I wanted to solve the following PDE with initial condition
$$ \left\{\begin{array}{c} xu_t+u_x=0\\ u(0,x)=f(x) \end{array}\right.$$
Proving that:
(i) if $f(x) = \sin(x),$ then it is impossible to find a solution which is valid for all point in $\mathbb{R}^2.$
(ii) if $f(x) = \cos(x),$ has infinite solutions defined in all $\mathbb{R}^2.$
I tried to solve this using the method of characteristics.
First of all the characteristic system is
$$ \left\{\begin{array}{c} \frac{d t}{d s} = x\\ \frac{d x}{d s}=1 \\ \frac{d u}{d s}=0 \end{array}\right.$$ with initial conditions $$ t(0,\tau)=0 $$ $$ x(0,\tau)=\tau $$ $$ u(0,\tau)=\sin \tau $$
Computing, and proceeding by an analogue method to $\cos(x),$ I find $u(t,x) = \sin (\sqrt{-2t + x^2})$ and $u(t,x) = \cos (\sqrt{-2t + x^2}),$ but both are not defined in $(1,0).$ There is a problem with my calculation, or the initial statement are wrong?
If you solve the characteristic equations, the families of characteristic curves are $\phi(x,y,u) = u$ and $\psi(x,y,u) = \frac{x^2}{2} - t $. Then the general solution is given by $$ h(\psi) = \phi$$ with $h$ an arbitrary differentiable function. Applying the Cauchy data you get
$$h\left(\frac{x^2}{2}\right) = f(x) $$ So, to define the solution for all points in $\mathbb{R}^2$, $f$ must be an even function.