Numerical Solution for the Laplace Equation

© Siamak Faridani, Oct 2006 

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Matlab Code

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Abstract:In this homework Laplace equation has been solved for a rectangle, all the values on the boundary has been set to zero except for one third of one of the sides

One third of the left side is set to 1

Laplace Equation:

02

2

2

2

y

u

x

u

Decartelization:

))(,)((0)(

2

)(

222

2

1,,1,

2

,1,,1 yxOy

UUU

x

UUU jijijijijiji

Approach:Since in the former homework I faced some difficulties with Fortran and I was afraid of using it again, I chose MATLAB to solve this problem

Inputs

Number of the nodes in x and y direction will be given to the program as inputsUser will be asked if he wants to set the boundary conditions on the middle of the line or on the cornerAnd the desired maximum error will be asked as well

N = input('Number of Nodes in Y '); M = input('Number of Nodes in X ');

% Boundary values can be set along the corner or in the middlereply = input('Corner or Middle boundary conditions C/M [C]: ', 's');

%Maximum Error abs(x2-x1)maxerr= input('Desired Maximum Error [.00001]: ');

For boundary conditions the default mode is on the corner and for the maximum error the default mode is .00001. User can change them if he wants

Matlab Code:

% Laplace Equation Solver

% Siamak Faridani

% Oct 2, 2006

clc;clear all;close all;

N = input('Number of Nodes in Y '); M = input('Number of Nodes in X ');

% Boundary values can be set along the corner or in the middle

reply = input('Corner or Middle boundary conditions C/M [C]: ', 's');

%Maximum Error abs(x2-x1)

maxerr= input('Desired Maximum Error [.00001]: ');if isempty(reply) reply = 'C';endif isempty(maxerr) maxerr = .00001;endmymatrix=zeros(M,N);reply=lower(reply);if reply=='c' for i=1:floor(M/3) mymatrix(i,1)=1;endelse for i=floor(M/3):floor(2*M/3) mymatrix(i,1)=1;endend

maxr=1;errormatrix=zeros(M,N);itteration=0;while maxr>maxerr for i=2:M-1 for j=2:N-1 tempval=mymatrix(i,j); mymatrix(i,j)=(mymatrix(i+1,j)+mymatrix(i-1,j)+mymatrix(i,j+1)+mymatrix(i,j-1))/4; errormatrix(i,j)=abs(mymatrix(i,j)-tempval); end

end maxr=max(max(errormatrix)); itteration=itteration+1;end

contourf (mymatrix, 'DisplayName', 'mymatrix'); figure(gcf)itteration

Test Cases:

A 10×10 grid with default error53 Iterations

The same problem, Elemental representation

A 100×100 grid with default error2257 Iterations

The same problem elemental representation

Surface View

Middle

A 10×10 grid with default error64 Iterations

The same problem, Elemental representation

A 100×100 grid

A 100×100 grid with default error2999 Iterations

Surface Mesh Representation

N (N×N grid) BC type Iterations 10 Corner 5320 Corner 18930 Corner 39450 Corner 866100 Corner 2267

Iteration VS Grid Size

53189

394

866

2257

0

500

1000

1500

2000

2500

0 20 40 60 80 100 120

N (NxN grid)

Iter

atio

ns