Prove that PDE $u_t+x^2u_x=0$ has infinite solutions the text of the exercise I am takling is this one.

Given
  \begin{cases} 
u_t+x^2u_x=0 \text{ in } (0, +\infty)\times \mathbf{R}\\
u(0,x)=0 \text{ in } \mathbf{R}
\end{cases}
  prove that for every $\delta>0$ there are infinite solutions in $[0,\delta]\times \mathbf{R}$. 

I tried using the method of characteristic. The system i solve is
\begin{cases} 
\dot{v}=v^2 \\
\dot{w}=0 \\
v(0)=r \\
w(0)=0
\end{cases}
Even before solving the system I see that there is a problem: I will obatin $w(t)\equiv0 $, and any solution of the type $u(t,x)=w(t,r(t,x))$ will be the one constantly zero. Drawing a picture I find out that the vector field $(1,x^2)$ in the plane $(t,x)$ is directed as the $x$-axe, which is the line where I have my initial value for the problem.
Anyway solving the system for $v$ I obtain $v(t)=\frac{1}{c-t}$ with $c=\frac{1}{r}$. In this way I can find a family of characteristic curves $x_r(t)$. I see that for every fixed $\delta$ I can choose $r$ small enough to make the curve exist until time $\delta$. 
My problem is how to define anon-zero solution on this curve.
Beyond this single problem, what is the reason for the difficuty of this exercise? Is there a general method to solve problems like this? 
 A: The problem is that when you consider the characteristics backward, not all of them intersect the line $t = 0$.
For $x > 0$, even though the characteristics curves blow up in finite time, this is not actually a problem. What we care is that we can extend the characteristic backwards in time. In order to not get confuse, lets use $s$ as the parameter for the curve
$$ x(s) = \frac{r}{1 - rs}$$
and lets take a fixed point $(t,x) \in [0,\delta]\times \mathbb R_+$. Then the curve that intersects $(t,x)$ satisfies
$$ x = x(t) = 
\frac{r}{1 - rt}$$
so
$$ r = \frac{x}{xt + 1}$$
This value is always well defined and more importantly, $x(s)$ is well defined in $[0,t]$ due to the fact that
$$ 1 - rs = 1 - \frac{xs}{xt + 1} = \frac{x(t - s) + 1}{xt + 1} > 0 $$
This forces then $u(t,x) = u_0(r) = 0$
For $x < 0$ on the other hand, we may encounter some problems. The first one is the possibility that $xt + 1 = 0$ in which case there characteristic curve is $x(s) = -1/s$ that never touches the $t = 0$ line. Even worse is the case $xt + 1 < -1$ where all the computation of the first part remain valid, except that the characteristic curve has a singularity at $s = \frac{xt + 1}{x} = t + \frac1x \in (0,t)$ (this value was always bigger than $t$ in the previous case)
The problem then is that in this case the characteristic comes from $x = -\infty$ rather than the line $t = 0$.
Finally if $xt + 1 > 0$ then all the analysis done in the first case is still valid so $$u(t,x) = 0$$ as well.
With all this analysis we can construct the non trivial solutions:
Lets take $\phi \in C^1(\mathbb R)$ such that
$$\lim_{y \to -\infty} y \phi(y) = 0$$
Then a solution is given by
$$ u(t,x) = \begin{cases}
0 & tx + 1\geq 0\\
\phi \left( \frac{x}{xt + 1}\right )& tx + 1 < 0
\end{cases}
$$
To prove that it is a solution, in the case $tx +1 \not=0$ is just direct computations. 
In the case $tx + 1 = 0$ you can prove that $\partial_t u = \partial_x u = 0$ using the limit condition on $\phi$.
As we mentioned earlier, this solutions correspond to a wave coming from $-\infty$ so we would need a condition in $-\infty$ to have uniqueness.
A: The characteristics $t\mapsto (t,x(t))$ are indeed the curves $x(t) = \frac{r}{1-rt}$ along which $u = 0$. These curves do not intersect given that $\frac{r_1}{1-r_1 t} = \frac{r_2}{1-r_2 t}$ for all $t$ implies $r_1 = r_2$. Now, let us plot these curves in the $x$-$t$ plane:


*

*Characteristic curves starting at $x=r\le 0$ exist until $t\to \infty$ where $x \to 0$. They remain within $-1/t< x\le 0$, without intersecting. Thus, the solution $u \equiv 0$ is uniquely defined over that domain. However, note that points located in the region $x<-1/t$ cannot be reached by any of these curves. Therefore the solution is not uniquely defined in the region $x<-1/t$ with $0 \leq t \leq \delta$.


*Characteristic curves starting at $x=r>0$ end at $t = 1/r$. Nevertheless, any point located in the region $x>0$ is reached by a single characteristic curve.
