[proofplan] We prove uniqueness by applying the standard $L^2$ energy method to the difference $w=u-v$. Define \begin{align*} E:[0,T] &\to [0,\infty), \qquad t \mapsto \frac{1}{2}\|w(t,\cdot)\|_{L^2(U)}^2. \end{align*} Since $w$ is $C^1$ as an $L^2(U)$-valued map, $E$ is differentiable on $(0,T)$ and satisfies \begin{align*} E'(t) = (\partial_t w(t,\cdot),w(t,\cdot))_{L^2(U)} = (\Delta w(t,\cdot),w(t,\cdot))_{L^2(U)} = -\|\nabla w(t,\cdot)\|_{L^2(U)}^2 \leq 0. \end{align*} Since $E(0)=0$ and $E(t)\geq 0$, the energy vanishes identically, hence $w=0$ in $C^0([0,T];L^2(U))$ and therefore $u=v$. [/proofplan]

[step:Define the $L^2$ energy and compute its derivative]Define the energy function \begin{align*} E:[0,T]&\to[0,\infty) \end{align*} \begin{align*} t&\mapsto \frac{1}{2}\|w(t,\cdot)\|_{L^2(U)}^2. \end{align*} The $L^2(U)$ [inner product](/page/Inner%20Product) used here is \begin{align*} (f,g)_{L^2(U)}=\int_U f(x)g(x)\,d\mathcal{L}^n(x) \end{align*} for real-valued functions $f,g\in L^2(U)$. Since $w\in C^1([0,T];L^2(U))$, the map $E$ is continuous on $[0,T]$ and differentiable on $(0,T)$. For each $t\in(0,T)$, \begin{align*} E'(t)=(\partial_t w(t,\cdot),w(t,\cdot))_{L^2(U)}. \end{align*} Indeed, for $h\neq 0$ with $t+h\in[0,T]$, define \begin{align*} d_h:[0,T]\supset\{t\}&\to L^2(U) \end{align*} \begin{align*} t&\mapsto \frac{w(t+h,\cdot)-w(t,\cdot)}{h}. \end{align*} Then \begin{align*} \frac{E(t+h)-E(t)}{h}=(d_h,w(t,\cdot))_{L^2(U)}+\frac{h}{2}\|d_h\|_{L^2(U)}^2. \end{align*} As $h\to0$, $d_h\to\partial_t w(t,\cdot)$ in $L^2(U)$, and the displayed formula gives the derivative identity.[/step]

[guided]We first isolate the quantity that the [heat equation](/page/Heat%20Equation) controls. Define \begin{align*} E:[0,T]&\to[0,\infty) \end{align*} \begin{align*} t&\mapsto \frac{1}{2}\|w(t,\cdot)\|_{L^2(U)}^2. \end{align*} The factor \begin{align*} \frac{1}{2} \end{align*} is chosen so that differentiating the square of the norm does not leave an extra factor of $2$. The inner product on $L^2(U)$ is \begin{align*} (f,g)_{L^2(U)}=\int_U f(x)g(x)\,d\mathcal{L}^n(x), \end{align*} for real-valued $f,g\in L^2(U)$. We now justify the differentiation of the squared norm. Fix $t\in(0,T)$, and let $h\neq0$ satisfy $t+h\in[0,T]$. Define the difference quotient \begin{align*} d_h=\frac{w(t+h,\cdot)-w(t,\cdot)}{h}\in L^2(U). \end{align*} Using the [Hilbert space](/page/Hilbert%20Space) identity \begin{align*} \|a+b\|_{L^2(U)}^2-\|a\|_{L^2(U)}^2=2(a,b)_{L^2(U)}+\|b\|_{L^2(U)}^2 \end{align*} with $a=w(t,\cdot)$ and $b=w(t+h,\cdot)-w(t,\cdot)=h d_h$, we obtain \begin{align*} \frac{E(t+h)-E(t)}{h}=(d_h,w(t,\cdot))_{L^2(U)}+\frac{h}{2}\|d_h\|_{L^2(U)}^2. \end{align*} Because $w\in C^1([0,T];L^2(U))$, the difference quotient $d_h$ converges in $L^2(U)$ to $\partial_t w(t,\cdot)$ as $h\to0$. In particular, the family $\|d_h\|_{L^2(U)}$ remains bounded for $h$ sufficiently small. Therefore the second term $\frac{h}{2}\|d_h\|_{L^2(U)}^2$ tends to $0$, while the first term tends to $(\partial_t w(t,\cdot),w(t,\cdot))_{L^2(U)}$ by continuity of the $L^2$ inner product. Hence \begin{align*} E'(t)=(\partial_t w(t,\cdot),w(t,\cdot))_{L^2(U)}. \end{align*}[/guided]

[step:Use the heat equation to rewrite the energy derivative] For every $t\in(0,T)$, the hypothesis \begin{align*} \partial_t w(t,\cdot)-\Delta w(t,\cdot)=0 \end{align*} holds in $L^2(U)$. Therefore \begin{align*} \partial_t w(t,\cdot)=\Delta w(t,\cdot) \end{align*} in $L^2(U)$, and the derivative identity becomes \begin{align*} E'(t)=(\Delta w(t,\cdot),w(t,\cdot))_{L^2(U)}. \end{align*} [/step]

[step:Apply the Sobolev Green identity using the zero boundary condition] Fix $t\in(0,T)$ and define \begin{align*} z:U&\to\mathbb{R} \end{align*} \begin{align*} x&\mapsto w(t,x). \end{align*} By the hypotheses, $z\in H_0^1(U)$ and $\Delta z\in L^2(U)$ in the weak sense. The definition of the weak Laplacian gives \begin{align*} (\Delta z,\phi)_{L^2(U)}=-\int_U \nabla z(x)\cdot\nabla \phi(x)\,d\mathcal{L}^n(x) \end{align*} for every $\phi\in H_0^1(U)$. Taking $\phi=z$ is admissible because $z\in H_0^1(U)$, and therefore \begin{align*} (\Delta z,z)_{L^2(U)}=-\int_U |\nabla z(x)|^2\,d\mathcal{L}^n(x). \end{align*} Thus \begin{align*} E'(t)=-\|\nabla w(t,\cdot)\|_{L^2(U)}^2\leq0. \end{align*} Here \begin{align*} \|\nabla w(t,\cdot)\|_{L^2(U)}^2=\int_U |\nabla w(t,x)|^2\,d\mathcal{L}^n(x). \end{align*} [/step]

[step:Conclude that the energy vanishes for all times] Since $E'(t)\leq0$ for every $t\in(0,T)$, the function $E$ is nonincreasing on $[0,T]$. Also, \begin{align*} E(0)=\frac{1}{2}\|w(0,\cdot)\|_{L^2(U)}^2=0 \end{align*} because $w(0,\cdot)=0$ in $L^2(U)$. For each $t\in[0,T]$, nonincreasingness gives $E(t)\leq E(0)=0$, while the definition of $E$ gives $E(t)\geq0$. Hence \begin{align*} E(t)=0 \end{align*} for every $t\in[0,T]$. Therefore \begin{align*} \|w(t,\cdot)\|_{L^2(U)}=0 \end{align*} for every $t\in[0,T]$, so $w(t,\cdot)=0$ in $L^2(U)$ for every $t\in[0,T]$. Since $w=u-v$ as a $C^0([0,T];L^2(U))$ map, this proves \begin{align*} u=v \end{align*} in $C^0([0,T];L^2(U))$. [/step]