Combinatorial Functional Equations: Advanced Theory 9783110627336, 9783110624359

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Yanpei Liu Combinatorial Functional Equations

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De Gruyter Series in Nonlinear Analysis and Applications Jürgen Appell et al. (Eds.) ISSN 0941-813X

Yanpei Liu

Combinatorial Functional Equations |

Volume 2: Advanced Theory

Author Prof. Yanpei Liu Dept. of Mathematics School of Science Beijing Jiaotong University 3 Shangyuancun Haidian District 100044 Beijing People’s Republic of China [email protected]

ISBN 978-3-11-062435-9 e-ISBN (PDF) 978-3-11-062733-6 e-ISBN (EPUB) 978-3-11-062480-9 Library of Congress Control Number: 2019951713 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2020 Walter de Gruyter GmbH, Berlin/Boston Cover image: Roberto Pangiarella/EyeEm/Getty Images Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

Preface This is Combinatorial Functional Equations II—Advanced Theory. We shall refer to it as Book II (or Volume 2). By advanced is meant equations such that all of them have the meson functional and without the specific case that is meaningful for either trees or near-trees on the plane because of higher complexity. On meson functional equations, my friends domestic and abroad used to tell me that they “look very simple in form and yet there is no way to solve them.” However, this book presents a universal approach to establishing a qualitative theory and to evaluating their solutions in the form of a recursion to an explicision (i. e., explicit expression) in an algebraic manner for passing three stages: systematization in theory, efficientization in running and intelligentization in usage. From the procedure used, an equation established for a set of maps considered with a given vertex or face partition vector always has a solution. However, any of its solutions is not always meaningful for the given set of maps. Attention should be paid to the condition for the well-definedness. In 1999, the first monograph Enumerative Theory of Maps (see Liu YP [56]) was published by Kluwer (Springer) and Science. A great number of decompositions for a variety of map sets, planar mainly and only a few nonplanar, were discovered for transformations on the extension of integral domain. Then in General Theory of Map Census (see Liu YP [61]) published by Science in 2009, nonplanar maps were extensively considered. However, none of them reaches a favorite explicision in the form of a finite sum with all terms positive, or summationfree, except for the two most simple cases of tree, or tree-like equations, shown in Chapter 8 through Chapter 10 of Book I (or Volume 1). In this book (i. e., Book II), the equations considered with all coefficient parameters and initial values constants are divided into three classes, based on three levels of our research on the specific case in this aspect. All are shown to be well-defined, i. e., we have existence and uniqueness of a solution, provided certain conditions are satisfied. All meson functional equations considered in this book are divided into three parts. Part one is of outer type. Four classic models are called the crackless outer model and superwheel outer model in Chapter 1; and the restricted outer model and ordinary outer model in Chapter 2. All solutions are extracted in an explicit form with each coefficient a finite sum with all terms positive, or one term, involving powers of a matrix of finite dimension for some suitable parameter given. Part two is of inner type. Eight classic models, called the general ordinary inner model and Eulerian ordinary inner model; the general loopless inner model and Eulerian loopless inner model; the general cutless inner model and Eulerian cutless inner model; the general simple inner model and bipartite simple inner model are, respectively, discussed in Chapter 3; Chapter 4; Chapter 5, and Chapter 6. All of their soluhttps://doi.org/10.1515/9783110627336-201

VI | Preface tions evaluated have a recursion in the form of a finite sum with all terms positive. For the specific case related to planar maps, an explicision has been extracted indirectly via enumeration of certain planar maps involving with another finite parameter vector introduced. They are all suitable for efficientization in running on computers and further for intelligentization in usages. Part three is of surface type. Five classic models called join-end surface model, bridgeless surface model, loopless surface model, Eulerian surface model and ordinary surface model are, respectively, discussed in Chapter 7, Chapter 8, Chapter 9, Chapter 10 and Chapter 11. All of their solutions have a recursion in the form of a finite sum with all terms positive. Moreover, their explicisions for the specific case of maps on surfaces are indirectly found to not make it necessary to introduce an additional parameter vector as in part two. In part three, all chapters except only for Chapter 9 have found a direct explicision via powers of a matrix. Particularly, the result in Chapter 11 enables us to establish a new method for extracting a genus series of combinatorial maps. In this manner, all solutions in this part enable us to get their genus series of corresponding maps. Before this book, only the genus series of ordinary maps with some of its relatives had been found: by employing the Young tableau in algebra as shown in Jackson DM [19] and Jackson DM, Visentin TI [20]. However, it does not seem to be a suitable idea yet to attach the other four types of maps for genus series in this manner. Although all restriction sections are meaningful in maps, the procedures accepted for theoretical usages in context are completely independent of their originals. However, the examples presented with figures of maps serve us for checking the correctness of pure theoretical results. Although all equations have combinatorial background as a specific case or one of their applications, the basic theoretical principles presented here can be found in pure mathematics, independent of combinatorics, graphs, hypergraphs, lattices, networks, matroids and particularly maps. For each meson functional equation, from certain conditions, a number of function equations, difference(straight and slope) equations, and differential (ordinary and partial) equations are also involved with an application for classifying varieties of combinatorial objects. From these, explicit expressions of the solution of corresponding meson equations are indirectly extracted in each of all chapters from Chapter 3 through Chapter 8 in Book I. Another important application besides maps, or the specific case of meson functionals, I have to mention is the shadow functional which is on function space in its own right going from the basis {xi | ∀i ≥ 1} to the basis of {(x)i | ∀i ≥ 1} where (x)i = ∏i−1 j=0 (x − j) for i ≥ 1 having a connection to combinatorial geometry, number theory and abstract algebras such as the Hopf algebra. This branch is due to Rota’s group such as in Rota GC [86], Rota GC, Taylor BD [87]. An attempt has been made to keep the presentation as self-contained as possible. Introduction is for briefly describing all the meson functional equations considered

Preface | VII

in the whole volume. Chapter 1 and Chapter 2 in Book I are on basic concepts and theoretical background for usage in the whole of this book as well. There are two more things I have to mention on the contents of this book. One is that Sections 3.3 through 11.3 and 3.7 through 6.7 reflect novel progress in quantitative theory. The other is that Sections 1.4 through 11.4 and 1.8 through 6.8 are completely different from the originals in reasoning without consideration of corresponding combinatorial maps. On this occasion, I would like to express my heartiest appreciation to all individuals who made contributions to this book, particularly to Junliang Cai, Rongxia Hao, Zhaoxiang Li, Erling Wei and Liangxia Wan for carefully reading the manuscript to point out errors and mistakes suggesting corrections. Of course, any error or mistake remaining is due to myself. Daotiancun, Beijing July, 2019

Y. P. Liu

Contents Preface | V Introduction | XIII 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

Outer equations first part | 1 Crackless outer model | 1 Solution crackless outer case | 3 Explicision crackless outer case | 8 Restrictions crackless outer case | 10 Super-wheels outer model | 12 Solution super-wheels outer case | 15 Explicision super-wheels outer case | 19 Restrictions super-wheels outer case | 23 Notes | 26

2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9

Outer equations second part | 31 Restricted outer model | 31 Solution restricted outer case | 33 Explicisions restricted outer case | 36 Restrictions restricted outer case | 40 Ordinary outer model | 43 Solution ordinary outer case | 45 Explicisions ordinary outer case | 48 Restrictions ordinary outer case | 50 Notes | 54

3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Inner equations first part | 57 General ordinary inner model | 57 Solution general ordinary inner case | 60 Explicitness general ordinary inner case | 68 Restrictions general ordinary inner case | 73 Eulerian ordinary inner model | 83 Solution Eulerian ordinary inner case | 87 Explicitness Eulerian ordinary inner case | 91 Restrictions Eulerian ordinary inner case | 96 Notes | 103

4 4.1 4.2

Inner equations second part | 105 General loopless inner model | 105 Solution general loopless inner case | 109

X | Contents 4.3 4.4 4.5 4.6 4.7 4.8 4.9

Explicitness general loopless inner case | 113 Restrictions general loopless inner case | 118 Eulerian loopless inner model | 126 Solution Eulerian loopless inner case | 133 Explicitness Eulerian loopless inner case | 134 Restrictions Eulerian loopless inner case | 139 Notes | 146

5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

Inner equations third part | 149 General cutless inner model | 149 Solution general cutless inner case | 155 Explicitness general cutless inner case | 160 Restrictions general cutless inner case | 164 Eulerian cutless inner model | 173 Solution Eulerian cutless inner case | 179 Explicitness Eulerian cutless inner case | 182 Restrictions Eulerian cutless inner case | 185 Notes | 192

6 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

Inner equations fourth part | 197 General simple inner model | 197 Solution general simple inner case | 201 Explicitness general simple inner case | 207 Restrictions general simple inner case | 212 Bipartite simple inner model | 218 Solution bipartite simple inner case | 224 Explicitness bipartite simple inner case | 225 Restrictions bipartite simple inner case | 230 Notes | 237

7 7.1 7.2 7.3 7.4 7.5

Surface equations first part | 241 Join-end surface model | 241 Solution join-end surface | 246 Explicitness join-end surface | 250 Restrictions join-end surface | 252 Notes | 259

8 8.1 8.2 8.3 8.4

Surface equations second part | 263 Bridgeless surface model | 263 Solution bridgeless surface | 267 Explicitness bridgeless surface | 270 Restrictions bridgeless surface | 272

Contents | XI

8.5

Notes | 278

9 9.1 9.2 9.3 9.4 9.5

Surface equations third part | 283 Loopless surface model | 283 Solution loopless surface | 286 Explicitness loopless surface | 292 Restrictions loopless surface | 295 Notes | 301

10 10.1 10.2 10.3 10.4 10.5

Surface equations fourth part | 305 Eulerian surface model | 305 Solution Eulerian surface | 310 Explicitness Eulerian surface | 313 Restrictions Eulerian surface | 315 Notes | 321

11 11.1 11.2 11.3 11.4 11.5

Surface equations fifth part | 325 Ordinary surface model | 325 Solution ordinary surface | 330 Explicitness ordinary surface | 336 Restrictions ordinary surface | 338 Notes | 344

Bibliography | 349 Index | 353

Introduction This book, Combinatorial Functional Equations II—Advanced Theory (Book II, or Volume 2), contains eleven chapters: Chapter 1 through Chapter 11. All equations considered are involved with the meson functional. One might prefer to call them advanced on comparing with those called basic in Chapter 3 through Chapter 10 of Combinatorial Functional Equations I—Basic theory (Book I, or Volume 1). Equations of this book are listed in Introduction II for an overall view as well. There is a division into three parts, concerned with outer type, inner type and surface type. Part one, on outer type, consists of Chapter 1 and Chapter 2 for four models of meson equations whose specific case is an application to combinatorial enumeration, particularly to combinatorial outer planar maps. In Chapter 1, two models of meson equations are covered. First, consider the meson equation for f ∈ ℛ{x, y} xy𝜕x,y f |x=u 2 { { {a2 ∫ 1 − xy = f − a1 x ; { { y { {f |x=0,y=0 = a0 ,

(1)

where a0 , a1 , a2 ∈ ℤ+ and f ∈ ℛ{x, y}. Because a solution of equation (1) when a0 = 0 and a1 = a2 = 1 is involved with outer-planar (outer) maps without crack vertex, this equation is called a crackless outer model, or in short, crackless outer. Second, consider the equation for f ∈ ℛ{x, y} as y 2 2 { {bx ∫ x − cyf = f − ax ; y { { f | { x=0⇒y=0 = d,

(2)

where a, b, c, d ∈ ℤ+ and y = (y2 , y3 , y4 , . . .). Because of the specific case when a = b = c = 1 and d = 0 determining the enufunction of asymmetric classes of super-wheels (non-separable outer planar maps in dual) with the root-vertex (root-face) valency and the vertex (face)-partition vector as parameters, by the equivalence between super-wheels the duals of non-separable outer planar maps, equation (2) is called a super-wheel model, or a non-separable outer model. In Chapter 2, two models of meson equations are covered. Third, consider the meson equation for f ∈ ℛ{x, y} 2 { {a2 x ∫ yδx,y (uf |x=u ) = (1 − a3 x )f − a1 ; y { { f | { x=0,y=0 = a0 ,

where a0 , a1 , a2 , a3 ∈ ℤ+ and f ∈ ℛ{x, y}. https://doi.org/10.1515/9783110627336-202

(3)

XIV | Introduction Because of the solution of equation (3) for a0 = a1 = a2 = a3 = 1 related to restricted outer planar maps shown in [36] (Liu YP, 1987), this equation is called a restricted outer model. Fourth, consider the meson equation for g ∈ ℛ{x, y} 2 { {a2 x ∫ yδx,y (ug|x=u ) = (1 − x φ(x))g − a1 ; y { { g| { x=0,y=0 = a0 ,

(4)

where a0 , a1 , a2 ∈ ℤ+ and φ(x) ∈ ℛ+ {x} such that, for any integer n ≥ 0, ϕm = 𝜕xm φ ∈ ℤ+ and ϕ2i+1 = 0, i ≥ 0. Because of the solution of equation (4) for a0 = a1 = a2 = 1 involved with ordinary outerplanar maps shown in [33] (Liu YP, 1986) and [36] (Liu YP, 1987), this equation is called an ordinary outer model. Part two, for inner type, consists of Chapter 3 through Chapter 6 for eight models of meson equations whose specific cases have applications to eight types of combinatorial enumeration, particularly of combinatorial planar (inner) maps. In Chapter 3, two models of meson equations are covered. First, consider the meson equation for g ∈ ℛ{x, y} 2 { {a2 x ∫ yδx,y (ug|x=u ) = (1 − x g)g − a1 ; y { { g| { y=0⇒x=0 = a0 ,

(5)

where a0 , a1 , a2 ∈ ℤ+ . Because of the solution of equation (5) for a0 = a1 = a2 = 1 involved with general ordinary planar maps shown in [57] (Liu YP, 1999, p. 204), and then [68] (Liu YP, 2013, p. 229), this equation is called a general ordinary inner equation. Second, consider the meson equation for f ∈ ℛ{x, y} 2 2 2 { {a2 x ∫ y δx2 ,y2 f |x2 =u = (1 − x f )f − a1 ; y { { f | { x=0⇒y=0 = a0 ,

(6)

where a0 , a1 , a2 ∈ ℤ+ and f = f (z, y) ∈ ℛ{z, y} for z = x2 . Because of the solution of equation (6) for a0 = a1 = a2 = 1 involved with Eulerian ordinary planar maps shown in [32] (Liu YP, 1986) and then [61] (Liu YP, 2009, p. 122), this equation is called an Eulerian ordinary inner equation. Chapter 4 covers two models of meson equations. Third, consider the meson equation for g ∈ ℛ+ {x, y} 1 { {a2 ∫ 1 − a 𝜕 (u2 g| ) = g; 1 x,y x=u (7) y { { g| = a , { x=0⇒y=0 0 where a0 , a1 , a2 ∈ ℤ+ , g ∈ ℛ+ {x, y} and 𝜕x,y is the slope difference operator.

Introduction

| XV

Because of the solution of equation (7) for a0 = a1 = a2 = 1 involved with general loopless planar maps shown in [61] (Liu YP, 2010, p. 179), this equation is called a general loopless inner equation. Fourth, we discuss the meson equation for f ∈ ℛ{x.y} 1 − a1 𝜕x2 ,y2 (uf |x2 =u ) { { = f; {a2 ∫ 1 − 2a1 𝜕x2 ,y2 (uf |x2 =u ) − x2 y2 δx2 ,y2 (uf |2x2 =u ) { { y { {f |x=0⇒y=0 = a0 ,

(8)

where a0 , a1 , a2 ∈ ℤ+ , f ∈ ℛ{x, y} and y = (y1 , y2 , y3 , . . .). When a0 = a1 = a2 , one might find the equation or its equivalent version in [32] (Liu YP, 1986) and [60] (Liu YP, 2008, p. 186). Because of the solution of equation (8) for a = b = c = 1 involved with Eulerian loopless planar maps, this is called an Eulerian loopless inner equation. In Chapter 5, two models of meson equations are covered. Fifth, consider the meson equation for f ∈ ℛ{x, y} y𝜕x,y f |x=u { { = f − a1 x2 ; {a2 x ∫ 1 − 𝜕 f | x,y x=u y { { { f | { x=0⇒y=0 = a0 ,

(9)

where a0 , a1 , a2 ∈ ℤ+ . Because of the solution of equation (9) for a = 0 and b = c = 1 involved with general cutless planar maps shown in [28] (Liu YP, 1985) and [56] (Liu YP, 1999, p. 187), this equation is then called a general cutless inner equation. Sixth, consider the meson equation for f ∈ ℛ{x, y} y2 δx2 ,y2 f |x2 =u 2 { { = f − a1 x 2 ; {a 2 x ∫ (1 − 𝜕x2 ,y2 f |x2 =u )2 − (xyδx2 ,y2 f |x2 =u )2 y { { { f | { x=0⇒y=0 = a0 ,

(10)

where a0 , a1 , a2 ∈ ℤ+ . Because of the solution of equation (10) for a0 = 0 and a1 = a2 = 1 involved with Eulerian cutless planar maps shown in [32] (Liu YP, 1986) and [60] (Liu YP, 2008, p. 165), this equation is called an Eulerian cutless inner equation. In Chapter 6, two models of meson equations are covered. Seventh, consider the meson equation for f ∈ ℛ{x, y} 2 2 { {a2 x ∫ yδx,y (uf |x=u ) = ∫((1 + xy)f − 1)f |x=y − a1 x f ; y y { { {f |x=0⇒y=0 = a0 (≠ 0),

where a0 , a1 , a2 ∈ ℤ+ .

(11)

XVI | Introduction Because of the solution of equation (11) for a0 = a1 = a2 = 1 involved with general simple planar maps shown in [47] (Liu YP, 1989) and [61] (Liu YP, 2009, p. 194), this equation is called a general simple inner equation. Eighth, consider the meson equation for f ∈ ℛ{x, y} 2 2 2 2 { {a2 x ∫ y δx2 ,y2 f |x2 =u = (f − 1) ∫ f |x=y − a1 x f ; y y { { {f |x=0⇒y=0 = a0 ,

(12)

where a0 , a1 , a2 ∈ ℤ+ . In [56] (Liu YP, 1999, p. 231) and [58] (Liu YP, 2001, p. 271), one might find the main equation of (12), or its equivalence when a0 = a1 = a2 = 1. However, attention should be paid to some errors in the symbols. Because of the solution of equation (12) for a0 = a1 = a2 = 1 involved with bipartite simple planar maps, this equation is called a bipartite simple inner equation. Part three, concerned with surface type equations, consists of Chapter 7 through Chapter 11 for five meson equations whose specific cases have applications to five types of combinatorial enumeration, particularly combinatorial maps on surfaces. In Chapter 7, one model of meson equations is covered. First, consider the meson equation for f ∈ ℛ{x, y} 2 3 𝜕f { { {a2 x ∫ y(1 + 𝜕x,y (f − 1)|x=u ) = (1 − x )f − x 𝜕x − a1 ; y { { { f { x=0,y=0 = a0 ,

(13)

where a0 , a1 , a2 ∈ ℤ+ are given. Because of the contribution from join-end maps on surfaces to the specific case of a0 = a1 = a2 = 1, i. e., 2 3 𝜕f { { {x ∫ y(1 + 𝜕x,y (f − 1)|x=u ) = (1 − x )f − x 𝜕x − 1; y { { { f { x=0,y=0 = 1,

(14)

equation (13) is called a join-end surface model. In Chapter 8, one model of meson equations is covered. Second, consider the meson equation for g ∈ ℛ{x, y} 2 3 𝜕g { { {a3 x ∫ y𝜕x,y (g − a0 )|x=u = (1 − x )g − a2 x 𝜕x − a1 ; y { { { g { x=0,y=0 = a0 ,

where a0 , a1 , a2 , a3 ∈ ℤ+ .

(15)

Introduction

| XVII

Because of the contribution from maps without cut-edge (i. e., bridge) on surfaces to the case of a0 = a1 = a2 = a3 = 1, i. e., 2 3 𝜕g { { {x ∫ y𝜕x,y (g − 1)|x=u = (1 − x )g − x 𝜕x − 1; y { { { g { x=0,y=0 = 1,

(16)

equation (15) is called a bridgeless surface model. In Chapter 9, one model of meson equations is covered. Third, consider the meson equation for f ∈ ℛ{x, y} 𝜕f |yi =yi 2 { { {a2 x ∑ ∫ y 𝜕y = f − a1 − a3 xf ∫ yf |x=y ; i≥1 y y { { { {f |x=0⇒y=0 = a0 ,

(17)

where a0 , a1 , a2 , a3 ∈ ℤ+ and y = (y1 , y2 , y3 , . . .). Because of the contribution from maps without selfloop on surfaces to the case of a0 = a1 = a2 = a3 = 1, i. e., 𝜕f |yi =yi 2 { { {x ∑ ∫ y 𝜕y = f − 1 − xf ∫ yf |x=y ; i≥1 y y { { { f | = 1, { x=0⇒y=0

(18)

which is found in [69](Liu YP, 2015, p. 398), equation (17) is called a loopless surface model. In Chapter 10, one model of meson equations is covered. Fourth, consider the meson equation for g ∈ ℛ{x, y} 2 2 4 𝜕g 2 { { {a2 x ∫ y δx2 ,y2 g|x2 =u = −a1 + (1 − x )g − 2a3 x 𝜕x 2 ; y { { { g| { x=0,y=0 = a0 ,

(19)

where a0 , a1 , a2 , a3 ∈ ℤ+ . Because of the contribution from Eulerian maps on surfaces to the case of a0 = a1 = a2 = a3 = 1, i. e., 2 2 4 𝜕g 2 { { {x ∫ y δx2 ,y2 g|x2 =u = −1 + (1 − x )g − 2x 𝜕x 2 ; y { { { g| = 1, x=0,y=0 {

(20)

found in [60](Liu YP,2008, p. 152), or [61](Liu YP, 2009, p. 142), equation (19) is called a Eulerian surface model.

XVIII | Introduction In Chapter 11, one model of meson equations is covered. Fifth, consider the equation for f ∈ ℛ{x, y} 2 3 𝜕f { { {a2 x ∫ yδx,y (uf |x=u ) = −a1 + (1 − x )f − a3 x 𝜕x ; y { { { f { x=0, y=0 = a0 ,

(21)

where a0 , a1 , a2 , a3 ∈ ℤ+ . Because of the contribution from ordinary maps on surfaces to the case of a0 = a1 = a2 = a3 = 1, i. e., 2 3 𝜕f { { {x ∫ yδx,y (uf |x=u ) = −1 + (1 − x )f − x 𝜕x ; y { { { f { x=0, y=0 = 1,

(22)

found in [67] (Liu YP, 2012), equation (21) is called an ordinary surface model. Note that on the basis of the whole qualitative theory established in this book all ℤ+ in the initiation of every meson functional equation are replaced by ℚ+ , or even ℝ+ , without loss of validity.

1 Outer equations first part 1.1 Crackless outer model Consider the equation for f xy𝜕x,y f |x=u 2 { { {a2 ∫ 1 − xy = f − a1 x ; { { y { {f |x=0,y=0 = a0 ,

(1.1.1)

where a0 , a1 , a2 ∈ ℤ+ and f ∈ ℛ{x, y}. This is equation (1) in Introduction. For a solution of equation (1.1.1) when a0 = 0 and a1 = a2 = 1 are involved with outerplanar maps without crack vertex, this equation is called a crackless outer model. By doing equivalent transformations for the part under the meson functional in equation (1.1.1) on ℛ{x, y}, the term is expanded. Let f = f (x) = ∑ Fa[j] xj

(1.1.2)

j≥0

where Fa[j] ∈ ℛ{y} and a = (a0 , a1 , a2 ). Because 𝜕x,y (f |x=u ) =

yf − xf |x=y x−y

= ∑ ∑ Fa[j] xk+1 yj−k−1 , k≥0 j≥k+1

we have xy𝜕x,y (f |x=u ) 1 − xy

=

Fa[j] xi yi +j−2k−2 , 󸀠

∑ i󸀠 ≥k+2,j≥k+1 k≥0

󸀠

and hence ∫ y

xy𝜕x,y (f |x=u ) 1 − xy

= ∑( ∑ + ∑ )Fa[j] yj+i−2k−2 xi . i≥2

0≤k≤j−2 1≤j≤i−1

0≤k≤i−2 j≥i

(1.1.3)

By substituting (1.1.3) for the part in (1.1.1), we have f = a1 x2 + a2 ∑ ( ∑ + ∑ )Fa[j] ym+j−2k−2 xm . m≥2

0≤k≤j−2 1≤j≤m−1

0≤k≤m−2 j≥m

(1.1.4)

Observation 1.1.1. If a0 ≠ 0, then equation (1.1.1) is not consistent. Proof. Since x is a factor on the right hand side of equation (1.1.4), we have f |x=0⇒y=0 = Fa[0] = 0. This should be the initial condition of equation (1.1.1), a contradiction to the given condition a0 ≠ 0. https://doi.org/10.1515/9783110627336-001

2 | 1 Outer equations first part This observation shows that if f is a solution of equation (1.1.1), then Fa[0] = 𝜕x0 f = 0 and hence a0 = 0. In what follows, whenever equation (1.1.1) occurs, a0 = 0 is always assumed. Thus, a is (a1 , a2 ) instead of (a0 , a1 , a2 ). Observation 1.1.2. If f is a solution of equation (1.1.1), then Fa[1] = 𝜕x1 f = 0. Proof. Since there is a common factor x2 on the right hand side of (1.1.4), the observation is true. The two observations above enable us only to determine Fa[m] , m ≥ 2, for getting a solution f of equation (1.1.1). On the basis of equation (1.1.4), one sees that + ∑ 0≤k≤m−2 )ym+j−2k−2 Fa[j] , {a1 + a2 (∑ 0≤k≤j−2 j≥m+1 2≤j≤m Fa[m] = { a2 (∑ 0≤k≤j−2 + ∑ 0≤k≤m−2 )ym+j−2k−2 Fa[j] , { j≥m+1 2≤j≤m

when m = 2; when m ≥ 3.

(1.1.5)

Lemma 1.1.3. For any integer m ≥ 2, Fa[m] is independent of y1 . Proof. In (1.1.5), m ≥ 2 and j ≥ 2. When m = 2 and j = 2, then k = 0. It is easily seen that j + m − 2k − 1 ≥ 2 + 2 − 0 − 1 ≥ 2. This implies that Fa[m] can only depend on yl , l ≥ 2. This lemma enables us to discuss only y = (y2 , y3 , y4 , . . .). For m = 2, let us observe the first equality of (1.1.5). In the former ∑, m = 2 󳨐⇒ j = 2 󳨐⇒ k = m − 2 = 0. In the latter ∑, k = m − 2 = 0 with Lemma 1.1.3 leads to Fa[2] = a1 + a2 y2 Fa[2] + a2 ∑ yj Fa[j] . j≥3

(1.1.6)

Thus, equation (1.1.5) becomes when m = 2; {a1 + a2 ∑j≥2 yj Fa[j] , Fa[m] = { a (∑ 0≤k≤j−2 + ∑ 0≤k≤m−2 )ym+j−2k−2 Fa[j] , when m ≥ 3. j≥m+1 2≤j≤m { 2

(1.1.7)

Lemma 1.1.4. Equation (1.1.1) for a0 = 0 is equivalent to fT = a2 Ya[clo] fT + a1 eT1

(1.1.8)

where f = (Fa[2] , Fa[3] , Fa[4] , . . .), Ya[clo] = (yi,j )i,j≥1 such that for t ≥ 1, ∑t−1 yt+1+2k , { { { k=0 yi,j = { (i, j) = (1, t) + k(1, 1), when j ≥ i; { { when j < i. {yj,i

(1.1.9)

1.2 Solution crackless outer case

| 3

Proof. For a solution of equation (1.1.1), the above procedure shows that equation (1.1.8) is satisfied. Conversely, for a solution of equation (1.1.8), then f = xfT is seen to be a solution of equation (1.1.1) for a0 = 0. This is the conclusion. tion

On the basis of Lemma 1.1.4, by an equivalent transformation on ℛ{x, y}, the equa(I − a2 Ya[clo] )fT = a1 eT1

(1.1.10)

is extracted. Theorem 1.1.5. Equation (1.1.1) is well defined on ℛ{x, y} if, and only if, a0 = 0. Proof. Sufficiency. Because of the existence and the uniqueness of the inverse for coefficient matrix of equation (1.1.10), this equation and hence equation (1.1.1) are well defined. Necessity is from Observation 1.1.1. In fact, it is easily checked that (I − a2 Ya[clo] )−1 = ∑ (a2 Ya[clo] )i . i≥0

(1.1.11)

Therefore, the solution of equation (1.1.10) is in the form of an explicision as fT = ∑ a1 (a2 Ya[clo] )i eT1 . i≥0

(1.1.12)

1.2 Solution crackless outer case In order to get a favorite expression of the solution, we have to observe its inner constructions. For determining Fa[m] , m ≥ 2, its terms are partitioned into parts for each of which all terms have the same degree n = |n| of yn , denoted by Fa[m,n⟩ = ⟨Fa[m] ⟩n ∈ ℛ{y}, so that Fa[m] = ∑ Fa[m,n⟩ . n≥0

(1.2.1)

Observation 1.2.1. When n = 0. For any integer m ≥ 2, a1 , when m = 2; Fa[m,0⟩ = { 0, when m ≥ 3. Proof. This is a direct result of (1.1.7).

(1.2.2)

4 | 1 Outer equations first part This observation enables us only necessary to consider the case of n ≥ 1, because Fa[1] = 0, it is only necessary to observe Fa[m] for m ≥ 2. Now, let us see what happens for m = 2 and 3 on the basis of (1.1.7). For integer n ≥ 0,

Fa[2,n⟩

a1 , when n = 0; { { { { { {a1 a2 y2 , when n = 1; ={ 2 2 {a1 a y + a2 ∑ yj Fa[j,1] , when n = 2; { j≥3 2 2 { { { n n n−2 l {a1 a2 y2 + a2 ∑j≥3 ∑l=0 y2 yj Fa[j,n−l−1] , when n ≥ 3.

(1.2.3)

For m = 3, because of the former ∑ in the second equality of equation (1.1.7) being limited by m = 3 󳨐⇒ 2 ≤ j ≤ m = 3 󳨐⇒ 0 ≤ k ≤ j − 2 and the latter ∑ being limited by 0 ≤ k ≤ 1, Lemma 1.1.3 leads to Fa[3] = a2 y3 Fa[2] + a2 ∑(yj−1 + yj+1 )Fa[j] 󳨐⇒ j≥3

Fa[3]

yj−1 + yj+1 a2 y3 Fa[2] = + a2 ∑ F . 1 − y2 − y4 1 − y2 − y4 a[j] j≥4

(1.2.4)

On this basis, for integer n ≥ 0,

Fa[3,n⟩

0, { { { { { {a1 a2 y3 , = { n−1 k { {∑k=0 a2 y3 (y2 + y4 ) Fa[2,n−k−1⟩ { { { αj,k Fa[j,n−k−1⟩ , { + ∑ 0≤k≤n−2 j≥4

when n = 0; when n = 1;

(1.2.5)

when n ≥ 2,

where αj,k = a2 (yj−1 + yj+1 )(y2 + y4 )k . Consider the general case, i. e., m ≥ 4. For any integer n ≥ 1, from (1.1.7), Fa[m,n⟩ = a2 ( ∑ + ∑ )yj+m−2k−2 Fa[j,n−1⟩ . 0≤k≤j−2 2≤j≤m

0≤k≤m−2 j≥m+1

(1.2.6)

Lemma 1.2.2. For any integer m ≥ 2,

Fa[m,1⟩

a1 a2 y2 , { { { = {a1 a2 y3 , { { {a1 a2 ym ,

when m = 2; when m = 3; when m ≥ 4.

(1.2.7)

1.2 Solution crackless outer case

| 5

Proof. When m = 2 and 3, the conclusions are, respectively, drawn from (1.2.3) and (1.2.5) for the case of n = 1. When m ≥ 4, from (1.2.6), (11.2.2) leads to 0

Fa[m,1⟩ = a2 ∑ ym−2k Fa[2,0⟩ = a1 a2 ym . k=0

Therefore, the conclusion is obtained. This lemma combined with (1.2.2)–(1.2.6) shows that ⟨f⟩n = (Fa[2,n⟩ , Fa[3,n⟩ , Fa[4,n⟩ , . . .) can be determined by ⟨f⟩n−1 for n ≥ 0. Theorem 1.2.3. The solution fa[rlo] = xT of equation (1.1.1) for a0 = 0 is determined by

Fa[m,n⟩

{a1 (m = 2), 0(m ≥ 3), { { { { {a1 a2 ym (m ≥ 2), ={ {a2 (∑ 0≤k≤j−2 + ∑ 0≤k≤m−2 )yj+m−2k−2 Fa[j,n−1⟩ { { j≥m+1 2≤j≤m { { (m ≥ 2), {

when n = 0; when n = 1;

(1.2.8)

when n ≥ 2.

Proof. The case of n = 0 is from (1.2.2). The case of n = 1 is from (1.2.7). In general, for n ≥ 2, the conclusion can be drawn by (1.2.3)–(1.2.6). Although this theorem provides the solution fa[clo] of equation (1.1.1) for a0 = 0, each case of n ≥ 2 is involved with an infinite series which is not favorite for realization. This motivates us to seek a new parameter so that whenever it is given, it is only necessary to evaluate a polynomial instead of a power series. Let s be defined as s = ∑(j + 1)nj (= π(yn ), or π(n)) j≥1

(1.2.9)

for n occurring in Fa[m,n] . Hence, Fa[m,n] can be partitioned as (m,n) Fa[m,n⟩ = ∑ Pa(s) s≥0

(1.2.10)

(m,n) = Fa[m,n⟩ |π(n)=s ∈ ℛ{y}. where Pa(s) For a term yn of the function f in ℛ{x, y}, π(n) is called the in-size of the term. The maximum among all in-sizes of terms in a function f is called the in-size of f . In fact, when n = 0, for m ≥ 2, there is only one term in Fa[m,0⟩ , i. e., Fa[2,0⟩ = a1 , such that π(Fa[2,0⟩ ) = 0. Moreover, for any n ≥ 0 and n ≠ 0, we have s ≥ 2. This enables us to discuss only s ≥ 2 without loss of generality. (m,n) Lemma 1.2.4. For any integer n ≥ 1, s = m (mod 2) in Pa(s) .

6 | 1 Outer equations first part Proof. First, when n = 1. From (1.2.7), Fa[m,1⟩ = a1 a2 ym and hence π(Fa[m,1⟩ ) = m is allowed for m ≥ 2. Thus, s = m. Of course, s = m (mod 2). Then, for n ≥ 2. We proceed by induction on n. Assume π(Fa[m,l⟩ ) = m (mod 2) allowable for 1 ≤ l ≤ n − 1 and m ≥ 2. We prove π(Fa[m,n⟩ ) = m (mod 2) to be allowable. By (1.2.6), the inductive assumption results in π(Fa[m,n⟩ ) = 2j + m − 2k − 2 (mod 2) = m (mod 2). This is the conclusion. a(0,n) a(1,n) = 0 and Pa(s) Because Fa(0) = 0 and Fa(1) = 0, Pa(s) = 0 for s ≥ 0 and n ≥ 0. From (1.2.2), for m ≥ 2,

a1 , when m = 2; (m,0) ={ Pa(s) 0, when m ≥ 3. (m,n) , we are allowed to discuss m ≥ 2, n ≥ 1 and s ≥ 2 On this basis, in order to get Pa(s) (Lemma 1.1.3). Let ℒ(Fa[m,n⟩ ) = {n | ∀n occurs in Fa[m,n⟩ }.

Lemma 1.2.5. For any integer m ≥ 2, (m,n) } = m = π(Fa[m,1⟩ ). min{s | ∀Pa(s) n≥0

Proof. Since for any integer m ≥ 2 given, (m,n) } = min{π(n) | n ∈ ℒ(Fa[m,n⟩ ), |n| = n} min{s | Pa(s) n≥1

n≥0

= min{π(n) | n ∈ ℒ(Fa[m,n⟩ ), |n| = 1} n≥0

= π(Fm,1 ) = m, the conclusion can be drawn. This lemma tells us that if Pa(s) is the sum of all terms of f , each of which has π = s, then, for any m ≥ 2, Pa(s) =

⌊(s−2)/2⌋

∑ l=0

(s−2l,1+l) xs−2l Pa(s)

(1.2.11)

is a polynomial on ℛ{x, y}. Lemma 1.2.6. The polynomial Pa(s) is independent of yl for l ≥ s + 1. Proof. We proceed by contradiction. Assume Pa(s) depends on one yl , l ≥ s + 1, then there exists a term with yn , where y = (y1 , y2 , y3 , . . .) such that ns+1 > 0. This leads to π(n) = ∑ jnj = ∑ jnj + (s + 1)ns+1 ≥ s + 1 > s. j≥1

A contradiction to s = π(n).

j=s+1 ̸ j≥2

1.2 Solution crackless outer case

| 7

In order to avoid casual confusion between vector y and its power n in the subindices, the form used in this lemma is always adopted throughout this chapter without declaration. Then, observe how to determine Pa(s) ∈ ℛ{x, y} for s ≥ 2. Theorem 1.2.7. The solution fa[clo] of equation (1.1.1) is determined by the following polynomials on ℛ{x, y}:

Pa(l)

a1 a2 y2 x2 , when s = 2; { { { { 3 { {a1 a2 y3 x , when s = 3; ={ 4 2 2 2 { a1 a2 y4 x + a1 a2 y2 x , when s = 4; { { { { ⌊(s−2)/2⌋ s−2l (s−2l,1+l) s , when s ≥ 5, x Pa(s) {a1 a2 ys x + ∑l=1

(1.2.12)

for 0 ≤ l ≤ s. Proof. On the basis of (1.2.11), when s = 2, from l = 0, (1.2.3) results in (2,1) = a1 a2 y2 x2 . Pa(2) = x2 Pa(2)

(1.2.13)

When s = 3, from l = 0, (1.2.5) results in (3,1) = a1 a2 y3 x3 . Pa(3) = x3 Pa(3)

(1.2.14)

When s = 4, from l in between 0 and 1, (1.2.7) and (1.2.3) result in (2,2) (4,1) = a1 a2 y4 x4 + a1 a22 y22 x2 ; + x2 Pa(4) Pa(4) = x4 Pa(4)

(1.2.15)

When s = 5, from 0 ≤ l ≤ 1, (1.2.7) and (1.2.5) result in (5,1) (3,2) Pa(5) = x5 Pa(5) + x3 Pa(5) = a1 a2 y5 x5 + a1 a2 (1 + a2 )y2 y3 x3 ;

(1.2.16)

When s = 6, from l in between 0 and 2, (1.2.7) and (1.2.3) result in (4,2) (2,3) (4,2) (6,1) + a1 a32 y23 x2 = a1 a2 y6 x6 + x4 Pa[6] + x2 Pa(6) + x4 Pa(6) Pa(6) = x6 Pa(6)

(1.2.17)

(4,2) where Pa(6) = (Fa[4,2⟩ )6 . By (1.1.7),

Fa[4] = y4 Fa[2] + (y3 + y5 )Fa[3] + (y2 + y4 + y6 )Fa[4] + ∑(yj−2 + yj + yj+2 )Fa[j] ,

(1.2.18)

j≥5

i. e., (1.2.7) results in (4,2) = (y4 Fa[2,1] )6 + (y3 Fa[3,1] )6 + (y2 Fa[4,1] )6 = 2a1 a2 y2 y4 + a1 a2 y32 . Pa(6)

(1.2.19)

8 | 1 Outer equations first part Therefore, Pa(6) = a1 a2 y6 x6 + (2a1 a2 y2 y4 + a1 a2 y32 )x4 + a1 a32 y23 x2 .

(1.2.20)

(s−2l,1+l) determined by Fa[s−2l,1+l⟩ which In general, for any integer s ≥ 7, we have Pa(s) is known from Fa[s−2l,j⟩ for j ≤ l. By the principle of induction and the equivalence between equations (1.1.1) and (1.1.7), the conclusion can be drawn.

Although (1.2.12) shows the solution fa[clo] in the form of a finite sum of all terms positive in its own right, the parity of the parameter s should be considered. In order to avoid such sophistication, another parameter k = (s + m)/2 ≥ 0 (an integer from Lemma 1.2.4!) can be introduced so that the solution of equation (1.1.1) is expressed as fa[clo] = ∑ Qa(k) k≥0

(1.2.21)

where Qa(k) ∈ ℛ{x, y} can still be shown as a polynomial with each term positive. Furthermore, for any integer k ≥ 0, Qa(k) is determined by Qa(l) , l < k, via suitable Fa[m,n⟩ .

1.3 Explicision crackless outer case Now, let us go bach to equations (1.1.8)–(1.1.9), to evaluate Pa(s) , s ≥ 2, for getting an explicision of the solution of equation (1.1.1) for a0 = 0. Observation 1.3.1. For an integer s ≥ 2 given, polynomial Pa(s) comes with the maximum degree s of x. Proof. On the basis of (1.2.11) and (1.2.12), by induction, the conclusion can be drawn. This observation enables us only to consider fm−1 = (Fa[2] , Fa[3] , . . . , Fa[m] ) instead of f in (1.1.8). Observation 1.3.2. For an integer s ≥ 2 given, the polynomial Pa(s) comes with the maximum degree n = |n| = ⌊s/2⌋ of y. Proof. On the basis of (1.2.11) and (1.2.12), by induction, the conclusion can be drawn. This observation enables us to omit all yi for |i| ≥ n + 1 in (1.1.12) Observation 1.3.3. For an integer s ≥ 2 given, the polynomial Pa(s) has y = (y2 , y3 , . . . , ys ), or write ys−1 instead of y. Proof. On the basis of (1.2.11) and (1.2.12), by induction, the conclusion can be drawn.

1.3 Explicision crackless outer case

| 9

This observation enables us to delete all yi for i ≥ s + 1 in (1.1.9). Theorem 1.3.4. If integer ŝ ≥ 2 is the in-size of the solution of equation (1.1.1) for a0 = 0, then the solution is determined by the (ŝ − 1)-dimensional vector equation as follows: T T (Is−1 ̂ − a2 {(Ya[rlo] )(s−1)×( ̂ ̂ }ŝ )fs−1 s−1) ̂ = a1 (e1 )s−1 ̂

(1.3.1)

where (Ya[clo] )(s−1)×( = (yi,j )1≤i,j≤s−1 ̂ ̂ ̂ , s−1) for 1 ≤ t ≤ ŝ − 1, {∑t−1 yt+1+2k }ŝ , { { { k=0 yi,j = { (i, j) = (1, t) + k(1, 1), when ŝ − 1 ≥ j ≥ i ≥ 1; { { when 1 ≤ j < i ≤ ŝ − 1. {yj,i

(1.3.2)

Proof. On the basis of (1.1.12), by employing Observation 1.3.1–Observation 1.3.3, the conclusion can be drawn. This is a consistent linear system of equations with ŝ − 1 unknowns. Observation 1.3.5. For given integer ŝ ≥ 2, ⌈log s⌉̂

i {∑ (Ya[rlo] )i(s−1)×( ̂ ̂ } = { ∑ (Ya[clo] )(s−1)×( ̂ ̂ } . s−1) s−1) ŝ

i≥0

i=0

ŝ

(1.3.3)

Proof. This is a direct result of Observation 1.3.2. This observation enables us to get the inverse of coefficient matrix in equation (1.3.1): ⌈log s⌉̂

i i {(Is−1 ̂ − a2 {(Ya[clo] )(s−1)×( ̂ ̂ ) }ŝ = ∑ a2 {(Ya[clo] )(s−1)×( s−1) ̂ ̂ }ŝ . s−1) −1

i=0

(1.3.4)

Theorem 1.3.6. For given in-size ŝ ≥ 2, the solution of equation (1.1.1) for a0 = 0 is determined by the explicision in the form of a finite sum with all terms positive: ⌈log s⌉̂

i i T fTs−1 ̂ = a1 ( ∑ a2 {(Ya[clo] )(s−1)×( ̂ ̂ }ŝ (e1 )s−1 ̂ ). s−1) i=0

(1.3.5)

Proof. From Theorem 1.3.4 and (1.3.4), the conclusion can be drawn. [k] [k] ̂ then all yi,j Let (Ya[clo] )k(s−1)×( = (yi,j )1≤i,j≤ŝ for 0 ≤ k ≤ ⌈log s⌉, can be done from ̂ ̂ s−1)

[1] yi,j = yi,j as shown in

yi,j , when k = 1; [k] yi,j = { ⌈log s⌉̂ [k−1] ∑l=1 yi,l yl,j , when 2 ≤ k ≤ s.̂

(1.3.6)

10 | 1 Outer equations first part

1.4 Restrictions crackless outer case In [37] (Liu YP, 1987), one can find the following equation for f without the initial condition: xy𝜕x,y f |x=u 2 { { {∫ 1 − xy = f − x ; y { { { f { |x=0,y=0 = 0,

(1.4.1)

where f ∈ ℛ{x, y}. As a specific case of a0 = 0 and a1 = a2 = 1 in equation (1.1.1), Theorem 1.1.5 leads to the well-definedness of equation (1.4.1). Theorem 1.4.1. Let f[clo] be the solution of equation (1.4.1) determined by Fm,n = 𝜕xm f[clo] |n=n ∈ ℛ{y} for m, n ≥ 0, then

Fm,n

1(m = 2), 0(m ≥ 3), { { { { { {ym (m ≥ 2), ={ {(∑ 0≤k≤j−2 + ∑ 0≤k≤m−2 )yj+m−2k−2 Fj,n−1 { { j≥m+1 2≤j≤m { { (m ≥ 2), {

when n = 0; when n = 1;

(1.4.2)

when n ≥ 2.

Proof. We have the specific case of Theorem 1.2.3 with a0 = 0 and a1 = a2 = 1. Because of the infinite summation involved in (1.4.2), another parameter s = π(n), called the in-size (i. e., the number of semi-edges independent of the root-vertex), has to be considered. On the basis of Section 1.2, let Ps be the sum of all terms of π(yn ) = s in f[clo] ∈ ℛ{x, y} for s ≥ 0, then f[clo] is the sum of all Ps over s ≥ 0. Obviously, Fm,n = Fa[m,n] |a0 =0,a1 =a2 =1 = ∑ Ps(m,n) . s≥0

(1.4.3)

When n = 0, for any integer m ≥ 2, only the term F2,0 = 1 ≠ 0 (s = 0) among all Fm,0 . For any n ≥ 0 and n ≠ 0, we have s ≥ 2. This enables us to discuss only s ≥ 2 without loss of generality. Observation 1.4.2. For any integer n ≥ 1, s = m (mod 2) in Ps(m,n) . Proof. This is the specific case of Lemma 1.2.4 with a0 = 0 and a1 = a2 = 1. Because F0 = 0 and F1 = 0, Ps(0,n) = 0 and Ps(1,n) = 0 for s ≥ 0 and n ≥ 0. From (1.2.2), for m ≥ 2, 1, when m = 2; Ps(m,0) = { 0, when m ≥ 3.

1.4 Restrictions crackless outer case

| 11

On this basis, in order to get Ps(m,n) , we are allowed to discuss only m ≥ 2, n ≥ 1 and s ≥ 2 (Lemma 1.1.3). Observation 1.4.3. For any integer m ≥ 2, min{s | ∀Ps(m,n) } = m = π(Fm,1 ). n≥1

(1.4.4)

Proof. We have the specific case of Lemma 1.2.5 with a0 = 0 and a1 = a2 = 1. Observation 1.4.4. The polynomial Ps is independent of yl for l ≥ s + 1. Proof. We have the specific case of Lemma 1.2.6 with a0 = 0 and a1 = a2 = 1.

s as

The last three observations tell us that Ps is a homogeneous polynomial of in-size

Ps =

⌊(s−2)/2⌋

∑ l=0

xs−2l Ps(s−2l,1+l) .

(1.4.5)

Theorem 1.4.5. The solution f[clo] of equation (1.4.1) is determined by the following polynomials on ℛ{x, y}: 2

x , { { { { { {y3 x3 , Pl = { 4 {y4 x + y2 x2 , { 2 { { { s ⌊(s−2)/2⌋ s−2l (s−2l,1+l) y x + x Ps , ∑ { s l=1

when s = 2; when s = 3; when s = 4;

(1.4.6)

when s ≥ 5,

for 0 ≤ l ≤ s. Proof. We have the specific case of Theorem 1.2.7 with a0 = 0 and a1 = a2 = 1. Example 1.4.1. Root-isomorphic classes of crackless outer planar maps with rootvertex valency and vertex-partition vector as parameters. A crack vertex of a map is such a vertex that its incident semi-edges can be partitioned into two parts as two new vertices in distinct connected components of the result map. A map is said to be crack if it has a crack vertex. Figure 1.4.1 provides the classes of crackless outer planar maps with the number not greater than 6 of semi-edges not incident to the root-vertex. For example, P0 = a = x2 , P1 = 0, P2 = b = y2 x2 , P3 = c = y3 x3 , P4 = d + e = y4 x4 + y22 x2 , P5 = f + g = y5 x5 + 2y2 y3 x3 and P6 = h + (i + j) + k = y6 x6 + (2y2 y4 + y32 )x4 + y23 x2 shown by (1.4.5) combined with (1.2.13), (1.2.14), (1.2.15), (1.2.16) and (1.2.20). In Figure 1.4.1, the power of x is the root-vertex valency and the power of yi , 2 ≤ 2 ≤ 6, is the number of non-rooted vertices of valency i. Ps is shown as in the proof of Theorem 1.4.5.

12 | 1 Outer equations first part

Figure 1.4.1: Classes of crackless outer planar rooted maps with in-size not greater than 6.

1.5 Super-wheels outer model In Liu YP [39] (Liu, Y. P., 1988), particularly [43] (Liu, Y. P., 1989), one can find the specific case when a = b = c = 1 and d = 0 without initiation in the equation for f ∈ ℛ{x, y}, y 2 2 { {bx ∫ x − cyf = f − ax ; y { { f | { x=0⇒y=0 = d,

(1.5.1)

i

i

i

where a, b, c, d ∈ ℤ+ , y = (y2 , y3 , y4 , . . .) and hence yi = y21 y32 y24 ⋅ ⋅ ⋅. This is equation (2) in Introduction. Because of the specific case determining the enufunction of asymmetric classes of super-wheels with the root-vertex valency and the vertex-partition vector as parameters and by the equivalence between super-wheels and dual of non-separable outer

1.5 Super-wheels outer model | 13

planar maps with the link map as only exception, equation (1.5.1) is called a superwheels outer model, or non-separable outer model. Observation 1.5.1. Let f ∈ ℛ{x, y} be a solution of equation (1.5.1), then f is independent of y1 and d = 0. Proof. Because ℛ{x, y} only for y = (y2 , y3 , y4 , . . .) independent of y1 , the first statement is done, because the constant term of f is 0, the second statement is done. This observation enables us to only consider equation (1.5.1) which is neither with y1 nor d ≠ 0 in spite of any particular declaration. From now on, a is denoted by (a, b, c) instead of (a, b, c, d). By the substitution on ℛ{x, y} f = xh,

(1.5.2)

the first equality in equation (1.5.1) becomes h = ax + b ∑ yi+1 (ch)i .

(1.5.3)

i≥1

Lemma 1.5.2. Equation (1.5.1) for f with d = 0 is equivalent to the following equation for h: i

{ {h = ax + b ∑ yi+1 (ch) ; i≥1 { { h = { x=0⇒y=0 0,

(1.5.4)

where h ∈ ℛ{x, y}. Proof. From (1.5.1) and (1.5.3), both first equations are equivalent on ℛ{x, y}. From x = 0 ⇒ y = 0, the two initiations are equivalent. Let h = ha be a solution of equation (1.5.4). For m ≥ 0, let [h]m = 𝜕xm h = Ha[m] , then ha = ∑ Ha[m] xm m≥0

[i] and hia = ∑ Ha[m] xm , m≥0

i ≥ 2,

(1.5.5)

[i] where Ha[m] , Ha[m] ∈ ℛ{y} for m ≥ 0. Moreover, for any integer i ≥ 1,

Ha[m] , [i] ={ m Ha[m] [i−1] ∑l=0 Ha[l] Ha[m−l]

when i = 1; when i ≥ 2.

By the initiation of (1.5.4), it is seen that [i] = 0, i ≥ 1. Ha[0] = 0 󳨐⇒ Ha[0]

Thus, for any integer m ≥ 1, Ha[m] , when i = 1; [i] = { m−1 Ha[m] [i−1] ∑l=1 Ha[l] Ha[m−l] , when i ≥ 2.

(1.5.6)

14 | 1 Outer equations first part Lemma 1.5.3. Given integer m ≥ 2, for any integer i ≥ m, we have Hm , [i] = { a[1] Ha[m] 0,

when i = m;

(1.5.7)

when i ≥ m + 1.

Proof. Let g ∈ ℛ{x, y}. Denote by dx g the minimum power of x in g. We have Ha[0] = 0, m dx (hia ) = i. When m ≤ i − 1, 𝜕xm hia = 0, and when m = i, 𝜕xm hm a = Ha[1] . This is the conclusion. This conclusion enables us to deal with the infinite sum in the first part of equation (1.5.4) as a finite sum for the integer m given. Lemma 1.5.4. Equation (1.5.4) is equivalent to the system of equations as a when m = 1; { 1−bcy2 , Ha[m] = { i m bc yi+1 [i] {∑i=2 1−bcy2 Ha[m] , when m ≥ 2,

(1.5.8)

for m ≥ 2 on ℛ{y}. Proof. On the basis of the main equation of equation (1.5.4), for integer m = 1, its initiation leads to [ha ]1 = a + bcy2 [ha ]1 ⇔ Ha[1] = a + bcy2 Ha[1] ⇒ Ha[1] =

a . 1 − bcy2

This is the first equation of equations (1.5.8). For integer m ≥ 2, from Lemma 1.5.2 and Lemma 1.5.3, m

[i] Ha[m] = bcy2 Ha[m] + b ∑ ci yi+1 Ha[m] i=2

⇒ Ha⟨m⟩

m b [i] . = ∑ ci yi+1 Ha[m] 1 − bcy2 i=2

This is the second equation of equations (1.5.8). Because of Lemma 1.5.3, m [m] = Ha[1] = Ha[m]

am . (1 − bcy2 )m

Then equation (1.5.8) becomes a { 1−bcy2 , Ha[m] = { m bc ym+1 m−1 { (1−bcy2 )m+1 + b ∑i=2

when m = 1; i

c yi+1 H [i] , 1−bcy2 a[m]

when m ≥ 2.

(1.5.9)

1.6 Solution super-wheels outer case

| 15

On this basis, explicisions of Ha[2] and Ha[3] can be, respectively, presented: Ha[2] =

bc2 y3 2+n = ∑( )(bcy2 )n y3 3 2 (1 − bcy2 ) n≥0

(1.5.10)

[2] = 2Ha[1] Ha[2] , and by considering Ha[3]

Ha[3] =

bc2 y3 [2] bc3 y4 H + (1 − bcy2 )4 1 − bcy2 a[3]

3+n 4+n = ∑ (( )bc3 y4 + 2( )ab2 c4 y32 )(bcy2 )n . 3 4 n≥0

(1.5.11)

Theorem 1.5.5. Equation (1.5.1) is well defined on ℛ{x, y} if, and only if d = 0. Proof. Sufficiency. From Lemma 1.5.2, equation (1.5.4) is equivalent to equation (1.5.1). From Lemma,1.5.4, equation (1.5.4) is equivalent to the system of equations (1.5.9). These enable us to discuss only the latter statement. On the basis of (1.5.10) and (1.5.11), for m ≤ 3, Ha[m] is determined by the system of equations (1.5.9). For m ≥ 4. By induction on m, assume that when l ≤ m − 1, all Ha[l] ∈ ℛ{y} are determined by (1.5.9). To evaluate Ha[m] . [i] Because all Ha[m] for i ≤ m − 1 only depend on Ha[l] (1 ≤ l ≤ m − 1) in (1.5.9), Ha[m] is determined by Ha[l] , 1 ≤ l ≤ m − 1 from the induction hypothesis. Therefore, equation (1.5.1) has a solution. Moreover, because of the uniqueness of the procedure for determining Ha[m] (m ≥ 1) by equations (1.5.9) from the initiation Ha[1] = 0, the solution of equation (1.5.1) is the only one. Necessity. This is a result of Observation 1.5.1.

1.6 Solution super-wheels outer case New, let us introduce another integer n ≥ 1 as the degree, i. e., the maximum power i = |i| over terms involving with yi among all available i in a function on ℛ{y}. For convenience, denote by ⟨f ⟩n the part of f ∈ ℛ{y} for which all terms are of degree n of y. This implies that ⟨f ⟩n is a homogeneous polynomial of y with degree n for integer n ≥ 0. Therefore, f is partitioned into homogeneous polynomials of y. Observation 1.6.1. Given Ha[m] determined by equation (1.5.9) for integer m ≥ 1. Let Ha[m,n⟩ = ⟨Ha[m] ⟩n for integer n ≥ 1, then all Ha[m,n⟩ are homogeneous polynomials of degree n. Proof. Because all Ha[m] in ℛ{y}, the conclusion is naturally drawn.

16 | 1 Outer equations first part On the basis of Observation 1.6.1, equation (1.5.9) enables us to determine all Ha[m,n⟩ for the solution ha of equation (1.5.4). Theorem 1.6.2. The solution ha is determined by Ha[m,n⟩ for integers m, n ≥ 1 as a(bcy2 )n , for 1 ≤ n and m = 1; Ha[m,n⟩ = { n m+n−1 m−1 [i] n−1 m+n−1 b c ( m )y2 ym+1 + ∑i=2 Aa[m,n⟩ , for 1 ≤ n and m ≥ 2,

(1.6.1)

where, for 2 ≤ i ≤ m − 1, n−i−1

[i] l+1 l+i l A[i] a[m,n⟩ = ∑ b c y2 yi+1 Ha[m,n−l−1⟩ . l=0

[i] =0 Proof. On the basis of the initiation of equation (1.5.4), Ha[0]⟩ = 0 and hence Ha⟨[0] for all i ≥ 1 are known. For m = 1, by the first equation of the system of equations (1.5.8),

Ha[1,n⟩ = ⟨

a ⟩ = a(bc)n y2n 1 − bcy2 n

where n ≥ 1. For m ≥ 2, on the basis of the second equation in equations (1.5.8), because ⟨

bcm ym+1 1 ⟩ = bcm ym+1 ⟨ ⟩ (1 − bcy2 )m+1 n (1 − bcy2 )m+1 n−1 m + n − 1 n−1 = bn cm+n−1 ( )y2 ym+1 m

and A[i] a[m,n⟩ = ⟨

n−i−1 bci yi+1 [i] [i] Ha[m] ⟩ = ∑ bl+1 cl+i y2l yi+1 Ha[m,n−l−1⟩ 1 − bcy2 n l=0

for 2 ≤ i ≤ m − 1, the conclusion is easily drawn. However, for n given, still infinitely many Ha[m,n⟩ for m ≥ 1 need to be evaluated. This suggests us to seek a new parameter s related to both m and n so that only a finite number of Ha[m,n⟩ have to be determined for the solution of equation (1.5.4). If i = (i1 , i2 , i3 , . . .) is a power vector of y = (y1 , y2 , y3 , . . .) in a term of Ha[m,n⟩ , then let σ(i) = ∑(j + 1)ij . j≥1

(1.6.2)

For convenience of description for P ∈ ℛ{y}, let σ(P) = σ(i) for the moment where i is the power vector of y in a term of P.

1.6 Solution super-wheels outer case

| 17

Observation 1.6.3. For any A, B ∈ ℛ{y}, σ(AB) = σ(A) + σ(B). i

i

Proof. Let σ1 (i1 ) = σ(A) and σ2 (i2 ) = σ(b), because y11 y22 = (y1 y2 )i1 +i2 , σ(AB) = σ(i1 + i2 ) = σ(i1 ) + σ(i2 ) = σ(A) + σ(B). This is the conclusion. The conclusion of Observation 1.6.3 is called the additivity of σ. [i] has σ(i) = 2n + m − i where Lemma 1.6.4. For any integer m ≥ 2, n ≥ 1 and i ≥ 1, Ha[m,n⟩ [i] . i is the power vector of y in a term of Ha[m,n⟩

Proof. When m = 1, from the first equation of the system of equations (1.5.9), for any integer n ≥ 1, Ha[1,n⟩ = y2n has σ(i) = 2n = 2n + m − 1. In fact, this is the conclusion for [i] ⟩n in the case of i = 1. ⟨Ha[1] For m ≥ 2. Assume the conclusion is true for m, n and i smaller. We show the con[i] [i] ⟩n , on the basis of (1.5.6), from the additivity = ⟨Ha[m] clusion for m, n and i. For Ha[m,n⟩ of σ, we have, for i ≥ 2, the inductive assumption process [i−1] [i] ) = σ(⟨Ha[l] ⟩k ) + σ(⟨Ha[m−l] ⟩n−k ) σ(Ha[m,n⟩

= (2k + l − 1) + (2(n − k) + (m − l) − (i − 1)) = 2n + m − i. This is the conclusion. The case of i = 1 in this lemma has to be mentioned in particular because of its fundamental importance. Corollary 1.6.5. For integers m ≥ 2 and n ≥ 1, Ha[m,n⟩ comes with σ(i) = 2n + m − 1. Proof. Only the case of i = 1 in Lemma 1.6.4. Because σ(Ha[m,n⟩ ) is independent of the choice of y in a term of Ha[m,n⟩ , σ(i) in Ha[m,n⟩ can always be replaced by σ(Ha[m,n⟩ ) without clarification. Observation 1.6.6. For any integers m, n ≥ 1, σ(Ha[m,n⟩ ) ≠ m(mod 2). Proof. On account of σ(Ha[m,n⟩ ) = 2n + m − 1 = m − 1(mod 2), from Corollary 1.6.5, the conclusion can be drawn. This observation enables us to introduce another parameter s as 2s = σ(Ha⟨m,n] ) + (m + 1), and hence 2s = (2n + m − 1) + (m + 1) = 2(n + m) ⇒ s = m + n.

(1.6.3)

18 | 1 Outer equations first part On the basis of (1.6.3), because m and n are both positive integers, s is a positive integer as well. Observation 1.6.7. When s ≥ 2 is given, we have 1 ≤ m, n ≤ s − 1. Proof. Because s = m + n for m, n ≥ 1, (1.6.3) leads to s − 1 being the greatest value of m and n according as 1 is the least value of n and m. This is the conclusion. Let Ha(s) = (ha )s be the part of the solution ha of equation (1.5.4) with all terms of s = m + n, then ha is determined by all H(s) ∈ ℛ{x, y} for s ≥ 0. Because Ha(0) = Ha⟨0,0] = 0 and Ha(1) = Ha⟨0,1] + Ha⟨1,0] = 0 + a = a from the initiation and (1.5.8), it is only necessary to consider s ≥ 2 in what follows. Lemma 1.6.8. All Ha(s) for s ≥ 2 are polynomials on ℛ{x, y}. Proof. On the basis of Theorem 1.6.2, all Ha⟨m,n] for integers m, n ≥ 1 given are polynomials in ℛ{y}, because Ha(s) = ∑ Ha[m,n⟩ xm 1≤n≤s−1 1≤m≤s−1

(1.6.4)

for any integer s ≥ 2, Ha(s) is a polynomial on ℛ{x, y}. On the basis of (1.6.4), because of the maximum of j for yj appearing in Ha[s−1,1⟩ among all Ha[m,n⟩ , 1 ≤ n ≤ s − 1, we have the maximum j = s. Observation 1.6.9. For any integer s ≥ 2 given, Ha(s) is independent of all yj , j ≥ s + 1. Proof. The reason is that the maximum j of yj appearing in a term of Ha[m,n⟩ for 1 ≤ m, n ≤ s − 1 is s = m + n. This observation enables us to only restrict ourselves ys−1 = (y2 , y3 , . . . , ys ) instead of y for integer s given in the solution of equation (1.5.4). Theorem 1.6.10. For integer s ≥ 0 given, the solution ha of equation (1.5.4) is determined by Ha(l) for 0 ≤ l ≤ s as 0, { { { Ha(l) = {a, { { l−1 m {∑m=1 Ha[m,l−m⟩ x ,

when l = 0; when l = 1;

(1.6.5)

when 2 ≤ l ≤ s,

where a = (a, b, c). Proof. On the basis of Lemma 1.6.8, the first two cases of l = 0 and 1 are done according as the initiation of equation (1.5.4) and the first equation of equations (1.5.8). The last case of 2 ≤ l ≤ s is from the second equation of equations (1.5.8). On account of Observation 1.6.7, Lemma 1.6.8 and Observation 1.6.9, this theorem shows that all Ha(s) for s ≥ 2 are polynomials on ℛ{x, ys−1 }.

1.7 Explicision super-wheels outer case

| 19

1.7 Explicision super-wheels outer case For a power vector n, it is seen that σ(n) is dependent on neither the choice of terms nor a in Ha[m,n⟩ . This enables us to write σm,n = σ(Hm,n ) = 2n + m − 1 where Hm,n = Ha[m,n⟩ |a=(1,1,1)=1 . For convenience of description, let ℋm,n = {n ≥ 0 | n is power of y in terms of Hm,n }.

(1.7.1)

On account of Corollary 1.6.5, denote ℒm,n = {n ≥ 0 | |n| = n, σ(n) = 2n + m − 1}.

(1.7.2)

Lemma 1.7.1. For integer m ≥ 1, Ha[m] is independent of all yj for j ≥ m + 2. Proof. First, for m ≤ 3. From the first line of (1.5.9), (1.5.10) and (1.5.11) show the conclusion. Then, for m ≥ 4. Assume that Ha[i] satisfies the conclusion when i ≤ m − 1. By induction on i = m, we show the conclusion. On the basis of (1.5.7) and (1.5.8), by the induction assumption, the conclusion can be drawn. This lemma enables us to consider Ha[m] ∈ ℛ{ym } instead of ℛ{y} for m ≥ 1 where ym = (y2 , y3 , y4 , . . . , ym+1 ).

(1.7.3)

Lemma 1.7.2. For any integers m ≥ 2 and n ≥ 1, ℋm,n = ℒm,n . Proof. Corollary 1.6.5 tells us that, for any n ∈ ℋm,n , n ∈ ℒm,n . This is ℋm,n ⊆ ℒm,n . Conversely, for any n ∈ ℒm,n , we prove n ∈ ℋm,n . When (m, n) = (1, 1), (2, 1), (1, 2) and (m, n) = (3, 1), (2, 2), (1, 3). From the first line of (1.5.9), (1.5.10) and (1.5.11), the conclusion is checked. For example, when (m, n) = (1, 1), from Lemma 1.7.1, it is only necessary to consider the power vector (i1 , i2 ). For integers i1 , i2 ≥ 0, the system of equations |(i1 , i2 )| = i1 + i2 = 1 and π(i1 , i2 ) = 2i1 + 3i2 = 2 has only one solution as (i1 , i2 ) = (1, 0). This leads to ℒ1,1 = {(1, 0)}. We have H1,1 = y2 , ℋ1,1 = {(1, 0)}. For another example, consider the system of equations i1 + i2 + i3 = 2; 2i1 + 3i2 + 4i3 = 6. The solution shows ℒ3,2 = {(0, 2, 0), (1, 0, 1)}. For (0, 2, 0), (y2 , y3 , y4 )(0,2,0) = y32 . For (1, 0, 1), (y2 , y3 , y4 )(1,0,1) = y2 y4 . We have H3,2 = 2y32 + 4y2 y4 , and it is seen that (0, 2, 0), (1, 0, 1) ∈ ℋ3,2 . These suggest us to assume that, for any integers (i, j) < (m, n), ℒi,j ⊆ ℋi,j , we prove ℒm,n ⊆ ℋm,n by the principle of induction. Let i ∈ ℒm,n . From (1.5.9) and (1.6.1), Ha[m,n⟩ =

m−1 m−1 1 bn cm+n−1 (m + n)! n−1 [i] ⟩ . y2 ym+1 + b ∑ ci yi+1 ⟨ ∑ Hl Hn−l (m + 1)!(n − 1)! 1 − bcy 2 i=2 l=1 n−1

(1.7.4)

By Observation 1.6.9, it is only necessary to consider i = im . If im ≠ 0, then we only have im = 1 and hence i1 = n − 1 such that i ∈ ℋm,n . Otherwise, there exists an i, 1 ≤ i ≤ m − 1,

20 | 1 Outer equations first part such that i = ei + i󸀠 , i󸀠 ∈ ℒs,t and (s, t) < (m, n) in which ei is the vector with its ith entry of value 1 and 0 for all other entries. Then, from (1.7.4), there is a term on the right hand side such that i is its power vector. Therefore, i ∈ ℋm,n and hence ℒm,n ⊆ ℋm,n . Now, let us introduce a new function: n

γn (x) = (∑ yj xj ) .

(1.7.5)

j≥2

Denote 2(n−1)+m

𝒢2(n−1)+m = {i ≥ 0 | power vector of a term in 𝜕x

γn (x)}.

(1.7.6)

Lemma 1.7.3. For integers m ≥ 2 and n ≥ 1, 𝒢2(n−1)+m = ℒm,n . Proof. For any integer i ≥ 0, i ∈ 𝒢2(n−1)+m . We have i as a power vector in 𝜕x2(n−1)+m γn (x), i is an integral solution of the system of equations ∑ ij = n; { j≥1 ∑j≥1 (j + 1)ij = 2n + m − 1, for ij ≥ 0, j ≥ 1.

(1.7.7)

This implies i ∈ ℒm,n . Conversely, for an integral vector i ≥ 0, i ∈ ℒm,n . We have i satisfying equation (1.7.7), and it is seen that i ∈ 𝒢2(n−1)+m . From Lemma 1.7.2 and Lemma 1.7.3, (1.7.8)

ℒm,n = 𝒢2(n−1)+m

for any integers m ≥ 2 and n ≥ 1, because 𝜕x2(n−1)+m γn (x) =

∑

n∈𝒢2(n−1)+m

n! i y, n!

(1.7.9)

Lemma 1.7.3 leads to the existence of βai ∈ ℛ+ for i ∈ 𝒢2(n−1)+m such that Hm,n =

∑

n∈𝒢2(n−1)+m

βan

n! n y . n!

(1.7.10)

Theorem 1.7.4. The solution of equation (1.5.8) is determined by the following summation-free explicision: n n+1 n−1 {∑n≥1 b c y2 y3 , Ha[m] = { n! n y , ∑ n∈𝒢2(n−1)+m βan n! { n≥1

where βan is given in (1.7.10).

when m = 2; when m ≥ 3,

(1.7.11)

1.7 Explicision super-wheels outer case

| 21

Proof. The first line of (1.5.9) leads to the first line of (1.7.11). Because Ha[m] = ∑ Ha[m,n⟩ , n≥1

(1.7.10) leads to the second line of (1.7.11). In principle, βan can be determined from Theorem 1.6.2 in a recursive manner, However, the best manner should be direct as follows. Observation 1.7.5. Equation (1.5.4) with y1 instead of x is the same as equation (8.1.6) for a0 = 0 in Book I. Proof. If (a1 , a2 , a3 ) = (a, b, c), then equation (8.1.6) becomes equation (1.5.4) with y1 instead of x. The observation enables us to extract an explicision of the solution of equation (1.5.4) for a3 = 1 such that each term with a coefficient summation-free. Theorem 1.7.6. In the solution h[a,b] of equation (1.5.4) for c = 1, we have swl 𝜕yi++ h[a,b] = swl

ai0 b|i+ |−i0 (|i+ | − 1)! i+ !

(1.7.12)

where y+ = (x, y) and i+ = (i0 , i). Proof. This is a direct result of equation (8.4.7) in Book I. Combined with (1.7.12) and (1.7.10), it is shown that βai+ =

ai0 b|i+ |−i0 . |i+ |

(1.7.13)

On the basis of Observation 1.7.5 and Observation 8.4.1 in Book I, let ha[swl] = (ha1 , ha2 , ha3 , . . .) be a solution of the equation hTa[swl] = bYa[swl] hTa[swl] + axeT1

(1.7.14)

where a = (a, b, c) ∈ ℤ3+ , hai ∈ ℛ{x, y} and Ya[swl] = (ya[i,j] )i,j≥0 such that

ya[i,j]

cj x, when j − i = −1; { { { j = {c yj−i+2 , when j − i ≥ 0; { { when j = i ≤ −2, {0,

(1.7.15)

for i, j ≥ 1. The well-definedness of equation (1.7.14) is a result of Theorem 8.4.2 in Book I.

22 | 1 Outer equations first part Observation 1.7.7. Let ha[swl] = (ha[swl]1 , ha[swl]2 , ha[swl]3 , . . .) be the solution of equation (1.7.14), then hTa[swl] = ax ∑ (bYa[swl] )k eT1 k≥0

(1.7.16)

[k] where (Ya[swl] )k = (ya[i,j] )i,j≥1 is determined by

δi,j , when k = 0; [k] ={ ya[i,j] [k−1] ∑l≥1 ya[i,l] ya[l,j] , when k ≥ 1,

(1.7.17)

for i, j ≥ 1. Proof. Because equation (1.7.14) is equivalent to (I − bYa[swl] )hTa[swl] = axeT1 and (I − bYa[swl] )−1 = ∑ (bYa[swl] )k , k≥0

equation (1.7.16) is true. From (1.7.15), it is shown that all Yka[swl] for k ≥ 1 are only dependent on c but also on both a and b. One might prefer to see the solution of an equation in an expression as a finite sum of terms, particularly ones all positive. On the basis of Observation 1.7.5. According to Theorem 8.4.5 in Book I, this objection can be attained. Let y󸀠 = (x, y2 , y3 , y4 , . . . , yn−1 ) where n is the greatest power of terms with y󸀠 in a function f ∈ ℛ{y󸀠 }. Theorem 1.7.8. For the integer n ≥ 1 given, the solution fan of equation (1.5.4) is ta1(n−1) |y1 =x , i. e., n−1

[i] han = ax(1 + ∑ bi ya[1,1] )|y1 =x i=1

(1.7.18)

[i] , 1 ≤ i ≤ n − 1, are known from (8.4.4) of Book I. where ya[1,1]

Proof. A result of Theorem 8.4.5 in Book I. Example 1.7.1. When n = 4, on the basis of (1.7.16), (1.7.17) and (1.7.18), because of the coefficient matrix with dimension 3 Ya[swl]

cy2 = ( cx 0

c 2 y3 c 2 y2 c2 x

c 3 y4 c 3 y3 ) , c 3 y2

[3] [1] [2] = c2 y22 + c3 xy3 and ya[1,1] = c3 y23 + (2 + c)c4 xy2 y3 + c6 x2 y4 . we have ya[1,1] = cy2 , ya[1,1]

1.8 Restrictions super-wheels outer case

| 23

By employing (1.7.18), han = ax(1 + bcy2 + (bc)2 y22 + b3 c3 y23 )

+ ax2 (b2 c3 y3 + b3 (2 + c)c4 y2 y3 ) + ax 3 (b3 c6 y4 )

= a(1 + bcy2 + (bc)2 y22 + (bc)3 y23 )x + a(b2 c3 y3 + b3 (2 + c)c4 y2 y3 )x2 + a(b3 c6 y4 )x 3 .

1.8 Restrictions super-wheels outer case In Liu YP [54] (1993), one can find see the equation as y 2 2 { { {f = x + x ∫( x − yf ); y { { { {f |x=0,y=0 = 0, i

i

(1.8.1)

i

where y = (y2 , y3 , y4 , . . .). Hence, yi = y21 y32 y43 ⋅ ⋅ ⋅. This is the case a = b = c = 1 and d = 0 of equation (1.5.1). For avoiding some complication in evaluation, let us still employ its equivalence as i

{ {h = x + ∑ yi+1 h ; i≥1 { { h { x=0⇒y=0 = 0,

(1.8.2)

which is the case a = b = c = 1 of equation (1.5.4). Theorem 1.8.1. Equation (1.8.2) and hence equation (1.8.1) are well defined on ℛ{x, y}. Proof. This is a direct result of Theorem 1.5.5. By equation (1.5.2), the solution f = zh of equation (1.8.1) is done from the solution h of equation (1.8.2). This enables us only to consider equation (1.8.2) instead of equation (1.8.1). From [54] (Liu, Y. P., 1993), the enufunction fswl = f of super-wheels is determined via hswl = h, the solution of equation (1.8.2). Theorem 1.8.2. The solution h is determined by Hm,n for integers m, n ≥ 1 as yn , for 1 ≤ n and m = 1; Hm,n = { 2 m−1 [i] n−1 A , for 1 ≤ n and m ≥ 2, (m+n−1 )y y + ∑ m+1 i=2 m,n 2 m where for 2 ≤ i ≤ m − 1, n−i−1

l [i] A[i] m,n = ∑ y2 yi+1 Hm,n−l−1 . l=0

(1.8.3)

24 | 1 Outer equations first part Proof. We have the specific case of Theorem 1.6.2 in a = b = c = 1. One might think of the parameter s = m + n such that h is partitioned into a sum of s polynomials as h(l) with 1 ≤ l = m + n ≤ s because of h(0) = 0 given by the initiation. Theorem 1.8.3. For integer s ≥ 0 given, the solution h(s) of equation (1.8.2) is determined by Ha(l) for 0 ≤ l ≤ s as 0, when l = 0; { { { Ha(l) = {1, when l = 1; { { l−1 m {∑m=1 Hm,l−m x , when 2 ≤ l ≤ s.

(1.8.4)

Proof. We have the specific case of Theorem 1.6.10 with a = b = c = 1. Above, we have seen recursions of two types each of which is in the form of a sum with all terms positive. However, in what follows, direct explicisions of two types are investigated. On the basis of (1.7.14) and (1.7.15), consider the solution hswl = h1 in ha[swl] |a=(1,1,1) = (h1 , h2 , h3 , . . .) for the maximum degree of n = |(m, i)| where n = (m, i) is the power vector of (x, y2 , y3 , y4 , . . .). Theorem 1.8.4. For the integer n ≥ 1 given, the solution h(n) of equation (1.8.2) is h1(n−1) |y1 =x , i. e., 󵄨󵄨 n−1 [i] 󵄨󵄨󵄨 )󵄨󵄨 h(n) = x(1 + ∑ y1,1 󵄨󵄨 i=1 󵄨y1 =x

(1.8.5)

[i] , 1 ≤ i ≤ n − 1, are known from (8.4.4) of Book I. where y1,1

Proof. This is a direct result of Theorem 1.7.8 for a = b = c = 1. Example 1.8.1. Vertex partition of super-wheels. A super-wheel is a rooted planar map such that its dual is a non-separable outer planar rooted map. Or in other words, the result of deleting the root-vertex with incident edges is a tree. Let Fm,n is the number of non-isomorphic super-wheels with n non-rooted vertices, root-vertex valency m and vertex-partition vector n as the power vector of y = (y2 , y3 , y4 , . . .), then Fm,n = Hm−1,n , m ≥ 2, n ≥ 0. From the first equality of (1.8.3), H1 = ∑ H1,n = ∑ y2n = 1 + y2 + y22 + y23 + y24 + y25 + y26 + ⋅ ⋅ ⋅ . n≥0

n≥0

This implies F2,0 = H1,0 = 1, F2,1 = H1,1 = y2 , F2,2 = H1,2 = y22 , F2,3 = H1,3 = y23 , F2,4 = H1,4 = y24 , F2,5 = H1,5 = y25 and F2,6 = H1,6 = y26 . In Figure 1.8.1, it is seen that a = 1 = F2,0 , b = y2 = F2,1 , c = y22 = F2,2 , d = y23 = F2,3 , e = y24 = F2,4 , f = y25 = F2,5 and g = y26 = F2,6 .

1.8 Restrictions super-wheels outer case

| 25

Figure 1.8.1: Isomorphic classes of super-wheels of root-vertex valency 2 and the number of nonrooted vertices not greater than 6.

From the second equality of (1.8.3), H2 = y3 + 3y2 y3 + 6y22 y3 + 10y23 y3 + ⋅ ⋅ ⋅ . This implies F3,1 = H2,1 = y3 , F3,2 = H2,2 = 3y2 y3 , F3,3 = H2,3 = 6y22 y3 and F3,4 = H2,4 = 10y23 y3 . In Figure 1.8.2, it is seen that a = y3 = F3,1 , b = 3y2 y3 = F3,2 , c+d = 3y22 y3 +3y22 y3 = 6y22 y3 = F3,3 and e + f + g = 3y23 y3 + 6y23 y3 + y23 y3 = F3,4 . Example 1.8.2. Face partition of non-separable outer planar maps. In the solution of equation (1.8.1), Fm,n = Hm−1,n provides the number of isomorphic classes of nonseparable outer planar maps with root-face valency m and non-root faces number n via a face partition vector. By (1.8.3), H3 = y4 + 4y2 y4 + 2y32 + 10y2 y32 + 10y22 y4 + 30y22 y32 + 20y23 y4 + ⋅ ⋅ ⋅ . This implies that F4,1 = H3,1 = y4 , F4,2 = H3,2 = 4y2 y4 + 2y32 , F4,3 = H3,3 = 10y2 y32 + 10y22 y4 and F4,4 = H3,4 = 30y22 y32 + 20y23 y4 . In Figure 1.8.3, it is seen that a = y4 = F4,1 , b + c = 2y32 + 4y2 y4 = F4,2 , (d + f ) + (e + g + h) = (2+8)y2 y32 +(4+2+4)y22 y4 = 10y2 y32 +10y22 y4 = F4,3 and (i+j+k+l+m+n)+(o+p+q+r) = (8 + 4 + 8 + 4 + 4 + 2)y22 y32 + (4 + 8 + 4 + 4)y23 y4 = 30y22 y32 + 20y23 y4 = F4,4 . Example 1.8.3. Face partition of non-separable outer planar simple maps. In Figure 1.8.3, only a = y4 and 2b = 2y32 .

26 | 1 Outer equations first part

Figure 1.8.2: Isomorphic classes of super-wheels with non-rooted vertex valency 3 and the number of non-rooted vertices not greater than 4.

Example 1.8.4. Face partition of non-separable outer planar bipartite maps. In Figure 1.8.3, only a = y4 , 4c = 4y2 y4 , 4e = 4y22 y4 , 2g = 2y22 y4 , 4h = y22 y4 , 4o = 4y23 y4 , 8p = 8y23 y4 , 4q = 4y23 y4 and 4r = 4y23 y4 . Example 1.8.5. Face partition of non-separable outer planar simple bipartite maps. In Figure 1.8.3, only one a = y4 .

1.9 Notes 1.9.1. It looks that the two equations (1.1.1) and (1.5.1) are very different because the former is linear and the latter non-linear in its expansion. However, the latter is transformed into a linear one. Both of them are, in this chapter, shown to have their coefficient matrices with the same full upper triangle sub-matrix. 1.9.2. Although both of equation (1.1.1) and equation (1.5.1) are equivalent to a linear system of infinite equations, the coefficient matrix is shown having the inverse such that all entries are in ℛ{y}. This enables us to evaluate powers of an infinite matrix and further those of a finite matrix whenever a suitable parameter is chosen. The simplest explicit form still needs to be further considered. 1.9.3. Is there one, or are there a series of transformations on ℛ{x, y} from one of equation (1.1.1) and equation (1.5.1), or equation (1.4.1) and equation (1.8.1), to another such that a favorite type of relationships among their solutions are established?

1.9 Notes | 27

Figure 1.8.3: Isomorphic classes of non-separable outer planar maps with root-face valency 4 and non-root vertices number 3.

1.9.4. On the basis of Theorem 1.2.7 and Theorem 1.6.10, or Theorem 1.4.5 and Theorem 1.8.4, a type of relationships among the functions determined on ℛ{y} is established to see what is meaningful in combinatorics particularly in number theory (par-

28 | 1 Outer equations first part

Figure 1.8.3: (continued)

titions) by Andrews GE [1] (Andrews, G. E., 1976), graph theory (combinatorial maps) by Liu YP [56] (Liu, Y. P., 1999), or umbral calculus (shadow functional) by Rota GC, Taylor BD [87] (Rota, G. C., Taylor, B. D., 1994). 1.9.5. It looks that (1.2.12) in Theorem 1.2.7 in general and hence (1.4.2) in Theorem 1.4.1 are meaningful in maps and are favorite to making efficient for running and further to intelligentization for usage. However, (1.3.5) in Theorem 1.3.6 in general and hence (1.4.6) in Theorem 1.4.5 specifically look available to extract an explicision. 1.9.6. Similarly, (1.6.5) in Theorem 1.6.10 in general and hence (1.8.4) in Theorem 1.8.3 are meaningful in maps and look favorite to making efficient for running and further to intelligentization for usage. However, (1.7.12) in Theorem 1.7.6 or (1.7.18) in Theorem 1.7.8 in general and hence (1.8.5) in Theorem 1.8.4 specifically seem to be available to extract an explicision. 1.9.7. As an application, (1.4.6) can be employed to enumerate the root-isomorphic classes of crackless outer planar maps with root-vertex valency m and size (m + s)/2, or order 1 + |n|, two parameters. For relevant results one is referred to Liu YP [33] (Liu, Y. P., 1986), [15] (Dong, F. M., Yan, J. Y., 1989). 1.9.8. As another application, (1.8.5) can be employed to enumerate the root-isomorphic classes of super-wheels(a type of non-separable outer planar maps) with root-vertex valency m and size (m + s)/2, or order 1 + |n|, two parameters. For relevant results one is referred to [34] (Liu, Y. P., 1986), [39] (1988), [45] (1989). 1.9.9. Asymptotic behavior (or asymptotics). Both of Note 1.9.7 and Note 1.9.8 remind us to investigate their asymptotic behavior as shown in [94] (Yan, J. Y., Liu, Y. P., 1991). 1.9.10. Stochastic behavior (or stochastics). On the basis of (1.4.6) and (1.8.5), the distributions of sizes, orders, semi-automorphic group orders and genera from, respectively, size polynomials, order polynomials, semi-automorphic group order polynomials and genus polynomials need to be investigated. For relevant results one is referred to [59] (Liu, Y. P., 2003, 221–230), [94], [11] (Chen, Y. C., et al., 2007), [92] (Wan, L. X., et al., 2008) etc.

1.9 Notes | 29

1.9.11. Equation (1.1.1) is from Program 85 in which equation (116) has constant coefficients c = a0 , a = a1 and b = a2 referred to in [70] (Liu, Y. P., 2015, Vol. 23, pp. 11696– 11697). Equation (1.5.1) is the same as equation (113) from Program 82 in [70] (Liu, Y. P., 2015, Vol. 23, pp. 11695–11696).

2 Outer equations second part 2.1 Restricted outer model Two models are investigated in this chapter. First, consider the equation for f 2 { {a2 x ∫ yδx,y (uf |x=u ) = (1 − a3 x )f − a1 ; y { { {f |x=0,y=0 = a0 ,

(2.1.1)

where a0 , a1 , a2 , a3 ∈ R+ and f ∈ ℛ{x, y}. This is equation (3) in Introduction. Because of the solution of equation (2.1.1) for a0 = a1 = a2 = a3 = 1 related to restricting outerplanar maps shown in [36] (Liu YP, 1987), this equation is called a restricted outer model. The first equality of equation (2.1.1) is equivalently transformed into as f = a1 + a3 x2 f + a2 x ∫ yδx,y (uf |x=u ).

(2.1.2)

y

Observation 2.1.1. If f = fa ∈ ℛ{x, y} where a = (a0 , a1 , a2 , a3 ) is a solution of equation (2.1.1), then a0 = a1 . Proof. Because of fa (0) = a1 from equation (2.1.2), the initial condition of equation (2.1.1) causes a0 = a1 . This observation enables us to only consider a0 = a1 in equation (2.1.1). Then equation (2.1.1) becomes 2 { {f = a1 + a3 x f + a2 x ∫ yδx,y (uf |x=u ); y { { f | = a . 1 { x=0,y=0

(2.1.3)

Let a = (a1 , a2 , a3 ) instead of (a0 , a1 , a2 , a3 ) from now on. For f ∈ ℛ{x, y}, write Fa[m] = 𝜕xm f ∈ ℛ{y}, m ≥ 0. Because of Fa[m] ∈ ℛ{y}, fa = fa (x) is determined by Fa[m] for m ≥ 0. It is well known that δx,y (uf |x=u ) = ∑ Fa[m] ( m≥0

xm+1 − ym+1 ) = ∑ xm ∑ yl−m Fa[l] x−y m≥0 l≥m

(2.1.4)

and the term with the meson functional in (2.1.2) is x ∫(yδx,y (uf |x=u )) = ∑ xm ∑ yl−m+2 Fa[l] . y

https://doi.org/10.1515/9783110627336-002

m≥1

l≥m−1

(2.1.5)

32 | 2 Outer equations second part Lemma 2.1.2. Equation (2.1.3) is equivalent to the system of equations

Fa[m]

a1 , { { { = {a2 ∑l≥0 yl+1 Fa[l] , { { {a3 Fa[m−2] + a2 ∑l≥m−1 yl−m+2 Fa[l] ,

when m = 0 (initial condition!); when m = 1;

(2.1.6)

when m ≥ 2,

on ℛ{y}. Proof. By substituting (2.1.5) into equation (2.1.3), equation (2.1.6) can then be obtained. The conclusion can be drawn. Let fa = (Fa[1] , Fa[2] , Fa[3] , . . .) and x = (x, x2 , x3 , . . .), then fa = a1 + xfTa . Lemma 2.1.3. Equation (2.1.6) is equivalent to the system of equations for fa ∈ ℛ{y}∞ fTa = Ya[rto] fTa + a2 y1 e1 + a3 e2

(2.1.7)

on ℛ{y}∞ where Ya[rto] = (ya[i,j] )i≥1,j≥1 such that

ya[i,j]

a2 yt+2 , when j − i = t ≥ 0; { { { { { {a2 y1 , when j − i = −1; ={ { a3 , when j − i = −2; { { { { when j − i ≤ −3. {0,

(2.1.8)

Proof. This is a direct result of Lemma 2.1.2. This lemma enables us to establish a qualitative theory on equation (2.1.7) and hence equation (2.1.3). Theorem 2.1.4. Equation (2.1.7) has, and is the only one to have, a solution on ℛ{y}. Proof. By the cancelation law, Ya[rto] fTa is removed from the right hand side of equation (2.1.7) to the left hand side leaving us with (I − Ya[rto] )fTa = (a2 y1 e1 + a3 e2 )T ,

(2.1.9)

which is equivalent to equation (2.1.7). Because of the existence and uniqueness for the inverse of the matrix (I − Ya[rto] ), equation (2.1.9) and hence equation (2.1.7) are well defined. This theorem enables us, directly, to get the solution of equation (2.1.9) in ℛ{y}∞ . Corollary 2.1.5. The solution of equation (2.1.7) is of the form fTa = ∑ (Ya[rto] )i (a2 y1 e1 + a3 e2 )T . i≥0

(2.1.10)

2.2 Solution restricted outer case

| 33

Proof. Because we checked that the inverse of I − Ya[rto] is ∑ (Ya[rto] )i ,

i≥0

it is seen that (2.1.10) is the solution of equation (2.1.9). From Lemma 2.1.3 and Lemma 2.1.2, the conclusion can be drawn. Then we are allowed to state our qualitative theorem for equation (2.1.1), or equivalently equation (2.1.3). Theorem 2.1.6. Equation (2.1.1) is well defined on ℛ{x, y} if, and only if, a0 = a1 . Proof. The sufficiency is from Theorem 2.1.4. The necessity is from Observation 2.1.1.

2.2 Solution restricted outer case Although Corollary 2.1.5 provides us with a form of the solution of equation (2.1.3), because of the infinity, the form is impracticable. Local structures of the solution have to be further investigated. On the basis of equation (2.1.6), by equivalent transformations on ℛ{y},

Fa[m]

a1 , { { { { { { 1−aa2 y (y1 + ∑l≥2 yl+1 Fa[l] ), = { a2 2 a { { 1−a3 y Fa[m−2] + 1−a2 y (y1 Fa[m−1] { 2 2 { { 22 y F + ∑ l−m+2 a[l] ), l≥m+1 {

when m = 0 (initial condition!); when m = 1;

(2.2.1)

when m ≥ 2,

is deduced. For convenience, let Fa[m,n] = [Fa[m] ]n = Fa[m] ||n|=n , i. e., Fa[m,n] = ∑ Fa[m,n] yn n≥0 |n|=n

(2.2.2)

where Fa[m,n] = 𝜕yn Fa[m] ∈ ℛ+ for m ≥ 0 and n ≥ 0. Observation 2.2.1. When m = 0, we have, for any integer n ≥ 0, a1 , when n = 0; Fa[0,n] = { 0, when n ≥ 1.

(2.2.3)

Proof. This is a direct result of (2.2.1). In other words, Fa[0,n] = a1 δ0,n where δ is the Kronecker symbol and hence Fa[0] = a1 .

34 | 2 Outer equations second part Observation 2.2.2. For any integer m ≥ 0, a, { { 1 Fa[m,0] = { 0, { m/2 { a1 a3 ,

m = 1(mod 2); }, m = 0(mod 2)

when m = 0; when m ≥ 1.

(2.2.4)

Proof. From (2.2.1), it follows that Fa[0,0] = a1 and Fa[1,0] = 0. These are the cases of m = 0 and m = 1 in the conclusion. For any integer m ≥ 2, we proceed by induction. Assume that, for any 2 ≤ l ≤ m − 1, 0, Fa[l,0] = { l/2 a1 a3 ,

when l = 1(mod 2); when l = 0(mod 2).

We prove the general case of l = m for the conclusion. On account of Fa[m,0] = a3 [Fa[m−2] ]0 = a3 Fa[m−2,0] from (2.2.1), by the assumption, the conclusion can be drawn. This observation shows that Fa[m,0] = a1 am/2 3 δm(mod 2),0 . Although (2.2.1) provides an explicit expression of the solution of equation (2.1.1), some complication might be involved with the evaluation for determining all Fa[m,n] when any m and n = |n| are chosen. When m = 1, from (2.2.3), it is only necessary to discuss n ≥ 1. By (2.1.6), Fa[1,n] = a2 [∑ yl+1 Fa[l] ] = a2 ∑ yl+1 Fa[l,n−1] . n

l≥0

l≥0

On the basis of this, Fa[1,1] via (2.2.4) can be found to be Fa[1,1] = a2 ∑ yl+1 Fa[l,0] = a1 a2 ∑ ai3 y2i+1 . i≥0

l≥0

However, in order to determine Fa[1,2] , we have to get all Fa[m,1] for m ≥ 2 beforehand. When m = 2, from (2.2.4), it is only necessary to discuss n ≥ 1 as well. By (2.1.6) with (2.2.3) and (2.2.4), Fa[2,1] = a3 Fa[0,1] + a2 ∑ yl Fa[l.0] = a1 a2 ∑ ai3 y2i . l≥1

i≥1

When m = 3, from (2.2.4), Fa[3,0] = 0. It is only necessary to consider n ≥ 1. By (2.1.6) and then (2.2.4), Fa[3,1] = a3 Fa[1,1] + a2 ∑ yi Fa[i+1,0] = 2a1 a2 a3 ∑ at3 y2t+1 . i≥1

t≥0

2.2 Solution restricted outer case

| 35

When m ≥ 4, by (2.1.6), it follows that Fa[m,1] = a3 Fa[m−2,1] + a2 ∑ yi Fa[i+m−2,0] .

(2.2.5)

i≥1

On the basis of (2.2.5), Fa[m,1] for n = 1 and for m ≥ 1 has the form of i

Fa[m,1]

a1 a2 ∑i≥1 a3 y2i+1 , { { { { { {a1 a2 ∑i≥1 ai3 y2i , ={ { {2a1 a2 a3 ∑t≥0 at3 y2t+1 , { { { {a3 Fa[m−2,1] + a2 ∑i≥1 yi Fa[i+m−2,0] ,

when m = 1; when m = 2; when m = 3;

(2.2.6)

when m ≥ 4.

Furthermore, a compact form of Fa[m,1] for m ≥ 1 can then be obtained. Lemma 2.2.3. For any integer m ≥ 1, a y2

Fa[m,1]

s−1 3 {sa1 a2 a3 ∫y 1−a3 y2 , ={ a3 y s−2 {sa a1 a2 a3 ∫y 1−a3 y2 ,

when m = 2s, s ≥ 1; when m = 2s − 1, s ≥ 1,

(2.2.7)

where sa = 2 + (s − 2)a3 . Proof. When m = 1, 2 and 3, Fa[m,1] is found from (2.2.6). Because of the results for Fa[2,1] and Fa[3,1] , respectively, Fa[2s,1] and Fa[2s−1,1] for s = 1 are checked to satisfy (2.2.7). For s ≥ 2, assume that, for any integer 1 ≤ l ≤ s − 1, the conclusion is true. When m = 2s, s ≥ 2, by (2.2.6), (2.2.4) and the inductive assumption, Fa[2s,1] = a3 Fa[2s−2,1] + a2 ∑ yi Fa[i+2s−2,0] = sa1 a2 as−1 3 ∫ i≥1

y

a3 y 2 . 1 − a3 y 2

When m = 2s − 1, s ≥ 2, by (2.2.6), (2.2.4) and the inductive assumption, Fa[2s−1,1] = a3 Fa[2s−3,1] + a2 ∑ yi Fa[i+2s−3,0] = sa a1 a2 as−1 3 ∫ i≥1

y

y 1 − a3 y 2

where sa = 2 + (s − 2)a3 . Therefore, the lemma is proved. From Fa[m,0] and Fa[m,1] , Fa[m,2] can be derived. For example, by (2.1.6) and (2.2.7), we have a y2

Fa[1,2] = ∑ yi+1 Fa[i,1] i≥0

s−1 3 {sa1 a2 ∑s≥0 y2s+1 a3 ∫y 1−a3 y2 , ={ a3 y s−2 {sa a1 a2 ∑s≥0 y2s+2 a3 ∫y 1−a3 y2 ,

where sa = 2 + (s − 2)a3 .

when i = 2s, s ≥ 0; when i = 2s + 1, s ≥ 0, (2.2.8)

36 | 2 Outer equations second part Theorem 2.2.4. The solution of equation (2.1.1) is determined by Fa[m,n] , m ≥ 0, for any integer n ≥ 0 given from 0 to n in the natural order:

Fa[m,n]

Fa[m,0] (= a1 ) shown by (2.2.4) { { { { { for m ≥ 0, { { { { { { F { a[m,1] shown by (2.2.7) for m ≥ 1 { { { { { and Fa[0,1] = 0 by (2.2.3), ={ { a3 Fa[m−2,n] + a2 ∑i≥1 yi Fa[i+m−2,n−1] { { { { { { provided from Fa[m−2,n] and Fa[i+m−2] { { { { { { known for all i ≥ 1 and Fa[0,n] = 0 { { { by (2.2.3), {

when n = 0; when n = 1;

(2.2.9)

when n ≥ 2.

Proof. From (2.2.3), all Fa[0,n] for n ≥ 0 are known. From (2.2.4), all Fa[m,0] for m ≥ 0 are known. In the natural order of n = 1, 2, . . ., we proceed by induction on n from n = 0. For n chosen, we proceed by induction on m from m = 0, to determine Fa[m,n] . On the basis of (2.2.6), the conclusion can be drawn.

2.3 Explicisions restricted outer case Although the solution of (2.1.1) can be evaluated by (2.2.9), because all Fm,n have the form of an infinite sum, some compact forms are still needed to seek for usage. In order to accomplish this task, a new parameter has to be introduced so that the solution fa can be partitioned into a series of polynomials. For any power vector n = (n1 , n2 , n3 , . . .) of y in a term of a function f ∈ ℛ{x, y}, let π(n) = ∑ ini , i≥1

(2.3.1)

which is called the inner size of the term. Observation 2.3.1. Let s and t be two power vectors of y, then we have π(s + t) = π(s) + π(t). Proof. Because of i(si + ti ) = isi + iti for i ≥ 1, the conclusion can be drawn from (2.3.1). This observation enables us to determine the inner size of each term in a function product fg for f , g ∈ ℛ{x, y}. For f ∈ ℛ{x, y}, let π(Fa[m,n] ) be the set of all π(n) for a power vector n of y in Fa[m,n] . Lemma 2.3.2. Let ℐm be the set of all power vectors in Fa[m] for m ≥ 1, then, for any n ∈ ℐm , π(n) = m(mod 2).

2.3 Explicisions restricted outer case

| 37

Proof. When n = 0, from (2.2.4), Fa[m,0] is constant for any m ≥ 0, a triviality. For n = 1, by (2.2.6), it can be checked that m = π(n)(mod 2). For any integer n ≥ 2, assume m = π(Fa[m,n−1] ), m ≥ 1, and i = π(Fa[i,n] )(mod 2), i ≤ m − 1. To prove m = π(Fa[m,n] )(mod 2), when m = 1. From (2.1.6), Observation 2.3.1 and the inductive assumption, π(Fa[1,n] ) = π(a2 ∑ yl+1 Fa[l,n−1] ) = (l + 1) + l = 1(mod 2). l≥0

This is the case m = 1 of the conclusion. In general, for m ≥ 2, by (2.1.6), π(Fa[m,n] ) = π(a3 Fa[m−2,n] + a2 ∑ yl−m+2 Fa[l,n−1] ). l≥m−1

By the assumption, π(Fm−2,n ) = m−2 = m(mod 2) and (l−m+2)+π(Fl,n−1 ) = (l−m+2)+l = m(mod 2). This is the conclusion. This lemma enables us to write s = (m + π(n))/2 such that, for any integer m ≥ 1, s (s ≥ 1) is also an integer. Denote Ha[s] = ⟨fa ⟩s = fa |m+π(n)=2s for the solution fa ∈ ℛ{x, y} of equation (2.1.1), then fa = a1 + ∑ Ha[s] . s≥1

(2.3.2)

Hence, whenever the Ha[s] for s ≥ 1 are known, fa is determined. Observation 2.3.3. Given s ≥ 1. The inner size π of a term of the function fa ∈ ℛ{x, y} is neither greater than 2s nor less than 1. Proof. It is seen that 2s = m + π, 1 ≤ π ≤ 2s from m, π ∈ ℤ+ . This is the conclusion. On the basis of Observation 2.3.3, from (2.1.6), 2s

Ha[s] = ∑ xm ⟨Fa[m] ⟩2s−m . m=1

(2.3.3)

Observation 2.3.4. For s ≥ 1, Ha[s] is independent of yi for i ≥ 2s. Proof. By contradiction. If Ha[s] has a yi with i ≥ 2s, then there is a power vector n ∈ ℐ (Ha[s] ) such that ni ≥ 1 and hence π(n) ≥ 2s. This is a contradiction to the fact that the power of x is not 0 in Ha[s] . Because of Observation 2.3.4, the infinite vector y may be replaced by the finite vector y2s−1 of 2s − 1 dimension. This implies that Ha[s] ∈ ℛ{x, y2s−1 } where y2s−1 = (y1 , y2 , . . . , y2s−1 ).

(2.3.4)

38 | 2 Outer equations second part Lemma 2.3.5. For any integer s ≥ 1, Hs is a polynomial of y2s−1 in ℛ{x, y2s−1 } with degree at most 2s. Proof. On account of n2s−1 ≥ 0, for any integer 1 ≤ m ≤ 2s, max{|n| | π(n) = 2s − m, n ∈ ℐ (Fa[m] )} 󵄨 󵄨 = 󵄨󵄨󵄨(2s − m, 0, 0, . . . , 0)2s−1 󵄨󵄨󵄨 = 2s − m. In (2.3.2), each term of xm ⟨Fm ⟩2s−m is a polynomial of y2s−1 with degree at most 2s. Therefore, the conclusion can be drawn. On the basis of (2.3.3) and (2.1.6), for any integer s ≥ 1, 2s−2

Ha[s] = a2 x ∑ yl+1 ⟨Fa[l] ⟩2s−l−2 l=0

2s

2s−2

m=2

l=m−1

+ a3 ∑ xm (⟨Fa[m−2] ⟩2s−m + a2 ∑ yl−m+2 ⟨Fa[l] ⟩2s−l−2 ). By (2.2.4), 2s−2

2s−1

l=0

m=2

Ha[s] = a2 x ∑ yl+1 ⟨Fa[l] ⟩2s−l−2 + a3 ∑ xm Aa[m,s] + a1 as3 x2s

(2.3.5)

where, for 2 ≤ m ≤ 2s − 1, 2s−2

Aa[m,s] = ⟨Fa[m−2] ⟩2s−m + a2 ∑ yl−m+2 ⟨Fa[l] ⟩2s−l−2 . l=m−1

On the basis of (2.3.5), Ha[s] , 0 ≤ s ≤ 2, can be determined as follows. From the initial condition of (2.1.1), Ha[0] = a1 . For s = 1, Ha[1] = a2 xy1 ⟨Fa[0] ⟩0 + a1 a3 x2 = a1 a2 xy1 + a1 a3 x2 .

(2.3.6)

For s = 2, 2

Ha[2] = a2 x(y1 ⟨F0 ⟩2 + y2 ⟨F1 ⟩1 + y3 ⟨F2 ⟩0 ) + a3 x2 (⟨F0 ⟩2 + a2 ∑ yl ⟨Fl ⟩2−l ) l=1

2

+ a3 x3 (⟨F1 ⟩1 + a2 ∑ yl−1 ⟨Fl ⟩2−l ) + a1 a23 x4 . l=2

By (2.2.4) and (2.2.7)–(2.2.9) with rearrangement of terms, we have Ha[2] = a2 (y1 y2 + a1 a3 y3 )x + a3 (a2 y12 + a1 a3 y2 )x2 + a3 (1 + a1 a2 a3 )y1 x3 + a1 a23 x4 .

