
Acta Appl Math (2013) 123:261284DOI
10.1007/s1044001297654
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 selfadjoint
elliptic Steklov eigenvalueproblems with periodic rapidly
oscillating coefficients and with indefinite (signchanging)density
function is investigated in periodically perforated domains. We
prove that the spectrum of this problem is discrete and consists
of two sequences, one tending to andanother to +. The limiting
behavior of positive and negative eigencouples depends crucially
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 twoscale convergence method, we investigate allthree
cases.
Keywords Homogenization Eigenvalue problems Perforated domains
Indefiniteweight function Twoscale 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,Swedenemail:
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, eigenvalue 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(signchanging) 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 eigenvalue 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 = 12 + + 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 socalled 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 Signchanging 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 pointwise 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
singchanging 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 studied 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 expansion 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 coefficient) but still with
constant density ( = 1). All the justcited works deal only with
onesequence of positive eigenvalues.
In this paper we take it to the general tricky step. We
investigate in periodically perforated domains the asymptotic
behavior of Steklov eigenvalue problems for periodic ellipticlinear
differential operators of order two in divergence form with L
coefficients and asingchanging 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 twoscale
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 pointwisepositive 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 twoscale 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 functions 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
accumulation 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 twoscale convergence 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 twoscale
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 twoscale convergence 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 twoscale converges inL2() to some u0 L2( Y ).

Steklov Eigenvalue Problems with Signchanging 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 variable 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) 1Y 
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 moreover 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 twoscale
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 twoscale 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 twoscale limit with its
surface twoscale 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 twoscale converges in the
sense of Definition 2.6 tosome u0 L2(;L2(S)) which is nothing but
the trace on S of the usual twoscale limit, afunction in L2(;H
1per (Y )). More precisely, as E 0

Steklov Eigenvalue Problems with Signchanging 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 justdefined
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 Signchanging 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
socalled homogenized coefficients. 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
semibounded selfadjointoperators 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 selfadjoint. Recall the gradient norm is
equivalent to the H 1()norm on V . Looking

Steklov Eigenvalue Problems with Signchanging 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 selfadjoint 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 orthonormalized 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 ,
u02dx = 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 Signchanging 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 RellichKondrachov 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 spectrum 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 Signchanging 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 Signchanging 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 eigencouple 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 ,
v02 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 orthonormalized 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 Signchanging 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 RellichKondrachovtheorem. 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 righthand 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 Signchanging 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 Signchanging Density Function
283
Where often the limit passage, we used the integration by part
formula, then the weak formulation 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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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