Characteristics and the Riccati equation for the spatial derivative

Part (i). With $a(x,t)=f^{\mathrm{\prime }}(u(x,t))$, [F2] gives $\frac{d}{dt}u(x(t),t)=u_t+u_x\dot{x}$ along the characteristic. Substituting the characteristic ODE $\dot{x}=f^{\mathrm{\prime }}(u(x(t),t))$ and then the pointwise equation of [F1] gives $\frac{d}{dt}u(x(t),t)=u_t+f^{\mathrm{\prime }}(u)u_x=u_t+(f(u){)}_x=0$. Hence $t\mapsto u(x(t),t)$ is constant along every characteristic.

Part (ii): differentiation in $x$. Assume now $u\in C^2$, so that $p=u_x$ is $C^1$ and $f^{\mathrm{\prime }}(u)p$ is $C^1$ by [F1] and [F3]. Differentiating the pointwise equation $u_t+(f(u){)}_x=0$ in $x$ gives $p_t+(f^{\mathrm{\prime }}(u)p{)}_x=0$, and the product and chain rules give $(f^{\mathrm{\prime }}(u)p{)}_x=f^{\mathrm{\prime }\mathrm{\prime }}(u)u_{x}p+f^{\mathrm{\prime }}(u)p_x=f^{\mathrm{\prime }\mathrm{\prime }}(u)p^2+f^{\mathrm{\prime }}(u)p_x$, so that $p_t+f^{\mathrm{\prime }}(u)p_x+f^{\mathrm{\prime }\mathrm{\prime }}(u)p^2=0$.

The Riccati equation along characteristics. Along the characteristic of step 1.1, [F2] applied to the $C^1$ function $p$ gives $\dot{p}=p_t+p_x\dot{x}=p_t+f^{\mathrm{\prime }}(u)p_x$. By step 1.2 this equals $-f^{\mathrm{\prime }\mathrm{\prime }}(u(x(t),t))p^2$, which is the asserted Riccati equation.

Monotonicity. By step 1.1, $u(x(t),t)\equiv u_{\ast }$ is constant along the characteristic, so along that curve $f^{\mathrm{\prime }\mathrm{\prime }}(u(x(t),t))=f^{\mathrm{\prime }\mathrm{\prime }}(u_{\ast })$ is a constant $c$ and step 2.1 reads $\dot{p}=-c{p}^2$. If $f^{\mathrm{\prime }\mathrm{\prime }}\ge 0$ on the range of $u$, then $c\ge 0$, so $\dot{p}\le 0$ and $p$ is nonincreasing along the characteristic.

Exact Riccati solution. Suppose $p_0=p(t_0)<0$ and $c:=f^{\mathrm{\prime }\mathrm{\prime }}(u(x_0,t_0))\ge \kappa >0$; by step 3.1 the constant is $c=f^{\mathrm{\prime }\mathrm{\prime }}(u_{\ast })$ along the whole characteristic. The scalar ODE $\dot{p}=-c{p}^2$ has a locally Lipschitz right-hand side, so uniqueness in [F4] and the zero solution imply that $p$ cannot reach zero on its interval of existence. Since $p(t_0)<0$, it remains negative there. Hence one has $\frac{d}{dt}\left(\frac{1}{p}\right)=-\dot{p}\mathrm{/}{p}^2=c$ by step 3.1, hence $\frac{1}{p(t)}=\frac{1}{p_0}+c(t-t_0)$ and therefore $p(t)=\frac{p_0}{1+c{p}_0(t-t_0)}=\frac{p_0}{1-c\mathrm{\mid }{p}_0\mathrm{\mid }(t-t_0)}$.

Blow-up and the persistence bound. Since $u(x(t),t)\equiv u_{\ast }$ by step 1.1, the speed $\dot{x}=f^{\mathrm{\prime }}(u_{\ast })$ is constant, so the characteristic is the straight line $x(t)=x_0+f^{\mathrm{\prime }}(u_{\ast })(t-t_0)$, on its maximal interval. If that interval ended at an interior time $t_1\in (0,T)$, the straight line would have a finite endpoint $x_1$; continuity would give $u(x_1,t_1)=u_{\ast }$, and [F4] would extend the characteristic with initial condition $x(t_1)=x_1$, contradicting maximality. Thus its interval is $(0,T)$. The denominator in step 4.1 is positive exactly for $t<t_{\ast }=t_0+1\mathrm{/}(c\mathrm{\mid }{p}_0\mathrm{\mid })\le t_0+1\mathrm{/}(\kappa \mathrm{\mid }{p}_0\mathrm{\mid })$ and tends to $0$ as $t\uparrow t_{\ast }$, while $p_0<0$, so $p(t)\to -\mathrm{\infty }$. Were a $C^2$ solution defined on $\mathbb{R}\times (0,T)$ with $T>t_{\ast }$, then $p=u_x$ would be continuous, hence finite, on the compact rectangle $[-A,A]\times [t_0,\frac{1}{2}(t_{\ast }+T)]$, with $A>1+\mathrm{\mid }{x}_0\mathrm{\mid }+\mathrm{\mid }{f}^{\mathrm{\prime }}(u_{\ast })\mathrm{\mid }(T-t_0)$, and along the straight characteristic it would equal the explicit solution of step 4.1, which is unbounded on that interval near $t_{\ast }$: a contradiction. Hence the $C^2$ solution cannot persist through $t_{\ast }$ along this characteristic.