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Consider $$ -y\frac{\partial F}{\partial x} +x\frac{\partial F}{\partial y} = G(x,y) $$

with the condition $F(x,0) = 0$ for all $x > 0$. How does one show that this initial-value problem has a single- value solution on $\mathbb{R} \backslash \{0\} $ if and only if $\int_0^{2\pi} G(A\cos(t),A\sin(t)) \, dt = 0$ for all $A$?

So far I know $x = A\cos(t)$ and $y = A\sin(t)$ but I am not sure where to go from there.

John Sweeney
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  • It's a consequence of the method of characteristics, which you've used for your $x(t)$ and $y(t)$, although your solution should be $x(t) = a\sin(t) + b\cos(t),\ y(t) = b\sin(t) - a\cos(t)$, although I would accept that the sum of two harmonics of the same phase is a phase-shifted harmonic of different amplitude, so our solutions are in a sense, equivalent. – bjorne Dec 10 '13 at 02:44
  • Essentially, your characteristics are circles, and since the solution grows according to $F_t=G$ along each circle it must be taking the integral over the entire circle is zero, otherwise we would reach a contradiction whereby the value at a point doesn't equal itself. – bjorne Dec 10 '13 at 02:48
  • @bjorne : consider making your comment an answer – Sergio Parreiras Dec 11 '13 at 02:54

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