " num_iter = 250\n", + " lmb = 0.01\n", + "\n", + " P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " ## Store the results\n", + " g_dnn_ag = np.zeros((Nx, Nt))\n", + " G_analytical = np.zeros((Nx, Nt))\n", + " for i,x_ in enumerate(x):\n", + " for j, t_ in enumerate(t):\n", + " point = np.array([x_, t_])\n", + " g_dnn_ag[i,j] = g_trial(point,P)\n", + "\n", + " G_analytical[i,j] = g_analytic(point)\n", + "\n", + " # Find the map difference between the analytical and the computed solution\n", + " diff_ag = np.abs(g_dnn_ag - G_analytical)\n", + " print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))\n", + "\n", + " ## Plot the solutions in two dimensions, that being in position and time\n", + "\n", + " T,X = np.meshgrid(t,x)\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))\n", + " s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Analytical solution')\n", + " s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Difference')\n", + " s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " ## Take some slices of the 3D plots just to see the solutions at particular times\n", + " indx1 = 0\n", + " indx2 = int(Nt/2)\n", + " indx3 = Nt-1\n", + "\n", + " t1 = t[indx1]\n", + " t2 = t[indx2]\n", + " t3 = t[indx3]\n", + "\n", + " # Slice the results from the DNN\n", + " res1 = g_dnn_ag[:,indx1]\n", + " res2 = g_dnn_ag[:,indx2]\n", + " res3 = g_dnn_ag[:,indx3]\n", + "\n", + " # Slice the analytical results\n", + " res_analytical1 = G_analytical[:,indx1]\n", + " res_analytical2 = G_analytical[:,indx2]\n", + " res_analytical3 = G_analytical[:,indx3]\n", + "\n", + " # Plot the slices\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t1)\n", + " plt.plot(x, res1)\n", + " plt.plot(x,res_analytical1)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t2)\n", + " plt.plot(x, res2)\n", + " plt.plot(x,res_analytical2)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t3)\n", + " plt.plot(x, res3)\n", + " plt.plot(x,res_analytical3)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Solving the wave equation with Neural Networks\n", + "\n", + "The wave equation is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\frac{\\partial^2 g(x,t)}{\\partial t^2} = c^2\\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "with $c$ being the specified wave speed.\n", + "\n", + "Here, the chosen conditions are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "\tg(0,t) &= 0 \\\\\n", + "\tg(1,t) &= 0 \\\\\n", + "\tg(x,0) &= u(x) \\\\\n", + "\t\\frac{\\partial g(x,t)}{\\partial t} \\Big |_{t = 0} &= v(x)\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $\\frac{\\partial g(x,t)}{\\partial t} \\Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions.\n", + "\n", + "\n", + "The wave equation to solve for, is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{equation} \\label{wave} \\tag{19}\n", + "\\frac{\\partial^2 g(x,t)}{\\partial t^2} = c^2 \\frac{\\partial^2 g(x,t)}{\\partial x^2}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $c$ is the given wave speed.\n", + "The chosen conditions for this equation are" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "g(0,t) &= 0, &t \\geq 0 \\\\\n", + "g(1,t) &= 0, &t \\geq 0 \\\\\n", + "g(x,0) &= u(x), &x\\in[0,1] \\\\\n", + "\\frac{\\partial g(x,t)}{\\partial t}\\Big |_{t = 0} &= v(x), &x \\in [0,1]\n", + "\\end{aligned} \\label{condwave} \\tag{20}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In this example, let $c = 1$ and $u(x) = \\sin(\\pi x)$ and $v(x) = -\\pi\\sin(\\pi x)$.\n", + "\n", + "\n", + "\n", + "Setting up the network is done in similar matter as for the example of solving the diffusion equation.\n", + "The only things we have to change, is the trial solution such that it satisfies the conditions from ([20](#condwave)) and the cost function.\n", + "\n", + "The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is\n", + "\n", + "$$\n", + "g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P)\n", + "$$\n", + "\n", + "where\n", + "\n", + "$$\n", + "h_1(x,t) = (1-t^2)u(x) + tv(x)\n", + "$$\n", + "\n", + "Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example.