Regularity of the free boundary in the biharmonic obstacle problem

Abstract

In this article we use a flatness improvement argument to study the regularity of the free boundary for the biharmonic obstacle problem with zero obstacle. Assuming that the solution is almost one-dimensional, and that the non-coincidence set is an non-tangentially accessible domain, we derive the $C^{1,\alpha }$-regularity of the free boundary in a small ball centred at the origin. From the $C^{1,\alpha }$-regularity of the free boundary we conclude that the solution to the biharmonic obstacle problem is locally $C^{3,\alpha }$ up to the free boundary, and therefore $C^{2,1}$. In the end we study an example, showing that in general $C^{2,\frac{1}{2}}$ is the best regularity that a solution may achieve in dimension $n \ge 2$.

1 Introduction

Let $\Omega \subset {\mathbb {R}}^n$ be a given domain, and $\varphi \in C^2 ({\overline{\Omega }}), \varphi \le 0 \text { on } \partial \Omega$ be a given function, called an obstacle. Then the minimiser to the following functional

$$\begin{aligned} J[u]= \int _\Omega \left( \Delta u(x) \right) ^2 dx, \end{aligned}$$

(1.1)

over all functions $u \in W_0^{2,2}(\Omega ),$ such that $u\ge \varphi$, is called the solution to the biharmonic obstacle problem with obstacle $\varphi$. The solution satisfies the following variational inequalities

$$\begin{aligned} \Delta ^2 u\ge 0, ~ u \ge \varphi , ~ \Delta ^2u \cdot (u-\varphi )=0. \end{aligned}$$

(1.2)

It has been shown in [1] that the solution $u \in W^{3,2}_{loc}(\Omega )$, and in [2] that $\Delta u \in L^\infty _{loc}(\Omega )$, and moreover $u\in W^{2,\infty }_{loc}(\Omega )$, see also [3]. Furthermore, in the paper [3], the authors show that in dimension $n=2$ the solution $u\in C^2(\Omega )$ and that the free boundary $\Gamma _u :=\partial \{ u= \varphi \}$ lies on a $C^1$-curve in a neighbourhood of the points $x_0\in \Gamma _u$, such that $\Delta u(x_0)> \Delta \varphi (x_0)$.

The setting of our problem is slightly different from the one in [1, 3] and [2]. We consider a zero-obstacle problem with general nonzero boundary conditions. Let $\Omega$ be a bounded domain in ${\mathbb {R}}^n$ with smooth boundary. We study the minimiser of the functional (1.1) over the admissible set

$$\begin{aligned} {\mathscr {A}}:=\left\{ u \in W^{2,2}(\Omega ),~ u \ge 0, ~u=g>0, \frac{\partial u}{\partial \nu }=f \text { on } \partial \Omega \right\} . \end{aligned}$$

The minimiser u exists, it is unique, and is called the solution to the biharmonic obstacle problem. We will denote the free boundary by $\Gamma _u := \partial \Omega _u \cap \Omega$, where $\Omega _u: = \{ u > 0\}$.

There are several important questions regarding the biharmonic obstacle problem that remain open. For example, the optimal regularity of the solution, the characterisation of blow-ups at free boundary points, etc. In this article we focus on the regularity of the free boundary for an n-dimensional biharmonic obstacle problem, assuming that the solution is close to the one-dimensional solution $\frac{1}{6}(x_n)_+^3$. In [4], using flatness improvement argument, the author, John Andersson, shows that the free boundary in the p-harmonic obstacle problem is a $C^{1,\alpha }$ graph in a neighbourhood of the points where the solution is almost one-dimensional. We apply the same technique in order to study the regularity of the free boundary in the biharmonic obstacle problem.

In Sect. 2 we study the basic properties of the solution, adapted to our setting. The existence, uniqueness as well as $W^{3,2}_{loc} \cap C^{1,\alpha }_{loc}$-regularity of the solution are known. For the sake of completeness statements and some sketched proofs are included, together with references.

In Sect. 3 we introduce the class ${\mathscr {B}}_\kappa ^\varrho (\varepsilon )$ of solutions to the biharmonic obstacle problem, which are close to the one-dimensional solution $\frac{1}{6} (x_n)_+^3$. Following the linearisation argument by John Andersson (see for instance [4]), we show that if $\varepsilon$ is small enough, then there exists a rescaling $u_s(x)= \frac{u(sx)}{s^3}$, such that

$$\begin{aligned} ||\nabla 'u_s ||_{W^{2,2}(B_2)} \le \gamma ||\nabla 'u ||_{W^{2,2}(B_2)} \le \gamma \varepsilon \end{aligned}$$

in a normalised coordinate system, where $\nabla '_\eta := \nabla -\eta (\eta \cdot \nabla ), \nabla ':= \nabla '_{e_n}$, and $\gamma <1$ is a constant. Repeating the argument for the rescaled solutions, $u_{s^k}$, we show that there exists a unit vector $\eta _0 \in {\mathbb {R}}^n$, such that

$$\begin{aligned} \frac{ ||\nabla '_{\eta _0} u_{s^k} ||_{W^{2,2}(B_2)}}{\Arrowvert D^3 u_{s^k}||_{L^2(B_1)}} \le \gamma ^k \varepsilon \end{aligned}$$

(1.3)

for some $0<s<\gamma <1$. Then the $C^{1,\alpha }$-regularity of the free boundary in a neighbourhood of the origin follows via a standard iteration argument.

From the $C^{1,\alpha }$-regularity of the free boundary it follows that $\Delta u \in C^{1,\alpha }$ up to the free boundary, furthermore, u is $C^{3,\alpha }$ up to the free boundary. Thus a solution $u\in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$ is locally $C^{2,1}$, which is the best possible regularity. We provide a two-dimensional counterexample to the $C^{2,1}$-regularity, showing that without our flatness assumptions there exists a solution that is $C^{2,\frac{1}{2}}$ but is not $C^{2,\alpha }$ for $\alpha >\frac{1}{2}$. Hence $C^{2,\frac{1}{2}}$ is the best regularity that a solution may achieve in dimension $n \ge 2$.

2 The obstacle problem for the biharmonic operator

This section can be viewed as a background text, where we present the known regularity theory for the biharmonic obstacle problem, with adaptations to our problem. The material in this section is known for the biharmonic obstacle problem with a general obstacle, and zero boundary conditions. Later on in Sect. 3 we will need quantitive estimates for $||u||_{W_{loc}^{3,2} }$ and for $||u ||_{C_{loc}^{1, \alpha }}$, hence we decided to include these estimates in the manuscript, with references to the original papers.

2.1 The known regularity theory for the solution

The existence and uniqueness of the minimiser of functional (1.1) over the admissible set ${\mathscr {A}}$ follows by standard arguments in the calculus of variations, see for instance [5]. We give a summary of the known regularity theory for the solution. The following result is due to Frehse, see [1].

Theorem 2.1

(J. Frehse) Let u be the solution to the biharmonic obstacle problem in $\Omega$, where $\Omega \subset {\mathbb {R}}^n$ is an open bounded set. Then

$$\begin{aligned} \Arrowvert u ||_{W^{3,2}(V)} \le C \Arrowvert u ||_{W^{2,2}(\Omega )}, \end{aligned}$$

for any given $V \subset \subset \Omega$, where $C>0$ is a universal constant, depending only on the space dimension and sets $V, \Omega$, but not on the solution u.

Now let us discuss further properties of the solution, referring to the paper by Caffarelli and Friedman [3].

Lemma 2.2

The solution to the biharmonic obstacle problem satisfies the following equation in the distribution sense

$$\begin{aligned} \Delta ^2u=\mu _u, \end{aligned}$$

(2.1)

where $\mu _u$ is a positive measure on $\Omega$.

Proof

For any nonnegative test function $\eta \in C_0^{\infty }(\Omega )$, the function $u+\varepsilon \eta$ is obviously admissible for any $\varepsilon >0$. Hence $J[u+\varepsilon \eta ]\ge J[u]$, consequently

$$\begin{aligned} \int {\varepsilon ^2(\Delta \eta )^2+2\varepsilon \Delta u \Delta \eta } \ge 0, \end{aligned}$$

and after dividing by $\varepsilon$ and letting $\varepsilon$ go to zero, we obtain

$$\begin{aligned} \int {\Delta u \Delta \eta } \ge 0, \end{aligned}$$

for all $\eta \in C_0^{\infty }(\Omega )$, $\eta \ge 0$, so $\Delta ^2 u \ge 0$ in the sense of distributions.

Let us study the following linear functional $\Lambda (\eta ):=\int _\Omega \Delta u \Delta \eta$, $\eta \in C_0^{\infty }(\Omega )$. By the Riesz theorem, $\Lambda$ is a positive measure, let us denote it by $\mu := \mu _u$. Then $\Delta ^2 u = \mu _u$ in the sense that

$$\begin{aligned} \int _\Omega \Delta u \Delta \eta = \int _\Omega \eta d\mu _u. \end{aligned}$$

for every $\eta \in C_0^{\infty }(\Omega )$. $\square$

Corollary 2.3

There exists an upper semicontinuous function $\omega$ in $\Omega$, such that $\omega = \Delta u \text { a.e. in } \Omega$.

Proof

For any fixed $x_0 \in \Omega$, the function

$$\begin{aligned} \omega _r(x_0):= \fint _{B_r(x_0)} \Delta u(x) dx \end{aligned}$$

is decreasing in $r>0$, since $\Delta u$ is subharmonic by Lemma 2.2. Define $\omega (x):= \lim _{r\rightarrow 0} \omega _r(x)$, then $\omega$ is an upper semicontinuous function. On the other hand $\omega _r(x) \rightarrow \Delta u (x)$ as $r \rightarrow 0$ a.e., hence $\omega = \Delta u$ a.e. in $\Omega$. $\square$

In [2] Frehse proved that $\Delta u \in L^\infty _{loc}$, later on L. Caffarelli and A. Friedman gave another proof to the same result, see Theorem 3.1 in [3]. Below we follow the proof in the latter paper in order to obtain a quantitative estimate on $\Arrowvert \Delta u ||_{L^\infty }$ in our setting. The next lemma is a restatement of the corresponding result in [3], Theorem 2.2.

Lemma 2.4

Let $\Omega \subset {\mathbb {R}}^n$ be a bounded open set with a smooth boundary, and let u be a solution to the biharmonic obstacle problem with zero obstacle. Denote by S the support of the measure $\mu _u = \Delta ^2 u$ in $\Omega$, then

$$\begin{aligned} \omega (x_0) \ge 0,~~ \text { for every }x_0 \in S . \end{aligned}$$

(2.2)

Proof

The detailed proof of Lemma 2.4 can be found in the original paper [3] and in the book [6, pp. 92–94], so we will provide only a sketch, showing the main ideas.

Extend u to a function in $W^{2,2}_{loc}({\mathbb {R}}^n)$, and denote by $u_\varepsilon$ the $\varepsilon$-mollifier of u. Let $x_0 \in \Omega$, assume that there exists a ball $B_r(x_0)$, such that $u_\varepsilon \ge \alpha >0$ in $B_r(x_0)$. Let $\eta \in C_0^{\infty }(B_r(x_0))$, $\eta \ge 0$ and $\eta = 1$ in $B_{r/2}(x_0)$. Then for any $\zeta \in C_0^\infty ( B_{r/2}(x_0))$ and $0< t < \frac{\alpha }{2 ||\zeta ||_\infty }$ the function

$$\begin{aligned} v= \eta u_\varepsilon +(1-\eta )u \pm t \zeta \end{aligned}$$

is nonnegative and it satisfies the same boundary conditions as u. Hence

$$\begin{aligned} \int (\Delta u)^2 \le \int \left( \Delta ( \eta u_\varepsilon +(1-\eta )u \pm t \zeta ) \right) ^2, \end{aligned}$$

after passing to the limit in the last inequality as $\varepsilon \rightarrow 0$, we obtain

$$\begin{aligned} \int (\Delta u)^2 \le \int ( \Delta u \pm t \Delta \zeta ) ^2, \end{aligned}$$

Therefore

$$\begin{aligned} \int {\Delta u \Delta \zeta }=0 , \end{aligned}$$

for all $\zeta \in C_0^{\infty }(B_{r/2}(x_0))$, hence $\Delta ^2 u=0$ in $B_{r/2}(x_0)$ and $x_0 \notin S$. It follows that if $x_0 \in S$, then there exists $x_m \in \Omega$, $x_m \rightarrow x_0$, and $\varepsilon _m \rightarrow 0$, such that

$$\begin{aligned} u_{\varepsilon _m}(x_m) \rightarrow 0, \text { as } m\rightarrow \infty . \end{aligned}$$

Then by Green’s formula,

$$\begin{aligned} u_{\varepsilon _m}(x_m)= \fint _{\partial B_\rho (x_m)} u_{\varepsilon _m} d {\mathcal {H}}^{n-1} -\int _{B_\rho (x_m)} \Delta u_{\varepsilon _m}(y) V(x_m-y)dy, \end{aligned}$$

where $\rho < dist(x_0, \partial \Omega )$ and $- V(z)$ is Green’s function for Laplacian in the ball $B_\rho (0)$. Hence

$$\begin{aligned} \liminf _{m\rightarrow \infty } \int _{B_\rho (x_m)} \Delta u_{\varepsilon _m}(y) V(x_m-y)dy \ge 0, \end{aligned}$$

Then it follows from the convergence of the mollifiers and the upper semicontinuity of $\omega$, that $\omega (x_0)\ge 0$, for any $x_0\in S$. $\square$

Knowing that $\Delta u$ is a subharmonic function, and $\omega \ge 0$ on the support of $\Delta ^2 u$, we can show that $\Delta u$ is locally bounded (Theorem 3.1 in [3]).

Theorem 2.5

Let u be the solution to the biharmonic obstacle problem with zero obstacle in $\Omega$, $B_1 \subset \subset \Omega$. Then

$$\begin{aligned} \Arrowvert \Delta u ||_{L^\infty (B_{1/3})} \le C \Arrowvert u ||_{W^{2,2}(\Omega )}, \end{aligned}$$

(2.3)

where the constant $C>0$ depends on the space dimension n and on $dist(B_1, \partial \Omega )$.

Proof

The detailed proof of the theorem can be found in the original paper [3], Theorem 3.1, and in the book [6, pp. 94–97]. Here we will only provide a sketch of the proof.

Let $\omega$ be the upper semicontinuous equivalent of $\Delta u$ and $x_0 \in B_{1/2}$, then

$$\begin{aligned} \omega (x_0) \le \fint _{B_{1/2}(x_0)} \Delta u(x) dx, \end{aligned}$$

since $\omega$ is a subharmonic function. Applying Hölder’s inequality, we obtain

$$\begin{aligned} \omega (x_0) \le |B_{1/2}|^{-\frac{1}{2}} ||\Delta u ||_{L^2(B_1)}. \end{aligned}$$

(2.4)

It remains to show that $\Delta u$ is bounded from below in $B_{1/2}$. Let $\zeta \in C_0^\infty (B_1)$, $\zeta =1$ in $B_{2/3}$ and $0 \le \zeta \le 1$ elsewhere. Referring to [6, p. 96], the following formula holds for any $x \in B_{1/2}$

$$\begin{aligned} \omega (x)= - \int _{B_{1/2}} V(x-y) d \mu - \int _{B_1{\setminus } B_{1/2}} \zeta (y) V(x-y) \Delta ^2 u dy + \delta (x), \end{aligned}$$

(2.5)

where V is Green’s function for the unit ball $B_1$, and $\delta$ is a bounded function,

$$\begin{aligned} \Arrowvert \delta ||_{L^\infty ( B_{1/2})} \le C_n \Arrowvert \Delta u ||_{L^2(B_1)}. \end{aligned}$$

(2.6)

Denote

$$\begin{aligned} {\tilde{V}}(x):= \int _{B_{1/2}} V(x-y) d\mu (y), \end{aligned}$$

then ${\tilde{V}}$ is a superharmonic function in ${\mathbb {R}}^n$, and the measure $\upsilon :=\Delta {\tilde{V}}$ is supported on $S_0:= B_{1/2}\cap S$, moreover according to Lemma 2.4, (2.2)

$$\begin{aligned} {\tilde{V}} (x) \le -\omega (x)+\delta (x)\le \delta (x) ~\text { on } S_0. \end{aligned}$$

Taking into account that ${\tilde{V}}(+\infty ) <\infty$, the authors in [3] apply Evans maximum principle, [7] to the superharmonic function ${\tilde{V}}-{\tilde{V}}(+\infty )$, and conclude that

$$\begin{aligned} {\tilde{V}}(x)\le \Arrowvert \delta ||_{L^\infty ( B_{1/2})}~ \text { in }~{\mathbb {R}}^n. \end{aligned}$$

(2.7)

It follows from equation (2.5) that

$$\begin{aligned} \omega (x) \ge - \Arrowvert \delta ||_{L^\infty ( B_{1/2})}-c_n \mu _u(B_1)+\delta (x), \end{aligned}$$

(2.8)

for any $x\in B_{1/3}$.

Let $\eta \in C_0^\infty (\Omega )$ be a nonnegative function, such that $\eta =1$ in $B_1$ and $0 \le \eta \le 1$ in $\Omega$. Then

$$\begin{aligned} \mu _u(B_1)\le \int _\Omega \eta d\mu _u =\int _\Omega \Delta u \Delta \eta \le \Arrowvert \Delta u ||_{L^2(\Omega )} \Arrowvert \Delta \eta ||_{L^2(\Omega )}, \end{aligned}$$

and $\eta$ can be chosen such that $\Arrowvert \Delta \eta ||_{L^2(\Omega )} \le C(dist(B_1, \partial \Omega ))$. Hence

$$\begin{aligned} \mu _u(B_1)\le C \Arrowvert \Delta u ||_{L^2(\Omega )}, \end{aligned}$$

(2.9)

where the constant $C>0$ depends on the space dimension and on $dist(B_1, \partial \Omega )$.

Combining the inequalities (2.4) and (2.8) together with (2.9), (2.6), we obtain (2.3). $\square$

Corollary 2.6

Let u be the solution to the biharmonic obstacle problem in $\Omega$. Then $u\in C^{1,\alpha }_{loc}$, for any $0<\alpha <1$, and

$$\begin{aligned} \Arrowvert u ||_{C^{1,\alpha }(K)}\le C\Arrowvert u ||_{W^{2,2}(\Omega )}, \end{aligned}$$

(2.10)

where the constant C depends on the space dimension and $dist(K, \partial \Omega )$.

Proof

It follows from Theorem 2.5 via a standard covering argument, that

$$\begin{aligned} \Arrowvert \Delta u ||_{L^\infty (K)}\le C \Arrowvert u ||_{W^{2,2}(\Omega )}. \end{aligned}$$

Then inequality (2.10) follows from the Calderón–Zygmund inequality and the Sobolev embedding theorem. $\square$

According to Corollary 2.6, u is a continuous function in $\Omega$, and therefore the noncoincidence set $\Omega _u:=\{u>0\}$ is an open subset of $\Omega$. Define the free boundary by

$$\begin{aligned} \Gamma _u = \partial \Omega _u \cap \Omega . \end{aligned}$$

(2.11)

By standard arguments in the theory of free boundary problems, we can see that u is a biharmonic function in the noncoincidence set $\Omega _u$. It follows from our discussion that $\mu _u= \Delta ^2 u$ is a positive measure supported on $\Gamma _u$, and the variational inequalities (1.2) holds with $\varphi \equiv 0$.

3 Regularity of the free boundary

In this section we investigate the regularity of the free boundary $\Gamma _u$, under the assumption that the solution to the biharmonic obstacle problem is close to the one-dimensional solution $\frac{1}{6}(x_n)^3_+$.

3.1 One-dimensional solutions

First we find the explicit solution to the biharmonic obstacle problem in the interval $(0,1) \subset {\mathbb {R}}$.

Example 3.1

The minimiser $u_0$ of the functional

$$\begin{aligned} J[u]=\int _0^1 (u''(x))^ 2 dx, \end{aligned}$$

(3.1)

over nonnegative functions $u \in W^{2,2}(0,1)$, with boundary conditions $u(0)=1,u'(0)=\lambda <-3$ and $u(1)=0,u'(1)=0$, is a piecewise 3-rd order polynomial,

$$\begin{aligned} u_0(x)= \frac{\lambda ^3}{3^3} \left( x+ \frac{3}{\lambda } \right) _-^3, \quad x \in (0,1), \end{aligned}$$

(3.2)

hence $u_0 \in C^{2,1}(0,1)$.

Proof

Let $u_0$ be the minimiser to the given biharmonic obstacle problem. If $0<x_0 <1$, and $u_0(x_0)>0$, then $\int u_0'' \eta '' = 0$, for all infinitely differentiable functions $\eta$ compactly supported in a small ball centered at $x_0$. Hence the minimiser $u_0$ has a fourth order derivative, $u_0^{(4)}(x) =0$ if $x \in \{ u_0>0 \}$. Therefore $u_0$ is a piecewise polynomial of degree less than or equal to three. Denote by $\gamma \in (0,1]$ the first point where the graph of $u_0$ hits the x-axes. Our aim is find the explicit value of $\gamma$. Then we can also compute the minimiser $u_0$.

Observe that $u_0(\gamma )=0$, and $u'_0(\gamma )=0$, since $u'_0$ is an absolutely continuous function in (0, 1). Taking into account the boundary conditions at the points 0 and $\gamma$, we can write $u_0(x)= ax^ 3+bx^2+\lambda x+1$ in $(0,\gamma )$, where

$$\begin{aligned} a=\frac{\lambda \gamma +2}{\gamma ^3}, ~~ b=-\frac{2\lambda \gamma +3}{\gamma ^2}. \end{aligned}$$

We see that the point $\gamma$ is a zero of second order for the third order polynomial $u_0$, and $u_0 \ge 0$ in $(0, \gamma ]$. That means the third zero is not on the open interval $(0, \gamma )$, hence $\gamma \le -\frac{3}{\lambda }$.

Consider the function

$$\begin{aligned} F(\gamma ):=\int _0^{\gamma }(u''(x))^ 2 dx, \end{aligned}$$

then $F(\gamma )=\frac{4}{\gamma ^3}(\lambda ^2\gamma ^2+3\lambda \gamma +3)$. Hence $F'(\gamma )=-\frac{4}{\gamma ^4}(\lambda \gamma +3)^2$, showing that the function F is decreasing, so it achieves minimum at the point $\gamma = -\frac{3}{\lambda }$. Therefore we may conclude that

$$\begin{aligned} u_0(x)= \frac{\lambda ^3}{3^3} \left( x+ \frac{3}{\lambda } \right) _-^3, \quad x \in (0,1), \end{aligned}$$

(3.3)

and $\gamma = -\frac{3}{\lambda }$ is a free boundary point. Observe that $u''(\gamma )=0$, and $u''$ is a continuous function, but $u'''$ has a jump discontinuity at the free boundary point $\gamma = -\frac{3}{\lambda }$. $\square$

The example above characterises one-dimensional solutions. It also tells us that one-dimensional solutions are $C^{2,1}$, and in general are not $C^3$.

3.2 The class ${\mathscr {B}}_\kappa ^\varrho (\varepsilon )$ of solutions to the biharmonic obstacle problem

Without loss of generality, we assume that $0\in \Gamma _u$, and study the regularity of the free boundary, when $u \approx \frac{1}{6}(x_n)^3_+$.

Let us start by recalling the definition of non-tangentially accessible domains [8].

Definition 3.2

A bounded domain $D \subset {\mathbb {R}}^n$ is called non-tangentially accessible (abbreviated NTA) when there exist constants M, $r_0$ and a function $l :{\mathbb {R}}_+ \mapsto {\mathbb {N}}$ such that

• 1. D satisfies the corkscrew condition; that is for any $x_0 \in \partial D$ and any $r<r_0$, there exists $P=P(r,x_0) \in D$ such that

  $$\begin{aligned} M^{-1}r< |P-x_0|<r ~ \text { and } ~dist(P, \partial D) > M^{-1}r. \end{aligned}$$

  (3.4)

• 2. $D^c:= {\mathbb {R}}^n {\setminus } D$ satisfies the corkscrew condition.

• 3. Harnack chain condition; if $\epsilon >0$ and $P_1, P_2 \in D$, $dist(P_i,\partial D)>\epsilon$, and $|P_1-P_2 |<C \epsilon$, then there exists a Harnack chain from $P_1$ to $P_2$ whose length l depends on C, but not on $\epsilon$, $l=l(C)$. A Harnack chain from $P_1$ to $P_2$ is a chain of balls $B_{r_k}(x^k)$, $k=1,\ldots ,l$ such that $P_1 \in B_{r_1}(x^1)$, $P_2 \in B_{r_l}(x^l)$, $B_{r_k}(x^k) \cap B_{r_{k+1}}(x^{k+1}) \ne \emptyset$, and

  $$\begin{aligned} M r_k> dist( B_{r_k}(x^k), \partial D)>M^{-1} r_k. \end{aligned}$$

  (3.5)

Let us define rigorously, what we mean by $u \approx \frac{1}{6}(x_n)^3_+$.

Definition 3.3

Let $u \ge 0$ be the solution to the biharmonic obstacle problem in a domain $\Omega$, $B_2 \subset \subset \Omega$ and assume that $0 \in \Gamma _u$ is a free boundary point. We say that $u \in {\mathscr {B}}^{\varrho }_{\kappa }(\varepsilon )$, if the following assumptions are satisfied:

• 1. u is almost one dimensional, that is

  $$\begin{aligned} \Arrowvert \nabla ' u||_{W^{2,2}(B_2)} \le \varepsilon , \end{aligned}$$

  where $\nabla ':= \nabla -e_n \frac{\partial }{\partial x_n}$.

• 2. The set $\Omega _u :=\{u > 0\}$ is an NTA domain with constants $r_0=M^{-1}= \varrho$, and with a function l, indicating the length of a Harnack chain.

• 3. There exists $2>t>0$, such that $u=0$ in $B_2 \cap \{ x_n < -t \}$.

• 4. We have the following normalisation

  $$\begin{aligned} ||D^3 u ||_{L^2(B_1)}= \frac{1}{6} \left\| D^3 (x_n)_+^3 \right\| _{L^2(B_1)} =\frac{|B_1|^\frac{1}{2}}{2^\frac{1}{2}}:= \omega _n, \end{aligned}$$

  (3.6)

  and we also assume that

  $$\begin{aligned} ||D^3 u ||_{L^2(B_2)} < \kappa , \end{aligned}$$

  (3.7)

  where $\kappa > \frac{1}{6} \left\| D^3 (x_n)_+^3 \right\| _{L^2(B_2)}= 2^\frac{n}{2}\omega _n$.

In the notation of the class ${\mathscr {B}}_\kappa ^\varrho (\varepsilon )$ we did not include the length function l, since later it does not appear in our estimates. For the rest of this paper we will assume that we have a fixed length function l. Later on in Corollary 3.6 we will see that the precise value of the parameter t in assumption 3 is not very important, and therefore we also omit the parameter t in our notation. From now on $\kappa >2^\frac{n}{2}\omega _n$ and $1>\varrho >0$ are fixed parameters.

Evidently $\frac{1}{6} (x_n)_+^3 \in {\mathscr {B}}^{\varrho }_{\kappa }(\varepsilon )$, for any $\varepsilon >0$ and $\varrho >0$. We show that if $u\in {\mathscr {B}}_\kappa ^{\varrho }(\varepsilon )$, with $\varepsilon >0$ small, then $u \approx \frac{1}{6} (x_n)_+^3$ in $W^{3,2}(B_1)$. First we need to prove the following easy lemma.

Lemma 3.4

Let u be the solution to the biharmonic obstacle problem in $\Omega$. Take $K \subset \subset \Omega$, and a function $\zeta \in C^\infty _0(K)$, $\zeta \ge 0$, then

$$\begin{aligned} \int _\Omega \Delta u_{x_i} \Delta ( \zeta u_{x_i}) \le 0, \end{aligned}$$

(3.8)

for all $i=1,2,\ldots , n$.

Proof

Fix $1 \le i \le n$, denote $u_{i,h}(x):= u(x+he_i)$, where $0< |h |< dist(K,\partial \Omega )$, hence $u_{i,h}$ is defined in K. Let us observe that the function $u+ t \zeta (u_{i,h}- u )$ is well defined and nonnegative in $\Omega$ for any $0<t< \frac{1}{ ||\zeta ||_{L^\infty }}$, and it satisfies the same boundary conditions as u. Therefore

$$\begin{aligned} \int _\Omega \left( \Delta (u+ t \zeta (u_{i,h}- u )) \right) ^2 \ge \int _\Omega (\Delta u )^2, \end{aligned}$$

after dividing the last inequality by t, and taking the limit as $t \rightarrow 0$, we obtain

$$\begin{aligned} \int _K \Delta u \Delta (\zeta ( u_{i,h}-u )) \ge 0. \end{aligned}$$

(3.9)

Note that $u_{i,h}$ is the solution to the biharmonic obstacle problem in K, and $u_{i,h}+ t \zeta (u-u_{i,h})$ is an admissible function, hence

$$\begin{aligned} \int _K \left( \Delta (u_{i,h}+ t \zeta (u-u_{i,h} )) \right) ^2 \ge \int _K (\Delta u_{i,h })^2, \end{aligned}$$

after dividing the last inequality by t, and taking the limit as $t \rightarrow 0$, we obtain

$$\begin{aligned} \int _K \Delta u_{i,h} \Delta (\zeta ( u- u_{i,h} )) \ge 0. \end{aligned}$$

(3.10)

Inequalities (3.9) and (3.10) imply that

$$\begin{aligned} \int _K (\Delta u_{i, h}- \Delta u ) \Delta ( \zeta ( u_{i,h}-u)) \le 0, \end{aligned}$$

(3.11)

dividing the last inequality by $h^2$, and taking into account that $u\in W^{3,2}_{loc}$, we may pass to the limit as $|h |\rightarrow 0$ in (3.11), and conclude that

$$\begin{aligned} \int _K \Delta u_{x_i} \Delta (\zeta u_{x_i}) \le 0. \end{aligned}$$

$\square$

Lemma 3.5

There exists a modulus of continuity $\sigma =\sigma (\varepsilon )\ge 0$, such that

$$\begin{aligned} \left\| u(x)-\frac{1}{6} (x_n)_+^3\right\| _{W^{3,2}(B_1)}\le \sigma (\varepsilon ), \end{aligned}$$

(3.12)

for any $u\in {\mathscr {B}}^{\varrho }_\kappa (\varepsilon )$.

Proof

We argue by contradiction. Assume that there exists $\sigma _0>0$ and a sequence of solutions, $u^j \in {\mathscr {B}}_\kappa ^{\varrho }({\varepsilon _j})$, such that

$$\begin{aligned} \Arrowvert \nabla ' u^j ||_{W^{2,2}(B_2)} = \varepsilon _j \rightarrow 0, \end{aligned}$$

but

$$\begin{aligned} \left\| u^j(x)-\frac{1}{6} (x_n)^3_+ \right\| _{W^{3,2}(B_1)}> \sigma _0 >0. \end{aligned}$$

(3.13)

According to assumption 4 in Definition 3.3, $\Arrowvert D^3 u^j ||_{L^2(B_2)} < \kappa$ and according to Assumption 2 the functions $u^j$ are vanishing on an open subset of $B_2$. Therefore it follows from the Poincaré inequality that $\Arrowvert u^j ||_{W^{3,2}(B_2)} \le C(\varrho , n)\kappa$. Hence up to a subsequence $u^j \rightharpoonup u^0$ weakly in $W^{3,2}(B_2)$, $u^j \rightarrow u^0$ strongly in $W^{2,2}(B_2)$ and according to Corollary 2.6 $u^j \rightarrow u^0$ in $C^{1,\alpha }(B_{3/2})$. Hence

$$\begin{aligned} \Arrowvert \nabla ' u^0 ||_{W^{1,2}(B_2)}= \lim _{j\rightarrow \infty } \Arrowvert \nabla ' u^j ||_{W^{1,2}(B_2)}\le \lim _{j\rightarrow \infty }\varepsilon _j =0. \end{aligned}$$

This implies that $u^0$ is a one-dimensional solution (depending only on the variable $x_n$). Example 3.1 tells us that one-dimensional solutions in the interval $(-2,2)$ have the form

$$\begin{aligned} u^0(x_n) = c_1(x_n-a_1)^3_- + c_2(x_n -a_2)^3_+, \end{aligned}$$

where $c_1, c_2 \ge 0$ and $-2 \le a_1 \le a_2 \le 2$ are constants. According to assumption 3 in Definition 3.3, $u^0= c (x_n-a)^3_+$. In order to obtain a contradiction to Assumption (3.13), we need to show that $u^j \rightarrow u_0= \frac{1}{6}(x_n)_+^3$ in $W^{3,2}(B_1)$. The proof of the last statement can be done in two steps.

Step 1 We show that

$$\begin{aligned} u^j \rightarrow c(x_n-a)_+^3~\text { in } W^{3,2}(B_1). \end{aligned}$$

(3.14)

Denote $u^j_n:= \frac{\partial u^j}{\partial x_n} \in W^{2,2}(B_2)$, $j \in {\mathbb {N}}_0$, and let $\zeta \in C_0^\infty (B_\frac{3}{2})$ be a nonnegative function, such that $\zeta \equiv 1$ in $B_1$. According to Lemma 3.4,

$$\begin{aligned} 0\ge \int _{B_2} \Delta (\zeta u^j_n) \Delta u^j_n = \int _{B_2} u^j_n \Delta \zeta \Delta u^j_n+ \int _{B_2}\zeta ( \Delta u^j_n )^2+ 2 \int _{B_2} \nabla \zeta \nabla u^j_n \Delta u^j_n, \end{aligned}$$

and therefore

$$\begin{aligned} \begin{aligned} \limsup _{j\rightarrow \infty } \int _{B_2} \zeta (\Delta u^j_n )^2&\le - \lim _{j\rightarrow \infty } \int _{B_2} u^j_n \Delta \zeta \Delta u^j_n- 2 \lim _{j\rightarrow \infty } \int _{B_2} \nabla \zeta \nabla u^j_n \Delta u^j_n \\&= -\int _{B_2} u^0_n \Delta \zeta \Delta u^0_n - 2 \int _{B_2} \nabla \zeta \nabla u^0_n \Delta u^0_n =\int _{B_2} \zeta (\Delta u^0_n)^2, \end{aligned} \end{aligned}$$

(3.15)

where in the last step we used integration by parts.

On the other hand, since $\Delta u^j_n \rightharpoonup \Delta u^0_n$ weakly in $L^2(B_2)$, it follows that

$$\begin{aligned} \liminf _{j\rightarrow \infty } \int _{B_2} \zeta (\Delta u^j_n)^2 \ge \int _{B_2} \zeta (\Delta u^0_n)^2. \end{aligned}$$

(3.16)

Therefore, we may conclude from (3.15) and (3.16) that

$$\begin{aligned} \lim _{j\rightarrow \infty } \int _{B_2} \zeta (\Delta u^j_n)^2 = \int _{B_2} \zeta (\Delta u^0_n)^2. \end{aligned}$$

Hence we obtain

$$\begin{aligned} \frac{\partial \Delta u^j}{\partial x_n} \rightarrow \frac{\partial \Delta u^0}{\partial x_n} \text { in } ~ L^2(B_1). \end{aligned}$$

Similarly $\frac{\partial \Delta u^j}{\partial x_i} \rightarrow 0$ in $L^2(B_1)$, for $i=1,\ldots , n-1$. Knowing that

$$\begin{aligned} \Arrowvert \nabla \Delta u^j -\nabla \Delta u^0||_{L^2(B_1)} \rightarrow 0, ~\text { and }~ \Arrowvert u^j - u^0||_{W^{2,2}(B_2)} \rightarrow 0, \end{aligned}$$

we may apply the Calderón–Zygmund inequality, and conclude (3.14). Recalling that $\Arrowvert D^3 u^j ||_{L^2(B_1)}=\omega _n$, we see that

$$\begin{aligned} \Arrowvert D^3 u^0 ||_{L^2(B_1)}=\omega _n. \end{aligned}$$

(3.17)

Since $u^0 = c (x_n-a)_+^3\ge 0$, it follows that

$$\begin{aligned} \Arrowvert D^3 u^0 ||^2_{L^2(B_1)}= c^2 {\mathcal {L}}^n( B_1 \cap \{x_n> a\} ) > 0, \end{aligned}$$

hence

$$\begin{aligned} c >0~ \text { and } ~ a<1. \end{aligned}$$

(3.18)

Step 2 We show that $a=0$ and $c=\frac{1}{6}$. Taking into account that $u^j\rightarrow u^0$ in $C^{1,\alpha }$ and $u^j(0)=0$, we conclude that $u^0(0)= 0$, thus $a\ge 0$. Assume that $a>0$. Since $0 \in \Gamma _j$, and $\Omega _j$ is an NTA domain, there exists $P_j=P(r,0) \in \Omega _j$, for $0<r<\min ( \varrho , a/2)$ as in the corkscrew condition,

$$\begin{aligned} \varrho r<|P_j|<r~ \text {and} ~dist(P_j, \partial \Omega _j)> \varrho r. \end{aligned}$$

Therefore up to a subsequence $P_j \rightarrow P_0$, hence $~r \varrho \le |P_0|\le r$, $~ B_{r'}(P_0) \subset \Omega _j,$ for all j large enough, where $0<r' <r \varrho$ is a fixed number. Since we have chosen $r < a/2$, we may conclude that

$$\begin{aligned} B_{r'}(P_0) \subset \{ x_n < a\} \cap \Omega _j . \end{aligned}$$

Thus $\Delta u^j$ is a sequence of harmonic functions in the ball $B_{r'}(P_0)$, and therefore

$$\begin{aligned} \Delta u^j \rightarrow 0 \text { locally uniformly in} ~B_{r'}(P_0), \end{aligned}$$

(3.19)

according to (3.14).

Let $Q:= e_n$, then $u^0(Q)= c (1-a)^3 >0$, since $u^j \rightarrow u^0$ uniformly in $B_{ 3/2}$, we see that $u^j(Q) >0$ for large j, and $Q \in \Omega _j$. Therefore there exists a Harnack chain connecting $P_0$ with Q; $\{ B_{r_1}(x^1), B_{r_2}(x^2),\ldots , B_{r_l}(x^l) \} \subset \Omega _j$, whose length l does not depend on j. Denote by $K^j := \cup _i B_{r_i}(x^i) \subset \subset \Omega _j$, and let $V^j \subset \subset K^j \subset \subset \Omega _j$ where $V^j$ is a regular domain, such that $dist(K^j,\partial V^j )$ and $dist(V^j,\partial \Omega _j )$ depend only on r and $\varrho$.

Let $w^j_+$ be a harmonic function in $V^j$, with boundary conditions $w^j_+= (\Delta u^j)_+ \ge 0$ on $\partial V^j$, then $w^j_+ -\Delta u^j$ is a harmonic function in $V^j$, and $w^j_+ -\Delta u^j = (\Delta u^j)_- \ge 0$ on $\partial V^j$ , hence

$$\begin{aligned} 0 \le w^j_+-\Delta u^j \le \Arrowvert (\Delta u^j)_-||_{L^\infty (V^j)} \text { in }~ V^j. \end{aligned}$$

(3.20)

Let us observe that $\Delta u ^j \rightarrow \Delta u^0 =6c (x_n-a)_+$ implies that $\Arrowvert (\Delta u ^j )_-||_{L^2(B_2)} \rightarrow 0$. Since $(\Delta u^j)_-$ is a subharmonic function in $\Omega _j$, and $V^j \subset \subset \Omega _j$ it follows that

$$\begin{aligned} \Arrowvert (\Delta u^j)_-||_{L^\infty (V^j)} \le C(n,r, \varrho ) \Arrowvert (\Delta u ^j )_-||_{L^2(B_2)} \rightarrow 0. \end{aligned}$$

So $w^j_+$ is a nonnegative harmonic function in $V^j$, and by the Harnack inequality

$$\begin{aligned} C_H \inf _{ B_{r_l}(x^l)} w^j_+ \ge \sup _{ B_{r_l}(x^l)} w^j_+ \ge w^j_+(e_n) \ge \Delta u^j(e_n) \ge \frac{1}{2} \Delta u_0(e_n)= 3c(1-a), \end{aligned}$$

if j is large, where $C_H$ is the constant in Harnack’s inequality, it depends on $\varrho$ and r but not on j. Denote $C(a, c):= 3c(1-a)>0$ by (3.18). Applying the Harnack inequality again, we see that

$$\begin{aligned} C_H \inf _{ B_{r_{l-1}}(x^{l-1})} w^j_+ \ge \sup _{B_{r_{l-1}}(x^{l-1}) } w^j_+ \ge \inf _{ B_{r_l}(x^l)} w^j_+ > \frac{C(a, c)}{C_H}. \end{aligned}$$

Inductively, we obtain that

$$\begin{aligned} C_H \inf _{ B_{r_1}(x^1)} w^j_+ \ge \sup _{ B_{r_{1}}(x^{1})} w^j_+ > \frac{C(a, c)}{C_H^{l-1}}, \end{aligned}$$

(3.21)

where l does not depend on j. Hence $w^j_+(P_0) \ge \frac{C(a, c)}{C_H^{l}}$ for all j large, and according to (3.20),

$$\begin{aligned} \lim _{j\rightarrow \infty }\Delta u^j (P_0) \ge \frac{C(a, c)}{C_H^{l}}>0, \end{aligned}$$

the latter contradicts (3.19). Therefore we may conclude that $a=0$.

Recalling that $\Arrowvert D^3 u^0 ||_{L^2(B_1)}=\omega _n$, we see that $c = \frac{1}{6}$, but then we obtain $u^j \rightarrow \frac{1}{6}(x_n)_+^3$ in $W^{3,2}(B_1)$ which is a contradiction, since we assumed (3.13). $\square$

Lemma 3.5 has an important corollary, which will be very useful in our later discussion.

Corollary 3.6

Let u be the solution to the biharmonic obstacle problem, $u \in {\mathscr {B}}^\varrho _\kappa (\varepsilon )$. Then for any fixed $t>0$ we have that $u(x)=0$ in $B_2 \cap \{ x_n <-t \}$, provided $\varepsilon =\varepsilon (t)>0$ is small.

Proof

Once again we argue by contradiction. Assume that there exist $t_0 >0$ and a sequence of solutions $u_j \in {\mathscr {B}}^\varrho _\kappa ({\varepsilon _j})$, $\varepsilon _j \rightarrow 0$, such that $x^j \in B_2 \cap \Gamma _j$, and $x^j_n < - t_0$. For $0<r<\min (\varrho , {t_0}/{2})$ choose $P^j=P(r,x^j) \in \Omega _j$ as in the corkscrew condition,

$$\begin{aligned} r \varrho< |x^j-P^j|<r, ~~B_{r\varrho }(P^j) \subset \Omega _j. \end{aligned}$$

Upon passing to a subsequence, we may assume that $P^j \rightarrow P^0$. Fix $0<r' <r \varrho$, then for large j

$$\begin{aligned} B_{r'}(P^0) \subset \subset \Omega _j \cap \{ x_n < 0\}. \end{aligned}$$

Hence $\Delta u ^j$ is a sequence of harmonic functions in $B_{r'}(P^0)$. According to Lemma 3.5, $u^j \rightarrow \frac{1}{6}(x_n)_+^3$, and therefore $\Delta u^j \rightarrow 0$ in $B_{r'}(P^0)$, and $\Delta u^j(e_n)\rightarrow 1$. Since $\Omega _j$ is an NTA domain, there exists a Harnack chain connecting $P^0$ with $Q:= e_n \in \Omega _j$; $\{ B_{r_1}(x^1), B_{r_2}(x^2),\ldots , B_{r_k}(x^k) \} \subset \Omega _j$, whose length does not depend on j. Arguing as in the proof of Lemma 3.5, we will obtain a contradiction to $\Delta u_j \rightarrow 0$ in $B_{r'}(P^0)$. $\square$

3.3 Linearisation

Let $\{ u^j\}$ be a sequence of solutions in $\Omega \supset \supset B_2$, $u^j \in {\mathscr {B}}^{\varrho }_\kappa ({ \varepsilon _j})$, and assume that $\varepsilon _j \rightarrow 0$ as $j\rightarrow \infty$. It follows from Lemma 3.5, that up to a subsequence

$$\begin{aligned} u^j \rightarrow \frac{1}{6}(x_n)_+^3 ~ \text { in }~~ W^{2,2}(B_2) \cap C_{loc}^{1,\alpha }(B_2). \end{aligned}$$

(3.22)

Let us denote

$$\begin{aligned} \delta ^j_i:= \left\| \frac{\partial u^j}{\partial x_i} \right\| _{W^{2,2}(B_2)}. \end{aligned}$$

Without loss of generality we may assume that $\delta ^j_i> 0$, for all $j \in {\mathbb {N}}$. Indeed, if $\delta ^j_i =0$ for all $j \ge J_0$ large, then $u^j$ does not depend on the variable $x_i$, and the problem reduces to a lower dimensional case. Otherwise we may pass to a subsequence satisfying $\delta ^j_i > 0$ for all j.

Denote

$$\begin{aligned} v^j_i: =\frac{1}{\delta ^j_i}\frac{\partial u^j}{\partial x_i}, \quad \text { for } i=1,\ldots ,n-1, \end{aligned}$$

(3.23)

then $\Arrowvert v^j_i ||_{W^{2,2}(B_2)}=1$. Therefore up to a subsequence $v^j_i$ converges to a function $v^0_i$ weakly in $W^{2,2}(B_2)$ and strongly in $W^{1,2}(B_2)$. For the further discussion we need strong convergence $v^j_i \rightarrow v^0_i$ in $W^{2,2}$, at least locally.

Lemma 3.7

Assume that $\{u^j\}$ is a sequence of solutions in $\Omega \supset \supset B_2$, $u^j \in {\mathscr {B}}^\varrho _\kappa ({\varepsilon _j})$, $\varepsilon _j\rightarrow 0$. Let $v^j_i$ be the sequence given by (3.23), and assume that $v^j_i \rightharpoonup v^0_i$ weakly in $W^{2,2}(B_2)$, strongly in $W^{1,2}(B_2)$, for $i=1,\ldots ,n-1$, then

$$\begin{aligned} \Delta ^2 v^0_i=0 \text { in } B_2^+, ~~v^0_i \equiv 0 \text { in } B_2 {\setminus } B_2^+. \end{aligned}$$

(3.24)

Furthermore, for any $0<R<2$

$$\begin{aligned} \Arrowvert v^j_i -v^0_i ||_{W^{2,2}(B_{R})} \rightarrow 0. \end{aligned}$$

(3.25)

Proof

Denote by $\Omega _j:= \Omega _{u^j},~\Gamma _j:= \Gamma _{u^j}$. It follows from Corollary 3.6 that $v^0_i \equiv 0$ in $B_2 {\setminus } B_2^+$, hence $v^0_i=|\nabla v^0_i |=0$ on $\{x_n=0\} \cap B_2$ in the trace sense. Moreover, if $K \subset \subset B_2^+$ is an open subset, then $K \subset \Omega _j$ for large j by (3.22). Hence $\Delta ^2 v^j_i =0$ in K, and therefore $\Delta ^2 v^0_i =0$ in $B_2^+$, and (3.24) is proved.

Now let us proceed to the proof of the strong convergence. Let $\zeta \in C_0^\infty (B_{2})$ be a nonnegative function, such that $\zeta \equiv 1$ in $B_R$ and $0\le \zeta \le 1$ in $B_2$. It follows from (3.24) that

$$\begin{aligned} 0= \int _{B_{{2}}} \Delta v_i^0 \Delta (\zeta v^0_i) = \int _{B_{{2}}} v^0_i \Delta \zeta \Delta v^0_i+ \int _{B_{{2}}}\zeta ( \Delta v^0_i )^2+ 2 \int _{B_{{2}}} \nabla \zeta \nabla v^0_i \Delta v^0_i. \end{aligned}$$

(3.26)

According to Lemma 3.4

$$\begin{aligned} 0 \ge \int _{B_{{2}}} \Delta (\zeta v^j_i) \Delta v^j_i = \int _{B_{{2}}} v^j_i \Delta \zeta \Delta v^j_i+ \int _{B_{{2}}}\zeta ( \Delta v^j_i )^2+ 2 \int _{B_{{2}}} \nabla \zeta \nabla v^j_i \Delta v^j_i, \end{aligned}$$

(3.27)

and therefore

$$\begin{aligned} \begin{aligned} \limsup _{j \rightarrow \infty } \int _{B_{{2}}} \zeta (\Delta v^j_i )^2&\le -\lim _{j \rightarrow \infty }\int _{B_{{2}}} v^j_i \Delta \zeta \Delta v^j_i -2 \lim _{j \rightarrow \infty }\int _{B_{{2}}} \nabla \zeta \nabla v^j_i \Delta v^j_i \\&= -\int _{B_{{2}}} v^0_i \Delta \zeta \Delta v^0_i- 2 \int _{B_{{2}}} \nabla \zeta \nabla v^0_i \Delta v^0_i, \end{aligned} \end{aligned}$$

where we used that $v^j_i \rightarrow v^0_i \text { in } W^{1,2}(B_2)$ and $\Delta v^j_i \rightharpoonup \Delta v^0_i \text { in } L^2(B_2)$.

From the last inequality and (3.26) we may conclude that

$$\begin{aligned} \limsup _{j \rightarrow \infty } \int _{B_{{2}}} \zeta (\Delta v^j_i )^2 \le \int _{B_{{2}}} \zeta (\Delta v^0_i )^2. \end{aligned}$$

(3.28)

On the other hand

$$\begin{aligned} \liminf _{j \rightarrow \infty } \int _{B_{{2}}} \zeta (\Delta v^j_i )^2 \ge \int _{B_{{2}}} \zeta (\Delta v^0_i )^2 \end{aligned}$$

(3.29)

follows from the weak convergence $\Delta v^j_i \rightharpoonup \Delta v^0_i$ in $L^2(B_2)$, and we may conclude from (3.28) and (3.29) that

$$\begin{aligned} \lim _{j \rightarrow \infty } \int _{B_{{2}}} \zeta (\Delta v^j_i )^2 = \int _{B_{{2}}} \zeta (\Delta v^0_i )^2. \end{aligned}$$

Hence we obtain $\Arrowvert \Delta v^j_i - \Delta v^0_i ||_{L^{2}(B_R)} \rightarrow 0$, and therefore $v^j_i \rightarrow v^0_i$ in $W^{2,2}_{loc}(B_2)$ according to the Calderón–Zygmund inequality. $\square$

3.4 Properties of solutions in a normalised coordinate system

Let us define

$$\begin{aligned} u_{r,x_0}(x):=\frac{u(rx+x_0)}{r^3}, \text { for } x_0 \in \Gamma _u, ~x \in B_2, ~ r \in (0,1), \end{aligned}$$

(3.30)

and $u_r := u_{r, 0}$. We would like to know how fast $\Arrowvert \nabla ' u_r ||_{W^{2,2}(B_2)}$ decays with respect to $\Arrowvert \nabla ' u ||_{W^{2,2}(B_2)}$, for $r<1$. In particular, the inequality

$$\begin{aligned} \Arrowvert \nabla ' u_s ||_{W^{2,2}(B_2)} \le \tau \Arrowvert \nabla ' u||_{W^{2,2}(B_2)} , \end{aligned}$$

(3.31)

for some $0<s,\tau <1$ would provide good decay estimates for $\Arrowvert \nabla ' u_{s^ k }||_{W^{2,2}(B_2)}$, $k\in {\mathbb {N}}$.

We show that the inequality (3.31) holds in a special coordinate system depending on the solution u and parameter $s>0$. Then iterating the inequality (3.31) and the coordinate system we obtain the existence of the unit normal vector to the free boundary at the origin.

Let us observe that $\frac{1}{6}(\eta \cdot x)_+^3 \in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$ if $|\eta - e_n |\le C_n \varepsilon$, for some dimensional constant $C_n$.

Definition 3.8

Let u be the solution to the biharmonic obstacle problem. We say that the coordinate system is normalised with respect to u, if

$$\begin{aligned}&\inf _{\eta \in {\mathbb {R}}^n,|\eta |=1 } \left\| \nabla '_\eta \left( u(x)-\frac{1}{6}(\eta \cdot x)_{+}^3 \right) \right\| _{L^{2}(B_{2})} \\&\quad = \left\| \nabla '_{e_n} \left( u(x)- \frac{1}{6}( x_n)_{+}^3 \right) \right\| _{L^{2}(B_{2})}, \end{aligned}$$

where $\nabla '_\eta := \nabla - (\eta \cdot \nabla )\eta$, and $\nabla ':=\nabla '_{e_n}$.

A minimiser $\eta$ always exists for a function $u\in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$, and since $\nabla '_{-\eta }= \nabla '_{\eta }$, $-\eta$ is also a minimiser, thus we always choose a minimiser satisfying the condition $e_n\cdot \eta \ge 0$. A normalised coordinate system always exists by choosing $\eta = e_n$ in the new coordinate system.

Lemma 3.9

Let u be the solution to the biharmonic obstacle problem in a normalised coordinate system with respect to u. Then

$$\begin{aligned} \int _{B_2} \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_n}dx=0, \quad \text { for all } 1 \le i\le n-1. \end{aligned}$$

(3.32)

Proof

Let us observe that for every $\eta \in {\mathbb {R}}^n$,

$$\begin{aligned} \nabla '_\eta \left( u(x)-\frac{1}{6}(\eta \cdot x)_{+}^3 \right) = \nabla '_\eta u(x) \end{aligned}$$

and

$$\begin{aligned} \left\| \nabla '_\eta u\right\| _{L^{2}(B_{2})}^2 = \left\| \nabla u\right\| _{L^{2}(B_{2})}^2 - \left\| \eta \cdot \nabla u\right\| _{L^{2}(B_{2})}^2 . \end{aligned}$$

For any fixed $1 \le i\le n-1$, and real number $-1<t <1$, let $\eta (t):= t e_i +\sqrt{1-t^2} e_n$. By the definition of a normalised coordinate system, the function $\varphi (t):=\left\| \eta \cdot \nabla u\right\| _{L^{2}(B_{2})}^2$, $t\in (-1,1)$ has a local maximum at the point $t=0$. Hence

$$\begin{aligned} \varphi ^\prime (0)=2\int _{B_2} \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_n}dx=0, \end{aligned}$$

(3.33)

which implies (3.32). $\square$

Lemma 3.10

Assume that $u\in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$ solves the biharmonic obstacle problem in a fixed coordinate system with basis vectors $\{ e_1,\ldots , e_n\}$. Let $\{ e_1^1,\ldots ,e^1_n \}$ be a normalised coordinate system with respect to u, and assume that $e_n^1 \cdot e_n \ge 0$. Then

$$\begin{aligned} |e_n-e_n^1|\le C(n) \Arrowvert \nabla ' u ||_{L^2(B_2)}\le C(n) \varepsilon , \end{aligned}$$

if $\varepsilon$ is small, where $C(n) >0$ is a dimensional constant.

Proof

According to Definition 3.8,

$$\begin{aligned} \Arrowvert \nabla '_{e_n^1} u ||_{L^{2}(B_2)}= \Arrowvert \nabla u -e_n^1 ( e_n^1 \cdot \nabla u)||_{L^2(B_2)}\le \Arrowvert \nabla ' u ||_{L^2(B_2)} . \end{aligned}$$

(3.34)

It follows from the triangle inequality that

$$\begin{aligned} \begin{aligned}&\left\| \frac{\partial u}{\partial x_n}- (e_n \cdot e_n^1)^2 \frac{\partial u}{\partial x_n} \right\| _{L^2(B_2)} \le \left\| \frac{\partial u}{\partial x_n}- (e_n \cdot e_n^1) ( e_n^1 \cdot \nabla u) \right\| _{L^2(B_2)} \\&\qquad +\left\| (e_n \cdot e_n^1 ) ( e_n^1 \cdot \nabla u) - (e_n \cdot e_n^1 )^2 \frac{\partial u}{\partial x_n} \right\| _{L^2(B_2)} \\&\quad \le \Arrowvert \nabla _{e_n^1}' u ||_{L^2(B_2)}+ ( e_n\cdot e_n^1 ) \Arrowvert e_n^1 \cdot \nabla ' u ||_{L^2(B_2)} \le 2 \Arrowvert \nabla ' u ||_{L^2(B_2)}, \end{aligned} \end{aligned}$$

(3.35)

according to (3.34), and taking into account that $0 \le e_n \cdot e_n^1 \le 1$.

Note that Lemma 3.5 implies that $\left\| \frac{\partial u }{\partial x_n} \right\| _{L^2(B_2)} \approx \left\| \frac{x_n^2 }{2} \right\| _{L^2(B_2^+)}$ is uniformly bounded from below by a dimensional constant if $\varepsilon >0$ is small. We may conclude from (3.35) that

$$\begin{aligned} 1-(e_n \cdot e_n^1)^2 \le \frac{2 \Arrowvert \nabla ' u ||_{L^2(B_2)} }{ \left\| \frac{\partial u }{\partial x_n} \right\| _{L^2(B_2)} } \le C(n) \Arrowvert \nabla ' u ||_{L^2(B_2)}. \end{aligned}$$

Since $0 \le e_n \cdot e_n^1 \le 1$, we get

$$\begin{aligned} 0\le 1-e_n \cdot e_n^1 \le 1-(e_n \cdot e_n^1)^2 \le C(n) \Arrowvert \nabla ' u ||_{L^2(B_2)}. \end{aligned}$$

(3.36)

Denote by $(e_n^1)':= e_n^1- e_n (e_n \cdot e_n^1)$. It follows from the triangle inequality and (3.34) that

$$\begin{aligned} \begin{aligned}&\Arrowvert (e_n^1)' (e_n^1 \cdot \nabla u) ||_{L^2(B_2)} \le \Arrowvert \nabla ' u -(e_n^1)' (e_n^1 \cdot \nabla u) ||_{L^2(B_2)} \\&\quad + \Arrowvert \nabla ' u ||_{L^2(B_2)} \le \Arrowvert \nabla '_{e_n^1} u ||_{L^2(B_2)}+\Arrowvert \nabla ' u ||_{L^2(B_2)} \le 2 \Arrowvert \nabla ' u ||_{L^2(B_2)} . \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} |(e_n^1)'|\le \frac{2 \Arrowvert \nabla ' u ||_{L^2(B_2)} }{ \left\| e_n^1 \cdot \nabla u \right\| _{L^2(B_2)} }. \end{aligned}$$

Let us choose $\varepsilon > 0$ small, then $\left\| e_n^1 \cdot \nabla u \right\| _{L^2(B_2)}$ is bounded from below by a dimensional constant according to Lemma 3.5 and inequality (3.36). Therefore we obtain

$$\begin{aligned} |(e_n^1)'|\le C(n) \Arrowvert \nabla ' u ||_{L^2(B_2)}. \end{aligned}$$

(3.37)

Note that

$$\begin{aligned} |e_n-e_n^1|\le |1-e_n\cdot e_n^1 |+ |(e_n^1)'|. \end{aligned}$$

Applying inequalities (3.36) and (3.37) we obtain the desired inequality,

$$\begin{aligned} |e_n-e_n^1|\le C(n) \Arrowvert \nabla ' u ||_{L^2(B_2)} \le C(n) \varepsilon , \end{aligned}$$

and the proof of the lemma is now complete. $\square$

Lemma 3.10 provides an essential estimate, which will be useful in our later discussion. Next we state another supporting lemma, the proof of which is quite standard, but we include it for our convenience.

Lemma 3.11

• 1. Let v be a biharmonic function in the ball $B_2$, then

  $$\begin{aligned} \Arrowvert \Delta v ||_{L^2(B_1)} \le C_n \Arrowvert v ||_{L^2(B_{2})}. \end{aligned}$$

  (3.38)

• 2. If v is a biharmonic function in the half-ball $B_2^+$, such that $v=|\nabla v|=0$ on $\{ x_n=0\} \cap B_2$, then

  $$\begin{aligned} \Arrowvert \Delta v ||_{L^2(B_1^+)} \le C_n \Arrowvert v ||_{L^2(B^+_{2})}. \end{aligned}$$

  (3.39)

Proof

Throughout $\zeta \in C_0^\infty (B_{2})$ is a fixed function, such that $\zeta \equiv 1$ in $B_1$, $0 \le \zeta \le 1$ in $B_{2}$.

• 1. If v is a biharmonic function in $B_2$, then

  $$\begin{aligned} 0&= \int _{B_{{2}}} \Delta v \Delta (\zeta ^4 v) = \int _{B_{{2}}} 2 v( \Delta \zeta ^2+4|\nabla \zeta |^2)\zeta ^2 \Delta v \\&\quad + \int _{B_{{2}}}\zeta ^4 ( \Delta v )^2+ 8 \int _{B_{{2}}} \zeta \nabla \zeta \nabla v \zeta ^2 \Delta v. \end{aligned}$$

  Hence by Cauchy’s inequality,

  $$\begin{aligned} \begin{aligned} \int _{B_{{2}}}\zeta ^4 ( \Delta v )^2&=- \int _{B_{{2}}} 2 v( \Delta \zeta ^2+4|\nabla \zeta |^2)\zeta ^2 \Delta v - 8 \int _{B_{{2}}} \zeta \nabla \zeta \nabla v \zeta ^2 \Delta v \\&\le \frac{1}{4}\int _{B_{{2}}} \zeta ^4 ( \Delta v)^2 +4 \int _{B_{{2}}} v^2 (\Delta \zeta ^2+4|\nabla \zeta |^2 )^2 + \frac{1}{4}\int _{B_{{2}}}\zeta ^4 (\Delta v )^2 \\&\quad + 64 \int _{B_{{2}}} (\zeta \nabla \zeta \nabla v)^2 \le \frac{1}{2}\int _{B_{{2}}} \zeta ^4 ( \Delta v)^2 + C_n \Arrowvert v ||^2_{L^2(B_{2})} + C_n \Arrowvert \zeta \nabla v ||^2_{L^2(B_{2})} . \end{aligned} \end{aligned}$$

  (3.40)

  On the other hand,

  $$\begin{aligned} \begin{aligned} \int _{B_{2}} \zeta ^2 | \nabla v |^2&= \int _{B_{2}} \nabla (\zeta ^2 v) \nabla v -2 \int _{B_{2}} \zeta v \nabla \zeta \nabla v\\&=- \int _{B_{2}} \zeta ^2 v \Delta v -2 \int _{B_{2}} \zeta v \nabla \zeta \nabla v \le \frac{1}{8C_n}\int _{B_2}\zeta ^4 (\Delta v )^2+ 2C_n \int _{B_{{2}}} v ^2 \\&\quad +\frac{1}{2} \int _{B_{2}} \zeta ^2 | \nabla v |^2+2 \int _{B_{2}} | \nabla \zeta |^2 v ^2, \end{aligned} \end{aligned}$$

  and therefore

  $$\begin{aligned} C_n \Arrowvert \zeta \nabla v ||^2_{L^2(B_2)} \le \frac{1}{4} \int _{B_2}\zeta ^4 ( \Delta v )^2+ {\tilde{C}}_n \int _{B_2} v ^2 \end{aligned}$$

  (3.41)

  Combining estimates (3.40) and (3.41), we obtain

  $$\begin{aligned} \int _{B_2} \zeta ^4 ( \Delta v )^2 \le \frac{3}{4} \int _{B_2} \zeta ^4 ( \Delta v )^2 + {\bar{C}}_n \Arrowvert v ||^2_{L^2(B_2)}, \end{aligned}$$

  which implies (3.38).

• 2. In order to prove the second part of the lemma, it is enough to observe that

  $$\begin{aligned} \int _{B_2^+} \Delta v \Delta (\zeta ^4 v) =0, \end{aligned}$$

  since $\zeta ^4 v\in W^{2,2}_0(B_2^+)$. The rest of the proof follows as in the first part.

$\square$

Proposition 3.12

For any small number $0< s<2^{-n-4}e^{-1}n^{-2}$, there exists $\varepsilon _0=\varepsilon _0(s)>0$ small, such that if $\varepsilon <\varepsilon _0$, then for any $u \in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$

$$\begin{aligned} \left\| \nabla ' u_{2s} \right\| _{L^{2}(B_2)} \le C_n \left\| \nabla ' u \right\| _{W^{2,2}(B_2)}, \end{aligned}$$

(3.42)

where $C_n$ is a dimensional constant, not depending on s. Furthermore, if the coordinate system is normalised with respect to $u_{2s}$, then

$$\begin{aligned} \left\| \nabla ' u_s \right\| _{W^{2,2}(B_2)} \le \tau \left\| \nabla ' u \right\| _{W^{2,2}(B_2)}, \end{aligned}$$

(3.43)

where $1>\tau >\Lambda _n s$ is a fixed number and $\Lambda _n$ is a dimensional constant to be specified.

Proof

The proof of inequalities (3.42) and (3.43) follows the exact same procedure, so we will mainly focus on the proof of the second one, since it is the core of the linearisation argument.

We want to show that the inequality (3.43) holds in a normalised coordinate system with respect to $u_{2s}$. By the Cauchy–Schwartz inequality, it is enough to show that the inequality

$$\begin{aligned} \left\| \frac{ \partial u_s }{\partial x_i}\right\| _{W^{2,2}(B_2)} \le \tau \left\| \frac{ \partial u}{\partial x_i} \right\| _{W^{2,2}(B_2)} \end{aligned}$$

holds for any $i\in \{1,\ldots ,n-1\}$, provided $\varepsilon$ is small enough. We argue by contradiction. Assume that there exists small numbers $0<s<2^{-n-4}e^{-1}n^{-2}$, $\Lambda _n s<\tau <1$ and a sequence of solutions $\{ u^j \} \subset {\mathscr {B}}_\kappa ^\varrho (\varepsilon _j)$, in a coordinate system normalised with respect to $u^j_{2s}$, such that $\varepsilon _j \rightarrow 0$, as $j\rightarrow \infty$, but for some $i \in \{ 1,2,\ldots ,n-1 \}$

$$\begin{aligned} \left\| \frac{ \partial u^j_{s} }{\partial x_i}\right\| _{W^{2,2}(B_2)} > \tau \left\| \frac{\partial u^j}{\partial x_i} \right\| _{W^{2,2}(B_2)}. \end{aligned}$$

(3.44)

Let $v^j_i$ be given by (3.23), then according to Lemma 3.7, $v^j_i\rightarrow v^0_i$ in $W^{2,2}_{loc}(B_2)$, where $v^0_i$ is a biharmonic function in the half-ball $\{ x_n > 0\} \cap B_2$, satisfying $v^0_i=|\nabla v^0_i|=0$ on $\{x_n=0\}\cap B_2$. Inequality (3.44) implies that

$$\begin{aligned} \left\| \frac{v_i^0(s\cdot )}{s^2}\right\| _{W^{2,2}(B_2)} \ge \tau . \end{aligned}$$

(3.45)

Lemma 3.11, part 2. and the Calderón–Zygmund inequality imply that

$$\begin{aligned} \left\| \frac{v_i^0(2s\cdot )}{4s^2}\right\| _{W^{2,2}(B_1)} \le C_n \left\| \frac{v_i^0(2s\cdot )}{4s^2}\right\| _{L^{2}(B_2)}, \end{aligned}$$

hence

$$\begin{aligned} \left\| \frac{v_i^0(2s\cdot )}{4s^2}\right\| _{L^{2}(B_2)} \ge \frac{1}{C_n} \left\| \frac{v_i^0(2s\cdot )}{4s^2}\right\| _{W^{2,2}(B_1)} \ge C_n \left\| \frac{v_i^0(s\cdot )}{s^2}\right\| _{W^{2,2}(B_2)}\ge C_n \tau , \end{aligned}$$

(3.46)

where $C_n$ represents a general dimensional constant, and it does not depend neither on the function $v_i^0$ nor on the parameter s. We will derive a contradiction to (3.45), if we show that $\left\| \frac{v_i^0(2s\cdot )}{4s^2}\right\| _{L^{2}(B_2)}$ can be made arbitrarily small by choosing $s>0$ small initially.

Since $v^0_i$ is a biharmonic function in the half-ball $\{ x_n > 0\} \cap B_2$ and $v^0_i=|\nabla v^0_i|= 0$ on $\{ x_n = 0\} \cap B_2$, we can apply the reflection principle for biharmonic functions, and extend $v^0_i$ to a biharmonic function in the ball $B_2$, see for instance [9] or [10]. Let ${\bar{v}}^0_i$ denote the extended function given by Duffin’s formula

$$\begin{aligned} {\bar{v}}^0_i(x',-x_n)=-v^0_i(x',x_n)+2x_n\frac{\partial v^0_i}{\partial {x_n}}(x',x_n)-x_n^2\Delta v^0_i(x',x_n), ~x_n >0. \end{aligned}$$

(3.47)

The formula (3.47) implies that

$$\begin{aligned} \Arrowvert {\bar{v}}^0_i ||_{L^2(B_{2})} \le c_n \Arrowvert v^0_i ||_{W^{2,2}(B_{2})}, \end{aligned}$$

(3.48)

where $c_n>0$ is yet another dimensional constant.

The function ${\bar{v}}^0_i$ is biharmonic in the ball $B_2$, therefore analytic and it may be written as a Taylor series

$$\begin{aligned} {\bar{v}}^0_i(x)= \sum _{|\alpha |=0}^\infty \frac{D^\alpha {\bar{v}}^0_i(0) }{\alpha !} x^\alpha =\sum _{k=0}^\infty b_k(x), \end{aligned}$$

(3.49)

where $\alpha$ is a multiindex, and $b_k$ is a homogeneous degree k biharmonic polynomial. It follows from boundary conditions for the function $v^0_i$ on $\{x_n=0\}$ that

$$\begin{aligned} b_0=b_1\equiv 0, \text { and }b_2(x)= \frac{\partial ^2 {\bar{v}}^0_i(0)}{\partial x_n^2} \frac{x_n^2}{2}. \end{aligned}$$

(3.50)

Lemma 3.5 implies that $\frac{\partial u^j}{\partial x_n} \rightarrow \frac{1}{2}(x_n^+)^2$ in ${L^2(B_2)}$, and according to Lemma 3.7, $\frac{v^j_i (2sx)}{4s^2} \rightarrow \frac{v^0_i(2sx)}{4s^2}$ in ${W^{2,2}(B_2)}$ as $j\rightarrow \infty$, and $v^0_i= 0$ in $B_2 {\setminus } B_2^+$. By Lemma 3.9,

$$\begin{aligned} \int _{B_2} v^j_i (2sx)\frac{\partial u^j}{\partial x_n} (2sx)dx=\frac{1}{\delta ^j_i}\int _{B_2} \frac{\partial u^j}{\partial x_i }(2sx)\frac{\partial u^j}{\partial x_n}(2sx)dx =0, \end{aligned}$$

and after passing to the limit as $j\rightarrow \infty$, we obtain that

$$\begin{aligned} \int _{B_2^+} v^0_i (2sx)x_n^2dx =0. \end{aligned}$$

(3.51)

Note that (3.51) implies that

$$\begin{aligned} \left\| \frac{v^0_i(2s\cdot )}{4s^2}-b_2 \right\| ^2_{L^2(B_2^+)} = \left\| \frac{v^0_i(2s\cdot )}{4s^2} \right\| ^2_{L^2(B_2^+)} + \left\| b_2 \right\| ^2_{L^2(B_2^+)} , \end{aligned}$$

hence

$$\begin{aligned} \left\| \frac{v^0_i(2s\cdot )}{4s^2} \right\| _{L^2(B_2^+)} \le \left\| \frac{v^0_i(2s\cdot )}{4s^2}-b_2 \right\| _{L^2(B_2^+)}. \end{aligned}$$

(3.52)

Next we show that $\left\| v^0_i(2s\cdot )-4s^2b_2 \right\| ^2_{L^2(B_2^+)}$ is of order $s^3$. By the triangle inequality,

$$\begin{aligned} \begin{aligned} \left\| \frac{v^0_i(2s\cdot )}{4s^2}-b_2 \right\| _{L^2(B_2^+)}&=\left\| \sum _{k=3}^\infty (2s)^{-2} b_k (2s\cdot ) \right\| _{L^2(B_2^+)} \\ \le \sum _{k=3}^\infty \left\| (2s)^{-2} b_k (2s\cdot ) \right\| _{L^2(B_2^+)}&=\sum _{k=3}^\infty (2s)^{k-2} \left\| b_k \right\| _{L^2(B_2^+)}. \end{aligned} \end{aligned}$$

(3.53)

Now it is time to refer to the estimates on derivatives for biharmonic functions (see “Appendix A”),

$$\begin{aligned} \begin{aligned} b_k(x)&=\sum _{|\alpha |=k}\frac{ D^\alpha {\bar{v}}^i_0(0)}{\alpha !} x^\alpha , \text { and }\\ |D^\alpha {\bar{v}}^i_0(0) |&\le \frac{(2^{n+1}nk)^k}{r^{n+k}} \left( \left\| {\bar{v}}^0_i \right\| _{L^1(B_{r})}+\frac{r^2}{2(n+2)} \left\| \Delta {\bar{v}}^0_i \right\| _{L^1(B_{r})} \right) . \end{aligned} \end{aligned}$$

(3.54)

Hence

$$\begin{aligned} \left\| b_k \right\| _{L^2(B_2^+)} \le&\sum _{|\alpha |=k}\frac{(2^{n+1}nk)^k 2^{n+k} }{|B_1|^\frac{1}{2}\alpha ! } \left( \left\| {\bar{v}}^0_i \right\| _{L^1(B_{1})} +\frac{1}{2(n+2)} \left\| \Delta {\bar{v}}^0_i \right\| _{L^1(B_{1})} \right) \\ \overset{(3.38)}{\le }&C_n {(2^{n+2}nk)^k } \sum _{|\alpha |=k}\frac{1}{\alpha ! } \left\| {\bar{v}}^0_i \right\| _{L^2(B_{2})} = C_n \frac{(2^{n+2}nk)^k n^k }{ k!} \left\| {\bar{v}}^0_i \right\| _{L^2(B_{2})} \\ \le&C_n{2^{k(n+2)}n^{2k} e^k } \left\| {\bar{v}}^0_i \right\| _{L^2(B_{2})} , \end{aligned}$$

where we used Stirling’s inequality in the last step.

Let $\lambda := 2^{n+2} e n^{2}$ be a fixed number, then by (3.53),

$$\begin{aligned} \begin{aligned} \left\| \frac{v^0_i(2s\cdot )}{4s^2}-b_2 \right\| _{L^2(B_2^+)}&\le C_n s^{-2}\sum _{k=3}^\infty (2s)^{k} \lambda ^{k} {\left\| {\bar{v}}^0_i \right\| _{L^2(B_{2})}} \\&= C_n \frac{\lambda ^3 s}{1-2\lambda s}\left\| {\bar{v}}^0_i \right\| _{L^2(B_{2})} \le {\tilde{C}}_n s\left\| {\bar{v}}^0_i \right\| _{L^2(B_{2})}, \end{aligned} \end{aligned}$$

(3.55)

where by assumption $2s \lambda <1/2$.

Finally, combining the inequalities (3.48) , (3.52) and (3.55), we obtain

$$\begin{aligned} \left\| \frac{v^0_i(2s\cdot )}{4s^2}\right\| _{L^2(B_2)} \le c_n {\tilde{C}}_n s ||v^0_i||_{W^{2,2}(B_{2})} \le A_n s, \end{aligned}$$

where $A_n >0$ is a dimensional constant, and s is fixed small number, $sA_n<1$. Let $\Lambda _n := A_n /C_n$, where $C_n$ is the dimensional constant in (3.46). Recalling that $1>\tau >s\Lambda _n$, we derive a contradiction to (3.46).

The proof of the inequality (3.42) is very similar. Any biharmonic function v in the half ball $B_2^+$, satisfying the boundary conditions $v=|\nabla v|=0$ on $\{x_n=0\}$ can be written as (3.49). Employing the estimates of derivatives of biharmonic functions, we can show that $\left\| \frac{v(s\cdot )}{s^2}\right\| _{L^2(B_2)}$ is bounded by a dimensional constant if $0< s<2^{-n-2} e^{-1}n^{-2}$, and (3.42) follows. $\square$

3.5 $C^{1,\alpha }$-regularity of the free boundary

In this section we perform an iteration argument, based on Proposition 3.12 and Lemma 3.10, that leads to the existence of the unit normal $\eta _0$ of the free boundary at the origin, and provides good decay estimates for $\Arrowvert \nabla '_{\eta _0} u_r ||_{W^{2,2}(B_2)}$.

First we would like to verify that $u \in {\mathscr {B}}_\kappa ^\varrho ( \varepsilon )$ imply that $u_s \in {\mathscr {B}}^\varrho _\kappa ({ \varepsilon })$. It is easy to check that the property of being an NTA domain is scaling invariant, in the sense that if D is an NTA domain and $0\in \partial D$, then for any $0<s<1$ the set $D_s := s^{-1}(D \cap B_s)$ is also an NTA domain with the same parameters as D.

Assumption 3 in Definition 3.3 holds for $u_s$ according to Corollary 3.6. Indeed, let $t=s$ in Corollary 3.6, then $u(sx)=0$ if $x_n < -1$ .

Thus $u_s$ satisfies  2, 3 in Definition 3.3, but it may not satisfy 4. Instead we consider rescaled solutions defined as follows

$$\begin{aligned} U_s(x):=\frac{ \omega _n u_s(x)}{\Arrowvert D^3 u_s ||_{L^2(B_1)}}, \end{aligned}$$

(3.56)

then assumption 4 also holds. Indeed, $\Arrowvert D^3 U_s||_{L^2(B_1)}= \omega _n$ by definition of $U_s$, and

$$\begin{aligned} \Arrowvert D^3 U_s||_{L^2(B_2)}&= \frac{\omega _n\Arrowvert D^3 u_s||_{L^2(B_2)}}{\Arrowvert D^3 u_s||_{L^2(B_1)}}= \frac{\omega _n \Arrowvert D^3 u ||_{L^2(B_{2s})}}{\Arrowvert D^3 u ||_{L^2(B_{s})}}\\&\le \omega _n \frac{ \omega _n (2s)^\frac{n}{2} + \sigma (\varepsilon )}{\omega _n (s)^\frac{n}{2}- \sigma (\varepsilon )} < \kappa , \end{aligned}$$

according to Lemma 3.5 provided $\varepsilon =\varepsilon (n, \kappa , s)$ is small.

In the next lemma we show that $U_s \in {\mathscr {B}}_\kappa ^\varrho ({\gamma \varepsilon })$ in a normalised coordinate system, then we argue inductively to show that $U_{s^k} \in {\mathscr {B}}_\kappa ^\varrho ({\gamma ^k \varepsilon })$, $\gamma <1$.

Lemma 3.13

Assume that $u \in {\mathscr {B}}^\varrho _{\kappa }(\varepsilon )$ solves the biharmonic obstacle problem in a normalised coordinate system $\{e_1, e_2,\ldots , e_n\}$. Then for any $0<\alpha <1$ there exist $r_0>0$ and a unit vector $\eta _0 \in {\mathbb {R}}^n$, such that $|\eta _0 -e_n |\le C\varepsilon$, and for any $0<r<r_0$

$$\begin{aligned} \frac{\Arrowvert \nabla '_{\eta _0} u_{r}||_{W^{2,2}(B_2)} }{\Arrowvert D^3 u_{r}||_{L^2(B_1)} }\le C r^\alpha \varepsilon , \end{aligned}$$

(3.57)

provided $\varepsilon =\varepsilon (n,\kappa ,\varrho , \alpha )$ is small enough. The constant $C>0$ depends only on the given parameters.

Proof

Throughout $\{e_1,\ldots ,e_n \}$ is a fixed coordinate system normalised with respect to the solution $u\in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$, and $\nabla 'u= \nabla '_{e_n}u$. We may renormalise the coordinate system with respect to $U_{2s}$ and denote by $\{e_1^1,\ldots ,e_n^1\}$ the set of basis vectors in the new system. Inductively, $\{e_1^k,\ldots ,e_n^k\}$, $k\in {\mathbb {N}}$ is a normalised system with respect to $U_{2s^k}$, and $e_i^0:=e_i$. According to Lemma 3.10,

$$\begin{aligned} |e_n^{k+1}-e_n^{k} |\le C(n) \frac{\Arrowvert \nabla '_{e_n^{k}} u_{2s^{k+1}}||_{L^2(B_2)}}{ \Arrowvert D^3 u_{s^k }||_{L^2(B_1)}} \overset{(3.42)}{\le } C(n) \frac{\Arrowvert \nabla '_{e_n^{k}} u_{s^{k}}||_{W^{2,2}(B_2)}}{ \Arrowvert D^3 u_{s^k }||_{L^2(B_1)}} , \end{aligned}$$

(3.58)

provided $\Arrowvert \nabla '_{e_n^{k}} U_{s^k} ||_{W^{2,2}(B_2)}$ is sufficiently small.

In the following discussion $0<s<\tau <1$ are small fixed numbers, satisfying the assumptions in Proposition 3.12.

Now let us consider the sequence of numbers $\{A_k\}_{k\in {\mathbb {N}}_0}$, defined as follows:

$$\begin{aligned} A_{k}:=\frac{\omega _n \Arrowvert \nabla '_{e_n^{k}} u_{s^k} ||_{W^{2,2}(B_2)}}{\Arrowvert D^3 u_{s^k}||_{L^2(B_1)}} , \text { for } k=0,1,2,\ldots . \end{aligned}$$

(3.59)

By definition, $A_0\le \varepsilon$, and

$$\begin{aligned} A_{1}=\frac{\omega _n \Arrowvert \nabla '_{e_n^1} u_{s} ||_{W^{2,2}(B_2)}}{\Arrowvert D^3 u_s||_{L^2(B_1)}}= \frac{\Arrowvert D^3 u||_{L^2(B_1)}}{\Arrowvert D^3 u_{s}||_{L^2(B_1)}} { \Arrowvert \nabla '_{e_n^1} u_{s} ||_{W^{2,2}(B_2)}}. \end{aligned}$$

(3.60)

Applying Proposition 3.12 and Lemma 3.10 for a function $u \in {\mathscr {B}}_\kappa ^\varrho ({ \varepsilon })$, we obtain

$$\begin{aligned} \begin{aligned}&\Arrowvert \nabla '_{e_n^{1}} u_{s}||_{W^{2,2}(B_2)} \le \tau \Arrowvert \nabla '_{e_n^1} u||_{W^{2,2}(B_2)} \\&\quad \le \tau \left( \Arrowvert \nabla ' u||_{W^{2,2}(B_2)}+2|e_n-e_n^1| \Arrowvert \nabla u||_{W^{2,2}(B_2)}\right) \\&\quad \overset{(3.58)}{\le } \tau \left( \Arrowvert \nabla ' u||_{W^{2,2}(B_2)}+2C(n)\frac{\Arrowvert \nabla ' u||_{W^{2,2}(B_2)}}{|| D^3u | |_{L^2(B_1)}} \Arrowvert \nabla u||_{W^{2,2}(B_2)} \right) \\&\quad \overset{(3.42)}{\le } \tau C(n,\kappa ) \Arrowvert \nabla ' u||_{W^{2,2}(B_2)}. \end{aligned} \end{aligned}$$

(3.61)

Let $\beta := \tau C(n,\kappa )$, and $\beta<\gamma <1$ be fixed numbers. Then

$$\begin{aligned} \frac{\Arrowvert D^3 u||_{L^2(B_1)}}{\Arrowvert D^3 u_{s}||_{L^2(B_1)}}&= \frac{s^\frac{n}{2} \Arrowvert D^3 u||_{L^2(B_1)}}{\Arrowvert D^3 u ||_{L^2(B_{s})}} \\&\le \frac{s^\frac{n}{2} \omega _n }{s^\frac{n}{2} \omega _n - \sigma ( A_0)} \le \frac{\gamma }{\beta }. \end{aligned}$$

according to Lemma 3.5, provided $A_0 \le \varepsilon$ is small depending on the parameter s and dimension n. The last inequality together with (3.60) and (3.61) implies that $A_1 \le \gamma \varepsilon$.

We use an induction argument to show that

$$\begin{aligned} A_{k} \le \gamma ^{k} \varepsilon , ~~~ \text { for all } k \in {\mathbb {N}}_0 \end{aligned}$$

(3.62)

for fixed 1$> \gamma>\beta>\tau>\Lambda _n s>0$. Assuming that (3.62) holds for $k\in {\mathbb {N}}$, we will show that $A_{k+1} \le \gamma A_k$.

By the induction assumption

$$\begin{aligned} \Arrowvert \nabla '_{e_n^{k}} U_{s^k }||_{W^{2,2}(B_2)} = \frac{\omega _n \Arrowvert \nabla '_{e_n^{k}} u_{s^k }||_{W^{2,2}(B_2)} }{\Arrowvert D^3 u_{s^k}||_{L^2(B_1)}} = A_k \le \gamma ^k \varepsilon . \end{aligned}$$

(3.63)

Hence

$$\begin{aligned} U_{s^k}= \frac{\omega _n u_{s^k}(x)}{\Arrowvert D^3 u_{s^k}||_{L^2(B_1)}} \in {\mathscr {B}}_\kappa ^\varrho ({\gamma ^k \varepsilon }) \end{aligned}$$

in the coordinate system $\{ e_1^{k}, \ldots , e_n^{k}\}$. By definition, $\{ e_1^{k+1}, \ldots , e_n^{k+1}\}$ is a normalised coordinate system with respect to $U_{2s^{k+1}}\in {\mathscr {B}}_\kappa ^\varrho ({\beta ^k \varepsilon })$, and by (3.43)

$$\begin{aligned} A_{k+1}{{\varvec{=}}}\frac{\omega _n \Arrowvert \nabla '_{e_n^{k+1}} u_{s^{k+1}} ||_{W^{2,2}(B_2)}}{\Arrowvert D^3 u_{s^{k+1}}||_{L^2(B_1)}} \le \frac{\omega _n \tau \Arrowvert \nabla '_{e_n^{k+1}} u_{s^{k}} ||_{W^{2,2}(B_2)}}{\Arrowvert D^3 u_{s^{k}}||_{L^2(B_1)}} \frac{\Arrowvert D^3 u_{s^{k}}||_{L^2(B_1)}}{\Arrowvert D^3 u_{s^{k+1}}||_{L^2(B_1)}} . \end{aligned}$$

(3.64)

First we observe that

$$\begin{aligned} \begin{aligned} \frac{\Arrowvert D^3 u_{s^{k}}||_{L^2(B_1)}}{\Arrowvert D^3 u_{s^{k+1}}||_{L^2(B_1)}}&= \frac{s^\frac{n}{2} \Arrowvert D^3 u_{s^{k}}||_{L^2(B_1)}}{\Arrowvert D^3 u_{s^{k}}||_{L^2(B_{s})}} \\&\le \frac{s^\frac{n}{2} \omega _n }{s^\frac{n}{2} \omega _n - \sigma (\gamma ^{k} A_0)} \le \frac{\gamma }{\beta }, \end{aligned} \end{aligned}$$

(3.65)

according to Lemma 3.5, since $U_{s^k} \in {\mathscr {B}}_\kappa ^\varrho (\gamma ^k \varepsilon )$ and $\gamma ^{k} \varepsilon < \varepsilon$ is small.

Next we estimate

$$\begin{aligned} \begin{aligned}&\Arrowvert \nabla '_{e_n^{k+1}} u_{s^k} ||_{W^{2,2}(B_2)}\le \Arrowvert \nabla '_{e_n^k} u_{s^k} ||_{W^{2,2}(B_2)}+2|e_n^{k+1}-e_n^k | \Arrowvert \nabla u_{s^k} ||_{W^{2,2}(B_2)}\\&\quad \overset{(3.58)}{ \le } \Arrowvert \nabla '_{e_n^k} u_{s^k} ||_{W^{2,2}(B_2)}+\frac{C(n) \Arrowvert \nabla '_{e_n^k} u_{s^{k}} ||_{W^{2,2}(B_2)} }{{\Arrowvert D^3 u_{s^{k}}||_{L^2(B_{1})}} } \Arrowvert \nabla u_{s^k} ||_{W^{2,2}(B_2)}\\&\quad \overset{(3.42)}{\le } C(n,\kappa ) \Arrowvert \nabla '_{e_n^k} u_{s^k} ||_{W^{2,2}(B_2)}. \end{aligned} \end{aligned}$$

(3.66)

Finally we obtain from (3.65) and (3.66) that

$$\begin{aligned} A_{k+1} \le \frac{ \omega _n \tau C(n,\kappa ) \gamma }{\beta } \frac{\Arrowvert \nabla '_{e_n^k} u_{s^k} ||_{W^{2,2}(B_2)}}{\Arrowvert D^3 u_{s^k}||_{L^2(B_1)}}= \gamma A_k \le \gamma ^{k+1} A_0, \end{aligned}$$

(3.67)

this completes the proof of inequality (3.62).

Next we show that $\{ e_n^k\}$ is a Cauchy sequence by using (3.58) and (3.62). Indeed for any $m, k \in {\mathbb {N}}$,

$$\begin{aligned} |e_n^{k+m}-e_n^k |\le&\sum _{l=1}^m |e_n^{k+l}-e_n^{k+l-1} |\le C(n) \sum _{l=1}^m \Arrowvert \nabla '_{e_n^{k+l-1}} U_{2s^{k+l}} ||_{L^2(B_2)} \\ \le&C(n) \sum _{l=1}^m A_{k+l-1} \le {C(n) A_0 } \sum _{l=1}^m \gamma ^{k+l-1} \le \frac{C(n) A_0 }{ (1-\gamma )} \gamma ^k, \end{aligned}$$

hence $e_n^k \rightarrow \eta _0$, as $k\rightarrow \infty$ for some $\eta _0 \in {\mathbb {R}}^n$, $|\eta _0|=1$ and

$$\begin{aligned} |\eta _0 -e_n^k |\le C'(n) A_0 \gamma ^k \le C'(n) \gamma ^k \varepsilon , \end{aligned}$$

(3.68)

in particular $|\eta _0 -e_n |\le C'(n)\varepsilon$.

Now the inequality (3.57) follows via a standard iteration argument. Let $0<\alpha <1$ be any number, choose $s=s(n,\alpha )$ small, satisfying the assumption in Proposition 3.12, and such that $\gamma = C_n s <s^\alpha$. If $0<r\le s$, then there exists $k\in {\mathbb {N}}_0$, such that $s^{k+1}\le r <s^k$. Hence

$$\begin{aligned} \begin{aligned}&\frac{\Arrowvert \nabla '_{\eta _0} u_{r}||_{W^{2,2}(B_2)} }{\Arrowvert D^3 u_{r}||_{L^2(B_1)} }\le C \frac{\Arrowvert \nabla '_{\eta _0} u_{s^k}||_{W^{2,2}(B_2)} }{\Arrowvert D^3 u_{s^k}||_{L^2(B_1)} } \\&\quad \le C \frac{\Arrowvert \nabla '_{e_n^k} u_{s^k}||_{W^{2,2}(B_2)} +2|e_n^k- \eta _0| \Arrowvert \nabla u_{s^k}||_{W^{2,2}(B_2)}}{\Arrowvert D^3 u_{s^k}||_{L^2(B_1)} } \le C \gamma ^k \varepsilon \le C s^{\alpha k} \varepsilon \le C r^\alpha \varepsilon , \end{aligned} \end{aligned}$$

(3.69)

where C depends on the space dimension and on the given parameters. $\square$

Now we are ready to prove the $C^{1,\alpha }$-regularity of the free boundary.

Theorem 3.14

Let $0<\alpha <1$ be a given number. Assume that $u \in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$, with an $\varepsilon > 0$ small, depending on $\alpha$ and the space dimension. Then there exists $0<r_0<1$ depending on the given parameters, such that $\Gamma _u \cap B_{r_0}$ is a $C^{1,\alpha }$-graph and the $C^{1,\alpha }$-norm of the graph is bounded by $C\varepsilon$.

Proof

Let $0<\alpha <1$ and fix $s=s(n, \alpha )>0$ small as in (3.69). It follows from Lemma 3.13 that for $u\in {\mathscr {B}}^\varrho _{\kappa }(\varepsilon )$

$$\begin{aligned} \frac{ \Arrowvert \nabla ' u_r ||_{W^{2,2}(B_2)}}{\Arrowvert D^3 u_r ||_{L^2(B_1)}} \le C r^\alpha \rightarrow 0 ~ \text { as } ~ r\rightarrow 0, \end{aligned}$$

after a change of variable, by choosing $e_n = \eta _0$, where $\eta _0$ is the same vector as in Lemma 3.13. Then

$$\begin{aligned} \frac{\omega _n u_r(x)}{\Arrowvert D^3u_r ||_{L^2(B_1)}} \rightarrow \frac{1}{6} (x_n)_+^3 \end{aligned}$$

according to Lemma 3.5.

So we have shown that in the initial coordinate system,

$$\begin{aligned} \frac{ \omega _n u(rx)}{ r^3 \Arrowvert D^3u_r||_{L^2(B_1)}} \rightarrow \frac{1}{6} (\eta _0 \cdot x )_+^3 \text { in } W^{3,2} (B_1)\cap C^{1,\alpha }(B_1), \text { as } r\rightarrow 0, \end{aligned}$$

(3.70)

and therefore $\eta _0$ is the measure theoretic normal to $\Gamma _u$ at the origin.

Now let $x_0 \in \Gamma _u \cap B_{s}$ be a free boundary point, and consider the function $u_{{x_0,{1/2}}}(x) = \frac{u(x/2+x_0)}{(1/2)^3}, x \in B_2$, then

$$\begin{aligned} U_{x_0}(x):= \frac{\omega _n u_{{x_0}, 1/2}(x)}{ \Arrowvert D^3 u_{x_0, 1/2}||_{L^2(B_1)}}\in {\mathscr {B}}^\varrho _\kappa ({ C(n) \varepsilon }). \end{aligned}$$

According to Lemma 3.13, $U_{x_0}$ has a unique blow-up

$$\begin{aligned} U_{r, x_0}(x):=\frac{U_{x_0}(rx)}{r^3}= \frac{\omega _n u_{{x_0}, 1/2}(rx)}{ r^3 \Arrowvert D^3 u_{x_0, 1/2}(rx)||_{L^2(B_1)}}\rightarrow \frac{1}{6}(\eta _{x_0}\cdot x)_+^3 . \end{aligned}$$

and therefore $\eta _{x_0}$ is the normal to $\Gamma _u$ at $x_0$.

Next we show that $\eta _x$ is a Hölder continuous function on $\Gamma _u \cap B_s$. If $x_0 \in \Gamma _u \cap B_s$, then $s^{k+1}< |x_0 |\le s^k$, for some $k\in {\mathbb {N}}_0$. Hence $\Arrowvert \nabla '_{\eta _0} U_{s^k, x_0}||_{W^{2,2}(B_2)} \le C \gamma ^k \varepsilon$, and $\Arrowvert \nabla '_{\eta _{x_0}}U_{r, x_0} ||_{W^{2,2}(B_1)} \rightarrow 0$ as $r\rightarrow 0$. Applying Lemma 3.13 for the function $U_{s^k,x_0} \in {\mathscr {B}}^\varrho _\kappa ({ C \gamma ^k \varepsilon })$, we obtain

$$\begin{aligned} |\eta _{x_0} -\eta _0 |\le C \gamma ^k \varepsilon \le \frac{C}{\gamma } |x_0 |^{\alpha } \varepsilon . \end{aligned}$$

(3.71)

Furthermore, the inequality

$$\begin{aligned} |\eta _x -\eta _y |\le C|x-y |^\alpha \varepsilon , ~~~ \text { for any } x,y \in \Gamma _u \end{aligned}$$

follows from (3.71). $\square$

4 On the regularity of the solution

In this section we study the regularity of the solution to the biharmonic obstacle problem. Assuming that $u\in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$, with $\varepsilon >0$ small, we derive from Theorem 3.14 that $u\in C^{2,1}_{loc} (B_1)$. In the end we provide an example showing that without the NTA domain assumption, there exist solutions, which are not $C^{2,1}$.

4.1 $C^{2,1}$-regularity of the solutions in ${\mathscr {B}}^\varrho _\kappa (\varepsilon )$

After showing the $C^{1,\alpha }$-regularity of the free boundary $\Gamma _u \cap B_{r_0}$, we may go further to derive improved regularity for the solution $u\in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$.

Theorem 4.1

Let $u\in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$ be the solution to the biharmonic obstacle problem in $\Omega \supset \supset B_2$, and let $0<\alpha <1$ be a fixed number. Then there exists $r_0>0$ such that $u\in C^{2,1}(B_{r_0})$, provided $\varepsilon =\varepsilon (\kappa ,\varrho , \alpha )$ is small. Furthermore, the following estimate holds

$$\begin{aligned} \Arrowvert u ||_{C^{3,\alpha }({\overline{\Omega }}_u \cap B_{r_0} )} \le C(n) \Arrowvert u ||_{W^{2,2}(B_2)}\le C(n) \kappa , \end{aligned}$$

where C(n) is just a dimensional constant.

Proof

According to Theorem 3.14, $\Gamma _u \cap B_s$ is a graph of a $C^{1,\alpha }$-function. We know that $\Delta u \in W^{1,2}(B_2)$ is a harmonic function in $\Omega _u:= \{u> 0\}$, and also $u \in W^{3,2}(B_2)$, $u \equiv 0$ in $\Omega {\setminus } \Omega _u$, hence $\Delta u =0$ on $\Gamma _u= \partial \Omega _u \cap B_2$ in the trace sense. Therefore we may apply Corollary 8.36 in [11], to conclude that $\Delta u \in C^{1,\alpha } ( (\Omega _u \cup \Gamma _u) \cap B_{3s/4})$, and

$$\begin{aligned} \Arrowvert \Delta u ||_{ C^{1,\alpha } ({\overline{\Omega }}_u \cap B_{3s/4}) } \le C(n) \Arrowvert \Delta u ||_{L^\infty ( B_1)}. \end{aligned}$$

(4.1)

It follows from the Calderón–Zygmund estimates that $u \in W^{3,p}(B_{s/2})$, for any $p < \infty$. According to the Sobolev embedding theorem, $u \in C^{2,\alpha }(B_{s/2})$, for all $\alpha < 1$, with the following estimate

$$\begin{aligned} \Arrowvert u ||_{C^{2,\alpha }(\overline{ \Omega _u } \cap B_{s/2}) } \le C(n) \left( \Arrowvert \Delta u ||_{C^{1,\alpha }({\overline{\Omega }}_u \cap B_{3s/4})}+ \Arrowvert u ||_{C^{1,\alpha }(B_{3/4})} \right) . \end{aligned}$$

(4.2)

Denote by $u_{ij}:= \frac{\partial ^2 u}{\partial x_i \partial x_j}$. Then $u_{ij} \in W^{1,2}(B_1)\cap C^\alpha (\Omega _u \cap B_{3s/4})$ is a weak solution of $\Delta u_{ij}= \frac{\partial f_j}{\partial x_i}$ in $\Omega _u \cap B_{3s/4}$, where $f_j:= \frac{\partial \Delta u }{\partial x_j} \in C^\alpha ( \Omega _u \cap B_{3/4} )$. Taking into account that $u_{ij}=0$ on $\partial \Omega _u \cap B_{1/2}$, we may apply Corollary 8.36 in [11] once again and conclude that

$$\begin{aligned} \Arrowvert u_{ij}||_{C^{1,\alpha }( {\overline{\Omega }}_u \cap B_{s/4}) } \le&C(n)\left( \Arrowvert u_{ij}||_{C^{0}( {\overline{\Omega }}_u \cap B_{s/2})}+ \Arrowvert \Delta u ||_{C^{1,\alpha }( {\overline{\Omega }}_u\cap B_{3s/4})} \right) , \end{aligned}$$

hence

$$\begin{aligned} \Arrowvert D^2 u||_{C^{1,\alpha }( {\overline{\Omega }}_u \cap B_{s/4})} \le C'_n \left( \Arrowvert \Delta u ||_{C^{1,\alpha }({\overline{\Omega }}_u \cap B_{3s/4})}+ \Arrowvert u ||_{C^{1,\alpha }(B_{3/4})} \right) , \end{aligned}$$

according to (4.2).

Therefore we obtain

$$\begin{aligned} \Arrowvert u ||_{C^{3,\alpha }({\overline{\Omega }}_u \cap B_{s/4} )}&\le \Arrowvert u ||_{C^{1,\alpha }(B_{3/4})}+ \Arrowvert D^2 u||_{C^{1,\alpha }( {\overline{\Omega }}_u \cap B_{s/4})} \\&\le C(n) \left( \Arrowvert \Delta u ||_{C^{1,\alpha }({\overline{\Omega }}_u \cap B_{3s/4})}+ \Arrowvert u ||_{C^{1,\alpha }(B_{3s/4})} \right) . \end{aligned}$$

Taking into account that

$$\begin{aligned} \Arrowvert D^3 u ||_{L^\infty (B_{s/4} ) } \le \Arrowvert D^3 u||_{C^{0,\alpha }( {\overline{\Omega }}_u \cap B_{s/4})}, \end{aligned}$$

we see that $u\in C^{2,1}( B_{s/4} )$. $\square$

4.2 In general the solutions are not better than $C^{2,\frac{1}{2}}$

Let us observe that the assumption $u \in {\mathscr {B}}_\kappa ^\varrho (\varepsilon )$ is essential in the proof of $u\in C^{2,1}(B_r)$. The next example shows that without our flatness assumptions there exists a solution to the biharmonic obstacle problem in ${\mathbb {R}}^2$, that do not possess $C^{2,1}$- regularity.

Example 4.2

Consider the following function given in polar coordinates in ${\mathbb {R}}^2$,

$$\begin{aligned} u(r,\varphi )= r^{\frac{5}{2}}\left( \cos \frac{\varphi }{2} -\frac{1}{5}\cos \frac{5 \varphi }{2} \right) , ~~ r \in [0,1), ~~ \varphi \in [-\pi , \pi ) \end{aligned}$$

(4.3)

then $u \in C^{2, \frac{1}{2}}$ is the solution to the biharmonic zero-obstacle problem in the unit ball $B_1 \subset {\mathbb {R}}^2$.

Proof

It is easy to check that $u \ge 0$, $u(x)=0$ if and only if $-1 \le x_1 \le 0$ and $x_2 =0$. Hence the set $\Omega _u = \{u >0\}$ is not an NTA domain, since the complement of $\Omega _u$ does not satisfy the corkscrew condition.

Let us show that $\Delta ^2 u$ is a nonnegative measure supported on $[-1,0]\times \{0 \}$. For any nonnegative $f \in C_0^\infty (B_1)$, we compute

$$\begin{aligned} \int _{B_1} \Delta u(x) \Delta f(x)dx&=\int _{0}^1 \int _{-\pi }^\pi r \Delta f(r,\varphi ) \Delta u(r, \varphi ) d \varphi d r \\&= 6 \int _{0}^1 \int _{-\pi }^\pi r^{\frac{3}{2}} \Delta f(r,\varphi ) \cos \frac{\varphi }{2} d \varphi d r =6 \int _0^1 {r^{- \frac{1}{2}}} {f(r,\pi )} dr \ge 0, \end{aligned}$$

where we used integration by parts, and that f is compactly supported in $B_1$.

We obtain that u solves the following variational inequality,

$$\begin{aligned} u \ge 0, ~ \Delta ^2 u \ge 0, ~ u \cdot \Delta ^2 u =0. \end{aligned}$$

(4.4)

Now we show that u is the unique minimiser to the following zero-obstacle problem: minimize the functional (1.1) over

$$\begin{aligned} {\mathscr {A}}:=\left\{ v \in W^{2,2}(B_1), v \ge 0, \text { s.t. } v= u, \frac{\partial v}{\partial n }= \frac{\partial u}{\partial n}, \text { on } \partial B_1 \right\} \ne \emptyset . \end{aligned}$$

The functional J admits a unique minimiser over ${\mathscr {A}}$, let us call it v. It follows from (4.4), that

$$\begin{aligned} \int _{B_1} \Delta u \Delta (v-u)= \int _{B_1} (v-u) \Delta ^2 u = \int _{B_1} v \Delta ^2 u \ge 0. \end{aligned}$$

Hence

$$\begin{aligned} \int _{B_1} (\Delta u)^2 \le \int _{B_1} \Delta u \Delta v \le \left( \int _{B_1} (\Delta u)^2 \right) ^\frac{1}{2} \left( \int _{B_1} (\Delta v)^2 \right) ^\frac{1}{2}, \end{aligned}$$

where we used the Hölder inequality in the last step. Therefore we obtain

$$\begin{aligned} \int _{B_1} (\Delta u)^2 \le \int _{B_1} (\Delta v)^2, \end{aligned}$$

thus $u \equiv v$, and u solves the biharmonic zero-obstacle problem in the unit ball.

However $\Delta u= 6 r^{\frac{1}{2}} \cos \frac{\varphi }{2}$, which implies that u is $C^{2,\frac{1}{2}}$, and that the exponent $\frac{1}{2}$ is optimal, in particular u is not $C^{2,1}$. $\square$

Estimates on derivatives of biharmonic functions

In this part of the paper estimates on derivatives of biharmonic functions are obtained. We believe that these estimates are known, but we could not find a reference, and therefore included in the paper.

Lemma A.1

Let v be a biharmonic function in the ball $B_1 \subset {\mathbb {R}}^n$, and assume that $B_r(x_0)\subset B_1$. Then

$$\begin{aligned} |D^\alpha v (x_0)|\le \frac{(2^{n+1}nk)^k }{|B_1| r^{n+k}} \left( \Vert v \Vert _{L^1(B_r(x_0))}+ \frac{r^2}{2(n+2)}\Vert \Delta v \Vert _{L^1(B_r(x_0))} \right) , \end{aligned}$$

(A.1)

where $\alpha$ is a multiindex, and $k=|\alpha |$.

Proof

The following mean value properties are known for biharmonic functions

$$\begin{aligned} v(x_0)=\fint _{\partial B_r(x_0)} vdS -\frac{r^2}{2n} \Delta v(x_0) \end{aligned}$$

(A.2)

and

$$\begin{aligned} v(x_0)=\fint _{B_r(x_0)} vdx -\frac{r^2}{2(n+2)} \Delta v(x_0). \end{aligned}$$

(A.3)

The proofs of (A.2) and (A.3) are similar to the proofs of the mean value properties for harmonic functions. For a fixed $x_0$, let $\phi (r):=\fint _{\partial B_r(x_0)} vdS$. It is easy to see by Green’s formula that

$$\begin{aligned} \phi ^\prime (r)=\frac{r}{n} \fint _{B_r(x_0)}\Delta v dx= \frac{r}{n} \Delta v(x_0), \end{aligned}$$

(A.4)

where we also used the mean value property for the harmonic function $\Delta v$. Hence (A.2) follows by integrating (A.4) in the interval (0, r).

Now (A.3) can be shown by using (A.2) and the co-area formula;

$$\begin{aligned} \begin{aligned} \fint _{B_r(x_0)}vdx&=|B_r|^{-1}\int _0^r \int _{\partial B_s(x_0)} vdS ds= \int _0^r \frac{n s^{n-1}}{r^n} \left( v(x_0)+\frac{s^2}{2n} \Delta v(x_0) \right) ds \\&=v(x_0) \frac{n}{r^n} \int _0^rs^{n-1} ds+\Delta v(x_0)\frac{ 1}{2r^n }\int _0^rs^{n+1} ds =v(x_0)+\Delta v(x_0)\frac{ r^2 }{2(n+2)}. \end{aligned} \end{aligned}$$

Let us proceed to the proof of (A.1). Estimate (A.1) is well known for harmonic functions, which will be used to show that it also holds for biharmonic functions. We follow the proof of estimates on derivatives of harmonic functions (see for instance [5]), and employ (A.2) and (A.3). The proof uses an argument of induction on $k=| \alpha |$. The formula (A.3) implies that

$$\begin{aligned} |v(x_0)|\le \frac{1}{r^n |B_1|}\left( \Vert v\Vert _{L^1(B_r(x_0))}+ \frac{r^2}{2(n+2)} \Vert \Delta v\Vert _{L^1(B_r(x_0))} \right) . \end{aligned}$$

Let $k=1$, then

$$\begin{aligned} \begin{aligned} v_{x_i}(x_0)&=\fint _{B_{r/2}(x_0)} v_{x_i}dx -\frac{r^2}{2^3(n+2)} \fint _{B_{r/2}(x_0)}\Delta v_{x_i} dx \\&=\frac{2^n}{|B_1| r^n} \int _{\partial B_{r/2}(x_0)} v{\nu _i}dS -\frac{r^22^{n-2}}{2(n+2)|B_1| r^n} \int _{\partial B_{r/2}(x_0)}{\nu _i} \Delta vdS. \end{aligned} \end{aligned}$$

hence

$$\begin{aligned} \begin{aligned} |v_{x_i}(x_0)|\le&\frac{2n \Vert v\Vert _{L^\infty (\partial B_{r/2}(x_0))}}{ r} + \frac{rn \Vert \Delta v\Vert _{L^\infty (\partial B_{r/2}(x_0))}}{2^2(n+2) } \\ \le&\frac{ 2^{n+1}n \Vert v\Vert _{L^1( B_{r}(x_0))}}{ |B_1| r^{n+1}} + \frac{n 2^{n-1}r^2\Vert \Delta v\Vert _{L^1( B_{r}(x_0))}}{ 2(n+2)|B_1| r^{n+1}} +\frac{r^2 n 2^{n-1}\Vert \Delta v\Vert _{L^1(B_{r}(x_0))}}{2(n+2) |B_1| r^{n+1}} \\ \le&\frac{n2^{n+1}}{|B_1| r^{n+1}}\left( \Vert v\Vert _{L^1(B_{r}(x_0))} +\frac{r^2}{ 2^2(n+2)}\Vert \Delta v\Vert _{L^1(B_{r}(x_0))} \right) . \end{aligned} \end{aligned}$$

Assuming that (A.1) is true for $k-1$, we will show that it is true for k. Let $|\alpha |=k$, and $D^\alpha v=(D^\beta v)_{x_i}$, where $|\beta |=k-1$. By (A.3)

$$\begin{aligned} \begin{aligned} D^\alpha v(x_0)&= \fint _{B_{r/k}(x_0)} D^\alpha v dx -\frac{r^2}{k^2 2(n+2)} D^\alpha \Delta v(x_0) \\&=\frac{k^n}{|B_1| r^n} \int _{\partial B_{r/k}(x_0)} D^\beta v {\nu _i}dS - \frac{r^2}{k^2 2(n+2)} D^\alpha \Delta v(x_0) \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} |D^\alpha v(x_0)| \le \frac{kn}{r} \Vert D^\beta v \Vert _{L^\infty (\partial B_{r/k(x_0)})} + \frac{r^2}{k^2 2(n+2)}| D^\alpha \Delta v(x_0)|. \end{aligned} \end{aligned}$$

If $x \in \partial B_{r/k}(x_0)$, then $B_{r(k-1)/k}(x) \subset B_r(x_0)$, and by induction assumption

$$\begin{aligned} \begin{aligned} |D^\alpha v (x_0)| \le&\frac{kn (k-1)^{k-1} (2^{n+1}n)^{k-1}k^{n+k-1}}{r |B_1| r^{n+k-1} (k-1)^{n+k-1}} \Vert v\Vert _{L^1(B_{r}(x_0))} \\&+\frac{ r^2(k-1)^2kn (k-1)^{k-1} (2^{n+1}n)^{k-1}k^{n+k-1}}{k^2 2(n+2)r |B_1| r^{n+k-1} (k-1)^{n+k-1}} \Vert \Delta v\Vert _{L^1(B_{r}(x_0))} \\&+\frac{r^2(2^{n+1}nk)^k}{|B_1| r^{n+k}k^22(n+2)} \Vert \Delta v\Vert _{L^1(B_{r}(x_0))} \le \frac{(kn2^{n+1})^k}{|B_1| r^{n+k}}\Vert v\Vert _{L^1(B_{r}(x_0))} \\&+\frac{(kn2^{n+1})^k((k-1)^2+1)}{2(n+2)|B_1| k^2 r^{n+k-2}}\Vert \Delta v\Vert _{L^1(B_{r}(x_0))}\\ \le&\frac{(kn2^{n+1})^k}{|B_1| r^{n+k}} \left( \Vert v\Vert _{L^1(B_{r}(x_0))}+\frac{r^2}{2(n+2)} \Vert \Delta v\Vert _{L^1(B_{r}(x_0))}\right) , \end{aligned} \end{aligned}$$

and the proof of the lemma is now complete. $\square$