Oleinik's one-sided estimate characterizes bounded entropy solutions

Shifting and positive-strip gradient bounds. Fix $c\in I$ and put $J:=I-c$, which contains $0$. Set $\tilde{f}(s):=f(s)-f(0)$ and $g(v):=\tilde{f}(v+c)-\tilde{f}(c)=f(v+c)-f(c)\mathrm{.}$ Then $g(0)=0$ and $g^{\mathrm{\prime }\mathrm{\prime }}\ge \kappa$ on $J$. For a smooth compactly supported datum $v_0^{\mathrm{s}}$ with values in $J$, apply [F2] to flux $g$ and datum $v_0^{\mathrm{s}}$, obtaining $v^{\epsilon }$ with values in $J$. The unshifted profile $U^{\epsilon }:=v^{\epsilon }+c$ has datum $v_0^{\mathrm{s}}+c$, equals $c$ outside a compact set, and solves the equation with normalized flux $\tilde{f}$; thus the compact datum to which [F2] is applied is $v_0^{\mathrm{s}}$, not the generally noncompact function $v_0^{\mathrm{s}}+c$ or a nonzero-tail shift. The Oleinik slope is unchanged by adding $c$. Set $M_v:=\mathrm{\parallel }{v}_0^{\mathrm{s}}{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $G_0:={\sup }_{v\in J}\mathrm{\mid }{g}^{\mathrm{\prime }}(v)\mathrm{\mid }<\mathrm{\infty }$. By variation of constants, for $0<d<t$, $v^{\epsilon }(t)=H_{\epsilon (t-d)}{v}^{\epsilon }(d)-{\int }_d^t{\mathrm{\partial }}_x{H}_{\epsilon (t-s)}g(v^{\epsilon }(s))ds\mathrm{.}$ Indeed, differentiating $H_{\epsilon (t-s)}{v}^{\epsilon }(s)$ for $d<s<t$ and using the viscous equation gives $-{\mathrm{\partial }}_x{H}_{\epsilon (t-s)}g(v^{\epsilon }(s))$; the integral converges at $s=t$ because $\mathrm{\parallel }{\mathrm{\partial }}_x{\mathrm{\Gamma }}_{\epsilon r}{\mathrm{\parallel }}_1\le C_{\epsilon }{r}^{-1\mathrm{/}2}$. The Gaussian derivative bounds [F2] also give $\mathrm{\parallel }{\mathrm{\partial }}_x{\mathrm{\Gamma }}_{\epsilon r}(\cdot +h)-{\mathrm{\partial }}_x{\mathrm{\Gamma }}_{\epsilon r}{\mathrm{\parallel }}_1\le \min (C_{\epsilon }{r}^{-1\mathrm{/}2},C_{\epsilon }\mathrm{\mid }h\mathrm{\mid }{r}^{-1}),\mathrm{\parallel }{\mathrm{\partial }}_r{\mathrm{\partial }}_x{\mathrm{\Gamma }}_{\epsilon r}{\mathrm{\parallel }}_1\le C_{\epsilon }{r}^{-3\mathrm{/}2}\mathrm{.}$ Splitting the spatial integral at $r=\mathrm{\mid }h{\mathrm{\mid }}^2$ and the time integral at $r=\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }$ in the restarted identity shows that, for every $0<\alpha <1$ and every $\tau >0$, $\mathrm{\mid }{v}^{\epsilon }(t,x+h)-v^{\epsilon }(t,x)\mathrm{\mid }\le C_{\tau ,T}\mathrm{\mid }h{\mathrm{\mid }}^{\alpha },\mathrm{\mid }{v}^{\epsilon }(t,x)-v^{\epsilon }(t^{\mathrm{\prime }},x)\mathrm{\mid }\le C_{\tau ,T}\mathrm{\mid }t-t^{\mathrm{\prime }}{\mathrm{\mid }}^{1\mathrm{/}2}$ on $[\tau \mathrm{/}2,T]\times \mathbb{R}$. Fix $\tau >0$, put $d=\tau \mathrm{/}2$, and differentiate the restarted identity for $t\ge \tau$, first omitting its final $\eta$-length of integration and then letting $\eta \downarrow 0$. Since $\int {\mathrm{\partial }}_{xx}{\mathrm{\Gamma }}_{\epsilon r}=0$, cancellation gives $p(t,x)={\mathrm{\partial }}_x{H}_{\epsilon (t-d)}{v}^{\epsilon }(d,x)-{\int }_0^{t-d}{\int }_{\mathbb{R}}{\mathrm{\partial }}_{xx}{\mathrm{\Gamma }}_{\epsilon r}(z)(g(v^{\epsilon }(t-r,x-z))-g(v^{\epsilon }(t,x)))dzdr\mathrm{.}$ The range bound and the positive-strip estimates imply $\mathrm{\mid }g(v^{\epsilon }(t-r,x-z))-g(v^{\epsilon }(t,x))\mathrm{\mid }\le G_0{C}_{\tau ,T}(\mathrm{\mid }z{\mathrm{\mid }}^{\alpha }+r^{1\mathrm{/}2})$. By scaling, $\int \mathrm{\mid }{\mathrm{\partial }}_{xx}{\mathrm{\Gamma }}_{\epsilon r}(z)\mathrm{\mid }(\mathrm{\mid }z{\mathrm{\mid }}^{\alpha }+r^{1\mathrm{/}2})dz\le C_{\epsilon ,\alpha }(r^{-1+\alpha \mathrm{/}2}+r^{-1\mathrm{/}2}),$ which is integrable at $r=0$; the first term is bounded by $C_{\epsilon ,\tau }{M}_v$ because $t-d\ge \tau \mathrm{/}2$. Thus ${\sup }_{(t,x)\in [\tau ,T]\times \mathbb{R}}\mathrm{\mid }p(t,x)\mathrm{\mid }<\mathrm{\infty }\mathrm{.}$ This bound is global in space and holds on every positive-time strip.

