I asked a somewhat similar question previously but perhaps it might have been too specific for anyone to really answer. Here is a bit more general of a question that I am struggling with. Consider the following system:
$$ -\nabla\cdot(D_{1}(u_{2})\nabla u_{1}) = \nabla\cdot\mathbf{f}_{1}(u_{2}) $$ $$ \frac{\partial u_{2}}{\partial t} + \nabla\cdot\mathbf{f}_{2}(u_{1},u_{2}) - \nabla\cdot(D_{2}(u_{2})\nabla u_{2}) = 0 $$
assuming a general set of BCs: $$ u_{i} = u_{i,D}, \;\;\;\text{on}\;\Gamma_{D} $$ $$ D_{i}\nabla u_{i}\cdot\mathbf{n} = u_{i,N}, \;\;\;\text{on} \;\Gamma_{N} $$
Using DGFEM for spatial discretizations and backwards Euler for the time derivative. We decouple like this:
Solve for $u_{1}^{k+1}$ using $u_{2}^{k}$: $$ -\nabla\cdot(D_{1}(u_{2}^{k})\nabla u_{1}^{k+1}) = \nabla\cdot\mathbf{f}_{1}(u_{2}^{k}) $$
Solve for $u_{2}^{k+1}$ using $u_{1}^{k+1}$: $$ \frac{\partial u_{2}}{\partial t} + \nabla\cdot\mathbf{f}_{2}(u_{1}^{k+1},u_{2}^{k},u_{2}^{k+1}) - \nabla\cdot(D_{2}(u_{2}^{k+1})\nabla u_{2}^{k+1}) = 0 $$
Newton's method is used to handle the non-linearity.
In the particular example I'm looking at, I have $$ \mathbf{f}_{2}(u_{1},u_{2}) = u_{2}(-\nabla u_{1} + u_{2}\nabla u_{2}) $$
so
$$ \mathbf{f}_{2}(u_{1}^{k+1},u_{2}^{k},u_{2}^{k+1}) = u_{2}^{k+1}(-\nabla u_{1}^{k+1} + u_{2}^{k}\nabla u_{2}^{k}) $$
The issue is that the evaluation of $\mathbf{f}_{2}$ at $u_{1}^{k+1}$ and $u_{2}^{k}$ is really messing up my solution of $u_{2}^{k+1}$ to the point where Newton's method is not converging quadratically and I'm not observing the correct convergence of the discretization errors. I originally wanted to evaluate $\mathbf{f}_{2}$ at $u_{2}^{k}$ so that the vector part, $-\nabla u_{1}^{k+1} + u_{2}^{k}\nabla u_{2}^{k}$ could be treated as a constant at each Newton iteration. Otherwise, it would have to be added to the second order term, which I wanted to avoid doing.
To test my code, I'm using method of manufactured solutions. As a test, I'm considering the second equation by itself and evaluating $\mathbf{f}_{2} = \mathbf{f}_{2}(u_{1,\text{exact}}^{k+1}, u_{2}^{k+1}, u_{2,\text{exact}}^{k+1}) = u_{2}^{k+1}(-\nabla u_{1,\text{exact}}^{k+1} + u_{2,\text{exact}}^{k+1}\nabla u_{2,\text{exact}}^{k+1})$ using the exact solutions and there's, no problem here. Newton's method converges quadratically, I'm observing correct convergence of discretization error, etc. so I'm pretty sure the error here lies with the decoupling and not the discretization.
Is this suboptimal performance in the first scenario expected given the decoupling? I know the decoupling will limit the time step, but I'm pretty sure I'm taking the time step sufficiently small.
The next step I was going to take was to split up $\mathbf{f}_{2}$ and toss the $u_{2}\nabla u_{2}$ into the second order term. This would have the advantage of making the formulation more implicit since the only explicit part would be the use of $u_{1}^{k+1}$ in the convective term.
I hadn't been able to find much literature on what effect decoupling would have on convergence, so if anyone could point me in the right direction or offer some advice, that would be great.