Citation: Thomas Bintsis. Foodborne pathogens[J]. AIMS Microbiology, 2017, 3(3): 529-563. doi: 10.3934/microbiol.2017.3.529
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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:Symn→R 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<λ≤Λ with
λ‖N‖≤F(M+N)−F(M)≤Λ‖N‖ for every M,N∈Symn with N≥0, |
where ‖M‖=max|x|=1|Mx| denotes the (L2,L2)-norm of the matrix M.
Let Ω⊂Rn be a bounded Lipschitz domain and f1,f2∈C(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)={u≤0}ou+ν(x)=G(u−ν,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
infx∈Ω,|ν|=1G(0,x,ν)>0. | (1.2) |
Moreover, u+ν and u−ν 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 x0∈F(u) is regular from the right (resp. left) if there exists a ball B⊂Ω+(u) (resp. B⊂Ω−(u)) such that ¯B∩F(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 x0∈F is regular from the right with touching ball B, then, near x0,
u+(x)≥α⟨x−x0,ν⟩++o(|x−x0|)in B, with α≥0 |
and
u−(x)≤β⟨x−x0,ν⟩−+o(|x−x0|)in Bc, with β≥0, |
with equality along every non-tangential direction, and
α≤G(β,x0,ν(x0)); |
(ⅱ) (subsolution condition) if x0∈F is regular from the left with touching ball B, then, near x0,
u+(x)≤α⟨x−x0,ν⟩++o(|x−x0|)in Bc, with α≥0 |
and
u−(x)≥β⟨x−x0,ν⟩−+o(|x−x0|)in B, with β≥0, |
with equality along every non-tangential direction, and
α≥G(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 w∈C(¯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 x0∈F(w) is regular from the left, with touching ball B, then
w+(x)≤α⟨x−x0,ν⟩++o(|x−x0|)in Bc, with α≥0 |
and
w−(x)≥β⟨x−x0,ν⟩−+o(|x−x0|)in B, with β≥0, |
with
α≤G(β,x0,ν(x0)); |
(c) if x0∈F(w) is not regular from the left then
w(x)=o(|x−x0|). |
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 x0∈F(u_) is regular from the right, with touching ball B, and near x0
u_+(x)≥α⟨x−x0,ν⟩++ω(|x−x0|)|x−x0|in B, with α>0, |
where ω(r)→0 as r→0+, and
u_−(x)≤β⟨x−x0,ν⟩−+o(|x−x0|)in Bc, with β≥0, |
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 {w∈class:w≥u_, w=g on ∂Ω} is not empty, then
u=inf{w:w∈class,w≥u_} |
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 (n−1)-dimensional Hausdorff measure. More precisely, there exists a universal constant r0>0 such that for every r<r0, for every x0∈F(u),
Hn−1(F(u)∩Br(x0))≤rn−1. |
Moreover, the reduced boundary F∗(u) of Ω+(u) has positive density in Hn−1 measure at any point of F(u), i.e. for r<r0, r0 universal
Hn−1(F∗(u)∩Br(x))≥crn−1, |
for every x∈F(u). In particular
Hn−1(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 Hn−1 a.e. point x0∈F(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 E⊂∂B+1∩{x1>0} be any subset such that there exists ˉx∈E with
E⊃∂B+1∩{x1>0}∩Bσ(ˉx). |
Let u be the solution to
{F(D2u)=rinB+1u=δgEon∂B+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)=n∑i,j=1aij(x)uxixj≡Luu, |
with (F=F(M))
aij=∫10∂F∂Mij(tD2u)dt. |
We have
λ|ξ|2≤aij(x)ξiξj≤Λ|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,
c0δx1≤v≤c1δx1. |
Put z(x)=12mina11(x1−x21)r. The function z is positive in B+1 and
Luz=−a11mina11r≤−r. |
Therefore Lu(w+z)≤0 in B+1 and w+z≥0 on ∂B+1. By the maximum principle w≥−c2rx1 in B+1, where c2=a11mina11>0.
Summing up we get, in B+1/2,
u=v+w≥(c0δ−c2r)x1≥c3x1 |
for r small enough, having c3>0.
