2
$\begingroup$

I've been trying to figure out how to prove that the following equation is conservative

$$\int_{D^k} \partial_t(u^k) v_j^k + \partial_x(f) v_j^k dx = [(f -f^*) v^k_j]_{x^k}^{x^{k+1}}$$

where $v_j^k$ is a test function for element $k$, $f$ is the flux and $f^*$ is the numerical flux.

Can anyone point me out to some reference?

$\endgroup$
1
  • $\begingroup$ I don't think this would be regarded as conservative unless the right hand side is zero. $\endgroup$
    – Bill Barth
    Jan 11 '15 at 21:09
1
$\begingroup$

What you've written is considered the "strong form" of DG. If you integrate by parts the spatial term and rearrange you get $$ \int_{D^k} \left(\partial_t(u^k)v^k - f \partial_x(v^k)\right) dx = -[f^* v^k]_{x^k}^{x^{k+1}} = 0. $$ Taking $v = 1$ then gives $$ \int_{D^k} \partial_t(u^k) dx = f^*(x^{k})-f^*(x^{k+1}). $$ This is a restatement of conservation - that change in the amount of $u^k$ over the element $D^k$ is equal to flux in minus flux out.

$\endgroup$
4
  • $\begingroup$ I agree with that result, however, it seems to me a little bit weird because from the original formulation (not shown in the question) I integrated by parts once to get the weak form, then again integrated by parts to get the strong form and now apply that same trick to show it is conservative. Perhaps I am missing something subtle in this manipulation. $\endgroup$
    – BRabbit27
    Jan 11 '15 at 21:42
  • $\begingroup$ You can just plug in $v^k = 1$ and then integrate by parts to get the same result as well. The "strong form" of DG is really unnecessary mathematically (hence we undo it to show conservation), but is convenient computationally. $\endgroup$
    – Jesse Chan
    Jan 11 '15 at 22:04
  • 1
    $\begingroup$ @JesseChan Why is it convenient computationally? $\endgroup$ May 15 '18 at 15:31
  • $\begingroup$ It can expose some structure in matrices. The matrices involved in the strong form (differentiation and lift matrices) tend to have more readily available structure than weak form matrices. $\endgroup$
    – Jesse Chan
    May 24 '18 at 2:11
1
$\begingroup$

The strong and weak form of the DG approximation are equivalent approximations (full details and analysis are given in this paper). As was pointed out by JLC the conservation of the DG approximation can be explicitly seen from the weak form and then, by equivalence, conservation of the strong form follows. The one-dimensional case is relatively straightforward, but in higher dimensions one must be concerned with the mapping of the approximation.

$\endgroup$

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.