Complex Delay-Differential Equations 311056016X, 9783110560169

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Kai Liu, Ilpo Laine, and Lianzhong Yang Complex Delay-Differential Equations

De Gruyter Studies in Mathematics

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Edited by Carsten Carstensen, Berlin, Germany Gavril Farkas, Berlin, Germany Nicola Fusco, Napoli, Italy Fritz Gesztesy, Waco, Texas, USA Niels Jacob, Swansea, United Kingdom Zenghu Li, Beijing, China Karl-Hermann Neeb, Erlangen, Germany

Volume 78

Kai Liu, Ilpo Laine, and Lianzhong Yang

Complex Delay-Differential Equations |

Mathematics Subject Classification 2010 30D35, 39A10, 34M55 Authors Prof. Kai Liu Nanchang University Department of Mathematics 330031 Nanchang People’s Republic of China [email protected]

Prof. Lianzhong Yang Shandong University School of Mathematics 250100 Shandong People’s Republic of China [email protected]

Prof. Ilpo Laine University of Eastern Finland Department of Physics and Mathematics PO Box 111 FIN-80101 Joensuu Finland [email protected]

ISBN 978-3-11-056016-9 e-ISBN (PDF) 978-3-11-056056-5 e-ISBN (EPUB) 978-3-11-056040-4 ISSN 0179-0986 Library of Congress Control Number: 2021932017 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. © 2021 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

Contents Preface | VII Introduction | IX 1 1.1 1.2 1.2.1 1.2.2 1.2.3

Introduction of Nevanlinna theory and its difference versions | 1 Basic notations and facts in the Nevanlinna theory | 1 Difference Nevanlinna theory | 11 Elementary difference considerations in the complex domain | 11 Difference variants of the logarithmic derivative lemma for meromorphic functions | 12 Key notions and results in difference Nevanlinna theory | 16

2 2.1 2.2 2.3 2.4 2.5

Value distribution of complex delay-differential polynomials | 21 Value distribution of difference polynomials | 21 Value distribution of delay-differential polynomials | 24 Borel exceptional values of delay-differential polynomials | 51 The zeros of [P(f )f (z + c)](k) − α(z) | 56 Zeros on delay-differential polynomials of exponential type | 59

3 3.1 3.2

Uniqueness of delay-differential polynomials | 67 Value sharing on f with its linear delay-differential operator | 67 Uniqueness of delay-differential polynomials | 78

4 4.1 4.2 4.3

Difference Wiman–Valiron theory | 87 Differential setting of the Wiman–Valiron theory | 87 Difference setting of the Wiman–Valiron theory | 88 Wiman–Valiron theorems for q-differences | 90

5 5.1 5.2 5.2.1

The linear complex delay-differential equations | 93 First-order delay-differential equations | 93 Higher-order linear complex delay-differential equations | 100 Higher-order homogeneous linear complex delay-differential equations | 100 Higher-order nonhomogeneous linear complex delay-differential equations | 106

5.2.2

6 6.1 6.2 6.3

Fermat-type delay-differential equations | 117 Fermat equations | 117 Fermat-type functional and differential equations | 117 Fermat difference equations | 122

VI | Contents 6.3.1 6.3.2 6.3.3

Entire solutions of Fermat difference equations | 122 Meromorphic solutions of Fermat difference equations | 128 Fermat delay-differential equations | 132

7 7.1 7.2

Delay-differential Riccati equations | 149 Complex differential and difference Riccati equations | 149 Complex delay-differential Riccati equations | 154

8 8.1 8.2

Malmquist-type delay-differential equations | 163 Malmquist-type difference equations | 163 Malmquist-type delay-differential equations | 164

9 9.1 9.2

Nonlinear complex delay-differential equations | 195 Background | 195 Nonlinear complex delay-differential equations | 195

10 10.1 10.2 10.3

Complex q-delay-differential equations | 213 Value distribution for q-delay-differential polynomials | 213 Uniqueness on q-delay-differential polynomials | 221 q-delay-differential equations | 226

11 11.1 11.2 11.3 11.4

Systems of complex delay-differential equations | 241 Background | 241 Malmquist-type delay-differential systems | 241 Fermat-type delay-differential systems | 245 Systems of q-delay-differential equations | 251

12 12.1 12.2

Periodicity of entire functions with delay-differential polynomials | 255 Basic results on periodic functions | 255 Periodicity of differential, difference, and delay-differential polynomials | 258

Bibliography | 279 Index | 289

Preface Nevanlinna theory, one of the great achievement in mathematical research in the 20th century, is an important part of modern function theory. Its basic ideas have been applied in several related fields, such as several complex variables, potential theory, minimal surfaces, and tropical mathematics. The background to this book is the applications of Nevanlinna theory to investigate meromorphic solutions to complex differential equations; see, e. g., the books by Jank and Volkmann [113] and by the second author [120]. Similarly, the uniqueness theory of meromorphic functions, see the book by Yang ad Yi [235], has been an important aspect behind this book. The development of difference Nevanlinna theory 15 years ago naturally prompted to investigating applications of Nevanlinna theory to meromorphic solutions to complex difference equations; see the recent book by Chen [29]. This book presents, in some sense, the next step: applications of Nevanlinna theory to meromorphic solutions to delay-differential equations; these are complex equations that include difference operators such as, e. g., f (z + c) and f (qz), and the derivatives of a meromorphic function f (z). These investigations abound during the last decade. The aim of this book is collecting together what we feel to be the key material within this topic. Perhaps, we should point out here that much of the material referred to in this book speak about difference-differential equations, whereas we prefer to call them as delay-differential equations. We also hope that this collection of recent results is interesting enough to prompt further research in this field, being certainly not yet completely explored. The authors would like to express their thanks to several colleagues whose discussions and concrete help has been of an utmost importance during our preparation of this book. In particular, we would like to mention Hongxun Yi, Zongxuan Chen, Peichu Hu, Tingbin Cao and Xiaoguang Qi, who read the manuscript and gave many valuable suggestions, and Z. Latreuch whose collaboration with the second author can be found in the contents of Chapter 2. Last but definitely not least, our warm thanks are due to our families for their support and understanding during the process of preparing. This book was supported by the National Natural Science Foundation of China (Nos. 11661052, 12061042) and the Natural Science Foundation of Jiangxi (No. 20202BAB201003).

https://doi.org/10.1515/9783110560565-201

Introduction This book may be understood as a continuation to previous expositions devoted to applications of the Nevanlinna theory to complex differential equations, see, e. g., the books by Jank and Volkmann [113] and by the second author [120], to the uniqueness theory of meromorphic functions, see the book by Yang and Yi [235], and to complex difference equations, see the book by Chen [29]. Looking at the combinations of various difference operators such as the shift operator f (z+c), the basic difference operator f (z + c) − f (z), and the q-difference operator f (qz), combined with the derivatives of f (z), this book is the first monograph introducing applications of Nevanlinna theory to the value distribution theory of complex delay-differential polynomials and complex delay-differential equations of various types. The book is organized in twelve chapters. Most of the results presented here may be found either in the respective original references or are to be found in a closely related form at least. The basic notations and facts of Nevanlinna theory are included in Chapter 1, most proofs and details being omitted. As to this background material, the reader may consult three key monographs in Nevanlinna theory by Gol’dberg and Ostrovskii [61], Hayman [90], and Jank and Volkmann [113]. Difference Nevanlinna theory, including difference variants of the logarithmic derivative lemma, the second main theorem in the difference setting, see, e. g., [83], and difference- and delay-differential forms of the Clunie lemma will also be introduced in Chapter 1. In the short Chapter 4, Wiman–Valiron-type results in the differential, difference, and q-difference setting are included for the convenience of the reader, the proofs again being omitted. Chapter 2 contains a number of difference- and delay-differential variants of some classical results treating zeros of complex differential polynomials, zeros of complex difference, and, in particular, delay-differential polynomials of various types. Chapter 3 then offers uniqueness results, related to value sharing, of meromorphic functions, presenting delay-differential variants of classical results. Chapter 5 then offers some results in first- and higher-order complex delaydifferential equations. Also, comparisons are given to similarities and nonsimilarities while looking at various properties of complex differential and complex delaydifferential equations. The next four chapters are then devoted to considering nonlinear complex delay-differential equations. Chapter 6 presents Fermat-type delaydifferential equations, first recalling Fermat-type functional and differential equations as the background. Chapter 7 is devoted to delay-differential Riccati equations, again related to essential properties of differential and difference Riccati equations. Chapter 8 then treats Malmquist-type delay-differential equations. Here some important equations have been singled out, with the existence of sufficiently many finite-order and hyper-order less than one meromorphic solutions, considered as a version of the https://doi.org/10.1515/9783110560565-202

X | Introduction Painlevé property. Based on some trigonometric formulas, some related nonlinear complex delay-differential equations are then treated in Chapter 9. The corresponding investigations related to the q-difference operator f (qz), such as Nevanlinna theory for f (qz), value distribution and uniqueness of q-delaydifferential polynomials and q-delay-differential equations, are presented in Chapter 10. Meromorphic solutions to systems of complex delay-differential equations are then included in Chapter 11. To close the contents of this book, some basic results on the periodicity of meromorphic functions are recalled, and the periodicity of complex delay-differential polynomials is considered by making use of the Nevanlinna theory and its difference versions in Chapter 12.

1 Introduction of Nevanlinna theory and its difference versions 1.1 Basic notations and facts in the Nevanlinna theory This preceding chapter is devoted to reviewing the basic facts in the Nevanlinna theory needed for the material included in this book. In addition, we also include some facts that may be useful in the future research. For more material, for proofs omitted here and for most of the notations, the reader may consult Gol’dberg and Ostrovskii [61], Hayman [90], and Jank and Volkmann [113], to mention three key monographs in the field. Before shortly reviewing the basic notations and key results in the Nevanlinna theory, exceptional sets are throughout needed therein, as well as in the present book. Typically, it means considering the linear measure m(E) := ∫E dt and the logarithmic

measure l(E) := ∫E∩[1,∞) dtt for a set E ⊂ [0, ∞); such a set E is always called an exceptional set if it is of finite linear measure or of finite logarithmic measure. Trivially, l(E) ≤ m(E). Hence E is of finite logarithmic measure whenever it is of finite linear 1 measure. However, the set E = ⋃+∞ n=1 [n, n + n ] shows that the converse is not true. To eliminate such exceptional sets, the following elementary lemma sometimes appears to be useful. Lemma 1.1.1. Let ϕ(r) and ψ(r) be increasing functions on [0, +∞) satisfying ϕ(r) ≤ ψ(r) outside an exceptional set of finite linear measure. Given α > 1, there exists a number r0 = r0 (α) > 0 such that ϕ(r) ≤ ψ(αr) on [r0 , +∞). Proving this lemma is nothing but an easy exercise; see also [120, Lemma 1.1.1], whereas the corresponding logarithmic measure variant, with the same conclusion, is less trivial. For the proof, see [72], Lemma 5. Lemma 1.1.2. Let ϕ(r) and ψ(r) be increasing functions on [0, +∞) satisfying ϕ(r) ≤ ψ(r) for all r ∉ E ∪ [0, 1], where E ⊂ (1, ∞) is a set of finite logarithmic measure. Given α > 1, there exists r0 = r0 (α) > 0 such that ϕ(r) ≤ ψ(αr) for all r ≥ r0 . To define the corresponding notations of densities, denote E(r) := E ∩ [1, r]. Then we define the upper linear, resp., logarithmic density as dens E = lim sup r→∞

m(E(r)) , r−1

resp., log dens E = lim sup r→∞

https://doi.org/10.1515/9783110560565-001

l(E(r)) , log r

2 | 1 Introduction of Nevanlinna theory and its difference versions and similarly the corresponding lower densities by using lim inf instead of lim sup. It is not difficult to see that 0 ≤ dens E ≤ log dens E ≤ log dens E ≤ dens E ≤ 1. Moreover, if l(E) < ∞, resp., m(E) < ∞, then E is of zero upper density. Let f (z) be an arbitrary meromorphic function in ℂ. For r > 0, denote by n(r, f ) the cardinal number of the poles of f (z) in the disk |z| ≤ r, each pole being counted according to its multiplicity. Then the counting function of f (z) is defined as r

N(r, f ) := ∫ 0

n(t, f ) − n(0, f ) dt + n(0, f ) log r, t

(1.1)

which measures the average frequency of poles in the disk |z| ≤ r. If each pole is counted just once, independently of multiplicity, we will use the corresponding notation N(r, f ). For x > 0, we write log+ x = max{log x, 0}. The proximity function of f (z) is defined as 2π

m(r, f ) :=

1 ∫ log+ |f (reiθ )|dθ, 2π

(1.2)

0

which measures the average magnitude of log+ |f | on the circle |z| = r. 1 1 1 Natural definitions then follow for N(r, f −a ), N(r, f −a ), m(r, f −a ), where a ∈ ℂ. Then we define T(r, f ) := m(r, f ) + N(r, f ),

(1.3)

1 the characteristic function of f (z), and similarly T(r, f −a ). We easily see the following inequalities for meromorphic functions f (z), g(z):

m(r, αf + βg) ≤ m(r, f ) + m(r, g) + O(1), m(r, fg) ≤ m(r, f ) + m(r, g), T(r, αf + βg) ≤ T(r, f ) + T(r, g) + O(1), T(r, fg) ≤ T(r, f ) + T(r, g), where α, β ∈ ℂ. Let (aj ) and (bk ), respectively, be the zeros and poles of g(z) in the disk |z| < r, each repeated according to its multiplicity. By the Poisson–Jensen formula, for every z satisfying |z| < r, 2π

r 2 − |z|2 1 dθ log |g(z)| = ∫ log |g(reiθ )| ⋅ iθ 2π |re − z|2 0

󵄨󵄨 r(z − a ) 󵄨󵄨 󵄨󵄨 r(z − b ) 󵄨󵄨 󵄨 j 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 − ∑ log 󵄨󵄨󵄨󵄨 + ∑ log 󵄨󵄨󵄨󵄨 2 󵄨. 󵄨󵄨 r 2 − b̄ z 󵄨󵄨󵄨 󵄨󵄨 r − ā j z 󵄨󵄨󵄨 |b | 0, then the deficient value a is also called a Nevanlinna exceptional value. For a transcendental meromorphic function f (z), by Theorem 1.1.3, if δ(a, f ) < 1, then f (z) admits infinitely many a-points. If f (z) admits a finite number of a-points in ℂ only, then δ(a, f ) = 1. To count the multiple poles, we put n1 (r, f ) = ∑|τ|≤r (μ(τ) − 1), where μ(τ) denotes the multiplicity of a pole τ, and ∑|τ|≤r denotes the summation over all poles in the disk |z| ≤ r. Then r

N1 (r, f ) := ∫ 0

n1 (t, f ) − n1 (0, f ) dt + n1 (0, f ) log r t

(1.10)

measures the frequency of the multiple poles. Then, for a ∈ ℂ, we define ϑ(a, f ) := lim inf r→∞

1 N1 (r, f −a )

T(r, f )

(1.11)

and ϑ(∞, f ) := lim inf r→∞

N1 (r, f ) , T(r, f )

(1.12)

1.1 Basic notations and facts in the Nevanlinna theory | 5

which are called the ramification indexes of a and ∞, respectively. If all a-points are simple, then ϑ(a, f ) = 0, and if they are double, then ϑ(a, f ) ≤ 1/2. Let ϕ(r) be a function defined on an interval [r0 , +∞), where r0 > 1. We write ϕ(r) = S(r, f ) if ϕ(r) = o(T(r, f )) as r → +∞ outside a possible exceptional set of finite linear measure. Applying the differential operator (𝜕/𝜕x − i𝜕/𝜕y) to (1.4) (with g = f ), we obtain 2π

reiθ f 󸀠 (z) 1 = ∫ log |f (reiθ )| ⋅ dθ iθ f (z) π (re − z)2 0

+ ∑ ( |aj | 0, and let A ⊂ ℂ be a finite set of complex numbers. Then (k − 1)N(r, f ) + ∑ N1 (r, a∈A

1 1 ) ≤ N (r, (k) ) + εT(r, f ) f −a f

for all r > e outside a set E ⊂ [e, +∞) of logarithmic density 0, depending on f , k, ε, and A.

1.1 Basic notations and facts in the Nevanlinna theory | 7

In making use of logarithmic derivatives in connection of differential, difference, and delay-differential equations, the Gundersen estimates [71], frequently appear to be most useful. For the convenience of applications below, we recall the following propositions. Let H := {(k1 , j1 ), . . . , (kq , jq )} be a finite set of distinct pairs of integers such that ki > ji ≥ 0 for i = 1, . . . , q. Proposition 1.1.15. Given H, let f be of finite order ρ, and let ε > 0 be given. Then there exists a set E ⊂ (0, 2π) of linear measure zero such that if ψ ∈ [0, 2π) \ E, then there is a constant R = R(ψ) > 1 such that for z satisfying arg z = ψ and |z| ≥ R and for all (k, j) ∈ H, we have 󵄨󵄨 (k) 󵄨󵄨 󵄨󵄨 f (z) 󵄨󵄨 (k−j)(ρ−1+ε) 󵄨 󵄨󵄨 . 󵄨󵄨 f (j) (z) 󵄨󵄨󵄨 ≤ |z| 󵄨 󵄨 Proposition 1.1.16. Given H, let f be of finite order ρ, and let ε > 0. Then there exists a set E ⊂ (1, ∞) of finite logarithmic measure such that for all z satisfying |z| ∉ E ∪ [0, 1] and for all (k, j) ∈ H, we have 󵄨󵄨 (k) 󵄨󵄨 󵄨󵄨 f (z) 󵄨󵄨 (k−j)(ρ−1+ε) 󵄨 󵄨󵄨 . 󵄨󵄨 f (j) (z) 󵄨󵄨󵄨 ≤ |z| 󵄨 󵄨 Proposition 1.1.17. Given H, let f be of finite order ρ, and let ε > 0. Then there exists a set E ⊂ [0, ∞) of finite linear measure such that for all z satisfying |z| ∉ E and for all (k, j) ∈ H, we have 󵄨󵄨 (k) 󵄨󵄨 󵄨󵄨 f (z) 󵄨󵄨 (k−j)(ρ+ε) 󵄨 󵄨󵄨 . 󵄨󵄨 f (j) (z) 󵄨󵄨󵄨 ≤ |z| 󵄨 󵄨 For more details and corresponding estimates, the reader may consult [71]. In the next chapter, we will need a simple variant of Proposition 1.1.16, which follows by a minor modification of the corresponding proof in [71]: Proposition 1.1.18. Let f be a transcendental meromorphic function of finite order ρ, and let g be a meromorphic function such that for some λ < ρ, T(r, g) = O(r λ+ε ) + S(r, f ) outside a possible exceptional set of finite logarithmic measure. Then for all z such that |z| ∉ E ∪ [0, 1], where E is a set of finite logarithmic measure, and for all k > j, 󵄨󵄨 (k) 󵄨󵄨 󵄨󵄨 g (z) 󵄨󵄨 (k−j)(λ−1+ε) 󵄨 󵄨󵄨 . 󵄨󵄨 g (j) (z) 󵄨󵄨󵄨 ≤ |z| 󵄨 󵄨 The second main theorem in the Nevanlinna theory is stated as follows. Theorem 1.1.19. For an arbitrary nonconstant meromorphic function f (z) and distinct points a1 , . . . , aq ∈ ℂ, q ∈ ℕ, we have q

m(r, f ) + ∑ m(r, 1/(f − aj )) + N(r, 1/f 󸀠 ) + N1 (r, f ) ≤ 2T(r, f ) + S(r, f ). j=1

8 | 1 Introduction of Nevanlinna theory and its difference versions From this theorem we immediately obtain q

δ(∞, f ) + ϑ(∞, f ) + ∑ (δ(aj , f ) + ϑ(aj , f )) ≤ 2. j=1

This implies that for each n ∈ ℕ, the number of the points a ∈ ℂ such that δ(a, f ) ≥ 1/n (resp., ϑ(a, f ) ≥ 1/n) does not exceed 2n, and hence all points with a nonzero deficiency (resp., a nonzero ramification index) constitute a countable set. Thus we have the following corollary, a far-reaching generalization of the classical Picard theorem. Corollary 1.1.20. For an arbitrary nonconstant meromorphic function f (z), ∑ (δ(a, f ) + ϑ(a, f )) ≤ 2,

̂ = ℂ ∪ {∞}, ℂ

̂ a∈ℂ

̂ such that δ(a, f ) > 0 or ϑ(a, f ) > 0. where the summation ranges over all points in ℂ The second main theorem has another frequently applied important form. Theorem 1.1.21. Let f be a nonconstant meromorphic function, let q ≥ 2, and let a1 , . . . , aq be distinct complex constants. Then q

(q − 1)T(r, f ) ≤ N(r, f ) + ∑ N (r, k=1

1 ) + S(r, f ). f − ak

We also remark that the second main theorem is also true when constants are replaced by small functions; see [228, Corollary 1] for the general case and [90, Theorem 2.5] for three small functions, which we also will frequently apply in this book. We recall [90, Theorem 2.5]. Theorem 1.1.22. Let f be a nonconstant meromorphic function, and let a1 , a2 , a3 be distinct small meromorphic functions with respect to f . Then 3

T(r, f ) ≤ ∑ N (r, k=1

1 ) + S(r, f ). f − ak

The following lemma is due to Clunie [41]; see also Laine [120, p. 39–44]. Lemma 1.1.23. Let f be a transcendental meromorphic function such that f p+1 = Q(z, f ), p ∈ ℕ, where Q(z, u) is a polynomial in u and its derivatives with meromorphic coefficients aμ (z), μ ∈ M. Suppose that the total degree of Q(z, u) as a polynomial in u and its derivatives does not exceed p. Then m(r, f ) = O( ∑ m(r, aμ )) + S(r, f ). μ∈M

(1.15)

Moreover, if ρ(f ) < +∞, then the error term S(r, f ) can be replaced by O(log r) without exceptional set.

1.1 Basic notations and facts in the Nevanlinna theory | 9

For a more general version of the Clunie lemma in the differential setting, see [237, Theorem 1]. For the corresponding versions in the difference- and delay-differential settings, see the next section. The following lemma, due to Mohon’ko and Mohon’ko [173], see also Laine [120, Proposition 9.2.3], gives an estimate for the proximity function of the reciprocal of f (z) − c. Lemma 1.1.24. Let F(z, u) be a polynomial in u and its derivatives with meromorphic coefficients bκ (z), κ ∈ K. Suppose that f is a transcendental meromorphic function satisfying F(z, f ) = 0 and that c is a complex number. If F(z, c) ≢ 0, then m (r,

1 ) = O( ∑ T(r, bκ )) + S(r, f ). f −c κ∈K

(1.16)

Moreover, if ρ(f ) < +∞, then the error term S(r, f ) can be replaced by O(log r) without exceptional set. Remark 1.1.25. In the preceding two lemmas, S(r, f ) can be replaced by O(log(rT(r, f ))) as r → +∞ outside a possible exceptional set of finite linear measure. Proposition 1.1.26. For any three distinct meromorphic functions f , g, h, T(r, fg + gh + hf ) ≤ T(r, f ) + T(r, g) + T(r, h) + O(1). The proof of this proposition does not usually appear in standard references. An interested reader may consult [64, p. 281]. More generally, we obtain the following: Theorem 1.1.27. Given distinct meromorphic functions f1 , . . . fn , let {J} denote the collection of all nonempty subsets of {1, . . . , n}, and suppose that αJ ∈ ℂ for each J ∈ {J}. Then n

T(r, ∑ αJ ( ∏ fj )) ≤ ∑ T(r, fk ) + O(1). J

j∈J

k=1

Remark 1.1.28. By a straightforward modification we obtain the corresponding estimate as in the preceding theorem if the constant coefficients αJ are replaced by small meromorphic functions αJ (z). Theorem 1.1.29. Let f be a meromorphic function, and let R(z, f ) =

a0 (z) + a1 (z)f (z) + ⋅ ⋅ ⋅ + ap (z)f (z)p b0 (z) + b1 (z)f (z) + ⋅ ⋅ ⋅ + bq (z)f (z)q

be an irreducible rational in f (z) with meromorphic coefficients aj (z), bk (z) of small growth S(r, f ). Then T(r, R(z, f (z))) = max(p, q)T(r, f (z)) + S(r, f ).

10 | 1 Introduction of Nevanlinna theory and its difference versions This theorem is usually called the Valiron–Mohon’ko theorem. We suggest the reader to see the proof of this theorem in, for example, Laine [120, Theorem 2.2.5]. We shortly recall the definition of the exponent of convergence for the zeros λ(f ) as follows; for more details, see [120, p. 7–9] or [235, p. 96–98]. Let (aj )j∈ℕ be the nonzero zeros of f (z). Denote n(t) := card ((aj )j∈ℕ ∩ {|z| ≤ t}) and r

N(r) := ∫ 0

n(t) dt. t

Thus n(r, f1 ) = n(r) + n(0, f1 ) and N(r, f1 ) = N(r) + n(0, f1 ) log r. Then the exponent of convergence for the zeros of f is defined as 󵄨󵄨 ∞ 1 󵄨󵄨 ∞ n(t) { } 󵄨󵄨 󵄨 λ(f ) : = inf {α > 0 󵄨󵄨 ∑ < ∞} = inf {α > 0 󵄨󵄨󵄨 ∫ α+1 dt < ∞} α 󵄨󵄨 󵄨 |a | t 󵄨 j j=1 0 { } log n(r) log N(r) = lim sup = lim sup log r log r r→∞ r→∞ = lim sup r→∞

log n (r, f1 ) log r

= lim sup r→∞

log N (r, f1 ) log r

.

Natural definitions then follow for the exponent of convergence for the poles λ( f1 ) and the exponent of convergence for a-points λ(f − a), where a is a constant or a small function. The Weierstrass factorization theorem can be stated as follows; see [120, p. 7]. Theorem 1.1.30. Let f be an entire function with a zero of multiplicity k ≥ 0 at z = 0. Then f (z) has the representation f (z) = z k P(z)eQ(z) , where P(z) is the canonical product of f (z) formed with the nonzero zeros of f (z), and Q(z) is an entire function. The corresponding version on entire functions of finite order is called the Hadamard factorization theorem; see [235, p. 101–103]. Theorem 1.1.31. Let f be an entire function of finite order ρ(f ) with a zero of multiplicity k ≥ 0 at z = 0. Then f (z) = z k P(z)eQ(z) , where P(z) is the canonical product of f (z) formed with the nonzero zeros of f (z), and Q(z) is a polynomial of deg(Q(z)) ≤ ρ(f ). To close this preliminary section, we recall three results on the combinations of meromorphic functions to be applied in subsequent chapters. The first one is due to Borel, which we further call the Borel theorem; see also [235, Theorem 1.52]. Theorem 1.1.32. Let fj (z), j = 1, 2, . . . , n (n ≥ 2) be meromorphic functions, and let gj (z), j = 1, 2, . . . , n, be entire functions satisfying

1.2 Difference Nevanlinna theory | 11

(i) ∑nj=1 fj (z)egj (z) ≡ 0, (ii) the orders of fj are less than those of egh −gk for 1 ≤ j ≤ n, 1 ≤ h ≤ k ≤ n. Then fj (z) ≡ 0, j = 1, 2, . . . , n. The following two theorems can be found in [235, Theorems 1.51 and 1.62]. Theorem 1.1.33. Let fj (z), j = 1, 2, . . . , n, n ≥ 2, be meromorphic functions, and let gj (z), j = 1, 2, . . . , n, be entire functions satisfying (i) ∑nj=1 fj (z)egj (z) ≡ 0, (ii) gj (z) − gk (z) are not constant for 1 ≤ j < k ≤ n, (iii) T(r, fj ) = o(T(r, egh −gk )) (r → ∞, r ∈ ̸ E) for 1 ≤ j ≤ n, 1 ≤ h < k ≤ n, where E ⊂ (1, ∞) is a set of finite linear measure or finite logarithmic measure. Then fj (z) ≡ 0, j = 1, 2, . . . , n. Theorem 1.1.34. Let fj (z) be meromorphic functions, and let fk (z), k = 1, 2, . . . , n − 1, be nonconstant and satisfy ∑nj=1 fj = 1 and n ≥ 3. If fn (z) ≢ 0 and n n 1 ∑ N (r, ) + (n − 1) ∑ N(r, fj ) < (λ + o(1))T(r, fk ), fj j=1 j=1

where λ < 1, then fn (z) ≡ 1.

1.2 Difference Nevanlinna theory In this section, we consider the difference variants of the Nevanlinna theory, including additional material pointing out how the difference variant results are to be modified if a delay-differential situation is to be considered instead of the difference situation. We divide this section into three subsections.

1.2.1 Elementary difference considerations in the complex domain Looking at the shifts and differences of meromorphic functions, as well as at their value distribution, we may frequently have to apply elementary difference considerations. For such elementary expressions and results, classical references are [114, 172, 179], and [219]. Let f (z) be a meromorphic function. Given a complex number c ≠ 0, the c-shift of f is the function obtained by composing f with the parallel c-shift z 󳨃→ z + c, i. e., f (z + c), whereas the c-difference of f is Δc f (z) := f (z + c) − f (z). If c = 1, then we use the notations Ef (z) := f (z + 1) and Δf (z) := f (z + 1) − f (z). As to the iterated differences,

12 | 1 Introduction of Nevanlinna theory and its difference versions first, recall the classical identities k k E k f = E(E k−1 f ) = ( ∑ ( ) Δk−j f (z)) j j=0

and k k f (z + k − j) Δk f = Δ(Δk−1 f ) = ( ∑ (−1)j ( ) ) f (z); j f (z) j=0

see, e. g., [114]. The following trivial identities may also be sometimes useful: Δs (Δt f (z)) = Δs+t f (z),

Δ(C1 f (z) + C2 g(z)) = C1 Δf (z) + C2 Δg(z),

(1.17) C1 , C2 ∈ ℂ,

Δ(f (z)g(z)) = f (z)Δg(z) + Eg(z)Δf (z), Δ(

g(z)Δf (z) − f (z)Δg(z) f (z) )= . g(z) g(z)Eg(z)

(1.18) (1.19) (1.20)

For a number of other potentially useful identities, see, e. g., [114], Chapter 2.

1.2.2 Difference variants of the logarithmic derivative lemma for meromorphic functions Value distribution of shifts and differences of meromorphic functions have been under frequent considerations during the last two decades. An apparent background to this interest has been the general interest to discrete Painlevé equations in mathematical physics and mathematics. Some of the first papers to consider value distribution of shifts and differences of meromorphic functions were [82] and [38]. Both these papers offer the key estimate to the proximity function m(r, f (z + c)/f (z)) of the difference quotient. Such an estimate, given in [82, Theorem 2.1], reads as follows. Proposition 1.2.1. Let f be a nonconstant meromorphic function, and let c ∈ ℂ, δ < 1, and ε > 0. Then m (r,

f (z + c) T(r + |c|, f )1+ε ) = o( ) f (z) rδ

for all r outside a possible exceptional set of finite logarithmic measure. For the convenience of the reader, we repeat the proof from [82], as the reasoning in this proof is the key to obtain a number of similar results corresponding in the difference setting to the lemma on the logarithmic derivative.

1.2 Difference Nevanlinna theory | 13

Proof. Let ξ (x) and ϕ(r) be positive nondecreasing continuous functions defined for e ≤ x < ∞ and r0 ≤ r < ∞, and assume that T(r + |c|, f ) ≥ e for all r ≥ r0 . Then, by a standard Borel lemma-type argument given in [36, Lemma 3.3.1], T (r + |c| +

ϕ(r) , f ) ≤ 2T(r + |c|, f ) ξ (T(r + |c|, f ))

for all r outside a set E satisfying ∫ E∩[r0 ,R]

1 1 dr ≤ + ϕ(r) ξ (e) log 2

T(R+|c|,f )

∫ e

dx xξ (x)

for R < ∞. Taking now ϕ(r) = r and ξ (x) = xε/2 , and fixing α := 1 +

r , (r + |c|)T(r + |c|, f )ε/2

we obtain T(α(r + |c|), f ) = T (r + |c| +

ϕ(r) , f ) ≤ 2T(r + |c|, f ). ξ (T(r + |c|, f ))

Applying now Lemma 2.3 in [82], see below, the claim follows whenever f (0) ≠ 0, ∞. As for the remaining case where either f (0) = 0 or f (0) = ∞, we may consider z p f (z) for a suitable integer p. Lemma 1.2.2. Let f be a meromorphic function, and let c ∈ ℂ. If f (0) ≠ 0, ∞, then for all α > 1, δ < 1, and r ≥ 1, m (r,

K(α, δ, c) 1 f (z + c) )≤ (T(α(r + |c|), f ) + log+ ), f (z) |f (0)| rδ

where K(α, δ, c) is a constant depending on α, δ, c only: K(α, δ, c) :=

8|c|(3α + 1) + 8α(α − 1)|c|δ . δ(1 − δ)(α − 1)2

On the other hand, [38, Theorem 2.4] gives a similar estimate, which may be shortly written as the following: Proposition 1.2.3. Let α, R, R󸀠 be real numbers such that 0 < α < 1 and 0 < R < R󸀠 , and let c be a nonzero complex number. Then there are positive constants C1 , C2 , depending on c and α only, such that for a meromorphic function f , we have m (r,

f (z + c) f (z) ) + m (r, ) f (z) f (z + c)

≤ C1 (m(R, f ) + m(R, 1/f )) + C2 (N(R󸀠 , f ) + N(R󸀠 , 1/f )) .

14 | 1 Introduction of Nevanlinna theory and its difference versions The proof of this proposition follows by a careful analysis of the Poisson–Jensen formulas for f (z) and f (z + c); see [38, p. 117–120]. Remark 1.2.4. If f is a meromorphic function of finite order ρ and 0 < δ < 1, then from Proposition 1.2.1 it immediately follows that m (r,

f (z + c) ) = o(r ρ−δ+ε ) f (z)

(1.21)

for all r outside an exceptional set of finite logarithmic measure. Similarly, m (r,

f (z + c) ) = O(r ρ−1+ε ) f (z)

(1.22)

without an exceptional set; see [38, Corollary 2.5]. | instead of Remark 1.2.5. Sometimes it may be useful to consider the quotient | f (z+c) f (z) the proximity function. As to this end, recall a double inequality in [38, Theorem 8.2]. Suppose that f is of finite order ρ and ε > 0. Then 󵄨󵄨 f (z + c) 󵄨󵄨 󵄨󵄨 ≤ exp(r ρ−1+ε ) exp(−r ρ−1+ε ) ≤ 󵄨󵄨󵄨󵄨 󵄨 󵄨 f (z) 󵄨󵄨

(1.23)

outside a possible exceptional set of finite logarithmic measure. Remark 1.2.6. A much more precise conclusion follows, provided that f is a transcendental meromorphic function of order ρ < 1; see [14]. To this end, recall that E is an ε-set if it is a countable union of discs not containing the origin, whose centers tend to infinity, whose radii have a finite sum, and the set of r ≥ 1 for which the circle S(0, r) meets E has finite logarithmic measure. Fix h > 0. Then there exists an ε-set E such that lim

z∈ℂ\E

f (z + c) =1 f (z)

uniformly in c for |c| ≤ h. If f is of infinite order, then the estimates in Remark 1.2.4 cannot be obtained in general. As a simple example, take f (z) := exp(exp(z)). Then m (r,

f (z + 1) ) = m(r, (e − 1)f (z)) = (e − 1)T(r, f ). f (z)

However, if the order of growth of f is suitably restricted, then estimates for the proximity functions of difference quotients may be obtained. In particular, if the hyperorder ρ2 = ρ2 (f ) < 1, then an estimate corresponding to (1.21) easily follows as shown by Halburd, Korhonen, and Tohge [87, Theorem 5.1]: m (r,

f (z + c) T(r, f ) ) = o ( 1−ρ −ε ) f (z) r 2

(1.24)

1.2 Difference Nevanlinna theory | 15

for all r outside an exceptional set of finite logarithmic measure. Another similar estimate appears in [118, Lemma 3.1]: If ν > 0, 0 < ε < ν, and log T(r, f ) ≤ r/(log r)2+ν for all r sufficiently large, then m (r,

T(r, f ) f (z + c) ) = o( ) f (z) (log r)ν−ε

(1.25)

for all r outside an exceptional set of finite logarithmic measure. A number of variants to these estimates may be found in the recent literature. As an example, recall Lemma 2.2 in [117]: Lemma 1.2.7. Let f be a nonrational meromorphic function, and let ω(z) := cz n + pn−1 z n−1 + ⋅ ⋅ ⋅ + p0 and φ(z) := cz n + qn−1 z n−1 + ⋅ ⋅ ⋅ + q0 be nonconstant polynomials. If lim sup r→∞

then m (r,

log log T(r, f ) 1 < 2, log r n

f ∘ω ) = o(T(|c|r n , f )) f ∘φ

outside a possible exceptional set of finite logarithmic measure. In particular, if n = c = 1, then we may write ω(z) = z + c1 and φ(z) = z + c2 , which results in f (z + c1 ) ) = S(r, f ) (1.26) m( f (z + c2 )

for all r outside an exceptional set of finite logarithmic measure, provided that the hyperorder ρ2 (f ) < 1, Finally, observe that delay-differential counterparts to the results described above easily follow from the obvious inequality m (r,

f (k) (z + c) f (k) (z + c) f (z + c) ) ≤ m (r, ) + m (r, ). f (z) f (z + c) f (z)

For our later use, recall two such consequences. Lemma 1.2.8. Let f be a transcendental meromorphic function of hyperorder ρ2 (f ) < 1. Then m (r,

f (k) (z + c) ) = S(r, f ) f (z)

(1.27)

outside a possible exceptional set of finite logarithmic measure. Lemma 1.2.9. Let f be a transcendental meromorphic function of hyperorder ρ2 (f ) < 1, and let P(z, f ) be a linear delay-differential polynomial in f with small coefficients. Then m (r,

P(z, f ) ) = S(r, f ) f (z)

outside a possible exceptional set of finite logarithmic measure.

(1.28)

16 | 1 Introduction of Nevanlinna theory and its difference versions 1.2.3 Key notions and results in difference Nevanlinna theory In this section, we collect some results of the difference Nevanlinna theory related to the two Nevanlinna main theorems and their basic applications. As to our knowledge, Dugué [46] was the first one to consider the dependence of Nevanlinna theory notions (in fact, dependence of defects) of shifts in the argument z. The example f (z) :=

exp(2πiez − 1) exp(2πie−z − 1)

proposed by Dugué, where the defects depend on the choice of the origin, is of infinite order (and, of course, of hyperorder ρ2 (f ) = 1). A complete difference variant of Nevanlinna theory was obtained, quite recently, by Halburd and Korhonen [83]. To explain the main theorems in the difference Nevanlinna theory, observe first that the difference variants to the first main theorem in fact are immediate consequences of the following real analysis lemma; see Lemma 8.3 in [87] and its proof therein. Lemma 1.2.10. Let T : [0, +∞) → [0, +∞) be nondecreasing continuous, and let c ∈ (0, +∞). If the hyperorder ρ2 = ρ2 (T) := lim sup r→∞

log log T(r) 0 such that λ+ε < ρ outside a possible exceptional set of finite logarithmic measure. Suppose that the total degree deg U(z, f ) = n and deg Q(z, f ) ≤ n, both total degrees in f and its shifts. Moreover, assume that U(z, f ) contains just one term of maximal total degree in f and its shifts. Then, for each ε > 0, m(r, P(z, f )) = O(r ρ−1+ε ) + S(r, f ). Looking at the proof of Lemma 1.2.15, see [123] and [125], note the key observation in the proof while using a completely similar reasoning to prove the two lemmas above: Denoting, as we may, U(z, f ) = ∑nj=0 c̃j (z)f (z)j , we have m(r, 1/c̃j ) = O(r ρ−1+ε ) + S(r, f ) in addition to our assumption that m(r, c̃j ) = O(r ρ−1+ε ) + S(r, f ) for all j = 0, . . . , n. As m(r, 1/f ) may not be small although m(r, f ) is small, we have to be careful when trying to construct delay-differential variants of the Clunie lemma. This obstacle may immediately be overcome if P(z, f ), Q(z, f ) are delay-differential polynomials, provided that U(z, f ) is a difference polynomial. As an example of such a delaydifferential variant of the Clunie lemma, recall the following [100]: Lemma 1.2.16. Let f be a transcendental meromorphic solution of hyperorder ρ2 < 1 of a delay-differential equation of the form U(z, f )P(z, f ) = Q(z, f ), where U(z, f ) is a difference polynomial, and P(z, f ), Q(z, f ) are delay-differential polynomials, all three with small coefficients such that the total degree deg U(z, f ) = n and deg Q(z, f ) ≤ n. Moreover, assume that U(z, f ) contains just one term of maximal total degree in f and its shifts. Then m(r, P(z, f )) = S(r, f ).

20 | 1 Introduction of Nevanlinna theory and its difference versions Observe that the proof of this lemma [100, p. 135–137] is, again, essentially the same as the proof of Lemma 1.2.14 in [123]. Another obstacle to obtain other variants of the difference Clunie lemma is that U(z, f ) contains more than one term of maximal total degree. A simple example provided in [123] is f (z) = sin z, which satisfies (f (z)2 + f (z + π/2)2 )f (z) = −f (z + π). We now have U(z, f ) = c̃2 (z)f (z)2 = (1 + (f (z +

2

π )/f (z)) )f (z)2 , 2

and hence m(r, 1/c̃2 ) = m(r, sin2 (z)) = O(r). If U(z, f ) contains more than one term of maximal total degree, Lemma 1.2.15 still holds, provided that the sum S(z) of terms in U(z, f ) of maximal total degree does not vanish identically and, additionally, N(r, f )+N(r, 1/f ) = O(r ρ−1+ε )+S(r, f ) (in the finiteorder case). For such a situation, recall the following [102]: Lemma 1.2.17. Let f be a transcendental meromorphic solution of finite order ρ of a difference equation of the form U(z, f )P(z, f ) = Q(z, f ), where U(z, f ), P(z, f ), Q(z, f ) are difference polynomials with small coefficients such that the total degree of deg U(z, f ) = n and deg Q(z, f ) ≤ n. Moreover, assume that the sum of terms of maximal total degree in U(z, f ) in f and its shifts does not vanish identically. Moreover, suppose that N(r, f ) + N(r, 1/f ) = O(r ρ−1+ε ) + S(r, f ). Then, for each ε > 0, m(r, P(z, f )) = O(r ρ−1+ε ) + S(r, f ). Again, under assumptions as in [102], a similar delay-differential variant may be obtained, provided that U(z, f ) is a difference polynomial, whereas P(z, f ), Q(z, f ) are allowed to be delay-differential polynomials.

2 Value distribution of complex delay-differential polynomials In this chapter, we present some difference and delay-differential variants of some classical results on the zeros of complex differential polynomials.

2.1 Value distribution of difference polynomials This section contains a number of difference versions of Hayman’s result in [89] and their generalizations. One of Hayman’s results ([89, Theorem 10]) therein was stated as follows. Theorem 2.1.1. If f (z) is a transcendental entire function and n ≥ 2 is a positive integer, then f (z)n f 󸀠 (z) − a has infinitely many zeros for any nonzero constant a. Theorem 2.1.1 was the starting point to considering value distribution of complex differential polynomials. Recall that Clunie [42] proved that Theorem 2.1.1 is also true for n = 1. If f (z) is a transcendental meromorphic function, the conclusion is valid for n ≥ 3 by Hayman [89, Corollary to Theorem 9], for n = 2 by Mues [175], and for n = 1 by Bergweiler and Eremenko [12, Theorem 2], Chen and Fang [21], and Zalcman [240]. Recall that some uniqueness problems [234] and normal family problems [128, 182] were also inspired by Theorem 2.1.1. The shift operators f (z + c) or difference operators Δc f := f (z + c) − f (z), where c is a nonzero constant, can be seen as discrete versions of f 󸀠 (z). Laine and Yang [122, Theorem 2] were the first who considered the zeros of f (z)n f (z + c) − a, where a is a nonzero constant, which can be viewed as a difference analogue to Theorem 2.1.1: Theorem 2.1.2. Let f (z) be a transcendental entire function of finite order, and let c be a nonzero complex constant. Then for n ≥ 2, f (z)n f (z + c) takes every nonzero value a ∈ C infinitely often. The following results are devoted to some improvements of Theorem 2.1.2, restricted to the difference framework, whereas the more general delay-differential case will be treated in the next section. The difference analogue of the logarithmic derivative lemma is of essential importance in these considerations. For the difference analogue of the logarithmic derivative lemma on meromorphic functions with finite order or hyperorder less than one, see Chapter 1.2.2. After that, many researchers started to consider the zeros of complex difference polynomials of different types. For some investigations on improving Theorem 2.1.2, see, e. g., [122, 140, 143, 152, 168, 244]. We start with a slight extension to Theorem 2.1.2. https://doi.org/10.1515/9783110560565-002

22 | 2 Value distribution of complex delay-differential polynomials Theorem 2.1.3. Let f (z) be a transcendental entire function of hyperorder ρ2 (f ) < 1. If n ≥ 2, then f (z)n f (z + c) − a(z) has infinitely many zeros for any nonzero small function a(z) with respect to f (z). Remark 2.1.4. Theorem 2.1.3 is not true when n = 1. We can see this by looking at izπ f (z) = a + e c with a(z) ≡ a2 , where a is a nonzero constant. Also, Theorem 2.1.3 is in general not valid for transcendental entire functions of hyperorder ρ2 (f ) = 1. As an z example, consider f (z) = ee , ec = −n, and a(z) = −p(z)ez + 1, where p(z) is a nonzero polynomial. Then f (z)n f (z + c) − a(z) = p(z)ez has finitely many zeros only. To prove Theorem 2.1.3, we need an estimate on the characteristic function of f (z)n f (z + c). Lemma 2.1.5. Let f (z) be a transcendental entire function of hyperorder ρ2 (f ) < 1. If n ≥ 1, then T(r, f (z)n f (z + c)) = (n + 1)T(r, f ) + S(r, f ).

(2.1)

Proof. Write first f (z)n+1 = f (z)n f (z + c)

f (z) . f (z + c)

Since f is a transcendental entire function, by standard arguments offered in Chapter 1 we see that (n + 1)T(r, f ) = (n + 1)m(r, f ) = m (r, f (z)n+1 ) + S(r, f ) f (z) ) + S(r, f ) f (z + c) ≤ T(r, f (z)n f (z + c)) + S(r, f ). ≤ m(r, f (z)n f (z + c)) + m (r,

(2.2)

On the other hand, T(r, f (z)n f (z + c)) ≤ nT(r, f ) + T(r, f (z + c)) + S(r, f ) ≤ (n + 1)T(r, f ) + S(r, f ).

(2.3)

Thus (2.1) follows from (2.2) and (2.3). Proof of Theorem 2.1.3. Denote G(z) := f (z)n f (z + c). By Lemma 2.1.5 G(z) is not a constant. Assume that G(z) − a(z) has finitely many zeros only. From the second main theorem for three small functions (Theorem 1.1.22) we get 1 1 ) + N (r, ) + S(r, G) G G − a(z) 1 1 ≤ N (r, ) + N (r, ) + S(r, f ) f f (z + c) ≤ 2T(r, f ) + S(r, f ).

T(r, G) ≤ N(r, G) + N (r,

If n ≥ 2, then we have a contradiction from Lemma 2.1.5, verifying the claim.

(2.4)

2.1 Value distribution of difference polynomials |

23

Next, let n be a positive integer, let P(z) = an z n + an−1 z n−1 + ⋅ ⋅ ⋅ + a1 z + a0 be a nonzero polynomial, where a0 , a1 , . . . , an (≠ 0) are complex constants, and let t be the number of distinct zeros of P(z). Luo and Lin [168, Theorem 1] improved Theorem 2.1.3 as follows. Theorem 2.1.6. Let f (z) be a transcendental entire function of hyperorder ρ2 (f ) < 1. If n > t, then P(f )f (z + c) − a(z) has infinitely many zeros for any nonzero small function a(z) with respect to f (z). Remark 2.1.7. The restriction n > t in Theorem 2.1.6 is indispensable. Indeed, the theorem is not true if n = t = 1; for example, if f (z) = ez +1 and ec = −1, then f (z)f (z+c)−1 = −e2z has no zeros. More generally, for any positive integers n, t satisfying n = t ≥ 2, Theorem 2.1.6 fails by taking f (z) = e1z +1, ec = −1, and P(z) = (z −1− d1 ) ⋅ ⋅ ⋅ (z −1− d1 ), where 1

n

di ≠ 1, i = 1, 2, . . . , n, are the distinct zeros of z n+1 − 1 = 0. Then P(f )f (z + c) − 1 = has no zeros.

−1 e(n+1)z

To extend Theorem 2.1.3 to the case of meromorphic functions, Liu, Liu, and Cao [142, Theorem 1.2] considered the case of meromorphic functions of finite order and n ≥ 6. With an idea of common zeros and poles, which will be given in detail in the next section, Wang and Ye [212] improved the condition n ≥ 6 to n ≥ 4 for meromorphic functions of finite order. Their results remain valid for the meromorphic functions of hyperorder less than one as well. We omit the proofs of the following results, as these theorems are just particular cases of the corresponding results for delay-differential polynomials; see the next section. Theorem 2.1.8. Let f (z) be a transcendental meromorphic function of hyperorder ρ2 (f ) < 1, and let c be a nonzero complex constant. If n ≥ 4, then the difference polynomial f (z)n f (z + c) − α(z) has infinitely many zeros for any nonzero small function α(z) with respect to f (z). Remark 2.1.9. The assumption ρ2 (f ) < 1 cannot be removed. To see this, take f (z) = 1 c z , e = −n (n ≥ 4), a nonconstant polynomial P(z), and a nonzero rational funcP(z)ee tion R(z). Then f (z) is of hyperorder ρ2 (f ) = 1 and has finitely many poles, whereas f (z)n f (z + c) − R(z) =

1 − P(z)n P(z + c)R(z) P(z)n P(z + c)

has finitely many zeros only. For n = 2 and n = 3, we can find counterexamples to Theorem 2.1.8. For example, z z take f (z) = eez −1 and c = iπ. Then the functions f (z)2 f (z + c) = eez −1 and f (z)3 f (z + c) = +1 +1 z

( eez −1 )2 avoid the value −1. +1

24 | 2 Value distribution of complex delay-differential polynomials If f (z) is a transcendental meromorphic function, then we can state a corresponding result to Theorem 2.1.6. Theorem 2.1.10. Let f (z) be a transcendental meromorphic function of hyper-order ρ2 (f ) < 1. If n ≥ t + 3, then P(f )f (z + c) − a(z) has infinitely many zeros for any nonzero small function a(z) with respect to f (z). Replacing f (z+c) in the preceding results by more general linear shift polynomials k

g(f ) := ∑ bj (z)f (z + cj ), j=1

where the coefficients b1 (z), . . . , bk (z) are small meromorphic functions with respect to f (z), and c1 , . . . , ck are nonzero distinct complex constants, the following result can be found in [124]. Theorem 2.1.11. Let f be a transcendental entire function of finite order ρ, let b0 be a nonvanishing small meromorphic function with respect to f , let the linear shift polynomial g(f ) be nonvanishing, and let n ≥ 2, s ≥ 1. Then F := f n g(f )s − b0 has sufficiently many zeros to satisfy λ(F) = ρ, where λ is the exponent of convergence for the zeros.

2.2 Value distribution of delay-differential polynomials For the convenience of the reader, we first look at some simple results in this direction, before proceeding to more general considerations. To start with, we consider the zeros of complex delay-differential polynomials of type f (z)n f (k) (z + c) − a(z), where k and n are positive integers. Liu, Liu, and Zhou [160] proved the following theorem. Theorem 2.2.1. Let f (z) be a transcendental meromorphic function of hyper-order ρ2 (f ) < 1, and a(z) be a nonzero small function with respect to f (z). If n ≥ 2k + 6 or if n ≥ 3 and f (z) is transcendental entire, then f (z)n f (k) (z + c) − a(z) has infinitely many zeros. To prove Theorem 2.2.1, we need a simple estimate of the characteristic function of f (z)n f (k) (z + c): Lemma 2.2.2. Let f (z) be a transcendental meromorphic function of ρ2 (f ) < 1 and denote F(z) := f (z)n f (k) (z + c). Then (n − k − 1)T(r, f ) + S(r, f ) ≤ T(r, F) ≤ (n + k + 1)T(r, f ) + S(r, f ).

(2.5)

Moreover, if f (z) is transcendental entire, then nT(r, f ) + S(r, f ) ≤ T(r, F) ≤ (n + 1)T(r, f ) + S(r, f ).

(2.6)

2.2 Value distribution of delay-differential polynomials | 25

Proof. Observe that 1 1 f (k) (z + c) = . f (z)n+1 F(z) f (z) Recalling Chapter1, by elementary reasoning, including the standard Valiron–Mohon’ko theorem and the fact that ρ2 (f ) < 1, we obtain (n + 1)T(r, f ) = T (r, ≤ T (r,

1 ) + S(r, f ) f (z)n+1

f (k) (z + c) 1 ) + T (r, ) + S(r, f ) F f (z)

f (k) (z + c) ) + S(r, f ) f (z) 1 ≤ T(r, F) + N(r, f ) + kN(r, f ) + N (r, ) + S(r, f ) f ≤ T(r, F) + (k + 2)T(r, f ) + S(r, f ).

≤ T(r, F) + N (r,

(2.7)

On the other hand, T(r, F) ≤ nT(r, f ) + T(r, f (k) (z + c)) + S(r, f ) ≤ nT(r, f ) + T(r, f ) + kN(r, f ) + S(r, f ) ≤ (n + k + 1)T(r, f ) + S(r, f ).

(2.8)

Thus (2.5) follows from (2.7) and (2.8). If f is transcendental entire with ρ2 (f ) < 1, then (n + 1)T(r, f ) = T (r, f (z)n+1 ) = m (r, f (z)n+1 ) ≤ m (r,

F(z)f (z) ) f (k) (z + c)

f (k) (z + c) ) + S(r, f ) f (z) 1 ≤ T(r, F) + N (r, ) + S(r, f ) f ≤ T(r, F) + T(r, f ) + S(r, f ). ≤ T(r, F) + T (r,

(2.9)

On the other hand, the estimate T(r, F) ≤ (n + 1)T(r, f ) + S(r, f ) trivially follows from (2.8). Proof of Theorem 2.2.1. Let F(z) = f (z)n f (k) (z + c), which is not a constant by Lemma 2.2.2. Assume that F(z) − a(z) has finitely many zeros only. From the second main theorem for three small functions (Theorem 1.1.22) we have T(r, F) ≤ N(r, F) + N (r,

1 1 ) + N (r, ) + S(r, F) F F − a(z)

26 | 2 Value distribution of complex delay-differential polynomials 1 1 ≤ N(r, f ) + N(r, f (k) (z + c)) + N (r, ) + N (r, (k) ) + S(r, f ) f f (z + c) 1 ) + S(r, f ). ≤ 3T(r, f ) + N (r, (k) f (z + c)

(2.10)

To estimate the second term on the right-hand side in (2.10), recalling Lemma 1.2.10 and Corollary 1.1.12, we obtain N (r,

1 1 ) ≤ N (r, ) + kN(r, f (z + c)) + S(r, f ) f (z + c) f (k) (z + c) 1 ≤ N (r, ) + kN(r, f ) + S(r, f ). f

(2.11)

Combining (2.10) and (2.11), we get T(r, F) ≤ (k + 4)T(r, f ) + S(r, f ). Since f (z) is a transcendental meromorphic function, by Lemma 2.2.2 we obtain (n − k − 1)T(r, f ) ≤ (k + 4)T(r, f ) + S(r, f ), contradicting with n ≥ 2k + 6. If f (z) is a transcendental entire function, then we can get nT(r, f ) ≤ T(r, F) ≤ 2T(r, f ) + S(r, f ), which is a contradiction to n ≥ 3. Remark 2.2.3. (1) The condition ρ2 (f ) < 1 cannot be removed in Theorem 2.2.1. This z can be seen by looking at f (z) = ee of ρ2 (f ) = 1. Then f (z)n f 󸀠 (z+c)+nez +P(z) = P(z) has finitely many zeros, provided that ec = −n and P(z) is a nonzero polynomial. In fact, for any integer k, we can choose an appropriate α(z) to make f (z)n f (k) (z + c) + α(z) + P(z) = P(z), where α(z) is a suitable polynomial in ez . (2) The condition that a(z) is a nonzero small function cannot be removed in Theorem 2.2.1. Indeed, take f (z) = ez and ec = 2. Then f (z)n f (k) (z + c) = 2e(n+1)z has no zeros. We return back to this situation later on in this section. The condition n ≥ 2k + 6 in Theorem 2.2.1 has been improved to n ≥ k + 4 by making use of common zeros and poles; see Wang and Ye [212]. We will first explain the basic idea of common zeros and poles, to be applied in the next theorem. Let f , g be nonconstant meromorphic functions. Denote by n0 (r), resp., n1 (r), be the number of the common zeros, resp., poles, of fg and g, ignoring multiplicities. Let p, q be positive integers. We assume that the Laurent series of f , g at z0 are as follows: f (z) =

1 f (z), (z − z0 )p 1

g(z) = (z − z0 )q g1 (z),

2.2 Value distribution of delay-differential polynomials | 27

where f1 (z) and g1 (z) are analytic functions at z0 such that f1 (z0 ) ≠ 0 and g1 (z0 ) ≠ 0. So z0 is a zero of g(z) with multiplicity q. If q > p, then z0 is a zero of f (z)g(z) with multiplicity q − p, and thus the contribution to n0 (r) is 1 at z0 . If q ≤ p, then z0 is a pole of f (z)g(z) with multiplicity p − q or an analytic point of f (z)g(z), and thus the contribution to n0 (r) is 0 at z0 . A similar method can be applied to n1 (r). The other cases can be discussed in a similar way. As usual, define by N 0 (r), resp., N 1 (r), the counting function of the common zeros, resp., poles, of fg and g, ignoring multiplicities. Thus we have N (r,

1 1 ) ≤ N (r, ) + N 0 (r) fg f

and N(r, fg) ≤ N(r, f ) + N 1 (r). Now we provide a delay-differential version of Theorem 2.1.10: Theorem 2.2.4. Let f (z) be a transcendental meromorphic function of hyperorder ρ2 (f ) < 1, and let a(z) be a nonzero small function with respect to f (z). Suppose that P(f ) is a polynomial in f of degree n, possessing t distinct zeros. If n ≥ k + t + 3, then P(f )f (k) (z + c) − a(z) has infinitely many zeros. Proof. Denote H(z) := P(f )f (k) (z + c) − a(z). Then H(z) + a(z) ) f (k) (z + c) 1 ≤ m(r, H + a) + m ( (k) ) + O(1), f (z + c) H(z) + a(z) ) nN(r, f ) + S(r, f ) = N(r, P(f )) = N ( (k) f (z + c) 1 ≤ N(r, H + a) + N ( (k) ) − N 0 (r) − N 1 (r), f (z + c)

nm(r, f ) + S(r, f ) = m(r, P(f )) = m (

(2.12)

(2.13)

where N 0 (r), resp., N 1 (r), stands for the distinct common zeros, resp., poles, of H + a and f (k) (z + c). Using (2.12) and the first main theorem, we get nT(r, f ) ≤ T(r, H + a) + T (r,

1 ) − N 0 (r) − N 1 (r) + S(r, f ). f (k) (z + c)

(2.14)

Moreover, we have N(r, H + a) ≤ N(r, P(f )) + N 1 (r) ≤ T(r, f ) + N 1 (r), 1 1 N (r, ) ≤ N (r, ) + N 0 (r). H +a P(f )

(2.15) (2.16)

28 | 2 Value distribution of complex delay-differential polynomials By the second main theorem for three small functions (Theorem 1.1.22), from (2.15) and (2.16) we obtain 1 1 ) + N (r, ) + S(r, H) H +a H 1 1 ≤ T(r, f ) + N 1 (r) + N (r, ) + N 0 (r) + N (r, ) + S(r, f ). P(f ) H

T(r, H + a) ≤ N(r, H + a) + N (r,

Combining (2.14) with this inequality, we first obtain (n − 1)T(r, f ) ≤ T (r,

1

f (k) (z

+ c)

) + N (r,

1 1 ) + N (r, ) + S(r, f ), P(f ) H

and hence (n − 1)T(r, f ) ≤ m(r, f (k) (z + c)) + N(r, f (k) (z + c)) + N (r, + N (r, Writing f (k) (z + c) =

1 ) + S(r, f ). H

f (k) (z+c) (k) f (z) f (k) (z)

1 ) P(f )

and using (1.24) and Lemma 1.2.10, we obtain

(n − 1)T(r, f ) ≤ m(r, f (k) (z)) + N(r, f (k) (z)) + N (r,

1 1 ) + N (r, ) + S(r, f ). P(f ) H

Hence, by Corollary 1.1.9, (n − 1)T(r, f ) ≤ (k + 1)T(r, f ) + N (r,

1 1 ) + N (r, ) + S(r, f ). P(f ) H

Since by assumption P(f ) has t distinct zeros, we now easily conclude that (n − k − t − 2)T(r, f ) ≤ N (r,

1 ) + S(r, f ). H

Hence, if n ≥ k + t + 3, then H must have infinitely many zeros. Thus n ≥ 2k + 6 in Theorem 2.2.1 can be improved to n ≥ k + 4. It remains open to further reduce n ≥ k + 4. If f is entire, then a related result easily follows: Theorem 2.2.5. Let f be a transcendental entire function of finite order, and let a(z) be a nonzero meromorphic function of order λ := ρ(a) < ρ(f ). If n ≥ 2, then f (z)n f (k) (z + c) − a(z) has infinitely many zeros. We omit the proof, as this is just a particular case of Theorem 2.2.6. To proceed, consider linear delay-differential polynomials of the form k

g(f ) := ∑ bj (z)f (kj ) (z + cj ), j=1

2.2 Value distribution of delay-differential polynomials | 29

where f is a meromorphic function of finite order ρ, the coefficients bj are small meromorphic functions with respect to f in the sense that for λ < ρ and all ε ∈ (0, ρ − λ), T(r, bj ) = O(r λ+ε ) + S(r, f ) outside a possible exceptional set of finite logarithmic measure, k is a positive integer, the shifts cj are nonzero complex numbers, and kj are nonnegative integers. In what follows, such small functions with respect to f are called λ-small, denoted as T(r, b) = Sλ (r, f ). Provided that f is a meromorphic function with few poles, we now prove some theorems that are substantial extensions of Theorem 2.2.1. Theorem 2.2.6. Let f be a transcendental meromorphic function of finite order ρ with N(r, f ) = Sλ (r, f ), let b0 be a nonvanishing λ-small function of f , let g(f ) be nonvanishing, and let n ≥ 2 and s ≥ 1. Then F := f n g(f )s − b0 has sufficiently many zeros to satisfy λ(F) = ρ. Remark 2.2.7. (1) The condition N(r, f ) = Sλ (r, f ) is necessary in Theorem 2.2.6. As an example, the function f (z) := tan z is of order 1, and N(r, f ) = O(r). On the other hand, if g(f )(z) = f (z + π/2) = − cot z, then F(z) = f 2 (z)(g(f ))2 (z) − 2 = −1 has no zeros. (2) Considering F := fg(f )s − b0 with s ≥ 2, the conclusion similar to that in Theorem 2.2.6 does not necessarily hold. As an example, suggested by Zhibo Huang (private communication), take f (z) := ez + z and g(f )(z) := 2f (z) − f (z + log 2) = z − log 2. Then, for every integer s ≥ 1, F(z) := f (z)g(f )s (z) − z(z − log 2)s = (z − log 2)s ez has finitely many zeros only. Now we give the next two results from [125] to point out what can be shown for n = 1 and s ≥ 2. Theorem 2.2.8. Let f be a transcendental meromorphic function of finite order ρ with N(r, f ) = Sλ (r, f ), let b0 be a nonvanishing λ-small function of f , and let g(f ) be nonvanishing. If s ≥ 2, then F := fg(f )s − b0 satisfies max{λ(F), λ(f )} = ρ. Moreover, if ρ ∉ ℕ, then λ(F) = λ(f ) = ρ. The example given in Remark 2.2.7(2) shows a case where λ(F) < ρ although λ(f ) = ρ. For the next result from [125], we slightly modify the expression for g(f ) as follows: n

m

̃ ) := ∑ ∑ bij (z)f (j) (z + ci ), g(f i=0 j=0

where each bij (z) (0 ≤ i ≤ n, 0 ≤ j ≤ m) is a λ-small meromorphic function, and bim (z) ≡ 1 for every 0 ≤ i ≤ n. In addition to [125, Theorem 2.7] (see also [3]), we have the following:

30 | 2 Value distribution of complex delay-differential polynomials Theorem 2.2.9. Let f be a transcendental meromorphic function of finite order ρ with ̃ ) be nonvanN(r, f ) = Sλ (r, f ), let b0 be a nonvanishing λ-small function of f , and let g(f ishing. Suppose that ̃ )) ≠ Sλ (r, f ) T(r, g(f

(2.17)

T(r, w) = Sλ (r, f )

(2.18)

and

̃ ̃ ))s − b0 satisfies for every meromorphic solution w of g(w) = 0. Then for s ≥ 2, F := f (g(f λ(F) = ρ. Again, Remark 2.2.7(2) shows that Theorem 2.2.9 can fail without hypothesis (2.17). Next, we see that without hypothesis (2.18), we may have λ(F) < ρ. Example 2.2.10. Let f (z) := ez − 32 e−z/2 and ̃ )(z) := f 󸀠 (z + 4πi) − f (z) = e−z/2 . g(f ̃ )) ≠ Sλ (r, f ), and w(z) = ez is a solution of g(w) ̃ Clearly, T(r, g(f = 0 not satisfying (2.18). Moreover, 2 2 ̃ ))2 (z) − 1 = (ez − e−z/2 )e−z − 1 = − e−3z/2 F(z) := f (z)(g(f 3 3 has no zeros. Returning to our original definition of g(f ), we now treat the case n = s = 1; see [125, Theorem 2.3]. Theorem 2.2.11. Let f be a transcendental meromorphic function of finite order ρ with N(r, f ) = Sλ (r, f ), let b0 be a nonvanishing λ-small function of f , and let g1 (f ) and g2 (f ) be nonvanishing linear delay-differential polynomials with λ-small coefficients of f such that g1 (f ) and g2 (f ) are not equivalent. Then the functions F1 := fg1 (f ) − b0 and F2 := fg2 (f ) − b0 satisfy max{λ(F1 ), λ(F2 )} = ρ, except when one of the following cases holds: (i) g1 (f ) = L1 (z)f +M(z)f 󸀠 and g2 (f ) = L2 (z)f −M(z)f 󸀠 , where L1 , L2 , M are non-vanishing λ-small functions of f , and L1 + L2 ≢ 0. (ii) Only one of gi (f ), i = 1, 2, is a λ-small function of f . Remark 2.2.12. We need some examples to illustrate the necessity of the assumptions in Theorem 2.2.11: (1) Two nonequivalent delay-differential polynomials g1 (f ) and g2 (f ) are indeed needed. As an example, take f (z) = ez + z and g(f )(z) = 2f (k) (z + πi) + f (z) for any integer k ≥ 2. Then f (z)g(f )(z) − z 2 = −e2z has no zeros. (2) For an example to case (ii) in Theorem 2.2.11, consider f (z) = ez + 1, g1 (f ) ≡ 1, and g2 (f ) = f (z + πi). Then both f (z)g1 (f )(z) − 1 = ez and f (z)g2 (f )(z) − 1 = −e2z have no zeros.

