PDE - Wave Equation 
Let $A(x) = (a^{ij}(x))_{ij}$, $i, j = 1, 2, 3$ be a smooth matrix-valued function on $\mathbb{R}^3$ such that $a^{ij} = a^{ji}$, and there exists constants $c, C$ such that $c|\zeta|^2 \le \langle A(x)\zeta, \zeta \rangle \le C|\zeta|^2$ for all $(x, \zeta) \in \mathbb{R}^3$ x $\mathbb{R}^3$.
Show that if $u$ is a smooth solution to the variable-coefficient wave equation:
$\partial_t ^2 u$ - div$(A \nabla u) = 0$,
such that $u$ and its derivatives decay rapidly as $|x| \rightarrow \infty$ (say, bounded by $|x|^{-3/2 - \epsilon}$ for some $\epsilon > 0$), then:
$E(t) := \frac{1}{2} \int_{\mathbb{R}^3} (|\partial_t u|^2 + \langle A \nabla_x u, \nabla_x u \rangle) dx$
is conserved.

This is what I've done so far:
$E(t) := \frac{1}{2} \int_{\mathbb{R}^3} (|d_t u|^2 + \langle A \nabla u, \nabla u \rangle) dx$
= $\int_{\mathbb{R}^3}(u_t u_{tt} + \frac{1}{2} [A \nabla u_t \cdot \nabla u + A \nabla u \cdot \nabla u_t])dx$
= $\int_{\mathbb{R}^3} (u_t \nabla \cdot (A \nabla u) + \frac{1}{2}[A \nabla u_t \cdot u + A \nabla u \cdot \nabla u_t])dx$
But I don't know where to go for the next step. I am honestly completely stuck on this part. Any help/guidance on where to go next would be great. Thank you.
 A: We are given the partial differential equation
$\partial_t^2 u = \text{div}(A \nabla u) = \nabla \cdot (A \nabla u), \tag 1$
for which we define the energy at time $t$ via an integral over $\Bbb R^3$:
$E(t) = \displaystyle \dfrac{1}{2} \int_{\Bbb R^3} ((\partial_t u)^2 + \langle A\nabla u, \nabla u \rangle) \; dx; \tag 2$
we wish to show that
$\dot E(t) = \partial_t E(t) = 0. \tag 3$
Before proceeding with the demonstration of (3), a few remarks are in order:
first, there is no need to take the absolute value of $\partial_t u$ when forming the term $(\partial_t u)^2$ appearing in (2), as our OP Axioms does in the text of his question.  Since we are taking the square of $\partial_t u$, the absolute value signs are unnecesary:  $\vert \partial_t u \vert^2 = (\partial_t u)^2$; this simplification makes taking the $t$-derivative of $\vert \partial_t u \vert^2 = (\partial_t u)^2$ somewhat more transparent, since it makes clear that we needn't be concerned with the failure of $\vert \cdot \vert$ to be differentiable when its argument vanishes; second, the hypothesis that $a^{ij}(x) = a^{ji}(x)$, i.e., that $A(x)$ is a symmetric matrix, $A^T(x) = A(x)$, for all $x \in \Bbb R^3$ implies, as is well known, that for any vectors $v, w$ we have
$\langle v, A(x) w \rangle = \langle A(x) v, w \rangle; \tag 4$
indeed,
$\langle v, A(x) w \rangle = \langle A^T(x) v, w \rangle = \langle A(x) v, w \rangle; \tag 5$
this property of the coefficient matrix $A(x)$ will enter into the discussion which follows; third, there is no need to write the subscript $x$ when taking $\nabla u = \nabla_x u$, since it is understood that $\nabla$ refers only to the spatial variables $x$; bearing these remarks in mind, we proceed.
