On the Two-Phase Membrane Problem With Coefficients Below the Lipschitz Threshold

On the two-phase membrane problem with coefficients below the Lipschitz threshold

Erik Lindgren, Henrik Shahgholian, Anders Edquist

Department of Mathematics, Royal Institute of Technology, 100 44 Stockholm, Sweden Received 4 November 2008; received in revised form 5 March 2009; accepted 23 March 2009

Available online 27 May 2009

Abstract

We study the regularity of the two-phase membrane problem, with coefficients below the Lipschitz threshold. For the Lipschitz coefficient case one can apply a monotonicity formula to prove theC^1,1-regularity of the solution and that the free boundary is, near the so-called branching points, the union of twoC¹-graphs. In our case, the same monotonicity formula does not apply in the same way. In the absence of a monotonicity formula, we use a specific scaling argument combined with the classification of certain global solutions to obtainC^1,1-estimates. Then we exploit some stability properties with respect to the coefficients to prove that the free boundary is the union of two Reifenberg vanishing sets near so-called branching points.

©2009 Elsevier Masson SAS. All rights reserved.

MSC:35J70; 35J60; 35J85

Keywords:Obstacle problem; Elliptic equation; Regularity

1. Introduction and main result

1.1. Problem

Given two strictly positive functionsλ₁andλ₂and boundary datag∈H¹(B₁)∩L^∞(B₁)we study the minimizer of the functional

J (u)=

B₁

|∇u|²

2 +λ1(x)u⁺+λ2(x)u⁻dx (1.1)

over the set{u: u−g∈H₀¹(B₁)}, with its corresponding Euler–Lagrange equation

u=λ₁(x)χ_{u>0}−λ₂(x)χ_{u<0} inB₁. (1.2)

✩ This research project is part of the ESF program Global. E. Lindgren has been supported by grant KAW 2005.0098 from the Knut and Alice Wallenberg foundation. H. Shahgholian is partially supported by the Swedish Research Council.

* Corresponding author.

E-mail addresses:eriklin@math.kth.se (E. Lindgren), henriksh@math.kth.se (H. Shahgholian), edquist@math.kth.se (A. Edquist).

0294-1449/$ – see front matter ©2009 Elsevier Masson SAS. All rights reserved.

doi:10.1016/j.anihpc.2009.03.006

Fig. 1.x¹a positive one-phase point,x²a negative one-phase point,x³a branching point andx⁴a non-branching two-phase point.

The existence and uniqueness of minimizers of (1.1) and solutions of (1.2) can be obtained by standard methods, see [16]. In general, if g attains both positive and negative values, u will have a jump where u changes sign, i.e. across the set ∂{u=0}. We call this set the free boundary, and also we denote its two parts ∂{u >0} and

∂{u <0}by Γ⁺(u) and Γ⁻(u) respectively (see Fig. 1). The main purpose of the present paper is to study the regularity of u andΓ (u). Note that in the sets{u >0}and{u <0}the regularity of uis completely determined by the regularity of the coefficients λ_i. The difficulty that arises, when we have coefficients that are not Lipschitz continuous, is that we cannot directly apply the monotonicity formula from [6]. In the absence of a monotonicity formula, we use a scaling argument combined with the classification of certain global solutions to obtain C^1,1- estimates. These arguments are elaborated versions of those introduced in [8]. Once the optimal regularity is settled we exploit some stability properties with respect to the coefficients to prove that the free boundary is the union of two Reifenberg vanishing sets near so-called branching points. The stability arguments here are based on the ideas in [3].

1.2. Known results

The one-phase case, i.e. when the minimizer is assumed not to change sign, is the classical obstacle problem and it has been well-studied before. In [5] it is proved that minimizers are locallyC^1,1and that the free boundary is locally aC^1,α-graph if the coefficients are Lipschitz and if the zero set satisfies a certain thickness assumption. Later, in [3]

Blank proved that minimizers are locallyC^1,1and that the free boundary is locally aC¹-graph under similar thickness assumptions when the coefficients are only assumed to be Dini continuous.

The two-phase case (when the minimizer is allowed to change sign) has also been studied before under stronger as- sumptions on the coefficients. In [15] theC^1,1-regularity of minimizers is proved when the coefficients are assumed to be constant. This result was extended in [12] to the case when the coefficients are assumed to be Lipschitz. Moreover, in [13] global solutions of (1.2) is considered and classified in the case of constant coefficients. This result is then later used in [14] to prove that in the case of Lipschitz coefficients, the free boundary is the union of twoC¹-graphs close to branching points, i.e. points close to the set∂{u >0} ∩∂{u <0} ∩ {|∇u| =0}.

