how to prove the uniqueness solution of the following PDE how to prove the uniqueness solution of the three dimensional wave:
$c^2\frac{\partial^2u}{\partial t^2}=\Delta u  $ 
with satisfy the boundary conditions :
$u(x,y,z,t) = F(x,y,z,t) $  on $S$
and the initial conditions:
$(u(x,y,z,0) = G(x,y,z)$ 
$\frac{\partial u}{\partial t}(x,y,z,0)=H(x,y,z)$
 A: Suppose that you have two solutions to the equation, $u(x,y,z)$ and $v(x,y,z)$. Then we must have that, for $w=v-u$,
$$
c^2\frac{\partial^2 w}{\partial t^2} = \Delta w
$$
with boundary condition
$$
w = 0 \text{ on }S
$$
and the initial conditions
$$
w(t=0) = 0
$$
and
$$
\left.\frac{\partial w}{\partial t}\right|_{t=0}=0
$$
So the problem reduces to proving that the only solution to this system of equations is $w=0$. Now, consider the volume integral
$$
E(t)=\frac{1}{2}\int_V (c^2w_t^2+(\nabla w)^2)dV
$$
Taking the derivative with respect to time, you get
$$
\frac{\partial E}{\partial t} = \int_V\left(c^2w_{tt}w_t+\nabla w\cdot \nabla \frac{\partial w}{\partial t}\right)dV
$$
Using the first Green's identity, we can write this as
$$\begin{align}
\frac{\partial E}{\partial t} &= \int_V c^2w_{tt}w_tdV+\int_S w_t\nabla w\cdot d\mathbf{S}-\int_V w_t \Delta w dV\\
&= \int_V w_T\left(c^2w_{tt}-\Delta w\right)dV+\int_S w_t\nabla w\cdot d\mathbf{S}\\
&= \int_S w_t\nabla w\cdot d\mathbf{S}
\end{align}$$
But $w=0$ on $S$ means that $w_t=0$ on $S$. Therefore, $E(t)=0$ (because $\nabla w=\mathbf{0}$ and $w_t=0$ when $t=0$). But if $w_t\neq 0$ or $\nabla w\neq \mathbf{0}$, $E(t)>0$. Therefore, $w_t=0$ and $\nabla w = \mathbf{0}$.
Therefore, $w=0$, and the solution must be unique.
