How to get single curve.

H=10;R=5;Pr=1;Q=H-(R/Pr);
xl=0; xr=5; J = 10; dx = (xr-xl) / J; tf = 01; Nt = 100; dt = tf/Nt; mu = dt/(dx)^2;
% Evaluate the initial conditions
x = xl : dx : xr; % generate the grid point
f = 0;    %%%I.C
u = zeros(J+1,Nt);
for n = 1:Nt
    t = n*dt; % current time
    % B.C at left side
    gl = t;
    % B.C at right side
    gr = 0;
    if n==1 % first time step
        for j=2:J % interior nodes
            u(j,n) = (1+dt*Q)*u(j) + (mu/Pr)*(u(j+1)-2*u(j)+u(j-1));
        end
        u(1,n) = gl;
        u(J+1,n) = gr;
    else
        for j=2:J % interior nodes
            u(j,n)= (1+dt*Q)*u(j,n-1)+ (mu/Pr)*(u(j+1,n-1)-2*u(j,n-1)+u(j-1,n-1));
        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)
surf(x,tt,u'); % 3-D surface plot
hold on
figure(2)
plot(x,u)
hold on

%%%This code is giving many curves in a single figure but I need only one

[Adam Danz]

Also see

https://www.mathworks.com/matlabcentral/answers/476590-error-came-as-index-exceeds-matrix-dimensions

Answer 1

[KALYAN ACHARJYA]

You defined f as scalar

f = 0;    %%%I.C

And in the following line, try to acess as a array.

u(j,n) = (1+dt*Q)*f(j) + (mu/Pr)*(f(j+1)-2*f(j)+f(j-1));
%...................^........................^.....^

When you iterate the for loop till J (10) iterartion, how can it get the value of f, as f=0, defined as earlier.

Please note , suppose f=[5 6 4 5 90 200], the

f(1) means 5

f(2) menas 6 ...so on

Alos please ensure that the loop iterartion does not exceded the vector length, those are tring to ecess with the loop.

[MINATI PATRA]

But my f=0 (initial condition u(x,0)=0) for all x

and also gr=0 (Boundary condition for x tends to infinity)

So how to code?