Optimal Regularity and Structure of the Free Boundary for Minimizers in Cohesive Zone Models

We study optimal regularity and free boundary for minimizers of an energy functional arising in cohesive zone models for fracture mechanics. Under smoothness assumptions on the boundary conditions and on the fracture energy density, we show that minimizers are \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^{1, 1/2}$$\end{document}C1,1/2, and that near non-degenerate points the fracture set is \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^{1, \alpha }$$\end{document}C1,α, for some \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha \in (0, 1)$$\end{document}α∈(0,1).


Introduction
In recent years, a variational formulation of fracture evolution has been proposed by Francfort and Marigo [21], and later developed by Dal Maso and Toader [17], and Dal Maso, Francfort, and Toader [15,16] (see also [22] and the references therein, for a variational theory of rate independent processes). Such evolution is based on the idea that at any given time the configuration of the elastic body is an absolute minimiser of the energy functional (see also [4,11,14,18] in the context of plasticity where, more in general, critical points of the energy are allowed).
In this paper we study optimal regularity and free boundary for minimizers of an energy functional arising in cohesive zone models for fracture mechanics. Such models describe the situation in which the energy density of the fracture depends on the distance between the lips of the crack (see for instance [4,[11][12][13]19]). We consider the energy functional associated to an elastic body occupying the open strip R n × (−A, A), with n ≥ 2 and A > 0. Denoting a generic point z ∈ R n × (−A, A) by (x, y), with x ∈ R n and y ∈ (−A, A), we shall consider deformations ensuring that cracks can only appear on the hyperplane {y = 0}. The assumption of confining fractures to a given hyperplane is a standard simplification that avoids some technical difficulties but does not prevent the crack set from being irregular, thus keeping the main features of the problem.
We consider the situation in which the elastic body can only undergo deformations that are parallel to a fixed given direction lying on {y = 0}. In this way, the displacement can be represented by a scalar function v : R n × (−A, A) → R. According to Barenblatt's cohesive zone model [7], the energy associated to a displacement v ∈ H 1 (R n × (−A, A)\{y = 0}) is given by where v RT and v LT are the right and left traces on {y = 0} of v | R n ×(0,A) and v | R n ×(−A,0) , respectively, and g ∈ C 2 [0, ∞)∩C 3 (0, ∞) is strictly increasing, bounded, with g(0) = 0 and g (0 + ) ∈ (0, +∞). The parameter g (0 + ) has an important physical meaning, and it can be identified with the maximal sustainable stress of the material along {y = 0}, see [11,Theorem 4.6]. A critical point u of (1.1) with boundary conditions u A , u −A satisfies (see [ where sgn(·) denotes the sign function. Note that, because ∂ y u RT = ∂ y u LT , we can use the notation ∂ y u to denote the y derivative of u on {y = 0} without paying attention to the side on which the derivative is computed. For simplicity, we will assume that u A (x) = −u −A (x) for every x ∈ R n , and we will focus on solutions that are odd with respect to the hyperplane {y = 0}. In this situation, our problem reduces to the study of a function u ∈ H 1 (R n × (0, A)) satisfying ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ u = 0 i n R n × (0, A), u = u A on {y = A}, |∂ y u| ≤ g (0 + ) on {y = 0}, ∂ y u = g (2|u|) sgn(u) on {y = 0} ∩ {u = 0}, (1.2) where we used the notation u(x, 0) = u RT (x, 0) for every x ∈ R n . In this setting, the crack K u is represented by the discontinuity set of u, and is given by We assume that the boundary condition u A satisfies the following: for some β ∈ (0, 1) and lim |x|→∞ u A (x) = 0.
( 1.4) Under these assumptions, we want to the study optimal regularity of the restriction of u to R n ×[0, A], and the regularity of the free boundary ∂ K u (where the boundary is defined in the topology of R n × {0}).
boundary point coming from the positive phase, we subtract from u the linear function g (0 + )y, and then we reflect evenly with respect to the hyperplane {y = 0}, defining v(x, y) := u(x, y) − g (0 + )y for every (x, y) ∈ R n × (0, A), v(x, −y) for every (x, y) ∈ R n × (−A, 0). (1.5) Then, inspired by [9], we prove a variant of Almgren's monotonicity formula. More precisely, suppose that (0, 0) ∈ ∂ K u , and set v (r ) := r d dr log max{F v (r ), r n+4 } , where B r is the ball of R n+1 centred at 0 with radius r , and H n denotes the Hausdorff n-dimensional measure. We show that there exists C > 0 such that for r sufficiently small the function r → v (r )e Cr is nondecreasing (see Proposition 5.1). This implies that v (0 + ) exists, and we can show that either v (0 + ) = n + 3, or v (0 + ) ≥ n + 4 (see Proposition 6.1). This allows us to classify subquadratic blow up profiles of v: more precisely, considering the family {v r } r >0 of functions we can classify the possible limits as r → 0 + provided d r r 2 → +∞. In other words, provided v decays slower than quadratic, we obtain the following theorem, which is the second main result of the paper: Theorem 1.2. Let the assumptions of Theorem 1.1 be satisfied, and let u ∈ H 1 (R n ×(0, A)) be a solution of (1.2). Suppose that (0, 0) ∈ ∂ K u , with u(·, 0) ≥ 0 near (0, 0), and let v be defined by (1.5). If lim inf r →0 + d r r 2 = +∞, (1.6) then the free boundary ∂ K u is of class C 1,α near (0, 0), for some α ∈ (0, 1).
To prove Theorem 1.2 we show that (1.6) implies that v attains its smallest possible value, namely v (0 + ) = n + 3, and that in this case blow up profiles of v are homogeneous solutions of the classical Signorini problem (that is the classical thin obstacle problem), with homogeneity degree 1/2( v (0 + ) − n). Thanks to this fact, the blow ups can be easily classified (see Proposition 6.2) and the result follows as in the classical theory. The paper is organised as follows. In Section 2 we introduce the notation and the setting of the problem. We show basic regularity properties of the solution u in Section 3, while Section 4 is devoted to the separation of phases and the optimal regularity. Frequency formula is the subject of Section 5, and in Section 6 we study blow up profiles. Finally, in Section 7 we prove the regularity of the free boundary.

