1. Not finding help here? Sign up for a free 30min 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 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

    User Avatar
    Staff Emeritus
    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: 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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laplace's equation w/ polar coordinates
Loading...