\n", + "\n", + "\n", + "The analytical solution for our specific problem, is\n", + "\n", + "$$\n", + "g(x,t) = \\sin(\\pi x)\\cos(\\pi t) - \\sin(\\pi x)\\sin(\\pi t)\n", + "$$" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import autograd.numpy as np\n", + "from autograd import hessian,grad\n", + "import autograd.numpy.random as npr\n", + "from matplotlib import cm\n", + "from matplotlib import pyplot as plt\n", + "from mpl_toolkits.mplot3d import axes3d\n", + "\n", + "## Set up the trial function:\n", + "def u(x):\n", + " return np.sin(np.pi*x)\n", + "\n", + "def v(x):\n", + " return -np.pi*np.sin(np.pi*x)\n", + "\n", + "def h1(point):\n", + " x,t = point\n", + " return (1 - t**2)*u(x) + t*v(x)\n", + "\n", + "def g_trial(point,P):\n", + " x,t = point\n", + " return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)\n", + "\n", + "## Define the cost function\n", + "def cost_function(P, x, t):\n", + " cost_sum = 0\n", + "\n", + " g_t_hessian_func = hessian(g_trial)\n", + "\n", + " for x_ in x:\n", + " for t_ in t:\n", + " point = np.array([x_,t_])\n", + "\n", + " g_t_hessian = g_t_hessian_func(point,P)\n", + "\n", + " g_t_d2x = g_t_hessian[0][0]\n", + " g_t_d2t = g_t_hessian[1][1]\n", + "\n", + " err_sqr = ( (g_t_d2t - g_t_d2x) )**2\n", + " cost_sum += err_sqr\n", + "\n", + " return cost_sum / (np.size(t) * np.size(x))\n", + "\n", + "## The neural network\n", + "def sigmoid(z):\n", + " return 1/(1 + np.exp(-z))\n", + "\n", + "def deep_neural_network(deep_params, x):\n", + " # x is now a point and a 1D numpy array; make it a column vector\n", + " num_coordinates = np.size(x,0)\n", + " x = x.reshape(num_coordinates,-1)\n", + "\n", + " num_points = np.size(x,1)\n", + "\n", + " # N_hidden is the number of hidden layers\n", + " N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer\n", + "\n", + " # Assume that the input layer does nothing to the input x\n", + " x_input = x\n", + " x_prev = x_input\n", + "\n", + " ## Hidden layers:\n", + "\n", + " for l in range(N_hidden):\n", + " # From the list of parameters P; find the correct weigths and bias for this layer\n", + " w_hidden = deep_params[l]\n", + "\n", + " # Add a row of ones to include bias\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)\n", + "\n", + " z_hidden = np.matmul(w_hidden, x_prev)\n", + " x_hidden = sigmoid(z_hidden)\n", + "\n", + " # Update x_prev such that next layer can use the output from this layer\n", + " x_prev = x_hidden\n", + "\n", + " ## Output layer:\n", + "\n", + " # Get the weights and bias for this layer\n", + " w_output = deep_params[-1]\n", + "\n", + " # Include bias:\n", + " x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)\n", + "\n", + " z_output = np.matmul(w_output, x_prev)\n", + " x_output = z_output\n", + "\n", + " return x_output[0][0]\n", + "\n", + "## The analytical solution\n", + "def g_analytic(point):\n", + " x,t = point\n", + " return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)\n", + "\n", + "def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):\n", + " ## Set up initial weigths and biases\n", + " N_hidden = np.size(num_neurons)\n", + "\n", + " ## Set up initial weigths and biases\n", + "\n", + " # Initialize the list of parameters:\n", + " P = [None]*(N_hidden + 1) # + 1 to include the output layer\n", + "\n", + " P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias\n", + " for l in range(1,N_hidden):\n", + " P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias\n", + "\n", + " # For the output layer\n", + " P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included\n", + "\n", + " print('Initial cost: ',cost_function(P, x, t))\n", + "\n", + " cost_function_grad = grad(cost_function,0)\n", + "\n", + " # Let the update be done num_iter times\n", + " for i in range(num_iter):\n", + " cost_grad = cost_function_grad(P, x , t)\n", + "\n", + " for l in range(N_hidden+1):\n", + " P[l] = P[l] - lmb * cost_grad[l]\n", + "\n", + "\n", + " print('Final cost: ',cost_function(P, x, t))\n", + "\n", + " return P\n", + "\n", + "if __name__ == '__main__':\n", + " ### Use the neural network:\n", + " npr.seed(15)\n", + "\n", + " ## Decide the vales of arguments to the function to solve\n", + " Nx = 10; Nt = 10\n", + " x = np.linspace(0, 1, Nx)\n", + " t = np.linspace(0,1,Nt)\n", + "\n", + " ## Set up the parameters for the network\n", + " num_hidden_neurons = [50,20]\n", + " num_iter = 1000\n", + " lmb = 0.01\n", + "\n", + " P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)\n", + "\n", + " ## Store the results\n", + " res = np.zeros((Nx, Nt))\n", + " res_analytical = np.zeros((Nx, Nt))\n", + " for i,x_ in enumerate(x):\n", + " for j, t_ in enumerate(t):\n", + " point = np.array([x_, t_])\n", + " res[i,j] = g_trial(point,P)\n", + "\n", + " res_analytical[i,j] = g_analytic(point)\n", + "\n", + " diff = np.abs(res - res_analytical)\n", + " print(\"Max difference between analytical and solution from nn: %g\"%np.max(diff))\n", + "\n", + " ## Plot the solutions in two dimensions, that being in position and time\n", + "\n", + " T,X = np.meshgrid(t,x)\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))\n", + " s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Analytical solution')\n", + " s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + "\n", + " fig = plt.figure(figsize=(10,10))\n", + " ax = fig.gca(projection='3d')\n", + " ax.set_title('Difference')\n", + " s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)\n", + " ax.set_xlabel('Time $t$')\n", + " ax.set_ylabel('Position $x$');\n", + "\n", + " ## Take some slices of the 3D plots just to see the solutions at particular times\n", + " indx1 = 0\n", + " indx2 = int(Nt/2)\n", + " indx3 = Nt-1\n", + "\n", + " t1 = t[indx1]\n", + " t2 = t[indx2]\n", + " t3 = t[indx3]\n", + "\n", + " # Slice the results from the DNN\n", + " res1 = res[:,indx1]\n", + " res2 = res[:,indx2]\n", + " res3 = res[:,indx3]\n", + "\n", + " # Slice the analytical results\n", + " res_analytical1 = res_analytical[:,indx1]\n", + " res_analytical2 = res_analytical[:,indx2]\n", + " res_analytical3 = res_analytical[:,indx3]\n", + "\n", + " # Plot the slices\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t1)\n", + " plt.plot(x, res1)\n", + " plt.plot(x,res_analytical1)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t2)\n", + " plt.plot(x, res2)\n", + " plt.plot(x,res_analytical2)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.figure(figsize=(10,10))\n", + " plt.title(\"Computed solutions at time = %g\"%t3)\n", + " plt.plot(x, res3)\n", + " plt.plot(x,res_analytical3)\n", + " plt.legend(['dnn','analytical'])\n", + "\n", + " plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Resources on differential equations and deep learning\n", + "\n", + "1. [Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al](https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf)\n", + "\n", + "2. [Neural networks for solving differential equations by A. Honchar](https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c)\n", + "\n", + "3. [Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener](http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf)\n", + "\n", + "4. [Introduction to Partial Differential Equations by A. Tveito, R. Winther](https://www.springer.com/us/book/9783540225515)" + ] + } + ], + "metadata": {}, + "nbformat": 4, + "nbformat_minor": 4 +} diff --git a/doc/LectureNotes/chapter12.ipynb b/doc/LectureNotes/chapter12.ipynb new file mode 100644 index 000000000..d88cef4d1 --- /dev/null +++ b/doc/LectureNotes/chapter12.ipynb @@ -0,0 +1,732 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Convolutional Neural Networks (recognizing images)\n", + "\n", + "\n", + "Convolutional neural networks (CNNs) were developed during the last\n", + "decade of the previous century, with a focus on character recognition\n", + "tasks. Nowadays, CNNs are a central element in the spectacular success\n", + "of deep learning methods. The success in for example image\n", + "classifications have made them a central tool for most machine\n", + "learning practitioners.\n", + "\n", + "CNNs are very similar to ordinary Neural Networks.\n", + "They are made up of neurons that have learnable weights and\n", + "biases. Each neuron receives some inputs, performs a dot product and\n", + "optionally follows it with a non-linearity. The whole network still\n", + "expresses a single differentiable score function: from the raw image\n", + "pixels on one end to class scores at the other. And they still have a\n", + "loss function (for example Softmax) on the last (fully-connected) layer\n", + "and all the tips/tricks we developed for learning regular Neural\n", + "Networks still apply (back propagation, gradient descent etc etc).\n", + "\n", + "What is the difference? **CNN architectures make the explicit assumption that\n", + "the inputs are images, which allows us to encode certain properties\n", + "into the architecture. These then make the forward function more\n", + "efficient to implement and vastly reduce the amount of parameters in\n", + "the network.**\n", + "\n", + "Here we provide only a superficial overview, for the more interested, we recommend highly the course\n", + "[IN5400 – Machine Learning for Image Analysis](https://www.uio.no/studier/emner/matnat/ifi/IN5400/index-eng.html)\n", + "and the slides of [CS231](http://cs231n.github.io/convolutional-networks/).\n", + "\n", + "Another good read is the article here . \n", + "\n", + "\n", + "As an example, consider\n", + "an image of size $32\\times 32\\times 3$ (32 wide, 32 high, 3 color channels), so a\n", + "single fully-connected neuron in a first hidden layer of a regular\n", + "Neural Network would have $32\\times 32\\times 3 = 3072$ weights. This amount still\n", + "seems manageable, but clearly this fully-connected structure does not\n", + "scale to larger images. For example, an image of more respectable\n", + "size, say $200\\times 200\\times 3$, would lead to neurons that have \n", + "$200\\times 200\\times 3 = 120,000$ weights. \n", + "\n", + "We could have\n", + "several such neurons, and the parameters would add up quickly! Clearly,\n", + "this full connectivity is wasteful and the huge number of parameters\n", + "would quickly lead to possible overfitting.