1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laplace's equation inside a circular annulus

  1. Oct 3, 2013 #1
    1. The problem statement, all variables and given/known data

    1eoviw.png

    2. Relevant equations

    General solution:
    1z23h4z.png

    Fourier series:
    2md14k1.png

    where [itex]r_{1}=a[/itex], [itex]r_{2}=b[/itex], [itex]f_{1}(\theta)=sin(\theta)[/itex], and [itex]f_{2}(\theta)=2sin(\theta)cos(\theta)[/itex].

    3. The attempt at a solution

    By evaluating the Fourier series shown above, I determined that [itex]a_{o}=b_{o}=a_{n}=b_{n}=c_{n}=d_{n}=0[/itex].

    Is this correct?
     
    Last edited: Oct 3, 2013
  2. jcsd
  3. Oct 3, 2013 #2
    Well, no. You could guess that for yourself because that would mean the answer is identically 0, which is not awfully likely for the diff eq is it?

    I looked at ##\int_0^{2\pi}sin\theta cos{n \theta}d\theta ## . The best way to evaluate it is to express both the sin and cos in exponential form: ##sinx = \frac{e^x-e^{-x}}{2i}## and similarly for the cosnx. Multiply it all out and regroup your terms. You'll have a sin and a cosh. The former is periodic in ##\pi## the latter is not.

    Do try again.
     
  4. Oct 4, 2013 #3
    I evaluated the integral you mentioned after expressing both the sin and cos in exponential form as follows:

    263ygd0.png

    Isn't this still periodic in [itex]\pi[/itex]?
     
  5. Oct 4, 2013 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    How about ##2 \sin \theta \cos n \theta = \sin ((n+1)\theta) - \sin ((n-1) \theta)##?
     
  6. Oct 4, 2013 #5
    Isn't that still always going to be zero at all n, where n is an integer and can't equal +-1, after integrating?
     
  7. Oct 4, 2013 #6
    ****
    Here's what I have:

    ##sinx \cdot cos(nx)##= ##\frac{1}{4i}[(e^{ix}-e^{-ix}) \cdot (e^{inx}+e^{-inx})] ## = ##\frac{1}{4i}[(e^{i(n+1)x} - e^{-i(n+1)x}) + (e^{(n+1)x} - e^{-(n+1)x})]##

    Integrating up the last expression and evaluating at 0 and ##2\pi ## gives

    ## \frac{1}{4(n+1)i}(e^{2(n+1)\pi} + e^{-2(n+1)\pi}) ##

    This is not zero.

    Try to be very careful when you do your integration.. Also, note that the product of two periodic functions is periodic, but not necessarily in the period of either.
     
  8. Oct 5, 2013 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Quite so, so maybe something else is wrong. But it's much easier than writing out all those exponentials.
     
  9. Oct 5, 2013 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not right. You should get some n-1 terms.
     
  10. Oct 5, 2013 #9
    You are right, and I'm the one who should be careful about integrals. I do not know by what slip of the mind I lost the i's in the exponents. The other integral is 0 also.

    So temaire is right, they are all 0, unless ##r_1 = r_2## (which isn't really an annulus,but whatever).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Laplace's equation inside a circular annulus
Loading...