# Homework Help: About Separation of Variables for the Laplace Equation

1. Sep 7, 2009

### hectoryx

1. The problem statement, all variables and given/known data

This is a try for the solution of Laplace Equation. We have to calculate the potential distribution in a cylinder coordinate. However, there is a step really bring us trouble. Please go to the detail. You can either read it in the related URL, or in my PDF attachment..
The uncompleted solution is:
http://i1021.photobucket.com/albums/af335/hectoryx/Bessel001.jpg

2. Relevant equations

The method on the book is that:
http://i1021.photobucket.com/albums/af335/hectoryx/Bessel002.jpg

3. The attempt at a solution

I really do not know what the basis of above equation is. Why can we get (2) from (1)? Does anyone give me any advice?
Thanks in advance.

Regards

Hector

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2. Sep 8, 2009

### gabbagabbahey

You do exactly what they say you should do:

$$U_0=\sum_{m=0}^{\infty}A_m\sinh\left(\frac{P_m h}{a}\right)J_0\left(\frac{P_m r}{a}\right)$$

$$\implies\int_0^a U_0 J_0\left(\frac{P_n r}{a}\right)rdr=\int_0^a \left[\sum_{m=0}^{\infty}A_m\sinh\left(\frac{P_m h}{a}\right)J_0\left(\frac{P_m r}{a}\right)\right] J_0\left(\frac{P_n r}{a}\right)rdr=\sum_{m=0}^{\infty}A_m\sinh\left(\frac{P_m h}{a}\right)\left[\int_0^aJ_0\left(\frac{P_m r}{a}\right) J_0\left(\frac{P_n r}{a}\right)rdr\right]$$

What does the orthoganality condition tell you about the integral on the RHS?

3. Sep 8, 2009

### hectoryx

Wo, Thanks for your reply so soon!

I understood your means.

About the orthogonality condition, actually, there is one of the charactrestics of Bessel function, isn't it?

we have:

$$\int _0^{\alpha }J_0\left(\frac{P_mr}{\alpha }\right)J_0\left(\frac{P_nr}{\alpha }\right)rdr=0$$ if $$m\neq n$$

where $$P_m$$ is the solution of Bessel Function $$J_0(x)=0$$

Regards

Hector

4. Sep 8, 2009

### gabbagabbahey

Right, so the only non-zero term in the sum

$$\sum_{m=0}^{\infty}A_m\sinh\left(\frac{P_m h}{a}\right)\left[\int_0^aJ_0\left(\frac{P_m r}{a}\right) J_0\left(\frac{P_n r}{a}\right)rdr\right]$$

will be the $m=n$ term.

$$\implies\int_0^a U_0 J_0\left(\frac{P_m r}{a}\right)rdr=A_m\sinh\left(\frac{P_m h}{a}\right)\int_0^a\left[J_0\left(\frac{P_m r}{a}\right)\right]^2 rdr$$

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