(2.3.7)

2.3 Explicisions restricted outer case

| 39

For the general case of s ≥ 3, it is only necessary to evaluate ⟨Fa[m] ⟩t (0 ≤ m, t ≤ 2(s − 1)) for determining Ha[s] from (2.3.5). Because of ⟨Fa[m] ⟩0 = Fa[m,0] and ⟨Fa[0] ⟩t = Fa[0,t] , respectively, from (2.2.4) and (2.2.3), it is only necessary to consider 1 ≤ m, t ≤ 2(s − 1) in what follows. On the basis of (2.1.6), when m = 1, ⟨Fa[1] ⟩t = a2 ∑ ⟨yl+1 Fa[l] ⟩t = a2 ∑ yl+1 ⟨Fa[l] ⟩t−l−1 . l≥0

l≥0

By considering t − l − 1 ≥ 0, we have t−1

(2.3.8)

⟨Fa[1] ⟩t = a2 ∑ yl+1 ⟨Fa[l] ⟩t−l−1 . l=0

When m ≥ 2, ⟨Fa[m] ⟩t = a3 ⟨Fm−2 ⟩t + a2 ∑ ⟨yl−m+2 Fl ⟩t l≥m−1

= a3 ⟨Fa[m−2] ⟩t + a2 ∑ yl−m+2 ⟨Fa[l] ⟩t−l+m−2 . l≥m−1

By considering t − l + m − 2 ≥ 0, it becomes t+m−2

a3 ⟨Fa[m−2] ⟩t + a2 ∑ yl−m+2 ⟨Fa[l] ⟩t−l+m−2 . l=m−1

Then, by i = l − m + 1, we have t−1

⟨Fa[m] ⟩t = a3 ⟨Fa[m−2] ⟩t + a2 ∑ yi+1 ⟨Fa[m+i−1] ⟩t−i−1 . i=0

(2.3.9)

Thus, for any integers m, t ≥ 1, t−1

⟨Fa[m] ⟩t = a3 ⟨Fa[m−2] ⟩t + a2 ∑ yl+1 ⟨Fa[m+l−1] ⟩t−l−1 . l=0

(2.3.10)

Notably, for any integer m < 0, ⟨Fa[m] ⟩t = 0 is shown. As a matter of fact, all ⟨Fa[m] ⟩t , m, t ≥ 0, can be evaluated by (2.3.10) starting from ⟨Fa[0] ⟩0 = a1 . This enables us to deal with all ⟨Fa[m] ⟩t , m, t ≥ 0, as is well known. Theorem 2.3.6. The solution of equation (2.1.1) is determined by Ha[s] for s ≥ 0 in the form of finite sums with all terms positive when a1 , a2 , a3 ∈ ℝ+ , a1 , Ha[s] = { s 2s m a1 a3 x + a3 ∑2s−1 m=1 (⟨Fa[m−2] ⟩2s−m + a2 Σa[m,s] )x , where, for 1 ≤ m ≤ 2s − 1, 2s−m−1

Σa[m,s] = ∑ yi+1 ⟨Fa[i+m−1] ⟩2s−i−m−1 . i=0

when s = 0; when s ≥ 1,

(2.3.11)

40 | 2 Outer equations second part Proof. From the initial condition of equation (2.1.1), the case of s = 0 in (2.3.11) is done. When s ≥ 1, on the basis of 2s−m−1

∑ ⟨yl−m+2 Fa[l] ⟩2s−m = ∑ yi+1 ⟨Fa[i+m−1] ⟩2s−i−m−1 = Σa[m,s] , i=0

l≥m−1

from (2.3.3), 2s−1

Ha[s] = a1 as3 x2s + ∑ a3 (⟨Fa[m−2] ⟩2s−m + a2 Σa[m,s] )x m . m=1

Therefore, from (2.3.8) and (2.3.9), the conclusion can be drawn. Then let us go back to (2.1.10) on the basis of Lemma 2.3.5. Because Hs is a polynomial of y2s−1 with term degrees between 1 and 2s for s given, the matrix Ysa[rto] is 2s × 2s with y = y2s−1 , denoted by Ya[s] = (ya[i,j] )2s×2s , where

ya[i,j]

a2 yt+2 , when 0 ≤ t = j − i ≤ 2s − 3; { { { { { {a2 y1 , when j − i = −1; ={ {a3 , when j − i = −2; { { { { when j − i ≤ −3. {0,

(2.3.12)

Theorem 2.3.7. The solution of equation (2.1.3) for size s ≥ 2 given is determined by an explicision of the form s−1

T

fTa[2s] = ∑ Yia[s] (a2 y1 (e2s )1 + a3 (e2s )2 ) . i=0

(2.3.13)

Proof. On the basis of (2.3.12), from 1 + π(y1 y2s ) = 2s + 2, Yta[s] = 0 when t ≥ s. On account of Observations 2.3.1, Observation 2.3.3 and Observation 2.3.4, from (2.1.10), the conclusion can be drawn.

2.4 Restrictions restricted outer case Because of the meaningfulness for a = (1, 1, 1, 1) of equation (2.1.1) in counting rootisomorphic classes of corresponding maps, the equation 2 { {x ∫ yδx,y (uf |x=u ) = (1 − x )f − 1; y { { f | { x=0,y=0 = 1,

(2.4.1)

is considered for f ∈ ℛ{x, y}. One might see the first equation in equation (2.4.1) from [36] (Liu YP, 1987) for determining the enufunction of restricted outer planar maps with the vertex-partition vector as parameter.

2.4 Restrictions restricted outer case

| 41

Theorem 2.4.1. Equation (2.4.1) is well defined on ℛ{x, y}. Proof. This is a direct result of Theorem 2.1.4 for a = (1, 1, 1, 1). This theorem tell us that whenever the initial condition is meaningful for such maps, the solution of equation (2.4.1) is just the enufunction of outer planar restrict maps with the vertex-partition vector as parameter. m,n Because of f ∈ ℛ{x, y} determined by Fm,n = 𝜕x,y f ||n|=n for integers m, n ≥ 0, we are allowed to only discuss all Fm,n ∈ ℛ{y} instead of f itself. Let 2s = m + π(n) for n = |n| be double the size of outer planar maps with rootvalency m and a number n of all non-root-vertices. Observation 2.4.2. For the size s ≥ 0 given, we have 0 ≤ m ≤ 2s, 0 ≤ n ≤ s

and

yi = 0

(i ≥ 2s).

(2.4.2)

Proof. Because of the vertex map and the loop-map of order 1 with all semi-edges incident to the vertex in restricted outer planar maps, the first inequality follows. Because of all non-root-vertices of valency 1 allowable in a restricted outer planar map, the second inequality is seen. Because of there being only one non-root-vertex of valency 2s − 1 available in a restricted outer planar map, the third inequality is done. On the basis of Theorem 2.2.4, the solution of equation (2.4.1) can be determined by a recursion in the form of a finite sum with all terms positive. Theorem 2.4.3. For any integer s ≥ 0, the solution of equation (2.4.1) is determined by

Fm,n

1, when m = 0; { { }, { { { { 0, when m ≥ 1 { { m s−1 { { when m = 0(mod 2); ∑i=1 y2i , { 2 { { }, { m+1 s = { 2 ∑i=1 y2i−1 , when m = 1(mod 2); { { { 0, when m = 0; { { { { 2s { { y F , when m = 1; ∑ { { i=0 i+1 i,n−1 { { 2s−m+2 yi Fi+m−2,n−1 , when m ≥ 2; { Fm−2,n + ∑i=1

when n = 0; when n = 1; } } } , } } } }

(2.4.3)

when n ≥ 2

for 0 ≤ m ≤ 2s, 0 ≤ n ≤ s and y2s−1 . Proof. By considering Observation 2.4.2 and (2.2.9) for a = (a1 , a2 , a3 ) = (1, 1, 1) and a0 = a1 , the conclusion can be drawn. Because of Theorem 2.3.6, it can be seen that, for any s ≥ 0 given, Hs = Ha[s] |a=1 is a polynomial of y2s−1 with degree 2s where 1 = (1, 1, 1).

42 | 2 Outer equations second part Theorem 2.4.4. The solution of equation (2.4.1) is determined by Ha[s] for s ≥ 0 in the form of a finite sum with all terms positive, 1, when s = 0; Hs = { 2s 2s−1 m x + ∑m=1 (⟨Fm−2 ⟩2s−m + Σm,s )x , when s ≥ 1,

(2.4.4)

where for 1 ≤ m ≤ 2s − 1, Σm,s is given in Theorem 2.3.6. Proof. This is a direct result of Theorem 2.3.6 when a1 = a2 = a3 = 1. On the basis of (2.3.12) and (2.3.13), an explicision of the solution of equation (2.4.1) can also deduced. Theorem 2.4.5. The solution of equation (2.4.1) for size s ≥ 2 given is determined by the explicision in the form of s−1

fT2s = ∑ Yis (y1 (e2s )1 + (e2s )2 )

T

(2.4.5)

i=0

where f2s = (F1 , F2 , . . . , F3s ) and Ys are given in (2.3.13). Proof. This is a direct result of Theorem 2.3.7 in the case of a1 = a2 = a3 = 1 and a0 = a1 . Example 2.4.1. Root-isomorphic classes of restricted outer planar maps of size at most 2. By a restricted outer planar map is meant that the interior of any self-loop without another self-loop and root-link is not on an at least 3-multiple edge. For a size not greater than 2, on the basis of Theorem 2.4.4, we have

0

1 F1 y2 [ [ F2 y 1 ( )=[ )+( 1 [( F3 1 1 [ F4 0 1 [ y1 y1 y2 + y3 1 y 2 + y2 =( )+( 1 ) 0 2y1 0 1 y1 + y1 y2 + y3 1 + y12 + y2 =( ). 2y1 1

0

y3 y2 y1 1

0 y3 y2 y1

y1 0 ] 1 0 ] )] ( ) 0 y3 ] ] 0 y2 ]

In Figure 2.4.1, it is seen that F1 = y1 + y1 y2 + y3 = F1,1 + F1,2 + F1,3 = (a) + (b) + (c) = (y1 ) + (y1 y2 ) + (y3 ), F2 = 1 + y2 + y12 = F2,0 + F2,1 + F2,2 = (d) + (f ) + e = (1) + (y2 ) + (y12 ), F3 = 2y1 = F3,1 = 2(g) = 2(y1 ), F4 = 1 = F4,0 = (h) = (1).

2.5 Ordinary outer model | 43

Figure 2.4.1: Classes of restricted outer planar maps with size at most 2.

2.5 Ordinary outer model Second, consider the equation for g ∈ ℛ{x, y} 2 { {a2 x ∫ yδx,y (ug|x=u ) = (1 − x φ(x))g − a1 ; y { { g| { x=0⇒y=0 = a0 ,

(2.5.1)

where a0 , a1 , a2 ∈ R+ with a = (a0 , a1 , a2 ) and φ(x) ∈ ℛ+ {x} such that, for any integer n ≥ 0, ϕm = 𝜕xm φ ∈ ℛ+ and ϕ2i+1 = 0, i ≥ 0. This is equation (4) in Introduction. Because of a solution of equation (2.5.1) for a0 = a1 = a2 = 1 involved with ordinary outerplanar maps shown in [33] (Liu YP, 1986) and [36] (Liu YP, 1987), this equation is called an ordinary outer model. On g ∈ ℛ{x, y} determined by Ga[m] = 𝜕xm g ∈ ℛ{y} for m ≥ 0, we have δx,y (ug|x=u ) = ∑ Ga[m] ( m≥0

xm+1 − ym+1 ) = ∑ xl ∑ ym−l Ga[m] . x−y l≥0 m≥l

The first equality of equation (2.5.1) is involved with the term x ∫(yδx,y (ug|x=u )) = ∑ xm ∑ yl−m+2 Ga[m] . y

m≥1

l≥m−1

Observation 2.5.1. If a1 ≠ a0 , then equation (2.5.1) has no solution.

(2.5.2)

44 | 2 Outer equations second part Proof. By contradiction. Assume g is a solution of equation (2.5.1). From the initiation, it follows that 0 = Ga[0] −a1 = a0 −a1 ⇒ a1 = a0 . A contradiction to the given condition occurs. This observation enables us to only address a0 = a1 and a1 a2 ≠ 0 without loss of generality for equation (2.5.1), with two constant parameters: a1 and a2 . Hence, we write a = (a1 , a2 ) instead of (a0 , a1 , a2 ). Lemma 2.5.2. Equation (2.5.1) on ℛ{x, y} is equivalent to the system a1 , when m = 0 (the initiation!); { { { Ga[m] = {a2 ∑l≥0 yl+1 Ga[l] , when m = 1; { { m−2 {∑i=0 ϕi Ga[m−i−2] + a2 ∑l≥m−1 yl−m+2 Ga[l] , when m ≥ 2, (2.5.3) on ℛ{y}. Proof. By substituting (2.5.2) for the first equation of equation (2.5.1) under Observation 2.5.1. Via equating the coefficients of the x-terms with the same power on two sides of equation (2.5.2), the equivalence between equation (2.5.1) and equation (2.5.3) under the initiation is seen. This is the conclusion. On the basis of the lemma above, we are allowed to evaluate Ga[m,n] = Ga[m] ||n|=n ∈ ℛ{y} for n and then m from 0 on by the order of natural numbers. Lemma 2.5.3. When n = 0, we have, for m ≥ 0,

Ga[m,0]

a1 , { { { { { { 0, { { { = {a1 ϕ0 , { { { { 0, { { { { p−1 {∑i=0 ϕ2i Ga[2p−2−2i,0] ,

if m = 0; if m = 1;

(2.5.4)

if m = 2; if m = 1(mod 2), m ≥ 3; if m = 2p, m ≥ 4.

Proof. By induction on m ≥ 0. Because of Ga[0,0] = a1 from the initiation, Ga[1,0] = 0 and Ga[2,0] = 0 are checked from (2.5.3). For m ≥ 3, assume that all cases for 1 ≤ i ≤ m−1 are known. We show that Ga[m,0] is determined as (2.5.4), by considering that Ga[m,0] is determined by Ga[i,0] (0 ≤ i ≤ m − 1). From (2.5.3), the conclusion can be drawn. On the basis of this lemma, by (2.5.3), Ga[m,1] for m ≥ 0 can be evaluated. Lemma 2.5.4. When n = 1, we have, for m ≥ 0,

Ga[m,1]

0, { { { { { {a1 a2 y1 + a2 ∑i≥1 Ga[2i,0] y2i+1 , { { = {a1 a2 ϕ0 y2 + a2 ∑i≥2 Ga[2i,0] y2i , { { { {ϕ1 Ga[1,1] + a2 ∑i≥2 y2i Ga[2i,0] , { { { m−3 {∑i=0 ϕi Ga[m−i−2,1] + a2 ∑i≥1 y2i−m−2 Ga[2i,0] ,

if m = 0; if m = 1; if m = 2; if m = 3; if m ≥ 4.

(2.5.5)

2.6 Solution ordinary outer case

| 45

Proof. By a similar procedure to the proof of Lemma 2.5.3. This lemma shows that Ga[m,1] for any integer m ≥ 1 can be evaluated recursively on the basis of the initial value of Ga[0,1] = 0. Theorem 2.5.5. Equation (2.5.1) for a0 , a1 , a2 ∈ ℝ+ is well defined on ℛ{x, y} if, and only if, a0 = a1 . Proof. From Observation 2.5.1, the necessity is done. For the sufficiency, we proceed by induction. On account of Lemma 2.5.2 and Lemma 2.5.3, for any n ≥ 2, assume all Ga[m,n−1] (m ≥ 0) and Ga[k,n] (0 ≤ k ≤ m − 1) are known while Ga[0,n] = 0. By (2.5.3), we evaluate a1 , when m = 0; { { { Ga[m,n] = {a2 ∑l≥0 yl+1 Ga[l,n−1] , when m = 1; { { m−2 ϕ G + a y G , when m ≥ 2. ∑ ∑ 2 l≥m−1 l−m+2 a[l,n−1] { i=0 i a[m−i−2,n]

(2.5.6)

Because of all Ga[m,n−1] (m ≥ 0) and Ga[k,n] (0 ≤ k ≤ m − 1) while Ga[0,n] = 0 is known from the assumption, Ga[m,n] for m ≥ 0 and n ≥ 0 are determined for a solution of equation (2.5.1). By considering the uniqueness of the procedure from the initiation, this solution is unique.

2.6 Solution ordinary outer case Although (2.5.6) provides us with a solution of equation (2.5.1) formally, it is still necessary to find a sum of finite terms for the solution when some parameter has suitably been chosen. For a term of xm yn in Ga[m,n] , let π(n) = ∑ ini i≥1

(2.6.1)

where n = |n| and n = (n1 , n2 , n3 , . . .). Observation 2.6.1. For any term of xm yn in Ga[m,n] , we have m = π(n)(mod 2). Proof. By induction on n ≥ 0. On the basis of Lemma 2.5.3 and Lemma 2.5.4, for n = 0 and n = 1, the conclusion is checked to be true for m = 0 and m = 1. For m ≥ 2, by induction on m, assume that the conclusion is true for any 0 ≤ t ≤ m − 1. From the additivity of π(n) and the assumption, the conclusion is true for any m ≥ 2. Then, for n ≥ 2, by assumption on n, for any 0 ≤ k ≤ n − 1, the conclusion is true. We prove the conclusion for k = n. Because Ga[0,n] = 0 from (2.5.6) and Ga[1,n] = a2 ∑ yl+1 Ga[1,n−1] l≥0

46 | 2 Outer equations second part from Lemma 2.5.2, the assumption tells us that, for any n in Ga[1,n−1] , π(n) = 1(mod 2). This is the conclusion for m = 1. For m ≥ 2, assume that the conclusion is true for all 0 ≤ t ≤ m − 1 while n ≥ 2. On the basis of this assumption and (2.5.3), the additivity of π leads to the conclusion for any n ≥ 2. This observation enables us to take 2s = m+π(n) for a term of Ga[m,n] in Ga[m,n] such that n = |n|, called semi-size, i. e., twice the size s ≥ 0. Obviously, s is a non-negative integer. On account of this, let us write the solution of equation (2.5.1) as ga = ∑ Ga⟨s⟩ = a1 + ∑ Ga⟨s⟩ s≥0

s≥1

(2.6.2)

where for Ga⟨s⟩ ∈ ℛ{x, y} all terms have the same size s. Observation 2.6.2. For the size s ≥ 1 given, we have 1 ≤ m ≤ 2s in Ga⟨s⟩ . Proof. On the basis of (2.5.6), by a similar procedure to the proof of Observation 2.6.1, the conclusion can be drawn for m in Ga⟨s⟩ . This observation enables us to only discuss Ga[m,n] on 0 ≤ m ≤ 2s for s ≥ 0 arbitrarily given. Observation 2.6.3. For the size s ≥ 1 given, we have 0 ≤ n ≤ s. Proof. On the basis of (2.5.6), by similar procedure to the proof of Observation 2.6.1, the conclusion can be drawn for n in Ga⟨s⟩ . This observation enables us to only discuss Ga[m,n] on 0 ≤ n ≤ s for s ≥ 0 arbitrarily given. Observation 2.6.4. For integer s ≥ 1 given, we see that Ga[m,n] is independent of yk for k ≥ 2s. Proof. On the basis of (2.5.6), by a similar procedure to the proof of Observation 2.6.1, the conclusion can be drawn for yk , k ≥ 2s, in Ga⟨s⟩ . This observation enables us only to discuss Ga[m,n] on yk , 0 ≤ k ≤ 2s − 1 for the size s ≥ 1 arbitrarily given. On the basis of (2.5.6), the three observations above enable us to evaluate the solution of equation (2.5.1) in the form of a finite sum with all terms positive. Theorem 2.6.5. For size s ≥ 0 given, the solution ga[odo] of equation (2.5.1) is determined [odo] = ga[odo] ||n2s−1 |=n for 0 ≤ m ≤ 2s and 0 ≤ n ≤ s, by Ga[m,n] = Ga[m,n]

Ga[m,n]

a1 , when m = 0 ⇒ n = 0 & s = 0; { { { 2s−2 = {a2 ∑l=0 yl+1 Ga[l,n−1] , when m = 1; { { 2s−m+1 m−2 yl+1 Ga[l+m−1,n−1] , when m ≥ 2. {∑i=0 ϕi Ga[m−i−2,n] + a2 ∑l=0

(2.6.3)

2.6 Solution ordinary outer case | 47

Proof. When m = 0, from (2.5.3), it is the initiation of equation (2.5.1). When m = 1, by (2.5.3) with Observation 2.6.3 and Observation 2.6.4, 2s−2

Ga[1,n] = a2 ∑ yl+1 Ga[l,n−1] = a2 ∑ yl+1 Ga[l,n−1] , l≥0

l=0

which is the same as shown in (2.6.3). Then, when m ≥ 2, on account of (2.5.3), m−2

Ga[m,n] = ∑ ϕi Ga[m−i−2,n] + a2 ∑ yl−m+2 Ga[l,n−1] i=0

l≥m−1

m−2

2s−m+1

i=0

l=0

= ∑ ϕi Ga[m−i−2,n] + a2 ∑ yl+1 Ga[l+m−1,n−1] , which is the case of m ≥ 2 in (2.6.3). In consequence, by employing Lemma 2.5.2, the conclusion can be drawn. The solution can also be expressed in the way shown by Ga⟨s⟩ for s ≥ 0 via (2.6.2). Lemma 2.6.6. For any integer s ≥ 1 given, Ga⟨s⟩ has x as a factor. Proof. On the basis of Theorem 2.6.5, from (2.5.4) and (2.5.5), the conclusion can be derived. This lemma enables us to realize that, for any integer s ≥ 1, all terms in Ga⟨s⟩ come with the minimum of x degrees at least 1. Lemma 2.6.7. For any integer s ≥ 1 given, Ga⟨s⟩ is a polynomial of y2s−1 with degree between 0 and s. Proof. On the basis of Theorem 2.6.5. From (2.5.4) and (2.5.5), the conclusion can be derived. This lemma enables us to only consider Ga[m,n] , 0 ≤ n ≤ s, for Ga⟨s⟩ when any s ≥ 1 is given. Lemma 2.6.8. For any integer s ≥ 0 given, Ga⟨s⟩ is a polynomial of x with degree between 0 and 2s. Proof. On the basis of Theorem 2.6.5. From (2.5.4) and (2.5.5), the conclusion can be derived. This lemma enables us to only consider Ga[m,n] , 1 ≤ m ≤ 2s, for Ga⟨s⟩ when any s ≥ 1 is given. Theorem 2.6.9. For any integer s ≥ 0 given, the solution ga[odo] of equation (2.5.1) is determined by

48 | 2 Outer equations second part

Ga⟨s⟩

a1 , { { { = {a1 a2 y1 x + a1 ϕ0 x2 , { { 2s m {∑m=1 x ⟨Ga[m] ⟩2s−m

when s = 0; when s = 1;

(2.6.4)

when s ≥ 2,

where s

Ga[m] = ∑ Ga[m,n] n=0

(2.6.5)

such that all Ga[m,n] for 0 ≤ n ≤ s are determined by Theorem 2.6.5. Proof. Because of 2s = m + π(n), from m = 0 and n = 0, it is seen that s = 0. Thus, Ga⟨0⟩ = a1 . This is the initiation of equation (2.5.1). When s = 1, by Observation 2.6.2, m is only allowed to be 1 and 2. From Lemmas 2.6.6–2.6.8 and (2.6.3), 2

1

Ga⟨1⟩ = ∑ ∑ ⟨Ga[m,n] ⟩2s−m = x(a2 a1 y1 ) + x2 (a1 ϕ0 ). m=1 n=0

This is the case of s = 1 in (2.6.4). When s ≥ 2, on the basis of Lemma 2.6.6 and Lemma 2.6.8, 2s

Ga⟨s⟩ = ∑ xm ⟨Ga[m] ⟩2s−m . m=1

Then, by Lemma 2.6.7, the conclusion can be drawn.

2.7 Explicisions ordinary outer case On the basis of (2.5.3), the following equation is established: (I − Ya[odo] )gT = a1 bT

(2.7.1)

where I is the unity matrix, g = (Ga[1] , Ga[2] , Ga[3] , . . .), b = (b1 , b2 , b3 , . . .), y1 , when i = 1; { { { bi = {ϕi , when i = 0(mod 2), i ≥ 2; { { {0, when i = 1(mod 2), i ≥ 2,

(2.7.2)

Ya[odo] = D + a2 U,

(2.7.3)

in which D = (di,j )i,j≥1 and U = (ui,j )i,j≥0 such that yk+2 , { { { di,j = {y1 , { { {0,

when j − i = k ≥ 0; when j − i = −1; when j − i ≤ −2,

(2.7.4)

2.7 Explicisions ordinary outer case

| 49

and 0, { { { ui,j = {ϕk−2 , { { {0,

when j − i ≥ −1; when j − i ≤ −k = 0(mod 2), k ≥ 2;

(2.7.5)

when j − i ≤ −k = 1(mod 2), k ≥ 2.

On the basis of Observations 2.6.2–2.6.4, for integer s ≥ 0 given as the size, it is only necessary to consider ga[2s] = (Ga[1] , Ga[2] , . . . , Ga[2s] ) ∈ ℛ{y}2s , ba[2s] and Ya[2s×2s] instead of, respectively, g, b and Ya[odo] in (2.7.1)–(2.7.5). Observation 2.7.1. For size s ≥ 0 given, we have (Ya[2s×2s] )k = 02s×2s

(2.7.6)

for k ≥ s. [s] ) ≥ 2s−1, Observation 2.6.3 leads to the desired conclusion. Proof. Because of π(y1,1

This observation combined with those in Section 2.6 enables us to evaluate all solutions for size less than and equal to s given arbitrarily. Observation 2.7.2. For size s ≥ 1 given, we have s−1

(I − Ya[2s×2s] )−1 = ∑ (Ya[2s×2s] )k . k=0

(2.7.7)

Proof. A result of Observation 2.7.1. On the basis of the above two observations, equation (2.7.1) becomes a system of finite equations for integer s ≥ 0 given. Observation 2.7.3. For an integer s ≥ 1 given, equation (2.5.1) is equivalent to the finite system of equations as gTa[2s] = Ya[2s×2s] gTa[2s] + bTa[2s] .

(2.7.8)

Proof. Because of the equivalence between equation (2.5.3) and equation (2.5.1), the conclusion can be drawn. On the basis of this observation, an explicision of the solution of equation (2.5.1) is then evaluated. Theorem 2.7.4. For size s ≥ 1 given, the solution ga[2s] ∈ ℛ{y}2s of equation (2.5.1) has an explicision s−1

gTa[2s] = ∑ (Ya[2s×2s] )k bTa[2s] . k=0

Proof. This is a direct result of (2.7.7).

(2.7.9)

50 | 2 Outer equations second part Let ga⟨s⟩ ∈ ℛ{x, y} be partitioned as s

ga⟨s⟩ = ∑ Ga⟨i⟩ i=0

(2.7.10)

for s ≥ 0 where all Ga⟨i⟩ ∈ ℛ{x, y} as well. Theorem 2.7.5. Given integer s ≥ 1, let ga⟨s⟩ be the solution of equation (2.5.1). For 1 ≤ t ≤ s, we have 2t

Ga⟨t⟩ = ∑⟨Ga[i] ⟩π=2s−i xi i=1

(2.7.11)

where (Ga[1] , Ga[2] , . . . , Ga[2t] , ) = ga[2t] , ga[i] ∈ ℛ{y} for 1 ≤ i ≤ 2t as follows from (2.7.9). Proof. Because all Ga[i] (1 ≤ i ≤ s) are known from (2.7.9), the conclusion can be drawn. On the basis of this theorem, all Ga[m,n] for m, n ≥ 1 in the solution of equation (2.5.1) can be found as another type of an explicision. Theorem 2.7.6. For any integer s ≥ 1 given, the solution of equation (2.5.1) determined by Ga[m,n] ∈ ℛ{y} has an explicision s

Ga[m,n] = ∑[Ga⟨t⟩ ]m ||n|=n,π(n)=2t−m t=1

(2.7.12)

where ga[m] for 2s ≥ m ≥ 1 and s ≥ n ≥ 1 are given by (2.7.9). Proof. Because all Ga⟨t⟩ (1 ≤ t ≤ s) are known from (2.7.11), the conclusion can be drawn.

2.8 Restrictions ordinary outer case Now, let us go back to the particular case of equation (2.5.1) for a0 = a1 = a2 = 1. Consider the meson functional equation for g ∈ ℛ{x, y} 2 { {x ∫ yδx,y (ug|x=u ) = (1 − x ϕ(x))g − 1; { y { {g|x=0⇒y=0 = 1.

(2.8.1)

Its first equality appears in [36] (Liu YP, 1987) with the initiation determined by what is meaningful for g as the function of enumerating root-isomorphic classes of ordinary outer planar maps with the valency (power m of x) of root-vertex and vertex-partition vector (power vector n of y) given.

2.8 Restrictions ordinary outer case

| 51

Theorem 2.8.1. Equation (2.8.1) is well defined on ℛ{x, y}. Proof. This is a direct result of Theorem 2.5.5 for a0 = a1 = a2 = 1. On the basis of this theorem, we are allowed to evaluate the solution of equation (2.8.1). Theorem 2.8.2. The solution of equation (2.8.1) is determined by Gm , m ≥ 0, for any integer n ≥ 0 given from 0 to n in the natural order as 1 when m = 0; { { { Gm = {∑l≥0 yl+1 Gl , when m = 1; { { m−2 {∑l=0 ϕl Gm−l−2 + ∑l≥m−1 yl−m+2 Gl , when n ≥ 2.

(2.8.2)

Proof. This is a direct result of (2.5.3) for a0 = a1 = a2 = 1 because of Theorem 2.8.1. This is in the form of an infinite sum with all terms positive. We are urged to extract a form of Gm,n as a finite sum with all terms positive by introducing the new parameter s ≥ 0, which is the size of the solution. Theorem 2.8.3. For size s ≥ 0 given, the solution g[odo] of equation (2.8.1) is determined by [odo] [odo] ||n2s−1 |=n ∈ ℛ{y} ) = Gm Gm,n (= Gm,n

for 0 ≤ m ≤ 2s and 0 ≤ n ≤ s as

Gm,n

1, { { { 2s−2 = {∑l=0 yl+1 Gl,n−1 , { { m−2 2s−m+1 yl+1 Gl+m−1,n−1 , {∑i=0 ϕi Gm−i−2,n + ∑l=0

when m = 0 ⇒ n = 0 & s = 0; when m = 1; when m ≥ 2.

(2.8.3)

Proof. The direct result of Theorem 2.6.5 for a0 = a1 = a2 = 1 Another expression of the solution of equation (2.8.1) by G⟨s⟩ ∈ ℛ{x, y} can also be obtained on the basis of Theorem 2.8.3. Theorem 2.8.4. For any integer s ≥ 0 given, the solution g[odo] of equation (2.8.1) is determined by G⟨s⟩

1, { { { = {y1 x + ϕ0 x2 , { { 2s m {∑m=1 x ⟨Gm ⟩2s−m

when s = 0; when s = 1; when s ≥ 2,

(2.8.4)

52 | 2 Outer equations second part where s

Gm = ∑ Gm,n n=0

(2.8.5)

such that all Gm,n for 0 ≤ n ≤ s are determined by Theorem 2.8.3. Proof. This is a direct result of Theorem 2.6.9 for a0 = a1 = a2 = 1. For s ≥ 1, let g2s , b2s and Y2s×2s be taken, instead of, respectively, ga[2s] ∈ ℛ{y}2s , ba[2s] and Ya[2s×2s] in (2.7.9). Then g2s = (G1 , G2 , . . . , G2s )

(2.8.6)

provides us with the solution g⟨s⟩ = x2s gT2s of equation (2.8.1) for integer s ≥ 1. Theorem 2.8.5. For s ≥ 1, the solution g2s of equation (2.8.1) has a direct explicision as s−1

gT2s = ∑ (Y2s×2s )k bT2s . k=0

(2.8.7)

Proof. This is a direct result of Theorem 2.7.4 for a0 = a1 = a2 = 1. Let g⟨s⟩ ∈ ℛ{x, y} be the solution of equation (2.8.1), then s

g⟨s⟩ = ∑ G⟨i⟩ (∈ ℛ{x, y}) i=0

(2.8.8)

as shown in (2.7.10) when a0 = a1 = a2 = 1. Theorem 2.8.6. For any integer s ≥ 1 given, the solution of equation (2.8.1) determined by Gm,n ∈ ℛ{y} has an explicision Gm,n = [Gm ]|n|=n,π(n)+m=2s

(2.8.9)

where Gm for 2s ≥ m ≥ 1 and s ≥ n ≥ 1 are given by (2.7.9) when a0 = a1 = a2 = 1. Proof. This is a direct result of Theorem 2.7.6 for a0 = a1 = a2 = 1. Example 2.8.1. Root-isomorphic classes of ordinary outer planar maps with rootvertex valency(or rooted semi-size) and vertex-partition vector as parameters. According to Liu YP [36] (1987), [54] (1993, Theorem 2.6), or [56] (1999, pp. 76–79), the solution of equation (2.8.1) is just the enufunction for enumerating root-isomorphic classes of ordinary outer planar maps with root-vertex valency and the vertex-partition vector given when ϕi = ((2n)!)/(n!(n + 1)!), if i = 2n; 0, otherwise where integer i ≥ 0. By Theorem 2.8.5, g2s = (G1 , G2 , G3 , G4 ) for s = 2 is shown as

2.8 Restrictions ordinary outer case

0

1 G1 y2 [ [ G2 y 1 ( )=[ )+( 1 ( [ G3 1 1 [ G4 0 1 [ y1 y1 y2 + y3 1 y 2 + y2 =( )+( 1 ) 0 2y1 + y3 1 1 + y2 y1 + y1 y2 + y3 1 + y12 + y2 ). =( 2y1 + y3 2 + y2

0

y3 y2 y1 1

0 y3 y2 y1

| 53

0 y1 ] 0 ] 1 )] ( ) ] y3 ] 0 y2 1 ]

On the basis of this, from (2.8.4), G⟨1⟩ and G⟨2⟩ are then obtained: G⟨1⟩ = x⟨y1 + y1 y2 + y3 ⟩π=1 + x2 ⟨1 + y12 + y2 ⟩π=0 = x(y1 ) + x2 (1) = y1 x + x2 and G⟨2⟩ = x⟨y1 + y1 y2 + y3 ⟩π=3 + x2 ⟨1+y12 +y2 ⟩π=2 +x3 ⟨2y1 +y3 ⟩π=1 +x4 ⟨2+y2 ⟩π=0 = x(y1 y2 +y3 )+x2 (y12 +y2 )+x3 (2y1 )+x4 (2) = (y1 y2 + y3 )x + (y12 + y2 )x2 + 2y1 x3 + 2x4 . Because G⟨2⟩ is known as the polynomial for enumerating root-isomorphic ordinary outer planar maps of size 2 with vertex-partition vector and the root-vertex valency given, Figure 2.8.1 shows G⟨2⟩ = (1a + 1b)x + (1c + 1d)x2 + (2e)x3 + (1f + 1h)x4 = (y1 y2 + y3 )x + (y12 + y2 )x2 + (2y1 )x3 + (2)x4 .

Figure 2.8.1: Classes of ordinary outer planar maps with size at most 2.

54 | 2 Outer equations second part

2.9 Notes 2.9.1. By observing the two equations (2.1.1) and (2.5.1) as regards their form, it is seen that if ϕ(x) is replaced by a3 , then equation (2.5.1) becomes equation (2.1.1). 2.9.2. Although both equation (2.1.1) and equation (2.5.1) are equivalent to a linear system of infinite equations, the coefficient matrix is shown to have the inverse such that all entries are in ℛ{y}. This enables us to evaluate powers of an infinite matrix and further those of a finite matrix whenever a suitable parameter is chosen. The simplest explicit form still needs to be further considered. 2.9.3. Is there one, or are there a series of transformations on ℛ{x, y} from equation (2.1.1) and equation (2.5.1), or equation (2.4.1) and equation (2.8.1), to the other such that a type of relationships among their solutions is established? 2.9.4. On the basis of Theorem 2.2.4 and Theorem 2.6.9, or Theorem 2.4.4 and Theorem 2.8.3, a type of relationships among the functions on ℛ{y} is established that is meaningful in combinatorics particularly in number theory (partitions) as in Andrews GE [1] (Andrews, G. E., 1976), graph theory (combinatorial maps) as in Liu YP [56] (Liu, Y. P., 1999), or umbral calculus (shadow functional) as in Rota GC, Taylor BD [87] (Rota, G. C., Taylor, B. D., 1994). 2.9.5. It seems that (2.2.9) in Theorem 2.2.4 in general, and hence (2.4.3) in Theorem 2.4.3, being meaningful in maps, is favorite to doing efficientization for running and further to doing intelligentization for usage. However, (2.3.13) in Theorem 2.3.7 in general and hence (2.4.5) in Theorem 2.4.5 looks favorite to extract an explicision. 2.9.6. Similarly, (2.6.4) in Theorem 2.6.5 in general, and hence (2.8.3) in Theorem 2.8.3, being meaningful in maps, looks favorite to doing efficientization for running and further to doing intelligentization for usage. However, (2.7.9) in Theorem 2.7.4 or (2.7.11) in Theorem 2.7.5 in general, and hence (2.8.7) in Theorem 2.8.5, looks favorite to extract an explicision. 2.9.7. As an application, (2.4.3), or (2.4.4), can be employed to enumerate the rootisomorphic classes of restricted outer planar maps with root-vertex valency m and size (m + s)/2, or order 1 + |n|, two parameters. Furthermore, one might address the genus series via determining the dual order. Relevant results are referred to in Liu YP [33] (Liu, Y. P., 1986), Dong FM, Yan JY [15] (Dong, F. M., Yan, J. Y., 1989). 2.9.8. As another application, (2.8.3), or (2.8.4)–(2.8.5), can be employed to enumerate the root-isomorphic classes of ordinary outer planar maps with root-vertex valency m and size (m + s)/2, or order 1 + |n|, with two parameters. Relevant results are referred to in [34] (Liu, Y. P., 1986), [39] (1988), [45] (1989), [15] (Dong, F. M., Yan, J. Y., 1989).

2.9 Notes | 55

2.9.9. Asymptotic behavior (or asymptotics). Both Note 2.9.7 and Note 2.9.8 remind us to investigate their asymptotic behavior as shown in [56] (Liu, Y. P., 1999, pp. 359–389), [94] (Yan, J. Y., Liu, Y. P., 1991). 2.9.10. Stochastic behavior (or stochastics). On the basis of (2.4.3) and (2.8.7), the distributions of sizes, orders, semi-automorphic group orders and genera from, respectively, size polynomials, order polynomials, semi-automorphic group order polynomials and genus polynomials need to be investigated. Relevant results are referred to in [56] (Liu, Y. P., 1999, pp. 359–389), [59] (Liu, Y. P., 2003, 221–230), [94], [11] (Chen, Y. C., et al., 2007), [92] (Wan, L. X., et al., 2008) etc. 2.9.11. These topics of equation (2.1.1) and equation (2.5.1) are, respectively, based on the foundation from Program 86 and Program 87 in [71] (Liu, Y. P., 2016, Vol. 22, p. 10740).

3 Inner equations first part 3.1 General ordinary inner model Two models are investigated in this chapter. First, consider the equation 2 { {a2 x ∫ yδx,y (ug|x=u ) = (1 − x g)g − a1 ; y { { {g|y=0⇒x=0 = a0 ,

(3.1.1)

where a = (a0 , a1 , a2 ) ∈ ℝ3+ and g ∈ ℛ{x, y} with y = (y1 , y2 , y3 , . . .). This is equation (5) in Introduction. Because of a solution of equation (3.1.1) for a0 = a1 = a2 = 1 is involved with general ordinary planar maps as shown in [57] (Liu YP, 1999, p. 204), and then [68] (Liu YP, 2013, p. 229), this equation is called a general ordinary inner model. Because of g ∈ ℛ{x, y}, g is determined by Ga[m] = 𝜕xm g ∈ ℛ{y} for m ≥ 0. From δx,y (ug|x=u ) = ∑ Ga[m] ( m≥0

xm+1 − ym+1 ) = ∑ xl ∑ ym−l Ga[m] , x−y l≥0 m≥l

we have ∫(yδx,y (zg)) = ∑ xl ∫ ∑ ym−l+1 Ga[m] = ∑ Λa[l] xl

(3.1.2)

Λa[l] = ∑ ym−l+1 Ga[m] .

(3.1.3)

l≥0

y

l≥0

y m≥l

where, for l ≥ 0, m≥l

[2] Because of g 2 ∈ ℛ{y} determined by Gm = 𝜕xm g 2 for m ≥ 0, we have m

[2] = ∑ Ga[m−j] Ga[j] . Ga[m] j=0

(3.1.4)

On account of the first equality in equation (3.1.1), [2] l g = a1 + x2 ∑ Ga[l] x + a2 x ∑ Λa[l] xl l≥0

= a1 + a2 Λa[0] x +

l≥0

[2] ∑ (Ga[m−2] m≥2

+ a2 Λa[m−1] )xm .

Observation 3.1.1. If equation (3.1.1) is consistent, then a1 = a0 . https://doi.org/10.1515/9783110627336-003

(3.1.5)

58 | 3 Inner equations first part Proof. Because of a solution ga = g ∈ ℛ{x, y} of equation (3.1.1) satisfying (3.1.5), its constant term is a1 . However, the initiation of equation (3.1.1) tells us that the constant term of any solution has to be a0 . Therefore, a1 = a0 . This observation enables us to discuss equation (3.1.1) only for a1 instead of a0 afterward. Thus, a = (a1 , a2 ) instead of (a0 , a1 , a2 ) without loss of generality. Theorem 3.1.2. Equation (3.1.1) on ℛ{x, y} is equivalent, for integer m ≥ 0, to

Ga[m]

a1 , when m = 0; { { { = {a2 Λa[0] , when m = 1; { { [2] {Ga[m−2] + a2 Λa[m−1] , when m ≥ 2,

(3.1.6)

[2] and Λa[m−1] are, respectively, given by (3.1.4) and (3.1.3). on ℛ{y} where Ga[m−2]

Proof. Because of there being only a cancelation law used from equation (3.1.1) to equations (3.1.6), the conclusion can be drawn. Because of the infinity of terms in Λa[m−1] , another parameter has to be sought for the availability in evaluation. Let 𝒥m be the set of all n which is a power vector of y involving a term of Ga[m] for any integer m ≥ 0. For any power vector n ∈ 𝒥m , denote by π(n) = inT the y-size where i = (1, 2, 3, . . .). Now, the y-size is chosen as the new parameter s = π(n). Let Ga[m,s] = Ga[m] |π(n)=s

(3.1.7)

for any integer s ≥ 0, then we have Ga[m] = ∑ Ga[m,s] ∈ ℛ{y}. s≥0

(3.1.8)

Lemma 3.1.3. For integer s ≥ 0, a1 , when s = 0; Ga[0,s] = { 0, when s ≥ 1.

(3.1.9)

Proof. From the initiation of equation (3.1.1) and Observation 3.1.1, the conclusion can be drawn. This is the case m = 0 of equation (3.1.6) and hence equation (3.1.1) in Theorem 3.1.2. Then we proceed on the basis of Theorem 3.1.2, let us observe the cases of m ≥ 1. Observation 3.1.4. When both integers m ≥ 0 and s ≥ 0 are given, then both m and s are, respectively, in the intervals [0, s] and [0, m].

3.1 General ordinary inner model | 59

Proof. Because of both m and s given are not less than 0, m + s is a constant. The conclusion is apparent. This apparent fact is easily seen to have a favorable usage for determining all Ga[m,s] as polynomials. Lemma 3.1.5. For integer s ≥ 0, 0, when s = 0; Ga[1,s] = { s−1 a2 ∑k=0 yk+1 Ga[k,s−k−1] , when s ≥ 1.

(3.1.10)

Proof. We proceed on the basis of Theorem 3.1.2, for the case of s = 0, the conclusion is from the second equation of equations (3.1.6) via (3.1.3). For the cases of s ≥ 1, from the second equation of equations (3.1.6), s−1

Ga[1,s] = a2 Λa[0] |s = a2 ∑ yk+1 Ga[k,s−k−1] . k=0

Hence, the conclusion can be drawn. This is the case m = 1 of equations (3.1.6) in Theorem 3.1.2. Then we proceed on the basis of Theorem 3.1.2, let us observe the general cases of m ≥ 2. Observation 3.1.6. For integer s ≥ 0 given, Λa[m−1,s] is independent of all Ga[k,l] for k+l ≥ m + s − 1, k, l ≥ 0. Proof. We proceed on the basis of (3.1.3), s+m−2

Λa[m−1,s] = ∑ yk−m+2 Ga[k,s−k+m−2] . k=m−1

(3.1.11)

Because of k + (s − k + m − 2) = s + m − 2, the conclusion can be drawn. This observation tells us that Λa[m−1,s] is only determined by some Ga[k,l] for k + l ≤ m + s − 1, k, l ≥ 0. By the substitution of j = k − m + 1 for k, s−1

Λa[m−1,s] = ∑ yj+1 Ga[j+m−1,s−j−1] . j=0

(3.1.12)

Lemma 3.1.7. Given integer s ≥ 0, for integer m ≥ 2, we have [2] + a2 Λa[m−1,s] Ga[m,s] = Ga[m−2,s]

where Λa[m−1,s] is known in the form of a sum of finite terms by (3.1.12). Proof. By employing Theorem 3.1.2, [2] [2] + a2 Λa[m−1,s] . + a2 Λa[m−1] )s = Ga[m−2,s] Ga[m,s] = (Ga[m−2]

This is (3.1.13).

(3.1.13)

60 | 3 Inner equations first part We proceed on the basis of Lemma 3.1.3, Lemma 3.1.5 and Lemma 3.1.7, a qualitative theorem of equations (3.1.6) and hence equation (3.1.1) can then be established. [2] is independent of Ga[k,l] for all k, l ≥ Observation 3.1.8. For any integers m, s ≥ 1, Ga[m,s] 0 and k + l ≥ m + s + 1.

Proof. We proceed on the basis of (3.1.8), by the additivity of π = s, [2] = ∑ Ga[i,j] Ga[m−i,s−j] . Ga[m,s] 0≤j≤s 0≤i≤m

(3.1.14)

Because of 0 ≤ i + j, m + s − i − j ≤ m + s, the conclusion can be drawn. [2] is determined only by Ga[m,j] for 0 ≤ j ≤ s from (3.1.14). In fact, Ga[m,s]

Theorem 3.1.9. Equation (3.1.1) is well defined on ℛ{x, y} for a ∈ ℝ3+ if, and only if, a1 = a0 . Proof. The necessity is from Observation 3.1.1. From Theorem 3.1.2, the sufficiency is from the well-definedness of equations (3.1.6) for a1 , a2 ∈ ℝ+ . First, we get a solution of equations (3.1.6). Because of Ga[0,0] = a1 and Ga[0,1] = Ga[1,0] = 0 as is known from (3.1.9) and (3.1.10), Ga[m,s] are determined for 0 ≤ m + s ≤ 1. In general, assume that all Ga[k,l] are determined for 1 ≤ k +l ≤ m+s−1 by the induction principle on m+s. From Observation 3.1.6 and Observation 3.1.8, it is seen that Ga[m,s] is determined by (3.1.13). A solution of equations (3.1.6) is obtained. Then, by the uniqueness of the procedure for evaluating the solution from the initiation Ga[0,0] = a1 , this solution is the only one.

3.2 Solution general ordinary inner case The structural properties of the solution of equations (3.1.6), and hence of equation (3.1.1) from Theorem 3.1.2, have to be further investigated. First, observe Ga[m,s] which determine the solution of equations (3.1.6) when s is smaller. Observation 3.2.1. For any integer m ≥ 1, Ga[m,0] = 0 if m = 1(mod 2). [2] = 0 for m = Proof. On account of (3.1.12), Λa[m−1,0] = 0 for any m ≥ 1. From Ga[m−2,0] 2k + 1 = 1 (i. e., k = 0), (3.1.13) leads to Ga[1,0] = 0. Assume Ga[2k−1,0] = 0 for all integer 1 ≤ 2k − 1 ≤ m − 1 by the induction principle on k for m = 2k + 1, k ≥ 1. We prove Ga[2k+1,0] = 0. We proceed on the basis of 2k−1

[2] Ga[2k−1,0] = ∑ Ga[i,0] Ga[2k−1−i,0] , i=0

3.2 Solution general ordinary inner case

| 61

because i + (2k − 1 − i) = 1(mod 2) and 0 ≤ i, 2k − 1 − i ≤ 2(k − 1) + 1 = 2k − 1, the [2] assumption leads to Ga[2k−1,0] = 0. By considering Λa[m−1,0] = 0, (3.1.13) leads to the conclusion. For determining all Ga[m,0] (m ≥ 1), this observation enables us to only discuss the case of m = 0(mod 2). Observation 3.2.2. For any integer t ≥ 1 and m = 2t, we have t−1

[2] = ∑ Ga[2j,0] Ga[2(t−1−j),0] . Ga[2t,0] j=0

Proof. We proceed on the basis of equations (3.1.6), (3.1.12) and Observation 3.2.1 we have t−1

[2] + a2 Λa[m−1,0] = ∑ Ga[2j,0] Ga[2(t−1−j),0] . Ga[m,0] = Ga[m−2,0] j=0

This is the conclusion. This observation shows that Ga[2t,0] is known whenever Ga[0,0] is given because Ga[2t,0] only depends on Ga[2l,0] for 0 ≤ l ≤ t − 1. Lemma 3.2.3. For any integer m ≥ 0, we have

Ga[m,0]

a1 , { { { = {0, { { at+1 1 (2t)! { t!(t+1)! ,

when m = 0; when m = 1(mod 2);

(3.2.1)

when m = 2t, t ≥ 1.

Proof. Because of Observation 3.2.1, it is only necessary to observe m = 2t, t ≥ 0. When t = 0, (3.1.9) tells us Ga[0,0] = a1 . Then, for t ≥ 1, assume that the conclusion is true for any i : 0 ≤ i ≤ t − 1. By induction on t, we prove that the conclusion is true as well. We proceed on the basis of Observation 3.2.2 and the inductive assumption, t−1

t−1

(2(t − 1 − j))! (2j)! . j!(j + 1)! (t − 1 − j)!(t − j)! j=0

Ga[2t,0] = ∑ Ga[2j,0] Ga[2(t−1−j),0] = at+1 1 ∑ j=0

By (3.1.8) in Book I, the conclusion can be drawn. Although the finiteness of the Ga[k,l] , which determine Ga[m,s] for m and s arbitrarily given, is seen, it is still necessary for us to observe the maximum degree of y and the minimum of i such that yi does not involve Ga[m,s] for i ≥ 1. Observation 3.2.4. For any integers m, s ≥ 0, Ga[m,s] is a polynomial of y in ℛ{y} with the maximum degree s.

62 | 3 Inner equations first part Proof. Let n ∈ 𝒥m,s , the set of power vectors in Ga[m,n] . Because of π(n) = s, s = ∑ ini = inT . i≥1

Hence, the degree of Ga[m,s] is d(Ga[m,s] ) ≤ max{|n| | inT , n ≥ 0} = |s11 | = s|11 | = s where 11 is n as n1 = 1 and ni = 0 for i ≥ 2. This is the conclusion. Then, how many of yi for i ≥ 1 are involved with Ga[m,s] for m, s ≥ 0, given? Is it a finite number? Observation 3.2.5. For integers m, s ≥ 0 given, Ga[m,s] is independent of yi for i ≥ s + 1. Proof. When s = 0, from (3.2.1), Ga[m,0] is independent of yl , l ≥ 1 = s+1, for any integer m ≥ 0. When s ≥ 1, assume for any integer l : 0 ≤ l ≤ s − 1, Ga[m,l] , m ≥ 0, is independent of yi , i ≥ l + 1. When m = 0, because of Ga[0,s] = 0 for s ≥ 1, Ga[0,s] is independent of yl , l ≥ s + 1. When m ≥ 1, assume for any integer k : 0 ≤ k ≤ m − 1, Ga[k,s] are all independent of yl for l ≥ s + 1. We prove that Ga[m,s] is independent of yl , l ≥ s + 1. Since for any integer k : 0 ≤ k ≤ m − 1, Ga[k,s] are all independent of yi , i ≥ s + 1, m−2

s

i=0

j=0

[2] = ∑ ( ∑ Ga[i,j] Ga[m−2−i,s−j] ) Ga[m−2,s]

is independent of yl , l ≥ s + 1. Since for any integer l : 0 ≤ l ≤ s − 1, Ga[m,l] are all independent of yi , i ≥ l + 1, from the assumption, s−1

∑ yl+1 Ga[l+m−1,s−l−1]

l=0

is independent of yl , l ≥ s + 1. Then from (3.1.13), the conclusion can be drawn. This observation enables us to only consider ys = (y1 , y2 , y3 , . . . , ys ) for determining Ga[m,s] when m, s ≥ 0 are given. Observation 3.2.6. If Ga[m,s] ≠ 0 for any m, s ≥ 0, then m = s(mod 2). Proof. When m + s = 0, from (3.2.1), Ga[0,0] = a1 . Thus, s = π(n) = 0 = m, n ∈ 𝒥0,0 . When m+s = 1, because of Ga[0,1] = Ga[1,0] = 0, s = π(n) ≠ m(mod 2), n ∈ 𝒥0,1 ∪ 𝒥1,0 . For any integers m, s ≥ 0, m + s = t ≥ 1. Assume Ga[i,j] = 0, i + j = 1(mod 2), whenever for any integers i, j ≥ 0, i + j ≤ t − 1. By induction on t, we prove that the conclusion is true for m + s = t.

3.2 Solution general ordinary inner case

| 63

We proceed by contradiction. If π(n) ≠ m(mod 2), then m + s = 1(mod 2). In Ga[m,s] , [2] only depends on Ga[i,j] for (3.1.13) is employed. Because of m + s = 1(mod 2), Ga[m−2,s] i + j ≤ m + s − 1, and i + j = 1(mod 2). By the assumption, all Ga[i,j] = 0, and hence [2] = 0. Similarly, Ga[m−2,s] s−1

Λa[m−1,s] = ∑ yl+1 Ga[l+m−1,s−l−1] = 0. l=0

A contradiction to the condition Ga[m,s] ≠ 0 occurs. Therefore s = π(n) = m(mod 2). According to the principle of induction, the conclusion can be drawn. This observation enables us to investigate all Ga[m,s] only for integers m, s ≥ 1 such that m + s = 0(mod 2). Lemma 3.2.7. For all integers m, s ≥ 0 such that m + s ≠ 0(mod 2), Ga[m,s] = 0. Proof. A direct result of Observation 3.2.6. This lemma enables us to discuss Ga[m,s] only for m + s = 0(mod 2) of integers m, s ≥ 0 because of all Ga[m,s] = 0 for m + s ≠ 0(mod 2). For s = 1, Observation 3.2.6 suggests us that only Ga[m,1] for m = 1(mod 2) are available. Observation 3.2.8. For m = 1, Ga[1,1] = a1 a2 y1 . Proof. From (3.1.10) with (3.1.9), Ga[1,1] = a2 y1 Ga[0,0] = a1 a2 y1 . The conclusion can be drawn. Then consider the case for m = 3 we proceed on the basis of (3.1.12) and (3.1.13). Observation 3.2.9. For m = 3, Ga[3,1] = 3a21 a2 y1 . Proof. On account of 1

1

[2] = ∑ ∑ Ga[i,j] Ga[1−i,1−j] = 2a21 a2 y1 , Ga[1,1] i=0 j=1

[2] + a2 y1 G[2,0] = 3a21 a2 y1 . The conclusion can be drawn. from (3.1.10), Ga[3,1] = G[1,1]

For an integer t ≥ 2 and hence m = 2t + 1 ≥ 5, because t−1

t

i=0

i=1

[2] Ga[2t−1,1] = ∑ Ga[2i,0] Ga[2t−1−2i,1] + ∑ Ga[2i−1,1] Ga[2(t−i),0] ,

(3.2.2)

(3.1.13) for m = 2t + 1 leads to [2] + Ga[2t+1,1] = Ga[2t−1,1] [2] where Ga[m−2,1] is shown in (3.2.2).

at+1 1 a2 (2t)! y t!(t + 1)! 1

(3.2.3)

64 | 3 Inner equations first part Lemma 3.2.10. For any integer m ≥ 0, we have

Ga[m,1]

0, { { { 2 = {3a1 a2 y1 , { { [2] {Ga[2t−1,1] +

when m = 2t, t ≥ 0; at+1 1 a2 (2t)! y, t!(t+1)! 1

when m = 3;

(3.2.4)

when m = 2t + 1, t ≥ 2,

[2] is given by (3.2.2). where Ga[2t−1,1]

Proof. The case of m = 2t is from Lemma 3.2.7. The case of m = 2t +1 is from (3.2.3). It seems to be more helpful to evaluate Ga[m,2] for integer m ≥ 0 by the procedure in general. Because Ga[0,2] = 0 and Ga[1,2] = 0, it is only necessary for us to evaluate Ga[m,2] for m ≥ 2. Because of Ga[0,2] = 0 (Lemma 3.1.3) and Ga[1,2] = 0 (Lemma 3.2.7), the Ga[m,2] for m ≤ 1 are known. When m = 2, from Ga[0,2] = Ga[0,1] = 0, 2 [2] = 2Ga[0,0] Ga[0,2] + Ga[0,1] = 0. Ga[0,2]

(3.2.5)

Then by employing (3.1.12) and (3.1.13), 1

Ga[2,2] = a2 ∑ yj+1 Ga[j+1|1−j] . j=0

(3.2.6)

Observation 3.2.11. For m = s = 2, we have Ga[2,2] = a1 a2 (a2 y12 + a1 y2 ).

(3.2.7)

Proof. By employing (3.2.6), Ga[2,2] = a2 (y1 Ga[1,1] + y2 Ga[2,0] ) = a1 a22 y12 + a21 a2 y2 . This is the conclusion. When m = 4, we proceed on the basis of Ga[0,0] = a1 (Lemma 3.2.3), Ga[1,1] = a1 a2 y1 (Observation 3.2.8), and the results for Ga[2,2] (Observation 3.2.11), Ga[3,1] (Observation 3.2.9) and Ga[4,0] (Lemma 3.2.3), it is seen that [2] = Ga[2,2]

∑ 0≤i,j≤2 i+j=0(mod 2)

Ga[i,j] Ga[2−i,2−j] = 3(a1 a2 y1 )2 + 2a31 a2 y2

(3.2.8)

and 1

∑ yj+1 Ga[j+3,1−j] = y1 Ga[3,1] + y2 Ga[4,0] = 3a21 a2 y12 + 2a31 y2 .

j=0

(3.2.9)

3.2 Solution general ordinary inner case

| 65

Lemma 3.2.12. For m = 4, Ga[4,2] = a21 a2 (6a2 y12 + 4a1 y2 ). Proof. Because of (3.2.8) and (3.2.9), 1

[2] Ga[4,2] = Ga[2,2] + a2 ∑ yj+1 Ga[j+3,1−j] = 6(a1 a2 y1 )2 + 4a31 a2 y2 . j=0

This is the conclusion. We proceed on the basis of Observation 3.2.6, for m ≥ 5, it is only necessary to observe Ga[m,2] for m = 2t, t ≥ 3. Observation 3.2.13. For m = 2t, t ≥ 3, we have t−1 t−i a (2(t − 1 − i))! ai+1 1 (2i)! Ga[2(t−1−i),2] + ∑ 1 G . i!(i + 1)! (t − 1 − i)!(t − i)! a[2i,2] i=1 i=0 t−2

[2] =∑ Ga[2t−2,2]

(3.2.10)

Proof. On account of (3.1.13) t−1

[2] = ∑ (Ga[2i,0] Ga[2(t−1−i),2] + Ga[2i,2] Ga[2(t−1−i),0] ). Ga[2t−2,2] i=0

By employing Lemma 3.2.3, the conclusion can be drawn. Since all Ga[2(t−1−i),2] for 0 ≤ i ≤ t − 2 and Ga[2i,2] for 1 ≤ i ≤ t − 1 have been [2] determined, so is Ga[2t−2,2] as well in (3.2.10). Observation 3.2.14. For m = 2t, t ≥ 3, we have Λa[m−1,2] =

(2t)! at1 a2 (2t − 2)! 2 at+1 y + 1 y. (t − 1)!t! 1 t!(t + 1)! 2

(3.2.11)

Proof. From (3.1.12), Λa[m−1,2] = y1 Ga[2t−1,1] + y2 Ga[2t,0] =

(2t)! at1 a2 (2t − 2)! 2 at+1 y1 + 1 y. (t − 1)!t! t!(t + 1)! 2

This is (3.2.11). We proceed on the basis of Observation 3.2.13 and Observation 3.2.14, all Ga[m,2] for m ≥ 0 can be determined. Lemma 3.2.15. For integer m ≥ 0,

Ga[m,2]

0, { { { { { {a1 a2 (a2 y12 + a1 y2 ), ={ 2 { {4a1 a2 (a2 y12 + a1 y2 ), { { { [2] {Ga[2t−2,2] + a2 Λa[2t−1,2] ,

when m = 0 or m = 1(mod 2); when m = 2; when m = 4; when m = 2t, t ≥ 3,

[2] where Ga[2t−2,2] and Λa[2t−1,2] are, respectively, given by (3.2.10) and (3.2.11).

(3.2.12)

66 | 3 Inner equations first part Proof. The case of m = 0 or m = 1(mod 2) is from, respectively, (3.1.9) or Lemma 3.2.7. The case of m = 2 is from (3.2.7). The case of m = 4 is from Lemma 3.2.12. The case of m = 2t, t ≥ 3 is from Observation 3.2.13 and Observation 3.2.14. Let d = m + s, called the double size, be given. Because of both m and s being nonnegative integers, only 0 ≤ m = d −s ≤ d are available while 0 ≤ s ≤ d for Ga[m,s] . When s = 2, m = d − 2, on account of Lemma 3.2.7, only even d are available for determining Ga[m,n] ; half of double size is the size of Ga[m,s] . Similarly, m and s are, respectively, the x-size and y-size. From Lemma 3.1.3 and Lemma 3.2.3, it is only necessary to consider Ga[m,s] for m, s ≥ 1. Furthermore, Lemma 3.1.5 and Lemma 3.2.10 enable us to only consider m, n ≥ 2 for Ga[m,s] . Observation 3.2.16. For any m, s ≥ 2,

[2] Ga[m−2,s]

∑ 0≤j≤t Ga[2i,2j] Ga[2(l−1−i),2j] , { { { 0≤i≤l−1 { { { { when m = 2l, l ≥ 1 ⇒ s = 2t, t ≥ 1; ={ { ∑ 0≤j≤t Ga[2i+1,2j+1] Ga[2(l−1−i),2(t−j)] , { { { 0≤i≤l−1 { { when m = 2l + 1 ⇒ s = 2t + 1, t ≥ 1. {

(3.2.13)

Proof. We proceed on the basis of Lemma 3.2.7, the conclusion can be deduced from (3.1.14). [2] only by some Ga[i,j] for 0 ≤ i+j ≤ This observation enables us to determine Ga[m−2,s] m + s − 2.

Observation 3.2.17. For m, s ≥ 2, 0, when m + s ≠ 0(mod 2); Λa[m−1,s] = { s−1 ∑j=0 yj+1 Ga[m+j−1,s−j−1] , otherwise.

(3.2.14)

Proof. On account of (m + j − 1) + (s − j − 1) = m + s − 2 = m + s(mod 2), the case of m + s ≠ 0(mod 2) is deduced from Lemma 3.2.7. The other case is from (3.1.12). We proceed on the basis of Observation 3.2.16 and Observation 3.2.17, the solution of equations (3.1.6) and hence equation (3.1.1) can be evaluated from Theorem 3.1.2. Theorem 3.2.18. The solution of equation (3.1.1) is determined by all Ga[m,s] for m, s ≥ 0 as Ga[m,s]

a1 , { { { = {0, { { [2] {Ga[m−2,s] + a2 Λa[m−1,s] ,

when m = s = 0; when m + s ≠ 0(mod 2); when m + s = 0(mod 2),

[2] where Ga[m−2,s] and Λa[m−1,s] are, respectively, given by (3.2.13) and (3.2.14).

(3.2.15)

3.2 Solution general ordinary inner case

| 67

Proof. The case of m = s = 0 is from (3.2.1). The case of m + s ≠ 0(mod 2) is from Lemma 3.2.7. The last case is from Observation 3.2.16 and Observation 3.2.17. We proceed on the basis of this theorem, the solution is evaluated by the procedure starting from Ga[0,0] of m + s = 0, increasingly two by two up to m + s = 2L, for getting Ga[2L,0] , Ga[2L−1,1] , Ga[2L−2,2] , . . . , Ga[0,2L] if L is the object, as shown in Table 3.2.1. In this table, all Ga[0,2l] = 0 for l ≥ 1 are omitted. Example 3.2.1. For integers m, s ≥ 0, let l = m + s and Γa[l] = ∑ Ga[m,s] xm .

(3.2.16)

m+s=l m,s≥0

From Lemma 3.2.7, it is only necessary to consider l = 0(mod 2). It is easily checked that Γa[l] for l ≥ 0 are all polynomials on ℛ[x, y] ⊆ ℛ{x, y}.

Table 3.2.1: Chart of determining all Ga[m,n] . m+s 0 2 4 6 .. .

Ga[m,s]

Ga[6,0]

Ga[4,0] Ga[5,1]

Ga[2,0] Ga[3,1] Ga[4,2]

Ga[0,0] Ga[1,1] Ga[2,2] Ga[3,3]

Ga[1,3] Ga[2,4]

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

Ga[1,5]

We proceed on the basis of Table 3.2.1, for 0 ≤ l ≤ 6, we have

Γa[l]

a1 , { { { { 2 2 { { {a1 x + a1 a2 y1 x, { { { { 2a31 x4 + 3a21 a2 y1 x3 + a1 a2 (a2 y12 + a1 y2 )x2 { { { { { { + a1 a2 (a2 y1 y2 + a1 y3 )x, { { { { 4 6 { {5a x + 10(a31 a2 y1 )x5 + a21 a2 (6a2 y12 ={ 1 { + 4a1 y2 )x4 + a1 a2 (a22 y13 + 6a1 a2 y1 y2 { { { { { { + 4a21 y3 )x3 + a1 (2a22 y12 y2 + 4a1 a2 y1 y3 { { { { { { + a1 a2 y22 + 2a21 y4 )x2 + a1 (a22 y1 y22 { { { { { + 3a1 a2 y1 y4 + a22 y12 y3 { { { { + 2a1 a2 y2 y3 + 2a21 y5 )x, {

when l = 0; when l = 2; when l = 4; (3.2.17)

when l = 6.

68 | 3 Inner equations first part

3.3 Explicitness general ordinary inner case Because no explicision has yet directly been deduced from Theorem 3.2.18, the relationship between the solutions of general equation (3.1.1) and its restriction for a0 = a1 = a2 = 1 has to be investigated by considering the latter to be meaningful in graph theory. Let ga[goi] and g[goi] ∈ ℛ{x, y} be the solutions of, respectively, equation (3.1.1) [goi] [goi] and Gm,s ∈ ℛ{y} for integers m, s ≥ 0. Deand its restriction determined by Ga[m,s]

[goi] [goi] note by 𝒩m,s the set of power vectors in Ga[m,s] and hence G[m,s] . For any n ∈ 𝒩m,s , Aa[m,n] |s=π(n) ∈ ℛa = ℛ{a} and Am,n |s=π(n) ∈ ℝ+ are the coefficients of yn in, respec[goi] [goi] tively, Ga[m,s] and Gm,s . Then [goi] n [goi] | = 𝜕yn Ga[m,s] A[goi] and A[goi] m,n |π(n)=s = 𝜕y Gm,s . a[m,n] π(n)=s

is a polyTheorem 3.3.1. For any integers m, s ≥ 1 and integral vector n ∈ 𝒩m,s , A[goi] a[m,n] = A[goi] nomial of a with degree at most l + 1 where l = (m + s)/2 such that A[goi] m,n . 1[m,n]

Proof. We proceed on the basis of Theorem 3.2.18. By induction, the conclusion can be drawn. For a planar embedding μ(G) of a general ordinary planar graph G, let n = (n1 , n2 , n3 , . . .) be the vertex-partition vector where ni is the number of y-vertices (or non-root-vertex) with valency i for integer i ≥ 1. A vertex v of G whose deletion makes G − v have at least one component greater than G does is called a cut-vertex of G. Now, G is always assumed to be connected. It is allowed to restrict a face of maximum length boundary at a cut-vertex as the outerface without loss of generality. The cut-valency of a cut-vertex v is the number of travels without repetition of v on outerface boundary. Each travel with its inner parts is call a cut-component. A cut-component is called a fat-component if it is not a path. Then the fat valency of a cut-vertex is known in its own right. A graph without cut-vertex is said to be nonseparable. A maximal nonseparable edge-induced subgraph is called a block of the graph. For a cut-vertex v, let a = ρcut (= av ), c(= cv ) and b = ρv (= bv )(= bv ) be, respectively, the cut-valency, the fat valency and the valency of v. Because of planarity, all cut-components at v have a rotation in the order as G1 , G2 , . . . , Ga as in clockwise. At cut-vertex v, let ti be the contribution of Gi to the valency of v, then t1 + t2 + ⋅ ⋅ ⋅ + ta = b. Three cases have to be considered at a cut-vertex v. Case I Reflection for fat-components; it turns out that we have [goi] = 2c−1 πvI

(topologically) distinct planar embeddings.

(3.3.1)

3.3 Explicitness general ordinary inner case

| 69

Case II For cut-rotation at v; it turns out that we have [goi] = (a − 1)! πvII

(3.3.2)

(topologically) distinct planar embeddings. Case III For an integer k ≥ 1 given, denote I = {i1 , i2 , . . . , ik } by distributing all Gj , j ∈ ̸ I, into s = (ti1 −1)+(ti2 −1)+⋅ ⋅ ⋅+(tik −1) angles on 𝒢I = {Gi |∀i ∈ I} at v as (x1 , x2 , . . . , xts ) in the s angles such that s

∑ xj = a(= ρvcut ) − k; j=1

0 ≤ xj ≤ a − k, 1 ≤ j ≤ s. Let us denote by a−k ⟨ ⟩ x1 , . . . , xts the number of ways for such distributions, then it turns out that we have s a−k πk[goi] = ⟨ ⟩ ∏ xj ! x1 , . . . , xts j=1

(topologically) distinct planar embeddings. From the disjointness of such embeddings, we have a−1 a [goi] = ∑ ( )πk[goi] πvIII k k=1

(3.3.3)

(topologically) distinct planar embeddings. [goi] Lemma 3.3.2. For a general ordinary planar graph G, let VCut be the set of all cutvertices. Then all cut-vertices turn out to involve [goi] [goi] [goi] n[goi] Cut (G) = ∏ (πvI πvII πvIII ) v∈VCut

(3.3.4)

(topologically) distinct planar embeddings. Proof. On account of the independency among Case I, Case II and Case III. From (3.3.1), (3.3.2) and (3.3.3), the conclusion can be drawn. As regards a planar embedding μ = μ(G) of graph G without cut-vertex, if two vertices u and v in a block have the property that G − {u, v} have at least two components, then {u, v} is called a splitting pair. Since an embedding is a point set in its own right on the plane, pu and pv are the corresponding points of, respectively, u and v.

70 | 3 Inner equations first part Because μ can be taken as a point set, any component of μ − {pu , pv } is called a splitting slice, or slice. A splitting slice that is not a path is called a splitting chunk, or chunk. Let S be the set of all splitting pairs consisting of {ui , vi } for 1 ≤ i ≤ s = |S| in μ = μ(G). All components of μ − S are splitting slices in μ(G). For a splitting pair, its slice valent is the number of splitting slices it is incident with. Similarly for its chunk valency. Observation 3.3.3. Let t and 1 ≤ p ≤ t be, respectively, the slice valency and chunk valency of a splitting pair on a general ordinary planar embedding of graph G, then the splitting pair produces [goi] = 2p−1 (t − 1)! πspp

(3.3.5)

(topologically) distinct planar embeddings of G. Proof. On account of the independency between a rotation on the slice valency and reflection on a chunk, by considering the connectivity of embedding, the conclusion can be drawn. Let pi be the number of splitting chunks and ti the slice valency at the splitting pair {ui , vi }, 1 ≤ i ≤ s on μ(G). Lemma 3.3.4. For a general ordinary planar graph G, the number of topologically distinct embeddings of G without consideration of a cut-vertex is l

p−s n[goi] ∏((j − 1)!) j Nct = 2 j≥2

(3.3.6)

where p = p1 + p2 + ⋅ ⋅ ⋅ + ps and lj is the number of splitting pairs with slice valency j ≥ 2. Proof. We proceed on the basis of Lemma 3.3.3, since s

s

pi −1 (ti − 1)!) = 2p−s ∏ ((t i − 1)!) n[goi] Nct = ∏ (2 i=1

=2

p−s

lj

i=1

∏((j − 1)!) . j≥2

This is the conclusion. This lemma is a refinement and a concise form of Theorem 10.4.3 with (10.4.13) in Liu YP [72] (2017, pp. 196–197), or Theorem 7.4.3 with (7.4.13) in Liu YP [55] (1994, p. 142). We proceed on the basis of Lemma 3.3.2 and Lemma 3.3.4, the result is useful in the context of determining the number of general ordinary planar embeddings of a general ordinary planar graph.

3.3 Explicitness general ordinary inner case | 71

Lemma 3.3.5. For a general ordinary planar graph G, the number of planar embeddings of G is [goi] n[goi] (G) = n[goi] Cut (G)nNct (G)

(3.3.7)

where nCut (G) and nNct (G) are, respectively, given by (3.3.4) and (3.3.6). Proof. We proceed on the basis of the uniqueness of a planar embedding of a 3-connected graph. Because a graph without vertex-cut set of one or two vertices has to be 3-connected, only the two independent cases with and without cut-vertex need to be considered. Lemma 3.3.2 and Lemma 3.3.4 lead to the conclusion. ̃ Denote by t = aut(G) the order of the semi-automorphism group of graph G. The relationship between embeddings and upper maps of G can be obtained via t. Lemma 3.3.6. Let ℰ [goi] (G) and ℳ[goi] (G) be, respectively, the sets of all distinct topological planar embeddings and root-isomorphic upper maps of G. Then 󵄨󵄨 [goi] 󵄨󵄨 2ϵ 󵄨󵄨 [goi] 󵄨󵄨 (G)󵄨󵄨 (G)󵄨󵄨 = 󵄨󵄨ℰ 󵄨󵄨ℳ t

(3.3.8)

where ϵ = ϵ(G) is the size of G. Proof. Refer to Liu YP [59] (2003, pp. 225–226). [goi] Let 𝒢m,s be the set of all general ordinary planar graphs with rooted semi-valency m (i. e., the number of semi-edges incident to the root-vertex) and the un-rooted semisize s (i. e., the number of semi-edges not incident to the root-vertex). [goi] [goi] Denote by 𝒯m,s the set of semi-automorphism group orders of graphs in 𝒢m,s . [goi] [goi] Let 𝒥m,s be the set of all partition vectors n in 𝒢m,s . [goi] Lemma 3.3.7. Let Em,n be the number of (topologically) distinct embeddings of the [goi] graphs in 𝒢m,n . Then we have [goi] Em,n =

∑ n[goi] (G)

(3.3.9)

[goi] G∈𝒢m,n

where n[goi] is given in (3.3.7). [goi] Proof. By the partition of all embeddings on 𝒢m,s from each graph, the conclusion can be drawn.

We proceed on the basis of Lemma 3.3.7. Let 𝒯m,n be the set of semi-automorphism [goi] [goi] [goi] group orders for graphs in 𝒢m,s and 𝒢m,s;t , the set of all graphs in 𝒢m,s with semĩ automorphism group order t. Denote by aut(G) the semi-automorphism group order of G.

72 | 3 Inner equations first part Lemma 3.3.8. The number of root-isomorphic classes of upper maps of all graphs in

[goi] 𝒢m,s is

m + π(n) [goi] Em,n;t t

(3.3.10)

∑ ngoi (G)|aut(G)=t ̃

(3.3.11)

A[goi] m,n = ∑

t∈𝒯m,n

[goi] where A[goi] m,n = Am,n (𝒢m,s ), s = π(n) and [goi] Em,n;t =

[goi] G∈𝒢m,s

is determined by (3.3.9). Proof. We proceed on the basis of Lemma 3.3.6. By considering m + π(n) = 2ϵ, the conclusion can be drawn. This lemma enables us to obtain the solution ga[goi] of equation (3.1.1) when a0 = [goi] a1 = a2 = 1. Denote g[goi] = ga[goi] |a=1 , it being determined by Gm,s = 𝜕xm g[goi] |π(y)=s for m, s ≥ 0. Lemma 3.3.9. For integers m, n ≥ 1, we have [goi] Gm,s =

∑ [goi] n∈𝒥m,s

n A[goi] m,n y

(3.3.12)

where A[goi] m,n is given by (3.3.10). Proof. From Lemma 3.3.8, the conclusion can be drawn. We proceed on the basis of Lemma 3.3.9 and Theorem 3.3.1; we present our main result in this section. [goi] we determine the solution of equation (3.1.1) as Theorem 3.3.10. For m, s ≥ 1, Ga[m,s] [goi] = Ga[m,s]

∑ [gop] n∈𝒥m,s

yn A[goi] a[m,n]

(3.3.13)

where A[goi] is a polynomial of a with degree at most 1 + (m + s)/2 such that A[goi] = a[m,n] 1[m,n]

A[goi] m,n .

Proof. A result of Theorem 3.3.1 and Lemma 3.3.9. From (3.3.4) and (3.3.6), nCut and nNct are summation-free. From (3.3.7), n[goi] is [goi] summation-free. Because A[goi] m,s is determined by n[goi] , Gm,s given by (3.3.12) is a direct [goi] explicision and, hence, so is Ga[m,s] in (3.3.13).

3.4 Restrictions general ordinary inner case | 73

3.4 Restrictions general ordinary inner case First, consider the equation 2 { { {x ∫ yδx,y (ug|x=u ) = (1 − x g)g − 1; y { { { g|y=0⇒x=0 = 1, {

(3.4.1)

where g ∈ ℛ{x, y} with y = (y1 , y2 , y3 , . . .). This is the equation obtained from equation (3.1.1) for a0 = a1 = a2 = 1. Equation (3.4.1) is also called a restriction of equation (3.1.1). One of the solutions of the first equality of equation (3.4.1) found in [57] (Liu YP, 1999, p. 204), and then [68] (Liu YP, 2013, p. 229) is just the enufunction of general ordinary planar maps with the vertexpartition vector as a parameter for root-isomorphic classes. Because of g ∈ ℛ{x, y}, g is determined by Gm = 𝜕xm g ∈ ℛ{y} for m ≥ 0. From δx,y (ug|x=u ) = ∑ Gm ( m≥0

xm+1 − ym+1 ) = ∑ xl ∑ ym−l Gm , x−y l≥0 m≥l

we have ∫(yδx,y (zg)) = ∑ xl ∫ ∑ ym−l+1 Gm = ∑ Λl xl

(3.4.2)

Λl = ∑ ym−l+1 Gm .

(3.4.3)

y

l≥0

y m≥l

l≥0

where, for l ≥ 0, m≥l

[2] Because g 2 ∈ ℛ{y} is determined by Gm = 𝜕xm g 2 for m ≥ 0, we have m

[2] Gm = ∑ Gm−j Gj . j=0

(3.4.4)

On account of the first equality in equation (3.4.1), g = 1 + x2 ∑ Gl[2] xl + x ∑ Λl xl l≥0

= 1 + Λ0 x +

l≥0

[2] ∑ (Gm−2 m≥2

+ Λm−1 )xm .

(3.4.5)

Observation 3.4.1. The initiation of equation (3.4.1) is consistent. Proof. A direct result of Observation 3.1.1 because of a0 = a1 = 1. This observation enables us to discuss how many solutions equation (3.4.1) has.

74 | 3 Inner equations first part Theorem 3.4.2. Equation (3.4.1) on ℛ{x, y} is equivalent, for integer m ≥ 0, to 1, { { { Gm = {Λ0 , { { [2] {Gm−2 + Λm−1 ,

when m = 0; when m = 1;

(3.4.6)

when m ≥ 2,

[2] on ℛ{y} where Gm−2 and Λm−1 are, respectively, given by (3.4.4) and (3.4.3).

Proof. Because only the cancelation law is used from equation (3.4.1) to equations (3.4.6), the conclusion can be drawn. Because of the infinity for terms in Λm−1 , another parameter has to be sought for the availability in evaluation. Let 𝒥m be the set of all n which is a power vector of y involved with a term of Gm for any integer m ≥ 0. For any power vector n ∈ 𝒥m , denote by π(n) = inT the y-size where i = (1, 2, 3, . . .). Now, the y-size is chosen as a new parameter s = π(n). Let Gm,s = Gm |π(n)=s

(3.4.7)

Gm = ∑ Gm,s ∈ ℛ{y}.

(3.4.8)

1, when s = 0; G0,s = { 0, when s ≥ 1.

(3.4.9)

for any integer s ≥ 0, then we have s≥0

Lemma 3.4.3. For integer s ≥ 0,

Proof. From the initiation of equation (3.4.1) and Observation 3.4.1, the conclusion can be drawn. This is the case m = 0 of equation (3.4.6), and hence equation (3.4.1), in Theorem 3.4.2. Then we proceed on the basis of Theorem 3.4.2; let us observe the cases of m ≥ 1. Observation 3.4.4. When both integers m ≥ 0 and s ≥ 0 are given, then both m and s are, respectively, in the intervals [0, s] and [0, m]. Proof. This is the case of Observation 3.1.4 for a0 = a1 = a2 = 1. This apparent fact is easily seen to be favorable for usage for determining all Gm|s as polynomials.

3.4 Restrictions general ordinary inner case | 75

Lemma 3.4.5. For integer s ≥ 0, 0, when s = 0; G1,s = { s−1 ∑k=0 yk+1 Gk,s−k−1 , when s ≥ 1.

(3.4.10)

Proof. See the proof of Lemma 3.1.5 for a = 1. This is the case m = 1 of equation (3.4.6), and hence equation (3.4.1), in Theorem 3.4.2. We proceed on the basis of Theorem 3.4.2; let us observe the general cases of m ≥ 2. Observation 3.4.6. For integer s ≥ 0 given, Λm−1,s is independent of all Gk,l for k + l ≥ m + s − 1, k, l ≥ 0. Proof. We proceed on the basis of (3.4.3), Λm−1|s is the summation of (yk−m+2 Gk )|s over k : k ≥ m − 1. By the additivity of π, it is the summation of yk−m+2 Gk|s−k+m−2 over k : k ≥ m − 1. Then, by s − k + m − 2 ≥ 0, we have s+m−2

Λm−1,s = ∑ yk−m+2 Gk,s−k+m−2 . k=m−1

(3.4.11)

Because k + (s − k + m − 2) = s + m − 2, the conclusion can be drawn. This observation tells us that Λm=1,s is only determined by some Gk,l for k + l ≤ m + s − 1, k, l ≥ 0. By the substitution of j = k − m + 1 for k, s−1

Λm−1,s = ∑ yj+1 Gj+m−1,s−j−1 . j=0

(3.4.12)

Lemma 3.4.7. Given integer s ≥ 0. For integer m ≥ 2, we have [2] Gm,s = Gm−2,s + Λm−1,s

(3.4.13)

where Λm−1,s is known in the form of a sum of finite terms by (3.4.12). [2] [2] + Λm−1,s . This is + Λm−1 )|s = Gm−2,s Proof. By employing Theorem 3.4.2, Gm,s = (Gm−2 (3.4.13).

We proceed on the basis of Lemma 3.4.3, Lemma 3.4.5 and Lemma 3.4.7; a qualitative theorem of equations (3.4.3), and hence equation (3.4.1), can be then established. [2] Observation 3.4.8. For any integers m, s ≥ 1, Gm,s is independent of Gk,l for all k, l ≥ 0 and k + l ≥ m + s + 1.

Proof. We proceed on the basis of (3.4.8). By the additivity of π = s, [2] Gm,s = ∑ Gi,j Gm−i,s−j . 0≤j≤s 0≤i≤m

Because of 0 ≤ i + j, m + s − i − j ≤ m + s, the conclusion can be drawn.

(3.4.14)

76 | 3 Inner equations first part [2] In fact, Gm,s is determined only by Gm,j for 0 ≤ j ≤ s from (3.4.14).

Theorem 3.4.9. Equation (3.4.1) is well defined on ℛ{x, y}. Proof. A direct result of Theorem 3.1.9 because of a0 = a1 = 1. We proceed on the basis of Theorem 3.4.2; some structural properties of the solution of equations (3.4.6), and hence of equation (3.4.1) have to be further investigated. First, we address all Gm|s which determine the solution of equations (3.4.6) when s is smaller. Observation 3.4.10. For any integer m ≥ 1, Gm,0 = 0 if m = 1(mod 2). Proof. See the proof of Observation 3.2.1 for a = 1. In order to determine all Ga[m,0] for m ≥ 1, this observation enables us to only discuss the case of m = 0(mod 2). Observation 3.4.11. For any integer t ≥ 1 and m = 2t, we have t−1

[2] G2t,0 = ∑ G2j,0 G2(t−1−j),0 . j=0

Proof. See the proof of Observation 3.2.2 for a = 1. This observation shows that G2t,0 is known whenever G0,0 is given because G2t,0 only depends on G2l,0 for 0 ≤ l ≤ t − 1. Lemma 3.4.12. For any integer m ≥ 0, we have

Gm,0

1, when m = 0; { { { = {0, when m = 1(mod 2); { { (2t)! { t!(t+1)! , when m = 2t, t ≥ 1.

(3.4.15)

Proof. See the proof of Lemma 3.2.3 for a = 1. Although we have the finiteness of the Gk,l which determine Ga[m,s] for m and s given, it is still necessary for us to observe the maximum degree of y and the minimum number of i(≥ 0) such that yi is involved with Ga[m,s] . Observation 3.4.13. For any integers m, s ≥ 0, Gm,s is a polynomial of y in ℛ{y} with the maximum degree s. Proof. See the proof of Observation 3.2.4 for a = 1. Then how many of yi for i ≥ 1 are involved with Gm,s for m, s ≥ 0, given? Is it a finite number? Observation 3.4.14. For integers m, s ≥ 0 given, Gm,s is independent of yi for i ≥ s + 1.

3.4 Restrictions general ordinary inner case | 77

Proof. See the proof of Observation 3.2.5 for a = 1. This observation enables us to only consider ys = (y1 , y2 , y3 , . . . , ys ) for determining Gm,s when m, s ≥ 0 are given. Observation 3.4.15. If Gm,s ≠ 0 for any m, s ≥ 0, then m = s(mod 2). Proof. See the proof of Observation 3.2.6 for a = 1. This observation enables us to investigate all Gm,s only for integers m, s ≥ 1 such that m + s = 0(mod 2). Lemma 3.4.16. For all integers m, s ≥ 0 such that m + s ≠ 0(mod 2), Gm|s = 0. Proof. A direct result of Observation 3.4.15. This lemma enables us to discuss Gm,s only for m+s = 0(mod 2) of integers m, s ≥ 0 because of all Gm,s = 0 for m + s ≠ 0(mod 2). For s = 1, Observation 3.4.15 suggests us that only the Gm,1 for m = 1(mod 2) are available. Observation 3.4.17. For m = 1, G1,1 = y1 . Proof. From (3.4.10), G1,1 = y1 G0,0 = y1 . The conclusion can be drawn. Then consider the case for m = 3 we proceed on the basis of (3.4.12) and (3.4.13). Observation 3.4.18. For m = 3, G3,1 = 3y1 . [2] [2] Proof. On account of G1,1 = 2G0,0 G1,1 , by (3.4.15) and Observation 3.4.17, we have G1,1 =

[2] + y1 G2,0 = 2y1 + y1 G2,0 . By Lemma 3.2.3, we have 2y1 . From (3.4.6) and (3.4.3), G3,1 = G1,1 G3,1 = 3y1 . The conclusion can be drawn.

For integer t ≥ 2 and hence m = 2t + 1 ≥ 5, on account of t−1

t

i=0

i=1

[2] G2t−1,1 = ∑ G2i|0 G2t−1−2i|1 + ∑ G2i−1|1 G2(t−i)|0 .

(3.4.16)

For m = 2t + 1, (3.4.13) leads to [2] [2] G2t+1,1 = G[2t−1,1] + y1 G2t|0 = G2t−1,1 +

(2t)! y t!(t + 1)! 1

(3.4.17)

[2] is shown in (3.4.16). where Gm−2,1

Lemma 3.4.19. For any integer m ≥ 0, we have Gm,1

0, { { { = {3y1 , { { [2] {G2t−1,1 +

[2] where G2t−1,1 is given by (3.4.16).

when m = 2t, t ≥ 0; when m = 3; (2t)! y, t!(t+1)! 1

when m = 2t + 1, t ≥ 2,

(3.4.18)

78 | 3 Inner equations first part Proof. The case of m = 2t is from Lemma 3.4.5. The case of m = 2t+1 is from (3.4.17). It looks more helpful to evaluate Gm,2 for integer m ≥ 0 by the general procedure. Because G0,2 = 0 and G1,2 = 0, it is only necessary for us to evaluate Gm,2 for m ≥ 2. Because G0,2 = 0 (Lemma 3.1.3) and G1,2 = 0 (Lemma 3.4.5), Gm,2 for m ≤ 1 are known. When m = 2, from G0,2 = G0,1 = 0 we have 2 [2] = 2G0,0 G0,2 + G0,1 = 0. G0,2

(3.4.19)

Then by employing (3.4.12) and (3.4.13), 1

G2,2 = ∑ yj+1 Gj+1,1−j . j=0

(3.4.20)

Observation 3.4.20. For m = s = 2, we have G2,2 = y12 + y2 .

(3.4.21)

Proof. By employing (3.4.20), G2,2 = y1 G1,1 + y2 G2,0 = y12 + y2 . This is the conclusion. When m = 4, we proceed on the basis of G0,0 = 1 (Lemma 3.2.3), G1,1 = y1 (Observation 3.2.8), G2,2 (Observation 3.4.20), G3,1 (Observation 3.2.9) and G4,0 (Lemma 3.2.3); it is seen that [2] G2,2 =

∑ 0≤i,j≤2 i+j=0(mod 2)

Gi,j G2−i,2−j = 3y12 + 2y2

(3.4.22)

and 1

∑ yj+1 Gj+3,1−j = y1 G3,1 + y2 G4,0 = 3y12 + 2y2 .

j=0

(3.4.23)

Lemma 3.4.21. For m = 4, G4,2 = 6y12 + 4y2 . Proof. Because of (3.4.22) and (3.4.23), 1

[2] G4,2 = G2,2 + ∑ yj+1 Gj+3,1−j = 6y12 + 4y2 . j=0

This is the conclusion. We proceed on the basis of Observation 3.2.6. For m ≥ 5, it is only necessary to address Gm,2 for m = 2t, t ≥ 3. Observation 3.4.22. For m = 2t, t ≥ 3, we have t−2

t−1 (2i)! (2(t − 1 − i))! G2(t−1−i),2 + ∑ G . i!(i + 1)! (t − 1 − i)!(t − i)! 2i,2 i=0 i=1

[2] G2t−2,2 =∑

(3.4.24)

3.4 Restrictions general ordinary inner case | 79

Proof. On account of (3.4.14), t−1

[2] G2t−2,2 = ∑ (G2i,0 G2(t−1−i),2 + G2i,2 G2(t−1−i),0 ). i=0

By employing Lemma 3.2.3, the conclusion can be drawn. Since all G2(t−1−i),2 for 0 ≤ i ≤ t − 2 and G2i,2 for 1 ≤ i ≤ t − 1 have been determined, [2] in (3.4.24). so is G2t−2,2 Observation 3.4.23. For m = 2t, t ≥ 3, we have Λm−1,2 =

(2t)! (2t − 2)! 2 y + y. (t − 1)!t! 1 t!(t + 1)! 2

(3.4.25)

Proof. From (3.4.12), Λm−1,2 = y1 G2t−1,1 + y2 G2t,0 =

(2t − 2)! 2 (2t)! y1 + y. (t − 1)!t! t!(t + 1)! 2

This is (3.4.25). We proceed on the basis of Observation 3.4.15 and Observation 3.4.23; all Gm,2 for m ≥ 0 are determined. Lemma 3.4.24. For integer m ≥ 0,

Gm,2

0, { { { { { { y 2 + y2 , ={ 12 {4y + y2 , { 1 { { { [2] G { 2t−2,2 + Λ2t−1,2 ,

when m = 0 or m = 1(mod 2); when m = 2; when m = 4;

(3.4.26)

when m = 2t, t ≥ 3,

[2] where G2t−2,2 and Λa[2t−1,2] are, respectively, given by (3.4.24) and (3.4.25).

Proof. The case of m = 0 or m = 1(mod 2) is from, respectively, (3.1.9) or Lemma 3.4.5. The case of m = 2 is from (3.2.7). The case of m = 4 is from Lemma 3.2.12. The case of m = 2t, t ≥ 3 is from Observation 3.4.22 and Observation 3.4.23. Let d = m + s, called the double size, be given. Because both m and s are nonnegative integers, only 0 ≤ m = d − s ≤ d are available while 0 ≤ s ≤ d for Gm,s . When s = 2, m = d − 2, on account of Lemma 3.4.16, only even d are available for determining Ga[m,n] . Half of the double size is the size of Gm,s . Similarly, m and s are, respectively, x-size and y-size. From Lemma 3.4.3 and Lemma 3.4.12, it is only necessary to consider Gm,s for m, s ≥ 1. Further, Lemma 3.4.5 and Lemma 3.4.19 enable us to only consider m, n ≥ 2 for Gm,s .

80 | 3 Inner equations first part Observation 3.4.25. For any m, s ≥ 2,

[2] Gm−2,s

∑ 0≤j≤t G2i,2j G2(l−1−i),2j , { { { 0≤i≤l−1 { { { { when m = 2l, l ≥ 1 ⇒ s = 2t, t ≥ 1; ={ { ∑ 0≤j≤t G2i+1,2j+1 G2(l−1−i),2(t−j) , { { { { 0≤i≤l−1 { when m = 2l + 1 ⇒ s = 2t + 1, t ≥ 1. {

(3.4.27)

Proof. We proceed on the basis of Lemma 3.4.5, the conclusion can then be deduced from (3.4.14). [2] This observation enables us to determine Gm−2,s only by some Gi,j for 0 ≤ i + j ≤ m + s − 2.

Observation 3.4.26. For m, s ≥ 2, 0, Λm−1,s = { s−1 ∑j=0 yj+1 Gm+j−1,s−j−1 ,

when m + s ≠ 0(mod 2); otherwise.

(3.4.28)

Proof. On account of (m + j − 1) + (s − j − 1) = m + s − 2 = m + s(mod 2), the case of m+s ≠ 0(mod 2) is deduced from Observation 3.4.15. The other case is from (3.4.12). We proceed on the basis of Observation 3.4.25 and Observation 3.4.26; the solution of equations (3.4.6), and hence equation (3.4.1) can be evaluated from Theorem 3.4.2. Theorem 3.4.27. The solution of equation (3.4.1) is determined by all Gm,s for m, s ≥ 0 as Gm,s

1, when m = s = 0; { { { = {0, when m + s ≠ 0(mod 2); { { [2] {Gm−2,s + Λm−1,s , when m + s = 0(mod 2),

(3.4.29)

[2] and Λm−1,s are, respectively, given by (3.4.27) and (3.4.28). where Gm−2,s

Proof. The case of m = s = 0 is from (3.4.15). The case of m + s ≠ 0(mod 2) is from Observation 3.4.15. The last case is from Observation 3.4.25 and Observation 3.4.26. We proceed on the basis of this theorem, the solution is evaluated by the procedure starting from G0,0 of m + s = 0 increasingly two by two up to m + s = 2L for getting G2L,0 , G2L−1,1 , G2L−2,2 , . . . , G0,2L if L is the object as shown in Table 3.4.1 where G0,2l = 0 are omitted for l ≥ 1 and t = (m + s)/2. Example 3.4.1. For integers m, s ≥ 0, let l = m + s and Γl = ∑ Gm,s xm . m+s=l m,s≥0

(3.4.30)

3.4 Restrictions general ordinary inner case | 81 Table 3.4.1: Procedure of determining all Ga[m,n] . m+s

Gt,t

0 2 4 6

G4,0 = 2 G6,0 = 5

.. .

G5,1 = 10y1

G4,2 = 6y12 + 4y2

G0,0 = 1 G1,1 = y1 G2,2 = y12 + y2 G3,3 = y13 + 6y1 y2 + 4y3

⋅⋅⋅

⋅⋅⋅

G2,0 = 1 G3,1 = 3y1

G1,3 = y1 y2 + y3 G2,4 = 2y12 y2 + 4y1 y3 + y22 + 2y4

G1,5 = y1 y22 + 3y1 y4 + y12 y3 + 2y2 y3 + 2y5

⋅⋅⋅

From Observation 3.4.15, it is only necessary to consider l = 0(mod 2). It is easily checked that Γl , l ≥ 0, are all polynomials on ℛ[x, y] ⊆ ℛ{x, y}. We proceed on the basis of Table 3.4.1; for 0 ≤ l ≤ 6, we have

1, { { { { { 2 { x { + a1 a2 y1 x, { { { { {2x 4 + 3y1 x3 + (y12 + y2 )x2 + (y1 y2 + y3 )x, Γl = { 6 { 5x + 10y1 x5 + (6y12 + 4y2 )x4 + (y13 + 6y1 y2 { { { { { { + 4y3 )x3 + (2y12 y2 + 4y1 y3 + y22 + 2y4 )x2 { { { { + (y1 y22 + 3y1 y4 + y12 y3 + 2y2 y3 + 2y5 )x, {

when l = 0; when l = 2; when l = 4;

(3.4.31)

when l = 6.

Example 3.4.2. Root-isomorphic classifications of general ordinary planar maps with

double size and vertex-partition vector as parameters. In Liu YP [56] (1999, pp. 236– 242), it is seen that Γl , l = 2t = m + s ≥ 0, are meaningful in combinatorial maps where

t, m and s are, respectively, the size, x-size and y-size. Stated precisely, the coefficient

of the term xm yn in Γl is the number of distinct root-isomorphic classes of general

ordinary planar maps with double size l = m + s. In Figure 3.4.1, all Gm,s for 2 ≤ l ≤ 6 are seen.

For example, Γ2 = (b)x +(a)x2 = y1 x +x2 ; Γ4 = (c +d)x +(e +f )x2 +(3g)x3 +(h+i)x4 =

(y1 y2 + y3 )x + (y12 + y2 )x2 + 3y1 x3 + 2x4 ; Γ6 = (y + (2z + z1) + z2 + (z3 + z4) + (z5 + z5))x + (2t + (2u + 2v) + w + 2x)x2 + (j + (3k + 3l) + (m + 3n))x3 + ((4q + 2r) + 4s)x4 + (5o + 5p)x 5 + 5x 6 .

Hence, Γ6 = (y1 y22 + 3y1 y4 + y12 y3 + 2y2 y3 + 2y5 )x + (2y12 y2 + 4y1 y3 + y22 + 2y4 )x2 + (y13 +

6y1 y2 + 4y3 )x3 + (6y12 + 4y2 )x4 + (10y1 )x5 + 5x 6 . The items that appeared in Γ2 , Γ4 and Γ6

are the same as those shown in (3.4.31).

82 | 3 Inner equations first part

Figure 3.4.1: Root-isomorphic classes of planar maps by double size and vertex-partition vector.

3.5 Eulerian ordinary inner model | 83

Figure 3.4.1: (continued).

3.5 Eulerian ordinary inner model Second, consider the equation 2 2 2 { {a2 x ∫ y δx2 ,y2 f |x2 =u = (1 − x f )f − a1 ; y { { {f |x=0⇒y=0 = a0 ,

(3.5.1)

where a0 , a1 , a2 ∈ ℝ+ and f = f (z, y) ∈ ℛ{z, y} for z = x2 . This is equation (6) in Introduction. Because a solution of equation (3.5.1) for a0 = a1 = a2 = 1 is involved with Eulerian ordinary planar maps shown in [32] (Liu YP, 1986) and then [61] (Liu YP, 2009, p. 122), this equation is called an Eulerian ordinary inner model. Observation 3.5.1. If f ∈ ℛ{z, y} is a solution of equation (3.5.1), then a1 = a0 .

84 | 3 Inner equations first part Proof. From the first equality of equation (3.5.1), 0 = f |x=0⇒y=0 − a1 is known. The initiation causes a1 = a0 . Let fa be a solution of equation (3.5.1). For fa ∈ ℛ{x, y}, fa is determined by Fa[m] = 𝜕x2m f ∈ ℛ{y}, m ≥ 0, where a = (a0 , a1 , a2 ). Because of δx2 ,y2 (fa |u=x2 ) = ∑ Fa[m] ( m≥0

we have

x2m − y2m ) = ∑ x2i y2(m−i−1) Fa[m] , x2 − y2 m≥i i≥0

y2 δx2 ,y2 = ∑ x2i y2(l−i) Fa[l] l≥i i≥0

and hence

∫ y2 δx2 ,y2 (fa |u=x2 ) = ∑ x2i y2(l−i) Fa[l] . l≥i i≥0

y

(3.5.2)

We proceed on the basis of equation (3.5.1) and Observation 3.5.1. For integer m ≥ 0, a0 = a1 , when m = 0; Fa[m] = { [2] Fa[m−1] + a2 ∑l≥m−1 y2(l−m+1) Fa[l] , when m ≥ 1,

(3.5.3)

[2] = 𝜕x2m fa2 , m ≥ 1. where Fa[m]

Theorem 3.5.2. Equation (3.5.1) is equivalent to the system of equations (3.5.3) for Fa[m] ∈ ℛ{y}, m ≥ 0. [2] Proof. Because fa ∈ ℛ{x, y}, Fa[m] ∈ ℛ{y} for m ≥ 0, from fa2 ∈ ℛ{y}, Fa[m−1] ∈ ℛ{y} is deduced. Since equations (3.5.2) are defined on ℛ{x, y}, equation (3.5.3) is transformed to equation (3.5.1) on ℛ{x, y} only by the cancelation law. The conclusion can be drawn.

Because there is no term with odd degree of x known from equation (3.5.1), f is an even function of x, i. e., f (−x) = f (x). Because there is no term with odd degree of y in equation (3.5.1), Theorem 3.5.2 enables us to see that, for any integer i ≥ 0, f is independent of y2i+1 . By the infinity of equations in equations (3.5.3), we are required to seek a new parameter, such that, for any value of the parameter given, it is only necessary to consider a system of finite equations. For any m ≥ 0 given, denote by yn for y = (y2 , y4 , . . .) and n = (n2 , n4 , . . .) as a term of Fa[m] . Let s= called the inner size.

π(n) 1 = ∑(2i)n2i = ∑ in2i , 2 2 i≥1 i≥1

(3.5.4)

3.5 Eulerian ordinary inner model |

85

Observation 3.5.3. The inner size s is a non-negative integer for any integer m ≥ 0 given. Proof. A direct result of (3.5.4). For convenience, let Fa[m,s] = [Fa[m] ]s for any integers m, s ≥ 0, i. e., the part of Fa[m] ∈ ℛ{y} such that π(n)/2 = s. Lemma 3.5.4. When m = 0. For any integer s ≥ 0, a1 , if s = 0; Fa[0,s] = { 0, if s ≥ 1.

(3.5.5)

Proof. From the first case of (3.5.3), the conclusion is derived. This lemma enables us to only consider what happens for m ≥ 1. Lemma 3.5.5. When s = 0, for any integer m ≥ 0, a1 , Fa[m,0] = { am+1 (2m)! 1

m!(m+1)!

if m = 0; ,

(3.5.6)

if m ≥ 1.

Proof. When m = 0, from the case of s = 0 in Lemma 3.5.4, Fa[m,0] = Fa[0,0] = a1 . When m ≥ 1, assume Fa[n,0] =

an+1 1 (2n)! n!(n + 1)!

for 1 ≤ n ≤ m − 1. By induction on m, we prove Fa[m,0] =

am+1 1 (2m)! . m!(m + 1)!

[2] 2 We proceed on the basis of (3.5.3). For m = 1, Fa[1,0] = Fa[0,0] = Fa[0,0] = a21 . For m ≥ 2, m−1

[2] + a2 ∑ y2(l−m+1) Fa[l] ] = ∑ Fa[i,0] Fm−1−i,0 . Fa[m,0] = [Fa[m−1] l≥m−1

0

i=0

Then, by the inductive assumption, m−1

Fa[m,0] = ∑

i=0

m−i ai+1 am+1 (2m)! 1 (2i)! a1 (2(m − 1 − i))! = 1 . i!(i + 1)! (m − 1 − i)!(m − i)! m!(m + 1)!

This is the conclusion. In what follows, only the case of s ≥ 1 is concerned. Because f 2 does not have a term of odd degree of x for m ≥ 1, we have m 2a F , [2] = ∑ Fa[i] Fa[m−i] = { 1 a[1] Fa[m] 2a1 Fa[m] + ∑m−1 i=1 Fa[i] Fa[m−i] , i=0

when m = 1; when m ≥ 2,

(3.5.7)

86 | 3 Inner equations first part and, for any integer s ≥ 1, s

[ ∑ y2(l−m+1) Fa[l] ] = ∑ y2k Fa[k+m−1,s−k] . s

l≥m−1

k=0

(3.5.8)

From (3.5.7) and (3.5.8), equations (3.5.3) become a , if s = 0; { } { 1 Fa[m,s] = { 0, if s ≥ 1, { [2] {Fa[m−1,s] + a2 Σa[m−1,s] , where

when m = 0; when s ≥ 0, m ≥ 1,

s

Σa[m−1,s] = ∑ y2k Fa[k+m−1,s−k] . k=0

(3.5.9)

(3.5.10)

Lemma 3.5.6. Equation (3.5.9) is equivalent to equation (3.5.3) on ℛ{y}. Proof. Because all transformations on ℛ{y} from equations (3.5.3) to equations (3.5.9) are equivalent, the conclusion can be drawn. For m + s ≤ 4, we see what happens in determining Fa[m,s] by (3.5.9). Whenever m + s = 0, Fa[0,0] = a1 is known from (3.5.6). When m + s = 1, from Lemma 3.5.4, it is only necessary to consider Fa[1,0] . From Lemma 3.5.5, Fa[1,0] = a21 .

