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Let $\Omega$ be an open, bounded, connected subset of $\mathbb R^n$ with smooth boundary. Does there always exist an almost everywhere solution $u \in W^{1, \infty}$ to the following system of PDE?

$$|Du| = 1$$ $$\lim_{\delta \to 0_+} \text{sup}_{y, z \in B_d (x)} \,|Du(y) - Du(z)| = 0$$ $$u = 0 \text{ on } \partial \Omega.$$

Note: We say $u \in W^{1, \infty}$ is an almost everywhere solution if the first two equations above are satisfied for Lebesgue almost every $x \in \mathbb R^n$.

In other words, is there a Lipschitz solution to the Eikonal equation with almost everywhere continuous derivative?

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    $\begingroup$ Not a real answer: unless I'm mistaken there always exists a (unique) viscosity solution. Such solutions are well-known to be semi concave (or semi convex, depending on your convention for visosity solutions). And such functions are in turn a.e. twice differentiable (Alexandrov's theorem). Maybe this is enough to conclude? I'm not an expert, though... $\endgroup$
    – leo monsaingeon
    Commented Mar 11 at 11:11
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    $\begingroup$ Another important piece is the well-known fact that iviscosity solutions satisfy the equation in the pointwise sense at any point of differentiability. $\endgroup$
    – leo monsaingeon
    Commented Mar 11 at 11:12

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If you take $u$ to be the distance from the boundary of $\Omega$ the two properties you look for are statisfied at every point of differentiability for $u$, indeed in any such point $x$ the gradient of $u$ is given by $$\frac{x-\pi(x)}{|x-\pi(x)|}$$ where $\pi(x)$ is the (unique) point realising the distance. Uniqueness follows from the assumption about differentiability and it holds a.e.

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  • $\begingroup$ Thanks for the answer! Is it obvious that $x - \pi(x)$ is continuous at any point of differentiability? Having trouble showing this formally. $\endgroup$
    – Nate River
    Commented May 27 at 20:46
  • $\begingroup$ It follows from uniqueness, if $x_k\to x$ and $\pi(x_k)\to y$, then it is easy to see that $y$ realizes the distance and so it has to coincide with $\pi(x)$. $\endgroup$
    – Guido De Philippis
    Commented May 28 at 5:23
  • $\begingroup$ Ah uniqueness would do it… $\endgroup$
    – Nate River
    Commented May 28 at 7:43

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