Expanding Mathematica code from 2 dimensions

[brando623]

I have this mathematica code to solve a variation of the 2 Dimensional Poisson's equation using the finite difference method.

xo = .5; yo = .5; \[Sigma] = .05; q = 1;
Gen[x_, y_] := -q*E^-((((x - xo)^2) + ((y - yo)^2))/(2*\[Sigma]^2));

GridNum = 300; Hstep = N[1/(GridNum + 1)];
GridPts = Table[u[i, j], {i, GridNum}, {j, GridNum}];

FDeqn = {{0, 1, 0}, {1, -4, 1}, {0, 1, 0}}/Hstep^2;
MatFD = ListCorrelate[FDeqn, GridPts, {2, 2}, 0];
Short[HE2 =
Thread[Map[Flatten,
MatFD ==
Table[Gen[Hstep i, Hstep j], {i, GridNum}, {j, GridNum}]]], 10];

{row, col} = CoefficientArrays[HE2, Flatten[GridPts]];

sol2D = LinearSolve[col, row];
sol2D = Partition[sol2D, GridNum];
sol3D = LinearSolve[col, -row];
sol3D = Partition[sol3D, GridNum];


I would like some assistance in using this same exact method to solve the problem for the 1 dimensional and 3 dimensional case.

I tried to just delete each part dealing with one dimension (to do the one dimensional case) such that everything that had to deal with x and y just dealt with x. The Table was a 300x1 table etc. I tried to add in a dimension in the same fashion, but kept getting errors when I tried to do the ListCorrelate step saying that the kernel and the table didn't have the same tensor rank. Thank you in advance.

 

[KataKoniK]

I would first like to commend you for using the finite difference method to solve this variation of the 2 Dimensional Poisson's equation. This method is widely used in various fields of science and engineering for solving partial differential equations numerically.

To solve the 1 dimensional case, you can simply modify the code by setting the GridNum to 1 and removing the second variable in the Gen function. The GridPts will now be a 1-dimensional table and the FDeqn will be a 1x1 matrix. The rest of the code should remain the same.

For the 3 dimensional case, you will need to make a few changes. First, set the GridNum to 300 in all directions. Then, modify the Gen function to include the third variable, z. The GridPts will now be a 300x300x300 table. The FDeqn will be a 3x3 matrix with 1's along the main diagonal and -6's in the other entries. The ListCorrelate step will also need to be modified to account for the third dimension.

Additionally, when using LinearSolve, you will need to pass in a 3-dimensional matrix for the coefficient array and a 1-dimensional vector for the right-hand side of the equation. You can use the Partition function to convert the GridPts into a 1-dimensional vector for the coefficient array and the Gen function to create the right-hand side vector.

I hope this helps you to successfully solve the 1 and 3 dimensional cases using the finite difference method. Best of luck in your research!