0
$\begingroup$

Suppose that we are given a numerical scheme.

In order to find the CFL condition , we set $U_j^n= \lambda ^ne^{ik x_j}$ and put it into the numerical scheme.

I have shown that the given method is unstable since $|\lambda|>1$.

In this case, how do we calculate the CFL condition?

I get $| \lambda |= \sqrt{1+\gamma^2 \nu^2 \sin^2{(kh)}}$. This can't be $<1$. What do we do in this case?

The given equation is $u_t+ \gamma u_x=0, \gamma>0$. We know that $k \in \mathbb{R}, \nu=\frac{\tau}{h}$ where $\tau$ is the step of the discretization of time.

enter image description here

EDIT: Also suppose that we have a numerical scheme , put into it $U_j^n= \lambda^n e^{ik x_j}$ and get $a_i^n \nu \leq 1$. Can we then say that the method is stable if and only if $a_i^n \nu \leq 1$ ?

Improve this question
$\endgroup$
16
  • 1
    $\begingroup$ choose |\lambda|<1? $\endgroup$
    – Michael Medvinsky
    Dec 1 '15 at 20:49
  • 1
    $\begingroup$ I get $| \lambda |= \sqrt{1+\gamma^2 \nu^2 \sin^2{(kh)}}$. This can't be <1. What do we do in this case? $$$$ The given equation is $u_t+ \gamma u_x=0, \gamma>0, x \in \mathbb{R}, t \in [0,T_f]$. @MichaelMedvinsky $\endgroup$
    – Mary Star
    Dec 1 '15 at 20:53
  • 3
    $\begingroup$ Add those details to your question $\endgroup$
    – nicoguaro ♦
    Dec 1 '15 at 20:56
  • $\begingroup$ also add what is $\nu$ and $k$ $\endgroup$
    – Michael Medvinsky
    Dec 1 '15 at 21:06
  • 1
    $\begingroup$ exactly! The central differencing scheme you use is known to be unstable. You may want to try upwind differencing: $$\frac{u_i^{n+1}-u_i^n}{\Delta t}=\gamma\frac{u_{i+1}^n-u_i^n}{\Delta x}$$ or some form of implicit methods. $\endgroup$
    – nluigi
    Dec 1 '15 at 21:57
3
$\begingroup$

You assert that for your scheme to be stable you need $|λ|≤1$ which is correct, however you find that all values for $c=γν$ will give $|λ|>1$. This means that the discretization you have applied is unstable and there is no CFL condition for stability. Consider using upwind differencing or an implicit method.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.