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!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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




Similar Discussions: Particular solution of a PDE
  1. Particular Solution (Replies: 17)

  2. Particular Solution (Replies: 14)

  3. Particular solutions (Replies: 1)

Loading...