Equivalence of the pointwise and distributional forms. Fix $t\in (0,T)$, assume first that (ii) holds at this $t$, and put $w(x)=u(t,x)-x\mathrm{/}(\kappa t)$, so that $w(y)\le w(x)$ for almost every $x<y$. For $n\ge 1$ set $a:=1\mathrm{/}n$ and $w_n(x)=n{\int }_x^{x+a}w(z)dz$. If $x<x^{\mathrm{\prime }}$, put $d:=x^{\mathrm{\prime }}-x>0$. When $d\le a$, cancellation of the overlap gives $w_n(x)-w_n(x^{\mathrm{\prime }})=n({\int }_x^{x+d}w(z)dz-{\int }_{x+a}^{x+a+d}w(\zeta )d\zeta )\ge 0,$ because the two intervals have equal length $d$ and every interior pair $z\in [x,x+d]$, $\zeta \in [x+a,x+a+d]$ satisfies $z<\zeta$; explicitly, $d({\int }_x^{x+d}w-{\int }_{x+a}^{x+a+d}w)={\int }_{[x,x+d]\times [x+a,x+a+d]}(w(z)-w(\zeta ))dzd\zeta \ge 0$ by the assumed a.e. pair inequality. When $d\ge a$, the intervals $[x,x+a]$ and $[x^{\mathrm{\prime }},x^{\mathrm{\prime }}+a]$ are ordered and have equal length, so $w_n(x)-w_n(x^{\mathrm{\prime }})=n({\int }_x^{x+a}w(z)dz-{\int }_{x^{\mathrm{\prime }}}^{x^{\mathrm{\prime }}+a}w(\zeta )d\zeta )\ge 0,$ since $a({\int }_x^{x+a}w-{\int }_{x^{\mathrm{\prime }}}^{x^{\mathrm{\prime }}+a}w)={\int }_{[x,x+a]\times [x^{\mathrm{\prime }},x^{\mathrm{\prime }}+a]}(w(z)-w(\zeta ))dzd\zeta \ge 0$. Thus $w_n$ is nonincreasing and $D_x{w}_n\le 0$ distributionally. Since $w_n\to w$ in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ as $n\to \mathrm{\infty }$, distributional differentiation passes to the limit, so $D_{x}w\le 0$, which is (iii) at this $t$. Conversely, assume (iii) at some $t$ and let ${\rho }_{\delta }$ be a nonnegative spatial mollifier; then $w_{\delta }=w\ast {\rho }_{\delta }$ is smooth with $D_x{w}_{\delta }=(D_{x}w)\ast {\rho }_{\delta }\le 0$, so $w_{\delta }$ is nonincreasing and $w_{\delta }(y)\le w_{\delta }(x)$ for all $x<y$; since $w_{\delta }\to w$ in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$, passing to an almost-everywhere convergent subsequence gives the two-point inequality of (ii) at this $t$ for almost every pair. Hence (ii) and (iii) hold for the same full-measure set of times, proving the equivalence.

