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, and let u be a continuous weak solution of Lu = 0 in {u (cid:2)= 0}. In this paper, we show that if φ satisfies a suitable differential inequality, then vφ(x) = sup Bφ(x)(x)u is a subsolution of Lu = 0 away from its zero set. We apply this result to prove C1,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 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 indeed C1,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 type F(D2u), and again Feldman [4] fully nonlinear operators, not necessary concave, of the type F(D2u,Du). In [5], Cerutti et al. consider variable coefficients operators in nondivergence form and Ferrari [6] a class of fully non- linear operators F(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 than C1,α, e.g.) is not included in the above results and it is
2 Boundary Value Problems
(cid:3)
(cid:2)
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 (cid:2)(λ,Λ,ω) of elliptic operators
(1.1) A(x)∇ L = div
defined in a domain Ω ⊂ Rn, with symmetric and uniformly elliptic matrix, that is,
(1.2) A(x) = A(cid:5)(x), λI ≤ As(x) ≤ ΛI
and modulus of continuity of the coefficients given by
(cid:4) (cid:4)A(x) − A(y)
(cid:4) (cid:4).
(1.3) ω(r) = sup |x−y|≤r
Theorem 1.1. Let u be a continuous function in Ω. Assume that in {u > 0} u is a C2-weak solution of Lu = 0, L ∈ (cid:2)(λ,Λ,ω), ω(r) ≤ c0r. Let φ be a positive C2-function such that 0 < φmin ≤ φ ≤ φmax and
(cid:3) (cid:2) x + φ(x)ν
(1.4) u u = sup |ν|=1 vφ(x) = sup Bφ(x)(x)
(cid:6) ,
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
(cid:5)(cid:4) (cid:4)∇φ(x)
(cid:4) (cid:4)2 + ω2
0
(1.5) φLφ ≥ C
then v is a weak subsolution of Lu = 0 in {v > 0}.
R(0) × (−R,R) we are given a
R(0) be the ball of radius R in Rn−1. In (cid:3)R = B(cid:9) loc function u satisfying the following.
= B(cid:9) Let B(cid:9) R continuous H 1 (i)
(cid:3)
= 0
We now introduce the class of free boundary problems we are going to study and the appropriate notion of weak solution.
(cid:2) A(x)∇u
(1.6) Lu = div
in Ω+(u) = {x ∈ (cid:3)R : u(x) > 0}, and in Ω−(u) = {x ∈ (cid:3)R : u(x) ≤ 0}0, in the weak sense. We call F(u) ≡ ∂Ω+(u) ∩ (cid:3)R the free boundary. We say that a point x0 ∈ F(u) is regular
(cid:3) ,
⊂ Ω−(u), resp. (cid:7) (cid:8) . x0
from the right (left) if there exists a ball B: (cid:2) B ⊂ Ω+(u) (1.7) B ∩ F(u) =
F. Ferrari and S. Salsa 3
(cid:10)−
(cid:4) (cid:3) (cid:4)
(ii) Along F(u) the following conditions hold: (a) if x0 ∈ F(u) is regular from the right, then, near x0,
(cid:10)+ − β
(cid:9) x − x0,ν u+(x) ≥ α
(cid:9) x − x0,ν
(cid:2)(cid:4) (cid:4)x − x0
, (1.8) + o
for some α > 0, β ≥ 0 with equality along every nontangential domain in both cases, and
(1.9) α ≤ G(β);
(cid:10)−
(cid:4) (cid:3) (cid:4)
(b) if x0 ∈ F(u) is regular from the left, then, near x0,
(cid:10)+ − β
(cid:9) x − x0,ν u+(x) ≤ α
(cid:9) x − x0,ν
(cid:2)(cid:4) (cid:4)x − x0
, (1.10) + o
for some α ≥ 0, β > 0 with equality along every nontangential domain in both cases, and
(1.11) α ≥ G(β).
ν = G(u−
× (−R,R).
The conditions (a) and (b), where ν denotes the unit normal to ∂B at x0, towards the ν ) in a weak sense; accord- positive phase, express the free boundary relation u+ ingly, we call u a weak solution of f.b.p. Via an approximation 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 free boundaries.
Theorem 1.2. Let u be a weak solution to f.b.p. in (cid:3)R = B(cid:9) R
R/2, f is a C1,γ function with γ = γ(n,l,N,λ,Λ,ω).
is a Lipschitz continuous function with Suppose that 0 ∈ F(u) and that (i) L ∈ (cid:2)(λ,Λ,ω); (ii) Ω+(u) = {(x(cid:9),xn) : xn > f (x(cid:9))} where f Lip( f ) ≤ l; (iii) G = G(z) is continuous, strictly increasing and for some N > 0, z−N G(z) is decreasing in (0,+∞). Then, on B(cid:9)
By using of the monotonicity formula in [7] we can prove the following.
(cid:3) ,
Corollary 1.3. In f.b.p. let
(cid:2) A(x,u)∇u
(1.12) 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 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(cid:9)) G = G(z,ν,x) is continuous strictly increasing in z and, for some N > 0 indepen- dent of ν and x, z−N G(z,ν,x) is decreasing in (0, ∞); (iii(cid:9)(cid:9)) logG is Lipschitz continuous with respect to ν, x, uniformly 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) in dyadic balls that corresponds to a C1,γ regularity of F(u) for a suitable γ.
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 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.
loc(Ω) be a weak solution
2. Monotonicity properties of weak solutions
(cid:11)
In this section we assume that ω(r) ≤ c0ra, 0 < a ≤ 1. Let u ∈ H 1 of Lu = 0 in Ω, that is,
(cid:9) A(x)∇u(x), ∇ϕ(x)
(cid:10) dx = 0,
Ω
(2.1)
(cid:7)(cid:2)
(cid:3)
for every test function ϕ supported in Ω. If L ∈ (cid:2)(λ,Λ,ω), u ∈ C1,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, the level sets of u are uniformly Lipschitz surfaces. Precisely, we consider domains of the form
(cid:8) ∈ R : |x(cid:9)| < s, f (x(cid:9)) < xn < 2ls
, (2.2) x(cid:9),xn Ts =
(cid:7)
(cid:8)
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 = {xn = f (x(cid:9))} ∩ ∂T4. Then, there exists η such that in
∩ T1,
(2.3) (cid:4)η(F) = f (x(cid:9)) < xn < f (x(cid:9)) + η
F. Ferrari and S. Salsa 5
u is increasing along the directions τ belonging to the cone Γ(en,θ), with axis en and opening θ = (1/2)cot−1 l. Moreover, in (cid:4)η(F),
≤ Dnu(x) ≤ c
, (2.4) c−1 u(x) dx u(x) dx
(cid:3)
where dx = dist(x,F).
