[asifnafi]

Hi,
I am trying to solve this problem numerically (Boundary Layer over a flat plate, see attached).
Matlab code:

nx=2001;ny=51;
xmax=1; ymax=0.25; vis = 1.4e-5;
dx = xmax/(nx-1);dy = ymax / (ny - 1);
%grid
u = nan (nx,ny); v = nan (nx,ny);

%Boundary condition
u(1, : ) = 0; v(1,: ) = 0; u(nx,: ) = 10;

%calc

for p = 1:nx-1
for k = 2:ny-1
u(p+1,k)= u(p,k) + (vis*dx / u(p,k)*(dy^2)) * (u(p,k+1) - 2*u(p,k) ...
+ u(p,k-1)) - (v(p,k)*dx / 2*u(p,k)*dy) * (u(p,k+1) - u(p,k-1));

v(p+1,k) = v(p+1,k-1) - 0.5* dy * ((u(p+1,k) - u(p,k))/dx + ...
(u(p+1,k-1) - u(p,k-1))/dx);

end
end

But i dont get any value of u and v. I would very much appreciate your help

[MangoNrFive]

Well you're dividing by u(p,k) in your formula for u(p+1,k) but your boundary condition is set to zero. As a result you get a NaN because the numerator is zero aswell (0/0 = NaN in Matlab). This NaN gets then propagated through your whole calculation because you always have NaN's from the row before (If you have a NaN in the formula the result will be a NaN aswell). Maybe you have the indices mixed up or something.

[FMDenaro]

You can solve this problem in a simplified equation using the von Mises transformation

[cothiele]

Originally Posted by MangoNrFive

Well you're dividing by u(p,k) in your formula for u(p+1,k) but your boundary condition is set to zero. As a result you get a NaN because the numerator is zero aswell (0/0 = NaN in Matlab). This NaN gets then propagated through your whole calculation because you always have NaN's from the row before (If you have a NaN in the formula the result will be a NaN aswell). Maybe you have the indices mixed up or something.

would you have to change your boundary condition or start at the next step?