[proofplan] We prove the comparison on an arbitrary compact time subinterval $[0,S] \subset I$, then let $S$ vary. The main idea is to compare $u$ with the strict lower barrier $\varphi(t) - \varepsilon e^{At}$, where $A$ dominates the Lipschitz constant of $F$ on the relevant compact range. If the strict comparison fails, compactness of $M$ gives a first spacetime contact point. At that point the spatial Laplacian has the favorable sign and the differential inequality forces the time derivative to be positive, contradicting the first-contact condition. [/proofplan] [step:Fix a compact time interval and choose a Lipschitz constant for the reaction term] Let $S \in I$ with $0 \le S \le T$ be fixed. If $S = 0$, the conclusion follows from the initial comparison, so assume $S > 0$. Define the compact set of values \begin{align*} K_S := \{u(x,t) : (x,t) \in M \times [0,S]\} \cup \{\varphi(t) : t \in [0,S]\} \subset \mathbb{R}. \end{align*} The set $K_S$ is compact by the Extreme Value Theorem, applied to the continuous maps $u: M \times [0,S] \to \mathbb{R}$ and $\varphi: [0,S] \to \mathbb{R}$ on compact domains. Since $F$ is locally Lipschitz in the first variable uniformly on compact subsets of $\mathbb{R} \times [0,T]$, there exists a constant $L_S \ge 0$ such that \begin{align*} |F(a,t) - F(b,t)| \le L_S |a-b| \end{align*} for all $a,b \in K_S$ and all $t \in [0,S]$. Choose a constant $A > L_S$. [guided] We first localize the argument to a fixed time interval $[0,S] \subset I$. This is necessary because the solution $\varphi$ may only be defined on the interval $I$, and the Lipschitz constant for $F$ is only assumed locally. We need a Lipschitz constant for $F$ on exactly the range where the comparison will take place. Define \begin{align*} K_S := \{u(x,t) : (x,t) \in M \times [0,S]\} \cup \{\varphi(t) : t \in [0,S]\}. \end{align*} The map $u: M \times [0,S] \to \mathbb{R}$ is continuous, and $M \times [0,S]$ is compact because $M$ is compact. By the Extreme Value Theorem, the image $u(M \times [0,S])$ is compact. Similarly, $\varphi([0,S])$ is compact because $\varphi$ is continuous on the compact interval $[0,S]$. Therefore $K_S$ is a compact subset of $\mathbb{R}$. The hypothesis on $F$ says that, on compact subsets of $\mathbb{R} \times [0,T]$, the dependence on the first variable is Lipschitz uniformly in the time variable. Applying this to $K_S \times [0,S]$, there is $L_S \ge 0$ such that \begin{align*} |F(a,t) - F(b,t)| \le L_S |a-b| \end{align*} for all $a,b \in K_S$ and all $t \in [0,S]$. We then choose $A > L_S$. This strict inequality is the quantitative ingredient that will make the exponential barrier produce a contradiction at the first contact point. [/guided] [/step] [step:Introduce a strict exponential barrier] Fix $\varepsilon > 0$. Define the function $w_\varepsilon: M \times [0,S] \to \mathbb{R}$ by \begin{align*} w_\varepsilon(x,t) := u(x,t) - \varphi(t) + \varepsilon e^{At}. \end{align*} Then $w_\varepsilon$ is smooth in $x$ and $C^1$ in $t$. At $t=0$, \begin{align*} w_\varepsilon(x,0) = u(x,0) - \varphi(0) + \varepsilon \ge \inf_{y \in M} u(y,0) - \varphi(0) + \varepsilon \ge \varepsilon > 0 \end{align*} for every $x \in M$. [guided] We now make the comparison strict. Fix $\varepsilon > 0$ and define the function $w_\varepsilon: M \times [0,S] \to \mathbb{R}$ by \begin{align*} w_\varepsilon(x,t) := u(x,t) - \varphi(t) + \varepsilon e^{At}. \end{align*} The added term $\varepsilon e^{At}$ is positive and will later produce the strictly positive error $(A-L_S)\varepsilon e^{At}$ at a possible first contact point. Since $u$ is smooth on spacetime, $\varphi$ is $C^1$ on $[0,S]$, and $t \mapsto \varepsilon e^{At}$ is smooth, the function $w_\varepsilon$ is smooth in the spatial variable and $C^1$ in time. At the initial time, the hypothesis $\varphi(0) \le \inf_M u(\cdot,0)$ gives, for every $x \in M$, \begin{align*} w_\varepsilon(x,0) = u(x,0) - \varphi(0) + \varepsilon \ge \inf_{y \in M} u(y,0) - \varphi(0) + \varepsilon \ge \varepsilon > 0. \end{align*} Thus the strict barrier starts strictly positive everywhere on $M$. [/guided] [/step] [step:Assume a first contact point exists and record its differential properties] Suppose, toward a contradiction, that $w_\varepsilon(x,t) < 0$ for some $(x,t) \in M \times [0,S]$. Define $m: [0,S] \to \mathbb{R}$ by \begin{align*} m(t) := \min_{x \in M} w_\varepsilon(x,t). \end{align*} The function $m$ is continuous because $w_\varepsilon$ is continuous and $M$ is compact. Since $m(0)>0$ and $m(t)<0$ for at least one $t \in [0,S]$, the set \begin{align*} E := \{t \in [0,S] : m(t) \le 0\} \end{align*} is nonempty. Define \begin{align*} t_0 := \inf E. \end{align*} Then $0 < t_0 \le S$. Since $m$ is continuous, $m(t_0)=0$: if $m(t_0)>0$, then continuity gives $m>0$ on a neighbourhood of $t_0$, contradicting the definition of $t_0$ as the infimum of $E$; if $m(t_0)<0$, then continuity gives a point of $E$ strictly smaller than $t_0$. By compactness of $M$, the Extreme Value Theorem gives $x_0 \in M$ such that $w_\varepsilon(x_0,t_0)=m(t_0)=0$. Also $m(t)>0$ for every $t \in [0,t_0)$, hence $w_\varepsilon(x,t)>0$ for every $x \in M$ and every $t \in [0,t_0)$. At the spatial minimum of $w_\varepsilon(\cdot,t_0)$, the second-derivative test on the Riemannian manifold $(M,g(t_0))$ gives \begin{align*} \Delta_{g(t_0)} w_\varepsilon(x_0,t_0) \ge 0. \end{align*} Define the one-variable function $h: [0,t_0] \to \mathbb{R}$ by $h(t):=w_\varepsilon(x_0,t)$. Then $h(t)>0$ for $0 \le t<t_0$ and $h(t_0)=0$, so the left endpoint derivative at $t_0$ satisfies \begin{align*} \partial_t w_\varepsilon(x_0,t_0) = h'(t_0) \le 0. \end{align*} [guided] Assume, for contradiction, that the strict comparison fails on $M \times [0,S]$. Thus there is a point $(x,t) \in M \times [0,S]$ with $w_\varepsilon(x,t)<0$. To turn this into a first-contact point, define the spatial minimum function $m: [0,S] \to \mathbb{R}$ by \begin{align*} m(t) := \min_{x \in M} w_\varepsilon(x,t). \end{align*} This minimum exists for each $t$ by the Extreme Value Theorem, because $M$ is compact and $w_\varepsilon(\cdot,t)$ is continuous. The function $m$ is continuous because a [continuous function](/page/Continuous%20Function) on the compact product $M \times [0,S]$ is uniformly continuous, so nearby times change all spatial values of $w_\varepsilon$ by a uniformly small amount. We know $m(0)>0$ from the strict initial comparison. The assumed negative value gives $m(t)<0$ for at least one time $t$. Therefore the set \begin{align*} E := \{t \in [0,S] : m(t) \le 0\} \end{align*} is nonempty. Define \begin{align*} t_0 := \inf E. \end{align*} Because $m(0)>0$, continuity gives a neighbourhood of $0$ on which $m>0$, so $t_0>0$; and since $E \subset [0,S]$ is nonempty, $t_0 \le S$. Continuity also gives $m(t_0)=0$. Indeed, $m(t_0)>0$ would keep $m$ positive on a neighbourhood of $t_0$, contradicting that $t_0$ is the infimum of times with $m \le 0$; while $m(t_0)<0$ would put earlier times in $E$ by continuity. By the Extreme Value Theorem, compactness of $M$ gives a point $x_0 \in M$ where the minimum at time $t_0$ is achieved, so $w_\varepsilon(x_0,t_0)=m(t_0)=0$. The definition of $t_0$ gives $m(t)>0$ for every $t \in [0,t_0)$, hence $w_\varepsilon(x,t)>0$ for every $x \in M$ and every $t \in [0,t_0)$. Now record the two differential consequences of being a first spacetime contact. First, $x_0$ is a spatial minimum of the smooth function $w_\varepsilon(\cdot,t_0)$ on the Riemannian manifold $(M,g(t_0))$. By the second-derivative test, the Hessian is nonnegative in every tangent direction at $x_0$, and taking the trace with respect to the positive definite metric $g(t_0)$ gives \begin{align*} \Delta_{g(t_0)} w_\varepsilon(x_0,t_0) \ge 0. \end{align*} Second, define the one-variable function $h: [0,t_0] \to \mathbb{R}$ by $h(t):=w_\varepsilon(x_0,t)$. Then $h(t)>0$ for $0 \le t<t_0$ and $h(t_0)=0$. Thus the left derivative at $t_0$ cannot be positive, and since $w_\varepsilon$ is $C^1$ in time, \begin{align*} \partial_t w_\varepsilon(x_0,t_0) = h'(t_0) \le 0. \end{align*} These are the only maximum-principle facts needed in the proof, and both have been derived from compactness, continuity, and the elementary second-derivative test. [/guided] [/step] [step:Use the differential inequality to contradict first contact] Because $\varphi'(t_0)=F(\varphi(t_0),t_0)$ and $\Delta_{g(t_0)}\varphi(t_0)=0$, we compute at $(x_0,t_0)$. Differentiating the definition of $w_\varepsilon$ in time gives \begin{align*} \partial_t w_\varepsilon(x_0,t_0) = \partial_t u(x_0,t_0) - \varphi'(t_0) + A\varepsilon e^{At_0}. \end{align*} Using the assumed differential inequality for $u$ and the ODE for $\varphi$ gives \begin{align*} \partial_t w_\varepsilon(x_0,t_0) \ge \Delta_{g(t_0)}u(x_0,t_0) + F(u(x_0,t_0),t_0) - F(\varphi(t_0),t_0) + A\varepsilon e^{At_0}. \end{align*} Since $\varphi(t_0)$ and $\varepsilon e^{At_0}$ are spatially constant, this is \begin{align*} \partial_t w_\varepsilon(x_0,t_0) \ge \Delta_{g(t_0)}w_\varepsilon(x_0,t_0) + F(u(x_0,t_0),t_0) - F(\varphi(t_0),t_0) + A\varepsilon e^{At_0}. \end{align*} At the contact point $w_\varepsilon(x_0,t_0)=0$, so \begin{align*} u(x_0,t_0) = \varphi(t_0) - \varepsilon e^{At_0}. \end{align*} Both $u(x_0,t_0)$ and $\varphi(t_0)$ belong to $K_S$, so the Lipschitz estimate gives \begin{align*} F(u(x_0,t_0),t_0) - F(\varphi(t_0),t_0) \ge -L_S |u(x_0,t_0)-\varphi(t_0)| = -L_S\varepsilon e^{At_0}. \end{align*} Using also $\Delta_{g(t_0)}w_\varepsilon(x_0,t_0)\ge 0$, we obtain \begin{align*} \partial_t w_\varepsilon(x_0,t_0) \ge (A-L_S)\varepsilon e^{At_0} > 0. \end{align*} This contradicts the first-contact inequality $\partial_t w_\varepsilon(x_0,t_0) \le 0$. Therefore \begin{align*} w_\varepsilon(x,t) \ge 0 \end{align*} for every $(x,t) \in M \times [0,S]$. [guided] At the first contact point, the differential inequality for $u$ must be incompatible with the