# Relation between zero stability and absolute stability of numerical methods for solving ODEs

I'm confused by the relationship between the notions of zero stability and absolute stability in numerical analysis. Consider the leapfrog method given by:

$$y_{k+2} =y_k + 2hf(t_{k+1},y_{k+1})$$

It is known that this method has stability region $$S=\{z \in \mathbb{C} | Re(z)=0, |Im(z)|<1\}$$, where $$z=\lambda h$$. Therefore no step size $$h>0$$, however small, will cause the numerical process to remain bounded when applied to the test problem $$y' = \lambda y$$, meaning:

$$max_{k\geq 0}|y_k - \tilde{y_k}|= \infty$$ for any $$h>0$$

However, the characteristic polynomial associated with this method satisfies the root condition, and therefore it must be zero stable, meaning there exist strictly positive constants $$h^*, K$$ such that:

$$max_{k\geq 0}|y_k - \tilde{y_k}| for $$0

Where $$w_k$$ are input errors at step $$k$$.

Given these definitions, it seems impossible for a method to be zero stable but not to remain bounded for any $$h>0$$. What am I missing?

• Are you sure it is supposed to be $y_{\color{red}{k+2}}$? Commented Apr 10, 2021 at 16:31
• Yes I made an error thank you for pointing it out. Commented Apr 10, 2021 at 16:32
• @mattos : This is the central Euler method or 2-step Nystrom method, with the cited weak stability. Commented Apr 10, 2021 at 16:33
• @LutzLehmann Yes, there was a factor of $2$ missing which made it confusing. Commented Apr 10, 2021 at 16:34
• Can you first see if page 9 of this review I wrote 13 years ago has any helpful hints? It's almost 2:30am here in Waterloo, Ontario and I can't write a full answer. Commented Apr 21, 2021 at 6:19

One thing you are missing in your post, is that the leapfrog method is not unstable when $$\lambda$$ is purely imaginary, which can be the case in many real-world applications, such as hyperbolic problems in fluid dynamics or electromagnetism.
The next thing is that the region you've given is not for stability but for A-stability. The leapfrog method actually can be stable in the case where $$h$$ is constant, so even when $$\textrm{Re}(\lambda) \ne 0$$ it still can converge, it just doesn't satisfy the criteria for absolute stability (it also says this on pg 53 of this PDF, and the difference between "stability" and "A-stability" can be seen on pg 36 of the same PDF).