Specific questions for 2-D Multigrid

Question

I am simulating $\nabla^{2}u=0$ with mixed Dirichlet-Neumann boundary conditions on 2-D using 2-Grid method. Dirichlet conditions are applied to left and bottom boundary with $u=1+x+y$ and Neuman conditions are applied to top and right boundary with $\frac{\partial u}{\partial n}=1$. The solution converges for 4x4 grid without using 2-grid. With 2-grid, please let me know if the following are correct (as I am unable to converge with 2-grid).

1. I do 2 iterations on the fine grid (using Pure Jacobi), calculate residual, use injection to restrict to Coarse grid (2x2). Is it okay to use Pure Jacobi or do we always use Damped (weighted) Jacobi ?

2. With error matrix $e=0$ as initial guess, I solve the residual equation $Ae =r$ till full convergence, then interpolate/prolongate the error back to fine grid. Should the residual equation be solved till convergence or would 2-3 iterations be sufficient ? ( The error equation does converge fully.)

3. Where exactly do I have to use the norm of an error/residual/whatever ? Since I know the true solution on the Fine Grid is $u = 1 + i*h + j*h$ where h=1/4for i,j vary from 0 to4, I am using it as convergence criterion on the fine grid i.e. till $error = u_{new} - (1 + i*h + j*h)$ is greater than the tolerance of $10^{-5}$ keep iterating.

Answer 1

Jacobi is usually not a great smoother, Gauss-Seidel is a better choice.

You might also want to use a larger initial grid so you see whether the defect is being generated at the boundary or in the interior. Your 2x2 coarse grid is essentially all boundary condition.

Mixed boundaries are tricky to implement. I'd get only Dirichlet boundaries working, and then only Neumann working, and then try to mix them. You have to think carefully about how to handle restriction at the corners.

This discussion might also be helpful: Multigrid stops converging when more grid levels are used

Answer 2

I used weighted Jacobi (doesn't work with pure Jacobi). In this problem the Neumann Boundary condition becomes $\frac{\partial e}{\partial n} = 0$ for the error matrix i.e. while solving $Ae = r$ if we take the central difference at the boundary of the coarse grid then $\frac{e_{i+1,j} - e_{i-1,j}}{2h_{c}} = 0$, where $e_{i,j}$ is the point on the Neumann boundary. Similarly $\frac{e_{i,j+1} - e_{i,j-1}}{2h_{c}} = 0$, where $e_{i,j}$ is a boundary point (and not the fictitious boundary point, of course !) on the coarse grid.