Q:

Uniqueness to the wave equation in $\mathbb R^n$ using the energy method

Let $u(x,t)\in{C^2}$ be a solution for the wave equation in $\mathbb{R}^n$.
\begin{align*}
u_{tt}-\Delta{u}&=0 ,\ \ \ \ \ \ \ x\in{\mathbb{R}^n}, t>0\\
u(x,0)&=f(x), \ x\in{\mathbb{R}^n}\\
u_t(x,0)&=g(x), \ x\in{\mathbb{R}^n}.
\end{align*}
Let $x_0\in{\mathbb{R}^n}$ and $T>0$, define for $0\leq t\leq T$ the energy integral:
$$e(t):=\frac{1}{2}\int_{B(x_0,T-t)} (u_t^2+|\nabla_xu|^2))\mathrm dx.$$

A: Prove for all $0\le t\le T , e'(t)\le0$ (Hint: Use the polar coordinates:
$$\int_{B(x_0,T-t)} (u_t^2+|\nabla_xu|^2)dx)=\int_{0}^{T-t}(\int_{\partial  B(x_0,r)}(u_t^2+|\nabla_xu^2|)dSx)dr$$
then use Green's formula and finally use the formula $xy\le 1/2(x^2+y^2)$).

B: Prove that there is only one solution to the wave equation in $\mathbb{R}^n$.

I tried many many times to solve part A according to the tips and using the derivation rules for integrals, but every time I was stuck.
In part B,by assuming that there are two solutions to the given problem $u_1 and u_2$, defining $u:=u_1-u_2$. Then $u$ solves the homogeneous problem. using A, $e(t)$ is an decreasing function. Note that $e(0)=0$, so $e(t)\leq 0$. On the other hand, clearly $e(t)\geq 0$ by definition of $e(t)$, finally $e(t)=0$. Thus both $u_t=0$ and $\nabla {u}=0$ , then $u$ is a constant, $u(x,0)=0$ leading to $u=0$ and $u_1=u_2$. That proves that there's only one solution to the wave equation $\forall x  \in \mathbb{R}^n$ and $t \in [0,T]$.
Is that right?
Can someone help me in part A and give me tips for part B?

A:

One has
$$2e(t):= \int_{B(x_0,T-t)} (u_t^2+|\nabla_xu|^2)\mathrm dx)=\int_{0}^{T-t}\left(\int_{\partial  B(x_0,r)}(u_t^2+|\nabla_xu^2|)\mathrm dS_x \right)dr$$
which gives
\begin{align*}
e'(t) &= -\frac{1}{2}\int_{\partial  B(x_0,T-t)}(u_t^2+|\nabla_xu^2|)\mathrm dS_x + \int_0^{T-t} \int_{\partial B(x_0,r)}(u_tu_{tt}+\langle \nabla_xu, \nabla_x u_t\rangle)\mathrm dS_x \mathrm dt\\
&=-\frac{1}{2}\int_{\partial  B(x_0,T-t)}(u_t^2+|\nabla_xu^2|)\mathrm dS_x + \int_{B(x_0, T-t)}(u_tu_{tt}+\langle \nabla_xu, \nabla_x u_t\rangle)\mathrm dx
\end{align*}
The Green's theorem gives
\begin{align*}
\int_{B(x_0,T-t)}\langle \nabla_xu, \nabla_x u_t\rangle\mathrm d x &= \int_{B(x_0,T-t)}(\nabla_x (u_t \nabla_x u) - u_t \Delta u)\mathrm d x \\
&= \int_{\partial B(x_0,T-t)} u_t \langle \nabla_x u, \vec n\rangle \ \mathrm dS_x-\int_{B(x_0, T-t)} u_t \Delta u\ \mathrm d x 
\end{align*}
where $\vec n$ is the unit normal vector of $\partial B(x_0, T-t)$. Using the equation $u_{tt} = \Delta u$, we have
\begin{align*}
e'(t) = \frac{1}{2}\int_{\partial  B(x_0,T-t)}(-u_t^2-|\nabla_xu|^2 + 2 u_t \langle \nabla_x u, \vec n\rangle )\mathrm dS_x 
\end{align*}
Using
$$  |u_t \langle \nabla_x u, \vec n\rangle | \le |u_t| |\nabla _x u| | \vec n| = |u_t| |\nabla _x u| \le \frac{1}{2}(u_t^2 + |\nabla _x u|^2),$$
we obtain that $e'(t) \le 0$.

