[proofplan] The difference $w = u - v$ of two solutions satisfies the homogeneous heat equation with zero parabolic boundary data. Applying the [Parabolic Strong Maximum Principle](/theorems/560) to both $w$ and $-w$ forces $w \leq 0$ and $w \geq 0$ everywhere, giving $w \equiv 0$. [/proofplan] [step:Reduce to a homogeneous problem by taking the difference of two solutions] Suppose $u$ and $v$ both solve the stated initial/boundary-value problem. Define $w := u - v$. By linearity of the heat operator: \begin{align*} \partial_t w - \Delta w = (\partial_t u - \Delta u) - (\partial_t v - \Delta v) = f - f = 0 \quad \text{in } \Omega_T. \end{align*} On the parabolic boundary $\Gamma_T$: $w = u - v = g - g = 0$. [/step] [step:Apply the maximum principle to $w$ and $-w$] The function $w$ solves the homogeneous heat equation $\partial_t w - \Delta w = 0$ in $\Omega_T$ with $w = 0$ on $\Gamma_T$. By the [Parabolic Strong Maximum Principle](/theorems/560): \begin{align*} \max_{\overline{\Omega}_T} w = \max_{\Gamma_T} w = 0. \end{align*} The function $-w$ also solves the homogeneous heat equation with $-w = 0$ on $\Gamma_T$. Applying the [Parabolic Strong Maximum Principle](/theorems/560) to $-w$: \begin{align*} \max_{\overline{\Omega}_T} (-w) = \max_{\Gamma_T} (-w) = 0. \end{align*} [/step] [step:Conclude $u \equiv v$] The two inequalities $\max_{\overline{\Omega}_T} w = 0$ and $\max_{\overline{\Omega}_T} (-w) = 0$ give $w \leq 0$ and $-w \leq 0$ on $\overline{\Omega}_T$. Therefore $w \equiv 0$, i.e., $u \equiv v$ on $\overline{\Omega}_T$. [/step]