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\title{A short proof of the $C^{1,1}$ regularity for the eikonal equation}
\alttitle{Une preuve simple de la r\'egularit\'e $C^{1,1}$ pour l'\'equation eikonale}

\author{\firstname{Radu} \lastname{Ignat}}
\address{Institut de Math\'ematiques de Toulouse, UMR 5219, Universit\'e de Toulouse, CNRS, UPS IMT, 31062 Toulouse Cedex 9, France}
\email{radu.ignat@math.univ-toulouse.fr}
\thanks{The author is partially supported by the ANR projects ANR-21-CE40-0004 and ANR-22-CE40-0006-01}
\CDRGrant[ANR]{ANR-21-CE40-0004}
\CDRGrant[ANR]{ANR-22-CE40-0006-01}

\dedicatory{In the memory of my PhD advisor Ha\"im Brezis}

\begin{abstract}
We give a short and self-contained proof of the interior $C^{1,1}$ regularity of solutions $\varphi\colon\Omega \to \mathbb{R}$ to the eikonal equation $\lvert\nabla\varphi\rvert=1$ in an open set $\Omega\subset \mathbb{R}^{N}$ in dimension $N\geq 1$ under the assumption that $\varphi$ is pointwise differentiable in $\Omega$.
\end{abstract}

\begin{altabstract}
Nous pr\'esentons une preuve courte et auto-contenue de la r\'egularit\'e int\'erieure $C^{1,1}$ des solutions $\varphi\colon\Omega \to \mathbb{R}$ de l'\'equation eikonale $\lvert\nabla\varphi\rvert=1$ dans un ouvert $\Omega\subset \mathbb{R}^{N}$ en toute dimension $N\geq 1$ sous l'hypoth\`ese que $\varphi$ est diff\'erentiable en tout point de $\Omega$.
\end{altabstract}

\keywords{\kwd{Eikonal equation} \kwd{characteristics} \kwd{regularity}}
\altkeywords{\kwd{\'Equation eikonale} \kwd{caract\'eristiques} \kwd{r\'egularit\'e}}

\editornote{Article submitted by invitation}
\alteditornote{Article soumis sur invitation}

\subjclass{35F21, 35B65}

\dateposted{2025-07-21}
\begin{document}

\maketitle

\section{Introduction}

Ha\"im Brezis always liked short and new proofs of known results. The aim of this note is to give such a short and self-contained proof of the following result known in the theory of Hamilton--Jacobi equations.
\begin{theo}\label{thm1}
Let $\Omega \subset \mathbb{R}^{N}$ be an open set in dimension $N\geq 1$ and $\varphi \colon \Omega \to \mathbb{R}$ be a pointwise differentiable solution to the eikonal equation $\abs{\nabla \varphi}=1$ in $\Omega$. Then $\nabla \varphi$ is locally Lipschitz in $\Omega$.
\end{theo}

The usual (standard) proof of this result is based on the following steps (see e.g.\ Lions~\cite{Lions}, Cannarsa--Sinestrari~\cite{CS}): first, one checks that $\f$ (and $-\f$) is a viscosity solution to the eikonal equation (see~\cite[Definition~5.2.1]{CS}); second, one proves that $\f$ is both semiconcave and semiconvex with linear modulus (see~\cite[Theorem~5.3.7]{CS}); third, one proves that $\f$ is $C^1$ (see~\cite[Theorem~3.3.7]{CS}) and finally, that $\f$ is locally $C^{1,1}$ in $\Omega$ (see~\cite[Corollary~3.3.8]{CS}).

Our approach is based on the geometry of characteristics associated to the eikonal equation. More precisely, if $x_{0} \in \Omega$, we say that $X\coloneqq X_{x_0}$ is a characteristic of a solution $\f$ passing through $x_0$ in some time interval $t \in[-T, T]$ if
\begin{equation}\label{cara}
\begin{cases}
\dot{X}(t)=\nabla \f\parens[\big]{X(t)} 	& \text{for $t\in [-T,T]$},\\
X(0)=x_{0}.
\end{cases}
\end{equation}
Then the beautiful proof of Caffarelli--Crandall~\cite[Lemma~2.2]{CC} shows in a short and self-contained manner that every point $x_0\in \Omega$ has a characteristic $X_{x_0}$ that is a straight line along which $\nabla \f$ is constant and $\f$ is affine. Finally, we give a geometric argument on the structure of characteristics yielding the locally Lipschitz regularity of $\nabla \f$ in $\Omega$.

