Getting same final value for various time-steps

Hi.

I am trying to modify my Crank Nicholson method code so that it breaks when 14.9<T(11,11,k+1)<15.1 condition is satisfied.

This is the equation

Since time interval is unknown,time variable t, time step variable p and variables including p and t are commented out. I arbitrarily picked p values to obtain the time when 14.9<T(11,11,k+1)<15.1 but whatever the p interval is, I always face with the same result ("final" matrix). What should be done to make the code work properly?

%t=4;                % Total time (s)
delta_t=0.05;        % Time step
L=2;                 % Length of each edge
delta_x=0.1;         % Spacing

% Boundary temperatures
T1=10;
T2=10;
T3=10;
T4=10;

% Initial temperature
T5=400;

n=((L/delta_x)+1)^2;    % Total no of nodes
m=sqrt(n);              % Number of nodes in each row and column
%r1=(t/delta_t)+1;       % Number of time steps
%p=round(r1);            % Number of time steps (rounded)
r=0.033;

% Placing initial and boundary conditions
%T=zeros(m,m,p);         % Pre-allocating

for k=1:8000          % Time-step loop
    for i=1:m           % x coord loop
        for j=1:m       % y coord loop
            if(i==1)&&(j==1)
                T(i,j,k)=(T4+T1)/2;
            elseif(i==1)&&(j==m)
                T(i,j,k)=(T2+T1)/2;
            elseif(i==m)&&(j==m)
                T(i,j,k)=(T2+T3)/2;
            elseif(i==m)&&(j==1)
                T(i,j,k)=(T4+T3)/2;
            elseif(i==1&&(j>1&&j<m))
                T(i,j,k)=T1;
            elseif(j==m&&(i>1&&i<m))
                T(i,j,k)=T2;
            elseif(i==m&&(j>1&&j<m))
                T(i,j,k)=T3;
            elseif(j==1&&(i>1&&i<m))
                T(i,j,k)=T4;
            else
                T(i,j,k)=T5;
            end
        end
    end
end

% Solution

keepgoing = true;
for k=1:7999
    for i=2:m-1
        for j=2:m-1
            
            T(i,j,k+1)=1/(1+2*r)*(r/2*(T(i-1,j,k+1)+ T(i+1,j,k+1)+ T(i,j+1,k+1)+ T(i,j-1,k+1))+(1-2*r)*T(i,j,k)+r/2*(T(i-1,j,k)+ T(i+1,j,k)+ T(i,j+1,k)+ T(i,j-1,k)));
            
            if (i==11 && j==11 && T(i,j,k+1)>14.9 && T(i,j,k+1)<15.1)
             keepgoing =false;
             break;
            end
        end
        if ~keepgoing; break; end
    end
    if ~keepgoing; break; end
end
final=T(:,:,k);
imagesc(T(:,:,k));
colorbar;
title('Temperature Profile')

Thanks.

[Walter Roberson]

is the breakpoint being encountered at all?

[Jan]

Easier to debug:

T=zeros(m,m, 8000);
% for k=1:8000   The body does not depend on k, so omit this loop!
for i=1:m           % x coord loop
    for j=1:m       % y coord loop
        if i==1
            if j==1
               T(i,j,:) = (T4+T1)/2;
            elseif j==m
               T(i,j,:) = (T2+T1)/2;
            else
               T(i,j,:) = T1;
            end
            
        elseif i==m
            if j==1
               T(i,j,:) = (T4+T3)/2;
            elseif j==m
               T(i,j,:) = (T2+T3)/2;
            else
               T(i,j,:) = T3;
            end
            
        else
            if j==1
               T(i,j,:) = T4;
            elseif j==m
               T(i,j,:) = T2;
            else
               T(i,j,:) = T5;
            end
        end
    end
end

Or what about:

T = zeros(m, m, 8000);
T(1,     1,     :) = (T4+T1) / 2;
T(1,     m,     :) = (T2+T1) / 2;
T(1,     2:m-1, :) = T1;
T(m,     1,     :) = (T4+T3) / 2;
T(m,     m,     :) = (T2+T3) / 2;
T(m,     2:m-1, :) = T3;
T(2:m-1, 1,     :) = T4;
T(2:m-1, m,     :) = T2;
T(2:m-1, 2:m-1, :) = T5;

Now, what is your problem? You get the same output final if you vary what?

[Turgut Ataseven]

@Jan Thanks for the reply.

I get the same output "final" as I vary ending value of k (number of time steps).

keepgoing = true;
for k=1:7999 % this value %%
    for i=2:m-1
        for j=2:m-1
            T(i,j,k+1)=1/(1+2*r)*(r/2*(T(i-1,j,k+1)+ T(i+1,j,k+1)+ T(i,j+1,k+1)+ T(i,j-1,k+1))+(1-2*r)*T(i,j,k)+r/2*(T(i-1,j,k)+ T(i+1,j,k)+ T(i,j+1,k)+ T(i,j-1,k)));
            
            if (i==11 && j==11 && T(i,j,k+1)>14.9 && T(i,j,k+1)<15.1)
             keepgoing =false;
             break;
            end
        end
        if ~keepgoing; break; end
    end
    if ~keepgoing; break; end
end
        end

T matrix converges to same values after approximately 260 steps. It is probably because both sides of the equation has "k+1".

Answer 1

[Torsten]

for k=1:7999
  for i=2:m-1
    for j=2:m-1         
      T(i,j,k+1)=1/(1+2*r)*(r/2*(T(i-1,j,k+1)+ T(i+1,j,k+1)+ T(i,j+1,k+1)+ T(i,j-1,k+1))+(1-2*r)*T(i,j,k)+r/2*(T(i-1,j,k)+ T(i+1,j,k)+ T(i,j+1,k)+ T(i,j-1,k)));  
    end       
  end
  if T(11,11,k+1)>14.9 && T(11,11,k+1)<15.1
    break
  end
end