Regularity of the free boundary in the biharmonic obstacle problem

In this article we use 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 (NTA) domain, we derive the $C^{1,\alpha}$-regularity of the free boundary in a small ball centered 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 $n \geq 2$.


Introduction
Let Ω ⊂ R n be a given domain, and ϕ ∈ C 2 (Ω), ϕ ≤ 0 on ∂Ω be a given function, called an obstacle. Then the minimizer to the following functional over all functions u ∈ W 2,2 0 (Ω), such that u ≥ ϕ, is called the solution to the biharmonic obstacle problem with obstacle ϕ. The solution satisfies the following variational inequality It has been shown in [1] and [6] that the solution u ∈ W 3,2 loc (Ω), ∆u ∈ L ∞ loc (Ω), and moreover u ∈ W 2,∞ loc (Ω). Furthermore, in the paper [1], the authors show that in dimension n = 2 the solution u ∈ C 2 (Ω) and that the free boundary Γ u := ∂{u = ϕ} lies on a C 1 -curve in a neighbourhood of the points x 0 ∈ Γ u , such that ∆u(x 0 ) > ∆ϕ(x 0 ).
The setting of our problem is slightly different from the one in [1] and [6]. We consider a zero-obstacle problem with general nonzero boundary conditions. Let Ω be a bounded domain in R n with smooth boundary. We consider the problem of minimizing the functional (1.1) over the admissible set A := u ∈ W 2,2 (Ω), u ≥ 0, u = g > 0, ∂u ∂ν = f on ∂Ω .
The minimizer u exists, it is unique. The minimizer is called the solution to the biharmonic obstacle problem. We will denote the free boundary by Γ u := ∂Ω u ∩ Ω, where Ω 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 characterization 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 1 6 (x n ) 3 + . In [4], using flatness improvement argument, the authors show that the free boundary in the p-harmonic obstacle problem is a C 1,α graph in a neighborhood of the points where the solution is almost one-dimenional. We apply the same technique in order to study the regularity of the free boundary in the biharmonic obstacle problem.
In Section 2 we study the basic properties of the solution in the new setting, and show that it is locally in W 3,2 ∩ C 1,α .
In Section 3 we introduce the class B ̺ κ (ε) of solutions to the biharmonic obstacle problem, that are close to the one-dimensional solution 1 6 (x n ) 3 + . Following [4], we show that if ε is small enough, then there exists a rescaling u s (x) = u(sx) s 3 , such that ∇ ′ u s W 2,2 (B1) ≤ γ ∇ ′ u W 2,2 (B2) ≤ γε in a normalized coordinate system, where ∇ ′ η := ∇ − η(η · ∇), ∇ ′ := ∇ ′ en , and γ < 1 is a constant. Repeating the argument for the rescaled solutions, u s k , we show that there exists a unit vector η 0 ∈ R n , such that ∇ ′ η0 u s k W 2,2 (B1) D 3 u s k L 2 (B1) ≤ Cβ k ε (1.2) for 0 < s < β < 1. Then the C 1,α -regularity of the free boundary in a neighborhood of the origin follows via a standard iteration argument. From the C 1,α -regularity of the free boundary it follows that ∆u ∈ C 1,α up to the free boundary. We move further and show that u is C 3,α up to the free boundary. Thus a solution u ∈ B ̺ κ (ε) is locally C 2,1 , which is the best regularity that a solution may achieve. 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, 1 2 but is not C 2,α for α > 1 2 . Hence C 2, 1 2 is the best regularity that a solution may achieve in dimension n ≥ 2.

The obstacle problem for the biharmonic operator
In this section we show that there exists a unique solution to the biharmonic obstacle problem. Furthermore we show that the solution is locally W 3,2 ∩ C 1,α . Proof. Here we use the standard terminology from [5]. Let us start with an observation that the functional J is weakly lower semicontinuous, i.e. given a sequence {u k } converging weakly to a function u ∈ W 2,2 (Ω), then

Existence, uniqueness and W 3,2 -regularity of the solution
(2.1) Upon passing to a subsequence, we may assume that According to the definition of weak convergence in W 2,2 (Ω), ∆u k converges to ∆u weakly in L 2 (Ω), hence lim k→∞ˆ∆ u k ∆u =ˆ(∆u) 2 , and the inequalitŷ after passing to a limit as k → ∞, we get the desired inequality, (2.1). Next we take a minimizing sequence {u k } ⊂ A , and show that it converges weakly to some function u in W 2,2 (Ω) through a subsequence, and that u is an admissible function. Define Let us note that J[u k ] = ∆u k 2 L 2 , so ∆u k is bounded in L 2 , and since u k −ω = 0 and ∂(u k −ω) ∂n = 0 on ∂Ω in the trace sense for any fixed ω ∈ A , the sequence is bounded in W 2,2 (Ω). Hence it has a subsequence which converges weakly in W 2,2 , we will keep the notation, call it {u k }. We want to show that the limit function u ∈ A . According to the Sobolev embedding theorem {u k } converges to u strongly in L 2 up to a subsequence, hence upon passing to a new subsequence u k → u a.e. in Ω. The latter proves that u ≥ 0 a.e..
It remains to show that u satisfies the boundary conditions. For any ω ∈ A , u k − ω ∈ W 2,2 0 (Ω), since W 2,2 0 (Ω) is a closed, linear subspace of W 2,2 (Ω), it is weakly closed, according to Mazur's theorem ( [5], pp. 471 and 723). This proves that u − ω ∈ W 2,2 0 (Ω) and therefore u ∈ A . According to (2.1), m ≥ J[u], but the reversed inequality is also true since u is admissible and according to our choice of the sequence {u k }. Thus m = J[u], and u is a minimizer.
The uniqueness of the minimizer follows from the convexity of the functional: assuming that both u and v are minimizers, it follows that u+v 2 is also admissible, so but the reversed inequality is also true with equality if and only if ∆u = ∆v. Thus if u and v are both minimizers in A then ∆(u−v) = 0 and u−v ∈ W 2,2 0 (Ω), which implies that u = v in Ω. Now we turn our attention to the regularity of the solution to the biharmonic obstacle problem. Proposition 2.2. Let u be the solution to the biharmonic obstacle problem in the unit ball B 1 , then where the constant C depends only on the space dimension.
Proof. The proof is based on a difference quotient method. Let {e 1 , e 2 , ..., e n } be the standard basis in R n . For a fixed i ∈ {1, 2, ..., n} denote Take a nonnegative function ζ ∈ C ∞ 0 (B 3 4 ), such that ζ ≡ 1 in B 1 2 . Then for small values of the parameter t > 0, the function u + tζ 2 (u i,h − u) is admissible for the biharmonic obstacle problem in is small, and obviously it satisfies the same boundary conditions as the minimizer u. Hencê Assuming that h < 1 4 , the inequality will still hold if we replace the integration over the ball B 1 by B 1−h , since ζ is zero outside the ball B 3 4 . It is clear that u i,h is the solution to the biharmonic obstacle problem in After dividing both sides of the inequalities (2.3) and (2.4) by t, and taking the limit as t → 0, we getˆB We rewrite inequalities (2.5) and (2.6) explicitly, that iŝ After summing the inequalities above, we obtain Dividing both sides of the last inequality by h 2 , we get First let us study the first integral on the right side of (2.7) 8) where we applied Hölder's inequality, and used the fact that the L 2 -norm of the first and second order difference quotients of a function u ∈ W 2,2 are uniformly bounded by its W 2,2 -norm.
Next we estimate the absolute value of the second integral in (2.7) where we applied Cauchy's inequality.
Combining inequalities (2.7), (2.8) and (2.9), we obtain According to our choice of function ζ, so the L 2 -norm of the difference quotients of ∆u is uniformly bounded in B 1 2 hence ∆u ∈ W 1,2 (B 1 2 ), and where the constant C depends only on the function ζ, and can be computed explicitly, depending only on the space dimension.
Corollary 2.3. Assume that Ω is a bounded open set in R n . Then the solution to the obstacle problem is in W 3,2 (K) for any K ⊂⊂ Ω, and where the constant C depends on the space dimension n and on dist(K, ∂Ω).
Let us observe that the function u + tζ(u i,h − u) is well defined and nonnegative in Ω for any 0 < t < 1 ζ L ∞ , and it satisfies the same boundary conditions as u. Thereforê after dividing the last inequality by t, and taking the limit as t → 0, we obtain K ∆u∆(ζ(u i,h − u)) ≥ 0. (2.12) Note that u i,h is the solution to the biharmonic obstacle problem in K, and after dividing the last inequality by t, and taking the limit as t → 0, we obtain Inequalities (2.12) and (2.13) imply that dividing the last inequality by h 2 , and taking into account that u ∈ W 3,2 loc , we may pass to the limit as |h| → 0 in (2.14), and conclude that K ∆u xi ∆(ζu xi ) ≤ 0.

