[proofplan] We write the local energy as an integral over a moving ball and compute its derivative for $0 < s < t_0$. The [wave equation](/page/Wave%20Equation) gives a local conservation law for the energy density, while the shrinking of the ball contributes a negative boundary term. On the boundary sphere, the remaining flux integrand is controlled by the elementary inequality $2ab \le a^2+b^2$, which gives $E'(s) \le 0$. Integrating this differential inequality gives the desired monotonicity on $[0,t_0]$. [/proofplan]

[step:Define the energy density and derive the local conservation law]Define the energy density as the map $e: [0,T] \times \mathbb{R}^n \to [0,\infty)$ given by \begin{align*} e(t,x) := \frac{1}{2}\left(|\partial_t u(t,x)|^2 + |\nabla u(t,x)|^2\right). \end{align*} Since $u \in C^2([0,T] \times \mathbb{R}^n)$, the functions $\partial_t u$ and $\nabla u$ are $C^1$, and $e$ is $C^1$ on $(0,T) \times \mathbb{R}^n$. For every $(t,x) \in (0,T) \times \mathbb{R}^n$, \begin{align*} \partial_t e(t,x) = \partial_t u(t,x)\partial_t^2u(t,x) + \nabla u(t,x)\cdot \nabla \partial_t u(t,x). \end{align*} Using the wave equation $\partial_t^2u=\Delta u$, we get \begin{align*} \partial_t e(t,x) = \partial_t u(t,x)\Delta u(t,x) + \nabla u(t,x)\cdot \nabla \partial_t u(t,x). \end{align*} Define the energy flux vector field as the map $F: (0,T) \times \mathbb{R}^n \to \mathbb{R}^n$ given by \begin{align*} F(t,x) := \partial_t u(t,x)\nabla u(t,x). \end{align*} Then $F$ is $C^1$, and the product rule gives \begin{align*} \nabla_x \cdot F(t,x) = \nabla \partial_t u(t,x)\cdot \nabla u(t,x) + \partial_t u(t,x)\Delta u(t,x). \end{align*} Therefore \begin{align*} \partial_t e(t,x) = \nabla_x \cdot F(t,x) \end{align*} for every $(t,x) \in (0,T) \times \mathbb{R}^n$.[/step]

[guided]We first isolate the pointwise identity that turns the wave equation into an energy statement. Define the map $e: [0,T] \times \mathbb{R}^n \to [0,\infty)$ by \begin{align*} e(t,x) := \frac{1}{2}\left(|\partial_t u(t,x)|^2 + |\nabla u(t,x)|^2\right). \end{align*} This is the energy density at time $t$ and position $x$. Because $u \in C^2([0,T]\times\mathbb{R}^n)$, both $\partial_t u$ and the spatial gradient $\nabla u$ are $C^1$, so differentiating $e$ with respect to $t$ is justified. Applying the chain rule gives \begin{align*} \partial_t e(t,x) = \partial_t u(t,x)\partial_t^2u(t,x) + \nabla u(t,x)\cdot \nabla \partial_t u(t,x). \end{align*} The wave equation says $\partial_t^2u(t,x)=\Delta u(t,x)$, hence \begin{align*} \partial_t e(t,x) = \partial_t u(t,x)\Delta u(t,x) + \nabla u(t,x)\cdot \nabla \partial_t u(t,x). \end{align*} Now define the flux vector field as the map $F: (0,T) \times \mathbb{R}^n \to \mathbb{R}^n$ given by \begin{align*} F(t,x) := \partial_t u(t,x)\nabla u(t,x). \end{align*} This vector field is $C^1$ because $u$ is $C^2$. The spatial divergence of $F$ is computed using the product rule: \begin{align*} \nabla_x \cdot F(t,x) = \nabla \partial_t u(t,x)\cdot \nabla u(t,x) + \partial_t u(t,x)\Delta u(t,x). \end{align*} This is exactly the expression obtained for $\partial_t e(t,x)$. Therefore \begin{align*} \partial_t e(t,x) = \nabla_x \cdot F(t,x) \end{align*} for every $(t,x) \in (0,T)\times\mathbb{R}^n$. This identity is the local conservation law: changes in energy density are represented as the divergence of a spatial flux.[/guided]

[step:Differentiate the energy over the shrinking ball]Let $\mathcal{H}^{n-1}$ denote $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on hypersurfaces in $\mathbb{R}^n$, and let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on intervals in $\mathbb{R}$. Define the radius map \begin{align*} r: (0,t_0) \to (0,\infty), \qquad r(s) := t_0-s. \end{align*} Define the local energy map \begin{align*} E: [0,t_0] \to [0,\infty), \qquad E(s) := \int_{K_s} e(s,x)\, d\mathcal{L}^n(x). \end{align*} Fix $s \in (0,t_0)$. Then $r(s)>0$ and $K_s=B(x_0,r(s))$. If $n\ge 2$, its boundary is the smooth sphere $\partial K_s = \partial B(x_0,r(s))$; if $n=1$, its boundary consists of the two endpoints $x_0-r(s)$ and $x_0+r(s)$. Let $\nu_s: \partial K_s \to \mathbb{R}^n$ denote the outward unit normal, given for every $x\in\partial K_s$ by \begin{align*} \nu_s(x) := \frac{x-x_0}{|x-x_0|}. \end{align*} Since $e$ is $C^1$ on $(0,T)\times\mathbb{R}^n$, the function $x\mapsto e(s,x)$ is continuous on the compact closure $\overline{K_s}$. The family $B(x_0,r(\sigma))$ has smooth moving boundary near $\sigma=s$ for $n\ge 2$ and moving endpoints for $n=1$, and its outward normal velocity is $r'(s)=-1$. Therefore the transport formula for a smoothly moving domain, with the standard one-dimensional endpoint interpretation when $n=1$, gives \begin{align*} E'(s) = \int_{K_s} \partial_t e(s,x)\, d\mathcal{L}^n(x) - \int_{\partial K_s} e(s,x)\, d\mathcal{H}^{n-1}(x). \end{align*} Using the conservation law $\partial_t e=\nabla_x\cdot F$ from the previous step and applying the [divergence theorem](/theorems/2754) on the ball $K_s$, whose boundary has outward unit normal $\nu_s$, yields \begin{align*} \int_{K_s} \partial_t e(s,x)\, d\mathcal{L}^n(x) = \int_{\partial K_s} F(s,x)\cdot \nu_s(x)\, d\mathcal{H}^{n-1}(x). \end{align*} Therefore \begin{align*} E'(s) = \int_{\partial K_s} \left(\partial_t u(s,x)\nabla u(s,x)\cdot \nu_s(x) - \frac{1}{2}\left(|\partial_t u(s,x)|^2+|\nabla u(s,x)|^2\right)\right)\, d\mathcal{H}^{n-1}(x). \end{align*}[/step]

[guided]We now differentiate the energy over a domain that changes with time, so every object in the moving-domain formula must be declared. Let $\mathcal{H}^{n-1}$ denote $(n-1)$-dimensional Hausdorff measure on hypersurfaces in $\mathbb{R}^n$ when $n\ge 2$, and let $\mathcal{H}^0$ denote counting measure on zero-dimensional boundaries when $n=1$. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on intervals in $\mathbb{R}$. Define the radius map $r: (0,t_0) \to (0,\infty)$ by \begin{align*} r(s) := t_0-s. \end{align*} Define the local energy map $E: [0,t_0] \to [0,\infty)$ by \begin{align*} E(s) := \int_{K_s} e(s,x)\, d\mathcal{L}^n(x). \end{align*} For a fixed $s\in(0,t_0)$ we have $r(s)>0$, so $K_s=B(x_0,r(s))$ is a genuine ball. If $n\ge 2$, $\partial K_s=\partial B(x_0,r(s))$ is a smooth sphere; if $n=1$, $\partial K_s$ consists of the two endpoints $x_0-r(s)$ and $x_0+r(s)$. Its outward unit normal is the map $\nu_s: \partial K_s \to \mathbb{R}^n$ given by \begin{align*} \nu_s(x) := \frac{x-x_0}{|x-x_0|}. \end{align*} The transport formula applies because the boundary is a smooth moving sphere for $n\ge 2$ and a pair of smoothly moving endpoints for $n=1$, the radius function $r$ is $C^1$, and $e$ is $C^1$ on $(0,T)\times\mathbb{R}^n$. The normal velocity of the moving boundary is $r'(s)=-1$, meaning the boundary moves inward with speed one in the outward-normal convention. Thus \begin{align*} E'(s) = \int_{K_s} \partial_t e(s,x)\, d\mathcal{L}^n(x) - \int_{\partial K_s} e(s,x)\, d\mathcal{H}^{n-1}(x). \end{align*} The minus sign is exactly the contribution of the shrinking domain. The local conservation law from the previous step gives $\partial_t e=\nabla_x\cdot F$. Since $F$ is $C^1$ and $K_s$ is a bounded smooth domain, the [divergence theorem](/theorems/3614) applies on $K_s$ and gives \begin{align*} \int_{K_s} \partial_t e(s,x)\, d\mathcal{L}^n(x) = \int_{K_s} \nabla_x\cdot F(s,x)\, d\mathcal{L}^n(x) = \int_{\partial K_s} F(s,x)\cdot \nu_s(x)\, d\mathcal{H}^{n-1}(x). \end{align*} Substituting the definition $F(s,x)=\partial_t u(s,x)\nabla u(s,x)$ and the definition of $e$ into the transport identity gives \begin{align*} E'(s) = \int_{\partial K_s} \left(\partial_t u(s,x)\nabla u(s,x)\cdot \nu_s(x) - \frac{1}{2}\left(|\partial_t u(s,x)|^2+|\nabla u(s,x)|^2\right)\right)\, d\mathcal{H}^{n-1}(x). \end{align*} This formula separates the change in local energy into the physical flux through the boundary and the energy lost because the ball itself shrinks.[/guided]

