Generalized Chessboard Structures Whose Effective Conductivities Are Integer Valued

Abstract

We consider generalized chessboard structures where the local conductivity takes two values 𝑎 and 𝑏. All integer combinations of 𝑎 and 𝑏 which make the components of effective conductivity matrix integer valued are found. Moreover, we discuss the problem of estimating the effective conductivity matrix by using the finite-element method.

1. Introduction

Let ℎ be a large positive integer and consider a periodic composite material with period equal to 1/ℎ. The stationary heat conduction problem can then be formulated by the following minimum principle:𝐸_ℎ=min_𝑢ℱ^ℎ(𝑢)−_Ω𝑢(𝑥)𝑔(𝑥)𝑑𝑥,(1.1) whereℱ_ℎ(𝑢)=_Ω12grad𝑢(𝑥)⋅𝐶^ℎ(𝑥)grad𝑢(𝑥)𝑑𝑥.(1.2) Here, 𝑢 is the temperature; the conductivity matrix 𝐶_ℎ(𝑥) is given by𝐶_ℎ(𝑥)=𝐶(ℎ𝑥),(1.3) where 𝐶(⋅) is periodic relative to the unit cube of ℝ^𝑚, 𝑔 is the source field, Ω is a bounded-open subset of ℝ^𝑚, and the minimization is taken over some Sobolev space depending on the boundary conditions. It is possible to prove that the “energy” 𝐸_ℎ converges to the so-called homogenized “energy” 𝐸_hom as ℎ→∞, defined by𝐸_hom=min_𝑢ℱ^hom(𝑢)−_Ω𝑢(𝑥)𝑔(𝑥)𝑑𝑥,(1.4) whereℱ_hom(𝑢)=_Ω12𝐷𝑢(𝑥)⋅𝜎^∗𝐷𝑢(𝑥)𝑑𝑥.(1.5) The matrix 𝜎^∗ is often called the homogenized or effective conductivity matrix and is found by solving a number of boundary value problems on the cell of periodicity. For an elementary introduction to the theory of homogenization, see, for example, the book Persson et al. [1].

There are very few microstructures where all elements of the effective conductivity matrix are known in terms of closed form explicit formulae. Laminates and chessboards are the most classical. Mortola and Steffé [2] studied in 1985 a chessboard structure consisting of four equally sized squares in each period with (isotropic) conductivities 𝑎, 𝑏, 𝑐, and 𝑑, respectively (see Figure 1). They conjectured that the corresponding effective conductivity matrix of this composite structure is given by𝜎^∗=𝜎¹¹0𝜎0₂₂,(1.6) where𝜎𝜎₁₁=𝐺(𝐻(𝐴(𝑎,𝑏),𝐴(𝑐,𝑑)),𝐴(𝐻(𝑎,𝑑),𝐻(𝑏,𝑐)))₂₂,=𝐺(𝐻(𝐴(𝑎,𝑑),𝐴(𝑏,𝑐)),𝐴(𝐻(𝑎,𝑏),𝐻(𝑐,𝑑))).(1.7) Here, 𝐴, 𝐺, and 𝐻 denote the arithmetic mean, the geometric mean and the harmonic mean, respectively, given by the formulae:𝐴(𝛼,𝛽)=𝛼+𝛽2,𝐺(𝛼,𝛽)=√𝛼𝛽,𝐻(𝛼,𝛽)=2𝛼𝛽𝛼+𝛽.(1.8) Mortola and Steffés conjecture was ultimately proved by Craster and Obnosov [3] and Milton [4] in 2001 (see also [5]).

Figure 1 

The four component chessboard structure.

In this paper, we consider the special cases when the local conductivity only takes two values (see Figure 2). One such case is when 𝑎=𝑐=𝑑, which will be referred to as the connected case. The remaining cases are the chessboard case and the laminate case. The first of these is characterized by the property 𝑎=𝑐 and 𝑏=𝑑, and the second is characterized by the property 𝑎=𝑑 and 𝑏=𝑐. For these three cases, we find all integer combinations of 𝑎 and 𝑏 which make the components of 𝜎^∗ integer valued. We also discuss the problem of estimating 𝜎^∗ by using the finite-element method.

(a)

(b)

(c)

Figure 2 

The three special cases when the local conductivity only takes two values.