(3.5.11)

When m + s = 2, from Lemma 3.5.4, it is necessary to consider Fa[2,0] and Fa[1,1] . From Lemma 3.5.5, Fa[2,0] = 2a31 .

(3.5.12)

From (3.5.9), (3.5.7) and Lemma 3.5.4, 1

[2] + a2 ∑ y2k Fa[k,1−k] = a21 a2 y2 . Fa[1,1] = Fa[0,1] k=0

(3.5.13)

When m + s = 3, from Lemma 3.5.4, it is only necessary to consider Fa[3,0] , Fa[2,1] and Fa[1,2] . From Lemma 3.5.5, Fa[3,0] = 5a41 . From (3.5.9), (3.5.7), (3.5.13) and (3.5.6), we have 1

[2] + a2 ∑ y2k Fa[k+2−1,1−k] = 4a31 a2 y2 . Fa[2,1] = Fa[1,1] k=0

(3.5.14)

From (3.5.9) and Lemma 3.5.4, 2

[2] Fa[1,2] = Fa[0,2] + a2 ∑ y2k Fa[k,2−k] = a21 a2 (a2 y22 + 2a1 y4 ). k=0

(3.5.15)

We proceed on the basis of Lemma 3.5.6; we are allowed to state the main theorem of this section.

3.6 Solution Eulerian ordinary inner case

| 87

Theorem 3.5.7. Equation (3.5.1) is well defined on ℛ{x, y} if, and only if, a0 = a1 . Proof. Sufficiency. Because equations (3.5.9) are equivalent to equations (3.5.3) on

ℛ{y}, Theorem 3.5.2 enables us to only discuss that, for integers m, s ≥ 0, and only for them, there is a set of Fa[m,s] satisfying equations (3.5.9) on ℛ{y}.

First, for integers m, s ≥ 0 (m + s ≤ 3), the Fa[m,s] satisfying equations (3.5.9) are known. Then, for other m, s, assume that, for i, j ≥ 0 and i+j ≤ m+s−1, all Fi,j satisfying equations (3.5.9) are known. By induction on m + n, we evaluate Fa[m,n] from equations (3.5.9). In equations (3.5.9) with (3.5.10), the first term is m−1

m−1 s

i=0

i=0 j=0

[2] = ∑ [Fa[i] Fa[m−1−i] ]s = ∑ ∑ Fa[i,j] Fa[m−1−i,s−j] . Fa[m−1,s]

Because of i + j ≤ m − 1 + j ≤ m + s − 1 and (m − 1 − i) + (s − j) ≤ (m − 1) + (s − j) ≤ [2] is deduced from the assumption. The last term is (m − 1) + s = m + s − 1, Fm−1,s s

∑ Fa[k+m−1,s−k] y2k .

k=0

Because (k + m − 1) + (s − k) = m + s − 1 ≤ m + s − 1, the summation is done as well. Therefore, all Fa[m,s] determine a solution of equation (3.5.1). Moreover, by considering the uniqueness of the procedure on equations (3.5.9) from the initiation of equation (3.5.1), the solution is the only one. The necessity is done from Observation 3.5.1.

3.6 Solution Eulerian ordinary inner case From the procedure presented in the proof of Theorem 3.5.7, it is seen that (3.5.9) is, in principle, an algorithm for evaluating the solution of equation (3.5.1). However, some inner structures of the solution of equation (3.5.1) still need to be clarified for simplifying the solution. Lemma 3.6.1. For any integers m, s ≥ 1, Fa[m,s] = 0 whenever s ≥ m + 1. Proof. On Fa[m,s] , for m + s ≤ 3, we have seen that when s ≥ m + 1, Fa[m,s] = 0. For the general case of m + s ≥ 4, assume that, for i + j ≤ m + s − 1, Fa[i,j] = 0 whenever j ≥ i + 1. By induction on m + s, we prove that, when s ≥ m + 1, Fa[m,s] = 0. For this purpose, from (3.5.9), it is only necessary to prove m−1 s

∑ ∑ Fa[k,l] Fa[m−1−k,s−l] = 0

k=0 l=0

and

s

∑ y2k Fa[k+m−1,s−k] = 0.

k=0

On the former, if l ≤ k, then by l ≤ k and s ≤ m + 1, (s − l) − (m − 1 − k) = s − m + 1 + k − l ≥ s − m + 1 ≥ 2, i. e., (s − l) ≥ (m − 1 − k) + 2 > (m − 1 − k) + 1. From the

88 | 3 Inner equations first part assumption, in the summation, all terms are 0. On the latter, because of s ≥ m + 1, (s − k) − (k + m − 1) = s − m + 1 ≥ 2 > 1 implies s − k ≥ (k + m − 1) + 1. Similarly by assumption, all terms in the summation are 0. Therefore, the conclusion can be drawn. This lemma enables us to reduce to half the amount of labor for evaluating Fa[m,s] . Lemma 3.6.2. For integers m, s ≥ 1, Fa[m,s] ∈ ℛ+ [y] whenever a ∈ ℝ3+ . Proof. From the results of calculation in the last section, it is checked that, for integers m, s ≥ 0 such that m + s ≤ 3, Fa[m,s] ∈ ℛ+ [y] whenever a ∈ ℝ3+ . In the general cases of m + s ≥ 4, assume that, for i + j < m + s, Fa[i,j] ∈ ℛ+ [y] (a polynomial!) whenever a ∈ ℝ3+ . By induction on m + s, we prove Fa[m,s] ∈ ℛ+ [y] whenever a ∈ ℝ3+ . We proceed on the basis of (3.5.9). As shown in the proof of Lemma 3.6.1, because k +l ≤ (m−1)+s = m+s−1 and (m−1+k)+(s−l) ≤ (m−1)+s = m+s−1, the assumption [2] ∈ ℛ+ [y] whenever a ∈ ℝ3+ . Because (k + m − 1) + (s − k) = enables us to have Fa[m−1,s] m + s − 1, the assumption enables us to have s

∑ Fk+m−1,s−k y2k ∈ ℛ+ {y}

k=0

whenever a ∈ ℝ3+ . Therefore, the conclusion can be drawn from (3.5.9)–(3.5.10). In fact, the coefficient of each term in Fa[m,s] is a non-negative integer if a ∈ ℤ+ . Lemma 3.6.3. For any integers m, s ≥ 1, Fa[m,s] is independent of yi for i ≥ s + 1. Proof. When 1 ≤ m+s ≤ 3. From (3.5.11)–(3.5.15), the conclusion is satisfied. In general, when m + s ≥ 4, assume, for any integer i, j ≥ 1 with i + j ≤ m + s − 1, that the conclusion is true. By induction on m + s, we prove the conclusion. By (3.5.3), [2] ]s + a2 [ ∑ y2(l−m+1) Fa[l] ] Fa[m,s] = [Fa[m−1] l≥m−1

s

s+m−1

[2] + a2 ∑ y2(l−m+1) Fa[l,s−l+m−1] . = Fa[m−1,s] l=m−1

From the assumption, the conclusion can be drawn. The three lemmas above enable us to refine the solution of equations (3.5.9) to more compact form. [Eoi] = Theorem 3.6.4. Let fa[Eoi] = fa be the solution of equation (3.5.1), determined by Fa[m,s] m Fa[m,s] = 𝜕x fa |s=π(y) for integers m, s ≥ 0. Then all of Fa[m,s] have the form of a finite sum

3.6 Solution Eulerian ordinary inner case

| 89

with all terms positive whenever a ∈ ℝ3+ as

Fa[m,s]

a1 , when s = 0; { }, { { { 0, when s ≥ 1 { { { { = { 0,m+1 { a1 (2m)! { { , { m!(m+1)! { { { [2] { Fa[m−1,s] + a2 Σa[m−1,s] ,

for m = 0; when s ≥ m + 1; } } } when s = 0 , } } } otherwise, }

for m ≥ 1,

(3.6.1)

[2] and Σa[m−1,s] are, respectively, given by (3.5.7) and (3.5.10). where Fm−1,s

Proof. When m = 0 and s = 0, this is the initiation of equation (3.5.1). When m = 0 and s ≥ 1, the case is given by Lemma 3.5.4. For others cases, when s ≥ m + 1, it is given by Lemma 3.6.1; otherwise, the case is determined by Lemma 3.5.5 and Lemma 3.6.2. We proceed on the basis of Theorem 3.6.4; let us continue to determine Fa[m,s] in the solution of equation (3.5.1) for m + s = 4. Here, it is only necessary to evaluate Fa[4,0] , Fa[3,1] , Fa[2,2] and Fa[1,3] . For F4,0 , from Lemma 3.6.1, Fa[4,0] =

a51 8! = 14a51 . 4!5!

(3.6.2)

[2] 2 = [2a1 Fa[2] + Fa[1] For Fa[3,1] , from (3.5.7), Fa[2,1] ]1 . By setting s = 1 to each term in square brackets, we find 2a1 F2,1 + 2F1,0 F1,1 . By (3.5.14), (3.5.11) and (3.5.13), we have [2] = 2a1 (4a31 a2 y2 ) + 2a21 (a21 a2 y2 ) = 10a41 a2 y2 . Fa[2,1]

(3.6.3)

From (3.5.10), Σa[2,1] = y2 Fa[3,0] . By (3.5.6), we have Σa[2,1] = 10a41 y2 .

(3.6.4)

[2] + a2 Σa[2,1] . By (3.6.3) and (3.6.4), we have Then, from (3.6.1), Fa[3,1] = Fa[2,1]

Fa[3,1] = 15a41 a2 y2 .

(3.6.5)

[2] = [2a1 Fa[1] ]2 = 2a1 Fa[1,2] . By (3.5.15), we have For Fa[2,2] , from (3.5.7), Fa[1,2] [2] = 2a21 a2 (a2 y22 + a1 y4 ). Fa[1,2]

(3.6.6)

From (3.5.10), Σa[1,2] = y2 Fa[2,1] + y4 Fa[3,0] . By (3.5.14) and Lemma 3.5.5, Σa[1,2] = 4a31 a2 y22 + 5a41 y4 .

(3.6.7)

[2] + a2 Σa[1,2] . By (3.6.6) and (3.6.7), we have 2a21 a2 (a2 y22 + Then, from (3.6.1), Fa[2,2] = Fa[1,2]

a1 y4 ) + a2 (4a31 a2 y22 + 5a41 y4 ). Then we have

Fa[2,2] = 2a21 a22 (1 + 2a1 )y22 + a31 a2 (2 + 5a1 )y4 .

(3.6.8)

90 | 3 Inner equations first part [2] For Fa[1,3] , from (3.5.7), Fa[0,3] = 0. From (3.5.10), Σa[0,3] = y2 Fa[1,2] + y4 Fa[2,1] +

y6 Fa[3,0] . By (3.5.15), (3.5.14) and Lemma 3.5.5, we have y2 (2a21 a2 (a2 y22 + a1 y4 )) + y4 (4a31 a22 y2 ) + y6 (5a41 ). Then we have Σa[0,3] = 4a31 a2 y22 + 5a41 y4 .

(3.6.9)

From (3.6.1), Fa[1,3] = a2 Σa[0,3] . By (3.6.9), we have a2 (y2 (2a21 a2 (a2 y22 +a1 y4 ))+y4 (4a31 a22 y2 )+ y6 (5a41 )). Then we have Fa[1,3] = a1 a32 y23 + 2a21 a22 (1 + 2a2 )y2 y4 + 5a41 a2 y6 .

(3.6.10)

We proceed on the basis of this section and the last; the solution of equation (3.5.1) is evaluated by the procedure starting from Fa[0,0] of m + s = 0 increasingly one by one up to m + s = 4 for getting Fa[L,0] , Fa[L−1,1] , Fa[L−2,2] , . . . , Fa[1,L] . L = 4 is the object shown in Table 3.6.1. Here, Fa[0,l] = 0 for L ≥ l ≥ 1 are omitted. Table 3.6.1: Chart of determining all Fa[m,s] . m+s 0 1 2 3 4 .. .

Fa[m,s]

Fa[4,0]

Fa[3,0] ⋅⋅⋅

Fa[2,0] Fa[3,1]

Fa[1,0] Fa[2,1]

Fa[0,0] Fa[1,1] Fa[2,2]

⋅⋅⋅

Fa[1,2]

Fa[1,3]

⋅⋅⋅

Example 3.6.1. For integers m, s ≥ 0, let l = m + s and Λa[l] = ∑ Fa[m,s] x2m . m+s=l m,s≥0

(3.6.11)

From Theorem 3.6.1, it is easily checked that Λa[l] , l ≥ 0, are all polynomials on ℛ[x, ys ] ⊆ ℛ{x, y}. We proceed on the basis of Table 3.6.1; for 0 ≤ l ≤ 4, we have

Λa[l]

{ a1 , when l = 0; { { { { 2 2 { a1 x , when l = 1; { { { { {(a3 )x4 + (a2 a y )x2 , { when l = 2; { 1 2 2 { { 1 = {(5a41 )x6 + (4a31 a2 y2 )x4 + (a21 a2 (a2 y22 + 2a1 y4 ))x2 , when l = 3; { { { { (14a51 )x8 + (15a41 a2 y2 )x6 { { { { { { + (2a21 a22 (1 + 2a1 a2 )y22 + a31 a2 (2 + 5a1 )y4 )x4 { { { { + (a1 a32 y23 + 2a21 a22 (1 + 2a2 )y2 y4 + 5a41 a2 y6 )x2 , when l = 4. {

(3.6.12)

3.7 Explicitness Eulerian ordinary inner case | 91

3.7 Explicitness Eulerian ordinary inner case Because of no explicision having yet directly been deduced from Theorem 3.6.4, the relationship between the solutions of the Eulerian ordinary inner equation (3.5.1) and its restriction for a0 = a1 = a2 = 1 has to be investigated on account of the latter being meaningful in graph theory. Let fa[Eoi] and f[Eoi] ∈ ℛ{x, y} be the solutions of, respectively, equation (3.5.1) and [Eoi] [Eoi] and Fm,s ∈ ℛ{y} for integers m, s ≥ 0. Denote by its restriction determined by Fa[m,s] [Eoi] [Eoi] [Eoi] ∈ and hence Fm,s . For any n ∈ 𝒩m,s , Ua[m,n] 𝒩m,s the set of power vectors in Fa[m,s]

[Eoi] [Eoi] |a=1 ∈ ℝ+ are the coefficients of yn in, respectively, ℛa = ℛ{a} and Um,n = Ua[m,n]

[Eoi] [Eoi] and Fm,s . Then Fa[m,s]

[Eoi] [Eoi] [Eoi] [Eoi] . = 𝜕yn Fm,s and Um,n = 𝜕yn Fa[m,s] Ua[m,n] [Eoi] is a Theorem 3.7.1. For any integers m, s ≥ 1 and integral vector n ∈ 𝒩m,s , Ua[m,n] polynomial of a1 = a0 and a2 with degree at most l + 1 where l = m + s such that [Eoi] [Eoi] |a=1 = Um,n . Ua[m,n]

Proof. We proceed on the basis of Theorem 3.6.4. By induction, the conclusion can be drawn easily. For a planar embedding μ(G) of a Eulerian ordinary planar graph G, let n = (n1 , n2 , n3 , . . .) be the vertex-partition vector where ni is the number of non-root-vertices of valency 2i for integer i ≥ 1. A vertex v is a point of μ(G) such that μ(G) − v has at least one component greater than μ(G); it is called a cut-vertex of G. Now, G is always assumed to be connected. It is allowed to restrict a face of maximum length boundary at a cut-vertex as the outerface without loss of generality. The cut-valency of a cut-vertex v is the number of components of μ(G) − v. The closure of such a component is called a cut-component. If a cut-component is not a path, then it is called a fat-component. The number of fatcomponents contributing to the cut-valency of v is called the fat valency of v. A graph without cut-vertex is said to be nonseparable. A maximal nonseparable edge-induced subgraph is called a block of the graph. For a cut-vertex v, let a = av = ρvcut , c = cv and b = bv = ρv be, respectively, the cutvalency, fat valency and the valency of v. Because of the planarity, all cut-components at v have a rotation in the order as cut-components G1 , G2 , ⋅ ⋅ ⋅, Ga . At cut-vertex v, let ti be the contribution of Gi to the valency of v, then t1 + t2 + ⋅ ⋅ ⋅ + ta = b. Three cases have to be considered at cut-vertex v for a Eulerian ordinary planar graph.

92 | 3 Inner equations first part Case I Reflection for Gi , 1 ≤ i ≤ c, it turns out that we have [Eoi] = 2c−1 πvI

(3.7.1)

(topologically) distinct planar embeddings. Case II For cut-rotation at v, it turns out that we have [Eoi] = (a − 1)! πvII

(3.7.2)

(topologically) distinct planar embeddings. Case III For an integer k ≥ 1 given, denote I = {i1 , i2 , . . . , ik } by distributing all Gj , j ∈ ̸ I, into s = (ti1 − 1) + (ti2 − 1) + ⋅ ⋅ ⋅ + (tik − 1) angles on 𝒢I = {Gi | ∀i ∈ I} at v as (x1 , x2 , . . . , xts ) in the s angles such that s

∑ xj = c − k; j=1

0 ≤ xj ≤ c − k, 1 ≤ j ≤ s. Let us denote by

c−k ⟨ ⟩ x1 , . . . , xts

the number of ways for such distributions; then it turns out that we have s c−k ⟩ ∏ xj ! πk[Eoi] = ⟨ x1 , . . . , xts j=1

(topologically) distinct planar embeddings. From the disjointness of the embeddings, we have c−1 c [Eoi] = ∑ ( )πk[Eoi] πvIII k k=1

(3.7.3)

(topologically) distinct planar embeddings. [Eoi] be the set of all cutLemma 3.7.2. For a Eulerian ordinary planar graph G, let VCut vertices. Then all cut-vertices turn out to involve [Eoi] [Eoi] [Eoi] n[Eoi] Cut (G) = ∏ (πvI πvII πvIII ) v∈VCut

(3.7.4)

(topologically) distinct planar embeddings. Proof. On account of the independency among Case I, Case II and Case III, from (3.7.1), (3.7.2) and (3.7.3), the conclusion can be drawn. On an embedding μ of Eulerian planar graph G without cut-vertex, if two vertices u and v have the property that G − {u, v} have at least two components, then {u, v} is

3.7 Explicitness Eulerian ordinary inner case | 93

called a splitting pair. Since an embedding is a point set in its own right on the plane, pu and pv are as the corresponding points of u and v. Because of μ as a point set, any component of μ − {pu , pv } is called a splitting slice, or slice. A splitting slice that is not a path is called a splitting chunk, or chunk. Let S be the set of all splitting pairs consisted of {ui , vi } for 1 ≤ i ≤ s = s(S) = |S| in μ = μ(G). All components of μ − S are all splitting slices in μ(G). For a splitting pair, its slice valent is the number of splitting slices it is incident with. Similarly, for its chunk valency. Observation 3.7.3. Let t and 1 ≤ p ≤ t be, respectively, the slice valency and chunk valency of a splitting pair on an embedding of the Eulerian ordinary planar graph G, then the splitting pair produces [Eoi] πspp = 2p−1 (t − 1)!

(3.7.5)

(topologically) distinct planar embeddings of G. Proof. On account of the independency between rotation on slice valency and reflection on a chunk, by considering the connectivity of embedding, the conclusion can be drawn. Let p be the number of all splitting chunks and ti the slice valency at the splitting pair {ui , vi }, 1 ≤ i ≤ s on μ(G). Lemma 3.7.4. For a Eulerian ordinary planar graph G, the number of topologically distinct embeddings of G generated by all splitting pairs is s

p−s n[Eoi] ∏(ti − 1)! Nct = 2 i=1

(3.7.6)

where pi and ti are, respectively, the number of chunk and slice valencies at splitting pair {ui , vi }, 1 ≤ i ≤ s = |S| such that p = p1 + p2 + ⋅ ⋅ ⋅ + ps . Proof. A result of Observation 3.7.3. This lemma is a refinement and a concise form of Theorem 10.4.3 with (10.4.13) in Liu YP [72] (2017, pp. 196–197), or Theorem 7.4.3 with (7.4.13) in Liu YP [55] (1994, p. 142). We proceed on the basis of Lemma 3.7.2 and Lemma 3.7.4; the result is found to be useful in the context for determining the number of planar embeddings of a Eulerian ordinary planar graph. Lemma 3.7.5. For a Eulerian ordinary planar graph G, the number of (topologically) distinct planar embeddings of G is [Eoi] n[Eoi] (G) = n[Eoi] Cut (G)nNct (G) [Eoi] where n[Eoi] Cut (G) and nNct (G) are, respectively, given by (3.7.4) and (3.7.6).

(3.7.7)

94 | 3 Inner equations first part Proof. We consider the uniqueness of 3-connected embedding of a graph. Because a graph without vertex-cut set of one or two vertices has to be 3-connected, only the two independent cases with and without cut-vertex should be considered. Lemma 3.7.2 and Lemma 3.7.4 lead to the conclusion. ̃ Denote by t = aut(G) the semi-automorphism group order of an Eulerian ordinary planar graph G. The relationship between embeddings and upper maps of G can be found via t. Lemma 3.7.6. Let ℰ [Eoi] (G) and ℳ[Eoi] (G) be, respectively, the sets of all planar embeddings and upper maps of an Eulerian ordinary planar graph G. Then 󵄨󵄨 [Eoi] 󵄨󵄨 2ϵ 󵄨󵄨 [Eoi] 󵄨󵄨 (G)󵄨󵄨 = 󵄨󵄨ℰ (G)󵄨󵄨 󵄨󵄨ℳ t

(3.7.8)

where ϵ = ϵ(G) is the size of G. Proof. Refer to Liu YP [59] (2003, pp. 225–226). [Eoi] Let 𝒢m,s be the set of all Eulerian ordinary planar graphs with rooted semi-valency 2m (i. e., the number of semi-edges incident to the root-vertex) and the un-rooted semisize 2s (i. e., the number of semi-edges not incident to the root-vertex). [Eoi] Denote by 𝒯m,s the set of semi-automorphism group orders of Eulerian ordinary [Eoi] [Eoi] [Eoi] planar graphs in 𝒢m,s . Let 𝒥m,s be the set of all partition vectors n in 𝒢m,s . [Eoi] be the number of (topologically) distinct embeddings of EuleLemma 3.7.7. Let Em,n [Eoi] rian ordinary planar graphs in 𝒢m,n . Then we have [Eoi] = Em,n

∑ n[Eoi] (G)

(3.7.9)

[Eoi] G∈𝒢m,n

where n[Eoi] is given in (3.7.7). Proof. Given the vertex-partition vector n, by considering all embeddings of each [Eoi] graph in 𝒢m,s with n, the conclusion is easily drawn. [Eoi] We proceed on the basis of Lemma 3.7.7. Let 𝒯m,n be the set of semi-automorphism [Eoi] [Eoi] [Eoi] group orders among graphs in 𝒢m,n and 𝒢m,n;t , the set of all graphs in 𝒢m,n with semĩ automorphism group order t. Denote by aut(G) the semi-automorphism group order of G.

Lemma 3.7.8. The number of root-isomorphic classes of upper maps of all graphs in

[Eoi] 𝒢m,n is

A[Eoi] m,n =

∑ [Eoi] t∈𝒯m,n

2(m + π(n)) [Eoi] Em,n;t t

(3.7.10)

3.7 Explicitness Eulerian ordinary inner case | 95

[Eoi] where A[Eoi] m,n = Am,n (ℳm|n ) and [Eoi] Em,n;t =

∑ n[Eoi] (G)|aut(G)=t ̃

(3.7.11)

[Eoi] G∈𝒢m,n

determined by (3.7.9). Proof. We proceed on the basis of Lemma 3.7.6. By considering m + π(n) = ϵ, the conclusion can easily be drawn. This lemma enables us to turn out the solution fa[Eoi] of equation (3.5.1) when a0 = [Eoi] a1 = a2 = 1. Let f[Eoi] = fa[Eoi] |a=1 determined by Fm,s = 𝜕xm f[Eoi] |π(y)=s for m, s ≥ 0. Lemma 3.7.9. For integers m, n ≥ 1, we have [Eoi] Fm,s =

∑ [Eoi] n∈𝒥m,s

n A[Eoi] m,n y

(3.7.12)

where A[Eoi] m,n is given by (3.7.10). Proof. From the set of all underline graphs of Eulerian ordinary planar maps with [Eoi] [Eoi] m and n being the same as 𝒢m,n , A[Eoi] m,n = Um,n is shown in Theorem 3.7.1. From Lemma 3.7.8, the conclusion can be drawn. We proceed on the basis of Lemma 3.7.9 and Theorem 3.7.1; we are allowed to illustrate our main result of this section. [Eoi] determine the solution of equation (3.5.1): Theorem 3.7.10. For m, s ≥ 1, all Fa[m,s] [Eoi] Fa[m,s] =

∑ [Eoi] n∈𝒥m,s

[Eoi] Ua[m,n] yn

(3.7.13)

[Eoi] is the polynomial of a with degree at most m + s + 1 given where a = (a1 , a2 ) and Ua[m,n] in Theorem 3.7.1.

Proof. A result of Theorem 3.7.1 and Lemma 3.7.9. [Eoi] is evaluated by the procedure shown in (3.6.1) or directly The polynomial Ua[m,n] from the inner structures of the solution on its own right. [Eoi] [Eoi] |a=1 determine the solution of equation Corollary 3.7.11. For m, s ≥ 1, Fm,s = Fa[m,s] (3.5.1) when a = 1 by [Eoi] Fm,s =

∑ [Eoi] n∈𝒥m,s

[Eoi] n Um,n y

(3.7.14)

[Eoi] where Um,n = A[Eoi] m,n is given in (3.7.10).

Proof. A direct result of Theorem 3.7.10. [Eoi] This is a direct explicision (explicit expression) of Fm,s for integers m, s ≥ 1.

96 | 3 Inner equations first part

3.8 Restrictions Eulerian ordinary inner case Consider the equation 2 2 2 { { {x ∫ y δx2 ,y2 f |x2 =u = (1 − x f )f − 1; y { { { f |x=0⇒y=0 = 1. {

(3.8.1)

This is equation (3.5.1) when a0 = a1 = a2 = 1. The first equality of equation (3.8.1) involved with root-classification of Eulerian ordinary planar maps under a vertex-partition vector as parameter appears in [32] (Liu YP, 1986) and then [61] (Liu YP, 2009, p. 122). Let f[Eoi] be a solution of equation (3.8.1). For f ∈ ℛ{x, y}, f is determined by Fm = 2m 𝜕x f ∈ ℛ{y}, m ≥ 0. Thus, δx2 ,y2 (f |u=x2 ) = ∑ Fm ( m≥0

x2m − y2m ) = ∑ x2i y2(m−i−1) Fm . x2 − y2 m≥i i≥0

Therefore, y2 δx2 ,y2 (f |u=x2 ) = ∑ x2i ∑ y2(m−i) Fm = ∑ x2i y2(l−i) Fl i≥0

m≥i

l≥i i≥0

and hence ∫ y2 δx2 ,y2 (f |u=x2 ) = ∑ x2i y2(l−i) Fl . y

l≥i i≥0

(3.8.2)

We proceed on the basis of equation (3.8.1). For integer m ≥ 0, 1, when m = 0; Fm = { [2] Fm−1 + ∑l≥m−1 y2(l−m+1) Fl , when m ≥ 1,

(3.8.3)

[2] where Fm = 𝜕x2m f 2 , m ≥ 1.

Theorem 3.8.1. Equation (3.8.1) is equivalent to the system of equations (3.8.3) for Fm ∈ ℛ{y}, m ≥ 0. [2] ∈ ℛ{y} Proof. Because of f ∈ ℛ{x, y}, Fm ∈ ℛ{y} for m ≥ 0. From f 2 ∈ ℛ{y}, Fm−1 is deduced. Because of equations (3.8.2) being defined on ℛ{x, y}, equation (3.8.3) is transformed to equation (3.8.1) on ℛ{x, y}. The conclusion can be drawn.

Because no term with an odd number as the degree of x is known from equation (3.8.1), f is an even function of x, i. e., f (−x) = f (x). There being no term with odd degree of y in equation (3.8.1), Theorem 3.8.1 enables us to see that, for any integer i ≥ 0, f is independent of y2i+1 .

3.8 Restrictions Eulerian ordinary inner case | 97

By the infinity of equations in equations (3.8.3), we are required to seek a new parameter, such that, for any value of the parameter given, it is only necessary to consider a system of finite equations. For any m ≥ 0 given, denote by yn for y = (y2 , y4 , . . .) and n = (n2 , n4 , . . .) a term of Fm . Let s=

π(n) 1 = ∑(2i)n2i = ∑ in2i , 2 2 i≥1 i≥1

(3.8.4)

called the inner size. Observation 3.8.2. The inner size s is a non-negative integer for any integer m ≥ 0 given. Proof. A direct result of (3.8.4). For convenience, let Fm,s = [Fm ]s for any integers m, s ≥ 0, i. e., the part of Fm such that π(n)/2 = s. Lemma 3.8.3. When m = 0. For any integer s ≥ 0, 1, if s = 0; F0,s = { 0, if s ≥ 1.

(3.8.5)

Proof. From the first case of (3.8.3), the conclusion is derived. This lemma enables us only consider what happens for m ≥ 1. Lemma 3.8.4. When s = 0, for any integer m ≥ 0, 1, Fm,0 = { (2m)!

m!(m+1)!

if m = 0; ,

if m ≥ 1.

(3.8.6)

Proof. See the proof of Lemma 3.5.5 for a = 1. In what follows, only the case of s ≥ 1 is addressed. f 2 does not have a term of degree odd of x. For m ≥ 1, Lemma 3.8.3 leads to 2Fa[1] , [2] ={ Fa[m] 2Fa[m] + ∑m−1 i=1 Fa[i] Fa[m−i] ,

when m = 1; when m ≥ 2.

(3.8.7)

For any integer s ≥ 1, [ ∑ y2(l−m+1) Fa[l] ] = ∑ [Fl y2(l−m+1) ]s . l≥m−1

s

l≥m−1

By distributing s to each factor of a term, we have the summation of y2(l−m+1) Fa[l,s−(l−m+1)] over l : l ≥ m − 1. By considering s − (l − m + 1) ≥ 0, it becomes the finite summation of

98 | 3 Inner equations first part y2(l−m+1) Fa[l,s−(l−m+1)] over l : m − 1 ≤ l ≤ s + m − 1. Via l being replaced by k = l − m + 1, we have s

[ ∑ y2(l−m+1) Fa[l] ] = ∑ y2k Fa[k+m−1,s−k] . s

l≥m−1

(3.8.8)

k=0

From (3.8.7) and (3.8.8), equation (3.8.3) becomes Fm,s

1 if s = 0; { }, { = { 0, if s ≥ 1 { [2] {Fm−1,s + Σm−1,s ,

when m = 0;

(3.8.9)

when m ≥ 1,

in which, for s ≥ 0 and m ≥ 1, s

Σm−1,s = ∑ y2k Fk+m−1,s−k . k=0

(3.8.10)

Lemma 3.8.5. Equation (3.8.9) is equivalent to (3.8.3) on ℛ{y}. Proof. Because all transformations from equations (3.8.3) to equations (3.8.9) are equivalent on ℛ{y}, the conclusion can be drawn. For m + s ≤ 4, we see what happens in determining Fm,s by (3.8.9). When m + s = 0, F0,0 = a1 . When m + s = 1, from Lemma 3.8.3, it is only necessary to consider F1,0 . From Lemma 3.8.4, F1,0 = 1.

(3.8.11)

When m + s = 2, from Lemma 3.8.3, it is necessary to consider F2,0 and F1,1 . First, from Lemma 3.8.4, F2,0 = 2.

(3.8.12)

Then, from (3.8.9) with (3.8.7) and (3.8.8), F1,1 = 2F0,1 + F0,0 y2 . By Lemma 3.8.3, we have F1,1 = y2 .

(3.8.13)

When m + s = 3, from Lemma 3.8.3, it is necessary to consider F3,0 , F2,1 and F1,2 . From Lemma 3.8.4, F3,0 = 5. From (3.8.9) with (3.8.7) and (3.8.8), F2,1 = 2F1,1 + F2,0 y2 . By (3.8.13) and (3.8.6), we have F2,1 = 2y2 + 2y2 = 4y2 .

(3.8.14)

From (3.8.9) with (3.8.7) and (3.8.8), F1,2 = F1,1 y2 + F2,0 y4 . By F1,1 and F2,0 , shown above, we have F1,2 = y22 + 2y4 .

(3.8.15)

We proceed on the basis of Lemma 3.8.5; we are allowed to state the main theorem of this section.

3.8 Restrictions Eulerian ordinary inner case | 99

Theorem 3.8.6. Equation (3.8.1) is well defined on ℛ{x, y}. Proof. A direct result of Theorem 3.5.7 for a = 1. From the procedure presented in the proof of Theorem 3.5.7, it is seen that (3.8.9) is, in principle, an algorithm for evaluating the solution of equation (3.8.1). However, some inner structures of the solution of equation (3.8.1) still need to be clarified. Lemma 3.8.7. For any integers m, s ≥ 1, Fm,s = 0 if s ≥ m + 1. Proof. See the proof of Lemma 3.6.1 for a = 1. This lemma enables us to reduce to half of the labor for evaluating Fm,s . Lemma 3.8.8. For integers m, s ≥ 1, Fm,s ∈ ℛ+ [y]. Proof. A direct result of Lemma 3.6.2 for a = 1. In fact, the coefficient of each term in Fm,s is a non-negative integer. Lemma 3.8.9. For any integers m, s ≥ 1, Fm,s is independent of yi for i ≥ s + 1. Proof. See the proof of Lemma 3.6.3 for a = 1. The three lemmas above enable us to refine the solution of equations (3.8.9) to much more compact form to a certain extent. Theorem 3.8.10. Let f[Eoi] be the solution of equation (3.8.1), determined by Fm,s = 𝜕xm f[Eoi] |s=π(y) for integers m, s ≥ 0. Then all of Fm,s are of the form of a finite sum with all terms positive:

Fm,s

1, when s = 0; { { }, { { { 0, when s ≥ 1 ={ { 0, when s ≥ m + 1; { { { [2] }, { Fm−1,s + Σm−1,s , otherwise,

for m = 0; for m ≥ 1,

(3.8.16)

[2] and Σm−1,s are, respectively, given by (3.8.7) and (3.8.10). where Fm−1,s

Proof. When m = 0 and s = 0, this is the initiation of equation (3.8.1). When m = 0 and s ≥ 1, the case is given by Lemma 3.8.3. For m ≥ 1, when s ≥ m + 1, the case given by Lemma 3.8.7; otherwise, it is determined by Lemma 3.8.9. We proceed on the basis of Theorem 3.8.10; let us continue to determine Fm,s in the solution of equation (3.8.1) for m + s = 4. Here, it is only necessary to evaluate F4,0 , F3,1 , F2,2 and F1,3 . For F4,0 , from Lemma 3.8.3, F4,0 =

8! = 14. 4!5!

(3.8.17)

100 | 3 Inner equations first part [2] = [2F2 + F12 ]1 . By considering the value 1 for each term For F3,1 , from (3.8.7), F2,1 in the square brackets, we have 2F2,1 + 2F1,0 F1,1 . By F2,1 , F1,0 and F1,1 shown above, we have [2] F2,1 = 2(4y2 ) + 2(y2 ) = 10y2 .

(3.8.18)

From (3.8.10), Σ2,1 = y2 F3,0 . By (3.8.6), we have Σ2,1 = 10y2 .

(3.8.19)

[2] Then from (3.8.16), F3,1 = F2,1 + Σ2,1 . By (3.8.18) and (3.8.19), we have

F3,1 = 10y2 + (5y2 ) = 15y2 .

(3.8.20)

[2] = [2F1 ]2 = 2F1,2 . By (3.8.15), we have For F2,2 , from (3.8.7), F1,2 [2] F1,2 = 2(y22 + y4 ).

(3.8.21)

From (3.8.10), Σ1,2 = y2 F2,1 + y4 F3,0 . By (3.8.14) and (3.8.6), we have Σ1,2 = y2 (4y2 ) + y4 (5) = 4y22 + 5y4 .

(3.8.22)

[2] Then, from (3.8.16), F2,2 = F1,2 + Σ1,2 . By (3.8.21) and (3.8.22), we have

F2,2 = 2(y22 + y4 ) + (4y22 + 5y4 ) = 6y22 + 7y4 .

(3.8.23)

[2] = 0. From (3.8.10), Σ0,3 = y2 F1,2 +y4 F2,1 +y6 F3,0 . By (3.8.15), For F1,3 , from (3.8.7), F0,3 (3.8.14) and F3,0 = 5, we have

Σ0,3 = y2 (y22 + 2y4 ) + y4 (4y2 ) + y6 (5) = y23 + 6y2 y4 + 5y6 .

(3.8.24)

[2] = 0, F1,3 = Σ0,3 . By (3.8.14), we have We proceed on the basis of (3.8.16). From F0,3

F1,3 = y23 + 6y2 y4 + 5y6 .

(3.8.25)

We proceed on the basis of what was mentioned above: the solution of equation (3.8.1) is evaluated by the procedure starting from F0,0 of m + s = 0 increasingly one by one up to m + s = 4 for getting FL,0 , FL−1,1 , FL−2,2 , . . . , F1,L if L = 4 is the object as shown in Table 3.8.1. Here, F0,l = 0 for L ≥ l ≥ 1 are omitted.

3.8 Restrictions Eulerian ordinary inner case

| 101

Table 3.8.1: Table Chart of determining all Fm,s . m+s 0 1 2 3 4 .. .

Fm,s

F4,0

F3,0 ⋅⋅⋅

F2,0 F3,1

F0,0

F1,0

F1,1

F2,1

F2,2

⋅⋅⋅

F1,2

F1,3

⋅⋅⋅

Example 3.8.1. For integers m, s ≥ 0, let l = m + s and Λl = ∑ Fm,s x2m .

(3.8.26)

m+s=l m,s≥0

From Theorem 3.8.5, it is easily checked that Λl , l ≥ 0, are all polynomials on ℛ[x, ys ] ⊆ ℛ{x, y}. We proceed on the basis of Table 3.8.1; for 0 ≤ l ≤ 4, we have 1, { { { { { {x2 , { { { { { {x4 + (y2 )x2 , Λl = { 6 { 5x + (4y2 )x4 + (y22 + 2y2 )x2 , { { { { {14x8 + (15y )x6 + (6y2 + 7y )x4 { { 2 4 2 { { { 3 2 + (y2 + 6y2 y4 + 5y6 )x , {

when l = 0; when l = 1; when l = 2; when l = 3;

(3.8.27)

when l = 4.

Example 3.8.2. Root-classes of Eulerian ordinary planar maps with vertex-partition. In Liu YP [58] (2001, pp. 174–177), the set of Eulerian ordinary planar maps rooted with vertex-partition is decomposed for determining Fm,s as shown by Theorem 3.8.10 where 2m and 2s are the numbers of semi-edges incident with, respectively, rooted and unrooted vertices. Thus, m + s is the edge number (or size) of each map. In Figure 3.8.1, rooted classes of such maps with size m + s ≤ 4 are provided. Because Fm,0 (m ≥ 2) is the Catalan number, no occurrences except only for a (F0,0 = 1) are in this figure. When m + s = 1, F1,0 = 1(1a), shown as a. When m + s = 2, F1,1 = 1(1b)y2 . When m + s = 3, F2,1 = 4(4c)y2 , F1,2 = 1(1d)y22 + 2(2e)y4 . When m + s = 4, F3,1 = 15(6q + 3r + 6s)y2 , F2,2 = 6(4f + 2i)y22 + 7(4g + 1h + 2j)y4 and F1,3 = 1(1k)y23 + 6(2l + 4m)y2 y4 + 5(2n + 2o + 1p)y6 .

102 | 3 Inner equations first part

Figure 3.8.1: Root-classes of Euler planar maps with vertex-partition.

3.9 Notes | 103

3.9 Notes 3.9.1. Equation coefficients a variable. The generalization of equation (3.1.1), or equation (3.5.1), with coefficients a variable might be extensively investigated for more combinatorial properties of partitions, probably with weights. 3.9.2. One way for explicision. An explicision of the solution of equation (3.1.1), or equation (3.5.1), still needs to be evaluated only by transformations on ℛ{x, y} directly for simplification. 3.9.3. Another way for explicision. Another manner for getting an explicision of the solution of equation (3.1.1), or equation (3.5.1), is in the case of a = (1, 1, 1), i. e., equation (3.4.1), or equation (3.8.1). That is, we observe the root-classifications of general, or Eulerian, ordinary planar maps with vertex-partition given directly without considering semi-automorphism group orders of underline graphs. 3.9.4. New results on embedding of general, or Eulerian, ordinary planar graphs. All results in Section 3.3, or Section 3.7, on embedding of general, or Eulerian, ordinary planar graphs for reaching the corresponding root-enumerations of general, or Eulerian, ordinary planar maps are a type of foundation for getting an explicision for the solution of equation (3.1.1), or equation (3.5.1). 3.9.5. Making efficient and then more intelligent (3.2.15) and (3.6.1), particularly (3.4.29) and (3.8.16) have to be further investigated for running on computers and then for practical usage. 3.9.6. Specification of parameters. For a small number of parameters given, more research can be found in Liu YP [24] (1983), [25] (1982), [26] (1983); Cai JL, Liu YP [8] (1997), [9] (2000), [6] (Cai, J. L., Hao, R. X., Liu, Y. P., 2001), [10] (2005), [95] (Zhang, Y. L., et al., 2010) etc. One can find some relevant results obtained in a different manner in [13] (Cori, R., Dulusq, S., Viennot, G., 1986). 3.9.7. Direct explicision without consideration of symmetry. A direct explicision of (3.7.13) for upper maps of general ordinary planar graphs without consideration of symmetry can be considered. If symmetry is considered for maps, the root-isomorphic maps distribution by their automorphism group orders has to be done as shown in Liu YP [59] (Liu, Y. P., 2003, pp. 221–230), or [73] (2017, pp. 188–194). 3.9.8. The relationship between equation (3.1.1) and equation (3.5.1), or equation (3.4.1) and equation (3.8.1), might be investigated via inner structures over general ordinary planar maps and Eulerian ordinary planar maps. Are there a series of transformations on ℛ{x, y} from one of equation (3.1.1) and equation (3.5.1), particularly, equation (3.4.1) and equation (3.8.1) to the other? 3.9.9. The Tutte formula for planar embeddings of Eulerian graphs with vertex-labels by given vertex-partition vector. Let vi (1 ≤ i ≤ k) be k ≥ 3 vertices labeled. Denote by 2ni

104 | 3 Inner equations first part the valency of vertex v1 for 1 ≤ i ≤ k. Then the size (edge number) is n = n1 +n2 +⋅ ⋅ ⋅+nk . Tutte had found in [89] (Tutte, W. T., 1962) that the number of label-isomorphic planar embeddings of such Eulerian graphs is k (2ni − 1)! (n − 1)! . ∏ (n − k + 2)! i=1 ni !(ni − 1)!

(3.9.1)

On this basis, one might find an explicision of the solution of equation (3.8.1) and then equation (3.5.1) as Theorem 3.7.10 by a similar procedure as shown in [76] (Mao, L. F., Liu, Y. P., Wei, E. L., 2006). 3.9.10. Asymptotic behavior (or asymptotics). Note 3.9.6 reminds us to investigate their asymptotic behavior as shown in [56] (Liu, Y. P., 1999, pp. 359–389), [94] (Yan, J. Y., Liu, Y. P., 1991). 3.9.11. Stochastic behavior (or stochastics). We proceed on the basis of (3.4.29) and (3.8.16); the distributions of sizes, orders, semi-automorphic group orders and genera from, respectively, size polynomials, order polynomials, semi-automorphic group order polynomials and genus polynomials need to be investigated. Relevant results are referred to in [56] (Liu, Y. P., 1999, pp. 359–389), [59] (Liu, Y. P., 2003, 221–230), [94], [11] (Chen, Y. C., et al., 2007), [92] (Wan, L. X., et al., 2008) etc. 3.9.12. Equation (3.1.1) is from Program 89 in [70] as equation (120) (Liu, Y. P., 2015, Vol. 23, p. 11698). Equation (3.5.1) is from Program 92 in [70] as equation (123) (Liu, Y. P., 2015, Vol. 23, p. 11699). The results in this chapter are in the stage of systematization in theory with certain consideration on other two stages: efficientization in running and intelligentization in usage.

4 Inner equations second part 4.1 General loopless inner model Two models are included in this chapter. First (Sections 4.1–4.4), we consider the equation 1 { { {a2 ∫ 1 − a 𝜕 (u2 g| ) = g; 1 x,y x=u y { { { = a0 , g| { x=0⇒y=0

(4.1.1)

where a0 , a1 , a2 ∈ ℝ+ , g ∈ ℛ{x, y} and 𝜕x,y is the slope difference operator. This is equation (7) in Introduction. Because a solution of equation (4.1.1) for a0 = a1 = a2 = 1 is involved with general loopless planar (inner) maps as shown in [61] (Liu YP, 2010, p. 179), this equation is called a general loopless inner model. Note that the original equation is from [36] (Liu YP, 1987). If ga where a = (a0 , a1 , a2 ) is a solution of equation (4.1.1), then ga ∈ ℛ{x, y} is determined by Ga[m] = 𝜕xm ga ∈ ℛ+ {y} for integer m ≥ 0. The relationship between 𝜕x,y (u2 ga |x=u ) and Ga[j] for j ≥ 0 has first to be investigated. Because of 𝜕x,y (u2 ga |x=u ) =

yx 2 ga − xy2 ga |x=y x−y

= xy ∑ Ga[m] ( m≥0

xm+1 − ym+1 ) x−y

and m

xm+1 − ym+1 = (x − y) ∑ xi ym−i , i=0

we have 𝜕x,y (u2 ga |x=u ) = ∑( ∑ Ga[l] yl−j+2 )xj . j≥1 l≥j−1

(4.1.2)

Then the relationship between (𝜕x,y (u2 ga |x=u ))k for k ≥ 2 and Ga[j] for j ≥ 0 can be recursively found. Denote Pa[j] = ∑ Ga[l] yl−j+2 = ∑ Ga[i+j−2] yi , l≥j−1

https://doi.org/10.1515/9783110627336-004

i≥1

(4.1.3)

106 | 4 Inner equations second part then, for any integer k ≥ 2, j

j−1

i=0

i=1

(4.1.4)

[k−1] [k−1] [k] Pa[j−i] . Pa[j−i] = ∑ Pa[i] = ∑ Pa[i] Pa[j] [1] = Pa[i] is guaranteed for 0 ≤ i ≤ j. When k = 1, Pa[i] On account of

1 k = ∑ (a 𝜕 (u2 ga |x=u )) , 1 − a1 𝜕x,y (u2 ga |x=u ) k≥0 1 x,y from equation (4.1.1) and (14.1.3), [k] Ga[m] = a2 ∑ ∫ ak1 Pa[m]

(4.1.5)

[k] Π[k] a[m] = ∫ Pa[m] .

(4.1.6)

k≥0 y

for integer m ≥ 0. For integers m, k ≥ 0, let

y

From (4.1.3) and (4.1.4),

Π[k] a[m]

1, when m = 0; { { } { { 0, when m ≥ 1 { { { { { { 0, when m = 0; } ={ i { G y , when m ≥ 1 ∑ i≥1 a[i+m−2] { { { { { 0, when 0 ≤ m ≤ k − 1; { { { m−1 [k−1] } { ∑i=k−1 Πa[i] ⊗ Πa[m−i] , when m ≥ k,

if k = 0; if k = 1;

(4.1.7)

if k ≥ 2.

From (4.1.5), we have Ga[m] = a2 ∑ ak1 Π[k] a[m] . k≥0

(4.1.8)

Observation 4.1.1. Let ga be a solution of equation (4.1.1), then a0 = a2 . Proof. We have ga ∈ ℛ{x, y}. Since a0 and a2 are the constant terms of, respectively, the left and right hand sides on the equation, the conclusion is true. The observation enables us to only consider a0 = a2 in equation (4.1.1) with a = (a0 , a1 ) instead of (a0 , a1 , a2 ). Theorem 4.1.2. On ℛ{x, y}, equation (4.1.1) is equivalent to the system of equations a0 ∑k≥0 ak1 Π[k] , for m ≥ 1; a[m] Ga[m] = { a0 (= a2 ), for m = 0.

(4.1.9)

4.1 General loopless inner model |

107

Proof. First, from (4.1.3), Pa[0] = 0, and hence Πa[0] = 0. Because Ga[0] = a0 + a1 Πa[0] = a0 , the initiation of equations (4.1.9) is the same of equation (4.1.1). Then, for integer m ≥ 1, from equation (4.1.1) and Observation 4.1.1, equation (4.1.8) is derived. Thus, by the cancelation law on ℛ{x, y}, the conclusion can be drawn. Unfortunately, in equations (4.1.8), the summation is not finite. This motivates us to seek a new parameter s such that the Ga[m,s] are all of the form of a finite sum for any integers m, s ≥ 0. If s is chosen to be the power |n| of y|n| , Ga[|n|] can be shown to be infinite as well. For any n ∈ 𝒥m , we address the set of all power vectors in Ga[m] . Let π(n) = inT , where i = (1, 2, 3, . . .). Because Ga[m] ∈ ℛ{y}, we have Ga[m] = ∑ Ga[m,n] yn n∈𝒥m

(4.1.10)

where 𝒥m is the set of power vectors of y which appear in Ga[m] . Of course, Ga[m,n] = 𝜕yn Ga[m] ∈ ℝ. For any integers m, s ≥ 0, denote Ga[m,s] = ∑ Ga[m,n] yn |π(n)=s = ∑ Ga[m,n] yn . n∈𝒥m

n∈𝒥m π(n)=s

(4.1.11)

Thus, π(Ga[m,s] ) = s is allowed to be called the y-semi-size of Ga[m,s] . Denote 𝒥m,s = {n ∈ 𝒥m |π(n) = s}. Because Ga[0] = a0 (= a2 ) from (4.1.9), Ga[0,s] = 0, s ≥ 1. That is, for s ≥ 0, a0 , when s = 0; Ga[0,s] = { 0, when s ≥ 1.

(4.1.12)

It is only necessary to consider the case of m ≥ 1. When k = 1, from (4.1.7), s−1

Πa[m,s] = ∑ yj+1 Ga[j+m−1,s−j−1] . j=0

(4.1.13)

Lemma 4.1.3. For any integer m ≥ 0 and s = 0, a0 , when m = 0; Ga[m,0] = { 0, when m ≥ 1.

(4.1.14)

Proof. When m = 0, from the initiation of (4.1.9), Ga[0,0] = a0 . When m ≥ 1. From (4.1.7) and (4.1.8), Π[k] = 0 for k ≥ 0 leads to the conclusion. a[0,0]

108 | 4 Inner equations second part In what follows, we address Ga[m,1] . We see how the case of s = 1 can be deduced from Lemma 4.1.3. Lemma 4.1.4. For any integer m ≥ 0,

Ga[m,1]

0, when m = 0; { { { 2 = {a0 a1 y1 , when m = 1; { { when m ≥ 2. {0,

(4.1.15)

Proof. From (4.1.9), Ga[0,0] = a0 and then Ga[0,s] = 0, s ≥ 1. This is the conclusion for m = 0. = Πa[1,1] = On the general case of m ≥ 1, for m = 1, (4.1.13) yields Π[1] a[1,1] ⟨y1 Ga[0,0] ⟩1 = a0 y1 . [k] Because Π[0] 0,1 = 0, (4.1.7) and (4.1.13) show that Πa[1,1] = 0 for k ≥ 2, from (4.1.8), [1] ) = a20 a1 y1 . Ga[1,1] = a0 (a1 Πa[1,1]

[1] 2 = 0 and Πa[2,1] From (4.1.7), Π[0] = ⟨Pa[1] ⟩1 = = y1 Ga[1,0] = 0. From (4.1.7), Π[2] a[2,1] a[2,1]

2 Pa[1,1]

= ⟨y12 ⟩1 = 0. Similarly, Π[k] = 0, k ≥ 3. Thus, (4.1.8) yields Ga[2,1] = 0. Furthera[2,1] more, Ga[m,1] = 0 for integer m ≥ 3.

We proceed on the basis of Ga[1,0] = Ga[0,1] = 0; it suggests to observe the distribution of 0 among all Ga[m,s] for m, s ≥ 0, so that the amount of labor can be reduced for their evaluation. Lemma 4.1.5. Given integers m, s ≥ 0. If n ∈ 𝒥m,s such that Ga[m,s] ≠ 0, then s = π(n) = m(mod 2). Proof. From the evaluation above, it is seen that when m + s ≤ 1, only Ga[0,0] ≠ 0. In this case, m = s(mod 2). The conclusion is true. For m + s ≥ 2, by induction on m + s. Assume Ga[i,j] ≠ 0 for i + j < m + s and i = j(mod 2) with i, j ≥ 0. We prove it for Ga[m,s] ≠ 0, s = m(mod 2). From (4.1.7) and (4.1.8), m

Ga[m,s] = a0 ∑ ak1 Π[k] a[m,s] . k=0

(4.1.16)

for k ≥ 2 are only dependent on Πa[i,j] for i + j ≤ m + s − 1. From From (4.1.7), Π[k] a[m,s] the assumption, all terms in Ga[m,s] with k ≥ 2 satisfy the conclusion. The only case remaining is for k = 1. Consider the terms in (4.1.13) as yj+1 Ga[j+m−1,s−j−1] , 0 ≤ j ≤ s − 1. We proceed on the basis of (j + m − 1) + (s − j − 1) = m + s − 2 ≤ m + s − 1. From the assumption, (j + m − 1) = (s − j − 1)(mod 2) 󳨐⇒ m − 1 = s − 1(mod 2) 󳨐⇒ m = s(mod 2). This is the conclusion. Because m = s(mod 2) is the same as m + s = 0(mod 2), this lemma enables us to see that m + π(n) = 0(mod 2) for n ∈ 𝒥m,s .

4.2 Solution general loopless inner case

| 109

Theorem 4.1.6. Equation (4.1.1) is well defined on ℛ{x, y} if, and only if, a0 = a2 . Proof. Given integers m, s ≥ 0. When m + s = 0, then m = s = 0. From (4.1.14), Ga[0,0] = a0 . This is the initiation of equation (4.1.1). When m + s = 1, this is to determine Ga[0,1] and Ga[1,0] . From Lemma 4.1.5, Ga[0,1] = Ga[1,0] = 0 satisfies equations (4.1.9). From Theorem 4.1.2, they also satisfy equation (4.1.1). On this basis, assume for any integers l, t ≥ 0, all Ga[l,t] satisfy equations (4.1.9) if l + t < m + s. By induction on m + s, we prove that the Ga[m,s] satisfy equations (4.1.9). From equation (4.1.13), Πa[m,s] only depends on Ga[j+m−1,s−j−1] . Because (j + m − 1) + (s−j−1) = m+s−2 < m+s, the assumption shows that Πa[m,s] is determined. From (4.1.7), Πka[m,s] is only determined by Πa[m,s] for 2 ≤ k ≤ m. Thus, Ga[m,s] is also determined by (4.1.16). This implies that equation (4.1.9) has a solution. From Theorem 4.1.2, equation (4.1.1) has a solution. Further, from the uniqueness of the solution of equations (4.1.9), equation (4.1.1) has only one solution. The sufficiency is proved. The necessity is done from Observation 4.1.1.

4.2 Solution general loopless inner case In order to simplify the solution as much as possible, the relevant structures of the solution have to be further investigated. Lemma 4.2.1. For any integer s ≥ 1 given, if the integer s ≤ m − 1, then Ga[m,s] = 0. Proof. When s = 0 and s = 1, (4.1.14) and (4.1.15) are, respectively, employed. If the positive integer s ≤ m − 1, then Ga[m,s] = 0. When s ≥ 2, assume that, for any integer t ≤ s − 1, Ga[m,t] = 0 for t ≤ m − 1. By induction on s, we prove the conclusion for t = s. From (4.1.14), it is seen that, for s ≥ 1, (4.1.16) becomes m

Ga[m,s] = a0 ∑ ak1 Π[k] a[m,s] . k=1

(4.2.1)

= 0 for integers 1 ≤ In order to show Ga[m,s] = 0, it is only necessary to show Π[k] a[m,s] k ≤ m. When k = 1, we proceed on the basis of (4.1.13). From s ≤ m−1, s−j −1 ≤ m−1−j −1. By s − 1 ≥ j ≥ 0, s − j − 1 ≤ (m + j − 1) − 1. On account of s − j − 1 ≤ s − 1, by the inductive assumption, Ga[m+j−1,s−j−1] = 0 for all integers j : s − 1 ≥ j ≥ 0. From (4.1.13), Πa[m,s] = Π[1] = 0. a[m,s] = 0 is found for k ≥ 2. Further from (4.1.7), Π[k] a[m,s]

This lemma enables us to evaluate Ga[m,s] for the solution of equations (4.1.9) without consideration on the area m ≥ s + 1 of m = s(mod 2) over m, s ≥ 1. Lemma 4.2.2. For any integers m, s ≥ 1, if m + s is odd, then Ga[m,s] = 0.

110 | 4 Inner equations second part Proof. In fact, this is a direct result of Lemma 4.1.5. The two lemmas above enable us to reduce by three-fourths amount the labor for solving the system of equations (4.1.9). Lemma 4.2.3. Given an integer s ≥ 1, for any integer m ≥ 1, Ga[m,s] ∈ ℛ{y} is independent of yl for l ≥ s + 1. is for 1 ≤ k ≤ m. From (4.1.7), Π[k] Proof. From (4.2.1), Ga[m,s] is determined by Π[k] a[m,s] a[m,s]

for 2 ≤ k ≤ m. From (4.1.13), Πa[m,s] is determined determined only by Πa[m,s] = Π[1] a[m,s] by Ga[j+m−1,s−j−1] for 0 ≤ j ≤ s − 1 and yl for 1 ≤ l ≤ s. For s = 0 and 1, the conclusion can be drawn from, respectively, (4.1.14) and (4.1.15). For s ≥ 2, we proceed by induction on s. Assume that, for integer t : 1 ≤ t ≤ s−1, Ga[m,t] is only dependent on yl , 1 ≤ l ≤ s − 1. We prove that Ga[m,s] is dependent on yl , 1 ≤ l ≤ s. Because s − j − 1 ≤ s − 1 for 0 ≤ j ≤ s − 1, from the assumption, it is seen that Ga[m,s] only depends on some yl , 1 ≤ l ≤ s. The conclusion can be drawn. This lemma suggests it is only necessary to consider ys for evaluating Ga[m,s] (m, s ≥ 1). That implies Ga[m,s] ∈ ℛ{ys }. Lemma 4.2.4. For two integers m, s ≥ 1, Ga[m,s] is a polynomial on ℛ{ys } with terms of minimum degree 1 and maximum degree s. Proof. From Lemma 4.2.3, we are allowed to only consider ns = (n1 , n2 , . . . , ns ) ∈ 𝒥m,s . The term of power vector ns has its degree |ns |. Denote by 𝒟 = {|ns | | ns ∈ 𝒥m,s } the finite set of all degrees among all terms in Ga[m,s] . Since 1 = |1[i] | = min |ns | ns ∈𝒟

(1 ≤ i ≤ s)

where 1[i] is the vector of all nj = δi,j for 1 ≤ j ≤ s and s = |s11 | = max |ns |, ns ∈𝒟

the conclusion can be drawn. This lemma tells us that, for any integers m, s ≥ 1, the evaluation of Ga[m,n] is running on a space of finite dimension. Lemma 4.2.5. The solution of equation (4.1.1) is determined by the following polynomials of all terms with positive coefficients:

Ga[m,s]

a0 , when m + s = 0, i. e., m = s = 0; { { { m [k] k = {a0 ∑k=1 a1 Πa[m,s] , when m + s ≥ 1, m, s ≥ 1, m = s(mod 2); { { otherwise, {0,

(4.2.2)

4.2 Solution general loopless inner case

| 111

{ ∑s−1 { j=0 yj+1 Ga[j+m−1,s−j−1] (= Πa[m,s] ), when k = 1; { { [s−1] for k = m = s; otherwise, = {∑1≤i,j≤s−1 (Πa[s−i,s−j] ⊗ Πa[i,j] ), { { {∑ 1≤j≤s−1 (Π[k−1] ⊗ Πa[i,j] ), when k ≥ 2. { k−1≤i≤m−1 a[m−i,s−j]

(4.2.3)

for integers m, s ≥ 0 where

Π[k] a[m,s]

Proof. From Lemma 4.2.1, Lemma 4.2.2 and (4.2.1), (4.2.2) is derived. Further, (4.2.3) is found. We proceed on the basis of this theorem, to evaluate Ga[m,s] for m + s ≤ 6. From (4.2.2), only the four cases: m + s = 0, 2, 4 and 6 need to be considered. When m + s = 0, only Ga[0,0] = a0 is considered for m = s = 0. Then for m, s ≥ 1. When m+s = 2, it is only necessary to evaluate Ga[1,1] from Ga[2,0] = 0 (Lemma 4.1.3) and Ga[0,2] = 0 by (4.1.12). We have Ga[1,1] = a0 a1 Πa[1,1] and 1−1

Πa[1,1] = ∑ yj+1 Ga[j,−j] = y1 Ga[0,0] . j=0

From Ga[0,0] = a0 , we have Ga[1,1] = a20 a1 y1 .

(4.2.4)

When m + s = 4. On account of Ga[4,0] = 0 (Lemma 4.1.3), Ga[3,1] = 0 (Lemma 4.1.4) and Ga[0,4] = 0 by (4.1.12), it is only necessary to evaluate Ga[2,2] and Ga[1,3] . [1] [2] +a21 Π[2] Because Ga[2,2] = a0 (a1 Π[1] ), Πa[2,2] = Πa[2,2] and Πa[2,2] = Πa[1,1] ⊗Πa[1,1] , a[2,2] a[2,2] we have [1] {Πa[2,2] = y1 Ga[1,1] + y2 Ga[2,0] = y1 Ga[1,1] ; { [2] 2 {Πa[2,2] = y1 Ga[0,0] ⊗ y1 Ga[0,0] = y2 Ga[0,0] .

By the second case of (4.2.2), we have [1] 2 [2] + a1 Πa[2,2] + a1 y1 Ga[1,1] ). ) = a0 (a21 y2 Ga[0,0] Ga[2,2] = a0 (a21 Πa[2,2]

By Ga[0,0] = a0 and Ga[1,1] = a20 a1 y1 , Ga[2,2] = a0 (a21 y2 a20 + a1 y1 a20 a1 y1 ) = a0 (a0 a1 )2 (y12 + y2 ).

(4.2.5)

We proceed on the basis of Ga[1,3] = a0 a1 Πa[1,3] and Πa[1,3] as the summation of yj+1 Ga[j,2−j] over 0 ≤ j ≤ 2. By (4.2.1), we have y1 Ga[0,2] + y2 Ga[1,1] + y3 Ga[2,0] . By Ga[0,2] = Ga[2,0] = 0 and (4.2.4), Πa[1,3] = y2 G1,1 = a20 a1 y1 y2 . From (4.2.1), we have Ga[1,3] = a0 a1 Πa[1,3] = a30 a21 y1 y2 .

(4.2.6)

112 | 4 Inner equations second part When m + s = 6, on account of Ga[6,0] = 0 (Lemma 4.1.3), Ga[5,1] = 0 (Lemma 4.1.4), Ga[4,2] = 0 (Lemma 4.2.1) and Ga[0,6] = 0 (by (4.1.12)), it is only necessary to evaluate Ga[3,3] , Ga[2,4] and Ga[1,5] . = . We have Π[1] + Π[3] + Π[2] We proceed on the basis of Ga[3,3] = Π[1] a[3,3] a[3,3] a[3,3] a[3,3]

= Πa[1,1] ⊗ Π[2] = a30 y3 . From the = 2a30 a1 y1 y2 and Π[3] a0 (a0 a1 )2 (y13 + y1 y2 ), Π[2] a[2,2] a[3,3] a[3,3] second case of (4.2.2), we have Ga[3,3] = a0 (a1 (a0 a1 )2 (y13 + y1 y2 ) + a21 (2a30 a1 y1 y2 ) + a31 (a30 y3 )) = (a0 a1 )3 y13 + (a0 a1 )3 (1 + 2a0 )y1 y2 + a40 a31 y3 .

(4.2.7)

+ a21 Π[2] We proceed on the basis of Ga[2,4] = a0 (a1 Π[1] = ). Because of Π[1] a[2,4] a[2,4] a[2,4]

a30 a21 (2y12 y2 + y22 ) and Π[2] = 2a0 (a20 a1 )y1 y3 , we have a[2,4]

Ga[2,4] = a0 (a1 (a30 a21 (2y12 y2 + y22 )) + a21 (2a0 (a20 a1 )y1 y3 )) = a40 a31 (2y12 y2 + 2y1 y3 + y22 ).

(4.2.8)

[1] We proceed on the basis of Ga[1,5] = a0 a1 Πa[1,5] = a0 a1 Πa[1,5] . Because of Πa[1,5] =

a30 a21 (y1 y22 + y12 y3 + y2 y3 ), we have

Ga[1,5] = a0 a1 (a30 a21 (y1 y22 + y12 y3 + y2 y3 )) = a40 a31 (y1 y22 + y12 y3 + y2 y3 ).

(4.2.9)

= as0 ys . Corollary 4.2.6. For any integer s ≥ 1, Π[s] a[s,s] Proof. We proceed on the basis of (4.2.3). When m = k = s, s−1 s−1

[s−1] Π[s] a[s,s] = ∑ ∑ (Πa[s−i,s−j] ⊗ Πa[i,j] ). i=1 j=1

If i ≠ j, then either i > j or s − i > s − j. From Lemma 4.2.1, only i = j is considered. This implies that [s−1] Π[s] a[s,s] = Πa[s−j,s−j] ⊗ Πa[j,j]

(4.2.10)

for j ≥ 1. If j ≠ 1, i. e., j ≥ 2, then s − j ≤ s − 2 < s − 1. From Lemma 4.2.1, Π[s−1] = 0. This s−j,s−j implies that [s−1] Π[s] a[s,s] = Πa[s−1,s−1] ⊗ Πa[1,1] .

Because Πa[1,1] = a0 y1 , the conclusion is true for s = 1. For s ≥ 2, we are allowed to assume that Π[s−1] = as−1 0 ys−1 by the inductive principle on n. On account a[s−1,s−1]

4.3 Explicitness general loopless inner case | 113

of (4.2.10), s−1 s Π[s] a[s,s] = a0 ys−1 ⊗ Πa[1,1] = ys−1 ⊗ a0 y1 = a0 ys .

The conclusion can be drawn. This corollary enables us to determine Π[s] s,s for s ≥ 1 in a very simple way in summation-free form. [gli] = Theorem 4.2.7. The solution ga[gli] of equation (4.1.1) is determined by Ga[m,s] m 𝜕x ga[gli] |π(n)=s for integers m, s ≥ 0,

[gli] Ga[m,s]

a0 , when m + s = 0, i. e., m = s = 0; { { { m [k] k = {a0 ∑k=1 a1 Πa[m,s] , when m + s ≥ 1, s ≥ m ≥ 1 and m = s(mod 2); (4.2.11) { { otherwise, {0,

where for s ≥ m ≥ 1 and m = s(mod 2),

Π[k] a[m,s]

{ as0 ys , when k = m = s; otherwise, { { { s−1 = {∑j=0 yj+1 Ga[j+m−1,s−j−1] (= Πa[m,s] ), when k = 1; { { {∑ 1≤j≤s−1 (Π[k−1] ⊗ Πa[i,j] ), when k ≥ 2. { k−1≤i≤m−1 a[m−i,s−j]

(4.2.12)

Proof. A direct result of Lemma 4.2.5 and Corollary 4.2.6.

4.3 Explicitness general loopless inner case Because no explicision has yet directly been deduced from Theorem 4.2.7, the relationship between the solutions of the general loopless inner equation (4.1.1) and its restriction for a0 = a1 = a2 = 1 has to be investigated by considering the latter to be meaningful in combinatorial maps. Let ga[gli] and g[gli] ∈ ℛ{x, y} be the solutions of, respectively, equation (4.1.1)

[gli] [gli] and Gm,s ∈ ℛ{y} for integers m, s ≥ 0. Deand its restriction determined by Ga[m,s] note by 𝒩m,s the set of all power vectors in Ga[m,s] and hence Gm,s . For any n ∈ 𝒩m,s , [gli] [gli] n | and A[gli] A[gli] m,n |s=π(n) are the coefficients of y in, respectively, Ga[m,s] and Gm,s . a[m,n] s=π(n) Then [gli] | = 𝜕yn Ga[m,s] A[gli] a[m,n] π(n)=s

n [gli] and A[gli] m,n |π(n)=s = 𝜕y Gm,s .

is a Theorem 4.3.1. For any integers m, s ≥ 1, and integral vector n ∈ 𝒩m,s , A[gli] a[m,n]

polynomial of a with degree at most l + 1 where l = m + s such that A[gli] = A[gli] m,n . 1[m,n]

Proof. We proceed on the basis of Theorem 4.2.7. By induction, the conclusion can be drawn.

114 | 4 Inner equations second part For a planar embedding μ(G) of a general loopless planar graph G, let n = (n1 , n2 , n3 , . . .) be the vertex-partition vector where ni is the number of y-vertices (or non-root-vertex) with valency i for integer i ≥ 1. A vertex v of G whose deletion makes G − v have at least one component greater than G is called a cut-vertex of G. Now, G is always assumed to be connected. It is allowed to restrict a face of maximum length boundary at a cut vertex as the outerface without loss of generality. The cut-valency of a cut-vertex v is the number of travels without repetition of v on outerface boundary. Each travel with its inner parts is called a cut-component. A cut-component is called a fat-component if it is not a path. Then the fat-valency of a cut-vertex is significant in its own right. A graph without cut-vertex is said to be nonseparable. A maximal nonseparable edge-induced subgraph is called a block of the graph. For a cut-vertex v, let a = ρcut (= av ), c(= cv ) and b = ρv (= bv ) be, respectively, the cut-valency, the fat-valency and the valency of v. Because of planarity, all cutcomponents at v have a rotation in the order as G1 G2 , . . . , Ga clockwise. At cut-vertex v, let ti be the contribution of Gi to the valency of v, then t1 + t2 + ⋅ ⋅ ⋅ + ta = b. Three cases have to be considered at a cut-vertex v. Case I Reflection for fat-components; it turns out that we have [gli] = 2c−1 πvI

(4.3.1)

(topologically) distinct planar embeddings. Case II For a cut-rotation at v, it turns out that we have [gli] = (a − 1)! πvII

(4.3.2)

(topologically) distinct planar embeddings. Case III For an integer k ≥ 1 given, denote I = {i1 , i2 , . . . , ik } by distributing all Gj , j ∈ ̸ I, into s = (ti1 −1)+(ti2 −1)+⋅ ⋅ ⋅+(tik −1) angles on 𝒢I = {Gi |∀i ∈ I} at v as (x1 , x2 , . . . , xts ) in the s angles such that s

∑ xj = a(= ρvcut ) − k; j=1

0 ≤ xj ≤ a − k, 1 ≤ j ≤ s. Let us denote by ⟨

a−k ⟩ x1 , . . . , xts

the number of ways for such distributions, then it turns out that we have s a−k πk[gli] = ⟨ ⟩ ∏(xj )! x1 , . . . , xts j=1

4.3 Explicitness general loopless inner case | 115

(topologically) distinct planar embeddings. From the disjointness of such embeddings, we have a−1 a [gli] = ∑ ( )πk[gli] πvIII k k=1

(4.3.3)

(topologically) distinct planar embeddings. [gli] Lemma 4.3.2. For a general loopless planar graph G, let VCut be the set of all cutvertices. Then all cut-vertices turn out to be [gli] [gli] [gli] n[gli] Cut (G) = ∏ (πvI πvII πvIII ) v∈VCut

(4.3.4)

(topologically) distinct planar embeddings. Proof. On account of the independency among Case I, Case II and Case III, from (4.3.1), (4.3.2) and (4.3.3), the conclusion can be drawn. On a planar embedding μ = μ(G) of graph G without cut-vertex, if two vertices u and v in a block have the property that ν(G) − {u, v} has at least two components, then {u, v} is called a splitting pair. Since an embedding is a point set in its own right on the plane, pu and pv are as the corresponding points of u and v. Because of μ as a point set, any component of μ − {pu , pv } is called a splitting slice, or slice. A splitting slice that is not a path is called a splitting chunk, or chunk. Let S be the set of all splitting pairs consisting of {ui , vi } for 1 ≤ i ≤ s = |S| in μ = μ(G). All components of μ − S are splitting slices in μ(G). For a splitting pair, its slice valency is the number of splitting slices incident with it. Similarly, for chunk valency. Observation 4.3.3. Let t and 1 ≤ p ≤ t be, respectively, the slice valency and chunk valency of a splitting pair on a general loopless planar embedding of graph G, then the splitting pair produces [gli] πspp = 2p−1 (t − 1)!

(4.3.5)

(topologically) distinct planar embeddings of G. Proof. On account of the independency between rotation on slice valency and reflection on a chunk, by considering the connectivity of embedding, the conclusion can be drawn. Let pi be the number of splitting chunks and ti , the slice valency at the splitting pair {ui , vi }, 1 ≤ i ≤ s on μ(G). Lemma 4.3.4. For a general loopless planar graph G, the number of topologically distinct embeddings of G without consideration of cut-vertex is

116 | 4 Inner equations second part l

p−s n[gli] ∏((j − 1)!) j Nct (G) = 2

(4.3.6)

j≥2

where p = p1 + p2 + ⋅ ⋅ ⋅ + ps and lj is the number of splitting pairs with slice valency j ≥ 2. Proof. We proceed on the basis of Observation 4.3.3. Since s

l

pi −1 (ti − 1)!) = 2p−s ∏((j − 1)!) j , n[gli] Nct (G) = ∏ (2 i=1

j≥2

the conclusion can be drawn. This lemma is a refinement and a concise form of Theorem 10.4.3 with (10.4.13) in Liu YP [72] (2017, pp. 196–197), or Theorem 7.4.3 with (7.4.13) in Liu YP [55] (1994, p. 142). We proceed on the basis of Lemma 4.3.2 and Lemma 4.3.4, the result being useful in the context of determining the number of general loopless planar embeddings of a graph. Lemma 4.3.5. For a general loopless planar graph G, the number of planar embeddings of G is [gli] n[gli] (G) = n[gli] Cut (G)nNct (G)

(4.3.7)

[gli] where n[gli] Cut (G) and nNct (G) are, respectively, given by (4.3.4) and (4.3.6).

Proof. We proceed on the basis of uniqueness of planar embedding of a 3-connected graph. Because a graph without vertex-cut set of one or two vertices has to be 3-connected, only the two independent cases with and without cut vertex need to be considered. Lemma 4.3.2 and Lemma 4.3.4 lead to the conclusion. ̃ Denote by t = aut(G) the order of the semi-automorphism of graph G. The relationship between embeddings and upper maps of G can be shown via t. Lemma 4.3.6. Let ℰ [gli] (G) and ℳ[gli] (G) be, respectively, the sets of all distinct topological planar embeddings and root-isomorphic upper maps of G. Then 󵄨󵄨 [gli] 󵄨󵄨 2ϵ 󵄨󵄨 [gli] 󵄨󵄨 󵄨󵄨ℳ (G)󵄨󵄨 = 󵄨󵄨ℰ (G)󵄨󵄨 t

(4.3.8)

where ϵ = ϵ(G) is the size of G. Proof. Refer to Liu YP [59] (2003, pp. 225–226). [gli] Let 𝒢m,s be the set of all general loopless planar graphs with rooted semi-valency m (i. e., the number of semi-edges incident to the root-vertex) and the un-rooted semisize s (i. e., the number of semi-edges not incident to the root-vertex). [gli] [gli] Denote by 𝒯m,s the set of semi-automorphism group orders of graphs in 𝒢m,s . Let [gli] [gli] 𝒥m,s be the set of all partition vectors n in 𝒢m,s .

4.3 Explicitness general loopless inner case | 117

[gli] Lemma 4.3.7. Let Em,n be the number of distinct (topologically) planar embeddings of [gli] graphs in 𝒢m,n . Then we have [gli] Em,n =

∑ n[gli] (G)

(4.3.9)

[gli] G∈𝒢m,n

where n[gli] is given in (4.3.7). [gli] Proof. By the partition of all embeddings on 𝒢m,s from each graph, the conclusion can be drawn.

We proceed on the basis of Lemma 4.3.7. Let 𝒯m,n be the set of semi-automorphism [gli] [gli] [gli] group orders for graphs in 𝒢m,s and 𝒢m,s;t the set of all graphs in 𝒢m,s with semĩ automorphism group order t. Denote by aut(G) the semi-automorphism group order of G. Lemma 4.3.8. The number of root-isomorphic classes of upper planar maps of all [gli] graphs in 𝒢m,s is A[gli] m,n = ∑

t∈𝒯n

m + π(n) [gli] Em,n;t t

(4.3.10)

[gli] where A[gli] m,n = Am,n (𝒢m,s ), s = π(n) and [gli] Em,n;t =

∑ ngoi (G)|aut(G)=t ̃

(4.3.11)

[gli] G∈𝒢m,s

as determined by (4.3.9). Proof. We proceed on the basis of Lemma 4.3.6. By considering m + π(n) = 2ϵ, the conclusion can be drawn. This lemma enables us to find the solution ga[gli] of equation (4.1.1) when a0 = a1 = [gli] a2 = 1. Denote g[gli] = ga[gli] |a=1 as determined by Gm,s = 𝜕xm g[gli] |π(y)=s for m, s ≥ 0. Lemma 4.3.9. For integers m, n ≥ 1, we have [gli] Gm,s =

n ∑ A[gli] m,n y

(4.3.12)

[gli] n∈𝒥m,s

where A[gli] m,n is given by (4.3.10). Proof. From Lemma 4.3.8, the conclusion can be drawn. We proceed on the basis of Lemma 4.3.9 and Theorem 4.3.1; we are allowed to describe our main result of this section.

118 | 4 Inner equations second part [gli] determines the solution of equation (4.1.1): Theorem 4.3.10. For m, s ≥ 1, Ga[m,s] [gli] = Ga[m,s]

yn ∑ A[gli] a[m,n]

(4.3.13)

[gli] n∈𝒥m,s

= A[gli] is a polynomial of a with degree at most 1+(m+s) such that A[gli] where A[gli] m,n . 1[m,n] a[m,n] Proof. This is a result of Theorem 4.3.1 and Lemma 4.3.9. From (4.3.4) and (4.3.6), known that nCut and nNct are summation-free. From [gli] (4.3.7), n[gli] is summation-free. Because A[gli] m,s is determined by n[gli] , Gm,s given by

[gli] in (4.3.13). (4.3.12) is a direct explicision and hence so is Ga[m,s]

4.4 Restrictions general loopless inner case An application or a restriction of equation (4.1.1) for a0 = a1 = a2 = 1 is illustrated in the theory of enumerating root-isomorphic classes of general loopless planar maps. Consider the equation for g 1 { { {∫ 1 − 𝜕 (u2 g| ) = g; x,y x=u y { { { = 1, g| { x=0⇒y=0

(4.4.1)

where g ∈ ℛ{x, y} and 𝜕x,y is the slope difference operator. This is the equation (4.1.1) for a0 = a1 = a2 = 1. The solution of equation (4.4.1) is involved with general loopless planar maps as shown in [61] (Liu YP, 2010, p. 179). Note that the original equation is from [36] (Liu YP, 1987). If g ∈ ℛ{x, y} is a solution of equation (4.4.1), then g is determined by Gm = 𝜕xm g ∈ ℛ+ {y} for integer m ≥ 0. In fact, all coefficients of y in Gm for m ≥ 0 are in ℤ+ . The relationship between 𝜕x,y (u2 g|x=u ) and Gj for j ≥ 0 have firstly to be investigated. Because of 𝜕x,y (u2 g|x=u ) = xy ∑ Gm ( m≥0

xm+1 − ym+1 ), x−y

we have 𝜕x,y (u2 g|x=u ) = ∑( ∑ Gl yl−j+2 )xj . j≥1 l≥j−1

(4.4.2)

Then the relationship between (𝜕x,y (u2 g|x=u ))k for k ≥ 2 and Gj for j ≥ 0 can be recursively found. Denote Pj = ∑ Gi+j−2 yi , i≥1

(4.4.3)

4.4 Restrictions general loopless inner case | 119

then, for any integer k ≥ 2, j−1

Pj[k] = ∑ Pi[k−1] Pj−i .

(4.4.4)

i=1

When k = 1, Pi[1] = Pi is guaranteed for 0 ≤ i ≤ j. We have 1 k = ∑ (𝜕 (u2 g|x=u )) . 1 − 𝜕x,y (u2 g|x=u ) k≥0 x,y From equation (4.4.1), [k] Gm = ∑ ∫ Pm

(4.4.5)

[k] Π[k] m = ∫ Pm .

(4.4.6)

k≥0 y

for integer m ≥ 0. For integers m, k ≥ 0, let y

From (4.4.3) and (4.4.4),

Π[k] m

1, when m = 0; { { } { { 0, when m ≥ 1 { { { { { { 0, when m = 0; } ={ i { G y , when m ≥ 1 ∑ { i≥1 i+m−2 { { { { 0, when 0 ≤ m ≤ k − 1; { { { m−1 [k−1] } Π ⊗ Π , when m ≥ k, ∑ m−i { i=k−1 i

if k = 0; if k = 1;

(4.4.7)

if k ≥ 2.

From (4.4.5), we have Gm = ∑ Π[k] m . k≥0

(4.4.8)

Observation 4.4.1. Let g be a solution of equation (4.4.1), then G0 = 1. Proof. This is the initiation of equation (4.4.1). on.

The observation enables us to only consider G0 = 1 in equation (4.4.1) from now

Theorem 4.4.2. On ℛ{x, y}, equation (4.4.1) is equivalent to the system of equations Gm = ∑k≥0 Π[k] m , for m ≥ 1; { G0 = 1, for m = 0.

(4.4.9)

120 | 4 Inner equations second part Proof. First, from (4.4.3), P0 = 0, and hence Π0 = 0. Because G0 = 1 + Π0 = 1, the initiation of equations (4.4.9) is the same as equation (4.4.1). Then, for integer m ≥ 1, from equation (4.4.1) and Observation 4.4.1, equation (4.4.8) is derived. Thus, by the cancelation law on ℛ{x, y}, the conclusion can be drawn. Unfortunately, in equations (4.4.8), the summation is not finite. This motivates us to seek a new parameter s such that Gm,s are all finite for any integers m, s ≥ 0. If s is chosen to be the power |n| of y|n| , Gm,|n| can be shown to be infinite as well. For any n ∈ 𝒥m , we address the set of all power vectors in Gm . Let π(n) = inT , where i = (1, 2, 3, . . .), be called the y-size. Because Gm ∈ ℛ{y}, we have Gm = ∑ Gm,n yn n∈𝒥m

(4.4.10)

where 𝒥m is the set of power vectors of y which appear in Gm . Of course, Gm,n ∈ ℝ. For any n ≥ 0, let π(n) = inT , called the y-semi-size. For any integers m, k ≥ 0, denote Gm,s = ∑ Gm,n yn |π(n)=s = ∑ Gm,n yn . n∈𝒥m

n∈𝒥m π(n)=s

(4.4.11)

Thus, π(Gm,s ) = s is allowed to be called the y-semi-size of Gm,s and we denote 𝒥m,s = {n ∈ 𝒥m |π(n) = s}. Because G0 = 1 from (4.4.9), G0,s = 0, s ≥ 1. That is, for s ≥ 0, 1, when s = 0; G0,s = { 0, when s ≥ 1.

(4.4.12)

It is only necessary to consider the case of m ≥ 1. When k = 1, from (4.4.7), s−1

Πm,s = ∑ yj+1 Gj+m−1,s−j−1 . j=0

(4.4.13)

Lemma 4.4.3. For any integer m ≥ 0 and s = 0, 1, when m = 0; Gm,0 = { 0, when m ≥ 1.

(4.4.14)

Proof. When m = 0. From the initiation of (4.1.9), G0,0 = 1. When m ≥ 1, from (4.4.7) and (4.4.8), Π[k] 0,0 = 0 for k ≥ 0 leads to the conclusion. In what follows, we address Gm,1 , i. e., we see how the case of s = 1 can be deduced from Lemma 4.4.3.

4.4 Restrictions general loopless inner case | 121

Lemma 4.4.4. For any integer m ≥ 0, in equations (4.4.9),

Gm,1

0, when m = 0; { { { = {y1 , when m = 1; { { {0, when m ≥ 2.

(4.4.15)

Proof. See the proof of Lemma 4.1.4 for a = 1. We proceed on the basis of G1,0 = G0,1 = 0; we are suggested to observe the distribution of 0 among all Gm,s for m, s ≥ 0, so that the amount of labor in evaluation of them is decreased. Lemma 4.4.5. Given integers m, s ≥ 0. If n ∈ 𝒥m,s such tat Gm,s ≠ 0, then s = π(n) = m(mod 2). Proof. From the evaluation above, it is seen that when m + s ≤ 1, only G0,0 ≠ 0. In this case, m = s(mod 2). The conclusion is true. For m + s ≥ 2, we proceed by induction on m + s. Assume Gi,j ≠ 0 for i + j < m + s with i, j ≥ 0 and i = j(mod 2). We prove it for Gm,s ≠ 0, s = m(mod 2). From (4.4.7) and (4.4.8), m

Gm,s = ∑ Π[k] m,s . k=0

(4.4.16)

From (4.4.7), Π[k] m,s for k ≥ 2 are only dependent on Πi,j for i + j ≤ m + s − 1. From the assumption, all terms in Gm,s with k ≥ 2 satisfy the conclusion. The only case remaining is for k = 1. Consider the terms in (4.4.13) as yj+1 Gj+m−1,s−j−1 , 0 ≤ j ≤ s − 1. We proceed on the basis of (j + m − 1) + (s − j − 1) = m + s − 2 ≤ m + s − 1. From the assumption, (j + m − 1) = (s − j − 1)(mod 2) 󳨐⇒ m − 1 = s − 1(mod 2) 󳨐⇒ m = s(mod 2). This is the conclusion. Because m = s(mod 2) is the same as m + s = 0(mod 2), this lemma enables us to see that m + π(n) = 0(mod 2) for n ∈ 𝒥m,s . Theorem 4.4.6. Equation (4.4.1) is well defined on ℛ{x, y}. Proof. This is a result of Theorem 4.1.6 for a = 1. In order to simplify the solution as much as possible, relevant structures of the solution have to be further investigated. Lemma 4.4.7. For any integer s ≥ 1 given, if the integer s ≤ m − 1, then Gm,s = 0. Proof. When s = 0 and s = 1, (4.4.14) and (4.4.15) are, respectively employed. If the positive integer s ≤ m − 1, then Gm,s = 0. When s ≥ 2, assume that, for any integer t ≤ s − 1, Gm,t = 0 for t ≤ m − 1. By induction on s, we prove the conclusion for t = s.

122 | 4 Inner equations second part From (4.4.14), it is seen that, for s ≥ 1, (4.4.16) becomes m

Gm,s = ∑ Π[k] m,s . k=1

(4.4.17)

In order to show Gm,s = 0, it is only necessary to show Π[k] m,s = 0 for integers 1 ≤ k ≤ m. When k = 1, we proceed on the basis of (4.4.13). From s ≤ m−1, s−j−1 ≤ m−1−j−1. By s−1 ≥ j ≥ 0, s−j−1 ≤ (m+j−1)−1. We have s−j−1 ≤ s−1. By the inductive assumption, Gm+j−1,s−j−1 = 0 for all integers j : s − 1 ≥ j ≥ 0. From (4.4.13), Πm,s = Π[1] m,s = 0. [k] Further, from (4.4.7), Πm,s = 0 is seen for k ≥ 2. This lemma enables us to evaluate Ga[m,s] for the solution of equations (4.4.9) without consideration of the area of m = s(mod 2) over m, s ≥ 1. Lemma 4.4.8. For any integers m, s ≥ 1. If m + s is odd, then Gm,s = 0. Proof. In fact, this is a direct result of Lemma 4.4.5. The two lemmas above enable us to reduce by three-fourth the amount of labor for solving the system of equations (4.4.9). Lemma 4.4.9. Given an integer s ≥ 1. For any integer m ≥ 1, Gm,s ∈ ℛ{y} is independent of yl for l ≥ s + 1. Proof. See the proof of Lemma 4.2.3 for a = 1. This lemma suggests us only necessary to consider ys for evaluating Gm,s (m, s ≥ 1). That implies Gm,s ∈ ℛ{ys }. Lemma 4.4.10. For two integers m, s ≥ 1, Gm,s is a polynomial on ℛ{ys } with terms of minimum degree 1 and maximum degree s. Proof. See the proof of Lemma 4.2.4 for a = 1. This lemma tells us that, for any integers m, s ≥ 1, the evaluation of Gm,n is defined on a space of finite dimension. Lemma 4.4.11. The solution of equation (4.1.1) is determined by the following polynomials of all terms with positive coefficients with integers m, s ≥ 0: Gm,s

1, when m + s = 0, i. e., m = s = 0; { { { m [k] = {∑k=1 Πa[m,s] , when m + s ≥ 1, m, s ≥ 1, m = s(mod 2); { { otherwise, {0,

(4.4.18)

{ ∑s−1 { j=0 yj+1 Gj+m−1,s−j−1 (= Πm,s ), when k = 1; { { [s−1] when k = m = s; = {∑1≤i,j≤s−1 (Πs−i,s−j ⊗ Πi,j ), { { {∑ 1≤j≤s−1 (Π[k−1] ⊗ Π ), when k ≥ 2. i,j { k−1≤i≤m−1 m−i,s−j

(4.4.19)

where Π[k] m,s

4.4 Restrictions general loopless inner case | 123

Proof. From Lemma 4.4.7, Lemma 4.4.8 and (4.4.17), (4.4.18) is derived. Further, (4.4.19) is found. We proceed on the basis of this theorem, to evaluate Gm,s for m + s ≤ 6. From (4.4.18), we have only the four cases: m + s = 0, 2, 4 and 6. When m + s = 0, only G0,0 = 1 is considered for m = s = 0. Then for m, s ≥ 1. When m + s = 2, it is only necessary to evaluate G1,1 from G2,0 = 0 (Lemma 4.4.3) and G0,2 = 0 by (4.4.12). We have of G1,1 = Π1,1 and Π1,1 = y1 G0,0 . From G0,0 = 1, we have G1,1 = y1 .

(4.4.20)

When m + s = 4,because of G4,0 = 0 (Lemma 4.4.3), G3,1 = 0 (Lemma 4.4.4) and G0,4 = 0 by (4.4.12), it is only necessary to evaluate G2,2 and G1,3 . [2] [1] [2] On account of G2,2 = (Π[1] 2,2 + Π2,2 ) where Π2,2 = Π2,2 and Π2,2 = Π1,1 ⊗ Π1,1 , we have

[2] 2 Π[1] 2,2 = y1 G1,1 + y2 G2,0 = y1 G1,1 and Π2,2 = y1 G0,0 ⊗ y1 G0,0 = y2 G0,0 . By G0,0 = 1 and G1,1 = y1 , we have

G2,2 = y12 + y2 .

(4.4.21)

This is (4.2.5) for a = 1. We have of G1,3 = Π1,3 where Π1,3 = y1 G0,2 + y2 G1,1 + y3 G2,0 = y2 G1,1 = y1 y2 . From (4.4.17), we have G1,3 = Π1,3 = y1 y2 .

(4.4.22)

This is (4.2.6) for a = 1. When m + s = 6, because of G6,0 = 0 (Lemma 4.4.3), G5,1 = 0 (Lemma 4.4.4), G4,2 = 0 (Lemma 4.4.7) and G0,6 = 0 (by (4.4.12)), it is only necessary to evaluate G3,3 , G2,4 and G1,5 . [2] [3] [1] 3 We proceed on the basis of G3,3 = Π[1] 3,3 +Π3,3 +Π3,3 . Because Π3,3 = y1 G2,2 = y1 +y1 y2 ,

[3] 3 Π[2] 3,3 = 2(y2 G1,1 G0,0 ) = 2y1 y2 and Π3,3 = y3 G0,0 = y3 , we have

G3,3 = y13 + 3y1 y2 + y3 .

(4.4.23)

This is (4.2.7) for a = 1. [1] [2] We proceed on the basis of G2,4 = (Π[1] 2,4 + Π2,4 ). Because of Π2,4 = y1 G1,3 + y2 G2,2 = 2y12 y2 + y22 and Π[2] 2,4 = 2y3 G0,0 G1,1 = 2y1 y3 , we have

G2,4 = (2y12 y2 + y22 ) + (2y1 y3 ) = 2y12 y2 + 2y1 y3 .

(4.4.24)

This is (4.2.8) for a = 1. We proceed on the basis of G1,5 = Π[1] 1,5 = Π1,5 . Because of Π1,5 = y2 G1,3 + y3 G2,2 = 2 y1 y2 + y12 y3 + y2 y3 , we have G1,5 = y1 y22 + y12 y3 + y2 y3 . This is (4.2.9) for a = 1.

(4.4.25)

124 | 4 Inner equations second part Corollary 4.4.12. For any integer s ≥ 1, Π[s] s,s = ys . Proof. We proceed on the basis of (4.4.19). When m = k = s, s−1 s−1

[s−1] Π[s] s,s = ∑ ∑ (Πs−i,s−j ⊗ Πi,j ). i=1 j=1

If i ≠ j, then either i > j or s − i > s − j. From Lemma 4.4.7, only i = j is considered. This implies that [s−1] Π[s] s,s = Πs−j,s−j ⊗ Πj,j

(4.4.26)

for j ≥ 1. If j ≠ 1, i. e., j ≥ 2, then s − j ≤ s − 2 < s − 1. From Lemma 4.4.7, Π[s−1] = 0. This s−j,s−j

[s−1] implies that Π[s] s,s = Πs−1,s−1 ⊗ Π1,1 . Because Π1,1 = y1 , the conclusion is true for s = 1. For s ≥ 2, we are allowed to assume that Π[s−1] s−1,s−1 = ys−1 by the inductive principle on n. On account of (4.4.26), [s] Πs,s = ys−1 ⊗ Π1,1 = ys−1 ⊗ y1 = ys . The conclusion can be drawn.

This corollary enables us to determine Π[s] s,s for s ≥ 1 in a very simple way in a summation-free form. [gli] Theorem 4.4.13. The solution g[gli] of equation (4.4.1) is determined by Gm,s m 𝜕x g[gli] |π(n)=s for integers m, s ≥ 0 as

1, { { { [gli] [k] Gm,s = {∑m k=1 Πm,s , { { {0,

when m + s = 0, i. e., m = s = 0; when m + s ≥ 1, s ≥ m ≥ 1 and m = s(mod 2);

=

(4.4.27)

otherwise,

where for s ≥ m ≥ 1 and m = s(mod 2),

Π[k] m,s

{ ys , { { { s−1 = {∑j=0 yj+1 Gj+m−1,s−j−1 (= Πm,s ), { { {∑ 1≤j≤s−1 (Π[k−1] ⊗ Π ), i,j { k−1≤i≤m−1 m−i,s−j

when k = m = s; otherwise, when k = 1;

(4.4.28)

when k ≥ 2.

Proof. A direct result of Lemma 4.4.11 and Corollary 4.4.12. This theorem is the case of a = 1 in Theorem 4.2.7. Example 4.4.1. Root-isomorphic classes of loopless planar maps with root-vertex valency m and vertex-partition vector n as parameters. In Liu YP [61], the equation shown as the first equality of equation (4.4.1) is satisfied by the enumerating function of root-isomorphic classes for general loopless planar maps with root-vertex valency m (power of x) and vertex-partition vector n (power vector of y) as parameters. Theorem 4.4.13 shows that the coefficient of yn in Gm,s is the number of root-isomorphic

4.4 Restrictions general loopless inner case | 125

Figure 4.4.1: Root-isomorphic classes of general loopless planar maps.

classes of such maps. In Figure 4.4.1, all root-isomorphic classes of loopless planar maps for m + s ≤ 6 are shown by a hole circle as root. From Figure 4.4.1, it is seen that G1,1 = (a) = y1 ; G2,2 = (b) + (c) = y12 + y2 ; G1,3 = (d) = y1 y2 ; G3,3 = (e) + 3(f ) + (g) = y13 + 3y1 y2 + y3 ; G2,4 = 2(h) + 2(i) + (j) = 2y12 + 2y1 y3 + y22 ; [5] G1,5 = (m) + (k) + (l) = y1 y22 + y12 y3 + y2 y3 . And, Π[4] 4,4 = (n) = y4 , Π5,5 = (o) = y5 ,

Π[6] = (p) = y6 . 6,6

126 | 4 Inner equations second part

4.5 Eulerian loopless inner model Second, we discuss the equation 1 − a1 𝜕x2 ,y2 (uf |x2 =u ) { { {a2 ∫ 1 − 2a 𝜕 2 2 (uf | 2 ) − x2 y2 δ 2 2 (uf |2 ) = f ; 1 x ,y x =u x ,y x2 =u y { { { {f |x=0⇒y=0 = a0

(4.5.1)

where a0 , a1 , a2 ∈ ℛ+ , f ∈ ℛ{x, y} and y = (y1 , y2 , y3 , . . .). This is equation (8) in Introduction. When a0 = a1 = a2 = 1, one might find the equation without the initiation or its equivalent version in [32] (Liu YP, 1986), and then [60] (Liu YP, 2008, p. 186). Because of a solution of equation (4.5.1) for a0 = a1 = a2 = 1 being involved with Eulerian loopless planar maps, this is called a Eulerian loopless inner model. There being no term with odd degree of x in equation (4.5.1), f = fa is an even function of x. There being no term with odd degree of y, f is independent of y2i+1 , for integer i ≥ 0. For convenience, let pa = 𝜕x2 ,y2 (ufa |u=x2 ); { qa = δx2 ,y2 (ufa2 |u=x2 ). Denote Λa = From (4.5.2),

1 − a1 𝜕x2 ,y2 (ufa |u=x2 )

1 − 2a1 𝜕x2 ,y2 (ufa |u=x2 ) − x2 y2 δx2 ,y2 (ufa |2u=x2 )

Λa =

(4.5.2)

.

1 − a1 pa 1 − 2a1 pa − x2 y2 qa

2i ai+1 1 k! 2(k−i) 2(k−i) i+1 k−i x y pa qa i!(k − i)! i=0 k

= 1 + ∑ (∑ k≥0

2i ai1 k! 2(k−i+1) 2(k−i+1) i k−i+1 x y pa qa ). i!(k − i)! i=0 k

+∑

In the first i-sum, the term of i = k is a1 (2a1 )k pk+1 = 2k (a1 pa )k+1 . In the second i-sum, a the term of i = 0 is x2(k+1) y2(k+1) qak+1 . The remainder of the second i-sum is k

(2a1 )i k! 2(k−i+1) 2(k−i+1) i k−i+1 x y pa qa i!(k − i)! i=1

∑

k−1

=∑

i=0

2i+1 k! k−i x2(k−i) y2(k−i) pi+1 a qa . (i + 1)!(k − i − 1)!

4.5 Eulerian loopless inner model | 127

By combining the two remainders, Λa = 1 + ∑ (2k (a1 pa )k+1 + x2(k+1) y2(k+1) qak+1 k≥0

k−1

k−i + ∑ αa[k,i] x2(k−i) y2(k−i) pi+1 a qa ) i=0

where

k k αa[k,i] = (2a1 )i ( ) + (2a1 )i+1 ( ) i i+1 (2a1 )i (2a1 k − (2a1 − 1)i + 1)k! = ∈ ℤ+ ⊆ ℝ+ . (i + 1)!(k − i)!

(4.5.3)

Observation 4.5.1. If fa is a solution of equation (4.5.1), then 𝜕x0 fa = a0 = a2 . Proof. We proceed on the basis of (4.5.1)–(4.5.3); the equation considered is { { {f = a2 (1 + ∫(Λa (x, y) − 1)); y { { { {f |x=0⇒y=0 = a0 .

(4.5.4)

Because of 𝜕x0 fa = a2 = fa |x=0⇒y=0 = a0 , the conclusion can be drawn. This observation enables us to only consider a2 = a0 and a = (a0 , a1 ) instead of (a0 , a1 , a2 ) from now on. Theorem 4.5.2. Equation (4.5.1) is equivalent to [k+1] f = a0 + a0 ∑ (2k ak+1 + x2(k+1) (y2(k+1) ⊗ Q[k+1] ) { 1 Pa a { { { k≥0 { { { k−1 { + ∑ αa[k,i] x2(k−i) (y2(k−i) ⊗ Pa[i+1] ⊗ Qa[k−i] )); { { { { i=0 { { f | { x=0⇒y=0 = a0 ,

(4.5.5)

on ℛ{x, y} where for any integer k ≥ 0, k − 1 ≥ i ≥ 0, αa[k,i] is given in (4.5.3) and, for integer l ≥ 1, { Pa[l] = Pa[l−1] ⊗ Pa , Pa[1] = P = ∫ pa ; { { { { y { [l] [l−1] [1] { {Qa = Qa ⊗ Qa , Qa = Q = ∫ qa , { { y { such that p and q are, respectively, given by the first and second rows of (4.5.2).

128 | 4 Inner equations second part Proof. We proceed on the basis of equation (4.5.1). By substituting (4.5.2) into the first equality of (4.5.1), f = a0 ∫ y

1 − a1 pa 1 − 2a1 pa − x2 y2 qa

[k+1] = a0 + a0 ∑ (2k ak+1 + x2(k+1) (y2(k+1) ⊗ Q[k+1] ) 1 Pa a k≥0

k−1

+ ∑ αa[k,i] x2(k−i) (y2(k−i) ⊗ Pa[i+1] ⊗ Qa[k−i] )). i=0

Thus, the conclusion can be drawn. We proceed on the basis of Theorem 4.5.2 the equation (4.5.1) for f ∈ ℛ{x, y} is transformed into the system of infinite equations for Fm ∈ ℛ{y} for m ≥ 0 shown in equations (4.5.5). For an even function g of x on ℛ{x, y}, denote by m Gm = [g]m x2 = 𝜕x2 g

(4.5.6)

the coefficient of x2m in g for m ≥ 0. Because f ∈ ℛ{x, y} is even for x, pa ∈ ℛ{x, y} and qa ∈ ℛ{x, y} are even for x as well. From (4.5.6), we see the meaning of Fa[m] , Pa[m] and Qa[m] . For any integer k ≥ 1, m

[k] = [g k ]x2 = 𝜕xm2 g k Ga[m]

(4.5.7)

[1] where Gm = Gm .

Lemma 4.5.3. Given integer k ≥ 1, if we have a non-negative integer m ≤ k − 1, then [k] = 0. Pa[m] Proof. We proceed on the basis of (4.5.2). When k = 1, Pa[0] = 0. For k ≥ 2, assume [k] [k−1] = 0 for m ≤ k − 1. = 0, m ≤ k − 2. By induction on k, we prove Pa[m] Pa[m] We have m

m

i=0

i=1

[k−1] [k−1] [k] Pa[i] Pa[i] = ∑ Pa[m−i] = ∑ Pa[m−i] Pa[m] [k−1] = 0 for m ≥ i ≥ 1, we and m ≤ k − 1 󳨐⇒ m − 1 ≤ k − 2. By the inductive principle, Pa[m−i] [k] have Pa[m] = 0 for m ≤ k − 1. This is the conclusion.

[k] This lemma enables us to only consider Pa[m] for k ≤ m.

Lemma 4.5.4. Given integer k ≥ 1. If we have anon-negative integer m ≤ k − 1, then Q[k] = 0. a[m]

4.5 Eulerian loopless inner model | 129

Proof. We proceed on the basis of (4.5.2). When k = 1, Qa[0] = 0. For k ≥ 2, assume = 0 for m ≤ k − 1. = 0 for m ≤ k − 2. By the inductive principle, we show Q[k] Q[k−1] a[m] a[m] We have m

m

i=0

i=1

[k−1] [k−1] Q[k] a[m] = ∑ Qa[m−i] Qa[i] = ∑ Qa[m−i] Qa[i]

= 0 for m ≥ i ≥ 1, and we have and m ≤ k − 1 󳨐⇒ m − 1 ≤ k − 2. By induction, Q[k−1] a[m−i]

= 0 for m ≤ k − 1. This is the conclusion. Q[k] a[m]

for k ≤ m. This lemma enables us to only consider Q[k] a[m] Theorem 4.5.5. The equation system for Fa[m] for m ≥ 0

Fa[m]

a0 , { { { { k k+1 [k+1] { {a0 ∑m−1 k=0 2 a1 Pa[m] ={ { + a0 ∑⌊m/2⌋−1 (y2(k+1) ⊗ Q[k+1] ) { { k=0 a[m−k−1] { { m−1 k−1 [k+1] + a0 ∑k=1 ∑i=0 αa[k,i] (y2(k−i) ⊗ Ra[m−k+i] ), {

when m = 0; (4.5.8) when m ≥ 1,

where αa[k,i] ∈ ℝ is given in Theorem 4.5.2 and R[k+1] = a[m−k+i]

m−2k−1

[k+1] [k−i] ) ⊗ Qa[j] ∑ (Pa[m−k+i−j]

j=k−i

is equivalent to equation (4.5.1) on ℛ{x, y}. Proof. We proceed on the basis of Theorem 4.5.2. From the initiation, Fa[0] = a0 when m = 0; when m ≥ 1, [k+1] [k+1] Fa[m] = a0 ∑ 2k ak+1 1 Pa[m] + a0 ∑ (y2(k+1) ⊗ Qa[m−k−1] ) k≥0

k≥0

k−1

m−k+i

+ a0 ∑ ∑ αa[k,i] (y2(k−i) ⊗ [Pa[i+1] ⊗ Qa[k−i] ]x2 k≥0 i=0

).

