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Journal of Applied Mathematics
VolumeΒ 2012, Article IDΒ 874239, 7 pages
Research Article

Generalized Chessboard Structures Whose Effective Conductivities Are Integer Valued

1Narvik University College, P.O. Box 385, 8505 Narvik, Norway
2Norut Narvik, P.O. Box 250, 8504 Narvik, Norway

Received 28 September 2011; Accepted 18 November 2011

Academic Editor: SrinivasanΒ Natesan

Copyright Β© 2012 Dag Lukkassen and Annette Meidell. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


We consider generalized chessboard structures where the local conductivity takes two values π‘Ž and 𝑏. All integer combinations of π‘Ž 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 β„Ž be a large positive integer and consider a periodic composite material with period equal to 1/β„Ž. The stationary heat conduction problem can then be formulated by the following minimum principle:πΈβ„Ž=minπ‘’ξ‚΅β„±β„Žξ€œ(𝑒)βˆ’Ξ©ξ‚Άπ‘’(π‘₯)𝑔(π‘₯)𝑑π‘₯,(1.1) whereβ„±β„Ž(ξ€œπ‘’)=Ξ©ξ‚€12grad𝑒(π‘₯)β‹…πΆβ„Ž(π‘₯)grad𝑒(π‘₯)𝑑π‘₯.(1.2) Here, 𝑒 is the temperature; the conductivity matrix πΆβ„Ž(π‘₯) is given byπΆβ„Ž(π‘₯)=𝐢(β„Žπ‘₯),(1.3) where 𝐢(β‹…) is periodic relative to the unit cube of β„π‘š, 𝑔 is the source field, Ξ© is a bounded-open subset of β„π‘š, and the minimization is taken over some Sobolev space depending on the boundary conditions. It is possible to prove that the β€œenergy” πΈβ„Ž converges to the so-called homogenized β€œenergy” 𝐸hom as β„Žβ†’βˆž, defined by𝐸hom=min𝑒ℱhomξ€œ(𝑒)βˆ’Ξ©ξ‚Άπ‘’(π‘₯)𝑔(π‘₯)𝑑π‘₯,(1.4) whereβ„±hom(ξ€œπ‘’)=Ξ©ξ‚€12𝐷𝑒(π‘₯)β‹…πœŽβˆ—ξ‚π·π‘’(π‘₯)𝑑π‘₯.(1.5) The matrix πœŽβˆ— 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 SteffΓ© [2] studied in 1985 a chessboard structure consisting of four equally sized squares in each period with (isotropic) conductivities π‘Ž, 𝑏, 𝑐, and 𝑑, respectively (see Figure 1). They conjectured that the corresponding effective conductivity matrix of this composite structure is given byπœŽβˆ—=ξ‚ƒπœŽ1100𝜎22ξ‚„,(1.6) where𝜎11,𝜎=𝐺(𝐻(𝐴(π‘Ž,𝑏),𝐴(𝑐,𝑑)),𝐴(𝐻(π‘Ž,𝑑),𝐻(𝑏,𝑐)))22=𝐺(𝐻(𝐴(π‘Ž,𝑑),𝐴(𝑏,𝑐)),𝐴(𝐻(π‘Ž,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)=2𝛼𝛽.𝛼+𝛽(1.8) Mortola and SteffΓ©s 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 π‘Ž=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 π‘Ž=𝑐 and 𝑏=𝑑, and the second is characterized by the property π‘Ž=𝑑 and 𝑏=𝑐. For these three cases, we find all integer combinations of π‘Ž and 𝑏 which make the components of πœŽβˆ— integer valued. We also discuss the problem of estimating πœŽβˆ— by using the finite-element method.

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:ξ‚™πœŽ=π‘Ž23𝑏+π‘Ž.3π‘Ž+𝑏(2.1) Assume that π‘Ž and 𝑏 are positive integers. In order to obtain an integer value of 𝜎, there must exist integers π‘Ÿ1, π‘Ÿ2, π‘˜, and 𝑙 such that3𝑏+π‘Ž=π‘˜π‘Ÿ21,3π‘Ž+𝑏=π‘˜π‘Ÿ22,π‘Ž=π‘™π‘Ÿ2,(2.2) whereξ€·π‘Ÿgcd1,π‘Ÿ2ξ€Έ=1.(2.3) We note that (2.2) is equivalent with3ξ€·π‘Ÿπ‘Ž=π‘˜2ξ€Έ2βˆ’ξ€·π‘Ÿ1ξ€Έ283ξ€·π‘Ÿ,𝑏=π‘˜1ξ€Έ2βˆ’ξ€·π‘Ÿ2ξ€Έ28,π‘Ž=π‘™π‘Ÿ2.(2.4) Hence,π‘˜ξ‚€3ξ€·π‘Ÿ2ξ€Έ2βˆ’ξ€·π‘Ÿ1ξ€Έ2=8π‘™π‘Ÿ2.(2.5) It is clear that gcd(3(π‘Ÿ2)2βˆ’(π‘Ÿ1)2,π‘Ÿ2)=1 (otherwise, 3(π‘Ÿ2)2βˆ’(π‘Ÿ1)2=𝑑𝑠 and π‘Ÿ2=𝑑𝑑 for some integers 𝑑𝑠 and 𝑑>1, which would mean that π‘Ÿ21=𝑑(3𝑑𝑑2βˆ’π‘ ), and since π‘Ÿ22=𝑑2𝑑2, we obtain that gcd(π‘Ÿ21,π‘Ÿ22)β‰₯𝑑, which contradicts (2.3)). According to (2.5), this gives thatπ‘˜=𝑠2π‘Ÿ2(2.6) for some integer 𝑠2. By checking the four combinations when π‘Ÿ1 and π‘Ÿ2 are odd/even, we can easily verify thatξ‚€3ξ€·π‘Ÿgcd2ξ€Έ2βˆ’ξ€·π‘Ÿ1ξ€Έ23ξ€·π‘Ÿ,8=gcd1ξ€Έ2βˆ’ξ€·π‘Ÿ2ξ€Έ2,8.(2.7) Hence, using (2.2), we obtain thatπ‘Ž=π‘ π‘Ÿ23ξ€·π‘Ÿ2ξ€Έ2βˆ’ξ€·π‘Ÿ1ξ€Έ2𝑑,𝑏=π‘ π‘Ÿ23ξ€·π‘Ÿ1ξ€Έ2βˆ’ξ€·π‘Ÿ2ξ€Έ2𝑑,(2.8) whereξ‚€π‘Ÿπ‘‘=gcd2ξ‚€3ξ€·π‘Ÿ2ξ€Έ2βˆ’ξ€·π‘Ÿ1ξ€Έ2,8,(2.9) and 𝑠 is some positive integer. The corresponding integer value of 𝜎 is then given byξ‚™πœŽ=π‘Ž3𝑏+π‘Ž3π‘Ž+𝑏=π‘ π‘Ÿ23ξ€·π‘Ÿ1ξ€Έ2βˆ’ξ€·π‘Ÿ2ξ€Έ2π‘‘ξ„Άξ„΅ξ„΅βŽ·π‘˜π‘Ÿ21π‘˜π‘Ÿ22=π‘ π‘Ÿ13ξ€·π‘Ÿ1ξ€Έ2βˆ’ξ€·π‘Ÿ2ξ€Έ2𝑑.(2.10) In order to have positive values of π‘Ž and 𝑏, (2.8) shows that we only can choose values of π‘Ÿ1 and π‘Ÿ2 such that13ξ€·π‘Ÿ2ξ€Έ2β‰€ξ€·π‘Ÿ1ξ€Έ2ξ€·π‘Ÿβ‰€32ξ€Έ2.(2.11)

Summing up, all integers π‘Ž and 𝑏 making 𝜎 integer valued are of the form (2.8) for some positive integers 𝑠, π‘Ÿ1 and π‘Ÿ2, satisfying (2.11), where 𝑑 is given by (2.9). The corresponding value of 𝜎 is𝜎=π‘ π‘Ÿ13ξ€·π‘Ÿ2ξ€Έ2βˆ’ξ€·π‘Ÿ1ξ€Έ2𝑑.(2.12) Conversely, all π‘Ž and 𝑏 of the form (2.8), satisfying (2.11), are positive integers and make 𝜎 integer valued.

For π‘Žβ‰ π‘, the smallest integer value of 𝜎 is obtained when π‘Ÿ1=5, π‘Ÿ2=3 (making 𝑑=2) and 𝑠=1, corresponding to the values π‘Ž=3, 𝑏=99, and 𝜎=5.

