[proofplan] We define the labelled characteristic projection $X: [0,\infty)\times\mathbb{R}\to\mathbb{R}$ by \begin{align*} X(t,x_0)=x_0+t f'(u_0(x_0)). \end{align*} Wherever the projection is single-valued and locally invertible, the corresponding characteristic representation satisfies \begin{align*} u(t,X(t,x_0))=u_0(x_0). \end{align*} The [Characteristic System](/theorems/49) transports the initial value $u_0(x_0)$ along the labelled curve $t \mapsto X(t,x_0)$. The whole criterion reduces to differentiating the projection $x_0 \mapsto X(t,x_0)$ and reading off when its one-dimensional Jacobian vanishes, which is exactly the nondegeneracy condition used by the [Inverse Function Theorem](/theorems/51). The infimum of the declared map $q: \mathbb{R}\to\mathbb{R}$ gives the earliest predicted degeneration time, and at a zero of \begin{align*} 1+tq(x_0) \end{align*} the transported spatial derivative formula forces gradient blow-up rather than merely a failure of an inverse-function-theorem hypothesis. [/proofplan]

[step:Differentiate the characteristic map with respect to the label]Define the characteristic map $X: [0,\infty) \times \mathbb{R} \to \mathbb{R}$ by \begin{align*} X(t,x_0)=x_0+t f'(u_0(x_0)). \end{align*} This is the labelled characteristic projection supplied by the [Characteristic System](/theorems/49); wherever $x_0 \mapsto X(t,x_0)$ is single-valued and locally invertible, the associated characteristic representation is the function $u$ determined by \begin{align*} u(t,X(t,x_0))=u_0(x_0). \end{align*} Since $f \in C^2(I)$ and $u_0 \in C^1(\mathbb{R};I)$, the composite map $a: \mathbb{R} \to \mathbb{R}$ defined by \begin{align*} a(x_0)=f'(u_0(x_0)) \end{align*} belongs to $C^1(\mathbb{R})$. By the ordinary $C^1$ chain rule applied to the composite $f'\circ u_0$, \begin{align*} a'(x_0)=f''(u_0(x_0))u_0'(x_0)=q(x_0). \end{align*} Therefore, for every $t \geq 0$ and every $x_0 \in \mathbb{R}$, \begin{align*} \partial_{x_0}X(t,x_0)=1+tq(x_0). \end{align*}[/step]

[guided]The label $x_0$ identifies the characteristic starting from the point $x_0$ at time $0$. The characteristic map $X: [0,\infty) \times \mathbb{R} \to \mathbb{R}$ is defined by \begin{align*} X(t,x_0)=x_0+t f'(u_0(x_0)). \end{align*} This is the labelled projection obtained from the [Characteristic System](/theorems/49). On any time interval where this projection is single-valued and locally invertible in the label, the characteristic representation defines $u$ by the identity $u(t,X(t,x_0))=u_0(x_0)$. To decide whether different labelled characteristics cross, we inspect the derivative of the projection $x_0 \mapsto X(t,x_0)$. Define $a: \mathbb{R} \to \mathbb{R}$ by \begin{align*} a(x_0)=f'(u_0(x_0)). \end{align*} Because $u_0$ is $C^1$ and takes values in $I$, and because $f'$ is $C^1$ on $I$, the composite $a=f'\circ u_0$ is $C^1$ on $\mathbb{R}$. Applying the chain rule to this composite gives \begin{align*} a'(x_0)=f''(u_0(x_0))u_0'(x_0). \end{align*} By the definition of $q$, this is exactly $q(x_0)$. Hence differentiating \begin{align*} X(t,x_0)=x_0+t a(x_0) \end{align*} with respect to the label $x_0$, again using the ordinary $C^1$ chain rule for the composite term already computed, gives \begin{align*} \partial_{x_0}X(t,x_0)=1+tq(x_0). \end{align*} This identity is the entire mechanism: local invertibility of the characteristic projection can only persist while this derivative avoids degenerating.[/guided]

[step:Use the finite negative infimum to force a negative derivative after $T_*$] Assume $-\infty < m < 0$ and define $T_*$ by \begin{align*} T_*=-\frac{1}{m}. \end{align*} Fix a time $t \in (T_*,\infty)$. Since $m<0$, the inequality \begin{align*} t>-\frac{1}{m} \end{align*} is equivalent to \begin{align*} m<-\frac{1}{t}. \end{align*} Because $m$ is the infimum of $q(\mathbb{R})$, there exists $x_0 \in \mathbb{R}$ such that \begin{align*} q(x_0)<-\frac{1}{t}. \end{align*} Using the derivative formula from the previous step, \begin{align*} \partial_{x_0}X(t,x_0)=1+tq(x_0)<1+t\left(-\frac{1}{t}\right)=0. \end{align*} Thus for every $t>T_*$ there is a characteristic label $x_0$ at which the label projection has negative derivative. [/step]

