**Proof plan.** We specialise the [method of characteristics](/page/Method%20of%20Characteristics) to the conservation law $\partial_t \rho + q'(\rho)\partial_x \rho = 0$. The characteristic ODE system simplifies: since the right-hand side $c = 0$, the solution value is constant along each characteristic, and the characteristics are straight lines with slope $q'(g(a))$. This gives $\rho$ implicitly as the solution of $H(t, x, \rho) = 0$. We apply the [Implicit Function Theorem](/theorems/52) to determine when this defines a $C^1$ [function](/page/Function), then verify by direct computation that the implicitly defined function satisfies the PDE.
**Step 1: Derive the characteristic equations.**
The conservation law $\partial_t \rho + q'(\rho)\partial_x \rho = 0$ is a quasilinear first-order PDE of the form $\partial_t \rho + b(t, x, \rho)\partial_x \rho = c(t, x, \rho)$ with $b(t, x, \rho) = q'(\rho)$ and $c(t, x, \rho) = 0$. Following the [method of characteristics](/page/Method%20of%20Characteristics), we write the characteristic ODE system with parameter $a \in \mathbb{R}$ labelling the initial point:
\begin{align*}
\frac{dX}{dt}(t, a) = q'(Z(t, a)), \quad X(0, a) = a, \qquad \frac{dZ}{dt}(t, a) = 0, \quad Z(0, a) = g(a).
\end{align*}
The second equation integrates immediately to $Z(t, a) = g(a)$ for all $t \geq 0$: the solution value is constant along each characteristic. Substituting into the first equation, $\frac{dX}{dt} = q'(g(a))$ is now a constant, so the characteristic from $a$ is the straight line:
\begin{align*}
X(t, a) = a + q'(g(a))t.
\end{align*}
**Step 2: Obtain the implicit formula.**
Along the characteristic from $a$, we have $\rho(t, X(t, a)) = Z(t, a) = g(a)$. To express $\rho$ as a function of $(t, x)$, we must invert $a \mapsto X(t, a)$ to recover $a$ from $(t, x)$, then set $\rho = g(a)$. Since $\rho = g(a)$ and $x = a + q'(g(a))t = a + q'(\rho)t$, we get $a = x - q'(\rho)t$. Substituting:
\begin{align*}
\rho = g(x - q'(\rho)t).
\end{align*}
Define $H: [0, \infty) \times \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ by:
\begin{align*}
H(t, x, \rho) := \rho - g(x - q'(\rho)t).
\end{align*}
**Step 3: Apply the [implicit function theorem](/page/Implicit%20Function%20Theorem).**
At $t = 0$: $H(0, x, g(x)) = g(x) - g(x) = 0$, so the zero set is non-empty. Since $q \in C^2(\mathbb{R})$ and $g \in C^1(\mathbb{R})$, the function $H$ is $C^1$ in all variables. By the [Implicit Function Theorem](/theorems/52), $H = 0$ defines $\rho$ as a $C^1$ function of $(t, x)$ in a neighbourhood of any point where $\partial_\rho H \neq 0$.
Computing by the chain rule:
\begin{align*}
\partial_\rho H(t, x, \rho) = 1 + t \, q''(\rho) \, g'(x - q'(\rho)t).
\end{align*}
[claim: Local Existence Time]
If $\sup_{r \in \mathbb{R}} |q''(r)| \leq M$ and $\sup_{r \in \mathbb{R}} |g'(r)| \leq M$, then $\partial_\rho H(t, x, \rho) > 0$ for all $(t, x, \rho)$ with $0 \leq t < M^{-2}$.
[/claim]
[proof]
For any $(t, x, \rho)$ with $t \geq 0$:
\begin{align*}
|\partial_\rho H(t, x, \rho) - 1| = |t \, q''(\rho) \, g'(x - q'(\rho)t)| \leq t \cdot |q''(\rho)| \cdot |g'(x - q'(\rho)t)| \leq t M^2.
\end{align*}
Therefore $\partial_\rho H \geq 1 - tM^2 > 0$ whenever $t < M^{-2}$.
[/proof]
[claim: Global Existence Under The Sign Condition]
If $q''(r)g'(s) \geq 0$ for all $r, s \in \mathbb{R}$, then $\partial_\rho H(t, x, \rho) \geq 1$ for all $t \geq 0$ and all $(x, \rho)$.
[/claim]
[proof]
Under the sign condition, $q''(\rho) \cdot g'(x - q'(\rho)t) \geq 0$ for every $\rho$ and every $x - q'(\rho)t$. Since $t \geq 0$:
\begin{align*}
\partial_\rho H(t, x, \rho) = 1 + t \, q''(\rho) \, g'(x - q'(\rho)t) \geq 1 > 0.
\end{align*}
The [Implicit Function Theorem](/theorems/52) applies for all $t \geq 0$.
[/proof]
**Step 4: Verify that the implicitly defined $\rho$ solves the PDE.**
Let $\rho(t, x)$ be the $C^1$ function defined implicitly by $H(t, x, \rho(t, x)) = 0$. We must show $\partial_t \rho + q'(\rho)\partial_x \rho = 0$. Differentiating $H(t, x, \rho(t,x)) = 0$ in $t$ and $x$ separately by the chain rule gives:
\begin{align*}
\partial_t H + \partial_\rho H \cdot \partial_t \rho = 0, \qquad \partial_x H + \partial_\rho H \cdot \partial_x \rho = 0.
\end{align*}
Writing $a := x - q'(\rho)t$ for the initial point of the characteristic, the partial derivatives of $H$ are:
\begin{align*}
\partial_t H &= -g'(a) \cdot (-q'(\rho)) = q'(\rho) \, g'(a), \\
\partial_x H &= -g'(a), \\
\partial_\rho H &= 1 + t \, q''(\rho) \, g'(a).
\end{align*}
From the implicit [differentiation](/page/Derivative) identities:
\begin{align*}
\partial_t \rho = -\frac{\partial_t H}{\partial_\rho H} = -\frac{q'(\rho) \, g'(a)}{1 + t \, q''(\rho) \, g'(a)}, \qquad \partial_x \rho = -\frac{\partial_x H}{\partial_\rho H} = \frac{g'(a)}{1 + t \, q''(\rho) \, g'(a)}.
\end{align*}
Therefore:
\begin{align*}
\partial_t \rho + q'(\rho) \, \partial_x \rho = \frac{-q'(\rho) \, g'(a) + q'(\rho) \, g'(a)}{1 + t \, q''(\rho) \, g'(a)} = 0.
\end{align*}
The initial condition holds since $H(0, x, \rho) = 0$ gives $\rho(0, x) = g(x)$.
**Step 5: Uniqueness.**
Let $\rho$ be any $C^1$ solution of the conservation law on $[0, T) \times \mathbb{R}$.
[claim: Any Solution Is Constant Along Characteristics]
For every $a \in \mathbb{R}$, $\rho(t, X(t, a)) = g(a)$ for all $t \in [0, T)$.
[/claim]
[proof]
Define $Y(t) := \rho(t, X(t, a))$ where $X(t, a) = a + q'(g(a))t$. Differentiating by the chain rule:
\begin{align*}
\frac{dY}{dt} = \partial_t \rho(t, X) + \partial_x \rho(t, X) \cdot q'(g(a)).
\end{align*}
At $t = 0$, $Y(0) = \rho(0, a) = g(a)$, so $q'(g(a)) = q'(Y(0))$. Since $\rho$ satisfies $\partial_t \rho + q'(\rho)\partial_x \rho = 0$ and $\rho(t, X(t,a)) = Y(t)$:
\begin{align*}
\frac{dY}{dt} = -q'(Y)\partial_x \rho + q'(g(a))\partial_x \rho = (q'(g(a)) - q'(Y))\partial_x \rho.
\end{align*}
This is an ODE for $Y$ with $Y(0) = g(a)$. The constant function $Y(t) = g(a)$ satisfies this ODE (since $q'(g(a)) - q'(g(a)) = 0$). By uniqueness of ODE solutions (the right-hand side is locally Lipschitz in $Y$ since $q' \in C^1$ and $\partial_x \rho$ is [continuous](/page/Continuity)), $Y(t) = g(a)$ is the only solution.
[/proof]
[claim: The Characteristics Cover All Of $[0, T) \times \mathbb{R}$]
For each $t \in [0, T)$, the map $a \mapsto X(t, a) = a + q'(g(a))t$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$.
[/claim]
[proof]
The derivative $\partial_a X(t, a) = 1 + q''(g(a))g'(a)t$ satisfies $\partial_a X \geq 1 - tM^2 > 0$ for $t < T = M^{-2}$ (by the same estimate as in the Local Existence Time claim). Therefore $a \mapsto X(t, a)$ is strictly increasing.
For surjectivity, the lower bound $\partial_a X \geq 1 - tM^2 =: \lambda > 0$ gives, by [integration](/page/Integral) from $0$ to $a$ (for $a > 0$):
\begin{align*}
X(t, a) = X(t, 0) + \int_0^a \partial_a X(t, a') \, d\mathcal{L}^1(a') \geq X(t, 0) + \lambda a \to +\infty \quad \text{as } a \to +\infty.
\end{align*}
Similarly, $X(t, a) \to -\infty$ as $a \to -\infty$. Since $a \mapsto X(t, a)$ is continuous, strictly increasing, and has range $\mathbb{R}$, it is a bijection.
[/proof]
Combining the two claims: for any $(t, x) \in [0, T) \times \mathbb{R}$, there exists a unique $a \in \mathbb{R}$ with $X(t, a) = x$, and any $C^1$ solution satisfies $\rho(t, x) = \rho(t, X(t, a)) = g(a)$. This is the same value produced by the construction in Step 2, so the solution is unique.