Mollification commutator. Assume (iii). Fix a nonnegative test function $\phi$ supported in $[a,b]\times [-R,R]$ with $0<a<b<T$, and let ${\rho }_{\delta }$ be a nonnegative unit-mass mollifier on ${\mathbb{R}}^2$ supported in the ball of radius $\delta <\frac{1}{2}\min \{a,T-b\}$. Extend $u$ boundedly to all of ${\mathbb{R}}^2$ by a fixed value in $I$ outside $(0,T)\times \mathbb{R}$, and set $u_{\delta }=u\ast {\rho }_{\delta }$, $F_{\delta }=f(u)\ast {\rho }_{\delta }$, $r_{\delta }=f(u_{\delta })-F_{\delta }$. By [F5], distributional derivatives commute with convolution, so on a neighbourhood of the support of $\phi$ the identity ${\mathrm{\partial }}_t{u}_{\delta }+{\mathrm{\partial }}_x{F}_{\delta }=0$ holds; since the mollification averages only over times $\ge a-\delta >a\mathrm{/}2$, the distributional bound (iii) gives ${\mathrm{\partial }}_x{u}_{\delta }\le 2\mathrm{/}(\kappa a)$ there; and $u_{\delta }$ still takes values in $I$. The tangent inequality $f(s)\ge f(u_{\delta })+f^{\mathrm{\prime }}(u_{\delta })(s-u_{\delta })$ for convex $f$, averaged against ${\rho }_{\delta }$, gives $r_{\delta }\le 0$. Finally, localizing $u$ and $f(u)$ by a cutoff equal to $1$ on a slightly larger compact set and applying approximate-identity convergence in $L^1$ [F5] gives $u_{\delta }\to u$, $F_{\delta }\to f(u)$ and $r_{\delta }\to 0$ in $L^1$ on the support of $\phi$, the last two also using that $f$ is Lipschitz on the bounded interval $I$.

A classical barrier after flux regularization. Take smooth normalized $g_l$ converging to $g$ in $C^2$ on the compact state interval $J$, with $g_l(0)=0$ and $g_l^{\mathrm{\prime }\mathrm{\prime }}\ge {\kappa }_l:=\kappa -1\mathrm{/}l>0$ there (mollify $g$ at sufficiently small scales). Choose $0<{\epsilon }_l\le 1$ tending to zero, and let $v^l$ have the fixed smooth datum of step 1.1 and flux $g_l$. Its range lies in $J$. By [F2], $p=v_x^l$ is classical at positive times, is bounded globally on positive strips by step 1.1, and satisfies $p_t+g_l^{\mathrm{\prime }}(v^l)p_x+g_l^{\mathrm{\prime }\mathrm{\prime }}(v^l)p^2={\epsilon }_l{p}_{xx}$. Fix $\tau >0$, put $q(t)=1\mathrm{/}({\kappa }_l(t-\tau ))$, $b=g_l^{\mathrm{\prime }}(v^l)$ and $W=p-q$. Where $W>0$, $({\mathrm{\partial }}_t+b{\mathrm{\partial }}_x-{\epsilon }_l{\mathrm{\partial }}_{xx})W<0$ since $p>q>0$. If $G_l={\sup }_J\mathrm{\mid }{g}_l^{\mathrm{\prime }}\mathrm{\mid }$, choose $A>G_l+2{\epsilon }_l$ and $\mathrm{\Phi }=e^{A(t-\tau )}(1+x^2)$, so the same operator applied to $\mathrm{\Phi }$ is strictly positive. For $\rho >0$, $Z=W-\rho \mathrm{\Phi }$ is negative at some $t_0>\tau$ sufficiently close to $\tau$, by the positive-strip bound on $p$, and negative on the sides of a sufficiently large rectangle. A positive maximum on that rectangle would have $Z_t\ge 0$, $Z_x=0$, $Z_{xx}\le 0$, contradicting the strict operator inequality there. Thus $Z\le 0$. Let $\rho \downarrow 0$ and then $\tau \downarrow 0$ to get $v_x^l\le 1\mathrm{/}({\kappa }_{l}t)$; integrating in $x$ gives $v^l(t,y)-v^l(t,x)\le (y-x)\mathrm{/}({\kappa }_{l}t)$. This uses a classical maximum argument, with no Sobolev positive-part test.

The entropy production tends to a nonpositive limit. For any convex $\eta \in C^2$ and $q^{\mathrm{\prime }}={\eta }^{\mathrm{\prime }}{f}^{\mathrm{\prime }}$, the chain and product rules applied to ${\mathrm{\partial }}_t{u}_{\delta }+{\mathrm{\partial }}_x(f(u_{\delta })-r_{\delta })=0$ give ${\mathrm{\partial }}_t\eta (u_{\delta })+{\mathrm{\partial }}_{x}q(u_{\delta })={\mathrm{\partial }}_x({\eta }^{\mathrm{\prime }}(u_{\delta })r_{\delta })-{\eta }^{\mathrm{\prime }\mathrm{\prime }}(u_{\delta })({\mathrm{\partial }}_x{u}_{\delta })r_{\delta }\le {\mathrm{\partial }}_x({\eta }^{\mathrm{\prime }}(u_{\delta })r_{\delta })+\frac{2\mathrm{\parallel }{\eta }^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I)}}{\kappa a}(-r_{\delta })$, using $r_{\delta }\le 0$ and ${\mathrm{\partial }}_x{u}_{\delta }\le 2\mathrm{/}(\kappa a)$ from step 1.3. Tested against the nonnegative $\phi$, the first term is bounded by $\mathrm{\parallel }{\eta }^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I)}\mathrm{\parallel }{\phi }_x{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{r}_{\delta }{\mathrm{\parallel }}_{L^1(\mathrm{supp}\phi )}$ after an integration by parts, and the second by $\frac{2\mathrm{\parallel }{\eta }^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}}{\kappa a}\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{r}_{\delta }{\mathrm{\parallel }}_{L^1(\mathrm{supp}\phi )}$; both tend to $0$ as $\delta \downarrow 0$. Since ${\mathrm{\partial }}_t\eta (u_{\delta })+{\mathrm{\partial }}_{x}q(u_{\delta })\to {\mathrm{\partial }}_t\eta (u)+{\mathrm{\partial }}_{x}q(u)$ distributionally by [F5] (local $L^1$ convergence of $u_{\delta }$ and continuity of $\eta ,q$), the limit satisfies $⟨{\mathrm{\partial }}_t\eta (u)+{\mathrm{\partial }}_{x}q(u),\phi ⟩\le 0$ for every nonnegative test function supported in $(0,T)\times \mathbb{R}$.