The regularity result in Theorem~\ref{thm1} is optimal: such solution $\f$ of the eikonal equation is not $C^2$ in general (see e.g.~\cite[Proposition~1]{Ign}). We mention that a more general regularizing effect (i.e., $\nabla \f$ is locally Lipschitz away from vortex point singularities) is proved under a weaker assumption $\nabla \f\in W^{1/p,p}$ for $p\in [1,3]$, see~\cite{Ign, DI}. Similar results are obtained in the context of the Aviles--Giga model which can be seen as a regularization of the eikonal equation (see~\cite{JOP, GL}).

\section{Proof of the main result}

The first step is to show that each point $x_0\in \Omega$ has a characteristic $X\coloneqq X_{x_0}$ that is a straight line in direction $\nabla \f(x_0)$. Moreover, $\nabla \f$ is constant while $\f$ is affine along this characteristic. This fact yields $\f\in C^1(\Omega)$. In order to have a self-contained proof of Theorem~\ref{thm1}, we repeat here the very nice argument of Caffarelli--Crandall~\cite[Lemma~2.2]{CC} based on a maximum type principle for the eikonal equation.

\begin{lemm}\label{lem1}
Let\/ $\Omega \subset \mathbb{R}^{N}$ be an open set and $\varphi \colon \Omega \to \mathbb{R}$ be a pointwise differentiable solution of the eikonal equation $\abs{\nabla \varphi}=1$ in $\Omega$. Then for every $x_0\in \Omega$, $X(t)=x_0+t\nabla \f(x_0)$ is a characteristic of~\eqref{cara} and
\[
\nabla \varphi\parens[\big]{X(t)}=\nabla \varphi\parens{x_{0}}, \quad \f\parens[\big]{X(t)}=\f(x_0)+t, \quad \forall t \in[-T, T],
\]
for some $T>0$. As a consequence, $\f\in C^1(\Omega)$.
\end{lemm}

\begin{proof}
This proof follows the lines of~\cite[Lemma~2.2]{CC}. Let $R>0$ be such that $\bar{B}_R(x_0)\subset \Omega$ and consider
\[
M_r=\max_{\bar{B}_r(x_0)} \f, \quad m_r=\min_{\bar{B}_r(x_0)} \f, \quad \forall r\in [0,R].
\]

\begin{enonce}{Claim}\label{ballext}
$M_r=\f(x_0)+r$ and $m_r=\f(x_0)-r$ for every $r\in [0,R]$.
\end{enonce}

\begin{proof}
For $r\in [0,R]$, we pick some maximum point $x_r^+\in \bar{B}_r(x_0)$ such that $\f(x_r^+)=M_r$. First, we show that $r\in [0,R]\mapsto M_r$ is a nondecreasing $1$-Lipschitz function. Indeed, for $R\geq r>\tilde r$, as $\abs{x_r^+-x_0}\leq r$, we can find a vector $e\in \R^N$ such that $\abs{e}\leq r-\tilde r$ and $\abs{x_r^++e-x_0}\leq \tilde r$, i.e., $x_r^++e\in \bar{B}_{\tilde r}(x_0)$; this yields
\[
0\leq M_r-M_{\tilde r}\leq \f(x_r^+)-\f(x_r^++e)\leq \abs{e}\leq r-\tilde r
\]
because $\f$ is $1$-Lipschitz. Second, we prove that $\frac{\dd M_r}{\dd r}=1$ a.e.\ in $(0,R)$ because for $r\in (0,R)$ and for small $h>0$, as $x_r^++ h\nabla \f(x_r^+)\in \bar{B}_{r+h}(x_0)$, we have
\[
\liminf_{h\to 0}\frac{M_{r+ h}-M_r}{h}\geq \liminf_{h\to 0} \frac{\f\parens[\big]{x_r^+ + h\nabla \f(x_r^+)}-\f(x_r^+)}{h}= \abs[\big]{\nabla \f(x_r^+)}^2=1.
\]
As $M_0=\f(x_0)$, we conclude $M_r=\f(x_0)+r$. Up to changing $\f$ in $-\f$, one also gets $m_r=\f(x_0)-r$ for $r\in [0,R]$.
\end{proof}

To conclude the proof of Lemma~\ref{lem1}, pick some minimum point $x_R^-\in \bar{B}_R(x_0)$ such that $\f(x_R^-)=m_R=\f(x_0)-R$ (by Claim~\ref{ballext}). As $\f$ is $1$-Lipschitz, we have, again by Claim~\ref{ballext}:
\[
2R=\f(x_R^+)-\f(x_R^-)\leq \abs{x_R^+-x_R^-}\leq 2R,
\]
which means that $[x_R^+, x_R^-]$ is a diameter in $ \bar{B}_R(x_0)$. Note that $x_R^+$ (resp.\ $x_R^-$) is the unique maximum (resp.\ minimum) of $\f$ in $\bar{B}_R(x_0)$ because if $\tilde x_R^+$ is another maximum, then it has to be antipodal to~$x_R^-$, that is, $\tilde x_R^+=x_R^+$ (the same for the uniqueness of $x_R^-$). In particular, $e_*=\frac{x_R^+-x_0}R\in \Ss^{N-1}$. Define $g \colon [-R,R]\to \R$ by $g(r)=\f(x_0+re_*)-\f(x_0)$. Then $g$ is $1$-Lipschitz and $g(\pm R)=\f(x_R^\pm)-\f(x_0)=\pm R$ (by Claim~\ref{ballext}). So $g(r)=r$ for every $r\in (-R,R)$ yielding $1=g'(r)=e_*\cdot \nabla \f(x_0+re_*)$ for every $r$. Thus, $\nabla \f(x_0+re_*)=e_*$ for every $r\in [-R,R]$, in particular, $e_*=\nabla \f(x_0)$, i.e., $X(r)=x_0+r\nabla \f(x_0)$ is a characteristic of~\eqref{cara} and
\[
\nabla \varphi\parens[\big]{X(r)}=\nabla \varphi\parens{x_{0}}, \quad \f\parens[\big]{x_0+r\nabla \f(x_0)}=\f(x_0)+r, \quad \forall r \in[-R, R].
\]
In particular, the (unique) maximum and minimum of $\f$ in
$\bar{B}_R(x_0)$ are achieved at the points $x_R^\pm=x_0\pm R\nabla \varphi(x_{0})$.

It remains to prove that $\nabla \f$ is continuous in $\Omega$. Indeed, let $x_n\to x_0$ in $\Omega$ and $\bar{B}_R(x_n)\subset \Omega$ for large $n$. Up to a subsequence, we may assume that $\nabla \f(x_n)\to e\in \Ss^{N-1}$. By above, we know that $\f\parens[\big]{x_n+R\nabla \f(x_n)}=\f(x_n)+R$. Passing to the limit, we obtain $\f(x_0+Re)=\f(x_0)+R$, meaning that $x_0+Re$ is the maximum of $\f$ in $\bar{B}_R(x_0)$. By uniqueness of the maximum point $x_R^+$, we conclude that $e=\nabla \f(x_0)$. The uniqueness of the limit $e$ for such subsequences yield the convergence of the whole sequence $\parens[\big]{\nabla \f(x_n)}_n$ to $\nabla \f(x_0)$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm1}] Let $B$ be a ball, $\bar{B}\subset \Omega$ and we consider $d \in \parens[\big]{0, \frac{\dist(B, \partial \Omega)}{5}}$. We will prove the following.
\begin{enonce}{Claim}\label{cl2}
There exists a universal constant $C>0$ such that
\[
\abs[\big]{\nabla \varphi(x)-\nabla \varphi(y)} \leqslant \frac{C}{d}\abs{x-y}, \quad \forall x, y \in B \ \text{with} \ \abs{x-y}<\frac d{10}.
\]
\end{enonce}

