# How to set the boundary conditions of 3D Poisson Equation

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Anthony Owusu-Mensah on 2 Mar 2020
Commented: David Martín Oliva Zavala on 12 Nov 2020 at 0:04
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);

#### 1 Comment

David Martín Oliva Zavala on 12 Nov 2020 at 0:04
Hey Anthony, have you finished solving poisson eqn in 3D? I´m just starting this project and I may need some help in the future jeje:)

Dinesh Yadav on 5 Mar 2020
Edited: Dinesh Yadav on 5 Mar 2020
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

Anthony Owusu-Mensah on 5 Mar 2020
Thanks Dinesh for reaching out however I don't understand the part that has to do with the Neumann boundary conditions. The section below
V(i,dx+1,k) = V(i,dx-1,k); % dv/dx = 0
V(dy+1,j,k) = V(dy-1,j,k); % dv/dy = 0
V(i,j,dz+1) = V(i,j,dz-1); % dv/dz = 0
Suppose I try running the code, it going to throw an exception because i,j and k have not been defined in your code. How do I compensate for that?.
Dinesh Yadav on 5 Mar 2020
Yes, replace i with dy j with dx and k with dz. Just to make it more readable. I have corrected the above code.
Anthony Owusu-Mensah on 5 Mar 2020
Thank you very much. But I guess changing the position of the x and y indices won't affect the solution because I intend on maitaining the indeces for x and y when solving for the internal nodes else per your submission I would have to also change the x and y indices for that too. Below is how I intend to slove for the internal nodes
l = 2:length(x1)-1;
m = 2:length(y1)-1;
n = 2:length(z1)-1;
for t = 1:1000
Vold = V;
%%Solve for the internal nodes
for i = 2:length(x1)-1
for j = 2:length(y1)-1
for k = 2:length(z1)-1
V(i,j,k) = 1/6*(Vold(i+1,j,k)+ Vold(i-1,j,k)+...
Vold(i,j+1,k)+ Vold(i,j-1,k)+...
Vold(i,j,k+1)+ Vold(i,j,k-1));
end
end
end
%Dirichlet Boundary Condtions
%Neumann Boundary Conditions
V(l+1,m,n) = V(l-1,m,n);
V(l,m+1,n) = V(l,m-1,n);
V(l,m,n+1) = V(l,m,n-1);
end