[proofplan] We first compute the characteristic curves and verify that any $C^1$ local inverse branch of the [characteristic map](/page/Method%20of%20Characteristics) produces a classical solution. The possible breakdown of this parametrisation is then reduced to the vanishing of the derivative $\partial_a F_t(a)=1+t u_0'(a)$. The lower bound $u_0'(a)\geq m$ gives positivity for every $t<T_*$, while attainment or non-attainment of the infimum distinguishes whether the threshold degeneracy occurs at a single label or only along a sequence. [/proofplan] [step:Recover the classical solution from a local inverse branch] For each $t \geq 0$, define $F_t: \mathbb{R} \to \mathbb{R}$ by \begin{align*} F_t(a) = a + t u_0(a). \end{align*} Fix a spacetime region $\Omega \subset [0,\infty) \times \mathbb{R}$ that is relatively open in $[0,\infty) \times \mathbb{R}$ and on which there is a $C^1$ local inverse branch $A: \Omega \to \mathbb{R}$ such that \begin{align*} F_t(A(t,x)) = x \end{align*} for every $(t,x) \in \Omega$. All differentiations below are taken at interior points with $t>0$; on points of the initial slice $t=0$, $\partial_t$ is interpreted as the right time derivative. Define $u: \Omega \to \mathbb{R}$ by \begin{align*} u(t,x) = u_0(A(t,x)). \end{align*} Here a classical solution means a function $u \in C^1(\Omega)$ satisfying the differential equation pointwise at interior points with $t>0$ and satisfying the initial condition on the initial slice where that slice is included in $\Omega$. Let $J: \Omega \to \mathbb{R}$ be defined by \begin{align*} J(t,x) = 1 + t u_0'(A(t,x)). \end{align*} Differentiating the identity \begin{align*} A(t,x) + t u_0(A(t,x)) = x \end{align*} with respect to $x$ gives \begin{align*} J(t,x)\partial_x A(t,x) = 1. \end{align*} Thus $J(t,x) \neq 0$ on $\Omega$, and \begin{align*} \partial_x A(t,x) = \frac{1}{J(t,x)}. \end{align*} Differentiating the same identity with respect to $t$ gives \begin{align*} \partial_t A(t,x) + u_0(A(t,x)) + t u_0'(A(t,x))\partial_t A(t,x) = 0, \end{align*} hence \begin{align*} \partial_t A(t,x) = -\frac{u_0(A(t,x))}{J(t,x)}. \end{align*} By the chain rule, \begin{align*} \partial_t u(t,x) = u_0'(A(t,x))\partial_t A(t,x) \end{align*} and \begin{align*} \partial_x u(t,x) = u_0'(A(t,x))\partial_x A(t,x). \end{align*} Substituting the formulas for $\partial_t A$ and $\partial_x A$ yields \begin{align*} \partial_t u(t,x) + u(t,x)\partial_x u(t,x)= u_0'(A(t,x))\left(-\frac{u_0(A(t,x))}{J(t,x)}\right)+ u_0(A(t,x))u_0'(A(t,x))\frac{1}{J(t,x)}. \end{align*} The two terms cancel, so \begin{align*} \partial_t u(t,x) + u(t,x)\partial_x u(t,x) = 0. \end{align*} At $t=0$, the characteristic map is $F_0(a)=a$, so $A(0,x)=x$ and therefore $u(0,x)=u_0(x)$ wherever the initial slice is included in $\Omega$. [guided] For each $t \geq 0$, define the characteristic map $F_t: \mathbb{R} \to \mathbb{R}$ by \begin{align*} F_t(a) = a + t u_0(a). \end{align*} The characteristic formula gives the label $a$ implicitly through \begin{align*} x = F_t(a) = a + t u_0(a). \end{align*} On a relatively open spacetime region $\Omega \subset [0,\infty) \times \mathbb{R}$ where this equation can be solved by a $C^1$ local inverse branch $A: \Omega \to \mathbb{R}$, the value $A(t,x)$ is the characteristic label that reaches the spacetime point $(t,x)$, and it satisfies \begin{align*} A(t,x) + t u_0(A(t,x)) = x. \end{align*} The differentiations below are ordinary partial derivatives at points with $t>0$; if the initial slice is included, the time derivative at $t=0$ is understood as a right derivative. We then define the candidate solution by transporting the initial value along the characteristic: $u: \Omega \to \mathbb{R}$ by \begin{align*} u(t,x) = u_0(A(t,x)). \end{align*} In this proof, a classical solution means a function $u \in C^1(\Omega)$ satisfying the differential equation pointwise at interior points with $t>0$ and satisfying the initial condition on the initial slice where that slice is included in $\Omega$. The key quantity controlling the inverse branch is the derivative of the characteristic map with respect to the label. Define $J: \Omega \to \mathbb{R}$ by \begin{align*} J(t,x) = 1 + t u_0'(A(t,x)). \end{align*} We now compute the derivatives of $A$ from the identity \begin{align*} A(t,x) + t u_0(A(t,x)) = x. \end{align*} Differentiating with respect to $x$ gives \begin{align*} \partial_x A(t,x) + t u_0'(A(t,x))\partial_x A(t,x) = 1. \end{align*} Factoring the left-hand side gives \begin{align*} J(t,x)\partial_x A(t,x) = 1, \end{align*} so $J(t,x) \neq 0$ at every point of the branch and \begin{align*} \partial_x A(t,x) = \frac{1}{J(t,x)}. \end{align*} Differentiating the same identity with respect to $t$ gives \begin{align*} \partial_t A(t,x) + u_0(A(t,x)) + t u_0'(A(t,x))\partial_t A(t,x) = 0. \end{align*} Again factoring the terms involving $\partial_t A$ gives \begin{align*} J(t,x)\partial_t A(t,x) + u_0(A(t,x)) = 0, \end{align*} and therefore \begin{align*} \partial_t A(t,x) = -\frac{u_0(A(t,x))}{J(t,x)}. \end{align*} Now apply the chain rule to $u(t,x)=u_0(A(t,x))$. We obtain \begin{align*} \partial_t u(t,x) = u_0'(A(t,x))\partial_t A(t,x) \end{align*} and \begin{align*} \partial_x u(t,x) = u_0'(A(t,x))\partial_x A(t,x). \end{align*} Substituting the formulas for $\partial_t A$ and $\partial_x A$ gives \begin{align*} \partial_t u(t,x) = -\frac{u_0'(A(t,x))u_0(A(t,x))}{J(t,x)} \end{align*} and \begin{align*} u(t,x)\partial_x u(t,x)= u_0(A(t,x))\frac{u_0'(A(t,x))}{J(t,x)}. \end{align*} Adding these two expressions gives \begin{align*} \partial_t u(t,x) + u(t,x)\partial_x u(t,x) = 0. \end{align*} Thus the parametrisation produces a classical solution wherever the inverse branch exists, with the one-sided time derivative convention on the initial boundary. At $t=0$, $F_0(a)=a$, so the inverse label is $A(0,x)=x$ and the initial condition becomes \begin{align*} u(0,x)=u_0(A(0,x))=u_0(x). \end{align*} [/guided] [/step] [step:Identify loss of local invertibility with vanishing of the label derivative] For each fixed $t \geq 0$, define $F_t: \mathbb{R} \to \mathbb{R}$ by \begin{align*} F_t(a) = a + t u_0(a). \end{align*} Since $u_0 \in C^1(\mathbb{R})$, we have $F_t \in C^1(\mathbb{R})$, and its derivative is \begin{align*} F_t'(a) = 1 + t u_0'(a). \end{align*} If $F_t'(a) \neq 0$, the [Inverse Function Theorem](/theorems/51) gives open intervals $I \subset \mathbb{R}$ containing $a$ and $V \subset \mathbb{R}$ containing $F_t(a)$ such that $F_t|_I: I \to V$ has a $C^1$ inverse. Conversely, if a $C^1$ inverse branch $B: V \to I$ satisfies $F_t(B(y))=y$ and $B(F_t(a))=a$, differentiating at $y=F_t(a)$ gives \begin{align*} F_t'(a)B'(F_t(a)) = 1. \end{align*} Therefore $F_t'(a)=0$ prevents a $C^1$ inverse branch at that point. Thus the characteristic parametrisation can lose $C^1$ local invertibility only at labels $a \in \mathbb{R}$ and times $t \geq 0$ satisfying \begin{align*} 1 + t u_0'(a) = 0. \end{align*} Equivalently, for $t>0$, such a degeneracy requires \begin{align*} u_0'(a) = -\frac{1}{t}. \end{align*} [guided] For a fixed time $t \geq 0$, the object whose invertibility matters is the label-to-position [characteristic map](/page/Method%20of%20Characteristics). Define $F_t: \mathbb{R} \to \mathbb{R}$ by \begin{align*} F_t(a) = a + t u_0(a). \end{align*} Because $u_0 \in C^1(\mathbb{R})$, the map $F_t$ is also $C^1$, and differentiating with respect to the label $a$ gives \begin{align*} F_t'(a) = 1 + t u_0'(a). \end{align*} We now connect this derivative to local invertibility. The [Inverse Function Theorem](/theorems/51) applies to the $C^1$ map $F_t: \mathbb{R} \to \mathbb{R}$. Its hypotheses are satisfied because $F_t$ is $C^1$ and $F_t'(a) \neq 0$. It gives open intervals $I \subset \mathbb{R}$ containing $a$ and $V \subset \mathbb{R}$ containing $F_t(a)$ such that $F_t|_I: I \to V$ is bijective and has a $C^1$ inverse. Thus nonvanishing of $F_t'(a)$ is enough to build the local inverse branch used in the characteristic parametrisation. The converse obstruction is just as important. Suppose $B: V \to I$ were a $C^1$ inverse branch near $F_t(a)$, with $F_t(B(y))=y$ for $y \in V$ and $B(F_t(a))=a$. Differentiating the identity $F_t(B(y))=y$ at $y=F_t(a)$ gives \begin{align*} F_t'(a)B'(F_t(a)) = 1. \end{align*} This equality cannot hold when $F_t'(a)=0$. Therefore vanishing of $F_t'(a)$ prevents a $C^1$ inverse branch at that label. Combining these two directions, the relevant degeneracy condition is \begin{align*} 1 + t u_0'(a) = 0. \end{align*} For $t>0$, this is equivalent to \begin{align*} u_0'(a) = -\frac{1}{t}. \end{align*} [/guided] [/step] [step:Use the lower bound on $u_0'$ to exclude degeneracy before $T_*$] [guided] Assume $-\infty < m < 0$, and define \begin{align*} T_* := -\frac{1}{m}. \end{align*} Because $m<0$, the number $T_*$ is positive. The infimum condition means that $m$ is a global lower bound for $u_0'$: for every label $a \in \mathbb{R}$, \begin{align*} u_0'(a) \geq m. \end{align*} Fix $t$ with $0 \leq t<T_*$. Multiplication by the non-negative number $t$ preserves the inequality, so \begin{align*} t u_0'(a) \geq t m. \end{align*} Adding $1$ gives \begin{align*} 1+t u_0'(a) \geq 1+t m. \end{align*} Since $t<T_*=-1/m$ and $m<0$, this last lower bound is strictly positive: \begin{align*} 1+t m>0. \end{align*} Therefore \begin{align*} F_t'(a)=1+t u_0'(a)>0 \end{align*} for every $a \in \mathbb{R}$ and every $0 \leq t<T_*$. The previous step identified possible $C^1$ local invertibility failure with the vanishing of $F_t'$, so no loss of local invertibility can occur before $T_*$. [/guided] Assume $-\infty < m < 0$, and define \begin{align*} T_* := -\frac{1}{m}. \end{align*} Since $m<0$, we have $T_*>0$. For every $a \in \mathbb{R}$, the definition of the infimum gives \begin{align*} u_0'(a) \geq m. \end{align*} If $0 \leq t < T_*$, then multiplying the inequality by $t \geq 0$ gives \begin{align*} t u_0'(a) \geq t m. \end{align*} Therefore \begin{align*} 1 + t u_0'(a) \geq 1 + t m. \end{align*} Because $m<0$ and $t<T_*=-1/m$, we have \begin{align*} 1 + t m > 0. \end{align*} Hence \begin{align*} F_t'(a)=1+t u_0'(a)>0 \end{align*} for every $a \in \mathbb{R}$ and every $0 \leq t<T_*$. Consequently no loss of local invertibility can occur before $T_*$. [/step] [step:Determine what happens at the threshold time] [guided] There are two different possibilities at the threshold. First suppose there exists $a_* \in \mathbb{R}$ such that $u_0'(a_*)=m$. Since $T_*=-1/m$, direct substitution gives \begin{align*} F_{T_*}'(a_*) = 1 + T_* u_0'(a_*) = 1 + \left(-\frac{1}{m}\right)m = 0. \end{align*} By the derivative criterion already proved, the characteristic map loses $C^1$ local invertibility at the label $a_*$ when $t=T_*$. Now suppose the infimum $m$ is not attained. The definition of infimum gives a sequence of labels $(a_k)_{k\in\mathbb{N}}$ with \begin{align*} m < u_0'(a_k) < m + \frac{1}{k}. \end{align*} The lower inequality is strict because no label attains the infimum. Hence $u_0'(a_k)\to m$. Since $m<0$, we may ignore finitely many initial terms and assume $u_0'(a_k)<0$ for all remaining $k$. Define $t_k \in (0,\infty)$ by \begin{align*} t_k := -\frac{1}{u_0'(a_k)}. \end{align*} This definition is chosen so that the derivative vanishes exactly at the pair $(t_k,a_k)$: \begin{align*} F_{t_k}'(a_k)=1+t_k u_0'(a_k)=0. \end{align*} Because $r\mapsto -1/r$ is continuous for $r<0$ and $u_0'(a_k)\to m<0$, these