Combining Theorem 2.3 and Corollary 2.5 of this paper gives that, for a strictly convex conservation law $$u_t + f(u)_x = 0,$$ satisfying the entropy condition $$\eta(u)_t + q(u)_x \le 0$$ in the sense of distribution for one entropy-entropy flux pair $(\eta,q)$ is equivalent to satisfying it for all entropy-entropy flux pair $(\eta,q)$.

On a high level, the key point of the paper is based on the fact that one entropy is enough to guarantee that the solution of the conservation law corresponds to the unique viscosity solution Hamilton-Jacobi equation $$h_t + f(h_x) = 0$$ (which is, indeed, the content of Theorem 2.3).

Question: Can you give me a high-level overview of this passage of the proof? Specifically, what goes on in Section 4.2 (that is, pages 7--8) of the paper? I feel like I'm loosing the intuition behind their argument and the motivation behind their estimates.

Let me ask more precisely what are some points of pag. 7--8 of the linked paper that I'd like to clarify.

  • We want to prove that, if $\zeta$ is a smooth function such that $h-\zeta$ has a minimum at some $(t, x) \in \Omega$, then $\left[\zeta_{t}+f\left(\zeta_{x}\right)\right](t, x) \geq 0$. For simplicity we assume that $(t, x)=(0,0)$ and $[h-\zeta](0,0)=0$. Moreover, we assume that the minimum is strict. Indeed, if we choose $\varepsilon>0$ and consider $\zeta^{\varepsilon}(t, x):=\zeta(t, x)+\varepsilon\left(t^{2}+x^{2}\right)$, then $\left[h-\zeta^{\varepsilon}\right]$ has a strict minimum at $(0,0)$ and $\left[\zeta_{t}^{\varepsilon}+f\left(\zeta_{x}^{\varepsilon}\right)\right](0,0)=\left[\zeta_{t}+f\left(\zeta_{x}\right)\right](0,0)$

For any $\delta>0$ consider \begin{align*} \Omega_{\delta}:=\text { connected component of }\{(t, x):[h-\zeta](t, x)<\delta\} \text { containing }(0,0) \text {. } \end{align*}

Since $h$ is continuous and the origin is a strict minimum, $\Omega_{\delta}$ is an open set and$\operatorname{diam}\left(\Omega_{\delta}\right) \downarrow 0$ as $\delta \downarrow 0$. We introduce the notation \begin{align*} \langle g\rangle_{\delta}:=\frac{1}{\left|\Omega_{\delta}\right|} \int_{\Omega_{\delta}} g(t, x) d t d x \end{align*}

Why, heuristically, are we considering this average?

  • By definition of (h),

\begin{align}\label{eq:19} \left\langle\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right)\right\rangle_{\delta}=-\left\langle\left(\begin{array}{c} (h-\zeta)_{t} \\ (h-\zeta)_{x} \end{array}\right)\right\rangle_{\delta}+\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right)\right\rangle_{\delta} \end{align}

For $\delta$ sufficiently small we have $\Omega_{\delta} \subset \subset B_{1}$. Thus \begin{align*} \left\langle(h-\zeta)_{t}\right\rangle_{\delta}=\int_{\Omega_{\delta}}(h-\zeta)_{t}=\frac{1}{\left|\Omega_{\delta}\right|} \int_{\Omega_{\delta}}(h-\zeta)_{t}=\frac{1}{\left|\Omega_{\delta}\right|} \int_{B_{1}}(\min \{h-\zeta-\delta, 0\})_{t} \end{align*}

Since the function $\min \{h-\zeta-\delta, 0\}$ is continuous and identically zero on a neighborhood of $\partial B_{1}$, the right hand side above vanishes. The same argument applies to $\left\langle(h-\zeta)_{x}\right\rangle_{\delta}$. Hence, from \eqref{eq:19} we get \begin{align*} \left\langle\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right)\right\rangle_{\delta}=\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right)\right\rangle_{\delta} \end{align*}

Why is it true that For $\delta$ sufficiently small we have $\Omega_{\delta} \subset \subset B_{1}$? Why exactly, by continuity, $\min \{h-\zeta-\delta, 0\}$ is identically zero on a neighborhood of $\partial B_{1}$?

On the other hand we have \begin{align*} \left\langle\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right) \cdot\left(\begin{array}{c} \eta(u) \\ q(u) \end{array}\right)\right\rangle_{\delta} \geq\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right) \cdot\left(\begin{array}{c} \eta(u) \\ q(u) \end{array}\right)\right\rangle_{\delta} \end{align*}

It seems that, up to this point, we stated an entropy inequality for the "averaged" equation. Is this a good intuition of what happens? Why does it help?

