Explore the Multiple Grid Method: Get Accurate Solution Faster

[I like Serena]

Do we not have the residue problem at the step 2 if we are not on the coarsest grid, i.e. if we don't have $H=2^jh$ and so we calculate $f_{2h}$ and $v_{2h}$? (Wondering)

Ah yes. I overlooked that. (Tmi)

Turns out that $v_{2h}$ is actually a residue instead of an actual approximation on the coarser grid.

Could you explain to me the natrices $I_h^{2h}$ and $I_{2h}^h$ and especially the graphs Fig. 25 and Fig. 26? (Wondering)

Consider the Fig. 25: Do we consider three points on the fine grid and we take from there the average and the result is then one point on the coarse grid, i.e. the first box? (Wondering)

In figure 25 we want to keep only about half of the points, so that the calculations become easier.
But we don't just want to discard the information that is contained in the points.
So we pick which points we want to keep, and we take a weighted average at a ratio of 1:2:1 with the neighboring points that are discarded.
In the example of figure 25 we keep points 2 and 4 while discarding points 1, 3, and 5.
Since we don't just want to throw the information away, we replace points 2 and 4 by a weighted average that includes points 1, 3, and 5. (Thinking)

In figure 26 we go into the other direction: coarser to finer grid.
In the example we have only 2 points, but we need to get back to 5 points.
To find the missing points we interpolate linearly between the points we have.
Points 2 and 4 remain the same, point 1 is found as the average of 0 and point 2, and point 3 is found by taking the average of point 2 and 4. (Thinking)

 

[mathmari]

Thank you so much for your help! (Bow) (Sun)