How to change diagonal, subdiagonal and superdiagonal values with respect time while using loop and conditional statement?
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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
4 Comments
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
B
C
Yanni
on 14 Mar 2023
Dyuman Joshi
on 14 Mar 2023
"You changing that matrix size only."
How else are you going to get the values on diagonals?
What are the expected outputs?
Dyuman Joshi
on 15 Mar 2023
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.
Accepted Answer
Dyuman Joshi
on 15 Mar 2023
Edited: Dyuman Joshi
on 15 Mar 2023
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
B
C
1 Comment
Dyuman Joshi
on 15 Mar 2023
So you need A, B, C and D to be vectors for executing the function TriDiag?
More Answers (1)
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
B
C
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