lars/FYS-STK4155
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity
Files
b5d6383ce78e48fb6fcac84d9059cac0e4d0ae61
FYS-STK4155/doc/LectureNotes/_build/html/chapter11.html
T
Morten Hjorth-Jensen b5d6383ce7 added p1
2025-09-02 06:34:11 +02:00

2761 lines
122 KiB
HTML
Raw Blame History

This file contains ambiguous Unicode characters
This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.
<!DOCTYPE html>
<html lang="en" data-content_root="./" >
<head>
<meta charset="utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1.0" /><meta name="viewport" content="width=device-width, initial-scale=1" />
<title>15. Solving Differential Equations with Deep Learning &#8212; Applied Data Analysis and Machine Learning</title>
<script data-cfasync="false">
document.documentElement.dataset.mode = localStorage.getItem("mode") || "";
document.documentElement.dataset.theme = localStorage.getItem("theme") || "";
</script>
<!-- Loaded before other Sphinx assets -->
<link href="_static/styles/theme.css?digest=dfe6caa3a7d634c4db9b" rel="stylesheet" />
<link href="_static/styles/bootstrap.css?digest=dfe6caa3a7d634c4db9b" rel="stylesheet" />
<link href="_static/styles/pydata-sphinx-theme.css?digest=dfe6caa3a7d634c4db9b" rel="stylesheet" />
<link href="_static/vendor/fontawesome/6.5.2/css/all.min.css?digest=dfe6caa3a7d634c4db9b" rel="stylesheet" />
<link rel="preload" as="font" type="font/woff2" crossorigin href="_static/vendor/fontawesome/6.5.2/webfonts/fa-solid-900.woff2" />
<link rel="preload" as="font" type="font/woff2" crossorigin href="_static/vendor/fontawesome/6.5.2/webfonts/fa-brands-400.woff2" />
<link rel="preload" as="font" type="font/woff2" crossorigin href="_static/vendor/fontawesome/6.5.2/webfonts/fa-regular-400.woff2" />
<link rel="stylesheet" type="text/css" href="_static/pygments.css?v=fa44fd50" />
<link rel="stylesheet" type="text/css" href="_static/styles/sphinx-book-theme.css?v=eba8b062" />
<link rel="stylesheet" type="text/css" href="_static/togglebutton.css?v=13237357" />
<link rel="stylesheet" type="text/css" href="_static/copybutton.css?v=76b2166b" />
<link rel="stylesheet" type="text/css" href="_static/mystnb.8ecb98da25f57f5357bf6f572d296f466b2cfe2517ffebfabe82451661e28f02.css?v=6644e6bb" />
<link rel="stylesheet" type="text/css" href="_static/sphinx-thebe.css?v=4fa983c6" />
<link rel="stylesheet" type="text/css" href="_static/sphinx-design.min.css?v=95c83b7e" />
<!-- Pre-loaded scripts that we'll load fully later -->
<link rel="preload" as="script" href="_static/scripts/bootstrap.js?digest=dfe6caa3a7d634c4db9b" />
<link rel="preload" as="script" href="_static/scripts/pydata-sphinx-theme.js?digest=dfe6caa3a7d634c4db9b" />
<script src="_static/vendor/fontawesome/6.5.2/js/all.min.js?digest=dfe6caa3a7d634c4db9b"></script>
<script src="_static/documentation_options.js?v=9eb32ce0"></script>
<script src="_static/doctools.js?v=9a2dae69"></script>
<script src="_static/sphinx_highlight.js?v=dc90522c"></script>
<script src="_static/clipboard.min.js?v=a7894cd8"></script>
<script src="_static/copybutton.js?v=f281be69"></script>
<script src="_static/scripts/sphinx-book-theme.js?v=887ef09a"></script>
<script>let toggleHintShow = 'Click to show';</script>
<script>let toggleHintHide = 'Click to hide';</script>
<script>let toggleOpenOnPrint = 'true';</script>
<script src="_static/togglebutton.js?v=4a39c7ea"></script>
<script>var togglebuttonSelector = '.toggle, .admonition.dropdown';</script>
<script src="_static/design-tabs.js?v=f930bc37"></script>
<script>const THEBE_JS_URL = "https://unpkg.com/thebe@0.8.2/lib/index.js"; const thebe_selector = ".thebe,.cell"; const thebe_selector_input = "pre"; const thebe_selector_output = ".output, .cell_output"</script>
<script async="async" src="_static/sphinx-thebe.js?v=c100c467"></script>
<script>var togglebuttonSelector = '.toggle, .admonition.dropdown';</script>
<script>const THEBE_JS_URL = "https://unpkg.com/thebe@0.8.2/lib/index.js"; const thebe_selector = ".thebe,.cell"; const thebe_selector_input = "pre"; const thebe_selector_output = ".output, .cell_output"</script>
<script>window.MathJax = {"options": {"processHtmlClass": "tex2jax_process|mathjax_process|math|output_area"}}</script>
<script defer="defer" src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js"></script>
<script>DOCUMENTATION_OPTIONS.pagename = 'chapter11';</script>
<link rel="index" title="Index" href="genindex.html" />
<link rel="search" title="Search" href="search.html" />
<link rel="next" title="16. Convolutional Neural Networks" href="chapter12.html" />
<link rel="prev" title="14. Building a Feed Forward Neural Network" href="chapter10.html" />
<meta name="viewport" content="width=device-width, initial-scale=1"/>
<meta name="docsearch:language" content="en"/>
</head>
<body data-bs-spy="scroll" data-bs-target=".bd-toc-nav" data-offset="180" data-bs-root-margin="0px 0px -60%" data-default-mode="">
<div id="pst-skip-link" class="skip-link d-print-none"><a href="#main-content">Skip to main content</a></div>
<div id="pst-scroll-pixel-helper"></div>
<button type="button" class="btn rounded-pill" id="pst-back-to-top">
<i class="fa-solid fa-arrow-up"></i>Back to top</button>
<input type="checkbox"
class="sidebar-toggle"
id="pst-primary-sidebar-checkbox"/>
<label class="overlay overlay-primary" for="pst-primary-sidebar-checkbox"></label>
<input type="checkbox"
class="sidebar-toggle"
id="pst-secondary-sidebar-checkbox"/>
<label class="overlay overlay-secondary" for="pst-secondary-sidebar-checkbox"></label>
<div class="search-button__wrapper">
<div class="search-button__overlay"></div>
<div class="search-button__search-container">
<form class="bd-search d-flex align-items-center"
action="search.html"
method="get">
<i class="fa-solid fa-magnifying-glass"></i>
<input type="search"
class="form-control"
name="q"
id="search-input"
placeholder="Search this book..."
aria-label="Search this book..."
autocomplete="off"
autocorrect="off"
autocapitalize="off"
spellcheck="false"/>
<span class="search-button__kbd-shortcut"><kbd class="kbd-shortcut__modifier">Ctrl</kbd>+<kbd>K</kbd></span>
</form></div>
</div>
<div class="pst-async-banner-revealer d-none">
<aside id="bd-header-version-warning" class="d-none d-print-none" aria-label="Version warning"></aside>
</div>
<header class="bd-header navbar navbar-expand-lg bd-navbar d-print-none">
</header>
<div class="bd-container">
<div class="bd-container__inner bd-page-width">
<div class="bd-sidebar-primary bd-sidebar">
<div class="sidebar-header-items sidebar-primary__section">
</div>
<div class="sidebar-primary-items__start sidebar-primary__section">
<div class="sidebar-primary-item">
<a class="navbar-brand logo" href="intro.html">
<img src="_static/logo.png" class="logo__image only-light" alt="Applied Data Analysis and Machine Learning - Home"/>
<script>document.write(`<img src="_static/logo.png" class="logo__image only-dark" alt="Applied Data Analysis and Machine Learning - Home"/>`);</script>
</a></div>
<div class="sidebar-primary-item">
<script>
document.write(`
<button class="btn search-button-field search-button__button" title="Search" aria-label="Search" data-bs-placement="bottom" data-bs-toggle="tooltip">
<i class="fa-solid fa-magnifying-glass"></i>
<span class="search-button__default-text">Search</span>
<span class="search-button__kbd-shortcut"><kbd class="kbd-shortcut__modifier">Ctrl</kbd>+<kbd class="kbd-shortcut__modifier">K</kbd></span>
</button>
`);
</script></div>
<div class="sidebar-primary-item"><nav class="bd-links bd-docs-nav" aria-label="Main">
<div class="bd-toc-item navbar-nav active">
<ul class="nav bd-sidenav bd-sidenav__home-link">
<li class="toctree-l1">
<a class="reference internal" href="intro.html">
Applied Data Analysis and Machine Learning
</a>
</li>
</ul>
<p aria-level="2" class="caption" role="heading"><span class="caption-text">About the course</span></p>
<ul class="nav bd-sidenav">
<li class="toctree-l1"><a class="reference internal" href="schedule.html">Course setting</a></li>
<li class="toctree-l1"><a class="reference internal" href="teachers.html">Teachers and Grading</a></li>
<li class="toctree-l1"><a class="reference internal" href="textbooks.html">Textbooks</a></li>
</ul>
<p aria-level="2" class="caption" role="heading"><span class="caption-text">Review of Statistics with Resampling Techniques and Linear Algebra</span></p>
<ul class="nav bd-sidenav">
<li class="toctree-l1"><a class="reference internal" href="statistics.html">1. Elements of Probability Theory and Statistical Data Analysis</a></li>
<li class="toctree-l1"><a class="reference internal" href="linalg.html">2. Linear Algebra, Handling of Arrays and more Python Features</a></li>
</ul>
<p aria-level="2" class="caption" role="heading"><span class="caption-text">From Regression to Support Vector Machines</span></p>
<ul class="nav bd-sidenav">
<li class="toctree-l1"><a class="reference internal" href="chapter1.html">3. Linear Regression</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapter2.html">4. Ridge and Lasso Regression</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapter3.html">5. Resampling Methods</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapter4.html">6. Logistic Regression</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapteroptimization.html">7. Optimization, the central part of any Machine Learning algortithm</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapter5.html">8. Support Vector Machines, overarching aims</a></li>
</ul>
<p aria-level="2" class="caption" role="heading"><span class="caption-text">Decision Trees, Ensemble Methods and Boosting</span></p>
<ul class="nav bd-sidenav">