1.3. Main result

The main result of this paper is that solutions to (1.2) are locallyC^1,1if the coefficientsλ_i are Hölder continuous and that the free boundary,Γ (u), is the union of two Reifenberg vanishing sets close to branching points under the weaker assumption that theλ_i’s are merely continuous. In order to state our main theorems in their precise forms we must define the class of solutions that we consider in this paper.

Definition 1.1.We sayu∈P1(M, M0, ω)if

(1) u=λ₁(x)χ_{u>0}−λ₂(x)χ_{u<0}inB₁, in the sense of distributions.

(2) sup_B1|u|M₀. (3) infλi1/M >0.

(4) supλiM.

(5) λ₁andλ₂are uniformly continuous withωas modulus of continuity.

(6) 0∈∂{u >0} ∩∂{u <0} ∩ {|∇u| =0}.

Our two main theorems are the following:

Theorem 1.2.Assume that u∈P1(M, M0, ω)where ω(r)Mr^α for some0< α <1. Then there arer0 and C depending onM,M0and the dimension such that

u_C^1,1_(Br

0)C.

Remark 1.3.The observant reader might ask if condition (3) is really necessary for theC^1,1-regularity. Indeed, if λ₁=λ₂=0 then the function would be harmonic and therefore analytic. However, the technique used in this paper relies heavily on a classification of global solution which would fail ifλ_i =0.

Theorem 1.4.Assumeu∈P1(M, M0, ω). Then there arer0andγdepending onM,M0andωsuch that|∇u(y)|γ and dist(y, Γ^±(u))γ implies thatΓ (u)⁺∩Br₀(y)and Γ (u)⁻∩Br₀(y)are both Reifenberg vanishing sets. In particular they both admit Hölder parameterizations.

Remark 1.5.Interesting here is to note that the assumptions onωare weaker in Theorem 1.4 than in Theorem 1.2.

The reason for that is that there are some technical difficulties in Proposition A.1 in Appendix A which we need in order to obtainC^1,1-estimates. See also Remark 2.9.

2. C^1,1-estimates

In this section we prove Theorem 1.2. This is done by first proving a weaker type of quadratic growth away from branching points. The method used here is a contradictory scaling argument very similar to the one used in [8]. After that we use some properties of coercive elliptic systems to obtainC^1,1-regularity close to points on the free boundary where the gradient does not vanish. Gluing these pieces together we finally obtainC^1,1-estimates close to branching points.

2.1. Non-degeneracy

Here we present the well known-result that the solutions do not grow too slow around free boundary points; the proof is standard.

Proposition 2.1.Letu∈P1(M, M0, ω). Then there is a constantcdepending on the dimension andMsuch that for ally∈Γ (u)∩B₁

sup

B_r(y)∩{u>0}ucr²

and

B_r(y)inf∩{u<0}u−cr², forr <dist(y, ∂B1).

Proof. Let y ∈ {u >0} ∩B₁ and w(x)=u(x)−t|x −y|² and r so small that B_r(y)⊂B₁. Then w0 in {u >0} ∩B₁ if t=2ninfλ₁. Since w(y) >0 andw is subharmonic, there isx_y∈∂(B_r(y)∩ {u >0}) such that w(x_y) >0. Now onΓ⁺(u)we havew0, hencex_y∈∂B_r(y)∩ {u >0}. In other words

sup

∂B_r(y)∩{u>0}w >0.

Takingz∈Γ⁺(u)we can find a sequence of pointsy_n∈ {u >0}such thaty_n→z, then by continuity we obtain sup

B_r(z)∩{u>0}

w0,

which implies the desired result. OnΓ⁻(u)we can argue similarly. 2 2.2. Classification of global solutions of a related problem

In order to go through with our scaling argument we will need a classification of global solutions to the equation u=λ1χ_{u>−ax₁}−λ2χ_{u<−ax₁},

whereλ1andλ2are constants. One major difference here is that one cannot apply the monotonicity formula from [1]

to the pair of functions(∂eu)^±for all directionse, only for those directions orthogonal toe1. Of course, translating the functionulinearly withax1yields a global solution to the two-phase obstacle problem. However the gradient will not vanish at the origin, so we cannot apply the classification from [13] directly.

