So, to prove the energy $ E ( t ) $ is constant along solutions, differentiate $ ( 15 ) $, $$ \begin {align*} \frac { \mathrm d } { \mathrm d t } E ( t ) & = \frac { \mathrm d } { \mathrm d t } \int _ { - \infty } ^ { + \infty } \left ( \frac 1 2 u _ t ^ 2 ( x , t ) + \frac 1 2 c ^ 2 u _ x ^ 2 ( x , t ) \right ) \, \mathrm d x \\ & = \int _ { - \infty } ^ { + \infty } \left ( \frac 1 2 \frac \partial { \partial t } u _ t ^ 2 ( x , t ) + \frac 1 2 c ^ 2 \frac \partial { \partial t } u _ x ^ 2 ( x , t ) \right ) \, \mathrm d x \\ & = \int _ { - \infty } ^ { + \infty } \left ( u _ t ( x , t ) u _ { t t } ( x , t ) + c ^ 2 u _ x ( x , t ) u _ { x t } ( x , t ) \right ) \, \mathrm d x \\ & = \int _ { - \infty } ^ { + \infty } \left ( u _ t ( x , t ) u _ { t t } ( x , t ) - c ^ 2 u _ { x x } ( x , t ) u _ t ( x , t ) \right ) \, \mathrm d x \\ & = \int _ { - \infty } ^ { + \infty } u _ t ( x , t ) \left ( u _ { t t } ( x , t ) - c ^ 2 u _ { x x } ( x , t ) \right ) \, \mathrm d x = 0 \text , \end{align*} $$
I don't understand the step from line 3 to 4, where the negative sign is first introduced, and would really appreciate it if someone could explain it. Just for a little background, this is the energy functional for the one-dimensional wave equation. Also, if you don't mind, can you very briefly show why "demonstrating that a functional is non-negative, $ 0 $ at $ t = 0 $, and non-increasing, implies that any solution to the one-dimensional wave equation is unique" (The working out above is showing that it's non-increasing I'm assuming)? Thank you in advance.
This step is integration by parts:
$\int_{-\infty}^\infty u_x(x,t)u_{xt}(x,t) \, dx = u_x(x,t)u_t(x,t)\big\vert_{x=-\infty}^{x=\infty} - \int_{-\infty}^\infty u_{xx}(x,t) u_t(x,t) \, dx.$
You have to assume that $\lim_{x\to \pm \infty} u_x(x,t) = 0$ and/or $\lim_{x\to \pm \infty} u_t(x,t) = 0$ to kill the first term. This is true, for example, when the initial data $u(x,0)$ and $u_t(x,0)$ for the wave equation vanish outside some interval $[-M,M]$. Then due to finite speed of propagation, the solution $u(x,t)$ will vanish outside the interval $[M-ct,M+ct]$, and the limits above will be zero.
This argument shows that the energy $E(t)$ is constant for any solution of the wave equation. So for all $t>0$
$\int_{-\infty}^\infty u_t(x,t)^2 + c^2u_x(x,t)^2 \, dx = \int_{-\infty}^\infty u_t(x,0)^2 + c^2u_x(x,0)^2 \, dx.$
If you have two solutions $u$ and $v$ of the wave equation with the same initial data $u(x,0)=v(x,0)$ and $u_t(x,0)=v_t(x,0)$, then set $w(x,t) = u(x,t) - v(x,t)$. Then $w$ solves the wave equation with zero initial data $w(x,0) = 0$ and $w_t(x,0) = 0$. In particular, we also have $w_x(x,0) = 0$. Applying conservation of energy to $w$ yields
$\int_{-\infty}^\infty w_t(x,t)^2 + c^2w_x(x,t)^2 \, dx = \int_{-\infty}^\infty w_t(x,0)^2 + c^2w_x(x,0)^2 \, dx = 0$
for all $t>0$. Since the integrand on the left hand side is always positive or zero, and its integral is zero, the integrand itself must vanish everywhere (provided it is continuous, but you probably assume all solutions are classical in your course anyway). Hence $w_t(x,t) = 0$ and $w_x(x,t) = 0$. This means $w$ is constant in $x$ and $t$, and so
$w(x,t) = w(x,0) = 0$.
Finally, since $w=u-v$, this means that $u=v$, so solutions of the wave equation are unique.