Finding a Particular Solution for a Non-Homogeneous PDE

Then use the boundary conditions to determine the f(y^2-x^2) term.In summary, the solution to the non-homogeneous equation is u(x,y) = e^{x-y} + (y^2-x^2)^2. To find a particular solution, we can use the boundary conditions to determine the f(y^2-x^2) term. The final solution is u(x,y) = e^{x-y} + (y^2-x^2)^2.
  • #1
roldy
237
2

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]
 
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  • #2
roldy said:

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
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
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##.
 

FAQ: Finding a Particular Solution for a Non-Homogeneous PDE

What is a particular solution of a PDE?

A particular solution of a PDE (partial differential equation) is a specific function that satisfies the given PDE and any associated boundary or initial conditions. It is a solution that is unique and fully satisfies the PDE, unlike a general solution which may contain arbitrary constants.

How do you find a particular solution of a PDE?

To find a particular solution of a PDE, one must first determine the type of PDE (e.g. elliptic, parabolic, hyperbolic) and then use appropriate techniques such as separation of variables, method of characteristics, or integral transforms to solve the PDE and obtain the particular solution.

Can a particular solution of a PDE be unique?

Yes, a particular solution of a PDE can be unique. This is because it must satisfy the given PDE and any associated boundary or initial conditions, leaving no room for arbitrary constants. However, it is possible for a particular solution to not exist if the PDE and boundary/initial conditions are not compatible.

What is the difference between a general solution and a particular solution of a PDE?

A general solution of a PDE is a solution that contains arbitrary constants and can satisfy the PDE, but may not satisfy any additional conditions. A particular solution, on the other hand, is a specific solution that fully satisfies the PDE and any associated conditions. It is unique and does not contain any arbitrary constants.

Can a particular solution of a PDE change over time?

It depends on the type of PDE and the nature of the problem being modeled. In some cases, the particular solution of a PDE may remain constant over time, while in others it may change as the problem evolves. For example, in heat conduction problems, the particular solution may change as heat is transferred over time.

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