3. The Chessboard-and Laminate Case

For the chessboard case, 𝜎11=𝜎22=𝜎, where√𝜎=𝐺(π‘Ž,𝑏)=π‘Žπ‘.(3.1) This case is very simple. We just use the representation:π‘Ž=π‘˜π‘Ÿ21,𝑏=π‘˜π‘Ÿ22,(3.2) where π‘Ÿ1, π‘Ÿ2, and π‘˜ are integers (giving 𝜎=π‘˜π‘Ÿ1π‘Ÿ2).

For the laminate case, we find that 𝜎11=𝐴(π‘Ž,𝑏) and 𝜎22=𝐻(π‘Ž,𝑏). It is possible to show that the integers π‘Ž and 𝑏 making the harmonic mean integer valued are precisely those on the form:π‘Ž=𝑑𝑝(𝑝+π‘ž),𝑏=π‘‘π‘ž(𝑝+π‘ž),(3.3) (in this case, 𝐻(π‘Ž,𝑏)=2π‘‘π‘žπ‘) and the form:π‘Ž=𝑑(2𝑝+1)(𝑝+π‘ž+1),𝑏=𝑑(2π‘ž+1)(𝑝+π‘ž+1),(3.4) (in this case, 𝐻(π‘Ž,𝑏)=𝑑(2π‘ž+1)(2𝑝+1)) where 𝑝, π‘ž, and 𝑑 are positive integers. For a proof of this fact, see [6]. Therefore, if 𝜎11=𝐴(π‘Ž,𝑏) and 𝜎22=𝐻(π‘Ž,𝑏) are integer valued, (π‘Ž,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 π‘ž are odd (for which (3.3) may be written on the form (3.4) with 𝑑 replaced by 2𝑑) or both even, or 𝑑 is even. In any case, if both 𝐻 and 𝐴 are integers, we end up with the form:π‘Ž=2𝑑𝑝(𝑝+π‘ž),𝑏=2π‘‘π‘ž(𝑝+π‘ž),(3.5) (in this case 𝜎11=𝑑(𝑝+q)2 and 𝜎22=4π‘‘π‘žπ‘) and the form:π‘Ž=𝑑(2𝑝+1)(𝑝+π‘ž+1),𝑏=𝑑(2π‘ž+1)(𝑝+π‘ž+1)(3.6) (in this case 𝜎11=𝑑(𝑝+π‘ž+1)2 and 𝜎22=𝑑(2π‘ž+1)(2𝑝+1)).

4. Calculating πœŽβˆ— 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:𝑒div(πœ†(π‘₯)grad𝑒)=0onπ‘Œ,0,π‘₯2ξ€Έξ€·=0,𝑒1,π‘₯2ξ€Έ=1,πœ•π‘’πœ•π‘₯2ξ€·π‘₯1ξ€Έ=,0πœ•π‘’πœ•π‘₯2ξ€·π‘₯1ξ€Έ,1=0.(4.1) Here, π‘Œ is the unit cell π‘Œ=[0,1]2 and the conductivity πœ† is defined byξ‚»[]πœ†(π‘₯)=π‘Žifπ‘₯βˆˆπ‘Œβ§΅0.25,0.752,[]𝑏ifπ‘₯∈0.25,0.752.(4.2) In addition, we must assume continuity of normal component of πœ†(π‘₯)grad𝑒 through the four surfaces where πœ†(π‘₯) changes its value from π‘Ž to 𝑏. The effective conductivity 𝜎 is then found by calculating the integral:ξ€œπœŽ=π‘Œ||||πœ†(π‘₯)grad𝑒2𝑑π‘₯.(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 π‘Ž=1 and 𝑏=0 turns out to be 0.5773, which is close to the exact value:1𝜎=√3β‰ˆ0.5773502692.(4.4) Using the class of integer values for π‘Ž 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 π‘Ž 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 πœ†(π‘₯) is defined byξ‚»[]πœ†(π‘₯)=π‘Ž,ifπ‘₯∈0,0.52βˆͺ[]0.5,12,𝑏,otherwise.(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 π‘Ž=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 πœŽβˆ— 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 πœŽπ‘–π‘–.


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


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