\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "Convolutional Neural Networks take advantage of the fact that the\n", + "input consists of images and they constrain the architecture in a more\n", + "sensible way. \n", + "\n", + "In particular, unlike a regular Neural Network, the\n", + "layers of a CNN have neurons arranged in 3 dimensions: width,\n", + "height, depth. (Note that the word depth here refers to the third\n", + "dimension of an activation volume, not to the depth of a full Neural\n", + "Network, which can refer to the total number of layers in a network.)\n", + "\n", + "To understand it better, the above example of an image \n", + "with an input volume of\n", + "activations has dimensions $32\\times 32\\times 3$ (width, height,\n", + "depth respectively). \n", + "\n", + "The neurons in a layer will\n", + "only be connected to a small region of the layer before it, instead of\n", + "all of the neurons in a fully-connected manner. Moreover, the final\n", + "output layer could for this specific image have dimensions $1\\times 1 \\times 10$, \n", + "because by the\n", + "end of the CNN architecture we will reduce the full image into a\n", + "single vector of class scores, arranged along the depth\n", + "dimension. \n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "A simple CNN is a sequence of layers, and every layer of a CNN\n", + "transforms one volume of activations to another through a\n", + "differentiable function. We use three main types of layers to build\n", + "CNN architectures: Convolutional Layer, Pooling Layer, and\n", + "Fully-Connected Layer (exactly as seen in regular Neural Networks). We\n", + "will stack these layers to form a full CNN architecture.\n", + "\n", + "A simple CNN for image classification could have the architecture:\n", + "\n", + "* **INPUT** ($32\\times 32 \\times 3$) will hold the raw pixel values of the image, in this case an image of width 32, height 32, and with three color channels R,G,B.\n", + "\n", + "* **CONV** (convolutional )layer will compute the output of neurons that are connected to local regions in the input, each computing a dot product between their weights and a small region they are connected to in the input volume. This may result in volume such as $[32\\times 32\\times 12]$ if we decided to use 12 filters.\n", + "\n", + "* **RELU** layer will apply an elementwise activation function, such as the $max(0,x)$ thresholding at zero. This leaves the size of the volume unchanged ($[32\\times 32\\times 12]$).\n", + "\n", + "* **POOL** (pooling) layer will perform a downsampling operation along the spatial dimensions (width, height), resulting in volume such as $[16\\times 16\\times 12]$.\n", + "\n", + "* **FC** (i.e. fully-connected) layer will compute the class scores, resulting in volume of size $[1\\times 1\\times 10]$, where each of the 10 numbers correspond to a class score, such as among the 10 categories of the MNIST images we considered above . As with ordinary Neural Networks and as the name implies, each neuron in this layer will be connected to all the numbers in the previous volume.\n", + "\n", + "CNNs transform the original image layer by layer from the original\n", + "pixel values to the final class scores. \n", + "\n", + "Observe that some layers contain\n", + "parameters and other don’t. In particular, the CNN layers perform\n", + "transformations that are a function of not only the activations in the\n", + "input volume, but also of the parameters (the weights and biases of\n", + "the neurons). On the other hand, the RELU/POOL layers will implement a\n", + "fixed function. The parameters in the CONV/FC layers will be trained\n", + "with gradient descent so that the class scores that the CNN computes\n", + "are consistent with the labels in the training set for each image.\n", + "\n", + "\n", + "\n", + "### CNNs in brief\n", + "\n", + "In summary:\n", + "\n", + "* A CNN architecture is in the simplest case a list of Layers that transform the image volume into an output volume (e.g. holding the class scores)\n", + "\n", + "* There are a few distinct types of Layers (e.g. CONV/FC/RELU/POOL are by far the most popular)\n", + "\n", + "* Each Layer accepts an input 3D volume and transforms it to an output 3D volume through a differentiable function\n", + "\n", + "* Each Layer may or may not have parameters (e.g. CONV/FC do, RELU/POOL don’t)\n", + "\n", + "* Each Layer may or may not have additional hyperparameters (e.g. CONV/FC/POOL do, RELU doesn’t)\n", + "\n", + "For more material on convolutional networks, we strongly recommend\n", + "the course\n", + "[IN5400 – Machine Learning for Image Analysis](https://www.uio.no/studier/emner/matnat/ifi/IN5400/index-eng.html)\n", + "and the slides of [CS231](http://cs231n.github.io/convolutional-networks/) which is taught at Stanford University (consistently ranked as one of the top computer science programs in the world). [Michael Nielsen's book is a must read, in particular chapter 6 which deals with CNNs](http://neuralnetworksanddeeplearning.com/chap6.html).