Does the Helmholtz equation on a square with constant but nonzero boundary conditions have a closed solution?
(One finds everywhere the solution for a zero boundary condition, but this is useless to me.)
Does the Helmholtz equation on a square with constant but nonzero boundary conditions have a closed solution?
(One finds everywhere the solution for a zero boundary condition, but this is useless to me.)
There can't be such a solution because these boundary conditions violate the equation at the corners. Both second derivatives are zero, hence the Laplacian is zero, but the function value isn't.
[Edit in response to comment:]
If you don't mind singular behaviour at the corners, you can regard the function along the edges as a square wave and expand it in sines that vanish at the corners. For each such sine, say, on the edge $y=0$, there are two linearly independent solutions in the $y$ direction that you can multiply it by to get a solution of the Helmholtz equation; you can choose the coefficients to match the value on the opposite edge $y=L$, and then sum over all the sines in the square wave. The result is constant on the edges in the $x$ direction and zero on the edges in the $y$ direction. Then you can do the same thing for the $y$ direction and add the two solutions to get a solution that's constant on all the edges. But I suspect the result won't look nice at the corners, and I doubt there's a closed form for it. (Of course this is just one particular solution, and you can add any solution for zero boundary conditions to it to get another solution.)
In fact this has solution, but the method is quite advanced:
http://eqworld.ipmnet.ru/en/solutions/lpde/lpde303.pdf#page=2