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Research article

Optimal dengue vaccination strategies of seropositive individuals

  • The dengue vaccine, CYD-TDV (Dengvaxia), has been licensed in 20 countries in Latin America and Southeast Asia beginning in 2015. In April 2018, the World Health Organization (WHO) advised that CYD-TDV should only be administered to individuals with a history of previous dengue virus infection. Using literature-based parameters, a mathematical model of dengue transmission and vaccination was developed to determine the optimal vaccination strategy while considering the effect of antibody-dependent enhancement (ADE). We computed the optimal vaccination rates under various vaccination costs and serological profiles. We observe that the optimal dengue vaccination rates for seropositive individuals are highest at the initial phase of a vaccination program, requiring intense effort at the early phase of an epidemic. The model shows that even in the presence of ADE, vaccination could reduce dengue incidence and provide population benefits. Specifically, optimal vaccination rates increase with a higher proportion of monotypic seropositive individuals, resulting in a higher impact of vaccination. Even in the presence of ADE and with limited vaccine efficacy, our work provides a population-level perspective on the potential merits of dengue vaccination.

    Citation: Eunha Shim. Optimal dengue vaccination strategies of seropositive individuals[J]. Mathematical Biosciences and Engineering, 2019, 16(3): 1171-1189. doi: 10.3934/mbe.2019056

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  • The dengue vaccine, CYD-TDV (Dengvaxia), has been licensed in 20 countries in Latin America and Southeast Asia beginning in 2015. In April 2018, the World Health Organization (WHO) advised that CYD-TDV should only be administered to individuals with a history of previous dengue virus infection. Using literature-based parameters, a mathematical model of dengue transmission and vaccination was developed to determine the optimal vaccination strategy while considering the effect of antibody-dependent enhancement (ADE). We computed the optimal vaccination rates under various vaccination costs and serological profiles. We observe that the optimal dengue vaccination rates for seropositive individuals are highest at the initial phase of a vaccination program, requiring intense effort at the early phase of an epidemic. The model shows that even in the presence of ADE, vaccination could reduce dengue incidence and provide population benefits. Specifically, optimal vaccination rates increase with a higher proportion of monotypic seropositive individuals, resulting in a higher impact of vaccination. Even in the presence of ADE and with limited vaccine efficacy, our work provides a population-level perspective on the potential merits of dengue vaccination.


    In the last years the regularity theory for two phase problems governed by uniformly elliptic equations with distributed sources has reached a considerable level of completeness (see for instance the survey paper [10]) extending the results in the seminal papers [2,4] (for the Laplace operator) and in [17,18] (for concave fully non linear operators) to the inhomogeneous case, through a different approach first introduced in [7].

    In particular the papers [15] and [8] provides optimal Lipschitz regularity for viscosity solutions and their free boundary for a large class of fully nonlinear equations.

    Existence of a continuous viscosity solution through a Perron method has been established for linear operators in divergence form in [3] (homogeneous case) and in [9] (inhomogeneous case), and for a class of concave operators in [19]. The main aim of this paper is to adapt the Perron method to extend the results of [19] to the inhomogeneous case. Although we are largely inspired by the papers [3] and [9], the presence of a right hand side and the nonlinearity of the governing equation presents several delicate points, significantly in Section 6, which require new arguments.

    We now introduce our class of free boundary problems and their weak (or viscosity) solutions.

    Let Symn denote the space of n×n symmetric matrices and let F:SymnR denote a positively homogeneous map of degree one, smooth except at the origin, concave and uniformly elliptic, i.e. such that there exist constants 0<λΛ with

    λNF(M+N)F(M)ΛN for every M,NSymn with N0,

    where M=max|x|=1|Mx| denotes the (L2,L2)-norm of the matrix M.

    Let ΩRn be a bounded Lipschitz domain and f1,f2C(Ω)L(Ω). We consider the following two-phase inhomogeneous free boundary problem (f.b.p. in the sequel).

    {F(D2u+)=f1in Ω+(u):={u>0}F(D2u)=f2χ{u<0}in Ω(u)={u0}ou+ν(x)=G(uν,x,ν)along F(u):={u>0}Ω. (1.1)

    Here ν=ν(x) denotes the unit normal to the free boundary F=F(u) at the point x, pointing toward Ω+, while the function G(β,x,ν) is Lipschitz continuous, strictly increasing in β, and

    infxΩ,|ν|=1G(0,x,ν)>0. (1.2)

    Moreover, u+ν and uν denote the normal derivatives in the inward direction to Ω+(u) and Ω(u) respectively.

    As we said, the main aim of this paper is to adapt Perron's method in order to prove the existence of a weak (viscosity) solution of the above f.b.p., with assigned Dirichlet boundary conditions

    For any u continuous in Ω we say that a point x0F(u) is regular from the right (resp. left) if there exists a ball BΩ+(u) (resp. BΩ(u)) such that ¯BF(u)=x0. In both cases, we denote with ν=ν(x0) the unit normal to B at x0, pointing toward Ω+(u).

    Definition 1.1. A weak (or viscosity) solution of the free boundary problem (1.1) is a continuous function u which satisfies the first two equality of (1.1) in viscosity sense (see Appendix A), and such that the free boundary condition is satisfied in the following viscosity sense:

    (ⅰ) (supersolution condition) if x0F is regular from the right with touching ball B, then, near x0,

    u+(x)αxx0,ν++o(|xx0|)in B, with α0

    and

    u(x)βxx0,ν+o(|xx0|)in Bc, with β0,

    with equality along every non-tangential direction, and

    αG(β,x0,ν(x0));

    (ⅱ) (subsolution condition) if x0F is regular from the left with touching ball B, then, near x0,

    u+(x)αxx0,ν++o(|xx0|)in Bc, with α0

    and

    u(x)βxx0,ν+o(|xx0|)in B, with β0,

    with equality along every non-tangential direction, and

    αG(β,x0,ν(x0));

    We will construct our solution via Perron's method, by taking the infimum over the following class of admissible supersolutions class.

    Definition 1.2. A locally Lipschitz continuous function wC(¯Ω) is in the class class if

    (a) w is a solution in viscosity sense to

    {F(D2w+)f1in Ω+(w)F(D2w)f2χ{u<0}in Ω(w);

    (b) if x0F(w) is regular from the left, with touching ball B, then

    w+(x)αxx0,ν++o(|xx0|)in Bc, with α0

    and

    w(x)βxx0,ν+o(|xx0|)in B, with β0,

    with

    αG(β,x0,ν(x0));

    (c) if x0F(w) is not regular from the left then

    w(x)=o(|xx0|).

    The last ingredient we need is that of minorant subsolution.

    Definition 1.3. A locally Lipschitz continuous function u_C(¯Ω) is a strict minorant if

    (a) u_ is a solution in viscosity sense to

    {F(D2u_+)f1in Ω+(u_)F(D2u_)f2χ{u_<0}in Ω(u_);

    (b) every x0F(u_) is regular from the right, with touching ball B, and near x0

    u_+(x)αxx0,ν++ω(|xx0|)|xx0|in B, with α>0,

    where ω(r)0 as r0+, and

    u_(x)βxx0,ν+o(|xx0|)in Bc, with β0,

    with

    α>G(β,x0,ν(x0)).

    Our main result is the following.

    Theorem 1.4. Let g be a continuous function on Ω. If

    (a) there exists a strict minorant u_ with u_=g on Ω and

    (b) the set {wclass:wu_, w=g on Ω} is not empty, then

    u=inf{w:wclass,wu_}

    is a weak solution of (1.1) such that u=g on Ω.

    Once existence of a solution is established, we turn to the analysis of the regularity of the free boundary.

    Theorem 1.5. The free boundary F(u) has finite (n1)-dimensional Hausdorff measure. More precisely, there exists a universal constant r0>0 such that for every r<r0, for every x0F(u),

    Hn1(F(u)Br(x0))rn1.

    Moreover, the reduced boundary F(u) of Ω+(u) has positive density in Hn1 measure at any point of F(u), i.e. for r<r0, r0 universal

    Hn1(F(u)Br(x))crn1,

    for every xF(u). In particular

    Hn1(F(u)F(u))=0.

    Using the results in [8] we deduce the following regularity result.

    Corollary 1.6. F(u) is a C1,γ surface in a neighborhood of Hn1 a.e. point x0F(u).

    Notation. Constants c, C and so on will be termed "universal" if they only depend on λ, Λ, n, Ω, fi and g.

    In this section we show that positive solutions of F(D2u)=f (with f continuous up to the boundary) have asymptotically linear behavior at any boundary point which admits a touching ball, either from inside or from outside the domain. We need the following preliminary result.

    Lemma 2.1. Let r>0, δ>0, σ>0, B+1:=B1{x1>0} and let EB+1{x1>0} be any subset such that there exists ˉxE with

    EB+1{x1>0}Bσ(ˉx).