[k+1] , it is only necessary to consider We proceed on the basis of Lemma 4.5.3. For Pa[m] m ≥ k + 1 󳨐⇒ k ≤ m − 1. Thus, m−1

k k+1 [k+1] [k+1] ∑ 2k ak+1 1 Pa[m] = ∑ 2 a1 Pa[m] .

k≥0

k=0

We proceed on the basis of Lemma 4.5.4. For Q[k+1] , it is only necessary to cona[m−k−1] sider m − k − 1 ≥ k + 1 󳨐⇒ 2k ≤ m − 2. Thus, )= ∑ (y2(k+1) ⊗ Q[k+1] a[m−k−1]

k≥0

⌊m/2⌋−1

). ∑ (y2(k+1) ⊗ Q[k+1] a[m−k−1]

k=0

130 | 4 Inner equations second part We have m−k+i

[Pa[i+1] ⊗ Qa[k−i] ]x2

m−k+i

[i+1] [k−i] = ∑ Pa[m−k+i−j] ⊗ Qa[j] j=0

m−k−1

[k+1] [i+1] = ∑ Pa[m−k+i−j] ⊗ Q[k−i] a[j] = Ra[m−k+i] . j=k−i

From m − k − 1 ≥ 0 󳨐⇒ k ≤ m − 1, the last infinite summation is m−1 k−1

). ∑ ∑ αa[k,i] (y2(k−i) ⊗ R[k+1] a[m−k+i]

k=1 i=0

Therefore, the conclusion can be drawn. In order to solve the system of equations (4.5.8), the relationship between Pa[m] and Qa[m] expressed by Fa[m] has to be considered beforehand. We proceed on the basis of (4.5.2). For given integer m ≥ 0, because pa = ∑ Fa[m] m≥0

y2 x2(m+1) − x2 y2(m+1) x2 − y2

= ∑ x2j ∑ Fa[i] y2(i−j+1) , j≥1

i≥j

we have 0, when m = 0; [pa ]m x2 = { 2(i+1) , when m ≥ 1. ∑i≥0 Fa[i+m] y From Theorem 4.5.2, it is seen that, for integers l, m ≥ 1, ∑ Fa[i+m] y2(i+1) (= Pa[m] ), [l] = { i≥0 Pa[m] [l−1] ∑m−1 i=1 Pa[m−i] Pa[i] ,

when l = 1; when l ≥ 2.

(4.5.9)

[2] = 𝜕xm2 fa2 , m ≥ 0, then Let Fa[m] [2] qa = ∑ Fa[m] m≥0

x2(m+1) − y2(m+1) [2] y2i . = ∑ x2j ∑ Fa[i+j] x2 − y2 i≥0 j≥0

From Theorem 4.5.2, it is seen that, for integers l ≥ 1 and m ≥ 0, [2] y2i (= Qa[m] ), when l = 1; ∑i≥0 Fa[i+m] = { Q[l] a[m] m [l−1] when l ≥ 2. ∑i=0 Qa[m−i] Qa[i] ,

(4.5.10)

In order to solve the system of equations (4.5.8), a new parameter s = π/2 is introduced so that, for any power vector n, π(y) = (11 + 212 + 313 + ⋅ ⋅ ⋅)nT . Let Fa[m,s] be the part of Fa[m] such that n satisfies π(n) = 2s.

4.5 Eulerian loopless inner model | 131

Lemma 4.5.6. When m = 0. For any integer s ≥ 0, a0 , when s = 0; Fa[0,s] = { 0, when s ≥ 1.

(4.5.11)

Proof. When s = 0. From the initiation of equation (4.5.1), Fa[0] = a0 and hence Fa[0,0] = a0 . Because Fa[0] is independent of y, the case of s ≥ 1 in (4.5.10) is found. From Lemma 4.5.6, we are allowed to only discuss m ≥ 1 without loss of generality. Lemma 4.5.7. When s = 0, for any integer m ≥ 0, a0 , Fa[m,0] = { 0,

when m = 0; when m ≥ 1.

(4.5.12)

Proof. When m = 0, from Lemma 4.5.6, Fa[0,0] = a0 . When m ≥ 1, from (4.5.7), m−1

[k+1] Fa[m,0] = ∑ 2k ak+1 1 Pa[m,0] , k=0

by Lemma 4.5.3,

= 0. Hence, the conclusion can be drawn. From Lemma 4.5.7, it is only necessary to discuss s ≥ 1. In order to determine [l] for some l ≥ 1. and Q[l] Fa[m,s] for m, s ≥ 1, Theorem 4.5.5 suggests us to find Pa[m,s] a[m,s]

[l] As for Pm,s , Lemma 4.5.3 enables us to only consider m ≥ l. We proceed on the basis of (4.5.8). For any integer s ≥ 1,

∑s−1 [l] i=0 Fa[i+m,s−i−1] y2(i+1) (= Pa[m,s] ), when l = 1; = { m−1 Pa[m,s] [l−1] Pa[i,j] , when l ≥ 2. ∑i=1 ∑sj=0 Pa[m−i,s−j]

(4.5.13)

Proposition 4.5.8. Let ℱm+s−1 = {Fa[r,t] | r + t ≤ m + s − 1, r, t ≥ 0}. For any integers [l] is determined by ℱm+s−1 . m ≥ l ≥ 1 and s ≥ 1, Pa[m,s] Proof. For any integers s−1 ≥ i ≥ 0 and (i+m)+(s−i−1) = m+s−1, Fa[i+m,s−i−1] ∈ ℱm+s−1 . [1] is determined by ℱm+s−1 . When l ≥ 2, From the first case of (4.5.13), Pa[m,s] = Pa[m,s]

[l] [l−1] is determined is determined by ℱr+t−1 . By induction, we prove that Pa[m,s] assume Pa[r,t] by ℱm+s−1 . As a matter of fact, by the second case of (4.5.13), the assumption leads to the conclusion.

. Lemma 4.5.4 enables us to only discuss m − 1 ≥ l ≥ 1. We Similarly, on Q[l] a[m,s] proceed on the basis of (4.5.10). For any integer s ≥ 1, [2] y2i (= Qa[m] ), when l = 1; ∑si=0 Fa[i+m,s−i] = { Q[l] a[m,s] m−1 s [l−1] ∑i=1 ∑j=0 Qa[m−i,s−j] Qa[i,j] , when l ≥ 2.

(4.5.14)

132 | 4 Inner equations second part Proposition 4.5.9. Let ℱm+s−1 = {Fa[r,t] | r + t ≤ m + s − 1, r, t ≥ 0}. For any integers is determined by Fa[r,t] , r + t ≤ m + s − 1. m ≥ l ≥ 1 and s ≥ 1, Q[l] a[m,s] Proof. For integers s−1 ≥ i ≥ 0 and (i+m)+(s−i−1) = m+s−1, Fi+m,s−i−1 ∈ ℱm+s−1 . When [1] is determined by ℱm+s−1 . When l ≥ 2, m = 1, by the first case of (4.5.13), Pa[m,s] = Pa[m,s]

[l−1] have been determined by ℱr+t−1 . By induction, we prove that assume that the Pa[r,t]

[l] is determined by ℱm+s−1 . We proceed on the basis of the second case of (4.5.13); Pa[m,s] the assumption leads to the conclusion.

We proceed on the basis of (4.5.8). For integers m ≥ 1 and s ≥ 1, m−1

⌊m/2⌋−1

k=0

k=0

[k+1] Fa[m,s] =a0 ∑ 2k ak+1 1 Pa[m,s] + a0 m−1 k−1

+ a0 ∑ ∑ αa[k,i] (y2(k−i) ⊗ k=1 i=0

) ∑ (y2(k+1) ⊗ Q[k+1] a[m−k−1,s−k−1]

(4.5.15)

R[k+1] ). a[m−k+i,s−k+i]

On account of Proposition 4.5.8 and Proposition 4.5.9, R[k+1] is determined by a[m−k+i,s+i−k] Fa[r,t] , r + t ≤ (m − k + i) + (s − k + i) − 1 = m + s − 2k + 2i − 1 ≤ m + s − 1. Theorem 4.5.10. Equation (4.5.1) is well defined on ℛ{x, y} if, and only if, a0 = a2 . Proof. We proceed on the basis of Theorem 4.5.5. In order to get a solution of equation (4.5.1), it is only necessary to get Fa[m,s] for any integers m, s ≥ 0 from the initiation. The procedure of getting Fm,s is in the order from 0 increasingly to m + s. First, for the sufficiency. When m + s = 0, we have Fa[0,0] = a0 = a2 . From Lemma 4.5.6 and Lemma 4.5.7, it is only necessary to discuss m, s ≥ 1. When m + s = 1, Fa[1,0] = Fa[0,1] = 0 are known. When m + s = 2, because Fa[2,0] = Fa[0,2] = 0, only Fa[1,1] has to be found. From (4.5.8) and (4.5.13), 1−1

[k+1] = a0 a1 Pa[1,1] y2 Fa[1,1] = a0 ∑ 20 a1 Pa[1,1] k=0

= a0 a1 Fa[0,0] y2 = a20 a1 y2 . When m + s ≥ 3 in general. By induction, assume all Fr,t have been found for r + t ≤ m + s − 1, and we determine Fa[m,s] from (4.5.8). From Proposition 4.5.8 and [k+1] , Q[k+1] and R[k+1] Proposition 4.5.9, the assumption shows to that Pa[m,s] a[m−k−1,s] a[m−k+i,s−k+i] are known by (4.5.15). Therefore, Theorem 4.5.5 yields a solution of equation (4.5.1). From the uniqueness of the procedure under the initiation on ℛ{y}, the solution of equation (4.5.1) is unique. This is the sufficiency. As regards necessity: this is This is a result of Observation 4.5.1.

4.6 Solution Eulerian loopless inner case

| 133

4.6 Solution Eulerian loopless inner case In order to simplifying the procedure illustrated in the last section, more structural properties of the solution of equation (4.5.1) have to be particularly investigated. Lemma 4.6.1. If fa ∈ ℛ{x, y} is a solution of (4.5.1), then, for any integers m, s ≥ 0, Fa[m,s] = [𝜕xm2 fa ]sy ∈ ℛ+ [y2s ] is a polynomial of y with degree s. Proof. For integers m + s ≥ 0 (m, s ≥ 0), we proceed by induction. When 0 ≤ m + s ≤ 2, only one of Fa[m,n] is not zero, i. e., Fa[1,1] = a20 a1 y2 is a polynomial of y with degree s = 1. When m + s ≥ 3, assume that, for any integers r, t ≥ 0 and r + t ≤ m + s − 1, Fa[r,t] ∈ ℛ+ [y2t ] is a polynomial of y with degree t. We prove that Fa[m,s] = [𝜕xm2 f ]sy ∈ ℛ+ [y2s ] is a polynomial of y with degree s. We proceed on the basis of Proposition 4.5.8 and Proposition 4.5.9. By (4.5.15), the assumption yields the conclusion. From this lemma, Fa[m,s] can be seen to have all coefficients of the form of a finite sum of positive terms if, and only if, a0 , a1 ∈ ℤ+ for m, s ≥ 0. Lemma 4.6.2. For any integers m, s ≥ 1 given, if m ≥ s + 1, then Fa[m,s] = 0. Proof. First, for m + s ≤ 2, it is seen that the conclusion is true. Then consider the general case m + s ≥ 3. Assume for any integers, r, t ≥ 0, r + t ≤ m + s − 1, Fa[r,t] = 0 for r ≥ t + 1. By induction, we prove Fa[m,s] = 0 for m ≥ s + 1. We proceed on the basis of the assumption; it is seen that, for any integer k, 0 ≤ [k+1] = 0, for any integer k, 0 ≤ k ≤ ⌊m/2⌋−1, whenever k ≤ m−1, whenever m ≥ s+1, Pa[m,s] [k+1] = 0, and, for any integers k(0 ≤ k ≤ m − 1), i(0 ≤ i ≤ k − 1), m ≥ s + 1, Qa[m−k−1.s−k−1]

whenever m ≥ s + 1, R[k+1] = 0. From (4.5.15), Fa[m,s] = 0 for m ≥ s + 1. a[m−k+i,s−k+i] This is the conclusion.

This lemma enables us to decrease by half the amount of labor for solving equation (4.5.1) by using the procedure shown in the proof of Theorem 4.5.10. Theorem 4.6.3. Let fa[Eli] be the solution of equation (4.5.1). For integers m, s ≥ 0, de[Eli] [Eli] [Eli] ∈ ℛ+ {y} is of the form of a finite sum with , then Fa[m,s] = [𝜕xm2 fa[Eli] ]sy = Fa[m,s] note Fa[m,s] all terms positive whenever a0 , a1 ∈ ℝ+ ,

[Eli] Fa[m,s]

a0 , { { { { { { 0, { { { { { { ={ k k+1 [k+1] { a0 ∑m−1 { k=0 2 a1 Pa[m,s] { { { { { +a0 ∑⌊m/2⌋−1 (y2(k+1) ⊗ Q[k+1] ) { k=0 a[m−k−1,s−k−1] { { { m−1 k−1 [k+1] +a0 ∑k=1 ∑i=0 αa[k,i] (y2(k−i) ⊗ Ra[m−k+i,s−k+i] ), {

when s = m = 0; when s ≥ 1, m = 0, or m ≥ s + 1;

otherwise,

(4.6.1)

134 | 4 Inner equations second part where R[k+1] = a[m−k+i,s−k+i]

m−2k−1 k−i

∑

[k+1] ⊗ Q[k−i] ∑ (Pa[m−k+i−j,s−k+i−n] a[j,n] ),

j=k−i n=0

(4.6.2)

[Eli] ∑s−1 [l] i=0 Fa[i+m,s−i−1] y2(i+1) (= Pa[m,s] ), when l = 1; = { m−1 Pa[m,s] [l−1] Pa[i,j] , when l ≥ 2, ∑i=1 ∑sj=0 Pa[m−i,s−j]

(4.6.3)

[Eli][2] y2i (= Qa[m] ), when l = 1; ∑si=0 Fa[m,s] = { Q[l] a[m,s] m−1 s [l−1] ∑i=1 ∑j=0 Qa[m−i,s−j] Qa[i,j] , when l ≥ 2,

(4.6.4)

and

in which m

s

[Eli][2] [Eli] [Eli] Fa[m,s] = ∑ ∑ Fa[m−i,s−j] . Fa[i,j] i=0 j=0

(4.6.5)

Proof. When s = m = 0. We have the result of the initiation of equation (4.5.1). When s ≥ 1, m = 0, or m ≥ s + 1, we have the results of, respectively, Lemma 4.5.6 or Lemma 4.6.2. Or otherwise, when m, s ≥ 2, m ≤ s, as shown in the proof of Theo[Eli] = a20 a1 y2 , determine rem 4.5.10. By the increasing order of m + s starting from F1,1

[Eli] [Eli] Fa[m,s] . By Lemma 4.6.1, all Fa[m,s] ∈ ℛ+ [y2s ] are of the form of a finite sum with terms positive whenever a0 , a1 ∈ ℝ+ .

4.7 Explicitness Eulerian loopless inner case Because no explicision has yet directly been deduced from Theorem 4.6.3, the relationship between the solutions of Eulerian loopless inner equation (4.5.1) and its restriction for a0 = a1 = a2 = 1 has to be investigated on account of the latter being meaningful in graph theory. Let fa[Eli] and f[Eli] ∈ ℛ{x, y} be the solutions of, respectively, equation (4.5.1) and [Eli] [Eli] and Fm,s ∈ ℛ{y} for integers m, s ≥ 0. Denote by its restriction determined by Fa[m,s] [Eli] [Eli] [Eli] ∈ and hence Fm,s . For any n ∈ 𝒩m,s , Ua[m,n] 𝒩m,s the set of power vectors in Fa[m,s]

[Eli] [Eli] and ℛa = ℛ{a} and Um,n | ∈ ℛ1 = ℛ are the coefficients of yn in, respectively, Fa[m,s] [Eli] Fm,s . Then

[Eli] [Eli] = 𝜕yn Fa[m,s] Ua[m,n]

[Eli] [Eli] = 𝜕yn Fm,s . and Um,n

[Eli] is a Theorem 4.7.1. For any integers m, s ≥ 1 and integral vector n ∈ 𝒩m,s , Ua[m,n] polynomial of a2 = a0 and a1 with degree at most l + 1 where l = m + s such that [Eli] [Eli] Ua[m,n] |a=1 = Um,n .

4.7 Explicitness Eulerian loopless inner case | 135

Proof. We proceed on the basis of Theorem 4.6.3. By induction, the conclusion can be easily drawn. For an embedding μ(G) of a Eulerian loopless planar graph G, let n = (n1 , n2 , n3 , . . .) be the vertex-partition vector where ni is the number of non-root-vertices of valency 2i for integer i ≥ 1. A vertex v is a point of μ(G) such that μ(G) − v has at least one component greater than μ(G); it is called a cut-vertex of G. Now, G is always assumed to be connected. It is allowed to restrict a face of maximum length boundary at a cut vertex as the outerface without loss of generality. The cut-valency of a cut-vertex v is the number of components of μ(G) − v. The closure of such a component is called a cut-component. If a cut-component is not a path, then it is called a fat-component. The number of fatcomponents contributing to the cut-valency of v is called the fat-valency of v. A graph without cut-vertex is said to be nonseparable. A maximal nonseparable edge-induced subgraph is called a block of the graph. For a cut-vertex v, let a = av = ρvcut , c = cv and b = bv = ρv be, respectively, the cut-valency, fat-valency and the valency of v. Because of planarity, all cut-components at v have a rotation in the order of cut-components G1 , G2 , . . . , Ga . At cut-vertex v, let ti be the contribution of Gi to the valency of v, then t1 + t2 + ⋅ ⋅ ⋅ + ta = b. Three cases have to be considered at cut-vertex v for a Eulerian loopless planar graph. Case I Reflection for Gi , 1 ≤ i ≤ c, it turns out that we have [Eli] = 2c−1 πvI

(4.7.1)

(topologically) distinct planar embeddings. Case II For a cut-rotation at v, it turns out that we have [Eli] = (a − 1)! πvII

(4.7.2)

(topologically) distinct planar embeddings. Case III For an integer k ≥ 1 given, denote I = {i1 , i2 , . . . , ik } by distributing all Gj , j ∈ ̸ I, into s = (ti1 − 1) + (ti2 − 1) + ⋅ ⋅ ⋅ + (tik − 1) angles on 𝒢I = {Gi | ∀i ∈ I} at v as (x1 , x2 , . . . , xts ) in the s angles such that s

∑ xj = c − k; j=1

0 ≤ xj ≤ c − k, 1 ≤ j ≤ s. Let us denote by c−k ⟨ ⟩ x1 , . . . , xts

136 | 4 Inner equations second part the number of ways for such distributions, then it turns out that we have s c−k πk[Eli] = ⟨ ⟩ ∏(xj )! x1 , . . . , xts j=1

(topologically) distinct planar embeddings. From the disjointness of embeddings, we have c−1 c [Eli] = ∑ ( )πk[Eli] πvIII k k=1

(4.7.3)

(topologically) distinct planar embeddings. Lemma 4.7.2. For a Eulerian loopless planar graph G, let VCut be the set of all cutvertices. Then all cut-vertices turn out to be [Eli] [Eli] [Eli] n[Eli] Cut (G) = ∏ (πvI πvII πvIII ) v∈VCut

(4.7.4)

(topologically) distinct planar embeddings. Proof. On account of the independency among Case I, Case II and Case III, from (4.7.1), (4.7.2) and (4.7.3), the conclusion can be drawn. We address an embedding μ of a Eulerian planar graph G without cut-vertex. If two vertices u and v have the property that G − {u, v} has at least two components, then {u, v} is called a splitting pair. Since an embedding is a point set in its own right on the plane, pu and pv are as the corresponding points of u and v. Because of μ as a point set, any component of μ − {pu , pv } is called a splitting slice, or slice. A splitting slice that is not a path is called a splitting chunk, or a chunk. Let S be the set of all splitting pairs consisting of {ui , vi } for 1 ≤ i ≤ s = s(S) = |S| in μ = μ(G). All components of μ − S are splitting slices in μ(G). For a splitting pair, its slice valent is the number of splitting slices incident with it. Similarly, for a chunk valency. Observation 4.7.3. Let t and 1 ≤ p ≤ t be, respectively, the slice valency and chunk valency of a splitting pair on an embedding of Eulerian loopless planar graph G, then the splitting pair produces [Eli] πspp = 2p−1 (t − 1)!

(4.7.5)

(topologically) distinct planar embeddings of G. Proof. On account of the independency between rotation on slice valency and reflection on a chunk, by considering the connectivity of an embedding, the conclusion can be drawn.

4.7 Explicitness Eulerian loopless inner case | 137

Let p be the number of all splitting chunks and ti , the slice valency at the splitting pair {ui , vi }, 1 ≤ i ≤ s on μ(G). Lemma 4.7.4. For a Eulerian loopless planar graph G, the number of topologically distinct embeddings of G generated by all splitting pairs is s

p−s n[Eli] ∏(ti − 1)! Nct (G) = 2 i=1

(4.7.6)

where pi and ti are, respectively, the number of chunks the slice valency at splitting pair {ui , vi }, 1 ≤ i ≤ s = |S| such that p = p1 + p2 + ⋅ ⋅ ⋅ + ps . Proof. This is a result of Observation 4.7.3. This lemma is a refinement and a concise form of Theorem 10.4.3 with (10.4.13) in Liu YP [72] (2017, pp. 196–197), or Theorem 7.4.3 with (7.4.13) in Liu YP [55] (1994, p. 142). We proceed on the basis of Lemma 4.7.2 and Lemma 4.7.4, the result being useful in the context of determining the number of planar embeddings of a Eulerian loopless graph. Lemma 4.7.5. For a Eulerian loopless planar graph G, the number of (topologically) distinct planar embeddings of G is [Eli] n[Eli] (G) = n[Eli] Cut (G)nNct (G)

(4.7.7)

[Eli] where n[Eli] Cut (G) and nNct (G) are, respectively, given by (4.7.4) and (4.7.6).

Proof. On the uniqueness of 3-connected embedding of a graph, because a graph without vertex-cut set of one or two vertices has to be 3-connected, only the two independent cases with and without cut vertex need to be considered. Lemma 4.7.2 and Lemma 4.7.4 lead to the conclusion. ̃ Denote by t = aut(G) the semi-automorphism group order of a Eulerian loopless planar graph G. The relationship between embeddings and upper maps of G can be shown via t. Lemma 4.7.6. Let ℰ [Eli] (G) and ℳ[Eli] (G) be, respectively, the sets of all planar embeddings and upper maps of a Eulerian ordinary planar graph G. Then 󵄨󵄨 [Eli] 󵄨󵄨 2ϵ 󵄨󵄨 [Eli] 󵄨󵄨 (G)󵄨󵄨 󵄨󵄨ℳ (G)󵄨󵄨 = 󵄨󵄨ℰ t where ϵ = ϵ(G) is the size of G. Proof. Refer to Liu YP [59] (2003, pp. 225–226).

(4.7.8)

138 | 4 Inner equations second part [Eli] Let 𝒢m,s be the set of all Eulerian loopless planar graphs with rooted semi-valency 2m (i. e., the number of semi-edges incident to the root-vertex) and the un-rooted semisize 2s (i. e., the number of semi-edges not incident to the root-vertex). [Eli] Denote by 𝒯m,s the set of semi-automorphism group orders of Eulerian loopless [Eli] [Eli] [Eli] planar graphs in 𝒢m,s . Let 𝒥m,s be the set of all partition vectors n in 𝒢m,s . [Eli] Lemma 4.7.7. Let Em,n be the number of (topologically) distinct embeddings of Eulerian [Eli] loopless planar graphs in 𝒢m,n . Then we have [Eli] = Em,n

∑ n[Eli] (G)

(4.7.9)

[Eli] G∈𝒢m,n

where n[Eli] is given in (4.7.7). [Eli] Proof. Given the partition n, by considering all embeddings of each graph in 𝒢m,s with n, the conclusion is easily drawn. [Eli] We proceed on the basis of Lemma 4.7.7. Let 𝒯m,n be the set of semi-automorphism [Eli] [Eli] [Eli] group orders among graphs in 𝒢m,n and 𝒢m,n;t , the set of all graphs in 𝒢m,n with semĩ automorphism group order t. Denote by aut(G) the semi-automorphism group order of G.

Lemma 4.7.8. The number of root-isomorphic classes of upper maps of all graphs in [Eli] 𝒢m,n is A[Eli] m,n = ∑

[Eli] t∈𝒯m,n

2(m + π(n)) [Eli] Em,n;t t

(4.7.10)

[Eli] where A[Eli] m,n = Am,n (ℳm,n ) and [Eli] Em,n;t =

∑ n[Eli] (G)|aut(G)=t ̃

(4.7.11)

[Eli] G∈𝒢m,n

as determined by (4.7.9). Proof. We proceed on the basis of Lemma 4.7.6. By considering m + π(n) = ϵ, the conclusion is easily drawn. This lemma enables us to show the solution fa[Eli] of equation (4.5.1) when a0 = [Eli] a1 = a2 = 1. Denote f[Eli] = fa |a0 =a1 =a2 =1 , which is determined by Fm,s = 𝜕xm f[Eli] |π(y)=s for m, s ≥ 0. Lemma 4.7.9. For integers m, n ≥ 1, we have [Eli] Fm,s =

where A[Eli] m,n is given by (4.7.10).

n ∑ A[Eli] m,n y

[Eli] n∈𝒥m,s

(4.7.12)

4.8 Restrictions Eulerian loopless inner case |

139

Proof. It being known that the set of all underline graphs of Eulerian ordinary planar [Eli] [Eli] maps with m and n is the same as 𝒢m,n , A[Eli] m,n = Um,n is shown in Theorem 4.7.1. From Lemma 4.7.8, the conclusion can be drawn. We proceed on the basis of Lemma 4.7.9 and Theorem 4.7.1; we are allowed to state our main result of this section. [Eli] determine the solution of equation (3.5.1) as Theorem 4.7.10. For m, s ≥ 1, Fa[m,s] [Eli] = Fa[m,s]

[Eli] yn ∑ Ua[m,n]

(4.7.13)

[Eli] n∈𝒥m,s

[Eli] is the polynomial of a with degree at most m + s + 1 given where a = (a0 , a1 ) and Ua[m,n] in Theorem 4.7.1.

Proof. This is a result of Theorem 4.7.1 and Lemma 4.7.9. [Eli] is evaluated by the procedure shown in (4.6.1) or directly The polynomial Ua[m,n] from the inner structures of the solution in its own right. [Eli] [Eli] |a=1 determine the solution of equation Corollary 4.7.11. For m, s ≥ 1, Fm,s = Fa[m,s] (4.5.1) when a = 1 as [Eli] = Fm,s

[Eli] n y ∑ Um,n

(4.7.14)

[Eli] n∈𝒥m,s

[Eli] where Um,n = A[Eli] m,n is given in (4.7.10).

Proof. A direct result of Theorem 4.7.10. [Eli] This is a direct explicision (explicit expression) of Fm,s for integers m, s ≥ 1.

4.8 Restrictions Eulerian loopless inner case Now, an application of equation (4.5.1) with a0 = a1 = a2 = 1 is considered, 1 − 𝜕x2 ,y2 (uf |x2 =u ) { { = f; {∫ 1 − 2𝜕x2 ,y2 (uf |x2 =u ) − x2 y2 δx2 ,y2 (uf |2x2 =u ) y { { { {f |x=0⇒y=0 = 1,

(4.8.1)

where f ∈ ℛ{x, y} and y = (y1 , y2 , y3 , . . .). This is the restriction of equation (4.5.1) for a = 1. One might find the equation without the initiation or its equivalent version in [32] (Liu YP, 1986), and [60] (Liu YP, 2008, p. 186).

140 | 4 Inner equations second part The solution of equation (4.8.1) for f is just the enufunction for root-classifications of Eulerian loopless planar maps with the root-vertex valency (power of x) and the vertex-partition vector as parameters (power of y). There being no term with odd degree of x in equation (4.8.1), f is an even function of x. There being no term with odd degree of y, f is independent of y2i+1 , for any integer i ≥ 0. For convenience, let p = pa |a=1 and q = qa |a=1 , i. e., p = 𝜕x2 ,y2 (uf |u=x2 ); { q = δx2 ,y2 (uf 2 |u=x2 ).

(4.8.2)

Denote by Λ the part under the meson functional in equation (4.8.1), then Λ = Λ1 in equation (4.5.1), i. e., Λ=

1 − 𝜕x2 ,y2 (uf |u=x2 )

1 − 2𝜕x2 ,y2 (uf |u=x2 ) − x2 y2 δx2 ,y2 (uf |2u=x2 )

.

From (4.8.2), Λ=

1−p p + x2 y2 q = 1 + , 1 − p − x2 y2 q 1 − 2p − x2 y2 q k

2i k! x2(k−i) y2(k−i) pi+1 qk−i i!(k − i)! i=0

= 1 + ∑ (∑ k≥0

k

2i k! x2(k−i+1) y2(k−i+1) pi qk−i+1 ). i!(k − i)! i=0

+∑

In the first i-sum, the term of i = k is (2p)k p. In the second i-sum, the term of i = 0 is x2(k+1) y2(k+1) qk+1 . The remainder of the second i-sum is k

2i k! x2(k−i+1) y2(k−i+1) pi qk−i+1 i!(k − i)! i=1

∑

k−1

=∑

i=0

2i+1 k! x2(k−i) y2(k−i) pi+1 qk−i . (i + 1)!(k − i − 1)!

By combining the two remainders, k−1

Λ = 1 + ∑ (2k pk+1 + x2(k+1) y2(k+1) qk+1 + ∑ α[k,i] x2(k−i) y2(k−i) pi+1 qk−i ) k≥0

i=0

where k k 2i (2k − i + 1)k! ∈ ℤ+ ⊆ ℝ+ . α[k,i] = 2i ( ) + 2i+1 ( )= (i + 1)!(k − i)! i i+1

(4.8.3)

4.8 Restrictions Eulerian loopless inner case

| 141

Observation 4.8.1. If f is a solution of equation (4.8.1), then 𝜕x0 f = 1. Proof. We proceed on the basis of (4.8.1)–(4.8.3), the equation considered is { {f = ∫ Λ(x, y); y { { f | { x=0⇒y=0 = 1.

(4.8.4)

Because of 𝜕x0 Λ = 1 = f |x=0⇒y=0 , the conclusion can be drawn. This observation enables us to only consider 𝜕xm f for m ≥ 1 from now on. Theorem 4.8.2. Equation (4.8.1) is equivalent to f = 1 + ∑ (2k P [k+1] + x2(k+1) (y2(k+1) ⊗ Q[k+1] ) { { { { k≥0 { { { k−1 { + ∑ αa[k,i] x2(k−i) (y2(k−i) ⊗ P [i+1] ⊗ Q[k−i] )); { { { i=0 { { { f | { x=0⇒y=0 = 1,

(4.8.5)

on ℛ{x, y} where, for any integer k ≥ 0, k − 1 ≥ i ≥ 0, α[k,i] is given in (4.8.3) and, for integer l ≥ 1, { P [l] = P [l−1] ⊗ P, P [1] = P = ∫ p; { { { { y { [l] [l−1] [1] { { Q = Q ⊗ Q, Q = Q = ∫ q, { { y { such that p and q are, respectively, given by the first and second cases of (4.8.2). Proof. This is a result of Theorem 4.5.2 for a = 1. We proceed on the basis of Theorem 4.8.2; equation (4.8.1) for f ∈ ℛ{x, y} is transformed into the system of infinite equations for Fm ∈ ℛ{y} for m ≥ 0 shown in equation (4.8.5). For convenience, for an even function g of x on ℛ{x, y}, denote by m Gm = [g]m x2 = 𝜕x2 g ∈ ℛ{y}

(4.8.6)

the coefficient of x2m in g for m ≥ 0. This is because f ∈ ℛ{x, y} is even for x, p ∈ ℛ{x, y} and q ∈ ℛ{x, y} are even for x as well. From (4.8.6), we see the meaning of Fm , Pm and Qm . For any integer k ≥ 1, m

[k] Gm = [g k ]x2 = 𝜕xn2 g k [1] where Gm = Gm .

(4.8.7)

142 | 4 Inner equations second part Lemma 4.8.3. Given integer k ≥ 1, if we have a non-negative integer m ≤ k − 1, then [k] Pm = 0. Proof. See the proof of Lemma 4.5.3 for a = 1. [k] This lemma enables us to only consider Pm for k ≤ m.

Lemma 4.8.4. Given an integer k ≥ 1, if we have a non-negative integer m ≤ k − 1, then = 0. Q[k] a[m] Proof. This is a result of Lemma 4.5.4 for a = 1. This lemma enables us to only consider Q[k] m for k ≤ m. Theorem 4.8.5. The system of equations for Fm for m ≥ 0 1, when m = 0; { { { m−1 k [k+1] ⌊m/2⌋−1 [k+1] Fm = {∑k=0 2 Pm + ∑k=0 (y2(k+1) ⊗ Qm−k−1 ) { { m−1 k−1 [k+1] when m ≥ 1, { + ∑k=1 ∑i=0 α[k,i] (y2(k−i) ⊗ Rm−k+i ),

(4.8.8)

where α[k,i] ∈ ℤ+ ⊆ ℝ+ is given by (4.5.3) for a = 1 and R[k+1] = m−k+i

m−2k−1

[k+1] ⊗ Qj[k−i] ) ∑ (Pm−k+i−j

j=k−i

is equivalent to equation (4.8.1) on ℛ{x, y}. Proof. See the proof of Theorem 4.5.5 for a = 1. In order to solve the system of equations (4.8.8), the relationship between Pm and Qm expressed by Fm has to be considered beforehand. We proceed on the basis of (4.8.2). For given integer m ≥ 0, because of p = ∑ Fm m≥0

y2 x2(m+1) − x2 y2(m+1) = ∑ x2j ∑ Fi y2(i−j+1) , x2 − y2 j≥1 i≥j

we have 0, when m = 0; [p]m x2 = { 2(i+1) , otherwise. ∑i≥0 Fi+m y From Theorem 4.8.2, it is seen that, for integers l, m ≥ 1, ∑ Fi+m y2(i+1) (= Pm ), when l = 1; [l] Pm = { i≥0 [l−1] when l ≥ 2. ∑m−1 i=1 Pm−i Pi ,

(4.8.9)

4.8 Restrictions Eulerian loopless inner case

| 143

[2] Let Fm = 𝜕xm2 f 2 , m ≥ 0, then [2] q = ∑ Fm m≥0

x2(m+1) − y2(m+1) [2] 2i = ∑ x2j ∑ Fi+j y . x2 − y2 i≥0 j≥0

From Theorem 4.8.2, it is seen that, for integers l ≥ 1 and m ≥ 0, [2] y2i (= Qm ), ∑i≥0 Fi+m Q[l] m ={ m [l−1] ∑i=0 Qm−i Qi ,

when l = 1; when l ≥ 2.

(4.8.10)

In order to solve the system of equations (4.8.8), a new parameter s = π/2 is introduced so that, for any power vector n, π(y) = (11 + 212 + 313 + ⋅ ⋅ ⋅)nT . Let Fm,s be the part of Fm such that n satisfies π(n) = 2s. Lemma 4.8.6. When m = 0, for any integer s ≥ 0, 1, when s = 0; F0,s = { 0, when s ≥ 1.

(4.8.11)

Proof. When s = 0. From the initiation of equation (4.8.1), F0 = 1 and hence F0,0 = 1. This is the case of s = 0. Because F0 is independent of y, the case of s ≥ 1 in (4.8.10) is found. From Lemma 4.8.6, we are allowed to only discuss m ≥ 1 without loss of generality. Lemma 4.8.7. When s = 0, for any integer m ≥ 0, 1, when m = 0; Fm,0 = { 0, when m ≥ 1.

(4.8.12)

Proof. When m = 0, from Lemma 4.8.6, F0,0 = 1. When m ≥ 1, from (4.8.7), m−1

[k+1] Fm,0 = ∑ 2k Pm,0 = 0. k=0

Hence, the conclusion can be drawn. From Lemma 4.8.7, it is only necessary to discuss s ≥ 1. In order to determine Fm,s [l] for m, s ≥ 1, Theorem 4.8.5 suggests us to find Pm,s and Q[l] m,s for some l ≥ 1. As regards [l] Pm,s , Lemma 4.8.3 enables to only consider, m ≥ l. We proceed on the basis of (4.8.8). For any integer s ≥ 1, ∑s−1 Fi+m,s−i−1 y2(i+1) , (= Pm,s ) when l = 1; [l] Pm,s = { i=0 s [l−1] when l ≥ 2. ∑m−1 i=1 ∑j=0 Pm−i,s−j Pi,j ,

(4.8.13)

144 | 4 Inner equations second part Proposition 4.8.8. Let ℱm+s−1 = {Fr,t | r + t ≤ m + s − 1, r, t ≥ 0}. For any integers [l] m ≥ l ≥ 1 and s ≥ 1, Pm,s is determined by ℱm+s−1 . Proof. For any integers s − 1 ≥ i ≥ 0 and (i + m) + (s − i − 1) = m + s − 1, Fi+m,s−i−1 ∈ ℱm+s−1 . [1] From the first case of (4.8.13), Pm,s = Pm,s is determined by ℱm+s−1 . When l ≥ 2, assume [l−1] [l] Pr,t is determined by ℱr+t−1 . By induction, we prove that Pm,s is determined by ℱm+s−1 . As a matter of fact, by the second case of (4.8.13), the assumption leads to the conclusion. Similarly, on Q[l] m,s . Lemma 4.8.4 enables us to only discus m − 1 ≥ l ≥ 1. We proceed on the basis of (4.8.10). For any integer s ≥ 1, [2] y2i (= Qm ), ∑si=0 Fi+m,s−i = { Q[l] m,s m−1 s Q , ∑i=1 ∑j=0 Q[l−1] m−i,s−j i,j

when l = 1; when l ≥ 2.

(4.8.14)

Proposition 4.8.9. Let ℱm+s−1 = {Fr,t | r + t ≤ m + s − 1, r, t ≥ 0}. For any integers m ≥ l ≥ 1 and s ≥ 1, Q[l] m,s is determined by Fr,t , r + t ≤ m + s − 1. Proof. See the proof of Proposition 4.5.9, where a = 1. We proceed on the basis of (4.8.8). For integers m ≥ 1 and s ≥ 1, m−1

⌊m/2⌋−1

k=0

k=0

[k+1] Fm,s = ∑ 2k Pm,s +

) ∑ (y2(k+1) ⊗ Q[k+1] m−k−1,s−k−1

m−1 k−1

+ ∑ ∑ α[k,i] (y2(k−i) ⊗ k=1 i=0

(4.8.15)

R[k+1] ). m−k+i,s−k+i

On account of Proposition 4.8.8 and Proposition 4.8.9, R[k+1] is determined by m−k+i,s+i−k Fr,t , r + t ≤ (m − k + i) + (s − k + i) − 1 = m + s − 2k + 2i − 1 ≤ m + s − 1. Theorem 4.8.10. Equation (4.8.1) is well defined on ℛ{x, y}. Proof. A direct result of Theorem 4.5.10 for a = 1. In order to simplify the procedure illustrated in the last section, more structural properties of the solution of equation (4.8.1) have to be particularly investigated. Lemma 4.8.11. If f ∈ ℛ{x, y} is a solution of (4.8.1), then, for any integers m, s ≥ 0, Fm,s = [𝜕xm2 f ]sy ∈ ℛ+ [y2s ] is a polynomial of y with degree s. Proof. See the proof of Lemma 4.6.1 for a = 1. From this lemma, Fm,s can be seen to have all coefficients of the form of a finite sum of terms positive for m, s ≥ 0. Lemma 4.8.12. For any integers m, s ≥ 1 given. If m ≥ s + 1, then Fm,s = 0. Proof. See the proof of Lemma 4.6.2 for a = 1.

4.8 Restrictions Eulerian loopless inner case | 145

This lemma enables us to reduce by half the amount of labor for solving equation (4.8.1) by using the procedure shown in the proof of Theorem 4.8.10. Theorem 4.8.13. Let f[Eli] be the solution of equation (4.8.1). For integers m, s ≥ 0, de[Eli] [Eli] [Eli] , then Fm,s ∈ ℛ+ {y} is of the form of a finite sum with all note Fm,s = [𝜕xm2 f[Eli] ]sy = Fm,s terms positive:

[Eli] Fm,s

1, { { { { { { 0, { { { m−1 k [k+1] = {∑k=0 2 Pm,s { { { { + ∑⌊m/2⌋−1 (y2(k+1) ⊗ Q[k+1] ) { { k=0 m−k−1,s−k−1 { { m−1 k−1 [k+1] + ∑k=1 ∑i=0 α[k,i] (y2(k−i) ⊗ Rm−k+i,s−k+i ), {

when s = m = 0; when s ≥ 1, m = 0, or m ≥ s + 1;

otherwise,

where R[k+1] = m−k+i,s−k+i

m−2k−1 k−i

∑

[k+1] [k−i] ⊗ Qj,n ), ∑ (Pm−k+i−j,s−k+i−n

j=k−i n=0

[Eli] y2(i+1) (= Pm,s ), ∑s−1 Fi+m,s−i−1 [l] Pm,s = { i=0 m−1 s [l−1] ∑i=1 ∑j=0 Pm−i,s−j Pi,j ,

when l = 1; when l ≥ 2,

(4.8.16) (4.8.17)

(4.8.18)

and [Eli][2] y2i (= Qm ), ∑si=0 Fm,s Q[l] = { m,s s [l−1] Q ∑m−1 ∑ i=1 j=0 m−i,s−j Qi,j ,

in which

m

when l = 1; when l ≥ 2,

s

[Eli] [Eli] [Eli][2] Fi,j . Fm,s = ∑ ∑ Fm−i,s−j i=0 j=0

(4.8.19)

(4.8.20)

Proof. A direct result of Theorem 4.6.3 for a = 1. Example 4.8.1. Root-isomorphic classes of Euler loopless planar maps with rootvertex valency and vertex-partition vector as parameters. Because the enufunction of [Eli] such maps satisfies the equation (4.8.1), Fm,s determined by (4.8.16)–(4.8.20) in Theorem 4.8.13 shows the number of root-isomorphic classes of Eulerian loopless planar maps with root-vertex valency 2m and inner semi-size 2s = π(n). Thus, m + s is the size of the map. When m + s = 2, only one map is considered, as shown by a in Figure 4.8.1 as [Eli] = x 2 y2 . x2 F1,1 [ELi] = x2 y22 . When m+s = 3, we also have one map, shown by b in Figure 4.8.1 as x2 F1,2

[Eli] When m+s = 4, we have five maps, shown by c, d and e in Figure 4.8.1 as x2 (F1,3 )+

[Eli] [Eli] [Eli] x4 (F2,2 ), where F1,3 = 2c + 1e = 2y2 y4 + y23 and F2,2 = 2c + d = 2y22 + y4 .

146 | 4 Inner equations second part

Figure 4.8.1: Root-isomorphic classes of Euler loopless planar maps with vertex partition. [Eli] )+ When m + s = 5, we have 17 maps, shown by f , g and h in Figure 4.8.1 as x2 (F1,4

[Eli] [Eli] [Eli] = 6f + 2g + h = 6y22 y4 + 2y42 + y24 and F2,3 = 4f + 4g = 4y23 + 4y2 y4 . x4 (F2,3 ), where F1,4

4.9 Notes 4.9.1. Equation coefficients variable. The generalization of equation (4.1.1), or equation (4.5.1), with variable equation coefficients a such that each entry of a has a linear form, a quadratic form or so is extensively investigated for combinatorial properties of partitions with a certain type of weights variable. 4.9.2. One way for explicision. An explicision of the solution of equation (4.1.1), or equation (4.5.1), still needs to be evaluated only by transformations on ℛ{x, y} to directly seek a favorite resolution. 4.9.3. Another way for explicision. Another manner for getting an explicision of the solution of equation (4.1.1), or equation (4.5.1), in the case of a = (1, 1, 1), i. e., equation (4.4.1), or equation (4.8.1). This is to observe the root-classifications of general, or Eulerian, loopless planar maps with vertex-partition given directly without considering semi-automorphism group orders of the underline graphs. 4.9.4. New results on embedding of general, Eulerian, loopless planar graphs. All results in Section 4.3 or Section 4.7 on embedding of general, or Eulerian, loopless planar graphs for reaching the corresponding root-enumerations of general, or Eule-

4.9 Notes | 147

rian, loopless planar maps are seen as a type of foundation for getting an explicision for the solution of equation (4.1.1), or equation (4.5.1). 4.9.5. The efficientization and then intelligentization of Theorem 4.2.7 and Theorem 4.6.3, particularly Theorem 4.4.13 and Theorem 4.8.13, have to be further investigated for running on computers and then for practical usage. 4.9.6. Specification of parameters. For a small number of parameters given, more research might be found in [27] (Liu, Y. P., 1983), [26] (1983), [51] (1989); [8] (Cai, J. L., Liu, Y. P., 1997); Cai JL, Hao RX, Liu YP [6] (2001) etc. 4.9.7. Direct explicision without consideration of symmetry. One might address a direct explicision of (4.7.13) for upper maps of general loopless planar graphs without consideration of symmetry. If symmetry is considered for maps, the root-isomorphic maps distribution by their automorphism group orders has to be done as shown in Liu YP [59] (Liu, Y. P., 2003, pp. 221–230), or [73] (2017, pp. 188–194). 4.9.8. The relationship between equation (4.1.1) and equation (4.5.1), or equation (4.4.1) and equation (4.8.1), is investigated via inner structures over general loopless planar maps and Eulerian loopless planar maps. Are there a series of transformations on ℛ{x, y} from one of equation (4.1.1) and equation (4.5.1), or particularly, from equation (4.4.1) and equation (4.8.1) to the other? 4.9.9. Theorem 4.2.7 and Theorem 4.6.3 can be, respectively, employed for evaluating of (a0 = a2 , a1 ) in Theorem 4.3.1 and Theorem 4.7.1. and A[Eli] the polynomials A[gli] a[m,n] a[m,n] 4.9.10. By noticing that equation (4.8.1) is equivalent to the equation as 1 { {∫ 1 − Γ(f , δ, 𝜕; y) = f ; y { { {f |x=0⇒y=0 = 1,

(4.9.1)

where Γ(f , δ, 𝜕; y) = (𝜕x2 ,y2 (uf |x2 =u ) + x2 y2 δx2 ,y2 (uf 2 |x2 =u ))/(1 − 𝜕x2 ,y2 (uf |x2 =u )). One might see that equation (4.8.1) has a similar form to equation (4.4.1). This suggests to seek a relationship between g in equation (4.4.1) and f in equation (4.8.1) via the corresponding two types of maps. 4.9.11. Asymptotic behavior (or asymptotics). Note 4.9.6 reminds us to investigate their asymptotic behavior as shown in [56] (Liu, Y. P., 1999, pp. 359–389), [94] (Yan, J. Y., Liu, Y. P., 1991). 4.9.12. Stochastic behavior (or stochastics). We proceed on the basis of (4.4.27)– (4.4.28) and (4.8.16)–(4.8.20); the distributions of sizes, orders, semi-automorphic group orders and genera from, respectively, size polynomials, order polynomials, semi-automorphic group order polynomials and genus polynomials need to be investigated. Relevant results are referred to in [56] (Liu, Y. P., 1999, pp. 359–389), [59] (Liu,

148 | 4 Inner equations second part Y. P., 2003, 221–230), [94] (Yan, J. Y., Liu, Y. P., 1991), [11] (Chen, Y. C., et al., 2007), [92] (Wan, L. X., et al., 2008) etc. 4.9.13. Equation (4.1.1) is from Program 90 as equation (121) in [70] (Liu, Y. P., 2015, Vol. 23, p. 11698). Equation (4.5.1) is from Program 96 as equation (127) in [70] (Liu, Y. P., 2015, Vol. 23, p. 11701).

5 Inner equations third part 5.1 General cutless inner model Two models are included in this chapter. First, consider the equation for f y𝜕x,y f |x=u { { = f − a1 x2 ; {a2 x ∫ 1 − 𝜕 f | x,y x=u y { { { f | { x=0⇒y=0 = a0

(5.1.1)

where a = (a0 , a1 , a2 ) ∈ ℝ3 , f ∈ ℛ{x, y} and y = (y1 , y2 , y3 , . . .). This is equation (9) in Introduction. Because of a solution of equation (5.1.1) for a0 = 0 and a1 = a2 = 1 is involved with general cutless planar maps as shown in [28] (Liu YP, 1985) and [56] (Liu YP, 1999, p. 187), this equation is called a general cutless inner model. From f = fa ∈ ℛ{x, y}, f is determined by Fa[m] = 𝜕xm f ∈ ℛ{y} for m ≥ 0. We proceed on the basis of 𝜕x,y f |u=x = xy ∑ Fa[l] ( l≥0

xl−1 − yl−1 ). x−y

By considering l−2

xl−1 − yl−1 = (x − y)( ∑ yi xl−2−i ), i=0

we have 𝜕x,y f |u=x = ∑(∑ Fa[l−1] yl−i−1 )xi . i≥1 l≥i

Therefore, y𝜕x,y f |u=x = ∑(∑ Fa[l−1] yl−i )xi . i≥1 l≥i

In the part under the meson functional, because of 1 = ∑ (𝜕 f | )j , 1 − 𝜕x,y f |u=x j≥0 x,y u=x we have

y𝜕x,y f |u=x

1 − 𝜕x,y f |u=x https://doi.org/10.1515/9783110627336-005

= ∑ y(𝜕x,y f |u=x )j . j≥1

(5.1.2)

150 | 5 Inner equations third part By (5.1.2), y𝜕x,y f |u=x

1 − 𝜕x,y f |u=x

j

= ∑ y(∑ ( ∑ Fa[l−1] yl−i−1 )xi ) . j≥1

i≥0 l≥i+1

(5.1.3)

For convenience, denote ∇a[m] = 𝜕xm (𝜕x,y f |u=x ) and 0, when k = 0; { { { [k] = {∇a[m] , ∇a[m] when k = 1; { { m [k−1] {∑j=0 ∇a[m−j] ∇a[j] , when k ≥ 2,

(5.1.4)

for any integers k ≥ 1 and m ≥ 0. [k] = 𝜕xm (𝜕x,y f |u=x )k . By It is easily seen that, for any integers k ≥ 1 and m ≥ 0, ∇a[m] (5.1.2), l−m−1 , {∇a[m] = ∑ Fa[l−1] y { { { l≥m+1 m { [k] [k−1] { { {∇a[m] = ∑ ∇a[m−j] ∇a[j] , j=0 {

when k = 1; when k ≥ 2,

(5.1.5)

for any integer m ≥ 1. For any integers k ≥ 1 and m ≥ 0, let {∫y (y∇a[m] ), when k = 1; △[k] a[m] = { ∫ (y∇[k] ), when k ≥ 2. { y a[m]

(5.1.6)

[k] = 0. = 0 ⇔ ∇a[m] Thus, for any integers k ≥ 1 and m ≥ 0, △[k] a[m]

Observation 5.1.1. If f is a solution of equation (5.1.1), then 𝜕x0 f = a0 = 0. Proof. Because of Fa[0] = 0 from the first equality of equation (5.1.1), the conclusion can be drawn. This observation enables us only to consider a0 = 0 for equation (5.1.1) without loss of generality. Here, a = (a1 , a2 ). Theorem 5.1.2. Equation (5.1.1) is equivalent to the system of equations for Fa[m] for m≥0

Fa[m]

on ℛ{x, y}.

0, when m = 0; { { { { [k] { {a2 ∑k≥1 △a[0] , when m = 1; ={ [k] { {a1 + a2 ∑k≥1 △a[1] , when m = 2; { { { [k] when m ≥ 3, {a2 ∑k≥1 △a[m−1] ,

(5.1.7)

5.1 General cutless inner model | 151

Proof. We proceed on the basis of (5.1.3) and (5.1.4), y𝜕x,y f |u=x

1 − 𝜕x,y f |u=x

j

= ∑ y(∑ ∇a[i] xi ) = ∑ (∑ y∇a[i] )xi . j≥1

[j]

i≥0 j≥1

i≥0

From (5.1.6), ∫ y

y𝜕x,y f |u=x

1 − 𝜕x,y f |u=x

= ∑ (∑ ∫(y∇a[i] ))xi = ∑ (∑ △a[i] )xi . [j]

[j]

i≥0 j≥1

i≥0 j≥1 y

This shows that (5.1.1) becomes f = a1 x2 + a2 ∑ (∑ △a[m−1] )xm . [j]

m≥0 j≥1

Then, for any integer m ≥ 0, a2 ∑k≥1 △a[−1] , { { { { { {a2 ∑ △[k] , { k≥1 a[0] ={ { { , a1 + a2 ∑k≥1 △[k] { a[1] { { { [k] {a2 ∑k≥1 △a[m−1] , [k]

Fa[m]

when m = 0; when m = 1; when m = 2; when m ≥ 3.

When m = 0, because △a[−1] = 0 ⇒ △a[−1] = 0 for j ≥ 2, Fa[0] = 0. This is the initiation of equation (5.1.1). When m ≥ 1, the results agrees with those shown in equations (5.1.7). The conclusion can be drawn. [j]

This theorem enables us to only consider the system of equations shown in equations (5.1.7) instead of equation (5.1.1) itself. Because each equation in equations (5.1.7) is infinite, we are required to seek surplus equations and some parameter other than m so that, for any specified value, the system of equations (5.1.7) becomes finite. = 0 for 0 ≤ m ≤ k. Lemma 5.1.3. Given any integer k ≥ 1, △[k] a[m] = 0 for k ≥ 1, Fa[1] = 0. Proof. When k = 1, because of Fa[0] = 0 ⇒ △[k] a[0]

= 0 (0 ≤ m ≤ t), for any 1 ≤ t ≤ k − 1. By induction on k, When k ≥ 2, assume △[t] a[m] we prove the conclusion for t = k. We proceed on the basis of (5.1.5) and (5.1.6). By the assumption, the conclusion can be easily drawn. for k ≥ m in the This lemma shows that it is unnecessary to consider terms △[k] a[m] system of equations (5.1.7) for any integer m ≥ 1. Thus, for any integer m ≥ 0, equation (5.1.7) becomes Fa[m]

0, { { { { = {a1 + a2 △a[1] , { { { m−1 [k] {a2 ∑k=1 △a[m−1] ,

when m ≤ 1; when m = 2; when m ≥ 3.

(5.1.8)

152 | 5 Inner equations third part From equations (5.1.8), all infinite sum in equations (5.1.7) become finite. For Fa[m] ∈ ℛ{y}, let 𝒥m be the set of power vectors of all terms in Fa[m] . For any n ∈ 𝒥m , denote by π(n) = (∑ in[i] )nT i≥1

the inner semi-size where n[i] is the vector of all entries 0 except only for the ith entry, which is 1. Because of y = (y1 , y2 , y3 , . . .), π(n) = ∑ ini .

(5.1.9)

i≥1

Observe the case of m = 2 for (5.1.8). From (5.1.5) and (5.1.6), △a[1] = ∫(∑ Fa[l−1] yl−2 ) = ∑ Fa[l−1] yl−2 . y

l≥2

l≥2

(5.1.10)

Because of the infinity of equation (5.1.10), equation (5.1.8) is infinite as well. Observation 5.1.4. For integer m ≥ 0, Fa[m] is independent of y0 and y1 . Proof. Because no y0 appears in y, Fa[m] is independent of y0 . We proceed on the basis of (5.1.5) and (5.1.6). By induction, the independency of y1 in Fa[m] can be seen. From this observation, △a[1] = ∑ Fa[l−1] yl−1 .

(5.1.11)

0, when m ≤ 1; { { { Fa[m] = {a1 + a2 ∑l≥3 Fa[l−1] yl−2 , when m = 2; { { m−1 [k] when m ≥ 3. {a2 ∑k=1 △a[m−1] ,

(5.1.12)

l≥3

Thus, (5.1.8) becomes

For any integer s ≥ 0, denote Fa[m,s] = ⟨Fa[m] ⟩s . Because of Fa[m] ∈ ℛ{y}, Fa[m,s] = ∑ Fa[m,n] yn n∈𝒥m π(n)=s

(5.1.13)

where Fa[m,n] ∈ ℛ. Observe Fa[2,s] . From the case of m = 2 in (5.1.12), a1 , Fa[2,s] = { a2 ∑sl=1 Fa[l,s−l] yl ,

when s = 0; when s ≥ 1.

(5.1.14)

5.1 General cutless inner model | 153

for In order to determine Fa[m,s] for m ≥ 3, it is only necessary to get △[k] a[m−1,s]

[k] . From (5.1.5), 0 ≤ k ≤ m − 1 from (5.1.7) and Lemma 5.1.3. By (5.1.6), observe ∇a[m] m

[k−1] [k] ( ∑ Fa[l−1] yl−j−1 ). = ∑ ∇a[m−j] ∇a[m] j=0

l≥j+1

(5.1.15)

In convenience, for integers k ≥ 1 and m ≥ 0, let (5.1.16)

[k] Π[k] a[m] = ∫ ∇a[m] . y

. Then we have From (5.1.5), denote Πa[m] = Π[1] a[m] ∑l≥m+1 Fa[l−1] yl−m−1 , when k = 1; Π[k] a[m] = { m ⊗ Πa[j] , when k ≥ 2. ∑j=0 Π[k−1] a[m−j]

(5.1.17)

By (5.1.6), △[k] m = y1 ⊗ Πa[m] . We proceed on the basis of (5.1.17). From the property of convolution, when k = 1; ∑l≥1 Fa[l+m−1] yl , △[k] a[m] = { m [k−1] ∑j=0 Πa[m−j] ⊗ (∑l≥j+1 Fa[l−j−1] yl ), when k ≥ 2. = 0 for k ≥ 1, From l − m − 1 ≥ 0, ∇a[0] = 0 ⇒ Πa[0] = 0 and Π[k] a[0] when k = 1; ∑l≥1 Fa[l+m−1] yl , △[k] a[m] = { m−1 [k−1] ∑j=1 Πa[m−j] ⊗ (∑l≥j+1 Fa[l−j−1] yl ), when k ≥ 2.

(5.1.18)

On this basis, when k = 1, for any integers m ≥ 2 and s ≥ 0, s

s

△a[m,s] = ⟨ ∑ Fa[l+m−1] yl ⟩ = ∑ Fa[l+m−1,s−l] yl . l=1

s

l=1

When k ≥ 2, for two integers m ≥ 2 and s ≥ 0, m−1 s−1

r

j=1 r=1

l=1

[k−1] △[k] a[m,s] = ∑ ∑ (Πa[m−j,s−r] ⊗ ∑ Fa[l+j−1,r−l] yl ).

By combining the two formulas above, ∑sl=m+1 Fa[l−m−1,s−l] yl , = { △[k] a[m,s] m−1 s−1 [k−1] ∑r=1 (Πa[m−j,s−r] ⊗ ∑rl=1 Fa[l+j−1,r−l] yl ), ∑j=1

when k = 1; when k ≥ 2.

(5.1.19)

154 | 5 Inner equations third part Theorem 5.1.5. Equation (5.1.1) for f ∈ ℛ{x, y} is equivalent to the system of equations for Fa[m,s] ∈ ℛ{y} for integers m ≥ 2 and s ≥ 0,

Fa[m,s]

a1 , when s = 0; { { }, { a ∑s F y , when s≥1 a[l,s−l] l 2 l=2 ={ { { r [k−1] a ∑ 1≤r≤s−1 ((∑m−1 k=2 Πa[m−j−1,s−r] ) ⊗ ∑l=1 Fa[l+j−1,r−l] yl ), { 2 1≤j≤m−2

when m = 2; when m ≥ 3.

(5.1.20)

Proof. From Theorem 5.1.2, it is only necessary to discuss the equivalence between equations (5.1.20) and equations (5.1.7) (or equations (5.1.12)). First, we proceed on the basis of the case of m = 2 for equations (5.1.12). From (5.1.14), the case of m = 2 for (5.1.20) is obtained. Then we proceed on the basis of the general case of m ≥ 3 for equations (5.1.12). !), From (5.1.19) (notice that Fa[m,s] is determined by △[k] a[m−1,s] m−1

r

k=2

l=1

Fa[m,s] = a2 ∑ ( ∑ Π[k−1] a[m−j−1,s−r] ) ⊗ ∑ Fa[l+j−1,r−l] yl . 1≤r≤s−1 1≤j≤m−2

This is the conclusion. We proceed on the basis of equations (5.1.20). For m ≥ 2 and s ≥ 0, all of Fa[m,s] = 0 are omitted. Lemma 5.1.6. Given any integers m ≥ 2 and s ≥ 0. If Fa[m,s] ≠ 0, then m = s(mod 2). Proof. When 0 ≤ m + s ≤ 2. Only for m = 2, s = 0, Fa[2,0] ≠ 0. From (5.1.14), Fa[2,0] = a1 . The conclusion is true. When n = m + s ≥ 3, assume i = j(mod 2), if Fa[i,j] ≠ 0 for any integers i ≥ 2 and j ≥ 0, i + j ≤ n − 1. By induction on n, we prove m + s = 0(mod 2) if Fa[m,s] ≠ 0 for any integers m ≥ 2 and s ≥ 0 such that m + s = n. We proceed on the basis of (5.1.19). Because (m−j−1)+(s−r) = m+s−j−r−1, by j ≥ 1 and r ≥ 1, (m−j−1)+(s−r) ≤ m+s−3 < m+s−1 = n−1. Because of (l+j−1)+r − l = r+j−2, by r ≤ s − 1 and j ≤ m, r + j − 2 ≤ (s − 1) + j − 2 ≤ m + s − 3 < m + s − 1 = n − 1, we have ((m−j−1)+(l+j−1))+((s−r)+(r −l)) = (m+l−2)+(s−l) = m+s−2 < m+s−1 = n−1. From the assumption, (m − j − 1) + (s − r) = 0(mod 2), (l + j − 1) + (r − l) = r + j − 1 = 0(mod 2), and then 0 = ((m − j − 1) + (s − r)) + (r + j − 1) = m + s − 2 = m + s(mod 2). Therefore, the conclusion can be drawn. Furthermore, it is still necessary to discuss, for any integers m, s ≥ 0, what y there are that Fa[m,s] depends on. Lemma 5.1.7. For any integers m ≥ 2 and s ≥ 0, Fa[m,s] (≠ 0) is independent of all yl , l ≥ s + 1.

5.2 Solution general cutless inner case

| 155

Proof. For any n ∈ 𝒥m,s , π(n) = s, if Fa[m,s] has some yl for l ≥ s + 1, then there exists n ∈ 𝒥m,s such that ns+1 ≥ 1. From this, s ≥ (s + 1)ni+1 ≥ s + 1. This is a contradiction to π(n) = s. This lemma enables us to restrict ourselves only on Fa[m,s] in the scope of y = ys = (y1 , y2 , . . . , ys ) for m ≥ 2 and s ≥ 1. Theorem 5.1.8. Equation (5.1.1) is well defined on ℛ{x, y} if, and only if, a0 = 0. Proof. We proceed on the basis of Theorem 5.1.5. It is only necessary to consider equations (5.1.20). First of all, we address the case of m + s ≤ 4 with m ≥ 2 and s ≥ 0. Because Fa[2,0] = a1 for m + s ≤ 2, from Lemma 5.1.7, it is only necessary to consider that m + s is even. When m+s = 4, we only have Fa[4,0] = 0, Fa[3,1] = 0 and Fa[2,2] = a2 F2,0 y2 = a2 a1 y2 . When m + s ≥ 6, assume all Fa[p,q] are determined by (5.1.20) for p ≥ 2, q ≥ 0, and p + q ≤ m + n − 1. By induction, we determine Fa[m,s] . are determined by Fa[l+j−1,r−l] from (5.1.20). We As a matter of fact, all Π[k−1] a[m−j−1,s−r] have (l+j−1)+(r −l) = r +j−1. By j ≤ m−1 and r ≤ s−1, (l+j−1)+(r −l) ≤ m+s−3 < m+s, by the assumption, Fm,s is obtained from (5.1.20). By considering the uniqueness for a0 = 1 given, we see the sufficiency. The necessity is seen from Observation 5.1.1.

5.2 Solution general cutless inner case We proceed on the basis of equations (5.1.20). In order to evaluate the solution of equation (5.1.1), some available structures of the solution have to be investigated. Lemma 5.2.1. Let fa[gci] ∈ ℛ{x, y} be the solution of equation (5.1.1). Denote [gci] (m,n) Sa[m,s] = ∑ (𝜕(x,y) f[gci] )yn . n∈𝒥m π(n)=s

(5.2.1)

If a1 , a2 ∈ ℝ+ , then Sm,s ∈ ℛ+ {y}. [gci] = Fa[m,s] satisfies the system of equations (5.1.20). By Proof. From Theorem 5.1.8, Sa[m,s] employing the procedure in the proof of Theorem 5.1.8, it is shown that all coefficients of yn are in ℝ+ if a1 , a2 ∈ ℝ+ . This is the conclusion.

This lemma enables us to obtain the solution with the coefficient of each term in the form of a sum of positive real numbers. = 0 for Lemma 5.2.2. Given any two integers m ≥ 1 and s ≥ 0. If m ≥ s + 1, then Π[k] a[m,s] any integer k ≥ 1 where Π[k] is shown in (5.1.16) and (5.1.17). a[m,s]

156 | 5 Inner equations third part Proof. When k = 1, from (5.1.17), s

Πa[m,s] = ∑ Fa[t+m,s−t] yt . t=0

(5.2.2)

When s = 0, by (5.2.2), Πa[m,0] = Fa[m,0] y0 for any integer m ≥ 2, because there is no y0 in y = (y1 , y2 , . . .), Πm,0 = 0 for m ≥ 1. When s = 1, by (5.2.2), Πa[m,1] = Fa[m,1] y0 + Fa[m+1,0] y1 = Fa[m+1,0] y1 . From (5.1.12), Fa[m,0] = 0 for m ≥ 1 except only for Fa[2,0] = a1 . Thus, a1 y1 , when m = 1; Πa[m,1] = { 0, when m ≥ 2.

(5.2.3)

That is, Πa[m,1] = 0 for m ≥ s + 1 = 2. When integer s ≥ 2, assume, for any integers m ≥ s, that Πa[m,r] = 0, and hence Fa[m,r] = 0 where 1 ≤ r ≤ s − 1. By induction, we prove Πm,s = 0. That is, Πa[m,s] = 0 ⇒ Fa[m,s] = 0.

(5.2.4)

This is the case of k = 1 for the conclusion. When k ≥ 2, assume for any integers 1 ≤ l ≤ k − 1, known that if m ≥ s + 1, then Π[l] = 0. By induction, we prove Π[k] = 0. a[m,s] a[m,s] From (5.1.17), m

m

[k−1] [k−1] Π[k] a[m,s] = [ ∑ Πa[m−j] ⊗ Πa[j] ] = ∑ [Πa[m−j] ⊗ Πa[j] ]s j=0

where

s

j=0

(5.2.5)

s

[k−1] [Π[k−1] a[m−j] ⊗ Πa[j] ]s = ∑ Πa[m−j,s−r] ⊗ Πa[j,r] . r=0

Because of m ≥ s + 1, we have, for any integers 0 ≤ r ≤ s, m − j ≤ s − r 󳨐⇒ −j ≤ −(r + 1) 󳨐⇒ j ≥ r + 1. = 0. Therefore, = 0 or Π[k−1] By the assumption, for any 0 ≤ r ≤ s, either Π[k−1] a[j] a[m−j]

= 0. This is the conclusion. Π[k] a[m,s]

This lemma also shows to us that, for any integers m ≥ s + 1, Fa[m,s] = 0. Corollary 5.2.3. For any integers m ≥ 2 and s ≥ 0. If m ≥ s + 1, then Fa[m,s] = 0 except only Fa[2,0] = a1 . Proof. On the basis of (5.1.12) and (5.1.18), the conclusion is a direct result of Lemma 5.2.2.

5.2 Solution general cutless inner case

| 157

We proceed on the basis of the last two lemmas. A compact form of the solution of equation (5.1.1) can be evaluated. In order to do this, the cases of m + s ≤ 8 have to be firstly observed. From Lemma 5.1.6, it is only necessary to observe the cases of m = s(mod 2). By (5.1.17), when k = 1, if m = 1 and 0 ≤ s ≤ 7, because of Πa[1,s] =

⌊(s+3)/2⌋

∑ l=2

Fa[l−1,s+2−l] yl−2 ,

we have

Πa[1,s]

Fa[2,0] y1 , when s = 1; { { { { { {Fa[2,2] y1 , when s = 3; ={ {Fa[2,4] y1 + Fa[3,3] y2 , when s = 5; { { { { {Fa[2,6] y1 + Fa[3,5] y2 + Fa[4,4] y3 , when s = 7.

(5.2.6)

If m = 2 and 0 ≤ s ≤ 7, because of Πa[2,s] =

⌊(s+4)/2⌋

∑ l=3

Fa[l−1,s+3−l] yl−3 ,

we have

Πa[2,s]

Fa[2,0] y0 , when s = 0; { { { { { {Fa[2,2] y0 , when s = 2; ={ {Fa[2,4] y0 + Fa[3,3] y1 , when s = 4; { { { { {Fa[2,6] y0 + Fa[3,5] y1 + Fa[4,4] y2 , when s = 6.

(5.2.7)

If m = 3 and 0 ≤ s ≤ 7, because of s+4

⌊(s+5)/2⌋

l=4

l=4

Πa[3,s] = ∑ Fa[l−1,s+4−l] yl−4 =

∑

Fa[l−1,s+4−l] yl−4 ,

we have

Πa[3,s]

0, { { { { { {Fa[3,3] y0 , ={ {Fa[3,5] y0 + Fa[4,4] y1 , { { { { {Fa[3,7] y0 + Fa[4,6] y1 + Fa[5,5] y2 ,

when s = 1; when s = 3; when s = 5; when s = 7.

If m = 4 and 0 ≤ s ≤ 7, because of s+6

Πa[4,s] = ∑ Fa[l−1,s+5−l] yl−5 = l=5

⌊(s+6)/2⌋

∑ l=5

Fa[l−1,s+5−l] yl−5 ,

(5.2.8)

158 | 5 Inner equations third part we have 0, when s = 0 and s = 2; { { { Πa[4,0] = {Fa[4,4] y0 , when s = 4; { { {Fa[4,6] y0 + Fa[5,5] y1 , when s = 6.

(5.2.9)

If m = 5 and 0 ≤ s ≤ 7, because of s+7

⌊(s+7)/2⌋

l=6

l=6

Πa[5,s] = ∑ Fa[l−1,s+6−l] yl−6 =

∑

Fa[l−1,s+6−l] yl−6 ,

we have Πa[5,s]

0, when s = 1 and s = 3 { { { = {Fa[5,5] y0 , when s = 5; { { {Fa[5,7] y0 + Fa[6,6] y1 , when s = 7.

(5.2.10)

If m = 6 and 0 ≤ s ≤ 7, because of s+8

Πa[6,s] = ∑ Fa[l−1,s+7−l] yl−7 = l=7

⌊(s+8)/2⌋

∑ l=7

Fa[l−1,s+7−l] yl−7 ,

we have 0, when s = 5 s = 0, 2 and 4; Πa[6,s] = { Fa[6,6] y0 , when s = 6.

(5.2.11)

If m = 7 and 0 ≤ s ≤ 7, because of s+9

⌊(s+9)/2⌋

l=8

l=8

Πa[7,s] = ∑ Fa[l−1,s+8−l] yl−8 =

∑

Fa[l−1,s+8−l] yl−8 ,

we have 0, when s = 1, 3 and 5; Πa[7,s] = { Fa[7,7] y0 , when s = 7.

(5.2.12)

We proceed on the basis of calculations above, all Fa[m,s] are evaluated for m+s ≤ 8 when m ≥ 2 and s ≥ 0. When m + s = 2, it is only necessary to do Fa[2,0] . From (5.1.10), Fa[2,0] = a1 . When m + s = 4, only we do Fa[4,0] and Fa[2,2] . From Corollary 5.2.3, Fa[4,0] = 0. Because of Πa[1,1] = Fa[2,0] y1 = a1 y1 , Fa[2,2] = a1 a2 y2 .

(5.2.13)

5.2 Solution general cutless inner case

| 159

When m + s = 6, only we do Fa[6,0] = 0, Fa[3,3] and Fa[2,4] . Because of Π[1] = a[2,2]

Fa[2,2] y0 = 0 and Π[2] = Πa[1,1] ⊗ Πa[1,1] = a1 y1 ⊗ a1 y1 = a21 y2 , a[2,2] Fa[3,3] = a2 a21 y2 ⊗ y1 = a21 a2 y3 .

(5.2.14)

Fa[2,4] = a1 a22 y22 .

(5.2.15)

Because of Πa[1,3] = Fa[2,2] y1 ,

When m + s = 8, from Corollary 5.2.3, it is only necessary to observe Fa[4,4] , Fa[3,5] and Fa[2,6] . Because of Π[1] = a31 y3 , we have = 0 and Π[3] = Fa[3,3] y0 = 0, Π[2] a[3,3] a[3,3] a[3,3] 3

3 Fa[4,4] = a2 ∑ Π[k] a[3,3] ⊗ y1 = a1 a2 y4 . k=1

(5.2.16)

[2] . BeIn order to determine Fa[3,5] , it is only necessary to do Π[1] = Πa[2,4] and Πa[2,4] a[2,4]

cause of Πa[2,4] = Fa[2,4] y0 + Fa[3,3] y1 = Fa[3,3] y1 and Π[2] = Πa[1,3] ⊗ Πa[1,1] + Πa[1,1] ⊗ a[2,4] Πa[1,3] = 2(Πa[1,3] ⊗ Πa[1,1] ) = 2Fa[2,2] Fa[2,0] y2 , we have 2 2 [2] Fa[3,5] = a2 (Π[1] a[2,4] + Πa[2,4] ) ⊗ y1 = 3a1 a2 y2 y3 .

(5.2.17)

In order to determine Fa[2,6] only necessary to do Πa[1,5] = Fa[2,4] y1 + Fa[3,3] y2 , we have Fa[2,6] = a2 (Fa[2,4] y1 + Fa[3,3] y2 ) ⊗ y1 = a1 a32 y23 + a21 a22 y32 .

(5.2.18)

Lemma 5.2.4. For an integer m ≥ 3, Π[m−1] = Π[m−2] ⊗ Πa[1,1] . a[m−1,m−1] a[m−2,m−2] Proof. For any integer m ≥ 3, Lemma 5.2.2 yields [m−2] Π[m−1] a[m−1,m−1] = Πa[m−2,m−2] ⊗ Πa[1,1] .

Therefore, the conclusion can be drawn. This lemma helps us to determine Fa[m,m] for m ≥ 1 in a much more simple way. Corollary 5.2.5. For any integer m ≥ 2, Fa[m,m] = am 1 ym . = 0. When k = 1, Proof. First, we prove that, for integers 1 ≤ k ≤ m − 2, Π[k] a[m−1,m−1] = 0. from Corollary 5.2.3, Πa[m−1,m−1] = 0. This implies Π[1] a[m−1,m−1]

= 0. = 0. By induction on k, we prove Π[k] When k ≥ 2, assume Π[k−1] a[m−1,m−1] a[m−1,m−1]

= 0. We proceed on the basis of As a matter of fact, Corollary 5.2.3 results in Π[k] a[m−1,m−1]

this and (5.1.12). For m ≥ 3, Lemma 5.2.4 yields Fa[m,m] = Π[m−1] ⊗y1 = (Π[m−2] ⊗ a[m−1,m−1] a[m−2,m−2] Πa[1,1] ) ⊗ y1 . Then, on account of Fa[2,2] = a1 a2 y2 , Fa[3,3] = a21 a2 y3 and Fa[4,4] = a31 a2 y4 . For any integer m ≥ 5, we prove Fa[m−1,m−1] = am−2 a2 ym−1 . By induction, we prove Fa[m,m] = 1 [m−2] m−2 = a y am−1 a y . Because of Π m−2 , the assumption results in Fa[m,m] = 2 m 1 1 a[m−2,m−2] a2 (Π[m−2] × Πa[1,1] ) ⊗ y1 = am−1 1 a2 ym . Therefore, the conclusion can be drawn. a[m−2,m−2]

160 | 5 Inner equations third part We proceed on the basis of what was just mentioned above; the solution of equation (5.1.1) can be further simplified to get the coefficient of each term in the form of a finite sum with all terms positive. [gci] = Theorem 5.2.6. Let f[gci] be the solution of equation (5.1.1) determined by Fa[m,s] m 𝜕x f[gci] |s=π(n) which is of the form of

[gci] Fm,s

a1 , when s = 0; { { }, {a Π ⊗ y , when s≥1 1 = { 2 a[1,s−1] { { ⊗ Πa[j,r] )) ⊗ y1 , a ∑m−1 (∑ 1≤r≤s−1 (Π[k−1] a[m−j,s−r] { 2 k=1 1≤j≤m−2

when m = 2; when m ≥ 3,

(5.2.19)

where Πa[j,r]

[gci] r when j = 1; {∑l=1 Fa[l,r−l] yl , ={ [gci] r {∑l=1 Fa[l−j,r−l] yl , when j ≥ 2,

(5.2.20)

for 1 ≤ j ≤ m − 2, 1 ≤ r ≤ s − 1, m ≥ 2 and s ≥ 0 such that m ≤ s and m = s(mod 2). [gci] ∈ Proof. When m+s is smaller, from (5.2.2)–(5.2.8), it is shown how to determine Sa[m,s] ℛ{y} from the initial condition of equation (5.1.1). Then, for general m + s greater, we have l + (r − l) = r ≤ s − 1 < m + s and (l − j) + (r − l) = r − j ≤ (s − 1) − 1 = s − 2 < m + s. [gci] [gci] By the inductive principle, all Sa[l,r−l] and Sa[l−j,r−l] for 1 ≤ l ≤ r, and hence Πa[j,r] and

are known. Thus, the conclusion can be drawn. Π[k−1] a[m−j,s−r]

This theorem enables us to find a number of corollaries. Only one is listed as an example. [gci] for m ≥ 2 and s ≥ 0 are polynomials with all Corollary 5.2.7. If a ∈ ℤ2+ , then all Fa[m,s] coefficients in ℤ+ .

Proof. By following the procedure in the proof of Theorem 5.2.6, the conclusion can be drawn.

5.3 Explicitness general cutless inner case Because no explicision has yet directly been deduced from Theorem 5.2.6, the relationship between the solutions of general equation (5.1.1) and its restriction for a0 = 0, a1 = a2 = 1, i. e., a = (a1 , a2 ) = (1, 1) = 1, has to be investigated by considering the latter to be meaningful in graph theory. Let fa[gci] and f[gci] ∈ ℛ{x, y} be the solutions of, respectively, equation (5.1.1) and [gci] [gci] and Fm,s ∈ ℛ{y} for integers m, s ≥ 0. Detheir restrictions determined by Fa[m,s] [gci] [gci] note by 𝒩m,s the set of power vectors in Fa[m,s] and hence Fm,s . For any n ∈ 𝒩m,s ,

5.3 Explicitness general cutless inner case

| 161

n | ∈ ℛa = ℛ{a} and A[gci] A[gci] m,n |s=π(n) ∈ ℛ are the coefficients of y in, respeca[m,n] s=π(n)

[gci] [gci] and Fm,s . Then tively, Fa[m,s]

[gci] | = 𝜕yn Fa[m,s] A[gci] a[m,n] π(n)=s

n [gci] and A[gci] m,n |π(n)=s = 𝜕y Fm,s .

is a polyTheorem 5.3.1. For any integers m, s ≥ 1 and integral vector n ∈ 𝒩m,s , A[gci] a[m,n]

= A[gci] nomial of a with degree at most l + 1 where 2l = m + s such that A[gci] m,n . 1[m,n]

Proof. We proceed on the basis of Theorem 5.2.6. By induction, the conclusion can be drawn. For an embedding μ(G) of a general cutless planar graph G, let n = (n1 , n2 , n3 , . . .) be the vertex-partition vector where ni is the number of y-vertices (or non-root-vertices) with valency i for integer i ≥ 1. A vertex v of G whose deletion makes that μ(G) − v has at least one component greater than G is called a cut-vertex of G. No cutless graph has any cut-vertex. Given a planar embedding μ = μ(G) of a general cutless graph G (connected!). If two vertices u and v have the property that μ(G)−{pu , pv } have at least two components, then {u, v} is called a splitting pair. Since an embedding is a point set in its own right on the plane, pu and pv are as the corresponding points of u and v. Because μ is a point set, any component of μ − {pu , pv } is called a splitting slice, or slice. A splitting slice that is not a path is called a splitting chunk, or chunk. Let S be the set of all splitting pairs consisting of {ui , vi } for 1 ≤ i ≤ s = |S| in μ = μ(G). It is known that all components of μ − S are splitting slices in μ(G). For a splitting pair, its slice valent is the number of splitting slices incident with. Similarly, for chunk valency. Observation 5.3.2. Let t and 1 ≤ p ≤ t be, respectively, the slice valency and chunk valency of a splitting pair on a general cutless planar embedding of graph G, then the splitting pair produces [gci] πspp = 2p−1 (t − 1)!

(5.3.1)

(topologically) distinct planar embeddings of G. Proof. We have the independency between rotation on slice valency and reflection on a chunk. By considering the connectivity of embedding, the conclusion can be drawn. Let ci be the number of splitting chunks and ti , the slice valency at the splitting pair {ui , vi }, 1 ≤ i ≤ s on μ(G). Lemma 5.3.3. For a general cutless planar graph G, the number of topologically distinct embeddings is l

c−s n[gci] ∏((j − 1)!) j Nct = 2 j≥2

(5.3.2)

162 | 5 Inner equations third part where c = c1 + c2 + ⋅ ⋅ ⋅ + cs and lj is the number of splitting pairs with slice valency j ≥ 2. Proof. We proceed on the basis of Lemma 5.3.2. Since s

s

i=1

i=1

l

ci −1 (ti − 1)!) = 2c−s ∏ ((t i − 1)!) = 2c−s ∏((j − 1)!) j , n[gci] Nct = ∏ (2 j≥2

this is the conclusion. This lemma is a refinement and a concise form of Theorem 10.4.3 with (10.4.13) in Liu YP [72] (2017, pp. 196–197), or Theorem 7.4.3 with (7.4.13) in Liu YP [55] (1994, p. 142). We proceed on the basis of Lemma 5.3.3 and Lemma 5.3.4, the result is useful in the context of determining the number of general cutless planar embeddings of a graph. Lemma 5.3.4. For a general cutless planar graph G, the number of planar embeddings of G is n[gci] (G) = n[gci] Nct (G)

(5.3.3)

where nNct (G) is given by (5.3.2). Proof. We consider the uniqueness of 3-connected embedding of a graph. Because of there being neither cut-vertex nor splitting pair of two vertices in a graph, the graph is 3-connected. All planar embeddings of a cutless graph are generated in the way shown in Lemma 5.3.3. ̃ Denote by t = aut(G) the order of the semi-automorphism of graph G. The relationship between embeddings and upper maps of G can be shown via t. Lemma 5.3.5. Let ℰ [gci] (G) and ℳ[gci] (G) be, respectively, the sets of all distinct topological planar embeddings and root-isomorphic upper maps of G. Then 󵄨󵄨 [gci] 󵄨󵄨 2ϵ 󵄨󵄨 [gci] 󵄨󵄨 (G)󵄨󵄨 (G)󵄨󵄨 = 󵄨󵄨ℰ 󵄨󵄨ℳ t

(5.3.4)

where ϵ = ϵ(G) is the size of G. Proof. Refer to Liu YP [59] (2003, pp. 225–226). [gci] be the set of all general cutless planar graphs with rooted semi-valency Let 𝒢m|s m (i. e., the number of semi-edges incident to the root-vertex) and the un-rooted semisize s (i. e., the number of semi-edges not incident to the root-vertex). [gci] [gci] Denote by 𝒯m,s the set of semi-automorphism group orders of graphs in 𝒢m,s . Let [gci] [gci] 𝒥m,s be the set of all partition vectors n in 𝒢m,s .

5.3 Explicitness general cutless inner case

| 163

[gci] Lemma 5.3.6. Let Em,n be the number of distinct (topologically) embeddings of graphs [gci] in 𝒢m,n . Then we have [gci] Em,n =

∑ n[gci] (G)

(5.3.5)

[gci] G∈𝒢m,n

where n[gci] given in (5.3.3). [gci] Proof. By the partition of all embeddings on 𝒢m,s from each graph, the conclusion can be drawn.

We proceed on the basis of Lemma 5.3.6. Let 𝒯m,n be the set of semi-automorphism [gci] [gci] [gci] and 𝒢m,s;t with semigroup orders for graphs in 𝒢m,s , the set of all graphs in 𝒢m,s ̃ automorphism group order t. Denote by aut(G) the semi-automorphism group order of G. Lemma 5.3.7. The number of root-isomorphic classes of upper maps of all graphs in

[gci] 𝒢m,s is

= ∑ A[gci] m|n

t∈𝒯n

m + π(n) [gci] Em,n;t t

(5.3.6)

[gci] where A[gci] m,n = Am,n (𝒢m,s ), s = π(n) and [gci] Em,n;t =

∑ ngoi (G)|aut(G)=t ̃

(5.3.7)

[gci] G∈𝒢m,s

as determined by (5.3.5). Proof. We proceed on the basis of Lemma 5.3.5. By considering m + π(n) = 2ϵ, the conclusion can be drawn. This lemma enables us to turn out the solution fa[gci] of equation (5.1.1) when [gci] a0 = 0, a1 = a2 = 1. Denote f[gci] = fa[gci] |a=1 as determined by Fm,s = 𝜕xm f[gci] |π(y)=s for m, s ≥ 0. Lemma 5.3.8. For integers m, n ≥ 1, we have [gci] Fm,s =

∑ [gci] n∈𝒥m,s

n A[gci] m,n y

(5.3.8)

where A[gci] m,n is given by (5.3.6). Proof. From Lemma 5.3.7, the conclusion can be drawn. We proceed on the basis of Lemma 5.3.8 and Theorem 5.3.1; we are allowed to state our main result of this section.

164 | 5 Inner equations third part [gci] determines the solution of equation (5.1.1), Theorem 5.3.9. For m, s ≥ 1, Ga[m,s] [gci] = Fa[m,s]

∑ [gci] n∈𝒥m,s

yn A[gci] a[m,n]

(5.3.9)

= is a polynomial of a with degree at most 1 + (m + s)/2 such that A[gci] where A[gci] 1[m,n] a[m,n]

A[gci] m,n .

Proof. This is a result of Theorem 5.3.1 and Lemma 5.3.8. From (5.3.2), nNct is summation-free. From (5.3.3), n[gci] is summation-free. Be[gci] cause A[gci] m,s is determined by n[gci] , Fm,s given by (5.3.8) is a direct explicision and [gci] hence so is Fa[m,s] in (5.3.9). [gci] [gci] |a=1 determine the solution of equation Corollary 5.3.10. For m, s ≥ 1, Fm,s = Fa[m,s] (5.1.1) when a = 1, or equation (5.4.1), as [gci] = Fm,s

∑ [gci] n∈𝒥m,s

[gci] n y Um,n

(5.3.10)

[gci] where Um,n = A[gci] m,n is given in (5.3.6).

Proof. This is a direct result of Theorem 5.3.9. [gci] This is a direct explicision (explicit expression) of Fm,s for integers m, s ≥ 1.

5.4 Restrictions general cutless inner case Consider the equation for f y𝜕x,y f |x=u { { = f − x2 ; {x ∫ 1 − 𝜕 f | x,y x=u y { { { {f |x=0⇒y=0 = 0,

(5.4.1)

where f ∈ ℛ{x, y} and y = (y1 , y2 , y3 , . . .). This is equation (5.1.1) when a0 = 0 and a1 = a2 = 1 for general cutless planar maps as shown in [28] (Liu YP, 1985) and [56] (Liu YP, 1999, p. 187). From f ∈ ℛ{x, y}, f is determined by Fm = 𝜕xm f ∈ ℛ{y} for m ≥ 0. We proceed on the basis of 𝜕x,y f |u=x = ∑(∑ Fl−1 yl−i−1 )xi . i≥1 l≥i

We have y𝜕x,y f |u=x = ∑(∑ Fl−1 yl−i )xi . i≥1 l≥i

(5.4.2)

5.4 Restrictions general cutless inner case

| 165

In the part under the meson functional, because of 1

1 − 𝜕x,y f |u=x we have

y𝜕x,y f |u=x

1 − 𝜕x,y f |u=x

= ∑ (𝜕x,y f |u=x )j , j≥0

= ∑ y(𝜕x,y f |u=x )j . j≥1

By (5.4.2), y𝜕x,y f |u=x

1 − 𝜕x,y f |u=x

j

= ∑ y(∑ ( ∑ Fl−1 yl−i−1 )xi ) . j≥1

i≥0 l≥i+1

(5.4.3)

For convenience, denote ∇m = 𝜕xm (𝜕x,y f |u=x ) and

∇m[k]

0, when k = 0; { { { = {∇m , when k = 1; { { m [k−1] {∑j=0 ∇m−j ∇j , when k ≥ 2,

(5.4.4)

for any integers k ≥ 1 and m ≥ 0. It is easily seen that, for any integers k ≥ 1 and m ≥ 0, ∇m[k] = 𝜕xm (𝜕x,y f |u=x )k . By (5.4.2), ∇m = ∑ Fl−1 yl−m−1 , { { { { l≥m+1 m { { [k−1] [k] { {∇m = ∑ ∇m−j ∇j , j=0 {

when k = 1; when k ≥ 2,

(5.4.5)

for any integer m ≥ 1. For any integers k ≥ 1 and m ≥ 0, let {∫y (y∇m ), when k = 1; △[k] m ={ ∫ (y∇[k] ), when k ≥ 2. { y m

(5.4.6)

[k] Thus, for any integers k ≥ 1 and m ≥ 0, △[k] m = 0 ⇔ ∇m = 0.

Theorem 5.4.1. Equation (5.4.1) is equivalent to the system of equations for Fm for m ≥ 0 0, { { { { { {∑ △[k] , Fm = { k≥1 0 [k] { 1 + ∑k≥1 △1 , { { { { [k] {∑k≥1 △m−1 , on ℛ{x, y}.

when m = 0; when m = 1; when m = 2; when m ≥ 3,

(5.4.7)

166 | 5 Inner equations third part Proof. See the proof of Theorem 5.1.2 for a = 1. This theorem enables us to only consider the system of equations shown in equations (5.4.7) instead of equation (5.4.1) itself. Because of each equation in equations (5.4.7) is infinite, we are required to seek surplus equations and some other parameter than m so that, for any specified value, the system of equations (5.4.7) becomes finite. Lemma 5.4.2. Given any integer k ≥ 1, △[k] m = 0 for 0 ≤ m ≤ k. Proof. When k = 1, because of F0 = 0 ⇒ △[k] 0 = 0 for k ≥ 1, F1 = 0. [t] When k ≥ 2, assume △m = 0 (0 ≤ m ≤ t), for any 1 ≤ t ≤ k − 1. By induction on k, we prove the conclusion for t = k. We proceed on the basis of (5.4.5) and (5.4.6). By the assumption, the conclusion can be easily drawn. This lemma enables us unnecessary to consider terms △[k] m for k ≥ m in the system of equations (5.4.7) for any integer m ≥ 1. This implies that equations (5.4.7) becomes, for any integer m ≥ 0, 0, when m ≤ 1; { { { Fm = {1 + △1 , when m = 2; { { m−1 [k] {∑k=1 △m−1 , when m ≥ 3.

(5.4.8)

From equations (5.4.8), it is seen that all infinite sum in equations (5.4.7) become finite. For Fm ∈ ℛ{y}, let 𝒥m be the set of power vectors of all terms in Fm . For any n ∈ 𝒥m , denote by π(n) = (∑ in[i] )nT i≥1

the inner semi-size where n[i] is the vector of all entries 0 except only for the ith entry 1. Because of y = (y1 , y2 , y3 , . . .), π(n) = ∑ ini . i≥1

(5.4.9)

Observe the case of m = 2 for (5.4.8). From (5.4.5) and (5.4.6), △1 = ∫(∑ Fl−1 yl−2 ) = ∑ Fl−1 yl−2 . y

l≥2

l≥2

(5.4.10)

Because of the infinity of equation (5.4.10), equation (5.4.8) is infinite as well. Observation 5.4.3. For integer m ≥ 0, Fm is independent of y0 and y1 . Proof. Because no y0 appears in y, Fm is independent of y0 . We proceed on the basis of (5.4.7). By induction, the independency of y1 in Fm can be seen.

5.4 Restrictions general cutless inner case

| 167

From this observation, △1 = ∑ Fl−1 yl−1 .

(5.4.11)

0, when m ≤ 1; { { { Fm = {1 + ∑l≥3 Fl−1 yl−2 , when m = 2; { { m−1 [k] when m ≥ 3. {∑k=1 △m−1 ,

(5.4.12)

l≥3

Thus, (5.4.8) becomes

For any integer s ≥ 0, denote Fm,s = ⟨Fm ⟩s . Because of Fm ∈ ℛ{y}, Fm,s = ∑ Fm,n yn n∈𝒥m π(n)=s

(5.4.13)

where Fm,n ∈ ℛ. Observe F2,s . From the case of m = 2 in (5.4.12), 1, when s = 0; F2,s = { s ∑l=1 Fl,s−l yl , when s ≥ 1.

(5.4.14)

In order to determine Fm,s for m ≥ 3, it is only necessary to get △[k] m−1,s for 0 ≤ k ≤

m − 1 from (5.4.7) and Lemma 5.4.2. By (5.4.6), observe ∇m[k] . From (5.4.5), m

[k−1] ∇m[k] = ∑ ∇m−j ( ∑ Fl−1 yl−j−1 ). j=0

l≥j+1

(5.4.15)

For convenience, k ≥ 1 and m ≥ 0 are considered. Let [k] Π[k] m = ∫ ∇m .

(5.4.16)

y

From (5.4.5), denote Πm = Π[1] m . Then we have ∑l≥m+1 Fl−1 yl−m−1 , when k = 1; Π[k] m ={ m ⊗ Πj , when k ≥ 2. ∑j=0 Π[k−1] m−j

(5.4.17)

By (5.4.6), △[k] m = y1 ⊗ Πm . We proceed on the basis of (5.4.17). From the property of convolution, when k = 1; ∑l≥1 Fl+m−1 yl , △[k] m ={ m [k−1] ∑j=0 Πm−j ⊗ (∑l≥j+1 Fl−j−1 yl ), when k ≥ 2.

168 | 5 Inner equations third part From l − m − 1 ≥ 0, ∇0 = 0 ⇒ Π0 = 0 and Π[k] 0 = 0 for k ≥ 1, when k = 1; ∑l≥1 Fl+m−1 yl , △[k] m = { m−1 [k−1] ∑j=1 Πm−j ⊗ (∑l≥j+1 Fl−j−1 yl ), when k ≥ 2.

(5.4.18)

On this basis, when k = 1, for any integers m ≥ 2 and s ≥ 0, s

△m,s = ∑ Fl+m−1,s−l yl . l=1

When k ≥ 2, for two integers m ≥ 2 and s ≥ 0, m−1 s−1

r

j=1 r=1

l=1

[k−1] △[k] m,s = ∑ ∑ n(Πm−j,s−r ⊗ ∑ Fl+j−1,r−l yl ).

By combining the two formulas above, when k = 1; ∑sl=m+1 Fl−m−1,s−l yl , △[k] = { m,s r m−1 s−1 [k−1] ∑j=1 ∑r=1 (Πm−j,s−r ⊗ ∑l=1 Fl+j−1,r−l yl ), when k ≥ 2.

(5.4.19)

Theorem 5.4.4. Equation (5.4.1) for f ∈ ℛ{x, y} is equivalent to the system of equations for Fm,s ∈ ℛ{y} for integers m ≥ 2 and s ≥ 0 as Fm,s

1, when s = 0; { { } { s F y , ∑ = { l=2 l,s−l l when s ≥ 1, { r {∑ 1≤r≤s−1 ((∑m−1 Π[k−1] k=2 m−j−1,s−r ) ⊗ ∑l=1 Fl+j−1,r−l yl ), { 1≤j≤m−2

when m = 2; when m ≥ 3.

(5.4.20)

Proof. See the proof of Theorem 5.1.5 for a = 1. We proceed on the basis of equations (5.4.20). For m ≥ 2 and s ≥ 0, all of Fm,s = 0 are omitted. Lemma 5.4.5. Given any integers m ≥ 2 and s ≥ 0. If Fm,s ≠ 0, then m = s(mod 2). Proof. This is a result of Lemma 5.1.6 for a = 1. Furthermore, it is still necessary to discuss that, for any integers m, s ≥ 0, what in y are there Fm,s depends on. Lemma 5.4.6. For any integers m ≥ 2 and s ≥ 0, Fm,s (≠ 0) is independent of all yl , l ≥ s + 1. Proof. This is a result of Lemma 5.1.7 for a = 1. This lemma enables us to restrict ourselves only on Fm,s in the scope of y = ys = (y1 , y2 , . . . , ys ) for m ≥ 2 and s ≥ 1.

5.4 Restrictions general cutless inner case

| 169

Theorem 5.4.7. Equation (5.4.1) is well defined on ℛ{x, y}. Proof. This is a direct result of Theorem 5.1.8 for a0 = 0. We proceed on the basis of equations (5.4.20). In order to evaluate the solution of equation (5.4.1), some available structures of the solution have to be investigated. Lemma 5.4.8. Let f[gci] ∈ ℛ{x, y} be the solution of equation (5.4.1). Denote [gci] (m,n) Fm,s = ∑ (𝜕(x,y) f[gci] )yn , n∈𝒥m π(n)=s

(5.4.21)

[gci] then Fm,s ∈ ℛ+ {y}. [gci] Proof. From Theorem 5.4.7, Fm,s = Fm,s satisfies equations (5.4.20). By employing the procedure in the proof of Theorem 5.4.7, it is shown that all coefficients of yn are in ℝ+ . This is the conclusion.

This lemma enables us to obtain the solution with the coefficient of each term in the form of a sum of positive real numbers. Lemma 5.4.9. Given any two integers m ≥ 1 and s ≥ 0. If m ≥ s + 1, then Π[k] m,s = 0 for [k] any integer k ≥ 1 where Πm,s is shown in (5.4.16) and (5.4.17). Proof. See the proof of Lemma 5.2.2 for a = 1. This lemma also shows us that, for any integers m ≥ s + 1, Fm,s = 0. Corollary 5.4.10. For any integers m ≥ 2 and s ≥ 0, if m ≥ s + 1, then Fm,s = 0 except only F2,0 = 1. Proof. On the basis of (5.4.12) and (5.4.18), the conclusion is a direct result of Lemma 5.4.9. We proceed on the basis of the last two lemmas. A compact form of the solution of equation (5.4.1) can be evaluated. In order to do this, the cases of m + s ≤ 8 need to be firstly observed. From Lemma 5.4.5, it is only necessary to observe the cases of m = s(mod 2). By (5.4.17), when k = 1, on m = 1 and 0 ≤ s ≤ 7, because of Π1,s =

⌊(s+3)/2⌋

∑ l=2

Fl−1,s+2−l yl−2 ,

we have

Π1,s

F2,0 y1 , { { { { { {F2,2 y1 , ={ {F2,4 y1 + F3,3 y2 , { { { { {F2,6 y1 + F3,5 y2 + F4,4 y3 ,

when s = 1; when s = 3; when s = 5; when s = 7.

(5.4.22)

170 | 5 Inner equations third part On m = 2 and 0 ≤ s ≤ 7, because of Π2,s =

⌊(s+4)/2⌋

∑ l=3

Fl−1,s+3−l yl−3 ,

we have

Π2,s

F2,0 y0 , when s = 0; { { { { { {F2,2 y0 , when s = 2; ={ {F2,4 y0 + F3,3 y1 , when s = 4; { { { { {F2,6 y0 + F3,5 y1 + F4,4 y2 , when s = 6.

(5.4.23)

On m = 3 and 0 ≤ s ≤ 7, because of Π3,s =

⌊(s+5)/2⌋

∑ l=4

Fl−1,s+4−l yl−4 ,

we have

Π3,s

0, { { { { { {F3,3 y0 , ={ { F3,5 y0 + F4,4 y1 , { { { { {F3,7 y0 + F4,6 y1 + F5,5 y2 ,

when s = 1; when s = 3; when s = 5;

(5.4.24)

when s = 7.

On m = 4 and 0 ≤ s ≤ 7, because of Π4,s =

⌊(s+6)/2⌋

∑ l=5

Fl−1,s+5−l yl−5 ,

we have 0, when s = 0 and s = 2; { { { Π4,0 = {F4,4 y0 , when s = 4; { { {F4,6 y0 + F5,5 y1 , when s = 6.

(5.4.25)

On m = 5 and 0 ≤ s ≤ 7, because of Π5,s =

⌊(s+7)/2⌋

∑ l=6

Fl−1,s+6−l yl−6 ,

we have Π5,s

0, { { { = {F5,5 y0 , { { {F5,7 y0 + F6,6 y1 ,

when s = 1 and s = 3 when s = 5; when s = 7.

(5.4.26)

5.4 Restrictions general cutless inner case | 171

On m = 6 and 0 ≤ s ≤ 7, because of Π6,s =

⌊(s+8)/2⌋

∑ l=7

Fl−1,s+7−l yl−7 ,

we have 0, when s = 5 s = 0, 2 and 4; Π6,s = { F6,6 y0 , when s = 6.

(5.4.27)

On m = 7 and 0 ≤ s ≤ 7, because of Π7,s =

⌊(s+9)/2⌋

∑ l=8

Fl−1,s+8−l yl−8 ,

we have 0, Π7,s = { F7,7 y0 ,

when s = 1, 3 and 5; when s = 7.

(5.4.28)

We proceed on the basis of calculations above; we address all Fm,s for m + s ≤ 8 when m ≥ 2 and s ≥ 0. When m + s = 2, it is only necessary to do F2,0 . From the initiation, F2,0 = 1. When m + s = 4, only we do F4,0 and F2,2 . From Corollary 5.4.10, F4,0 = 0. Because of Π1,1 = F2,0 y1 = y1 , F2,2 = y1 ⊗ y1 = y2 . Π[2] 2,2

(5.4.29)

When m + s = 6, only we do F6,0 = 0, F3,3 and F2,4 . Because of Π[1] 2,2 = F2,2 y0 = 0 and = Π1,1 ⊗ Π1,1 = y1 ⊗ y1 = y2 ,

Because of Πa[1,3] = Fa[2,2] y1 ,

F3,3 = y2 ⊗ y1 = y3 .

(5.4.30)

F2,4 = F2,2 y2 = y22 .

(5.4.31)

When m + s = 8, from Corollary 5.4.10, it is only necessary to observe F4,4 , F3,5 [2] [3] and F2,6 . Because of Π[1] 3,3 = F3,3 y0 = 0, Π3,3 = Π1,1 ⊗ Π2,2 = 0 and Π3,3 = y1 ⊗ y1 ⊗ y1 = y3 , we have 3

F4,4 = ∑ Π[k] 3,3 ⊗ y1 = y3 ⊗ y1 = y4 . k=1

(5.4.32)

[2] In order to determine F3,5 , it is only necessary to do Π[1] 2,4 = Π2,4 and Π2,4 . Because of

Π2,4 = F2,4 y0 +F3,3 y1 = F3,3 y1 and Π[2] 2,4 = Π1,3 ⊗Π1,1 +Π1,1 ⊗Π1,3 = 2(Π1,3 ⊗Π1,1 ) = 2F2,2 F2,0 y2 , we have [2] F3,5 = (Π[1] 2,4 + Π2,4 ) ⊗ y1 = 3y2 y3 .

(5.4.33)

172 | 5 Inner equations third part In order to determine F2,6 , it is only necessary to do Π1,5 = F2,4 y1 + F3,3 y2 ; we have F2,6 = (F2,4 y1 + F3,3 y2 ) ⊗ y1 = y23 + y32 .

(5.4.34)

[m−2] Lemma 5.4.11. For an integer m ≥ 3, Π[m−1] m−1,m−1 = Πm−2,m−2 ⊗ Π1,1 .

Proof. For any integer m ≥ 3, [m−2] Π[m−1] m−1,m−1 = Πm−2,m−2 ⊗ Π1,1 .

