Volume 2007, Article ID 57049,21pages doi:10.1155/2007/57049
Research Article
Subsolutions of Elliptic Operators in Divergence Form and
Application to Two-Phase Free Boundary Problems
Fausto Ferrari and Sandro Salsa
Received 29 May 2006; Accepted 10 September 2006
Recommended by Jos´e Miguel Urbano
LetLbe a divergence form operator with Lipschitz continuous coefficients in a domain
Ω, and letube a continuous weak solution ofLu=0 in{u=0}. In this paper, we show that ifφsatisfies a suitable differential inequality, thenvφ(x)=supBφ(x)(x)uis a subsolution
ofLu=0 away from its zero set. We apply this result to proveC1,γregularity of Lipschitz free boundaries in two-phase problems.
Copyright © 2007 F. Ferrari and S. Salsa. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction and main results
In the study of the regularity of two-phase elliptic and parabolic problems, a key role is played by certain continuous perturbations of the solution, constructed as supremum of the solution itself over balls of variable radius. The crucial fact is that if the radius satisfies a suitable differential inequality, modulus a small correcting term, the perturbations turn out to be subsolutions of the problem, suitable for comparison purposes.
This kind of subsolutions have been introduced for the first time by Caffarelli in the classical paper [1] in order to prove that, in a general class of two-phase problems for the laplacian, Lipschitz free boundaries are indeedC1,α.
This result has been subsequently extended to more general operators: Feldman [2] considers linear anisotropic operators with constant coefficients, Wang [3] a class of con-cave fully nonlinear operators of the typeF(D2u), and again Feldman [4] fully nonlinear operators, not necessary concave, of the typeF(D2u,Du). In [5], Cerutti et al. consider variable coefficients operators in nondivergence form and Ferrari [6] a class of fully non-linear operatorsF(D2u,x), H¨older continuous in the space variable.
precisely the subject of this paper. Once again, the key point is the construction of the previously mentioned family of subsolutions. Unlike the case of nondivergence or fully nonlinear operators, in the case of divergence form operators, the construction turns out to be rather delicate due to the fact that in this case not only the quadratic part of a function controls in average the action of the operator but also the linear part has an equivalent influence. Here we require Lipschitz continuous coefficients.
To state our first result we introduce the classᏸ(λ,Λ,ω) of elliptic operators
L=divA(x)∇ (1.1)
defined in a domainΩ⊂Rn, with symmetric and uniformly elliptic matrix, that is,
A(x)=A(x), λI≤As(x)≤ΛI (1.2)
and modulus of continuity of the coefficients given by
ω(r)= sup
|x−y|≤r
A(x)−A(y). (1.3)
Theorem1.1. Letube a continuous function inΩ. Assume that in{u >0}uis aC2-weak
solution ofLu=0,L∈ᏸ(λ,Λ,ω),ω(r)≤c0r. Letφbe a positive C2-function such that 0< φmin≤φ≤φmaxand
vφ(x)= sup Bφ(x)(x)
u=sup
|ν|=1
ux+φ(x)ν (1.4)
is well defined inΩ. There exist positive constantsμ0=μ0(n,λ,Λ)andC=C(n,λ,Λ), such
that, if|∇φ| ≤μ0,ω0=ω(φmax), and
φLφ≥C∇φ(x)2+ω2 0
, (1.5)
thenvis a weak subsolution ofLu=0in{v >0}.
We now introduce the class of free boundary problems we are going to study and the appropriate notion of weak solution.
LetBR=BR(0) be the ball of radiusRinRn−1. InᏯR=BR(0)×(−R,R) we are given a continuousH1
locfunctionusatisfying the following. (i)
Lu=divA(x)∇u=0 (1.6)
in Ω+(u)= {x∈Ꮿ
R:u(x)>0}, and in Ω−(u)= {x∈ᏯR:u(x)≤0}0, in the weak sense.
We callF(u)≡∂Ω+(u)∩Ꮿ
Rthefree boundary. We say that a pointx0∈F(u) isregular from the right (left) if there exists a ballB:
B⊂Ω+(u) ⊂Ω−(u), resp., B∩F(u)=x0
(ii) AlongF(u) the following conditions hold:
(a) ifx0∈F(u) is regular from the right, then, nearx0,
u+(x)≥α x−x 0,ν
+
−β x−x0,ν
−+ox−x
0, (1.8)
for someα >0,β≥0 with equality along every nontangential domain in both cases, and
α≤G(β); (1.9)
(b) ifx0∈F(u) is regular from the left, then, nearx0,
u+(x)≤α x−x 0,ν
+
−β x−x0,ν
−
+ox−x0, (1.10)
for someα≥0,β >0 with equality along every nontangential domain in both cases, and
α≥G(β). (1.11)
The conditions (a) and (b), whereνdenotes the unit normal to∂Batx0, towards the positive phase, express the free boundary relationu+
ν =G(u−ν) in a weak sense; accord-ingly, we callua weak solution off.b.p.
Via an approximation argument it is possible to show thatTheorem 1.1holds for the positive and negative parts of a solution of our f.b.p.
Here are our main results concerning the regularity of Lipschitz free boundaries.
Theorem1.2. Letube a weak solution to f.b.p. inᏯR=BR×(−R,R).
Suppose that0∈F(u)and that
(i)L∈ᏸ(λ,Λ,ω);
(ii)Ω+(u)= {(x,x
n) :xn> f(x)} where f is a Lipschitz continuous function with
Lip(f)≤l;
(iii)G=G(z)is continuous, strictly increasing and for someN >0,z−NG(z)is decreasing
in(0, +∞).
Then, onBR/2, f is aC1,γfunction withγ=γ(n,l,N,λ,Λ,ω).
By using of the monotonicity formula in [7] we can prove the following.
Corollary1.3. In f.b.p. let
Lu=divA(x,u)∇u, (1.12)
whereLis a uniformly elliptic divergence form operator. Assume (ii) and (iii) inTheorem 1.2
hold and replace (i) with the assumption thatAis Lipschitz continuous with respect toxand
u. Then the same conclusion holds.
(iii)G=G(z,ν,x) is continuous strictly increasing inzand, for someN >0 indepen-dent ofνandx,z−NG(z,ν,x) is decreasing in (0,∞);
(iii ) logGis Lipschitz continuous with respect toν,x, uniformly with respect to its first argumentz∈[0,∞).
The proof ofTheorem 1.2goes along well-known guidelines and consists in the fol-lowing three steps: to improve the Lipschitz constant of the level sets ofufar fromF(u), to carry this interior gain to the free boundary, to rescale and iterate the first two steps. This procedure gives a geometric decay of the Lipschitz constant ofF(u) in dyadic balls that corresponds to aC1,γregularity ofF(u) for a suitableγ.
The first step follows with some modifications [5, Sections 2 and 3] and everything works with H¨older continuous coefficients. We will describe the relevant differences in
Section 2.
The second step is the crucial one. At difference with [5] we use the particular struc-ture of divergence and the fact that weak sub- (super-) solutions of operators in diver-gence form with H¨older coefficients can be characterized pointwise, through lower (su-per) mean properties with respect to a base of regular neighborhoods of a point, involving theL-harmonic measure.Section 3contains the proof of the main result,Theorem 1.1, and some consequences.
InSection 4the above results are applied to our free boundary problem, preparing the
necessary tools for the final iteration.
The third step can be carried exactly as in [5, Sections 6 and 7], since here the particular form of the operator does not play any role anymore. Actually the linear modulus of continuity allows some simplifications.
2. Monotonicity properties of weak solutions
In this section we assume thatω(r)≤c0ra, 0< a≤1. Letu∈Hloc1 (Ω) be a weak solution ofLu=0 inΩ, that is,
Ω A(x)∇u(x),∇ϕ(x)
dx=0, (2.1)
for every test functionϕsupported inΩ. IfL∈ᏸ(λ,Λ,ω),u∈C1,a(Ω).
In this section we prove that if the domainΩis Lipschitz anduvanishes on a relatively open portionF⊂∂Ω, then, nearF, the level sets ofuare uniformly Lipschitz surfaces.
Precisely, we consider domains of the form
Ts=
x,xn
∈R:|x|< s, f(x)< xn<2ls
, (2.2)
where f is a Lipschitz function with constantl.
Theorem2.1. Letube a positive solution toLu=0inT4, vanishing onF= {xn= f(x)} ∩
∂T4. Then, there existsηsuch that in
ᏺη(F)=
f(x)< xn< f(x) +η
uis increasing along the directionsτbelonging to the coneΓ(en,θ), with axisenand opening
θ=(1/2) cot−1l. Moreover, inᏺ η(F),
c−1u(x)
dx ≤Dnu(x)≤c
u(x)
dx , (2.4)
wheredx=dist(x,F).
Proof ofTheorem 2.1. Letzbe the solution of the Dirichlet problem
divA(0)∇z(x)=0, T2, z=g, ∂T2,
(2.5)
wheregis a smooth function vanishing onᏲand equal to 1 at pointsxwithdx>1/10. Then, see [1],Dnz >0 inQρ, withρ=ρ(n,l). By rescaling the problem (if necessary), we may assumeρ=3/2. Sincez(en)≥c >0, by Harnack inequality we have that, ify∈T1, dy≥η0,
z(y)∼cη0
, Dnz(y)∼
z(y)
dy ∼
cη0
. (2.6)
Clearlyz,u∈C0,a(T
2).
Lemma2.2. Forr >0, letwrbe theC1,a(T2)weak solution to
divA(rx)∇wr(x)
=0, T2, wr=z, ∂T2.
(2.7)
Then, givenη0>0, there existsr0=r0(η0), such that ifr≤r0,
Dnwr(y)≥0 (2.8)
for everyy∈T1, withdy≥η0.
Proof. Let
divA(rx)∇wr(x)− ∇z(x)
=divA(rx)−A(0)∇z(x). (2.9)
For everyσ >0, let
Qσ2=
x∈T2, distx,∂T2> σ. (2.10)
Notice thathr=wr−z∈C0,a(T2), and moreoverhr∈C1,a(Q2σ). Notice that ((A(rx)− A(0))∇z(x))i∈L∞(Q2σ), and from standard estimates we have
sup Qσ
2
wr−z≤sup
∂Qσ
2
hr+CQσ 2
1/n
ω(r)∇zL∞(Qσ
Hence
sup Qσ
2
|hr| ≤c
σa+r
σ
. (2.12)
Choosingr=σ1+awe get that for everyy∈Qσ 2,
hr(y)≤crβ, (2.13)
whereβ=a/(1 +a).
Lety∈T1, withdy≥η0,r0≤(1/3)η10/(1+a), andρ=η0/3. It follows that
ρDnhr(y)≤C
raz(y) +rz
L∞(Bρ(y))≤Crβ+rz(y)
=Cρrβ+rz(y)
ρ ≤Cρr
βD nz(y).
(2.14)
Hence ifr≤r0=min{(2c(η0))−1/β, (1/3)ηa0+1}, we get
1
2Dnz(y)≤Dnwr(y)≤ 3
2Dnz(y). (2.15)
The following two lemmas are similar to [5, Lemmas 2 and 3], respectively.
Lemma2.3. Letη0>0be fixed andwandzas inLemma 2.2. Then there existr0=r0(η0)
andt0=t0(λ,Λ,n)>1such that, ifr≤r0,
c−1 w(y)
dy ≤Dnw(y)≤c
w(y)
dy (2.16)
for everyy∈T1,dy≥t0η0.
Proof. The right-hand side inequality follows Schauder’s estimates and Harnack inequal-ity. Let nowy∈T1, withdy≥t0η0,t0to be chosen. We may assumey=tη0en. From the boundary Harnack principle (see, e.g., [8] or [9]) ify=η0en, then
z(y)≤ct−az(y) (2.17)
and, ift≥(2c)1/a≡t 0, then
z(y)≤1
2z(y). (2.18)
On the other hand, ifdy≥t0η0andr≤r0(η0), from (2.6), (2.13), and (2.15) we have
w(y)≤3
2z(y), Dnw(y)≥ 1
Thus, ift0η0≤dy≤10t0η0, applying Harnack inequality toDnz, we get
w(y)≤3
2z(y)≤3
z(y)−z(y)=3
1
0 d dsz
sy+ (1−s)yds
≤ct0η0Dnz(y)≤cDnz(y)dy≤cDnw(y)dy.
(2.20)
Repeating the argument with y=10t0η0, we get that (2.18) holds for 10t0η0≤dy≤ 20t0η0. After a finite number of steps, (2.18) follows fordy≥t0η0,y∈T1.
Lemma2.4. Letube as inTheorem 2.1. Then there exists a positiveη, such that for every
x∈T1,dx≤η,
Dnu(x)≥0. (2.21)
Moreover, in the same set
c−1u(x) dx ≤
Dnu(x)≤cud(x) x
. (2.22)
Proof. Lett0 be as inLemma 2.3, andη0 small to be chosen later. Setη1=2η0t0. It is enough to show that ifx=η1renandr≤r0(η0), thenDnu(x)≥0. Consider a small box
T2r and defineu(y)=u(r y). Thenusatisfies div(A(x)∇u(x))≡div(A(rx)∇u(x))=0 in
T2, where f is replaced by fr(y)= f(r y)/r.
We will show thatDnu(y)>0, where y=η1en, by comparinguwith the functionw constructed in Lemma 2.2, normalized in order to getu(y)=w(y). Notice that if we chooser0=r0(η0) according toLemma 2.3, we have
c−1w(y)
dy ≤Dnw(y)≤c
w(y)
dy . (2.23)
Ifdy≥η1. From the comparison theorem (see [8] or [9]), we know thatu/w ∈C0,a(T3/2) so that inBη0(y)
wu((yy))−1
≤cηa0, (2.24)
which implies
u(y)−w(y)≤cηa
0w(y)≤cηa0w(y). (2.25)
Moreover, sinceη0∼dy,
Dnu(y)−Dnw(y)≤cηa−1
0 w(y)≤cηa0Dnw(y), (2.26) from which we get
Dnu(y)≥
1−cηa 0
Dnw(y), (2.27)
and (2.21) holds ifη0is sufficiently small. Inequality (2.22) is now a consequence of (2.23)
To complete the proof ofTheorem 2.1, it is enough to observe that the above lemmas hold if we replaceenby any unit vectorτsuch that the angle betweenτandenis less than
θ=1/2 cot−1l.
Thus, we obtain a coneΓ(en,θ) of monotonicity foru. ApplyingTheorem 2.1to the positive and negative parts of the solutionuof our free boundary problem, we conclude that in aη-neighborhood ofF(u) the functionuis increasing along the direction of a cone
Γ(en,θ). Far from the free boundary, the monotonicity cone can be enlarged improving the Lipschitz constant of the level sets ofu.
This is a consequence of the following strong Harnack principle, where the cone
Γ(en,θ) is obtained fromΓ(en,θ) by deleting the “bad” directions, that is, those in a neighborhood of the generatrix opposite to∇u(en). Precisely, ifτ∈Γ(en,θ), denote by
ωτthe solid angle between the planes span{en,∇}and span{en,τ}. Delete fromΓ(en,θ) the directionsτsuch that (say)ωτ≥(99/100)π and callΓ(en,θ) the resulting set of di-rections. Ifτ∈Γ(en,θ), then
∇,τ ≥c3δ, (2.28)
whereδ=π/2−θ. We callδthedefect angle.
