PDE with initial value Problem is that
Solve $xu_t+uu_x=0 $ with $u(x,0)=x$ 
Hint is in the textbook that replacing  $x$ into $x^2$
I am stuck with it.
Would you help me?
 A: Some version of the characteristics method stabilishes to solve first an auxiliar system of ode's, this:
$\dfrac{dt}{x}=\dfrac{dx}{u}=\dfrac{du}{0}$
The last ratio means that $u=c_1$, so is $u$ is constant along the characteristics. With the first proportion we get the second constant:
$dt=\dfrac{xdx}{c_1}\implies c_1t+c_2=x^2/2\implies ut+c_2=x^2/2$
For the solution we have to impose some relation between the constants, so is, we have to impose that $c_2=f(c_1)$ for some single argument, differentiable function $f$.
$ut+f(c_1)=x^2/2\;;ut+f(u)=x^2/2$ is the general solution.
Imposing the initial condition $u(x,0)=x$ we can get $f$:
$f(x)=x^2/2$ that substituting in the general solution gives (maybe here is where the hint given fits):
$ut+u^2/2=x^2/2$ as the requested solution in implicit form.
A: $xu_t+uu_x=0$
$u_t+\dfrac{u}{x}u_x=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$\dfrac{dx}{ds}=\dfrac{u}{x}=\dfrac{u_0}{x}$ , we have $x^2=2u_0s+f(u_0)=2ut+f(u)$ , i.e. $u=F(x^2-2ut)$
$u(x,0)=x$ :
$F(x^2)=x$
$F(x)=\pm\sqrt x$
$\therefore u=\pm\sqrt{x^2-2ut}$
$u^2=x^2-2tu$
$u^2+2tu-x^2=0$
$u(x,t)=\dfrac{-2t+\sqrt{4t^2+4x^2}}{2}=-t+\sqrt{x^2+t^2}$
