# ERROR came as "Index exceeds matrix dimensions".

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MINATI PATRA le 18 Août 2019
Commenté : MINATI PATRA le 24 Août 2019
%% 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
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
madhan ravi le 18 Août 2019
Modifié(e) : madhan ravi le 18 Août 2019

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### Réponses (3)

Adam Danz le 18 Août 2019
Modifié(e) : Adam Danz le 18 Août 2019
f1 = 0; % 12th line of your code; so, f1 only has 1 element
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.
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Adam Danz le 19 Août 2019
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 le 19 Août 2019
% 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) ?

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MINATI PATRA le 20 Août 2019
Dear Walter
Please see Equation (12), (Attached pdf)
2nd line is for gl
3rd line is for gr
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Adam Danz le 23 Août 2019
Kudos to Walter for persistence!
MINATI PATRA le 24 Août 2019
Its not working

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MINATI PATRA le 21 Août 2019
u, θand ϕ are coded in u, v and w respectively.
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