# How can I solve this PDE system by discontinuous Galerkin method?

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?

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.
• I don't see why $v$ can't depend on time. I think it would depend on the boundary conditions for $v$. Jan 19, 2023 at 21:22
• @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? Jan 26, 2023 at 3:32