Lemma 2.2. Let Ω1 be a bounded domain with 0∈∂Ω1 and
B+1:=B1∩{x1>0}⊂Ω1. |
Let u be non-negative and Lipschitz in ¯Ω1∩B2, such that F(D2u)=f in Ω1∩B2 and that u=0 in ∂Ω1∩B2 . Then there exists α≥0 such that
u(x)=αx+1+o(|x|)asx→0, x1>0. |
Proof. Let αk=sup{β:u(x)≥βx1 in B+1/k} for k≥1. Then the sequence {αk}k is increasing and αk≤L 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)}k⊂B+1, with |xk|=rk→0, such that
u(xk)≥αx1,k+δ0rk. |
Since u is Lipschitz, a simple computation implies that
u(x)≥αx1+δ02rk≥αkx1+δ02rkin {x:|x|=rk,|x−xk|≤δ0rk4L}. |
Let
uk(x)=u(rkx)rk−αkx1. |
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)≤rk‖f‖∞. |
Moreover uk(x)≥0 on ∂B+1 and uk≥δ0/2 in Ek={x:x∈∂B+1,x1>0,|x−xk|≤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)−αkx1≥Cx1in 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 B−1:=B1∩{x1<0},
¯B−1∩¯Ω1={0}. |
Let u be non-negative and Lipschitz in ¯Ω1∩B2(0), such that F(D2u)=f in Ω1∩B2(0) and that u=0 in ∂Ω1∩B2(0) . Then there exists α≥0 such that
u(x)=αx+1+o(|x|)asx→0, x∈Ω1. |
Proof. By assumption, we have that
Ω1∩B1⊂B+1. |
Then we can extend u as the zero function on B+1∖Ω1 so that it is a Lipschitz, non-negative solution to
F(D2u)≥−‖f‖∞in B+1. |
Reasoning in a similar way as in Lemma 2.2, we define αk=inf{β:u(x)≤βx1 in B1/k}, k≥1. Then 0≤αk<+∞ (u is Lipschitz), and αk↘α≥0, with u(x)≤αx1+o(|x|) in B+1. Again, let us suppose by contradiction that
u(xk)≤αx1,k−δ0rk. |
where δ0>0 and {xk}k={(x1,k,x2,k,...,xn,k)}k⊂B+1, is such that |xk|=rk→0. As before, such inequality propagates by Lipschitz continuity:
u(x)≤αx1−δ02rk≤αkx1−δ02rkin {x:|x|=rk,|x−xk|≤δ0rk4L}. |
Defining the elliptic, homogeneous operator F∗(M)=−F(−M), we have that the functions
uk(x)=αkx1−u(rkx)rk |
solve
F∗(D2uk(x))≤rk‖f‖∞in B+1, |
with uk(x)≥0 on ∂B+1 and uk≥δ0/2 in Ek={x:x∈∂B+1,x1>0,|x−xk|≤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 0∈∂Ω1 and
eitherB1(e1)⊂Ω1or¯B1(−e1)∩¯Ω1={0}. |
Let u be non-negative and Lipschitz in ¯Ω1∩B2(0), such that F(D2u)=f in Ω1∩B2(0) and that u=0 in ∂Ω1∩B2(0). Then there exists α≥0 such that
u(x)=αx1+o(|x|) |
as x→0 and either x∈B1(e1) or x∈Ω.
Proof. In both cases, we use the smooth change of variable
{y1=x1−ψ(x′)y′=x′, |
where ψ(x′) is smooth, with ψ(x′)=1−√1−|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 u∈C(¯Br(re1)), r≤1, be such that
{F(D2u)=finBr(re1),u≥0,u(0)=0. |
Moreover, assume that u(re1)≥Cr, for some C>0. Then
u(x)≥αx1+o(|x|),whereα≥c1u(re1)r−c2r‖f‖∞, |
as x→0, for r≤ˉr, where c1,c2 and ˉr only depend on λ,Λ,n.
Proof. Let
ur(x)=u(r(e1+x))r,x∈B1(0). |
Then
{F(D2ur)=rfin B1,ur≥0,ur(−e1)=0. |
By Harnack's inequality [5,Theorem 4.3] we have that
inf∂B1/2ur≥c(ur(0)−r‖f‖∞)=: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 α≥c1a−c2r‖f‖∞=c′1ur(0)−c′2r‖f‖∞, |
as x→−e1, 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 Δui≥−1, u1⋅u2=0 in B1. Then there exist universal constants C0 and r0, such that the functional
Φ(r):=1r4∫Br|∇u1|2rn−2∫Br|∇u2|2rn−2 |
satisfies, for 0<r≤r0,
Φ(r)≤C0(1+‖u1‖2L2(B1)+‖u2‖2L2(B1))2. |
Lemma 3.2. Let w∈class. There exists ˜w∈class such that
1. F(D2˜w)=f1 in Ω+(˜w),
2. ˜w≤w, ˜w−=w−, and
3. ˜w≥u_ in Ω.
