ASYMPTOTIC ANALYSIS OF AN ARRAY OF CLOSELY SPACED ABSOLUTELY CONDUCTIVE INCLUSIONS

We consider the conductivity problem in an array structure with square closely spaced absolutely conductive inclusions of the high concentration, i.e. the concentration of inclusions is assumed to be close to 1. The problem depends on two small parameters: ε, the ratio of the period of the micro-structure to the characteristic macroscopic size, and δ, the ratio of the thickness of the strips of the array structure and the period of the microstructure. The complete asymptotic expansion of the solution to problem is constructed and justified as both ε and δ tend to zero. This asymptotic expansion is uniform with respect to ε and δ in the area {ε = O(δ), δ = O(ε)} for any positive α, β.

1. Introduction: statement of the problem.A lot of engineering problems lead to the PDE's stated in some domains of a small measure.One of such examples is the so called array structures (Fig. 1) introduced in [11], [12] and then studied by several authors in [1], [5], [13], [14], [15].
These array structures are presented by domains in R s (s = 2, 3) depending on two small parameters ε and δ.Here ε stands for a period of the microstructure (while the macroscopic characteristic size is taken equal to 1), and every periodic cell consists of thin strips (rods in 3-dimensional case) of thickness εδ, i.e. δ is the ratio of the thickness to the length of each rod.
As mentioned above, the PDE's modeling the physical field or process, are set in this array structure, and at the boundary Neumann or Dirichlet conditions are prescribed (cf.[11], [12], [1], [5], [13], [14], [15], [9], [10]).In [12] the complete asymptotic expansion was constructed for a solution of the conductivity equation, and in [13] for a solution of the elasticity equations.In particular, in [13] it was e d e Figure 1.Rectangular array proved that if ε δ = const or ε δ → ∞ then there is no convergence to the solution of formally homogenized problem; the leading term of the solution was constructed; it may be not bounded.For the details we refer to the book [14].
In the present paper we consider a composite rod which consists of one layer of the array structure Ω εδ and infinitely conductive inclusions occupying the "holes" G i of such structure (Fig. 2).

Figure 2. Domain Ω εδ
The leading term of the effective conductivity of such structure (as ε → 0 followed by δ → 0) was obtained in [6] by the network approximation technique (see [2], [3], [4]).Below we construct an asymptotic expansion of the solution independently of the order of passage to the limit as ε → 0, δ → 0.
More precisely, we consider a domain Ω εδ = Π ε i∈Z G i εδ , where in which the Laplace equation is set with the Neumann boundary condition at the boundary ∂Π ε : and with conditions of infinitely conductive inclusions at the boundary of each G i εδ , that is where C i is unknown constant, f ∈ C ∞ (R) is a T -periodic function of x 1 independent on ε (the number T is divisible by ε), such that (4) Our goal is to study an asymptotic of the solution u εδ to the problem (1)÷(3) as ε → 0 + and δ → 0 + .Extend u εδ by constant C i on every G i εδ .The existence and the uniqueness of solution u εδ of this problem, such that u εδ (x) dx = 0 (5) is proved in Appendix 1.
Remark 1. Assume that f and f (x 1 − T 2 ) are odd functions, T 2 is divisible by ε.Then there exists a unique solution of problem (1)÷( 3), ( 5) such that Indeed, we observe that in our assumption u εδ (•, x 2 ) is odd, too.In fact if we consider −u εδ (−x 1 , x 2 ) , we have 3) and (5).Then it is also a solution of problem (1)÷(3), (5).By uniqueness of solution we have ) is odd and by periodicity we see that So, the T -periodic in x 1 solution of problem (1)÷(3), ( 5) is also a solution of more usual boundary value problem (1)÷( 3), ( 6), (7).All results of Appendix 1 are still valid for solution of this problem (Fig. 3).

Figure 3. Finite bar
The paper is organized as follows.First we recall the asymptotic expansion technique in the case of finitely conductive inclusions (we follow [14] section 2.2.2).Then in section 2.1, we consider the simplified one-dimensional problem with absolutely conductive inclusions.This auxiliary problem helps to understand the behavior of the solutions of cell problems inside the inclusions and in the vertical strips of the domain Ω εδ .These solutions are used further in section 2.2 for the construction of the complete asymptotic expansion of the solution of problem (1)÷(3), (5).In particular, the solutions of one-dimensional cell problems of section 2.1 are modified in the neighborhoods of the corners of the periodic cell outside inclusions, as well as in the horizontal strips of Ω εδ .The "corner correctors" are functions of the boundary layer type: they are exponentially decaying as the distance from the corner divided by εδ tends to infinity.
Finally a priori estimate for (1)÷(3), ( 5) is applied to prove the estimate of order (ε √ ε in H 1 −norm for the K−th partial sum of the truncated asymptotic expansion (it is uniform with respect to ε and δ such that ε = O(δ α ) and δ = O(ε β ) for any positive α, β).This a priori estimate (as well as the existence and uniqueness of solution of problem (1)÷( 3), ( 5)) is proved in Appendix 1.The exponential decaying of the solutions of the corner boundary layer problems (58), (59) is proved in Appendix 2: for each infinite branch the problem is periodically extended and reduced to the case [8].
Recall the asymptotic expansion method for the case of finite conductivity of inclusions described in [14] section 2.2.2.That is, consider the problem, which is similar to (1)÷(3), ( 5) but with finite conductivity X ( x ε ) (since there is no dependence on the parameter δ here we drop such a subscript): where (the differentiability condition here is not important: see [14]).Suppose that f (x 1 ) is T -periodic in x 1 and T 0 f (x 1 ) dx 1 = 0; we seek for a Tperiodic in x 1 solution u ε of problem (8), (9) with vanishing average on . Assume that X (ξ 1 , ξ 2 ) = X (ξ 2 , ξ 1 ) .We look for the asymptotic solution u ε in the following form: for some K > 0, where and , where δ kj is the Kronecker symbol.Substituting (10) into (8) yields: this relation is supposed to be true up to the terms of order ε K , here the discrepancy function can be estimated by r ε L2,C ≤ C, (see below) and with operator L ξξ = div ξ (X (ξ) ∇ ξ ).Hereafter we set N m = 0 for m < 0. Substituting (10) into the boundary condition (9) gives us: We require that (a) where h ℓ is a constant, defined below in Remark 2. Note that for L ξξ = △ (that is, A kj = δ kj ) the equality (13) becomes: Remark 2. (Solvability of the problem (15)) There exists (up to a constant) a solution to problem (15) if and only if For ℓ = 0, 1, 2 we have h 0 = 0, h 1 = 0, h 2 = 2 j=1 A 1j ∂N1 ∂ξj + A 11 = 0, respectively, and when ℓ = 1 the problem (15) is the standard cell problem: where which is called the higher order homogenized equation.
Next we find coefficients of the expansion (11) by substituting it into (18): This relation is supposed to be true up to the terms of order ε K , and the second remainder is . Thus, for any m = 0, . . ., K the function v j (x 1 ) satisfies the following equation: with periodic boundary conditions.In particular, for m = 0: for m = 1: for 1 < m ≤ K: The solvability condition for m = 0 : is satisfied due to the assumption made above, while as well as is satisfied automatically due to the periodicity of the function v j .After this step, the equation ( 8) is satisfied up to a remainder ε K r ε (x)+ε K+1 r 1 ε (x 1 ).Applying the standard a priori estimate, we obtain which justifies the above procedure.
In the next section we apply the same methods to the problem (1)÷( 3), (5).It should be modified in order to take into account the new boundary conditions (3) and the dependency of the domain Ω εδ on the second small parameter δ.Remark 3. Extending problem (1)÷( 3), (5) ε−periodically with respect to x 2 , one can see that problem (8), ( 9) set in R 2 has a solution which tends to v 0 , and v 0 satisfies equation ( 21) equivalent to the equation −h 2 ∆v 0 = f (x 1 ) (because v 0 depends only on x 1 ).It means that h 2 is the effective conductivity of the periodic medium in x 1 direction.Here the effective conductivity is defined as the conductivity of an homogeneous medium mechanically equivalent to the heterogeneous one.It means that for any smooth right hand side f the solution of the conductivity equation for the heterogeneous medium is close to the solution of the conductivity problem of the effective homogeneous medium.We recall that if X = +∞ in the inclusions and X = 1 out of inclusions, then it is proved in [14], p. 316 that h 2 = 1 δ + o 1 δ as δ → 0. Let us mention here that if we consider a perforated medium and the right hand side has a support out of the holes then the effective conductivity is defined as h 2 multiplied by the measure of the periodic cell without the hole in dilated ξ− variables, that is, one minus the volume concentration of the holes.This factor can be explained by the following reason: in the mechanically equivalent homogeneous medium the macroscopically equivalent right hand side is "diffused" everywhere, even inside the holes.It means that it is equal to the original f multiplied by the mentioned above factor.
2. Asymptotic expansion of the problem for a strip with infinitely conductive inclusions.We apply now the technique presented in the previous section to the case of the infinitely conductive inclusions, that is, to problem (1)÷( 3), (5).
For ε → 0 + and δ → 0 + we are looking for the solution of (1)÷( 3), (5) in the form of an asymptotic expansion: for some K > 0, where where, as before, N ℓ is 1-periodic continuous function of 2.1.Simplified one-dimensional problem.First, we consider a simplified onedimensional problem for a strip with vertical infinitely conductive inclusions (Fig. 4): Then the solution of (1)÷(3), (5) in such a domain depends on x 1 only and such problem can be rewritten as (x 1 is denoted by x) where with boundary conditions or due to periodicity where C i is an unknown constant.As before, we seek for a T -periodic solution u εδ .Substitution of the expansion (26) into equation ( 28) yields (ξ 1 is denoted by ξ): with h ℓ = where −δ/2 •dξ.Note that N 0 = 1 and N ℓ = 0 for ℓ < 0.
We remark that condition (30a) means that du εδ dx = 0 in δ 2 ε, (1 − δ 2 )ε , then after substituting (26) into this equation we have: up to a remainder dx K+1 from which we obtain the following equations for N ℓ : Remark 4. The remainder r i,εδ shows that the solution does not belong to the subspace of functions equal to constant on the inclusions.So we will have to think about the construction of a special corrector, equal to − r (3) i,εδ dx 1 on inclusions placing the asymptotic expansion into the space of functions, constant on inclusions.Condition (30b) implies: The function v εδ is expanded into the Taylor series around a point x 0 ∈ δε 2 , (1 − δ 2 )ε : for some y ∈ (x 0 , x).For the point x 0 = ε 2 we have the following: Hence, where After substituting (35), (36), (37) into (34) we obtain: up to remainders r where x 0 = ε 2 .Thus, due to periodicity of N ℓ (38) can be rewritten as follows: (also up to above remainders) from what we obtain Now slightly move coordinate system so that its origin is in the middle of the inclusion.Taking into account the periodicity of the function N ℓ and (39) we obtain the following problem for N ℓ , ℓ ≥ 2: where (41) In particular, the first three problems for N 0 , N 1 and N 2 are as follows. (a) hence, hence (see Fig. 5), (44) Also consider problem for N 2 : (a) (45) For simplicity we move the system of coordinates so that perfectly conducting inclusion occupies the interval (− 1 2 , − δ 2 ) ( δ 2 , 1 2 ).Thus, two functions N 1 (ξ) and N 2 (ξ) would be (47) (see Fig. 7, 8).
The goal of this section was to obtain problem (41) for the functions N ℓ .They can be constructed as some piecewise polynomial functions.The coefficients h l can be expanded in powers of δ: Then we seek for the solution v εδ in the form of series ( 27) and obtain the chain of problems for v jr in the same way as in Introduction: , where are the right hand sides defined by v ′′ l1j1 such that l 1 ≤ l and j 1 < j or l 1 < l and j 1 ≤ j; f 00 = f 01 = 0, f 02 = f.3), (5).In this subsection we modify functions N ℓ constructed for the one-dimensional case in order to obtain solutions of cell problems for two-dimensions.For this end we construct some special correctors in horizontal strips of Ω εδ and some exponentially decaying boundary layer type correctors in the neighborhoods of the corners of Ω εδ .The values of constants h ℓ will be also modified because the measure of the periodic cell is greater in the 2-dimensional case, and the functions N ℓ are not the same.

