[proofplan] We first rewrite the conservation law in quasilinear form using the chain rule. Then we follow the solution along a characteristic curve whose velocity is $f'(u)$ evaluated on the solution. Along such a curve, the total derivative of $u$ vanishes, so $u$ is constant on the characteristic. Once $u$ is constant, the characteristic velocity is constant as well, and integrating the velocity gives the formula $x=x_0+t f'(u_0(x_0))$. Local invertibility of the footpoint map then permits the same relation to be written as a [representation formula](/theorems/39) in terms of the inverse footpoint. [/proofplan]

[step:Rewrite the conservation law in quasilinear form] Since $f \in C^2(I)$ and $u \in C^1([0,T)\times \mathbb{R};I)$, the composition \begin{align*} f \circ u : [0,T)\times \mathbb{R} \to \mathbb{R} \end{align*} is $C^1$. Applying the chain rule to the spatial derivative gives, for every $(t,x) \in (0,T)\times \mathbb{R}$, \begin{align*} \partial_x(f(u(t,x))) = f'(u(t,x))\,\partial_x u(t,x). \end{align*} Hence the conservation law is equivalent on $(0,T)\times \mathbb{R}$ to \begin{align*} \partial_t u(t,x) + f'(u(t,x))\,\partial_x u(t,x) = 0. \end{align*} [/step]

[step:Show that the solution is constant along each characteristic]Fix $x_0 \in \mathbb{R}$ and $t \in [0,T)$. Define the characteristic velocity field $a:[0,T)\times \mathbb{R}\to \mathbb{R}$ by \begin{align*} a(s,x)=f'(u(s,x)). \end{align*} Since $f\in C^2(I)$ and $u\in C^1([0,T)\times\mathbb{R};I)$, the map $a$ is $C^1$. The Picard-Lindelöf local existence and uniqueness theorem for ordinary differential equations therefore gives a number $\varepsilon>0$ and a $C^1$ map $\gamma:[0,\varepsilon)\to\mathbb{R}$ such that \begin{align*} \gamma(0)=x_0 \end{align*} and \begin{align*} \gamma'(s)=a(s,\gamma(s))=f'(u(s,\gamma(s))) \end{align*} for $s\in(0,\varepsilon)$. Define the curve-value map $w:[0,\varepsilon)\to I$ by \begin{align*} w(s)=u(s,\gamma(s)). \end{align*} By the one-variable chain rule, for $s\in(0,\varepsilon)$, \begin{align*} w'(s)=\partial_t u(s,\gamma(s))+\gamma'(s)\,\partial_x u(s,\gamma(s)). \end{align*} Substituting the defining equation for $\gamma'$ and using the quasilinear form of the PDE on $(0,T)\times\mathbb{R}$ gives $w'(s)=0$ for $s\in(0,\varepsilon)$. Hence $w$ is constant on $[0,\varepsilon)$ by continuity, and \begin{align*} u(s,\gamma(s))=u_0(x_0) \end{align*} for $s\in[0,\varepsilon)$. Consequently the ODE becomes $\gamma'(s)=f'(u_0(x_0))$ on $(0,\varepsilon)$, so the local characteristic agrees with the affine curve $s\mapsto x_0+s f'(u_0(x_0))$. Define the set of reached times $A\subset[0,t]$ by \begin{align*} A=\{\tau\in[0,t]: u(s,x_0+s f'(u_0(x_0)))=u_0(x_0) \text{ for every } s\in[0,\tau]\}. \end{align*} The local argument above shows that $A$ is nonempty. The set $A$ is closed in $[0,t]$ because $u$ is continuous and the defining identity is preserved under limits of times. It remains to prove openness relative to $[0,t]$. Let $r\in A$ with $r<t$, and set \begin{align*} z_r=x_0+r f'(u_0(x_0)). \end{align*} Starting the same ODE at time $r$ from the point $z_r$, the Picard-Lindelöf local existence and uniqueness theorem for ordinary differential equations gives a local characteristic. Since $u(r,z_r)=u_0(x_0)$, the same chain-rule computation along this characteristic shows that $u$ remains equal to $u_0(x_0)$ on a short interval after $r$, and uniqueness forces the characteristic segment to be \begin{align*} s\mapsto x_0+s f'(u_0(x_0)). \end{align*} Thus $A$ is open in $[0,t]$. Since $[0,t]$ is connected and $A$ is nonempty, open, and closed in $[0,t]$, we have $A=[0,t]$. Therefore \begin{align*} u(s,x_0+s f'(u_0(x_0)))=u_0(x_0) \end{align*} for every $s\in[0,t]$.[/step]