Therefore, the conclusion can be drawn. This lemma helps us to determine Fm,m for m ≥ 1 in a much more simple way. Corollary 5.4.12. For any integer m ≥ 2, Fm,m = ym . Proof. See the proof of Corollary 5.4.12 for a = 1. We proceed on the basis of what was just mentioned above; the solution of equation (5.4.1) can be further simplified to get the coefficient of each term in the form of a finite sum with all terms positive. [gci] = Theorem 5.4.13. Let f[gci] be the solution of equation (5.4.1) determined by Sm,s m 𝜕x f[gci] |s=π(n) which is of the form of

[gci] Sm,s

1, when s = 0; { { }, { Π ⊗ y , when s≥1 1 = { 1,s−1 { {∑m−1 (∑ [k−1] ⊗ Πj,r )) ⊗ y1 , 1≤r≤s−1 (Π m−j,s−r 1≤j≤m−2 { k=1

when m = 2; when m ≥ 3,

(5.4.35)

where [gci] r {∑l=1 Sl,r−l yl , Πa[j,r] = { [gci] r {∑l=1 Sl−j,r−l yl ,

when j = 1; when j ≥ 2,

(5.4.36)

for 1 ≤ j ≤ m − 2, 1 ≤ r ≤ s − 1, m ≥ 2 and s ≥ 0 such that m ≤ s and m = s(mod 2). Proof. This is a direct result of Theorem 5.2.6 for a = 1. This theorem enables us to find a number of corollaries. Only one is listed as an example. [gci] Corollary 5.4.14. For any two integers m ≥ 2 and s ≥ 0, Sm,s is a polynomial with all coefficients in ℤ+ .

Proof. By following the procedure in the proof of Theorem 5.4.13, the conclusion can be drawn.

5.5 Eulerian cutless inner model | 173

Figure 5.4.1: Root-isomorphic classes of general cutless planar maps.

Example 5.4.1. Root-isomorphic classification of general cutless planar maps by rootvertex valency and vertex-partition vector given. A map is called cutless (or nondetachable) if it has a vertex v = (x1 , x2 , x3 , . . .) with i < j such that v1 = (xi , xi+1 , . . . , xj−1 ) and v2 = (xj , xj+1 , . . . , xi−1 ) generate two maps. The vertex map is guaranteed as an exclusion of such maps. Figure 5.4.1 shows the root-isomorphic classes of general cutless planar maps with size not greater than 4. For instance, a = 1 = F2,0 , b = y2 = F2,2 , c = y22 = F2,4 , d = y3 = F3,3 , e + f = y23 + y32 = F2,6 , g = 3y2 y3 = F3,5 and h = y4 = F4,4 .

5.5 Eulerian cutless inner model Second, consider the following equation for f : y2 δx2 ,y2 f |x2 =u 2 { { = f − a1 x 2 ; {a 2 x ∫ (1 − 𝜕x2 ,y2 f |x2 =u )2 − (xyδx2 ,y2 f |x2 =u )2 y { { { f | { x=0⇒y=0 = a0 ,

(5.5.1)

where a0 , a1 , a2 ∈ ℛ+ , f = f (x2 , y) ∈ ℛ{x, y} and y = (y2 , y4 , y6 , . . .). This is equation (10) in Introduction. Because a solution of equation (5.5.1) for a0 = 0 and a1 = a2 = 1 is involved with Eulerian cutless planar maps as shown in [32] (Liu YP, 1986), and then [60] (Liu YP, 2008, p. 165), this equation is called a Eulerian cutless inner model.

174 | 5 Inner equations third part Since no term has odd degree of x in the first equality of equation (5.5.1), f is an even function of x. This enables us to convert f into a function of u = x2 . For any integer m ≥ 0, denote Fa[m] = 𝜕um f = 𝜕x2m f . Observation 5.5.1. For any solution f of equation (5.5.1), 𝜕x0 f = a0 = 0. Proof. On the right hand side of the first equality in equation (5.5.1), no constant term occurs. This implies 𝜕x0 f = Fa[0] = 0. However, the initial condition shows Fa[0] = a0 . Therefore, a0 = 0. This observation enables us to consider a = (a1 , a2 ) instead of (a0 , a1 , a2 ). For convenience, equation (5.5.1) has to be transformed into a suitable form. Let ⊤a = δx2 ,y2 f |u=x2 and ∇a = 𝜕x2 ,y2 f |u=x2 , then δx2 ,y2 f |u=x2

(1 − 𝜕x2 ,y2 f |u=x2 )2 − (xyδx2 ,y2 f |u=x2 )2 We have

=

⊤a . (1 − ∇a )2 − (xy⊤a )2

(5.5.2)

2xy⊤a 1 1 − = , 1 − ∇a − xy⊤a 1 − ∇a + xy⊤a (1 − ∇a )2 − (xy⊤a )2

we have δx2 ,y2 f |u=x2

(1 − 𝜕x2 ,y2 f |u=x2 )2 − (xyδx2 ,y2 f |u=x2 )2 1 = ∑ ((∇ + xy⊤a )k − (∇a − xy⊤a )k ). 2xy k≥1 a

(5.5.3)

Because of (∇a + xy⊤a )k − (∇a − xy⊤a )k k =2 ∑ ( )(xy⊤a )2r+1 ∇ak−2r−1 , 2r + 1 r=0 ⌊k/2⌋

we have δx2 ,y2 f |u=x2

(1 − 𝜕x2 ,y2 f |u=x2 )2 − (xyδx2 ,y2 f |u=x2 )2 k k−2r−1 )(xy)2r ⊤2r+1 . =∑ ∑ ( a ∇a 2r +1 k≥1 r=0 ⌊k/2⌋

We proceed on the basis of ⊤a = ∑ Fa[i] i≥0

x2i − y2i = ∑ ∑ Fa[i] x2j y2(i−1)−2j x2 − y2 j≥0 i≥j+1

(5.5.4)

5.5 Eulerian cutless inner model | 175

and ∇a = ∑ Fa[i] i≥0

y2 x2i − x2 y2i = ∑ ∑ Fa[i] x2(j+1) y2(i−1)−2j . x2 − y2 j≥0 i≥j+2

For integer m ≥ 1 (note that m = 0 makes no sense in δx,y and 𝜕x,y !), let Ta[m] = 𝜕x2m ⊤a = 2m 2m [⊤a ]2m x and Pa[m] = 𝜕x ∇a = [∇a ]x , then from the initial condition Fa[0] = 0 of equation (5.5.1), T = ∑ F y2(i−m−1) ; { { { a[m] i≥m+1 a[i] { 2(i−m) { . {Pa[m] = ∑ Fa[i] y i≥m+1 {

(5.5.5)

Furthermore, for any integer l ≥ 1, denote La[l] = ∫ ⊤la y

and

Δa[l] = ∫ ∇al .

(5.5.6)

y

Theorem 5.5.2. Equation (5.5.1) is equivalent to k { { )x2r (y2r+2 ⊗ La[2r+1] ⊗ Δa[k−2r−1] ); {f = a1 x2 + a2 x2 ∑ ∑ ( 2r + 1 r=0 k≥1 { { { {f |x=0⇒y=0 = 0, ⌊k/2⌋

(5.5.7)

on ℛ{x, y} where ⨂ is the convolution of two vectors in 𝒱 (Section 2.5 of Book I). Proof. We proceed on the basis of Observation 5.5.1. By substituting (5.5.4) for the first equality of equation (5.5.1), k k−2r−1 )x2r y2(r+1) ⊤2r+1 f = a1 x2 + a2 x2 ∫ ∑ ∑ ( a ∇a 2r +1 k≥1 r=0 ⌊k/2⌋

y

k = a1 x2 + a2 x2 ∑ ∑ ( )x2r (y2(r+1) ⊗ La[2r+1] ⊗ Δa[k−2r−1] ). 2r +1 k≥1 r=0 ⌊k/2⌋

This is the first equality in equation (5.5.7). Therefore, the conclusion can be drawn. This theorem suggests us to evaluate La[2r+1] and Δa[k−2r−1] in ℛ{x, y} before determining Fa[m] for m ≥ 1. Because of [⊤a ]2m when r = 0; 2m x , [⊤2r+1 ] = { a x m 2r 2(m−i) 2i [⊤a ]x , when r ≥ 1, ∑i=0 [⊤a ]x

(5.5.8)

176 | 5 Inner equations third part and [∇a ]2m , when r = ⌊k/2⌋; 2m [∇ak−2r−1 ]x = { m x k−2r−2 2(m−i) 2i ]x [∇]x , when 0 ≤ r ≤ ⌊k/2⌋ − 1, ∑i=0 [∇a

(5.5.9)

for any integers r ≥ 0 and m ≥ 0. From (5.5.5), we have, for 0 ≤ r ≤ ⌊k/2⌋, Ta[m] , when r = 0; 2m { { [⊤2r+1 { a ]x = { m 2r 2(m−i) { { Ta[i] , otherwise, ∑i=0 [⊤a ]x { { and { { { { P , when r = ⌊k/2⌋; { 2m { {[∇ak−2r−1 ]x = { a[m] m k−2r−2 2(m−i) ]x Pa[i] , otherwise. ∑i=0 [∇a {

(5.5.10)

Theorem 5.5.3. Equation (5.5.7) for f ∈ ℛ{x, y} is equivalent to the system of equations for Fm = 𝜕xm f ∈ ℛ{y} for m ≥ 0 Fa[m]

0, when m = 0; { { { k = {a1 δ1,m + a2 ∑ 0≤r≤⌊k/2⌋ (2r+1) k≥1 { { × (y2r+2 ⊗ [La[2r+1] ⊗ Δa[k−2r−1] ]2(m−1−r) ), when m ≥ 1, { x

(5.5.11)

where δ1,m is Kronecker index, La[2r+1] and Δa[k−2r−1] are given in (5.5.6). Proof. From equation (5.5.7), Fa[m]

0, when m = 0; { { { 0 0 = {a1 + a2 ∑k≥1 k(y2 ⊗ [La[1] ]x ⊗ [Δa[k−1] ]x ), when m = 1; { { ⌊k/2⌋ k 2(m−1−r) ), when m ≥ 2. {a2 ∑k≥1 ∑r=0 (2r+1)(y2r+2 ⊗ [L2r+1 ⊗ Δk−2r−1 ]x

Thus, for integer m ≥ 2, k Fa[m] = ∑ ∑ ( )(y2r+2 ⊗ [L2r+1 ⊗ Δk−2r−1 ]2(m−1−r) ). x 2r + 1 r=0 k≥1 ⌊k/2⌋

When m = 1 and hence r = 0, k ) )(y ⊗ [La[2r+1] ⊗ Δa[k−2r−1] ]2(m−1−r) ∑ ∑ ( x 2r + 1 2r+2 k≥1 r=0 ⌊k/2⌋

= ∑ k(y2 ⊗ [La[1] ]0x ⊗ [Δa[k−1] ]0x ). k≥1

Therefore, the conclusion can be drawn. In order to determine Fa[2m] for m ≥ 1, it is necessary to introduce s, such that, for any integers m and s ≥ 0, Fa[m,s] ∈ ℛ[y] is a polynomial with Fa[m] = ∑ Fa[m,s] . s≥0

(5.5.12)

5.5 Eulerian cutless inner model | 177

For any power vector n of Fa[m] , let s = π(n)/2 where π(n) is the inner semi-size of n. When s = 0, because of the initial condition of equation (5.5.1), Fa[0] = 0 󳨐⇒ Fa[0,s] = 0 for any integer s ≥ 0. It is only necessary to consider the case of m ≥ 1. Lemma 5.5.4. When s = 0, for any integer m ≥ 1, a1 , when m = 1; Fa[m,0] = { 0, otherwise.

(5.5.13)

Proof. For any integer m ≥ 1, ) ≥ π(y2r+2 ), π(y2r+2 ⊗ [L2r+1 ⊗ Δk−2r−1 ]2(m−1−r) x

by r ≥ 0,

≥ π(y2 ) = 2,

i. e., s ≥ 1. From (5.5.11), the conclusion can be drawn. This lemma enables us to only consider s ≥ 1 from now on without loss of generality. Lemma 5.5.5. When m = 1, for any integer s ≥ 0, a1 , when s = 0; Fa[1,s] = { s 0 s−1 a2 (y2 ⊗ ∑k=1 k[[La[1] ⊗ Δa[k−1] ]x ]y ), otherwise.

(5.5.14)

Proof. By the second equality of equation (5.5.11), Fa[1,s] = [a1 + a2

s

k )] )(y2r+2 ⊗ [La[2r+1] ⊗ Δa[k−2r−1] ]2(−r) ∑ ( x 2r + 1 y 0≤r≤⌊k/2⌋ k≥1

s−1

= a1 δ0,s + a2 (y2 ⊗ ∑ k[[La[1] ⊗ Δa[k−1] ]0x ]y ). k≥1

Because of π([y2 ⊗ [La[1] ⊗ Δa[k−1] ]0x ]sy ) ≥ 2, s = π/2 is not 0. Thus when s = 0, Fa[1,0] = a1 δ0,0 = a1 . When s ≥ 1, because of δ0,s = 0, s−1

Fa[1,s] = a2 (y2 ⊗ ∑ k[[La[1] ⊗ Δa[k−1] ]0x ]y ). k≥1

We have s−1

s−1

l

s−1−l

[[La[1] ⊗ Δa[k−1] ]0x ]y = ∑ [[La[1] ]0x ]y ⊗ [[Δa[k−1] ]0x ]y

.

l=0

We consider that k − 1 ≥ (s − 1) + 1 ⇒ k ≥ s + 1 yields the part of k ≥ s + 1 that in the summation disappeared. This is the conclusion.

178 | 5 Inner equations third part Now, observe what happens on Fa[m,s] for m + s ≤ 2. When 0 ≤ m + s ≤ 1, the three possibilities Fa[0,0] = 0, Fa[0,1] = 0 and Fa[1,0] = a1 are known from Lemma 5.5.4 and Lemma 5.5.5. When m + s = 2, it is only necessary to determine Fa[2,0] and Fa[1,1] . For the former, Lemma 5.5.4 tells us Fa[2,0] = 0. For the latter, by Lemma 5.5.5, 0

0

Fa[1,1] = ([[La[1] ]0x ]y ⊗ [[Δa[0] ]0x ]y ) ⊗ y2 .

(5.5.15)

From (5.5.6) and (5.5.5), La[1] = ∑ ( ∑ Fa[i] y2(i−1)−2m )x2m , m≥0 i≥m+1

and hence 0

[L[1] ]0x = ∑ Fa[i] y2(i−1) 󳨐⇒ [[La[1] ]0x ]y = Fa[1,0] y0 = 0. i≥1

(5.5.16)

Similarly from (5.5.6) and (5.5.5), Δa[0] = ∫ ∇a0 = 1, y

and hence 0

[Δa[0] ]0x = 0 󳨐⇒ [[Δa[0] ]0x ]y = 0.

(5.5.17)

Therefore, by substituting (5.5.16) and (5.5.17) into (5.5.15), Fa[1,1] = a2 y2 is obtained. Lemma 5.5.6. For any integers m ≥ 2 and s ≥ 1, Fa[m,s] =

k ( )a (y 2r + 1 2 2r+2 0≤r≤⟨⌊k/2⌋,s−1⟩ ∑

1≤k≤⟨s,2m−1⟩

⊗ [[La[2r+1] ⊗

s−r−1 Δa[k−2r−1] ]2(m−1−r) ]y ), x

(5.5.18)

where for integers a, b ∈ ℝ+ , ⟨a, b⟩ = min{a, b}. Proof. We proceed on the basis of (5.5.11). For any integers m ≥ 2 and s ≥ 1, Fa[m,s] = a2

k s )]y . )[(y2r+2 ⊗ [La[2r+1] ⊗ Δa[k−2r−1] ]2(m−1−r) ∑ ( x 2r + 1 0≤r≤⌊k/2⌋ k≥1

Similarly to Lemma 5.5.5, when k ≥ s + 1, s

)]y = 0 󳨐⇒ k ≤ s. [(y2r+2 ⊗ [La[2r+1] ⊗ Δa[k−2r−1] ]2(m−1−r) x By considering that m − 1 − r ≥ 0 ⇒ m − 1 − ⌊k/2⌋ ≥ 0 ⇒ k ≤ 2m − 1 and s − r − 1 ≥ 0 ⇒ r ≤ s − 1, the conclusion can be drawn.

5.6 Solution Eulerian cutless inner case

| 179

We proceed on the basis of last three lemmas; the qualitative theorem of equation (5.5.1) can be illustrated. Theorem 5.5.7. Equation (5.5.1) is well defined on ℛ{x, y} − {x2l+1 , y2l+1 |l ≥ 0} if, and only if, a0 = 0. Proof. We proceed on the basis of Theorem 5.5.3. For any integers m, s ≥ 0, Fa[m,s] is determined in the increased order of m + s one by one from 0. For m + s ≤ 2, all Fa[m,n] have been shown above. For m + s ≥ 3 in general, assume that, for any integers i and j, Fa[i,j] is known whenever i+j ≤ m+s−1. By induction, we evaluate Fa[m,n] . From (5.5.14) and (5.5.18), it is seen that Fa[m,n] is only dependent on Fa[i,j] for i + j ≤ m + s − 1. By the assumption, Fa[m,n] is obtained. We find all Fa[m,n] as a solution of equations (5.5.11). From Theorem 5.5.2 and Theorem 5.5.3, by noticing that Fm,n ∈ ℛ{x, y}−{x 2l+1 , y2l+1 | l ≥ 0}, equation (5.5.1) with a0 = 0 has a solution. By considering the uniqueness of the procedure for determining Fa[m,n] under the given initiation, the solution is the only one. This is the sufficiency. The necessity is from Observation 5.5.1.

5.6 Solution Eulerian cutless inner case In order to simplify the solution of equation (5.5.1), some useful structures need to be further investigated. Lemma 5.6.1. Given any integer s ≥ 1. For any integer m ≥ s + 1, Fa[m,s] = 0. Proof. When m + s ≤ 2. From (5.5.15), (5.5.16) and (5.5.17) in Section 5.5, the conclusion can be drawn. For m + s ≥ 3 in general, assume that, for any integers i ≥ 2 and j ≥ 1, Fa[i,j] = 0 whenever i + j ≤ m + s − 1, and i ≥ j + 1. By induction, we prove that Fa[m,n] = 0 whenever m ≥ s + 1. We proceed on the basis of (5.5.18). If m ≥ s + 1, by the assumption similarly to the proof of Lemma 5.5.6, s−r−1

y2r+2 ⊗ [[La[2r+1] ⊗ Δa[k−2r−1] ]2(m−1−r) ]y x

=0

leads to Fa[m,s] = 0. This lemma suggests us that, for any integer k ≥ 1, it is not necessary to evaluate all {Fa[m,s] | ∀m, s ≥ 0, m + s = k} by starting from the initiation of equation (5.5.1) and Lemma 5.6.1. As a matter of fact, only half, {Fm+s−i,s+i | ⌈(m + s)/2⌉ ≤ i ≤ m − 1}, is enough. We proceed on the basis of Theorem 5.5.7. Because it is known that the case of a0 = 0 and a1 = a2 = 1 in equation (5.5.1) has its solution as the enufunction of rootisomorphic classes for Eulerian cutless planar maps by root-vertex valency (m) and vertex-partition vector as parameters (s = π(n)/2). Two lemmas in what follows are

180 | 5 Inner equations third part shown in this sense instead of induction to avoid more space occupied. For instance, any cutless planar maps of size not less than 2 is without an edge as self-loop and the number (2s) of semi-edges not incident to the root-vertex is not greater than the rootvertex valency (m) in such a map. These two facts are useful in the present context. This is why Fm,s = 0 whenever m ≥ s + 1 as shown in Lemma 5.6.1. α (m) α (m)

Lemma 5.6.2. For any integer m ≥ 1, Fa[m,m] = a1 1 a2 2 are the occurrence indices of, respectively, a1 and a2 .

y2m where α1 (m) and α2 (m)

Proof. Only combinatorial explanation is accepted for saving space although it can be done from structures of the equation in its own right via (5.5.15), because there is only one Eulerian cutless planar map of size 2m, i. e., the planar link bundle of size 2m. This leads to Fa[m,m] |a=(0,1,1) = y2m . The cases of α1 (m) and α2 (m) can be done by the procedure to determine Fa[m,m] , and the conclusion can be drawn. When m = 1, for s ≥ 0, although Lemma 5.5.5 provides an expression of Fa[1,s] , the following lemma presents a very simple coefficient of a specific term in Fa[1,s] . Lemma 5.6.3. Let Fa[1,s] |sy2 be the coefficient of term yn = y2s in F1,s . Then, for any integer α (s) α (s)

s ≥ 1, F1,s |sy2 = a1 1 a2 2 a1 and a2 .

where α1 (s) and α2 (s) are the occurrence index of, respectively,

Proof. Because we have only 2-valent vertices and connectedness, the Eulerian cutless α (s) α (s) planar map is an s-circuit. From the uniqueness, Fa[1,s] |sy2 = a1 1 a2 2 . Now, we are allowed to illustrate the solution f = fa[Eci] of equation (5.5.1), deter[Eci] = 𝜕xm fa[Eci] in the form of a finite sum with all terms positive. mined by Fa[m,n] Theorem 5.6.4. For any integers s ≥ m ≥ 0, we have

[Eci] Fa[m,s]

0, { { { { { { { a1 , when s = 0; { { } { } { α1 (s) α2 (s) s { a1 a2 y2 , when yn = y2s ; } , { { { } s = { a2 y2 ⊗ ∑k=1 k[[La[1] ⊗ Δa[k−1] ]0x ]s−1 y } { { { α1 (m) α2 (m) { { a1 a2 y2m , { { { { k )(y { a2 ∑ 0≤r≤⟨⌊k/2⌋,s−1⟩ (2r+1 { 2r+2 ⊗ [[La[2r+1] { 1≤k≤⟨s,2m−1⟩ { { 2(m−1−r) s−r−1 ⊗ Δa[k−2r−1] ]x ]y ), {

when s ≥ 0, m = 0; when m = 1;

(5.6.1)

when m = s ≥ 2;, otherwise,

where L and Δ are given in (5.5.6). Proof. When s ≥ 0, m = 0 or s ≤ m − 1, m ≥ 2, the results are determined by, respectively, the initiation of equation (5.5.1) or Lemma 5.5.4. When m = 1, the results follow by Lemma 5.5.5 and Lemma 5.6.3. When m = s ≥ 2, the result follows by Lemma 5.6.2. The others follow by (5.5.18).

5.6 Solution Eulerian cutless inner case

| 181

Further, the cases of m + s = 3 are observed. When m + s = 3, it is only necessary to consider Fa[3,0] Fa[2,1] and Fa[1,2] . By Lemma 5.5.4, Fa[3,0] = 0. By (5.5.11), 1

Fa[2,1] = a2 [y2 ⊗ [La[1] ⊗ Δa[0] ]2x ]y . Because of La[1] = ∑ ∑ Fa[i] y2(i−1)−2j x2j j≥0 i≥j+1

and Δa[0] = 0, [La[1]a[ ] ⊗ Δa[0] ]2x = [La[1] ]2x . As y0 disappeared, [La[1] ]2x = ∑ Fa[i] y2i−4 . i≥3

We proceed on the basis of y2 ⊗ (∑ Fa[i] y2i−4 ) = ∑ Fa[i] y2i−2 . i≥3

i≥3

By noticing (2i − 2)/2 = 2 > 1 for i = 3, 1

[∑ Fa[i] y2i−2 ] = 0. y

i≥3

Therefore, Fa[2,1] = 0. On Fa[1,2] , from (5.5.14), 2

1

Fa[1,2] = a2 ∑ k[[La[1] ⊗ Δa[k−1] ]0x ]y ⊗ y2 k=1

1

1

= a2 ([[La[1] ⊗ Δa[0] ]0x ]y ⊗ y2 ) + 2a2 ([[La[1] ⊗ Δa[1] ]0x ]y ⊗ y2 ). By considering that [La[1] ⊗ Δa[0] ]0x = [La[1] ]0x ⊗ [Δa[0] ]0x , [La[1] ⊗ Δa[1] ]0x = [La[1] ]0x ⊗ [Δa[1] ]0x and [La[1] ]0x = ∑ Fa[i] y2(i−1) ; { { { { i≥1 { { [Δa[0] ]0x = 0; { { { { 0 { {[Δa[1] ]x = ∑ Fa[i] y2(i−1) , i≥2 { we address [[La[1] ]0x ]0y (non-existing, because of y0 ), [[L1 ]0x ]1y = F1,0 y2 = a1 y2 (by F1,0 = a1 ) and the others either non-existing or 0. Hence, Fa[1,2] = a1 a2 y2 ⊗ y2 = a1 a2 y22 .

182 | 5 Inner equations third part

5.7 Explicitness Eulerian cutless inner case Because no explicision has yet directly been deduced from Theorem 5.6.4, the relationship between the solutions of the Eulerian cutless inner equation (5.5.1) and its restriction for a0 = 0, a1 = a2 = 1 has to be investigated on account of the latter being meaningful in graph theory. Let fa[Eci] and f[Eci] ∈ ℛ{x, y} be the solutions of, respectively, equation (5.5.1) and [Eci] [Eci] and Fm,s ∈ ℛ{y} for integers m, s ≥ 0. Denote by its restriction determined by Fa[m,s] [Eci] [Eci] [Eci] ∈ . For any n ∈ 𝒩m,s , Ua[m,n] and hence Fm,s 𝒩m,s the set of power vectors in Fa[m,s]

[Eci] [Eci] and ℛa = ℛ{a} and Um,n | ∈ ℛ1 = ℛ are the coefficients of yn in, respectively, Fa[m,s] [Eci] Fm,s ∈ ℛ{y}. Then

[Eci] [Eci] = 𝜕yn Fa[m,s] Ua[m,n]

[Eci] [Eci] . = 𝜕yn Fm,s and Um,n

[Eci] Theorem 5.7.1. For any integers m, s ≥ 1 and integral vector n ∈ 𝒩m,s , Ua[m,n] is a polynomial of a1 = a0 and a2 with degree at most l + 1 where l = m + s such that [Eci] [Eci] Ua[m,n] |a=1 = Um,n .

Proof. We proceed on the basis of Theorem 5.6.4. By induction, the conclusion can easily be drawn. For an embedding μ(G) of a Eulerian ordinary planar graph G, let n = (n1 , n2 , n3 , . . .) be the vertex-partition vector where ni is the number of non-root-vertices of valency 2i for integer i ≥ 1. A vertex v as a point of μ(G) such that μ(G) − v has at least one component greater than μ(G) is called a cut-vertex of G. Now, G is always assumed to be connected. Because there is no cut-vertex in a Eulerian cutless graph, all graphs (connected!) here are always guaranteed without cut-vertex. On an embedding μ of a Eulerian planar graph G without cut-vertex. If two vertices u and v have the property that μ(G)−{pu , pv } has at least two components, then {u, v} is called a splitting pair. Since an embedding is a point set in its own right on the plane, pu and pv are the corresponding points of u and v. Because of μ as a point set, any component of μ − {pu , pv } is called a splitting slice, or slice. A splitting slice that is not a path is called a splitting chunk, or chunk. Let S be the set of all splitting pairs consisting of {ui , vi } for 1 ≤ i ≤ s = s(S) = |S| in μ = μ(G). It is known that all components of μ − S are all splitting slices in μ(G). For a splitting pair, its slice valent is the number of splitting slices it is incident with. Similarly, for chunk valency. Observation 5.7.2. Let t and 1 ≤ p ≤ t be, respectively, the slice valency and chunk valency of a splitting pair on an embedding of a Eulerian ordinary planar graph G, then the splitting pair produces [Eci] πspp = 2p−1 (t − 1)!

(topologically) distinct planar embeddings of G.

(5.7.1)

5.7 Explicitness Eulerian cutless inner case | 183

Proof. We have the independency between rotation on slice valency and reflection on a chunk. By considering the connectivity of embedding, the conclusion can be drawn. Let p be the number of all splitting chunks and ti , the slice valency at the splitting pair {ui , vi }, 1 ≤ i ≤ s on μ(G). Lemma 5.7.3. For a Eulerian cutless planar graph G, the number of topologically distinct embeddings of G generated by all splitting pairs is s

p−s n[Eci] ∏(ti − 1)! Nct = 2 i=1

(5.7.2)

where pi and ti are, respectively, the number of chunks with the slice valency at splitting pair {ui , vi }, 1 ≤ i ≤ s = |S| such that p = p1 + p2 + ⋅ ⋅ ⋅ + ps . Proof. This is a result of Observation 5.7.2. This lemma is a refinement and a concise form of Theorem 10.4.3 with (10.4.13) in Liu YP [72] (2017, pp. 196–197), or Theorem 7.4.3 with (7.4.13) in Liu YP [55] (1994, p. 142). We proceed on the basis of Lemma 5.7.3 and Lemma 5.7.4, the result being useful in the context of determining the number of planar embeddings of a Eulerian cutless graph. Lemma 5.7.4. For a Eulerian cutless planar graph G, the number of (topologically) distinct planar embeddings of G is n[Eci] (G) = n[Eci] Nct (G)

(5.7.3)

where n[Eci] Nct (G) is given by (5.7.2). Proof. On the uniqueness of 3-connected embedding of a graph. Because a graph without vertex-cut set of one or two vertices has to be 3-connected, only the two independent cases with and without cut-vertex need to be considered. Observation 5.7.2 and Lemma 5.7.3 yield the conclusion. ̃ Denote by t = aut(G) the semi-automorphism group order of the Eulerian cutless planar graph G. The relationship between embeddings and upper maps of G can be shown via t. Lemma 5.7.5. Let ℰ [Eci] (G) and ℳ[Eci] (G) be, respectively, the sets of all planar embeddings and upper maps of a Eulerian cutless planar graph G. Then 󵄨󵄨 [Eci] 󵄨󵄨 2ϵ 󵄨󵄨 [Eci] 󵄨󵄨 (G)󵄨󵄨 (G)󵄨󵄨 = 󵄨󵄨ℰ 󵄨󵄨ℳ t where ϵ = ϵ(G) is the size of G. Proof. Refer to Liu YP [59] (2003, pp. 225–226).

(5.7.4)

184 | 5 Inner equations third part [Eci] be the set of all Eulerian cutless planar graphs with rooted semi-valency Let 𝒢m|s 2m (i. e., the number of semi-edges incident to the root-vertex) and the un-rooted semisize 2s (i. e., the number of semi-edges not incident to the root-vertex). [Eci] the set of semi-automorphism group orders of Eulerian cutless Denote by 𝒯m|s [Eci] [Eci] [Eci] . be the set of all partition vectors n in 𝒢m|s . Let 𝒥m|s planar graphs in 𝒢m|s

[Eci] be the number of (topologically) distinct embeddings of EuleLemma 5.7.6. Let Em|n

[Eci] . Then we have rian cutless planar graphs in 𝒢m|n [Eci] = Em|n

∑ n[Eci] (G)

(5.7.5)

[Eci] G∈𝒢m|n

where n[Eci] is given in (5.7.7). [Eci] Proof. Given the partition n, by considering all embeddings of each graph in 𝒢m|s with n, the conclusion is easily drawn. [Eci] be the set of semi-automorphism We proceed on the basis of Lemma 5.7.6. Let 𝒯m|n

[Eci] [Eci] [Eci] with , the set of all graphs in 𝒢m|n and 𝒢m|n,t group orders among the graphs in 𝒢m|n ̃ semi-automorphism group order t. Denote by aut(G) the semi-automorphism group

order of G.

Lemma 5.7.7. The number of root-isomorphic classes of upper maps of all graphs in [Eci] is 𝒢m|n A[Eci] m|n =

∑ [Eci] t∈𝒯m|n

2(m + π(n)) [Eci] Em|n,t t

(5.7.6)

) and = Am|n (ℳ[Eci] where A[Eci] m|n m|n [Eci] = Em|n,t

∑ n[Eci] (G)|aut(G)=t ̃

(5.7.7)

[Eci] G∈𝒢m|n

as determined by (5.7.5). Proof. We proceed on the basis of Lemma 5.7.6. By considering m + π(n) = ϵ, the conclusion is easily drawn. This lemma enables us to find the solution fa[Eci] of equation (5.5.1) when a0 = a1 = [Eci] a2 = 1. Denote f[Eci] = fa |a1 =a2 =1 as determined by Fm,s = 𝜕xm f[Eci] |π(y)=s for m, s ≥ 0. Lemma 5.7.8. For integers m, n ≥ 1, we have [Eci] Fm,s =

where A[Eci] is given by (5.7.6). m|n

∑ [Eci] n∈𝒥m|s

n A[Eci] m,n y

(5.7.8)

5.8 Restrictions Eulerian cutless inner case

| 185

Proof. It is known that the set of all underline graphs of Eulerian cutless planar maps [Eci] [Eci] = Um,n as shown in Theorem 5.7.1. From with m and n is the same as 𝒢m,n , A[Eci] m|n Lemma 5.7.7, the conclusion can be drawn. We proceed on the basis of Lemma 5.7.6 and Theorem 5.7.1; we are allowed to illustrate our main result of this section. [Eci] determine the solution of equation (5.5.1), Theorem 5.7.9. For m, s ≥ 1, the Fa[m,s] [Eci] = Fa[m,s]

∑ [Eci] n∈𝒥m|s

[Eci] yn Ua[m,n]

(5.7.9)

[Eci] is the polynomial of a with degree at most m + s + 1 given where a = (a1 , a2 ) and Ua[m,n] in Theorem 5.7.1.

Proof. This is a result of Theorem 5.7.1 and Lemma 5.7.8. [Eci] is evaluated by the procedure shown in (5.6.1) or directly The polynomial Ua[m,n] from the inner structures of the solution in its own right. [Eci] [Eci] |a=1 determine the solution of equation Corollary 5.7.10. For m, s ≥ 1, Fm,s = Fa[m,s] (5.5.1) when a = 1 by [Eci] Fm,s =

∑ [Eci] n∈𝒥m|s

[Eci] n Um,n y

(5.7.10)

[Eci] where Um,n = A[Eci] m,n is given in (5.7.6).

Proof. This is a direct result of Theorem 5.7.9. [Eci] This is a direct explicision (explicit expression) of Fm,s for integers m, s ≥ 1.

5.8 Restrictions Eulerian cutless inner case Consider the following equation for f : y2 δx2 ,y2 f |x2 =u 2 { { = f − x2 ; {x ∫ (1 − 𝜕x2 ,y2 f |x2 =u )2 − (xyδx2 ,y2 f |x2 =u )2 { { y { {f |x=0⇒y=0 = 0,

(5.8.1)

where f = f (x2 , y) ∈ ℛ{x, y} and y = (y2 , y4 , y6 , . . .). This is the case of equation (5.5.1) with a0 = 0 and a1 = a2 = 1. The first equality in equation (5.8.1) is involved with Eulerian cutless planar maps shown in [32] (Liu YP, 1986) and [60] (Liu YP, 2008, p. 165). Since no term has odd degree of x in the first equality of equation (5.8.1), f is an even function of x. This enables us to convert f into a function of u = x2 . For any integer m ≥ 0, denote Fm = 𝜕um f = 𝜕x2m f .

186 | 5 Inner equations third part Observation 5.8.1. For equation (5.8.1), the initiation F0 = 0 is consistent. Proof. Because of a factor x occurring on the left hand side of the first equality in equation (5.8.1), the conclusion can be drawn. For convenience, equation (5.8.1) has to be transformed into a suitable form. Let ⊤ = δx2 ,y2 f |u=x2 and ∇ = 𝜕x2 ,y2 f |u=x2 , then δx2 ,y2 f |u=x2

(1 − 𝜕x2 ,y2 f |u=x We have

2 2)

− (xyδx2 ,y2 f |u=x

2 2)

=

(1 −

∇)2

⊤ . − (xy⊤)2

(5.8.2)

1 1 2xy⊤ − = , 1 − ∇ − xy⊤ 1 − ∇ + xy⊤ (1 − ∇)2 − (xy⊤)2

we have

δx2 ,y2 f |u=x2

(1 − 𝜕x2 ,y2 f |u=x2 )2 − (xyδx2 ,y2 f |u=x2 )2 1 = ∑ ((∇ + xy⊤)k − (∇ − xy⊤)k ). 2xy k≥1

(5.8.3)

Because of k (∇ + xy⊤)k − (∇ − xy⊤)k = 2 ∑ ( )(xy⊤)2r+1 ∇k−2r−1 , 2r + 1 r=0 ⌊k/2⌋

we have

δx2 ,y2 f |u=x2

(1 − 𝜕x2 ,y2 f |u=x2 )2 − (xyδx2 ,y2 f |u=x2 )2 k )(xy)2r ⊤2r+1 ∇k−2r−1 . =∑ ∑ ( 2r + 1 k≥1 r=0 ⌊k/2⌋

(5.8.4)

We proceed on the basis of ⊤ = ∑ Fi i≥0

and ∇ = ∑ Fi i≥0

x2i − y2i = ∑ ∑ Fi x2j y2(i−1)−2j x2 − y2 j≥0 i≥j+1

y2 x2i − x2 y2i = ∑ ∑ Fi x2(j+1) y2(i−1)−2j . x2 − y2 j≥0 i≥j+2

For integer m ≥ 1 (note that m = 0 makes no sense in δx,y and 𝜕x,y !). Let Tm = 𝜕x2m ⊤ = 2m 2m [⊤]2m x and Pm = 𝜕x ∇ = [∇]x , then from the initial condition F0 = 0 of equation (5.8.1), T = ∑ F y2(i−m−1) ; { { { m i≥m+1 i { 2(i−m) { . {Pm = ∑ Fi y i≥m+1 {

(5.8.5)

5.8 Restrictions Eulerian cutless inner case | 187

Furthermore, for any integer l ≥ 1, denote L = ∫ ⊤l

and

Δ = ∫ ∇l .

(5.8.6)

y

y

Theorem 5.8.2. Equation (5.8.1) is equivalent to k { { )x2r (y2r+2 ⊗ L2r+1 ⊗ Δk−2r−1 ); {f = x 2 + x 2 ∑ ∑ ( 2r + 1 r=0 k≥1 { { { f | = 0, x=0⇒y=0 { ⌊k/2⌋

(5.8.7)

on ℛ{x, y} where ⨂ is the convolution of two vectors in 𝒱 (Section 2.5 of Book I). Proof. We proceed on the basis of Observation 5.8.1. By substituting (5.8.4) into the first equality of equation (5.8.1), we have k )x2r y2(r+1) ⊤2r+1 ∇k−2r−1 f = a1 x2 + a2 x2 ∫ ∑ ∑ ( 2r + 1 k≥1 r=0 ⌊k/2⌋

y

k = x2 + x2 ∑ ∑ ( )x2r (y2(r+1) ⊗ L2r+1 ⊗ Δk−2r−1 ). 2r + 1 k≥1 r=0 ⌊k/2⌋

Therefore, the conclusion can be drawn. This theorem suggests us to evaluate L2r+1 and Δk−2r−1 in ℛ{x, y} before determining Fm for m ≥ 1. We have [⊤]2m , when r = 0; 2m [⊤2r+1 ]x = { mx 2r 2(m−i) 2i [⊤]x , when r ≥ 1, ∑i=0 [⊤ ]x

(5.8.8)

and [∇]2m , when r = ⌊k/2⌋; 2m [∇k−2r−1 ]x = { mx k−2r−2 2(m−i) 2i ]x [∇]x , when 0 ≤ r ≤ ⌊k/2⌋ − 1, ∑i=0 [∇

(5.8.9)

for any integers r ≥ 0 and m ≥ 0. From (5.8.5), we have, for 0 ≤ r ≤ ⌊k/2⌋, Tm , when r = 0; 2m { { [⊤2r+1 ]x = { m { 2r 2(m−i) { { Ti , otherwise, ∑i=0 [⊤ ]x { { and { { { { when r = ⌊k/2⌋; { {[∇k−2r−1 ]2m = {Pm , { x m k−2r−2 2(m−i) [∇ ] P , otherwise. ∑ { i i=0 x

(5.8.10)

188 | 5 Inner equations third part Theorem 5.8.3. Equation (5.8.7) for f ∈ ℛ{x, y} is equivalent to the system of equations for Fm = 𝜕xm f ∈ ℛ{y} for m ≥ 0 0, { { { k ) Fm = {δ1,m + ∑ 0≤r≤⌊k/2⌋ (2r+1 k≥1 { { × (y2r+2 ⊗ [L2r+1 ⊗ Δk−2r−1 ]2(m−1−r) ), { x

when m = 0; (5.8.11) when m ≥ 1,

where δ1,m is the Kronecker symbol, L2r+1 and Δk−2r−1 are given in (5.8.6). Proof. See the proof of Theorem 5.5.3 for a = 1. In order to determine Fm for m ≥ 1, it is necessary to introduce s, such that for any integers m and s ≥ 0, Fm,s ∈ ℛ[y] is a polynomial with Fm = ∑ Fm,s .

(5.8.12)

s≥0

For any power vector n of Fm , let s = π(n)/2 where π(n) is the inner semi-size of n. When s = 0, because of the initial condition of equation (5.8.1), F0 = 0 󳨐⇒ F0,s = 0 for any integer s ≥ 0. It is only necessary to consider the case of m ≥ 1. Lemma 5.8.4. When s = 0, for any integer m ≥ 1, 1, when m = 1; Fm,0 = { 0, otherwise.

(5.8.13)

Proof. For any integer m ≥ 1 and r ≥ 0, π(y2r+2 ⊗ [L2r+1 ⊗ Δk−2r−1 ]x2(m−1−r) ) ≥ π(y2r+2 ) ≥ π(y2 ) = 2, i. e., s ≥ 1. From (5.8.11), the conclusion can be drawn. This lemma enables us to only consider s ≥ 1 from now on without loss of generality. Lemma 5.8.5. When m = 1, for any integer s ≥ 0, 1, when s = 0; F1,s = { , otherwise. y2 ⊗ ∑sk=1 k[[L1 ⊗ Δk−1 ]0x ]s−1 y

(5.8.14)

Proof. See the proof of Lemma 5.5.5 for a = 1. Now, observe what happens on Fm,s for m + s ≤ 2. When 0 ≤ m + s ≤ 1, the three possibilities F0,0 = 0, F0,1 = 0 and F1,0 = a1 are known from Lemma 5.8.4 and Lemma 5.8.5. When m + s = 2, it is only necessary to determine F2,0 and F1,1 . For the former, Lemma 5.8.4 tells us F2,0 = 0. For the latter, by Lemma 5.8.5, 0

0

F1,1 = ([[L1 ]0x ]y ⊗ [[Δ0 ]0x ]y ) ⊗ y2 .

(5.8.15)

5.8 Restrictions Eulerian cutless inner case | 189

From (5.8.6) and (5.8.5), L1 = ∑ ( ∑ Fi y2(i−1)−2m )x2m m≥0 i≥m+1

and hence 0

(5.8.16)

0

(5.8.17)

[L1 ]0x = ∑ Fi y2(i−1) 󳨐⇒ [[L1 ]0x ]y = F1,0 y0 = 0. i≥1

Similarly from (5.8.6) and (5.8.5), Δ0 = ∫ ∇0 = 1 y

and hence [Δ0 ]0x = 0 󳨐⇒ [[Δ0 ]0x ]y = 0.

Therefore, by substituting (5.8.16) and (5.8.17) into (5.8.15), F1,1 = y2 is obtained. Lemma 5.8.6. For any integers m ≥ 2 and s ≥ 1, Fm,s =

k ( )(y 2r + 1 2r+2 0≤r≤⟨⌊k/2⌋,s−1⟩ ∑

1≤k≤⟨s,2m−1⟩

⊗ [[L2r+1 ⊗

s−r−1 ]y ), Δk−2r−1 ]2(m−1−r) x

(5.8.18)

where for integers a, b ∈ ℝ+ , ⟨a, b⟩ = min{a, b}. Proof. See the proof of Lemma 5.5.6 for a = 1. We proceed on the basis of the last three lemmas; the qualitative theorem of equation (5.8.1) can be illustrated. Theorem 5.8.7. Equation (5.8.1) is well defined on ℛ{x, y} − {x 2l+1 , y2l+1 | l ≥ 0}. Proof. This is a direct result of Theorem 5.5.7 for a = 1. In order to simplifying the solution of equation (5.8.1), some useful structures need to be further discussed. Lemma 5.8.8. Given any integer s ≥ 1, for any integer m ≥ s + 1, Fm,s = 0. Proof. See the proof of Lemma 5.6.1 for a = 1. This lemma suggests us that, for any integer k ≥ 1, it is not necessary to evaluate all {Fm,s | ∀m, s ≥ 0, m + s = k} by starting the initiation of equation (5.8.1) and Lemma 5.8.8. As a matter of fact, only its half, {Fm+s−i,s+i | for ⌈(m + s)/2⌉ ≤ i ≤ m − 1}, are enough.

190 | 5 Inner equations third part We proceed on the basis of Theorem 5.8.7. It is known that equation (5.8.1) has its solution as the enufunction of root-isomorphic classes for Eulerian cutless planar maps by root-vertex valency (m) and vertex-partition vector as parameters (s = π(n)/2). Two lemmas in what follows are shown in this sense instead of induction to avoid using too much space. For instance, any cutless planar maps of size not less than 2 is without an edge as self-loop and the number (2s) of semi-edges not incident to the root-vertex is not greater than the root-vertex valency (2m) in such a map. These two fact are useful in the present context. This is why Fm,s = 0 whenever m ≥ s+1 as shown in Lemma 5.8.8. Lemma 5.8.9. For any integer m ≥ 1, Fm,m = y2m . Proof. Only a combinatorial explanation is accepted for saving space although it can be done from structures of the equation in its own right via (5.8.18), because there is only one Eulerian cutless planar map, i. e., the planar link bundle of size 2m. This leads to Fm,m = y2m . When m = 1, for s ≥ 0, although Lemma 5.8.5 provides an expression for F1,s , the following lemma presents a much more simple coefficient of a specific term in F1,s . Lemma 5.8.10. For any integer s ≥ 1, F1,s |sy2 = y2s . Proof. Because there are only 2-valent vertices and we have connectedness, the Eulerian cutless planar map is an s-circuit. From the uniqueness, F1,s |sy2 = y2s . Now, we are allowed to illustrate the solution f = f[Eci] of equation (5.8.1), deter[Eci] mined by Fm,n = 𝜕xm fEci in the form of a finite sum with all terms positive. Theorem 5.8.11. For any integers s ≥ m ≥ 0, we have

[Eci] Fm,s

{0, { { { 1, when s = 0; { { } { } { s n s { y , when y = y ; { , 2 2 { } { } { { y2 ⊗ ∑sk=1 k[[L1 ⊗ Δk−1 ]0x ]s−1 y } ={ { {y2m , { { { k { { { )(y ⊗ [[L2r+1 ∑ 0≤r≤⟨⌊k/2⌋,s−1⟩ ( { { 1≤k≤⟨s,2m−1⟩ 2r + 1 2r+2 { { 2(m−1−r) s−r−1 { ), ⊗ Δk−2r−1 ]x ]y {

when s ≥ 0, m = 0; when m = 1; when m = s ≥ 2;,

(5.8.19)

otherwise,

where L and Δ are given in (5.8.6). Proof. When s ≥ 0, m = 0 or s ≤ m − 1, m ≥ 2, the results are determined by, respectively, the initiation of equation (5.8.1) and Lemma 5.8.8. When m = 1, the results are by Lemma 5.8.5 and Lemma 5.8.10. When m = s ≥ 2, the result is by Lemma 5.8.9. The others follow by (5.8.18).

5.8 Restrictions Eulerian cutless inner case | 191

Furthermore, the cases on m + s = 3 are observed. When m + s = 3, it is only [Eci] [Eci] [Eci] [Eci] = 0. By (5.8.11), . By Lemma 5.8.4, F3,0 F2,1 and F1,2 necessary to consider F3,0 1

[Eci] F2,1 = [y2 ⊗ [L1 ⊗ Δ0 ]2x ]y .

Because of L1 = ∑ ∑ Fi y2(i−1)−2j x2j j≥0 i≥j+1

and Δ0 = 0, [L1 ⊗ Δ0 ]2x = [L1 ]2x . As y0 disappeared, [L1 ]2x = ∑ Fi[Eci] y2i−4 . i≥3

We proceed on the basis of y2 ⊗ (∑ Fi[Eci] y2i−4 ) = ∑ Fi y2i−2 . i≥3

i≥3

By noticing (2i − 2)/2 = 2 > 1 for i = 3, 1

[∑ Fi[Eci] y2i−2 ] = 0. y

i≥3

Therefore, F2,1 = 0. [Eci] , from (5.8.14), On F1,2 1

1

[Eci] F1,2 = ([[L1 ⊗ Δ0 ]0x ]y ⊗ y2 ) + 2([[L1 ⊗ Δ1 ]0x ]y ⊗ y2 ).

By considering that [L1 ⊗ Δ0 ]0x = [L1 ]0x ⊗ [Δ0 ]0x , [L1 ⊗ Δ1 ]0x = [L1 ]0x ⊗ [Δ1 ]0x and [L1 ]0x = ∑ Fi[Eci] y2(i−1) ; { { { { i≥1 { { [Δ0 ]0x = 0; { { { { [Eci] 0 { {[Δ1 ]x = ∑ Fi y2(i−1) , i≥2 { [Eci] [Eci] we address [[L1 ]0x ]0y (non-existing because of y0 ), [[L1 ]0x ]1y = F1,0 y2 = y2 (by F1,0 = 1)

[Eci] = y2 ⊗y2 = y22 . Or, we may consider and the others either non-existing or 0. Hence, F1,2 the specific case of s = 2 in Lemma 5.8.10.

192 | 5 Inner equations third part Example 5.8.1. Root-isomorphic classification of Eulerian cutless planar maps by root-vertex valency and vertex-partition vector as parameters. In Liu YP [48] (1989), one might find the discussion of the enumeration of distinct root-isomorphic classes of bipartite cutless planar maps by the root-face valency and the size as parameters. [Eci] (y) the enufunction of such maps of size i, then Denoting by F⟨i⟩ [Eci] [Eci] [Eci] (y) = y2 ; (y) = F⟨3⟩ (y) = F⟨2⟩ F⟨1⟩ { { { { { [Eci] [Eci] (y) = 2y2 + 4y4 ; (y) = y2 + y4 ; F⟨5⟩ F { { ⟨4⟩ { { { [Eci] 2 4 6 {F⟨6⟩ (y) = 6y + 12y + y ,

are found where the power of y is the root-face valency. From the planar duality, [Eci] (y) is also the enufunction of Eulerian cutless planar maps of size i. In FigF⟨i⟩

[Eci] ure 5.8.1, one might like to check that Fm,s is the number of distinct root-isomorphic classes of such maps with size m + s, root-vertex valency 2m and vertex-partition vector. The results for size up to 6 shown in Figure 5.8.1 are as follows:

When m + s = 1, { { { { { { { { When m + s = 2, { { { { { { { When m + s = 3, { { { { { { { { When m + s = 4, { { { { { { { { { { { When m + s = 5, { { { { { { { { { { { { { { { When m + s = 6, { { { { { { { { { { { { { {

[Eci] = 1(a) = 1; F1,0 [Eci] F2,0 =0

[Eci] = 1(b) = y2 ; and F1,1

[Eci] [Eci] =0 = F2,1 F3,0 [Eci] [Eci] F4,0 = F3,1 = 0,

[Eci] = 1(c) = y22 ; and F1,2 [Eci] F2,2 = 1(e) = y4

[Eci] and F1,3 = 1(d) = y23 ; [Eci] [Eci] [Eci] = 0, = F3,2 = F4,1 F5,0

[Eci] = 4(h) = 4y2 y4 F2,3

[Eci] and F1,4 = 1(f ) + 1(g) = y24 + y42 ; [Eci] [Eci] [Eci] = 0, = F4,2 F6,0 = F5,1

[Eci] F3,3 = 1(p) = y6 ,

[Eci] = 4(m) + 4(n) + 4(o) = 12y22 y4 F2,4

and

[Eci] = 1(i) + 2(j) + 1(k) + 2(l) = 1y25 + 5y2 y42 . F1,5

5.9 Notes 5.9.1. Equation coefficients variable. The generalization of equation (5.1.1), or equation (5.5.1), with coefficients a variable might be extensively investigated for combinatorial properties of partitions with certain type of weights variable.

5.9 Notes | 193

Figure 5.8.1: Root-isomorphic classes of Eulerian cutless planar maps.

5.9.2. One way of explicision. An explicision of the solution of equation (5.1.1), or equation (5.5.1), is still necessary to evaluate only by transformations on ℛ{x, y} directly for seeking a favorable resolution. 5.9.3. Another way for explicision. This manner for getting an explicision of the solution of equation (5.1.1), or equation (5.5.1), in the case of a = (0, 1, 1), i. e., equation (5.4.1), or equation (5.8.1), is to observe the root-classifications of general, or Eu-

194 | 5 Inner equations third part lerian, cutless planar maps with vertex-partition given directly without considering semi-automorphism group orders of the underline graphs. 5.9.4. New results on embedding of general, or Eulerian, cutless planar graphs. All results in Section 5.3, or Section 5.7, on embedding of general, or Eulerian, cutless planar graphs for reaching the corresponding root-enumerations of general, or Eulerian, cutless planar maps are new. As seen, they form a type of foundation for getting an explicision for the solution of equation (5.1.1), or equation (5.5.1). 5.9.5. Making more efficient and then more intelligent Theorem 5.2.6 and (5.6.1), particularly Theorem 5.4.13 and (5.8.19) have to be further investigated for running on computers and for theoretical and/or practical usage. 5.9.6. Specification of parameters. For a small number of parameters given, more research might be found in Tutte WT [90] (1963), Liu YP [25] (1982), [28] (1985), [30] (1986), [32] (1986), [35] (1986), [37] (1987), [50] (1990), [54] (1993), [8] (Cai, J. L., Liu, Y. P., 1997), [5] (Cai, J. L., Gao, C. L., Liu, Y. P., 2002), [74] (Long, S. D., Cai, J. L., 2010) etc. 5.9.7. Direct explicision without consideration of symmetry. A direct explicision of (5.3.10), or (5.7.10), is for upper maps of general, or Eulerian, cutless planar graphs without consideration of symmetry. If symmetry is considered for maps, the rootisomorphic maps distribution by their automorphism group orders has to be done as shown in Liu YP [59] (Liu, Y. P., 2003, pp. 221–230), or [73] (2017, pp. 188–194). 5.9.8. The relationship between equation (5.1.1) and equation (5.5.1), or equation (5.4.1) and equation (5.8.1), is investigated via inner structures over general cutless planar maps and Eulerian cutless planar maps. Are there a series of transformations on ℛ{x, y} from one of equation (5.1.1) and equation (5.5.1), or particularly, equation (5.4.1) and equation (5.8.1) to the other? 5.9.9. One might wish to find if there a suitable way to label vertices of a Eulerian cutless planar graph to generate all its planar embeddings or directly to get the number of all its embeddings so that an explicision of the solution of equation (5.8.1) and then equation (5.5.1) as Theorem 5.7.9 can be extracted by a similar procedure to [76] (Mao, L. F., Liu, Y. P., Wei, E. L., 2006). 5.9.10. Asymptotic behavior (or asymptotics). Note 5.9.6 reminds us to investigate their asymptotic behavior as shown in [56] (Liu, Y. P., 1999, pp. 359–389), [94] (Yan, J. Y., Liu, Y. P., 1991). 5.9.11. Stochastic behavior (or stochastics). On the basis of (5.4.35)–(5.4.36) and (5.8.19), the probability of orders, semi-automorphic group order etc. for given sizes of, respectively, general and Eulerian cutless planar maps via size polynomials, order polynomials and semi-automorphic group order polynomials are investigated. Rele-

5.9 Notes | 195

vant results are referred to in [56] (Liu, Y. P., 1999, pp. 359–389), [59] (Liu, Y. P., 2003, 221–230), [94] (Yan, J. Y., Liu, Y. P., 1991) etc. 5.9.12. Distributions. Theorem 5.4.13 and Theorem 5.8.11 enable us to determine the distributions of orders on given sizes for, respectively, general cutless planar maps and Eulerian cutless planar maps and then to estimate their corresponding moments including the average values (i. e., mathematical expectations). 5.9.13. Equation (5.1.1) is from Program 91 as equation (122) for c = 1 in [71] (Liu, Y. P., 2016, p. 10742). Equation (5.5.1) is from Program 95 as equation (126) in [71] (Liu, Y. P., 2016, p. 10743).

6 Inner equations fourth part 6.1 General simple inner model Two models are included in this chapter. First, consider the equation for f ∈ ℛ{x, y} 2 2 { {a2 x ∫ yδx,y (uf |x=u ) = ∫((1 + xy)f − 1)f |x=y − a1 x f ; y y { { f | = a = ̸ 0, 0 { x=0⇒y=0

(6.1.1)

where a0 , a1 , a2 ∈ ℝ and y = (y1 , y2 , y3 , . . .) ∈ 𝒱 . This is equation (11) in Introduction. Because a solution of equation (6.1.1) for a0 = a1 = a2 = 1 is involved with general simple planar maps as shown in [47] (Liu YP, 1989) and [61] (Liu YP, 2009, p. 194), this equation is called a general simple inner model. For convenience, equation (6.1.1) is transformed to its equivalence on ℛ{x, y} as f = 1 + a1 x2 f 2 + a2 x ∫(yδx,y (uf |x=u )) { { { { { y { { − xf ∫(yf |x=y ) − (f − 1) ∫(f |x=y − 1); { { { { { y y { { f | = a = ̸ 0. 0 { x=0⇒y=0

(6.1.2)

[2] = 𝜕xm f 2 the coeffiFor any integer m ≥ 0, let Fa[m] = 𝜕xm f (or 𝜕m f ). Denote by Fa[m]

cient of xm in f 2 , then

m

[2] = ∑ Fa[i] Fa[m−i] . Fa[m] i=0

(6.1.3)

From the initiation of equation (6.1.1), Fa[0] = a0 . Thus, for any integer m ≥ 0,

[2] Fa[m]

a0 , when m = 0; { { { = {2a0 Fa[1] , when m = 1; { { m−1 {2a0 Fa[m] + ∑i=1 Fa[i] Fa[m−i] , when m ≥ 2.

(6.1.4)

Observation 6.1.1. If fa is a solution of equation (6.1.1), then a0 (1 − a0 ) = 0. Proof. We have Fa[0] = a0 ≠ 0 by the initiation of equation (6.1.1). However, Fa[0] = 2 1 − (Fa[0] − 1) ⇒ a0 (1 − a0 ) = 0 by the first equality of equation (6.1.2). Therefore, a0 = 1. This observation makes it only necessary to consider a0 = 1 for equation (6.1.1) and hence (6.1.2). https://doi.org/10.1515/9783110627336-006

198 | 6 Inner equations fourth part Let Aa[m] = 𝜕m (f ∫y yf |x=y ) for integer m ≥ 0. Because of f = ∑ Fa[m] xm , m≥0

Observation 6.1.1 yields Aa[m] = Fa[m] (y1 + ∑ Fa[i−1] yi ).

(6.1.5)

i≥2

For any integer m ≥ 0, let Ba[m] = 𝜕m ((f − 1) ∫y (f |x=y − 1)). Because of f = 1 + ∑ Fa[m] xm , m≥1

we have 0, Ba[m] = { Fa[m] (∑i≥1 Fa[i] yi ),

when m = 0; when m ≥ 1,

(6.1.6)

for integer m ≥ 0. According to the evaluation in Section 3.1, δx,y (uf |x=u ) = ∑ xm ∑ yl Fa[l+m] . m≥0

Let ∇a[m] =

𝜕xm (δx,y (uf |x=u )),

l≥0

then ∇a[m] = ∑ yl Fa[l+m]

(6.1.7)

l≥0

for any integer m ≥ 0. Furthermore, let △a[m] = ∫y (y∇a[m] ), then, by (6.1.7), △a[m] = ∑ Fa[l+m] yl+1

(6.1.8)

l≥0

for any integer m ≥ 0. Theorem 6.1.2. Any solution of equation (6.1.1) is determined by the system of equations for Fa[m] ∈ ℛ{y} for m ≥ 0 Fa[m]

1 − Ba[0] , { { { = {a2 △a[0] −Aa[0] − Ba[1] , { { [2] {a1 Fa[m−2] + a2 △a[m−1] −Aa[m−1] − Ba[m] ,

when m = 0; when m = 1;

(6.1.9)

when m ≥ 2.

Proof. If f is a solution of equation (6.1.1), we have f ∈ ℛ{x, y} and Fa[m] = 𝜕xm f for m ≥ 0. It is seen that the Fa[m] for m ≥ 0 form a solution of equations (6.1.9) on the basis of (6.1.2) through (6.1.6). Conversely, if Fa[m] ∈ ℛ{y} for m ≥ 0 form a solution of (6.1.9), then (6.1.3) through (6.1.6) are checked for f ∈ ℛ{x, y} determined by Fa[m] = 𝜕xm f for m ≥ 0 satisfying equation (6.1.2) and hence equation (6.1.1). The conclusion can be drawn.

6.1 General simple inner model | 199

We proceed on the basis of this theorem and equations (6.1.9), in order to determine Fa[m] ∈ ℛ{y} for m ≥ 0; it is still necessary to introduce another parameter s ≥ 0, so that Fa[m] = ∑ Fa[m,s]

(6.1.10)

s≥0

and so that, for any non-negative integers m and s given, Fa[m,s] is a polynomial on ℛ{y}.

Let 𝒥m be the set of power vectors n = (n1 , n2 , n3 , . . .) with term yn in Fa[m] where y = (y1 , y2 , y3 , . . .), and 𝒥m,s , i. e., the subset of 𝒥m such that π(n) = ∑ ini = s. i≥1

From (6.1.9),

Fa[m,s]

1, { { { = {a2 △a[0,s] −Aa[0,s] − Ba[1,s] , { { [2] {a1 Fa[m−2,s] + a2 △a[m−1,s] −Aa[m−1,s] − Ba[m,s] ,

when m = s = 0; when m = 1;

(6.1.11)

when m ≥ 2,

for any integer s ≥ 0. On this basis, Fa[0] , Fa[1] and Fa[m,0] for m ≥ 0 can be determined. Lemma 6.1.3. For Fa[0] , we have 1, when s = 0; Fa[0,s] = { 0, when s ≥ 1.

(6.1.12)

Proof. This is a result of the initiation of equation (6.1.1). This lemma shows that Fa[0] is determined. Lemma 6.1.4. For any integer m ≥ 0, 1, when m = 0; for m ≥ 1, { { { Fa[m,0] = {0, when m = 1(mod 2); { { m−1 F F , a ∑ { 1 i=0 i,0 m−2−i,0 when m = 0(mod 2). Proof. When m = 0, from Lemma 6.1.3, Fa[0,0] = 1. When m ≥ 1, on account of △a[m−1,0] = ∑ Fa[l+m−1,0−l−1] yl+1 = 0, l≥0

Aa[m−1,0] = Fa[m−1,0] [∑ Fa[i−1] yi ] = 0, i≥2

0

(6.1.13)

200 | 6 Inner equations fourth part and Ba[m,0] = Fa[m,0] [∑ Fa[i] yi ] = 0, 0

i≥1

from (6.1.11), 0, Fa[m,0] = { [2] , a1 Fa[m−2,0]

when m = 1; when m ≥ 2.

We proceed on the basis of m−2

[2] = ∑ Fa[i,0] Fa[m−2−i,0] . Fa[m,0] i=0

From Fa[0,0] = 1 and Fa[1,0] = 0, we are allowed to assume that, for any integers 0 ≤ l ≤ m − 1, Fa[l,0] satisfies (6.1.13). By induction, we prove that Fa[m,0] satisfies (6.1.13). From 0 ≤ i, m − 2 − i ≤ m − 1 and from the fact that both i and m + 2 − i are even only if m is even, by the assumption, Fa[m,0] = 0 when m = 1(mod 2). This is the conclusion. This lemma enables us to determine all Fa[m,0] from the initiation Fa[0,0] = 1 by the following procedure. Corollary 6.1.5. For any integer m ≥ 0, {0, Fa[m,0] = { ak m! 1 , { k!(k+1)!

when m = 2k − 1, k ≥ 1; when m = 2k, k ≥ 0.

(6.1.14)

Proof. We proceed on the basis of Lemma 6.1.4. By induction, the conclusion is easily drawn. In the sense of combinatorial maps, the number F1[m,0] shows how many distinct root-isomorphic classes of plane trees with size k where m = 2k. Lemma 6.1.6. For any integer s ≥ 1, (a2 − 1)y1 , Fa[1,s] = { (a2 − 1) ∑sl=2 Fa[l−1,s−l] yl ,

when s = 1; when s ≥ 2.

(6.1.15)

Fa[1,s] = 0 if, and only if, a2 = 1. Proof. When s = 0, from Lemma 6.1.4, Fa[1,0] = 0. For any integer s ≥ 1 given, according to the case of m = 1 in Theorem 6.1.2, we have Fa[1] = a2 △a[0] −Aa[0] − Ba[1] . On account of (6.1.5), (6.1.6) and (6.1.8), y1 , when s = 1; △a[0,s] = { s−1 ∑l=0 Fa[l,s−l−1] yl+1 , when s ≥ 2,

6.2 Solution general simple inner case

| 201

and y1 , when s = 1; Aa[0,s] = { s ∑l=2 Fa[l−1,s−l] yl , when s ≥ 2. By the inductive assumption, s−1

Ba[1,s] = ∑[Fa[1] Fa[l] ]s−l yl = 0. l=1

Thus, when s = 1; (a2 − 1)y1 , Fa[1,s] = a2 △a[0,s] −(Aa[0,s] + Ba[1,s] ) = { s (a2 − 1) ∑l=2 Fa[l−1,s−l] yl , when s ≥ 2. The first statement of (6.1.15) is done. The second statement follows from the first. This lemma tells us that, for any integer s ≥ 1, Fa[1,s] is determined by some Fi,j for i + j ≤ s − 1 and i, j ≥ 0 starting from Fa[1,0] = 0. Because of the importance of a2 = 1 for the simplicity on maps from Lemma 6.1.6, in what follows, equation (6.1.1) is under the condition a2 = 1 to avoid sophistication. Theorem 6.1.7. Equation (6.1.1) is well defined on ℛ{x, y} if, and only if, a0 = 1. Proof. We proceed on the basis of Lemma 6.1.3, Lemma 6.1.4 and Lemma 6.1.6. From Fa[0,0] = 1 and Fa[0,1] = Fa[1,0] = 0, we are allowed to assume that, for any integers p, q ≥ 0, p + q ≤ t − 1, t ≥ 2, all Fa[p,q] are known. By induction on m + s, we determine Fa[m,s] for m + s = t. Fa[0,t] and Fa[t,0] for t ≥ 0 are known from, respectively, Lemma 6.1.3 and Lemma 6.1.4. It is only necessary to consider 1 ≤ m ≤ t − 1. From Lemma 6.1.6, only m ≥ 2 has to be considered. [2] , △a[m−1,t−m] , Aa[m−1,s] When m ≥ 2, we proceed on the basis of (6.1.11). Fa[m−2,t−m] and Ba[m,t−m] are known from the assumption by, respectively, (6.1.3), (6.1.8), (6.1.5) and (6.1.6). By (6.1.11), Fa[m,s] = Fa[m,t−m] is determined. Thus, a solution of (6.1.9), and hence equation (6.1.1) from Theorem 6.1.2, is obtained. Furthermore, this solution is the only one by the uniqueness of the procedure starting from the initiation. The sufficiency is done. The necessity of the theorem follows from Observation 6.1.1.

6.2 Solution general simple inner case In order to evaluate the solution of equation (6.1.1) in the form of a finite sum with all terms positive, further constructions of the solution have to be investigated.

202 | 6 Inner equations fourth part According to the last section, one might see that, for any integers m, s ≥ 1, s−1

{ { Δa[m,s] = ∑ Fa[l+m,s−l−1] yl+1 ; { { { { l=0 { { s { { A = [Fa[m] Fa[l−1] ]s−l yl ; ∑ a[m,s] { { { l=1 { { s { { { { {Ba[m,s] = ∑[Fa[m] Fa[l] ]s−l yl . l=1 {

(6.2.1)

Lemma 6.2.1. For any integers m, s ≥ 0, if m + s = 1(mod 2), then Fa[m,s] = 0. Proof. When m + s = 1, from (6.1.12) and (6.1.13), Fa[0,1] = 0 and Fa[1,0] = 0 are known. When m+s = k ≥ 2, one might assume that, for any two integers p+q ≤ k −1, Fa[p,q] = 0 whenever p + q = 1(mod 2). By induction on k, we prove Fa[m,s] = 0 whenever k is odd. We proceed on the basis of (6.1.11) and (6.2.1). First, we show that Δa[m−1,s] = 0, Aa[m−1,s] = 0 and Ba[m,s] = 0. We take into account (l+m−1)+(s−l−1) = m−1+s−1 = m+s−2 = k−2 ≤ k−1 for any integer 0 ≤ l ≤ s − 1 and k − 2 odd as well as k odd. By the assumption, Fa[l+m,s−l−1] = 0. Thus by the first row of (6.2.1), Δa[m−1,s] = 0. We take into account that ((m − 1) + i) + ((l − 1) + (s − l − i)) = m + s − 2 = k − 2 ≤ k − 1 for any 1 ≤ l ≤ s and 0 ≤ i ≤ s − l. At least one of (m − 1) + i and (l − 1) + (s − l − i) is odd whenever k odd. By the assumption, [Fa[m−1] Fa[l−1] ]s−l = 0. By the second row of (6.2.1), Aa[m−1,s] = 0. We take into account that (m + i) + (l + (s − l − i)) = m + s = k for 1 ≤ l ≤ s and 0 ≤ i ≤ s − l. We have m + i ≤ k − 1 and l + (s − l − i) ≤ k − 1 for m ≥ 1 and l ≥ 1. By k = 1(mod 2), at least one of m + i and l + (s − l − i) is odd. By the assumption, [Fa[m] Fa[l] ]s−l = 0. From the third row of (6.2.1), Ba[m,s] = 0. [2] = 0 can be done. Therefore, (6.1.11) leads to Fm,s = 0 for m + s = Similarly, Fa[m−2,s] 1(mod 2). This is the conclusion. This lemma enables us to determine Fa[m,s] it is only necessary for m+s = 0(mod 2). Because of Lemma 6.1.3, Lemma 6.1.4 and Lemma 6.1.6, it is only necessary to discuss Fa[m,s] for m ≥ 2 and s ≥ 1. Thus, all Fa[m,s] for m + s ≤ 3 are done. When m + s = 4. Only Fa[3,1] and Fa[2,2] need to be determined. For simplicity, the case of a2 = 1 is recalled in what follows. We consider Fa[3,1] . We proceed on the basis of Corollary 6.1.5 and Lemma 6.2.1, 0−1

Δa[2,1] = ∑ Fa[l+2,1−l−1] yl+1 = a1 y1 , l=0

Aa[2,1] + Ba[3,1] = [Fa[2] Fa[0] ]0 y1 = a1 y1 , and [2] = 2[Fa[0] Fa[1] ]1 = 2F0,0 F1,1 = 0. Fa[1,1]

6.2 Solution general simple inner case

| 203

By employing (6.1.11), [2] + a2 Δa[2,1] − (Aa[2,1] + Ba[3,1] ) Fa[3,1] = a1 Fa[1,1]

= 0 + a1 y1 − a1 y1 = 0. We consider Fa[2,2] . We take into account that Δa[1,2] = Fa[1,1] y1 + Fa[2,0] y2 = a1 y2 . From Lemma 6.1.6 and Corollary 6.1.5, Aa[1,2] + Ba[2,2] = Fa[2,0] Fa[2,0] y2 = a21 y2 and [2] = [Fa[0] Fa[0] ]2 = 2Fa[0,0] Fa[0,2] + Fa[0,1] Fa[0,1] = 0. By employing (6.1.11), Fa[0,2] [2] + a2 Δa[1,2] − (Aa[1,2] + Ba[2,2] ) Fa[2,2] = a1 Fa[0,2]

= 0 + a2 a1 y2 − a21 y2 = a1 (a2 − a1 )y2 .

Then, for a2 = a1 , Fa[2,2] = 0. The above two results show that when m+s = 4, a2 Δa[m−1,s] −(Aa[m−1,s] +Ba[m,s] ) ≥ 0 if a2 ≥ a1 . Lemma 6.2.2. For any integers m, s ≥ 1, Δm−1,s ≥ Am−1,s + Bm,s if a1 = a2 = 1. Proof. We proceed on the basis of Theorem 6.1.7 (a0 = 1!). Equation (6.1.3) has, and is the only one to have, a solution. In combinatorial maps, Δa[m−1,s] |a1 =a2 =1 − (Aa[m−1,s] + Ba[m,s] ) is the number of distinct root-isomorphic classes in a type of general simple planar maps. The conclusion holds. This lemma tells us that, for any integers m, s ≥ 1, Γm,s ∈ ℛ+ {y} is not always true except for a1 = a2 = 1. For any integers m ≥ 2 and s ≥ 0, let Γa[m,s] = a2 Δa[m−1,s] − (Aa[m−1,s] + Ba[m,s] ), then from (6.2.1), s

Γa[m,s] = ∑[a2 Fa[l+m−2] − Fa[m−1] Fa[l−1] − Fa[m] Fa[l] ]s−l yl . l−1

(6.2.2)

For convenience, for integers 1 ≤ l ≤ s, denote Γ(m,s) = [a2 Fa[l+m−2] − Fa[m−1] Fa[l−1] − Fa[m] Fa[l] ]s−l . a[l]

(6.2.3)

From Lemma 6.2.1 and Lemma 6.2.2, = 0, when m ≠ s(mod 2); Γ(m,s) { a[l] ∈ ℝ, when m = s(mod 2).

(6.2.4)

From Lemma 6.1.3, Lemma 6.1.6 and (6.1.11),

Fa[m,s]

1, { { { { { {0, ={ { { { { { s (m,s) [2] {a1 Fm−2,s + ∑l=1 Γa[l] yl ,

for any integers m, s ≥ 0.

when m = s = 0 or m = 2, s = 0; when m = 0, s ≥ 1 or m = 1, s ≥ 0, and m ≠ s(mod 2); otherwise,

(6.2.5)

204 | 6 Inner equations fourth part Lemma 6.2.3. For any integers m, s ≥ 0, Fa[m,s] ∈ ℛ+ {y} whenever a1 = a2 = 1. Proof. Similar reasoning to the proof of Lemma 6.2.2. One might like to see whether, or not, Fa[m,s] is determined by a finite number of entries in y for any integers m, s ≥ 0 given. Lemma 6.2.4. For any integers m, s ≥ 0 given, Fa[m,s] is independent of y1 , y2 and all yl , l ≥ s + 1. Proof. As we have seen, when m + s ≤ 4, the lemma is true. For m + s ≥ 6, m ≥ 3, assume that the Fa[p,q] satisfy the conclusion of the lemma for p + q ≤ m + s − 1. By induction, we prove that Fm,s satisfies the conclusion. We take into account (6.1.4). Because of (m−2)+s = m+s−2 < m+s, the assumption [2] is independent of y1 , y2 and all yl , l ≥ s + 1. We take into account shows that Fa[m−2,s] (6.2.3). For any integer 1 ≤ l ≤ s, s−l

s−l−1

j=1

j=1

Γ(m,s) = a2 Fa[l+m−2,s−l] − ∑ Fa[m−1,j] Fa[l−1,s−l−j] − ∑ Fa[m,j] Fa[l,s−l−j] . a[l] We have (l + m − 2) + (s − l) = m + s − 2, (m − 1) + j ≤ m + s − l − 1 ≤ m + s − 2, (l − 1) + (s − l − j) = s − j − 1 ≤ s − 2 ≤ m + s − 1 (m ≥ 3), m + j ≤ m + (s − l) ≤ m + s − 1 and (m,s) l + (s − l − j) = s − 1 ≤ m + s − 1(m ≥ 3). From the assumption, Γa[l] is independent of y1 , y2 and all yl , l ≥ s + 1. Therefore, from (6.2.5), Fa[m,s] is independent of y1 , y2 and all yl , l ≥ s + 1. This lemma tells us that, for any integers m, s ≥ 0 given, Fa[m,s] ∈ ℛ{ys }, where ys = (y1 , y2 , . . . , ys ), i. e., a vector of s dimensions. Lemma 6.2.5. When m = 2. For any integer s ≥ 1, Fa[2,s] = 0 if a1 = a2 . Proof. From the evaluation above, Fa[2,1] = 0 and Fa[2,2] = a1 (a2 −a1 )y2 . The conclusion is true for s ≤ 2. In the general case of s ≥ 3, one might think of the assumption that Fa[2,l] = 0 for 1 ≤ l ≤ s − 1 when a2 = a1 . By induction on s, we prove Fa[2,s] = 0. [2] = 0 for = 0 and Γ(2,s) From (6.2.5), it is only necessary to prove the two facts: Fa[0,s] a[l] 1 ≤ l ≤ s. The first fact follows from (6.1.4). As regards the second fact, from (6.2.3), s−l−1

Γ(2,s) = a2 Fa[l,s−l] − ∑ Fa[2,j] Fa[l,s−l−j] − Fa[2,s−l] Fa[l,0] . a[l] j=0

Furthermore by assumption from (2 + (s − l) ≤ 2 + s − 1), s−l−1

Γ(2,s) = a2 Fa[l,s−l] − ∑ Fa[2,j] Fa[l,s−l−j] − Fa[2,0] Fa[l,s−l] = 0. a[l] j=1

Therefore the conclusion can be drawn.

6.2 Solution general simple inner case

| 205

This lemma enables us to only discuss m ≥ 3 without loss of generality for a2 = a1 . Lemma 6.2.6. For any integer m ≥ 3 given, if integer s ≥ m + 1, then Fa[m,s] = 0 under the condition of a2 = a1 . Proof. The conclusion is true when m + s ≤ 4. In general for m + s ≥ 5, the assumption can be made that, for any integers p + q ≤ m + s − 1, Fp,q = 0 whenever q ≥ p + 1 for p, q ≥ 0. By induction on m + s, we show Fa[m,s] = 0 for s ≥ m + 1. From (6.2.5), it is only necessary to prove the two facts: fact 1, that, for s ≥ m + 1, [2] = 0, and fact 2, that, for any integer 1 ≤ l ≤ s, Γ(m,s) = 0. Fa[m−2,s] a[l] We consider fact 1. Consider m−2 i

[2] = ∑ ∑ Fa[i,j] Fa[m−2−i,s−j] . Fa[m−2,s] i=0 j=0

By assumption, only when s − j ≤ m − 2 − i, i. e., j ≥ i + (s − m) + 2, Fm−2−i,s−j ≠ 0 is a possibility. However, s ≥ m + 1 leads to j ≥ i + 3 ≥ i + 1. By assumption Fa[i,j] = 0. [2] = 0. Therefore, Fa[m−2,s] We consider fact 2. From (6.2.3), Γ(m,s) = [a2 Fa[l+m−2] − Fa[m−1] Fa[l−1] − Fa[m] Fa[l] ]s−l . a[l] By assumption, Γ(a[m,s]) = 0 for 1 ≤ l ≤ s. a[l] In consequence, the conclusion can be drawn. Furthermore, for m + s ≥ 6, we evaluate Fa[m,s] . When m + s = 6, consider Fa[6,0] = 4a1 + a21 (Corollary 6.1.5), Fa[5,1] = Fa[4,2] = 0 (Lemma 6.2.4), F3,3 and Fa[2,4] = 0 (by (6.2.5)). Only Fa[3,3] is unknown. We proceed on the basis of (6.2.5). Fa[1] = 0 leads to, for [2] [k] = 0. Then, what remains is the summation = 0 and hence Fa[1,3] any integer k ≥ 2, Fa[1] y over l : 1 ≤ l ≤ 3. By (6.2.3) and Lemma 6.2.4, of Γ(3,3) a[l] l 3

yl = a1 (2a2 − a1 )y3 . ∑ Γ(3,3) l l=1

From (6.2.5), Fa[3,3] = a1 (2a2 − a1 )y3 . Lemma 6.2.7. For integer m ≥ 3 given, let Fa[m,m] (ym ) be the part of Fa[m,m] with ym , then y3 , when m = 3; Fa[m,m] (ym ) = { (m,m) Γa[m] , when m ≥ 4,

(6.2.6)

where, for a1 = a2 = 1, Γ(m,m) 1[m] =

(2m − 2)! (2⌊m/2⌋)! − . (m − 1)!m! ⌊m/2⌋!(⌊m/2⌋ − 1)!

(6.2.7)

206 | 6 Inner equations fourth part Proof. We proceed on the basis of m > (m−2)+1 = m−1. From the proof of Lemma 6.2.6, [2] = 0. From (6.2.5) and Lemma 6.2.4, Fa[m−2,m] m

m

l=1

l=3

yl . yl = ∑ Γ(m,m) Fa[m,m] = ∑ Γ(m,m) a[l] a[l] . By s = m. There is no ym in for 3 ≤ l ≤ m − 1 and hence Fa[m,m] (ym ) = Γ(m,m) a[m] Furthermore, from (6.2.3) and (6.1.14), Γ(m,m) a[l]

Γ(m,m) 1[m] =

(2m − 2)! − (F1[m−1,0] F1[m−1,0] + F1[m,0] F1[m,0] ). (m − 1)!m!

From (6.2.5), F1[m,0] F1[m,0] F1[m−1,0] F1[m−1,0] + F1[m,0] F1[m,0] = { F1[m−1,0] F1[m−1,0]

when m = 2k; when m = 2k + 1.

Thus, from (6.1.14), F1[m−1,0] F1[m−1,0] + F1[m,0] F1[m,0] =

(2k)! . k!(k + 1)!

Because of k = ⌊m/2⌋, (6.2.7) is done. Now, we are allowed to find the solution of equation (6.1.1) in the form of a polynomial in ℛ{x, ys } whenever s is given in any way. Theorem 6.2.8. Let fa[gsi] be the solution of equation (6.1.3). For any integers m, s ≥ 0, [gsi] = [𝜕xm fa[gsi] ]s , then denote Fa[m,s] 1, when m = s = 0; { { { { 0, when m ≠ s(mod 2), or 0 ≤ m ≤ 1, { { { { { or m = 2, s ≥ 1, or s ≥ m + 1, { { { { or 1 ≤ s ≤ 2; { { { { at1 m! { { , when s = 0 and m = 2t, t ≥ 1; { t!(t+1)! { { { { { { { 0, when m = 3, s = 1; [gsi] ={ Fa[m,s] } when m + s = 4; { 0, when m = s = 2 { { { { { 0, when (m, s) = (5, 1), (4, 2); { { { } when m + s = 6; { { a1 (2a2 − a1 )y3 , when m = s = 3, { { { { { { 0, when 1 ≤ s ≤ 2; { } { } { } gsi[2] s (m,s) { { Γ y , when 3 ≤ s ≤ m without y ; + a F ∑ { m } 1 a[m−2,s] otherwise, l=3 a[l] l { { } { (m,m) } when s = m with ym { Γa[m] ym , } (6.2.8) [gsi] is given by (6.2.3) when F = F . where Γ(m,s) a[l] Proof. We proceed on the basis of Lemma 6.2.1–Lemma 6.2.7; the conclusion can be directly obtained.

6.3 Explicitness general simple inner case

| 207

6.3 Explicitness general simple inner case Because no explicision has yet directly been deduced from Theorem 6.2.8, the relationship between the solutions of general simple inner equation (6.1.1), or equivalently equation (6.1.2) and its restriction for a0 = a1 = a2 = 1 has to be investigated on account of the latter being meaningful in graph theory. Let fa[gsi] and f[gsi] ∈ ℛ{x, y} be the solutions of, respectively, equation (6.1.1) and [gsi] [gsi] and Fm,s ∈ ℛ{y} for integers m, s ≥ 0. Denote by its restriction determined by Fa[m,s]

[gsi] . For any n ∈ 𝒩m,s , Ua[m,n] ∈ 𝒩m,s the set of power vectors in Fa[m,s] and hence Fm,s [gsi]

[gsi]

[gsi] ℛa = ℛ{a} and Um,n | ∈ ℛ1 = ℛ are the coefficients of yn in, respectively, Fa[m,s] and [gsi]

[gsi] Fm,s .

Then,

[gsi] [gsi] = 𝜕yn Fa[m,s] Ua[m,n]

[gsi] [gsi] . = 𝜕yn Fm,s and Um,n

[gsi] is a Theorem 6.3.1. For any integers m, s ≥ 1 and integral vector n ∈ 𝒩m,s , Ua[m,n] polynomial of a0 = 1, a1 and a2 with degree at most l + 1 where 2l = m + s such that [gsi] [gsi] . |a=1 = Um,n Ua[m,n]

Proof. We proceed on the basis of Theorem 6.2.8. By induction, the conclusion can easily be drawn. For an embedding μ(G) of a general simple planar graph G, let n = (n1 , n2 , n3 , n4 , . . .) be the face-partition vector where ni is the number of non-root faces of valency i for integer i ≥ 1. A vertex v as a point of μ(G) such that μ(G) − v that has at least one component greater than μ(G) is called a cut-vertex of G. Now, G is always assumed to be connected. It is allowed to restrict a face of maximum length boundary at a cut-vertex as the outerface without loss of generality. The cut-valency of a cut-vertex v is the number of components of μ(G) − v. The closure of such a component is called a cut-component. If a cut-component is not a path, then it is called a fat-component. The number of fatcomponents contributing to the cut-valency of v is called the fat-valency of v. A graph without cut-vertex is said to be nonseparable. A maximal nonseparable edge-induced subgraph is called a block of the graph. For a cut-vertex v, let a = av = ρvcut , c = cv and b = bv = ρv be, respectively, the cutvalency, fat-valency and the valency of v. Because of the planarity, all cut-components at v have a rotation in the order of the cut-components G1 , G2 , . . . , Ga . At cut-vertex v, let ti be the contribution of Gi to the valency of v, then t1 + t2 + ⋅ ⋅ ⋅ + ta = b. Three cases have to be considered at cut-vertex v for the dual embedding μ(G). Case I Reflection for Gi , 1 ≤ i ≤ c, it turns out that we have [gsi] πvI = 2c−1

(topologically) distinct planar embeddings.

(6.3.1)

208 | 6 Inner equations fourth part Case II For cut rotation at v, it turns out that we have [gsi] = (a − 1)! πvII

(6.3.2)

(topologically) distinct planar embeddings. Case III For an integer k ≥ 1 given, denote I = {i1 , i2 , . . . , ik } by distributing all Gj , j ∈ ̸ I, into s = (ti1 − 1) + (ti2 − 1) + ⋅ ⋅ ⋅ + (tik − 1) angles on 𝒢ℐ = {Gi | ∀i ∈ I} at v as (x1 , x2 , . . . , xts ) in the s angles such that s

∑ xj = c − k; j=1

0 ≤ xj ≤ c − k, 1 ≤ j ≤ s. Let us denote by c−k ⟨ ⟩ x1 , . . . , xts the number of ways for such distributions, then it turns out that we have s c−k ⟩ ∏ xj ! πk[gsi] = ⟨ x1 , . . . , xts j=1

(topologically) distinct planar embeddings. From the disjointness of the embeddings, we have c−1 c [gsi] = ∑ ( )πk[gsi] πvIII k k=1

(6.3.3)

(topologically) distinct planar embeddings. Lemma 6.3.2. For μ(G), let VCut be the set of all cut-vertices. Then, all cut-vertices turn out to be [gsi] [gsi] [gsi] n[gsi] Cut (G) = ∏ (πvI πvII πvIII ) v∈VCut

(6.3.4)

(topologically) distinct planar embeddings. Proof. We take into account the independency among Case I, Case II and Case III; from (6.3.1), (6.3.2) and (6.3.3), the conclusion can be drawn. We consider a dual embedding μ(G) of general simple graph G without cut-vertex. If two vertices u and v have the property that μ(G) − {pu , pv } have at least two components, then {u, v} is called a splitting pair. Since an embedding is a point set in its own right on the plane, pu and pv are the corresponding points of u and v.

6.3 Explicitness general simple inner case

| 209

Because of μ as a point set, any component of μ − {pu , pv } is called a splitting slice, or slice. A splitting slice that is not a path is called a splitting chunk, or chunk. Let S be the set of all splitting pairs consisting of {ui , vi } for 1 ≤ i ≤ s = s(S) = |S| in μ = μ(G). All components of μ − S are splitting slices in μ(G). For a splitting pair, its slice valency is the number of splitting slices incident with it. Similarly, for chunk valency. Observation 6.3.3. Let t and 1 ≤ p ≤ t be, respectively, the slice valency and chunk valency of a splitting pair on a dual embedding of general simple planar graph G; then the splitting pair produces [gsi] = 2p−1 (t − 1)! πspp

(6.3.5)

(topologically) distinct planar embeddings of G. Proof. We take into account the independency between rotation on slice valency and reflection on a chunk. By considering the connectivity of the embedding, the conclusion can be drawn. Let p be the number of all splitting chunks and ti the slice valency at the splitting pair {ui , vi }, 1 ≤ i ≤ s on μ(G). Lemma 6.3.4. For a general simple planar graph G, the number of topologically distinct dual embeddings of G generated by all splitting pairs is s

p−s n[gsi] ∏(ti − 1)! Nct = 2 i=1

(6.3.6)

where pi and ti are, respectively, the number of chunks the slice valency at splitting pair {ui , vi }, 1 ≤ i ≤ s = |S| such that p = p1 + p2 + ⋅ ⋅ ⋅ + ps . Proof. This is a result of Lemma 6.3.2. This lemma is a refinement and a concise form of Theorem 10.4.3 with (10.4.13) in Liu YP [72] (2017, pp. 196–197), or Theorem 7.4.3 with (7.4.13) in Liu YP [55] (1994, p. 142). We proceed on the basis of Lemma 6.3.2 and Lemma 6.3.4; the result is useful in the context of determining the number of planar embeddings of a Eulerian ordinary graph. Lemma 6.3.5. For a general simple planar graph G, the number of (topologically) distinct planar embeddings of G is [gsi] n[gsi] (G) = n[gsi] Cut (G)nNct (G) [gsi] where n[gsi] Cut (G) and nNct (G) are, respectively, given by (6.3.4) and (6.3.6).

(6.3.7)

210 | 6 Inner equations fourth part Proof. We consider the uniqueness of 3-connected embedding of a graph. Because a graph without vertex-cut set of one or two vertices has to be 3-connected, only the two independent cases with and without cut-vertex need to be considered. Lemma 6.3.2 and Lemma 6.3.4 yield the conclusion. ̃ Denote by t = aut(G) the semi-automorphism group order of general simple planar graph G. The relationship between embeddings and upper dual maps of G can be shown via t. Lemma 6.3.6. Let ℰ [gsi] (G) and ℳ[gsi] (G) be, respectively, the sets of all planar embeddings and upper maps of general simple planar graph G. Then, 󵄨󵄨 [gsi] 󵄨󵄨 2ϵ 󵄨󵄨 [gsi] 󵄨󵄨 (G)󵄨󵄨 󵄨󵄨ℳ (G)󵄨󵄨 = 󵄨󵄨ℰ t

(6.3.8)

where ϵ = ϵ(G) is the size of G. Proof. Refer to Liu YP [59] (2003, pp. 225–226). [gsi] be the set of all general simple planar graphs with rooted semi-valency Let 𝒢m|s 2m (i. e., the number of semi-edges incident to the root-vertex) and the un-rooted semisize 2s (i. e., the number of semi-edges not incident to the root-vertex). [gsi] the set of semi-automorphism group orders of general simple plaDenote by 𝒯m|s [gsi] [gsi] [gsi] . be the set of all partition vectors n in 𝒢m|s . Let 𝒥m|s nar graphs in 𝒢m|s

[gsi] be the number of (topologically) distinct embeddings of general Lemma 6.3.7. Let Em|n

[gsi] . Then we have simple planar graphs in 𝒢m|n [gsi] = Em|n

∑ n[gsi] (G)

(6.3.9)

[gsi] G∈𝒢m|n

where n[gsi] is given in (6.3.7). [gsi] Proof. Given the partition n. By considering all embeddings of each graph in 𝒢m|s with n, the conclusion is easily drawn. [gsi] be the set of semi-automorphism We proceed on the basis of Lemma 6.3.7. Let 𝒯m|n

[gsi] [gsi] [gsi] with semi, the set of all graphs in 𝒢m|n and 𝒢m|n,t group orders among graphs in 𝒢m|n ̃ automorphism group order t. Denote by aut(G) the semi-automorphism group order

of G.

Lemma 6.3.8. The number of root-isomorphic classes of upper maps of all graphs in [gsi] 𝒢m|n is = A[gsi] m|n

∑ [gsi] t∈𝒯m|n

m + π(n) [gsi] Em|n,t t

(6.3.10)

6.3 Explicitness general simple inner case | 211

) and = Am|n (ℳ[gsi] where A[gsi] m|n m|n [gsi] = Em|n,t

∑ n[gsi] (G)|aut(G)=t ̃

(6.3.11)

[gsi] G∈𝒢m|n

as determined by (6.3.9). Proof. We proceed on the basis of Lemma 6.3.6. By considering m + π(n) = 2ϵ, the conclusion is easily drawn. This lemma enables us to turn out the solution fa[gsi] of equation (6.1.1) when a0 = [gsi] a1 = a2 = 1. Denote f[gsi] = fa |a0 =a1 =a2 =1 as determined by Fm,s = 𝜕xm f[gsi] |π(y)=s for m, s ≥ 0. Lemma 6.3.9. For integers m, n ≥ 1, we have [gsi] Fm,s =

∑ [gsi] n∈𝒥m|s

yn A[gsi] m|n

(6.3.12)

is given by (6.3.10). where A[gsi] m|n Proof. The set of all underline graphs of general simple planar maps with m and n is [gsi] [gsi] = Um,n is shown in Theorem 6.3.1. From Lemma 6.3.8, the the same as 𝒢m,n , A[gsi] m|n conclusion can be drawn. We proceed on the basis of Lemma 6.3.9 and Theorem 6.3.1; we are allowed to illustrate our main result of this section. [gsi] determine the solution of equation (6.1.1) as Theorem 6.3.10. For m, s ≥ 1, Fa[m,s] [gsi] Fa[m,s] =

∑ [gsi] n∈𝒥m|s

[gsi] Ua[m,n] yn

(6.3.13)

[gsi] where a = (1, a1 , a2 ) and Ua[m,n] is the polynomial of a with degree at most m + s + 1 given in Theorem 6.3.1.

Proof. This is a result of Theorem 6.3.1 and Lemma 6.3.9. [gsi] is evaluated by the procedure shown in (6.2.8) or directly The polynomial Ua[m,n] from the inner structures of the solution on its own right. [gsi] [gsi] |a=1 determine the solution of equation Corollary 6.3.11. For m, s ≥ 1, Fm,s = Fa[m,s] (3.5.1) when a = 1 as [gsi] Fm,s =

[gsi] where Um,n = A[gsi] is given in (6.3.10). m|n

∑ [gsi] n∈𝒥m|s

[gsi] n Um,n y

(6.3.14)

212 | 6 Inner equations fourth part Proof. This is a direct result of Theorem 6.3.10. [gsi] This is a direct explicision (explicit expression) of Fm,s for integers m, s ≥ 1.

6.4 Restrictions general simple inner case Consider the equation for f ∈ ℛ{x, y} 2 2 { { {x ∫ yδx,y (uf |x=u ) = ∫((1 + xy)f − 1)f |x=y − x f ; y y { { { f |x=0⇒y=0 = 1, {

(6.4.1)

where y = (y1 , y2 , y3 , . . .) ∈ 𝒱 . A solution of equation (6.4.1) is involved with general simple planar maps as shown in [47] (Liu YP, 1989) and [61] (Liu YP, 2009, p. 194). This equation is shown to be equation (6.1.1) when a = 1. For convenience, equation (6.4.1) is transformed to its equivalence on ℛ{x, y} as { f = 1 + x2 f 2 + x ∫(yδx,y (uf |x=u )) { { { { { y { { (yf | − xf ∫ { x=y ) − (f − 1) ∫(f |x=y − 1); { { { y y { { { { f| = 1. { x=0⇒y=0

(6.4.2)

[2] = 𝜕xm f 2 the For any integer m ≥ 0, let Fm = 𝜕xm f (or 𝜕m f ) ∈ ℛ{y}. Denote by Fa[m]

coefficient of xm in f 2 , then

m

[2] Fm = ∑ Fi Fm−i . i=0

(6.4.3)

From the initiation of equation (6.4.1), Fa[0] = 1. Thus, for any integer m ≥ 0, 1, when m = 0; { { { [2] Fm = {2Fa[1] , when m = 1; { { m−1 {2Fm + ∑i=1 Fi Fm−i , when m ≥ 2.

(6.4.4)

Observation 6.4.1. If f is a solution of equation (6.4.1), then F0 = 1. Proof. We have Fa[0] = a0 by the initiation of equation (6.4.1). However, Fa[0] = 1 by the first equality of equation (6.4.2). Therefore, a0 = 1. This observation shows us that the initiation of equation (6.4.1) is consistent.

6.4 Restrictions general simple inner case | 213

Let Am = 𝜕m (f ∫y yf |x=y ) for integer m ≥ 0. Because of f = ∑ Fm x m , m≥0

we have Am = Fm (y1 + ∑ Fi−1 yi ). i≥2

(6.4.5)

For any integer m ≥ 0, let Bm = 𝜕m ((f − 1) ∫y (f |x=y − 1)). Because of f = 1 + ∑ Fm xm , m≥1

we have 0, when m = 0; Bm = { Fm (∑i≥1 Fi yi ), when m ≥ 1,

(6.4.6)

for integer m ≥ 0. According to the evaluation in Section 3.1, δx,y (uf |x=u ) = ∑ xm ∑ yl Fl+m . m≥0

l≥0

Let ∇m = 𝜕xm (δx,y (uf |x=u )), then ∇m = ∑ yl Fl+m l≥0

(6.4.7)

for any integer m ≥ 0. Furthermore, let △m = ∫y (y∇m ), then, by (6.4.7), △m = ∑ Fl+m yl+1 l≥0

(6.4.8)

for any integer m ≥ 0. Theorem 6.4.2. Any solution of equation (6.4.1) is determined by the system of equations for Fa[m] ∈ ℛ{y} for m ≥ 0 1 − B0 , when m = 0; { { { Fm = {△0 − A0 − B1 , when m = 1; { { [2] {Fm−2 + △m−1 − Am−1 − Bm , when m ≥ 2.

(6.4.9)

Proof. If f is a solution of equation (6.4.1), we have f ∈ ℛ{x, y} and Fm = 𝜕xm f for m ≥ 0. It is seen that the Fm for m ≥ 0 form a solution of equations (6.4.9) from (6.4.2) through (6.4.6). Conversely, if Fm ∈ ℛ{y} for m ≥ 0 form a solution of (6.4.9), then (6.4.3) through (6.4.6) are checked for f ∈ ℛ{x, y} as determined by Fm = 𝜕xm f for m ≥ 0 satisfying equation (6.4.2) and hence equation (6.4.1). The conclusion can be drawn.

214 | 6 Inner equations fourth part We proceed on the basis of this theorem and equations (6.4.9). In order to determine Fm ∈ ℛ{y} for m ≥ 0, it is still necessary to introduce another parameter s ≥ 0, so that Fm = ∑ Fm,s s≥0

(6.4.10)

and, for any non-negative integers m and s given, Fm,s is a polynomial on ℛ{y}. Let 𝒥m be the set of power vectors n = (n1 , n2 , n3 , . . .) with term yn in Fm where y = (y1 , y2 , y3 , . . .), and 𝒥m,s , i. e., the subset of 𝒥m,s of 𝒥m such that π(n) = ∑i≥1 ini = s. From (6.4.9),

Fm,s

1, when m = s = 0; { { { = {△0,s − A0,s − B1,s , when m = 1; { { [2] {Fm−2,s + a2 △m−1,s −Am−1,s − Bm,s , when m ≥ 2,

(6.4.11)

for any integer s ≥ 0. On this basis, F0 , F1 and Fm,0 for m ≥ 0 are determined. Lemma 6.4.3. For F0 , we have 1, when s = 0; F0,s = { 0, when s ≥ 1.

(6.4.12)

Proof. This is a result of the initiation of equation (6.4.1). This lemma shows us that Fa[0] is determined. Lemma 6.4.4. We consider Fm,0 . For any integer m ≥ 0, 1, when m = 0; for m ≥ 1, { { { Fm,0 = {0, when m = 1(mod 2); { { m−1 {∑i=0 Fi,0 Fm−2−i,0 , when m = 0(mod 2).

(6.4.13)

Proof. See the proof of Lemma 6.1.4 for a = 1. This lemma enables us to determine all Fm,0 from the initiation F0,0 = 1 in any way by following the procedure. Corollary 6.4.5. For any integer m ≥ 0. Let Fm,0 = Fa[m,0] |a=1 , then 0, Fm,0 = { (2k)!

k!(k+1)!

when m = 2k − 1, k ≥ 1; ,

when m = 2k, k ≥ 0.

Proof. This is the Catalan number Ck from the recursion in (6.4.13).

(6.4.14)

6.4 Restrictions general simple inner case | 215

In the sense of combinatorial maps, the number shows how many distinct rootisomorphic classes of plane trees with size k. Lemma 6.4.6. We consider F1 . For any integer s ≥ 0, F1,s = 0,

i. e.,

F1 = 0.

(6.4.15)

Proof. See the proof of Lemma 6.1.6 for a = 1. The last three lemmas enable us to only observe m ≥ 2 and s ≥ 1 in what follows. Theorem 6.4.7. Equation (6.4.1) is well defined on ℛ{x, y}. Proof. This is a direct result of Theorem 6.1.7 for a = 1. In order to evaluate the solution of equation (6.4.1) in the form of a finite sum with all terms positive, further constructions of the solution have to be investigated. According to what was discussed in Section 6.2, one might see that, for any integers m, s ≥ 1, s−1

{ { Δm,s = ∑ Fl+m,s−l−1 yl+1 ; { { { { l=0 { { s { { A = [Fm Fl−1 ]s−l yl ; ∑ m,s { { { l=1 { { s { { { { {Bm,s = ∑[Fm Fl ]s−l yl . l=1 {

(6.4.16)

Lemma 6.4.8. For any integers m, s ≥ 0, if m + s = 1(mod 2), then Fm,s = 0. Proof. See the proof of Lemma 6.2.1 for a = 1. This lemma shows that to determine Fm,s it is only necessary to address m + s = 0(mod 2). Because of Lemma 6.4.3, Lemma 6.4.4 and Lemma 6.4.6, it is only necessary to discuss Fm,s for m ≥ 2 and s ≥ 1. Thus, all Fm,s for m + s ≤ 3 are done. When m + s = 4, only F3,1 and F2,2 need to be determined. We consider F3,1 . We proceed on the basis of Corollary 6.4.5 and Lemma 6.4.8, Δ2,1 = F2,0 y1 = y1 ,

A2,1 + B3,1 = [F2 F0 ]0 y1 = y1 [2] F1,1

and

= 2[F0 F1 ]1 = 0.

By employing (6.4.11), [2] + Δ2,1 − (A2,1 + B3,1 ) = 0 + y1 − y1 = 0. F3,1 = a1 F1,1

We consider F2,2 . On account of Lemma 6.4.4 and Lemma 6.4.5, Δ1,2 = F1,1 y1 + F2,0 y2 = y2 ,

216 | 6 Inner equations fourth part A1,2 + B2,2 = F2,0 F2,0 y2 = y2 [2] F0,2

and

= 2F0,0 F0,2 + F0,1 F0,1 = 0.

By employing (6.4.11), [2] + Δ1,2 − (A1,2 + B2,2 ) = 0 + y2 − y2 = 0. F2,2 = F0,2

The above two results show that when m + s = 4, Δm−1,s − (Am−1,s + Bm,s ) ≥ 0. Lemma 6.4.9. For any integers m, s ≥ 1, Δm−1,s ≥ Am−1,s + Bm,s . Proof. We proceed on the basis of Theorem 6.4.7. Equation (6.4.3) has, and is the only one to have, a solution. In map theory, Δm−1,s − (Am−1,s + Bm,s ) is the number of distinct root-isomorphic classes in a type of general simple planar maps. The conclusion holds. This lemma tells us that, for any integers m, s ≥ 1, Γm,s ∈ ℛ+ {y}. For any integers m ≥ 2 and s ≥ 0, let Γm,s = Δm−1,s − (Am−1,s + Bm,s ), then from (6.4.16), s

Γm,s = ∑[Fl+m−2 − Fm−1 Fl−1 − Fm Fl ]s−l yl . l−1

(6.4.17)

For convenience, for integers 1 ≤ l ≤ s, denote Γ(m,s) = [Fl+m−2 − Fm−1 Fl−1 − Fm Fl ]s−l . l

(6.4.18)

From Lemma 6.4.8 and Lemma 6.4.9, = 0, when m ≠ s(mod 2); Γ(m,s) { l ≥ 0, when m = s(mod 2).

(6.4.19)

From Lemma 6.4.3, Lemma 6.4.6 and (6.4.11),

Fm,s

1, when m = s = 0 or m = 2, s = 0; { { { { { {0, when m = 0, s ≥ 1 or m = 1, s ≥ 0, ={ { and m ≠ s(mod 2); { { { { [2] s (m,s) {Fm−2,s + ∑l=1 Γl yl , otherwise,

(6.4.20)

for any integers m, s ≥ 0. Lemma 6.4.10. For any integers m, s ≥ 0, Fm,s ∈ ℛ+ {y}. Proof. Similar reasoning to the proof of Lemma 6.4.9. One might like to see whether, or not, Fm,s is determined by a finite number of entries in y for any integers m, s ≥ 0 given.

