**Proof plan.** If $u_1, u_2$ are two solutions, their difference $w := u_1 - u_2$ solves the homogeneous problem with zero initial and [boundary](/page/Boundary) data. Multiply the PDE by $w$, integrate over $\Omega$, and use Green's identity to convert the Laplacian term into $-\int |\nabla w|^2$. The boundary term vanishes for both Dirichlet and Neumann conditions. This gives $\frac{d}{dt} E \leq 0$ where $E(t) = \frac{1}{2}\int |w|^2$, and since $E(0) = 0$, we conclude $w \equiv 0$. **Step 1 (Reduction to the homogeneous problem).** Let $u_1, u_2$ be two solutions to the same problem, sharing the same data. Define $w := u_1 - u_2$. Then $w$ satisfies \begin{align*} \begin{cases} \partial_t w - \kappa \Delta w = 0, & (t,x) \in Q_T, \\ w(0,x) = 0, & x \in \Omega, \end{cases} \end{align*} with homogeneous boundary conditions (of either Dirichlet or Neumann type). **Step 2 (Energy identity).** Multiply the equation $\partial_t w = \kappa \Delta w$ by $w$ and integrate over $\Omega$: \begin{align*} \int_\Omega \partial_t w \, w \, d\mathcal{L}^n = \kappa \int_\Omega \Delta w \, w \, d\mathcal{L}^n. \end{align*} The left-hand side equals $\frac{1}{2} \frac{d}{dt} \int_\Omega |w|^2 \, d\mathcal{L}^n$. For the right-hand side, apply [Green's first identity](/page/Laplace's%20Equation): \begin{align*} \kappa \int_\Omega \Delta w \, w \, d\mathcal{L}^n = -\kappa \int_\Omega |\nabla w|^2 \, d\mathcal{L}^n + \kappa \int_{\partial \Omega} w \, \partial_\nu w \, d\mathcal{H}^{n-1}. \end{align*} **Step 3 (Boundary term vanishes).** If homogeneous Dirichlet conditions hold, then $w = 0$ on $\partial\Omega$, so $\int_{\partial\Omega} w \, \partial_\nu w \, d\mathcal{H}^{n-1} = 0$. If homogeneous Neumann conditions hold, then $\partial_\nu w = 0$ on $\partial\Omega$, so the boundary [integral](/page/Integral) again vanishes. **Step 4 (Energy decay and conclusion).** Define the energy $E(t) := \frac{1}{2}\int_\Omega |w(t,x)|^2 \, d\mathcal{L}^n(x)$. From Steps 2–3: \begin{align*} \frac{d}{dt} E(t) = -\kappa \int_\Omega |\nabla w|^2 \, d\mathcal{L}^n \leq 0. \end{align*} Therefore $E(t) \leq E(0) = 0$ (since $w(0,x) = 0$). Since $E(t) \geq 0$ by definition, we conclude $E(t) = 0$ for all $t$, which forces $w \equiv 0$ and hence $u_1 = u_2$.