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Solve $-yu_x+xu_y = u$, with the initial condition $u(x,0)= F(x)$

My attempt:

The characteristic ODE are

$\dfrac{d x}{ d t} = -y$, $\dfrac{d y}{ d t} = x$, $\dfrac{d u}{ d t} = u$, with initial conditions $x(0)=x_0$, $y(0)=0$, $u(x(0),0)=u_0 = F(x_0)$.

The solutions of the ODE are

$x(t) = x_0 \cos(t)$, $y(t) =x_0 \sin(t) $, $u = F(x_0)\exp(t)$

(Is my solution right so far?)

We have $t= \arctan \big(\frac{y}{x}\big)$ and $x_0 ^2 = \sqrt{x^2+y^2}$.

It is not clear to me if I should choose $x_0 = \sqrt{x^2+y^2}$ or $x_0 = -\sqrt{x^2+y^2}$. Which one is right?

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  • $\begingroup$ The point is that the characteristics are circles being traversed counterclockwise, with the boundary data kicking in at the line $ y=0$. This can create an ambiguity in the solution; the BC should be either only on the positive $x $ axis or the negative one but not both. $\endgroup$ – Ian Sep 8 '16 at 19:00
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You approach is correct. So, one have to chose between $x_0 = \sqrt{x^2+y^2}$ and $x_0 = -\sqrt{x^2+y^2}$

$u(x,y)=\exp(t)F(x_0)\quad$ with condition $\quad u(x,0)=F(x)\quad$ at $\quad t=0$

$u(x,0)=F(x)=e^0 F\left(\pm \sqrt{x^2+0}\right)$ which implies : $\begin{cases} F(x)=F\left(\sqrt{x^2}\right)\quad \text{ if}\quad x>0\\ F(x)=F\left(-\sqrt{x^2}\right)\quad \text{ if}\quad x<0 \end{cases}$

Hence the solution is :

$$\begin{cases} u(x,y)=\exp{\left(\tan^{-1}\left(\frac{y}{x}\right)\right)}F\left(\sqrt{x^2+y^2}\right)\quad \text{ if}\quad x>0\\ u(x,y)=\exp{\left(\tan^{-1}\left(\frac{y}{x}\right)\right)}F\left(-\sqrt{x^2+y^2}\right)\quad \text{ if}\quad x<0 \end{cases}$$

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