[step:Show that the boundary flux is non-positive]For every $x \in \partial K_s$, the [Cauchy-Schwarz inequality](/theorems/432) in $\mathbb{R}^n$ and the fact that $|\nu_s(x)|=1$ give \begin{align*} \partial_t u(s,x)\nabla u(s,x)\cdot \nu_s(x) \le |\partial_t u(s,x)|\,|\nabla u(s,x)|. \end{align*} Applying $2ab \le a^2+b^2$ with $a=|\partial_t u(s,x)|$ and $b=|\nabla u(s,x)|$, we obtain \begin{align*} |\partial_t u(s,x)|\,|\nabla u(s,x)| \le \frac{1}{2}\left(|\partial_t u(s,x)|^2+|\nabla u(s,x)|^2\right). \end{align*} Thus the boundary integrand in the formula for $E'(s)$ is everywhere non-positive on $\partial K_s$, and hence \begin{align*} E'(s) \le 0 \end{align*} for every $s \in (0,t_0)$.[/step]

[guided]The only remaining issue is the sign of the boundary integral. At a boundary point $x \in \partial K_s$, the outward normal vector $\nu_s(x)$ has Euclidean norm $|\nu_s(x)|=1$. We estimate the flux term by Cauchy-Schwarz in the Euclidean [inner product](/page/Inner%20Product) on $\mathbb{R}^n$: \begin{align*} \partial_t u(s,x)\nabla u(s,x)\cdot \nu_s(x) \le |\partial_t u(s,x)|\,|\nabla u(s,x)|\,|\nu_s(x)|. \end{align*} Since $|\nu_s(x)|=1$, this becomes \begin{align*} \partial_t u(s,x)\nabla u(s,x)\cdot \nu_s(x) \le |\partial_t u(s,x)|\,|\nabla u(s,x)|. \end{align*} Now apply the elementary inequality $2ab \le a^2+b^2$ with $a=|\partial_t u(s,x)|$ and $b=|\nabla u(s,x)|$. This gives \begin{align*} |\partial_t u(s,x)|\,|\nabla u(s,x)| \le \frac{1}{2}\left(|\partial_t u(s,x)|^2+|\nabla u(s,x)|^2\right). \end{align*} Combining the two estimates, \begin{align*} \partial_t u(s,x)\nabla u(s,x)\cdot \nu_s(x) - \frac{1}{2}\left(|\partial_t u(s,x)|^2+|\nabla u(s,x)|^2\right) \le 0. \end{align*} This is the key sign: the possible outward flux through the spherical boundary is never larger than the energy removed by the inward motion of the boundary. Therefore the integral of this boundary integrand is non-positive, so \begin{align*} E'(s) \le 0 \end{align*} for every $s \in (0,t_0)$.[/guided]