2.2 Value distribution of delay-differential polynomials | 31

Lemma 2.2.13. Let f be a transcendental meromorphic function of finite order ρ such that N(r, f ) = Sλ (r, f ) for λ < ρ, and such that the linear delay-differential polynomial g(f ) with λ-small coefficients is not vanishing. Suppose n ≥ 1, s ≥ 1. Then we have ρ(f n g(f )s ) = ρ. Proof. Suppose that ρ(f n g(f )s ) = λ < ρ. Writing k

g(f )(z) = ∑ bj (z) j=1

f (kj ) (z + cj ) f (z + cj ) f (z + cj )

f (z)

and using the logarithmic derivative lemma and Lemma 1.2.11, we obtain T(r, g(f )) = O(r ρ−1+ε ) + O(r λ+ε ) + S(r, f ). Of course, the same estimate applies to (g(f ))s . Therefore T(r, f n ) ≤ T(r, f n g(f )s ) + T(r, 1/g(f )s ) = O(r ρ−1+ε ) + O(r λ+ε ) + S(r, f ), which leads to the contradiction ρ(f ) < ρ. Remark 2.2.14. Observe that the same contradiction follows if we assume in the proof that T(r, f n g(f )s ) = Sλ (r, f ) instead of ρ(f n g(f )s ) = λ < ρ. Proof of Theorem 2.2.6. Suppose, on the contrary, that λ(F) = λ < ρ. Since N(r, f ) = Sλ (r, f ), we have N(r, F) = Sλ (r, f ). By the standard Hadamard representation we may write f n g(f )s = b0 + βeh ,

(2.19)

where β is a λ-small function of f , and h is a polynomial of degree ≤ ρ. Actually, deg h = ρ. Indeed, if deg h ≤ μ < ρ, then T(r, f n g(f )s ) = O(r μ+ε ) + Sλ (r, f ), leading to ρ(f n g(f )s ) ≤ max(μ, λ) < ρ, a contradiction with Lemma 2.2.13. Now differentiating (2.19) and eliminating eh , we obtain n

f󸀠 g(f )󸀠 β󸀠 A +s − − h󸀠 = n , f g(f ) β f (g(f ))s

β󸀠

(2.20)

where A := b󸀠0 − b0 β − b0 h󸀠 cannot vanish identically. Indeed, if A = 0, then b0 = Cβeh for some constant C ≠ 0. If C = −1, then f = 0 follows by (2.19), whereas if C ≠ −1, then f n (g(f ))s = C+1 b is a λ-small function of f , contradicting Lemma 2.2.13. C 0 Since n ≥ 2 and N(r, f ) = Sλ (r, f ), equation (2.20) gives N(r, 1/f ) = Sλ (r, f ). If s ≥ 2, then we similarly observe that N(r, 1/g(f )) = Sλ (r, f ). By the second main theorem of Nevanlinna (Theorem 1.1.22) we get T(r, f n g(f )s ) ≤ N(r, f n g(f )s ) + N (r, = Sλ (r, f ),

1 1 ) + N (r, ) + S(r, f ) f n g(f )s F

32 | 2 Value distribution of complex delay-differential polynomials contradicting Lemma 2.2.13. It remains to consider the case of n ≥ 2 and s = 1. Since f is a meromorphic function of finite order ρ such that max {N(r, f ), N(r, 1/f )} = Sλ (r, f ), it may be represented as f (z) = γ(z)eQ(z) , where T(r, γ) = Sλ (r, f ), and Q is a polynomial of deg Q = ρ. Write now g(f ) = G(f )(z)eQ(z) , where k

G(f )(z) := ∑ bj (z)

f (kj ) (z + cj ) f (z + cj )

j=1

f (z + cj )

f (z)

γ(z).

By the logarithmic derivative lemma (Proposition 1.1.8) and its difference analogue (1.22) we have T(r, G(f )) = O(r ρ−1+ε ) + Sλ (r, f ). Hence N (r,

1 1 ) = N (r, ) = O(r ρ−1+ε ) + Sλ (r, f ). g(f ) G(f )

Recalling that N(r, f ) = Sλ (r, f ) and again applying Theorem 1.1.22, we obtain 1 1 1 T(r, f n g(f )) ≤ nN (r, ) + N (r, ) + N (r, ) + N(r, f n g(f )) + S(r, f ) f g(f ) F = Sλ (r, f ).

Thus ρ(f n g(f )) ≤ max(ρ − 1, λ), contradicting Lemma 2.2.13. Proof of Theorem 2.2.8. Suppose that max{λ(F), λ(f )} < ρ. We may write f (g(f ))s = b0 + βeh ,

(2.21)

where β is a nonvanishing λ-small function of f , and h is a polynomial of degree ρ. Differentiating (2.21) and eliminating eh as before, we obtain b󸀠

β󸀠

b0 ( b0 − β − h󸀠 ) f󸀠 g(f )󸀠 β󸀠 󸀠 0 +s − −h = . f g(f ) β f (g(f ))s

(2.22)

Since s ≥ 2, from (2.22) we conclude that N (r,

1 ) = Sλ (r, f ). g(f )

(2.23)

By this and (2.22) we conclude again that N (r,

1 1 ) ≤ N (r, ) + Sλ (r, f ). f (g(f ))s f

(2.24)

2.2 Value distribution of delay-differential polynomials | 33

Again by Theorem 1.1.22 we get T(r, f (g(f ))s ) ≤ N(r, f (g(f ))s ) + N (r,

1 ) f (g(f ))s

1 ) + S(r, f ) f (g(f ))s − b0 1 ≤ N (r, ) + Sλ (r, f ). f

+ N (r,

(2.25)

From (2.25) and Lemma 2.2.13 we obtain ρ = ρ(f (g(f ))s ) ≤ λ(f ) < ρ, a contradiction. Now suppose that ρ ∉ N. Then, clearly, λ(f ) = ρ. On the other hand, by Lemma 2.2.13 again we have ρ(F) = ρ ∉ N. Hence λ(F) = ρ. Proof of Theorem 2.2.9. Suppose, on the contrary, that λ(F) < ρ. On the other hand, since b0 is a nonvanishing λ-small function of f , F = b0 F ∗ , where F ∗ :=

1 ̃ )s − 1. f g(f b0

Clearly, λ(F) = λ(F ∗ ) and ρ(F) = ρ(F ∗ ) = ρ. So, we further only consider F ∗ . Put bi,−1 = bi,m+1 = 0. Since bim (z) ≡ 1 for every 0 ≤ i ≤ n, we have n

m

n m+1

̃ ) = ∑ ∑ bi,j f (j) (z + ci ) = ∑ ∑ bi,j f (j) (z + ci ). g(f i=0 j=0

i=0 j=−1

Since b󸀠i,m+1 = 0, we have n

m

n

m

̃ )󸀠 = ∑ ∑ b󸀠i,j f (j) (z + ci ) + ∑ ∑ bi,j f (j+1) (z + ci ) g(f i=0 j=0

i=0 j=0

n m+1

= ∑ ∑ (b󸀠i,j + bi,j−1 )f (j) (z + ci ). i=0 j=0

(2.26)

(j) ̃ ) = ∑ni=0 ∑m+1 Since g(f j=0 bi,j f (z + ci ), we find that w = f solves the delay-differential equation n m+1

∑ ∑ di,j w(j) (z + ci ) = 0,

i=0 j=0

where di,j = b󸀠i,j + bi,j−1 − for all 0 ≤ i ≤ n, 0 ≤ j ≤ m + 1.

̃ )󸀠 g(f b , ̃ ) i,j g(f

and d0,m+1 = 1

(2.27)

34 | 2 Value distribution of complex delay-differential polynomials ̃ )s . By using Leibniz’ rule in (2.27) and the conDenote w := uv, where v = b0 /g(f k vention that Cj = 0 for k > j, we get j

n m+1

n m+1

i=0 j=0

i=0 j=0

k=0

n m+1

m+1

i=0 j=0

k=0

∑ ∑ di,j (uv)(j) (z + ci ) = ∑ ∑ di,j ∑ Cjk u(k) (z + ci )v(j−k) (z + ci ) = ∑ ∑ di,j ∑ Cjk u(k) (z + ci )v(j−k) (z + ci ) = 0.

Dividing the right-hand side by v(z), we obtain n m+1

m+1

i=0 k=0

j=0

0 = ∑ ∑ u(k) (z + ci ) ∑ Cjk di,j

n m+1 v(j−k) (z + ci ) = ∑ ∑ Ai,k (z)u(k) (z + ci ), v(z) i=0 k=0

where again, since Cjk = 0 for k > j, m+1

Ai,k (z) = ∑ Cjk di,j j=k

v(j−k) (z + ci ) . v(z)

(2.28)

In particular, this gives n

n m+1

∑ Ai,0 (z) = ∑ ∑ di,j

i=0

i=0 j=0

v(j) (z + ci ) v(z)

n m+1

= ∑ ∑ (b󸀠i,j + bi,j−1 − i=0 j=0

=

̃ 󸀠− g(v)

̃ )󸀠 g(f ̃ ̃ ) g(v) g(f

v(z)

̃ )󸀠 v(j) (z + ci ) g(f bi,j ) ̃ ) g(f v(z)

.

(2.29)

We now claim that n

∑ Ai,0 (z) ≢ 0.

i=0

To prove this, we suppose the contrary. Using (2.29), we get ̃ 󸀠= g(v)

̃ )󸀠 g(f ̃ g(v). ̃ ) g(f

We consider two cases. ̃ Case 1: If g(v) ≢ 0, then by simple integration of the above equation we get ̃ ̃ ), g(v) = cg(f

(2.30)

2.2 Value distribution of delay-differential polynomials | 35

where c is a nonzero constant. Defining H := v − cf , the linearity of g̃ implies that ̃ g(H) = 0. By assumption T(r, H) = Sλ (r, f ). Further, defining G := f + c1 H, we see that ̃ ) = g(f ̃ + H) = g(G). ̃ v = cG and g(f ̃ )s , we get Therefore T(r, G) = T(r, f ) + Sλ (r, f ). On the other hand, since v = b0 /g(f 1=

c c ̃ )s = ̃ s, Gg(f Gg(G) b0 b0

contradicting Lemma 2.2.13. ̃ Case 2: If g(v) ≡ 0, then ̃ )) + Sλ (r, f ) Sλ (r, f ) = T(r, v) = sT(r, 1/g(f ̃ )) + Sλ (r, f ), = sT(r, g(f

̃ )) ≠ Sλ (r, f ). a contradiction with the hypothesis that T(r, g(f Now returning to our proof, we have u=

w 1 ̃ )s = F ∗ + 1. = f g(f v b0

So F ∗ + 1 solves the delay-differential equation n m+1

∑ ∑ Ai,k (z)u(k) (z + ci ) = 0.

i=0 k=0

Hence n m+1

n

i=0 k=0

i=0

∑ ∑ Ai,k (z)F ∗(k) (z + ci ) = − ∑ Ai,0 (z).

(2.31)

From (2.28) and (2.23) we deduce that T(r, Ai,k (z)) = O(r ρ−1+ε ) + Sλ (r, f ) for all 0 ≤ i ≤ n and 0 ≤ k ≤ m + 1. Since λ(F ∗ ) < ρ, we may write F ∗ (z) = β∗ (z)eh(z) ,

(2.32)

where T(r, β∗ ) = Sλ (r, f ), and h is polynomial of degree ρ. Obviously, F ∗(k) (z + ci ) = ψk (z + ci )eh(z+ci ) ,

(2.33)

where ψk (k = 0, . . . , m + 1) are differential polynomials in β and h. Substituting (2.32) and (2.33) into (2.31), since ∑ni=0 Ai,0 ≢ 0, we get n m+1

∑∑

i=0 k=0

Ai,k (z)ψk (z + ci ) ∑ni=0 Ai,0 (z)

eh(z+ci )−h(z) = −e−h(z) .

Hence T(r, eh ) = O(r ρ−1+ε ) + Sλ (r, f ), resulting in the contradiction deg h = ρ(f ) ≤ max{ρ − 1, λ} < ρ. This completes the proof of the theorem.

36 | 2 Value distribution of complex delay-differential polynomials We next proceed to proving Theorem 2.2.11. As a preparation to this proof, we need the following lemma, whose proof closely follows the idea given in the proof of Proposition 4.1 in [124, p. 814–817]; see also [125]. Lemma 2.2.15. Let f be a transcendental meromorphic function of finite order ρ such that N(r, f ) = Sλ (r, f ), let b0 be a nonvanishing λ-small function of f , and let g1 (f ) and g2 (f ) be nonvanishing, not equivalent linear delay-differential polynomials in f with λ-small coefficients of f . Suppose that fgj (f ) = b0 + βj ehj

(j = 1, 2),

(2.34)

where β1 and β2 are λ-small meromorphic functions of f , and h1 and h2 are polynomials. Then deg h1 = deg h2 = ρ. Moreover, if deg(h1 + h2 ) < ρ, then the delay-differential polynomials g1 (f ) and g2 (f ) reduce to g1 (f ) = L1 (z)f + M(z)f 󸀠 and g2 (f ) = L2 (z)f − M(z)f 󸀠 , where L1 , L2 , M are nonvanishing λ-small functions of f such that L1 + L2 ≢ 0. Proof. First, observe that the polynomials h1 and h2 are nonconstant. Indeed, if h1 , say, is constant, then T(r, fg1 (f )) = Sλ (r, f ). This immediately implies the contradiction ρ(f ) < ρ; see Lemma 2.2.13 and the remark complementing this lemma. (1) Next, we clearly have deg h1 ≤ ρ, deg h2 ≤ ρ, and deg(h1 ± h2 ) ≤ ρ. Therefore we may assume that ρ > 0. Suppose first that deg h1 < ρ. By (1.29) in Section 1.2 we have T(r, g1 (f )) = O(r ρ−1+ε ) + Sλ (r, f ), and hence T(r, f ) = O(r ρ−1+ε ) + Sλ (r, f ) as well, and so, ρ(f ) < ρ, a contradiction. Similarly, we obtain that deg h2 = ρ. (2) Suppose next that deg(h1 + h2 ) < ρ. Some preparations are now needed before we are able to deduce a contradiction. Note that these preparations are also needed in the actual proof of Theorem 2.2.11. Differentiating (2.34) and eliminating exponentials, we obtain, as before, 󸀠 βj󸀠 f 󸀠 gj (f ) 1 + − − h󸀠j = Aj , f gj (f ) βj f (gj (f )) β󸀠

(2.35)

where Aj := b󸀠0 −b0 βj −b0 h󸀠j for j = 1, 2. Moreover, A1 and A2 are not vanishing identically j

as known by recalling the proof of Theorem 2.2.6. By (2.35), for j = 1, 2, m (r,

1 ) = Sλ (r, f ) fgj (f )

2.2 Value distribution of delay-differential polynomials | 37

and N(2 (r,

1 ) = Sλ (r, f ), fgj (f )

where N(2 (r, ⋅) is the integrated counting function for the nonsimple zeros. Therefore T(r, f (gj (f ))) = N1) (r,

1 ) + Sλ (r, f ), fgj (f )

where N1) (r, ⋅) now counts the simple zeros only. Clearly, we also have N(2 (r, f1 ) = Sλ (r, f ), and hence 1 N(r, 1/f ) = N1) (r, ) + Sλ (r, f ). f Using the identity

1 f2

=

gj (f ) 1 , f f (gj (f ))

we have

1 m (r, ) = O(r ρ−1+ε ) + Sλ (r, f ). f Assuming, as we may, that ρ − 1 < λ < ρ, we obtain 1 T(r, f ) = N1) (r, ) + Sλ (r, f ). f

(2.36)

Writing now (2.35) in the form f 󸀠 (gj (f )) + f (gj (f ))󸀠 − (

βj󸀠 βj

+ h󸀠j ) (f (gj (f ))) = Aj ,

(2.37)

we observe that f 󸀠 (z0 )(gj (f ))(z0 ) − Aj (z0 ) = 0, provided that z0 is a simple zero of f , outside of all possible zeros and poles of Aj , βj . Since (2.37) holds for both j = 1, 2, we easily see that all possible poles of H :=

A1 (g2 (f )) − A2 (g1 (f )) f

(2.38)

are multiple except perhaps at the shift value f (z0 ) + cj , and hence N(r, H) = Sλ (r, f ). Moreover, m(r, H) = Sλ (r, f ). Hence H is a small function with respect to f in the sense of T(r, H) = Sλ (r, f ). (3) Since deg(h1 + h2 ) ≤ ρ − 1 < λ, we have T(r, eh1 +h2 ) = O(r λ+ε ). Moreover, T(r, φ) = Sλ (r, f ) for φ := β1 β2 eh1 +h2 . Looking now at simple zeros, say z0 , of f , we may assume at the same time that b0 , β1 , β2 , φ and all coefficients of g1 (f ) and g2 (f ) are nonzero and finite at z0 . Write now (2.34) in the form fg1 (f ) = b0 + β1 eh1 ,

fg2 (f ) = b0 +

φ −h1 e . β1

38 | 2 Value distribution of complex delay-differential polynomials Thus we obtain eh1 = −

b0 φ =− , β1 b0 β1

and so b20 = φ at z0 . If b20 − φ is not vanishing identically, then we conclude that N1) (r, 1/f ) ≤ Sλ (r, f ) = O(r λ+ε ) + S(r, f ). Hence T(r, f ) = Sλ (r, f ), resulting in the contradiction ρ(f ) < ρ. Therefore it remains to consider the case b20 ≡ φ. Then b20 = β1 β2 eh1 +h2 = (fg1 (f ) − b0 )(fg2 (f ) − b0 ), resulting in fg1 (f )g2 (f ) = b0 (g1 (f ) + g2 (f )).

(2.39)

But now from g1 (f )g2 (f ) = b0 (

g1 (f ) g2 (f ) + ) f f

we see that m(r, g1 (f )g2 (f )) = Sλ (r, f ), and hence T(r, g1 (f )g2 (f )) = Sλ (r, f ) as well. Now denote ψ := g1 (f )g2 (f ). Using (2.39), we get b0 + β1 eh1 = fg1 (f ) =

b b0 g (f )(g1 (f ) + g2 (f )) = 0 g1 (f )2 + b0 . ψ 1 ψ

Thus we obtain g1 (f )2 =

β1 ψ h1 e . b0

(2.40)

g2 (f )2 =

β2 ψ h2 e , b0

(2.41)

Similarly,

and, further, (

g1 (f ) 2 β1 h1 −h2 ) = e . g2 (f ) β2

Recalling identity (2.38), we now consider A1 g2 (f ) − A2 g1 (f ) = Hf .

(2.42)

2.2 Value distribution of delay-differential polynomials | 39

If H vanishes identically, then we see that (A1 /A2 )2 = (

g1 (f ) 2 β1 h1 −h2 ) = e , g2 (f ) β2

and hence T(r, eh1 −h2 ) = Sλ (r, f ).

(2.43)

Therefore deg(h1 − h2 ) ≤ ρ − 1. Combining this with deg(h1 + h2 ) ≤ ρ − 1, we obtain deg h1 < ρ, contradicting to deg h1 = ρ. Therefore we need to consider (2.42) assuming that H does not vanish identically. Squaring (2.42) and using (2.40) and (2.41), we get f2 =

A21 β2 ψ h2 A22 β1 ψ h1 2A1 A2 e + 2 e − ψ. H 2 b0 H b0 H2

Multiplying by g1 (f )2 , using (2.40) and (2.41) again, and recalling that f 2 g1 (f )2 = (b0 + β1 eh1 )2 , by an elementary computation we get (β12 −

A22 β1 ψ 2 2h1 A21 2 2A1 A2 β1 2 h1 2 ) ) e + (2b β − ψ ) e + (b − ( ψ ) = 0. 0 1 0 H 2 b0 H 2 b0 H2

(2.44)

Then we have T(r, eh1 ) = Sλ (r, f ), resulting in the contradiction deg h1 < ρ by Lemma 2.2.13, provided that not all coefficients in (2.44) are vanishing identically. Finally, suppose that all coefficients in (2.44) vanish. Then b20 =

A21 2 A22 2 A1 A2 2 ψ = 2ψ = ψ H2 H H2

implies that A1 = −A2 , and (2.42) takes the form g1 (f ) = κ(z)f − g2 (f ),

(2.45)

where κ = H/A1 . Differentiating (2.45) and substituting this and (2.45) into (2.35) for j = 1, we obtain (κ󸀠 − κ (

β1󸀠 β󸀠 + h󸀠1 )) f 2 + 2κf 󸀠 f + ( 1 + h󸀠1 ) fg2 (f ) − (g2 (f ))󸀠 f − g2 (f )f 󸀠 = A1 . β1 β1

(2.46)

Adding (2.46) to equation (2.35) for j = 2 and keeping in mind that A1 = −A2 , we get −(

β1󸀠 β2󸀠 β󸀠 − + (h1 − h2 )󸀠 ) g2 (f ) = (κ󸀠 − κ ( 1 + h󸀠1 )) f + 2κf 󸀠 . β1 β2 β1

(2.47)

40 | 2 Value distribution of complex delay-differential polynomials β󸀠

β󸀠

β󸀠

We easily see that the coefficients β1 − β2 +(h1 −h2 )󸀠 , resp., κ󸀠 −κ( β1 +h󸀠1 ) above are nonva1 2 1 nishing. Indeed, if not, then we get deg(h1 − h2 ) < ρ, resp., deg h1 < ρ, a contradiction. Therefore (2.47) can be rewritten in the form g2 (f ) = L2 (z)f − M(z)f 󸀠 ,

(2.48)

where β󸀠

L2 = −

β1󸀠 β1

κ󸀠 − κ( β1 + h󸀠1 ) −

β2󸀠 β2

1

+ (h1 − h2

)󸀠

,

M(z) =

β1󸀠 β1

−

β2󸀠 β2

2κ + (h1 − h2 )󸀠

.

Similarly, we obtain g1 (f ) = L1 (z)f + M(z)f 󸀠 ,

(2.49)

where β󸀠

L1 =

β1󸀠 β1

κ󸀠 − κ( β2 + h󸀠2 ) −

β2󸀠 β2

2

+ (h1 − h2 )󸀠

.

Proof of Lemma 2.2.15 is now completed. Proof of Theorem 2.2.11. Clearly, we have to show that either case (i) or case (ii) occurs in the theorem whenever max{λ(F1 ), λ(F2 )} < ρ. Therefore now suppose that max{λ(F1 ), λ(F2 )} < λ for some λ < ρ. Then, as before, fgj (f ) − b0 = βj ehj ,

j = 1, 2,

(2.50)

where the βj are λ-small functions of f , and hj are polynomials. By Lemma 2.2.15 we have deg h1 = deg h2 = ρ. If now deg(h1 + h2 ) < ρ, then we have the exceptional case (i) by Lemma 2.2.15. Hence we may now proceed under the assumption that deg(h1 + h2 ) = ρ, and we have to show that the exceptional case (ii) is the only possibility. To prove this, we need to show that the following two cases are impossible: (a) both g1 (f ) and g2 (f ) are λ-small functions of f , and (b) both g1 (f ) and g2 (f ) are not λ-small functions of f . (a) Suppose that both g1 (f ) and g2 (f ) are λ-small functions of f . Applying the second main theorem, we conclude that T(r, f ) ≤ N(r, f ) + N (r,

1 f−

b0 g1 (f )

) + N (r,

1 f−

b0 g2 (f )

= N(r, 1/β1 ) + N(r, 1/β2 ) + S(r, f ) = Sλ (r, f ), which is impossible.

) + S(r, f )

2.2 Value distribution of delay-differential polynomials | 41

(b) Suppose now that both g1 (f ) and g2 (f ) are not λ-small functions of f . To start this case, we first show that deg(h1 − h2 ) = ρ. If not, i. e., if deg(h1 − h2 ) < ρ, then we immediately obtain form (2.50) that f (g1 (f ) −

β1 h1 −h2 β e g2 (f )) = b0 (1 − 1 eh1 −h2 ) . β2 β2

(2.51)

By a modification of the Clunie lemma, see Lemma 4.1 in [125], we get m (g1 (f ) −

β1 h1 −h2 e g2 (f )) = O(r ρ−1+ε ) + Sλ (r, f ). β2

(2.52)

Assuming now, as we may, that ρ − 1 < λ < ρ, we further see that T (r, g1 (f ) −

β1 h1 −h2 e g2 (f )) = Sλ (r, f ). β2

β

(2.53)

β

If now g1 (f )− β1 eh1 −h2 g2 (f ) vanishes, then β1 eh1 −h2 ≡ 1 follows by (2.51), and thus g2 (f ) ≡ 2 2 g1 (f ), contradicting the assumption that g2 (f ) ≢ g1 (f ). Therefore we have that g1 (f ) − β1 h1 −h2 e g2 (f ) is not vanishing identically. But then β 2

T(r, f ) = T (r,

b0 (1 − g1 (f ) −

β1 h1 −h2 e ) β2

β1 h1 −h2 e g2 (f ) β2

) = Sλ (r, f ),

a contradiction. Therefore we now proceed under the assumption that deg(h1 − h2 ) = ρ. Recall the notations in the proof of Lemma 2.2.15, where we had fgj (f ) − b0 = βj ehj ,

j = 1, 2,

(2.54)

where T(r, βj ) = Sλ (r, f ) for j = 1, 2, and h1 , h2 are polynomials of degree ρ. Moreover, f 󸀠 (gj (f )) + f (gj (f ))󸀠 − (

βj󸀠 βj

+ h󸀠j ) (f (gj (f ))) = Aj ,

j = 1, 2.

(2.55)

Finally, recall the λ-small function H :=

A1 (g2 (f )) − A2 (g1 (f )) , f

(2.56)

where A1 , A2 are λ-small functions of f as well. If now H vanishes, then (2.54) implies that −β1 A2 eh1 + β2 A1 eh2 + (A1 − A2 )b0 = 0. Since deg(h1 − h2 ) = ρ, we may apply Theorem 1.1.32 to see that all coefficients in this equation are vanishing identically, a contradiction.

42 | 2 Value distribution of complex delay-differential polynomials Therefore we may now proceed under the assumption that the λ-small function H is not vanishing identically. By (2.56) we have g1 (f ) =

A1 H g2 (f ) − f . A2 A2

(2.57)

Differentiating (2.57), substituting g1 (f ) and (g1 (f ))󸀠 into (2.55), and adding identity (2.55) multiplied by −A1 /A2 in the case j = 2, we obtain (

B1 H H 󸀠 −2H 󸀠 − ( ) )f + ( ) f = Dg2 (f ), A2 A2 A2

(2.58)

where Bj := βj󸀠 /βj + h󸀠j , j = 1, 2, and D :=

B1 A1 A 󸀠 AB −( 1) − 1 2. A2 A2 A2

(2.59)

The coefficients in (2.58), denoted as T1 D for f and T2 D for f 󸀠 (and D for g2 (f )) in what follows, are not vanishing identically. Indeed, suppose first that T1 D = BA1 H − ( AH )󸀠 2

2

vanishes identically. Then by elementary integration, β1 eh1 = CH/A2 , where C is a constant, implying that T(r, eh1 ) = Sλ (r, f ). As before, we see that ρ(eh1 ) < ρ, a contradiction. Clearly, T2 D is not vanishing identically. Finally, if −D : −

A 󸀠 AB B1 A1 + ( 1 ) + 1 2 ≡ 0, A2 A2 A2

then integration results in β1 h1 −h2 e =: γ, β2 where γ satisfies T(r, γ) = Sλ (r, f ). We may now use representations (2.54) to obtain f (g1 (f ) − γg2 (f )) = (1 − γ)b0 . If now γ = 1, then the immediate contradiction g1 (f ) = g2 (f ) follows. Otherwise, we may use the standard Clunie reasoning to obtain T(r, g1 (f ) − γg2 (f )) = m(r, g1 f − γg2 (f )) + Sλ (r, f ) = Sλ (r, f ). Therefore T(r, f ) = T (f ,

(1 − γ)b0 ) = Sλ (r, f ), g1 (f ) − γg2 (f )

and the contradiction ρ(f ) < ρ follows as before.

2.2 Value distribution of delay-differential polynomials | 43

Therefore we now obtain g2 (f ) = T1 f + T2 f 󸀠

(2.60)

with λ-small coefficients. Differentiating (2.60) and substituting this expression and (2.60) into (2.55) with j = 2 results in (T1󸀠 − B2 T1 )f 2 + (2T1 + T2󸀠 − B2 T2 )ff 󸀠 + T2 (f 󸀠 )2 + T2 ff 󸀠󸀠 = A2 .

(2.61)

If z0 is now a simple zero of f such that, simultaneously, all coefficients of (2.61) take a nonzero finite value at z0 , then we conclude, by a careful analysis of (2.61), that T2 (f 󸀠 )2 = A2

(2.62)

at z0 . Differentiating now (2.61) and taking into account (2.62) result in (T1󸀠 − B2 T1 )󸀠 f 2 + (4T1 + T2󸀠󸀠 − B󸀠2 T2 − B2 T2󸀠 − 2B2 T1 )ff 󸀠

+ (2T1 + 2T2󸀠 − B2 T2 )(f 󸀠 )2 + 3T2 f 󸀠 f 󸀠󸀠 + (2T1 + 2T2󸀠 − B2 T2 )ff 󸀠󸀠 + T2 ff (3) = A󸀠2 .

Considering now a simple zero, say z0 , of f that is neither a zero nor a pole of the coefficients of this expression, we obtain (2T1 + 2T2󸀠 − B2 T2 )f 󸀠 + 3T2 f 󸀠󸀠 −

A󸀠2 =0 f󸀠

(2.63)

at z0 . Eliminating 1/f 󸀠 from (2.62) and (2.63) implies that (2A2 T1 + 2A2 T2 − A󸀠2 T2 − A2 B2 T2 )f 󸀠 + 3A2 T2 f 󸀠󸀠 vanishes at z0 . Hence 󸀠 󸀠 󸀠󸀠 ̃ := (2A2 T1 + 2A2 T2 − A2 T2 − A2 B2 T2 )f + 3A2 T2 f H f

is a λ-small function. Writing now f 󸀠󸀠 =

̃ 2A T + 2A2 T2 − A󸀠2 − A2 B2 T2 󸀠 H f− 2 1 f 3A2 T2 3A2 T2

(2.64)

and substituting (2.64) into (2.61), we see that Q1 f 2 + Q2 ff 󸀠 + T2 (f 󸀠 )2 = A2

(2.65)

with λ-small coefficients Q1 := T1󸀠 − B2 T1 +

̃ H 3A2

(2.66)

44 | 2 Value distribution of complex delay-differential polynomials and Q2 :=

A󸀠 1 (4T1 + T2󸀠 − 2B2 T2 + 2 T2 ) . 3 A2

(2.67)

For an argument further, first, observe that Q2 does not vanish identically. Indeed, if Q2 vanishes, then 4

A󸀠 T1 T2󸀠 + − 2B2 + 2 = 0. T2 T2 A2

Recalling the definitions of T1 D, T2 D, A2 , B2 , a simple computation from 4T1 D +

T2󸀠 A󸀠 T2 D − 2B2 T2 D + 2 T2 D = 0 T2 A2

results in e2(h1 +h2 ) = C

H T β−2 β−2 . A2 2 1 2

Since T2 =

−1

A 󸀠 AB 2H B1 A1 ( −( 1) − 1 2) , A2 A2 A2 A2

we see that T(r, e2(h1 +h2 ) ) = Sλ (r, f ), and hence deg(h1 + h2 ) ≤ λ < ρ. This is a contradiction to our assumption for this key part of the proof that deg(h1 + h2 ) = ρ. We now proceed by differentiating (2.65) to obtain Q󸀠1 f 2 + (2Q1 + Q󸀠2 )ff 󸀠 + (Q2 + T2󸀠 )(f 󸀠 )2 + Q2 ff 󸀠󸀠 + 2T2 f 󸀠 f 󸀠󸀠 = A󸀠2 . Considering simple zeros z0 of f as before, we obtain (T2 f 󸀠 −

A2 ) (z0 ) = 0 f󸀠

and ((Q2 + T2󸀠 )f 󸀠 + 2T2 f 󸀠󸀠 −

A󸀠2 ) (z0 ) = 0. f󸀠

Eliminating 1/f 󸀠 , we conclude that (A2 (Q2 + T2󸀠 ) − A󸀠2 T2 )f 󸀠 + 2A2 T2 f 󸀠󸀠 vanishes at z0 . Therefore 󸀠 󸀠 󸀠 󸀠󸀠 ̃ := (A2 (Q2 + T2 ) − A2 T2 )f + 2A2 T2 f R f

(2.68)

2.2 Value distribution of delay-differential polynomials | 45

̃ = Sλ (r, f ), and we may write satisfies T(r, R) f 󸀠󸀠 =

̃ A󸀠 T − A2 (Q2 + T2󸀠 ) 󸀠 R f+ 2 2 f . 2A2 T2 2A2 T2

(2.69)

Substitute now (2.69) into (2.68), which results in (Q󸀠1 +

󸀠 ̃ ̃ Q2 R R 1 Q2 1 A2 ) f 2 + (2Q1 + Q󸀠2 − (Q2 + T2󸀠 ) + Q2 + ) ff 󸀠 2A2 T2 2 T2 2 A2 A2

+

A󸀠2 T (f 󸀠 )2 = A󸀠2 . A2 2

(2.70)

Comparing now (2.65) and (2.70) implies (Q󸀠1 +

̃ A󸀠 Q2 R − 2 Q1 ) f 2A2 T2 A2

+ (2Q1 + Q󸀠2 −

󸀠 ̃ Q 1 A2 1 R Q2 − (Q2 + T2󸀠 ) 2 + ) f 󸀠 = 0. 2 A2 2 T2 A2

(2.71)

Considering again simple zeros of f , we obtain, similarly as before, a contradiction T(r, f ) = O(r λ+ε ) + S(r, f ), unless we have Q󸀠1 +

̃ A󸀠 Q2 R − 2 Q1 = 0 2A2 T2 A2

(2.72)

and 2Q1 + Q󸀠2 −

󸀠 ̃ Q 1 A2 1 R Q2 − (Q2 + T2󸀠 ) 2 + = 0. 2 A2 2 T2 A2

(2.73)

̃ 2 from (2.72), substituting into (2.73), and simplifying, we obtain Solving R/A T2 (4Q1 T2 − Q22 )

A󸀠2 + Q2 (4Q1 T2 − Q22 ) − T2 (4Q1 T2 − Q22 )󸀠 + T2󸀠 (4Q1 T2 − Q22 ) = 0. A2

(2.74)

Our final reasoning now divides in two cases. Suppose first that 4Q1 T2 − Q22 does not vanish identically. Then by (2.74) we have Q2 (4Q1 T2 − Q22 )󸀠 A󸀠2 T2󸀠 = − − . T2 A2 T2 4Q1 T2 − Q22

(2.75)

Comparing (2.75) with (2.67) and recalling the definitions of A2 , B1 , B2 , T1 , T2 , Q1 , Q2 , we obtain h󸀠1 + h󸀠2 = −

2 󸀠 A󸀠2 T2󸀠 (H/A2 )󸀠 β1󸀠 β2󸀠 3 (4Q1 T2 − Q2 ) + 2 + 2 + − − . 2 4Q1 T2 − Q22 A2 T2 H/A2 β1 β2

46 | 2 Value distribution of complex delay-differential polynomials Direct integration now results in the contradiction T(r, eh1 +h2 ) = Sλ (r, f ) to Lemma 2.2.15. It remains to consider the final case where 4Q1 T2 = Q22 . In this case, first, observe that by (2.72) equation (2.69) may be rewritten in the form f 󸀠󸀠 = (

󸀠 T󸀠 A󸀠2 Q1 Q󸀠1 Q 1 A − ) f + ( 2 − 2 − 2 ) f 󸀠. A2 Q2 Q2 2 A2 T2 T2

(2.76)

Comparing with (2.64) and considering simple zeros of f again, we can avoid a contradiction only if ̃ A󸀠2 Q1 Q󸀠1 H − = . A2 Q2 Q2 3A2 T2 On the other hand, by (2.66) we have ̃ Q1 T1󸀠 − B2 T1 H − = . T2 T2 3A2 T2 Therefore A󸀠2 Q1 Q󸀠1 Q1 T1󸀠 T1 T − = − + B2 1 , A2 Q2 Q2 T2 T1 T2 T2 and so 󸀠 󸀠 󸀠 T 1 A2 Q1 Q2 1 Q 2 T T = ( 2 ) − 1 1 + B2 1 . ( − ) 4 A2 Q1 T2 4 T2 T1 T2 T2

Using (2.67), we have 󸀠 󸀠 T 󸀠 A󸀠 T 1 A2 Q1 ( − ) (4 1 + 2 + 2 − 2B2 ) 12 A2 Q1 T2 T2 A2 2

=

T 󸀠 A󸀠 T󸀠 T T T 1 (4 1 + 2 + 2 − 2B2 ) − 1 1 + B2 1 . 36 T2 T2 A2 T1 T2 T2

Now recall that T1 1 (H/A2 )󸀠 = ( − B1 ) . T2 2 H/A2 Therefore, also using Bj = βj󸀠 /βj + h󸀠j for j = 1, 2, we get 󸀠 󸀠 β󸀠 β󸀠 (H/A2 )󸀠 T2󸀠 A󸀠2 1 A2 Q1 ( − ) (2 + + − 2 1 − 2 2 − 2(h󸀠1 + h󸀠2 )) 12 A2 Q1 H/A2 T2 A2 β1 β2

=

β󸀠 β󸀠 (H/A2 )󸀠 T2󸀠 A󸀠2 1 (2 + + − 2 1 − 2 2 − 2(h󸀠1 + h󸀠2 )) 36 H/A2 T2 A2 β1 β2

2

2.2 Value distribution of delay-differential polynomials | 47

+

󸀠 󸀠 (H/A2 )󸀠 β1󸀠 1 β2 T1 ( − + h󸀠2 ) ( − − h󸀠1 ) , 2 β2 T1 H/A2 β1

and hence (

β󸀠 β󸀠 A󸀠2 Q󸀠1 (H/A2 )󸀠 T2󸀠 A󸀠2 − ) (2 + + − 2 1 − 2 2 − 2(h󸀠1 + h󸀠2 )) A2 Q1 H/A2 T2 A2 β1 β2 −

β󸀠 β󸀠 (H/A2 )󸀠 T2󸀠 A󸀠2 1 (2 + + − 2 1 − 2 2 − 2(h󸀠1 + h󸀠2 )) 3 H/A2 T2 A2 β1 β2

= 6(

2

β2󸀠 T1󸀠 (H/A2 )󸀠 β1󸀠 − )( − ) β2 T1 H/A2 β1

− 6(

β2󸀠 T1󸀠 󸀠 (H/A2 )󸀠 β1󸀠 − ) h1 + 6 ( − ) h󸀠2 − 6h󸀠1 h󸀠2 . β2 T1 H/A2 β1

By Proposition 1.1.16 in Section 1.1 all logarithmic derivatives here are dominated by |z|λ−1+ε outside a possible exceptional set of finite logarithmic measure, whereas the derivatives h󸀠1 , h󸀠2 dominate 21 |z|ρ−1 for all |z| large enough. Denoting h1 (z) = αz ρ + ⋅ ⋅ ⋅ and h2 (z) = βz ρ + ⋅ ⋅ ⋅, we may let |z| → ∞ above, outside an exceptional set of finite logarithmic measure, to obtain h󸀠1 h󸀠2 αβ 2 = = . 󸀠 󸀠 2 2 |z|→∞ (h + h ) 9 (α + β) 1 2 lim

Hence we must have either α/β = 2 or α/β = 1/2. We proceed to considering the case α = 2β. We may write ρ

eh1 (z) = e2βz eP1 (z) ,

ρ

eh2 (z) = eβz eP2 (z) ,

where P1 (z) and P2 (z) are polynomials of degree at most ρ − 1. Recall that we have g1 (f ) =

A1 A A H H g2 (f ) − f = ( 1 T1 − ) f + 1 T2 f 󸀠 A2 A2 A2 A2 A2

and g2 (f ) = T1 f + T2 f 󸀠 . Therefore ρ

b0 + β2 eP2 (z) eβz = fg2 (f ) = T1 f 2 + T2 ff 󸀠 and ρ

b0 + β1 eP1 (z) e2βz = fg1 (f ) = (

A1 A H T − ) f 2 + 1 T2 ff 󸀠 . A2 1 A2 A2

48 | 2 Value distribution of complex delay-differential polynomials By a simple computation, T1 f + T2 f 󸀠 =

1 b0 1 A H A1 f 󸀠 b0 + β2 eP2 − 2 P1 ( ( 1 T1 − + T − 2 )) f β1 A2 A2 A2 2 f f

1/2

.

(2.77)

Note that the square-root in (2.77) is a meromorphic function. We next show that g2 (f ) is a λ-small function by computing T(r, T1 f + T2 f 󸀠 ). Indeed, T(r, T1 f + T2 f 󸀠 ) = m(r, T1 f + T2 f 󸀠 ) + O(r λ+ε ) + S(r, f ) 1 1 = ∫ log+ |T1 f + T2 f 󸀠 |dθ + ∫ log+ |T1 f + T2 f 󸀠 |dθ + Sλ (r, f ), 2π 2π E1

E2

where E1 , resp., E2 , means the part on the circle of radius r such that |f | ≤ 1, resp., |f | > 1. Then we first see that 󵄨󵄨 f 󸀠 󵄨󵄨 1 1 󵄨 󵄨 ∫ log+ |T1 f + T2 f 󸀠 |dθ ≤ m(r, T1 ) + m(r, T2 ) + ∫ log+ 󵄨󵄨󵄨 󵄨󵄨󵄨 dθ + Sλ (r, f ) 󵄨󵄨 f 󵄨󵄨 2π 2π E1

E1

= Sλ (r, f ).

To compute the E2 -part, we use (2.77) to obtain 1 ∫ log+ |T1 f + T2 f 󸀠 |dθ 2π E2

≤

1 1 1 1 ∫ log+ |b0 |dθ + ∫ log+ |β2 |dθ + ∫ log+ |eP2 − 2 P1 |dθ 2π 2π 2π

E2

E2

E2

󵄨󵄨 1 󵄨󵄨 󵄨󵄨 A 1 H 󵄨󵄨 1 + ∫ log+ 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 dθ + ∫ log+ 󵄨󵄨󵄨󵄨 1 T1 − 󵄨󵄨󵄨󵄨 dθ 4π 4π A2 󵄨 󵄨 β1 󵄨 󵄨 A2 E2

E2

󵄨󵄨 A 1 f 󸀠 󵄨󵄨󵄨 1 󵄨 + ∫ log+ 󵄨󵄨󵄨 1 T2 󵄨󵄨󵄨 dθ + ∫ log+ |b0 |dθ = Sλ (r, f ). 󵄨 󵄨 4π 4π 󵄨 A2 f 󵄨 E2

E2

Thus we have obtained T(r, g2 (f )) = T(r, T1 f + T2 f 󸀠 ) = Sλ (r, f ), a contradiction in the case α/β = 2. To arrive at our final contradiction, we now apply a completely similar reasoning in the case α/β = 1/2 to obtain T(r, g1 (f )) = Sλ (r, f ), and this contradiction now completes the proof of Theorem 2.2.11. In the remaining part of the present section, we shortly consider delay-differential products of type f n (g(f ))s , where f is a meromorphic function of finite order ρ, and k

g(f )(z) := ∑ bj (z)f (kj ) (z + cj ) j=1

2.2 Value distribution of delay-differential polynomials | 49

is a linear delay-differential polynomial with small meromorphic coefficients bj such that T(r, bj ) = Sλ (r, f ). So, we will consider what happens for the results of the first part of this section if b0 vanishes. The reader may refer to [125] for the results below and for some more results of this type. Theorem 2.2.16. Let f be a transcendental meromorphic function of finite order ρ such that N(r, f ) = Sλ (r, f ). Suppose n ≥ 1 and s ≥ 1 and define F := f n (g(f ))s . If λ(f ) = ρ, resp., λ(f ) < ρ, then λ(F) = ρ, resp., λ(F) < ρ. Proof. (1) Suppose first that λ(F) < ρ. Writing F = f n (g(f ))s = βeh , we have that T(r, β) = Sλ (r, f ) and h is a polynomial of degree ρ. Then n

f󸀠 (g(f ))󸀠 β󸀠 +s − = h󸀠 . f g(f ) β

If f vanishes at z0 , then F(z0 ) = 0, unless g(f ) has a pole at z0 . This may happen at the coefficients of g(f ) and the poles of f (z + cj ) only, contributing at most by Sλ (r, f ). Therefore λ(f ) < ρ, and the claim follows. ρ

(2) In this case, we may write f (z) = τ(z)eαz , where α ≠ 0 is a constant, and ρ τ is small. Then, of course, f (z + cj ) = τ(z + cj )τj (z)eαz , where τj is a meromorphic ρ

function of order ρ − 1. Therefore it is not difficult to see that g(f ) = T(z)eαz , where T(z) is a differential polynomial of τ, of its shifts, and of its derivatives. Then (g(f ))s = ρ (T(z))s esαz , and hence ρ(T) ≤ λ, implying that λ(F) < ρ. Theorem 2.2.17. Let f be a transcendental meromorphic function of finite order ρ such that N(r, f ) = Sλ (r, f ). Suppose n ≥ 1 and define F := f n g(f ). Then (i) If λ(f ) ≤ ρ − 1, λ < ρ − 1, and ρ ≠ 1, then λ(F) = ρ − 1. (ii) If ρ − 1 < λ(f ) = λ∗ < ρ and λ < λ∗ , then λ(F) = λ∗ . (iii) If λ(f ) = λ = 0 and ρ = 1, then λ(F) = 0. Proof. We divide the proof of claim (i) into three parts. (a) If λ(f ) < ρ − 1, then ρ(F) = ρ by Lemma 2.2.13. Write now f (z) = τ(z)eh(z) , where h is a polynomial of deg(h) = ρ ≥ 2, and τ is a λ-small function, where λ < ρ − 1. Recalling that k

g(f ) := ∑ bj (z)f (kj ) (z + cj ) j=1

with k ≥ 2, we may write k

k

j=1

j=1

g(f ) = ∑ τj (z)f (z + cj ) := ∑ bj (z)

f (kj ) (z + cj ) f (z + cj )

f (z + cj ).

50 | 2 Value distribution of complex delay-differential polynomials Suppose first, contrary to the claim, that λ(F) < ρ − 1. Writing, as we may, F(z) = σ(z)eQ(z) , where σ is a λ-small function, where λ < ρ − 1, and Q is a polynomial of deg(Q) = ρ. Therefore we now have k

σ(z) = ∑ αj (z)eβj (z) , j=1

where αj (z) = τn (z)τj (z)τ(z + cj ) are λ-small functions, where again λ < ρ − 1 and βj (z) = nh(z) + h(z + cj ) − Q(z) for every 1 ≤ j ≤ k. Also, set h(z) := bm z m + bm−1 z m−1 + ⋅ ⋅ ⋅ + b0 , bm ≠ 0, where bm , bm−1 , . . . , b0 are constants, and m = ρ ≥ 2. Hence, for every i ≠ j, βi (z) − βj (z) = h(z + ci ) − h(z + cj ) = mbm (ci − cj )z m−1 + ⋅ ⋅ ⋅ . If deg βj (z) ≥ ρ−1 for all j, then by Theorem 1.1.32 σ(z) ≡ 0, and αj (z) ≡ 0 for j = 1, . . . , k. Hence F vanishes, a contradiction. If this is not the case, then deg βj (z) < ρ−1 for some 1 ≤ j0 ≤ k. Therefore k

σ(z) − αj0 (z)eβj0 (z) = ∑ αj (z)eβj (z) . j=1,j=j̸ 0

Since deg(βi − βj ) = ρ − 1 for every 1 ≤ i ≠ j ≤ k, then again by Theorem 1.1.32 αj (z) ≡ 0 for all j ≠ j0 , and σ(z)e−βj0 (z) (z) ≡ αj0 (z). This implies that τj (z)τ(z + cj ) vanishes for all j ≠ j0 , and hence g(f ) includes just one term, a contradiction. We point out to the reader that it is impossible to have deg βj < ρ−1 for all 1 < j < k, since deg(βi − βj ) = ρ − 1 for every 1 ≤ i ≠ j ≤ k. (b) Suppose next that λ(F) > ρ − 1. We may write g(f ) as k

g(f ) = (∑ bj (z)τ(z + cj )eh(z+cj )−h(z) ) eh(z) . j=1

(2.78)

Since deg (h(z + cj ) − h(z)) = ρ − 1 and τj (z) (1 ≤ j ≤ k), τ(z) are λ-small functions, where λ < ρ − 1, we have λ(g(f )) ≤ ρ − 1. This inequality, together with λ(f ) < ρ − 1, implies the contradiction ρ − 1 < λ(F) = λ(f n g(f )) ≤ ρ − 1. (c) Finally, if λ(f ) = ρ − 1, then, from (2.78), λ(F) ≤ ρ − 1. On the other hand, we have 1 g(f ) 1 nN (r, ) = N (r, ) ≤ N (r, ) + Sλ (r, f ). f F F

(2.79)

2.3 Borel exceptional values of delay-differential polynomials | 51

By (2.79) and λ < ρ − 1, we obtain ρ − 1 = λ(f ) ≤ λ(F), which completes the proof of claim (i). To prove claim (ii), we may write f (z) = τ(z)eh(z) , where h is a polynomial of degree ρ ≥ 1, and τ := ττ1 is a meromorphic function, where τ1 and τ2 are the canonical 2 products of zeros and poles, respectively. Since λ(f ) = λ∗ (ρ − 1 < λ∗ < ρ) and λ < λ∗ , ρ(τ1 ) = λ∗ and ρ(τ2 ) < λ∗ . This leads to ρ(τ) = λ∗ . By this and (2.78) we deduce that λ(g(f )) ≤ λ∗ , and hence λ(F) ≤ λ∗ . On the other hand, since λ < λ∗ , from (2.79) we deduce that λ(F) ≤ λ∗ . To the final claim (iii), we may apply the same reasoning as in the proof of Theorem 2.2.16 to obtain that g(f ) = T(z)eαz , where α is a constant, and T is a meromorphic function of order zero. Therefore λ(F) = 0.

2.3 Borel exceptional values of delay-differential polynomials In this section, we add a few results about the connection of Borel exceptional values of f and of F := fg(f ), where g(f ) = ∑kj=1 bj (z)f (kj ) (z + cj ). Recall that the Borel exceptional value for small functions d of f (z) satisfies λ(f (z) − d) < ρ(f ), where λ(f (z) − d) is the exponent of convergence of zeros of f (z) − d. Theorem 2.3.1. Let f be a finite-order transcendental entire function with nonvanishing Borel exceptional small entire function d(z), and let b0 be a nonvanishing small meromorphic function. If n ≥ 1 and g(f ) is not a small function relative to f , then f n g(f ) − b0 has sufficiently many zeros to satisfy λ(F) = λ(f n g(f )) = ρ(f ). Proof. If n ≥ 2, the statement is trivial by Theorem 2.2.6. Therefore we may further assume that n = 1. Since d(z) is a nonvanishing Borel exceptional small entire function of f (z), the value 1 is a Borel exceptional value of df . By Corollary in [235, p. 107] f is of positive integer order, say ρ(f ) = ρ. We may now write ρ

f (z) = d(z) + h(z)eαz , where α ≠ 0 is a constant, and h is a nonvanishing entire function such that λ(h) ≤ ρ(h) < ρ. We may next write f (z + c) = d(z + c) + h(z + c)eα(z+c) = d(z + c) + h(z + c)e

ρ

α(z ρ +cCρ1 z ρ−1 +⋅⋅⋅+cρ )

(2.80)

ρ

= d(z + c) + h0 (z)eαz , 1 ρ−1

ρ

where h0 (z) := h(z + c)eα(cCρ z +⋅⋅⋅+c ) is a small function relative to f . Differentiating iteratively, we obtain by elementary computation that ρ

f (kj ) (z + c) = d(kj ) (z + c) + hkj (z)eαz ,

52 | 2 Value distribution of complex delay-differential polynomials where hkj is still a small function. Suppose now, contrary to the statement, that ρ

f (z)(g(f ))(z) − b0 (z) = A(z)eβz . Clearly, the small function A is nonvanishing by recalling Lemma 2.2.13. Therefore ρ

A(z)eβz = f (z)(g(f ))(z) − b0 (z)

k

ρ

ρ

= (d(z) + h(z)eαz ) (∑ bj (z) (d(kj ) (z + c) + hkj (z)eαz )) − b0 (z) j=1

k

k

= d(z) ∑ bj (z)d(kj ) (z + c) − b0 (z) + (h(z) ∑ bj (z)hkj (z)) e2αz j=1

ρ

j=1

k

k

j=1

j=1

ρ

+ (d(z) ∑ bj (z)hkj (z) + h(z) ∑ bj (z)d(kj ) (z + c)) eαz .

(2.81)

If now β ≠ α, 2α, then we may apply Theorem 1.1.32 to conclude that the coefficients in this exponential identity vanish. In particular, A = 0, a contradiction. Suppose next that β = 2α. Then ρ

2

f (z)(g(f ))(z) − b0 (z) = A(z) (eαz ) = A(z) (

f (z) − d(z) 2 ) , h(z)

and hence h2

g(f ) 1 h2 b0 + d2 A . − A + 2Ad = f f f2

Since A, b0 , d, h are small functions and N(r, g(f )) is small, we may immediately conclude that N(r, 1/f ) is small. By the second main theorem for three small functions, 1 1 ) + S(r, f ) T(r, f ) ≤ N(r, f ) + N (r, ) + N (r, f f −d 1 = N(r, 1/f ) + N (r, ) + S(r, f ) h is small as well, a contradiction. It remains to get a contradiction in the case β = α. In this case, applying Theorem 1.1.32 again, we get k

∑ bj (z)hkj (z) ≡ 0 j=1

and k

∑ bj (z)d(kj ) (z + c) ≡ j=1

b0 (z) . d(z)

2.3 Borel exceptional values of delay-differential polynomials | 53

Recalling the definition of g(f ), we obtain k

k

j=1

j=1

g(f ) = ∑ bj (z)f (kj ) (z + cj ) = ∑ bj (z)d(kj ) (z + cj ) =

b0 , d

meaning that g(f ) is a small function, contradicting the assumption. Theorem 2.3.2. Let f be a transcendental meromorphic function of finite order ρ with finite Borel exceptional value d and N(r, f ) = Sλ (r, f ), and let g(f ) be a nonconstant delay-differential polynomial with λ-small coefficients. Defining F := fg(f ), we have the following statements: (i) If d ≠ 0 and k

∑ bj (z) ≠ 0,

j=1,kj =0

then F(z) has at most one finite Borel exceptional value d∗ ≠ 0, which satisfies k d∗ d∗ − F(z) = = b (z). ∑ (d − f (z))2 d2 j=1,k =0 j j

(ii) If d ≠ 0 and k

∑ bj (z) = 0,

j=1,kj =0

then F(z) has no finite Borel exceptional values. (iii) If d = 0, then z = 0 is a Borel exceptional value of F(z) as well. Proof. (i) Suppose that d ≠ 0 is the Borel exceptional value of f (z) and k

∑ bj (z) ≢ 0.

j=1,kj =0

Clearly, the order ρ is an integer, and f (z) can be written in the form ρ

f (z) = d + π(z)eαz ,

(2.82)

where α ≠ 0 is a constant, and π(z) is a nonvanishing meromorphic function satisfying ρ(π) < ρ. Thus ρ

f (z + cj ) = d + π(z + cj )πj (z)eαz ,

(2.83)

54 | 2 Value distribution of complex delay-differential polynomials where πj is a meromorphic function of order ρ − 1. On the other hand, we may write g(f ) as g(f ) = ∑ bj (z)f (z + cj ) + ∑ bj (z)f (kj ) (z + cj ), j∈I1

(2.84)

j∈I2

where I1 = {1 ≤ j ≤ k : kj = 0} and I2 = {1 ≤ j ≤ k : kj > 0}. Hence, for every j ∈ I2 , ρ

f (kj ) (z + cj ) = Qkj (z)eαz ,

(2.85)

where Qkj is a meromorphic function of order less than ρ. By substituting (2.83) and (2.85) into (2.84) we get ρ

ρ

F(z) = d2 A(z) + d (B(z) + C(z) + π(z)A(z)) eαz + π(z) (B(z) + C(z)) e2αz ,

(2.86)

where A(z) = ∑ bj (z) (≢ 0), j∈I1

B(z) = ∑ π(z + cj )πj (z)bj (z) j∈I1

and C(z) = ∑ bj (z)Qkj (z). j∈I2

By Lemma 2.2.13, ρ(F) = ρ. If F(z) has a Borel exceptional value d∗ , then ρ

F(z) = d∗ + π ∗ (z)eβz ,

(2.87)

where β ≠ 0 is a constant, and π ∗ (z) (≢ 0) is a meromorphic function of order less than ρ. By (2.86) and (2.87) we have ρ

ρ

ρ

d(B(z) + C(z) + π(z)A(z))eαz + π(z)(B(z) + C(z))e2αz − π ∗ (z)eβz = d∗ − d2 A(z). (2.88) Case 1. If β ≠ α and β ≠ 2α, then by (2.88) and Theorem 1.1.32 in Section 1.1 we get π ∗ (z) ≡ 0, a contradiction. Case 2. If β = α and β ≠ 2α, then equation (2.88) may written as ρ

ρ

(d(B(z) + C(z) + π(z)A(z)) − π ∗ (z)) eαz + π(z)(B(z) + C(z))e2αz = d∗ − d2 A(z). By this and Theorem 1.1.32 again we get π ∗ (z) = and combining the result with (2.82), we obtain f (z)g(f )(z) = F(z) = d∗ + Thus g(f )(z) =

d∗ , d

d∗ π(z). d

Substituting this into (2.87)

ρ d∗ d∗ π(z)eαz = f (z). d d

contradicting the assumption that g(f ) is nonconstant.

2.3 Borel exceptional values of delay-differential polynomials | 55

Case 3. If β ≠ α and β = 2α, then equation (2.88) may written as ρ

ρ

d(B(z) + C(z) + π(z)A(z))eαz + (π(z)(B(z) + C(z)) − π ∗ (z)) e2αz = d∗ − d2 A(z). By this and Theorem 1.1.32 once more we get π ∗ (z) = − dd2 π 2 (z). Substituting this into (2.87) and combining the result with (2.82), we get ∗

F(z) = d∗ −

d∗ (f (z) − d)2 . d2

Clearly, d∗ ≠ 0 as we would have g(f ) ≡ 0 otherwise, a contradiction. Hence d∗ − F(z) d∗ = . (d − f (z))2 d2

(2.89)

We now prove the second identity. From (2.89) we get ∑ bj (z)

j∈I1

f (z + cj ) − d f (z) − d

+ ∑ bj (z)

f (kj ) (z + cj )

j∈I2

f (z) − d

d∗ + 2 = d

d∗ d

− d ∑j∈I1 bj (z) f (z) − d

.

Suppose, contrary to the claim, that d∗ − d2 ∑j∈I1 bj (z) ≢ 0. Then from the above equation we deduce 1 m (r, ) = O(r ρ−1+ϵ ) + Sλ (r, f ). f −d Since d is a Borel exceptional value of f , from this we obtain T (r,

1 ) = O(r ρ−1+ϵ ) + Sλ (r, f ), f −d

a contradiction. (ii) Suppose that d ≠ 0 and k

∑ bj (z) ≡ 0.

j=1,kj =0

Suppose that F has a finite Borel exceptional value d∗ . By the same proof as in (i), we obtain (2.88) as ρ

ρ

ρ

d(B(z) + C(z))eαz + π(z)(B(z) + C(z))e2αz − π ∗ (z)eβz = d∗ . If β ≠ α and β ≠ 2α, β = α and β ≠ 2α, or β ≠ α and β = 2α, then using Theorem 1.1.32 again, we get π ∗ (z) ≡ 0 in all three cases, a contradiction. (iii) Suppose that d = 0 is the Borel exceptional value of f . Using the same method as before, we obtain (2.86) with d = 0, ρ

F(z) = π(z)(B(z) + C(z))e2αz . Since ρ(F) = ρ, π(z)(B(z) + C(z)) ≢ 0, and since ρ(π(B + C)) < ρ, we deduce that d = 0 is the Borel exceptional value of f .

56 | 2 Value distribution of complex delay-differential polynomials Corollary 2.3.3. Under the hypotheses of Theorem 2.3.2, F(z) has no Borel exceptional value b such that k

b − d2 ∑ bj ≠ 0. j=1,kj =0

2.4 The zeros of [P(f )f (z + c)](k) − α(z) Observing that [f n+1 ]󸀠 = (n + 1)f n f 󸀠 , Wang and Fang [215, Corollary 1] extended Theorem 2.1.1 as follows. Theorem 2.4.1. Let f be a transcendental meromorphic function, and let n, k be two positive integers with n ≥ k + 1. Then (f n )(k) − 1 has infinitely many zeros. In this section, we consider the zeros of complex delay-differential polynomials [P(f )f (z +c)](k) −α(z) as delay-differential counterparts to Theorem 2.4.1. We first recall a result due to Liu, Liu, and Cao [144]: Theorem 2.4.2. Let f be a transcendental entire function of finite order, and let α(z) be a nonzero small function with respect to f (z). If n ≥ k + 2, then [f (z)n f (z + c)](k) − α(z) has infinitely many zeros. If f is not a periodic function with period c and n ≥ k + 3, then [f (z)n Δc f (z)](k) − α(z) has infinitely many zeros. Liu, Liu, and Yang [148] further considered the case with f (z)n replaced by P(f ). Here and in what follows, P(z) is a polynomial in z of deg(P) = n with t distinct zeros. Theorem 2.4.3. Let f be a transcendental entire function of hyperorder ρ2 (f ) < 1. If n ≥ t(k + 1) + 1, then [P(f )f (z + c)](k) − a(z) has infinitely many zeros for any nonzero small function a(z) with respect to f (z). Remark 2.4.4. (1) Theorem 2.4.3 is an improvement of Theorem 2.4.2 in the case t = 1 and an improvement of Theorem 2.1.6 in the case k = 0. z (2) Theorem 2.4.3 does not remain valid if ρ2 (f ) = 1. Indeed, take f (z) = ee , P(z) = z n , k ≥ 1, ec = −n, and a nonconstant polynomial a(z). Then [P(f )f (z + c)](k) − a(z) = −a(z) has finitely many zeros. (3) The condition a(z) ≠ 0 cannot be removed. Let f (z) = ez , P(z) = z n , and ec = −1. Then [P(f )f (z + c)](k) = −(n + 1)k e(n+1)z has no zeros. Theorem 2.4.5. Let f be a transcendental entire function of hyperorder ρ2 (f ) < 1, not periodic with period c. If n ≥ (t + 1)(k + 1) + 1, then [P(f )(Δc f )s ](k) − a(z) has infinitely many zeros for any nonzero small function a(z) with respect to f (z). Remark 2.4.6. The condition a(z) ≠ 0 cannot be removed in Theorem 2.4.5. This can be seen by taking f (z) = ez , P(z) = z n , and ec = 2. Then [P(f )Δc f ](k) = (n + 1)k e(n+1)z has no zeros.

2.4 The zeros of [P(f )f (z + c)](k) − α(z)

| 57

Proceeding similarly as in the proof of Lemma 2.1.5, we have the following three lemmas. We leave the detailed proofs as exercises for the reader. Lemma 2.4.7. Let f (z) be a transcendental meromorphic function of hyperorder ρ2 (f ) < 1. If F = P(f )f (z + c), then (n − 1)T(r, f ) + S(r, f ) ≤ T(r, F) ≤ (n + 1)T(r, f ) + S(r, f ).

(2.90)

If f (z) is a transcendental entire function of hyperorder ρ2 (f ) < 1, then T(r, F) = (n + 1)T(r, f ) + S(r, f ).

(2.91)

Lemma 2.4.8. Let f (z) be a transcendental entire function of hyperorder ρ2 (f ) < 1, and let s be a nonnegative integer. Then nT(r, f ) + S(r, f ) ≤ T(r, P(f )[f (z + c) − f (z)]s ) ≤ (n + s)T(r, f ) + S(r, f ).

(2.92)

Remark 2.4.9. Equalities in (2.92) may appear. To this end, consider the following two examples: If f (z) = ez and ec = 2, then T(r, f (z)n [f (z + c) − f (z)]s ) = T(r, e(n+s)z ) = (n + s)T(r, f ) + S(r, f ). If f (z) = ez + z and c = 2πi, then T(r, f (z)n [f (z + c) − f (z)]s ) = T(r, [ez + z]n (2πi)s ) = nT(r, f ) + S(r, f ). Lemma 2.4.10. Let f (z) be a transcendental meromorphic function of hyperorder ρ2 (f ) < 1. Then (n − s)T(r, f ) + S(r, f ) ≤ T(r, P(f )[f (z + c) − f (z)]s ) ≤ (n + 2s)T(r, f ) + S(r, f ).