Proof. Let w∈class and Ω+=Ω+(w). We define
V:={v∈C(¯Ω+):F(D2v)≥f1χ{v>0} in Ω+, v≥0 in Ω+, v=w on ∂Ω+} |
and
˜w(x):={sup{v(x):v∈V}x∈Ω+w(x)elsewhere. |
Since u_+∈V we obtain that V is not empty and that u_≤˜w≤w. Moreover ˜w is a solution of the obstacle problem (see [13])
{F(D2˜w)=f1in {˜w>0}˜w≥0in 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 ˜w∈class, we need to show that the free boundary conditions in Definition 1.2 hold true. Let x0∈F(˜w): if x0∈F(w) too, then the free boundary condition follows from the fact that ˜w≤w; otherwise, x0∈Ω+ is an interior zero of ˜w, and the free boundary condition follows by the C1,1 regularity of ˜w.
Lemma 3.3. Let w∈class with F(D2w)=f1 in Ω+(w), and let x0∈F(w) be regular from the right. Then u admits developments
w+(x)=α⟨x−x0,ν⟩++o(|x−x0|),w−(x)≥β⟨x−x0,ν⟩−+o(|x−x0|), |
with 0≤α≤G(β,x0,ν(x0)), and
αβ≤C0(1+‖u1‖22+‖u2‖22). |
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±≥−c‖f‖∞. |
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)||x−y|≤LD |
for every x,y∈D, x≠y, and for every w∈class with F(D2w)=f1 in Ω+(w).
Proof. Let x0∈Ω+(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)r≤M, |
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)≥αM⟨x−x0,ν⟩++o(|x−x0|), |
where αM=c1M−c2r‖f‖∞>0 for M sufficiently large. Then x0 is regular from the right and Lemma 3.3 applies, with αM≤α≤G(β,x0,ν(x0)), providing
αMG−1(αM)≤C0(1+‖u1‖22+‖u2‖22), |
where G−1(α):=infx,νG−1(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,w2∈class 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 w∈class 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, v≥0 in Ω1, v=w on ∂Ω1∖∂B, 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},w1≥0 in Ω1. | (4.1) |
On the other hand, working on B, let
V2:={v:F(D2v)≥f2χ{v>0} in B, v≥0 in B, v=w− on ∂B} |
(which is non empty, as w−∈V2). Again,
w2:=supV2 |
solves the obstacle problem
F(D2w2)≥f1 in {w2>0},w2≥0 in B. | (4.2) |
Under the above notation, the double replacement ˜w of w, relative to B, is defined as
˜w:={w1in Ω1−w2in Bwotherwise. |
Lemma 4.2. Let w∈class 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_≤˜w≤w in Ω;
2. ˜w<0 and F(D2˜w)=f2 in Bεh(x0), with
|∇˜w|≤Cε+εC‖f2‖∞inBεh/2(x0); |
3. ˜w∈class.
Proof. The inequality w1≤w in Ω1 follows by the maximum principle, while w2≥−w in B because w−∈V2. 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 u∗∈V1, so that w1≥u_; finally, by the maximum principle in {w2>0}, also −w2≥u_, 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 ˉx∈F(˜w), three possibilities may occur. If ˉx∈F(w) then, since ˜w≤w, 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 ˉx∈∂B∩Ω+(w). By Proposition 3.4, let us denote with L the Lipschitz constant of w in Bdist(x0,∂Ω)/2(x0). Then
˜w≤w+≤Lεhin B2εh(x0). |
Defining
˜wε(x):=˜w(x0+εhx)εh, |
we have that
{F(D2˜w+ε)=εhf1in (B2∖¯B1)∩Ω+(wε)˜w+ε≤Lon ∂B2˜w+ε≤0on ∂B1∩∂Ω+(wε). |
Then Lemma A.1 applies, yielding
˜w+ε≤α⟨x−ˉxε,ν(ˉxε)⟩++o(|x−ˉxε|), |
where
α≤c1L+c2εh‖f1‖∞,ˉxε:=ˉx−x0εh, |
for universal c1, c2. Going back to ˜w we obtain
˜w+≤α⟨x−ˉx,ν⟩++o(|x−ˉx|),α≤ˉL | (4.3) |
where ν=(ˉx−x0)/|ˉx−x0|.
On the other hand, we can apply Lemma 2.5 to (−w2)ε, obtaining
˜w−ε=−(w2)ε≥β⟨x−ˉxε,ν(ˉxε)⟩++o(|x−ˉxε|), |
where
β≥c′1ε−c′2εh‖f1‖∞, |
for universal c′1, c′2, and thus
˜w−≥β⟨x−ˉx,ν⟩−+o(|x−ˉx|),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, ˜wk≥u_, 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˜w−k)=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. ˜wk↘u uniformly on Bhεilon/4
Proof. Let u(x0)=−h<0, {wk}⊂class be such that wk↘u 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 ˜wk↘u uniformly on Bhεilon/4. By equicontinuity, ˜wk→˜w in Bεilonh/2(x0), and suppose by contradiction that ˜w(x1)>u(x1) for some x1∈Bεilonh/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 ¯D⊂Ω there exists {wk}k⊂class such that wk↘u uniformly in ¯D. Furthermore, if ¯D⊂Ω−(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 wk↘u uniformly in ¯B2ε(x0)⊂Ω−(u), and let
wεk(x)=wk(x0+εx)ε↘uεin B2. |
Let ϕ be such that
{Δϕ=−cε‖f‖∞in 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 uε≤0 in ¯B2, for k sufficiently large wk≤a/2 in ¯B2. Let us define
ˉwεk={min{wεk,ϕ}in ¯B2,wεotherwise. |
Then, by Lemma 4.1, the function
ˉwk(x)=εˉwεk(x−x0ε) |
satisfies ˉwk∈class, ˉwk≤0 in ¯Bε(x0) and ˉwk↘u 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 x0∈F(u) and let A be a connected component of Ω+(u)∩(Br(x0)∖¯Br/2(x0)) satisfying
¯A∩∂Br/2(x0)≠∅,¯A∩∂Br(x0)≠∅, |
for r≤r0 universal. Then
supAu≥Cr. |
Moreover
|A∩Br(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 L1loc.
The proof of the above result will follow by the two following lemmas.
Lemma 5.3. Let u be a Lipschitz function in ¯Ω∩B1(0), with 0∈∂Ω, satisfying
{F(D2u)=finΩ∩B1u=0on∂Ω∩B1. |
If there exists c>0 such that
u(x)≥cdist(x,∂Ω)foreveryx∈Ω∩B1/2 | (5.1) |
then there exists a constant C>0 such that
supBr(0)u≥Cr, |
for all r≤r0 universal.
Proof. Let x0∈Ω∩B1, ε=dist(x0,∂Ω), and let L denote the Lipschitz constant of u. Then
cε≤u(x0)≤Lε. |
We will show that, for δ>0 to be fixed, there exists x1∈Bε(x0) such that
u(x1)≥(1+δ)u(x0). | (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+δ)u(x0)−u(x)>0in Bε(x0)satisfies F∗(D2v)=−f. |
Let r(L)=1−c/(4L); using the Harnack inequality we have that there exists C(L) such that
v≤C(L)(δu(x0)+ε2‖f‖∞)≤12u(x0)in ¯Br(L)ε(x0), |
provided both δ and ε are sufficiently small (depending on c, L and ‖f‖∞). In terms of u, the previous inequality writes as
u≥cε2in ¯Br(L)ε(x0). |
On the other hand, there exists y0∈∂Br(L)ε(x0) such that dist(y0,∂Ω)=(1−r(L))ε and hence
min¯Br(L)ε(x0)u≤u(y0)≤Ldist(y0,∂Ω)=cε4. |
This is a contradiction, therefore (5.2) holds true.
Lemma 5.4. There exist universal constants ˉr, ˉC such that
u(x0)≥ˉCdist(x0,F(u))foreveryx0∈{x∈Ω+(u):dist(x,F(u))≤ˉr}. |
Proof. Let x0∈{x∈Ω+(u):dist(x,F(u))≤ˉr}, with ˉr universal to be specified later, and let r:=dist(x0,F(u)). We distinguish two cases.
First let us assume that
dist(x0,Ω+(u_))≤r2. |
In this case, for any x∈F(u_) we define
ρ(x):=max{r>0:for some z, x∈∂Br(z) and Br(z)⊂Ω+(u_)}. |
Notice that ρ(x)>0 for every x, since any point in F(u_) is regular from the right by assumption. Thus, recalling that u_+ has linear growth bounded below by infx,νG(0,x,ν), and noticing that B3r/4(x0)∩Ω+(u_) contains a ball of radius comparable with r (at least for a suitable choice of ˉr):
supB3r/4(x0)u+≥supB3r/4(x0)u_+≥ˉCr, |
where ˉC only depends on u_.
On the other hand, in case
dist(x0,Ω+(u_))≥r2, |
we have u_≤0 in Br/2(x0). By Corollary 4.4 we can find {wk}k⊂class converging uniformly to u on some D⊃Br(x0). By scaling
ur(x)=u(x0+rx)r,wrk(x)=wk(x0+rx)r, |
we need to find ˉC universal such that ur(0)≥ˉC. Let us assume by contradiction that
ur(0)<ˉC. |
Then by Harnack inequality
ur≤C(ˉC+r‖f1‖∞)in B1/2 |
and, for k sufficiently large,
wrk≤C′(ˉC+r‖f1‖∞)in B1/2. |
Now, reasoning as in the proof of Corollary 4.4, let ϕ be such that
{Δϕ=−cr‖f‖∞in B1/2∖¯B1/4ϕ=aon ∂B1/2ϕ=0on ∂B1/4, |
with a and r positive and sufficiently small so that ∇ϕ(e1/4)⋅e1<infx,νG(0,x,ν), in such a way that ϕ, extended to zero in B1/4, is a supersolution in B1/2. Then, choosing ˉC<(a−r‖f1‖∞)/C′ we obtain that wrk<ϕ on ∂B1/2 and then the functions
ˉwrk={0in ¯B1/4,min{wrk,ϕ}in ¯B1/2∖B1/4,wrkotherwise |
are continuous, while
ˉwk(x)=rˉwrk(x−x0r) |
satisfy ˉwk∈class, ˉwk≡0 in ¯Br/4(x0). This is in contradiction with the fact that u(x0)>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 x0∈F(u), with coefficients α and β, then α≤G(β,x0,νx0). To do that, we need to distinguish the two cases β>0 and β=0.
Lemma 6.1. Let x0∈F(u), and
u+(x)=α⟨x−x0,ν⟩++o(|x−x0|),u−(x)=β⟨x−x0,ν⟩−+o(|x−x0|), |
with
β>0. |
Then α≤G(β,x0,νx0).
Proof. Since β>0, then F(u) is tangent at x0 to the hyperplane
π:⟨x−x0,ν⟩=0 |
in the following sense: for any point x∈F(u), dist(x,F(u))=o(|x−x0|). Otherwise we get a contradiction to the asymptotic development of u.
Let {wk}k⊂class be uniformly decreasing to u, as in Corollary 4.4. By the non-degeneracy of u+ we have that, for k large, wk can not remain strictly positive near x0. Let dk=dH(F(wk),F(u)) be the Hausdorff distance between the two free boundaries. In the ball B2√dk(x0), F(u) is contained in a strip parallel to π of width o(√dk) and, since dk→0, F(wk) is contained in a strip Sk of width dk+o(√dk)=o(√dk).
Consider now the points xk=x0−√dkν and let Bk=Brk(xk) be the largest ball contained in Ω−(wk) with touching point zk∈F(wk). Then zk∈Sk and, since wk≥u, from the asymptotic developments of wk and u we have
β√dk+o(√dk)=u−(xk)≥w−(xk)=βkrk+o(rk), |
since
√dk+o(√dk)≤rk≤√dk. |
Passing to the limit we infer
limsupβk≤β. |
Reasoning in the same way on the other side th the points yk=x0+√dk (and the same zk, which are regular from the left), we get
α≤liminfαk. |
From αk≤G(βk,zk,νk), where νk=(xk−zk)/|xk−zk|, we get α≤G(β,x0,νx0).
To treat the case β=0 we need the following preliminary lemma.
Lemma 6.2. Let v≥0 continous in B1(x0) be such that Δv≥−M. Let
Ψr(x0,v)=1r2∫Br(x0)|∇v|2|x−x0|n−2dx. |
Then, for r small,
Ψr(x0,v)≤c(n){supB2r(x0)(vr)2+MsupB2r(x0)v}. | (6.1) |
Proof. We may assume x0=0 and write Ψr(0,v)=Ψr(v). Rescale setting vr(x)=v(rx)/r; we have Δvr≥−rM and
Ψr(v)=Ψ1(vr). |
Let η∈C∞0(B2), η=1 in B1. Since 2|∇vr|2≤2rMvr+Δv2r, we have:
Ψ1(vr)≤C∫B2η|∇vr|2|x|n−2≤C∫B2η2Mvr+Δv2r|x|n−2=C∫B2[2Mvr|x|n−2+v2rΔ(η|x|n−2)], |
so that
Ψr(v)=Ψ1(vr)≤c(n)(|vr|2L∞(B2)+rM|vr|L∞(B2))=c(n){supB2r(vr)2+MsupB2rv}, |
which is (6.1).
Lemma 6.3. Let x0∈F(u), and
u+(x)=α⟨x−x0,ν⟩++o(|x−x0|),u−(x)=o(|x−x0|). |
Then α≤G(β,x0,νx0).
Proof. As before, let {wk}k⊂class be uniformly decreasing to u, with wk that is not strictly positive near x0, 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
Bm,k=Bλm,k(x0+1mν) |
the largest ball centered at x0+ν/m contained in Ω+(wk), touching F(wk) at xm,k where νm,k is the unit inward normal of F(wk) at xm,k. Then up to proper subsequences we deduce that
λm,k→λm,xm,k→xm,νm,k→νm |
and Bλm(x0+ν/m) touches F(u) at xm, with unit inward normal νm. From the behavior of u+, we get that
|xm−x0|=o(1m), |
1m+o(1m)≤λm≤1m |
and
|νm−ν|=o(1). |
Now since wk∈F, near xm,k in Bm,k:
w+k≤αm,k⟨x−xm,k,νm,k⟩++o(|x−xm,k|) |
and in Ω∖Bm,k
w−k≥βm,k⟨x−xm,k,νm,k⟩−+o(|x−xm,k|) |
with
0≤αm,k≤G(βm,k,xm,k,νm,k), |
(by Lemma 2.5 the touching occurs at a regular point, for m,k large.) We know that
w+k≥u+≥α⟨x−x0,ν⟩++o(|x−x0|), |
hence
α_m=lim infk→∞αm,k≥α−εilonm |
and εilonm→0, as m→∞. We have to show that
β_=lim infm,k→+∞βm,k=0. |
We assume by contradiction that ˉβ>0. Acting as in [9,Lemma 6.3] we obtain, for r small,
(1+ω(r))Φr(xm,k,wk)+Cω(r)≥cnα2m,kβ2m,k, | (6.2) |
where
Φr(xm,k,wk)=Ψr(xm,k,w+k)Ψr(xm,k,w−k). |
By concavity we have that Δw±k≥−M where M=cmin(‖f1‖∞,‖f2‖∞). Lemma 6.2 implies
cnα2m,kβ2m,k≤(1+ω(r))Ψr(xm,k,w+k)Ψr(xm,k,w−k)+Cω(r)≤c2(n)(1+ω(r)){supB2r(xm,k)(w+kr)2+MsupB2r(xm,k)w+k}××{supB2r(xm,k)(w−kr)2+MsupB2r(xm,k)w−k}+Cω(r)≤C1(n,M,L)){supB2r(xm,k)(w−kr)2+MsupB2r(xm,k)w−k}+Cω(r), |
where L is the uniform Lipschitz constant of {w+k}k (recall Lemma 3.4). Taking the lim inf as m,k→∞ and using the uniform convergence of wk to u we infer
0<cnα2ˉβ2≤C1(n,M,L){supB2r(x0)(u−r)2+MsupB2r(x0)u−}+Cω(r). |
Recalling that, by assumption, u−(x)=o(|x−x0|) as x→x0, we have
supB2r(0)(u−r)2=o(1)as r→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 x0∈F(u) is a regular point from the left with touching ball B⊂Ω−(u), then near to x0
u−(x)=β⟨x−x0,ν⟩−+o(|x−x0|),β≥0, |
in B, and
u+(x)=α⟨x−x0,ν⟩++o(|x−x0|),α≥0 |
in Ω∖B. Indeed, even if β=0, then Ω+(u) and Ω−(u) are tangent to {⟨x−x0,ν⟩=0} at x0 since u+ is non-degenerate. Thus u has a full asymptotic development as in the next lemma. We want to show that α≥G(β,x0,ν). We follow closely [3] and [9].
Lemma 7.1. Assume that near x0∈F(u),
u(x)=α⟨x−x0,ν⟩+−β⟨x−x0,ν⟩−+o(|x−x0|), |
with α>0, β≥0. Then
α≥G(β,x0,ν). |
Proof. Assume by contradiction that α<G(β,x0,ν). We construct a supersolution w∈class which is strictly smaller than u at some point, contradicting the minimality of u. Let u0 be the two-plane solution, i.e.
u0(x):=limr→0u(x0+rx)r=α⟨x,ν⟩+−β⟨x,ν⟩−. |
Suppose that α≤G(β,x0,ν)−δ0 with δ0>0. Fix ζ=ζ(δ0), to be chosen later. By Corollary 4.4, we can find wk∈F↘u locally uniformly and, for r small, k large, the rescaling wk,r satisfies the following conditions:
if β>0, then
wk,r(x)≤u0+ζmin{α,β} on ∂B1; |
if β=0, then
wk,r(x)≤u0+αζ on ∂B1 |
and
wk,r(x)≤0,in{⟨x,ν⟩<−ζ}∩¯B1. |
In particular,
wk,r(x)≤u0(x+ζν)on∂B1. |
If β>0, let v satisfy
{F(D2v)=rfr1,in {⟨x,ν⟩>−ζ+εilonϕ(x)}F(D2v−)=rfr2,in {⟨x,ν⟩<−ζ+εilonϕ(x)}v(x)=0,on {⟨x,ν⟩=−ζ+εilonϕ(x)}v(x)=u0(x+ζν),on ∂B1, | (7.1) |
where ϕ≥0 is a cut-off function, ϕ≡0 outside B1/2, ϕ≡1 inside B1/4.
