Obstacle Problem: $C^{1,\alpha}$-Regularity of Lipschitz FB with continuous right hand side.


Posted online: 2018-12-11 12:45:15Z by Henrik Shahgholian43

Cite as: P-181211.1

  • Analysis of PDEs
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Problem's Description

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.

  1. Book Regularity of free boundaries in obstacle-type problems

    pp. x+221, year of publication: 2012

  2. Article A theorem of real analysis and its application to free boundary problems

    Communications on Pure and Applied Mathematics 38, 499-502, 1985fulltext

  3. Article A New Boundary Harnack Principle (equations with right hand side)

  4. Article Sharp results for the regularity and stability of the free boundary in the obstacle problem

    Indiana University Mathematics Journal 50, 1077-1112, 2001fulltext

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  • Created at: 2018-12-11 12:45:15Z