C 1,α -regularity of the solution
It has been shown in [1], Theorem 3.1 that ∆u ∈ L ∞ loc for the solution to the biharmonic obstacle problem with nonzero obstacle and zero boundary conditions. In this section we show that the statement remains true in our setting, with a quantitative estimate of ∆u L ∞ .
Lemma 2.5. The solution to the biharmonic obstacle problem satisfies the following equation in the distribution sense where µ u is a positive measure on Ω.
The next lemma is a restatement of the corresponding result in [1], Theorem 2.2.

Lemma 2.7.
Let Ω ⊂ 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 µ u = ∆ 2 u in Ω, then Proof. The detailed proof of Lemma 2.7 can be found in the original paper [1] and in the book [2](pp. 92-94), so we will provide only a sketch, showing the main ideas. Extend u to a function in W 2,2 loc (R n ), and denote by u ε the ε-mollifier of u.
is nonnegative and it satisfies the same boundary conditions as u. Hencê after passing to the limit in the last inequality as ε → 0, we obtain Then by Green's formula, Then it follows from the convergence of the mollifiers and the upper semicontinuity of ω, that ω(x 0 ) ≥ 0, for any x 0 ∈ S.
Knowing that ∆u is a subharmonic function, and ω ≥ 0 on the support of ∆ 2 u, we can show that ∆u is locally bounded (Theorem 3.1 in [1]).
Theorem 2.8. Let u be the solution to the biharmonic obstacle problem with zero obstacle in Ω, B 1 ⊂⊂ Ω. Then where the constant C > 0 depends on the space dimension n and on dist(B 1 , ∂Ω).
Proof. The detailed proof of the theorem can be found in the original paper [1], Theorem 3.1, and in the book [2], pp. 94-97. Here we will only provide a sketch of the proof.
Corollary 2.9. Let u be the solution to the biharmonic obstacle problem in Ω.
Then u ∈ C 1,α loc , for any 0 < α < 1, and where the constant C depends on the space dimension and dist(K, ∂Ω).
Proof. It follows from Theorem 2.8 via a standard covering argument, that Then inequality (2.24) follows from the Calderón-Zygmund inequality and the Sobolev embedding theorem.
According to Corollary 2.9, u is a continuous function in Ω, and therefore Ω u := {u > 0} is an open subset of Ω. We define the free boundary It follows from our discussion that the measure µ u = ∆ 2 u is supported on Γ u .

Regularity of the free boundary
In this section we investigate the regularity of the free boundary Γ u , under the assumption that the solution to the biharmonic obstacle problem is close to the one-dimensional solution 1 6 (x n ) 3 + .

One-dimensional solutions
Here we study the solution to the biharmonic obstacle problem in the interval (0, 1) ⊂ R.
Proof. Let u 0 be the minimizer to the given biharmonic obstacle problem. If 0 < x 0 < 1, and u 0 (x 0 ) > 0, then´u ′′ 0 η ′′ = 0, for all infinitely differentiable functions η compactly supported in a small ball centered at x 0 . Hence the minimizer u 0 has a fourth order derivative, u Therefore u 0 is a piecewise polynomial of degree less than or equal to three. Denote by γ ∈ (0, 1] the first point where the graph of u 0 hits the x-axes. Our aim is find the explicit value of γ. Then we can also compute the minimizer u 0 . Observe that u 0 (γ) = 0, and u ′ 0 (γ) = 0, since u ′ 0 is an absolutely continuous function in (0, 1). Taking into account the boundary conditions at the points 0 and γ, we can write u 0 (x) = ax 3 We see that the point γ is a zero of second order for the third order polynomial u 0 , and u 0 ≥ 0 in (0, γ]. That means the third zero is not on the open interval showing that the function F is decreasing, so it achieves minimum at the point γ = − 3 λ . Therefore we may conclude that and γ = − 3 λ is a free boundary point. Observe that u ′′ (γ) = 0, and u ′′ is a continuous function, but u ′′′ has a jump discontinuity at the free boundary point γ = − 3 λ .
The example above characterizes one-dimensional solutions. It also tells us that one-dimensional solutions are C 2,1 , and in general are not C 3 .

The class B ̺ κ (ε) of solutions to the biharmonic obstacle problem
Without loss of generality, we assume that 0 ∈ Γ u , and study the regularity of the free boundary, when u ≈ 1 6 (x n ) 3 + . Let us start by recalling the definition of non-tangentially accessible domains, [3]. 3. Harnack chain condition; if ǫ > 0 and P 1 , P 2 ∈ D, dist(P i , ∂D) > ǫ, and |P 1 − P 2 | < Cǫ, then there exists a Harnack chain from P 1 to P 2 whose length l depends on C, but not on ǫ, l = l(C). A Harnack chain from Let us define rigorously, what we mean by u ≈ 1 6 (x n ) 3 + .
Definition 3.3. Let u ≥ 0 be the solution to the biharmonic obstacle problem in a domain Ω, B 2 ⊂⊂ Ω and assume that 0 ∈ Γ u is a free boundary point. We say that u ∈ B ̺ κ (ε), if the following assumptions are satisfied: and with a function l, indicating the length of a Harnack chain.

There exists
4. We have the following normalization

6)
and we also assume that
In the notation of the class B ̺ κ (ε) 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 3.5 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.
Proof. We argue by contradiction. Assume that there exist σ 0 > 0 and a sequence of solutions, u j ∈ B ̺ κ (ε j ), such that According to assumption 4 in Definition 3.3, D 3 u j L 2 (B2) < κ 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 u j W 3,2 (B2) ≤ C(̺, n)κ. Hence up to a subsequence u j ⇀ u 0 weakly in W 3,2 (B 2 ), u j → u 0 strongly in W 2,2 (B 2 ) and according to Corollary 2.9 u j → u 0 in C 1,α (B 3/2 ). Hence This implies that u 0 is a 1-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 where c 1 , c 2 ≥ 0 and −2 ≤ a 1 ≤ a 2 ≤ 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 (3.9), we need to show that u j → u 0 = 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 Denote u j n := ∂u j ∂xn ∈ W 2,2 (B 2 ), j ∈ N 0 , and let ζ ∈ C ∞ 0 (B 3 2 ) be a nonnegative function, such that ζ ≡ 1 in B 1 . According to Lemma 2.4, and therefore lim sup where in the last step we used integration by parts.
Therefore up to a subsequence P j → P 0 , hence r̺ ≤ |P 0 | ≤ r, B r ′ (P 0 ) ⊂ Ω j , for all j large enough, where 0 < r ′ < r̺ is a fixed number. Since we have chosen r < a/2, we may conclude that Thus ∆u j is a sequence of harmonic functions in the ball B r ′ (P 0 ), and therefore ∆u j → 0 locally uniformly in B r ′ (P 0 ), (3.15) according to (3.10). Let Q := e n , then u 0 (Q) = c(1 − a) 3 > 0, since u j → u 0 uniformly in B 3/2 , we see that u j (Q) > 0 for large j, and Q ∈ Ω j . Therefore there exists a Harnack chain connecting P 0 with Q; {B r1 (x 1 ), B r2 (x 2 ), ..., B r l (x l )} ⊂ Ω j , whose length l does not depend on j. Denote by K j := ∪ i B ri (x i ) ⊂⊂ Ω j , and let V j ⊂⊂ K j ⊂⊂ Ω j where V j is a regular domain, such that dist(K j , ∂V j ) and dist(V j , ∂Ω j ) depend only on r and ̺.
