Distinct states with equal flux give a stationary weak discontinuity

The jump speed vanishes. With $u_L=1$, $u_R=-1$, $f(u)=u^2$: $f(u_L)=f(u_R)=1$, so [F1] gives $s=(f(u_R)-f(u_L))\mathrm{/}(u_R-u_L)=0\mathrm{/}(-2)=0$.

The stationary jump is a weak solution with the stated datum. Since $u(t,x)=\pm 1$ takes only the values $\pm 1$, one has $f(u(t,x))=u(t,x{)}^2=1$ almost everywhere, and $u$ is independent of $t$. Hence ${\int }_{{\mathrm{\Pi }}_T}(u{\phi }_t+f(u){\phi }_x)dxdt={\int }_{\mathbb{R}}u(x)[{\int }_0^T{\phi }_t(t,x)dt]dx+{\int }_0^T[{\int }_{\mathbb{R}}{\phi }_x(t,x)dx]dt=0+0=0$ for every $\phi \in C_c^{\mathrm{\infty }}({\mathrm{\Pi }}_T)$, because the inner $t$-integral of ${\phi }_t$ vanishes by compact support in time and the inner $x$-integral of ${\phi }_x$ vanishes by compact support in space. Since $u(t,\cdot )=u_0(\cdot )$ identically, the strong local $L^1$ initial trace condition holds with vanishing error. Thus $u$ is a distributional weak solution with Riemann datum $u_0$, and by [F4] it is the stationary self-similar profile of that Riemann problem.

Chord check. For the jump $u^{-}=u_L=1>u^{+}=u_R=-1$ with speed $s=0$, [F1] gives $F(z)=z^2-1-0\cdot (z-1)=z^2-1\le 0$ for $z\in [-1,1]$ by [F3], while $u_R-u_L=-2<0$; hence $F(z)(u_R-u_L)\ge 0$ for every $z$ between the states, and the chord condition holds. Equivalently, the chord through $(-1,1)$ and $(1,1)$ is the constant line $1$ and the parabola $z^2$ lies below it on $[-1,1]$.

The Kruzhkov inequalities. For general $k\in \mathbb{R}$, both ${\eta }_k(u)$ and $q_k(u)$ are piecewise constant with a single jump at $x=0$, and $u$ does not depend on $t$, so ${\mathrm{\partial }}_t{\eta }_k(u)=0$ and ${\mathrm{\partial }}_x{q}_k(u)=(q_k(u_R)-q_k(u_L)){\delta }_0$ in distributions. With $f(s)=s^2$ one computes $q_k(u_R)-q_k(u_L)=(1-k^2)[\mathrm{sgn}(-1-k)-\mathrm{sgn}(1-k)]=-(1-k^2)[\mathrm{sgn}(1+k)+\mathrm{sgn}(1-k)]\le 0$, because $1-k^2\ge 0$ exactly when $\mathrm{\mid }k\mathrm{\mid }\le 1$, where $\mathrm{sgn}(1+k)+\mathrm{sgn}(1-k)\ge 0$, and for $\mathrm{\mid }k\mathrm{\mid }>1$ the last bracket vanishes. Hence all Kruzhkov entropy inequalities hold with a nonpositive measure.

Conclusion. The jump has speed $0$ by step 1.1, the nonzero difference of states produces a genuine discontinuity, and the entropy inequalities hold for all Kruzhkov pairs by step 3.1 (with the smooth-pair check of step 2.1 as the geometric form of the same condition), so $u$ is a bounded Kruzhkov entropy solution whose flux values at the two states coincide. The equal flux values were responsible for the vanishing speed, while admissibility had to be verified separately through the chord condition.