The method of characteristics for $u_t+u{u}_x=0$ and the finite blow-up time of the gradient

Fix $t\in [0,T_0)$. If $a<b$ and ${\phi }_t(a)={\phi }_t(b)$, then $b-a=-t(f(b)-f(a))\le tK(b-a)<b-a$ by [F1] and $tK<1$, a contradiction; hence ${\phi }_t$ is injective. Moreover ${\phi }_t(a)=a+tf(a)\to \pm \mathrm{\infty }$ as $a\to \pm \mathrm{\infty }$ because $f$ is bounded and $a\to \pm \mathrm{\infty }$; by [F2] it is surjective. Finally ${\phi }_t^{\mathrm{\prime }}(a)=1+t{f}^{\mathrm{\prime }}(a)\ge 1-tK>0$, so [F2] makes ${\phi }_t$ a $C^1$ diffeomorphism onto $\mathbb{R}$.

Endpoint level set. Let $F_0(a):=a+T_{0}f(a)$. Its derivative is $1+T_0{f}^{\mathrm{\prime }}(a)\ge 0$, so [F1] makes it nondecreasing. Since $f$ is bounded, $F_0(a)\to \pm \mathrm{\infty }$ as $a\to \pm \mathrm{\infty }$. Its level set $S=F_0^{-1}(x_{\ast })$ is therefore a nonempty compact interval $[\alpha ,\beta ]$, containing $a_{\ast }$. If $\alpha <\beta$, then $F_0$ is constant on this interval, so $f^{\mathrm{\prime }}=-K$ in its interior and, by continuity, at its endpoints. If $\alpha =\beta =a_{\ast }$ the same derivative value follows from the hypothesis. Thus $f^{\mathrm{\prime }}=-K$ on all of $S$.

Fix $(t_0,x_0)$ with $0\le t_0<T_0$ and put $a_0:=a(t_0,x_0)$. By step 1.1, $1+t_0{f}^{\mathrm{\prime }}(a_0)>0$. Continuity gives an open rectangle $Q$ around $(t_0,a_0)$ on which $1+t{f}^{\mathrm{\prime }}(a)>0$. For each fixed $t$ in this rectangle, [F4] shows that $a\mapsto {\phi }_t(a)$ is strictly increasing on its $a$-interval. Thus $\mathrm{\Phi }(t,a):=(t,{\phi }_t(a))$ is injective on $Q$, and its derivative has determinant $1+t{f}^{\mathrm{\prime }}(a)>0$. The injective-regular corollary therefore makes $\mathrm{\Phi }{\mathrm{\mid }}_Q$ a $C^1$ diffeomorphism onto its open image. This also applies at $t_0=0$, because $Q$ may extend to small negative times and the determinant remains positive locally. On the portion with $t\ge 0$, the local inverse agrees with the globally defined $a(t,x)={\phi }_t^{-1}(x)$ by uniqueness of the inverse. Hence $a(t,x)$ is $C^1$ in a relative neighbourhood of every point of $[0,T_0)\times \mathbb{R}$, and differentiating the identity ${\phi }_t(a(t,x))=x$ is now legitimate: ${\mathrm{\partial }}_{x}a(t,x)=\frac{1}{1+t{f}^{\mathrm{\prime }}(a(t,x))},{\mathrm{\partial }}_{t}a(t,x)=-\frac{f(a(t,x))}{1+t{f}^{\mathrm{\prime }}(a(t,x))}\mathrm{.}$ By [F3], $u=f\circ a$ is $C^1$ jointly, $u(0,x)=f(x)$, and $u_t=-f^{\mathrm{\prime }}(a)\frac{f(a)}{1+t{f}^{\mathrm{\prime }}(a)},u_x=\frac{f^{\mathrm{\prime }}(a)}{1+t{f}^{\mathrm{\prime }}(a)},u_t+u{u}_x=\frac{f^{\mathrm{\prime }}(a)}{1+t{f}^{\mathrm{\prime }}(a)}(-f(a)+f(a))=0.$ This proves (i).

Label limit. The identity $a(t)+tf(a(t))=x_{\ast }$ and boundedness of $f$ keep $a(t)$ in a fixed compact interval. Every subsequential limit $a_0$ as $t\to T_0$ from below satisfies $F_0(a_0)=x_{\ast }$, hence belongs to $S$. For $a\in S$, $f(a)=K(x_{\ast }-a)$, so ${\phi }_t(a)=(1-tK)a+tK{x}_{\ast }$. If $x_{\ast }<\alpha$, this gives ${\phi }_t(\alpha )>x_{\ast }$ and strict monotonicity of ${\phi }_t$ from step 1.1 forces $a(t)<\alpha$; its only possible limit in $S$ is $\alpha$. Similarly, if $x_{\ast }>\beta$, $a(t)>\beta$ and its only possible limit in $S$ is $\beta$. If $x_{\ast }\in S$, then $f(x_{\ast })=0$ and ${\phi }_t(x_{\ast })=x_{\ast }$, so uniqueness of the inverse gives $a(t)=x_{\ast }$. Thus every subsequential limit as $t\to T_0$ from below equals $a^{†}$. If $a(t)$ did not converge to $a^{†}$, there would be $\epsilon >0$ and a sequence $t_m\to T_0$ from below with $\mathrm{\mid }a(t_m)-a^{†}\mathrm{\mid }\ge \epsilon$. Boundedness and [F5] give a convergent subsequence of labels, contradicting the identified limit. Therefore $a(t)\to a^{†}$, and by step 1.2 and continuity, $f^{\mathrm{\prime }}(a(t))\to -K$.

Fix $a_0\in \mathbb{R}$ and put $x(t):={\phi }_t(a_0)=a_0+tf(a_0)$; then $a(t,x(t))=a_0$, so $u(t,x(t))=f(a_0)$ is constant. By step 2.1, $q(t)=u_x(t,x(t))=f^{\mathrm{\prime }}(a_0)\mathrm{/}(1+t{f}^{\mathrm{\prime }}(a_0))$, and differentiating this quotient with [F3], $q^{\mathrm{\prime }}(t)=-\frac{f^{\mathrm{\prime }}(a_0{)}^2}{(1+t{f}^{\mathrm{\prime }}(a_0){)}^2}=-q(t{)}^2\mathrm{.}$ This proves (ii).

Fixed-spatial gradient blow-up. Step 2.1 gives $u_x(t,x_{\ast })=f^{\mathrm{\prime }}(a(t))\mathrm{/}(1+t{f}^{\mathrm{\prime }}(a(t)))$. Its denominator is positive for $t<T_0$ by step 1.1 and tends to $1-T_{0}K=0$, while its numerator tends to $-K\mathrm{\ne }0$ by step 2.2. Hence $\mathrm{\mid }{u}_x(t,x_{\ast })\mathrm{\mid }\to +\mathrm{\infty }$, and the supremum over $x$ diverges as well. Also $\mathrm{\mid }u(t,x_{\ast })\mathrm{\mid }=\mathrm{\mid }f(a(t))\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Independently, along the specified characteristic ${\phi }_t(a_{\ast })$, step 3.1 gives $u_x(t,{\phi }_t(a_{\ast }))=-K\mathrm{/}(1-tK)$. Thus the fixed-spatial and material-characteristic gradient blow-up claims both hold even when the endpoint map has a flat level interval; no injectivity at $T_0$ was assumed.