On the regularity of weak solutions to Burgers’ equation with finite entropy production

Abstract

Bounded weak solutions of Burgers’ equation $\partial _tu+\partial _x(u^2/2)=0$ that are not entropy solutions need in general not be BV. Nevertheless it is known that solutions with finite entropy productions have a BV-like structure: a rectifiable jump set of dimension one can be identified, outside which u has vanishing mean oscillation at all points. But it is not known whether all points outside this jump set are Lebesgue points, as they would be for BV solutions. In the present article we show that the set of non-Lebesgue points of u has Hausdorff dimension at most one. In contrast with the aforementioned structure result, we need only one particular entropy production to be a finite Radon measure, namely $\mu =\partial _t (u^2/2)+\partial _x(u^3/3)$. We prove Hölder regularity at points where $\mu$ has finite $(1+\alpha )$-dimensional upper density for some $\alpha >0$. The proof is inspired by a result of De Lellis, Westdickenberg and the second author : if $\mu _+$ has vanishing 1-dimensional upper density, then u is an entropy solution. We obtain a quantitative version of this statement: if $\mu _+$ is small then u is close in $L^1$ to an entropy solution.

Appendix: Lipschitz estimate for the viscosity solution

Let $h\in W^{1,\infty }(Q)$ solve (6) almost everywhere. The viscosity solution $\bar{h}\in W^{1,\infty }(Q)$ of (6) with $\bar{h}=h$ on $\partial _0 Q$ is given [25, §11] by the Hopf-Lax formula

$$\begin{aligned} \bar{h}(t,x)=\inf \left\{ h(s,y)+\frac{(x-y)^2}{2(t-s)}:(s,y)\in \partial _0 Q,\; s<t \right\} . \end{aligned}$$

Note that for $(t,x)\in Q$ the infimum is attained. Let $L:={\left\| \partial _x h\right\| }_{L^\infty (Q)}$, so that the initial data $h(0,\cdot )$ has Lipschitz constant $\le L$ and the boundary data $h(\cdot ,0)$ and $h(\cdot ,1)$ have Lipschitz constants $\le L^2/2$.

Lemma 10

It holds ${\left| \partial _x\bar{h}\right| }\le L$ a.e.

Proof

Let $(t_0,x_0)\in Q$ and denote by $(s_0,y_0)$ a point at which the infimum defining $\bar{h}(t_0,x_0)$ is attained. Then for any small x it holds

$$\begin{aligned} \bar{h}(t_0,x_0+x)-\bar{h}(t_0,x_0)&=\bar{h}(t_0,x_0+x)-h(s_0,y_0)-\frac{(x_0-y_0)^2}{2(t_0-s_0)}\\&\le \frac{(x_0+x-y_0)^2}{2(t_0-s_0)} -\frac{(x_0-y_0)^2}{2(t_0-s_0)}\\&= \frac{x_0-y_0}{t_0-s_0} x + \frac{1}{2(t_0-s_0)}x^2, \end{aligned}$$

so that ${\left| \partial _ x\bar{h}(t_0,x_0)\right| }\le {\left| x_0-y_0\right| }/(t_0-s_0)$ and to prove (7) it suffices to show that the infimum defining $\bar{h}(t_0,x_0)$ is attained at some $(s_0,y_0)$ with

$$\begin{aligned} \frac{{\left| x_0-y_0\right| }}{t_0-s_0}\le L. \end{aligned}$$

(28)

We show that for any $(s,y)\in \partial _0 Q\cap \lbrace s<t_0\rbrace$ with

$$\begin{aligned} \frac{{\left| x_0-y\right| }}{t_0-s} > L, \end{aligned}$$

(29)

there exists $(\tilde{s} ,\tilde{y})\in \partial _0 Q\cap \lbrace s<t_0\rbrace$ satisfying

$$\begin{aligned}&\frac{{\left| x_0-\tilde{y}\right| }}{t_0-\tilde{s}} < \frac{{\left| x_0-y\right| }}{t_0-s}\end{aligned}$$

(30)

$$\begin{aligned} \text {and}\quad&h(s_0,\tilde{y})+\frac{(x_0-\tilde{y})^2}{2(t_0-\tilde{s})} \le h(s_0, y)+\frac{(x_0-y)^2}{2(t_0-s)}, \end{aligned}$$

(31)

which proves (28).

There are two cases to consider, depending on which part of the parabolic boundary (s, y) belongs to.

Case 1 $(s,y)\in \lbrace 0 \rbrace \times [0,1]$. We look for $(\tilde{s},\tilde{y})$ defined through $\tilde{s}=0$ and

$$\begin{aligned} \frac{x_0-\tilde{y}}{t_0}=(1-\varepsilon )D,\quad D:=\frac{x_0-y}{t_0}, \end{aligned}$$

for some small $\epsilon >0$, so that (30) is satisfied. On the other hand since $h(0,\cdot )$ has Lipschitz constant $\le L$, to show (31) it suffices to establish

$$\begin{aligned} L{\left| \tilde{y}-y\right| } \le \frac{(x_0-y)^2-(x_0-\tilde{y})^2}{2t_0} \quad \Longleftrightarrow \quad \frac{L}{D} \le (1- \frac{1}{2} \varepsilon ), \end{aligned}$$

which is satisfied for small enough $\varepsilon$ since (29) amounts to ${\left| D\right| }>L$.

Case 2 $(s,y)\in (0,1)\times \lbrace 0,1 \rbrace$. We assume $y=0$, the case $y=1$ being similar. We look for $(\tilde{s},\tilde{y})$ defined through $\tilde{y}=0$ and

$$\begin{aligned} \frac{x_0}{t_0-\tilde{s}}=(1-\varepsilon ) D,\quad D:=\frac{x_0}{t_0-s}, \end{aligned}$$

for some small $\epsilon >0$, so that (30) is satisfied. On the other hand since $h(1,\cdot )$ has Lipschitz constant $\le L^2/2$, to show (31) it suffices to establish

$$\begin{aligned} \frac{L^2}{2}{\left| s-\tilde{s}\right| }\le \frac{x_0^2}{2}\left( \frac{1}{t_0-s}-\frac{1}{t_0-\tilde{s}}\right) \quad \Longleftrightarrow \quad \frac{L^2}{D^2}\le 1-\varepsilon , \end{aligned}$$

which is satisfied for small enough $\varepsilon$ since (29) amounts to ${\left| D\right| }>L$. $\square$