Báo cáo hóa học: " Research Article Subsolutions of Elliptic Operators in Divergence Form and Application to Two-Phase Free Boundary Problems" pdf

Hindawi Publishing Corporation
Boundary Value Problems
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
Let L be a divergence form operator with Lipschitz continuous coefficients in a domain
Ω,andletu be a continuous weak solution of Lu
= 0in{u = 0}. In this paper, we show
that if φ satisfies a suitable differential inequality, then v
φ
(x) = sup
B
φ(x)
(x)
u is a subsolution
of Lu
= 0 away from its zero set. We apply this result to prove C
1,γ
regularity 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 par abolic problems, a key role is
played by certain continuous per turbations of the solution, constructed as supremum of
the solution itself over balls of variable radius. The cr ucial fact is that if the radius satisfies
asuitabledifferential 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 indeed C
1,α
.
This result has been subsequently extended to more general operators: Feldman [2]
considers linear anisotropic operators with constant coefficients, Wang [3]aclassofcon-
cave fully nonlinear operators of the typ e F(D
2
u), and again Feldman [4 ] fully nonlinear
operators, not necessary concave, of the type F(D
2
u,Du). In [5], Cerutti e t al. consider
variable coefficients operators in nondivergence form and Ferrari [6] a class of fully non-
linear operators F(D
2
u,x), H
¨
older continuous in the space variable.
The important case of linear or semilinear operators in divergence form with non-
smooth coefficients (less than C
1,α
, e.g.) is not included in the above results and it is
2 Boundary Value Problems
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 class ᏸ(λ,Λ,ω)ofellipticoperators
L
= div

A(x)∇

(1.1)
defined in a domain Ω
⊂ R
n
, 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. Let u beacontinuousfunctioninΩ. Assume that in
{u>0} u is a C
2
-weak
solution of Lu
= 0, L ∈ ᏸ(λ,Λ,ω), ω(r) ≤ c
0
r.Letφ be a positive C
2
-function such that
0 <φ
min
≤ φ ≤ φ
max
and
v
φ
(x) = sup
B
φ(x)
(x)
u = sup
|ν|=1
u

x + φ(x)ν

(1.4)

is well defined in Ω. There exist positive constants μ
0
= μ
0
(n,λ,Λ) and C = C(n,λ,Λ),such
that, if
|∇φ|≤μ
0
, ω
0
= ω(φ
max
),and
φLφ
≥ C



∇
φ(x)


2
+ ω
2
0

, (1.5)
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 the
appropriate notion of weak solution.
Let B

R
= B

R
(0) be the ball of radius R in R
n−1
.InᏯ
R
= B

R
(0) × (−R,R)wearegivena
continuous H
1
loc
function u satisfying the following.
(i)
Lu
= div

A(x)∇u

=
0 (1.6)
in Ω
+

(u) ={x ∈ Ꮿ
R
: u(x) > 0},andinΩ
−
(u) ={x ∈ Ꮿ
R
: u(x) ≤ 0}
0
,inthe
weak sense.
We cal l F(u)
≡ ∂Ω
+
(u) ∩ Ꮿ
R
the free boundary. We say that a point x
0
∈ F(u)isregular
from the right (left) if there exists a ball B:
B
⊂ Ω
+
(u)

⊂
Ω
−
(u), resp.

,

B
∩ F(u) =

x
0

.
(1.7)
F. Ferra r i an d S. Sa lsa 3
(ii) Along F(u) the following conditions hold:
(a) if x
0
∈ F(u) is regular from the right, then, near x
0
,
u
+
(x) ≥ α

x − x
0
,ν

+
− β

x − x
0
,ν


−
+ o



x − x
0



, (1.8)
for some α>0, β
≥ 0 with equality along every nontangential domain in both
cases, and
α
≤ G(β); (1.9)
(b) if x
0
∈ F(u) is regular from the left, then, near x
0
,
u
+
(x) ≤ α

x − x
0
,ν

+

− β

x − x
0
,ν

−
+ o



x − x
0



, (1.10)
for some α
≥ 0, β>0 with equality along every nontangential domain in
both cases, and
α
≥ G(β). (1.11)
The conditions (a) and (b), where ν denotes the unit normal to ∂B at x
0
,towardsthe
positive phase, express the free boundary relation u
+
ν
= G(u
−

ν
) in a weak sense; accord-
ingly, we call u a weak solution of f.b.p.
Via an a pproximation argument it is possible to show that Theorem 1.1 holds for the
positive and negative parts of a solution of our f.b.p.
Here are our main results concerning the regularity of Lipschitz f ree boundaries.
Theorem 1.2. Let u be a weak solution to f.b.p. in Ꮿ
R
= 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, st rictly increasing and for some N>0, z
−N
G(z) is decreasing
in (0,+
∞).
Then, on B

R/2
, f is a C
1,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

A(x, u)∇u

, (1.12)
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 a l low a dependence on x and ν in the free boundary condition for G
= G(β, x,ν)
assuming instead of (iii) in Theorem 1.1
4 Boundary Value Problems
(iii

) G = G(z,ν,x) is continuous strictly increasing in z and, for s ome N>0indepen-
dent of ν and x, z

−N
G(z, ν, x) is decreasing in (0, ∞);
(iii

)logG is Lipschitz continuous with respect to ν, x, unifor mly with respect to its
first argument z
∈ [0,∞).
The proof of Theorem 1.2 goes along well-known guidelines and consists in the fol-
lowing three steps: to improve the Lipschitz constant of the level sets of u far from F(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 of F(u)indyadicballs
that corresponds to a C
1,γ
regularity of F(u)forasuitableγ.
The first step follows with some modifications [5, S ections 2 and 3] and ever ything
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
the L-harmonic measure. Section 3 contains the proof of the main result, Theorem 1.1,
and some consequences.
In Section 4 the 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 that ω(r)
≤ c
0
r
a
,0<a≤ 1. Let u ∈ H
1
loc
(Ω)beaweaksolution
of Lu
= 0inΩ, that is,

Ω

A(x)∇u( x),∇ϕ(x)

dx = 0, (2.1)
for every test function ϕ supported in Ω.IfL
∈ ᏸ(λ,Λ,ω), u ∈ C
1,a
(Ω).
In this section we prove that if the domain Ω is Lipschitz and u vanishes on a relatively
open portion F
⊂ ∂Ω, then, near F,thelevelsetsofu are uniformly Lipschitz surfaces.
Precisely, we consider domains of the form
T
s

=

x

,x
n

∈ R
: |x

| <s, f (x

) <x
n
< 2ls

, (2.2)
where f is a Lipschitz function with constant l.
Theorem 2.1. Let u be a positive solution to Lu
= 0 in T
4
, vanishing on F ={x
n
= f (x

)}∩
∂T
4
. Then, there exists η such that in
ᏺ

η
(F) =

f (x

) <x
n
<f(x

)+η

∩
T
1
, (2.3)
F. Ferra r i an d S. Sa lsa 5
u is increasing along the directions τ belonging to the cone Γ(e
n
,θ),withaxise
n
and opening
θ
= (1/2)cot
−1
l.Moreover,inᏺ
η
(F),
c
−1
u(x)

d
x
≤ D
n
u(x) ≤ c
u(x)
d
x
, (2.4)
where d
x
= dist(x,F).
Proof of Theorem 2.1. Let z be the solution of the Dirichlet problem
div

A(0)∇z(x)

=
0, T
2
,
z
= g, ∂T
2
,
(2.5)
where g 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>0inQ
ρ
,withρ = ρ(n, l). By rescaling the problem (if necessary),
we may assume ρ
= 3/2. Since z(e
n
) ≥ c>0, by Harnack inequality we have that, if y ∈ T
1
,
d
y
≥ η
0
,
z(y)
∼ c

η
0

, D
n
z(y) ∼
z(y)
d
y
∼ c

η

0

. (2.6)
Clearly z,u
∈ C
0,a
(T
2
). 
Lemma 2.2. For r>0,letw
r
be the C
1,a
(T
2
) weak solution to
div

A(rx)∇w
r
(x)

=
0, T
2
,
w
r
= z, ∂T
2

.
(2.7)
Then, given η
0
> 0,thereexistsr
0
= r
0
(η
0
), such that if r ≤ r
0
,
D
n
w
r
(y) ≥ 0 (2.8)
for every y
∈ T
1
,withd
y
≥ η
0
.
Proof. Let
div

A(rx)


∇
w
r
(x) −∇z(x)

=
div

A(rx) − A(0)

∇
z(x)

. (2.9)
For every σ>0, let
Q
σ
2
=

x ∈ T
2
,dist

x, ∂T
2

>σ


. (2.10)
Notice that h
r
= w
r
− z ∈ C
0,a
(T
2
), and moreover h
r
∈ C
1,a
(Q
σ
2
). Notice that ((A(rx) −
A(0))∇z(x))
i
∈ L
∞
(Q
σ
2
), and from standard estimates we have
sup
Q
σ
2



w
r
− z


≤
sup
∂Q
σ
2


h
r


+ C


Q
σ
2


1/n
ω(r)


∇

z


L
∞
(Q
σ
2
)
. (2.11)
6 Boundary Value Problems
Hence
sup
Q
σ
2
|h
r
|≤c

σ
a
+
r
σ

. (2.12)
Choosing r
= σ
1+a

wegetthatforeveryy ∈ Q
σ
2
,


h
r
(y)


≤
cr
β
, (2.13)
where β
= a/(1 +a).
Let
y ∈ T
1
,withd
y
≥ η
0
, r
0
≤ (1/3)η
1/(1+a)
0
,andρ = η

0
/3. It follows that
ρ


D
n
h
r
(y)


≤
C

r
a
z(y)+rz
L
∞
(B
ρ
(
y
))

≤
C

r

β
+ r

z(y)
= Cρ

r
β
+ r

z(y)
ρ
≤ Cρr
β
D
n
z(y).
(2.14)
Hence if r
≤ r
0
= min{(2c(η
0
))
−1/β
,(1/3)η
a+1
0
},weget
1

2
D
n
z(y) ≤ D
n
w
r
(y) ≤
3
2
D
n
z(y). (2.15)

The following two lemmas are similar to [5, Lemmas 2 and 3], respectively.
Lemma 2.3. Let η
0
> 0 be fixed and w and z as in Lemma 2.2. Then there exist r
0
= r
0
(η
0
)
and t
0
= t
0
(λ,Λ,n) > 1 such that, if r ≤ r
0

,
c
−1
w(y)
d
y
≤ D
n
w(y) ≤ c
w(y)
d
y
(2.16)
for every y
∈ T
1
, d
y
≥ t
0
η
0
.
Proof. The right-hand side inequality follows Schauder’s estimates and Harnack inequal-
ity. Let now y
∈ T
1
,withd
y
≥ t

0
η
0
, t
0
to be chosen. We may assume y = tη
0
e
n
.Fromthe
boundary Harnack principle (see, e.g., [8]or[9]) if
y = η
0
e
n
,then
z(
y) ≤ ct
−a
z(y) (2.17)
and, if t
≥ (2c)
1/a
≡ t
0
,then
z(
y) ≤
1
2

z(y). (2.18)
On the other hand, i f d
y
≥ t
0
η
0
and r ≤ r
0
(η
0
), from (2.6), (2.13), and (2.15)wehave
w(y)
≤
3
2
z(y), D
n
w(y) ≥
1
2
D
n
z(y). (2.19)
F. Ferra r i an d S. Sa lsa 7
Thus, if t
0
η
0
≤ d

y
≤ 10t
0
η
0
, applying Harnack inequality to D
n
z,weget
w(y)
≤
3
2
z(y)
≤ 3

z(y) − z(y)

=
3

1
0
d
ds
z

sy+(1− s)y

ds
≤ ct

0
η
0
D
n
z(y) ≤ cD
n
z(y)d
y
≤ cD
n
w(y)d
y
.
(2.20)
Repeating the argument with y = 10t
0
η
0
,wegetthat(2.18)holdsfor10t
0
η
0
≤ d
y
≤
20t
0
η
0

. After a finite number of steps, (2.18)followsford
y
≥ t
0
η
0
, y ∈ T
1
. 
Lemma 2.4. Let u be as in Theorem 2.1. Then the re exists a positive η, such that for every
x
∈ T
1
, d
x
≤ η,
D
n
u(x) ≥ 0. (2.21)
Moreover, in the same set
c
−1
u(x)
d
x
≤ D
n
u(x) ≤ c
u(x)
d

x
. (2.22)
Proof. Let t
0
be as in Lemma 2.3,andη
0
small to be chosen later. Set η
1
= 2η
0
t
0
.Itis
enough to show that if
x = η
1
re
n
and r ≤ r
0
(η
0
), then D
n
u(x) ≥ 0. Consider a small box
T
2r
and define u(y) = u(ry). Then u satisfies div(

A(x)∇u(x)) ≡ div ( A(rx)∇u(x)) = 0in

T
2
,where f is replaced by f
r
(y

) = f (ry

)/r.
We will show that D
n
u(y) > 0, where y = η
1
e
n
,bycomparingu with the function w
constructed in Lemma 2.2, normalized in order to get
u(y) = w(y). Notice that if we
choose r
0
= r
0
(η
0
)accordingtoLemma 2.3,wehave
c
−1
w(y)
d
y

≤ D
n
w(y) ≤ c
w(y)
d
y
. (2.23)
If d
y
≥ η
1
. From the comparison theorem (see [8]or[9]), we know that u/w ∈ C
0,a
(T
3/2
)
so that in B
η
0
(y)






u(y)
w(y)
− 1






≤
cη
a
0
, (2.24)
which implies



u(y) − w(y)


≤
cη
a
0
w(y) ≤ cη
a
0
w(y). (2.25)
Moreover, since η
0
∼ d
y
,



D
n
u(y) − D
n
w(y)


≤
cη
a−1
0
w(y) ≤ cη
a
0
D
n
w(y), (2.26)
from which we get
D
n
u(y) ≥

1 − cη
a
0

D
n
w(y), (2.27)

and (2.21)holdsifη
0
is sufficiently small. Inequality (2.22) is now a consequence of (2.23)
and the fact that w(
y) = u(y). 
8 Boundary Value Problems
To complete the proof of Theorem 2.1, it is enough to observe that the above lemmas
hold if we replace e
n
by any unit vector τ such that the angle between τ and e
n
is less than
θ
= 1/2cot
−1
l.
Thus, we obtain a cone Γ(e
n
,θ) of monotonicity for u.ApplyingTheorem 2.1 to the
positive and negative parts of the solution u of our free boundary problem, we conclude
that in a η-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 of u.
This is a consequence of the follow ing strong Harnack principle, where the cone
Γ

(e
n

,θ)isobtainedfromΓ(e
n
,θ) by deleting the “ bad” directions, that is, those in a
neighborhood of the generatrix opposite to
∇u(e
n
). Precisely, if τ ∈ Γ(e
n
,θ), denote by
ω
τ
the solid angle between the planes span{e
n
,∇} and span{e
n
,τ}.DeletefromΓ(e
n
,θ)
the directions τ such that (say) ω
τ
≥ (99/100)π and call Γ

(e
n
,θ) the resulting set of di-
rections. If τ
∈ Γ

(e
n

,θ), then
∇,τ≥c
3
δ, (2.28)
where δ
= π/2 − θ.Wecallδ the defect angle.
Lemma 2.5. Suppose u is a positive solution of div(A(rx)
∇u(x)) = 0 in T
4
vanishing on
F
={x
n
= f (x

)}, increasing along every τ ∈ Γ(e
n
,θ). Assume furthermore that (2.4)holds
in T
4
. The re exist p ositive r
0
and h, depending only on n, l,andλ, Λ, such that if r ≤ r
0
,for
eve ry small vector τ, τ
∈ Γ

(e
n

,θ/2),andforeveryx ∈ B
1/8
(e
n
),
sup
B
(1+hδ)
(x)
u(y − τ) ≤ u(x) − Cδu

e
n

, (2.29)
where
 =|τ|sin(θ/2).
For the proof see [5,Section3].
Corollary 2.6. In B
1/8
(x
0
), u is inc reasing along every τ ∈ Γ(τ
1
,θ
1
) with
δ
1
≤ bδ,


δ
1
=
π
2
− θ
1

,


ν
1
− e
1


≤
Cδ,
(2.30)
where
b = 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
2
min


r
0
,η

, (2.31)
with η as in Theorem 2.1 and
r
0
as in Lemma 2.5.Ifwedefine
u
s
(x) =
u(sx)
s
, (2.32)
F. Ferra r i an d S. Sa lsa 9
then u
+
s
satisfies L
s
u
+
s
≡ Lu
s
(sx) = 0inT
4
and falls under the hypothesis of Lemma 2.5.

Therefore, rescaling back we get the following result.
Theorem 2.7. Let u be a solution of our free boundary problem. Then in B
s/8
(se
n
),
sup
B
(1+hδ)
(x)
u(y − τ) ≤ u(x) − cδu

e
n

(2.33)
for every τ
∈ Γ

(e
n
,θ/2), |τ|s. As a consequence, in B
s/8
(se
n
), u is monotone along every
τ
∈ Γ(ν
1
,θ

1
),whereν
1
, θ
1
satisfy (2.30).
3. Proof of the main theorem
Before proving Theorem 1.1, we need to introduce some notations and to recall a point-
wise characterization of weak subsolutions.
If ᏻ
⊂ Ω is an open set, regular for the Dirichlet problem, we denote by G
ᏻ
= G
ᏻ
(x, y)
the Green function associated with the operator L in ᏻ and by ω
x
ᏻ
the L-harmonic mea-
sure for L in ᏻ. In this way,
w(x)
=

ᏻ
gdω
x
ᏻ
−

ᏻ

G
ᏻ
(x, y)h(y)dy (3.1)
is the u nique weak solution of Lu
= h in ᏻ, h = 0on∂ᏻ.
A function v
∈ H
1
(Ω)isaweaksubsolutioninΩ if

Ω

A(x)∇u( x),∇φ(x)

dx ≤ 0 (3.2)
for every nonnegative test function ϕ 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]forthedetails,we
say that a function v : Ω
→ R is L-subharmonic in a set Ω if it is upper semicontinuous in
Ω, locally upper bounded in Ω,and
(S) for every x
0
∈ Ω there exists a basis of regular neighbor hood Ꮾ
x
0
associated with
v such that for every B
∈ Ꮾ

x
0
,
v

x
0

≤

∂B
v(σ)dω
x
0
B
. (3.3)
A function v is L-superharmonic if
−v is L-subharmonic. Thus u is L-harmonic, or sim-
ply harmonic, whenever it is both L-subharmonic and L-superharmonic.
With such pointwise characterization, the definition of the Perron-Wiener-Brelot so-
lution of the D irichlet 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, if v is locally Lips-
chitz, v is L-subharmonic if and only if v is locally a subsolution.
10 Boundary Value Problems
Precisely, see [12, 13], if f is the trace on ∂Ω of a function

f ∈ C(Ω) ∩ H
1

(Ω), then
the weak solution of the Dirichlet problem (even if L has just bounded measurable coef-
ficients)
Lu
= 0inΩ,
u
= f on ∂Ω
(3.4)
and the parallel Perron-Wiener-Brelot one coincide. Moreover, in [15]Herv
´
ealsoproved
that the same result holds when

f is L-subharmonic and

f ∈ H
1
loc
(Ω).
Lemma 3.1. Let C>2 and φ be a C
2
weak solution of
div

A(x)∇φ(x)

≥
C



∇
φ(x)


2+ω
2
0
φ(x)
≡ Φ(x) (3.5)
in Ω, 0 <φ
min
≤ φ ≤ φ
max
. Then for any x ∈ Ω there exists a positive number r(x,φ
max
,φ
min
,
C) such that, for every r<
r(x) and every ball B
r
= B
r
(x) ⊂ Ω,

∂B
r


σ − x,∇φ(x)


+
1
2

Ᏸ
2
φ(x)(σ − x),(σ − x)

−
Φ(σ)

dω
x
B
(σ) ≥ 0. (3.6)
Proof. From Lemma A.5,foreveryballB
r
= B
r
(x) ⊂ Ω,
∇φ(x)

∂B
r
(σ − x)dω
x
B
(σ)+
1

2
n

i, j=1
D
ij
φ(x)

∂B
r

σ
i
− x
i

σ
j
− x

dω
x
B
(σ)
≥

B
r
G
B

r
(x, y)Φ(y)dy+ o

r
2

,
(3.7)
the proof follows easily.

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)

, (3.8)
for some η(x), where
|η(x)|=1. To prove that v
φ
is an L-subsolution we just check con-
dition (S), since by str aightforward calculations v
φ
is locally Lipschitz continuous. In par-
ticular we will prove that for every x
∈ Ω
+
(v) there exists a positive constant r

0
= r
0
(x)
such that for every ball B
r
≡ B
r
(x) ⊂ Ω
+
(v), r ≤ r
0
,andforeveryx
0
∈ B
r
,

∂B
r
v
φ
(σ)dω
x
0
B
r
(σ) ≥ v
φ


x
0

. (3.9)
Let
{e
1
, ,e
n
} be an orthonormal basis of R
n
where e
n
= η(x)andletξ be the following
vectorfield:
ξ(h)
= e
n
+
n−1

i=1

V
i
,h

e
i
, (3.10)

F. Ferra r i an d S. Salsa 11
where
{V
1
, ,V
n−1
}⊂R
n
will be chosen later. Let ν(h) = ξ(h)/|ξ(h)|,sothat
ν(h)
= e
n
+
n−1

i=1

V
i
,h

e
i
−
1
2
n−1

i=1


V
i
,h

2
e
n
+ o

|
h|
2

. (3.11)
Let x
0
∈ B
r
(x)andh = σ − x
0
. Then, letting
φ
0
= φ

x
0

, ∇φ
0

=∇φ

x
0

, Ᏸ
2
φ
0
= Ᏸ
2
φ

x
0

, (3.12)
we have
φ(σ)
= φ
0
+

∇
φ
0
,h

+
1

2

Ᏸ
2
φ
0
h,h

+ o

|
h|
2

(3.13)
as h
→ 0, uniformly in a neighborhood of x. As a consequence,
σ + φ(σ)ν

σ − x
0

=
y
∗
+ J
1
+ J
2
+ J

3
, (3.14)
where y
∗
= x
0
+ φ(x
0
)e
n
,
J
1
=


∇
φ
0
,h

e
n
+ h + φ
0
n
−1

i=1


V
i
,h

e
i

,
J
2
=


∇
φ
0
,h

n−1

i=1

V
i
,h

e
i
+
1

2

Ᏸ
2
φ
0
h,h

e
n
−
φ
0
2
n−1

i=1

V
i
,h

2
e
n

,
J
3
= o


|
h|
2

(3.15)
uniformly as h
→ 0.
Let J
= J
1
+ J
2
+ J
3
.Thenforeveryσ ∈ ∂B
r
(x
0
),
v(σ)
≥ u

y
∗
+ J

=
u


y
∗

+

∇
u

y
∗

,J

+
1
2

Ᏸ
2
u

y
∗

J,J

+ o

|
h|

2

, (3.16)
as h
→ 0, uniformly in a neighborhood of y
∗
.Wehave

∇
u

y
∗

,J
1

=


∇
u

y
∗




h,e

n

+

h,∇φ
0

,

∇
u

y
∗

,J
2

=


∇
u

y
∗





−
φ
0
2
n−1

i=1

V
i
,h

2
+
1
2

Ᏸ
2
φ
0
h,h


.
(3.17)
12 Boundary Value Problems
As a consequence,

∂B

r
v(σ)dω
x
0
B
r
(σ) ≥ v

x
0

+


∇
u

y
∗




∂B
r


h,∇φ
0


−
φ
0
2
n−1

i=1

V
i
,h

2
+
1
2

Ᏸ
2
φ
0
h,h


dω
x
0
B
r
+


∂B
r


∇
u

y
∗

,h

+
1
2

Ᏸ
2
u

y
∗

J,J

+ o




h
2




dω
x
0
B
r
= v

x
0

+


∇
u

y
∗




∂B
r



h,∇φ
0

−
φ
0
2
n−1

i=1

V
i
,h

2
+
1
2

Ᏸ
2
φ
0
h,h


dω

x
0
B
r
+
1
2

∂B
r

Ᏸ
2
u

y
∗

J
1
,J
1

dω
x
0
B
r
+ ∇u


y
∗


∂B
r
hdω
x
0
B
r
+ o

r
2

,
(3.18)
uniformly with respect to x
0
in a neighborhood of x.
Let

∂B

Ᏸ
2
u

y

∗

J
1
,J
1

dω
x
0
B
r
=
n

i, j
D
ij
u

y
∗


∂B
r
a
i
a
j

dω
x
0
B
r
(3.19)
with
a
i
= φ
0

V
i
,h

+

h,e
i

, i = 1, ,n, (3.20)
where the V
i
are still to be chosen, and
a
n
=

∇

φ
0
,h

+

h,e
n

. (3.21)
For i
= 1, ,n and j = 1, ,n,let
d
ij
= d
ij

x
0
,x
0

=

∂B
r
h
i
h
j

dω
x
0
B
r
(x
0
)
, d
∗
ij
= d
ij

y
∗
, y
∗

=

∂B
r
h
i
h
j
dω
y
∗

B
r
(y
∗
)
(3.22)
be the entries, of the matr ix of the moments (see the appendix), respectively, evaluated in
x
0
and y
∗
.
For i
= 1, ,n,and j = 1, ,n − 1, let
m
ij
=

∂B
r
a
i
a
j
dω
x
0
B
r
,

m
nn
=
n

p,q=1
D
p
φ
0
D
q
φ
0
d
pq
+2
n

p=1
D
p
φ
0
d
pn
+ d
nn
.
(3.23)

F. Ferra r i an d S. Salsa 13
Then
m
ij
=
n

p,q=1
φ
2
0
V
p
i
V
q
j
d
pq
+ φ
0
n

p=1
V
p
i
d
jp
+ φ

0
n

q=1
V
q
j
d
iq
+ d
ij
, (3.24)
m
in
= m
ni
=

∂B
r


∇
φ
0
,h

+ h
n



φ
0
n

p=1
V
p
i
h
p
+ h
i

dω
x
0
B
r
(σ)
= φ
0
n

p,q=1
V
p
i
D
p

φ
0
d
pq
+
n

p=1
D
q
φ
0
d
pi
+ φ
0
n

p=1
V
p
i
d
pn
+ d
in
.
(3.25)
SupposenowwecanfindV
1

, ,V
n−1
and a real number κ
0
,suchthatforeveryi =
1, ,n − 1andforeveryj = 1, ,n,
m
i, j
=

1+κ
0

d
∗
i, j
, m
nn
=

1+κ
0

d
∗
nn
.
(3.26)
Then
n−1


i, j=1
D
ij
u

y
∗

m
ij
+2
n−1

i=1
D
in
u

y
∗

m
in
+ D
nn
u

y
∗


m
nn
=

1+κ
0

n

i, j=1
D
ij
u

y
∗

d
∗
ij
.
(3.27)
In particular this means that V
1
, ,V
n−1
,andk
0
must solve the following system, for

i
= 1, ,n − 1andj = 1, ,n − 1,
φ
0
n

p
V
p
i
d
jp
+ φ
0
n

q=1
V
q
j
d
iq
+ φ
2
0
n

p,q=1
d
p,q

V
p
i
V
q
j
=−d
i, j
+

1+κ
0

d
∗
i,1
,
φ
0
n

p=1
V
p
i
d
pn
+ φ
0
n


p,q=1
V
p
i
D
p
φ
0
d
pq
=−d
i,n
+
n

p=1
D
q
φ
0
d
ip
+

1+κ
0

d
∗

i,n
,
n

p,q=1
D
p
φ
0
D
q
φ
0
d
pq
+2
n

p=1
D
q
φ
0
d
pn
+ d
nn
=

1+κ

0

d
∗
n,n
.
(3.28)
From the last equations and Lemma A.3, since d
∗
nn
>cλr
2
,forsmallr and |∇φ
0
|,there
exists a positive constant C
= C(λ,Λ)suchthat


κ
0


≤
C


∇
φ
0



+
d
nn
− d
∗
nn
d
∗
n,n
≤ C



∇
φ
0


+ φ
max

. (3.29)
We now start an iteration process to solve the above system (see [4, 6]).
14 Boundary Value Problems
Let (ᐂ
i
)
0

= 0, i = 1, ,n − 1, and for l ≥ 0, define recursively (ᐂ
i
)
(l+1)
as the solution
of the system (i
= 1, ,n − 1; j = 1, ,n − 1):
φ
0
n

p

ᐂ
p
i

(l+1)
d
jp
+ φ
0
n

q=1

ᐂ
p
j


(l)
d
iq
+ φ
2
0
n

p,q=1
d
p,q

ᐂ
p
i

(l)

