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


1 Answer 1


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 $\Delta t$.

Stability ensures that numerical computations remain stable, round-off errors do not grow, and you do not get infinities.

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.

  • $\begingroup$ 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. $\endgroup$ Commented Mar 16, 2020 at 13:24
  • $\begingroup$ 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 $[0,T]$". $\endgroup$
    – cfdlab
    Commented Mar 16, 2020 at 13:54

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