[proofplan] We derive the Rankine--Hugoniot jump condition by splitting the weak formulation integral across the shock curve, applying integration by parts (divergence theorem) on each half-domain where $u$ is $C^1$, computing the boundary integrals along the shock using the outward normal and arc-length parametrisation, and then combining the two contributions. The pointwise PDE holds on each side (so the bulk terms vanish), and the arbitrariness of the test function forces the jump condition. [/proofplan] [step:Split the weak form integral across the shock curve] Let $\phi \in C_c^\infty(\mathbb{R} \times (0,\infty))$ be a test function supported in a compact set that does not meet $\{t = 0\}$. The initial data term vanishes, so the weak form reduces to \begin{align*} 0 = \iint_{\Omega^-} \bigl(u\,\phi_t + f(u)\,\phi_x\bigr)\,d\mathcal{L}^2 + \iint_{\Omega^+} \bigl(u\,\phi_t + f(u)\,\phi_x\bigr)\,d\mathcal{L}^2. \end{align*} [/step] [step:Apply integration by parts on $\Omega^-$ using the pointwise PDE] On $\Omega^-$, $u$ is $C^1$ and satisfies $u_t + f(u)_x = 0$ pointwise. By the product rule: \begin{align*} u\,\phi_t + f(u)\,\phi_x = \partial_t(u\phi) + \partial_x(f(u)\phi) - \bigl(u_t + f(u)_x\bigr)\phi = \partial_t(u\phi) + \partial_x(f(u)\phi). \end{align*} By the divergence theorem applied to the vector field $(x,t) \mapsto (f(u)\phi, u\phi)$ on the region $\Omega^-$: \begin{align*} \iint_{\Omega^-} \bigl(u\,\phi_t + f(u)\,\phi_x\bigr)\,d\mathcal{L}^2 = \int_{\partial \Omega^-} \bigl(f(u)\phi,\, u\phi\bigr) \cdot n^-\,d\mathcal{H}^1, \end{align*} where $n^-$ is the outward unit normal to $\Omega^-$. Since $\phi$ has compact support away from $\{t=0\}$ and spatial infinity, only the shock curve contributes. [/step] [step:Compute the boundary integral along the shock for $\Omega^-$] The shock curve is parametrised by $t \mapsto (\xi(t), t)$, with arc-length element $d\mathcal{H}^1 = \sqrt{1 + \dot\xi(t)^2}\,d\mathcal{L}^1(t)$. The outward unit normal to $\Omega^-$ at the shock (pointing into $\Omega^+$) is \begin{align*} n^-(t) = \frac{(1,\,-\dot\xi(t))}{\sqrt{1+\dot\xi(t)^2}}, \end{align*} where the first component is in the $x$-direction and the second in the $t$-direction. Evaluating the dot product and cancelling $\sqrt{1+\dot\xi^2}$: \begin{align*} \int_{\partial\Omega^-} \bigl(f(u^-)\phi,\, u^-\phi\bigr) \cdot n^-\,d\mathcal{H}^1 = \int_0^\infty \phi(\xi(t),t)\,\bigl(f(u^-(t)) - \dot\xi(t)\,u^-(t)\bigr)\,d\mathcal{L}^1(t). \end{align*} [/step] [step:Perform the same computation on $\Omega^+$ with reversed normal] The outward normal to $\Omega^+$ at the shock is $n^+ = -n^-$. An identical argument gives: \begin{align*} \iint_{\Omega^+} \bigl(u\,\phi_t + f(u)\,\phi_x\bigr)\,d\mathcal{L}^2 = -\int_0^\infty \phi(\xi(t),t)\,\bigl(f(u^+(t)) - \dot\xi(t)\,u^+(t)\bigr)\,d\mathcal{L}^1(t). \end{align*} [/step] [step:Combine and extract the pointwise jump condition] Adding the two contributions and using the fact that the total is zero: \begin{align*} \int_0^\infty \phi(\xi(t),t)\Bigl[\bigl(f(u^-) - \dot\xi\,u^-\bigr) - \bigl(f(u^+) - \dot\xi\,u^+\bigr)\Bigr]\,d\mathcal{L}^1(t) = 0. \end{align*} Since $\phi \in C_c^\infty(\mathbb{R} \times (0,\infty))$ is arbitrary (we can localise its support to any neighbourhood of any point on the shock), the integrand must vanish for $\mathcal{L}^1$-almost every $t > 0$: \begin{align*} f(u^-(t)) - f(u^+(t)) = \dot\xi(t)(u^-(t) - u^+(t)), \end{align*} which is the Rankine--Hugoniot condition $\dot\xi(t)\,[u]_\xi = [f(u)]_\xi$. [/step]