Lemma 2.2.Letube a solution of the following problem

⎧⎨

⎩

u=λ₁χ_{u>−ax1}−λ₂χ_{u<−ax1} inRⁿ, u(0)=∇u(0)=0,

0∈∂{u=0},

(2.1)

wherea >0andλ₁andλ₂are constants. Assume also that for someC >0 sup

B_r

|u|Cr² (2.2)

wheneverr >1. Then one of the following holds (1) u(x)=^λ₂¹(x₁⁺)²−^λ₂²(x₁⁻)²,

(2) u(x)=^λ₂¹(x₁⁺)², (3) u(x)= −^λ₂²(x₁⁻)². In particular

sup

B₁

|u|1

2max(λ1, λ₂).

Remark 2.3. A more general form of this lemma with the right-hand side equal to λ1χ_{u>−L(x)}−λ2χ_{u<−L(x)} withLbeing a linear function can easily be obtained by a change of coordinates.

Proof of Lemma 2.2. We observe that (∂_eu)=∂_eu= λ₁+λ₂

|∇(u+ae1)|(∂_eu+ae·e₁)Hⁿ⁻¹{u= −ax₁}. Therefore withv^±=(∂eu)^±fore·e1=0 we have

(1) v⁺·v⁻=0, (2) (v^±)0, (3) v^±(0)=0.

Hence, the Alt–Caffarelli–Friedman monotonicity formula applies tov^±. It states that with φ(r, v)= 1

r⁴

B_r

|∇v⁺|²

|x|ⁿ⁻² dx

B_r

|∇v⁻|²

|x|ⁿ⁻² dx

we haveφ(r, v)0 for allr. Moreover, ifφ(r, v)=0 for allrthen one of the functionsv^±vanishes.

Defineu_r(x)=u(rx)/r². Now we wish to study the behavior ofu_r as rtends to∞and 0. Since the Laplacian is invariant under the quadratic scaling,u_r is uniformly bounded independently ofr. We observe that the function v=u+ax₁ is a solution to the usual two-phase membrane problem. Thereforev∈C_loc^1,1(Rⁿ)by the result in [12]

and [15]. Together with the hypotheses this implies that we have uniform quadratic growth asr→0 and asr→ ∞.

This allows us to extract a subsequencer_j tending either to 0 or to∞such thatu_rj converges to a limit inC_loc^1,α∩W_loc^2,p for any 0< α <1 and 1< p <∞.

We first consider the case whenr→ ∞: We claim thatu_rj tends tou_∞which is a global solution to the ordinary two-phase obstacle problem.

Indeed, in{u_∞>0}we haveu_∞=λ₁and in{u_∞<0},u_∞= −λ₂. Therefore it remains to determine what happens in the set{u_∞=0}. Now we can use the implicit function theorem to conclude that{u_∞=0} ∩ {|∇u_∞| =0}

is locally aC¹-surface and thus it has zero measure. When|∇u_∞| =0 we use thatu_∞∈W_loc^2,1and obtainu_∞=0 a.e. in this set.

By theC^1,α-convergence it follows thatu_∞(0)= |∇u_∞(0)| =0 and also by Proposition 2.1 the origin must be a two-phase point also foru_∞. Hence, by the classification done in [13]u_∞is a one-dimensional solution in some directionν and must be of form (1). Therefore, for any directionewe haveφ(1, ∂eu_∞)=0 which together with the ACF monotonicity formula mentioned above implies thatφ(r, ∂_ev)=0 for allr >0 andeorthogonal toe₁. Hence

∂_eu0 (or0) for any directionesuch thate·e₁=0. There are(n−1)such directions, and thus by calculus we can reduce the dependence ofu in those directions to one. Soucan only depend on two directions, saye₁ande₂. Moreover we know that∂_e2u0.

Now we consider instead the case whenr→0: We know thatu_rj solves u_rj=λ₁χ_{{urj>−(a/rj)x1}}−λ₂χ_{{urj<−(a/rj)x1}}.

In{ur_j >−(a/rj)x1},ur_j=λ1and we can use standard estimates to conclude that thereur_j ∈C^∞uniformly. This is also true in the set{ur_j<−(a/rj)x1}.

