
Extra 'boundary' conditions
Dear Dr. Sewell,
I wonder if you could tell me how to introduce extra boundary conditions in pde2d.
Concretely I have the following equation to solve
Uxx + Uyy = F
with boundary conditions
Ux( x=0, y ) = 0
Uy( x, y=0 ) = 0
U ( x=infinity, y ) = 0
U ( x, y=infinity ) = 0
but also,
U(0, 1) = 1.
I have not found a way to introduce this last extra condition into pde2d.
The solution is seek in a rectangular domain as shown:
U=0

 
Ux=0  
 
+ >U=1  U=0
 
 

Uy=0
Thanks in advance,
Lazaro Calderin
calderin@physics.queensu.ca

Lazaro,
I think you need to solve 2 separate problems, on different rectangles, each with Uxx+Uyy=F, the first on with BC:
Ux = 0 at x=0
U = 0 at x=B (B = "infinity")
Uy = 0 at y=0
U = 1 at y=1
and the other with
Ux = 0 at x=0
U = 0 at x=B
U = 1 at y=1
U = 0 at y=B
I don't see any other way to do this, since the 4 original BCs should
result in a unique solution, so adding U=1 at y=1 will contradict that
solution. This way the solution is not smooth at y=1, but will satisfy
Uxx+Uyy=F everywhere else.
Does that sound reasonable?
Granville

Hi Granville,
Thanks for your quick answer.
What you said does sound reasonable .. I should have mentioned that the equation is nonlinear so I guess that uniqueness of the solution is not guaranteed by the boundary conditions. Linear superposition should not be valid neither. But what you have suggested may work any way.
Let me take a step back and explain a little bit more. The full equation is
Uxx + Uyy = 4 (U  1)/x d(x) d(y1) + F(x,y,U,Ux,Uy) where d=Dirac delta function.
with the boundary condition Ux(0,y)=0, U(B,y)=0, Uy(x,0)=0, U(x,B)=0.
So, in principle we thought of two options:
1) Solve that equation just off the x=0 and y=0 boundaries using the extra condition U(0,1)=1 (which must be obey). In that case de delta functions are zero and we get the equation with the conditions as stated in the first email. This solution has been used many years ago by other people using pde2d.
2) Keep the delta functions, that is .. solve the problem as stated above in this email. But it seems that pde2d can not handle coefficients of the delta functions which are a function of U and x.
Thanks again,
Lazaro
calderin@physics.queensu.ca

Lazaro,
Oh, ignore the email I just sent you directly, I guess it is true that
d(x)*d(y1) = d(x,y1), so if the coefficient of this were a constant,
PDE2D could handle it. And even if it were a function of x,y, you'd just
have g(x,y)*d(x,y1) = g(0,1)*d(x,y1) and PDE2D could handle it.
But with the U as a coefficient, I guess not; and anyway, your "g(x,y)"
has x in the denominator, so g(0,1) = infinity!
Granville
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