`International Journal of Mathematics and Mathematical SciencesVolume 2012, Article ID 856927, 14 pageshttp://dx.doi.org/10.1155/2012/856927`
Research Article

## On Heat Conduction in Domains Containing Noncoaxial Cylinders

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

Received 28 November 2011; Accepted 16 January 2012

Copyright © 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.

#### Abstract

We consider heat conduction in domains containing noncoaxial cylinders. In particular, we present some regularity results for the solution and consider criteria which ensure the single valueness of the corresponding complex potential. Examples are discussed. In addition, we present some classes of cases where the parameters describing the solution are rational. Alternative ways of calculating the heat flux are also discussed.

#### 1. Introduction

Let be the unit-disk and let be some disk of the form . Moreover, let the conductivity be defined on the unit-disk as follows: where is some constant. We focus on the stationary heat conduction problems of the form: Here, is a region with smooth boundary , is a fixed smooth function defined on , is the temperature, and is the outward unit normal of the surface of . Since is piecewise harmonic (see Proposition 2.1), it is the real part (or imaginary part) of some piecewise analytic function , the associated complex potential Figure 1. Using a particular Möbius transformation of the type which maps onto and onto a disk with center at 0 and some radius (for a particular choice of the constant ,), we obtain that , where is the complex potential associated with the problem: Here, and , and denotes the function whenever . Of symmetry reasons, the auxiliary problem (1.4) is usually much easier to solve. Once this problem is solved, we obtain the solution of our original problem (1.2) by setting

Figure 1: The Möbius transformation.

In this paper, we discuss problems of this type when the boundary may be a little more general than that above. In Section 2, we recall the simplest possible example and present some regularity results for the solution. Moreover, we consider criteria which ensure the single valueness of the corresponding complex potential. An interesting example referred to as the Hashin-Shtrikman problem is discussed in Section 3. In Section 4, we describe methods, involving the so-called PQ-algorithm, for generating all integer values of and making , , and rational. Finally, in Section 5, we briefly discuss alternative ways of calculating the heat flux.

#### 2. Single-Valued Complex Potentials

The simplest example of a problem of the form (1.2) is when and for some constants and . In this case, it is easily seen that , where the complex potential associated with the auxiliary problem is given by In this case, is multiple connected and the harmonic conjugate of (given by ) is clearly multiple valued. The following results show in particular that this is not the case when is simply connected.

Proposition 2.1. Assume that the boundary is Lipschitz continuous and let be the solution of the weak formulation corresponding to (1.2). Then is harmonic in the interior of each region where is constant. Moreover, if is simply connected, then the harmonic conjugate of is single valued in these regions.

Proof. Let be a weak solution of (1.2), that is, belongs to the Sobolev space with trace satisfying for all . Assume that is constant in a disk with centre at some point . Then, (2.3) gives that for all , that is, is (by definition) a generalized solution of the Laplace equation in . Hence, by standard regularity results for elliptic partial differential equations, this gives that is a solution of the Laplace equation in in the classical sense (this is, e.g., a consequence of the regularity result stated in [1] Theorem 8.8; see also [2]). Hence, we conclude that is harmonic in the interior of each region where is constant.
By (2.3), where When is simply connected, it holds that (see e.g., [3, page 467]) where denotes the vector-function defined by . Hence, for some function . Thus, that is, putting , which clearly is single valued, we obtain that on every domain where is constant. This shows that is a harmonic conjugate of on these domains, since (2.9) is nothing but the Cauchy-Riemann equations.

#### 3. On the Hashin-Shtrikman Problem

In the case when , where and are constant real numbers, the problem (1.2) becomes identical with that used in the proof of the attainability of the famous Hashin and Shtrikman bounds [4] and is, therefore, called the Hashin-Shtrikman problem. These bounds are important in the homogenization theory for composite structures. For more general information, we refer to the literature, see, for example, [510]. We also like to mention that an elementary introduction to the homogenization method can be found in the book of Persson et al. [11].

It will be clear from the arguements below that the solution of the Hashin-Shtrikman problem is given by where (see, e.g., [12]). The constants and will determined below. It is interesting to note that is the real part of the complex potential: where and . This follows from the fact that Note also that , where is the well-known Joukowsky transformation: The fact that is analytic in the regions and shows that is harmonic in these regions. Hence, the first condition of (1.2) is satisfied. The unit normal vector on the boundaries and can be represented by the complex number (still denoted : Using the fact that (by Cauchy.Riemann equations), we find that that is, If , then Hence, multiplying with and using the formula yield which gives that If , then clearly Thus, in order to fulfill the continuity of , we obtain from (3.14) and (3.15) that Moreover, the continuity of the temperature on the circle gives that (c.f. (3.3)) Hence, In particular, this gives that at the boundary , where is the so-called Hashin-Shtrikman bound given by Moreover, noting that on on the unit circle , Greens formula, gives that Here, is the angle between and .

We can transform the solution on the unit-disk to a general disk with center at and radius where the conductivity is given by . It is easy to see that the solution of the problem, is given by where is given by (3.3). Moreover, similarly as above, we find that

Now, let and consider the interior of the square centered at 0 whose boundary consists of the four line segments , and . We can extend the conductivity function to by setting on . Thanks to the above results, it is now easy to see that the solution of the Dirichlet/Neumann problem, is the real part of the complex potential given by (see Figures 2 and 3)

Figure 2: The solution of (3.26) for the case when the conductivity of the disk is .
Figure 3: The solution of (3.26) for the cases when the conductivity of the disk is , and , respectively.

The inverse of the Möbius transformation, is of the same type, namely, For any fixed real value , we can transform the above problem on to a corresponding problem on the domain as follows (see Figure 4): Here, for an arbitrary set denotes the set

Figure 4: Dirichlet/Neumann problem on and the domain .

We obtain a more complex structure than that described on if we cover by an infinite sequence of nonintersecting discs and define the conductivity on each disk as we did in the definition of above. This function can also be extended by periodicity to the whole plane. The underlying structure is called the Hashin, coated cylinder assemblage. The Dirichlet/Neumann problem, is easily solved by using the above results. By (3.25) and the fact that , we obtain that for . Due to symmetry, we obtain the same result for if we replace with and with in (3.31).

#### 4. Rational Parameters

In order to find the value of and , we utilize the fact that maps the real line onto itself. In particular, this gives that and , where and . Hence, Provided , we find that since this solution satisfies the criteria while the the other solution, does not. This follows by the following inequalities: Hence, Moreover, if , then Similarly, we obtain that if .

The -value corresponding to is given by

Now, assume that and where and are integers. Then, It is clear that the pairs making and rational numbers are precisely those of the form: where , and are integers. It is possible to show that the integer solutions of the generalized Pell's equation, for a given integer which is not a square, are precisely those of the form where is odd, is even, and are the integers obtained from the following algorithm (called the PQ-algorithm, see [13, pages 346–348 and page 358] and [14, pages 125–128]).

Set For , set For , set Hence, by (4.10) all pairs making and rational numbers are precisely those of the form where and are odd, and and are even.

#### 5. Calculating the Heat Flux

If is a simple contour in , the corresponding heat flux is given by If is the real part of some complex potential , then when denotes the unit tangential vector. Hence,

As an example, consider the simplest case discussed at the beginning of Section 2 for which In this case, the derivative of the complex potential is given by By the residue theorem, we obtain that which gives that

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