[proofplan] We rewrite the scalar conservation law as the vanishing spacetime divergence of the flux $F=(u,f(u))$. Testing the weak formulation against functions supported in a small neighbourhood of the hypersurface and integrating by parts separately on the two sides leaves only boundary terms on $\Sigma$, because the classical equation cancels the interior terms. The two boundary contributions have opposite orientations and combine into the jump of $F$ across $\Sigma$ dotted with the spacetime normal. Since the trace of the [test function](/page/Test%20Function) on $\Sigma$ is arbitrary locally, that jump coefficient must vanish pointwise. [/proofplan] [step:Introduce the spacetime flux and localize near the hypersurface] Set $Q := (0,T)\times U$. Let $Q_-$ and $Q_+$ denote the two open sides of $Q\setminus\Sigma$, labelled so that the one-sided traces of $u$ from $Q_-$ and $Q_+$ are $u_-$ and $u_+$, respectively. Define the spacetime flux \begin{align*} F: Q\setminus \Sigma \to \mathbb{R}^{n+1}, \qquad F(t,x) := (u(t,x), f(u(t,x))). \end{align*} On each side $Q_\pm$, the chain rule and the classical conservation law give \begin{align*} \operatorname{div}_{t,x} F = \partial_t u+\operatorname{div}_x f(u) = 0. \end{align*} Fix a point $p \in \Sigma$. It suffices to prove the jump relation in a neighbourhood of $p$. Choose an open neighbourhood $V \subset Q$ of $p$ such that $\Sigma \cap V$ separates $V$ into two smooth open sets $V_- := V \cap Q_-$ and $V_+ := V \cap Q_+$. Let \begin{align*} N:\Sigma\cap V \to \mathbb{R}^{n+1}\setminus\{0\}, \qquad N(t,x):=(n_t(t,x),n_x(t,x)) \end{align*} denote the stated nonzero normal field. Let $\nu:\Sigma\cap V\to\mathbb{R}^{n+1}$ denote the unit normal oriented from $V_-$ toward $V_+$. Since $N$ and $\nu$ are normal to the same one-dimensional normal space at each point, there is a nonvanishing scalar function \begin{align*} \lambda:\Sigma\cap V \to \mathbb{R}\setminus\{0\} \end{align*} such that $N=\lambda\nu$ on $\Sigma\cap V$. We choose $V$ small enough that every test function used below has compact support in $V$, so no boundary term appears from $\partial V$. [/step] [step:Integrate by parts separately on the two sides of the hypersurface] Let $\varphi \in C_c^1(V)$. Since $u$ is a weak solution on $Q$, extending $\varphi$ by zero outside $V$ gives \begin{align*} 0 = \int_Q \left(u\,\partial_t \varphi+f(u)\cdot \nabla_x \varphi\right)\,d\mathcal{L}^{n+1}(t,x). \end{align*} Because $\varphi$ is supported in $V$ and $\Sigma$ has zero $\mathcal{L}^{n+1}$ measure, this becomes \begin{align*} 0 = \int_{V_-} F\cdot \nabla_{t,x}\varphi\,d\mathcal{L}^{n+1}(t,x) + \int_{V_+} F\cdot \nabla_{t,x}\varphi\,d\mathcal{L}^{n+1}(t,x). \end{align*} On each open side $V_\pm$, the vector field $F$ is $C^1$ and $\operatorname{div}_{t,x}F=0$. The product rule gives \begin{align*} \operatorname{div}_{t,x}(\varphi F) = F\cdot \nabla_{t,x}\varphi+\varphi\,\operatorname{div}_{t,x}F = F\cdot \nabla_{t,x}\varphi. \end{align*} Let $K:=\operatorname{supp}\varphi$. Choose a smooth [open set](/page/Open%20Set) $V'\Subset V$ with $K\subset V'$. We use the [Gauss-Green Theorem](/theorems/28) on the one-sided domains $V'\cap V_-$ and $V'\cap V_+$ for the vector field $\varphi F$, in the standard one-sided trace form. This form is obtained by applying Gauss-Green on smooth subdomains at positive distance from $\Sigma$ and then letting the subdomains exhaust $V'\cap V_\pm$; the boundary limits exist because $u_\pm$ are the one-sided traces of $u$ on $\Sigma$ and $f\in C^1(\mathbb{R};\mathbb{R}^n)$, so $\varphi F$ has one-sided boundary traces $\varphi F_\pm$ on $\Sigma\cap V'$. The boundary portions lying in $\partial V'$ give no contribution because $\varphi=0$ there, so the only boundary contributions are on $\Sigma \cap V'$, equivalently on $\Sigma\cap V$ after extending the integrand by zero outside $K$. The outward unit normal to $V_-$ along $\Sigma$ is $\nu$, while the outward unit normal to $V_+$ along $\Sigma$ is $-\nu$. Hence \begin{align*} 0=\int_{\Sigma\cap V} \varphi\,F_- \cdot \nu\,d\mathcal{H}^{n}(t,x)-\int_{\Sigma\cap V} \varphi\,F_+ \cdot \nu\,d\mathcal{H}^{n}(t,x), \end{align*} where \begin{align*} F_-:\Sigma\cap V \to \mathbb{R}^{n+1}, \qquad F_-(t,x):=(u_-(t,x),f(u_-(t,x))), \end{align*} and \begin{align*} F_+:\Sigma\cap V \to \mathbb{R}^{n+1}, \qquad F_+(t,x):=(u_+(t,x),f(u_+(t,x))). \end{align*} Therefore \begin{align*} \int_{\Sigma\cap V} \varphi\,\bigl(F_+-F_-\bigr)\cdot \nu\,d\mathcal{H}^{n}(t,x)=0. \end{align*} [guided] The weak formulation contains exactly the spacetime pairing of the flux with the spacetime gradient of the test function. To make this explicit, define \begin{align*} F: Q\setminus \Sigma \to \mathbb{R}^{n+1}, \qquad F(t,x):=(u(t,x),f(u(t,x))). \end{align*} Then \begin{align*} F\cdot \nabla_{t,x}\varphi = u\,\partial_t\varphi+f(u)\cdot \nabla_x\varphi. \end{align*} Fix $\varphi \in C_c^1(V)$. Since $u$ is a weak solution and $\varphi$ is supported in $V$, we have \begin{align*} 0 = \int_{V_-} F\cdot \nabla_{t,x}\varphi\,d\mathcal{L}^{n+1}(t,x) + \int_{V_+} F\cdot \nabla_{t,x}\varphi\,d\mathcal{L}^{n+1}(t,x). \end{align*} The hypersurface itself does not contribute to this volume integral because a smooth hypersurface has zero $(n+1)$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure). Now use the classical equation on each side. Since $u$ is $C^1$ on $V_\pm$ up to the interface from the corresponding side and $f \in C^1(\mathbb{R};\mathbb{R}^n)$, the map $F$ is $C^1$ on each side. The chain rule gives \begin{align*} \operatorname{div}_{t,x}F = \partial_tu+\operatorname{div}_x f(u) = 0 \end{align*} on $V_\pm$. Therefore the product rule gives \begin{align*} \operatorname{div}_{t,x}(\varphi F) = F\cdot \nabla_{t,x}\varphi+\varphi\,\operatorname{div}_{t,x}F = F\cdot \nabla_{t,x}\varphi. \end{align*} We apply the [Gauss-Green Theorem](/theorems/28) separately on truncated one-sided domains. Let $K:=\operatorname{supp}\varphi$, and choose a smooth open set $V'\Subset V$ with $K\subset V'$. The theorem applies on smooth subdomains lying inside $V'\cap V_-$ and $V'\cap V_+$ because $F$ is $C^1$ in the interior of each side. Letting those subdomains exhaust the two sides gives the boundary terms on $\Sigma$ with the one-sided traces, since $u_\pm$ are the one-sided traces of $u$ and $f\in C^1(\mathbb{R};\mathbb{R}^n)$. Define \begin{align*} F_-:\Sigma\cap V \to \mathbb{R}^{n+1}, \qquad F_-(t,x):=(u_-(t,x),f(u_-(t,x))) \end{align*} and \begin{align*} F_+:\Sigma\cap V \to \mathbb{R}^{n+1}, \qquad F_+(t,x):=(u_+(t,x),f(u_+(t,x))). \end{align*} The boundary portions lying in $\partial V'$ give no contribution because $\varphi=0$ there. Along $\Sigma$, the outward unit normal of $V_-$ is $\nu$ because $\nu$ is oriented from the minus side to the plus side. The outward unit normal of $V_+$ is therefore $-\nu$. Hence the two integrations by parts give \begin{align*} \int_{V_-} F\cdot \nabla_{t,x}\varphi\,d\mathcal{L}^{n+1}(t,x) = \int_{\Sigma\cap V} \varphi\,F_-\cdot \nu\,d\mathcal{H}^{n}(t,x), \end{align*} and \begin{align*} \int_{V_+} F\cdot \nabla_{t,x}\varphi\,d\mathcal{L}^{n+1}(t,x) = -\int_{\Sigma\cap V} \varphi\,F_+\cdot \nu\,d\mathcal{H}^{n}(t,x). \end{align*} Adding these identities and using the weak formulation yields \begin{align*} 0 = \int_{\Sigma\cap V} \varphi\,F_-\cdot \nu\,d\mathcal{H}^{n}(t,x) - \int_{\Sigma\cap V} \varphi\,F_+\cdot \nu\,d\mathcal{H}^{n}(t,x). \end{align*} Equivalently, \begin{align*} \int_{\Sigma\cap V} \varphi\,\bigl(F_+-F_-\bigr)\cdot \nu\,d\mathcal{H}^{n}(t,x)=0. \end{align*} This identity says that every possible trace of a test function on the interface is orthogonal, in the integral sense, to the scalar jump coefficient $(F_+-F_-)\cdot \nu$. [/guided] [/step] [step:Use arbitrary traces of test functions to force the jump coefficient to vanish] Define \begin{align*} g:\Sigma\cap V \to \mathbb{R}, \qquad g(t,x):=\bigl(F_+(t,x)-F_-(t,x)\bigr)\cdot \nu(t,x). \end{align*} The one-sided traces $u_\pm$, the flux $f$, and the normal field $\nu$ are continuous on $\Sigma\cap V$, so $g$ is continuous. The preceding step shows that \begin{align*} \int_{\Sigma\cap V} \varphi\,g\,d\mathcal{H}^{n}(t,x)=0 \end{align*} for every $\varphi \in C_c^1(V)$. We justify the trace-extension step. Since $\Sigma$ is a smooth hypersurface, after possibly shrinking $V$ there is a $C^1$ coordinate chart \begin{align*} \Theta: W \subset \mathbb{R}^{n+1} \to V \end{align*} whose image contains $p$ and in which $\Sigma\cap V$ is represented by $\{y_{n+1}=0\}$. For any $\psi \in C_c^1(\Sigma\cap V)$ with support inside this chart, choose a function $\rho \in C_c^1((-\varepsilon,\varepsilon))$ with $\rho(0)=1$. Define the ambient extension as the map \begin{align*} \widetilde{\varphi}:\Theta(W) \to \mathbb{R}, \qquad \widetilde{\varphi}(\Theta(y',y_{n+1})):=\psi(\Theta(y',0))\rho(y_{n+1}). \end{align*} Multiplying by a compactly supported cutoff inside $V$ and extending by zero outside the chart gives $\widetilde{\varphi}\in C_c^1(V)$ and $\widetilde{\varphi}|_{\Sigma\cap V}=\psi$ on the support under consideration. Using a finite [Partition of Unity](/page/Partition%20of%20Unity) on the compact support gives the same conclusion for every $\psi\in C_c^1(\Sigma\cap V)$. Therefore \begin{align*} \int_{\Sigma\cap V} \psi\,g\,d\mathcal{H}^{n}(t,x)=0 \end{align*} for every $\psi \in C_c^1(\Sigma\cap V)$. If $g(p)\neq 0$ at some point $p\in \Sigma\cap V$, continuity gives a smaller neighbourhood $W\subset \Sigma\cap V$ on which $g$ has one sign and is bounded away from $0$ in absolute value. Choosing a nonnegative function $\psi\in C_c^1(W)$ that is not identically zero gives \begin{align*} \int_{\Sigma\cap V} \psi\,g\,d\mathcal{H}^{n}(t,x)\neq 0, \end{align*} contradicting the preceding identity. Hence $g=0$ on $\Sigma\cap V$. [/step] [step:Pass from the unit-normal jump relation to the stated normal field] For every $(t,x)\in \Sigma\cap V$, \begin{align*} 0=\bigl(F_+(t,x)-F_-(t,x)\bigr)\cdot \nu(t,x). \end{align*} Because $N=\lambda\nu$ and $\lambda(t,x)\neq 0$, multiplying by $\lambda(t,x)$ gives \begin{align*} 0=\bigl(F_+(t,x)-F_-(t,x)\bigr)\cdot N(t,x). \end{align*} Using the definitions of $F_\pm$ and $N=(n_t,n_x)$, this is exactly \begin{align*} 0=(u_+(t,x)-u_-(t,x))\,n_t(t,x)+(f(u_+(t,x))-f(u_-(t,x)))\cdot n_x(t,x). \end{align*} Since $p\in \Sigma$ was arbitrary, the identity holds at every point of $\Sigma$. This proves the multidimensional Rankine-Hugoniot condition for any consistently oriented nonzero spacetime normal field. [/step]