Energy method in higher dimensions for waves confusion Background Information:
$$\begin{cases}
u_{tt} - \Delta u = f(x,t), \ \ x\in\mathbb{R}^n, t> 0 \\
u(x,0) = g(x)\\
u_t(x,0) = h(x)
\end{cases}$$
Theorem - If $g(x) = h(x) = 0$ for $|x - x_0|\leq t$, and $f(x,t) = 0$ for $(x,t)\in K(x_0,t_0),$ then we have that $u(x_0,t_0) = 0$.
The proof proceeds like this:
We define $$e(t) = \frac{1}{2}\int_{|x-x_0|\leq t_0 - t}u_t^{2}(x,t) + |\nabla u(x,t)|^2 dx$$
Then
\begin{align*}
\frac{de}{dt} &= \int_{|x-x_0|\leq t_0 - t}u_t u_{tt} + \nabla u \nabla u _t - \frac{1}{2}\int_{|x-x_0| = t_0 - t}u_t^{2} |\Delta u |^2 dS\\
&= \int_{|x-x_0|\leq t_0 - t}(u_t u_{tt} - \Delta u u_t)dx + \int_{|x-x_0| = t_0 - t}(u_t\nabla u\cdot n - \frac{1}{2}u_t^{2} - \frac{1}{2}|\nabla u|^2)dS
\end{align*}
I don't understand how we go from 
$$\int_{|x-x_0|\leq t_0 - t}u_t u_{tt} + \nabla u \nabla u _t - \frac{1}{2}\int_{|x-x_0| = t_0 - t}u_t^{2} |\Delta u |^2 dS$$
to
$$\int_{|x-x_0|\leq t_0 - t}(u_t u_{tt} - \Delta u u_t)dx + \int_{|x-x_0| = t_0 - t}(u_t\nabla u\cdot n - \frac{1}{2}u_t^{2} - \frac{1}{2}|\nabla u|^2)dS$$
I know this is from Green's theorem but I am not seeing it, specifically how do we get the $-\Delta u u_t$ term? Why is it negative? Also from the $-\frac{1}{2}\int$ part how does it become positive in the next line? Where does the $\frac{1}{2}$ come from?
Please try to explain carefully in detail so that I understand.
 A: The product rule for divergences (see for example here) gives the identity
$$\nabla \cdot (u_t\nabla u) = \nabla u\cdot \nabla u_t + u_t\Delta u$$
i.e.
$$\nabla u\cdot \nabla u_t = - u_t\Delta u + \nabla \cdot (u_t\nabla u). \tag{*}$$
On the other hand, the divergence theorem gives
$$\int_{|x-x_0|\leq t_0 - t} (\nabla \cdot (u_t\nabla u))dx = \int_{|x-x_0|= t_0 - t} (u_t\nabla u \cdot n)dS.$$
Integrating (*) over $|x-x_0|\leq t_0 - t$ and applying this form of the divergence theorem gives
$$\int_{|x-x_0|\leq t_0 - t}(\nabla u\cdot \nabla u_t)dx =
 - \int_{|x-x_0|\leq t_0 - t}(u_t\Delta u)dx +
\int_{|x-x_0|= t_0 - t}(u_t\nabla u \cdot n)dS.$$
Correcting the typos in the expression for $de/dt$,
\begin{align*}
\frac{de}{dt} &= \int_{|x-x_0|\leq t_0 - t}(u_t u_{tt} + \nabla u \cdot\nabla u _t)dx - \frac{1}{2}\int_{|x-x_0| = t_0 - t}(u_t^{2} +|\nabla u |^2) dS\\
 &= \int_{|x-x_0|\leq t_0 - t}(u_t u_{tt} + \nabla u \cdot\nabla u _t)dx + \int_{|x-x_0| = t_0 - t}(-\frac{1}{2}u_t^{2} -\frac{1}{2}|\nabla u |^2) dS\\
&= \int_{|x-x_0|\leq t_0 - t}(u_t u_{tt} - u_t\Delta u)dx + \int_{|x-x_0| = t_0 - t}(u_t\nabla u\cdot n - \frac{1}{2}u_t^{2} - \frac{1}{2}|\nabla u|^2)dS.
\end{align*}