Notation
In this brief section we introduce the notation that will be used, and we give the main assumptions. Throughout the paper, we fix n ∈ N, with n ≥ 2, and A > 0. For every point z ∈ R n × [−A, A] we will write z = (x, y), with x ∈ R n and y ∈ [−A, A]. The canonical basis of R n+1 is denoted by e 1 , . . . , e n+1 . For a, b ∈ R n+1 , a · b denotes the Euclidean scalar product between a and b, and | · | denotes both the absolute value in R and the Euclidean norm in R n or R n+1 , depending on the context. For every k ∈ N, H k stands for the Hausdorff k-dimensional measure. If z = (x, y) ∈ R n+1 and r > 0, we will denote by B r (z) the ball of R n+1 centered at z with radius r : and with B n r (x) the ball of R n centered at x with radius r : We will write B r and B n r for B n r (0) and B n r (0), respectively, and we will use the notation S n := ∂ B 1 and S n−1 := ∂ B n 1 , while ω n+1 denotes the (n +1)-dimensional Lebesgue measure of B 1 .
Throughout all the paper, C will denote a universal constant, possibly different from line to line. For any function v ∈ H 1 (R n ×(−A, A)\{y = 0}), we will denote by v RT and v LT the right and left traces on {y = 0} of v | R n ×(0,A) and v | R n ×(−A,0) , respectively, while we set v + := max{v, 0} and v − := min{v, 0}, When v is sufficiently regular, ∇v and D 2 v stand for the gradient and the Hessian of v, while ∇ x v and D 2 x x v are the gradient and the Hessian of the function x → v(x, y). We will say that v is homogeneous of degree μ if v can be written as v(z) = |z| μ h z |z| , Also, f is said to be semiconvex, with semiconvexity constant D 0 , if for every x, h ∈ R n . Similarly, we say that f is semiconcave, with semiconcavity constant D 0 , if for every x, h ∈ R n . We are now ready to state our assumptions. In the following, g ∈ C 2 [0, ∞) ∩ C 3 (0, ∞) is strictly increasing and bounded, with g(0) = 0 and g (0 + ) ∈ (0, +∞). We assume, in addition, that 2 g L ∞ < 1/A and g L ∞ < ∞, where g L ∞ and g L ∞ denote the L ∞ -norms of g and g , respectively. Moreover, we assume that u A : R n → R satisfies (1.4), that is Remark 2.1. The assumptions above imply, in particular, that u A is Lipschitz continuous with Lipschitz constant L A := ∇u A L ∞ . Moreover, denoting by λ min (x) and λ max (x) the smallest and largest eigenvalue of D 2 u A (x), respectively, we have that u A is semiconvex with semiconvexity constant D A := (λ min ) − L ∞ , and is semiconcave with semiconcavity constant C A := (λ max ) + L ∞ .
We will study optimal regularity and free boundary for a function u ∈ H 1 (R n × (0, A)) solving Equation (1.2): Note that the equation above implies that Also, by the maximum principle, In the next section we prove some basic regularity properties of u.

