Acta Appl Math (2013) 123:261–284 DOI 10.1007/s10440-012-9765-4 Homogenization of Steklov Spectral Problems with Indeﬁnite Density Function in Perforated Domains Hermann Douanla Received: 25 July 2011 / Accepted: 25 May 2012 / Published online: 14 June 2012 © Springer Science+Business Media B.V. 2012 Abstract The asymptotic behavior of second order self-adjoint elliptic Steklov eigenvalue problems with periodic rapidly oscillating coefﬁcients and with indeﬁnite (sign-changing) density function is investigated in periodically perforated domains. We prove that the spec- trum of this problem is discrete and consists of two sequences, one tending to −∞ and another to +∞. The limiting behavior of positive and negative eigencouples depends cru- cially on whether the average of the weight over the surface of the reference hole is positive, negative or equal to zero. By means of the two-scale convergence method, we investigate all three cases. Keywords Homogenization · Eigenvalue problems · Perforated domains · Indeﬁnite weight function · Two-scale convergence Mathematics Subject Classiﬁcation 35B27 · 35B40 · 45C05 1 Introduction In 1902, with a motivation coming from Physics, Steklov [33] introduced the following problem u = 0 in Ω, ∂u ∂n = ρλu on ∂Ω, (1.1) where λ is a scalar and ρ is a density function. The function u represents the steady state temperature on Ω such that the ﬂux on the boundary ∂Ω is proportional to the temperature. In two dimensions, assuming ρ = 1, problem (1.1) can also be interpreted as a membrane H. Douanla ( ) Department of Mathematical Sciences, Chalmers University of Technology, Gothenburg, 41296, Sweden e-mail: douanla@chalmers.se
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• Acta Appl Math (2013) 123:261284DOI 10.1007/s10440-012-9765-4

Homogenization of Steklov Spectral Problemswith Indefinite Density Function in Perforated Domains

Hermann Douanla

Received: 25 July 2011 / Accepted: 25 May 2012 / Published online: 14 June 2012 Springer Science+Business Media B.V. 2012

Abstract The asymptotic behavior of second order self-adjoint elliptic Steklov eigenvalueproblems with periodic rapidly oscillating coefficients and with indefinite (sign-changing)density function is investigated in periodically perforated domains. We prove that the spec-trum of this problem is discrete and consists of two sequences, one tending to andanother to +. The limiting behavior of positive and negative eigencouples depends cru-cially on whether the average of the weight over the surface of the reference hole is positive,negative or equal to zero. By means of the two-scale convergence method, we investigate allthree cases.

Keywords Homogenization Eigenvalue problems Perforated domains Indefiniteweight function Two-scale convergence

Mathematics Subject Classification 35B27 35B40 45C05

1 Introduction

In 1902, with a motivation coming from Physics, Steklov [33] introduced the followingproblem

u = 0 in ,u

n= u on ,

(1.1)

where is a scalar and is a density function. The function u represents the steady statetemperature on such that the flux on the boundary is proportional to the temperature.In two dimensions, assuming = 1, problem (1.1) can also be interpreted as a membrane

H. Douanla ()Department of Mathematical Sciences, Chalmers University of Technology, Gothenburg, 41296,Swedene-mail: douanla@chalmers.se

• 262 H. Douanla

with whole mass concentrated on the boundary. This problem has been later referred to asSteklov eigenvalue problem (Steklov is often transliterated as Stekloff). Moreover, eigen-value problems also arise from many nonlinear problems after linearization (see e.g., thework of Hess and Kato [15, 16] and that of de Figueiredo [13]). This paper deals with thelimiting behavior of a sequence of second order elliptic Steklov eigenvalue problems withindefinite(sign-changing) density function in perforated domains.

Let be a bounded domain in RNx (the numerical space of variables x = (x1, . . . , xN)),with C 1 boundary and with integer N 2. We define the perforated domain asfollows. Let T Y = (0,1)N be a compact subset of Y with C 1 boundary T ( S) andnonempty interior. For > 0, we define

t = {k ZN : (k + T ) },T =

kt(k + T )

and

= \ T .In this setup, T is the reference hole whereas (k + T ) is a hole of size and T is thecollection of the holes of the perforated domain . The family T is made up with a finitenumber of holes since is bounded. In the sequel, Y stands for Y \ T and n = (ni)Ni=1denotes the outer unit normal vector to S with respect to Y .

We are interested in the spectral asymptotics (as 0) of the following Steklov eigen-value problem

N

i,j=1

xi

(

aij

(x

)u

xj

)

= 0 in ,

N

i,j=1aij

(x

)u

xjni

(x

)

= (

x

)

u on T,

u = 0 on ,

(1.2)

where aij L(RNy ) (1 i, j N ), with the symmetry condition aji = aij , the Y -periodicityhypothesis: for every k ZN one has aij (y + k) = aij (y) almost everywhere in y RNy , andfinally the (uniform) ellipticity condition: there exists > 0 such that

N

i,j=1aij (y)j i | |2 (1.3)

for all RN and for almost all y RNy , where | |2 = |1|2 + + |N |2. The densityfunction Cper (Y ) changes sign on S, that is, both the set {y S,(y) < 0} and {y S,(y) > 0} are of positive N 1 dimensional Hausdorf measure (the so-called surfacemeasure). This hypothesis makes the problem under consideration nonstandard. We will see(Corollary 2.15) that under the preceding hypotheses, for each > 0 the spectrum of (1.2)is discrete and consists of two infinite sequences

0 < 1,+ 2,+ n,+ , limn+

n,+ = +

• Steklov Eigenvalue Problems with Sign-changing Density Function 263

and

0 > 1, 2, n, , limn+

n, = .

