TWO PHASE PROBLEMS FOR LINEAR ELLIPTIC OPERATORS WITH ...

TWO PHASE PROBLEMS FOR LINEAR ELLIPTIC OPERATORS WITH VARIABLE COEFFICIENTS: LIPSCHITZ FREE BOUNDARIES ARE C 1,γ M. CRISTINA CERUTTI, FAUSTO FERRARI, SANDRO SALSA

Abstract. We prove C 1,γ regularity of Lipschitz free boundaries of two fase problems for linear elliptic operators with H¨ older continuous coefficients.

1. Introduction and main results In this paper we start the study of the regularity of the free boundary in two phase problems for uniformly elliptic operators with variable coefficients. The theory for the Laplace operator has been developed by L. Caffarelli in the well known and by now classical papers [3], [4]. These results have¡been¢extended by P. Wang to a class of concave fully non linear operators of the type F D2 u in [10]. M. Feldman considers in [7] anisotropic operators with constant and in [8] a class of non ¡ 2 coefficients ¢ concave fully non linear operators of the type F D u, Du . He shows that Lipschitz free boundaries are smooth, thus extending the results in [3]. We are interested in the following free boundary problem and in its weak solutions in the sense of [3]. Definition 1.1. In the unit cylinder C1 = B1n−1 × (−1, 1) ⊂ Rn we are given a continuous function u satisfying (1)

(2)

L1 u =

2

L u=

n X

a1ij (x) Dij u = 0

in Ω+ (u) = {x ∈ Ω : u (x) > 0}

a2ij (x) Dij u = 0

in Ω− (u) = {x ∈ Ω : u (x) ≤ 0}0

i,j=1 n X

i,j=1

Moreover, if x0 ∈ F (u) ≡ ∂Ω+ (u) ∩ C1 and there exists a ball B% (y) such that x0 ∈ ∂B% (y) and B% (y) is contained either in Ω+ (u) or in Ω− (u), then (3)

u (x) = α < x − x0 , ν >+ −β < x − x0 , ν >− +o (|x − x0 |)

as x → x0

where ν is the unit unit normal to ∂B% (y) at x0 pointing inward to Ω+ (u) and (4)

α = G (β) .

The set F (u) is the free boundary and (4) expresses the free boundary condition − u+ ν = G (uν ) in a weak sense. Date: 10/07/2002. 1991 Mathematics Subject Classification. 35R35, 35R60,35B05. Key words and phrases. Free boundaries, elliptic equations, two fase problems. We thank Professor Luis Caffarelli for some useful discussions. 1

2

M. CRISTINA CERUTTI, FAUSTO FERRARI, SANDRO SALSA

L1 , L2 are uniformly elliptic operators with ellipticity constant Λ that is, ³ symmetric, ´ s s if A (x) = aij (x) , s = 1, 2, then

(5)

asij (x) = asji (x), ΛI ≤ As (x) ≤ Λ−1 I

x ∈ C1

where I is the identity matrix. Our purpose is to show that if F is a Lipschitz surface then it is C 1,γ . The natural hypothesis on the coefficients is then H¨older continuity. Our main result is the following. Theorem 1.2. Let u be a weak solution of the free boundary problem (1)-(4) in C1 . Suppose 0 ∈ F (u) and that i) As ∈ C 0,a (C1 ),s = 1, 2 0 < a ≤ 1 and (5) holds. ii) Ω+ (u) = {(x0 , xn ) : xn > f (x0 )} where f is a Lipschitz continuous function with Lip(f ) ≤ L. iii) G = G(z) is continuous, strictly increasing and for some N > 0, z −N G(z) is den−1 creasing in (0, +∞). Then, on B1/2 , f is a C 1,γ function with γ = γ(n, N, L, Λ, a).

The strategy of the proof follows the guidelines in [3] and consists of the following three main steps: to improve the Lipschitz constant of the level sets of u away from the free boundary, to carry this interior gain to the free boundary, to rescale and iterate the first two steps. Let us briefly describe below the new features and difficulties that arise in the case of operators with variable coefficients. In the case of operators independent of x, the proofs in the first step heavily rely on the fact that the directional derivatives of a solution are still solutions of the same equation. Clearly this is not the case for variable coefficients operators. Via an approximation argument it is possible to bypass this obstacle (see sections 2 and 3). The second step (the key one) requires the construction of a C 2 function φ such that the function v (x) = supBφ(x) (x) u is L−subharmonic whenever Lu = 0. Our technique, that, we believe, gives to the construction a new geometrical insight, is based on a decomposition of the operator L in the sum of a term acting tangentially to the level sets of u, plus a term that can be easily handled (sections 4 and 5). Unlike the constant coefficients case, the condition that φ has to satisfy (see inequality (30)) turns out to be non-homogeneous w.r.t. φ itself, with the non-homogeneity controlled by the anisotropy of the matrices As . The main consequence is the impossibility to carry at once to the free boundary the full gain in the flatness of the level sets of u obtained in the second step. What one can do is to carry ε−flatness, which amounts to saying that F (u) is trapped in a ε−strip between two Lipschitz functions with Lipschitz constant L0 < L. To achieve C 1,γ regularity of F (u) one has to set up an iteration procedure balancing rescaling and flatness improvements at each step (sections 6 and 7). Various more or less immediate extensions of the above theorem hold. In particular the same conclusions can be reached with the same technique if one considers operators with first order terms, Ls u =

n X

i,j=1

asij (x) Dij u +

n X

bsi (x) Di u

i=1

with bi H¨ older continuous (see remark 4.3) and therefore also divergence form operators with smooth coefficients. Fully non linear operators with x−dependence will be considered in a forthcoming paper.

TWO PHASE PROBLEMS FOR ELLIPTIC OPERATORS WITH VARIABLE COEFFICIENTS

3

It is also possible to introduce a dependence on x and ν in the free boundary condition, through a relation of the type α = G (β, ν, x) assuming, instead of iii) in Theorem 1.2 iii’) G = G (z, ν, x) is continuous, strictly increasing in z and, for some N > 0 independent of ν and x, z −N G (z, ν, x) is decreasing in (0, +∞) . and iii”) log G is Lipschitz continuous w.r.t. ν, x, uniformly w.r.t. its first argument z ∈ [0, +∞).

Also, conormal derivatives can be taken into account, replacing ν by A1 (x0 )−1 ν and A2 (x0 )−1 ν on both sides, respectively, in (3). The various constants c that will appear in the sequel may vary from formula to formula. If for a constant c, η, µ etc., we don’t give any explicit dependence we mean that it depends only on some of the relevant parameters n, L, N, Λ, a. 2. Positive L-harmonic functions in Lipschitz domains In this section we prove that positive solutions to Lu = 0 in Lipschitz domains, vanishing on a portion F of the boundary, are monotone along a cone of directions in a neighborhood of F . Here L is a uniformly elliptic operator in the same class of L1 and L2 . Let Ts = {(x0 , xn ) ∈ Rn : |x0 | < s, f (x0 ) < xn < 2Ls}, where f is a Lipschitz function with constant L. © Theorem 2.1. Let u be a positive solution to Lu = 0 in T4 , vanishing on F = xn = ª f (x0 ) . Then, there exists η such that in © ª Nη (F ) = f (x0 ) < xn < f (x0 ) + η ∩ T1

u is increasing along the directions τ belonging to the cone Γ(en , θ), with axis en and opening θ = 21 cot−1 L. Moreover, in Nη (F ), (6)

c−1

u(x) u(x) ≤ Dn u(x) ≤ c dx dx

where dx = dist(x, F ). In order to prove the above theorem, we need some approximation lemmas. Let z be a solution to ½ ∆z = 0 in T2 z=g on ∂T2 , where g is a smooth function vanishing on F and equal to 1 at points x with dx > 1/10. Then by Lemma 5 in [3], Dn z > 0 in Tρ , with ρ = ρ(n, L). By rescaling the problem (if necessary), we may assume ρ = 3/2. Since z(en ) ≥ c > 0, by Harnack inequality we have that, if y ∈ T1 , dy ≥ η0 (7) (8)

z(y) ∼ c(η0 ), Dn z(y) ∼

z(y) ∼ c(η0 ) dy

4

M. CRISTINA CERUTTI, FAUSTO FERRARI, SANDRO SALSA

and |Dij z(y)| ≤

(9)

z(y) ≤ c(η0 ) . dy 2

¡ ¢ Clearly z, u ∈ C 0, a T 2 , lowering, if necessary, the H¨older exponent a in condition n P aij (ry)Dij w(y). In this section, without loss of i) of Theorem 1.2. Let now Lr w = i,j=1

generality, we may assume L0 = ∆.

Lemma 2.2. For r > 0, let w be the solution to ½ Lr w = 0 in T2 w=z on ∂T2

Then, given η0 > 0, there exists r0 = r0 (η0 ), such that if r ≤ r0 Dn w(y) ≥ 0

for every y ∈ T1 , with dy ≥ η0 .

¡ ¢ Proof. First observe that w ∈ C 0,a T 2 and let h(y) = z(y)−w(y); then also h ∈ C 0,a (T 2 ) and it satisfies  n  L h(y) = P (a (ry) − δ )D z(y) in T ij

r



ij

ij

2

i,j=1

h=0

on ∂T2 .

σ , Now, for σ > 0, let T2σ ≡ {y ∈ T2 : dist(y, ∂T2 ) > σ}; then on ∂T2/δ

|h(y)| ≤ cσ a .

Using the maximum principle in T2σ and Schauder estimates, we get |h(y)| ≤ c σ a + cra sup |Dij z(y)| ≤ c(σ a + ra σ −2 ), T2σ

since |Dij z(y)| ≤ c σ −2 sup z ≤ c σ −2 and z ≤ 1. Bσ/2 (y)

a

Choosing σ = cargmin(σ a + ra σ −2 ) = r a+2 , we have that a2

|h(y)| ≤ c r a+2 ≡ c rβ .

(10)

a+2

Consider now y ∈ T1 , with dy ≥ η0 and let r0 ≤ 31 η0 a . Then, using Schauder estimates in Bη0 /3 (y) together with (8) and (9) we get |Dn h(y)| ≤ cd−1 sup |h| + cra sup |Dij z| ≤ c(η0 )rβ Dn z(y) y Bη0 /3 (y)

Bη0 /3 (y)

¾ 1 a+2 1 a , η , we conclude: Therefore, if r ≤ r0 = min [2c (η0 )]1/β 3 0 ³ ´ 1 (11) Dn w(y) ≥ 1 − c(η0 )rβ Dn z(y) > Dn z(y). 2 ½

Lemma 2.3. Let η0 > 0 be fixed ad w and z as in Lemma 2.2. Then there exist r0 = r0 (η0 ) and t0 = t0 (L, n) > 1 such that, if r ≤ r0 c−1 for every y ∈ T1 , dy ≥ t0 η0 .

w(y) w(y) ≤ Dn w(y) ≤ c dy dy

TWO PHASE PROBLEMS FOR ELLIPTIC OPERATORS WITH VARIABLE COEFFICIENTS

5

Proof. The right hand side inequality follows from Schauder estimates and Harnack inequality. Let now y ∈ T1 , with dy ≥ t0 η0 , t0 to be chosen. We may assume y = tη0 en . From the boundary Harnack principle (see e.g. [5] or [9]) if y˜ = η0 en , then z(˜ y ) ≤ ct−a z(y)

and, if t ≥ (2 c)1/a ≡ t0 , then

1 z(˜ y ) ≤ z(y). 2 On the other hand, if dy ≥ t0 η0 and r ≤ r0 (η0 ), from (7), (10) and (11) we have 1 3 w(y) ≤ z(y) and Dn w(y) ≥ Dn z(y). 2 2 Thus, if t0 η0 ≤ dy ≤ 10t0 η0 , applying Harnack inequality to Dn z, we get Z 1 ¡ ¢ 3 d w(y) ≤ z(y) ≤ 3 z(y) − z(˜ y) = 3 z(sy + (1 − s)˜ y )ds 2 0 ds ≤ ct0 η0 Dn z(y) ≤ cDn z(y)dy ≤ cDn w(y)dy . (12)

Repeating the argument with y˜ = 10t0 η0 , we get that (12) holds for 10t0 η0 ≤ dy ≤ 20t0 η0 . After a finite number of steps, (12) follows for dy ≥ t0 η0 , y ∈ T1 . Lemma 2.4. Let u be as in theorem 2.1. Then there exists η positive, such that for every x ∈ T1 , dx ≤ η. Dn u(x) ≥ 0.

(13)

Moreover, in the same set c−1

(14)

u(x) u(x) ≤ Dn u(x) ≤ c . dx dx

Proof. Let t0 be as in Lemma 2.3, and η0 small to be chosen later. Set η1 = 2η0 t0 . It is enough to show that if x = η1 ren and r ≤ r0 (η0 ), then Dn u(x) ≥ 0. Consider a small box T2r and define u ˜(y) = u(ry). Then u ˜ satisfies Lr u ˜ = 0 in T2 , where f is replaced by fr (y 0 ) = f (ry 0 )/r. We will show that Dn u ˜(y) > 0, where y = η1 en , by comparing u ˜ with the function w constructed in Lemma 2.2, normalized in order to get u ˜(y) = w(y). Notice that, if we choose r0 = r0 (η0 ) according to Lemma 2.3, we have c−1

(15) if dy ≥ η1 . From [6] we know that

which implies

w(y) w(y) ≤ Dn w(y) ≤ c . dy dy

u ˜ ∈ C 0,a (T 3 ) so that in Bη0 (y) 2 w ¯ ¯ ¯ ¯u ¯ ˜(y) − 1¯ ≤ cη0a , ¯ ¯ w(y)

|˜ u(y) − w(y)| ≤ cη0a w(y) ≤ cη0a w(y).

From Schauder estimates, being η0 ∼ dy¯, from which, we get (16)

|Dn u ˜(y) − Dn w(y)| ≤ cη0a−1 w(y) ≤ cη0a Dn w(y), Dn u ˜(y) ≥ (1 − cη0a )Dn w(y),

6

M. CRISTINA CERUTTI, FAUSTO FERRARI, SANDRO SALSA

and (13) holds if η0 is sufficiently small. Inequality (14) is now a consequence of (15) and the fact that w(y) = u(y). To complete the proof of Theorem 2.1, it is enough to observe that the above lemmas hold if we replace en by any unit vector τ such that the angle between τ and en is less than θ = 1/2 cot−1 L. Thus, we obtain a cone Γ(en , θ) of monotonicity for u. 3. Interior gain Applying Theorem 2.1 to the positive and negative part of the solution u of our free boundary problem, we conclude that in a η-neighborhood of F (u) the function u is increasing along the direction of a cone Γ(en , θ). In this section we prove that, far from the free boundary, the monotonicity cone can be enlarged improving the Lipschitz constant of the level sets of u. Again, to control the errors due to the anisotropy of the matrices As (x), s = 1, 2, we need another approximation lemma. We still assume L(0) = ∆. Lemma 3.1. Suppose u is a positive solution of Lr u = 0 in T4 vanishing on F = {xn = f (x0 )}, increasing along every τ ∈ Γ(en , θ). Assume furthermore that (6) holds in T4 . Then, in B1/8 (en ), Dτ u(x) ≥ (c1 < ∇, τ > −c2 rβ )u(en )

(17)

for every τ ∈ Γ(en , θ), | τ |= 1, where ∇ =

∇u(en ) |∇u(en )|

and r is small enough.

Proof. Let v be harmonic in T2 , v = u on ∂T2 , and put h = v − u. Then Lr h(y) = P σ i,j (aij (ry) − δij )Dij v(y) in T2 , h = 0 on ∂T2 . Arguing as in Lemma 2.2, on ∂T2 , with a

σ = r a+2 , we get, by the boundary Harnack principle,

| h(y) |≤ cσ a max u ≤ cσ a u(en )

(18)

T2

and, since sup | Dij v |≤ cσ T2σ

−2

u(en ) we obtain, for every y ∈ T2σ ,

a2 . a+2 In particular, thanks to Harnack inequality, for every y ∈ B1/4 (en ) |h(y)| ≤ crβ u(en ), β =

(19)

(C − crβ )u(en ) ≤ v(y) ≤ c(1 + rβ )u(en ).

By Schauder inequality in B1/4 (en ), for every τ ∈ Γ(en , θ), | τ |= 1, we conclude (20) In particular, this implies (21)

| Dτ v − Dτ u |≤ crβ u(en ).

| ∇v(en ) |≥| ∇u(en ) | −crβ u(en ) ≥ (C − crβ )u(en )

thanks to (6). From (19) we infer that Dτ v + crβ u(en ) is a positive harmonic function in B1/4 (en ), hence, by Harnack inequality in B1/8 (en ), Thus, (22)

Dτ v(y) + crβ u(en ) ≥ C(Dτ v(en ) + crβ u(en )). Dτ v ≥ CDτ v(en ) − crβ u(en ) ­ ® = C | ∇v(en ) | · ∇v , τ − crβ u(en ),

TWO PHASE PROBLEMS FOR ELLIPTIC OPERATORS WITH VARIABLE COEFFICIENTS

where ∇v =

∇v(en ) |∇v(en )| .

