inhomogeneous global minimizers to the one-phase …

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INHOMOGENEOUS GLOBAL MINIMIZERS TO THE

ONE-PHASE FREE BOUNDARY PROBLEM

DANIELA DE SILVA, DAVID JERISON, AND HENRIK SHAHGHOLIAN

Abstract. Given a global 1-homogeneous minimizer U0 to the Alt-Caffarellienergy functional, with sing(F (U0)) = 0, we provide a foliation of the half-space R

n × [0,+∞) with dilations of graphs of global minimizers U ≤ U0 ≤ U

with analytic free boundaries at distance 1 from the origin.

1. Introduction

1.1. Background. In this paper we are interested in minimizers to the energyfunctional,

(1.1) J(u) = JΩ(u) =

ˆ

Ω

(|∇u|2 + χu>0) dx,

where Ω is a domain in Rn (n ≥ 2) and u ≥ 0.Minimizers of J were first investigated

systematically by Alt and Caffarelli. Two fundamental properties are proved inthe pioneering article [AC], namely, the Lipschitz regularity of minimizers and theregularity of “flat” free boundaries. These in turn, give the almost-everywhereregularity of minimizing free boundaries. The viscosity approach to the regularityof the associated free boundary problem

(1.2)

∆u = 0 in Ω+(u) := Ω ∩ u > 0

|∇u| = 1 on F (u) := ∂u > 0 ∩ Ω,

was later developed by Caffarelli in [C1,C2]. There is a wide literature on this prob-lem and the corresponding two-phase problem, and for a comprehensive treatmentwe refer the reader to [CS] and the references therein.

Here we are interested in one-phase global minimizers, that is, minimizers u ≥ 0of J over all balls BR ⊂ R

n among functions that agree with u on ∂BR. Indimension n = 2, Alt and Caffarelli [AC] showed that (up to rotation) the onlyglobal minimizer of J is x+n . The same result was obtained by Caffarelli, Jerison andKenig in dimension n = 3 [CJK], and by Jerison and Savin in dimension n = 4 [JS].The classification of global minimizers implies the smoothness of minimizing freeboundaries in both the one-phase and two-phase problem in dimension n ≤ 4.These results rely on the flatness theorem and on the Weiss Monotonicity formula(see Section 2 for the precise statement) which allows one to consider only the caseof 1-homogeneous minimizers. It is conjectured that the results above remain trueup to dimension n ≤ 6. On the other hand De Silva and Jerison provided in [DJ] anexample of a singular 1-homogeneous minimal solution in dimension n = 7. In [H],Hong studied the stability of Lawson-type cones for (1.2) in low dimensions and

DD is supported by a NSF grant (RTG 1937254). DJ is supported by a Simons FoundationGrant (601948 DJ). HS is supported by Swedish Research Council.

1

http://arxiv.org/abs/2106.14576v1

2 DANIELA DE SILVA, DAVID JERISON, AND HENRIK SHAHGHOLIAN

showed that in dimension n = 7 there is another stable cone besides the minimizerin [DJ].

1.2. Main result. Let U0 ≥ 0 be a global energy minimizer to J , and assumethat U0 is homogeneous of degree 1, and F (U0) \ 0 is an analytic hypersurface,while 0 is a singular point (hence, by the discussion above, n ≥ 5). In this paper weconstruct smooth inhomogeneous global minimizers asymptotic to U0 at infinity. Infact, U0 is trapped between two global smooth inhomogeneous minimizers obtainedby perturbing away the singularity, and whose dilations foliate the whole space.The existing theory of one-phase minimizers establishes a strong resemblance be-tween the theory of free boundaries and the theory of minimal surfaces. Our resultreaffirms this similarity, providing an analogue of Hardt and Simon’s result in thecontext of area minimizing cones [HS].

In order to state our main theorem precisely, we recall some known facts andrefer to [JS] for further details on the linearized problem (see also Section 3). LetH > 0 denote the mean curvature of ∂U0 > 0 oriented towards the complementof the connected set U0 > 0, and consider the linearized problem associated toU0:

(1.3)

∆w = 0 in U0 > 0

∂νw +Hw = 0 on F (U0) \ 0,

with ν the interior unit normal to F (U0). Let w(x) := |x|−γ v(θ) with v the firsteigenfunction of the Laplacian on S

n−1 ∩ U0 > 0,

∆Sn−1 v = λv, on Sn−1 ∩ U0 > 0,

with Neumann boundary condition

∂ν v +Hv = 0, on F (U0) ∩ Sn−1.

Then v > 0 on Sn−1∩U0 > 0 and λ > 0. It is easily computed (see Section 4) that

w solves (1.3) as long as γ = γ± ∈ R, γ+ ≥ γ− > 0, satisfy γ2 − (n − 2)γ + λ = 0.If γ− 6= γ+ we call

(1.4) Vγ±(x) := |x|−γ± v.

By abuse of notation, if γ− = γ+ and we call γ0 this common value, we set

(1.5) Vγ−(x) := |x|−γ0(ln |x|+ 1) v, Vγ+

(x) = |x|−γ0 v.

Our main result reads as follow. As usual, BR(x0) ⊂ Rn denotes the ball of

radius R > 0 and center x0, and when x0 = 0 we drop the dependence on it. Also,given a function u ≥ 0 defined on R

n, we define for t > 0

ut(x) =1

tu(tx), Γ(ut) := (x, ut(x)) : x ∈ ut > 0.

Finally, constants that depend only on the ingredients i.e. n, U0, are called univer-sal.

Theorem 1.1. Let U0 be as above. There exist U ,U global minimizers to (1.1),such that

(i) U ≤ U0 ≤ U , and dist(F (U), 0) = dist(F (U), 0) = 1;

(ii) F (U), F (U) are analytic hypersurfaces;

INHOMOGENEOUS GLOBAL MINIMIZERS 3

(iii) there exist universal constants R > 0 large, α′ > 0 small, such that ifγ− 6= γ+,

U(x) = U0(x) + aVγ−(x) +O(|x|−γ−−α′

), in U0 > 0 \BR,

or

U(x) = U0(x) + aVγ+(x) +O(|x|−γ+−α′

), in U0 > 0 \BR,

for some a > 0 universal. The analogous statement for U holds in U >0 \BR with a < 0. If γ− = γ+, then

U(x) = U0(x) + aVγ−(x) + bVγ+

+O(|x|−γ0−α′

), in U0 > 0 \BR,

with a ≥ 0 and if a = 0 then b > 0. The analogous statement for U holdsin U > 0 \BR with a ≤ 0 and b < 0 whenever a = 0;

(iv) for any x ∈ U0 = 0o, the ray tx, t > 0, intersects F (U) in a singlepoint and the intersection is transverse; similarly, for any x ∈ U0 > 0,the ray tx, t > 0, intersects F (U) in a single point and the intersectionis transverse;

(v) the graphs Γt := Γ(Ut),Γt := Γ(Ut) foliate the half-space Rn × [0,+∞),

i.e.,

Rn × [0,+∞) =

⋃

t∈(0,∞)

(Γt ∪ Γt);

(vi) if V is a global minimizer to J and V ≥ U0 (resp. V ≤ U0), then

V ≡ Ut, (resp. V ≡ U t),

for some t > 0, unless V ≡ U0.

