Poissons Equation Problem

[hc91]

Homework Statement

First part of the question is to obtain a Fourier series of the function f(theta)= 1 0=<theta<pi
-1 pi=<theta<2pi
Then to find the solution of Poissons equation inside the unit disk r=<1
∇^2(P) = (1/r)d/dr(r(dP/dr))+ (1/r^2)d^2P/d^theta = f(theta) with Boundary Condition that when r=1, P=0. Given HINT that general solution of (r^2)R'' + rR' -(n^2)R = r^2 is R= a(r^n) + b(r^-n) - (r^2)/(n^2-4)

Then required to show that the integral of f(theta)*P dA =
(-4/pi) sum over n odd (1/(n^2)((n+2)^2))

Homework Equations

The Attempt at a Solution

I think I have got the Fourier series correct as f = sum over n odd ((4/n*pi)sin(n*theta))

I have then tried separation of variables letting P=R(r)THETA(theta), and got
R''(THETA) + (1/r)R'(THETA) + (1/r^2)R(THETA)'' = f(theta)

and then said as r=1 when P=0, R(r)=0 at r=1, so THETA''=f(theta) and so by integrating, THETA = (-1/n^2)THETA'', which can be subbed back into get the form for the HINT. This means I can construct P =R(r)THETA(theta) but then when I go ahead and integrate this, I a) cannot get rid of the a_n, b_n that occur from summing the HINT, as only one BC
b) cannot get t the right form for the final integral, but I do have a lot of powers of n, n-2, n+2 so feel I am on the right sort of lines.

Sorry this is so messy, any help would be much appreciated.

 

[mighty2000]

Hello,

Thank you for your post. It seems like you are on the right track with your approach to finding the solution for Poisson's equation. However, there are a few things that need to be clarified.

Firstly, for the Fourier series, the coefficients should be (2/n*pi)sin(n*theta) instead of (4/n*pi)sin(n*theta). Also, keep in mind that the Fourier series will have different forms for the even and odd terms, so you will need to split the sum into two parts accordingly.

For the solution of Poisson's equation, you are correct in using separation of variables. However, the boundary condition at r=1 should be P=1 instead of P=0. This is because the function f(theta) has a value of 1 for 0<=theta<=pi and a value of -1 for pi<=theta<=2pi. So, when r=1, the value of P should be 1. This will give you the correct form for the HINT and the final solution for P.

As for the final integral, you will need to use the Fourier series that you obtained for f(theta) and substitute it into the integral. Then, you can use the orthogonality property of sine functions to simplify the integral and eventually arrive at the desired form.

I hope this helps clarify some of the steps for your solution. Good luck with your work!