From (2) we have
$\dot E(t) = \partial_t E(t) =  \displaystyle \dfrac{1}{2} \partial_t \int_{\Bbb R^3} ((\partial_t u)^2 + \langle A\nabla u, \nabla u \rangle) \; dx$
$= \displaystyle \dfrac{1}{2} \int_{\Bbb R^3} \partial_t ((\partial_t u)^2 + \langle A\nabla u, \nabla u \rangle) \; dx = \dfrac{1}{2} \int_{\Bbb R^3} (\partial_t (\partial_t u)^2 + \partial_t \langle A\nabla u, \nabla u \rangle) \; dx; \tag 6$
now,
$\partial_t (\partial_t u)^2 = 2 \partial_t u \; \partial_t^2 u; \tag 7$
also,
$\partial_t \langle A\nabla u, \nabla u \rangle  = \langle \partial_t(A \nabla u), \nabla u \rangle + \langle A \nabla u, \partial_t \nabla u \rangle, \tag 8$
and
$\partial_t \nabla u = \nabla \partial_t u, \tag 9$
since the $t$- and $x$-derivatives commute, and
$\partial_t A \nabla u = A \partial_t \nabla u, \tag{10}$
since $A(x)$ is constant with respect to $t$; combining (8), (9) and (10) we see that
$\partial_t \langle A\nabla u, \nabla u \rangle  = \langle A \nabla \partial_t u, \nabla u \rangle + \langle A \nabla u,  \nabla \partial_t u \rangle, \tag{11}$
and using the symmetric property (4) of $A$ and of the inner product ($\langle v, w \rangle = \langle w, v \rangle$), 
$\partial_t \langle A\nabla u, \nabla u \rangle  = \langle \nabla \partial_t u, A \nabla u \rangle + \langle \nabla \partial_t u, A \nabla u \rangle = 2\langle \nabla \partial_t u, A \nabla u \rangle; \tag{12}$
using (7) and (12) we return to (6) and write it in the form
$\dot E(t) = \partial_t E(t) = \displaystyle \dfrac{1}{2} \int_{\Bbb R^3} (\partial_t (\partial_t u)^2 + \partial_t \langle A\nabla u, \nabla u \rangle) \; dx$
$= \displaystyle \int_{\Bbb R^3} (\partial_t u \; \partial_t^2u + \langle \nabla \partial_t u, A \nabla u \rangle) \; dx, \tag{13}$
where the factors of $2$ occurring in (7) and (12) have been cancelled against the $1/2$ occurring in (6).
Now
$\nabla \cdot (\partial_t u (A \nabla u)) = \langle \nabla \partial_t u, A \nabla u) \rangle + \partial u_t \nabla \cdot (A \nabla u), \tag{14}$
which follows from the standard formula from vector calculus
$\nabla \cdot (fX) = \langle \nabla f, X \rangle + f \nabla \cdot X \tag{15}$
applied with $f = \partial_t u$ and $X = A \nabla u$; re-arranging (14), 
$\langle \nabla \partial_t u, A \nabla u) \rangle = \nabla \cdot (\partial_t u (A \nabla u)) - \partial_t u \nabla \cdot (A \nabla u); \tag{16}$
we integrate (16) over $\Bbb R^3$, as follows:
$\displaystyle \int_{\Bbb R^3} \langle \nabla \partial_t u, A \nabla u \rangle \; dx =  \int_{\Bbb R^3}  \nabla \cdot (\partial_t u (A \nabla u)) \; dx -  \int_{\Bbb R^3} \partial_t u \nabla \cdot (A \nabla u) \; dx; \tag{17}$
we evaluate the first integral on the right of (17) by first taking it over a ball $B(R)$, of radius $R$ and centered at $(0, 0, 0)$, and then letting $R \to \infty$; we have, using the divergence theorem,
$\displaystyle \int_{B(R)} \nabla \cdot (\partial_t u (A \nabla u)) \; dx  = \int_{\partial B(R)} \partial_t u (A \nabla u) \cdot n \; dA, \tag{18}$ 
where $n$ is the outward normal on the sphere $\partial S(R)$, the boundary of $B(R)$, and $dA$ is its area element; under the assumption that $u$ and its derivatives fall off sufficiently rapidly as $R \to \infty$, we have
$\displaystyle \int_{\Bbb R^3} \nabla \cdot (\partial_t u (A \nabla u)) \; dx$
$ = \displaystyle \lim_{R \to \infty} \int_{B(R)} \nabla \cdot (\partial_t u (A \nabla u)) \; dx  = \lim_{R \to \infty} \int_{\partial B(R)} \partial_t u (A \nabla u) \cdot n \; dA = 0, \tag{19}$
so that (17) yields
$\displaystyle \int_{\Bbb R^3} \langle \nabla \partial_t u, A \nabla u \rangle \; dx =  -\int_{\Bbb R^3} \partial_t u \nabla \cdot (A \nabla u) \; dx; \tag{20}$
using (20), we may re-assemble (13) to find
$\dot E(t) = \displaystyle \int_{\Bbb R^3} (\partial_t u \; \partial_t^2u - \partial u_t \nabla \cdot (A \nabla u)) \; dx = \int_{\Bbb R^3} \partial_t u (\partial_t^2u - \nabla \cdot (A \nabla u)) \; dx = 0, \tag{21}$
by virtue of (1).  The conservation of the energy (2) is thus estsblished.
In closing, we note that there is nothing inherently $3$-dimensional about this derivation; everything can carried into $\Bbb R^n$ provided the conditions on $u$ and $\nabla u$ as $R \to \infty$ are appropriately modified.
