Solution for u_x+2xy^2u_y=0 with Initial Condition u(x,0)=\phi (x)

[oyolasigmaz]

PDE--initial condition

Homework Statement

Solve the equation u_x+2xy^2u_y=0 with u(x,0)=\phi (x). Strauss PDE 2nd ed., chapter 1.5, exercise 6.

Homework Equations

The question is under "Well-Posed Problems" section, so this might be about existence, uniqueness, or stability.

The Attempt at a Solution

I can easily solve the first order PDE without the constaint, and get the solution as u(x,y)=f(x^2+\frac{1}{y}). However, when I try to apply the initial condition, since y=0 there, I got stuck. I cannot really interpret this situation, so I need your help.

 

[oyolasigmaz]

Bumping, could not find the answer yet.

 

[Dick]

I THINK it's fairly simple. To determine u(x,0) from your general solution you need to take the limit as y->0. That's either the limit f(z) as z goes to plus or minus infinity (depending on which side of the x-axis you are on). That means you need an f that has that limit at infinity. Once you get that what kind of function must u(x,0) be? Finally, ask yourself about uniqueness. I think that's it.

 

[oyolasigmaz]

Alright, I believe I got it and it seems quite simple now. So the function is not unique and that was what we were looking for I guess, conceptually.

Thanks for your help.