\n", + "\n", + "\n", + "\n", + "\n", + "## CNNs in more detail, building convolutional neural networks in Tensorflow and Keras\n", + "\n", + "\n", + "As discussed above, CNNs are neural networks built from the assumption that the inputs\n", + "to the network are 2D images. This is important because the number of features or pixels in images\n", + "grows very fast with the image size, and an enormous number of weights and biases are needed in order to build an accurate network. \n", + "\n", + "As before, we still have our input, a hidden layer and an output. What's novel about convolutional networks\n", + "are the **convolutional** and **pooling** layers stacked in pairs between the input and the hidden layer.\n", + "In addition, the data is no longer represented as a 2D feature matrix, instead each input is a number of 2D\n", + "matrices, typically 1 for each color dimension (Red, Green, Blue). \n", + "\n", + "\n", + "\n", + "It means that to represent the entire\n", + "dataset of images, we require a 4D matrix or **tensor**. This tensor has the dimensions:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "(n_{inputs},\\, n_{pixels, width},\\, n_{pixels, height},\\, depth) .\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The MNIST dataset consists of grayscale images with a pixel size of\n", + "$28\\times 28$, meaning we require $28 \\times 28 = 724$ weights to each\n", + "neuron in the first hidden layer.\n", + "\n", + "If we were to analyze images of size $128\\times 128$ we would require\n", + "$128 \\times 128 = 16384$ weights to each neuron. Even worse if we were\n", + "dealing with color images, as most images are, we have an image matrix\n", + "of size $128\\times 128$ for each color dimension (Red, Green, Blue),\n", + "meaning 3 times the number of weights $= 49152$ are required for every\n", + "single neuron in the first hidden layer.\n", + "\n", + "\n", + "\n", + "Images typically have strong local correlations, meaning that a small\n", + "part of the image varies little from its neighboring regions. If for\n", + "example we have an image of a blue car, we can roughly assume that a\n", + "small blue part of the image is surrounded by other blue regions.\n", + "\n", + "Therefore, instead of connecting every single pixel to a neuron in the\n", + "first hidden layer, as we have previously done with deep neural\n", + "networks, we can instead connect each neuron to a small part of the\n", + "image (in all 3 RGB depth dimensions). The size of each small area is\n", + "fixed, and known as a [receptive](https://en.wikipedia.org/wiki/Receptive_field).\n", + "\n", + "\n", + "\n", + "The layers of a convolutional neural network arrange neurons in 3D: width, height and depth. \n", + "The input image is typically a square matrix of depth 3. \n", + "\n", + "A **convolution** is performed on the image which outputs\n", + "a 3D volume of neurons. The weights to the input are arranged in a number of 2D matrices, known as **filters**.\n", + "\n", + "\n", + "Each filter slides along the input image, taking the dot product\n", + "between each small part of the image and the filter, in all depth\n", + "dimensions. This is then passed through a non-linear function,\n", + "typically the **Rectified Linear (ReLu)** function, which serves as the\n", + "activation of the neurons in the first convolutional layer. This is\n", + "further passed through a **pooling layer**, which reduces the size of the\n", + "convolutional layer, e.g. by taking the maximum or average across some\n", + "small regions, and this serves as input to the next convolutional\n", + "layer.\n", + "\n", + "\n", + "\n", + "By systematically reducing the size of the input volume, through\n", + "convolution and pooling, the network should create representations of\n", + "small parts of the input, and then from them assemble representations\n", + "of larger areas. The final pooling layer is flattened to serve as\n", + "input to a hidden layer, such that each neuron in the final pooling\n", + "layer is connected to every single neuron in the hidden layer. This\n", + "then serves as input to the output layer, e.g. a softmax output for\n", + "classification.