2. The Connected Case

For the connected case, it is easily verified that 𝜎₁₁=𝜎₂₂=𝜎, where 𝜎 is given by the formula:𝜎=𝑎²3𝑏+𝑎3𝑎+𝑏.(2.1) Assume that 𝑎 and 𝑏 are positive integers. In order to obtain an integer value of 𝜎, there must exist integers 𝑟₁, 𝑟₂, 𝑘, and 𝑙 such that3𝑏+𝑎=𝑘𝑟²¹,3𝑎+𝑏=𝑘𝑟²²,𝑎=𝑙𝑟₂,(2.2) wheregcd𝑟₁,𝑟₂=1.(2.3) We note that (2.2) is equivalent with𝑎=𝑘3𝑟²²−𝑟₁²8,𝑏=𝑘3𝑟¹²−𝑟₂²8,𝑎=𝑙𝑟².(2.4) Hence,𝑘3𝑟₂²−𝑟₁²=8𝑙𝑟².(2.5) It is clear that gcd(3(𝑟₂)²−(𝑟₁)²,𝑟₂)=1 (otherwise, 3(𝑟₂)²−(𝑟₁)²=𝑑𝑠 and 𝑟₂=𝑑𝑡 for some integers 𝑡𝑠 and 𝑑>1, which would mean that 𝑟²¹=𝑑(3𝑑𝑡²−𝑠), and since 𝑟²²=𝑑²𝑡², we obtain that gcd(𝑟²¹,𝑟²²)≥𝑑, which contradicts (2.3)). According to (2.5), this gives that𝑘=𝑠₂𝑟₂(2.6) for some integer 𝑠₂. By checking the four combinations when 𝑟₁ and 𝑟₂ are odd/even, we can easily verify thatgcd3𝑟₂²−𝑟₁²,8=gcd3𝑟₁²−𝑟₂²,8.(2.7) Hence, using (2.2), we obtain that𝑎=𝑠𝑟₂3𝑟₂²−𝑟₁²𝑑,𝑏=𝑠𝑟²3𝑟₁²−𝑟₂²𝑑,(2.8) where𝑑=gcd𝑟₂3𝑟²²−𝑟₁²,8,(2.9) and 𝑠 is some positive integer. The corresponding integer value of 𝜎 is then given by𝜎=𝑎3𝑏+𝑎3𝑎+𝑏=𝑠𝑟²3𝑟₁²−𝑟₂²𝑑⎷𝑘𝑟²¹𝑘𝑟²²=𝑠𝑟₁3𝑟₁²−𝑟₂²𝑑.(2.10) In order to have positive values of 𝑎 and 𝑏, (2.8) shows that we only can choose values of 𝑟₁ and 𝑟₂ such that13𝑟²²≤𝑟₁²≤3𝑟₂².(2.11)

Summing up, all integers 𝑎 and 𝑏 making 𝜎 integer valued are of the form (2.8) for some positive integers 𝑠, 𝑟₁ and 𝑟₂, satisfying (2.11), where 𝑑 is given by (2.9). The corresponding value of 𝜎 is𝜎=𝑠𝑟₁3𝑟₂²−𝑟₁²𝑑.(2.12) Conversely, all 𝑎 and 𝑏 of the form (2.8), satisfying (2.11), are positive integers and make 𝜎 integer valued.

For 𝑎≠𝑏, the smallest integer value of 𝜎 is obtained when 𝑟₁=5, 𝑟₂=3 (making 𝑑=2) and 𝑠=1, corresponding to the values 𝑎=3, 𝑏=99, and 𝜎=5.

3. The Chessboard-and Laminate Case

For the chessboard case, 𝜎₁₁=𝜎₂₂=𝜎, where𝜎=𝐺(𝑎,𝑏)=√𝑎𝑏.(3.1) This case is very simple. We just use the representation:𝑎=𝑘𝑟²¹,𝑏=𝑘𝑟²²,(3.2) where 𝑟₁, 𝑟₂, and 𝑘 are integers (giving 𝜎=𝑘𝑟₁𝑟₂).

For the laminate case, we find that 𝜎₁₁=𝐴(𝑎,𝑏) and 𝜎₂₂=𝐻(𝑎,𝑏). It is possible to show that the integers 𝑎 and 𝑏 making the harmonic mean integer valued are precisely those on the form:𝑎=𝑡𝑝(𝑝+𝑞),𝑏=𝑡𝑞(𝑝+𝑞),(3.3) (in this case, 𝐻(𝑎,𝑏)=2𝑡𝑞𝑝) and the form:𝑎=𝑡(2𝑝+1)(𝑝+𝑞+1),𝑏=𝑡(2𝑞+1)(𝑝+𝑞+1),(3.4) (in this case, 𝐻(𝑎,𝑏)=𝑡(2𝑞+1)(2𝑝+1)) where 𝑝, 𝑞, and 𝑡 are positive integers. For a proof of this fact, see [6]. Therefore, if 𝜎₁₁=𝐴(𝑎,𝑏) and 𝜎₂₂=𝐻(𝑎,𝑏) are integer valued, (𝑎,𝑏) must belong to the class (3.3) or (3.4). The latter is directly seen to generate integer values also for 𝐴. However, (3.3) gives integer values of 𝐴 only if both 𝑝 and 𝑞 are odd (for which (3.3) may be written on the form (3.4) with 𝑡 replaced by 2𝑡) or both even, or 𝑡 is even. In any case, if both 𝐻 and 𝐴 are integers, we end up with the form:𝑎=2𝑡𝑝(𝑝+𝑞),𝑏=2𝑡𝑞(𝑝+𝑞),(3.5) (in this case 𝜎₁₁=𝑡(𝑝+q)² and 𝜎₂₂=4𝑡𝑞𝑝) and the form:𝑎=𝑡(2𝑝+1)(𝑝+𝑞+1),𝑏=𝑡(2𝑞+1)(𝑝+𝑞+1)(3.6) (in this case 𝜎₁₁=𝑡(𝑝+𝑞+1)² and 𝜎₂₂=𝑡(2𝑞+1)(2𝑝+1)).

4. Calculating 𝜎^∗ by Numerical Methods

As mentioned in the introduction, the effective conductivity matrix is found by solving a number of boundary value problems on the cell of periodicity. For the connected case, the effective conductivity 𝜎 can be found by solving the following boundary value problem:𝑢di0v,(𝑥𝜆₂(𝑥)grad𝑢)=0on𝑌,𝜕𝑢=0,𝑢1,𝑥₂=1,𝜕𝑥₂𝑥₁,0=𝜕𝑥₂𝜕𝑥𝑢₁,1=0.(4.1) Here, 𝑌 is the unit cell 𝑌=[0,1]² and the conductivity 𝜆 is defined by𝜆(𝑥)=𝑎if𝑥∈𝑌⧵[0.25,0.75]²𝑏if𝑥∈[0.25,0.75]₂,.(4.2) In addition, we must assume continuity of normal component of 𝜆(𝑥)grad𝑢 through the four surfaces where 𝜆(𝑥) changes its value from 𝑎 to 𝑏. The effective conductivity 𝜎 is then found by calculating the integral:𝜎=_𝑌𝜆(𝑥)||grad𝑢||²𝑑𝑥.(4.3) The above boundary value problem can be solved numerically by using the finite-element method with relatively good accuracy. The hardest case is assumed to be the one when 𝑏=0, but even in this case, we obtain a numerical value close to the exact one. Using a couple of thousand elements (built up by second-order polynomials), the numerical value of 𝜎 for the case 𝑎=1 and 𝑏=0 turns out to be 0.5773, which is close to the exact value:𝜎=1√3≈0.5773502692.(4.4) Using the class of integer values for 𝑎 and 𝑏 giving integer valued effective conductivity, we may relatively easily make internet-based student projects where the students themselves are supposed to train their skills in using FEM programs, simply by randomly generating 𝑎 and 𝑏 from (2.8) and ask the students to find the integer which is the closest to their numerical estimate. Without knowing the exact formula, there are no way that they can guess the correct value without doing the numerical FEM calculation. The actual evaluation can be done by a simple Java-script or other programs which can register the students progress.

For the chessboard structure, we can use the same method with the only difference that 𝜆(𝑥) is defined by𝜆(𝑥)=𝑎,if𝑥∈[0,0.5]²𝑏,otherwise.∪[0.5,1]²,(4.5) However, the numerical estimation of the effective conductivity turns out to be significantly more difficult. In order to illustrate, we have made a numerical estimation for the case when 𝑎=1 and 𝑏=10000. Even with about 10000 quadrilateral 8-node elements (with increasing number of elements close to the midpoint of the unit-cell), our numerical value turned out to be as high as 557, which is more than 5 times higher than the actual value which is 𝜎=100. There exists a numerical method which is much more efficient than the finite-element method in such problems. Concerning this, we refer to the paper [7].

The numerical calculation of 𝜎^∗ for the laminate case turns out to be trivial. In fact, we only need two elements to obtain a numerical value which is exactly equal to 𝜎_𝑖𝑖.

Acknowledgment

The authors are grateful to two anonymous reviewers for some helpful comments which have improved the presentation of this paper.