Plotting a variable that changes in a loop iteration vs the iteration number

I am trying to plot the change in u (diff_u) agaist the iteration to prove that diff_u converges. What is the best way to do this?

Below is my code, with z being the iteration counter:

%% Parameters
rho = 1.225; % kg/m^3 - Air sea level standard day
U = 10; % m/s - freestream
mu = 1.789e-5; % kg/(m*s) - Air sea level standard day
nu = mu/rho; % m^2/s
L = 0.25; % m
H = 0.05; % m
 
% nx and ny defines matrix grid, dx and dy for step size
nx = 50;
ny = 10;
dx = .005; % 50 grid points in x-dir - .005m spacing
dy = .005; % 10 grid points in y-dir - .005m spacing
 
% Re based on length
Re_L = U*L/nu;
 
% Filling mesh with 0 to begin
v = zeros(ny,nx); % All initial v are 0
u = ones(ny,nx)*U; % Filling the entire mesh with all u = 10
 
% Boundary Conditions
% Goes (j,i) each segment, fills mesh with known values
u(:,1) = U; % Inlet
v(:,1) = 0; % Inlet
u(1,:) = 0; % No Slip wall
v(1,:) = 0; % No Slip wall
u(ny,:) = U; % Free stream (top of mesh)

%% Solving
% Guess values of alpha
alphau = 0.001; %guess, will need to figure out best value for this 
alphav = 0.1;
 
% Differences
diff_u=[];
diff_v=[];
diff_u_max=[];
diff_v_max=[];
diff_both=[];
 
% Tolerance and convergence for main while loop
tol = .001;
converge = false;
z = 0; % Starting the index for iterations

while z < 20 % This is number of iterations it performs
    z = z + 1;
    for j = 2:1:ny-1
        for i = 2:1:nx-1
            u_i = u(j,i+1);
            u(j,i+1) = (2*dx/u(j,i)) * (((mu/rho)*((u(j-1,i) + 2*u(j,i) + u(j+1,i))/(dy^2))) - (v(j,i)*((u(j-1,i) - u(j+1,i))/(2*dy)))) + u(j,i-1);
            change_u = u_i - u(j,i+1);
            u(j,i+1) = u_i + (alphau * change_u);
            v_j = v(j+1,i);       
            v(j,i+1) = v(j-1,i+1) - dy/2/dx*(u(j,i+1)-u(j,i)+u(j-1,i+1)-u(j-1,i));
            change_v = v_j - v(j,i+1);
            v(j,i+1)= v_j + (alphav * change_v);
            diff_u = [diff_u change_u];
            diff_v = [diff_v change_v];
        end
    end
    diff_U = max(abs(diff_u))
    diff_V = max(abs(diff_v))
    diff_u_max = [diff_u_max diff_U];
    diff_v_max = [diff_v_max diff_V];
    diff = [diff_U diff_V];
    max_diff = max(diff);
    
    %if max_diff < tol
    %    z = 0; % I dont think this is right. Want max_diff < tol then iterations = 0 so it stops iterating
    %end
    
    % WANT TO PLOT THE CHANGE IN u TO SHOW IT CONVERGES
   
end

Answer 1

[Geoff Hayes]

Colin - outside of the while loop, just do

plot(diff_u);

But is that what you really want to plot? Your comment

%if max_diff < tol
%    z = 0; % I dont think this is right. Want max_diff < tol then iterations = 0 so it stops iterating
%end
    

where

diff = [diff_U diff_V];
max_diff = max(diff);

also includes a v. Do you want to plot diff_v instead? Or max_diff? Also, if you want to exit the while loop, just do

if abs(max_diff) < tol
    break;
end

[Colin Thiele]

Thanks Geoff!

I actually will need 2 difference lines plotted. One that shows the differnce in v and one for difference in u to show they are both converging throughout the iteration. I figured I would plot diff_u and diff_v on the same plot, since max_diff takes into account both differnces (of u and v) and combines them. When the iteration finsihed max_diff is just one number.

Also thank you for the help with the if statment at the end of the while loop!

while
 % Eqns and such 
    if abs(max_diff) < tol
        break
    end
    z = z + 1;
end
figure
hold on
plot(diff_u)
plot(diff_v)
hold off
xlabel('Iterations');
ylabel('Chnage in u and v');
title('Convergence of u and v Throughout Iterations');
legend('u difference', 'v difference');