\n", + "\n", + "\n", + "\n", + "### Prerequisites: Collect and pre-process data" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "%matplotlib inline\n", + "\n", + "# import necessary packages\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "from sklearn import datasets\n", + "\n", + "\n", + "# ensure the same random numbers appear every time\n", + "np.random.seed(0)\n", + "\n", + "# display images in notebook\n", + "%matplotlib inline\n", + "plt.rcParams['figure.figsize'] = (12,12)\n", + "\n", + "\n", + "# download MNIST dataset\n", + "digits = datasets.load_digits()\n", + "\n", + "# define inputs and labels\n", + "inputs = digits.images\n", + "labels = digits.target\n", + "\n", + "# RGB images have a depth of 3\n", + "# our images are grayscale so they should have a depth of 1\n", + "inputs = inputs[:,:,:,np.newaxis]\n", + "\n", + "print(\"inputs = (n_inputs, pixel_width, pixel_height, depth) = \" + str(inputs.shape))\n", + "print(\"labels = (n_inputs) = \" + str(labels.shape))\n", + "\n", + "\n", + "# choose some random images to display\n", + "n_inputs = len(inputs)\n", + "indices = np.arange(n_inputs)\n", + "random_indices = np.random.choice(indices, size=5)\n", + "\n", + "for i, image in enumerate(digits.images[random_indices]):\n", + " plt.subplot(1, 5, i+1)\n", + " plt.axis('off')\n", + " plt.imshow(image, cmap=plt.cm.gray_r, interpolation='nearest')\n", + " plt.title(\"Label: %d\" % digits.target[random_indices[i]])\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Importing Keras and Tensorflow" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "from tensorflow.keras import datasets, layers, models\n", + "from tensorflow.keras.layers import Input\n", + "from tensorflow.keras.models import Sequential #This allows appending layers to existing models\n", + "from tensorflow.keras.layers import Dense #This allows defining the characteristics of a particular layer\n", + "from tensorflow.keras import optimizers #This allows using whichever optimiser we want (sgd,adam,RMSprop)\n", + "from tensorflow.keras import regularizers #This allows using whichever regularizer we want (l1,l2,l1_l2)\n", + "from tensorflow.keras.utils import to_categorical #This allows using categorical cross entropy as the cost function\n", + "#from tensorflow.keras import Conv2D\n", + "#from tensorflow.keras import MaxPooling2D\n", + "#from tensorflow.keras import Flatten\n", + "\n", + "from sklearn.model_selection import train_test_split\n", + "\n", + "# representation of labels\n", + "labels = to_categorical(labels)\n", + "\n", + "# split into train and test data\n", + "# one-liner from scikit-learn library\n", + "train_size = 0.8\n", + "test_size = 1 - train_size\n", + "X_train, X_test, Y_train, Y_test = train_test_split(inputs, labels, train_size=train_size,\n", + " test_size=test_size)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "def create_convolutional_neural_network_keras(input_shape, receptive_field,\n", + " n_filters, n_neurons_connected, n_categories,\n", + " eta, lmbd):\n", + " model = Sequential()\n", + " model.add(layers.Conv2D(n_filters, (receptive_field, receptive_field), input_shape=input_shape, padding='same',\n", + " activation='relu', kernel_regularizer=regularizers.l2(lmbd)))\n", + " model.add(layers.MaxPooling2D(pool_size=(2, 2)))\n", + " model.add(layers.Flatten())\n", + " model.add(layers.Dense(n_neurons_connected, activation='relu', kernel_regularizer=regularizers.l2(lmbd)))\n", + " model.add(layers.Dense(n_categories, activation='softmax', kernel_regularizer=regularizers.l2(lmbd)))\n", + " \n", + " sgd = optimizers.SGD(lr=eta)\n", + " model.compile(loss='categorical_crossentropy', optimizer=sgd, metrics=['accuracy'])\n", + " \n", + " return model\n", + "\n", + "epochs = 100\n", + "batch_size = 100\n", + "input_shape = X_train.shape[1:4]\n", + "receptive_field = 3\n", + "n_filters = 10\n", + "n_neurons_connected = 50\n", + "n_categories = 10\n", + "\n", + "eta_vals = np.logspace(-5, 1, 7)\n", + "lmbd_vals = np.logspace(-5, 1, 7)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "CNN_keras = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n", + " \n", + "for i, eta in enumerate(eta_vals):\n", + " for j, lmbd in enumerate(lmbd_vals):\n", + " CNN = create_convolutional_neural_network_keras(input_shape, receptive_field,\n", + " n_filters, n_neurons_connected, n_categories,\n", + " eta, lmbd)\n", + " CNN.fit(X_train, Y_train, epochs=epochs, batch_size=batch_size, verbose=0)\n", + " scores = CNN.evaluate(X_test, Y_test)\n", + " \n", + " CNN_keras[i][j] = CNN\n", + " \n", + " print(\"Learning rate = \", eta)\n", + " print(\"Lambda = \", lmbd)\n", + " print(\"Test accuracy: %.3f\" % scores[1])\n", + " print()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Final visualization" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# visual representation of grid search\n", + "# uses seaborn heatmap, could probably do this in matplotlib\n", + "import seaborn as sns\n", + "\n", + "sns.set()\n", + "\n", + "train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "\n", + "for i in range(len(eta_vals)):\n", + " for j in range(len(lmbd_vals)):\n", + " CNN = CNN_keras[i][j]\n", + "\n", + " train_accuracy[i][j] = CNN.evaluate(X_train, Y_train)[1]\n", + " test_accuracy[i][j] = CNN.evaluate(X_test, Y_test)[1]\n", + "\n", + " \n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(train_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Training Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()\n", + "\n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Test Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## The CIFAR01 data set\n", + "\n", + "The CIFAR10 dataset contains 60,000 color images in 10 classes, with\n", + "6,000 images in each class. The dataset is divided into 50,000\n", + "training images and 10,000 testing images. The classes are mutually\n", + "exclusive and there is no overlap between them." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "import tensorflow as tf\n", + "\n", + "from tensorflow.keras import datasets, layers, models\n", + "import matplotlib.pyplot as plt\n", + "\n", + "# We import the data set\n", + "(train_images, train_labels), (test_images, test_labels) = datasets.cifar10.load_data()\n", + "\n", + "# Normalize pixel values to be between 0 and 1 by dividing by 255. \n", + "train_images, test_images = train_images / 255.0, test_images / 255.0" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "To verify that the dataset looks correct, let's plot the first 25 images from the training set and display the class name below each image." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "class_names = ['airplane', 'automobile', 'bird', 'cat', 'deer',\n", + " 'dog', 'frog', 'horse', 'ship', 'truck']\n", + "​\n", + "plt.figure(figsize=(10,10))\n", + "for i in range(25):\n", + " plt.subplot(5,5,i+1)\n", + " plt.xticks([])\n", + " plt.yticks([])\n", + " plt.grid(False)\n", + " plt.imshow(train_images[i], cmap=plt.cm.binary)\n", + " # The CIFAR labels happen to be arrays, \n", + " # which is why you need the extra index\n", + " plt.xlabel(class_names[train_labels[i][0]])\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The 6 lines of code below define the convolutional base using a common pattern: a stack of Conv2D and MaxPooling2D layers.\n", + "\n", + "As input, a CNN takes tensors of shape (image_height, image_width, color_channels), ignoring the batch size. If you are new to these dimensions, color_channels refers to (R,G,B). In this example, you will configure our CNN to process inputs of shape (32, 32, 3), which is the format of CIFAR images. You can do this by passing the argument input_shape to our first layer." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "model = models.Sequential()\n", + "model.add(layers.Conv2D(32, (3, 3), activation='relu', input_shape=(32, 32, 3)))\n", + "model.add(layers.MaxPooling2D((2, 2)))\n", + "model.add(layers.Conv2D(64, (3, 3), activation='relu'))\n", + "model.add(layers.MaxPooling2D((2, 2)))\n", + "model.add(layers.Conv2D(64, (3, 3), activation='relu'))\n", + "\n", + "# Let's display the architecture of our model so far.\n", + "\n", + "model.summary()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "You can see that the output of every Conv2D and MaxPooling2D layer is a 3D tensor of shape (height, width, channels). The width and height dimensions tend to shrink as you go deeper in the network. The number of output channels for each Conv2D layer is controlled by the first argument (e.g., 32 or 64). Typically, as the width and height shrink, you can afford (computationally) to add more output channels in each Conv2D layer.\n", + "\n", + "\n", + "\n", + "To complete our model, you will feed the last output tensor from the\n", + "convolutional base (of shape (4, 4, 64)) into one or more Dense layers\n", + "to perform classification. Dense layers take vectors as input (which\n", + "are 1D), while the current output is a 3D tensor. First, you will\n", + "flatten (or unroll) the 3D output to 1D, then add one or more Dense\n", + "layers on top. CIFAR has 10 output classes, so you use a final Dense\n", + "layer with 10 outputs and a softmax activation." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "model.add(layers.Flatten())\n", + "model.add(layers.Dense(64, activation='relu'))\n", + "model.add(layers.Dense(10))\n", + "Here's the complete architecture of our model.\n", + "\n", + "model.summary()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "As you can see, our (4, 4, 64) outputs were flattened into vectors of shape (1024) before going through two Dense layers.\n", + "\n", + "Compile and train the model." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "model.compile(optimizer='adam',\n", + " loss=tf.keras.losses.SparseCategoricalCrossentropy(from_logits=True),\n", + " metrics=['accuracy'])\n", + "​\n", + "history = model.fit(train_images, train_labels, epochs=10, \n", + " validation_data=(test_images, test_labels))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Finally, we evaluate the model." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "plt.plot(history.history['accuracy'], label='accuracy')\n", + "plt.plot(history.history['val_accuracy'], label = 'val_accuracy')\n", + "plt.xlabel('Epoch')\n", + "plt.ylabel('Accuracy')\n", + "plt.ylim([0.5, 1])\n", + "plt.legend(loc='lower right')\n", + "\n", + "test_loss, test_acc = model.evaluate(test_images, test_labels, verbose=2)\n", + "\n", + "print(test_acc)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Recurrent neural networks: Overarching view\n", + "\n", + "Till now our focus has been, including convolutional neural networks\n", + "as well, on feedforward neural networks. The output or the activations\n", + "flow only in one direction, from the input layer to the output layer.\n", + "\n", + "A recurrent neural network (RNN) looks very much like a feedforward\n", + "neural network, except that it also has connections pointing\n", + "backward. \n", + "\n", + "RNNs are used to analyze time series data such as stock prices, and\n", + "tell you when to buy or sell. In autonomous driving systems, they can\n", + "anticipate car trajectories and help avoid accidents. More generally,\n", + "they can work on sequences of arbitrary lengths, rather than on\n", + "fixed-sized inputs like all the nets we have discussed so far. For\n", + "example, they can take sentences, documents, or audio samples as\n", + "input, making them extremely useful for natural language processing\n", + "systems such as automatic translation and speech-to-text.\n", + "\n", + "\n", + "\n", + "\n", + "### A simple example" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "# Start importing packages\n", + "import pandas as pd\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "import tensorflow as tf\n", + "from tensorflow.keras import datasets, layers, models\n", + "from tensorflow.keras.layers import Input\n", + "from tensorflow.keras.models import Model, Sequential \n", + "from tensorflow.keras.layers import Dense, SimpleRNN, LSTM, GRU\n", + "from tensorflow.keras import optimizers \n", + "from tensorflow.keras import regularizers \n", + "from tensorflow.keras.utils import to_categorical \n", + "\n", + "\n", + "\n", + "# convert into dataset matrix\n", + "def convertToMatrix(data, step):\n", + " X, Y =[], []\n", + " for i in range(len(data)-step):\n", + " d=i+step \n", + " X.append(data[i:d,])\n", + " Y.append(data[d,])\n", + " return np.array(X), np.array(Y)\n", + "\n", + "step = 4\n", + "N = 1000 \n", + "Tp = 800 \n", + "\n", + "t=np.arange(0,N)\n", + "x=np.sin(0.02*t)+2*np.random.rand(N)\n", + "df = pd.DataFrame(x)\n", + "df.head()\n", + "\n", + "plt.plot(df)\n", + "plt.show()\n", + "\n", + "values=df.values\n", + "train,test = values[0:Tp,:], values[Tp:N,:]\n", + "\n", + "# add step elements into train and test\n", + "test = np.append(test,np.repeat(test[-1,],step))\n", + "train = np.append(train,np.repeat(train[-1,],step))\n", + " \n", + "trainX,trainY =convertToMatrix(train,step)\n", + "testX,testY =convertToMatrix(test,step)\n", + "trainX = np.reshape(trainX, (trainX.shape[0], 1, trainX.shape[1]))\n", + "testX = np.reshape(testX, (testX.shape[0], 1, testX.shape[1]))\n", + "\n", + "model = Sequential()\n", + "model.add(SimpleRNN(units=32, input_shape=(1,step), activation=\"relu\"))\n", + "model.add(Dense(8, activation=\"relu\")) \n", + "model.add(Dense(1))\n", + "model.compile(loss='mean_squared_error', optimizer='rmsprop')\n", + "model.summary()\n", + "\n", + "model.fit(trainX,trainY, epochs=100, batch_size=16, verbose=2)\n", + "trainPredict = model.predict(trainX)\n", + "testPredict= model.predict(testX)\n", + "predicted=np.concatenate((trainPredict,testPredict),axis=0)\n", + "\n", + "trainScore = model.evaluate(trainX, trainY, verbose=0)\n", + "print(trainScore)\n", + "\n", + "index = df.index.values\n", + "plt.plot(index,df)\n", + "plt.plot(index,predicted)\n", + "plt.axvline(df.index[Tp], c=\"r\")\n", + "plt.show()" + ] + } + ], + "metadata": {}, + "nbformat": 4, + "nbformat_minor": 4 +}