ISSN 1064-5624, Doklady Mathematics, 2017, Vol. 96, No. 3, pp. 601–606. © Pleiades Publishing, Ltd., 2017. Published in Russian in Doklady Akademii Nauk, 2017, Vol. 477, No. 5, pp. 516–522.
MATHEMATICS
Homogenization of the Boundary Value Problem for the Laplace Operator in a Domain Perforated along (n – 1)-Dimensional Manifold with Nonlinear Robin Type Boundary Condition on the Boundary of Arbitrary Shaped Holes: Critical Case1 A. V. Podolskiy* and T. A. Shaposhnikova** Presented by Academician of the RAS V.V. Kozlov June 26, 2017 Received July 3, 2017
Abstract—The asymptotic behavior of the solution to the boundary value problem for the Laplace operator in a domain perforated along an (n − 1) -dimensional manifold is studied. A nonlinear Robin-type condition is assumed to hold on the boundary of the holes. The basic difference of this work from previous ones concerning this subject is that the domain is perforated not by balls, but rather by sets of arbitrary shape (more precisely, by sets diffeomorphic to the ball). A homogenized model is constructed, and the solutions of the original problem are proved to converge to the solution of the homogenized one. DOI: 10.1134/S1064562417060229
In the present paper we study the asymptotic behavior, as ε → 0, of the solution uε to the boundary value problem for the Laplace operator in a domain Ωε ⊂ Rn, n ≥ 3 , perforated by sets Gε with the diameter equal to aε = C0εα , α > 1. It is supposed that the cavities that form the perforation have an arbitrary shape and are ε -periodically distributed along an (n − 1)-dimensional manifold γ . The nonlinear Robintype condition ∂ νuε + ε−k σ(uε ) = 0 , where k ∈ R , is specified on the boundary of the holes. Thereby, the problem has several parameters: ε , αε , k ; different relations between these parameters lead to a different asymptotic behavior of the solution (see [1–8]). There exists a set of parameter values such that, under homogenization, the character of nonlinearity changes (see [1, 2, 5–10, 12]); this is the so-called critical case. We note that, in all the above papers, the critical case was studied only for domains perforated by balls. In the current paper, we consider the critical set of parameters, namely, α = k = n − 1 and assume n−2 1 The article was translated by the authors.
Faculty of Mechanics and Mathematics, Moscow State University, Moscow, 119992 Russia * e-mail:
[email protected] ** e-mail:
[email protected]
that the holes forming the perforation have an arbitrary shape. It is proved that the homogenized problem contains a new nonlinear transmission condition on the manifold along which the perforation was made. Note that the case n = 2 for the nonlinear Robin-type boundary condition and the general geometry of the perforations was studied in [9]. Let Ω be a bounded domain in Rn , n ≥ 3 , with a smooth boundary ∂Ω and γ = Ω ∩ {x1 = 0} ≠ φ be a n set on the plane {x1 = 0}. Define Y = − 1 , 1 . Con2 2 sider a set G0 such that G0 ⊂ T1/ 4 and G0 is diffeomorphic to the ball, where Tr denotes the ball of radius r centered at the origin. We define δB = {x: δ−1 x ∈ B},
( )
δ > 0, Ω+ = Ω ∩ {x1 > 0} , and Ω− = Ω ∩ {x1 < 0}. Let
Gε =
∪ (a G
z∈Z '
ε
0
+ εz ) =
∪G,
z∈Z '
z ε
(1)
where Z' is a set of vectors of the form z = (0, z2, … , zn ) with integer coefficients zi , i = 2, 3, … , n, aε = C0εk , k = n − 1 , C0 > 0 , Gεz = aεG0 + εz. n−2 The set Gε is defined as the union of sets Gεj ⊂ G ε
such that Gεj ⊂ Ω and ρ(∂Ω, Gεj ) ≥ 2ε , i.e., 601
602
PODOLSKIY, SHAPOSHNIKOVA
Gε =
∪G, z ε
z∈ϒε
ϒε = {j ∈ Z', ρ(∂Ω, Gεj ) ≥ 2ε} .
where d = const > 0 , and
⊂ Taε ⊂
Gεj
j
Tε/j 4
Note that |ϒε | ≅ d ε1− n ,
⊂ Yε ,
Sε = ∂G ε ,
∂Ωε = Sε ∪ ∂Ω.
∂ νuε + ε σ(uε ) = 0, x ∈ Sε , uε = 0, x ∈ ∂Ω,
ε
Ωε
+ε−k (3)
where f ∈ L2(Ω), ν is the outward normal unit vector to Sε , and ∂ νu is the normal derivative of u. We assume that the function σ: R → R is continuously differentiable, σ(0) = 0 , and there exists positive constants k1 , k2 such that
k1 ≤ ∂uσ(u) ≤ k2
∀u ∈ R.
k1u ≤ uσ(u) ≤ k2u
2
∀u ∈ R.
∫ ∇u ∇ϕdx + ε ∫ σ(u )ϕds = ∫ ε
Ωε
ε
Sε
f ϕdx,
(5)
Ωε
where ϕ is an arbitrary function from H 1(Ωε , ∂Ω) . Let H 1(Ωε , ∂Ω) denote the closure in H 1(Ωε ) of the set of infinitely differentiable functions in Ωε vanishing near the boundary ∂Ω . It is well known that problem (3) has a unique weak solution uε ∈ H 1(Ωε , ∂Ω). Let uε be an H1-extension of the function uε to Ω with the properties
||uε ||H 1(Ω) ≤ K ||uε ||H 1(Ω ), ε
ε−k ||uε ||2L2(Sε ) ≤ K ,
(9)
f (ϕ − uε )dx.
Ωε
Sε
We need to introduce some auxiliary functions and lemmas for proving a homogenization theorem. We define the function wεj (u, x) as a solution of the boundary value problem
Δwεj = 0,
x ∈ Tε/j 4 \Gεj ,
∂ ν wεj − ε−k σ(u − wεj ) = 0, j wε
= 0,
x∈
x ∈ ∂Gεj ,
(10)
j ∂Tε/ 4 ,
⎧wεj (u, x), x ∈ Tε/j 4 \Gεj , ⎪ Wε (u, x) = ⎨ n Tε/j 4 . ⎪0, x ∈ R \ j∈ϒε ⎩ The following lemma is valid.
(7)
here and below, the constants K and K1 are independent of ε .
j ∈ ϒε (11)
∪
Lemma 1. For all u ∈ R , we have Wε (u, .) ∈ H (Ωε , ∂Ω) and the following estimates are valid: 1
−k
||∇Wε ||L2(Ω ) + ε ||Wε ||L2(S ) ≤ Ku , 2
2
ε
2
ε
(12)
||Wε ||L2(Ω ) ≤ K ε|u|, ε
where K is a constant independent of u and ε . Proof. The integral identity for this problem yields
||∇uε ||L2(Ω) ≤ K ||∇uε ||L2(Ωε ). (6)
Here and below, K is a constant independent of ε. An extension with such properties was constructed in [3]. Setting ϕ = uε as a test function in the integral identity (5) and using the conditions imposed on σ(u) , we get the estimates
||∇uε ||L2(Ωε ) ≤ K ,
ε
where u ∈ R is a parameter, j ∈ ϒε . Then we consider
The weak solution of problem (3) is defined as a function uε ∈ H 1(Ωε , ∂Ω) satisfying the integral identity −k
∫ σ(ϕ)(ϕ − u )ds ≥ ∫
(4)
Therefore, we have 2
(8)
∫ ∇ϕ∇(ϕ − u )dx
(2)
x ∈ Ωε ,
−k
1
By using the properties of the function σ(u) , we can switch from the consideration of integral identity (5) to the consideration of an integral inequality of the form
We consider the boundary value problem
−Δuε = f ( x),
weakly in H 0(Ω), strongly in L2(Ω).
uε u0 uε → u0
j
where Tr j = Tr + εj and Yε j = εY + εj for j ∈ ϒε . Define
Ω ε = Ω \ Gε ,
From estimates (6) and (7), we conclude that there exists a subsequence (denoted as the original sequence uε ) such that, as ε → 0, we have
j
∫
|∇wε | dx − ε j2
−k
∫
σ(u − wε )wε ds = 0. j
j
∂Gεj
j
Tε / 4 \Gε
Then this relation can be rewritten as
j
∫
|∇wε | dx + ε j2
−k
∫
σ(u − wε )(u − wε )ds j
j
∂G ε
j
Tε / 4 \Gε
=ε
−k
∫
j
σ(u − wεj )uds.
∂G ε
j
Using the properties of the function σ, we derive the estimate DOKLADY MATHEMATICS
Vol. 96
No. 3
2017
HOMOGENIZATION OF THE BOUNDARY VALUE PROBLEM −k
||∇wε ||L2(T j j 2
j ε / 4 \Gε )
≤ε
−k
∫
+ k1ε ||u − wε ||L (∂G j ) j 2
ε
2
|σ(u − wε )||u|ds ≤ k2 |u|ε j
−k
∂Gεj
∫
|u − wε |ds. j
∂Gεj
Then we use Young’s inequality with parameter δ = k1 to get k2 |u| −k
||∇wε ||L2(T j j 2
j
ε / 4 \Gε )
+ k1ε ||u − wε ||L2(∂G j ) j 2
ε
−k
k1ε ||u − wεj ||2L2(∂G j ) + K |u|2εn−1. ε 2 Hence, we obtain ≤
j ε / 4 \Gε )
Lemma 2. The function wˆ(u, y) is Lipschitz continuous and monotonic with respect to the parameter u ; moreover, the following estimate is valid:
|wˆ(u1, y) − wˆ(u2, y)| ≤ |u1 − u2 |,
+
k1 −k ε ||Wε ||2L2(S ) ≤ Ku2. ε ε 2 Friedrichs’ inequality implies that j j ε / 4 \Gε )
≤ Kε
2
j 2 ||∇wε ||L2(T
j j ε / 4 \Gε )
This equation was widely considered in the literature [2, 6–9, 12], in which domains perforated by balls were considered. Remark 3. We define the function
H (u) = = C0
.
This estimate completes the proof of the lemma. Thereby, as ε → 0, we have 1
(13)
where W ε is an H 1-extension of Wε satisfying inequalities (6). Let wˆ(u, y) be a solution to the problem
Δwˆ = 0, y ∈ Rn \G0, ∂ ν y wˆ − C0σ(u − wˆ) = 0, y ∈ ∂G0, wˆ → 0 as y → ∞,
(14)
Vol. 96
No. 3
(16)
∀u ∈ R.
If G0 is the unit ball, then, by using the previous remark, we get
H (u) = C0 =
∫
σ(u − wˆ(u, y))ds y
∫
(n − 2)*(u)ds = (n − 2)ωn*(u),
∂ G0
where ωn is the surface area of the unit sphere in Rn . Lemma 3. The function H (u) defined by formula (16) is monotone and Lipschitz continuous. The following lemma gives an estimate for the proximity of the functions wεj and wˆ . Lemma 4. For the introduced functions wεj (u, x) and wˆ(u, y) , the following estimates are true:
+ k1ε−k ||vεj ||2L2(∂G j ) ≤ K |u|2εn,
||∇vεj ||2L2(T j
j ε / 4 \G ε )
where u ∈ R is a parameter. Remark 1. The existence and uniqueness of a weak solution wˆ(u, y) ∈ H 1(Rn \G0 ) of problem (14) are proved by using standard methods, taking into account the monotonicity of the function σ(u) . In addition, since wˆ(u, y) is a harmonic function, we have wˆ ∈ H 1(Rn \G0 ) ∩ C ∞ (Rn \G0 ) . The smoothness of the function wˆ(u, y) up to the boundary ∂Ω for a smooth boundary ∂Ω and a sufficiently smooth function σ follows from [11]. The following lemma establishes certain properties of wˆ(u, y) . DOKLADY MATHEMATICS
∫
σ(u − wˆ(u, y))ds y
∂G 0
ε
weakly in H 0(Ω), strongly in L2(Ω),
∂ ν y wˆ(u, y)ds y
∂G 0
||Wε ||2L2(Ω ) ≤ K ε2 ||∇Wε ||2L2(Ω ) ≤ K ε2u2.
W ε 0 W ε → 0
∫
∂G 0
Summing this result over all cells and using the obtained estimates, we derive ε
(15)
Remark 2. Let us consider the case when G0 is the unit ball. Then problem (14) has the solution wˆ(u, y) = *(u) , where the function *(u) satisfies the functional n −2 r equation
||∇Wε ||2L2(Ω ) + j 2 ||wε ||L2(T
∀u1, u2 ∈ R.
(n − 2)*(u) = C0σ(u − *(u)).
k1 −k ε ||u − wεj ||2L2(∂G j ) ≤ Ku2εn−1. ε 2 Summing the result over all cells j ∈ ϒε , we get ||∇wεj ||2L2(T j
603
2017
ε
||vεj ||2L2(T j
j ε / 4 \G ε )
≤ K |u|2εn+ 2,
(17) (18)
⎛ x − Pε j ⎞ where vεj (u, x ) = wεj (u, x ) − wˆ ⎜ u, ⎟. aε ⎠ ⎝ Proof. The function vεj is a solution of the problem
Δvε = 0, j
x ∈ Tε/ 4 \Gε , j
j
∂ νvεj − ε−k (σ(u − wεj ) − σ(u − wˆ)) = 0, ⎛ x − Pε j ⎞ vεj = −wˆ ⎜ u, ⎟, aε ⎠ ⎝
x ∈ ∂Gεj ,
x ∈ ∂Tε/j 4 .
604
PODOLSKIY, SHAPOSHNIKOVA
⎛ x − Pε j ⎞ j j Note that |vεj | ≤ wˆ ⎜ u, ⎟ in Tε/ 4 \Gε . In the intea ε ⎝ ⎠ gral identity for the above problem, we use vεj as a test function: ||∇vεj ||2L2(T j
j ε / 4 \Gε )
− ε −k
∫
Lemma 5. Let H (u) be the function defined in (16), ϕ ∈ C0∞ (Ω), hε , h ∈ H 01(Ω), and hε h in H 01(Ω) as ε → 0. Then ϒε
∑ ∫
(σ(u − wεj )
j =1 ∂T j ε/4
∂ Gε
− σ(u − wˆ))vεj ds = −
j
∫
j ⎛ x − Pε ⎞ ∂ ν x wˆ ⎜ ϕ(Pε j ), ⎟ hε ( x)ds aε ⎠ ⎝
n−2 + C0
ˆ . ∂ νvεj wds
The integral on the right-hand side of the equation is transformed into the form
−
∫
∫
=−
∂Tε j/ 4
j ˆ ∇vε ∇wdx
+
∫
j ˆ . ∂ νvε wds
∂Tε j/ 8
j j Tε / 4 \Tε / 8
= Kn ε
∫
∫
j ⎛ j j x − Pε ⎞ ∂ νθε = ∂ ν wˆ ⎜ ϕ(Pε ), ⎟, aε ⎠ ⎝
j
∂Tε / 8
j
j
Tε / 4 \Tε / 8
For θεj , the following estimates are valid:
∑ ||∇θ ||
j 2 ε L2 (Yε j \Tε j/ 4 )
j∈ϒε
j 2 ||∇vε ||L2(T
j∈ϒε ∂T j ε/4
j j ε / 4 \Gε )
+ k1ε
j 2 ||vε ||L2(∂G j ) ε
≤ K |u| ε . 2 n
This estimate yields the inequality
||vε ||L2(T j j 2
j ε / 4 \Gε )
Thus, the lemma is proved.
2 n+ 2
≤ K |u| ε
.
j∈ϒε
n−2
+C0
j 2 ε L2 (Yε j \Tε j/ 4 )
≤ Kε . 3
∫ H (ϕ(0, x'))h dx' ε
γ
≤
2 n
−k
∑ ||θ ||
j ⎛ x − Pε ⎞ ∂ ν wˆ ⎜ ϕ(Pε j ), ⎟ hεds aε ⎠ ⎝
∑ ∫
≤ 1 ||∇vε ||L2(T j \G j ) + K |u| ε . ε/2 ε 2 Combining the estimates obtained and using the properties of the function σ, we derive
≤ K ε,
From the integral identity for the solution θεj , it follows that
j 2 2 ≤ 1 ||∇vε ||L2(T j \G j ) + K ||∇wˆ||L2(T j \T j ) ε/4 ε ε/4 ε/8 2
j 2
}
μεj = −C0n−2H (ϕ(Pε j )).
⎛ x − Pε j ⎞ Having ∇wˆ ⎜ u, ⎟ ≤ K |u| , we get aε ⎠ ⎝
∫
{
ε
ˆ ≤ K |u| max ∂ νvεj wds |wˆ||∂Tε/j 8 | ≤ K |u|2εn. j
j ˆ ∇vε ∇wdx
j
x ∈ ∂Yε j ∩ x|x1 = − ε = γ −j,ε , 2 j j − ∂ νθε = 0, x ∈ ∂Yε \γ j,ε ,
∂ x1θεj = μεj ,
∂Tε /16 ( x )
j
x ∈ ∂Tε/ 4 ,
where ν is the unit normal vector to the boundary of the domain Yε j \Tε/4j . The constant μεj is defined by the solvability condition for this problem:
vε νi dx ≤ K |u|,
∂Tε / 8
j j x ∈ Yε \T ε/ 4,
Δθε = 0,
j
where Tr ( x ) is the ball of radius r with the center at the point x . It follows immediately from the last estimate that |∇vεj ( x )| ≤ K |u| for x ∈ ∂Tε/j 8 . Thus, we obtain an estimate of the second integral:
∫
Proof. Let ϕ ∈ C0∞ (Ω) be an arbitrary function. We introduce the function θεj as a solution to the problem
〈θε 〉Y j = 0,
∂vεj ≤ dx |Tε/16( x )| Tε /16( x ) ∂xi 1
as ε → 0, where ν x is the unit outward normal to ∂Tε/j 4 .
j
Let us estimate the integrals on the right-hand side of this equality. Let x ∈ ∂Tε/ 8 . On ∂Tε/j16( x ) from the maximum principle, we conclude that |vεj | ≤ |wˆ| ≤ K |u| ≤ K |u|ε2 − naεn−2 = K |u|ε2 − nεn−1 = K |u|ε. j n−2 |( x − Pε )/aε | Therefore, j |∂ xi vε ( x )|
∫ H (ϕ(0, x'))hdx' → 0 γ
∂Tε j/ 4
j ˆ ∂ νvε wds
(19)
∑ ∫
j∈ϒε Y j \ j ε T ε/4
+
∑μ ∫ j∈ϒε
j ε
∇θε ∇hεdx j
hεdx' + C0n−2 H (ϕ(0, x'))hεdx' .
