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Homework Help: Linear 1st order PDE (boundary conditions)

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve the equation [tex]u_{x}+2xy^{2}u_{y}=0[/tex] with [tex]u(x,0)=\phi(x)[/tex]

    2. Relevant equations
    Implicit function theorem
    [tex]\frac{dy}{dx}=-\frac{\partial u/\partial x}{\partial u/\partial y}[/tex]


    3. The attempt at a solution
    [tex]-\frac{u_x}{u_y}=\frac{dy}{dx}=2xy^2[/tex]
    Separating variables
    [tex]\frac{dy}{y^2}=2xdx[/tex]
    [tex]\frac{-1}{y}=x^2+c[/tex]
    [tex]C=x^2+\frac{1}{y}[/tex]
    So [tex]u(x,y)=f(x^2+\frac{1}{y})[/tex]
    The boundary condition is given as evaluating at [tex]y=0[/tex] which doesn't seem to make sense. Any thoughts? Thanks!
     
    Last edited: Feb 24, 2010
  2. jcsd
  3. Feb 24, 2010 #2

    Dick

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    Wouldn't be boundary condition mean you need to pick an f such that f(u)->0 as u->infinity?
     
  4. Feb 24, 2010 #3
    Since [tex]u(x,y)=f(x^2+\frac{1}{y})[/tex] and [tex]u(x,0)=\phi(x) [/tex],
    we have [tex]u(x,0)=f(x^2+\frac{1}{\epsilon})=\phi(x)[/tex] where [tex]\epsilon\rightarrow 0[/tex]. This would imply that [tex]\lim_{u\rightarrow \infty}f(u)=\phi(x)[/tex]. Then again, that can't be right because the limit as [tex]u\rightarrow \infty[/tex] should not depend on x. I'm not sure if I see why you have "f(u)->0".
     
  5. Feb 24, 2010 #4

    Dick

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    I have "f(u)->0" because I was reading the boundary condition as u(x,0)=0, not phi(x). My mistake.
     
  6. Feb 24, 2010 #5
    So then if [tex]\phi(x)[/tex] were not a constant, would the question even make sense?
     
  7. Feb 24, 2010 #6

    Dick

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    I'm having trouble seeing how. But then maybe I'm missing something...
     
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