In fact, the uniqueness property (vi) holds more generally for global criticalpoints that satisfy a uniform non-degeneracy condition (see Section 2 for this no-tion). Furthermore the expansion in (iii) implies the same expansion for the freeboundaries F (U), F (U), as graphs over the cone F (U0) in the outer normal direc-tion −ν (see Remark 3.3).

1.3. Further background. In the context of critical point, Hauswirth, Helein,and Pacard [HHP] discovered an explicit family of simply-connected planar regions(so-called exceptional domains)

Ωa := (x1, x2) ∈ R2 : |x1/a| < π/2 + cosh(x2/a), a > 0

whose boundary consists of two curves (hairpins), and a positive harmonic functionHa(x) = aH1(x/a) on Ωa that satisfies the free boundary conditions Ha = 0 and|∇Ha| = 1 on ∂Ωa. Extending Ha to be zero in the complement of Ωa, we havea non-trivial entire solution to (1.2) (an explicit formula for Ha is given usingconformal mappings). The graphs of the functions Ha give a foliation of spaceusing dilates of an unstable critical point of the functional. In [JK], Jerison andKamburov characterized blow-up limits of classical solutions to (1.2) in the disk,with simply-connected positive phase, as either (a) half-plane, (b) two-plane or(c) hairpin solutions. This relates to previous results of classifications of entiresolutions with simply-connected positive phase due to Khavinson, Lundberg and


Teodorescu [KLT] and Traizet [T]. Traizet showed that classical entire solutionssatisfying that no connected component of F (u) is compact in the open disk mustbe one of the forms (a), (b), or (c). Khavinson et al. (2013) showed that the sameconclusion is true under a natural, weak regularity assumption on the free boundaryknown as the Smirnov property.

1.4. Organization of the paper. The paper is organized as follows. In Section2 we provide some preliminary properties of minimizers. The following sectionis devoted to the proof of Theorem 1.1. The core of the proof is the analysis ofthe asymptotic behaviour of a solution of a perturbation of the linearized problem(1.3), using appropriate families of subsolutions and supersolutions. This analysisis carried on in Section 4. Finally, the Appendix contains a technical result (relyingon the Hodograph transform) needed in the proof of the main theorem.

When we announced this result, M. Engelstein and coauthors informed us thatthey have obtained what appear to be essentially similar results to the ones in thisarticle in a paper under preparation [ESV].

2. Preliminaries

In this section we recall some basic properties of minimizers and we obtain a fewpreliminary results which we use in Section 3 towards the proof of Theorem 1.1.Throughout the paper, for a non-negative function u on a domain Ω, we use thenotation

F (u) := Ω ∩ ∂u > 0.

We start by a quick recap of the classical theory for minimizers of (1.1), which wewill be using throughout the paper (see [AC]). First of all, we say that u ∈ H1

loc(Ω)minimizes J in Ω if on any smooth compact set Ω′ ⊂ Ω,

J(u) ≤ J(u+ v), ∀v ∈ H10 (Ω

′).

For a given boundary data ϕ ≥ 0 with finite energy, there exists a non-negativeminimizer u ∈ H1(Ω) of J such that u − ϕ ∈ H1

0 (Ω). Moreover, u ∈ C0,1(Ω) andsatisfies (1.2) in the viscosity sense (see [C2] for the definition of viscosity solutionand a proof of this claim). Finally, u satisfies a strong non-degeneracy property,that is for any x0 ∈ F (u), supBr(x0) u ≥ cr for all balls Br(x0) ⊂ Ω. Lipschitzcontinuity and non-degeneracy are the key to the following compactness property(see Lemma 4.7 and Lemma 5.4 in [AC]).

Lemma 2.1. Let uk ∈ H1loc(Ω) be a sequence of minimizers to J in Ω with

uk → u uniformly on compact subsets, then

• uk > 0 → u > 0 and F (uk) → F (u) locally in the Hausdorff distance;• χuk>0 → χu>0 in L1

loc(Ω);• ∇uk → ∇u a.e. in Ω.

Moreover, u minimizes J in Ω.

We also recall Weiss Monotonicity Formula [W].

Theorem 2.2. If u is a minimizer to J in BR, then

Φu(r) := r−nJBr (u)− r−n−1

ˆ

∂Br

u2, 0 < r ≤ R,

is increasing in r. Moreover Φu is constant if and only if u is homogeneous of degree1.


We now notice that,

J(minu,w) + J(maxu,w) = J(u) + J(w),

from which we deduce that if u,w minimize J in Ω and u ≥ w on ∂B1, thenminu,w,maxu,w minimize J in Ω as well (with boundary data w and u re-spectively.). We can then provide a comparison principle.

Proposition 2.3. Let u,w minimize J in B1 and assume that for 0 < ǫ < 1,

u ≥ w on B1 \B1−ǫ.

Then u ≥ w in B1.

Proof. Since u ≥ w on ∂B1, we have that v := minu,w minimizes J amongcompetitors with boundary value w. However, by our assumptions, v ≡ w inB1\B1−ǫ, hence by the harmonicity of minimizers in their positivity set, we concludethat v ≡ w in D ∩B1 for any connected component D of w > 0 which intersectthe annulus. By the maximum principle, any connected component of w > 0 willintersect the annulus, hence v ≡ w in B1.

Proposition 2.4. Let u,w minimize J in B1 and assume that

u ≥ w on ∂B1, u(x0) > w(x0) at x0 ∈ ∂B1.

If w > 0 ∩ ∂B1 is connected, then u ≥ w in B1.

Proof. We argue as in the proof of the lemma above, however in this case weonly have that minu,w ≡ w in a neighborhood of x0. Thus they coincide inthe connected component of w > 0 which includes this neighborhood, and theconclusion follows from the connectivity assumption.

We now obtain a uniqueness result. Recall that U0 > 0 is connected [CJK].

Lemma 2.5 (Uniqueness). Let w be a minimizer of (1.1) in B1 such that w = U0

on ∂B1. Then w ≡ U0 in B1.

Proof. LetW = w in B1, W = U0 in B2 \B1.

Since w − U0 ∈ H10 (B1), we have W − U0 ∈ H1

0 (B2) and W and U0 both minimizeJ in B2 among competitors with boundary data U0. Hence,

∆W = 0 in B2 ∩ W > 0, ∆U0 = 0 in B2 ∩ U0 > 0.

In particular, sinceW = U0 in B2\B1, by unique continuation w and U0 must agree

on B1∩U0 > 0. If w(x) > 0 at a point x ∈ B1∩U0 = 0o, then JB1(w) > JB1

(U0)in B1, contradicting minimality.

Finally, we prove a strict separation lemma.

Lemma 2.6. Let w be a minimizer to (1.1) in B1 such that w ≥ U0 on ∂B1.

Then w ≥ U0 in B1. Moreover, w > U0 in B1 ∩ U0 > 0, unless w ≡ U0 in

B1 ∩ U0 > 0.

Proof. Since w ≥ U0 on ∂B1, minw,U0 is also a minimizer to J in B1 withboundary data U0. By the previous uniqueness result, we deduce that minw,U0 ≡U0 in B1 and the first part of our lemma is proved. Assume w 6≡ U0 in B1 ∩U0 > 0. By the maximum principle, we immediately deduce that w > U0 in


B2∩U0 > 0 Moreover, if w touches U0 by above at a free boundary point x0 6= 0,since ∂νU0(x0) = ∂νw(x0) = 1, we contradict Hopf’s boundary point lemma. Inparticular,

w > U0 in (B3/4 \B1/2) ∩ U0 > 0,

implying that for δ small,

w(x) ≥ U0(x+ y), |y| ≤ δ, in B3/4 \B1/2,

and by Proposition 2.3 this is true in B3/4. Choosing y so that U0(y) > 0, we getthat w(0) > 0, concluding the proof of the desired claim.