    Let u be the solution to

    {F(D2u)=rinB+1u=δgEonB+1, (2.1)

    where gE is a cut-off function, gE=1 on E. If r is sufficiently small then there exists a positive constant C=C(δ,σ) such that

    u(x)Cx1inB+1/2.

    Proof. We write

    F(D2u)=ni,j=1aij(x)uxixjLuu,

    with (F=F(M))

    aij=10FMij(tD2u)dt.

    We have

    λ|ξ|2aij(x)ξiξjΛ|ξ|2.

    Denote u=v+w with Luv=0, v=δχE on B+1 and Luw=r, w=0 on B+1. By [11] we have that v(e1/2)Cδ, for some constant C=C(n,λ,Λ,σ), and by the Boundary Harnack principle applied to v and u1(x)=x1 we get that, in B+1/2, for some positive constants c0 and c1,

    c0δx1vc1δx1.

    Put z(x)=12mina11(x1x21)r. The function z is positive in B+1 and

    Luz=a11mina11rr.

    Therefore Lu(w+z)0 in B+1 and w+z0 on B+1. By the maximum principle wc2rx1 in B+1, where c2=a11mina11>0.

    Summing up we get, in B+1/2,

    u=v+w(c0δc2r)x1c3x1

    for r small enough, having c3>0.

    Lemma 2.2. Let Ω1 be a bounded domain with 0Ω1 and

    B+1:=B1{x1>0}Ω1.

    Let u be non-negative and Lipschitz in ¯Ω1B2, such that F(D2u)=f in Ω1B2 and that u=0 in Ω1B2 . Then there exists α0 such that

    u(x)=αx+1+o(|x|)asx0, x1>0.

    Proof. Let αk=sup{β:u(x)βx1 in B+1/k} for k1. Then the sequence {αk}k is increasing and αkL for any k, where L is the Lipschitz constant of u. Let α=limkαk. By definition, u(x)αx1+o(|x|) in B+1, where x=(x1,x2,...,xn).

    Suppose by contradiction that u(x)αx1+o(|x|) in B+1. Then there exist a constant δ0>0 and a sequence {xk}k={(x1,k,x2,k,...,xn,k)}kB+1, with |xk|=rk0, such that

    u(xk)αx1,k+δ0rk.

    Since u is Lipschitz, a simple computation implies that

    u(x)αx1+δ02rkαkx1+δ02rkin {x:|x|=rk,|xxk|δ0rk4L}.

    Let

    uk(x)=u(rkx)rkαkx1.

    The functions uk are defined in B+1 and, by assumption of homogeneity on F, we have

    F(D2uk(x))=F(rkD2u(rkx))=rkF(D2u(rkx))=rkf(rkx)rkf.

    Moreover uk(x)0 on B+1 and ukδ0/2 in Ek={x:xB+1,x1>0,|xxk|δ04L}. We deduce that uk is a supersolution of (2.1). By comparison and Lemma 2.1, there exists C>0, not depending on k, such that

    uk(x)=1rku(rkx)αkx1Cx1in B+1/2.

    Writing z=rkx we obtain u(z)(αk+C)z1 in B+rk/2. Choosing k, k in such a way that αk+C>α and k>2/rk we obtain

    αk>α,

    a contradiction.

    Lemma 2.3. Let Ω1 be a bounded domain such that, writing B1:=B1{x1<0},

    ¯B1¯Ω1={0}.

    Let u be non-negative and Lipschitz in ¯Ω1B2(0), such that F(D2u)=f in Ω1B2(0) and that u=0 in Ω1B2(0) . Then there exists α0 such that

    u(x)=αx+1+o(|x|)asx0, xΩ1.

    Proof. By assumption, we have that

    Ω1B1B+1.

    Then we can extend u as the zero function on B+1Ω1 so that it is a Lipschitz, non-negative solution to

    F(D2u)fin B+1.

    Reasoning in a similar way as in Lemma 2.2, we define αk=inf{β:u(x)βx1 in B1/k}, k1. Then 0αk<+ (u is Lipschitz), and αkα0, with u(x)αx1+o(|x|) in B+1. Again, let us suppose by contradiction that

    u(xk)αx1,kδ0rk.

    where δ0>0 and {xk}k={(x1,k,x2,k,...,xn,k)}kB+1, is such that |xk|=rk0. As before, such inequality propagates by Lipschitz continuity:

    u(x)αx1δ02rkαkx1δ02rkin {x:|x|=rk,|xxk|δ0rk4L}.

    Defining the elliptic, homogeneous operator F(M)=F(M), we have that the functions

    uk(x)=αkx1u(rkx)rk

    solve

    F(D2uk(x))rkfin B+1,

    with uk(x)0 on B+1 and ukδ0/2 in Ek={x:xB+1,x1>0,|xxk|δ04L}. As a consequence, a contradiction can be obtained by reasoning as in Lemma 2.2.

    Lemma 2.4. Let Ω1 be bounded domain with 0Ω1 and

    eitherB1(e1)Ω1or¯B1(e1)¯Ω1={0}.

    Let u be non-negative and Lipschitz in ¯Ω1B2(0), such that F(D2u)=f in Ω1B2(0) and that u=0 in Ω1B2(0). Then there exists α0 such that

    u(x)=αx1+o(|x|)

    as x0 and either xB1(e1) or xΩ.

    Proof. In both cases, we use the smooth change of variable

    {y1=x1ψ(x)y=x,

    where ψ(x) is smooth, with ψ(x)=11|x|2 for |x| small. Then, by direct calculations, the function ˜u(y)=u(y1+ψ(y),y) satisfies

    ˜F(D2˜u,˜u,y)=F(D2u),

    where ˜F is still a uniformly elliptic operator. As a consequence the lemma follows by arguing as in the proofs of Lemmas 2.2, 2.3, with minor changes.

    We conclude this section by providing a uniform estimate from below of the development coefficient α, in case the touching ball is inside the domain.

    Lemma 2.5. Let uC(¯Br(re1)), r1, be such that

    {F(D2u)=finBr(re1),u0,u(0)=0.

    Moreover, assume that u(re1)Cr, for some C>0. Then

    u(x)αx1+o(|x|),whereαc1u(re1)rc2rf,

    as x0, for rˉr, where c1,c2 and ˉr only depend on λ,Λ,n.

    Proof. Let

    ur(x)=u(r(e1+x))r,xB1(0).

    Then

    {F(D2ur)=rfin B1,ur0,ur(e1)=0.

    By Harnack's inequality [5,Theorem 4.3] we have that

    infB1/2urc(ur(0)rf)=:a,

    where c only depends on λ,Λ,n. We are in a position to apply Lemma A.2, which provides

    ur(x)α(x1+1)+o(|x+e1|),with αc1ac2rf=c1ur(0)c2rf,

    as xe1, and the lemma follows.

    Remark 2.6. Notice that the above results can be applied both to F(D2u+)=f1 in Ω+(u) and to F(D2u)=f2χ{u<0} in Ω(w).

    In this section we adapt the strategy developed in [3], in order to show that u+ is locally Lipschitz. To this aim we need to use the following almost-monotonicity formula, provided in [6,14].

    Proposition 3.1. Let ui, i=1,2 be continuous, non-negative functions in B1, satisfying Δui1, u1u2=0 in B1. Then there exist universal constants C0 and r0, such that the functional

    Φ(r):=1r4Br|u1|2rn2Br|u2|2rn2

    satisfies, for 0<rr0,

    Φ(r)C0(1+u12L2(B1)+u22L2(B1))2.

    Lemma 3.2. Let wclass. There exists ˜wclass such that

    1. F(D2˜w)=f1 in Ω+(˜w),

    2. ˜ww, ˜w=w, and

    3. ˜wu_ in Ω.

    Proof. Let wclass and Ω+=Ω+(w). We define

    V:={vC(¯Ω+):F(D2v)f1χ{v>0} in Ω+, v0 in Ω+, v=w on Ω+}

    and

    ˜w(x):={sup{v(x):vV}xΩ+w(x)elsewhere.

    Since u_+V we obtain that V is not empty and that u_˜ww. Moreover ˜w is a solution of the obstacle problem (see [13])

    {F(D2˜w)=f1in {˜w>0}˜w0in Ω+˜w=won Ω+.

    In particular, regularity results for the obstacle problem for fully nonlinear equations imply that ˜w is C1,1 in Ω+ (see [13]). To conclude that ˜wclass, we need to show that the free boundary conditions in Definition 1.2 hold true. Let x0F(˜w): if x0F(w) too, then the free boundary condition follows from the fact that ˜ww; otherwise, x0Ω+ is an interior zero of ˜w, and the free boundary condition follows by the C1,1 regularity of ˜w.

