Analyzing specific qualities of the Wave Equation Let $u$ be a solution of the wave equation $u_{tt} = c^2u_{xx}$ on the domain $\Omega=(0,1)$, subject to homogeneous Neumann boundary conditions.
(a) Define $$E(t)=\int_{0}^{1}\big(u_t(t,x)^2+c^2u_x(t,x)^2\big)dx$$ Show that $E(t)$ is constant, i.e., it does not depend on the time $t$. That is, show that the derivative satisfies $dE(t)/dt = 0$. 
(b) Use (a) to show that the only differentiable function satisfying the wave equation $u_{tt }= c^2u_{xx}$ such that $u(0,x) = 0$ and $u_t(0,x) = 0$ for all $x\in (0,1)$, as well as $u_x(t,0) = u_x(t,1) = 0$ for all $t > 0$ is the trivial function $u(t,x) \equiv 0$. Hint: Find $E(0)$ for a function with these initial conditions
 A: In what follows, I assume without further comment that $u(t, x)$ is "sufficiently differentiable"; enough so to make the logic fly.
For part (a):
With
$u_{tt} = c^2 u_{xx} \tag 1$
and
$E(t)  = \displaystyle \int_0^1 (u_t^2(t, x) + c^2 u_x^2(t, x))dx, \tag 2$
we may perform the "lite" computation of $\dot E(t) = E_t(t) = dE(t)/dt$, where I use the word "lite" since I am not going to worry about or otherwise vindicate differentiating under the integral sign with respect to $t$, an operation which would probably merit some justifying analysis and/or remarks in a more thorough treatment, as follows:
$\dfrac{dE(t)}{dt} = \dfrac{d}{dt}\displaystyle \int_0^1 (u_t^2(t, x) + c^2 u_x^2(t, x))dx= \int_0^1 \dfrac{d}{dt}(u_t^2(t, x) + c^2 u_x^2(t, x))dx$
$= \displaystyle \int_0^1 (2u_t(t, x)u_{tt}(t, x) + 2c^2u_x(t, x)u_{xt}(t, x))dx$
$ = \displaystyle 2\int_0^1 (u_t(t, x)u_{tt}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx; \tag 3$
using (1), the rightmost integral in (3) becomes
$\displaystyle \int_0^1 (u_t(t, x)u_{tt}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx$ $= \displaystyle \int_0^1 (c^2u_t(t, x)u_{xx}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx; \tag 4$
furthermore,
$\displaystyle \int_0^1 (c^2u_t(t, x)u_{xx}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx$
$= \displaystyle \int_0^1c^2 u_t(t, x)u_{xx}(t, x)dx + \int_0^1 c^2u_x(t, x)u_{xt}(t, x)dx; \tag 5$
as for the second integral on the right of (5), consider the identity
$(u_t(t, x) u_x(t, x))_x = u_{tx}(t, x) u_x(t, x) + u_t(t, s) u_{xx}(t, x), \tag 6$
which results from a simple application of the Leibniz rule for derivatives of products to $u_t(t, x) u_x(t, x)$; (6) may be re-arranged to yield
$u_{tx}(t, x) u_x(t, x) = (u_t(t, x) u_x(t, x))_x -  u_t(t, s) u_{xx}(t, x), \tag 7$
which we integrate in $x$ over $[0, 1]$:
$\displaystyle \int_0^1 u_{tx}(t, x) u_x(t, x)dx = \int_0^1(u_t(t, x) u_x(t, x))_x dx -  \int_0^1 u_t(t, s) u_{xx}(t, x)dx; \tag 8$
when $t \ge 0$, the leftmost integral on the right-hand side of (8) may be explicitly found:
$\displaystyle \int_0^1(u_t(t, x) u_x(t, x))_x dx= u_t(t, 1)u_x(t, 1) - u_t(t, 0)u_x(t, 0) = 0, \tag 9$
by virtue of the homogeneous Neumann boundary conditions $u_x(t, 1) = u_x(t, 0) = 0$, $t \ge 0$; thus, (8) becomes
$\displaystyle \int_0^1 u_{tx}(t, x) u_x(t, x)dx = -  \int_0^1 u_t(t, s) u_{xx}(t, x)dx, \tag {10}$
in the light of which (5) yields
$\displaystyle \int_0^1 (c^2u_t(t, x)u_{xx}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx$
$= \displaystyle \int_0^1 c^2 u_t(t, x)u_{xx}(t, x)dx - \int_0^1 c^2 u_t(t, x) u_{xx}(t, x)dx = 0; \tag {11}$
now following the logic back from (5) to (3) we see that
$\dfrac{dE(t)}{dt} = \dfrac{d}{dt}\displaystyle \int_0^1 (u_t^2(t, x) + c^2 u_x^2(t, x))dx, \tag{12}$
for $t \ge 0$; the requisite result.
Now as for part (b):
Since $u(0, x) = 0$ we have
$u_x(0, x) = 0, \tag{13}$
and since we are given that
$u_t(0, x) = 0, \tag{14}$
we see that
$E(0) = 0; \tag{15}$
then by the result of part (a) we find
$E(t) = 0 \tag{16}$
for all $t \ge 0$.  Since we (though perhaps tacitly) assume $u_t(t, x)$ and $u_x(t, x)$ are continuous, this implies
$u_t(t, x) = u_x(t, x) = 0 \tag{17}$
for  $t \ge 0$; since the derivatives of $u(t, x)$ are both $0$, $u(t, x)$ is constant in both time and space for $t \ge 0$; since $u(x, 0) = 0$ it follows that
$u(x, t) = 0 \; \text{for} \; (x, t) \in [0, 1] \times [0, \infty), \tag{18}$
and the hypothesis proposed in part (b) is affirmed.
