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?