Asymptotic expansion of solution to problem (1)÷(
For the correspondent two-dimensional problem we work with the solution u (K) εδ of the form (26) where v εδ are sought in the form (27), satisfying (12), with H ℓ , defined by ( 15) and ( 16). Denote shown in Fig. 9, that is, Γ ± q = x q = i + 1 2 ± 1−δ 2 ∩ Γ i , q = 1, 2. Since N ℓ is 1-periodic in ξ 1 we drop the index i and work with the periodicity cell S i δ .3), ( 5) we obtain, as in the previous subsections, the following chain of cell problems for N ℓ (recalling that L ξξ = △): where N in ℓ (ξ 1 ) is defined in the inclusion by relations (33) with N in 0 (ξ 1 ) = 1, and dξ and the normal vector ν is directed inside the inclusion .We find the surface integral of the right hand side of (50) from the following integral condition: for r = ℓ.The surface integral of the right hand side of (50) corresponds to j = 0 in (51) taking into account the direction of ν.Therefore, Note that where two last parentheses in (53) are equal zero on the inclusions.We decompose the solution N ℓ of ℓ th -cell problem (53) as follows: where 2 ) (see Fig. 10, 11).ℓ in each region would be constructed separately.So, constant h ℓ is fixed by the integral condition.
First, consider the vertical strip V .Then as mentioned above: We choose the function M v ℓ to satisfy the following problem: Since the constructed function M v ℓ does not depend on ℓ we drop this subscript hereafter.
Thus, for function N v ℓ (ξ 1 ) we have the following problem where N in ℓ is taken to be equal to the solution of the corresponding one-dimensional ℓ th -cell problem considered above in (33).
In horizontal strips H 1 and H 2 we decompose the function N h ℓ (ξ 1 , ξ 2 ) as follows: Since the constructed function M h ℓ does not depend on ℓ we drop this subscript hereafter.
We choose N h ℓ in the form: We extend both N v ℓ and N h ℓ in the periodic cell with zero where they are not defined.
For the corresponding solution N c ℓ (ξ 1 , ξ 2 ) in the half-cross C 1 ∪V ∪H 1 we consider similar decomposition (the other half-cross C 2 ∪ V ∪ H 2 is treated similarly): , where the function N c ℓ satisfies the following problem: (59) where Σ ± i , i = 1, 2 are shown in Fig. 12.Note that the problem for M c ℓ also does not depend on ℓ so we drop this subscript.Let us translate the origin of the coordinates to point (0, −1/2) and extend this problem to the infinite half-cross (−∞, +∞) × (0, δ 2 ) ∪ (− δ 2 , δ 2 ) × (0, +∞).Here we remark that if one denotes then M c does not depend on δ and satisfies: 0, otherwise with the boundary conditions generated by the above boundary conditions for M c .For such a function there exists an unique solution and it satisfies the following estimate: with some positive constants c 1 , c 2 due to Phrägmen-Lindelöf type theorem (see Appendices 1 and 2).The functions N c ℓ and N c ℓ also decay exponentially as translated ξ δ → +∞ (cf.Appendix 2).Therefore, we multiply these functions by a cutting function that vanishes at the distance |ξ| ≥ 2  3 and that is equal to 1 if |ξ| ≤ 1 3 .This multiplication will produce an error of order O e − c δ that is negligible in comparison with the desired error estimate O ε K−1 + δ K−1 √ ε if δ and ε are related by some bounds with some α and β from (0, +∞) .Thus, we have to add one more remainder in equation (1), that is, r εδ (x) such that r Consider now a periodicity cell − 1 2 , 1 2 × − 1 2 , 1 2 and denote by a j (j = 1, 2)"the corner points" a 1 = 0, − 1 2 and a 2 = 0, 1 2 ; denote by N c ℓ (a j ; ξ) and M c (a j ; ξ) the solutions of the above boundary layer problems corresponding to the half-crosses containing point a j .Set and where a j (j = 1, .., 4) is the corner of the unit square and χ(ξ) is defined by: (60) Note that functions N ℓ satisfy the cell problem (49) up to a remainder r (4) By induction it can be seen that with N v ℓj (η), N h ℓj (η) independent of δ, the sign ± is taken with respect to the domain H 1 or H 2 where the solution is sought; and r v ℓ K , r h ℓ K are bounded in the same sense, and that with N c ℓj (η) independent of δ and exponentially decaying, and r c δℓ K is exponentially decaying and bounded in L ∞ norm.
