[proofplan] We prove the identity by the standard $L^2$ energy method. First we define the scalar energy $E(r)=\|u(r,\cdot)\|_{L^2(U)}^2$ and differentiate it using the Hilbert-space chain rule, which follows directly from the assumed $C^1$ regularity into $L^2(U)$. Then we pair the equation $\partial_tu=\Delta u$ with $u$ and use the zero-trace condition to justify the Sobolev integration-by-parts identity. Integrating the resulting differential identity over $[s,t]$ gives the stated equality. [/proofplan]

[step:Define the energy and differentiate the $L^2$ norm]Let $(\cdot,\cdot)_{L^2(U)}$ denote the real Hilbert-space [inner product](/page/Inner%20Product) \begin{align*} (f,g)_{L^2(U)}:=\int_U f(x)g(x)\,d\mathcal{L}^n(x) \end{align*} for $f,g\in L^2(U)$. Define the energy function \begin{align*} E:[0,T]&\to\mathbb{R} \end{align*} \begin{align*} r&\mapsto \|u(r,\cdot)\|_{L^2(U)}^2. \end{align*} Since $u\in C^1([0,T];L^2(U))$, the map $E$ is $C^1$ on $[0,T]$, and for every $r\in[0,T]$, \begin{align*} E'(r)=2(\partial_tu(r,\cdot),u(r,\cdot))_{L^2(U)}. \end{align*}[/step]

[guided]We first isolate the quantity whose evolution we want to control. The relevant [Hilbert space](/page/Hilbert%20Space) is $L^2(U)$ with inner product \begin{align*} (f,g)_{L^2(U)}:=\int_U f(x)g(x)\,d\mathcal{L}^n(x). \end{align*} Define \begin{align*} E:[0,T]&\to\mathbb{R} \end{align*} \begin{align*} r&\mapsto \|u(r,\cdot)\|_{L^2(U)}^2. \end{align*} Because $u\in C^1([0,T];L^2(U))$, the difference quotient \begin{align*} \frac{u(r+h,\cdot)-u(r,\cdot)}{h} \end{align*} converges to $\partial_tu(r,\cdot)$ in $L^2(U)$ as $h\to0$, for each $r\in[0,T]$ where the quotient is taken inside the interval. Using the polarization of the norm, \begin{align*} \frac{E(r+h)-E(r)}{h}=\left(\frac{u(r+h,\cdot)-u(r,\cdot)}{h},u(r+h,\cdot)\right)_{L^2(U)}+\left(u(r,\cdot),\frac{u(r+h,\cdot)-u(r,\cdot)}{h}\right)_{L^2(U)}. \end{align*} The continuity of $u:[0,T]\to L^2(U)$ gives $u(r+h,\cdot)\to u(r,\cdot)$ in $L^2(U)$, and the $C^1$ assumption gives convergence of the difference quotient to $\partial_tu(r,\cdot)$ in $L^2(U)$. Passing to the limit in the two inner products yields \begin{align*} E'(r)=2(\partial_tu(r,\cdot),u(r,\cdot))_{L^2(U)}. \end{align*} This is the Hilbert-space chain rule specialized to the squared $L^2$ norm.[/guided]

[step:Use the heat equation to rewrite the time derivative of the energy] Fix $r\in[0,T]$. The hypothesis states that \begin{align*} \partial_tu(r,\cdot)=\Delta u(r,\cdot) \end{align*} in $L^2(U)$. Therefore the preceding identity gives \begin{align*} E'(r)=2(\Delta u(r,\cdot),u(r,\cdot))_{L^2(U)}. \end{align*} [/step]

[step:Prove the zero-boundary integration by parts identity]For every $v\in H^2(U)\cap H_0^1(U)$, \begin{align*} (\Delta v,v)_{L^2(U)}=-\|\nabla v\|_{L^2(U)}^2. \end{align*} Indeed, since $v\in H_0^1(U)$, there exists a sequence $(\varphi_k)_{k=1}^{\infty}$ in $C_c^\infty(U)$ such that $\varphi_k\to v$ in $H^1(U)$. For each coordinate index $i\in\{1,\dots,n\}$, the [weak derivative](/page/Weak%20Derivative) identity applied to the compactly supported smooth [test function](/page/Test%20Function) $\varphi_k$ gives \begin{align*} \int_U \varphi_k(x)\partial_{x_i}^2v(x)\,d\mathcal{L}^n(x)=-\int_U \partial_{x_i}\varphi_k(x)\partial_{x_i}v(x)\,d\mathcal{L}^n(x). \end{align*} Since $\partial_{x_i}^2v\in L^2(U)$ and $\partial_{x_i}v\in L^2(U)$, passing to the limit $k\to\infty$ in $L^2(U)$ gives \begin{align*} \int_U v(x)\partial_{x_i}^2v(x)\,d\mathcal{L}^n(x)=-\int_U |\partial_{x_i}v(x)|^2\,d\mathcal{L}^n(x). \end{align*} Summing over $i=1,\dots,n$ yields \begin{align*} (\Delta v,v)_{L^2(U)}=-\sum_{i=1}^n\int_U |\partial_{x_i}v(x)|^2\,d\mathcal{L}^n(x)=-\|\nabla v\|_{L^2(U)}^2. \end{align*}[/step]