3. The proof of Theorem 1.1

This section is devoted to the proof of our main Theorem 1.1. The first subsectiondeals with the existence, while the second one handles the regularity.

3.1. Existence and basic properties. In this subsection we prove part (i) ofTheorem 1.1, that is the existence of U , U , and provide some key technical lemmasabout their vertical distance from U0. We start with the existence.

Proof of Theorem 1.1-(i). First, we construct the global minimizer U . Letx0 ∈ ∂B1 ∩ U0 > 0 and let g ≥ 0 be a smooth function compactly supportedin a neighborhood of x0 on ∂B1. For ǫ > 0, we define the boundary value gǫ :=U0 + ǫg ≥ U0 on ∂B1, gǫ(x0) > U0(x0), and call uǫ a minimizer to JB1

with thisboundary value. Notice that since gǫ is Holder continuous, uǫ is uniformly Holdercontinuous up to the boundary and it achieves the boundary data continuously.By Lemma 2.6, 0 6∈ F (uǫ) and uǫ ≥ U0 for all ǫ > 0. Let ǫj ց 0 (as j → ∞),and for each j, let Brj ⊂ uǫj > 0 be the largest ball included in the positivephase of uǫj . Notice that by Proposition 2.4, uǫj is a decreasing sequence. Bythe Lipschitz continuity and the compactness Lemma 2.1, we conclude that up toextracting a subsequence, uǫj converges to a minimizer v of J in B1, and sincegǫj → U0, we conclude by the uniqueness Lemma 2.5 that v = U0 and in particular

rj → 0 and j → ∞. Consider now for each j the rescaling Uj(x) = uǫj(rjx)/rj .

Then Uj is a minimizer in B1/rj , B1 ⊂ Uj > 0, and ∂B1 ∩F (Uj) 6= ∅. Moreover,

Uj(x) ≤ C(1+ |x|) in B 12rj

, where C is independent of j, since each uǫj is uniformly

Lipschitz in B1/2. Again by compactness, up to extracting a subsequence, Uj → U

and U is a global minimizer with B1 ⊂ U > 0, and universal Lipschitz bound.Finally by construction U ≥ U0 as desired.

Similarly, U can be build starting from a family of minimizers with boundarydata, 0 ≤ gǫ := U0 − ǫg ≤ U0.

Next, we prove the following asymptotic behavior for U , U.

Lemma 3.1. U , U are asymptotic to U0 at infinity, i.e.,

limt→∞

Ut(x) = limt→∞

U t(x) = U0(x),

uniformly on compact sets in Rn.

Proof. We prove the statement for U . The same argument also applies to U . For

tk → ∞ as k → ∞, consider the sequence of rescalings, Utk(x) :=U(tkx)

tkwhich in

view of the equi-Lipschitz continuity and the compactness Lemma 2.1, will converge


to a global minimizer V ≥ U0 (up to extracting a subsequence). On the other hand,again by the Lipschitz continuity,

ΦV (r) = limk→∞

ΦUtk(r) = lim

k→∞ΦU (tkr) = a ∈ R, ∀r > 0,

and by Theorem 2.2, V is homogeneous of degree 1.Thus, the restrictions of V, U0 to the unit sphere solve the eigenvalue problem,

∆Sn−1u = −λ1u on u > 0 ∩ Sn−1,

for λ1 = n− 1. However, since U0 > 0 ⊂ V > 0 and eigenvalues are monotonewith respect to the inclusion of domains, we deduce from standard arguments thatU0 ≡ V.

Finally, using Lemma 3.1 and Proposition 5.1 in the Appendix, we can deducethe following result, which is key to our strategy. Here and henceforth, we denoteby Ec := R

n \ E. Also, constants depending on n, U0 will be called universal andmay vary from one appearance to another in the body of the proofs. For notationalsimplicity, the positivity set of U0 is called Ω0 := U0 > 0, and as pointed out inthe introduction, −(∂ννU0) = H > 0 is the mean curvature of ∂Ω0 oriented towardsthe complement of the connected set Ω0.

To fix ideas, from now on, we prove the statements involving U .

Lemma 3.2. There exists universally large R > 0, such that v := U − U0 > 0satisfies

(3.1)

∆v = 0 in Ω0 ∩BcR,

∂νv +H(1 +O( v|x|))v = 0 on ∂Ω0 ∩Bc

R,

with ν the interior unit normal to ∂Ω0. Moreover,

O

(

v

|x|

)

= o(1), as |x| → ∞.

Proof. Let Us(x) := s−1U(sx), s ≥ 1. We prove that vs := Us − U0 satisfies

(3.2)

∆vs = 0 in Ω0 ∩A,

∂νvs +Hvs = O(|vs|2) on ∂Ω0 ∩A,

with ν the interior unit normal to ∂Ω0, the constant in O(|vs|2) universal, and Aan open annulus included in B2 \ B1. Below Al := B2−l \ B1+l, l < 1.

Since Us, U0 are harmonic in their positivity set, and Us ≥ U0, the harmonicityof vs in Ω0 is obvious. In order to prove the boundary condition we proceed asfollows. First, in view of Lemma 3.1

‖Us − U0‖L∞(B2\B1) → 0, as s→ ∞,

thus, by the flatness implies regularity results [C2], there exists ǫ0 universal suchthat if s = s(ǫ0) is large enough so that

(3.3) ‖Us − U0‖L∞(B2\B1) ≤ ǫ0

then in A1/5 ∩ Us > 0, ‖Us‖C3 ≤ C with C > 0 universal. Moreover, in a

sufficiently small ball Bρ(x), x ∈ F (U0) ∩ A1/5, Us and U0 satisfy Proposition 5.1

(in view of the convergence of Us to U0 and non-degeneracy), hence by a coveringargument we obtain that in A2/5 ∩ Ω0,

(3.4) ‖Us − U0‖2,α ≤ C‖Us − U0‖∞, Us − U0 ∼ ǫ := ‖Us − U0‖∞,


s large, and ǫ = ǫ(s) ≪ ǫ0.Let us define the function ψs > 0, such that

(3.5) x ∈ F (U0) ∩ A2/5 −→ x− ψs(x)νx ∈ F (Us).

In view of (3.4), since ∂νU0(x) = 1 on F (U0), we have that ∂νUs(x) = 1 + O(ǫ)on F (U0) ∩ A2/5. Hence, F (Us) is a graph over F (U0) locally in A2/5, ψs is welldefined, it is bounded by Cǫ, and in fact

(3.6) ψs(x) = vs(x) +O(ǫ2).

Then,

∇Us(x− ψs(x)νx) = ∇Us(x)− vs(x)D2Us(x)νx +O(ǫ2)

= νx − vs(x)D2Us(x)νx +∇vs(x) +O(ǫ2).

Hence, again using (3.4), and the free boundary condition for Us, we get

1 = |∇Us(x− ψsνx)|2 = 1 + 2(∂νvs − vs (∂ννU0)) + O(ǫ2),

which gives the second condition in (3.2). Rescaling back we get that v satisfies(3.1), as desired. Moreover, by Lemma 3.1,

O

(

v

|x|

)

= o(1), as |x| → ∞.