(2.93)

Proofs of Theorems 2.4.3 and 2.4.5. Let F(z) = P(f )f (z + c). From Lemma 2.4.7 we know that F(z) is not a constant, and S(r, F) = S(r, F (k) ) = S(r, f ) follows. Assume that F(z)(k) − α(z) has finitely many zeros only. Combining the second main theorem for three small functions (Theorem 1.1.22) for a transcendental entire function f and Theorem 1.1.13, we get T(r, F (k) ) ≤ N(r, F (k) ) + N (r,

F (k)

) + N (r,

F (k)

1 ) + S(r, F (k) ) − α(z)

1 ) + S(r, F (k) ) − α(z) 1 ≤ T(r, F (k) ) − T(r, F) + N≤(k+1) (r, ) + S (r, F (k) ) . F

≤ N≤1 (r,

1

1

F (k)

) + N (r,

F (k)

Combining (2.91) with (2.94), we obtain (n + 1)T(r, f ) + S(r, f ) = T(r, F) ≤ N≤(k+1) (r,

1 ) + S(r, f ) F

(2.94)

58 | 2 Value distribution of complex delay-differential polynomials 1 1 ) + S(r, f ) ≤ t(k + 1)N (r, ) + N (r, f f (z + c) ≤ [t(k + 1) + 1]T(r, f ) + S(r, f ),

(2.95)

which is a contradiction to n ≥ t(k + 1) + 1. Thus Theorem 2.4.3 is proved. To prove Theorem 2.4.5, define G(z) := P(f )[Δc f ]s . Suppose that G(z)(k) − α(z) has only finitely many zeros. Proceeding similarly as before and using Lemma 2.4.8, we obtain 1 ) + S(r, f ) G 1 1 ≤ t(k + 1)N (r, ) + (k + 1)N (r, ) + S(r, f ) f f (z + c) − f (z) ≤ (t + 1)(k + 1)T(r, f ) + S(r, f ),

nT(r, f ) + S(r, f ) ≤ T(r, G) ≤ N≤(k+1) (r,

(2.96)

a contradiction to n ≥ (t + 1)(k + 1) + 1. For the case that f (z) is a transcendental meromorphic function in Theorems 2.4.3 and 2.4.5, we obtain the following results. Theorem 2.4.11. Let f be a transcendental meromorphic function of hyperorder ρ2 (f ) < 1. If n ≥ t(k + 1) + 5, then [P(f )f (z + c)](k) − a(z) has infinitely many zeros. Remark 2.4.12. Theorem 2.4.11 is a partial answer to a question raised by Luo and Lin [168, p. 448]. Theorem 2.4.13. Let f be a transcendental meromorphic function of hyperorder ρ2 (f ) < 1. If n ≥ (t + 2)(k + 1) + 3 + s, then [P(f )(Δc f )s ](k) − a(z) has infinitely many zeros. Proofs of Theorems 2.4.11 and 2.4.13. Let F(z) = P(f )f (z + c). From Lemma 2.4.7 we know that F(z) is not a constant, and S(r, F) = S(r, F (k) ) = S(r, f ) follows. Assume that F(z)(k) − α(z) has finitely many zeros only. Combining Theorem 1.1.22 again and (1.13) with the fact that f is a transcendental meromorphic function, we have 1 1 ) + N (r, (k) ) + S(r, F (k) ) F (k) F − α(z) 1 ≤ N(r, f ) + N(r, f (z + c)) + N≤1 (r, (k) ) F 1 (k) + N (r, (k) ) + S(r, F ) F − α(z) 1 ≤ 2T(r, f ) + T(r, F (k) ) − T(r, F) + N≤(k+1) (r, ) + S(r, F (k) ). F

T(r, F (k) ) ≤ N(r, F (k) ) + N (r,

Combining this with (2.90), we obtain (n − 1)T(r, f ) + S(r, f ) ≤ T(r, F) ≤ 2T(r, f ) + N≤(k+1) (r,

1 ) + S(r, f ) F

1 1 ≤ t(k + 1)N (r, ) + N (r, ) + 2T(r, f ) + S(r, f ) f f (z + c)

2.5 Zeros on delay-differential polynomials of exponential type

| 59

≤ [t(k + 1) + 3]T(r, f ) + S(r, f ), which is a contradiction to n ≥ t(k + 1) + 5. Thus Theorem 2.4.11 is proved. To prove Theorem 2.4.13, define G(z) := P(f )[Δc f ]s . Suppose that G(z)(k) − α(z) has finitely many zeros only. Using a similar method as before and Lemma 2.4.8, we get (n − s)T(r, f ) + S(r, f ) ≤ T(r, G) ≤ 2T(r, f ) + N≤(k+1) (r,

1 ) + S(r, f ) G

1 1 ) + S(r, f ) ≤ 2T(r, f ) + t(k + 1)N (r, ) + (k + 1)N (r, f f (z + c) − f (z) ≤ [(t + 2)(k + 1) + 2]T(r, f ) + S(r, f ), which is a contradiction to n ≥ (t + 2)(k + 1) + 3 + s, completing the proof.

Remark 2.4.14. We remark that n in the theorems of this section is probably not sharp, perhaps being deserved to be reduced somehow.

2.5 Zeros on delay-differential polynomials of exponential type To close this chapter, we give some results on the zeros of delay-differential polynomials of exponential type. Exponential-type functions are an important subclass of transcendental entire functions. Consider the exponential-type functions f (z) = s1 (z)et1 (z) + ⋅ ⋅ ⋅ + sn (z)etn (z) ,

(2.97)

where si (z) and ti (z) (i = 1, 2, . . . , n) are entire functions. If all si (z) are constants and all ti (z) are linear functions in z, then (2.97) is called an exponential sum. Considerations of exponential sums trace back to Ritt [194]. If all si (z) and ti (z) are nonzero polynomials in z, then (2.97) is called an exponential polynomial. Rewrite an exponential polynomial as f (z) = P1 (z)eQ1 (z) + ⋅ ⋅ ⋅ + Pk (z)eQk (z) ,

(2.98)

where Pj (z) and Qj (z) (j = 1, 2, . . . , k) are polynomials in z. Let q = max{deg Qj (z) : Qj (z) ≢ 0}, and let ω1 , . . . , ωm (1 ≤ m ≤ k) be pairwise different leading coefficients of the polynomials Qj (z) of maximum degree q. Following Steinmetz [202], (2.98) can be rewritten in the form q

q

f (z) = H0 (z) + H1 (z)ew1 z + ⋅ ⋅ ⋅ + Hm (z)ewm z ,

(2.99)

where Hj (z) ≢ 0 (j = 1, 2, . . . , m) are either exponential polynomials of degrees ≤ q − 1 or ordinary polynomials in z. The convex hull of a set W ⊂ ℂ, denoted by co(W), is the intersection of all convex sets containing W. If W contains only finitely many elements, then co(W) is obtained

60 | 2 Value distribution of complex delay-differential polynomials as an intersection of finitely many closed half-planes, and hence co(W) is either a compact polygon (with a nonempty interior) or a line segment. We denote the perimeter of co(W) by C(co(W)). If co(W) is a line segment, then C(co(W)) equals twice the length of this line segment. We fix the notations W := {w1 , . . . , wm } and W0 := {0, w1 , . . . , wm }. We recall the following two results, which are important when considering zeros of exponential polynomials; see [202, Satz 1, 2]. Lemma 2.5.1. Let f be given by (2.99). Then T(r, f ) = C(co(W0 ))

rq + o(r q ). 2π

(2.100)

Lemma 2.5.2. Let f be given by (2.99). If H0 (z) ≢ 0, then 1 m(r, ) = o(r q ), f that is, 1 rq N(r, ) = C(co(W0 )) + o(r q ), f 2π whereas if H0 (z) ≡ 0, then 1 rq N(r, ) = C(co(W)) + o(r q ). f 2π By the fact that the exponential polynomials are of finite order and by previous discussions, if looking at zeros of expressions of type f (z)n f (z + c), it is sufficient to consider the case n = 1 only. The following results can be found in [157]. Theorem 2.5.3. Let f be an exponential polynomial given by (2.99), let c ∈ ℂ\{0}, and let p(z) be a nonzero polynomial. (i) If m = 1 and ρ(f ) ≥ 2 or if m ≥ 2, then f (z)f (z + c) − p(z) has infinitely many zeros. (ii) If m = 1 and ρ(f ) = 1, then f (z)f (z + c) − p(z) has infinitely many zeros, except when f (z) = H0 (z)(1 + λeω1 z ) and p(z) = H0 (z)H0 (z + c), where eω1 c = −1, λ is a nonzero constant, and H0 (z) is a nonzero polynomial. Proof. We divide the proof into three cases. q

Case 1. If m = 1 and ρ(f ) = q ≥ 2, then f (z) = H0 (z) + H1 (z)eω1 z from (2.99), where H0 (z), H1 (z)(≢ 0) are exponential polynomials of degree ≤ q − 1 or ordinary polynomials in z. Thus q

q

f (z)f (z + c) − p(z) = T1 (z) + A1q (z)eω1 z + B1q (z)e2ω1 z ,

(2.101)

2.5 Zeros on delay-differential polynomials of exponential type

| 61

where T1 (z) = H0 (z)H0 (z + c) − p(z), { { { { q q A1q (z) = H0 (z)H1 (z + c)eω1 (z+c) −ω1 z + H0 (z + c)H1 (z), { { { { ω1 (z+c)q −ω1 z q . {B1q (z) = H1 (z)H1 (z + c)e

(2.102)

We first consider the case H0 (z) ≡ 0. Thus T1 (z) = −p(z) ≢ 0, A1q (z) ≡ 0, and B1q (z) ≢ 0. Lemma 2.5.2 yields N (r,

2|w1 | q 1 )= r + o(r q ). f (z)f (z + c) − p(z) π

Lemmas 2.5.1 and 2.1.5 imply that T(r, f (z)f (z + c)) = 2T(r, f ) + S(r, f ) =

2|w1 | q r + o(r q ). π

Hence f (z)f (z + c) − p(z) must have infinitely many zeros. If H0 (z) ≢ 0, then we state that A1q (z) ≢ 0. Otherwise, assume that q

A1q (z) = H0 (z)H1 (z + c)eω1 (z+c)

−ω1 z q

+ H0 (z + c)H1 (z) ≡ 0.

The difference analogue of logarithmic derivative lemma, see (1.22) in Section 1.2.2, implies that q

m (eω1 (z+c)

−ω1 z q

) = m(

H0 (z + c)H1 (z) ) = O(r q−2+ε ), H0 (z)H1 (z + c)

which is a contradiction. Thus A1q (z) ≢ 0. Obviously, B1q (z) ≢ 0. By Lemma 2.5.2, if T1 (z) ≡ 0, then N (r,

rq 1 ) = C(co(W)) + o(r q ), f (z)f (z + c) − p(z) 2π

where W = {ω1 , 2ω1 }, and if T1 (z) ≢ 0, then N (r,

1 rq ) = C(co(W0 )) + o(r q ), f (z)f (z + c) − p(z) 2π

where W0 = {0, ω1 , 2ω1 }. We easily see that both C(co(W)) and C(co(W0 )) are not equal to zero. Hence f (z)f (z + c) − p(z) must have infinitely many zeros by the above two equations and Lemma 2.5.1. Case 2. If m = 1 and ρ(f ) = q = 1, then using similar discussions as in Case 1 when H0 (z) ≡ 0, we get that f (z)f (z + c) − p(z) must have infinitely many zeros. According to

62 | 2 Value distribution of complex delay-differential polynomials Lemma 2.5.2 and similarly as in Case 1, when H0 (z) ≢ 0, we see that f (z)f (z + c) − p(z) must have infinitely many zeros except when T1 (z) = H0 (z)H0 (z + c) − p(z) ≡ 0,

{

(2.103)

A11 (z) = H0 (z)H1 (z + c)eω1 c + H0 (z + c)H1 (z) ≡ 0,

where H0 (z) and H1 (z) are nonzero polynomials, since p(z) is a nonzero polynomial. The second equation of (2.103) shows that eω1 c = −1 and H1 (z) ≡ λH0 (z), where λ is a nonzero constant. Case 3. If m ≥ 2, substitute (2.99) into f (z)f (z + c) − p(z). Then G1 (z) := f (z)f (z + c) − p(z) =

q q 1 m m ∑ ∑ (H (z)Hj (z + c)eωj (z+c) −ωj z 2 i=0 j=0 i q

+ Hj (z)Hi (z + c)eωi (z+c)

−ωi z q

q

)e(ωi +ωj )z − p(z),

where ω0 = 0. If i ≠ j (i, j ≥ 1), then ωi ≠ ωj . Using the difference analogue of logarithmic derivative lemma again, we get q

Hi (z)Hj (z + c)eωj (z+c)

−ωj z q

q

+ Hj (z)Hi (z + c)eωi (z+c)

−ωi z q

≢ 0.

If i = j (i, j ≥ 1), then this inequality is obviously valid. Denote G

W0 1 := {0, ωi , ωi + ωj },

i, j = 1, 2, . . . , m,

and W G1 := {ωi + ωj },

i, j = 1, 2, . . . , m.

G

It is not difficult to see that W0 1 and W G1 include at least two nonzero elements by G comparing the moduli of the elements. Then C(co(W0 1 )) and C(co(W G1 )) are not equal to zero no matter what H0 (z) equals to. Thus f (z)f (z + c) − p(z) must have infinitely many zeros by Lemma 2.5.2. Theorem 2.5.4. Let f be an exponential polynomial given by (2.99), let c ∈ ℂ\{0}, and let p(z) be a nonzero polynomial. (i) If m = 1 and ρ(f ) ≥ 2 or if m ≥ 2, then f (z)Δf (z) − p(z) has infinitely many zeros. (ii) If m = 1 and ρ(f ) = 1, then f (z)Δf (z) − p(z) has infinitely many zeros, except when f (z) = H0 (z) + H1 eω1 z or p(z) = H0 (z)(H0 (z + c) − H0 (z)), where eω1 c = 1, H0 (z) is a nonzero polynomial, and H1 is a nonzero constant. Proof. The proof can again be divided by three cases. q

Case 1. If m = 1 and ρ(f ) = q ≥ 2, substituting f (z) = H0 (z) + H1 (z)eω1 z into f (z)Δc f (z) − p(z), we get q

q

f (z)Δc f (z) − p(z) = T2 (z) + A2q (z)eω1 z + B2q (z)e2ω1 z ,

2.5 Zeros on delay-differential polynomials of exponential type

| 63

where T2 (z) = H0 (z)(H0 (z + c) − H0 (z)) − p(z), { { { { q q A2q (z) = H0 (z)[H1 (z + c)eω1 (z+c) −ω1 z − H1 (z)] + H1 (z)(H0 (z + c) − H0 (z)), { { { { ω1 (z+c)q −ω1 z q − H1 (z)]. {B2q (z) = H1 (z)[H1 (z + c)e If H0 (z) ≡ 0, then T2 (z) = −p(z) ≢ 0, A2q (z) ≡ 0, and B2q (z) ≢ 0 from (1.22) in Section 1.2.2. Again, Lemma 2.5.2 yields N (r,

2|w1 | q 1 )= r + o(r q ). f (z)Δc f (z) − p(z) π

Lemmas 2.5.1 and 2.4.8 imply that T(r, f (z)Δc f (z)) ≥ T(r, f ) + S(r, f ) =

2|w1 | q r + o(r q ). π

Hence f (z)Δc f (z) − p(z) has infinitely many zeros. If H0 (z) ≢ 0, then A2q (z) ≢ 0 and B2q (z) ≢ 0 from (1.22) in Section 1.2.2. Since ω1 ≠ 2ω1 , Lemma 2.5.2 implies that f (z)Δc f (z) − p(z) must have infinitely many zeros no matter what T2 (z) equals to. Case 2. If m = 1 and ρ(f ) = q = 1, we have f (z)Δc f (z) − p(z) = T2 (z) + A21 (z)eω1 z + B21 (z)e2ω1 z ,

(2.104)

where T2 (z) = H0 (z)(H0 (z + c) − H0 (z)) − p(z), { { { A21 (z) = H0 (z)[H1 (z + c)eω1 c − H1 (z)] + H1 (z)(H0 (z + c) − H0 (z)), { { { ω1 c {B21 (z) = H1 (z)[H1 (z + c)e − H1 (z)],

(2.105)

and H0 (z) and H1 (z) are polynomials. The case of H0 (z) ≡ 0 can be considered by a similar method as in Case 1. If f (z)Δc f (z) − p(z) has finitely many zeros only, then only two of T2 (z), A21 (z), B21 (z) vanish. The case where all T2 (z), A21 (z), and B21 (z) vanish at the same time is impossible by Lemma 2.4.8. Subcase 1. From T2 (z) = 0, B21 (z) = 0, and A21 (z) ≠ 0 we obtain that eω1 c = 1 and H1 (z) reduces to a nonzero constant H1 . Thus f (z) = H0 (z) + H1 eω1 z and H0 (z)(H0 (z + c) − H0 (z)) = p(z). Subcase 2. From T2 (z) = 0, A21 (z) = 0, and B21 (z) ≠ 0 we obtain that H1 (z) and H0 (z) are nonzero constants and eω1 c = 1. Thus B21 (z) = 0, a contradiction. Subcase 3. From A21 (z) = 0, B21 (z) = 0, and T2 (z) ≠ 0, we obtain that H1 (z) and H0 (z) are nonzero constants and eω1 c = 1. Then f (z) = H0 + H1 eω1 z .

64 | 2 Value distribution of complex delay-differential polynomials Case 3. If m ≥ 2, then we have G2 (z) = f (z)Δf (z) − p(z) =

q q 1 m m ∑ ∑ (Hi (z)[Hj (z + c)eωj (z+c) −ωj z − Hj (z)] 2 i=0 j=0 q

+ Hj (z)[Hi (z + c)eωi (z+c)

−ωi z q

q

− Hi (z)])e(ωi +ωj )z − p(z).

(2.106)

Using a similar method as in Case 3 of the proof of Theorem 2.5.3, we obtain that f (z)Δc f (z) − p(z) has infinitely many zeros. Remark 2.5.5. The exceptional cases in Theorems 2.5.3 and 2.5.4 may in fact appear. For example, if f (z) = zez −z, ec = −1, and a = z(z +c), then f (z)f (z +c)−a = −z(z +c)e2z has finitely many zeros. If f (z) = ez + z, ec = 1, and a = cz, then f (z)Δc f − a = cez has no zeros. Theorem 2.5.6. Let f be an exponential polynomial given by (2.99), let c ∈ ℂ\{0}, let p(z) be a nonzero polynomial, and let k be a positive integer. Then f (z)f (k) (z + c) − p(z) has infinitely many zeros. Proof. We consider the case of k = 1 only in some detail; in the case of k ≥ 2, we only recall a similar reasoning in the proof of Theorem 2.5.3. q

Case 1. If k = 1, then substituting f (z) = H0 (z) + H1 (z)eω1 z into f (z)f 󸀠 (z + c) − p(z), we have q

q

f (z)f 󸀠 (z + c) − p(z) = T3 (z) + A3q (z)eω1 z + B3q (z)e2ω1 z , where 󸀠

T3 (z) = H0 (z)H0 (z + c) − p(z), { { { { q q { { {A3q (z) = [H0 (z)H1󸀠 (z + c) + H0 (z)H1 (z + c)ω1 q(z + c)q−1 ]eω1 (z+c) −ω1 z { { + H0󸀠 (z + c)H1 (z), { { { { { 󸀠 q−1 ω1 (z+c)q −ω1 z q . {B3q (z) = [H1 (z)H1 (z + c) + H1 (z)H1 (z + c)ω1 q(z + c) ]e

(2.107)

We easily see that B3q (z) ≢ 0, since otherwise H1 (z) would be of order q. The case of H0 (z) ≡ 0 can be considered in a similar way as in the proof of Theorem 2.5.4. Assume that H0 (z) ≢ 0. We also state that A3q (z) ≢ 0. Otherwise, if q ≥ 2, then [

q q H 󸀠 (z + c) H1󸀠 (z + c) H1 (z + c) + ω1 q(z + c)q−1 ]eω1 (z+c) −ω1 z = − 0 , H1 (z) H1 (z) H0 (z)

which is a contradiction to (1.22) in Section 1.2.2 and the logarithmic derivative lemma for finite-order meromorphic functions. If q = 1 and if H0 (z) and H1 (z) are nonzero polynomials, then A31 (z) ≢ 0 from the expression of A31 (z). Hence we see that at least

2.5 Zeros on delay-differential polynomials of exponential type

| 65

two of T3 (z), A3 (z), and B3 (z) are not equal to zero. Using Lemma 2.5.2 again, we have that f (z)f 󸀠 (z + c) − p(z) must have infinitely many zeros. Case 2. If k ≥ 2, then f (z)f (k) (z + c) − p(z) =

q q 1 k k ∑ ∑ (Hi (z)[Hj󸀠 (z + c) + Hj (z + c)ωj q(z + c)q−1 ]eωj (z+c) −ωj z 2 i=0 j=0 q

+ Hj (z)[Hi󸀠 (z + c) + Hi (z + c)ωi q(z + c)q−1 ]eωi (z+c)

−ωi z q

)e(ωi +ωj )z

q

− p(z), where ω0 = 0. Using the similar method as in the case m ≥ 2 in Theorem 2.5.3, we obtain that f (z)f (k) (z + c) − p(z) has infinitely many zeros. To close this section, recall that the difference analogues of the logarithmic derivative lemma are valid for the meromorphic functions of finite order or of hyperorder less z than one. However, observe that if f (z) = zee and ec = −n, then f (z)n f (z + c) − a = z z n (z + c) − a has finitely many zeros, and f (z)n Δc f − z n (z + c) = −z n+1 e(n+1)e has finitely z many zeros. If f (z) = ee + 1, then f (z)n f (z + c) − a and f (z)n Δc f − a have infinitely many zeros, where a is a nonzero constant, and ec = −n. Hence it seems that we can raise the following question. Question 2.5.7. Let f (z) be an entire function of infinite order with infinitely many zeros, let a(z) be a nonzero polynomial, and let c ∈ ℂ\{0}. Under what kind of conditions, f (z)n f (z + c) − a(z) may have infinitely many zeros for n ≥ 2? We give a preliminary result for some entire functions of infinite order. z

Theorem 2.5.8. Let f (z) = p(z)ee + q(z), where p(z)(≢ 0), a(z)(≢ 0), and q(z) are polynomials, and n ≥ 2. If f (z)n f (z + c) − a(z) has finitely many zeros, then q(z) ≡ 0 and ec = −n. z

Remark 2.5.9. (1) Theorem 2.5.8 implies that if f (z) = p(z)ee +q(z) has infinitely many zeros and ec ≠ −n, then f (z)n f (z + c) − a(z) has infinitely many zeros, where p(z), q(z), and a(z) are nonzero polynomials. (2) Question 2.5.7 is not valid when n = 1 or the polynomial a(z) is replaced by z a small function with respect to f (z). This can be seen by taking f (z) = (1 + ez )ee . Obviously, f (z) has infinitely many zeros, but f (z)f (z + iπ) − 1 = −e2z has no zeros, and f (z)n f (z + c) − α(z) = −ne(n+1)z has no zeros, where ec = −n, and α(z) = (1 + ez )n (1 − nez ) + ne(n+1)z . Proof of Theorem 2.5.8. If f (z)n f (z+c)−a(z) has finitely many zeros, then we may apply the Hadamard factorization theorem to write f (z)n f (z+c)−a(z) = H(z)eh(z) , where H(z)

66 | 2 Value distribution of complex delay-differential polynomials is a nonzero polynomial, and h(z) is an entire function with ρ(h) ≤ 1. If q(z) ≢ 0, then z

[p(z)ee + q(z)]n [p(z + c)ee

z+c

+ q(z + c)] − a(z) = H(z)eh(z) .

Hence we have p(z)n p(z + c)e(n+e n

+ q(z) p(z + c)e +

Cn1 q(z

ec ez

c

)ez

+ Cn1 p(z)n−1 q(z)p(z + c)e(n−1+e n nez

+ q(z + c)p(z) e n−1 ez

+ c)p(z)q(z)

n

+

Cn1 q(z

c

)ez

n−1

+ c)p(z)

e + q(z) q(z + c) − a(z) = H(z)e

+ ⋅⋅⋅ q(z)e

h(z)

(n−1)ez

(2.108) + ⋅⋅⋅

.

Since n ≥ 2, at least three elements in the set I := {n + ec , (n − 1) + ec , . . . , ec , n, n − 1, . . . , 1} are different. This observation is important when using Theorems 1.1.32 and 1.1.34. If ec = ±1, then the cardinality of I is n + 1. If ec ≠ ±1, then the cardinality of I is greater than n + 1. We proceed to discussing the following four cases. Without loss of generality, we assume that the elements in I are different from each other. Case 1. h(z) is a polynomial, and H(z)eh(z) − q(z)n q(z + c) + a(z) ≡ 0. In this case, using Theorem 1.1.32, we get a contradiction. Case 2. h(z) is a polynomial, and H(z)eh(z) −q(z)n q(z+c)+a(z) ≢ 0, which are small functions with respect to f (z). Dividing both sides of (2.108) by H(z)eh(z) − q(z)n q(z + c) + a(z), we get a contradiction by using Theorem 1.1.32, since any terms cannot equal one. Case 3. h(z) is a transcendental entire function, and q(z)n q(z + c) − a(z) ≡ 0. In this case, h(z) may take one of the elements in the set {(n + ec )ez , ((n − 1) + ec )ez , . . . , ec ez , nez , (n − 1)ez , . . . , ez }. Anyway, we also get a contradiction from Theorem 1.1.32. Case 4. h(z) is a transcendental entire function, and q(z)n q(z + c) − a(z) ≢ 0. We can get a contradiction with Theorem 1.1.32. Thus we get q(z) ≡ 0, and we have z

p(z)n ene p(z + c)ee n nez

ez+c

z+c

h(z)

− a(z) = H(z)eh(z) .

p(z+c)e and f2 := −H(z)e . Thus we have f1 + f2 = 1. To avoid a conLet f1 := p(z) e a(z) a(z) tradiction to the second main theorem for three small functions, f1 and f2 should be constants, and thus ec = −n, and h(z) should be a constant.

3 Uniqueness of delay-differential polynomials In this chapter, we mainly present uniqueness results for meromorphic functions concerning difference or delay-differential variants on classical results.

3.1 Value sharing on f with its linear delay-differential operator Recall the standard notation in the uniqueness theory of meromorphic functions. Sup̂ = C ⋃{∞}, resp., pose that f and g are nonconstant meromorphic functions and a ∈ C a is a small meromorphic function in the usual Nevanlinna theory sense. We denote by E(a, f ) the set of points z ∈ C where f (z) = a, resp., f (z) = a(z). We say that f and g share a IM (ignoring multiplicities) if E(a, f ) = E(a, g). If E(a, f ) = E(a, g) and the multiplicities of the zeros of f (z) − a and g(z) − a are the same at each z ∈ C, then f and g share a CM (counting multiplicities). The classical results in the uniqueness theory of meromorphic functions are five IM and four CM theorems due to Nevanlinna [178]; see also [90, 235]. The assumption of 4 CM was improved to 3 CM + 1 IM and 2 CM + 2 IM by Gundersen [69, 70], whereas the case 1 CM + 3 IM still remains an open problem. Heittokangas et al. [98, 97] considered the value sharing problem between f (z) and f (z + c). These results may be viewed as 3 CM and 2 CM + 1 IM in difference. Denote ̂ ) = S(f ) ∪ {∞}, where S(f ) is the set of small functions with respect to f . S(f Theorem 3.1.1 ([97]). Let f be a transcendental meromorphic function of finite order, let ̂ ) be three distinct periodic functions with period c. If c ∈ ℂ\{0}, and let a1 , a2 , a3 ∈ S(f f (z) and f (z + c) share a1 , a2 CM and a3 IM, then f (z) = f (z + c) for all z ∈ ℂ. Many authors have attempted to relax the value sharing conditions; we suggest the reader to see, e. g., [20, 141, 183, 242]. For the case 1 CM + 2 IM in Theorem 3.1.1, the question remains open. Note that Bergweiler and Langley [14, Lemma 3.5] proved that there exists an ε-set E such that f (z + c) − f (z) = cf 󸀠 (z)(1 + o(1)) as z → ∞

in ℂ \ E

uniformly in c for |c| ≤ h, where f (z) is a transcendental meromorphic function of order ρ(f ) < 1, and h > 0 is a constant. Qi, Li, and Yang [187] considered the uniqueness in the case where f 󸀠 (z) shares values with Δf as follows. Theorem 3.1.2. Let f (z) be a meromorphic function of finite order. Suppose that f 󸀠 and Δf share four distinct finite values a1 , a2 , a3 , a4 IM. Then f = Δf . The proof of Theorem 3.1.2 depends on the following two results on the uniqueness of meromorphic functions. https://doi.org/10.1515/9783110560565-003

68 | 3 Uniqueness of delay-differential polynomials Lemma 3.1.3 ([235, Theorem 2.16]). Let f (z) and g(z) be two nonconstant rational functions, and let a1 , a2 , a3 , a4 be four distinct values. If f (z) and g(z) share a1 , a2 , a3 , a4 IM, then f (z) = g(z). Lemma 3.1.4 ([73, Lemma 1]). Let f (z) and g(z) be two distinct nonconstant meromorphic functions, and let a1 , a2 , a3 , a4 ∈ ℂ ∪ {∞} be four distinct values. If f (z) and g(z) share a1 , a2 , a3 , a4 IM, then (1) T(r, f ) = T(r, g) + O(log(rT(r, f ))); 1 ) + O(log(rT(r, f ))) as E ∋ r → ∞, where E ⊂ (1, ∞) is of (2) 2T(r, f ) = ∑4i=1 N(r, f −a i finite linear measure. Proof of Theorem 3.1.2. Suppose, on the contrary, that f 󸀠 ≠ Δf . If f (z) is a rational function, then the conclusion follows by Lemma 3.1.3. Assume that f (z) is transcendental. Then f 󸀠 (z) must be transcendental. Case 1. If f 󸀠 is transcendental and Δf is a rational function, then Lemma 3.1.4(1) implies that T(r, f 󸀠 ) = T(r, Δf ) + O(log(rT(r, f 󸀠 ))) = O(log(rT(r, f 󸀠 ))), which is a contradiction. Case 2. If both f and Δf are transcendental, then applying the second main theorem (Theorem 1.1.21) and Lemma 3.1.4(2), we obtain 4

3T(r, f 󸀠 ) ≤ N(r, f 󸀠 ) + ∑ N (r, i=1

1 ) + S(r, f 󸀠 ) f 󸀠 − ai

≤ N(r, f 󸀠 ) + 2T(r, f 󸀠 ) + S(r, f 󸀠 ) ≤ 3T(r, f 󸀠 ) + S(r, f 󸀠 ), which implies that N(r, f 󸀠 ) = N(r, f 󸀠 ) + S(r, f 󸀠 ), that is, N(r, f ) + N(r, f ) = N(r, f ) + S(r, f 󸀠 ). Hence N(r, f ) = S(r, f 󸀠 ), which means that N(r, f 󸀠 ) = S(r, f 󸀠 ). From Lemma 3.1.4(1) we have T(r, f 󸀠 ) = T(r, Δf ) + S(r, f 󸀠 ).

(3.1)

3.1 Value sharing on f with its linear delay-differential operator

| 69

Combining this equality and (3.1), it follows that 4

3T(r, f 󸀠 ) ≤ N(r, f 󸀠 ) + ∑ N (r, i=1

󸀠

≤ N(r, f ) + N (r,

1 ) + S(r, f 󸀠 ) f 󸀠 − ai

1 ) + S(r, f 󸀠 ) f 󸀠 − Δf

≤ T(r, f 󸀠 ) + T(r, Δf ) + S(r, f 󸀠 ) ≤ 2T(r, f 󸀠 ) + S(r, f 󸀠 ), which is a contradiction. The conclusion holds.

Theorem 3.1.5. Let f (z) be a meromorphic function such that its order of growth is neither an integer nor infinite, and let a1 , a2 be two distinct values. If f 󸀠 and Δf share a1 , a2 , ∞ CM, then f 󸀠 = Δf . Proof. If f (z) is a finite-order meromorphic function, then ρ(Δf ) ≤ ρ(f ) by equation (1.29). Since f 󸀠 and Δf share the values a1 , a2 , and ∞ CM, then we have f (z + 1) − f (z) − a1 = eP(z) f 󸀠 (z) − a1

(3.2)

f (z + 1) − f (z) − a2 = eQ(z) , f 󸀠 (z) − a2

(3.3)

and

where P(z) and Q(z) are polynomials such that max{deg P(z), deg Q(z)} ≤ ρ(f ). Since ρ(f ) is not an integer, we see that max{deg P(z), deg Q(z)} < ρ(f ).

(3.4)

It follows from (3.2) and (3.3) that (eQ(z) − eP(z) )f 󸀠 (z) = a1 − a2 + a2 eQ(z) − a1 eP(z) .

(3.5)

To avoid a contradiction in (3.5), it follows that eQ(z) = eP(z) . From (3.5) we obtain (a1 − a2 )(1 − eQ(z) ) = 0. Since a1 ≠ a2 , we get eQ(z) = 1, and hence we finish our proof from (3.3). Qi, Li, and Yang [187] also considered the case where f 󸀠 (z) and f (z + c) share two values, where c is a nonzero complex number.

70 | 3 Uniqueness of delay-differential polynomials Theorem 3.1.6. Let f (z) be a transcendental entire function of finite order, and let a ≠ 0. If f 󸀠 (z) and f (z + c) share 0, a CM, then f 󸀠 (z) = f (z + c). Proof. Similarly as in the proof of Theorem 3.1.5, we get f 󸀠 (z) = eP(z) f (z + c)

(3.6)

f 󸀠 (z) − a = eQ(z) , f (z + c) − a

(3.7)

and

where P(z) and Q(z) are polynomials such that max{deg P(z), deg Q(z)} ≤ ρ(f ). If eP(z) = eQ(z) , then f 󸀠 (z) = f (z + c) from (3.6) and (3.7). Next, we consider the case where eP(z) ≠ eQ(z) . Rewrite (3.6) as follows: f 󸀠 (z) f (z) = eP(z) . f (z) f (z + c) Applying the logarithmic derivative lemma (Proposition 1.1.8) and its difference version (1.22), we obtain T(r, eP(z) ) = m(r, eP(z) ) = O(r ρ(f )−1+ε ) + O(log r). This implies that ρ(eP(z) ) < ρ(f ). If ρ(eQ(z) ) < ρ(f ), then we get max{deg P(z), deg Q(z)} < ρ(f ). Using a similar reasoning as in proof of Theorem 3.1.5, we have f 󸀠 (z) = f (z + c). If ρ(eQ(z) ) = ρ(f ), then T(r, eP(z) ) = S(r, eQ(z) ). From (3.6) and (3.7) it follows that f 󸀠 (z) =

aeQ(z) − a P(z) e . eQ(z) − eP(z)

(3.8)

From (3.8) and the assumption that f (z) is entire we know that eP(z) − eQ(z) = 0, and hence eP(z) = eQ(z) = 1. Thus N (r,

1 1 ) ≤ N (r, P(z) ) = S(r, eQ(z) ). eQ(z) − eP(z) e −1

According to the second main theorem for three small functions (Theorem 1.1.22), we get T(r, eQ(z) ) = S(r, eQ(z) ), which is a contradiction.

3.1 Value sharing on f with its linear delay-differential operator

| 71

Remark 3.1.7. Theorem 3.1.6 is also related to the classical result given by Rubel and Yang [197]: if a transcendental entire function f (z) shares two values a, b CM with its first derivative f 󸀠 , then f = f 󸀠 . This result was improved to 2 IM by Mues and Steinmetz [176], to f and f (k) sharing a, b CM by Yang [239], and to f and f (k) sharing a, b IM by Li and Yang [133]. For the details on these results, also see [235, Chapter 8]. The Brück conjecture [17], related to an entire function f (z) that shares one value or one small function with its derivative, has many investigations of itself and improvements. Brück Conjecture. Let f be a nonconstant entire function of hyperorder ρ2 (f ) < +∞, where ρ2 (f ) is not a positive integer. If f and f 󸀠 share one finite value a CM, then f − a = c(f 󸀠 − a) for some constant c ≠ 0. We recall some partial results on the Brück conjecture. – The cases a = 0 or a ≠ 0 with the condition N(r, f1󸀠 ) = S(r, f ) were proved by Brück [17]. – The case where f is of finite order was proved by Gundersen and Yang [74] using the properties of transcendental entire solutions of complex differential equations. – For entire functions of infinite order, Chen and Shon [32] proved that the conjecture holds if ρ2 (f ) < 21 . – Cao [18] settled the case ρ2 (f ) = 21 by studying the infinite hyperorder solutions of the linear differential equation f (k) + A(z)f = Q(z), where Q(z) is an entire function of finite order, and A(z) is a transcendental entire function. The difference version of Brück conjecture considered by Heittokangas et al. [97] can be stated as follows. Theorem 3.1.8 ([97, Theorem 1]). Let f be an entire function of hyperorder ρ(f ) < 2. If f (z) and f (z + c) share the value a CM, then f (z + c) − a =τ f (z) − a

(3.9)

for a nonzero constant τ. 2

Remark 3.1.9. (1) Heittokangas et al. [97] also provided the function f (z) = ez + 1 to show that ρ(f ) < 2 cannot be reduced to ρ(f ) ≤ 2. Obviously, f (z + c) and f (z) share the value 1 CM, and 2 f (z + c) − 1 = e2cz+c . f (z) − 1

Note that the shared value 1 is also the Borel exceptional value of f . (2) Equation (3.9) can also be written as a first-order linear difference equation f (z + c) − τf (z) = a(1 − τ).

72 | 3 Uniqueness of delay-differential polynomials Log τ

Then f (z) = d(z)e c z + a, where Log denotes the principal branch of the logarithm, and d(z) is a periodic function with period c. Obviously, if τ = 1, then f is a periodic function with period c. Furthermore, Li and Gao [136, Theorem 1.6] showed that if f (z) is a transcendental entire function of finite order and f (z + c) and f (z) share the value a CM, then f can be described as follows. Theorem 3.1.10. Let f (z) be a nonperiodic transcendental entire function of finite order ρ(f ) < +∞, and let c be a nonzero constant. If f (z) and f (z + c) share a finite value a CM, then ρ(f ) ≥ 1 and f (z) satisfies exactly one of the following cases: (i) ρ(f ) ≤ λ(f − a) + 1, and f (z) is of the form f (z) = P(z)eQ(z) + a, where P(z) is an entire function such that ρ(P) = λ(f − a), and Q(z) is a polynomial such that deg(Q) ≤ ρ(P) + 1; (ii) ρ(f ) > λ(f − a) + 1, and f (z) is of the form f (z) = P(z)eQ(z) + a, where P(z) is an entire function such that ρ(P) = λ(f − a), and Q(z) is a polynomial such that deg(Q) = ρ(P) ≥ 3; (iii) ρ(f ) > λ(f − a) + 1, and f (z) is of the form f (z) = eQ(z) + a, where Q(z) is a polynomial such that deg(Q) ≥ 2. Replacing f (z +c) with Δη f (z) := f (z +η)−f (z) in Theorem 3.1.8, Chen [30, Theorem 1.1] and Chen and Yi [27] proved the following theorem. Theorem 3.1.11. Let f (z) be a transcendental entire function of finite order with finite Borel exceptional value α, and let c ∈ ℂ be a constant such that f (z + c) ≢ f (z). If Δc f (z) and f (z) share the value a ≠ α CM, then Δc f (z) − a a = . f (z) − a a−α If a = α, then a = 0, and f (z + c) = (A + 1)f (z), where A ≠ 0, −1 is a constant. Some results on difference analogues of the Brück conjecture also can be found in [138, 137] and [29, Chapter 11]. It should be of interest to explore the delay-differential analogue of the Brück conjecture by the following example. z

Example 3.1.12. Take f (z) = ee and ec = 1. Then

f (z+c) f 󸀠 (z)

= e−z , where f 󸀠 (z) and f (z + c)

share 0 CM. If f (z) = ez − 1 and ec = 2, then f 󸀠 (z) and f (z + c) share the value 1 CM, and f 󸀠 (z)−1 = 21 . f (z+c)−1 Liu and Dong [149, Theorem 4.3] obtained the following result.

Theorem 3.1.13. Let f (z) be a transcendental entire function of finite order. Suppose that f (z) has a Picard exceptional value d and that f 󸀠 (z) and f (z + η) share the con-

3.1 Value sharing on f with its linear delay-differential operator

| 73

stant a CM. Then f 󸀠 (z) − a = A, f (z + η) − a where A is a nonzero constant. Furthermore, if a ≠ 0 and a ≠ d, then A =

a . a−d

Proof. Since f (z) has a Picard exceptional value d, we have f (z) = d + eh(z) , where ρ(f ) = deg(h(z)), and h(z) is a nonconstant polynomial. Thus f (z + c) = d + eh(z+c) and f 󸀠 (z) = h󸀠 (z)eh(z) . Since f 󸀠 (z) and f (z + c) share the constant a CM, we have a polynomial, and deg p(z) ≤ deg(h(z)). Thus we have

f 󸀠 (z)−a f (z+c)−a

= ep(z) , where p(z) is

h󸀠 (z)eh(z) − (d − a)ep(z) − ep(z)+h(z+c) = a.

(3.10)

Case 1: a = 0. Then (3.10) implies that h󸀠 (z)eh(z) − dep(z) − ep(z)+h(z+c) = 0.

(3.11)

If d = 0, then h󸀠 (z)eh(z) = ep(z)+h(z+c) . Thus h󸀠 (z) has no zeros, so h(z) is a polynomial with deg(h(z)) = 1. Then p(z) must be a constant. If d ≠ 0, then (3.11) changes into h󸀠 (z)eh(z)−p(z) eh(z+c) − = 1. d d h(z)−p(z)

Define f1 (z) = h (z)ed and f2 (z) = e main theorem of Nevanlinna, we have 󸀠

T(r, f2 ) ≤ N (r,

h(z+c)

d

(3.12)

. Hence f1 (z) − f2 (z) = 1. Using the second

1 1 ) + N (r, ) + N(r, f2 ) + S(r, f2 ) = S(r, f2 ), f2 f2 + 1

which is a contradiction. Thus, if a = 0, then d = 0, and p(z) reduces to a constant. Case 2: a ≠ 0 and a ≠ d. From Lemma 1.1.34 and (3.10) it follows that −ep(z)+h(z+c) = a and h󸀠 (z)eh(z) − (d − a)ep(z) = 0 or −(d − a)ep(z) = a. In the former case, we have that h(z) is a polynomial of degree one and p(z) + h(z + c) and p(z) − h(z) are constants simultaneously, thus p(z) must be a constant. Then h(z) also must be a constant. This is a contradiction since h(z) is a polynomial of degree one. In the second case, ep(z) = a , and hence p(z) is a constant. If a ≠ 0 and a = d, we easily get a contradiction a−d from (3.10). Thus we have completed the proof of Theorem 3.1.13.

74 | 3 Uniqueness of delay-differential polynomials Look next for complex homogeneous linear delay-differential polynomials of f (z): ψ(f ) := ∑ aj f (kj ) (z + ηj ) + ∑ bi f (z + ζi ), j∈J

i∈I

(3.13)

where kj (j ∈ J) > 0 are positive integers, ηj and ζi are complex constants, and aj and bi are nonzero constants, where I and J are two finite index sets. A general result can be stated as follows; see [44, Theorem 1.1]. Theorem 3.1.14. Let f (z) be a transcendental entire function of finite order, and let ψ(f ) be defined in (3.13) with ∑i∈I bi = 0. Suppose that ψ(f ) and f (z) share a finite value a CM and f (z) has a finite exceptional value α ≠ a. (i) If a ≠ 0 and α is a Nevanlinna exceptional value, then ψ(f ) − a = τ(≠ 0); f (z) − a (ii) if α is a Borel exceptional value, then ψ(f ) − a a = . f (z) − a a−α Example 3.1.15. Let ψ(f ) = 5f 󸀠󸀠 (z + log 3) − 3f 󸀠 (z + log 2) − 3f (z + log 4) + 4f (z), and let f (z) = ez . Then f (z) has a Borel exceptional value 0. Thus ψ(f ) and f (z) share the value 1 and satisfy 5f 󸀠󸀠 (z + log 3) − 3f 󸀠 (z + log 2) − 3f (z + log 4) + 4f (z) − 1 = 1. f (z) − 1 If ψ(z) = [f (z + η) − f (z)](n) , then we have an improvement of Theorem 3.1.11. Corollary 3.1.16. Let f (z) be a transcendental entire function of finite order such that (n) f (z) ≢ f (z + η), where η ∈ ℂ\{0}. Suppose that Δ(n) and f (z) η f (z) = [f (z + η) − f (z)] share a finite value a CM and that f (z) has a Borel exceptional value α ≠ a. Then, we have a ≠ 0 and Δ(n) η f (z) − a f (z) − a

=

a , a−α

where n is a nonnegative integer. Example 3.1.17. Suppose that f (z) = ez + 1 and η = log 3. Then f (z) has a Borel exceptional value 1. Thus Δ(n) η f (z) and f (z) share the value 2 for every n ≥ 0 and satisfy Δ(n) η f (z) − 2 f (z) − 2

=

2 = 2. 2−1

3.1 Value sharing on f with its linear delay-differential operator

| 75

Proof of Theorem 3.1.14. We first prove case (i). Since f is a transcendental entire function of finite order, we have that ψ(f ) is of finite order as well. Since ψ(f ) and f (z) share the value a CM, we obtain ψ(f ) − a = eϕ(z) , f (z) − a

(3.14)

where ϕ(z) is a polynomial. We proceed to show that ϕ(z) is a constant. If not, then assume that deg(ϕ(z)) ≥ 1. Differentiating (3.14) and eliminating eϕ(z) , we obtain ϕ󸀠 (z) =

ψ󸀠 (f ) f 󸀠 (z) − . ψ(f ) − a f (z) − a

(3.15)

Obviously, m(r, ϕ󸀠 (z)) = S(r, f ). Rewriting (3.15), we have ϕ󸀠 (z) = (f (z) − α){

ψ󸀠 (f ) 1 1 f 󸀠 (z) − }. f (z) − α ψ(f ) − a f (z) − α f (z) − a

(3.16)

Furthermore, from (3.15) we also obtain f (z) − α f (z) − α Φ(f ) + Ψ(f ), a α−a

(3.17)

ψ󸀠 (f ) ψ󸀠 (f )ψ(f ) − , (f (z) − α)(ψ(f ) − a) f (z) − α

(3.18)

ϕ󸀠 (z) = where Φ(f ) = Ψ(f ) =

f 󸀠 (z) f 󸀠 (z) − . f (z) − α f (z) − a

(3.19)

Note that ϕ󸀠 (z) ≢ 0. Thus by (3.17) we obtain 1 = f (z) − α

1 Φ(f ) a

+

1 Ψ(f ) α−a

ϕ󸀠 (z)

.

(3.20)

Since ϕ(z) is a polynomial, we have m (r,

1 ) ≤ m(r, Φ(f )) + m(r, Ψ(f )) + S(r, f ). f (z) − α

Considering that f (z) is of finite order ρ(f ) and ∑i∈I bi = 0, we obtain by the logarithmic derivative lemma and its difference counterpart (Proposition 1.1.8 and (1.22)) that ψ(f ) ψ󸀠 (f ) ψ󸀠 (f ) ) + m (r, ) + m (r, ) + O(1) f (z) − α ψ(f ) − a f (z) − α ψ(f ) = m (r, ) + S(r, f ) f (z) − α

m(r, Φ(f )) ≤ m (r,

76 | 3 Uniqueness of delay-differential polynomials

= m (r, ∑ aj j∈J

≤ ∑ m (r,

f

= m (r, ∑ bi ≤ ∑ m (r, i∈I

f (z) − α

(kj )

j∈J

i∈I

f (kj ) (z + ηj ) (z + ηj )

f (z) − α

+ ∑ bi i∈I

f (z + ζi ) ) + S(r, f ) f (z) − α

) + m (r, ∑ bi i∈I

f (z + ζi ) ) + S(r, f ) f (z) − α

f (z + ζi ) − α ) + S(r, f ) f (z) − α

f (z + ζi ) − α ) + S(r, f ) = S(r, f ). f (z) − α

Using a similar method, we also have m(r, Ψ(f )) = S(r, f ). 1 Thus we obtain m(r, f (z)−α ) = S(r, f ), that is, δ(α, f ) = 0. Hence we get that α is not a Nevanlinna exceptional value of f (z), which is a contradiction. Thus ϕ is a constant, and hence (i) holds.

(ii) We first prove that a ≠ 0. Contrary to our statement, we assume that ψ(f ) and f (z) share the value 0 CM. Then we obtain ψ(f ) = eh(z) , f (z)

(3.21)

where h(z) is a polynomial. Since f (z) has a finite Borel exceptional value α ≠ 0, f (z) can be written in the form f (z) = A(z)eP(z) + α,

(3.22)

where A(z) is a nonzero entire function, P(z) is a polynomial such that deg(P(z)) ≥ 1, and A(z) and P(z) satisfy λ(A) = ρ(A) = λ(f − α) < ρ(f ) = deg(P(z)).

(3.23)

For every integer kj > 0 (j ∈ J), we have (A(z)eP(z) )(kj ) = φj (z)eP(z) ,

(3.24)

where φj (z) is a differential polynomial formed by A(z) and P(z) and their derivatives. By (3.23) we get T(r, φ(z)) = o{T(r, eP(z) )} and ρ(φ(z)) < ρ(f ) = deg(P(z)). Substituting (3.22) and (3.24) into (3.21), we obtain ∑j∈J aj φj (z)eP(z+ηj ) + ∑i∈I bi [A(z + ζi )eP(z+ζi ) + α] A(z)eP(z) + α

= eh(z) .

(3.25)

3.1 Value sharing on f with its linear delay-differential operator

| 77

By the condition ∑i∈I bi = 0, (3.25) can be rewritten as ∑j∈J aj φj (z)eP(z+ηj )−P(z) + ∑i∈I bi A(z + ζi )eP(z+ζi )−P(z) A(z) + αe−P(z)

= eh(z) .

(3.26)

Considering ρ(φj (z)) < ρ(f ), deg(P(z +ηj )−P(z)) < ρ(f ), and deg(P(z +ζi )−P(z)) < ρ(f ), we obtain λ(Γ) ≤ ρ(Γ) < ρ(f ),

(3.27)

where Γ = ∑j∈J aj φj (z)eP(z+ηj )−P(z) + ∑i∈I bi A(z + ζi )eP(z+ζi )−P(z) . 1 ) = O(r ρ−1+ε ). Since α ≠ 0 is a Borel exceptional value of f (z), we have N(r, f −α Applying the second main theorem to the entire function f (z), we obtain 1 T(r, f ) ≤ N (r, ) + O(r ρ−1+ε ) + S(r, f ), f

(3.28)

and thus ρ(f ) ≤ λ(f ). Since λ(f ) ≤ ρ(f ) again, we get that λ(A(z) + αe−P(z) ) = λ(f ) = ρ(f ).

(3.29)

It is a contradiction to (3.26). Hence we have a ≠ 0. Continue now assuming that a ≠ 0. By the condition that α ≠ a is a Borel exceptional value, α must be a Nevanlinna exceptional value, and so we get that ψ(f ) − a = τ(≠ 0). f (z) − a

(3.30)

Since α is a Borel exceptional value of f (z), we see that f (z) can be written as f (z) = A(z)eP(z) + α,

(3.31)

where A(z) is a nonzero entire function, P(z) is a polynomial such that deg(P(z)) ≥ 1, and A(z), P(z) satisfy λ(A) = ρ(A) = λ(f − α) < ρ(f ) = deg(P(z)).

(3.32)

Substituting (3.31) and (3.24) into (3.30), we obtain ∑j∈J aj φj (z)eP(z+ηj ) + ∑i∈I bi [A(z + ζi )eP(z+ζi ) + α] − a A(z)eP(z) + α − a

= τ.

(3.33)

By the condition ∑i∈I bi = 0, (3.33) can be rewritten as ∑ aj φj (z)eP(z+ηj )−P(z) + ∑ bi A(z + ζi )eP(z+ζi )−P(z) j∈J

i∈I

= (τ(α − a) − a)e

−P(z)

+ τA(z).

(3.34)

It is easy to infer that ρ(aj φj (z)) < ρ(f ) and ρ(bi A(z + ζi )) < ρ(f ) by (3.32), and thus we get that the order of the left-hand side of (3.34) is less than ρ(f ). But on the right-hand a side of (3.34), we see that ρ(e−P(z) ) = ρ(f ), and thus τ(α − a) − a = 0. Hence τ = a−α . The proof of Theorem 3.1.14 is completed.

78 | 3 Uniqueness of delay-differential polynomials

3.2 Uniqueness of delay-differential polynomials Uniqueness theory of differential polynomials is related to Hayman’s result in Chapter 2. Yang and Hua [234] considered the case where f n f 󸀠 and g n g 󸀠 share one value, and Fang [48, Theorem 1] obtained an improved the result as follows. Theorem 3.2.1. Let f and g be two nonconstant entire functions, and let n > 2k + 4. If (f n )(k) and (g n )(k) share the value 1 CM, then either f (z) = c1 ecz or g(z) = c2 e−cz , where c1 , c2 , and c are constants satisfying (−1)k (c1 c2 )n (nc)2k = 1 or f = tg, where t n = 1. We will also consider the uniqueness of difference products of entire functions sharing a common value. The main purpose is to obtain relationships between f and g when P(f )f (z + c) and P(g)g(z + c) share a common value. In fact, some authors always considered two particular types P(z) = z n and P(z) = z n (z m − 1), see, e. g., [142, 144, 168, 185, 244], where m, n are positive integers. Liu, Liu, and Cao [144, Theorem 1.5] considered the uniqueness of delay-differential polynomials on [f (z)n f (z + c)](k) and [g(z)n g(z + c)](k) sharing one common value. Their result can be stated as follows. Theorem 3.2.2. Let f (z) and g(z) be transcendental entire functions of finite order, and let n ≥ 2k + 6. If [f (z)n f (z + c)](k) and [g(z)n g(z + c)](k) share the value 1 CM, then either f (z) = c1 eCz or g(z) = c2 e−Cz , where c1 , c2 , and C are constants satisfying (−1)k (c1 c2 )n [(n+ 1)C]2k = 1 or f = tg, where t n+1 = 1. Furthermore, Liu, Liu, and Yang [148] considered the uniqueness problems of entire functions of hyperorder less than one. Theorem 3.2.3. Let f (z) and g(z) be transcendental entire functions of hyperorder less than one, and let n ≥ 2k+m+6. If [f (z)n (f (z)m −1)f (z+c)](k) and [g(z)n (g(z)m −1)g(z+c)](k) share the value 1 CM, then f = tg, where t n+1 = t m = 1. Theorem 3.2.4. The conclusion of Theorem 3.2.3 is also valid if n ≥ 5k + 4m + 12 and [f (z)n (f (z)m − 1)f (z + c)](k) and [g(z)n (g(z)m − 1)g(z + c)](k) share the value 1 IM. For the proof of Theorems 3.2.3 and 3.2.4, we need the following lemma. Lemma 3.2.5. Let f and g be transcendental entire functions of hyperorder ρ2 (f ) < 1, and let c be a nonzero constant. If n ≥ m + 5 and [f (z)n (f (z)m − 1)f (z + c)](k) ≡ [g(z)n (g(z)m − 1)g(z + c)](k) , then f = tg, where t

n+1

m

= t = 1.

Proof. From (3.35) we get f (z)n (f (z)m − 1)f (z + c) ≡ g(z)n (g(z)m − 1)g(z + c) + Q(z), where Q(z) is a polynomial of degree at most k − 1. If Q(z) ≢ 0, then f (z)n (f (z)m − 1)f (z + c) g(z)n (g(z)m − 1)g(z + c) ≡ + 1. Q(z) Q(z)

(3.35)

3.2 Uniqueness of delay-differential polynomials | 79

From the second main theorem of Nevanlinna theory, Lemma 2.4.7, and Lemma 1.2.10 we conclude that f (z)n (f (z)m − 1)f (z + c) ) + S(r, f ) Q(z) f (z)n (f (z)m − 1)f (z + c) Q(z) ≤ N (r, ) + N (r, ) Q(z) f (z)n (f (z)m − 1)f (z + c) Q(z) + N (r, ) + S(r, f ) g(z)n (g(z)m − 1)g(z + c) 1 1 1 ) + N (r, ) + N (r, n m ) ≤ N (r, n m f (f − 1) f (z + c) g (g − 1) 1 + N (r, ) + S(r, f ) g(z + c) ≤ (m + 2)T(r, f ) + (m + 2)T(r, g) + S(r, f ) + S(r, g).

(n + m + 1)T(r, f ) = T (r,

(3.36)

(3.37)

Similarly, we obtain (n + m + 1)T(r, g) ≤ (m + 2)T(r, f ) + (m + 2)T(r, g) + S(r, f ) + S(r, g). Thus we get (n + m + 1)[T(r, f ) + T(r, g)] ≤ 2(m + 2)[T(r, f ) + T(r, g)] + S(r, f ) + S(r, g), which is a contradiction to n ≥ m + 5. Hence we get Q(z) ≡ 0. This implies that f (z)n (f (z)m − 1)f (z + c) ≡ g(z)n (g(z)m − 1)g(z + c). Define G(z) :=

f (z) . g(z)

(3.38)

Assume that G(z) is not a constant. From (3.38) we have g(z)m ≡

G(z)n G(z + c) − 1 . G(z)n+m G(z + c) − 1

(3.39)

If the value 1 is a Picard exceptional value of G(z)n+m G(z + c), applying the second main theorem of Nevanlinna theory and Lemma 1.2.10, we get T(r, G(z)n+m G(z + c)) ≤ N(r, G(z)n+m G(z + c)) + N (r,

1 ) G(z)n+m G(z + c)

1 ) + S(r, G) G(z)n+m G(z + c) − 1 ≤ 2T(r, G(z)) + 2T(r, G(z + c)) + S(r, G)

+ N (r,

≤ 4T(r, G) + S(r, G). Combining (3.40) with Lemma 2.4.7, we obtain (n + m − 1)T(r, G) ≤ 4T(r, G) + S(r, G),

(3.40)

80 | 3 Uniqueness of delay-differential polynomials which is a contradiction to n ≥ m + 5. Therefore the value 1 is not a Picard exceptional value of G(z)n+m G(z + c). Thus there exists z0 such that G(z0 )n+m G(z0 + c) = 1. We now consider two cases. Case 1. G(z)n+m G(z + c) ≢ 1. Since g(z) is an entire function, from (3.39) we get G(z0 )n G(z0 + c) = 1. Thus G(z0 )m = 1. Therefore N (r,

1 1 ) ≤ N (r, m ) G(z)n+m G(z + c) − 1 G −1 ≤ mT(r, G) + S(r, G).

(3.41)

By (3.41) and Lemma 1.2.10, applying the second main theorem of Nevanlinna theory, we get T(r, G(z)n+m G(z + c)) ≤ N(r, G(z)n+m G(z + c)) + N (r, + N (r,

1

G(z)n+m G(z

+ c) − 1

1 ) G(z)n+m G(z + c)

) + S(r, G)

≤ (m + 2)T(r, G) + 2T(r, G(z + c)) + S(r, G) ≤ (m + 4)T(r, G) + S(r, G). On the other hand, we have (n + m)T(r, G) = T(r, Gn+m )

≤ T(r, G(z)n+m G(z + c)) + T(r, G(z + c)) + O(1)

≤ (m + 5)T(r, G) + S(r, G),

(3.42)

a contradiction to n ≥ m + 5 ≥ 6. Case 2. G(z)n+m G(z + c) ≡ 1. Now (n + m)T(r, G) = T(r, G(z + c)) + S(r, G) = T(r, G) + S(r, G),

(3.43)

which is a contradiction to n ≥ m + 5. Thus G must be a constant, and hence f = tg, where t is a nonzero constant. Then t m = t n+1 = 1 follows from f (z)n (f (z)m − 1)f (z + c) ≡ g(z)n (g(z)m − 1)g(z + c). Lemma 3.2.6. If n ≥ k + 1, then there are no transcendental entire functions f and g of hyperorder less than one satisfying [f (z)n (f (z)m − 1)f (z + c)](k) ⋅ [g(z)n (g(z)m − 1)g(z + c)](k) ≡ 1.

(3.44)

Proof. Let f and g be transcendental entire functions of hyperorder less than one satisfying (3.44). By (3.44) and n ≥ k + 1 both f and g have no zeros. Let f (z) = eb(z) and

3.2 Uniqueness of delay-differential polynomials | 81

g(z) = ed(z) , where b(z) and d(z) are entire functions of order less than one. Substituting f and g into (3.44), we get [enb(z) (emb(z) − 1)eb(z+c) ](k) [end(z) (emd(z) − 1)ed(z+c) ](k) = 1.

(3.45)

Let (n+m)b(z)+b(z+c) =: B1 (z), nb(z)+b(z+c) =: B2 (z) and (n+m)d(z)+d(z+c) =: D1 (z), nd(z) + d(z + c) =: D2 (z). We easily to see that B1 (z) and B2 (z) are not constants at the same time. Indeed, if so, then b(z) would be a constant, hence f (z) as well. We next proceed to show that one of B1 (z) and B2 (z) must be a constant for any positive integer k. Equation (3.45) can be written as (eB1 (z) − eB2 (z) )(k) (eD1 (z) − eD2 (z) )(k) = 1.

(3.46)

Thus we obtain (eB1 − eB2 )(k) = (B󸀠1 k + Mk )eB1 − (B󸀠2 k + Nk )eB2

= [(B󸀠1 k + Mk )eB1 −B2 − (B󸀠2 k + Nk )]eB2 ,

(k) 󸀠 where Mk = Mk (B󸀠1 , B󸀠󸀠 1 , . . . , B1 ) is a differential polynomial of B1 of degree k − 1, and (k) 󸀠 󸀠󸀠 󸀠 Nk = Nk (B2 , B2 , . . . , B2 ) is a differential polynomial of B2 of degree k − 1. Observing that 0 is the only Picard exceptional value of eB1 (z)−B2 (z) , we get B󸀠1 k + (k) (k) (k) 󸀠k 󸀠 󸀠󸀠 󸀠k 󸀠 󸀠󸀠 Mk (B󸀠1 , B󸀠󸀠 1 , . . . , B1 ) = 0 or B2 + Nk (B2 , B2 , . . . , B2 ) = 0. If B1 + Mk (B1 , B1 , . . . , B1 ) = 0, from the Clunie lemma [41] we get that m(r, B󸀠1 ) = S(r, B󸀠1 ). This implies that the entire function B1 (z) must reduce to a constant. Similarly, we see that B2 is a constant (k) if B󸀠2 k + Nk (B󸀠2 , B󸀠󸀠 2 , . . . , B2 ) = 0. Thus we see that one of B1 (z) and B2 (z) must be a constant. If B1 (z) ≡ B1 is a constant, then f (z)n+m f (z + c) = eB1 . From Lemma 2.4.7 we obtain T(r, f ) = S(r, f ), a contradiction. If B2 (z) ≡ B2 is a constant, then f (z)n f (z + c) = eB2 . From Lemma 2.4.7 we now get T(r, f ) = S(r, f ), a contradiction. The proof of Lemma 3.2.6 is completed.

For the proof of the theorems, we need the following lemmas. Lemma 3.2.7 ([234, Lemma 3]). Let F and G be nonconstant meromorphic functions. If F and G share the value 1 CM, then one of the following three cases holds: (i) max{T(r, F), T(r, G)} ≤ N≤2 (r, F1 ) + N≤2 (r, F) + N≤2 (r, G1 ) + N≤2 (r, G) + S(r, F) + S(r, G), (ii) F = G, (iii) F ⋅ G = 1. Lemma 3.2.8 ([222, Lemma 2.3]). Let F and G be nonconstant meromorphic functions sharing the value 1 IM. Define H :=

F 󸀠󸀠 F󸀠 G󸀠󸀠 G󸀠 − 2 + 2 − . F󸀠 F − 1 G󸀠 G−1

82 | 3 Uniqueness of delay-differential polynomials If H ≠ 0, then 1 1 ) + N≤2 (r, F) + N≤2 (r, ) + N≤2 (r, G)) F G 1 1 + 3 (N(r, F) + N(r, ) + N(r, G) + N(r, )) F G + S(r, F) + S(r, G).

T(r, F) + T(r, G) ≤ 2(N≤2 (r,

(3.47)

Proof of Theorem 3.2.3. Let F(z) := [f (z)n (f (z)m −1)f (z+c)](k) and G(z) := [g(z)n (g(z)m − 1)g(z + c)](k) . From the conditions of Theorem 3.2.3 we get that F and G share the value 1 CM. Since f is a transcendental entire function, T(r, F) ≤ T(r, f (z)n (f (z)m − 1)f (z + c)) + S(r, f ).

(3.48)

Combining (3.48) with Lemma 2.4.7, we have S(r, F) = S(r, f ). We also have S(r, G) = S(r, g). From (1.13) we obtain 1 1 ) ) = N≤2 (r, F [f (z)n (f (z)m − 1)f (z + c)](k)

N≤2 (r,

≤ T(r, F) − T(r, f (z)n (f (z)m − 1)f (z + c)) 1 + N≤(k+2) (r, ) + S(r, f ). f (z)n (f (z)m − 1)f (z + c)

(3.49)

Combining Lemma 2.4.7 with (3.49), we get (n + m + 1)T(r, f ) = T (r, f (z)n (f (z)m − 1)f (z + c)) + S(r, f ) 1 1 ≤ T(r, F) − N≤2 (r, ) + N≤(k+2) (r, ) F f (z)n (f (z)m − 1)f (z + c) + S(r, f ).

(3.50)

From (1.14) we obtain N≤2 (r,

1 1 ) ≤ N≤(k+2) (r, ) + S(r, f ) F f (z)n (f (z)m − 1)f (z + c) 1 1 1 ≤ (k + 2)N (r, ) + N (r, m ) + N (r, ) + S(r, f ) f f −1 f (z + c) ≤ (k + m + 3)T(r, f ) + S(r, f ).

(3.51)

Similarly, we also obtain (n + m + 1)T(r, g) ≤ T(r, G) − N≤2 (r, + N≤(k+2) (r,

1 ) G

1 ) + S(r, g) g n (z)(g m (z) − 1)g(z + c)

(3.52)

and N≤2 (r,

1 ) ≤ (k + m + 3)T(r, g) + S(r, g). G

(3.53)

3.2 Uniqueness of delay-differential polynomials | 83

If (i) of Lemma 3.2.7 is satisfied, then we get max{T(r, F), T(r, G)} ≤ N≤2 (r,

1 1 ) + N≤2 (r, ) + S(r, F) + S(r, G). F G

Thus, combining this with (3.50)–(3.53), we obtain (n + m + 1)[T(r, f ) + T(r, g)] ≤ 2N≤(k+2) (r,

1 ) f (z)n (f (z)m − 1)f (z + c)

1 ) + S(r, f ) + S(r, g) − 1)g(z + c) ≤ 2(k + m + 3)[T(r, f ) + T(r, g)] + S(r, f ) + S(r, g),

+ 2N≤(k+2) (r,

g(z)n (g(z)m

which is a contradiction to n ≥ 2k + m + 6. Hence F = G or F ⋅ G = 1. From Lemmas 3.2.5 and 3.2.6 we get f = tg for t m = t n+1 = 1. This completes the proof of Theorem 3.2.3. Proof of Theorem 3.2.4. Let F(z) := [f (z)n (f (z)m −1)f (z+c)](k) and G(z) := [g(z)n (g(z)m − 1)g(z + c)](k) . We proceed to show that F = G or F ⋅ G = 1 under the conditions of Theorem 3.2.4. Assume that H ≠ 0, where H is defined in Lemma 3.2.8. Then from (3.47) we get 1 1 1 1 ) + N≤2 (r, )) + 3 (N(r, ) + N(r, )) F G F G + S(r, F) + S(r, G). (3.54)

T(r, F) + T(r, G) ≤ 2 (N≤2 (r,

Combining (3.54) with (3.50)–(3.53) and (1.14), we obtain (n + m + 1)(T(r, f ) + T(r, g)) ≤ T(r, F) + T(r, G) 1 + N≤(k+2) (r, ) f (z)n (f (z)m − 1)f (z + c) 1 ) + N≤(k+2) (r, g(z)n (g(z)m − 1)g(z + c) 1 1 − N≤2 (r, ) − N≤2 (r, ) + S(r, f ) + S(r, g) F G 1 ≤ 2N≤(k+2) (r, ) f (z)n (f (z)m − 1)f (z + c) 1 + 2N≤(k+2) (r, ) g(z)n (g(z)m − 1)g(z + c) 1 1 + 3 (N(r, ) + N(r, )) + S(r, f ) + S(r, g) F G ≤ (5k + 5m + 12)[T(r, f ) + T(r, g)] + S(r, f ) + S(r, g), which is a contradiction to n ≥ 5k + 4m + 12. Thus we get H = 0. Integrating H twice, we obtain F=

(b + 1)G + (a − b − 1) (a − b − 1) − (a − b)F ,G = . bG + (a − b) Fb − (b + 1)

(3.55)

84 | 3 Uniqueness of delay-differential polynomials This implies that T(r, F) = T(r, G) + O(1). We now consider three cases. Case 1. b ≠ 0, −1. If a − b − 1 ≠ 0, then by (3.55) we get N(r,

1 1 ) = N (r, ). F G − a−b−1

(3.56)

b+1

By the second main theorem of Nevanlinna theory, (1.13), and (1.14), we obtain (n + m + 1)T(r, g) ≤ T(r, G) + N≤k (r,

1 ) g(z)n (g(z)m − 1)g(z + c)

1 ) + S(r, g) G 1 1 ) + S(r, g) ≤ N≤k (r, ) + N (r, n m g(z) (g(z) − 1)g(z + c) G − a−b−1 − N (r,

b+1

≤ (k + m + 1)T(r, g) + (k + m + 2)T(r, f ) + S(r, f ) + S(r, g).

(3.57)

Similarly, we get (n + m + 1)T(r, f ) ≤ (k + m + 1)T(r, f ) + (k + m + 2)T(r, g) + S(r, f ) + S(r, g). Thus we see that (n + m + 1)[T(r, f ) + T(r, g)] ≤ (2k + 2m + 3)[T(r, f ) + T(r, g)] + S(r, f ) + S(r, g), which is a contradiction to n ≥ 5k + 4m + 12. Thus a − b − 1 = 0, meaning that F=

(b + 1)G . bG + 1

Since F is an entire function, (3.58) implies that N(r, ing, we have (n + m + 1)T(r, g) ≤ T(r, G) + N≤k (r,

(3.58) 1 G+ b1

) = 0. Using a similar reason-

1 ) g(z)n (g(z)m − 1)g(z + c)

1 ) + S(r, g) G 1 1 ≤ N≤k (r, ) + N (r, n m g(z) (g(z) − 1)g(z + c) G+ − N (r,

≤ (k + m + 1)T(r, g) + S(r, g),

1 b

) + S(r, g) (3.59)

which is a contradiction. Case 2. b = 0, a ≠ 1. From (3.55) we obtain F=

G+a−1 . a

(3.60)

3.2 Uniqueness of delay-differential polynomials | 85

As before, a contradiction again follows. Thus a = 1. This implies that F = G. Case 3. b = −1, a ≠ −1. From (3.55) we obtain F=

a . a+1−G

(3.61)

Similarly as before, we again get a contradiction, and hence a = −1. Thus we get F ⋅ G = 1. From Lemmas 3.2.5 and 3.2.6 we get f = tg with t m = t n+1 = 1, completing the proof of Theorem 3.2.4. To close this chapter, we recall a result given by Luo and Lin [168, Theorem 2] considering the case where f n or f n (f m − 1) is replaced by a polynomial P(f ) of f . Definition 3.2.9. Let P(z) = an z n + an−1 z n−1 + ⋅ ⋅ ⋅ + a1 z + a0 be a nonzero polynomial, where a0 , a1 , . . . , an (an ≠ 0) are complex constants. Denote Γ0 := m1 + 2m2 , where m1 is the number of the simple zeros of P(z), and m2 is the number of multiple zeros of P(z), and denote d := GCD{λ0 , λ1 , . . . , λn }, where i + 1,

λi = {

n + 1,

ai = 0, ai ≠ 0.

(3.62)

Theorem 3.2.10. Let f and g be transcendental entire functions of finite order, let c be a nonzero complex constant, and let P(z) be as above. If n > 2Γ0 + 1 and P(f )f (z + c) and P(g)g(z + c) share 1 CM, then one of the following cases holds: (1) f ≡ tg for a constant t such that t d = 1, where d is defined above, (2) f and g satisfy the algebraic equation R(f , g) ≡ 0, where R(ω1 , ω2 ) = P(ω1 )ω1 (z + c) − P(ω2 )ω2 (z + c), (3) f (z) = eα(z) and g(z) = eβ(z) , where α(z) and β(z) are two polynomials, b is a constant satisfying α + β ≡ b, and a2n e(n+1)b = 1. Remark 3.2.11. The reader may find some results on the uniqueness of [P(f )f (z + c)](k) and [P(g)g(z + c)](k) under the condition of sharing a common value or a small function. Here we omit such considerations.

4 Difference Wiman–Valiron theory Wiman–Valiron theory is an important device to considering the value distribution of entire solutions of complex differential equations. Some of such applications can be found in [120]. Extending the Wiman–Valiron-type reasoning to complex differences, resp., q-differences, the corresponding difference, resp., q-difference variants of Wiman–Valiron theory can be constructed. In this short chapter, we present the basic notations and key results of this theory.

4.1 Differential setting of the Wiman–Valiron theory For the basic notations and results of Wiman–Valiron theory in the differential setting, we refer to [113] and [94]. To fix basic notations, let f (z) be an entire function with Taylor expansion +∞

f (z) = ∑ an z n . n=0

For r > 0, define the maximum modulus M(r, f ) and maximum term μ(r, f ) of f by M(r, f ) := max |f (z)| |z|=r

and μ(r, f ) := max |an |r n , n≥0

respectively. Recall that the power series of f is absolutely convergent in compact subsets of ℂ, and hence |an |r n → 0 as n → ∞. This observation makes us possible to define the central index ν(r, f ) of f as the largest exponent m such that |am |r m = μ(r, f ). In [113, p. 31–38], we can find a clear exposition of basic properties of the maximum term and central index. We list here some of them: (1) μ(r, f ) is strictly increasing for all r sufficiently large, continuous, and tending to +∞ as r → +∞; (2) ν(r, f ) is increasing, piecewise constant, right continuous, and also tending to +∞ as r → +∞; (3) the order ρ(f ) of f can be expressed in terms of the central index or of the maximum term: ρ(f ) = lim sup r→∞

log+ log+ μ(r, f ) log+ ν(r, f ) = lim sup . log r log r r→∞

In addition, recall [120, Proposition 3.3], where it is shown that (4) if f (z) is a transcendental entire function with ρ(f ) = 0 and k, l ∈ N, then ν(r, f )k = 0. r→∞ rl lim

https://doi.org/10.1515/9783110560565-004

88 | 4 Difference Wiman–Valiron theory The key result in Wiman–Valiron theory typically applied to complex differential equations can be stated as follows ([113, p. 187–199] and [94, p. 28–30]): Theorem 4.1.1. Let f be a transcendental entire function, and let 0 < δ < chosen such that |z| = r

and

1 8

and z be

1

|f (z)| > M(r, f )ν(r, f )− 8 +δ .

Then f (n) (z) ν(r, f ) n =( ) (1 + Rn (z)), f (z) z where 1

Rn (z) = O((ν(r, f ))− 8 +δ ) for all n ∈ ℕ and r ∈ ̸ E ∪ [0, 1]. Here E ⊂ (1, ∞) is a set of finite logarithmic measure. In case the reader wants to consider more details of Wiman–Valiron theory, we recommend the reader to the extensive presentation by Hayman [92].

4.2 Difference setting of the Wiman–Valiron theory For the difference variants of Wiman–Valiron theory, we point out the corresponding researches due to Chiang and Feng [39, 40] and Ishizaki and Yanagihara [106]. Observe that [106] depends on the factorial series (or binomial series), that is, series written in terms of the basis {z(n) = z(z + 1) ⋅ ⋅ ⋅ (z + n − 1)}, which is different from the usual basis {z n }. In what follows, we mainly rely on the difference versions of the Wiman–Valiron theorem given by Chiang and Feng [39, 40]. To start with, recall first [39, Theorem 5.2]: Theorem 4.2.1. Let f be a meromorphic function of order ρ(f ) = ρ < 1, and let c be a nonzero constant. For any ε > 0 and n ∈ N, there exists a set E ⊂ (1, +∞), which depends on f and has finite logarithmic measure, such that for all z satisfying |z| = r ∈ E ∪ [0, 1], Δnc f (z) f (n) (z) = cn + O(r (n+1)(ρ−1)+ε ). f (z) f (z)

(4.1)

Using this theorem, we obtain the following difference version of the Wiman– Valiron theorem. Theorem 4.2.2 ([39, Theorem 6.1]). Let f be a transcendental entire function of order ρ(f ) = ρ < 1, and let 0 < ε < 81 and z be chosen such that |z| = r

1

and |f (z)| > M(r, f )ν(r, f )− 8 +ε .

(4.2)

4.2 Difference setting of the Wiman–Valiron theory | 89

Then for each positive integer n, there exists a set E ⊂ (1, ∞) of finite logarithmic measure such that for all r ∈ ̸ E ∪ [0, 1], if ρ = 0, then 1 Δnc f (z) ν(r, f ) n = cn ( ) (1 + O(ν(r, f )− 8 +ε )). f (z) z

(4.3)

Δnc f (z) ν(r, f ) n = cn ( ) + O(r nρ−n−γ+ε ), f (z) z

(4.4)

If 0 < ρ < 1, then

where γ = min{ 81 ρ, 1 − ρ}. The remainder in (4.4) is not sharp for entire functions of order less than one. For a sharp error bound, recall Theorem 3 in [40]: Theorem 4.2.3. Let f be a meromorphic function with order ρ(f ) = ρ < 1, 0 < ε < min{ 81 , 1 − ρ}, and let z satisfy (4.2). Then for each positive integer n, there exists a set E ⊂ [1, +∞) of finite logarithmic measure such that for all |z| = r ∈ ̸ E ∪ [0, 1], we have Δnc f (z) ν(r, f ) n =( ) (1 + Rn (z)), f (z) z

(4.5)

where Rn (z) = O((ν(r, f ))−κ+ε ) and κ = min{ 81 , 1 − ρ}. Another possibility to considering Wiman–Valiron-type estimates is looking at the formal expansion +∞

Δnc f (z) = n! ∑

k=n

ℑk(n) k!

f (n) (z)

(4.6)

in terms of the Stirling numbers of the second kind ℑk(n) ; see [2, p. 825]. Observe that these Stirling numbers ℑk(n) count the number of different ways to partition a set of n objects into m nonempty subsets. In particular, these may be computed by the generating function n

xn = ∑ ℑk(n) x(x − 1) ⋅ ⋅ ⋅ (x − m + 1); m=0

see more details in [2, p. 822–825]. We shortly add here a key application of Theorem 4.2.3 to linear difference equations. To this end, consider linear difference equations of the form an (z)Δnc f (z) + ⋅ ⋅ ⋅ + a1 (z)Δc f (z) + a0 (z)f (z) = 0, A

(4.7)

j where aj (z) = ∑k=0 ak z k , Aj = deg aj (z), and an (z) ≢ 0. Write

(j)

ℜj := {(x, y); x ≥ j, y ≤ An−j − (n − j)} for 0 ≤ j ≤ n. The Newton polygon for (4.7) is defined as the convex hull of ℜj = ⋃nj=0 ℜj .

90 | 4 Difference Wiman–Valiron theory Theorem 4.2.4 ([40, Theorem 4]). Let f be a transcendental entire solution of (4.7) with ρ(f ) = ρ < 1. Then ρ is a rational number among the slopes of the Newton polygon of equation (4.7). In particular, log M(r, f ) = Lr ρ (1 + o(1)), where L > 0, ρ > 0, and M(r, f ) = max|z|=r |f (z)|. For more related such results, we recommend the reader to look at [39] and [106].

4.3 Wiman–Valiron theorems for q-differences Let q ∈ ℂ \ {0, 1} be a complex number. Counterpart of Wiman–Valiron theory for q-differences follows by considering the relationship between the logarithmic 󸀠 q-difference log f (qz) − log f (z) and ff . This was recently obtained by Wen and Ye [218] for meromorphic functions of order less than 21 . Recalling [218, Theorem 2.1], we first obtain the following: Theorem 4.3.1. If f (z) is a transcendental meromorphic function of order strictly less than 21 , then log

f 󸀠 (z) f (qz) = (q − 1)z + O(1) f (z) f (z)

(4.8)

or, equivalently, f 󸀠 (z) f (qz) = e(q−1)z f (z) +O(1) f (z)

(4.9)

for any r outside an exceptional set of finite logarithmic measure. Remark 4.3.2. The restriction on order ρ(f ) < 21 cannot be relaxed in Theorem 4.3.1. For example, the transcendental entire function f (z) = cos √z is of order ρ(f ) = 21 . Moreover, estimates (4.8)–(4.9) fail for this function; see [218, Theorem 2]. The q-difference analogue for the Wiman–Valiron theorem for entire functions of order less than 21 can be proved by using Theorem 4.3.1; see [218, Theorem 2.3]: Theorem 4.3.3. Suppose that m is any positive integer and q is a complex number with qm ∈ ℂ \ {0, 1}. Let f (z) be a transcendental entire function of order strictly less than 21 , and let E ⊂ R+ be a set of finite logarithmic measure. Then for any 0 < δ < 41 and any z 1

with |z| = r ∈ ̸ E satisfying |f (z)| > M(r, f )ν(r, f )δ− 4 , we have m f (qm z) = e(q −1)ν(r,f )(1+o(1)) . f (z)

4.3 Wiman–Valiron theorems for q-differences | 91

Remark 4.3.4. To close this chapter, we remark that there are other interesting variants of Wiman–Valiron theory. To give some examples, we may recall the Wiman– Valiron theory in (1) the unit disc [50–52, 129], (2) simply connected domains [53], and (3) several complex variables [49, 54]. Moreover, it is worth mentioning, in addition, the Wiman–Valiron theory for Wilson operators [22] and Jackson difference operators [19].

5 The linear complex delay-differential equations 5.1 First-order delay-differential equations We first recall a few basic observations on the first-order complex difference equation f (z + 1) = α(z)f (z),

(5.1)

where α(z) is a nonzero meromorphic function. We may use the gamma function to express the solutions of (5.1) in explicit form, see, e. g., [114, p. 48] and [116, p. 115–116], provided that α(z) is a rational function. Indeed, we have the following: Theorem 5.1.1. Let α(z) := R(z) = ρ

∏nk=1 (z − αk ) ∏m j=1 (z − βj )

be a rational function, where ρ ≠ 0, αk , k = 1, 2, . . . , n, and βj , j = 1, 2, . . . , m, are complex numbers. Then (5.1) can be solved as f (z) = Q(z)ρz

∏nk=1 Γ(z − αk ) , ∏m j=1 Γ(z − βj )

where Q(z) is an arbitrary periodic function of period 1, and Γ(z) is the gamma function. Remark 5.1.2. (1) It is well known that Γ(z) is a particular meromorphic solution with ρ(Γ) = 1 of the linear difference equation y(z + 1) = zy(z). There exist several formulas for the gamma function, each of which can be regarded as the definition of the gamma function; see [116, p. 114–115]. Hölder [99] proved that the gamma function cannot solve an algebraic differential equation with rational coefficients. (2) Theorem 5.1.1 shows that the transcendental meromorphic solution f (z) of (5.1) is of order ρ(f ) ≥ 1. We cannot find a uniform upper bound for the growth of meromorphic solutions of (5.1), since f (z) contains arbitrary periodic functions Q(z). This also happens for the growth of meromorphic solutions of most linear difference equations. (3) Let α(z) be a polynomial of degree deg α(z) ≥ 1 in (5.1). Chen, Huang, and Zhang [23, Theorem 3] showed that any transcendental meromorphic solution of (5.1) has at most one Borel exceptional value. If α(z) is an entire function or a meromorphic function of order ρ(α), then Whittaker [219, p. 30–31] states that (5.1) has a meromorphic solution of order ≤ ρ(α) + 1. However, the solution cannot be explicitly expressed in terms of the given coefficient α(z). The above Whittaker’s result shows us the existence theorem for meromorphic solutions. Chiang and Feng [38, Theorems 9.2 and 9.4] revealed more on the meromorphic solutions of (5.1), provided that α(z) is entire: https://doi.org/10.1515/9783110560565-005

94 | 5 The linear complex delay-differential equations Theorem 5.1.3. Let α(z) be an entire function of order ρ(α). Then all meromorphic solutions f satisfy ρ(f ) ≥ ρ(α) + 1, and (5.1) admits a meromorphic solution of order ρ(f ) = ρ(α) + 1. Remark 5.1.4. If α(z) is a meromorphic function, then the first statement of Theorem 5.1.3 is not valid. For example, α(z) = − tan πz and f (z) = cos πz solve (5.1). The 2 2 solutions with minimal growth have been found in [37, 198] under certain conditions. Of course, what is said above applies to slightly more general equations of the form f (z + c) = α(z)f (z).

(5.2)

By the change of variables z = cz̄ and f ̄(z)̄ = f (cz)̄ equation (5.2) reduces to f ̄(z̄ + 1) = ̄ α(̄ z)̄ f ̄(z). We next proceed to looking at first-order linear complex difference equations of the form f (z + 1) − f (z) = R1 (z),

(5.3)

where R1 (z) is a nonzero meromorphic function. A more general first-order linear complex difference equation related to (5.3) is P(z)f (z + 1) + Q(z)f (z) = R(z),

(5.4)

where P(z), Q(z), R(z) are nonzero meromorphic functions. Equation (5.4) can be solved in the following way: Let s(z) be a solution of the equation s(z + 1) = −

Q(z) s(z), P(z)

(5.5)

and let h(z) be a solution of the equation h(z + 1) − h(z) = −

R(z) . Q(z)s(z)

(5.6)

Then f (z) = s(z)h(z) solves (5.4). Recall the method to construct the solutions of (5.3) in [116, p. 80–83], provided that R1 (z) is rational. Define the operator S that is the inverse of the difference operator △, just like the integral ∫ is the inverse of the differential operator, that is, Δ(Sa(z)) = a(z). So the general solution of (5.3) is written as f (z) = S(R1 (z)) + Q(z), where Q(z) is an arbitrary periodic function of period 1. Define the symbol z(0) := 1 and z(n) := z(z−1)⋅⋅⋅(z−n+1) ; we may also use the symbol (nz ). It is easy to derive the formulas n! S1 = z,

Sz(n) = z(n + 1),

n = 1, 2, . . . .

5.1 First-order delay-differential equations | 95

Theorem 5.1.5. Let M

l

N nj

R1 (z) = ∑ bl z + ∑ ∑ l=0

cjk

k j=1 k=1 (z − αj )

be a rational function, where bj , βj , j = 1, 2, . . . , N, and cjk are complex numbers. Then (5.3) can be solved as n

j M N (−1)k−1 dk−1 Ψ(z − αj ), f (z) = Q(z) + ∑ bl̃ z(l + 1) + ∑ ∑ (k − 1)! dz k−1 j=1 k=1 l=0

where Q(z) is an arbitrary periodic function of period 1, Ψ(z) = function, and M

M

l=0

l=0

Γ󸀠 (z) Γ(z)

is the Gauss psi-

S ( ∑ bl z l ) = ∑ bl̃ z(l + 1). Bank and Kaufman [5] showed the existence of finite-order meromorphic solutions of (5.3): Theorem 5.1.6. For any rational function R1 (z), equation (5.3) always admits a meromorphic solution f (z) satisfying T(r, f ) = O(r) as r → ∞. Chen [29, Chapter 5] also presents some results on the value distribution of meromorphic solutions of first-order difference equations of the form (5.3). We recall the following result due to Chen, Huang, and Zhang [23, Theorem 1]. Theorem 5.1.7. For any rational function R1 (z) ≢ 0, the following statements hold for (5.3): (1) Equation (5.3) has a meromorphic solution f that satisfies λ(f ) = ρ(f ) = 1. Moreover, all finite-order transcendental meromorphic solutions to (5.3) satisfy λ(f ) = ρ(f ) ≥ 1. (2) Every finite-order transcendental meromorphic solution f (z) to (5.3) has at most one Borel exceptional value. If f (z) has infinitely many poles, then λ( f1 ) ≥ 1. (3) If R1 (z) ≢ 1 and f (z) is a finite-order meromorphic solution, then λ(f − z) = ρ(f ). As our final remark on first-order linear difference equations, we recall the following existence result in Whittaker [219, Theorems 3, 4]. Theorem 5.1.8. (1) If g(z) is an entire function, then there is an entire function f such that ρ(f ) = ρ(g) and f (z + 1) − f (z) = g(z). (2) If g(z) is a meromorphic function, then there exists a meromorphic solution f to f (z + 1) − f (z) = g(z) such that ρ(f ) ≤ ρ(g) + 1. The simplest delay-differential equation f (x − k) = f 󸀠 (x) with (k > 0) has been extensively studied in real analysis, including numerous applications ranging from cell

96 | 5 The linear complex delay-differential equations growth models to current collection systems for an electric locomotive to wavelets; see, e. g., the book [11]. However, it seems that there are a few considerations only on meromorphic solutions of such complex delay-differential equations using Nevanlinna theory. We begin by looking at the complex delay-differential equation f 󸀠 (z) = f (z + c),

(5.7)

where c is a nonzero complex constant. Remark 5.1.9. (1) Nonvanishing meromorphic solutions f (z) to (5.7) must be transcendental entire. We immediately see that nonvanishing polynomial solutions cannot exist. Moreover, if f (z) has a pole, then f (z) has infinitely many poles by (5.7). Let E be the set of the multiplicities of all poles of f (z). We call the minimum mf (z) (E) of this set the index of the function f (z). Clearly, mf (z+c) (E) = mf (z) (E), whereas mf 󸀠 (z) (E) = 1 + mf (z) (E), contradicting equation (5.7). (2) Transcendental entire solutions to (5.7) exist. As a trivial example, f (z) = ez is an entire solution of (5.7), provided that c = 2kiπ, k ∈ ℤ \ {0}. In addition, f (z) = sin z and f (z) = cos z are solutions to (5.7) for suitable c. More generally, entire functions of the form f (z) = αj ∑nj=1 eλj z are also solutions to (5.7), provided that eλj c = λj and αj are constants. (3) The order of entire solutions of (5.7) must be satisfy ρ(f ) ≥ 1. This immediately follows by Lemma 5.1.10. (4) Liu and Dong [149, Theorem 3.6] showed that f (z) admits no nonzero Borel exceptional polynomials, provided that f (z) is a finite-order entire solution to (5.7). (5) It remains an open question on the existence of entire solutions to (5.7) of infinite order or of finite order ρ(f ) > 1. Lemma 5.1.10 ([14, Lemma 3.3]). Let f (z) be a transcendental meromorphic function of order ρ(f ) < 1, h > 0. Then there exists an ε-set E such that f 󸀠 (z + c) →0 f (z + c)

and

f (z + c) → 1 as f (z)

z → ∞ in ℂ \ E,

uniformly in c for |c| ≤ h. Further, E may be chosen so that for large z ∈ ̸ E, the function f has no zeros or poles in |ζ − z| ≤ h. Liu and Song [153] stated two results on the entire solutions of equations, which are more general than (5.7). As a preparation, we first prove the following: Lemma 5.1.11. Let L(f ) := f (k) (z) + ak−1 (z)f (k−1) (z) + ⋅ ⋅ ⋅ + a1 (z)f 󸀠 (z) + a0 (z)f (z),

5.1 First-order delay-differential equations | 97

where a0 (z), . . . , ak−1 (z) are polynomials, and p(z) is a nonconstant polynomial of deg(p(z)) = n. If f (z) is an entire nonvanishing solution to f (z + c) = ep(z) L(f ),

(5.8)

then ρ(f ) ≥ n + 1. Proof. The solution f is transcendental. Indeed, if f is a polynomial, then f (z + c)/L(f ) is rational, whereas ep(z) is of order n ≥ 1, a contradiction. The claim is trivial if f is of infinite order. Therefore we now assume that f is of finite order. From (5.8) we obtain that f (z + c) L(f ) = ep(z) [ ]. f (z) f (z) Since m (r,

L(f ) ) = O(log r) f (z)

and m (r,

f (z) ) = O(r ρ(f )−1+ε ), f (z + c)

we have that m(r, ep(z) ) =

|bn | n r + o(r n ) = m(r, e−p(z) ) + O(1) ≤ O(r ρ(f )−1+ε ) + O(log r), π

where bn is the coefficient of z n in p(z). Therefore ρ(f ) ≥ n + 1. Remark 5.1.12. Observe that the claim of Lemma 5.1.11 fails in general if p(z) is constant. For example, f 󸀠 (z) + f (z) = f (z + 1) has an entire solution f (z) = z. Theorem 5.1.13. If f (z) is an entire solution of (5.7) and λ(f ) < ρ(f ) < +∞, then ρ(f ) = 1. Furthermore, one of the following three cases holds: (1) f (z) = (b1 z + b0 )eez+B , where b1 is a nonzero constant. A+2kiπ) (2) f (z) = b0 eAz+B , where c = log |A|+i(arg , and A is a nonzero constant. A Az+B (3) f (z) = g(z)e , where g(z) is a transcendental entire function such that g 󸀠 (z) = A[g(z + c) − g(z)] and ρ(g) < 1, where A is a nonzero constant. Proof. Since λ(f ) < ρ(f ) and f (z) has a Borel exceptional value z = 0, f (z) must be of positive integer order [235, p. 106, Corollary]. Denote q := ρ(f ). Then, by the Hadamard representation, f (z) = g(z)eQ(z) ,

98 | 5 The linear complex delay-differential equations where Q(z) is a polynomial of deg(Q(z)) = q, and ρ(g) = λ(f ) < ρ(f ) = q < +∞. From (5.7) we may write [g 󸀠 (z) + g(z)Q󸀠 (z)]eQ(z)−Q(z+c) = g(z + c).

(5.9)

If Q(z) − Q(z + c) is nonconstant, then Lemma 5.1.11 implies that ρ(g) ≥ q, since deg(Q(z) − Q(z + c)) = q − 1, a contradiction. Therefore Q(z + c) − Q(z) = D is a constant, and Q(z) = Az + B, where A ≠ 0 and B are constants. Thus Ac = D, ρ(g) < 1, and ρ(f ) = 1. From (5.9) we have g 󸀠 (z) + Ag(z) = eD g(z + c).

(5.10)

g(z + c) g 󸀠 (z) + A = eD , g(z) g(z)

(5.11)

Since

Lemma 5.1.10 implies that A = eD . Thus (5.10) changes to g 󸀠 (z) − A[g(z + c) − g(z)] = 0.

(5.12)

If now g is transcendental, then claim (3) follows. Now suppose that g(z) is a polynomial, say, g(z) = bn z n + bn−1 z n−1 + ⋅ ⋅ ⋅ + b1 z + b0 . To avoid a contradiction to (5.12), we must have g(z) = b1 z + b0 and Ac = 1, where b1 is a nonzero constant, or g(z) = b, where b is a nonzero constant. If g(z) = b1 z + b0 , then D = 1, A = e, and c = e1 , and thus f (z) = (b1 z + b0 )eez+B . If g(z) = b0 , then eAc = A, and thus c =

log |A|+i(arg A+2kiπ) , A

and f (z) = b0 eAz+B .

Theorem 5.1.14. If f (z) is an exponential polynomial solution to (5.7) of the form k

f (z) = ∑ Pj (z)eQj (z) , j=1

then ρ(f ) = 1, and Pj (z) is a linear polynomial for j = 1, 2, . . . , k. Proof. Substituting f (z) = ∑kj=1 Pj (z)eQj (z) into (5.7), we obtain [P1󸀠 (z) + P1 (z)Q󸀠1 (z)]eQ1 (z) + ⋅ ⋅ ⋅ + [Pk󸀠 (z) + Pk (z)Q󸀠k (z)]eQk (z) = P1 (z + c)eQ1 (z+c) + ⋅ ⋅ ⋅ + Pk (z + c)eQk (z+c) .

Defining ϕk (z) = Pk󸀠 (z) + Pk (z)Q󸀠k (z), it follows that [ϕ1 (z) − P1 (z + c)eQ1 (z+c)−Q1 (z) ]eQ1 (z) + ⋅ ⋅ ⋅

5.1 First-order delay-differential equations | 99

+ [ϕk (z) − Pk (z + c)eQk (z+c)−Qk (z) ]eQk (z) = 0. Thus by Theorem 1.1.32 we have ϕj (z) − Pj (z + c)eQj (z+c)−Qj (z) ≡ 0 for j = 1, 2, . . . , k. By Lemma 5.1.11 we see that Qk (z) must be a linear polynomial Qk (z) = Ak z +Bk . Thus ρ(f ) = 1 follows. We next see that all Pj (z) are linear polynomials as well. Without loss of generalization, take j = 1 and write P1 (z) = an z n + an−1 z n−1 + ⋅ ⋅ ⋅ + a1 z + a0 , where n ≥ 2 and an ≠ 0. Thus P1 (z) satisfies P1󸀠 (z) + P1 (z)Q󸀠1 (z) − P1 (z + c)eA1 c ≡ 0. Therefore (A1 − eA1 c )an z n + (nan + A1 an−1 − eA1 c an−1 − eA1 c an nc)z n−1 + [(n − 1)an−1 + A1 an−2 − eA1 c (an

n(n − 1) 2 c + an−1 (n − 1)c + an−2 )]z n−2 + ⋅ ⋅ ⋅ = 0. 2

We now get the following equations by comparing the coefficients above: A1 − eA1 c = 0,

nan − eA1 c an nc = 0, and (n − 1)an−1 + A1 an−2 − eA1 c (an From these identities we conclude that eA1 c an which is impossible. Thus n ≤ 1.

n(n − 1) 2 c + an−1 (n − 1)c + an−2 ) = 0. 2

n(n − 1) 2 c = 0, 2

Bélair and Giroux [10, Theorem 2] considered f 󸀠 (z) + af (z + c) = g(z).

(5.13)

Of course, this is slightly more general than (5.7). They obtained the existence of entire solutions of (5.13), provided that g(z) is an entire function of exponential type τ, meaning that its growth is bounded by the exponential function eτ|z| for some constants τ as r → ∞: Theorem 5.1.15. Let the type τ of g(z) satisfy τ < lution to (5.13) of exponential type τ.

|a| . 1+|ac|

Then there exists an entire so-

Remark 5.1.16. As the proof of this theorem needs q-difference considerations, we advice the reader to look at Chapter 10 to complete understanding this result.

100 | 5 The linear complex delay-differential equations

5.2 Higher-order linear complex delay-differential equations 5.2.1 Higher-order homogeneous linear complex delay-differential equations We first recall some results for higher-order homogeneous linear complex differential equations of the form f (k) (z) + ak−1 (z)f (k−1) (z) + ⋅ ⋅ ⋅ + a0 (z)f (z) = 0,

(5.14)

where the coefficients a0 (z)(≢ 0), . . . , ak−1 (z) are entire functions. It is well known that all meromorphic solutions of (5.14) are entire functions. A classical result concerning the growth of solutions compared with the coefficients in (5.14) is due to Wittich; see [220] and [120, Theorem 4.1]. Note that the original proof depends on the Wiman– Valiron theory. Theorem 5.2.1. The coefficients a0 (z), . . . , ak−1 (z) are polynomials if and only if all solutions of (5.14) are entire functions of finite order. Considering relations between the exponent of convergence of the zeros with the growth of entire solutions, a well-known result may be stated as follows; see, e. g., [120, Theorem 4.3]. Theorem 5.2.2. If the coefficients a0 (z), . . . , ak−1 (z) are polynomials and f is an entire function, then λ(f − α) = ρ(f ), where α ≠ 0. n

Remark 5.2.3. Theorem 5.2.2 may fail for α = 0. As an example, f (z) = e−z is an entire solution to f 󸀠 + nz n−1 f = 0. To recall more detailed information on the growth of entire solution f of (5.14) with polynomials coefficients, the reader may consult the important paper [75] by Gundersen, Steinbart, and Wang. Their key result may be described as follows under the assumption that the coefficients a0 (z)(≢ 0), . . . , ak−1 (z) are polynomials with dj = deg(aj (z)) if aj (z) ≢ 0 and dj = −∞ if aj (z) ≡ 0. First, define the integers s1 , s2 , . . . , sp as follows: s1 := min {j :

dj

n−j

= max

0≤k≤n−1

dk } n−k

(5.15)

and, given sj , j ≥ 1, sj+1 := min {i :

di − dsj n−j

= max

0≤k −1} .

(5.16)

For a certain p, sp will be defined, but not sp+1 , meaning that the sequence s1 , s2 , . . . , sp terminates with sp . Clearly, p ≤ n, and s1 > s2 > ⋅ ⋅ ⋅ > sp ≥ 0.

5.2 Higher-order linear complex delay-differential equations |

101

We next define, for j = 1, . . . , p, αj := 1 +

dsj − dsj−1 sj−1 − sj

(5.17)

,

where s0 := n and ds0 = dn := 0. By the definitions above αj > 0 for all 1 ≤ j ≤ p. We are now ready to state the following: Theorem 5.2.4. For equation (5.14) with polynomial coefficients, we have: (i) If f is a transcendental solution to (5.14), then ρ(f ) = αj for all j, 1 ≤ j ≤ p. (ii) If s1 ≥ 1 and p ≥ 2, then α1 > α2 > ⋅ ⋅ ⋅ > αp ≥

1 1 1 ≥ ≥ . sp−1 − sp s1 − sp s1

(iii) If s1 = 0, then all nontrivial solutions f to (5.14) satisfy ρ(f ) = 1 +

d0 . n

We now proceed to considering some of the corresponding results for complex difference equations αn (z)f (z + cn ) + ⋅ ⋅ ⋅ + α1 (z)f (z + c1 ) + α0 (z)f (z + c0 ) = 0,

(5.18)

where α0 (z), . . . , αn (z)(≢ 0) are entire functions and ci , i = 0, 1, . . . , n are distinct constants. Chiang and Feng [38, Theorems 9.4 and 9.2] obtained estimates for the growth of meromorphic solution f (z) of (5.18): Theorem 5.2.5. If α0 (z), . . . , αn (z)(≢ 0) are polynomials, resp., transcendental entire functions and there exists an integer l, 0 ≤ l ≤ n, such that deg(αl ) > max{deg(αj )}, j=l̸

resp., ρ(αl ) > max{ρ(αj )}, j=l̸

then all nontrivial meromorphic solutions to (5.18) satisfy ρ(f ) ≥ 1, resp., ρ(f ) ≥ ρ(αl ) + 1. The same conclusion follows when equation (5.18) admits a dominating coefficient in the following sense [123, Theorem 5.2]: Theorem 5.2.6. Let α0 (z), . . . , αn (z)(≢ 0) be transcendental entire functions such that among those having the maximal order ρ(αl ) > max{ρ(αj )}, j=l̸

exactly one of them has its type strictly greater than the others. Then ρ(f ) ≥ ρ(αl ) + 1.

102 | 5 The linear complex delay-differential equations Chen [25, 26] further studied relations between the growth with the exponent of convergence of the zeros on (5.18) and obtained the following difference counterparts to Theorem 5.2.2, complementing Theorem 5.2.5 as well. Theorem 5.2.7. Let αj (z), j = 1, . . . , n, be polynomials such that αn α0 ≢ 0. Suppose that f (z) is a finite-order transcendental meromorphic solution to (5.18). Then (i) λ(f ) ≥ ρ(f ) − 1, (ii) if αn + ⋅ ⋅ ⋅ + α0 ≢ 0 or deg(α1 (z) + ⋅ ⋅ ⋅ + αn (z)) = max{deg αj (z), j = 1, 2, . . . , n} ≥ 1, then f (z) takes every nonzero finite value a infinitely often, λ(f − a) = ρ(f ), and ρ(f ) ≥ 1. We now proceed to considering the corresponding delay-differential equations. We start by looking at equations of the form αn (z)f (n) (z + cn ) + ⋅ ⋅ ⋅ + α1 (z)f 󸀠 (z + c1 ) + α0 (z)f (z + c0 ) = 0,

(5.19)

where α0 (z), α1 (z), . . . , αn (z)(≢ 0) are polynomials, and c1 , . . . , cn are complex constants. We first prove the following: Theorem 5.2.8. Let f be a transcendental meromorphic solution to (5.19) and suppose that deg(α0 ) > maxj≥1 {deg(αj )}. Then ρ(f ) ≥ 1. Proof. Suppose, on the contrary, that ρ(f ) < 1 and rewrite (5.19) in the form αn (z)

f (n) (z + cn ) f 󸀠 (z + c1 ) + ⋅ ⋅ ⋅ + α1 (z) + α0 (z) = 0. f (z + c0 ) f (z + c0 )

Writing now f (j) (z + cj ) f (z + c0 )

=

f (j) (z + cj ) f (z + cj )

f (z + cj ) f (z + c0 )

for j = 1, . . . , n, we see that f (j) (z + cj )/f (z + cj ) approaches zero as z → ∞ by Proposition 1.1.16 and f (z + cj )/f (z + c0 ) approaches one as r → ∞ by Lemma 5.1.10, both outside of a possible exceptional set of finite logarithmic measure. But then, with b := maxj=0̸ {deg(αj )}, 󵄨󵄨 (j) 󵄨 n 󵄨󵄨 f (z + cj ) 󵄨󵄨󵄨 󵄨󵄨󵄨 f (z + cj ) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ≤ O(r b ) ≤ 1 r deg(α0 ) |α0 (z)| ≤ ∑ |αj (z)| 󵄨󵄨󵄨 󵄨󵄨 f (z + cj ) 󵄨󵄨󵄨 󵄨󵄨󵄨 f (z + c0 ) 󵄨󵄨󵄨 2 j=1 󵄨 󵄨 outside a possible exceptional set of finite logarithmic measure. This is a contradiction as |α0 (z)| > 21 r deg(α0 ) for all |z| = r large enough.

5.2 Higher-order linear complex delay-differential equations | 103

We next present the following result concerning relations between the growth of solutions and their exponents of convergence of the zeros [153, Theorem 2]: Theorem 5.2.9. Let α0 (z), α1 (z), . . . , αn (z)(≢ 0) be polynomials in (5.19). Suppose that f (z) is a transcendental entire solution of finite order to (5.19). Then λ(f − a) ≥ ρ(f ) − 1 for every finite value a. Proof. The claim is trivial if ρ(f ) ≤ 1. Therefore assume, contrary to the statement, that ρ(f ) > 1 and λ(f − a) < ρ(f ) − 1 =: k − 1. By the Hadamard representation we may write f (z) = H(z)eg(z) + a,

(5.20)

where H(z) is the canonical product formed by the zeros of f (z), so that λ(H) = ρ(H) < ρ(f ) − 1 = k − 1, and g(z) is a polynomial of deg g(z) = k, to be written as g(z) = bk z k + bk−1 z k−1 + ⋅ ⋅ ⋅ + b0 ,

bk ≠ 0.

(5.21)

Since λ(f − a) < ρ(f ) − 1, we have k ≥ 2. Observe now that (H(z)eg(z) + a)(t) = (H(z)eg(z) )(t) = φt (z)eg(z) ,

(5.22)

where 1 ≤ t ≤ n, and φt (z) is a polynomial formed by H(z), g(z), and their derivatives. Since ρ(H) < ρ(f ) − 1, we conclude that ρ(φt ) < k − 1. Substituting now (5.20), (5.21), and (5.22) into (5.19), we obtain αn (z)φn (z + cn )eg(z+cn ) + ⋅ ⋅ ⋅ + α0 (z)H(z + c0 )eg(z+c0 ) + aα0 (z) = 0.

(5.23)

Since g(z + cj ) = bk z k + (kbk cj + bk−1 )z k−1 + ⋅ ⋅ ⋅ and the shifts cj are distinct, we observe that, for j ≠ s, g(z + cj ) − g(z + cs ) = kbk (cj − cs )z k−1 + ⋅ ⋅ ⋅

(5.24)

is a nonconstant polynomial of degree k−1 ≥ 1. Since ρ(φj ) < k−1 and the exponentials exp(g(z +cj )−g(z +cs )) are of regular growth, we obtain for all 1 ≤ j ≤ n and 1 ≤ j < s ≤ n that T(r, αj (z)φj (z + cj )) = o(T(r, exp(g(z + cj ) − g(z + cs )))). If a = 0, then we may use Theorem 1.1.33 to conclude that αn (z)φn (z + cn ) = ⋅ ⋅ ⋅ = α1 (z)φ1 (z + c1 ) = α0 (z)H(z + c0 ) = 0. This is clearly a contradiction. If a ≠ 0, then we obtain αn (z)φn (z + cn ) g(z+cn ) α (z)φ1 (z + c1 ) g(z+c1 ) e + ⋅⋅⋅ + 1 e aα0 (z) aα0 (z)

(5.25)

104 | 5 The linear complex delay-differential equations

+

H(z + c0 ) g(z+c0 ) e = −1. a

Differentiating this identity, we may again apply Theorem 1.1.33 to conclude that the coefficients of this differentiated exponential identity vanish. In particular, we see that H 󸀠 (z+c0 )/H(z+c0 ) = −g 󸀠 (z+c0 ), and hence H(z+c0 ) = Ce−g(z+c0 ) for a nonzero constant C. But now ρ(H(z + c0 )) < k − 1, whereas ρ(e−g(z+c0 ) ) = k, a contradiction, completing the proof. Remark 5.2.10. For the convenience of the reader, we included the above proof here. However, the result above is nothing but a particular case of Theorem 5.2.14. Remark 5.2.11. The conclusion λ(f − a) ≥ ρ(f ) − 1 cannot be improved to λ(f − a) = ρ(f ) in Theorem 5.2.9, as shown by the following two examples. The first example shows that λ(f ) = ρ(f ) − 1 may happen, whereas in Example 2, we have λ(f − 1) = ρ(f ) − 1. We could not to find an example such that ρ(f ) − 1 < λ(f ) < ρ(f ). Example 1. The entire function f (z) = zez satisfies the homogeneous linear delaydifferential equation (z + 2πi + 1)f 󸀠󸀠 (z) − (z + 2)f 󸀠 (z + 2πi) = 0.

(5.26)

Here λ(f ) = ρ(f ) − 1 and λ(f − a) = ρ(f ) whenever a ≠ 0. Example 2. The entire function f (z) = ez + 1 satisfies the homogeneous linear delaydifferential equation f 󸀠󸀠 (z + 2πi) − f 󸀠 (z) = 0. Here λ(f − 1) = ρ(f ) − 1 and λ(f ) = ρ(f ). In the case where the coefficients of (5.19) are meromorphic functions other than polynomials, Qi and Yang [188, Theorem 1.1] obtained the following result. Theorem 5.2.12. Suppose αj (z) (j = 0, 1, . . . , n) are meromorphic functions in (5.19), and set σ := maxj {ρ(αj )}. Let f (z) be a transcendental meromorphic solution of finite order ρ = ρ(f ) to (5.19). Denoting λf := max{λ(f ), λ( f1 )}, we have: (1) if σ ≥ λf , then σ ≥ ρ(f ) − 1; (2) if σ < λf , then λf ≥ ρ(f ) − 1. Proof. (1) The proof of this case is very much the same as that of Theorem 5.2.9. For the convenience of the reader, we sketch the proof. Contrary to the statement, suppose H1 (z) g(z) that σ ≥ λf and σ < ρ − 1. By the Hadamard representation, f (z) = H e , where 2 (z) H1 (z), resp., H2 (z), is the canonical product formed by the zeros, resp., the poles, of f (z). Then, of course, λ(H1 ) = λ(f ) < ρ − 1, and similarly for H2 . As in the proof of Theorem 5.2.9, g(z) is a polynomial of degree k = ρ, and, in fact, k ≥ 2, since λf
σ = 1 and λf > ρ(f ) − 1 = 1. This is an example to Theorem 5.2.12(2). Theorem 5.2.14. Suppose f (z) is a meromorphic solution of finite order ρ(f ) = ρ < ∞ to the homogeneous linear delay-differential equation n

∑ bj (z)f (kj ) (z + cj ) = 0, j=1

(5.27)

where kj are nonnegative integers, cj are complex constants, and the nonvanishing meromorphic coefficients bj are small in the sense of T(r, bj ) = O(r λ+ε ) + S(r, f ), λ + ε < ρ − 1. Then 1 max {λ( ), λ(f )} ≥ ρ − 1. f

106 | 5 The linear complex delay-differential equations Remark 5.2.15. For this result, see a special case previously offered in [167]. Proof. First, observe that we may assume that ρ > 1, as otherwise the claim is trivial. We may write equation (5.27) in the form n

∑ (bj (z)

f (kj ) (z + cj )

j=1

f (z + cj )

) f (z + cj ) = 0.

(5.28)

Assume, contrary to the statement, that max{λ( f1 ), λ(f )} < ρ − 1. By the Hadamard rep-

resentation we may write f (z) = Q(z)eg(z) , where Q(z) is formed with the Weierstrass products of zeros and poles of f . Clearly, then we have 1 1 max {λ ( ) , λ(Q)} = max {λ ( ) , λ(f )} < ρ − 1, Q f and g is a polynomial of degree ρ =: k. Clearly, Q must be of order < k − 1. Writing now (5.28) in the form n

∑ βj (z)eg(z+cj ) = 0, j=1

(5.29)

where βj (z) = bj (z)

f (kj ) (z + cj ) f (z + cj )

Q(z + cj ),

(5.30)

we may use Proposition 1.1.8 and Lemma 1.2.11 to conclude that each βj (z) is of order < k − 1. Since each g(z + cj ) − g(z + ci ) for j ≠ i is a polynomial of degree k − 1, we may apply Theorem 1.1.32 to see that each coefficient βj (z) vanishes. This means that each f (kj ) (z + cj ) vanishes, and hence f must be a polynomial, contradicting the fact that ρ(f ) = k > 1. 5.2.2 Higher-order nonhomogeneous linear complex delay-differential equations In this section, we first shortly recall nonhomogeneous complex differential equations of the form f (k) + ak−1 (z)f (k−1) + ⋅ ⋅ ⋅ + a0 (z)f = F(z),

(5.31)

where aj (z) (j = 0, 1, . . . , k − 1) and F(z)(≢ 0) are entire functions. It is well known that all solutions of (5.31) are entire functions; see [120, Proposition 8.1]. Moreover, if aj (z)(j = 0, 1, . . . , k−1) are polynomials and F(z) ≢ 0 is an entire function of finite order, then all solutions of (5.31) are of finite order; see [120], Corollary 8.2. Furthermore, [120, Theorem 8.3] shows the following:

5.2 Higher-order linear complex delay-differential equations | 107

Theorem 5.2.16. If a0 (z), . . . , ak−1 (z), and F(z) ≢ 0 are polynomials, then every solution f (z) to (5.31) satisfies λ(f ) = ρ(f ).

(5.32)

For more basic results concerning solutions to (5.31), see [76] and [72]. For the corresponding nonhomogeneous complex difference equations of the form Pn (z)f (z + cn ) + ⋅ ⋅ ⋅ + P1 (z)f (z + c1 ) + P0 (z)f (z + c0 ) = F(z),

(5.33)

where P0 (z), . . . , Pn (z), F(z) ≢ 0 are polynomials, we recall, as an example, the following result due to Chen [25, Theorem 1.1]. Theorem 5.2.17. Let P0 (z), . . . , Pn (z), F(z) be polynomials such that deg(P0 (z) + ⋅ ⋅ ⋅ + Pn (z)) = max{deg Pj (z)} ≥ 1. j

Then every finite-order transcendental meromorphic solution f (z) to (5.33) satisfies ρ(f ) ≥ 1 and λ(f ) = ρ(f ). For a recent presentation treating nonhomogeneous complex difference equations in more detail, see [29, Chapter 5]. We now proceed to considering nonhomogeneous delay-differential equations of the form Pn (z)f (n) (z + cn ) + ⋅ ⋅ ⋅ + P1 (z)f 󸀠 (z + c1 ) + P0 (z)f (z + c0 ) = F(z),

(5.34)

where c0 , c1 , . . . , cn are distinct complex constants, P0 (z), . . . , Pn (z) are polynomials, and F(z) is a nonzero polynomial. In this case, we first recall an observation due to Wu and Zheng [221, Theorem 1.9]. Theorem 5.2.18. All transcendental meromorphic solutions to (5.34) are of order ρ(f ) ≥ 1. Proof. Suppose, contrary to the statement, that ρ(f ) < 1 and first consider the case that f has finitely many poles at most. We may use, again, Proposition 1.1.16 and Lemma 5.1.10 to see that f (j) (z + cj ) = o(1)f (z) and f (z + c0 ) = (1 + o(1))f (z) outside a possible exceptional set of finite logarithmic measure for j = 1, . . . , n. Therefore f (z) =

F(z) (1 + o(1))P0 (z) + o(1)P1 (z) + ⋅ ⋅ ⋅ + o(1)Pn (z)

outside this exceptional set, say H. Since N(r, f ) = O(log r), we easily see that T(r, f ) = O(log r) outside H. By a standard reasoning it follows that T(r, f ) = O(log r), a contradiction.

108 | 5 The linear complex delay-differential equations Assume next that f has infinitely many poles. We may consider these poles being sufficiently large, say in |z| > R, so that P0 , . . . , Pn , and F have no zeros in |z| > R. Increasing R, if needed, we may assume that all z + c0 , . . . , z + cn also are in |z| > R. Let z + c0 be a pole of f . Then there is at least one j1 such that f (j1 ) (z + cj1 ) = ∞, and hence f (z + cj1 ) = f (z + (cj1 − c0 ) + c0 ) = ∞ as well. Similarly, z + (cj2 − c0 ) + (cj1 − c0 ) + c0 is a pole of f . Proceeding inductively, we may find poles of f at z + m(cjk − c0 ) + c0 for some k = 1, . . . , n and for all m ∈ ℕ, that is, λ( f1 ) ≥ 1, so ρ(f ) ≥ 1 follows. We next consider the zeros of solutions to (5.34). Liu and Song [153, Theorem 3] obtained the following result. Theorem 5.2.19. Let F(z) and Pj (z) (j = 0, 1, 2, . . . , n) be polynomials such that Pn (z)F(z) ≢ 0 in (5.34). If f (z) is a transcendental entire solution of finite order to (5.34), then λ(f ) = ρ(f ). If F(z) ≢ dP0 (z), then λ(f − d) = ρ(f ), where d ≠ 0 is a constant. We omit the proof, as this result is a particular case of Theorem 5.2.21. Before proceeding, we add the following remarks. Remark 5.2.20. (1) We see that λ(f ) = ρ(f ) in the preceding theorem cannot be improved to λ(f − a) = ρ(f ) for an arbitrary finite value a unless we assume that F(z) ≢ aP0 (z). As an example, f (z) = ez + a solves f 󸀠 (z + 2πi) − f (z) = −a, and we have λ(f − a) = ρ(f ) − 1. (2) Observe that complex difference equations of type (5.33) may admit entire solutions of infinite order. As an example, consider f (z) = esin(zπ) + z, a solution of f (z + 2) − f (z) = 2. However, we have no example of a proper delay-differential equations of type (5.34) with polynomial coefficients that admit entire solutions of infinite order. Proceeding now to the case of transcendental meromorphic coefficients, we write equation (5.34) in the form αn (z)f (n) (z + cn ) + ⋅ ⋅ ⋅ + α1 (z)f 󸀠 (z + c1 ) + α0 (z)f (z + c0 ) = Q(z),

(5.35)

where α0 , . . . , αn , and Q ≠ 0 are meromorphic functions. Qi and Yang proved an extension to Theorem 5.2.19; see [188, Theorem 1.2]: Theorem 5.2.21. Suppose that f (z) is a transcendental meromorphic solution of finite order to (5.35) in the sense that T(r, αj ) = S(r, f ) and T(r, Q) = S(r, f ), i. e., f is an admissible solution to (5.35). Then λ(f ) = ρ(f ). Moreover, if Q(z) ≢ dα0 (z), then λ(f − d) = ρ(f ), provided that d is a constant. Proof. Substituting g(z) := f (z) − d, we may write H(z) + dα0 (z) − Q(z) = 0,

5.2 Higher-order linear complex delay-differential equations | 109

where H(z) := αn (z)g (n) (z + cn ) + ⋅ ⋅ ⋅ + α1 (z)g 󸀠 (z + c1 ) + α0 (z)g(z + c0 ). Since S(r, g) = S(r, f ), using Lemma 1.2.9, we conclude that m (r,

H ) = S(r, g). g

Therefore 1 H 1 H m (r, ) ≤ m (r, ) + m (r, ) = m (r, ) + S(r, g) = S(r, g). g g H g 1 ) = S(r, f ) implies that λ(f − d) = ρ(f ). Hence m(r, f −d

Corollary 5.2.22. The same conclusion follows if f (z) is a transcendental meromorphic solution of finite order to (5.35) such that α1 , . . . , αn , Q all are of order < ρ(f ). Proof. Repeating, essentially, the proof of Theorem 5.2.21, we see that m(r, g1 ) =

O(r ρ(f )−1+ε ). Hence from T(r, g) = N(r, g1 ) + m(r, g1 ) + O(1) it follows that λ(g) = ρ(f ), proving the claim. Remark 5.2.23. As an example to illustrate Theorem 5.2.21, consider the equation 1 1 f 󸀠󸀠 (z + 2) + (ez + e−z )f 󸀠 (z + 1) − (e4z+4 + e3z+1 + ez+1 )f (z) 2z + 2 4(z + 2)2 + 2 1 1 2 = (e4z+4 + e3z+1 + ez+1 ) + (ez + e−z ) − . z 2(z + 1)3 4(z + 2)5 + 2(z + 2)3 2

This equation has a solution f (z) = ez − z1 . Here Q(z) =

1 4z+4 1 2 (e + e3z+1 + ez+1 ) + (ez + e−z ) − , z 2(z + 1)3 4(z + 2)5 + 2(z + 2)3

so that T(r, Q) = S(r, f ). Moreover, we easily verify that α0 (z) = −(e4z+4 + e3z+1 + ez+1 ) satisfies Q(z) ≢ dα0 (z). Clearly, λ(f ) = λ(f − d) = ρ(f ) = 2. In addition, the next result is closely related to the preceding theorem; see [188], Theorem 1.3. Theorem 5.2.24. Suppose that f (z) is an admissible transcendental meromorphic solution of finite order to equation (5.35). If, as before, Q ≠ 0 and ρ(f ) > ρ(Q), then λf = max{λ(1/f ), λ(f )} = ρ(f ). Proof. Suppose, contrary to the statement, that λf =: λ < ρ(f ) =: ρ, and denote ρ(Q) =: σ < ρ. Write equation (5.35) in the form αn (z)

f (z + c0 ) Q(z) f (n) (z + cn ) f 󸀠 (z + c1 ) + ⋅ ⋅ ⋅ + α1 (z) + α0 (z) = . f (z) f (z) f (z) f (z)

110 | 5 The linear complex delay-differential equations Recalling the logarithmic derivative lemma, see Proposition 1.1.8 and (1.22), we conclude that Q 1 1 m (r, ) ≤ m (r, ) + m (r, ) = O(r ρ−1+ε ) + O(r σ+ε ) + S(r, f ). f f Q Since N(r, f1 ) = O(r λ+ε ), it is obvious that ρ(f ) ≤ max{ρ−1, σ, λ} < ρ, a contradiction. A fairly immediate consequence is the following: Theorem 5.2.25. Suppose that f (z) is an admissible transcendental meromorphic solution of finite order to equation (5.35). If Q ≠ 0, ρ(f ) > ρ(Q), and φ is a meromorphic function satisfying ρ(φ) < ρ(f ) such that φ is not a solution to (5.35), then max{λ(1/(f − φ)), λ(f − φ)} = ρ(f ). Proof. Denoting g(z) := f (z) − φ(z), we can write (5.35) in the form ̃ αn (z)g (n) (z + cn ) + ⋅ ⋅ ⋅ + α1 (z)g 󸀠 (z + c1 ) + α0 (z)g(z + c0 ) = Q(z), ̃ where Q(z) = Q(z) − (αn (z)φ(n) (z + cn ) + ⋅ ⋅ ⋅ + α1 (z)φ󸀠 (z + c1 ) + α0 (z)φ(z + c0 )). Since ̃ < ρ(f ), the claim is an immediate corollary of Theorem 5.2.24. ρ(Q) We next proceed to considering delay-differential equations of the form f (z + c) − eP(z) f (k) (z) = Q(z).

(5.36)

For such equations, Qi and Yang [188] obtained the following two results. Theorem 5.2.26. Let P(z) and Q(z) ≢ 0 be polynomials, and let P(z) be nonconstant. If f (z) is a finite-order transcendental entire solution to (5.36), then ρ(f ) ≥ deg(P(z)) and λ(f ) = ρ(f ). Proof. Writing equation (5.36) in the form f (z + c) − Q(z) = eP(z) , f (k) (z) it immediately follows that deg(P(z)) ≤ max{ρ(f (z + c)), ρ(f (k) (z))} = ρ(f ). It remains to show that λ(f ) = ρ(f ). Suppose this is not the case, and so we have λ(f ) =: λ < ρ(f ). Therefore 1 N (r, ) = O(r λ+ε ). f

5.2 Higher-order linear complex delay-differential equations | 111

By Corollary 1.1.12 N (r,

1

f (k)

1 ) ≤ N (r, ) + S(r, f ) = O(r λ+ε ) + S(r, f ), f

and by Lemma 1.2.10 N (r,

1 1 ) ≤ N (r, ) + S(r, f ) = O(r λ+ε ) + S(r, f ) f (z + c) f

as well. Now we immediately obtain that N (r,

1 1 ) = N (r, P(z) (k) ) = O(r λ+ε ) + S(r, f ). f (z + c) − Q(z) e f (z)

By the second main theorem T(r, f (z + c)) ≤ N (r,

1 1 ) + N (r, ) = O(r λ+ε ) + S(r, f ). f (z + c) f (z + c) − Q(z)

Recalling (1.22), we conclude that T(r, f ) = T(r, f (z + c)) + O(r ρ−1+ε ) + O(log r) = O(r λ+ε ) + O(r ρ−1+ε ) + S(r, f ). This now easily implies that ρ(f ) ≤ max{ρ − 1, λ} < ρ, a contradiction. On the other hand, if we replace e−P(z) by certain polynomials of exponential functions, that is, if we consider delay-differential equations of the form n−1

f (n) (z + cn ) + ∑ {Pj (eA(z) ) + Qj (e−A(z) )}f (j) (z + cj ) = 0, j=0

(5.37)

where Pj (z) and Qj (z) are polynomials, and A(z) = am z m + ⋅ ⋅ ⋅ + a0 is a polynomial of degree m, then we can obtain the following result; see [188], Theorem 1.6. Theorem 5.2.27. Let Pj (z) and Qj (z) (j = 0, 1, . . . , n − 1) be polynomials, and let A(z) = am z m + am−1 z m−1 + ⋅ ⋅ ⋅ + a0 be a nonconstant polynomial of degree m. If deg (P0 ) > deg (Pj )

or

deg (Q0 ) > deg (Qj ),

j = 1, . . . , n − 1,

then all meromorphic solutions f (z)(≢ 0) of finite order to (5.37) satisfy ρ(f ) ≥ m + 1. Moreover, we have max {λ(f − a), λ ( for all a ≠ 0.

1 )} = ρ(f ) f −a

112 | 5 The linear complex delay-differential equations Proof. Letting f be a nonvanishing solution to (5.37), we may assume that f is of finite order ρ. To prove first that ρ ≥ 1, suppose that ρ < 1. Using Remark 1.2.5 and Proposition 1.1.16, we find an exceptional set in (1, ∞) of finite logarithmic measure such that for all |z| = r outside this exceptional set, 󵄨󵄨 (k) 󵄨 󵄨 󵄨 󵄨󵄨 f (z + ck ) 󵄨󵄨󵄨 󵄨󵄨󵄨 f (k) (z + ck ) 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 f (z + ck ) 󵄨󵄨󵄨󵄨 ρ−1+ε k(ρ−1+ε) 󵄨󵄨 󵄨󵄨 = 󵄨󵄨 󵄨 }r 󵄨󵄨 f (z + c0 ) 󵄨󵄨 󵄨󵄨 f (z + ck ) 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 f (z + c0 ) 󵄨󵄨󵄨󵄨 ≤ exp{r 󵄨 󵄨 󵄨 󵄨

(5.38)

for k = 1, . . . , n. We now proceed under the assumption that deg(P0 ) > deg(Pj ) for j = 0, 1, . . . , n − 1 and using the notation Pj (z) = aj,pj z pj + aj,pj−1 z pj−1 + ⋅ ⋅ ⋅ + aj,0 for j = 1, . . . , n − 1. Proceeding now outside the possible exceptional set and letting z be large enough, we see that m

|P0 (z)eA(z) + Q0 (z)e−A(z) | = |a0,p0 |ep0 |am |r (1 + o(1)) 󵄨󵄨 (n) 󵄨 󵄨 󵄨󵄨 (n−1) (z + cn−1 ) 󵄨󵄨󵄨 󵄨 f (z + cn ) 󵄨󵄨󵄨 󵄨f 󵄨󵄨 + |Pn−1 (eA(z) ) + Qn−1 (e−A(z) )| 󵄨󵄨󵄨 󵄨 ≤ 󵄨󵄨󵄨󵄨 󵄨󵄨 f (z + c0 ) 󵄨󵄨󵄨 + ⋅ ⋅ ⋅ 󵄨󵄨 f (z + c0 ) 󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󸀠 A(z) −A(z) 󵄨󵄨󵄨 f (z + c1 ) 󵄨󵄨󵄨 + |P1 (e ) + Q1 (e )| 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (z + c0 ) 󵄨󵄨 ≤ er

ρ−1+ε

m

r n(ρ−1+ε) + |an−1,pn−1 |epn−1 |am |r er m

+ ⋅ ⋅ ⋅ + |a1,p1 |ep1 |am |r er

ρ−1+ε

ρ−1+ε

r (n−1)(ρ−1+ε) (1 + o(1))

r ρ−1+ε (1 + o(1)).

Now by (5.38) and the temporary assumption that ρ < 1, we conclude that |f (k) (z + ck )/f (z + c0 )| = o(1) as r → ∞ outside the exceptional set. Therefore we now see that m

m

|a0,p0 |ep0 |am |r (1 + o(1)) ≤ 1 + |an−1,pn−1 |epn−1 |am |r (1 + o(1)) + ⋅ ⋅ ⋅ m

+ |a1,p1 |ep1 |am |r (1 + o(1)) m

≤ nMemax{p1 ,...,pn−1 }|am |r (1 + o(1)), where M = max{|an−1,pn−1 |, . . . , |a1,p1 |, 1}. Denoting N := max{p1 , . . . , pn−1 }, we have p0 > N by our assumption that deg (P0 ) > max{deg (P1 ), . . . , deg (Pn−1 )}. By the preceding inequality we obtain that |a0,p0 | nM

m

e(p0 −N)|am |r (1 + o(1)) ≤ 1,

a contradiction. Having now proved that ρ(f ) ≥ 1, we now easily arrive at the original claim ρ(f ) ≥ m + 1. Indeed, observing that |a0,p0 | nM

m

e(p0 −N)|am |r (1 + o(1)) ≤ er

ρ−1+ε

,

5.2 Higher-order linear complex delay-differential equations | 113

we may use the standard reasoning to remove the exceptional set, and hence ρ(f ) ≥ m + 1. By a completely similar reasoning we may treat the case with the alternative assumption deg (Q0 ) > deg (Qj ), j = 1, . . . , n − 1. Suppose that max{λ(f − a), λ(1/(f ))} < ρ(f ). Clearly, for each j = 0, . . . , n − 1, ρ(Pj (eA(z) ) + Qj (e−A(z) )) ≤ m < m + 1 ≤ ρ(f ). Since, moreover, a(P0 (eA(z) ) + Q0 (e−A(z) )) ≢ 0, the contradiction λ(f − a) = ρ(f ) immediately follows by Corollary 5.2.22. As a delay-differential version of [60, Theorem 2] (see also [120, Theorem 8.8]), Wang, Han, and Hu [210, Theorem 1.2] proposed to consider hn (z)f (n) (z + cn ) + ⋅ ⋅ ⋅ + h1 (z)f 󸀠 (z + c1 ) + h0 (z)f (z + c0 ) = P1 (z)eP0 (z) f (k) (z + c), (5.39) where k > 0 is an integer, P0 is entire, P1 , h0 ≢ 0, h1 , . . . , hn are small functions to f , and c, c0 , c1 , . . . , cn are pairwise distinct constants. For equations of this type, the following result follows. Theorem 5.2.28. Let f be a transcendental entire solution to (5.39) of hyperorder ρ2 (f ) < 1. Then we have: (i) ρ(f ) ≥ ρ(eP0 ). (ii) If f has a finite Borel exceptional value b and either ρ(f ) > ρ(eP0 ) or ρ(f ) = ρ(eP0 ) while ρ2 (f ) > ρ(P0 ), then b = 0. (ii) If f is of finite order such that max{ρ(eP0 ), ρ(P1 ), ρ(hj )} < ρ(f ) − 1, j = 1, . . . , n, then λ(f − a) ≥ ρ(f ) − 1 for every finite value a ∈ ℂ. Proof. To prove the first claim, rewrite (5.39) as hn (z)f (n) (z + cn ) + ⋅ ⋅ ⋅ + h1 (z)f 󸀠 (z + c1 ) + h0 (z)f (z + c0 ) = eP0 (z) . P1 (z)f (k) (z + c) Since f is of hyperorder < 1, we may combine Lemma 1.2.10 with standard Nevanlinna theory reasoning to obtain n

T(r, eP0 ) ≤ T(r, ∑ hj (z)f (j) (z + cj )) + T (r, n

j=0

1

f (k) (z

+ c)

) + T (r,

1 ) P1 (z)

n

≤ ∑ T(r, f (j) (z + cj )) + T(r, f (k) (z + c)) + ∑ T(r, hj (z)) + S(r, f ) j=0 n

j=0

≤ ∑ T(r, f (j) (z)) + T(r, f (k) (z)) + S(r, f ) ≤ (n + 2)T(r, f ) + S(r, f ). j=0

Then it immediately follows that ρ(eP0 ) ≤ ρ(f ). As for the second claim, suppose b ≠ 0 is a Borel exceptional value of f . By the Hadamard representation, f (z) = g(z)eQ(z) + b, where g is an entire function, so that

114 | 5 The linear complex delay-differential equations T(r, g) = S(r, f ) and Q(z) is an entire function, and thus T(r, eQ ) = T(r, f ) + S(r, f ). Clearly, f (t) (z) = (g(z)eQ(z) + b)(t) = (g(z)eQ(z) )(t) = φt (z)eQ(z) , where each φt (z) is a polynomial in g, Q, and of their derivatives. Hence T(r, φt ) = S(r, f ). Substituting now these representations into (5.39), we have to look at hn (z)φn (z + cn )eQ(z+cn ) + ⋅ ⋅ ⋅ + h1 (z)φ1 (z + c1 )eQ(z+c1 )

+ h0 (z)g(z + c0 )eQ(z+c0 ) + bh0 (z) = P1 (z)φk (z + c)eP0 (z)+Q(z+c) .

If n = 0, then we just have h0 (z)g(z + c0 )eQ(z+c0 ) + bh0 (z) = P1 (z)φk (z + c)eP0 (z)+Q(z+c) . Looking at the entire function g(z + c0 )eQ(z+c0 ) , we immediately see that 1 1 N(r, g(z+c )e Q(z+c0 ) ) = S(r, f ) and N(r, g(z+c )eQ(z+c0 ) +b ) = S(r, f ). An obvious contradiction 0 0 now follows by the second main theorem. Therefore we now proceed by assuming that n ≥ 1. Denoting fj (z) :=

hj (z)φj (z + cj )eQ(z+cj ) −bh0 (z)

fn+1 (z) :=

,

j = 1, . . . , n,

g(z + c0 )(z)eQ(z+c0 ) , −b

and fn+2 (z) :=

P1 (z)φk (z + c)eP0 (z)+Q(z+c) , bh0 (z)

we have f1 + ⋅ ⋅ ⋅ + fn+2 = 1. Using the condition that ρ(f ) > ρ(eP0 ) or ρ(f ) = ρ(eP0 ) while ρ2 (f ) > ρ(P0 ), we obtain that T(r, fj ) = T(r, eQ(z) ) + S(r, f ) = T(r, f ) + S(r, f ) for j = 1, . . . , n + 1, whereas fn+2 cannot be a constant. A contradiction follows immediately by Theorem 1.1.34. Thus b = 0, and claim (ii) is proved. To complete the proof, suppose that a ∈ ℂ is such that λ(f − a) < ρ(f ) − 1. Then a is a Borel value of f , and we may write f (z) = g(z)eQ(z) + a, where, as before, g is an entire function with ρ(g) = λ(g) = λ(f − a) < ρ(f ) − 1, and Q is a polynomial such that deg Q = ρ(f ) := q. Using the same notations as in the previous part of the proof, we also have ρ(φt ) < ρ(f ) − 1 for t = 1, . . . , n, k. Substitution into (5.39) again results in hn (z)φn (z + cn )eQ(z+cn ) + ⋅ ⋅ ⋅ + h1 (z)φ1 (z + c1 )eQ(z+c1 )

+ h0 (z)g(z + c0 )eQ(z+c0 ) + ah0 (z) = P1 (z)φk (z + ck )eP0 (z)+Q(z+c) .

(5.40)

Since ρ(f ) > ρ(eP0 ) + 1 > 1, we see that ρ(f ) = deg Q = q ≥ 2, and therefore ρ(e ) < q − 1. Hence P0

ρ(eQ(z+cj )−Q(z+ci ) ) = ρ(eQ(z+cj )−P0 (z)−Q(z+c) ) = q − 1

5.2 Higher-order linear complex delay-differential equations | 115

for all 0 ≤ j ≠ i ≤ n. Since here the exponential functions are of regular growth and max{ρ(hj φj ), ρ(P1 φk )} < q − 1 for j = 0, . . . , n (here φ0 stands for g(z + c0 ) for brevity), we clearly obtain, for j ≠ i, max{T(r, hj φj ), T(r, P1 φk )} = S(r, eQ(z+cj )−Q(z+ci ) ) and max{T(r, hj φj ), T(r, P1 φk )} = S(r, eQ(z+cj )−P0 (z)−Q(z+c) ). If a = 0, then we may apply Theorem 1.1.33 to (5.40) to conclude that h0 (z)g(z + c0 ) identically vanishes, a contradiction. Assume next that a ≠ 0 and n = 0. Now we may use the same reasoning as in the previous part of the proof to obtain a contradiction by using the second main theorem. Finally, if a ≠ 0 and n ≥ 1, again the same argument as in the preceding part of the proof brings a contradiction, completing the proof. The next examples show that the results in Theorem 5.2.28 are, essentially, best possible. Example 5.2.29. Observe that f (z) = ez sin z is a transcendental entire solution of finite order to the equation f 󸀠󸀠 (z + 2πi) + 2f (z) = 2f 󸀠 (z). In this case, ρ(f ) > ρ(eP0 ). We have not found an example to illustrate the case ρ(f ) = ρ(eP0 ) Example 5.2.30. Next, f (z) = ez is a transcendental entire solution having finite order to the equation f 󸀠󸀠 (z + 2πi) + f (z) = 2f 󸀠 (z), illustrating case (ii) in Theorem 5.2.28. 3

Example 5.2.31. In the case where f (z) = ez is a transcendental entire solution of 2 2 3 finite order to the equation 3z 2 f (z+η0 ) = e3η0 z +3η0 z+η0 f 󸀠 (z) for P0 (z) = 3η0 z 2 +3η20 z+η30 , we have ρ(f ) = ρ(eP0 ) + 1, whereas λ(f ) < ρ(f ) − 1. Hence the condition ρ(f ) > ρ(eP0 ) + 1 cannot be further reduced.

6 Fermat-type delay-differential equations 6.1 Fermat equations As it is well known, the Fermat equations xn +yn = 1 do not admit nontrivial rational solutions by the Fermat last theorem whenever n ≥ 3, whereas it trivially admits nontrivial rational solutions in the Pythagorean case n = 2. Recently, Fermat-type equations, e. g., ∑nj=1 fj (z)k = 1, have been treated in various functional fields, including Fermattype differential equations, difference equations, and delay-differential equations. In this chapter, our main emphasis is on Fermat-type delay-differential equations, including short introductory remarks to Fermat-type functional equations and differential equations plus recalling some key results about Fermat-type difference equations. In what follows, we use the notations ℒ for linear polynomials, 𝒫 for polynomials, ℰ for entire functions, ℛ for rational functions, and ℳ for meromorphic functions.

6.2 Fermat-type functional and differential equations Classical results on meromorphic solutions of Fermat-type functional equations with two terms f (z)n + g(z)n = 1

(6.1)

may shortly be stated as follows: – Montel [174] proved that (6.1) has no transcendental entire solutions whenever n ≥ 3. – Gross [66] proved that if n = 2, then the entire solutions of (6.1) are f (z) = sin(h(z)) and g(z) = cos(h(z)), where h(z) is any entire function. – Gross [65] also proved that (6.1) has no transcendental meromorphic solutions whenever n ≥ 4. – Baker [4] showed that if n = 3, then the nonconstant meromorphic solutions of (6.1) are of the form f (z) = F(ω(z)) and g(z) = cG(ω(z)), where c3 = 1, ω(z) is an 1

1+3− 2 φ󸀠 (z)

1

1−3− 2 φ󸀠 (z)

entire function, F(z) = , G(z) = , and φ(z) denotes the Weier2φ(z) 2φ(z) strass function with periods ω1 and ω2 defined as φ(z) :=

–

1 1 1 + { − }, ∑ 2 2 z (z + μω1 + νω2 ) (μω1 + νω2 )2 μ,ν∈ℤ;μ2 +ν2 =0 ̸

satisfying the differential equation (φ󸀠 )2 = 4φ3 − 1. Gross [65] showed that if n = 2, then all meromorphic solutions of (6.1) are of the 2β 1−β2 forms f (z) = 1+β2 and g = 1+β2 , where β is any meromorphic function.

https://doi.org/10.1515/9783110560565-006

118 | 6 Fermat-type delay-differential equations We next recall some results on the existence of meromorphic solutions of Fermat-type functional equations with three terms f (z)n + g(z)n + h(z)n = 1.

(6.2)

First, recall nonexistence results concerning the function classes ℳ, ℛ, ℰ , 𝒫 for equation (6.2); see [93] and [79]. Theorem 6.2.1. (1) For n ≥ 9, there do not exist f , g, h ∈ ℳ satisfying (6.2), (2) For n ≥ 8, there do not exist f , g, h ∈ ℛ satisfying (6.2), (3) For n ≥ 7, there do not exist f , g, h ∈ ℰ satisfying (6.2), (4) For n ≥ 6, there do not exist f , g, h ∈ 𝒫 satisfying (6.2). As to the existence of meromorphic solutions to (6.2), some special solutions are easy found, letting a(z) ≠ 0 be a meromorphic function: For n = 3, f (z) = 21 (1 + a(z)3 ), g(z) = 21 (1 − a(z)3 ), h(z) = √3 −3(a(z)2 ); for n = 4, f (z) =

1 3 4 (a(z) √ 8

+

1 ), a(z)

g(z) =

for n = 5, f (z) = 31 ((2 − √6)a(z) + 1 +

1 (a(z)3 4 √ −8

−

1 ), a(z)

and h(z) = √4 −1a(z)2 ;

2+√6 ), a(z)

g(z) =

1 (√6 + 2) − (3√2 + 2√3i) (((√6 − 2) + (3√2 − 2√3i))a(z) + 2 − ), 3 a(z)

h(z) =

1 (√6 + 2) + (3√2 + 2√3i) (((√6 − 2) − (3√2 − 2√3i))a(z) + 2 − ). 3 a(z)

and

In the case n = 6, Gundersen [77] constructed the transcendental meromorphic μ , g(z) = ω+b , and h(z) = i ν , where solutions f (z) = ω−b ν ν ω=a

(4 + 2i)F(z) − (3 + 4i)A − (1 − 2i)B , (4 + 2i)F(z) − 5A + (1 − 2i)B

ν6 = (1 − 112 i)(ω2 − β2 )3 , μ6 = (1 + 112 i)(ω2 − α2 )3 , and F(z) is an elliptic function that satisfies (F 󸀠 )2 = k(F − A)(F − B)(F − C), C = 41 (5A − B), A and B being arbitrary distinct constants. This example was further considered by Tohge [204] by means of the theta function. In the case n ≤ 5, nonconstant rational solutions to equation (6.2) are immediately found by choosing a(z) above as a rational function. Similarly, nonconstant entire solutions to equation (6.2) are obtained by choosing a(z) = eα(z) , where α(z) is an entire function. Note that in the cases n = 6, 7 the existence of nonconstant rational solutions, in the cases n = 4, 5 the existence of nonconstant polynomial solutions, in the

6.2 Fermat-type functional and differential equations | 119

cases n = 6 the existence of nonconstant entire solutions, and in the cases n = 7, 8 the existence of nonconstant meromorphic functions still remain open. However, for n = 8, Ishizaki [105] showed that if f , g, h are nonconstant meromorphic solutions of (6.2), then there must exist a small function a(z) with respect to f , g, h such that the Wronskian determinant satisfies W(f 8 , g 8 , h8 ) = a(z)(f (z)g(z)h(z))6 . A more careful analysis of this situation was made in [109] for the cases n = 7, 8. Denote F := f n , G := g n , H := hn , and Δ := W(F, G, H)/(FGH). Then a possible meromorphic solution f , g, h to (6.2) satisfies the differential equation W(F, G, H) = f n g n hn Δ. For more considerations of this situation, the reader may consult [109]. To close these introductory remarks, we shortly consider general Fermat-type functional equations n

∑ fj (z)k = 1,

(6.3)

j=1

where n, k are positive integers. Let F𝒮 (k) denote the smallest number n such that (6.3) admits nonconstant solutions in 𝒮 , where 𝒮 is one of function classes ℒ, 𝒫 , ℰ , ℛ, ℳ. Clearly, F𝒮 (k) depends on k and S. As a key reference to this notion, the reader may consult [93], [80], and [115]. We collect some estimates for F𝒮 in the following: Theorem 6.2.2. Let k ≥ 2 and n ≥ 2. Let f1 (z), f2 (z), . . . , fn (z) be nonconstant solutions in 𝒮 satisfying (6.3). Then we have the following lower estimates for (6.3): Fℒ (k) = k + 1,

F𝒫 (k) >

1 1 + √k + , 2 4

Fℛ (k) > √k + 1,

Fℰ (k) ≥

FM (k) ≥ √k + 1.

1 1 + √k + , 2 4

We next recall some results related to Fermat-type differential equations. First, look at equation (6.1), where g(z) is a linear differential polynomial L(f ) of f (z). The following observation is a particular case from Yang and Li [236]. Theorem 6.2.3. Let n be a positive integer, let b0 , b1 , . . . , bn−1 , bn (≠ 0) be constants, and let L(f ) = ∑nk=0 bk f (k) . Then the transcendental meromorphic solutions to f 2 + L(f )2 = 1

(6.4)

must be of the form f (z) = 21 (Peλz + P1 e−λz ) = cosh(λz + A) = cos(λzi + Ai), where eA = P, P is a nonzero constant, and λ satisfies the following equations: n 1 ∑ bk λk = , i k=0

n 1 ∑ bk (−λ)k = − . i k=0

(6.5)

Corollary 6.2.4. The transcendental meromorphic solutions of the Fermat-type differential equation f (z)2 + f 󸀠 (z)2 = 1

(6.6)

120 | 6 Fermat-type delay-differential equations must satisfy f (z) =

1 1 (Pe−iz + eiz ) = cos (z + Ai) , 2 P

where eA = P, and P is nonzero constant. The basic idea to considering possible meromorphic solutions of (6.4) is factoring the left-hand side of (6.4) (or related equations of similar type) to work out expressions of f (z). For the existence of meromorphic solutions, this depends, of course, on the concrete form of the equation under consideration. Here we give examples of the existence and nonexistence cases. First, consider f (z)2 + R(z)(f (k) (z))2 = Q(z).

(6.7)

Looking at this Fermat-type differential equation, Zhang and Liao [249] showed that by reducing (6.7) to some linear differential equations the existence of transcendental meromorphic solutions to (6.7) implies that R(z) and Q(z) must be of special type: Theorem 6.2.5. Let f (z) be a transcendental meromorphic solution to (6.7). (i) If k = 1, then Q(z) must be a constant, say C, and the zeros of R(z) must be at most of 1 double multiplicity. Then f (z) = √C cos α(z), where α(z) is a primitive of √R(z) such

that √C cos α(z) is a transcendental meromorphic function. (ii) If k ≥ 2, then k is an odd integer, Q(z) and R(z) are constants, say C and A, respectively. Then f (z) = √C cos(az + b), where a2k = A1 . On the other hand, if L(f ) = bn f (n) + bn+1 f (n+1) in (6.4), then Yang and Li [236, Theorem 2] showed the nonexistence of meromorphic solutions: Theorem 6.2.6. Let bn and bn+1 be nonzero constants. Then 2

f 2 + (bn f (n) + bn+1 f (n+1) ) = 1

(6.8)

has no transcendental meromorphic solutions. However, simple examples of the existence situations are easily constructed: Example 6.2.7. The function f (z) = cos z satisfies 2 3 1 f 2 + ( f 󸀠 + f 󸀠󸀠󸀠 ) = 1, 2 2

and f (z) = sin z solves 2 1 1 f 2 + ( f 󸀠 − f 󸀠󸀠󸀠 ) = 1. 2 2

6.2 Fermat-type functional and differential equations | 121

Looking at situations where L(f ) consists of several terms, we give two notations for more convenient statements. Denote by D(L(f )) the number of differential operator terms in L(f ) and by W(L(f )) the sum of orders of all differential operators in L(f ). For if L(f ) = ∑nk=0 bk f (k) and bk ≠ 0, k = 0, 1, . . . , n. So far, it is example, W(L(f )) = n(n+1) 2 known that if D(L(f )) is odd, resp., even, then a necessary condition of the existence of transcendental solutions of (6.4) is that W(L(f )) is odd, resp., even, provided that L(f ) has no more than three terms. However, this is not true if L(f ) has four terms: Example 6.2.8. The function f (z) = cos(−√ bb2 z + Ai) is an entire solution of 4

2

f 2 + (b1 f 󸀠 + b2 f 󸀠󸀠 + b3 f 󸀠󸀠󸀠 + b4 f (4) ) = 1, where b1 b4 + b2 b3 = b4 √ bb2 . Here D(L(f )) = 4, and W(L(f )) = 10. The function f (z) =

sin z solves the equation

4

2 1 3 f 2 + (f 󸀠 + f 󸀠󸀠 + 2f (4) + f (6) ) = 1. 2 2

Here D(L(f )) = 4, and W(L(f )) = 13. Of course, the situation becomes more complicated if f (z)2 is replaced by an expression with more than one term in (6.4). As an example, let us consider (f + f 󸀠 )2 + (f + f 󸀠󸀠 )2 = 1.

(6.9)

In [150] the following result is proved. Theorem 6.2.9. Equation (6.9) has no transcendental meromorphic solutions. On the other hand, considering a more general equation (af + bf 󸀠 )2 + (cf + df 󸀠󸀠 )2 = 1,

(6.10)

Liu and Dong [150] showed that meromorphic solutions may exist in certain special cases for constant coefficients a, b, c, d. Again, the reasoning works by using the factorization: Theorem 6.2.10. If equation (6.10) admits transcendental meromorphic solutions f (z), then we have one of the following cases: (i) a = 0, c = 0, b ≠ 0, d ≠ 0, and f (z) = bid2 sinh( bi z − A) + B, where A, B are constants. d 2p

a

p

(ii) a, b, c, d are nonzero constants and ac = 1−e i, and then f (z) = De− b z + e +e , where 2a 1+e2p p is a constant, and D is a nonzero constant. c dc (iii) a = 0, b ≠ 0, c ≠ 0, and f (z) = bi+d sinh( bi+d z − A) + B, where d + b2 i+bd = 0, and bc A, B are constants. −p

122 | 6 Fermat-type delay-differential equations

6.3 Fermat difference equations Equations of type (6.1) where g(z) is a linear difference polynomial of f (z) are called Fermat difference equations. We include here some considerations of such equations, as the reasoning applied may be useful to considering Fermat delay-differential equations, which we treat in more detail in the next section. 6.3.1 Entire solutions of Fermat difference equations Liu, Cao, and Cao [145] obtained a detailed expression for entire solutions of finite order to f (z)2 + f (z + c)2 = 1.

(6.11)

Theorem 6.3.1. Transcendental entire solutions of finite order to (6.11) must satisfy , and k is an integer. f (z) = sin(Az + B), where B is a constant, A = (4k+1)π 2c Proof. Let f (z) be a transcendental entire solution of finite order to (6.11). Recall that all entire solutions to f (z)2 + f (z + c)2 = 1 must be of the form f (z) = sin(h(z)), f (z + c) = cos(h(z)), where h(z) is an entire function; see [66]. Thus h(z +c) = h(z)+ π2 +2kπ, where k is an integer. Writing h(z) = (4k+1)π z + H(z), we see that H(z) is a periodic function 2c with period c. Hence f (z) = sin (

(4k + 1)π z + H(z)) . 2c

Since f (z) is a transcendental entire function of finite order, by periodicity H(z) must be a constant, say B, by Pólya’s theorem; see [90, Theorem 2.9]. For the more general case of f (z)n + f (z + c)n = 1,

(6.12)

see Section 6.3.2. On the other hand, replacing f (z + c) by f (z + c) − f (z) in (6.11), that is, f (z)2 + [f (z + c) − f (z)]2 = 1,

(6.13)

Liu [141, Proposition 5.3] proved the following: Theorem 6.3.2. There exist no transcendental entire solutions of finite order to equation (6.13). Looking at the more general case of f (z)2 + (a0 f (z) + a1 f (z + c1 ) + ⋅ ⋅ ⋅ + an f (z + cn ))2 = 1,

(6.14)

6.3 Fermat difference equations | 123

where a0 , a1 , . . . , an are nonzero constants, it is not surprising that equations of type (6.14) admit transcendental solutions in special cases only, although examples of such cases are not difficult to find: Example 6.3.3. The function f (z) = sin z is a solution to f (z)2 + (2f (z) + f (z +

2

π ) + f (z + π)) = 1 2

and to f (z)2 + (2f (z) + f (z +

2

1 3 π ) + f (z + π) + f (z + 3π)) = 1. 2 2 2

As a concrete example of special cases related to (6.13), we may consider D2 f (z)2 + (Af (z + c) + Bf (z))2 = 1;

(6.15)

see [150, Theorem 1.13]: Theorem 6.3.4. Let A, B be constants. If there are transcendental entire solutions of finite order to (6.15), then A2 = B2 + D2 . Proof. Let f (z) be a transcendental entire solution. By Gross [66] we may write Df (z) = sin(h(z)),

Af (z + c) + Bf (z) = cos(h(z)).

Therefore A sin(h(z + c)) + B sin(h(z)) = D cos(h(z)), and hence A sin(h(z + c)) = √D2 + B2 sin(h(z) + φ(z)), where tan φ = −D/B. Recalling elementary expressions of the sine function, we get eih(z+c) − e−ih(z+c) √D2 + B2 eih(z)+iφ(z) − e−ih(z)−iφ(z) = . 2i A 2i Now writing this as e2ih(z+c) −

√D2 + B2 ih(z+c)+ih(z)+iφ(z) √D2 + B2 ih(z+c)−ih(z)−iφ(z) e + e = 1, A A

we may use Theorem 1.1.34 to conclude that √D2 + B2 ih(z+c)−ih(z)−iφ(z) e =1 A

124 | 6 Fermat-type delay-differential equations and e2ih(z+c) −

√D2 + B2 A

eih(z+c)+ih(z)+iφ(z) = 0

simultaneously. To avoid a contradiction, we easily conclude that h(z + c) − h(z) − φ(z) vanishes identically, and hence A2 = D2 + B2 . Zhang [243] considered an improvement of (6.13) as f (z)2 + [f (z + c) − f (z)]2 = β2 ,

(6.16)

where β is a nonzero small function of f . The following result is an interesting situation where a Fermat difference equation can be transformed to a differential equation. Theorem 6.3.5. Let f be a transcendental entire function of finite order. Suppose that f is a solution to (6.16). Then, after the transformation f = βg, (6.16) reduces to a differential equation of g such that g 2 + (αg 󸀠 )2 = 1, where α is a small meromorphic function of g and has one zero at least. Proof. Differentiating both sides of (6.16) results in ff 󸀠 + △c f △c f 󸀠 = ββ󸀠 .

(6.17)

Multiplying by △c f the both sides of (6.17), together with (6.16), leads to ff 󸀠 △c f + △c f 󸀠 (β2 − f 2 ) = ββ󸀠 △c f ,

(6.18)

f (f 󸀠 △c f − △c f 󸀠 f ) = ββ󸀠 △c f − β2 △c f 󸀠 .

(6.19)

that is,

Set φ := f 󸀠 △c f − △c f 󸀠 f . Then it follows from equations (6.18) and (6.19) that fφ = ββ󸀠 △c f − β2 △c f 󸀠 .

(6.20)

Applying Lemma 1.2.14 to (6.20), we get T(r, φ) = m(r, φ) = S(r, f ), that is, φ is a small entire function of f . If φ = 0, then it follows from (6.20) that △c f = c0 β, where c0 is a nonzero constant. By substituting △c f = c0 β into (6.16) we see that f 2 = (1 − c02 )β2 ,

6.3 Fermat difference equations | 125

which is impossible. Therefore we may further assume that φ ≢ 0. On the other hand, by eliminating β, β󸀠 from the right sides of (6.16) and (6.17) we get β󸀠 (f 2 + (△c f )2 ) − β(ff 󸀠 + △c f △c f 󸀠 ) = 0. This leads to f (β󸀠 f − βf 󸀠 ) = β △c f △c f 󸀠 − β󸀠 (△c f )2 = −

(6.21)

△c ffφ . β

(6.22)

Therefore it follows from (6.22) that △c f = Set f = gβ and α =

β2 . φ

g 󸀠 β3 . φ

(6.23)

Then it follows from equation (6.23) that g󸀠 =

and hence △c f =

β 󸀠 (β f − βf 󸀠 ). φ

β󸀠 f − βf 󸀠 φ △c f = , β3 β2

(6.24)

Substituting this into equation (6.16) leads to g 2 + (αg 󸀠 )2 = 1.

(6.25)

Finally, if α has no zeros, then we see that g is an entire function by comparing the order of possible poles on both sides of (6.25), Then we can rewrite (6.25) into the form g 2 + (αg 󸀠 )2 = (g + iαg 󸀠 )(g − iαg 󸀠 ) = 1.

(6.26)

This means g + iαg 󸀠 , resp., g − iαg 󸀠 has no poles and zeros. Recalling that g is of finite order, we get g + iαg 󸀠 = eγ and g − iαg 󸀠 = e−γ , respectively, where γ is a nonconstant polynomial. Therefore we obtain g = cos p, where p = iγ is a polynomial as well. Substituting g = cos p and f = βg into (6.16) leads to β(z + c) cos p ± sin p = = β(z) cos(p + △p)

1∓i 2ip e 2 cos △p+i sin △p 2ip e 2

+ +

1±i 2 cos △p−i sin △p 2

.

(6.27)

Observe that the coefficients cos △p, sin △p, and cos △p ± i sin △p are just small funcβ(z+c) tions of ep . Applying Theorem 1.1.29, we obtain T(r, e2ip ) = T(r, β(z) )+S(r, g) = S(r, g), which is impossible. Thus α must have at least one zero. Slightly more generally, related to equation (6.16), Liu and Yang [147, Theorem 2.3] also considered Fermat-difference equations f (z)2 + P(z)2 (△c f )2 = Q(z)

(6.28)

and proved that there are no transcendental entire solutions of finite order if P(z) and Q(z) are nonzero polynomials. Returning to equation (6.16), Liu et al. [154] proved the following nonexistence result.

126 | 6 Fermat-type delay-differential equations Theorem 6.3.6. Equation (6.16) admits no transcendental entire solutions of finite order, provided that the small function β satisfies one the following conditions: (1) β is a nonzero constant; (2) β is a nonzero periodic function with period c; (3) β is an entire function without zeros. Proof. Let f (z) be a finite-order transcendental entire solution of (6.16). Case (1): If β is a nonzero constant, then by (6.16) (△c f )2 = −[f (z)2 − β2 ]. Thus △c f must have infinitely many zeros, since otherwise both f − β and f + β would have finitely many zeros, which is a contradiction to the fact f (z) is an entire function. So, we have that at least one of f − β and f + β has infinitely many zeros. If both f − β and f + β have infinitely many zeros, then the zeros should have multiplicities at least two. Thus we have N (r,

1 1 1 1 ) ≤ N (r, ) ≤ T(r, f ). f ±β 2 f ±β 2

Shifting backward equation (6.16), we have 2f (z − c)2 − 2f (z)f (z − c) = β2 − f (z)2 = (△c f )2 , and hence 2f (z − c)[f (z − c) − f (z)] = [f (z + c) − f (z)]2 .

(6.29)

If f (z − c) has finitely many zeros, combining the second main theorem with Lemma 1.2.10 results in 1 1 1 ) + N (r, ) + S(r, f ) 2T(r, f ) ≤ N(r, f ) + N (r, ) + N (r, f f +β f −β 1 1 1 ≤ N (r, ) + N (r, ) + N (r, ) + S(r, f ) f (z − c) f +β f −β ≤ T(r, f (z)) + S(r, f ),

(6.30)

which is a contradiction. If one of f − β and f + β has infinitely many zeros, then inequality (6.30) also holds. Thus f (z − c) must have infinitely many zeros. Furthermore, from (6.29) we have that f (z), f (z − c), and f (z + c) − f (z) share 0 IM. Indeed, if f (z) and f (z − c) have no common zeros, then the multiplicities of the zeros of f (z − c) should be at least two, and then (6.30) is replaced by 2T(r, f ) ≤ 32 T(r, f )+S(r, f ), which is impossible. Moreover, the case where f (z) and f (z + c) − f (z) share the value 0 IM is impossible from (6.16) as well.

6.3 Fermat difference equations | 127

Case (2): Using a similar reasoning, we see that if β is a nonzero small periodic function with respect to f (z), then we also get that f (z), f (z − c), and f (z + c) − f (z) 1 ) ≤ N(r, β1 ) + S(r, f ), and we also get a contradiction from share 0 IM. Thus N(r, f (z) (6.30). Case (3): We suppose that β is a transcendental entire function without zeros. Thus β(z) = eh(z) . It is easy to see that (6.16) can be written as f (z + c)2 − 2f (z + c)f (z) + 2f (z)2 = β2 .

(6.31)

Equation (6.31) implies that f (z + c) =

β(z) (U + V), √2

√2f (z) = β(z) (U − V), √2

where U, V are entire functions, since β has no zeros and satisfies the Fermat equation (1 −

√2 √2 t)U 2 + (1 + t)V 2 = 1 2 2

(6.32)

with t 2 = 1. So we have U=

sin(α(z)) √1 −

√2 t 2

= t1 sin(α(z)),

V=

cos(α(z)) √1 +

√2 t 2

= t2 cos(α(z)).

Then f (z) =

β(z) β(z) (t sin(α(z)) − t2 cos(α(z))) = (sin(α(z) + B)) 2 1 2

and f (z + c) =

β(z) β(z) (t sin(α(z)) + t2 cos(α(z))) = (sin(α(z) + D)). √2 1 √2

Thus we obtain β(z + c) β(z) (sin(α(z + c) + B)) = (sin(α(z) + D)). √2 2 Furthermore, we get β(z + c) i(α(z+c)+α(z)+B+D) (e − e−i(α(z+c)−α(z)+B−D) ) − √2e2i(α(z)+D) β(z) = −√2. Applying Theorem 1.1.34 to this equation, we see that α(z) must be a linear polynomial. Since β(z) is a small function with respect to f (z), β(z) is a constant, a contradiction. Thus we have completed the proof of Theorem 6.3.6.

128 | 6 Fermat-type delay-differential equations 6.3.2 Meromorphic solutions of Fermat difference equations Looking at the solutions to (6.6), we immediately see that there are no nonentire meromorphic functions satisfying (6.6). However, equation (6.12) can admit nonentire meromorphic solutions, at least for the case n = 2; see the next theorem and examples below. The following result can be found in [151]; see also [231]. Theorem 6.3.7. The meromorphic solutions to (6.11) should satisfy f (z) = where h(z) is a meromorphic function satisfying one of the following two cases: (i) h(z + c) = −ih(z); (ii) h(z + c)h(z) = i.

1 h(z)+ h(z) 2

,

z

Remark 6.3.8. (1) If h(z + c) = −ih(z), then h(z) = (−i) c d(z), where d(z) is a periodic meromorphic function with period c. Using a result given by Ozawa [180], there exists a prime periodic entire function h(z) of order ρ(h) ≥ 1. This implies that there do not exist meromorphic solutions to (6.11) of order less than one in this case. (2) If h(z + c)h(z) = i, then h(z) is a periodic meromorphic function with period 2c. In this case, h(z) cannot be an entire function of finite order. This can be easily seen by the difference analogue of the logarithmic derivative lemma in Section 1.2.2. (3) Theorem 6.3.7 actually displays how to construct non-entire meromorphic solutions of (6.11). The following Examples 6.3.9 and 6.3.11 show that the first case of Theorem 6.3.7 can appear. Examples 6.3.10 and 6.3.12 show that the second case of Theorem 6.3.7 may also appear. In addition, there may also exist meromorphic solutions of infinite order to (6.11); see Examples 6.3.11 and 6.3.12. Example 6.3.9. Let ec = −i, that is c = 2kiπ −

f (z) =

ez sin 2zπ c

iπ 2

with integer k. The function

+

sin 2zπ c ez

2

is a finite-order meromorphic solution to f (z)2 + f (z + 2kiπ − iπ2 )2 = 1. Here h(z) = satisfies h(z + c) = −ih(z).

ez sin 2zπ c

Example 6.3.10. Let c = π2 . The function f (z) =

1 √i tan z

+ √i tan z 2

is a finite-order meromorphic solution to f (z)2 + f (z + satisfies h(z + c)h(z) = i.

π 2 ) 2

= 1. Here h(z) =

1 √i tan z

6.3 Fermat difference equations | 129

Example 6.3.11. Let ec = 1, that is c = 2kiπ with nonzero integer k. The function f (z) = sin(ez +

z ) 4ki

is an infinite-order entire solution to f (z)2 + f (z + 2kiπ)2 = 1. Here h(z) = ei(e satisfies h(z + c) = −ih(z).

z

z − π2 ) + 4ki

Example 6.3.12. Let c = π2 . The function f (z) =

√−i tan(e4zi + z) + 2

1 √−i tan(e4zi +z)

is an infinite-order meromorphic solution to f (z)2 + f (z + √−i tan(e4zi + z) satisfies h(z + c)h(z) = i.

, π 2 ) 2

= 1. Here h(z) =

We next consider Fermat cubic difference equations of the form f (z)3 + f (z + c)3 = 1.

(6.33)

All possible meromorphic solutions of this equation must be of hyperorder ρ2 (f ) ≥ 1, see [165] and [119]. For the proof of this result, see [165, p. 131–133] and [119]. Theorem 6.3.13. There are no transcendental meromorphic solutions to (6.33) of hyperorder less than one. Proof. Suppose f is a meromorphic solution to equation (6.33). Recalling [4], we observe that f (z) = H(φ(z)) and f (z + c) = ηG(φ(z)) = ηH(−φ(z)) = H(−η2 φ(z)), where φ(z) is an entire function, η3 = 1, H(z) :=

1 + 𝒫 󸀠 (z)/√3 , 2𝒫 (z)

G(z) :=

1 − 𝒫 󸀠 (z)/√3 , 2𝒫 (z)

and 𝒫 (z) is the Weierstrass elliptic function satisfying the differential equation 𝒫 󸀠 (z)2 = 4𝒫 (z)3 − 1. First, assume that f is of finite order. Using the differential equation 𝒫 󸀠 (z)2 = 1+𝒫 󸀠 (φ)/√3 4𝒫 (z)3 − 1 and expression f (z) = , by an elementary computation we see 2𝒫(φ) that

3f 2 (z)𝒫 2 (φ(z)) − 3f (z)𝒫 (φ(z)) + 1 = 𝒫 3 (φ(z)). Therefore 3T(r, 𝒫 (φ)) ≤ 2T(r, f ) + 2T(r, 𝒫 (φ)) + O(1), so 𝒫 (φ) ≤ 2T(r, f ) + O(1), and thus 𝒫 (φ) must be of finite order. Since 𝒫 is of finite order, it immediately follows by [47,

Corollary 1.2] that φ is a polynomial. On the other hand, T(r, f ) ≤ T(r, 𝒫 (φ)) + T(r, 𝒫 󸀠 (φ)) + O(1)

130 | 6 Fermat-type delay-differential equations

= T(r, 𝒫 (φ)) + N(r, 𝒫 (φ)) + m (r, ≤ 3T(r, 𝒫 (φ)) + O(log r),

𝒫 󸀠 (φ) ) + m(r, 𝒫 (φ)) + O(1) 𝒫 (φ)

and hence ρ(f ) = ρ(𝒫 (φ)) and S(r, f ) = S(r, 𝒫 (φ)). By the expressions for f (z) and f (z + c) we immediately conclude that η (1 −

𝒫 󸀠 (φ(z))

√3

) 𝒫 (φ(z + c)) = (1 +

𝒫 󸀠 (φ(z + c))

√3

) 𝒫 (φ(z)).

Differentiating this identity and looking at [165, p. 132], we conclude that if 𝒫 (φ(z)) = 0, then 𝒫 (φ(z + c)) = ∞, except for finitely many such points, as φ(z) is a polynomial, and N(r, 1/𝒫 (φ(z))) ≤ N(r, 𝒫 (φ(z + c))) + S(r, 𝒫 (φ)).

(6.34)

Next, from −f (z+c)3 = f (z)3 −1 = (f −1)(f −η)(f −η2 ) we see that all zeros of f −1, f −η, f −η2 are at least of triple multiplicity. Therefore by basic Nevanlinna theory 3

2T(r, f ) ≤ ∑ N (r, j=1

≤

1 ) + N(r, f ) + S(r, f ) f − ηj

1 3 1 ) + N(r, f ) + S(r, f ) ∑ N (r, 3 j=1 f − ηj

≤ T(r, f ) + N(r, f ) + S(r, f ). This means that m(r, f ) = S(r, f ) = S(r, 𝒫 (φ)).

Now using the expression for f , we may write m (r,

1 𝒫 (φ)

1 1 2 𝒫(φ)

=f −

√3 𝒫 󸀠 (φ) . 6 𝒫(φ)

) ≤ m(r, f ) + O(log r) = S(r, f ) = S(r, 𝒫 (φ)).

Hence (6.35)

Now since all poles of 𝒫 are double, all poles of 𝒫 (φ) are at least of double multiplicity, and we obtain that 1 N(r, 𝒫 (φ(z + c))) ≤ T(r, 𝒫 (φ(z + c))) + S(r, 𝒫 (φ)). 2 Combining this inequality with (6.34), (6.35), and (1.29), we get 1 T(r, 𝒫 (φ)) = S(r, 𝒫 (φ)) + O(r ρ(𝒫(φ))−1+ε ), 2 a contradiction. We next proceed by assuming that a solution f to (6.33) must be of infinite order by the reasoning in [119]. Again recalling the expressions for f (z) and f (z + c), f (z) = H(φ) and f (z + c) = H(−η2 φ), it follows that φ(z + c) = −η2 φ + ω, where ω is a period of

6.3 Fermat difference equations | 131

𝒫 (φ). Then φ is a transcendental entire function of order at least 1 because φ can be

ω expressed by φ = (−η2 )z ν + 1+η 2 , where ν is a periodic function with period ω. Recall the expressions for f (z) and f (z + c) again. We also get ηf (z) + f (z + c) = η/𝒫 (φ). By logarithmic differentiation of this identity we obtain

φ󸀠 = −

(ηf (z) + f (z + c))󸀠 𝒫 (φ) , ηf (z) + f (z + c) 𝒫 󸀠 (φ)

(6.36)

and thus (φ󸀠 )2 = (

2

𝒫 (φ)2 (ηf (z) + f (z + c))󸀠 ) . ηf (z) + f (z + c) 4𝒫 (φ)3 − 1

(6.37)

Thus we see that the poles and ϑi -points of 𝒫 (φ) are of double multiplicity, where ϑi are the three cubic roots of 41 . Using the first main theorem, we see that 1 1 3 1 2T(r, 𝒫 (φ)) = m(r, 𝒫 (φ)) + ∑ m (r, ) 2 2 i=1 𝒫 (φ) − ϑi

1 1 3 1 + N(r, 𝒫 (φ)) + ∑ N (r, ) + O(1). 2 2 i=1 𝒫 (φ) − ϑi

(6.38)

Using the second main theorem, we see that 3

2T(r, 𝒫 (φ)) ≤ N(r, 𝒫 (φ)) + ∑ N (r, i=1

1

𝒫 (φ) − ϑi

) + S(r, 𝒫 (φ))

1 1 3 1 ≤ N(r, 𝒫 (φ)) + ∑ N (r, ) + S(r, 𝒫 (φ)). 2 2 i=1 𝒫 (φ) − ϑi

(6.39)

Combining (6.38) and (6.39), we have 3

m(r, 𝒫 (φ)) + ∑ m (r, i=1

1

𝒫 (φ) − ϑi

) = O(log rT(r, 𝒫 (φ))) = O(log rT(r, f )).

Obviously, T(r, f (z + c)) = T(r, f ) + O(1) follows (6.33). Moreover, by (6.37) and the logarithmic derivative lemma we next obtain that T(r, φ󸀠 ) = m(r, φ󸀠 ) ≤ O(log rT(r, f )). By elementary computation, ρ2 (f ) ≥ 1, and we are done. Remark 6.3.14. We remark that (6.33) admits transcendental meromorphic solutions. Indeed, we may take c = iπ, φ(z) = ez , f (z) = H(φ(z)), and f (z + c) = G(φ(z)) in the proof of Theorem 6.3.13. In addition, we may also take any entire function φ satisfying φ(z + c) = −φ(z).

132 | 6 Fermat-type delay-differential equations 6.3.3 Fermat delay-differential equations We concentrate in this section on discussing entire solutions only. For Fermat delaydifferential equations of the form (f (k) (z))n + f (z + c)m = 1, we may first point out [145, Theorem 1.2] to recall the following result. Theorem 6.3.15. Equation (f (k) (z))n + f (z + c)m = 1 admits no transcendental entire solutions of finite order whenever the positive integers m, n satisfy m ≠ n. Proof. Suppose f is a transcendental entire solution of finite order ρ(f ) = ρ, and first assume that m > n. Then by (1.22) we observe at once that mT(r, f (z + c)) + O(r ρ−1+ε ) + O(log r) = mT(r, f (z)) = nT(r, f (k) (z)) + O(1) ≤ nT(r, f (z)) + O(log r), and hence (m − n)T(r, f ) = O(r ρ−1+ε ) + O(log r), a contradiction. Next, assume that n > m ≥ 2. Then by the second main theorem 1 1 ) + N (r, (k) n ) + S(r, f (k) ) f (k) (f ) − 1 1 ) + S(r, f (k) ), ≤ T(r, f (k) ) + N (r, (k) n (f ) − 1

nT(r, f (k) ) = T(r, (f (k) )n ) ≤ N (r,

and hence 1 ) + S(r, f (k) ) (f (k) )n − 1 1 = N (r, ) + S(r, f (k) ) f (z + c)m 1 ) + S(r, f ) ≤ N (r, f (z + c)

(n − 1)T(r, f (k) ) ≤ N (r,

≤ T(r, f (z + c)) + S(r, f ) = T(r, f ) + O(r ρ−1+ε ) + S(r, f ).

Therefore 2T(r, f ) + O(r ρ−1+ε ) + S(r, f ) = 2T(r, f (z + c)) ≤ mT(r, f (z + c)) = nT(r, f (k) ) + O(1) n ≤ T(r, f ) + O(r ρ−1+ε ) + S(r, f ) n−1 3 ≤ T(r, f ) + O(r ρ−1+ε ) + S(r, f ), 2 resulting in T(r, f ) = O(r ρ−1+ε ) + S(r, f ), a contradiction. It remains to consider the case n > m = 1. As we now have (f 󸀠 (z))n + f (z + c) = 1,

6.3 Fermat difference equations | 133

we obtain n(f 󸀠 (z))n−1 f 󸀠󸀠 (z) = −f 󸀠 (z + c). Therefore by Lemma 1.2.14 T(r, f 󸀠󸀠 (z)) = S(r, f 󸀠 ), which is a contradiction. This completes the proof. Making first the easy observation that there cannot be nonconstant polynomial solutions, Liu, Cao, and Cao [145] also considered transcendental entire solutions of finite order to f 󸀠 (z)2 + f (z + c)2 = 1.

(6.40)

Theorem 6.3.16. Transcendental entire solutions to (6.40) must satisfy f (z) = sin(z ± Bi), where B is a constant, c = 2kπ or c = 2kπ + π with integer k. Example 6.3.17. Obviously, the transcendental entire function f (z) = sin z is a solution to f 󸀠 (z)2 + f (z + 2kπ + π)2 = 1, and f (z) = cos z = sin(z + π2 ) is a solution to f 󸀠 (z)2 + f (z + 2kπ)2 = 1, where k is an integer. We omit the proof to Theorem 6.3.16, as Liu and Yang [147] proved a more general case by replacing f 󸀠 with f (k) : Theorem 6.3.18. The transcendental entire solutions of finite order to the delay-differential equation f (k) (z)2 + f (z + c)2 = 1 must satisfy one of the following two cases: (i) if k is odd, then f (z) = ∓ sin(Aiz + Bi), c = (ii) if k is even, then f (z) = ± cos(Aiz +Bi), c =

kπi , and Ak = A kπi+ πi2 , and Ak A

(6.41)

±i, = ±1, where B is a constant.

Proof. Assume that f (z) is a finite-order transcendental entire solution to equation (6.41). Then [f (k) (z) + if (z + c)][f (k) (z) − if (z + c)] = 1.

(6.42)

Thus both f (k) (z) + if (z + c) and f (k) (z) − if (z + c) have no zeros. Combining (6.42) with the Hadamard factorization theorem, we have f (k) (z) + if (z + c) = ep(z) and f (k) (z) − if (z + c) = e−p(z) . Thus we get f (k) (z) =

ep(z) + e−p(z) 2

(6.43)

134 | 6 Fermat-type delay-differential equations and f (z + c) =

ep(z) − e−p(z) , 2i

(6.44)

where p(z) is a nonconstant polynomial. Combining (6.43) with (6.44), we get f (k) (z + c) =

p1 (z)ep(z) − p2 (z)e−p(z) ep(z+c) + e−p(z+c) = , 2i 2

(6.45)

where p1 (z) = p󸀠 k + Mk (p(k) , . . . , p󸀠 ) and p2 (z) = (−1)k p󸀠 k + Nk (p(k) , . . . , p󸀠 ), and Mk and Nk are polynomials of p(k) , . . . , p󸀠 with degree less than k. Thus from (6.45) we get iep(z+c)−p(z) ie−p(z)−p(z+c) p2 (z)e−2p(z) + + = 1. p1 (z) p1 (z) p1 (z) p2 (z)e−2p(z) p1 (z) p(z+c)−p(z)

It is easy to see that neither

nor

i e−p(z+c)−p(z) p1 (z)

(6.46)

is a constant. By Theo-

rem 1.1.34 we obtain that ie ≡ p1 (z). This implies that p(z) = Az + B, where A is a nonzero constant, and B is a constant. Thus p1 (z) = Ak and p2 (z) = (−A)k . From (6.45) we get Ak eAz+B − (−A)k e−Az−B eA(z+c)+B + e−A(z+c)−B = . 2i 2 Therefore ieAc = Ak and ie−Ac = (−1)k+1 Ak . Thus A2k = (−1)k and e2Ac = (−1)k+1 . From (6.43) we also obtain f (z) =

1 Az+B e Ak

+

1 e−Az−B (−A)k

2

.

If k is odd, then Ak = ±i. If Ak = i, then f (z) = sin(−Aiz − Bi), where c = kπi . If Ak = −i, A kπi k k then f (z) = sin(Aiz + Bi), where c = A . If k is even, then A = ±1. If A = 1, then f (z) = cos(Aiz + Bi), where c = c=

kπi+ πi2 A

kπi+ πi2 A

. If Ak = −1, then f (z) = − cos(Aiz + Bi), where

. Thus the proof is completed.

Examples of delay-differential equations that are slightly more general than those above appear when replacing f (z + c) with Δc f (z) := f (z + c) − f (z). Recalling [145], we first observe the following: Theorem 6.3.19. Delay-differential equations of the form (f 󸀠 (z))n + (f (z + c) − f (z))m = 1 have no transcendental entire solutions of finite order, provided that the positive integers n, m satisfy m ≠ n > 1.

6.3 Fermat difference equations | 135

Before giving the proof of Theorem 6.3.19, we recall an important result, which is also related to Fermat-type functional equations; see [232], Theorem 1. Lemma 6.3.20. Let m, n be positive integers satisfying nonconstant entire solutions f (z) and g(z) that satisfy

1 m

+

1 n

< 1. Then there are no

a(z)f (z)n + b(z)g(z)m = 1, where a(z) and b(z) are small functions with respect to f (z). Proof of Theorem 6.3.19. By Lemma 6.3.20 it is sufficient to consider the case where n > 1 = m. By differentiation we obtain that n(f 󸀠 (z))n−1 f 󸀠󸀠 (z) = −f 󸀠 (z + c) + f 󸀠 (z). We may also use Lemma 1.2.14 to see that m(r, f 󸀠󸀠 ) = S(r, f 󸀠 ), again contradicting to the fact that f (z) is an entire function. Theorem 6.3.19 is thus proved. If n = m = 2, we obtain the following [145]: Theorem 6.3.21. Transcendental entire solutions of finite order to (f 󸀠 (z))2 + (f (z + c) − f (z))2 = 1 satisfy f (z) =

1 2

sin(2z + Bi), where c = (k + 21 )π, k ∈ ℤ, and B is a constant.

Proof. Suppose, contrary to the statement, that f (z) is a transcendental entire solution of finite order. Similarly as in the proof of Theorem 6.3.18, we may factorize to obtain (f 󸀠 (z) + ig(z))(f 󸀠 (z) − ig(z)) = 1, where g(z) := f (z +c)−f (z). This means that f 󸀠 (z)+ig(z) = ep(z) and f 󸀠 (z)−ig(z) = e−p(z) , where p(z) is a polynomial. Therefore f 󸀠 (z) =

ep(z) + e−p(z) 2

and g(z) = f (z + c) − f (z) =

ep(z) − e−p(z) . 2i

An immediate conclusion is that p(z) cannot be a constant, as f is transcendental. Now we can compute f 󸀠 (z + c) =

ep(z+c) + e−p(z+c) (p󸀠 (z) + i)(ep(z) + e−p(z) ) = . 2 2i

136 | 6 Fermat-type delay-differential equations Therefore (1 − p󸀠 (z))(ep(z+c)+p(z) + ep(z+c)−p(z) ) − e2p(z+c) = 1. It now easy to see that we may apply Theorem 1.1.34 to obtain that (1−p󸀠 (z))ep(z+c)−p(z) ≡ 1. This means that p󸀠 is a constant, and thus p(z) = Az + B, where A, B are constants, and A ≠ 0. Therefore (1 − p󸀠 (z))ep(z+c)−p(z) = (1 − Ai)eAc = 1. Returning to the expression of f 󸀠 (z + c) above, we get f 󸀠 (z + c) =

(A + i)eAz+B + (A + i)e−Az−B eA(z+c)+B + e−A(z+c)−B = . 2i 2

We now immediately observe that A = −2i and e−2ic = −1, c = (k + 21 )π with integer k. Hence f (z) =

1 e2iz−B − e−2iz+B 1 = sin(2z + Bi), 2 2i 2

and we are done. Proceeding to more general delay-differential equations, Fermat equations of type 3

3

{a0 f (z) + a1 f (z + c) + a2 f 󸀠 (z)} + {b0 f (z) + b1 f (z + c) + b2 f 󸀠 (z)} = eαz+β

(6.47)

were considered by Hu and Wang [101] as follows. See also some related results in Wang, Chen, and Hu [209]. Theorem 6.3.22. Let α, β, c ≠ 0 and ai , bi (i = 0, 1, 2) be complex numbers, and assume that rank (

a0 b0

a1 b1

a2 ) = 2. b2

If equation (6.47) has meromorphic solutions of finite order, then all such solutions are of the form f (z) = Ae

αz+β 3

+ CeDz ,

where A, C, D are constants. Moreover, if we define constants c0 , c1 by c03 + c13 = 1, then A, D are completely determined by ai , bi , α ≠ 0, c0 , c1 as follows: b c −a c (1) A = a 2b0−a 2b1 , C = 0 if a0 b1 − a1 b0 = 0, b1 a2 − a1 b2 = 0, (2) A = (3) A =

0 2

2 0

b1 c0 −a1 c1 , C = 0 if a0 b1 − a1 b0 ≠ 0, b1 a2 − a1 b2 = 0, a0 b1 −a1 b0 a b −a b 3(b1 c0 −a1 c1 ) , C ∈ ℂ, D = b1 a0 −a0b 1 if b1 a2 − a1 b2 ≠ (b1 a2 −a1 b2 )(α−3D) 1 2 1 2

0.

6.3 Fermat difference equations | 137

Remark 6.3.23. It should be pointed out that the simple delay-differential equation (f 󸀠 (z))3 + f 3 (z + c) = 1, a particular case of (6.47), may have no transcendental meromorphic solutions. To prove this, we may apply the reasoning used in Remark 5.1.9. Before proceeding now to prove Theorem 6.3.22, we remark that obtaining more general results on the existence of meromorphic solutions to delay-differential equations, even of linear ones, remains largely an unexplored topic. For example, we are not aware of such results for delay-differential equations of type (f 󸀠 (z))n + g(f )(z) = 1, where g(f ) is a linear shift polynomial of f . Proof of Theorem 6.3.22. First, rewrite (6.47) in the form F(z)3 + G(z)3 = 1,

(6.48)

where F and G are defined by F(z)e G(z)e

αz+β 3

= a0 f (z) + a1 f (z + c) + a2 f 󸀠 (z),

αz+β 3

(6.49)

= b0 f (z) + b1 f (z + c) + b2 f 󸀠 (z).

Suppose we were able to prove that F(z), G(z) are constants, say F(z) ≡ c0 , G(z) ≡ c1 . By the rank assumption we have three cases to consider. (1) If a0 b1 − a1 b0 = 0 and b1 a2 − a1 b2 = 0, then a0 b2 − b0 a2 ≠ 0. Solving equations (6.49), it follows that f (z) =

b c − a2 c1 αz+β b2 F(z) − a2 G(z) αz+β e 3 = 2 0 e 3 , a0 b2 − b0 a2 a0 b2 − b0 a2

(6.50)

proving case (1) of Theorem 6.3.22. (2) Similarly, if a0 b1 − a1 b0 ≠ 0 and b1 a2 − a1 b2 = 0, then f (z) =

b c − a1 c1 αz+β b1 F(z) − a1 G(z) αz+β e 3 = 1 0 e 3 , a0 b1 − b0 a1 a0 b1 − b0 a1

(6.51)

which is case (2) of the theorem. (3) The final case is b1 a2 − a1 b2 ≠ 0. Eliminating f (z + c) from (6.49), we obtain (b1 a2 − a1 b2 )f 󸀠 (z) + (a0 b1 − a1 b0 )f (z) = (b1 c0 − a1 c1 )e

αz+β 3

.

By the elementary theory of differential equations we now get, with C being the constant of integration: b0 a1 −a0 b1

z

1

f (z) = e b1 a2 −a1 b2 (C + Ae 3

(β+

3(b1 a0 −a1 b0 )+α(b1 a2 −a1 b2 ) z) a2 b1 −a1 b2

),

(6.52)

138 | 6 Fermat-type delay-differential equations 3(a c −b c )

1 1 0 . where A = 3(b a −a b1 )+α(a 0 1 0 1 1 b2 −a2 b1 ) Therefore it remains to show that F, G are constants. Proceed now under the assumption that this is not the case. By [4] we have

F(z) =

1+

℘󸀠 (h(z)) √3

2℘(h(z))

,

G(z) =

1−

℘󸀠 (h(z)) √3

2℘(h(z))

η.

(6.53)

Looking back at the proof of Theorem 6.3.13, we observe that h(z) is a polynomial. Moreover, as F and G are nonconstants by our provisional assumption, (6.53) implies that h(z) is nonconstant. By the rank assumption we have to distinguish three cases to obtain a contradiction from our provisional assumption. Case 1: a0 b1 − a1 b0 = 0, b1 a2 − a1 b2 = 0. Then we have a0 b2 − b0 a2 ≠ 0 by the rank assumption. Assume first that a2 = 0. Then a1 = 0 as well. Indeed, if a1 ≠ 0, then we have b2 = 0 by the assumption b1 a2 − a1 b2 = 0. This contradicts the rank assumption b2 a0 − a2 b0 ≠ 0. Since now a1 = a2 = 0, we have b2 a0 ≠ 0 from b2 a0 − a2 b0 ≠ 0 and a0 b1 = 0 from the assumption a0 b1 − a1 b0 = 0. Thus it follows that a0 ≠ 0, b1 = 0, and b2 ≠ 0. By using (6.53) we can rewrite equations (6.49) as follows: f (z) =

1+

℘󸀠 (h) √3

2a0 ℘(h)

e

αz+β 3

b0 f (z) + b2 f 󸀠 (z) =

,

1−

℘󸀠 (h) √3

2℘(h)

ηe

αz+β 3

(6.54)

.

Differentiating the first relation of (6.54) and recalling that (℘󸀠 )2 = 4℘3 −1 and ℘󸀠󸀠 = 6℘2 , it follows that f (z) = 󸀠

2 3 ℘ (h)h󸀠 √3

− ℘󸀠 (h)h󸀠 +

2a0

℘2 (h)

h󸀠 √3

e

αz+β 3

+

α (1 +

℘󸀠 (h) ) √3

6a0 ℘(h)

e

αz+β 3

.

(6.55)

Substituting f and f 󸀠 into the second equation of (6.54), we obtain 2b b bα ( 2 ℘3 (h)h󸀠 − (a0 η − b0 − 2 )℘(h) + 2 h󸀠 ) √3 √3 3 3

=(4℘ (h) − 1)(b2 h − 󸀠

a0 η + b0 + √3

b2 α 3

2

2

℘(h)) .

Therefore by Theorem 1.1.29 we get 6T(r, ℘(h)) = 5T(r, ℘(h)) + O(log r). Therefore T(r, ℘(h)) = O(log r), which means that ℘(h) is a rational function, a contradiction.

6.3 Fermat difference equations | 139

Therefore, to proceed in case 1, we may now assume that a2 ≠ 0. Solving equations (6.49) and noting (6.53), it follows that f (z) =

b +a η

b2 − a2 η + 2√32 ℘󸀠 (h) αz+β b2 F(z) − a2 G(z) αz+β e 3 = e 3 . a0 b2 − b0 a2 2(a0 b2 − b0 a2 )℘(h)

(6.56)

By differentiating (6.56) and recalling that (℘󸀠 )2 = 4℘3 − 1 and ℘󸀠󸀠 = 6℘2 , we have f 󸀠 (z) =

2(b2 +a2 η) 3 ℘ (h)h󸀠 √3

− (b2 − a2 η)℘󸀠 (h)h󸀠 +

2(a0 b2 − b0 a2

+

α (b2 − a2 η +

)℘2 (h)

b2 +a2 η 󸀠 ℘ (h)) √3

6(a0 b2 − b0 a2 )℘(h)

e

αz+β 3

b2 +a2 η 󸀠 h √3

e

αz+β 3

(6.57)

.

Substituting (6.56) and (6.57) into the first equation of (6.49), we have ((a0 b2 − b0 a2 )(1 +

℘󸀠 (h) a + a2 α b + a2 η 󸀠 )℘(h) − 0 (b2 − a2 η + 2 ℘ (h))℘(h) √3 √3 3

b + a2 η 󸀠 2(b2 + a2 η) 3 ℘ (h)h󸀠 − (b2 − a2 η)℘󸀠 (h)h󸀠 + 2 h ))℘(h(z + c)) √3 √3 αc b + a2 η 󸀠 = e 3 a1 (b2 − a2 η + 2 ℘ (h(z + c))) ℘2 (h). √3

− a2 (

(6.58)

Let {zj }∞ j=1 be the zeros of ℘ satisfying |zj | → ∞ as j → ∞. Then for each j, there exist finitely many complex numbers aj,k such that h(aj,k ) = zj . Moreover, we see that (℘󸀠 )2 (h(aj,k )) = (℘󸀠 )2 (zj ) = −1, since ℘(zj ) = 0. We next show that ℘(h(aj,k + c)) = 0 for at most finitely many aj,k . Assume, on the contrary, that there exists an infinite subsequence of {aj,k } such that ℘(h(aj,k + c)) = 0. Without loss of generality, we may consider {aj,k } itself. Thus we have (℘󸀠 )2 (h(aj,k +c)) = −1. Differentiating (6.58) and then setting z = aj,k , we obtain the identity a2 h󸀠 (aj,k )h󸀠 (aj,k + c) ((a2 η − b2 )℘󸀠 (h(aj,k )) +

b2 + a2 η ) = 0. √3

(6.59)

Clearly, ℘󸀠 (h(aj,k )) = ±i. Since a2 ≠ 0, equation (6.59) becomes h󸀠 (aj,k )h󸀠 (aj,k + c) (±(a2 η − b2 )i +

b2 + a2 η ) = 0. √3

Since h is a nonconstant polynomial and {aj,k } is an infinite sequence, there are infinitely many points aj,k such that h󸀠 (aj,k )h󸀠 (aj,k + c) ≠ 0. It follows that we have either (a2 η − b2 )i +

2πi b2 + a2 η 1 − √3i = (b2 + e 3 ηa2 ) = 0 √3 √3

140 | 6 Fermat-type delay-differential equations or (b2 − a2 η)i +

2πi b2 + a2 η 1 + √3i = (b2 + e− 3 ηa2 ) = 0. √3 √3

Hence b2 +e 3 ηa2 = 0. We may write b2 = −ka2 , where k = e 3 η. Since a2 ≠ 0, we have b1 = −ka1 by the assumption b1 a2 −a1 b2 = 0. Recalling the assumption a0 b1 −a1 b0 = 0, we get ka0 a1 + a1 b0 = 0. If a1 ≠ 0, then ka0 a1 + a1 b0 = 0 implies b0 = −ka0 , and hence bi = −kai for i = 0, 1, 2, contradicting the rank assumption. Therefore we must have a1 = 0. Using (6.53), we may rewrite the first equation of (6.49) in the form ±2πi

±2πi

℘󸀠 (h(z))

1 + √3 αz+β a f 󸀠 (z) = − 0 f (z) + e 3 . a2 2a2 ℘(h(z))

(6.60)

By elementary theory of linear differential equations the solutions to (6.60) are f (z) = e

a

− a0 z 2

(∫

1+

℘󸀠 (h(z)) √3

2a2 ℘(h(z))

e

3a0 +αa2 3a2

β

z+ 3

dz + C) ,

(6.61)

where C is the constant of integration. Let tj (j ≥ 1) be the poles of ℘(ζ ). As they are all double poles, we have ℘(ζ ) = gj (ζ )(ζ − tj )−2 , where gj is a holomorphic function in a neighborhood of tj such that gj (tj ) ≠ 0. Therefore 1+

℘󸀠 (ζ ) √3

2a2 ℘(ζ )

=−

gj󸀠 (ζ ) (ζ − tj )2 1 1 + + =− + O(1) √3a2 (ζ − tj ) 2a2 gj (ζ ) 2√3a2 gj (ζ ) √3a2 (ζ − tj )

as ζ → tj . Setting h(z) =: ζ for a while, we conclude that 1+

℘󸀠 (h(z)) √3

2a2 ℘(h(z))

=−

1 + O(1) √3a2 (h(z) − tj )

as h(z) → tj . Observe that equation h󸀠 (z) = 0 has finitely many solutions only, whereas −1 󸀠 −1 ⋃∞ j=1 h (tj ) is an infinite set. Thus there exist an integer j and a point z ∈ h (tj ) such 󸀠 󸀠 󸀠 that h (z ) ≠ 0, meaning that z is a simple zero of h(z) − tj . Therefore f (z) has a logarithmic singularity at z 󸀠 , which contradicts the assumption that f (z) is a meromorphic function. Since ℘(h(aj,k + c)) = 0 for at most finitely many aj,k only and h is a polynomial, there exists a positive integer J such that ℘(h(aj,k + c)) ≠ 0,

h󸀠 (aj,k ) ≠ 0,

j > J, k = 1, . . . , deg(h).

6.3 Fermat difference equations |

141

Returning to equation (6.58) and recalling that ℘(h(z)) = 0 at z = aj,k , we see that the coefficient of ℘(h(z + c)) at aj,k takes a nonzero value a2 h󸀠 (aj,k ) ((a2 η − b2 )℘󸀠 (h(aj,k )) +

b2 + a2 η ) ≠ 0, √3

for j > J, k = 1, . . . , deg(h), because (a2 η − b2 )℘󸀠 (h(aj,k )) +

b2 + a2 η ≠ 0. √3

Therefore equation (6.58) valued at aj,k immediately yields ℘(h(aj,k + c)) = ∞,

j > J, k = 1, . . . , deg(h),

which further yields N (r,

1 ) ≤ N (r, ℘(h(z + c))) + S(r, ℘(h)). ℘(h)

Note that the multiple zeros of ℘(h) occur at zeros of its derivative {℘(h)}󸀠 = ℘󸀠 (h)h󸀠 , that is, at the zeros of h󸀠 , since ℘󸀠 (h(aj,k )) = ±i. Hence we obtain the estimate N (r,

1 1 ) ≤ N (r, ) + O(log r) ≤ N (r, ℘(h(z + c))) + S(r, ℘(h)). ℘(h) ℘(h)

(6.62)

The following part is similar to Theorem 6.3.13. However, we give enough details to prove m(r, F) = S(r, ℘(h)). In fact, the expression of F in (6.53) yields T(r, F) ≤ T(r, ℘(h)) + T(r, ℘󸀠 (h)) + O(1) = O(T(r, ℘(h))). Based on the factorization −G3 = F 3 − 1 = (F − 1)(F − η)(F − η2 ), where η ≠ 1, we see that all zeros of F − 1, F − η, and F − η2 are of multiplicity ≥ 3. Hence the second main theorem gives 3

2T(r, F) ≤ ∑ N (r, m=1

≤

1 ) + N(r, F) + S(r, F) F − ηm

1 1 3 ) + N(r, F) + S(r, F) ∑ N (r, 3 m=1 F − ηm

≤ T(r, F) + N(r, F) + S(r, ℘(h)), which immediately implies the claim m(r, F) = S(r, ℘(h)).