    Lemma 3.3. Let wclass with F(D2w)=f1 in Ω+(w), and let x0F(w) be regular from the right. Then u admits developments

    w+(x)=αxx0,ν++o(|xx0|),w(x)βxx0,ν+o(|xx0|),

    with 0αG(β,x0,ν(x0)), and

    αβC0(1+u122+u222).

    Proof. If x0 is not regular from the left, then by definition of class the asymptotic developments hold with α=β=0 and there is nothing to prove. On the other hand, if x0 is also regular from the left, then the asymptotic developments and the free boundary condition hold true by definition of class and by Lemma 2.4. Also in this case, if α=0 then there is nothing else to prove, thus we are left to deal with the case α>0.

    Reasoning as in [3,Lemma 3], see also [19,Lemma 4.3], one can show that

    Φ(r)C(n)(α+o(1))2(β+o(1))2 (3.1)

    (recall that Φ(r) is defined in Proposition 3.1). On the other hand, since F is concave,

    Δw±cf.

    The conclusion follows by combining Proposition 3.1 with (3.1).

    Proposition 3.4. For every D⊂⊂Ω there exists a constant LD, depending only on D, G, u_ and class, such that

    |w+(x)w+(y)||xy|LD

    for every x,yD, xy, and for every wclass with F(D2w)=f1 in Ω+(w).

    Proof. Let x0Ω+(w)D such that

    r:=dist(x0,F(w))<12dist(¯D,Ω).

    We will show that there exists M>0, not depending on w, such that

    w(x0)rM,

    and the lemma will follow by Schauder estimates and Harnack inequality. By contradiction, let M large to be fixed and let as assume that

    w(x0)r>M.

    Then Lemma 2.5 applies and we obtain

    w(x)αMxx0,ν++o(|xx0|),

    where αM=c1Mc2rf>0 for M sufficiently large. Then x0 is regular from the right and Lemma 3.3 applies, with αMαG(β,x0,ν(x0)), providing

    αMG1(αM)C0(1+u122+u222),

    where G1(α):=infx,νG1(α,x,ν). This provides a contradiction for M sufficiently large.

    Corollary 3.5. u+ is locally Lipschitz and satisfies F(D2u)=f1 in Ω+(u).

    Now we turn to the Lipschitz continuity of u.

    Lemma 4.1. If w1,w2class then min{w1,w2}class.

    Proof. This follows by standard arguments, see e.g. [9,Lemma 4.1].

    To prove that u is Lipschitz continuous we use the double replacement technique introduced in [9]. Let wclass with w(x0)<0 and

    B:=BR(x0),Ω1:=Ω+(w)¯B.

    Working on Ω1, we define

    V1:={v:F(D2v)f1χ{v>0} in Ω1, v0 in Ω1, v=w on Ω1B, v=0 on B}

    (which is non empty, for R sufficiently small, as u_+V1). Then

    w1:=supV1

    solves the obstacle problem (see [13])

    F(D2w1)f1 in {w1>0},w10 in Ω1. (4.1)

    On the other hand, working on B, let

    V2:={v:F(D2v)f2χ{v>0} in B, v0 in B, v=w on B}

    (which is non empty, as wV2). Again,

    w2:=supV2

    solves the obstacle problem

    F(D2w2)f1 in {w2>0},w20 in B. (4.2)

    Under the above notation, the double replacement ˜w of w, relative to B, is defined as

    ˜w:={w1in Ω1w2in Bwotherwise.

    Lemma 4.2. Let wclass with w(x0)=h<0. There exists ε>0 (depending on dist(x0,Ω) and u_) such that:

    1. the double replacement ˜w of w, relative to Bεh(x0), satisfies u_˜ww in Ω;

    2. ˜w<0 and F(D2˜w)=f2 in Bεh(x0), with

    |˜w|Cε+εCf2inBεh/2(x0);

    3. ˜wclass.

    Proof. The inequality w1w in Ω1 follows by the maximum principle, while w2w in B because wV2. On the other hand, provided ε is sufficiently small (depending on the Lipschitz constant of u_), we have that u_<0 in B:=Bεh(x0) and uV1, so that w1u_; finally, by the maximum principle in {w2>0}, also w2u_, and part 1. follows.

    Turning to part 2., assume by contradiction that {w2>0}Bεh(x0). Then, by the regularity properties of the obstacle problem (4.2) (see [13]), we obtain that

    w2(x0)C(εh)2.

    Since w2(x0)w(x0)=h, we obtain a contradiction for ε sufficiently small. Then w2>0 in Bεh(x0), w2 solves the equation by (4.2), and the remaining part of 2. follows by standard Schauder estimates and Harnack inequality.

    Coming to part 3., the fact that ˜w satisfies (a) in Definition 1.2 follows by equations (4.1), (4.2) and by part 2. above, and we are left to check the free boundary conditions. For ˉxF(˜w), three possibilities may occur. If ˉxF(w) then, since ˜ww, then ˜w has the correct asymptotic behavior both when ˉx is regular and when it is not (recall that G(0,,)>0. If ˉx{w1>0}Ω1, then we can use again the regularity of the obstacle problem]4.1 to obtain the correct asymptotic behavior. We are left to the final case, when ˉxBΩ+(w). By Proposition 3.4, let us denote with L the Lipschitz constant of w in Bdist(x0,Ω)/2(x0). Then

    ˜ww+Lεhin B2εh(x0).

    Defining

    ˜wε(x):=˜w(x0+εhx)εh,

    we have that

    {F(D2˜w+ε)=εhf1in (B2¯B1)Ω+(wε)˜w+εLon B2˜w+ε0on B1Ω+(wε).

    Then Lemma A.1 applies, yielding

    ˜w+εαxˉxε,ν(ˉxε)++o(|xˉxε|),

    where

    αc1L+c2εhf1,ˉxε:=ˉxx0εh,

    for universal c1, c2. Going back to ˜w we obtain

    ˜w+αxˉx,ν++o(|xˉx|),αˉL (4.3)

    where ν=(ˉxx0)/|ˉxx0|.

    On the other hand, we can apply Lemma 2.5 to (w2)ε, obtaining

    ˜wε=(w2)εβxˉxε,ν(ˉxε)++o(|xˉxε|),

    where

    βc1εc2εhf1,

    for universal c1, c2, and thus

    ˜wβxˉx,ν+o(|xˉx|),β ˉcε. (4.4)

    Comparing (4.3) and (4.4) we have that, choosing ε small so that

    ˉL<infx,νG(ˉc/ε,x,ν),

    the free boundary condition holds true.

    Corollary 4.3. Let u(x0)=h<0. There exist an non-increasing sequence {˜wk}class, ˜wku_, and ε>0, depending on dist(x0,Ω) and u_, such that the following hold:

    1. ˜wk(x0)u(x0);

    2. ˜wk<0 and F(D2˜wk)=f2 in Bεh(x0);

    3. the sequence {˜wk} is uniformly Lipschitz in Bεh/2(x0), with Lipschitz constant L0 depending on dist(x0,Ω).

    4. ˜wku uniformly on Bhεilon/4

    Proof. Let u(x0)=h<0, {wk}class be such that wku in some neighborhood of x0 and {˜wk}class be the corresponding double replacements, as in Lemma 4.2. Then first three points are direct consequence of the lemma above, and we are left to prove that ˜wku uniformly on Bhεilon/4. By equicontinuity, ˜wk˜w in Bεilonh/2(x0), and suppose by contradiction that ˜w(x1)>u(x1) for some x1Bεilonh/4(x0). Then consider a new sequence {vk}k converging to u at x1 and define {˜uk}k as the double replacement of {min{˜vk,˜wk}}k in Bεilonh/2(x0). Then ˜uk˜u, ˜u˜w in Bεilonh/2(x0), ˜u(x0)=˜w(x0) and ˜u(x1)<˜w(x1). Since F(D2˜w)=F(D2˜u)=f2 in Bεilonh/2(x0), this contradicts the strong maximum principle.

    Corollary 4.4. For any ¯DΩ there exists {wk}kclass such that wku uniformly in ¯D. Furthermore, if ¯DΩ(u), then each wk may be taken non-positive in ¯D.

    Proof. The first part follows from the previous corollary. By compactness, it is enough to prove the second part for balls ¯Bε(x0)Ω(u), with ε small. Let wku uniformly in ¯B2ε(x0)Ω(u), and let

    wεk(x)=wk(x0+εx)εuεin B2.

    Let ϕ be such that

    {Δϕ=cεfin B2¯B1ϕ=aon B2ϕ=0on B1,

    with a and ε positive and sufficiently small so that

    ϕ(e1)e1<infx,νG(0,x,ν)

    (this is possible by explicit calculations, see for instance Lemma A.1); notice that this condition insure that ϕ, extended to zero in B1, is a supersolution in B2 (when c universal is suitably chosen). Since uε0 in ¯B2, for k sufficiently large wka/2 in ¯B2. Let us define

    ˉwεk={min{wεk,ϕ}in ¯B2,wεotherwise.

