Consider the Eikonal equation \begin{align*} \begin{cases}\left|D u\right|^{2}=1 & \text { on } \Omega \\ u \equiv 0 & \text { on } \partial \Omega\end{cases} \end{align*} and the viscous regularization \begin{align*} \begin{cases}\varepsilon \Delta u_{\varepsilon}+\left|D u_{\varepsilon}\right|^{2}=1 & \text { on } \Omega \\ u_{\varepsilon} \equiv 0 & \text { on } \partial \Omega\end{cases} \end{align*} It is well-known that $u_\varepsilon$ converges uniformly to the unique viscosity solution $u$ of the Eikonal equation. Does the error estimate $$\|u_\varepsilon - u\|_{\infty} \lesssim \sqrt{\varepsilon}$$ also holds?


1 Answer 1


Yes, this is a standard result. You can see sections 5.2 and 5.3 in these notes: https://www-users.cse.umn.edu/~jwcalder/viscosity_solutions.pdf

In fact, the rate is $O(\varepsilon)$ in one direction, when the boundary is $C^2$, since in this case $u$ is semiconcave.

  • $\begingroup$ In your notes, a lot of heavy lifting is one by the fact that the equation is of the form $u+H(x,\nabla u) = 0$, i.e. there's a zero-order term appearing with the right sign. Is it actually possible to prove the same result for $H(x,\nabla u)=0$? $\endgroup$
    – user140746
    Jul 7, 2022 at 22:26
  • $\begingroup$ Yes, have a look at Theorem 9.17 in my notes. This is for monotone finite difference schemes, but the proof is very similar to that of viscous approximations and this is for $H(x,\nabla u)=0$. You need some additional structure conditions on $H$ in this case to allow perturbation to strict sub/super solutions. $\endgroup$
    – Jeff
    Jul 8, 2022 at 0:29
  • $\begingroup$ The eikonal equations satisfies the structure condition I mentioned. $\endgroup$
    – Jeff
    Jul 8, 2022 at 0:30

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