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Volume 2007, Article ID 57049, 21 pagesdoi:10.1155/2007/57049 Research Article Subsolutions of Elliptic Operators in Divergence Form and Application to Two-Phase Free Boundary Problems F

Volume 2007, Article ID 57049, 21 pages

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

LetL be a divergence form operator with Lipschitz continuous coefficients in a domain

Ω, and let u be a continuous weak solution of Lu =0 in{u =0} In this paper, we showthat ifφ satisfies a suitable di fferential inequality, then v φ(x) =supB φ(x)(x) u is a subsolution

ofLu =0 away from its zero set We apply this result to proveC1,γregularity of Lipschitzfree boundaries in two-phase problems

Copyright © 2007 F Ferrari and S Salsa This is an open access article distributed underthe 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 isplayed by certain continuous perturbations of the solution, constructed as supremum ofthe 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 turnout to be subsolutions of the problem, suitable for comparison purposes

This kind of subsolutions have been introduced for the first time by Caffarelli in theclassical paper [1] in order to prove that, in a general class of two-phase problems for thelaplacian, 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 nonlinearoperators, not necessary concave, of the typeF(D2u, Du) In [5], Cerutti et al considervariable coefficients operators in nondivergence form and Ferrari [6] a class of fully non-linear operatorsF(D2u, x), H¨older continuous in the space variable.

The important case of linear or semilinear operators in divergence form with smooth coefficients (less than C1,α, e.g.) is not included in the above results and it is

non-precisely the subject of this paper Once again, the key point is the construction of thepreviously mentioned family of subsolutions Unlike the case of nondivergence or fullynonlinear 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 afunction controls in average the action of the operator but also the linear part has anequivalent influence Here we require Lipschitz continuous coefficients

To state our first result we introduce the classᏸ(λ,Λ,ω) of elliptic operators

L =div

defined in a domainΩ⊂ R n, with symmetric and uniformly elliptic matrix, that is,

and modulus of continuity of the coefficients given by

is well defined in Ω There exist positive constants μ0= μ0(n, λ, Λ) and C = C(n, λ, Λ), such

that, if |∇φ| ≤ μ0, ω0= ω(φmax), and

φLφ ≥ C

∇ φ(x) 2

+ω2 

then v is a weak subsolution of Lu = 0 in {v > 0}.

We now introduce the class of free boundary problems we are going to study and theappropriate notion of weak solution

LetB R = B R(0) be the ball of radiusR inRn −1 InᏯR = B R(0)×(−R, R) we are given a

We callF(u) ≡ ∂Ω+(u) ∩ᏯR the free boundary We say that a point x0∈ F(u) is regular

from the right (left) if there exists a ballB:

(ii) AlongF(u) the following conditions hold:

(a) ifx0∈ F(u) is regular from the right, then, near x0,

for someα ≥0,β > 0 with equality along every nontangential domain in

both cases, and

Here are our main results concerning the regularity of Lipschitz free boundaries

Theorem 1.2 Let u be a weak solution to f.b.p inᏯR = B R ×(−R, R).

Suppose that 0 ∈ F(u) and that

Then, on B R/2 , f is a C1,γ function with γ = γ(n, l, N, λ, Λ,ω).

By using of the monotonicity formula in [7] we can prove the following

Corollary 1.3 In f.b.p let

Lu =div

where L is a uniformly elliptic divergence form operator Assume (ii) and (iii) in Theorem 1.2

hold and replace (i) with the assumption that A is Lipschitz continuous with respect to x and

u Then the same conclusion holds.

We can allow a dependence onx and ν in the free boundary condition for G = G(β, x,ν)

assuming instead of (iii) inTheorem 1.1

(iii )G = G(z,ν,x) is continuous strictly increasing in z and, for some N > 0

indepen-dent ofν and x, z − N G(z,ν,x) is decreasing in (0,∞);

(iii ) logG is Lipschitz continuous with respect to ν, x, uniformly with respect to its

that corresponds to aC1,γregularity ofF(u) for a suitable γ.

The first step follows with some modifications [5, Sections 2 and 3] and everythingworks 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 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, involvingtheL-harmonic measure.Section 3contains the proof of the main result,Theorem 1.1,and some consequences

struc-InSection 4the above results are applied to our free boundary problem, preparing thenecessary tools for the final iteration

The third step can be carried exactly as in [5, Sections 6 and 7], since here the particularform of the operator does not play any role anymore Actually the linear modulus ofcontinuity allows some simplifications

2 Monotonicity properties of weak solutions

In this section we assume thatω(r) ≤ c0r a, 0< a ≤1 Letu ∈ H1

loc(Ω) be a weak solution

for every test functionϕ supported in Ω If L ∈ ᏸ(λ,Λ,ω), u ∈ C1,a(Ω)

In this section we prove that if the domainΩ is Lipschitz and u vanishes on a relatively

open portionF ⊂ ∂ Ω, then, near F, the level sets of u are uniformly Lipschitz surfaces.

Precisely, we consider domains of the form

T s =x ,x n



∈ R:|x | < s, f (x )< x n < 2ls

where f is a Lipschitz function with constant l.

Theorem 2.1 Let u be a positive solution to Lu = 0 in T4, vanishing on F = {x n = f (x )} ∩

∂T4 Then, there exists η such that in

ᏺη(F) =f (x )< x n < f (x ) +η

u is increasing along the directions τ belonging to the cone Γ(e n,θ), with axis e n and opening

whereg is a smooth function vanishing on Ᏺ and equal to 1 at points x with d x > 1/10.

Then, see [1],D n z > 0 in Q ρ, withρ = ρ(n, l) By rescaling the problem (if necessary),

we may assumeρ =3/2 Since z(e n)≥ c > 0, by Harnack inequality we have that, if y ∈ T1,

whereβ = a/(1 + a).

Lety ∈ T1, withd y ≥ η0,r0≤(1/3)η10/(1+a), andρ = η0/3 It follows that

2D n z(y) ≤ D n w r(y) ≤3



The following two lemmas are similar to [5, Lemmas 2 and 3], respectively

Lemma 2.3 Let η0> 0 be fixed and w and z as in Lemma 2.2 Then there exist r0= r0(η0)

and t0= t0(λ, Λ,n) > 1 such that, if r ≤ r0,

c −1 w(y)

d y ≤ D n w(y) ≤ c w(y)

for every y ∈ T1, d y ≥ t0η0.

Proof The right-hand side inequality follows Schauder’s estimates and Harnack

inequal-ity Let nowy ∈ T1, withd y ≥ t0η0,t0to be chosen We may assumey = tη0e n From theboundary Harnack principle (see, e.g., [8] or [9]) ify= η0e n, then

Thus, ift0η0≤ d y ≤10t0η0, applying Harnack inequality toD n z, we get

Repeating the argument with y =10t0η0, we get that (2.18) holds for 10t0η0≤ d y ≤

20t0η0 After a finite number of steps, (2.18) follows ford y ≥ t0η0,y ∈ T1 

Lemma 2.4 Let u be as in Theorem 2.1 Then there exists a positive η, such that for every

We will show thatD n u(y) > 0, where y = η1e n, by comparingu with the function w

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

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 lemmashold if we replacee nby any unit vectorτ such that the angle between τ and e nis less than

θ =1/2 cot −1l.

Thus, we obtain a coneΓ(e n,θ) of monotonicity for u ApplyingTheorem 2.1to thepositive and negative parts of the solutionu of our free boundary problem, we conclude

that in aη-neighborhood of F(u) the function u is increasing along the direction of a cone

Γ(e n,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 ( n,θ) is obtained from Γ(e n,θ) by deleting the “bad” directions, that is, those in a

neighborhood of the generatrix opposite to∇u(e n) Precisely, ifτ ∈ Γ(e n,θ), denote by

ω τthe solid angle between the planes span{e n,∇}and span{e n,τ} Delete fromΓ(e n,θ)

the directionsτ such that (say) ω τ ≥(99/100)π and callΓ ( n,θ) the resulting set of

di-rections Ifτ ∈Γ ( n,θ), then

whereδ = π/2 − θ We call δ the defect angle.

Lemma 2.5 Suppose u is a positive solution of div(A(rx)∇u(x)) = 0 in T4 vanishing on

F = {x n = f (x )}, increasing along every τ ∈ Γ(e n,θ) Assume furthermore that ( 2.4 ) holds

in T4 There exist positive r0and h, depending only on n, l, and λ, Λ, such that if r ≤ r0, for every small vector τ, τ ∈Γ ( n,θ/2), and for every x ∈ B1/8( n ),

For the proof see [5, Section 3]

Corollary 2.6 In B1/8(x0), u is increasing along every τ ∈ Γ(τ1,θ1) with

3 Proof of the main theorem

Before provingTheorem 1.1, we need to introduce some notations and to recall a wise characterization of weak subsolutions

point-Ifᏻ⊂ Ω is an open set, regular for the Dirichlet problem, we denote by Gᏻ= Gᏻ(x, y)

the Green function associated with the operatorL in ᏻ and by ω x

ᏻtheL-harmonic

mea-sure forL inᏻ In this way,

is the unique weak solution ofLu = h in ᏻ, h =0 on∂ᏻ

A functionv ∈ H1(Ω) is a weak subsolution in Ω if

Ω

for every nonnegative test functionϕ supported in Ω, while u is a weak supersolution in

Ω if−u is a weak subsolution.

We need to recall a pointwise characterization Indeed, see [10–14] for the details, wesay that a functionv :Ω→ RisL-subharmonic in a setΩ if it is upper semicontinuous in

Ω, locally upper bounded in Ω, and

(S) for everyx0∈Ω there exists a basis of regular neighborhood Ꮾx0associated with

v such that for every B ∈Ꮾx0,

A functionv is L-superharmonic if −v is L-subharmonic Thus u is L-harmonic, or

sim-ply harmonic, whenever it is bothL-subharmonic and L-superharmonic.