    Then, by Lemma 4.1, the function

    ˉwk(x)=εˉwεk(xx0ε)

    satisfies ˉwkclass, ˉwk0 in ¯Bε(x0) and ˉwku in ¯Bε(x0), as required.

    Corollary 4.5. u is locally Lipschitz in Ω, continuous in ¯Ω, u=g on Ω. Moreover u solves

    Lu=f2χ{u<0},inΩ(u).

    In this section we will show that u+ is non-degenerate, in the sense of the following result.

    Lemma 5.1. Let x0F(u) and let A be a connected component of Ω+(u)(Br(x0)¯Br/2(x0)) satisfying

    ¯ABr/2(x0),¯ABr(x0),

    for rr0 universal. Then

    supAuCr.

    Moreover

    |ABr(x0)||Br(x0)|C>0,

    where all the constants C depend on d(x,Ω) and on u_.

    Corollary 5.2. F(wk)F(u) locally in Hausdorff distance and χ{wk>0}χ{u>0} in L^1_{\mathit{\boldsymbol{loc}}}.

    The proof of the above result will follow by the two following lemmas.

    Lemma 5.3. Let u be a Lipschitz function in \overline{\Omega}\cap B_1(0), with 0\in\partial \Omega, satisfying

    \begin{cases} F(D^2 u) = f & in \; \Omega\cap B_1\\ u = 0 & on \;\partial\Omega\cap B_1. \end{cases}

    If there exists c>0 such that

    u(x)\geq c{\rm{dist}}(x, \partial \Omega) \qquad for\; every \;x\in \Omega\cap B_{1/2} (5.1)

    then there exists a constant C>0 such that

    \sup\limits_{B_r(0)}u\geq Cr,

    for all r \leq r_0 universal.

    Proof. Let x_0 \in \Omega\cap B_1, \varepsilon = {\rm{dist}}(x_0, \partial\Omega), and let L denote the Lipschitz constant of u. Then

    c\varepsilon\le u(x_0) \le L \varepsilon.

    We will show that, for \delta>0 to be fixed, there exists x_1\in B_\varepsilon(x_0) such that

    u(x_1) \ge (1+\delta) u(x_0). (5.2)

    Then, iterating the procedure, one can conclude as in [9,Lemma 5.1].

    Assume by contradiction that (5.2) does not hold. Then, defining the elliptic, homogeneous operator F^*(M) = - F(-M), we infer that

    v(x): = (1+\delta)u(x_0) - u(x) \gt 0 \quad \text{in }B_\varepsilon(x_0) \qquad\text{satisfies }F^*(D^2 v) = -f.

    Let r(L) = 1-c/(4L); using the Harnack inequality we have that there exists C(L) such that

    v\le C(L) (\delta u(x_0) + \varepsilon^2\|f\|_\infty) \le \frac12 u(x_0)\qquad \text{in }\overline B_{r(L)\varepsilon}(x_0),

    provided both \delta and \varepsilon are sufficiently small (depending on c, L and \|f\|_\infty). In terms of u, the previous inequality writes as

    u \ge \frac{c\varepsilon}{2}\quad \text{in }\overline B_{r(L)\varepsilon}(x_0).

    On the other hand, there exists y_0\in\partial B_{r(L)\varepsilon}(x_0) such that {\rm{dist}}(y_0, \partial\Omega) = (1-r(L))\varepsilon and hence

    \min\limits_{\overline B_{r(L)\varepsilon}(x_0)} u \le u(y_0) \le L {\rm{dist}}(y_0, \partial\Omega) = \frac{c\varepsilon}{4}.

    This is a contradiction, therefore (5.2) holds true.

    Lemma 5.4. There exist universal constants \bar r, \bar C such that

    u(x_0) \ge \bar C {\rm{dist}}(x_0, {\mathcal{F}}(u)) \qquad for \;every \; x_0\in\{x\in\Omega^+(u):{\rm{dist}}(x, {\mathcal{F}}(u))\le \bar r\}.

    Proof. Let x_0\in\{x\in\Omega^+(u):{\rm{dist}}(x, {\mathcal{F}}(u))\le \bar r\}, with \bar r universal to be specified later, and let r: = {\rm{dist}}(x_0, {\mathcal{F}}(u)). We distinguish two cases.

    First let us assume that

    {\rm{dist}}(x_0, \Omega^+(\underline u))\le \frac{r}{2}.

    In this case, for any x\in{\mathcal{F}}(\underline u) we define

    \rho(x): = \max\{r \gt 0:\text{for some }z, \ x\in\partial B_r(z)\text{ and }B_r(z)\subset\Omega^+ (\underline u)\}.

    Notice that \rho(x)>0 for every x, since any point in {\mathcal{F}}(\underline u) is regular from the right by assumption. Thus, recalling that \underline u^+ has linear growth bounded below by \inf_{x, \nu}G(0, x, \nu), and noticing that B_{3r/4}(x_0)\cap \Omega^+(\underline u) contains a ball of radius comparable with r (at least for a suitable choice of \bar r):

    \sup\limits_{B_{3r/4}(x_0)} u^+ \ge \sup\limits_{B_{3r/4}(x_0)} \underline u^+ \ge \bar C r,

    where \bar C only depends on \underline u.

    On the other hand, in case

    {\rm{dist}}(x_0, \Omega^+(\underline u))\ge \frac{r}{2},

    we have \underline u\le 0 in B_{r/2}(x_0). By Corollary 4.4 we can find \{w_k\}_k\subset{\rm{class}} converging uniformly to u on some D\supset B_{r}(x_0). By scaling

    u_r(x) = \frac{u(x_0+rx)}{r}, \qquad w_k^r(x) = \frac{w_k(x_0+rx)}{r},

    we need to find \bar C universal such that u_r(0)\ge \bar C. Let us assume by contradiction that

    u_r(0) \lt \bar C.

    Then by Harnack inequality

    u_r \le C(\bar C + r \|f_1\|_\infty)\qquad\text{in }B_{1/2}

    and, for k sufficiently large,

    w_k^r \le C'(\bar C + r \|f_1\|_\infty)\qquad\text{in }B_{1/2}.

    Now, reasoning as in the proof of Corollary 4.4, let \phi be such that

    \begin{cases} \Delta \phi = -c r \|f\|_\infty & \text{in }B_{1/2}\setminus \overline{B}_{1/4}\\ \phi = a & \text{on }\partial B_{1/2}\\ \phi = 0 & \text{on }\partial B_{1/4}, \end{cases}

    with a and r positive and sufficiently small so that \nabla\phi(e_1/4) \cdot e_1 < \inf_{x, \nu}G(0, x, \nu), in such a way that \phi, extended to zero in B_{1/4}, is a supersolution in B_{1/2}. Then, choosing \bar C < (a-r \|f_1\|_\infty)/C' we obtain that w_k^r < \phi on \partial B_{1/2} and then the functions

    \bar{w}_k^r = \begin{cases} 0 &\text{in }\overline B_{1/4}, \\ \min\{w_k^r, \phi\}&\text{in }\overline B_{1/2}\setminus B_{1/4}, \\ w^r_k &\text{otherwise} \end{cases}

    are continuous, while

    \bar w_k(x) = r\bar w_k^r\left(\frac{x-x_0}{r}\right)

    satisfy \bar w_k \in {\rm{class}}, \bar w_k \equiv 0 in \overline B_{r/4}(x_0). This is in contradiction with the fact that u(x_0)>0, and the lemma follows.

    This section is devoted to the proof that u satisfies the supersolution condition (ⅰ) in Definition 1.1. Thanks to Lemma 2.4 we only need to prove that, whenever u admits asymptotic developments at x_0\in{\mathcal{F}}(u), with coefficients \alpha and \beta, then \alpha\le G(\beta, x_0, \nu_{x_0}). To do that, we need to distinguish the two cases \beta>0 and \beta = 0.

    Lemma 6.1. Let x_0\in{\mathcal{F}}(u), and

    \begin{split} u^+(x)& = \alpha \left\langle x-x_0, \nu \right\rangle^+ + o(|x-x_0|), \\ u^-(x)& = \beta \left\langle x-x_0, \nu \right\rangle^- + o(|x-x_0|), \end{split}

    with

    \beta \gt 0.

    Then \alpha\le G(\beta, x_0, \nu_{x_0}).

    Proof. Since \beta >0, then {\mathcal{F}}\left(u\right) is tangent at x_{0} to the hyperplane

    \pi :\left\langle x-x_{0}, \nu \right\rangle = 0

    in the following sense: for any point x\in {\mathcal{F}}\left(u\right) , dist\left(x, {\mathcal{F}}\left(u\right) \right) = o\left(\left\vert x-x_{0}\right\vert \right). Otherwise we get a contradiction to the asymptotic development of u.

