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PDE: Laplace's Equation solutions

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that u(x,y) is a solution of Laplace's equation. If [itex]\theta[/itex] is a fixed real number, define the function v(x,y) = u(xcos[itex]\theta[/itex] - ysin[itex]\theta[/itex], xsin[itex]\theta[/itex] + ycos[itex]\theta[/itex]). Show that v(x,y) is a solution of Laplace's equation.

    2. Relevant equations

    Laplace's equation: uxx + uyy = 0.

    Separated solutions


    X''(x) - [itex]\lambda[/itex]X(x)=0.


    Y''(y) - [itex]\lambda[/itex]Y(y)=0.

    Solutions for [itex]\lambda[/itex] > 0


    X(x) = A1ekx + A2e-kx


    Y(y) = A3cosky + A4sinky.

    Solutions for [itex]\lambda[/itex] < 0


    Y(y) = A1ekx + A2e-kx


    X(x) = A3cosky + A4sinky.

    3. The attempt at a solution

    I began by trying to analyze each of the cases ([itex]\lambda[/itex]>0, [itex]\lambda[/itex]<0, [itex]\lambda[/itex]=0) for the solution. But working these out would take forever and I know it isn't the most elegant way of doing it. My thinking is that I can somehow just differentiate the arguments for v(x,y) so I would get a factor of k2sin2[itex]\theta[/itex]cos2[itex]\theta[/itex] for uxx and -k2sin2[itex]\theta[/itex]cos2[itex]\theta[/itex] for uyy. Would this be a valid way of proving the statement to avoid doing all the work? Or is there a better way? Thanks.
     
  2. jcsd
  3. Sep 16, 2011 #2

    Hootenanny

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    That is perfectly fine and the way I would do it. It is also likely to be the way intended by whomever wrote the problem.
     
  4. Jan 7, 2012 #3
    Separated solutions
    X''(x) - λX(x)=0.
    Y''(y) + λY(y)=0.


    there must be "+" , not "-"
     
  5. Jan 7, 2012 #4
    by the way

    therefore other equation will be changed
     
  6. Jan 7, 2012 #5
    Your second strategy would perhaps work best. Since you are given that u satisfies Laplace's equation, it is perhaps a good idea to plug v into Laplace's equation and try to get it in terms of u. Your main tool for that will be the chain rule for scalar fields.
     
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