Subscripted assignment dimension mismatch.

function main
D=1; %L=0;
Pr=1;R=0.1;Sc=1;
xa=0;xb=6;
Lv = [-2.5:0.025:0];
p = []; 
for i=1:length(Lv)
    L = Lv(i);
    fODE = @(x,y) [y(2);  y(3);  y(2)^2-y(3)*y(1)-1;  y(5);  -3*Pr*y(1)*y(5)/(3+4*R);  y(7); -Sc*y(1)*y(7)];  
    BCres= @(ya,yb)[ya(1); ya(2)-L-D*ya(3); ya(4)-1; ya(6)-1; yb(2)-1; yb(4);yb(6)]; 
    xint=linspace(xa,xb,101);
    solinit=bvpinit(xint,[0 1 0 1 0 1 0]);
    sol=bvp4c(fODE,BCres,solinit);
    sxint=deval(sol,xint);
       %%WE NEED TO PLOT for
    S(i,1)=sxint(3,:);
      end
   figure(1)
    plot(Lv,S,'-','Linewidth',1.5); 
    xlabel('\bf \lambda');
    ylabel('\bf C_{f}');
    hold on
end

%%While running the code following ERROR occurs:
Subscripted assignment dimension mismatch.
Error in  (line 17)
    S(i,1)=sxint(3,:);

Answer 1

[Walter Roberson]

function all_sxint = main
    D=1; %L=0;
    Pr=1; R=0.1; Sc=1;
    xa=0;xb=6;
    Lv = [-2.5:0.025:0];
    nLv = length(Lv);
    all_sxint = cell(nLv, 1);
    S = zeros(nLv, 7);
    for i=1:nLv
        L = Lv(i);
        fODE = @(x,y) [y(2);  y(3);  y(2)^2-y(3)*y(1)-1;  y(5);  -3*Pr*y(1)*y(5)/(3+4*R);  y(7); -Sc*y(1)*y(7)];  
        BCres= @(ya,yb)[ya(1); ya(2)-L-D*ya(3); ya(4)-1; ya(6)-1; yb(2)-1; yb(4);yb(6)]; 
        xint=linspace(xa,xb,101);
        solinit=bvpinit(xint,[0 1 0 1 0 1 0]);
        sol=bvp4c(fODE,BCres,solinit);
        sxint=deval(sol,xint);
        all_sxint{i} = sxint; 
        S(i,:) = sxint(3,:);
    end
  
    figure(1)
    plot(Lv, S, '-', 'Linewidth', 1.5); 
    xlabel('\bf \lambda');
    ylabel('\bf C_{f}');
    legend({'bc1', 'bc2', 'bc3', 'bc4', 'bc5', 'bc6', 'bc7'})
end

Assign the output to a variable so that you can examine all of the time points for all of the Lv values afterwards, as you indicate that you need to be able to do that.

[MINATI]

Still unsolved

[Walter Roberson]

Please confirm that what you want to plot is all derivatives for the third time point, not the third derivative for all time points.

function all_sxint = main
    D=1; %L=0;
    Pr=1; R=0.1; Sc=1;
    xa=0;xb=6;
    Lv = [-2.5:0.025:0];
    nLv = length(Lv);
    all_sxint = cell(nLv, 1);
    S = zeros(nLv, 7);
    for i=1:nLv
        L = Lv(i);
        fODE = @(x,y) [y(2);  y(3);  y(2)^2-y(3)*y(1)-1;  y(5);  -3*Pr*y(1)*y(5)/(3+4*R);  y(7); -Sc*y(1)*y(7)];  
        BCres= @(ya,yb)[ya(1); ya(2)-L-D*ya(3); ya(4)-1; ya(6)-1; yb(2)-1; yb(4);yb(6)]; 
        xint=linspace(xa,xb,101);
        solinit=bvpinit(xint,[0 1 0 1 0 1 0]);
        sol=bvp4c(fODE,BCres,solinit);
        sxint=deval(sol,xint);
        all_sxint{i} = sxint; 
        S(i,:) = sxint(:,3);
    end
  
    figure(1)
    plot(Lv, S, '-', 'Linewidth', 1.5); 
    xlabel('\bf \lambda');
    ylabel('\bf C_{f}');
    legend({'bc1', 'bc2', 'bc3', 'bc4', 'bc5', 'bc6', 'bc7'})
end

[MINATI]

Its third derivative for all time points.