Passage to the smooth-datum entropy solution. The varying-flux family $v^l$ of step 2.1 satisfies the common range and derivative bounds of the compactness lemma [F3]. Extract a locally $L^1$ and almost-everywhere convergent subsequence. The existence proof passes its weak and entropy identities to the limit because $g_l\to g$ in $C^1$ on the range; the uniform local time modulus supplies the initial trace. Uniqueness identifies the limit as the entropy solution $v$ for $g$ with this smooth datum. Fubini gives slicewise convergence at almost every time, and passage to the bound in step 2.1, with ${\kappa }_l\to \kappa$, gives $v(t,y)-v(t,x)\le (y-x)\mathrm{/}(\kappa t)$ for almost every time and almost every pair $x<y$.

Approximation for general data: (i) implies (ii). Now let $u$ be the given entropy solution with datum $u_0$ and range in $I$, and fix $c\in I$. For $j\ge 1$ set $w_j=((u_0-c){\mathbf{1}}_{[-j,j]})\ast {\rho }_{{\delta }_j}$, where ${\rho }_{{\delta }_j}$ is a nonnegative mollifier of radius ${\delta }_j\downarrow 0$: then $w_j\in C_c^{\mathrm{\infty }}$, its values lie in $I-c$ (a convex combination of values of $u_0-c$), and $w_j\to u_0-c$ in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{R})$. Let $v_j$ be the entropy solution with datum $w_j$ for the flux $g$: by step 3.1 each $v_j$ satisfies the two-point estimate, and by [F4] applied to $v_j$ and $\tilde{u}=u-c$, the differences converge to $0$ in $L^1$ on every compact cylinder; a diagonal subsequence converges almost everywhere on ${\mathrm{\Pi }}_T$. Passing the two-point estimate to that almost-everywhere limit proves (ii) for $u$.

Kruzhkov pairs and conclusion of (iii) implies (i). For $k\in \mathbb{R}$ and $m\ge 1$ put ${\eta }_m(s)=\sqrt{(s-k{)}^2+m^{-2}}$ and $q_m(s)={\int }_k^s{\eta }_m^{\mathrm{\prime }}(z)f^{\mathrm{\prime }}(z)dz$, so $({\eta }_m,q_m)$ is a smooth convex entropy pair. Step 2.2 gives ${\mathrm{\partial }}_t{\eta }_m(u)+{\mathrm{\partial }}_x{q}_m(u)\le 0$ for every $m$; letting $m\to \mathrm{\infty }$, ${\eta }_m\to \mathrm{\mid }s-k\mathrm{\mid }$ and $q_m\to \mathrm{sgn}(s-k)(f(s)-f(k))$ uniformly on the bounded interval $I$ by dominated convergence, so the distributional inequality passes to the limit and every Kruzhkov inequality holds. Together with the weak equation and the strong local $L^1$ trace (hypotheses), $u$ is a Kruzhkov entropy solution, proving (iii) implies (i); the reverse implication (i) implies (ii) is step 4.1, and the equivalence of (ii) and (iii) is step 1.2.

The piecewise $C^1$ remark. If $u$ is piecewise $C^1$ with a single jump at a curve $x=s(t)$ and satisfies the hypotheses, then (ii) forces the right trace not to exceed the left trace across an upward jump: taking $x\uparrow s(t)$, $y\downarrow s(t)$ in the two-point inequality and letting $x,y\to s(t)$ gives $u^{+}-u^{-}\le 0$, so an upward jump $u^{+}>u^{-}$ is excluded; for the remaining jumps with $u^{-}>u^{+}$ the Rankine--Hugoniot condition and the chord criterion [F6] make them entropy-admissible. This shows how the slope bound encodes admissibility in the piecewise smooth class, while the equivalence itself was proved for all bounded weak solutions.