Subsolutions of Elliptic Operators in Divergence Form and Application to Two-Phase Free Boundary Problems

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)=sup_Bφ(x)(x)uis a subsolution ofLu=0 away from its zero set. We apply this result to proveC^1,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 indeedC^1,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.

The important case of linear or semilinear operators in divergence form with non- smooth coefficients (less thanC^1,α, e.g.) is not included in the above results and it is

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≤A^s(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)

Theorem 1.1. 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+ω₀², (1.5)

thenvis a weak subsolution ofLu=0 in{v >0}.

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

LetB_R=B_R(0) be the ball of radiusRinRⁿ⁻¹. InᏯR=B_R(0)×(−R,R) we are given a continuousH_loc¹ functionusatisfying the following.

(i)

Lu=divA(x)∇u=0 (1.6)

in Ω⁺(u)= {x∈ᏯR:u(x)>0}, and in Ω⁻(u)= {x∈ᏯR:u(x)≤0}⁰, in the weak sense.

We callF(u)≡∂Ω⁺(u)∩ᏯRthe free boundary. We say that a pointx0∈F(u) is regular from the right (left) if there exists a ballB:

B⊂Ω⁺(u) ⊂Ω⁻(u), resp., B∩F(u)=

x0

. (1.7)

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

(a) ifx0∈F(u) is regular from the right, then, nearx0,

u⁺(x)≥α x−x0,ν⁺−β x−x0,ν⁻+ox−x0, (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−x0,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 of f.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.

Theorem 1.2. Letube a weak solution to f.b.p. inᏯR=B_R×(−R,R).

Suppose that 0∈F(u) and that (i)L∈ᏸ(λ,Λ,ω);

(ii)Ω⁺(u)= {(x,x_n) :x_n> 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, onB_R/2, f is aC^1,γfunction withγ=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=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.

We can allow a dependence onxandνin the free boundary condition forG=G(β,x,ν) assuming instead of (iii) inTheorem 1.1

(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 aC^1,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)≤c0r^a, 0< a≤1. Letu∈H_loc¹ (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∈C^1,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.

Theorem 2.1. Letube a positive solution toLu=0 inT4, vanishing onF= {x_n= f(x)} ∩

∂T4. Then, there existsηsuch that in

ᏺη(F)=

f(x)< x_n< f(x) +η∩T1, (2.3)

uis increasing along the directionsτbelonging to the coneΓ(en,θ), with axisenand opening θ=(1/2) cot⁻¹l. Moreover, inᏺη(F),

c⁻¹u(x)

dx ≤Dnu(x)≤cu(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],D_nz >0 inQ_ρ, 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

, D_nz(y)∼z(y) d_y ^∼cη0

. (2.6)

Clearlyz,u∈C^0,a(T2).

Lemma 2.2. Forr >0, letwrbe theC^1,a(T2) weak solution to divA(rx)∇wr(x)=0, T2,

w_r=z, ∂T2. (2.7)

Then, givenη0>0, there existsr0=r0(η0), such that ifr≤r0,

D_nw_r(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

Q^σ₂=

x∈T2, distx,∂T2

> σ. (2.10)

Notice thathr=wr−z∈C^0,a(T2), and moreoverhr∈C^1,a(Q₂^σ). Notice that ((A(rx)− A(0))∇z(x))_i∈L^∞(Q2^σ), and from standard estimates we have

sup

Q^σ2

w_r−z≤sup

∂Q2^σ

h_r+CQ^σ₂^1/nω(r)∇z_L∞(Q^σ2). (2.11)

Hence

sup

Q^σ2

|h_r| ≤c

σ^a+r σ

. (2.12)

Choosingr=σ^1+awe get that for everyy∈Q^σ₂,

hr(y)≤cr^β, (2.13)

whereβ=a/(1 +a).

Lety∈T1, withd_y≥η0,r0≤(1/3)η^1/(1+a)₀ , andρ=η0/3. It follows that ρD_nh_r(y)≤Cr^az(y) +rzL^∞(Bρ(y))

≤Cr^β+rz(y)

=Cρr^β+rz(y)

ρ ^≤Cρr^βD_nz(y).

(2.14)

Hence ifr≤r0=min{(2c(η0))^−1/β, (1/3)η^a+1₀ }, we get 1

2D_nz(y)≤D_nw_r(y)≤3

2D_nz(y). (2.15)

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

Lemma 2.3. Letη0>0 be fixed andwandzas inLemma 2.2. Then there existr0=r0(η0) andt0=t0(λ,Λ,n)>1 such that, ifr≤r0,

c⁻¹ w(y)

dy ≤Dnw(y)≤cw(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=η0e_n, then

z(y)≤ct^−az(y) (2.17)

and, ift≥(2c)^1/a≡t0, 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

2Dnz(y). (2.19)

Thus, ift0η0≤dy≤10t0η0, applying Harnack inequality toDnz, we get w(y)≤3

2z(y)≤3z(y)−z(y)=3 1

0

d

dszsy+ (1−s)yds

≤ct0η0D_nz(y)≤cD_nz(y)d_y≤cD_nw(y)d_y.

(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 ford_y≥t0η0,y∈T1. Lemma 2.4. Letube as inTheorem 2.1. Then there exists a positiveη, such that for every x∈T1,dx≤η,

D_nu(x)≥0. (2.21)

Moreover, in the same set

c⁻¹u(x)

d_x ^≤D_nu(x)_≤cu(x)

d_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=η1re_nandr≤r0(η0), thenD_nu(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 thatD_nu(y) >0, where y=η1e_n, 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)≤cw(y)

dy . (2.23)

Ifdy≥η1. From the comparison theorem (see [8] or [9]), we know thatu/w ∈C^0,a(T3/2) so that inBη0(y)

u(y)

w(y)⁻1≤cη^a₀, (2.24)

which implies

u(y)−w(y)≤cη₀^aw(y)≤cη^a₀w(y). (2.25) Moreover, sinceη0∼dy,

Dnu(y) −Dnw(y)≤cη^a₀⁻¹w(y)≤cη^a₀Dnw(y), (2.26) from which we get

D_nu(y) _≥1−cη^a₀D_nw(y), (2.27) and (2.21) holds ifη0is sufficiently small. Inequality (2.22) is now a consequence of (2.23)

and the fact thatw(y)=u(y).

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 ω_τthe solid angle between the planes span{e_n,∇}and span{e_n,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 callδthe defect angle.

Lemma 2.5. Supposeuis a positive solution of div(A(rx)∇u(x))=0 inT4 vanishing on F= {x_n= 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].

Corollary 2.6. InB_1/8(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

2minr0,η, (2.31)

withηas inTheorem 2.1andr0as inLemma 2.5. If we define u_s(x)₌u(sx)

s , (2.32)

thenu⁺_s satisfiesLsu⁺_s ≡Lus(sx)=0 inT4and falls under the hypothesis ofLemma 2.5.

Therefore, rescaling back we get the following result.

Theorem 2.7. 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 byG_ᏻ=G_ᏻ(x,y) the Green function associated with the operatorLinᏻand byω_ᏻ^x theL-harmonic mea- sure forLinᏻ. In this way,

w(x)=

ᏻgdω_ᏻ^x−

ᏻG_ᏻ(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^x_B⁰. (3.3)

A functionvisL-superharmonic if−visL-subharmonic. ThusuisL-harmonic, or sim- ply harmonic, whenever it is bothL-subharmonic andL-superharmonic.

With such pointwise characterization, the definition of the Perron-Wiener-Brelot so- 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, with the solution of the Dirichlet given by the variational approach. In general,L-subharmonic functions and such subsolutions do not coincide. On the other hand, ifvis locally Lips- chitz,visL-subharmonic if and only ifvis locally a subsolution.

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∈H_loc¹ (Ω).

Lemma 3.1. LetC >2 andφbe aC²weak solution of divA(x)∇φ(x)≥C^∇φ(x)2 +ω²₀

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 Ᏸ²φ(x)(σ−x), (σ−x)−Φ(σ)

dω^x_B(σ)≥0. (3.6) Proof. FromLemma A.5, for every ballBr=Br(x)⊂Ω,

∇φ(x)

∂Br

(σ−x)dω^x_B(σ) +1 2

n i,j=1

Di jφ(x)

∂Br

σi−xi

σj−xdω^x_B(σ)

≥

Br

GBr(x,y)Φ(y)d y+or²,

(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)=ux+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 calculationsv_φis locally Lipschitz continuous. In par- ticular we will prove that for everyx∈Ω⁺(v) there exists a positive constantr0=r0(x) such that for every ballBr≡Br(x)⊂Ω⁺(v),r≤r0, and for everyx0∈Br,

∂Br

vφ(σ)dω^x_B⁰_r(σ)≥vφ

x0

. (3.9)

Let{e1,. . .,en}be an orthonormal basis ofRⁿwhereen=η(x) and letξbe the following vectorfield:

ξ(h)=e_n+

n−1 i=1

V_i,he_i, (3.10)

where{V1,. . .,Vn−1} ⊂Rⁿwill be chosen later. Letν(h)=ξ(h)/|ξ(h)|, so that

ν(h)=en+

n−1 i=1

Vi,hei−1 2

n−1 i=1

Vi,h²en+o|h|²

. (3.11)

Letx0∈Br(x) andh=σ−x0. Then, letting φ0=φx0

, ∇φ0= ∇φx0

, Ᏸ²φ0=Ᏸ²φx0

, (3.12)

we have

φ(σ)=φ0+ ∇φ0,h+1

2 Ᏸ²φ0h,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,hei+1

2 Ᏸ²φ0h,hen−φ0

2

n−1 i=1

Vi,h²en

, 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 Ᏸ²uy^∗J,J+o|h|²

, (3.16)

ash→0, uniformly in a neighborhood ofy^∗. We have

∇uy^∗,J1

=∇uy^∗ h,e_n+ h,∇φ0

,

∇uy^∗,J2

=∇uy^∗

−φ0

2

n−1 i=1

V_i,h²+1

2 Ᏸ²φ0h,h

. (3.17)

As a consequence,

∂Br

v(σ)dω_B^x0_r(σ)≥vx0

+∇uy^∗

∂Br

h,∇φ0

−φ0

2

n−1 i=1

V_i,h²+1

2 Ᏸ²φ0h,h

dω^x_B⁰_r

+

∂Br

∇uy^∗,h+1

2 Ᏸ²uy^∗J,J+oh²

dω^x_B⁰_r=vx0 +∇uy^∗

∂Br

h,∇φ0

−φ0

2

n−1 i=1

V_i,h²+1

2 Ᏸ²φ0h,h

dω^x_B⁰_r +1

2

∂Br

Ᏸ²uy^∗J1,J1

dω_B^x0_r+∇uy^∗

∂Br

hdω^x_B⁰_r+or²,

(3.18)

uniformly with respect tox0in a neighborhood ofx.

Let

∂B Ᏸ²uy^∗J1,J1

dω^x_B⁰_r= n

i,j

D_{i j}uy^∗

∂Br

a_ia_jdω_B^x0_r (3.19)

with

ai=φ0 Vi,h+ h,ei

, i=1,. . .,n, (3.20)

where theV_iare 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

hihjdω^x_B⁰_r(x0), d_{i j}^∗=di j

y^∗,y^∗=

∂Br

hihjdω_B^y∗_r(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

aiajdω^x_B⁰_r, mnn=

n p,q=1

Dpφ0Dqφ0dpq+ 2 n p=1

Dpφ0dpn+dnn.

(3.23)

Then

mi j= n p,q=1

φ²₀V_i^pV_j^qdpq+φ0

n p=1

V_i^pdj p+φ0

n q=1

V^q_jdiq+di j, (3.24) m_in=m_ni=

∂Br

∇φ0,h+h_nφ0

n p=1

V_i^ph_p+h_i

dω^x_B⁰_r(σ)

=φ0

n p,q=1

V_i^pDpφ0dpq+ n p=1

Dqφ0dpi+φ0

n p=1

V_i^pdpn+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 +κ0

d^∗_i,j, mnn= 1 +κ0

d^∗_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,. . .,V_n−1, and k0 must solve the following system, for i=1,. . .,n−1 andj=1,. . .,n−1,

φ0

n p

V_i^pdj p+φ0

n q=1

V^q_jdiq+φ²₀ n p,q=1

dp,qV_i^pV^q_j = −di,j+1 +κ0 d^∗_i,1,

φ0

n p=1

V_i^pdpn+φ0

n p,q=1

V_i^pDpφ0dpq= −di,n+ n p=1

Dqφ0dip+1 +κ0

d_i,n^∗,

n p,q=1

D_pφ0D_qφ0d_pq+ 2 n p=1

D_qφ0d_pn+d_nn= 1 +κ0

d^∗_n,n.

(3.28)

From the last equations andLemma A.3, sinced^∗_nn> cλr², for smallr and|∇φ0|, there exists a positive constantC=C(λ,Λ) such that

κ0≤C∇φ0+dnn−d^∗_nn

d^∗_n,n ^≤C∇φ0+φmax

. (3.29)

We now start an iteration process to solve the above system (see [4,6]).