ERROR came as "Index exceeds matrix dimensions".

%% When I incorporate 3 eqns, ERROR came as "Index exceeds matrix dimensions".
%%Here I have initial condition f=[0 0 0 ], BC: gl=[coswt t t], gr =[0 0 0 ]. 

What next?

Here is my trial:

H=10;R=5;Pr=1;G1=5;G2=5;Kc=1;Sc=0.22;wt=pi/2;Q=H-(R/Pr);
xl=0; xr=5; % x domain [xl,xr]
J = 10; % J: number of division for x
dx = (xr-xl)/ J; % dx: mesh size
tf = 01; % final simulation time
Nt = 100; % Nt: number of time steps
dt = tf/Nt;
mu = dt/(dx)^2;
% Evaluate the initial conditions
x = xl : dx : xr; % generate the grid point
% f(1:J+1) since array index starts from 1
f1 = 0;f2 = 0;f3 = 0;    %%%I.C
% store the solution at all grid points for all time steps
u = zeros(J+1,Nt);
v = zeros(J+1,Nt);
w = zeros(J+1,Nt);
U=[u; v; w];
% Find the approximate solution at each time step
for n = 1:Nt
    t = n*dt; % current time
    % boundary condition at left side
    gl = [cos(wt); t; t];
    % boundary condition at right side
    gr = [0; 0; 0];
    if n==1 % first time step
        for j=2:J % interior nodes
            u(j,n) = (1+dt*Q)*f1(j) + dt*(G1*f2(j)+G2*f3(j))+mu*(f1(j+1)-2*f1(j)+f1(j-1));
            v(j,n) = (1+dt*Q)*f2(j) + (mu/Pr)*(f2(j+1)-2*f2(j)+f2(j-1));
            w(j,n) = (1-dt*Kc)*f3(j) + (mu/Sc)*(f3(j+1)-2*f3(j)+f3(j-1));
            U=[u(j,n) v(j,n) w(j,n)];
        end
        U(1,n) = gl; % the left-end point
        U(J+1,n) = gr; % the right-end point
    else
        for j=2:J % interior nodes
            u(j,n)= (1+dt*Q)*u(j,n-1)+ dt*(G1*v(j,n-1)+G2*w(j,n-1))+ mu*(u(j+1,n-1)-2*u(j,n-1)+u(j-1,n-1));
            v(j,n)= (1+dt*Q)*v(j,n-1)+ (mu/Pr)*(v(j+1,n-1)-2*v(j,n-1)+v(j-1,n-1));
            w(j,n)= (1+dt*Q)*w(j,n-1)+ (mu/Sc)*(w(j+1,n-1)-2*w(j,n-1)+w(j-1,n-1));
            U=[u(j,n) v(j,n) w(j,n)];
        end
        U(1,n) = gl; % the left-end point
        U(J+1,n) = gr; % the right-end point
    end
end
% Plot the results
tt = dt : dt : Nt*dt;
figure(1)
plot(x,u)
hold on

[madhan ravi]

Isn't this the same as your previous questions https://in.mathworks.com/matlabcentral/answers/476579-how-to-get-single-curve ?

Answer 1

[Adam Danz]

f1 = 0;    % 12th line of your code;  so, f1 only has 1 element 

% (later in your code...)

for j=2:J % interior nodes
      u(j,n) = (1+dt*Q)*f1(j) + dt*(G1*f2(j)+G2*f3(j))+mu*(f1(j+1)-2*f1(j)+f1(j-1));
      %                 ^^^^

When j = 2, you're trying to reference the 2nd element of f1 but f1 only has 1 element.

[MINATI PATRA]

But my f=0 for all the three variables (initial condition u(x,0)=0=v(x,0)=w(x,0)) for all x

and also gr=0 for all the three variables (Boundary condition for x tends to infinity)

So how to code?

[Adam Danz]

There is not a variable named "f" in the code you shared.

The problem is that when you try to do "f1(j)", and j = 2, f1 only has 1 value.

f1 = 0; 
f1(2)   % ERROR

[MINATI PATRA]

f=[f1 f2 f3]=[0 0 0]

[Adam Danz]

The variable "f" does not exist in your code.

This line below creates 3 variables named f1, f2 and f3.

f1 = 0;f2 = 0;f3 = 0;    %%%I.C

% same as 
f1 = 0;
f2 = 0;
f3 = 0;    %%%I.C

[MINATI PATRA]

ok

with this f1 = 0;f2 = 0;f3 = 0;

how to proceed

[Adam Danz]

If you didn't write this code and do not understand what it is suppose to be doing, maybe you could consult with whoever wrote it.

I can help explain why you're getting the error but I have no idea what the variables are supposed to be.

There's no way I could know what f1, f2, or f3 should be. This isn't my code.

[MINATI PATRA]

But still this is unsolved

Thanks

[Adam Danz]

Well, the error you were getting is explained. I hope you understand why you were getting the error so you can figure out what needs fixed.

[Walter Roberson]

Look at the lines

        for j=2:J % interior nodes
            u(j,n) = (1+dt*Q)*f1(j) + dt*(G1*f2(j)+G2*f3(j))+mu*(f1(j+1)-2*f1(j)+f1(j-1));

f1(j+1) is used, and j can be up to J. Therefore f1 must be at least length J+1 . That can be done by

f1 = zeros(1,J+1); f2 = zeros(1,J+1); f3 = zeros(1,J+1);    %%%I.C

[MINATI PATRA]

Thanks Walter

After modification again ERROR came as

Assignment has more non-singleton rhs dimensions than non-singleton subscripts

(line 36)

U(1,n) = gl; % the left-end point

Means some modification needed in gi and gr line:

Can I change gr = [ zeros(J+1, n); zeros(J+1, n); zeros(J+1, n) ]; and what for gl

gl = [

Please add to complete it.

[Walter Roberson]

No those will not help. You assign all of gl and all of gr to scalar locations, so they have to be scalars for your code to work.

[Adam Danz]

@MINATI ,

Often times when people are troubleshooting errors, they just try things that alleviate the error and lose sight of the bigger picture: what the algorithm is supposed to be doing and what each variable is supposed to represent. Changing a variable to avoid an error is much more simple than investing time into dissecting the code and understanding each line. Just because an error message no longer appears doesn't mean the algorithm is behaving as it should. That approach is actually dangerous because the lack of an error message can provide false security that the code "is correct".

I know it may be frustrating and it will be a large investment of time, but I suggest starting at the top of your code and understanding what each line does and what each variable represents. What shape and size should the variable take? Do the values in each variable look right? You're going to learn so much in that process and more importantly, you'll become independent in solving these types of errors.

Walter is the best and I've learned so much from him and others here but I've learned (and am still learning) the most by going through the troubleshooting process. Just my 2 cents.

[Walter Roberson]

    % boundary condition at left side
    gl = [cos(wt); t; t];

What is that intended to designate?

Is the intention that the first row should be assigned the boundary condition cos(wt) (the first entry) and that the last row should be assigned the boundary condition t (the last entry), and that all other rows should be assigned the boundary condition t (the middle entry) ?

Answer 2

[MINATI PATRA]

Dear Walter

Please see Equation (12), (Attached pdf)

2nd line is for gl

3rd line is for gr

[Walter Roberson]

That gives boundary information for three different variables, u, θand ϕ. I see a u variable in your code, but no θ and no ϕ.

[MINATI PATRA]

u, θand ϕ are coded in u, v and w respectively.

Give a last try

Dear Walter

[Walter Roberson]

One of your gl entries should be going into u, one should be going into v, one should be going into w.

[MINATI PATRA]

yes

gl is defined in that way perhaps but how to call ?

[Walter Roberson]

gl(1) into u, gl(2) into v, gl(3) into w.

[Walter Roberson]

It hardly seems worth assigning

    gl = [cos(wt); t; t];
    u(1,N) = gl(1);
    v(1,N) = gl(2);
    w(1,N) = gl(3);
    

when you could just do

    u(1,N) = cos(w*t);
    v(1,N) = t;
    w(1,N) = t;
    

[Adam Danz]

Kudos to Walter for persistence!

[MINATI PATRA]

Its not working

Answer 3

[MINATI PATRA]

u, θand ϕ are coded in u, v and w respectively.