6.4 Restrictions general simple inner case | 217

Lemma 6.4.11. For any integers m, s ≥ 0 given, Fm,s is independent of y1 , y2 and all yl , l ≥ s + 1. Proof. See the proof of Lemma 6.2.4 for a = 1. This lemma tells us that, for any integers m, s ≥ 0 given, Fa[m,s] ∈ ℛ{ys } where ys = (y1 , y2 , . . . , ys ), i. e., a vector of s dimensions. Lemma 6.4.12. When m = 2, for any integer s ≥ 1, F2,s = 0. Proof. See the proof of Lemma 6.2.5 for a = 1. This lemma enables us to only discuss m ≥ 3 without loss of generality. Lemma 6.4.13. For any integer m ≥ 3 given, if integer s ≥ m + 1, then Fa[m,s] = 0. Proof. See the proof of Lemma 6.2.6 for a = 1. Furthermore, for m + s ≥ 6, we evaluate Fm,s . When m + s = 6. Consider F6,0 = 2 (Corollary 6.4.5), F5,1 = F4,2 = 0 (Lemma 6.4.11), F3,3 and F2,4 = 0 (by (6.4.20)). Only F3,3 is unknown. We proceed on the basis of (6.4.20). Because of F1 = 0 show[2] ing that, for any integer k ≥ 2, F1[k] = 0 and hence F1,3 = 0. Then, is remains to consider 1 ≤ l ≤ 3. By (6.4.18) and Lemma 6.4.11, 3

3

l=1

l=1

yl = ∑[Fl+1 − F2 Fl−1 − F3 Fl ]3−l yl = y3 . ∑ Γ(3,3) l From (6.4.20), F3,3 = y3 . Lemma 6.4.14. For integer m ≥ 3 given, let Fm,m (ym ) be the part of Fm,m with the term ym , then y3 , when m = 3; Fm,m (ym ) = { (m,m) Γm ym , when m ≥ 4,

(6.4.21)

where Γ(m,m) = m

(2m − 2)! (2⌊m/2⌋)! − . (m − 1)!m! ⌊m/2⌋!(⌊m/2⌋ − 1)!

(6.4.22)

Proof. See the proof of Lemma 6.2.7 for a = 1. Now, we are allowed to find the solution of equation (6.4.1) in the form of a polynomial in ℛ{x, ys } whenever s is given in any way. Theorem 6.4.15. Let f[gsi] be the solution of equation (6.4.1). For any integers m, s ≥ 0, [gsi] denote Fm,s = [𝜕xm fgsi ]s , then

218 | 6 Inner equations fourth part

[gsi] Fm,s

1, when m = s = 0; { { { { { 0, when m ≠ s(mod 2), or 0 ≤ m ≤ 1, { { { { or m = 2, s ≥ 1, or s ≥ m + 1, { { { { { or 1 ≤ s ≤ 2; { { { (m)! { { , when s = 0 and m = 0(mod 2); { ⌊m/2⌋!(⌊m/2⌋+1)! { { { { { { 0, when m = 3, s = 1; { } when m + s = 4; ={ { 0, when m = s = 2; { { { { { 0, when (m, s) = (5, 1), (4, 2); { { { } when m + s = 6; { { y3 , when m = s = 3; { { { { { { 0, when 1 ≤ s ≤ 2; } { } { } { [gsi][2] s (m,s) { { Γ y , when 3 ≤ s ≤ m without y ; F + ∑ l m { l=3 m−2,s l } { } { } { (m,m) Γ y , when s = m with y m m m } {

otherwise,

(6.4.23) (m,m) [gsi] where Γ(m,s) and Γ are, respectively, given by (6.4.18) where F = F and (6.4.22). m l Proof. We proceed on the basis of Lemma 6.4.8–Lemma 6.4.14; the conclusion can be directly obtained. [gsi] We proceed on the basis of Theorem 6.4.15. In the example below, Fm,s of m+s = 8 [gsi] [gsi] [gsi] [gsi] [gsi] are provided. They are F8,0 = 14, F7,1 = F6,2 = 0, F5,3 = 8y3 and F4,4 = y4 .

Example 6.4.1. Root-isomorphic classes of general simple planar maps by root-face valency and face-partition vector. In Figure 6.4.1, the classes of such maps are shown [gsi] for size 4. By noticing that m+s is 2 times the size, F2,0 = 1 is m+s = 2, i. e., the case of size 1. We consider only one such map. Its root-face valency is 2 without face-partition vector. As shown in the figure a = x2 . Similarly for size two, i. e., m + s = 4, shown as [gsi] b = 2x 4 , i. e., F4,0 = 2. For size 3, i. e., m + s = 6, there are c + d + e = (3x6 ) + (2x6 ) + [gsi] gsi (x3 y3 ), i. e., F6,0 = 3(c) + 2(d) = 5 and F3,3 = y3 . For size 4, i. e., m + s = 8, there are

[gsi] f + g + h + i + j = (4x8 ) + (8x) + (2x8 ) + (8x5 y3 ) + (x4 y4 ), i. e., F8,0 = 4(f ) + 8(g) + 2(h) = 14,

[gsi] [gsi] [gsi] F5,3 (y5 ). = 8(i) = 8x5 y3 and F4,4 = y4 . In addition, k + l + m + n + o + p = 10x5 y5 = F5,5

6.5 Bipartite simple inner model Second, we discuss the equation for f ∈ ℛ{x, y} 2 2 2 2 { {a2 x ∫ y δx2 ,y2 f |x2 =u = (f − 1) ∫ f |x=y − a1 x f ; y y { { {f |x=0⇒y=0 = a0 ,

where a0 , a1 , a2 ∈ ℛ and f ∈ ℛ{x, y}. This is equation (12) in Introduction.

(6.5.1)

6.5 Bipartite simple inner model | 219

Figure 6.4.1: Root-isomorphic classes of general simple planar maps by face-partition vector.

In [56] (Liu YP, 1999, p. 231) and [58] (Liu YP, 2001, p. 271), one might see the main equation of equation (6.5.1), or its equivalence when a0 = a1 = a2 = 1. However, some errors in symbols should be paid attention to. Because of a solution of equation (6.5.1) for a0 = a1 = a2 = 1 is involved with bipartite simple planar maps, this equation is called a bipartite simple inner model.

220 | 6 Inner equations fourth part The first equality of equation (6.5.1) is transformed on ℛ{x, y} as follows. By adding (1 − f ) to the both sides of the equality, (1 − f ) + a2 x2 ∫ y2 δx2 ,y2 f |x2 =u = (1 − f ) + (f − 1) ∫ f |x=y − a1 x2 f 2 . y

y

By taking out the factor (f − 1) of the first two terms on the right hand side, (1 − f ) + a2 x2 ∫ y2 δx2 ,y2 f |x2 =u = (f − 1)(∫ f |x=y − 1) − a1 x2 f 2 . y

y

By adding f + a1 x2 f 2 on the both sides, 1 + a1 x2 f 2 + a2 x2 ∫ y2 δx2 ,y2 f |x2 =u = f + (f − 1)(∫ f |x=y − 1). y

y

Then, by adding −(f − 1)(∫y f |x=y − 1) to its two sides, we have f = 1 + a1 x2 f 2 + a2 x2 ∫ y2 δx2 ,y2 f |x2 =u − (f − 1)(∫ f |x=y − 1). y

y

Theorem 6.5.1. Equation (6.5.1) is equivalent to 2 2 2 2 { { {f = 1 + a1 x f + a2 x ∫(y δx2 ,y2 f |u=x2 ) − (f − 1)(∫ f |x=y − 1); y y { { { f | = a , 0 { x=0⇒y=0

(6.5.2)

on ℛ{x, y}. Proof. Because all transformations mentioned above are equivalent on ℛ{x, y}, the first equality of equation (6.5.1) is equivalent to the first equality of equation (6.5.2), on ℛ{x, y}. By considering that their second equalities are the same, the conclusion can be drawn. Because of the favorableness for evaluation, equation (6.5.2) is accepted in what follows instead of equation (6.5.1). By considering that no odd power of x occurs in equation (6.5.2), the function f ∈ ℛ{x, y} is even on x. Similarly, because no odd power of y occurs, y2i+1 for i ≥ 0 never occurs in f . In other words, f ∈ ℛ{x2 , (y2 , y4 , y6 , . . .)} ⊆ ℛ{x, y}. On account of f being determined by the coefficients of x2m , m ≥ 0, let f = fa be a solution of equation (6.5.2); then fa is determined by Fa[m] = 𝜕x2m fa for m ≥ 0. Lemma 6.5.2. When m = 0, Fa[m] = 1. Proof. From Fa[0] being independent of y in equation (6.5.2), f |x=0⇒y=0 = 1, and hence Fa[m] = 1 when m = 0.

6.5 Bipartite simple inner model | 221

By considering the initial condition of equation (6.5.1), a basic necessary condition of a solution of equation (6.5.1) can be found. Observation 6.5.3. If fa is a solution of equation (6.5.1), then a0 = 1. Proof. We proceed on the basis of Lemma 6.5.2; this is a direct result. We proceed on the basis of Lemma 6.5.2 and Observation 6.5.3; we are allowed to only consider Fa[m] for m ≥ 1 without loss of generality. From (6.5.2), observe, for any integer m ≥ 1, how the coefficients of x2m in x2 f 2 , 2 x ∫y (y2 δx2 ,y2 f |u=x2 ) and (f − 1)(∫y f |x=y − 1), denoted by, respectively, Aa[m] , Ba[m] and Ca[m] , are determined. We consider Aa[m] ∈ ℛ{y} for m ≥ 0. Because f is an even function of x, f 2 is even [2] [2] = 𝜕x2m f 2 , then, for = (Fa[0] )2 = 1 for m = 0. Let Fa[m] in x as well. by Lemma 6.5.2, Fa[0] m ≥ 1, m

[2] = ∑ Fa[m−i] Fa[i] . Fa[m] i=0

[2] , From Aa[m] = Fa[m−1]

Aa[m]

0, when m = 0; { { { = {1, when m = 1; { { m−1 {∑i=0 Fa[m−1−i] Fa[i] , otherwise.

(6.5.3)

We consider Ba[m] ∈ ℛ{y} for m ≥ 0. First, da = 𝜕x2 ,y2 f |u=x2 as da = ∑ Fa[i] i≥0

x2i − y2i = ∑ x2k (∑ Fa[k+j+1] y2j ). x2 − y2 j≥0 k≥0

Because of Ba[m] = x2 ∫y (y2 da ), 0, when m = 0; Ba[m] = { ∑j≥1 Fa[m+j−1] y2j , when m ≥ 1,

(6.5.4)

for m ≥ 0 We consider Ca[m] ∈ ℛ{y} for m ≥ 0. From Lemma 6.5.2, Fa[0] = 1. Then, f − 1 = ∑ Fa[m] x2m m≥1

and

∫ f |x=y − 1 = ∑ Fa[m] y2m . y

m≥1

Thus, for m ≥ 0, 0, Ca[m] = { Fa[m] ∑j≥1 Fa[j] y2j ,

when m = 0; when m ≥ 1.

(6.5.5)

222 | 6 Inner equations fourth part Theorem 6.5.4. Equation (6.5.2) is equivalent the system of equations for Fa[m] for m ≥ 0

Fa[m]

1, when m = 0; { { { m−2 = {2a1 Fa[m−1] + a1 ∑i=1 Fa[m−1−i] Fa[i] { { + ∑j≥1 (a2 Fa[m+j−1] − Fa[m] Fa[j] )y2j , when m ≥ 1, {

(6.5.6)

on ℛ{y}. Proof. We proceed on the basis of (6.5.2). For any integer m ≥ 0, Fa[m] = 1 + a1 Aa[m] + a2 Ba[m] − Ca[m] . From (6.5.3)–(6.5.5),

Fa[m]

when m = 0; {1, { { m−1 = {a1 ∑i=0 Fa[m−1−i] Fa[i] { { + ∑j≥1 (a2 Fa[m+j−1] − Fa[m] Fa[j] )y2j , when m ≥ 1. {

This is (6.5.6). Let us turn to the system of equations (6.5.6). Because of the infinite number of both unknowns and variables, there is no way available to evaluate the solution directly. Further constructive properties need to be investigated. For any integer m ≥ 1, a new parameter is introduced as s ∈ ℤ+ , such that, for any lower vector n of y, Fa[m] s = π(n)/2. Such a part of Fa[m] is denoted by Fa[m,s] . In what follows, it will be seen that Fa[m,s] ∈ ℛ[x, y], i. e., we have a polynomial on ℛ{x, y}. We proceed on the basis of (6.5.6). For any integers m ≥ 1 and s ≥ 1, m−2

s

i=1

j=1

Fa[m,s] = 2a1 Fa[m−1,s] + a1 ∑ Da[i,s] + ∑ Ea[j,s−j] y2j

(6.5.7)

where s

{ { Da[i,s] = ∑ Fa[m−1−i,s−j] Fa[i,j] ; { { { j=0 s−j { { { { {Ea[j,s−j] = a2 Fa[m+j−1,s−j] − ∑ Fa[m,s−j−k] Fa[j,k] . { k=0

(6.5.8)

First of all, observe some cases of non-negative integers m and/or s smaller. Lemma 6.5.5. When s = 0, for any integer m ≥ 1, 1, Fa[m,0] = { am (2m)! 1

m!(m+1)!

when m = 0; ,

when m ≥ 1.

6.5 Bipartite simple inner model | 223

Proof. By employing (6.5.7)–(6.5.8), for m ≥ 1, 1, Fa[m,0] = { a1 ∑m−1 i=0 Fa[m−1−i,0] Fa[i,0]

when m = 0; when m ≥ 1.

We have Fa[0,0] = 1 and Fa[1,0] = a1 . From (3.1.7) in Book I, Fa[m,0] |a1 =1 = Cm . By considering that a1 occurs m times for determining Fa[m,0] recursively, the conclusion can be drawn. Of course, the conclusion of Lemma 6.5.5 can also be proved by induction independently. Lemma 6.5.6. Given integers m ≥ 3, s ≥ 1. For any integer 1 ≤ i ≤ m − 2, all Da[i,s] are determined by Fa[l,t] for l + t ≤ m + s − 1 and l, t ≥ 0. Proof. We proceed on the basis of the first equality in (6.5.8). Because of 1 ≤ i ≤ m − 2, 0 ≤ j ≤ s, we have (m−1−i)+(s−j) = m+s−(i+j)−1 ≤ (m+s)−1 and i+j ≤ m−2+s ≤ m+s−1. Thus, all Fa[m−1−i,s−j] and Fa[i,j] for 0 ≤ j ≤ s are known and hence Da[i,s] is determined. The conclusion can be drawn. This lemma tells us that all Da[i,s] for 1 ≤ i ≤ m − 2 are determined by ℱm+s−1 = {Fa[l,t] |l + t ≤ m + s − 1, l, t ≥ 0}. Lemma 6.5.7. Given integers m ≥ 3, 1 ≤ s ≤ m. For any integer 1 ≤ j ≤ s, all Ea[j,s−j] are determined by Fa[l,t] for l + t ≤ m + s − 1 and l, t ≥ 0. Proof. We proceed on the basis of the second equality in (6.5.8). We have 1 ≤ j ≤ s, (m + j − 1) + (s − j) = m + s − 1. This implies Fa[m+j−1,s−j] ∈ ℱm+s−1 for any j : 1 ≤ j ≤ s. By j ≥ 1 and k ≥ 0, m + (s − j − k) ≤ m + s − 1. Thus, Fa[m,s−j−k] ∈ ℱm+s−1 . By 1 ≤ j ≤ s and k ≤ s − j, j + k ≤ 2s − j ≤ 2s − 1. We have s ≤ m, j + k ≤ m + s − 1. Thus, Fa[j,k] ∈ ℱm+s−1 . Therefore, the conclusion can be drawn. The last two lemmas tell us that, as seen from (6.5.7), all Fa[m,s] are determined by

ℱm+s−1 for integers m ≥ s ≥ 1.

Theorem 6.5.8. Equation (6.5.2) is well defined on ℛ{x, y} if, and only if a0 = 1. Proof. Sufficiency. We prove the well-definedness of equation (6.5.2) on ℛ{x, y} whenever a0 = 1. Because any solution of equation (6.5.2) with all Fa[m,s] determined by Lemma 6.5.6 and Lemma 6.5.7 in companion with the analogy of Lemma 6.2.4 for m, s ≥ 0 satisfy (6.5.7)–(6.5.8), by considering the uniqueness of Fa[m,s] for m, s ≥ 0 starting from the initiation, the well-definedness of equation (6.5.2) is done. Necessity. This is a result of Observation 6.5.3.

224 | 6 Inner equations fourth part

6.6 Solution bipartite simple inner case In this section, let us turn to equation (6.5.2) for observing the inner structural properties of its solution further. Lemma 6.6.1. Given integer s ≥ 1, for any integer 1 ≤ j ≤ s, s−j

F1[m+j−1,s−j] − ∑ F1[m,s−j−k] F1[j,k] ≥ 0. k=0

(6.6.1)

Proof. According to what was done above, the inequality (6.6.1) is checked for m+s ≤ 5. For the general case of m + s ≥ 6, by induction, assume that, for any two integers l, t ≥ 0, the inequality is shown to be true when l + t ≤ m + s − 1; t−j

F1[l+j−1,t−j] − ∑ F1[l,t−j−k] F1[j,k] ≥ 0. k=0

We prove the inequality for l = m and t = s. From j ≥ 1 and k ≥ 0, we have (m + j − 1) + (s − j) = m + s − 1 ≤ m + s − 1, m + (s − j − k) ≤ m + s − 1 − k ≤ m + s − 1. Furthermore, from m ≥ 1 and k ≤ s − j, j + k ≤ s ≤ m + s − 1. By the assumption, (6.6.1) is done. This lemma tells us that, for any integer 1 ≤ j ≤ s, E1[j,s−j] ∈ ℛ+ {y} in (6.5.7). Lemma 6.6.2. For integers m, s ≥ 0, Fa[m,s] is independent of y2 and y2i for i ≥ s + 1. Proof. From what was mentioned in last section, it is seen that, for m + s ≤ 5 with m, s ≥ 0, the conclusion is true. For the general case of m + s ≥ 6, we proceed on the basis of (6.5.7) with (6.5.8). The Fa[m−1,s] with Da[i,s] (1 ≤ i ≤ m − 2) and Ea[j,s−j] (1 ≤ j ≤ s) are only dependent on Fa[r,t] for 0 ≤ r + t ≤ m + s − 1 (r, t ≥ 0). By the principle of induction, the conclusion can be drawn. This lemma enables us to restrict ourselves to only the equation (6.5.2) on ℛ{x, y2s } for integer s ≥ 0 given. As a matter of fact, its solution is only dependent on y2i for 2 ≤ i ≤ s. Theorem 6.6.3. Let fa[bsi] be the solution of equation (6.5.1). For any integers m, s ≥ 0, [bsi] [bsi] = 𝜕x2m denote Fa[m,s] 2 f[bsi] |π(yn )=2s . Then Fa[m,s] in ℛ{y2s } are all polynomials,

[bsi] Fa[m,s]

1, { { { { { { 0, { { { { ={ { { am 1 (2m)! { , { { m!(m+1)! { { { { s [bsi][2] {a1 Fa[m−1,s] + ∑j=1 Aa[j,s−j] y2j ,

when m = s = 0; when s ≥ 1, m = 0; or s ≥ 1, m = 1; s = 1, m ≥ 0; when s = 0, m ≥ 1; otherwise,

(6.6.2)

6.7 Explicitness bipartite simple inner case | 225

where [bsi] [bsi][2] [bsi] ; = ∑ Fa[m−1−i,s−j] Fa[i,j] Fa[m−1,s] { { { 0≤j≤s { { 0≤i≤m−1 s−j { { { [bsi] [bsi] [bsi] { {Aa[j,s−j] = a2 Fa[m+j−1,s−j] − ∑ Fa[m,s−j−k] Fa[j,k] . k=0 {

(6.6.3)

Proof. We proceed on the basis of Theorem 6.5.8 for any two integers m, s ≥ 0. When m = s = 0 and s ≥ 1, m = 0. The conclusion is given by the initiation of equation (6.5.1) with Observation 6.5.3. When s ≥ 1, m = 1, it is given by Lemma 6.5.7. When s = 1, m ≥ 0, it is given by Lemma 6.5.6. When s = 0, m ≥ 1, it is given by Lemma 6.5.5. Otherwise, i. e., m, s ≥ 2, it is given by (6.5.7)–(6.5.8). As for 1 ≤ j ≤ 3, Aa[j,s−j] ∈ ℛ{y2s }. It is given by Lemma 6.6.2. [bsi] for One more thing that has to be mentioned is that if a0 = a1 = a2 = 1, all Fa[m,s] m, s ≥ 0 are shown to be a polynomial with each coefficient in the form of a finite sum with all terms positive. Direct proofs of them will be presented in Section 6.8.

6.7 Explicitness bipartite simple inner case Because no explicision has yet directly been deduced from Theorem 6.6.3, the relationship between the solutions of bipartite simple inner equation (6.5.1), or equivalently equation (6.5.2), and its restriction for a0 = a1 = a2 = 1 has to be investigated on account of the latter being meaningful in graph theory. Let fa[bsi] and f[bsi] ∈ ℛ{x, y} be the solutions of, respectively, equation (6.5.1), or [bsi] [bsi] and Fm,s equivalently equation (6.5.2), and its restriction determined by Fa[m,s] ∈ ℛ{y} [bsi] [bsi] and hence Fm,s . for integers m, s ≥ 0. Denote by 𝒩m,s the set of power vectors in Fa[m,s]

[bsi] [bsi] ∈ ℛa = ℛ{a} and Um,n | ∈ ℛ1 = ℛ are the coefficients of yn For any n ∈ 𝒩m,s , Ua[m,n]

[bsi] [bsi] and Fm,s . Then, in, respectively, Fa[m,s] [bsi] [bsi] = 𝜕yn Fa[m,s] Ua[m,n]

[bsi] [bsi] and Um,n = 𝜕yn Fm,s .

[bsi] is a Theorem 6.7.1. For any integers m, s ≥ 1 and integral vector n ∈ 𝒩m,s , Ua[m,n] polynomial of a1 = a0 and a2 with degree at most l + 1 where l = m + s such that [bsi] [bsi] |a=1 = Um,n . Ua[m,n]

Proof. We proceed on the basis of Theorem 6.6.3. By induction, the conclusion can easily be drawn. For an embedding μ(G) of a bipartite simple planar graph G, let n = (n1 , n2 , n3 , n4 , . . .) be the face-partition vector where ni is the number of non-root faces of valency 2i for integer i ≥ 1.

226 | 6 Inner equations fourth part A vertex v is a point of μ(G) such that μ(G) − v has at least one component greater than μ(G) and is called a cut-vertex of G. Now, G is always assumed to be connected. It is allowed to restrict a face of maximum length boundary at a cut-vertex as the outerface without loss of generality. The cut-valency of a cut-vertex v is the number of components of μ(G) − v. The closure of such a component is called a cut-component. If a cut-component is not a path, then it is called a fat-component. The number of fatcomponents contributing to the cut-valency of v is called the fat-valency of v. A graph without cut-vertex is said to be nonseparable. A maximal nonseparable edge-induced subgraph is called a block of the graph. For a cut-vertex v, let a = av = ρvcut , c = cv and b = bv = ρv be, respectively, the cut-valency, fat-valency and the valency of v. Because of planarity, all cut-components at v have a rotation in the order as cut-components G1 , G2 , . . . , Ga . At cut-vertex v, let ti be the contribution of Gi to the valency of v, then t1 + t2 + ⋅ ⋅ ⋅ + ta = b. Three cases have to be considered at cut-vertex v for a bipartite simple planar graph. Case I Reflection for Gi , 1 ≤ i ≤ c, it turns out that we have [bsi] = 2c−1 πvI

(6.7.1)

(topologically) distinct planar embeddings. Case II For cut rotation at v, it turns out that we have [bsi] = (a − 1)! πvII

(6.7.2)

(topologically) distinct planar embeddings. Case III For an integer k ≥ 1 given, denote I = {i1 , i2 , . . . , ik } by distributing all Gj , j ∈ ̸ I, into s = (ti1 − 1) + (ti2 − 1) + ⋅ ⋅ ⋅ + (tik − 1) angles on 𝒢ℐ = {Gi | ∀i ∈ I} at v as (x1 , x2 , . . . , xts ) in the s angles such that s

∑ xj = c − k; j=1

0 ≤ xj ≤ c − k, 1 ≤ j ≤ s. Let us denote by c−k ⟨ ⟩ x1 , . . . , xts the number of ways for such distributions, then it turns out that we have s c−k πk[bsi] = ⟨ ⟩ ∏ xj ! x1 , . . . , xts j=1

(topologically) distinct planar embeddings. From the disjointness of the embeddings, we have

6.7 Explicitness bipartite simple inner case | 227 c−1 c [bsi] = ∑ ( )πk[bsi] πvIII k k=1

(6.7.3)

(topologically) distinct planar embeddings. Lemma 6.7.2. For a bipartite simple planar graph G, let VCut be the set of all cut-vertices. Then, all cut-vertices turn out to be [bsi] [bsi] [bsi] n[bsi] Cut (G) = ∏ (πvI πvII πvIII ) v∈VCut

(6.7.4)

(topologically) distinct planar embeddings. Proof. On account of the independency among Case I, Case II and Case III, from (6.7.1), (6.7.2) and (6.7.3), the conclusion can be drawn. We consider an embedding μ of a bipartite simple planar graph G without cutvertex. If two vertices u and v have the property that μ − {pu , pv } has at least two components, then {u, v} is called a splitting pair. Since an embedding is a point set in its own right on the plane, pu and pv are the corresponding points of u and v. Because of μ as a point set, any component of μ − {pu , pv } is called a splitting slice, or slice. A splitting slice that is not a path is called a splitting chunk, or chunk. Let S be the set of all splitting pairs consisting of {ui , vi } for 1 ≤ i ≤ s = s(S) = |S| in μ = μ(G). All components of μ − S are splitting slices in μ(G). For a splitting pair, its slice valency is the number of splitting slices incident with. Similarly, for chunk valency. Observation 6.7.3. Let t and 1 ≤ p ≤ t be, respectively, the slice valency and chunk valency of a splitting pair on an embedding of bipartite simple planar graph G, then the splitting pair produces [bsi] πspp = 2p−1 (t − 1)!

(6.7.5)

(topologically) distinct planar embeddings of G. Proof. On account of the independency between a rotation on slice valency and reflection on a chunk, by considering the connectivity of the embedding, the conclusion can be drawn. Let p be the number of all splitting chunks, and ti the slice valency at the splitting pair {ui , vi }, 1 ≤ i ≤ s on μ(G). Lemma 6.7.4. For a bipartite simple planar graph G, the number of topologically distinct embeddings of G generated by all splitting pairs is s

p−s n[bsi] ∏(ti − 1)! Nct = 2 i=1

(6.7.6)

228 | 6 Inner equations fourth part where pi and ti are, respectively, the number of chunks the slice valency at splitting pair {ui , vi }, 1 ≤ i ≤ s = |S| such that p = p1 + p2 + ⋅ ⋅ ⋅ + ps . Proof. This is a result of Lemma 6.7.2. This lemma is a refinement and a concise form of Theorem 10.4.3 with (10.4.13) in Liu YP [72] (2017, pp. 196–197), or Theorem 7.4.3 with (7.4.13) in Liu YP [55] (1994, p. 142). We proceed on the basis of Lemma 6.7.2 and Lemma 6.7.4; the result is useful in the context of determining the number of planar embeddings of a bipartite simple graph. Lemma 6.7.5. For a bipartite simple planar graph G, the number of (topologically) distinct planar embeddings of G is [bsi] n[bsi] (G) = n[bsi] Cut (G)nNct (G)

(6.7.7)

[bsi] where n[bsi] Cut (G) and nNct (G) are, respectively, given by (6.7.4) and (6.7.6).

Proof. We consider the uniqueness of 3-connected embedding of a graph. Because a graph without vertex-cut set of one or two vertices has to be 3-connected, only the two independent cases with and without cut-vertex need to be considered. Lemma 6.7.2 and Lemma 6.7.4 yield the conclusion. ̃ Denote by t = aut(G) the semi-automorphism group order of bipartite simple planar graph G. The relationship between embeddings and upper maps of G can be shown via t. Lemma 6.7.6. Let ℰ [bsi] (G) and ℳ[bsi] (G) be, respectively, the sets of all planar embeddings and upper maps of bipartite simple planar graph G. Then, 󵄨󵄨 [bsi] 󵄨󵄨 2ϵ 󵄨󵄨 [bsi] 󵄨󵄨 (G)󵄨󵄨 = 󵄨󵄨ℰ (G)󵄨󵄨 󵄨󵄨ℳ t

(6.7.8)

where ϵ = ϵ(G) is the size of G. Proof. Refer to Liu YP [59] (2003, pp. 225–226). [bsi] be the set of all bipartite simple planar graphs with rooted semi-valency Let 𝒢m|s 2m (i. e., the number of semi-edges incident to the root-vertex) and the un-rooted semisize 2s (i. e., the number of semi-edges not incident to the root-vertex). [bsi] the set of semi-automorphism group orders of bipartite simple Denote by 𝒯m|s

[bsi] [bsi] [bsi] . be the set of all partition vectors n in 𝒢m|s . Let 𝒥m|s planar graphs in 𝒢m|s

[bsi] be the number of (topologically) distinct embeddings of biparLemma 6.7.7. Let Em|n [bsi] . Then we have tite simple planar graphs in 𝒢m|n [bsi] = Em|n

where n[bsi] is given in (6.7.7).

∑ n[bsi] (G)

[bsi] G∈𝒢m|n

(6.7.9)

6.7 Explicitness bipartite simple inner case | 229

[bsi] Proof. Given the partition n. By considering of all embeddings of each graph in 𝒢m|s with n, the conclusion is easily drawn. [bsi] be the set of semi-automorphism We proceed on the basis of Lemma 6.7.7. Let 𝒯m|n

[bsi] [bsi] [bsi] with , the set of all graphs in 𝒢m|n and 𝒢m|n,t group orders among the graphs in 𝒢m|n ̃ semi-automorphism group order t. Denote by aut(G) the semi-automorphism group

order of G.

Lemma 6.7.8. The number of root-isomorphic classes of upper maps of all graphs in [bsi] is 𝒢m|n A[bsi] m|n =

∑ [bsi] t∈𝒯m|n

2(m + π(n)) [bsi] Em|n,t t

(6.7.10)

) and = Am|n (ℳ[bsi] where A[bsi] m|n m|n [bsi] = Em|n,t

∑ n[bsi] (G)|aut(G)=t ̃

(6.7.11)

[bsi] G∈𝒢m|n

determined by (6.7.9). Proof. We proceed on the basis of Lemma 6.7.6. By considering m + π(n) = ϵ, the conclusion is easily drawn. This lemma enables us to find the solution fa[bsi] of equation (6.5.1) when a0 = a1 = [bsi] a2 = 1. Denote f[bsi] = fa |a0 =a1 =a2 =1 as determined by Fm,s = 𝜕xm f[bsi] |π(y)=s for m, s ≥ 0. Lemma 6.7.9. For integers m, n ≥ 1, we have [bsi] Fm,s =

∑ [bsi] n∈𝒥m|s

n A[bsi] m|n y

(6.7.12)

where A[bsi] is given by (6.7.10). m|n Proof. It being known that the set of all underline graphs of bipartite simple planar [bsi] [bsi] maps with m and n is the same as 𝒢m,n , A[bsi] = Um,n is shown in Theorem 6.7.1. From m|n Lemma 6.7.8, the conclusion can be drawn. We proceed on the basis of Lemma 6.7.9 and Theorem 6.7.1, we are allowed to illustrate our main result of this section. [bsi] determine the solution of equation (6.5.1) as Theorem 6.7.10. For m, s ≥ 1, the Fa[m,s] [bsi] = Fa[m,s]

∑ [bsi] n∈𝒥m|s

[bsi] yn Ua[m,n]

(6.7.13)

[bsi] where a = (a1 , a2 ) and Ua[m,n] is the polynomial of a with degree at most m + s + 1 given in Theorem 6.7.1.

230 | 6 Inner equations fourth part Proof. This is a result of Theorem 6.7.1 and Lemma 6.7.9. [bsi] is evaluated by the procedure shown in (6.6.1) or directly The polynomial Ua[m,n] from the inner structures of the solution on its own right. [bsi] [bsi] |a=1 determine the solution of equation Corollary 6.7.11. For m, s ≥ 1, Fm,s = Fa[m,s] (3.5.1) when a = 1, [bsi] Fm,s =

∑ [bsi] n∈𝒥m|s

[bsi] n Um,n y

(6.7.14)

[bsi] is given in (6.7.10). where Um,n = A[bsi] m|n

Proof. This is a direct result of Theorem 6.7.10. [bsi] This is a direct explicision (explicit expression) of Fm,s for integers m, s ≥ 1.

6.8 Restrictions bipartite simple inner case Consider the equation for f 2 2 2 2 { {x ∫ y δx2 ,y2 f |x2 =u = (f − 1) ∫ f |x=y − x f ; y y { { f | = 1, { x=0⇒y=0

(6.8.1)

where f ∈ ℛ{x, y}. In [56] (Liu YP, 1999, p. 231) and [58] (Liu YP, 2001, p. 271), one might find the main equation (6.8.1), or its equivalence. However, some errors in symbols should be paid attention to. A solution of equation (6.8.1) is involved with bipartite simple planar maps. The first equality of equation (6.8.1) is transformed on ℛ{x, y} as follows. By adding (1 − f ) to the both sides of the equality, (1 − f ) + x2 ∫ y2 δx2 ,y2 f |x2 =u = (1 − f ) + (f − 1) ∫ f |x=y − x2 f 2 . y

y

By taking out the common factor (f − 1) of the first two terms on the right hand side, (1 − f ) + x2 ∫ y2 δx2 ,y2 f |x2 =u = (f − 1)(∫ f |x=y − 1) − x2 f 2 . y

y

By adding f + x2 f 2 on the both sides, 1 + x2 f 2 + x2 ∫ y2 δx2 ,y2 f |x2 =u = f + (f − 1)(∫ f |x=y − 1). y

y

6.8 Restrictions bipartite simple inner case | 231

Then, by adding −(f − 1)(∫y f |x=y − 1) to its two sides, we have f = 1 + x2 f 2 + x2 ∫ y2 δx2 ,y2 f |x2 =u − (f − 1)(∫ f |x=y − 1). y

y

Theorem 6.8.1. Equation (6.8.1) is equivalent to 2 2 2 2 { { {f = 1 + x f + x ∫(y δx2 ,y2 f |u=x2 ) − (f − 1)(∫ f |x=y − 1); y y { { { {f |x=0⇒y=0 = a0 ,

(6.8.2)

on ℛ{x, y}. Proof. Because all transformations mentioned above are equivalent on ℛ{x, y}, the first equality of equation (6.8.1) is equivalent to the first equality of equation (6.8.2), on ℛ{x, y}. By considering that their second equalities are the same, the conclusion can be drawn. Because of the favorableness for evaluation, equation (6.8.2) is accepted in what follows instead of equation (6.8.1). By considering that no odd power of x occurs in equation (6.8.2), the function f ∈ ℛ{x, y} is even on x. Similarly, because no odd power of y occurs, y2i+1 for i ≥ 0 never occur in f . In other words, f ∈ ℛ{x2 , (y2 , y4 , y6 , . . .)} ⊆ ℛ{x, y}. On account of f being determined by the coefficients of x2m , m ≥ 0, if f is a solution of equation (6.8.2), then f is determined by Fm = 𝜕x2m f for m ≥ 0. Lemma 6.8.2. When m = 0, Fm = 1. Proof. From F0 being independent of y in equation (6.8.2), f |x=0⇒y=0 = 1, and hence Fm = 1 when m = 0. By considering the initial condition of equation (6.8.1), a basic necessary condition of a solution of equation (6.8.1) can be found. Observation 6.8.3. If f is a solution of equation (6.8.1), then F0 = 1. Proof. We proceed on the basis of Lemma 6.8.2; this is a direct result. We proceed on the basis of Lemma 6.8.2 and Observation 6.8.3; we are allowed to only consider Fm for m ≥ 1 without loss of generality. From (6.8.2), observe, for any integer m ≥ 1, what happens for the coefficients of x2m in x2 f 2 , x2 ∫y (y2 δx2 ,y2 f |u=x2 ) and (f −1)(∫y f |x=y −1). They are denoted by, respectively, Am , Bm and Cm for m ≥ 1 because of A0 = B0 = C0 = 0. We consider Aa[m] ∈ ℛ{y} for m ≥ 0. Because f is an even function of x, f 2 is even [2] in x as well. By Lemma 6.8.2, F0[2] = (F0 )2 = 1 for m = 0. Let Fm = 𝜕x2m f 2 , then, for

232 | 6 Inner equations fourth part m ≥ 1,

m

[2] Fm = ∑ Fm−i Fi . i=0

[2] , From Aa[m] = Fm−1

0, when m = 0; { { { Am = {1, when m = 1; { { m−1 {∑i=0 Fa[m−1−i] Fa[i] , otherwise.

(6.8.3)

We consider Bm ∈ ℛ{y} for m ≥ 0. First, d = 𝜕x2 ,y2 f |u=x2 is known, d = ∑ Fi i≥0

x2i − y2i = ∑ x2k (∑ Fk+j+1 y2j ). x2 − y2 j≥0 k≥0

Because of Bm = x2 ∫y (y2 da ), 0, when m = 0; Bm = { ∑j≥1 Fm+j−1 y2j , when m ≥ 1,

(6.8.4)

for m ≥ 0 We consider Cm ∈ ℛ{y} for m ≥ 0. From Lemma 6.8.2, F0 = 1 is known. Then, f − 1 = ∑ Fm x2m

and

∫ f |x=y − 1 = ∑ Fm y2m . y

m≥1

0, Cm = { Fm ∑j≥1 Fj y2j ,

when m = 0;

m≥1

Thus, for m ≥ 0, when m ≥ 1.

(6.8.5)

Theorem 6.8.4. Equation (6.8.2) is equivalent to the system of equations for Fm for m ≥ 0, 1, when m = 0; Fm = { m−2 2Fm−1 + ∑i=1 Fm−1−i Fi + ∑j≥1 (Fm+j−1 − Fm Fj )y2j , when m ≥ 1,

(6.8.6)

on ℛ{y}. Proof. We proceed on the basis of (6.8.2). For any integer m ≥ 0, Fm = 1 + Am + Bm − Cm . From (6.8.3)–(6.8.5), 1, when m = 0; Fm = { m−1 ∑i=0 Fm−1−i Fi + ∑j≥1 (Fm+j−1 − Fm Fj )y2j , when m ≥ 1. This is (6.8.6).

6.8 Restrictions bipartite simple inner case | 233

Let us turn to the system of equations (6.8.6). Because of there being an infinite number of both unknowns and variables, there is no way available to evaluate the solution directly. Further constructive properties need to be investigated. For any integer m ≥ 1, a new parameter is introduced as s ∈ ℛ, such that, for any lower vector n of y, Fm s = π(n)/2. Such a part of Fm is denoted by Fm,s . In what follows, it will be seen that Fm,s ∈ ℛ[x, y], i. e., a polynomial on ℛ{x, y}. We proceed on the basis of (6.8.6). For any integers m ≥ 1 and s ≥ 1, m−2

s

i=1

j=1

Fm,s = 2Fm−1,s + ∑ Di,s + ∑ Ej,s−j y2j where

(6.8.7)

s

{ { Di,s = ∑ Fm−1−i,s−j Fi,j ; { { { j=0 s−j { { { { {Ej,s−j = Fm+j−1,s−j − ∑ Fm,s−j−k Fj,k . { k=0

(6.8.8)

First of all, observe some cases of non-negative integers m and/or s that are smaller. Lemma 6.8.5. When s = 0, for any integer m ≥ 1, Fm,0 =

(2m)! . m!(m+1)!

Proof. By employing (6.8.7)–(6.8.8), for m ≥ 1, m−2

m−1

i=1

i=0

Fm,0 = 2Fm−1,0 + ∑ Fm−1−i,0 Fi,0 = ∑ Fm−1−i,0 Fi,0 . We have F0,0 = 1. From (3.1.7) in Book I, Fm,0 = Cm , i. e., the Catalan number. therefore, the conclusion can be drawn. This lemma enables us to only observe Fm,s for integer s ≥ 1. Lemma 6.8.6. For any integer m ≥ 0, Fm,1 = 0. Proof. By employing (6.8.7)–(6.8.8). For any integer m ≥ 1, m−2

Fm,1 = 2Fm−1,1 + ∑ (Fm−1−i,1 Fi,0 + Fm−1−i,0 Fi,1 ) + (Fm,0 − Fm,0 F1,0 )y2 . i=1

Because of F1,0 = 1 (Lemma 6.8.5!), m−2

Fm,1 = 2Fm−1,1 + ∑ (Fm−1−i,1 Fi,0 + Fm−1−i,0 Fi,1 ). i=1

On this basis, we proceed by induction. When m = 0, Lemma 6.8.2 tells us F0,1 = 0. When m ≥ 1, assume that, for any integer k ≤ m − 1, the Fk,0 = 0 are known. We prove

234 | 6 Inner equations fourth part Fm,1 = 0. As a matter of fact, we work from the assumption that Fm−1,1 = 0 and that, for 1 ≤ i ≤ m − 1, Fm−1−i,1 = 0 and Fi,1 = 0. Hence, Fm,1 = 0 + 0 = 0. This is the conclusion. The two lemmas above enable us to only observe Fm,s for integer s ≥ 2. Lemma 6.8.7. For any integer s ≥ 1, F1,s = 0. Proof. For s ≥ 1, by (6.8.7)–(6.8.8), s

s−j

j=1

k=0

F1,s = ∑(Fj,s−j − ∑ F1,s−j−k Fj,k )y2j . On this basis, by induction, when s = 1, F1,1 = 0 is known. When s ≥ 2, assume that the F1,k = 0 are known for any integer 1 ≤ k ≤ s − 1. We prove F1,s = 0. As a matter of fact, by the assumption, Fm−1,1 = 0, and, for 1 ≤ i ≤ m − 1, we have Fm−1−i,1 = 0 and Fi,1 = 0. Hence, Fm,1 = 0 + 0 = 0. This is the conclusion. Then, observe what happens for determining Fm,s as m + s is increased one by one starting from F0,0 = 1. When m + s = 0, only F0,0 = 1 (Lemma 6.8.5). When m + s = 1, F1,0 = 1 and F0,1 = 0. When m + s = 2, F2,0 = 2, F1,1 = F0,2 = 0. When m + s = 3, F3,0 = 5, F2,1 = F1,2 = F0,3 = 0. When m + s = 4, F4,0 = 14, F3,1 = F1,3 = F0,4 = 0. Only F2,2 needs to be addressed. [2] = 0, from (6.8.7)–(6.8.8) and the independency of y2 , Because of F1,2 F2,2 = (F3,0 − F2,0 F2,0 )y4 = (5 − 22 )y4 = y4 . When m + s = 5, F5,0 = 42 (Lemma 6.8.5!), F4,1 = F1,4 = F0,5 = 0. Only F3,2 and F2,3 need to be done. [2] = 2y4 . By (6.8.7)–(6.8.8) with Lemmas 6.8.5–6.8.7, For F3,2 , we use F2,2 2

2−j

j=1

k=0

F3,2 = 2y4 + ∑(F1+j,2−j − ∑ F3,2−j−k Fj,k )y2j = 6y4 . [2] For F2,3 , we use F1,3 = 2F1,3 = 0. From (6.8.7)–(6.8.8) with Lemmas 6.8.5–6.8.7, 3

3−j

j=1

k=0

F2,3 = ∑(F1+j,3−j − ∑ F2,3−j−k Fj,k )y2j = 4y6 . When m+s = 6, we proceed on the basis of F6,0 = 132 (Lemma 6.8.5), F5,1 = F1,5 = 0 (Lemma 6.8.6 and Lemma 6.8.7). It is only necessary to evaluate F4,2 , F3,3 and F2,4 . Similarly, F4,2 = 28y4 , F3,3 = 25y6 and F2,4 = 14y8 + 2y42 are done.

6.8 Restrictions bipartite simple inner case | 235

Theorem 6.8.8. Equation (6.8.2) is well defined on ℛ{x, y}. Proof. We prove that equation (6.8.2) has a solution on ℛ{x, y}. We address any solution of equation (6.8.2) whose Fm,s for m, s ≥ 0 satisfy (6.8.7)–(6.8.8). The procedure described above is to determine Fm,s in the increase order of m + s one by one, from m + s = 0 i. e., F0,0 = 1, known as the initiation. So, a solution is obtained. By considering the uniqueness of Fm,s for m, s ≥ 0 from the initiation, the solution of equation (6.8.2) is unique. Now, let us turn to equation (6.8.2) for observing more inner structural properties of its solution further. Lemma 6.8.9. Given integer s ≥ 1. For any integer 1 ≤ j ≤ s, s−j

Fm+j−1,s−j − ∑ Fm,s−j−k Fj,k ≥ 0. k=0

(6.8.9)

Proof. According to what has been done above, the inequality (6.8.9) is checked for m + s ≤ 5. For the general case of m + s ≥ 6, by induction, assume that for any two integers l, t ≥ 0, the inequality is shown to be true when l + t ≤ m + s − 1, t−j

Fl+j−1,t−j − ∑ Fl,t−j−k Fj,k ≥ 0. k=0

We prove the inequality for l = m and t = s. From j ≥ 1 and k ≥ 0, we have (m + j − 1) + (s − j) = m + s − 1 ≤ m + s − 1, m + (s − j − k) ≤ m + s − 1 − k ≤ m + s − 1. Furthermore, from m ≥ 1 and k ≤ s − j, j + k ≤ s ≤ m + s − 1, by the assumption, (6.8.9) is done. This lemma tells us that, for any integer 1 ≤ j ≤ s, Ej,s−j ∈ ℛ+ {y} as appearing in (6.8.7). Lemma 6.8.10. For integers m, s ≥ 0, Fm,s is independent of y2 and y2i for i ≥ s + 1. Proof. From what was mentioned in last section, it is seen that for m + s ≤ 5 with m, s ≥ 0, the conclusion is true. For the general case of m + s ≥ 6, we proceed on the basis of (6.8.7) with (6.8.8). Fm−1,s with Di,s (1 ≤ i ≤ m − 2) and Ej,s−j (1 ≤ j ≤ s) is only dependent on Fr,t for 0 ≤ r + t ≤ m + s − 1 (r, t ≥ 0). By the principle of induction, the conclusion can be drawn. This lemma enables us to restrict ourselves to only considering equation (6.8.2) on ℛ{x, y2s } for integer s ≥ 0 given. As a matter of fact, its solution is only dependent on y2i for 2 ≤ i ≤ s. Theorem 6.8.11. Let f[bsi] be the solution of equation (6.8.1). For any integers m, s ≥ 0, [bsi] [bsi] denote Fm,s = 𝜕x2m 2 f[bsi] |π(yn )=2s . Then Fm,s in ℛ{y2s } are all polynomials,

236 | 6 Inner equations fourth part

[bsi] Fm,s

1, when m = s = 0; { { { { { { 0, when s ≥ 1, m = 0; { { { { or s ≥ 1, m = 1; s = 1, m ≥ 0; ={ { { (2m)! { , when s = 0, m ≥ 1; { { { m!(m+1)! { { { [bsi][2] s {Fm−1,s + ∑j=1 Aj,s−j y2j , otherwise,

(6.8.10)

[bsi][2] [bsi] [bsi] = ∑ Fm−1−i,s−j Fm−1,s Fi,j ; { { { 0≤j≤s { { 0≤i≤m−1 s−j { { { [bsi] [bsi] [bsi] { {Aj,s−j = Fm+j−1,s−j − ∑ Fm,s−j−k Fj,k . k=0 {

(6.8.11)

where

Proof. We proceed on the basis of Theorem 6.8.8 for any two integers m, s ≥ 0. When m = s = 0 and s ≥ 1, m = 0, the conclusion is given by the initiation of equation (6.8.1) with Observation 6.8.3. When s ≥ 1, m = 1, it is given by Lemma 6.8.7. When s = 1, m ≥ 0, it is given by Lemma 6.8.6. When s = 0, m ≥ 1, it is given by Lemma 6.8.5. Otherwise, i. e., m, s ≥ 2, it is given by (6.8.7)–(6.8.8). As for 1 ≤ j ≤ 3, Aj,s−j ∈ ℛ{y2s }, it is given by Lemma 6.8.9. [bsi] for m, s ≥ 0 can be One more thing that has to be clarified is whether all Fa[m,s] shown to be a polynomial with each coefficient in the form of a finite sum with all terms positive.

Example 6.8.1. Root-isomorphic classes of bipartite simple planar maps by root-face [bsi] valency and face-partition vector. In fact, Fm,s provides the number of such classes of bipartite simple planar maps with root-face valency 2m and face-partition vector n such that π(n) = 2s (the number of semi-edges not in the root-face boundary). Their classes are shown in Figure 6.8.1 distinguished by tiny hollow circles. Here, a circuit is included without length 2. Because 2m + 2s is doubling the size, m + s is the size of the map. In this figure, a = x4 y4 is the case of size m + s = 4; b = 6x 6 y4 + 4x4 y6 is of size m + s = 5; the others are of size m + s = 6. Thus, one might see that a = x4 y4 = x4 F2,2 , i. e., F2,2 = y4 ; b = 4x4 y6 + 6x 6 y4 = x4 F2,3 + x6 F3,2 , i. e., F2,3 = 4y6 and F3,2 = 6y4 ; c(x6 y6 ) + d(4x 4 y8 + 8x 8 y4 ) + e(6x6 y6 ) + f (4x 4 y8 + 8x 8 y4 ) + g(12x 6 y6 ) + h(2x4 y8 + 4x8 y4 ) + i(6x 6 y6 ) + j(4x4 + 8x8 y4 ) + k(2x4 y42 ). By merging common terms, x4 (4y8 (d) + 4y8 (f ) + 2y8 (h) + 4y8 (j) + 2y42 (k)) = x4 (14y8 + 2y42 ) = x4 (F2,4 ); x6 (y6 (c) + 6y6 (e) + 12y6 (g) + 6y6 (i)) = x6 (25y6 ) = x6 (F3,3 ); x8 (8y4 (d) + 8y4 (f ) + 4y4 (h) + 8y4 )(j) = x8 (28y4 ) = x8 (F4,2 ).

6.9 Notes | 237

Figure 6.8.1: Root-isomorphic classes of bipartite simple planar maps with sizes from 4 through 6.

6.9 Notes 6.9.1. Consider the equation for f ∈ ℛ{x, y} 2 2 { {a2 x ∫ yδx,y (uf |x=u ) = ∫((a3 + xy)f − a3 )f |x=y − a1 x f ; y y { { {f |x=0⇒y=0 = a0 ,

(6.9.1)

where a0 , a1 , a2 , a3 ∈ ℛ and y = (y1 , y2 , y3 , . . .) ∈ 𝒱 . For this equation, one might see the following qualitative theorem. Theorem 6.9.1. Equation (6.9.1) is well defined if, and only if, a3 = a0 = 1. Proof. The proof is similar to the proof of the qualitative theorem (Theorem 6.1.7) of equation (6.1.1).

238 | 6 Inner equations fourth part In fact, equation (6.1.1) is a specific case of a3 = 1 in equation (6.9.1). 6.9.2. Equation coefficients variable. The generalization of equation (6.1.1), or equation (6.5.1), with coefficients a variable such that each entry of a has a linear form, or a quadratic form or so in x and/or yi (i ≥ 1), has been extensively investigated for combinatorial properties of partitions with certain type of weights variable. 6.9.3. One way for explicision. An explicision of the solution of equation (6.1.1), or equation (6.5.1), still needs to be evaluated only by transformations on ℛ{x, y} directly for seeking a favorite resolution. 6.9.4. Another way for explicision. A manner for getting an explicision of the solution of equation (6.1.1), or equation (6.5.1), in the case of a = (1, 1, 1), i. e., equation (6.4.1), or equation (6.8.1) is to observe the root-classifications of general, or bipartite, simple planar maps with face-partition vector given directly without considering semi-automorphism group orders of the underline graphs. 6.9.5. New results on embedding of general, or bipartite, simple planar graphs. All results in Section 6.3, or Section 6.7, on embedding of general, or bipartite, simple planar graphs are available for reaching the corresponding root-enumerations of general, or bipartite, simple planar maps. As seen, this is a type of foundation for getting an explicision of the solution of equation (6.1.1), or equation (6.5.1). 6.9.6. Making efficient and more intelligent (6.2.1) and (6.6.1), particularly (6.4.3) and (6.8.4) have to be further investigated for running on computers and then for theoretical and/or practical usage. 6.9.7. Specification of parameters. For a small number of parameters given, more research might be found in [41] (Liu, Y. P., 1987), [26] (1983), [40] (1987), [38] (1988), [47] (1989), [48] (1989), [51] (1989), [52] (1992), [8] (Cai, J. L., Liu, Y. P., 1997), [85] (Ren, H., Liu, Y. P., 2002) etc. 6.9.8. Direct explicision without consideration of symmetry. A direct explicision of (6.7.13) for upper maps of general, or bipartite, simple planar graphs without consideration of symmetry. If symmetry is considered for maps, the root-isomorphic maps distribution by their automorphism group orders has to be done as shown in Liu YP [59] (Liu, Y. P., 2003, pp. 221–230), or [73] (2017, pp. 188–194). 6.9.9. The relationship between equation (6.1.1) and equation (6.5.1), or equation (6.4.1) and equation (6.8.1), might be known via inner structures over general simple planar maps and bipartite simple planar maps. Are there a series of transformations on ℛ{x, y} from one of equation (6.1.1) and equation (6.5.1), particularly equation (6.4.1) and equation (6.8.1), to the other?

6.9 Notes | 239

6.9.10. Theorem 6.2.8 and Theorem 6.6.3 might be, respectively, employed for evaluof (a0 = 1, a1 , a2 ∈ ℝ) in Theorem 6.3.1 and and A[bsi] ating the polynomials A[gsi] a[m,n] a[m,n] Theorem 6.7.1. 6.9.11. Asymptotic behavior (or asymptotics). Note 6.9.7 reminds us to investigate corresponding asymptotics as shown in [56] (Liu, Y. P., 1999, pp. 359–389), [94] (Yan, J. Y., Liu, Y. P., 1991). Relevant results are in [40] (1987), [38] (1988), [41] (1989), [48] (1989), [8] (Cai, J. L., Liu, Y. P., 1997), [85] (Ren, H., Liu, Y. P., 2002) look available for this topic. 6.9.12. Stochastic behavior (or stochastics). On the basis of (6.4.23) and (6.8.10)– (6.8.11), the distributions of sizes, orders, semi-automorphic group orders and genera from, respectively, size polynomials, order polynomials, semi-automorphic group order polynomials and genus polynomial of general or bipartite simple planar maps are estimated. Relevant results are referred to in [56] (Liu, Y. P., 1999, pp. 359–389), [59] (Liu, Y. P., 2003, 221–230), [94] (Yan, J. Y., Liu, Y. P., 1991) etc. 6.9.13. Distributions. Theorem 6.4.15 and Theorem 6.8.11 enable us to determine the distributions of dual orders on given sizes for, respectively, general and bipartite simple planar maps and to estimate their corresponding moments including the average values (i. e., mathematical expectations). 6.9.14. Equation (6.1.1) might be found in Program 93 as equation (124) in the part with constant coefficients which occurs on page 10743, Vol. 22 of [71] (Liu, Y. P., 2016). Equation (6.5.1) occurs on page 10743, Vol. 22 in Program 94 as equation (125) in the part with constant coefficients.

7 Surface equations first part 7.1 Join-end surface model Observe the equation for f ∈ ℛ{x, y} 2 3 𝜕f { { {a2 x ∫ y(1 + 𝜕x,y (f − a0 )|x=u ) = (1 − x )f − a3 x 𝜕x − a1 ; y { { { f { x=0,y=0 = a0 ,

(7.1.1)

where a = (a0 , a1 , a2 , a3 ) ∈ ℝ4+ is given. This is equation (13) in Introduction. Because of the contribution from joint-end maps on surfaces as the special case of a0 = a1 = a2 = 1, equation (7.1.1) is then called a join-end surface model. From [66] (Liu YP, 2012), one might see that the present equation shown is a special case. However, f under 𝜕x,y should be replaced by f − 1. One might also observe the equation for f ∈ ℛ{x, y} only with a non-negative integer constant a, 2 3 𝜕f { { {x ∫ y(1 + 𝜕x,y (f − a)|u=x ) = (1 − x )f − x 𝜕x − a; y { { { {fx=0,y=0 = a,

(7.1.2)

where a ∈ ℛ+ . In Liu YP [73] (2015, pp. 379–389), one might consider equation (7.1.2) and the like. In order to evaluate a solution of equation (7.1.1) in the form of a sum with all terms positive, the equation is equivalently transformed on ℛ{x, y} to 2 3 𝜕f { { {f = a1 + x f + a3 x 𝜕x + a2 x ∫ y(1 + 𝜕x,y (f − a1 )|u=x ); y { { { f = a . x=0,y=0 0 {

(7.1.3)

m For integer m ≥ 0, let Fa[m] = [f ]m x = 𝜕x f , then

0, m [x2 f ]x = { Fa[m−2] ,

when 0 ≤ m ≤ 1; otherwise.

(7.1.4)

Lemma 7.1.1. For any integer m ≥ 0, if Fa[l] ∈ ℛ+ {y}, 0 ≤ l ≤ m, then Fa[m−2] ∈ ℛ+ {y}. Proof. Because of m − 2 ≤ m, the given condition implies the conclusion. 𝜕f 󸀠 For integer m ≥ 0, let Fa[m] = 𝜕xm 𝜕x , then

Fa[1] , 󸀠 Fa[m] ={ (m + 1)Fa[m+1] , https://doi.org/10.1515/9783110627336-007

when m = 0; otherwise.

(7.1.5)

242 | 7 Surface equations first part 󸀠 Lemma 7.1.2. For any integer m ≥ 0, if Fa[l] ∈ ℛ+ {y}, 0 ≤ l ≤ m, then Fa[m−1] ∈ ℛ+ {y}. 󸀠 Proof. From (7.1.5), Fa[m−1] = mFa[m] . By the given condition, Fa[m] ∈ ℛ+ {y}. Therefore, 󸀠 Fa[m−1] ∈ ℛ+ {y}. This is the conclusion.

Let ∇i = 𝜕xi (𝜕x,y f |u=x ), i ≥ 0. For any integer k ≥ 2, k−1

𝜕x,y uk = ∑ xj yk−j . j=1

Incidentally, 𝜕x,y u0 = −1 and 𝜕x,y u = 0 are guaranteed. From the linearity of 𝜕x,y , we see that, for integer m ≥ 0, a0 − Fa[0] , 𝜕xm (y𝜕x,y (f − a1 )|u=x ) = { ∑k≥m+1 Fa[k] yk−m+1 ,

when m = 0; when m ≥ 1.

(7.1.6)

Thus, from the linearity of the meson functional, we see that, for any integer m ≥ 0, m

y1 (1 + (a1 − Fa[0] )), when m = 0; [∫ y(1 + 𝜕x,y (f − a1 ))] = { ∑k≥m+1 Fa[k] yk−m+1 , when m ≥ 1. x y

(7.1.7)

Observation 7.1.3. For any solution fa of equation (7.1.1), fa |m=0⇒y=0 = a0 = a1 . Proof. On the left hand side of the first equality in equation (7.1.1), a common factor x occurs. As a result its constant term is 0. On the right hand side of the equality, the constant term is Fa[0] − a1 . Thus, Fa[0] − a1 = 0 ⇒ Fa[0] = a1 . However, the initiation of equation (7.1.1) is Fa[0] = a0 . Therefore, a0 = a1 . This observation enables us to only need to observe a0 instead of a1 in what follows without loss of generality. Thus, a = (a0 , a2 , a3 ) but not (a0 , a1 , a2 , a3 ). Theorem 7.1.4. Equation (7.1.3) for f on ℛ{y} is equivalent to the system of equations for Fa[m] on ℛ{y} for m ≥ 0

Fa[m]

a0 (= a1 ), when m = 0; { { { { { {a2 (y1 + ∑k≥1 Fa[k] yk+1 ), when m = 1; ={ {a0 + a2 ∑ Fa[k] yk , when m = 2; { k≥2 { { { {(a3 (m − 2) + 1)Fa[m−2] + a2 ∑k≥m Fa[k] yk−m+2 , when m ≥ 3.

(7.1.8)

Proof. From x | (f − a1 ) in equation (7.1.3), F0 = a1 = a0 (Observation 7.1.3). This is the initiation. When m = 0, (7.1.8) holds, for the general case of m ≥ 1, on the basis of (7.1.4)–(7.1.7). From (7.1.3), (7.1.8) is then equivalently deduced.

7.1 Join-end surface model | 243

In order to solve the system of equations with an infinite number of unknowns Fa[m] for m ≥ 0, a new parameter s has to be introduced on Fa[m] so that its terms are classified by s as polynomials. Let 𝒥m be the set of power vectors of n = (n1 , n2 , n3 , . . .) such that yn is in a term of Fa[m] . Denote 𝒥m,s = {n | π(n) = s} ⊆ 𝒥m , then (7.1.9)

𝒥m = ∑ 𝒥m,s s≥0

where π(n) = i nT and i = (1, 2, 3, . . . , i, . . .). Observation 7.1.5. For any vector n ∈ ̸ 𝒥m,s , s ≠ m(mod 2) where integers m, s = π(n) ≥ 0. Proof. Because of Fa[0,0] = a0 and Fa[1,0] = Fa[0,1] = 0, the conclusion is clear for m + s ≤ 1. We address the general case of m + s ≥ 2. Assume that, for any integers i, j ≥ 0 and i + j ≤ m + s − 1, Fa[i,j] = 0, i. e., n ∈ ̸ 𝒥i,j where π(n) = j. By induction on m + s, we prove Fa[m,s] = 0 when s ≠ m(mod 2). We proceed on the basis of (7.1.8). When m = 1 and s ≠ 1(mod 2), because s

s−1

y

k=1

[∑ Fa[k] yk+1 ] = ∑ Fa[k,s−k−1] yk+1 k≥1

and k + (s − k − 1) = s − 1 < s with k ≠ s − k − 1(mod 2), the assumption leads to all Fa[k,s−k−1] = 0 and hence we have the conclusion. Similarly for the cases of m = 2 and m ≥ 3. The conclusion can be drawn. This observation enables us to only consider Fa[m.s] such that m = s(mod 2) for m, s ≥ 0 without loss of generality. Lemma 7.1.6. For any integer s ≥ 0, a0 , when s = 0; Fa[0,s] = { 0, otherwise.

(7.1.10)

Proof. Because of Fa[0] = a0 ∈ ℛ+ is a constant, Fa[0] is independent of s ≥ 1. Hence, Fa[0,s] = 0 for s ≥ 1. By the Kronecker symbol δs,t = 1, s = t; 0, s ≠ t, the lemma tells us Fa[0,s] = δ0,s for s ≥ 0. Lemma 7.1.7. For any integer s ≥ 0, 0, Fa[1,s] = { a2 y1 δ1,s + a2 ∑s−1 k=1 Fa[k,s−k−1] yk+1 ,

when s = 0(mod 2); when s = 1(mod 2).

(7.1.11)

244 | 7 Surface equations first part Proof. From the second equation in (7.1.8), s

Fa[1,s] = a2 [y1 + ∑ Fa[k] yk+1 ] = { k≥1

y

a2 y1 , a2 ∑s−1 k=1 Fa[k,s−k−1] yk+1 ,

when s = 1; when s ≥ 2.

When s = 0(mod 2), it is a result of Observation 7.1.5. On the basis of this theorem, let us observe Fa[1,1] , Fa[1,3] and Fa[1,5] . We have Fa[1,1] = a2 y1 from (7.1.11). Then Fa[1,3] and Fa[1,5] are analogous. Although (7.1.11) shows a finite sum of terms to evaluate Fa[1,s] whenever s is given, only Fa[0,0] = a0 and Fa[1,1] = a2 y1 are directly determined by Lemma 7.1.6 and Lemma 7.1.7, excluding trivial cases. However, the last two lemmas enable us to only discuss integer m ≥ 2 in what follows without loss of generality. From (7.1.8), we see that, for any integer m ≥ 2, when m = 2; a0 δs,0 + a2 ∑sk=2 Fa[k,s−k] yk , Fa[m,s] = { F y , when m ≥ 3, α(a3 , m)Fa[m−2,s] + a2 ∑s+m−2 a[k,s−k+m−2] k−m+2 k=m

(7.1.12)

where α(a3 , m) = a3 (m − 2) + 1. Lemma 7.1.8. For any integers m ≥ 2, a0 , when m = 2t, t = 0 and 1; { { { Fa[m,0] = {a0 ∏t−1 (2ia + 1), when m = 2t, t ≥ 2; 3 i=1 { { when m = 2t + 1, {0,

(7.1.13)

where t ≥ 1. Proof. We proceed on the basis of (7.1.11). Because 0

∑ Fa[k,s−k] yk = 0

k=2

and

m−2

∑ Fa[k,s−k+m−2] yk−m+2 = 0,

k=m

we see that, for m ≥ 2, a0 , Fa[m,0] = { (a3 (m − 2) + 1)Fm−2,0 ,

when m = 2; when m ≥ 3.

By (7.1.10), Fa[0,0] = a0 . From Lemma 7.1.7, Fa[1,0] = 0. For the case of m = 2t + 1 for t ≥ 1, we have a direct result of Observation 7.1.5. For m = 2t, t ≥ 1. We proceed on the basis of Fa[2,0] = Fa[0,0] = a0 and Fa[4,0] = (2a3 + 1)Fa[2,0] = (2a3 + 1)a0 . For any integer t ≥ 2 in general, assume Fa[2(t−1),0] = a0 ∏t−2 i=1 (2ia3 + 1). By the inductive assumption, t−2

Fa[2t,0] = (2(t − 1)a3 + 1)Fa[2(t−1),0] = (2(t − 1)a3 + 1)(a0 ∏(2ia3 + 1)). i=1

Therefore, the conclusion can be drawn.

7.1 Join-end surface model | 245

From this lemma, one might see that, for a3 = 1, Fa[m,0] ∈ ℛ+ {y} if, and only if, a0 ∈ ℛ+ . Lemma 7.1.9. For any integer m ≥ 0, 0, when m = 2t, t ≥ 0; { { { Fa[m,1] = {a2 y1 , when m = 1; { { t−1 {a2 ∏i=0 ((2i + 1)a3 + 1)y1 , when m = 2t + 1, t ≥ 1.

(7.1.14)

Proof. Lemma 7.1.6 and Lemma 7.1.7 lead, respectively, to Fa[0,1] = 0 and Fa[1,1] = a2 y1 . We proceed on the basis of (7.1.11). Because 1

∑ Fa[k,s−k] yk = 0

k=2

and

m−1

∑ Fa[k,s−k+m−2] yk−m+2 = 0,

k=m

we see that, for any integer m ≥ 2, 0, Fa[m,1] = { ((m − 2)a3 + 1)Fa[m−2,1] ,

when m = 2; when m ≥ 3.

On account of Fa[0,1] = 0, we have Fa[1,1] = a2 y1 and Fa[2,1] = 0. When m = 2t + 1 ≥ 3 for t ≥ 1, assume for any integer 2(t − 1) + 1 that we have t−2

Fa[2(t−1)+1,1] = a2 ∏((2i + 1)a3 + 1)y1 . i=0

By induction, we show that Fa[m,1] satisfies the conclusion. When m = 0(mod 2), the conclusion is a direct result of Observation 7.1.5. When m = 2t + 1, because of Fa[2t+1,1] = ((2t − 1)a3 + 1)Fa[2t−1,1] , t−2

Fa[2t+1,1] = ((2t − 1)a3 + 1)(a2 ∏((2i + 1)a3 + 1)y1 ). i=0

This leads to the conclusion. Lemma 7.1.8 and Lemma 7.1.9 enable us to only discuss s ≥ 2, without loss of generality of s ≥ 0. Lemma 7.1.10. Given two integers m ≥ 2 and s ≥ 2. If for integers l, r ≥ 0, Fa[l,r] ∈ ℛ+ {y} whenever l + r ≤ m + s − 1, then s+m−2

∑ Fa[k,s−k+m−2] yk−m+2 ∈ ℛ+ {y}.

k=m

(7.1.15)

Proof. We have k + (s − k + m − 2) = m + s − 2 ≤ m + s − 1 for k : m ≤ k ≤ m + s − 2. By the given condition, Fa[k,s−k+m−2] ∈ ℛ+ {y}. Thus, by considering yk−m+2 ∈ ℛ+ {y}, the conclusion can be drawn.

246 | 7 Surface equations first part Now, let us observe how to determine Fa[m,s] by the procedure starting from m+s = 0 in increasing order one by one (in increment 1) on the basis of (7.1.11). When m + s = 0, only Fa[0,0] = a0 , given by Lemma 7.1.6. When m + s = 1, we have only two cases: Fa[1,0] = 0 and Fa[0,1] = 0, as given by Observation 7.1.3. When m + s = 2, we have Fa[2,0] = a2 a0 and Fa[0,1] = 0 by, respectively, Lemma 7.1.8 and Lemma 7.1.6. Finally, Fa[1,1] = a2 y1 by Lemma 7.1.9. When m + s = 3, Fa[3,0] = 0 (Lemma 7.1.8) and Fa[0,3] = 0 (Observation 7.1.5). Only two equations remain: Fa[2,1] = 0 and Fa[1,2] = 0 as given by Observation 7.1.5. Theorem 7.1.11. Equation (7.1.3) is well defined on ℛ+ {x, y} if, and only if, a0 = a1 . Proof. Sufficiency. We proceed on the basis of a0 = a1 . By the procedure mentioned above, starting from Fa[0,0] = a0 , all Fa[i,j] are determined for i + j ≤ 3. For the general case of i + j ≤ m + s − 1 in m + s ≥ 4, assume all Fa[i,j] are known. By induction, we determine Fa[m,n] . On the basis of what mentioned above, we are allowed to only discuss m ≥ 2 without loss of generality. When m = 2, from the first case of (7.1.12), one might see that, for any integer k ≥ 0, k + (s − k) = s ≤ 2 + s − 1 = s + 1 < 2 + s. By the assumption, Fa[k,s−k] , 2 ≤ k ≤ s, are determined and hence Fa[2,s] for any integer s ≥ 0 and 2 + s = m + s. When m ≥ 3, we take into account (m − 2) + s ≤ m + s − 1 and k + (s − k + m − 2) = s + m − 2 ≤ m + s − 1 for m ≤ k ≤ m + s − 2. By the assumption, Fa[m−2,s] and all Fa[k,s−k+m−2] for m ≤ k ≤ s + m − 2 are determined. Thus, all Fa[m,s] for m ≥ 3 and s ≥ 0 are determined from the second case of (7.1.12). Because these Fa[m,s] for integers m, s ≥ 0 satisfy equations (7.1.8), equation (7.1.2) has, and is the only one to have, a solution from Theorem 7.1.4. Necessity. This is a result of Observation 7.1.3.

7.2 Solution join-end surface In order to evaluate the solution of equation (7.1.1) in the form of a finite sum with all terms positive, more combinatorial structures have to be investigated further. Observation 7.2.1. For any two integers m, s ≥ 0, Fa[m,s] are determined by some Fa[i,j] for non-negative integers i and j such that i + j ≤ m + s − 1. Proof. The proof results from the procedure in the proof of Theorem 7.1.11. This observation suggests that the solution of equation (7.1.1) can always be extracted recursively from its initial condition. Lemma 7.2.2. For any two integers m, s ≥ 0 given. If m + s = 1(mod 2), then Fa[m,s] = 0.

7.2 Solution join-end surface | 247

Proof. It is easily seen that the conclusion is true when m + s ≤ 2. When m + s ≥ 3, assume Fa[l,t] = 0 for any integers l, t ≥ 0 such that l + t ≤ m + s − 1 and m + s = 1(mod 2). By induction, we prove Fa[m,s] = 0 for any integers m, s ≥ 0 with m + s = 1(mod 2). For the reason mentioned above, only m, s ≥ 2 are considered on the basis of (7.1.12). When m = 2, given s ≥ 2 such that 2+s = s = 1(mod 2). For the reason of k +(s−k) = s ≠ 0(mod 2) and k +(s−k) = s ≤ 2+s−1 = s+1 < s+2 for any k ≥ 0, by the assumption, Fa[k,s−k] = 0. Hence, by the first case of (7.1.12), Fa[2,s] = 0. When m ≥ 3. We take into account m + s = 1(mod 2) and (m − 2) + s = m + s(mod 2), k + (s − k + m − 2) = s + m − 2 = m + s(mod 2). We have (m − 2) + s = m + s − 2 ≤ m + s − 1 and k + (s − k + m − 2) = s + m − 2 ≤ m + s − 1. By the assumption, Fa[m−2,s] = 0 and Fa[k,s−k+m−2] = 0. By the second case of (7.1.12), Fa[m,s] = 0. This lemma enables us to reduce half the amount of labor for evaluating the solution of equation (7.1.1). Lemma 7.2.3. Given two integers m, s ≥ 0. If m + s = 0(mod 2), then Fa[m,s] is independent of yi for i ≥ s + 1. Proof. Given integer s ≥ 2. For any integer m ≥ 2 and vector n ∈ 𝒥m , s = ∑ ini , i≥1

ni ∈ ℤ+ .

We proceed by contradiction. If there is yj such that j ≥ s + 1, or say nj ≥ 1, then s ≥ jnj ≥ (s + 1). This is a contradiction. Therefore, no vector which has an entry ni > 0 with i ≥ s + 1 occurs in 𝒥m,s . This is the conclusion. This lemma tells us that the variable vector is y = ys = (y1 , y2 , y3 , . . . , ys ) in Fa[m,s] . Lemma 7.2.4. Given two integers m, s ≥ 0. If m + s = 0(mod 2), then Fa[m,s] is a polynomial of ys with degree at most s. Proof. As a matter of fact, we need to show that, for any n ∈ 𝒥m,s , |n| ≤ s. Let ℋ = {n ≥ 0 | inT = s}. We have max{|n| | ∀n ∈ ℋ} = is 1Ts = s and 𝒥m,s ⊆ ℋ, Fa[m,s] is a polynomial of y with degree at most s. As is known, 1s is the s-dimensional vector of all entries 0 except for only the sth, which is 1. Lemma 7.2.5. Given two integers m, s ≥ 0. Fa[m,s] ∈ ℛ+ {y} if, and only if, a0 = a1 , a2 , a3 ∈ ℝ+ . Proof. When m + s ≤ 3. It can be checked that Fa[m,s] ∈ ℛ+ {y} if, and only if, a0 = a1 , a2 , a3 ∈ ℛ+ .

248 | 7 Surface equations first part When m + s ≥ 4, assume Fa[l,t] ∈ ℛ+ {y} for integers l, t ≥ 0, l + t ≤ m + s − 1 if, and only if, a ∈ ℛ+ . By induction, we prove that Fa[m,s] satisfies the conclusion. We proceed on the basis of (7.1.12), Lemma 7.1.1, Lemma 7.1.2 and Lemma 7.1.10. By the assumption, Fa[m,s] satisfies the conclusion. The last four lemmas tell us that, for any two integers m, s ≥ 0, Fa[m,s] is a polynomial of ys with degree at most s and that Fa[m,s] ∈ ℛ+ [ys ] if, and only if, a0 = a1 , a2 , a3 ∈ ℛ+ . Theorem 7.2.6. Let f[jes] be the solution of equation (7.1.1). Given any two integers

m, s ≥ 0. Denote Fa[m,n] = [f[jes] ]sy (= Ea[m,s] ), then Ea[m,s] has the form of a finite sum with all terms positive, [jes]

Ea[m,s]

a0 δ0,s , { { { { { { 0, { { { { { { { { { { { a0 ∏t−1 { i=1 (2ia3 + 1), { { { { { a2 y1 , { { { { { {a2 ∏t−1 i=0 ((2i + 1)a3 + 1)y1 , ={ s−1 {a2 ∑ Ea[k,s−k−1] yk+1 , { k=1 { { { { { { { { { { { a2 ∑sk=2 Ea[k,s−k] yk , { { { { { { { { { { { (a3 (m − 2) + 1)Ea[m−2,s] { { { { s+m−2 { + a2 ∑k=m Ea[k,s−k+m−2] yk−m+2 ,

when s ≥ 0, m = 0; when m = 0, s ≥ 1; m = 1, s ≥ 2, s = 0(mod 2), or s ≥ 0, m ≠ s(mod 2); when s = 0 and m = 2t, t ≥ 1; when s = 1, m = 1; when s = 1, m = 2t + 1, t ≥ 1; when s ≥ 3 (s = 1(mod 2)), m = 1; when s ≥ 2 (s = 0(mod 2)), m = 2; when m ≥ 3, s ≥ 2, m = s(mod 2).

(7.2.1)

Proof. We may consider s ≥ 0, m = 0, by Lemma 7.1.6. When m = 0, s ≥ 1; m = 1, s ≥ 2, s = 0(mod 2); or s ≥ 0, m ≠ s(mod 2), by, respectively, Lemma 7.1.6; Lemma 7.1.8; or Lemma 7.2.2. We may proceed with s = 0 and m = 2t, t ≥ 1, by Lemma 7.1.8. When s = 1, m = 2t + 1, t ≥ 0, by Lemma 7.1.10. When s ≥ 3 (s = 1(mod 2)), m = 1, by Lemma 7.1.7. When s ≥ 2 (s = 0(mod 2)), m = 2 and when m ≥ 3, s ≥ 2, m = s(mod 2), by (7.1.12). [jes]

Now, we look at what happens when 4 ≤ m + s ≤ 6. By (7.2.1), we evaluate Fa[m,s] (= Ea[m,s] ). When m + s = 4, because of Fa[4,0] = a0 ∏2−1 i=1 (2ia3 + 1) = a0 (2a3 + 1) and Fa[0,4] = 0, it is only necessary to evaluate Fa[3,1] , Fa[2,2] and Fa[1,3] . We consider Fa[1,3] . We have 3−1

Fa[1,3] = a2 ∑ Fk,2−k yk+1 = a2 (F1,1 y2 + F2,0 y3 ). k=1

By Lemma 7.1.8, we have Fa[1,3] = a22 y1 y2 + a0 a2 y3 .

(7.2.2)

7.2 Solution join-end surface | 249

Next F2,2 . We have 2

Fa[2,2] = a2 ∑ Fa[k,2−k] yk = a2 Fa[2,0] y2 . k=2

(7.2.3)

By Lemma 7.1.8, we have Fa[2,2] = a0 a2 y2 . On Fa[3,1] , from 1+3−2

Fa[3,1] = (a3 + 1)Fa[1,1] + a2 ∑ Fa[k,2−k] yk = (a3 + 1)Fa[1,1] , k=3

(7.2.4)

by Lemma 7.1.9, we have Fa[3,1] = a2 (a3 + 1)y1 . When m + s = 5, because of m ≠ s(mod 2), Fa[m,s] = 0. When m+s = 6, because of Fa[6,0] = a0 (2a3 +1)(4a3 +1), Fa[5,1] = a2 (a3 +1)(3a3 +1)y1 and Fa[0,6] = 0, it is only necessary to evaluate Fa[4,2] , Fa[3,3] , Fa[2,4] and Fa[1,5] . Furthermore, 5−1

Fa[1,5] = a2 ∑ Fa[k,4−k] yk+1 = a32 y1 y22 + a0 a22 y2 y3 k=1

+

a0 a22 y2 y3

+

a22 (a3

(7.2.5)

+ 1)y1 y4 + a0 a2 (2a3 + 1)y5 ,

we have Fa[1,5] = a32 y1 y22 + 2a0 a22 y2 y3 + a22 (a3 + 1)y1 y4 + a0 a2 (2a3 + 1)y5 . Then we consider Fa[2,4] . From 4

Fa[2,4] = a2 ∑ Fa[k,4−k] yk = a2 (a2 a0 y2 )y2 k=2

(7.2.6)

+ a2 (a2 (a3 + 1)y1 )y3 + a2 (a0 (2a3 + 1))y4 , we have Fa[2,4] = a0 a22 y22 + a22 (a3 + 1)y1 y3 + a0 a2 (2a3 + 1)y4 . On Fa[3,3] , from 3+1

Fa[3,3] = 2a3 Fa[1,3] + a2 ∑ Fa[k,3−k+1] yk−1 =

2a3 (a22 y1 y2

k=3

+ a2 a0 y3 ) + a2 (a2 (a3 + 1)y1 )y2

(7.2.7)

+ a2 (a0 (2a3 + 1))y3 ,

we have Fa[3,3] = a22 (3a3 + 1)y1 y2 + a0 a2 (4a3 + 1)y3 . From 6−2

Fa[4,2] = 3a3 Fa[2,2] + a2 ∑ Fa[k,4−k] yk−2 k=4

= 3a3 (a0 a2 y2 ) + a2 (a0 (2a3 + 1))y2 , we have Fa[4,2] = a0 a2 (5a3 + 1)y2 .

(7.2.8)

250 | 7 Surface equations first part

7.3 Explicitness join-end surface [jes]

Let ℳm,n be the set of all root-isomorphic classes of join-end maps on all orientable surfaces with root-vertex valency m ≥ 0 and the vertex-partition vector n ≥ 0. For [jes] any integer s ≥ 0, denote by ℳ[jes] m,s the union of ℳm,n for n ≥ 0 such that s = π(n). [jes]

Because it has been shown that the enufunction of ℳm,n satisfies equation (7.1.1) for a0 = a1 = a2 = a3 = 1, from Theorem 7.1.11, its solution fa[jes] ∈ ℛ{x, y} is determined by Fa[m,s] = 𝜕xm fa[jes] |π(n)=s ∈ ℛ{y} for m ≥ 0. [jes]

Observation 7.3.1. For any integers m, s ≥ 0, 󵄨󵄨 󵄨 [jes] 󵄨 F1[m,s] 󵄨󵄨󵄨y=1 = 󵄨󵄨󵄨ℳ[jes] m,s 󵄨󵄨

(7.3.1)

where a = (a0 , a1 , a2 , a3 ) and 1 = (1, 1, 1, 1), or (1, 1, 1, . . .) according as a or y. Proof. This case is seen to be the specific case of equation (7.1.1) when a = 1. [jes]

[jes] Let 𝒢m,s and 𝒢m,n be the sets of all join-end graphs of root-vertex valency m ≥ 0 with, respectively, s = π(n) for all n ≥ 0 and an n ≥ 0 given. [jes]

Observation 7.3.2. Let 𝒢 (ℳ[jes] m,s ) and 𝒢 (ℳm,n ) be the sets of all underline graphs of maps in, respectively, ℳ[jes] m,s and

[jes] ℳm,n .

[jes]

[jes]

𝒢m,n = 𝒢 (ℳm,n )

Then

[jes] = 𝒢 (ℳ[jes] and hence 𝒢m,s m,s ).

(7.3.2)

Proof. The proof is by showing a bijection between two sides of an equality. [jes]

[jes] Let Fm,s |y=1 or Fm,n be the numbers of distinct root-isomorphic classes of join-end maps on all orientable surfaces of root-vertex valency m with, respectively, s = π(n) for all n ≥ 0 or a vertex-partition vector n given. [jes]

Lemma 7.3.3. Given G ∈ 𝒢m,n , denote by μG the set of topologically non-equivalent embeddings of G. Then |μG | = (m − 1)! ∏(i − 1)!ni = (m − 1)!(i − 1)!n i≥1

(7.3.3)

where i = (1, 2, 3, . . .) and 1 = (1, 1, 1, . . .). Proof. By the existence of a bijection between an embedding and a vertex-rotation system of a graph. [jes]

[jes]

[jes]

Let 𝒢m,n = 𝒢 (ℳm,n ) be the set of all underline graphs of map M ∈ ℳm,n and [jes] [jes] hence ℳm,n = ℳ(𝒢m,n ). [jes] Denote by t = aut[sm] (G) the order of the semi-automorphic group of G ∈ 𝒢m,n . Let [jes]

[jes]

ℐm,n = {t | ∃G ∈ 𝒢m,n , t = aut[sm] (G)}. [jes](t)

[jes]

(7.3.4)

Denote by 𝒢m,n the subset of 𝒢m,n in which every graph has t as the order of its semi-automorphic group.

7.3 Explicitness join-end surface

| 251

Lemma 7.3.4. For integer m ≥ 0 and integral vector n ≥ 0, F1[m,n] = ∑ Λ1[m,n] yn ,

(7.3.5)

m + π(n) 󵄨󵄨 n [jes](t) 󵄨 󵄨󵄨μ(𝒢m,n )󵄨󵄨󵄨(m − 1)!(i − 1)! t

(7.3.6)

[jes](t)

[jes]

[jes] t∈ℐm,n

where [jes](t)

Λ1[m,n] =

with π(n)(= inT ), i = (1, 2, 3, . . .) and n = (n1 , n2 , n3 , . . .). Proof. See Liu YP [59] (2003, Theorem 4.1, p. 211) and particularly Mao LF, Liu YP [75] (2003). [jes]

[jes]

By (7.2.1) and Lemma 7.3.4, E1[m,s] = F1[m,s] and then E1[m,n] = F1[m,n] has the following explicit expression: ∑ F1[m,n] yn = [jes]

[jes]

F1[m,s] =

π(n)=s [jes] n∈𝒥m,s

∑ ( ∑ Λ1[m,n] )yn , [jes](t)

(7.3.7)

[jes] [jes] n∈𝒥m,s t∈ℐm,n

[jes](t)

where Λ1[m,n] is given by (7.3.6). Lemma 7.3.5. For an integer m ≥ 0 and an integral vector n ≥ 0, [jes]

[jes](t)

Fa[m,n] = ∑ Λa[m,n]

(7.3.8)

[jes] t∈ℐm,n

[jes](t)

where Λa[m,n] ∈ ℝ is a polynomial of a with degree at most π(n) + 1, as extracted from the procedure in the proof of Theorem 7.2.6 such that [jes](t)

[jes](t)

Λa[m,n] |a=1 = Λ1[m,n] ,

(7.3.9)

which is just given by (7.3.6). Proof. This is a result of Observation 7.3.1 and Observation 7.3.2. Now, we are allowed to illustrate an explicit solutions of equation (7.1.1) via its specific case of a = 1 in the combinatorial sense. [jes]

Theorem 7.3.6. Let fa[jes] be the solution of equation (7.1.1) determined by Fa[m,s] for in[jes]

tegers m, s ≥ 0. Then Fa[m,s] has an explicit expression, [jes]

Fa[m,s] = [jes](t)

where Λa[m,n] is given in (7.3.8).

∑ ( ∑ Λa[m,n] )yn [jes](t)

[jes] n∈𝒥m,s

[jes] t∈ℐm,n

(7.3.10)

252 | 7 Surface equations first part Proof. On the basis of Lemma 7.3.5 and Observation 7.3.1 we obtain the conclusion. This theorem enables us directly to deduce an explicit expression of the number of root-isomorphic classes of join-end maps with root-vertex valency m and vertexpartition vector n on all orientable surfaces. Corollary 7.3.7. Given integer m ≥ 0 and integral vector n ≥ 0, the solution f[jes] of equation (7.1.1) under a = 1 is determined by m,n f[jes] = ∑ Λ1[m,n] 𝜕x,y

[jes](t)

(7.3.11)

[jes] t∈ℐm,s

[jes](t)

where Λ1[m,n] is given by (7.3.6). Proof. This is a result of Theorem 7.3.6.

7.4 Restrictions join-end surface Consider the equation for f 2 3 𝜕f { { {x ∫ y(1 + 𝜕x,y (f − 1)|x=u ) = (1 − x )f − x 𝜕x − 1; y { { { f { x=0,y=0 = 1,

(7.4.1)

where f ∈ ℛ{x, y}. This equation is the specific case of a0 = a1 = a2 = 1 in equation (7.1.1). From [66] (Liu YP, 2012), one might see the first equality in equation (7.4.1). However, f under 𝜕x,y should be replaced by f − 1. One might think of the equation for f ∈ ℛ{x, y} as only having one non-negative integer constant a, 2 3 𝜕f { { {x ∫ y(1 + 𝜕x,y (f − a)|u=x ) = (1 − x )f − x 𝜕x − a; { { y { {fx=0,y=0 = a,

(7.4.2)

where a ∈ ℤ+ . In Liu YP [73] (2012, pp. 379–389), equation (7.4.2) and the like occur in first time. Equation (7.4.1) is also a specific case of equation (7.4.2) with a = 1. For convenience in evaluating a solution of equation (7.4.1) in the form of a sum with all terms positive, the equation is equivalently transformed on ℛ{x, y} into 2 3 𝜕f { { {f = 1 + x f + x 𝜕x + x ∫ y(1 + 𝜕x,y (f − 1)|u=x ); y { { { f = 1. { x=0,y=0

(7.4.3)

7.4 Restrictions join-end surface

| 253

m For integer m ≥ 0, let Fm = [f ]m x = 𝜕x f , then

0, m [x2 f ]x = { Fm−2 ,

when 0 ≤ m ≤ 1; otherwise.

(7.4.4)

Lemma 7.4.1. For any integer m ≥ 0. If Fl ∈ ℛ+ {y}, 0 ≤ l ≤ m, then Fm−2 ∈ ℛ+ {y}. Proof. Because of m − 2 ≤ m, the given condition implies the conclusion. 𝜕f 󸀠 , then For integer m ≥ 0, let Fm = 𝜕xm 𝜕x

F1 , 󸀠 ={ Fm (m + 1)Fm+1 ,

when m = 0;

(7.4.5)

otherwise.

󸀠 Lemma 7.4.2. For any integer m ≥ 0, if Fl ∈ ℛ+ {y}, 0 ≤ l ≤ m, then Fm−1 ∈ ℛ+ {y}. 󸀠 󸀠 ∈ = mFm . By the given condition, Fm ∈ ℛ+ {y}. Therefore, Fm−1 Proof. From (7.4.5), Fm−1 ℛ+ {y}. This is the conclusion.

Let ∇i = 𝜕xi (𝜕x,y f |u=x ), i ≥ 0. For any integer k ≥ 1, k−1

𝜕x,y uk = ∑ xj yk−j . j=1

From the linearity of 𝜕x,y , we see that, for integer m ≥ 0, 0, when m = 0; 𝜕xm (y𝜕x,y (f − a1 )|u=x ) = { k−m+1 , when m ≥ 1. ∑k≥m+1 Fk y

(7.4.6)

Thus, from the linearity of the meson functional, we see that, for any integer m ≥ 0, m

y1 , when m = 0; [∫ y(1 + 𝜕x,y (f − 1))] = { ∑k≥m+1 Fk yk−m+1 , when m ≥ 1. x y

(7.4.7)

Observation 7.4.3. For a solution f of the first equality in equation (7.4.1), the initiation is consistent. Proof. Let F0 be the constant term of f ∈ ℛ{x, y}. Because of a common factor x on the left hand side of the first equality in equation (7.4.1), its constant term is 0. On the other hand, the constant term on the right hand side results in F0 − 1. Thus F0 − 1 = 0 ⇒ F0 = 1. This is just the initiation F0 = fx=0, y=0 = 1. This observation tells us that this equation is worth our while to investigate further.

254 | 7 Surface equations first part Theorem 7.4.4. Equation (7.4.3) for f on ℛ{x, y} is equivalent to the system of equations for Fm on ℛ{y} for m ≥ 0 1, { { { { { {y1 + ∑k≥1 Fk yk+1 , Fm = { {1 + ∑ Fk yk , { k≥2 { { { (m − 1)F m−2 + ∑k≥m Fk yk−m+2 , {

when m = 0; when m = 1; when m = 2;

(7.4.8)

when m ≥ 3.

Proof. This is a direct result of Theorem 7.1.4 for a = 1. In order to solve the system of equations with an infinite number of unknowns Fm , m ≥ 0, a new parameter s has to be introduced on Fm so that its terms are classified by s as polynomials. Let 𝒥m be the set of power vectors n = (n1 , n2 , n3 , . . .) of y such that yn is in a term of Fm . Denote 𝒥m,s = {n|π(n) = s} ⊆ 𝒥m , then (7.4.9)

𝒥m = ∑ 𝒥m,s s≥0

where π(n) = i nT and i = (1, 2, 3, . . . , i, . . .). Observation 7.4.5. For any vector n ∈ 𝒥m,s , s = m(mod 2) where integers m, s = π(n) ≥ 0. Proof. See the proof of Observation 7.1.5 for a = 1. This observation enables us to only consider Fm.s such that m = s(mod 2) for m, s ≥ 0 without loss of generality. Lemma 7.4.6. For any integer s ≥ 0, 1, when s = 0; F0,s = { 0, otherwise.

(7.4.10)

Proof. Because F0 = 1 ∈ ℝ+ is a constant, F0 is independent of s ≥ 1. Hence, F0,s = 0 for s ≥ 1. By the Kronecker symbol δs,t = 1, s = t; 0, s ≠ t, the lemma shows F0,s = δ0,s for s ≥ 0. Lemma 7.4.7. For any integer s ≥ 0, 0, F1,s = { y1 δ1,s + ∑s−1 k=1 Fk,s−k−1 yk+1 , Proof. See the proof of Lemma 7.1.7 for a = 1.

when s = 0(mod 2); when s = 1(mod 2).

7.4 Restrictions join-end surface

| 255

On the basis of this lemma, let us observe F1,1 , F1,3 and F1,5 . Because of 1 ≤ k ≤ s − 1, we have F1,1 = y1 , F1,3 = F1,1 y2 + F2,0 y3 = y1 y2 + y3 and F1,5 = y1 y22 + 2y2 y3 + 2y1 y4 + 2ay5 . The last two lemmas enable us to only discuss m ≥ 2 in what follows without loss of generality. From (7.4.8), we see that, for any integer m ≥ 2, when m = 2; δs,0 + ∑sk=2 Fk,s−k yk , Fm,s = { s+m−2 (m − 1)Fm−2,s + ∑k=m Fk,s−k+m−2 yk−m+2 , when m ≥ 3.

(7.4.11)

Lemma 7.4.8. For any integers m ≥ 2, m! t

Fm,0 = { 2 t! 0,

,

when m = 2t; when m = 2t + 1,

(7.4.12)

where t ≥ 1. Proof. A direct result of Lemma 7.1.8 for a = 1. From this lemma, one finds that Fm,0 ∈ ℛ+ {y}. Lemma 7.4.9. For any integer m ≥ 0,

Fm,1

0, when m = 2t, t ≥ 0; { { { = { y1 , when m = 1; { { t {2 t!y1 , when m = 2t + 1, t ≥ 1.

(7.4.13)

Proof. A direct result of Lemma 7.1.9 for a = 1. Lemma 7.4.8 and Lemma 7.4.9 enable us to only discuss s ≥ 2, without loss of the generality of s ≥ 0. Lemma 7.4.10. Given two integers m ≥ 2 and s ≥ 2. If for integers l, r ≥ 0, Fl,r ∈ ℛ+ {y} whenever l + r ≤ m + s − 1, then s+m−2

∑ Fk,s−k+m−2 yk−m+2 ∈ ℛ+ {y}.

k=m

Proof. We take into account m ≤ k ≤ m + s − 2 and k + (s − k + m − 2) = m + s − 2 ≤ m + s − 1. By the given condition, Fk,s−k+m−2 ∈ ℛ+ {y}. Thus, by considering yk−m+2 ∈ ℛ+ {y}, the conclusion can be drawn.

256 | 7 Surface equations first part Now, let us observe how to determine Fm,s by the procedure as starting from m+s = 0 in the increasing order one by one with increment 1 on the basis of (7.4.11). When m + s = 0, we have only one case: F0,0 = 1. Given by Lemma 7.4.6. When m + s = 1, we have only two cases: F1,0 = 0 and F0,1 = 0. This is given by Observation 7.4.3. When m+s = 2, F2,0 = 1 and F0,2 = 0 by, respectively, Lemma 7.4.8 and Lemma 7.4.6. The remaining case is F1,1 = y1 by Lemma 7.4.9. When m + s = 3. F3,0 = 0 (Lemma 7.4.8) and F0,3 = 0 (Observation 7.4.5), we have only two remaining cases: F2,1 = 0 and F1,2 = 0 as given by Observation 7.4.5. Theorem 7.4.11. Equation (7.4.3) is well defined on ℛ{x, y}. Proof. A direct result of Theorem 7.1.11 for a = 1. In order to evaluate the solution of equation (7.4.1) in the form of a finite sum with all terms positive, more combinatorial structures have to be investigated further. Observation 7.4.12. For any two integers m, s ≥ 0, the Fm,s are determined by some Fi,j for non-negative integers i and j such that i + j ≤ m + s − 1. Proof. This is a direct result of Observation 7.2.1 for a = 1. This observation suggests us that the solution of equation (7.4.1) can always be extracted recursively from its initial condition. Lemma 7.4.13. For any two integers m, s ≥ 0 given, if m + s = 1(mod 2), then Fa[m,s] = 0. Proof. See the proof of Lemma 7.2.2 for a = 1. This lemma enables us to reduce to half the amount of labor for obtaining the solution of equation (7.4.1). Lemma 7.4.14. Given two integers m, s ≥ 0. If m+s = 0(mod 2), then Fm,s is independent of yi for i ≥ s + 1. Proof. This is a result of Lemma 7.2.3 for a = 1. This lemma tells us that the variable vector is y = ys = (y1 , y2 , y3 , . . . , ys ) of s dimensions in Fa[m,s] . Lemma 7.4.15. Given two vectors m, s ≥ 0. If m + s = 0(mod 2), then Fm,s is a polynomial of ys with degree at most s. Proof. This is a result of Lemma 7.2.4 for a = 1. Lemma 7.4.16. Given two integers m, s ≥ 0, we have Fm,s ∈ ℛ+ {y} Proof. When m + s ≤ 3, it can be checked that Fm,s ∈ ℛ+ {y}. When m + s ≥ 4, assume Fl,t ∈ ℛ+ {y} for integers l, t ≥ 0, l + t ≤ m + s − 1. By induction, we prove that Fm,s satisfies the conclusion.

7.4 Restrictions join-end surface

| 257

We proceed on the basis of (7.4.11), Lemma 7.4.1, Lemma 7.4.2 and Lemma 7.4.8. By the assumption, Fm,s satisfies the conclusion. The last four lemmas tell us that, for any two integers m, s ≥ 0, Fm,s is a polynomial of ys with degree at most s and that Fm,s ∈ ℛ+ [ys ]. Theorem 7.4.17. Let f[jes] be the solution of equation (7.4.1). Given any two integers m, s ≥ 0. Denote [jes] Fm,s = [f[jes] ]sy (= Em,s ),

then Em,s is in the form of a finite sum with all terms positive,

Em,s

δ0,s , when s ≥ 0, m = 0; { { { { { { 0, when m = 0, s ≥ 1; m = 1, s ≥ 2, ; { { { { { s = 0(mod 2) or s ≥ 0, m ≠ s(mod 2); { { { { { m! { , when s = 0 and m = 2t, t ≥ 1; { { 2t t! { { { {y1 , when s = 1, m = 1; (7.4.14) ={ t {2 t!y1 , when s = 1, m = 2t + 1, t ≥ 1; { { { { { {∑s−1 when s ≥ 3 (s = 1(mod 2)), m = 1; { k=1 Ek,s−k−1 yk+1 , { { { s { { when s ≥ 2 (s = 0(mod 2)), m = 2; ∑k=2 Ek,s−k yk , { { { { { (m − 1)Em−2,s { { { { s+m−2 { + ∑k=m Ek,s−k+m−2 yk−m+2 , when m ≥ 3, s ≥ 2, m = s(mod 2).

Proof. When s ≥ 0, m = 0, by Lemma 7.4.6. When m = 0, s ≥ 1; m = 1, s ≥ 2, s = 0(mod 2); or s ≥ 0, m ≠ s(mod 2), by, respectively, Lemma 7.4.6, Lemma 7.4.8, or Lemma 7.4.13. When s = 0 and m = 2t, t ≥ 1, by Lemma 7.4.8. When s = 1, m = 2t + 1, t ≥ 0, by Lemma 7.4.10. When s ≥ 3 (s = 1(mod 2)), m = 1, by Lemma 7.4.7. When s ≥ 2 (s = 0(mod 2)), m = 2 and when m ≥ 3, s ≥ 2, m = s(mod 2), by (7.4.11). Now, we look at what happens when 4 ≤ m + s ≤ 6. By (7.4.13), we evaluate Fm,s (= Em,s ). When m + s = 4, because of F4,0 = 24!t t! = 3 and F0,4 = 0, it is only necessary to evaluate F3,1 , F2,2 and F1,3 . From F1,3 = F1,1 y2 +F2,0 y3 = y1 y2 +y3 , we have F1,3 = y1 y2 +y3 . From F2,2 = F2,0 y2 = y2 , we have F2,2 = y2 . From F3,1 = 2F1,1 = 2y1 , we have F3,1 = 2y1 . When m + s = 5, because of m ≠ s(mod 2), Fm,s = 0. 2 When m + s = 6, because of F6,0 = 26! 3 3! = 15, F5,1 = 2 2!y1 = 8y1 and F0,6 = 0, it is only necessary to evaluate F4,2 , F3,3 , F2,4 and F1,5 . From F1,5 = y1 y22 + 2y2 y3 + 2y1 y4 + 2y5 , we have F1,5 = y1 y22 + 2y2 y3 + 2y1 y4 + 2y5 . From F2,4 = y22 + 2y1 y3 + 3y4 , we have F2,4 = y22 + 2y1 y3 + 3y4 . From F3,3 = 2(y1 y2 + y3 ) + (2y1 )y2 + (3)y3 , we have F3,3 = 4y1 y2 + 5y3 . From F4,2 = 3y2 + F4,0 y2 , we have F4,2 = (3 + 3)y2 = 6y2 .

258 | 7 Surface equations first part

Figure 7.4.1: Root-isomorphic classes of join-end maps on surfaces of size up to 3.

Example 7.4.1. Root-isomorphic classes of join-end maps on all orientable surfaces. A map without end on an inner (non-root) face is said to be join-end. The number of root-isomorphic classes of join-end maps with given root-vertex valency m ≥ 0 and [jes] vertex-partition vector n such that π(n) = s ≥ 0 is just Fm,s = Fm,s as shown in Theorem 7.4.17. For example, F1,1 = y1 , F1,3 = y1 y2 + y3 and F1,5 = y1 y22 + 2y2 y3 + 2y1 y4 + 2y5 . Take a0 = a1 = a2 = a3 = 1, i. e., a = 1, F1[1,1] = y1 , F1[1,3] = y1 y2 + y3 and F1[1,5] = y1 y22 + 2y2 y3 + 2y1 y4 + 2y5 . In Figure 7.4.1, a = y1 = F1[1,1] , b+c = y1 y2 +y3 = F1[1,3] and d+(e+f )+2g +(i +j +k) = y1 y22 + (2y1 y3 ) + 2y1 y4 + (3y5 ) = F1[1,5] . Note that, although h = y1 y4 , because an end not rooted occurs, it is not in F1[1,5] .

7.5 Notes | 259

7.5 Notes 7.5.1. In this chapter, only orientable surfaces are considered. All surfaces orientable and non-orientable still need to be considered in a similar manner via decompositions as presented in Liu YP [61] (Liu, Y. P., 2009, pp. 231–236) for some types of subsets of maps join-end. [jes]

7.5.2. Although the coefficients of Fa[m,s] are determined by following the procedure of the evaluation shown in Section 7.2 as polynomials of a, all these polynomials still need to be extracted as explicit expressions as simple as possible. 7.5.3. On the basis of (7.1.8) or (7.4.8), an infinite system of linear equations can be established and its solution determined by powers of a matrix related to, respectively, a and y, or only y. In each case, an explicit expression of the solution can be directly obtained as in the following. Theorem 7.5.1. Let f = (Fa[1] , Fa[2] , Fa[1] , . . .) and ca = (a2 y1 , a0 , 0, . . .), the following equation: fT = Ya[jes] f + cTa[jes]

(7.5.1)

where Ya[jes] = (ya[i,j] )i,j≥1 such that

ya[i,j]

a2 yj−i+2 , when j − i = k ≥ 0; { { { { { {0, when j − i = −1; ={ { ja3 + 1, when j − i = −2; { { { { when j − i ≤ −3, {0,

(7.5.2)

is equivalent to equation (7.1.8). Proof. We need to only employ an equivalent operation on ℛ{y}∞ . Because of the existence of the inverse of I − Ya (I − Ya[jes] )−1 = ∑ Yka[jes] , k≥0

(7.5.3)

equation (7.5.1) is well defined on ℛ{y}. [k] )i,j≥1 for k ≥ 0 where Theorem 7.5.2. Let Yka[jes] = (ya[i,j] [k−1] [k] , = ∑ ya[i,l] ya[l,j] ya[i,j] l≥1

(7.5.4)

[1] = ya[i,j] for i, j ≥ 1. Then the solution of equation (7.5.1) has the explicit in which ya[i,j] expression

260 | 7 Surface equations first part [l] fT = cTa[jes] + ∑(a2 y1 y[l] a[∗,1] + a0 y∗,2 )

(7.5.5)

l≥1

j

and y[l] where y[l] ∗,2 are, respectively, the first and second columns of Ya[jes] for l ≥ 1. a[∗,1] Proof. We prove this by only considering operations on ℛ{y}∞ . 7.5.4. If parameter π(n) is given, the infinite summation in (7.5.5) becomes finite because of the limitation restricted in the s-dimensional system. This guarantees that the solution of equation (7.1.1) can always be evaluated whenever s = π(n) ≥ 0 is given. The form of solution is particularly favorite to use for making things more efficient and more intelligent. 7.5.5. Consider some types of linear transformations from y to z for getting favorite expression so that only arithmetics needs to be employed on the extension of the integral domain. 7.5.6. For a as a functional vector of x and/or y, we can find some meaningful objects except only for maps in combinatorics because of the universality of partitions of a set for classifications everywhere. 7.5.7. Equation (7.1.1) provides a theoretical bases on meson equations of join-end surface type for extending a certain number of other equations of surface types to reflect the theoretical ideas of tree-like meson equations for extending a certain numbers of outer types. Almost all of them have linearity. About these topics, one might like to read the original articles such as in Liu YP [31] (Liu, Y. P., 1986), [33] (1986), [34] (1986), [36] (1987), [37] (1987), [43] (1989), [45] (1989), [54] (1993) etc. only on the sphere and [61] (2009, equation (5.5.11) on p. 142; equation (6.5.11) on p. 171), [66] (2012), [67] (2012) etc. on all orientable surfaces. 7.5.8. One might think of the extension from only all orientable surfaces to all surfaces including both orientable and non-orientable surfaces. It looks reasonable that the equation is of the same type with distinction only in the coefficients. This idea reminds us to see that the three pairs of equations, equation (1.1) and equation (1.2); equation (1.3) and equation (1.4), equation (1.5) and equation (1.6), are, respectively, on only orientable surfaces and on all surfaces as shown in [65] (Liu, Y. P., 2012). 7.5.9. A genus series (or genus polynomial) of rooted join-end maps can be extracted on the basis of (7.3.11) by introducing the dual order (or face number) as a parameter. It seems no result has yet been reached by the technique of Jackson et al. via Young tableaux as shown in [19] (Jackson, D. M., 1995) and [20] (Jackson, D. M., Visentin, T. I., 1990) etc. for rooted ordinary maps. However, we seem to have the existence of a relationship between join-end maps and ordinary maps, so that one of the two genus series can be converted to the other.

7.5 Notes | 261

7.5.10. Asymptotics. Note 7.5.4 enables us to investigate the corresponding asymptotics as shown in [56] (Liu, Y. P., 1999, pp. 359–389), [94] (Yan, J. Y., Liu, Y. P., 1991). 7.5.11. Stochastic behavior (or stochastics). We proceed on the basis of (7.4.14), the probability of order, co-order, semi-automorphic group order or genus for join-end orientable maps with size given can be found via order polynomials, co-order polynomials, semi-automorphic group order polynomials or genus polynomials. 7.5.12. Distributions. Theorem 7.4.17 enables us to determine the distribution of order, co-order (equivalent to genus) for join-end orientable maps with size given. Then all statistical moments, including the mathematical expectation, can be established. 7.5.13. The topic in this chapter is from Program 97 in [71] (Liu, Y. P., Vol. 22, 2016, p. 10744) as a primary part for systematization in theory.

8 Surface equations second part 8.1 Bridgeless surface model For a function g ∈ ℛ{x, y}, consider the equation 2 3 𝜕g { { {a2 x ∫ y𝜕x,y (g − a0 )|x=u = (1 − x )g − a3 x 𝜕x − a1 ; y { { { g { x=0,y=0 = a0 ,

(8.1.1)

where a0 , a1 , a2 , a3 ∈ ℤ+ . This is equation (15) in Introduction. Because of the contribution from maps without cut-edge (i. e., bridge) on all orientable surfaces in the case of a0 = a1 = a2 = a3 = 1, equation (8.1.1) is then called a bridgeless surface model. In [60] (Liu YP, 2008, p. 180), one might find the first equality of equation (8.1.1) when a0 = a1 = a2 = a3 = 1. However, the coefficient of the term with partial differentiation is x3 instead of x2 , and f under 𝜕x,y should be replaced by g − 1. In [69] (Liu YP, 2015, p. 389), the following equation for g ∈ ℛ{x, y}: 2 3 𝜕g { { {x ∫ y(𝜕x,y (g − 1)|u=x ) = (1 − x )g − x 𝜕x − b; y { { { {gx=0,y=0 = a,

(8.1.2)

where a, b ∈ ℤ+ ⊆ ℝ+ occurs. Although the specific case of equation (8.1.2) with a = b = 1 is the same as that of equation (8.1.1) with a0 = a1 = a2 = a3 = 1, equation (8.1.2) itself is also a specific case of equation (8.1.1) with a0 = a, a1 = b and a2 = a3 = 1. For integer m ≥ 0, let Ga[m] = 𝜕xm f where a = (a0 , a1 , a2 , a3 ). Observation 8.1.1. If a0 ≠ a1 , then (8.1.1) has no solution. Proof. Let Ga[0] be the constant term of g. For the reason of there being a common factor x on the left hand side of the first equality in (8.1.1), the constant term is 0 on this side. By the constant term G0 − a1 on the right hand side, we have G0 − a1 = 0 󳨐⇒ G0 = a1 . However, the initiation tells us that G0 = a0 . Therefore, equation (8.1.1) is not consistent whenever a1 ≠ a0 . This lemma enables us to only consider equation (8.1.1) in a0 = a1 , 2 3 𝜕g { { {a2 x ∫ y(𝜕x,y (g − a0 )|u=x ) = (1 − x )g − a3 x 𝜕x − a0 ; y { { { g { x=0,y=0 = a0 . https://doi.org/10.1515/9783110627336-008

(8.1.3)

264 | 8 Surface equations second part For the convenience of evaluating a solution via the forms where each is a finite sum of terms all positive, equation (8.1.1) on ℛ{x, y} is transformed into 2 3 𝜕g { { {g = a0 + x g + a3 x 𝜕x + a2 x ∫ y(𝜕x,y (g − a0 )|u=x ); y { { { {gx=0,y=0 = a0 .

(8.1.4)

󸀠 Given an integer m ≥ 0. Let Ga[m] = 𝜕xm 𝜕g , then 𝜕x

Ga[1] , when m = 0; 󸀠 ={ Ga[m] (m + 1)Ga[m+1] , otherwise,

(8.1.5)

where a = (a0 , a2 , a3 ) is instead of (a0 , a1 , a2 , a3 ) from Observation 8.1.1. 󸀠 Lemma 8.1.2. Given an integer m ≥ 0. If Ga[l] ∈ ℛ+ {y}, 0 ≤ l ≤ m, then Ga[m−1] ∈ ℛ+ {y}. 󸀠 Proof. By (8.1.5), Ga[m−1] = mGa[m] . From the known condition, Ga[m] ∈ ℛ+ {y}. There󸀠 fore, Ga[m−1] ∈ ℛ+ {y}. This is the conclusion.

Let ∇a[i] = 𝜕xi (𝜕x,y g|u=x ), i ≥ 0. For integer k ≥ 1, 𝜕x,y uk =

yx k − xyk k−1 j k−j = ∑xy . x−y j=1

Note that 𝜕x,y 1 = −1 is guaranteed. We proceed on the basis of linearity of the operator 𝜕x,y . For integer m ≥ 0, 0, when m = 0, 1; 𝜕xm (y𝜕x,y (g − a0 )|u=x ) = { ∑k≥m+1 Ga[k] yk−m+1 , when m ≥ 2.

(8.1.6)

Thus, on the linearity of the meson functional, for m ≥ 0, m

0, [∫ y(𝜕x,y (g − a))] = { ∑k≥m+1 Ga[k] yk−m+1 , x y

when m = 0, 1; when m ≥ 2.

(8.1.7)

Theorem 8.1.3. Equation (8.1.3) for g ∈ ℛ{x, y} is equivalent to the system of equations for Ga[m] ∈ ℛ{y} for m ≥ 0

Ga[m]

a0 , { { { { { {0, ={ {a0 + a2 ∑ Ga[k] yk , { k≥2 { { { a (m − 1)G a[m−2] + a2 ∑k≥m Ga[k] yk−m+2 , { 3

when m = 0; when m = 1; when m = 2; when m ≥ 3.

(8.1.8)

8.1 Bridgeless surface model | 265

Proof. Because of x| (g − a) in equation (8.1.3), Ga[0] = a0 . This is the initiation. When m = 0, equation (8.1.8) holds. By (8.1.7), 󵄨󵄨 󵄨 x2 󵄨󵄨󵄨 ∫ y(𝜕x,y (g − a0 )). 󵄨󵄨 y From equation (8.1.3), x2 | (g − a0 ). This implies that Ga[1] = [g]1x = a0 . When m = 1, (8.1.8) holds. For m ≥ 2, we proceed on the basis of (8.1.4)–(8.1.7). By equation (8.1.3), the system of equations (8.1.8) is equivalently transformed. In order to solve the system of infinite equations with the infinite unknowns Ga[m] for m ≥ 0, a new parameter s needs to be introduced such that all terms of Ga[m] can be partitioned with each part containing only finite terms upon s. Let 𝒥m be the set of all power vectors n = (n1 , n2 , n3 , . . .) such that Ga[m] contains a term with yn . Denote 𝒥m,s = {n|π(n) = s} ⊆ 𝒥m , then 𝒥m = ∑ 𝒥m,s s≥0

(8.1.9)

where π(n) = i nT and i = (1, 2, 3, . . . , i, . . .). Lemma 8.1.4. For any integer s ≥ 0, a0 , when s = 0; Ga[0,s] = { 0, otherwise.

(8.1.10)

Proof. Ga[0] = a0 ∈ ℛ+ is a constant; Ga[0] is independent of all s ≥ 1. Therefore Ga[0,s] = 0 for any s ≥ 1. By use of the Kronecker symbol δs,t = 1, s = t; 0, s ≠ t, this lemma is Ga[0,s] = a0 δ0,s . Lemma 8.1.5. For any integer s ≥ 0, Ga[1,s] = 0. Proof. From the second equality of equation (8.1.7), the conclusion can be drawn. The two lemmas above enable us to only discuss Ga[m,s] for m ≥ 2 without loss of generality. We proceed on the basis of equation (8.1.8). For integers m ≥ 2 and s ≥ 0, when m = 2; a0 δs,0 + a2 ∑sk=2 Ga[k,s−k] yk , Ga[m,s] = { s+m−2 a3 (m − 1)Ga[m−2,s] + a2 ∑k=m Ga[k,s−k+m−2] yk−m+2 , when m ≥ 3.

(8.1.11)

Lemma 8.1.6. For any integer m ≥ 2, Ga[m,0] = where the integer t ≥ 1.

a0 at3 m! , t { 2 t!

0,

when m = 2t; when m = 2t + 1,

(8.1.12)

266 | 8 Surface equations second part Proof. On account of (8.1.11), because of 0

∑ Ga[k,s−k] yk = 0

k=2

and

m−2

∑ Ga[k,s−k+m−2] yk−m+2 = 0,

k=m

we see that, for any integer m ≥ 2, a0 , when m = 2; Ga[m,0] = { a3 (m − 1)Ga[m−2,0] , when m ≥ 3. From (8.1.10), Ga[0,0] = a0 . From Lemma 8.1.5, Ga[1,0] = 0. For integers m ≥ 2, assume all Ga[i,0] for 0 ≤ i ≤ m − 1 are known. By induction, we prove the case of i = m. We proceed on the basis of Ga[m,0] = a3 (m − 1)Ga[m−2,0] . From m = m − 2(mod 2), the assumption shows that when m = 1(mod 2), Ga[m,0] = 0 and that when m = 2t(t ≥ 1), Ga[m,s] = a3 (2t − 1)(

a0 at−1 a0 at3 (2t − 1)! a0 at3 (2t)! 3 (2t − 2)! ) = = . 2t t! 2t−1 (t − 1)! 2t−1 (t − 1)!

Therefore, the conclusion can be drawn. This lemma tells us that Ga[m,0] ∈ ℛ+ {y} if, and only if, a0 , a3 ∈ ℝ+ . Lemma 8.1.7. For any integer m ≥ 2, Ga[m,1] = 0. Proof. By Lemma 8.1.4 and Lemma 8.1.5, Ga[0,1] = Ga[1,1] = 0. When m = 2, by (8.1.11), because of 1

∑ Ga[k,1−k] yk = 0

k=2

and

m−1

∑ Ga[k,1−k+m−2] yk−m+2 = 0,

k=m

we see that Ga[2,1] = 0 and Ga[m,1] = a3 (m − 1)Gm−2,1 for, respectively, m = 2 and m ≥ 3. Assume all Ga[l,1] = 0 for 0 ≤ l ≤ m − 1. By induction, we prove Ga[m,1] = 0. By the assumption, Ga[m−2,1] = 0. Therefore, (m−1)Ga[m−2,1] = 0, i. e., Ga[m,1] = 0. The conclusion can be drawn. Lemma 8.1.6 and Lemma 8.1.7 enable us to only discuss s ≥ 2 without loss of generality for s ≥ 0. Lemma 8.1.8. Given two integers m ≥ 2 and s ≥ 2. If Ga[l,r] ∈ ℛ+ {y} whenever l + r ≤ m + s − 1 for any two integers l, r ≥ 0, then s+m−2

∑ Ga[k,s−k+m−2] yk−m+2 ∈ ℛ+ {y}.

k=m

Proof. We take into account k + (s − k + m − 2) = m + s − 2 ≤ m + s − 1 for any integers m ≤ k ≤ m + s − 2. By the known condition, Ga[k,s−k+m−2] ∈ ℛ+ {y}. Therefore, the conclusion can be drawn by considering yk−m+2 ∈ ℛ+ {y}.

8.2 Solution bridgeless surface |

267

Now, let us have a look at what happens to determine Ga[m,s] in the order of increasing m + s one by one starting from m + s = 0 by (8.1.10). When m + s = 0, only Ga[0,0] = a0 as given by Lemma 8.1.4. When m + s = 1, only Ga[1,0] = 0 and Ga[0,1] = 0, given by, respectively, Lemma 8.1.5 and Lemma 8.1.4. When m + s = 2, in spite of Ga[2,0] = a0 a3 and Ga[0,1] = 0, only Ga[1,1] has to be found. By Lemma 8.1.5, Ga[1,1] = 0. When m + s = 3, in spite of Ga[3,0] = 0 (by Lemma 8.1.5) and Ga[0,3] = 0 (by Lemma 8.1.4), only Ga[2,1] and Ga[1,2] have to be known. From (8.1.11), Ga[2,1] = 0. From Lemma 8.1.7, Ga[1,2] = 0. Theorem 8.1.9. Equation (8.1.1) is well defined on ℛ{x, y} if, and only if, a0 = a1 . Proof. Sufficiency. Because of a0 = a1 , it is only necessary to evaluate a solution of equation (8.1.3) under the initiation. When m + s is smaller, the Ga[m,s] are determined from (8.1.10). When m + s is greater, assume all Ga[l,t] determined for integers, l, t ≥ 0 such that l + t ≤ m + s − 1. By induction, we evaluate Ga[m,s] . When m = 2, because of k + (s − k) = s ≤ 2 + s − 1 = s + 1 for any integers k ≥ 0 from the first equality of (8.1.11), Ga[k,s−k] are determined for 2 ≤ k ≤ s. Thus, Ga[2,s] can be determined for any integer s ≥ 0, 2 + s = m + s. When m ≥ 3, because of (m−2)+s ≤ m+s−1 and k+(s−k+m−2) = s+m−2 ≤ m+s−1 for any integer k ≥ 0, by the assumption, Ga[m−2,s] and all Ga[k,s−k+m−2] are determined for m ≤ k ≤ s + m − 2. Thus, all Ga[m,s] for m ≥ 3, s ≥ 0 are determined from the second equality of (8.1.11). Since all such Ga[m,s] satisfy the system of equations (8.1.8), Theorem 8.1.3 tells us that equation (8.1.3) has a solution on ℛ{x, y}. By considering the uniqueness of the procedure used above from the initiation, there is only one solution. Necessity is a result of Observation 8.1.1.

8.2 Solution bridgeless surface In order to extract the solution of equation (8.1.3) partitioned by each part in the form of a finite sum with all terms positive that is as simple as possible, more investigation of its inner structures needs to be done. Lemma 8.2.1. For any integers m, s ≥ 0, if m + s = 1(mod 2), then Ga[m,s] = 0. Proof. From the evaluation above, it is seen that, for m + s small, all Ga[m,s] = 0 whenever m ≠ s(mod 2). For m + s greater, assume for any integers l, t ≥ 0, Ga[l,t] = 0 if l + t ≤ m + s − 1 and m ≠ s(mod 2). By induction, we prove that, for any integers m, s ≥ 0, Ga[m,s] = 0 whenever m ≠ s(mod 2). From Lemma 8.1.4—Lemma 8.1.7, we are allowed only to discuss m, s ≥ 2.

268 | 8 Surface equations second part When m = 2, for s ≥ 2, s ≠ 0(mod 2), because of k + (s − k) = s ≠ 0(mod 2) and k+(s−k) = s ≤ 2+s−1 = s+1 for any integer k ≥ 0, the assumption leads to Ga[k,s−k] = 0. Therefore, by the first equality of (8.1.11), Fa[2,s] = 0, s = 1(mod 2). When m ≥ 3, we have m + s = 1(mod 2) and (m − 2) + s = m + s(mod 2), k + (s − k + m − 2) = s + m − 2 = m + s(mod 2). From (m − 2) + s = m + s − 2 ≤ m + s − 1 and k + (s − k + m − 2) = s + m − 2 ≤ m + s − 1, the assumption leads to Ga[m−2,s] = 0 and Ga[k,s−k+m−2] = 0. By the second equality of (8.1.11), Ga[m,s] = 0. The conclusion can be drawn. This lemma enables us to reduce to half the amount of labor in evaluating the solution of equation (8.1.3). Lemma 8.2.2. Given two integers m, s ≥ 0, if m + s = 0(mod 2), then Ga[m,s] is independent of y1 and yi for i ≥ s + 1. Proof. From Lemma 8.1.7, the Ga[m,s] are all independent of y1 . Given s ≥ 2, for any integer m ≥ 2 and integral vector n ∈ 𝒥m , we have s = ∑ ini , i≥1

ni ∈ ℤ+ , i ≥ 0.

If there exists an yi , i ≥ s + 1, such that ni ≥ 1, then s ≥ ini ≥ (s + 1). This is a contradiction. Thus, 𝒥m,s has no vector containing ni > 0 such that i ≥ s + 1. This is the conclusion. This lemma tells us that Ga[m,s] is only a function of ys = (0, y2 , y3 , . . . , ys ). Lemma 8.2.3. Given integers m, s ≥ 0, if m + s = 0(mod 2), then Ga[m,s] is a polynomial of ys with degree at most s. Proof. As a matter of fact, we are to prove this for any n ∈ 𝒥m,s , |n| ≤ s. Let ℋ = {n | n ≥ 0, inT = s}, because of max{|n| | ∀n ∈ ℋ} = |s11 | = s and 𝒥m,s ⊆ ℋ, Ga[m,s] is a polynomial of ys with degree at most s. The last two lemmas imply that, for any integers m, s ≥ 0, Ga[m,s] is of the form of a sum of finite terms. Lemma 8.2.4. Given two integers m, s ≥ 0. Ga[m,s] ∈ ℛ+ {y} if, and only if, a ∈ ℝ3+ . Proof. When m + s is small, the Ga[m,s] are checked for Ga[m,s] ∈ ℛ+ {y} if, and only if, a ∈ ℝ4+ . When m + s is greater, assume for any integers l, t ≥ 0 such that for l + t ≤ m + s − 1, we have Ga[l,t] ∈ ℛ+ {y} if, and only if, a ∈ ℝ3+ . By induction, we prove that Ga[m,s] satisfies the conclusion. We proceed on the basis of (8.1.11). From Lemma 8.1.2, Lemma 8.1.6 and Lem ma 8.1.8, the assumption shows that Ga[m,s] satisfies the conclusion.

8.2 Solution bridgeless surface |

269

The last three lemmas tell us that, for any two integers m, s ≥ 0, Ga[m,s] is a polynomial of ys with degree at most s and Fa[m,s] ∈ ℛ+ {y} if, and only if a ∈ ℝ3+ . Theorem 8.2.5. Let ga[bls] be the solution of equation (8.1.3). For any integers m, s ≥ 0, [bls] [bls] denote Ga[m,s] = [ga[bls] ]sy , then Ga[m,s] has the form of a finite sum with all terms positive:

[bls] Ga[m,s]

a0 δ0,s , { { { { { { 0, { { { { { { { { t { { { a0 at 3 m! , 2 = { t! { { [bls] { { yk , a2 ∑sk=2 Ga[k,s−k] { { { { { { [bls] { a3 (m − 1)Ga[m−2,s] { { { { { s+m−2 [bls] { + a2 ∑k=m Ga[k,s−k+m−2] yk−m+2 ,

when s ≥ 0, m = 0; when m = 1, s ≥ 1; s = 1; m ≥ 1; or s ≥ 0, m ≠ s(mod 2); when s = 0 and m = 2t, t ≥ 1; when s ≥ 2 (s = 0(mod 2)), m = 2; when m ≥ 3, s ≥ 2 (m = s(mod 2)). (8.2.1)

[bls] = Ga[m,s] . When s ≥ 0, m = 0, by (8.1.8). Proof. From Theorem 8.1.9, Ga[m,s] When m = 1, s ≥ 1; s = 1, m ≥ 1; or s ≥ 0, m ≠ s(mod 2), by, respectively, Lemma 8.1.5; Lemma 8.1.7; or Lemma 8.2.1. When s = 0 and m = 2t, t ≥ 1, by Lemma 8.1.6. When s ≥ 2 (s = 0(mod 2)), m = 2 and when m ≥ 3, s ≥ 2, m = s(mod 2), by, respectively, the first and the second equalities of (8.1.11).

Let us have a look at what happens to Ga[m,s] when 4 ≤ m + s ≤ 6 by employing (8.2.1). a a2 4!

When m + s = 4, because of Ga[4,0] = 02t t!3 = 3a0 a23 , Ga[3,1] = 0, Ga[1,3] = 0 and Ga[0,4] = 0, it is only necessary to evaluate Ga[2,2] . For Ga[2,2] , because of 2

Ga[2,2] = a2 ∑ Ga[k,2−k] yk , k=2

(8.2.2)

we have Ga[2,2] = a0 a2 a3 y2 . When m + s = 5, because of m ≠ s(mod 2), Ga[m,s] = 0. a a3 6!

When m + s = 6, because of Ga[6,0] = 023 3!3 = 15a0 a33 , Ga[5,1] = 0, Ga[1,5] = 0 and Ga[0,6] = 0, it is only necessary to evaluate Ga[4,2] , Ga[3,3] and Ga[2,4] . For Ga[2,4] , because of 4

Ga[2,4] = a2 ∑ Ga[k,4−k] yk = (a0 a22 a3 y2 )y2 + (3a0 a2 a23 )y4 , k=2

we have Ga[2,4] = a0 a22 a3 y22 + 3a0 a2 a23 y4 .

(8.2.3)

270 | 8 Surface equations second part For Ga[3,3] , because of 3+1

Ga[3,3] = 2a3 Ga[1,3] + a2 ∑ Ga[k,3−k+1] yk−1 = (3a0 a2 a23 )y3 , k=3

(8.2.4)

we have Ga[3,3] = 3a0 a2 a23 y3 . For Ga[4,2] , because of 6−2

Ga[4,2] = 3a3 Ga[2,2] + a2 ∑ Ga[k,4−k] yk−2 k=4

(8.2.5)

= 3a3 (a0 a2 a3 y2 ) + a2 Ga[4,0] y2 , we have Ga[4,2] = (3a0 a2 a23 + 3a0 a2 a23 )y2 = 6a0 a2 a23 y2 .

8.3 Explicitness bridgeless surface Let ℳ[bls] m,n be the set of all root-isomorphic classes of bridgeless maps with root-vertex valency m ≥ 0 and the vertex-partition vector n ≥ 0 on all orientable surfaces. For any [bls] integer s ≥ 0, denote by ℳ[bls] m,s the union of all ℳm,n such that s = π(n) for n ≥ 0. Be[bls] cause the enufunction of ℳm,n satisfies equation (8.1.1) for a0 = a1 = a2 = a3 = 1, from [bls] [bls] )= (Ga[m,s] Theorem 8.2.5, its solution fa[bls] (ga[bls] ∈ ℛ{x, y}) is determined by Fa[m,s] m 𝜕x fa[bls] |π(n)=s ∈ ℛ{y} for m, s ≥ 0. Observation 8.3.1. For any integers m, s ≥ 0, 󵄨 [bls] 󵄨 [bls] 󵄨󵄨 F1[m,s] 󵄨󵄨y=1 = 󵄨󵄨󵄨ℳm,s 󵄨󵄨󵄨

(8.3.1)

where a = (a0 , a2 , a3 ) and 1 = (1, 1, 1), or (1, 1, 1, . . .) according as a or y. Proof. The result follows from the specific case of equation (8.1.1) when a = 1. [bls] [bls] Let 𝒢m,s and 𝒢m,n be the sets of all bridgeless graphs of root-vertex valency m ≥ 0 with, respectively, s = π(n) for all n ≥ 0 and the vertex-partition vector n ≥ 0. [bls] Observation 8.3.2. Let 𝒢 (ℳ[bls] m,s ) and 𝒢 (ℳm,n ) be the sets of all underline graphs of [bls] maps in, respectively, ℳ[bls] m,s and ℳm,n . Then [bls]

[bls]

𝒢m,n = 𝒢 (ℳm,n )

[bls] and hence 𝒢m,s = 𝒢 (ℳ[bls] m,s ).

(8.3.2)

Proof. The proof is found by showing a bijection between two sides of each equality. [bls] [bls] Let Fm,s |y=1 or Fm,n be the numbers of distinct root-isomorphic classes of bridgeless maps of root-vertex valency m with, respectively, s = π(n) for all n ≥ 0 or the vertex-partition vector n given on all orientable surfaces.

8.3 Explicitness bridgeless surface

| 271

[bls] Lemma 8.3.3. Given G ∈ 𝒢m,n , denote by μG the set of topologically non-equivalent embeddings of G. Then

|μG | = (m − 1)! ∏(i − 1)!ni = (m − 1)!(i − 1)!n i≥1

(8.3.3)

where i = (1, 2, 3, . . .) and 1 = (1, 1, 1, . . .). Proof. The result follows by the existence of a bijection between an embedding and a vertex-rotation system of a graph. [bls] [bls] Let 𝒢m,n = 𝒢 (ℳ[bls] m,n ) be the set of all underline graphs of map M ∈ ℳm,n and [bls] [bls] hence ℳm,n = ℳ(𝒢m,n ) from Observation 8.3.2. [bls] Denote by t = aut[sm] (G) the order of the semi-automorphic group of G ∈ 𝒢m,n . Let [bls]

[bls]

ℐm,n = {t | ∃G ∈ 𝒢m,n , t = aut[sm] (G)}.

(8.3.4)

[bls] [bls](t) in which every graph has t as the order of its Denote by 𝒢m,n the subset of 𝒢m,n semi-automorphic group.

Lemma 8.3.4. For integer m ≥ 0 and integral vector n ≥ 0, [bls] n F1[m,n] = ∑ Λ[bls](t) 1[m,n] y ,

(8.3.5)

m + π(n) 󵄨󵄨 n [bls](t) 󵄨 󵄨󵄨μ(𝒢m,n )󵄨󵄨󵄨(m − 1)!(i − 1)! t

(8.3.6)

[bls] t∈ℐm,n

where Λ[bls](t) 1[m,n] =

with π(n)(= inT ), i = (1, 2, 3, . . .) and n = (n1 , n2 , n3 , . . .). Proof. See Liu YP [59] (2003, Theorem 4.1, p. 211) and particularly Mao LF, Liu YP [75]. [bls] [bls] has the foland then E1[m,n] = F1[m,n] By (8.3.1) and Lemma 8.3.4, E1[m,s] = F1[m,s] lowing explicit expression: [bls] F1[m,s] =

n ∑ ( ∑ Λ[bls](t) 1[m,n] )y ,

[bls] [bls] t∈ℐm,n n∈𝒥m,s

(8.3.7)

where Λ[bls](t) is given by (8.3.6). 1[m,n] Lemma 8.3.5. For an integer m ≥ 0 and an integral vector n ≥ 0, [bls](t) n [bls] y = ∑ Λa[m,n] Fa[m,n] [bls] t∈ℐm,n

(8.3.8)

272 | 8 Surface equations second part where Λ[bls](t) ∈ ℛ is a polynomial of a with degree at most π(n) + 1, as extracted from a[m,n] the procedure in the proof of Theorem 7.2.6 such that 󵄨󵄨 [bls](t) Λ[bls](t) a[m,n] 󵄨󵄨a=1 = Λ1[m,n] ,

(8.3.9)

which is just given by (8.3.6). Proof. This is a result of Observation 8.3.1 and Observation 8.3.2. Now, we are allowed to illustrate explicit solutions of equation (8.1.1) via its specific case of a = 1 in the combinatorial sense. [bls] for Theorem 8.3.6. Let fa[bls] be the solution of equation (8.1.1) determined by Fa[m,s] [bls] has the explicit expression integers m, s ≥ 0. Then Fa[m,s] [bls] = Fa[m,s]

n ∑ ( ∑ Λ[bls](t) a[m,n] )y

(8.3.10)

[bls] [bls] t∈ℐm,n n∈𝒥m,s

is given in (8.3.8). where Λ[bls](t) a[m,n] Proof. Based on Lemma 8.3.5, Observation 8.3.1 yields the conclusion. This theorem enables us directly to deduce the explicit expression of the number of root-isomorphic classes of bridgeless maps with root-vertex valency m and vertexpartition vector n on all orientable surfaces. Corollary 8.3.7. Given integer m ≥ 0 and integral vector n ≥ 0, the solution f[bls] of equation (7.1.1) under a = 1 is determined by m,n f[bls] = ∑ Λ[bls](t) 𝜕x,y 1[m,n]

(8.3.11)

[bls] t∈ℐm,s

is given by (8.3.6). where Λ[bls](t) 1[m,n] Proof. This is a direct result of Theorem 8.3.6.

8.4 Restrictions bridgeless surface For function g ∈ ℛ{x, y}, observe the equation 2 3 𝜕g { { {x ∫ y𝜕x,y (g − 1)|x=u = (1 − x )g − x 𝜕x − 1; y { { { g { x=0,y=0 = 1.

This is equation (8.1.1) in the case of a0 = a1 = a2 = a3 = 1.

(8.4.1)

8.4 Restrictions bridgeless surface

| 273

In [60] (Liu YP, 2008, p. 180), one might refer to the first equality of equation (8.4.1). However, the coefficient of the term with partial differentiation is x3 instead of x2 , and f under 𝜕x,y should be replaced by g − 1. In [69] (Liu YP, 2015, p. 389), one might refer to the following equation for g ∈ ℛ{x, y}: 2 3 𝜕g { { {x ∫ y(𝜕x,y (g − a)|u=x ) = (1 − x )g − x 𝜕x − b; y { { { g { x=0,y=0 = a,

(8.4.2)

where a, b ∈ ℤ+ ⊆ ℝ+ . Although the specific case of equation (8.4.2) with a = b = 1 is the same as that of equation (8.1.1) with a0 = a1 = a2 = a3 = 1, equation (8.4.2) itself is also a specific case of equation (8.1.1) with a0 = a, a1 = b and a2 = a3 = 1. For integer m ≥ 0, let Ga[m] = 𝜕xm g. Observation 8.4.1. The initiation gx=0,y=0 = 1 is consistent in equation (8.4.1). Proof. Let G0 be the constant term of g ∈ ℛ{x, y}. Because of a common factor x on the left hand side of the first equality in equation (8.4.1), its constant term is 0. On the other hand, the constant term on the right hand side results in −1 + G0 . Thus −1 + G0 = 0 ⇒ G0 = 1. This is just the initiation G0 = gx=0, y=0 = 1. For the convenience of evaluating a solution via forms each as a finite sum of terms all positive, equation (8.4.1) on ℛ{x, y} is, by the cancelation law, transformed into 2 3 𝜕g { { {g = 1 + x g + x 𝜕x + x ∫ y(𝜕x,y (g − 1)|u=x ); y { { { {gx=0,y=0 = 1.

(8.4.3)

󸀠 Given an integer m ≥ 0. Let Gm = 𝜕xm 𝜕g , then 𝜕x

G1 , when m = 0; 󸀠 Gm ={ (m + 1)Gm+1 , otherwise.

(8.4.4)

󸀠 Lemma 8.4.2. Given an integer m ≥ 0. If Gl ∈ ℛ+ {y}, 0 ≤ l ≤ m, then Gm−1 ∈ ℛ+ {y}. 󸀠 Proof. By (8.4.4), Gm−1 = mGm . From the known condition, Gm ∈ ℛ+ {y}. Therefore, 󸀠 Gm−1 ∈ ℛ+ {y}. This is the conclusion.

Let ∇i = 𝜕xi (𝜕x,y g|u=x ), i ≥ 0. For integer k ≥ 1, k−1

𝜕x,y uk = ∑ xj yk−j . j=1

274 | 8 Surface equations second part Note that 𝜕x,y 1 = −1. We proceed on the basis of linearity of the operator 𝜕x,y . For integer m ≥ 0, 0, 𝜕xm (y𝜕x,y (g − 1)|u=x ) = { ∑k≥m+1 Gk yk−m+1 ,

when m = 0, 1; when m ≥ 2.

(8.4.5)

Thus, we address the linearity of a meson functional. For m ≥ 0, m

0, when m = 0, 1; [∫ y(𝜕x,y (g − 1))] = { ∑k≥m+1 Gk yk−m+1 , when m ≥ 2. x y

(8.4.6)

Theorem 8.4.3. Equation (8.4.3) for g ∈ ℛ{x, y} is equivalent to the system of equations for Gm ∈ ℛ{y} for m ≥ 0 1, when m = 0; { { { { { {0, when m = 1; Gm = { { 1 + ∑ Gk yk , when m = 2; { k≥2 { { { {(m − 1)Gm−2 + ∑k≥m Gk yk−m+2 , when m ≥ 3.

(8.4.7)

Proof. A direct result of Theorem 8.1.3 for a = 1. In order to solve the system of infinite equations with the infinite unknowns Gm for m ≥ 0, a new parameter s needs to be introduced such that all terms of Gm can be partitioned with each part containing only finite terms by s. Let 𝒥m be the set of all power vectors n = (n1 , n2 , n3 , . . .) such that Gm contains a term with yn . Denote 𝒥m,s = {n|π(n) = s} ⊆ 𝒥m , then 𝒥m = ∑ 𝒥m,s s≥0

(8.4.8)

where π(n) = i nT and i = (1, 2, 3, . . . , i, . . .). Lemma 8.4.4. For any integer s ≥ 0, we have 1, when s = 0; G0,s = { 0, otherwise.

(8.4.9)

Proof. Because G0 = 1 ∈ ℝ+ is a constant, G0 is independent of all s ≥ 1. This is that G0,s = 0, for any s ≥ 1. δ0,s .

By use of the Kronecker symbol δs,t = 1, s = t; 0, s ≠ t, this lemma shows G0,s =

Lemma 8.4.5. For any integer s ≥ 0, G1,s = 0.

8.4 Restrictions bridgeless surface

| 275

Proof. From the first equality of equation (8.4.7), the conclusion can be drawn. The two lemmas above enable us to discuss only Gm,s for m ≥ 2 without loss of generality. We proceed on the basis of equation (8.4.7). For integers m ≥ 2 and s ≥ 0, when m = 2; δs,0 + ∑sk=2 Gk,s−k yk , Gm,s = { s+m−2 (m − 1)Gm−2,s + ∑k=m Gk,s−k+m−2 yk−m+2 , when m ≥ 3.

(8.4.10)

Lemma 8.4.6. For any integer m ≥ 2, m! t

Gm,0 = { 2 t! 0,

,

when m = 2t; when m = 2t + 1,

(8.4.11)

where the integer t ≥ 1. Proof. See the proof of Lemma 8.1.6 for a = 1. This lemma tells us that Gm,0 ∈ ℛ+ {y} for m ≥ 0. Lemma 8.4.7. For any integer m ≥ 0, Gm,1 = 0. Proof. See the proof of Lemma 8.1.7 for a = 1. Lemma 8.4.6 and Lemma 8.4.7 enable us to discuss only s ≥ 2 without loss of generality for s ≥ 0. Lemma 8.4.8. Given two integers m ≥ 2 and s ≥ 2. If Gl,r ∈ ℛ+ {y} whenever l+r ≤ m+s−1 for any two integers l, r ≥ 0, then s+m−2

∑ Gk,s−k+m−2 yk−m+2 ∈ ℛ+ {y}.

k=m

Proof. On account of k + (s − k + m − 2) = m + s − 2 ≤ m + s − 1 for any integers k : m ≤ k ≤ m + s − 2. By the known condition, Gk,s−k+m−2 ∈ ℛ+ {y}. Therefore, the conclusion can be drawn by considering yk−m+2 ∈ ℛ+ {y}. Now, let us have a look at what happens to determine Gm,s in the order of increasing m + s one by one starting from m + s = 0 by (8.4.9). When m + s = 0, only G0,0 = 1 given by Lemma 8.4.4. When m + s = 1, only G1,0 = 0 and G0,1 = 0, given by, respectively, Lemma 8.4.5 and Lemma 8.4.4. When m + s = 2, in spite of Ga[2,0] = 1 and G0,1 = 0, only G1,1 has to be found. By Lemma 8.4.5, G1,1 = 0. When m+s = 3, in spite of G3,0 = 0 (by Lemma 8.4.5) and G0,3 = 0 (by Lemma 8.4.4), only G2,1 and G1,2 have to be found. From (8.4.10), G2,1 = 0. From Lemma 8.4.7, G1,2 = 0.

276 | 8 Surface equations second part Theorem 8.4.9. Equation (8.4.1) is well defined on ℛ{x, y}. Proof. A direct result of Theorem 8.1.9 for a = 1. In order to extract the solution of equation (8.4.3) partitioned by each part in the form of a finite sum with all terms positive as simple as possible, more investigation of its inner structures need to be performed. Lemma 8.4.10. For any integers m, s ≥ 0, if m + s = 1(mod 2), then Ga[m,s] = 0. Proof. See the proof of Lemma 8.2.1 for a = 1. This lemma enables us to reduce half amount of labor in evaluating the solution of equation (8.4.3). Lemma 8.4.11. Given two integers m, s ≥ 0. If m+s = 0(mod 2), then Gm,s is independent of y1 and yi for i ≥ s + 1. Proof. See the proof of Lemma 8.2.2 for a = 1. This lemma tells us that Gm,s is only a function of ys = (0, y2 , y3 , . . . , ys ). Lemma 8.4.12. Given integers m, s ≥ 0, if m + s = 0(mod 2), then Gm,s is a polynomial of ys with degree at most s. Proof. As a matter of fact, we are to prove that, for any n ∈ 𝒥m,s , |n| ≤ s. Let ℋ = {n | n ≥ 0, inT = s}, because of max{|n| | ∀n ∈ ℋ} = |s11 | = s and 𝒥m,s ⊆ ℋ, we find that Gm,s is a polynomial of ys with degree at most s. This lemma with the next implies that, for any integers m, s ≥ 0, Gm,s is of the form of a sum of finite terms all positive. Lemma 8.4.13. Given two integers m, s ≥ 0, Gm,s ∈ ℛ+ {y}. Proof. When m + s is small, the Ga[m,s] are checked for Ga[m,s] ∈ ℛ+ {y}. When m + s is greater, assume for any integers l, t ≥ 0 such that l + t ≤ m + s − 1, we have Gl,t ∈ ℛ+ {y}. By induction, we prove that Gm,s satisfies the conclusion. We proceed on the basis of (8.4.7). From Lemma 8.4.2, Lemma 8.4.4 and Lemma 8.4.8, the assumption shows that Gm,s satisfies the conclusion. The last three lemmas tell us that, for any two integers m, s ≥ 0, Gm,s is a polynomial of ys with degree at most s and Fm,s ∈ ℛ+ {y}. Theorem 8.4.14. Let g[bls] be the solution of equation (8.4.3). For any integers m, s ≥ 0, [bls] [bls] denote Gm,s = [g[bls] ]sy , then Gm,s has the form of a finite sum with all terms positive:

8.4 Restrictions bridgeless surface

[bls] Gm,s

δ0,s , { { { { { 0, { { { { { { { { { { m! = { 2t t! , { { { [bls] { { yk , ∑sk=2 Gk,s−k { { { { [bls] { { (m − 1)Gm−2,s { { { s+m−2 [bls] { + ∑k=m Gk,s−k+m−2 yk−m+2 ,

| 277

when s ≥ 0, m = 0; when m = 1, s ≥ 1; s = 1; m ≥ 1; or s ≥ 0, m ≠ s(mod 2); when s = 0 and m = 2t, t ≥ 1;

(8.4.12)

when s ≥ 2 (s = 0(mod 2)), m = 2; when m ≥ 3, s ≥ 2 (m = s(mod 2)).

[bls] Proof. From Theorem 8.4.9, Gm,s = Gm,s . When s ≥ 0, m = 0, by (8.4.9). When m = 1, s ≥ 1; s = 1, m ≥ 1; or s ≥ 0, m ≠ s(mod 2), by, respectively, Lemma 8.4.5; Lemma 8.4.7; or Lemma 8.4.8. When s = 0 and m = 2t, t ≥ 1, by Lemma 8.4.6. When s ≥ 2 (s = 0(mod 2)), m = 2 and when m ≥ 3, s ≥ 2, m = s(mod 2), by, respectively, the first and the second equalities of (8.4.10).

Let us have a look at what happens to Gm,s when 4 ≤ m + s ≤ 6 by employing (8.4.12). When m + s = 4, because of G4,0 = 24! 2 2! = 3, G3,1 = 0, G1,3 = 0 and G0,4 = 0, it is only necessary to evaluate G2,2 . For G2,2 , because of 2

G2,2 = ∑ Gk,2−k yk , k=2

(8.4.13)

we have G2,2 = y2 . When m + s = 5, because of m ≠ s(mod 2), Gm,s = 0. When m + s = 6, because of G6,0 = 26! 3 3! = 15, G5,1 = 0, G1,5 = 0 and G0,6 = 0, it is only necessary to evaluate G4,2 , G3,3 and G2,4 . For G2,4 , because of 4

G2,4 = ∑ Gk,4−k yk = (y2 )y2 + (3)y4 , k=2

we have G2,4 = y22 + 3y4 . For G3,3 , because of

3+1

G3,3 = 2G1,3 + ∑ Gk,3−k+1 yk−1 = (3)y3 , k=3

we have G3,3 = 3y3 . For G4,2 , because of

6−2

G4,2 = 3G2,2 + ∑ Gk,4−k yk−2 = 3(y2 ) + G4,0 y2 , k=4

we have G4,2 = (3 + 3)y2 = 6y2 .

(8.4.14)

(8.4.15)

(8.4.16)

278 | 8 Surface equations second part Example 8.4.1. Root-isomorphic classes of bridgeless maps with given root-vertex valency and vertex-partition vector on all orientable surfaces. By bridgeless of a map is meant without cut-edge in a map. A solution of equation (8.4.1), i. e., equation (8.1.1) under a = 1, is just the enufunction of such maps with root-vertex valency and vertexpartition vector as parameters. [bls] As shown in (8.4.12), the coefficient of the term yn (π(n) = s) in Gm,s is just the number of root-isomorphic classes of bridgeless maps on all orientable surfaces with root-vertex valency m and vertex-partition vector n where (m + s)/2 is the size of the map. [bls] [bls] are, respectively, known from Section 8.1 and and Gm,s For 0 ≤ m + s ≤ 6, Ga[m,s] Section 8.4. [bls] [bls] . When m + s = 0, i. e., size is 0, only G0,0 = 1 = G1[0,0] [bls] [bls] When m + s = 2, i. e., size is 1, only G2,0 = 1 = G1[2,0] .

[bls] [bls] [bls] [bls] and G2,2 = 3 = G1[4,0] When m + s = 4, i. e., size is 2 only G4,0 . = y2 = G1[2,2]

[bls] [bls] [bls] [bls] [bls] , G4,2 When m + s = 6, i. e., only G6,0 = 15 = G1[6,0] , G3,3 = 3y3 = = 6y2 = G1[4,2]

[bls] [bls] [bls] and G2,4 G1[3,3] . = y22 + 3y4 = G1[2,4] In Figure 8.4.1, root-isomorphic classes of bridgeless maps of size 2 and size 3 on all orientable surfaces are shown. For instance, [bls] 2a + b = 3 = G4,0 , [bls] G2,2 = y2

(obviously—it is not in the figure),

[bls] 2c + 3d + 4e + 3f + 3g = 15 = G6,0 , [bls] 4h + i = 4y2 + 2y2 = 6y2 = G4,2 , [bls] j + 2k = y3 + 2y3 = 3y3 = G3,3

and

[bls] l + 2m + n = y22 + 2y4 + y4 = y22 + 3y4 = G2,4 .

8.5 Notes 8.5.1. If y2 is missing from the solution of equation (8.4.1), then the result is a solution of equation (6.511) in Liu YP [61] (2009, pp. 168–173). It is just meaningful for bridgeless maps without vertex of valency 2 on all orientable surfaces. 8.5.2. We proceed on the basis of (8.1.8) and (8.4.7); an infinite system of linear equations can be established and their solutions are determined by powers of a matrix related to, respectively, a and y (acceptable as an example!), or only y. In each case, an explicit expression of the solutions can be directly obtained as in Section 8.3 or in Liu YP [29] (1985), or [31] (1986) for seeking the origin of such an approach.

8.5 Notes | 279

Figure 8.4.1: Root-isomorphic classes of bridgeless maps of size 2 and size 3 on all orientable surfaces.

Theorem 8.5.1. Let g = (Ga[2] , Ga[3] , Ga[4] , . . .) and da[bls] = (a0 , 0, 0, . . .); the following equation for g: gT = Ya[bls] gT + dTa[blg]

(8.5.1)

280 | 8 Surface equations second part where Ya[bls] = (ya[i,j] )i,j≥1 such that

ya[i,j]

a2 yj−i+2 , when j − i = k ≥ 0; { { { { { {0, when j − i = −1; ={ {(j + 2)a3 , when j − i = −2; { { { { when j − i ≤ −3, {0,

(8.5.2)

is equivalent to equations (8.1.8). Proof. We only need to employ the equivalent operation on ℛ{y}∞ . Because of the existence of the inverse of I − Ya[bls] , (I − Ya[bls] )−1 = ∑ Yka[bls] , k≥0

(8.5.3)

equation (8.5.1) is well defined on ℛ{y}∞ . [k] )i,j≥1 for k ≥ 0 where Theorem 8.5.2. Let Yk = (ya[i,j] [k] [k−1] = ∑ ya[i,l] ya[l,j] ya[i,j] l≥1

(8.5.4)

[1] = ya[i,j] for i, j ≥ 1. Then the solution ga[bls] of equation (8.5.1) has the in which ya[i,j] explicit expression

gTa[bls] = dTa[bls] + ∑(a0 y[l] a[∗,1] ) l≥1

(8.5.5)

j

is the first column of Ya[bls] for l ≥ 1. where y[l] a[∗,1] Proof. We need only consider operations on ℛ{y}∞ . Three things have to be mentioned. The first is that g = (Ga[2] , Ga[3] , . . .) instead of (G1 , G2 , . . .) is from Ga[1] = 0 for equation (8.1.1). Second, whenever ga[bls] is obtained by (8.5.5), the solution of equation (8.1.1) is ga[bls] = a0 +ga[bls] xT where x = (x2 , x3 , . . .). Third, whenever s = π(n) is given, g becomes gs = (G2 , G3 , . . . , Gs ), which only depends on ys . Hence, gs ∈ ℛ{ys }s−1 . 8.5.3. On the linearity of equation (8.1.1), the infinite coefficient matrix has its full upper triangle sub-matrix in common with that of equation (7.1.1). This suggests us to observe the common inner constructions of their corresponding combinatorial maps. 8.5.4. If parameter π(n) is given, the infinite summation in (8.5.5) becomes finite because of the limitation restricted in the s-dimensional system. This enables us to consider a variety of polynomials, such as order polynomials, size polynomials, face polynomials, genus polynomials and so forth.

8.5 Notes | 281

8.5.5. Consider a linear transformation from y to z for getting more simple explicisions of a solution of equation (8.1.1) or equation (8.4.1). 8.5.6. For a as a functional vector of x and/or y, we may wish to find some meaningful objects except only for maps in combinatorics because of the universality of partitions as classifications everywhere. 8.5.7. Equation (8.1.1) provides a theoretical base on meson equations of bridgeless surface type for extending a certain number of other equations of surface types to reflect the original ideas of tree-like meson equations for extending a certain number of outer types. Almost all of them have linearity. About these topics, one might like to refer to the more original statements such as in [31] (Liu, Y. P., 1986). 8.5.8. One might think of the extension from only orientable surfaces to all surfaces including both orientable and non-orientable for the bridgeless case. It seems reasonable in that the equation is of the same type with distinction only in coefficients. This idea reminds us of [65] (Liu Y. P., 2012; or [70], Vol. 20, 2015, pp. 9532–9539) for the equations themselves, or [61] (Liu, Y. P., 2009, pp. 139–142, 168–172, 202–205, 231–236) for the extraction of equations via decompositions on certain infinite sets of maps. 8.5.9. Asymptotics. Note that 8.5.4 enables us to investigate the corresponding asymptotics from known polynomials as shown in [56] (Liu, Y. P., 1999, pp. 359–389), [94] (Yan, J. Y., Liu, Y. P., 1991). 8.5.10. Stochastics. On the basis of (8.3.11), (8.4.12) or (8.5.5) for a = 1, the probability of order, co-order (or face number), semi-automorphic group order and genus from, respectively, order polynomials, co-order polynomials, semi-automorphic group order polynomials and genus polynomials of bridgeless maps on all orientable surfaces can be estimated for given size. 8.5.11. Distributions. Theorem 8.4.14 enables us to determine the distribution of order, co-order, semi-automorphic group order, or genus for bridgeless maps on all orientable surfaces via, respectively, order polynomials, co-order polynomials, semiautomorphic group order polynomials, or genus polynomials with given size. Then the moments including the mathematical expectation can be estimated. 8.5.12. The topic in this chapter is from Program 98 in [71] (Liu, Y. P., Vol. 22, 2016, pp. 10744–10745) as a primary part for systematization in theory.

9 Surface equations third part 9.1 Loopless surface model For the function f ∈ ℛ{x, y}, observe the equation 𝜕f |yi =yi 2 { { {a2 x ∑ ∫ y 𝜕y = f − a1 − a3 xf ∫ yf |x=y ; i≥1 y y { { { f | = a , x=0⇒y=0 0 {

(9.1.1)

where a = (a0 , a1 , a2 , a3 ) ∈ ℝ4+ and y = (y1 , y2 , y3 , . . .). This is equation (17) in Introduction. Because of the contribution from maps without self-loop on surfaces in the case of a0 = a1 = a2 = a3 = 1, which is found in [70] (Liu YP, 2015, p. 398), equation (9.1.1) is said to be in loopless surface model. In Liu YP [61] (2009, p. 204), one might also refer to the following meson equation for f ∈ ℛ{x, y}: { { {f = b + xf ∫ yf |x=y + x ∑ ∫ y(y i≥1 y y { { { {f |x=0⇒y=0 = a,

𝜕f |yi =yi 𝜕y

);

(9.1.2)

where y = (y1 , y2 , y3 , . . .) and a, b ∈ ℝ+ as two constant coefficients. This is also a specific case of equation (9.1.1) for a2 = a3 = 1, a = a0 and b = a1 in ℝ+ . Lemma 9.1.1. If a1 ≠ a0 , then equation (9.1.1) has no solution. Proof. Since f − a1 has a common factor x, the constant term of f − a1 is Fa[0] − a1 = 0 where Fa[0] is the constant term of f . This implies Fa[0] = a1 . However, the initial condition of equation (9.1.1) is Fa[0] = a0 . Thus, a0 = a1 , a contradiction to the given condition. The conclusion can be drawn. This lemma enables us to only consider equation (9.1.1) with a0 = a1 without loss of generality. Hence, a = (a0 , a2 , a3 ) instead of (a0 , a1 , a2 , a3 ). For convenience, equation (9.1.1) is equivalently transformed to { { {f = a0 + a3 xf ∫ yf |x=y + a2 x ∑ ∫ y(y i≥1 y y { { { {f |x=0⇒y=0 = a0 ∈ ℛ.

𝜕f |yi =yi 𝜕y

);

(9.1.3)

Because of f ∈ ℛ{x, y}, f is determined by Fa[m] = 𝜕xm f = [f ]m x ∈ ℛ{y} for integer m ≥ 0. Let αa = xf ∫ yf |x=y y

https://doi.org/10.1515/9783110627336-009

and βa = x ∑ ∫ y(y i≥1 y

𝜕f |yi =yi 𝜕y

),

(9.1.4)

284 | 9 Surface equations third part then because of αa , βa ∈ ℛ{x, y}, it is only necessary to determine Aa[m] = 𝜕xm αa for m ≥ 0 and Ba[m] = 𝜕xm βa for m ≥ 0. In order to evaluate them, certain relations among Fa[m] for m ≥ 0 have to be firstly investigated. On account of that, for any integer m ≥ 0, m

Aa[m] = [x ∫ yf |x=y ] = Fa[m−1] ∫ yf |x=y . x

y

y

The expansion of the meson functional leads to 0, when m = 0; Aa[m] = { Fa[m−1] ∑i≥1 Fa[i−1] yi , when m ≥ 1.

(9.1.5)

Because of 𝜕f |yi =yi 𝜕y

=

𝜕f |yi =yi dyi 𝜕f , = iyi−1 𝜕yi dy 𝜕yi

we have, by B0 = B1 = 0, for any integer m ≥ 2, Ba[m] = [∑ ∫ y(y

𝜕f |yi =yi 𝜕y

i≥1 y

m−1

)]

x

= ∑ iyi+1 i≥1

𝜕Fa[m−1] . 𝜕yi

(9.1.6)

Theorem 9.1.2. Equation (9.1.3) for f on ℛ{x, y} is equivalent to the system of equations Fa[m]

a0 , when m = 0; { { { = {a2 ∑i≥1 Fa[i−1] yi , when m = 1; { { 𝜕Fa[m−1] {a3 Fa[m−1] ∑i≥1 Fa[i−1] yi + a2 ∑i≥1 iyi+1 𝜕yi , when m ≥ 2,

(9.1.7)

for Fa[m] , m ≥ 0, on ℛ{y} for m ≥ 0. Proof. From the first equality of equation (9.1.1) and equation (9.1.4), f = a0 + a3 αa + a2 βa . Thus, for any integer m ≥ 0, Fa[m] = a0 δ0,m + a3 Am + a2 Bm . When m = 0, we have Aa[0] = Ba[0] = 0 and δ0,0 = 1, Fa[0] = a0 . This is the initiation of equation (9.1.1). When m = 1, from (9.1.5), we have Aa[1] = Fa[0] ∑ Fa[i−1] yi = a0 ∑ Fa[i−1] yi . i≥1

i≥1

From (9.1.6), we have Ba[1] = ∑ iyi+1 i≥1

𝜕Fa[0] = 0. 𝜕yi

Hence, the second equality of (9.1.7) is done. When m ≥ 2, by (9.1.5) and (9.1.6), the third equality of (9.1.7) is done. Therefore, the conclusion is proved.

9.1 Loopless surface model | 285

Because of the infinity of Fa[m] for m ≥ 1, a difficulty is involved in solving the system of equations (9.1.7). How should we introduce a new parameter s such that Fa[m,s] becomes a finite sum? This is the problem we have to address in the following. s For any integers m ≥ 0 and s ≥ 0, let Fa[m,s] = [f ]m x |π(y)=s = [Fa[m] ]y where s = π(y), T or π(n) = i n for n as the power vector of y. Because of Fa[m] ∈ ℛ{y}, Fa[m,s] ∈ ℛ{y : π(y) = s}. In order to clarify Fa[m,s] , Aa[m,s] and Ba[m,s] have to be investigated beforehand. As regards Aa[m,s] , from the first equality of (9.1.5), for any integer s ≥ 0, Aa[0,s] = 0. It is only necessary to consider m ≥ 1. From the second equality of (9.1.5), s

s s−i

y

i=1 j=0

Aa[m,s] = [Fa[m−1] ∑ Fa[i−1] yi ] = ∑ ∑ Fa[m−1,s−i−j] Fa[i−1,j] yi . i≥1

(9.1.8)

Lemma 9.1.3. For two integers m, s ≥ 0, if Fa[r,t] for r + t ≤ m + s − 1 and r, t ≥ 0 are known, then Aa[m,s] is determined. Proof. From (9.1.5) the result is easily seen. By reason of Ba[0] = Ba[1] = 0, for any integer s ≥ 0, Ba[0,s] = Ba[1,s] = 0. It is only necessary to consider m ≥ 2. From (9.1.6), for integer s ≥ 0, Ba[m,s] = [∑ iyi+1 i≥1

𝜕Fa[m−1] s s−1 𝜕Fa[m−1,s−1] ] = ∑ iyi+1 . 𝜕yi 𝜕yi y i=1

(9.1.9)

Lemma 9.1.4. For integers m, s ≥ 0, if Fa[r,t] for r + t ≤ m + s − 1 and r, t ≥ 0 are known, then Ba[m,s] is determined. Proof. From (9.1.6), the result is easily seen. We proceed on the basis of (9.1.8) and (9.1.9). For integer m ≥ 1, from Fa[m] = a3 Aa[m] + a2 Ba[m] , the system of equations (9.1.7) becomes, for two integers m, s ≥ 0,

Fa[m,s]

a0 δ0,s , { { { { { {a3 ∑si=1 Fi−1,s−i yi , ={ { a3 ∑si=1 ∑s−i { j=0 Fa[m−1,s−i−j] Fa[i−1,j] yi { { { 𝜕Fa[m−1,s−1] s−1 { + a2 ∑i=1 iyi+1 𝜕yi ,

when m = 0; when m = 1;

(9.1.10)

when m ≥ 2.

On this basis, we evaluate Fa[m,s] for m + s ≤ 3 and m, s ≥ 0. When m + s = 0, we have only m = s = 0. From the first equality of (9.1.10), Fa[0,0] = a0 . When m + s = 1, because of Fa[0,1] = 0, it is only necessary to evaluate Fa[1,0] . From the second equality of (9.1.10), Fa[1,0] = 0. When m + s = 2, only Fa[2,0] , Fa[1,1] and Fa[0.2] have to be evaluated. From the first equality of (9.1.10), Fa[0,2] = 0. From the third equality of (9.1.10), Fa[2,0] = 0. Only Fa[1,1] remains. From the second equality of (9.1.10), Fa[1,1] = a3 Fa[0,0] y1 = a0 a3 y1 .

286 | 9 Surface equations third part When m + s = 3, only Fa[3,0] , Fa[2,1] , Fa[1,2] and Fa[0,3] are available. From the first equality of (9.1.10), Fa[0,3] = 0. From the second equality of (9.1.10), Fa[1,2] = Fa[0,1] y1 + Fa[1,0] y2 = 0. From the third equality of (9.1.10), Fa[2,1] = Fa[1,0] Fa[0.0] y1 = 0 and Fa[3,0] = 0. Theorem 9.1.5. Equation (9.1.1) is well defined on ℛ{x, y} if, and only if, a0 = a1 . Proof. As for sufficiency, first, we find a solution of the system of equations (9.1.10), then that of equation (9.1.3), and hence equation (9.1.2), starting from a0 = a1 . By following the procedure for evaluating Fa[m,s] described above, all Fm,s are found when m + s ≤ 3. When m + s ≥ 4, assume that all Fr,t for r + t ≤ m + s − 1 and m, s ≥ o are already known. By induction, we prove that Fa[m,s] can be evaluated. When m = 0, by the first equality of equation (9.1.10). Because of s ≥ 4, Fa[0,s] = 0. When m = 1, we have (i − 1) + (s − i) = s − 1 ≤ s = (1 + s) − 1 = m + s − 1 for any integers 1 ≤ i ≤ 5. By the assumption, Fa[1,s] are determined by the second equality of equation (9.1.10). When m ≥ 2, we take into account that (m − 1) + (s − i − j) = m + s − i − j − 1 ≤ m + s − i − 1 ≤ m + s − 2 ≤ m + s − 1 and (i − 1) + j ≤ (i − 1) + s − i = s − 1 ≤ m + s − 1 for 1 ≤ i ≤ s and 0 ≤ j ≤ s − i. By the assumption, the first term in the third equality of (9.1.10) is determined. From (m − 1) + (s − 1) = m + s − 2 ≤ m + s − 1, the assumption shows that the second term in the third equality of (9.1.10) is determined. Thus, Fa[m,s] is determined. From the arbitrariness of m and s, a solution of the system of equations (9.1.7) is settled. By Theorem 9.1.2, a solution of equation (9.1.1) is done. By considering the uniqueness of the procedure for the given initiation, the solution is the only one. Necessity is a direct result of Lemma 9.1.1.

9.2 Solution loopless surface Now, let us have a look at how to simplify the procedure of evaluating each of Fa[m,s] in the form of a finite sum with all terms positive via its inner constructions explored further. Lemma 9.2.1. For any integer s ≥ 0, a0 , when s = 0; Fa[0,s] = { 0, when s ≥ 1.

(9.2.1)

Proof. This is the first equality of (9.1.10). This lemma enables us to only discuss m ≥ 1 without loss of generality where m ≥ 0.

9.2 Solution loopless surface

| 287

Lemma 9.2.2. For any integer m ≥ 0, a0 , Fa[m,0] = { 0,

when m = 0; when m ≥ 1.

(9.2.2)

Proof. From Lemma 9.2.1, it is only necessary to discuss m ≥ 1. When m = 1, by the second equality of (9.1.10), Fa[1,0] = 0. When m ≥ 2. Because of two sums being 0 in the third equality of (9.1.10), Fa[m,0] = 0. The last two lemmas enable us to only discuss m ≥ 1 and s ≥ 1 without loss of generality. Lemma 9.2.3. Given two integers m, s ≥ 0, Fa[m,s] = 0 whenever m + s = 1(mod 2). Proof. When m = 0, from Lemma 9.2.1, Fa[0,2t+1] = 0 for t ≥ 0. When s = 0, from Lemma 9.2.2, Fa[2t+1,0] = 0 for t ≥ 0. For m + s ≤ 3, the conclusion is checked to be true. For m + s ≥ 4 in general, assume Fa[r,t] = 0 for r + t = 1(mod 2), r + t ≤ m + s − 1 and r, t ≥ 0. By induction, we prove Fa[m,s] = 0 for m + s = 1(mod 2). When m = 1, we take into account of 1 + s = 1(mod 2) ⇒ s = 0(mod 2). Let s = 2l for l ≥ 1. We proceed on the basis of the second equality of (9.1.10). From (i − 1) + (s − i) = s − 1 = 1(mod 2) ⇒ s = 0(mod 2) and s − 1 ≤ (1 + s) − 1, the assumption leads to that, for any integers 1 ≤ i ≤ s, Fa[i−1,s−i] = 0 and hence Fa[1,s] = Fa[1,2l] = 0. When m ≥ 2. We proceed on the basis of the third equality of (9.1.10). For integers 0 ≤ j ≤ s − i and 1 ≤ i ≤ s, ((m − 1) + (s − i − j)) + ((i − 1) + j) = m + s − 2, m + s = 1(mod 2) ⇒ m + s − 2 = 1(mod 2) ⇒ if not (m − 1) + (s − i − j) = 1(mod 2) then (i − 1) + j = 1(mod 2). By reason of (m − 1) + (s − i − j) ≤ m + s − 1 and (i − 1) + j ≤ m + s − 1, the assumption leads to Fa[m−1,s−i−j] Fa[i−1,j] = 0. This shows that the first sum in the equality is 0. By (m − 1) + (s − 1) = m + s − 2, m + s = 1(mod 2) ⇒ (m − 1) + (s − 1) = 1(mod 2) and (m − 1) + (s − 1) ≤ m + s − 1, the assumption results in Fa[m−1,s−1] = 0. This shows that the second sum in the equality is 0. Therefore, Fa[m,s] = 0. This lemma enables us to reduce by half the amount of labor for evaluating the solution of equation (9.1.1). Lemma 9.2.4. For any two integers m, s ≥ 0, Fa[m,s] = 0 whenever s ≤ m − 1. Proof. When m = 1, Fa[1,0] = 0 and when m = 2, Fa[2,0] = Fa[2,1] = 0 follows. When m ≥ 3 in general, assume Fa[r,0] = Fa[r,1] = ⋅ ⋅ ⋅ = Fa[r,r−1] = 0 for 0 ≤ r ≤ m − 1. By induction, we prove Fa[m,0] = Fa[m,1] = ⋅ ⋅ ⋅ = Fa[m,m−1] = 0, i. e., Fa[m,s] = 0 for s ≤ m − 1. Because of m ≥ 3, by the third equality of (9.1.10), from m − s ≥ 1, (m − 1) − (s − i − j) = m − s + i + j − 1 ≥ i + j for 0 ≤ j ≤ s − i, 1 ≤ i ≤ s. By the assumption, Fa[m−1,s−i−j] = 0 for 0 ≤ j ≤ s − i, 1 ≤ i ≤ s. Therefore, s s−i

∑ ∑ Fm−1,s−i−j Fi−1,j yi = 0. i=1 j=0

288 | 9 Surface equations third part This is the first sum in the third equality of (9.1.10) being 0. By the assumption, Fa[m−1,s−1] = 0. This is the second sum in the third equality of (9.1.10) being 0. In consequence, Fa[m,s] = 0 for m − s ≥ 1. We proceed on the basis of Lemma 9.2.3; this lemma further reduces by another half the amount of labor in evaluating Fa[m,s] for m, s ≥ 0. Lemma 9.2.5. For two integers m, s ≥ 0, Fa[m,s] is independent of yl for l ≥ s + 1. Proof. When m + s ≤ 3, the conclusion is checked to be true. When m + s ≥ 4 in general, assume Fa[r,t] depends on at most yl for 2 ≤ l ≤ t whenever r + t ≤ m + s − 1 and r, t ≥ 0. By induction, we prove that Fa[m,s] depends on at most yl for 2 ≤ l ≤ s. When m = 0, it is trivial because of the initiation. When m = 1, by the second equality of (9.1.10), we have (i − 1) + (s − i) = s − 1 ≤ (1 + s) − 1 = s for 1 ≤ i ≤ s, all Fi−1,s−i depend on at most yl for 2 ≤ l ≤ s − 1. However, s

Fa[1,s] = a3 ∑ Fi−1,s−i yi i=1

involves ys . That implies Fa[1,s] is dependent on at most yl for 2 ≤ l ≤ s. When m ≥ 2, by employing the third equality of (9.1.10), because of (m − 1) + (s − i − j) = m + s − i − j − 1 ≤ m + s − 1 and (i − 1) + j ≤ (i − 1) + (s − i) = s − 1 ≤ m + s − 1, the assumption shows that in Fa[m,s] , the first sum s s−i

Σ1 = ∑ ∑ Fm−1,s−i−j Fi−1,j yi i=1 j=0

has each coefficient of yi dependent on at most yl for 2 ≤ l ≤ s − 1. Because of yi with i we have at most s, Σ1 depends on at most yl for 2 ≤ l ≤ s. Because of (m − 1) + (s − 1) = m + s − 2 ≤ m + s − 1, the assumption shows that Fa[m−1,s−1] depends on at most ys−1 , and further 𝜕Fm−1,s−1 𝜕yi

depends on at most

ys−1 { ys

when i ≠ s − 1; when i = s − 1.

Therefore, Fa[m,s] has its second sum s−1

Σ2 = ∑ iyi+1 i=1

𝜕Fm−1,s−1 𝜕yi

depending on at most ys . Because of Fa[m,s] = a3 Σ1 + a2 Σ2 , Fa[m,s] depends on at most yl for 2 ≤ l ≤ s. The conclusion can be drawn. This lemma tells us that, for any two integers m, s ≥ 0, Fa[m.s] is a function of at most s variables as ys = (y1 , y2 , . . . , ys ).

9.2 Solution loopless surface

| 289

Lemma 9.2.6. For two integers m, s ≥ 0, Fa[m,s] is a polynomial of ys with degree at most s. Proof. When m + s ≤ 3, the conclusion is checked that all Fa[m,s] are a polynomial of ys with degree at most s. When m + s ≥ 4 in general, assume all Fr,t are a polynomial of yt with degree at moat t for r +t ≤ m+s−1 and r, t ≥ 0. By induction, we prove that Fa[m,s] is a polynomial of ys with degree at most s. When m = 0, from the initiation it is trivial. When m = 1, by the second equality of (9.1.10), because of (i − 1) + (s − i) = s − 1 ≤ (1 + s) − 1 = s for 1 ≤ i ≤ s, the assumption shows that all Fa[i−1,s−i] are a polynomial of y with degree at most s − 1. However, for s

Fa[1,s] = a3 ∑ Fa[i−1,s−i] yi i=1

all coefficients of yi including ys are polynomial of ys−1 with degree at most s − 1. Thus, Fa[1,s] is a polynomial of ys with degree at most s. When m ≥ 2, by the third equality of (9.1.10), because of (m − 1) + (s − i − j) = m + s − i − j − 1 ≤ m + s − 1 and (i − 1) + j ≤ (i − 1) + (s − i) = s − 1 ≤ m + s − 1, the assumption as a result shows that Fa[m,s] has as its first sum s s−i

Σa[1] = ∑ ∑ Fa[m−1,s−i−j] Fa[i−1,j] yi i=1 j=0

with each coefficient of yi including ys as a polynomial of ys−1 in degree at most s − 1. Because of (m − 1) + (s − 1) = m + s − 2 ≤ m + s − 1, the assumption as a result shows that Fa[m−1,s−1] is a polynomial of ys−1 with degree at most s − 1, and then the second sum s−1

Σa[2] = ∑ iyi+1 i=1

𝜕Fm−1,s−1 𝜕yi

is a polynomial of ys with degree at most s. Thus, Fa[m,s] = a3 Σ1 +a2 Σa[2] is a polynomial of ys with degree at most s. This is the conclusion. This lemma with Lemma 9.2.5 shows that, for any two integers m, s ≥ 0, Fa[m.s] is a polynomial of s indeterminates with degree at most s. Lemma 9.2.7. For two integers m, s ≥ 0, Fa[m,s] ∈ ℛ+ {y} if, and only if, a0 = a1 , a2 , a3 ∈ ℝ+ . Proof. From the initiation of equation (9.1.1) and Lemma 9.1.1, a0 = a1 ∈ ℝ+ . This leads to the necessity. Conversely, by starting from a0 = a1 ∈ ℝ+ , we proceed on the basis of (9.1.10), and the conclusion is checked for m + s ≤ 3, i. e., Fa[m,s] ∈ ℛ+ {y}.

290 | 9 Surface equations third part For m + s ≥ 4 in general, assume all Fa[r,t] ∈ ℛ+ {y} for r + t ≤ m + s − 1 and r, t ≥ 0. By induction, we prove Fa[m,s] ∈ ℛ+ {y}. When m = 0, from the initiation, it is trivial. When m = 1, by the second equality of (9.1.10), because of (i − 1) + (s − i) = s − 1 ≤ (1 + s) − 1 = s for 1 ≤ i ≤ s, the assumption yields Fi−1,s−i ∈ ℛ+ {y}. Thus, from a3 ∈ ℝ+ , s

Fa[1,s] = a3 ∑ Fa[i−1,s−i] yi ∈ ℛ+ {y}. i=1

When m ≥ 2, by the third equality of (9.1.10), because of (m − 1) + (s − i − j) = m + s − i − j − 1 ≤ m + s − 1 and (i − 1) + j ≤ (i − 1) + (s − i) = s − 1 ≤ m + s − 1, the assumption shows that Fm,s has its first sum s s−i

Σa[1] = ∑ ∑ Fa[m−1,s−i−j] Fa[i−1,j] yi i=1 j=0

with each coefficient of yi in ℛ+ {y}. Thus, Σa[1] ∈ ℛ+ {y}. Because of (m − 1) + (s − 1) = m + s − 2 ≤ m + s − 1, the assumption shows that Fa[m−1,s−1] ∈ ℛ+ {y}, and further the second sum s−1

Σa[2] = ∑ iyi+1 i=1

𝜕Fa[m−1,s−1] ∈ ℛ+ {y}. 𝜕yi

Thus from a2 ∈ ℝ+ , Fa[m,s] = a3 Σa[1] + a2 Σa[2] ∈ ℛ+ {y}. This shows sufficiency. This lemma and Theorem 9.1.5 provide a theoretical basis for evaluating the solution of equation (9.1.1) in the form of a finite sum with all terms positive. Theorem 9.2.8. Let fa[lls] be the solution of equation (9.1.1). For any two integers [lls] [lls] has the form of a finite sum with all m, s ≥ 0, denote Fa[m,s] = 𝜕xm fa[lls] |π(n)=s , then Fa[m,s] terms positive when a ∈ ℝ3+ ,

[lls] Fa[m,s]

{ a0 , { { { { { 0, { { { { { { { { = {a ∑s F [lls] { 3 i=1 a[i−1,s−i] yi , { { { { [lls] [lls] { { a3 ∑si=1 ∑s−i { j=0 Fa[m−1,s−i−j] Fa[i−1,j] yi { { { { 𝜕F [lls] { , + a2 ∑s−1 iyi+1 a[m−1,s−1] i=1 𝜕yi {

when m = s = 0; when m = 0, s ≥ 1, m ≥ 1, s = 0, or m + s = 1(mod 2), or s ≤ m − 1; when m = 1, s = 2l(l ≥ 1);

otherwise,

where {otherwise} = {m ≥ 2, m = s(mod 2) and s ≥ m}.

(9.2.3)

Proof. We have m = s = 0, by (9.2.1). When m = 0, s ≥ 1; m ≥ 1, s = 0 or m+s = 1(mod 2) or s ≤ m − 1, by, respectively, (9.2.1); (9.2.2), or Lemma 9.2.3, or Lemma 9.2.4. When

9.2 Solution loopless surface

| 291

m = 1, s = 2l (l ≥ 1), by the second equality of (9.1.10). Otherwise, we have the third equality of (9.1.10). [lls] = We proceed on the basis of Theorem 9.1.5. For any integers m, s ≥ 0, Fa[m,s] Fa[m,s] . By employing Theorem 9.2.8, let us evaluate Fa[m,s] for 4 ≤ m + s ≤ 6. When m + s = 5, because of Fa[m,s] = 0, only m + s = 4 and m + s = 6 need to be considered. When m + s = 4, because of Fa[4,0] = Fa[3,1] = Fa[0,4] = 0, it is only necessary to consider Fa[2,2] and Fa[1,3] . For Fa[1,3] , from the third equality of (9.2.3), 3

Fa[1,3] = a3 ∑ Fa[i−1,3−i] = a3 Fa[1,1] y2 . i=1

By Fa[1,1] = a0 a3 y1 ,

Fa[1,3] = a0 a23 y1 y2 .

(9.2.4)

For Fa[2,2] , from the fourth equality of (9.2.3), 2 2−i

1

i=1 j=0

i=1

Fa[2,2] = a3 ∑ ∑ Fa[2−1,2−i−j] Fa[i−1,j] yi + a2 ∑ iyi+1

𝜕Fa[1,s−1] . 𝜕yi

On the first sum, by expanding for i, 1

∑ Fa[1,1−j] y1 = (Fa[1,1] + Fa[1,0] )y1 = Fa[1,1] y1 = a0 a3 y12 .

j=0

On the second sum, by i = 1 and y2

𝜕Fa[1,1] = a0 a3 y2 . 𝜕y1

Thus, Fa[2,2] = a3 (a0 a3 y12 ) + a2 (a0 a3 y2 ) = a0 a3 (a3 y12 + a2 y2 ).

(9.2.5)

When m + s = 6, from Fa[6,0] = Fa[5,1] = Fa[4,2] = Fa[0,6] = 0, we only need to consider Fa[3,3] , Fa[2,4] and Fa[1,5] . On Fa[1,5] , by the third equality of (9.2.3), 5

Fa[1,5] = a3 ∑ Fa[i−1,5−i] yi = a0 a23 (a3 y1 y22 + a3 y12 y3 + a2 y2 y3 ). i=1

On Fa[2,4] , by the fourth equality of (9.2.3), 4 4−i

3

i=1 j=0

i=1

Fa[2,4] = a3 ∑ ∑ Fa[1,4−i−j] Fa[i−1,j] yi + a2 ∑ iyi+1

𝜕Fa[1,4−1] . 𝜕yi

(9.2.6)

292 | 9 Surface equations third part For the first term over sum, denoted by Σa[1] , because both coefficients of y3 and y4 are 0, Σa[1] = Fa[1,3] Fa[0,0] y1 + Fa[1,1] Fa[1,1] y2 = 2(a0 a3 )2 y12 y2 . For the second term over sum, denoted by Σa[2] , because no y3 appears in F1,3 , Σa[2] = y2

𝜕Fa[1,3] 𝜕F1,3 + 2y3 = a0 a23 (2y1 y3 + y22 ). 𝜕y1 𝜕y2

Thus, Fa[2,4] = a3 Σa[1] + a2 Σa[2] = a0 a23 (2a0 a3 y12 y2 + 2a2 y1 y3 + a2 y22 ).

(9.2.7)

On Fa[3,3] , by the fourth equality of (9.2.3), 3 3−i

2

i=1 j=0

i=1

Fa[3,3] = a3 ∑ ∑ Fa[2,s−i−j] Fa[i−1,j] yi + a2 ∑ iyi+1

𝜕Fa[2,2] . 𝜕yi

For the first term over sum, denoted by Λa[1] , because the coefficient of y3 is 0, 2

1

j=0

j=0

Λa[1] = ∑ Fa[2,2−j] Fa[0,j] y1 + ∑ F2,1−j F1,j y2 = a20 a3 (a3 y13 + a2 y1 y2 ). On the second term over sum, denoted by Λa[2] , because of Fa[2,2] = a1 a3 (a3 y12 + a2 y2 ), Λa[2] = y2

𝜕Fa[2,2] 𝜕Fa[2,2] + 2y3 = 2a0 a3 (a3 y1 y2 + a2 y3 ). 𝜕y1 𝜕y2

Therefore, F3,3 = a3 Λa[1] + a2 Λa[2] = a20 a23 y13 + a0 a2 a23 (2 + a0 )y1 y2 + 2a0 a22 a3 y3 .

(9.2.8)

9.3 Explicitness loopless surface Let ℳ[lls] m,n be the set of all root-isomorphic classes of loopless maps with root-vertex valency m ≥ 0 and vertex-partition vector n ≥ 0. For any integer s ≥ 0, denote by [lls] [lls] ℳ[lls] m,s the union of ℳm,n over all π(n) = s. Because the enufunction of ℳm,n satisfies equation (9.1.1) for a0 = a1 = a2 = a3 = 1, from Theorem 9.1.5, its solution fa[lls] ∈ [lls] = 𝜕xm fa[lls] |π(n)=s ∈ ℛ{y} for m ≥ 0. ℛ{x, y} is determined by Fa[m,s] Observation 9.3.1. For any integers m, s ≥ 0, 󵄨 [lls] 󵄨 [lls] 󵄨󵄨 F1[m,s] 󵄨󵄨y=1 = 󵄨󵄨󵄨ℳm,s 󵄨󵄨󵄨 where a = (a0 , a2 , a3 ) and 1 = (1, 1, 1), or (1, 1, 1, . . .) according as a, or y.

(9.3.1)

9.3 Explicitness loopless surface

| 293

Proof. This is shown from the specific case of equation (9.1.1) when a = 1. [lls] [lls] Let 𝒢m,s and 𝒢m,n be the sets of all loopless graphs of root-vertex valency m ≥ 0 with, respectively, s = π(n) for all n ≥ 0 and the vertex-partition vector n ≥ 0. [lls] Observation 9.3.2. Let 𝒢 (ℳ[lls] m,s ) and 𝒢 (ℳm,n ) be the sets of all underline graphs of [lls] [lls] maps in, respectively, ℳm,s and ℳm,n . Then [lls]

[lls]

𝒢m,n = 𝒢 (ℳm,n )

[lls] and hence 𝒢m,s = 𝒢 (ℳ[lls] m,s ).

(9.3.2)

Proof. By showing a bijection between two sides of an equality. [lls] [lls] Let Fm,s |y=1 or Fm,n be the number of distinct root-isomorphic classes of loopless maps on all orientable surfaces of root-vertex valency m with, respectively, s = π(n) for all n ≥ 0 or the vertex-partition vector n given. [lls] Lemma 9.3.3. Given G ∈ 𝒢m,n , denote by μG the set of topologically non-equivalent embeddings of G. Then

|μG | = (m − 1)! ∏(i − 1)!ni = (m − 1)!(i − 1)!n i≥1

(9.3.3)

where i = (1, 2, 3, . . .) and 1 = (1, 1, 1, . . .). Proof. One proves this by the existence of a bijection between an embedding and a vertex-rotation system of a graph. [lls] [lls] Let 𝒢m,n = 𝒢 (ℳ[lls] m,n ) be the set of all underline graphs of map M ∈ ℳm,n and [lls] [lls] hence ℳm,n = ℳ(𝒢m,n ). [lls] Denote by t = aut[sm] (G) the order of the semi-automorphic group of G ∈ 𝒢m,n . Let

ℐm,n = {t | ∃G ∈ 𝒢m,n , t = aut[sm] (G)}. [lls]

[lls]

(9.3.4)

[lls](t) [lls] Denote by 𝒢m,n the subset of 𝒢m,n in which every graph has t as the order of its semi-automorphic group.

Lemma 9.3.4. For integer m ≥ 0 and integral vector n ≥ n, n [lls] = ∑ Λ[lls](t) F1[m,n] 1[m,n] y ,

(9.3.5)

[lls] t∈ℐm,n

where Λ[lls](t) 1[m,n] =

m + π(n) 󵄨󵄨 n [lls](t) 󵄨 󵄨󵄨μ(𝒢m,n )󵄨󵄨󵄨(m − 1)!(i − 1)! t

(9.3.6)

with π(n)(= inT ), i = (1, 2, 3, . . .) and n = (n1 , n2 , n3 , . . .). Proof. See Liu YP [59] (2003, Theorem 4.1, p. 211) and particularly Mao LF-Liu YP [75].

294 | 9 Surface equations third part [lls] [lls] have the following explicit expression: and thus F1[m,n] By (9.2.3), F1[m,s] [lls] = F1[m,s]

[lls] yn = ∑ F1[m,n]

π(n)=s [lls] n∈𝒥m,s

n ∑ ( ∑ Λ[lls](t) 1[m,n] )y ,

[lls] [lls] t∈ℐm,n n∈𝒥m,s

(9.3.7)

is given by (9.3.6). where Λ[lls](t) 1[m,n] Lemma 9.3.5. For an integer m ≥ 0 and an integral vector n ≥ 0, [lls] n Fa[m,n] = ∑ Λ[lls](t) a[m,n] y

(9.3.8)

[lls] t∈ℐm,n

∈ ℛ is a polynomial of a with degree at most π(n) + 1, as extracted from where Λ[lls](t) a[m,n] the procedure in the proof of Theorem 9.2.8 such that m + π(n) 󵄨󵄨 󵄨󵄨 n [lls](t) 󵄨 Λ[lls](t) 󵄨󵄨μ(𝒢m,n )󵄨󵄨󵄨(m − 1)!(i − 1)! , a[m,n] 󵄨󵄨a=1 = t

(9.3.9)

which is just given by (9.3.6). Proof. This is a result of Observation 9.3.1 and Observation 9.3.2. Now, we are allowed to illustrate an explicit solution of equation (7.1.1) via its specific case of a = 1 in the combinatorial sense. [lls] Theorem 9.3.6. Let fa[lls] be the solution of equation (7.1.1) determined by Fa[m,s] for in-

[lls] has the explicit expression tegers m, s ≥ 0. Then Fa[m,s] [lls] = Fa[m,s]

n ∑ ( ∑ Λ[lls](t) a[m,n] )y

(9.3.10)

[lls] [lls] t∈ℐm,n n∈𝒥m,s

is given in (9.3.8). where Λ[lls](t) a[m,n] Proof. We proceed on the basis of Lemma 9.3.5. This theorem enables us to directly deduce the explicit expression of the number of root-isomorphic classes of loopless maps with root-vertex valency m and vertexpartition vector n on all orientable surfaces. Corollary 9.3.7. Given the integer m ≥ 0 and integral vector n ≥ 0, the solution f[lls] of equation (9.1.1) under a = 1 is determined by m,n 𝜕x,y f[lls] = ∑ Λ[lls](t) 1[m,n] [lls] t∈ℐm,s

where Λ[lls](t) is given by (9.3.6). 1[m,n] Proof. This is a direct result of Theorem 9.3.6.

(9.3.11)

9.4 Restrictions loopless surface

| 295

9.4 Restrictions loopless surface Consider the equation for f ∈ ℛ{x, y} 𝜕f |yi =yi 2 { { {x ∑ ∫ y 𝜕y = f − 1 − xf ∫ yf |x=y ; i≥1 y y { { { f | = 1, { x=0⇒y=0

(9.4.1)

where y = (y1 , y2 , y3 , . . .). This is the equation obtained from equation (9.1.1) in the case of a0 = a1 = a2 = a3 = 1, which is found in [70] (Liu YP, 2015, p. 398). In Liu YP [61] (2009, p. 204), one might find the meson equation (9.1.2) for f ∈ ℛ{x, y} with two constant coefficients a = a0 and b = a1 . It is the specific case of equation (9.1.3) in a2 = a3 = 1. Also, its specific case with a = b = 1 is { { {f = 1 + xf ∫ yf |x=y + x ∑ ∫ y(y i≥1 y y { { { {f |x=0⇒y=0 = 1.

𝜕f |yi =yi 𝜕y

);

(9.4.2)

Lemma 9.4.1. Equation (9.4.2) is equivalent to equation (9.4.1) on ℛ{x, y}. Proof. By the cancelation law, one can be seen from the other on ℛ{x, y}. This lemma enables us to only consider equation (9.4.2) instead of equation (9.4.1) without loss of generality. Because of f ∈ ℛ{x, y}, f is determined by Fm = 𝜕xm f = [f ]m x ∈ ℛ{y} for integer m ≥ 0. Let and β = x ∑ ∫ y(y

α = xf ∫ yf |x=y

i≥1 y

y

𝜕f |yi =yi 𝜕y

),

(9.4.3)

then, because of α, β ∈ ℛ{x, y}, it is only necessary to determine Am = 𝜕xm α for m ≥ 0 and Bm = 𝜕xm β for m ≥ 0. In order to evaluate them, certain relations among Fm for m ≥ 0 have to be investigated first. On account of that, for any integer m ≥ 0, m

Am = [x ∫ yf |x=y ] = Fm−1 ∫ yf |x=y . y

x

y

The expansion of the meson functional leads to 0, when m = 0; Am = { Fm−1 ∑i≥1 Fi−1 yi , when m ≥ 1.

(9.4.4)

296 | 9 Surface equations third part Because of 𝜕f |yi =yi 𝜕y

=

𝜕f |yi =yi dyi 𝜕f , = iyi−1 𝜕yi dy 𝜕yi

we have, by B0 = B1 = 0, for any integer m ≥ 2, Bm = [∑ ∫ y(y

𝜕f |yi =yi 𝜕y

i≥1 y

m−1

)]

x

= ∑ iyi+1 i≥1

𝜕Fm−1 . 𝜕yi

(9.4.5)

Theorem 9.4.2. Equation (9.4.2) for f on ℛ{x, y} is equivalent to the system of equations 1, { { { Fa[m] = {∑i≥1 Fa[i−1] yi , { { 𝜕Fm−1 {Fm−1 ∑i≥1 Fi−1 yi + ∑i≥1 iyi+1 𝜕yi ,

when m = 0; when m = 1;

(9.4.6)

when m ≥ 2,

for Fm on ℛ{y} for m ≥ 0. Proof. See the proof of Theorem 9.1.2 for a = 1. Because of the infinity of Fm for m ≥ 1, difficulties are involved with solving the system of equations (9.4.6). How should we introduce a new parameter s such that Fm,s becomes a finite sum? This is the problem we have to address in what follows. s For any integers m ≥ 0 and s ≥ 0, let Fm,s = [f ]m x |π(y)=s = [Fm ]y where s = π(y), or π(n) = i nT for n as the power vector of y. We have Fm ∈ ℛ{y}, Fm,s ∈ ℛ{y : π(y) = s}. In order to clarify Fm,s , Am,s and Bm,s have to be investigated beforehand. On Am,s , from the first equality of (9.4.4), for any integer s ≥ 0, A0,s = 0, it is only necessary to consider m ≥ 1. From the second equality of (9.4.4), s s−i

Am,s = ∑ ∑ Fm−1,s−i−j Fi−1,j yi . i=1 j=0

(9.4.7)

Lemma 9.4.3. For two integers m, s ≥ 0, if Fr,t for r + t ≤ m + s − 1 and r, t ≥ 0 are known, then Am,s is determined. Proof. From (9.4.4), it is easily seen. On Bm,s , by reason of B0 = B1 = 0, for any integer s ≥ 0, B0,s = B1,s = 0. It is only necessary to consider m ≥ 2. On (9.4.5), for integer s ≥ 0, s−1

Bm,s = ∑ iyi+1 i=1

𝜕Fm−1,s−1 . 𝜕yi

(9.4.8)

Lemma 9.4.4. For integers m, s ≥ 0, if Fr,t for r + t ≤ m + s − 1 and r, t ≥ 0 are known, then Bm,s is determined.

9.4 Restrictions loopless surface

| 297

Proof. From (9.4.5), it is easily seen. We proceed on the basis of (9.4.7) and (9.4.8). For integer m ≥ 1, from Fm = Am +Bm , the system of equations (9.4.6) becomes, for two integers m, s ≥ 0,

Fm,s

δ0,s , when m = 0; { { { s = {∑i=1 Fi−1,s−i yi , when m = 1; { { s 𝜕Fm−1,s−1 s−1 s−i {∑i=1 ∑j=0 Fm−1,s−i−j Fi−1,j yi + ∑i=1 iyi+1 𝜕yi , when m ≥ 2.

(9.4.9)

On this basis, we evaluate Fm,s for m + s ≤ 3 and m, s ≥ 0. When m + s = 0, we have only m = s = 0, from the first equality of (9.4.9), F0,0 = 1. When m + s = 1, because of F0,1 = 0, it is only necessary to evaluate F1,0 . From the second equality of (9.4.9), F1,0 = 0. When m + s = 2, only F2,0 , F1,1 and F0.2 have to be evaluated. From the first equality of (9.4.9), F0,2 = 0. From the third equality of (9.4.9), F2,0 = 0. Only F1,1 remains. From the second equality of (9.4.9), F1,1 = F0,0 y1 = y1 . When m + s = 3, only F3,0 , F2,1 , F1,2 and F0,3 are available. From the first equality of (9.4.9), F0,3 = 0. From the second equality of (9.4.9), F1,2 = F0,1 y1 + F1,0 y2 = 0. From the third equality of (9.4.9), F2,1 = F1,0 F0.0 y1 = 0 and F3,0 = 0. Theorem 9.4.5. Equation (9.4.1) is well defined on ℛ{x, y}. Proof. This is a direct result of Theorem 9.1.5 when a = 1. Now, let us have a look at how to simplify the procedure of evaluating each of Fm,s in the form of a finite sum with all terms positive via its inner constructions explored further. Lemma 9.4.6. For any integer s ≥ 0, 1, when s = 0; F0,s = { 0, when s ≥ 1.

(9.4.10)

Proof. This is the first equality of (9.4.9). This lemma enables us to only discuss m ≥ 1 without loss of generality for m ≥ 0. Lemma 9.4.7. For any integer m ≥ 0, 1, when m = 0; Fm,0 = { 0, when m ≥ 1.

(9.4.11)

Proof. This is a direct result of Lemma 9.2.2 for a = 1. The last two lemmas enable us to only discuss m ≥ 1 and s ≥ 1 without loss of generality.

298 | 9 Surface equations third part Lemma 9.4.8. Given two integers m, s ≥ 0, Fm,s = 0 whenever m + s = 1(mod 2). Proof. See the proof of Lemma 9.2.3 for a = 1. This lemma enables us to reduce by half the amount of labor for evaluating the solution of equation (9.4.1). Lemma 9.4.9. For any two integers m, s ≥ 0, Fm,s = 0 whenever s ≤ m − 1. Proof. See the proof of Lemma 9.2.4 for a = 1. We proceed on the basis of Lemma 9.4.8; this lemma further reduces by another half the amount of labor in evaluating all Fm,s for m, s ≥ 0. Lemma 9.4.10. For two integers m, s ≥ 0, Fm,s is independent of yl for l ≥ s + 1. Proof. See the proof of Lemma 9.2.5 for a = 1. This lemma tells us that, for any two integers m, s ≥ 0, Fm.s is a function of at most s variables, ys = (y1 , y2 , . . . , ys ). Lemma 9.4.11. For two integers m, s ≥ 0, Fm,s is a polynomial of ys with degree at most s. Proof. See the proof of Lemma 9.2.6 for a = 1. This lemma with Lemma 9.4.10 shows that, for any two integers m, s ≥ 0, Fm.s is a polynomial of s indeterminates with degree at most s. Lemma 9.4.12. For two integers m, s ≥ 0, Fm,s ∈ ℛ+ {y}. Proof. See the proof of Lemma 9.2.7 for a = 1. This lemma and Theorem 9.4.5 provide a theoretical basis for evaluating the solution of equation (9.4.1) in the form of a finite sum with all terms positive. Theorem 9.4.13. Let f[lls] be the solution of equation (9.4.1). For any two integers m, s ≥ [lls] [lls] has the form of a finite sum with all terms 0, denote Fm,s = 𝜕xm f[lls] |π(n)=s , then Fm,s positive,

[lls] Fm,s

{1, { { { { 0, { { { { { { { { = {∑s F [lls] y , { i=1 i−1,s−i i { { { { [lls] { { yi Fi−1,j ∑s ∑s−i F [lls] { { { i=1 j=0 m−1,s−i−j { [lls] { 𝜕F s−1 m−1,s−1 { + ∑i=1 iyi+1 𝜕y , i

when m = s = 0; when m = 0, s ≥ 1, m ≥ 1, s = 0, or m + s = 1(mod 2), or s ≤ m − 1; when m = 1, s = 2l − 1 (l ≥ 1);

otherwise,

where {otherwise} = {m ≥ 2, m = s(mod 2) and s ≥ m}.

(9.4.12)

9.4 Restrictions loopless surface

| 299

Proof. We have m = s = 0, by (9.4.10). When m = 0, s ≥ 1; m ≥ 1, s = 0 or m + s = 1(mod 2), or s ≤ m−1, by, respectively, (9.4.10); (9.4.11), or Lemma 9.4.8, or Lemma 9.4.9. When m = 1, s = 2l − 1 (l ≥ 1), by the second equation of equations (9.4.9). Otherwise, we have the third equation of equations (9.4.9). [lls] We proceed on the basis of Theorem 9.4.5. For any integers m, s ≥ 0, Fm,s = Fm,s . By employing Theorem 9.4.13, let us continue to evaluate Fm,s for 4 ≤ m + s ≤ 6. Apart from m + s = 5, because of Fm,s = 0, only m + s = 4 and m + s = 6 need to be considered. When m + s = 4, because of F4,0 = F3,1 = F0,4 = 0, it is only necessary to consider F2,2 and F1,3 . For F1,3 , from the third equality of (9.4.12), 3

F1,3 = ∑ Fi−1,3−i = F1,1 y2 . i=1

By F1,1 = y1 , F1,3 = y1 y2 .

(9.4.13)

For F2,2 , from the fourth equality of (9.4.12), 2 2−i

1

i=1 j=0

i=1

F2,2 = ∑ ∑ F2−1,2−i−j Fi−1,j yi + ∑ iyi+1

𝜕F1,s−1 . 𝜕yi

On the first sum, by expanding for i, 1

∑ F1,1−j y1 = (F1,1 + F1,0 )y1 = F1,1 y1 = y12 .

j=0

On the second sum, by i = 1 we have y2

𝜕F1,1 = y2 . 𝜕y1

Thus, F2,2 = (y12 ) + (y2 ) = y12 + y2 .

(9.4.14)

When m + s = 6, from F6,0 = F5,1 = F4,2 = F0,6 = 0, we only need to consider F3,3 , F2,4 and F1,5 . On F1,5 , by the third equality of (9.4.12), 5

F1,5 = ∑ Fi−1,5−i yi = y1 y22 + y12 y3 + y2 y3 . i=1

(9.4.15)

300 | 9 Surface equations third part On F2,4 , by the fourth equality of (9.4.12), 4 4−i

3

i=1 j=0

i=1

F2,4 = ∑ ∑ F1,4−i−j Fi−1,j yi + ∑ iyi+1

𝜕F1,4−1 . 𝜕yi

For the first term over sum, denoted by Σ1 , because both coefficients of y3 and y4 are 0, Σ1 = F1,3 F0,0 y1 + F1,1 F1,1 y2 = 2y12 y2 . For the second term over sum, denoted by Σ2 , because no y3 appears in F1,3 , Σ2 = y2

𝜕F1,3 𝜕F1,3 + 2y3 = 2y1 y3 + y22 . 𝜕y1 𝜕y2

Thus, F2,4 = Σa[1] + Σa[2] = 2y12 y2 + 2y1 y3 + y22 .

(9.4.16)

On F3,3 , by the fourth equality of (9.4.12), 3 3−i

2

i=1 j=0

i=1

F3,3 = ∑ ∑ F2,s−i−j Fi−1,j yi + ∑ iyi+1

𝜕F2,2 . 𝜕yi

For the first term over sum, denoted by Λ1 , because the coefficient of y3 is 0, 2

1

Λ1 = ∑ F2,2−j F0,j y1 + ∑ F2,1−j F1,j y2 j=0

= F2,2 F0,0 y =

(y13

j=0

+ y1 y2 ).

On the second term over sum, denoted by Λ2 , because F2,2 = y12 + y2 , Λ2 = y2

𝜕F2,2 𝜕F2,2 + 2y3 = 2(y1 y2 + y3 ). 𝜕y1 𝜕y2

Therefore, F3,3 = Λ1 + Λ2 = y13 + 3y1 y2 + 2y3 .

(9.4.17)

Example 9.4.1. Root-isomorphic classes of loopless maps on all orientable surfaces by root-vertex valency and vertex-partition vector as parameters. When we consider equation (9.1.1) under a = (1, 1, 1, 1), then it becomes equation (9.4.1), whose solution provides the number of root-isomorphic classes of loopless maps on all orientable surfaces for root-vertex valency and vertex-partition vector arbitrarily given. [lls] Let f[lls] ∈ ℛ{x, y} be the solution of equation (9.4.1) and Fm,s = 𝜕xm f[lls] |π(n)=s . The m n coefficient of x y for π(n) = s is the number of root-isomorphic classes of loopless

9.5 Notes | 301

maps on all orientable surfaces by root-vertex valency m and vertex-partition vector n. Equation (9.4.1) is extracted from the decomposition principle provided in Liu YP [62] [lls] [lls] [lls] [lls] |a=1 . F1,3 = y1 = Fa[1,1] (2009, pp. 202—203). For instance, F1,1 |a=1 , = y1 y2 = Fa[1,3]

[lls] [lls] [lls] [lls] [lls] |a=1 , F2,4 = y1 y22 + y12 y3 + y2 y3 = Fa[1,5] = 2y1 y3 + y22 + |a=1 . F1,5 = y12 + y2 = Fa[2,2] F2,2 [lls] [lls] [lls] 2a3 y12 y2 = Fa[2,4] |a=1 , F3,3 = y13 + 3y1 y2 + 2y3 = Fa[3,3] |a=1 .

[lls] [lls] In Figure 9.4.1, one can see the following: F1,3 = y1 y2 = a, F2,2 = y12 (= a) + y2 (= b).

[lls] [lls] [lls] = d + c + e = y1 y22 + y12 y3 + y2 y3 , F2,4 = 2d + 2e + f = 2y12 y2 + 2y1 y3 + y22 , F3,3 = F1,5 3e + (g + h) = 3y1 y2 + 2y3 .

Figure 9.4.1: Root-isomorphic classes of loopless maps on orientable surfaces.

9.5 Notes 9.5.1. We proceed on the basis of (9.1.7). Because of the occurrence of a term with degree 2 for f , equation (9.1.1) cannot be transformed into a linear system of equations. Is it possible for us to seek a new type of decomposing the set of loopless maps so that a linear meson equation can be established accompanied by some known relations and/or functions? 9.5.2. On the linearization of nonlinear meson equation, it might be possible by the realization of the nonlinear part via a linear function or a known function extracted

302 | 9 Surface equations third part from some other equation. For equation (9.1.1), one might think of root-vertex separable maps as shown in [25] (Liu, Y. P., 1982) on surfaces instead of on the sphere (i. e., the plane). 9.5.3. Because of equation (9.1.1) accompanied by loopless maps on surfaces, this topic has its origin in a variety of investigations on a variety of loopless maps. The earliest involvement of the author was with enumeration of planar loopless maps with one or two parameters in [27] (Liu, Y. P., 1983), and in [42] (Liu, Y. P., 1987), [46] (Liu, Y. P., 1989), [49] (Liu, Y. P., 1990) etc. all with the vertex-partition vector as an infinite parameter on the sphere via a meson functional, particularly in [61] (Liu, Y. P., 2009, pp. 202–205) on all surfaces (orientable and/or non-orientable). However, only [69] (2015, pp. 398–409) addresses the meson functional under an infinite summation. 9.5.4. Two topics are considered. The first is to establish differential equations satisfied by the enufunction of loopless maps on all orientable surfaces with two parameters: the semi-rooted size and the semi-nonrooted size via a procedure as shown in [61] (Liu, Y. P., 2009, pp. 202–205). The second is to establish a linear meson functional equation by considering the vertex-partition vector as in the following. Let f = (Fa[1] , Fa[2] , Fa[3] , . . .), then f = Fa[0] + fxT occurs in equation (9.1.3) as the unknown. For m ≥ 1, because of Fa[0] = a0 , denote a0 a2 y1 , when m = 1; ψa[m] = { [lls] a0 a3 Fa[m−1] y1 + a2 λa[m−1] , when m ≥ 2,

(9.5.1)

where λa[m−1] = ∑ iyi+1 i≥1

𝜕Fa[m−1] . 𝜕yi

(9.5.2)

Theorem 9.5.1. Equation (9.1.3) and hence equation (9.1.1) are equivalent to the infinite system of equations for f fT = Ta fT + dTa

(9.5.3)

a2 (y2 , y3 , y4 , . . .), when i = 1; ti,⋅ = { [lls] (y2 , y3 , y4 , . . .), when i ≥ 2, a3 Fa[i−1]

(9.5.4)

da = (ψa[1] , ψa[2] , ψa[3] , . . .).

(9.5.5)

where the ith row vector of Ta is

for i ≥ 1 and

Proof. This is a direct result of Theorem 9.1.2.

9.5 Notes | 303

Although the well-definedness of equations (9.5.3) can be formally done, the infinity shows that the evaluation of a solution is not executable. This stimulates us to consider the parameter s = π(n) as shown in Section 9.2. Theorem 9.5.2. If integer s ≥ 1 is given, then equation (9.5.3) is equivalent to the following finite system of equations: fTs = Ta[s×s] fTs + dTa[s] .

(9.5.6)

Proof. A result of Lemmas 9.2.4–9.2.6. This theorem enables us to evaluate the solution by the procedure of taking powers of the matrix Ya[s×s] in the increasing order of s starting from s = 0, i. e., the initiation Fa[0] = a0 . This is just the procedure shown by (9.2.3). The approach presented in this note seems to be available to solve all nonlinear meson functional equations encountered so far. 9.5.5. The decomposition of loopless maps on all orientable surfaces or all surfaces including both orientable and non-orientable referred to in [61] (Liu, Y. P., 2009, pp. 201–203) might be helpful for us to discover other types of decomposition for such maps so that a linear meson functional equation is extracted to produce an explicision of the solution directly. 9.5.6. Specification. Only two types are mentioned as specific cases of equation (9.1.1). One is related to a variety of restrictions on the powers of yi , i ≥ 2. For example, the case d = ni for i ≥ 2, is called near d-regular, or y = yi (i ≥ 2) etc. When d = 3 or 4 we have, respectively, near triangulation, or near quadrangulation. This is referred to in Tutte WT [88] (1962), [77] (Mullin, R. C., Nemeth, E., Schellenberg, P. J., 1970), [53] (Liu, Y. P., 1992), [14] (Dong, F. M., Liu YP., 1993), [7] (Cai, J. L., Liu, X. Y., et al., 2006), etc. The other is on surfaces of given small genus. This is referred to in [4] (Brown, W. G., 1965), [9] (Cai, J. L., Liu Y. P., 2000), [81] (Ren, H., Liu, Y. P., 1999), [83] (Ren, H., Liu, Y. P., 2000), [84] (Ren, H., Liu, Y. P., 2001), [21] (Li, Z. X., Liu, Y. P., 2002), [79] (Ren, H., Deng, M., et al., 2005), [93] (Xu, Y., et al., 2007), [63], [64] (Liu, Y. P., 2012), [78] (Pan, L. Y., Liu, Y. P., 2013) etc. 9.5.7. Asymptotic estimation. Theorem 9.4.13 suggests us to investigate the corresponding asymptotics as shown in [56] (Liu, Y. P., 1999, pp. 359–389), [94] (Yan, J. Y., Liu, Y. P., 1991). 9.5.8. Stochastic behavior. On the basis of (9.4.12), the probability of order, co-order, semi-automorphic group order, or genus for all loopless orientable maps with size given can be done via order polynomials, co-order polynomials, semi-automorphic group order polynomials or genus polynomials. This shows that it is time to investigate their random chains in dependence on the size.

304 | 9 Surface equations third part 9.5.9. Distributions. On the basis of the above note, one addresses the distribution of order, co-order, semi-automorphic group order, or genus for all loopless orientable maps via, respectively, order polynomials, co-order polynomials, semi-automorphic group order polynomials, or genus polynomials with size given. Then the moments including the mathematical expectation can further be established. 9.5.10. The topic in this chapter is from Program 99 in [71] (Liu, Y. P., 2015, Vol. 22, p. 10745) as the primary stage for systematization in theory. Two other stages remain: making it more efficient in running and more intelligent in usage are still left to address.

10 Surface equations fourth part 10.1 Eulerian surface model Consider the equation for g ∈ ℛ{x, y} 2 2 2 4 𝜕g { { {a2 x ∫ y δx2 ,y2 g|x2 =u = −a1 + (1 − x )g − 2a3 x 𝜕x 2 ; y { { { {g|x=0,y=0 = a0 ,

(10.1.1)

where a0 , a1 , a2 , a3 ∈ ℝ+ . This is equation (19) in Introduction. The contribution from Eulerian maps on surfaces in the case of a1 = a2 = a3 = 1 without the initial condition, i. e., x2 ∫ y2 δx2 ,y2 g|x2 =u = −1 + (1 − x2 )g − 2x4 y

𝜕g 𝜕x 2

(10.1.2)

is found in [60] (Liu YP, 2008, p. 152). Equation (10.1.1) is then called a Eulerian surface model. In Liu YP [61] (p. 142), one finds the following equation for g ∈ ℛ{x, y}: 2 2 2 4 𝜕g { { {2x 𝜕x 2 = −b + (1 − x )g − x ∫ y δx2 ,y2 g|u=x2 ; y { { { g|x=0,y=0 = a, {

(10.1.3)

where a, b ∈ ℤ+ ⊆ ℝ+ . This equation is also a specific case of equation (10.1.1) with a0 = a, a1 = b and a2 = a3 = 1. Lemma 10.1.1. If a1 ≠ a0 , then neither equation (10.1.1) nor equation (10.1.3) has a solution. Proof. Denote by Ga[0] the constant term of g satisfying equation (10.1.1). As is seen, x2 is a common factor on the left hand side of the first equality in equation (10.1.1) and hence, there is no constant term. However, there is a constant term −a1 + Ga[0] on the right hand side. Thus, −a1 + Ga[0] = 0 ⇒ Ga[0] = a1 . By the initiation, a0 = a1 . This is a contradiction to the given condition. This lemma enables us to only consider equation (10.1.1) with a0 = a1 without loss of generality. Thus, a = (a0 , a2 , a3 ) instead of (a0 , a1 , a2 , a3 ). For convenience, equation (10.1.1) is equivalently by the cancelation law on ℛ{x, y}, transformed into the equation

https://doi.org/10.1515/9783110627336-010

306 | 10 Surface equations fourth part 2 4 𝜕g 2 2 { { {g = a0 + x g + 2a3 x 𝜕x 2 + a2 x ∫ y δx2 ,y2 g|u=x2 ; y { { { g|x=0,y=0 = a0 , {

(10.1.4)

where a0 , a2 , a3 ∈ ℤ+ ⊆ ℛ+ . From Lemma 10.1.1, equation (10.1.1) and equation (10.1.4) are equivalent on ℛ{x, y}. This enables us to only consider equation (10.1.4) instead of equation (10.1.1). From equation (10.1.4), it is seen that g is an even function of x because there is no term in equation (10.1.4) involving x with odd power. This enables us to treat g as a function of x2 . ,0≤ Because of g ∈ ℛ{x, y}, g is determined by the infinite set {Ga[m] | Ga[m] = [g]m x2 2m m g, or say, we have the coefficient of x in g. = 𝜕 m ∈ 𝒵+ } where [g]m x2 x2 𝜕g i 󸀠 For any integer i ≥ 0, let Ga[i] = [ 𝜕x 2 ]x 2 . Because of [x2

i

𝜕g ] = iGa[i] , 𝜕x2 x2

we have 󸀠 = (m + 1)Ga[m+1] , Ga[m]

(10.1.5)

where the integer m ≥ 0. 󸀠 Lemma 10.1.2. If Ga[i] ∈ ℛ+ {y} for integer 0 ≤ i ≤ l, then Ga[j] ∈ ℛ+ {y} for integer 0 ≤ j ≤ l − 1. 󸀠 󸀠 Proof. Because of Ga[l] ∈ ℛ+ {y}, by (10.1.5), Ga[l−1] = lGa[l] . Thus, Ga[l−1] is found. This is the conclusion.

for integer m ≥ 0. For any integer i ≥ 0, Let δ = δx2 ,y2 g|u=x2 and Δa[m] = [δ]m x2 i−1

δx2 ,y2 ui = ∑ x2(i−1−j) y2j , j=0

where, when i = 0, its value is guaranteed to be 0. Then 0, when m = 0; Δa[m] = { 2(i−m−1) , when m ≥ 1. ∑i≥m+1 Ga[i] y

(10.1.6)

Denote ∇a[m] = ∫y y2 Δa[m] , then, for integer m ≥ 1, ∇a[m] = ∑ Ga[i] y2(i−m) . i≥m+1

Lemma 10.1.3. Function g is independent of all y2l+1 for l ≥ 0.

(10.1.7)

10.1 Eulerian surface model | 307

Proof. From equation (10.1.4), all entries of y are generated by a function of the variable y instead of x in g(x). Because g(x) is even for x, g is only a function of y2i for i ≥ 1. This is the conclusion. This lemma tells us that g is an even function of x only in dependence of on y = (0, y2 , 0, y4 , . . .) = (y2 , y4 , y6 , . . .). Hence, for any integer m ≥ 0, Ga[m] is a function of y2i for i ≥ 1. We take into account (10.1.6). In order to determine g, a new parameter available only related to y needs to be introduced. For any integer m ≥ 0, denote 𝒥m = {n | power vector of a term of y in Ga[m] }. For any integer s ≥ 0, let s = π(n)/2 where π(n) = inT . Denote 𝒥m,s = {n | n ∈ 𝒥m , π(n) = s}, then we have (10.1.8)

𝒥m = ∑ 𝒥m,s . s≥0

Given two integers m, s ≥ 0, let Ga[m,s] = Gm |s=π(n)/2 . It consists of all terms yn in Ga[m] for π(n)/2 = s. From (10.1.8), Ga[m] = ∑ Ga[m,s] .

(10.1.9)

s≥0

We proceed on the basis of (10.1.7). For any integer s ≥ 0, s

∇a[m,s] = ∑ Ga[j+m,s−j] y2(j) .

(10.1.10)

j=1

Lemma 10.1.4. Given two integers m, s ≥ 0. If all Ga[r,t] are determined for integers r, t ≥ 0 and r + t ≤ m + s, ∇a[m,s] is then determined. Proof. Because of (j+m)+(s−j) = m+s for any integers 1 ≤ j ≤ s, Ga[j+m,s−j] is determined by Ga[r,t] for r + t ≤ m + s and r, t ≥ 0. By (10.1.10), ∇a[m,s] is then determined. We proceed on the basis of (10.1.4). From (10.1.5) and (10.1.7), an infinite system of equations for Ga[m] for m ≥ 0 can be found. Theorem 10.1.5. Equation (10.1.4) for g ∈ ℛ{x, y} is equivalent to the system of equations for Ga[m] ∈ ℛ{y} for m ≥ 0 a0 , { { { Ga[m] = {a0 + a2 ∑i≥1 Ga[i] y2(i) , { { {(2a3 m − 1)Ga[m−1] + a2 ∑i≥m Ga[i] y2(i−m+1) ,

when m = 0; when m = 1;

(10.1.11)

when m ≥ 2.

Proof. When m = 0, from the first equality of equation (10.1.4), x| (g − a0 ). This implies Ga[0] = a0 . We have g|x=0,y=0 = Ga[0] . This is the initiation of equation (10.1.4).

308 | 10 Surface equations fourth part When m = 1, from the first equality of equation (10.1.4), Ga[1] = Ga[0] + a2 ∇a[0] = a0 + a2 ∑ Ga[i] y2(i) . i≥1

This is the second equality of equations (10.1.11). When m ≥ 2, from the first equality of equation (10.1.4), Ga[m] = Ga[m−1] + 2a3 [

m−2

𝜕g ] 𝜕x 2 x2

m−1

+ a2 [∫ y2 δx2 ,y2 g|u=x2 ] y

x2

= (2a3 m − 1)Ga[m−1] + a2 ∑ Ga[i] y2(i−m+1) . i≥m

This is the third equality of equations (10.1.11). By considering that all operations employed above are equivalences on ℛ{y}, the conclusion can be drawn. There seems to be no simple way to solve the infinite system of equations (10.1.11) for Ga[m] , m ≥ 0. It is only necessary to consider how Ga[m,s] are determined for m, s ≥ 0. Let us have a look at what happens for smaller integers m and/or s. Lemma 10.1.6. For any integer s ≥ 0, a0 , when s = 0; Ga[0,s] = { 0, when s ≥ 1.

(10.1.12)

Proof. We proceed on the basis of Theorem 10.1.5. By employing the first equality of equations (10.1.11), the conclusion can be drawn. This lemma enables us to only discuss m ≥ 1 without loss of the generality as for m ≥ 0. Lemma 10.1.7. For any integer m ≥ 0, a0 , when m = 0; Ga[m,0] = { m a0 ∏i=1 (2ia3 − 1), when m ≥ 1.

(10.1.13)

Proof. When m = 0, from Lemma 10.1.6, the first equality of (10.1.13) is done. When m = 1, from the second equality of equations (10.1.11), Ga[1,0] = a0 (2a3 − 1). For m ≥ 2 in general, assume that m−1

Ga[m−1,0] = a0 (2ma3 − 1) ∏ (2ia3 − 1) i=1

is known. We work by induction, to determine Gm,0 , from the third equality of equations (10.1.11). Because of 0

[ ∑ Ga[i] y2(i−m+1) ] = 0, i≥m

y

10.1 Eulerian surface model | 309

we have m

Ga[m,0] = (2ma3 − 1)Ga[m−1,0] = a0 ∏(2ia3 − 1). i=1

This is the conclusion. The two lemmas above enable us to only discuss m, s ≥ 1 without loss of generality. We proceed on the basis of (10.1.11). For integers m, s ≥ 1, a2 ∑si=1 Ga[i,s−i] y2(i) , Ga[m,s] = { (2ma3 − 1)Ga[m−1,s] + a2 ∑s−1 j=0 Ga[j+m,s−j−1] y2(j+1) ,

when m = 1; when m ≥ 2.

(10.1.14)

Now, according to the order of m + s increasingly one by one starting from Ga[1,1] for m + s = 2 we evaluate Ga[m,s] for 2 ≤ m + s ≤ 3 and m, s ≥ 1. When m + s = 2, because Ga[2,0] = a0 (4a3 − 1)(2a3 − 1) (by Lemma 10.1.7) and Ga[0,2] = 0 (by Lemma 10.1.6), it is only necessary to evaluate Ga[1,1] . By the first equality of (10.1.14), Ga[1,1] = Ga[1,0] y2 = a0 (2a3 − 1)y2 .

(10.1.15)

That is, Ga[1,1] = a0 (2a3 − 1)y2 . When m + s = 3, we have Ga[3,0] = a0 (6a3 − 1)(4a3 − 1)(2a3 − 1) (Lemma 10.1.7) and Ga[0,3] = 0 (Lemma 10.1.6). It is only necessary to evaluate Ga[2,1] and Ga[1,2] . On Ga[2,1] , from the second equality of (10.1.14), Ga[2,1] = (4a3 − 1)a0 (2a3 − 1)y2 + a0 (4a3 − 1)(2a3 − 1)y2 .

(10.1.16)

That is, Ga[2,1] = 2a0 (4a3 − 1)(2a3 − 1)y2 . On Ga[1,2] , from the first equality of (10.1.14), Ga[1,2] = a0 (2a3 − 1)y22 + a0 (4a3 − 1)(2a3 − 1)y4 .

(10.1.17)

That is, Ga[1,2] = a0 (2a3 − 1)(y22 + (4a3 − 1)y4 ). Theorem 10.1.8. Equation (10.1.4) is well defined on ℛ{x, y} if, and only if, a0 = a1 . Proof. Sufficiency. When m + s ≤ 3, the conclusion is checked for all Ga[m,s] . Here, only we discuss m + s ≥ 4. For m + s = n ≥ 4, assume Ga[r,t] satisfies the conclusion whenever the integers r + t ≤ n − 1 and r, t ≥ 1. We work by induction, to evaluate Ga[m,s] = Ga[m,n−m] via (10.1.14). When m = 1, Ga[1,s] = Ga[1,n−1] . From the first equality of (10.1.14), n−1

Ga[1,n−1] = a2 ∑ Ga[i,n−1−i] y2(i) . i=1

310 | 10 Surface equations fourth part Because of i + (n − 1 − i) = n − 1 ≤ n − 1 for integers 1 ≤ i ≤ n − 1, the assumption shows that all Ga[i,n−1−i] are known for 1 ≤ i ≤ n − 1. Thus, Ga[1,n−1] is determined. When m ≥ 2, Ga[m,s] = Ga[m,n−m] . From the second equality of equations (10.1.14), n−m−1

Ga[m,n−m] = (2ma3 − 1)Ga[m−1,n−m] + a2 ∑ Ga[j+m,n−m−j−1] y2(j+1) . j=0

Because of (m − 1) + (n − m) = n − 1 ≤ n − 1, the assumption shows that Ga[m−1,n−m] are known, the first term of Ga[m,n−m] is evaluated. Because of (j +m)+(n−m−j −1) = n−1 ≤ n − 1 for any integers 0 ≤ j ≤ n − m − 1, the assumption shows that Ga[j+m,n−m−j−1] is determined. Thus, the second term of Ga[m,n−m] is evaluated. Hence Ga[m,n−m] = Ga[m,s] is evaluated. Based on the arbitrariness of n, for any integer m ≥ 0, Ga[m] = ∑ Ga[m,s] ∈ ℛ{y}, s≥0

s = n − m,

is determined. We proceed on the basis of Theorem 10.1.5. All Ga[m] together form a solution of equation (10.1.4). By considering only the solution is obtained on ℛ{x, y} from the initiation. The sufficiency is done. Necessity is a result of Lemma 10.1.1.

10.2 Solution Eulerian surface In what follows, Ga[m,s] for m, s ≥ 0 are investigated further for some favorite inner properties. Lemma 10.2.1. For integers m, s ≥ 0, Ga[m,s] is only dependent on y2i for 1 ≤ i ≤ s. Proof. From Lemma 10.1.3, Ga[m,s] is known in dependency on y2i for i ≥ 0. Here, i ≤ s is shown to be enough. For m + s ≤ 3, it is easy to check the case of i ≤ s. For m + s ≥ 4 in general, assume Ga[r,t] is dependent on y2i for i ≤ t whenever r + t ≤ m + s − 1. We work by induction, and we prove that Ga[m,s] is dependent on only y2i for i ≤ s. We proceed on the basis of (10.1.14). When m = 1, because of i +(s−i) = s ≤ m+s−1 for any integer 1 ≤ i ≤ s, the assumption shows that Ga[i,s−i] depends on at most y2(s−i) . Because of i ≥ 1, Gi,s−i depends at most on y2(s−1) . By considering i ≤ s, y2i depends on at most y2s . Thus, by the first equality of equations (10.1.14), Ga[1,s] depends on at most y2s . When m ≥ 2, by the second equality of equations (10.1.14), because of (m − 1) + s ≤ m+s−1, the assumption shows that Ga[m−1,s] depends on at most y2s . The first term has

10.2 Solution Eulerian surface

| 311

no problem. In the second term, because of (j + m) + (s − j − 1) ≤ m + s − 1 for any integer j : 0 ≤ j ≤ s − 1, the assumption shows that Ga[j+m,s−j−1] depends on at most y2(s−j−1) . Because of j ≥ 0, all Gj+m,s−j−1 depend on at most y2(s−1) . By considering j ≤ s − 1, y2(j+1) is at most y2s . Thus, the second term depends on at most y2s . In consequence, the conclusion can be drawn. This lemma tells us that, for any integers m and s, Ga[m,s] is a function of at most s variables in y2s . By noticing that s is independent of m, the determination of Ga[m,s] can be done by the order of s increasingly one by one starting from s = 0. Lemma 10.2.2. For any integers m, s ≥ 0, Ga[m,s] is a polynomial of y2s with degree at most s. Proof. When m + s ≤ 3. It is easily checked that Ga[m,s] is a polynomial of y2s with degree at most s. When m + s ≥ 4 in general, assume Ga[r,t] is a polynomial of y2t with degree at most t whenever r + t ≤ m + s − 1. We work by induction, to prove that Ga[m,s] is a polynomial of y2s with degree at most s. We proceed on the basis of equations (10.1.14). When m = 1, because of i + (s − i) = s ≤ m+s−1 for any integer 1 ≤ i ≤ s, the assumption shows that Ga[i,s−i] is a polynomial of y2(s−i) with degree at most s − i. Because of i ≥ 1, Gi,s−i is a polynomial of y2(s−1) with degree at most s − 1. By considering that y2i for i ≤ s, they are polynomials of at most y2s with degree s. Thus, by the first equality of equations (10.1.14), Ga[1,s] is a polynomial of y2s with degree s. When m ≥ 2, by the second equality of equations (10.1.14). Because of (m − 1) + s ≤ m+s−1, the assumption shows that Ga[m−1,s] is a polynomial of y2s with degree at most s. The first term poses no problem. In the second term, because of (j + m) + (s − j − 1) ≤ m + s − 1 for any integer j : 0 ≤ j ≤ s − 1, the assumption shows that Ga[j+m,s−j−1] is a polynomial of y2(s−j−1) with degree at most s − j − 1. Because of j ≥ 0, all Ga[j+m,s−j−1] are polynomials of at most y2(s−1) with degree s − 1. By considering j ≤ s − 1, they are polynomials of at most y2s with degree s. Thus, the second term is a polynomial of at most y2s with degree s. In consequence, the conclusion can be drawn. This lemma with last but one shows that, for any two integers m, n ≥ 0, Ga[m,s] is a polynomial of finite indeterminates. Lemma 10.2.3. For two integers m, s ≥ 0, Ga[m,s] ∈ ℛ+ {y} if, and only if, a ∈ ℝ3+ . Proof. If for two integers m, s ≥ 0, Ga[m,s] ∈ ℛ+ {y} but a ∈ ̸ ℝ3+ for instant a0 (others are analogous!), then because of Ga[0,0] = a0 , we have a contradiction to the initiation of equation (10.1.1) that a0 ∈ ℝ+ . This shows the necessity. Conversely, if a ∈ ℝ3+ , we have Gm,s ∈ ℛ+ {y} for 0 ≤ m + s ≤ 3. On m + s ≥ 4 in general, assume that, for any two integers r, t ≥ 0, Ga[r,t] ∈ ℛ+ {y} whenever r + t ≤ m + s − 1. We work by induction, to prove Ga[m,s] ∈ ℛ+ {y}.

312 | 10 Surface equations fourth part We proceed on the basis of equations (10.1.14). When m = 1, because of i + (s − i) = s ≤ m + s − 1 for any integer i : 1 ≤ i ≤ s, the assumption shows that Ga[i,s−i] ∈ ℛ+ {y}. Thus, by the first equation of equations (10.1.14), Ga[1,s] ∈ ℛ+ {y}. When m ≥ 2, by the second equation of equations (10.1.14), because of (m − 1) + s ≤ m + s − 1, the assumption shows that Ga[m−1,s] ∈ ℛ+ {y}. The first term is in ℛ+ {y}. In the second term, because of (j + m) + (s − j − 1) ≤ m + s − 1 for any integer j : 0 ≤ j ≤ s − 1, Ga[j+m,s−j−1] ∈ ℛ+ {y}. Thus, the second term is in ℛ+ {y}. In consequence, the conclusion can be drawn. Because of a ∈ ℝ4+ in equation (10.1.1), this lemma tells us that the solution is always in ℛ+ {x, y}. Theorem 10.2.4. Let ga[Ers] be the solution of equation (10.1.1). For any two integers [Ers] [Ers] = [𝜕x2m g[Ers] ]2s m, s ≥ 0, denote Ga[m,s] y = Ga[m,s] , then Ga[m,s] ∈ ℛ+ {y} is of the form of finite sum with all terms positive as

[Ers] Ga[m,s]

a0 δ0,s , { { { { { {a0 ∏m i=1 (2ia3 − 1), ={ s [Ers] { y2(i) , a2 ∑i=1 Gi,s−i { { { { s−1 [Ers] [Ers] {(2ma3 − 1)Ga[m−1,s] + a2 ∑j=0 Ga[j+m,s−j−1] y2(j+1) ,

when s ≥ 0, m = 0; when s = 0, m ≥ 1; when s ≥ 1, m = 1; when s ≥ 1, m ≥ 2. (10.2.1)

Proof. When s ≥ 0, m = 0, from Lemma 10.1.6. When s = 0, m ≥ 1, from Lemma 10.1.7. When s ≥ 1, m = 1, from the second equality of equations (10.1.14). Otherwise, from the second equality of equations (10.1.14). We proceed on the basis of (10.2.1). We continue to evaluate Ga[m,s] for m + s = 4. Because Ga[4,0] = a0 ∏4i=1 (2ia3 − 1) (the second equality!) and Fa[0,4] = 0 (the first equality of Lemma 10.1.7!), it is only necessary to address Ga[3,1] , Ga[2,2] and Ga[1,3] . On Ga[3,1] , by the fourth equality of (10.2.1), 3

3

i=1

i=1

Ga[3,1] = 2a0 ∏(2ia3 − 1)y2 + a0 ∏(2ia3 − 1)y2 .

(10.2.2)

That is, Ga[3,1] = 3a0 ∏3i=1 (2ia3 − 1)y2 . On Ga[2,2] , by the fourth equality of (20.2.1), Ga[2,2] = 3a0 (2a3 − 1)(y22 + (4a3 − 1)y4 ) + (2a0 (4a3 − 1)(2a3 − 1)y2 )y2

+ (a0 (2a3 − 1))y4 ,

i. e., Ga[2,2] = a0 (2a3 − 1)((8a3 + 1)y22 + (12a3 − 2)y4 ).

(10.2.3)

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On Ga[1,3] , by the third equality of (10.2.1), Ga[1,3] = a0 (2a3 − 1)(y22 + (4a3 − 1)y4 )y2 + (2a0 (4a3 − 1)(2a3 − 1)y2 )y4 3

(10.2.4)

+ (a0 ∏(2ia3 − 1))y6 , i=1

i. e., Ga[1,3] = a0 (2a3 − 1)(y23 + 3(4a3 − 1)y2 y4 + (6a3 − 1)(4a3 − 1)y6 ).

10.3 Explicitness Eulerian surface Let ℳ[Ers] m,n be the set of all root-isomorphic classes of Eulerian maps on all orientable surfaces with root-vertex valency 2m ≥ 0 and the vertex-partition vector 2n ≥ 0. For [Ers] any integer s ≥ 0, denote by ℳ[Ers] m,s the union of ℳm,n over n such that π(n) = s. [Ers] Because the enufunction of ℳm,n satisfies the equation (10.1.1) for a0 = a1 = a2 = [Ers] a3 = 1. From Theorem 10.1.8, its solution fa[Ers] ∈ ℛ{x, y} is determined by Fa[m,s] = m 𝜕x fa[Ers] |π(n)=s ∈ ℛ{y} for m ≥ 0. Observation 10.3.1. For any integers m, s ≥ 0, 󵄨 [Ers] 󵄨 [Ers] 󵄨󵄨 F1[m,s] 󵄨󵄨y=1 = 󵄨󵄨󵄨ℳm,s 󵄨󵄨󵄨

(10.3.1)

where a = (a0 , a1 , a2 , a3 ) and 1 = (1, 1, 1, 1), or (1, 1, 1, . . .) according as a or y. Proof. The result follows from the specific case of equation (10.1.1) when a = 1. [Ers] [Ers] Let 𝒢m,s and 𝒢m,n be the sets of all Eulerian graphs of root-vertex valency 2m ≥ 0 with, respectively, 2s = π(n) for all n ≥ 0 and the vector n ≥ 0. [Ers] Observation 10.3.2. Let 𝒢 (ℳ[Ers] m,s ) and 𝒢 (ℳm,n ) be the sets of all underline graphs of [Ers] [Ers] maps in, respectively, ℳm,s and ℳm,n . Then [Ers]

[Ers]

𝒢m,n = 𝒢 (ℳm,n )

[Ers] and hence 𝒢m,s = 𝒢 (ℳ[Ers] m,s ).

(10.3.2)

Proof. We see the result by showing a bijection between two sides of an equality. [Ers] [Ers] Let Fm,s |y=1 or Fm,n be the numbers of distinct root-isomorphic classes of Eulerian maps of root-vertex valency m with, respectively, s = π(n) for all n ≥ 0 or the vertex-partition vector n given on all orientable surfaces. [Ers] Lemma 10.3.3. Given G ∈ 𝒢m,n , denote by μG the set of topologically non-equivalent embeddings of G. Then

|μG | = (m − 1)! ∏(i − 1)!ni = (m − 1)!(i − 1)!n i≥1

where i = (1, 2, 3, . . .) and 1 = (1, 1, 1, . . .).

(10.3.3)

314 | 10 Surface equations fourth part Proof. The result follows by the existence of a bijection between an embedding and a vertex-rotation system of a graph. [Ers] [Ers] Let 𝒢m,n = 𝒢 (ℳ[Ers] m,n ) be the set of all underline graphs of map M ∈ ℳm,n and [Ers] [Ers] hence ℳm,n = ℳ(𝒢m,n ). [Ers] Denote by t = aut[sm] (G) the order of the semi-automorphic group of G ∈ 𝒢m,n . Let [Ers]

[Ers]

ℐm,n = {t | ∃G ∈ 𝒢m,n , t = aut[sm] (G)}.

(10.3.4)

[Ers](t) [Ers] Denote by 𝒢m,n the subset of 𝒢m,n in which every graph is with t as the order of its semi-automorphic group.

Lemma 10.3.4. For integer m ≥ 0 and integral vector n ≥ 0, n [Ers] = ∑ Λ[Ers](t) F1[m,n] 1[m,n] y ,

(10.3.5)

m + π(n) 󵄨󵄨 n [Ers](t) 󵄨 󵄨󵄨μ(𝒢m,n )󵄨󵄨󵄨(m − 1)!(i − 1)! t

(10.3.6)

[Ers] t∈ℐm,n

where Λ[Ers](t) 1[m,n] =

with π(n)(= inT ), i = (1, 2, 3, . . .) and n = (n1 , n2 , n3 , . . .). Proof. See Liu YP [59] (2003, Theorem 4.1, p. 211) and particularly Mao LF-Liu YP [75]. [Ers] [Ers] [Ers] [Ers] , has the following explicit = F1[m,n] , and thus G1[m,n] = F1[m,s] By (10.2.1), G1[m,s] expression: [Ers] = F1[m,s]

[Ers] n y = ∑ F1[m,n]

π(n)=s [Ers] n∈𝒥m,s

n ∑ ( ∑ Λ[Ers](t) 1[m,n] )y ,

[Ers] [Ers] n∈𝒥m,s t∈ℐm,n

(10.3.7)

is given by (10.3.6). where Λ[Ers](t) 1[m,n] Lemma 10.3.5. For an integer m ≥ 0 and an integral vector n ≥ 0, [Ers] n = ∑ Λ[Ers](t) Fa[m,n] a[m,n] y

(10.3.8)

[Ers] t∈ℐm,n

∈ ℛa is a polynomial of a with degree at most π(n) + 1, extracted from the where Λ[Ers](t) a[m,n] procedure in the proof of Theorem 10.2.4 such that Λ[Ers](t) a[m,n] |a=1 = which is just given by (10.3.6).

m + π(n) 󵄨󵄨 n [Ers](t) 󵄨 󵄨󵄨μ(𝒢m,n )󵄨󵄨󵄨(m − 1)!(i − 1)! , t

(10.3.9)

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| 315

Proof. A result of Observation 10.3.1 and Observation 10.3.2. Now, we are allowed to illustrate an explicit solutions of equation (10.1.1) via its specific case of a = 1 in combinatorial sense. [Ers] for Theorem 10.3.6. Let fa[Ers] be the solution of equation (10.1.1) determined by Fa[m,s] [Ers] has the explicit expression integers m, s ≥ 0. Then Fa[m,s] [Ers] = Fa[m,s]

n ∑ ( ∑ Λ[Ers](t) a[m,n] )y

(10.3.10)

[Ers] [Ers] n∈𝒥m,s t∈ℐm,n

is shown in Lemma 10.3.5. where Λ[Ers](t) a[m,n] Proof. We proceed on the basis of Lemma 10.3.5; Observation 10.3.1 results in the conclusion. This theorem enables us directly to deduce the explicit expression of the number of root-isomorphic classes of join-end maps with root-vertex valency m and vertexpartition vector n on all orientable surfaces. Corollary 10.3.7. Given integer m ≥ 0 and integral vector n ≥ 0, the solution f[Ers] of equation (10.1.1) under a = 1 is determined by m,n f[Ers] = ∑ Λ[Ers](t) 𝜕x,y 1[m,n]

(10.3.11)

[Ers] t∈ℐm,s

is given by (10.3.6). where Λ[Ers](t) 1[m,n] Proof. This is a direct result of Theorem 10.3.6.

10.4 Restrictions Eulerian surface Consider the equation for g ∈ ℛ{x, y} 2 2 4 𝜕g 2 { { {x ∫ y δx2 ,y2 g|x2 =u = −1 + (1 − x )g − 2x 𝜕x 2 ; y { { { g| { x=0,y=0 = 1.

(10.4.1)

This is equation (10.1.1) with a0 = a1 = a2 = a3 = 1, or equation (10.1.3) in the case of a = b = 1. The first equality of equation (10.4.1) is seen in [60] (Liu YP, 2008, p. 152) for enumerating the root-isomorphic classes of Eulerian maps on all orientable surfaces with given root-vertex valency and vertex-partition vector. Lemma 10.4.1. The initiation of equation (10.4.1) is consistent.

316 | 10 Surface equations fourth part Proof. Let G0 be the constant term of g ∈ ℛ{x, y}. Because of a common factor x2 on the left hand side of the first equality in equation (10.4.1), its constant term is 0. On the other hand, the constant term on the right hand side results in −1 + G0 . Thus G0 = 1. This is just the initiation G0 = gx=0, y=0 = 1. For convenience, equation (10.4.1) is, equivalently by the cancelation law on ℛ{x, y}, transformed into the equation by 2 4 𝜕g 2 2 { { {g = 1 + x g + 2x 𝜕x 2 + x ∫ y δx2 ,y2 g|u=x2 ; y { { { {g|x=0,y=0 = 1.

(10.4.2)

From Lemma 10.4.1, equation (10.4.1) and equation (10.4.2) are equivalent on

ℛ{x, y}. This enables us to only consider equation (10.4.2) instead of equation (10.4.1).

From observing equation (10.4.2), it is seen that g is an even function of x because there is no term in equation (10.4.2) involving x of odd power. This enables us to treat g as a function of x2 . for 0 ≤ Because of g ∈ ℛ{x, y}, g is determined by the infinite set {Gm | Gm = [g]m x2 2m m g, or, say, the coefficient of x in g. = 𝜕 m ∈ ℤ+ } where [g]m x2 x2 𝜕g i For any integer i ≥ 0, let Gi󸀠 = [ 𝜕x 2 ]x 2 . Because of [x 2

i

𝜕g ] = iGi , 𝜕x 2 x2

we have 󸀠 = (m + 1)Gm+1 , Gm

(10.4.3)

where the integer m ≥ 0. Lemma 10.4.2. If Gi ∈ ℛ+ {y} for integer 0 ≤ i ≤ l, then Gj󸀠 ∈ ℛ+ {y} for integer 0 ≤ j ≤ l − 1. 󸀠 󸀠 Proof. Gl ∈ ℛ+ {y} is known. By (10.4.3), Gl−1 = lGl . Thus, Gl−1 is found. This is the conclusion.

for integer m ≥ 0. For any integer i ≥ 0, Let δ = δx2 ,y2 g|u=x2 and Δm = [δ]m x2 i−1

δx2 ,y2 ui = ∑ x2(i−1−j) y2j , j=0

where, when i = 0, its value is guaranteed to be 0. Then 0, Δm = { ∑i≥m+1 Gi y2(i−m−1) ,

when m = 0; when m ≥ 1.

(10.4.4)

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| 317

Denote ∇m = ∫y y2 Δm , then, for integer m ≥ 1, ∇m = ∑ Gi y2(i−m) .

(10.4.5)

i≥m+1

Lemma 10.4.3. The function g is independent of all y2l+1 for l ≥ 0. Proof. From equation (10.4.2), all entries of y are generated by a function of a variable y instead of x in g(x). Because of g(x) being even for x, g is only a function of y2i for i ≥ 1. This is the conclusion. This lemma tells us that g is an even function of x only dependent on y = (0, y2 , 0, y4 , . . .) = (y2 , y4 , y6 , . . .). Hence, for any integer m ≥ 0, Gm is a function of y2i for i ≥ 1. We take into account (10.4.4). In order to determine g, a new parameter only related to y needs to be introduced. For any integer m ≥ 0, denote 𝒥m = {n | a power vector of a term of y in Gm }. For any integer s ≥ 0, let s = π(n)/2 where π(n) = inT . Denote 𝒥m,s = {n | n ∈ 𝒥m , π(n) = s}, then we have (10.4.6)

𝒥m = ∑ 𝒥m,s . s≥0

Given two integers m, s ≥ 0, let Gm,s = Gm |s=π(n)/2 . It consists of all terms yn in Gm for π(n)/2 = s. From (10.4.6), Gm = ∑ Gm,s .

(10.4.7)

s≥0

We proceed on the basis of (10.4.5). For any integer s ≥ 0, s

∇m,s = ∑ Gj+m,s−j y2j .

(10.4.8)

j=1

Lemma 10.4.4. Given two integers m, s ≥ 0. If all Gr,t are determined for integers r, t ≥ 0 and r + t ≤ m + s, then ∇m,s is determined. Proof. Because (j + m) + (s − j) = m + s for any integers 1 ≤ j ≤ s, Gj+m,s−j is determined by Gr,t for r + t ≤ m + s and r, t ≥ 0. By (10.4.8), ∇m,s is then determined. We proceed on the basis of (10.4.2). From (10.4.3) and (10.4.5), an infinite system of equations for Gm for m ≥ 0 can be found. Theorem 10.4.5. Equation (10.4.2) for g ∈ ℛ{x, y} is equivalent to the system of equations for Gm ∈ ℛ{y} for m ≥ 0 1, { { { Gm = {1 + ∑i≥1 Gi y2(i) , { { {(2m − 1)Gm−1 + ∑i≥m Gi y2(i−m+1) ,

when m = 0; when m = 1; when m ≥ 2.

(10.4.9)

318 | 10 Surface equations fourth part Proof. See the proof of Theorem 10.1.5 for a = 1. There seems to be no favorite way to directly solve the system of equations (10.4.9) for Gm , m ≥ 0. It is only necessary to consider how to determine Gm,s for m, s ≥ 0. Let us have a look at what happens for smaller integers m and/or s. Lemma 10.4.6. For any integer s ≥ 0, 1, when s = 0; G0,s = { 0, when s ≥ 1

(10.4.10)

Proof. We proceed on the basis of Theorem 10.4.5. By employing the first equation of equations (10.4.9), the conclusion can be drawn. This lemma enables us to only discuss m ≥ 1 without loss of generality on m ≥ 0. Lemma 10.4.7. For any integer m ≥ 0, 1, Gm,0 = { (2m)! 2m m!

when m = 0; ,

when m ≥ 1.

(10.4.11)

Proof. See the proof of Lemma 10.1.7 for a = 1. The last two lemmas enable us to only discuss m, s ≥ 1 without loss of generality. We proceed on the basis of (10.4.9). For integers m, s ≥ 1, ∑s Gi,s−i y2(i) , Gm,s = { i=1 (2m − 1)Gm−1,s + ∑s−1 j=0 Gj+m,s−j−1 y2(j+1)

when m = 1; otherwise.

(10.4.12)

Now, according to the order of m + s increasing one by one starting from G1,1 for m + s = 2, we evaluate Gm,s for 2 ≤ m + s ≤ 3 and m, s ≥ 1. When m + s = 2, because of G2,0 = 3 (by Lemma 10.4.7) and G0,2 = 0 (by Lemma 10.4.6), it is only necessary to evaluate G1,1 . By the first equality of (10.4.12), G1,1 = G1,0 y2 .

(10.4.13)

By G1,0 = 1 (Lemma 10.4.7), G1,1 = y2 . When m + s = 3, we take into account G3,0 = 15 (by Lemma 10.4.7) and G0,3 = 0 (by Lemma 10.4.6). It is only necessary to evaluate G2,1 and G1,2 . On G2,1 , from the second equality of (10.4.12), G2,1 = 3G1,1 + G2,0 y2 = 3y2 + 3y2 .

(10.4.14)

That is, G2,1 = 6y2 . On G1,2 , from the first equality of (10.4.12), G1,2 = G1,1 y2 + G2,0 y4 = y22 + 3y4 . That is, G1,2 = y22 + 3y4 .

(10.4.15)

10.4 Restrictions Eulerian surface

| 319

Theorem 10.4.8. Equation (10.4.2) is well defined on ℛ{x, y}. Proof. This is a direct result of Theorem 10.1.8 for a = 1. In what follows, the Gm,s for m, s ≥ 0 are investigated further for some favorite inner properties. Lemma 10.4.9. For integers m, s ≥ 0, Gm,s is only dependent on y2i for 1 ≤ i ≤ s. Proof. See the proof of Lemma 10.2.1 for a = 1. This lemma tells us that, for any integers m and s, Gm,s is a function of at most s variables in y2s . By noticing that s is independent of m, the determination of Gm,s can be done by the order of s increasing one by one starting from s = 0. Lemma 10.4.10. For any integers m, s ≥ 0, Gm,s is a polynomial of y2s with degree at most s. Proof. See the proof of Lemma 10.2.2 for a = 1. This lemma with the penultimate shows that, for any two integers m, n ≥ 0, Gm,s is a polynomial of finite indeterminates. Lemma 10.4.11. For two integers m, s ≥ 0, Gm,s ∈ ℛ+ {y}. Proof. See the proof of Lemma 10.2.3 for a = 1. All lemmas above in this section enable us to illustrate a compact form for the solution of equation (10.4.1). Theorem 10.4.12. Let g[Ers] be the solution of equation (10.4.1). For any two integers [Ers] [Ers] m, s ≥ 0, denote Gm,s = [𝜕x2m g[Ers] ]2s y = Gm,s , then Gm,s ∈ ℛ+ {y} is of the form of a finite sum with all terms positive,

[Ers] Gm,s

δ0,s , when s ≥ 0, m = 0; { { { { (2m)! { { m , when s = 0, m ≥ 1; = { 2 sm! [Ers] { when s ≥ 1, m = 1; ∑i=1 Gi,s−i y2(i) , { { { { s−1 [Ers] [Ers] {(2m − 1)Gm−1,s + ∑j=0 Gj+m,s−j−1 y2(j+1) , when s ≥ 1, m ≥ 2.

(10.4.16)

Proof. When s ≥ 0, m = 0, from Lemma 10.4.6. When s = 0, m ≥ 1, from Lemma 10.4.7. When s ≥ 1, m = 1, from the second equality of equations (10.4.12). Otherwise, the result follows from the second equality of equations (10.4.12). We proceed on the basis of (10.4.16). We continue to evaluate Gm,s for m + s = 4. Because G4,0 = 248!4! = 105 (the second case!) and F0,4 = 0 (the first case!), it is only necessary to do G3,1 , G2,2 and G1,3 .

320 | 10 Surface equations fourth part On G3,1 , by the fourth case of (10.4.16), G3,1 = 5G2,1 + G3,0 y2 = 30y2 + 15y2 .

(10.4.17)

That is, G3,1 = 45y2 . On G2,2 , by the fourth case, G2,2 = 3G1,2 + (G2,1 y2 + G1,0 y4 ) = 3(y22 + 3y4 ) + 6y22 + y4 .

(10.4.18)

That is, G2,2 = 9y22 + 10y4 . On G1,3 , by the third case, G1,3 = G1,2 y2 + G2,1 y4 + G3,0 y6 = (y22 + 3y4 )y2 + 6y2 y4 + 15y6 .

(10.4.19)

That is, G1,3 = y23 + 9y2 y4 + 15y6 . Example 10.4.1. Root-isomorphic classes of Eulerian maps on all orientable surfaces [Ers] with given root-vertex valency and vertex-partition vector. Denote Fm,s = Gm,s , then Fm,s provides the number of root-isomorphic classes for Eulerian maps on all orientable surfaces by root-vertex valency 2m and vertex-partition vector 2n such that π(n)/2 = s over n ≥ 0. For instance, as above, F1,1 = G1,1 = y2 , F2,1 = G2,1 = 6y2 and F1,2 = G1,2 = y22 + 3y4 .

Figure 10.4.1: Root-isomorphic classes of Euler maps of size 3.

In Figure 10.4.1, F1,1 stands for a map of root-vertex valency 2 (2m) and only one non-root-vertex of valency 2 (y2 ). This is 2-link-bundle(size 2). F1,2 is for a map of rootvertex valency 2 (2m) and two non-rooted vertices each of which has valency 2 (y22 ). Others include F2,1 = 4a + 2b = 6y2 and 3y4 = 2c + d in F1,2 . Known above as G3,1 = 45y2 , G2,2 = 9y22 + 10y4 , and G1,3 = y23 + 9y2 y4 + 15y6 . Seen F3,1 = 45y2 , F2,2 = 9y22 + 10y4 , and F1,3 = y23 + 9y2 y4 + 15y6 . In Figure 10.4.2, for save space, only two terms in F1,3 , i. e., y23 and 9y2 y4 , are illustrated. Others, the like. The former shows that the number of root-isomorphic classes of Euler maps of size 4 with vertex-partition vector n = 312 is 1, as a. The number of classes of Euler maps of size 4 with vertex-partition n = 12 + 14 is 9, as b, c and d.

10.5 Notes | 321

Figure 10.4.2: Root-isomorphic classes of Euler maps of size 4.

10.5 Notes 10.5.1. On the basis of (10.1.11), or (10.4.9) in the specific case of a = 1, an infinite system of linear equations for g = (Ga[1] , Ga[2] , Ga[3] , . . .) can be established and its solution determined by powers of a matrix related to, respectively, a and y, or only y. In each case, an explicit expression of the solution is directly obtained. Theorem 10.5.1. Let da[Ers] = (a0 , 0, 0, . . .); we consider the following equation: gT = Ya[Ers] gT + dTa[Ers]

(10.5.1)

where Ya[Ers] = (ya[i,j] )i,j≥1 such that

ya[i,j]

a2 yj−i+2 , { { { = {2(j + 1)a3 − 1, { { {0,

when j − i = k ≥ 0; when j − i = −1;

(10.5.2)

when j − i ≤ −2,

is equivalent to equations (10.1.11). Proof. It is easily seen from equations (10.1.11). Because of the existence of the inverse of I − Ya , (I − Ya[Ers] )−1 = ∑ Yka[Ers] , k≥0

(10.5.3)

equation (10.5.1) is well defined on ℛ{y}. [k] )i,j≥1 for k ≥ 0 where Theorem 10.5.2. Let Yka[Ers] = (ya[i,j] [k−1] [k] = ∑ ya[i,l] ya[l,j] ya[i,j] l≥1

(10.5.4)

322 | 10 Surface equations fourth part [1] = ya[i,j] for i, j ≥ 1. Then the solution of equation (10.5.1) has the explicit in which ya[i,j] expression

gTa[Ers] = dTa[Ers] + ∑ (a0 y[k] a[∗,1] ) k≥1

(10.5.5)

is for the first columns of Yka[Ers] for k ≥ 1. where y[k] a[∗,1] Proof. This result is directly evaluated from equation (10.5.1). 10.5.2. On the linearity of equation (10.1.3): the infinite coefficient matrix has the full upper triangle sub-matrix shown in (10.5.2) is in common with the full upper triangle sub-matrix of equation (7.1.1) shown in (7.5.2). This suggests us to observe the inner common structures between the join-end model and the Eulerian model of surface equations. 10.5.3. If the parameter π(n) is given, the infinite summation in (10.5.5) becomes finite because of the limitation in the s-dimensional system restricted from Theorem 10.2.4 and Theorem 10.4.8. 10.5.4. Consider a linear transformation from y to z for getting a simple explicision. We establish the reducibility for the solution of equation (10.1.1) and hence equation (10.4.1) on the basis of transformations. 10.5.5. For a as a functional vector of x and/or y, we may wish to find some meaningful objects except only for maps in combinatorics because of the universality of partitions as classification everywhere. 10.5.6. Equation (10.1.3) provides a theoretical base on the meson equation of Eulerian surface type for extending a certain number of other equations of surface types to reflect the original ideas of tree-like meson equations for extending a certain number of outer types. Almost all of them show linearity. About these topics, one might like to read the more original statements such as in [60] (Liu YP, 2008, pp. 149–152) and [61] (Liu, Y. P., 2009, pp. 139–142) etc. 10.5.7. One might think of the extension from only all orientable surfaces to all surfaces including both orientable and non-orientable for Eulerian. It looks reasonable in that the equation is of the same type with distinction only in the coefficients. This idea reminds us of Liu YP [59] (2003, 183–214), [61] (Liu, Y. P., 2009, pp. 231–236), [65] (2012) and [69] (2015, 164–165, 198–209) etc. 10.5.8. Specification of parameters or surfaces. For surfaces of small genus with small number of parameters, more research is found in [51] (Liu, Y. P., 1992), [8] (Cai, J. L., Liu, Y. P., 1997), [80, 81, 82, 83] (Ren, H., Liu, Y. P., in 1999–2000), etc. 10.5.9. Asymptotic estimation. Theorem 10.4.12 enables us to estimate the asymptotic [Ers] values of the sums of Gm,s over all m : 1 ≤ m ≤ s and certain relatives.

10.5 Notes | 323

10.5.10. Stochastic behavior. On the basis of the solution of equation (10.4.1), the probability of order, co-order, semi-automorphic group order, or genus for all Eulerian orientable maps with size given can be done via, respectively, order polynomials, co-order polynomials, semi-automorphic group order polynomials, or genus polynomials. It seems that the time is coming to investigate their random chains dependent on size. 10.5.11. Distributions. On the basis of the above note, we might address the distribution of order, co-order, semi-automorphic group order, or genus for all Eulerian orientable maps via, respectively, order polynomials, co-order polynomials, semiautomorphic group order polynomials, or genus polynomials with size given. Then the moments including the mathematical expectation can further be estimated. 10.5.12. The topic of this chapter is closely related to Program 100 in [71] (Liu, Y. P., Vol. 22, 2016, p. 10745) as a primary part of the stage for systematization in theory. Furthermore, there are two stages: making it more efficient in running and more intelligent in usage are still necessary.

11 Surface equations fifth part 11.1 Ordinary surface model Consider the equation for f ∈ ℛ{x, y} 2 3 𝜕f { { {a2 x ∫ yδx,y (uf |x=u ) = −a1 + (1 − x )f − a3 x 𝜕x ; y { { { f { x=0, y=0 = a0 ,

(11.1.1)

where a0 , a1 , a2 , a3 ∈ ℝ+ . This is equation (21) in Introduction. Because of the contribution to ordinary maps on orientable surfaces in the case of a0 = a1 = a2 = a3 = 1, from [67] (Liu YP, 2012), equation (11.1.1) is called an ordinary surface model. In Liu YP [67] (2012), one can also find the equation for f ∈ ℛ{x, y} 3 𝜕f 2 { { {x 𝜕x = −b + (1 − x )f − x ∫ yδx,y (uf |u=x ); y { { { f = a, { x=0, y=0

(11.1.2)

where a, b ∈ ℤ+ ⊆ ℝ+ which is also the specific case of equation (11.1.1) with a2 = a3 = 1. Lemma 11.1.1. If a1 ≠ a0 , then equation (11.1.1) has no solution. Proof. Let Fa[0] be the constant term of f . Because of a common factor x on the left hand side of the first equality in equation (11.1.1), its constant term is 0. On the other hand, the constant term on the right hand side results in −a1 + Fa[0] . Thus Fa[0] − a1 = 0 󳨐⇒ Fa[0] = a1 . However, from the initiation of equation (11.1.1), Fa[0] = a0 ⇒ a0 = a1 . Therefore, if a1 ≠ a0 , equation (11.1.2) has no solution. This lemma enables us to only consider equation (11.1.1) and equation (11.1.2) always with a0 = a1 without loss of generality. Thus, a = (a0 , a2 , a3 ) instead of (a0 , a1 , a2 , a3 ) is accepted in what follows. For convenience, equation (11.1.1) is, by the cancelation law, transformed into the form 2 3 𝜕f { { {f = a0 + x f + a3 x 𝜕x + a2 x ∫ yδx,y (uf |u=x ); y { { { f | = a , 0 { x=0,y=0

(11.1.3)

where a0 , a2 , a3 ∈ ℤ+ ⊆ ℝ+ . From Lemma 11.1.1, equation (11.1.1) or equation (11.1.2) is equivalent to equation (11.1.3) on ℛ{x, y}. This enables us to only discuss equation (11.1.3) instead of equation (11.1.1) or equation (11.1.2). https://doi.org/10.1515/9783110627336-011

326 | 11 Surface equations fifth part

[f ]m x

f is determined by the infinite set {Fa[m] | Fa[m] = [f ]m x , m ∈ ℤ+ , m ≥ 0} where = 𝜕xm f , or, say, the coefficient of xm in f . 𝜕f i 󸀠 For any integer i ≥ 0, denote Fa[i] = [ 𝜕x ]x , then because of i

[x

𝜕f ] = iFa[i] , 𝜕x x

we have 󸀠 = (m + 1)Fa[m+1] Fa[m]

(11.1.4)

where the integer m ≥ 0. 󸀠 ∈ ℛ+ {y} for any Lemma 11.1.2. If Fa[i] ∈ ℛ+ {y} for any integer i : 0 ≤ i ≤ l, then Fa[j] integer j : 0 ≤ j ≤ l − 1. 󸀠 Proof. We take into account that Fa[l] ∈ ℛ+ {y} is determined. By (11.1.4), Fa[l−1] = lFa[l] , 󸀠 and hence Fa[l−1] is determined. This is the conclusion.

Let δ = δx,y (uf |u=x ) and, for integer m ≥ 0, Δa[m] = [δ]m x . From i

δx,y ui+1 = ∑ xi−j yj j=0

for any integer i ≥ 0, Fa[0] = a0 , Δa[m] = { ∑i≥m Fa[i] yi−m ,

when m = 0; when m ≥ 1.

(11.1.5)

Denote ∇a[m] = ∫y yΔa[m] . Then, for m ≥ 1, ∇a[m] = ∑ Fa[i] yi−m+1 .

(11.1.6)

i≥m

Theorem 11.1.3. Equation (11.1.3) for f ∈ ℛ{x, y} is equivalent to the following system of equations for Fa[m] ∈ ℛ{y} for integer m ≥ 0:

Fa[m]

a0 , { { { { { {a2 ∑i≥0 Fa[i] yi+1 , ={ { a0 + a2 ∑i≥1 Fa[i] yi , { { { { {(a3 (m − 2) + 1)Fa[m−2] + a2 ∑i≥m−1 Fa[i] yi−m+2 ,

when m = 0; when m = 1; when m = 2;

(11.1.7)

when m ≥ 3.

Proof. When m = 0, from the first equality of (11.1.3), x| (f −a0 ). This implies Fa[0] = a0 . We take into account that f |x=0⇔⇒y=0 = Fa[0] . This is the initiation of equation (11.1.3).

11.1 Ordinary surface model | 327

When m = 1, from the first equality of equation (11.1.3), Fa[1] = ∇a[0] = ∑ Fa[i] yi+1 . i≥0

This is the second equation of equations (11.1.7). When m = 2, by the first equality of equation (11.1.3), Fa[2] = Fa[0] + a2 ∇a[1] = a0 + a2 ∑ Fa[i] yi . i≥1

This is the third equation of equations (11.1.7). When m ≥ 3, from the first equality of equation (11.1.3), Fa[m] = (a3 (m − 2) + 1)Fa[m−2] + a2 ∑ Fa[i] yi−m+2 . i≥m−1

This is the fourth equation of equations (11.1.7). In consequence, the conclusion can be drawn. We proceed on the basis of equations (11.1.7). To consider only the parameter relevant to x is not enough to determine f practically. Another new parameter relevant to y has to be introduced. For any integer m ≥ 0, denote 𝒥m = {n | a power vector of y in Fa[m] }. For any integer s ≥ 0, let s = π(n) where π(n) = i nT . Denote 𝒥m,s = {n | n ∈ 𝒥m , π(n) = s}, then 𝒥m = ∑ 𝒥m,s . s≥0

(11.1.8)

For any two integers m, s ≥ 0, denote by Fa[m,s] = Fa[m] |π(n)=s the sum of all terms of yn in Fa[m] such that π(n) = s. By (11.1.8), Fa[m] = ∑ Fa[m,s] . s≥0

(11.1.9)

Lemma 11.1.4. Given two integers m, s ≥ 0. If all Fa[r,t] are determined for r, t ≥ 0 and r + t < m + s, then ∇a[m,s] is determined. Proof. From (11.1.6), s−1

∇a[m,s] = ∑ Fa[j+m,s−j−1] yj+1 . j≥0

All Fa[j+m,s−j−1] are determined for 0 ≤ j ≤ s − 1 and (j + m) + (s − j − 1) = m + s − 1 < m + s. From the result above, ∇a[m,s] is determined.

328 | 11 Surface equations fifth part This lemma shows that Fa[m,s] can be evaluated from some Fa[r,t] for r + t < m + s. Lemma 11.1.5. The system of equations for Fa[m,s] for m, s ≥ 0 on ℛ{y}

Fa[m,s]

a0 δ0,s , { { { { { { a2 ∑s−1 { i=0 Fa[i,s−i−1] yi+1 , { { = {a0 δ0,s + a2 ∑si=1 Fa[i,s−i] yi , { { { { (a3 (m − 2) + 1)Fa[m−2,s] { { { { s { + a2 ∑i=1 Fa[i+m−2,s−i] yi ,

s ≥ 0, m = 0; s ≥ 0, m = 1; s ≥ 0, m = 2;

(11.1.10)

s ≥ 0, m ≥ 3,

is equivalent to the system of equations (11.1.7) for Fm ∈ ℛ{y} for m ≥ 0. Proof. When m = 0, for any integer s ≥ 0, the conclusion can be drawn from the first equation of equations (11.1.7). When m = 1, for any integer s ≥ 0, from the second equation of equations (11.1.7), s−1

Fa[1,s] = a2 ∑ Fa[i,s−i−1] yi+1 . i=0

This is the second equation of equations (11.1.10). When m = 2, for any integer s ≥ 0, from the third equation of equations (11.1.7), s

Fa[2,s] = a0 δ0,s + ∑ Fa[i,s−i] yi . i=1

This is the third equation of equations (11.1.10). When m ≥ 3, for any integer s ≥ 0, from the fourth equation of equations (11.1.7), s

Fa[m,s] = (a3 (m − 2) + 1)Fa[m−2,s] + ∑ Fa[j+m−2,s−j] yj . j=1

This is the fourth equation of equations (11.1.10). Now, let us have a look at what happens for m + s ≤ 2 starting from m + s = 0. We proceed on the basis of equations (11.1.10), according to the order of increasingly one by one from m + s = 0, to evaluate Fa[m,s] . When m + s = 0, because of m, s ≥ 0, only Fa[0,0] is considered. From the first equation of equations (11.1.10), Fa[0,0] = a0 . When m + s = 1, two candidates are Fa[1,0] and Fa[0,1] . They are determined by, respectively, the second and first equations of equations (11.1.10), Fa[1,0] = 0 and Fa[0,1] = 0. When m + s = 2, three candidates are Fa[2,0] , Fa[1,1] and Fa[0,2] . They are, respectively, determined by the third, second and first equations of equations (11.1.10). Then

11.1 Ordinary surface model | 329

we have Fa[2,0] = a0 , Fa[0,2] = 0 and 1−1

Fa[1,1] = a2 ∑ Fa[i,1−i−1] yi+1 = a2 Fa[0,0] y1 = a2 a0 y1 . i=0

(11.1.11)

Theorem 11.1.6. Equation (11.1.1) on ℛ{x, y} is well defined if, and only if, a0 = a1 . Proof. For sufficiency, when m+s ≤ 2, it can be checked that all Fa[m,s] are determined. Only m + s ≥ 3 is considered. When m+s = n ≥ 3, assume all Fa[r,t] are known for integers r, t ≥ 0 and r+t ≤ n−1. By induction on n, we evaluate Fa[m,s] = Fa[m,n−m] via equations (11.1.10). When m = 0, all Fa[0,s] for s ≥ 0 are known from the first equation of (11.1.10). When m = 1, Fa[1,s] = Fa[1,n−1] . From the second equation of equations (11.1.10) n−2

Fa[1,n−1] = a2 ∑ Fa[i,n−2−i] yi+1 . i=0

All Fa[i,n−2−i] are determined for 0 ≤ i ≤ n − 2 and i + (n − 2 − i) = n − 2 ≤ n − 1 from the assumption. Thus, Fa[1,n−1] is evaluated. When m = 2, Fa[2,s] = Fa[2,n−2] . From the third equation of equations (11.1.10), it is only necessary to discuss s ≥ 1; we have n−2

Fa[2,n−2] = a2 ∑ Fa[i,n−i−2] yi . i=1

Because i + (n − i − 2) = n − 2 ≤ n − 1, the assumption shows that Fa[i,n−i−2] is evaluated. Thus, Fa[2,n−2] is found. When m ≥ 3, Fa[m,s] = Fa[m,n−m] . From the fourth equation of equations (11.1.10), n−m

Fa[m,n−m] = (a3 (m − 2) + 1)Fa[m−2,n−m] + a2 ∑ Fa[i+m−2,n−m−i] yi . i=0

Because of (m−1)+(n−m) = n−1 ≤ n−1, the assumption yields Fa[m−1,n−m] is evaluated. Thus, the first term is done. Because of (j + m) + (n − m − j − 1) = n − 1 ≤ n − 1 for all integers j : 0 ≤ j ≤ n−m−1, the assumption shows that all Fa[j+m,n−m−j−1] are evaluated. Thus, the second term is done. This result shows that Fa[m,n−m] = Fa[m,s] is determined. From Theorem 11.1.3 and Lemma 11.1.5, all Fa[m,s] are formed as a solution of equation (11.1.3). By considering the uniqueness of the procedure under the initiation, the solution is unique on ℛ{x, y}. Necessity is a result of Lemma 11.1.1.

330 | 11 Surface equations fifth part

11.2 Solution ordinary surface In order to simplify the procedure for the solution of equation (11.1.1), some favorite inner structures of the solution have to be further investigated. Lemma 11.2.1. Given two integers m s ≥ 0. If m + s = 1(mod 2), then Fa[m,s] = 0. Proof. When m + s ≤ 2, only m + s = 1 is available. It is found that Fa[m,s] = 0. This is the conclusion. When m + s ≥ 3 in general, assume all Fa[r,t] = 0 for r + t = 1(mod 2) whenever r + t ≤ m + s − 1 for m, s ≥ 0. By induction, we prove Fa[m,s] = 0 for m + s = 1(mod 2). We proceed on the basis of equations (11.1.10). When m = 0, from the first equation of equations (11.1.10), the conclusion is seen. When m = 1, by the second equation of equations (11.1.10), because of i+(s−i−1) = s − 1 ≤ (1 + s) − 1 and 1 + s = s − 1(mod 2) for any integer i : 0 ≤ i ≤ s − 1, the assumption shows that s−1

Fa[1,s] = a2 ∑ Fa[i,s−i−1] yi+1 = 0 i=0

whenever 1 + s = 1(mod 2). When m = 2. By the third equation of equations (11.1.10). Because of i + (s − i) = s ≤ (2 + s) − 1 and s = s + 21(mod 2) for any integer i : 1 ≤ i ≤ s, the assumption shows that s

Fa[2,s] = a2 ∑ Fa[i,s−i] yi = 0 i=1

whenever 2 + s = 1(mod 2). When m ≥ 3 in general, by the fourth equation of equations (11.1.10), we have (m−2)+s = m+s−2 ≤ m+s−1 and m+s−2 = m+s(mod 2), and the assumption shows that Fa[m−2,s] = 0 whenever m+s = 1(mod 2). Because of (i+m−2)+(s−i) = m+s−2 ≤ (m+s)−1 and m + s − 2 = m + s(mod 2) for any integer i : 1 ≤ i ≤ s, the assumption shows that s

∑ Fa[i+m−2,s−i] yi = 0 i=1

whenever m + s = 1(mod 2). Thus, s

Fa[m,s] = (a3 (m − 2) + 1)Fa[m−2,s] + ∑ Fa[i+m−2,s−i] yi = 0 i=1

whenever m + s = 1(mod 2). In consequence, the conclusion can be drawn. This lemma enables us to reduce by half the amount of labor in the procedure of evaluating the solution of equation (11.1.1).

11.2 Solution ordinary surface

| 331

Lemma 11.2.2. For any integer m ≥ 0,

Fa[m,0]

a0 , when m = 0; { { { = {0, when m = 2t + 1, t ≥ 0; { { 2 t {a0 ∏i=1 (2(i − 1)a3 + 1), when m = 2t, t ≥ 1.

(11.2.1)

Proof. When m = 0, it follows from the first equation of equations (11.1.10). When m = 1(mod 2), the conclusion is given from Lemma 11.2.1. When m = 0(mod 2), it follows by equations (11.1.10). First, Fa[0,0] = a0 and Fa[2,0] = a20 are, respectively, given by the first and second equations. For m = 2t (t ≥ 2), assume t−1

Fa[2(t−1),0] = a20 ∏ (((2i − 2)a3 + 1)) i=1

when m = 2(t − 1). By induction on t, the conclusion is shown. From the fourth equation of equations (11.1.10), t

F2t,0 = a20 ∏ ((2i − 2)a3 + 1). i=1

This is the conclusion. This lemma tells us that the Fa[m,s] for m ≥ 0 are, in increasing order of s, done one by one starting from s = 0. Lemma 11.2.3. For any integer m ≥ 0,

Fa[m,1]

0, when m = 2t, t ≥ 0; { { { { { {a0 a2 y1 , when m = 1; ={ { {(a3 (m − 2) + 1)Fa[m−2,1] { { { t 2 { + a0 a2 ∏i=1 (2(i − 1)a3 + 1)y1 , when m = 2t + 1, t ≥ 1.

(11.2.2)

Proof. When m = 0(mod 2), the conclusion is given by Lemma 11.2.1. When m = 1, by the second equation of equations (11.1.10), Fa[1,1] = a2 Fa[0,0] y1 = a2 a0 y1 .

(11.2.3)

This is the second case of (11.2.2). When m = 2t + 1, t ≥ 1 in general. By the fourth equation of equations (11.1.10), t

Fa[2t+1,1] = (a3 (2t − 1) + 1)Fa[2t−1,1] + a20 a2 ∏ ((2i − 2)a3 + 1)y1 . i=1

This is the third case of (11.2.2).

332 | 11 Surface equations fifth part This lemma provides all cases of Fa[m,s] for s = 1 given at a time. Lemma 11.2.4. For integer m ≥ 0,

Fa[m,2]

0, { { { { { {a0 a2 (a2 y12 + a0 y2 ), ={ {(a3 (m − 2) + 1)Fa[m−2,2] + a2 Fa[m−1,1] y1 { { { { t 2 { + a0 a2 ∏i=1 (2(i − 1)a3 + 1)y2 ,

when m = 0, or m = 1(mod 2); when m = 2; when m = 2t, t ≥ 2,

where Fa[m−1,1] is given in (11.2.2).

(11.2.4)

Proof. The cases m = 0, or m = 1(mod 2), are, respectively, given by Lemma 11.2.2, or Lemma 11.2.1. When m = 2, from the third equation of equations (11.1.10), Fa[2,2] = a2 (Fa[1,1] y1 + Fa[2,0] y2 ) = a0 a2 (a2 y12 + a0 y2 ).

(11.2.5)

This is the second case of (11.2.4). When m = 2t for t ≥ 2 in general, by the fourth equation of (11.1.10), t

Fa[2t,2] = (a3 (2t − 2) + 1)Fa[2t−2,2] + F2t−1,1 y1 + a20 a2 ∏((2i − 2)a3 + 1)y2 . i=1

This is the third case of (11.2.4). The three lemmas above tell us that the procedure works well for going on to evaluate Fm,s by increasing the order one by one starting from s = 0. Lemma 11.2.5. For integers m, s ≥ 0, Fa[m,s] is independent of all yl for l ≥ s + 1. Proof. When m + s ≤ 3, the conclusion is checked to be true. When m+s ≥ 4 in general, assume that any Fa[r,t] is independent of all yl for l ≥ t+1 whenever r + t ≤ m + s − 1 and m, s ≥ 0. By induction, we show that Fa[m,s] satisfies the conclusion. When m = 0, from the first equation of equations (11.1.10), the result is trivial. When m = 1, we proceed by the second equation of equations (11.1.10). Because of i + (s − i − 1) = s − 1 ≤ s = m + s − 1 for any integer i : 0 ≤ i ≤ s − 1, the assumption shows that Fi,s−i−1 is independent of all yl for l ≥ s − i. From i ≥ 0, {Fi,s−i−1 : 0 ≤ i ≤ s − 1} is independent of yl for l ≥ s. Thus, s−1

Fa[1,s] = a2 ∑ Fa[i,s−i−1] yi+1 i=0

is independent of yl for l ≥ s + 1. When m = 2, by the third equation of equations (11.1.10), we have i + (s − i) = s ≤ s + 1 = m + s − 1 for any integer i : 1 ≤ i ≤ s, and the assumption shows that Fa[i,s−i] is

11.2 Solution ordinary surface

| 333

independent of yl for l ≥ s − i + 1. From i ≥ 1, {Fa[i,s−i] : 1 ≤ i ≤ s} is independent of yl for l ≥ s. Thus, s

Fa[2,s] = a0 δ0,s + a2 ∑ Fa[i,s−i] yi i=1

is independent of yl for l ≥ s + 1. When m ≥ 3 in general, by the fourth equation of equations (11.1.10), from the assumption, Fa[m−2,s] is independent of yl for l ≥ s + 1. Because of (i + m − 2) + (s − i) = m + s − 2 ≤ m + s − 1 for any integer i : 1 ≤ i ≤ s, the assumption shows that Fa[i+m−2,s−i] is independent of yl for l ≥ s − i + 1. From i ≥ 1, {Fa[i+m−2,s−i] : 1 ≤ i ≤ s} is independent of yl for l ≥ s. Thus, s

∑ Fa[i+m−2,s−i] yi i=1

is independent of yl for l ≥ s + 1. This results in s

Fa[m,s] = (a3 (m − 2) + 1)Fa[m−2,s] + a2 ∑ Fa[i+m−2,s−i] yi i=1

being independent of yl for l ≥ s + 1. This lemma enables us to deal with Fa[m,s] as a function of s variables as ys = (y1 , y2 , . . . , ys ). Lemma 11.2.6. For any integers m, s ≥ 0, Fa[m,s] is a polynomial of ys with degree at most s. Proof. For m + s ≤ 3, it can be checked that Fa[m,s] is a polynomial of ys with degree at most s. For m + s ≥ 4 in general, assume Fa[r,t] is a polynomial of yt with degree at most t whenever r + t ≤ m + s − 1. By induction, we prove that Fa[m,s] is a polynomial of ys with degree at most s. We proceed on the basis of equations (11.1.10). When m = 0, it is trivial. When m = 1, because of i+(s−i−1) = s−1 ≤ m+s−1 for any integer i : 0 ≤ i ≤ s−1, the assumption shows that Fa[i,s−i−1] is a polynomial of ys−i−1 with degree at most s − i − 1. Because of i ≥ 0, Fa[i,s−i−1] is a polynomial of ys−i−1 with degree at most s−1. We consider yi for i ≤ s − 1, a polynomial of ys with degree at most s. Thus, by the second equation of equations (11.1.10), Fa[1,s] is a polynomial of ys with degree at most s. When m = 2, by the third equation of equations (11.1.10), because of i + (s − i) = s ≤ m + s − 1 for any integer i : 1 ≤ i ≤ s, the assumption shows that Fa[i,s−i] is a polynomial of ys−i with degree at most s − i. Because of i ≥ 1, all Fa[i,s−i] are polynomials of ys−1 with degree at most s − 1. By i ≤ s, of ys we have a degree of at most s. Thus, Fa[2,s] is a polynomial of ys with degree at most s.

334 | 11 Surface equations fifth part When m ≥ 3 in general, by the fourth equation of equations (11.1.10), from the assumption, Fa[m−2,s] is a polynomial of ys with degree at most s. Because of (i + m − 2) + (s − i) = m + s − 2 ≤ m + s − 1 for any integer i : 1 ≤ i ≤ s, the assumption shows that Fa[i,s−i] is a polynomial of ys−i with degree at most s − i. Because of i ≥ 1, all Fa[i,s−i] are dependent on ys−1 with degree at most s − 1. By i ≤ s, of ys we have a degree of at most s. Thus, Fa[m,s] is a polynomial of ys with degree at most s. In consequence, the conclusion can be drawn. This lemma shows that, for any integer s ≥ 0, all Fa[m,s] have finite terms with an upper bound independent of m. Lemma 11.2.7. For any integers m, s ≥ 0, Fa[m,s] ∈ ℛ+ {y} if, and only if, a = (a0 = a1 , a2 , a3 ) ∈ ℝ3+ . Proof. For sufficiency: if for any two integers m, s ≥ 0, Fa[m,s] ∈ ℛ+ {y} and a ∈ ̸ ℝ+ , then because of the contradiction Fa[0,0] = a0 ∈ ̸ ℝ+ to the initiation of (11.1.3) with a0 ∈ ℝ+ , the sufficiency is done. Necessity: we proceed on the basis of Fa[m,s] ∈ ℛ+ {y} for m + s ≥ 0, to show a ∈ ℝ+ . When m + s ≤ 3, the conclusion is checked. When m + s ≥ 4 in general, assume all Fa[r,t] ∈ ℛ+ {y} for r, t ≥ 0 and r + t ≤ m + s − 1 result in the conclusion. By induction, we show Fa[m,s] ∈ ℛ+ {y} results in the conclusion. We proceed on the basis of equations (11.1.10). When m = 0, it is trivial. When m = 1, because of i + (s − i − 1) = s − 1 ≤ s =≤ m + s − 1 for any integer i : 0 ≤ i ≤ s − 1, the assumption shows Fa[i,s−i−1] ∈ ℛ+ {y} ⇒ a ∈ ℝ+ . By the second equation of equations (11.1.10), F1,s ∈ ℛ+ {y} ⇒ a ∈ ℝ+ . When m = 2, by the third equation of (11.1.10), because of i + (s − i) = s ≤ m + s − 1, the assumption shows Fa[i,s−i] ∈ ℛ+ {y} ⇒ a ∈ ℝ+ and hence Fa[2,s] ∈ ℛ+ {y} ⇒ a ∈ ℝ+ When m ≥ 3 in general, by the fourth equation of equations (11.1.10), the assumption results in Fa[m−2,s] ∈ ℛ+ {y} ⇒ a ∈ ℝ+ . This belongs to the first term. In the second term, because of (i + m − 2) + (s − i) = m + s − 2 ≤ m + s − 1 for any integer i : 1 ≤ i ≤ s, the assumption shows Fa[i+m−2,s−i] ∈ ℛ+ {y} ⇒ a ∈ ℝ+ and hence we have the second term in ℛ+ {y} ⇒ a ∈ ℝ+ . Thus, Fa[m,s] ∈ ℛ+ {y} ⇒ a ∈ ℝ+ . In consequence, the conclusion can be drawn. All lemmas above in this section enable us to illustrate a compact form for the solution of equation (11.1.1). Theorem 11.2.8. Let fa[ons] be the solution of equation (11.1.1). For any two integers m, s ≥ 0, denote s

a[ons] = [𝜕xm fa[ons] ]y (= Ua[m,s] ), Fa[m,s]

then Ua[m,s] is of the form of a finite sum with all terms positive,

11.2 Solution ordinary surface

Ua[m,s]

a0 , { { { { { { 0, { { { { { { { { { { {a20 ∏ti=1 (2(i − 1)a3 + 1), { { { { { {a2 a0 y1 , ={ { (a3 (m − 2) + 1)Ua[m−2,1] { { { { { { + a2 a20 ∏ti=1 (2(i − 1)a3 + 1)y1 , { { { { { { a2 ∑s−1 { i=0 Ua[i,s−i−1] yi+1 , { { { { (a3 (m − 2) + 1)Ua[m−2,s] { { { { s { + a2 ∑i=1 Ua[i+m−2,s−i] yi ,

| 335

when m = s = 0; when m = 0, s ≥ 1, or m ≠ s(mod 2), m ≥ 1, s ≥ 0; when s = 0, m = 2t ≥ 1 when m = s = 1;

(11.2.6)

when s = 1, m = 2t + 1, t ≥ 1; when m = 1, s = 2t + 1, t ≥ 1; otherwise,

where {otherwise} = {s ≥ 2, m ≥ 2, m = s(mod 2)}. Proof. When m = s = 0, we proceed by the initiation of equation (11.1.1). When m = 0, s ≥ 1, or m ≠ s(mod 2), m ≥ 1, s ≥ 0, by, respectively, the first equation of equation (11.1.10), or Lemma 11.2.1. When s = 0, m = 2t ≥ 1, by Lemma 11.2.2. When m = s = 1, we proceed by the second equation of equation (11.2.2). When s = 1, m = 2t + 1, t ≥ 1, by the third equation of equation (11.2.2). When m = 1, s = 2t + 1, t ≥ 1, by the second equation of equation (11.1.10). When s ≥ 2, m ≥ 2, m = s(mod 2) (i. e., otherwise), by the fourth equation of equation (11.2.2). [ons] ) for We continue by having a look at what happens with Ua[m,s] (= Fa[m,s] 3 ≤ m + s ≤ 6. From Lemma 11.2.1, it is only necessary to discuss m+s = 4 and m+s = 6. When m + s = 4, because of Ua[4,0] = (2a3 + 1)a20 and Ua[0,4] = 0, it is only necessary to evaluate Ua[3,1] , Ua[2,2] and Ua[1,3] . On Ua[3,1] , by (11.2.6),

Ua[3,1] = (a3 + 1)Ua[1,1] + a20 a2 y1 = (a3 + 1)(a0 a2 y1 ) + a20 a2 y1 .

(11.2.7)

This is Ua[3,1] = a0 a2 (a3 + 2)y1 . On Ua[2,2] , by (11.2.6), 2

Ua[2,2] = a3 Ua[2−2,2] + a2 ∑ Ua[i,2−i] yi = a2 Ua[1,1] y1 + a2 Ua[2,0] y2 . i=1

(11.2.8)

This is Ua[2,2] = a0 a2 (a2 y12 + a0 y2 ). On Ua[1,3] , by (11.2.6), 2

Ua[1,3] = a2 ∑ Ua[i,2−i] yi+1 = a2 (Ua[1,1] y2 + Ua[2,0] y3 ). i=0

This is Ua[1,3] = a0 a2 (a2 y1 y2 + a0 y3 ).

(11.2.9)

336 | 11 Surface equations fifth part When m + s = 6. In spite of Ua[6,0] = a20 (2a3 + 1)(4a3 + 1) (Lemma 11.2.2), Ua[0,6] = 0 and Ua[5,1] = (3a3 + 1)Ua[3,1] + a20 a2 (2a3 + 1)y1 (Lemma 11.2.3). On Ua[1,5] , by (11.2.6), Ua[1,5] = a0 a32 y1 y22 + a0 a22 (a3 + 2)y1 y4 + a0 a32 y12 y3 + 2a20 a22 y2 y3 + a20 a2 (2a3 + 1)y5 .

(11.2.10)

Then Ua[4,2] , Ua[3,3] and Ua[2,4] are omitted for saving space.

11.3 Explicitness ordinary surface Let ℳ[ons] m,n be the set of all root-isomorphic classes of ordinary maps with root-vertex valency m ≥ 0 and the vertex-partition vector n ≥ 0. For any integer s ≥ 0, denote by [ons] ℳ[ons] m,s the union of ℳm,n over n such that π(n) = s. It was shown that the enufunc[ons] tion of ℳm,n satisfies equation (11.1.1) for a0 = a1 = a2 = a3 = 1. From Theorem 11.2.8, [ons] = 𝜕xm fa[ons] |π(n)=s ∈ ℛ{y} for inteits solution fa[ons] ∈ ℛ{x, y} is determined by Fa[m,s] gers m, s ≥ 0. Observation 11.3.1. For any integers m, s ≥ 0, 󵄨 [ons] 󵄨 [ons] 󵄨󵄨 F1[m,s] 󵄨󵄨y=1 = 󵄨󵄨󵄨ℳm,s 󵄨󵄨󵄨

(11.3.1)

where a = (a0 , a1 , a2 , a3 ) and 1 = (1, 1, 1, 1), or (1, 1, 1, . . .) according as a or y. Proof. We have the specific case of equation (11.1.1) when a = 1. [ons] [ons] Let 𝒢m,s and 𝒢m,n be the sets of all ordinary graphs of root-vertex valency m ≥ 0 with, respectively, s = π(n) for all n ≥ 0 and the vertex-partition vector n(≥ 0). [ons] Observation 11.3.2. Let 𝒢 (ℳ[ons] m,s ) and 𝒢 (ℳm,n ) be the sets of all underline graphs of [ons] [ons] maps in, respectively, ℳm,s and ℳm,n . Then [ons]

[ons]

𝒢m,n = 𝒢 (ℳm,n )

[ons] and hence 𝒢m,s = 𝒢 (ℳ[ons] m,s ).

(11.3.2)

Proof. The result follows by showing a bijection between two sides of an equality. [ons] [ons] Let Fm,s |y=1 or Fm,n be the numbers of distinct root-isomorphic classes of ordinary maps on all orientable surfaces of root-vertex valency m with, respectively, s = π(n) for all n ≥ 0 or the vertex-partition vector n given. [ons] Lemma 11.3.3. Given G ∈ 𝒢m,n , denote by μG the set of topologically non-equivalent embeddings of G. Then

|μG | = (m − 1)! ∏(i − 1)!ni = (m − 1)!(i − 1)!n i≥1

where i = (1, 2, 3, . . .) and 1 = (1, 1, 1, . . .).

(11.3.3)

11.3 Explicitness ordinary surface

| 337

Proof. The result follows from the existence of a bijection between an embedding and a vertex-rotation system of graph G. [ons] [ons] Let 𝒢m,n = 𝒢 (ℳ[ons] m,n ) be the set of all underline graphs of map M ∈ ℳm,n and [ons] [ons] hence ℳm,n = ℳ(𝒢m,n ) from Observation 11.3.2. [ons] Denote by t = aut[sm] (G) the order of the semi-automorphic group of G ∈ 𝒢m,n . Let [ons]

[ons]

ℐm,n = {t | ∃G ∈ 𝒢m,n , t = aut[sm] (G)}.

(11.3.4)

[ons](t) [ons] Denote by 𝒢m,n the subset of 𝒢m,n in which every graph is with t as the order of its semi-automorphic group.

Lemma 11.3.4. For integer m ≥ 0 and integral vector n ≥ 0, n ∑ Λ[ons](t) 1[m,n] y ,

(11.3.5)

m + π(n) 󵄨󵄨 n [ons](t) 󵄨 󵄨󵄨μ(𝒢m,n )󵄨󵄨󵄨(m − 1)!(i − 1)! t

(11.3.6)

[ons] = F1[m,n]

where Λ[ons](t) 1[m,n] =

[ons] t∈ℐm,n

with π(n)(= inT ), i = (1, 2, 3, . . .) and n = (n1 , n2 , n3 , . . .). Proof. See Liu YP [59] (2003, Theorem 4.1, p. 211) and particularly Mao LF,Liu YP [75]. [ons] [ons] has the following explicit and then U1[m,n] = F1[m,n] By (11.2.6), U1[m,s] = F1[m,s] expression: [ons] = F1[m,s]

∑ π(n)=s [ons] n∈𝒥m,s

[ons] n y = F1[m,n]

n ∑ ( ∑ Λ[ons](t) 1[m,n] )y ,

[ons] [ons] n∈𝒥m,s t∈ℐm,n

(11.3.7)

where Λ[ons](t) is given by (11.3.6). 1[m,n] Lemma 11.3.5. For an integer m ≥ 0 and an integral vector n ≥ 0, [ons] = Fa[m,n]

n ∑ Λ[ons](t) a[m,n] y

(11.3.8)

[ons] t∈ℐm,n

∈ ℛ is a polynomial of a with degree at most π(n) + 1, extracted from the where Λ[ons](t) a[m,n] procedure in the proof of Theorem 11.2.8 such that Λ[ons](t) a[m,n] |a=1 = which is just given by (11.3.6).

m + π(n) 󵄨󵄨 n [ons](t) 󵄨 󵄨󵄨μ(𝒢m,n )󵄨󵄨󵄨(m − 1)!(i − 1)! , t

(11.3.9)

338 | 11 Surface equations fifth part Proof. A result of Observation 11.3.1 and Observation 11.3.2. Now, we are allowed to illustrate an explicit solution of equation (11.1.1) via its specific case of a = 1 in the combinatorial sense. [ons] for Theorem 11.3.6. Let fa[ons] be the solution of equation (11.1.1) determined by Fa[m,s]

[ons] has the explicit expression integers m, s ≥ 0. Then Fa[m,s] [ons] = Fa[m,s]

n ∑ ( ∑ Λ[ons](t) a[m,n] )y

(11.3.10)

[ons] [ons] n∈𝒥m,s t∈ℐm,n

where Λ[ons](t) is given in (11.3.8). a[m,n] Proof. We proceed on the basis of Lemma 11.3.5; Observation 11.3.1 results in the conclusion. This theorem enables us directly to deduce the explicit expression of the number of root-isomorphic classes of ordinary maps with root-vertex valency m and vertexpartition vector n on all orientable surfaces. Corollary 11.3.7. Given the integer m ≥ 0 and integral vector n ≥ 0, the solution f[ons] of equation (11.1.1) under a = 1 is determined by m,n f[ons] = 𝜕x,y

∑ Λ[ons](t) 1[m,n]

(11.3.11)

[ons] t∈ℐm,s

where Λ[ons](t) is given by (11.3.6). 1[m,n] Proof. The result follows as a direct result of Theorem 11.3.6.

11.4 Restrictions ordinary surface Consider the equation for f ∈ ℛ{x, y} 2 3 𝜕f { { {x ∫ yδx,y (uf |x=u ) = −1 + (1 − x )f − x 𝜕x ; y { { { f { x=0, y=0 = 1.

(11.4.1)

This is the case of a0 = a1 = a2 = a3 = 1 in equation (11.1.1) and the case of a = b = 1 in equation (11.1.2) as well. Because it is meaningful in the enumerative theory of combinatorial maps, in order to refine the original ideas and procedures, some details and applications are shown. Lemma 11.4.1. The initiation fx=0, y=0 = 1 is consistent for the equation.

11.4 Restrictions ordinary surface

| 339

Proof. Let F0 be the constant term of f ∈ ℛ{x, y}. Because of a common factor x on the left hand side of the first equality in equation (11.4.1), its constant term is 0. On the other hand, the constant term on the right hand side results in −1 + F0 . Thus F0 = 1. This is just the initiation F0 = fx=0, y=0 = 1. For convenience, equation (11.4.1) is, by the cancelation law, transformed into the form 2 3 𝜕f { { {f = 1 + x f + x 𝜕x + x ∫ yδx,y (uf |u=x ); y { { { f | = 1. { x=0,y=0

(11.4.2)

From Lemma 11.4.1, equation (11.4.1) is equivalent to equation (11.4.2) on ℛ{x, y}. This enables us to only discuss equation (11.4.2) instead of equation (11.4.1). Here f is determined by the infinite set {Fm | Fm = [f ]m x , 0 ≤ m ∈ 𝒵+ , m ≥ 0} where m m m [f ]x = 𝜕x f , or, say, the coefficient of x in f . 𝜕f i ]x , then because of For any integer i ≥ 0, denote Fi󸀠 = [ 𝜕x i

[x

𝜕f ] = iFi , 𝜕x x

we have 󸀠 Fm = (m + 1)Fm+1 ,

(11.4.3)

where the integer m ≥ 0. Lemma 11.4.2. If Fi ∈ ℛ+ {y} for any integer i : 0 ≤ i ≤ l, then Fj󸀠 ∈ ℛ+ {y} for any integer j : 0 ≤ j ≤ l − 1. 󸀠 Proof. We take into account that Fl ∈ ℛ+ {y} is determined. By (11.4.3), Fl−1 = lFl , and 󸀠 hence Fl−1 is determined. This is the conclusion.

Let δ = δx,y (uf |u=x ) and, for integer m ≥ 0, Δm = [δ]m x . We take into account that i

δx,y ui+1 = ∑ xi−j yj . j=0

For any integer i ≥ 0, F0 = 1, when m = 0; Δm = { i−m ∑i≥m Fi y , when m ≥ 1.

(11.4.4)

Denote ∇m = ∫y yΔm . Then, for m ≥ 1, ∇m = ∑ Fi yi−m+1 . i≥m

(11.4.5)

340 | 11 Surface equations fifth part Theorem 11.4.3. Equation (11.4.2) for f ∈ ℛ{x, y} is equivalent to the following system of equations for Fm ∈ ℛ{y} for integer m ≥ 0: 1, { { { { { {∑ Fi yi+1 , Fm = { i≥0 { 1 + ∑i≥1 Fi yi , { { { { {(m − 1)Fm−2 + ∑i≥m−1 Fi yi−m+2 ,

when m = 0; when m = 1; when m = 2;

(11.4.6)

when m ≥ 3.

Proof. See the proof of Theorem 11.1.3 for a = 1. We proceed on the basis of equations (11.4.6). Only the parameter relevant to x is not enough to determine f practically. Another new parameter relevant to y has to be introduced. For any integer m ≥ 0, denote 𝒥m = {n | a power vector of y in Fm }. For any integer s ≥ 0, let s = π(n) where π(n) = i nT . Denote 𝒥m,s = {n | n ∈ 𝒥m , π(n) = s}, then (11.4.7)

𝒥m = ∑ 𝒥m,s . s≥0

For any two integers m, s ≥ 0, denote by Fm,s = Fm |π(n)=s the sum of all terms of yn over n in Fm such that π(n) = s. By (11.4.7), Fm = ∑ Fm,s .

(11.4.8)

s≥0

Lemma 11.4.4. Given any two integers m, s ≥ 0. If all Fr,t are determined for r, t ≥ 0 and r + t ≤ m + s, then ∇m,s is determined. Proof. From (11.4.5), s−1

∇m,s = ∑ Fj+m,s−j−1 yj+1 . j≥0

All Fj+m,s−j−1 are determined for 0 ≤ j ≤ s − 1 and (j + m) + (s − j − 1) = m + s − 1 ≤ m + s. From the result above, ∇m,s is determined. This lemma shows that ∇m,s can be evaluated from some Fr,t for r + t ≤ m + s − 1. Lemma 11.4.5. The system of equations for Fm,s for m, s ≥ 0 on ℛ{y} as

Fm,s

δ0,s , { { { { { {∑s−1 Fi,s−i−1 yi+1 , = { i=0 s { δ + { 0,s ∑i=1 Fi,s−i yi , { { { s {(m − 1)Fm−2,s + ∑i=1 Fi+m−2,s−i yi ,

s ≥ 0, m = 0; s ≥ 0, m = 1; s ≥ 0, m = 2; s ≥ 0, m ≥ 3,

is equivalent to the system of equations (11.4.6) for Fm ∈ ℛ{y} for m ≥ 0.

(11.4.9)

11.4 Restrictions ordinary surface |

341

Proof. See the proof of Lemma 11.1.5 for a = 1. Now, let us have a look at what happens for m + s ≤ 2. Starting from m + s = 0. We proceed on the basis of equations (11.4.9). According to the order of increasingly one by one from m + s = 0, we evaluate Fm,s . When m+s = 0, because of m, s ≥ 0, only F0,0 is considered. From the first equation of equations (11.4.9), F0,0 = 1. When m + s = 1, two candidates are F1,0 and F0,1 . They are determined by, respectively, the second and first equations of equations (11.4.9) as F1,0 = 0 and F0,1 = 0. When m + s = 2, three candidates are F2,0 , F1,1 and F0,2 . They are, respectively, determined by the third, second and first equations of equations (11.4.9) as F2,0 = 1 F0,2 = 0 and 1−1

F1,1 = ∑ Fi,1−i−1 yi+1 = F0,0 y1 = y1 . i=0

(11.4.10)

Theorem 11.4.6. Equation (11.4.1) on ℛ{x, y} is well defined. Proof. A direct result of Theorem 11.1.6 for a = 1. In order to simplify the procedure for getting the solution of equation (11.4.1), some favorite inner structures of the solution have to be further investigated. Lemma 11.4.7. Given two integers m s ≥ 0. If m + s = 1(mod 2), then Fm,s = 0. Proof. See the proof of Lemma 11.2.1 for a = 1. This lemma enables us to reduce by half the amount of labor in the procedure of evaluating the solution of equation (11.4.1). Lemma 11.4.8. For any integer m ≥ 0,

Fm,0

1, when m = 0; { { { = {0, when m = 2t + 1, t ≥ 0; { { m! { 2t t! , when m = 2t, t ≥ 1.

(11.4.11)

Proof. This follows as a direct result of Lemma 11.2.2 for a = 1. This lemma tells us that the Fm,s for m ≥ 0 are directly done when s = 0 is given. Lemma 11.4.9. For any integer m ≥ 0, 0, { { { Fm,1 = {y1 , { { {(m − 1)Fm−2,1 +

when m = 2t, t ≥ 0; when m = 1; (m−1)! y, 2t t! 1

when m = 2t + 1, t ≥ 1.

Proof. This follows as a direct result of Lemma 11.2.3 for a = 1.

(11.4.12)

342 | 11 Surface equations fifth part This lemma provides all cases of Fm,s for s = 1 given at a time. Lemma 11.4.10. For integer m ≥ 0,

Fm,2

0, { { { { { {y2 + ay2 , ={ 1 { {(m − 1)Fm−2,2 { { { { + Fm−1,1 y1 +

when m = 0, or m = 1(mod 2); when m = 2; (m)! y, 2t t! 2

(11.4.13)

when m = 2t, t ≥ 2,

where Fm−1,1 is given in (11.4.12). Proof. This follows as a direct result of Lemma 11.2.4 for a = 1. The three lemmas above tell us that the procedure works for going on to evaluate Fm,s via the order of increasingly one by one starting from s = 0. Lemma 11.4.11. For integers m, s ≥ 0, Fm,s is independent of all yl for l ≥ s + 1. Proof. See the proof of Lemma 11.2.5 for a = 1. This lemma enables us to deal with Fm,s as a function of s variables as ys = (y1 , y2 , . . . , ys ). Lemma 11.4.12. For any integers m, s ≥ 0, Fm,s is a polynomial of ys with degree at most s. Proof. See the proof of Lemma 11.2.6 for a = 1. This lemma shows that, for any integer s ≥ 0, all Fm,s have finite terms with a upper bound independent of m. Lemma 11.4.13. For any integers m, s ≥ 0, Fm,s ∈ ℛ+ {y}. Proof. This follows as a direct result of Lemma 11.2.7 for a = 1. All lemmas above in this section enable us to illustrate a compact form for the solution of equation (11.4.1). Theorem 11.4.14. Let f[ons] be the solution of equation (11.4.1). For any two integers m, s ≥ 0, denote s

[ons] Fm,s = [𝜕xm f[ons] ]y (= Um,s ),

then Um,s is of the form of a finite sum with all terms positive,

11.4 Restrictions ordinary surface

Um,s

1, { { { { { 0, { { { { { { { { { { { { m! t , = { 2 t! { y1 , { { { { { { (m − 1)Um−2,1 + (m−1)! y, { 2t t! 1 { { s−1 { { { ∑i=0 Ui,s−i−1 yi+1 , { { { s {(m − 1)Um−2,s + ∑i=1 Ui+m−2,s−i yi ,

| 343

when m = s = 0; when m = 0, s ≥ 1, or m ≠ s(mod 2), m ≥ 1, s ≥ 0; when s = 0, m = 2t ≥ 1 when m = s = 1;

(11.4.14)

when s = 1, m = 2t + 1, t ≥ 1; when m = 1, s = 2t + 1, t ≥ 1; otherwise,

where {otherwise} = {s ≥ 2, m ≥ 2, m = s(mod 2)}. Proof. When m = s = 0, this follows by the initiation of equation (11.4.1). When m = 0, s ≥ 1, or m ≠ s(mod 2), m ≥ 1, s ≥ 0, by, respectively, the first equation of equation (11.4.9), or Lemma 11.4.8. When s = 0, m = 2t ≥ 1, by Lemma 11.4.9. When m = s = 1, by the second equation of equation (11.4.12). When s = 1, m = 2t + 1, t ≥ 1, by the third equation of equation (11.4.12). When m = 1, s = 2t + 1, t ≥ 1, by the second equation of equation (11.4.9). When s ≥ 2, m ≥ 2, m = s(mod 2) (i. e., otherwise), by the fourth equation of equation (11.4.9). We have a look at what happens with Um,s (= Fm,s ) for 3 ≤ m + s ≤ 6. From Lemma 11.4.8, it is only necessary to discuss m + s = 4 and m + s = 6. When m + s = 4, because U4,0 = 3 and U0,4 = 0, it is only necessary to evaluate U3,1 , U2,2 and U1,3 . On U3,1 , by (11.4.14), U3,1 = (3 − 1)U1,1 +

2! y = 2y1 + y1 . 2 1

(11.4.15)

This is U3,1 = 3y1 . On U2,2 , by (11.4.14), 2

U2,2 = (2 − 1)U2−2,2 + ∑ Ui,2−i yi = (y1 )y1 + y2 . i=1

(11.4.16)

This is U2,2 = y12 + y2 . On U1,3 , by (11.4.14), 2

U1,3 = ∑ Ui,2−i yi+1 = U0,2 y1 + U1,1 y2 + U2,0 y3 = (y1 )y2 + y3 . i=0

(11.4.17)

This is U1,3 = y1 y2 + y3 . When m+s = 6, in spite of U6,0 = 15, U0,6 = 0, U5,1 = 4U3,1 +3y1 = 15y1 and U0,6 = 0, it is only necessary to evaluate U1,5 , U2,4 , U3,3 and U4,2 . On U1,5 , by (11.4.14),

344 | 11 Surface equations fifth part 5−1

U1,5 = ∑ Ui,4−i yi+1 = U0,4 y1 + U1,3 y2 + U2,2 y3 + U3,1 y4 + U4,0 y5 i=0

=

y1 y22

+ 3y1 y4 +

y12 y3

(11.4.18)

+ 2y2 y3 + 3y5 .

Then U4,2 , U3,3 and U2,4 are omitted for saving space. Example 11.4.1. Root-isomorphic classes of ordinary maps on all orientable surfaces by root-vertex valency and vertex-partition vector as parameters. The solution of equation (11.4.1) is shown in (11.4.14).

Figure 11.4.1: Root-isomorphic classes of ordinary maps on all orientable surfaces.

For instance, U1,5 = y1 y22 +3y1 y4 +y12 y3 +2y2 y3 +3y5 . In the five terms of U1,5 , the first term y1 y22 shows that a map whose underline graph is a path of size 3 = (m+s)/2, root-vertex of valency m = 1, vertex-partition vector n = (1, 2) of s = π(n) = 1 + 2 × 2 = 5. It has only 1 class. In Figure 11.4.1, one sees the second term as 3y1 y4 = 2a + b, the third term as y12 y3 = c and the fourth term as 2y2 y3 = d + e and the fifth term as 3y5 = f + g + h.

11.5 Notes 11.5.1. The origin of equation (11.1.1) is equation (11.4.1), which occurs in [67] (Liu, Y. P., 2012, equation (13)) where one shows its extraction from establishing a suitable theory for decomposing certain infinite sets of ordinary combinatorial maps.

11.5 Notes | 345

11.5.2. On the basis of (11.1.7) or (11.4.6), an infinite system of linear equations for f = (Fa[1] , Fa[2] , Fa[3] , . . .) is established. Its solution is determined by powers of a matrix related to, respectively, a and y, or only y. In each case, an explicit expression of the solution is directly obtained. Theorem 11.5.1. Let da = (a2 a0 y1 , a0 , 0, . . .), the following equation: fT = Ya f + dTa

(11.5.1)

where Ya = (ya[i,j] )i,j≥1 such that

ya[i,j]

a2 yj−i+2 , { { { { { {a2 y1 , ={ { ja3 + 1, { { { { {0,

when j − i = k ≥ 0; when j − i = −1;

(11.5.2)

when j − i = −2; when j − i ≤ −3,

is equivalent to equations (11.1.7). Proof. This is easily seen from equations (11.1.7). Because of the existence of the inverse of I − Ya , (I − Ya )−1 = ∑ Yka ,

(11.5.3)

k≥0

equation (11.5.1) is well defined on ℛ{y}. [k] )i,j≥1 for k ≥ 0 where Theorem 11.5.2. Let Yka = (ya[i,j] [k−1] [k] = ∑ ya[i,l] ya[l,j] ya[i,j]

(11.5.4)

l≥1

[1] = ya[i,j] for i, j ≥ 1. Then the solution of equation (11.5.1) has the explicision and ya[i,j] [l] fTa = dTa + ∑(a0 a2 y1 y[l] a[∗,1] + a0 ya[∗,2] )

(11.5.5)

l≥1

j

[l] are, respectively, the first and second columns of Ya for l ≥ 1. and ya[∗,2] where y[l] a[∗,1]

Proof. This is directly evaluated from equation (11.5.1). Let fa[ons] and fa[ons] be, respectively, the solutions of equation (11.1.1) and equations (11.5.1). Then fa[ons] = a0 + fa[ons] xT where x = (x, x2 , x3 , . . .). 11.5.3. Equation (11.1.1) provides a theoretical base on meson equations of ordinary surface type for extending a certain number of other equations of surface types to reflect the original ideas of tree-like meson equations for extending a certain meson equations of outer type. Almost all of them show linearity. About these topics, one might like to read the more original statements such as in [29] (Liu, Y. P., 1985), [31] (Liu, Y. P., 1986), [44] (Liu, Y. P., 1989) etc.

346 | 11 Surface equations fifth part 11.5.4. One might think of the extension from only all orientable surfaces to all surfaces including both orientable and non-orientable ones for ordinary surface type. It seems reasonable in that the equation is of the same type with the distinction only in the coefficients. This idea reminds us of [61] (Liu, Y. P., 2009, Theorem 8.5.1 p. 233 and Theorem 8.5.2, p. 236), [65] (Liu, Y. P., 2012) etc. 11.5.5. The explicision of the solution of equation (11.4.1), as a specific case of equation (11.1.1), enables us to determine the genus series for all ordinary maps on orientable surfaces. The only thing we have to do is to determine the face distribution of such maps. Now, this is a simple way by determining all faces from a map with given order and size. On the genus series, the reader is referred to Jackson DM [19] (Jackson, D. M., 1995) and Jackson DM, Visentin TI [20] (Jackson, D. M., Visentin, T. I., 1990) via algebra without being concerned with the equation. However, it seems that this chapter as a specific case provides a type of realization. 11.5.6. Another type of realization is to attach genus series of maps via usage on computers as in Walsh THS, Lahman AB [91] (Walsh, T. H. S., Lahman, A. B., 1972) with a few parameters, still without consideration of the meson equation. 11.5.7. It seems more or less surprising that the solutions of equation (7.1.1) and equation (11.1.1) are nearly the same even if their appearances different. However, their specific cases in combinatorics should reflect a reasonable illustration. 11.5.8. Specification of parameters or surfaces. For surfaces of small genus with small number of parameters, more research might be found in [2] (Arques, D., 1987), [17] (Gao, Z. C., 1993), [6] (Cai, J. L., Hao, R. X., Liu, Y. P., 2001), [18] (Hao, R. X., Liu, Y. P., 2002), [23] (Liu, W. Z., Liu, Y. P., 2007) etc. 11.5.9. Asymptotic estimation. Theorem 11.4.14 provides a new theoretical base for [ons] estimating an asymptotic value of the sum of Fm,s over m : 0 ≤ m ≤ s when the size is given. However, one can find a number of relevant results extracted from another manner in [3] (Bender, E. A., Canfield, E. R., Richmond, L. B., 1986), [16] (Drmota, M., 1994), [12] (Conder, M., 1996), [22] (Liskovets, V. A., 1996) etc. 11.5.10. Stochastic behavior. On the basis of (11.4.14), the probability of order, coorder, semi-automorphic group order, or genus for all ordinary orientable maps with size given can be done via order polynomials, co-order polynomials, semiautomorphic group order polynomials or genus polynomials. This shows that it is time to investigate their random chains in dependence on the size. Moreover, the probability of an order of join-end maps, bridgeless maps, loopless maps, or Eulerian maps occurring in ordinary orientable maps with size given can also be done for relevant estimations. 11.5.11. Distributions. On the basis of the above note, one may consider the distribution of order, co-order, semi-automorphic group order, or genus for all ordinary

11.5 Notes | 347

orientable maps via, respectively, order polynomials, co-order polynomials, semiautomorphic group order polynomials, or genus polynomials with size given. Then the moments including the mathematical expectation can further be estimated. 11.5.12. The topic of this chapter is closely related to Program 101 [71] (Liu, Y. P., 2016, Vol. 22, p. 10746) as the part with four constant coefficients of the equation on the research stage of systematization in theory with consideration of two other remaining stages: making it more efficient in running and intelligent in usage.

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Index additivity 17 Andrews, G. E. 28, 54 Arques, D. 346 asymptotics 28, 55, 104, 147, 194, 239 Bender, E. A. 346 bipartite simple inner XVI bipartite simple inner model 219 block 68, 91, 114, 135, 207, 226 bridge 263 bridgeless map 278 bridgeless surface model XVII, 263 Brown, W. G. 303 Cai, J. L. 103, 147, 194, 238, 239, 303, 322, 346 Canfield, E. R. 346 Chen, Y. C. 28, 55, 104, 148 chunk 70, 93, 115, 136, 161, 182, 209, 227 chunk valency 70, 93, 115, 136, 161, 182, 209, 227 co-order 281 Conder, M. 346 Cori, R. 103 crack map 11 crack vertex 11 crackless outer XIII crackless outer model XIII, 1 cut-component 68, 91, 114, 135, 207, 226 cut-edge 263 cut-valency 68, 91, 114, 135, 207, 226 cut-vertex 68, 91, 114, 135, 161, 182, 207, 226 cutless 173 Deng, M. 303 Dong, F. M. 28, 54, 303 double size 66, 79 Drmota, M. 346 dual order 260 Dulusq, S. 103 Eulerian cutless inner XV Eulerian cutless inner model 173 Eulerian loopless inner XV Eulerian loopless inner model 126 Eulerian ordinary inner XIV Eulerian ordinary inner model 83 Eulerian surface model XVII, 305

fat valency 68 fat-component 68, 91, 114, 135, 207, 226 fat-valency 91, 114, 135, 207, 226 Gao, C. L. 194 Gao, Z. C. 346 general cutless inner XV general cutless inner model 149 general loopless inner XV general ordinary inner XIV general ordinary inner model 57 general simple inner XVI general simple inner model 197 genus series 260 Hao, R. X. 103, 147, 346 in-side 10 in-size 5 inner semi-size 152, 166 inner size 36, 84, 97 Jackson, D. M. 260, 346 join-end map 258 join-end surface model XVI, 241 Lahman, A. B. 346 Li, Z. X. 303 Liskovets, V. A. 346 Liu, W. Z. 346 Liu, X. Y. 303 Liu, Y. P. XIV–XVIII, 10, 12, 23, 28, 29, 31, 40, 43, 50, 54, 55, 57, 70, 73, 81, 83, 93, 96, 103–105, 116, 118, 126, 137, 139, 147–149, 162, 164, 173, 183, 185, 194, 195, 197, 209, 210, 212, 219, 228, 230, 238, 239, 241, 251, 252, 259–261, 263, 271, 273, 278, 281, 283, 293, 295, 302–305, 314, 315, 322, 323, 325, 337, 344–347 Long, S. D. 194 loopless surface model XVII, 283 Mao, L. F. 104, 194, 251, 271, 293, 314, 337 Mullin, R. C. 303 near d-regular 303 near quadrangulation 303 near triangulation 303 Nemeth, E. 303

354 | Index

non-detachable 173 non-root vertex 68, 114, 161 non-separable outer model XIII, 13 nonseparable 68, 91, 114, 135, 207, 226 ordinary outer model XIV, 43 ordinary surface model XVIII, 325 Pan, L. Y. 303 Ren, H. 238, 239, 303, 322 restricted outer model XIV, 31 restricted outer planar map 42 restriction 73 Richmond, L. B. 346 rooted semi-size 52 rooted semi-valency 71, 94, 116, 138, 162, 184, 210, 228 Rota, G. C. 28, 54 Schellenberg, P. J. 303 semi-size 46 size 101 slice 70, 93, 115, 136, 161, 182, 209, 227 slice valency 70, 93, 115, 136, 161, 182, 209, 227 slice valent 70, 93, 115, 136, 161, 182, 209, 227 splitting chunk 70, 93, 115, 136, 161, 182, 209, 227 splitting pair 69, 93, 115, 136, 161, 182, 208, 227

splitting slice 70, 93, 115, 136, 161, 182, 209, 227 stochastics 28, 55, 104, 147, 194, 239, 261 super-wheel 24 super-wheel model XIII super-wheels outer model 13 Taylor, B. D. 28, 54 Tutte, W. T. 104, 194, 303 un-rooted semi-size 71, 94, 116, 138, 162, 184, 210, 228 Viennot, G. 103 Visentin, T. I. 260, 346 Walsh, T. H. S. 346 Wan, L. X. 28, 55, 104, 148 Wei, E. L. 104, 194 x-size 66, 79 Xu, Y. 303 y-semi-size 107, 120 y-size 58, 66, 74, 79, 120 y-vertex 68, 114, 161 Yan, J. Y. 28, 54, 55, 104, 147, 148, 194, 195, 239, 261, 281, 303 Zhang, Y. L. 103