Open

Posted online: 2018-12-11 12:45:15Z by Henrik Shahgholian56

Cite as: P-181211.1

Let $v$ be a solution to the obstacle problem $$ \Delta v = h\chi_{\{v > 0\}}, \qquad v \geq 0 \qquad \hbox{in } B_1. $$ We assume $h \geq c_0 >0 $ is Lipschitz, and a Dirichlet data on $\partial B_1$ has been prescribed.

For $h$ constant, I. Athanasopoulos and L.A. Caffarelli [2], using the standard Boundary Harnack Principle, could deduce $C^{1,\alpha}$-regularity of the free boundary for the obstacle problem.

Using a recent Boundary Harnack Principle [3] the remarkably simple and novice proof of [2] can be adapted to prove similar statement when $h$ is Lipschitz.

**Here is how it works:**

Suppose that the free boundary is a Lipschitz graph (with small Lipschitz norm) in $B_1(0) $ and in direction $e_1$, with the origin being a free boundary point. For a large positive constant $C$, set $H(x) := v_{e_1} - C v $. Then one can show that $H$ satisfies $$ H > 0 , \qquad \Delta H = h_{e_1} - C h \leq 0 \qquad \hbox{in } \{v >0\} \cap B_{1/2}. $$

Apply the BHP with r.h.s. [3] to $H=v_{e_1} -Cv,$ and $ v_e,$ where $e \perp e_1$, and $\gamma=0$. This implies \begin{equation}\tag{1} \frac{ v_e }{ v_{e_1} - Cv } = \frac{v_e}{H} \ \in C^\alpha ( B_{1/4}). \end{equation}

Next fix a level surface $v=l$, and denote this surface by $x_1=G(x')$. The level surface is smooth since $v_{e_1} > 0$ there. Differentiating both side of $$v( G(x') , x') = l$$ gives $$ G_e = \frac{- v_e }{ v_{e_1} } . $$

We want to show $G_e$ is $C^\alpha$ for all directions $e \in \mathbb R^{n-1}$.

Rephrasing equation (1) and inserting this gives us $$ \frac{ -v_e }{ v_{e_1} - Cv } = \frac{ \frac{- v_e}{ v_{e_1} } }{ 1 - \frac{ Cv}{ v_{e_1} }} = \frac{G_e }{ 1 - \frac{ Cv}{ v_{e_1} } } , $$ is $C^\alpha$, independent of $l$. Since $v=l$ and $v_{e_1} \approx \sqrt{l}$ we have that $$ \frac{- v_e }{ v_{e_1} - Cv } = \frac{G_e }{ 1 - \frac{ Cv}{ v_{e_1} } } = \frac{G_e }{ 1 - \frac{ Cl}{ \sqrt l } } \to G_e, \qquad \hbox{as } l \to 0. $$ Hence $G_e$ is $C^\alpha$.

**Question:** Does continuous RHS $h$, and FB Lipschitz (small Lip. norm) imply FB is $C^{1,\alpha}$? (*Prove or disprove it.*)

**Remark: ** Observe that this does not violate Ivan Blank's result [4] since we assume FB is initially Lipschitz.

A strategy to prove this may work as follows: Take $h^\epsilon$ be an Lipschitz approximation of $h$. Suppose we can prove that the approximate problem also has Lipschitz FB (small Lip. norm). Then one deduces as above that FB is $C^{1,\alpha}$. The $C^{1,\alpha}$-norm however blows-up as $\epsilon$ tends to zero. Not withstanding this it should be possible to prove that the $C^{1,\alpha}$ actually is uniform on (exactly) the approximate FB. This would then mean that we can allow $\epsilon$ tend to zero and keep the $C^{1,\alpha}$-norm intact.

I believe if the conjecture is true, the strategy above should work.

No solutions added yet

Created at: 2018-12-11 12:45:15Z

No remarks yet