Journal of Mathematical Sciences, Vol. 154, No. 1, 2008
HOMOGENIZATION OF A BOUNDARY VALUE PROBLEM IN A THICK CASCADE JUNCTION T. A. Mel’nik National Taras Shevchenko University of Kyiv 64, Volodymyrska St., Kyiv 01033, Ukraine
[email protected]
G. A. Chechkin Lomonosov Moscow State University Vorob’evy Gory, Moscow 119991, Russia Narvik University College Postboks 385, Narvik 8505, Norway
[email protected]
UDC 517.956.225 & 517.956.8
We consider a homogenization problem in a singularly perturbed two-dimensional domain of a new type that consists of a junction body and many alternating thin rods of two classes. One of the classes consists of rods of finite length, whereas the other contains rods of small length, and inhomogeneous Fourier boundary conditions (the third type boundary conditions) with perturbed coefficients are imposed on boundaries of thin rods. Homogenization theorems are proved. Bibliography: 38 titles. Illustrations: 2 figures.
Introduction In this paper, we begin to study asymptotic properties of solutions to the boundary value problems in singularly perturbed domains of a new type (regarding methods for studying solutions in some known singularly perturbed domains, cf., for example, [1]–[7]). We assume that a domain consists of a junction body, independent of a small parameter, and thin rods of two classes (cf. Fig. 1).
Translated from Problemy Matematicheskogo Analiza, No. 37, 2008, pp. 47–72.
c 2008 Springer Science+Business Media, Inc. 1072-3374/08/1541-0050
50
% Figure 1. The thick cascade junction Ωε .
The first class consists of thin rods of finite length located periodically with period ε, where ε = a/N is a small discrete parameter, a = const. The second class consists of thin rods of small length (of order ε) that are also located periodically with period ε. Rods of these two classes ε-periodically alternate, the lengths of rods are considerably different while changing the class. The boundaries of rods of the second class are not necessarily described by functions. In fact, the domains under consideration are a symbiosis of domains with thick junction of rods of finite length (boundary value problems in such domains were considered in [1] and [8]— [20]) and domains with rapidly oscillating boundary (boundary value problems in such domains were considered in [21]–[35]). Note that the methods presented in the above-mentioned papers (except for [33]) for domains with rapidly oscillating boundary essentially use the fact that the oscillation is determined by some smooth function, whereas, in the present paper, we deal with some other geometry of oscillations of boundary which cannot be described by a function. This fact leads to additional difficulties which we overcome in this work. On the other hand, unlike the case of multilevel thick junctions of finite length [9, 16, 18], new effects appear in the problem under consideration. In this paper, we assume that different inhomogeneous Fourier boundary conditions (the third kind boundary conditions) are imposed on the boundaries of thin rods of different classes, and these conditions involve additional parameters. The asymptotic behavior of a solution to the initial problem essentially depends on these parameters. Such thick cascade junctions are prototypes of many physical and biological systems. In Sec. 1, we describe some known results and state the problem. We also formulate and discuss the main results of this paper. In Sec. 2, we prove auxiliary results on asymptotics. In Sec. 3, we prove homogenization theorems. 51
1. Statement of the Problem and Formulation of the Main Results Let a, b1 , b2 , h1 , h2 be positive real numbers such that 0 < b1 < b2
0, and many thin rods [0,a]
(1) Gj (dk , ε)
(2) Gj (ε)
εh1 , = x ∈ R2 : |x1 − ε(j + dk )| < 2
εh2 1 , = x ∈ R2 : |x1 − ε(j + )| < 2 2
x2 ∈ (−εl1 , 0] ,
x2 ∈ (−l2 , 0] ,
k = 1, . . . , 4,
j = 0, 1, . . . , N − 1,
where d1 = b1 , d2 = b2 , d3 = 1 − b2 , d4 = 1 − b1 . Thus, (2) Ωε = Ω0 ∪ G(1) ε ∪ Gε ,
where G(1) ε =
N −1 4 j=0
k=1
(1) Gj (dk , ε) ,
G(2) ε =
N −1 j=0
(2)
Gj (ε); (1)
(2)
the number of thin rods is equal to 5N and the rods are divided into two classes Gε and Gε depending on length. The first class contains rod of length εl1 , whereas the second class consists 52
of rods of lenfth l2 . So, the length of a rod considerably changes while passing from one class to the other. The parameter ε characterizes the distance between neighboring thin rods and also their thickness, equal to εh1 for rods of the first class and to εh2 for rods of the second class. Thin rods of different classes ε-periodically alternate along the segment I0 = {x : x1 ∈ [0, a], x2 = 0}. In Ωε , we consider the boundary value problem x ∈ Ωε ,
− ∆x uε (x) = fε (x),
∂ν uε (x) + ετ k1 uε (x) = pε (x),
x ∈ Υ(1) ε ,
∂ν uε (x) + εµ k2 uε (x) = εβ gε (x),
x ∈ Υ(2) ε ,
(1.1)
x ∈ Γ1 ,
uε (x) = 0,
x ∈ Γε ,
∂ν uε (x) = 0,
where ∂ν = ∂/∂ν is the derivative with respect to the outward normal, the constants k1 and k2 (i) are positive parameters, β 1, µ, τ ∈ R, Υε is the union of lateral sides and lower bases of (1) (2) rods of the ith class, i = 1, 2, Γ1 = {x : x2 = γ(x1 ), x1 ∈ [0, a]}, Γε = ∂Ωε \ Υε ∪ Υε ∪ Γ1 . Note that (2) [∂x2 uε ] = 0 on I0 ∩ G(1) ∪ G , (1.2) [uε ] = 0, ε ε where the square brackets mean the jump of a function. Without loss of generality, we assume that fε ∈ L2 (Ω2 ), where Ω2 = Ω0 ∪ D2 , D2 = (0, a) × (−l2 , 0) is a rectangle occupied by thin rods of the second class, and L2 (Ω2 )
fε →f0
in
as ε → 0;
(1.3)
gε g0
in H 1 (D2 ) as ε → 0.
(1.4)
gε ∈ H 1 (D2 ), and
(1)
(1)
Let pε ∈ H 1 (Dε ) and Dε = (0, a) × (−εl1 , 0) satisfy the following conditions: 1) there exists a constant c > 0 such that √ pε H 1 (D(1) ) c ε, ε
(1.5)
2) the traces converge: pε (·, 0)→p0
in L2 (0, a)
as ε → 0.
(1.6)
Remark 1.1. Hereinafter, the constants ci Ci are independent of ε. We recall that uε ∈ H 1 (Ωε , Γ1 ) = {v ∈ H 1 (Ωε ) : v|Γ1 = 0} is a weak solution to the problem (1.1) if uε satisfies the integral identity 53
∇x uε · ∇x ψdx + ετ k1
Ωε
uε ψ dσx + εµ k2
(1)
uε ψ dσx (2)
Υε
Υε
=
β
fε ψ dx + Ωε
pε ψ dσx + ε (1)
gε ψ dσx
∀ ψ ∈ H 1 (Ωε , Γ1 ).
