0
$\begingroup$

I want to solve the advection equation: $u_t+au_x=0, a > 0$. Here is our method:

$$U_j^{n+1}=U_{j-1}^n-\left(\frac{ak}{h}-1\right)(U_{j-1}^n - U_{j-2}^n)$$

I am trying to answer the following question: "Say we fix $h$, then for which values of $k$ is this method exact? Note that there's more than one value."

I know that the following is exact: $U_j^{n+1}=U_{j-1}^n$, which corresponds to $k=\frac{h}{a}$. But what is the other value of $k$?

Thanks.

$\endgroup$
2
  • $\begingroup$ What, precisely, do you mean by "exact"? Do you have any requirements on the order of accuracy of your finite difference scheme? $\endgroup$
    – Charles
    Jun 25, 2016 at 8:10
  • $\begingroup$ @Charlie Based on his example exact solution, I'd guess that he's interested in a solution where, given exact arithmetic, the analytical solution is reproduced. I'm not aware of any besides the "magic time step" in the first-order scheme. $\endgroup$ Jun 26, 2016 at 16:00

1 Answer 1

1
$\begingroup$

Let $c=a k/h$ denote so called Courant number. Your scheme is reasonable only for $c \in [1,2]$. It is exact for $c=1$, as you mentioned, and for $c=2$ in the sense that if the values at time level $n$ are exact, they will be exact also at time level $n+1$. It is unusual to use a scheme that is inprecise (and very likely unstable) for small time step $k$, i. e. when $c \in (0,1)$, you may include standard first order upwind scheme for this case. There exist schemes that solve your advection equation for arbitrary time step $k$ and even for variable $a$, e. g. modified method of characteristics.

$\endgroup$
1
  • $\begingroup$ Thanks! I actually reached the same conclusion over the weekend, and your answer agrees with what I was thinking. $\endgroup$
    – Kurt
    Jun 27, 2016 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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