  • Using the Proposition 3.2 (already discussed in the answer below), we obtain \begin{align*} \begin{aligned} &\frac{c^{2}}{3}\left\langle\left(u-\langle u\rangle_{\delta}\right)^{4}\right\rangle_{\delta} \\ &\quad \leq\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right) \cdot\left(\begin{array}{c} \eta(u) \\ q(u) \end{array}\right)\right\rangle_{\delta}-\left\langle\left(\begin{array}{c} -f(u) \\ u \end{array}\right)\right\rangle_{\delta} \cdot\left\langle\left(\begin{array}{c} \eta(u) \\ q(u) \end{array}\right)\right\rangle_{\delta} \\ &\leq \quad C \sup _{\Omega_{\delta}}\left|\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right)-\left\langle\left(\begin{array}{c} \zeta_{t} \\ \zeta_{x} \end{array}\right)\right\rangle_{\delta}\right| \end{aligned} \end{align*}

And, since $\zeta$ is smooth, we deduce

\begin{align*} \lim _{\delta \downarrow 0}\left\langle\left(u-\langle u\rangle_{\delta}\right)^{4}\right\rangle_{\delta}=0 \end{align*} and, assuming $f$ Lipschitz, \begin{align*} \left|\langle f(u)\rangle_{\delta}-f\left(\langle u\rangle_{\delta}\right)\right| \leq C\left\langle\left|u-\langle u\rangle_{\delta}\right|\right\rangle_{\delta} \leq C\left\langle\left(u-\langle u\rangle_{\delta}\right)^{4}\right\rangle_{\delta}^{1 / 4} \longrightarrow 0 \text{ as \delta \to 0} \end{align*}

Are we here only applying Lebesgue's dominated convergence theorem? The final conclusion seems to be that the average of the flux is equal to the flux of the average of $u$; where does the intuition for this come from and lead to ?

  • Plugging this back into the equations above gives \begin{align*} \lim _{\delta \downarrow 0}\left|-\left\langle\zeta_{t}\right\rangle_{\delta}-f\left(\left\langle\zeta_{x}\right\rangle_{\delta}\right)\right|=0 \end{align*} and, since $\zeta$ is smooth, \begin{align*} -\zeta_{t}(0,0)-f\left(\zeta_{x}(0,0)\right)=0 \end{align*} which is the conclusion.

What is the final message of this proof? Can it be streamlined/simplified a bit?


1 Answer 1


My impression is the following. In order that $u$ be an entropy solution, it must not have increasing shocks (because $f$ is strictly convex). Thus if $u$ does not experience at $(t,x)$ a decreasing discontinuity, the PDE should be satisfied pointwise there. At the level of $h$, the assumption amounts to saying that there exists a smooth $\zeta$ such that $h-\zeta$ has a strict local minimum at $(t,x)$. Therefore the goal is to prove that at every such point, the PDE is satisfied pointwisely in a reasonable way. This reasonnable way is the viscosity sense $\zeta_t+f(\zeta_x)=0$.

Remark. The case of points $(t,x)$ belonging to shock curves is not covered by the super-solution condition, because there does not exist such a smooth $\zeta$. Instead one may have a local maximum at $(t,x)$, and this is covered by the sub-solution condition, which is covered in paragraph 4.1. The paper hides the difficulty of this case by refereing to the paper by Crandall & Lions.

Edit. Here is an alternate proof of Prop 3.2.b, perhaps more illuminating. If $s\mapsto m(s)$ is a numerical function, define $[m](s,t):=m(t)-m(s)$. Then on the one hand $$B(f,\eta)=\frac12\langle\mu\otimes\mu,[{\rm id}][q]-[f][\eta]\rangle,$$ where $\mu\otimes\mu$ acts on functions of two variables. On the other hand $$([{\rm id}][q]-[f][\eta])(s,t)=\frac12\int_s^t\int_s^t(f'(b)-f'(a))(\eta'(t)-\eta'(a))da db$$ is non-negative because both $f$ and $\eta$ are convex. Since $\mu\otimes\mu$ is non-negative, $B(f,\eta)\ge0$.

  • $\begingroup$ Thanks! The new proof of Prop. 3.2 was indeed very illuminating for me. I've edited the question to outline my questions (and lack of intuition) about the rest of the steps in the proof of the main theorems. If you have some time, could you take a look? Thanks again $\endgroup$
    – user139844
    Oct 12, 2022 at 15:29
  • $\begingroup$ Hello. Did you have any occasion to look into the further questions? $\endgroup$
    – user139844
    Oct 24, 2022 at 7:43

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