Now consider the remaining set,{ur_j= −(a/rj)x1}. We know that we have quadratic growth and thereforeur_j is bounded on every compact set. Thus in every compact subset of{x1>0}we have thatur_j >−a/rjx1forrj small enough. Using the same argument in{x1<0}with{ur_j <−(a/rj)x1}we can therefore conclude thatur_j →u0, where

u0=λ1χ_{x₁>0}−λ2χ_{x₁<0}.

We also observe that by the reasoning above we have that u_rj →u₀ inC^∞(K) for any compact K contained in {x1=0}. Lethbe the one-dimensional solution in thex1-direction given by

h(x)=λ1

2 x₁⁺2

−λ2

2 x₁⁻2

.

Then(u₀−h)=0. Moreoveru₀ has quadratic growth at infinity, 0= |∇u₀(0)| =u₀(0)and 0∈∂{u₀=0}. This implies that u₀−his a quadratic harmonic polynomial of two variables P (x₁, x₂)and thereforeP is of the form C(x₁²−x₂²)+Ax₁x₂. Since we also know that∂_e2u₀0 we must have

0∂_e2P (x₁, x₂)= −2Cx2+Ax₁

which can only be true ifC=A=0. Thus we can conclude thatP =P (e₁)soP=0 since it must be quadratic and harmonic. Moreover, the fact thatu₀only depends onx₁and thatu_rj→u₀inC_loc² (Rⁿ\ {x₁=0})implies

0=∂1∂2u0(x)= lim

r_j→0∂1∂2u(rjx), (2.3)

for anyx /∈ {x₁=0}. Assume now thatudoes really depend onx₂. By the implicit function theorem, the free boundary is near the origin at the graph of aC^1,α-function. Hence we can apply the Hopf’s lemma from [17]. We observe that

∂2u(0)=0 and∂2u0, moreover∂2uis harmonic away from the free boundary, which has normale1. Hopf’s lemma says that

lim inf

x→0 ∂₁∂₂u(x) >0

which contradicts (2.3). Thereforeuis one-dimensional and the following hold:

(1) u(x₁)=^λ₂¹(x₁⁺)²−^λ₂²(x₁⁻)², (2) u(x1)=^λ₂¹(x₁⁺)²,

(3) u(x1)= −^λ₂²(x₁⁻)².

This is a consequence of solving (2.1) in one dimension:

u=λ₁χ_{u>−ax1}−λ₂χ_{u<−ax1} inR.

In{u >−ax₁}every solution is of the formu(x₁)=λ₁x₁²/2+Ax+B. Using the continuity we know thatuandu tend to zero as x1→0⁺, thusA=B=0 andu(x1)=λ1x₁²/2 is the only solution. The same arguments applied in{u <−ax₁}imply thatu(x₁)= −λ₂x₁²/2 is the only possible solution. In the remaining set{u= −ax₁}we only have the zero solution. The zero solution itself does not contain a free boundary point at the origin and thus, as stated above we need to combine the zero solution with at least one of the first two solutions to obtain a global solution. 2 2.3. Quadratic growth

Now we are ready to prove a certain type of quadratic growth which validity depends on the modulus of the gradient. Before proving that we need a result that says that the gradient have to vanish on the free boundary for global solutions having one-phase points.

Lemma 2.4.Letλ1andλ2be constants. Supposeusatisfies (1) sup_Bρ|u|Cρ²forρ >1,

(2) u=λ₁χ_{u>0}−λ₂χ_{u<0}inRⁿ, (3) 0∈Γ⁺(u)\Γ⁻(u),

for some constantC. Then|∇u| =0onΓ (u).

Proof. If the statement of the lemma does not hold then there must be a pointy∈Γ (u)where the gradient does not vanish. Theny must be a two-phase point. Since we have a solution to the two-phase problem with constant coeffi- cients the Alt–Caffarelli–Friedman monotonicity formula applies tow^±_e =(∂ev₀^C)^±as in [13]. Now as in Lemma 2.2 we study the limits of the scalings

vj(x)=u(rjx)/r_j²

for sequencesr_j→ ∞. The functionsv_j satisfy

(1) sup_B1|v_j|cfrom Proposition 2.1, (2) sup_Bρ|vj|Cρ²forρ >1,

(3) v_j=λ₁χ_{vj>0}−λ₂χ_{vj<0}inRⁿ, (4) 0∈Γ⁺(v_j)\Γ⁻(v_j).

