**Proof plan.** We verify three things: (i) in each of the three regions (left constant, fan, right constant), $\rho$ satisfies the PDE $\partial_t \rho + q'(\rho)\partial_x \rho = 0$ classically; (ii) $\rho$ is continuous across the boundaries of the fan, so no shock conditions are needed; (iii) $\rho$ satisfies the weak formulation including the initial data. **Step 1: PDE in the constant regions.** For $x < q'(\rho_L)t$: $\rho(t, x) = \rho_L$ is constant, so $\partial_t \rho = 0$ and $\partial_x \rho = 0$. The PDE $\partial_t \rho + q'(\rho)\partial_x \rho = 0 + q'(\rho_L) \cdot 0 = 0$ holds. For $x > q'(\rho_R)t$: identical argument with $\rho_R$. **Step 2: PDE in the fan region.** For $q'(\rho_L)t < x < q'(\rho_R)t$ and $t > 0$, define $\xi := x/t$ and write $\rho(t, x) = (q')^{-1}(\xi)$. Computing the partial [derivatives](/page/Derivative) by the chain rule: \begin{align*} \partial_t \rho = \frac{d(q')^{-1}}{d\xi} \cdot \partial_t\!\left(\frac{x}{t}\right) = \frac{1}{q''(\rho)} \cdot \left(-\frac{x}{t^2}\right) = -\frac{\xi}{t \, q''(\rho)}, \end{align*} \begin{align*} \partial_x \rho = \frac{d(q')^{-1}}{d\xi} \cdot \partial_x\!\left(\frac{x}{t}\right) = \frac{1}{q''(\rho)} \cdot \frac{1}{t} = \frac{1}{t \, q''(\rho)}, \end{align*} where we used $\frac{d(q')^{-1}}{d\xi} = 1/q''((q')^{-1}(\xi)) = 1/q''(\rho)$ by the inverse [function](/page/Function) rule (valid since $q'' > 0$). Substituting into the PDE and using $q'(\rho) = \xi$ (which is exactly the definition $(q')^{-1}(\xi) = \rho$): \begin{align*} \partial_t \rho + q'(\rho)\partial_x \rho = -\frac{\xi}{t \, q''(\rho)} + \xi \cdot \frac{1}{t \, q''(\rho)} = 0. \end{align*} [claim: [Continuity](/page/Continuity) Across The Fan Boundaries] $\rho$ is continuous at $x = q'(\rho_L)t$ and $x = q'(\rho_R)t$ for every $t > 0$. [/claim] [proof] At the left [boundary](/page/Boundary) $x = q'(\rho_L)t$: the fan formula gives $\rho = (q')^{-1}(x/t) = (q')^{-1}(q'(\rho_L)) = \rho_L$, which matches the constant value on the left. At the right boundary $x = q'(\rho_R)t$: the fan formula gives $(q')^{-1}(q'(\rho_R)) = \rho_R$, matching the constant value on the right. [/proof] **Step 3: Weak formulation.** Since $\rho$ is $C^1$ on each of the three open regions and continuous across the two boundaries, it is a classical solution on $\{t > 0\} \setminus \{x = q'(\rho_L)t\} \cup \{x = q'(\rho_R)t\}$, with no jumps. To verify the weak formulation, take any $\varphi \in C_c^\infty(\mathbb{R} \times [0, \infty))$ and integrate by parts on each smooth region. Since the PDE holds classically on each region (Steps 1–2), the volume [integrals](/page/Integral) produce zero. The boundary integrals along the fan edges involve the jumps $[\rho]$ and $[q(\rho)]$, but these are zero by continuity (the claim above). The only remaining boundary contribution is from $\{t = 0\}$, which gives: \begin{align*} \int_{\mathbb{R}} g(x) \, \varphi(0, x) \, d\mathcal{L}^1(x) = \int_{-\infty}^0 \rho_L \, \varphi(0, x) \, d\mathcal{L}^1(x) + \int_0^\infty \rho_R \, \varphi(0, x) \, d\mathcal{L}^1(x). \end{align*} This is the correct initial data term, so $\rho$ satisfies the weak formulation.