Abstract
We consider generalized chessboard structures where the local conductivity takes two values and . All integer combinations of and 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 be a large positive integer and consider a periodic composite material with period equal to . The stationary heat conduction problem can then be formulated by the following minimum principle: where Here, is the temperature; the conductivity matrix is given by where is periodic relative to the unit cube of , is the source field, is a bounded-open subset of , and the minimization is taken over some Sobolev space depending on the boundary conditions. It is possible to prove that the βenergyβ converges to the so-called homogenized βenergyβ as , defined by where The matrix 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 , , , and , respectively (see Figure 1). They conjectured that the corresponding effective conductivity matrix of this composite structure is given by where Here, , , and denote the arithmetic mean, the geometric mean and the harmonic mean, respectively, given by the formulae: Mortola and SteffΓ©s conjecture was ultimately proved by Craster and Obnosov [3] and Milton [4] in 2001 (see also [5]).
In this paper, we consider the special cases when the local conductivity only takes two values (see Figure 2). One such case is when , 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 and , and the second is characterized by the property and . For these three cases, we find all integer combinations of and which make the components of integer valued. We also discuss the problem of estimating by using the finite-element method.
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2. The Connected Case
For the connected case, it is easily verified that , where is given by the formula: Assume that and are positive integers. In order to obtain an integer value of , there must exist integers , , , and such that where We note that (2.2) is equivalent with Hence, It is clear that (otherwise, and for some integers and , which would mean that , and since , we obtain that , which contradicts (2.3)). According to (2.5), this gives that for some integer . By checking the four combinations when and are odd/even, we can easily verify that Hence, using (2.2), we obtain that where and is some positive integer. The corresponding integer value of is then given by In order to have positive values of and , (2.8) shows that we only can choose values of and such that
Summing up, all integers and making integer valued are of the form (2.8) for some positive integers , and , satisfying (2.11), where is given by (2.9). The corresponding value of is Conversely, all and of the form (2.8), satisfying (2.11), are positive integers and make integer valued.
For , the smallest integer value of is obtained when , (making ) and , corresponding to the values , , and .
3. The Chessboard-and Laminate Case
For the chessboard case, , where This case is very simple. We just use the representation: where , , and are integers (giving .
For the laminate case, we find that and . It is possible to show that the integers and making the harmonic mean integer valued are precisely those on the form: (in this case, ) and the form: (in this case, ) where , , and are positive integers. For a proof of this fact, see [6]. Therefore, if and are integer valued, must belong to the class (3.3) or (3.4). The latter is directly seen to generate integer values also for . However, (3.3) gives integer values of only if both and are odd (for which (3.3) may be written on the form (3.4) with replaced by ) or both even, or is even. In any case, if both and are integers, we end up with the form: (in this case and ) and the form: (in this case and ).
4. Calculating 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 can be found by solving the following boundary value problem: Here, is the unit cell and the conductivity is defined by In addition, we must assume continuity of normal component of through the four surfaces where changes its value from to . The effective conductivity is then found by calculating the integral: 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 , 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 for the case and turns out to be 0.5773, which is close to the exact value: Using the class of integer values for and 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 and 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 is defined by 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 and . 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 . 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 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 .
Acknowledgment
The authors are grateful to two anonymous reviewers for some helpful comments which have improved the presentation of this paper.