Then (52) yields: where h ℓj are independent of parameters.In particular Taking into account that the right hand side support is the domain Ω εδ , we can calculate the leading term of the effective conductivity of the strip multiplying this value h 2 by the measure of the periodic cell in the extended variables ξ (see Remark 3).We get then that the leading term for the effective conductivity is ε δ .If we extend the problem (1)÷(3), (5) ε−periodically with respect to x 2 , then it will model the conductivity of the periodic medium with absolutely conductive inclusions.Its effective conductivity has the leading term 1 δ .It corresponds to the asymptotic analysis of [14], p. 316, Theorem 4.10.1.Substituting now expansion (27) into (18) allows us to obtain the set of equations for v jr : v ′′ jr = f jr (x 1 ) where f jr are the right hand sides defined by v j1r1 , such that j 1 ≤ j and r 1 < r or j 1 < j and r 1 ≤ r; f 00 = f 01 = 0, f 02 = f .These equations can be solved successfully by induction in j, r with an additional condition T 0 v jr (x 1 ) dx 1 = 0. Finally, we obtain the equation (1) satisfied up to the remainders which are: • the remainder analogous to ε K r ε of Introduction, that is, εδ (x 1 ) ; • the remainder analogous to ε K+1 r ε (x 1 ) of Introduction, that is, where v jδ (x 1 ) = K r=0 δ r v jr (x 1 ); • the remainder of Taylor formula: of relations (34), (38); • the remainder r that should be compensated by a corrector equal to a primitive of the function −r i,εδ (x 1 ) extended by 0 outside the domain iε . This corrector will place an asymptotic solution into the space of functions equal to constants on the inclusions (Remark 4).
More precisely, this means that to compensate the last remainder r i,εδ given by (61), in the equation ∂u ∂x1 = 0 on the infinitely conductive inclusion, we add a corrector Φ εδ (x 1 ) to the asymptotic solution u + εZ such that Φ εδ (0) = 0.This corrector places the asymptotic solution to the space H 1 per,ε (U ε ) where an a priori estimate is proved (cf.Appendix 1), but it generates a new remainder in the right hand side of the Laplace equation, which is the Laplacian of the corrector Φ εδ or simply the derivative of the remainder r (3) i,εδ ; it is of order O ε K−1 in L ∞ norm.This adds a complementary term of such an order to the error estimate.
Finally, taking into account estimates for all the remainders and applying a priori estimate of Appendix 1, we obtain So we have proved the following result: Theorem 1.Let for any α, β > 0, ε = O (δ α ) and δ = O ε β .Then estimate (62) holds.
Theorem 4 justifies the asymptotic expansion of the solution of problem (1)÷( 3), (5) constructed in section 2. It gives the complete analysis of the conductivity of the periodic strip with infinitely conductive inclusions of the high concentration uniformly with respect to small parameters ε and δ such that ε = O (δ α ) and δ = O ε β for any α, β > 0. It justifies the existence of the effective conductivity of the strip, that is not evident for array structures (cf.[13], where it is not the case for the elasticity equations).The leading term of this macroscopic conductivity coincides with calculated in [6].Variational formulation of problem (1)÷( 4) is as follows: Find u εδ ∈ H  Note that the same result can be obtained easily directly from (68) by the Fourier expansion of f and u in ξ 2 .
Note that the Agmon-Duglas-Nirenberg theory gives estimate (70) with some constants for the derivatives of u if the derivatives of the right hand side f decay exponentially.

Figure 9 .
Figure 9.The boundary of an inclusion

Figure 13 .
Figure 13.The half-cross domain of definition of M c

( 4 )
K related to the multiplication of boundary layer functions N c ℓ and M c by the function χ given by (60); • the remainder related to the truncation of the expansions in δ of N v ℓ , N h ℓ , N c ℓ and h ℓ at the terms of order δ K ; • the remainder in (32):