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Homework Help: Laplace's equation w/ polar coordinates

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data

    The lecture notes say that ∇ = urr + (1/r)ur + (1/r2)uθθ. I'm not sure how this comes about. The notes never explain it.

    2. Relevant equations

    (???)

    3. The attempt at a solution

    No attempts on the actual homework problem until this ∇ thing is cleared up.
     
  2. jcsd
  3. May 6, 2012 #2

    HallsofIvy

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    Science Advisor

    That is the "Laplacian" operator, [itex]\partial^2/\partial x^2+ \partial^2/\partial x^2[/itex] in two dimensional xy- coordinates, in polar coordinates.

    For [itex]\phi[/itex] any differentiable function of x and y, by the chain rule,
    [tex]\frac{\partial \phi}{\partial x}= \frac{\partial \phi}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial \phi}{\partial \theta}\frac{\partial \theta}{\partial x}[/tex]

    Since [itex]r= (x^2+ y^2)^{1/2}[/itex], [itex]\frac{\partial r}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= (1/2)2rcos(\theta)/r= cos(\theta)[/itex]. Since [itex]\theta= artctan(y/x)[/itex],
    [tex]\frac{\partial \theta}{\partial x}= (1./(1+ y^2/x^2))(-(y/x^2)= -y/(x^2+ y^2)= -rsin(\theta)/r^2= -sin(\theta)/r[/tex].

    So
    [tex]\frac{\partial \phi}{\partial x}= cos(\theta)\frac{\partial \phi}{\partial r}- \frac{sin(\theta)}{r} \frac{\partial \phi}{\partial \theta}[/tex]

    Then
    [tex]\frac{\partial^2\phi}{\partial x^2}= cos(\theta)\frac{\partial}{\partial r}\left(cos(\theta)\frac{\partial \phi}{\partial r}- \frac{sin(\theta)}{r} \frac{\partial \phi}{\partial \theta}\right)[/tex][tex]+ \frac{sin(\theta)}{r}\left(cos(\theta)\frac{\partial \phi}{\partial r}- \frac{sin(\theta)}{r} \frac{\partial \phi}{\partial \theta}\right)[/tex]

    Complete that then do the same for [itex]\partial^2\phi/\partial \theta^2[/itex].
     
    Last edited by a moderator: May 6, 2012
  4. May 6, 2012 #3
    Bingo. Got it.
     
  5. May 6, 2012 #4
    HallsofIvy, I have another question if you have the time. I'm trying to solve

    2u=0 ​

    with the boundary conditions

    u(x,0)=0, u(0,y)=0, u(x,1)=x, u(1,y)=y.​

    I've found that u(x,0)=0 and u(0,y)=0 imply

    u(x,y)= D * sinh(√kx) * sin(√ky)​

    where k was my separation constant.

    I then did

    u(x,1)=x → uxx(x,1)=0=D * sinh(√kx) * sin(√ky)​

    to conclude that √k = n∏ and that our solution is of the form

    Ʃn≥1 Dn * sinh(n∏x) * sin(n∏y).​

    Finally, I solved for Dn by

    u(1,y)=y=Ʃn≥1 Dn * sinh(n∏) * sin(n∏y)
    → Dn * sinh(n∏) = (2/∏)∫[0,∏] y * sin(n∏y) dy.​

    Since that gives me a very messy solution for u, I'm assuming it's wrong. Thoughts?
     
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