On a free boundary problem

Nonlinear Analysis 68 (2008) 2328–2348 www.elsevier.com/locate/na

On a free boundary problem Sabri Bensid, S.M. Bouguima ∗ Department of Mathematics, Faculty of Sciences, University of Tlemcen, B.P. 119, Tlemcen 13000, Algeria Received 20 October 2006; accepted 25 January 2007

Abstract In this paper, we will be concerned with the existence of positive solutions of the following discontinuous problem:  −∆u = f (u)H (u − µ) in Ω , u=h on ∂Ω ,

(P)

where Ω is the unit ball of Rn (n ≥ 3) and H is the Heaviside function. The problem is written in an equivalent integral equation form. The shape of the free boundary problem {x ∈ Ω /u(x) = µ} is studied for when µ is a positive real parameter, f and h are given functions. c 2007 Elsevier Ltd. All rights reserved.

MSC: 34R35; 35J25 Keywords: Green function; Maximum principle; Implicit function theorem; Free boundary problem

1. Introduction Some free boundaries are curves or surfaces with a priori unknown location which separate geometric regions with different characteristics. An example is the equilibrium shape of a confined plasma (see [23,9]) where the location of the boundary of the region occupied by the plasma (separating it from the vacuum) is not given a priori and changes during the process. This lack of knowledge requires a major study of the free boundary. Many of these free boundary problems can be regarded as partial differential equations with discontinuous nonlinearities which serve as models in many concrete problems in mathematical physics like those of combustion theory, vortex flow, porous media and plasma physics. (See for instance [8,15,24].) The purpose of this work is to investigate the following problem:  −∆u = f (u)H (u − µ) in Ω , (P) u=h on ∂Ω , where Ω is the unit ball of Rn (n ≥ 3) and H is the Heaviside unit step function, i.e. H (t) = 1

if t > 0

and

H (t) = 0

if t ≤ 0

f, h are given functions and µ is a real parameter. ∗ Corresponding author. Tel.: +213 75 02 93 47.

E-mail addresses: Bensid [email protected] (S. Bensid), [email protected] (S.M. Bouguima). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.01.047

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Problem (P) represents an electric arc experiencing a balance between the energy generated by the arc current and the energy carried out by conduction and radiation (see [11] for more details). Because the nonlinear term in (P) is discontinuous at u = µ, we require a suitable concept of solution. We say that a function u ∈ W 2, p (Ω )( p > 1) is a solution of problem (P) if −∆u = f (u)H (u − µ)

a.e. in Ω

and the trace of u on ∂Ω is equal to h. When h = 0, problem (P) is called the reduced problem. This case is studied in [4]. The treatment of a more general case with a p-Laplacian operator ( p > 1) can be found in [7]. When h does not vanish identically, problem (P) is called the perturbed problem. The existence of a positive solution and the behavior of the corresponding free boundary with respect to h seem to pose a difficult question. For the case when f (u) = λ is a real parameter, problem (P) is considered in [1,2]. By using local methods, the authors obtained at least two positive solutions. With the same methods, problem (P) has been investigated in [9,10] for the two-dimensional case with some restrictions on the functions f and h. In this paper, we will consider the case n ≥ 3 with more general conditions on the nonlinearity f which are more realistic for applications. The following assumptions will be needed throughout this work. Let λ1 be the first eigenvalue of −∆ in H01 (Ω ) with the corresponding eigenfunction ϕ1 > 0 in Ω . We will assume: (H1 ) f (u) ≥ 0, ∀u ≥ 0, f is k-Lipschitzian with k ∈ (0, λ1 ) and nondecreasing. (H2 ) Let n ≥ 3; there exists µ > 0 such that ! kϕ1 k L 1 f (µ) max Mn , 2λ1 < . µ kϕ1 k2 2 L

Here Mn =   2 n

n(n − 2) 2   n . n−2 n−2 − n2

2− 1 , p

(H3 ) h ∈ W p (∂Ω , R), for a certain p > n and h is small enough with 0 ≤ khk∞ < µ where khk∞ = maxx∈∂ Ω |h(x)|. Remark 1.1. (i) h can be seen as the trace of a W 2, p function defined on Ω . (ii) Let f (u) = λ, a positive real parameter. For fixed µ > 0 we obtain ! λ kϕ1 k L 1 f (µ) = > max Mn , 2λ1 µ µ kϕ1 k2L 2 for λ large enough. (iii) Now, let f (u) = αu + β with 0 < α < λ1 and β > 0; then it is easy to see that conditions (H1 ), (H2 ) are satisfied. Let us check hypothesis (H2 ). We have ! β kϕ1 k L 1 f (µ) = α + > max Mn , 2λ1 µ µ kϕ1 k2L 2 for µ small enough. (iv) When Ω = B(0, R) is a ball of radius R > 0, the condition in hypothesis (H2 ) becomes ! f (µ) kϕ1 k L 1 < max Mn (R), 2λ1 (R) 2 µ kϕ1 k 2

(C)

L

where Mn (R) = R12 Mn . We can see that Mn (R) and λ1 (R) tend to zero as R −→ +∞ (see Appendix B). Hence for fixed µ > 0, the condition (C) is satisfied for R large enough.

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The main result of this paper is the following theorem that gives conditions under which problem (P) has positive solutions. Theorem 1.1. Under hypotheses (H1 ), (H2 ) and (H3 ), the problem (P) has at least two positive solutions u and v such that their corresponding free boundaries {x ∈ Ω : u(x) = µ}

and

{x ∈ Ω : v(x) = µ}

are C 1,α -hypersurfaces in Rn for some α ∈ (0, 1). Remark 1.2. This theorem remains true in the case n = 2; the proof is similar to that for the case of higher dimensions. Since Ω is supposed a ball of Rn , Schwarz symmetrization (see [19]) implies that the positive solutions of the reduced problem are radials and their corresponding free boundaries are spheres with some radius r0 ∈ (0, 1). When h 6= 0, the basic idea of the proof consists in the parametrization of the free boundary in the form r0 + b(θ ) where b is the amplitude of the perturbation caused by h. To our knowledge the case of a general domain Ω is far from being solved. Further works on elliptic problems with discontinuous nonlinearities may be found in [17,18,20], and the references given there. This paper is organized as follows. In Section 2, we discuss the case of the reduced problem. In Section 3, using a parametrization of the unknown free boundary, we will see that (P) is equivalent to a nonlinear integral equation. In Section 4, the implicit function theorem is used to analyze the integral equation. Section 5 is devoted to the regularity of the free boundary. Finally the last section is devoted to an appendix which contains some useful results. 2. The reduced problem We consider the following problem:  −∆u = f (u)H (u − µ) in Ω , u=h on ∂Ω = S,

(1)

where Ω is the unit ball of Rn , n ≥ 3; f : R → R, h : Rn → R are given functions H : R → R is the Heaviside step function i.e. H (t) = 1 if t > 0 and H (t) = 0 if t ≤ 0 ∆ is the Laplace operator in Rn and µ is a real parameter. When h ≡ 0, the problem (1) is called the reduced problem, i.e.  −∆u = f (u)H (u − µ) in Ω , u=0 on ∂Ω .