7

If r is small enough inequality (21) implies that, | ∇v(en ) |≥ Cu(en )

and also that | ∇v(en ) − ∇u(en ) | | (| ∇v(en ) | − | ∇u(en ) |) | + | ∇v(en ) | | ∇u(en ) |

| ∇v − ∇ | ≤

≤ crβ . Hence

­

® ­ ® ­ ® ­ ® ∇v , τ = ∇v − ∇, τ + ∇, τ ≥ ∇, τ − crβ .

Inserting in (22) we get

­ ® Dτ v(y) ≥ (C1 ∇, τ − crβ )u(en )

and the lemma follows.

We now control the error term in (17) by deleting from the original cone Γ(en , θ) the ”bad” directions, that is those in a neighborhood of the generatrix opposite to ∇u(en ). Precisely, if τ ∈ Γ(en , θ), denote by ωτ the solid angle between the planes span{en , ∇} 99 π and and span{en , τ }. Delete from Γ(en , θ) the directions τ such that (say) ωτ ≥ 100 0 0 call Γ (en , θ) the resulting set of directions. If τ ∈ Γ (en , θ), then ­ ® (23) ∇, τ ≥ c3 δ where δ =

π 2

− θ. We call δ the defect angle.

Lemma 3.2. Let u be as in Lemma 3.1. There exist positive r0 and h, depending only on n, L and Λ, such that if r ≤ r0 , for every small vector τ , τ ∈ Γ0 (en , θ/2), and for every x ∈ B1/8 (en ): (24)

sup B(1+hδ)² (x)

u(y − τ ) ≤ u(x) − C²δu(en ),

where ² =| τ | sin 2θ .

Proof. Let ν be a unit vector in Rn . We have ¡ ¢ u(x) − u x − τ + (1 + hδ)²ν = [u(x) − u(x − τ + ²ν)] £ ¡ ¢¤ + u(x − τ + ²ν) − u x − τ + (1 + hδ)²ν = W1 + W2 . ³ ´1/β and Put τ = τ − ²ν; then | τ |≥| τ | −² ≥ c². From (17) and (23), if r0 = c21cc23 δ r ≤ r0 we have, in B1/8 (en ), Dτ u ≥ cδu(en ).

(25) Therefore,

W1 = Dτ u(x) ≥ C²δu(en ) and |W2 | ≤| ∇u(y) | hδ² ≤ c²hδu(en ). Thus,

and (24) follows if h
0}. Observe that the function u ˜(x) = u(A1/2 (0)x)

is a solution of Tr(A˜0 (x)D2 u ˜(x)) = 0, ˜ in {˜ u > 0}, where A˜0 (z) = A1/2 (0)A(A1/2 (0)x)([A1/2 (0)]−1 )T . Note that A(0) = I. 1/2 1/2 ˜ Moreover if φ(x) = φ(A (0)x) and v˜(x) = v(A (0)x) then ˜ v˜(x) = sup (32) u ˜(x + φ(x)ν) = v(x). |A1/2 (0)ν|=1

and, if φ satisfies (30), then φ˜ satisfies ˜ ¡ ¢ ˜ − C | ∇φ˜ |2 +ω 2 ≥ 0 (33) ∆φ(0) ˜ φ(0)

with C˜ = ΛC. Thus, we must show that, if (33) holds, v˜ is subharmonic at the origin, in the viscosity sense. It is enough to prove that (34)

˜ r := M

Z

∂B1

v˜(ry) dHn−1 (y) ≥ v˜(0) + o(r2 ),

as r → 0, r > 0. ∗ where ˜ Let P a maximum point in (32), i.e., v˜(0) = u ˜(P ), and suppose P = φ(0)aν ∗ 1/2 ∗ −1 ν is a unit vector and a =| A (0)ν | . ¡ ¢ ˜ ¿From the definition of v˜ it is enough to estimate u ˜ ry + φ(ry)ν(ry) from below, where ν is a suitable smooth vector field with ν(0) = a−1 ν ∗ . To simplify a little the ˜ ˜ notations let φ(0) = φ˜0 , ∇φ(0) = ∇φ0 . We now divide the proof in several steps. Step 1. We select an orthonormal basis in Rn as follows. We choose en = ∇u(P )/ | ∇u(P ) |. Consider now the matrix Pn = I − en ⊗ en and DT = DT (P ) = Pn D2 u(P )Pn .

The symmetric matrix DT has en as an eigenvector corresponding to the eigenvalue βn (P ) = 0. The other eigenvectors belong to the tangent space at P to the level set ΣP = {˜ u = u ˜(P )}. We choose e1 , . . . , en−1 as an orthonormal basis for this tangent eigenspace and we call β1 (P ), . . . , βn−1 (P ) the corresponding eigenvalues. Incidentally, notice that the numbers λj = −βj (P )/ | ∇˜ u(P ) | represent the principal curvatures of ΣP at P .

