Summer Bradford

2022-06-26

Consider the IVP

$\begin{array}{r}{y}^{\u2033}=-y\end{array}$

for $t\ge 0$, and $y(0)=1$, ${y}^{\prime}(0)=2$.

I have rewritten this differential equation as a system of first-order ODE's such that

$\begin{array}{r}{u}^{\prime}=v\\ {v}^{\prime}=-u\end{array}$

with $u(0)=1,v(0)=2$.

The solution is $y=2\mathrm{sin}(t)+\mathrm{cos}(t)$, ${y}^{\prime}=2\mathrm{cos}(t)-\mathrm{sin}(t)$

I am asked to perform one step of Euler's method with $h=0.5$, and determine if Euler's method is stable for this problem.

For the first part, I find that one step of Euler's method yields

$\begin{array}{r}{y}_{1}={y}_{0}+hf({t}_{0},{y}_{0})=1+(0.5)(2\mathrm{cos}(0)-\mathrm{sin}(0))=2.\end{array}$

But how do I determine if Euler's method is stable? I know that for the equation ${y}^{\prime}=\lambda y$, Euler's method is stable for $|1+h\lambda |\le 1$, but since this problem is in a different form I'm not sure what I need to do here.

$\begin{array}{r}{y}^{\u2033}=-y\end{array}$

for $t\ge 0$, and $y(0)=1$, ${y}^{\prime}(0)=2$.

I have rewritten this differential equation as a system of first-order ODE's such that

$\begin{array}{r}{u}^{\prime}=v\\ {v}^{\prime}=-u\end{array}$

with $u(0)=1,v(0)=2$.

The solution is $y=2\mathrm{sin}(t)+\mathrm{cos}(t)$, ${y}^{\prime}=2\mathrm{cos}(t)-\mathrm{sin}(t)$

I am asked to perform one step of Euler's method with $h=0.5$, and determine if Euler's method is stable for this problem.

For the first part, I find that one step of Euler's method yields

$\begin{array}{r}{y}_{1}={y}_{0}+hf({t}_{0},{y}_{0})=1+(0.5)(2\mathrm{cos}(0)-\mathrm{sin}(0))=2.\end{array}$

But how do I determine if Euler's method is stable? I know that for the equation ${y}^{\prime}=\lambda y$, Euler's method is stable for $|1+h\lambda |\le 1$, but since this problem is in a different form I'm not sure what I need to do here.

Bornejecbo

Beginner2022-06-27Added 19 answers

Applying the Euler iteration procedure we have

$\{\begin{array}{l}{u}_{k}={u}_{k-1}+h{v}_{k-1}\\ {v}_{k}={v}_{k-1}-h{u}_{k-1}\end{array}$

or

$\left(\begin{array}{c}{u}_{k}\\ {v}_{k}\end{array}\right)=\left(\begin{array}{cc}1& h\\ -h& 1\end{array}\right)\left(\begin{array}{c}{u}_{k-1}\\ {v}_{k-1}\end{array}\right)$

or

${U}_{k}={M}^{k}{U}_{0}$

this sequence converges as long as the eigenvalues of $M$ have absolute value less than 1. Here the $M$ eigenvalues are $1\pm ih$ with absolute value $\sqrt{1+{h}^{2}}>1$ so the Euler procedure diverges.

$\{\begin{array}{l}{u}_{k}={u}_{k-1}+h{v}_{k-1}\\ {v}_{k}={v}_{k-1}-h{u}_{k-1}\end{array}$

or

$\left(\begin{array}{c}{u}_{k}\\ {v}_{k}\end{array}\right)=\left(\begin{array}{cc}1& h\\ -h& 1\end{array}\right)\left(\begin{array}{c}{u}_{k-1}\\ {v}_{k-1}\end{array}\right)$

or

${U}_{k}={M}^{k}{U}_{0}$

this sequence converges as long as the eigenvalues of $M$ have absolute value less than 1. Here the $M$ eigenvalues are $1\pm ih$ with absolute value $\sqrt{1+{h}^{2}}>1$ so the Euler procedure diverges.

Layla Velazquez

Beginner2022-06-28Added 11 answers

You apply the Euler step to the first-order system.

${u}_{1}={u}_{0}+h{u}_{0}^{\prime}={u}_{0}+h{v}_{0},\phantom{\rule{0ex}{0ex}}{v}_{1}={v}_{0}+h{v}_{0}^{\prime}={v}_{0}-h{u}_{0}.$

As the Euler step is tangential to the convex solution curve, it will always move outwards, away from the center of the concentric exact solution curves.

${u}_{1}={u}_{0}+h{u}_{0}^{\prime}={u}_{0}+h{v}_{0},\phantom{\rule{0ex}{0ex}}{v}_{1}={v}_{0}+h{v}_{0}^{\prime}={v}_{0}-h{u}_{0}.$

As the Euler step is tangential to the convex solution curve, it will always move outwards, away from the center of the concentric exact solution curves.

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