Lemma2.5. Supposeuis a positive solution ofdiv(A(rx)∇u(x))=0inT4 vanishing on F= {xn= f(x)}, increasing along everyτ∈Γ(en,θ). Assume furthermore that (2.4) holds
inT4. There exist positiver0andh, depending only onn,l, andλ,Λ, such that ifr≤r0, for
every small vectorτ,τ∈Γ(en,θ/2), and for everyx∈B1/8(en),
sup B(1+hδ)(x)
u(y−τ)≤u(x)−Cδuen
, (2.29)
where= |τ|sin(θ/2).
For the proof see [5, Section 3].
Corollary2.6. InB1/8(x0),uis increasing along everyτ∈Γ(τ1,θ1)with
δ1≤bδ,
δ1=π
2−θ1
,
ν1−e1≤Cδ, (2.30)
whereb=b(n,a,l,λ,Λ)<1.
We now apply the above results to the solution of our free boundary problem in a properly chosen neighborhood of the origin. Precisely, set for the moment
s=1
2min
r0,η, (2.31)
withηas inTheorem 2.1andr0as inLemma 2.5. If we define
thenu+
s satisfiesLsu+s ≡Lus(sx)=0 inT4and falls under the hypothesis ofLemma 2.5. Therefore, rescaling back we get the following result.
Theorem2.7. Letube a solution of our free boundary problem. Then inBs/8(sen),
sup B(1+hδ)(x)
u(y−τ)≤u(x)−cδuen (2.33)
for everyτ∈Γ(en,θ/2),|τ| s. As a consequence, inBs/8(sen),uis monotone along every
τ∈Γ(ν1,θ1), whereν1,θ1satisfy (2.30).
3. Proof of the main theorem
Before provingTheorem 1.1, we need to introduce some notations and to recall a point-wise characterization of weak subsolutions.
Ifᏻ⊂Ωis an open set, regular for the Dirichlet problem, we denote byGᏻ=Gᏻ(x,y) the Green function associated with the operatorLinᏻand byωᏻx theL-harmonic mea-sure forLinᏻ. In this way,
w(x)=
ᏻgdω
x
ᏻ−
ᏻGᏻ(x,y)h(y)d y (3.1)
is the unique weak solution ofLu=hinᏻ,h=0 on∂ᏻ. A functionv∈H1(Ω) is a weak subsolution inΩif
Ω A(x)∇u(x),∇φ(x)
dx≤0 (3.2)
for every nonnegative test functionϕsupported inΩ, whileuis a weak supersolution in
Ωif−uis a weak subsolution.
We need to recall a pointwise characterization. Indeed, see [10–14] for the details, we say that a functionv:Ω→RisL-subharmonic in a setΩif it is upper semicontinuous in
Ω, locally upper bounded inΩ, and
(S) for everyx0∈Ωthere exists a basis of regular neighborhoodᏮx0associated with vsuch that for everyB∈Ꮾx0,
vx0
≤
∂Bv(σ)dω x0
B. (3.3)
A functionvisL-superharmonic if−visL-subharmonic. ThusuisL-harmonic, or sim-ply harmonic, whenever it is bothL-subharmonic andL-superharmonic.
Precisely, see [12,13], if f is the trace on∂Ωof a function f∈C(Ω)∩H1(Ω), then the weak solution of the Dirichlet problem (even ifLhas just bounded measurable coef-ficients)
Lu=0 inΩ,
u=f on∂Ω (3.4)
and the parallel Perron-Wiener-Brelot one coincide. Moreover, in [15] Herv´e also proved that the same result holds when fisL-subharmonic and f∈Hloc1 (Ω).
Lemma3.1. LetC >2andφbe aC2weak solution of
divA(x)∇φ(x)≥C∇φ(x)2 +ω 2 0
φ(x) ≡Φ(x) (3.5)
inΩ,0< φmin≤φ≤φmax. Then for anyx∈Ωthere exists a positive numberr(x,φmax,φmin, C)such that, for everyr < r(x)and every ballBr=Br(x)⊂Ω,
∂Br
σ−x,∇φ(x)+1
2 Ᏸ
2φ(x)(σ−x), (σ−x)−Φ(σ)dωx
B(σ)≥0. (3.6)
Proof. FromLemma A.5, for every ballBr=Br(x)⊂Ω,
∇φ(x)
∂Br
(σ−x)dωxB(σ) + 1 2
n
i,j=1 Di jφ(x)
∂Br
σi−xi
σj−x
dωxB(σ)
≥
Br
GBr(x,y)Φ(y)d y+o
r2,
(3.7)
the proof follows easily.
We are now ready for the proof of the main theorem.
Proof ofTheorem 1.1. We have
vφ(x)=u
x+φ(x)η(x), (3.8)
for someη(x), where|η(x)| =1. To prove thatvφis anL-subsolution we just check con-dition (S), since by straightforward calculationsvφis locally Lipschitz continuous. In par-ticular we will prove that for everyx∈Ω+(v) there exists a positive constantr
0=r0(x) such that for every ballBr≡Br(x)⊂Ω+(v),r≤r0, and for everyx0∈Br,
∂Br
vφ(σ)dωxB0r(σ)≥vφ
x0
. (3.9)
Let{e1,. . .,en}be an orthonormal basis ofRnwhereen=η(x) and letξbe the following vectorfield:
ξ(h)=en+ n−1
i=1 Vi,h
where{V1,. . .,Vn−1} ⊂Rnwill be chosen later. Letν(h)=ξ(h)/|ξ(h)|, so that
ν(h)=en+ n−1
i=1 Vi,h
ei−1
2 n−1
i=1 Vi,h
2 en+o
|h|2. (3.11)
Letx0∈Br(x) andh=σ−x0. Then, letting
φ0=φ
x0
, ∇φ0= ∇φ
x0
, Ᏸ2φ0=Ᏸ2φ
x0
, (3.12)
we have
φ(σ)=φ0+ ∇φ0,h
+1
2 Ᏸ
2φ 0h,h
+o|h|2 (3.13)
ash→0, uniformly in a neighborhood ofx. As a consequence,
σ+φ(σ)νσ−x0
=y∗+J1+J2+J3, (3.14)
wherey∗=x0+φ(x0)en,
J1=
∇φ0,hen+h+φ0 n−1
i=1
Vi,hei
,
J2=
∇φ0,h
n−1
i=1 Vi,h
ei+1
2 Ᏸ
2φ 0h,h
en−φ0
2 n−1
i=1 Vi,h
2 en
,
J3=o
|h|2
(3.15)
uniformly ash→0.
LetJ=J1+J2+J3. Then for everyσ∈∂Br(x0),
v(σ)≥uy∗+J=uy∗+ ∇uy∗,J+1
2 Ᏸ
2uy∗J,J+o|h|2, (3.16)
ash→0, uniformly in a neighborhood ofy∗. We have
∇uy∗,J1
=∇uy∗ h,en
+ h,∇φ0
,
∇uy∗,J2
=∇uy∗
−φ0 2
n−1
i=1 Vi,h
2 +1
2 Ᏸ
2φ 0h,h
As a consequence,
∂Br
v(σ)dωx0
Br(σ)≥v
x0
+∇uy∗
∂Br
h,∇φ0
−φ0
2 n−1
i=1 Vi,h
2 +1
2 Ᏸ
2φ 0h,h
dωx0
Br
+
∂Br
∇uy∗,h+1
2 Ᏸ
2uy∗J,J+oh2
dωx0
Br=v
x0
+∇uy∗
∂Br
h,∇φ0
−φ0
2 n−1
i=1 Vi,h
2 +1
2 Ᏸ
2φ 0h,h
dωx0
Br
+1 2
∂Br
Ᏸ2uy∗J
1,J1
dωx0
Br+∇u
y∗
∂Br
hdωx0
Br+o
r2,
(3.18)
uniformly with respect tox0in a neighborhood ofx. Let
∂B Ᏸ 2uy∗J
1,J1
dωx0
Br= n
i,j
Di ju
y∗
∂Br
aiajdωBx0r (3.19)
with
ai=φ0 Vi,h
+ h,ei
, i=1,. . .,n, (3.20)
where theViare still to be chosen, and
an= ∇φ0,h
+ h,en
. (3.21)
Fori=1,. . .,nandj=1,. . .,n, let
di j=di j
x0,x0
=
∂Br
hihjdωxB0r(x0), d ∗
i j=di j
y∗,y∗=
∂Br
hihjdωy ∗ Br(y∗)
(3.22)
be the entries, of the matrix of the moments (see the appendix), respectively, evaluated in
x0andy∗.
Fori=1,. . .,n, and j=1,. . .,n−1, let
mi j=
∂Br
aiajdωxB0r,
mnn= n
p,q=1
Dpφ0Dqφ0dpq+ 2 n
p=1
Dpφ0dpn+dnn.
Then
mi j= n
p,q=1 φ2
0V p i V
q jdpq+φ0
n
p=1
Vipdj p+φ0 n
q=1
Vqjdiq+di j, (3.24)
min=mni=
∂Br
∇φ0,h
+hn
φ0
n
p=1
Viphp+hi
dωx0
Br(σ)
=φ0 n
p,q=1
VipDpφ0dpq+ n
p=1
Dqφ0dpi+φ0 n
p=1
Vipdpn+din.
(3.25)
Suppose now we can findV1,. . .,Vn−1and a real numberκ0, such that for everyi= 1,. . .,n−1 and for everyj=1,. . .,n,
mi,j=1 +κ0d∗i,j, mnn=1 +κ0d∗nn. (3.26)
Then
n−1
i,j=1
Di juy∗mi j+ 2 n−1
i=1
Dinuy∗min+Dnnuy∗mnn=1 +κ0
n
i,j=1
Di juy∗d∗i j.
(3.27)
In particular this means thatV1,. . .,Vn−1, and k0 must solve the following system, for i=1,. . .,n−1 andj=1,. . .,n−1,
φ0 n
p
Vipdj p+φ0 n
q=1
Vqjdiq+φ20 n
p,q=1
dp,qVipV q
j = −di,j+1 +κ0d∗i,1,
φ0 n
p=1
Vipdpn+φ0 n
p,q=1
VipDpφ0dpq= −di,n+ n
p=1
Dqφ0dip+1 +κ0
di∗,n,
n
p,q=1
Dpφ0Dqφ0dpq+ 2 n
p=1
Dqφ0dpn+dnn=
1 +κ0
d∗n,n.
(3.28)
From the last equations andLemma A.3, sinced∗nn> cλr2, for smallr and|∇φ
0|, there exists a positive constantC=C(λ,Λ) such that
κ0≤C∇φ0+dnn−d∗nn
d∗
n,n
≤C∇φ0+φmax
. (3.29)
Let (ᐂi)0=0,i=1,. . .,n−1, and forl≥0, define recursively (ᐂi)(l+1)as the solution of the system (i=1,. . .,n−1; j=1,. . .,n−1):
φ0 n p ᐂp i
(l+1) dj p+φ0
n
q=1
ᐂp
j
(l) diq+φ20
n
p,q=1 dp,q
ᐂp
i
(l)
ᐂq j
(l)
= −di,j+
1 +κ0
di∗,1,
φ0 n
p=1
ᐂp
i
(l+1) dpn+φ0
n
p,q=1
ᐂp
j
(l)
Dpφ0dpq= −di,n+ n
p=1
Dqφ0dip+
1 +κ0
di∗,n.
(3.30)
Notice that the sequence is well defined, since the matrixD(x0,x0) is nonsingular (Lemma
A.3in the appendix). Moreover, if|∇φ(x0)|is kept small, denoting bydiandd∗i the vec-tors (di1,. . .,din) and (di∗1,. . .,d∗in), from the estimates inLemma A.3, we get, by induction,
ᐂ(l+1)
i ≤C
di−d∗i
r2φ
0 +
∇φ0 φ
d∗
i
r2 +φ a 0
≤Cφ−1
0
φ0+∇φ0 (3.31)
withC=C(n,Λ,λ). Since the sequences (ᐂi(l))l∈Nare bounded for everyi∈ {1,. . .,n−1},
there exist subsequences (that we still call) (ᐂ(il))l∈N, converging toᐂiwith
ᐂi≤C(n,Λ,λ)φ−1 0
φ0+∇φ0. (3.32)
Now, from (3.18), (1.11), andLemma A.5, we get
∂Bv(σ)dω x0
Br(σ)≥v
x0
+∇uy∗
∂Br
h,∇φ0
−φ0
2 n−1
i=1 Vi,h
2 +1
2 Ᏸ
2φ 0h,h
dωx0
Br
+
∂Br
∇uy∗,hdωx0
Br+
1 2
∂Br
Ᏸ2uy∗J
1,J1
dωx0
Br+o
r2=vx 0
+∇uy∗
∂Br
h,∇φ0−φ0 2
n−1
i=1
Vi,h2+1
2 Ᏸ
2φ 0h,h
dωx0
Br
+
∂Br
∇uy∗,hdωx0
Br+
1 +κ0 2
∂Br(y∗)
Ᏸ2uy∗h∗,h∗dωy∗ Br(y∗)+o
r2
=vx0
+∇uy∗
∂Br
hdωx0
Br−
1 +κ0
∂Br(y∗)
h∗dωBy∗r(y∗)
+∇uy∗
∂Br
h,∇φ0−φ0 2
n−1
i=1
Vi,h2+1
2 Ᏸ
2φ
0h,hdωBx0r+or2. (3.33)
Consider
T=
∂Br
hdωx0
B(σ)−
1 +κ0
∂Br(y∗)
hdωBy∗r(y∗)(σ)
FromLemma A.3and (3.29), we get
|T| ≤Kr2. (3.35)
Thus, from (3.32), we deduce that
∂Br
v(σ)dωx0
B(σ)
≥vx0+∇uy∗
∂Br
h,∇φ0−φ0 2
n−1
i=1
Vi,h2+1
2 Ᏸ
2φ
0h,h−Kr2
dωx0
Br
≥vx0+∇uy∗
∂Br
h,∇φ0+1
2 Ᏸ
2φ
0h,h−C
∇φ02+φ02
φ0 r 2−Kr2
dωx0
Br.
(3.36)
FromLemma 2.5, ifris small, andClarge depending onx0andφ0, we have
∂Br
h,∇φ0
+1
2 Ᏸ
2φ 0h,h
−C
∇φ0 2
+φ2 0
φ0
r2−Kr2
dωx0
Br≥0, (3.37)
so thatvφis a weakL-subsolution in its positivity set.
Remark 3.2. We emphasize that the construction of the vectorsVi,i=1,. . .,n−1, involves only the Lipschitz continuity ofA.
4. Construction of the family of subsolutions and application to the free boundary problem
For the application to our free boundary problem we need a slightly different version of
Theorem 1.1. Indeed consider a small vectorτand the function
vτ(x)= sup Bφ(x)(x)
u(y−τ)=sup
|ν|=1
ux−τ+φ(x)ν. (4.1)
The proof ofTheorem 1.1holds, with minor changes, also in this case. In particular the following result holds.
Corollary4.1. Letube a continuous function inΩ. Assume that in{u >0}uis aC2-weak
solution ofLu=0,L∈ᏸ(λ,Λ,ω). For any vectorτletφbe a positiveC2-function such that 0< φmin≤φ≤φmax,
vτ(x)= sup Bφ(x)(x)
u(y−τ)=sup
|ν|=1
ux−τ+φ(x)ν, (4.2)
is well defined inΩ. There exist positive constantsρ0,μ0=μ0(n,λ,Λ)andC=C(n,λ,Λ),
such that if|∇φ| ≤μ0,|τ|< ρ0,ω0=ω(φmax), and
φLφ≥C∇φ(x)2+ω02
, (4.3)
Remark 4.2. The key point inCorollary 4.1 is that the estimates (3.29) and (3.32) for the vectorsVi,i=1,. . .,n−1, andk0depend only on the distance between the matrices D(x0,x0) andD(y∗,y∗).
We now construct a family of radii, with the right properties to be used in the final comparison theorem.
LetD=B2(0)\B1/8(en). We may assume with out loss of generality thatA(0)=Iand that
sup B3
A(x)−I≤ω11. (4.4)
By a slight modification of [5, Lemma 7] we can construct a family of functions satisfying the properties expressed in the following lemma.
Lemma4.3. LetC > 0. There exist positive numbersc,η,μ,ω < ημ/2and a family of func-tionsφt,0≤t≤1, such thatgt∈C2(D)and
(i) 0<1−ω≤φt≤1 +μt, (ii)φt≤1−ωinB2\B5/3, (iii)φt≥1−ω+ημtinB1/2, (iv)|∇φt| ≤c(μt+ω),
(v)
φtLφt≥C
∇φt 2
+ωmaxφt
2
. (4.5)
We are now in position to proveTheorem 1.2.
Proof ofTheorem 1.2. We first observe that Theorem 1.1 (andCorollary 4.1) holds for weak solutions, not necessarilyC2. In fact letu±
j be the functions constructed as solu-tions of the following problems:
Lju±j =0 inΩ±(u),
u±j =uj onΩ±(u),
(4.6)
and setuj=u+j −u−j. Thenujconverges locally inC1,a(Ω±(u)) touand it is not difficult to check that (suppressing for clarity the indext)
vj(x)= sup Bφ(x)(x)
uj (4.7)
converges locally inC1,a(Ω±(u)∩D) to
vφ(x)= sup Bφ(x)(x)
u. (4.8)
FromTheorem 1.1,vj is a weak subsolution forLjinΩ±(uj)∩D. But thenvφis a weak
L-subsolution inΩ±(u)∩D.
inequality (4.5) alsoφt satisfies the same inequality for every>0. Therefore, we can simplify the proof given in [5] avoiding, in the iteration process, to go through the im-provement of the-monotonicity and prove directly that in a sequence of dyadic balls
B4−kuis monotone along everyτ∈Γ(νk,θk) with
δk+1≤bδk
δ0=δ,δk=π 2 −θk
, νk+1−νk≤cδk. (4.9)
These conditions imply thatF(u) isC1,γ,γ=γ(b), at the origin.
Proof ofCorollary 1.3. SinceF(u) is Lipschitz,uis H¨older continuous inᏯ1. We only need to show thatuis Lipschitz inᏯ2/3 across the free boundary. This follows from a simple application of the monotonicity formula in [16, Lemma 1] and a barrier argument. Pre-cisely, letx0∈Ω+(u)∩Ꮿ2/3,d0=dist(x0,F(u)), andu(x0)=λ. From Harnack inequality
u(x)∼λ (4.10)
inBd0/2(x0). Letwbe the solution of
divA(x,u)∇w=0 (4.11)
inBd0(x0)\Bd0/2(x0) such thatw=0 on∂Bd0(x0),w=λon∂Bd0/2(x0). By maximum
prin-ciple
u≥cw inBd0
x0
\Bd0/2
x0
(4.12)
and, from theCanature ofAandC1,aestimates, ify
0∈∂Bd0(x0)∩F(u),
w(x)≥cλ d0
x−y0,ν
+
(4.13)
withν=(x0−y0)/|x0−y0|. Thus, neary0,uhas the asymptotic behavior
u(x)≥α x−y0,ν
+
−β x−y0,ν
−+ox−y
0 (4.14)
with
cλ
d0 ≤α≤G(β). (4.15)
Then, the monotonicity formula gives
λ d0G
−1
cλ
d0
≤Cu2
L∞(Ꮿ3/4) (4.16)
so that, from interior estimates,
∇u+x 0G−1
This gives the Lipschitz continuity ofu+. Similarly, we get
G∇u−x0∇u−
x0≤C1u2L∞(Ꮿ3/4) (4.18)
and the proof is complete.
Appendix Auxiliary lemmas
We collect here some estimates on theL-harmonic measure and its moments that are used in the paper. Hereω(r)≤c0ra, 0< a≤1.
Definition A.1. For anyx0,x,y∈Ω, andr >0,Br(x0)⊂Ω, letdi(x0,y) be, fori=1,. . .,n,
di(x0,y)=
∂Br(x0)
σi−x0i
dωByr(x0)(σ) (A.1)
and letdi j(x0,y) be, for everyi,j, 1≤i,j≤n,
di j
x0,y
=
∂Br(x0)
σi−x0i
σj−x0j
dωByr(x0)(σ). (A.2)
We call, respectively, (di(x0,y))1≤i≤n the vector of the first moment of theL-harmonic measure inBr(x), andD(x0,y)=(di j(x0,y))1≤i,j≤nthe matrix of the second moment of theL-harmonic measure inBr(x).
Denote byL0=div(A(x0)∇) and byG0r=G0r(x,y) the Green function forL0inBr=
Br(x0). We have the following.
LemmaA.2. LetL0wr= −1inBr(x0),wr=0on∂Br(x0). Then
wrx0
= r2
2TrAx0
. (A.3)
Proof. Suppose x0=0. Let gi j(x)=xixj and let vi j be the solution ofL0vi j=0 in Br,
vi j=gi jon∂Br. SinceL0gi j=2ai j(0) andgi j(0)=0, we have
vi j(0)=2ai j(0)wr(0). (A.4)
On the other hand,ni=1vii(0)=r2and (A.3) follows.
LemmaA.3. LetB2r(x0)⊂Ω. Then: (1)for everyi=1,. . .,n,
sup Br(0)
d x0,y
−yi≤C(λ,Λ,n)r1+a; (A.5)
(2)for everyi,j=1,. . .,n, di j
x0,x0
−2wr
x0
ai j
x0≤Cr2+a, (A.6)
Proof. Letx0=0 and
di(y)=di(0,y)=
∂Br
σidωByr(σ). (A.7)
Thendi(y)−yi=0 on∂Brand
L0
di(y)−yi
=divA(0)−A(y)ei
inBr. (A.8)
From standard estimates, we get
di−yi
L∞(B
r)≤CrA(0)−A(y)
eiL∞(B
r)≤Cr
1+a. (A.9)
Consider now
di j(y)=di j(0,y)=
∂Br
σiσjdωByr(σ). (A.10)
Ifvi jis as inLemma 2.2, we havehi j−vi j=0 on∂Brand
L0
di j(y)−vi j(y)
=A(0)−A(y)∇vi j
inBr. (A.11)
Therefore,
di j−vi j
L∞(B
r)≤CrA(0)−A(y)
∇vi jL∞(B
r)≤C∇vi jLs(Br)r
1+a≤Cr2+a. (A.12)
Hence, fromLemma 2.2, we get
di j(0)−2ai j(0)wr(0)=di j(0)−vi j(0)≤Cr2+a. (A.13)
CorollaryA.4. Forr≤r0(n,λ,Λ,a), the matrix(di j(0)/r2)is nonsingular.
LemmaA.5. Letwbe a weak solution of
divA(x)∇w(x)= f (A.14)
inΩ, where f is continuous. Then for everyx∈Br(x0)⊂Ω,
∇w(x)·
∂Br(x0)
(σ−x)dωBxr(x0)(σ) +
Br(x0)
GBr(x0)(x,y)f(y)d y=R
x0,x
, (A.15)
where
Rx0,x
=
∂Br
x0
1
0 ∇w
x+s(σ−x)− ∇w(x),σ−xds
Moreover, ifu∈C2(Ω),
∇w(x)·
∂Br(x0)
(σ−x)dωBxr(x0)(σ) +1 2
∂Br(x0)
D2w(x)(σ−x), (σ−x)dωx Br(x0)(σ)
+
Br(x0)
GBr(x0)(x,y)f(y)d y=o
r2.
(A.17)
Proof. Let div(A(x)∇w(x))=f, thenw∈C1,a(Ω) and for anyσ,x∈B
r(x0)⊂Ω,
w(σ)=w(x) +
1
0 ∇w
x+s(σ−x),σ−xds=w(x) + ∇w(x),σ−x
+
1
0 ∇w
x+s(σ−x)− ∇w(x),σ−xds.
(A.18)
On the other hand,
w(x)=
∂Br(x0)
w(σ)dωx
Br(x0)(σ)−
Br(x0)
GBr(x0)(x,y)f(y)d y, (A.19)
hence
∇w(x)·
∂Br(x0)
(σ−x)dωBxr(x0)(σ) +
Br(x0)
GBr(x0)(x,y)f(y)d y=R
x0,x
. (A.20)
The rest of the proof follows from Taylor expansion.
CorollaryA.6. Letu∈C2(Ω)be a weak solution of
divA(x)∇u(x)=0 (A.21)
inΩ. Then
∇ux0
·
∂Br(x0)
σ−x0
dωx0
Br(x0)(σ)=O
r2. (A.22)
Proof. It is enough to observe that
∂Br(x0)
σi−x0iσj−x0jdωxB0r(x0)(σ)=O
r2. (A.23)
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Fausto Ferrari: Dipartimento di Matematica, Universit`a di Bologna, Piazza di Porta S. Donato 5, 40126 Bologna, Italy; C.I.R.A.M., Via Saragozza 8, 40123 Bologna, Italy
Email address:[email protected]
Sandro Salsa: Dipartimento di Matematica, Politecnico di Milano, Via Bonardi 7, 20133 Milano, Italy
