1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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 equation in polars

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Laplace's equation in 2 dimensions may be written, using plane polar coordinates r, θ, as

    Find all separable solutions of this equation which have the form V(r, θ)=R(r)S(θ), which are single valued for all r, θ. What property of the equation makes any linear combination of such solutions also a solution?

    2. Relevant equations

    3. The attempt at a solution
    i get how to separate the variables and i am left with to expressions equaling a constant. but from there it doesnt make sense to me how to end up with 2 solutions
  2. jcsd
  3. Apr 1, 2009 #2
    assume a seperable solution then sub it in and do the derivatives. then multiply thorugh by [itex]\frac{r^2}{RS}[/itex]

    if you make your constant [itex]\lambda[/itex] you should get a nice solution for [itex]\lambda>0[/itex] hint : let [itex]R=r^m[/itex] with m a constant. the others will be less pretty unless you have simplifying boundary conditions.
  4. Apr 1, 2009 #3
    ok, got it thanks. the second part of the question which i didnt include before is:
    A continuous potential V(r, θ) satisfies Laplace's equation everywhere except on the concentric circles r=a, r=b where b>a.
    (i) Given that V(r=a, θ)=Vo(1+cos θ), and that V is finite as r-->infinity , find V in the region r less than or equal to a
    (ii) given, separately, that V(r=0, θ)=2Vo and V is finite as r--> ∞, find V for r≥b

    for (i), i dont really know how to select solutions for that V. my solutions arecombinations of sinmtheta cosmtheta r^m and r^-m. is there a way to do this through an expansion?
  5. Apr 1, 2009 #4
    ok i think that the periodicity of the cos term means that the solutions for [itex]\lambda<0[/itex] are useless here and the [itex]\lambda=0[/itex] solutions will be trivial after you apply the b.c. that it must be finit as r goes to infinity.

    so for the [itex]\lambda>0[/itex] solutions i have

    [itex]V=(Ar^n+Br^{-n})(C \cos{n \theta} + D \sin{n \theta})[/itex]

    for this to be finite at infinity what can you say about the coefficient of [itex]r^n[/itex]
  6. Apr 1, 2009 #5
    then for finding V in the region r<a, your right to think of expansions, consider

    [itex]\frac{1}{2}c_0 + \sum_{n=0}^{\infty} r^{-m} (c_n \cos{n \theta} + d_n \sin{n \theta})[/itex]

    equate this to the boundary condition you have and you should be able to work out the relative non-zero coefficients.

    my answer : [itex]V(r<a)=2V_0+\frac{aV_0}{r}[/itex]
  7. Apr 1, 2009 #6
    Well Br^-n goes to 0 so A should be a constant or 1 so that the whole expression doesn't go to zero?
  8. Apr 1, 2009 #7
    zero is finite though. so that's allowed. what is not allowed is have [itex]\infty^n[/itex]. so what can you say about A and B now?
  9. Apr 1, 2009 #8
    ok so should i then just throw out the solutions r^n and r^-n since they don't meet the initial condition?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook