# Showing stability of numerical scheme: Decaying norm implies stability?

I have a little trouble formulating my question since I am not really sure what conclusions I am supposed to draw from the results I have obtained. I am sorry for the long problem formulation below, but I will keep it as short as possible. I was given a general scheme,

$$U^{n+1} = Q(t_{n})U^{n} + \Delta t F^{n},\\[3mm] U^{0} = g,$$

where $$Q$$ represents some discrete differential operator (I think) and where $$F$$ is an inhomogeneous term and $$g$$ is the initial data and $$t_{n} = n\Delta t$$. Then by the discrete form of Duhamel's Principle it holds that the solution to this scheme at any time level is given by

$$U^{n} = S_{\Delta t}(t_{n},0)g + \Delta t \sum_{\nu = 0}^{n-1}S_{\Delta t}(t_{n},t_{\nu + 1})F^{\nu},$$

where $$S_{\Delta t}(t,\tau)$$ is the solution operator (also called transition matrix in system theory) with the following properties:

$$S_{\Delta t}(t,t) = I,\quad t\in \mathbb{R}, \\[3mm] S_{\Delta t}(t_{n+1},t_{\mu}) = Q(t_{n})S_{\Delta t}(t_{n},t_{\mu}).$$

Let us agree that all this holds (it is not hard to prove Duhamel's Principle in this case) and let us assume that

$$\vert\vert S_{\Delta t}(t_{\kappa},t_{\nu}) \vert\vert \leq K e^{\alpha(t_{\kappa}-t_{\nu})},$$

where $$\vert\vert \cdot \vert\vert_{h}$$ is some discrete norm, then it can be shown that

$$\vert\vert U^{n} \vert\vert_{h} \leq K\left(e^{\alpha t_{n}}\vert\vert g \vert\vert_{h} + \int_{0}^{t_{n}}e^{\alpha(t_{n}-s)}ds\; \max_{0\leq \nu \leq n-1}\vert\vert F^{\nu}\vert\vert_{h}\right).$$

Here they presume that $$\alpha$$ does not depend on $$\Delta t$$ or $$t_{\kappa}-t_{\nu}$$ (but what would change if it did?).

Now, I interpret this result as a proof that the numerical general scheme above is stable since the upperbound decays as $$\Delta t$$ becomes smaller (because then $$t_{n} = n\Delta t$$ becomes smaller and the integral goes towards 0 and what remains in the limit is the initial data times some factor $$K$$). But what I do not understand is how to interpret the norm of the solution $$U^{n}$$: What does this measure exactly? Is it the error?

Furthermore, as I wrote in paratheses above, what does it matter if the $$\alpha$$ depends on e.g. $$\Delta t$$? Why do we have to point that out in this case?

All discussion is welcome and appreciated! \Best regards

If $$\alpha$$ depends on $$\Delta t$$ then it might blow up as $$\Delta t \to 0$$. Hence it is useful to put the requirement that it is independent of $$\Delta t$$.
Finally, you want to show convergence, for which both consistency and stability are required. The exact solution satisfies $$u^{n+1} = Q(t_n) u^n + \Delta t F^n + \Delta t \tau^n$$ where $$\tau^n$$ is local truncation error which is small by consistency, that is $$\tau^n \to 0$$ as $$\Delta t \to 0$$. Then the error $$e = u - U$$ satisfies $$e^{n+1} = Q(t_n) e^n + \Delta t \tau^n, \qquad e^0 = 0$$ Because you have stability with $$\alpha$$ independent of $$\Delta t$$, you can conclude that $$\|e^n\|_h \le K \int_0^{t_n} e^{\alpha(t_n-s)} ds \max_{0 \le \nu \le n-1}\| \tau^\nu \|_h \to 0, \qquad\textrm{as}\qquad \Delta t \to 0$$ with $$n\Delta t = t_n$$, i.e, for some fixed time.
If you allow $$\alpha$$ to depend on $$\Delta t$$, then you have to do extra work to somehow prove that the exponential term containing $$\alpha$$ is bounded, so that the error can be shown to go to zero.
• Thank you! That clears some things up for me. I had some problem understanding why this dependence was important since $t_{n}$ already is dependent on $\Delta t$, but it seems $t_{n}$ is actually just a constant. Mar 16 '20 at 13:24
• Yes, we assume some fixed time interval in this definition. The notation is not very nice, it should be stated as "for some time interval $[0,T]$". Mar 16 '20 at 13:54