Answer to the Question 07/01

CONDUCTING CHESSBOARD

The question was:





Small square metal sheets are made from two different metals (green and yellow in the picture) and combined into a single very large "chessboard". The two-dimensional conductivity of one metal is S1 and of the other - is S2. Calculate the effective two-dimensional conductivity of this "chessboard" by following steps:

(a) Show that the electrostatic potential inside every, say, green square is given by the same harmonic function f1(x,y) (up to a constant which needs to be added depending on the detailed position of the square). Similar function g1(x,y) describes yellow squares.

(b) Functions f1(x,y) and g1(x,y) can be treated as real parts of ANALYTIC functions F(x,y)=f1(x,y)+if2(x,y) and G(x,y)=g1(x,y)+ig2(x,y) of a complex variable z=x+iy. Show that f2(x,y) and g2(x,y) also solve (a different) chessboard conductivity problem.

(c) By comparing the solutions of the conductivity problems described in (b) find the effective conductivity of this "chessboard".


Answer (to part (c)): The effective conductivity of this "chessboard" is S=sqrt{S1*S2}.

Historical note: The problem of conductivity of two-dimensional composites was originally considered by A.M. Dykhne, JETP 7, 110 (1970). In particular, using his theory it can be shown that a symmetric mixture of two components has conductivity which is geometrical average of the conductivities of constituent materials. This applies also to the chessboard problem, i.e. the effective conductivity can be found without actually finding the potentials! Nevertheless, the potentials themselves are known: they have been found by V.L. Berdichevskii and published in Vestnik Moskovskogo Universiteta, 40(4), 56 (1985). For additional (more recent) references see the pdf file mentioned in the discussion of a modified problem by K. T. McDonald below.


(9/98) The problem has been solved (19/9/02) by Jared Kaplan (e-mail jaredk@stanford.edu), an undergraduate student at Stanford, CA (USA), and by Dan Gulotta (e-mail dgulotta@imsa.edu), a high school senior at Illinois Mathematics and Science Academy in Aurora, IL (USA).

The solution:

The detailed solution provided by Kaplan and Gulotta can be found in the following pdf file. Here, we only mention few essential points:

(a) In potentials of the problem in each type of the squares are assumed to be real parts of complex functions, then the imaginary parts represent field rotated by 90 degrees.

(b) By examining the boundary conditions between the grains one notices that the fields generated by the imaginary parts are currents in a conductivity problem in which conductivity Si of each square has been replaced by 1/Si.

(c) By writing down the expressions for the effective conductivities of the original and the "rotated" problems, we notice that they are simply related (one the the inverse of the other). By further using the isotropy of the problem, homogeneity of the solutions, and the symmetry between the two components, one arrives at the expression for conductivity.

This problem is highly non-trivial, and you are advised to read the detailed solution mentioned above. It should be noticed that the solution is valid not only for checkerboard, but also for random two-dimensional 50%-50% mixtures which are symmetric (i.e. both components have the same type of geometry). Over the years, the solutions were slightly extended. In particular a three component mixture can be solved, under some (very restrictive) conditions on geometry and conductivity.


(9/2002) Y. Kantor: If you are sure that you understood the solution of the continuous problem here is another challenge for you: show that the same result is valid for a two-dimensional square lattice whose bonds are randomly chosen resistors of two types. Each bond has a 50%-50% chance to have either of two selected values. The effective resistivity of such random lattice is given by the geometrical mean of two resistor values. While the solution of this discrete problem resembles, in its spirit, the continuous solution, you will discover many challenging obstacles that did not appear in continuum problem.

Below we present some additional approaches and related problems.



"Heuristic" approach: Now that we know what the answer is, we can try to see if we can get it easier, by using some general properties of conductivity and some guesses. The effective conductivity S is a function of constituent conductivities S1 and S2: S=H(S1,S2). Obviously (because of linearity of equations of conductivity), the increase of both constituent conductivities by factor k, will increase the effective conductivity by the same factor. Therefore H(ka,kb)=k*H(a,b). By choosing k=1/S2, we get (S/S2)=H((S1/S2),1). So now we see that (S/S2) is only a function of (S1/S2). What kind of function it can be? The function is equal to 1, when (S1/S2)=1. The function obviously vanishes if the argument vanishes (i.e. if one of the material is non-conducting), and the function is infinite when the argument is infinite (i.e. first material has infinite conductivity, while the second material has finite conductivity). [The latter property is not self-evident; one really needs to look deeper into a problem of a contact between two corners, to understand what is happening to currents.] All these properties lead us to a guess that the function is a simple power law. [We do not know simple rigorous argument that this should be so.] Consequently,

(S/S2)=(S1/S2)p,

or

S=S1p*S21-p.

However, the value of S cannot change if we exchange locations of green and yellow squares, i.e. exchange S1 and S2 in the above formula. Consequently, p=1/2, and the answer S=sqrt{S1*S2} follows.

We should note that this "heuristic" approach involved guesses regarding the shape of the function, and explicitly used the symmetry of the system (interchange two types of materials). In three dimensions, the simple power law assumption is incorrect, and, consequently, there the solution is not known.



Kirk T. McDonald from Princeton (e-mail mcdonald@puphed.princeton.edu suggested (22/10/2001) a related simpler problem, which allows an exact solution which includes finding the fields! He considered a circular (two-dimensional) region, divided into four parts by two straight lines passing through its center and forming pre-defined angle between them. The conductivities of 4 regions alternate between two possible conductivities. The conductance of such system can be found exactly. The modified problem and its solution are presented in the following pdf file.


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