Basic Properties of the Solution
We study in this section the basic regularity properties of a solution u of Equation (1.2). We start by showing that condition 2 g L ∞ < 1/A implies uniqueness. Lemma 3.1. Let u A satisfy (1.4), and let g ∈ C 2 [0, ∞) be strictly increasing and bounded, with g(0) = 0 and g (0 + ) ∈ (0, +∞). If 2 g L ∞ < 1/A, then there exists a unique u ∈ H 1 (R n × (0, A)) solving (1.2). In particular, there is a unique critical point of (1.1) that coincides with the global minimizer.
Proof. Suppose, by contradiction, that there exist u 1 , u 2 ∈ H 1 (R n × (0, A)) solutions of (1.2), with u 1 ≡ u 2 . In particular, since u 1 = u 2 on {y = A}, this implies We will prove the statement into two steps.
Step 1: We show that Using the weak formulation of the equation (see [11,Proposition 3.1]) we havê Analogously, using the weak formulation of the equation for u 2 , with test function Adding together the last two relations, we obtain We now observe that where we also used the fact that |u 2 − u 1 | = | |u 2 | − |u 1 | | whenever u 1 u 2 > 0.
First of all, note that for every x ∈ R n and i = 1, 2.
We now show that x −→ u(x, 0) is infinitesimal as |x| → ∞. Define now, for every k ∈ N, the function u k : Since u k is harmonic for every k ∈ N and {u k } k∈N is uniformly bounded in R n × [0, A] and u k H 1 (R n ×(0,A)) ≤ C, up to subsequences we have for some harmonic function u : where we also used the fact that u k u weakly in H 1 loc (R n × (0, A)). Since u(·, A) ≡ 0 this implies u ≡ 0, which contradicts the fact that u(0, 0) = a = 0.
We now prove that the crack set K u defined in (1.3) is bounded. Proposition 3.3. Let u A and g be as in Theorem 1.1, and let u ∈ H 1 (R n × (0, A)) be a solution of (1.2). Then, u(·, 0) has compact support. Proof. We start by showing that there exist positive constants R = R(g, A), c = c(g, A), and r = r (g, A) ∈ (0, 1) with the following property: if x 1 ∈ R n is such that |x 1 | > R and u(x 1 , 0) = 0, then Before proving the claim, let us show that this implies the conclusion. Indeed, suppose by contradiction that the support of u(·, 0) is not bounded. Then, there exists a sequence {x k } k∈N ⊂ R n with |x k | → ∞ such that |x k | > R and u(x k , 0) = 0 for every k ∈ N. By (3.5), for every k ∈ N there exists z k ∈ B n 1 (x k ) such that Without loss of generality, we can assume that |x j − x k | ≥ 4 for every j = k, so that the balls {B n r (z k )} k∈N are pairwise disjoint. Therefore, Let us now show the claim. By Lemma 3.2, Let V : B n 1 × [0, A] → R be the solution of the following problem: and let a = a(g, A) > 0 be so small that (3.7) By (3.6), there exists a constant R = R(g, u A ) > 2 such that , for every x with |x| > R − 2. (3.8) Let x 1 ∈ R n be such that |x 1 | > R and u(x 1 , 0) > 0 (the case u(x 1 , 0) < 0 can be treated in the same way). We will show that there exist z 1 ∈ B n 1 (x 1 ), c > 0, and r ∈ (0, 1) such that (3.5) holds true.
We then have only two possibilities.
Case i: y = 0. Let us show that this is not possible. First of all, note that in this case it must be |x − thanks to (3.8). Thus, using (3.7) and the fact that u( which gives a contradiction.
Case ii: 0 < y < A and |x − x 1 | = 1. Let us show that, for a sufficiently small, there exists a positive constant From (1.2) and (1.4) it follows that for every x ∈ R n for some positive constant C 0 > g (0 + ). Therefore, setting C 1 := (C 0 −g (0 + ))/A, by the maximum principle (note that ∂ y u is harmonic) where we used (3.8). The above inequality implies Analogously, we have We can now show (3.5). At the contact point, we have Then, by Harnack inequality and by (3.9), there exists a radius r = r (c 1 ) ∈ (0, 1) such that The inequality above implies that for every Integrating with respect to x, we obtain . Setting the claim follows.
We now show that, under the assumption 2 A g L ∞ < 1, the Lipschitz continuity of u A implies the Lipschitz continuity of u(·, y), uniformly with respect to y.

