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Laplace's equation

  1. Jan 14, 2012 #1
    1. The problem statement, all variables and given/known data
    f = r^-n-1 * cos(n+1) θ satisfies laplace's equation.
    r^2 [itex]\partial^2 f / \partial r^2[/itex] + r[itex] \partial f / \partial r[/itex] + [itex]\partial^2 f / \partial θ^2[/itex] = 0

    2. Relevant equations
    P.D.E

    3. The attempt at a solution
    [itex]\partial f / \partial r[/itex] = nr^n-1 * sin (nθ)
    [itex]\partial f/ \partial θ [/itex] = ncos(nθ)*r^n
    Are these derivatives right ?
     
    Last edited: Jan 14, 2012
  2. jcsd
  3. Jan 14, 2012 #2
    What happened to partial differential equation?
     
  4. Jan 14, 2012 #3
    Sorry , I have fixed my latex commands.Refresh the webpage.
     
  5. Jan 14, 2012 #4
    The function is :


    f = [itex]r^{-n-1}[/itex] * cos (n+1)[itex]θ[/itex]

    Now I have been trying to figure out what bits are exactly related to variable 'r' and 'theta'...
    [itex]\frac {\partial f}{\partial θ}[/itex] = {sin (n+1)θ + cos (n+1) } * r^-n-1 or
    sin (n+1) * r^-n-1.
    Can someone clarify this for me . Thanks!
     
  6. Jan 14, 2012 #5
    If [itex]f = cos[(n+1)\theta] r^{-(n+1)}[/itex]

    then [tex]\frac{\partial f}{\partial r} = -(n+1)cos[(n+1)\theta] r^{-(n+2)}[/tex]
    and [tex]\frac{\partial f}{\partial \theta} = -(n+1)sin[(n+1)\theta] r^{-(n+1)}[/tex]
     
    Last edited: Jan 14, 2012
  7. Jan 14, 2012 #6
    Thanks for the help. This makes sense.. I was being sloppy, I didn't realize the fact that differentiating sin nθ is no different than sin (n+1)θ ah....
     
  8. Jan 14, 2012 #7
    Ha yeah, no problem.
     
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