How to obtain convergence of the curve to my boundary condtion?

I want a curve in each 'j'th iteration. But, I didn't get convergence of the curve to plot each 'j'th iterations between limit 1 to 0. Because, my boundary condition i.e.

T(1,:)=1;T(end,:)=0

doesn't fit with my code. how to use code for convergence condition?

Further, My data values of each 'j'the iterations are not gradual values along my boundary condition and the data values in 'j' th iteration are increasing only.,

clc; close all; clear all;
ymax=20; m=80; dy=ymax/m;                       %'i'th row
xmax=1; n=20;  dx=xmax/n;                       %'j'th column   
tmax=100; nt=500; dt=tmax/nt;       
tol=1e-5; max_difference=1; 
%====================INTIALIZATION=================================%
UOLD=zeros(m,n); VOLD=zeros(m,n); 
TNEW=0; TOLD=ones(m,n); 
A=zeros([1,m]); 
B=A;
C=A;
D=A;
T=TOLD; 
tic
   for j=1:n
        for i=1:m
            if j>i
                C(i)=(dt*VOLD(i,j)/4*dy)-(dt/(2*dy^2));
            elseif i>j
                A(i)=(-dt*VOLD(i,j)/4*dy)-(dt/(2*dy^2));
            elseif i==j
                B(i)=1+(dt*UOLD(i,j)/2*dx)+(dt/(dy^2));
            end
        end
    end
    for j=2:n
       
            for i=1:m
                    D(i)=(-dt*UOLD(i,j)*(-TNEW+TOLD(i,j)-TNEW)/2*dx)+(dt/(2*dy^2)*(TOLD(i+1,j)-2*TOLD(i,j)+TNEW))-(dt*VOLD(i,j)/4*dy*(TNEW-TOLD(i+1,j)-TOLD(i,j)));
            end
        end
        T(:,j)=TriDiag(A,B,C,D);   %call tridiagonal
        dt=0.2+dt;
      
        %T(1,:)=1;T(end,:)=0;  % how to use thsi to get perfect curve

Thanks for advance!

Answer 1

[LeoAiE]

It seems like you are trying to enforce the boundary condition T(1,:) = 1 and T(end,:) = 0 after the solution is computed in each iteration. However, this approach won't necessarily converge to the correct solution.

To incorporate the boundary conditions correctly, you need to modify the system of equations itself. In your case, the boundary conditions should be applied during the construction of the A, B, C, and D matrices.

clc; close all; clear all;
ymax=20; m=80; dy=ymax/m;                       %'i'th row
xmax=1; n=20;  dx=xmax/n;                       %'j'th column   
tmax=100; nt=500; dt=tmax/nt; t=0:dt:tmax;      % time at 'j'
tol=1e-2; max_difference(1)=1; 
%====================INTIALIZATION=================================%
UOLD=zeros(m,nt); VOLD=zeros(m,nt); 
TNEW=0; TOLD=TNEW*ones(m,nt); TWALL=ones(1,length(t));
A=zeros([1,m]); 
B=A;
C=A;
D=A;
T=TOLD; 
tic
for j=1:nt
    for i=1:m
        if i == 1  % Top boundary condition
            A(1) = 0;
            B(1) = 1;
            C(1) = 0;
            D(1) = 1;
        elseif i == m  % Bottom boundary condition
            A(m) = 0;
            B(m) = 1;
            C(m) = 0;
            D(m) = 0;
        elseif j>i
            C(i)=(dt*VOLD(i,j)/4*dy)-(dt/(2*dy^2));
        elseif i>j
            A(i)=(-dt*VOLD(i,j)/4*dy)-(dt/(2*dy^2));
        elseif i==j
            B(i)=1+(dt*UOLD(i,j)/2*dx)+(dt/(dy^2));
        end
    end

