[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.
[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]
[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]