The asymptotic behavior of the eigencouples depends crucially on whether the average ofthe density over S, MS() =

S(y) d(y), is positive, negative or equal to zero. All three

cases are carefully investigated in this paper.The homogenization of spectral problems has been widely explored. In a fixed domain,

homogenization of spectral problems with point-wise positive density function goes back toKesavan [18, 19]. Spectral asymptotics in perforated domains was studied by Vanninathan[35] and later in many other papers, including [9, 12, 17, 28, 29, 32] and the referencestherein. Homogenization of elliptic operators with sing-changing density function in a fixeddomain with Dirichlet boundary conditions has been investigated by Nazarov et al. [2224]via a combination of formal asymptotic expansion with Tartars energy method. In porousmedia, spectral asymptotics of elliptic operator with sign changing density function is stud-ied in [11] with the two scale convergence method.

The asymptotics of Steklov eigenvalue problems in periodically perforated domains wasstudied in [35] for the Laplace operator and constant density ( = 1) using asymptotic ex-pansion and Tartars test function method. The same problem for a second order periodicelliptic operator has been studied in [29] (with C coefficients) and in [9] (with L coef-ficient) but still with constant density ( = 1). All the just-cited works deal only with onesequence of positive eigenvalues.

In this paper we take it to the general tricky step. We investigate in periodically perfo-rated domains the asymptotic behavior of Steklov eigenvalue problems for periodic ellipticlinear differential operators of order two in divergence form with L coefficients and asing-changing density function. We obtain accurate and concise homogenization results inall three cases: MS() > 0 (Theorem 3.1 and Theorem 3.3), MS() = 0 (Theorem 3.5),MS() < 0 (Theorem 3.1 and Theorem 3.3), by using the two-scale convergence method[1, 21, 25, 36] introduced by Nguetseng [25] and further developed by Allaire [1]. In short;

(i) If MS() > 0, then the positive eigencouples behave like in the case of point-wisepositive density function, i.e., for k 1, k,+ is of order and 1 k,+ converges as 0 to the kth eigenvalue of the limit Dirichlet spectral problem, correspondingextended eigenfunctions converge along subsequences.

As regards the negative eigencouples, k, converges to at the rate 1 andthe corresponding eigenfunctions oscillate rapidly. We use a factorization technique[20, 35] to prove that

k, =11 + k, + o(1), k = 1,2, . . . ,

where (1 , 1 ) is the first negative eigencouple to the following local Steklov spectral

problem

div(a(y)Dy) = 0 in Y ,

a(y)Dy n = (y) on S, Y -periodic,

(1.4)

and {k, }k=1 are eigenvalues of a Steklov eigenvalue problem similar to (1.2). Wethen prove that { k,

1

2} converges to the kth eigenvalue of a limit Dirichlet spec-

tral problem which is different from that obtained for positive eigenvalues. As regards

• 264 H. Douanla

eigenfunctions, extensions of { uk,(1 )

}Ewhere (1 )(x) = 1 ( x )converge alongsubsequences to the kth eigenfunctions of the limit problem.

(ii) If MS() = 0, then the limit spectral problem generates a quadratic operator pencil andk, converges to the (k,)th eigenvalue of the limit operator, extended eigenfunctionsconverge along subsequences as well. This case requires a new convergence result asregards the two-scale convergence theory, Lemma 2.9.

(iii) The case when MS() < 0 is equivalent to that when MS() > 0, just replace with .

Unless otherwise specified, vector spaces throughout are considered over R, and scalar func-tions are assumed to take real values. We will make use of the following notations. LetF(RN) be a given function space. We denote by Fper(Y ) the space of functions in Floc(RN)(when it makes sense) that are Y -periodic, and by Fper (Y )/R the space of those functionsu Fper(Y ) with

Yu(y)dy = 0. We denote by H 1per (Y ) the space of functions in H 1(Y )

assuming same values on the opposite faces of Y and H 1per (Y )/R stands for the subset ofH 1per (Y

) made up of functions u H 1per (Y ) verifying

Y u(y)dy = 0. Finally, the letter Edenotes throughout a family of strictly positive real numbers (0 < < 1) admitting 0 as ac-cumulation point. The numerical space RN and its open sets are provided with the Lebesguemeasure denoted by dx = dx1, . . . , dxN . The usual gradient operator will be denoted by D.For the sake of simple notations we hide trace operators. The rest of the paper is organizedas follows. Section 2 deals with some preliminary results while homogenization processesare considered in Sect. 3.

2 Preliminaries

We first recall the definition and the main compactness theorems of the two-scale conver-gence method. Let be a smooth open bounded set in RNx (integer N 2) and Y = (0,1)N ,the unit cube.

Definition 2.1 A sequence (u)E L2() is said to two-scale converge in L2() tosome u0 L2( Y ) if as E 0,

u(x)

(

x,x

)

dx

Yu0(x, y)(x, y)dxdy (2.1)

for all L2(; Cper (Y )).

Notation We express this by writing u2s u0 in L2().

The following compactness theorems [1, 25, 27] are cornerstones of the two-scale con-vergence method.

Theorem 2.2 Let (u)E be a bounded sequence in L2(). Then a subsequence E canbe extracted from E such that as E 0, the sequence (u)E two-scale converges inL2() to some u0 L2( Y ).

• Steklov Eigenvalue Problems with Sign-changing Density Function 265

Theorem 2.3 Let (u)E be a bounded sequence in H 1(). Then a subsequence E canbe extracted from E such that as E 0

u u0 in H 1()-weak,u u0 in L2(),

u

xj

2s u0xj

+ u1yj

in L2() (1 j N),

where u0 H 1() and u1 L2(;H 1per (Y )). Moreover, as E 0 we have

u(x)

(

x,x

)

dx

Yu1(x, y)(x, y)dx dy (2.2)

for D() (L2per (Y )/R).

Remark 2.4 In Theorem 2.3 the function u1 is unique up to an additive function of vari-able x. We need to fix its choice according to our future needs. To do this, we introduce thefollowing space

H 1,per (Y ) ={

u H 1per (Y ) :

Y u(y)dy = 0

}

.

This defines a closed subspace of H 1per (Y ) as it is the kernel of the bounded linear functionalu

Y u(y)dy defined on H1per (Y ). It is to be noted that for u H 1,per (Y ), its restriction to

Y (which will still be denoted by u in the sequel) belongs to H 1per (Y )/R.

We will use the following version of Theorem 2.3.

Theorem 2.5 Let (u)E be a bounded sequence in H 1(). Then a subsequence E canbe extracted from E such that as E 0

u u0 in H 1()-weak, (2.3)u u0 in L2(), (2.4)

u

xj

2s u0xj

+ u1yj

in L2() (1 j N), (2.5)

where u0 H 1() and u1 L2(;H 1,per (Y )). Moreover, as E 0 we have

u(x)

(

x,x

)

dx

Yu1(x, y)(x, y)dx dy (2.6)

for D() (L2per (Y )/R).

Proof Let u1 L2(;H 1per (Y )) be such that Theorem 2.3 holds with u1 in place of u1. Put

u1(x, y) = u1(x, y) 1|Y |

Y u1(x, y)dy, (x, y) Y,

• 266 H. Douanla

where |Y | stands for the Lebesgue measure of Y . Then u1 L2(;H 1,per (Y )) and more-over Dyu1 = Dyu1 so that (2.5) holds.

In the sequel, S stands for T and the surface measures on S and S are denotedby d(y) (y Y ), d(x) (x , E), respectively. The space of squared integrablefunctions, with respect to the previous measures on S and S are denoted by L2(S) andL2(S) respectively. Since the volume of S grows proportionally to 1

as 0, we endow

L2(S) with the scaled scalar product [3, 30, 31]

(u, v)L2(S) =

Su(x)v(x)d(x)

(u,v L2(S)).

Definition 2.1 and Theorem 2.2 then generalize as

Definition 2.6 A sequence (u)E L2(S) is said to two-scale converge to some u0 L2( S) if as E 0,

Su(x)

(

x,x

)

d(x)

Su0(x, y)(x, y)dxd(y)

for all C(; Cper (Y )).

Theorem 2.7 Let (u)E be a sequence in L2(S) such that

S

u(x)

2d(x) C,

where C is a positive constant independent of . There exists a subsequence E of E suchthat (u)E two-scale converges to some u0 L2(;L2(S)) in the sense of Definition 2.6.

In the case when (u)E is the sequence of traces on S of functions in H 1(), onecan link its usual two-scale limit with its surface two-scale limits. The following propositionwhose proof can be found in [3] clarifies this.

Proposition 2.8 Let (u)E H 1() be such that

uL2() + DuL2()N C,

where C is a positive constant independent of and D denotes the usual gradient. Thesequence of traces of (u)E on S satisfies

S

u(x)

2d(x) C ( E)

and up to a subsequence E of E, it two-scale converges in the sense of Definition 2.6 tosome u0 L2(;L2(S)) which is nothing but the trace on S of the usual two-scale limit, afunction in L2(;H 1per (Y )). More precisely, as E 0

• Steklov Eigenvalue Problems with Sign-changing Density Function 267

Su(x)

(

x,x

)

d(x)

Su0(x, y)(x, y)dxd(y),

u(x)

(

x,x

)

dxdy

Yu0(x, y)(x, y)dxdy,

for all C(; Cper (Y )).

In our homogenization process, more precisely in the case when MS() = 0, we willneed a generalization of (2.2) to periodic surfaces. Notice that (2.2) was proved for the firsttime in a deterministic setting by Nguetseng and Woukeng in [27] but to the best of ourknowledge its generalization to periodic surfaces is not yet available in the literature. Westate and prove it below.

Lemma 2.9 Let (u)E H 1() be such that as E 0

u2s u0 in L2(), (2.7)

u

xj

2s u0xj

+ u1yj

in L2() (1 j N) (2.8)

for some u0 H 1() and u1 L2(;H 1per (Y )). Then

lim0

Su(x)(x)

(x

)

d(x) =

Su1(x, y)(x)(y)dxd(y) (2.9)

for all D() and Cper (Y ) with

S(y)d(y) = 0.

Proof We first transform the above surface integral into a volume integral by adapting thetrick in [7, Sect. 3]. By the mean value zero condition over S for we conclude that thereexists a unique solution H 1per (Y )/R to

{ y = 0 in Y ,Dy(y) n(y) = (y) on S,

(2.10)

where n = (ni)Ni=1 stands for the outward unit normal to S with respect to Y . Put = Dy .We get

Dxu(x)(x) Dy

(x

)

dx

=

Su(x)(x)Dy

(x

)

n(

x

)

d(x)

u(x)Dx(x) Dy

(x

)

dx 1

u(x)(x)y

(x

)

dx

=

Su(x)(x)

(x

)

d(x)

u(x)Dx(x)

(x

)

dx. (2.11)

• 268 H. Douanla

Next, sending to 0 yields

lim0

Su(x)(x)

(x

)

d(x) =

Y [Dxu0(x) + Dyu1(x, y)

](x) (y)dxdy

+

Y u0(x)Dx(x) (y)dxdy

=

Y Dyu1(x, y)(x) (y)dxdy.

We finally have

Y Dyu1(x, y)(x) (y)dxdy =

Y u1(x, y)(x)y(y)dxdy

+

Su1(x, y)(x)(y) n(y)dxd(y)

=

Su1(x, y)(x)(y) dxd(y).

The proof is completed.

We now gather some preliminary results. We introduce the characteristic function G of

G = RNy \ with

=

kZN(k + T ).

It is clear that G is an open subset of RNy . Next, let E be arbitrarily fixed and defineV =

{u H 1() : u = 0 on }.

We equip V with the H 1()-norm which makes it a Hilbert space. We recall the followingclassical extension result [8].

Proposition 2.10 For each E there exists an operator P of V into H 10 () with thefollowing properties: P sends continuously and linearly V into H 10 (). (Pv)| = v for all v V . D(Pv)L2()N cDvL2()N for all v V , where c is a constant independent of .

In the sequel, we will explicitly write the just-defined extension operator everywhereneeded but we will abuse notations on the local extension operator (see [8] for its definition):the extension to Y of u H 1per (Y )/R will still be denoted by u (this extension is an elementof H 1,per (Y )).

Now, let Q = \(). This defines an open set in RN and \Q is the intersection of with the collection of the holes crossing the boundary . The following result impliesthat the holes crossing the boundary are of no effects as regards the homogenizationprocess.

• Steklov Eigenvalue Problems with Sign-changing Density Function 269

Lemma 2.11 [26] Let K be a compact set independent of . There is some 0 > 0 suchthat \ Q \ K for any 0 < 0.

We introduce the space

F10 = H 10 () L2

(;H 1,per (Y )

)

and endow it with the following norm

vF

10= Dxv0 + Dyv1L2(Y)

(v = (v0, v1) F10

),

which makes it a Hilbert space admitting F0 = D() [D() C,per (Y )] (whereC,per (Y ) = {u Cper (Y ) :

Y u(y)dy = 0}) as a dense subspace. For (u,v) F10 F10, let

a(u,v) =N

i,j=1

Y aij (y)

(u0

xj+ u1

yj

)(v0

xi+ v1

yi

)

dxdy.

This define a symmetric, continuous bilinear form on F10 F10. We will need the followingresults whose proof can be found in [9].

Lemma 2.12 Fix = (0,1) F0 and define : R ( > 0) by

(x) = 0(x) + 1(

x,x

)

(x ).

If (u)E H 10 () is such thatu

xi

2s u0xi

+ u1yi

in L2() (1 i N)

as E 0 for some u = (u0, u1) F10, then

a(u,) a(u,)

as E 0, where

a(u,) =N

i,j=1

aij

(x

)u

xj

xidx.

We now construct and point out the main properties of the so-called homogenized coef-ficients. Put

a(u, v) =N

i,j=1

Y aij (y)

u

yj

v

yidy,

lj (v) =N

k=1

Y akj (y)

v

ykdy (1 j N)

• 270 H. Douanla

and

l0(v) =

S

(y)v(y)d(y),

for u,v H 1per (Y )/R. Equipped with the norm

uH 1per (Y )/R = DyuL2(Y )N(u H 1per

(Y

)/R

), (2.12)

H 1per (Y)/R is a Hilbert space.

Proposition 2.13 Let 1 j N . The local variational problems

u H 1per(Y

)/R and a(u, v) = lj (v) for all v H 1per

(Y

)/R (2.13)

and

u H 1per(Y

)/R and a(u, v) = l0(v) for all v H 1per

(Y

)/R (2.14)

admit each a unique solution, assuming for (2.14) that MS() = 0.

Let 1 i, j N . The homogenized coefficients read

qij =

Y aij (y)dy

N

l=1

Y ail(y)

j

yl(y)dy, (2.15)

where j (1 j N) is the solution to (2.13). We recall that qji = qij (1 i, j N) andthere exists a constant 0 > 0 such that

N

i,j=1qij j i 0| |2

for all RN (see e.g., [4]).We now visit the existence result for (1.2). The weak formulation of (1.2) reads: Find

(, u) C V , (u = 0) such that

a(u, v) = (u, v

)

S, v V, (2.16)

where

(u, v)S =

Suvd(x).

Since changes sign, the classical results on the spectrum of semi-bounded self-adjointoperators with compact resolvent do not apply. To handle this, we follow the ideas in [24].The bilinear form (u, v)S defines a bounded linear operator K : V V such that

(u, v

)

S= a(Ku,v) (u, v V).

The operator K is symmetric and its domains D(K) coincides with the whole V , thus itis self-adjoint. Recall the gradient norm is equivalent to the H 1()-norm on V . Looking

• Steklov Eigenvalue Problems with Sign-changing Density Function 271

at Ku as the solution to the boundary value problem

div(

a

(x

)

Dx(Ku

))

= 0 in ,

a

(x

)

DxKu n

(x

)

= u on S,

Ku(x) = 0 on ,

(2.17)

we get a constant C > 0 such that KuV CuL2(S). But the trace operator V L2(S) is compact. The compactness of K follows thereby. We can rewrite (2.16) as follows

Ku = u, = 1

.

We recall that (see e.g., [5]) in the case 0 on S, the operator K is positive and itsspectrum (K) lies in [0,K] and = 0 belongs to the essential spectrum e(K). LetL be a self-adjoint operator and let p (L) and c(L) be its set of eigenvalues of infinitemultiplicity and its continuous spectrum, respectively. We have e(L) = p (L) c(L) bydefinition. The spectrum of K is described by the following proposition whose proof issimilar to that of [24, Lemma 1].

Lemma 2.14 Let Cper (Y ) be such that the sets {y S : (y) < 0} and {y S :(y) > 0} are both of positive surface measure. Then for any > 0, we have (K) [K,K] and = 0 is the only element of the essential spectrum e(K). Moreover,the discrete spectrum of K consists of two infinite sequences

1,+ 2,+ k,+ 0+,1, 2, k, 0.

Corollary 2.15 The hypotheses are those of Lemma 2.14. Problem (1.2) has a discrete setof eigenvalues consisting of two sequences

0 < 1,+ 2,+ k,+ +,0 > 1,+ 2, k, .

We may now address the homogenization problem for (1.2).

