How to change diagonal, subdiagonal and superdiagonal values with respect time while using loop and conditional statement?

xmax=1; ymax=7; m=20; n=29; 
dx=xmax/m; dy=ymax/n; dt=0.2; 
UOLD=zeros(m,n);VOLD=zeros(m,n);
A=zeros([1,m]);
B=A;
C=A;
while dt<tmax
    for j=1:n
        for i=1:m
            if j>i
                C(i)=((dt*VOLD(i,j))/(4*dy))-(dt/(2*(dy^2)));  %superdiagonal
            elseif i>j
                A(i)=((-dt*VOLD(i,j))/(4*dy))-(dt/(2*(dy^2)));  %sub diagonal 
            elseif i==j
                B(i)=1+((dt*UOLD(i,j))/(2*dx))+(dt/(dy^2));     %main diagonal
            end
        end
    end
    dt=0.2+dt;
end

[Dyuman Joshi]

Is this what you are looking for?

xmax=1; ymax=7; m=20; n=29; tmax=100; nt=500; 
dx=xmax/m; dy=ymax/n; dt=tmax/nt; 
UOLD=zeros(m,n);VOLD=zeros(m,n);
A=zeros([m,m]);
B=A;
C=A;
while dt<tmax
    for j=1:n
        for i=1:m
            if j>i
                C(j-1,j)=((dt*VOLD(i,j))/(4*dy))-(dt/(2*(dy^2)));  %superdiagonal
            elseif i>j
                A(i,i-1)=((-dt*VOLD(i,j))/(4*dy))-(dt/(2*(dy^2)));  %sub diagonal 
            elseif i==j
                B(i,j)=1+((dt*UOLD(i,j))/(2*dx))+(dt/(dy^2));     %main diagonal
            end
        end
    end
dt=0.2+dt;
end
A

A = 20×20

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0

B

B = 20×20

1.0e+03 * 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0

C

C = 28×29

0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

[Yanni]

You changing that matrix size only. but i need distinct main daigonal values and other two as well. here, main diagonal values are same. so, how to get distinct values with respect to time?

[Dyuman Joshi]

"You changing that matrix size only."

How else are you going to get the values on diagonals?

What are the expected outputs?

[Dyuman Joshi]

I didn't understand what you meant by your comment below my response.

Also, you did not answer my questions - How else are you going to get the values on "diagonals"?

What are the expected outputs? Please provide exactly what the values of A, B and C you want to get.

Answer 1

[Dyuman Joshi]

xmax=1; ymax=7; m=20; n=29; tmax=100; nt=500; 
dx=xmax/m; dy=ymax/n; dt=tmax/nt; 
UOLD=zeros(m,n);VOLD=zeros(m,n);
A=zeros(m,m);
B=A;
C=A;
while dt<tmax
    for j=1:n
        for i=1:m
            if j>i
                C(j-1,j)=((dt*VOLD(i,j))/(4*dy))-(dt/(2*(dy^2)));  %superdiagonal
            elseif i>j
                A(i,i-1)=((-dt*VOLD(i,j))/(4*dy))-(dt/(2*(dy^2)));  %sub diagonal 
            elseif i==j
                B(i,j)=1+((dt*UOLD(i,j))/(2*dx))+(dt/(dy^2));     %main diagonal
            end
        end
        dt=0.2+dt;
    end
end
A

A = 20×20

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -847.8653 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -849.5816 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -851.2980 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -853.0143 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -854.7306 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -858.1633 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -859.8796 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -861.5959 0 0 0 0 0 0 0 0 0 0 0

B

B = 20×20

1.0e+03 * 1.6967 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7002 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7036 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7070 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7105 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7173 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7208 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7242 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7276 0 0 0 0 0 0 0 0 0 0

C

C = 28×29

0 -849.5816 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -851.2980 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -853.0143 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -854.7306 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -858.1633 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -859.8796 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -861.5959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -863.3122 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -865.0286 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

[Dyuman Joshi]

So you need A, B, C and D to be vectors for executing the function TriDiag?

Answer 2

[Oguz Kaan Hancioglu]

As @Dyuman Joshi mentioned, the problem statement is not clear. However vectors don't have diagonal. Please change the A,B, and C as a matrix and use correct indeces. I think it solves your problem.

xmax=1; ymax=7; m=20; n=29; tmax=100; nt=500; 
dx=xmax/m; dy=ymax/n; dt=tmax/nt; 
UOLD=zeros(m,n);VOLD=zeros(m,n);
A=zeros([m,n]);
B=A;
C=A;
while dt<tmax
    for j=1:n
        for i=1:m
            if j>i
                C(i,j)=((dt*VOLD(i,j))/(4*dy))-(dt/(2*(dy^2)));  %superdiagonal
            elseif i>j
                A(i,j)=((-dt*VOLD(i,j))/(4*dy))-(dt/(2*(dy^2)));  %sub diagonal 
            elseif i==j
                B(i,j)=1+((dt*UOLD(i,j))/(2*dx))+(dt/(dy^2));     %main diagonal
            end
        end
    end
    dt=0.2+dt;
end
A

A = 20×29

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

B

B = 20×29

1.0e+03 * 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.7139 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

C

C = 20×29

0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 0 0 0 0 0 0 0 0 0 0 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469 -856.4469

[Yanni]

In particular, you got the main diagonal value is same. but I want DIFFERENT main diagonal values.