Let w j + be a harmonic function in V j , with boundary conditions w j + = (∆u j ) + ≥ 0 on ∂V j , then w j + − ∆u j is a harmonic function in V j , and w j Let us observe that ∆u j → ∆u 0 = 6c(x n − a) + implies that (∆u j ) − L 2 (B2) → 0. Since (∆u j ) − is a subharmonic function in Ω j , and V j ⊂⊂ Ω j it follows that So w j + is a nonnegative harmonic function in V j , and by the Harnack inequality if j is large, where C H is the constant in Harnack's inequality, it depends on ̺ and r but not on j. Denote C(a, c) := 3c(1 − a) > 0 by (3.14). Applying the Harnack inequality again, we see that Inductively, we obtain that where l does not depend on j. Hence w j + (P 0 ) ≥ C(a,c) C l H for all j large, and according to (3.16), the latter contradicts (3.15). Therefore we may conclude that a = 0. Recalling that D 3 u 0 L 2 (B1) = ω n , we see that c = 1 6 , but then we obtain u j → 1 6 (x n ) 3 + in W 3,2 (B 1 ) which is a contradiction, since we assumed (3.9). Lemma 3.4 has an important corollary, which will be very useful in our later discussion.
Proof. Once again we argue by contradiction. Assume that there exist t 0 > 0 and a sequence of solutions u j ∈ B ̺ κ (ε j ), ε j → 0, such that x j ∈ B 2 ∩ Γ j , and x j n < −t 0 . For 0 < r < min(̺, t 0 /2) choose P j = P (r, x j ) ∈ Ω j as in the corkscrew condition, Upon passing to a subsequence, we may assume that P j → P 0 . Fix 0 < r ′ < r̺, then for large j B r ′ (P 0 ) ⊂⊂ Ω j ∩ {x n < 0}.
Hence ∆u j is a sequence of harmonic functions in B r ′ (P 0 ). According to Lemma 3.4, u j → 1 6 (x n ) 3 + , and therefore ∆u j → 0 in B r ′ (P 0 ), and ∆u j (e n ) → 1. Since Ω j is an NTA domain, there exists a Harnack chain connecting P 0 with Q := e n ∈ Ω j ; {B r1 (x 1 ), B r2 (x 2 ), ..., B r k (x k )} ⊂ Ω j , whose length does not depend on j. Arguing as in the proof of Lemma 3.4, we will obtain a contradiction to ∆u j → 0 in B r ′ (P 0 ).

Linearization
Let {u j } be a sequence of solutions in Ω ⊃⊃ B 2 , u j ∈ B ̺ κ (ε j ), and assume that ε j → 0 as j → ∞. It follows from Lemma 3.4, that up to a subsequence Let us denote Without loss of generality we may assume that δ j i > 0, for all j ∈ N. Indeed, if δ j i = 0 for all j ≥ 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 δ j i > 0 for all j. Denote then v j i W 2,2 (B2) = 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 → v 0 i in W 2,2 , at least locally.
Let v j i be the sequence given by (3.19), and assume that v j i ⇀ v 0 i weakly in W 2,2 (B 2 ), strongly in W 1,2 (B 2 ), for i = 1, ..., n − 1, then Proof. Denote by Ω j := Ω u j , Γ j := Γ u j . It follows from Corollary 3.5 that v 0 2 is an open subset, then K ⊂ Ω j for large j by (3.18). Hence ∆ 2 v j i = 0 in K, and therefore ∆ 2 v 0 i = 0 in B + 2 , and (3.20) is proved. Now let us proceed to the proof of the strong convergence. Let ζ ∈ C ∞ 0 (B 3 2 ) be a nonnegative function, such that ζ ≡ 1 in B 1 . It follows from (3.20 (3.23) and therefore lim sup . From the last inequality and (3.22) we may conclude that lim sup On the other hand lim inf follows from the weak convergence ∆v j i ⇀ ∆v 0 i in L 2 (B 2 ), and we may conclude from (3.24) and (3.25) that Hence we obtain ∆v j i − ∆v 0 i L 2 (B1) → 0, and therefore v j i → v 0 i in W 2,2 (B 1 ) according to the Calderón-Zygmund inequality.
Let v j i be the corresponding sequence, given by (3.19). The inequality (3.27) implies that After passing to the limit in (3.28), we obtain