10

M. CRISTINA CERUTTI, FAUSTO FERRARI, SANDRO SALSA

According to the above choice, denote by U the orthonormal matrix that diagonalizes DT and set ˜ )U, D∗ = U T (D2 u(P ) − DT (P ))U A∗ = U T A(P N

d∗ij

∗ , we have d∗ = 0 for i, j = 1, . . . , n − 1. In Observe that, if are the elements of DN ­ 2 ® ij this orthogonal frame the bilinear form D u(P )ξ, ξ has the following expression:

(35)

while, if

DT∗

­

®

2

D u(P )ξ, ξ =

n−1 X

ξi2 βi (P )

+2

i=1

n−1 X

d∗in ξi ξn + ξn2 d∗nn ,

i=1

= diag(β1 , · · · , βn−1 , 0),

∗ ˜u(P ) = Tr(AD ˜ 2u ) L˜ ˜) = Tr(A∗ DT∗ ) + Tr(A∗ DN

(36)

=

n−1 X

∗ αjj βj (P ) + 2

j=1

Step 2. Introduce the vector field

V (y) = ν +

n−1 X i=1

where the vectors V

i,

We have, uniformly in y,

ν(ry) = a



­

® V i , y ei

will be chosen later, and define ν(y) =

 

∗ ∗ ∗ αjn djn + αnn d∗nn .

j=1

∗

(37)

n−1 X

ν∗ + r

n−1 X i=1

­

V (y) |

A1/2 (0)V

® V i , y ei −

r2 2

 

(y) |

n−1 X

i,j=1

­ ® where µij = A(0)ei , ej . From Taylor expansion,

­

.



®­ ® V i , y V j , y a2 µij  ν ∗ + o(r2 )

  

,

˜ φ(ry)ν(ry) = P + rl(y) + r2 q(y) + o(r2 ),

where l(y) =

n−1 X i=1

and

® ­ ® ¢ ¡ ­ a φ˜0 V i , y ei + ∇φ˜0 , y ν ∗ ,

n−1 ´ X­ X­ ®­ ® ® ­ ® n−1 ® 1 ³­ V i , y V j , y a2 µij ν ∗ , V i , y ei − φ˜0 q(y) = a D2 φ˜0 y, y ν ∗ + 2 ∇φ˜0 , y 2 i=1

i,j=1

Therefore

˜ u ˜(ry + φ(ry)ν(ry)) =u ˜(P ) + rL(y) + r2 (M (y) + N (y)) + o(r2 ), where

­ ® L(y) = ∇˜ u(P ), (l(y) + y) , ­ ® M (y) = ∇˜ u(P ), q(y) ,

TWO PHASE PROBLEMS FOR ELLIPTIC OPERATORS WITH VARIABLE COEFFICIENTS

N (y) =

n n E X X ® 1D 2 1­ 2 D u ˜(P ) D u ˜(P )(l(y) + y), (l(y) + y) = Ni (y), Ni (y) 2 2 i=1

=

11

1 D¡

2

2 DN u ˜(P ) + DT2 u ˜(P )

with (for i = 1, . . . , n) Ni (y) = a

³­

¢

n X

Ni (y),

i=1

n X

i=1

E

Ni (y) ,

i=1

® ­ ®´ φ˜0 V i + pi ∇φ˜0 , y + a−1 y, ei ei ,

­ ® pi = ν ∗ , ei and V n = 0. Observing that Z L(y)dHn−1 (y) = 0 ∂B1

we have

Z

˜ u ˜(ry + φ(ry)h(ry))dH n−1 (y) ∂B1 Z 2 (N (y) + M (y))dHn−1 (y) + o(r2 ). =u ˜(P ) + r

˜r ≥ M

(38)

∂B1

Now, since ∇˜ u(P ) is parallel to A(0)ν ∗ , we have Z n−1 X ®n ­ ®o a­ ∗ ˜ ˜ ∇˜ u(P ), ν ∆φ(0) − φ0 µij a2 V i , V j . (39) M (y)dHn−1 (y) = 2n ∂B1 i,j=1

Set, for i = 1, . . . , n,

wi = api ∇φ˜0 + ei ­ ® so that Ni (y) = aφ˜0 V i + wi , y ei and Nn (y) = wn , y en . From (35), we get ­

(40)

®

o X­ ®2 ®­ ® ­ ® 1n X ­ ˜ i aφ0 V + wi , y βi (P ) + 2 aφ˜0 V i + wi , y wn , y d∗in + ( wn , y )2 d∗nn . N (y) = 2 n−1

n−1

i=1

i=1

Averaging over ∂B1 gives Z n−1 n−1 X X (41) N dHn−1 = c2i βi (P ) + 2 cin d∗in + c2n d∗nn , ∂B1

i=1

i=1

where, for i = 1, . . . , n − 1 ­ ® 2nc2i =| wi + aφ˜0 V i |2 , 2nc2n =| wn |2 and ncin = aφ˜0 V i + wi , wn . Step 3. We now choose the vectors V i , i = 1, . . . , n − 1 as follows. Set ki =

∗ ∗ αin αii , and γ = . i ∗ ∗ αnn αnn

Observe that (42)

| ki − 1 |≤ cω(m0 /Λ) and | γi |≤ cω(m0 /Λ).

For each i = 1, · · · , n − 1, minimize h(y) =| y |2 under the constraints (43)

| y + w i |2 = ki | w n |2

12

M. CRISTINA CERUTTI, FAUSTO FERRARI, SANDRO SALSA

and

­

(44)

® y + w i , w n = γi | w n |2 .

If | ∇φ˜0 | and ω(m0 /Λ) are small, Lagrange multiplier method gives the minimizer i ymin = η1i wi + η2i wn

where η1i η2i

n 3

¡

i

n 2

­

i

=| w | | w | w | − w , w

n

®

w

n

¢

|

−1

q

ki2 − γi2 − 1

q ¡ ­ ® ¢ ­ ® = γi + ki2 − γi2 | wn | · | wi | wn |2 − wi , wn wn |−1 wi , wn .

