Cauchy problem/characteristics method with initial condition on ellipse

[math2011]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

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

How do I invert these to get u(x,y)?

 

[jackmell]

Since u(t,s)=e^{4t}, you need only solve for t=T(x,y) in the system:

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

then:

u(x,y)=u(T(x,y))=e^{4T(x,y)}

I'll start it for you:

y^2=e^{2t}(1-4s^2)

but from the first equation:

s^2=x^2e^{-2t}

now you finish it to find t=T(x,y). Not sure about the \pm though on y.

 

[math2011]

Thank you. I got t = \frac{1}{2} \ln (y^2 + 4x^2) and hence u = e^{2 \ln(y^2 + 4x^2)}.