Fourier COSINE Transform (solving PDE - Laplace Equation) I'm trying to solve Laplace equation using Fourier Cosine Transform (I have to use that), but I don't know if I'm doing everything OK.
NOTE: $U(\cdot)$ is the Fourier transform of $u(\cdot)$
This are the equations (Laplace, boundary, etc.):
$$u_{xx}+u_{yy} = 0 \text{ with } y>0,\  0<x<a$$
$$u_y(x,0) = u(0, y) =0$$
$$u(a,y) = g(y)$$
$$|u(x,y)|<M$$
I used "Transform Methods for Solving PDE", from G. Duffy and this is what I'm doing (maybe you have a better way):
Now, since $x$ is between from $0$ to $a$ and $y$ is between $0$ and $\infty$, I use the definition of Fourier Cosine Transform and:
$$\int_{0}^{\infty} u_{xx} \cos{(w y)} \ dy + \int_{0}^{\infty} u_{yy} \cos{(w y)} \ dy = 0$$
where:
$$\int_{0}^{\infty} u_{xx}.cos(w.y) dy = U_{yy}(x,w)$$
$$\int_{0}^{\infty} u_{yy}.cos(w.y) dy = [u_y(x,y).cos(w.y)] - w.[u(x,y).sin(w.y)] - w.\int_{0}^{\infty}u(x,y).cos(w.y) dy$$
Note: I don't know how to write the Barrow Rule in LaTeX. Where it says [...] it's Barrow from $0$ to $\infty$
Now, I know that:
$[u_y(x,y).cos(w.y)] = 0$ because of the conditions: $u_y(x,0) =0$ and $|u(x,y)|<M$ (is that ok?)
But now, I want to solve this: $[u(x,y).sin(w.y)]$ and I don't know why, because I don't have any condition for $u(x,0)$ (but I have a condition for $u(0,y)$.
What's wrong? I searched everywhere but I couldn't find anything that helps me. Thanks!!!
Note 2: using Fourier Cosine Transform definition, I know that:
$$\int_{0}^{\infty}u(x,y).cos(w.y) dy = U(x,w).$$ That's correct, isn't it?
 A: It usually helps to follow Fourier's prescription rather than jump to the solution. For Laplace's equation, the separated solutions $X(x)Y(y)$ satisfy
$$
                 X''Y+XY'' = 0, \\
                 \frac{X''}{X} = -\frac{Y''}{Y} = \lambda,
$$
where $\lambda$ is a separation parameter. So the Fourier ODEs are
$$
\begin{align}
                               X''=\lambda X, &\;\;\;\;\; Y''=-\lambda Y \\
                                0 < x < a, & \;\;\;\;\;0 < y < \infty \\
                                  X(0) = 0, & \;\;\;\;\; Y'(0) = 0.
\end{align}
$$
Fourier always assumed the separated solutions were bounded on any unbounded domain. The conditions for $X$ do not determine parameters $\lambda$; however the boundedness of $Y$ in conjunction with $Y'(0)=0$ forces $\lambda \ge 0$. (If $\lambda < 0$ then $Y(y)=\cosh(\sqrt{\lambda}y)$ is unbounded.) Therefore, $\lambda \ge 0$ is assumed, and the separated solutions for $\lambda=\mu^{2}$ satisfying the above conditions are
$$
        X_{\mu}(x)Y_{\mu}(y) = C(\mu)\sinh(\mu x)\cos(\mu y).
$$
A sum of these must be an integral sum:
$$
                u(x,y) = \int_{0}^{\infty}C(\mu)\sinh(\mu x)\cos(\mu y)dy.
$$
The coefficient function $C(\mu)$ is the only missing component, and must be chosen so that
$$
                 g(y) = u(a,y) = \int_{0}^{\infty}C(\mu)\sinh(\mu a)\cos(\mu y)dy.
$$
Writing $g$ as a Fourier cosine transform gives
$$
                      g(y) = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}\cos(\mu u)g(u)du\right)\cos(\mu y)d\mu,
$$
which leads to
$$
                 C(\mu)\sinh(\mu a) = \frac{2}{\pi}\int_{0}^{\infty}\cos(\mu u)g(u)du \\
         C(\mu) = \frac{2}{\pi\sinh(\mu a)}\int_{0}^{\infty}\cos(\mu u)g(u)du.
$$
The final Fourier solution is
$$
       u(x,y) = \frac{2}{\pi}\int_{0}^{\infty}\left(\frac{1}{\sinh(\mu a)}
                  \int_{0}^{\infty}\cos(\mu u)g(u)du\right)\sinh(\mu x) \cos(\mu y)d\mu.
$$
Sanity Check: Note that the expression in parentheses is a function of $\mu$ only--it is $C(\mu)$. Visual inspection of the final, proposed solution shows that $u(a,y)$ is the inverse Fourier cosine transform of the Fourier cosine transform of $g$. So $u(a,y)=g(y)$ under suitable smoothness conditions on $g$. Cleary $u(0,y)=0$. Under suitable assumptions to ensure the convergence of the differentiated integral expression, $u_{y}(x,0)=0$ follows. All of the required conditions check, and this does give a solution of Laplace's equation given sufficient convergence of the different derived series.
