# On Heat Conduction in Domains Containing Noncoaxial Cylinders

Authors: Dag Lukkassen; Annette Meidell
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## 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.


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## Body

## 1. Introduction

LetD be the unit-disk D={(x,y):x2+y2≤1} and let B⊂D be some disk of the form B={(x,y):(x-x0)2+y2≤R2}. Moreover, let the conductivity λ=λ(x,y) be defined on the unit-disk as follows:(1.1)λ(x,y)={kif(x,y)∈B,1if(x,y)∈D∖B,
where k>0 is some constant. We focus on the stationary heat conduction problems of the form:(1.2)div⁡(λgradu)=0on⁡Ω∖∂B,u=gon⁡∂Ω,uandλ∂u∂narecontinuousthrough∂B.
Here, Ω⊂D is a region with smooth boundary ∂Ω, g is a fixed smooth function defined on ∂Ω, u is the temperature, and n is the outward unit normal of the surface ∂B of B. Since u(x,y) is piecewise harmonic (see Proposition 2.1), it is the real part (or imaginary part) of some piecewise analytic function F(z),z=x+iy, the associated complex potential Figure 1. Using a particular Möbius transformation  z=x+iy→z*=x*+iy* of the type(1.3)z*=f(z)=z-bbz-1,
which maps D onto D and B onto a disk B*⊂D with center at 0 and some radius R* (for a particular choice of the constant b,-1<b<1), we obtain that F(z)=F*(z*), where F*(z*) is the complex potential associated with the problem:(1.4)div⁡(λ*gradu*)=0on⁡Ω*∖∂B*,u*=g*on⁡∂Ω*,λ*∂u*∂niscontinuousthrough∂B*.
Here, Ω*=f(Ω)and B*=f(B), and h*:D→ℝ denotes the function h*(x*,y*)=h(x,y) whenever h:D→ℝ. 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(1.5)u(x,y)=u*(x*,y*)=u*(Ref(x+iy),Im⁡f(x+iy)).Figure 1
The Möbius transformation.In this paper, we discuss problems of this type when the boundary∂Ω 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 (x0-R)-1 and (x0+R)-1 making x0, R, and b 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 Ω=D∖B and(2.1)g={u0on⁡∂D,u1on⁡∂B,
for some constants u0 and u1. In this case, it is easily seen that u(x,y)=Ref(z)=Ref*(f(z)), where the complex potential associated with the auxiliary problem F* is given by(2.2)F*(z*)=u1-u0ln⁡R*log⁡z*+u0.
In this case, Ω is multiple connected and the harmonic conjugate of u (given by v(x,y)=Ref(z)) is clearly multiple valued. The following results show in particular that this is not the case when Ω is simply connected.Proposition 2.1.
Assume that the boundary∂Ω is Lipschitz continuous and let u be the solution of the weak formulation corresponding to (1.2). Then u is harmonic in the interior of each region where λ is constant. Moreover, if Ω is simply connected, then the harmonic conjugate v of u is single valued in these regions.Proof.
Letu be a weak solution of (1.2), that is, u belongs to the Sobolev space W1(Ω) with trace u|∂Ω=g satisfying
(2.3)∫Ωλ(gradu)⋅gradφdxdy=0,
for all φ∈C0∞(Ω). Assume that λ is constant in a disk O with centre at some point (x0,y0). Then, (2.3) gives that
(2.4)∫O(∂u∂x∂φ∂x+∂u∂y∂φ∂y)dxdy=0,
for all φ∈W01,2(O), that is, u is (by definition) a generalized solution of the Laplace equation in O. Hence, by standard regularity results for elliptic partial differential equations, this gives that u is a solution of the Laplace equation in O 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 u is harmonic in the interior of each region where λ is constant.
By (2.3),
(2.5)p=λ(gradu)∈Lsol2(Ω),
where
(2.6)Lsol2(Ω)={q∈L2(Ω):∫Ωq⋅gradφdxdy=0∀φ∈C0∞(Ω)}.
When Ω is simply connected, it holds that (see e.g., [3, page 467])
(2.7)Lsol2(Ω)={curl⁡φ:φ∈W1,2(Ω)},
where curlφ denotes the vector-function defined by curlφ=(-∂φ/∂y,∂φ/∂x). Hence, p=curlψ for some function ψ∈W1(Ω). Thus,
(2.8)λ-1(curl⁡ψ)=λ-1p=gradu,
that is, putting v=-λ-1ψ, which clearly is single valued, we obtain that
(2.9)-curl⁡v=gradu,
on every domain where λ is constant. This shows that v is a harmonic conjugate of u on these domains, since (2.9) is nothing but the Cauchy-Riemann equations.


## 3. On the Hashin-Shtrikman Problem

In the case whenx0=0,(3.1)Ω=D,g(x,y)=(ξ1x+ξ2y),
where ξ1 and ξ2 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, [5–10]. 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(3.2)u(x,y)=w(x,y)+ξ1x+ξ2y,
where(3.3)w(x,y)={l1(ξ1x+ξ2y)if(x,y)∈B,l(ξ1x+ξ2y-ξ1x+ξ2yx2+y2)if(x,y)∈D∖B
(see, e.g., [12]). The constants l1 and l will determined below. It is interesting to note that u is the real part of the complex potential:(3.4)F(z)={(l1+1)ξ¯zifz∈B,l|ξ|h(ξ¯|ξ|-1z)+ξ¯zifz∈D∖B,
where(3.5)h(z)=z-1z,
and ξ=ξ1+iξ2. This follows from the fact that(3.6)g(z)=|ξ|h(ξ¯|ξ|-1z)=l(ξ¯z-ξz¯|z|2).
Note also that h(z)=-id(iz), where d is the well-known Joukowsky transformation:(3.7)d(z)=z+1z.
The fact that F(z) is analytic in the regions B and D∖B shows that u=ReF(z) is harmonic in these regions. Hence, the first condition of (1.2) div(λgradu)=0 is satisfied. The unit normal vector n=(n1,n2) on the boundaries ∂B and ∂D can be represented by the complex number (still denoted n):(3.8)n=n1+in2=z|z|.
Using the fact that(3.9)F′(z)=∂u∂x-i∂u∂y
(by Cauchy.Riemann equations), we find that(3.10)∂u∂n=gradu⋅n=∂u∂xn1+∂u∂yn2=Re((n1+in2)(∂u∂x-i∂u∂y))=Re(nF′(z)),
that is,(3.11)∂u∂n=Re(nF′(z)).
If z∈D∖B, then(3.12)F′(z)=lξ¯h′(ξ¯|ξ|-1z)+ξ¯=lξ¯(1+1(ξ¯|ξ|-1z)2)+ξ¯=lξ¯(1+ξz¯ξz¯(ξξ¯|ξ|-1zz¯)2)+ξ¯=lξ¯(1+ξz¯ξz¯|ξ|2|z|4)+ξ¯.
Hence, multiplying with n=z/|z| and using the formula |q|2=qq¯ yield(3.13)nF′(z)=l(ξ¯n+z|z|ξ¯ξz¯ξz¯|ξ|2|z|4)+nξ¯=l(nξ¯+z¯|z|ξ¯ξzz¯ξ¯ξ|z|4ξ)+nξ¯=l(nξ¯+1|z|2n¯ξ)+nξ¯,
which gives that(3.14)∂u∂n=Re(nF′(z))=(n1ξ1+n2ξ2)(l(1+1|z|2)+1).
If z∈B, then clearly(3.15)∂u∂n=gradu⋅n=(l1+1)(n1ξ1+n2ξ2).
Thus, in order to fulfill the continuity of λ∂u/∂n, we obtain from (3.14) and (3.15) that(3.16)l(1+1R2)+1=k(l1+1).
Moreover, the continuity of the temperature u on the circle x2+y2=R gives that (c.f. (3.3))(3.17)l(1-1R2)+1=(l1+1).
Hence,(3.18)l=R2(k-1)2+(1-R2)(k-1),l1=(k-1)(R2-1)2+(1-R2)(k-1).
In particular, this gives that(3.19)λ∂u∂n=k̂(n1ξ1+n2ξ2)=k̂n⋅ξ
at the boundary ∂D, where k̂ is the so-called Hashin-Shtrikman bound given by(3.20)k̂=(R2(k-1)2+(1-R2)(k-1)+1).
Moreover, noting that u=ξ1x+ξ2y=n1ξ1+n2ξ2=n·ξ on on the unit circle ∂D, Greens formula,(3.21)∫Dudiv⁡(λgradu)︸=0dxdy=-∫Dλ|gradu|2dxdy+∫∂Duλ∂u∂nds,
gives that(3.22)1|D|∫Dλ|gradu|2dxdy=1|D|∫∂Duλ∂u∂nds=1|D|∫∂Dk̂(n⋅ξ)2ds=1|D|∫02πk̂(|ξ|cos⁡θ)2dθ=k̂|ξ|2.
Here, θ is the angle between nand ξ.We can transform the solution on the unit-diskD to a general disk D̃ with center at c0 and radius ɛ where the conductivity λ̃ is given by λ̃(x)=λ((x-c0)/ɛ),x=(x,y). It is easy to see that the solution ũ of the problem,(3.23)div⁡(λ̃gradũ)=0on⁡D̃,ũ(x)=ξxon⁡∂D̃,ũandλ∂ũ∂narecontinuousthrough∂Band∂D,
is given by(3.24)ũ(x)=ɛw(x-c0ɛ)+ξx,
where w is given by (3.3). Moreover, similarly as above, we find that(3.25)1|D̃|∫D̃λ|gradũ|2dxdy=k̂|ξ|2.Now, letξ=(0,1) and consider the interior of the square □=[-1,1]2 centered at 0 whose boundary consists of the four line segments Γ1,Γ2,Γ3, and Γ4. We can extend the conductivity function λ to □ by setting λ(x)=k̂ on □∖D. Thanks to the above results, it is now easy to see that the solution u of the Dirichlet/Neumann problem,(3.26)div⁡(λgradu)=0on⁡□∖(∂B∪∂D),∂u∂n=0on⁡Γ1,Γ3,u=-1on⁡Γ2,u=1on⁡Γ4,u,λ∂u∂narecontinuousthrough∂Band∂D,
is the real part of the complex potential F*(z) given by (see Figures 2 and 3)(3.27)F*(z)={(l1+1)ξ¯zifz∈B,l|ξ|h(ξ¯|ξ|-1z)+ξ¯zifz∈D∖B,ξ¯zifz∈□∖D.Figure 2
The solution of (3.26) for the case when the conductivity of the disk B is k=5.Figure 3
The solution of (3.26) for the cases when the conductivity of the disk B is k=0,k=1, and k=5, respectively.The inverse of the Möbius transformation,(3.28)w=f(z)=z-bbz-1,
is of the same type, namely,(3.29)z=f-1(w)=w-bbw-1.
For any fixed real value b,|b|<1, we can transform the above problem on □ to a corresponding problem on the domain □′=f-1(□) as follows (see Figure 4):(3.30)div⁡(λgradu)=0on⁡□′∖(∂B′∪∂D),∂u∂n=0on⁡Γ1′,Γ3′,u=-1on⁡Γ2′,u=1on⁡Γ4′,u,λ∂u∂narecontinuousthrough∂B′and∂D.
Here, for an arbitrary set A⊂□,A′ denotes the set A′=f-1(A)Figure 4
Dirichlet/Neumann problem on□ and the domain □′=f-1(□).We obtain a more complex structure than that described on□ if we cover □ by an infinite sequence of nonintersecting discs {D̃i},D̃i⊂□ and define the conductivity λ on each disk as we did in the definition of λ̃ 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,(3.31)div⁡(λgradu)=0on⁡D̃i,∂u∂n=0on⁡Γ1,Γ3,u=-1on⁡Γ2,u=1on⁡Γ4,u,λ∂u∂narecontinuousinthen-direction,
is easily solved by using the above results. By (3.25) and the fact that |□|=∑i=1∞|D̃i|, we obtain that(3.32)1|□|∫|□|λ|gradu|2dxdy=1|□|∑i=1∞∫Diλ|gradu|2dxdy=1|□|∑i=1∞|D̃i|k̂|ξ|2=k̂|ξ|2
for ξ=(0,1). Due to symmetry, we obtain the same result for ξ=(1,0) if we replace Γ1 with Γ2 and Γ3 with Γ4 in (3.31).