142 | 6 Fermat-type delay-differential equations We now rewrite (6.53) in the form ℘󸀠 (h) 1 = 2F − . √3℘(h) ℘(h) This means that m (r,

℘󸀠 (h) 1 ) ≤ m(r, F) + m (r, ) + O(1). ℘(h) ℘(h)

Applying the logarithmic derivative lemma, we get m (r,

℘󸀠 (h) ℘󸀠 (h)h󸀠 1 ) ≤ m (r, ) + m (r, 󸀠 ) = S(r, ℘(h)), ℘(h) ℘(h) h

and hence m (r,

1 ) = S(r, ℘(h)). ℘(h)

(6.63)

Combining (6.62) with (6.63) and recalling that all poles of ℘(z) are double, by the first Nevanlinna main theorem we get 1 1 1 ) + O(1) = m (r, ) + N (r, ) + O(1) ℘(h) ℘(h) ℘(h) 1 ≤ N(r, ℘(h(z + c))) + S(r, ℘(h)) ≤ N(r, ℘(h(z + c))) + S(r, ℘(h)) 2 1 1 ≤ T(r, ℘(h(z + c))) + S(r, ℘(h)) ≤ T(r, ℘(h)) + S(r, ℘(h)), 2 2

T(r, ℘(h)) = T (r,

a contradiction. Case 2: a0 b1 − a1 b0 ≠ 0, b1 a2 − a1 b2 = 0. Solving equations (6.49) and noting (6.53), it follows that b +a η

b1 − a1 η + 1√31 ℘󸀠 (h) αz+β b F(z) − a1 G(z) αz+β f (z) = 1 e 3 = e 3 . a0 b1 − b0 a1 2(a0 b1 − b0 a1 )℘(h)

(6.64)

By differentiating (6.64) and noting that (℘󸀠 )2 = 4℘3 − 1 and ℘󸀠󸀠 = 6℘2 , we obtain f (z) = 󸀠

2(b1 +a1 η) 3 ℘ (h)h󸀠 √3

− (b1 − a1 η)℘󸀠 (h)h󸀠 +

2(a0 b1 − b0 a1

+

α (b1 − a1 η +

)℘2 (h)

b1 +a1 η 󸀠 ℘ (h)) √3

6(a0 b1 − b0 a1 )℘(h)

e

αz+β 3

.

b1 +a1 η 󸀠 h √3

e

αz+β 3

(6.65)

6.3 Fermat difference equations |

143

Substituting (6.64) and (6.65) into the first equation of (6.49), we have ((a0 b1 − b0 a1 )(1 +

℘󸀠 (h) aα b + a1 η 󸀠 )℘(h) − (a0 + 2 )(b1 − a1 η + 1 ℘ (h))℘(h) √3 √3 3

b + a1 η 󸀠 2(b1 + a1 η) 3 ℘ (h)h󸀠 − (b1 − a1 η)℘󸀠 (h)h󸀠 + 1 h ))℘(h) √3 √3 αc b + a1 η 󸀠 = e 3 a1 (b1 − a1 η + 1 ℘ (h(z + c))) ℘2 (h). √3

(6.66)

− a2 (

Similarly to case 1, we again proceed to show that ℘(h(aj,k + c)) = 0 for at most finitely many aj,k . Assume for a while that there exists an infinite subsequence of {aj,k }, denoting by {aj,k } itself without loosing generality, such that ℘(h(aj,k + c)) = 0. Thus we have (℘󸀠 )2 (h(aj,k +c)) = −1. Differentiating (6.66) and then setting z = aj,k , we derive the relation a2 h󸀠 (aj,k )h󸀠 (aj,k + c) ((a1 η − b1 )℘󸀠 (h(aj,k )) +

b1 + a1 η ) = 0. √3

(6.67)

Note that ℘󸀠 (h(aj,k )) = ±i. If a2 ≠ 0, then equation (6.67) becomes h󸀠 (aj,k )h󸀠 (aj,k + c) (±(a1 η − b1 )i +

b1 + a1 η ) = 0. √3

Since h is a nonconstant polynomial and {aj,k } is an infinite sequence, there are infinitely many aj,k such that h󸀠 (aj,k )h󸀠 (aj,k + c) ≠ 0. It follows that ± (a1 η − b1 )i +

b1 + a1 η = 0. √3

(6.68)

Let tj (j ≥ 1) be the poles of ℘(z) satisfying |tj | → ∞ as j → ∞ and take bj,k ∈ ℂ satisfying h(bj,k ) = tj for k = 1, . . . , deg(h). Then there exists an integer j0 such that for j > j0 , bj,k are simple zeros of h(z) − tj and h(z + c) − tj has only simple zeros. Thus the unique term 2(b1 + a1 η)℘3 (h(z))h󸀠 (z)/√3 with poles of multiplicity 6 at bj,k (j > j0 ) must vanish, that is, b1 + a1 η = 0. Combining this with (6.68), we obtain a1 = b1 = 0. This contradicts the assumption a0 b1 − a1 b0 ≠ 0. If a2 = 0, then we have a1 b2 = 0 by the assumption b1 a2 − a1 b2 = 0. Hence we distinguish two cases depending on whether a1 is zero or not as follows. If a1 = 0, then we have a0 b1 ≠ 0 by the assumption a0 b1 − a1 b0 ≠ 0. Now by (6.53) we can rewrite equations (6.49) as follows: f (z) =

1+

℘󸀠 (h(z)) √3

2a0 ℘(h(z))

e

αz+β 3

,

b0 f (z) + b1 f (z + c) + b2 f (z) = 󸀠

1−

℘󸀠 (h(z)) √3

2℘(h(z))

ηe

αz+β 3

.

(6.69)

Differentiating the first relation of (6.69) and noting that (℘󸀠 )2 = 4℘3 − 1 and ℘󸀠󸀠 = 6℘2 , it follows that f (z) = 󸀠

2℘3 (h)h󸀠 √3

− ℘󸀠 (h)h󸀠 + 2a0

℘2 (h)

h󸀠 √3

e

αz+β 3

+

α(1 +

℘󸀠 (h) ) √3

6a0 ℘(h)

e

αz+β 3

.

(6.70)

144 | 6 Fermat-type delay-differential equations Substituting f and f 󸀠 into the second equation of (6.69), we have (a0 η (1 − − b2 (

℘󸀠 (h) ℘󸀠 (h) bα ) ℘(h) − (b0 + 2 ) (1 + ) ℘(h) √3 √3 3

2℘3 (h)h󸀠 h󸀠 − ℘󸀠 (h)h󸀠 + ))℘(h(z + c)) √3 √3

αc

=e 3 b1 (1 +

(6.71)

℘󸀠 (h(z + c)) ) ℘2 (h). √3

Differentiating (6.71) and then setting z = aj,k , we derive the relation b2 h󸀠 (aj,k )h󸀠 (aj,k + c) (℘󸀠 (h(aj,k )) −

1 ) = 0. √3

(6.72)

Note that ℘󸀠 (h(aj,k )) = ±i. If b2 ≠ 0, then (6.72) gives a contradiction. Hence we have b2 = 0. Now we can rewrite the second equation of (6.69) as follows: b0 f (z) + b1 f (z + c) =

1−

℘󸀠 (h(z)) √3

2℘(h(z))

ηe

αz+β 3

.

Substituting the first relation of (6.69) into the above equation, we have αc

b1 e 3 (1 +

℘󸀠 (h(z + c)) a η + b0 󸀠 )℘(h(z)) = (a0 η − b0 − 0 ℘ (h(z)))℘(h(z + c)). √3 √3

(6.73)

Differentiating this equation and then setting z = aj,k , we derive the relation αc

b1 e 3 (1 +

℘󸀠 (h(aj,k + c))

)℘󸀠 (h(aj,k ))h󸀠 (aj,k ) √3 a η + b0 󸀠 =(a0 η − b0 − 0 ℘ (h(aj,k )))℘󸀠 (h(aj,k + c))h󸀠 (aj,k + c). √3 Noting that ℘󸀠 (h(aj,k )) = ±i and ℘󸀠 (h(aj,k + c)) = ±i, this equation immediately implies one and only one of the following four situations: A1 h󸀠 (aj,k ) = B1 h󸀠 (aj,k + c),

A󸀠1 h󸀠 (aj,k ) = B2 h󸀠 (aj,k + c),

(6.74)

A2 h󸀠 (aj,k ) = B1 h󸀠 (aj,k + c),

A󸀠2 h󸀠 (aj,k ) = B2 h󸀠 (aj,k + c),

where αc

A1 = −A󸀠1 = b1 e 3 (1 +

i ), √3

B1 = a0 η − b0 −

a0 η + b0 i, √3

6.3 Fermat difference equations | 145

αc

A2 = −A󸀠2 = −b1 e 3 (1 −

i ), √3

B2 = a0 η − b0 +

a0 η + b0 i. √3

Obviously, A1 and A2 are nonzero constants by the assumption a0 b1 ≠ 0. Since h is a nonconstant polynomial and {aj,k } is an infinite sequence, relations (6.74) immediately yield the functional equations A1 h󸀠 (z) = B1 h󸀠 (z + c),

A󸀠1 h󸀠 (z) = B2 h󸀠 (z + c),

(6.75)

A2 h󸀠 (z) = B1 h󸀠 (z + c),

A󸀠2 h󸀠 (z) = B2 h󸀠 (z + c).

These equations mean that one of the four equalities A1 = B1 ,

−A1 = B2 ,

A2 = B1 ,

−A2 = B2

holds by comparing the leading coefficients. Thus we obtain h󸀠 (z) = h󸀠 (z +c), which implies h(z) = az +b, where a ≠ 0 and b are constants. Now the equations ℘(h(aj,k )) = 0 and ℘(h(aj,k +c)) = 0 become ℘(aaj,k +b) = 0 and ℘(aaj,k + b + ac) = 0, that is, {aaj,k + b, aaj,k + b + ac} ⊂ {zj }∞ j=1 . Hence we have (aaj,k + b + ac) − (aaj,k + b) = ac ∈ Pa ∪ {mω1 + nω2 : m, n ∈ ℤ}, since ℘ has only two distinct zeros in the parallelogram Pa and ℘ is a function of double periods ω1 and ω2 . If ac = mω1 + nω2 for some m, n ∈ ℤ, then equation (6.73) becomes αc

αc a0 η + b0 + b1 e 3 󸀠 ℘ (az + b) = a0 η − b0 − b1 e 3 √3

(6.76)

αc

αc

since ℘(az + b) = ℘(az + b + ac). Hence a0 η + b0 + b1 e 3 = 0 and a0 η − b0 − b1 e 3 = 0, because ℘󸀠 (az +b) is a transcendental meromorphic function. We obtain a0 = 0, which is a contradiction. When ac ∈ Pa , we rewrite (6.73) in the form a0 η − b0 −

a0 η+b0 󸀠 ℘ (az √3

℘(az + b)

+ b)

αc

=

b1 e 3 (1 +

℘󸀠 (az+b+ac) ) √3

℘(az + b + ac)

.

(6.77)

It is obvious that the function on the left-hand side of (6.77) has a pole z = − ba , whereas the function on the right-hand side of (6.77) takes a finite value. This leads to a contradiction.

146 | 6 Fermat-type delay-differential equations Therefore we must have a1 ≠ 0. By our assumptions a2 = 0 and b1 a2 − b2 a1 = 0 we see that b2 = 0. Now using (6.53), we can rewrite equations (6.49) as follows: ℘󸀠 (h(z))

αz+β 1 1 + √3 e 3 , a0 f (z) + a1 f (z + c) = 2 ℘(h(z))

η1−

b0 f (z) + b1 f (z + c) =

℘󸀠 (h(z)) √3

2 ℘(h(z))

e

αz+β 3

(6.78)

.

Solving equations (6.78), it follows that f (z) =

b1 − a1 η +

(b1 +a1 η)℘󸀠 (h) √3

2(a0 b1 − a1 b0 )℘(h)

f (z + c) =

b0 − a0 η +

e

αz+β 3

(b0 +a0 η)℘󸀠 (h) √3

2(a1 b0 − a0 b1 )℘(h)

, (6.79) e

αz+β 3

.

This implies the equation b0 + a0 η 󸀠 ℘ (h)) ℘(h(z + c)) √3 b + a1 η 󸀠 ℘ (h(z + c))) ℘(h). (b1 − a1 η + 1 √3

(b0 − a0 η + αc

= −e 3

(6.80)

Differentiating this equation and then setting z = aj,k , we derive the relation b0 + a0 η 󸀠 ℘ (h(aj,k ))) ℘󸀠 (h(aj,k + c))h󸀠 (aj,k + c) √3 b + a1 η 󸀠 (b1 − a1 η + 1 ℘ (h(aj,k + c))) ℘󸀠 (h(aj,k ))h󸀠 (aj,k ). √3

(b0 − a0 η + αc

=−e3

Noting that ℘󸀠 (h(aj,k )) = ±i and ℘󸀠 (h(aj,k + c)) = ±i, this equation immediately implies one and only one of (6.74), where αc b0 + a0 η b + a1 η i, B1 = −e 3 (b1 − a1 η + 1 i) , √3 √3 αc b + a0 η b + a1 η A2 = A󸀠2 = −(b0 − a0 η − 0 i), B2 = e 3 (b1 − a1 η − 1 i) . √3 √3

A1 = A󸀠1 = b0 − a0 η +

Note that h is a nonconstant polynomial and {aj,k } is an infinite sequence. This immediately yields the corresponding functional equations (6.75). Therefore one of the four equalities A1 = B1 ,

A1 = B2 ,

A2 = B1 ,

holds by comparing the leading coefficients.

A2 = B2

(6.81)

6.3 Fermat difference equations |

147

We now consider the case A1 = B1 in (6.81) only, omitting the other cases due to the similarity of the arguments. Note that ℘󸀠 (h(aj,k )) = ±i. Without loss of generality, we may assume that ℘󸀠 (h(aj,k )) = i. Suppose that A1 = 0. Moreover, aj,k is also a zero of b0 − a0 η +

b0 + a0 η 󸀠 ℘ (h(z)). √3

Assume that aj,k is a zero of ℘(h(z)) with multiplicity l. It follows from (6.79) that aj,k may be a pole of f (z) and f (z + c) of multiplicity ≤ l − 1. However, it follows from (6.78) that aj,k is a pole of a0 f (z) + a1 f (z + c) of multiplicity l. This is a contradiction. Hence we must have A1 ≠ 0. Similarly, if ℘󸀠 (h(aj,k )) = −i, then we also obtain A2 ≠ 0. We also have B1 ≠ 0, B2 ≠ 0. Thus we obtain h󸀠 (z) = h󸀠 (z + c), again from (6.75), Hence h(z) = az + b, where a ≠ 0 and b are constants. Next, we derive a contradiction according to the arguments between (6.75) and (6.77). Therefore we now have that ℘(h(aj,k + c)) = 0 only for at most finitely many aj,k . To complete the proof for case 2, we may now proceed as in the proof of case 1. Case 3: b1 a2 − a1 b2 ≠ 0. From (6.49) and (6.53) we have (b1 a2 − a1 b2 )f 󸀠 (z) + (a0 b1 − a1 b0 )f (z) = (b1 F(z) − a1 G(z))e

αz+β 3

.

Solving this first-order linear differential equation, we get the solution f (z) = e

a b −a b

− b0 a1 −a1 b0 z 1 2

1 2

(∫

b1 − a1 η +

b1 +a1 η 󸀠 ℘ (h(z)) ( a0 b1 −a1 b0 + α )z+ β √3 b a −a b 3 3

2(b1 a2 − a1 b2 )℘(h(z))

e

1 2

1 2

dz + C) ,

(6.82)

where C is a constant. If b1 + a1 η ≠ 0, then letting tj (j ≥ 1) be the poles of ℘(z), we have ℘(z) = gj (z)(z − tj )−2 , where gj is a holomorphic function in a neighborhood of tj with gj (tj ) ≠ 0. Hence b1 − a1 η +

b1 +a1 η 󸀠 ℘ (z) √3

2(b1 a2 − a1 b2 )℘(z)

=−

(b1 − a1 η)(z − tj )2 b1 + a1 η 1 + √3(b1 a2 − a1 b2 ) z − tj 2(b1 a2 − a1 b2 )gj (z)

gj󸀠 (z) b1 + a 1 η 2√3(b1 a2 − a1 b2 ) gj (z) b1 + a1 η 1 + O(1) =− √3(b1 a2 − a1 b2 ) z − tj +

as z → tj . Setting z = h(z), we derive the relation b1 − a1 η +

b1 +a1 η 󸀠 ℘ (h(z)) √3

2(b1 a2 − a1 b2 )℘(h(z))

=−

b1 + a1 η 1 + O(1) √3(b1 a2 − a1 b2 ) h(z) − tj

(6.83)

as h(z) → tj . Note that the equation h󸀠 (z) = 0 has only finitely many solutions and −1 󸀠 −1 ⋃∞ j=1 h (tj ) is an infinite set. There exist an integer j and a point z ∈ h (tj ) such that

148 | 6 Fermat-type delay-differential equations h󸀠 (z 󸀠 ) ≠ 0, that is, z 󸀠 is a simple zero of h(z) − tj . It follows from (6.83) that f (z) has a logarithmic singular point z 󸀠 , a contradiction, since f (z) is a meromorphic function. If b1 + a1 η = 0, then we have b1 − a1 η ≠ 0. Indeed, if b1 − a1 η = 0, then b1 = 0 and a1 = 0, contradicting the assumption b1 a2 − a1 b2 ≠ 0. Now we can rewrite equation (6.82) in the form f (z) = e

a b −a b

− b0 a1 −a1 b0 z 1 2

1 2

(∫

a b −a b β b1 − a1 η ( 0 1 1 0 + α )z+ e b1 a2 −a1 b2 3 3 dz + C) . 2(b1 a2 − a1 b2 )℘(h(z))

Similarly as before, there exist an integer j and a point z 󸀠󸀠 ∈ h−1 (zj ) such that h󸀠 (z 󸀠󸀠 ) ≠ 0. Hence f (z) has a logarithmic singular point z 󸀠󸀠 , since ℘ has only simple zeros, again a contradiction. Hence F = c0 and G = c1 are constants such that c03 + c13 = 1. Thus (6.82) admits a solution f (z) = where D =

a1 b0 −a0 b1 b1 a2 −a1 b2

αz+β 3(b1 c0 − a1 c1 ) e 3 + CeDz , (b1 a2 − a1 b2 )(α − 3D)

is a constant. This is just case 3 in Theorem 6.3.22.

7 Delay-differential Riccati equations This chapter is devoted to delay-differential Riccati equations in the complex plane. As a background, we shortly recall some essential properties of differential and difference Riccati equations. For more details on differential Riccati equations, the reader may consult [6] and [120, Chapter 9 ].

7.1 Complex differential and difference Riccati equations The classical Malmquist theorem, originally published in [170], states that a differential equation of the form y󸀠 (z) = R(z, y(z)),

(7.1)

where R(z, y) is a rational function in z and y, which admits a transcendental meromorphic solution, reduces to a differential Riccati equation y󸀠 (z) = a0 (z) + a1 (z)y(z) + a2 (z)y(z)2 ,

(7.2)

where ai (z) (i = 0, 1, 2) are rational functions. If a2 ≠ 0, then the elementary transformation y=

a󸀠 a u − 1 − 22 a2 2a2 2a2

transforms (7.2) into the special case u󸀠 = A(z) + u2 ,

(7.3)

where A(z) = a0 a2 −

2

a21 a󸀠1 3 a󸀠2 a a󸀠 1 a󸀠󸀠 2 + − ( ) − 1 2 + . 4 2 4 a2 2 a2 2 a2

The difference counterpart of the Malmquist theorem was obtained by Yanagihara [230]. Suppose that a difference equation of the form f (z + 1) = R(z, f (z)) =

P(z, f (z)) Q(z, f (z))

(7.4)

with polynomial coefficients and P, Q being irreducible in f , possesses a transcendental meromorphic solution. Denoting q := degf (R(z, f )), Yanagihara [230] proved the following: https://doi.org/10.1515/9783110560565-007

150 | 7 Delay-differential Riccati equations Theorem 7.1.1. Let f be a transcendental meromorphic solution to equation (7.4). If q ≥ 2, then T(r, f ) ≥ Kqr , where K is a nonzero constant. It is not difficult to show that Theorem 7.1.1 remains valid if the difference equation (7.4) has meromorphic coefficients and the solution f is admissible. Therefore, if (7.4) admits an admissible meromorphic solution f of hyperorder ρ2 (f ) < 1, then q = degf (R(z, f )) ≤ 1. Because of this result, equations of the form f (z + 1) =

a(z) + b(z)f (z) , c(z) + d(z)f (z)

(7.5)

where a(z), b(z), c(z), d(z) are small functions with respect to f (z), and a(z)d(z) − b(z)c(z) ≢ 0, are called difference Riccati equations. Note the following particular cases of equation (7.5): If d(z) ≡ 0, then (7.5) is a linear difference equation. This equation is integrable; see [219], also see Chapter 5. If d(z) ≢ 0, then (7.5) can be written as f (z + 1) =

a(z) + b(z)f (z) , c(z) − f (z)

a(z) + b(z)c(z) ≢ 0,

(7.6)

without restricting generality. If now b(z) ≡ −c(z + 1), then a(z) + b(z)c(z) = a(z) − c(z + 1)c(z) ≢ 0. Setting 1 f (z) = g(z) + c(z) in (7.6) and redenoting g as f again, we obtain f (z + 1) = where α(z) =

α(z) , f (z)

1 . c(z)c(z+1)−a(z)

If b(z) ≢ −c(z + 1), we may use the transformation f (z) = Again redenoting g as f , (7.6) changes to f (z + 1) =

A(z) + f (z) , 1 − f (z)

c(z)+b(z−1) g(z) + c(z)−b(z−1) . 2 2

A(z) ≢ −1,

(7.7)

where A(z) =

4a(z) − b(z)b(z − 1) + 3b(z)c(z) − b(z − 1)c(z + 1) − c(z)c(z + 1) . (b(z) + c(z + 1))(b(z − 1) + c(z))

Note that equation (7.7) may also be written as △ f (z) =

A(z) + f (z)2 , 1 − f (z)

A(z) ≢ −1,

(7.8)

7.1 Complex differential and difference Riccati equations | 151

where △f = f (z + 1) − f (z). The passage from (7.7) or (7.8) to a linear difference equation of second order Δ2 y(z) + A(z)y(z) = 0

(7.9)

was considered by Ishizaki [107]. For a nontrivial meromorphic solution y(z) of (7.9), we set f (z) := −

Δy(z) . y(z)

(7.10)

Then Δ2 y(z) = −(Δf (z))y(z) − f (z + 1)Δy(z) = −(Δf (z))y(z) + f (z + 1)f (z)y(z).

(7.11)

Combining (7.11) and (7.9), we have − (Δf (z))y(z) + f (z + 1)f (z)y(z) + A(z)y(z) = −(f (z + 1) − f (z))y(z) + f (z + 1)f (z)y(z) + A(z)y(z) = 0. This implies that (−1 + f (z))f (z + 1) + f (z) + A(z) = 0, and (7.7) or (7.8) follows. Conversely, if (7.8) has a meromorphic solution f (z), then a meromorphic solution y(z) of first-order difference equation (7.10) satisfies Δ2 y(z) = (−f (z + 1) + f (z) + f (z)f (z + 1))y(z)

−A(z) − f (z) f (z) − f (z)2 A(z)f (z) + f (z)2 + + ) y(z) 1 − f (z) 1 − f (z) 1 − f (z) A(z)(1 − f (z)) y(z) = −A(z)y(z). =− 1 − f (z) =(

Recalling that whenever f1 (z), f2 (z), f3 (z) are distinct meromorphic solutions of (7.3), then (7.3) possesses a one-parameter family of meromorphic solutions (fc )c∈C , see [6], p. 371–373. Difference analogues of this property are due to Ishizaki [107] and Chen and Shon [32]: Theorem 7.1.2 ([107, Proposition 2.1]). Suppose that (7.8) possesses three distinct meromorphic solutions f1 , f2 , f3 , and A(z) is a meromorphic function. Then all meromorphic solutions f of (7.8) can be represented as f =

f1 f2 − f2 f3 − f1 f2 Q + f1 f3 Q , f1 − f3 − f2 Q + f3 Q

(7.12)

where Q is a periodic function of period 1. Conversely, if Q is a periodic function of period 1, then f defined by (7.12) is a meromorphic solution to (7.8).

152 | 7 Delay-differential Riccati equations Theorem 7.1.3 ([32, Theorem 1.1]). Suppose that (7.5) possesses three distinct meromorphic solutions f0 , f1 , f2 . Then all meromorphic solutions f of (7.5) constitute a oneparameter family H(f (z)) = {f0 , f =

(f1 − f0 )(f2 − f0 ) + f }, Q(f2 − f1 ) + (f2 − f0 ) 0

(7.13)

where Q is a periodic function of period 1. If Q = 0, then f = f1 , and if Q = −1, then f = f2 . Remark 7.1.4. (1) If f0 , f1 , f2 are three distinct rational functions, then (7.5), resp., (7.8), admits infinitely many rational solutions by choosing different constants Q; see (7.12), resp., (7.13). (2) Equation (7.5) may admit a meromorphic solution f of any order ρ(f ) ≥ 1 by taking a suitable transcendental periodic function Q. This can be proved by (7.13) but not by (7.12). Recall the following observations on differential Riccati equations: If (7.3) has two distinct rational solutions f1 (z) and f2 (z), then it may happen that f1 , f2 are the only meromorphic solutions; see [6, p. 396]. If f3 is the third meromorphic solution that is rational, then all meromorphic solutions (7.3) are rational, whereas if f3 (z) is transcendental, then there are no other rational solutions than f1 (z) and f2 (z); see [6, p. 393–394]. For difference counterparts to these observations, we refer to Ishizaki [107, Proposition 2.2] and Chen and Shon [32, Theorem 1.2]: Theorem 7.1.5. Suppose that (7.8) possesses two distinct rational solutions a1 , a2 and A(z) is a meromorphic function. Then there exists a meromorphic solution a3 , distinct from a1 , a2 , such that any meromorphic solution f (z) of (7.8) can be represented in the form (7.12). Theorem 7.1.6. Let a(z), b(z), c(z), d(z) be polynomials, ac ≢ 0, ad − bc ≢ 0, and deg(c) > max{deg(a), deg(b)}. If the difference equation (7.5) has a rational solution f0 such that f0 ↛ 0 as z → ∞, then (7.5) has at most two distinct rational solutions. Theorem 7.1.6 and Remark 7.1.4(1) imply that (7.5) either admits at most two distinct rational solutions or admits infinitely rational solutions. Chen and Shon [24, Theorem 1] also considered the existence of rational solutions in the case of rational A(z) in (7.7). Theorem 7.1.7. Let A(z) = m(z) be an irreducible rational function, where m(z) and n(z) n(z) are polynomials with deg(m(z)) = m and deg(n(z)) = n. (1) Suppose that m ≥ n and m − n is an even number or zero. If (7.7) has an irreducible P(z) rational solution f (z) = Q(z) , where P(z) and Q(z) are polynomials with deg(P(z)) = p and deg(Q(z)) = q, then p − q = m−n . 2 (2) Suppose that m < n and n−m = k ≥ 2, where k is an integer. If (7.7) has an irreducible P(z) rational solution f (z) = Q(z) , then q − p = k − 1 or q − p = 1.

7.1 Complex differential and difference Riccati equations | 153

(3) Suppose that either m > n and m − n is an odd number or m < n and n − m = 1. Then (7.7) has no rational solutions. Suppose that (7.7) has a transcendental meromorphic solution f of finite order. Ishizaki [107, Theorem 3.1] showed that ρ(f ) ≥ 21 . Chen and Shon [31, Theorem 1.2] improved Ishizaki’s result and considered the value distribution of transcendental meromorphic solutions. More related results can also be found in [111, 112, 181, 246]. Theorem 7.1.8. Suppose that A(z) is a rational function in (7.7) and suppose that (7.7) possesses a rational solution f0 . Then every transcendental meromorphic solution f of finite order satisfies (i) λ(f ) = λ( f1 ) = ρ(f ) ≥ 1;

(ii) If A(z) = a(z)2 , where a(z) is a nonconstant rational function, then λ(Δf ) = λ( Δf1 ) = ρ(f ) ≥ 1, and λ( Δff ) = λ( Δff ) = ρ(f ) ≥ 1.

Remark 7.1.9. It remains open whether the condition that (7.7) possesses a rational solution may be removed in Theorem 7.1.8. We next want to point out a similarity, resp., nonsimilarity, between differential equation (7.3) and difference equation (7.7). First, assume that w(z) is a transcendental solution to (7.3), resp., to (7.7), whereas α(z) is a rational function that is not a solution to (7.3), resp., to (7.7). Then, in both cases, w − α has infinitely many zeros by Lemma 1.1.24 and its difference version. To point out a corresponding nonsimilarity, suppose that α is a rational solution to (7.3). Then ν := w − α satisfies ν󸀠 = ν(ν + 2α). This readily implies that ν = w − α has finitely many zeros only by the rationality of α. However, this property does not hold for (7.7). Indeed, there may exist two transcendental meromorphic solutions f1 , f2 to (7.7) such that f1 − α(z) admits infinitely many zeros, whereas f2 − α(z) admits finitely many zeros only, as shown by the following example due to Ishizaki [107, Example 3.1]. Example 7.1.10. Consider the difference Riccati equation △f (z) =

A(z) + f (z)2 , 1 − f (z)

A(z) = −

(z 2

z 4 − z 2 + 2z + 3 , + z − 1)(z 2 + 3z + 1)

which possesses the rational solution α(z) =

z2 − z + 1 . z2 + z − 1 2

1 α(z−1)−1 +z−1 Set f (z) = α(z) + β(z)k(z) with β(z) = A(z−1)+1 = − z 2z . Then k(z) satisfies the nonho2 mogeneous first-order difference equation

k(z + 1) =

z−1 k(z) + 1 z2

(7.14)

154 | 7 Delay-differential Riccati equations with the associated homogenous difference equation kH (z + 1) = Obviously, k0 (z) = fore be written as

1 Γ(z)(z−1)

z−1 k (z). z2 H

(7.15)

solves (7.15). The general solution k(z) of (7.14) may therek(z) = Q(z)k0 (z) +

z , z−1

1 where Q(z) is a periodic function with period 1. Since Γ(z) is an entire function, choosing an entire periodic function Q(z), f (z) − α(z) has finitely many zeros only, whereas when Q(z) is a meromorphic function with infinitely many poles, then f (z) − α(z) has infinitely many zeros.

As a completion to this example, we recall the following more complicated result due to Ishizaki [108, Theorem 5.1]. Theorem 7.1.11. Let A(z) be a rational function, and let f be a transcendental meromorphic solution of (7.7). Then either f has infinitely many simple poles, or f (z) is a periodic function of period 3 or 4.

7.2 Complex delay-differential Riccati equations Replacing the left-hand side in (7.1) and (7.4) with the shift of derivative, we consider f 󸀠 (z + c) = R(z, f ) =

P(z, f ) , Q(z, f )

(7.16)

where P(z, f ) and Q(z, f ) are polynomials irreducible in f and with rational coefficients. Of course, we may assume that the coefficients are meromorphic functions, letting f be admissible meromorphic solutions to (7.16). Looking at a slightly more general case, we easily obtain a delay-differential counterpart to the classical Malmquist theorem: Theorem 7.2.1. If (f 󸀠 (z + c))n =

a0 (z) + a1 (z)f (z) + ⋅ ⋅ ⋅ + ap (z)f (z)p b0 (z) + b1 (z)f (z) + ⋅ ⋅ ⋅ + bq (z)f (z)q

=:

P(z, f ) =: R(z, f ) Q(z, f )

(7.17)

admits an admissible transcendental meromorphic solution of hyperorder ρ2 (f ) < 1, then the degrees of P and Q relative to f satisfy max{p, q} ≤ 2n. Proof. Since f (z) is a transcendental meromorphic function, T(r, R(z, f )) = max{p, q}T(r, f ) + S(r, f )

7.2 Complex delay-differential Riccati equations |

155

by the Valiron–Mohon’ko theorem; see Theorem 1.1.29. Therefore, using Corollary 1.1.9 and Lemma 1.2.10, we obtain max{p, q}T(r, f ) + S(r, f ) = T(r, R(z, f ))

= T(r, (f 󸀠 (z + c))n ) = nT(r, f 󸀠 (z + c))

≤ n(T(r, f (z + c)) + N(r, f (z + c)) + S(r, f (z + c)) ≤ 2nT(r, f ) + S(r, f ) outside a possible exceptional set of finite logarithmic measure. To avoid a contradiction, we get max{p, q} ≤ 2n. Considering possible entire solutions to (7.17), we obtain the following: Theorem 7.2.2. If (7.17) admits an admissible transcendental entire solution of finite order, then q = 0 and p = n. Proof. Suppose that f is an entire solution. By the proof of Theorem 7.2.1, max{p, q} ≤ n. Since f cannot have two Picard values, we easily to conclude that q ≤ 1. We consider two cases. Case 1: If q = 1, then b0 (z) + b1 (z)f (z) has no zeros. Hence f (z) can be written as eh(z) −b (z)

0 f (z) = b1 (z) changes into

=: a(z)eh(z) + b(z), where h(z) is a nonzero polynomial. Thus (7.17) n

([a󸀠 (z + c) + a(z + c)h󸀠 (z + c)]eh(z+c) + b󸀠 (z + c)) =

a0 (z) + ⋅ ⋅ ⋅ + ap (z)(a(z)eh(z) + b(z))p eh(z)

.

Thus Tn (z)enh(z+c)+h(z) + ⋅ ⋅ ⋅ + T1 (z)eh(z+c)+h(z) + Sp (z)eph(z) + ⋅ ⋅ ⋅ + S1 (z)eh(z) = H(z) for suitable small coefficients. Since p ≤ n, by standard application of Theorem 1.1.34 we can avoid a contradiction only by Tn (z) = 0. Thus we have a(z) = e−h(z) , so f (z) does not admit transcendental entire solutions. Case 2: If q = 0, then p ≤ n. Assume that p < n. Then (7.17) reduces to [f 󸀠 (z + c)]n = a0 (z) + a1 (z)f (z) + ⋅ ⋅ ⋅ + an−1 (z)f (z)n−1 .

(7.18)

Writing this in the form f (z)n (

n

f 󸀠 (z + c) ) = a0 (z) + a1 (z)f (z) + ⋅ ⋅ ⋅ + an−1 (z)f (z)n−1 , f (z)

(7.19)

by Lemma 1.2.8 and the Clunie lemma [120, Lemma 2.4.2] we have that m(r, f ) = S(r, f ), a contradiction. Thus p = n.

156 | 7 Delay-differential Riccati equations By the two results above we may call this type of reduced form of equation (7.16) delay-differential-type Riccati equations. In what follows, we may assume that f (z) is not a periodic function with period c. Indeed, otherwise, we would return to classical Riccati differential equations. The following four examples show that Riccati-type delay-differential equations can be more complicated than in the situation of the classical differential equations. Example 7.2.3. The meromorphic function f (z) = cot z solves the equation f 󸀠 (z +

−f (z)2 − 1 π )= . 2 f (z)2

(7.20)

Example 7.2.4. The meromorphic function f (z) = cot z − 1 solves the equation f 󸀠 (z +

π −f (z)2 − 2f (z) + 2 . )= 2 [f (z) + 1]2

Example 7.2.5. The meromorphic function f (z) = f 󸀠 (z + c) =

ez ez −1

(7.21)

solves the equation

f (z)2 − f (z) , [2f (z) − 1]2

(7.22)

where ec = −1. The following example, due to Ishizaki [108, Example 1.1], presents a delaydifferential equation where the power on the left-hand side is > 1, see (7.23): Example 7.2.6. Let ℘(z) be the Weierstrass elliptic function constructed by the lattice {2n + mω | n, m ∈ ℤ}, where ω is a nonreal complex number. Write ℘(1) = e1 , ℘( ω2 ) = e2 , ℘(1 + ω2 ) = e3 . Then ℘(z) satisfies the equation (℘󸀠 (z))2 = 4(℘(z) − e1 )(℘(z) − e2 )(℘(z) − e3 ). Thus all zeros of ℘(z) − e1 , ℘(z) − e2 , ℘(z) − e3 are not simple. By the properties of ℘(z), since e1 + e2 + e3 = 0 and ℘󸀠 (1) = 0, we have ℘(z + 1) =

−(e2 e3 + e12 ) − e1 ℘(z) . e1 − ℘(z)

Taking the first derivative of this equation, we obtain ℘󸀠 (z + 1) =

−(e2 e3 + 2e12 )℘󸀠 (z) , [e1 − ℘(z)]2

and thus [℘󸀠 (z + 1)]2 =

(e2 e3 + 2e12 )2 (℘󸀠 (z))2 [e1 − ℘(z)]4

7.2 Complex delay-differential Riccati equations |

= =

157

(e2 e3 + 2e12 )2 4(℘(z) − e1 )(℘(z) − e2 )(℘(z) − e3 ) [e1 − ℘(z)]4 4(e2 e3 + 2e12 )2 (℘(z) − e2 )(℘(z) − e3 ) . [℘(z) − e1 ]3

(7.23)

Example 7.2.7. Note that Theorem 7.2.1 does not hold in general for meromorphic solutions of infinite order, even for solutions of hyperorder = 1. For example, f (z) := ez1 e −1 such that ec = 2 satisfies f 󸀠 (z + c) =

−2ez (f (z) − 1)2 . f (z)2 (f (z) + 2)2

(7.24)

It is not surprising to expect that results similar to Theorem 7.2.1 may be obtained to equations slightly more general than equation (7.16). As an example, define the following delay-differential polynomials with meromorphic coefficients small with respect to f : n

Ω1 (z, f ) := ∑ ai (z)(f 󸀠 (z + c1 ))li,1 ⋅ ⋅ ⋅ (f 󸀠 (z + cn ))li,n i=1

and n

Ω2 (z, f ) := ∑ bj (z)(f 󸀠 (z + c1 ))mj,1 ⋅ ⋅ ⋅ (f 󸀠 (z + cn ))mj,n . j=1

We now immediately get the following result [214, Theorem 2.1]. Theorem 7.2.8. Let P(z, f ) and Q(z, f ) be relatively prime polynomials in f with small meromorphic coefficients. If a delay-differential equation of type Ω1 (z, f ) P(z, f ) = R(z, f ) := Ω2 (z, f ) Q(z, f ) admits an admissible meromorphic solution f of hyperorder ς = ρ2 (f ) < 1, then max{p, q} ≤ 2λ, where λ := ∑nk=1 λk := ∑nk=1 maxi,j {li,k , mj,k }. Proof. As in the proof of Theorem 7.2.1, we readily obtain max{p, q}T(r, f ) + S(r, f ) = T(r, R(z, f )) = T (r, n

Ω1 (z, f ) ) Ω2 (z, f )

≤ ∑ λk T(r, f 󸀠 (z + ck )) + S(r, f ) k=1 n

≤ ∑ 2λk T(r, f (z + ck )) + S(r, f ) k=1 n

≤ ∑ 2λk T(r, f (z)) + S(r, f ) k=1

158 | 7 Delay-differential Riccati equations ≤ (2λ + o(1))T(r, f ) + S(r, f ), again outside a possible exceptional set of finite logarithmic measure. The statement follows. Using the same notations as in the preceding theorem, we present another additional result; see [214], Theorem 2.3. Theorem 7.2.9. Suppose f is a transcendental meromorphic solution to Ω1 (z, f ) = f (p(z)), Ω2 (z, f ) where p is a polynomial of degree d ≥ 2. Assuming, moreover, that 2λ ≥ d, the order of f satisfies T(r, f ) = O(logα+ε r), where α := log 2λ/ log d, and ε is arbitrarily small. Proof. Recalling [62, p. 363], denoting μ := |ad | − ε for p(z) = ad z d + ⋅ ⋅ ⋅, and using the same reasoning as in the proof of Theorem 7.2.8, we obtain (1 − ε) T (μr d , f ) ≤ T(r, f (p(z))) = T (r, n

Ω1 (z, f ) ) Ω2 (z, f ) n

≤ ∑ λk T(r, f 󸀠 (z + ck )) + S(r, f ) ≤ ∑ 2λk T(r, f (z + ck )) + S(r, f ) k=1 n

k=1

≤ ∑ 2λk T(r + c, f (z)) + S(r, f ) = 2λT(r + c, f (z)) + S(r, f ), k=1

where c := max{|c1 |, . . . , |cn |}. Of course, for each β > 1 and r large enough, we have T(r + c, f ) ≤ T(βr, f ), and further, outside a possible exceptional set of finite linear measure, (1 − ε)T(μr d , f ) ≤ 2λ(1 + ε)T(βr, f ). We may eliminate the exceptional set by a standard argument by fixing γ > 1 and taking r large enough to obtain (1 − ε)T(μr d , f ) ≤ 2λ(1 + ε)T(βγr, f ). Denoting now t := βγr, we may write T(

μ d 2λ(1 + ε) t ,f) ≤ T(r, f ). d 1−ε (βγ)

Recalling [63, Lemma 3], we now have T(r, f ) = O(logs r), where s := log 2λ(1+ε) / log d = 1−ε log 2λ/ log d + o(1).

7.2 Complex delay-differential Riccati equations | 159

Looking next back at Examples 7.2.3, 7.2.4, and 7.2.5, observe that the denominators on the right-hand side of these equations are squares of the expression f (z) − b(z). Wang [214, Theorem 2.2] considered the poles of f (z + c) and f (z) for more general delay-differential equations. For this result, take distinct non-zero complex numbers c1 , c2 , . . . , cn and look at the delay-differential polynomials Ω1 (z, f ) and Ω2 (z, f ) as defined for Theorem 7.2.8. We now prove the following: Theorem 7.2.10. Suppose that f (z) is a transcendental meromorphic solution to p Ω1 (z, f ) P(z, f ) a0 (z) + a1 (z)f (z) + ⋅ ⋅ ⋅ + ap (z)f (z) , = R(z, f ) = = Ω2 (z, f ) Q(z, f ) b0 (z) + b1 (z)f (z) + ⋅ ⋅ ⋅ + bq (z)f (z)q

(7.25)

where P(z, f ) and Q(z, f ) are relatively prime polynomials in f over the field of small meromorphic functions relative to f . Also, assume that Q(z, f ) is a monic polynomial. Moreover, suppose that q > 0 and max{p, q} = 2λ, where λ := ∑nk=1 λk and λk = maxi∈I,j∈J {li,k , mj,k }. If there exists α ∈ [0, 2λ) such that for all r sufficiently large, N (r,

Ω1 (z, f ) ) ≤ αN(r + c, f ) + S(r, f ) Ω2 (z, f )

(7.26)

and n

∑ 2λk N(r, f (z + ck )) ≤ αN(r + c, f ) + S(r, f ),

k=1

(7.27)

where c = max{|c1 |, |c2 |, . . . , |cn |}, then either ρ(f ) = ∞, or Q(z, f (z)) = (f (z) + h(z))q , where h is a small meromorphic function relative to f . Proof. Suppose that f (z) is a transcendental meromorphic solution of (7.25) and the second alternative of the conclusion is not true. Combining (7.25) and (7.26) with the Tumura–Clunie theorem in [217] and Lemma 2.3 in [121], we obtain 1 ) + N(r, f ) + S(r, f ) Q(z, f ) P(z, f ) ≤ N (r, ) + N(r, f ) + S(r, f ) Q(z, f ) Ω (z, f ) ≤ N (r, 1 ) + N(r, f ) + S(r, f ) Ω2 (z, f )

T(r, f ) ≤ N (r,

≤ αN(r + c, f ) + N(r, f ) + S(r, f ),

160 | 7 Delay-differential Riccati equations where c = max{|c1 |, |c2 |, . . . , |cn |}. Therefore we have T(r, f ) − N(r, f ) ≤ αN(r + c, f ) + S(r, f ),

(7.28)

where α ∈ [0, 2λ). Suppose, moreover, that ρ(f ) < ∞. Then S(r, f (z + ck )) = S(r, f ),

k = 1, 2, . . . , n.

From (7.28) we have T(r, f (z + ck )) − N(r, f (z + ck )) ≤ αN(r + c, f (z + ck )) + S(r, f ). Then max{p, q}T(r, f ) = T (r, n

Ω1 (z, f ) ) + S(r, f ) Ω2 (z, f ) n

≤ ∑ λk T(r, f 󸀠 (z + ck )) + S(r, f ) ≤ ∑ 2λk T(r, f (z + ck )) + S(r, f ) k=1 n

k=1

n

≤ ∑ 2λk [T(r, f (z + ck )) − N(r, f (z + ck ))] + ∑ 2λk N(r, f (z + ck )) + S(r, f ) k=1 n

k=1

≤ ∑ 2λk [αN(r + c, f (z + ck ))] + αN(r + c, f (z)) + S(r, f ) k=1 n

≤ ∑ 2λk αN(r + 2c, f ) + αN(r + 2c, f ) + S(r, f ) k=1

= 2λαN(r + 2c, f ) + αN(r + 2c, f ) + S(r, f ) = 2λαN(r + 2c, f ) + αN(r, f ) + S(r, f ). Because max{p, q} = 2λ, we have T(r, f ) ≤ αN(r + 2c, f ) +

α N(r, f ) + S(r, f ). 2λ

Therefore T(r, f ) − N(r, f ) ≤ αN(r + 2c, f ) +

α N(r, f ) − N(r, f ) + S(r, f ). 2λ

(7.29)

We next prove the following inequality by induction: T(r, f ) − N(r, f ) ≤ αN(r + 2mc, f ) +

mα N(r, f ) − mN(r, f ) + S(r, f ). 2λ

(7.30)

The case m = 1 being already verified, we may assume that this inequality holds for m = t: max{p, q}T(r, f ) n

≤ ∑ 2λk [T(r, f (z + ck )) − N(r, f (z + ck ))] + αN(r + c, f ) + S(r, f ) k=1

7.2 Complex delay-differential Riccati equations | 161 n

≤ ∑ 2λk [αN(r + 2tc, f (z + ck )) + k=1

+ αN(r + c, f ) + S(r, f ) n

tα N(r + c, f ) − tN(r − c, f )] 2λ

≤ ∑ 2λk [αN(r + 2tc + c, f ) + k=1

+ αN(r, f ) + S(r, f )

≤ 2λ[αN(r + (2t + 1)c, f ) +

tα N(r, f (z + ck )) − tN(r, f (z + ck ))] 2λ

tα N(r, f ) − tN(r − c, f )] + αN(r, f ) + S(r, f ). 2λ

Thus we have T(r, f ) ≤ αN(r + (2t + 1)c, f ) +

tα α N(r, f ) − tN(r − c, f ) + N(r, f ) + S(r, f ). 2λ 2λ

By Lemma 1.2.11 we obtain T(r, f ) ≤ αN(r + (2t + 2)c, f ) +

tα α N(r, f ) − tN(r, f ) + N(r, f ) + S(r, f ). 2λ 2λ

Further, we obtain T(r, f ) − N(r, f ) ≤ αN(r + (2t + 2)c, f ) +

(t + 1)α N(r, f ) − (t + 1)N(r, f ) + S(r, f ). 2λ

This inequality shows that inequality (7.30) holds for m = t + 1, completing induction. Recalling again Lemma 1.2.11 and combining with (7.30), we obtain α α N(r + 2mc, f ) + N(r, f ) + S(r, f ) m 2λ 1 1 = α ( + ) N(r, f ) + S(r, f ). m 2λ

N(r, f ) ≤

(7.31)

Noting that α ∈ [0, 2λ), let m be large enough such that α(

1 1 + ) < 1. m 2λ

It follows from (7.31) and (7.32) that (1 − α (

1 1 + )) N(r, f ) ≤ S(r, f ). m 2λ

This implies that N(r, f ) = S(r, f ). By (7.29), using once more Lemma 1.2.11, we have T(r, f ) = S(r, f ), a contradiction. Therefore ρ(f ) = ∞, completing the proof.

(7.32)

162 | 7 Delay-differential Riccati equations The following result is now a direct consequence of Theorem 7.2.10. Corollary 7.2.11. Suppose that f (z) is a transcendental meromorphic solution to f 󸀠 (z + c) =

a0 (z)(f (z) − a1 (z))(f (z) − a2 (z)) (f (z) − b1 (z))(f (z) − b2 (z))

(7.33)

with rational coefficients. If there exists α ∈ [0, 1) such that for all r sufficiently large, N(r, f (z + c)) ≤ αN(r + c, f (z)) + S(r, f ), then either ρ(f ) = ∞ or b1 (z) = b2 (z). However, we have not found a concrete example satisfying condition (7.34).

(7.34)

8 Malmquist-type delay-differential equations 8.1 Malmquist-type difference equations In Section 7.2, we already recalled the Malmquist theorem to introduce Riccati equations in the differential setting. For more detailed theorems of Malmquist–Yosida– Steinmetz type, see [203], [192], and [120, Chapter 10]. Proceeding to Malmquist-type theorems in the difference setting, we first recall Shimomura [199, Theorem 2.5], who proved that first-order difference equations of the form f (z + 1) = P(f (z)) have nontrivial entire solutions if P(f (z)) is a nonconstant polynomial in f (z). Yanagihara [230, Corollary 6] also proved that the first-order difference equations f (z + 1) = R(f (z)) have nontrivial meromorphic solutions if R(f (z)) is a nonconstant rational function of f (z). The following result due to Yanagihara [230, Theorem 1’] can now be understood as a basic difference counterpart to the Malmquist theorem: Theorem 8.1.1. If the first-order difference equation f (z + 1) = R(z, f ),

(8.1)

where R(z, f ) is rational in both arguments, admits a transcendental meromorphic solution of finite order, then degf (R) = 1. Yanagihara’s work prompted a number of possible generalizations to secondorder difference equations of the forms f (z + 1) + f (z − 1) = R(z, f )

(8.2)

f (z + 1)f (z − 1) = R(z, f ),

(8.3)

and

where R(z, f ) is rational in both arguments. We first recall a result of Ablowitz, Halburd, and Herbst [1, Theorems 3 and 5] who obtained the following: Theorem 8.1.2. If the second-order difference equation (8.2) or (8.3) admits a transcendental meromorphic solution of finite order, then degf (R(z, f )) ≤ 2. The class of equations of the form (8.2) with degf (R(z, f )) ≤ 2 contains equations that are Painlevé difference equations. Recall that continuous limits of these https://doi.org/10.1515/9783110560565-008

164 | 8 Malmquist-type delay-differential equations equations reduce to Painlevé differential equations; for see more details, see [85]. Halburd and Korhonen [84, Theorem 1.1] obtained a list of possible equations, provided that (8.2) admits an admissible meromorphic solution of finite order. In this list, difference Painlevé I and Painlevé II equations, difference Riccati equations, and some linear difference equations appear. In addition, difference Painlevé III equations appear by classifying (8.3); see [85]. It was also shown in [87] and [195] that more or less the same lists result under the weaker assumption of a meromorphic solution of hyperorder less than one. Hence the existence of sufficiently many meromorphic solutions of finite-order or of hyperorder less than one can be considered as a version of the Painlevé property. For higher-order difference equations of type ⋁ f (z + ci ) = R(z, f ),

(8.4)

where the symbol ⋁ stands for either ∑ni=1 or ∏ni=1 , Heittokangas et al. [95, Propositions 8 and 9] obtained the following: Theorem 8.1.3. If the higher-order difference equation (8.4) admits a transcendental meromorphic solution of finite order, then degf (R(z, f )) ≤ n. More generally, Theorems 8.1.1 to 8.1.3 remain valid if R(z, f ) is with small coefficients (in the usual Nevanlinna theory sense) and the conditions of finite order is being replaced by ρ2 (f ) < 1. For more results on complex difference equations, we refer the readers to Chen [29, Chapters 7 and 9].

8.2 Malmquist-type delay-differential equations Proceeding to Malmquist-type delay-differential equations, we first give a couple of simple results for equations, see [201], that are of first order in the difference sense. Theorem 8.2.1. Delay-differential equations of type c(z)f (z + 1) + a(z)

f 󸀠 (z) = b(z), f (z)

(8.5)

where a, b, c are rational functions, admit no transcendental entire solutions. If f (z) is a transcendental meromorphic solution of hyperorder ρ2 (f ) < 1 to equation (8.5), then λ( f1 ) = ρ(f ). Proof. Suppose first that f (z) is a transcendental entire solution to (8.5). If f (z0 ) = 0, whereas the coefficients a, b, c have neither a zero nor a pole at z0 , then f (z) must have a pole at z0 + 1, a contradiction. Therefore f (z) has at most finitely many zeros, and we may write, by the Hadamard factorization theorem, f (z) = P(z)eg(z) , where P(z) is

8.2 Malmquist-type delay-differential equations | 165

a nonvanishing polynomial, and g(z) is entire. Substituting f (z) into (8.5), we get eg(z+1) =

P 󸀠 (z) 1 (b(z) − a(z) ( + g 󸀠 (z))) . c(z)P(z + 1) P(z)

Now by comparing the characteristic functions of above equation we get T(r, eg(z+1) ) = T(r, g(z)) + S(r, g(z)), which is impossible. To prove the second claim, we write (8.5) in the form f (z) = (b(z) − a(z)

f 󸀠 (z) f (z + 1) ) . f (z) c(z)f (z)

Now by using the standard logarithmic derivative lemma and its difference version, see Proposition 1.1.8 and (1.24), we conclude that m(r, f ) = S(r, f ), and hence λ( f1 ) = ρ(f ). The subsequent lemma is a delay-differential version of the Mohon’ko theorem for meromorphic solutions f of hyperorder strictly less than one to P(z, f ) = 0. Recall that a delay-differential polynomial in f (z) is of the form P(z, f ) = ∑ bl (z)f (z)l0,0 f (z + c1 )l1,0 ⋅ ⋅ ⋅ f (z + cν )lν,0 f 󸀠 (z)l0,1 ⋅ ⋅ ⋅ f (μ) (z + cν )lν,μ , l∈L

where c1 , . . . , cν are distinct complex constants, L is a finite index set consisting of elements of the form l = (l0,0 , . . . , lν,μ ), and the coefficients bl (z) are rational functions of z for all l ∈ L. Lemma 8.2.2. Let f (z) be a nonrational meromorphic solution of hyperorder ρ2 (f ) < 1 to P(z, f ) = 0,

(8.6)

let P(z, f ) be a delay-differential polynomial in f (z) with rational coefficients, and let a be a rational function satisfying P(z, a) ≢ 0. Then m(

1 ) = S(r, f ) f −a

(8.7)

for all r outside a possible exceptional set of finite logarithmic measure. Proof. For this proof, see the proof of [88, Lemma 2.1]. Substituting f (z) = g(z) + a into (8.6), it follows that Q(z, g) + R(z) = 0,

(8.8)

where R(z) ≢ 0 is a rational function, since P(z, a) ≢ 0. Moreover, Q(z, g) = ∑ bl (z)Gl (z, g) l∈L

(8.9)

166 | 8 Malmquist-type delay-differential equations is a delay-differential polynomial in g such that for all l ∈ L, Gl (z, g) is a nonconstant product of derivatives and shifts of g(z). The coefficients bl (z) in (8.9) are all rational. Now, defining E1 := {θ ∈ [0, 2π) : |g(reiθ )| ≤ 1} and E2 := [0, 2π)\E1 , we have m(

󵄨󵄨 1 󵄨󵄨 dθ 1 1 󵄨 󵄨󵄨 ) = m (r, ) = ∫ log+ 󵄨󵄨󵄨 󵄨 . 󵄨󵄨 g(reiθ ) 󵄨󵄨󵄨 2π f −a g

(8.10)

θ∈E1

Moreover, for all z = reiθ such that θ ∈ E1 , 󵄨󵄨 ∑ b (z)G (z, g) 󵄨󵄨 󵄨󵄨 l∈L l 󵄨󵄨 l 󵄨󵄨 󵄨󵄨 g 󵄨󵄨 󵄨󵄨 󵄨󵄨 ∑ b (z)g(z)l0,0 g(z + c )l1,0 ⋅ ⋅ ⋅ g(z + c )lν,0 g 󸀠 (z)l0,1 ⋅ ⋅ ⋅ g (μ) (z + c )lν,μ 󵄨󵄨 󵄨󵄨 󵄨 1 ν ν = 󵄨󵄨󵄨 l∈L l 󵄨󵄨 󵄨󵄨 󵄨󵄨 g(z) l l l l (μ) 1,0 ν,0 0,1 ν,μ 󸀠 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 g(z + c1 ) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ⋅ ⋅ ⋅ 󵄨󵄨󵄨 g(z + cν ) 󵄨󵄨󵄨 󵄨󵄨󵄨 g (z) 󵄨󵄨󵄨 ⋅ ⋅ ⋅ 󵄨󵄨󵄨 g (z + cν ) 󵄨󵄨󵄨 , ≤ ∑ bl (z)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 g(z) 󵄨󵄨󵄨 󵄨󵄨󵄨 g(z) 󵄨󵄨󵄨 󵄨󵄨󵄨 g(z) 󵄨󵄨󵄨 g(z) 󵄨 󵄨 l∈L since degg (Gl ) ≥ 1 for all l ∈ L with l = (l0,0 , . . . , lν,μ ). By (8.8) and (8.9) we see that 󵄨 󵄨 󵄨 󵄨 󵄨󵄨 1 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨 󵄨 + 󵄨󵄨 R(z) 󵄨󵄨󵄨 + 󵄨󵄨 1 󵄨󵄨󵄨 + 󵄨󵄨 Q(z, g) 󵄨󵄨󵄨 + 󵄨󵄨 1 󵄨󵄨󵄨 log+ 󵄨󵄨󵄨 󵄨󵄨 ≤ log 󵄨󵄨󵄨 󵄨󵄨 + log 󵄨󵄨󵄨 󵄨󵄨 = log 󵄨󵄨󵄨 󵄨󵄨 + log 󵄨󵄨󵄨 󵄨 󵄨󵄨 g(z) 󵄨󵄨 󵄨󵄨 g(z) 󵄨󵄨 󵄨󵄨 R(z) 󵄨󵄨 󵄨󵄨 g(z) 󵄨󵄨 󵄨󵄨 R(z) 󵄨󵄨󵄨 󵄨 󵄨 󵄨󵄨 ∑ b (z)G (z, g) 󵄨󵄨 󵄨󵄨 󵄨 l + 󵄨󵄨 1 󵄨󵄨󵄨 ≤ log+ 󵄨󵄨󵄨 l∈L l 󵄨󵄨 + log 󵄨󵄨󵄨 󵄨. 󵄨󵄨 󵄨󵄨 R(z) 󵄨󵄨󵄨 󵄨󵄨 g(z) Defining c0 = 0, we obtain by integration 󵄨󵄨 ∑ b (z)G (z, g) 󵄨󵄨 dθ 󵄨󵄨 󵄨 l ∫ log+ 󵄨󵄨󵄨 l∈L l 󵄨󵄨 󵄨󵄨 2π 󵄨󵄨 g(z)

θ∈E1 ν

μ

≤ ∑ ∑ ln,m m (r, n=0 m=0

g (m) (z + cn ) ) + O(log r). g(z)

The claim now follows by applying the delay-differential counterparts of the logarithmic derivative lemma; see Lemma 1.2.8. Before proceeding to the next result, we recall [88, Lemma 2.1], which appears to have a key role in proving the next theorem (and later on in this chapter): Lemma 8.2.3. Let f (z) be a nonrational meromorphic solution to P(z, f ) = 0, where P(z, f ) is a delay-differential polynomial in f (z) with rational coefficients, and let a1 , . . . , ak be rational functions satisfying P(z, aj ) ≢ 0 for all j ∈ {1, . . . , k}. If there exist s > 0 and τ ∈ (0, 1) such that k

∑ n (r, j=1

then ρ2 (f ) ≥ 1.

1 ) ≤ kτn(r + s, f ) + O(1), f − aj

(8.11)

8.2 Malmquist-type delay-differential equations | 167

Proof. First, observe by integration that k

∑ N (r, j=1

1 ) ≤ (τ + ε)kN(r + s, f ) + O(log r), f − aj

(8.12)

where ε > 0 is chosen so that τ + ε < 1. The first main theorem of Nevanlinna theory now yields k

kT(r, f ) = ∑ (m (r, j=1

1 1 ) + N (r, )) + O(log r). f − aj f − aj

(8.13)

If ρ2 (f ) < 1 (recall Lemma 1.2.10), then using (8.7), (8.12), and (8.13), we conclude that kT(r, f ) ≤ (τ + ε)kN(r + s, f ) + S(r, f ) ≤ (τ + ε)kT(r + s, f ) + S(r, f ) = (τ + ε)kT(r, f ) + S(r, f ) Thus we have T(r, f ) = S(r, f ) since τ + ε < 1, a contradiction. We now proceed to looking at delay-differential equations of the form c(z)f (z + 1) + a(z)

f 󸀠 (z) P(z, f (z)) = R(z, f (z)) = , f (z) Q(z, f (z))

(8.14)

where a(z) and c(z) are rational functions, P(z, f (z)) is a polynomial in f with rational coefficients, and Q(z, f (z)) is a monic polynomial in f with the roots that are nonvanishing rational functions in z that are not roots of P(z, f (z)). We now get the following result and supply a modified proof; see [201, Theorem 1.2]. Theorem 8.2.4. Let f (z) be a transcendental meromorphic solution of (8.14) of hyperorder ρ2 (f ) < 1, where P(z, f ) and Q(z, f ) satisfy the above conditions. Then degf (P) = degf (Q) + 1 = 2, or degf (R) ∈ {0, 1}. We first give the following lemma for the proof of Theorem 8.2.4. We omit the proof of the lemma, as we may just apply in the present first-order case the proof of Lemma 8.2.7 for second-order delay-differential equations. Lemma 8.2.5. Let f (z) be a nonrational meromorphic solution of (8.14), where a(z), resp., R(z, f ) = P(z, f )/Q(z, f ), are rational functions in one, resp., two variables. Moreover, we suppose to hold the assumptions given on P(z, f ) and Q(z, f ) in the preceding theorem. If ρ2 (f ) < 1, then degf (R(z, f )) ≤ 2.

168 | 8 Malmquist-type delay-differential equations Proof of Theorem 8.2.4. We consider the situation according to the number of roots of Q(z, f (z)). Case 1. Suppose first that Q(z, f (z)) has two distinct nonzero rational roots, say b1 (z) and b2 (z). Write equation (8.14) in the form c(z)f (z + 1) + a(z)

P(z, f (z)) f 󸀠 (z) = , f (z) (f (z) − b1 (z))(f (z) − b2 (z))

(8.15)

where degf (P) ≤ 2. Recall that P(z, f (z)) and Q(z, f (z)) have no common (rational) roots. Moreover, clearly, neither b1 (z) nor b2 (z) is a solution to (8.15). Consider now points z ∈ ℂ such that f (z)̂ = bj (z)̂ (say, of multiplicity p), j = 1, 2, assuming (here and later on in this proof) that ẑ is neither a zero nor a pole of the coefficients of (8.15) and of their shifts to be considered, as well as of b1 (z) and b2 (z) and their shifts needed. ̂ ≠ 0. Of course, the total number of such exMoreover, we may assume that P(z,̂ f (z)) ceptional points is finite, and hence their contribution to the nonintegrated counting function is O(1). Here and later in this proof, the notation ẑ means to be a generic zero of f (z) just described. Continuing now to look at nonexceptional points ẑ only, we observe that ẑ + 1 is a pole of f of multiplicity at least p. Therefore n (r,

1 1 ) + n (r, ) ≤ n(r + 1, f ) + O(1). f − b1 f − b2

By Lemma 8.2.3 we conclude that ρ2 (f ) ≥ 1, a contradiction. Case 2. Assume now that Q(z, f (z)) has exactly one nonzero rational root, say b1 (z). Now, equation (8.14) may be written in the form c(z)f (z + 1) + a(z)

f 󸀠 (z) P(z, f (z)) = , f (z) (f (z) − b1 (z))2

(8.16)

where degf (P) ≤ 2. Let ẑ be a zero of order p of f (z) − b1 (z), and hence f (ẑ + 1) is a pole of f of order ≥ 2p. Shifting by one, we conclude that f (ẑ + 2) is a simple pole of f . Fixing 2p now τ so that 2p+1 ≤ τ < 1, we see that n (r,

1 ) ≤ τn(r + 2, f ) + O(1). f − b1

Therefore, by Lemma 8.2.3, ρ2 (f ) ≥ 1, a contradiction. Case 3. Suppose now that Q(z, f (z)) has exactly one simple zero at b1 (z). Write now (8.14) in the form c(z)f (z + 1) + a(z)

f 󸀠 (z) P(z, f (z)) = . f (z) f (z) − b1 (z)

(8.17)

8.2 Malmquist-type delay-differential equations | 169

If degf (P) = 2, then degf (P) = degf (Q) + 1, which is just the first statement. If degf (P) = 1, then degf (R) = 1 as well, as claimed in the second statement. Case 4. It remains to consider the final case where R(z, f (z)) is a polynomial in f , and hence (8.14) takes the form c(z)f (z + 1) + a(z)

f 󸀠 (z) = P(z, f (z)), f (z)

(8.18)

where degf (P) ≤ 1. If degf (P) = 0 or degf (P) = 1, we have the second statement. Hence we assume that degf (P) = 2. If we first have that f (z) has finitely many zeros and poles only and ρ2 (f ) < 1, we may apply the standard Clunie reasoning in the difference setting, see Lemma 1.2.14, to get m(r, f ) = S(r, f ), and hence T(r, f ) = S(r, f ), a contradiction. Therefore we may assume that f (z) has infinitely many zeros; the case with infinitely many poles may be carried through completely similarly. Let now ẑ be a zero of f (z). Then shifting by one, f (z) has a pole of order 1 at ẑ + 1, a pole of order 2 at ẑ + 2, and, by repeating the shifting, of order 2d−1 at ẑ + d. Therefore n(ẑ + |d|, f ) ≥ 2d for all d ∈ ℕ. Therefore we immediately compute that the hyperexponent of convergence satisfies λ2 (1/f ) = lim sup r→∞

≥ lim sup r→∞

log log n(r, f ) log r log log n(|z|̂ + d, f ) log(|z|̂ + d)

≥ lim sup r→∞

log log 2d = 1. log(|z|̂ + d)

Therefore ρ2 (f ) ≥ λ2 (1/f ) ≥ 1, a contradiction, completing the proof. Concerning the Malmquist-type theorems for delay-differential equations of second order, Halburd and Korhonen [88, Theorem 1.1] obtained the following result to point at some equations that may allow meromorphic solutions of hyper-order less than one. Theorem 8.2.6. Let f (z) be a transcendental meromorphic solution to f (z + 1) − f (z − 1) + a(z)

f 󸀠 (z) P(z, f (z)) = R(z, f (z)) = , f (z) Q(z, f (z))

(8.19)

where a(z) is rational, P(z, f ) is a polynomial in f having rational coefficients in z, and Q(z, f ) is a polynomial in f (z) with the roots that are nonzero rational functions of z and not roots of P(z, f ). If ρ2 (f ) < 1, then degf (Q) + 1 = degf (P) ≤ 3 or

deg(R(z, f )) ≤ 1.

