# Explore the Multiple Grid Method: Get Accurate Solution Faster

• MHB
• mathmari
In summary, the conversation is about the multiple grid method and its applications in solving linear systems of differential equations. The method uses a combination of fine and coarse grids to achieve both accuracy and fast convergence. The conversation also discusses the use of the finite difference method to solve the linear system and the application of Fourier analysis in approximating the solution. The concept of spectrum is also mentioned, which refers to the decomposition of a function into its frequencies.
mathmari said:
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.

mathmari said:
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)

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

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