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PDE-initial condition

  • #1
PDE--initial condition

Homework Statement


Solve the equation [tex]u_x+2xy^2u_y=0[/tex] with [tex]u(x,0)=\phi (x)[/tex]. 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 [tex]u(x,y)=f(x^2+\frac{1}{y})[/tex]. 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.
 

Answers and Replies

  • #2


Bumping, could not find the answer yet.
 
  • #3
Dick
Science Advisor
Homework Helper
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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.
 
  • #4


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.
 

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