[step:Use the first vanishing of the characteristic Jacobian to force gradient blow-up]Fix $t>T_*$ and choose $x_0 \in \mathbb{R}$ such that $\partial_{x_0}X(t,x_0)<0$. Define $\gamma_{x_0}: [0,t] \to \mathbb{R}$ by \begin{align*} \gamma_{x_0}(s)=\partial_{x_0}X(s,x_0)=1+sq(x_0). \end{align*} The map $\gamma_{x_0}$ is continuous on $[0,t]$, with $\gamma_{x_0}(0)=1$ and $\gamma_{x_0}(t)<0$. Since $\gamma_{x_0}(t)<0$, the formula \begin{align*} \gamma_{x_0}(t)=1+tq(x_0) \end{align*} implies \begin{align*} q(x_0)<-\frac{1}{t}. \end{align*} Define $s_*\in\mathbb{R}$ by \begin{align*} s_*=-\frac{1}{q(x_0)}. \end{align*} Then $q(x_0)<0$ gives $s_*>0$, and $q(x_0)<-1/t$ gives $s_*<t$. Therefore $s_*\in(0,t)$ and \begin{align*} \gamma_{x_0}(s_*)=1+s_*q(x_0)=0. \end{align*} Since \begin{align*} q(x_0)<-\frac{1}{t}, \end{align*} we have $q(x_0)\neq 0$, hence $u_0'(x_0)\neq 0$ because \begin{align*} q(x_0)=f''(u_0(x_0))u_0'(x_0). \end{align*} Let $u: [0,s_*] \times \mathbb{R} \to I$ denote the single-valued function represented by the characteristic formula on the time interval under consideration. The hypotheses needed for the [Characteristic System](/theorems/49) are satisfied here: $f\in C^2(I)$ implies $f'\in C^1(I)$, and $u_0\in C^1(\mathbb{R};I)$ supplies the $C^1$ initial trace. Hence, for $0\leq s<s_*$, the characteristic system gives the transported-value identity \begin{align*} u(s,X(s,x_0))=u_0(x_0). \end{align*} If this representation were a $C^1$ single-valued classical solution up to and beyond $s_*$, then differentiating the identity with respect to $x_0$ at times $s<s_*$ would give \begin{align*} u_x(s,X(s,x_0))\partial_{x_0}X(s,x_0)=u_0'(x_0). \end{align*} Therefore, for $0\leq s<s_*$, \begin{align*} u_x(s,X(s,x_0))=\frac{u_0'(x_0)}{1+sq(x_0)}. \end{align*} As $s\to s_*$ from below, the denominator tends to $0$ while the numerator is nonzero, so $|u_x(s,X(s,x_0))|\to\infty$. If $u$ were $C^1$ on a time interval containing $s_*$, then $u_x$ would be continuous at the point $(s_*,X(s_*,x_0))$. Since $(s,X(s,x_0))\to(s_*,X(s_*,x_0))$ as $s\to s_*$ from below, continuity would make $u_x(s,X(s,x_0))$ finite and locally bounded along this approaching curve, contradicting the blow-up just obtained. Since $s_*<t$ for every $t>T_*$ chosen above, the characteristic formula cannot remain a $C^1$ single-valued classical solution on any time interval extending beyond $T_*$.[/step]