<li class="toctree-l1"><a class="reference internal" href="chapter6.html">9. Decision trees, overarching aims</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapter7.html">10. Ensemble Methods: From a Single Tree to Many Trees and Extreme Boosting, Meet the Jungle of Methods</a></li>
</ul>
<p aria-level="2" class="caption" role="heading"><span class="caption-text">Dimensionality Reduction</span></p>
<ul class="nav bd-sidenav">
<li class="toctree-l1"><a class="reference internal" href="chapter8.html">11. Basic ideas of the Principal Component Analysis (PCA)</a></li>
<li class="toctree-l1"><a class="reference internal" href="clustering.html">12. Clustering and Unsupervised Learning</a></li>
</ul>
<p aria-level="2" class="caption" role="heading"><span class="caption-text">Deep Learning Methods</span></p>
<ul class="current nav bd-sidenav">
<li class="toctree-l1"><a class="reference internal" href="chapter9.html">13. Neural networks</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapter10.html">14. Building a Feed Forward Neural Network</a></li>
<li class="toctree-l1 current active"><a class="current reference internal" href="#">15. Solving Differential Equations with Deep Learning</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapter12.html">16. Convolutional Neural Networks</a></li>
<li class="toctree-l1"><a class="reference internal" href="chapter13.html">17. Recurrent neural networks: Overarching view</a></li>
</ul>
<p aria-level="2" class="caption" role="heading"><span class="caption-text">Weekly material, notes and exercises</span></p>
<ul class="nav bd-sidenav">
<li class="toctree-l1"><a class="reference internal" href="exercisesweek34.html">Exercises week 34</a></li>
<li class="toctree-l1"><a class="reference internal" href="week34.html">Week 34: Introduction to the course, Logistics and Practicalities</a></li>
<li class="toctree-l1"><a class="reference internal" href="exercisesweek35.html">Exercises week 35</a></li>
<li class="toctree-l1"><a class="reference internal" href="week35.html">Week 35: From Ordinary Linear Regression to Ridge and Lasso Regression</a></li>
<li class="toctree-l1"><a class="reference internal" href="exercisesweek36.html">Exercises week 36</a></li>
<li class="toctree-l1"><a class="reference internal" href="week36.html">Week 36: Linear Regression and Gradient descent</a></li>
<li class="toctree-l1"><a class="reference internal" href="exercisesweek37.html">Exercises week 37</a></li>
</ul>
<p aria-level="2" class="caption" role="heading"><span class="caption-text">Projects</span></p>
<ul class="nav bd-sidenav">
<li class="toctree-l1"><a class="reference internal" href="project1.html">Project 1 on Machine Learning, deadline October 6 (midnight), 2025</a></li>
</ul>
</div>
</nav></div>
</div>
<div class="sidebar-primary-items__end sidebar-primary__section">
</div>
<div id="rtd-footer-container"></div>
</div>
<main id="main-content" class="bd-main" role="main">
<div class="sbt-scroll-pixel-helper"></div>
<div class="bd-content">
<div class="bd-article-container">
<div class="bd-header-article d-print-none">
<div class="header-article-items header-article__inner">
<div class="header-article-items__start">
<div class="header-article-item"><button class="sidebar-toggle primary-toggle btn btn-sm" title="Toggle primary sidebar" data-bs-placement="bottom" data-bs-toggle="tooltip">
<span class="fa-solid fa-bars"></span>
</button></div>
</div>
<div class="header-article-items__end">
<div class="header-article-item">
<div class="article-header-buttons">
<div class="dropdown dropdown-download-buttons">
<button class="btn dropdown-toggle" type="button" data-bs-toggle="dropdown" aria-expanded="false" aria-label="Download this page">
<i class="fas fa-download"></i>
</button>
<ul class="dropdown-menu">
<li><a href="_sources/chapter11.ipynb" target="_blank"
class="btn btn-sm btn-download-source-button dropdown-item"
title="Download source file"
data-bs-placement="left" data-bs-toggle="tooltip"
>
<span class="btn__icon-container">
<i class="fas fa-file"></i>
</span>
<span class="btn__text-container">.ipynb</span>
</a>
</li>
<li>
<button onclick="window.print()"
class="btn btn-sm btn-download-pdf-button dropdown-item"
title="Print to PDF"
data-bs-placement="left" data-bs-toggle="tooltip"
>
<span class="btn__icon-container">
<i class="fas fa-file-pdf"></i>
</span>
<span class="btn__text-container">.pdf</span>
</button>
</li>
</ul>
</div>
<button onclick="toggleFullScreen()"
class="btn btn-sm btn-fullscreen-button"
title="Fullscreen mode"
data-bs-placement="bottom" data-bs-toggle="tooltip"
>
<span class="btn__icon-container">
<i class="fas fa-expand"></i>
</span>
</button>
<script>
document.write(`
<button class="btn btn-sm nav-link pst-navbar-icon theme-switch-button" title="light/dark" aria-label="light/dark" data-bs-placement="bottom" data-bs-toggle="tooltip">
<i class="theme-switch fa-solid fa-sun fa-lg" data-mode="light"></i>
<i class="theme-switch fa-solid fa-moon fa-lg" data-mode="dark"></i>
<i class="theme-switch fa-solid fa-circle-half-stroke fa-lg" data-mode="auto"></i>
</button>
`);
</script>
<script>
document.write(`
<button class="btn btn-sm pst-navbar-icon search-button search-button__button" title="Search" aria-label="Search" data-bs-placement="bottom" data-bs-toggle="tooltip">
<i class="fa-solid fa-magnifying-glass fa-lg"></i>
</button>
`);
</script>
<button class="sidebar-toggle secondary-toggle btn btn-sm" title="Toggle secondary sidebar" data-bs-placement="bottom" data-bs-toggle="tooltip">
<span class="fa-solid fa-list"></span>
</button>
</div></div>
</div>
</div>
</div>
<div id="jb-print-docs-body" class="onlyprint">
<h1>Solving Differential Equations with Deep Learning</h1>
<!-- Table of contents -->
<div id="print-main-content">
<div id="jb-print-toc">
<div>
<h2> Contents </h2>
</div>
<nav aria-label="Page">
<ul class="visible nav section-nav flex-column">
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#example-exponential-decay">15.1. Example: Exponential decay</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#reformulating-the-problem">15.2. Reformulating the problem</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#gradient-descent">15.3. Gradient descent</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#the-code-for-solving-the-ode">15.4. The code for solving the ODE</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#the-network-with-one-input-layer-specified-number-of-hidden-layers-and-one-output-layer">15.5. The network with one input layer, specified number of hidden layers, and one output layer</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#example-population-growth">15.5.1. Example: Population growth</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#using-forward-euler-to-solve-the-ode">15.6. Using forward Euler to solve the ODE</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solving-the-one-dimensional-poisson-equation">15.7. Solving the one dimensional Poisson equation</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#comparing-with-a-numerical-scheme">15.7.1. Comparing with a numerical scheme</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#partial-differential-equations">15.8. Partial Differential Equations</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#type-of-problem">15.8.1. Type of problem</a></li>
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#network-requirements">15.8.2. Network requirements</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#example-the-diffusion-equation">15.9. Example: The diffusion equation</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#setting-up-the-network-using-autograd-the-full-program">15.9.1. Setting up the network using Autograd; The full program</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solving-the-wave-equation-with-neural-networks">15.10. Solving the wave equation with Neural Networks</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#resources-on-differential-equations-and-deep-learning">15.11. Resources on differential equations and deep learning</a></li>
</ul>
</nav>
</div>
</div>
</div>
<div id="searchbox"></div>
<article class="bd-article">
<!-- HTML file automatically generated from DocOnce source (https://github.com/doconce/doconce/)
doconce format html chapter11.do.txt --><section class="tex2jax_ignore mathjax_ignore" id="solving-differential-equations-with-deep-learning">
<h1><span class="section-number">15. </span>Solving Differential Equations with Deep Learning<a class="headerlink" href="#solving-differential-equations-with-deep-learning" title="Link to this heading">#</a></h1>
<p>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.</p>
<p>An ordinary differential equation (ODE) is an equation involving functions having one variable.</p>
<p>In general, an ordinary differential equation looks like</p>
<!-- Equation labels as ordinary links -->
<div id="ode"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{ode} \tag{1}
f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0
\end{equation}
\]</div>
<p>where <span class="math notranslate nohighlight">\(g(x)\)</span> is the function to find, and <span class="math notranslate nohighlight">\(g^{(n)}(x)\)</span> is the <span class="math notranslate nohighlight">\(n\)</span>-th derivative of <span class="math notranslate nohighlight">\(g(x)\)</span>.</p>
<p>The <span class="math notranslate nohighlight">\(f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)\)</span> is just a way to write that there is an expression involving <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)\)</span> on the left side of the equality sign in (<a class="reference internal" href="#ode"><span class="xref myst">1</span></a>).
The highest order of derivative, that is the value of <span class="math notranslate nohighlight">\(n\)</span>, determines to the order of the equation.
The equation is referred to as a <span class="math notranslate nohighlight">\(n\)</span>-th order ODE.
Along with (<a class="reference internal" href="#ode"><span class="xref myst">1</span></a>), some additional conditions of the function <span class="math notranslate nohighlight">\(g(x)\)</span> are typically given
for the solution to be unique.</p>
<p>Let the trial solution <span class="math notranslate nohighlight">\(g_t(x)\)</span> be</p>
<!-- Equation labels as ordinary links -->
<div id="_auto1"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation}
g_t(x) = h_1(x) + h_2(x,N(x,P))
\label{_auto1} \tag{2}
\end{equation}
\]</div>
<p>where <span class="math notranslate nohighlight">\(h_1(x)\)</span> is a function that makes <span class="math notranslate nohighlight">\(g_t(x)\)</span> satisfy a given set
of conditions, <span class="math notranslate nohighlight">\(N(x,P)\)</span> a neural network with weights and biases
described by <span class="math notranslate nohighlight">\(P\)</span> and <span class="math notranslate nohighlight">\(h_2(x, N(x,P))\)</span> some expression involving the
neural network. The role of the function <span class="math notranslate nohighlight">\(h_2(x, N(x,P))\)</span>, is to
ensure that the output from <span class="math notranslate nohighlight">\(N(x,P)\)</span> is zero when <span class="math notranslate nohighlight">\(g_t(x)\)</span> is
evaluated at the values of <span class="math notranslate nohighlight">\(x\)</span> where the given conditions must be
satisfied. The function <span class="math notranslate nohighlight">\(h_1(x)\)</span> should alone make <span class="math notranslate nohighlight">\(g_t(x)\)</span> satisfy
the conditions.</p>
<p>But what about the network <span class="math notranslate nohighlight">\(N(x,P)\)</span>?</p>
<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.</p>
<p>For the minimization to be defined, we need to have a cost function at hand to minimize.</p>
<p>It is given that <span class="math notranslate nohighlight">\(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\)</span> should be equal to zero in (<a class="reference internal" href="#ode"><span class="xref myst">1</span></a>).
We can choose to consider the mean squared error as the cost function for an input <span class="math notranslate nohighlight">\(x\)</span>.
Since we are looking at one input, the cost function is just <span class="math notranslate nohighlight">\(f\)</span> squared.
The cost function <span class="math notranslate nohighlight">\(c\left(x, P \right)\)</span> can therefore be expressed as</p>
<div class="math notranslate nohighlight">
\[
C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2
\]</div>
<p>If <span class="math notranslate nohighlight">\(N\)</span> inputs are given as a vector <span class="math notranslate nohighlight">\(\boldsymbol{x}\)</span> with elements <span class="math notranslate nohighlight">\(x_i\)</span> for <span class="math notranslate nohighlight">\(i = 1,\dots,N\)</span>,
the cost function becomes</p>
<!-- Equation labels as ordinary links -->
<div id="cost"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{cost} \tag{3}
C\left(\boldsymbol{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2
\end{equation}
\]</div>
<p>The neural net should then find the parameters <span class="math notranslate nohighlight">\(P\)</span> that minimizes the cost function in
(<a class="reference internal" href="#cost"><span class="xref myst">3</span></a>) for a set of <span class="math notranslate nohighlight">\(N\)</span> training samples <span class="math notranslate nohighlight">\(x_i\)</span>.</p>
<p>To perform the minimization using gradient descent, the gradient of <span class="math notranslate nohighlight">\(C\left(\boldsymbol{x}, P\right)\)</span> is needed.
It might happen so that finding an analytical expression of the gradient of <span class="math notranslate nohighlight">\(C(\boldsymbol{x}, P)\)</span> from (<a class="reference internal" href="#cost"><span class="xref myst">3</span></a>) gets too messy, depending on which cost function one desires to use.</p>
<p>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.</p>
<section id="example-exponential-decay">
<h2><span class="section-number">15.1. </span>Example: Exponential decay<a class="headerlink" href="#example-exponential-decay" title="Link to this heading">#</a></h2>
<p>An exponential decay of a quantity <span class="math notranslate nohighlight">\(g(x)\)</span> is described by the equation</p>
<!-- Equation labels as ordinary links -->
<div id="solve_expdec"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{solve_expdec} \tag{4}
g'(x) = -\gamma g(x)
\end{equation}
\]</div>
<p>with <span class="math notranslate nohighlight">\(g(0) = g_0\)</span> for some chosen initial value <span class="math notranslate nohighlight">\(g_0\)</span>.</p>
<p>The analytical solution of (<a class="reference internal" href="#solve_expdec"><span class="xref myst">4</span></a>) is</p>
<!-- Equation labels as ordinary links -->
<div id="_auto2"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation}
g(x) = g_0 \exp\left(-\gamma x\right)
\label{_auto2} \tag{5}
\end{equation}
\]</div>
<p>Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (<a class="reference internal" href="#solve_expdec"><span class="xref myst">4</span></a>).</p>
<p>The program will use a neural network to solve</p>
<!-- Equation labels as ordinary links -->
<div id="solveode"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{solveode} \tag{6}
g'(x) = -\gamma g(x)
\end{equation}
\]</div>
<p>where <span class="math notranslate nohighlight">\(g(0) = g_0\)</span> with <span class="math notranslate nohighlight">\(\gamma\)</span> and <span class="math notranslate nohighlight">\(g_0\)</span> being some chosen values.</p>
<p>In this example, <span class="math notranslate nohighlight">\(\gamma = 2\)</span> and <span class="math notranslate nohighlight">\(g_0 = 10\)</span>.</p>
<p>To begin with, a trial solution <span class="math notranslate nohighlight">\(g_t(t)\)</span> must be chosen. A general trial solution for ordinary differential equations could be</p>
<div class="math notranslate nohighlight">
\[
g_t(x, P) = h_1(x) + h_2(x, N(x, P))
\]</div>
<p>with <span class="math notranslate nohighlight">\(h_1(x)\)</span> ensuring that <span class="math notranslate nohighlight">\(g_t(x)\)</span> satisfies some conditions and <span class="math notranslate nohighlight">\(h_2(x,N(x, P))\)</span> an expression involving <span class="math notranslate nohighlight">\(x\)</span> and the output from the neural network <span class="math notranslate nohighlight">\(N(x,P)\)</span> with <span class="math notranslate nohighlight">\(P \)</span> 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.</p>
<p>In this network, there are no weights and bias at the input layer, so <span class="math notranslate nohighlight">\(P = \{ P_{\text{hidden}}, P_{\text{output}} \}\)</span>.
