Consider the PDE $$\partial_t u + a\partial_x u=0, -\infty<x<\infty, t>0$$ $$u(x,0)=u_0(x),$$

where $a>0$ is a constant.

I want to show that the difference scheme

$$\frac{U_j^{n+1}-U_j^n}{\Delta t} + a\left( \frac{U_{j+1}^n-U_{j-1}^n}{2\Delta x} \right)=0$$

is not stable. Unfortunately, I did not completely understand what exactly I need to show in order to conclude that this scheme is not stable. I think the stability is related to convergence. From my notes and the textbook, however, it is not clear how to show stability. There is a lot of overloaded notation and I can't find my way into understanding how this works. I'm more of a pure math type person, and computational math sometimes seems quite vague for me. Can someone please help me?

• Regardless of the spatial differencing, you are doing simple forward differencing in time. This is bad, for reasons that can be more readily understood in the ODE setting (where this difference scheme is called the forward Euler method or just Euler's method). I suggest that you look up a proof that forward Euler is not stable and try to adapt it. – Ian Apr 1 '17 at 2:55
• @Ian Unfortunately I don't see how the Forward Euler method for ODEs relates to the PDE above. – sequence Apr 1 '17 at 4:44
• They are in fact one and the same: for fixed $\Delta x$, you really have the (infinite) ODE system $\frac{du_j}{dt}+a \frac{u_{j+1}-u_{j-1}}{2\Delta x}=0$. This scheme (called the forward in time, centered in space scheme, or FTCS) then attempts to solve this system by forward Euler. It then suffers from basically the same stability issues as forward Euler. But as Giovanni said, the best way to prove that is von Neumann analysis. – Ian Apr 1 '17 at 10:38

The standard machinery to prove stability is the so called Von Neumann analysis. Let $\Delta x = h$ and let $$U_j^n = \frac{1}{\sqrt{2\pi}}\int^{\pi/h}_{-\pi/h}e^{-ijh \xi}\widehat{U^n}(\xi)\,d\xi. \tag 1$$ Write the scheme in update form and use $(1)$ to find the relation $$\widehat{U^n}(\xi) = (g(h\xi,\Delta t,\Delta x))^n\widehat{U^0}(\xi).$$Then one can prove the following result.
A one-step method is stable in the region $\Lambda$ if and only if $$|g(\theta,\Delta t,\Delta x)| \le 1 + C\Delta t\quad \forall(\Delta t,\Delta x) \in \Lambda,$$ where $\theta = h\xi$.
If you carry on the computation for FTCS you'll get $$g(\theta) = 1 - ia\frac{\Delta t}{\Delta x}\sin\theta.$$ As you can easily verify, $|g(\theta)| > 1$ for $\theta \neq 0$.
• Thanks. Unfortunately, i didn't understand much from your answer, but your Wikipedia link was helpful. Can you please clarify why the error $\epsilon(x,t)$ is the exact solution to the numerical approximation? They say that it is because the exact solution satisfies the discretized method exactly, the error must also satisfy it. But to me this is far from being obvious reasoning. Anyway, if I'm correct, then what I need to do is to substitute for the error in the numerical scheme and then derive an expression and show that it does not converge? – sequence Apr 1 '17 at 5:35
• There are two types of errors here. One is called the local truncation error which is the difference between the exact solution of the difference equation (numerical scheme implemented in infinite precision arithmetic) and the analytical solution of the pde. Another is round-off error (what is called $\epsilon$ here) which is the difference between the exact solution of the difference equation, which is the same as the solution of the numerical scheme in infinite precision and the numerical solution of the scheme in finite precision arithmetic. – rivendell Jul 27 '17 at 16:15