Poisson Equation on a Square Domain, Dirichlet Boundary Conditions

Introduction to the problem

Consider the boundary-value problem
$\frac{{\partial}^{2}u}{{\partial}x^{2}}+\frac{{\partial}^{2}u}{{\partial}y^{2}}=x+y$
Boundary condition:
$u(0,y)=0\\u(1,y)=1\\u(x,0)=0 \\u(x,1)=1$

$j \rightarrow$ gird in x direction
$k \rightarrow$ gird in y direction
$h \rightarrow$ length if the grid
$\left.\frac{{\partial}^{2}u}{{\partial}x^{2}}\right|_{x_{j},y_{k}} = \frac{u_{j-1,k}-2u_{j,k}+u_{j+1,k}}{h^{2}}$

$\left.\frac{{\partial}^{2}u}{{\partial}y^{2}}\right|_{x_{j},y_{k}} = \frac{u_{j,k-1}-2u_{j,k}+u_{j,k+1}}{h^{2}}$
$-u_{j-1,k}-u_{j+1,k}-u_{j,l-1}-u_{j,k+1}+4u_{j,k}=-h^{2}f_{j,k}$
These equations can be arranged in the form
$Au = b$
Where, for $N_{x}$ intervals in $x$ and $N_{y}$, $A$ is an $(N_{x}-1)(N_{y}-1)$-by-$(N_{x}-1)(N_{y}-1)$ matrix of the form

$A = \begin{bmatrix} D & -I & & & & & \\ -I & D & -I & & & & \\ & & . & & & & \\ & & & . & & & \\ & & & & . & & \\ & & & -I & D & -I \\ & & & & -I & D \\ \end{bmatrix}$
where $D$ is in $(N_{y}-1)$-by-$(N_{y}-1)$ matrix of the form
$D= \begin{bmatrix} 4 & -1 & & & & & \\ -1 & 4 & -1 & & & & \\ & & . & & & & \\ & & & . & & & \\ & & & & . & & \\ & & & -1 & 4 & -1 \\ & & & & -1 & 4 \\ \end{bmatrix}$
This is solved using the finite difference method with $N=10*10$ subintervals

Code

%實在沒找到matlab代碼的插入方法，就插入了python的代碼塊，其實用的還是matlab，代碼複製粘貼後換行有點問題
%應該不能直接運行，把回車刪一刪應該就能用
Nx    =                        10; % Number of sub-segments in x

Ny    =                        10; % Number of sub-segments in y

a     =                         0; % Location of boundary 'a' for 'x'

b     =                         1; % Location of boundary 'b' for 'x'

c     =                         0; % Location of boundary 'c' for 'y'

d     =                         1; % Location of boundary 'd' for 'y'

f     = @(x,y)              x + y; % Defining RH-side function 'f(x,y)'

 

function [gxy] = g(x,y,a,b,c,d)    % Defining boundary condition 'g(x,y)'

 

 gxy = 0;   % Default value

 

 if (x==a)  % g(a,y), Left BC

  gxy = 0;

 end

 

 if (x==b)  % g(b,y), Right BC

  gxy = 1;

 end

 

 if (y==c)  % g(x,y), Bottom BC

  gxy = 0;

 end

 

 if (y==d)  % g(x,y), Top BC

  gxy = 1;

 end

 

end

 

%%% Setting up System of Eq's Au=B: %%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

x = linspace(a,b,Nx+1); % Defining locations 'x'

y = linspace(c,d,Ny+1); % Defining locations 'x'

h =         (b-a)/Nx;
% Sub-interval size, assumed hx=hy=h

 

% Defining "D" Block

D(1) = 4;

for ii = 2:(Nx-1)

 D(ii,ii-1) = -1;

 D(ii-1,ii) = -1;

 D(ii,ii)   =  4;

end

 

% Defining "I" Block

I = -eye([Nx-1,Nx-1]);

 

A(1:(Nx-1),1:(Nx-1)) = D;

for ii = 2:(Ny-1)

 A((ii-1)*(Nx-1)+1:(ii-1)*(Nx-1)+Nx-1,...

  
(ii-1)*(Nx-1)+1:(ii-1)*(Nx-1)+Nx-1) = D;

 A((ii-2)*(Nx-1)+1:(ii-2)*(Nx-1)+Nx-1,...

  
(ii-1)*(Nx-1)+1:(ii-1)*(Nx-1)+Nx-1) = I;

 A((ii-1)*(Nx-1)+1:(ii-1)*(Nx-1)+Nx-1,...

  
(ii-2)*(Nx-1)+1:(ii-2)*(Nx-1)+Nx-1) = I; 

end

 

% Defining RHS

B = zeros((Ny-1)*(Nx-1),1);

for ii = 1:(Ny-1)

 for jj = 1:(Nx-1)

  

  B((ii-1)*(Nx-1)+jj)
= -h^2*f(x(jj+1),y(ii+1));

  

  % Adding BC to 'B'

  if (ii==1)

   B((ii-1)*(Nx-1)+jj)
= B((ii-1)*(Nx-1)+jj) + g(x(jj+1),y(ii),a,b,c,d);

  end

  

  % Adding BC to 'B'

  if (ii==(Ny-1))

   B((ii-1)*(Nx-1)+jj)
= B((ii-1)*(Nx-1)+jj) + g(x(jj+1),y(ii+2),a,b,c,d);

  end

  

  % Adding BC to 'B'

  if (jj==1)

   B((ii-1)*(Nx-1)+jj)
= B((ii-1)*(Nx-1)+jj) + g(x(jj),y(ii+1),a,b,c,d);

  end

  

  % Adding BC to 'B'

  if (jj==(Nx-1))

   B((ii-1)*(Nx-1)+jj)
= B((ii-1)*(Nx-1)+jj) + g(x(jj+2),y(ii+1),a,b,c,d);

  end

  

 end

end

 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

 

u = A\B; % Solving Au=b for unknowns'u'

 

% Placing solution 'u' inside larger array with BCs

uu = zeros(Ny+1,Nx+1);%

for ii = 1:(Ny+1)

 for jj = 1:(Nx+1)

  uu(ii,jj) =
g(x(jj),y(ii),a,b,c,d);

 end

end

uu(2:(end-1),2:(end-1)) = reshape(u,(Nx-1),(Ny-1))';

Iteration method

clear; % Clear stored variables
h = 0.1
Nx    =                        10; % Number of sub-segments in x
Ny    =                        10; % Number of sub-segments in y
a     =                         0; % Location of boundary 'a' for 'x'
b     =                         1; % Location of boundary 'b' for 'x'
c     =                         0; % Location of boundary 'c' for 'y'
d     =                         1; % Location of boundary 'd' for 'y'
f     = @(x,y)              x + y; % Defining RH-side function 'f(x,y)'
u = zeros(Nx,Ny);
u(1,:) = a;
u(Nx,:) = b;
u(:,1) = c;
u(:,Ny) = d;
for i = 1:1000
   for x = 2:Nx-1
       for y = 2:Ny-1
           u(x,y) = 0.25*(u(x-1,y)+u(x+1,y)+u(x,y+1)+u(x,y-1))-h^2*f(x,y);
       end
   end
end

Two method has different result. BUT why?
I will explain it in the future.