(2)

This section gives the existence of solutions of (2). Definition 2.1. A solution of (2) is a function u ∈ W 2, p (Ω ) distribution.

T

1, p

W0 (Ω )( p > 1) satisfying (2) in the sense of

Let λ1 be the first eigenvalue of −∆ in H01 (Ω ) with the corresponding eigenfunction ϕ1 > 0 in Ω . Suppose that: (H1 ) f (u) ≥ 0, ∀u ≥ 0, f is k-Lipschitzian with k ∈ (0, λ1 ) and nondecreasing. (H02 ) There exists µ > 0 such that for n ≥ 3, we have 2λ1

kϕ1 k L 1 f (µ) < 2 µ kϕ1 k L 2

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The classical variational method cannot be applied directly to these problems since the Euler–Lagrange functionals associated with them are not of class C 1 . In this case, the dual variational method will be employed (see [4] for more details). Let p(u) := f (u)H (u − µ) and m > 0 (large enough) such that Pm (u) = p(u) + mu is strictly increasing. Then equation (2) can be written as  −∆u + mu = Pm (u) in Ω , u=0 on ∂Ω . Let b p (s) :=



Pm (s) T = [mµ, f (µ) + mµ]

if s = 6 µ, if s = µ,

and p ∗ be the inverse function of b p , i.e. p ∗ (σ ) = s ⇔ σ ∈ b p (s). The function p ∗ is defined on R, continuous and verifies p ∗ (σ ) = µ ⇔ mµ ≤ σ ≤ f (µ) + mµ. Let P ∗ (σ ) =

σ

Z

p ∗ (τ )dτ ∈ C 1 (R, R).

0

We define the Green operator G ∈ l(E, E) where E = L 2 (Ω ) as follows: \ 1,2 G(σ ) = u ⇔ (−∆ + m)u = σ, u ∈ W 2,2 (Ω ) W0 (Ω ). Let us denote by J :E →R the functional  1 J (σ ) = P (σ ) − σ G(σ ) dx. 2 Ω Z 

∗

Note that J ∈ C 1 (E, R). Lemma 2.1 ([4]). Let σ ∈ E be such that J 0 (σ ) = 0; then u = G(σ ) is a solution of (2). By using the same technique as in [4], we should have that the problem (2) has at least two solutions. The first solution u is a global minimum; the second solution v is a critical point obtained from the Ambrosetti–Rabinowitz theorem [5]. More precisely, we have the following result: Proposition 2.1. Suppose that f verifies the hypotheses (H1 ), (H02 ). Then the problem (2) has two solutions u, v ∈ T W 2,2 (Ω ) W01,2 (Ω ). Moreover u, v are positive and radial and the free boundaries {x ∈ Ω : u(x) = µ} and {x ∈ Ω : v(x) = µ} are respectively two spheres of Rn centered at the origin with radius r1 , r2 ∈ (0, 1). The proof of Proposition 2.1 is essentially contained in [3], p. 46. See also [6] and [7], p. 1066. 3. The perturbed problem In this section, h is supposed not identically zero (h 6= 0). Let r0 denote one of the values r1 and r2 of Proposition 2.1. As h is small, we will look for the free boundary in the form r0 + b(θ ), where b is a smooth function, which is supposed small in some sense.

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Now, the boundary value function h will be taken in the following set: A = {h ∈ W

2− 1p , p

(S, R) : p > 1}.

Consider B = {b ∈ C(S, R) and 0 ≤ r0 + b(θ ) < 1, ∀θ ∈ S}. And we define w = {x ∈ Ω : u(x) > µ}

which is an open and bounded set.

We need the following result Proposition 3.1. Assume that the condition (H1 ) is satisfied; then the problem  −∆u = f (u) in w, u=µ on ∂w

(3)

has a unique solution u ∗ ∈ H 1 (w), ∀µ > 0. Proof. (i) Existence. By the assumption (H1 ) there exists k ∈ (0, λ1 ), β > 0 such that 0 ≤ f (u) ≤ k|u| + β. We claim that the problem (3) has a supersolution u ∈ C 2,α (w) ∩ C 1,α (w) for a certain α ∈ (0, 1). In fact, as k ∈ (0, λ1 ), then the linear problem  −∆u = ku + β in Ω (3i) u=µ on ∂Ω has a unique solution u ∈ C 2,α (Ω ) ∩ C 1,α (Ω ) for a certain α ∈ (0, 1). See [16], p. 107. From a maximum principle [21], Theorem 2.5, we deduce that u > µ in Ω . Consequently u > µ in w ⊂ Ω . Then 

−∆u = ku + β u>µ

in w on ∂w.

This means that u is a supersolution for the problem (3). We see also that u = µ is a subsolution of problem (3). Now a classical result (see [14]) shows that (3) has a solution u ∗ with µ ≤ u ∗ ≤ u. (ii) Uniqueness. Suppose that v1 and v2 are two solutions of (3); then  −∆vi = f (vi ) in w, vi = µ on ∂w for i = 1, 2. Define u = v1 − v2 and 1 f (v1 ) − f (v2 ) if v1 6= v2 ρ := k v1 − v2 We see that the function u verifies  −∆u = kρu in w, u=0 on ∂w.

and

ρ=0

if v1 = v2 .