Lemma 3.2 remains valid if v := U0−U , after extending U analytically in Ω0∩BcR

(for R large enough), and moreover v > 0 in this region. Indeed, in the proof above,(3.3) holds for Us. Thus Us can be extended analytically in Ω0 ∩ (B2 \ B1), and allthe estimates following (3.3) remain valid in Ω0.

Remark 3.3. Notice that, from (3.6), we deduce the expansion (R large)

ψ(x) = v(x) +O

(

v2

|x|

)

, x ∈ F (U0) ∩BcR.

While v solves the perturbed linearized equation above in a subset of Ω0, attimes our analysis leads to estimates that must be extended outside of Ω0. Forthat purpose we use the following remark. Here Al is the annulus defined in theprevious Lemma.

Remark 3.4. If u1, u2 are critical points to J in A0, with ui ≥ U0 and

‖ui − U0‖L∞ ≤ ǫ, u1 ≥ u2 + cǫ in Ω0 ∩ A0,

for some c > 0 and ǫ > 0 small depending on c, then

u1 ≥ u2 in A1/5.

Indeed, if we argue as for Us above, we obtain that if ψi is the associated functiondefining the free boundary of ui as in (3.5), then for x ∈ A1/5 ∩ F (U0),

ui(x − tν) = ui(x)− t+O(ǫ2), t ∈ [0, ψi(x)].

Hence, for all t’s for which this expansion holds and ǫ small,

u2(x− tν) ≤ u1(x)− cǫ− t+O(ǫ2) ≤ u1(x − tν)− cǫ+O(ǫ2) < u1(x − tν),

which gives the desired result.


3.2. Regularity. This subsection is devoted to the proof of parts (ii) − (vi) ofTheorem 1.1. It relies on the following asymptotic expansion, which will be derivedin the next section. Here we use Vγ±

, which have been defined in the introduction(see (1.4),(1.5)).

Proposition 3.5. Let v > 0 be a classical solution to (3.1). If γ− 6= γ+, there existα′ > 0 small universal and R0 > 0 large universal, such that in Bc

R0∩ Ω0, either

(3.7) v(x) = a−Vγ−(x) +O(|x|−γ−−α′

), a− > 0,

or

(3.8) v(x) = a+Vγ+(x) +O(|x|−γ+−α′

), a+ > 0.

If γ− = γ+, then in BcR0

∩ Ω0

(3.9) v(x) = aVγ−(x) + bVγ+

+O(|x|−γ0−α′

), a ≥ 0, b ∈ R, maxa, b > 0.

With this proposition at hands, we can now obtain (ii) in Theorem 1.1, that isthe following result.

Theorem 3.6. F (U) is analytic.

Proof. Towards the proof of analyticity, we wish to obtain the following claim.

Claim 1. There is a universal (large) R0 > 0 such that

∇U(x) · x− U(x) < 0, in U > 0 ∩BcR0.

In order to prove Claim 1, we set v := U − U0. Recall that, in view of Lemma3.2, v > 0 satisfies (3.1), and hence the asymptotic in Proposition 3.5. Notice thatsince U0 is homogeneous of degree 1,

∇U · x− U = ∇v · x− v, in Ω0.

On the other hand, by Proposition 3.5,

(3.10) v = Zγ0+O(|x|−γ0−α′

) in BcR0

∩ Ω0,

with Zγ0given by

Zγ0= aVγ0

if γ− 6= γ+, a > 0,

and γ0 either γ− or γ+, or

Zγ0= aVγ−

+ bVγ+if γ− = γ+ = γ0, a ≥ 0, b ∈ R, maxa, b > 0.

Now, let vs(x) = v(sx)/s and similarly (Zγ0)s(x) = Zγ0

(sx)/s. Then, for s large,according to (3.10),

vs(x) = (Zγ0)s(x) + s−γ0−α′−1O(1), in (B2 \ B1) ∩ Ω0,

and from the formula for Zγ0(see also (1.4)-(1.5)),

ǫ = ǫ(s) :=

‖(Zγ0)s‖∞ ∼ s−γ0−1, if γ− 6= γ+,

‖(Zγ0)s‖∞ ∼ s−γ0−1 ln s, if γ− = γ+.

From this we deduce that, for σ := α′

2(γ0+1) ,

(3.11) vs(x) = (Zγ0)s(x) +O(ǫ1+σ), in (B2 \ B1) ∩Ω0,


and hence for universally large s, we have ‖vs‖∞ ≤ Cǫ in the annulus (B2\B1)∩Ω0.Next, in view of the first inequality in (3.4) we have (with the notation for annuliin the proof of Lemma 3.2),

∥

∥

∥

∥

vs(x) − (Zγ0)s(x)

ǫ

∥

∥

∥

∥

2,α

≤ C, in A2/5 ∩ Ω0,

while from the expansion above,∥

∥

∥

∥

vs(x)− (Zγ0)s(x)

ǫ

∥

∥

∥

∥

∞

≤ Cǫσ, in A2/5 ∩ Ω0.

We can interpolate to obtain (because ∂Ω0 \ 0 is smooth),∥

∥

∥

∥

vs(x) − (Zγ0)s(x)

ǫ

∥

∥

∥

∥

0,1

≤ Cǫσ′

, in A7/15 ∩Ω0.

Combining (3.11) with the estimate above, we now compute (for s large)

∇vs · x− vs = ∇(Zγ0)s · x− (Zγ0

)s +O(ǫ1+σ′

)

≤ −Cǫ(γ0 + 1)(1 + o(1)) +O(ǫ1+σ′

) ≤ −cǫ < 0 in A7/15 ∩ Ω0,

where the first inequality follows from the formula for Zγ0(see also (1.4)-(1.5)).

Unraveling the scaling in the inequality above, we obtain Claim 1 in BcR0

∩ Ω0.

However, we need to extend this inequality to Us > 0. Let us take a point x0 ∈F (U0) and argue as in Lemma 3.2, that is, in view of the C2,α estimate on Us andthe fact that

Us(x0) = O(ǫ), ∂ν Us(x0) = 1 +O(ǫ),

we conclude that F (Us) is included in an ǫ-neighborhood of F (U0). On the otherhand, using that the C2,α norm of vs is controlled by its L∞ norm, and that byhomogeneity ∇U0 · x− U0 ≡ 0 in A7/15 ∩ Ω0, we get

(∇Us · x− Us)(x0) = (∇vs · x− vs)(x0) ≤ −Cǫ,

∂ν(∇Us · x− Us)(x0) = ∂ν(∇vs · x− vs)(x0) = O(ǫ),

from which we obtain (using the fact that the C2,α norm of Us is universallybounded) that for s large universal,

∇Us · x− Us < 0 in A7/15 ∩ Us > 0,

as desired.Claim 1 implies,

d

dtUt(x) < −

c

t2for tx ∈ U > 0 ∩ (B4R0

\BR0), (c = c(R0)).

Given x ∈ U > 0 let t = t(x) be the smallest value for which the ray tx ∈

U > 0 ∩ (B4R0\BR0

) for all t < t, then

(3.12) Ut(x) − U(x) = −

ˆ 1

t

d

dtUt(x) dt ≥ c(1− t).

From this we deduce that Ut(x) > 0, hence Ut(x) is strictly decreasing in t ∈ (1/2, 1]

when x ∈ U > 0∩(B3R0\B2R0

). Now, for δ small and x ∈ U > 0∩(B3R0\B2R0

),the same computation as in (3.12) gives that

U1−δ(x)− U(x) ≥ cδ.