degeneracy times satisfy \begin{align*} t_k\to -\frac{1}{m}=T_*. \end{align*} Thus the threshold is approached by actual possible degeneracies, even though at time $T_*$ itself no label is forced to degenerate. Indeed, if $1+T_*u_0'(a)=0$, then $u_0'(a)=m$, contradicting the assumption that the infimum is not attained. [/guided] Assume first that there exists $a_* \in \mathbb{R}$ such that $u_0'(a_*)=m$. Then \begin{align*} F_{T_*}'(a_*) = 1 + T_* u_0'(a_*) = 1 + \left(-\frac{1}{m}\right)m = 0. \end{align*} Thus the characteristic map loses local invertibility at the label $a_*$ when $t=T_*$. Now assume that the infimum $m$ is not attained. By the defining property of the infimum, for each $k \in \mathbb{N}$ there exists $a_k \in \mathbb{R}$ such that \begin{align*} m < u_0'(a_k) < m + \frac{1}{k}. \end{align*} The lower inequality is strict because no label attains the infimum. Since $m<0$, after discarding finitely many terms we may assume $u_0'(a_k)<0$ for every remaining $k$. For those $k$, define the degeneracy time $t_k \in (0,\infty)$ by \begin{align*} t_k := -\frac{1}{u_0'(a_k)}. \end{align*} Then \begin{align*} F_{t_k}'(a_k)=1+t_k u_0'(a_k)=0. \end{align*} Because $u_0'(a_k) \to m<0$ and the map $r \mapsto -1/r$ is continuous on $(-\infty,0)$, we have \begin{align*} t_k \to -\frac{1}{m}=T_*. \end{align*} Thus $T_*$ is approached by actual degeneracy times along a sequence of labels. However, because no label satisfies $u_0'(a)=m$, the equality \begin{align*} 1+T_*u_0'(a)=0 \end{align*} does not occur at any single label $a \in \mathbb{R}$ solely from this criterion. [/step] [step:Record the monotone and unbounded-slope alternatives] [guided] If $u_0'(a)\geq 0$ for every $a \in \mathbb{R}$, then for every finite time $t\geq 0$ and every label $a\in\mathbb{R}$, \begin{align*} F_t'(a)=1+t u_0'(a)\geq 1. \end{align*} Thus the label derivative never vanishes, and the derivative criterion predicts no finite loss of local invertibility. If instead \begin{align*} \inf_{a\in\mathbb{R}} u_0'(a)=-\infty, \end{align*} then for every $t>0$ there exists a label $a\in\mathbb{R}$ with \begin{align*} u_0'(a)<-\frac{1}{t}. \end{align*} For this label, \begin{align*} F_t'(a)=1+t u_0'(a)<0. \end{align*} This shows that the uniform positive lower-bound argument fails immediately after $t=0$. A negative nonzero derivative does not itself prove local non-invertibility at that label, because the [Inverse Function Theorem](/theorems/51) still gives a local inverse when the derivative is nonzero. Therefore the precise conclusion is that arbitrarily steep negative slopes predict immediate failure of the positive-slope criterion, but without an additional sign-change hypothesis they do not by themselves prove a zero of $F_t'$ at a specified label. [/guided] If $u_0'(a)\geq 0$ for every $a \in \mathbb{R}$, then for every $t\geq 0$ and every $a\in\mathbb{R}$, \begin{align*} F_t'(a)=1+t u_0'(a)\geq 1. \end{align*} Therefore this derivative never vanishes at finite time, so the criterion predicts no finite loss of local invertibility. If \begin{align*} \inf_{a\in\mathbb{R}} u_0'(a)=-\infty, \end{align*} then for every $t>0$ there exists $a\in\mathbb{R}$ such that \begin{align*} u_0'(a)<-\frac{1}{t}. \end{align*} For that label, \begin{align*} F_t'(a)=1+t u_0'(a)<0. \end{align*} A negative nonzero derivative at this label still gives local invertibility at that label by the [Inverse Function Theorem](/theorems/51). Thus the uniform positive lower-bound argument used for $t<T_*$ fails immediately after $t=0$. This proves the final criterion in its precise form: arbitrarily steep negative slopes predict immediate failure of the positive-slope criterion, but without an additional sign-change hypothesis they do not by themselves prove a zero of $F_t'$ at a specified label. [/step]