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Cauchy problem/characteristics method with initial condition on ellipse

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the PDE [itex]xu_x + y u_y = 4 u, -\infty < x < \infty, -\infty < y < \infty[/itex]. Find an explicit solution that satisfies [itex]u = 1[/itex] on the ellipse [itex]4x^2 + y^2 = 1[/itex].

    2. Relevant equations



    3. The attempt at a solution

    The characteristic curves are
    [itex]x(t,s) = f_1(s) e^t[/itex]
    [itex]y(t,s) = f_2(s) e^t[/itex]
    [itex]u(t,s) = f_3(s) e^{4t}[/itex].

    The initial conditions are
    [itex]x(0,s) = s[/itex]
    [itex]y(0,s) = \pm \sqrt{1 - 4s^2}[/itex]
    [itex]u(0,s) = 1[/itex].

    Parametric representation of the integral surface is then
    [itex]x(t,s) = s e^t[/itex]
    [itex]y(t,s) = \pm \sqrt{1 - 4s^2} e^t[/itex]
    [itex]u(t,s) = e^{4t}[/itex].

    How do I invert these to get [itex]u(x,y)[/itex]?
     
  2. jcsd
  3. Oct 29, 2011 #2
    Since u(t,s)=e^{4t}, you need only solve for t=T(x,y) in the system:

    [tex]x=se^t[/tex]
    [tex]y=e^t\sqrt{1-4s^2}[/tex]

    then:

    [tex]u(x,y)=u(T(x,y))=e^{4T(x,y)}[/tex]

    I'll start it for you:

    [tex]y^2=e^{2t}(1-4s^2)[/tex]

    but from the first equation:

    [tex]s^2=x^2e^{-2t}[/tex]

    now you finish it to find t=T(x,y). Not sure about the [itex]\pm[/itex] though on y.
     
  4. Oct 30, 2011 #3
    Thank you. I got [itex]t = \frac{1}{2} \ln (y^2 + 4x^2)[/itex] and hence [itex]u = e^{2 \ln(y^2 + 4x^2)}[/itex].
     
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