MATLAB: How to set the boundary conditions of 3D Poisson Equation

3d finite difference methodboundary conditions of 3d poisson equationmolecular dynamics simulation

I am trying to compute the electric potential at point (x,y,z) by solving the 3D Poisson equation below using finite difference method.
I have the charge densities at various positions of (x,y,z).
Below are the boundary condtions;
  1. Dirichlet boundary condition are applied at the top and bottom of the planes of the rectangular grid.
  2. Electric potential is to be incorporated by setting and , where h is the height of the simulation box.
  3. Neumann boundary conditions are also enforced at the remaining box interfaces by setting at faces with constant x , at faces with constant y, and at faces with constant z.
I did the implementation of the boundary conditions with this code and I would want to ascertain if
the implementation of the boundary condtions is right.
x1 = linspace(0,10,20);
y1 = linspace(0,10,20);
z1 = linspace(0,10,20);
V = zeros(length(x1),length(y1),length(z1)); 
%Dirichlet Boundary Conditions
%Top plane
V(:,end,end) = 0;
V(end,:,end) = 0;
V(end,end,:) = 0;
% Bottom plane
V(:,1,1) = 0;
V(1,:,1) = 0;
V(1,1,:) = 0;
%Incoporated electric potential
V(:,:,1)  = 0; 
V(:,:,end) = -40*z1(end);
%Neumann Boundary Condition
i  = 2:length(x1)-1;
j = 2:length(y1)-1;
k = 2:length(z1)-1;
V(i+1,j,k) = V(i-1,j,k);
V(i,j+1,k) = V(i,j-1,k);
V(i,j,k+1) = V(i,j,k-1);

Best Answer

Hi Anthony I think you have wrongly defined the top and bottom planes.A plane is a 2-D sheet structure however the line of code 
V(:,end,end);
represent a single row vector, similar to a line and not a plane. As V is a 20x20x20 3-D matrix, lets assume the first plane facing us is top plane i.e (assuming x direction increases no of columns and y direction increases no of rows, i.e origin is at (20,1,1))
V(:,:,1);    % top plane        face with constant z
V(:,:,end);   % bottom plane    face with constant z
V(:,1,:);    % face with constant x

V(:,end,:)   % face with constant x
V(1,:,:);    % face with constant y

V(end,:,:)   % face with constant y
dx  = 2:length(x1)-1;
dy = 2:length(y1)-1;
dz = 2:length(z1)-1;
V(dy,dx+1,dz) = V(dy,dx-1,dz);   % dv/dx = 0
V(dy+1,dx,dz) = V(dy-1,dx,dz);   % dv/dy = 0
V(dy,dx,dz+1) = V(dy,dx,dz-1);   % dv/dz = 0
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