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Laplaces equation in polar coordinates

  • #1
The function [tex] u(r,\theta) [/tex]

satisfies Laplace's equation in the wedge [tex] 0 \leq r \leq a, 0 \leq \theta \leq \beta [/tex]

with boundary conditions [tex] u(r,0) = u(r,\beta) =0, u_r(a,\theta)=h(\theta) [/tex]. Show that

[tex] u(r,\theta) = \sum_{n=0}^\infty A_nr^{n\pi/\beta}sin(\frac{n\pi\theta}{\beta}) [/tex]

[tex]A_n=a^{1-\frac{n\pi}{\beta}\frac{2}{n\pi}\int_{0}^{\beta}h(\theta)sin\frac{n\pi\theta}{\beta}d\theta [/tex]
 
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Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,805
932
You have posted here before so surely you know the basic rules!

No one is going to do your homework for you and it wouldn't help you if they did! Show us what you have tried so we can see where you went wrong or got stuck.
 
  • #3
I saw the question, realised I was incapable and thought I'd put it up for some hints... I do have a similar problem viz laplace in the unit circle.....


[tex] \nabla^2U=0 [/tex]

Boundary conditions are 1) U=0 at r=0

2) [tex] U(1,\theta)=2cos\theta [/tex]

now I have quoted from the notes that the general solution is

[tex] U(r,\theta) =C_0lnr + D_0 + \sum_{0}^\infty(C_0r+\frac{D_0}{r^n}).(A_ncosn\theta + B_nsinn\theta) [/tex]

now I am told that B.C 1 implies [tex] D_0 = 0=C_n [/tex] for n=0,1,2,3,4....

I am immediately confused why it is necessary to have these two coeffiecient set to zero, surely we could have some situation whereby the three terms could cancel to zero without insisting their coefficients are zero??
 
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  • #4
Dick
Science Advisor
Homework Helper
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618
The sines, cosines and the constant function are linearly independent on the unit circle.
 

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