[guided]The sign change of $\partial_{x_0}X$ proves that the one-dimensional characteristic Jacobian must vanish at an intermediate time. That is precisely the loss of the nondegeneracy condition used by the [Inverse Function Theorem](/theorems/51). To prove the stronger classical-breakdown conclusion, we use the transported derivative formula. Choose $x_0\in\mathbb{R}$ with $\partial_{x_0}X(t,x_0)<0$ and define $\gamma_{x_0}: [0,t]\to\mathbb{R}$ by \begin{align*} \gamma_{x_0}(s)=\partial_{x_0}X(s,x_0)=1+sq(x_0). \end{align*} This is a continuous real-valued function on $[0,t]$, but in this affine case we can locate the zero directly. The inequality $\gamma_{x_0}(t)<0$ says \begin{align*} 1+tq(x_0)<0, \end{align*} so \begin{align*} q(x_0)<-\frac{1}{t}. \end{align*} Define $s_*\in\mathbb{R}$ by \begin{align*} s_*=-\frac{1}{q(x_0)}. \end{align*} Because $q(x_0)<0$, we have $s_*>0$. Because $q(x_0)<-1/t$, we also have $s_*<t$. Hence $s_*\in(0,t)$ and \begin{align*} 1+s_*q(x_0)=\gamma_{x_0}(s_*)=0. \end{align*} The same inequality shows $q(x_0)\neq 0$. Since \begin{align*} q(x_0)=f''(u_0(x_0))u_0'(x_0), \end{align*} this also forces $u_0'(x_0)\neq 0$. Now assume, for contradiction, that the characteristic formula remains a $C^1$ single-valued classical solution through this time, and let $u: [0,s_*] \times \mathbb{R} \to I$ denote that represented function. The [Characteristic System](/theorems/49) applies because $f\in C^2(I)$ gives $f'\in C^1(I)$ and the initial trace satisfies $u_0\in C^1(\mathbb{R};I)$. Therefore, along the labelled characteristic, the transported value is constant: \begin{align*} u(s,X(s,x_0))=u_0(x_0). \end{align*} For each $s<s_*$, differentiating this identity with respect to the label $x_0$ is valid because $u$ is assumed $C^1$, $X$ is $C^1$, and $u_0$ is $C^1$. The chain rule gives \begin{align*} u_x(s,X(s,x_0))\partial_{x_0}X(s,x_0)=u_0'(x_0). \end{align*} Using $\partial_{x_0}X(s,x_0)=1+sq(x_0)$, we obtain \begin{align*} u_x(s,X(s,x_0))=\frac{u_0'(x_0)}{1+sq(x_0)}. \end{align*} The numerator is nonzero, while the denominator tends to $0$ as $s\to s_*$ from below. Hence $|u_x(s,X(s,x_0))|\to\infty$. On the other hand, if $u$ were $C^1$ on an interval containing $s_*$, then $u_x$ would be continuous at the point $(s_*,X(s_*,x_0))$. The points $(s,X(s,x_0))$ approach $(s_*,X(s_*,x_0))$ as $s\to s_*$ from below, so continuity would force the values $u_x(s,X(s,x_0))$ to remain finite and locally bounded near that limiting point. This contradicts the blow-up. Therefore the characteristic formula cannot remain a $C^1$ single-valued classical solution through $s_*$. This proves the claimed breakdown before the chosen time $t$, and since $t>T_*$ was arbitrary, no interval extending beyond $T_*$ can support the characteristic formula as a $C^1$ single-valued classical solution.[/guided]

[step:Show that a nonnegative infimum prevents characteristic crossing by this criterion]Assume $m \geq 0$. Since $m$ is a lower bound for $q(\mathbb{R})$, for every $x_0 \in \mathbb{R}$, \begin{align*} q(x_0)\geq m\geq 0. \end{align*} For every $t \geq 0$, the derivative formula gives \begin{align*} \partial_{x_0}X(t,x_0)=1+tq(x_0)\geq 1. \end{align*} Thus the characteristic projection is locally increasing at every label for every nonnegative time, and this characteristic-crossing mechanism predicts no shock.[/step]

[guided]In this case the infimum condition says that $q$ never takes negative values. More precisely, since $m$ is a lower bound for the set $q(\mathbb{R})$, every label $x_0 \in \mathbb{R}$ satisfies \begin{align*} q(x_0)\geq m\geq 0. \end{align*} The derivative formula proved in the first step then gives, for every $t\geq 0$, \begin{align*} \partial_{x_0}X(t,x_0)=1+tq(x_0)\geq 1. \end{align*} Thus the projection from labels to positions is increasing at each label for every time satisfying \begin{align*} t\geq 0. \end{align*} The criterion under discussion detects shock formation through characteristic crossing, and this derivative computation shows that such crossing is not predicted when \begin{align*} m\geq 0. \end{align*}[/guided]

[step:Show that an infinite negative infimum destroys every positive uniform lifespan]Assume $m=-\infty$ and fix $T>0$. Since $q$ is unbounded below, there exists $x_0 \in \mathbb{R}$ such that \begin{align*} q(x_0)<-\frac{1}{T}. \end{align*} Using the derivative formula, \begin{align*} \partial_{x_0}X(T,x_0)=1+Tq(x_0)<1+T\left(-\frac{1}{T}\right)=0. \end{align*} Since this holds for every $T>0$, no positive time interval can serve as a uniform interval on which the characteristic projection remains locally increasing for all labels. Hence the characteristic formula has no positive uniform classical lifespan predicted by this crossing criterion.[/step]

[guided]The case $m=-\infty$ means that the values of $q$ have no finite lower bound. Fix an arbitrary positive time $T>0$. By unboundedness below, there is a label $x_0\in\mathbb{R}$ such that \begin{align*} q(x_0)<-\frac{1}{T}. \end{align*} Using the derivative identity for the characteristic projection, we obtain \begin{align*} \partial_{x_0}X(T,x_0)=1+Tq(x_0)<1+T\left(-\frac{1}{T}\right)=0. \end{align*} Thus even at the arbitrarily chosen positive time $T$, some labelled characteristic has already passed into the negative-Jacobian regime. Since the same labelled derivative equals $1$ at time $0$ and depends continuously on time, the affine-zero computation from the finite-infimum case gives a time in $(0,T)$ at which the characteristic Jacobian vanishes. Because $T>0$ was arbitrary, there is no positive uniform time interval on which all labels keep a locally increasing characteristic projection. This is exactly the statement that the crossing criterion supplies no positive uniform classical lifespan when $m=-\infty$.[/guided]