[proofplan] We prove that the closed interval $[a,b]$ is invariant under the parabolic flow. The constants $a$ and $b$ are barriers, but because the reaction term is only locally Lipschitz we encode the barrier comparison by writing the deviation from the barrier as a linear parabolic inequality with a bounded zero-order coefficient. A direct maximum-principle argument for this inequality gives first $u \geq a$ and then $u \leq b$. The resulting uniform $L^\infty$ bound contradicts any continuation criterion whose only finite-time obstruction is blow-up of the $L^\infty$ norm. [/proofplan]

[step:Prove a maximum principle for a linear inequality with bounded zero-order coefficient]Let $Q_T := (0,T) \times \Omega$ and let $\partial_p Q_T$ denote the parabolic boundary \begin{align*} \partial_p Q_T := \bigl(\{0\} \times \overline{\Omega}\bigr) \cup \bigl([0,T] \times \partial\Omega\bigr). \end{align*} We first record the elementary comparison fact used below. [claim:Parabolic upper bound for a bounded zero-order coefficient] Let \begin{align*} w \in C^{1,2}(Q_T) \cap C([0,T] \times \overline{\Omega}) \end{align*} and let $c: Q_T \to \mathbb{R}$ be a bounded real-valued function. Suppose \begin{align*} \partial_t w - \Delta w + c w \leq 0 \quad \text{in } Q_T \end{align*} and $w \leq 0$ on $\partial_p Q_T$. Then $w \leq 0$ on $[0,T] \times \overline{\Omega}$. [/claim] [proof] Choose a constant $M > \|c\|_{L^\infty(Q_T)}$ and define the function \begin{align*} v: [0,T] \times \overline{\Omega} \to \mathbb{R}, \qquad v(t,x) := e^{-Mt} w(t,x). \end{align*} Then $v \in C^{1,2}(Q_T) \cap C([0,T] \times \overline{\Omega})$, and $v \leq 0$ on $\partial_p Q_T$. For $(t,x) \in Q_T$, the product rule gives \begin{align*} \partial_t v(t,x) - \Delta v(t,x) = e^{-Mt}\bigl(\partial_t w(t,x) - \Delta w(t,x) - M w(t,x)\bigr). \end{align*} Since $\partial_t w - \Delta w \leq -cw$, we obtain \begin{align*} \partial_t v(t,x) - \Delta v(t,x) \leq -(M+c(t,x))v(t,x). \end{align*} Assume for contradiction that $v$ is positive somewhere on $[0,T] \times \overline{\Omega}$. Let \begin{align*} A := \max_{[0,T]\times\overline{\Omega}} v. \end{align*} Then $A>0$. Choose a number $\alpha$ with $0<\alpha<A$. Define the first hitting time \begin{align*} t_\alpha := \inf\{t\in[0,T] : \max_{x\in\overline{\Omega}} v(t,x) \geq \alpha\}. \end{align*} Continuity of the map $t\mapsto \max_{x\in\overline{\Omega}}v(t,x)$ gives $t_\alpha\in(0,T)$ and \begin{align*} \max_{x\in\overline{\Omega}}v(t_\alpha,x)=\alpha. \end{align*} Indeed, $v\leq0$ at $t=0$, while the level $\alpha<A$ is reached before any time at which the maximum $A$ is attained. Choose $x_\alpha\in\overline{\Omega}$ with $v(t_\alpha,x_\alpha)=\alpha$. The lateral boundary condition $v\leq0$ on $[0,T]\times\partial\Omega$ implies $x_\alpha\in\Omega$. By the definition of $t_\alpha$, for every $0\leq s<t_\alpha$ and every $x\in\overline{\Omega}$ one has $v(s,x)<\alpha$. Hence the time derivative at the interior point satisfies $\partial_t v(t_\alpha,x_\alpha)\geq0$. Since $x_\alpha$ is a spatial maximum point of $v(t_\alpha,\cdot)$ inside $\Omega$, the second-derivative test gives $\Delta v(t_\alpha,x_\alpha)\leq0$. Therefore \begin{align*} \partial_t v(t_\alpha,x_\alpha) - \Delta v(t_\alpha,x_\alpha) \geq 0. \end{align*} On the other hand, $v(t_\alpha,x_\alpha)=\alpha>0$ and $M+c(t_\alpha,x_\alpha)>0$, hence the differential inequality gives \begin{align*} \partial_t v(t_\alpha,x_\alpha) - \Delta v(t_\alpha,x_\alpha) \leq -(M+c(t_\alpha,x_\alpha))v(t_\alpha,x_\alpha) < 0, \end{align*} a contradiction. Therefore $v \leq 0$ on $[0,T]\times\overline{\Omega}$, and since $e^{-Mt}>0$, also $w \leq 0$. [/proof][/step]

[guided]The comparison argument we need is not the pure heat-equation maximum principle, because after subtracting a constant barrier the nonlinear term produces a bounded coefficient multiplying the error. We therefore prove the exact linear estimate needed in the sequel. Let $Q_T := (0,T) \times \Omega$ and define its parabolic boundary by \begin{align*} \partial_p Q_T := \bigl(\{0\} \times \overline{\Omega}\bigr) \cup \bigl([0,T] \times \partial\Omega\bigr). \end{align*} Suppose \begin{align*} w \in C^{1,2}(Q_T) \cap C([0,T] \times \overline{\Omega}) \end{align*} and $c: Q_T \to \mathbb{R}$ is a bounded real-valued function such that \begin{align*} \partial_t w - \Delta w + c w \leq 0 \quad \text{in } Q_T \end{align*} and $w \leq 0$ on $\partial_p Q_T$. We want to show that $w$ cannot become positive inside the cylinder. The coefficient $c$ may have either sign, so we remove it by exponential damping. Choose $M > \|c\|_{L^\infty(Q_T)}$ and define \begin{align*} v: [0,T] \times \overline{\Omega} \to \mathbb{R}, \qquad v(t,x) := e^{-Mt}w(t,x). \end{align*} The factor $e^{-Mt}$ is positive, so $v$ and $w$ have the same sign. Also $v \leq 0$ on $\partial_p Q_T$. For $(t,x)\in Q_T$, differentiating the product gives \begin{align*} \partial_t v(t,x)-\Delta v(t,x) = e^{-Mt}\bigl(\partial_t w(t,x)-\Delta w(t,x)-Mw(t,x)\bigr). \end{align*} Using $\partial_t w-\Delta w\leq -cw$ and $w=e^{Mt}v$, this becomes \begin{align*} \partial_t v(t,x)-\Delta v(t,x) \leq -(M+c(t,x))v(t,x). \end{align*} The choice of $M$ ensures $M+c(t,x)>0$ for every $(t,x)\in Q_T$. Now assume $v$ is positive somewhere. Since $[0,T]\times\overline{\Omega}$ is compact and $v$ is continuous, the maximum \begin{align*} A := \max_{[0,T]\times\overline{\Omega}} v \end{align*} is positive. The possible difficulty is that the maximum of $v$ might occur at the final time $T$, where the hypotheses do not give a two-sided time derivative. To avoid using any derivative at $T$, choose $\alpha$ with $0<\alpha<A$ and define the first time at which the spatial maximum reaches $\alpha$: \begin{align*} t_\alpha := \inf\{t\in[0,T] : \max_{x\in\overline{\Omega}} v(t,x)\geq\alpha\}. \end{align*} Because $v\leq0$ at $t=0$ and because $\alpha$ is strictly below the global maximum $A$, continuity gives $t_\alpha\in(0,T)$ and \begin{align*} \max_{x\in\overline{\Omega}}v(t_\alpha,x)=\alpha. \end{align*} Choose $x_\alpha\in\overline{\Omega}$ such that $v(t_\alpha,x_\alpha)=\alpha$. The lateral boundary condition excludes $x_\alpha\in\partial\Omega$, since $v\leq0$ there and $\alpha>0$; hence $x_\alpha\in\Omega$. At the point $(t_\alpha,x_\alpha)$ we may use the classical derivatives of $v$, because $t_\alpha<T$ and $x_\alpha\in\Omega$. For every $s<t_\alpha$ and every $x\in\overline{\Omega}$, the definition of the first hitting time gives $v(s,x)<\alpha=v(t_\alpha,x_\alpha)$. Therefore the time derivative satisfies $\partial_t v(t_\alpha,x_\alpha)\geq0$. Also $x_\alpha$ is a spatial maximum point of $v(t_\alpha,\cdot)$ in the [open set](/page/Open%20Set) $\Omega$, so the second-derivative test gives $\Delta v(t_\alpha,x_\alpha)\leq0$. Combining these two facts, \begin{align*} \partial_t v(t_\alpha,x_\alpha)-\Delta v(t_\alpha,x_\alpha)\geq0. \end{align*} But the differential inequality gives \begin{align*} \partial_t v(t_\alpha,x_\alpha)-\Delta v(t_\alpha,x_\alpha) \leq -(M+c(t_\alpha,x_\alpha))v(t_\alpha,x_\alpha)<0, \end{align*} because $M+c(t_\alpha,x_\alpha)>0$ and $v(t_\alpha,x_\alpha)=\alpha>0$. This contradiction proves $v\leq0$, and the positivity of $e^{-Mt}$ gives $w\leq0$.[/guided]

[step:Compare $u$ with the lower constant barrier $a$] Define \begin{align*} w_a: [0,T]\times\overline{\Omega} \to \mathbb{R}, \qquad w_a(t,x) := a-u(t,x). \end{align*} Then $w_a \in C^{1,2}(Q_T)\cap C([0,T]\times\overline{\Omega})$. On the parabolic boundary, the initial condition $u(0,\cdot)=u_0$ and the hypotheses $u_0\geq a$ and $g\geq a$ give \begin{align*} w_a(0,x) = a-u(0,x)=a-u_0(x) \leq 0 \quad \text{for } x\in\overline{\Omega}, \end{align*} and \begin{align*} w_a(t,x) = a-g(t,x) \leq 0 \quad \text{for } (t,x)\in [0,T]\times\partial\Omega. \end{align*} Because $u$ is continuous on the compact set $[0,T]\times\overline{\Omega}$, its range is compact. Since $f$ is locally Lipschitz, there is a constant $L_a\geq0$ such that \begin{align*} |f(r)-f(s)|\leq L_a|r-s| \end{align*} for all $r,s$ in that range together with the point $a$. Define the function $c_a: Q_T \to \mathbb{R}$ by If $u(t,x)\neq a$, set \begin{align*} c_a(t,x)=\dfrac{f(u(t,x))-f(a)}{a-u(t,x)}. \end{align*} If $u(t,x)=a$, set \begin{align*} c_a(t,x)=0. \end{align*} Then $c_a\in L^\infty(Q_T)$ and $|c_a(t,x)|\leq L_a$ where $u(t,x)\neq a$. For $(t,x)\in Q_T$, \begin{align*} \partial_t w_a(t,x)-\Delta w_a(t,x) = -\partial_t u(t,x)+\Delta u(t,x) = -f(u(t,x)). \end{align*} Using the definition of $c_a$ where $u\neq a$, and observing that the same inequality below is immediate when $u=a$, we obtain \begin{align*} \partial_t w_a(t,x)-\Delta w_a(t,x)+c_a(t,x)w_a(t,x) = -f(a) \leq 0. \end{align*} The maximum principle from the previous step gives $w_a\leq0$ on $[0,T]\times\overline{\Omega}$. Hence \begin{align*} u(t,x)\geq a \end{align*} for every $(t,x)\in[0,T]\times\overline{\Omega}$. [/step]

[step:Compare $u$ with the upper constant barrier $b$] Define \begin{align*} w_b: [0,T]\times\overline{\Omega} \to \mathbb{R}, \qquad w_b(t,x) := u(t,x)-b. \end{align*} Then $w_b \in C^{1,2}(Q_T)\cap C([0,T]\times\overline{\Omega})$. On the parabolic boundary, the initial condition $u(0,\cdot)=u_0$ and the hypotheses $u_0\leq b$ and $g\leq b$ give \begin{align*} w_b(0,x) = u(0,x)-b=u_0(x)-b \leq 0 \quad \text{for } x\in\overline{\Omega}, \end{align*} and \begin{align*} w_b(t,x) = g(t,x)-b \leq 0 \quad \text{for } (t,x)\in [0,T]\times\partial\Omega. \end{align*} Since $u([0,T]\times\overline{\Omega})\cup\{b\}$ is compact and $f$ is locally Lipschitz, there exists $L_b\geq0$ such that \begin{align*} |f(r)-f(s)|\leq L_b|r-s| \end{align*} for all $r,s$ in this compact set. Define the function $c_b: Q_T \to \mathbb{R}$ by If $u(t,x)\neq b$, set \begin{align*} c_b(t,x)=-\dfrac{f(u(t,x))-f(b)}{u(t,x)-b}. \end{align*} If $u(t,x)=b$, set \begin{align*} c_b(t,x)=0. \end{align*} Then $c_b\in L^\infty(Q_T)$ and $|c_b(t,x)|\leq L_b$ where $u(t,x)\neq b$. For $(t,x)\in Q_T$, \begin{align*} \partial_t w_b(t,x)-\Delta w_b(t,x) = \partial_t u(t,x)-\Delta u(t,x) = f(u(t,x)). \end{align*} Using the definition of $c_b$ where $u\neq b$, and observing that the same inequality below is immediate when $u=b$, we get \begin{align*} \partial_t w_b(t,x)-\Delta w_b(t,x)+c_b(t,x)w_b(t,x) = f(b) \leq 0. \end{align*} The maximum principle from the first step gives $w_b\leq0$ on $[0,T]\times\overline{\Omega}$. Hence \begin{align*} u(t,x)\leq b \end{align*} for every $(t,x)\in[0,T]\times\overline{\Omega}$. [/step]

[step:Combine the barrier bounds and rule out finite-time breakdown] The two comparison estimates give \begin{align*} a \leq u(t,x)\leq b \end{align*} for every $(t,x)\in[0,T]\times\overline{\Omega}$. Consequently \begin{align*} \|u(t,\cdot)\|_{L^\infty(\Omega)} \leq \max\{|a|,|b|\} \end{align*} for every $t\in[0,T]$. Now suppose that $u$ is the associated maximal mild solution in $C(\overline{\Omega})$ with maximal existence time $T_{\max}\in(0,\infty]$, that $u$ satisfies the preceding classical regularity and boundary conditions on every compact time interval $[0,T]\subset[0,T_{\max})$, and that its continuation criterion states that $T_{\max}<\infty$ can occur only if \begin{align*} \limsup_{t\uparrow T_{\max}}\|u(t,\cdot)\|_{L^\infty(\Omega)}=\infty. \end{align*} Applying the invariant-bound estimate on every interval $[0,T]\subset[0,T_{\max})$ gives \begin{align*} \|u(t,\cdot)\|_{L^\infty(\Omega)} \leq \max\{|a|,|b|\} \end{align*} for all $0\leq t<T_{\max}$. Therefore the blow-up alternative cannot occur. Hence $T_{\max}=\infty$, so the solution exists for all $t\geq0$. [/step]