Implicit time stepping discretization

[jlperla]

This is a continuation of #31

The trickiest part of this is figuring out what to do at the $t=0$ and $t=T$ steps. On the right hand side, we want this to nest solving the stationary distribution. On the left hand side, we need to remember that there is no initial condition.

[stevenzhangdx]

I typed the algebra for implicit time stepping. A couple of time to point out: For the right end, the backward time difference generally allows us to formulate the boundary in the same way as what we do for interiors. When the steady state condition is applied, the right end is similar with the one we had with explicit time stepping. For the left end, since we do not have "ghost" point $t_0$ and corresponding $vi^0$ for all $i$, my guess is that we can assume $h{1, -} = h{1, +}$, which is similar with the logic we used to deal with $\Delta{1, -} = \Delta_{1, +}$ and also assume $v_i^1 - v_i^0 = v_i^2 - v_i^1$. Then the matrix $A$ can be formulated. Maybe my guess is wrong. Please take a look at it.

[jlperla]

So basically, this assumes that $\frac{\partial v(t_0, x}{\partial t} = \frac{\partial v(t_1, x}{\partial t}$

I am not sure if that is an acceptable assumption? Is there any example you can find that shows this is the right approach? Take a look at http://www.princeton.edu/~moll/HACTproject.htm and see if you can find precedent.

[jlperla]

One other consideration here: there is a chance that the implicit method here is simply not appropriate for this type of PDE. Typically it is the preferred method for stability, but those are for PDEs with initial conditions.

If we can't find other examples of code which solves using an implicit time scheme, and you can't figure out how to do it without assuming an additional condition, then we should rethink whether it is possible.

[stevenzhangdx]

I could not find example to show that the approach is correct. I tried to search in the PDE literature and I think this implicit stepping time method may require initial conditions to deterministically solve PDEs.

[jlperla]

OK, the evidence is building that this is simply not the right approach in this case. I am going to leave this issue open for a little final research in the new year (or if the current method seems unstable). But lets leave it for now, and I will comment out the latex code.