About Separation of Variables for the Laplace Equation

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hectoryx
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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
 

Attachments

on Phys.org
You do exactly what they say you should do:

[tex]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)[/tex]

[tex]\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][/tex]

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

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

where [tex]P_m[/tex] is the solution of Bessel Function [tex]J_0(x)=0[/tex]

Regards

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

[tex]\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][/tex]

will be the [itex]m=n[/itex] term.

[tex]\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[/tex]