Uniqueness of the solution of a PDE I have the following initial-value problem: $$ \begin{cases} u_t + c u_x = 0, \quad x \in \mathbb{R}, \ t > 0 \\ u(x,0) = g(x) \end{cases} \qquad [1] $$where $u_t$ and $u_x$ are the partial derivatives. It has been asked me to prove that the problem is well-posed if $g \in C_b ^1 (\mathbb{R})$ - id est to prove that a solution exists and it is unique, and that the solution depends continuously on the initial data in the norm $ \sup_{x \in \mathbb{R}} |g(x)|$.
Now, the existence and the continuous dependence are pretty straightforward: it is sufficient to notice that $u(x,t)=g(x-ct)$ is a solution of $[1]$. But what about the uniqueness? I couldn't figure out how to reach a contraddiction by assuming the existence of another solution $v(x,t)$... I would conclude by myself, so just give me an (eventual) hint.
Thanks in advance.
 A: Hint
There is at least two methods that can be used to show uniqueness. One of them is Maximum Principle (if holds for the equation), and another one is Energy Integral (google for one of them). 
In both of them you pretty much assume two different solutions $u_1$ and $u_2$ and need to show that the new function $u = u_1 - u_2 \equiv 0$ which is of course imply uniqueness. Note that the function $u$ will satisfy 
\begin{cases} u_t + c u_x  = (u_1-u_2)_t + c(u_1-u_2)_x= 0, \qquad\quad x \in \mathbb{R}, \ t > 0 \\ u(x,0) = u_1(x,0)-u_2(x,0) = g(x)-g(x) =0 \end{cases} 
Energy Integral (assuming $c>0$):
$$E(t) = \int_{-\infty}^\infty u^2(x,t)dx$$
Note that $E(0)=0$.
Now
$$\frac{d E}{d t}(t) = \int_{-\infty}^\infty 2 u u_t dx = \int_{-\infty}^\infty 2 u (-c u_x) dx
=-c \int_{-\infty}^\infty  2 u u_x dx =
 -c u^2\big|_{-\infty}^\infty$$
One writes
$$
 -c u^2\big|_{-\infty}^\infty =-c\lim_{r\to \infty} (u^2(r,t)-u^2(-r,t))
$$
given the limit exists.
From this point you want either $\frac{d E}{d t}(t)=0$ or  $\frac{d E}{d t}(t)$ never change sign. In both cases this is enough for $E(t)$ to be 0 which will imply that $u$ is 0.
However, neither obvious here. If you can claim a finite flow (which is vanishes at infinity) it will imply $\frac{d E}{d t}(t)=0$. A symmetric solution will make it $0$ again. But I don't know how to use this method for a general case.
Another method, using a solution
Since this is a first order PDE it is easy to obtain solution which is of course depends on initial data $g$. More precisely it is $g(x-c t)$. Assuming two different solutions like before and defining $u=u_1-u_2$ we note that $u$ satisfies the transport equation with zero initial data, i.e. $g=0$ which immediately implies that $u$ is zero, which mean there is only one unique solution.
