Wave Equation, Energy methods. I am reading the book of Evans, Partial differential Equations ... wave equation section 2.4; subsection 2.4.3: Energy methods. Arriving at the theorem:
Theorem 5 (Uniqueness for wave equation). There exists at most one function $u \in C^{2}(\overline{U}_{T})$ solving 
$u_{tt} -\Delta u=f $ in $ U_{T}$
$u=g $ on $ \Gamma_{T}$
$u_{t}=h$ on $U \times \{t=0\}.$  
Proof. If $\tilde{u}$ is another such solution, then $ w:=u-\tilde{u}$ solves 
$w_{tt} -\Delta w=0 $ in $ U_{T}$
$w=0 $ on $ \Gamma_{T}$
$w_{t}=0$ on $U \times \{t=0\}.$ 
Define the "energy" 
$e(t):=\frac{1}{2} \int_{U} w^{2}_{t}(x,t)+ \mid Dw(x,t)\mid ^{2} dx (0\leq t \leq T).$ 
We compute 
$\dot{e}(t)=\int_{U} w_{t}w_{tt}+ Dw \cdot Dw_{t}dx (\cdot = \frac{d}{dt})$ 
$=\int_{U}w_{t}(w_{tt} - \Delta w)dx=0$. 
There is no boundary term since $w=0$, and hence $w_{t}=0$, on $\partial U \times [0,T].$ Thus for all $0\leq t \leq T, e(t)= e(0)=0$, and so $w_{t}, Dw \equiv 0$ within $U_{T}$. Since $w \equiv 0$ on$ U \times \{t=0\}$, we conclude $w=u-\tilde {u}\equiv 0$ in $U_{T}$. 
I have two questions:
1) What is the motivation for the definition of $e(t)$
2)$\int_{U} w_{t}w_{tt}+ Dw \cdot Dw_{t}dx $ 
$=\int_{U}w_{t}(w_{tt} - \Delta w)dx$. How to justify this equality? 
Thank very much.

This quantity, the definition of $e(t)$, can be easily recognized as the Hamiltonian (basically another name of energy) of the system by a physics student. Let's explore more details.
The PDE $w_{tt}-\Delta w=0$ is equivalent to a variation problem (under some boundary conditions, maybe), whose Lagrangian (or Lagrangian density) is
$$
\mathcal L = \frac{1}{2}(w_t^2-(\nabla w)\cdot(\nabla w))
$$
There is a standard process obtaining the Hamiltonian from the Lagrangian. First, calculate the canonical momentum $\pi$:
$$
\pi = \frac{\partial \mathcal L}{\partial w_t}=w_t.
$$
Then use the Legendre transformation:
$$
\mathcal H = \pi w_t - \mathcal L = \frac{1}{2} (w_t^2+(\nabla w)^2),
$$
which is the integrand in the definition of $e(t)$. This is actually the Hamitonian density and should be integrated over space to give the total energy.

P.S.
As a somewhat more complicated case, consider the following equation:
$$
u_{tt} - \nabla \cdot (c^2(x) \nabla u)+q(x)u = 0
$$
with some appropriate boundary conditions and initial conditions imposed. The two given functions $c,q\ge 0$ and depend on $x$ only. You can check that the Lagrangian is
$$
\mathcal L = \frac{1}{2}(u_t^2 - c^2 (\nabla u)^2-qu^2),
$$
which gives the Hamiltonian
$$
\mathcal H = \frac{1}{2} (u_t^2+c^2(\nabla u)^2+qu^2)
$$
