# Particular solution of a PDE

1. Aug 29, 2012

### roldy

1. The problem statement, all variables and given/known data
Find the solution of
$$yu_x + xu_y = (y-x)e^{x-y}$$

that satisfies the auxiliary condition
$$u(x,0) = x^4 + e^x$$

2. Relevant equations
Given in question

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

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

This results in
$$x^4 + e^x = f(-x^2)$$

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:
$$x^4 = f(-x^2)$$
Re-writing this would give
$$(-x^2)^2 = f(-x^2)$$

2. Aug 29, 2012

### LCKurtz

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
3. Aug 29, 2012

### roldy

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
$$u(x,y) = u_p(x,y) + u_c(x,y)$$
$$u(x,y) = e^{x-y} + f(y^2-x^2)$$
$$x^4 + e^x = u(x,0) = e^{x-0} + f(0^2-x^2)$$
$$x^4 + e^x = e^x + f(-x^2)$$
$$x^4 = f(-x^2)$$
$$(-x^2)^2 = f(-x^2)$$

Solution is

$$u(x,y) = (y^2-x^2)^2$$

4. Aug 30, 2012

### LCKurtz

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 $u$.