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

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Posts: 1,577
Registered: 11/8/10
Re: Solving PDE (how to introduce an specific Boundary condition)
Posted: Aug 22, 2013 2:27 AM
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"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*(2-y)*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 x-derivative at the diagonal of the first quadrant is equal to -1
(so the derivative is not given in normal direction, but parallel to the x-axis).
Furthermore, the region where the PDE is to be solved is infinite (x>=0, y>=x).

Best wishes

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