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I am recently interested in the following special case of Eikonal equation: Let $U$ be an open subset of $\mathbb{R}^{n}$ and consider the equation

$$ \begin{cases} \left\| Du(x) \right\| =1, & \text{on $U$}; \\ u(x)=0, & \text{on $\mathbb{R}^{n}\setminus U$}. \end{cases} $$ My question is: Must $u$ be a distance function? (or do we need more assumption to conclude this?)

Any help or further referenece would be highly appreciated.

PS: I am not so familar with PDE and it's not my homework :)

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  • $\begingroup$ Well, since $u=0$ on the boundary, the gradient at the boundary has to be normal to the boundary. And it has to have magnitude one. So one can imagine that in a tiny strip around the boundary it has to be the distance function. Then one can solve the eikonal equation in the domain minus this strip, and use the same argument to march the solution into the interior. This is how the problem is solved numerically -- it is called the fast marching method. Of course singularities will form (even if $U$ has a smooth boundary), but this is a way of thinking about the problem. $\endgroup$ – felipeh Dec 2 '17 at 7:49
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Yes, if $U$ is bounded, then the unique viscosity solution of this Eikonal equation is exactly the distance function to $\mathbb{R}^n\setminus U$ (actually you just need the boundary condition $u=0$ on $\partial U$). If $U$ is unbounded, you need to additionally assume $\lim_{|x|\to \infty} u(x) = \infty$. This is proved in many places; for example the book by Bardi and Capuzzo-Dolcetta on Optimal Control and Viscosity Solutions is good start. You can also find a detailed study in the notes below:

http://www-users.math.umn.edu/~jwcalder/222BS16/viscosity_solutions.pdf

See Sections 1.1, 1.2, and Exercise 6.3.

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