Let $u \in C^2(\mathbb R^n \times [0, \infty))$ be a real solution of the wave equation
$$\begin{cases} \partial_t^2 u - \Delta_x u = 0 & \text{in } \mathbb R^n \times (0, \infty) \\ u(x, 0) = f(x), \quad u_t(x, 0) = g(x) & \text{in } x \in \mathbb R^n, f \in C_c^2(\mathbb R^n), g \in C_c^1(\mathbb R^n) \end{cases}$$
We define $k(t) := \frac 1 2 u_t(x, t)^2 d x$ as the kinetic energy, $p(t) := \frac 1 2 |\nabla_x u(x, t)^2| d x$ as the potential energy, and $e(t) := p(t) + k(t)$ as the total energy. I now want to show that $e(t)$ is constant on $[0, \infty)$.
Now I did find some threads like this one where this question is explored and answered for the case $n = 1$, but I'm not sure if the argument carries over the same for higher dimensions. Like in the answer given there, I would start with $\frac d{dt} e(t)$ and then use the Leibniz rule/differential rule for parameter integrals to pull the $\frac d{dt}$ into the integral, i.e.
$$\frac d{dt} e(t) = \frac 1 2 \int_{\mathbb R^n} \frac d{dt} \left( u_t(x, t)^2 d x + |\nabla_x u(x, t)|^2 \right) d x$$
which would get me
$$= \frac 1 2 \int_{\mathbb R^n} \left( 2 u_t u_{t t} d x + \frac d{dt} |\nabla_x u(x, t)|^2 \right) d x $$
but my problem is, how do I now handle the $\frac d{dt} |\nabla_x u(x, t)|^2$-term? Unlike in the other answer where $x$ has only one component, I'm not sure where to go from here or what to make of it.
What confuses me is that I'm also given the hint that I might first want to show: for any $t_0 > 0$, there exists a compact set $K \subseteq \mathbb R^n$ so that $\mathrm{supp} u(\cdot, t) \subseteq K$ for all $t \in [0, t_0]$. But I'm not sure what for I would actually need this.
The correct replacement for $(u^2_x)_t = 2u_xu_{xt}$ is $$ \partial_t |\nabla u|^2 = 2\nabla u\cdot\nabla u_t $$ In fact this follows from the 1D rule because for any vector valued function $f=(f_1(t),\dots,f_n(t))$ $$ \partial_t |f|^2 =\partial_t \sum_{i}f_i^2 = \sum_i \partial_t (f_i^2)=2∑f_i\partial_tf_i = 2f\cdot\partial_t f$$ One concludes what you want by integration by parts, since $\int_{ℝ^n} \nabla a \cdot \vec{b} = -\int_{ℝ^n} a\nabla\cdot \vec{b}$.