About Separation of Variables for the Laplace Equation

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Homework Help Overview

The discussion revolves around the application of the separation of variables method to solve the Laplace equation in cylindrical coordinates. Participants are examining the potential distribution and addressing specific steps that are causing confusion, particularly regarding the derivation of certain equations and the properties of Bessel functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the basis for transitioning between equations in the context of the Laplace equation. There is a focus on the orthogonality condition of Bessel functions and its implications for integrals involving these functions.

Discussion Status

Some participants have provided insights into the orthogonality of Bessel functions and its relevance to the problem. There is an ongoing exploration of how these properties affect the solution process, with no explicit consensus reached yet.

Contextual Notes

Participants are working with specific equations and properties related to Bessel functions, and there are references to external resources for further detail. The discussion is framed within the constraints of a homework assignment, which may limit the depth of exploration.

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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

Physics news 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]
 

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