Volume 2012, Article ID 856927,14pages doi:10.1155/2012/856927
Research Article
On Heat Conduction in Domains Containing Noncoaxial Cylinders
Dag Lukkassen
1, 2and Annette Meidell
1, 21Narvik University College, P.O. Box 385, 8505 Narvik, Norway
2Norut Narvik, P.O. Box 250, 8504 Narvik, Norway
Correspondence should be addressed to Dag Lukkassen,dl@hin.no Received 28 November 2011; Accepted 16 January 2012
Academic Editor: Adolfo Ballester-Bolinches
Copyrightq2012 D. Lukkassen and A. 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 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
LetD be the unit-diskD {x, y : x2y2 ≤ 1}and letB ⊂ Dbe some disk of the form B{x, y:x−x02y2≤R2}. Moreover, let the conductivityλλx, ybe defined on the unit-disk as follows:
λ x, y
k if x, y
∈B, 1 if
x, y
∈D\B, 1.1
wherek > 0 is some constant. We focus on the stationary heat conduction problems of the form:
div
λgradu
0 on Ω\∂B, ug on ∂Ω,
uand λ∂u
∂n are continuous through∂B.
1.2
1 0.8 0.6 0.4 0.2
−0.2
−0.4
−0.6
−0.8
−1
1 0.8 0.6 0.4
−0.2 0.2
−0.4
−0.6
−0.8
−1 1
0.8 0.6 0.4 0.2
−0.2
−0.4
−0.6
−0.8
−1
1 0.8 0.6 0.4
−0.2 0.2
−0.4
−0.6
−0.8
−1 x
y
x∗ y∗
f(z)
Figure 1:The M ¨obius transformation.
Here,Ω ⊂ D is a region with smooth boundary ∂Ω,g is a fixed smooth function defined on∂Ω,uis the temperature, andnis the outward unit normal of the surface∂BofB. Since ux, yis piecewise harmonicseeProposition 2.1, it is the real partor imaginary partof some piecewise analytic functionFz, zxiy, the associatedcomplex potentialFigure 1.
Using a particular M ¨obius transformationzxiy → z∗x∗iy∗of the type z∗fz z−b
bz−1, 1.3
which mapsDontoDandBonto a diskB∗ ⊂Dwith center at 0 and some radiusR∗for a particular choice of the constantb,−1< b <1, we obtain thatFz F∗z∗, whereF∗z∗is the complex potential associated with the problem:
div
λ∗gradu∗
0 on Ω∗\∂B∗, u∗g∗ on ∂Ω∗,
λ∗∂u∗
∂n is continuous through∂B∗.
1.4
Here,Ω∗ fΩandB∗ fB, andh∗ :D → Rdenotes the functionh∗x∗, y∗ hx, y wheneverh : D → R. Of symmetry reasons, the auxiliary problem1.4is usually much easier to solve. Once this problem is solved, we obtain the solution of our original problem 1.2by setting
u x, y
u∗ x∗, y∗
u∗ Ref
xiy ,Imf
xiy
. 1.5
In this paper, we discuss problems of this type when the boundary ∂Ω may be a little more general than that above. InSection 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 ofx0−R−1 andx0R−1makingx0,R, andbrational. Finally, inSection 5, we briefly discuss alternative ways of calculating the heat flux.
2. Single-Valued Complex Potentials
The simplest example of a problem of the form1.2is whenΩ D\Band
g
u0 on ∂D,
u1 on ∂B, 2.1
for some constantsu0andu1. In this case, it is easily seen thatux, y Refz Ref∗fz, where the complex potential associated with the auxiliary problemF∗is given by
F∗z∗ u1−u0
lnR∗ logz∗u0. 2.2
In this case,Ωis multiple connected and the harmonic conjugate ofugiven byvx, y Refzis clearly multiple valued. The following results show in particular that this is not the case whenΩis simply connected.
Proposition 2.1. Assume that the boundary∂Ωis Lipschitz continuous and letube the solution of the weak formulation corresponding to1.2. Thenuis harmonic in the interior of each region where λis constant. Moreover, ifΩis simply connected, then the harmonic conjugatevofuis single valued in these regions.
Proof. Letube a weak solution of1.2, that is,ubelongs to the Sobolev spaceW1Ωwith traceu|∂Ωgsatisfying
Ωλ gradu
·gradϕ dx dy0, 2.3
for allϕ ∈C∞0 Ω. Assume thatλis constant in a diskOwith centre at some pointx0, y0. Then,2.3gives that
O
∂u
∂x
∂ϕ
∂x ∂u
∂y
∂ϕ
∂y
dx dy0, 2.4
for allϕ∈W01,2O, that is,uisby definitiona 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 thatuis harmonic in the interior of each region whereλis constant.
By2.3,
pλ gradu
∈L2solΩ, 2.5
where
L2solΩ
q∈L2Ω:
Ωq·gradϕ dx dy0∀ϕ∈C0∞Ω . 2.6 WhenΩis simply connected, it holds thatsee e.g.,3, page 467
L2solΩ
curlϕ:ϕ∈W1,2Ω
, 2.7
where curlϕdenotes the vector-function defined by curlϕ −∂ϕ/∂y, ∂ϕ/∂x. Hence,p curlψfor some functionψ∈W1Ω. Thus,
λ−1 curlψ
λ−1pgradu, 2.8
that is, puttingv−λ−1ψ, which clearly is single valued, we obtain that
−curlvgradu, 2.9
on every domain whereλis constant. This shows thatvis a harmonic conjugate ofuon these domains, since2.9is nothing but the Cauchy-Riemann equations.
3. On the Hashin-Shtrikman Problem
In the case whenx00,
Ω D, g x, y
ξ1xξ2y
, 3.1
where ξ1 andξ2 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
u x, y
w x, y
ξ1xξ2y, 3.2
where
w x, y
⎧⎪
⎨
⎪⎩ l1
ξ1xξ2y
if x, y
∈B, l
ξ1xξ2y−ξ1xξ2y x2y2
if
x, y
∈D\B 3.3
see, e.g.,12 . The constantsl1andlwill determined below. It is interesting to note thatuis the real part of the complex potential:
Fz
⎧⎨
⎩
l11ξz ifz∈B, l|ξ|h
ξ|ξ|−1z
ξz ifz∈D\B, 3.4
where
hz z−1
z, 3.5
andξξ1iξ2. This follows from the fact that gz |ξ|h
ξ|ξ|−1z l
ξz− ξz
|z|2
. 3.6
Note also thathz −idiz, wheredis the well-known Joukowsky transformation:
dz z1
z. 3.7
The fact thatFzis analytic in the regionsBandD\Bshows thatuReFzis harmonic in these regions. Hence, the first condition of1.2divλgradu 0 is satisfied. The unit normal vectorn n1, n2on the boundaries∂Band∂Dcan be represented by the complex number still denotedn:
nn1in2 z
|z|. 3.8
Using the fact that
Fz ∂u
∂x−i∂u
∂y 3.9
by Cauchy.Riemann equations, we find that
∂u
∂n gradu·n ∂u
∂xn1∂u
∂yn2Re
n1in2 ∂u
∂x−i∂u
∂y
Re nFz
, 3.10
that is,
∂u
∂n Re nFz
. 3.11
Ifz∈D\B, then
Fz lξh ξ|ξ|−1z
ξlξ
⎛
⎜⎝1 1 ξ|ξ|−1z2
⎞
⎟⎠ξ
lξ
⎛
⎜⎝1 ξzξz ξξ|ξ|−1zz2
⎞
⎟⎠ξlξ
1 ξzξz
|ξ|2|z|4
ξ.
3.12
Hence, multiplying withnz/|z|and using the formula|q|2qqyield
nFz l
ξn z
|z|
ξξzξz
|ξ|2|z|4
nξl
nξ z
|z|
ξξzz ξξ|z|4ξ
nξl
nξ 1
|z|2nξ
nξ, 3.13
which gives that
∂u
∂n Re nFz
n1ξ1n2ξ2
l
1 1
|z|2
1
. 3.14
Ifz∈B, then clearly
∂u
∂n gradu·n l11n1ξ1n2ξ2. 3.15 Thus, in order to fulfill the continuity ofλ∂u/∂n, we obtain from3.14and3.15that
l
1 1 R2
1kl11. 3.16
Moreover, the continuity of the temperatureuon the circlex2y2Rgives thatc.f.3.3
l
1− 1 R2
1 l11. 3.17
Hence,
l R2k−1 2 1−R2k−1, l1 k−1
R2−1 2 1−R2k−1.
3.18
In particular, this gives that
λ∂u
∂n kn 1ξ1n2ξ2 kn·ξ 3.19
at the boundary∂D, wherekis the so-called Hashin-Shtrikman bound given by
k
R2k−1
2 1−R2k−11
. 3.20
Moreover, noting thatu ξ1xξ2y n1ξ1 n2ξ2 n·ξ on on the unit circle ∂D, Greens formula,
D
udiv
λgradu
0
dx dy−
D
λgradu2 dx dy
∂D
uλ∂u
∂nds, 3.21
gives that 1
|D|
D
λgradu2 dx dy 1
|D|
∂D
uλ∂u
∂nds 1
|D|
∂D
kn ·ξ2ds
1
|D|
2π
0
k|ξ| cosθ2dθk|ξ| 2.
3.22
Here,θis the angle betweennandξ.
We can transform the solution on the unit-diskDto a general diskDwith center atc0 and radiusεwhere the conductivityλis given byλx λx−c0/ε, x x, y. It is easy to see that the solutionuof the problem,
div
λgradu
0 on D, ux ξx on ∂D, uand λ∂ u
∂n are continuous through∂B and∂D,
3.23
is given by
ux εw x−c0
ε
ξx, 3.24
wherewis given by3.3. Moreover, similarly as above, we find that
1 D
D
λgradu2dx dyk|ξ| 2. 3.25
Now, let ξ 0,1and consider the interior of the square −1,1 2 centered at 0 whose boundary consists of the four line segmentsΓ1, Γ2, Γ3, andΓ4. We can extend the conductivity functionλtoby settingλx kon\D. Thanks to the above results, it is now easy to see that the solutionuof the Dirichlet/Neumann problem,
div
λgradu
0 on \∂B∪∂D,
∂u
∂n 0 on Γ1,Γ3, u−1 on Γ2, u1 on Γ4, u,λ∂u
∂n are continuous through∂Band∂D,
3.26
is the real part of the complex potentialF∗zgiven bysee Figures2and3
F∗z
⎧⎪
⎪⎨
⎪⎪
⎩
l11ξz ifz∈B, l|ξ|h
ξ|ξ|−1z
ξz ifz∈D\B,
ξz ifz∈\D.
3.27
The inverse of the M ¨obius transformation,
wfz z−b
bz−1, 3.28
is of the same type, namely,
zf−1w w−b
bw−1. 3.29
Figure 2:The solution of3.26for the case when the conductivity of the diskBisk5.
k=0 k=1 k=5
Figure 3:The solution of3.26for the cases when the conductivity of the diskBisk0, k1, andk5, respectively.
For any fixed real valueb, |b|<1, we can transform the above problem onto a correspond- ing problem on the domain f−1as followsseeFigure 4:
div
λgradu
0 on \
∂B ∪∂D ,
∂u
∂n 0 on Γ1,Γ3, u−1 on Γ2, u1 on Γ4, u,λ∂u
∂n are continuous through∂B and∂D.
3.30
Here, for an arbitrary setA⊂, A denotes the setA f−1A
We obtain a more complex structure than that described on if we cover by an infinite sequence of nonintersecting discs{Di}, Di⊂and define the conductivityλon each
1 0.8 0.6 0.4 0.2
−0.2
−0.4
−0.6
−0.8
−1
0.5 1
−0.5
−1
1.6 1.4 1.2
−1.6
−1.4
−1.2 1
0.8 0.6 0.4 0.2
−0.2
−0.4
−0.6
−0.8
− 1
0.5 1
−0.5
−1
1.6 1.4 1.2
−1.6
−1.4
−1.2
Γ3
Γ1
Γ4
Γ2
y y
x x
Γ′2
Γ′3 Γ′1
Γ′4
Figure 4:Dirichlet/Neumann problem onand the domain f−1.
disk as we did in the definition ofλabove. 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, div
λgradu
0 on Di,
∂u
∂n 0 on Γ1,Γ3, u−1 on Γ2, u1 on Γ4, u,λ∂u
∂n are continuous in then-direction,
3.31
is easily solved by using the above results. By3.25 and the fact that || !∞
i1|Di|, we obtain that
1
||
||λgradu2dxdy 1
||
"∞ i1
Di
λgradu2dxdy 1
||
"∞ i1
Dik|ξ|2k|ξ|2
3.32
forξ 0,1. Due to symmetry, we obtain the same result forξ 1,0if we replaceΓ1with Γ2andΓ3withΓ4in3.31.
