I understand to purpose of the fourier transformation is to transfer a PDE into an ODE. By solving the ODE and then take the inverse transform, I can get the answer.
I have the following PDE ($\phi $ is a function of x and y, $\eta$ is a function of x):
$$ \nabla^2 \phi = 0,\quad 0<y<1, \quad(1)$$
$$ \frac{\partial \phi}{\partial y} = \frac{d \eta}{dx}\quad on \quad y=1, \quad (2)$$
$$\frac{\partial \phi}{\partial x}+\frac{\eta}{F^2}+P(x)=0 \quad on \quad y =1, \quad (3)$$
$$ \frac{\partial \phi}{\partial y}=0 \quad on \quad y=0, \quad (4)$$
$$P(x)=e^{-x^2/l^2}$$
$F$ and $l$ are constant.
Here is what I did: I tried to fourier transform equation (1), thus I get
$$\frac{\partial^2 \tilde\phi}{\partial y}-k^2 \tilde\phi=0$$
which is an ODE that I can solve, I get
$$\tilde\phi(k,y)=Ae^{ky}+Be^{-ky}$$
By plugging in my initial condition (4), I get $$A=-B$$ That means
$$\tilde\phi(k,y)=2A\sinh(ky)$$
Here is the part I got stuck: when I am using my boundary conditions (2) and (3), there are $\eta$ in it. I get two equations:
$$2A\cosh(k)=i\tilde\eta,$$
$$2Aik\sinh(k)+\tilde\eta / F^2+ \tilde P(k)=0$$
which I cannot solve for A
What did I do wrong? And I'm confused because the answer gives me solution of $\eta$ not $\phi$.
The answer is $$\eta =-F^2P+\frac{F^2}{2 \pi}\ \int^{\infty}_{-\infty} \frac{k\tilde P(k)}{k-\tanh(k)/F^2} e^{ikx}\,dk.\ $$
$\tilde P(k)$ is the fourier transform of $P(x)$