For context of this question, please refer to page 689 of PDE Evans, 2nd edition. My question is at the bottom.
As an example, consider this initial-value problem in three space dimensions: $$\begin{cases}u_{tt}-\Delta u - |u|^p = 0 & \text{in }\mathbb{R}^3 \times (0,\infty) \\ \qquad \, \, \, \, \,u=g,u_t = h & \text{on }\mathbb{R}^2 \times \{t=0\}. \end{cases}\tag{8}$$ Take any smooth initial data with $$\int_{\mathbb{R}^3} g \, dx > 0, \quad \int_{\mathbb{R}^3} h \, dx > 0 \tag{9}$$ and $\operatorname{spt}g,\operatorname{spt}h \subset B(0,R)$. We will deduce from these conditions alone that there is no solution, provided the exponent $p > 1$ is small enough.
In case $\text{spt}g$ is unfamiliar to some, it means "support", which is also written conventionally as $\operatorname{supp}g$, for example.
THEOREM 2 (Nonexistence for small data). Assume that $$1 < p < 1+\sqrt{2}.$$ Then under the above conditions on $g$ and $h$, the initial-value problem $(8)$ does not have a smooth solution $u$ existing for all times $t \ge 0$.
Here is a fragment of the textbook's proof:
Proof. 1. We assume to the contrary that $u$ is in fact a solution and derive a contradiction. Since the initial data have support within $B(0,R)$, Theorem 3 from §12.1.2 implies that $u(\cdot,t)$ is supported within the ball $B(0,R+t)$.
$\quad$ Put $$I(t) = \int_{\mathbb{R}^3} u \, dx.$$ Then $$I'' = \int_{\mathbb{R}^3} u_{tt} \, dx = \int_{\mathbb{R}^3} |u|^p \, dx.$$
This proof continues, but my question concerns the expression $I''$. I understand the first equality, but how did we obtain the second equality? Namely, why is $$\int_{\mathbb{R}^3} u_{tt} \, dx = \int_{\mathbb{R}^3} |u|^p \, dx$$ when the PDE in $(8)$ gives $u_{tt}=|u|^p+\Delta u$? Because integrating this gives $$I'' = \int_{\mathbb{R}^3} u_{tt} \, dx = \int_{\mathbb{R}^3} |u|^p + \Delta u \, dx.$$ For some reason, however, the textbook "omits" the $\Delta u$ term...
I think it is just an application of (one of) Green's formulas.
As $u(\cdot, t)$ has support in $B(0,R+t)$, then $u\equiv 0$ outside $B(0,R+t)$; in particular $\nabla u \equiv 0$ on $\partial B(0,r)$ for large enough $r > R + t$.
So \begin{equation} \int_{B(0,r)}\Delta u d x = \int_{\partial B(0,r)}\nabla u \cdot \hat{n}\,d S = 0. \end{equation} Therefore \begin{equation} \int_{\mathbb{R}^3}\Delta u\, dx = \lim_{r \to \infty}\int_{B(0,r)}\Delta u\, d x = \lim_{r \to \infty} \int_{\partial B(0,r)}\nabla u \cdot \hat{n}\,d S = 0. \end{equation}