Find cfl condition

We have the advection equation $u_t+a u_x=0, a>0, 0<t<T_f, x \in \mathbb{R}$ with initial condition $u(0,x)=u_0(x)$.

Suppose that we have the following sheme:

I want to find the CFL condition using the domain of dependence of the finite difference method.

To calculate $U_j^{n+1}$ we need the values $U_{j-1}^n$ and $U_{j+1}^n$.

So we get the following domain of dependence:

Is this correct?

The slope of the left line is $\frac{\tau}{h}$ and the slope of the right line is $-\frac{\tau}{h}$, right?

To find the CFL condition do we require that the left slope is smaller than the right slope?

Or have I understood it wrong?

• What do you mean by domain of dependence? Dec 13 '15 at 20:24
• It is described here: pastebin.com/5F1gAR17 at page 74 but I haven't understood it. Dec 13 '15 at 20:31
• @nluigi I edited my post... Dec 13 '15 at 22:11

Note that the analytic solution of $u_t+au_x=0$ is $u(t,x)=u_0(x-at)$. So the slope of the actual dependence is $a$. The CFL condition just says that the numerical dependence must include the actual dependence, which means $-\frac{h}{\tau}\leq a \leq \frac{h}{\tau}$ in your case.
• @ThomasKlimpel I think $-\frac{h}{\tau}\leq a \leq \frac{h}{\tau}$ this is valid for cfl number less than or equal to one. Since he didn't provide the value of $\nu$, its not possible to find the cfl number, so cfl may not be always equal to one, at that time numerical scheme stability may change and numerical area of influence may change. Am i right? I'm assuming $h$ is space step ans $\tau$ is time step and $cfl =c*\tau/h$
• @ArunGovindNeelanA Yes, $-\frac{h}{\tau}\leq a\leq\frac{h}{\tau}$ expresses the condition that the cfl number is less than or equal to one. I assumed $\nu=\frac{t}{h}$, but this doesn't enter anywhere, because the scheme is not consistent for $t\to 0$ anyway. The stability will certainly change with the cfl number, but the formula for the numerical area of influence is not affected by this. Dec 15 '15 at 8:44