So as before we have, passing to a subsequence if necessary, thatv_j→v₀inC_loc^1,α(Rⁿ)wherev₀satisfies (1) sup_B1|v0|c,

(2) sup_Bρ|v₀|Cρ²forρ >1, (3) v₀(0)= |∇v₀(0)| =0,

(4) v0=λ1χ_{v0>0}−λ2χ_{v0<0}inRⁿ.

Also, since there is a two-phase pointy, any such limitv0must by Proposition 2.1 have a two-phase point at the origin.

Hence we can apply the classification of global solutions from [13] and conclude that it must be one-dimensional. This in turn implies that the functionsφ(r, w_e)in the monotonicity formula vanishe for allr and all directionse, which as in Lemma 2.2 gives thatumust be one-dimensional. Ifu is one-dimensional it is easy to see that∇uvanishes onΓ (u). 2

Now we can proceed with the growth result.

Proposition 2.5.Letu∈P₁(M, M₀, ω). ForA >0there are anε=ε_Aand anr₀=r₀(A)such that ifsupω < εthen for anyy∈B_1/4∩∂{u=0} ∩ {0<|∇u|< Ar}we have

S_r(y, u)CM₀r², for allr < r₀. Here

Sr(y, u)=sup

x∈Br

u(y+x)− ∇u(y)·x

andCdepends onMand the dimension.

Proof. We argue by contradiction. Assume thatuj∈P1(M, M0, ω)with supω < εj→0 so that the statement of the proposition fails foru_j. Then for a subsequence, again labeledu_j, we know thatu_j converges to a solutionu₀ of the two-phase obstacle problem with constant coefficients. For this problem there are localC^1,1-estimates (see [12]

and [15]). ThereforeS_r(y, u₀)C₁(λ_i(0))M0r²for anyy∈B1

4 and anyr1/4. This implies

S_r(y, u_j)C₁M₀r²+τ (j ). (2.4)

In particular, sinceτ→0 asj→ ∞we have S_r(y, u_j)2C1M₀r²

forj large enough at least whenr=1/4. For any sequencey_j∈B1

4 ∩∂{u_j =0}and for a constantC > C₁to be chosen later define

rj=sup r: Sr(yj, uj) >2CM0r² .

If the assertion fails forCthen there is a convergent sequencey_j→y₀such thatr_j=r_j(y_j)→0, since we otherwise would have a contradiction to (2.4). Moreover, we know thatr_j1/4. Let

v_j(x)=uj(rjx+yj)− ∇uj(yj)·rjx

r_j² .

Then

(1) sup_B1|vj| =2CM0,

(2) sup_Bρ|v_j|2CM0ρ²forρ >1,

(3) v_j(0)= |∇v_j(0)| =0,

(4) v_j=λ₁(r_jx+y_j)χ_{{vj>−∇u(yj)·x/rj}}−λ₂(r_jx+y_j)χ_{{vj<−∇u(yj)·x/rj}}inB_1/rj.

Note that|∇u_j(y_j)|/r_j< A, which means that for a subsequence∇u_j(y_j)/r_j→νwhere|ν|A. We consider two cases:ν=0 orν=0.

Case 1.ν=0. Since|v_j|is uniformly bounded onB_1/rj this implies that for a subsequencev_j→v₀inC^1,α_loc(Rⁿ)∩ W_loc^2,p(Rⁿ)for some functionv₀which satisfies

(1) sup_B1|v0| =2CM0,

(2) sup_Bρ|v₀|2CM0ρ²forρ >1, (3) v₀(0)= |∇v₀(0)| =0,

(4) v₀=λ₁(y₀)χ_{{v0>−ν·x}}−λ₂(y₀)χ_{{v0<−ν·x}}inRⁿ. By Lemma 2.2

sup

B₁

|v₁|1 2max

λ₁(y₀), λ₂(y₀) ,

which will contradict (1) if 4CM0>max(λ1(y₀), λ₂(y₀)).

The second case needs to be split into two subcases: 1) In the limitv₀has a two-phase point at the origin or 2) in the limitv₀has a one-phase point at the origin.

Case 2a.ν=0 and in the limit we have a two-phase point at the origin. Since|v_j|is uniformly bounded onB_1/rj this implies that for a subsequencev_j→v₀inC_loc^1,α(Rⁿ)for some functionv₀which satisfies

(1) sup_B1|v₀| =2CM0,

(2) sup_Bρ|v₀|2CM0ρ²forρ >1, (3) v₀(0)= |∇v₀(0)| =0,

(4) v₀=λ₁(y₀)χ_{v0>0}−λ₂(y₀)χ_{v0<0}inRⁿ, (5) 0∈Γ⁺(v₀)∩Γ⁻(v₀).