On the other hand, for any unit vector τ , since U1−δ is Lipschitz, with Lipschitzconstant in the annulus independent of δ, it follows from the inequality above that

for x ∈ U > 0 ∩ (B3R0\B2R0

),

U1−δ(x+ c1δτ) ≥ U1−δ(x) − Lc1δ ≥ U(x),

as long as c1 (being universal) is small enough. Hence,

U1−δ(x+ c1δτ) ≥ U(x), 2R0 ≤ |x| ≤ 3R0,

and by the comparison principle Proposition 2.3, we conclude that this inequalityholds for all x’s. Therefore F (U) is a Lipschitz radial graph, and by standardarguments it is also locally Lipschitz. By the classical regularity theory for onephase free boundaries [AC,KN,C1], we conclude that F (U) is analytic.

We can now deduce the proof of parts (iii)− (vi) in Theorem 1.1.

Proof of Theorem 1.1 (iii)-(vi). Claim 1 in the proof of Theorem 3.6 and Proposition3.5, provide the statements in part (iii) − (iv) of Theorem 1.1. The statement in(v) follows immediately from (iv). We are left with the proof of (vi), from whichwe also deduce the universality of the coefficient in the expansion in (iii). AssumeV is not identically equal to U0. Then the same arguments in Lemma 3.1, Lemma3.2 and Proposition 3.5 can be applied to V giving that V is asymptotic to U0, and

(3.13) V − U0 = Vγ1(d+ o(1)) in Bc

R0∩ Ω0,

with γ1 either γ− or γ+, and d > 0.On the other hand,

(3.14) U − U0 = Vγ0(a+ o(1)) in Bc

R0∩ Ω0,

with γ0 either γ− or γ+, and a > 0.If U and V both have expansions in terms of Vγ−

(resp. Vγ+), then we can argue

as follows. Define t0 by,

at−γ−−10 = d.

Using the expansions above for a given t > t0, we conclude that for |x| large, x ∈ Ω0,

V − Ut

Vγ−

= (d− at−γ−−1 + o(1)), d− at−γ−−1 > 0.

Thus, for any t > t0, there exists Rt large such that in an annulus (B2Rt \BRt)∩Ω0,

V ≥ c(t)ǫ + Ut, ǫ = ‖Vγ−‖∞,

where we have used that Vγ−∼ ‖Vγ−

‖∞ in (B2Rt \BRt)∩Ω0. By a rescaled version

of Remark 3.4 we get V ≥ Ut in the annulus B2Rt \ BRt , and the comparisonProposition 2.3 gives,

V ≥ Ut, ∀t > t0, in Rn.

Similarly,

V ≤ Ut, ∀t < t0, in Rn,

and the claim follows letting t → t0. If the expansions are different, say U andVγ−

have a expansions in terms of V respectively Vγ+, then arguing as above we

conclude that Ut ≥ V for all t’s, and by letting t → ∞ we obtain U0 ≥ V , henceU0 ≡ V a contradiction.


4. The perturbed linearized problem

This section is intended for the proof of Proposition 3.5. For convenience, werecall some notation from the introduction and refer the reader to [JS] for furtherdetails on the discussion below. Consider the problem,

(4.1)

∆w = 0 in Ω0

∂νw +Hw = 0 on ∂Ω0 \ 0,

with ν the interior unit normal to ∂Ω0, and −(∂ννU0) = H > 0 the mean curvatureof ∂Ω0 oriented towards the complement of the connected set Ω0. Let w := f(r)v(θ),with f ≥ 0 a radial function, r := |x|, and v the corresponding first eigenfunctionof the Laplacian on S

n−1 ∩Ω0, i.e.,

∆Sn−1 v = λv, on Sn−1 ∩Ω0,

satisfying the Neumann condition,

(4.2) ∂ν v +Hv = 0, on ∂Ω0 ∩ Sn−1.

Then v > 0 on Sn−1 ∩ Ω0 and λ > 0. We compute,

∆w = v∆f + 2∇v · ∇f + f∆v = v

(

f ′′ + (n− 1)f ′

r+ λ

f

r2

)

,

thus, for f = r−γ , we obtain that w solves (4.1) as long as γ = γ± satisfy γ2 −(n − 2)γ + λ = 0. The stability of U0 is equivalent to the fact that this quadraticequation must have real roots i.e. (n − 2)2 − 4λ ≥ 0. Moreover, λ > 0, thusγ = γ± ∈ R, γ+ ≥ γ− > 0.

If γ− 6= γ+ we callVγ±

(x) := |x|−γ± v,

while by abuse of notation, if γ− = γ+ and we call γ0 this common value, we set

(4.3) Vγ−(x) := |x|−γ0(ln |x|+ 1) v, Vγ+

(x) := |x|−γ0 v.

Next, we build the following special family of functions, which will play an es-sential role in the proof of Theorem 1.1. Call u0 := |x|−1U0, and define for realnumbers γ, β, A = A(γ)

(4.4) W βγ (x) =

|x|−γ(Av + u0) A > 0 if 0 < γ < γ−,

|x|−γ(Av + u0) A < 0 if γ− < γ < γ+,

|x|−γ(Av + u0) + βVγ+A > 0 if γ > γ+.

In what follows, when γ < γ+, we drop the dependence on β. Notice that ourdefinition also includes the case γ− = γ+. Then, using the above formula,

∆W βγ = |x|−γ−2A[γ(γ − n+ 2) + λ]v + u0[γ(γ − n+ 2) + (1− n)]

and for |A| large enough,

∆W βγ ≥ 0, in Ω0.

Moreover, using that v solves (4.2), while u0 = 0, (u0)ν = 1/r on ∂Ω0 \ 0, we get

(4.5) ∂νWβγ +HW β

γ = |x|−γ−1 on ∂Ω0 \ 0.

We remark that Wγ > 0 in Ω0 when γ < γ−, and choosing |A| large, Wγ < 0 in Ω0

when γ− < γ < γ+. The sign of W βγ depends on the choice of β.

Having introduced this family of functions, we can now provide the proof ofProposition 3.5. The proof relies on a comparison principle for solutions to (3.1)


in an annulus (BR \ B1) ∩ Ω0, which we prove in the next lemma. Consider theproblem,

(4.6)

∆V = 0 in D,

∂νV = h(x)V on T ⊂ ∂D,

with D a Lipschitz domain in Rn, T a smooth open subset of ∂D, and ν the inward

unit normal to T .

Lemma 4.1. If there exists a classical supersolution to (4.6) (continuous up to∂D) such that V > 0 on D, then the comparison principle holds i.e. if z, w arerespectively a classical supersolution and a classical subsolution to (4.6), with z ≥ von ∂D \ T , then z ≥ v in D.

Proof. Let w := v − z, and set M := maxDwV = w

V (x0) for some x0 ∈ D. If x0 ∈∂D\T , then by our assumptionsM ≤ 1, which implies our claim. Similarly, if x0 ∈D, thenMV −w attains a minimum at x0 and by the maximum principleMV ≡ w,which contradicts out assumptions. Finally, x0 cannot occur on T . Indeed, againMV −w ≥ 0 attains a minimum at x0 and by Hopf Lemma, ∂ν(MV −w)(x0) > 0.On the other hand,

0 < M∂νV (x0)− ∂νw(x0) ≤ h(x0)(MV − w)(x0) = 0,

a contradiction.