If there are <span class="math notranslate nohighlight">\(N_{\text{hidden} }\)</span> neurons in the hidden layer, then <span class="math notranslate nohighlight">\(P_{\text{hidden}}\)</span> is a <span class="math notranslate nohighlight">\(N_{\text{hidden} } \times (1 + N_{\text{input}})\)</span> matrix, given that there are <span class="math notranslate nohighlight">\(N_{\text{input}}\)</span> neurons in the input layer.</p>
<p>The first column in <span class="math notranslate nohighlight">\(P_{\text{hidden} }\)</span> 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 <span class="math notranslate nohighlight">\(N_{\text{output} }\)</span> neurons in the output layer, then <span class="math notranslate nohighlight">\(P_{\text{output}} \)</span> is a <span class="math notranslate nohighlight">\(N_{\text{output} } \times (1 + N_{\text{hidden} })\)</span> matrix.</p>
<p>Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.</p>
<p>It is given that <span class="math notranslate nohighlight">\(g(0) = g_0\)</span>. The trial solution must fulfill this condition to be a proper solution of (<a class="reference internal" href="#solveode"><span class="xref myst">6</span></a>). A possible way to ensure that <span class="math notranslate nohighlight">\(g_t(0, P) = g_0\)</span>, is to let <span class="math notranslate nohighlight">\(F(N(x,P)) = x \cdot N(x,P)\)</span> and <span class="math notranslate nohighlight">\(A(x) = g_0\)</span>. This gives the following trial solution:</p>
<!-- Equation labels as ordinary links -->
<div id="trial"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{trial} \tag{7}
g_t(x, P) = g_0 + x \cdot N(x, P)
\end{equation}
\]</div>
</section>
<section id="reformulating-the-problem">
<h2><span class="section-number">15.2. </span>Reformulating the problem<a class="headerlink" href="#reformulating-the-problem" title="Link to this heading">#</a></h2>
<p>We wish that our neural network manages to minimize a given cost function.</p>
<p>A reformulation of out equation, (<a class="reference internal" href="#solveode"><span class="xref myst">6</span></a>), must therefore be done,
such that it describes the problem a neural network can solve for.</p>
<p>The neural network must find the set of weights and biases <span class="math notranslate nohighlight">\(P\)</span> such that the trial solution in (<a class="reference internal" href="#trial"><span class="xref myst">7</span></a>) satisfies (<a class="reference internal" href="#solveode"><span class="xref myst">6</span></a>).</p>
<p>The trial solution</p>
<div class="math notranslate nohighlight">
\[
g_t(x, P) = g_0 + x \cdot N(x, P)
\]</div>
<p>has been chosen such that it already solves the condition <span class="math notranslate nohighlight">\(g(0) = g_0\)</span>. What remains, is to find <span class="math notranslate nohighlight">\(P\)</span> such that</p>
<!-- Equation labels as ordinary links -->
<div id="nnmin"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{nnmin} \tag{8}
g_t'(x, P) = - \gamma g_t(x, P)
\end{equation}
\]</div>
<p>is fulfilled as <em>best as possible</em>.</p>
<p>The left hand side and right hand side of (<a class="reference internal" href="#nnmin"><span class="xref myst">8</span></a>) must be computed separately, and then the neural network must choose weights and biases, contained in <span class="math notranslate nohighlight">\(P\)</span>, 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 <span class="math notranslate nohighlight">\(P\)</span> of the neural network.</p>
<p>This gives the following cost function our neural network must solve for:</p>
<div class="math notranslate nohighlight">
\[
\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\}
\]</div>
<p>(the notation <span class="math notranslate nohighlight">\(\min_{P}\{ f(x, P) \}\)</span> means that we desire to find <span class="math notranslate nohighlight">\(P\)</span> that yields the minimum of <span class="math notranslate nohighlight">\(f(x, P)\)</span>)</p>
<p>or, in terms of weights and biases for the hidden and output layer in our network:</p>
<div class="math notranslate nohighlight">
\[
\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\}
\]</div>
<p>for an input value <span class="math notranslate nohighlight">\(x\)</span>.</p>
<p>If the neural network evaluates <span class="math notranslate nohighlight">\(g_t(x, P)\)</span> at more values for <span class="math notranslate nohighlight">\(x\)</span>, say <span class="math notranslate nohighlight">\(N\)</span> values <span class="math notranslate nohighlight">\(x_i\)</span> for <span class="math notranslate nohighlight">\(i = 1, \dots, N\)</span>, then the <em>total</em> error to minimize becomes</p>
<!-- Equation labels as ordinary links -->
<div id="min"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{min} \tag{9}
\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\}
\end{equation}
\]</div>
<p>Letting <span class="math notranslate nohighlight">\(\boldsymbol{x}\)</span> be a vector with elements <span class="math notranslate nohighlight">\(x_i\)</span> and <span class="math notranslate nohighlight">\(C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2\)</span> denote the cost function, the minimization problem that our network must solve, becomes</p>
<div class="math notranslate nohighlight">
\[
\min_{P} C(\boldsymbol{x}, P)
\]</div>
<p>In terms of <span class="math notranslate nohighlight">\(P_{\text{hidden} }\)</span> and <span class="math notranslate nohighlight">\(P_{\text{output} }\)</span>, this could also be expressed as</p>
<div class="math notranslate nohighlight">
\[
\min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\boldsymbol{x}, \{P_{\text{hidden} }, P_{\text{output} }\})
\]</div>
<p>For simplicity, it is assumed that the input is an array <span class="math notranslate nohighlight">\(\boldsymbol{x} = (x_1, \dots, x_N)\)</span> with <span class="math notranslate nohighlight">\(N\)</span> elements. It is at these points the neural network should find <span class="math notranslate nohighlight">\(P\)</span> such that it fulfills (<a class="reference internal" href="#min"><span class="xref myst">9</span></a>).</p>
<p>First, the neural network must feed forward the inputs.
This means that <span class="math notranslate nohighlight">\(\boldsymbol{x}s\)</span> 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 <span class="math notranslate nohighlight">\(N_{\text{input} }\)</span> neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be <span class="math notranslate nohighlight">\(N_{\text{hidden} }\)</span>.</p>
<p>For the <span class="math notranslate nohighlight">\(i\)</span>-th in the hidden layer with weight <span class="math notranslate nohighlight">\(w_i^{\text{hidden} }\)</span> and bias <span class="math notranslate nohighlight">\(b_i^{\text{hidden} }\)</span>, the weighting from the <span class="math notranslate nohighlight">\(j\)</span>-th neuron at the input layer is:</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
z_{i,j}^{\text{hidden}} &amp;= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\
&amp;=
\begin{pmatrix}
b_i^{\text{hidden}} &amp; w_i^{\text{hidden}}
\end{pmatrix}
\begin{pmatrix}
1 \\
x_j
\end{pmatrix}
\end{aligned}
\end{split}\]</div>
<p>The result after weighting the inputs at the <span class="math notranslate nohighlight">\(i\)</span>-th hidden neuron can be written as a vector:</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
\boldsymbol{z}_{i}^{\text{hidden}} &amp;= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\
&amp;=
\begin{pmatrix}
b_i^{\text{hidden}} &amp; w_i^{\text{hidden}}
\end{pmatrix}
\begin{pmatrix}
1 &amp; 1 &amp; \dots &amp; 1 \\
x_1 &amp; x_2 &amp; \dots &amp; x_N
\end{pmatrix} \\
&amp;= \boldsymbol{p}_{i, \text{hidden}}^T X
\end{aligned}
\end{split}\]</div>
<p>The vector <span class="math notranslate nohighlight">\(\boldsymbol{p}_{i, \text{hidden}}^T\)</span> constitutes each row in <span class="math notranslate nohighlight">\(P_{\text{hidden} }\)</span>, which contains the weights for the neural network to minimize according to (<a class="reference internal" href="#min"><span class="xref myst">9</span></a>).</p>
<p>After having found <span class="math notranslate nohighlight">\(\boldsymbol{z}_{i}^{\text{hidden}} \)</span> for every <span class="math notranslate nohighlight">\(i\)</span>-th neuron within the hidden layer, the vector will be sent to an activation function <span class="math notranslate nohighlight">\(a_i(\boldsymbol{z})\)</span>.</p>
<p>In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron:</p>
<div class="math notranslate nohighlight">
\[
f(z) = \frac{1}{1 + \exp{(-z)}}
\]</div>
<p>It is possible to use other activations functions for the hidden layer also.</p>
<p>The output <span class="math notranslate nohighlight">\(\boldsymbol{x}_i^{\text{hidden}}\)</span> from each <span class="math notranslate nohighlight">\(i\)</span>-th hidden neuron is:</p>
<div class="math notranslate nohighlight">
\[
\boldsymbol{x}_i^{\text{hidden} } = f\big( \boldsymbol{z}_{i}^{\text{hidden}} \big)
\]</div>
<p>The outputs <span class="math notranslate nohighlight">\(\boldsymbol{x}_i^{\text{hidden} } \)</span> are then sent to the output layer.</p>
<p>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 <span class="math notranslate nohighlight">\(w_i^{\text{output}}\)</span>
and biases <span class="math notranslate nohighlight">\(b_i^{\text{output}}\)</span>. In this case,
it is assumes that the number of neurons in the output layer is one.</p>
<p>The procedure of weighting the output neuron <span class="math notranslate nohighlight">\(j\)</span> in the hidden layer to the <span class="math notranslate nohighlight">\(i\)</span>-th neuron in the output layer is similar as for the hidden layer described previously.</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
z_{1,j}^{\text{output}} &amp; =
\begin{pmatrix}
b_1^{\text{output}} &amp; \boldsymbol{w}_1^{\text{output}}
\end{pmatrix}
\begin{pmatrix}
1 \\
\boldsymbol{x}_j^{\text{hidden}}
\end{pmatrix}
\end{aligned}
\end{split}\]</div>
<p>Expressing <span class="math notranslate nohighlight">\(z_{1,j}^{\text{output}}\)</span> as a vector gives the following way of weighting the inputs from the hidden layer:</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\boldsymbol{z}_{1}^{\text{output}} =
\begin{pmatrix}
b_1^{\text{output}} &amp; \boldsymbol{w}_1^{\text{output}}
\end{pmatrix}
\begin{pmatrix}
1 &amp; 1 &amp; \dots &amp; 1 \\
\boldsymbol{x}_1^{\text{hidden}} &amp; \boldsymbol{x}_2^{\text{hidden}} &amp; \dots &amp; \boldsymbol{x}_N^{\text{hidden}}
\end{pmatrix}
\end{split}\]</div>
<p>In this case we seek a continuous range of values since we are approximating a function. This means that after computing <span class="math notranslate nohighlight">\(\boldsymbol{z}_{1}^{\text{output}}\)</span> the neural network has finished its feed forward step, and <span class="math notranslate nohighlight">\(\boldsymbol{z}_{1}^{\text{output}}\)</span> is the final output of the network.</p>
<p>The next step is to decide how the parameters should be changed such that they minimize the cost function.</p>
<p>The chosen cost function for this problem is</p>
<div class="math notranslate nohighlight">
\[
C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2
\]</div>
<p>In order to minimize the cost function, an optimization method must be chosen.</p>
<p>Here, gradient descent with a constant step size has been chosen.</p>
</section>
<section id="gradient-descent">
<h2><span class="section-number">15.3. </span>Gradient descent<a class="headerlink" href="#gradient-descent" title="Link to this heading">#</a></h2>
<p>The idea of the gradient descent algorithm is to update parameters in
a direction where the cost function decreases goes to a minimum.</p>
<p>In general, the update of some parameters <span class="math notranslate nohighlight">\(\boldsymbol{\omega}\)</span> given a cost
function defined by some weights <span class="math notranslate nohighlight">\(\boldsymbol{\omega}\)</span>, <span class="math notranslate nohighlight">\(C(\boldsymbol{x},
\boldsymbol{\omega})\)</span>, goes as follows:</p>
<div class="math notranslate nohighlight">
\[
\boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})
\]</div>
<p>for a number of iterations or until <span class="math notranslate nohighlight">\( \big|\big| \boldsymbol{\omega}_{\text{new} } - \boldsymbol{\omega} \big|\big|\)</span> becomes smaller than some given tolerance.</p>
<p>The value of <span class="math notranslate nohighlight">\(\lambda\)</span> decides how large steps the algorithm must take
in the direction of <span class="math notranslate nohighlight">\( \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})\)</span>.