    for j=2:nt
        if j==1
            for i=1:m
                if i==1
                    D(i)=(-dt*UOLD(i,j)*(-TNEW+TOLD(i,j)-TNEW)/2*dx)+(dt/(2*dy^2)*(TOLD(i+1,j)-2*TOLD(i,j)+TNEW))-(dt*VOLD(i,j)/4*dy*(TNEW-TOLD(i+1,j)-TOLD(i,j)))-(dt/4*dy*VOLD(i,j))-(dt/2*dy^2*TNEW);
                elseif i==m
                    D(i)=(-dt*UOLD(i,j)*(-TNEW+TOLD(i,j)-TNEW)/2*dx)+(dt/(2*dy^2)*(TWALL(j)-2*TOLD(i,j)+TOLD(i-1,j)))-(dt*VOLD(i,j)/4*dy*(TOLD(i-1,j)-TWALL(j)+TOLD(i,j)))-((-dt)/4*dy*VOLD(i,j))-(dt/2*dy^2*TWALL);
                else
                    D(i)=(-dt*UOLD(i,j)*(-TNEW+TOLD(i,j)-TNEW)/2*dx)+(dt/(2*dy^2)*(TOLD(i+1,j)-2*TOLD(i,j)+TOLD(i-1,j)))-(dt*VOLD(i,j)/4*dy*(TOLD(i-1,j)-TOLD(i+1,j)+TOLD(i,j)));
                end
            end
        else
            for i=1:m
                if i==1
                    D(i)=(-dt*UOLD(i,j)*(-T(i,j-1)+TOLD(i,j)-TOLD(i,j-1))/2*dx)+(dt/(2*dy^

[Yanni]

@Aladdin I'm trying to include your idea in my code,. but it's not working. It's starting with 0 instead of 1 and I couldn't see another BC i.e. 1 at 'm' the iteration.

I want my data values start with 1 and gradually will increase and then gradually decrease to 0. This process should be fall in each column of T. I already shared one column for sample.

Moreover, if we can't use the BC T(1,:)=1;T(end,:)=0, then please consider the initialization

TWALL=ones(1,length(t)) i.e.,1 will fall in first row and no.of time steps.,It means TWALL at Y=0(top BC) and TNEW=0 i.e., 0 will in last row., it means TNEW at Y=m (bottom BC). and interior values should be increase gradually and left over, then reduce to 0. here, " How to use TWALL and TNEW values to fall in T?

[LeoAiE]

Maybe I have an idea of what you are looking for here. Try this

clc; close all; clear all;
ymax=20; m=80; dy=ymax/m;                       %'i'th row
xmax=1; n=20;  dx=xmax/n;                       %'j'th column   
tmax=100; nt=500; dt=tmax/nt; t=0:dt:tmax;      % time at 'j'
tol=1e-2; max_difference(1)=1; 
%====================INTIALIZATION=================================%
UOLD=zeros(m,nt); VOLD=zeros(m,nt); 
TNEW=0; TOLD=TNEW*ones(m,nt); TWALL=ones(1,length(t));
A=zeros([1,m]); 
B=A;
C=A;
D=A;
T=TOLD; 
tic
for j=1:nt
    for i=1:m
        if i == 1  % Top boundary condition
            A(1) = 0;
            B(1) = 1;
            C(1) = 0;
            D(1) = TWALL(j);
        elseif i == m  % Bottom boundary condition
            A(m) = 0;
            B(m) = 1;
            C(m) = 0;
            D(m) = TNEW;
        elseif j>i
            C(i)=(dt*VOLD(i,j)/4*dy)-(dt/(2*dy^2));
        elseif i>j
            A(i)=(-dt*VOLD(i,j)/4*dy)-(dt/(2*dy^2));
        elseif i==j
            B(i)=1+(dt*UOLD(i,j)/2*dx)+(dt/(dy^2));
        end
    end
    D(1) = TWALL(j); % Top boundary condition
    D(m) = TNEW; % Bottom boundary condition
    T(:,j)=TriDiag(A,B,C,D);   %call tridiagonal
    dt=0.2+dt;
    TOLD=T;
    max_difference(j) = max(abs(T(:,j)-T(:,j-1))./max(1,abs(T(:,j-1))));
    if max_difference(j)<tol
        break
    end
end