Lemma 3.4. Let u A and g be as in Theorem
Proof. Let h ∈ R n \{0}, α > 0, and define for every C > 0 Let us first show that the claim proves the lemma. Indeed, if (3.11) is true then for Since x, y and h are arbitrary, from the last inequality and letting α thus concluding. Let us now prove the claim. By maximum principle and thanks to (3.6), there In the following we assume C α h > 0, since otherwise (3.11) is trivially satisfied. We have two possibilities.
Case 2: y = 0. We conclude the proof of the lemma, showing that for α sufficiently large this case is impossible. At the contact point, the following equality holds true: We consider now three possible subcases, in which we will always reach a contradiction.
This follows as in case 2a: This proves the claim and, in turn, the lemma.
We now show a property that implies the semiconvexity of u + (·, y), for any y ∈ [0, A].
Lemma 3.5. Let u A and g be as in Theorem 1.1, and let u ∈ H 1 (R n × (0, A)) be a solution of (1.2). Then, there exists D > 0 such that for every (x, y) ∈ R n ×[0, A], In particular, for every y ∈ [0, A] the function u + (·, y) is semiconvex, with semiconvexity constant D.
An analogous result holds true for u − . Lemma 3.6. Let u A and g be as in Theorem 1.1, and let u ∈ H 1 (R n × (0, A)) be a solution of (1.2). Then, there exists C > 0 such that for every (x, y) ∈ R n × [0, A], In particular, for every y ∈ [0, A] the function u − (·, y) is semiconcave, with semiconcavity constant C.
The following remark will be useful in the proof of Proposition 4.1.
Remark 3.7. Combining Lemmata 3.5 and 3.6 , we obtain and c g > 0 is a positive constant such that We only give the proof of Lemma 3.5, since that one of Lemma 3.6 is analogous.

Proof of Lemma 3.5.
For every h ∈ R n \{0}, α > 0, ε > 0, and C > 0, we define the function where , and the constants c A , B A,g , and c g are defined in Remark 3.8. Before proving the claim, let us show how this will imply the lemma. Setting from (3.14) and by definition of C α,ε h it follows that .
From this, it follows that for every ε ∈ (0, with D defined in Remark 3.8. Therefore, minimizing in α the left hand side of (3.15) we obtain for every (x, y) ∈ R n × [0, A], and ε ∈ (0, D A /2). Taking the limit as ε → 0 + we conclude.
Let us now show (3.14). By definition of C α,ε h , the maximum principle, and thanks to (3.6), there exists (x, y) ∈ R n × {0, A} such that (3. 16) In what follows we assume that C α,ε h > 0, since otherwise (3.14) is trivially satisfied. We have two possibilities.
From the fact that the right hand side in the above expression is positive, it follows that Moreover, identity (3.18) becomes Observing now that the role played by u(x + h, 0) and u(x − h, 0) is symmetric, we only need to consider three subcases.
Let us now show that for every a, b ≥ 0 Summing up the last two relations we obtain the claim. Applying (3.23) with a = 2u(x + h, 0) and b = 2u(x − h, 0), and using (3.21), relation (3.22) gives By Lemma 3.4 it follows that u(·, 0) is Lipschitz continuous, with Lipschitz constant 2u(x, 0)) The inequality above is only possible if the last term in the right hand side is non-negative, that is, if .

(3.24)
Case 2b: y = 0 with u(x + h, 0) ≥ 0 and u(x − h, 0) < 0. In this case, recalling (2.1), (3.20) implies that (3.25) By Lemma 3.4, u(·, 0) is Lipschitz continuous, with Lipschitz constant L A /c A . Therefore, the right hand side of the above expression can be estimated as follows: On the other hand, thanks to (3.21) we can estimate the left hand side of (3.25) as Combining the last two inequalities and (3.25) we obtain We now distinguish two subcases. Case 2bi: Small values of |h|. Let c g > 0 be defined by (3.13). From (3.26) it follows that for every |h| ∈ [0, 1/c g ) we have which is impossible. Therefore, (3.26) can only be satisfied for |h| ≥ 1/c g .
Case 2d: Proof of (3.14). From the previous steps it follows that at least one among inequalities (3.17), (3.24), and (3.27) has to be satisfied. Recalling the definition of f ε (α), this concludes the proof of (3.14) and, in turn, of the lemma.

