[proofplan] We prove both implications by converting between the flux through $\partial V$ and the spatial divergence over $V$. In the integral-to-differential direction, the integral conservation law and the [divergence theorem](/theorems/2754) imply that the integral of the [continuous function](/page/Continuous%20Function) $\partial_t u+\operatorname{div}_x(f(u))$ over every smooth subdomain is zero; testing on small balls forces that function to vanish pointwise. In the differential-to-integral direction, we integrate the pointwise equation over $V$, justify differentiating under the integral sign from the $C^1$ regularity, and apply the [divergence theorem](/theorems/3614) in reverse. [/proofplan] [step:Define the flux field and justify the time derivative under the integral] Define the spatial flux field \begin{align*} F: (0,T) \times U \to \mathbb{R}^n,\qquad F(t,x) = f(u(t,x)). \end{align*} Since $f \in C^1(\mathbb{R};\mathbb{R}^n)$ and $u \in C^1((0,T)\times U)$, the chain rule gives $F \in C^1((0,T)\times U;\mathbb{R}^n)$, and \begin{align*} \operatorname{div}_x F(t,x) = \sum_{i=1}^n \partial_{x_i}(f_i(u(t,x))). \end{align*} Thus $\operatorname{div}_x F=\operatorname{div}_x(f(u))$. Let $V \Subset U$ be bounded with $C^1$ boundary. For each $t_0 \in (0,T)$, choose $\varepsilon>0$ such that $(t_0-\varepsilon,t_0+\varepsilon)\subset (0,T)$. Since $\partial_t u$ is continuous on the compact set $[t_0-\varepsilon/2,t_0+\varepsilon/2]\times \overline V$, it is bounded there. The one-dimensional [fundamental theorem of calculus](/theorems/632) gives, for $|h|<\varepsilon/2$, \begin{align*} \frac{u(t_0+h,x)-u(t_0,x)}{h} = \frac{1}{h}\int_{t_0}^{t_0+h}\partial_t u(s,x)\,d\mathcal{L}^1(s). \end{align*} The right-hand side converges uniformly in $x\in \overline V$ to $\partial_t u(t_0,x)$ by [uniform continuity](/page/Uniform%20Continuity) of $\partial_t u$ on the same compact set. Therefore \begin{align*} \frac{d}{dt}\bigg|_{t=t_0}\int_V u(t,x)\,d\mathcal{L}^n(x) = \int_V \partial_t u(t_0,x)\,d\mathcal{L}^n(x). \end{align*} [guided] We first isolate the object whose divergence appears in the conservation law. Define \begin{align*} F: (0,T) \times U \to \mathbb{R}^n,\qquad F(t,x) = f(u(t,x)). \end{align*} This is a legitimate $C^1$ map because $u: (0,T)\times U \to \mathbb{R}$ is $C^1$ and $f:\mathbb{R}\to\mathbb{R}^n$ is $C^1$. The chain rule then gives $F\in C^1((0,T)\times U;\mathbb{R}^n)$, so the classical divergence theorem may be applied to the map $x\mapsto F(t,x)$ on any smooth bounded subdomain $V\Subset U$. We also need to justify that differentiating the mass over $V$ gives the integral of $\partial_t u$. Fix a bounded [open set](/page/Open%20Set) $V\Subset U$ with $C^1$ boundary and fix $t_0\in(0,T)$. Choose $\varepsilon>0$ such that $(t_0-\varepsilon,t_0+\varepsilon)\subset(0,T)$. Since $\partial_t u$ is continuous on the compact cylinder $[t_0-\varepsilon/2,t_0+\varepsilon/2]\times\overline V$, it is bounded and uniformly continuous there. For $|h|<\varepsilon/2$, the fundamental theorem of calculus in the time variable gives, for every $x\in \overline V$, \begin{align*} \frac{u(t_0+h,x)-u(t_0,x)}{h} = \frac{1}{h}\int_{t_0}^{t_0+h}\partial_t u(s,x)\,d\mathcal{L}^1(s). \end{align*} Uniform continuity of $\partial_t u$ implies that this difference quotient converges uniformly in $x\in\overline V$ to $\partial_t u(t_0,x)$ as $h\to 0$. Because $V$ is bounded, $\mathcal{L}^n(V)<\infty$, so [uniform convergence](/page/Uniform%20Convergence) permits passage of the limit through the integral: \begin{align*} \frac{d}{dt}\bigg|_{t=t_0}\int_V u(t,x)\,d\mathcal{L}^n(x) = \int_V \partial_t u(t_0,x)\,d\mathcal{L}^n(x). \end{align*} This is the analytic justification behind the formal operation of moving $\partial_t$ inside the spatial integral. [/guided] [/step] [step:Derive the pointwise equation from the integral conservation law] Assume the integral conservation law holds. Fix $t\in(0,T)$ and a bounded open set $V\Subset U$ with $C^1$ boundary. By the previous step, \begin{align*} \frac{d}{dt}\int_V u(t,x)\,d\mathcal{L}^n(x) = \int_V \partial_t u(t,x)\,d\mathcal{L}^n(x). \end{align*} Applying the classical divergence theorem to the $C^1$ vector field $F(t,\cdot):\overline V\to\mathbb{R}^n$ gives \begin{align*} \int_{\partial V} F(t,x)\cdot \nu_V(x)\,d\mathcal{H}^{n-1}(x) = \int_V \operatorname{div}_x F(t,x)\,d\mathcal{L}^n(x) \end{align*} (citing a result not yet in the wiki: Classical Divergence Theorem). Hence the integral conservation law becomes \begin{align*} \int_V \left(\partial_t u(t,x)+\operatorname{div}_x(f(u(t,x)))\right)\,d\mathcal{L}^n(x)=0. \end{align*} Define \begin{align*} g_t: U \to \mathbb{R},\qquad g_t(x)=\partial_t u(t,x)+\operatorname{div}_x(f(u(t,x))). \end{align*} The function $g_t$ is continuous because $u$ and $F$ are $C^1$. Suppose there exists $x_0\in U$ with $g_t(x_0)>0$. By continuity, there is $r>0$ such that $B(x_0,r)\Subset U$ and $g_t(x)>g_t(x_0)/2$ for all $x\in B(x_0,r)$. Since $B(x_0,r)$ is a bounded smooth domain, \begin{align*} 0=\int_{B(x_0,r)} g_t(x)\,d\mathcal{L}^n(x) > \frac{g_t(x_0)}{2}\mathcal{L}^n(B(x_0,r))>0, \end{align*} a contradiction. The case $g_t(x_0)<0$ is identical after applying the previous argument to $-g_t$. Therefore $g_t(x)=0$ for every $x\in U$. Since $t\in(0,T)$ was arbitrary, \begin{align*} \partial_t u(t,x)+\operatorname{div}_x(f(u(t,x)))=0 \end{align*} for every $(t,x)\in(0,T)\times U$. [/step] [step:Recover the integral conservation law from the pointwise equation] Assume \begin{align*} \partial_t u(t,x)+\operatorname{div}_x(f(u(t,x)))=0 \end{align*} for every $(t,x)\in(0,T)\times U$. Let $V\Subset U$ be a bounded open set with $C^1$ boundary and fix $t\in(0,T)$. Integrating the pointwise identity over $V$ with respect to $\mathcal{L}^n$ gives \begin{align*} \int_V \partial_t u(t,x)\,d\mathcal{L}^n(x)+\int_V \operatorname{div}_x(f(u(t,x)))\,d\mathcal{L}^n(x)=0. \end{align*} By the first step, \begin{align*} \int_V \partial_t u(t,x)\,d\mathcal{L}^n(x)=\frac{d}{dt}\int_V u(t,x)\,d\mathcal{L}^n(x). \end{align*} Applying the classical divergence theorem to $F(t,\cdot)$ on $V$ gives \begin{align*} \int_V \operatorname{div}_x(f(u(t,x)))\,d\mathcal{L}^n(x)=\int_{\partial V} f(u(t,x))\cdot\nu_V(x)\,d\mathcal{H}^{n-1}(x) \end{align*} (citing a result not yet in the wiki: Classical Divergence Theorem). Substituting these two identities yields \begin{align*} \frac{d}{dt}\int_V u(t,x)\,d\mathcal{L}^n(x)+\int_{\partial V} f(u(t,x))\cdot\nu_V(x)\,d\mathcal{H}^{n-1}(x)=0. \end{align*} Since $V$ and $t$ were arbitrary, the integral conservation law holds for every bounded smooth $V\Subset U$ and every $t\in(0,T)$. [/step]