Does Laplace's Equation Apply to Infinite Boundary Conditions and Fourier Transforms?

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Homework Statement


Consider Laplace's equation uxx + uyy = 0 on the region -inf <= x <= inf, 0 <= y <= 1 subject to the boundary conditions u(x,0) = 0, u(x,1) = f(x), limit as x tends to inf of u(x,y) = 0.
Show that the solution is given by u(x,y) = F-1(sinh(wy)f(hat)/sinh(wy))


Homework Equations





The Attempt at a Solution


I used Fourier transforms in x.
I got u(hat)(w,y) = Aeky + Be-ky
In Fourier space:
u(hat)(w,0) = F(0) = 0
u(hat)(w,1) = f(hat)(w)
But u(hat) is a function of y. My question is how do I apply the 3rd boundary condition (as this is the limit as x(not y) tends to inf) to u(hat)
 
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Is it meant to be limit as y tends to inf not x.
 
squenshl said:
I used Fourier transforms in x.
I got u(hat)(w,y) = Aeky + Be-ky

You mean \hat{u}(\omega,y)=Ae^{\omega y}+Be^{-\omega y}, right? :wink:

My question is how do I apply the 3rd boundary condition (as this is the limit as x(not y) tends to inf) to u(hat)

You already used it. If \lim_{x\to\pm\infty} u(x,y)\neq 0[/itex], its Fourier transform (from x to \omega) might not exist (the integral could diverge).
 
Very true. That is what I meant.
But when did I use this boundary condition?
 
squenshl said:
But when did I use this boundary condition?

When you took the FT of u(x,y), and hence assumed that it existed.
 
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