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Trying to plot number of iterations vs gridsize for steady state

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Jayden Cavanagh
Jayden Cavanagh 2021 年 6 月 5 日
編集済み: SALAH ALRABEEI 2021 年 6 月 5 日
I need to compute the number of iterations taken to reach the steady-state within a given tolerance. I then need to plot the final number of iterations against the grid size but I cannot for the life of me work out how I am supposed to do that. How am I suppose to set up and then plot those two things?
Code:
n=2;
nx=2^n;
nz=nx;
a=25;
b=25;
x = linspace(0, a, nx);
z = linspace(0, b, nz);
[X, Z] = meshgrid(x,z);
Tnp1 = zeros(nx, nz);
Tnp1(:,1) = 20;
Tnp1(:,end) = 20;
Tnp1(1,:) = 20+380*sin((X.*pi)/25)+205*sin((X.*5*pi)/25);
Tnp1(end,:) = 20;
err = 1;
tol = 1e-8;
k=0;
while err > tol
Tn = Tnp1;
k=k+1;
for i = 2:nx-1
for j = 2:nz-1
Tnp1(i,j) = (1/4)*(Tn(i+1,j)+Tn(i-1,j)+Tn(i,j+1)+Tn(i,j-1));
end
end
err = max(abs(Tnp1(:) - Tn(:)));
end
T = Tnp1;
plot(X,k)
end
end

回答 (1 件)

SALAH ALRABEEI
SALAH ALRABEEI 2021 年 6 月 5 日
編集済み: SALAH ALRABEEI 2021 年 6 月 5 日
Hi, this looks like a diffusion equation solved by FDM. you had to mistakes, I already corrected them, n is the step size or grid size, and the u should have used x not X in the boundary condition conditions. See the code below
clear N = [2,4]; for kk = 1: length(N) n=N(kk); nx=2^n; nz=nx; a=25; b=25; x = linspace(0, a, nx); z = linspace(0, b, nz); [X, Z] = meshgrid(x,z); Tnp1 = zeros(nx, nz); Tnp1(:,1) = 20; Tnp1(:,end) = 20; Tnp1(1,:) = 20+380*sin((x.*pi)/25)+205*sin((x.*5*pi)/25); Tnp1(end,:) = 20; err = 1; tol = 1e-8; k=0; while err > tol Tn = Tnp1; k=k+1; for i = 2:nx-1 for j = 2:nz-1 Tnp1(i,j) = (1/4)*(Tn(i+1,j)+Tn(i-1,j)+Tn(i,j+1)+Tn(i,j-1)); end end err = max(abs(Tnp1(:) - Tn(:))); end T = Tnp1; %subplot(2,2,kk) %surf(X,Z,T),grid %title([' Profile at grid = ' num2str(kk)])
Num_iter(kk) = k; end
figure(2) plot(N,Num_iter),grid
  2 件のコメント
Jayden Cavanagh
Jayden Cavanagh 2021 年 6 月 5 日
編集済み: Jayden Cavanagh 2021 年 6 月 5 日
It is a steady-state temperature field in a 2D rectangle solved using G-S method. So it isn't exactly what I am looking for. For my problem I have to choose the number of divisions in the domain to be ndiv = 2^n for n = 2 : 9 in both the x and z directions. n=2 was just a way to reduce processing power when testing. While your code works for N=[2,4] I get error code
Requested array exceeds the maximum possible variable size.
Error in linspace (line 44)
y = d1 + (0:n1).*(d2 - d1)./n1;
Error in hope (line 9)
x = linspace(0, a, nx);
when I try and run it for the maximum value of n=8 and nx=2^8. Any way the code could be changed to be able to composate for that?
SALAH ALRABEEI
SALAH ALRABEEI 2021 年 6 月 5 日
編集済み: SALAH ALRABEEI 2021 年 6 月 5 日
%
clear
N = [2,4];
for kk = 1: length(N)
n=N(kk);
nx=2^n;
nz=nx;
a=25;
b=25;
x = linspace(0, a, nx);
z = linspace(0, b, nz);
[X, Z] = meshgrid(x,z);
Tnp1 = zeros(nx, nz);
Tnp1(:,1) = 20;
Tnp1(:,end) = 20;
Tnp1(1,:) = 20+380*sin((x.*pi)/25)+205*sin((x.*5*pi)/25);
Tnp1(end,:) = 20;
err = 1;
tol = 1e-8;
k=0;
while err > tol
Tn = Tnp1;
k=k+1;
for i = 2:nx-1
for j = 2:nz-1
Tnp1(i,j) = (1/4)*(Tn(i+1,j)+Tn(i-1,j)+Tn(i,j+1)+Tn(i,j-1));
end
end
err = max(abs(Tnp1(:) - Tn(:)));
end
T = Tnp1;
%subplot(2,2,kk)
%surf(X,Z,T),grid
%title([' Profile at grid = ' num2str(kk)])
Num_iter(kk) = k;
end
figure(2)
plot(N,Num_iter),grid

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