Phases Separation and Optimal Regularity
As already mentioned in the Introduction, the main problem in establishing optimal regularity is that one cannot exclude a priori the existence of free boundary points where the function u changes sign. Indeed, at such points ∂ y u(·, 0) would be discontinuous, with a jump of 2g (0 + ). This is ruled out by the next proposition, which shows that the two "phases" {x ∈ R n : u(x, 0) > 0} and {x ∈ R n : u(x, 0) < 0} are well separated.  × (0, A)) be a solution of (1.2), and let x ∈ ∂ K u , where K u is defined by (1.3). Then, there exists Before proving Proposition 4.1, we show how this allows us to prove Theorem 1.1.
Proof of Theorem 1.1. Let x ∈ ∂ K u . Without any loss of generality, thanks to Proposition 4.1, we can assume that where r 0 is given by Proposition 4.1. We claim that there exists 0 < r ≤ r 0 and D > 0, such that Indeed, let us write u = u 1 + u 2 + u 3 , where u 1 , u 2 , and u 3 are the harmonic functions in R n × (0, A) with the following boundary conditions: Note now that u 3 is C ∞ in a neighborhood of (x, 0), since u − = 0 in B n r 0 (x). Analogously, u 1 is also C ∞ in a neighborhood of (x, 0). On the other hand, by maximum principle u 2 ≥ 0. Therefore, an argument similar to the one used in the proof of Lemma 3.5 shows that, for every y ∈ [0, A], u 2 (·, y) is semiconvex. Therefore, Then, using the fact that u 1 and u 3 are smooth, (4.1) follows. We now note that which is just a minor variation of the classical Signorini problem v∂ y v = 0 [5,6]. Thus, the remaining part of the proof of Theorem 1.1 can easily be obtained by repeating (with the needed minor modifications) the arguments used in [5,8].
We now give the proof of Proposition 4.1.
Proof of Proposition 4.1. Without any loss of generality, we can assume x = 0. We will argue by contradiction, assuming that We divide the proof into two steps.
Step 1a: We show that

Suppose, by contradiction, that (4.3) is satisfied but
(4.6) By (4.2), we can find a sequence {x k } k∈N ⊂ R n \{0} with x k → 0 such that u(x k , 0) < 0 for every k ∈ N. Setting h k := 2|x k |e 1 , thanks to (4.6) we have where the last equality follows from our choice of the sequence {h k } k∈N . On the other hand, by (4.4) it follows that for k sufficiently large. Thanks to Remark 3.7, combining (4.7), (4.9), and (4.9), we have that, for k large enough, which is impossible.
Step 1b: We conclude the proof of Step 1. By Step 1a, there exists d > 0 and e ∈ S n ∩ {y = 0} such that de ∈ ∂ + x u − (0, 0). Since, by Lemma 3.6, u − (·, 0) is semiconcave with semiconcavity constant C, by repeating the same argument used to show (4.4) we have that where we set x d := d 2C e. Taking into account (4.5), this implies e = −e 1 , thus We will now show that ∂ x 1 u(·, 0) is unbounded, against Lemma 3.4. To this aim, for every ε > 0 we set w ε := −εe 1 . In this way, w ε → 0 as ε → 0 + and w ε ∈ B n |x d | (x d ) for ε sufficiently small, so that u(w ε , 0) < 0. We claim that for some positive dimensional constant C n . Sinceũ − u vanishes on {y = 0} and is harmonic in R n × (0, A), we have x → ∂ y (ũ − u)(x, 0) ∈ C ∞ (R n ). Therefore, to prove our claim it will be sufficient to show that the last integral diverges as ε → 0 + . Let f : R n → R be defined as In what follows, we set Given r > 0 with 0 < r < λ/4, we split the integral under consideration as follows: We can disregard the first integral, which is bounded for ε small enough. Concerning the second integral, using the fact that u(w ε , 0) < 0, we havê Since f is Lipschitz continuous, I ε 2 is uniformly bounded in ε. By definition of f and by (4.5), we can split I ε 1 in the following way: We claim that I ε 1,2 is uniformly bounded in ε. Indeed, first of all we observe that, thanks to (2.1) and Lemma 3.4, for every z ∈ B n r ∩ {u ≥ 0}. Then, recalling (4.10), we have where we set ρ := ω ∈ S n−1 : |ω 1 | < ρ λ , and λ is defined by (4.12). Since w ε := −εe 1 , we note that for every ρ ∈ (0, r ) and ω ∈ ρ for ε sufficiently small. Therefore, for ε sufficiently small we obtain where we used the fact that H n−1 ( ρ ) ≤ Cρ for some positive constant C = C(n).
Step 2: We show that there exist positive constants γ , η, and r such that (4.13) By Step 1 we know that u(·, 0) is differentiable at x = 0 with ∇ x u(0, 0) = 0, hence there exists a continuous function σ : Note that, with no loss of generality, we can assume that σ (r ) ≥ r for all r .
Let M, C, η be positive constants to be chosen later, and for every r ∈ (0, 1) set We consider the harmonic function V + : r → R defined as where x + 0 ∈ R n is such that u(x + 0 , 0) > 0 and |x + 0 | = c 0 r with 0 < c 0 1 (note that that such a point x + 0 exists, because we are assuming, by contradiction, that (0, 0) is a boundary point both for {u > 0} and for {u < 0}). Since V + is harmonic, we have where the positivity comes from the fact that V + (x + 0 , 0) = u(x + 0 , 0) > 0. We now have several possibilities.