[MINATI]

for ref.

We need

Fig. 1 Only solid lines

[Walter Roberson]

function main2
    D=1; %L=0;
    Pr=1; R=0.1; Sc=1;
    xa=0; xb=6;
    xint=linspace(xa,xb,101);
    nxint = length(xint);
    Lv = [-2.5:0.025:0];
    nLv = length(Lv);
    S = zeros(nLv, nxint);
    for i=1:nLv
        L = Lv(i);
        fODE = @(x,y) [y(2);  y(3);  y(2)^2-y(3)*y(1)-1;  y(5);  -3*Pr*y(1)*y(5)/(3+4*R);  y(7); -Sc*y(1)*y(7)];  
        BCres = @(ya,yb)[ya(1); ya(2)-L-D*ya(3); ya(4)-1; ya(6)-1; yb(2)-1; yb(4);yb(6)]; 
        solinit = bvpinit(xint,[0 1 0 1 0 1 0]);
        sol = bvp4c(fODE,BCres,solinit);
        sxint = deval(sol,xint);
        S(i,:) = sxint(3,:);
    end
  
    figure(1)
    plot(Lv, S, '-', 'Linewidth', 1.5); 
    xlabel('\bf \lambda');
    ylabel('\bf C_{f}');
end

However I think you are plotting the wrong thing. Your Y axis is one of the boundary conditions, but the Y axis in the paper is . The paper looks like it is plotting for 3 different δ values, but you are only using one δ value. You asked to plot for all values for all timepoints; in the paper the x axis is and it is not obvious to me that time comes into the plot at all.

It is not obvious to me that you are even calculating the right thing.

[MINATI]

At least for \delta=1 (D=1), the curve should match.

That Rex^(1/2)Cf = sxint(3,1) = f"(0)(in the PAPER), for which straight line will come. But in the paper perhaps they have taken sxint(3,:), which are all values of f"(\eta) of the paper, thats why a curve came.

[Walter Roberson]

The paper says there is the boundary condition . If your D represents delta, then your D should be appearing in your boundary conditions.

[MINATI]

In boundary condition

ya(2)-L-D*ya(3)

i.e, D is present

[Walter Roberson]

I noticed some oddities in the output. I used options to push the permitted mesh points way up, and zoomed in closer. It turns out there is something unusual going on at -2.4648 .

Warning: Unable to meet the tolerance without using more than 500000 mesh points.

The last mesh of 292855 points and the solution are available in the output argument.

The maximum residual is 0.204915, while requested accuracy is 0.001.

The above was for Lv = linspace(-2.475, -2.425, 50);

The residual is staying pretty much the same as I increase the number of mesh points and zoom in more closely, which is potentially hinting at a singularity.

Different lines correspond to different xint.

[MINATI]

No we cant go beyond \lambda = - 2

[Walter Roberson]

Lv = [-2.5:0.025:0]; is in your existing code

Answer 2

[Matt J]

sxint(3,:) is not a scalar, but the left hand side S(i,1) is a scalar location.

[MINATI]

@ Matt J

So what to do?

[Walter Roberson]

You are looping over several Lv values, running a boundary value problem for each. What would you like plotted from each of those runs?

I suspect you want

S(i,1) = deval(sol, xa, 3);

[MINATI]

@Walter Roberson

Actually we need for all values of sxint[3] i.e, f "(eta)

[Walter Roberson]

sxint(3) is a scalar. It does not make sense to talk about "all values".

There is no eta defined in your question, so I do not know what f''(eta) refers to.

Perhaps you want

S(i,:)=sxint(3,:);

However if you only want the information for the third time point, xint(3), then it is a waste of time to compute sxint for all the time points, and might as well use

sxint = deval(sol, xint(3));
S(i,:) = sxint;

[Matt J]

@MINATI

So what to do?

You must tell us what the sizes of the left and right hand side of the assigment are intended to be.

[MINATI]

@Matt J

For all values of sxint(3,:), what changes should be made in S(i,1) OR in Lv

i.e, we need for all values

[Walter Roberson]

Is there a reason you need to calculate at all of the time points, and then to store data for only the third time point? Is there a particular reason why my suggestion to calculate only at the third time point will not work for you?

[MINATI]

@Walter Roberson

Yes, Actually I need to validate a previous study with mine but unable. Since they have drawn with all points, I am trying in that way.