On Heat Conduction in Domains Containing Noncoaxial Cylinders

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 𝐷 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:𝜆(𝑥,𝑦)={𝑘if(𝑥,𝑦)∈𝐵,1if(𝑥,𝑦)∈𝐷⧵𝐵,(1.1) where 𝑘>0 is some constant. We focus on the stationary heat conduction problems of the form:div(𝜆grad𝑢)=0onΩ⧵𝜕𝐵,𝑢=𝑔on𝜕Ω,𝑢and𝜆𝜕𝑢𝜕𝑛arecontinuousthrough𝜕𝐵.(1.2) 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 type𝑧∗=𝑓(𝑧)=𝑧−𝑏𝑏𝑧−1,(1.3) 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:div(𝜆∗grad𝑢∗)=0onΩ∗⧵𝜕𝐵∗,𝑢∗=𝑔∗on𝜕Ω∗,𝜆∗𝜕𝑢∗𝜕𝑛iscontinuousthrough𝜕𝐵∗.(1.4) 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 setting𝑢(𝑥,𝑦)=𝑢∗(𝑥∗,𝑦∗)=𝑢∗(Re𝑓(𝑥+𝑖𝑦),Im𝑓(𝑥+𝑖𝑦)).(1.5)

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 and𝑔={𝑢0on𝜕𝐷,𝑢1on𝜕𝐵,(2.1) 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 by𝐹∗(𝑧∗)=𝑢1−𝑢0ln𝑅∗log𝑧∗+𝑢0.(2.2) 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. Let 𝑢 be a weak solution of (1.2), that is, 𝑢 belongs to the Sobolev space 𝑊1(Ω) with trace 𝑢|𝜕Ω=𝑔 satisfying ∫Ω𝜆(grad𝑢)⋅grad𝜑𝑑𝑥𝑑𝑦=0,(2.3) for all 𝜑∈𝐶∞0(Ω). Assume that 𝜆 is constant in a disk 𝑂 with centre at some point (𝑥0,𝑦0). Then, (2.3) gives that ∫𝑂(𝜕𝑢𝜕𝑥𝜕𝜑𝜕𝑥+𝜕𝑢𝜕𝑦𝜕𝜑𝜕𝑦)𝑑𝑥𝑑𝑦=0,(2.4) for all 𝜑∈𝑊1,20(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), 𝑝=𝜆(grad𝑢)∈𝐿2sol(Ω),(2.5) where 𝐿2sol(Ω)={𝑞∈𝐋2(Ω)∶∫Ω𝑞⋅grad𝜑𝑑𝑥𝑑𝑦=0∀𝜑∈𝐶∞0(Ω)}.(2.6) When Ω is simply connected, it holds that (see e.g., [3, page 467]) 𝐿2sol(Ω)={curl𝜑∶𝜑∈𝑊1,2(Ω)},(2.7) where curl𝜑 denotes the vector-function defined by curl𝜑=(−𝜕𝜑/𝜕𝑦,𝜕𝜑/𝜕𝑥). Hence, 𝑝=curl𝜓 for some function 𝜓∈𝑊1(Ω). Thus, 𝜆−1(curl𝜓)=𝜆−1𝑝=grad𝑢,(2.8) that is, putting 𝑣=−𝜆−1𝜓, which clearly is single valued, we obtain that −curl𝑣=grad𝑢,(2.9) 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 when 𝑥0=0,Ω=𝐷,𝑔(𝑥,𝑦)=(𝜉1𝑥+𝜉2𝑦),(3.1) 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 