[guided]The [integration by parts](/theorems/210) step is where the boundary condition is used. We prove the precise Sobolev identity needed here: for every $v\in H^2(U)\cap H_0^1(U)$, \begin{align*} (\Delta v,v)_{L^2(U)}=-\|\nabla v\|_{L^2(U)}^2. \end{align*} The point of working with $H_0^1(U)$ is that functions in this space can be approximated in $H^1(U)$ by compactly supported smooth functions. Thus there exists a sequence $(\varphi_k)_{k=1}^{\infty}$ with $\varphi_k\in C_c^\infty(U)$ such that \begin{align*} \varphi_k\to v \end{align*} in $H^1(U)$. In particular, $\varphi_k\to v$ in $L^2(U)$ and $\partial_{x_i}\varphi_k\to\partial_{x_i}v$ in $L^2(U)$ for each coordinate index $i\in\{1,\dots,n\}$. Fix such an index $i$. Since $v\in H^2(U)$, the second weak derivative $\partial_{x_i}^2v$ belongs to $L^2(U)$. The definition of weak derivative, tested against the compactly supported smooth function $\varphi_k$, gives \begin{align*} \int_U \varphi_k(x)\partial_{x_i}^2v(x)\,d\mathcal{L}^n(x)=-\int_U \partial_{x_i}\varphi_k(x)\partial_{x_i}v(x)\,d\mathcal{L}^n(x). \end{align*} No boundary term appears because $\varphi_k$ has compact support in $U$. Now pass to the limit. Since $\partial_{x_i}^2v\in L^2(U)$ and $\varphi_k\to v$ in $L^2(U)$, the left-hand side converges to \begin{align*} \int_U v(x)\partial_{x_i}^2v(x)\,d\mathcal{L}^n(x). \end{align*} Since $\partial_{x_i}\varphi_k\to\partial_{x_i}v$ in $L^2(U)$ and $\partial_{x_i}v\in L^2(U)$, the right-hand side converges to \begin{align*} -\int_U |\partial_{x_i}v(x)|^2\,d\mathcal{L}^n(x). \end{align*} Therefore \begin{align*} \int_U v(x)\partial_{x_i}^2v(x)\,d\mathcal{L}^n(x)=-\int_U |\partial_{x_i}v(x)|^2\,d\mathcal{L}^n(x). \end{align*} Summing this identity over all coordinate indices gives \begin{align*} (\Delta v,v)_{L^2(U)}=-\sum_{i=1}^n\int_U |\partial_{x_i}v(x)|^2\,d\mathcal{L}^n(x). \end{align*} By definition, the vector-valued gradient norm is \begin{align*} \|\nabla v\|_{L^2(U)}^2=\sum_{i=1}^n\int_U |\partial_{x_i}v(x)|^2\,d\mathcal{L}^n(x), \end{align*} so \begin{align*} (\Delta v,v)_{L^2(U)}=-\|\nabla v\|_{L^2(U)}^2. \end{align*}[/guided]

[step:Apply integration by parts at each time] For each $r\in[0,T]$, the hypothesis gives $u(r,\cdot)\in H^2(U)\cap H_0^1(U)$. Applying the preceding identity with $v=u(r,\cdot)$ gives \begin{align*} (\Delta u(r,\cdot),u(r,\cdot))_{L^2(U)}=-\|\nabla u(r,\cdot)\|_{L^2(U)}^2. \end{align*} Combining this with the differential energy identity gives, for every $r\in[0,T]$, \begin{align*} E'(r)+2\|\nabla u(r,\cdot)\|_{L^2(U)}^2=0. \end{align*} [/step]

[step:Integrate the differential identity from $s$ to $t$] Because $u\in C^0([0,T];H^2(U)\cap H_0^1(U))$, the map \begin{align*} r\mapsto \|\nabla u(r,\cdot)\|_{L^2(U)}^2 \end{align*} is continuous on $[0,T]$. Hence it is integrable with respect to $\mathcal{L}^1$ on $[s,t]$. Integrating \begin{align*} E'(r)+2\|\nabla u(r,\cdot)\|_{L^2(U)}^2=0 \end{align*} over $[s,t]$ and using the [fundamental theorem of calculus](/theorems/632) for the $C^1$ function $E:[0,T]\to\mathbb{R}$ yields \begin{align*} E(t)-E(s)+2\int_s^t \|\nabla u(r,\cdot)\|_{L^2(U)}^2\,d\mathcal{L}^1(r)=0. \end{align*} Substituting the definition of $E$ gives \begin{align*} \|u(t,\cdot)\|_{L^2(U)}^2+2\int_s^t \|\nabla u(r,\cdot)\|_{L^2(U)}^2\,d\mathcal{L}^1(r)=\|u(s,\cdot)\|_{L^2(U)}^2. \end{align*} This is the desired energy identity for all $0\le s\le t\le T$. [/step]