# Fourier COSINE Transform (solving PDE - Laplace Equation)

1. Feb 8, 2015

### juan.

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 Fourier Transform 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 infinity, 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 FTC in LaTeX. Where it says [...] it's FTC from 0 to infinity

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).sen(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:
$$\int_{0}^{\infty}u(x,y).cos(w.y) dy = U(x,w)$$

Last edited: Feb 8, 2015
2. Feb 9, 2015

### Svein

That is the definition of a harmonic function.