It is straight-forward to prove convergence of a numerical method given consistency and stability, but when does this break down? The proof for convergence says: \begin{equation} |y^{n} - \hat{y}^{n}| \leq \left(\sum_{p=0}^{n-1} \sigma ^{p} \right) (k \tau_{k}) \end{equation} where $\hat{y}$ is the exact solution, $\sigma^{p}$ is the stability condition: \begin{equation} \sigma = \dfrac{y^{n+1}}{y^{n}} \leq 1 \end{equation} $\tau_{k}$ is the truncation error (time step k): \begin{equation} \dfrac{\hat{y}^{n} - y^{n}}{k} \end{equation} Convergence says that $n\rightarrow \infty$; $k \rightarrow 0$; $nk =t$: \begin{equation} \lim |y^{n} - \hat{y}^{n}| \rightarrow 0 \end{equation} From this proof, it looks as it $k\tau_{k}$ will always approach zero as $k \rightarrow 0$, so why do we require stability? My professor said that there is a power $q$ in the stability condition where convergence is no longer held ($\sigma \leq 1+ k^{q}$), and challenged us to figure out the value of power $q$... but I'm not seeing it! Any help would be greatly appreciated.
In class, we solved for the summation term (setting $\sigma = 1+k$) and proved that it approached zero using L'Hôpital's rule and making the substitution ($k=t/n$), but then said "it all goes to zero anyways because it's being multiplied by zero". Also, since convergence holds with $\sigma \leq (1+k)$, isn't this telling us that stability isn't necessary? I'm missing something, thanks again!

  • $\begingroup$ How do you know the stability condition doesn't approach $\infty$ faster than $k \to 0$? $\endgroup$ – mattos Nov 23 '15 at 4:54
  • $\begingroup$ We don't. Im guessing that is why this breaks down, but how do you mathematically explain this? I apologize, I'm a physicist by trade, this is definitely not my specialty! $\endgroup$ – mfordcc7 Nov 23 '15 at 5:10

Your first equation does not look like a "consistency" check. There's no approximate solution in the consistency check, only the true solution. Consistency is the check of how far is the true solution from satisfying the discrete scheme. The "consistency error" is $\tau_k$. Your first equation looks more like a proof for convergence.

Anyhow, the point is that if $\sigma>1$, then the summation in parenthesis diverges (for $n\to\infty$), and it diverges much faster than how $k\to 0$. For instance, if $\sigma=2$, you have

$$ |y^n-\hat{y}^n|\leq 2^n \frac{t}{n}\tau $$

which does not yield convergence.

  • $\begingroup$ Thanks for catching that error, I meant to say "convergence" for the first eqn. You say, though, that convergence isn't held when $\sigma$>1, however we showed in class that it is held when sigma is 1+k, the question is how much bigger than 1 can you get? $\endgroup$ – mfordcc7 Nov 23 '15 at 5:36
  • $\begingroup$ Sure, you don't need "uniform" stability. What you need, as you already guessed, is a proper balance between stability constant and consistency. In the case $\sigma=1+k$, then the sum in parentheses adds up to $((1+k)^n-1)/k$. So, multiplying by $k\tau$, and assuming $\tau\to 0$, then you still achieve convergence. This is because $k\tau\to 0$ "faster" than $k$ (but you don't have any hint on how much faster), so you can accept a stability constant growing at most linearly with $k$. $\endgroup$ – bartgol Nov 23 '15 at 15:51
  • $\begingroup$ This is pretty common in numerical analysis. When the stability constant is independent on the discretization parameters, it's great! You only need stability to converge (for linear methods). Otherwise, you need to prove that the degeneration of the stability constant (as the discretization parameter goes to 0) is not too bad, meaning that the "good" properties of the consistency can make up for it. $\endgroup$ – bartgol Nov 23 '15 at 15:54

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