How to use Fourier Transform with non-trivial boundary conditions such as in potential flow around a plate? I'd specifically like to be able to solve this PDE with boundary conditions corresponding to flow around a line (plate cross-section), otherwise known as flow-tangency, with integral transforms.
$$(1) \quad {{\partial^2 \psi} \over {\partial x^2}}+{{\partial^2 \psi} \over {\partial y^2}}=0$$
Where the boundary condition satisfies that the streamline is constant at the boundary of this line.
$$(2) \quad \psi(x,0)=\mu \ , \quad |x| \lt r$$
Fourier Transform (1) with respect to x, and simplify using the derivative rules. We get.
$$(3) \quad {{d^2 \Psi} \over {d y^2}}-\omega^2 \cdot \Psi=0$$
Solve (3) up to additive constants to get...
$$(4) \quad \Psi=A \cdot e^{-|\omega| \cdot y}+B \cdot e^{|\omega| \cdot y}$$
I've taken the absolute value because I feel as though the square root of something squared should be positive. However, this could very well be the problem with my derivation. Feedback would be welcome. I could derive the rest of my wrong result, but at this point I think the error in my reasoning has become clear enough. How do I fix it?
If it helps, I eliminate the right side of (4) to avoid an infinite potential, apply an IFT, then apply the convolution theorem and end up with...
$$\psi(x,y)={1 \over \pi} \cdot \int_{-\infty}^{\infty}\psi(x,0) \cdot {y \over {(x-\omega)^2+y^2}} \ d\omega$$
My Thoughts: I think that perhaps the boundary conditions need to be expanded to enforce a constant velocity as the distance tends to infinity. I'm also pretty sure that putting absolute values on the solution to the ODE was a mistake. Edit: It seems as though the problem I solve is correct, but isn't the problem I wanted to solve. Perhaps someone could show me where my boundary conditions went wrong?
Questions: I'd like some help deriving flow around a plate cross-section using Integral Transforms. I'm not interested in using other methods that don't involve a Fourier Transform or at least an integral transform of one kind or another. Basically the point is to use integral transforms not separation by variables or some other technique. Below is what I am trying to describe, I've just flipped the axis in the above "derivation"...
Also, how does one generally use an integral transform if the boundary conditions involve 0 around a body and a constant at infinity or other region? For instance, how do you solve the particle in a box problem if $\Psi$ must be 0 beyond the boundary's of the box? As another question, how do you use the Fourier transform in polar coordinates? Is it legal to change the 0 to $2\pi$ angular region to an infinite bound? How about changing the semi-infinite region to an infinite region? For instance, flow around a cylinder would combine 0 valued boundary conditions, and all the above problems.
I'm having a hard time asking this question, but I've tried my best. I'll summarize. How do you use the Fourier transform for boundary conditions and coordinate systems similar to the above, such as flow tangency?

 A: Trying this a little differently, using Laplace Transforms (since I'm used to those).  I'm using the graph's coordinate system, so this is in terms of y, not x (since the graph helps).
$$\mathcal L(\psi(x,y)) = \Psi(y,s) $$
$$\frac{d^2\Psi}{dy^2} + s^2\Psi = 0$$
$$ \Psi(y,s) = C_1(s)\sin(sy) + C_2(s)\cos(sy) $$
We know immediately that, in terms of y, $\Psi(-y,s) = \Psi(y,s)$, so that means $C_1(s) = 0$  This solves the "absolute value" problem from earlier.
This leaves the boundary conditions: $\lim_{y \to \infty} \psi(x,y) = 0$, which requires a negative exponential in $C_2(s)$, and $\lim_{x \to \infty} \psi(x,y) = 0$  Now the tricky boundary condition of when x = 0, for some |y| < r, the function goes to a constant.  This is best handled with the derivative, not a constant:
$$\frac{d\psi(x,y<a)}{dx} = 0 $$
Which means breaking up the discontinuity into different plates.  In fact, I'd use four plates, as I've shown here:

This then turns into a system of partial differential equations, each one of the functions convoluting the output.  Ultimately, it's a tricky problem and that's when FEA usually gets involved.
