searching for a better approach for solving a 3-D matrix

Hey guys,

I need to solve the following equation.

u(i,j,k+1)=(2*dt/3).*(u(i,j,k).*(u(i+1,j,k)-u(i-1,j,k))./(2*dr)+u(i,j,k-1).*(u(i,j+1,k-1)-u(i,j-1,k-1))./(2*dz));

where dt and dr are some known constants.

The approach I can think of is given below:

for i=2:n, for j=2:n, for k=2:n 

u(i,j,k+1)=(2*dt/3).*(u(i,j,k).*(u(i+1,j,k)-u(i-1,j,k))./(2*dr)+u(i,j,k-1).*(u(i,j+1,k-1)-u(i,j-1,k-1))./(2*dz)); 

end, end, end

Is there any better approach to solve the same equation?

Please help me with this problem and suggest me a better approach, for my approach is not good enough for very long equations including 3 subscripted independent variables i, j, k where each denote r, z and time(t) components respectively, working in cylindrical coordinate system.

Thank you.

Udit Srivastava.

Answer 1

[KSSV]

How about?

i = 2:n ;
j = 2:n ;
for k=1:n
    u(i,j,k+1)=(2*dt/3).*(u(i,j,k).*(u(i+1,j,k)-u(i-1,j,k))./(2*dr)+u(i,j,k-1).*(u(i,j+1,k-1)-u(i,j-1,k-1))./(2*dz)); 
end

[Udit Srivastava]

Sir, I actually want to remove the 'k' term from my equation and adjust it somewhere else so that the equation works like a 2-D matrix, making it look less cumbersome, like putting an outer loop on 'k' or something like that. Is it anyway possible?

[KSSV]

It is possible...you have to save only two steps of k i.e u(i,j,1) and u(i,j,2). Every time you have to save the result into a file.