    Let \{w_k\}_k\subset {\rm{class}} be uniformly decreasing to u, as in Corollary 4.4. By the non-degeneracy of u^+ we have that, for k large, w_k can not remain strictly positive near x_0. Let d_{k} = d_{H}\left({\mathcal{F}}\left(w_{k}\right), {\mathcal{F}}\left(u\right) \right) be the Hausdorff distance between the two free boundaries. In the ball B_{2\sqrt{d_{k}}}\left(x_{0}\right) , {\mathcal{F}}\left(u\right) is contained in a strip parallel to \pi of width o\left(\sqrt{d_{k}} \right) and, since d_{k}\rightarrow 0, {\mathcal{F}}\left(w_{k}\right) is contained in a strip S_{k} of width d_{k}+o\left(\sqrt{d_{k}}\right) = o\left(\sqrt{d_{k}}\right) .

    Consider now the points x_{k} = x_{0}-\sqrt{d_{k}}\nu and let B_{k} = B_{r_{k}}\left(x_{k}\right) be the largest ball contained in \Omega ^{-}\left(w_{k}\right) with touching point z_{k}\in {\mathcal{F}}\left(w_{k}\right) . Then z_{k}\in S_{k} and, since w_{k}\geq u, from the asymptotic developments of w_{k} and u we have

    \beta \sqrt{d_{k}}+o\left( \sqrt{d_{k}}\right) = u^{-}\left( x_{k}\right) \geq w^{-}\left( x_{k}\right) = \beta _{k}r_{k}+o(r_{k}),

    since

    \sqrt{d_{k}}+o\left( \sqrt{d_{k}}\right) \leq r_{k}\leq \sqrt{d_{k}}.

    Passing to the limit we infer

    \lim \sup \beta _{k}\leq \beta .

    Reasoning in the same way on the other side th the points y_{k} = x_{0}+\sqrt{ d_{k}} (and the same z_{k}, which are regular from the left), we get

    \alpha \leq \lim \inf \alpha _{k}\text{.}

    From \alpha _{k}\leq G\left(\beta _{k}, z_k, \nu_{k}\right) , where \nu _{k} = \left(x_{k}-z_{k}\right) /\left\vert x_{k}-z_{k}\right\vert , we get \alpha \leq G\left(\beta, x_0, \nu_{x_0} \right) .

    To treat the case \beta = 0 we need the following preliminary lemma.

    Lemma 6.2. Let v\geq 0 continous in B_1(x_0) be such that \Delta v\geq -M. Let

    \Psi _{r}\left( x_0, v\right) = \frac{1}{r^{2}} \int_{B_{r}(x_0)}\frac{\left\vert \nabla v\right\vert ^{2}}{\left\vert x-x_0\right\vert ^{n-2}}dx.

    Then, for r small,

    \Psi _{r}\left( x_0, v\right) \leq c\left( n\right) \left\{ \sup\limits_{B_{2r}(x_0)}\left( \frac{v}{r}\right) ^{2}+M\sup\limits_{B_{2r}(x_0)}v\right\} . (6.1)

    Proof. We may assume x_0 = 0 and write \Psi _{r}\left(0, v\right) = \Psi _{r}\left(v\right). Rescale setting v_{r}\left(x\right) = v\left(rx\right) /r; we have \Delta v_{r}\geq -rM and

    \Psi _{r}\left( v\right) = \Psi _{1}\left( v_{r}\right) \text{.}

    Let \eta \in C_{0}^{\infty }\left(B_{2}\right) , \eta = 1 in B_{1}. Since 2\left\vert \nabla v_{r}\right\vert ^{2}\leq 2rMv_{r}+\Delta v_{r}^{2}, we have:

    \begin{eqnarray*} \Psi _{1}\left( v_{r}\right) &\leq &C\int_{B_{2}}\eta \frac{\left\vert \nabla v_{r}\right\vert ^{2}}{\left\vert x\right\vert ^{n-2}}\leq C\int_{B_{2}}\eta \frac{2Mv_{r}+\Delta v_{r}^{2}}{\left\vert x\right\vert ^{n-2}} \\ & = &C\int_{B_{2}}\left[\frac{2Mv_{r}}{\left\vert x\right\vert ^{n-2}} +v_{r}^{2}\Delta \left( \frac{\eta }{\left\vert x\right\vert ^{n-2}}\right)\right], \end{eqnarray*}

    so that

    \begin{eqnarray*} \Psi _{r}\left( v\right) & = &\Psi _{1}\left( v_{r}\right) \leq c\left( n\right) \left( \left\vert v_{r}\right\vert _{L^{\infty }\left( B_{2}\right) }^{2}+rM\left\vert v_{r}\right\vert _{L^{\infty }\left( B_{2}\right) }\right) \\ & = &c\left( n\right) \left\{ \sup\limits_{B_{2r}}\left( \frac{v}{r}\right) ^{2}+M\sup\limits_{B_{2r}}v\right\} , \end{eqnarray*}

    which is (6.1).

    Lemma 6.3. Let x_0\in{\mathcal{F}}(u), and

    \begin{split} u^+(x)& = \alpha \left\langle x-x_0, \nu \right\rangle^+ + o(|x-x_0|), \\ u^-(x)& = o(|x-x_0|). \end{split}

    Then \alpha\le G(\beta, x_0, \nu_{x_0}).

    Proof. As before, let \{w_k\}_k\subset {\rm{class}} be uniformly decreasing to u, with w_k that is not strictly positive near x_0, for k large. The first part of the proof is exactly as in Lemma 6.3 of [9], until equation (6.2) below. For the reader's convenience, we recall such argument here.

    For each k we denote with

    B_{m, k} = B_{\lambda _{m, k}}\left(x_{0}+\frac{1}{m}\nu \right)

    the largest ball centered at x_{0}+\nu/{m} contained in \Omega ^{+}(w_{k}), touching {\mathcal{F}}(w_{k}) at x_{m, k} where \nu _{m, k} is the unit inward normal of {\mathcal{F}}(w_{k}) at x_{m, k}. Then up to proper subsequences we deduce that

    \lambda _{m, k}\rightarrow \lambda _{m}, \quad x_{m, k}\rightarrow x_{m}, \quad \nu _{m, k}\rightarrow \nu _{m}

    and B_{\lambda _{m}}(x_{0}+\nu/m) touches {\mathcal{F}}(u) at x_{m}, with unit inward normal \nu _{m}. From the behavior of u^{+}, we get that

    |x_{m}-x_{0}| = o\left(\frac{1}{m}\right),
    \frac{1}{m}+o\left(\frac{1}{m}\right)\leq \lambda _{m}\leq \frac{1}{m}

    and

    |\nu _{m}-\nu | = o(1).

    Now since w_{k}\in \mathcal{F}, near x_{m, k} in B_{m, k}:

    w_{k}^{+}\leq \alpha _{m, k}\langle x-x_{m, k}, \nu _{m, k}\rangle ^{+}+o(|x-x_{m, k}|)

    and in \Omega \setminus B_{m, k}

    w_{k}^{-}\geq {\beta _{m, k}}\langle x-x_{m, k}, \nu _{m, k}\rangle ^{-}+o(|x-x_{m, k}|)

    with

    0\leq \alpha _{m, k}\leq G(\beta _{m, k}, x_{m, k}, \nu _{m, k}),

    (by Lemma 2.5 the touching occurs at a regular point, for m, k large.) We know that

    w_{k}^{+}\geq u^{+}\geq \alpha \langle x-x_{0}, \nu \rangle ^{+}+o(|x-x_{0}|),

    hence

    \underline{\alpha }_{m} = \liminf\limits_{k\rightarrow \infty }\alpha _{m, k}\geq \alpha -\varepsilon ilon _{m}

    and \varepsilon ilon _{m}\rightarrow 0, as m\rightarrow \infty. We have to show that

    \underline{\beta } = \liminf\limits_{m, k\rightarrow +\infty }\beta _{m, k} = 0.

    We assume by contradiction that \bar\beta>0. Acting as in [9,Lemma 6.3] we obtain, for r small,

    (1+\omega (r))\Phi _{r}(x_{m, k}, w_{k})+C\omega (r)\geq c_{n}\;\alpha _{m, k}^{2}\beta _{m, k}^{2}, (6.2)

    where

    \Phi _{r}\left( x_{m, k}, w_{k}\right) = \Psi _{r}\left( x_{m, k}, w_{k}^{+}\right) \Psi _{r}\left( x_{m, k}, w_{k}^{-}\right) .

    By concavity we have that \Delta w_{k}^{\pm }\geq -M where M = c\min \left(\| f_{1}\|_{\infty }, \|f_{2}\|_{\infty }\right) . Lemma 6.2 implies

    \begin{split} c_n\alpha_{m, k} ^{2}{\beta}^{2}_{m, k} &\leq \left( 1+\omega \left( r\right) \right) \Psi _{r}\left( x_{m, k}, w_{k}^{+}\right) \Psi _{r}\left( x_{m, k}, w_{k}^{-}\right) +C\omega \left( r\right) \\ &\leq c^{2}\left( n\right) \left( 1+\omega \left( r\right) \right) \left\{ \sup\limits_{B_{2r}\left( x_{m, k}\right) }\left( \frac{w_{k}^{+}}{r}\right) ^{2}+M\sup\limits_{B_{2r}\left( x_{m, k}\right) }w_{k}^{+}\right\} \times \\ & \qquad \times\left\{ \sup\limits_{B_{2r}\left( x_{m, k}\right) }\left( \frac{w_{k}^{-}}{r}\right) ^{2}+M\sup\limits_{B_{2r}\left( x_{m, k}\right) }w_{k}^{-}\right\} +C\omega \left( r\right) \\ &\leq C_{1}\left( n, M, L)\right) \left\{ \sup\limits_{B_{2r}\left( x_{m, k}\right) }\left( \frac{w_{k}^{-}}{r}\right) ^{2}+M\sup\limits_{B_{2r}\left( x_{m, k}\right) }w_{k}^{-}\right\} +C\omega \left( r\right) , \end{split}

    where L is the uniform Lipschitz constant of \{w_k^+\}_k (recall Lemma 3.4). Taking the \liminf as m, k\to\infty and using the uniform convergence of w_{k} to u we infer

    0 \lt c_n \alpha ^{2}\bar{\beta}^{2}\leq C_{1}\left( n, M, L\right) \left\{ \sup\limits_{B_{2r}\left( x_0\right) }\left( \frac{u^{-}}{ r}\right) ^{2}+M\sup\limits_{B_{2r}\left( x_0\right) }u^{-}\right\} +C\omega \left( r\right) .

    Recalling that, by assumption, u^-(x) = o(|x-x_0|) as x\to x_0, we have

    \sup\limits_{B_{2r}\left( 0\right) }\left( \frac{u^{-}}{r}\right) ^{2} = o \left(1\right) \qquad\text{as }r\rightarrow 0,

    and we get a contradiction.

    In this section we want to show that u is a subsolution according to Definition 1.1. Note that, if x_{0}\in {\mathcal{F}}(u) is a regular point from the left with touching ball B\subset \Omega ^{-}(u), then near to x_{0}

    u^{-}(x) = \beta \langle x-x_{0}, \nu \rangle ^{-}+o(| x-x_{0}| ), \quad \beta \geq 0,

    in B, and

    u^{+}(x) = \alpha \langle x-x_{0}, \nu \rangle ^{+}+o(| x-x_{0}| ), \quad \alpha \geq 0

    in \Omega \backslash B. Indeed, even if \beta = 0, then \Omega ^{+}(u) and \Omega ^{-}(u) are tangent to \{\langle x-x_{0}, \nu \rangle = 0\} at x_{0} since u^{+} is non-degenerate. Thus u has a full asymptotic development as in the next lemma. We want to show that \alpha \geq G(\beta, x_{0}, \nu). We follow closely [3] and [9].

    Lemma 7.1. Assume that near x_{0}\in {\mathcal{F}}(u),

    u(x) = \alpha \langle x-x_{0}, \nu \rangle ^{+}-\beta \langle x-x_{0}, \nu \rangle ^{-}+o(| x-x_{0}|),

    with \alpha >0, \beta \geq 0. Then

    \alpha \geq G(\beta , x_{0}, \nu ).

    Proof. Assume by contradiction that \alpha < G(\beta, x_{0}, \nu). We construct a supersolution w\in {\rm{class}} which is strictly smaller than u at some point, contradicting the minimality of u. Let u_{0} be the two-plane solution, i.e.

    u_{0}(x): = \lim\limits_{r\rightarrow 0}\frac{u(x_{0}+rx)}{r} = \alpha \langle x, \nu \rangle ^{+}-\beta \langle x, \nu \rangle ^{-}.

    Suppose that \alpha \leq G(\beta, x_{0}, \nu)-\delta _{0} with \delta _{0}>0. Fix \zeta = \zeta (\delta _{0}), to be chosen later. By Corollary 4.4, we can find w_{k}\in F\searrow u locally uniformly and, for r small, k large, the rescaling w_{k, r} satisfies the following conditions:

    if \beta >0, then

    w_{k, r}(x)\leq u_{0}+\zeta \min \{\alpha , \beta \}\ \text{on }\partial B_{1};

    if \beta = 0, then

    w_{k, r}(x)\leq u_{0}+\alpha \zeta \;\text{ on }\partial B_{1}

    and

    w_{k, r}(x)\leq 0, \quad \text{in}\; \{\langle x, \nu \rangle \lt -\zeta \}\cap \overline{B}_{1}.

    In particular,

    w_{k, r}(x)\leq u_{0}(x+\zeta \nu )\quad \text{on}\; \partial B_{1}.

    If \beta >0, let v satisfy

    \begin{cases} {F(D}^{2}v) = rf_{1}^{r}, &\text{in } \{\langle x, \nu \rangle \gt -\zeta +\varepsilon ilon \phi (x)\} \\ {F(D}^{2}v^{-}) = rf_{2}^{r}, &\text{in } \{\langle x, \nu \rangle \lt -\zeta +\varepsilon ilon \phi (x)\} \\ v(x) = 0, &\text{on } \{\langle x, \nu \rangle = -\zeta +\varepsilon ilon \phi (x)\} \\ v(x) = u_{0}(x+\zeta \nu ), &\text{on } \partial B_{1}, \end{cases} (7.1)

    where \phi \geq 0 is a cut-off function, \phi \equiv 0 outside B_{1/2}, \phi \equiv 1 inside B_{1/4}.

    For \beta = 0, replace the second equation with v = 0.

    Along the new free boundary, {\mathcal{F}}(v) = \{\langle x, \nu \rangle = -\zeta +\varepsilon ilon \phi (x)\} we have the following estimates:

    |v_{\nu }^{+}-\alpha |\leq c(\varepsilon +\zeta )+Cr, \quad |v_{\nu }^{-}-\beta |\leq c(\varepsilon +\zeta )+Cr,

    with c, C universal.

    Indeed,

    v^{+}-\alpha \langle x, \nu \rangle ^{+}

    is a solution of

    {F}(D^2(v-\alpha \langle x, \nu \rangle ^{+})) = rf_{1}^{r}.

    Thus, by standard C^{1, \gamma } regularity estimates (see [16,Theorem 1.1])

    |v_{\nu }^{+}-\alpha |\leq C\left( \Vert v-\alpha \langle x, \nu \rangle ^{+}\Vert _{\infty }+[-\zeta +\varepsilon ilon \phi ]_{1, \gamma }+r\Vert f_{1}\Vert _{\infty }\right) ,

    which gives the desired bound. Similarly, one gets the bound for v_{\nu }^{-}.

    Hence, since \alpha \leq G(\beta, x_{0}, \nu (x_{0}))-\delta _{0}, say for \varepsilon = 2\zeta and \zeta, r small depending on \delta _{0}

    v_{\nu }^{+} \lt G(v_{\nu }^{-}, x_{0}, \nu ),

    and the function,

    \bar{w}_{k} = \left\{ \begin{array}{l} \min \{w_{k}, \lambda v(\frac{x-x_{0}}{\lambda })\}\quad \text{ in } B_{\lambda }(x_{0}), \\ w_{k}\quad \text{in }\Omega \setminus B_{\lambda }(x_{0}), \end{array} \right.

    is still in {\rm{class}}. However, the set

    \{\langle x, \nu \rangle \leq -\zeta +\varepsilon ilon \phi \}

    contains a neighborhood of the origin, hence rescaling back x_{0}\in \Omega ^{-}(\bar{w}_{k}). We get a contradiction since x_{0}\in F(u) and \Omega ^{+}(u)\subseteq \Omega ^{+}(\bar{w}_{k}).

    In this section we prove the weak regularity properties of the free boundary. Both statements and proofs are by now rather standard and follows the papers [3] and [9] for problems governed by homogeneous and inhomogeneous divergence equations, respectively. Thus we limit ourselves to the few points in which differences from the previous cases emerge. Denote by \mathcal{N}_{\varepsilon }\left(A\right) an \varepsilon-neighborhood of the set A. The following lemma provides a control of the {\mathcal{H}}^{n-1} measure of {\mathcal{F}}\left(u\right) and implies that \Omega ^{+}\left(u\right) is a set of finite perimeter.

    Lemma 8.1. Let u be our Perron solution. Let x_{0}\in {\mathcal{F}}\left(u\right) \cap B_{1}. There exists a positive universal \delta_{0} < 1 such that, for every 0 < \varepsilon < \delta \leq \delta _{0}, the following quantities are comparable:

    1. \frac{1}{\varepsilon }\left\vert \left\{ 0 < u < \varepsilon\right\} \cap B_{\delta }\left(x_{0}\right) \right\vert ,

    2. \frac{1}{\varepsilon }\left\vert \mathcal{N}_{\varepsilon }\left({\mathcal{F}}\left(u\right) \right) \cap B_{\delta }\left(x_{0}\right) \right\vert ,

    3. N\varepsilon ^{n-1}, where N is the number of any family of balls of radius \varepsilon, with finite overlapping, covering {\mathcal{F}}\left(u\right) \cap B_{\delta }\left(x_{0}\right) ,

    4. {\mathcal{H}}^{n-1}\left({\mathcal{F}}\left(u\right) \cap B_{\delta }\left(x_{0}\right) \right).

    Proof. From [3], it is sufficient to prove the following two equivalences:

    c_{1}\varepsilon ^{n}\leq \int_{B_{\varepsilon }\left( x_{0}\right) }\left\vert \nabla u\right\vert ^{2}\leq C_{1}\varepsilon ^{n} (8.1)

    and

    c_{3}\varepsilon \delta ^{n-1}\leq \int_{\left\{ 0 \lt u \lt \varepsilon \right\} \cap B_{\delta }\left( x_{0}\right) }\left\vert \nabla u\right\vert ^{2}\leq C_{2}\varepsilon \delta ^{n-1} (8.2)

    with universal constants c_{1}, c_{2}, C_{1}, C_{2}.

    Since F \left(D^{2}u\right) = \inf_{\alpha }L_{\alpha }u where L_{\alpha } is a uniformly elliptic operator with constant coefficients and ellipticity constant \lambda, \Lambda , we have L_{\alpha }u^{+}\geq f_{1} in \Omega ^{+}\left(u\right) . Fix \alpha = \alpha _{0} and set

    L_{\alpha _{0}} = L = \sum\limits_{i, j = 1}^{n}a_{ij}\partial _{_{i}{}_{j}}\text{, }\quad A = \left( a_{ij}\right) .

    The upper bound in (8.1) follows by the Lipschitz continuity of u. The lower bound follows from \sup_{B_{\varepsilon }\left(x_{0}\right) }u^{+}\geq c\varepsilon , c universal, \inf_{B_{\varepsilon }\left(x_{0}\right) }u^{+} = 0, the Lipschitz continuity of u, and the Poincaré inequality (see [1,Lemma 1.15]).

    To prove (8.2), rescale by setting

    u_{\delta }\left( x\right) = \frac{u\left( x_{0}+\delta x\right) }{\delta }, \text{ }f_{1}^{\delta }\left( x\right) = f_{1}\left( x_{0}+\delta x\right) \quad x\in B_{1} = B_{1}\left( 0\right) .

    Then Lu_{\delta }\geq \delta f_{1}^{\delta } in \Omega ^{+}\left(u^{\delta }\right) \cap B_{1}. For 0 < \varepsilon < \delta, let

    u_{\delta, s, \varepsilon } = u_{s, \varepsilon }: = \max \left\{ s/\delta , \min \left\{ u_{\delta }, \varepsilon /\delta \right\} \right\} .

    We have:

    \begin{split} & -\delta \int_{B_{1}}f_{1}^{\delta }u_{\varepsilon , s} = -\int_{B_{1}}u_{\varepsilon , s}Lu_{\delta }^{+} \\ & = \int_{B_{1}}\langle A\nabla u_{\delta }^{+}, \nabla u_{\varepsilon , s}^{+}\rangle dx-\int_{\partial B_{1}}\langle A\nabla u_{\delta }^{+}, \nu \rangle u_{\varepsilon , s}d\mathcal{H}^{n-1} \\ & = \int_{B_{1}\cap \{0 \lt s/\delta \lt u_{\delta } \lt \varepsilon /\delta \}}\langle A\nabla u_{\delta }, \nabla u_{\delta }\rangle dx-\int_{\partial B_{1}}\langle A\nabla u_{\delta }^{+}, \nu \rangle u_{\varepsilon , s}d \mathcal{H}^{n-1} \end{split}

    since \nabla u_{\varepsilon, s} = \nabla u_{\delta }\cdot \chi _{\{s/\delta < u_{\delta } < \varepsilon /\delta \}}.

    By uniform ellipticity, since u^{+} is Lipschitz and f_{1} is bounded, we get (\delta < 1)

    \int_{B_{1}\cap \{0 \lt s/\delta \lt u_{\delta } \lt \varepsilon /\delta \}}| \nabla u_{\delta }| ^{2}dx\leq C\frac{\varepsilon }{\delta },

    with C universal. Letting s\rightarrow 0 and rescaling back, we obtain the upper bound in (8.2).

    For the lower bound, let V be the solution to

    \left\{ \begin{array}{l} LV = -\frac{\chi _{B_{\sigma }}}{| B_{\sigma }| }, \quad \mbox{in}\quad B_{1} \\ V = 0, \quad \mbox{on}\quad \partial B_{1} \end{array} \right. (8.3)

    with \sigma to be chosen later. By standard estimates, see for example [12], V\leq C\sigma ^{2-n} and -\langle A\nabla V, \nu \rangle \sim C^{\ast } on \partial B_{1}, with C^{\ast } independent of \sigma. By Green's formula

    \int_{B_{1}}(LV)\frac{u_{\delta }^{+}u_{\varepsilon , 0}}{\varepsilon }-\left(L \frac{u_{\delta }^{+}u_{\varepsilon , 0}}{\varepsilon }\right)V = \int_{\partial B_{1}}\frac{u_{\delta }^{+}u_{\varepsilon , 0}}{\varepsilon }\langle A\nabla V, \nu \rangle d\mathcal{H}^{n-1} (8.4)

    since V = 0 on \partial B_{1}. We estimate

    \delta \left| \int_{B_{1}}(LV)\frac{u_{\delta }^{+}u_{\varepsilon }}{ \varepsilon }dx\right| = \frac{\delta }{| B_{\sigma }| } \left|\int_{B_{\sigma }}\frac{u_{\delta }^{+}u_{\varepsilon }}{\varepsilon }dx\right| \leq \bar{C}\sigma , (8.5)

    since u is Lipschitz and 0\leq u_{\varepsilon, 0}\leq \varepsilon /\delta . From (8.4) and (8.5) and the fact that \langle A_{\delta }\nabla V, \nu \rangle \sim -C^{\ast } on \partial B_{1} we deduce that

    \begin{split} \delta \int_{B_{1}}\left(L\frac{u_{\delta }^{+}u_{\varepsilon , 0}}{\varepsilon } \right)Vdx& \geq -\bar{C}\sigma -\delta \int_{\partial B_{1}}\frac{u_{\delta }^{+}u_{\varepsilon , 0}}{\varepsilon }\langle A\nabla V, \nu \rangle d \mathcal{H}^{n-1} \\ & \geq -\bar{C}\sigma +C^{\ast }\delta \int_{\partial B_{1}}\frac{u_{\delta }^{+}u_{\varepsilon , 0}}{\varepsilon }d\mathcal{H}^{n-1}. \end{split}

    Thus using that u^{+} is non-degenerate and choosing \sigma small enough (universal) we get that (\delta >\varepsilon )

    \delta \int_{B_{1}}\left(L\frac{u_{\delta }^{+}u_{\varepsilon , 0}}{\varepsilon } \right)Vdx\geq \tilde{C}. (8.6)

    On the other hand in \{0 < u_{\delta }^{+} < \varepsilon /\delta \}\cap B_{1},

    Lu_{\delta }^{+}u_{\varepsilon , 0} = 2\delta u_{\varepsilon }f_{1}^{\delta }+\langle A\nabla u_{\delta }, \nabla u_{\delta }\rangle . (8.7)

    Combining (8.6), (8.7) and using the ellipticity of A we get that

    \frac{2\delta ^{2}}{\varepsilon }\int_{B_{1}}u_{\varepsilon }f_{1}^{\delta }V+\frac{\delta \Lambda }{\varepsilon }\int_{B_{1}}|\nabla u_{\delta }|^{2}V\geq \bar{C}.

    From the estimate on V we obtain that for \delta small enough

    \frac{\delta }{\varepsilon }\int_{B_{1}}|\nabla u_{\delta }|^{2}\geq C

    for some C universal. Rescaling, we obtain the desired lower bound.

    Lemma 8.1 implies that \Omega ^{+}(u)\cap B_{r}(x), x\in {\mathcal{F}}(u), is a set of finite perimeter. Next we show that in fact this perimeter is of order r^{n-1}.

    Theorem 8.2. Let u be our Perron solution. Then, the reduced boundary of \Omega ^{+}(u) has positive density in \mathcal{H}^{n-1}-measure at any point of {\mathcal{F}}(u), i.e. for r < r_{0}, r_{0} universal,

    \mathcal{H}^{n-1}({\mathcal{F}}^{\ast }(u)\cap B_{r}(x))\geq cr^{n-1}

    for every x\in {\mathcal{F}}(u).

