The strategy is to restrict to test functions supported away from $\{t = 0\}$, split the weak formulation integral across $\Gamma$ into integrals over $\Omega_L$ and $\Omega_R$, use the product rule and the classical PDE to convert each integrand into a divergence (Claim 1), apply the [Gauss-Green Theorem](/theorems/28) to reduce to boundary integrals along $\Gamma$ (Claim 2), compute these boundary integrals by parametrising $\Gamma$ and explicitly evaluating the outward normal and $\mathcal{H}^1$-Jacobian (Claim 3), and localise via the [Fundamental Lemma of Calculus of Variations](/theorems/45) to extract the pointwise jump condition (Claim 4).

**Step 1: Restrict to test functions supported away from $\{t = 0\}$.**

Let $\varphi \in C_c^\infty(\mathbb{R} \times (0, T))$ — note the open interval, so $\varphi$ vanishes near $t = 0$ and $t = T$. The initial data term drops out and the weak formulation becomes: \begin{align*} \int_0^T \int_{\mathbb{R}} \left(u \, \partial_t \varphi + f(u) \, \partial_x \varphi\right) d\mathcal{L}^1(x) \, d\mathcal{L}^1(t) = 0. \end{align*}

**Step 2: Split across $\Gamma$ and reduce to a divergence.**

[claim:Divergence Reduction] On each of $\Omega_L$ and $\Omega_R$, \begin{align*} u\,\partial_t\varphi + f(u)\,\partial_x\varphi &= \partial_t(u\varphi) + \partial_x(f(u)\varphi). \end{align*} [/claim]

[proof] Expanding the right-hand side by the product rule: \begin{align*} \partial_t(u\varphi) + \partial_x(f(u)\varphi) &= u\,\partial_t\varphi + \varphi\,\partial_t u + f(u)\,\partial_x\varphi + \varphi\,\partial_x f(u) \\ &= \bigl(u\,\partial_t\varphi + f(u)\,\partial_x\varphi\bigr) + \varphi\,\bigl(\partial_t u + \partial_x f(u)\bigr). \end{align*} On $\Omega_L$ and $\Omega_R$, $u$ is $C^1$ and satisfies $\partial_t u + \partial_x f(u) = 0$ classically, so the second term vanishes. [/proof]

Splitting the domain and applying the claim: \begin{align*} 0 &= \iint_{\Omega_L} \bigl(\partial_t(u\varphi) + \partial_x(f(u)\varphi)\bigr)\,d\mathcal{L}^2 + \iint_{\Omega_R} \bigl(\partial_t(u\varphi) + \partial_x(f(u)\varphi)\bigr)\,d\mathcal{L}^2. \end{align*}

**Step 3: Apply the divergence theorem.**

Define the vector field $\mathbf{F} := (f(u)\varphi,\, u\varphi)$ in the $(x, t)$-plane, so that $\partial_x F_1 + \partial_t F_2 = \nabla \cdot \mathbf{F}$. By the [Gauss-Green Theorem](/theorems/28) applied to $\Omega_L$ (and similarly to $\Omega_R$): \begin{align*} \iint_{\Omega_L} \nabla \cdot \mathbf{F}\,d\mathcal{L}^2 &= \int_{\partial\Omega_L} \mathbf{F} \cdot \mathbf{n}_L\,d\mathcal{H}^1, \end{align*} where $\mathbf{n}_L$ is the outward unit normal to $\Omega_L$. Since $\varphi \in C_c^\infty(\mathbb{R} \times (0, T))$, the function $\varphi$ vanishes on every part of $\partial\Omega_L$ except possibly the shock curve $\Gamma$: on the portions of $\partial\Omega_L$ at spatial infinity $\varphi = 0$ because $\varphi$ has compact support in $x$, and near $t = 0$ and $t = T$ because $\varphi$ is supported in $\mathbb{R} \times (0, T)$. The boundary integral therefore reduces to an integral along $\Gamma$: \begin{align*} \int_{\partial\Omega_L} \mathbf{F} \cdot \mathbf{n}_L\,d\mathcal{H}^1 &= \int_\Gamma \mathbf{F}\big|_L \cdot \mathbf{n}_L\,d\mathcal{H}^1, \end{align*} where $\mathbf{F}\big|_L$ denotes $\mathbf{F}$ evaluated using the left-hand limit $u_L$.

**Step 4: Compute the boundary integrals along $\Gamma$.**

[claim:Boundary Integral Computation] Parametrise $\Gamma$ by $\gamma: (0, T) \to \mathbb{R}^2$, $\gamma(\tau) := (s(\tau), \tau)$, where $\tau$ is a curve parameter ranging over the same interval as the time coordinate (the second component of $\gamma(\tau)$ is $\tau$, so the parameter value coincides with the time coordinate along $\Gamma$, but we use a separate letter to distinguish the roles). Then \begin{align*} \int_\Gamma \mathbf{F}\big|_L \cdot \mathbf{n}_L\,d\mathcal{H}^1 &= \int_0^T \varphi(s(\tau), \tau)\bigl(f(u_L(\tau)) - \dot{s}(\tau)\,u_L(\tau)\bigr)\,d\mathcal{L}^1(\tau), \end{align*} and the corresponding integral for $\Omega_R$ gives \begin{align*} \int_\Gamma \mathbf{F}\big|_R \cdot \mathbf{n}_R\,d\mathcal{H}^1 &= -\int_0^T \varphi(s(\tau), \tau)\bigl(f(u_R(\tau)) - \dot{s}(\tau)\,u_R(\tau)\bigr)\,d\mathcal{L}^1(\tau). \end{align*} [/claim]

[proof] The parametrisation $\gamma(\tau) = (s(\tau), \tau)$ has tangent vector $\gamma'(\tau) = (\dot{s}(\tau), 1)$ and speed $|\gamma'(\tau)| = \sqrt{\dot{s}(\tau)^2 + 1}$. The outward unit normal to $\Omega_L = \{(x, t) : x < s(t)\}$ points into $\Omega_R$, i.e. in the direction of increasing $x$ relative to $\Gamma$. Rotating $\gamma'(\tau)$ clockwise by $\pi/2$ gives the vector $(1, -\dot{s}(\tau))$, which points into $\{x > s(t)\}$ (since the $x$-component is positive). Normalising: \begin{align*} \mathbf{n}_L(\gamma(\tau)) &= \frac{(1, -\dot{s}(\tau))}{\sqrt{1 + \dot{s}(\tau)^2}}. \end{align*} The $\mathcal{H}^1$ measure on $\Gamma$ is related to the parameter $\tau$ by the [Substitution Formula for Hausdorff Integrals](/theorems/868) (with $m = 1$, $\sigma = \gamma$, $\Psi = \mathrm{id}$): for any integrable $h: \Gamma \to \mathbb{R}$, \begin{align*} \int_\Gamma h\,d\mathcal{H}^1 &= \int_0^T h(\gamma(\tau))\,|\gamma'(\tau)|\,d\mathcal{L}^1(\tau) = \int_0^T h(\gamma(\tau))\,\sqrt{1 + \dot{s}(\tau)^2}\,d\mathcal{L}^1(\tau). \end{align*} Applying this with $h = \mathbf{F}\big|_L \cdot \mathbf{n}_L$: \begin{align*} \int_\Gamma \mathbf{F}\big|_L \cdot \mathbf{n}_L\,d\mathcal{H}^1 &= \int_0^T \bigl(\mathbf{F}\big|_L \cdot \mathbf{n}_L\bigr)(\gamma(\tau))\,\sqrt{1 + \dot{s}(\tau)^2}\,d\mathcal{L}^1(\tau). \end{align*} Now compute the integrand. The point $\gamma(\tau) = (s(\tau), \tau)$ has $(x, t)$-coordinates $(s(\tau), \tau)$, so evaluating $\mathbf{F}\big|_L$ at this point gives: \begin{align*} \mathbf{F}\big|_L(\gamma(\tau)) &= \bigl(f(u_L(\tau))\,\varphi(s(\tau), \tau),\;\; u_L(\tau)\,\varphi(s(\tau), \tau)\bigr), \end{align*} where $u_L(\tau) := \lim_{x \to s(\tau)^-} u(x, \tau)$. The dot product with $\mathbf{n}_L$ is: \begin{align*} \bigl(\mathbf{F}\big|_L \cdot \mathbf{n}_L\bigr)(\gamma(\tau)) &= \frac{f(u_L(\tau))\,\varphi(s(\tau), \tau) \cdot 1 + u_L(\tau)\,\varphi(s(\tau), \tau) \cdot (-\dot{s}(\tau))}{\sqrt{1 + \dot{s}(\tau)^2}} = \frac{\varphi(s(\tau), \tau)\bigl(f(u_L(\tau)) - \dot{s}(\tau)\,u_L(\tau)\bigr)}{\sqrt{1 + \dot{s}(\tau)^2}}. \end{align*} The factors $\sqrt{1 + \dot{s}(\tau)^2}$ from the normal and from the substitution formula cancel: \begin{align*} \int_\Gamma \mathbf{F}\big|_L \cdot \mathbf{n}_L\,d\mathcal{H}^1 &= \int_0^T \frac{\varphi(s(\tau), \tau)\bigl(f(u_L(\tau)) - \dot{s}(\tau)\,u_L(\tau)\bigr)}{\sqrt{1 + \dot{s}(\tau)^2}} \cdot \sqrt{1 + \dot{s}(\tau)^2}\,d\mathcal{L}^1(\tau) \\ &= \int_0^T \varphi(s(\tau), \tau)\bigl(f(u_L(\tau)) - \dot{s}(\tau)\,u_L(\tau)\bigr)\,d\mathcal{L}^1(\tau). \end{align*} For $\Omega_R = \{(x, t) : x > s(t)\}$, the outward normal points into $\Omega_L$, so $\mathbf{n}_R(\gamma(\tau)) = -\mathbf{n}_L(\gamma(\tau)) = (-1, \dot{s}(\tau))/\sqrt{1 + \dot{s}(\tau)^2}$. The same computation gives \begin{align*} \int_\Gamma \mathbf{F}\big|_R \cdot \mathbf{n}_R\,d\mathcal{H}^1 &= -\int_0^T \varphi(s(\tau), \tau)\bigl(f(u_R(\tau)) - \dot{s}(\tau)\,u_R(\tau)\bigr)\,d\mathcal{L}^1(\tau). \end{align*} [/proof]

**Step 5: Extract the jump condition.**

[claim:Pointwise Jump Condition] For every $t \in (0, T)$, \begin{align*} f(u_L(t)) - f(u_R(t)) &= \dot{s}(t)\bigl(u_L(t) - u_R(t)\bigr). \end{align*} [/claim]

[proof] Adding the two boundary contributions from Claim 3 (reverting from the curve parameter $\tau$ to the time coordinate $t$, which are identified along $\Gamma$), the weak formulation $0 = \int_{\partial\Omega_L} \mathbf{F}\big|_L \cdot \mathbf{n}_L\,d\mathcal{H}^1 + \int_{\partial\Omega_R} \mathbf{F}\big|_R \cdot \mathbf{n}_R\,d\mathcal{H}^1$ becomes \begin{align*} \int_0^T \varphi(s(t), t)\bigl[\bigl(f(u_L) - f(u_R)\bigr) - \dot{s}(t)\bigl(u_L - u_R\bigr)\bigr]\,d\mathcal{L}^1(t) &= 0. \end{align*} Define $\psi: (0, T) \to \mathbb{R}$ by $\psi(t) := \varphi(s(t), t)$. The freedom in $\varphi$ allows $\psi$ to range over all of $C_c^\infty((0, T))$: given any $\psi_0 \in C_c^\infty((0, T))$, choose $\varphi(x, t) := \psi_0(t)\,\eta(x - s(t))$ where $\eta \in C_c^\infty(\mathbb{R})$ satisfies $\eta(0) = 1$; then $\varphi \in C_c^\infty(\mathbb{R} \times (0, T))$ and $\varphi(s(t), t) = \psi_0(t)$. The integral condition therefore reads \begin{align*} \int_0^T \psi(t)\,h(t)\,d\mathcal{L}^1(t) &= 0 \quad \text{for all } \psi \in C_c^\infty((0, T)), \end{align*} where $h(t) := (f(u_L) - f(u_R)) - \dot{s}(u_L - u_R)$ is continuous on $(0, T)$. By the [Fundamental Lemma of Calculus of Variations](/theorems/45), $h(t) = 0$ for $\mathcal{L}^1$-a.e. $t \in (0, T)$. Since $h$ is continuous (as $u_L$, $u_R$, $f$, and $\dot{s}$ are all continuous), $h$ vanishes everywhere on $(0, T)$. [/proof]