Boundary condition

You have to set boundary conditions on all the boundaries of your domain.

Quote:
Thank you for your reply. I agree with your first point. I think I don't have to specify the y boundary condition for B. The second equation will evolve by itself as you point out. But, is coupling a problem? What about I guess I still don't have to specify the y boundary condition in B, is that? Because there is no ydirection propagation in the second equation. 
Quote:
But this second equation is now different ...it contains A and is coupled to the first. In any case it is valid for any y... I suggest to classify the system.... Then, at a steady state, if A <> 0 you see that B must be a function only of y, then from the first equation you have dA/dx = f(y) > A(x,y) = f(y) (xx0) + A0 
Quote:
I would say that I agree with Aeronautics. When you will discretize the first equation for A, you will need to know B at the boundary since you have the operator dB/dy involved in the equation. However if you use left sided FD scheme instead of central scheme you could avoid to specify B on the boundary. Indeed you don't have y propagation of B in the B equation, but you have it on the A equation. So everything will depend on the scheme you use to discretize dB/dy in the A equation. 
All times are GMT 4. The time now is 09:27. 