ᐂ
q
j

(l)
=−d
i, j
+

1+κ
0


d
∗
i,1
,
φ
0
n

p=1

ᐂ
p
i

(l+1)
d
pn
+ φ
0
n

p,q=1

ᐂ
p
j

(l)
D
p

φ
0
d
pq
=−d
i,n
+
n

p=1
D
q
φ
0
d
ip
+

1+κ
0

d
∗
i,n
.
(3.30)
Notice that the sequence is well defined, since the matrix D(x
0
,x
0

) is nonsingular (Lemma
A.3 in the appendix). Moreover, if
|∇φ(x
0
)| is kept small, denoting by d
i
and d
∗
i
the vec-
tors (d
i1
, ,d
in
)and(d
∗
i1
, ,d
∗
in
), from the estimates in Lemma A.3,weget,byinduction,



ᐂ
(l+1)
i




≤
C



d
i
− d
∗
i


r
2
φ
0
+


∇
φ
0


φ


d
∗
i



r
2
+ φ
a
0

≤
Cφ
−1
0

φ
0
+


∇
φ
0



(3.31)
with C
= C(n,Λ,λ). Since the sequences (ᐂ
(l)
i
)

l∈N
are bounded for every i ∈{1, ,n − 1 },
there exist subsequences (that we still call) (ᐂ
(l)
i
)
l∈N
,convergingtoᐂ
i
with


ᐂ
i


≤
C(n,Λ,λ)φ
−1
0

φ
0
+


∇
φ
0




. (3.32)
Now, from (3.18), (1.11), and Lemma A.5 ,weget

∂B
v(σ)dω
x
0
B
r
(σ) ≥ v

x
0

+


∇
u

y
∗




∂B
r



h,∇φ
0

−
φ
0
2
n−1

i=1

V
i
,h

2
+
1
2

Ᏸ
2
φ
0
h,h


dω

x
0
B
r
+

∂B
r

∇
u

y
∗

,h

dω
x
0
B
r
+
1
2

∂B
r

Ᏸ

2
u

y
∗

J
1
,J
1

dω
x
0
B
r
+ o

r
2

=
v

x
0

+



∇
u

y
∗




∂B
r


h,∇φ
0

−
φ
0
2
n−1

i=1

V
i
,h

2
+

1
2

Ᏸ
2
φ
0
h,h


dω
x
0
B
r
+

∂B
r

∇
u

y
∗

,h

dω
x

0
B
r
+
1+κ
0
2

∂B
r
(y
∗
)

Ᏸ
2
u

y
∗

h
∗
,h
∗

dω
y
∗
B

r
(y
∗
)
+ o

r
2

=
v

x
0

+ ∇u

y
∗



∂B
r
hdω
x
0
B
r
−


1+κ
0


∂B
r
(y
∗
)
h
∗
dω
y
∗
B
r
(y
∗
)

+


∇
u

y
∗





∂B
r


h,∇φ
0

−
φ
0
2
n−1

i=1

V
i
,h

2
+
1
2

Ᏸ
2
φ

0
h,h


dω
x
0
B
r
+ o

r
2

.
(3.33)
Consider
T
=


∂B
r
hdω
x
0
B
(σ) −

1+κ

0


∂B
r
(y
∗
)
hdω
y
∗
B
r
(y
∗
)
(σ)

. (3.34)
F. Ferra r i an d S. Salsa 15
From Lemma A.3 and (3.29), we get
|T|≤Kr
2
. (3.35)
Thus, from (3.32), we deduce that

∂B
r
v(σ)dω
x

0
B
(σ)
≥ v

x
0

+


∇
u

y
∗




∂B
r


h,∇φ
0

−
φ
0

2
n−1

i=1

V
i
,h

2
+
1
2

Ᏸ
2
φ
0
h,h

−
Kr
2

dω
x
0
B
r
≥v


x
0

+


∇
u

y
∗




∂B
r


h,∇φ
0

+
1
2

Ᏸ
2
φ

0
h,h

−
C



∇
φ
0


2
+ φ
2
0

φ
0
r
2
− Kr
2

dω
x
0
B
r

.
(3.36)
From Lemma 2.5,ifr is small, and C large depending on x
0
and φ
0
,wehave

∂B
r


h,∇φ
0

+
1
2

Ᏸ
2
φ
0
h,h

−
C




∇
φ
0


2
+ φ
2
0

φ
0
r
2
− Kr
2

dω
x
0
B
r
≥ 0, (3.37)
so that v
φ
is a weak L-subsolution in its positivity set. 
Remark 3.2. We emphasize that the construction of the vectors V
i
, i = 1, ,n − 1, involves
only the Lipschitz continuity of A.

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 vector τ and the function
v
τ
(x) = sup
B
φ(x)
(x)
u(y − τ) = sup
|ν|=1
u

x − τ + φ(x)ν

. (4.1)
The proof of Theorem 1.1 holds, with minor changes, also in this case. In particular the
following result holds.
Corollary 4.1. Let u be a continuous function in Ω. Assume that in
{u>0} u is a C
2
-weak
solution of Lu
= 0, L ∈ ᏸ(λ,Λ,ω). For any vector τ let φ beapositiveC
2
-function such that
0 <φ
min
≤ φ ≤ φ

max
,
v
τ
(x) = sup
B
φ(x)
(x)
u(y − τ) = sup
|ν|=1
u

x − τ + φ(x)ν

, (4.2)
is well defined in Ω. There exist positive constants ρ
0
, μ
0
= μ
0
(n,λ,Λ) and C = C(n,λ,Λ),
such that if
|∇φ|≤μ
0
, |τ| <ρ
0
, ω
0
= ω(φ

max
),and
φLφ
≥ C



∇
φ(x)


2
+ ω
2
0

, (4.3)
then v
τ
is a weak subsolution of Lu = 0 in {v
τ
> 0}.
16 Boundary Value Problems
Remark 4.2. The key point in Corollary 4.1 is that the estimates (3.29)and(3.32)for
the vectors V
i
, i = 1, ,n − 1, and k
0
depend only on the distance between the matrices
D( x

0
,x
0
)andD(y
∗
, y
∗
).
We now construct a family of r adii, with the right properties to be used in the final
comparison theorem.
Let D
= B
2
(0)\B
1/8
(e
n
). We may assume with out loss of generality that A(0) = I and
that
sup
B
3


A(x) − I


≤
ω
1

 1. (4.4)
By a slig ht modification of [5, Lemma 7] we can construct a family of functions satisfying
the properties expressed in t he following lemma.
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
∈ C
2
(D) and
(i) 0 < 1
− ω ≤ φ
t
≤ 1+μt,
(ii) φ
t
≤ 1 − ω in B
2
\B
5/3
,
(iii) φ
t
≥ 1 − ω + ημt in B
1/2
,
(iv)

|∇φ
t
|≤c(μt + ω),
(v)
φ
t
Lφ
t
≥

C



∇
φ
t


2
+ ω

maxφ
t

2

. (4.5)
We are now in position to prove Theorem 1.2.
Proof of Theorem 1.2. We first observe that Theorem 1.1 (and Corollary 4.1)holdsfor

weak solutions, not necessarily C
2
.Infactletu
±
j
be the functions constructed as solu-
tions of the following problems:
L
j
u
±
j
= 0inΩ
±
(u),
u
±
j
= u
j
on Ω
±
(u),
(4.6)
and set u
j
= u
+
j
− u

−
j
.Thenu
j
converges locally in C
1,a
(Ω
±
(u)) to u and it is not difficult
to check that (suppressing for clarit y the index t)
v
j
(x) = sup
B
φ(x)
(x)
u
j
(4.7)
converges locally in C
1,a
(Ω
±
(u) ∩ D)to
v
φ
(x) = sup
B
φ(x)
(x)

u. (4.8)
From Theorem 1.1, v
j
isaweaksubsolutionforL
j
in Ω
±
(u
j
) ∩ D.Butthenv
φ
isaweak
L-subsolution in Ω
±
(u) ∩ D.
With this result at hand, the proof goes as in [5, Section 7]. Indeed, the particular
form of the operator does not play any role anymore. Actually observe that if φ
t
satisfies
F. Ferra r i an d S. Salsa 17
inequality (4.5)also
φ
t
satisfies the s ame 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
B
4
−k

u is monotone along every τ ∈ Γ(ν
k
,θ
k
)with
δ
k+1
≤ bδ
k

δ
0
= δ,δ
k
=
π
2
− θ
k

,


ν
k+1
− ν
k


≤

cδ
k
. (4.9)
These conditions imply that F(u)isC
1,γ
, γ = γ(b), at the origin. 
Proof of Corollary 1.3. Since F(u) is Lipschitz, u is H
¨
older continuous in Ꮿ
1
. We only need
to show that u is Lipschitz in Ꮿ
2/3
across the free boundar y. This follows from a simple
application of the monotonicity formula in [16, Lemma 1] and a barrier argument. Pre-
cisely, let x
0
∈ Ω
+
(u) ∩ Ꮿ
2/3
, d
0
= dist(x
0
,F(u)), and u(x
0
) = λ. From Harnack inequality
u(x)
∼ λ (4.10)

in B
d
0
/2
(x
0
). Let w be the solution of
div

A(x, u)∇w

=
0 (4.11)
in B
d
0
(x
0
)\B
d
0
/2
(x
0
)suchthatw = 0on∂B
d
0
(x
0
), w = λ on ∂B

d
0
/2
(x
0
). By maximum prin-
ciple
u
≥ cw in B
d
0

x
0

\
B
d
0
/2

x
0

(4.12)
and, from the C
a
nature of A and C
1,a
estimates, if y

0
∈ ∂B
d
0
(x
0
) ∩ F(u),
w(x)
≥ c
λ
d
0

x − y
0
,ν

+
(4.13)
with ν
= (x
0
− y
0
)/|x
0
− y
0
|. Thus, near y
0

, u has the asymptotic behavior
u(x)
≥ α

x − y
0
,ν

+
− β

x − y
0
,ν

−
+ o



x − y
0



(4.14)
with
c
λ
d

0
≤ α ≤ G(β). (4.15)
Then, the monotonicity formula gives
λ
d
0
G
−1

c
λ
d
0

≤
Cu
2
L
∞
(Ꮿ
3/4
)
(4.16)
so that, from interior estimates,


∇
u
+


x
0



G
−1



∇
u
+

x
0




≤
C
1
u
2
L
∞
(Ꮿ
3/4
)

. (4.17)
18 Boundary Value Problems
This gives the Lipschitz continuity of u
+
. Similarly, we get
G



∇
u
−

x
0






∇
u
−

x
0




≤
C
1
u
2
L
∞
(Ꮿ
3/4
)
(4.18)
and the proof is complete.

Appendix
Auxiliary lemmas
WecollectheresomeestimatesontheL-harmonic measure and its moments that are
used in the paper. Here ω(r)
≤ c
0
r
a
,0<a≤ 1.
Definit ion A.1. For any x
0
,x, y ∈ Ω,andr>0, B
r
(x
0
) ⊂ Ω,letd
i

(x
0
, y)be,fori = 1, ,n,
d
i
(x
0
, y) =

∂B
r
(x
0
)

σ
i
− x
0i

dω
y
B
r
(x
0
)
(σ)(A.1)
and let d
ij

(x
0
, y)be,foreveryi, j,1≤ i, j ≤ n,
d
ij

x
0
, y

=

∂B
r
(x
0
)

σ
i
− x
0i

σ
j
− x
0j

dω
y

B
r
(x
0
)
(σ). (A.2)
We call, respectively, (d
i
(x
0
, y))
1≤i≤n
the vector of the first moment of the L-harmonic
measure in B
r
(x), and D(x
0
, y) = (d
ij
(x
0
, y))
1≤i, j≤n
the matrix of the second moment of
the L-harmonic measure in B
r
(x).
Denote by L
0
= div(A(x

