# Forward Euler Adaptive Step Size Stability

Given

with a generalization using adaptive times-stepping as

then is it still reasonable to assume that to ensure stability of the Euler’s forward method we need the growth factor for all n to be obey

• While I know where the question originates, the way you ask it assumes that someone knows what $y_n$, $y_0$, and $\lambda$ are. You might do well to add some more context. In fact, if you do that, you might find that you already have the answer :-) Mar 18 at 19:50
• See your answer here: scicomp.stackexchange.com/a/33604/28872 Mar 18 at 23:59
• So this is my first post on this site and I'm learning the ropes, Giving an answer / comment without a score would have worked just fine for me. I'll put more info next time. Notwithstanding, thank you for the information. Mar 19 at 1:55

I assume you solve the linear ODE $$y'=\lambda y$$, with $$\lambda \in \mathbb{C},~\mathrm{Re}(\lambda)<0$$.
Otherwise if $$\mathrm{Re}(\lambda)>0$$, the true solution diverges and, if $$\mathrm{Re}(\lambda)=0$$, forward Euler cannot be stable as the eigenvalue $$\lambda$$ will always be outside of its stability domain.
If you have taken a time step such that the stability criterion is not fulfilled, then you locally let your solution diverge. You can still recover a stable solution (although not an accurate one) by lowering $$\Delta t$$ such that the stability criterion is fulfilled for the next steps. This will stabilise the solution. For nonlinear problems it may be more difficult to assess.
In the case of adaptative time stepping based on an error estimate, this estimate will generally blow up if the eigenvalues of your problem lie outside of the stability domain of your temporal integrator. Thus the numerical instability will be corrected by choosing a lower $$\Delta t$$.