The notation <span class="math notranslate nohighlight">\(\nabla_{\boldsymbol{\omega}}\)</span> express the gradient with respect
to the elements in <span class="math notranslate nohighlight">\(\boldsymbol{\omega}\)</span>.</p>
<p>In our case, we have to minimize the cost function <span class="math notranslate nohighlight">\(C(\boldsymbol{x}, P)\)</span> with
respect to the two sets of weights and biases, that is for the hidden
layer <span class="math notranslate nohighlight">\(P_{\text{hidden} }\)</span> and for the output layer <span class="math notranslate nohighlight">\(P_{\text{output}
}\)</span> .</p>
<p>This means that <span class="math notranslate nohighlight">\(P_{\text{hidden} }\)</span> and <span class="math notranslate nohighlight">\(P_{\text{output} }\)</span> is updated by</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
P_{\text{hidden},\text{new}} &amp;= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\boldsymbol{x}, P) \\
P_{\text{output},\text{new}} &amp;= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\boldsymbol{x}, P)
\end{aligned}
\end{split}\]</div>
</section>
<section id="the-code-for-solving-the-ode">
<h2><span class="section-number">15.4. </span>The code for solving the ODE<a class="headerlink" href="#the-code-for-solving-the-ode" title="Link to this heading">#</a></h2>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span>%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(&#39;Initial cost: %g&#39;%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(&#39;Final cost: %g&#39;%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__ == &#39;__main__&#39;:
# 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(&#39;Max absolute difference: %g&#39;%np.max(np.abs(res - res_analytical)))
# Plot the results
plt.figure(figsize=(10,10))
plt.title(&#39;Performance of neural network solving an ODE compared to the analytical solution&#39;)
plt.plot(x, res_analytical)
plt.plot(x, res[0,:])
plt.legend([&#39;analytical&#39;,&#39;nn&#39;])
plt.xlabel(&#39;x&#39;)
plt.ylabel(&#39;g(x)&#39;)
plt.show()
</pre></div>
</div>
</div>
</div>
</section>
<section id="the-network-with-one-input-layer-specified-number-of-hidden-layers-and-one-output-layer">
<h2><span class="section-number">15.5. </span>The network with one input layer, specified number of hidden layers, and one output layer<a class="headerlink" href="#the-network-with-one-input-layer-specified-number-of-hidden-layers-and-one-output-layer" title="Link to this heading">#</a></h2>
<p>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.</p>
<p>The number of neurons within each hidden layer are given as a list of integers in the program below.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span>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(&#39;Initial cost: %g&#39;%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(&#39;Final cost: %g&#39;%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__ == &#39;__main__&#39;:
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(&#39;Performance of a deep neural network solving an ODE compared to the analytical solution&#39;)
plt.plot(x, res_analytical)
plt.plot(x, res[0,:])
plt.legend([&#39;analytical&#39;,&#39;dnn&#39;])
plt.ylabel(&#39;g(x)&#39;)
plt.show()
</pre></div>
</div>
</div>
</div>
<section id="example-population-growth">
<h3><span class="section-number">15.5.1. </span>Example: Population growth<a class="headerlink" href="#example-population-growth" title="Link to this heading">#</a></h3>
<p>A logistic model of population growth assumes that a population converges toward an equilibrium.
The population growth can be modeled by</p>
<!-- Equation labels as ordinary links -->
<div id="log"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{log} \tag{10}
g'(t) = \alpha g(t)(A - g(t))
\end{equation}
\]</div>
<p>where <span class="math notranslate nohighlight">\(g(t)\)</span> is the population density at time <span class="math notranslate nohighlight">\(t\)</span>, <span class="math notranslate nohighlight">\(\alpha &gt; 0\)</span> the growth rate and <span class="math notranslate nohighlight">\(A &gt; 0\)</span> is the maximum population number in the environment.
Also, at <span class="math notranslate nohighlight">\(t = 0\)</span> the population has the size <span class="math notranslate nohighlight">\(g(0) = g_0\)</span>, where <span class="math notranslate nohighlight">\(g_0\)</span> is some chosen constant.</p>
<p>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.</p>
<p>Here, we will model a population <span class="math notranslate nohighlight">\(g(t)\)</span> in an environment having carrying capacity <span class="math notranslate nohighlight">\(A\)</span>.
The population follows the model</p>
<!-- Equation labels as ordinary links -->
<div id="solveode_population"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{solveode_population} \tag{11}
g'(t) = \alpha g(t)(A - g(t))
\end{equation}
\]</div>
<p>where <span class="math notranslate nohighlight">\(g(0) = g_0\)</span>.</p>
<p>In this example, we let <span class="math notranslate nohighlight">\(\alpha = 2\)</span>, <span class="math notranslate nohighlight">\(A = 1\)</span>, and <span class="math notranslate nohighlight">\(g_0 = 1.2\)</span>.</p>
<p>We will get a slightly different trial solution, as the boundary conditions are different
compared to the case for exponential decay.</p>
<p>A possible trial solution satisfying the condition <span class="math notranslate nohighlight">\(g(0) = g_0\)</span> could be</p>
<div class="math notranslate nohighlight">
\[
h_1(t) = g_0 + t \cdot N(t,P)
\]</div>
<p>with <span class="math notranslate nohighlight">\(N(t,P)\)</span> being the output from the neural network with weights and biases for each layer collected in the set <span class="math notranslate nohighlight">\(P\)</span>.</p>
<p>The analytical solution is</p>
<div class="math notranslate nohighlight">
\[
g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)}
\]</div>
<p>The network will be the similar as for the exponential decay example, but with some small modifications for our problem.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span>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&#39;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(&#39;Initial cost: %g&#39;%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(&#39;Final cost: %g&#39;%cost_function_deep(P, x))
return P
if __name__ == &#39;__main__&#39;:
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(&quot;The max absolute difference between the solutions is: %g&quot;%diff_ag)
plt.figure(figsize=(10,10))
plt.title(&#39;Performance of neural network solving an ODE compared to the analytical solution&#39;)
plt.plot(t, g_analytical)
plt.plot(t, g_dnn_ag[0,:])
plt.legend([&#39;analytical&#39;,&#39;nn&#39;])
plt.xlabel(&#39;t&#39;)
plt.ylabel(&#39;g(t)&#39;)
plt.show()
</pre></div>
</div>
</div>
</div>
</section>
</section>
<section id="using-forward-euler-to-solve-the-ode">
<h2><span class="section-number">15.6. </span>Using forward Euler to solve the ODE<a class="headerlink" href="#using-forward-euler-to-solve-the-ode" title="Link to this heading">#</a></h2>
<p>A straightforward way of solving an ODE numerically, is to use Euler’s method.</p>
<p>Euler’s method uses Taylor series to approximate the value at a function <span class="math notranslate nohighlight">\(f\)</span> at a step <span class="math notranslate nohighlight">\(\Delta x\)</span> from <span class="math notranslate nohighlight">\(x\)</span>:</p>
<div class="math notranslate nohighlight">
\[
f(x + \Delta x) \approx f(x) + \Delta x f'(x)
\]</div>
<p>In our case, using Euler’s method to approximate the value of <span class="math notranslate nohighlight">\(g\)</span> at a step <span class="math notranslate nohighlight">\(\Delta t\)</span> from <span class="math notranslate nohighlight">\(t\)</span> yields</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
g(t + \Delta t) &amp;\approx g(t) + \Delta t g'(t) \\
&amp;= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big)
\end{aligned}
\end{split}\]</div>
<p>along with the condition that <span class="math notranslate nohighlight">\(g(0) = g_0\)</span>.</p>
<p>Let <span class="math notranslate nohighlight">\(t_i = i \cdot \Delta t\)</span> where <span class="math notranslate nohighlight">\(\Delta t = \frac{T}{N_t-1}\)</span> where <span class="math notranslate nohighlight">\(T\)</span> is the final time our solver must solve for and <span class="math notranslate nohighlight">\(N_t\)</span> the number of values for <span class="math notranslate nohighlight">\(t \in [0, T]\)</span> for <span class="math notranslate nohighlight">\(i = 0, \dots, N_t-1\)</span>.</p>
<p>For <span class="math notranslate nohighlight">\(i \geq 1\)</span>, we have that</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
t_i &amp;= i\Delta t \\
&amp;= (i - 1)\Delta t + \Delta t \\
&amp;= t_{i-1} + \Delta t
\end{aligned}
\end{split}\]</div>
<p>Now, if <span class="math notranslate nohighlight">\(g_i = g(t_i)\)</span> then</p>
<!-- Equation labels as ordinary links -->
<div id="odenum"></div>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{equation}
\begin{aligned}
g_i &amp;= g(t_i) \\
&amp;= g(t_{i-1} + \Delta t) \\
&amp;\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\
&amp;= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big)
\end{aligned}
\end{equation} \label{odenum} \tag{12}
\end{split}\]</div>
<p>for <span class="math notranslate nohighlight">\(i \geq 1\)</span> and <span class="math notranslate nohighlight">\(g_0 = g(t_0) = g(0) = g_0\)</span>.</p>
<p>Equation (<a class="reference internal" href="#odenum"><span class="xref myst">12</span></a>) could be implemented in the following way,
extending the program that uses the network using Autograd:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span># Assume that all function definitions from the example program using Autograd
# are located here.
if __name__ == &#39;__main__&#39;:
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(&quot;The max absolute difference between the solutions is: %g&quot;%diff_ag)
plt.figure(figsize=(10,10))
plt.title(&#39;Performance of neural network solving an ODE compared to the analytical solution&#39;)
plt.plot(t, g_analytical)
plt.plot(t, g_dnn_ag[0,:])
plt.legend([&#39;analytical&#39;,&#39;nn&#39;])
plt.xlabel(&#39;t&#39;)
plt.ylabel(&#39;g(t)&#39;)
## 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(&#39;Max absolute difference between Euler method and analytical: %g&#39;%diff1)
print(&#39;Max absolute difference between deep neural network and analytical: %g&#39;%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([&#39;euler&#39;,&#39;analytical&#39;,&#39;dnn&#39;])
plt.xlabel(&#39;Time t&#39;)
plt.ylabel(&#39;g(t)&#39;)
plt.show()
</pre></div>
</div>
</div>
</div>
</section>
<section id="solving-the-one-dimensional-poisson-equation">
<h2><span class="section-number">15.7. </span>Solving the one dimensional Poisson equation<a class="headerlink" href="#solving-the-one-dimensional-poisson-equation" title="Link to this heading">#</a></h2>
<p>The Poisson equation for <span class="math notranslate nohighlight">\(g(x)\)</span> in one dimension is</p>
<!-- Equation labels as ordinary links -->
<div id="poisson"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{poisson} \tag{13}
-g''(x) = f(x)
\end{equation}
\]</div>
<p>where <span class="math notranslate nohighlight">\(f(x)\)</span> is a given function for <span class="math notranslate nohighlight">\(x \in (0,1)\)</span>.</p>
<p>The conditions that <span class="math notranslate nohighlight">\(g(x)\)</span> is chosen to fulfill, are</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{align*}
g(0) &amp;= 0 \\
g(1) &amp;= 0
\end{align*}
\end{split}\]</div>
<p>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.</p>
<p>Here, the function <span class="math notranslate nohighlight">\(g(x)\)</span> to solve for follows the equation</p>
<div class="math notranslate nohighlight">
\[
-g''(x) = f(x),\qquad x \in (0,1)
\]</div>
<p>where <span class="math notranslate nohighlight">\(f(x)\)</span> is a given function, along with the chosen conditions</p>
<!-- Equation labels as ordinary links -->
<div id="cond"></div>
<div class="math notranslate nohighlight">
\[
\begin{aligned}
g(0) = g(1) = 0
\end{aligned}\label{cond} \tag{14}
\]</div>
<p>In this example, we consider the case when <span class="math notranslate nohighlight">\(f(x) = (3x + x^2)\exp(x)\)</span>.</p>
<p>For this case, a possible trial solution satisfying the conditions could be</p>
<div class="math notranslate nohighlight">
\[
g_t(x) = x \cdot (1-x) \cdot N(P,x)
\]</div>
<p>The analytical solution for this problem is</p>
<div class="math notranslate nohighlight">
\[
g(x) = x(1 - x)\exp(x)
\]</div>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span>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(&#39;Initial cost: %g&#39;%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(&#39;Final cost: %g&#39;%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__ == &#39;__main__&#39;:
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(&quot;The max absolute difference between the solutions is: %g&quot;%max_diff)
plt.figure(figsize=(10,10))
plt.title(&#39;Performance of neural network solving an ODE compared to the analytical solution&#39;)
plt.plot(x, g_analytical)
plt.plot(x, g_dnn_ag[0,:])
plt.legend([&#39;analytical&#39;,&#39;nn&#39;])
plt.xlabel(&#39;x&#39;)
plt.ylabel(&#39;g(x)&#39;)
plt.show()
</pre></div>
</div>
</div>
</div>
<section id="comparing-with-a-numerical-scheme">
<h3><span class="section-number">15.7.1. </span>Comparing with a numerical scheme<a class="headerlink" href="#comparing-with-a-numerical-scheme" title="Link to this heading">#</a></h3>
<p>The Poisson equation is possible to solve using Taylor series to approximate the second derivative.</p>
<p>Using Taylor series, the second derivative can be expressed as</p>
<div class="math notranslate nohighlight">
\[
g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x)
\]</div>
<p>where <span class="math notranslate nohighlight">\(\Delta x\)</span> is a small step size and <span class="math notranslate nohighlight">\(E_{\Delta x}(x)\)</span> being the error term.</p>
<p>Looking away from the error terms gives an approximation to the second derivative:</p>
<!-- Equation labels as ordinary links -->
<div id="approx"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{approx} \tag{15}
g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2}
\end{equation}
\]</div>
<p>If <span class="math notranslate nohighlight">\(x_i = i \Delta x = x_{i-1} + \Delta x\)</span> and <span class="math notranslate nohighlight">\(g_i = g(x_i)\)</span> for <span class="math notranslate nohighlight">\(i = 1,\dots N_x - 2\)</span> with <span class="math notranslate nohighlight">\(N_x\)</span> being the number of values for <span class="math notranslate nohighlight">\(x\)</span>, (<a class="reference internal" href="#approx"><span class="xref myst">15</span></a>) becomes</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
g''(x_i) &amp;\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\
&amp;= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2}
\end{aligned}
\end{split}\]</div>
<p>Since we know from our problem that</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
-g''(x) &amp;= f(x) \\
&amp;= (3x + x^2)\exp(x)
\end{aligned}
\end{split}\]</div>
<p>along with the conditions <span class="math notranslate nohighlight">\(g(0) = g(1) = 0\)</span>,
the following scheme can be used to find an approximate solution for <span class="math notranslate nohighlight">\(g(x)\)</span> numerically:</p>
<!-- Equation labels as ordinary links -->
<div id="odesys"></div>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{equation}
\begin{aligned}
-\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &amp;= f(x_i) \\
-g_{i+1} + 2g_i - g_{i-1} &amp;= \Delta x^2 f(x_i)
\end{aligned}
\end{equation} \label{odesys} \tag{16}
\end{split}\]</div>
<p>for <span class="math notranslate nohighlight">\(i = 1, \dots, N_x - 2\)</span> where <span class="math notranslate nohighlight">\(g_0 = g_{N_x - 1} = 0\)</span> and <span class="math notranslate nohighlight">\(f(x_i) = (3x_i + x_i^2)\exp(x_i)\)</span>, which is given for our specific problem.</p>
<p>The equation can be rewritten into a matrix equation:</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
\begin{pmatrix}
2 &amp; -1 &amp; 0 &amp; \dots &amp; 0 \\
-1 &amp; 2 &amp; -1 &amp; \dots &amp; 0 \\
\vdots &amp; &amp; \ddots &amp; &amp; \vdots \\
0 &amp; \dots &amp; -1 &amp; 2 &amp; -1 \\
0 &amp; \dots &amp; 0 &amp; -1 &amp; 2\\
\end{pmatrix}
\begin{pmatrix}
g_1 \\
g_2 \\
\vdots \\
g_{N_x - 3} \\
g_{N_x - 2}
\end{pmatrix}
&amp;=
\Delta x^2
\begin{pmatrix}
f(x_1) \\
f(x_2) \\
\vdots \\
f(x_{N_x - 3}) \\
f(x_{N_x - 2})
\end{pmatrix} \\
\boldsymbol{A}\boldsymbol{g} &amp;= \boldsymbol{f},
\end{aligned}
\end{split}\]</div>
<p>which makes it possible to solve for the vector <span class="math notranslate nohighlight">\(\boldsymbol{g}\)</span>.</p>
<p>We can then compare the result from this numerical scheme with the output from our network using Autograd:</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span>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(&#39;Initial cost: %g&#39;%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(&#39;Final cost: %g&#39;%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__ == &#39;__main__&#39;:
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(&#39;Performance of neural network solving an ODE compared to the analytical solution&#39;)
plt.plot(x, g_analytical)
plt.plot(x, g_dnn_ag[0,:])
plt.legend([&#39;analytical&#39;,&#39;nn&#39;])
plt.xlabel(&#39;x&#39;)
plt.ylabel(&#39;g(x)&#39;)
## 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(&quot;The max absolute difference between the analytical solution and DNN Autograd: %g&quot;%max_diff1)
print(&quot;The max absolute difference between the analytical solution and numerical scheme: %g&quot;%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([&#39;numerical scheme&#39;,&#39;analytical&#39;,&#39;dnn&#39;])
plt.show()
</pre></div>
</div>
</div>
</div>
</section>
</section>
<section id="partial-differential-equations">
<h2><span class="section-number">15.8. </span>Partial Differential Equations<a class="headerlink" href="#partial-differential-equations" title="Link to this heading">#</a></h2>
<p>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.</p>
<p>In general, a partial differential equation for a function <span class="math notranslate nohighlight">\(g(x_1,\dots,x_N)\)</span> with <span class="math notranslate nohighlight">\(N\)</span> variables may be expressed as</p>
<!-- Equation labels as ordinary links -->
<div id="PDE"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{PDE} \tag{17}
f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0
\end{equation}
\]</div>
<p>where <span class="math notranslate nohighlight">\(f\)</span> is an expression involving all kinds of possible mixed derivatives of <span class="math notranslate nohighlight">\(g(x_1,\dots,x_N)\)</span> up to an order <span class="math notranslate nohighlight">\(n\)</span>. In order for the solution to be unique, some additional conditions must also be given.</p>
<section id="type-of-problem">
<h3><span class="section-number">15.8.1. </span>Type of problem<a class="headerlink" href="#type-of-problem" title="Link to this heading">#</a></h3>
<p>The problem our network must solve for, is similar to the ODE case.
We must have a trial solution <span class="math notranslate nohighlight">\(g_t\)</span> at hand.</p>
<p>For instance, the trial solution could be expressed as</p>
<div class="math notranslate nohighlight">
\[
\begin{align*}
g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))
\end{align*}
\]</div>
<p>where <span class="math notranslate nohighlight">\(h_1(x_1,\dots,x_N)\)</span> is a function that ensures <span class="math notranslate nohighlight">\(g_t(x_1,\dots,x_N)\)</span> satisfies some given conditions.
The neural network <span class="math notranslate nohighlight">\(N(x_1,\dots,x_N,P)\)</span> has weights and biases described by <span class="math notranslate nohighlight">\(P\)</span> and <span class="math notranslate nohighlight">\(h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))\)</span> is an expression using the output from the neural network in some way.</p>
<p>The role of the function <span class="math notranslate nohighlight">\(h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))\)</span>, is to ensure that the output of <span class="math notranslate nohighlight">\(N(x_1,\dots,x_N,P)\)</span> is zero when <span class="math notranslate nohighlight">\(g_t(x_1,\dots,x_N)\)</span> is evaluated at the values of <span class="math notranslate nohighlight">\(x_1,\dots,x_N\)</span> where the given conditions must be satisfied. The function <span class="math notranslate nohighlight">\(h_1(x_1,\dots,x_N)\)</span> should alone make <span class="math notranslate nohighlight">\(g_t(x_1,\dots,x_N)\)</span> satisfy the conditions.</p>
</section>
<section id="network-requirements">
<h3><span class="section-number">15.8.2. </span>Network requirements<a class="headerlink" href="#network-requirements" title="Link to this heading">#</a></h3>
<p>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 <span class="math notranslate nohighlight">\(P\)</span> such that the expression <span class="math notranslate nohighlight">\(f\)</span> in (<a class="reference internal" href="#PDE"><span class="xref myst">17</span></a>) is as
close to zero as possible.</p>
<p>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</p>
<div class="math notranslate nohighlight">
\[
C\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2
\]</div>
<p>If we let <span class="math notranslate nohighlight">\(\boldsymbol{x} = \big( x_1, \dots, x_N \big)\)</span> be an array containing the values for <span class="math notranslate nohighlight">\(x_1, \dots, x_N\)</span> respectively, the cost function can be reformulated into the following:</p>
<div class="math notranslate nohighlight">
\[
C\left(\boldsymbol{x}, P\right) = f\left( \left( \boldsymbol{x}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}) }{\partial x_N^n} \right) \right)^2
\]</div>
<p>If we also have <span class="math notranslate nohighlight">\(M\)</span> different sets of values for <span class="math notranslate nohighlight">\(x_1, \dots, x_N\)</span>, that is <span class="math notranslate nohighlight">\(\boldsymbol{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)\)</span> for <span class="math notranslate nohighlight">\(i = 1,\dots,M\)</span> being the rows in matrix <span class="math notranslate nohighlight">\(X\)</span>, the cost function can be generalized into</p>
<div class="math notranslate nohighlight">
\[
C\left(X, P \right) = \sum_{i=1}^M f\left( \left( \boldsymbol{x}_i, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}_i) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}_i) }{\partial x_N^n} \right) \right)^2.
\]</div>
</section>
</section>
<section id="example-the-diffusion-equation">
<h2><span class="section-number">15.9. </span>Example: The diffusion equation<a class="headerlink" href="#example-the-diffusion-equation" title="Link to this heading">#</a></h2>
<p>In one spatial dimension, the equation reads</p>
<div class="math notranslate nohighlight">
\[
\frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2}
\]</div>
<p>where a possible choice of conditions are</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{align*}
g(0,t) &amp;= 0 ,\qquad t \geq 0 \\
g(1,t) &amp;= 0, \qquad t \geq 0 \\
g(x,0) &amp;= u(x),\qquad x\in [0,1]
\end{align*}
\end{split}\]</div>
<p>with <span class="math notranslate nohighlight">\(u(x)\)</span> being some given function.</p>
<p>For this case, we want to find <span class="math notranslate nohighlight">\(g(x,t)\)</span> such that</p>
<!-- Equation labels as ordinary links -->
<div id="diffonedim"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation}
\frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2}
\end{equation} \label{diffonedim} \tag{18}
\]</div>
<p>and</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{align*}
g(0,t) &amp;= 0 ,\qquad t \geq 0 \\
g(1,t) &amp;= 0, \qquad t \geq 0 \\
g(x,0) &amp;= u(x),\qquad x\in [0,1]
\end{align*}
\end{split}\]</div>
<p>with <span class="math notranslate nohighlight">\(u(x) = \sin(\pi x)\)</span>.</p>
<p>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.</p>
<p>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 <span class="math notranslate nohighlight">\(t\)</span>
and position <span class="math notranslate nohighlight">\(x\)</span>. The variables will be represented by a
one-dimensional array in the program. The program will evaluate the
network at each possible pair <span class="math notranslate nohighlight">\((x,t)\)</span>, given an array for the desired
<span class="math notranslate nohighlight">\(x\)</span>-values and <span class="math notranslate nohighlight">\(t\)</span>-values to approximate the solution at.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span>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]
</pre></div>
</div>
</div>
</div>
<p>The cost function must then iterate through the given arrays
containing values for <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(t\)</span>, defines a point <span class="math notranslate nohighlight">\((x,t)\)</span> the deep
neural network and the trial solution is evaluated at, and then finds
the Jacobian of the trial solution.</p>
<p>A possible trial solution for this PDE is</p>
<div class="math notranslate nohighlight">
\[
g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P)
\]</div>
<p>with <span class="math notranslate nohighlight">\(A(x,t)\)</span> being a function ensuring that <span class="math notranslate nohighlight">\(g_t(x,t)\)</span> satisfies our given conditions, and <span class="math notranslate nohighlight">\(N(x,t,P)\)</span> being the output from the deep neural network using weights and biases for each layer from <span class="math notranslate nohighlight">\(P\)</span>.</p>
<p>To fulfill the conditions, <span class="math notranslate nohighlight">\(A(x,t)\)</span> could be:</p>
<div class="math notranslate nohighlight">
\[
h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x)
\]</div>
<p>since <span class="math notranslate nohighlight">\((0) = u(1) = 0\)</span> and <span class="math notranslate nohighlight">\(u(x) = \sin(\pi x)\)</span>.</p>
<p>The Jacobian is used because the program must find the derivative of
the trial solution with respect to <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(t\)</span>.</p>
<p>This gives the necessity of computing the Jacobian matrix, as we want
to evaluate the gradient with respect to <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(t\)</span> (note that the
Jacobian of a scalar-valued multivariate function is simply its
gradient).</p>
<p>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 <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(t\)</span>, the Jacobian is calculated using
the values of <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(t\)</span>.</p>
<p>To find the second derivative with respect to <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(t\)</span>, 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 <span class="math notranslate nohighlight">\(g(x,t)\)</span>.</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span># 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
</pre></div>
</div>
</div>
</div>
<section id="setting-up-the-network-using-autograd-the-full-program">
<h3><span class="section-number">15.9.1. </span>Setting up the network using Autograd; The full program<a class="headerlink" href="#setting-up-the-network-using-autograd-the-full-program" title="Link to this heading">#</a></h3>
<p>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.</p>
<p>The analytical solution of our problem is</p>
<div class="math notranslate nohighlight">
\[
g(x,t) = \exp(-\pi^2 t)\sin(\pi x)
\]</div>
<p>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.</p>
<p>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!</p>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span>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(&#39;Initial cost: &#39;,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(&#39;Final cost: &#39;,cost_function(P, x, t))
return P
if __name__ == &#39;__main__&#39;:
### 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(&#39;Max absolute difference between the analytical solution and the network: %g&#39;%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=&#39;3d&#39;)
ax.set_title(&#39;Solution from the deep neural network w/ %d layer&#39;%len(num_hidden_neurons))
s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel(&#39;Time $t$&#39;)
ax.set_ylabel(&#39;Position $x$&#39;);
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection=&#39;3d&#39;)
ax.set_title(&#39;Analytical solution&#39;)
s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel(&#39;Time $t$&#39;)
ax.set_ylabel(&#39;Position $x$&#39;);
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection=&#39;3d&#39;)
ax.set_title(&#39;Difference&#39;)
s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel(&#39;Time $t$&#39;)
ax.set_ylabel(&#39;Position $x$&#39;);
## 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(&quot;Computed solutions at time = %g&quot;%t1)
plt.plot(x, res1)
plt.plot(x,res_analytical1)
plt.legend([&#39;dnn&#39;,&#39;analytical&#39;])
plt.figure(figsize=(10,10))
plt.title(&quot;Computed solutions at time = %g&quot;%t2)
plt.plot(x, res2)
plt.plot(x,res_analytical2)
plt.legend([&#39;dnn&#39;,&#39;analytical&#39;])
plt.figure(figsize=(10,10))
plt.title(&quot;Computed solutions at time = %g&quot;%t3)
plt.plot(x, res3)
plt.plot(x,res_analytical3)
plt.legend([&#39;dnn&#39;,&#39;analytical&#39;])
plt.show()
</pre></div>
</div>
</div>
</div>
</section>
</section>
<section id="solving-the-wave-equation-with-neural-networks">
<h2><span class="section-number">15.10. </span>Solving the wave equation with Neural Networks<a class="headerlink" href="#solving-the-wave-equation-with-neural-networks" title="Link to this heading">#</a></h2>
<p>The wave equation is</p>
<div class="math notranslate nohighlight">
\[
\frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2}
\]</div>
<p>with <span class="math notranslate nohighlight">\(c\)</span> being the specified wave speed.</p>
<p>Here, the chosen conditions are</p>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{align*}
g(0,t) &amp;= 0 \\
g(1,t) &amp;= 0 \\
g(x,0) &amp;= u(x) \\
\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &amp;= v(x)
\end{align*}
\end{split}\]</div>
<p>where <span class="math notranslate nohighlight">\(\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}\)</span> means the derivative of <span class="math notranslate nohighlight">\(g(x,t)\)</span> with respect to <span class="math notranslate nohighlight">\(t\)</span> is evaluated at <span class="math notranslate nohighlight">\(t = 0\)</span>, and <span class="math notranslate nohighlight">\(u(x)\)</span> and <span class="math notranslate nohighlight">\(v(x)\)</span> being given functions.</p>
<p>The wave equation to solve for, is</p>
<!-- Equation labels as ordinary links -->
<div id="wave"></div>
<div class="math notranslate nohighlight">
\[
\begin{equation} \label{wave} \tag{19}
\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2}
\end{equation}
\]</div>
<p>where <span class="math notranslate nohighlight">\(c\)</span> is the given wave speed.