Notice that, if | ∇φ˜0 | is small, wi is ”almost” orthogonal to wn , therefore ¯ wn ­ i n ®¯¯ ¯ i w , w ¯ ≥ c > 0. ¯w − | w n |2 1 i We choose V i = y . It is not difficult to prove that, if | ∇φ˜0 | and ω(m0 /Λ) are aφ˜0 min small q ¯ ¯ ¡ ¢ ¯ ¯ i 2 2 ˜ | V | ≤ C | ∇φ 0 | + ¯1 − k i − γ i ¯ + γ i (45) ¢ C¡ | ∇φ˜0 | +ω(m0 /Λ) . ≤ φ˜0

Inserting Vi in (41) and taking in account (36) we have: Z c2 ˜ u(P ) = 0. N dHn−1 = ∗n L˜ αnn ∂B1 ­ ® Moreover, from (39), since ∇˜ u(P ), ν ∗ ≥ c | ∇˜ u(P ) |, (46)

Z

M dHn−1 ∂B1

ca | ∇˜ u(P ) | ≥ 2n

¾ ½ ¢ C ¡ 2 2 ˜ ˜ | ∇φ0 | +ω(m0 /Λ) ∆φ(0) − ˜ φ(0)

and from (38), if (33) holds, we conclude the proof.

Remark 4.2. In the case of the Laplace operator or of an operator with constant coefficients the above proof gives an approach alternative to the one in [3] and [7]. In this case, for every i = 1, . . . , n − 1, we have A∗ = I, ki = 1, γi = 0, pi = 0 while pn = 1. The choice of V i is ½ ¾ w0i 1 n i −ei + | w | , V = φ0 | w0i | ® n ­ where w0i = ei − ∇φ˜0 , ei |wwn |2 .

Remark 4.3. Lemma 4.1 holds for uniformly elliptic operators of the form n n X X L= aij (x) Dij + bi (x) Di + c (x) i,j=1

i=1

with c, b = (b1 , ...bn ) H¨ older continuous and c ≤ 0. Condition (30) has to be replaced in this case by ) ( |∇φ|2 + ω 2 + (|b| + |c|) (|∇φ| + ω) Lφ ≥C φ

TWO PHASE PROBLEMS FOR ELLIPTIC OPERATORS WITH VARIABLE COEFFICIENTS

13

We will actually apply Lemma 4.1 to u(x − τ ), where τ ∈ Rn is a vector belonging to the monotonicity cone Γ0 (en , θ) of u, so that we need a slightly different version of that lemma, expressed in the following corollary. Notice that condition (30) remains unchanged. Corollary 4.4. Let u, φ be as in Lemma 4.1, with φ, in particular, satisfying (30). Let v¯τ,φ (x) := sup u(y − τ ). Bφ(x)

where τ ∈ Rn is a fixed small vector. Then Lvτ,φ ≥ 0 in {vτ,φ > 0}. Proof. Let Mr,A,vτ,φ (x) =

Z

vτ,φ (x + rA(x)1/2 y)dHn−1 (y) ∂B1

Define B(z) = A(z + τ ), ψ(z) = φ(z + τ ). Then Z ³ ´ Mr,A,vτ,φ (0) = sup u −τ + rB 1/2 (−τ )y + ψ(−τ + rB 1/2 (−τ )ν) dHn−1 (y) ∂B1 |ν|=1

= Mr,B,v0,ψ (−τ ).

Since −τ is an arbitrary (small) vector, it is enough to control from below Mr,B,v0,ψ (0). This can be done exactly as in the proof of Lemma 4.1 replacing φ0 with ψ0 = ψ(0) and A0 with B0 = B(0). The conclusion immediately follows.

5. A family of subsolutions If φ satisfies (30) and we apply Lemma 4.1 to the positive and negative parts of the solution u of our free boundary problem, we conclude that the function (47)

vφ (x) =

sup

u

Bφ(x) (x)

¡ ¢ ¡ ¢ is L-subharmonic in Ω+ vφ ∪ Ω− vφ . Moreover, if |∇φ| is small enough, i.e. |∇φ| ≤ c < 1, the conclusions in Lemma 10 and Lemma 11 in [3] hold for vφ , with identical proofs. For the reader’s convenience, we list them below: ¡ ¢ a) F vφ is a Lipschitz graph with Lipschitz constant L0 ≤ L + C sup |∇φ|.

¡ ¢ ¡ ¢ b) Every point of F vφ is regular. In particular, if x1 ∈ Ω+ vφ and y1 ∈ Ω+ (u) ∩ ∂Bφ(x1 ) (x1 ), then x1 is regular from the right. Furthermore, if u has the following asymptotic behavior near y1 ¡ ¢ u(y) ≥ α < y − y1 , ν >+ − β < y − y1 , ν >− + o |y − y1 | then, near x1

¡ ¢ vφ (x) ≥ α ¯ < x − x1 , ν¯ >+ − β¯ < x − x1 , ν¯ >− + o |x − x1 | , ¡ ¢ where ν¯ = ν + ∇φ(x1 ) /|ν + ∇φ(x1 )| and µ ¶ α ¯ β¯ ≥ G (48) 1 − |∇φ(x1 )| 1 + |∇φ(x1 )|

14

M. CRISTINA CERUTTI, FAUSTO FERRARI, SANDRO SALSA

Using (47) and (48) we now construct a family of subsolutions. Let x0 = 14 en and, as in [7], let M be a large positive number and (49)

ψ(x) =

¡

16M − 2M

¢−4

© ª4 max |x − x0 |−M − 2M , 0 ;

then φ ∈ C 2 , φ ≡ 0 outside B1/2 (x0 ), 0 ≤ ψ ≤ 1 and |∇ψ| ≤ C(M ) in B1 \ B1/16 (x0 ), ψ ≥ η¯ > 0 in B1/8 and Lψ ≥ C M |∇ψ| .

Lemma 5.1. Let C¯ > 0. There exist positive numbers η¯, µ, ω0 , ω0 0 and ²2 ≥ Cω 6. ²-monotonicity improvement at the free boundary We now use the family of L-subharmonic functions constructed in Lemma 5.1 and Corollary 5.2 to get an improvement in the opening of the monotonicity cone up to the free boundary. At variance with the constant coefficients case, it is not possible possible to carry at once to the free boundary the full monotonicity gain expressed in Theorem 3.4. Instead we will show that if ² is a given small positive number, a suitable blow up of the solution u is ²-monotone up to the free boundary along the direction of a larger cone. We achieve the C 1,γ regularity of F (u) through an iteration process consisting in balancing ²-monotonicity, cone opening, and rescaling. We recall that a function is ²-monotone in a domain D, along a direction τ if u(x) − u(x − ²0 τ ) ≥ 0,

for every x ∈ D and every ²0 ≥ ². We start with the following extension of Lemma 14 in [3]. Lemma 6.1. Let u1 ≤ u2 be two solutions of our free boundary problem in B1 , with ¯ η¯, µ as in Lemma (5.1). Assume that, in B1 : 0 ∈ F (u2 ). Let ² > 0 and C, 2

i) || As (x) − As (0) ||≤ ω0 , with ω0 ≤ min{ ²C¯ , 12 η¯µ}; ii) v² (x) = supB² (x) u1 ≤ u2 (x), in B1−² ; iii) for σ > 0, small and x0 = 41 en , B1/16 (x0 ) ⊂ Ω+ (u1 ) and

v² (x0 ) ≤ (1 − σ²)u2 (x0 ). ¯ = h(n, ¯ µ, L, Λ) such that in B1/8 Then, for ² small enough, there exists h

(51)

v(1+hσ)² (x) ≤ u2 (x). ¯

TWO PHASE PROBLEMS FOR ELLIPTIC OPERATORS WITH VARIABLE COEFFICIENTS

15

Proof. Let φt , 0 ≤ t ≤ 1 be the family constructed in Lemma 5.1 and let

(52)

vt (x) =

sup u1 + cσ²wσt (x)

B²φσt (x)

¯9/10 \ B1/16 (x0 ) defined by the following where wt is the continuous function in Ω = B set of conditions:  2  L wt = 0, x ∈ Ωt = Ω+ (vt ) \ B1/8 (x0 ), wt = 0, x ∈ Ω− (vt ) ∪ ∂B9/10 ,  wt = u2 (x0 ), x ∈ ∂B1/16 .

If ² and µ in Lemma 5.1 are small enough, vt is a family of subsolutions admissible for the comparison Lemma 7 in [3]. The rest of the proof then follows as in Lemma 14 of [3]. Remark 6.2. The proof of Lemma 6.1 gives actually uλ0 (x) − uλ0 (x − τ ) ≤ c0 ²δu(x),

in B1/2 , for every τ ∈ Γ(ν1 , θ1 ), since by the boundary Harnack principle (see [1] or [6]), wt (x) ≤ Cu(x) in B1/2 . To apply Lemma 6.1 in our situation we first fix ² = ²0 small and choose a vector τ ∈ Γ0 (en , θ/2) such that, ²0 ≥| τ | sin 2θ . Then we set u1 (x) = uλ0 (x − τ ), u2 (x) = uλ0 (x) changing the starting rescaling factor 2/β from λ to λ0 ≤ min{²0 , λ}, where β is defined in (10). Moreover let σ = cδ (see (18)). In this way i) is satisfied with ω0 = cλβ0 and the other hypothesis are satisfied thanks to Theorem 2.1 and Theorem 3.4. Corollary 5.2 assures that vt in (52) is L2 -subharmonic in Ωσt , although u1 is a solution of a free boundary problem corresponding to shifted operators. Therefore, in B1/8 , Rescaling back to u we get:

v(1+chσ)² (x) ≤ u(x). ¯

Lemma 6.3. In Bλ0 /8 u is λ0 ²0 -monotone along every τ ∈ Γ(ν1 , θ1 ) with π (53) δ1 ≤ bδ (δ1 = − θ1 ) 2 and (54) where b = b(n, Λ, a, L) < 1. Corollary 6.4. Let ²k =

²0 , 2k

| ν1 − en |≤ Cδ, ¯ 2/β , λ}. Then, in B λk u λk = λk+1 with λ0 ≤ min{(²/2C) 0

is λk ²k -monotone along every τ ∈ Γ(ν1 , θ1 ).

8

7. Final iteration Corollary 6.4 says that we can improve the ²-monotonicity (by decreasing ²) along the directions of the enlarged cone, in smaller and smaller neighborhoods of the origin. To prove C 1,γ regularity of the free boundary we need to show that along the sequence of balls Bλk /2 not only we improve the ²-monotonicity, but also the opening of the monotonicity cone, with a geometric decay of the defect angle δk . We first need to observe that, as in the constant coefficient case, ²-monotonicity forces full monotonicity M ² away from the free boundary (see [4], Lemma 1). This is consequence of the following lemma. Lemma 7.1. Let u > 0, Lu = 0 in T1 . Assume that

16

M. CRISTINA CERUTTI, FAUSTO FERRARI, SANDRO SALSA

i) u is ²-monotone along a direction τ and moreover, if x ∈ T2/3 , u(x) − u(x − τ ) ≥ c0 ²δu(x)

(55)

¯ 1 c0 ²δ}; ii) || A(x) − A(0) ||≤ crβ ≤ ω, with ω ≤ min{²2 /C, 3 iii) B2M ² (x0 ) ⊂ T1/2 .

