As is known to all, the discontinuous Galerkin method (DG) was first used to solve the equation $u_t+u_x=0$. Now I have the following system of PDEs:
$$
\begin{cases}
u_t=f(u,v,\nabla{v}),\\
\Delta{v}=0.
\end{cases}$$
I want to approach the equation $u_t=f(u,v,\nabla{v})$, where $f$ doesn't depend on $\nabla{u}$, by DG while for the equation $\Delta{v}=0$ I'd chose any FEM.
Question. Since $u_t=f(u,v,\nabla{v})$ isn’t the standard form of $u_t+u_x=0$, how can I implement the DG scheme?
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1 Answer
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If $f$ does not depend on $\nabla u$, then the equation is purely local: For every $\mathbf x$, you have the ODE $$ \frac{d}{dt}u(\mathbf x,t) = f(u(\mathbf x,t), v(\mathbf x), \nabla v(\mathbf x)). $$ Furthermore, because $v$ does not depend on time, it is really just a constant-in-time coefficient in the equation, and so you end up with an equation that in reality is just $$ \frac{d}{dt}u(\mathbf x,t) = g(u(\mathbf x)), $$ where $g(u(\mathbf x)) = f(u(\mathbf x,t), v(\mathbf x), \nabla v(\mathbf x))$.
If you want to represent $u$ as a finite element field, you could just solve this ODE in time for each node point one after the other. You can of course choose a discontinuous finite element field for $u$, and in that case you would just solve the ODE above for the node points of that discontinuous element.
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$\begingroup$ I don't see why $v$ can't depend on time. I think it would depend on the boundary conditions for $v$. $\endgroup$ Jan 19 at 21:22
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$\begingroup$ @wolfgang many thanks Prof Bangerth. If this u ODE is purely local, can we conclude that DG is more accurate/converges faster than CG, because DG has more and independent DOFs across elements? $\endgroup$– feynmanJan 20 at 9:04
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1$\begingroup$ CG and DG differ in accuracy by some factor, but they generally have the same convergence order. DG may be slightly more accurate. $\endgroup$ Jan 20 at 18:11
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$\begingroup$ @WolfgangBangerth I wonder for such a local equation $u_t=f(u,v,\nabla{v})$ with an IC of u(x) and v(x), is there any reference showing that how DG is better than CG? I guess DG is esp superior when the IC has discontinuities in x? $\endgroup$– feynmanJan 26 at 3:32