    Proof. The proof follows the lines of Corollary 4 in [3] and Theorem 8.2 in [9]. Let w_{k}\in {\rm{class}}, w_{k}\searrow u in \overline{ B}_{1} and L as in Lemma 8.1. Then \Omega ^{+}\left(u\right) \subset \subset \Omega ^{+}\left(w_{k}\right) and Lw_{k}\geq F\left(D^{2}w_{k}\right) = f_{1} in \Omega ^{+}(u). Let x_{0}\in {\mathcal{F}}(u). We rescale by setting

    u_{r}(x) = \frac{u(x_{0}+rx)}{r}, \quad w_{k, r} = \frac{w_{k}(x_{0}+rx)}{r}\quad x\in B_{1}.

    Let V be the solution to (8.3). Since \nabla w_{k, r} is a continuous vector field in \overline{\Omega _{r}^{+}(u_{r})\cap B_{1}}, we can use it to test for perimeter. We get

    \begin{split} & \int_{B_{1}\cap \Omega _{r}^{+}(u_{r})}\left( Vrf_{1}^{r}-w_{k, r}LV\right) \leq \int_{B_{1}\cap \Omega _{r}^{+}(u_{r})}\left( VLw_{kr}-w_{k, r}LV\right) \\ & = \int_{{\mathcal{F}}^{\ast }(u_{r})\cap B_{1}}\left( V\langle A\nabla w_{k, r}, \nu \rangle -w_{kr}\langle A\nabla V, \nu \rangle \right) d\mathcal{H} ^{n-1}-\int_{\partial B_{1}\cap \Omega _{r}^{+}(u_{r})}w_{kr}\langle A\nabla V, \nu \rangle d\mathcal{H}^{n-1}. \end{split} (8.8)

    Using the estimates for V and the fact that the w_{k} are uniformly Lipschitz, we get that

    \left| \int_{{\mathcal{F}}^{\ast }(u_{r})\cap B_{1}}V\langle A\nabla w_{k, r}, \nu \rangle d \mathcal{H}^{n-1}\right| \leq C(\sigma )\mathcal{H}^{n-1}({\mathcal{F}}^{\ast }(u_{r})\cap B_{1}). (8.9)

    As in [3] we have, as k\rightarrow \infty ,

    \int_{{\mathcal{F}}^{\ast }(u_{r})\cap B_{1}}w_{k, r}\langle A\nabla V, \nu \rangle d \mathcal{H}^{n-1}\rightarrow 0,
    \int_{\partial B_{1}\cap \Omega _{r}^{+}(u_{r})}w_{kr}\langle A\nabla V, \nu \rangle d\mathcal{H}^{n-1}\rightarrow \int_{\partial B_{1}}u_{r}^{+}\langle A\nabla V, \nu \rangle d\mathcal{H}^{n-1}

    and

    -\int_{B_{1}\cap \Omega _{r}^{+}(u_{r})}w_{k, r}LV\rightarrow {{\rlap{-} \smallint }_{B_{\sigma }}u_{r}^{+}}.

    Passing to the limit in (8.8) and using all of the above we get

    \begin{split} &\left |r\int_{B_{1}\cap \Omega ^{+}(u_{r})}Vf_{1}^{r}+{{\rlap{-} \smallint }_{B_{\sigma }}u_{r}^{+}}+\int_{\partial B_{1}}u_{r}^{+}\langle A\nabla V, \nu \rangle d \mathcal{H}^{n-1}\right| \\ & \leq C(\sigma )\mathcal{H}^{n-1}({\mathcal{F}}^{\ast }(u_{r})\cap B_{1}). \end{split} (8.10)

    Since u is Lipschitz and non-degenerate, for \sigma small

    \frac{1}{\left\vert B_{\sigma }\right\vert }\int_{B_{\sigma }}u_{r}^{+}\leq \bar{C}\sigma,

    and using the estimate for \langle A\nabla V, \nu \rangle

    -\int_{\partial B_{1}}u_{r}^{+}\langle A\nabla V, \nu \rangle d\mathcal{H} ^{n-1}\geq \bar{c} \gt 0.

    Also, since f_{1}^{r} is bounded,

    \int_{B_{1}\cap \Omega _{r}^{+}(u_{r})}Vf_{1}^{r}\leq \bar{C}(\sigma ).

    Hence choosing first \sigma and then r sufficiently small we get that

    \mathcal{H}^{n-1}({\mathcal{F}}^{\ast }(u_{r})\cap B_{1})\geq \tilde{C},

    \tilde{C} universal.

    For the reader's convenience we collect here some explicit barrier functions which arise frequently in our arguments. Their proof is based on comparison arguments, together with the well known chain of inequalities

    \mathcal{P}^-_{\lambda/n, \Lambda} u \le F(D^2u) \le c \Delta u, (A.1)

    where \mathcal{P}^-_{\lambda/n, \Lambda} denotes the lower Pucci operator, and c = c(\lambda, \Lambda, n)> 0 since F is concave (see [5] for further details).

    Lemma A.1 (Barrier for subsolutions). Let u satisfy

    \begin{cases} F(D^2u)\ge f & in \;B_{2}(0)\setminus \overline{B}_{1}(0)\\ u\le a & on \;\partial B_{2}(0)\\ u\le 0 & on \;\partial B_{1}(0). \end{cases}

    Then

    u(x)\le \alpha(x_1-1) + o(|x-e_1|)\qquad where \;\alpha\le c_1 a + c_2\|f\|_\infty,

    as x\to e_1, where the positive constants c_1, c_2 only depend on \lambda, \Lambda, n.

    Proof. By comparison and (A.1) we infer that u\ge \phi in B_{2} \setminus \overline{B}_{1}, where \phi solves

    \begin{cases} \Delta \phi = -c\|f\|_\infty & \text{in }B_{2}\setminus \overline{B}_{1}\\ \phi = a & \text{on }\partial B_{2}\\ \phi = 0 & \text{on }\partial B_{1}, \end{cases}

    for a universal c. Then direct calculations show that, for n\ge3,

    \phi(x) = A(|x|^2-1)+B(|x|^{-n+2}-1),

    where

    A = -\frac{c}{2n}\|f\|_\infty, \qquad B = \frac{3}{1-2^{-n+2}} A - \frac{1}{1-2^{-n+2}} a.

    Then the Lemma follows by choosing

    \alpha : = \nabla \phi (e_1) \cdot e_1 = 2 A - (n-2) B.

    The proof in dimension n = 2 is analogous.

    Lemma A.2 (Barrier for supersolutions). Let u satisfy

    \begin{cases} F(D^2u)\le r f & in \;B_{2}(0)\setminus B_{1}(0)\\ u\ge 0 & on \;\partial B_{2}(0)\\ u\ge a \gt 0 & on \;\partial B_{1}(0). \end{cases}

    Then

    u(x)\ge \alpha(x_1+2) + o(|x+2e_1|)\qquad \text{where }\alpha\ge c_1 a - c_2r\|f\|_\infty,

    as x\to -2e_1, whenever r\le \bar r, where the positive constants c_1, c_2 and \bar r only depend on \lambda, \Lambda, n.

    Proof. By comparison and (A.1) we infer that u\ge \phi in B_{2} \setminus \overline{B}_{1}, where \phi solves

    \begin{cases} \mathcal{P}^-_{\lambda/n, \Lambda} \phi = r\|f\|_\infty & \text{in }B_{2}\setminus \overline{B}_{1}\\ \phi = 0 & \text{on }\partial B_{2}\\ \phi = a & \text{on }\partial B_{1}. \end{cases}

    Then direct calculations show that

    \phi(x) = A(|x|^2-4)+B(|x|^{-\gamma}-2^{-\gamma}), \qquad\text{where } \gamma = \frac{\Lambda n (n-1)}{\lambda} -1 \ge 1

    and

    A = \frac{n}{2(\gamma+2)\lambda}r\|f\|_\infty \gt 0, \qquad B = \frac{1}{1-2^{-\gamma}} a + \frac{3}{1-2^{-\gamma}} A \gt 0.

    To check this, one needs to choose r\le\bar r = \bar r(\gamma), in such a way that D^2\phi(x) has exactly one positive eigenvalue, for 1\le|x|\le2. Then the Lemma follows by choosing

    \alpha : = \nabla \phi (-2e_1) \cdot e_1 = -4 A + \gamma 2^{-\gamma-1} B.

    Work partially supported by the INDAM-GNAMPA group. G. Verzini is partially supported by the project ERC Advanced Grant 2013 n. 339958: "Complex Patterns for Strongly Interacting Dynamical Systems - COMPAT"and by the PRIN-2015KB9WPT Grant: "Variational methods, with applications to problems in mathematical physics and geometry".

    The authors declare no conflict of interest.



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