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Cauchy problem/characteristics method with initial condition on ellipse

  • Thread starter math2011
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  • #1
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Homework Statement



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].

Homework Equations





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]?
 

Answers and Replies

  • #2
1,796
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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.
 
  • #3
10
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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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