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

[squenshl]

Homework Statement

Consider Laplace's equation u_xx + u_yy = 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) = Ae^ky + 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)

 

[squenshl]

Is it meant to be limit as y tends to inf not x.

 

[gabbagabbahey]

I used Fourier transforms in x.
I got u(hat)(w,y) = Ae^ky + Be^-ky

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

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 [tex]\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).[/tex]

 

[squenshl]

Very true. That is what I meant.
But when did I use this boundary condition?

 

[gabbagabbahey]

But when did I use this boundary condition?

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