(1.7)
(2)
Υε
Υε
According to the theory of boundary value problems (cf., for example, [36]), for any fixed ε > 0 there exists a unique weak solution to the problem (1.1). Our goal is to study the asymptotic behavior of the weak solution to the problem (1.1) as ε → 0, i.e., when the number of thin joined rods of every class unboundedly increases, whereas the thickness tends to zero. Note that the limit passage goes with perturbations of the coefficients in the boundary conditions by the parameters β, τ , and µ. We also study the influence of these perturbations on the asymptotic behavior of the solution. 1.1. The Main Results and Discussion. We introduce the zero extension operation for (2) functions in the spaces H 1 Gε :
y ε (x) =
⎧ ⎨yε , x ∈ G(2) ε , ⎩0,
(2)
x ∈ D2 \ Gε ,
where D2 = (0, a) × (−l2 , 0) is a rectangle occupied by thin rods of the second class as ε → 0. Theorem 1.1 (case τ 0 and µ 1). The solutions uε to the problem (1.1) satisfies the following relations:
uε v0+ in u ε h2 v0− in
H 1 (Ω0 , Γ1 ), L2 (D2 ),
⎪ L2 (D2 ),⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ 2 L (D2 ),
− ∂ x2 uε h2 ∂x2 v0 in
∂ x1 uε 0 in
⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ as
ε → 0,
(1.8)
where the function v0 (x) =
v0+ (x), v0− (x),
x ∈ Ω0 , x ∈ D2 ,
is a unique solution to the following homogenization problem for (1.1): 54
(1.9)
− ∆x v0+ (x) = f0 (x), v0+ (x) = 0,
x ∈ Ω0
x ∈ Γ1 x ∈ ∂Ω0 \ Γ1 ∪ I0 ,
∂ν v0+ (x) = 0,
− h2 ∂x22 x2 v0− (x) + 2δµ,1 k2 v0− (x) = h2 f0 (x) + δβ,1 g0 (x), v0+ (x1 , 0) = v0− (x1 , 0),
(1.10)
(x1 , 0) ∈ I0 ,
(h2 ∂x2 v0− − ∂x2 v0+ + δτ,0 k1 Ξv0+ )(x1 , 0) = Ξp0 (x1 ), ∂x2 v0− (x1 , −l2 ) = 0,
x ∈ D2 ,
(x1 , 0) ∈ I0 ,
(x1 , −l2 ) ∈ Il2 .
Here, Il2 = {x : x2 = −l2 , x1 ∈ (0, a)}, δα,k is the Kronecker symbol, Ξ := (4h1 + 8l1 ) is the length of the boundary of “small” rods on the periodicity cell in the stretching coordinates. Furthermore, the energy integrals converge as ε → 0 :
2
τ
u2ε dσx
|∇x uε | dx + ε k1
Eε (uε ) : = Ωε
→
µ
+ ε k2
(1)
Ω0
+
u2ε dσx
(2)
Υε
|∇v0+ |2 dx
Υε
h2 |∂x2 v0− |2 + 2δµ,1 k2 |v0− |2 dx + δτ,0 k1 Ξ
D2
|v0+ (x1 , 0)|2 dx1
I0
=: E0 (v0 ).
(1.11)
Theorem 1.2 (case τ < 0 and µ 1). The solutions uε to the problem (1.1) satisfy the relations uε v0+ in u ε h2 v0− in
⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
H 1 (Ω0 , Γ1 ), L2 (D2 ),
⎪ L2 (D2 ),⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ 2 L (D2 ),
− ∂ x2 uε h2 ∂x2 v0 in
∂ x1 uε 0 in
as
ε → 0,
(1.12)
where the functions v0+ and v0− are solutions to the following homogenization problems for the problem (1.1): − ∆x v0+ (x) = f0 (x), v0+ (x) = 0, ∂ν v0+ (x) = 0,
x ∈ Ω0
x ∈ Γ1 ∪ I0
x ∈ ∂Ω0 \ Γ1 ∪ I0 ,
(1.13)
55
− h2 ∂x22 x2 v0− (x) + 2δµ,1 k2 v0− (x) = h2 f0 (x) + δβ,1 g0 (x), v0− (x1 , 0) = 0,
x ∈ D2 ,
(x1 , 0) ∈ I0 ,
∂x2 v0− (x1 , −l2 ) = 0,
(1.14)
(x1 , −l2 ) ∈ Il2 .
Furthermore, the energy integrals converge as ε → 0 :
Eε (uε ) →
|∇v0+ |2 dx
+ h2
Ω0
|∂x2 v0− |2 dx
+ 2δµ,1 k2
D2
|v0− |2 dx
D2
=: E0 (v0+ ) + E0 (v0− ).
(1.15)
Theorem 1.3 (case µ < 1). The solutions uε to the problem (1.1) satisfy the relations uε v0+ in in u ε → 0
H 1 (Ω0 , Γ1 ), L2 (D2 ),
as
ε → 0,
(1.16)
where the function v0+ is a solution to the problem − ∆x v0+ (x) = f0 (x), v0+ (x) = 0,
x ∈ Ω0
x ∈ Γ1 ∪ I0
∂ν v0+ (x) = 0,
x ∈ ∂Ω0 \ Γ1 ∪ I0 .
(1.17)
Furthermore, the energy integrals converge as ε → 0 :
Eε (uε ) →
|∇v0+ |2 dx =: E0 (v0+ ).
(1.18)
Ω0
1.2. Discussion. The above results show that the boundary conditions essentially affect the asymptotic behavior of solutions to the problem (1.1) (there are three qualitatively different cases). The third type boundary condition is imposed on the walls of thin rods of the second class: ∂ν uε + εµ k2 uε = εβ gε . At the first glance, it seems that there are no differences between this condition and the homogeneous Neumann condition since the term k2 uε is multiplied by εµ and gε is multiplied by εβ , µ 1, β 1. However, this assertion is valid only if µ > 1 and β > 1. But, in the case β = 1 and µ 1, the term εgε is transformed to the right-hand side of the averaged differential equation in the domain D2 (cf. (1.10) and (1.14)). If µ = 1, then we find a new term 2k2 v0− in this averaged equation in D2 which describes the local extraction of a quantity with density v0− . 56
We emphasize the qualitative delicate influence of the parameters τ and µ. First, for τ 0 and µ 1 we find a nonstandard inhomogeneous consistency condition h2 ∂x2 v0− (x1 , 0) − ∂x2 v0+ (x1 , 0) + δτ,0 k1 Ξv0+ (x1 , 0) = Ξp0 (x1 ),
(x1 , 0) ∈ I0 ,
(1.19)
which takes into account the geometry of thin rods of the first class and interaction of walls of these rods with a medium. Second, in fact, for τ < 0 and µ 1 the problem (1.1), is divided into two independent problems (1.13) and (1.14) after the limit passage. A similar phenomenon is observed [18] in the case of multilevel thick junctions. However, in [18], such a division was caused by the homogeneous Dirichlet conditions on the walls of thin cylinders of the second level, whereas, in our problem (1.1), the reason is an interaction of walls of “small” rods with a medium. Moreover, this depends on pε . Note that the function pε can coincide with the function gε . Then the trace of g0 appears on the right-hand side of (1.19). In the case µ < 1, interactions between walls of “long” rods and the medium plays a dominant role in the asymptotic behavior of solutions. Note that this interaction is not necessarily too large locally for 0 µ < 1. However, such an effect takes place because of the total length of the boundaries of “long” rods. In this case (µ < 1), there are no dependence on gε , pε , β, and τ. As was noted in [4], for functionals that are defined on reflexive spaces and grow faster than the norm, only one natural definition of homogenization seems to be reasonable; namely, in terms of the energy convergence. Therefore, it is important to prove the convergence of energy integrals, which gives us the possibility to study variational problems and problems of optimal control in thick cascade junctions.
2. Auxiliary Assertions on Asymptotics Note that there are specific difficulties in the study of of boundary value problems in thick junctions with inhomogeneous Neumann boundary conditions, Fourier conditions, or nonlinear boundary conditions on the boundaries of thin domains. For homogenization of such problems a new approach was suggested in [15, 16, 17]. This approach is based on the use of special integral identities (the type of a thick junction is determined by its form). In our case, such an integral identity has the form (cf. [16]) εh2 2
vdx − ε
vdx2 = (2)
Sε
(2)
Y2
x 1
ε
∂x1 vdx
∀v ∈ H 1 G(2) . ε
(2.1)
(2)
Gε
Gε
(i)
Here, Y2 (ξ) = −ξ + [ξ] + 12 , [ξ] is the integer part of ξ, Sε is the union of lateral sides of thin (i)
(i)
(i)
rods of the ith class, i = 1, 2. We denote by Qε := Υε \ Sε the union of the lower bases of the thin rods of the ithe class, i = 1, 2. Taking into account that max |Y2 | 1, from (2.1) we obtain the inequality R
1
vL2 (S (2) ) C2 ε− 2 vH 1 (G(2) ) . ε
ε
(2.2)
By standard methods (cf. [37, 38]), we find 57
vL2 (Q(2) ) C3 vH 1 (G(2) ) . ε
(2.3)
ε
Lemma 2.1. Any function v ∈ H 1 (Ωε , Γ1 ) satisfies the estimates
2
v (x) dσx C1 ε (1)
and
Gε
|∇v| dx + 4 (1)
Υε
2
v 2 (x1 , 0) dx1
(2.4)
I0
v 2 (x) dσx C2 v2H 1 (Ωε ) .
(2.5)
(1)
Υε
Proof. We represent the values of v on the vertical part of the boundary of a “small” branch, (1) which form Sε as follows: (˘ x1 ,0)
v(x1 , y) = (x1 ,y)
where x ˘1 = x1 + (cf. Fig. 2).
∂v ds + v(˘ x1 , 0), ∂s
(2.6)
h1 (εl1 − y), and s is the natural parameter of the segment [(x1 , y) ; (˘ x1 , 0)] 4l1
(1)
Figure 2. The “small” branch of class Gε . We square both sides of the equality and, using the Cauchy–Bunyakovskii inequality and the inequality (a + b)2 2a2 + 2b2 , write 58
(˘ x
1 ,0)
2
v (x1 , y) 2C3 ε (x1 ,y)
∂v ∂s
2
ds + 2v 2 (˘ x1 , 0).
Then we integrate with respect to y from −εl1 to 0 and summarize over all the lateral sides of “small” branches. Then
v 2 (x1 , x2 ) dx2 2C3 ε
(1)
|∇v|2 dx + 2
v 2 (x1 , 0) dx1 .
(2.7)
I0
(1)
Sε
Gε
Similarly, the value of v on the horizontal part of the boundary of a “small’ branch, which (1) form Qε are represented as
0
∂v dx2 + v(x1 , 0). ∂x2
v(x1 , −εl1 ) = − −εl1
Making transformations similar to those made in the case of (2.6), we find
0
2
v (x1 , −εl1 ) 2l1 ε −εl1
∂v ∂x2
2
dx2 + 2v 2 (x1 , 0).
Further, we integrate with respect to x1 over the base of the “small” branch and summarize over all branches. Then
2
2
v (x1 , −εl1 ) dx1 2l1 ε (1)
|∇v| dx + 2
v 2 (x1 , 0) dx1 .
(2.8)
I0
(1)
Qε
Gε
Finally, from (2.7) and (2.8) we obtain (2.4) which implies (2.5). Remark 2.1. Similarly, we for any v ∈ H 1 (Ωε , Γ1 ) we can establish the estimate
⎛ ⎜ 2 χ(1) ε (x1 )v (x1 , 0) dx1 C4 ⎝ε
I0
|∇v|2 dx +
(1)
Gε
⎞ ⎟ v 2 (x) dσx ⎠ ,
(2.9)
(1)
Υε
(1)
where χε (x1 ) = χh1 (x1 /ε), and χh1 (ξ), ξ ∈ R, is a 1-periodic function defined on [0, 1] by the formula
χh1 (ξ) =
⎧ ⎨1,
ξ ∈ Ih1 := Ih1 (b1 ) ∪ Ih1 (b2 ) ∪ Ih1 (1 − b2 ) ∪ Ih1 (1 − b1 ),
⎩0,
ξ ∈ [0, 1] \ Ih1 .
(2.10)
(1)
It is obvious that χε → 4h1 weakly in L2 (0, a) as ε → 0. 59
SEtting ϕ = uε in the integral identity (1.7) and taking into account the inequalities (2.2), (2.3), and (2.5), we find
β |∇uε | dx fε uε dx + pε uε dσx + ε gε uε dσx 2
Ωε
Ωε
(1)
(2)
Υε
Υε
fε L2 (Ωε ) uε L2 (Ωε ) + pε L2 (Υ(1) ) uε L2 (Υ(1) ) + εβ gε L2 (Υ(2) ) uε L2 (Υ(2) ) ε
ε
ε
ε
1 C4 fε L2 (Ωε ) + pε L2 (Υ(1) ) + εβ− 2 gε L2 (Υ(2) ) uε H 1 (Ωε ) ε
ε
or, by the conditions on fε , pε , gε , and the values of the parameter β, uε H 1 (Ωε ) C5 .
(2.11)
Lemma 2.2. Under the above assumptions on the function gε , the following convergence holds as ε → 0:
gε ψ dσx −→ 2
ε
∀ψ ∈ H 1 (D2 ).
g0 ψdx
D2
(2)
Υε
Proof. By the identity (2.1), for any ψ ∈ H 1 (D2 ) we have
ε (2)
2 gε ψ dσx = h2
Υε
χ(2) ε (x1 )gε ψdx
D2
2 −ε h2
Y2
x 1
ε
∂x1 gε ψ dx + ε
(2)
gε ψdx1 ,
(2.12)
(2)
Gε
Qε
(2)
where χε (x1 ) = χh2 t(x1 /ε) and χh2 (ξ), ξ ∈ R, is a 1-periodic function defined on [0, 1] by the formula
χh2 (ξ) =
1, |ξ − 1/2| h2 /2, 0, h2 /2 < |ξ − 1/2| 1.
(2.13)
(2)
As is known, χε → h2 weakly in L2 (0, a) as ε → 0. Using (1.4), we find the limit of the first term on the right-hand side of (2.12): 2 lim ε→0 h2
χ(2) ε (x1 )gε ψdx
=2
D2
g0 ψdx.
D2
Since max |Y2 (ξ1 )| 1 ξ1 ∈R
and the norms of gε are uniformly bounded in H 1 (D2 ) with respect to ε, we have 60
(2.14)
2
x 1 Y2 ∂x1 gε ψ dx εC1 ψH 1 (D2 ) . ε h2 ε (2)
Gε
Taking into account the conditions on gε and the inequality (2.3), we find
gε ψdx1 εgε L2 (Q(2) ) ψL2 (Q(2) ) εC2 ψH 1 (D2 ) . ε ε ε (2)
Qε
The obtained estimates and limit (2.14) allow us to pass to the limit on the right-hand side of (2.12) and thereby prove the lemma. (1)
Let ϕ ∈ H 1 (Dε ). Then for almost all t ∈ [−εl1 , 0]
0 1 ϕ(·, x2 )dx2 ϕ(·, t) − εl1
L2 (0,a)
−εl1
√ l1 ε∂x2 ϕL2 (D(1) ) . ε
(2.15)
Indeed,
a
0 2 1 ϕ(x1 , x2 )dx2 dx1 ϕ(x1 , t) − εl1 0
−εl1
1 = (εl1 )2 1 = εl1
a 0 0
−εl1
2 1 ϕ(x1 , t) − ϕ(x1 , x2 ) dx2 dx1 εl1
a 0 t ∂x2 ϕ(x1 , y)dy 0 −εl1
= εl1
2
a 0 0 dx2 dx1
x2
a 0
2 ϕ(x1 , t) − ϕ(x1 , x2 ) dx2 dx1
0 −εl1
2 ∂x2 ϕ(x1 , y) dydx2 dx1
0 −εl1 −εl1
2 ∂x2 ϕ(x1 , x2 ) dx.
(1)
Dε
From (1.5), (1.6), and (2.15) it follows that
1 εl1
0 pε (x1 , x2 )dx2 →p0 (x1 ) in
L2 (0, a)
as
ε → 0.
(2.16)
−εl1
Lemma 2.3. Suppose that (1.5) and (1.6) hold. Then the following convergence holds as ε → 0: 61
pε (x)ψ(x1 )dσx −→ (4h1 + 8l1 )
∀ ψ ∈ H 1 (0, a).
p0 (x1 )ψ(x1 )dx1
(2.17)
I0
(1)
Υε
Proof. In this case, we use a trick similar to that used for “long” rods (cf. (2.1)) by introducing a special function Y1 of the form ⎧ −t + b1 , ⎪ ⎪ ⎪ ⎪ ⎨−t + b , 2 Y1 (t) = ⎪ −t + 1 − b2 , ⎪ ⎪ ⎪ ⎩ −t + 1 − b1 ,
t ∈ [0, δ0 ), t ∈ [δ0 , 12 ),
t ∈ [ 12 , 1 − δ0 ), t ∈ [1 − δ0 , 1),
where δ0 = (b1 + b2 )/2. We have the integral identity (similar to (2.1))
(1)
2 pε ϕdx2 = h1 ε
Sε
(1)
2 pε ϕdx − h1
Gε
Y1
x 1
ε
∀ϕ ∈ H 1 (G(1) ε )
∂x1 (pε ϕ) dx
(2.18)
(1)
Gε
which implies
(1)
2l1 pε ψdx2 = h1
Sε
a
χ(1) ε (x1 )
0
1 εl1
0 pε dx2 ψ(x1 ) dx1 − −εl1
2 h1
Y1
x 1
ε
∂x1 (pε ψ) dx,
(2.19)
(1)
Gε
(1)
where χε (x1 ) is defined in (2.10). Taking into account the boundedness of (1.5) and the convergence (2.16), from (2.19) we find
pε ψdx2 −→ 8l1
p0 ψdx1 .
(2.20)
I0
(1)
Sε
By (1.6), (2.15), and (2.16),
a
pε ψdx1 = (1)
Qε (l1 )
χ(1) ε (x1 )pε (x1 , −εl1 )ψ(x1 )dx1
0
−→ 4h1
p0 ψdx1 .
(2.21)
I0
Thus, combining (2.20) and (2.21), we obtain (2.17) (1)
Corollary 2.1. Suppose that ϕε ∈ H 1 (Dε ), the norms ϕε H 1 (D(1) ) are uniformly bounded ε with respect to ε, and 62
ϕε (x1 , 0)→ϕ0 (x1 )
in
L2 (0, a)
as
ε → 0.
(2.22)
Then the following convergence holds as ε → 0:
pε (x)ϕε (x)dσx −→ (4h1 + 8l1 )
p0 (x1 )ϕ0 (x1 )dx1 . I0
(1)
Υε
Proof. It is easy to verify that
pε (x)ϕε (x)dσx = (1)
(1)
Υε
ϕε (x)p0 (x1 )dσx +
pε (x) − p0 (x1 ) ϕε (x)dσx .
(1)
Υε
Υε
Taking into account the conditions on ϕε and (2.15), we have
1 εl1
0
in L2 (0, a)
ϕε (x1 , x2 )dx2 →ϕ0 (x1 )
as
ε → 0.
(2.23)
−εl1
From (1.5) and (2.16) it follows that p0 ∈ H 1 (0, a). As in Lemma 2.3, we show that
ϕε (x)p0 (x1 )dσx = (4h1 + 8l1 ) p0 (x1 )ϕ0 (x1 )dx1 . lim ε→0
I0
(1)
Υε
It is obvious that
pε (x1 , −εl1 ) − p0 (x1 ) ϕε (x1 , −εl1 )dx1 −→ 0 as
ε → 0.
(1)
Qε (l1 )
Using (2.18), we find
(1)
2 pε (x) − p0 (x1 ) ϕε (x)dx2 = εh1
Sε
pε (x) − p0 (x1 ) ϕε (x)dx
(1)
Gε
2 − h1
Y1
x 1
ε
∂x1
pε (x) − p0 (x1 ) ϕε (x) dx.
(2.24)
(1)
Gε
By (1.5) and the uniform boundedness of ϕε H 1 (D(1) ) , the limit of the last integral in (2.24) ε vanishes. For the first integral on the right-hand side we make the following transformations: 63
2 εh1
(1)
2 pε (x) − p0 (x1 ) ϕε (x)dx = εh1
Gε
+
(1)
Gε
2 εh1
pε (x) − pε (x1 , 0) ϕε (x) − ϕε (x1 , 0) dx
(1)
Gε
2 + εh1
2 pε (x) − pε (x1 , 0) ϕε (x1 , 0)dx + εh1
pε (x1 , 0) − p0 (x1 ) ϕε (x1 , 0)dx
(1)
Gε
pε (x1 , 0) − p0 (x1 ) ϕε (x) − ϕε (x1 , 0) dx.
(2.25)
(1)
Gε
By (1.6), (2.16), (2.22), and (2.23), the limit of the last three integrals in (2.25) vanishes. For the first integral on the right-hand side of (2.24) we have the following inequalities: 1 (x) − p (x , 0) ϕ (x) − ϕ (x , 0) dx p ε ε 1 ε ε 1 ε (1)
Gε
0 0 2 2 1 ∂ p (x , y)dy dx ∂ ϕ (x , y)dy dx x ε 1 x ε 1 2 2 ε (1)
Gε
x2
(1)
Gε
x2
3
Cε∂x2 pε L2 (G(1) ) ∂x2 ϕε L2 (G(1) ) Cε 2 . ε
ε
The corollary is proved.
3. Proof of Theorems Proof of Theorem 1.1. 1. We extend by zero the solution uε bounded on a rod of the second class. Since boundaries of thin rods are linear, we have ∂ x2 uε = ∂x2 u ε
almost everywhere in D2 .
Therefore, u ε belongs to the anisotropic Sobolev space W 0,1 (D2 ) = v ∈ L2 (D2 ) : the derivative ∂x2 v ∈ L2 (D2 ) exists .
(3.1)
(3.2)
From (2.11) it follows that uε H 1 (Ω0 ) and u ε W 0,1 (D2 ) are bounded uniformly with respect to ε. Then there exists a subsequence {ε } ⊂ {ε}, denoted again by ε, such that uε v0+ in
H 1 (Ω0 , Γ1 ),
u ε v =: h2 v0− in ∂ xi uε ζi in 64
L2 (D2 ),
L2 (D2 ), i = 1, 2,
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
as ε → 0,
(3.3)
where v0+ , v0− , ζ1 , ζ2 are some functions which will be specified below. 2. Let us find ζ2 . Consider an arbitrary function ψ ∈ C0∞ (D2 ). By (3.1), we have
∂ x2 uε ψdx =
D2
∂x2 u ε ψdx = −
D2
u ε ∂x2 ψdx.
D2
Passing to the limit as ε → 0, we find
ζ2 ψdx = −h2 D2
v0− ∂x2 ψdx
∀ψ ∈ C0∞ (D2 ),
D2
which implies ζ2 = h2 ∂x2 v0− almost everywhere in D2 . Now, we find ζ1 . We consider the test function ⎧ ⎨0, Φε (x) =
(1)
⎩εY x1 ψ, 2
ε
x ∈ Ω0 ∪ Gε , x∈
(2) Gε ,
∀ψ ∈ C0∞ (D2 ),
where the function Y2 is defined in (2.1). It is easy to see that Φε ∈ H 1 (Ωε , Γ1 ) and x x 1 1 ∂x1 ψ, εY2 ∂x2 ψ , ∇Φε = −ψ + εY2 ε ε
x ∈ G(2) ε .
Substituting Φε into the integral identity (1.7), we find
x x x 1 1 1 β+1 ∇uε · ∇ψ dx = ε ψdx + ε ψdσx . f ε Y2 gε Y2 −∂x1 uε ψ + εY2 ε ε ε (2)
(2)
Gε
(2)
Gε
Υε
Taking into account the properties of Y2 and the relations (2.2), (2.3), (1.3), (1.4), and the fact that β 1, from the previous equality we derive the inequality
∂x1 uε ψdx εc1 (∇uε L2 (G(2) ) ∇ψL2 (G(2) ) ε ε (2)
Gε
√ √ + fε L2 (G(2) ) ψL2 (G(2) ) + εβ−1 εgε L2 (Υ(2) ) εψL2 (Υ(2) ) ) ε
ε
ε
ε
εc1 fε L2 (Ωε ) + εβ−1 gε H 1 (D2 ) ψH 1 (D2 ) εc2 ψH 1 (D2 ) . Passing to the limit in this inequality as ε → 0, we obtain the identity
ζ1 ψdx = 0
∀ψ ∈ C0∞ (D2 ),
D2
which implies that ζ1 = 0 almost everywhere in D2 . 65
3. It remains to find v0+ and v0− . First, we find the traces of these functions on I0 = {(x1 , 0) : 0 < x1 < a}. By the compactness of the trace operator and the first relation in (3.3), we have uε (x1 , 0)→v0+ (x1 , 0)
in
L2 (0, a)
as
ε → 0.
(3.4)
We consider the equality
u ε (x1 , 0 − 0) = χ(2) ε (x1 )uε (x1 , 0 + 0)
for almost all x1 ∈ (0, a),
(3.5)
(2)
where the function χε is defined in (2.13). It is clear that the right-hand side of (3.5) weakly converges to h2 v0+ (x1 , 0) in L2 (0, a) as ε → 0 in view of (3.4). On the other hand,
a
χ2ε (x1 )uε (x1 , 0
0
=
1 l2
a u ε (x1 , 0 − 0)ψ(x1 )dx1
+ 0)ψ(x1 )dx1 = 0
u ε ψ(x1 )dx1 dx2 + D2
1 l2
(x2 + l2 )∂x2 u ε ψ(x1 )dx1 dx2
∀ψ ∈ C0∞ (0, a). (3.6)
D1
Passing to the limit in (3.6) as ε → 0, we find
a h2 0
h2 v (x1 , 0)ψ(x1 )dx1 = l2 +
v0− ψ(x1 )dx1 dx2
D2
a = h2
h2 + l2
v0− (x1 , 0)ψ(x1 )dx1
(x2 + l2 )∂x2 v0− ψ(x1 )dx1 dx2
D2
∀ψ ∈ C0∞ (0, a),
0
which implies that v0+ (x1 , 0) = v0− (x1 , 0)
for almost all x1 ∈ (0, a).
(3.7)
! ∈ H 1 (Ωε , Γ1 ) as follows: 4. Consider a function ϕ ∈ C ∞ (Ω2 ), ϕ|Γ1 = 0, and define ϕ ⎧ (2) ⎨ ϕ(x), x ∈ Ω0 ∪ Gε ; ϕ(x) ! = ⎩ ϕ(x , 0), x ∈ G(1) . ε 1 Using the identity (2.1), we can write the integral identity (1.7) with test function ϕ ! in the form 66
∇x uε · ∇x ϕdx +
Ω0
∇ x uε · ∇x ϕdx +
D2 µ−1 2k2
+ε
h2
u ε ϕdx − ε D2
Ωε
Y2
h2
χ2ε (x1 )fε ϕ dx +
D2
x 1
ε
∂x1
uε ϕ dx + εµ
Υε
uε ϕ dx1 (2)
Qε
fε ϕ(x1 , 0)dx (1)
β
Gε
pε ϕ(x1 , 0) dσx + ε
+
uε ϕ(x1 , 0) dσx (1)
(2)
Gε
∂x1 uε ∂x1 ϕ(x1 , 0)dx + ε k1 Gε
µ 2k2
fε ϕ dx +
τ
(1)
=
(1)
gε ϕ dσx
(3.8)
(2)
Υε
Υε
Now, we pass to the limit in this identity. Since
√ ∂x1 uε ∂x1 ϕ(x1 , 0)dx ∂x1 uε 2 (1) ∂x1 ϕ(x1 , 0) 2 (1) C ε, L Gε L Gε (1)
Gε
where the limit of the first line in (3.8) is equal (by (3.3)) to the expression
∇x v0+
· ∇x ϕdx + h2
Ω0
∂x2 v0− ∂x2 ϕdx.
(3.9)
D2
Arguing in the same way and using the condition (1.3), we find the limit of the fourth line in (3.8):
f0 ϕdx + h2 Ω0
f0 ϕdx.
D2
By Lemmas 2.2 and 2.3, the limit of the last line in (3.8) is equal to the expression
(4h1 + 8l1 )
p0 (x1 )ϕ(x1 , 0)dx1 + δβ,1 2
I0
g0 ϕdx.
D2
Arguing in the same way as in Lemma 2.2, we find the limit of the third line in (3.8):
δµ,1 2k2
v0− ϕdx.
(3.10)
D2
It remains to find the limit of the second line in (3.8), i.e., the term
τ uε ϕ(x1 , 0)dσx . ε k1 (1)
Υε
67
For this purpose, we argue in the same way as in Lemma 2.3, take into account the limit in (3.3) and the result of item 3 (the traces of v0+ and v0− on I0 are equal (cf. (3.8)). We briefly represent these arguments and emphasize the differences in our situation. We have
uε ϕ dσx = k1
k1 (1)
uε ϕ dx2 + k1
(1)
Υε
uε ϕ dx1 . (1)
Sε
Qε
Let us show that
v0+ (x1 , 0)ϕ dx1
(3.11)
v0+ (x1 , 0)ϕ dx1 .
(3.12)
uε ϕ dx1 −→ 4h1 I0
(1)
Qε
and
uε ϕ dx1 −→ 8l1 I0
(1) Sε
(1)
To prove (3.11), we subtract and add from the left the term χε uε (x1 , 0)ϕ(x1 , 0) in the integrands. Then
(uε − uε (x1 , 0)) ϕ dx1 +
uε ϕ dx1 = (1)
Qε
I0
(1)
Qε
+
(1) + u (x , 0) − χ v (x , 0) ϕ dx1 χ(1) ε 1 1 ε ε 0
+ χ(1) ε v0 (x1 , 0)
4h1 v0+ (x1 , 0)
−
ϕ dx1 + 4h1
I0
v0+ (x1 , 0)ϕ dx1 ,
(3.13)
I0
The second integral on the right-hand side of (3.13) converges to zero because of the convergence of traces (3.4), whereas the third integral converges to zero by the properties of the characteristic function. Let us show that the first integral converges to zero. Applying twice the Cauchy– Bunyakovskii inequality, we find
0
(uε − uε (x1 , 0)) ϕ dx1 = (1)
Qε
∂x2 uε dx2 ϕ dx1 (1)
Qε
−εl1
0
2 ∂x2 uε dx2
2
dx1
−εl1
(1)
Qε
⎛ ⎜ ⎝εl1
⎛ ⎜ ⎝
⎞1 2
⎟ ϕ2 (x1 , 0) dx1 ⎠
(1)
Qε
⎞1 ⎛ 2
⎟ ⎜ (∂x2 uε )2 dx⎠ ⎝
(1)
Gε
√ C εuε H 1 (Ωε ) −→ 0 68
1
⎞1 2
⎟ ϕ2 (x1 , 0) dx1 ⎠
(1)
Qε
as ε → 0.
Now, we prove (3.12). We have
(1) Sε
2 uε ϕdx2 = h1 ε
(1) Gε
a
2l1 = h1
2 uε ϕdx − h1
χ(1) ε (x1 )
0
1 εl1
Y1
x 1
ε
∂x1 (uε ϕ) dx
(1) Gε
0 uε dx2 ϕ(x1 ) dx1 − −εl1
2 h1
Y1
x 1
ε
∂x1 (uε ϕ) dx.
(1)
Gε
(1)
As above, we subtract and add the term χε uε (x1 , 0)ϕ(x1 , 0). Then
(1)
a
2l1 uε ϕdx2 = h1
χ(1) ε (x1 )
0
Sε
2l1 = h1
(1)
⎛ ⎝ 1 εl1
Qε
2l1 + h1 2l1 + h1
1 εl1
0
0 uε dx2 ϕ(x1 , 0) dx1 − −εl1
2 h1
Y1
x 1
ε
∂x1 (uε ϕ) dx
(1)
Gε
⎞ uε dx2 − uε (x1 , 0)⎠ ϕ dx1
−εl1
(1) + u (x , 0) − χ v (x , 0) ϕ dx1 χ(1) ε 1 1 ε ε 0 I0
+ + χ(1) ε v0 (x1 , 0) − 4h1 v0 (x1 , 0) ϕ dx1 I0
+ 8l1
v0+ (x1 , 0)ϕ dx1 −
I0
2 h1
Y1
x 1
ε
∂x1 (uε ϕ) dx.
(3.14)
(1)
Gε
The second integral on the right-hand side of (3.14) converges to zero in view of the convergence of traces (3.4), and the third integral on the right-hand side converges to zero because of the properties of the characteristic function. Let us show that the first integral converges to zero. Using the Cauchy–Bunyakovskii inequality, we find
0 1 2 2l1 uε dx2 −uε (x1 , 0) ϕ dx1 = (uε (x1 , x2 ) − uε (x1 , 0)) ϕ dx1 h1 εl1 εh1 (1)
Qε
−εl1
(1)
Gε
0 2 ∂ u dx ϕ dx = x ε 2 1 2 εh1 (1)
Gε
x2
69
2 εh1
0
2 ∂x2 uε dx2
1 2
dx1
⎛ ⎜ ⎝
x2
(1)
Gε
⎞1
2
⎟ ϕ2 (x1 , 0) dx1 ⎠
(1)
Gε
#1"
" #1
2 2 2l1 (∂x2 uε )2 dx ϕ2 (x1 , 0) dx1 εl1 h1 (1)
(1)
Gε
Qε
√ C εuε H 1 (Ωε ) −→ 0
as ε → 0.
It remains to show that the last integral on the right-hand side of (3.14) converges to zero. Taking into account the boundedness of Y1 and using the Cauchy–Bunyakovskii inequality, we find
x 2 1 ∂ Y (u ϕ) dx C ∂ (u ) ϕdx + C u ∂ ϕdx 1 x1 ε x1 ε ε x1 h1 ε (1)
(1)
Gε
(1)
Gε
#1 "
"
(∂x1 uε )2 dx
2
ϕ2 (x1 , 0) dx1
εl1
(1)
"
#1
2
Gε
2
εl1
(1)
Qε
√ C εuε H 1 (Ωε ) −→ 0
(uε )2 dx
+
(1)
Gε
#1 "
#1 (∂x1 ϕ)2 dx1
2
(1)
Gε
Qε
as ε → 0.
Thus,
τ
uε ϕ(x1 , 0)dσx −→ δτ,0 k1 (4h1 + 8l1 )
ε k1
v0+ (x1 , 0)ϕ(x1 , 0)dx1
as ε → 0.
(3.15)
I0
(1)
Υε
Taking into account the obtained results (cf. (3.9)–(3.10), (3.15)), the integral identity (3.8) is transformed in limit as follows:
∇x v0+ · ∇x ϕdx + h2
Ω0
∂x2 v0− ∂x2 ϕdx + δµ,1 2k2
D2
+ δτ,0 k1 (4h1 + 8l1 )
=
f0 ϕdx + h2
Ω0
+ δβ,1 2 D2
70
g0 ϕdx
v0− ϕdx
D2
v0+ (x1 , 0)ϕ(x1 , 0)dx1
I0
f0 ϕdx + (4h1 + 8l1 )
D2
p0 (x1 )ϕ(x1 , 0)dx1 I0
∀ϕ ∈ C ∞ (Ω2 ), ϕ|Γ1 = 0.
(3.16)
5. Since the set {v ∈ C ∞ (Ω2 ) : v|Γ1 = 0} is everywhere dense in the anisotropic Sobolev space H = {v ∈ L2 (Ω2 ) : there exists a weak derivative ∂x2 v ∈ L2 (Ω2 ) and v|Ω0 ∈ H 1 (Ω0 , Γ1 )} equipped with the inner product
∇x u · ∇x vdx + h2
(u, v)H := Ω0
∂x2 u∂x2 vdx + δµ,1 2k2
D2
uvdx + δτ,0 k1 Ξ
D2
u(x1 , 0)v(x1 , 0)dx1 , I0
where Ξ := (4h1 + 8l1 ), we conclude that the function v0+ (x), x ∈ Ω0 , v0 (x) = v0− (x), x ∈ D2 , which, as follows from items 1–3 of the proof, belongs to the space H, is a solution to the integral identity (3.16) with an arbitrary test function ϕ ∈ H. The uniqueness of such a solution is established in a standard way. Furthermore, it is easy to see that v0 is a weak solution to the problem (1.10) in the sense of the integral identity (3.16) in the space H. By the uniqueness of a solution to the homogenization problem (1.10), the above arguments in this proof remain valid in the case of an arbitrary subsequence of the sequence {ε} chosen at the beginning of the proof. Therefore, (1.8) also holds for the entire sequence {ε}. 6. We prove the convergence (1.11) for the energy integral Eε (uε ). It is obvious that
Eε (uε ) =
pε uε dσx + εβ
fε uε dx + Ωε
(1)
Ω0
fε u ε dx +
fε uε dx + D2 µ−1 2k2
+ε
h2
gε uε dσx (2)
Υε
=
Υε
(1)
Gε
(fε − f0 )uε dx +
f0 uε dx +
µ 2k2
gε u ε dx − ε
h2
D2
(1)
(1)
Gε
x 1
Y2
ε
pε uε dσx Υε
∂x1 gε uε dx + εµ
(2)
gε uε dx1 .
(3.17)
(2)
Gε
Qε
By (1.3), (1.8), (2.11), and the absolute continuity of the Lebesgue integral, the limit of the second line in (3.17) is equal to the expression
f0 v0+ dx + h2
Ω0
f0 v0− dx.
D2
Since µ 1, the limit of the last two terms in (3.17) vanishes in view of the conditions on the function gε and the inequalities (2.3), (2.11). By (1.4) and (1.8), we have µ−1 2k2
lim ε
ε→0
h2
gε u ε dx = δµ,1 2k2
D2
g0 v0− dx.
D2
Combining the calculations in Corollary 2.1 with the arguments in the proof of (3.15), we find 71
lim
ε→0
I0
(1)
Υε
Thus,
lim Eε (uε ) =
ε→0
p0 (x1 )v0+ (x1 , 0)dx1 .
pε uε dσx = (4h1 + 8l1 )
f0 v0+ dx
+ h2
Ω0
f0 v0− dx
g0 v0− dx
+ δβ,1 2
D2
p0 (x1 )v0+ (x1 , 0)dx1 .
+Ξ
D2
I0
On the other hand, since (v0 , v0 )H = E0 (v0 ) and the integral identity (3.16) with the test function ϕ = v0 has the form
(v0 , v0 )H =
f0 v0+ dx
f0 v0− dx
+ h2
Ω0
g0 v0− dx
+ δβ,1 2
D2
D2
+Ξ
p0 (x1 )v0+ (x1 , 0)dx1 ,
I0
we obtain (1.11). Theorem 1.1 is proved.
Proof of Theorem 1.2. The convergence (1.12) on a subsequence is proved almost wordby-word as in items 1 and 2 of the proof of Theorem 1.1. Let us show that the traces of v0+ and v0− on I0 = {(x1 , 0) : 0 < x1 < a} vanish. The integral identity (1.7) implies the inequality
ετ k1
(1)
Υε
u2ε dσx fε uε dx + pε uε dσx + εβ gε uε dσx Ωε
(1)
(2)
Υε
Υε
C4 (fε L2 (Ωε ) + pε 2 L
(1) Υε
+ εβ− 12 gε
(2)
L2 Υε
)uε
H 1 (Ωε ) .
Taking into account (2.11), we conclude that uε 2 L
(1)
Υε
C7 ε− τ2 −→ 0
as ε → 0.
(1)
Taking into account (2.9) and the fact that χε → 4h1 weakly in L2 (0, a) as ε → 0, we conclude that v0− (x1 , 0) vanishes. As in the case τ 0, we can prove (3.7). Thus, v0+ (x1 , 0) = v0− (x1 , 0) = 0 for almost all x1 ∈ (0, a).