= 0, ∂T2,
div T2, (2.5) Proof of Theorem 2.1. Let z be the solution of the Dirichlet problem (cid:2) A(0)∇z(x) z = g,
where g is a smooth function vanishing on (cid:5) and equal to 1 at points x with dx > 1/10.
(cid:2)
(cid:3) ,
Then, see [1], Dnz > 0 in Qρ, with ρ = ρ(n,l). By rescaling the problem (if necessary), we may assume ρ = 3/2. Since z(en) ≥ c > 0, by Harnack inequality we have that, if y ∈ T1, dy ≥ η0,
∼ c
(cid:3) .
(cid:2) η0
(cid:2)
(2.6) z(y) ∼ c η0 Dnz(y) ∼ z(y) dy
Clearly z,u ∈ C0,a(T 2).
(cid:3)
Lemma 2.2. For r > 0, let wr be the C1,a(T2) weak solution to
= 0, ∂T2.
(cid:2) A(rx)∇wr(x) wr = z,
div T2, (2.7)
Then, given η0 > 0, there exists r0 = r0(η0), such that if r ≤ r0,
(2.8) Dnwr(y) ≥ 0
for every y ∈ T1, with dy ≥ η0.
(cid:2)
(cid:3)(cid:3)
(cid:2)(cid:2)
= div
Proof. Let
(cid:2) A(rx)
(cid:3) A(rx) − A(0) ∇z(x)
(cid:3) .
∇wr(x) − ∇z(x)
(2.9) div
(cid:3)
(cid:8)
=
For every σ > 0, let
(cid:7) x ∈ T2,dist
(cid:2) x,∂T2
2 ). Notice that ((A(rx) −
(2.10) > σ . Qσ 2
(cid:12) (cid:12)
Notice that hr = wr − z ∈ C0,a(T 2), and moreover hr ∈ C1,a(Qσ A(0))∇z(x))i
(cid:4) (cid:4)1/nω(r)
(cid:12) (cid:12)∇z
(cid:4) (cid:4)Qσ
∈ L∞(Qσ 2 ), and from standard estimates we have (cid:4) (cid:4) (cid:4) + C (cid:4)wr − z
(cid:4) (cid:4)hr
2
2 ). L∞(Qσ
(cid:4) (cid:4) ≤ sup ∂Qσ 2
(2.11) sup Qσ 2
6 Boundary Value Problems
(cid:13)
(cid:14)
Hence
|hr | ≤ c
(2.12) σ a + . r σ sup Qσ 2
Choosing r = σ 1+a we get that for every y ∈ Qσ 2 ,
(cid:4) (cid:4) ≤ crβ,
(cid:4) (cid:4)hr(y)
(2.13)
0
(cid:2)
(cid:3)
(cid:3)
(cid:4) (cid:4) ≤ C
≤ C
(cid:4) (cid:4)Dnhr(y)
where β = a/(1 + a). Let y ∈ T1, with dy ≥ η0, r0 ≤ (1/3)η1/(1+a) , and ρ = η0/3. It follows that (cid:2) ρ rβ + r z(y) raz(y) + r(cid:13)z(cid:13)L∞(Bρ(y))
(cid:2)
= Cρ
≤ CρrβDnz(y).
(cid:3) z(y) ρ
}, we get
(2.14) rβ + r
0
Hence if r ≤ r0 = min{(2c(η0))−1/β,(1/3)ηa+1
Dnz(y). 1 2 Dnz(y) ≤ Dnwr(y) ≤ 3 2 (2.15) (cid:2)
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,
≤ Dnw(y) ≤ c
(2.16) c−1 w(y) dy w(y) dy
for every y ∈ T1, dy ≥ t0η0.
Proof. The right-hand side inequality follows Schauder’s estimates and Harnack inequal- ity. Let now y ∈ T1, with dy ≥ t0η0, t0 to be chosen. We may assume y = tη0en. From the boundary Harnack principle (see, e.g., [8] or [9]) if (cid:15)y = η0en, then
(2.17) z((cid:15)y) ≤ ct−az(y)
and, if t ≥ (2c)1/a ≡ t0, then
(2.18) z(y). z((cid:15)y) ≤ 1 2
On the other hand, if dy ≥ t0η0 and r ≤ r0(η0), from (2.6), (2.13), and (2.15) we have
(2.19) z(y), Dnz(y). w(y) ≤ 3 2 Dnw(y) ≥ 1 2
F. Ferrari and S. Salsa 7
(cid:11)
1
(cid:3)
= 3
(cid:2) z(y) − z((cid:15)y)
(cid:2) sy + (1 − s)(cid:15)y
(cid:3) ds
Thus, if t0η0 ≤ dy ≤ 10t0η0, applying Harnack inequality to Dnz, we get
0
≤ ct0η0Dnz(y) ≤ cDnz(y)dy ≤ cDnw(y)dy.
z(y) ≤ 3 z w(y) ≤ 3 2 d ds (2.20)
Repeating the argument with (cid:15)y = 10t0η0, we get that (2.18) holds for 10t0η0 ≤ dy ≤ (cid:2) 20t0η0. After a finite number of steps, (2.18) follows for dy ≥ t0η0, y ∈ T1.
Lemma 2.4. Let u be as in Theorem 2.1. Then there exists a positive η, such that for every x ∈ T1, dx ≤ η,
(2.21) Dnu(x) ≥ 0.
Moreover, in the same set
≤ Dnu(x) ≤ c
(2.22) . c−1 u(x) dx u(x) dx
Proof. Let t0 be as in Lemma 2.3, and η0 small to be chosen later. Set η1 = 2η0t0. It is enough to show that if x = η1ren and r ≤ r0(η0), then Dnu(x) ≥ 0. Consider a small box T2r and define (cid:15)u(y) = u(r y). Then (cid:15)u satisfies div( (cid:15)A(x)∇(cid:15)u(x)) ≡ div(A(rx)∇(cid:15)u(x)) = 0 in T2, where f is replaced by fr(y(cid:9)) = f (r y(cid:9))/r.
We will show that Dn (cid:15)u(y) > 0, where y = η1en, by comparing (cid:15)u with the function w constructed in Lemma 2.2, normalized in order to get (cid:15)u(y) = w(y). Notice that if we choose r0 = r0(η0) according to Lemma 2.3, we have
≤ Dnw(y) ≤ c
(2.23) . c−1 w(y) dy w(y) dy
− 1
If dy ≥ η1. From the comparison theorem (see [8] or [9]), we know that (cid:15)u/w ∈ C0,a(T 3/2) so that in Bη0(y)
≤ cηa 0,
(cid:4) (cid:4) (cid:4) (cid:4) (cid:4)
(cid:4) (cid:4) (cid:4) (cid:4) (cid:4)
(cid:15)u(y) w(y)
(2.24)
which implies
(cid:4) (cid:4)(cid:15)u(y) − w(y)
(cid:4) (cid:4) ≤ cηa
0w(y) ≤ cηa
0w(y).
(2.25)
Moreover, since η0 ∼ dy,
(cid:4) (cid:4) ≤ cηa−1
(cid:4) (cid:4)Dn (cid:15)u(y) − Dnw(y)
0 w(y) ≤ cηa
0Dnw(y),
(2.26)
from which we get
(cid:3) Dnw(y),
(cid:2) 1 − cηa 0
(2.27) Dn (cid:15)u(y) ≥
and (2.21) holds if η0 is sufficiently small. Inequality (2.22) is now a consequence of (2.23) (cid:2) 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 en by any unit vector τ such that the angle between τ and en is less than θ = 1/2cot−1 l.
Thus, we obtain a cone Γ(en,θ) of monotonicity for u. Applying Theorem 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(en,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 following strong Harnack principle, where the cone Γ(cid:9)(en,θ) is obtained from Γ(en,θ) by deleting the “bad” directions, that is, those in a neighborhood of the generatrix opposite to ∇u(en). Precisely, if τ ∈ Γ(en,θ), denote by ωτ the solid angle between the planes span{en, ∇} and span{en,τ}. Delete from Γ(en,θ) the directions τ such that (say) ωτ ≥ (99/100)π and call Γ(cid:9)(en,θ) the resulting set of di- rections. If τ ∈ Γ(cid:9)(en,θ), then
(cid:14)∇,τ(cid:15) ≥ c3δ,
(2.28)
where δ = π/2 − θ. We call δ the defect angle.
(cid:2) en
Lemma 2.5. Suppose u is a positive solution of div(A(rx)∇u(x)) = 0 in T4 vanishing on F = {xn = f (x(cid:9))}, increasing along every τ ∈ Γ(en,θ). Assume furthermore that (2.4) holds in T4. There exist positive r0 and h, depending only on n, l, and λ, Λ, such that if r ≤ r0, for every small vector τ, τ ∈ Γ(cid:9)(en,θ/2), and for every x ∈ B1/8(en), (cid:3) , (2.29) u(y − τ) ≤ u(x) − C(cid:2)δu sup B(1+hδ)(cid:2)(x)
where (cid:2) = |τ|sin(θ/2).
For the proof see [5, Section 3].
(cid:6) ,
Corollary 2.6. In B1/8(x0), u is increasing along every τ ∈ Γ(τ1,θ1) with
− θ1
(cid:5) δ1 = π 2 (cid:4) (cid:4) ≤ Cδ,
(cid:4) (cid:4)ν1 − e1
δ1 ≤ bδ, (2.30)
where b = b(n,a,l,λ,Λ) < 1.
(cid:7)
(cid:8)
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
min , (2.31) r0,η s = 1 2
with η as in Theorem 2.1 and r0 as in Lemma 2.5. If we define
, (2.32) us(x) = u(sx) s
≡ Lus(sx) = 0 in T4 and falls under the hypothesis of Lemma 2.5.
s satisfies Lsu+ s
F. Ferrari and S. Salsa 9
then u+ Therefore, rescaling back we get the following result.
(cid:3)
Theorem 2.7. Let u be a solution of our free boundary problem. Then in Bs/8(sen),
(cid:2) en
(2.33) u(y − τ) ≤ u(x) − c(cid:2)δu sup B(1+hδ)(cid:2)(x)
for every τ ∈ Γ(cid:9)(en,θ/2), |τ| (cid:16) s. As a consequence, in Bs/8(sen), u is monotone along every τ ∈ Γ(ν1,θ1), where ν1, θ1 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.
(cid:11)
(cid:11)
If (cid:6) ⊂ Ω is an open set, regular for the Dirichlet problem, we denote by G(cid:6) = G(cid:6)(x, y) (cid:6) the L-harmonic mea- the Green function associated with the operator L in (cid:6) and by ωx sure for L in (cid:6). In this way,
(cid:6) −
(cid:6)
(cid:6)
(3.1) gdωx G(cid:6)(x, y)h(y)d y w(x) =
is the unique weak solution of Lu = h in (cid:6), h = 0 on ∂(cid:6). A function v ∈ H 1(Ω) is a weak subsolution in Ω if (cid:11)
(cid:9) A(x)∇u(x), ∇φ(x)
(cid:10) dx ≤ 0
Ω
(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.
(cid:11)
(cid:3)
≤
We need to recall a pointwise characterization. Indeed, see [10–14] for the details, 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 x0 ∈ Ω there exists a basis of regular neighborhood (cid:7)x0 associated with v such that for every B ∈ (cid:7)x0,
(cid:2) x0
∂B
(3.3) v v(σ)dωx0 B .
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 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, 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 (cid:15)f ∈ C(Ω) ∩ H 1(Ω), then the weak solution of the Dirichlet problem (even if L has just bounded measurable coef- ficients)
(3.4) Lu = 0 u = f in Ω, on ∂Ω
loc(Ω).
(cid:3)
and the parallel Perron-Wiener-Brelot one coincide. Moreover, in [15] Herv´e also proved that the same result holds when (cid:15)f is L-subharmonic and (cid:15)f ∈ H 1
0
(cid:2) A(x)∇φ(x)
≥ C
≡ Φ(x)
(cid:11)
(cid:16)
Lemma 3.1. Let C > 2 and φ be a C2 weak solution of (cid:4) (cid:4) (cid:4)2 + ω2 (cid:4)∇φ(x) (3.5) div φ(x)
(cid:10)
(cid:10)
(cid:17) dωx
− Φ(σ)
(cid:9) σ − x, ∇φ(x)
B(σ) ≥ 0.
∂Br
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 Br = Br(x) ⊂ Ω, (cid:9) (3.6) + (cid:8)2φ(x)(σ − x),(σ − x) 1 2
(cid:11)
(cid:11)
n(cid:18)
(cid:3)(cid:2)
∇φ(x)
Proof. From Lemma A.5, for every ball Br = Br(x) ⊂ Ω,
(cid:3) dωx
(cid:2) σi − xi
B(σ) +
B(σ)
∂Br
i, j=1
∂Br (cid:11)
≥
(cid:3) ,
(cid:2) GBr (x, y)Φ(y)d y + o r2
Br
(cid:2)
(σ − x)dωx Di jφ(x) σ j − x 1 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
(cid:3) (cid:2) , x + φ(x)η(x)
(3.8) vφ(x) = u
(cid:11)
for some η(x), where |η(x)| = 1. To prove that vφ is an L-subsolution we just check con- dition (S), since by straightforward calculations vφ is locally Lipschitz continuous. In par- ticular we will prove that for every x ∈ Ω+(v) there exists a positive constant r0 = r0(x) such that for every ball Br ≡ Br(x) ⊂ Ω+(v), r ≤ r0, and for every x0 ∈ Br,
(cid:3) .
(cid:2) x0
Br (σ) ≥ vφ
∂Br
(3.9) vφ(σ)dωx0
n−1(cid:18)
(cid:10)
Let {e1,...,en} be an orthonormal basis of Rn where en = η(x) and let ξ be the following vectorfield:
(cid:9) Vi,h
i=1
(3.10) ξ(h) = en + ei,
F. Ferrari and S. Salsa 11
n−1(cid:18)
n−1(cid:18)
(cid:9)
(cid:10)
(cid:9)
(cid:3)
(cid:10)2
where {V1,...,Vn−1} ⊂ Rn will be chosen later. Let ν(h) = ξ(h)/|ξ(h)|, so that
(cid:2) |h|2 en + o
i=1
i=1
(3.11) . ν(h) = en + Vi,h Vi,h ei − 1 2
(cid:3) ,
(cid:3) ,
(cid:3) ,
Let x0 ∈ Br(x) and h = σ − x0. Then, letting
(cid:2) x0
∇φ0 = ∇φ
(cid:2) x0
(cid:2) x0
(3.12) φ0 = φ (cid:8)2φ0 = (cid:8)2φ
(cid:9)
(cid:10)
(cid:10)
(cid:3)
(cid:9)
we have
(cid:2) |h|2 + o
∇φ0,h
+ (3.13) φ(σ) = φ0 + (cid:8)2φ0h,h 1 2
(cid:3)
as h → 0, uniformly in a neighborhood of x. As a consequence,
(cid:2) σ + φ(σ)ν σ − x0
= y∗ + J1 + J2 + J3,
(3.14)
(cid:19)
(cid:20)
n−1(cid:18)
(cid:9)
(cid:10)
(cid:10)
where y∗ = x0 + φ(x0)en,
∇φ0,h
(cid:9) Vi,h
i=1
(cid:19)
(cid:20)
n−1(cid:18)
(cid:9)
(cid:10) n−1(cid:18)
(cid:9)
(cid:10)
(cid:9)
, J1 = en + h + φ0 ei
∇φ0,h
(cid:9) (cid:8)2φ0h,h
(cid:10)2en
(cid:10) en − φ0 2
i=1
i=1
(cid:3)
(cid:2) |h|2 J3 = o
(3.15) , J2 = Vi,h ei + Vi,h 1 2
(cid:2)
(cid:3)
(cid:2)
(cid:3)
(cid:9)
(cid:2)
(cid:10)
(cid:2)
(cid:10)
(cid:9)
uniformly as h → 0. Let J = J1 + J2 + J3. Then for every σ ∈ ∂Br(x0),
(cid:3) ,
= u
∇u
(cid:3) ,J
(cid:3) J,J
(cid:2) + o
|h|2
+ + (3.16) y∗ + J y∗ y∗ (cid:8)2u y∗ v(σ) ≥ u 1 2
(cid:2)
(cid:10)
(cid:2)(cid:9)
(cid:10)
(cid:9)
(cid:2)
(cid:3)(cid:4) (cid:4)
=
(cid:10)(cid:3) ,
(cid:9) ∇u
as h → 0, uniformly in a neighborhood of y∗. We have
(cid:3) ,J1
(cid:4) (cid:4)∇u (cid:21)
(cid:22)
n−1(cid:18)
(cid:9)
(cid:10)
(cid:9)
(cid:9)
(cid:2)
(cid:10)
(cid:2)
y∗ y∗ h, ∇φ0 h,en
(cid:3)(cid:4) (cid:4)
=
∇u
(cid:4) (cid:4)∇u
(cid:3) ,J2
− φ0 2
i=1
(3.17) + (cid:10)2 + . y∗ y∗ (cid:8)2φ0h,h Vi,h 1 2
12 Boundary Value Problems
(cid:11)
(cid:3)
As a consequence,
(cid:2) x0
∂Br
(cid:19)
Br (σ) ≥ v (cid:11)
n−1(cid:18)
(cid:2)
(cid:10)
(cid:9)
(cid:10)
(cid:3)(cid:4) (cid:4)
v(σ)dωx0
(cid:10)2 +
(cid:4) (cid:4)∇u
(cid:9) h, ∇φ0
(cid:9) (cid:8)2φ0h,h
(cid:20) dωx0 Br
− φ0 2
∂Br
(cid:11)
(cid:16)
(cid:17) (cid:3)
(cid:9)
(cid:2)
(cid:10)
(cid:2)
(cid:9)
(cid:10)
(cid:3)
(cid:4) (cid:4)
+ y∗ Vi,h 1 2
(cid:3) ,h
∇u
i=1 (cid:3) J,J
(cid:2)(cid:4) (cid:4)h2
= v
(cid:2) x0
∂Br
(cid:19)
(cid:11)
n−1(cid:18)
(cid:10)
(cid:9)
(cid:2)
(cid:10)
(cid:3)(cid:4) (cid:4)
+ + y∗ (cid:8)2u y∗ + o (3.18) dωx0 Br 1 2
(cid:10)2 +
(cid:9) h, ∇φ0
(cid:9) (cid:8)2φ0h,h
(cid:20) dωx0 Br
− φ0 2
∂Br
i=1
(cid:11)
(cid:4) (cid:4)∇u (cid:11)
(cid:2)
(cid:23)(cid:9)
(cid:10)(cid:24)
(cid:3)
(cid:2)
+ y∗ Vi,h 1 2
(cid:3) ,
(cid:3) J1,J1
Br + ∇u
(cid:2) r2 Br + o
∂Br
∂Br
+ (cid:8)2u y∗ dωx0 y∗ hdωx0 1 2
(cid:11)
(cid:11)
n(cid:18)
(cid:23)(cid:9)
(cid:2)
(cid:10)(cid:24)
(cid:3)
(cid:2)
=
uniformly with respect to x0 in a neighborhood of x. Let
(cid:3) J1,J1
∂B
∂Br
i, j
(3.19) (cid:8)2u y∗ y∗ Di ju dωx0 Br aia jdωx0 Br
(cid:10)
(cid:9)
(cid:10)
with
(cid:9) Vi,h
+ , (3.20) i = 1,...,n, ai = φ0 h,ei
(cid:9)
(cid:10)
(cid:9)
(cid:10)
where the Vi are still to be chosen, and
∇φ0,h
+ (3.21) . an = h,en
(cid:11)
(cid:11)
(cid:3)
(cid:2)
(cid:3)
=
=
For i = 1,...,n and j = 1,...,n, let
(cid:2) x0,x0
= di j
Br (x0),
Br (y∗)
∂Br
∂Br
y∗, y∗ di j = di j hih jdωx0 hih jdωy∗ d∗ i j
(3.22)
be the entries, of the matrix of the moments (see the appendix), respectively, evaluated in x0 and y∗.
For i = 1,...,n, and j = 1,...,n − 1, let (cid:11)
∂Br
n(cid:18)
n(cid:18)
mi j = aia jdωx0 Br ,
p=1
p,q=1
(3.23) mnn = Dpφ0Dqφ0dpq + 2 Dpφ0dpn + dnn.
F. Ferrari and S. Salsa 13
n(cid:18)
n(cid:18)
n(cid:18)
Then
i V q
j dpq + φ0
p,q=1
q=1
j diq + di j, (cid:22)
(3.24) V q mi j =
0V p φ2 (cid:11)
p=1 (cid:6)
n(cid:18)
(cid:5)(cid:9)
(cid:10)
V p i d j p + φ0 (cid:21)
∇φ0,h
i hp + hi
Br (σ)
∂Br
p=1
V p dωx0 φ0 min = mni = + hn
n(cid:18)
n(cid:18)
n(cid:18)
(3.25)
= φ0
i Dpφ0dpq +
i dpn + din.
p=1
p,q=1
p=1
V p V p Dqφ0dpi + φ0
(cid:2)
Suppose now we can find V1,...,Vn−1 and a real number κ0, such that for every i = 1,...,n − 1 and for every j = 1,...,n,
(cid:2) 1 + κ0
(cid:3) d∗ i, j,
(cid:3) d∗ nn.
1 + κ0 mi, j = mnn = (3.26)
n−1(cid:18)
n−1(cid:18)
(cid:2)
(cid:2)
(cid:2)
(cid:3)
(cid:3) n(cid:18)
(cid:2)
Then
(cid:2) 1 + κ0
(cid:3) mi j + 2
(cid:3) min + Dnnu
(cid:3) d∗ i j .
i, j=1
i=1
i, j=1
y∗ y∗ y∗ y∗ Di ju Dinu mnn = Di ju
(3.27)
n(cid:18)
n(cid:18)
n(cid:18)
In particular this means that V1,...,Vn−1, and k0 must solve the following system, for i = 1,...,n − 1 and j = 1,...,n − 1,
(cid:2) 1 + κ0
= −di, j +
(cid:3) d∗ i,1,
i d j p + φ0
j diq + φ2 0
i V q
j
p
q=1
p,q=1
n(cid:18)
n(cid:18)
n(cid:18)
V p V q φ0 dp,qV p
(cid:2) 1 + κ0
(cid:3) d∗ i,n,
i dpn + φ0
i Dpφ0dpq = −di,n +
p,q=1
p=1
p=1
n(cid:18)
n(cid:18)
(cid:3)
(cid:2) 1 + κ0
V p V p φ0 Dqφ0dip + (3.28)
p,q=1
p=1
nn > cλr2, for small r and |∇φ0|, there
Dpφ0Dqφ0dpq + 2 Dqφ0dpn + dnn = d∗ n,n.
(cid:4) (cid:4) +
From the last equations and Lemma A.3, since d∗ exists a positive constant C = C(λ,Λ) such that
(cid:4) (cid:4) ≤ C
(cid:3) .
≤ C
(cid:4) (cid:4)κ0
(cid:4) (cid:4)∇φ0
(cid:2)(cid:4) (cid:4)∇φ0
(cid:4) (cid:4) + φmax
(3.29) dnn − d∗ nn d∗ n,n
We now start an iteration process to solve the above system (see [4, 6]).
14 Boundary Value Problems
n(cid:18)
n(cid:18)
n(cid:18)
(cid:3)(l+1)
(cid:3)(l)
(cid:3)(l)(cid:2)
Let ((cid:9)i)0 = 0, i = 1,...,n − 1, and for l ≥ 0, define recursively ((cid:9)i)(l+1) as the solution of the system (i = 1,...,n − 1; j = 1,...,n − 1):
(cid:2) 1 + κ0
(cid:3)(l) = −di, j +
(cid:3) d∗ i,1,
(cid:2) (cid:9)p i
(cid:2) (cid:9)p j
(cid:2) (cid:9)p i
p
q=1
p,q=1
n(cid:18)
n(cid:18)
n(cid:18)
(cid:3)(l+1)
(cid:3)(l)
(cid:2) 1 + κ0
φ0 d j p + φ0 dp,q diq + φ2 0 (cid:9)q j
(cid:3) d∗ i,n.
(cid:2) (cid:9)p i
(cid:2) (cid:9)p j
p,q=1
p=1
p=1
φ0 dpn + φ0 Dpφ0dpq = −di,n + Dqφ0dip +
(cid:22)
(cid:4) (cid:4)
(cid:4) (cid:4)
(cid:3)
(cid:4) (cid:4)
i
i
(3.30)
(cid:4) (cid:4) (cid:4) ≤ C
(cid:4) (cid:4)∇φ0
(cid:2) φ0 +
≤ Cφ−1 0
0
(cid:4) (cid:4) (cid:4)(cid:9)(l+1) i
(cid:4) (cid:4)d∗ r2 + φa
(cid:4) (cid:4)∇φ0 φ
i )l∈N are bounded for every i ∈ {1,...,n − 1},
(3.31) + Notice that the sequence is well defined, since the matrix D(x0,x0) is nonsingular (Lemma A.3 in the appendix). Moreover, if |∇φ(x0)| is kept small, denoting by di and d∗ i the vec- i1,...,d∗ tors (di1,...,din) and (d∗ in), from the estimates in Lemma A.3, we get, by induction, (cid:21) (cid:4) (cid:4) (cid:4)di − d∗ (cid:4) r2φ0
i )l∈N, converging to (cid:9)i with (cid:4) (cid:3) (cid:4) .
(cid:4) (cid:4)∇φ0
(cid:2) φ0 +
(cid:4) (cid:4)(cid:9)i
0
(3.32) with C = C(n,Λ,λ). Since the sequences ((cid:9)(l) there exist subsequences (that we still call) ((cid:9)(l) (cid:4) (cid:4) ≤ C(n,Λ,λ)φ−1
(cid:11)
(cid:3)
Now, from (3.18), (1.11), and Lemma A.5, we get
∂B
(cid:2) x0 (cid:19)
Br (σ) ≥ v (cid:11)
n−1(cid:18)
(cid:9)
(cid:10)
(cid:2)
(cid:9)
(cid:9)
(cid:10)
(cid:3)(cid:4) (cid:4)
v(σ)dωx0
(cid:10)2 +
(cid:20) dωx0 Br
− φ0 2
∂Br
i=1
(cid:11)
(cid:4) (cid:4)∇u (cid:11)
(cid:9)
(cid:2)
(cid:10)
(cid:23)(cid:9)
(cid:2)
(cid:10)(cid:24)
(cid:3)
(cid:3)
+ y∗ h, ∇φ0 (cid:8)2φ0h,h Vi,h 1 2
(cid:3) ,h
∇u
= v
(cid:3) J1,J1
(cid:2) x0
Br +
∂Br
∂Br
(cid:19)
(cid:11)
n−1(cid:18)
(cid:9)
(cid:10)
(cid:2)
(cid:10)
(cid:9)
(cid:9)
(cid:3)(cid:4) (cid:4)
+ y∗ dωx0 (cid:8)2u y∗ dωx0 1 2
(cid:10)2 +
(cid:2) r2 Br + o (cid:20) dωx0 Br
− φ0 2
∂Br
i=1
(cid:4) (cid:4)∇u (cid:11)
(cid:11)
(cid:9)
(cid:2)
(cid:10)
(cid:23)(cid:9)
(cid:2)
(cid:10)(cid:24)
(cid:3)
+ y∗ h, ∇φ0 (cid:8)2φ0h,h Vi,h 1 2
(cid:3) ,h
∇u
Br +
(cid:2) r2 Br (y∗) + o
(cid:3) h∗,h∗ (cid:17)
∂Br (cid:3)
(cid:16) (cid:11) (cid:3)
(cid:2)
∂Br (y∗) (cid:11) (cid:3)
−
= v
+ y∗ dωx0 (cid:8)2u y∗ dωy∗ 1 + κ0 2
(cid:2) x0
(cid:2) 1 + κ0
Br (y∗)
∂Br
∂Br (y∗)
(cid:11)
(cid:16)
(cid:17)
n−1(cid:18)
(cid:2)
(cid:10)
(cid:9)
(cid:10)
(cid:3)(cid:4) (cid:4)
+ ∇u y∗ h∗dωy∗ hdωx0 Br
(cid:10)2 +
(cid:4) (cid:4)∇u
(cid:3) .
(cid:9) h, ∇φ0
(cid:9) (cid:8)2φ0h,h
(cid:2) r2 Br + o
− φ0 2
∂Br
i=1
+ y∗ dωx0 Vi,h 1 2 (3.33)
(cid:11)
(cid:16) (cid:11)
(cid:3)
Consider
(cid:17) .
(cid:2) 1 + κ0
B (σ) −
Br (y∗)(σ)
∂Br
∂Br (y∗)
(3.34) hdωx0 hdωy∗ T =
F. Ferrari and S. Salsa 15
From Lemma A.3 and (3.29), we get
|T| ≤ Kr2.
(3.35)
Thus, from (3.32), we deduce that (cid:11)
B (σ)
∂Br
(cid:19)
(cid:11)
n−1(cid:18)
(cid:9)
(cid:3)
(cid:2)
(cid:10)
(cid:9)
(cid:10)
(cid:9)
(cid:3)(cid:4) (cid:4)
v(σ)dωx0
(cid:10)2 +
(cid:4) (cid:4)∇u
≥ v
(cid:2) x0
(cid:20) dωx0 Br
− φ0 2
∂Br
i=1
− Kr2 (cid:6)
(cid:19)
(cid:11)
(cid:9)
(cid:3)
(cid:2)
(cid:10)
(cid:10)
(cid:9)
0
(cid:3)(cid:4) (cid:4)
+ y∗ h, ∇φ0 Vi,h
(cid:4) (cid:4)∇u
≥v
−C
(cid:2) x0
(cid:20) dωx0 Br .
∂Br
+ + y∗ r2 − Kr2 h, ∇φ0 (cid:8)2φ0h,h 1 2 1 (cid:8)2φ0h,h 2 (cid:5)(cid:4) (cid:4) (cid:4)2 + φ2 (cid:4)∇φ0 φ0 (3.36)
(cid:6)
(cid:19)
(cid:11)
(cid:9)
(cid:10)
(cid:10)
0
From Lemma 2.5, if r is small, and C large depending on x0 and φ0, we have
≥ 0,
− C
(cid:9) (cid:8)2φ0h,h
(cid:20) dωx0 Br
(cid:5)(cid:4) (cid:4) (cid:4)2 + φ2 (cid:4)∇φ0 φ0
∂Br
(cid:2)
+ (3.37) r2 − Kr2 h, ∇φ0 1 2
so that vφ is a weak L-subsolution in its positivity set.
Remark 3.2. We emphasize that the construction of the vectors Vi, 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
(cid:3) (cid:2) x − τ + φ(x)ν .
(4.1) u u(y − τ) = sup |ν|=1 vτ(x) = sup Bφ(x)(x)
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 C2-weak solution of Lu = 0, L ∈ (cid:2)(λ,Λ,ω). For any vector τ let φ be a positive C2-function such that 0 < φmin ≤ φ ≤ φmax,
(cid:3) (cid:2) x − τ + φ(x)ν ,
(4.2) u u(y − τ) = sup |ν|=1 vτ(x) = sup Bφ(x)(x)
(cid:6) ,
(cid:5)(cid:4) (cid:4)∇φ(x)
0
(4.3) 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 (cid:4) (cid:4)2 + ω2 φLφ ≥ C
then vτ is a weak subsolution of Lu = 0 in {vτ > 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 Vi, i = 1,...,n − 1, and k0 depend only on the distance between the matrices D(x0,x0) and D(y∗, y∗).