170 | 8 Malmquist-type delay-differential equations The proof of this theorem is rather lengthy. To help the reader, we first prove some lemmas. Our first lemma is slightly different from the corresponding lemma in [88, Lemma 3.1]. Lemma 8.2.7. Let f (z) be a nonrational meromorphic solution to (8.19), where a(z), resp., R(z, f ) = P(z, f )/Q(z, f ), are rational functions in one, resp., two variables. Moreover, we suppose that the assumptions given on P(z, f ) and Q(z, f ) in the preceding theorem hold. If ρ2 (f ) < 1, then degf (R(z, f )) ≤ 3. Proof. Let f (z) be a transcendental meromorphic solution of (8.19) with ρ2 (f ) < 1. Recalling the assumptions given for P(z, f ) and Q(z, f ) in Theorem 8.2.6, it follows that if P(z,f ) f = 0 is not a root of Q(z, f ), then f = 0 is not a pole of Q(z,f . From the left-hand side ) of (8.19) we see that either f has finitely many zeros that are zeros of a(z) or f (z) has infinitely many zeros that are poles of f (z + 1) or f (z − 1). Therefore N (r, f (z + 1) − f (z − 1) + a(z)

f 󸀠 (z) ) ≤ N(r, f (z + 1)) + N(r, f (z − 1)) f (z) + N(r, f ) + S(r, f ).

Combining with the logarithmic derivative lemma and its difference version, see Proposition 1.1.8 and (1.22), we obtain f 󸀠 (z) ) f (z) f (z − 1) f (z + 1) ) + m(r, ) ≤ m(r, f (z)) + m(r, f (z) f (z) + 3N(r, f (z)) + S(r, f )

T(r, f (z + 1) − f (z − 1) + a(z)

≤ 3T(r, f (z)) + S(r, f ). Thus degf (R(z, f )) ≤ 3 immediately follows by the Valiron–Mohon’ko theorem; see Theorem 1.1.29. For the proof of Theorem 8.2.6, we first separately consider particular cases. We first assume that R(z, f ) reduces to a polynomial P(z, f ) in f . For this particular case, see [88, Lemma 3.2]. Lemma 8.2.8. Let f be a nonrational meromorphic solution to the equation f (z + 1) − f (z − 1) + a(z)

f 󸀠 (z) = P(z, f (z)), f (z)

(8.20)

where a(z) is rational in z, and P(z, f ) is a polynomial in f with rational coefficients. If the hyperorder ρ2 (f ) < 1, then degf (P) ≤ 1. Proof. Assume, contrary to the statement, that degf (P) ≥ 2, and first suppose that f (z) has either infinitely many zeros or infinitely many poles (or both). Let z = ẑ be either a

8.2 Malmquist-type delay-differential equations | 171

zero or a pole of f (z). Then either there is a cancelation with a zero or a pole by some of the coefficients in (8.20), or f (z) has a pole of order at least 1 either at z = ẑ + 1 or at z = ẑ − 1. Since the coefficients of (8.20) are rational, we can always choose a zero or a pole of f (z) such that there is no cancelation with the coefficients. Suppose, without loss of generality, that there is a pole of f (z) at z = ẑ + 1. Shifting (8.20) by one, we obtain f (z + 2) − f (z) + a(z + 1)

f 󸀠 (z + 1) = P(z + 1, f (z + 1)), f (z + 1)

(8.21)

from which it follows that f (z) has a pole at z = ẑ +2 of order at least degf (P) and a pole of order at least (degf (P))2 at z = ẑ + 3. Continuing shifting, we obtain a string of poles with exponential growth of multiplicity. The only way that such a string either terminates, or that there appears a drop in the order of poles, is that there is a cancelation with a suitable zero or pole of a coefficient of (8.20). Since the coefficients are rational and f (z) has infinitely many zeros or poles, we can choose the starting point ẑ of the iteration from outside a sufficiently large disc in such a way that no such cancelation occurs. Therefore ̂ f ) ≥ (degf (P))d n(d + |z|, for all d ∈ ℕ. Then it follows that 1 log log n(r, f ) λ2 ( ) = lim sup f log r r→∞ ̂ f) log log n(d + |z|, ≥ lim sup ̂ log(d + |z|) d→∞ ≥ lim sup d→∞

log log(degf (P))d = 1. ̂ log(d + |z|)

Therefore ρ2 (f ) ≥ λ2 ( f1 ) ≥ 1, a contradiction. Suppose next that f (z) has finitely many poles and zeros only and that ρ2 (f ) < 1. Of course, we then have N(r, f1 ) = S(r, f ) and N(r, f ) = S(r, f ). From (8.20) we obtain f (z)[f (z + 1) − f (z − 1)] + a(z)f 󸀠 (z) = P(z, f (z))f (z).

(8.22)

If degf (P) ≥ 2, then it is immediate that the total degree of the right-hand side of (8.22) in f is at least 3. Using the delay-differential Clunie lemma, see Lemma 1.2.16, to (8.22), it follows that m(r, f ) = S(r, f ). This leads to T(r, f ) = S(r, f ), which is a contradiction. Thus ρ2 (f ) ≥ 1. To continue, we next consider the case that Q(z, f ) has a repeated root as a polynomial in f ; see [88, Lemma 3.3].

172 | 8 Malmquist-type delay-differential equations Lemma 8.2.9. Let f be a transcendental meromorphic solution to the equation f (z + 1) − f (z − 1) + a(z)

f 󸀠 (z) P(z, f (z)) = , ̃ f (z)) f (z) (f (z) − b1 (z))κ Q(z,

(8.23)

̃ f (z)) are polynomials in f and where a(z) and b1 (z) are rational in z, P(z, f ) and Q(z, ̃ f (z)) are pairwise rational in z, and κ ≥ 2. Assume that f (z) − b1 (z), P(z, f ), and Q(z, coprime. Then ρ2 (f ) ≥ 1. Proof. Notice that b1 (z) is not a solution to (8.23) even if b1 ≡ 0, and thus the first condition of Lemma 8.2.3 is satisfied for b1 . Suppose that ẑ is a zero of f (z) − b1 (z) of order ̃ f (z)), have a zero or p and that neither a(z), b1 (z) nor the coefficients of P(z, f ) and Q(z, a pole at z.̂ We will also require that the coefficients have no zeros or poles at points of the form ẑ + j for a finite number of integers j (in this particular case, we may assume that −4 < j < 4). We will again call such a point ẑ a generic zero of order p. We further assume, if needed, that in similar situations, we are considering generic zeros only. Since the coefficients are rational, when estimating the corresponding nonintegrated counting functions, the contribution from the nongeneric zeros can be included in a bounded error term, leading to an error term of the type O(log r) in the integrated estimates involving T(r, f ). Now either f (z + 1) or f (z − 1) has a pole of order q ≥ κp at z = z,̂ and we may assume, without loss of generality, that ẑ + 1 is such a pole. Suppose that ̃ degf (P) ≤ κ + degf (Q).

(8.24)

Then f (z) has a pole of order one at ẑ +2 and a pole of order q at ẑ +3. By continuing the iteration it follows that f (z) has either a simple pole or a finite value at ẑ + 4. Therefore it may be that f (ẑ +4) = b1 (ẑ +4), and so it is at least in principle possible that f (ẑ +5) is of finite value. This can only happen if the order of f (z) − b1 (z) at z = ẑ + 4 is p󸀠 = qκ ≥ p. But even in such a case, by considering the multiplicities of zeros and poles of f − b1 in the set {z,̂ . . . , ẑ + 4} we find that there are 2q + 1 > 2q ≥ κp + kp󸀠 poles of f for p + p󸀠 zeros of f − b1 . This is the “worst-case scenario” in the sense that if f (ẑ + 4) ≠ b1 (ẑ + 4) ̃ f (z)), then ẑ + 5 is a pole of f of order q ≥ κp, and we have even more or a zero of Q(z, poles for every zero of f − b1 . By adding up the contribution from all points ẑ to the corresponding counting functions it follows that n (r,

1 1 ) ≤ n(r + 4, f ) + O(1). f − b1 κ

Thus the assumptions of Lemma 8.2.3 are satisfied, and so ρ2 (f ) ≥ 1. ̃ + 1. Suppose that ẑ is a generic zero of Assume now that degf (P) ≥ κ + degf (Q) f (z) − b1 (z) of order p. Then, as in the previous case, f (z) has a pole of order q ≥ κp at either ẑ + 1 or ẑ − 1, say at ẑ + 1. This implies that f (z) has a pole of order q󸀠 ≥ q at ẑ + 2, and so the only way that f (ẑ + 4) can be finite is that f (z) − b1 (z) has a zero at

8.2 Malmquist-type delay-differential equations | 173

̃ f ). Even if this were the case, we ẑ + 3 of multiplicity p󸀠 = qκ or f (ẑ + 3) is a zero of Q(z, 󸀠 have found at least κp + κp poles, taking into account multiplicities, that correspond uniquely to at most p + p󸀠 zeros of f − b1 . Therefore we have 󸀠

n (r,

1 1 ) ≤ n(r + 3, f ) + O(1) f − b1 κ

(8.25)

by going through all zeros of f − b1 in this way. Lemma 8.2.3 implies that ρ2 (f ) ≥ 1. Now we proceed to complete the proof of Theorem 8.2.6. From Lemmas 8.2.7, 8.2.8, and 8.2.9 it follows that degf (P) ≤ 3 and 1 ≤ degf (Q) ≤ 3 and that Q(z, f ) has simple nonzero roots only as a polynomial in f . Now first consider the case degf (Q) = 1. We may assume that Q(z, f ) = f (z) − b1 (z) without restricting generality. Equation (8.19) takes the form f (z + 1) − f (z − 1) + a(z)

f 󸀠 (z) P(z, f (z)) = . f (z) f (z) − b1 (z)

(8.26)

Suppose degf (P) = 3. Let ẑ be a generic zero of f (z) − b1 (z) of order p. Then f (z) has a pole of order at least p at either ẑ + 1 or ẑ − 1, say ẑ + 1 is such a pole. Shifting equation (8.26) by one, we see that f (z) has a pole of order at least 2p at ẑ + 2. Therefore n (r,

1 1 ) ≤ n(r + 2, f ) + O(1). f − b1 3

(8.27)

By Lemma 8.2.3, ρ2 (f ) ≥ 1, a contradiction. If now degf (P) ≤ 2, then we must have degf (P) = degf (Q) + 1 = 2, whereas if degf (P) ≤ 1, then degf (R) ≤ 1, according to the statement. Suppose now that Q(z, f ) has at least two simple nonzero rational roots for f , say b1 (z) ≢ 0 and b2 (z) ≢ 0. Then we may write (8.19) in the form f (z + 1) − f (z − 1) + a(z)

f 󸀠 (z) P(z, f (z)) = ̃ f (z)) f (z) (f (z) − b1 (z))(f (z) − b2 (z))Q(z,

(8.28)

̃ f (z)) ≢ 0 are polynomials in f (z) of degrees at most 3 where P(z, f (z)) ≢ 0 and Q(z, and 1, respectively, with no common factors. Then neither b1 (z) nor b2 (z) is a solution to (8.28), and the key assumption of Lemma 8.2.3 holds. Let z = ẑ be a generic zero of order p of f − b1 . Now by (8.28) it follows that either f (z + 1) or f (z − 1) has a pole at z = ẑ of order at least p. We may assume that f (z + 1) has such a pole at z = z.̂ Then by shifting equation (8.28) forward we obtain f 󸀠 (z + 1) f (z + 1) P(z + 1, f (z + 1))

f (z + 2) − f (z) + a(z + 1) =

̃ + 1, f (z + 1)) (f (z + 1) − b1 (z + 1))(f (z + 1) − b2 (f + 1))Q(z

.

(8.29)

174 | 8 Malmquist-type delay-differential equations This implies that f (z + 2) has a pole of order one at z = z,̂ provided that ̃ + 2. degf (P) ≤ degf (Q)

(8.30)

Suppose now that (8.30) holds. Iterating (8.28) forward by two, we obtain f 󸀠 (z + 2) f (z + 2) P(z + 2, f (z + 2))

f (z + 3) − f (z + 1) + a(z + 2) =

̃ + 2, f (z + 2)) (f (z + 2) − b1 (z + 2))(f (z + 2) − b2 (z + 2))Q(z

.

(8.31)

Now if p > 1, then f must be a pole of order at least p at ẑ + 3. Hence in this case, we can pair up the zeros of f − b1 at z = ẑ together with the poles of f at ẑ + 1 without the possibility of a similar sequence of iterates starting from another point, say z = ẑ + 3, and resulting in pairing the pole at ẑ +1 with another zero of f −b1 or of f −b2 . Therefore we have found a pole of order at least p, which can be uniquely associated with the zero of f − b1 at z.̂ If, on the other hand, p = 1, it may in principle be possible that there is another zero of f − b1 or of f − b2 at z = ẑ + 3, which needs to be paired with the pole of f at z = ẑ + 2. But since now all the poles in the iteration are simple, we may still pair up the zero of f − b1 at z = ẑ and the pole of f at z = ẑ + 1. If there is another zero of, say f − b1 , at z = ẑ + 3 such that f (ẑ + 4) is finite, we can pair it up with the pole of f at z = ẑ + 2. Thus for any p ≥ 1, there is a pole of multiplicity at least p, which can be paired up with the zero of f − b1 at z = z.̂ This argument can now be repeated for zeros of w−b2 without any possible overlap in the pairing of poles with the zeros of f − b1 and f − b2 . By considering all generic zeros of f − b1 and f − b2 it follows that n (r,

1 1 ) + n (r, ) ≤ n(r + 1, f ) + O(1). f − b1 f − b2

(8.32)

Therefore the assumptions of Lemma 8.2.3 are satisfied, and so ρ2 (f ) ≥ 1, a contradic̃ tion. Hence we must now have that degf (P) > degf (Q)+2. This implies that degf (P) = 3, completing the proof of Theorem 8.2.6. Remark 8.2.10. Observe that if the condition “Q(z, f ) has roots that are non-zero rational functions of z” is dropped, then Theorem 8.2.6 may fail, as can be seen by the following example. Example 8.2.11. The transcendental meromorphic function f (z) = tan( π4 z) with ρ2 (f ) = 0 solves f (z + 1) − f (z − 1) +

4 f 󸀠 (z) 1 + 4f (z)2 − f (z)4 = . π f (z) f (z)(f (z) + 1)(1 − f (z))

We next consider (8.19) independently of the assumption that Q(z, f ) has roots that are nonzero rational functions of z; see Zhang and Huang [248, Theorem 1.1]. We

8.2 Malmquist-type delay-differential equations | 175

want to see what happens when a solution to this delay-differential equation is an entire function. In certain sense, this result may be understood as a delay-differential variant of the Malmquist theorem. Theorem 8.2.12. Let R(z, f ) be an irreducible rational function in f (z) with rational coefficients, and let a(z) be a rational function. If equation (8.19) admits a transcendental entire solution f (z) with ρ2 (f ) < 1, then (8.19) reduces either to f 󸀠 (z) = a1 (z)f (z) + a0 (z) f (z)

(8.33)

f 󸀠 (z) a2 (z)f (z)2 + a1 (z)f (z) + a0 (z) = . f (z) f (z)

(8.34)

f (z + 1) − f (z − 1) + a(z) or to f (z + 1) − f (z − 1) + a(z)

To prove Theorem 8.2.12, we first give two lemmas; see [248, Lemmas 2.5 and 2.6]. Lemma 8.2.13. Let R(z, f ) be a nonvanishing irreducible rational function in f (z) with rational coefficients such that degf (R) ≤ 2, let a(z) be a nonvanishing rational function, and let f (z) be a transcendental entire solution to (8.19). If ρ2 (f ) < 1 and f (z) has finitely many zeros only, then (8.19) takes the form (8.33) such that a1 (z) and a0 (z) are rational functions with at least one of them being nonvanishing. Proof. By the Hadamard factorization theorem we may write f (z) = H(z)eg(z) , where H(z) is a nonvanishing polynomial, and g(z) is a nonconstant entire function such that ρ2 (f ) = ρ2 (eg ) = ρ(g) < 1. Substituting this representation into (8.19) and defining s(z) := H(z + 1)eg(z+1)−g(z) − H(z − 1)eg(z−1)−g(z) , we obtain s(z)eg(z) + a(z) (

P(z, f (z)) H 󸀠 (z) )= . H(z) + g 󸀠 (z) Q(z, f (z))

(8.35)

If now s(z) vanishes, then we immediately see by (8.35) that T(r, R(z, f (z))) = S(r, eg ) = S(r, f ). By the Valiron–Mohon’ko theorem, see Theorem 1.1.29, it follows that degf (Q) = degf (P) = 0. This means that (8.19) takes the form (8.33), where a1 (z) vanishes, and a0 (z) is a nonvanishing rational function. Suppose now that s(z) is nonvanishing. We may now apply (1.24) to conclude that T(r, s) = m(r, s) ≤ m (r,

eg(z+1) eg(z−1) ) + m (r, g(z) ) + O(log r) = S(r, eg ). g(z) e e

176 | 8 Malmquist-type delay-differential equations By (8.35) we now see that T(r, R(z, f )) ≤ T(r, eg ) + S(r, eg ) = T(r, f ) + S(r, f ). By the Valiron–Mohon’ko theorem again, degf (Q) ≤ 1 and degf (P) ≤ 1. Therefore (8.35) may be written in the form s(z)eg(z) + a(z) (

ã 1 (z)H(z)eg(z) + ã 0 (z) H 󸀠 (z) , ) = H(z) + g 󸀠 (z) b̃ 1 (z)H(z)eg(z) + b̃ 0 (z)

(8.36)

where ã 0 (z), ã 1 (z), b̃ 0 (z), b̃ 1 (z) are rational functions. We may now write (8.36) in the form H (z) (b̃ 0 (z)s(z) + b̃ 1 (z)H(z)a(z) ( + g 󸀠 (z)) ã 1 (z)H(z)) eg(z) H(z) H 󸀠 (z) + s(z)b̃ 1 (z)H(z)e2g(z) + b̃ 0 (z)a(z) ( + g 󸀠 (z)) − ã 0 (z) = 0. H(z) 󸀠

Applying the Borel theorem (Theorem 1.1.32) to this identity, we conclude that b̃ 1 (z) vanishes. Therefore (8.19) takes the form (8.33) such that ã 1 (z) is nonvanishing and both ã 1 (z) and ã 0 (z) are rational functions. Lemma 8.2.14. Let R(z, f ) be a nonvanishing irreducible rational function in f (z) with rational coefficients such that degf (R) ≤ 2, let a(z) be a nonvanishing rational function, and let f (z) be a transcendental entire solution of equation (8.19). If ρ2 (f ) < 1 and there exists a nonvanishing rational function r(z) such that f (z) + r(z) has finitely many zeros only, then (8.19) reduces to the form (8.34), where a2 (z), a1 (z), a0 (z) are rational functions, and a0 (z) is nonvanishing. Proof. By the Hadamard factorization theorem, f (z) may be written in the form f (z) = H(z)eg(z) + r(z), where H(z) is a nonvanishing polynomial, and g(z) is a nonconstant entire function such that ρ2 (f ) = ρ2 (eg ) = ρ(g) < 1. Defining s(z) := H(z + 1)eg(z+1)−g(z) − H(z − 1)eg(z−1)−g(z) , we obtain f (z + 1) − f (z − 1) = s(z)

f (z) + r(z) − r(z + 1) + r(z − 1) H(z)

(8.37)

and f 󸀠 (z) = (

H 󸀠 (z) H 󸀠 (z) r 󸀠 (z) + g 󸀠 (z)) f (z) + r(z) ( + g 󸀠 (z) − ). H(z) H(z) r(z)

(8.38)

Since degf (R) ≤ 2, substituting (8.37) and (8.38) into (8.19), we obtain s(z) f (z)2 H(z)

(z) + g 󸀠 (z) − + t(z)f (z) + a(z)r(z) ( HH(z) 󸀠

f (z) ã (z)f (z) + ã 1 (z)f (z) + ã 0 (z) = 2 , b̃ (z)f (z)2 + b̃ (z)f (z) + b̃ (z) 2

2

1

0

r 󸀠 (z) ) r(z)

(8.39)

8.2 Malmquist-type delay-differential equations | 177

where t(z) :=

s(z)r(z) H 󸀠 (z) − r(z + 1) + r(z − 1) + a(z) ( + g 󸀠 (z)) , H(z) H(z)

and ã j (z), b̃ j (z), j = 0, 1, 2, are rational functions. Since ρ2 (eg ) < 1, we may use (1.24) to verify that T(r, s) = S(r, f ). Hence all coefficients in (8.39) are small functions relative to f . If now H 󸀠 /H + g 󸀠 = r 󸀠 /r, then f would be rational, contradicting the assumption that f is transcendental. Therefore ar(h󸀠 /h + g 󸀠 − r 󸀠 /r) does not vanish. If now s ≠ 0, then multiplying both sides of (8.39) by f (b̃ 2 f 2 + b̃ 1 f + b̃ 0 ), we get s b̃ 2 f 4 + t3 f 3 + t2 f 2 + t1 f + b̃ 0 ar(H 󸀠 /H + g 󸀠 − r 󸀠 /r) = 0, H

(8.40)

where s t3 := b̃ 2 t + b̃ 1 − ã 2 , H s t2 := b̃ 2 ar(H 󸀠 /H + g 󸀠 − r 󸀠 /r) + b̃ 1 t + b̃ 0 − ã 1 , H and t1 := b̃ 1 ar(H 󸀠 /H + g 󸀠 − r 󸀠 /r) + b̃ 0 t − ã 0 all are small functions relative to f . By (8.40) we get a contradiction T(r, f ) = S(r, f ) unless all coefficients of (8.40) vanish. But then b̃ 2 = b̃ 0 = 0, meaning that (8.19) reduces to the form (8.34), where a0 , a1 , a2 are rational. Since R(z, f ) is irreducible, we must have a0 ≠ 0. It remains to consider the case s = 0. By (8.39) we conclude that degf (R) ≤ 1. Then (8.39) reduces to (z) t(z)f (z) + a(z)r(z) ( HH(z) + g 󸀠 (z) − 󸀠

f (z)

r 󸀠 (z) ) r(z)

=

ã 1 (z)f (z) + ã 0 (z) . b̃ (z)f (z) + b̃ (z) 1

0

(8.41)

Similarly as in the preceding case, we conclude that (8.19) reduces to (8.34), where a2 = 0, and a1 , a0 are rational with a0 ≠ 0. We are now ready to complete the proof of Theorem 8.2.12: Proof of Theorem 8.2.12. Case 1. Suppose first that a(z) ≡ 0. Therefore T(r, R(z, f (z))) = T(r, f (z + 1) − f (z − 1)) = m(r, f (z + 1) − f (z − 1)) f (z + 1) f (z − 1) ≤ m(r, f (z)) + m (r, ) + m (r, ) f (z) f (z) = m(r, f ) + S(r, f ).

178 | 8 Malmquist-type delay-differential equations By the Valiron–Mohon’ko theorem, degf (R) ≤ 1, and equation (8.19) takes the form f (z + 1) − f (z − 1) =

ã 1 (z)f (z) + ã 0 (z) b̃ (z)f (z) + b̃ (z) 1

0

(8.42)

with rational coefficients. Suppose first that b̃ 1 ≠ 0. Again by the Valiron–Mohon’ko theorem, (8.42) implies that T(r, f (z + 1) − f (z − 1)) = T(r, f ) + S(r, f ). We may now write (8.42), by simple computation, in the form f (z)(f (z + 1) − f (z − 1)) = −

ã (z) b̃ 0 (z) ã (z) (f (z + 1) − f (z − 1)) + 1 f (z) + 0 . ̃b (z) ̃b (z) b̃ 1 (z) 1 1

Applying now the delay-differential Clunie lemma (Lemma 1.2.16) we deduce that T(r, f (z + 1) − f (z − 1)) = m(r, f (z + 1) − f (z − 1)) = S(r, f ), a contradiction. Therefore we must have b̃ 1 = 0, and equation (8.19) takes the form (8.33) with rational coefficients. Case 2. We now continue by assuming that a ≠ 0. Recalling the assumption, we now have degf (P) ≤ 2 and degf (Q) ≤ 2, and we may write P(z, f (z)) ã 2 (z)f (z)2 + ã 1 (z)f (z) + ã 0 (z) = Q(z, f (z)) b̃ 2 (z)f (z)2 + b̃ 1 (z)f (z) + b̃ 0 (z)

(8.43)

with rational coefficients. If now f has finitely many zeros only, we see by Lemma 8.2.13 that (8.19) takes the form (8.33). Moreover, if there exists a rational function r ≠ 0 such that f + r has finitely many zeros only, then Lemma 8.2.14 implies that (8.19) reduces to the form (8.34). Therefore we may now assume that f has infinitely many zeros and that f + r has infinitely many zeros for an arbitrary rational function r ≠ 0. Take now, as we may, f (z0 ) = 0 so that neither a(z) nor any coefficient of R(z, f ) has a zero or pole at z0 . If 󸀠 (z) b̃ 0 ≠ 0, then z0 is a simple pole of f (z + 1) − f (z − 1) + a(z) ff (z) , whereas R(z0 , f (z0 )) is ̃ finite, a contradiction. Hence we must have b0 = 0. Consider now the case where b̃ ≠ 0 and b̃ = 0. Then (8.19) takes the form 2

f (z + 1) − f (z − 1) + a(z)

1

f 󸀠 (z) ã 2 (z)f (z)2 + ã 1 (z)f (z) + ã 0 (z) = . f (z) b̃ 2 (z)f (z)2

(8.44)

Since the right-hand side of (8.44) is irreducible, we have ã 0 ≠ 0. Taking a zero z0 as above, we observe that the left-hand side of (8.44) has there a simple pole, whereas the right-hand side has there a multiple pole, a contradiction.

8.2 Malmquist-type delay-differential equations | 179

Next, we consider the situation where b̃ 2 ≠ 0 and b̃ 1 ≠ 0. Then we consider f (z + 1)f (z) − f (z − 1)f (z) + a(z)f (z) = 󸀠

ã 2 (z) f (z)2 b̃ 2 (z)

ã 1 (z) f (z) b̃ 2 (z) ̃ f (z) + b̃ 1 (z) b (z)

+

+

ã 0 (z) b̃ 2 (z)

(8.45)

.

2

Now again, the right-hand side of (8.45) is irreducible, meaning that

ã 2 (z) f (z)2 b̃ 2 (z)

+

ã 0 (z) and f (z) + may have at most finitely many common zeros. Moreb̃ 2 (z) 2 ̃ ̃b (z) over, f (z) + ̃ 1 has infinitely many zeros. So, we may take a zero z1 of f (z) + b̃ 1 (z) so b2 (z) b2 (z) ̃ ̃ ã (z) that neither a(z) nor ã 2 (z) f (z)2 + ã 1 (z) f (z) + ̃0 has a zero or a pole at z1 . Hence z1 is b2 (z) b2 (z) b2 (z) ã 1 (z) f (z) b̃ 2 (z)

+

b̃ 1 (z) b̃ (z)

a pole of the right-hand side of (8.45), whereas its left-hand side takes a finite value, a contradiction. Therefore it remains to consider the case b̃ 0 = b̃ 2 = 0. This means that equation (8.19) reduces to (8.34), completing the proof.

The following examples 8.2.15 and 8.2.16 show that the case (8.33) in Theorem 8.2.12 may appear. Example 8.2.15. The transcendental entire function w(z) = zez solves w(z + 1) − w(z − 1) + a(z)

w󸀠 (z) e(z + 1) − e−1 (z − 1) z+1 = w(z) + a(z) , w(z) z z

where a(z) is any rational function. Example 8.2.16. The transcendental entire function w(z) = e2πiz solves w(z + 1) − w(z − 1) + a(z)

w󸀠 (z) = 2πia(z), w(z)

where a(z) is any rational function. Moreover, the following three examples 8.2.17, 8.2.18, and 8.2.19 show that the case (8.34) in Theorem 8.2.12 also may appear. Example 8.2.17. The transcendental entire function w(z) = z + ez solves w(z + 1) − w(z − 1) + a(z)

w󸀠 (z) (e − e−1 )w(z)2 + a(z)w(z) + a0 (z) = , w(z) w(z)

where a(z) is any rational function, and a0 (z) = −z(e − e−1 ) + 2 + a(z)(1 − z). Example 8.2.18. The transcendental entire function w(z) = e2πiz + 1 solves w(z + 1) − w(z − 1) + a(z) where a(z) is any rational function.

w󸀠 (z) 2πiza(z)w(z) − 2πia(z) = , w(z) w(z)

180 | 8 Malmquist-type delay-differential equations Example 8.2.19. The transcendental entire function w(z) = e2πiz + z solves w(z + 1) − w(z − 1) −

1 1 w󸀠 (z) 2z − πi = , πi w(z) w(z)

where a(z) is any rational function. We next complete considerations of entire solutions to (8.19) by the following result; see Theorems 1.2 and 1.3 in [248]. Theorem 8.2.20. (1) Suppose a, a0 , a1 are rational functions such that a0 ≠ 0 or a1 ≠ 0. If f is a transcendental entire solution to (8.33) with ρ2 (f ) < 1, then ρ(f ) ≥ 1 if a = 0 and f (z) = H(z)edz if a ≠ 0. In the latter case, H is a nonvanishing polynomial, and d ≠ 0 is a complex constant. (2) Suppose a, a2 , a1 , a0 ≠ 0 are rational functions. If f is a transcendental entire solution to (8.34) with ρ2 (f ) < 1, then ρ(f ) ≥ 1. Proof. (1) If a = 0, then (8.33) may be written in the form f (z + 2) − a1 (z + 1)f (z + 1) − f (z) = a0 (z + 1). If now a0 ≠ 0, then [28, Theorem 4] implies that ρ(f ) ≥ 1, whereas if a0 = 0, then a1 ≠ 0, and [28, Theorem 3] results in ρ(f ) ≥ 1. Suppose now that a ≠ 0. From (8.33) we see that f has at most finitely many zeros, and by the Hadamard factorization theorem, f (z) = H(z)eg(z) , where H ≠ 0 is a polynomial, and g is a nonconstant entire function such that ρ2 (f ) = ρ2 (eg ) = ρ(g) < 1. Substituting now f (z) = H(z)eg(z) into (8.33), we obtain −a1 (z)H(z)eg(z) + H(z + 1)eg(z+1) − H(z − 1)eg(z−1) = a0 (z) − a(z) (

H 󸀠 (z) + g 󸀠 (z)) . H(z)

We now easily see that the three entire functions g(z + 1) − g(z), g(z + 1) − g(z − 1), and g(z − 1) − g(z) are at the same time either transcendental or polynomials of the same degree, say d, in which case, deg(g) = d + 1. If we have the transcendental case or the polynomial case with d ≥ 1, then we may apply Theorem 1.1.32 to conclude that H ≡ 0, a contradiction. Therefore deg(g) = 1 means that f must be of the form f (z) = H(z)ebz , where b ≠ 0. (2) If a = 0 in (8.34), then we have T (r,

a2 f 2 + a1 f + a0 ) = T(r, f (z + 1) − f (z − 1)) f = m(r, f (z + 1) − f (z − 1)) ≤ m(r, f ) + S(r, f ).

8.2 Malmquist-type delay-differential equations | 181

By the Valiron–Mohon’ko theorem again, a2 = 0, and (8.34) takes the form f (z)(f (z + 1) − f (z − 1)) = a1 (z)f (z) + a0 (z).

(8.46)

Applying Clunie-type reasoning, see Lemma 1.2.14, we conclude that a1 f + a0 ) = T(r, f (z + 1) − f (z − 1)) f = m(r, f (z + 1) − f (z − 1)) = S(r, f ),

T(r, f ) + S(r, f ) = T (r,

a contradiction. Therefore we must have a ≠ 0. Write now (8.34) in the form f (z)(f (z + 1) − f (z − 1) − a2 (z)f (z) − a1 (z)) = −a(z)f 󸀠 (z) + a0 (z).

(8.47)

Since f is transcendental and a ≠ 0, we see that −a(z)f 󸀠 (z)+a0 (z) is nonvanishing, and hence f (z)(f (z + 1) − f (z − 1) − a2 (z)f (z) − a1 (z)) as well. Applying Clunie-type reasoning again to (8.47), we see that T(r, f (z + 1) − f (z − 1) − a2 (z)f (z) − a1 (z)) = O(log r). Therefore f (z + 1) − f (z − 1) − a2 (z)f (z) is a rational function, say r(z) = p(z)/q(z). Denoting now a2 (z) = P(z)/Q(z), we obtain q(z)Q(z)f (z + 1) − q(z)Q(z)f (z − 1) − P(z)q(z)f (z) = p(z)Q(z). Then by [28, Theorem 4] we conclude that ρ(f ) ≥ 1. We next proceed to considering (8.19) in the case where degf (R(z, f )) = 0. Therefore we look at f (z + 1) − f (z − 1) + a(z)

f 󸀠 (z) = b(z), f (z)

(8.48)

where a(z) and b(z) are rational. We first look at (8.48) by introducing an additional assumption that a meromorphic solution has sufficiently many simple zeros. In this case the rational coefficients a(z) and b(z) reduce to constants; see [88, Theorem 1.2]. Theorem 8.2.21. Let f (z) be a nonrational meromorphic solution to equation (8.48), where a(z)(≢ 0) and b(z) are rational. If ρ2 (f ) < 1 and for any ε > 0, 1 3 N (r, ) ≥ ( + ε) T(r, f ) + S(r, f ), f 4 then the coefficients a(z) and b(z) are both constants.

(8.49)

182 | 8 Malmquist-type delay-differential equations Proof. Let z = ẑ be a generic zero of f (z). Then by (8.48) there is a pole of f (z) at z = ẑ +1 or at z = ẑ − 1 (or at both points). We need to consider two cases. Suppose first that there are poles of f (z) at both z = ẑ + 1 and z = ẑ − 1. Then from (8.48) it follows that there are poles of f (z) at z = ẑ + 2 and z = ẑ − 2. In principle, we may have zeros of f at ẑ ± 3. Therefore, recalling Lemma 1.2.10, we obtain 4N(r, 1/f ) + S(r, f ) = 4N(r + 3, 1/f ) ≤ 3N(r + 2, f ) ≤ 3T(r, f ) + S(r, f ), contradicting (8.49). Therefore we may now assume that there is a pole of f (z) at only one of the points z = ẑ + 1 and z = ẑ − 1. Without loss of generality, we may suppose that f (z) has a pole at z = ẑ +1. The case where the pole is at z = ẑ −1 is completely analogous. We begin by showing that we need to consider simple generic zeros of f (z) only. Let N1 (r, f1 ) denote the integrated counting function for the simple zeros of f , and let N(p (r, f1 ) denote the counting function for the zeros of f of multiplicity ≥ p. Then 1 1 1 N (r, ) = N1 (r, ) + N(2 (r, ) f f f and 1 1 1 1 1 1 N (r, ) = N1 (r, ) + N (2 (r, ) ≤ N1 (r, ) + N(2 (r, ) f f f f 2 f 1 1 1 1 ≤ N1 (r, ) + N (r, ) . 2 f 2 f

(8.50)

Hence, using assumption (8.49), we get 1 1 1 3 1 N1 (r, ) ≥ 2N (r, ) − N (r, ) ≥ ( + ε) T(r, f ) − N (r, ) f f f 2 f 1 ≥ ( + ε) T(r, f ) + S(r, f ) 2

(8.51)

to estimate the number of simple zeros of f . So, if we consider the case where ẑ is a simple zero of f , we have that for K ∈ ℂ, α ∈ ℂ\{0}, ̂ f (z − 1) = K + O(z − z),

f (z) = α(z − z)̂ + O((z − z)̂ 2 ), a(z) f (z + 1) = − + O(1), z − ẑ a(z + 1) f (z + 2) = + O(1), z − ẑ a(z + 2) − a(z) f (z + 3) = + O(1) z − ẑ in a neighborhood of z.̂

8.2 Malmquist-type delay-differential equations | 183

̂ Therefore either + O(z − z). If now a(ẑ + 2) − a(z)̂ ≠ 0, then f (z + 4) = a(z+3)+a(z+1) z−ẑ we have infinitely many points such that a(z + 2) = a(z), and so the rational function a is a constant, or we can find at least four poles for every two simple zeros of f . In the second case, it follows that T(r, f ) ≤

1 2

1

1 N1 (r, ) + O(log r) f +ε

2 1 N (r + 2, f ) + O(log r) 1 + 2ε 2 1 1 ≤ T(r, f ) + S(r, f ). 1 + 2ε ≤

This implies that T(r, f ) = S(r, f ), a contradiction. Thus a(z) must be a constant. Furthermore, a ̂ + b(z) + K + O(z − z), z − ẑ a ̂ f (z + 2) = + b(z + 1) + b(z) + K + O(z − z), z − ẑ ̂ f (z + 3) = b(z + 2) − b(z + 1) + O(z − z).

f (z + 1) = −

If f (ẑ + 3) ≠ 0, then there are two poles (z = ẑ + 1 and z = ẑ + 2) in this sequence that can be uniquely associated with the zero of f (z) at z = z.̂ This can only be avoided if b(z) is a constant as well. Liu and Song [153, Theorem 5] obtained the following result on the meromorphic solutions to (8.48). Theorem 8.2.22. Transcendental entire solutions to (8.48) must be of the form f (z) = Cepπiz and b(z) ≡ kiπa(z), where k is an integer. If f (z) is a nonentire transcendental meromorphic solution of finite order to (8.48), then max{λ(f ), λ( f1 )} = ρ(f ). Proof. Let f (z) be a transcendental entire solution to (8.48). Rewrite (8.48) as follows: f (z)[f (z + 1) − f (z − 1)] = b(z)f (z) − a(z)f 󸀠 (z).

(8.52)

Clearly, f (z) cannot have infinitely many multiple zeros. Assume that f (z) has infinitely many simple zeros. Then we can choose a zero z0 such that b(z0 )f (z0 )−a(z0 )f 󸀠 (z0 ) ≠ 0, whereas the left-hand side of (8.52) is equal to zero at z0 , a contradiction. Therefore f (z) has at most finitely many zeros. We may now write f (z) = p(z)eg(z) , where p(z) is a nonzero polynomial, and g(z) is an entire function. Thus p(z + 1)eg(z+1) − p(z − 1)eg(z−1) = b(z) − a(z) (

p󸀠 (z) + g 󸀠 (z)) . p(z)

Here we have b(z) − a(z) (

p󸀠 (z) + g 󸀠 (z)) = 0, p(z)

(8.53)

184 | 8 Malmquist-type delay-differential equations since otherwise a contradiction with the second main theorem for three small functions follows from (8.53). Furthermore, g(z) must be a polynomial to avoid a contradiction with the above equation. From p(z + 1)eg(z+1) = p(z − 1)eg(z−1) we obtain g(z) = kπiz + B, where k is an integer, and p(z) is a constant. Hence b(z) ≡ kiπa(z). Let now f (z) be a nonentire transcendental meromorphic solution to (8.48) of fi󸀠 󸀠 (z) (z) ≢ 0. Indeed, if b(z) − a(z) ff (z) ≡ 0, then f (z) has nite order ρ. Thus b(z) + a(z) ff (z) finitely many poles only, whereas f (z + 1) = f (z − 1) implies that f (z) has infinitely many poles since f (z) has at least one pole. Now rewrite (8.48) as ( f (z+1) − f (z−1) ) 1 f (z) f (z) = . 󸀠 f (z) b(z) − a(z) f (z) f (z) Due to the logarithmic derivatives lemma and its difference counterpart, we have 1 f (z + 1) f (z − 1) f 󸀠 (z) T (r, ) ≤ T (r, − ) + T (r, b(z) − a(z) ) f f (z) f (z) f (z) ≤ N (r,

f 󸀠 (z) f (z + 1) f (z − 1) − ) + N (r, b(z) − a(z) ). f (z) f (z) f (z)

If we now assume that 1 max{λ(f ), λ( )} < ρ(f ), f then T(r, f1 ) = O(r ρ(f )−ε ), a contradiction. Hence max{λ(f ), λ( f1 )} = ρ(f ). Before proceeding to the situation where f (z + 1) − f (z − 1) in Theorem 8.2.6 is replaced by the corresponding product f (z + 1)f (z − 1), we shortly consider the corresponding higher-order situation proposed by Wang, Han, and Hu [210]: Theorem 8.2.23. Let f be a transcendental meromorphic solution to n

∑ aj (z)f (z + cj ) + a(z) j=1

f 󸀠 (z) P(z, f (z)) = , H(z, f (z)) Q(z, f (z))

(8.54)

where a, a1 , . . . , an are rational functions, and c1 , . . . , cn are distinct non-zero constants, H(z, f (z)) and P(z, f (z)) are polynomials in f with rational coefficients, and Q(z, f (z)) is a polynomial in f having zeros that are nonzero rational functions in z that in turn are not zeros of P(z, f (z)). If ρ2 (f ) < 1 and degf (H) ≥ 1, then degf (P) ≤ degf (Q) + 1.

8.2 Malmquist-type delay-differential equations | 185

Proof. Assume, contrary to the statement, that degf (P) = p ≥ degf (Q) + 2, and denote h := degf (H) ≥ 1. Moreover, write H(z, f (z)), P(z, f (z)) explicitly as H(z, f (z)) = dh (z)f (z)h + ⋅ ⋅ ⋅ + d1 (z)f (z) + d0 (z), P(z, f (z)) = bp (z)f (z)p + ⋅ ⋅ ⋅ + b1 (z)f (z) + b0 (z),

where d0 , . . . , dh and b0 , . . . , bp are rational functions. We may now write (8.54) in the form n

((∑ aj (z)f (z + cj ))H(z, f (z)) + a(z)f 󸀠 (z)) Q(z, f (z)) j=1

− (bp−1 (z)f (z)p−1 + ⋅ ⋅ ⋅ + b1 (z)f (z) + b0 (z))H(z, f (z))

− (dh−1 (z)f (z)h−1 + ⋅ ⋅ ⋅ + d1 (z)f (z) + d0 (z))bp (z)f (z)p = bp (z)dh (z)f (z)p+h . The total degree of the left-hand side as a polynomial in f , f 󸀠 and in shifts of f is at most p + h − 1, whereas the right-hand side is of degree p + h in f . An application of the delay-differential version of the Clunie lemma (Lemma 1.2.14) now implies that m(r, f ) = S(r, f ). Therefore f has infinitely many poles. Let now z1 be a generic pole of f with multiplicity k1 ≥ 1. If now degf (H) = 1, then z1 is a simple pole of a(z)f 󸀠 (z)/H(z, f (z)), whereas if degf (H) ≥ 2, then z1 is a regular point of a(z)f 󸀠 (z)/H(z, f (z)). In both cases the assumption degf (P) − degf (Q) ≥ 2 implies that f has a pole at least at one of the points z + c1 , . . . , z + cn . Denote by z2 such a pole of f of multiplicity k2 ≥ k1 (degf (P) − degf (Q)). Repeating the same process, starting with z2 , we obtain a pole z3 , one of the points z + c1 , . . . , z + cn , of f with multiplicity k3 ≥ k2 (degf (P) − degf (Q)) ≥ k1 (degf (P) − degf (Q))2 . Continuing the iteration, we construct a sequence zm of poles of f with corresponding multiplicities km ≥ k1 (degf (P) − degf (Q))m−1 . We may now proceed to estimate the counting function of f . Denote rm := |z1 |+(m−1)α, where α := max{|c1 |, . . . , |cn |}. Then by a simple geometric reasoning we see that zm ∈ B(0, rm ). Therefore n(rm , f ) ≥ k1 (degf (P) − degf (Q))m−1 . This now implies that log log n(rm , f ) log log n(r, f ) 1 ≥ lim sup λ2 ( ) ≥ lim sup f log r log rm r→∞ m→∞ ≥ lim sup m→∞

log log k1 (degf (P) − degf (Q))m−1 log rm

= 1,

and hence ρ2 (f ) ≥ λ2 ( f1 ) ≥ 1, contradicting the assumption that ρ2 (f ) < 1.

186 | 8 Malmquist-type delay-differential equations We now consider the product variant of equation (8.19); see Halburd and Korhonen [85, Theorem 5.5]. Such nonlinear difference equations f (z + 1)f (z − 1) = R(z, f (z)) are closely related to the difference Painlevé III equations. It is natural to ask what may happen when combining such an equation with Malmquist–Yosida-type reasoning. Liu and Song [155] considered this problem and obtained the following theorem. Theorem 8.2.24. Let f (z) be a nonrational meromorphic solution to f (z + 1)f (z − 1) + a(z)

P(z, f (z)) f 󸀠 (z) = R(z, f (z)) = , f (z) Q(z, f (z))

(8.55)

where a(z) is rational, P(z, f ) is a polynomial in f with rational coefficients, Q(z, f ) is a polynomial in f with degf (Q(z, f )) ≥ 1, and the roots of Q(z, f ) are nonzero rational functions of z and not roots of P(z, f ). If ρ2 (f ) < 1, then either degf (Q) + 1 = degf (P) ≤ 3,

(8.56)

or degf (R) = 1. The following remark on the poles of f (z + 1), f (z − 1), and f (z + 1)f (z − 1) will play an important part in the proof of Theorem 8.2.24. Remark 8.2.25. Assume that f (z + 1)f (z − 1) has a pole at z = ẑ of multiplicity λp. Then three cases may occur: Case 1: f (ẑ + 1) = ∞, and ẑ + 1 is a pole of f of multiplicity k, where λp > k > 0, f (ẑ − 1) = ∞, and ẑ − 1 is a pole of f of multiplicity λp − k. Case 2: f (ẑ +1) = ∞, where ẑ +1 is a pole of f of multiplicity k ≥ λp, and f (ẑ −1) = 0, where ẑ − 1 is a zero of f of multiplicity k − λp > 0. It may happen that k = λp, which means that f (ẑ − 1) is nonzero and finite. Case 3: f (ẑ −1) = ∞, where ẑ −1 is a pole of f of multiplicity k ≥ λp, and f (ẑ +1) = 0, where ẑ + 1 is a zero of f of multiplicity k − λp > 0. It may happen that k = λp, which means that f (ẑ + 1) is nonzero and finite. These will be called as cases 1, 2, 3 in the proof of Theorem 8.2.24. Proof of Theorem 8.2.24. Completely similarly as in Lemma 8.2.7, we conclude that degf (R) ≤ 3, i. e., degf (P) ≤ 3 and degf (Q) ≤ 3. We proceed to proving that Q(z, f ) is not of the form (f (z) − b1 (z))λ Q1 (z, f ), where λ ≥ 2, except if degf (P) = 3, λ = 2, and degf (Q1 ) = 0. In this case, degf (P) = 3 = degf (Q) + 1, which is the first statement. We may write f (z + 1)f (z − 1) + a(z)

f 󸀠 (z) P(z, f (z)) = , f (z) (f (z) − b1 (z))λ Q1 (z, f (z))

(8.57)

8.2 Malmquist-type delay-differential equations | 187

where Q1 (z, b1 (z)) ≢ 0. According to the conditions of Theorem 8.2.24, b1 (z) is a nonzero rational function. For the delay-differential polynomial P1 (z, f ) = (f (z)−b1 (z))λ Q1 (z, f (z))[f (z+1)f (z−1)f (z)+a(z)f 󸀠 (z)]−f (z)P(z, f (z)),

(8.58)

we have P1 (z, b1 (z)) ≢ 0. So, the first condition of Lemma 8.2.3 is satisfied. If there 1 ̂ then we have n(r, f −b ) = O(1). are finitely many points ẑ only satisfying f (z)̂ = b1 (z), 1 Therefore the second condition of Lemma 8.2.3 is satisfied as well. Thus ρ2 (f ) ≥ 1. ̂ Hence we may now assume that there are infinitely many ẑ satisfying f (z)̂ = b1 (z). ̂ Take now a generic z ∈ C that satisfies ̂ f (z)̂ = b1 (z),

(8.59)

generic in the sense that none of the rational coefficients and their shifts of (8.57) have ̂ ≠ 0. Let p denote the multiplicity of the zero of a zero or a pole at ẑ and that P(z,̂ f (z)) f − b1 at z = z.̂ In what follows, of course, the contributions from the nongeneric roots lead to an error term of the type O(log r) in the integrated estimate involving T(r, f ). By (8.57) it follows that f (z + 1)f (z − 1) has a pole at z = ẑ of order λp. This implies that cases 1, 2, 3 in Remark 8.2.25 may appear. Moreover, we assume that degf (P) ≤ degf (Q1 ) + λ.

(8.60)

In case 1, by shifting the equation (8.57) forward we have f 󸀠 (z + 1) f (z + 1) P(z + 1, f (z + 1)) = . (f (z + 1) − b1 (z + 1))λ Q1 (z + 1, f (z + 1))

f (z + 2)f (z) + a(z + 1)

(8.61)

Equation (8.61) means that f (z + 2) has a simple pole at z = z.̂ By shifting equation (8.57) backward we obtain f 󸀠 (z − 1) f (z − 1) P(z − 1, f (z − 1)) = . (f (z − 1) − b1 (z − 1))λ Q1 (z − 1, f (z − 1))

f (z)f (z − 2) + a(z − 1)

(8.62)

Equation (8.62) implies that f (z − 2) has a simple pole at z = z.̂ Thus z = ẑ satisfies ̂ and ẑ + 1, ẑ + 2, ẑ − 1, ẑ − 2 are poles of f (z) with multiplicities k, 1, λp − k, f (z)̂ = b1 (z), 1, respectively. Hence we have n (r,

1 p )≤ n(r + 2, f ) + O(1). f − b1 λp + 2

(8.63)

p Clearly, λp+2 ≤ 21 (λ ≥ 2). Therefore the second condition of Lemma 8.2.3 is satisfied, and so ρ2 (f ) ≥ 1.

188 | 8 Malmquist-type delay-differential equations In case 2, equation (8.61) implies that f (z + 2) has a simple pole at z = z.̂ Thus we have n (r,

p 1 )≤ n(r + 2, f ) + O(1). f − b1 k+1

(8.64)

p Since k ≥ λp, we have k+1 ≤ 21 (λ ≥ 2), and so Lemma 8.2.3 implies that ρ2 (f ) ≥ 1. In case 3, equation (8.62) implies that f (z − 2) has a pole of order one at z = z.̂ Thus inequality (8.64) is valid, and Lemma 8.2.3 implies that ρ2 (f ) ≥ 1. Hence we have

degf (P) > degf (Q1 ) + λ, which means that degf (P) = 3, λ = 2, and degf (Q1 (z, f )) = 0, that is, degf (P) = degf (Q) + 1 = 3. We further discuss three cases for Q(z, f ). Case I: Suppose first that Q(z, f ) = (f (z) − b1 (z))(f (z) − b2 (z))(f (z) − b3 (z)), where b1 (z), b2 (z), b3 (z) are distinct nonzero rational functions. By Lemma 8.2.2, f (z) − bj (z), j = 1, 2, 3, must have infinitely many zeros. Let pj denote the multiplicity of a generic zero of f (z) − bj (z) at z = z,̂ j = 1, 2, 3. Consider cases 1, 2, 3 with λ = 1 for f (z + 1) and f (z − 1) at z.̂ Shifting equation (8.57) forward, we get f 󸀠 (z + 1) f (z + 1) P(z + 1, f (z + 1)) = . (f (z + 1) − b1 (z + 1))(f (z + 1) − b2 (z + 1))(f (z + 1) − b3 (z + 1)) f (z + 2)f (z) + a(z + 1)

(8.65)

In case 1, the right-hand side of (8.65) takes a zero or a finite value at z = z,̂ depending on degf (P(z, f )). Thus f (z + 2) has a simple pole at z = z.̂ Using the same reasoning for the backward shift, we see that f (z − 2) has a simple pole at z = z.̂ Thus we obtain n (r,

pj 1 )≤ n(r + 2, f ) + O(1). f − bj pj + 2

(8.66)

In case 2, f (z + 2) must have a simple pole at z = z.̂ Thus n (r,

pj 1 )≤ n(r + 2, f ) + O(1). f − bj pj + 1

(8.67)

In case 3, using the same reasoning for the backward shift, we see that f (z − 2) has a simple pole at z = z.̂ Thus n (r,

pj 1 )≤ n(r + 2, f ) + O(1). f − bj pj + 1

8.2 Malmquist-type delay-differential equations | 189

Hence, no matter on which of cases 1, 2, 3 it appears, we have n (r,

1 1 1 ) + n (r, ) + n (r, ) ≤ 3τn(r + 2, f ) + O(1) f − b1 f − b2 f − b3

for some τ ∈ (0, 1). Lemma 8.2.3 now implies that ρ2 (f ) ≥ 1. Case II: Suppose now that Q(z, f ) = (f (z)−b1 (z))(f (z)−b2 (z)), where b1 (z) and b2 (z) are distinct nonzero rational functions. Lemma 8.2.2 again shows that f (z) − bj (z), j = 1, 2, have infinitely many zeros. Let pj denote the multiplicity of the zero of f (z) − bj (z), j = 1, 2, at a generic point z = z.̂ Cases 1, 2, 3 may again appear for f (z + 1) and f (z − 1) at z.̂ Shifting equation (8.57) forward, we obtain f (z + 2)f (z) + a(z + 1)

f 󸀠 (z + 1) P(z + 1, f (z + 1)) = . f (z + 1) (f (z + 1) − b1 (z + 1))(f (z + 1) − b2 (z + 1))

(8.68)

If degf (P(z, f )) = 3, then we then obtain degf (P) = degf (Q) + 1. Assume that degf (P(z, f )) ≤ 2. In case 1, we see that the right-hand side of (8.68) takes a zero or a finite value at z,̂ and thus (8.66) is also valid in this case. In case 2 or 3, (8.67) holds. Thus we have n (r,

1 1 ) + n (r, ) ≤ 2τn(r + 2, f ) + O(1) f − b1 f − b2

for some τ ∈ (0, 1). By Lemma 8.2.3, ρ2 (f ) ≥ 1. Case III: In this final case, we have Q(z, f ) = f (z) − b1 (z), where b1 (z) is a nonzero rational function. Let p denote the multiplicity of the zero of f (z) − b1 (z) at a generic point z = z.̂ Then cases 1, 2, 3 may again appear for f (z + 1) and f (z − 1) at z.̂ Shifting equation (8.57) forward, we have f (z + 2)f (z) + a(z + 1)

P(z + 1, f (z + 1)) f 󸀠 (z + 1) = . f (z + 1) f (z + 1) − b1 (z + 1)

(8.69)

If degf (P(z, f )) = 2, then degf (P) = degf (Q) + 1. We proceed to discuss the case degf (P(z, f )) = 3. In case 1, the right-hand side of (8.69) has a pole at ẑ with multiplicity 2k, and thus ẑ + 2 is a pole of f (z) of multiplicity 2k. Furthermore, ẑ − 2 is a pole of f (z) of multiplicity 2(p − k). Thus ẑ + 1, ẑ + 2, ẑ − 1, ẑ − 2 are poles of f (z) of multiplicities k, 2k, p − k, 2(p − k), respectively. Thus we have n (r,

1 1 ) ≤ n(r + 2, f ) + O(1). f − b1 3

In cases 2 and 3, we also get n (r,

1 p )≤ n(r + 2, f ) + O(1). f − b1 3k

Lemma 8.2.3 now implies that ρ2 (f ) ≥ 1. If degf (P(z, f )) = 1, then the degree of R(z, f ) as a rational function in f is one. Thus the proof of Theorem 8.2.24 is completed.

190 | 8 Malmquist-type delay-differential equations Remark 8.2.26. Recall the assumption in Theorem 8.2.24 that Q(z, f ) is a polynomial in f (z) with the roots that are nonzero rational functions of z and not the roots of P(z, f ). If Q(z, 0) ≡ 0, then the claim of Theorem 8.2.24 may fail. Examples 8.2.27 and 8.2.28 show that inequality (8.56) in Theorem 8.2.24 may hold. However, in Example 8.2.29, inequality (8.56) in Theorem 8.2.24 fails. Example 8.2.27. The meromorphic function f (z) = c1 tan π2 z, c ≠ 0, with ρ2 (f ) = 0 solves the equation 1 3 f (z) c13

2 f 󸀠 (z) f (z + 1)f (z − 1) + = π f (z)

+

1 f (z) c1

+ c13

f (z)2

,

where degf (P) = degf (Q) + 1 = 3. Example 8.2.28. The meromorphic function f (z) = the equation

f (z + 1)f (z − 1) +

π

e 2 iz −α π

e 2 iz +α

, α ≠ 0, with ρ2 (f ) = 0 solves

1 f 󸀠 (z) −f 2 (z) + 4f (z) + 1 = , iπ f (z) 4f (z)

where degf (P) = degf (Q) + 1 = 2. Example 8.2.29. The entire function f (z) = αez + β, α, β ≠ 0, with ρ2 (f ) = 0 solves f (z + 1)f (z − 1) + =

f (z)3 + (eβ +

β e

f 󸀠 (z) f (z)

− 2β)f (z)2 − (eβ2 + f (z)

β2 e

− 3β2 )f (z) − β3

.

In this case, degf (P) = 3 > degf (Q) + 1. For the case of Q(z, f ) = f μ , μ ≥ 2, we have the following theorem. Note that by Example 8.2.27 the condition degf (P(z, f )) ≠ 3 in this result cannot be removed. Theorem 8.2.30. Let f be a nonrational meromorphic solution to the equation f (z + 1)f (z − 1) + a(z)

f 󸀠 (z) P(z, f (z)) = , f (z) f (z)μ

(8.70)

where μ ≥ 2, degf (P(z, f )) ≠ 3, a(z) is a rational function in z, and P(z, f ) is a polynomial in f and rational in z. Then ρ2 (f ) ≥ 1. Proof. While considering equation (8.70) instead of (8.57), Lemma 8.2.2 again shows that f (z) has infinitely many zeros. A generic zero ẑ of f (z) of multiplicity p is a pole of f (z + 1)f (z − 1) of multiplicity μp (3p ≥ μp ≥ 2p). Thus ẑ is a pole of f (z + 1) or f (z − 1) of multiplicity at least p. Shifting equation (8.70) forward or backward, we obtain that ẑ is a pole of f (z + 2) or of f (z − 2) of multiplicity at least p, except the case

8.2 Malmquist-type delay-differential equations | 191

of degf (P(z, f )) = 3 (there is a possibility that p = 1 and degf (P(z, f )) = 3; in this case, ẑ may not be a pole of f (z + 2) or of f (z − 2)). From the assumptions we have 1 1 n(r, ) < n(r + 2, f ) + O(1). f 2 Lemma 8.2.3 now implies that ρ2 (f ) ≥ 1. For the case where Q(z, f ) is a constant, we have the following theorem. Theorem 8.2.31. Let f be a nonrational meromorphic solution of the equation f (z + 1)f (z − 1) + a(z)

f 󸀠 (z) = P(z, f (z)), f (z)

(8.71)

where a(z) is rational in z, and P(z, f ) is a polynomial in f and rational in z. If ρ2 (f ) < 1, then degf (P(z, f )) ≤ 2. Proof. Contrary to the statement, suppose that degf (P) = 3. Assume first that f (z) has either infinitely many poles, or infinitely many zeros, or both. Consider a generic pole of f (z) of multiplicity p at z = z.̂ Then ẑ is a pole of f (z + 1)f (z − 1) of multiplicity 3p by (8.71). Case A: Suppose that ẑ + 1 is a pole of f (z) of multiplicity k (3p > k > 0) and ẑ − 1 is a pole of f (z) of multiplicity 3p − k. Shifting (8.71) forward and backward, respectively, i. e., f (z + 2)f (z) + a(z + 1)

f 󸀠 (z + 1) = P(z + 1, f (z + 1)), f (z + 1)

f (z)f (z − 2) + a(z − 1)

f 󸀠 (z − 1) = P(z − 1, f (z − 1)), f (z − 1)

and

we have that ẑ is a pole of f (z + 2)f (z) of multiplicity 3k and ẑ is a pole of f (z − 2)f (z) of multiplicity 3(3p − k). This implies that ẑ + 2 is a pole of f (z) of multiplicity 3k − p and ẑ − 2 is a pole of f (z) of multiplicity 32 p − 3k − p. Continuing the iteration, it follows that ẑ + 2 is a pole of f (z) of multiplicity 32 k − 3p − k, ẑ − 3 is a pole of f (z) of multiplicity 33 p − 32 k − 6p + k, ẑ + 4 is a pole of f (z) of multiplicity 33 k − 32 p − 6k + p, ẑ − 4 is a pole of f (z) of multiplicity 34 p − 33 k − 32 p + 6k, and so on. Note that we started with a generic pole. Therefore cancelation with the poles considered above cannot happen, in particular, if ẑ was chosen outside a sufficiently large disc, as all coefficients here are rational functions. Applying now induction to poles of f (z) for all d ∈ ℕ, it follows that ̂ w) ≥ 3d p. n(d + |z|,

(8.72)

192 | 8 Malmquist-type delay-differential equations Case B: Suppose that ẑ + 1 is a pole of f (z) of multiplicity k (k > 3p) and ẑ − 1 is a zero of f (z) of multiplicity k − 3p. Then ẑ + 2 is a pole of f (z) of multiplicity 3k − p, ẑ + 3 is a pole of f (z) of multiplicity 3(3k − p) − k, ẑ + 4 is a pole of f (z) of multiplicity 4(3k − p) − 4k, and so on. Thus (8.72) remains valid. Case C: Suppose that ẑ − 1 is a pole of f (z) of multiplicity k (k > 3p) and ẑ + 1 is a zero of f (z) of multiplicity k − 3p. Then ẑ − 2 is a pole of f (z) of multiplicity 3k − p, ẑ − 3 is a pole of f (z) of multiplicity 3(3k − p) − k, ẑ − 4 is a pole of f (z) of order 4(3k − p) − 4k, and so on. Thus (8.72) again holds. Suppose now that ẑ is a zero of f (z) of multiplicity p. Then we have that ẑ is a simple pole of f (z + 1)f (z − 1). Now ẑ + 1 is a pole of f (z), or ẑ − 1 is a pole of f (z). Without loss of generality, we may discuss the case where ẑ + 1 is a pole of f (z). By iterating (8.71) we see that f (z) has a pole of multiplicity at least 3 + p at z = ẑ + 2, a pole of multiplicity at least 32 + 3p − 1 at z = ẑ + 3, a pole of multiplicity at least 33 + 32 p − 6 − p at z = ẑ + 4, and so on. By induction, for all d ≥ 2 ∈ N, we obtain ̂ f ) ≥ 3d−1 + 3d−2 p. n(d + |z|,

(8.73)

Thus by (8.72) and (8.73) we may compute the hyperexponent of convergence of poles to obtain log log n(r, f ) 1 λ2 ( ) = lim sup f log r r→∞ ̂ f) log log n(d + |z|, ≥ lim sup ̂ log(d + | z|) d→∞ ≥ lim sup d→∞

log log 3d−1 = 1. ̂ log(d + |z|)

Therefore ρ2 (f ) ≥ λ2 ( f1 ) ≥ 1, contradicting with our assumption. Finally, we assume that f (z) has finitely many poles and zeros and ρ2 (f ) < 1. In this case, from (8.71) we get f (z)f (z + 1)f (z − 1) + a(z)f 󸀠 (z) = P(z, f (z))f (z).

(8.74)

We easily see that the total degree of the left-hand side of (8.74) in f and its shifts is 3. If degf (P) = 3, then applying Lemma 1.2.14 to (8.74), we get that m(r, f ) = S(r, f ). This leads to T(r, f ) = S(r, f ), a contradiction, completing the proof of Theorem 8.2.31. To complete the preceding theorem, we give two examples to show that equation (8.71) can admit meromorphic solutions when degf (P(z, f )) = 2:

8.2 Malmquist-type delay-differential equations | 193

Example 8.2.32. (1) The function f (z) = αzez is a nonrational entire solution with ρ2 (f ) = 0 to f (z + 1)f (z − 1) +

z f 󸀠 (z) z 2 − 1 2 f (z) + 1, = z + 1 f (z) z2

where α is a nonzero constant. (2) The function f (z) =

β ez

is a nonrational entire solution with ρ2 (f ) = 0 to

f (z + 1)f (z − 1) + a(z)

f 󸀠 (z) = f 2 (z) − a(z), f (z)

where a(z) is rational, and β is a nonzero constant. To complete this chapter, we add a recent result. We recall [226, Theorem 1.6]. The original result in [226] is slightly more general than our variant below. For the proof in [226], see p. 9–16 therein. Theorem 8.2.33. Let c1 , . . . , ck be distinct complex numbers, let b1 (z), . . . , bk (z) be rational functions, and let R(z, f ) := P(z, f )/Q(z, f ), where P(z, f ) and Q(z, f ) are polynomials in f with rational coefficients with no common factors. If f (z) is a meromorphic solution of hyperorder ρ2 (f ) < 1 to k

f 󸀠 (z) = f (z)(R(z, f (z)) + ∑ bj (z)f (z − cj )), j=1

(8.75)

then degf (R) ≤ k + 2. Moreover, if degf (Q) = 0, and hence R(z, f (z)) is a polynomial in f , then degf (R) ≤ 1. Proof. For the first statement, by elementary Nevanlinna theory we obtain T(r, R(z, f )) ≤ T (r, ≤ N (r,

k f 󸀠 (z) ) + ∑ T(r, f (z − cj )) + O(log r) f (z) j=1 k k f 󸀠 (z) ) + ∑ N(r, f (z − cj )) + ∑ m(r, f (z − cj )) + O(log r) f (z) j=1 j=1 k

≤ N(r, f ) + N(r, 1/f ) + ∑ N(r, f (z − cj )) + m(r, f ) + S(r, f ) j=1

≤ N(r, f ) + N(r, 1/f ) + kN(r, f ) + m(r, f ) + S(r, f ) ≤ (k + 2)T(r, f ) + S(r, f ). By the Valiron–Mohon’ko theorem we have T(r, R(z, f )) = (degf (R))T(r, f ) + S(r, f ).

