0
$\begingroup$

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 !

$\endgroup$
5
  • $\begingroup$ can be typo for $ v=y^{'} ?$ $\endgroup$
    – Narasimham
    Mar 23 '20 at 17:19
  • $\begingroup$ I let $u=y, v=y'$, so that $u'=y'=v$, and $v'=y''=-y=-u$. $\endgroup$
    – mXdX
    Mar 23 '20 at 17:22
  • 1
    $\begingroup$ 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 $y$. $\endgroup$ Mar 23 '20 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
    Mar 23 '20 at 17:31
  • $\begingroup$ How can the method be applied to the system? $\endgroup$
    – mXdX
    Mar 23 '20 at 17:32
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Do you mind explaining a bit more about "the Euler step is tangential to the convex solution curve"? $\endgroup$
    – mXdX
    Mar 23 '20 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. $\endgroup$ Mar 23 '20 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.