Then if M is large enough, Dτ u(x0 ) > 0.

Proof. We may assume A(0) = I. Let v be harmonic in BM ² (x0 ) with u = v on ∂BM ² (x0 ). Then, as in Lemma 3.1, | u − v |≤ crβ u(x0 ) in BM ²/2 (x0 ). Therefore, in BM ²/2 (x0 ), using Harnack inequality, v(x) − v(x − ²τ ) ≥ u(x) − u(x − ²τ ) − 2crβ u(x0 ) 2 ≥ (c0 ²δ − c0 ²δ)u(x0 ) ≥ c0 ²δu(x0 ) 3 and v is ²-monotone along τ in BM ²/2 (x0 ). ¿From Lemma 1 in [4] we have, if M is large, c Dτ v(x0 ) ≥ (v(x0 ) − v(x0 − ²τ )) ≥ c²u(x0 ). ² ¿From Shauder estimates we get Dτ v(x0 ) ≤ Dτ u(x0 ) +

crβ c u(x0 ) ≤ Dτ u(x0 ) + δu(x0 ). M² M

In conclusion we obtain, Dτ u(x0 ) ≥ Dτ v(x0 ) −

c1 c δu(x0 ) ≥ δ(c − )u(x0 ), M M

which is positive if M is large enough. End of proof of Theorem 1.2. We apply Lemma 7.1 to the positive part of uλ1 (x) = u(λλ11 x) . Notice that Bλ1 ⊂ Bλ0 /8 and that, in Bλ1 , u is λ1 ²1 -monotone, so that uλ1 is ²1 -monotone in B1 along any τ ∈ Γ(ν1 , θ1 ). The validity of (55) is assured by Remark 6.2. By Lemma 7.1, uλ1 is fully monotone M ²1 away from F (u) and we perform the interior gain process described in Section 3. In order to assure the validity of Lemma 3.2 we choose λ0 ≤ min{( 2²C0¯ )2/β , λ, b1/β }, where b is defined in (53). From the propagation Lemma 6.1, applied to u1 (x) = uλ1 (x − τ ) and to u2 (x) = uλ1 (x), we deduce, after rescaling back, that u is λ1 ²1 -monotone in Bλ1 /8 along any τ ∈ Γ(ν2 , θ2 ) with | ν2 − ν1 |≤ cδ2 and δ2 ≤ bδ1 ≤ b2 δ, (δ2 = π2 − θ2 ). Iterating the process we construct a sequence of balls Bλk /8 in which u is λk ²k -monotone along every τ ∈ Γ(νk , θk ) with δk+1 ≤ bδk (δ0 = δ, δk =

π − θk ) 2

and | νk+1 − νk |≤ cδk . These conditions imply that F (u) is C 1,γ , γ = γ(b), at the origin.

TWO PHASE PROBLEMS FOR ELLIPTIC OPERATORS WITH VARIABLE COEFFICIENTS

17

References [1] A. Ancona Principe de Harnack ` a la fronti`ere et th´eor`eme de Fatou pour un op´erateur elliptique dans un domaine lipschitzien. Ann. Inst. Fourier (Grenoble) 28 (1978), no. 4, 169–213, x. [2] I. Athanasopoulos; L. Caffarelli; S. Salsa, Regularity of the free boundary in parabolic phasetransition problems, Acta Math. 176 (1996), no. 2, 245–282. [3] L. Caffarelli, A Harnack inequality approach to the regularity of free boundaries, Part 1: Lipschitz free boundaries are Cα1 , Revista Matematica Iberoamericana, 3 (1987), 139–162. [4] L. Caffarelli A Harnack inequality approach to the regularity of free boundaries. II. Flat free boundaries are Lipschitz, Comm. Pure Appl. Math. 42 (1989), no. 1, 55–78. [5] L. Caffarelli; E., Fabes; S., Mortola; S., Salsa, Boundary behavior of nonnegative solutions of elliptic operators in divergence form, Indiana Univ. Math. J. 30 (1981), no. 4, 621–640. [6] E. Fabes;N. Garofalo; S. Mar´in-Malave; S., Salsa, Fatou theorems for some nonlinear elliptic equations. Rev. Mat. Iberoamericana 4 (1988), no. 2, 227–251. [7] M. Feldman, Regularity for nonisotropic two-phase problems with Lipschitz free boundaries. Differential Integral Equations 10 (1997), no. 6, 1171–1179. [8] M. Feldman, Regularity of Lipschitz free boundaries in two-phase problems for fully nonlinear elliptic equations, Indiana Univ. Math. J. 50 (2001), no. 3, 1171–1200. [9] D.S.,Jerison;C.E. Kenig, Boundary behavior of harmonic functions in nontangentially accessible domains. Adv. in Math. 46 (1982), no. 1, 80–147. [10] P.Y, Wang Regularity of free boundaries of two-phase problems for fully nonlinear elliptic equations of second order. I. Lipschitz free boundaries are C 1,α , Comm. Pure Appl. Math. 53 (2000), no. 7, 799–810.

Maria Cristina Cerutti Dipartimento di Matematica del Politecnico Piazza Leonardo da Vinci, 32, 20133 Milano, Italy. E-mail: [email protected]

Fausto Ferrari Dipartimento di Matematica dell’Universit`a Piazza di Porta S. Donato, 5, 40126 Bologna, Italy and C.I.R.A.M. Via Saragozza, 8, 40123 Bologna, Italy. E-mail: [email protected]

Sandro Salsa Dipartimento di Matematica del Politecnico Piazza Leonardo da Vinci, 32, 20133 Milano, Italy. E-mail: [email protected]