Case 2a:
We show that, if C is sufficiently large, then there exists no x ∈ ∂ r ∩{y = 0} such that Suppose that such x exists. Then, it cannot be u(x, 0) ≤ 0, since in that case Therefore, u(x, 0) > 0, and If we choose C large enough we obtain ∂ y V + (x, 0) > 0, which is impossible.

Case 2b:
We show that, if M is sufficiently large and η is sufficiently small, then there exists no point (x, y) with |x| = r and y ∈ [0, ηr ] such that Indeed, suppose that such a point (x, y) exists. Then, since |x| = r and |x + 0 | = c 0 r , we have Thus, so that (recall that c 0 1) for η small enough. Thanks to (3.10), this last estimate gives which is impossible for M sufficiently large.

Case 2c:
We show that, if M is sufficiently large and η is sufficiently small, then there exists no point x ∈ R n with r/2 ≤ |x| ≤ r such that This case can be treated as Case 2b.
Thanks to (3.10), the function g (0 + ) + 2C 1 ηr − ∂ y u is harmonic and nonnegative in 2r . Thus, by (4.14) and Harnack inequality, there exists a constant C γ > 0 such that From the previous chain of inequalities we obtain Therefore, there exists r = r (γ ) such that thus showing (4.13).
Step 3: We conclude. An argument analogous to that one used in Step 2 can be applied to the harmonic function V − : r → R defined as Comparing the inequality above to (4.13), we obtain the desired contradiction.

Frequency Formula
In this section we prove a frequency formula, which will allow us to study the blow up profiles of solutions u of (1.2). To this purpose, assuming that (0, 0) is a free boundary point for u, and that u(x, 0) ≥ 0 in a neighborhood of 0 (cf. Proposition 4.1), we investigate the regularity properties of the function v : R n × [−A, A] → R defined by (1.5).
Throughout this section we assume that the hypotheses of Theorem 1.1 are satisfied, that v is given by (1.5) where u is a solution of (1.2), that (0, 0) is a free boundary point for v, and that v(x, 0) ≥ 0 for every x ∈ B n r 0 , where r 0 is given by Proposition 4.1. Therefore, v satisfies: Since v is even with respect to the hyperplane {y = 0}, we have First of all we observe that for every x ∈ B n r 0 , where we used the definition of v, Lemma 3.4, and the fact that (0, 0) is a free boundary point. From the above inequality it follows that the function Then, using the fact that v is harmonic in B r 0 \{y = 0} where we used (5.1). We can now state the main result of the section. Then, there exist 0 < r 0 ≤ r 0 , and a positive constant C, such that the function r → v (r )e Cr is monotone nondecreasing for r ∈ (0, r 0 ). In particular, there exists v (0 + ) = lim r →0 + v (r ). Before giving the proof, we need several auxiliary lemmas. When integrating along the boundary of a smooth (n + 1)-dimensional set, we will denote by ν the outer unit normal, and by v ν the derivative of v along ν. We will denote the tangential gradient of v by ∇ τ v, so that ∇ τ v = ∇v − v ν ν.
The next lemma is an adaptation of [9, Lemma 7.8].
Lemma 5.2. For every r ∈ (0, r 0 ) Then, Recalling that z = r ν on ∂ B r , we get Similarly, Combining (5.3), (5.4), and (5.5) we obtain Then, thanks to (5.1) and the equation satisfied by v, we conclude that We will also need the following lemma:

Proof.
Since v is harmonic in B r \{y = 0}, On the other hand, applying the divergence theorem in each half-spherê where we also used (5.1). Comparing the last two chains of equalities we conclude.
We now start by differentiating F v (r ).
Lemma 5.4. For every r ∈ (0, r 0 ) Proof. Writing the integral in polar coordinates and differentiating we obtain where the last equality follows by Lemma 5.3.
We now state a trace inequality, whose proof can be found in [20].
Lemma 5.5. For any r > 0 and any function w ∈ W 1,2 (B r ) we havê and C is a constant depending only on the dimension n.
In the following we will need an improvement of Lemma 5.5, which can be obtained using the fact that v is superharmonic.
Lemma 5.6. There exists a constant C, depending only on n, such that for any r ∈ (0, r 0 ) Proof. Let us start by proving the first inequality. Since v ∈ W 1,2 (B r ), by Lemma 5.5 Let now v be given by (5.2). Since v is superharmonic, The above inequality implies that Since v(x, 0) ≥ 0, we have v − (x, 0) = 0. Thus, by rescaling, where the positivity of c n follows by a standard compactness argument (actually, by spectral analysis theory, the minimum is attained by w(x, y) = y, thus c n = 1). Hence, by Hölder inequality, for every 0 < r < r 0 .
Proof. Suppose that F v (r ) > r n+4 . Then, by Lemma 5.6, Thanks to Lemma 5.3 and Lemma 5.6, for r sufficiently small we havê This shows the first inequality which, together with (5.8), allows us to conclude.
We are now ready to prove Proposition 5.1.
Proof of Proposition 5.1. Since r → max{F v (r ), r n+4 } is a semiconvex function (being the maximum between two smooth functions) and v (r ) = n + 4 on the region where max{F v (r ), r n+4 } = r n+4 , it suffices to prove the monotonicity of v (r )e Cr in the open set {r : F v (r ) > r n+4 }. Note that, thanks to Lemma 5.4, Setting v (r ) := v (r ) − n, the logarithmic derivative of v is given by where we used again Lemma 5.4. We now divide the remaining part of the proof into several steps. In the following, it will be convenient to define Step 1: We show that Indeed, thanks to Lemma 5.3, |∇v| 2 dH n + I 1 (r ). (5.12) Using first Lemma 5.2 and then Lemma 5.3, the first integral in the last expression can be written as where we used that, by the optimal regularity of v (see Theorem 1.1), |v| ≤ Cr 3/2 and |∇v| ≤ Cr 1/2 . Combining (5.13)-(5.15), for r sufficiently small the claim follows.
Step 3: where in the last step we used Hölder inequality and the fact that, by Lemma 5.7, the integral at the denominator is positive. Then, thanks to Step 2 and Lemma 5.7 again, we obtain v (r ) v (r ) The previous chain of inequalities gives Recalling that v (r ) = v (r ) − n, this shows that r → ( v (r ) − n)e Cr is increasing, and thus the conclusion.