0
)∇)andbyG
0
r
= G
0
r
(x, y) the Green function for L
0
in B
r
=
B
r
(x
0
). We have the following.
Lemma A.2. Let L
0
w
r
=−1 in B
r
(x
0
), w
r
= 0 on ∂B
r
(x

0
). Then
w
r

x
0

=
r
2
2TrA

x
0

. (A.3)
Proof. Suppose x
0
= 0. Let g
ij
(x) = x
i
x
j
and let v
ij
be the solution of L
0
v

ij
= 0inB
r
,
v
ij
= g
ij
on ∂B
r
.SinceL
0
g
ij
= 2a
ij
(0) and g
ij
(0) = 0, we have
v
ij
(0) = 2a
ij
(0)w
r
(0). (A.4)
On the other hand,

n
i

=1
v
ii
(0) = r
2
and (A.3)follows. 
Lemma A.3. Let B
2r
(x
0
) ⊂ Ω.Then:
(1) for every i
= 1, ,n,
sup
B
r
(0)


d

x
0
, y

−
y
i



≤
C(λ,Λ,n)r
1+a
;(A.5)
(2) for every i, j
= 1, ,n,


d
ij

x
0
,x
0

−
2w
r

x
0

a
ij

x
0




≤
Cr
2+a
,(A.6)
where w
r
is defined in Lemma A.2.
F. Ferra r i an d S. Salsa 19
Proof. Let x
0
= 0and
d
i
(y) = d
i
(0, y) =

∂B
r
σ
i
dω
y
B
r
(σ). (A.7)
Then d
i
(y) − y

i
= 0on∂B
r
and
L
0

d
i
(y) − y
i

=
div

A(0) − A(y)

e
i

in B
r
. (A.8)
From standard estimates, we get


d
i
− y
i



L
∞
(B
r
)
≤ Cr



A(0) − A(y)

e
i


L
∞
(B
r
)
≤ Cr
1+a
. (A.9)
Consider now
d
ij
(y) = d
ij

(0, y) =

∂B
r
σ
i
σ
j
dω
y
B
r
(σ). (A.10)
If v
ij
is as in Lemma 2.2,wehaveh
ij
− v
ij
= 0on∂B
r
and
L
0

d
ij
(y) − v
ij
(y)


=

A(0) − A(y)

∇
v
ij

in B
r
. (A.11)
Therefore,


d
ij
− v
ij


L
∞
(B
r
)
≤ Cr




A(0) − A(y)

∇
v
ij


L
∞
(B
r
)
≤ C


∇
v
ij


L
s
(B
r
)
r
1+a
≤ Cr
2+a
.

(A.12)
Hence, from Lemma 2.2,weget


d
ij
(0) − 2a
ij
(0)w
r
(0)


=


d
ij
(0) − v
ij
(0)


≤
Cr
2+a
. (A.13)

Corollary A.4. For r ≤ r
0

(n,λ,Λ,a), the matrix (d
ij
(0)/r
2
) is nonsingular.
Lemma A.5. Let w be a weak solution of
div

A(x)∇w(x)

=
f (A.14)
in Ω,where f is continuous. Then for every x
∈ B
r
(x
0
) ⊂ Ω,
∇w(x) ·

∂B
r
(x
0
)
(σ − x)dω
x
B
r
(x

0
)
(σ)+

B
r
(x
0
)
G
B
r
(x
0
)
(x, y) f (y)dy = R

x
0
,x

, (A.15)
where
R

x
0
,x

=


∂B
r

x
0



1
0

∇
w

x + s(σ − x)

−∇
w(x), σ − x

ds

dω
x
B
r
(x
0
)
(σ). (A.16)

20 Boundary Value Problems
Moreover, if u
∈ C
2
(Ω),
∇w(x) ·

∂B
r
(x
0
)
(σ − x)dω
x
B
r
(x
0
)
(σ)+
1
2

∂B
r
(x
0
)

D

2
w(x)(σ − x),(σ − x)

dω
x
B
r
(x
0
)
(σ)
+

B
r
(x
0
)
G
B
r
(x
0
)
(x, y) f (y)dy = o

r
2

.

(A.17)
Proof. Let div(A(x)
∇w(x)) = f ,thenw ∈ C
1,a
(Ω)andforanyσ,x ∈ B
r
(x
0
) ⊂ Ω,
w(σ)
= w(x)+

1
0

∇
w

x + s(σ − x)

,σ − x

ds = w(x)+

∇
w(x), σ − x

+

1

0

∇
w

x + s(σ − x)

−∇
w(x), σ − x

ds.
(A.18)
On the other hand,
w(x)
=

∂B
r
(x
0
)
w(σ)dω
x
B
r
(x
0
)
(σ) −


B
r
(x
0
)
G
B
r
(x
0
)
(x, y) f (y)dy, (A.19)
hence
∇w(x) ·

∂B
r
(x
0
)
(σ − x)dω
x
B
r
(x
0
)
(σ)+

B

r
(x
0
)
G
B
r
(x
0
)
(x, y) f (y)dy = R

x
0
,x

. (A.20)
The rest of the proof follows from Taylor expansion.

Corollary A.6. Let u ∈ C
2
(Ω) beaweaksolutionof
div

A(x)∇u( x)

=
0 (A.21)
in Ω. Then
∇u


x
0

·

∂B
r
(x
0
)

σ − x
0

dω
x
0
B
r
(x
0
)
(σ) = O

r
2

. (A.22)
Proof. It is enough to observe that


∂B
r
(x
0
)

σ
i
− x
0i

σ
j
− x
0j

dω
x
0
B
r
(x
0
)
(σ) = O

r
2


. (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:
Sandro Salsa: Dipartimento di Matematica, Politecnico di Milano, Via Bonardi 7,
20133 Milano, Italy
Email address: