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.):

$$u_{xx}+u_{yy} = 0 \text{ with } y>0,\ 0<x<a$$

$$u_y(x,0) = u(0, y) =0$$

$$u(a,y) = g(y)$$


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:

$$\int_{0}^{\infty} u_{xx} \cos{(w y)} \ dy + \int_{0}^{\infty} u_{yy} \cos{(w y)} \ dy = 0$$


$$\int_{0}^{\infty} u_{xx}.cos(w.y) dy = U_{yy}(x,w)$$

$$\int_{0}^{\infty} u_{yy}.cos(w.y) dy = [u_y(x,y).cos(w.y)] - w.[u(x,y).sin(w.y)] - w.\int_{0}^{\infty}u(x,y).cos(w.y) dy$$

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?


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.


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