(4)

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Let λi (ρ), i = 1, 2, . . ., denote the eigenvalues of (4). Then λ1 (ρ) ≥ λ1 (1) = λ1 . Since ρ ≤ 1 and k < λ1 , we have 0 ≤ kρ < λ1 (ρ), which implies that (4) has only the trivial solution. This contradiction shows that v1 ≡ v2 . Therefore (3) has a unique solution. This completes the proof of Proposition 3.1.  We denote by χw the characteristic function of w; then we have the following lemma: Lemma 3.1. Assume (H1 ), and suppose that hypothesis (H3 ) holds for some µ > 0. Let  ∗ u in w, v= µ in Ω \ w Then the problem  −∆u = f (v)χw (r, θ ) in Ω , u=h on ∂Ω

(5)

has a unique solution U0 ∈ C 1,α (Ω , R) with α = 1 − np Moreover if U0 (r0 + b(θ ), θ ) = µ with 0 ≤ |h|∞ < µ then U0 is a solution of (1). Proof. First let us observe that f (v)χw ∈ L p (Ω ) for p > 1; since | f (v)χw | ≤ | f (v)| ≤ k|v| + β and v is bounded in Ω , we will have f (v)χw ∈ L p (Ω ) for all p > 1 and from the Theorem 9.15, p. 241 of [16], there exists a solution U0 of problem (5) in W 2, p (Ω ). For p > n, W 2, p (Ω ) ⊂ C 1,α (Ω , R) with α = 1 − np . We remark also that U0 verifies  −∆U0 = f (v) in w, −∆U0 = 0 in Ω \ w  U0 = h on ∂Ω . Moreover, if there exists b ∈ B such that U0 (r0 + b(θ ), θ ) = µ, then U0 will be a solution of  −∆U0 = f (v) in w,    U0 = µ on ∂w in Ω \ w −∆U0 = 0   U0 = h on ∂Ω . The uniqueness of the solution in w implies that U0 = u ∗

in w.

Hence U0 satisfies  −∆U0 = f (U0 ) U0 = µ

in w, on ∂w.

By the maximum principle min U0 = min U0 = µ. w

∂w

Consequently, we will have U0 > µ in w. Similarly, we will obtain  −∆U0 = 0 in Ω \ w, U0 = h on ∂Ω  U0 = µ on ∂w.

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As 0 ≤ |h|∞ < µ, then max U0 = max U0 = µ ∂w

Ω \w

and consequently U0 < µ in Ω \ w. Therefore the function U0 verifies  −∆U0 = f (U0 )H (U0 − µ) in Ω , U0 = h on ∂Ω .  4. Resolution of the integral equation In this section, we will show that the relation U0 (r0 + b(θ ), θ ) = µ has a solution in B. For that we need the following result which gives an integral representation for the solution U0 . Theorem 4.1. The solution U0 of problem (5) has the following representation: Z Z U0 (x) = P(x, y)h(y)dS − G(x, y) f (v(y))χw dy ∂Ω

Ω

where P is the Poisson kernel and G is the Green function. (See Appendix A.) Remark 4.1. The solution U0 of the corresponding problem (5) does not belong to the class C 2 ; then we cannot apply Theorem 12, p. 35 of [14]. Proof of Theorem 4.1. The proof is based on the following lemmas. Lemma 4.1. If a function is defined by Z P(x, y)h(y)dS v1 (x) = S

then v1 satisfies the following problem:  ∆v1 = 0 in Ω , v1 = h on ∂Ω = S. Proof of Lemma 4.1 (See Theorem 2.6 p. 20 of [16]). Lemma 4.2. Let v2 (x) be the function defined by Z v2 (x) = − G(x, y) f (v(y))χw dy. Ω

Then v2 verifies the Poisson’s equation −∆v2 = f χw in Ω Proof of Lemma 4.2. We know that the Green function can be written as G(x, y) = F(x, y) − φ(x, y)

(See [13], p. 58.)

where F(x, y) is the fundamental solution of the Laplacian and φ(x, y) satisfies  ∆ y φ(x, .) = 0 in Ω , φ(x, .) = F(x, .) on ∂Ω = S. We can write v2 (x) as v2 (x) = v3 (x) − v4 (x) where v3 (x) = −

Z Ω

F(x, y) f (v(y))χw dy

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and v4 (x) = −

Z Ω

φ(x, y) f (v(y))χw dy.

Firstly, by using the fact that φ(x, y) is harmonic, we have  Z  φ(x, y) f (v(y))χw dy ∆v4 (x) = ∆ − Ω Z ∆φ(x, y) f (v(y))χw dy =− Ω

= 0. Now, for v3 (x), let g(v) = f (v)χw which is in L p (Ω ), for p > 1. It remains to prove that −∆v3 = g; for that, we have the following result: Proposition 4.1. The function v3 is in Beppo–Levi space B 2, p (Ω ) and verifies −∆v3 = g. Remark 4.2. Beppo–Levi space is defined as follows:   ∂ 2u p B 2, p (Ω ) = u ∈ L loc (Ω ), ∈ L p (Ω ), i, j = 1, . . . , n . ∂ xi x j Proof of Proposition 4.1. Firstly, we know that for every regular open set Ω , we have Z Z ∂ 2 F(x, y) ∂F ∂ 2 v3 (x) =− [g(v(y)) − g(v(x))]dy + g(v(x)) (x, y)n j (y)dS ∂ xi x j ∂ x x ∂ i j Ω ∂ Ω xi (See [12], vol. 1, p. 292.) We know also that a fundamental solution is given by F(x, y) =

1 |x − y|2−n , n(n − 2)wn

∀n ≥ 3

where wn is the volume of the unit ball in Rn ; this implies that n X ∂F ∂F (x, y) = (x, y)n i (y) ∂n x ∂ xi i=1

∂F (x, y)n(y) ∂x (x − y) n(y). =− nwn |x − y|n

=

x−y For y ∈ ∂Ω , the unit normal is given by n(y) = |x−y| . Therefore ∂F 1 (x, y) = − , ∀n ≥ 3. ∂n x nwn |x − y|n−1

Now, since v3 ∈ B 2, p (Ω ) (see [12], Vol. 4, corollary 2, p. 413), then Z ∂F ∆v3 = g(v(x)) dS ∂n x Z ∂Ω −g(v) dS = nwn ∂ Ω = −g(v) R where |x − y| = 1 and ∂ Ω dS = nwn . 

a.e. in Ω .

(∗)

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Now, we pass to polar coordinates, we find Z 1 Z Z (r 0 )n−1 f (v(r 0 , θ 0 ))χw (r 0 , θ 0 )G(r, θ, r 0 , θ 0 )dr 0 . P(r, θ, θ 0 )h(θ 0 )dθ 0 − dθ 0 U0 (r, θ ) = S

S

0

We define the following operator: F : A × R+ × B → C(S, R)

by : F(h, µ, b)(θ ) = U0 (r0 + b(θ ), θ ) − µ,

i.e. F(h, µ, b)(θ ) =

Z

P(r0 + b(θ ), θ, θ 0 )h(θ 0 )dθ 0 S

Z

dθ 0

−

r0 +b(θ 0 )

Z

(r 0 )n−1 f (v(r 0 , θ 0 ))G(r0 + b(θ ), θ, r 0 , θ 0 )dr 0 − µ.

0

S

Then, we have the following theorem which characterizes the free boundary. Theorem 4.2. Under the assumptions (H1 ), (H2 ), (H3 ), there exists a neighborhood V of 0 in A and a unique application b : V → B which is continuously differentiable and such that: (i) b(0) = 0. (ii) F(h, µ, b(h)) = 0 for h ∈ V . The proof of Theorem 4.2 is based on the implicit function theorem [14] and the following lemmas. Lemma 4.3. The operator F defined from A × R+ × B into C(S, R) is continuously differentiable at (0, µ, 0) ∈ A × R+ × B. Lemma 4.4. Under hypothesis (H2 ), the value r0 corresponding to the free boundary of the reduced problem (h = 0) 1 1 is different from ( n2 ) n−2 , i.e. r0 6= ( n2 )( n−2 ) . Let D j F denote the Frechet derivative of F, with respect to the variable of order j ( j = 1, 2, 3). Lemma 4.5. The operator D3 F(0, µ, 0) ∈ L(C(S)) has a bounded inverse. Proof of Lemma 4.3. We prove that F is continuously differentiable with respect to the variable b. We shall prove that D3 F(h, µ, b) is given by "Z ∂P D3 F(h, µ, b)β(θ ) = dθ 0 (r0 + b(θ ), θ, θ 0 )h(θ 0 ) ∂r S # Z Z r0 +b(θ 0 ) 0 n−1 0 0 0 ∂G 0 0 0 − dθ (r ) f (v(r , θ ))dr (r0 + b(θ ), θ, r , θ ) β(θ ) ∂r S 0 Z − dθ 0 (r0 + b(θ 0 ))n−1 f (v(r0 + b(θ 0 ), θ 0 ))G(r0 + b(θ ), θ, r0 + b(θ 0 ), θ 0 )β(θ 0 ). S

Let L = D3 F. Then F(h, µ, b + β)(θ ) − F(h, µ, b)(θ ) − Lβ(θ ) = I1 + I2 + I3 . Here Z

I1 =

P(r0 + b(θ ) + β(θ ), θ, θ 0 )h(θ 0 )dθ 0 − SZ − Pr h(r0 + b(θ ), θ, θ 0 )h(θ 0 )dθ 0 β(θ ) S

Z

P(r0 + b(θ ), θ, θ 0 )h(θ 0 )dθ 0 S

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Z

dθ 0

I2 = − ZS

dθ

+

0

r0 +b(θ 0 )+β(θ 0 )

Z

Z0 r0 +b(θ 0 ) 0

S

Z dθ

+

0

Z

r0 +b(θ 0 )

(r 0 )n−1 dr 0 f (v(r 0 , θ 0 ))G(r0 + b(θ ) + β(θ ), θ, r 0 , θ 0 )

(r 0 )n−1 dr 0 f (v(r 0 , θ 0 ))G(r0 + b(θ ), θ, r 0 , θ 0 ) (r 0 )n−1 dr 0 f (v(r 0 , θ 0 ))G r (r0 + b(θ ), θ, r 0 , θ 0 )β(θ )

0

S

and Z

dθ 0 [r0 + b(θ 0 )]n−1 f (v(r0 + b(θ 0 ), θ 0 ))G(r0 + b(θ ), θ, r0 + b(θ 0 ), θ 0 )β(θ 0 )

I3 = S

where Pr =

∂P ∂r

Gr =

∂G ∂r

are respectively the partial derivatives of P and G with respect to r . First, let Z ψh(r, θ ) := P(r, θ, θ 0 )h(θ 0 )dθ 0 , (r, θ ) ∈ (0, 1) × S. S

By Taylor’s theorem, we have ψh(r0 + b(θ ) + β(θ ), θ ) − ψh(r0 + b(θ ), θ ) −

∂ψ h(r0 + b(θ ), θ )β(θ ) = o(|β|∞ ) ∂r

Then I1 = o(|β|∞ )

when |β|∞ → 0.

Let Z

dθ 0

T =

r0 +b(θ 0 )

Z

(r 0 )n−1 dr 0 f (v(r 0 , θ 0 ))G(r0 + b(θ ) + β(θ ), θ, r 0 , θ 0 ).

0

S

Adding the term T to I2 and subtracting it, we find I2 = J1 + J2 with Z Z r0 +b(θ 0 ) J1 = − dθ 0 (r 0 )n−1 dr 0 f (v(r 0 , θ 0 ))G(r0 + b(θ ) + β(θ ), θ, r 0 , θ 0 ) ZS Z0 r0 +b(θ 0 ) 0 (r 0 )n−1 dr 0 f (v(r 0 , θ 0 ))G(r0 + b(θ ), θ, r 0 , θ 0 ) + dθ 0 S Z Z r0 +b(θ 0 ) 0 (r 0 )n−1 dr 0 f (v(r 0 , θ 0 ))G r (r0 + b(θ ), θ, r 0 , θ 0 , θ )β(θ ) + dθ 0

S

and Z

dθ 0

J2 = −

Z

S

r0 +b(θ 0 )+β(θ 0 )

r0 +b(θ 0 )

(r 0 )n−1 f (v(r 0 , θ 0 ))G(r0 + b(θ ) + β(θ ), θ, r 0 , θ 0 )dr 0 .

By similar arguments, we obtain that J1 = o(|β|∞ )

when |β|∞ → 0.

Thus, it remains to estimate J2 + I3 . We add the following term to J2 + I3 and subtract it: Z Z r0 +b(θ 0 )+β(θ 0 ) 0 (r 0 )n−1 f (v(r 0 , θ 0 ))G(r0 + b(θ ), θ, r 0 , θ 0 )dr 0 dθ S

r0 +b(θ 0 )

when |β|∞ → 0.

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and we find Z dθ

J2 + I3 = −

0

S

r0 +b(θ 0 )+β(θ 0 )

Z

r0 +b(θ 0 )

(r 0 )n−1 f (v(r 0 , θ 0 ))dr 0 [G(r0 + b(θ ) + β(θ ), θ, r 0 , θ 0 )

− G(r0 + b(θ ), θ, r 0 , θ 0 )] ! " Z Z r0 +b(θ 0 )+β(θ 0 ) 0 n−1 0 0 0 0 0 0 (r ) f (v(r , θ ))G(r0 + b(θ ), θ, r , θ )dr − dθ r0 +b(θ 0 )

S

# − (r0 + b(θ 0 ))n−1 f (v(r0 + b(θ 0 ), θ 0 ))G(r0 + b(θ ), θ, r0 + b(θ 0 ), θ 0 )β(θ 0 ) . We will estimate only the second integral; the first integral can be treated similarly. We know that the Green function can be decomposed into two parts, the regular and singular parts. (See Appendix A.) We try to estimate the singular part. Let " Z r0 +b(θ 0 )+β(θ 0 ) Z 0 0 n−1 0 K := − dθ (r ) dr f (v(r 0 , θ 0 ))G(r0 + b(θ ), θ, r 0 , θ 0 ) S

 −

r0 +b(θ 0 )

r0 + b(θ 0 ) r0

n−1

# f (v(r0 + b(θ ), θ ))G(r0 + b(θ ), θ, r0 + b(θ ), θ ) . 0

0

0

0

Let D be the ball of radius γ ∈ (0, 1) about (r0 + b(θ ), θ ): Z Z Z K = Q= Q+ Q E

E∩D

E∩D c

where E denotes the region of integration and Q the integrand. Then there exist two constants C, C 0 such that Z Z γ 0 Q ≤ C r 2−n r n−1 dr ≤ Cγ 2 . E∩D

0

We use the fact that f is bounded since it is continuous on a compact set. Outside of D and using the mean value formula, we will deduce that Z Z ≤ C0 Q r 1−n |β|∞ dr dθ ≤ C|β|n∞ γ 1−n . c c E∩D

E∩D

Combining the two results gives Z Q ≤ C1 γ 2 + C2 γ 1−n |β|n . ∞ E

Choosing γ = |β|1−τ ∞ , τ > 0. It is easy to see that Z Q = o(|β|∞ ) E

when β → 0. Then finally, F is continuously differentiable at (0, µ, 0) ∈ A × R+ × B. Now, the operator L : U → L(C(S)) is a continuous mapping in a neighborhood U of (0, µ, 0). In fact L can be written as ∂u (r0 + b(θ ), θ )β(θ ) − (φβ)(θ ) ∂r R where (φβ)(θ ) = S dθ 0 (r0 + b(θ 0 ))n−1 f (v(r0 + b(θ 0 ), θ 0 ))G(r0 + b(θ ), θ, r0 + b(θ 0 ), θ 0 )β(θ 0 ). φ is a continuous mapping, since the singularity of G is integrable and ∂u ∂r depends continuously on (h, µ, b) in a neighborhood of (0, µ, 0). 

S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348

2339

Proof of Lemma 4.4. According to the Proposition 2.1, we see that in the case of the reduced problem (h = 0), the solution U0 is radial and w = {(r, θ ) ∈ Ω : U0 (r, θ ) > µ} = {(r, θ ) : 0 < r < r0 , θ ∈ S}. From 

−∆U0 = 0 U0 = µ

in Ω \ w, on ∂w,

we obtain  (n − 1) 0  −U000 − U0 = 0 r U (1) = 0   0 U0 (r0 ) = µ.

in r0 < r < 1 (R1)

The solution of (R1) is given by U0 (r ) =

µr 2−n r02−n

−

−1

µ r02−n

−1

.

Then ∂U0 + (2 − n)µ . (r ) = ∂r 0 r0 − r0n−1 Now, consider the problem  −∆U0 = f (U0 ) in w, U0 = µ on ∂w. This implies that  1−n ∂/∂r (r n−1 ∂U0 /∂r ) = f (U0 ) −r 0 U (0) = 0  0 U0 (r0 ) = µ.

in 0 < r < r0 (R2)

Consequently, we obtain that Z r0 ∂U0 − 1−n (r ) = −r0 s n−1 f (U0 (s))ds. ∂r 0 0 Since U0 ∈ C 1,α (Ω ), then ∂U0 + ∂U0 − (r0 ) = (r ), we obtain that ∂r ∂r 0 Z r0 (2 − n)µ 1−n = −r s n−1 f (U0 (s))ds 0 r0 − r0n−1 0 Z r0 −r0 1−n ≤ −r0 s n−1 f (µ)ds = f (µ). n 0 Since by the maximum principle, U0 > µ in w, this implies that f (U0 ) ≥ f (µ) ( f is nondecreasing); then we have two possibilities. −r0 (a) (2−n)µ n−1 = n f (µ). r0 −r0

In this case, we will obtain that Let ψ(r ) :=

n(n−2) ; r 2 −r n

f (µ) µ

=

n(n−2) . r02 −r0n

this function has a minimum value Mn =

From hypothesis (H2 ), we have

f (µ) µ

1

> Mn ; then r0 6= ( n2 ) n−2 .

n(n−2) 2 n ( n2 ) n−2 −( n2 ) n−2

1

reached at the point r0 = ( n2 ) n−2 .

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(b)

S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348 (2−n)µ r0 −r0n−1

<

In this case

−r0 n

f (µ).

n(2−n) r02 −r0n

<

− f (µ) µ

which implies that ψ(r0 ) >

f (µ) µ .

1

If r0 = ( n2 ) n−2 , then Mn >

contradiction with hypothesis (H2 ). This completes the proof of Lemma 4.4.

f (µ) µ

which is a



Proof of Lemma 4.5. We know that Z ∂u n−1 D3 F(0, µ, 0)β(θ ) = f (µ)G(r0 , θ, r0 , θ 0 )β(θ 0 )dθ 0 . (r0 , θ )β(θ ) − r0 ∂r S This implies that D3 F(0, µ, 0)β(θ ) =

−r01−n

r0

Z

s

n−1

0

f (u(s))dsβ(θ ) − r0n−1

Z

f (µ)G(r0 , θ, r0 , θ 0 )β(θ 0 )dθ 0 . S

To prove that D3 F is invertible, let us first show the following results. Let K be the integral operator defined by Z Kβ(θ ) = G(r0 , θ, r0 , θ 0 )β(θ 0 )dθ 0 S

considered from the space C(S) into C(S). Then we have Claim 4.1. The integral operator K is continuous and compact in C(S). Claim 4.2. D3 F(0, µ, 0)β(θ ) = −r01−n

r0

Z 0

s n−1 f (u(s))dsβ(θ ) − r0n−1 f (µ)Kβ(θ )

is invertible if Z r0 1−n r0 s n−1 f (u(s))ds + r0n−1 f (µ)σ 6= 0 0

for any σ an eigenvalue of K . Proof of Claim 4.1. Using Cartesian coordinates, we can rewrite as follows: Z ˜ ˜ K β(x) = G(x, y)β(y)dw n (y) ∂ B(0,r0 )

where dwn (y) is the surface element of the sphere ∂ B(0, r0 ). • Let M be any set of measurable and equibounded functions. There exists a positive constant c1 > 0 such that ˜ ≤ c1 , for all β˜ ∈ M. We prove that K M is equibounded and equicontinuous. |β| ˜ ≤ c1 |K β|

Z ∂ B(0,r0 )

Z ≤ c1

∂ B(0,r0 )

|G(x, y)|dwn (y) |G 0 (|x − y|)|dwn (y) + c2 .

Recall that G 0 (|y||x − |y|y 2 |) is a smooth function on ∂ B and bounded by a constant c2 ; it remains to estimate the first integral. For this we fix ε > 0 sufficiently small such that Bε (x) is a ball of radius ε centered at x. We have Z ˜ ≤ c1 |K β| |x − y|2−n dwn (y) + c2 ∂ B(0,r0 ) Z Z = c1 |x − y|2−n dwn (y) + c1 |x − y|2−n dwn (y) + c2 . ∂ B∩Bε (x)

The function |x −

y|2−n

is bounded on ∂ B \ Bε (x).

∂ B\Bε (x)

S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348

Now, we prove that Z |x − y|2−n dwn (y) ∂ B(0,r0 )∩Bε (x)

2341

is bounded.

Let χ∂ B(0,r0 )∩Bε (x) be the characteristic function of ∂ B(0, r0 ) ∩ Bε (x); then we have Z Z |x − y|2−n χ∂ B(0,r0 )∩Bε (x) dwn (y) |x − y|2−n dwn (y) = ∂ B(0,r0 ) ∂ B(0,r0 )∩Bε (x) Z Z |z 0 − z|2−n dwn (z) |x − y|2−n dwn (y) = r0 ≤ ∂ B(0,1)

∂ B(0,r0 )

where x = r0 z 0 and y = r0 z. The last integral is uniformly bounded. (See Lemma 6.3, Appendix C.) • Now, we show that K (M) is equicontinuous. We have Z ˜ ˜ ˜ |(K β)(x1 ) − (K β)(x2 )| = [G(x1 , y) − G(x2 , y)]β(y)dwn (y) ∂B Z ≤ c1 |G(x1 , y) − G(x2 , y)|dwn (y). ∂B

Let Bε (x1 ), Bε (x2 ) be two balls centered respectively at x1 , x2 with radius ε > 0 sufficiently small such that Bε (x1 ) ∩ Bε (x2 ) = ∅ and let η > 0 such that |x1 − x2 | ≤ η; then Z Z c1 |G(x1 , y) − G(x2 , y)|dwn (y) = c1 ∂B

∂ B∩(Bε (x1 )∪Bε (x2 ))

|G(x1 , y) − G(x2 , y)|dwn (y)

Z + c1

∂ B\(Bε (x1 )∪Bε (x2 ))

|G(x1 , y) − G(x2 , y)|dwn (y)

= I1 + I2 .

The term I2 is bounded since the Green function is uniformly continuous on ∂ B \ (Bε (x1 ) ∪ Bε (x2 )). For the term I1 , we have Z I 1 = c1 |(G 0 (|x1 − y|) − G 0 (|x2 − y|)) + ψ(x1 , x2 , y)|dwn (y) ∂ B∩(Bε (x1 )∪Bε (x2 ))

where ψ is the analytical part of the Green function. Now, we estimate Z c1 |G 0 (|x1 − y|) − G 0 (|x2 − y|)dwn (y) := I ∂ B∩(Bε (x1 )∪Bε (x2 )) Z c1 I = ||x1 − y|2−n − |x2 − y|2−n |dwn (y) n(2Z − n)wn ∂ B∩(Bε (x1 )∪Bε (x2 )) ≤ c4 |(x1 − y)2−n − (x2 − y)2−n |dwn (y). ∂ B∩(Bε (x1 )∪Bε (x2 ))

Using the algebraic identity a n − bn = (a − b)

n X j=1

We conclude that

a n− j b j−1 .

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S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348

I ≤

≤

=

≤

n−2− j   j−1 Z n−2  X (x1 − x2 ) 1 1 c4 dwn (y) (x − y)(x − y) (x − y) (x − y) 1 2 1 2 ∂ B∩(Bε (x1 )∪Bε (x2 )) j=1 n−1− j  j Z n−2  X 1 1 c4 |x1 − x2 | dwn (y) (x − y) (x − y) 1 2 ∂ B∩(Bε (x1 )∪Bε (x2 )) j=1 n−1− j  j Z n−2  X 1 1 c4 |x1 − x2 | dwn (y) (x − y) (x − y) 1 2 ∂ B∩Bε (x1 ) j=1 n−1− j   j Z n−2  X 1 1 + c4 |x1 − x2 | dwn (y) (x − y) (x − y) 1 2 ∂ B∩Bε (x2 ) j=1 " Z # n−1− j  j Z n−2  n−2  X X 1 1 |x1 − x2 | c5 dwn (y) + c6 dwn (y) . ∂ B∩Bε (x1 ) j=1 (x 1 − y) ∂ B∩Bε (x2 ) j=1 (x 2 − y)

We can show by the previous method that the last two integrals are bounded. Finally, we obtain ˜ 1 ) − (K β)(x ˜ 2 )| ≤ C|x1 − x2 | |(K β)(x e is a compact operator. And by the Arzela–Ascoli theorem, we affirm that K β Proof of Claim 4.2. We have ( G(r0 , θ, r0 , θ 0 ) = r02−n

∞ X 1 − r02l+n−2 r0n−2 − 1 − (n − 2)N wn 2l + n − 2 l=1

!

NX (n,l)



) Ylm (θ )Ylm (θ 0 )

m=1

where N (n, l) = (2l + n − 2)

(l + n − 3)! l!(n − 2)!

∀l ∈ N, ∀n ≥ 3

Ylm are the spherical harmonic functions of degree l and wn denotes the volume of the unit ball in Rn (see Appendix A). From this expression, we read the eigenvalues σlm and the eigenfunctions of K to be r0n−2 σlm = −

1 − r02l+n−2 2l + n − 2

φlm = Ylm , l 6= 0.

For l = 0, we have r0n−2 σ00 = −

1 − r0n−2 n−2

φ00 = √

1 . nwn

Now, we show that Z r0 1−n s n−1 f (u(s))ds 6= −r0n−1 f (µ)σlm . r0 0

We have r01−n

r0

Z 0

s n−1 f (u(s))ds ≥ r01−n =

r0

Z

s n−1 f (µ)ds

0

r0 f (µ). n

It follows that ! Z r0 1 1 − r02l+n−2 r0 1−n n−1 n−1 n−1 f (µ) + r0 f (µ)σlm = r0 f (µ) − . r0 s f (u(s))ds + r0 f (µ)σlm ≥ n n 2l + n − 2 0

S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348

2343

For l ≥ 1, 1 1 − r02l+n−2 − ≥ n 2l + n − 2



 1 1 1 2l+n−2 − (1 − r0 ) = r02l+n−2 > 0. n n n

For l = 0, 1 1 − r0n−2 − = 0. n n−2 1

This implies r0 = (2/n) n−2 which is impossible by the result of Lemma 4.4.



