For the backward Euler discretization in time: $$ \left( \frac{u^{(k)}-u^{(k-1)}}{\Delta t}, v\right) + a(u^{(k)},v) = \ell(v) $$
where $a(\cdot,\cdot)$ is the bilinear operator associated with the discretization in space. I’m considering linear elliptic problems in 2D. So I rearrange the above equation: $$ \left(\frac{1}{\Delta t}u^{(k)},v\right) + a(u^{(k)},v) = \ell(v) + \left(\frac{1}{\Delta t}u^{(k-1)},v\right) $$ since $u^{(k-1)}$ is a known quantity from the previous time step. For non-time dependent problems, the convergence in space is typically on the order of $\mathcal{O}(h^{p+1})$, with $p$ being the basis degree. Now if I have a time-dependent problem as given above, how small does my time step have to be in order to test for convergence in space? If I take a large time step, the method is stable but yields error that is large, on the order of $10^{3}$.
Taking my time step extremely small yields proper convergence rates, but I don’t think this tells me anything since for small $\Delta t$, the time-dependent terms will dominate and I’m essentially solving $$ \left(\frac{1}{\Delta t}u^{(k)},v\right) = \left(\frac{1}{\Delta t}u^{(k-1)},v\right) $$ and if the time step is small, we can most likely expect $u^{(k)} \approx u^{(k-1)}$, so solving the system will yield the “correct solution”.
I’m just wondering if my implementation error of Backwards euler is incorrect. I know typically we solve the linear system: $$ (\mathbf{M} + \Delta t\mathbf{K})u^{(k)} = \mathbf{M}u^{(k-1)} + \Delta t\mathbf{f} $$
with $\mathbf{M}$ and $\mathbf{K}$ being the mass and stiffness matrices and $\mathbf{f}$ being the source term. But is the implementation I gave above incorrect? I originally implemented the first way I listed, then tried the second for comparison but they yield the same result in both cases, so I'm guessing that's not the issue.