[step:Integrate the differential inequality to obtain monotonicity]Let $0<s_1<s_2<t_0$. The moving-boundary formula above expresses $E'(s)$ as an integral over the smoothly varying boundary $\partial B(x_0,t_0-s)$, with the one-dimensional endpoint interpretation when $n=1$; since the integrand is continuous, $E'$ is continuous on every compact subinterval of $(0,t_0)$. Hence $E$ is $C^1$ on $(0,t_0)$. Since $E'(s)\le 0$ there, the [fundamental theorem of calculus](/theorems/632) gives \begin{align*} E(s_2)-E(s_1) = \int_{s_1}^{s_2} E'(s)\, d\mathcal{L}^1(s) \le 0. \end{align*} Hence $E(s_2)\le E(s_1)$ whenever $0<s_1<s_2<t_0$. It remains to include the endpoints. We first justify continuity of $E$ on $[0,t_0]$. Fix $s_*\in[0,t_0)$ and choose $\delta>0$ such that $s\in[0,t_0]$ and $|s-s_*|<\delta$ imply $t_0-s\le t_0-s_*+1$. On the compact cylinder $[0,t_0]\times\overline{B}(x_0,t_0-s_*+1)$, the [continuous function](/page/Continuous%20Function) $e$ is bounded by a constant $M_{s_*}>0$. [Uniform continuity](/page/Uniform%20Continuity) of $e$ on this compact cylinder and the fact that the symmetric difference of the balls $B(x_0,t_0-s)$ and $B(x_0,t_0-s_*)$ has $\mathcal{L}^n$-measure tending to $0$ as $s\to s_*$ imply $E(s)\to E(s_*)$. At $s=t_0$, for $s<t_0$ the bound of $e$ on $[0,t_0]\times\overline{B}(x_0,1)$ gives \begin{align*} 0 \le E(s) \le M_{t_0}\mathcal{L}^n(B(x_0,t_0-s)), \end{align*} and the right-hand side tends to $0$ as $s\uparrow t_0$. Since $K_{t_0}=B(x_0,0)=\varnothing$, we have \begin{align*} E(t_0)=\int_{\varnothing} e(t_0,x)\, d\mathcal{L}^n(x)=0. \end{align*} Thus $E$ is continuous on $[0,t_0]$. Taking limits as $s_1 \downarrow 0$ or $s_2 \uparrow t_0$ in the inequality for interior times gives \begin{align*} E(s_2)\le E(s_1) \end{align*} for all $0\le s_1\le s_2\le t_0$. Therefore the localized energy map $s\mapsto E(s)$ is non-increasing on $[0,t_0]$.[/step]

[guided]For interior times, the sign estimate already gives a differential inequality. Let $0<s_1<s_2<t_0$. The formula for $E'(s)$ obtained from the transport formula is an integral over the smoothly varying boundary $\partial B(x_0,t_0-s)$, with the one-dimensional endpoint interpretation when $n=1$. Because $u$ is $C^2$, the boundary integrand is continuous in both $s$ and $x$, so this boundary integral depends continuously on $s$ on every compact subinterval of $(0,t_0)$. Thus $E$ is $C^1$ on $(0,t_0)$. Since $E'(s)\le 0$ for every $s\in(0,t_0)$, the fundamental theorem of calculus gives \begin{align*} E(s_2)-E(s_1) = \int_{s_1}^{s_2} E'(s)\, d\mathcal{L}^1(s) \le 0. \end{align*} Thus $E(s_2)\le E(s_1)$ for all strictly interior times $0<s_1<s_2<t_0$. To pass to the endpoints, we need continuity of $E$ on the closed interval. Fix $s_*\in[0,t_0)$ and restrict attention to $s$ close to $s_*$. Then all balls $B(x_0,t_0-s)$ lie inside a single compact ball, for example $\overline{B}(x_0,t_0-s_*+1)$ after taking $s$ sufficiently close to $s_*$. Since $e$ is continuous, it is bounded and uniformly continuous on the compact cylinder $[0,t_0]\times\overline{B}(x_0,t_0-s_*+1)$. The difference $E(s)-E(s_*)$ is controlled by two effects: the integrand $e(s,x)$ converges uniformly to $e(s_*,x)$ on the common compact region, and the symmetric difference of the balls $B(x_0,t_0-s)$ and $B(x_0,t_0-s_*)$ has $\mathcal{L}^n$-measure tending to $0$ as $s\to s_*$. These two facts imply $E(s)\to E(s_*)$. At the terminal time $s=t_0$, the domains collapse. For $s<t_0$ close to $t_0$, continuity of $e$ gives a bound $M_{t_0}>0$ on $[0,t_0]\times\overline{B}(x_0,1)$, and therefore \begin{align*} 0 \le E(s) \le M_{t_0}\mathcal{L}^n(B(x_0,t_0-s)). \end{align*} The volume of $B(x_0,t_0-s)$ tends to $0$ as $s\uparrow t_0$, so $E(s)\to 0$. Since $K_{t_0}=B(x_0,0)=\varnothing$, we also have \begin{align*} E(t_0)=\int_{\varnothing} e(t_0,x)\, d\mathcal{L}^n(x)=0. \end{align*} Hence $E$ is continuous on $[0,t_0]$. Taking endpoint limits in the interior inequality yields \begin{align*} E(s_2)\le E(s_1) \end{align*} for all $0\le s_1\le s_2\le t_0$. This is exactly the assertion that the localized energy map $s\mapsto E(s)$ is non-increasing on $[0,t_0]$.[/guided]