1. Limited time only! Sign up for a free 30min personal 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!

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