Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find all solutions of u_(xx) + u_(yy) = 0, form u(x,y) = f(x^2 + y^2)

  1. Dec 20, 2009 #1
    1. Find all solutions of u_(xx) + u_(yy) = 0 which have the form u(x,y) = f(x^2 + y^2).




    3. General solution of laplace is u(x,y)= Eexp(ky)Coskx + Fexp(-ky)Coskx + Gexp(ky)Sinkx + Hexp(-ky)Sinkx, where E,F,G,H are arbitrary constants don't see how this can be linked to f(x^2+y^2)?
     
  2. jcsd
  3. Dec 20, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Re: laplace

    I suppose the idea is, that you can write [itex]u_{xx} = \partial^2 u/\partial x^2[/itex] and [itex]u_{yy} = \partial^2 u/\partial y^2[/itex] in terms of the "total" derivative [itex]df(r)/dr[/itex], making it an ordinary differential equation for f.
     
    Last edited: Dec 20, 2009
  4. Dec 20, 2009 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Re: laplace

    Why no attempt at all? Suppose u(x,y)= f(x^2+ y^2). What are u_xx and u_yy in terms of f? If you let z= x^2+ y^2, that reduces to a fairly simple d.e. for f(z).


    No, that is NOT the general solution to Laplace's equation. You could also have an infinite sum of such things- i.e. any function that has a Fourier series could be a solution.
     
  5. Dec 20, 2009 #4
    Re: laplace

    Hi Halls

    Yes I was thinking of doing that initially.

    u_xx=2f'(x^2+y^2)+4(x^2)f''(x^2+y^2)
    u_yy=2f'(x^2+y^2)+4(y^2)f''(x^2+y^2)

    Are you saying the addition of these is the answer!?
     
  6. Dec 20, 2009 #5

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Re: laplace

    What is the new diff.equation you get?
    Hint: write it in terms of z = x^2 + y^2 as Halls suggested.

    The answer is what you get when you solve this equation.

    PS Final hint for solving: consider the derivative of z f'(z)
     
  7. Dec 20, 2009 #6

    HallsofIvy

    User Avatar
    Science Advisor

    Re: laplace

    Well, no, adding those is not "the answer"- but gives, as I said, a simple equation to solve to get the answer.
    Adding those gives [itex]u_{xx}+ u_{yy}= 4f '(x^2+ y^2)+ 4(x^2+ y^2)f"(x^2+ y^2)= 0[/itex]. Setting [itex]z= x^2+ y^2[/itex], as I suggested, gives the ordinary differential equation [itex]4f '(z)+ 4z^2f"(z)= 0[/itex]. Letting [itex]g(z)= f'(z)[/itex] reduces that to a simple first order equation for g. After finding g, integrate to get f.
     
  8. Dec 21, 2009 #7

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Re: laplace

    Except that it is [itex]
    4f'(z)+ 4zf"(z)= 0
    [/itex], not [itex]
    4f'(z)+ 4z^2f"(z)= 0
    [/itex]
     
  9. Dec 21, 2009 #8

    HallsofIvy

    User Avatar
    Science Advisor

    Re: laplace

    You are right. I guess my mind shifted to "[itex]r= \sqrt{x^2+ y^2}[/itex]" while I was writing!
     
  10. Dec 21, 2009 #9
    Re: laplace

    Thanks guys

    The answer appears to be f(x^2+y^2)=Aexp(-(x^2+y^2)) where A is an arbitrary constant

    Thanks for your help
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook