The question is about the first-order exponential integration method described in this article.Consider a system of ordinary differential equations$$y'(t) = -A\,y(t) + \mathcal{N}(t, y), \qquad y(0) = y_0,$$where $\mathcal{N}$ is a nonlinear function. The solution can be approximated using the exponential (explicit) Euler method given by$$y_{n+1} = e^{-Ah}y_n + A^{-1}(1-e^{-Ah}) \mathcal{N}(t_n, \, y(t_n)).$$The question is: What can be said about the stability of the above integration scheme provided that A is a real, symmetric matrix? What if A is also positive semi-definite? I am aware that the ordinary (explicit) Euler scheme is known to be unstable, but what about its exponential analogue?Thank you!Regards,Ivan
Stability of the first-order exponential integrator method
ode;stability;time integration
It depends on your nonlinear function $\mathcal{N}.$  The stability means the solution continuously depends on the data. In particular, for discretised ODEs, it means if you give a perturbation to $y_n,$ the perturbation propagating to $y_{n+m}$ is bounded uniformly w.r.t. $m.$ In your case, we have$$|\tilde{y}_{n+1}-y_{n+1}|\leq |e^{-Ah}|\cdot|\tilde{y}_n-y_n| + (h + \frac{1}{2}|A|h^2 + ..)|\mathcal{N}(t_n,\tilde{y}_n)-\mathcal{N}(t_n,{y}_n)|.$$Assuming $\mathcal{N}$ is Lipschitz w.r.t. the second argument, you will need something like$$|e^{-Ah}| + Ch = 1 + \mathcal{O}(h)$$for asymptotic stability (stable as $h\rightarrow 0$) or if you want absolute stability$$|e^{-Ah}| + Ch<1,$$where $C$ depends on $|A|.$