Now we can apply the classification of global solutions done in [13] and obtain sup

B₁

|v₀|1 2max

λ₁(y₀), λ₂(y₀) ,

which will contradict (1) if 4CM0>max(λ1(y₀), λ₂(y₀)).

Case 2b.ν=0 and in the limit we have a one-phase point at the origin. We treat the case when this is a positive one-phase point. As before we havev_j→v0inC_loc^1,α(Rⁿ)for some functionv0which now will satisfy

(1) sup_B1|v0| =2CM0,

(2) sup_Bρ|v₀|2CM0ρ²forρ >1, (3) v₀(0)= |∇v₀(0)| =0,

(4) v₀=λ₁(y₀)χ_{v0>0}−λ₂(y₀)χ_{v0<0}inRⁿ, (5) 0∈Γ⁺(v₀)\Γ⁻(v₀).

By Lemma 2.4 this implies that∇v0vanishes onΓ (v0). Hencev^±₀ are solutions to the one-phase problem, i.e. to the obstacle problem, which in turn implies that we have

v⁺₀ =λ₁(y₀)χ_{v+

0>0} inRⁿ,

and

v₀⁻=λ2(y0)χ_{v−

0>0} inRⁿ.

Moreover, we know from (5) that 0∈Γ (v₀). Then Lemma 2.2 from [3] applies when|v₀|is considered to be a solution inB₂. This gives

sup

B1

|v0|C

λ1(y0)+λ2(y0) ,

whereCdepends only on the dimension. From (1) we know sup

B1

|v₀| =2CM0.

This is contradiction ifCis chosen big enough. 2

Proposition 2.5 only gives estimates for points outside a special neighborhood of the free boundary. In order to control the growth around free boundary where the gradient does not vanish, we need the following proposition.

Proposition 2.6.Letu∈P1(M, M0, ω)whereω(r)Mr^α. Then there are constantst0andCdepending onM,M0

and the dimension such that u_C^1,1_{(Bt0|∇u(y)|(y))}C

for ally∈Γ (u)∩ {0<|∇u(y)|r₀}, wherer₀=r₀(1)is from Proposition2.5.

Proof. Takey∈Γ (u)∩ {|∇u|>0}and definer_y= |∇u(y)|. Let v(x)=u(ryx+y)

r_y² .

Then Proposition 2.5 along with simple calculations give (1) v(x)=λ1(ryx+y)χ_{v>0}−λ2(ryx+y)χ_{v<0}, (2) sup_B1|v|CM0forry< r0,

(3) |∇v(0)| =1, (4) 0∈Γ⁺(v)∩Γ⁻(v).

By Proposition A.1 in Appendix A this implies thatvC^1,1(Bt0)C(M, M0), which after rescaling implies the desired result. 2

2.4. C^1,1-estimates

The result in the previous section actually implies localC^1,1-estimates under the assumption that the functionsλ_i areC^1,1-potentials, that is, there are functionsf_i∈C^1,1such thatλ_i=f_iwhich is of course the case if theλ_i’s are supposed to beC^α. First we need the following lemma.

Lemma 2.7.Suppose thatv=ψinB_2R andψ∈C^1,1(B_2R), then D²v

L^∞(BR)C

vL^∞(B2R)

R² +D²ψ

L^∞(B2R)

,

whereCdepends only on the dimension.

Proof. Putw=v−ψ−x· ∇ψ (0)−ψ (0). Thenwis harmonic inB_2Rand from classical estimates for harmonic functions we have

sup

B_R

D²wCsup_B2R|w|

R² C

sup_B2R|v|

R² +4D²ψ

L^∞(B_2R)

.

Butv=w+ψand thus|D²v||D²w| + |D²ψ|which leads to D²v

L^∞(BR)D²w

L^∞(BR)+D²ψ

L^∞(BR)C

wL^∞(B2R)

R² +D²ψ

L^∞(B2R)

. 2

Proposition 2.8.Letu∈P₁(M, M₀, ω)withω(r)Mr^α. Then there are constantsε, r₁such thatsupω < εimplies u∈C^1,1(B_r1/2)with

uC^1,1(B_r

1/2)C,

whereCdepends onM,M₀and the dimension.

Proof. Let ε be so small that Proposition 2.5 holds with r₀=r₀(1). Take x₀∈B_r1/2 with r₁< r₀ so small that

|∇u|< r0inBr₁, and letd=dist(x0, Γ (u)). Takey∈∂Bd(x0)∩Γ (u). Letv(x)=u(x)− ∇u(y)·x. Taket0as in Proposition 2.6 and apply Proposition 2.5 to obtain

sup

B2d/t0(y)

|v|CM₀4d²/t₀², whenever|∇u(y)|2d/t0and thus also

sup

B_d(y)

|v|CM₀4d²/t₀=CM₀d² (2.5)

forC=4C/t₀²whenever|∇u(y)|2d/t0. Now if|∇u(y)|>2d/t0then since|∇(y)|< r₀, Proposition 2.6 implies thatu∈C^1,1(B_2d(y))and therefore we also have (2.5) in this case.

Sincevequals eitherλ₁orλ₂inB_d(x)we can use Lemma 2.7 to obtain D²v

L^∞(B_d/2(x₀))C

vL^∞(B_d(x₀))

d² +M

.

Using (2.5) and observing thatD²v=D²uwe get D²u

L^∞(Bd/2(x0))C(CM₀+M). 2

Rescaling this result we obtain the correct regularity without the restriction on the oscillation and we can then prove Theorem 1.2.

Proof of Theorem 1.2. Takeρso small that sup

s<ρ

ω(s) < ε,

whereεis taken so that Proposition 2.8 is valid. Define v(x)=u(ρx)

ρ² .

Thenv∈P1(M, M0/ρ², ω(ρ·)). Therefore we can apply Proposition 2.8 to get vC^1,1(Br1/2)C/ρ²,

which is the same as

uC^1,1(Br1ρ/2)C/ρ².

Obviouslyρdoes only depend on the modulus of continuity of theω, the dimension,MandM0. 2

Remark 2.9.We wish to mention here that the authors strongly believe that Theorem 1.2 is true under the weaker assumption thatωsatisfies the Dini condition. The problem here is the proof of Proposition 2.6 which uses Proposi- tion A.1. There we invoke Theorem 9.3 in [2] which in turn requiresC^α-regularity of the coefficients. Probably this theorem is possible to extend to the case where the coefficients are only Dini continuous, which then would imply that Theorem 1.2 is true under the weaker assumption.

3. Partial regularity of the free boundary

In this section we prove that the free boundary is the union of two Reifenberg vanishing sets near so-called branch- ing points. We use arguments similar to those in [3]. First we start out with some comparison results which allows us to estimate distances between different free boundaries.

3.1. Stability results

Proposition 3.1. Letu, v∈P1(M, M0, ω)with coefficientsλi respectivelyγi andu=von∂B1. Assume moreover thatλ₁γ₁andλ₂γ₂. ThenuvinB₁.

Proof. We compareuandvin the set{u < v}.

(1) Whenv <0 then alsou <0. Thereforeu= −λ2−γ2=v.

(2) Whenv >0 thenv=γ₁λ₁=max(0, λ1,−λ₂)u.

(3) Whenv=0 thenu <0 and sov=0−λ₂=u.

Henceuvin{u < v} ∩B₁and therefore{u < v} = ∅. 2 Proposition 3.2.Letuanduεbe the solutions of

u=λ₁χ_{u>0}−λ₂χ_{u<0} inB₁, and

u_ε=(λ₁+ε)χ_{uε>0}−(λ₂−ε)χ_{uε<0} inB₁,

withu=uε=gon∂B1whereΓ^±(u)∩B1andΓ^±(uε)∩B1are allC¹-graphs. Then dist

Γ^±(u)∩B₁, Γ^±(u_ε)∩B₁

C√

ε,

whereCdepends on theC¹-norms ofΓ^±(u)∩B1andΓ^±(uε)∩B1.

Proof. We treat the caseε >0 and forΓ⁺(u)andΓ⁺(u_ε), the other cases can be treated similarly.

By Proposition 3.1 we haveu_εu. Now we claim thatu+εvu_εinB₁wherevis the solution tov=1 inB₁ with zero boundary data on∂B₁. Assume thatg(x₀)=u(x₀)+εv(x₀)−u_ε(x₀) >0 for somex₀∈B₁. Then we have the three different cases:

(1) u(x₀) >0 theng(x₀)=λ₁+ε−u_ε(x₀)0.