For β=0, replace the second equation with v=0.
Along the new free boundary, F(v)={⟨x,ν⟩=−ζ+εilonϕ(x)} we have the following estimates:
|v+ν−α|≤c(ε+ζ)+Cr,|v−ν−β|≤c(ε+ζ)+Cr, |
with c,C universal.
Indeed,
v+−α⟨x,ν⟩+ |
is a solution of
F(D2(v−α⟨x,ν⟩+))=rfr1. |
Thus, by standard C1,γ regularity estimates (see [16,Theorem 1.1])
|v+ν−α|≤C(‖v−α⟨x,ν⟩+‖∞+[−ζ+εilonϕ]1,γ+r‖f1‖∞), |
which gives the desired bound. Similarly, one gets the bound for v−ν.
Hence, since α≤G(β,x0,ν(x0))−δ0, say for ε=2ζ and ζ,r small depending on δ0
v+ν<G(v−ν,x0,ν), |
and the function,
ˉwk={min{wk,λv(x−x0λ)} in Bλ(x0),wkin Ω∖Bλ(x0), |
is still in class. However, the set
{⟨x,ν⟩≤−ζ+εilonϕ} |
contains a neighborhood of the origin, hence rescaling back x0∈Ω−(ˉwk). We get a contradiction since x0∈F(u) and Ω+(u)⊆Ω+(ˉwk).
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 Nε(A) an ε-neighborhood of the set A. The following lemma provides a control of the Hn−1 measure of F(u) and implies that Ω+(u) is a set of finite perimeter.
Lemma 8.1. Let u be our Perron solution. Let x0∈F(u)∩B1. There exists a positive universal δ0<1 such that, for every 0<ε<δ≤δ0, the following quantities are comparable:
1. 1ε|{0<u<ε}∩Bδ(x0)|,
2. 1ε|Nε(F(u))∩Bδ(x0)|,
3. Nεn−1, where N is the number of any family of balls of radius ε, with finite overlapping, covering F(u)∩Bδ(x0),
4. Hn−1(F(u)∩Bδ(x0)).
Proof. From [3], it is sufficient to prove the following two equivalences:
c1εn≤∫Bε(x0)|∇u|2≤C1εn | (8.1) |
and
c3εδn−1≤∫{0<u<ε}∩Bδ(x0)|∇u|2≤C2εδn−1 | (8.2) |
with universal constants c1,c2,C1,C2.
Since F(D2u)=infαLαu where Lα is a uniformly elliptic operator with constant coefficients and ellipticity constant λ,Λ, we have Lαu+≥f1 in Ω+(u). Fix α=α0 and set
Lα0=L=n∑i,j=1aij∂ij, A=(aij). |
The upper bound in (8.1) follows by the Lipschitz continuity of u. The lower bound follows from supBε(x0)u+≥cε, c universal, infBε(x0)u+=0, the Lipschitz continuity of u, and the Poincaré inequality (see [1,Lemma 1.15]).
To prove (8.2), rescale by setting
uδ(x)=u(x0+δx)δ, fδ1(x)=f1(x0+δx)x∈B1=B1(0). |
Then Luδ≥δ fδ1 in Ω+(uδ)∩B1. For 0<ε<δ, let
uδ,s,ε=us,ε:=max{s/δ,min{uδ,ε/δ}}. |
We have:
−δ∫B1fδ1uε,s=−∫B1uε,sLu+δ=∫B1⟨A∇u+δ,∇u+ε,s⟩dx−∫∂B1⟨A∇u+δ,ν⟩uε,sdHn−1=∫B1∩{0<s/δ<uδ<ε/δ}⟨A∇uδ,∇uδ⟩dx−∫∂B1⟨A∇u+δ,ν⟩uε,sdHn−1 |
since ∇uε,s=∇uδ⋅χ{s/δ<uδ<ε/δ}.
By uniform ellipticity, since u+ is Lipschitz and f1 is bounded, we get (δ<1)
∫B1∩{0<s/δ<uδ<ε/δ}|∇uδ|2dx≤Cεδ, |
with C universal. Letting s→0 and rescaling back, we obtain the upper bound in (8.2).