[guided]Fix $x_0\in\mathbb{R}$ and $t\in[0,T)$. Before differentiating along a characteristic on the whole interval, we must justify that such a curve exists. Define the velocity field $a:[0,T)\times\mathbb{R}\to\mathbb{R}$ by \begin{align*} a(s,x)=f'(u(s,x)). \end{align*} The hypothesis $f\in C^2(I)$ is used here: since $f'$ is $C^1$ and $u$ is $C^1$, the composition $a=f'\circ u$ is a $C^1$ velocity field. Thus the Picard-Lindelöf local existence and uniqueness theorem for ordinary differential equations applies to the initial value problem \begin{align*} \gamma(0)=x_0 \end{align*} and \begin{align*} \gamma'(s)=a(s,\gamma(s))=f'(u(s,\gamma(s))). \end{align*} It gives a number $\varepsilon>0$ and a $C^1$ characteristic curve $\gamma:[0,\varepsilon)\to\mathbb{R}$ solving this equation for $s\in(0,\varepsilon)$. Now define the solution value along the local characteristic by $w:[0,\varepsilon)\to I$ with \begin{align*} w(s)=u(s,\gamma(s)). \end{align*} Because $u$ and $\gamma$ are $C^1$, the one-variable chain rule gives, for $s\in(0,\varepsilon)$, \begin{align*} w'(s)=\partial_t u(s,\gamma(s))+\gamma'(s)\,\partial_x u(s,\gamma(s)). \end{align*} Substituting the characteristic equation makes this derivative equal to the left-hand side of the quasilinear PDE: \begin{align*} w'(s)=\partial_t u(s,\gamma(s))+f'(u(s,\gamma(s)))\,\partial_x u(s,\gamma(s)). \end{align*} The PDE holds on $(0,T)\times\mathbb{R}$, so this equality proves $w'(s)=0$ for $s\in(0,\varepsilon)$. We do not claim the PDE at $s=0$; instead, continuity of $w$ extends the constancy to the endpoint $0$. Therefore \begin{align*} u(s,\gamma(s))=w(s)=w(0)=u(0,x_0)=u_0(x_0) \end{align*} for $s\in[0,\varepsilon)$. Once the solution value is constant, the characteristic velocity is also constant: \begin{align*} \gamma'(s)=f'(u_0(x_0)) \end{align*} for $s\in(0,\varepsilon)$. Hence the local characteristic is the affine curve $s\mapsto x_0+s f'(u_0(x_0))$. To reach an arbitrary later time $t<T$, we make the continuation argument explicit. Define $A\subset[0,t]$ by \begin{align*} A=\{\tau\in[0,t]: u(s,x_0+s f'(u_0(x_0)))=u_0(x_0) \text{ for every } s\in[0,\tau]\}. \end{align*} The local calculation near $0$ proves that $A$ is nonempty. The set $A$ is closed in $[0,t]$: if $\tau_k\in A$ and $\tau_k\to\tau$, then continuity of the map $s\mapsto u(s,x_0+s f'(u_0(x_0)))$ preserves the identity at the limiting time, and the identity on the whole interval $[0,\tau]$ follows by approximating each $s\leq\tau$ with times lying below some $\tau_k$. Now let $r\in A$ with $r<t$. Put \begin{align*} z_r=x_0+r f'(u_0(x_0)). \end{align*} Since $r\in A$, we have $u(r,z_r)=u_0(x_0)$. Start the characteristic ODE at time $r$ from the point $z_r$. The velocity field is still $C^1$, so the Picard-Lindelöf local existence and uniqueness theorem for ordinary differential equations gives a local characteristic through $(r,z_r)$. Repeating the chain-rule computation on this shifted time interval shows that the solution value remains $u_0(x_0)$ along that local characteristic. Therefore the velocity on the new segment is again $f'(u_0(x_0))$, and uniqueness forces the segment to be the same affine curve \begin{align*} s\mapsto x_0+s f'(u_0(x_0)). \end{align*} Thus $A$ is open in $[0,t]$. Since $[0,t]$ is connected and $A$ is nonempty, open, and closed in $[0,t]$, it follows that $A=[0,t]$. Hence \begin{align*} u(s,x_0+s f'(u_0(x_0)))=u_0(x_0) \end{align*} for every $s\in[0,t]$.[/guided]

[step:Integrate the characteristic velocity after constancy is known] For the fixed time $t\in[0,T)$, define the characteristic endpoint map $X_t:\mathbb{R}\to\mathbb{R}$ by \begin{align*} X_t(y_0)=y_0+t f'(u_0(y_0)). \end{align*} Since $f\in C^2(I)$ and $u_0\in C^1(\mathbb{R};I)$, the map $f'\circ u_0$ is $C^1$, and hence $X_t$ is $C^1$. From the previous step applied with the initial point $x_0$, \begin{align*} u(s,x_0+s f'(u_0(x_0)))=u_0(x_0) \end{align*} for every $s\in[0,t]$. In particular, at time $t$, \begin{align*} X_t(x_0)=x_0+t f'(u_0(x_0)). \end{align*} Thus, if $x=X_t(x_0)$, then \begin{align*} u(t,x)=u_0(x_0). \end{align*} [/step]

[step:Use local invertibility to express the footpoint as a function of the endpoint] Assume that $X_t$ is locally invertible at $x_0$, as in the hypothesis; no derivative criterion such as $X_t'(x_0)\neq 0$ is needed here. Then there is a neighbourhood $V \subset \mathbb{R}$ of $x_0$ and an inverse map \begin{align*} Y_t:X_t(V) &\to V \end{align*} such that \begin{align*} Y_t(X_t(y_0))=y_0 \end{align*} for every $y_0 \in V$. Applying the characteristic formula with $y_0=Y_t(x)$ gives, for every $x \in X_t(V)$, \begin{align*} u(t,x)=u_0(Y_t(x)). \end{align*} This is exactly the claimed representation of the classical solution while the characteristic footpoint can be recovered locally from the endpoint. [/step]