## 4.  Rational Parameters

In order to find the value ofR* and b, we utilize the fact that f(z) maps the real line onto itself. In particular, this gives that |f(x1)|=|f(x2)|=R* and f(x1)=-f(x2), where x1=x0-R and x2=x0+R. Hence,(4.1)x1-bbx1-1=-x2-bbx2-1.
Provided x0≠0, we find that(4.2)b=x1x2+1-(1-x12)(1-x22)x1+x2,
since this solution satisfies the criteria |b|<1 while the the other solution,(4.3)x1x2+1+(1-x12)(1-x22)x1+x2,
does not. This follows by the following inequalities:(4.4)x1=x1x2+1-(1-x12)x1+x2<x1x2+1-(1-x12)(1-x22)x1+x2<x1x2+1-(1-x22)x1+x2=x2.
Hence,(4.5)-1≤x1<b<x2≤1.
Moreover, if x0>0, then(4.6)x1x2+1+(1-x12)(1-x22)x1+x2>x1x2+1+(1-x22)x1+x2=x02-R2+2-(x0+R)22x0=-2R2-2x0R+22x0=1-R2-x0Rx0=(1-R)(1+R)-x0Rx0≥x0(1+R)-x0Rx0=1.
Similarly, we obtain that(4.7)x1x2+1+(1-x12)(1-x22)x1+x2<-1
if x0<0.TheR*-value corresponding to b is given by(4.8)R*=|1-x1x2-(1-x12)(1-x22)x2-x1|.Now, assume thatx1=1/m and x2=1/n where m and n are integers. Then,(4.9)b=1+mn-(m2-1)(n2-1)m+n,R=|1m-n(-(m2-1)(n2-1)+mn-1)|.
It is clear that the pairs (m,n) making b and R rational numbers are precisely those of the form:(4.10)m2-1=dr2,n2-1=ds2,
where d, r and s are integers. It is possible to show that the integer solutions of the generalized Pell's equation,(4.11)x2-dy2=1,
for a given integer d which is not a square, are precisely those of the form x=Gkl-1,y=Bkl-1 where l is odd, k is even, and Gi,Bi 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(4.12)B-2=1,B-1=0,G-2=0,G-1=1,P0=0,Q0=1.
For i≥0, set(4.13)ai=(Pi+d)Qi,Bi=aiBi-1+Bi-2,Gi=aiGi-1+Gi-2.
For i≥1, set(4.14)Pi=ai-1Qi-1-Pi-1,Qi=(d-Pi2)Qi-1.
Hence, by (4.10) all pairs (m,n) making b and r rational numbers are precisely those of the form (m,n)=(Gk1l1-1,Gk2l2-1) where l1 and l2 are odd, and k1 and k2 are even.


## 5.  Calculating the Heat Flux

IfC is a simple contour in Ω¯, the corresponding heat flux is given by(5.1)∫Cλ∂u∂nds=∫Cλgradu⋅nds.
If u is the real part of some complex potential F(z), then(5.2)∫CλF′(z)dz=∫Cλ(ux-iuy)dz=∫Cλ(ux-iuy)(dx+idy)=(∫Cλuxdx+∫Cλuydy)+i(∫Cλuxdy-∫Cλuydx)=∫Cλ(ux,uy)⋅tds+i∫Cλ(ux,uy)⋅nds=∫Cλgradu⋅tds+i∫Cλgradu⋅nds,
when t denotes the unit tangential vector. Hence,(5.3)∫Cλ∂u∂nds=Im⁡∫CλF′(z)dz.As an example, consider the simplest case discussed at the beginning of Section2 for which(5.4)F*(z*)=u1-u0ln⁡R*log⁡z*+u0.
In this case, the derivative of the complex potential F(z) is given by(5.5)F′(z)=f′(z)F*′(f(z))=(b2-1)(bz-1)2(u1-u0ln⁡R*bz-1z-b)=u1-u0ln⁡R*b2-1(bz-1)(z-b).
By the residue theorem, we obtain that(5.6)∫CF′(z)dz=2πiu1-u0ln⁡R*,
which gives that(5.7)∫Cλ∂u∂nds=2πu1-u0ln⁡R*.



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