Gradient catastrophe before shock formation

The data. $u_0(y)=-\arctan y$ is smooth and bounded, with $u_0^{\mathrm{\prime }}(y)=-1\mathrm{/}(1+y^2)\in [-1,0)$ and ${\min }_y{u}_0^{\mathrm{\prime }}=-1$ attained only at $y=0$. The flux $f(u)=\frac{1}{2}{u}^2$ is $C^{\mathrm{\infty }}$ with $f^{\mathrm{\prime }}(u)=u$.

The characteristic map is a diffeomorphism for $t<1$. $X_t^{\mathrm{\prime }}(y)=1+t{u}_0^{\mathrm{\prime }}(y)=1-\frac{t}{1+y^2}=\frac{1+y^2-t}{1+y^2}>0$ for $0\le t<1$ and all $y$, so by the mean value theorem [F2] $X_t$ is strictly increasing; moreover $X_t(y)=y+O(1)$ tends to $\pm \mathrm{\infty }$ as $y\to \pm \mathrm{\infty }$, so $X_t$ maps $\mathbb{R}$ onto $\mathbb{R}$. A strictly increasing surjection is a homeomorphism, and since $X_t^{\mathrm{\prime }}$ never vanishes, the inverse function theorem [F2] makes the inverse $y(\cdot ,t)=X_t^{-1}$ smooth with $y_x(x,t)=1\mathrm{/}{X}_t^{\mathrm{\prime }}(y(x,t))$ and, differentiating $X_t(y(x,t))=x$ in $t$, $y_t(x,t)=-u_0(y(x,t))\mathrm{/}{X}_t^{\mathrm{\prime }}(y(x,t))$.

The outer branches for $t>1$. The function $h(a)=t\arctan a-a$ increases up to $\sqrt{t-1}$ and then decreases to $-\mathrm{\infty }$, so its unique positive zero $a(t)$ satisfies $a(t)>\sqrt{t-1}$. Hence $X_t^{\mathrm{\prime }}>0$ on $[a(t),\mathrm{\infty })$, and $X_t$ maps this interval bijectively onto $[0,\mathrm{\infty })$; by oddness it maps $(-\mathrm{\infty },-a(t)]$ bijectively onto $(-\mathrm{\infty },0]$. For $x>0$ choose the unique $y>a(t)$ with $X_t(y)=x$, and for $x<0$ choose the unique $y<-a(t)$; define $u(x,t)=-\arctan y$. These branches are smooth by the inverse function theorem, solve Burgers directly: implicit differentiation gives $y_x=1\mathrm{/}{X}_t^{\mathrm{\prime }}(y)$, $y_t=-u_0(y)\mathrm{/}{X}_t^{\mathrm{\prime }}(y)$, hence $u_t+u{u}_x=0$, and have traces $u^{-}=\arctan a(t)$, $u^{+}=-\arctan a(t)$ at $x=0$. Their fluxes agree, so the stationary jump satisfies Rankine--Hugoniot and is entropy-admissible by [F3].

The ansatz is a classical solution. Put $u(x,t)=u_0(y(x,t))$ for $0\le t<1$, which is smooth in $(x,t)$. By the chain rule and step 1.2, $u_t=u_0^{\mathrm{\prime }}(y)y_t=-u_0^{\mathrm{\prime }}(y)u_0(y)\mathrm{/}{X}_t^{\mathrm{\prime }}(y)$ and $u_x=u_0^{\mathrm{\prime }}(y)y_x=u_0^{\mathrm{\prime }}(y)\mathrm{/}{X}_t^{\mathrm{\prime }}(y)$. Hence $u_t+f(u{)}_x=u_t+f^{\mathrm{\prime }}(u)u_x=u_0^{\mathrm{\prime }}(y)[-u_0(y)+u_0(y)]\mathrm{/}{X}_t^{\mathrm{\prime }}(y)=0$, so $u$ solves $u_t+f(u{)}_x=0$ classically on $\mathbb{R}\times (0,1)$. Since $X_t$ satisfies ${\dot{X}}_t(y)=u_0(y)=f^{\mathrm{\prime }}(u(X_t(y),t))$, this is exactly the family of characteristics of [F1], along which $u$ is the constant $u_0(y)$.

The gradient formula. Differentiating $u(X_t(y),t)=u_0(y)$ in $y$ and using $u_x{X}_t^{\mathrm{\prime }}=u_0^{\mathrm{\prime }}$ gives $u_x(X_t(y),t)={\displaystyle \frac{u_0^{\mathrm{\prime }}(y)}{X_t^{\mathrm{\prime }}(y)}}={\displaystyle \frac{u_0^{\mathrm{\prime }}(y)}{1+t{u}_0^{\mathrm{\prime }}(y)}}=-{\displaystyle \frac{1}{1+y^2-t}}$. At $y=0$, where $X_t(0)=0$, this reads $u_x(0,t)=-1\mathrm{/}(1-t)$ for $t<1$.

Catastrophe at $T_{\ast }=1$. For each $t<1$, ${\sup }_x\mathrm{\mid }{u}_x(x,t)\mathrm{\mid }={\sup }_y\frac{1}{1+y^2-t}=\frac{1}{1-t}\to \mathrm{\infty }$ as $t\uparrow 1$, the supremum being attained at $y=0$; the classical solution exists for every $t<1$ by step 2.1, and its slope becomes unbounded as $t\uparrow 1$. Hence the first gradient catastrophe occurs at $T_{\ast }=1$.

The limit profile at $t=1$. $X_1^{\mathrm{\prime }}(y)=\frac{y^2}{1+y^2}\ge 0$ with equality only at $y=0$, so $X_1(y)=y-\arctan y$ is strictly increasing with range $\mathbb{R}$; its inverse is continuous, and the profile $u(x,1)=u_0(X_1^{-1}(x))$ is continuous. For $y\mathrm{\ne }0$, implicit differentiation as in step 2.2 with $t=1$ gives $u_x(X_1(y),1)=-1\mathrm{/}{y}^2$, which tends to $-\mathrm{\infty }$ as $y\to 0$, i.e. as the corresponding point $x=X_1(y)\to 0$. Thus at $t=1$ the solution is still continuous but has unbounded slope at $x=0$: a gradient catastrophe, not a jump.

Entropy continuation and uniqueness. The branches of step 1.3 give a bounded piecewise smooth profile for $t>1$. Its only jump is the descending stationary shock, so graph integration and the chord criterion give the weak equation and all smooth convex entropy inequalities; smooth convex approximation gives the Kruzhkov inequalities. For $t<1$ the smooth solution of step 2.1 has zero entropy production and attains $u_0$ locally uniformly. As $t\to 1$ from either side, the selected feet converge for every $x\mathrm{\ne }0$ to $X_1^{-1}(x)$; boundedness and dominated convergence give matching local $L^1$ traces to the continuous profile of step 3.2. Integrating separately below and above $t=1$ and taking these traces cancels the time-interface terms in both weak and entropy pairings. Thus this is a global entropy solution with the stated datum. Its uniqueness follows from [F3], even though $-\arctan x$ is not integrable. The first slope blow-up is at $t=1$, and a nonzero shock is present for every $t>1$.