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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.)

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up vote 0 down vote accepted

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.)

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But doesn't the differential equation have to hold only in the interior? Wouldn't your statment just mean that the second derivative cannot be continous at the corners? – Arnold Neumaier Mar 21 '12 at 20:21
Thanks. This construction is good enough for me. – Arnold Neumaier Mar 22 '12 at 17:07
Why the downvote? – joriki Oct 13 '12 at 4:18
Because you overlook so that you get the wrong conclusion that has no solution. – doraemonpaul Oct 15 '12 at 12:41

In fact this has solution, but the method is quite advanced:

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