(3.18)
Let ϕ ∈ C ∞ (Ω2 ), ϕ|Γ1 ∪I0 = 0, be an arbitrary function. Using ϕ, we define the function ϕˇ ∈ H 1 (Ωε , Γ1 ) as follows: ⎧ ⎨ ϕ(x), x ∈ Ω0 ∪ G(2) ε ; ϕ(x) ˇ = (1) ⎩ 0, x ∈ Gε and substitute it into the integral identity (1.7). Taking into account (2.1), we write the identity as follows: 72
∇ x uε · ∇x ϕdx
∇x uε · ∇x ϕdx + Ω0
+ εµ−1
2k2 h2
D2
u ε ϕdx − εµ
D2
=
fε ϕ dx + Ωε
2k2 h2
Y2
x 1
ε
(2)
uε ϕ dx1 (2)
Gε
χ2ε (x1 )fε ϕ dx + εβ
D2
∂x1 uε ϕ dx + εµ
Qε
gε ϕ dσx . (2)
Υε
Now, we pass to the limit in this identity. As in Theorem 1.1, we find
∇x v0+ · ∇x ϕdx + h2
Ω0
f0 ϕdx + h2
= Ω0
∂x2 v0− ∂x2 ϕdx + δµ,1 2k2
D2
D2
v0− ϕdx
D2
f0 ϕdx + δβ,1 2
∀ϕ ∈ C ∞ (Ω2 ), ϕ|Γ1 ∪I0 = 0.
g0 ϕdx
D2
By (3.18), this identity is equivalent to the following two identities:
∇x v0+
· ∇x ϕdx =
f0 ϕdx
Ω0
and
h2
∂x2 v0− ∂x2 ψdx
∀ϕ ∈ H 1 (Ω0 , Γ1 ∪ I0 )
(3.19)
Ω0
+ δµ,1 2k2
D2
v0− ψdx
D2
= h2 D2
f0 ψdx + δβ,1 2
g0 ψdx
(3.20)
D2
for any functions ψ ∈ W 0,1 (D2 ), ψ|I0 = 0, where the space W 0,1 (D2 ) is defined in (3.2). It is obvious that (3.19) is the integral identity corresponding to the weak solution to the problem (1.13), whereas the identity (3.20) corresponds to the problem (1.14). The existence and uniqueness of weak solutions to these problems are obvious. Therefore, (1.12) holds for the entire sequence {ε}. We prove the convergence (1.15) for the energy integral Eε (uε ), This integral admits the representation (3.17) and the limits of all its terms, except for
pε uε dσx , (1)
Υε
are equal to the same quantities as in item 6 of the proof of Theorem t1. Since uε (x1 , 0) → 0 in L2 (0, a), from Corollary 2.1 (cf. also Theorem 1.1,5 and (3.18)) we find
pε uε dσx = 0.
lim
ε→0
(3.21)
(1)
Υε
73
Then
lim Eε (uε ) =
ε→0
f0 v0+ dx + h2
Ω0
=
f0 v0− dx + δβ,1 2
D2
|∇v0+ |2 dx + h2
Ω0
g0 v0− dx
D2
|∂x2 v0− |2 dx + δµ,1 2k2
D2
|v0− |2 dx
D2
= E0 (v0+ ) + E0 (v0− ), where E0 (v0+ )
|∇v0+ |2 dx,
:=
E0 (v0− )
Ω0
:= h2
|∂x2 v0− |2 dx
+ δµ,1 2k2
D2
|v0− |2 dx.
D2
Theorem 1.2 is proved.
Proof of Theorem 1.3. From (2.11) we obtain the first convergence in (1.16) on some subsequence. From the integral identity (1.7) and the estimate (2.11) we obtain the inequality εµ k2
u2ε dσx C1 .
(3.22)
# "
2 2 2 C2 ε uε dσx + ε |∂x1 uε | dx ,
(3.23)
(2)
Υε
From (2.1) we find
(2)
Gε
u2ε dx
(2)
Sε
(2)
Gε
which implies, in view of (3.22) and (2.11),
u2ε dx C3 εϑ ,
(3.24)
(2)
Gε
where ϑ = min(1 − µ, 2). Thus, the second convergence in (1.16) also holds. Let us show that the trace of v0+ on I0 = {(x1 , 0) : 0 < x1 < a} vanishes. By (3.1), from (3.24) it follows that ∂x2 u ε weakly converge to zero in L2 (D2 ). Therefore, passing to the limit in (3.6) as ε → 0, we find v0+ (x1 , 0) = 0 for almost all x1 ∈ (0, a).
(3.25)
Consider an arbitrary function ϕ ∈ C ∞ (Ω0 ), ϕ|Γ1 ∪I0 = 0, and construct the test function ϕ ∈ H 1 (Ωε , Γ1 ) as follows: ϕ(x), x ∈ Ω0 , ϕ(x) = (1) (2) 0, x ∈ Gε ∪ Gε , 74
and substitute it into the integral identity (1.7). Then
∇x uε · ∇x ϕdx = Ω0
fε ϕ dx. Ω0
Passing to the limit in this identity, we find
∇x v0+ · ∇x ϕdx =
Ω0
f0 ϕdx
∀ϕ ∈ C ∞ (Ω0 ), ϕ|Γ1 ∪I0 = 0.
(3.26)
Ω0
The identity (3.26) means that v0+ is a unique weak solution to the problem (1.17). Therefore, (1.16) also holds for the entire sequence {ε}. Now, we prove the convergence (1.18) for the energy integral Eε (uε ). Taking into account (3.22), (3.24), (1.3), (1.4), (2.2), and (2.3), we find
1−µ ϑ ϑ β β−1 √ f u dx + ε g u dσ εgε L2 (Υ(2) ) C5 ε 2 C6 ε 2 . ε ε ε ε x fε L2 (G(2) ) C4 ε 2 + ε ε ε (2)
(2)
Gε
Υε
Since uε (x1 , 0) → 0 in L2 (0, a), the limit (3.21) holds. Thus, passing to the limit in the equality
Eε (uε ) =
fε uε dx + Ω0
pε uε dσx + εβ
fε uε dx + (2)
(1)
Gε
Υε
gε uε dσx , (2)
Υε
by (3.26), we find
lim Eε (uε ) =
ε→0
Ω0
f0 v0+
dx =
|∇v0+ |2 dx = E0 (v0+ ).
Ω0
Theorem 1.3 is proved.
Acknowledgement. The work of the first author is partially supported by the Alexander von Humboldt Foundation. The work of the second author is partially supported by the Russian Foundation for Basic Research (grant no. 06-01-00138). The major and final part of this work was made in Kyiv, autumn of 2007 within the framework of Agreement on scientific cooperation between the National Taras Shevchenko University of Kyiv and the Lomonosov Moscow State University. The second author deeply grateful for hospitality and support. References 1. V. A. Marchenko and E. Ya. Khruslov, Boundary Value Problems in Domains with a FineGrained Boundary [in Russian], Naukova Dumka, Kiev, 1974. 2. A. Bensoussan, J.-L. Lions, and G. Papanicolau, Asymptotic Analysis for Periodic Structures, North-Holland, Amsterdam, 1978. 3. N. S. Bakhvalov and G. P. Panasenko, Homogenization: Averaging Processes [in Russian], Nauka, Mowcow, 1984; English transl.: Kluwer Academic Publisher, Dordrecht, 1990. 75
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Submitted date: April 7, 2008 77
