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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
    cramster-equation-20093301451186337.gif

    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?
     
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