For the lower bound, let V be the solution to
{LV=−χBσ|Bσ|,inB1V=0,on∂B1 | (8.3) |
with σ to be chosen later. By standard estimates, see for example [12], V≤Cσ2−n and −⟨A∇V,ν⟩∼C∗ on ∂B1, with C∗ independent of σ. By Green's formula
∫B1(LV)u+δuε,0ε−(Lu+δuε,0ε)V=∫∂B1u+δuε,0ε⟨A∇V,ν⟩dHn−1 | (8.4) |
since V=0 on ∂B1. We estimate
δ|∫B1(LV)u+δuεεdx|=δ|Bσ||∫Bσu+δuεεdx|≤ˉCσ, | (8.5) |
since u is Lipschitz and 0≤uε,0≤ε/δ. From (8.4) and (8.5) and the fact that ⟨Aδ∇V,ν⟩∼−C∗ on ∂B1 we deduce that
δ∫B1(Lu+δuε,0ε)Vdx≥−ˉCσ−δ∫∂B1u+δuε,0ε⟨A∇V,ν⟩dHn−1≥−ˉCσ+C∗δ∫∂B1u+δuε,0εdHn−1. |
Thus using that u+ is non-degenerate and choosing σ small enough (universal) we get that (δ>ε)
δ∫B1(Lu+δuε,0ε)Vdx≥˜C. | (8.6) |
On the other hand in {0<u+δ<ε/δ}∩B1,
Lu+δuε,0=2δuεfδ1+⟨A∇uδ,∇uδ⟩. | (8.7) |
Combining (8.6), (8.7) and using the ellipticity of A we get that
2δ2ε∫B1uεfδ1V+δΛε∫B1|∇uδ|2V≥ˉC. |
From the estimate on V we obtain that for δ small enough
δε∫B1|∇uδ|2≥C |
for some C universal. Rescaling, we obtain the desired lower bound.
Lemma 8.1 implies that Ω+(u)∩Br(x), x∈F(u), is a set of finite perimeter. Next we show that in fact this perimeter is of order rn−1.
Theorem 8.2. Let u be our Perron solution. Then, the reduced boundary of Ω+(u) has positive density in Hn−1-measure at any point of F(u), i.e. for r<r0, r0 universal,
Hn−1(F∗(u)∩Br(x))≥crn−1 |
for every x∈F(u).
Proof. The proof follows the lines of Corollary 4 in [3] and Theorem 8.2 in [9]. Let wk∈class, wk↘u in ¯B1 and L as in Lemma 8.1. Then Ω+(u)⊂⊂Ω+(wk) and Lwk≥F(D2wk)=f1 in Ω+(u). Let x0∈F(u). We rescale by setting
ur(x)=u(x0+rx)r,wk,r=wk(x0+rx)rx∈B1. |
Let V be the solution to (8.3). Since ∇wk,r is a continuous vector field in ¯Ω+r(ur)∩B1, we can use it to test for perimeter. We get
∫B1∩Ω+r(ur)(Vrfr1−wk,rLV)≤∫B1∩Ω+r(ur)(VLwkr−wk,rLV)=∫F∗(ur)∩B1(V⟨A∇wk,r,ν⟩−wkr⟨A∇V,ν⟩)dHn−1−∫∂B1∩Ω+r(ur)wkr⟨A∇V,ν⟩dHn−1. | (8.8) |
Using the estimates for V and the fact that the wk are uniformly Lipschitz, we get that
|∫F∗(ur)∩B1V⟨A∇wk,r,ν⟩dHn−1|≤C(σ)Hn−1(F∗(ur)∩B1). | (8.9) |
As in [3] we have, as k→∞,
∫F∗(ur)∩B1wk,r⟨A∇V,ν⟩dHn−1→0, |
∫∂B1∩Ω+r(ur)wkr⟨A∇V,ν⟩dHn−1→∫∂B1u+r⟨A∇V,ν⟩dHn−1 |
and
−∫B1∩Ω+r(ur)wk,rLV→−∫Bσu+r. |
Passing to the limit in (8.8) and using all of the above we get
|r∫B1∩Ω+(ur)Vfr1+−∫Bσu+r+∫∂B1u+r⟨A∇V,ν⟩dHn−1|≤C(σ)Hn−1(F∗(ur)∩B1). | (8.10) |
Since u is Lipschitz and non-degenerate, for σ small
1|Bσ|∫Bσu+r≤ˉCσ, |
and using the estimate for ⟨A∇V,ν⟩
−∫∂B1u+r⟨A∇V,ν⟩dHn−1≥ˉc>0. |
Also, since fr1 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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