# is Euler's method stable for this problem?

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 !

• can be typo for $v=y^{'} ?$ Commented Mar 23, 2020 at 17:19
• I let $u=y, v=y'$, so that $u'=y'=v$, and $v'=y''=-y=-u$.
– mXdX
Commented Mar 23, 2020 at 17:22
• 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 $y$. Commented Mar 23, 2020 at 17:27
• 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.
– mXdX
Commented Mar 23, 2020 at 17:31
• How can the method be applied to the system?
– mXdX
Commented Mar 23, 2020 at 17:32

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.

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.

• Do you mind explaining a bit more about "the Euler step is tangential to the convex solution curve"?
– mXdX
Commented Mar 23, 2020 at 18:15
• 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. Commented Mar 23, 2020 at 18:32