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Particular solution of a PDE

  • Thread starter roldy
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  • #1
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Homework Statement


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]


Homework Equations


Given in question


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]
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


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]


Homework Equations


Given in question


The Attempt at a Solution


The general solution to this is [tex]u(x,y) = f(y^2-x^2)[/tex]
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:
  • #3
232
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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]
 
  • #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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