−
γ j ,ε
∫ γ
Using estimates for θεj and the definition of the constant μεj , we derive that the right-hand side of the last inequality tends to zero as ε → 0. This implies the assertion of the lemma. DOKLADY MATHEMATICS
Vol. 96
No. 3
2017
HOMOGENIZATION OF THE BOUNDARY VALUE PROBLEM
We now state and prove the homogenization theorem. Theorem 1. Let n ≥ 3 , α = k = n − 1 , and uε be a n−2 weak solution of problem (3). Then the function u0 defined in (8) is a weak solution of the problem
−Δu0 = f , x ∈ Ω− ∪ Ω+ , [u0 ] = 0, x ∈ γ,
−ε−k
j∈ϒε ∂G j ε
+ε−k
∑ ∫ σ(ϕ − w (ϕ(P ), x))(ϕ − w (ϕ(P ), x) − u )ds j ε
j∈ϒε ∂G j ε
≥
n−2
∪
j∈ϒε
ε −k
∫
ε
ε
−
x) − uε )ds
j
j∈ϒε ∂T j \G j ε/4 ε
∑ ∫ σ(ϕ − w (ϕ(P ), x))(ϕ − w (ϕ(P ), x) − u )ds j∈ϒε ∂G j
j
j ε
lim ε→0
f (ϕ − Wε (ϕ, x) − uε )dx,
ε
j
ε
∫ ∇ϕ∇(ϕ − W (ϕ, x) − u )dx ε
ε
Ωε
=
∫
∇ϕ∇(ϕ − u0 )dx,
(23)
Ω ε→0
∇(wε (ϕ(Pε ), x))∇(ϕ − wε (ϕ(Pε ), x ) − uε )dx ε
(22)
where Z ε → 0 as ε → 0. Using estimates (12), we can prove the relations
lim
ε
Ωε
j ε
∫
j
Ωε
∫ ∇ϕ∇(ϕ − W (ϕ, x) − u )dx
+ ε −k
ε
∂ ν wε (ϕ(Pε ), x)(ϕ − uε )ds + Z ε j
≥
Estimates (12) imply that, for an arbitrary function ϕ ∈ C0∞ (Ω), we have W ε (ϕ, x) 0 in H 01(Ω) as ε → 0, where W ε is an H 1-extension of the function Wε (ϕ, x) to the domain Ω satisfying estimates (6). By using estimates (20) and the definition of the function Wε (ϕ, x) , we obtain
∑ ∫
∑ ∫
j∈ϒε ∂T j ε/4
(21)
j
(σ(ϕ(Pε j ) − wεj (ϕ(Pε j ), x))
as ε → 0, we have the inequality
σ(ϕ − wεj (ϕ(Pε j ), x))
j
∑ ∫
∫ ∇ϕ∇(ϕ − W (ϕ, x) − u )dx
f (ϕ − Wε (φ, x) − uε )dx.
j
ε
Ωε
Ωε
−
j
×(ϕ − wεj (ϕ(Pε j ), x) − uε )ds → 0,
ε
ε
ε
−σ(ϕ( x) − wεj (ϕ(Pε j ), x )))
Ωε
≥
j ε
f (ϕ − Wε (ϕ, x) − uε )dx.
j∈ϒε ∂G j ε
∫ ∇(ϕ − W (ϕ, x))∇(ϕ − W (ϕ, x) − u )dx j∈ϒε ∂G j ε ×(ϕ − wεj (ϕ(Pε j ),
j
Given that
Wε (ϕ, x) = wεj (ϕ(Pε j ), x) on Tε/j 4 \Gεj for j ∈ ϒε , and Wε (ϕ, x ) = 0 for x ∈ Ω\ Tε/j4 . We get
∑ ∫
∫
ε
Ωε
where H (u) is defined in (16). Proof. We set ϕ − Wε (ϕ, x) as a test function in integral identity (9) for problem (3), where ϕ ∈ C0∞ (Ω),
+ε−k
σ(ϕ(Pε j ) − wεj (ϕ(Pε j ), x))
×(ϕ − wεj (ϕ(Pε j ), x) − uε )ds
(20)
[∂ x1u0 ] = C0 H (u0 ), x ∈ γ, u0 = 0, x ∈ ∂Ω,
ε
∑ ∫
605
∫
f (ϕ − Wε (ϕ, x) − uε )dx =
Ωε
∫
f (ϕ − u0 )dx. (24)
Ω
Let us find the limit of the remaining integral of ⎛ x − Pε j ⎞ inequality (22). We set vεj = wˆ ⎜ ϕ(Pε j ), ⎟ – aε ⎠ ⎝ wεj (ϕ(Pε j ), x) . Green’s formula and Lemma 4 imply
ε
≥
∫
∑ ∫
f (ϕ − Wε (ϕ, x) − uε )dx.