Besides the above comparison principle we also use the Harnack inequality forinhomogeneous Neumann problems, see for example [L]. The precise statement isprovided in the Appendix.

We are now ready to prove Proposition 3.5.

Proof of Proposition 3.5. The proof is divided into 5 steps. For simplicity, in eachstep we consider first the case when γ− 6= γ+, and then point out the modificationsneeded for the case γ− = γ+.

Since U0 is homogeneous of degree 1, after a dilation x→ ρx, ρ = ρ(δ) large, wemay assume that

(4.7)

∆v = 0 in Ω0 ∩Bc1

∂νv +H(1 + o(1))v = 0 on ∂Ω0 ∩Bc1,

with

(4.8) |o(1)| ≤ δ,

for δ > 0 to be made precise later. By abuse of notation, in what follows all dilationsof v will still be denoted by v.

Step 1: Decay at infinity. Let v satisfy (4.7)-(4.8). Given 0 < γ < γ−, thereexists M =M(γ, v) > 0 large such that

v ≤M |x|−γ on Ω0 ∩Bc1.

For this step, we do no need to distinguish whether γ− and γ+ coincide or not.Indeed, let γ be given and the constants below possibly depend on γ. To prove thedesired bound, we use the function Wγ defined in (4.4) (we drop the dependence


on β in the regime γ < γ−), and we construct a subsolution W = −CVγ−+ cWγ

to (4.7), such that for large C and small c,

(4.9) W ≤ R−γ on ∂BR ∩Ω0, R large, W ≤ 0 on ∂B1 ∩ Ω0.

This is possible because γ < γ−, hence Wγ > 0 is the leading term. For theboundary condition to yield a subsolution we need,

∂νW +H(1 + o(1))W ≥ 0 on ∂Ω0 ∩Bc1,

which in view of (4.5) and (4.8) holds as long as (recall that Vγ−solves (4.1))

−CHδVγ−+ c|x|−γ−1 − cδHWγ ≥ 0, on ∂Ω0 ∩B

c1.

This can be achieved by choosing δ small.Moreover, again since Wγ is the leading term, we can pick R < R large, such

that

(4.10) R−γ− ≤W on ∂BR ∩Ω0.

Now, for M large to be specified later, let us assume by contradiction thatv(x) > MR−γ at some point on ∂BR∩Ω0. Then by the interior Harnack inequalityand the Harnack inequality for the Neumann problem (3.1) (see Theorem 5.2), wehave that v ≥ cMR−γ on ∂BR ∩ Ω0, with c > 0 universal. Thus, using (4.9), andthe fact that v ≥ 0, we have that

v ≥ cMW on (∂BR ∪ ∂B1) ∩ Ω0.

By Lemma 4.1, which can be applied because v itself is a solution to (4.7) which isstrictly positive up to the boundary, we get

v ≥ cMW in (BR \B1) ∩ Ω0.

In particular, by (4.10) above,

v ≥ c(R)M on ∂BR ∩ Ω0,

hence we reach a contradiction if M is large enough, and Step 1 is proved.

Now, choosing γ = γ−/2, for a universal α0 = γ−/2 + 9/10 we have by Step 1that,

v(x)

|x|= o(|x|−α0 ), |x| → ∞

hence, again after a dilation, we can assume that v satisfies:

(4.11)

∆v = 0 in Ω0 ∩Bc1,

∂νv +H(1 + o(|x|−α0 ))v = 0 on ∂Ω0 ∩Bc1,

with

(4.12) |o(|x|−α0 )| ≤ δ|x|−α0

and δ small universal to be chosen later.

Step 2: Improved Decay at infinity. Let v satisfy (4.11)-(4.12). There existsM− > 0 large, such that

v

Vγ−

≤M− on Ω0 ∩Bc1.


The argument follows the lines of Step 1, but we need to distinguish whether γ−and γ+ coincide or not. If the roots are distinct, fix γ ∈ (γ−, γ+). We construct asubsolution W = Vγ−

+Wγ to (4.11), such that (R = R(γ) large),

(4.13) W ≤3

2Vγ−

on ∂BR ∩ Ω0, W ≤ 0 on ∂B1 ∩ Ω0.

The desired inequalities can be achieved as W/Vγ−→ 1 as r → ∞ and by choosing

|A| possibly larger. Moreover, for the same reason, we can pick a large R < R suchthat

(4.14)1

2Vγ−

≤W on ∂BR ∩ Ω0.

Now, for the boundary inequality to be satisfied, we need to choose (see (4.12)),

γ− < γ ≤ γ− + α0,

and δ small enough, so that

(4.15) − δH |x|−α0Vγ−+ |x|−γ−1 − δH |x|−α0Wγ ≥ 0, on ∂Ω0 ∩B

c1.

The argument is now the same as in Step 1. Given M large, to be specifiedlater, let us assume by contradiction that v

Vγ−≥M at some point at some point on

∂BR ∩ Ω0. Then by the Harnack inequality (interior and for a Neumann problemas in Theorem 5.2), we have that v

Vγ−≥ cM on ∂BR ∩ Ω0. This combined with

(4.13) and Lemma 4.1 (again since v > 0) gives that,

v

Vγ−

≥ cMW

Vγ−

in (BR \B1) ∩ Ω0.

Hence, in view of (4.14),

v ≥ c(R)M on ∂BR ∩ Ω0,

which is a contradiction if M is large enough.In the case when γ− = γ+, we need to choose W = Vγ−

+W βγ and let β be

negative and |β| large, so that the second inequality in (4.13) holds (see (4.4) forthe definition of W β

γ ).

Step 3: Limit at infinity and expansion. Let v satisfy (4.11)-(4.12). Then,

lim|x|→∞

v

Vγ−

= a− ≥ 0.

Moreover, if γ− 6= γ+,

(4.16) v(x) = a−Vγ−(x) +O(|x|−γ−−α′

), in Ω0 ∩Bc1,

while if γ− = γ+,

(4.17) v(x) = a−Vγ−(x) +O(Vγ+

), in Ω0 ∩Bc1.

We consider first that case when γ− 6= γ+. For ρ ≥ 1, let

a(ρ) := supa ≥ 0 | v ≥ aVγ−in Bc

ρ ∩ Ω0.

This is an increasing function which in view of Step 2 is bounded by M−. Thus,there exists

a− := limρ→∞

a(ρ) ≥ 0,


and by definition of a(ρ),

(4.18) lim inf|x|→∞

v

Vγ−

≥ a−.

We wish to show that

lim sup|x|→∞

v

Vγ−

≤ a−,

from which our claim will follow. Assume by contradiction that for a small η > 0along a sequence of points xk ∈ Ω0 with ρk := |xk| → ∞,

(4.19)v

Vγ−

(xk) ≥ a− + η,

while in view of (4.18), for k large and ǫ≪ η,

(4.20)v

Vγ−

≥ a− − ǫ in Bcρk/2

∩ Ω0.

Thus, by the Harnack inequality (Theorem 5.2) applied to v − (a− − ǫ)Vγ−we

conclude that for k large,

(4.21)v

Vγ−

≥ a− + cη, on ∂Bρk∩ Ω0.

Indeed, let us call w := v − (a− − ǫ)Vγ−and consider for 1

2 < |x| < 2 the functions

wk(x) = ργ−

k v(ρkx) − (a− − ǫ)Vγ−(x).

In view of (4.20), wk ≥ 0, and by (4.19), wk(x) ≥ c′η at some point x ∈ ∂B1 ∩Ω0.On the other hand, vk(x) = ρ

γ−

k v(ρkx) satisfies

∆vk = 0 in Ω0 ∩ (B2 \ B1/2),

∂νvk +H(1 + o(1))vk = 0 on ∂Ω0 ∩ (B2 \ B1/2),

and in view of Step 2, it is uniformly bounded. Thus by standard regularity esti-mates, the vk’s are uniformly Holder continuous in (B15/8 \B5/8) ∩ Ω0 and, up toextracting a subsequence, wk converges uniformly to a nonnegative limiting functionw which solves the unperturbed problem (4.1) in (B7/4 \ B3/4)∩Ω0 and w(x) ≥ c′ηat some point on ∂B1∩Ω0. By the Harnack inequality and the uniform convergencewe get, wk ≥ cη on ∂B1 ∩ Ω0 for c universal, and unraveling the scaling this givesthe desired claim (4.21).

As in Step 2, we now build a subsolution

W = (a− +c

2η)Vγ−

+Wγ , γ− < γ ≤ γ− + α0, γ < γ+,

such that for k ≥ k0 large

W ≤ (a− + cη)Vγ−on ∂Bρk

∩ Ω0.

Thus by Lemma 4.1, for k large,

v ≥W in (Bρk\Bρk0

) ∩ Ω0,

hence outside a very large ball

v

Vγ−

≥ a− +c

4η, in Bc

R ∩ Ω0,


contradicting the definition of a−. Finally, in order to obtain the expansion, weuse a similar argument as above to trap v in any annulus BR \ B1 with R large,between a subsolution and a supersolution, respectively,

(4.22) (a− − ǫ)Vγ−+ CWγ , (a− + ǫ)Vγ−

− CWγ ,

by choosing C large enough. Letting ǫ go to zero we get the desired expansion withα′ = γ. Rescaling back, we obtain the desired expansion.

There are two modifications needed in the case γ− = γ+. The first is the defini-tion of wk and vk, that is

wk(x) =1

ln ρk + 1ργ−

k v(ρkx) − (a− − ǫ)[Vγ+(x) +

1

ln ρk + 1Vγ−

(x)],

and

vk(x) =1

ln ρk + 1ργ−

k v(ρkx).

The second one is in the construction of the subsolution/supersolution in thelast step of the previous argument. Indeed the subsolution and the supersolutionin (4.22) must be replaced by

(4.23) (a− − ǫ)Vγ−+ CW β

γ , (a− + ǫ)Vγ−− CW β

γ ,

with β < 0 and |β| large.

Step 4: The case γ− 6= γ+. (Comparison with Vγ+.) Let v satisfy (4.11)-

(4.12) and assume that a− = 0 (from Step 3) and

v

Vγ−

= O(|x|−α′

), in Bc1, γ− < α′ ≤ γ− + α0, α′ < γ+.

There exists M+, R > 0 large, such that

v

Vγ+

≤M+ on BcR ∩ Ω0.

Since a− = 0, we can repeat the same arguments as above with α0 replaced byα1 = α′ + γ− + 9/10 > α0, that is a dilation of v satisfies (4.11) with α0 replacedby α1. If γ− + α1 < γ+, we use Lemma 4.1 and compare v with the supersolution(choose γ ≤ γ− + α1 according to a similar computation as in (4.15))

Z = ǫVγ−− CWγ .

We check that

v ≤ Z on ∂B1 ∩ Ω0, for C large

and

v ≤ Z on ∂BR ∩ Ω0, for R large.

The first inequality holds asWγ < −C < 0. The second inequality follows as a− = 0

andWγ

Vγ−→ 0 as |x| → ∞. As ǫ→ 0 we get that

v ≤ −CWγ in Bc1 ∩ Ω0.

We repeat the same argument with α2 = γ + 9/10 > α1, and we continue till wereach the first l ≥ 1 for which γ− + αl > γ+. Then, via Lemma 4.1, we compare vwith the supersolution (γ+ < γ ≤ γ− + αl)

Z = ǫVγ−− CW β

γ ,


(again see (4.4) for the definition of W βγ ) and choose β < 0 and |β| large enough

to guarantee that v ≤ Z on ∂B1 ∩ Ω0 (a similar computation as in (4.15) can becarried out again). Therefore we obtain that

v ≤ −CW βγ in Bc

1 ∩ Ω0,

and the claim follows from the definition of W βγ .

Step 5: Separation from 0 and Expansion. Let v solve (4.11)-(4.12) andwhenever γ− 6= γ+ replace α0 by αl from Step 4. Then,

(4.24) v ≥ cVγ+, in Bc

2 ∩ Ω0, c > 0.

To see this, we let a = v(x) > 0 for some x ∈ ∂B2 ∩Ω0. By the Harnack inequalityin (B3 \ B1) ∩Ω0, v ∼ a on ∂B2 ∩Ω0. Define (the subsolution) W := −ǫVγ−

+W 1γ

(with γ > γ+),1 such that W ≤ C on ∂B2 ∩ Ω0 and W ≤ 0 for R large. As usual,

we are using the function W βγ defined in (4.4). Then v ≥ caW in (BR \ B2) ∩ Ω0

and by letting ǫ→ 0 and using the definition of W 1γ , we get the desired statement.

To conclude the proof of the expansions (3.8)-(3.9), we prove the following im-provement of oscillation. First we consider the case of distinct roots.

Improvement of oscillation. There exist positive sequences ak increasing andbk decreasing with

bk+1 − ak+1 = (1 − c)(bk − ak)

such that

(4.25) ak ≤v

Vγ+

≤ bk, in Bc2k ∩ Ω0.

Let (4.25) hold for some k ≥ 1, call ρk := 2k, and for µ > 0 small, write bk − ak =

ρ−µk . Define,

w(x) := ρµ+γ+

k (v − akVγ+)(ρkx), x ∈ Bc

1 ∩ Ω0.

Then w is harmonic in Bc1 ∩ Ω0 and it satisfies the boundary condition

wν +Hw = o(|x|−αl−1−γ+) on ∂Ω0 ∩Bc1,

with |o(|x|−αl−1−γ+)| ≤ δ|x|−αl−1−γ+ . Moreover, in view of (4.25),

0 ≤w

Vγ+

≤ 1, in Ω0 ∩Bc1.

Assume that w ≥ 12 at some point x ∈ ∂B3/2 ∩ Ω0. Then by the interior Harnack

inequality and the Harnack inequality for a Neumann problem (see Theorem 5.2)w

Vγ+≥ c on ∂B3/2 ∩Ω0. We build a subsolution W (to the problem satisfied by w),

such that

(4.26) W ≤ c, on ∂B3/2 ∩ Ω0, W ≤ 0 on ∂BR ∩ Ω0, R large.

SetW = c1Vγ+

+ c2W0γ − ǫVγ−

, γ > γ+.

Then, the first inequality in (4.26) is satisfied for c1, c2 small, while the second onefollows as W/Vγ+

→ −∞ as |x| → ∞. We need to verify that

Wν +HW ≥ δ(|x|−αl−1−γ+) on ∂Ω0 ∩Bc3/2,

1In view of the iteration argument at the end of the previous step, this also works in the caseof distinct roots.


or equivalently,

c2|x|−γ−1 ≥ δ|x|−αl−1−γ+ on ∂Ω0 ∩B

c3/2.

