Numerical Solution of ODE’s

\(\require{cancel}\newcommand{\dee}[1]{\mathrm{d}#1} \newcommand{\half}{ \frac{1}{2} } \newcommand{\ds}{\displaystyle} \newcommand{\ts}{\textstyle} \newcommand{\es}{ {\varnothing}} \newcommand{\st}{ {\mbox{ s.t. }} } \newcommand{\pow}[1]{ \mathcal{P}\left(#1\right) } \newcommand{\set}[1]{ \left\{#1\right\} } \newcommand{\lin}{{\text{LIN}}} \newcommand{\quot}{{\text{QR}}} \newcommand{\simp}{{\text{SMP}}} \newcommand{\diff}[2]{ \frac{\mathrm{d}#1}{\mathrm{d}#2}} \newcommand{\bdiff}[2]{ \frac{\mathrm{d}}{\mathrm{d}#2} \left( #1 \right)} \newcommand{\ddiff}[3]{ \frac{\mathrm{d}^#1#2}{\mathrm{d}{#3}^#1}} \renewcommand{\neg}{ {\sim} } \newcommand{\limp}{ {\;\Rightarrow\;} } \newcommand{\nimp}{ {\;\not\Rightarrow\;} } \newcommand{\liff}{ {\;\Leftrightarrow\;} } \newcommand{\niff}{ {\;\not\Leftrightarrow\;} } \newcommand{\De}{\Delta} \newcommand{\bbbn}{\mathbb{N}} \newcommand{\bbbr}{\mathbb{R}} \newcommand{\bbbp}{\mathbb{P}} \newcommand{\cA}{\mathcal{A}} \newcommand{\cI}{\mathcal{I}} \newcommand{\cR}{\mathcal{R}} \newcommand{\cV}{\mathcal{V}} \newcommand{\Si}{\Sigma} \newcommand{\arccsc}{\mathop{\mathrm{arccsc}}} \newcommand{\arcsec}{\mathop{\mathrm{arcsec}}} \newcommand{\arccot}{\mathop{\mathrm{arccot}}} \newcommand{\erf}{\mathop{\mathrm{erf}}} \newcommand{\smsum}{\mathop{{\ts \sum}}} \newcommand{\atp}[2]{ \genfrac{}{}{0in}{}{#1}{#2} } \newcommand{\ave}{\mathrm{ave}} \newcommand{\llt}{\left \lt } \newcommand{\rgt}{\right \gt } \newcommand{\YEaxis}[2]{\draw[help lines] (-#1,0)--(#1,0) node[right]{$x$};\draw[help lines] (0,-#2)--(0,#2) node[above]{$y$};} \newcommand{\YEaaxis}[4]{\draw[help lines] (-#1,0)--(#2,0) node[right]{$x$};\draw[help lines] (0,-#3)--(0,#4) node[above]{$y$};} \newcommand{\YEtaxis}[4]{\draw[help lines] (-#1,0)--(#2,0) node[right]{$t$};\draw[help lines] (0,-#3)--(0,#4) node[above]{$y$};} \newcommand{\YEtaaxis}[4]{\draw[help lines] (-#1,0)--(#2,0) node[right]{$t$}; \draw[help lines] (0,-#3)--(0,#4) node[above]{$y$};} \newcommand{\YExcoord}[2]{\draw (#1,.2)--(#1,-.2) node[below]{$#2$};} \newcommand{\YEycoord}[2]{\draw (.2,#1)--(-.2,#1) node[left]{$#2$};} \newcommand{\YEnxcoord}[2]{\draw (#1,-.2)--(#1,.2) node[above]{$#2$};} \newcommand{\YEnycoord}[2]{\draw (-.2,#1)--(.2,#1) node[right]{$#2$};} \newcommand{\YEstickfig}[3]{ \draw (#1,#2) arc(-90:270:2mm); \draw (#1,#2)--(#1,#2-.5) (#1-.25,#2-.75)--(#1,#2-.5)--(#1+.25,#2-.75) (#1-.2,#2-.2)--(#1+.2,#2-.2);} \newcommand{\IBP}[7]{ \begin{array}{|c | l | l |} \hline \color{red}{\text{Option 1:}} & u=#2 &\color{red}{\dee{u}=#3 ~ \dee{#1}} \\ & \dee{v}=#5~\dee{#1} &\color{red}{v=#7} \\ \hline \color{blue}{\text{Option 2:}} & u=#5 &\color{blue}{\dee{u}=#6 ~ \dee{#1}} \\ &\dee{v}=#2 \dee{#1} &\color{blue}{v=#4} \\ \hline \end{array} } \renewcommand{\textcolor}[2]{{\color{#1}{#2}}} \newcommand{\trigtri}[4]{ \begin{tikzpicture} \draw (-.5,0)--(2,0)--(2,1.5)--cycle; \draw (1.8,0) |- (2,.2); \draw[double] (0,0) arc(0:30:.5cm); \draw (0,.2) node[right]{$#1$}; \draw (1,-.5) node{$#2$}; \draw (2,.75) node[right]{$#3$}; \draw (.6,1.1) node[rotate=30]{$#4$}; \end{tikzpicture}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \definecolor{fillinmathshade}{gray}{0.9} \newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} \)

Appendix D Numerical Solution of ODE’s

In Section 2.4 we solved a number of inital value problems of the form

\begin{align*} y'(t)&=f\big(t,y(t)\big) \\ y(t_0)&=y_0 \end{align*}

Here $f(t,y)$ is a given function, $t_0$ is a given initial time and $y_0$ is a given initial value for $y\text{.}$ The unknown in the problem is the function $y(t)\text{.}$ There are a number of other techniques for analytically solving some problems of this type. However it is often simply not possible to find an explicit solution. This appendix introduces some simple algorithms for generating approximate numerical solutions to such problems.

🔗

Prev Top Next