by𝑢(𝑥,𝑦)=𝑤(𝑥,𝑦)+𝜉1𝑥+𝜉2𝑦,(3.2) where𝑤(𝑥,𝑦)={{{𝑙1(𝜉1𝑥+𝜉2𝑦)if(𝑥,𝑦)∈𝐵,𝑙(𝜉1𝑥+𝜉2𝑦−𝜉1𝑥+𝜉2𝑦𝑥2+𝑦2)if(𝑥,𝑦)∈𝐷⧵𝐵(3.3) (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:𝐹(𝑧)={(𝑙1+1)𝜉𝑧if𝑧∈𝐵,𝑙|𝜉|ℎ(𝜉|𝜉|−1𝑧)+𝜉𝑧if𝑧∈𝐷⧵𝐵,(3.4) whereℎ(𝑧)=𝑧−1𝑧,(3.5) and 𝜉=𝜉1+𝑖𝜉2. This follows from the fact that𝑔(𝑧)=|𝜉|ℎ(𝜉|𝜉|−1𝑧)=𝑙(𝜉𝑧−𝜉𝑧|𝑧|2).(3.6) Note also that ℎ(𝑧)=−𝑖𝑑(𝑖𝑧), where 𝑑 is the well-known Joukowsky transformation:𝑑(𝑧)=𝑧+1𝑧.(3.7) 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):𝑛=𝑛1+𝑖𝑛2=𝑧|𝑧|.(3.8) Using the fact that𝐹′(𝑧)=𝜕𝑢𝜕𝑥−𝑖𝜕𝑢𝜕𝑦(3.9) (by Cauchy.Riemann equations), we find that𝜕𝑢𝜕𝑛=grad𝑢⋅𝑛=𝜕𝑢𝜕𝑥𝑛1+𝜕𝑢𝜕𝑦𝑛2=Re((𝑛1+𝑖𝑛2)(𝜕𝑢𝜕𝑥−𝑖𝜕𝑢𝜕𝑦))=Re(𝑛𝐹′(𝑧)),(3.10) that is,𝜕𝑢𝜕𝑛=Re(𝑛𝐹′(𝑧)).(3.11) If 𝑧∈𝐷⧵𝐵, then𝐹′(𝑧)=𝑙𝜉ℎ′(𝜉|𝜉|−1𝑧)+𝜉=𝑙𝜉(1+1(𝜉|𝜉|−1𝑧)2)+𝜉=𝑙𝜉(1+𝜉𝑧𝜉𝑧(𝜉𝜉|𝜉|−1𝑧𝑧)2)+𝜉=𝑙𝜉(1+𝜉𝑧𝜉𝑧|𝜉|2|𝑧|4)+𝜉.(3.12) Hence, multiplying with 𝑛=𝑧/|𝑧| and using the formula |𝑞|2=𝑞𝑞 yield𝑛𝐹′(𝑧)=𝑙(𝜉𝑛+𝑧|𝑧|𝜉𝜉𝑧𝜉𝑧|𝜉|2|𝑧|4)+𝑛𝜉=𝑙(𝑛𝜉+𝑧|𝑧|𝜉𝜉𝑧𝑧𝜉𝜉|𝑧|4𝜉)+𝑛𝜉=𝑙(𝑛𝜉+1|𝑧|2𝑛𝜉)+𝑛𝜉,(3.13) which gives that𝜕𝑢𝜕𝑛=Re(𝑛𝐹′(𝑧))=(𝑛1𝜉1+𝑛2𝜉2)(𝑙(1+1|𝑧|2)+1).(3.14) If 𝑧∈𝐵, then clearly𝜕𝑢𝜕𝑛=grad𝑢⋅𝑛=(𝑙1+1)(𝑛1𝜉1+𝑛2𝜉2).(3.15) Thus, in order to fulfill the continuity of 𝜆𝜕𝑢/𝜕𝑛, we obtain from (3.14) and (3.15) that𝑙(1+1𝑅2)+1=𝑘(𝑙1+1).(3.16) Moreover, the continuity of the temperature 𝑢 on the circle 𝑥2+𝑦2=𝑅 gives that (c.f. (3.3))𝑙(1−1𝑅2)+1=(𝑙1+1).(3.17) Hence,𝑙=𝑅2(𝑘−1)2+(1−𝑅2)(𝑘−1),𝑙1=(𝑘−1)(𝑅2−1)2+(1−𝑅2)(𝑘−1).(3.18) In particular, this gives that𝜆𝜕𝑢𝜕𝑛=̂‌𝑘(𝑛1𝜉1+𝑛2𝜉2)=̂‌𝑘𝑛⋅𝜉(3.19) at the boundary 𝜕𝐷, where ̂‌𝑘 is the so-called Hashin-Shtrikman bound given bŷ‌𝑘=(𝑅2(𝑘−1)2+(1−𝑅2)(𝑘−1)+1).(3.20) Moreover, noting that 𝑢=𝜉1𝑥+𝜉2𝑦=𝑛1𝜉1+𝑛2𝜉2=𝑛⋅𝜉 on on the unit circle 𝜕𝐷, Greens formula,∫𝐷𝑢div(𝜆grad𝑢)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟=0𝑑𝑥𝑑𝑦=−∫𝐷𝜆|grad𝑢|2𝑑𝑥𝑑𝑦+∫𝜕𝐷𝑢𝜆𝜕𝑢𝜕𝑛𝑑𝑠,(3.21) gives that1|𝐷|∫𝐷𝜆|grad𝑢|2𝑑𝑥𝑑𝑦=1|𝐷|∫𝜕𝐷𝑢𝜆𝜕𝑢𝜕𝑛𝑑𝑠=1|𝐷|∫𝜕𝐷̂‌𝑘(𝑛⋅𝜉)2𝑑𝑠=1|𝐷|∫2𝜋0̂‌𝑘(|𝜉|cos𝜃)2𝑑𝜃=̂‌𝑘|𝜉|2.(3.22) Here, 𝜃 is the angle between 𝑛and 𝜉.

We can transform the solution on the unit-disk 𝐷 to a general disk ̃‌𝐷 with center at 𝐜0 and radius 𝜀 where the conductivity ̃‌𝜆 is given by ̃‌𝜆(𝐱)=𝜆((𝐱−𝐜0)/𝜀),𝐱=(𝑥,𝑦). It is easy to see that the solution ̃𝑢 of the problem,div(̃‌𝜆grad̃𝑢)=0oñ‌𝐷,̃𝑢(𝐱)=𝜉𝐱on𝜕̃‌𝐷,̃𝑢and𝜆𝜕̃𝑢𝜕𝑛arecontinuousthrough𝜕𝐵and𝜕𝐷,(3.23) is given bỹ𝑢(𝐱)=𝜀𝑤(𝐱−𝐜0𝜀)+𝜉𝐱,(3.24) where 𝑤 is given by (3.3). Moreover, similarly as above, we find that1|̃‌𝐷|∫̃‌𝐷𝜆|grad̃𝑢|2𝑑𝑥𝑑𝑦=̂‌𝑘|𝜉|2.(3.25)

Now, let 𝜉=(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)=̂‌𝑘 on □⧵𝐷. Thanks to the above results, it is now easy to see that the solution 𝑢 of the Dirichlet/Neumann problem,div(𝜆grad𝑢)=0on□⧵(𝜕𝐵∪𝜕𝐷),𝜕𝑢𝜕𝑛=0onΓ1,Γ3,𝑢=−1onΓ2,𝑢=1onΓ4,𝑢,𝜆𝜕𝑢𝜕𝑛arecontinuousthrough𝜕𝐵and𝜕𝐷,(3.26) is the real part of the complex potential 𝐹∗(𝑧) given by (see Figures 2 and 3)𝐹∗(𝑧)={{{{{(𝑙1+1)𝜉𝑧if𝑧∈𝐵,𝑙|𝜉|ℎ(𝜉|𝜉|−1𝑧)+𝜉𝑧if𝑧∈𝐷⧵𝐵,𝜉𝑧if𝑧∈□⧵𝐷.(3.27)

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,𝑤=𝑓(𝑧)=𝑧−𝑏𝑏𝑧−1,(3.28) is of the same type, namely,𝑧=𝑓−1(𝑤)=𝑤−𝑏𝑏𝑤−1.(3.29) 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):div(𝜆grad𝑢)=0on□′⧵(𝜕𝐵′∪𝜕𝐷),𝜕𝑢𝜕𝑛=0onΓ′1,Γ′3,𝑢=−1onΓ′2,𝑢=1onΓ′4,𝑢,𝜆𝜕𝑢𝜕𝑛arecontinuousthrough𝜕𝐵′and𝜕𝐷.(3.30) 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 {̃‌𝐷𝑖},̃‌𝐷𝑖⊂□ and define the conductivity 𝜆 on each disk as we did in the definition of ̃‌𝜆 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,div(𝜆grad𝑢)=0oñ‌𝐷𝑖,𝜕𝑢𝜕𝑛=0onΓ1,Γ3,𝑢=−1onΓ2,𝑢=1onΓ4,𝑢,𝜆𝜕𝑢𝜕𝑛arecontinuousinthe𝑛-direction,(3.31) is easily solved by using the above results. By (3.25) and the fact that |□|=∑∞𝑖=1|̃‌𝐷𝑖|, we obtain that1|□|∫|□|𝜆|grad𝑢|2𝑑𝑥𝑑𝑦=1|□|∞∑𝑖=1∫𝐷𝑖𝜆|grad𝑢|2𝑑𝑥𝑑𝑦=1|□|∞∑𝑖=1|̃‌𝐷𝑖|̂‌𝑘|𝜉|2=̂‌𝑘|𝜉|2(3.32) 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 of 𝑅∗ 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,𝑥1−𝑏𝑏𝑥1−1=−𝑥2−𝑏𝑏𝑥2−1.(4.1) Provided 𝑥0≠0, we find that𝑏=𝑥1𝑥2+1−√(1−𝑥21)(1−𝑥22)𝑥1+𝑥2,(4.2) since this solution satisfies the criteria |𝑏|<1 while the the other solution,𝑥1𝑥2+1+√(1−𝑥21)(1−𝑥22)𝑥1+𝑥2,(4.3) does not. This follows by the following inequalities:𝑥1=𝑥1𝑥2+1−(1−𝑥21)𝑥1+𝑥2<𝑥1𝑥2+1−√(1−𝑥21)(1−𝑥22)𝑥1+𝑥2<𝑥1𝑥2+1−(1−𝑥22)𝑥1+𝑥2=𝑥2.(4.4) Hence,−1≤𝑥1<𝑏<𝑥2≤1.(4.5) Moreover, if 𝑥0>0, then𝑥1𝑥2+1+√(1−𝑥21)(1−𝑥22)𝑥1+𝑥2>𝑥1𝑥2+1+(1−𝑥22)𝑥1+𝑥2=𝑥20−𝑅2+2−(𝑥0+𝑅)22𝑥0=−2𝑅2−2𝑥0𝑅+22𝑥0=1−𝑅2−𝑥0𝑅𝑥0=(1−𝑅)(1+𝑅)−𝑥0𝑅𝑥0≥𝑥0(1+𝑅)−𝑥0𝑅𝑥0=1.(4.6) Similarly, we obtain that𝑥1𝑥2+1+√(1−𝑥21)(1−𝑥22)𝑥1+𝑥2<−1(4.7) if 𝑥0<0.

The 𝑅∗-value corresponding to 𝑏 is given by𝑅∗=|||1−𝑥1𝑥2−√(1−𝑥21)(1−𝑥22)𝑥2−𝑥1|||.(4.8)

Now, assume that 𝑥1=1/𝑚 and 𝑥2=1/𝑛 where 𝑚 and 𝑛 are integers. Then,𝑏=1+𝑚𝑛−√(𝑚2−1)(𝑛2−1)𝑚+𝑛,𝑅=|1𝑚−𝑛(−√(𝑚2−1)(𝑛2−1)+𝑚𝑛−1)|.(4.9) It is clear that the pairs (𝑚,𝑛) making 𝑏 and 𝑅 rational numbers are precisely those of the form:𝑚2−1=𝑑𝑟2,𝑛2−1=𝑑𝑠2,(4.10) where 𝑑, 𝑟 and 𝑠 are integers. It is possible to show that the integer solutions of the generalized Pell's equation,𝑥2−𝑑𝑦2=1,(4.11) 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]).

Set𝐵−2=1,𝐵−1=0,𝐺−2=0,𝐺−1=1,𝑃0=0,𝑄0=1.(4.12) For 𝑖≥0, set𝑎𝑖=(𝑃𝑖+√𝑑)𝑄𝑖,𝐵𝑖=𝑎𝑖𝐵𝑖−1+𝐵𝑖−2,𝐺𝑖=𝑎𝑖𝐺𝑖−1+𝐺𝑖−2.(4.13) For 𝑖≥1, set𝑃𝑖=𝑎𝑖−1𝑄𝑖−1−𝑃𝑖−1,𝑄𝑖=(𝑑−𝑃2𝑖)𝑄𝑖−1.(4.14) 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

If 𝐶 is a simple contour in Ω, the corresponding heat flux is given by∫𝐶𝜆𝜕𝑢𝜕𝑛𝑑𝑠=∫𝐶𝜆grad𝑢⋅𝑛𝑑𝑠.(5.1) If 𝑢 is the real part of some complex potential 𝐹(𝑧), then∫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,(5.2) when 𝑡 denotes the unit tangential vector. Hence,∫𝐶𝜆𝜕𝑢𝜕𝑛𝑑𝑠=Im∫𝐶𝜆𝐹′(𝑧)𝑑𝑧.(5.3)

As an example, consider the simplest case discussed at the beginning of Section 2 for which𝐹∗(𝑧∗)=𝑢1−𝑢0ln𝑅∗log𝑧∗+𝑢0.(5.4) In this case, the derivative of the complex potential 𝐹(𝑧) is given by𝐹′(𝑧)=𝑓′(𝑧)𝐹∗′(𝑓(𝑧))=(𝑏2−1)(𝑏𝑧−1)2(𝑢1−𝑢0ln𝑅∗𝑏𝑧−1𝑧−𝑏)=𝑢1−𝑢0ln𝑅∗𝑏2−1(𝑏𝑧−1)(𝑧−𝑏).(5.5) By the residue theorem, we obtain that∫𝐶𝐹′(𝑧)𝑑𝑧=2𝜋𝑖𝑢1−𝑢0ln𝑅∗,(5.6) which gives that∫𝐶𝜆𝜕𝑢𝜕𝑛𝑑𝑠=2𝜋𝑢1−𝑢0ln𝑅∗.(5.7)