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Consider the IVP \begin{align*} y''=-y \end{align*} for $t \geq 0$, and $y(0)=1$, $y'(0)=2$.

I have rewritten this differential equation as a system of first-order ODE's such that \begin{align*} u'=v\\ v'=-u\\ \end{align*} with $u(0)=1, v(0)=2$.

The solution is $y=2\sin(t)+\cos(t)$, $y'=2\cos(t)-\sin(t)$.

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{align*} y_1=y_0+hf(t_0, y_0)=1+(0.5)(2\cos(0)-\sin(0))=2. \end{align*} But how do I determine if Euler's method is stable? I know that for the equation $y'=\lambda y$, Euler's method is stable for $|1+h\lambda| \leq 1$, but since this problem is in a different form I'm not sure what I need to do here.

Thanks !

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  • $\begingroup$ can be typo for $ v=y^{'} ?$ $\endgroup$
    – Narasimham
    Commented Mar 23, 2020 at 17:19
  • $\begingroup$ I let $u=y, v=y'$, so that $u'=y'=v$, and $v'=y''=-y=-u$. $\endgroup$
    – mXdX
    Commented Mar 23, 2020 at 17:22
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    $\begingroup$ What does stability mean for a system with eigenvalues $\pm i$? That the numerical method gives a bounded result? How can you observe that in just one step? // You need to apply the method to the first order system $u'=v$, $v'=-u$ that you constructed. Please correct the question to the correct first order system without $y$. $\endgroup$ Commented Mar 23, 2020 at 17:27
  • $\begingroup$ I was asked in a previous question if the system is stable. Since the eigenvalues are $\pm i$, i.e., Re($\lambda_1$)=0 and Re($\lambda_2$)=0, the system is stable. But now the question is whether or not the system is stable particularly for Euler's method, and I'm not sure how to determine that. $\endgroup$
    – mXdX
    Commented Mar 23, 2020 at 17:31
  • $\begingroup$ How can the method be applied to the system? $\endgroup$
    – mXdX
    Commented Mar 23, 2020 at 17:32

2 Answers 2

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Applying the Euler iteration procedure we have

$$ \cases{ u_k = u_{k-1}+h v_{k-1}\\ v_k = v_{k-1}-h u_{k-1} } $$

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 $1\pm i h$ with absolute value $\sqrt{1+h^2} > 1$ so the Euler procedure diverges.

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You apply the Euler step to the first-order system. $$u_1=u_0+hu_0'=u_0+hv_0,\\ v_1=v_0+hv_0'=v_0-hu_0.$$ 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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  • $\begingroup$ Do you mind explaining a bit more about "the Euler step is tangential to the convex solution curve"? $\endgroup$
    – mXdX
    Commented Mar 23, 2020 at 18:15
  • $\begingroup$ The solutions in phase space are concentric circles, the Euler step is a line segment in tangential direction from the current circle. Draw yourself a picture to see that you get an outward spiral. $\endgroup$ Commented Mar 23, 2020 at 18:32

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