1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook