Solution of the Cauchy problem $xu_x + yu_y = 0$ $u(x,y) = x$ on $x^2 +y^2 =1$ The Cauchy problem $xu_x + yu_y = 0$
$u(x,y) = x$ on $x^2 +y^2 =1$ has 


*

*a solution for $x ,y \in \Bbb R$

*a unique bounded solution in [$x ,y| (x,y)  \neq (0, 0)$]

*an unique solution in [$x ,y| (x,y)  \neq (0, 0)$] but not bounded.
I have tried to solve this problem. Can anyone please check and tell me how to proceed further and which option(s) will be right?

 A: Let $$v(r,\theta) = u(r\cos \theta, r\sin \theta)$$
Then $$u_x = v_r\cos\theta - v_\theta r \sin \theta \\
 u_y = v_r\sin\theta + v_\theta r \cos \theta $$
$$xu_x + yu_y = v_r r\cos^2\theta - v_\theta r^2 \sin \theta \cos\theta \\
 +  v_r r\sin^2\theta + v_\theta r^2 \cos \theta \sin \theta$$
$$
0 = xu_x + yu_y = r v_r
$$
Therefore $v_r = 0$ everywhere except possibly at the origin, and 
$$
v(r, \theta) = f(\theta)
\\ u(x,y) = f(\mbox{atan2 } (y/x) ) $$
where the atan2 function is the usual quadrant-aware arctangent (atan2 is in the third quadrant if $x$ and $y$ are both negative, etc).
Now if we impose the boundary conditions, the unique solution is 
$$
u(x,y) = f(\cos\left(\mbox{atan2 } (y/x) \right)
$$
This is a bounded function, but it is not defined at $(x,y) = (0,0)$ because on an arbitrarily small circle around the origin, ($u(x,y)$ takes on all values in $[-1,1]$.
So the answer is 2 - bounded everywhere but domain does not include the origin.
A: From your notes, it's not quite clear (to me) what you have actually done, but what the PDE says is that $u$ must be constant on rays (half-lines) from the origin. Together with the given values on the unit circle, that should tell you anything you need to know about $u$.
A: Taking 
$$
xu_{x}+yu_{y}=0\text{ with } u(x,y)=x \text{ on }x^{2}+y^{2}=1
$$
Due to the contour conditions we will consider the change of variables
$$
\begin{array}{rcl}
x & = & r\cos\theta\\
y & = & r\sin\theta
\end{array}
$$
with
$$
\begin{array}{rcl}
dx & = & dr\cos\theta-r\sin\theta d\theta\\
dy & = & dr\sin\theta+r\cos\theta d\theta
\end{array}
$$
we have 
$$
\begin{array}{rcl}
u_{x} & = & u_{r}\frac{dr}{dx}+u_{\theta}\frac{d\theta}{dx}\\
u_{y} & = & u_{r}\frac{dr}{dy}+u_{\theta}\frac{d\theta}{dy}
\end{array}
$$
Here from the characteristic curves
$$
\frac{dx}{x}=\frac{dy}{y}\Rightarrow\frac{dy}{dx}=\frac{y}{x}=\tan\theta
$$
so we obtain
$$
xu_{x}+yu_{y}=0\Longleftrightarrow2ru_{r}=0\Rightarrow u(r,\theta)=C+\Phi(\theta)
$$
Note that
$$
\begin{array}{rcl}
\frac{dr}{dx} & = & \frac{1}{\cos\theta}\\
\frac{dr}{dy} & = & \frac{1}{\sin\theta}\\
\frac{d\theta}{dx} & = & 0\\
\frac{d\theta}{dy} & = & 0
\end{array}
$$
Now with the boundary conditions
$$
u(1,\theta)=\cos\theta\Rightarrow\Phi(\theta)=\cos\theta\text{ and }C=0
$$
and finally
$$
u(r,\theta)=\cos\theta
$$
or in $(x,y)$ coordinates, $u(x,y)=\frac{x}{\sqrt{x^{2}+y^{2}}}$ 