With such pointwise characterization, the definition of the Perron-Wiener-Brelot lution of the Dirichlet problem can be stated as usual, see [10] or [11] The Perron-Wiener-Brelot solution of the Dirichlet problem coincides, in any reasonable case, withthe solution of the Dirichlet given by the variational approach In general,L-subharmonic

so-functions and such subsolutions do not coincide On the other hand, ifv is locally

Lips-chitz,v is L-subharmonic if and only if v is locally a subsolution.

Precisely, see [12,13], if f is the trace on ∂Ω of a function f ∈ C(Ω)∩ H1(Ω), thenthe weak solution of the Dirichlet problem (even ifL has just bounded measurable coef-

in Ω, 0 < φmin≤ φ ≤ φmax Then for any x ∈ Ω there exists a positive number r(x,φmax,φmin,

C) such that, for every r < r(x) and every ball B r = B r(x) ⊂ Ω,

We are now ready for the proof of the main theorem

Proof of Theorem 1.1 We have

v φ(x) = u

x + φ(x)η(x)

for someη(x), where |η(x)| =1 To prove thatv φis anL-subsolution we just check

con-dition (S), since by straightforward calculationsv φis locally Lipschitz continuous In ticular we will prove that for everyx ∈Ω+(v) there exists a positive constant r0= r0(x)

par-such that for every ballB r ≡ B r(x) ⊂Ω+(v), r ≤ r0, and for everyx0∈ B r,

where{V1, , V n −1} ⊂ R nwill be chosen later Letν(h) = ξ(h)/|ξ(h)|, so that

Suppose now we can findV1, , V n −1and a real numberκ0, such that for everyi =

1, , n −1 and for everyj =1, , n,

Let (ᐂi)0=0,i =1, , n −1, and forl ≥0, define recursively (ᐂi)(l+1)as the solution

 (l)

ᐂq j

withC = C(n, Λ,λ) Since the sequences (ᐂ(l)

i )l ∈Nare bounded for everyi ∈ {1, , n −1},there exist subsequences (that we still call) (ᐂ(l)

FromLemma A.3and (3.29), we get

Remark 3.2 We emphasize that the construction of the vectors V i, =1, , n −1, involvesonly 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

is well defined in Ω There exist positive constants ρ0, μ0= μ0(n, λ, Λ) and C = C(n, λ, Λ),

such that if |∇φ| ≤ μ0, |τ| < ρ0, ω0= ω(φmax), and

Remark 4.2 The key point inCorollary 4.1 is that the estimates (3.29) and (3.32) forthe vectorsV i, =1, , n −1, andk0depend only on the distance between the matrices

Lemma 4.3 Let C > 0 There exist positive numbers c, η, μ, ω < ημ/2 and a family of func-

tions φ t , 0 ≤ t ≤ 1, such that g t ∈ C2(D) and

We are now in position to proveTheorem 1.2

Proof of Theorem 1.2 We first observe that Theorem 1.1 (andCorollary 4.1) holds forweak solutions, not necessarilyC2 In fact letu ± j be the functions constructed as solu-tions of the following problems:

L j u ± j =0 inΩ±(u),

and setu j = u+j − u − j Thenu jconverges locally inC1,a(Ω±(u)) to u and it is not difficult

to check that (suppressing for clarity the indext)

inequality (4.5) alsoφ t 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 provement of the-monotonicity and prove directly that in a sequence of dyadic balls

im-B4− k u is monotone along every τ ∈ Γ(ν k,θ k) with

δ k+1 ≤ bδ k δ0= δ, δ k = π

2 − θ k



, ν k+1 − ν k  ≤ cδ k (4.9)These conditions imply thatF(u) is C1,γ,γ = γ(b), at the origin. 

Proof of Corollary 1.3 Since F(u) is Lipschitz, u is H¨older continuous inᏯ1 We only need

to show thatu is Lipschitz inᏯ2/3across the free boundary This follows from a simpleapplication of the monotonicity formula in [16, Lemma 1] and a barrier argument Pre-cisely, letx0∈Ω+(u) ∩Ꮿ2/3,d0=dist(x0,F(u)), and u(x0)= λ From Harnack inequality

This gives the Lipschitz continuity ofu+ Similarly, we get

We collect here some estimates on theL-harmonic measure and its moments that are

used in the paper Hereω(r) ≤ c0r a, 0< a ≤1

Definition A.1 For any x0,x, y ∈ Ω, and r > 0, B r(x0)⊂ Ω, let d i(x0,y) be, for i =1, , n,

We call, respectively, (d i(x0,y))1≤ i ≤ n the vector of the first moment of theL-harmonic

measure inB r(x), and D(x0,y) =(d i j(x0,y))1≤ i, j ≤ nthe matrix of the second moment oftheL-harmonic measure in B r(x).

Denote byL0=div(A(x0)∇) and byG0

r = G0

r(x, y) the Green function for L0inB r =

B r(x0) We have the following

Lemma A.2 Let L0w r = − 1 in B r(x0), w r = 0 on ∂B r(x0) Then

Proof Suppose x0=0 Let g i j(x) = x i x j and let v i j be the solution ofL0v i j =0 in B r,

v i j = g i jon∂B r SinceL0g i j =2a i j(0) andg i j(0)=0, we have

On the other hand,n

Lemma A.3 Let B2r(x0)⊂ Ω Then: