[proofplan] We establish log-convexity of the $L^2$ energy $e(t) = \int_\Omega w^2 \, d\mathcal{L}^n$ for the difference $w = u - v$, which solves the homogeneous heat equation with zero boundary data and vanishes at $t = T$. The Cauchy--Schwarz inequality applied to the first and second derivatives of $e$ yields $e \cdot e'' \geq (e')^2$, equivalent to convexity of $\log e$. A convex function that tends to $-\infty$ at one endpoint must be identically $-\infty$, forcing $e \equiv 0$. [/proofplan] [step:Set up the difference and compute the first two derivatives of the $L^2$ energy] Define $w := u - v$. Then $w$ solves $\partial_t w - \Delta w = 0$ in $\Omega_T$ with $w = 0$ on $\partial\Omega \times [0, T]$ and $w(\cdot, T) = 0$. Define $e(t) := \int_\Omega w(x, t)^2 \, d\mathcal{L}^n(x)$. By the [Energy Dissipation](/theorems/564) computation (differentiating $\frac{1}{2}e(t)$ and applying Green's first identity with the zero boundary condition): \begin{align*} e'(t) = 2\int_\Omega w\,\partial_t w \, d\mathcal{L}^n = 2\int_\Omega w\,\Delta w \, d\mathcal{L}^n = -2\int_\Omega |\nabla w|^2 \, d\mathcal{L}^n. \end{align*} Differentiating again: \begin{align*} e''(t) = -4\int_\Omega \nabla w \cdot \nabla(\partial_t w) \, d\mathcal{L}^n = -4\int_\Omega \nabla w \cdot \nabla(\Delta w) \, d\mathcal{L}^n. \end{align*} Integrating by parts twice (the boundary terms vanish because $w = 0$ on $\partial\Omega$ implies $\Delta w = \partial_t w$ has the required boundary regularity): \begin{align*} e''(t) = 4\int_\Omega (\Delta w)^2 \, d\mathcal{L}^n. \end{align*} [/step] [step:Establish the log-convexity inequality $e \cdot e'' \geq (e')^2$ via Cauchy--Schwarz] [claim:The energy satisfies $e(t) \cdot e''(t) \geq (e'(t))^2$ whenever $e(t) > 0$] [/claim] [proof] Apply the Cauchy--Schwarz inequality in $L^2(\Omega)$ to the functions $w$ and $\Delta w$. First note that integration by parts gives $\int_\Omega |\nabla w|^2 \, d\mathcal{L}^n = -\int_\Omega w\,\Delta w \, d\mathcal{L}^n$ (the boundary term vanishes since $w = 0$ on $\partial\Omega$). Therefore: \begin{align*} \left(\int_\Omega |\nabla w|^2 \, d\mathcal{L}^n\right)^2 = \left(\int_\Omega (-w)(-\Delta w) \, d\mathcal{L}^n\right)^2 \leq \left(\int_\Omega w^2 \, d\mathcal{L}^n\right)\left(\int_\Omega (\Delta w)^2 \, d\mathcal{L}^n\right). \end{align*} Substituting $e'(t) = -2\int |\nabla w|^2$ and $e''(t) = 4\int(\Delta w)^2$: \begin{align*} \left(\frac{-e'(t)}{2}\right)^2 \leq e(t) \cdot \frac{e''(t)}{4}, \end{align*} which rearranges to $e(t) \cdot e''(t) \geq (e'(t))^2$. [/proof] This inequality is equivalent to $\frac{d^2}{dt^2}\log e(t) \geq 0$ whenever $e(t) > 0$: by the quotient rule, $(\log e)'' = (e'' e - (e')^2)/e^2 \geq 0$. Therefore $\log e(t)$ is a convex function of $t$ on any interval where $e > 0$. [/step] [step:Conclude $e \equiv 0$ from convexity and the vanishing at $t = T$] Suppose for contradiction that $e(t_0) > 0$ for some $t_0 \in [0, T)$. On the interval $[t_0, T]$, $\log e$ is convex (where $e > 0$) and satisfies $e(T) = 0$, i.e., $\log e(t) \to -\infty$ as $t \to T^-$. For any $t_0 < t_1 < T$, convexity of $\log e$ gives: \begin{align*} \log e(t_1) \leq \frac{T - t_1}{T - t_0}\log e(t_0) + \frac{t_1 - t_0}{T - t_0}\log e(T). \end{align*} Since $\log e(T) = -\infty$, the right side is $-\infty$, so $\log e(t_1) = -\infty$, meaning $e(t_1) = 0$. This holds for every $t_1 \in (t_0, T)$. By continuity, $e(t_0) = \lim_{t_1 \to t_0^+} e(t_1) = 0$, contradicting $e(t_0) > 0$. Therefore $e(t) = 0$ for all $t \in [0, T]$, giving $w \equiv 0$ and hence $u \equiv v$. [/step]