time-derivative sign obtained above. Since $\varphi$ solves the scalar ODE, $\varphi'(t_0)=F(\varphi(t_0),t_0)$. Since $\varphi(t_0)$ is constant as a function of the spatial variable $x$, its Laplacian with respect to $g(t_0)$ is zero. Therefore, at $(x_0,t_0)$, differentiating the definition of $w_\varepsilon$ in time gives \begin{align*} \partial_t w_\varepsilon(x_0,t_0) = \partial_t u(x_0,t_0) - \varphi'(t_0) + A\varepsilon e^{At_0}. \end{align*} The assumed parabolic differential inequality for $u$ and the identity $\varphi'(t_0)=F(\varphi(t_0),t_0)$ imply \begin{align*} \partial_t w_\varepsilon(x_0,t_0) \ge \Delta_{g(t_0)}u(x_0,t_0) + F(u(x_0,t_0),t_0) - F(\varphi(t_0),t_0) + A\varepsilon e^{At_0}. \end{align*} Finally, $\Delta_{g(t_0)}\varphi(t_0)=0$ and $\Delta_{g(t_0)}(\varepsilon e^{At_0})=0$ because both terms are spatially constant, so \begin{align*} \partial_t w_\varepsilon(x_0,t_0) \ge \Delta_{g(t_0)}w_\varepsilon(x_0,t_0) + F(u(x_0,t_0),t_0) - F(\varphi(t_0),t_0) + A\varepsilon e^{At_0}. \end{align*} The contact identity $w_\varepsilon(x_0,t_0)=0$ gives \begin{align*} u(x_0,t_0) = \varphi(t_0) - \varepsilon e^{At_0}. \end{align*} Both values $u(x_0,t_0)$ and $\varphi(t_0)$ lie in the compact set $K_S$ by its definition. Hence the Lipschitz estimate for $F$ on $K_S \times [0,S]$ gives \begin{align*} F(u(x_0,t_0),t_0) - F(\varphi(t_0),t_0) \ge -L_S |u(x_0,t_0)-\varphi(t_0)| = -L_S\varepsilon e^{At_0}. \end{align*} Combining this lower bound with the spatial minimum inequality $\Delta_{g(t_0)}w_\varepsilon(x_0,t_0)\ge 0$ yields \begin{align*} \partial_t w_\varepsilon(x_0,t_0) \ge (A-L_S)\varepsilon e^{At_0} > 0, \end{align*} because $A>L_S$ and $\varepsilon e^{At_0}>0$. This contradicts the first-contact consequence $\partial_t w_\varepsilon(x_0,t_0)\le 0$. Therefore the assumed negative value cannot exist, and \begin{align*} w_\varepsilon(x,t) \ge 0 \end{align*} for every $(x,t) \in M \times [0,S]$. [/guided] [/step] [step:Let the strict barrier vanish and conclude the comparison] From $w_\varepsilon \ge 0$ on $M \times [0,S]$, we have \begin{align*} u(x,t) \ge \varphi(t) - \varepsilon e^{At} \end{align*} for every $(x,t) \in M \times [0,S]$ and every $\varepsilon > 0$. Letting $\varepsilon \downarrow 0$ gives \begin{align*} u(x,t) \ge \varphi(t) \end{align*} for every $(x,t) \in M \times [0,S]$. Since $S \in I$ was arbitrary, the same inequality holds for every $x \in M$ and every $t \in I$. This proves the theorem. [guided] The strict barrier estimate says that, for every fixed $\varepsilon>0$, \begin{align*} w_\varepsilon(x,t) \ge 0 \end{align*} for every $(x,t) \in M \times [0,S]$. Substituting the definition of $w_\varepsilon$ gives \begin{align*} u(x,t) \ge \varphi(t) - \varepsilon e^{At} \end{align*} for every $(x,t) \in M \times [0,S]$. Now fix an arbitrary point $(x,t) \in M \times [0,S]$. The constant $e^{At}$ is finite and positive, and the preceding inequality holds for every $\varepsilon>0$. Letting $\varepsilon \downarrow 0$ gives \begin{align*} u(x,t) \ge \varphi(t). \end{align*} Because $(x,t)$ was arbitrary, the comparison holds on $M \times [0,S]$. Finally, $S \in I$ was arbitrary, so the same conclusion holds for every $x \in M$ and every time $t \in I$ for which $\varphi$ is defined. This is exactly the asserted comparison. [/guided] [/step]