Ωε
j∈ϒε ∂T j ε/4
By the definition of the functions rewrite this inequality in the form
∫
wεj (u,
x) , we can =
∑
j∈ϒε
∇ϕ∇(ϕ − Wε (ϕ, x) − uε )dx
∂ νvεj (ϕ − uε )ds
⎛ ⎞ j j ⎜ ∇vε ∇(ϕ − uε )dx − ∂ νvε (ϕ − uε )ds ⎟ ⎜ j j ⎟ ∂Gεj ⎝ Tε / 4 \Gε ⎠
∫
∫
Ωε
−
∑ ∫
j∈ϒε ∂T j ε/4
∂ ν wε (ϕ(Pε ), x)(ϕ − uε )ds j
≤
j
DOKLADY MATHEMATICS
Vol. 96
No. 3
∑ ∫ j∈ϒε
2017
j j Tε / 4 \Gε
∇vε ∇(ϕ − uε )dx j
(25)
606
+ε
PODOLSKIY, SHAPOSHNIKOVA −γ
∑∫
(σ(ϕ(Pε ) − wεj ) − σ(ϕ(Pε j ) − wˆ ))(ϕ − uε )ds j
j∈ϒε ∂G j
.
ε
Let us estimate the terms on the right-hand side of (25). Using the Cauchy and Young inequalities and applying Lemma 4 and estimates (7), we get the inequalities
≤
∑
j∈ϒε
1/ 2
≤ε
∇vεj ∇(ϕ
j
∑ ||∇(ϕ − u )|| j∈ϒε
− uε )dx
j∈ϒε ∂G j ε
+
2 j
j
L2 (Tε / 4 \Gε )
− uε )||L (T j 2
∑ε
j ε / 4 \Gε )
− 1/ 2
j∈ϒε
||∇vε ||L2(Ωε ) j 2
∑ ∫
−k
j∈ϒε ∂G j ε
≤ K1ε−1/ 2ε−k 1/ 2 − k
+K1ε ε
|vε ||ϕ − uε |ds j
∑ ||v ||
j∈ϒε 2 (||ϕ||L2(Sε )
j 2 ε L2 (∂Gεj )
+ ||uε ||2L2(Sε ) )
≤ K1ε−1/ 2εnε1− n + K1ε1/ 2 ≤ K ε. Combining the obtained estimates, we derive
∑ ∫
j∈ϒε ∂T j ε/4
j ⎛ x − Pε ⎞ (∂ ν wˆ ⎜ ϕ(Pε j ), ⎟ aε ⎠ ⎝
−∂ ν wε (ϕ(Pε ), x))(ϕ − uε )ds ≤ K ε. j
j
From this relation and Lemma 5, we deduce
− lim ε→0
∑ ∫
j∈ϒε ∂T j ε/4
∫ H (ϕ)(ϕ − u )dx'. 0
γ
From (21)–(26), it follows that
∫ ∇ϕ∇(ϕ − u )dx ∫ H (ϕ)(ϕ − u )dx' ≥ ∫ f (ϕ − u )dx,
∂ ν wεj (ϕ(Pε j ), x)(ϕ − uε )ds
0
0
Ω
where ϕ is an arbitrary function from H 01(Ω). This implies the assertion of the theorem. REFERENCES
(σ(ϕ(Pε j ) − wεj ) − σ(ϕ(Pε j ) − wˆ))(ϕ − uε )ds
≤ K1ε
n−2
= C0
γ
Then we estimate the second term of (25). Lemma 4 and inequalities (7) imply
∑ ∫
j∈ϒε ∂T j ε/4
+C0n−2
≤ K1(ε1/ 2 + ε−1/ 2εnε1− n ) = K ε.
ε −k
∑ ∫
Ω
j
Tε / 4 \Gε j ||∇vε ||L (T j \G j )||∇(ϕ 2 ε/4 ε
ε
ε→0
j ⎛ j x − Pε ⎞ ∂ ν wˆ ⎜ ϕ(Pε ), ⎟ (ϕ − uε )ds aε ⎠ ⎝
0
∑ ∫ j∈ϒε
= − lim
(26)
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DOKLADY MATHEMATICS
Vol. 96
No. 3
2017
