# General stability of implicit methods

given an initial value problem for an ordinary differential equation (ODE), we consider a method A-stable, if the approximation converges to $0$ for the test equation

$u(t)=e^{-\lambda t},\space\space\space for\space \lambda>0$

for any positive step size $\Delta t$.

But what if we consider same test equation with a positive exponent:

$u(t)=e^{\lambda t},\space\space\space for\space \lambda>0$.

According to my calculation the implicit Euler is now not unconditionally stable anymore for any $\Delta t$ while the explicit Euler is.

Does that imply that the explicit Euler should be preferred over the implicit one for any positive lambda?

Why would we not define A-Stability with a positive lambda then? How would I choose the method then if my solution is periodic (e.g. Lotka Volterra Equations), which means we have increasing and decreasing behavior?

Does that also mean that implicit methods are generally not unconditionally stable?

I am very confused and greatly appreciate any help.

Thank you!

• If $\lambda > 0$, the solution should not approach $0$ as $t \to +\infty$. Commented Apr 7, 2017 at 16:53
• That is correct but only the explicit Euler unconditionally exhibits this behavior for any step size $\Delta t$ according to my calculations if $\lambda>0$. Commented Apr 7, 2017 at 17:01

You can take for instance a look at the standard book by Butcher. Therein, stability is motivated by the demand that a numerical scheme should return a bounded approximation $$u_n$$ whenever the analytical solution $$u$$ is bounded, even if the number of timesteps $$n$$ goes arbitrarily large.
The analytical solution of the test equation $$u'(t) = \lambda u(t)$$ is given by $$u(t) = e^{\lambda t}$$ which is unbounded for $$\text{Re}(\lambda > 0), \: t \to \infty$$, so the whole notion of stability is simply not applicable.
I am not sure what you are trying to say about the explicit Euler method being stable, but the stability polynomial for the explicit Euler method is given by $$P(z) = 1 + z, \: z = \Delta t \lambda$$ and thus the method is only stable on a disk in the left half of the complex plane.
• $(1+λΔt)^n$ diverges less fast than $e^{λ(nΔt)}$, while $(1-λΔt)^{-n}$ has an almost immediate singularity for $λΔt\to 1$, in general if will diverge faster than the exact solution. As you said, in practical situations the exact as well as the numerical solution will rapidly leave any region-of-interest, how fast this happens is less interesting than that it happens at all. Commented Dec 1, 2022 at 11:06