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Topic:
Solving PDE (how to introduce an specific Boundary condition)
Replies:
3
Last Post:
Aug 22, 2013 2:27 AM



Torsten
Posts:
1,717
Registered:
11/8/10


Re: Solving PDE (how to introduce an specific Boundary condition)
Posted:
Aug 22, 2013 2:27 AM


"Bill Greene" wrote in message <kv2laa$j55$1@newscl01ah.mathworks.com>... > Hi, > > > I have some troubles when trying to solve for a second order PDE. Would be wonderful if you can help me. > > > > The PDE I want to solve is the following: > > > > uxx + (2*y^2)*uyy + (1+0.3*x*ux+0.5*(2y)*uy) = 0.3*u > > > > defined for x>=0 and y>=0 > > > > and with boundaries: u(0,y)=0 and ux(y,y)=1 > > > > It looks like it might be possible to solve this with PDE Toolbox. But, I am > unclear what this BC: > > ux(y,y)=1 > > means? Is that a typo? > > Bill
It means that the xderivative at the diagonal of the first quadrant is equal to 1 (so the derivative is not given in normal direction, but parallel to the xaxis). Furthermore, the region where the PDE is to be solved is infinite (x>=0, y>=x).
Best wishes Torsten.



