Method of characteristics for the PDE $xu_x +y u_y = 0$ with an initial condition on a circle How to solve this Cauchy problem?
$$xu_x +y u_y = 0$$
$$u(x,y) = x\quad \text{on}\quad x^2 + y^2 = 1$$
My attempt:
$$\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{du}{0}$$
Using this we have $\dfrac{y}{x}=c_1 $ and $u=c_2$
Now Consider $c_2=G(c_1)$ $\implies$ $u=G(\dfrac{y}{x})$
My doubt is that how to use initial condition in this type of question?
I am confused with $x^2 + y^2 = 1$. How to use it here?
 A: Functions of the form $u(x,y)=G(y/x)$ satisfy the PDE, but there are others. For one thing, there is nothing wrong with $x=0$ here... so I would rather use notation $u(x,y)=g(\theta)$ where $\theta$ is the polar coordinate. After all, the PDE $xu_x+yu_y = 0$ simply says  that $\partial u/\partial r = 0$ in polar coordinates. Any function that is independent of radial coordinate $r$ solves the PDE. And not every such function is of the form $G(y/x)$ since this form requires $u(-x,-y)=u(x,y)$ which need not be the case. 
When $r=1$, you want $u(x,y) = x$, which is $\cos\theta$. So, $u(x,y)=\cos \theta$ it is. It can be expressed in $x,y$ coordinates, of course.

A related question is The integral surface of the pde $xu_x+yu_y=0$ satisfying the condition $u(1,y)=y$ is given by: 
A: After finding the general solution $u=G\left(\frac{y}{x}\right)$ or equvalent $u=H\left(\frac{y^2}{x^2}\right)$  the condition $u(x,y)=x$ on $x^2+y^2=1$ must be applied :
$$x=H\left(\frac{1-x^2}{x^2}\right)$$
Let $\frac{1-x^2}{x^2}=t$
$x^2=\frac{1}{t+1}$
$$\sqrt{\frac{1}{t+1}}=H(t)$$
Now, the function $H$ is known.
$$u=H\left(\frac{y^2}{x^2}\right)=\sqrt{\frac{1}{\frac{y^2}{x^2}+1}}$$ 
$$u=\frac{x}{\sqrt{y^2+x^2}}$$
Note : The sign $\pm$ is omitted behind the radicals because only the signe $+$ is consistent with the condition $u(x,y)=x$ on the circle.
A: i want to see if changing to polar coordinates helps. we will make a change of variable $$x = r\cos \theta, y = r \sin \theta,  \frac{d}{dx} = \cos \theta\frac{d}{dr} -\frac{\sin \theta}r\frac{d}{d\theta}, \frac{d}{dy} = \sin \theta\frac{d}{dr} +\frac{\cos \theta}r\frac{d}{d\theta} $$
we can transform $$0=xu_x + yu_y = r\cos \theta\left(\cos \theta u_r-\frac{\sin \theta}ru_{\theta}\right)+r\sin \theta\left(\sin \theta u_r+\frac{\cos \theta}ru_{\theta}\right) = ru_r$$ to $$u_r = 0, u(1, \theta) = \cos \theta  \tag 1$$ the solution to $(1)$ is $$u(r, \theta) = \cos \theta = \frac{x}{\sqrt{x^2+y^2}}.$$