4. Rational Parameters
In order to find the value ofR∗andb, we utilize the fact thatfzmaps the real line onto itself.
In particular, this gives that|fx1||fx2|R∗andfx1 −fx2, wherex1x0−Rand x2x0R. Hence,
x1−b
bx1−1 − x2−b
bx2−1. 4.1
Providedx0/0, we find that
b x1x21−#
1−x21 1−x22
x1x2 , 4.2
since this solution satisfies the criteria|b|<1 while the the other solution,
x1x21#
1−x21 1−x22
x1x2 , 4.3
does not. This follows by the following inequalities:
x1 x1x21− 1−x21 x1x2
< x1x21−#
1−x12 1−x22 x1x2
< x1x21− 1−x22 x1x2 x2.
4.4
Hence,
−1≤x1< b < x2 ≤1. 4.5
Moreover, ifx0>0, then
x1x21#
1−x21 1−x22
x1x2 > x1x21 1−x22 x1x2 x02−R22−x0R2
2x0 −2R2−2x0R2
2x0 1−R2−x0R x0 1−R1R−x0R
x0 ≥ x01R−x0R
x0 1.
4.6
Similarly, we obtain that
x1x21#
1−x21 1−x22 x1x2
<−1 4.7
ifx0<0.
TheR∗-value corresponding tobis given by
R∗
1−x1x2−#
1−x12 1−x22 x2−x1
. 4.8
Now, assume thatx11/mandx21/nwheremandnare integers. Then,
b 1mn−$
m2−1n2−1
mn ,
R 1
m−n
−#
m2−1n2−1 mn−1 .
4.9
It is clear that the pairsm, nmakingband Rrational numbers are precisely those of the form:
m2−1dr2, n2−1ds2, 4.10
whered,randsare integers. It is possible to show that the integer solutions of the generalized Pell’s equation,
x2−dy21, 4.11
for a given integerdwhich is not a square, are precisely those of the formxGkl−1, yBkl−1 wherelis odd,kis even, andGi, Biare the integers obtained from the following algorithm called the PQ-algorithm, see13, pages 346–348 and page 358 and14, pages 125–128 .
Set
B−21, B−10, G−20, G−11, P00, Q01.
4.12
Fori≥0, set
ai
Pi√ d Qi , BiaiBi−1Bi−2, GiaiGi−1Gi−2.
4.13
Fori≥1, set
Piai−1Qi−1−Pi−1, Qi
d−Pi2
Qi−1 . 4.14
Hence, by4.10all pairsm, nmakingbandr rational numbers are precisely those of the formm, n Gk1l1−1, Gk2l2−1wherel1andl2are odd, andk1andk2are even.
5. Calculating the Heat Flux
IfCis a simple contour inΩ, the corresponding heat flux is given by
C
λ∂u
∂nds
C
λgradu·nds. 5.1
Ifuis the real part of some complex potentialFz, then
C
λFzdz
C
λ
ux−iuy dz
C
λ
ux−iuy
dxidy
C
λuxdx
C
λuydy
i
C
λuxdy−
C
λuydx
C
λ ux, uy
·tdsi
C
λ ux, uy
·nds
C
λgradu·tdsi
C
λgradu·nds, 5.2 whentdenotes the unit tangential vector. Hence,
C
λ∂u
∂ndsIm
C
λFzdz. 5.3
As an example, consider the simplest case discussed at the beginning ofSection 2for which
F∗z∗ u1−u0
lnR∗ logz∗u0. 5.4
In this case, the derivative of the complex potentialFzis given by
Fz fzF∗ fz
b2−1 bz−12
u1−u0
lnR∗ bz−1
z−b
u1−u0
lnR∗
b2−1
bz−1z−b. 5.5 By the residue theorem, we obtain that
C
Fzdz2πiu1−u0
lnR∗ , 5.6
which gives that
C
λ∂u
∂nds2πu1−u0
lnR∗ . 5.7
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