Properties of solutions in a normalized coordinate system
Let us define , for x 0 ∈ Γ u , x ∈ B 2 , r ∈ (0, 1), (3.29) and u r := u r,0 . First we would like to know how fast ∇ ′ ∆u r L 2 (B1) decays with respect to ∇ ′ ∆u L 2 (B1) , for r < 1. In particular, it is well known that an inequality for some 0 < s, τ < 1 would provide good decay estimates for ∇ ′ ∆u(s k x) L 2 (B1) , k ∈ N. By choosing a suitable coordinate system, we succeed to show a weaker version of the desired inequality. This weaker version of (3.30) is good enough to perform an iteration argument to prove the regularity of the free boundary. First let us observe that 1 6 (η · x) 3 + ∈ B ̺ κ (ε) if |η − e n | ≤ C n ε, for some dimensional constant C n .
A minimizer η always exists for a function u ∈ B ̺ κ (ε), and since ∇ ′ −η = ∇ ′ η , −η is also a minimizer, thus we always choose a minimizer satisfying the condition e n · η ≥ 0. A normalized coordinate system always exists by choosing η = e n in the new coordinate system.
Let us also observe that for every η ∈ R n , and Lemma 3.9. Assume that u ∈ B ̺ κ (ε) solves the biharmonic obstacle problem in a fixed coordinate system with basis vectors {e 1 , ..., e n }. Let {e 1 1 , ..., e 1 n } be a normalized coordinate system with respect to u, and assume that e 1 n · e n ≥ 0. Then Proof. According to Definition 3.8, It follows from the triangle inequality that ∂∆u ∂x n − (e n · e 1 n ) 2 ∂∆u ∂x n L 2 (B1) ≤ ∂∆u ∂x n − (e n · e 1 n )(e 1 n · ∇∆u) + (e n · e 1 n )(e 1 n · ∇∆u) − (e n · e 1 n ) 2 ∂∆u ∂x n L 2 (B1) according to (3.31), and taking into account that 0 ≤ e n · e 1 n ≤ 1. Note that Lemma 3.4 implies that ∂∆u ∂xn L 2 (B1) is uniformly bounded from below, and therefore by choosing ε > 0 small, we may conclude from (3.32) that Since 0 ≤ e n · e 1 n ≤ 1, we get Denote by (e 1 n ) ′ := e 1 n − e n (e n · e 1 n ). It follows from the triangle inequality and (3.31) that .
Proof. According to the Cauchy-Schwarz inequality, it is enough to show that the inequality holds for any i ∈ {1, ..., n − 1}, provided ε is small enough. We argue by contradiction. Assume that there exist 0 < s < τ < 1 and δ 0 > 0 for which there exists a sequence of solutions {u j } ⊂ B ̺ κ (ε j ) in a normalized coordinate system, such that ε j → 0, as j → ∞, but for some i ∈ {1, 2, ..., n − 1} Let v j i be given by (3.19), then according to Lemma 3.
and after passing to the limit as j → ∞, we obtain ∆v 0 i L 2 (B1) ≥ s n 2 δ 0 . According to Lemma 3.6, ∆v 0 i is a harmonic function in {x n > 0} ∩ B 1 . Therefore it may be written as a sum of homogeneous orthogonal harmonic polynomials where m indicates the degree of the polynomial a m i . Next we show that a 0 i = 0, using the definition of a normalized coordinate system. Consider the following direction where λ j > 0, is chosen so that |η j i | = 1, and since δ j i → 0, we may conclude that λ j → 1.
By the normalization of the coordinate system, Definition 3.8, it follows that or equivalently Expanding (3.37) in coordinates, and taking into account that ∂∆u j ∂xi = δ j i ∆v j i , we get .
The substitution .
Dividing the last inequality by (δ j i ) 2 , and taking into account . Hence .
(3.38) Lemma 3.4 implies that ∂∆u j ∂xn → χ B + 1 in L 2 (B 1 ), and according to Lemma 3.6, ∆v j i → ∆v 0 i in L 2 (B 1 ) as j → ∞, and v 0 i = 0 in B 1 \ B + 1 . Therefore we may pass to the limit in the inequality (3.38), and obtain and therefore for any 0 < s < 1 Then according to the strong convergence ∆v j i L 2 (B1) → ∆v 0 i L 2 (B1) , and taking into account that ∆v 0 for j > 1 large enough. Hence we may conclude that contradicting (3.36).

C 1,α -regularity of the free boundary
In this section we perform an iteration argument, based on Proposition 3.7, Proposition 3.10, and Lemma 3.9, that leads to the existence of the unit normal η 0 of the free boundary at the origin, and provides good decay estimates for ∇ ′ η0 u r W 2,2 (B1) . First we would like to verify that u ∈ B ̺ κ (ε) imply that u s ∈ B ̺ κ (Cε). 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 ∈ ∂D, then for any 0 < s < 1 the set D s := s −1 (D ∩ 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 3.5. Indeed, let t = s in Corollary 3.5, 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 then assumption 4 also holds. Indeed, D 3 U s L 2 (B1) = ω n by definition of U s , and according to Lemma 3.4 provided ε = ε(n, κ, s) is small.
In the next lemma we show that U s ∈ B ̺ κ (τ ε) in a normalized coordinate system, then we argue inductively to show that U s k ∈ B ̺ κ (Cβ k ε), β < 1.
(3.47) Now let us consider the sequence of numbers {A k } k∈N0 , defined as follows: and The assumption u ∈ B ̺ κ (ε) implies that Let us also observe that according to Lemma 3.4, since A 0 is small. Hence the inequality (3.47) implies that Thus A 1 ≤ βA 0 . We use induction to show that for a fixed γ < β < 1/2. Assuming that (3.49) holds up to k ∈ N, we will show that A k+1 ≤ β k+1 A 0 . The proof is quite long and technical. Recalling our notation (3.40) for U s k , and notation (3.48) for A k , we see that , ..., e k−1 n }, and ε 0 = (λ + 1)ε is small. By definition, {e k 1 , ..., e k n } is a normalized coordinate system with respect to U s k ∈ B ̺ κ (β k ε 0 ). The definition of a normalized coordinate system, and inequality (3.43) imply that (3.51) Applying Proposition 3.7 for the function U s k ∈ B ̺ κ (β k ε 0 ), we obtain It follows from Proposition 3.10 and our choice of the coordinate system {e k 1 , ..., e k n }, that where δ, c(δ) and δ 0 are the same numbers as before. Combining inequalities (3.51), (3.52) and (3.53), we derive the following estimate for A k+1 , It follows from the triangle inequality and Lemma 3.9 that The last inequality and (3.54) imply that In order to complete the induction argument, we observe that according to Lemma 3.4, since U s k ∈ B ̺ κ (β k ε 0 ) and β k ε 0 < ε 0 is small. Finally we obtain from (3.55) and (3.56) that this completes the proof of inequality (3.49).

On the regularity of the solution
In this section we study the regularity of the solution to the biharmonic obstacle problem. Assuming that u ∈ B ̺ κ (ε), with ε > 0 small, we derive from Theorem 3.12 that u ∈ C 2,1 loc (B 1 ). In the end we provide an example showing that without the NTA domain assumption, there exist solutions, that are not C 2,1 .

4.1
C 2,1 -regularity of the solutions in B ̺ κ (ε) After showing the C 1,α -regularity of the free boundary Γ u ∩ B 1 , we may go further to derive improved regularity for the solution u ∈ B ̺ κ (ε).

4.2
In general the solutions are not better that C 2, 1 2 Let us observe that the assumption u ∈ B ̺ κ (ε) is essential in the proof of u ∈ C 2,1 (B r ). The next example shows that without our flatness assumptions there exists a solution to the biharmonic obstacle problem in R 2 , that do not possess C 2,1 -regularity. then u ∈ C 2, 1 2 is the solution to the biharmonic zero-obstacle problem in the unit ball B 1 ⊂ R 2 .