2
$\begingroup$

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 (if I'm doing everything OK, the exercise is wrong and I don't think so).

NOTE: U(..) is the transform of Fourier of u(..)

This are the equations (Laplace, boundary, etc.):

$u_{xx}+u_{yy} = 0$, 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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.