# Viscous approximation of Eikonal equation

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?

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.

• 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 $H(x,\nabla u)=0$?
– user140746
Jul 7, 2022 at 22:26
• 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.
– Jeff
Jul 8, 2022 at 0:29
• The eikonal equations satisfies the structure condition I mentioned.
– Jeff
Jul 8, 2022 at 0:30