3 Homogenization Results

In this section we state and prove homogenization results for both cases MS() > 0 andMS() = 0. The homogenization results in the case when MS() < 0 can be deduced fromthe case MS() > 0 by replacing with . We start with the less technical case.

3.1 The Case MS() > 0

We start with the homogenization result for the positive part of the spectrum (k,+ , uk,+ )E .

• 272 H. Douanla

3.1.1 Positive Part of the Spectrum

We assume (this is not a restriction) that the corresponding eigenfunctions are orthonormal-ized as follows

S

(x

)

uk,+ ul,+ d(x) = k,l k, l = 1,2, . . . (3.1)

and the homogenization results states as

Theorem 3.1 For each k 1 and each E, let (k,+ , uk,+ ) be the kth positive eigencoupleto (1.2) with MS() > 0 and (3.1). Then, there exists a subsequence E of E such that

1k,+ k0 in R as E 0, (3.2)

Puk,+ uk0 in H 10 ()-weak as E 0, (3.3)

Puk,+ uk0 in L2() as E 0, (3.4)

Puk,+

xj

2s uk0

xj+ u

k1

yjin L2() as E 0 (1 j N), (3.5)

where (k0, uk0) R H 10 () is the kth eigencouple to the spectral problem

N

i,j=1

xi

(1

MS()qij

u0

xj

)

= 0u0 in ,

u0 = 0 on ,

|u0|2dx = 1MS()

,

(3.6)

and where uk1 L2(;H 1,per (Y )). Moreover, for almost every x the following hold true:(i) The restriction to Y of uk1(x) is the solution to the variational problem

uk1(x) H 1per(Y

)/R,

a(uk1(x), v

) = N

i,j=1

uk0xj

Y aij (y)

v

yidy,

v H 1per(Y

)/R;

(3.7)

(ii) We have

uk1(x, y) = N

j=1

uk0xj

(x)j (y) a.e. in (x, y) Y , (3.8)

where j (1 j N) is the solution to the cell problem (2.13).

Proof We present only the outlines since this proof is similar but less technical to that of thecase MS() = 0.

• Steklov Eigenvalue Problems with Sign-changing Density Function 273

Fix k 1. By means of the minimax principle, as in [35], one easily proves the existenceof a constant C independent of such that 1

k,+ < C. Clearly, for fixed E > 0, uk,+ lies

in V , and

N

i,j=1

aij

(x

)uk,+xj

v

xidx =

(1k,+

)

S

(x

)

uk,+ v d(x) (3.9)

for any v V . Bear in mind that

S( x

)(uk,+ )2dx = 1 and choose v = uk,+ in (3.9).

The boundedness of the sequence ( 1k,+ )E and the ellipticity assumption (1.3) imply at

once by means of Proposition 2.10 that the sequence (Puk,+ )E is bounded in H 10 ().Theorem 2.5 and Proposition 2.8 apply simultaneously and gives us uk = (uk0, uk1) F10 suchthat for some k0 R and some subsequence E E we have (3.2)(3.5), where (3.4) is adirect consequence of (3.3) by the Rellich-Kondrachov theorem. For fixed E, let beas in Lemma 2.12. Multiplying both sides of the first equality in (1.2) by and integratingover leads us to the variational -problem

N

i,j=1

aij

(x

)Pu

k,+

xj

xidx =

(1k,+

)

S

(Pu

k,+

)

(x

)

d(x). (3.10)

Sending E to 0, keeping (3.2)(3.5) and Lemma 2.12 in mind, we obtainN

i,j=1

Y aij (y)

(uk0xj

+ uk1

yj

)(0

xi+ 1

yi

)

dxdy = k0

Suk00(x)(y)dxd(y).

Therefore, (k0,uk) R F10 solves the following global homogenized spectral problem:

Find (,u) C F10 such thatN

i,j=1

Y aij (y)

(u0

xj+ u1

yj

)(0

xi+ 1

yi

)

dxdy = MS()

u00 dx

for all F10

(3.11)

which leads to the macroscopic and microscopic problems (3.6)(3.7) without any majordifficulty. As regards the normalization condition in (3.6), we fix k, l 1 and recall that thefollowing holds for any D() (Proposition 2.8)

limE0

S

(Pu

k,+

)(x)

(x

)

d(x) =

Suk0(x)(x)(y) dxd(y). (3.12)

But (3.12) still holds for any H 10 (). This being so, we write

S

(Pu

k,+

)(Pu

l,+

)

(x

)

d(x) MS()

uk0ul0 dx

=

S

(Pu

k,+

)(Pu

l,+ ul0

)

(x

)

d(x) +

S

(Pu

k,+

)ul0

(x

)

d(x)

MS()

uk0ul0 dx. (3.13)

• 274 H. Douanla

According to (3.12) the sum of the last two terms on the right hand side of (3.13) goes tozero with E. As the remaining term on the right hand side of (3.13) is concerned, wemake use of the Hlder inequality to get

S

(Pu

k,+

)(Pu

l,+ ul0

)

(x

)

d(x)

(

S

Pu

k,+

2d(x)

) 12(

S

Pu

l,+ ul0

2d(x)

) 12.

Next the trace inequality (see e.g., [29]) yields

S

Pu

k,+

2d(x) c

(

Pu

k,+

2dx + 2

D

(Pu

k,+

)2dx

)

(3.14)

S

Pu

l,+ ul0

2d(x) c

(

Pu

l,+ ul0

2dx + 2

D

(Pu

l,+ ul0

)2dx

)

,

(3.15)

for some positive constant c independent of . But the right hand side of (3.14) is boundedfrom above whereas that of (3.15) converges to zero. This concludes the proof.

Remark 3.2

The eigenfunctions {uk0}k=1 are in fact orthonormalized as follows

uk0ul0dx =

k,l

MS(), k, l = 1,2,3, . . . .

If k0 is simple (this is the case for 10), then by Theorem 3.1, k,+ is also simple, forsmall , and we can choose the eigenfunctions uk,+ such that the convergence results(3.3)(3.5) hold for the whole sequence E.

Replacing with in (1.2), Theorem 3.1 also applies to the negative part of the spec-trum in the case MS() < 0.

3.1.2 Negative Part of the Spectrum

We now investigate the negative part of the spectrum (k, , uk, )E . Before we can do thiswe need a few preliminaries and stronger regularity hypotheses on T , and the coefficients(aij )

Ni,j=1. We assume in this subsection that T is C2, and and the coefficients (aij )Ni,j=1

are -Hlder continuous (0 < < 1).The following spectral problem is well posed

Find (, ) C H 1per(Y

)

N

i,j=1

yj

(

aij (y)

yi

)

= 0 in Y ,

N

i,j=1aij (y)

yij = (y)(y) on S

(3.16)

• Steklov Eigenvalue Problems with Sign-changing Density Function 275

and possesses a spectrum with similar properties to that of (1.2), two infinite (one positiveand another negative) sequences. We recall that since we have MS() > 0, problem (3.16)admits a unique nontrivial eigenvalue having an eigenfunction with definite sign, the firstnegative one (see e.g., [34]). In the sequel we will only make use of (1 , 1 ), the firstnegative eigencouple to (3.16). After proper sign choice we assume that

1 (y) > 0 in y Y . (3.17)

We also recall that 1 is -Hlder continuous (see e.g., [14]), hence can be extended to afunction living in Cper (Y ) still denoted by 1 . Notice that we have

S

(y)(1 (y)

)2d(y) < 0, (3.18)

as is easily seen from the variational equality (keep the ellipticity hypothesis (1.3) in mind)

N

i,j=1

Y aij (y)

1yj

1yi

dy = 1

S

(y)(1 (y)

)2d(y).

Bear in mind that problem (3.16) induces by a scaling argument the following equalities:

N

i,j=1

xj

(

aij

(x

)

xi

)

= 0 in Q,

N

i,j=1aij

(x

)

xij

(x

)

= 1

(x

)

(x

)

on Q,

(3.19)

where (x) = ( x). However, is not zero on . We now introduce the following

Steklov spectral problem (with an indefinite density function)

Find (, v) C V

N

i,j=1

xj

(

aij

(x

)v

xi

)

= 0 in ,

N

i,j=1aij

(x

)v

xij

(x

)

= (

x

)

v on T,

v(x) = 0 on

(3.20)

with new spectral parameters (, v) C V , where aij (y) = (1 )2(y)aij (y) and (y) =(1 )

2(y)(y). Notice that aij (y) Lper (Y ) and (y) Cper (Y ). As 0 < c 1 (y) c+ 10 > 20 30 j0 as j .

Making use of (3.19) when following the same line of reasoning as in [35, Lemma 6.1], weobtain that the negative spectral parameters of problems (1.2) and (3.20) verify:

uk, =(1

)vk, ( E, k = 1,2, . . .) (3.24)

and

k, =11 + k, + o(1) ( E, k = 1,2, . . .). (3.25)

The presence of the term o(1) is due to integrals over \Q , which converge to zero with ,remember that (3.19) holds in Q but not . This trick, known as factorization principlewas introduced by Vaninathan [35] and has been used in many other works on averaging, seee.g., [2, 20, 23] just to cite a few. As will be seen below, the sequence ( k, )E is boundedin R. In another words, k, is of order 1/ and tends to as goes to zero. It is nowclear why the limiting behavior of negative eigencouples is not straightforward as that ofpositive ones and requires further investigations, which have just been made.

• Steklov Eigenvalue Problems with Sign-changing Density Function 277

Indeed, as the reader might be guessing now, the suitable orthonormalization conditionfor (3.20) is

S

(x

)

vk, vl, d(x) = k,l , k, l = 1,2, . . . (3.26)

which by means of (3.24) is equivalent to

S

(x

)

uk, ul, d(x) = k,l , k, l = 1,2, . . . . (3.27)

We may now state the homogenization theorem for the negative part of the spectrum of (1.2).

Theorem 3.3 For each k 1 and each E, let (k, , uk, ) be the kth negative eigen-couple to (1.2) with MS() > 0 and (3.27). Then, there exists a subsequence E of E suchthat

k,

1

2 k0 in R as E 0, (3.28)

Pvk, vk0 in H 10 ()-weak as E 0, (3.29)

Pvk, vk0 in L2() as E 0, (3.30)

Pvk,

xj

2s vk0

xj+ v

k1

yjin L2() as E 0 (1 j N), (3.31)

where ( k0 , vk0) R H 10 () is the kth eigencouple to the spectral problem

N

i,j=1

xi

(1

MS()qij

v0

xj

)

= 0v0 in ,

v0 = 0 on ,

|v0|2 dx = 1MS()

,

(3.32)

and where vk1 L2(;H 1,per (Y )). Moreover, for almost every x the following hold true:(i) The restriction to Y of vk1(x) is the solution to the variational problem

vk1(x) H 1per(Y

)/R,

a(vk1(x), u

) = N

i,j=1

vk0xj

Y aij (y)

u

yidy,

u H 1per(Y

)/R;

(3.33)

(ii) We have

vk1(x, y) = N

j=1

vk0xj

(x) j (y) a.e. in (x, y) Y , (3.34)

where j (1 j N) is the solution to the cell problem (3.23).

• 278 H. Douanla

Remark 3.4

The eigenfunctions {vk0}k=1 are orthonormalized by

vk0vl0dx =

k,lMS()

, k, l = 1,2,3, . . . .

If k0 is simple (this is the case for 10 ), then by Theorem 3.3, k, is also simple forsmall , and we can choose the eigenfunctions vk, such that the convergence results(3.29)(3.31) hold for the whole sequence E.

Replacing with in (1.2), Theorem 3.3 adapts to the positive part of the spectrum inthe case MS() < 0.

3.2 The Case MS() = 0

We prove an homogenization result for both the positive part and the negative part of thespectrum simultaneously. We assume in this case that the eigenfunctions are orthonormal-ized as follows

S

(x

)

uk, ul, d(x) = k,l , k, l = 1,2, . . . . (3.35)

Let 0 be the solution to (2.14) and put

2 =N

i,j=1

Y aij (y)

0

yj

0

yidy. (3.36)

Indeed, the right hand side of (3.36) is positive. We recall that the following spectral problemfor a quadratic operator pencil with respect to ,

N

i,j=1

xj

(

qiju0

xi

)

= 202u0 in ,

u0 = 0 on ,(3.37)

has a spectrum consisting of two infinite sequences

0 < 1,+0 < 2,+0 k,+0 , lim

n+k,+0 = +

and

0 > 1,0 > 2,0 k,0 , lim

n+k,0 =

with k,+0 = k,0 , k = 1,2, . . . and with the corresponding eigenfunctions uk,+0 = uk,0 .We note by passing that 1,+0 and

1,0 are simple. We are now in a position to state the

homogenization result in the present case.

Theorem 3.5 For each k 1 and each E, let (k, , uk, ) be the (k,)th eigencoupleto (1.2) with MS() = 0 and (3.35). Then, there exists a subsequence E of E such that

k, k,0 in R as E 0, (3.38)

• Steklov Eigenvalue Problems with Sign-changing Density Function 279

Puk, uk,0 in H 10 ()-weak as E 0, (3.39)

Puk, uk,0 in L2() as E 0, (3.40)

Puk,

xj

2s uk,0

xj+ u

k,1

yjin L2() as E 0 (1 j N), (3.41)

where (k,0 , uk,0 ) R H 10 () is the (k,)th eigencouple to the following spectral prob-

lem for a quadratic operator pencil with respect to ,

N

i,j=1

xi

(

qiju0

xj

)

= 202u0 in ,

u0 = 0 on ,(3.42)

and where uk,1 L2(;H 1,per (Y )). We have the following normalization condition

u

k,0

2 dx = 1

2k,0 2, k = 1,2, . . . . (3.43)

Moreover, for almost every x the following hold true:(i) The restriction to Y of uk,1 (x) is the solution to the variational problem

uk,1 (x) H 1per

(Y

)/R,

a(u

k,1 (x), v

) = k,0 uk,0 (x)

S

(y)v(y) d(y) N

i,j=1

uk,0

xj(x)

Y aij (y)

v

yidy

v H 1per(Y

)/R;

(3.44)(ii) We have

uk,1 (x, y) = k,0 uk,0 (x)0(y)

N

j=1

uk,0

xj(x)j (y) a.e. in (x, y) Y ,

(3.45)where j (1 j N) and 0 are the solutions to the cell problems (2.13) and (2.14),respectively.

Proof Fix k 1, using the minimax principle, as in [35], we get a constant C independentof such that |k, | < C. We have uk, V and

N

i,j=1

aij

(x

)uk,xj

v

xidx = k,

S

(x

)

uk, v d(x) (3.46)

for any v V . Bear in mind that

S( x

)(uk, )2 d(x) = 1 and choose v = uk,

in (3.46). The boundedness of the sequence (k, )E and the ellipticity assumption (1.3)imply at once by means of Proposition 2.10 that the sequence (Puk, )E is boundedin H 10 (). Theorem 2.5 and Proposition 2.8 apply simultaneously and gives us uk, =

• 280 H. Douanla

(uk,0 , u

k,1 ) F10 such that for some k,0 R and some subsequence E E we have

(3.38)(3.41), where (3.40) is a direct consequence of (3.39) by the Rellich-Kondrachovtheorem. For fixed E, let be as in Lemma 2.12. Multiplying both sides of the firstequality in (1.2) by and integrating over leads us to the variational -problem

N

i,j=1

aij

(x

)Pu

k,

xj

xidx = k,

S

(Pu

k,

)

(x

)

d(x).

Sending E to 0, keeping (3.38)(3.41) and Lemma 2.12 in mind, we obtain

a(uk,,

) = k,0

S

(u

k,1 (x, y)0(x)(y) + uk,0 1(x, y)(y)

)dxd(y). (3.47)

The right-hand side follows as explained below. we have

S

(Pu

k,

)

(x

)

d(x) =

S

(Pu

k,

)0(x)

(x

)

d(x)

+

S

(Pu

k,

)1

(

x,x

)

(x

)

d(x).

On the one hand we have

limE0

S

(Pu

k,

)1

(

x,x

)

(x

)

dx =

Su

k,0 1(x, y)(y) dxd(y).

On the other hand, owing to Lemma 2.9, the following holds:

limE0

S

(Pu

k,

)0(x)

(x

)

d(x) =

Su

k,1 (x, y)0(x)(y) dxd(y).

We have just proved that (k,0 ,uk,) R F10 solves the following global homogenizedspectral problem:

Find (,u) C F10 such that

a(u,) =

S

(u1(x, y)0(x) + u0(x)1(x, y)

)(y)dxd(y)

for all F10.

(3.48)

To prove (i), choose = (0,1) in (3.47) such that 0 = 0 and 1 = v1, where D() and v1 H 1per (Y )/R to get

(x)

[N

i,j=1

Y aij (y)

(u

k,0

xj+ u

k,1

yj

)v1

yidy

]

dx

=

(x)

[

k,0 u

k,0 (x)

S

v1(y)(y) d(y)

]

dx.

• Steklov Eigenvalue Problems with Sign-changing Density Function 281

Hence by the arbitrariness of , we have a.e. in

N

i,j=1

Y aij (y)

(u

k,0

xj+ u

k,1

yj

)v1

yidy = k,0 uk,0 (x)

S

v1(y)(y)d(y)

for any v1 in H 1per (Y )/R, which is nothing but (3.44).Fix x , multiply both sides of (2.13) by u

k,0

xj(x) and sum over 1 j N . Adding

side by side to the resulting equality that obtained after multiplying both sides of (2.14)by k,0 u

k,0 (x), we realize that z(x) =

Nj=1

uk,0

xj(x)j (y) + k,0 uk,0 (x)0(y) solves

(3.44). Hence

uk,1 (x, y) = k,0 uk,0 (x)0(y)

N

j=1

uk,0

xj(x)j (y) a.e. in Y , (3.49)

by uniqueness of the solution to (3.44). Thus (3.45). But (3.49) still holds almost everywherein (x, y) S as S is of class C 1. Considering now = (0,1) in (3.47) such that0 D() and 1 = 0 we get

N

i,j=1

Y aij (y)

(u

k,0

xj+ u

k,1

yj

)0

xidxdy

= k,0

Su

k,1 (x, y)(y)0(x) dxd(y),

which by means of (3.49) leads toN

i,j=1

qiju

k,0

xj

0

xidx + k,0

N

i,j=1

uk,0 (x)

0

xi

(

Y aij (y)

0

yj(y)dy

)

dx

= k,0N

j=1

uk,0

xj0(x)

(

S

(y)j (y) d(y)

)

dx

+ (k,0)2

uk,0 (x)0(x)

(

S

(y)0(y) d(y)

)

dx. (3.50)

Choosing l (1 l N) as test function in (2.14) and 0 as test function in (2.13) weobserve that

N

j=1

Y alj (y)

0

yj(y)dy =

S

(y)l(y) d(y) = a(l,0) (l = 1, . . .N).

Thus, in (3.50), the second term in the left hand side is equal to the first one in the right handside. This leaves us with

qiju

k,0

xj

0

xidx = (k,0

)2

uk,0 (x)0(x)dx

(

S

(y)0(y) d(y)

)

. (3.51)

• 282 H. Douanla

Choosing 0 as test function in (2.14) reveals that

S

(y)0(y) d(y) = a(0, 0) = 2.

HenceN

i,j=1

qiju

k,0

xj

0

xidx = (k,0

)22

uk,0 (x)0(x)dx,

and

N

i,j=1

xi

(

qiju

k,0

xj(x)

)

= (k,0)2

2uk,0 (x) in .

Thus, the convergence (3.38) holds for the whole sequence E. We now address (3.43). Fixk, l 1 and let H 1per (Y )/R be the solution to (2.10) where is replaced with ourdensity function . As in (2.11), we transform the surface integral into a volume integral

S

(Pu

k,

)(Pu

l,

)

(x

)

d(x)

=

(Pu

k,

)Dx

(Pu

l,

) Dy(

x

)

dx

+

Dx

(Pu

k,

)(Pu

l,

) Dy(

x

)

dx. (3.52)

A limit passage in (3.52) as E 0 yields

limE0

S

(Pu

k,

)(Pu

l,

)

(x

)

d(x)

=

Y u

k,0

(Dxu

l,0 + Dyul,1

) Dydxdy

+

Y (Dxu

k,0 + Dyuk,1

)u

l,0 Dydxdy

=

uk,0

(

Y Dyu

l,1 (x, y) Dy(y)dy

)

dx

+

ul,0

(

Y Dyu

k,1 (x, y) Dy(y)dy

)

dx

=

Su

k,0 (x)u

l,1 (x, y)(y)dxd(y) +

Su

l,0 (x)u

k,1 (x, y)(y)dxd(y)

= l,0 2

uk,0 (x)u

l,0 (x)dx + k,0 2

ul,0 (x)u

k,0 (x)dx

= (k,0 + l,0)2

uk,0 (x)u

l,0 (x)dx.

• Steklov Eigenvalue Problems with Sign-changing Density Function 283

Where often the limit passage, we used the integration by part formula, then the weak for-mulation of (2.10) and finally (3.45) and integration by part. If k = l, the above limit passageand (3.35) lead to the desired result, (3.43), completing thereby the proof.

Remark 3.6

The eigenfunctions {uk,0 }k=1 are in fact orthonormalized as follows

ul,0 (x)u

k,0 (x)dx =

k,l2(

l,0 + k,0 )

, k, l = 1,2, . . . .

If k,0 is simple (this is the case for 1,0 ), then by Theorem 3.5, k, is also simple,for small , and we can choose the eigenfunctions uk, such that the convergence results(3.39)(3.41) hold for the whole sequence E.

Final Remark After this paper was completed (see [10]) and submitted, we learned aboutan independent and similar work [6].

Acknowledgements The author is grateful to Dr. Jean Louis Woukeng for helpful discussions.

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34. Torne, O.: Steklov problem with an indefinite weight for the p-Laplacian. Elecron. J. Differ. Equ. 87,18 (2005)

35. Vanninathan, M.: Homogenization of eigenvalue problems in perforated domains. Proc. Indian Acad.Sci. Math. Sci. 90, 239271 (1981)

36. Zhikov, V.V.: On two-scale convergence. Tr. Semin. im. I. G. Petrovskogo No. 23, 149187, 410 (2003)(Russian); translation in J. Math. Sci. (N. Y.) 120(3), 13281352 (2004)

Homogenization of Steklov Spectral Problems with Indefinite Density Function in Perforated DomainsAbstractIntroductionPreliminariesHomogenization ResultsThe Case MS(rho)>0Positive Part of the SpectrumNegative Part of the Spectrum

The Case MS(rho)=0Final Remark

AcknowledgementsReferences

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