about Separation of Variables for the Laplace Equation

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/.../Bessel001.jpg

2. Relevant equations

The method on the book is that:
http://i1021.photobucket.com/albums/.../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

Attached Files

This is a try for the solution of Laplace Equation.pdf (44.6 KB, 1 views)

gabbagabbahey

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?

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:

[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

gabbagabbahey

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]