This holds for δ small, as long as γ < αl + γ+. By Lemma 4.1, which can beapplied because Vγ+

is a solution to (4.1), which is strictly positive on the closure

of D := (BR \ B3/2) ∩Ω0, while W −w is a subsolution to (4.1) in D, non-positiveon (∂BR ∪ ∂B3/2) ∩ Ω0, thus

w ≥W in (BR \B3/2) ∩Ω0

and by letting ǫ→ 0 we deduce,

w ≥ c1Vγ+in Bc

2 ∩Ω0,

which after unraveling what w is, gives the desired improvement.When γ− = γ+ the only modification is that, using the expansion in Step 3,

(4.25) will be satisfied by v − a−Vγ−, and

(4.27) w := w(x) := ρµ+γ+

k (v − a−Vγ−− akVγ+

)(ρkx).

The rest of the proof works in the same way.The proof of the proposition is now complete. Indeed, in the case γ− 6= γ+,

(4.16) immediately gives the desired expansions (3.7) if a− 6= 0. Otherwise, bystandard arguments, the improvement of oscillation in Step 5 leads to (3.8) (afterrescaling back), with a > 0 in view of (4.24). If γ− = γ+, again the improvementof oscillation in Step 5 and (4.24) give (3.9) (see (4.27)).

5. Appendix

In this Appendix, we provide a C2,α estimate for the difference of two nearbysmooth solutions of (1.2) in terms of the L∞ norm. This estimate is essential inour proof of Theorem 1.1, and it relies on the use of the hodograph transform. Werefer to [KN] for further details on this important tool. Here universal constantsonly depend on dimension.

Proposition 5.1. Let u1, u2 be classical solutions to (1.2) in B1, with

(5.1) (xn − ǫ1)+ ≤ u1 ≤ u2 ≤ (xn + ǫ1)

+,

for ǫ1 small universal. Then, v := u2 − u1 satisfies

(5.2) ‖v‖C2,α(B1/2∩u1>0) ≤ C‖v‖L∞(B1),v(x)

v(y)≤ C, x, y ∈ B1/2 ∩ u1 > 0,

with C > 0 universal.

Proof. Let k = 1, 2. In view of the flatness assumption (5.1), if ǫ1 is chosen small(depending on n) such that the improvement of flatness Theorem in [C2] holds,then ‖uk‖C3(B3/4) ≤ C, with C universal. Thus, for c > 0 universal, (again by

(5.1)),1− c ≤ ∂nuk ≤ 1 + c, in B3/4 ∩ uk > 0.

Then, we can perform the partial hodograph transform y′ = x′, yn = uk(x), withHuk

the inverse mapping given by the partial Legendre transform, and obtain

(5.3)

F(D2Huk,∇Huk

) = 0 in B1/2 ∩ yn > 0,

g(∇Huk) = 0, on B1/2 ∩ yn = 0.


The free boundary of uk is given by the graph of the trace of Hukon yn = 0.

The difference ψ := Hu2−Hu1

≥ 0 will then satisfy an equation of the form

aij∂ijψ + bi∂iψ = 0 in B1/2 ∩ yn > 0

with boundary condition,

ψn = di∂iψ on B1/2 ∩ yn = 0,

and aij , bi Holder continuous and di ∈ C1,α with norms depending on the C2,α

norm of Hukwhich in turn depends on the C2,α norm of uk (hence is bounded by

a universal constant). By the standard regularity estimates in Theorem 5.2, theC2,α norm of ψ is controlled by its L∞ norm. Moreover by the Harnack inequality,Theorem 5.2, ψ ∼ ‖ψ‖∞ in B1/4 ∩ yn > 0. Thus,

u1(x) = u2(x′, xn − ψ(x′, u1(x))), x ∈ B1/8 ∩ u1 > 0,

for ψ as above, from which our claim follows.

We conclude the Appendix by stating the Schauder estimates and the Harnackinequality that we need for boundary problems.

Theorem 5.2. Let v satisfy,

aij∂ijv + bi∂iv + cv = f in B1 ∩Ω,

β · ∇v + hv = g on B1 ∩ ∂Ω,

with aij uniformly elliptic, and β · ν ≥ c0 > 0, and

Ω := xn > φ(x′), φ(0) = 0.

(i) If aij , bi, c, f ∈ C0,α, β, h, g ∈ C1,α, φ ∈ C2,α, then

‖v‖C2,α(B1/2∩Ω) ≤ C(‖v‖∞ + ‖f‖0,α + ‖g‖1,α);

(ii) if v ≥ 0, aij , bi, c, β, h ∈ C0,α, φ ∈ C1, then

supB1/2∩Ω

v ≤ C( infB1/2∩Ω

v + ‖f‖∞ + ‖g‖∞),

for C > 0 depending on n, the coefficients, the ellipticity constants, andΩ.

Theorem 5.2 is contained for example in [L]. However (ii) is stated under a signassumption on the lower order coefficients, that is c, h ≤ 0, which is needed forthe existence theory. We point out that these assumptions are not needed for theHarnack inequality, since after a dilation it suffices to prove the result when thelower order coefficients are arbitrarily small, and the standard arguments continueto hold. Since we could not find an explicit reference, we provide below a sketch ofthe proof of the one-sided Harnack inequality which we used in Proposition 3.5.

Lemma 5.3. Let v ≥ 0 satisfy

∆v = 0 in B1 ∩Ω,

∂νv + hv = g, in B1 ∩ ∂Ω,

with Ω as in Theorem 5.2, and ν the inner unit normal to ∂Ω. If v(x) ≥ 1 atx = 1

4en, theninf

B1/2∩Ω

v ≥ c− C‖g‖∞,

for c, C > 0 depending on n, ‖h‖∞,Ω.


Proof. After a dilation, we may reduce to the case when ‖h‖∞, ‖g‖∞, ‖φ‖C1 ≤ ǫ0for some ǫ0 small, and we need to show that

v ≥c

2, in Br0 ∩ Ω, for some small r0.

By the interior Harnack inequality, v ≥ c0 on B1/8(x). Let

V := c1(|x− x|2−n − (1

2)2−n)+, in B1/2(x)

with c1 := c0((1/8)2−n − (1/2)2−n)−1. We claim that,

v ≥ V on B1/2(x) ∩ Ω,

from which the desired bound follows. Indeed, let w := v−V and assume w(x0) :=minB1/2(x)∩Ωw < 0. By the maximum principle, x0 cannot occur in (B1/2(x) \

B1/8(x))∩Ω. Thus, we only need to rule out that x0 ∈ B1/2(x)∩∂Ω. On the otherhand, at such point wν(x0) ≥ 0, hence vν(x0) ≥ Vν(x0) ≥ c2 > 0, if ǫ0 is chosensmall. Therefore,

ǫ0 + ǫ0V (x0) ≥ ǫ0 + ǫ0v(x0) ≥ g(x0)− h(x0)v(x0) = ∂νv(x0) ≥ c2,

and we reach a contradiction by choosing ǫ0 small enough.

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Department of Mathematics, Barnard College, Columbia University, New York, NY,

10027

Email address: [email protected]

Department of Mathematics, Massachusetts Institute of Technology, Cambridge

MA,


Department of Mathematics, KTH Royal Institute of Technology, Stockholm, Swe-

den


inhomogeneous global minimizers to the one-phase …

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