We apply the implicit function theorem to complete the proof of Theorem 4.2; therefore we have shown that when r0 takes the value r1 or r2 , the problem (P) has a solution. Hence, (P) has two solutions u and v corresponding respectively to the values r1 and r2 . This completes the proof of Theorem 1.1. 5. Regularity of the free boundary We prove in this section that the free boundary has a regularity C 1,α for some 0 < α < 1. Proposition 5.1. Let b ∈ C(S); if u is a solution of (P) such that u(r0 + b(θ ), θ ) = µ, then b ∈ C 1,α (S), for some α ∈ (0, 1). Proof. Let F(θ, g) = u(r0 + g, θ ) − µ = 0, First, note that the radial solution u ∈

θ ∈ S, g ∈ R.

W 2, p

⊂ C 1,α , with α = 1 − np , p > n; hence F is of class C 1,α . When g is

small, u is close in C 1,α to the solution of the reduced problem; hence Let θ0 ∈ S, and suppose g0 = b(θ0 ); we have

∂u ∂r (r, θ )

6= 0 for g small.

F(θ0 , g0 ) = u(r0 + b(θ0 ), θ0 ) − µ = 0. and Dg F(θ0 , g0 ) =

∂u (r0 + g0 , θ0 ). ∂g

We know that ∂u ∂u (r0 , θ0 ) = (r0 , θ0 ) 6= 0. ∂g ∂r Now, by the classical implicit function theorem, we deduce that there exists V (θ0 ), a neighborhood of θ0 , and V (g0 ), a neighborhood of g0 , and a unique function g in C 1,α such that g : V (θ0 ) → V (g0 )

and

g(θ0 ) = g0 .

We know that u(r0 + b(θ ), θ ) − µ = 0, ∀θ ∈ V (θ0 ); then by the uniqueness, b(θ ) = g(θ ), ∀θ ∈ V (θ0 ), which implies that b ∈ C 1,α (V (θ0 )). Since θ0 ∈ S is arbitrary, then we will have that b ∈ C 1,α (S).  Acknowledgments The authors thank the anonymous referee for valuable suggestions and comments. The second author expresses his grateful thanks to the I.C.T.P. of Trieste in Italy for kind hospitality during his visit as a regular associate. Appendix A. Spherical harmonics We denote by       ∂ 2 ∂ 2 ∂ 2 + + ··· + ∆q = ∂ x1 ∂ x2 ∂ xq the Laplace operator.

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S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348

Definition 6.1 (See [22]). Let Hn (x) be a homogeneous polynomial of degree n in q dimensions, which satisfies ∆q Hn (x) = 0. Then Sn (ξ ) =

1 Hn (τ ξ ) = Hn (ξ ) τn

is called a (regular) spherical harmonic of order n in q dimensions. Theorem 6.1 (See [22]). Let Sn, j be an orthonormal set of N (q, n) spherical harmonics of order n and dimension q. Then NX (q,n)

Sn, j (ξ )Sn, j (η) =

j=1

N (q, n) Pn (ξ, η) wq q

2 . where Pn (t) is the Legendre polynomial of degree n and dimension q, wq = q2π Γ ( q2 )

This theorem is called the addition theorem as it reduces to the addition theorem for the function cos ϕ in the twodimensional case (see Muller [22]). Lemma 6.1 (See [22]). For q ≥ 3, 0 ≤ x < 1, and −1 ≤ t ≤ 1 ∞ X

Cn (q)x n Pn (q, t) =

n=0

1 (1 + x 2 − 2xt)

q−2 2

.

Here Cn (q) =

Γ (n + q − 2) . Γ (q − 2)Γ (n + 1)

Expansion of the Green’s function in spherical harmonics A fundamental solution for Laplace’s equation is G 0 (r ) =

1 r 2−n , n(n − 2)wn

∀n ≥ 3

where wn is the volume of the unit ball in R n . Then Green’s function for the sphere is   y G(x, y) = G 0 (|x − y|) − G 0 |y| x − 2 see [13]. |y| On passing to polar coordinates, we find (C0)

q p G(r0 , θ, r0 , θ 0 ) = G 0 (r0 2(1 − cos γ )) − G 0 ( 1 + r04 − 2r02 cos γ ).

Here cos γ = θ.θ 0 . Now expand the second term on the right in Legendre polynomials (see [22]): q ∞ X 1 frac(l + n − 3)!l!(n − 3)!(r02 )l Pl (N , cos γ ) (C1) G 0 ( 1 + r04 − 2r02 cos γ ) = n(n − 2) l=0 and apply the Theorem 6.1 (addition theorem); we find (C2)

Pl (N , cos γ ) =

X,l) N wn n(N Ylm (θ )Ylm (θ 0 ) n(N , l) m=1

S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348

2345

with Ylm an orthonormal basis for the spherical harmonics of degree l and dimension N , and for each l the number of Ylm is (C3) n(N , l) = (2l + n − 2)

(l + n − 3)! . l!(n − 2)!

Using (C1), (C2) and (C3) we have (C4)

q G 0 ( 1 + r04 − 2r02 cos γ ) =

inf ty n(N X X,l) r02l 1 − Ylm (θ )Ylm (θ 0 ). (2 − n)N wn 2l + n − 2 l=1 m=1

For the first term on the right in (C0), we use Lemma 6.1, and the addition theorem again, obtaining ) (  n(N ∞  X,l) X r0n−2 1 2−n 2−n 0 0 − r0 Ylm (θ )Ylm (θ ) . (C5) G 0 (r0 , θ, r0 , θ ) = r0 (n − 2)N wn 2l + n − 2 m=1 l=1 The series converges uniformly in |θ − θ 0 | ≥ η > 0. Combining (C4) and (C5) gives finally the expansion of our kernel in spherical harmonics: ! ) ( ∞ X,l) X 1 − r02l+n−2 n(N r0n−2 − 1 2−n 0 0 Ylm (θ )Ylm (θ ) . G(r0 , θ, r0 , θ ) = r0 − (n − 2)N wn 2l + n − 2 l=1 m=1 Appendix B In this section, we will give a proof of the statements given in Remark 1.1(iv). More precisely, we will see that λ1 (R), Mn (R) tend to zero as R −→ +∞. In fact, let Ω = B(0, R), a ball of radius R > 0, and ψ1 the eigenfunction in Ω corresponding to λ1 (R). From the characterization of the first eigenvalue of −∆ with Dirichlet conditions, we have R 2 B(0,R) |∇ψ1 | dx . λ1 (R) = R 2 B(0,R) ψ1 dx Let x = Ry for y ∈ B(0, 1); then dx = R n dy and ψ1 (x) = ϕ1 (y) This implies that   n  n  X ∂ϕ1 ∂ xi 2 ∂ϕ1 2 X 2 = |∇ y ϕ1 | = ∂ yi ∂ xi ∂ yi i=1 i=1   n 2 X ∂ψ1 = R 2 |∇ψ1 |2 . = R ∂ x i i=1 Hence, 1 λ1 (R) = 2 R

