oopsteekwe

2022-10-21

Consider the initial value problem
$\left\{\begin{array}{l}{y}^{\prime }=1-{y}^{2},\phantom{\rule{1em}{0ex}}t>0\\ y\left(0\right)=\frac{e-1}{e+1}.\end{array}$
Is there a way I find a limit on the mesh-size $h$ such that the explicit Euler method is stable? Do I need to solve the IVP to find it?
Edit: After looking through a number of textbooks and online resources I can see that they apply a method to a linear test problem and then solve $|1+h\lambda |⩽1$. Is there any way I can do something similar to solve for $h$ in this nonlinear case.

Do you have a similar question?

elulamami

Expert

If you have an explicit global Lipschitz constant, then you can go through the linear development in more or less the same manner, but when you don't that won't work. And here there is no global Lipschitz constant at all, much less one you can compute.

That said here you can view things in a rather elementary manner. The point is that the true IVP stays in $\left(-1,1\right)$ if it begins there, so you'll have stability if your method does the same. Since ${y}^{\prime }>0$ while you are in $\left(-1,1\right)$, that means that as you go forward in time you want $y+h\left(1-{y}^{2}\right)<1$ when $y$ is any number between −1 and 1. From here you can readily infer a condition on $h$. If you want backward-in-time stability (which the true IVP also has) then you want $y-h\left(1-{y}^{2}\right)>-1$, which winds up resulting in the same condition on h in the end.

This is not much different from the stability analysis of ${y}^{\prime }=-y$ with explicit Euler, where everything is OK going forward in time as long as you avoid crossing the $x$-axis. Note that like that simpler problem, what I've said here does not imply you will have blowup if the method exits $\left(-1,1\right)$. In particular, you could have oscillation around $y=1$, or you could just decay to $y=1$ from above instead of below. In other words the condition we generate here is sufficient, not necessary, for stability.

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