We now construct a family of radii, with the right properties to be used in the final comparison theorem. Let D = B2(0)\B1/8(en). We may assume with out loss of generality that A(0) = I and that
(cid:4) (cid:4)A(x) − I
(cid:4) (cid:4) ≤ ω1 (cid:16) 1.
(4.4) sup B3
By a slight modification of [5, Lemma 7] we can construct a family of functions satisfying the properties expressed in the following lemma. Lemma 4.3. Let (cid:15)C > 0. There exist positive numbers c,η,μ,ω < ημ/2 and a family of func- tions φt, 0 ≤ t ≤ 1, such that gt ∈ C2(D) and
(cid:2)
(cid:3)2
(i) 0 < 1 − ω ≤ φt ≤ 1 + μt, (ii) φt ≤ 1 − ω in B2\B5/3, (iii) φt ≥ 1 − ω + ημt in B1/2, (iv) |∇φt| ≤ c(μt + ω), (v)
(cid:4) (cid:4)2 + ω
(cid:6) .
(cid:5)(cid:4) (cid:4)∇φt
(4.5) φtLφt ≥ (cid:15)C max φt
We are now in position to prove Theorem 1.2.
= 0
Proof of Theorem 1.2. We first observe that Theorem 1.1 (and Corollary 4.1) holds for weak solutions, not necessarily C2. In fact let u± j be the functions constructed as solu- tions of the following problems:
− u−
j . Then u j converges locally in C1,a(Ω±(u)) to u and it is not difficult
(4.6) in Ω±(u), on Ω±(u), L ju± j u± = u j j
and set u j = u+ j to check that (suppressing for clarity the index t)
(4.7) u j v j(x) = sup Bφ(x)(x)
converges locally in C1,a(Ω±(u) ∩ D) to
(4.8) u. vφ(x) = sup Bφ(x)(x)
From Theorem 1.1, v j is a weak subsolution for L j in Ω±(u j) ∩ D. But then vφ is a weak 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 φt satisfies
(cid:13)
F. Ferrari and S. Salsa 17
(cid:4) (cid:4)νk+1 − νk
(cid:4) (cid:4) ≤ cδk.
− θk
(cid:2)
inequality (4.5) also (cid:2)φt satisfies the same inequality for every (cid:2) > 0. Therefore, we can simplify the proof given in [5] avoiding, in the iteration process, to go through the im- provement of the (cid:2)-monotonicity and prove directly that in a sequence of dyadic balls B4−k u is monotone along every τ ∈ Γ(νk,θk) with (cid:14) , (4.9) δk+1 ≤ bδk δ0 = δ,δk = π 2
These conditions imply that F(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 in (cid:3)1. We only need to show that u is Lipschitz in (cid:3)2/3 across the free boundary. This follows from a simple application of the monotonicity formula in [16, Lemma 1] and a barrier argument. Pre- cisely, let x0 ∈ Ω+(u) ∩ (cid:3)2/3, d0 = dist(x0,F(u)), and u(x0) = λ. From Harnack inequality
(4.10) u(x) ∼ λ
(cid:3)
in Bd0/2(x0). Let w be the solution of
= 0
(cid:2) A(x,u)∇w
div (4.11)
(cid:2)
(cid:3)
in Bd0(x0)\Bd0/2(x0) such that w = 0 on ∂Bd0(x0), w = λ on ∂Bd0/2(x0). By maximum prin- ciple
(cid:2) x0
(cid:3) \Bd0/2
(4.12) x0 u ≥ cw in Bd0
(cid:10)+
and, from the Ca nature of A and C1,a estimates, if y0 ∈ ∂Bd0(x0) ∩ F(u),
(cid:9) x − y0,ν
(4.13) w(x) ≥ c λ d0
(cid:3)
(cid:10)−
(cid:4) (cid:4)
with ν = (x0 − y0)/|x0 − y0|. Thus, near y0, u has the asymptotic behavior
(cid:10)+ − β
(cid:9) x − y0,ν u(x) ≥ α
(cid:9) x − y0,ν
(cid:2)(cid:4) (cid:4)x − y0
(4.14) + o
with
≤ α ≤ G(β).
(4.15) c λ d0
(cid:14)
Then, the monotonicity formula gives (cid:13)
≤ C(cid:13)u(cid:13)2
L∞((cid:3)3/4)
(cid:6)
(cid:3)(cid:4) (cid:4)
(4.16) G−1 c λ d0 λ d0
(cid:4) (cid:4)∇u+
(cid:5)(cid:4) (cid:4)∇u+
(cid:2) x0
(cid:2) x0
≤ C1(cid:13)u(cid:13)2
L∞((cid:3)3/4).
(4.17) so that, from interior estimates, (cid:3)(cid:4) (cid:4)G−1
18 Boundary Value Problems
(cid:3)(cid:4) (cid:4)
This gives the Lipschitz continuity of u+. Similarly, we get
(cid:5)(cid:4) (cid:4)∇u−
(cid:6)(cid:4) (cid:4)∇u−
(cid:2) x0
(cid:2) x0
(cid:3)(cid:4) (cid:4) ≤ C1(cid:13)u(cid:13)2
L∞((cid:3)3/4)
(cid:2)
(4.18) G
and the proof is complete.
Appendix
(cid:11)
Auxiliary lemmas
(cid:3) dωy
Br (x0)(σ)
∂Br (x0)
(cid:11)
(cid:3)
(cid:3)(cid:2)
=
(A.1) We collect here some estimates on the L-harmonic measure and its moments that are used in the paper. Here ω(r) ≤ c0ra, 0 < a ≤ 1. Definition A.1. For any x0,x, y ∈ Ω, and r > 0, Br(x0) ⊂ Ω, let di(x0, y) be, for i = 1,...,n, (cid:2) σi − x0i di(x0, y) =
(cid:3) dωy
(cid:2) σi − x0i
Br (x0)(σ).
∂Br (x0)
(A.2) and let di j(x0, y) be, for every i, j, 1 ≤ i, j ≤ n, (cid:2) x0, y di j σ j − x0 j
= G0
We call, respectively, (di(x0, y))1≤i≤n the vector of the first moment of the L-harmonic measure in Br(x), and D(x0, y) = (di j(x0, y))1≤i, j≤n the matrix of the second moment of the L-harmonic measure in Br(x).
r (x, y) the Green function for L0 in Br =
Denote by L0 = div(A(x0)∇) and by G0 r Br(x0). We have the following.
(cid:3)
=
Lemma A.2. Let L0wr = −1 in Br(x0), wr = 0 on ∂Br(x0). Then
(cid:3) .
(cid:2) x0
(A.3) wr r2 (cid:2) 2TrA x0
Proof. Suppose x0 = 0. Let gi j(x) = xix j and let vi j be the solution of L0vi j = 0 in Br, vi j = gi j on ∂Br. Since L0gi j = 2ai j(0) and gi j(0) = 0, we have
(cid:25)
(cid:2)
(A.4) vi j(0) = 2ai j(0)wr(0).
n i=1 vii(0) = r2 and (A.3) follows.
On the other hand,
(cid:2)
(cid:3)
Lemma A.3. Let B2r(x0) ⊂ Ω. Then: (1) for every i = 1,...,n,
(cid:4) (cid:4)d
(cid:4) (cid:4) ≤ C(λ,Λ,n)r1+a;
− yi
(cid:3)
(A.5) x0, y sup Br (0)
(cid:3)(cid:4) (cid:4) ≤ Cr2+a,
(cid:2) x0,x0
(cid:2) x0
(cid:2) x0
− 2wr
(cid:3) ai j
(A.6) (2) for every i, j = 1,...,n, (cid:4) (cid:4)di j
where wr is defined in Lemma A.2.
F. Ferrari and S. Salsa 19
(cid:11)
Proof. Let x0 = 0 and
Br (σ).
∂Br
(A.7) di(y) = di(0, y) = σidωy
(cid:3)
(cid:2)(cid:2)
(cid:3)
= div
Then di(y) − yi = 0 on ∂Br and
(cid:2) di(y) − yi
(cid:3) ei
(A.8) A(0) − A(y) L0 in Br.
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
From standard estimates, we get
(cid:2) A(0) − A(y)
≤ Cr
≤ Cr1+a.
(cid:12) (cid:12)di − yi
(cid:3) ei
L∞(Br )
L∞(Br )
(A.9)
(cid:11)
Consider now
Br (σ).
∂Br
(cid:2)(cid:2)
(cid:3)
(cid:3)
=
(A.10) di j(y) = di j(0, y) = σiσ jdωy
(cid:2) di j(y) − vi j(y)
(A.11) A(0) − A(y) L0 If vi j is as in Lemma 2.2, we have hi j − vi j = 0 on ∂Br and (cid:3) ∇vi j in Br.
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:2) A(0) − A(y)
≤ Cr
≤ C
(cid:12) (cid:12)di j − vi j
(cid:3) ∇vi j
(cid:12) (cid:12)∇vi j
L∞(Br )
L∞(Br )
Ls(Br )r1+a ≤ Cr2+a.
Therefore,
(A.12)
(cid:4) (cid:4) =
(cid:4) (cid:4) ≤ Cr2+a.
(cid:4) (cid:4)di j(0) − 2ai j(0)wr(0)
(cid:4) (cid:4)di j(0) − vi j(0)
Hence, from Lemma 2.2, we get
(A.13) (cid:2)
Corollary A.4. For r ≤ r0(n,λ,Λ,a), the matrix (di j(0)/r2) is nonsingular.
(cid:3)
Lemma A.5. Let w be a weak solution of
(cid:2) A(x)∇w(x)
= f
(A.14) div
(cid:11)
(cid:11)
(cid:3)
in Ω, where f is continuous. Then for every x ∈ Br(x0) ⊂ Ω,
∇w(x) ·
(cid:2) x0,x
Br (x0)(σ) +
∂Br (x0)
Br (x0)
, (A.15) (σ − x)dωx GBr (x0)(x, y) f (y)d y = R
(cid:22)
(cid:11)
(cid:21) (cid:11)
1
(cid:3)
(cid:9)
(cid:3)
(cid:10)
(cid:3)
=
where
(cid:2) x + s(σ − x)
∇w
− ∇w(x),σ − x
(cid:2) x0,x
Br (x0)(σ).
0
(cid:2) x0
∂Br
(A.16) ds dωx R
20 Boundary Value Problems
(cid:11)
(cid:11)
(cid:9)
(cid:10)
∇w(x) ·
Moreover, if u ∈ C2(Ω),
Br (x0)(σ) +
Br (x0)(σ)
∂Br (x0)
(cid:11)
(σ − x)dωx D2w(x)(σ − x),(σ − x) dωx
∂Br (x0) (cid:3) .
Br (x0)
+ 1 2 (cid:2) r2 GBr (x0)(x, y) f (y)d y = o
(A.17)
(cid:11)
1
(cid:10)
(cid:10)
(cid:9)
(cid:2) x + s(σ − x)
(cid:3) ,σ − x
∇w
Proof. Let div(A(x)∇w(x)) = f , then w ∈ C1,a(Ω) and for any σ,x ∈ Br(x0) ⊂ Ω,
(cid:9) ∇w(x),σ − x
0
(cid:11)
1
(cid:3)
(cid:10)
(cid:9)
ds = w(x) + w(σ) = w(x) + (A.18)
(cid:2) x + s(σ − x)
∇w
− ∇w(x),σ − x
0
+ ds.
(cid:11)
(cid:11)
On the other hand,
Br (x0)(σ) −
∂Br (x0)
Br (x0)
(A.19) w(σ)dωx w(x) = GBr (x0)(x, y) f (y)d y,
(cid:11)
(cid:11)
hence
∇w(x) ·
(cid:3) .
(cid:2) x0,x
Br (x0)(σ) +
∂Br (x0)
Br (x0)
(cid:2)
(A.20) (σ − x)dωx GBr (x0)(x, y) f (y)d y = R
The rest of the proof follows from Taylor expansion.
(cid:3)
Corollary A.6. Let u ∈ C2(Ω) be a weak solution of
= 0
(cid:2) A(x)∇u(x)
div (A.21)
(cid:11)
(cid:3)
(cid:2)
·
in Ω. Then
∇u
(cid:3) dωx0
(cid:3) .
(cid:2) x0
(cid:2) σ − x0
Br (x0)(σ) = O
∂Br (x0)
(A.22) r2
(cid:11)
(cid:3)(cid:2)
(cid:3) dωx0
(cid:2) r2
(cid:3) .
(cid:2) σi − x0i
Proof. It is enough to observe that
Br (x0)(σ) = O
∂Br (x0)
σ j − x0 j (A.23) (cid:2)
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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: ferrari@dm.unibo.it
Sandro Salsa: Dipartimento di Matematica, Politecnico di Milano, Via Bonardi 7, 20133 Milano, Italy Email address: sansal@mate.polimi.it
F. Ferrari and S. Salsa 21