R

2 n B(0,1) |∇ y ϕ1 | R dy R 2 n B(0,1) ϕ1 R dy

=

1 λ1 R2

where λ1 = λ1 (R = 1). Note also that R R B(0,R) ϕ1 (x)dx B(0,1) ϕ1 (y)dy R R = 2 2 B(0,R) ϕ1 (x)dx B(0,1) ϕ1 (y)dy is bounded by a constant independent of R. kϕ k Hence, we obtain that 2λ1 (R) 1 2L 1 tends to zero as R −→ +∞. kϕ1 k

L2

In order to prove that Mn (R) goes to zero when R −→ +∞ and for the convenience of the reader, we will do some calculations similar to those given in Proof of Lemma 4.4. If u is a solution of the following problem:  −∆u = f (u)H (u − µ) in B(0, R), u=0 on ∂ B(0, R).

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S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348

Then u verifies  −∆u = 0 u=µ

in B(0, R) \ w, on ∂w.

Here w = {x ∈ B(0, R) : u(x) > µ}. From Proposition 2.1, the solution u is radial and w = {(r, θ ) ∈ B(0, R) : u(r, θ) > µ} = {(r, θ ) : 0 < r < r0 , θ ∈ S}. This implies that  (n − 1) 0  −u 00 − u =0 r u(R) = 0   u(r0 ) = µ.

in r0 < r < R (I)

The solution of (I) is given by u(r ) =

µr 2−n r02−n − R 2−n

−

µR 2−n r02−n − R 2−n

.

Then (2 − n)µr01−n ∂u + (r0 ) = 2−n . ∂r r0 − R 2−n Now, we consider the problem  −∆u = f (u) in w, u=µ on ∂w. We obtain that  1−n ∂/∂r (r n−1 ∂u/∂r ) = f (u) −r 0 u (0) = 0  u(r0 ) = µ.

in 0 < r < r0 (II)

Consequently, we have ∂u − (r ) = −r01−n ∂r 0 Since u ∈

C 1,α (Ω ),

r0

Z

s n−1 f (u(s))ds.

0

then

∂u + ∂u − (r ) = (r ), we obtain that ∂r 0 ∂r 0 Z r0 (2 − n)µr01−n 1−n = −r0 s n−1 f (u(s))ds r02−n − R 2−n 0 Z ≤ −r01−n

r0

s n−1 f (µ)ds =

0

−r0 f (µ). n

Since by the maximum principle, u > µ in w, this implies that f (u) ≥ f (µ) ( f is nondecreasing); then we have that f (µ) n(n − 2) ≤ 2 . µ r0 − r0n R 2−n Let ψ(r ) =

n(n−2) . r 2 −r n R 2−n 1

This function has a minimum value Mn (R) =

r0 = ( n R22−n ) n−2 . We see that Mn (R) =

1 R2

1 R2

n(n−2) 2

n

( n2 ) n−2 −( n2 ) n−2

Mn which tends to zero as R −→ +∞.

reached at the point

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Appendix C Lemma 6.2. Let x be a fixed vector of the unit sphere in n dimensions ∂ Bn ; then there exists an orthonormal system {ε1 , . . . , εn } such that the points of ∂ Bn can be represented by ηn = cos θ x + sin θ ηn−1 ,

for θ ∈ [0, π].

Here (x, ηn ) = cos θ . Here also ηn−1 is a unit vector in the space spanned by ε1 , . . . , εn−1 , and (. , .) denotes the scalar product of x and ηn . Proof of Lemma 6.2. Since x1 = x is different from 0, then there exists (n − 1) vectors x2 , . . . , xn such that S = {x1 , x2 , . . . , xn } is a basis for the Euclidean space Rn . Now, we use the Gram–Schmidt orthogonalization process in the following way: yn = x1 = x yn−1 = x2 − c1 x1 y1 = xn − cn−1 xn−1 − · · · − c1 x1 where the ci are chosen for orthogonalization. Divide each yi by its norm kyi k to get εi = Now, since εn = x, then, it follows that the points of ∂ Bn can be represented by

yi kyi k , i

= 1, 2, . . . , n.

ηn = cos θ x + sin θ ηn−1 with θ ∈ [0, π].



Lemma 6.3. Let Bn (0, 1) be the unit ball in n dimensions. Then for fixed x ∈ ∂ Bn (0, 1), Z |x − y|2−n dwn (y) ≤ 2 meas(∂ Bn−1 (0, 1)) ∂ Bn (0,1)

where meas(.) is the Lebesgue measure. Proof of Lemma 6.3. According to the previous lemma, we can write y = cos θ x + sin θ ηn−1 . (See Lemma 6.2.) Here θ ∈ [0, π]. We have |x − y| = = =

p q p

(x − y, x − y) kxk2 + kyk2 − 2(x, y) 2 − 2(x, y).

Then Z I :=

∂ Bn (0,1)

|x − y|2−n dwn (y) =

π

Z

=2

Z

√ ( 2 − 2 cos θ )2−n (sin θ )n−2 dwn−1 (y)dθ

0

∂ Bn−1 (0,1)

2−n 2

meas(∂ Bn−1 (0, 1))

π

Z

(1 − cos θ )

0

Let cos θ = t, with t ∈ [−1, +1]. Then Z +1 2−n 2−n p dt 2 I =2 meas(∂ Bn−1 (0, 1)) (1 − t) 2 ( 1 − t 2 )n−2 √ 1 − t2 −1 Z +1 2−n −1 n−3 = 2 2 meas(∂ Bn−1 (0, 1)) (1 − t) 2 (1 + t) 2 dt −1

2−n 2

(sin θ )n−2 dθ.

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S. Bensid, S.M. Bouguima / Nonlinear Analysis 68 (2008) 2328–2348

≤2

−1 2

meas(∂ Bn−1 (0, 1))

+1

Z

(1 − t)

−1 2

dt

−1

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