(2) u(x₀)=0 then we must haveu_ε(x₀) <0 and sog(x₀)=λ₂.

(3) u(x₀) <0 then againu_ε(x₀) <0 and sog(x₀)= −λ₂+ε+λ₂−ε=0.

So a maximum must be attained atx₀∈∂B₁but thereg=0. Henceg0. Since v(x)=ε|x|²−1

2n

this implies that u_ε −Cε +u. Also, due to the fact that Γ⁺(u) is a C¹-graph we know that u(x) C(dist(x, Γ⁺(u)))²forx∈ {u >0}which givesu_ε(x)−Cε+Cdist(x, Γ⁺(u))²forx∈ {u >0}and thusu_εis positive if dist(x, Γ⁺(u))C√

ε. Obviouslyu_ε(x)0 for anyx such thatu(x)0. This yields the desired re- sult. 2

Remark 3.3.The result corresponding to Theorem 5.4 in [3] we cannot match. There one obtains linear stability but we can only prove stability of order√

ε.

3.2. Main result

Definition 3.4 (Reifenberg-flatness).A compact setS inRⁿ is said to beδ-Reifenberg flat if for any compact set K⊂Rⁿthere exists anRK>0 such that for everyx∈K∩S and everyr∈(0, RK]we have a hyperplaneL(x, r) such that

dist

L(x, r)∩B_r(x), S∩B_r(x)

2rδ.

We define the modulus of flatness as θ_K(r)= sup

0<ρr

sup

x∈S∩K

dist(L(x, ρ)∩B_ρ(x), S∩B_ρ(x)) ρ

.

A set is called Reifenberg vanishing if

rlim→0θ_K(r)=0.

Then we have the following result:

Proposition 3.5.There is aρ >0such that Γ⁺(u)∪Γ⁻(u)

∩ ∇u(y)< σρ/2

∩ max dist

y, Γ⁺(u) ,

y, Γ⁻(u)

< σρ/2 are Reifenberg vanishing. Hereσ is taken as in Theorem1.1in[14].

Proof. Take a sequence of pointsy_k→y₀such that|∇u(y_k)|σ r_k/2 and dist(yk, Γ^±(u)) < σ r_k/2 withr_k→0.

Consider the functions v_k(x)=u(r_kx+y_k)

r_k² .

Then from Lemma 2.2 and Proposition 2.5 we know thatv_k is bounded, and therefore for a subsequence it converges to a limit inC_loc^1,α(Rⁿ).

Now letf_k¹=sup_B1λ1(rkx+yk),f_k²=infB₁λ2(rkx+yk),g¹_k=infB₁λ1(rkx+yk)andg_k²=sup_B1λ2(rkx+yk).

Takeh_k to solve

h_k=f_k¹χ_{hk>0}−f_k²χ_{hk<0} inB₁, andl_kto solve

l_k=g¹_kχ_{lk>0}−g²_kχ_{lk<0} inB₁,

where l_k =h_k =v_k on ∂B₁. Then |f_kⁱ −g_kⁱ| 2ω(rk) and by Proposition 3.1 h_k v_k l_k. Moreover, since h_k−v_k →0 in C_loc^1,α(B₁), |∇v_k(0)|σ/2 and dist(0, Γ^±(v_k))σ/2 we have for k large enough say k > K₀, that |∇h_k(0)|σ and dist(0, Γ^±(h_k))σ. Now we can use Theorem 1.1 in [14] to say that Γ^±(h_k) is uni- formlyC¹for k > K₀ inB_r0. The same reasoning applies to l_k as well. This means that there is a planeP_k such that dist(Γ^±(h_k), P_k)ρ(r_k)r_k inB_r0 and the corresponding forl_k. Hereρ depends on theC¹-norm ofΓ^±(l_k)and Γ^±(h_k)respectively. Moreover, by Proposition 3.2 we have

dist

Γ^±(h_k)∩B_r0, Γ^±(l_k)∩B_r0

C

ω(r_k).

Therefore by rescaling we obtain that dist(Γ^±(u), P_k)Cmax(√

ω(r_k), ρ(r_k)) in B_r0rk(y_k)for r_k < r_K0 which implies the result. 2

Now we can prove Theorem 1.4.