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Homework Help: Particular solution of a PDE

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the solution of
    [tex]yu_x + xu_y = (y-x)e^{x-y}[/tex]

    that satisfies the auxiliary condition
    [tex]u(x,0) = x^4 + e^x[/tex]


    2. Relevant equations
    Given in question


    3. The attempt at a solution
    The general solution to this is [tex]u(x,y) = f(y^2-x^2)[/tex]

    Applying the auxiliary condition I get
    [tex]x^4 + e^x = u(x,0) = f(0^2-x^2)[/tex]

    This results in
    [tex]x^4 + e^x = f(-x^2)[/tex]

    This is where I'm getting stuck. I need to "make" something on the left side that resembles what is shown in the parenthesis.

    For example:
    [tex]x^4 = f(-x^2)[/tex]
    Re-writing this would give
    [tex](-x^2)^2 = f(-x^2)[/tex]
     
  2. jcsd
  3. Aug 29, 2012 #2

    LCKurtz

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    That is apparently the general solution to the homogeneous equation. Let's call it ##u_c(x,y)##

    I'm not a PDE expert, but I think what you need here is to find a particular solution ##u_p(x,y)## that solves the NH that you can add to your ##u_c(x,y)##. After that you can apply your boundary conditions to the general solution ##u(x,y)=u_c(x,y)+u_p(x,y)##. In ordinary DE you have techniques like undetermined coefficients and variation of parameters to help you with such tasks. You must have some corresponding techniques for PDE's, eh?

    [Edit, Added later]: It isn't difficult to find a ##u_p(x,y)## by inspection.
     
    Last edited: Aug 29, 2012
  4. Aug 29, 2012 #3
    I think I figured out the answer and I think what threw me off was that the professor didn't have any work regarding the particular solution.

    Full solution
    [tex]u(x,y) = u_p(x,y) + u_c(x,y)[/tex]
    [tex]u(x,y) = e^{x-y} + f(y^2-x^2)[/tex]
    [tex]x^4 + e^x = u(x,0) = e^{x-0} + f(0^2-x^2)[/tex]
    [tex]x^4 + e^x = e^x + f(-x^2)[/tex]
    [tex]x^4 = f(-x^2)[/tex]
    [tex](-x^2)^2 = f(-x^2)[/tex]

    Solution is

    [tex]u(x,y) = (y^2-x^2)^2[/tex]
     
  5. Aug 30, 2012 #4

    LCKurtz

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    But that is a solution to the homogeneous equation, not the non-homogeneous equation you are given. You are close. Read my post #2 again to see the proper form for ##u##.
     
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