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 $u = e^{2 \ln(y^2 + 4x^2)}$.