# Convergence of numerical method without stability?

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: $$|y^{n} - \hat{y}^{n}| \leq \left(\sum_{p=0}^{n-1} \sigma ^{p} \right) (k \tau_{k})$$ where $\hat{y}$ is the exact solution, $\sigma^{p}$ is the stability condition: $$\sigma = \dfrac{y^{n+1}}{y^{n}} \leq 1$$ $\tau_{k}$ is the truncation error (time step k): $$\dfrac{\hat{y}^{n} - y^{n}}{k}$$ Convergence says that $n\rightarrow \infty$; $k \rightarrow 0$; $nk =t$: $$\lim |y^{n} - \hat{y}^{n}| \rightarrow 0$$ 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!

• How do you know the stability condition doesn't approach $\infty$ faster than $k \to 0$? – mattos Nov 23 '15 at 4:54
• 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! – 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
$$|y^n-\hat{y}^n|\leq 2^n \frac{t}{n}\tau$$
• 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? – mfordcc7 Nov 23 '15 at 5:36
• 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 $k$. – bartgol Nov 23 '15 at 15:51