Blow Up Profiles and Regularity of the Free Boundary
We are now going to study the blow up profiles of v and the regularity of the free boundary. As in the previous section, with no loss of generality we will assume that (0, 0) is a free boundary point and that v(x, 0) ≥ 0 for every x ∈ B n r 0 , where r 0 is given by Proposition 5.1.
We define, for every r ∈ (0, r 0 ), the function v r : where F v is as in Proposition 5.1. Note that 2) The next proposition shows which are the possible values of v (0 + ).
We first prove the proposition above in the case lim inf r →0 + d r r 2 < +∞.
We will show that in this case we always have v (0 + ) ≥ n + 4. Indeed, by assumption there exists C > 0 such that We then have two possibilities (see also the second part of the proof of [9, Lemma 6.1]). Case 1: there exists a sequence (r j ) j∈N with r j → 0 such that for j sufficiently large.
Then, v (r j ) = n + 4 for j sufficiently large, and therefore v (0 + ) = n + 4. Case 2: for r sufficiently small Then, Suppose now, by contradiction, that there exists η > 0 such that v (r ) ≤ n + 4 − η for r sufficiently small, and let (r j ) j∈N be a strictly decreasing sequence with r j → 0. Then, thanks to (6.3), for every k, l ∈ N with k < l we have Therefore, by the definition of v , which is impossible if we choose log r k − log r l → ∞.
In the next proposition we consider the case lim inf r →0 + d r r 2 = +∞. Proposition 6.2. Let the assumptions of Theorem 1.1 be satisfied, suppose that (0, 0) is a free boundary point for u, and that u(x, 0) ≥ 0 in B n r 0 . Let v be given by (1.5), and let F v and v be as in Proposition 5.1. Define v r as in (6.1), and assume that lim inf r →0 + d r r 2 = +∞. Then, there exists a sequence (r k ) k∈N with r k → 0, and a homogeneous function v ∞ ∈ W 1,2 (B 1 ) with homogeneity degree 1/2 and v r k → v ∞ in C 1,γ on compact subsets of B 1 ∩ {y ≥ 0}, (6.4) for any γ ∈ (0, 1/2). Moreover, v ∞ satisfies the classical Signorini problem in B 1 and is even with respect to y: y) in B 1 .
Proof of Proposition 6.2. and conclusion of proof of Proposition 6.1. Since d r /r 2 → ∞, for r sufficiently small we have F v (r ) > r n+4 . Then, thanks to Proposition 5.1 and by (5.10), for r < r 0 we have Therefore, by (5.9) and the definition of v r , for r sufficiently small. Consider now a sequence r k → 0. By the previous inequality and thanks to (6.2), the sequence (v r k ) k∈N is bounded in W 1,2 (B 1 ). Thus, up to subsequences, for some v ∞ ∈ W 1,2 (B 1 ). Thanks to the uniform C 1,1/2 regularity for solutions, we also have that (6.4) holds. Let us show that v ∞ ∂ y v ∞ = 0 on B n 1 , since the other conditions in (6.5) are a direct consequence of (6.4). Recalling the definition of v r k , from the identity v(r x, 0) ∂ y v(r x, 0) − g (2v(r x, 0)) + g (0 + ) = 0 for every x ∈ B n 1 it follows that, for every k ∈ N, Thanks to (6.4), since taking the limit in (6.7) we obtain To show that v ∞ is homogeneous, let us first prove that v ∞ is constant for r sufficiently small. Indeed, let r < s 1. A direct calculation shows that v r k (r ) − v r k (s) = v (r k r ) − v (r k s) for every k ∈ N.

C 1,α Regularity of the Free Boundary for μ = 3/2.
We now study the regularity of the free boundary in the special case in which v (0 + ) = n+3. Note that, by the argument in the previous section, this corresponds to the case lim inf r →0 + d r r 2 = +∞.
We start by proving the C 1 regularity. Lemma 7.1. Let the assumptions of Theorem 1.1 be satisfied, suppose that (0, 0) is a free boundary point for u, and that u(x, 0) ≥ 0 in B n r 0 . Let v be given by (1.5), and let v be as in Proposition 5.1. Assume that v (0 + ) = n + 3, and choose a coordinate system in R n such that (6.6) holds true. Then, for every c > 0 there exists ρ = ρ(c) > 0 with the following property: For every τ ∈ S n ∩ {y = 0} with τ · e n ≥ c we have ∂ τ v(z) ≥ 0, for every z ∈ B ρ . (7.1) In addition, near the origin the free boundary of v is the graph of a C 1 function x n = f (x 1 , . . . , x n−1 ).
We are now ready to prove that the free boundary is C 1 .
Proof of Lemma 7.1. Applying Lemma 7.2 to the functions h k introduced in (7.2) we obtain (7.1). As a consequence, for every L > 0 there exists r = r (L) > 0 such that (recall that K u is defined in (1. Consider now a pointx ∈ ∂ K u ∩ B n r and define the function vx (x, y) := v(x −x, y). Note that we can repeat the same argument (frequency formula and blow-up procedure) with vx in place of v. Also, observe that since the function h r → ∂ x n v ∞ uniformly in B n 2/3 , as r → 0 + . Recalling (6.6), it follows that for r small enough Let now i ∈ {1, . . . , n − 1}, τ i := e n +e i