The chosen conditions for this equation are</p>
<!-- Equation labels as ordinary links -->
<div id="condwave"></div>
<div class="math notranslate nohighlight">
\[\begin{split}
\begin{aligned}
g(0,t) &amp;= 0, &amp;t \geq 0 \\
g(1,t) &amp;= 0, &amp;t \geq 0 \\
g(x,0) &amp;= u(x), &amp;x\in[0,1] \\
\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &amp;= v(x), &amp;x \in [0,1]
\end{aligned} \label{condwave} \tag{20}
\end{split}\]</div>
<p>In this example, let <span class="math notranslate nohighlight">\(c = 1\)</span> and <span class="math notranslate nohighlight">\(u(x) = \sin(\pi x)\)</span> and <span class="math notranslate nohighlight">\(v(x) = -\pi\sin(\pi x)\)</span>.</p>
<p>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 (<a class="reference internal" href="#condwave"><span class="xref myst">20</span></a>) and the cost function.</p>
<p>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 <span class="math notranslate nohighlight">\(g_t(x,t)\)</span> is</p>
<div class="math notranslate nohighlight">
\[
g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P)
\]</div>
<p>where</p>
<div class="math notranslate nohighlight">
\[
h_1(x,t) = (1-t^2)u(x) + tv(x)
\]</div>
<p>Note that this trial solution satisfies the conditions only if <span class="math notranslate nohighlight">\(u(0) = v(0) = u(1) = v(1) = 0\)</span>, which is the case in this example.</p>
<p>The analytical solution for our specific problem, is</p>
<div class="math notranslate nohighlight">
\[
g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t)
\]</div>
<div class="cell docutils container">
<div class="cell_input docutils container">
<div class="highlight-none notranslate"><div class="highlight"><pre><span></span>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(&#39;Initial cost: &#39;,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(&#39;Final cost: &#39;,cost_function(P, x, t))
return P
if __name__ == &#39;__main__&#39;:
### 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(&quot;Max difference between analytical and solution from nn: %g&quot;%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=&#39;3d&#39;)
ax.set_title(&#39;Solution from the deep neural network w/ %d layer&#39;%len(num_hidden_neurons))
s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel(&#39;Time $t$&#39;)
ax.set_ylabel(&#39;Position $x$&#39;);
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection=&#39;3d&#39;)
ax.set_title(&#39;Analytical solution&#39;)
s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel(&#39;Time $t$&#39;)
ax.set_ylabel(&#39;Position $x$&#39;);
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection=&#39;3d&#39;)
ax.set_title(&#39;Difference&#39;)
s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel(&#39;Time $t$&#39;)
ax.set_ylabel(&#39;Position $x$&#39;);
## 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(&quot;Computed solutions at time = %g&quot;%t1)
plt.plot(x, res1)
plt.plot(x,res_analytical1)
plt.legend([&#39;dnn&#39;,&#39;analytical&#39;])
plt.figure(figsize=(10,10))
plt.title(&quot;Computed solutions at time = %g&quot;%t2)
plt.plot(x, res2)
plt.plot(x,res_analytical2)
plt.legend([&#39;dnn&#39;,&#39;analytical&#39;])
plt.figure(figsize=(10,10))
plt.title(&quot;Computed solutions at time = %g&quot;%t3)
plt.plot(x, res3)
plt.plot(x,res_analytical3)
plt.legend([&#39;dnn&#39;,&#39;analytical&#39;])
plt.show()
</pre></div>
</div>
</div>
</div>
</section>
<section id="resources-on-differential-equations-and-deep-learning">
<h2><span class="section-number">15.11. </span>Resources on differential equations and deep learning<a class="headerlink" href="#resources-on-differential-equations-and-deep-learning" title="Link to this heading">#</a></h2>
<ol class="arabic simple">
<li><p><a class="reference external" href="https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf">Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al</a></p></li>
<li><p><a class="reference external" href="https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c">Neural networks for solving differential equations by A. Honchar</a></p></li>
<li><p><a class="reference external" href="http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf">Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener</a></p></li>
<li><p><a class="reference external" href="https://www.springer.com/us/book/9783540225515">Introduction to Partial Differential Equations by A. Tveito, R. Winther</a></p></li>
</ol>
</section>
</section>
<script type="text/x-thebe-config">
{
requestKernel: true,
binderOptions: {
repo: "binder-examples/jupyter-stacks-datascience",
ref: "master",
},
codeMirrorConfig: {
theme: "abcdef",
mode: "python"
},
kernelOptions: {
name: "python3",
path: "./."
},
predefinedOutput: true
}
</script>
<script>kernelName = 'python3'</script>
</article>
<footer class="prev-next-footer d-print-none">
<div class="prev-next-area">
<a class="left-prev"
href="chapter10.html"
title="previous page">
<i class="fa-solid fa-angle-left"></i>
<div class="prev-next-info">
<p class="prev-next-subtitle">previous</p>
<p class="prev-next-title"><span class="section-number">14. </span>Building a Feed Forward Neural Network</p>
</div>
</a>
<a class="right-next"
href="chapter12.html"
title="next page">
<div class="prev-next-info">
<p class="prev-next-subtitle">next</p>
<p class="prev-next-title"><span class="section-number">16. </span>Convolutional Neural Networks</p>
</div>
<i class="fa-solid fa-angle-right"></i>
</a>
</div>
</footer>
</div>
<div class="bd-sidebar-secondary bd-toc"><div class="sidebar-secondary-items sidebar-secondary__inner">
<div class="sidebar-secondary-item">
<div class="page-toc tocsection onthispage">
<i class="fa-solid fa-list"></i> Contents
</div>
<nav class="bd-toc-nav page-toc">
<ul class="visible nav section-nav flex-column">
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#example-exponential-decay">15.1. Example: Exponential decay</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#reformulating-the-problem">15.2. Reformulating the problem</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#gradient-descent">15.3. Gradient descent</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#the-code-for-solving-the-ode">15.4. The code for solving the ODE</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#the-network-with-one-input-layer-specified-number-of-hidden-layers-and-one-output-layer">15.5. The network with one input layer, specified number of hidden layers, and one output layer</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#example-population-growth">15.5.1. Example: Population growth</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#using-forward-euler-to-solve-the-ode">15.6. Using forward Euler to solve the ODE</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solving-the-one-dimensional-poisson-equation">15.7. Solving the one dimensional Poisson equation</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#comparing-with-a-numerical-scheme">15.7.1. Comparing with a numerical scheme</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#partial-differential-equations">15.8. Partial Differential Equations</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#type-of-problem">15.8.1. Type of problem</a></li>
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#network-requirements">15.8.2. Network requirements</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#example-the-diffusion-equation">15.9. Example: The diffusion equation</a><ul class="nav section-nav flex-column">
<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#setting-up-the-network-using-autograd-the-full-program">15.9.1. Setting up the network using Autograd; The full program</a></li>
</ul>
</li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solving-the-wave-equation-with-neural-networks">15.10. Solving the wave equation with Neural Networks</a></li>
<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#resources-on-differential-equations-and-deep-learning">15.11. Resources on differential equations and deep learning</a></li>
</ul>
</nav></div>
</div></div>
</div>
<footer class="bd-footer-content">
<div class="bd-footer-content__inner container">
<div class="footer-item">
<p class="component-author">
By Morten Hjorth-Jensen
</p>
</div>
<div class="footer-item">
<p class="copyright">
© Copyright 2023.
<br/>
</p>
</div>
<div class="footer-item">
</div>
<div class="footer-item">
</div>
</div>
</footer>
</main>
</div>
</div>
<!-- Scripts loaded after <body> so the DOM is not blocked -->
<script src="_static/scripts/bootstrap.js?digest=dfe6caa3a7d634c4db9b"></script>
<script src="_static/scripts/pydata-sphinx-theme.js?digest=dfe6caa3a7d634c4db9b"></script>
<footer class="bd-footer">
</footer>
</body>
</html>
Reference in New Issue View Git Blame Copy Permalink