Question

对于我的特定二维亥姆霍兹方程，我无法使用Gauss-Seidel算法得到收敛值：

以下是代码：

%Final Project Helmhotz 2-D:              
%d^2u/dx^2+d^2u/dy^2+Lambda*u=F(x,y)       
%Over domain of ax < x < bx and ay < y < by                                                 
%Using Gauss-Seidel

clc              
clear;

%Given variables         
lamda1=0.5;             %Given         
ax=0;                   %Given       
ay=0;                   %Given        
Pi=4*atan(1);           %Given        
bx=2*Pi;                %Given     
by=2*Pi;                %Given       
v=0;                    %Given (du/dy @y=by = 0)    

%Initial Problem Setup    
Nx=20;                  %Initial number of points in x   
Ny=Nx;                  %Initial number of points in y     
Lx=bx;                  %Length of x      
Ly=by;                  %Length of y    
deltax=Lx/(Nx+1);           %Step size in x       
deltay=Ly/(Ny+1);           %Step size in y      


%Constants     
fbay = (by-ay)*(by-ay)*cos(Pi*ay/by);    
gbay = ay*(by-ay)*(by-ay);        
cons1 = bx-ax;                 
dy2=deltay*deltay;          
dx2=deltax*deltax;           
dx2dy2=deltax*deltax*deltay*deltay;        
denominator=2*deltay*deltay+2*deltax*deltax;     
denoma=denominator-dx2dy2*lamda1;         
x=0:deltax:2*pi;               
y=0:deltay:2*pi;

%Solution Matrix          
U1 = zeros(Nx+2,Ny+2);  % Create solution matix with initial guess of ZERO

%Boundary Conditions
for j = 1:Ny+2                      %u(x=ax,y)=fb(y) boundary condition        
    U1(1,j)=(by-deltay*(j-1))*(by-deltay*(j-1))*cos(Pi*deltay*(j-1)/by);      
end

for j=1:Ny+2                        %u(x=bx,y)=gb(y) boundary condition           
    U1(Nx+2,j)=(deltay*(j-1))*(by-deltay*(j-1))*(by-deltay*(j-1));      
end

for i=1:Nx+2                        %u(x,y=ay)  boundary conditoin
    U1(i,1)=fbay+(deltax*(i-1)-ax)/cons1*(gbay-fbay);     
end

%Part 1 Using Gauss Siedel            
F1=zeros(Nx+2,Ny+2);

for i=1:Nx+2             
    for j=1:Ny+2                   
        F1(i,j)=cos(Pi/2*(2*(deltax*(i-1)-ax)/(bx-ax)+1))*sin(Pi*(deltay*(j-1)-ay)/(by-ay));
    end
end


for z=1:100
    for i=2:Nx+1
        for j=2:Ny+1
            %%%%% Different Equations for U if wanted %%%%%%%%%%%%%%%%%%%%
            %U1(i,j)=(deltax*deltax*F1(i,j)-(U1(i-1,j)+U1(i+1,j)+U1(i,j-1)+U1(i,j+1)))/(-4+deltax*deltax*lamda1);
            %U1a(i,j)=(Cons4*F1(i,j)-dy2*U1(i+1,j)-dy2*U1(i-1,j)-dx2*U1(i,j+1)-dx2*U1(i,j-1))/(Cons5a);
            U1(i,j)=((U1(i-1,j)+U1(i+1,j)+U1(i,j-1)+U1(i,j+1))-deltax*deltax*F1(i,j))/(4-deltax*deltax*lamda1);
            %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
            %U1(i,j)=(dy2*U1(i-1,j)+dy2*U1(i+1,j)+dx2*U1(i,j-1)+dx2*U1(i,j+1))/(denoma)-(dx2dy2*F1(i,j))/(denoma);
        end
        %%%%%%%%%%%% Different Neumann Conditions if wanted %%%%%%%%%%%%%%
        %U1(i,Ny+2)=(deltax*deltax*F1(i,Ny+2)-(U1(i-1,Ny+2)+U1(i+1,Ny+2)+U1(i,Ny+1)+(U1(i,Ny+1)-2*deltax*v)))/(-4+deltax*deltax*lamda1);
        %U1a(i,Ny+2)=(Cons4*F1(i,Ny+2)-dy2*U1(i+1,Ny+2)-dy2*U1(i-1,Ny+2)-dx2*U1(i,Ny+1)-dx2*U1(i,Ny+1))/(Cons5a);
        U1(i,Ny+2)=((U1(i-1,Ny+2)+U1(i+1,Ny+2)+U1(i,Ny+1)+U1(i,Ny+1))-deltax*deltax*F1(i,Ny+2))/(4-deltax*deltax*lamda1);
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        %U1(i,Ny+2)=((dy2)*U1(i-1,Ny+2)+(dy2)*U1(i+1,Ny+2)+(dx2)*U1(i,Ny+1)+(dx2)*U1(i,Ny+1))/(denoma)-(dx2dy2*F1(i,Ny+2))/(denoma);
    end

end

figure()
surf(x,y,U1);

每当我运行此代码超过100次迭代时，值开始偏离边界条件。我不确定为什么，也不是我的教授。

我还发布了我给出的问题，边界条件以及显示我离散化的另一张图片的照片。

Shows my discretization Shows my given problem and boundary conditions

Gauss-Seidel代码没有收敛于2-D亥姆霍兹的解

0 个答案: