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The van der Pauw method is a well-known experimental technique in the applied sci- ences for measuring physical quantities such as the electrical conductivity or the Hall coefﬁcient of a given sample. Its popularity is attributable to its ﬂexibility: the same method works for planar samples of any shape provided they are simply connected. Mathematically, the method is based on the cross-ratio identity. Much recent work has been done by applied scientists attempting to extend the van der Pauw method to samples with holes (“holey samples”). In this article we show the relevance of two new function theoretic ingredients to this area of application: the prime function associated with the Schottky double of a multiply connected planar domain and the Fay trisecant identity involving that prime function. We focus here on the single-hole (doubly connected, or genus one) case. Using these new theoretical ingredients we are able to prove several mathematical conjectures put forward in the applied science literature. Keywords Cross ratio · Prime function · Fay trisecant identity · van der Pauw This article is dedicated to our former colleague, Professor Walter Hayman FRS, for his constant support over the years and his unwavering interest in what function theory can do for applications. His stories, anecdotes and songs, his career, and his fascinating life, remain an inspiration to us all at Imperial College London. Communicated by James K. Langley. Hiroyuki Miyoshi hiroyuki.miyoshi17@imperial.ac.uk Department of Mathematics, Imperial College London, 180 Queen’s Gate, London SW7 2AZ, UK Department of Earth Science and Engineering, Royal School of Mines, Prince Consort Rd, South Kensington, London SW7 2BP, UK 123 708 H. Miyoshi et al. Mathematics Subject Classiﬁcation 30Cxx · 14Qxx · 31Axx 1 Introduction One of the most prevalent and successful measurement techniques in the applied sciences is the four-point probe method [1,2]. The van der Pauw method [3,4]isan example of such a method used to measure the resistivity (or its inverse quantity, the conductivity) of a conducting laminate sample. Van der Pauw’s name is associated to this method because he was the ﬁrst to point out that, owing to the conformal invariance of the underlying boundary value problem for the electrical voltage potential, the basic idea of the four-point method works for determining the resistivity of uniform two- dimensional samples of any shape provided the contacts are placed at the edges of the sample [3,4]. Samples must have a ﬂat shape of unit thickness, be isotropic of uniform resistivity, and be simply connected. That is, they must not have holes. Figure 1 shows a set-up for the original van der Pauw measurement: four electrical contacts ( , , , ) are placed on the perimeter of a test sample. If and a b z w a represent a point source and sink of current J respectively, then the potential b ab difference V between two other points and can be measured while this current zw z w zw is ﬂowing. The resistance R = V / J is then a measured quantity; a second zw ab ab zb resistance R can be measured in exactly the same way. Van der Pauw [4]showed aw that for any arrangement of four electrical contacts, and given these two resistance zw zb measurements R and R , the speciﬁc resistivity σ can be found by solving the ab aw non-linear equation zw zb R R ab aw exp − + exp − =1(1) λ λ with λ ≡ (2) π d and where d is the uniform thickness of the two-dimensional sample. We will refer to this as the classical van der Pauw equation and it provides the basis for the van der Pauw method, the aim of which is to determine λ and, hence, the sample resistivity. Because this method needs only two resistance measurements, the method is widely applied for measuring the resistivity of superconductors or Hall coefﬁcients of mate- rials in laboratory experiments [3,5]. An efﬁcient numerical method to determine λ is discussed in [6]. While powerful, the van der Pauw technique is limited by its applicability only to simply connected samples. In reality, a given sample may well have holes, and these will affect the voltage measurements. Consequently, in recent years the van der Pauw method for samples with a single hole, or several holes, has been studied extensively [7–16]. It has been reported that the van der Pauw equation (1) is inaccurate in samples with several holes [17]. This is not surprising given that the formula (1) takes no account of the presence of those holes. 123 The Prime Function, the Fay Trisecant Identity... 709 x x Fig. 1 The original van der Pauw set-up. The ﬁrst measurement is the voltage difference between z and w with a point source of current at a and a point sink of current at b. A second measurement is of the voltage difference between z and b with a source a and a sink w. The associated resistances satisfy the van der Pauw equation (1) which can be solved for the sought-after parameter λ The present paper has been inspired by several conjectures in the literature that have arisen in the aforementioned attempts to generalize the van der Pauw methodology to holey samples. Given that the original van der Pauw method rests on the cross-ratio identity, these conjectures represent an interesting challenge for function theorists. As we show here, two new mathematical tools turn out to be relevant to the van der Pauw problem as it pertains to holey samples: the theory of the so-called prime function on the Schottky double of a multiply connected planar domain [18] and the Fay trisecant identities also associated with those same compact Riemann surfaces and which involve the prime function. This article focusses on the function theory aspects. A companion article [19] explores some of the practical implications of our results. In this paper, we use these mathematical tools—both of which are new to this application area—to prove two recent conjectures made in the applied science literature [7,12]. These conjectures concern a sample with an isolated hole, which is the natural ﬁrst case to study. We now outline the nature of these conjectures. It is a well-known consequence of an extension of the Riemann mapping theorem due to Koebe (see [20]) that any 2D sample with a single isolated hole can be trans- planted conformally into an annulus ρ< |ζ | < 1 in a complex ζ plane, say, where the radius of the inner circle of the annulus depends on the shape of sample [18,21]. By conducting both numerical and actual experiments Szymanski ´ et al. showed [7] that the van der Pauw equation (1) does not hold for a sample with a hole but they conjectured that the data instead satisﬁes the inequality zw zb ab aw exp − + exp − ≤ 1. (3) λ λ The same inequality has been mentioned in a series of papers [8,9]. It is one of the objectives of this paper to prove that (3) does indeed hold. zw zb In [8] Szymanski ´ et al. also ﬁnd that the pair of measured resistances (R , R ) aw ab satisﬁes another inequality which they dubbed a “lower envelope” and they proposed a method to measure λ based on the existence of this envelope. To explain this, Fig. 2 shows the results of a repeat of the same numerical experiment conducted in [8]. It 123 710 H. Miyoshi et al. Fig. 2 (i) Evidence for the two “envelopes”. Numerical experiments for pairs (X , Y ) where X ≡ zw zb exp(−R /λ) and Y ≡ exp(−R /λ) following [8]. The point z = 1isﬁxed, but (a, b,w) are picked at aw ab random with the ordering 0 = arg[z] < arg[w] < arg[a] < arg[b] < 2π.When ρ = 0, all pairs (X , Y ) are on the line X + Y = 1 which is (1). However, when ρ> 0, all points (X , Y ) lie in the gray-shaded region bounded by the straight line X + Y = 1 and a “lower envelope” which is curved. As shown in the center and right, the size of the gray-shaded area increases with ρ. (ii) The upper envelope (red line) and the lower envelope (blue line). The envelope is deﬁned by its set of tangents. The gray region is above the tangent line for each point on the lower envelope. The ﬁgure to the left shows what we mean by the “symmetric” choice of contact points in which the sector formed by the pair (a, b) subtends the same angle θ at the origin as that formed by the pair (z,w) shows the data from 40,000 pairs (X , Y ), where zw zb X ≡ exp(−R /λ), Y ≡ exp(−R /λ). (4) aw ab Three different samples are used with ρ = 0, 0.2 and 0.5. The contact points z, w, a, and b are chosen at random but always such that they retain the ordering 0 = arg[z] < arg[w] < arg[a] < arg[b] < 2π. The data is found to fall in the gray-shaded regions shown in Fig. 2. With no hole, which means ρ = 0, the pair satisﬁes X + Y = 1 which is just the original van der Pauw equation (1) in evidence. However, if ρ> 0, this is no longer true and the data (X , Y ) “ﬁlls in” a crescent-shaped domain shown shaded in Fig. 2. Szymanski ´ et al. [8] conjecture that the data (X , Y ) always lies in such a domain bounded by the upper envelope X + Y ≤ 1 and some lower envelope which depends only on the conformal modulus ρ. They also conjecture, again without a rigorous mathematical proof, that the lower envelope might correspond to the pair of 123 The Prime Function, the Fay Trisecant Identity... 711 (X , Y ), where (X , Y ) are measurements with the four electrical contacts having θ θ θ θ a certain symmetry shown on the lower left of Fig. 2(ii). The angle θ is deﬁned as θ ≡ arg[w/z]= arg[b/a]. In such a symmetric conﬁguration the sector formed by the pair (a, b) subtends the same angle θ at the origin as that formed by the pair (z,w). If these conjectures hold then the form of the upper and lower envelopes can be expressed mathematically as ∂Y /∂θ X + Y ≤ 1, Y − Y ≥ (X − X ), for 0 <θ <π. (5) θ θ ∂ X /∂θ Two examples of the tangent line (5) on the lower envelope are shown on the right of Fig. 2(ii). These inequalities have been put forward in several papers [7–10], but no rigorous proof of them has yet been obtained. This paper aims to understand the envelope structure mathematically and to offer proofs of the aforementioned conjectures. This is done by introducing, for the ﬁrst time, two important tools into this area of investigation: (i) use of the prime function, denoted by ω(ζ, c), associated with the concentric annulus ρ< |ζ | < 1; (ii) use of the Fay trisecant identity satisﬁed by this prime function. Using these tools we can gain insights into the two envelopes associated with the resistance measurements (4). 2 The Van der Pauw Equation and the Cross Ratio Identity While applied scientists are familiar with the van der Pauw equation (1) mathemati- cians are perhaps more familiar with a similar-looking cross-ratio identity given by p (z,w; b, a) + p (z, b; w, a) = 1, (6) 0 0 where the classical cross-ratio is deﬁned by ω(z, a)ω(w, b) p (z,w; a, b) ≡ , (7) ω(z, b)ω(w, a) and where, in a step usually not carried out, we introduce the prime function associated with the unit disc [18] as the simple monomial function ω(ζ, c) = ζ − c. (8) Establishing the identity (6) is a simple exercise. In complex analysis the cross-ratio [18,22] is most commonly encountered in a geometrical context as the Möbius mapping that provides a conformal mapping, as a function of the variable z, between 3 arbitrary complex points (a,w, b) in the complex z plane and the canonical choice of points (0, 1, ∞). Concerning the prime function (8), it is so simple in this case that it is usually not even given the designation “prime function”. However, a recent monograph [18] makes the case that recognizing it as the simplest instance of a more general notion of a prime 123 712 H. Miyoshi et al. function is important for generalizing many known results for simply connected planar geometries to multiply connected cases in a natural way. The van der Pauw problem of interest here is no exception: it will be shown later that the natural way to extend the classical van der Pauw method to multiply connected geometries is to treat the problem using the prime function—that is, the multiply connected generalization of (8)—and to make use of some important identities satisﬁed by those functions. There is a connection between (1) and (6). It arises by considering the complex potential H () of the complex variable = x + iy, where (x , y) denotes Cartesian coordinates in the physical plane, whose real part is the harmonic voltage potential V (x , y) in the sample: H () = V (x , y) + iχ(x , y), (9) where we have introduced χ(x , y), the harmonic conjugate of V (x , y) in the physical (x , y) plane. V (x , y) is harmonic in the sample and its normal derivative vanishes on the sample boundary; equivalently, by the Cauchy-Riemann equations, the harmonic conjugate χ(x , y) is constant on the boundary. By the well-known Riemann mapping theorem, and the conformal invariance of the boundary value problem for the potential V (x , y) driven by a source at and a compensating sink at [22], the complex potential h (ζ ) ≡ H () is given by the b 0 0 explicit formula σ J ω(ζ, a) ab h (ζ ) = log , (10) π d ω(ζ, b) where, once again, we notice the appearance of the prime function (8) and = f (a), = f (b), (11) a b where = f (ζ ) is the conformal mapping between the unit disc D in the ζ plane and the given laminate sample in the complex plane. The potential difference V zw is therefore given by ω(z, a)ω(w, b) V ≡ Re[h (z)]− Re[h (w)]= λ J log zw 0 0 ab ω(z, b)ω(w, a) (12) = λ J log | p (z,w; a, b)| , ab 0 where z and w are the preimages of the two measurement contact points and : z w = f (z), = f (w). (13) z w It follows that zw zw R ≡ = λ log p (z,w; a, b), (14) ab ab 123 The Prime Function, the Fay Trisecant Identity... 713 where we have removed the modulus symbols because the cross-ratio is real and positive when all the points z,w, a, and b are on the unit circle in the ζ plane, and 0 ≤ arg[z] < arg[w] < arg[a] < arg[b] < 2π. On combining (14) with (6), we arrive at the van der Pauw equation (1). 3 The Prime Function for the Annulus < || < 1 Now consider the van der Pauw method for a doubly connected sample. Let D denote a bounded sample with an isolated hole. Let ∂ D be the outer boundary of the sample and ∂ D the boundary of the hole. Similar to the original van der Pauw method, we assume that the sample thickness is d. We assume that the hole in the sample carries no net charge. The set-up is illustrated in Fig. 3. Suppose that 4 point contacts ( , , , ), of inﬁnitesimal width, are placed a b z w on ∂ D . It is known, by an extension of the Riemann mapping theorem [18,20], that any such domain is conformally equivalent to a concentric annulus ρ< |ζ | < 1 with circular boundaries C and C and 0 ≤ ρ< 1. The circle C is the unit circle; C is 0 1 0 1 the circle |ζ|= ρ. In other words, there exists an analytic function = f (ζ ) (15) that transplants the annulus ρ< |ζ | < 1 to the domain D with C being transplanted to ∂ D and C to ∂ D . 0 1 1 Let the required complex potential, as a function of = x + iy,be H () = V (x , y) + iχ(x , y). (16) We can still exploit the conformal invariance for the problem of determining the potential V (x , y) in this doubly connected domain. Crowdy [18,23] has shown that the complex potentials for any point source/sink driven harmonic ﬁeld in a multiply connected domain can be written down explicitly in terms of the prime function asso- ciated with that domain. It is important to emphasize, for possible future generalization of this work, that this fact holds for domains of any ﬁnite connectivity not just the doubly connected case of interest here. The prime function ω(ζ, c) for the annulus ρ< |ζ | < 1 can be deﬁned explicitly in this case by the formula [18] c ζ ω(ζ, c) =− P , (17) P(1) where 2n 2n ˆ ˆ P(ζ ) ≡ (1 − ζ) P(ζ ), P(ζ ) ≡ (1 − ρ ζ) 1 − . (18) n=1 123 714 H. Miyoshi et al. x x Fig. 3 The van der Pauw set-up for a sample with an isolated hole. The ﬁrst measurement is the voltage difference between z and w with a source of current a and a sink of current b. The second measurement is the difference between z and b with a source a and a sink w. Similar to the original van der Pauw method, the measured resistances are denoted by the logarithm of the Schottky–Klein prime function It is easy to show, directly from these deﬁnitions via an inﬁnite product, that −1 −1 2 −1 P(ζ ) =−ζ P(ζ ), P(ρ ζ) =−ζ P(ζ ). (19) These identities, which are the fundamental properties of the prime function in this case (see [18, Ch. 5] which is devoted to this function and its properties), will be useful later. For notational brevity, we write P(ζ ) even though this function also depends on the parameter ρ as is clear from its deﬁnition (18). The reader should bear in mind this additional parametric dependence. It is important to mention that while, merely for convenience and brevity of exposi- tion, we have here deﬁned the prime function for the concentric annulus in terms of an inﬁnite product (18) the monograph [18] gives a more intrinsic derivation of the prime function for multiply connected planar domains using the ﬁrst-type Green’s function as the starting point. Interested readers are encouraged to refer there for a much more general mathematical perspective. We will also need the functions K (ζ ) and L(ζ ) deﬁned as ∂ ∂ K (ζ ) K (ζ ) ≡ ζ log P(ζ ), L(ζ ) ≡ ζ . (20) ∂ζ ∂ζ The logarithmic derivative of the prime function K (ζ ) and its derivative L(ζ ) are also important functions in the general function theory on multiply connected domains [18]. For the concentric annulus it is easily shown, on use of (19), that these functions satisfy the functional relations −1 2 K (ζ ) = 1 − K (ζ ), K (ρ ζ) = K (ζ ) − 1, (21) −1 2 L(ζ ) = L(ζ ), L(ρ ζ) = L(ζ ). Actually, the function K (ζ ) can be related to the Weierstrass zeta function and L(ζ ) to the Weierstrass ℘-function [18] but those theoretical connections will not be needed here and this remark is made merely as a point of interest. 123 The Prime Function, the Fay Trisecant Identity... 715 Following [18,23] the complex potential h(ζ ) ≡ H ( f (ζ )) is given by −1 σ J ab ¯ ω(ζ, a)ω(ζ, a ¯ ) a P(ζ /a) ab h(ζ ) = log = λ J log , (22) ab −1 2π d |ab| b P(ζ /b) ω(ζ, b)ω(ζ, b ) where the ﬁrst equality is derived in [18,23] and follows from the general properties of the prime function, and where we have used (17) in the second equality. The relations (11) and (13) give the relationship between (a, b, z,w) and ( , , , ) a b z w although it is understood that the mapping f (ζ ) is now the new mapping from the concentric annulus to the holey sample. Note that, because two electrical contacts a −1 −1 and b are on C , a ¯ = a and b = b . The voltage difference between z and w is given by P(z/a) P(w/b) zw V ≡ Re[h(z)]− Re[h(w)]= λ J log . (23) ab ab P(z/b) P(w/a) Since all contacts are located on the same edge of the annulus, then Im[h(z)]− zw Im[h(w)]= 0. The measured resistance R is deﬁned as ab V P(z/a) P(w/b) zw zw R ≡ = λ log . (24) ab J P(z/b) P(w/a) ab In the same way, V P(z/a) P(b/w) zb zb R ≡ = λ log . (25) aw J P(z/w) P(b/a) aw It is straightforward to check that the formulas (24) and (25) above are equivalent to those given in [7]. More precisely, it can be checked that the function G(φ) used in [7] is related to P(ζ )—and hence to the prime function for the concentric annulus (17)—by the formula iφ iφ 4n 2n iφ/2 P(e ) = 1 − e 1 + ρ − 2ρ cos φ =−2ie P(i)G(φ), (26) n=1 where the function G(φ) introduced in [7] is deﬁned by φ cos φ G(φ) ≡ sin 1 − , h ≡ 2log ρ. (27) cosh hn n=1 Although (26) shows that our new expressions (24)–(25) coincide with those of [7], which is reassuring, there is much signiﬁcance in having recognized that the resistances can be written in terms of this special transcendental function known as the prime function [18] of the preimage concentric annulus. 123 716 H. Miyoshi et al. First, the notion of a prime function extends to a planar domain of any ﬁnite con- nectivity [18] which means that we have already provided a route to generalizing all the ideas in this paper presented here for the annulus to any higher connected domain, that is, to a sample with more than one hole. Crowdy [23] was the ﬁrst to show how the complex potentials for source/sink driven harmonic ﬁelds in multiply connected domains of arbitrary connectivity can be written explicitly in terms of the associated prime functions. His treatment uses irrotational ﬂuid mechanics as the physical context but, mathematically, the problem is identical to the electrical conduction problems of interest here. Second, it is known [18] that prime functions, including those associated with domains of connectivity higher than one, satisfy a so-called Fay trisecant identity. This identity can be viewed as an analogue of the cross-ratio identity (6) on a higher genus Riemann surface [24] and is the topic of the next section. 4 The Fay Trisecant Identity for the Annulus < || < 1 It is useful to introduce the function ω(z, a)ω(w, b) p(z,w; a, b) ≡ . (28) ω(z, b)ω(w, a) This formula is identical to that deﬁning the cross-ratio (6) but this quantity is no longer a cross-ratio because the deﬁnition of the prime function has changed. On use of (17)formula (28) can be written in terms of P(ζ ) as P(z/a) P(w/b) p(z,w; a, b) = . (29) P(z/b) P(w/a) From (29) and (24)–(25) we see that zw zb exp(−R /λ) = p(z,w; b, a), exp(−R /λ) = p(z, b; w, a). (30) ab aw The Fay trisecant identity associated with this prime function is P(kz/w) P(ka/b) P(kz/b) P(ka/w) p(z,w; b, a) + p(z, b; w, a) = P(k), (31) P(kza/wb) P(kza/wb) where k is an arbitrary complex number. This statement (31) of the genus-1 Fay trisecant identity expressed purely in terms of the prime function of the concentric annulus has been taken from Exercise 8.9 of Chapter 8 of the monograph [18] which asks the reader to prove it using the properties of so-called loxodromic functions. While (31) is a particular form of the more general Fay trisecant identity [24,25] the authors have not found it written in the form (31) anywhere else in the literature (besides [18]). This form (31) of the genus-one Fay trisecant identity will be a crucial tool in what follows. 123 The Prime Function, the Fay Trisecant Identity... 717 Although it is well established [24,25] from more general arguments it is instructive to sketch a proof this form (31) of the genus-one Fay trisecant identity. First we consider the left hand side of (31) as a function of z with all other quantities being treated as parameters. Let P(kz/w) P(ka/b) P(kz/b) P(ka/w) J (z) ≡ p(z,w; b, a) + p(z, b; w, a). (32) P(kza/wb) P(kza/wb) The properties (19) of the function P(ζ ) can be used to show that J (ρ z) = J (z). (33) Hence, since it is also meromorphic as a function of z, it is a loxodromic function of z;see [18, Ch. 8]. A loxodromic function is the name of an automorphic function on the Schottky double of the concentric annulus; it is a meromorphic function of that surface satisfying the functional identity (33). If we write N (z) J (z) = (34) P(kza/wb) P(z/a) P(w/b) so that N (z) = P(kz/w) P(ka/b) P(z/b) P(w/a) (35) − (w/b) P(kz/b) P(ka/w) P(z/w) P(b/a) then, it can be veriﬁed, again using the properties (19)of P(ζ ), that N (a) = N (wb/ka) = 0. (36) Since J (z) is a loxodromic function with removable poles at z = a and z = wb/ka— and, therefore, having no poles on the surface— it must be independent of z, which means it is a constant when considered as a function of z. We are employing a Liouville- type theorem on this genus-one Schottky double: any meromorphic function on it having no poles must be constant. Such results will be used extensively throughout this paper. We can also consider the left hand side of (31) as a function for w and write P(kz/w) P(ka/b) P(kz/b) P(ka/w) J (w) ≡ p(z,w; b, a) + p(z, b; w, a). (37) P(kza/wb) P(kza/wb) This can also be shown to be loxodromic, i.e., ˜ ˜ J (ρ w) = J (w) (38) and to have removable poles at w = b and w = kza/b. It is therefore independent of w. 123 718 H. Miyoshi et al. By similar arguments, considering the left hand side of (31) successively as a function of a and b it can be shown to be independent of those variables too. Putting all these facts together, we conclude that P(kz/w) P(ka/b) P(kz/b) P(ka/w) p(z,w; b, a) + p(z, b; w, a) = C (k), (39) P(kza/wb) P(kza/wb) where C (k) is a function to be determined. It can be found by matching to the limit of the left hand side of (39) in the double limit w → z and b → a which yields C (k) = P(k). (40) Thus we have established the Fay trisecant identity (31). On substituting (24) and (25)into(31), we obtain zw zb P(kz/w) P(ka/b) P(kz/b) P(ka/w) R ab aw exp − + exp − = 1 . (41) P(k) P(kza/wb) λ P(k) P(kza/wb) λ When ρ → 0, so that there is no hole in a sample, it is straightforward to check that P(kz/w) P(ka/b) P(kz/b) P(ka/w) = =1(42) P(k) P(kza/wb) P(k) P(kza/wb) because P(ζ ) = 1−ζ when ρ = 0. The original van der Pauw equation (1) is therefore retrieved from (41). All dependence on the new parameter k disappears in this simply connected limit. It is clear that (41) opens up new perspectives: that it reduces, as ρ → 0, to the original van der Pauw equation (1) is tantalizing. It also makes it a natural candidate, at least from the mathematical point of view, to ﬁnd ways to extend the van der Pauw method to holey samples. In contrast to the original van der Pauw equation, the zw zb coefﬁcients of exp(−R /λ) and exp(−R /λ) in (41) now depend not only on the ab aw electrical contact locations z, w, a, b but also on a ﬁfth complex parameter k. It should be emphasized that (41) holds for arbitrary choices of a, b, z,w and k even though, for present purposes, we will assume that a, b, z and w lie on C . The freedom in the choice of k will be exploited in the next section to gain insights into the envelope structure evident in Fig. 2. 5 Analysis of the Envelopes: The Integrated Fay Identity In Sect. 1 we discussed the existence of two envelopes, an “upper” and a “lower” envelope with the mathematical deﬁnitions given in (5). It is now shown how the new tools introduced in the previous two sections allow us to prove the conjecture enshrined in (5). 123 The Prime Function, the Fay Trisecant Identity... 719 x x Fig. 4 Illustration of the complex coordinates. The parameters θ and θ are seen as displacements from 1 3 the symmetric points where z is ﬁxed to 1 For arbitrary z,w, a, and b on C we can introduce new angular coordinates θ, θ 0 1 and θ : z = 1,w = exp(i(θ + θ )), a = exp(i(θ + θ )), b = exp(i(θ + θ )). (43) 1 1 3 3 Because 0 < arg[w] < arg[a] < arg[b] < 2π, the range of θ ,θ,θ are given by 1 3 − θ< θ <θ, θ <θ < 2π − θ, 0 <θ <π. (44) 1 3 It is important to mention that the symmetry case shown in Fig. 2 corresponds to the choices θ = 0 and θ = π. An illustration of the complex coordinates is shown in 1 3 Fig. 4. The parameters θ and θ are seen as displacements from the symmetric points 1 3 with parameter θ. The Fay trisecant identity (41) can be written ˆ ˆ A(θ ,θ, k)X + B(θ ,θ, k)Y = 1, (45) 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 where zw −i(θ +θ ) i(θ −θ ) 3 3 P(e ) P(e ) ab X ≡ exp − = , θ ,θ ,θ 1 3 −i(θ +θ ) i(θ −θ ) 1 3 1 3 λ P(e ) P(e ) (46) zb −i(θ +θ ) i(θ −θ ) 1 1 R P(e ) P(e ) aw Y ≡ exp − = , θ ,θ ,θ 1 3 −i(θ +θ ) i(θ −θ ) 1 3 3 1 λ P(e ) P(e ) 123 720 H. Miyoshi et al. and the coefﬁcient functions are −i(θ +θ ) i(−θ +θ ) 1 1 P(kz/w) P(ka/b) P(ke ) P(ke ) A(θ ,θ, k) ≡ = −2iθ P(k) P(kza/wb) P(k) P(ke ) −iθ iθ ˆ 1 ˆ 1 P(ke ) P(ke ) = , −iθ iθ ˆ ˆ P(ke ) P(ke ) (47) −i(θ +θ ) i(−θ +θ ) 3 3 P(kz/b) P(ka/w) P(ke ) P(ke ) B(θ ,θ, k) ≡ = −2iθ P(k) P(kza/wb) P(k) P(ke ) −iθ iθ ˆ 3 ˆ 3 P(ke ) P(ke ) = , −iθ iθ ˆ ˆ P(ke ) P(ke ) iθ and where we have set k = ke because k is arbitrary. Notice that, with this choice, the resistance measurements have the following symmetries: X = X = X = X , θ ,2π −θ ,θ −θ ,θ ,θ −θ ,2π −θ ,θ θ ,θ ,θ 1 3 1 3 1 3 1 3 (48) Y = Y = Y = Y . θ ,2π −θ ,θ −θ ,θ ,θ −θ ,2π −θ ,θ θ ,θ ,θ 1 3 1 3 1 3 1 3 ˆ ˆ It is also important to note that A(θ ,θ, k) is independent of θ and B(θ ,θ, k) is 1 3 3 independent of θ . A crucial next step is to consider an integrated form of the Fay trisecant identity ˆ ˆ ˆ (41). Consider integrals of A(θ ,θ, k) and B(θ ,θ, k) with respect to k around the 1 3 circle |k|= ρ.From(45), we obtain α(θ ,θ)X + β(θ ,θ)Y = 1, (49) 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 where 2π 2π i(φ−θ ) i(φ+θ ) 1 1 1 1 P(ρe ) P(ρe ) iφ α(θ ,θ) ≡ A(θ ,θ,ρe )dφ = dφ, 1 1 i(φ−θ) i(φ+θ) 2π 2π P(ρe ) P(ρe ) 0 0 (50) 2π 2π i(φ−θ ) i(φ+θ ) 3 3 1 1 P(ρe ) P(ρe ) iφ β(θ ,θ) ≡ B(θ ,θ,ρe )dφ = dφ. 3 3 i(φ−θ) i(φ+θ) 2π 2π P(ρe ) P(ρe ) 0 0 The integrated Fay’s trisecant identity (49) turns out to be essential for proving the conjectures about the envelope structure. On taking a derivative of (49) with respect to θ, ∂α(θ ,θ) ∂β(θ ,θ) 1 3 X + Y θ ,θ ,θ θ ,θ ,θ 1 3 1 3 ∂θ ∂θ ∂ X ∂Y θ ,θ ,θ θ ,θ ,θ 1 3 1 3 + α(θ ,θ) + β(θ ,θ) = 0. 1 3 ∂θ ∂θ (51) 123 The Prime Function, the Fay Trisecant Identity... 721 The sum of the ﬁrst two terms is zero because ∂α(θ ,θ) ∂β(θ ,θ) 1 3 X + Y θ ,θ ,θ θ ,θ ,θ 1 3 1 3 ∂θ ∂θ 2π iφ iφ = [ A(θ ,θ,ρe )X + B(θ ,θ,ρe )Y ] 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 2π i(φ−θ) i(φ+θ) ×[K (ρe ) − K (ρe )]dφ 2π i(φ−θ) i(φ+θ) = [K (ρe ) − K (ρe )]dφ = 0, 2π (52) where we used Fay’s identity (45) in the second equality. On use of (52)in(51), it is found that ∂ X ∂Y θ ,θ ,θ θ ,θ ,θ 1 3 1 3 α(θ ,θ) + β(θ ,θ) = 0. (53) 1 3 ∂θ ∂θ Suppose now that we ﬁx the two parameters θ and θ . Then (X , Y ) 1 3 θ ,θ ,θ θ ,θ ,θ 1 3 1 3 depends only on the single parameter θ and the tangent at (X , Y ) when θ ,θ ,θ θ ,θ ,θ 1 3 1 3 viewed as a function of θ is deﬁned as the set of (X , Y ) satisfying ∂Y /∂θ θ ,θ ,θ 1 3 Y − Y = (X − X ). (54) θ ,θ ,θ θ ,θ ,θ 1 3 1 3 ∂ X /∂θ θ ,θ ,θ 1 3 If we now make use of both (53) and (49) the tangent line (54) is equivalent to α(θ ,θ)X + β(θ ,θ)Y = 1. (55) 1 3 It is helpful to visualize this: Fig. 5 shows examples of these tangent lines. The red line shows the collection of points (X , Y ) where both θ and θ are ﬁxed θ ,θ ,θ θ ,θ ,θ 1 3 1 3 1 3 but where θ is varying; the blue lines in Fig. 5, given by (55), are tangent to those red lines. Because of the angular conditions (44), θ varies between |θ | and π −|π − θ |. 1 3 The important point, however, is that because the symmetry case shown in Fig. 2 corresponds to θ = 0 and θ = π, the two inequalities in (5) are equivalent to the 1 3 two statements X + Y ≤ 1, (56) α X + β Y ≥ 1,α ≡ α(0,θ), β ≡ β(π, θ), (57) θ θ θ θ for all pairs (X , Y ) and 0 <θ <π. The ﬁrst statement is that all pairs (X , Y ) for ρ ≥ 0 lie on or below the line X + Y = 1 relevant to the classical van der Pauw case ρ = 0; the second statement is that all pairs (X , Y ) for ρ ≥ 0 lie above a θ-parametrized curve 123 722 H. Miyoshi et al. Fig. 5 The red lines are collections of data points (X , Y ) with both θ and θ ﬁxed and only θ ,θ ,θ θ ,θ ,θ 1 3 1 3 1 3 parameter θ changed. The blue lines, given by (55), are tangent to the red lines at (X , Y ). θ ,θ ,θ θ ,θ ,θ 1 3 1 3 When θ = 0and θ = π, the red line is a tangent to a lower envelope 1 3 deﬁned for θ = 0 and θ = π where the parameter θ corresponds geometrically to 1 3 points arranged with the symmetry shown in Fig. 2. Thestrategyistoprove (56) and (57) for all (X , Y ) by considering the maximum and minimum values of the coefﬁcient functions α(θ ,θ) and β(θ ,θ) subject to the 1 3 condition (44). From the integrated Fay identity (49), min (α(θ ,θ))X + min (β(θ ,θ))Y 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 −θ<θ <θ θ<θ <2π −θ 1 3 ≤ α(θ ,θ)X + β(θ ,θ)Y = 1, (58) 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 max (α(θ ,θ))X + max (β(θ ,θ))Y 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 −θ<θ <θ θ<θ <2π −θ 1 3 ≥ α(θ ,θ)X + β(θ ,θ)Y = 1. (59) 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 The idea is to show that (58) is equivalent to (56), and (59) is equivalent to (57). To study the extrema of α(θ ,θ) and β(θ ,θ), we deﬁne the function 1 1 2π 2π i(φ−η) i(φ+η) 1 P(ρe ) P(ρe ) 1 iφ g (η) ≡ dφ = G (ρe )dφ, (60) θ θ,η i(φ−θ) i(φ+θ) 2π P(ρe ) P(ρe ) 2π 0 0 where 0 ≤ η ≤ 2π and the integrand is deﬁned as P(ζ /μ) P(ζ μ) iη iθ G (ζ ) ≡ ,μ ≡ e ,ν ≡ e . (61) θ,η P(ζ /ν) P(ζ ν) 123 The Prime Function, the Fay Trisecant Identity... 723 Fig. 6 Visualizing the relationship between tangent lines and the function g (η) deﬁned in (60). Red lines show the curves produced by changing only θ, and blue lines are tangents to the envelope at (X , Y ).When α(θ ,θ) and β(θ ,θ) are minimised, the tangent line corresponds to X +Y = 1. θ ,θ ,θ θ ,θ ,θ 1 3 1 3 1 3 When α(θ ,θ) and β(θ ,θ) are maximised, the line becomes a tangent line to the lower envelope at (X , Y ), 1 3 θ θ where X ≡ X and Y ≡ Y θ 0,π,θ θ 0,π,θ 123 724 H. Miyoshi et al. From the deﬁnitions (50) it is clear that the two coefﬁcients functions in (49) can be written in terms of this single function: α(θ ,θ) = g (θ ), β(θ ,θ) = g (θ ). (62) 1 θ 1 3 θ 3 Analysis of this function g (η) provides the key to the proofs of the conjectures. Before presenting the details, the strategy explained in (58)–(59) is illustrated in Fig. 6. By minimising both α(θ ,θ) and β(θ ,θ), the upper envelope X + Y = 1is 1 3 obtained. In contrast, by maximising both α(θ ,θ) and β(θ ,θ), a tangent line to the 1 3 lower envelope is obtained. It is important to mention that because of the deﬁnition of P(ζ ), iφ 4n−2 2n−1 P(ρe ) = 1 + ρ − 2ρ cos φ > 0, (63) n=1 iφ which means G (ρe ) is real and positive. By a log-sum and sum-log inequality, we θ,η obtain ⎡ ⎤ 2π i(φ−η) i(φ+η) 1 P(ρe ) P(ρe ) ⎣ ⎦ log g (η) = log dφ i(φ−θ) i(φ+θ) 2π P(ρe ) P(ρe ) 2π i(φ−η) i(φ+η) 1 P(ρe ) P(ρe ) ≥ log dφ i(φ−θ) i(φ+θ) 2π P(ρe ) P(ρe ) 2π i(φ−η) i(φ+η) = [log( P(ρe )) + log( P(ρe )) 2π i(φ−θ) i(φ+θ) − log( P(ρe ) − log( P(ρe ))]dφ = 0. (64) Thus, we can conclude that g (η) ≥ 1 for all θ and η. We will use this in the next section to analyse the behavior of the function g (η). @g 6 Analysis of the Functions g and () This section studies the behavior of the function g (η). First, we propose an alternative expression of g (η) to be: P(ν/μ) P(μν) g (η) = [K (μ/ν) − K (μν)], (65) ˆ 2 P(1) P(ν ) 123 The Prime Function, the Fay Trisecant Identity... 725 iη iθ where we recall that μ = e and ν = e . This expression is useful for the analysis of g (η). To explain the derivation of (65) we note that the integrand G (ζ ) deﬁned in (61) θ,η has two simple poles at ζ = ν and ζ = 1/ν and is readily conﬁrmed, on use of (19), to be loxodromic. The function G (ζ ) can therefore also be written as θ,η P(ζ /μ) P(ζ μ) G (ζ ) = = c [K (ζ /ν) − K (ζ ν)]+ c , (66) θ,η 1 2 P(ζ /ν) P(ζ ν) where c , c ∈ C. Since K (ζ ) is a logarithmic derivative of P(ζ ) and has functional 1 2 properties (21), the right hand side of (66) is also a loxodromic function and has two simple poles at ζ = ν and ζ = 1/ν. The coefﬁcients c and c are determined by 1 2 considering the limits ζ → ν and ζ → μ, P(ν/μ) P(μν) c =−c [K (μ/ν) − K (μν)], c = [K (μ/ν) − K (μν)]. (67) 2 1 2 P(1) P(ν ) A Liouville-type argument then conﬁrms the equivalence of the two expressions for G (ζ ) in (66). We therefore conclude by using (60) and (66) that θ,η 2π iφ g (η) = G (ρe )dφ θ θ,η 2π 2π i(φ−θ) i(φ+θ) = c [K (ρe ) − K (ρe )]+ c dφ = c , 1 2 2 2π (68) which is precisely (65). Now, we consider the behavior of the function g (η) by exploiting the derivative of g (η): ∂ g ∂μ ∂ g P(ν/μ) P(μν) θ θ = = i (L(μ/ν) − L(μν)) + ig [K (μν) − K (ν/μ)]. ∂η ∂η ∂μ ˆ P(1) P(ν ) (69) The function L(μ/ν) − L(μν) is a loxodromic function as a function of μ, and has two second-order poles at μ = ν and μ = 1/ν. Similar to the expression of g (η),we propose another expression of L(μ/ν) − L(μν) given by 2 2 2 P(1) P(μ ) P(ν ) L(μ/ν) − L(μν) = . (70) P(ν/μ) P(μ/ν) P(μν) 123 726 H. Miyoshi et al. To establish this expression, consider a new function 2 2 2 P(1) P(μ ) P(ν ) L(μ) ≡ L(μ/ν) − L(μν) − . (71) P(ν/μ) P(μ/ν) P(μν) The function L(μ) is shown to be a loxodromic function by using the functional properties (19) and (21)asfollows: 2 4 2 2 P(1) P(ρ μ ) P(ν ) 2 2 2 L(ρ μ) = L(ρ μ/ν) − L(ρ μν) − 2 2 2 2 P(ν/ρ μ) P(ρ μ/ν) P(ρ μν) 2 2 4 −1 2 2 P(1) (ρ μ ) P(μ ) P(ν ) = L(μ/ν) − L(μν) − 2 −1 −1 −2 2 (ρ μ/ν) P(ν/μ)(μ/ν) P(μ/ν)(μν) P(μν) = L(μ). (72) Furthermore, the asymptotic expansions of each term at μ = ν and μ = 1/ν are 1 1 L(μ/ν) − L(μν) ∼− + + O(1) at μ = ν, (1 − μ/ν) 1 − μ/ν 1 1 L(μ/ν) − L(μν) ∼ − + O(1) at μ = 1/ν, (1 − μν) 1 − μν 2 2 2 P(1) P(μ ) P(ν ) 1 ∼ + O(1) (73) P(ν/μ)P(μ/ν) P(μν) (1 − ν/μ)(1 − μ/ν) 1 1 =− + + O(1) at μ = ν, (1 − μ/ν) 1 − μ/ν 2 2 2 P(1) P(μ ) P(ν ) 1 1 ∼ − + O(1) at μ = 1/ν. 2 2 P(ν/μ)P(μ/ν) P(μν) (1 − μν) 1 − μν Thus the two poles at ν and 1/ν are removable. We conclude that L(μ) is a constant function, i.e., it is independent of μ. In addition, we deﬁne another function 2 2 2 P(1) P(μ ) P(ν ) L(ν) ≡ L(μ/ν) − L(μν) − . (74) P(ν/μ) P(μ/ν) P(μν) By similar arguments, the function L(ν) can be seen to be a loxodromic function of ν with removable poles ν = μ and ν = 1/μ. Thus, L(μ) is independent of both μ and ν. On use of the functional property (21)of L(ζ ), it can be veriﬁed that L(ρ) = 0, which means L(μ) vanishes everywhere leading to expression (70). 123 The Prime Function, the Fay Trisecant Identity... 727 Fig. 7 Typical behavior of the function g (η), k (η),and ∂ g /∂η. A local maximum of g (η) is obtained θ θ θ θ when η = 0or η = π. A local minimum of g(η) is obtained when η = θ or 2π − θ Expression (70) can now be used to analyse the behaviour of g (η). By substituting (70)into(69), ∂ g P(1) P(μ ) = i + g (η)(K (μν) − K (ν/μ)) ∂η P(μ/ν) P(μν) 1 P(1) P(μ ) = i(K (μν) − K (ν/μ)) g (η) − K (ν/μ) − K (μν) P(μ/ν) P(μν) = i(K (μν) − K (ν/μ)) g (η) − . g (θ ) (75) Since, from (64), g (η) ≥ 1 and g (θ ) ≥ 1 then θ η g (η) − ≥ 0. (76) g (θ ) The sign of the derivative (75) is therefore determined by the function k (η) ≡ i(K (μν) − K (ν/μ)). (77) The derivative of k (η) with respect to η is ∂k =−(L(μν) + L(ν/μ)). (78) ∂η iφ But it can also be shown that L(e ) for 0 <φ < 2π is real and positive—the proof is given in Appendix A. This means that ∂k < 0, 0 <η < 2π. (79) ∂η 123 728 H. Miyoshi et al. It is straightforward to check that k (0) = k (π ) = 0 and consequently, using conti- θ θ nuity arguments, lim k (η) =−∞, lim k (η) =+∞, (80) θ θ η→θ −0 η→θ +0 lim k (η) =−∞, lim k (η) =+∞. (81) θ θ η→2π −θ −0 η→2π −θ +0 With this information we can determine the maxima and the minima of g (η).The local minima occur when η = θ or η = 2π − θ. By the deﬁnition (60), g (θ ) = g (2π − θ) = 1. (82) θ θ On the contrary, local maxima occur when η = 0 and η = π. Recall that this case corresponds to the symmetric choice of points with parameter θ. Figure 7 shows the typical behaviour of the function g (η) for a range of ρ values; recall that P(ζ ), and hence g (η) and k (η) depend on the parameter ρ although this dependence is hidden θ θ in our notation. 7 Proof of the Two Conjectures (56)–(57) We will now prove the two conjectures (56) and (57) using what we have established about the behavior of g (η).From(58) and (82), we can conclude that X + Y ≤ 1 (83) θ ,θ ,θ θ ,θ ,θ 1 3 1 3 for all (X , Y ). This is equivalent to the conjecture (56). θ ,θ ,θ θ ,θ ,θ 1 3 1 3 The maxima of α(θ ,θ) and β(θ ,θ) occur when θ = 0 and θ = π. Hence, 1 3 1 3 from (59), we conclude that α X + β Y ≥ 1. (84) θ θ ,θ ,θ θ θ ,θ ,θ 1 3 1 3 The inequality (84) is not, however, equivalent to the inequality (57). What has been proven in (84)isthat for0 <θ <π, the pair (X , Y ) is above the tangent at θ ,θ ,θ θ ,θ ,θ 1 3 1 3 (X , Y ) on the lower envelope. Thus, for the ﬁnal step, we need to prove that for 0,π,θ 0,π,θ another θ = θ, the pair (X , Y ) lies above the tangent at (X , Y ). θ ,θ ,θ θ ,θ ,θ 0,π,θ 0,π,θ 1 3 1 3 The condition is equivalent to α X + β Y ≥ 1. (85) θ θ ,θ ,θ θ θ ,θ ,θ 1 3 1 3 Notice that because of the choice of angular coordinates (44) the range of θ and θ 1 3 are related to θ,but θ should be chosen independently of θ and θ since we need it 1 3 to parametrize the whole of the lower envelope. The procedure to prove (85) can be divided into two steps. ˜ ˜ In the ﬁrst step (Step 1) we ﬁnd a speciﬁc θ such that α ≥ α(θ , θ) and β ≥ θ 1 θ β(θ , θ). This step can be viewed as a movement from a point on the lower envelope 123 The Prime Function, the Fay Trisecant Identity... 729 (the set of 3 blue curves in Fig. 8) to a point on another curve (the red curve in Fig. 8) on which the point (X , Y ) lies. θ ,θ ,θ θ ,θ ,θ 1 3 1 3 In the next step (Step 2) we make a use of a log-sum and sum-log inequality to prove that ˜ ˜ α(θ , θ)X + β(θ , θ)Y ≥ 1. (86) 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 First let us explain Step 1. For an arbitrary value of θ with 0 <θ <π, there are three possible cases for the set (θ ,θ ,θ ). The three cases are illustrated in Fig. 8: 1 3 they can be understood as corresponding to the lower envelope being split into three parts: the three blue curve segments sitting on the lower envelope in Fig. 8 as shown in top, middle, and bottom of this ﬁgure. These will be called cases 1, 2a and 2b, respectively. The red solid curve on each Fig. 8 is drawn by changing θ from |θ | to π −|π − θ | while θ and θ are ﬁxed. The range of θ comes from the condition of 3 1 3 angular coordinates (44). The reason for the separation into three cases is that when θ < |θ | or π −|π − θ | <θ , an auxiliary point (X , Y ) does not lie on 1 3 θ ,θ ,θ θ ,θ ,θ 1 3 1 3 the red curve. Based on the argument above, we therefore assign a value θ to a value of θ according to the following three conditions: – Case 1: When |θ |≤ θ ≤ π −|π − θ |, which means that the point (X , Y ) 1 3 θ θ marked as “+” in the top right of Fig. 8 lies on the blue solid curve, an auxiliary point (X , Y ) also lies on the red curve marked as “x”. From the behavior θ ,θ ,θ θ ,θ ,θ 1 3 1 3 of g (η), we can see that α ≥ α(θ ,θ ) and β ≥ β(θ ,θ ). In this case, we θ θ 1 θ 3 choose θ = θ . – Case 2a: When θ < |θ | as shown in the middle of Fig. 8, an auxiliary point (X , Y ) does not lie on the red curve. In this case, the end point of the red θ ,θ ,θ θ ,θ ,θ 1 3 1 3 curve is chosen, that is, we make the choice θ =|θ |. From the behavior of g (η),it 1 θ is apparent that α ≥ α(θ , |θ |) = 1. Because of the angular condition (44) such θ 1 1 that |θ | <θ < 2π −|θ | and because β is a monotonically decreasing function 1 3 1 θ with respect to θ for 0 <θ <π (Appendix B), β >β = β(π, |θ |) ≥ θ |θ | 1 β(θ , |θ |). 3 1 – Case 2b: This is the same as case 2a but refers to the other end of the red curve. When π −|π −θ | <θ as shown in the bottom of Fig. 8, a point (X , Y ) θ ,θ ,θ θ ,θ ,θ 1 3 1 3 does not lie on the red curve (cf: Case 2a). In this case, the opposite boundary point of the red curve is now chosen, which means that we choose θ = π −|π −θ |.From the behavior of g (η), it is apparent that β ≥ β(θ ,π −|π − θ |) = 1. Because θ 3 3 π −|π −θ | <θ , |θ | <π −|π −θ |, and because α is a monotonically increasing 3 1 3 θ function with respect to θ for 0 <θ <π (Appendix B), α >α ≥ θ π −|π −θ | α(θ ,π −|π − θ |). 1 3 ˜ ˜ In all cases, it is possible to ﬁnd a speciﬁc θ such that α ≥ α(θ , θ) and β ≥ θ θ β(θ , θ). Thus, we obtain the important result that ˜ ˜ α X + β Y ≥ α(θ , θ)X + β(θ , θ)Y . (87) θ θ ,θ ,θ θ θ ,θ ,θ 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 1 3 1 3 This facilitates Step 2 where we use a log-sum and sum-log inequality for the right hand side of (87)asfollows: 123 730 H. Miyoshi et al. Fig. 8 Three possible cases for the set (θ ,θ ,θ ).(i) Case 1: α ≥ α(θ ,θ ) and β ≥ β(θ ,θ ). (ii) Case 1 3 θ 1 θ 3 2a: Because |θ | >θ , we choose θ =|θ |. We can see that α ≥ α(θ , |θ |) = 1and β >β(θ , |θ |). (iii) 1 1 1 1 3 1 θ θ Case 2b: Because |π −θ | >π −θ , we choose θ = π −|π −θ |. We can see that α ≥ α(θ ,π −|π −θ |) 3 3 1 3 and β >β(θ ,π −|π − θ |) = 1 θ 3 3 ˜ ˜ log α(θ , θ)X + β(θ , θ)Y 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 ⎡ ⎤ 2π iφ iφ ⎣ ˜ ˜ ⎦ = log ( A(θ , θ, ρe )X + B(θ , θ, ρe )Y )dφ 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 2π 123 The Prime Function, the Fay Trisecant Identity... 731 2π iφ iφ ˜ ˜ ≥ log A(θ , θ, ρe )X + B(θ , θ, ρe )Y dφ. (88) 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 2π However the right hand side can be written 2π 2π iφ 1 B(θ , θ, ρe ) 1 iφ log X + Y dφ + log A(θ , θ, ρe )dφ θ ,θ ,θ θ ,θ ,θ 1 1 3 1 3 iφ 2π 2π A(θ , θ, ρe ) 0 0 2π 2π iφ 1 B(θ ,θ,ρe ) 1 iφ = log X + Y dφ + log A(θ ,θ,ρe )dφ θ ,θ ,θ θ ,θ ,θ 1 1 3 1 3 iφ 2π A(θ ,θ,ρe ) 2π 0 0 2π iφ iφ = log A(θ ,θ,ρe )X + B(θ ,θ,ρe )Y dφ = 0, 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 2π (89) iφ where we used the fact that the integral of log A(θ ,θ,ρe ) is 2π 2π i(φ−θ ) i(φ+θ ) 1 1 1 1 P(ρe ) P(ρe ) iŒ log A(θ ,θ,ρe )dφ = log dφ i(φ−θ) i(φ+θ) 2π 2π P(ρe ) P(ρe ) 0 0 2π i(φ−θ ) i(φ+θ ) 1 1 = [log P(ρe ) + log P(ρe ) 2π i(φ−θ) i(φ+θ) − log P(ρe ) − log P(ρe )]dφ = 0 (90) and the fact that iφ iφ i(φ−θ ) i(φ+θ ) ˜ 3 3 B(θ , θ, ρe ) B(θ ,θ,ρe ) P(ρe ) P(ρe ) 1 1 = = (91) iφ i(φ−θ ) i(φ+θ ) ˜ iφ 1 1 A(θ ,θ,ρe ) P(ρe ) P(ρe ) A(θ , θ, ρe ) is independent of θ and θ. In the last line, we have used the Fay trisecant identity (45). Putting all this together it has been shown that ˜ ˜ log α(θ , θ)X + β(θ , θ)Y ≥ 0. (92) 1 θ ,θ ,θ 3 θ ,θ ,θ 1 3 1 3 From (92) it follows that α X + β Y ≥ 1, (93) θ θ ,θ ,θ θ θ ,θ ,θ 1 3 1 3 which means for 0 <θ <π, α X + β Y ≥ 1 for all (X , Y ). We have proved all the θ θ conjectured features of the envelope structure observed by previous authors. 123 732 H. Miyoshi et al. We now summarize steps 1 and 2 geometrically with the aid of Fig. 8.Instep1 we found a choice of an auxiliary tangent line (the red line) to the pink curve on which (X , Y ) sits (and which corresponds to a ﬁxed θ and θ ) above the θ ,θ ,θ θ ,θ ,θ 1 3 1 3 1 3 tangential line on the lower envelope (the blue line) for the three segments of the lower envelope into which it naturally divides for any ﬁxed θ and θ (i.e., cases 1, 2a and 2b). 1 3 In step 2 the point (X , Y ) is then shown to be above that red tangent line θ ,θ ,θ θ ,θ ,θ 1 3 1 3 by using the log-sum and sum-log inequalities. We thus prove that (X , Y ) θ ,θ ,θ θ ,θ ,θ 1 3 1 3 is above any point on the lower envelope. 8 Discussion By introducing the prime function on the Schottky double of a conformally equivalent concentric annulus to any given doubly connected sample with a single uncharged hole, and by using an integrated form of an associated Fay trisecant identity expressed in terms of that prime function, we have shown how to prove some conjectures made in the literature on an observed envelope structure associated with pairs of 4-point resistance measurements in a van der Pauw procedure. More practical implications of our results are discussed elsewhere [19]. The van der Pauw method for a sample with two or more holes is, of course, of interest and several studies have recently appeared [9,11]. The theory of the prime function on domains of general ﬁnite connectivity has been expounded in a recent monograph [18] and the relevance of the prime function to expressing the complex potential for source/sink ﬂows in such domains is already known [23]. At the same time, the Fay trisecant identity also carries over to higher genus Riemann surfaces, where the Fay identity can be written down in terms of the so-called prime form on those surfaces and associated theta functions [24]. Therefore, while we have focused here on what is the simplest non-trivial case of a holey sample, we expect all the ideas to be extendible to the case of a sample with any ﬁnite number of holes. The details, however, remain to be worked out and constitute an interesting future challenge. 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To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. 123 The Prime Function, the Fay Trisecant Identity... 733 Appendix A The Positiveness of L(e ) for 0 < < 2 First, we consider the derivative of L(ζ ) deﬁned in (20)asfollows: ∂ L M (ζ ) ≡ ζ . (94) ∂ζ Because we have another representation of L(μ/ν) − L(μν) as stated in (70), we can consider a limit of each side: L(μ/ν) − L(μν) 1 L(μ/ν) − L(μ) L(μ) − L(μν) lim = μ lim + = 2M (μ). ν→1 1 − ν ν→1 ν μ/ν − μ μ − μν (95) The right hand side is, from the deﬁnition in (18), 2 2 2 3 2 ˆ ˆ 1 P(1) P(μ ) P(ν ) P(1) P(μ ) lim = 2 . (96) 2 3 ν→1 1 − ν P(ν/μ) P(μ/ν) P(μν) P(1/μ) P(μ) We therefore ﬁnd an alternative representation of M (ζ ): 3 2 P(1) P(ζ ) M (ζ ) = . (97) P(1/ζ ) P(ζ ) iφ Consequently M (e ) is found to be 3 2iφ P(1) P(e ) i sin φ iφ M (e ) = = C (φ) , (98) iφ iφ 3 P(1/e ) P(e ) 8 sin where 4n 2n (1 + ρ − 2ρ cos 2φ) C (φ) = P(1) > 0. (99) 4n 2n 4 (1 + ρ − 2ρ cos φ) n=1 iφ iφ iφ Therefore, because ∂ L/∂φ (e ) = iM (e ), L(e ) decreases when 0 <φ <π, and iφ iφ iπ L(e ) increases when π< φ < 2π. The minimum of L(e ) is L(e ) = L(−1) where L(−1) can be calculated explicitly to be 2n 1 ρ L(−1) = + 2 > 0. (100) 2n 2 4 (1 + ρ ) n=1 iφ Therefore L(e ) is positive over this range of φ. 123 734 H. Miyoshi et al. B Proof of the Properties of ˛ and ˇ Here we will prove that for 0 <θ <π, α is a monotonically increasing function and β is a monotonically decreasing function with respect to θ, that is, ∂α ∂β θ θ > 0, < 0. (101) ∂θ ∂θ We will prove it using the integrated Fay trisecant identity. From (52), ∂α ∂β θ θ X + Y = 0, (102) 0,π,θ 0,π,θ ∂θ ∂θ which means the sign of ∂α /∂θ is opposite to ∂β /∂θ. By a log-sum and sum-log θ θ inequality, log[α X + β Y ] θ 0,π,θ θ 0,π,θ 2π iφ iφ ≥ log[ A(0,θ ,ρe )X + B(π, θ ,ρe )Y ]dφ 0,π,θ 0,π,θ 2π 2π iφ iφ = log[ A(0,θ,ρe )X + B(π,θ,ρe )Y ]dφ = 0, 0,π,θ 0,π,θ 2π (103) whereweused(90) and (91). This means α X + β Y ≥ 1. (104) θ 0,π,θ θ 0,π,θ From the Fay trisecant identity, α X + β Y = 1, so θ 0,π,θ θ 0,π,θ (α − α )X + (β − β )Y ≥ 0 (105) θ θ 0,π,θ θ θ 0,π,θ for 0 <θ <π. Hence, together with (102), we can obtain an inequality for the second derivatives of α and β as follows: θ θ 2 2 ∂ α ∂ β θ θ X + Y ≥ 0. (106) 0,π,θ 0,π,θ 2 2 ∂θ ∂θ When we take the derivative of (102) with respect to θ, we get 2 2 ∂α ∂ X ∂β ∂Y ∂ α ∂ β θ 0,π,θ θ 0,π,θ θ θ + + X + Y = 0. (107) 0,π,θ 0,π,θ 2 2 ∂θ ∂θ ∂θ ∂θ ∂θ ∂θ Thus, from (106), 2 2 ∂α ∂ X ∂β ∂Y ∂ α ∂ β θ 0,π,θ θ 0,π,θ θ θ + =− X − Y ≤ 0. (108) 0,π,θ 0,π,θ 2 2 ∂θ ∂θ ∂θ ∂θ ∂θ ∂θ 123 The Prime Function, the Fay Trisecant Identity... 735 The signs of ∂ X /∂θ and ∂Y /∂θ can be determined for 0 <θ <π, that is, 0,π,θ 0,π,θ ∂ X 0,π,θ iθ −iθ iθ = iX [K (−e ) − K (−e )]= iX [2K (−e ) − 1] < 0, 0,π,θ 0,π,θ ∂θ (109) ∂Y 0,π,θ iθ −iθ iθ = iY [K (e ) − K (e )]= iY [2K (e ) − 1] > 0, 0,π,θ 0,π,θ ∂θ iφ iφ where we used the positiveness of L(e ) to state that the imaginary part of K (e ) is iπ monotonically increasing and K (e ) = 1/2. Because the sign of ∂α /∂θ is opposite to ∂β /∂θ as deduced from (102), we can conclude that ∂α ∂β θ θ > 0, < 0. (110) ∂θ ∂θ References 1. Miccoli, I., Edler, F., Pfnür, H., Tegenkamp, C.: 2015 The 100th anniversary of the four-point probe technique: the role of probe geometries in isotropic and anisotropic systems. J. Phys. Condens. Matter 27, 223201 2. Singh, Y.: 2013 Electrical resistivity measurements: a review. In: Int. J. Mod. Phys. Conf. Ser., vol. 22, pp. 745–756. World Scientiﬁc 3. 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Computational Methods and Function Theory – Springer Journals
Published: Dec 1, 2021
Keywords: Cross ratio; Prime function; Fay trisecant identity; van der Pauw; 30Cxx; 14Qxx; 31Axx
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