[proofplan] We differentiate the squared $L^2$ norm in time and use the transport equation to replace $\partial_t u$ by $-b\cdot\nabla u$. The transport term is converted into a divergence by the identity $b\cdot\nabla(u^2)=2u\,b\cdot\nabla u$, and the boundary contribution vanishes because $b\cdot\nu=0$ on $\partial U$. The resulting differential identity gives an inequality for the energy $E(t)=\|u(\cdot,t)\|_{L^2(U)}^2$, and the exponential estimate follows by differentiating $e^{-Mt}E(t)$. [/proofplan] [step:Differentiate the squared $L^2$ energy in time] Define the energy function $E:[0,T]\to[0,\infty)$ by \begin{align*} E(t)=\int_U |u(x,t)|^2\,d\mathcal{L}^n(x). \end{align*} Since $u\in C^1(\overline{U}\times[0,T])$ and $U$ is bounded, differentiation under the integral sign gives, for every $t\in(0,T)$, \begin{align*} E'(t)=2\int_U u(x,t)\,\partial_t u(x,t)\,d\mathcal{L}^n(x). \end{align*} Using the equation $\partial_t u=-b\cdot\nabla u$ on $U\times(0,T)$, we obtain \begin{align*} \frac{1}{2}E'(t)=-\int_U u(x,t)\,b(x)\cdot\nabla u(x,t)\,d\mathcal{L}^n(x). \end{align*} [guided] We first isolate the quantity whose evolution we want to control. Define $E:[0,T]\to[0,\infty)$ by \begin{align*} E(t)=\int_U |u(x,t)|^2\,d\mathcal{L}^n(x). \end{align*} This is the square of the $L^2(U)$ norm of the time slice $u(\cdot,t)$. Because $u\in C^1(\overline{U}\times[0,T])$ and $U$ is bounded, the function $(x,t)\mapsto |u(x,t)|^2$ has a continuous time derivative on the compact set $\overline{U}\times[0,T]$. Therefore differentiation under the integral sign is valid, and for $t\in(0,T)$, \begin{align*} E'(t)=\int_U \partial_t\bigl(|u(x,t)|^2\bigr)\,d\mathcal{L}^n(x). \end{align*} By the chain rule, \begin{align*} \partial_t\bigl(|u(x,t)|^2\bigr)=2u(x,t)\,\partial_t u(x,t), \end{align*} so \begin{align*} E'(t)=2\int_U u(x,t)\,\partial_t u(x,t)\,d\mathcal{L}^n(x). \end{align*} The transport equation says exactly that $\partial_t u(x,t)=-b(x)\cdot\nabla u(x,t)$ for $(x,t)\in U\times(0,T)$. Substituting this into the preceding identity gives \begin{align*} \frac{1}{2}E'(t)=-\int_U u(x,t)\,b(x)\cdot\nabla u(x,t)\,d\mathcal{L}^n(x). \end{align*} This reduces the problem to computing the spatial integral of the transport term. [/guided] [/step] [step:Convert the transport term into a divergence and cancel the boundary flux] Fix $t\in(0,T)$. Define $w_t:\overline{U}\to\mathbb{R}$ by $w_t(x)=u(x,t)^2$. Let $\mathcal{H}^{n-1}$ denote the $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $\partial U$. Then $w_t\in C^1(\overline{U})$, and the product rule gives \begin{align*} \operatorname{div}(b\,w_t)(x)=(\operatorname{div} b)(x)w_t(x)+b(x)\cdot\nabla w_t(x). \end{align*} Since $\nabla w_t(x)=2u(x,t)\nabla u(x,t)$, we have \begin{align*} u(x,t)\,b(x)\cdot\nabla u(x,t) = \frac{1}{2}b(x)\cdot\nabla w_t(x). \end{align*} Therefore \begin{align*} \int_U u(x,t)\,b(x)\cdot\nabla u(x,t)\,d\mathcal{L}^n(x) = \frac{1}{2}\int_U \operatorname{div}(b\,w_t)(x)\,d\mathcal{L}^n(x) - \frac{1}{2}\int_U(\operatorname{div}b)(x)w_t(x)\,d\mathcal{L}^n(x). \end{align*} By the [Gauss-Green Theorem](/theorems/28) for the $C^1$ vector field $b\,w_t:\overline{U}\to\mathbb{R}^n$ on the smooth bounded domain $U$, \begin{align*} \int_U \operatorname{div}(b\,w_t)(x)\,d\mathcal{L}^n(x) = \int_{\partial U} w_t(x)\,b(x)\cdot\nu(x)\,d\mathcal{H}^{n-1}(x). \end{align*} The boundary integral is zero because $b(x)\cdot\nu(x)=0$ for every $x\in\partial U$. Hence \begin{align*} \int_U u(x,t)\,b(x)\cdot\nabla u(x,t)\,d\mathcal{L}^n(x) = -\frac{1}{2}\int_U(\operatorname{div}b)(x)|u(x,t)|^2\,d\mathcal{L}^n(x). \end{align*} [guided] Fix $t\in(0,T)$. The goal is to rewrite the integral involving $b\cdot\nabla u$ so that the boundary condition $b\cdot\nu=0$ can be used. Define $w_t:\overline{U}\to\mathbb{R}$ by $w_t(x)=u(x,t)^2$. Let $\mathcal{H}^{n-1}$ denote the $(n-1)$-dimensional Hausdorff measure on $\partial U$. Since $u\in C^1(\overline{U}\times[0,T])$, the time slice $x\mapsto u(x,t)$ belongs to $C^1(\overline{U})$, and hence $w_t\in C^1(\overline{U})$. The product rule for divergence gives, for every $x\in U$, \begin{align*} \operatorname{div}(b\,w_t)(x)=(\operatorname{div} b)(x)w_t(x)+b(x)\cdot\nabla w_t(x). \end{align*} By the chain rule applied to $w_t(x)=u(x,t)^2$, \begin{align*} \nabla w_t(x)=2u(x,t)\nabla u(x,t). \end{align*} Therefore \begin{align*} u(x,t)\,b(x)\cdot\nabla u(x,t)=\frac{1}{2}b(x)\cdot\nabla w_t(x). \end{align*} Substituting this into the product-rule identity and integrating over $U$ with respect to $\mathcal{L}^n$ gives \begin{align*} \int_U u(x,t)\,b(x)\cdot\nabla u(x,t)\,d\mathcal{L}^n(x) = \frac{1}{2}\int_U \operatorname{div}(b\,w_t)(x)\,d\mathcal{L}^n(x) - \frac{1}{2}\int_U(\operatorname{div}b)(x)w_t(x)\,d\mathcal{L}^n(x). \end{align*} We now apply the [Gauss-Green Theorem](/theorems/28). Its hypotheses are satisfied because $U$ is bounded with smooth boundary and $b\,w_t:\overline{U}\to\mathbb{R}^n$ is a $C^1$ vector field. The theorem gives \begin{align*} \int_U \operatorname{div}(b\,w_t)(x)\,d\mathcal{L}^n(x) = \int_{\partial U} w_t(x)b(x)\cdot\nu(x)\,d\mathcal{H}^{n-1}(x). \end{align*} The boundary condition says $b(x)\cdot\nu(x)=0$ for every $x\in\partial U$, so the boundary integral is zero. Since $w_t(x)=|u(x,t)|^2$, the transport integral becomes \begin{align*} \int_U u(x,t)\,b(x)\cdot\nabla u(x,t)\,d\mathcal{L}^n(x) = -\frac{1}{2}\int_U(\operatorname{div}b)(x)|u(x,t)|^2\,d\mathcal{L}^n(x). \end{align*} [/guided] [/step] [step:Derive the exact energy identity] Substituting the preceding identity into \begin{align*} \frac{1}{2}E'(t)=-\int_U u(x,t)\,b(x)\cdot\nabla u(x,t)\,d\mathcal{L}^n(x) \end{align*} gives \begin{align*} \frac{1}{2}E'(t) = \frac{1}{2}\int_U(\operatorname{div}b)(x)|u(x,t)|^2\,d\mathcal{L}^n(x). \end{align*} Since $E(t)=\int_U |u(x,t)|^2\,d\mathcal{L}^n(x)$, this is precisely \begin{align*} \frac{d}{dt}\frac{1}{2}\int_U |u(x,t)|^2\,d\mathcal{L}^n(x) = \frac{1}{2}\int_U(\operatorname{div}b)(x)|u(x,t)|^2\,d\mathcal{L}^n(x). \end{align*} [/step] [step:Apply the divergence bound to obtain the exponential estimate] Assume that $\|\operatorname{div}b\|_{L^\infty(U)}\leq M$ for some $M\geq 0$. From the energy identity, \begin{align*} E'(t)=\int_U(\operatorname{div}b)(x)|u(x,t)|^2\,d\mathcal{L}^n(x). \end{align*} Using $(\operatorname{div}b)(x)\leq |\operatorname{div}b(x)|\leq M$ for $\mathcal{L}^n$-a.e. $x\in U$, we get \begin{align*} E'(t)\leq M E(t) \end{align*} for every $t\in(0,T)$. Define $F:[0,T]\to[0,\infty)$ by \begin{align*} F(t)=e^{-Mt}E(t). \end{align*} Then, for $t\in(0,T)$, \begin{align*} F'(t)=e^{-Mt}\bigl(E'(t)-ME(t)\bigr)\leq 0. \end{align*} Since $F$ is continuous on $[0,T]$ and differentiable on $(0,T)$, the [mean value theorem](/theorems/186) implies that $F$ is nonincreasing on $[0,T]$. Thus $F(t)\leq F(0)$ for every $t\in[0,T]$, so \begin{align*} E(t)\leq e^{Mt}E(0). \end{align*} Taking square roots and using $E(t)=\|u(\cdot,t)\|_{L^2(U)}^2$ yields \begin{align*} \|u(\cdot,t)\|_{L^2(U)}\leq e^{Mt/2}\|u(\cdot,0)\|_{L^2(U)}. \end{align*} This proves the stated estimate. [guided] Assume that $\|\operatorname{div}b\|_{L^\infty(U)}\leq M$ for some $M\geq 0$. The exact identity already proved gives, for every $t\in(0,T)$, \begin{align*} E'(t)=\int_U(\operatorname{div}b)(x)|u(x,t)|^2\,d\mathcal{L}^n(x). \end{align*} The bound on $\operatorname{div}b$ means that $|\operatorname{div}b(x)|\leq M$ for $\mathcal{L}^n$-almost every $x\in U$. Since $|u(x,t)|^2\geq 0$, multiplication by $|u(x,t)|^2$ preserves the inequality, and therefore \begin{align*} E'(t)\leq M\int_U |u(x,t)|^2\,d\mathcal{L}^n(x)=ME(t). \end{align*} To integrate this differential inequality without invoking a separate Gronwall theorem, define $F:[0,T]\to[0,\infty)$ by \begin{align*} F(t)=e^{-Mt}E(t). \end{align*} The functions $t\mapsto e^{-Mt}$ and $E$ are differentiable on $(0,T)$, so the product rule gives \begin{align*} F'(t)=e^{-Mt}\bigl(E'(t)-ME(t)\bigr). \end{align*} Using $E'(t)\leq ME(t)$, we obtain $F'(t)\leq 0$ for every $t\in(0,T)$. Since $F$ is continuous on $[0,T]$ and differentiable on $(0,T)$, the mean value theorem implies that $F$ is nonincreasing on $[0,T]$. Hence $F(t)\leq F(0)$ for every $t\in[0,T]$. Written out, this says \begin{align*} e^{-Mt}E(t)\leq E(0). \end{align*} Multiplying by $e^{Mt}$ gives \begin{align*} E(t)\leq e^{Mt}E(0). \end{align*} Finally, $E(t)=\|u(\cdot,t)\|_{L^2(U)}^2$ and both sides are nonnegative, so taking square roots yields \begin{align*} \|u(\cdot,t)\|_{L^2(U)}\leq e^{Mt/2}\|u(\cdot,0)\|_{L^2(U)}. \end{align*} This proves the stated estimate. [/guided] [/step]