[proofplan] We first absorb the bounded zeroth-order term into an exponential weight, reducing the inequality to one with a non-positive zeroth-order coefficient. Since multiplication by a positive factor preserves the zero set, it is enough to prove the vanishing conclusion for the weighted function. The local strong parabolic maximum principle, applied in time-dependent coordinate charts, propagates a zero at positive time backward through a parabolic neighbourhood. Compactness and connectedness then let this local propagation reach every point of $M$ at every time not exceeding $t_0$. [/proofplan]

[step:Absorb the bounded zeroth-order term by an exponential weight]Since $a$ is bounded on $M \times [0,T]$, define \begin{align*} A := \sup_{(x,t) \in M \times [0,T]} |a(x,t)| < \infty. \end{align*} Define the smooth map $v: M \times [0,T] \to \mathbb{R}$ by sending each $(x,t)$ to $e^{-At}u(x,t)$. Then $v$ is nonnegative on $M \times [0,T]$, and $v(x_0,t_0)=0$. Since $e^{-At}$ is independent of $x$, the Laplace-Beltrami operator satisfies \begin{align*} \Delta_{g(t)}v(x,t)=e^{-At}\Delta_{g(t)}u(x,t) \end{align*} for every $(x,t) \in M \times [0,T]$. Also, the product rule gives \begin{align*} \partial_t v(x,t)=e^{-At}\partial_t u(x,t)-Ae^{-At}u(x,t). \end{align*} Using the assumed differential inequality for $u$, we obtain \begin{align*} \partial_t v(x,t)\geq e^{-At}\Delta_{g(t)}u(x,t)+e^{-At}a(x,t)u(x,t)-Ae^{-At}u(x,t). \end{align*} Substituting the definition of $v$ and the spatial Laplacian identity gives \begin{align*} \partial_t v(x,t)\geq \Delta_{g(t)}v(x,t)+(a(x,t)-A)v(x,t). \end{align*} Because $a(x,t)-A \leq 0$, the function $v$ satisfies \begin{align*} \partial_t v(x,t)-\Delta_{g(t)}v(x,t)-(a(x,t)-A)v(x,t)\geq 0 \end{align*} with non-positive zeroth-order coefficient $a-A$.[/step]

[guided]The zeroth-order term is harmless, but it is convenient to put it into the standard form required by the local strong parabolic maximum principle. Since $a$ is bounded on the compact spacetime cylinder $M \times [0,T]$, the number \begin{align*} A := \sup_{(x,t) \in M \times [0,T]} |a(x,t)| \end{align*} is finite. We multiply $u$ by a positive time-dependent factor and define the smooth map $v: M \times [0,T] \to \mathbb{R}$ by sending each $(x,t)$ to $e^{-At}u(x,t)$. The factor $e^{-At}$ is strictly positive, so $v \geq 0$ exactly when $u \geq 0$, and $v(x_0,t_0)=0$ exactly when $u(x_0,t_0)=0$. Now compute the differential inequality for $v$. Because the factor $e^{-At}$ does not depend on the spatial variable, the Laplace-Beltrami operator acts only on $u$: \begin{align*} \Delta_{g(t)}v(x,t)=e^{-At}\Delta_{g(t)}u(x,t). \end{align*} For the time derivative, the product rule gives \begin{align*} \partial_t v(x,t) &= e^{-At}\partial_t u(x,t)-Ae^{-At}u(x,t). \end{align*} Using the assumed inequality for $u$ gives \begin{align*} \partial_t v(x,t)\geq e^{-At}\Delta_{g(t)}u(x,t)+e^{-At}a(x,t)u(x,t)-Ae^{-At}u(x,t). \end{align*} The identity for the spatial Laplacian and the definition of $v$ then give \begin{align*} \partial_t v(x,t)\geq \Delta_{g(t)}v(x,t)+(a(x,t)-A)v(x,t). \end{align*} The choice of $A$ ensures $a(x,t)-A \leq 0$ everywhere. This is the useful sign: at a zero of the nonnegative function $v$, the lower-order term vanishes, and away from the zero it has the sign compatible with the strong maximum principle.[/guided]

[step:Apply the local strong parabolic maximum principle in coordinate cylinders]Fix a coordinate chart $(U,\varphi)$ on $M$, where $\varphi: U \to \varphi(U) \subset \mathbb{R}^n$ is a smooth coordinate map. Define the coordinate representative $\tilde v: \varphi(U) \times [0,T] \to \mathbb{R}$ by $\tilde v(z,t)=v(\varphi^{-1}(z),t)$. In these coordinates the operator $\Delta_{g(t)}$ has the uniformly parabolic local form \begin{align*} (\Delta_{g(t)}v)(\varphi^{-1}(z),t)=\sum_{i,j=1}^{n} \alpha_{ij}(z,t)\partial_{z_i}\partial_{z_j}\tilde v(z,t)+\sum_{i=1}^{n} \beta_i(z,t)\partial_{z_i}\tilde v(z,t), \end{align*} where $\alpha_{ij}:\varphi(U)\times[0,T]\to\mathbb{R}$ are the inverse-metric coefficient functions written with subscript indices, and $\beta_i:\varphi(U)\times[0,T]\to\mathbb{R}$ are smooth functions determined by $g(t)$ and its first spatial derivatives in the chart. On every compact coordinate subcylinder $K \times [s_1,s_2] \subset \varphi(U) \times (0,T]$, compactness and smoothness imply that the matrix $(\alpha_{ij}(z,t))_{i,j=1}^n$ is uniformly positive definite and that all coefficients are bounded. Thus the coordinate expression is a uniformly parabolic operator on each such subcylinder. We use the local strong parabolic maximum principle for uniformly parabolic operators in the following form. Let $W \subset \mathbb{R}^n$ be a connected open ball with compact closure in a coordinate domain, let $\sigma<\tau$, and let $q:W\times[\sigma,\tau]\to\mathbb{R}$ be a smooth nonnegative supersolution \begin{align*} \partial_t q-Lq-cq\geq 0 \end{align*} on $W\times(\sigma,\tau]$, where $L$ is a uniformly parabolic second-order operator with bounded smooth coefficients and $c\leq0$. If $q(z_*,\tau)=0$ for some $z_*\in W$, then $q(z,t)=0$ for every $(z,t)\in W\times[\sigma,\tau]$. The hypotheses apply to $\tilde v$ on every connected coordinate ball whose closure is compactly contained in $\varphi(U)$ because $\tilde v\geq0$, the coefficient $a-A$ is nonpositive, and the preceding paragraph verifies uniform parabolicity and boundedness of the local coefficients on its compact spacetime closure. Choose a coordinate ball $B\subset U$ with $x_0\in B$ and compact closure in $U$, and choose $\varepsilon_0\in(0,t_0]$ such that $[t_0-\varepsilon_0,t_0]\subset[0,T]$. Applying the local principle to $\varphi(B)\times[t_0-\varepsilon_0,t_0]$ gives \begin{align*} v(x,t)=0 \end{align*} for every $(x,t)\in B\times[t_0-\varepsilon_0,t_0]$.[/step]

[guided]We first reduce the geometric operator to the standard local parabolic form. Fix a chart $(U,\varphi)$, where $\varphi:U\to\varphi(U)\subset\mathbb{R}^n$ is smooth. Define the coordinate representative $\tilde v: \varphi(U) \times [0,T] \to \mathbb{R}$ by $\tilde v(z,t)=v(\varphi^{-1}(z),t)$. In these coordinates, \begin{align*} (\Delta_{g(t)}v)(\varphi^{-1}(z),t)=\sum_{i,j=1}^{n} \alpha_{ij}(z,t)\partial_{z_i}\partial_{z_j}\tilde v(z,t)+\sum_{i=1}^{n} \beta_i(z,t)\partial_{z_i}\tilde v(z,t). \end{align*} Here $\alpha_{ij}$ are the inverse-metric coefficients written with subscript indices, and $\beta_i$ are the first-order coefficient functions coming from the Christoffel-symbol terms in the Laplace-Beltrami operator. Since $g(t)$ is a smooth Riemannian metric and the coordinate cylinder is compact, the matrix $(\alpha_{ij})$ has eigenvalues bounded above and below by positive constants, and the coefficient functions are bounded. This is exactly the uniform parabolicity condition needed by the local theorem. The local result we use is the local strong parabolic maximum principle: a smooth nonnegative supersolution of $\partial_t q-Lq-cq\geq0$ on a connected coordinate ball with compact spacetime closure, with $L$ uniformly parabolic and with bounded smooth coefficients and $c\leq0$, that vanishes at an interior spatial point at the top time must vanish on the whole backward cylinder. In the present situation, $q=\tilde v$, the operator $L$ is the coordinate expression of the Laplace-Beltrami operator $\Delta_{g(t)}$, and $c=a-A\leq0$. The zero $v(x_0,t_0)=0$ becomes the zero $\tilde v(\varphi(x_0),t_0)=0$. Choosing a coordinate ball $B\subset U$ with $x_0\in B$ and compact closure in $U$, and choosing $\varepsilon_0\in(0,t_0]$ so that $[t_0-\varepsilon_0,t_0]\subset[0,T]$, the theorem gives \begin{align*} v(x,t)=0 \end{align*} for every $(x,t)\in B\times[t_0-\varepsilon_0,t_0]$. This gives a genuine neighbourhood in space and a genuine interval backward in time, which is the local propagation mechanism used next.[/guided]

[step:Propagate vanishing first across space and then down to time zero]For $s\in[0,t_0]$, define the zero set at time $s$ by \begin{align*} Z_s:=\{x\in M:v(x,s)=0\}. \end{align*} Continuity of $v(\cdot,s):M\to\mathbb{R}$ implies that $Z_s$ is closed. The previous step gives the following local implication: if $x\in Z_s$ and $s>0$, then there are an open neighbourhood $U_x\subset M$ of $x$ and a number $\varepsilon_x\in(0,s]$ such that \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in U_x\times[s-\varepsilon_x,s]$. In particular, $U_x\subset Z_s$, so $Z_s$ is open whenever it is nonempty and $s>0$. Since $x_0\in Z_{t_0}$ and $M$ is connected, the set $Z_{t_0}$ is nonempty, open, and closed in $M$. Hence $Z_{t_0}=M$, and therefore \begin{align*} v(x,t_0)=0 \end{align*} for every $x\in M$. Define \begin{align*} \tau:=\inf\{s\in[0,t_0]:v(x,t)=0\text{ for every }(x,t)\in M\times[s,t_0]\}. \end{align*} The set in this definition is nonempty because $t_0$ belongs to it. By continuity of $v$ on the compact set $M\times[0,t_0]$, the defining property is closed under decreasing limits, so $v(x,t)=0$ for every $(x,t)\in M\times[\tau,t_0]$. If $\tau>0$, apply the local implication at time $\tau$ to every point $x\in M=Z_\tau$. The resulting neighbourhoods $U_x$ cover the compact manifold $M$, so choose finitely many points $x_1,\dots,x_N\in M$ such that $M=\bigcup_{k=1}^N U_{x_k}$. Let \begin{align*} \varepsilon:=\min_{1\leq k\leq N}\varepsilon_{x_k}>0. \end{align*} Then $v(y,t)=0$ for every $(y,t)\in M\times[\tau-\varepsilon,\tau]$, after replacing $\tau-\varepsilon$ by $0$ if necessary. Together with the already known vanishing on $M\times[\tau,t_0]$, this gives vanishing on a strictly larger interval $M\times[\max\{0,\tau-\varepsilon\},t_0]$, contradicting the definition of $\tau$ unless $\tau=0$. Therefore \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in M\times[0,t_0]$.[/step]

[guided]The local maximum principle gives more than a single zero: if $x\in M$ is a zero at a positive time $s$, then the zero spreads to a whole spatial neighbourhood and to a short interval of earlier times. Formally, for each $s\in[0,t_0]$ define \begin{align*} Z_s:=\{x\in M:v(x,s)=0\}. \end{align*} Because $v(\cdot,s)$ is continuous, $Z_s$ is closed. If $x\in Z_s$ and $s>0$, the coordinate-cylinder result supplies an [open set](/page/Open%20Set) $U_x\subset M$ and a number $\varepsilon_x\in(0,s]$ such that \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in U_x\times[s-\varepsilon_x,s]$. Taking $t=s$ shows $U_x\subset Z_s$, so every point of $Z_s$ is an interior point. Thus $Z_s$ is open whenever $s>0$ and $Z_s$ is nonempty. At the time $t_0$, the set $Z_{t_0}$ contains $x_0$, so it is nonempty. It is also open and closed in the connected manifold $M$. Connectedness forces the only nonempty clopen subset to be the whole space, hence \begin{align*} Z_{t_0}=M. \end{align*} Equivalently, \begin{align*} v(x,t_0)=0 \end{align*} for every $x\in M$. Now we must push this vanishing all the way down to time zero. Define the earliest time from which full-slice vanishing is known by \begin{align*} \tau:=\inf\{s\in[0,t_0]:v(x,t)=0\text{ for every }(x,t)\in M\times[s,t_0]\}. \end{align*} The set is nonempty because $s=t_0$ belongs to it. Since $v$ is continuous on $M\times[0,t_0]$, if a decreasing sequence of such times tends to $\tau$, then the vanishing passes to the limit. Hence $v=0$ on $M\times[\tau,t_0]$. Suppose for contradiction that $\tau>0$. Since $v=0$ on $M\times\{\tau\}$, every point $x\in M$ lies in $Z_\tau$. Applying the local implication at each such $x$ gives open sets $U_x$ and backward time lengths $\varepsilon_x>0$. Compactness of $M$ is what prevents these time lengths from shrinking to zero globally: choose a finite subcover $U_{x_1},\dots,U_{x_N}$ of $M$ and set \begin{align*} \varepsilon:=\min_{1\leq k\leq N}\varepsilon_{x_k}>0. \end{align*} Then every point of $M$ lies in one of the finitely many $U_{x_k}$, so \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in M\times[\tau-\varepsilon,\tau]$, with the lower endpoint replaced by $0$ if $\tau-\varepsilon<0$. Combining this with the vanishing on $M\times[\tau,t_0]$ gives full vanishing starting before $\tau$, contradicting the definition of $\tau$. Therefore $\tau=0$, and \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in M\times[0,t_0]$.[/guided]

[step:Undo the exponential weight] For every $(x,t) \in M \times [0,t_0]$, we have proved \begin{align*} v(x,t)=0. \end{align*} By definition of $v$, \begin{align*} u(x,t)=e^{At}v(x,t). \end{align*} Since $e^{At}>0$ for every $t \in [0,t_0]$, it follows that \begin{align*} u(x,t)=0 \end{align*} for every $(x,t) \in M \times [0,t_0]$. This is the desired conclusion. [/step]