Dependence of the existence of solutions to a first-order linear PDE on the initial conditions. I have the following problem:
Show that there is no solution to the problem
$x\partial_tu+\partial_xu = 0 \\
u(x,0) = \sin(x)$, 
but that there are infinitely many solutions to the problem
$x\partial_tu+\partial_xu = 0 \\
u(x,0) = \cos(x)$
I know that the method of characteristics cannot be used since the line $x =0$ is characteristic in the sense that all $(0, a)$ for $a \in \mathbb{R}$ is in $char_x(L)$, but other then that I have no clue on how to proceed. Any hints will be the most appreciated.
Thank you very much. 
 A: The characteristics method works well: the charateristics curves for the second problem are:
$$\begin{cases}
 u=c_1\\
 x^2/2-t=c_2
\end{cases}$$
Being the general solution:
$$u=f(x^2/2-t)$$
with $f$ some function.
Imposig the initial conditions, 
$$f(x^2/2)=\cos x$$
$$f(y)=\cos\sqrt{2y}$$
Then, the particular solution must fulfill the condition:
$$u(x,t)=\cos\sqrt{x^2-2t}$$
But observe that the square root is not defined for values such that $2t\gt x^2$. This gives us a hint about what's going on here. The characteristics are parabolas, $x^2/2-t=c_2$ for which $u$ is constant and the boundary conditions we are imposing ($u(x,0)=\cos x$) say nothing about the parabolas not crosing the $x$ axis, e.g. $x^2/2-t=-1$, so, we can give any value to $u$ along such parabolas and we will not contradict the initial conditions. Further, such arbitrary assignation for $u$ along that family of parabolas can be done in a continuous and differentiable way, so, there are infinitely many solutions.
I don't get why you say that the line $x=0$ is characteristic. It is not. In fact, the line along which the boundary conditions are given is not a characteristic curve, and this is a necessary condition for the problem to have a single solution, but as we can see, it is not suficient.
Added
The general solution for first problem would have to be very similar to the found for the second,
$$u(x,t)=\sin\sqrt{x^2-2t}$$
and it miss too determine the value of $u$ for curves with $2t\gt x^2$, but the point to consider here is different: the solution cannot be differentiable at $(0,0)$.
Consider the partial derivative along the line $x=0$ that, because the boundary conditions, for any solution it must coincide with the partial derivative for the function we have found:
$$\left.\frac{\partial u}{\partial t}\right|_{x=0}=\frac{t\cos\sqrt{-2t}}{\sqrt{2}(-t)^{3/2}}$$
that is not defined at $t=0$.
