About Separation of Variables for the Laplace Equation

[hectoryx]

Homework Statement

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

Homework Equations

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

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

 

[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?

 

[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

 

[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