What is the PDE for the general solution U(x,y) = Phi(x+y) + Psi(x-2y)?

[shortwave]

Homework Statement

Find the PDE for this general solution:
U(x,y) = Phi(x+y) + Psi(x-2y)

Homework Equations

The Attempt at a Solution

I let my xi = x+y and my eta = x-2y and found that both roots are {-1,1/2}. From that I multiplied: (dy/dx - root1)*(dy/dx - root2) to give me the function that will give me my little a,b,c for the quadratic equation. But now I'm stuck and am not sure where to go. Any hints or ideas would be appreciated.

 

[Dick]

I don't know what the specific technique you are trying to use is. But one PDE is clearly [tex]\frac{\partial^2 F}{\partial \xi \partial \eta}=0[/tex], right? Now just change variables to x and y.

 

[shortwave]

Sorry, I should clarify. The PDE that I'm trying to find from the general solution should look like 2U_xx + U_xy - U_yy = 0.

 

[Dick]

Ok. Then you just want to factor [tex]2 \partial_x^2+\partial_x \partial_y-\partial_y^2[/tex] into linear factors, right?

 

[shortwave]

Since this is a hyperbolic pde (discriminant > 0) I would find the characteristics by using a form of the quadratic equation, which would give me two roots, xi and eta. But I'm working backwards; I have the roots but not I need to get back to the hyperbolic pde. Does this help?

 

[Dick]

Since this is a hyperbolic pde (discriminant > 0) I would find the characteristics by using a form of the quadratic equation, which would give me two roots, xi and eta. But I'm working backwards; I have the roots but not I need to get back to the hyperbolic pde. Does this help?

Not exactly. You are basically given xi and eta as functions of x and y, right? I don't see where any quadratic roots come into it. Go back to the PDE in my post #2. All you have to do is change the partial derivatives wrt xi and eta to partial derivatives wrt x and y (use the chain rule) - then you get a PDE in x and y.

 

[shortwave]

Thanks Dick! I managed to get a similar pde.