\begin{proof}
Let $X_x$ and $X_y$ be the characteristics passing through $x$ and $y$ constructed in Lemma~\ref{lem1} (that are lines in direction $\nabla \f(x)$ and $\nabla \f(y)$). If $X_x$ and $X_y$ coincide inside $B$ (in particular, $\nabla \f(x)=\pm \nabla \f(y)$ by Lemma~\ref{lem1}), then Lemma~\ref{lem1} implies that $\nabla\f(x)=\nabla \f(y)$ and the claim is trivial in that case. Otherwise, $\nabla \f(x)\neq \pm \nabla \f(y)$ and $X_x$ and $X_y$ cannot intersect inside $\Omega$ (as $\nabla \f$ is continuous in $\Omega$ by Lemma~\ref{lem1}). Let $\abs{x-y}<\frac d{10}$ and $x'$ be the projection of $y$ on $X_x$. Clearly,
\[
\ell=\abs{x'-y}\leq \abs{x-y}<\frac{d}{10},
\]
$\dist(x', \partial \Omega)>5d-\ell$ and $\nabla \f(x')=\nabla \f(x)$. Up to changing $\f$ in $-\f$, we may assume that $\f(x')\leq \f(y)$ and up to an additive constant, we also may assume that $0=\f(x')\leq \f(y)=a$. As $\abs{\nabla \f}=1$ on the segment $[x'y]$, we deduce that $0\leq a=\f(y)-\f(x')\leq \abs{x'-y}=\ell$.
Let $z$ be the point on the characteristic $X_y$ reached at time $t=d$ in direction $\nabla \f(y)$, i.e., $\abs{z-y}=d$, $\nabla \f(y)=\frac{z-y}d$ and $\f(z)=a+d$; in particular, $z\in \Omega$. Let $w$ be the point on the characteristic $X_{x'}$ reached at time $t=-d$ in direction $\nabla \f(x')$, i.e., $\abs{x'-w}=d$, $\nabla \f(x')=\frac{x'-w}d$ and $\f(w)=-d$; in particular, $w \in \Omega$. We deduce that
\[
\abs{z-w}=\int_{[zw]} \abs{\nabla \f}\,\dd\h^1\geq \abs*{ \int_{[zw]} \nabla \f \cdot \frac{z-w}{\abs{z-w}}\,\dd\h^1}=\abs[\big]{\f(z)-\f(w)}=2d+a\geq 2d.
\]
By Pythagoras's formula in the triangle $x'yw$ we have $\abs{y-w}^2=\ell^2+d^2$. Denoting $\alpha$ the angle between $\overrightarrow{wy}$ and $\nabla \f(y)$, the cosine formula in the triangle $wyz$ yields
\[
\begin{split}
-\cos \alpha
	& = \frac{\abs{y-w}^2+\abs{y-z}^2-\abs{z-w}^2}{2\abs{y-z}\cdot \abs{y-w}}
\\	& = \frac{2d^2+\ell^2-\abs{z-w}^2}{2d\sqrt{d^2+\ell^2}}
\\	& \leq -\frac{2d^2-\ell^2}{2d\sqrt{d^2+\ell^2}}
\\	& = -1+O\parens[\Bigg]{\frac{\ell^2}{d^2}}.
\end{split}
\]

\begin{figure}
\includegraphics[width=0.5\textwidth]{figure-eiko.pdf}
\caption{Geometry of characteristics $X_x$ (containing $x'$) and $X_y$.}
\end{figure}

As $\alpha\in \parens[\big]{0, \frac\pi2}$, it yields $\sin^2 \alpha=1-\cos^2 \alpha\leq 2(1-\cos \alpha)=O\parens[\big]{\frac{\ell^2}{d^2}}$. So, $\sin \alpha=O\parens[\big]{\frac{\ell}{d}}$. Denoting $\beta\in \parens[\big]{0, \frac\pi2}$ the angle between $\overrightarrow{wy}$ and $\nabla \f(x)$, we compute in the triangle $x'yw$:
\[
\sin \beta\leq \tan \beta=\frac{\abs{y-x'}}{\abs{x'-w}}=\frac{\ell}d.
\]
In particular, $0\leq \alpha+ \beta\leq \frac\pi2$ if $\ell<\frac{d}{10}$. Finally, denoting $\gamma$ the angle between $\nabla \f(x)$ and $\nabla \f(y)$, the triangle inequality yields
\[
\sin \gamma\leq \sin (\alpha+\beta)\leq \sin \alpha+\sin \beta=O\parens[\Bigg]{\frac{\ell}{d}}.
\]
We conclude
\[
\abs[\big]{\nabla \varphi(x)-\nabla \varphi(y)}=2\sin \frac\gamma2=O(\sin \gamma)=O\parens[\Bigg]{\frac{\ell}{d}}\leq \frac{C}{d}\abs{x-y}
\]
for some universal $C>0$.
\end{proof}
The conclusion of Theorem~\ref{thm1} follows.
\end{proof}

\section*{Acknowledgments}

The author thanks Xavier Lamy for very useful discussions.

\section*{Declaration of interests}

The author does not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and has declared no affiliations other than their research organizations.

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