194 | 8 Malmquist-type delay-differential equations Therefore (degf (R) − (k + 2))T(r, f ) ≤ S(r, f ), and hence degf (R) ≤ k + 2. Proceeding to the second statement, we may write (8.75) in the form f 󸀠 (z) k + ∑ b (z)f (z − cj ) = P(z, f (z)). f (z) j=1 j

(8.76)

Suppose, contrary to the statement, that degf (P) ≥ 2, and hence k + 2 ≥ degf (P) ≥ 2. Assume first that f has infinitely many poles. Let ẑ be such a pole of multiplicity t, and assume without loss of generality that ẑ is generic. Then, of course, ẑ is a simple pole 󸀠 (z) of ff (z) . By (8.76) ẑ is a pole of multiplicity t degf (P) ≥ 2t of P(z, f (z)). Then at least one of f (z − c1 ), . . . , f (z − ck ), say f (z − c1 ), has a pole of multiplicity ≥ t degf (P) ≥ 2t at z.̂ Shifting (8.76) by −c1 , we obtain f 󸀠 (z − c1 ) k + ∑ b (z − c1 )f (z − cj − c1 ) = P(z − c1 , f (z − c1 )). f (z − c1 ) j=1 j

(8.77)

Therefore ẑ is pole of multiplicity t(degf (P))2 ≥ 22 t of P(z − c1 , f (z − c1 )). Hence we find at least one cj1 , j1 ∈ {1, . . . , k} such that ẑ is a pole of multiplicity ≥ 22 t of f (z−cj1 −c1 ). Shifting by −c1 d times, we see that ẑ is a pole of multiplicity ≥ 2d t of f (z − cj1 − cj2 − ⋅ ⋅ ⋅ − cjd ), where j1 , . . . , jd ∈ {1, . . . , k}. This means that n(|z|̂ + d(|c1 | + ⋅ ⋅ ⋅ + |ck |), f ) ≥ 2d t for all positive integers d. Estimating now the hyperexponent of convergence of λ2 (1/f ), we get log log n(r, f ) log r r→∞ log log n(|z|̂ + d(|c1 | + ⋅ ⋅ ⋅ + |ck |) + O(1) ≥ lim sup log(|z|̂ + d(|c1 | + ⋅ ⋅ ⋅ + |ck |)) d→∞ log(d log 2 + log t + O(1)) = 1. ≥ lim sup d→∞ log(|z|̂ + d(|c1 | + ⋅ ⋅ ⋅ + |ck |))

λ2 (1/f ) ≥ lim sup

Therefore ρ2 (f ) ≥ λ2 (1/f ) ≥ 1, a contradiction. It remains to consider the case where f has finitely many poles only. Writing f (z) = h(z)eg(z) by the Hadamard factorization we have N(r, h) = O(log r) and T(r, eg ) = T(r, f ) + O(log r). Substituting into (8.75), we get k

(h󸀠 (z)/h(z) + g 󸀠 (z)) + (∑ bj (z)eg(z−cj )−g(z) ) eg(z) = P(z, h(z)eg(z) ). j=1

Recalling the Valiron–Mohon’ko theorem again, we immediately see that degf (P)T(r, f ) = T(r, P(z, heg )) + S(r, f ) = T(r, f ) + S(r, f ), contradicting the assumption that degf (P) ≥ 2, completing the proof.

9 Nonlinear complex delay-differential equations In this chapter, we shortly treat nonlinear complex differential, difference, and delaydifferential equations based on trigonometric formulas. We will also treat meromorphic solutions of more general complex delay-differential equations.

9.1 Background We begin this short section by recalling the trigonometric formula sin 3z = 3 sin z − 4 sin3 z. Thus f (z) = sin z solves the nonlinear differential equation 4f 3 + 3f 󸀠󸀠 = − sin 3z.

(9.1)

Heittokangas et al. [96] pointed out that f (z) = − 23 cos z − 21 sin z also solves (9.1). Furthermore, Li and Yang [236, Theorem 4] showed that (9.1) admits exactly three dis√ tinct transcendental entire solutions: f1 (z) = sin z, f2 (z) = − 23 cos z − 21 sin z, and √

f3 (z) = 23 cos z − 21 sin z. Some results on transcendental meromorphic solutions to certain nonlinear differential equations can be found in [139]. √ Note that the transcendental entire functions f1 (z) = sin πz, f2 (z) = 23 cos πz − √

1 2

sin πz, and f3 (z) = −

√3 2

cos πz − 21 sin πz also solve the nonlinear difference equation 4f (z)3 + 3f (z + 1) = − sin 3πz

(9.2)

and the delay-differential equation 4f (z)3 +

1 3 󸀠 f (z + ) = − sin 3πz. π 2

(9.3)

From this point of view, trigonometric functions may always solve some nonlinear differential, difference, or delay-differential equations.

9.2 Nonlinear complex delay-differential equations First, we recall a simple nonlinear difference type result by Yang and Laine [238, Theorem 2.5]. Theorem 9.2.1. A nonlinear difference equation f (z)3 + q(z)f (z + 1) = c sin bz, https://doi.org/10.1515/9783110560565-009

(9.4)

196 | 9 Nonlinear complex delay-differential equations where q(z) is a nonconstant polynomial, and b, c ∈ ℂ are nonzero constants, does not admit entire solutions of finite order. If q(z) = q is a nonzero constant, then (9.4) possesses c2 three distinct entire solutions of finite order, provided that b = 3πn and q3 = (−1)n+1 27 4 for a nonzero integer n. We now proceed to considering some basic nonlinear complex delay-differential equations. As our first example, we recall Theorem 1.2 in [35]. Theorem 9.2.2. Suppose p1 , p2 , λ, c are nonzero constants, k ≥ 0 is an integer, and a(z) is a nonvanishing polynomial. If f (z) is a transcendental entire solution of finite order to equation f (z)2 + a(z)f (k) (z + c) = p1 eλz + p2 e−λz , then a(z) reduces to a constant. Moreover: (i) If k is odd, then f (z) = ± 2i a(λ/2)k + c1 eλz/2 + c2 e−λz/2 , eλc = −1. (ii) If k is even, then f (z) = ± 2i a(λ/2)k + c1 eλz/2 + c2 e−λz/2 , eλc = 1. 1 4 In cases (i) and (ii), c1 , c2 are constants, 64 a (λ/2)4k = p1 p2 , and cj2 = pj , j = 1, 2.

(iii) If k = 0, then f (z) = ± 21 a + c1 eλz/2 + c2 e−λz/2 , eλc = 1. In this case, c1 , c2 are constants 1 4 9 4 again, 64 a = p1 p2 or 64 a = p1 p2 , and cj2 = pj , j = 1, 2. We omit the proof of this result; see [35, p. 144–146]. Next, we proceed to a more general theorem [196, Theorem 7]. Theorem 9.2.3. Let α1 ≠ α2 , p1 , p2 , and c ≠ 0 be constants. Suppose that k ≥ 0 and n ≥ 2 are integers and q(z) is a nonvanishing polynomial. If f (z) is a transcendental entire solution of hyperorder ρ2 (f ) < 1 to f (z)n + q(z)f (k) (z + c) = p1 eα1 z + p2 eα2 z ,

(9.5)

then q(z) reduces to a constant q, and ρ(f ) = 1. Moreover, the following possibilities may appear: (1) f (z) = c1 eα1 z/n , qc1 (α1 /n)k eα1 c/n = p2 , α1 = nα2 , c1n = p1 ; (2) f (z) = c2 eα2 z/n , qc2 (α2 /n)k eα2 c/n = p1 , α2 = nα1 , c2n = p2 . (3) If n = 2, then T(r, f ) ≤ N1) (r, 1/f )+T(r, φ)+S(r, f ), where N1) (r, 1/f ) counts the simple zeros of f , and φ = α1 α2 f 2 − n(α1 + α2 )ff 󸀠 + n(n − 1)(f 󸀠 )2 + nff 󸀠󸀠 , and if n = 3, then T(r, f ) ≤ N1) (r, 1/f ) + S(r, f ). Proof. We first show that ρ(f ) = 1. Using elementary Nevanlinna theory and the delaydifferential version of the logarithmic derivative lemma (Lemma 1.2.8), we first observe that T(r, p1 eα1 z + p2 eα2 z ) = T(r, f (z)n + q(z)f (k) (z + c))

≤ T(r, f n ) + T(r, f (k) (z + c)) + O(log r)

9.2 Nonlinear complex delay-differential equations | 197

f (k) (z + c) ) + m(r, f ) + O(log r) f (z) ≤ (n + 1)T(r, f ) + S(r, f ). ≤ T(r, f n ) + m (r,

On the other hand, T(r, p1 eα1 z + p2 eα2 z ) = T(r, f (z)n + q(z)f (k) (z + c))

≥ T(r, f n ) − T(r, f (k) (z + c)) + O(log r)

f (k) (z + c) ) − m(r, f ) + O(log r) f (z) ≥ (n − 1)T(r, f ) + S(r, f ). ≥ T(r, f n ) − m (r,

Hence we obtain (n − 1)T(r, f ) + S(r, f ) ≤ T(r, p1 eα1 z + p2 eα2 z ) ≤ (n + 1)T(r, f ) + S(r, f ). This readily implies that ρ(f ) = 1. Now denote g(z) := q(z)f (k) (z + c) for short and differentiate (9.5) to obtain nf n−1 f 󸀠 + g 󸀠 = α1 p1 eα1 z + α2 p2 eα2 z .

(9.6)

Eliminating eα2 z from this and (9.5), we get α2 f n − nf n−1 f 󸀠 + α2 g − g 󸀠 = (α2 − α1 )p1 eα1 z .

(9.7)

Differentiating this results in nα2 f n−1 f 󸀠 − n(n − 1)f n−2 (f 󸀠 )2 − nf n−1 f 󸀠󸀠 + α2 g 󸀠 − g 󸀠󸀠 = α1 (α2 − α1 )p1 eα1 z . By (9.7) and (9.8) we now get f n−2 φ = Q, where φ := α1 α2 f 2 − n(α1 + α2 )ff 󸀠 + n(n − 1)(f 󸀠 )2 + nff 󸀠󸀠 and Q := −α1 α2 g + (α1 + α2 )g 󸀠 − g 󸀠󸀠 . Note that Q is a delay-differential polynomial of f of degree one. Consider next the case where φ vanishes. Using the definition of φ, we get 0 = α1 α2 − n(α1 + α2 ) (

2

f󸀠 f󸀠 f 󸀠󸀠 ) + n(n − 1) ( ) + n ( ) f f f

(9.8)

198 | 9 Nonlinear complex delay-differential equations

= α1 α2 − n(α1 + α2 ) (

2

󸀠

f󸀠 f󸀠 f󸀠 ) + n2 ( ) + n ( ) . f f f

Recalling that f is entire of order ρ(f ) = 1 and n ≥ 2, we easily to see that f has no zeros. Therefore f 󸀠 /f is a constant, meaning that we have either f 󸀠 /f = α1 /n or f 󸀠 /f = α2 /n, and hence f (z) = c1 eα1 z/n or f (z) = c2 eα2 z/n . Looking at f (z) = c1 eα1 z/n , we first observe that c1n = p1 . Substituting this into (9.5), we have q(z)c1 (

α1 k α1 c/n α1 z/n ) e e = p2 eα2 z . n

Therefore α1 = nα2 , and q(z)c1 ( αn1 )k eα1 c/n = p2 . As we can see, q(z) has no zeros, implying that q(z) is constant, say q. This means that with f (z) = c1 eα1 z/n we obtain claim (1). Completely similarly, f (z) = c2 eα2 z/n results in claim (2). We now proceed to looking at the case where φ is nonvanishing. We first observe that a contradiction follows for n ≥ 4. Indeed, recalling that Q is a delay-differential polynomial of degree one in f , we may apply the Clunie lemma-type reasoning to conclude from f n−2 φ = Q that m(r, φ) = T(r, φ) = S(r, f ). Similarly, writing f n−3 (fφ) = Q, we see that m(r, fφ) = S(r, f ). Therefore T(r, f ) = m(r, f ) ≤ m(r, 1/φ) + m(r, fφ) = S(r, f ), a contradiction. Suppose now n = 3 and φ is nonvanishing. We again have T(r, φ) = m(r, φ) = S(r, f ). Now from 󸀠

2

1 1 f󸀠 f󸀠 f 󸀠󸀠 = [α α − n(α + α ) ( ) + n(n − 1) ( ) + n ( )] 1 2 1 2 f f f f2 φ we immediately obtain that m(r, 1/f ) = S(r, f ). Moreover, since each multiple zero of f is a zero of φ as well, we see that N(2 (r, 1/f ) = N(r, 1/φ) = S(r, f ). Therefore by the first fundamental theorem T(r, f ) ≤ N1) (r, 1/f ) + S(r, f ). Finally, suppose that n = 2 and φ is nonvanishing. Similarly as in the case n = 3, we get m(r, 1/f ) = m(r, 1/φ) + S(r, f ) and N(2 (r, 1/f ) = N(r, 1/φ) + S(r, f ). Therefore the claim for the case n = 2 readily follows. We next proceed to considering the case n ≥ 4 in a more general situation; see Yang and Laine [238, Theorem 2.6]. Theorem 9.2.4. Let n ≥ 4 be an integer, let M(z, f ) be a nonzero linear delay-differential polynomial of f with small coefficients with respect to f , and let h be a meromorphic function of finite order. Then the delay-differential equation f n + M(z, f ) = h

(9.9)

9.2 Nonlinear complex delay-differential equations | 199

possesses at most one admissible transcendental entire solution of finite order. If such a solution f exists, then f is of the same order as h. Proof. We first observe that ρ(h) = ρ(f ) for all entire solutions of finite order of (9.9). Since the inequality ρ(h) ≤ ρ(f ) trivially holds, suppose for a while that ρ(h) < σ < ρ(f ) := ρ and write (9.9) in the form f n−1 =

h M(z, f ) − . f f

(9.10)

By the delay-differential version of logarithmic derivatives lemma (Lemma 1.2.9), since M(z, f ) is a nonzero linear delay-differential polynomial, we conclude that (n − 1)T(r, f ) = (n − 1)m(r, f ) = m(r, f n−1 )

≤ T(r, h) + T(r, f ) + O(r ρ−1+ε ) + S(r, f )

≤ T(r, f ) + O(r σ ) + O(r ρ−1+ε ) + S(r, f )

for all r sufficiently large, outside an exceptional set of finite logarithmic measure. For ε chosen small enough, removing the exceptional set by standard reasoning, we obtain ρ(f ) ≤ max{ρ − 1 + 2ε, σ + ε} < ρ,

(9.11)

a contradiction. Assume now, contrary to the statement, that f , g are two distinct finite-order transcendental entire solutions of (9.9) and write f n + M(z, f ) = g n + M(z, g).

(9.12)

Clearly, ρ(f ) = ρ(g). From (9.12) we obtain f n − g n = M(z, g) − M(z, f ) = M(z, g − f ).

(9.13)

Let F :=

f n − g n n−1 M(z, f − g) = ∏(f − ηj g) = − . f −g f −g j=1

(9.14)

Thus F is an entire function, and ηj (≠ 1) (j = 1, . . . , n − 1) are the distinct roots of z n = 1. From this and from the delay-differential version of the logarithmic derivative lemma (Lemma 1.2.9) we conclude that T(r, F) = m(r, F) = m (r,

M(z, g − f ) ) f −g

= O(r ρ(f −g)−1+ε ) + S(r, f ) + S(r, g)

200 | 9 Nonlinear complex delay-differential equations ≤ O(r ρ(f )−1+ε ) + S(r, f ) =: Sρ (r, f ). Here ε > 0 is arbitrary and sufficiently small. An immediate observation results in n−1

∑ N (r, j=1

1 1 ) = N (r, ) = Sρ (r, f ), f − ηj g F

and therefore 1 ) = Sρ (r, f ) f − ηj g

N (r, for all j = 1, 2, . . . , n − 1. Since f g

1

− ηj

=g

1 , f − ηj g

we conclude that N (r,

f g

1

− ηj

) = Sρ (r, f )

for all j = 1, 2, . . . , n − 1. Assuming now that n ≥ 4, the second main theorem implies for ψ := gf that f T(r, ψ) = T (r, ) = Sρ (r, f ) g and T(r, f ) = T(r, g) + Sρ (r, f ). Using (9.14), we infer that n−1

n−1

j=1

j=1

F = ∏(f − ηj g) = g n−1 ∏(ψ − ηj ). If ψ is not identically equal to ηj (j = 1, 2, . . . , n − 1), then we see that (n − 1)T(r, f ) = (n − 1)T(r, g) + Sρ (r, f ) n−1

≤ T(r, F) + T (r, ∏(ψ − ηj )−1 ) + Sρ (r, f ) = Sρ (r, f ),

j=1

which is a contradiction. Therefore ψ = ηj for some j = 1, 2, . . . , n − 1. Then f = ηj g, f n = g n , and M(r, f ) = M(r, g). By the linearity of the delay-differential polynomial M we obtain M(r, f ) = ηj M(r, g). Since ηj ≠ 1, a contradiction again follows.

9.2 Nonlinear complex delay-differential equations | 201

Remark 9.2.5. (1) Qi and Yang [186, Theorem 1.2] considered the meromorphic case with few poles in Theorem 9.2.4 by a similar reasoning. (2) It remains open to improve f (z)n to a difference polynomial such as ∏nj=1 f (z + cj ) or to improve the linear delay-differential polynomial M to more general delaydifferential polynomials in Theorem 9.2.4. Yang and Laine proposed the following conjecture [238, Conjecture 2]. Conjecture. Let f be an entire function of infinite order, and let n ≥ 2 be an integer. Then the delay-differential polynomial of the form f n + Pn−1 (z, f ) cannot be a nonconstant entire function of finite order if Pn−1 (z, f ) is a delay-differential polynomial in f of total degree at most n − 1 in f , its derivatives, and its shifts, with entire functions of finite order as coefficients. Moreover, all terms of Pn−1 (z, f ) are supposed to have the total degree ≥ 1. Li and Yang [131, Theorem 1.7] showed that this conjecture holds for entire functions of infinite order such that ρ2 (f ) < 1: Theorem 9.2.6. Let f be an entire function of infinite order such that ρ2 (f ) < 1, and let n ≥ 2 be an integer. Then the delay-differential polynomial f n + Pn−1 (z, f ) cannot be a nonconstant entire function of finite order if Pn−1 (z, f ) is defined as in the conjecture. Proof. Set n−1

Pn−1 (z, f ) = ∑ Qj (z, f ), j=1

(9.15)

where Qj (z, f ) is a homogeneous delay-differential polynomial of degree j in f , with entire functions aλ of finite order as coefficients. Let f be an entire function of infinite order with ρ2 (f ) < 1. Suppose that the delay-differential polynomial of the form f n + Pn−1 (z, f ) is a nonconstant entire function of finite order h(z), that is, f n + Pn−1 (z, f ) = h(z).

(9.16)

We now immediately obtain 󵄨󵄨 󵄨󵄨 h(z) − Pn−1 (z, f ) 󵄨󵄨 󵄨󵄨 h(z) 󵄨󵄨 󵄨󵄨󵄨󵄨 ∑n−1 󵄨󵄨 ≤ 󵄨󵄨 󵄨󵄨 + 󵄨󵄨 j=1 Qj (z, f ) 󵄨󵄨󵄨󵄨 |f | = 󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 n−1 󵄨󵄨 󵄨󵄨 󵄨󵄨 f n−1 f n−1 󵄨 󵄨 f 󵄨 󵄨󵄨 󵄨 󵄨󵄨 n−1 󵄨󵄨 h(z) 󵄨󵄨 󵄨󵄨󵄨 Qj (z, f ) 󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨󵄨 n−1 󵄨󵄨󵄨󵄨 + ∑ 󵄨󵄨󵄨 n−1 󵄨󵄨󵄨 . 󵄨 f 󵄨 j=1 󵄨󵄨 f 󵄨󵄨

(9.17)

202 | 9 Nonlinear complex delay-differential equations Define E1 := {θ ∈ [0, 2π) : |f (reiθ )| > 1} and E2 := {θ ∈ [0, 2π) : |f (reiθ )| ≤ 1}. Then, for θ ∈ E1 , n−1 󵄨󵄨󵄨 Q (reiθ , f ) 󵄨󵄨󵄨 n−1 󵄨󵄨󵄨 Q (reiθ , f ) 󵄨󵄨󵄨 󵄨 j 󵄨 j 󵄨 󵄨󵄨 |f | ≤ |h(reiθ )| + ∑ 󵄨󵄨󵄨 j n−1−j 󵄨󵄨󵄨 ≤ |h(reiθ )| + ∑ 󵄨󵄨󵄨 󵄨󵄨 . j 󵄨󵄨 󵄨󵄨 f ⋅ f 󵄨󵄨 󵄨󵄨 f j=1 󵄨 j=1 󵄨 󵄨 󵄨

(9.18)

Thus 󵄨󵄨 󵄨 n−1 󵄨󵄨 Qj (reiθ , f ) 󵄨󵄨󵄨 󵄨󵄨 + log n, log+ |f | ≤ log+ |h(reiθ )| + ∑ log+ 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 fj 󵄨󵄨 j=1 󵄨

θ ∈ E1 .

(9.19)

1 Trivially, 2π ∫E log+ |f (reiθ )|dθ = 0. Therefore by the delay-differential version of the 2 logarithmic derivative lemma we conclude that

T(r, f ) = m(r, f ) =

1 ∫ log+ |f |dθ 2π E1

≤

󵄨󵄨 Qj (z, f ) 󵄨󵄨 1 1 n−1 󵄨 󵄨󵄨 ∑ ∫ log+ 󵄨󵄨󵄨 ∫ log+ |h(z)|dθ + 󵄨 dθ + O(1) 󵄨󵄨 f j 󵄨󵄨󵄨 2π 2π j=1 E1

E1

n−1

≤ m(r, h(z)) + ∑ m (r, j=1

Qj (z, f ) fj

) + O(1)

≤ m(r, h(z)) + ∑ m(r, aλ ) + S(r, f )

≤ T(r, h(z)) + ∑ T(r, aλ ) + S(r, f )

(9.20)

for all r sufficiently large outside an exceptional set of finite logarithmic measure. Removing the exceptional set by standard reasoning, we obtain ρ(f ) ≤ max{ρ(h), ρ(aλ )}.

(9.21)

This is a contradiction since h(z) and all aλ are finite-order entire functions and f (z) is an infinite-order entire function. Remark 9.2.7. (1) If n = 2 and ρ2 (f ) = 1, then conjecture does not hold in general. The following two examples show that f n + Pn−1 (z, f ) may be a nonzero constant or a nonconstant entire function: z (a) Taking f (z) = 1ez + ee of hyperorder ρ2 (f ) = 1 and ec = −2, we have e

f (z)2 −

1 󸀠 f (z + c) = 2. 2ez

z

(b) Similarly, let f (z) = ee + ez of hyperorder ρ2 (f ) = 1, and let ec = 2. Then f (z)2 − (f (z + c) + 2f 󸀠 (z)) = e2z − 4ez .

9.2 Nonlinear complex delay-differential equations | 203

(2) Zhang, Gao and Zhang [250, Example 5.2] gave an example showing that the conjecture does not hold in general when f (z) is of hyperorder ρ2 (f ) > 1. To this end, consider f (z) = eg(z) + g 󸀠 (z), where g(z) = Π(z)ez log 2 , and Π(z) is a periodic function with period 1 such that 1 < ρ(Π) < ∞, and thus g is of finite order. Then ρ(Π(z)ez log 2 ) > 1 and g(z + 1) = 2g(z), and thus ρ2 (f ) > 1. Now f (z) is of infinite order and satisfies equation f (z)2 − (f (z + 1) + 2f 󸀠 (z)) = h(z) := (g 󸀠 (z))2 − 2g 󸀠 (z) − 2g 󸀠󸀠 (z). (3) Zhang, Gao, and Zhang [250, Theorem 1.6] also showed that if Pd (z, f ) is a difference polynomial of degree at most n − 1, n ≥ 2, then all meromorphic solutions to f n + Pd (z, f ) = 0 are of infinite order. Recalling Theorem 9.2.2, we proceed to considering situations where the righthand side of the delay-differential equation has more exponential function terms, such as β1 eα1 z + β2 eα2 z + ⋅ ⋅ ⋅ + βs eαs z , where β1 , β2 , . . . , βs , α1 , α2 , . . . , αs are nonzero constants. Recall that the basic computations in Theorem 9.2.2 eliminate eα1 z , resp., eα2 z , from the right-hand by differentiating both sides of the delay-differential equation. However, the above replacement makes more complicated computations. Using linear algebra, Zhang and Huang [247, Theorem 1.1] obtained a difference result under the following assumption: (A). Fix c ∈ ℂ\{0} and positive integers n, s with n ≥ s + 2. Let β1 , β2 , . . . , βs be α nonzero constants, and let α1 , α2 , . . . , αs be distinct nonzero constants satisfying αi ≠ j

n for all i, j ∈ {1, 2, . . . , s}. If s ≥ 5, then assume that nαl (5 ≤ l ≤ s) are not linear combinations of α1 , . . . , αs with the weight n over {0, 1, . . . , n − 1}, that is, s

̂ , α⟩ = ∑ mj αj , nαl ≠ ⟨m j=1

l = 5, . . . , s,

̂ = (m1 , m2 , . . . , ms ) ∈ {0, 1, . . . , n − 1}s , |m ̂ | = m1 + ⋅ ⋅ ⋅ + ms = n, and α = where m (α1 , α2 , . . . , αs ). Theorem 9.2.8. Let p(z) be a nonzero polynomial, and let assumption (A) be satisfied. Then all meromorphic solutions f (z) to the difference equation f (z)n + p(z)f (z + c) = β1 eα1 z + β2 eα2 z + ⋅ ⋅ ⋅ + βs eαs z

(9.22)

satisfy ρ2 (f ) ≥ 1. Using a similar method as in [247], Wang, Zhan, and Hu [213, Theorem 1.1] (see also [100, Theorem 2.1]), treated delay-differential equations of the form p

s

i=1

l=1

∑ f (z)ni f (k) (z) + p(z)f (z + c) = ∑ βl eαl z

(9.23)

204 | 9 Nonlinear complex delay-differential equations under the following assumption: (B) Fix c ∈ ℂ\{0} and positive integers n1 , . . . , np , s, k with np > np−1 > ⋅ ⋅ ⋅ > n1 ≥ s + 2. Let β1 , β2 , . . . , βs be nonzero distinct constants as well as α1 , α2 , . . . , αs . For fixed i ∈ {1, . . . , p}, suppose that (ni + 1)αj ≠ αl for all j, l ∈ {1, 2, . . . , s} and (ni + 1)αj ≠ (nt + 1)αk for all 1 ≤ t(≠ i) ≤ p, 1 ≤ k(≠ j) ≤ s. If s ≥ 5, then assume that (ni + 1)αl , 5 ≤ l ≤ s, are not linear combinations of α1 , . . . , αs with the weight ni + 1 over {0, 1, . . . , ni }, that is, s

(ni + 1)αl ≠ ⟨mi , α⟩ = ∑ mi,j αj , j=1

l = 5, . . . , s,

where mi = (mi,1 , mi,2 , . . . , mi,s ) ∈ {0, 1, . . . , ni }s , |mi | = mi,1 + ⋅ ⋅ ⋅ + mi,s = ni + 1, and α = (α1 , α2 , . . . , αs ). In fact, Wang, Zhan, and Hu [213, Theorem 1.1] obtained the following result. Theorem 9.2.9. Let p(z) be a nonzero polynomial, and let assumption (B) be satisfied. Then all meromorphic solutions f of the delay-differential equation (9.23) satisfy ρ2 (f ) ≥ 1. Remark 9.2.10. (1) If n1 < s + 1, then Theorem 9.2.9 is not true. The delay-differential equation π f (z)4 f 󸀠 (z) − 2f (z + ) = ie5iz + 3ie3iz − 3ie−3iz − ie−5iz 2 has an entire solution f (z) = eiz + e−iz of hyperorder ρ2 (f ) = 0. In this case, we have 4 = n1 < s + 1 = 5. (2) The delay-differential equation (f 6 + f 5 )f 󸀠 − f (z + π/2) = −ieiz + ie6iz + ie7iz has a transcendental entire solution f (z) = eiz such that ρ2 (f ) = 0. This example shows that the condition “for a fixed i ∈ {1, . . . , p}, (ni + 1)αj ≠ αl for all j, l ∈ {1, 2, . . . , s} and (ni + 1)αj ≠ (nt + 1)αk for all 1 ≤ t(≠ i) ≤ p, 1 ≤ k(≠ j) ≤ s” is necessary in Theorem 9.2.9. To prove Theorem 9.2.9, we first need the following technical lemma [213, Lemma 2.2] related to the sums of exponential polynomials. In fact, it may be understood as a Borel-type lemma; see [16] on the sum of transcendental entire functions: Lemma 9.2.11. Let n1 , . . . , np , k, and s be positive integers such that np > np−1 > ⋅ ⋅ ⋅ > n1 ≥ s + 2, let cl , 1 ≤ l ≤ s, be constants, and let bj (z), 1 ≤ j ≤ 4, be rational functions. Suppose that for fixed i ∈ {1, . . . , p}, there exist distinct nonzero constants αl , 1 ≤ l ≤ s, such that (ni + 1)αj ≠ αl for all j, l ∈ {1, 2, . . . , s}, (ni + 1)αj ≠ (nt + 1)αk for all 1 ≤ t(≠ i) ≤ p, 1 ≤ k(≠ j) ≤ s, and p

4

i=1

j=1

ni 4

s

∑ (∑ bj (z)eαj z ) ∑ αjk bj (z)eαj z = ∑ cl eαl z . Then bj (z) ≡ 0, 1 ≤ j ≤ 4.

j=1

l=1

(9.24)

9.2 Nonlinear complex delay-differential equations | 205

Proof. By an elementary computation from (9.24) we have s

p

4

⟨mi ,̃ α⟩z ] , ∑ cl eαl z = ∑ [∑ αjk bj (z)ni +1 e(ni +1)αj z + ∑ cm ̃i (z)e ̃i |=ni +1 i=1 [ j=1 l=1 |m ] ̃

(9.25)

̃ i := (mi,1 , mi,2 , mi,3 , mi,4 ) ∈ {0, 1, . . . , ni }4 , cm where m ̃i (z) are rational functions, and 4

̃i , α ̃ ⟩ := ∑ mi,j αj . ⟨m

(9.26)

j=1

We first claim that there exists j ∈ {1, 2, 3, 4} such that (ni + 1)αj is not a linear combination of α1 , . . . , α4 with the weight ni + 1 over {0, 1, . . . , ni }. Otherwise, for any k ∈ {1, 2, 3, 4}, there would exist nonnegative integers dkj ∈ {0, 1, . . . , ni } (j = 1, 2, 3, 4) with dk1 + dk2 + dk3 + dk4 = ni + 1 such that (ni + 1)α1 = d11 α1 + d12 α2 + d13 α3 + d14 α4 , { { { { { { (ni + 1)α2 = d21 α1 + d22 α2 + d23 α3 + d24 α4 , { { { (ni + 1)α3 = d31 α1 + d32 α2 + d33 α3 + d34 α4 , { { { {(ni + 1)α4 = d41 α1 + d42 α2 + d43 α3 + d44 α4 .

(9.27)

We proceed to deduce a contradiction from (9.27). If there exists k ∈ {1, 2, 3, 4} such that only one of three integers dkj (j ≠ k) is greater than zero, say d21 > 0, so that d2j = 0 (j ≠ 1, 2), then the second equation of (9.27) implies that (ni + 1 − d22 )α2 = d21 α1 . Since d21 + d22 = ni + 1 and d21 > 0, we get α1 = α2 , which is a contradiction. Hence, for each k ∈ {1, 2, 3, 4}, at least two of three integers dkj (j ≠ k) are > 0, and we have dkj < ni + 1 − dkk for j ≠ k. Rewrite now (9.27) as (d11 − ni − 1)α1 + d12 α2 + d13 α3 + d14 α4 { { { { { { d21 α1 + (d22 − ni − 1)α2 + d23 α3 + d24 α4 { { d31 α1 + d32 α2 + (d33 − ni − 1)α3 + d34 α4 { { { { {d41 α1 + d42 α2 + d43 α3 + (d44 − ni − 1)α4

= 0, = 0, = 0, = 0,

and consider the matrix of coefficients of system (9.28) d11 − ni − 1 d21 B=( d31 d41

d12 d22 − ni − 1 d32 d42

d13 d23 d33 − ni − 1 d43

d14 d24 ). d34 d44 − ni − 1

(9.28)

206 | 9 Nonlinear complex delay-differential equations We need to show that rank(B) = 3. Adding the columns 2, 3, and 4 to column 1 and noting that dk1 + dk2 + dk3 + dk4 = ni + 1, k = 1, 2, 3, 4, we obtain that det(B) = 0. Next, we discuss the minor determinants of order 3 in B by distinguishing two cases. Case 1. d13 = d23 = d43 = 0. In this case, we have d12 > 0, d14 > 0, d21 > 0, d24 > 0, d41 > 0, d42 > 0. For the 3 × 3 minor B1 = (

d11 − ni − 1 d31 d41

d12 d32 d42

d14 ), d34 d44 − ni − 1

we have det(B1 ) = (d11 − ni − 1)d32 (d44 − ni − 1) + d12 d34 d41 + d14 d31 d42

− d14 d32 d41 − d12 d31 (d44 − ni − 1) − (d11 − ni − 1)d34 d42 .

If d32 > 0, then since d14 < ni + 1 − d11 and d41 < ni + 1 − d44 , we have det(B1 ) > d12 d34 d41 + d14 d31 d42 + d12 d31 d41 + d14 d34 d42 ≥ 0. If d32 = 0, then we have d31 > 0 and d34 > 0, and thus det(B1 ) > d12 d34 d41 + d14 d31 d42 + d12 d31 d41 + d14 d34 d42 > 0. Thus we have proved det(B1 ) > 0 in this case. Case 2. At least one of d13 , d23 , d43 is > 0. Now we consider the 3 × 3 minor B2 = (

d11 − ni − 1 d21 d41

d12 d22 − ni − 1 d42

d13 d23 ) d43

of the matrix B. Then det(B2 ) = (d11 − ni − 1)(d22 − ni − 1)d43 + d12 d23 d41 + d13 d21 d42

− d13 (d22 − ni − 1)d41 − d12 d21 d43 − (d11 − ni − 1)d23 d42 .

Hence if d43 > 0, then it follows that det(B2 ) > d12 d23 d41 + d13 d21 d42 + d13 d21 d41 + d12 d23 d42 ≥ 0 since d12 < ni + 1 − d11 and d21 < ni + 1 − d22 . However, if d43 = 0, then d41 > 0 and d42 > 0, and therefore det(B2 ) > d12 d23 d41 + d13 d21 d42 + d13 d21 d41 + d12 d23 d42 > 0 since at least one of d13 , d23 is > 0. Therefore we have proved that rank(B) = 3.

9.2 Nonlinear complex delay-differential equations | 207

Using elementary row transformations, we can transform the matrix B into the form 1 0 D=( 0 0

0 1 0 0

0 0 1 0

−1 −1 ). −1 0

(9.29)

Combining (9.28) with (9.29), it follows that α1 = α2 = α3 = α4 , a contradiction. Hence (9.27) does not hold. Without loss of generality, we may assume that (ni + 1)α4 is not a linear combination of α1 , . . . , α4 with weight ni + 1 over {0, 1, . . . , ni }, that is, (ni + 1)α4 ≠ mi,1 α1 + mi,2 α2 + mi,3 α3 + mi,4 α4

(9.30)

for all mi,1 , mi,2 , mi,3 , mi,4 ∈ {0, 1, . . . , ni } such that mi,1 + mi,2 + mi,3 + mi,4 = ni + 1. Multiplying (9.25) by e−(ni +1)α4 z , we obtain 3

s

∑ cl eαl z−(ni +1)α4 z = α4k b4 (z)ni +1 + ∑ αjk bj (z)ni +1 e(ni +1)(αj −α4 )z j=1

l=1

4

+

∑ αjk bj (z)ni +1 e(nt +1)αj z−(ni +1)α4 z

∑

j=1 l≤t(=i)≤p ̸ p

+∑

∑

̃i |=ni +1 i=1 |m

⟨mi ,̃ α⟩ cm z − (ni + 1)α4 z. ̃i (z)e ̃

Note that for fixed i ∈ {1, . . . , p}, there exists a nonzero constant αl , 1 ≤ l ≤ s, such that (ni + 1)α4 ≠ αl , 1 ≤ l ≤ s, and (ni + 1)α4 ≠ (nt + 1)αj for all 1 ≤ t(≠ i) ≤ p, 1 ≤ j ≤ 4. We may now combine this with (9.30) to see that α4k b4 (z)ni +1 is a linear combination of exponential functions. Therefore b4 = 0 by growth comparison. Thus (9.24) becomes p

3

i=1

j=1

∑ (∑ bj (z)e

αj z

ni 3

s

) ∑ αjk bj (z)eαj z = ∑ cl eαl z . j=1

l=1

(9.31)

Repeating the above argument, we may show that one of {b1 , b2 , b3 }, say b3 , is zero, so that (9.24) further becomes p

2

i=1

j=1

∑ (∑ bj (z)e

αj z

ni 2

s

) ∑ αjk bj (z)eαj z = ∑ cl eαl z . j=1

l=1

(9.32)

We may now deduce b2 = b1 = 0 similarly, completing the proof of Lemma 9.2.11. We are now ready to prove Theorem 9.2.9.

208 | 9 Nonlinear complex delay-differential equations Proof of Theorem 9.2.9. Let f be a meromorphic solution of (9.23) with ρ2 (f ) < 1. (1) First, note that f (z) cannot be a polynomial. Indeed, if so, then the left-hand side of equation (9.23) would be a polynomial, whereas the right-hand side would be of order one. (2) We next prove that f (z) is a transcendental entire function. Otherwise, assume that f has at least one pole z0 of multiplicity q ≥ 1. Since c ≠ 0, then z0 + c is also a pole of f of multiplicity ≥ (np + 1)q + k by (9.23). By shifting (9.23) forward we get p

s

i=1

l=1

∑ f (z + c)ni f (k) (z + c) + p(z + c)f (z + 2c) = ∑ βl eαl (z+c) .

(9.33)

By (9.33) z0 + 2c is also a pole of f of multiplicity ≥ (np + 1)2 q + k(np + 1) + k since z0 + c is a pole of f np f (k) of multiplicity ≥ (np + 1)2 q + k(np + 1) + k. Using induction, we see that for each integer j ≥ 1, the point z0 + jc is a pole of f of multiplicity ≥ (np + 1)j q + k[(np + 1)j−1 + (np + 1)j−2 + ⋅ ⋅ ⋅ + 1]. Hence for each integer m ≥ 1, we get the following estimate on the number n(r, f ) of poles of f in the disc |z| ≤ r: m

n(rm , f ) ≥ q + ∑(np + 1)j q + k [(np + 1)j−1 + (np + 1)j−2 + ⋅ ⋅ ⋅ + 1] , j=1

(9.34)

where rm = m|c| + |z0 | + 1. Thus we have log log n(r, f ) 1 ρ2 (f ) ≥ λ2 ( ) = lim sup f log r r→∞ log log n(rm , f ) ≥ lim sup log rm m→∞ log log(np + 1)m ≥ lim sup ≥1 log m m→∞ since n ≥ s + 2 ≥ 3. This is a contradiction to ρ2 (f ) < 1. Thus f must be a transcendental entire function. (3) Denote G(z) := ∑pi=1 f (z)ni f (k) (z) + p(z)f (z + c). Differentiating (9.23), we get G (z) = ∑sl=1 αl βl eαl z . Eliminating eα1 z from this and equation (9.23), we obtain 󸀠

s

α1 G − G󸀠 = ∑ βl (α1 − αl )eαl z . l=2

Differentiating again now results in s

α1 G󸀠 − G󸀠󸀠 = ∑ βl (α1 − αl )αl eαl z . l=2

9.2 Nonlinear complex delay-differential equations | 209

Therefore, eliminating eα2 z , we get s

α1 α2 G − (α1 + α2 )G󸀠 + G󸀠󸀠 = ∑ βl (α1 − αl )(α2 − αl )eαl z . l=3

Repeating the elimination inductively, we conclude that s

∑ (−1)j es−j (α1 , . . . , αs )G(j) = 0,

(9.35)

j=0

where ej (α1 , . . . , αs ), j = 0, . . . , s, are elementary symmetric polynomials in s variables α1 , . . . , αs ; see [169]. Consider now the linear differential operator s

L(w) := ∑ (−1)j es−j (α1 , . . . , αs )w(j) .

(9.36)

j=0

Recalling the definition of G, from (9.35) we obtain that p

L (∑ f (z)ni f (k) (z)) = −L(p(z)f (z + c)).

(9.37)

i=1

We deduce inductively that (m)

(f n f (k) )

m

(m−i)

= ∑ (mi ) (f n ) (f (k) ) i=0

(i)

i−1

m

= ∑ (mi ) (f (k) )

(m−i)

i=1

λj1

λj2

⋅ [nf n−1 f (i) + ∑ ∑ γjλ f n−j (f 󸀠 ) (f 󸀠󸀠 ) j=2 λ

⋅ ⋅ ⋅ (f (i−1) )

λj,i−1

(9.38)

i

+ n(n − 1) ⋅ ⋅ ⋅ (n − (i − 1))f n−i (f 󸀠 ) ] + f n f (k+m) for m = 1, 2, . . . , s, where γjλ are positive integers, λj1 , λj2 , . . . , λj,i−1 are non-negative integers, and the sum ∑λ is carried out over λj1 + λj2 + ⋅ ⋅ ⋅ + λj,i−1 = j and λj1 + 2λj2 + ⋅ ⋅ ⋅ + (i − 1)λj,i−1 = i. By (9.36) and (9.38) we get p

L (∑ f ni f (k) ) = f n1 −s ψ, i=1

(9.39)

where ψ is a differential polynomial in f of degree np − n1 + s + 1 with constant coefficients. From (9.36), (9.37), and (9.39) we obtain f (z)n1 −s ψ(z) = −L(p(z)f (z + c)),

(9.40)

where L(p(z)f (z + c)) is a delay-differential polynomial in f of degree one with polynomial coefficients.

210 | 9 Nonlinear complex delay-differential equations If ψ ≠ 0, then by (9.40) and the delay-differential version of the Clunie lemma (Lemma 1.2.16) T(r, ψ) = m(r, ψ) = S(r, f ), T(r, fψ) = m(r, fψ) = S(r, f ).

(9.41)

These two equalities give T(r, f ) ≤ T(r, fψ) + T (r,

1 ) = S(r, f ). ψ

(9.42)

This is a contradiction. If ψ = 0, then we have L(∑pi=1 f ni f (k) ) = 0 and L(p(z)f (z + c)) = 0. Using (9.36), we get s

L(p(z)f (z + c)) = ∑ (−1)j es−j (p(z)f (z + c))(j) = 0. j=0

(9.43)

The characteristic equation of (9.43) is (−1)s λs + (−1)s−1 e1 λs−1 + ⋅ ⋅ ⋅ + (−1)s−j ej λs−j + ⋅ ⋅ ⋅ + es = 0.

(9.44)

Since (9.44) has s distinct roots α1 , α2 , . . . , αs , we get that p(z)f (z + c) takes the form ̃ eα1 z + b ̃ eα2 z + ⋅ ⋅ ⋅ + b ̃ eαs z , p(z)f (z + c) = b 1 2 s

(9.45)

̃ (j = 1, 2, . . . , s) are constants. Hence where b j ̂ (z)eα1 z + b ̂ (z)eα2 z + ⋅ ⋅ ⋅ + b ̂ (z)eαs z , f (z) = b 1 2 s ̂ (z) = where b j

̃ e−αj c b j (j p(z−c)

(9.46)

= 1, 2, . . . , s) are rational functions.

Similarly, from L(∑pi=1 f ni f (k) ) = 0 we deduce that p

∑ f (z)ni f (k) (z) = c̃1 eα1 z + c̃2 eα2 z + ⋅ ⋅ ⋅ + c̃s eαs z , i=1

(9.47)

where c̃j (j = 1, 2, . . . , s) are constants. From (9.46) and (9.47) we obtain s

p

s

̂ (z)ni +1 e(ni +1)αl z + ∑ c̃l eαl z = ∑ (∑ αlk b l l=1

i=1

l=1

∑ |mi |=ni +1

cmi (z)e⟨mi ,α⟩z ) ,

(9.48)

where mi = (mi,1 , mi,2 , . . . , mi,s ) ∈ {0, 1, . . . , n}s , cmi (z) are rational functions, and s

⟨mi , α⟩ = ∑ mi,j αj . j=1

(9.49)

9.2 Nonlinear complex delay-differential equations | 211

For fixed i ∈ {1, . . . , p}, multiplying equation (9.48) by e−(ni +1)αs z , we obtain s

s−1

̂ ni +1 (z) + ∑ αk b ̂ ni +1 (z)e(n1 +1)(αl −αs )z ∑ c̃l eαl z−(ni +1)αs z = αsk b s l l l=1

s

+

l=1

̂ ni +1 (z)e(nt +1)αl z−(ni −1)αs z ∑ αlk b l

∑

1≤t(=i)≤p ̸ l=1 p

+ ∑∑ i=1

∑ |mi |=ni +1

cmi (z)e⟨mi ,α⟩z−(ni +1)αs z .

Since (ni + 1)αs ≠ αl for all l ∈ {1, . . . , s} and (ni + 1)αs ≠ (nl + 1)αl for all 1 ≤ t(≠ i) ≤ p, 1 ≤ l ≤ s, combining this with the fact that (ni + 1)αl , 5 ≤ l ≤ s, are not linear combinations ̂ ni +1 is a linear of α1 , . . . , αs with weight n1 + 1 over {0, 1, . . . , ni } for s ≥ 5, we see that αsk b s ̂ = 0 by growth comparison. combination of exponential functions, and hence b s Repeating above arguments, we conclude that ̂ =b ̂ = ⋅⋅⋅ = b ̂ = 0. b 5 6 s−1

(9.50)

̂ (z)eα1 z + b ̂ (z)eα2 z + b ̂ (z)eα3 z + b ̂ (z)eα4 z . f (z) = b 1 2 3 4

(9.51)

Then f becomes

̂ =b ̂ =b ̂ =b ̂ = 0, and hence From (9.47), (9.51), and Lemma 9.2.11 we obtain that b 1 2 3 4 f = 0, a contradiction, completing the proof of Theorem 9.2.9. After having completed the proof of Theorem 9.2.9, we remark that a slightly more general delay-differential equations of the form p

q

s

i=1

j=1

l=1

∑ f (z)ni f (k) (z) + ∑ pj (z)f (mj ) (z + cj ) = ∑ βl eαl z

(9.52)

was treated by Wang and Wang [211]. As an example, we include a result whose proof can be found in [211, Theorem 1.5]. Theorem 9.2.12. Suppose p1 (z), . . . , pq (z) are polynomials, c1 , . . . , cq , β1 , . . . , βs are nonzero constants, and α1 , . . . , αs are distinct nonzero constants. Moreover, suppose that s ≥ 1, k ≥ 0, mq > ⋅ ⋅ ⋅ > m1 ≥ 0, np > ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ > n1 ≥ s + 2. If f is an entire solution to (9.52) of finite order such that λ(f ) < ρ(f ) and ∑qj=1 pj (z)f (mj ) (z + cj ) does not vanish identically, then s = p + 1, and f is of the form f (z) = Ceαk z , where C ≠ 0 is constant, and αk ∈ {α1 , . . . , αs }. We close the section by a remark that more general equations of the form p

q

s

j=1

j=1

l=1

∑ cj (z)f (z + τj )nj + ∑ aj (z)f (mj ) (z + cj ) = ∑ βl eαl z under additional conditions may also be treated; see [45].

(9.53)

10 Complex q-delay-differential equations 10.1 Value distribution for q-delay-differential polynomials We start by looking at the Nevanlinna functions for the expression f (qz), where f is a meromorphic function. In this chapter, we always assume that q ≠ 0. For the first proposition, see [61, Theorem 1.6.7]. For the convenience of the reader, we include a short proof. Proposition 10.1.1. Let f be a meromorphic function. Then m(r, f (qz)) = m(|q|r, f (z)), and N(r, f (qz)) = N(|q|r, f (z)) − n(0, f (z)) log |q|. Thus T(r, f (qz)) = T(|q|r, f (z)) − n(0, f (z)) log |q| = T(|q|r, f (z)) + O(1).

(10.1)

Proof. By an elementary observation, 2π

m(r, f (qz)) =

1 ∫ log+ |f (qreiθ )|dθ 2π 0

2π

1 = ∫ log+ |f (|q|rei(θ+α) )|dθ = m(|q|r, f (z)), 2π 0

where arg q = α. Next, the number of poles of f (z) in the disc |z| < |q|r equals the number of poles of f (qz) in |z| < r. Therefore n(r, f (qz)) = n(|q|r, f (z)) and n(0, f (qz)) = n(0, f (z)). Thus |q|r

N(|q|r, f (z)) = ∫ 0 r

=∫ 0

n(t, f ) − n(0, f ) dt + n(0, f ) log |q|r t

n(s, f (qz)) − n(0, f (qz)) ds + n(0, f (qz))(log |q| + log r) s

= N(r, f (qz)) + n(0, f (qz)) log |q| = N(r, f (qz)) + n(0, f (z)) log |q|. By Proposition 10.1.1 we immediately get the relation for the order, resp., type, of f (z) and f (qz); see [61, Theorem 2.1.6]. https://doi.org/10.1515/9783110560565-010

214 | 10 Complex q-delay-differential equations Theorem 10.1.2. Let f be a meromorphic function, and let g(z) := f (qz). Then ρ(g) = ρ(f ) and τ(g) = |q|ρ(f ) τ(f ). In what follows, we need the notion of logarithmic density of a set E, defined as lim

r→∞

1 log r

∫ [1,r]∩E

1 dt. t

Treating meromorphic functions of order zero and small quantities related to such functions, by S∗ (r, f ) we denote any quantity satisfying S∗ (r, f ) = o(T(r, f )) for all r outside a set with logarithmic density 0. In this direction, we first recall the q-difference analogue of the logarithmic derivative lemma given by Barnett et al. [7, Theorem 1.1] for meromorphic functions of order zero. Theorem 10.1.3. Let f (z) be a nonconstant meromorphic function of order zero, and let q ∈ ℂ \ {0, 1}. Then m(

f (qz) ) = S∗ (r, f ) f (z)

(10.2)

on a set of logarithmic density one. Remarks. The restriction to order zero for q-difference is corresponding to the restriction to finite order for the ordinary shift case. The zero order cannot be replaced by small order in the statement of Theorem 10.1.3 as the following examples show. Clearly, the exponential function f (z) = ez does not satisfy (10.2). The entire function f (z) = √z sin √z is of order ρ(f ) = 21 . Then ff(4z) = 4 cos √z, and thus (z) m(

f (4z) ) = T(r, f ) + S∗ (r, f ). f (z)

(10.3)

If f (z) is an entire function of order ρ ∈ (0, 21 ) with real negative zeros, f (0) = 1, and n(r, f1 ) ∼ r ρ , then there is a positive constant c, independent of θ, such that

log |f (reiθ )| ∼ cr ρ cos ρθ on the ray θ = θ0 , where −π < θ0 < π; see [15, Theorem 4.1.1]. Thus 󵄨󵄨 f (qreiθ ) 󵄨󵄨 󵄨 󵄨󵄨 ρ ρ log 󵄨󵄨󵄨 󵄨󵄨 ∼ c(q − 1)r cos ρθ, 󵄨󵄨 f (reiθ ) 󵄨󵄨

meaning that (10.2) is not valid for q > 1. Concerning relations between N(r, f (qz)) and N(r, f (z)) and between T(r, f (qz)) and T(r, f (z)), using a result given by Hayman [91, Lemma 4], Zhang and Korhonen [245, Theorems 1.3 and 1.1] also obtained the following: Theorem 10.1.4. Let f (z) be a nonconstant meromorphic function of order zero, and let q ∈ ℂ \ {0}. Then N(r, f (qz)) = (1 + o(1))N(r, f ),

(10.4)

10.1 Value distribution for q-delay-differential polynomials | 215

and T(r, f (qz)) = (1 + o(1))T(r, f )

(10.5)

on a set of lower logarithmic density one. Remark 10.1.5. The assumption in Theorem 10.1.4 that ρ(f ) = 0 cannot be extended to ρ(f ) > 0. Indeed, take any meromorphic function f (z) such that T(r, f (z)) = r ρ + o(r ρ ), where ρ > 0. If q ∈ ℂ and |q| ≠ 1, then we may use (10.1) in Proposition 10.1.1 to obtain T(r, f (qz)) = |q|ρ r ρ + o(r ρ ) = |q|ρ (1 + o(1))T(r, f (z)). Barnett et al. [7, Theorem 3.1] presented the following q-difference analogue to the second main theorem of Nevanlinna. Theorem 10.1.6. Let f (z) be a nonconstant meromorphic function of order zero, let q ≠ 0, 1, and let a1 , a2 , . . . , ap ∈ ℂ be distinct constants with p ≥ 2. Then p

m(r, f ) + ∑ m ( k=1

1 ) ≤ 2T(r, f ) − Npair (r, f ) + S∗ (r, f ), f − ak

where Npair (r, f ) = 2N(r, f ) − N(r, Δq f ) + N (

1 ), Δq f

and Δq f = f (qz) − f (z). A q-difference analogue of the Clunie lemma is given in [7, Theorem 2.1]; an improvement to this appears in [123, Theorem 2.5] for general q-difference polynomials. We also get a q-delay-differential analogue of the Clunie lemma by using a similar method as in [123, Theorem 2.5]; see below. A q-delay-differential polynomial in f (z) can be expressed as P(z, f ) = ∑ bl (z)f (z)l0,0 f (c1 z)l1,0 ⋅ ⋅ ⋅ f (cν z)lν,0 f 󸀠 (z)l0,1 ⋅ ⋅ ⋅ f (μ) (cν z)lν,μ , l∈L

where the coefficients bl (z) are small meromorphic functions with respect to f (z) in the sense that their Nevanlinna characteristic functions are o(T(r, f )) on a set of logarithmic density one. Theorem 10.1.7. Let f be a transcendental meromorphic solution of order zero of a q-delay-differential equation of the form Uq (z, f )Pq (z, f ) = Qq (z, f ),

216 | 10 Complex q-delay-differential equations where Uq (z, f ), Pq (z, f ), Qq (z, f ) are q-delay-differential polynomials such that the total degree deg Uq (z, f ) = n in f (z) and its q-differences and of the derivatives of its q-difference, whereas deg Qq (z, f ) ≤ n. Moreover, assume that Uq (z, f ) contains just one term of maximal total degree in f (z) and its q-differences. Then m(r, Pq (z, f )) = o(T(r, f )) on a set of logarithmic density one. In addition, a q-difference analogue of the Mohon’ko lemma appears in [7, Theorem 2.2]. A corresponding q-delay-differential version may be stated as follows. Theorem 10.1.8. Let f be a nonconstant meromorphic solution of order zero to P(z, f ) = 0, where P(z, f ) is a q-delay-differential polynomial in f (z). If P(z, a) ≢ 0 for a small function a(z) of f (z), then m (r,

1 ) = o(T(r, f )) f −a

on a set of logarithmic density one. More generally, let now L(f ) be a linear q-delay-differential polynomial. It is now easy to get the following result. Theorem 10.1.9. Let f (z) be a nonconstant meromorphic function of order zero, and let q ∈ ℂ \ {0}. Then m(

L(f ) ) = S∗ (r, f ) f (z)

(10.6)

on a set of logarithmic density one. For a general linear operator L : ℳ → ℳ, where ℳ is the field of meromorphic functions, we have the following presentation Halburd and Korhonen [86, Theorem 4.1]. Theorem 10.1.10. Let 𝒩 be a subfield of ℳ, and let f ∈ 𝒩 \ ker(L), where L : ℳ → ℳ is a linear operator such that m (r,

L(f ) ) = o(T(r, f )) f

as r → ∞ outside an exceptional set E ⊂ (0, +∞). If a1 , a2 , . . . , aq are q ≥ 1 different elements of ker(L) ∩ 𝒮 (f ), then q

(q − 1)T(r, f ) + NL(f ) (r, f ) ≤ N(r, f ) + ∑ N (r, j=1

1 ) + S(r, f ), f − aj

(10.7)

10.1 Value distribution for q-delay-differential polynomials | 217

where NL(f ) (r, f ) = 2N(r, f ) − N(r, L(f )) + N (

1 ), L(f )

and S(r, f ) = o(T(r, f )) as r → ∞ outside E. Remark 10.1.11. By choosing different 𝒩 and L(f ), Theorem 10.1.10 incorporates and generalizes several different versions of the second main theorem. (1) Theorem 10.1.10 implies the classical Nevanlinna second main theorem in the complex plane by choosing L(f ) = f 󸀠 and 𝒩 = ℳ. (2) Theorem 10.1.10 also implies the difference analogue of the second main theorem by choosing L(f ) = f (z + 1) − f (z) and 𝒩 the field of meromorphic functions of hyperorder strictly less than one. (3) Theorem 10.1.10 similarly implies the q-difference analogue of the second main theorem by choosing L(f ) = f (qz) − f (z) and 𝒩 the field of meromorphic functions of order zero. We next proceed to considering zeros of q-difference polynomials and q-delaydifferential polynomials. We first state a result due to Fletcher, Langley, and Meyer [55, Theorem 1.5]. Theorem 10.1.12. Let q ∈ ℂ with |q| > 1. Let f be a transcendental meromorphic function with ς(f ) = lim inf r→∞

Then at least one of f (qz) − f (z) and

T(r, f ) = 0. (log r)2

f (qz)−f (z) f (z)

(10.8)

has infinitely many zeros.

Remark 10.1.13. (1) Hypothesis (10.8) is sharp in Theorem 10.1.12. Let q ∈ ℂ with |q| > 1 and write ∞

f (z) = ∏ (1 − n=0

z −1 ) , qn

∞

f (qz) = ∏ (1 − n=0

z

qn−1

−1

)

=

f (z) . 1 − qz

(z) qz qz Hence, neither f (qz)−f = 1−qz nor f (qz) − f (z) = f (z) 1−qz has infinitely many zeros, f (z) whereas ς(f ) is positive but finite. Indeed, this follows by [13, Theorems 1.1 and 1.2] from (1 − qz)f (qz) − f (z) = 0 by using the transformation z → z/q.

(2) If f (z) is a transcendental meromorphic solution to a q-difference Riccati equation of the form f (qz) =

a(z)f (z) + b(z) , f (z) + c(z)

(10.9)

218 | 10 Complex q-delay-differential equations where a(z), b(z), c(z) are rational functions, then f (qz) − f (z) and 2

f (qz)−f (z) f (z)

both have

infinitely many zeros if a(z) = c(z) and b(z) = (s(z)) , where s(z) ≢ a(z); see [110]. Observe that the transcendental meromorphic solutions f (z) to (10.9) satisfy T(r, f ) = O((log r)2 ), although they may not satisfy (10.8); see [29, Remark 10.4.1] or [78]. An interested reader may consult Chen [29, Chapter 10.4] and Huang and Zhang [104] to see more results on q-difference Riccati equations. Related to the q-difference analogue of Hayman conjecture, Zhang and Korhonen [245, Theorem 4.1] proved that q-difference polynomials f (z)n f (qz) − α have infinitely many zeros when n ≥ 6 for the meromorphic case and when n ≥ 2 for the entire case of order zero. Recalling the ideas applied before in Theorem 2.2.4, a better result below follows. It remains open to improve the assumption n ≥ 4. Theorem 10.1.14. Let f be a transcendental meromorphic function of order zero, and let q ∈ ℂ \ {0}. If n ≥ 4, then f (z)n f (qz) − α(z) has infinitely many zeros whenever α(z) is a nonzero small function with respect to f (z). Proof. Recalling the proof of Theorem 2.2.4, denote H(z) := f (z)n f (qz) − α(z). Then, clearly, nm(r, f ) = m(r, f n ) ≤ m(r, H + α) + m(r, 1/f (qz)). Denote now by N0 (r), resp., N1 (r), the distinct common zeros, resp., poles, of H +α and f (qz). We obtain nN(r, f ) = N(r, f n ) ≤ N(r, H + α) + N(r, 1/f (qz)) − N 0 (r) − N 1 (r). Therefore nT(r, f ) = T(r, f n ) ≤ T(r, H + α) + T(r, 1/f (qz)) − N 0 (r) − N 1 (r). By simple observation N(r, H + α) ≤ N(r, f n ) + N 1 (r) ≤ T(r, f ) + N 1 (r) and N(r, 1/(H + α)) ≤ N(r, 1/f n ) + N 0 (r) ≤ T(r, f ) + N 0 (r). We may now apply the second main theorem to obtain T(r, H + α) ≤ N(r, H + α) + N(r, 1/(H + α)) + N(r, 1/H) + S(r, H) ≤ T(r, f ) + N 1 (r) + T(r, f ) + N 0 (r) + N(r, 1/H) + S∗ (r, f ). Using (10.10), we get (n − 2)T(r, f ) ≤ T(r, 1/f (qz)) + N(r, 1/H) + S∗ (r, f ).

(10.10)

10.1 Value distribution for q-delay-differential polynomials | 219

By (10.5) we further conclude that (n − 3)T(r, f ) ≤ N(r, 1/H) + S∗ (r, f ). If now H has finite many zeros only, an immediate contradiction follows, since n ≥ 4. Corollary 10.1.15. Let f be a transcendental entire function of order zero, and let q ∈ ℂ \ {0}. If n ≥ 1, then f (z)n f (qz) − α(z) has infinitely many zeros whenever α(z) is a nonzero constant. Proof. Suppose the claim fails. Then f n (z)f (qz) must be a polynomial by the Hadamard representation. This implies that f has finite many zeros only, contradicting the fact that f is transcendental and of order zero. Remark 10.1.16. An example given in [103, Example 2.1] shows that if α = 0, then the claim of Theorem 10.1.14 may fail: Let 0 < |q| < 1. The q-gamma function Γq (x) is defined by Γq (x) :=

(q; q)∞ (1 − q)1−x , (qx ; q)∞

k x−1 where (a; q)∞ = ∏∞ Γq (x), z = qx , and γq (0) := k=0 (1 − aq ). By defining γq (z) := (1 − q) (q; q)∞ We see that γq (z) is a meromorphic function of order zero with no zeros. Then, clearly, γq (z)n γq (qz) has no zeros as well.

We now continue to consider zeros of q-delay-differential polynomials of type f (qz)n f 󸀠 (z) − a(z), where a(z) is a small function with respect to f . Before giving the result, we need an estimate on the characteristic function of f (qz)n f 󸀠 (z), which also can be proved by a similar method as in Lemma 2.2.2. Lemma 10.1.17. Let f (z) be a nonconstant zero-order meromorphic function, and let q ∈ ℂ \ {0} and n ≥ 2. Then (n − 2)T(r, f ) ≤ T(r, f (qz)n f 󸀠 (z)) + S∗ (r, f ) ≤ (n + 2)T(r, f ).

(10.11)

If f (z) is a nonconstant zero-order entire function, then nT(r, f ) ≤ T(r, f (qz)n f 󸀠 (z)) + S∗ (r, f ) ≤ (n + 1)T(r, f ).

(10.12)

Proof. Denoting F(z) := f (qz)n f 󸀠 (z), we clearly see that T(r, F) ≤ nT(r, qz) + 2T(r, f ) + S∗ (r, f ) = nT(r, f ) + 2T(r, f ) + S∗ (r, f ). To obtain the reversed inequality, we may write (n + 1)T(r, f ) + S∗ (r, f ) = T(r,

1 f (qz)n+1

=

1 f 󸀠 (z) . F f (qz)

Therefore

1 ) ≤ T(r, 1/F) + T(r, 1/f (qz)) + T(r, f 󸀠 ) f (qz)n+1

220 | 10 Complex q-delay-differential equations ≤ T(r, F) + T(r, f ) + S∗ (r, f ) + 2T(r, f ), and hence T(r, F) ≥ (n − 2)T(r, f ) + S∗ (r, f ). The remaining claim concerning the entire case is obvious. Observe now that f (qz)n f 󸀠 (z) − a(z) may have finitely many zeros for finite-order entire functions f (z). We can see this by taking f (z) = ez , q = − n1 , and a nonconstant polynomial a(z). For meromorphic functions f (z) and n ≥ 9, f (qz)n f 󸀠 (z) − a(z) has infinitely many zeros. This follows by (10.11) and the second main theorem. We leave this as an exercise for the reader. Applying now the idea of common zeros and common poles similarly as in Theorem 10.1.14, we obtain the following theorem. Theorem 10.1.18. Let f (z) be a transcendental meromorphic function of order zero, and let q ∈ ℂ \ {0} and n ≥ 5. Then f (qz)n f 󸀠 (z) − a(z) has infinitely many zeros, provided that a(z) is a nonzero small function with respect to f (z). Proof. Proceeding as in the proof of Theorem 10.1.15, we denote H(z) := f (qz)n f 󸀠 (z) − a(z). Then nm(r, f ) = m(r, f (qz)n ) + S∗ (r, f ) ≤ m(r, H + a) + m(r, 1/f 󸀠 ) + S∗ (r, f ) and nN(r, f ) ≤ N(r, f (qz)n ) + S∗ (r, f ) ≤ N(r, H + a) + N(r, 1/f 󸀠 ) − N 0 (r) − N 1 (r) + S∗ (r, f ), where, as before, N 0 (r), resp., N 1 (r), stands for the distinct common zeros, resp., poles, of H + a and f 󸀠 . Therefore nT(r, f ) ≤ T(r, H + a) + T(r, f 󸀠 ) − N 0 (r) + N 1 (r) + S∗ (r, f ). Obviously, we have N(r, H + a) ≤ N(r, f (qz)) + N 1 (r) ≤ T(r, f ) + N 1 (r) + S∗ (r, f ) and N(r, 1/(H + a)) ≤ N(r, 1/f (qz)) + N 0 (r) ≤ T(r, f ) + N 0 (r) + S∗ (r, f ). Therefore by the second main theorem T(r, H + a) ≤ N(r, H + a) + N(r, 1/(H + a)) + N(r, 1/H) + S(r, H) ≤ 2T(r, f ) + N 1 (r) + N 0 (r) + N(r, 1/H) + S∗ (r, f ).

(10.13)

10.2 Uniqueness on q-delay-differential polynomials | 221

Combining this estimate for T(r, H + a) with (10.13), we get nT(r, f ) ≤ 4T(r, f ) + S∗ (r, f ). If now H(z) = f (qz)n f 󸀠 (z) − a(z) has finitely many zeros only, then a contradiction follows from n ≥ 5. Remark. If we consider f (qz)n f (k) (z) − a(z) instead of f (qz)n f 󸀠 (z) − a(z), then the same reasoning implies that f (qz)n f (k) (z) − a(z) has infinitely many zeros if n ≥ k + 4.

10.2 Uniqueness on q-delay-differential polynomials We start this section by recalling a classical uniqueness result due to Rubel and Yang [197]. Theorem 10.2.1. Let f (z) be a nonconstant entire function, and let a, b ∈ ℂ. If f (z) and f 󸀠 (z) share a, b CM, then f 󸀠 (z) = f (z). Many improvements of Theorem 10.2.1 have been achieved since 1976. For example, f 󸀠 (z) was improved to f (k) (z) or to differential polynomials of f (z), the condition a, b CM was reduced to a, b IM, and the entire function f (z) was extended to a meromorphic function; see these results in [235, Chapter 8]. Furthermore, we also recall the following result [235, Theorem 2.17]. Lemma 10.2.2. Let f and g be nonconstant meromorphic functions of order ρ(f ) < 1. If f and g share 0 and ∞ CM, then there exists a nonzero constant K satisfying f = Kg. Using Lemma 10.2.2, Qi, Liu, and Yang [184, Theorem 1.1] considered the problem of f (z) and f (qz) sharing common values when |q| = 1. In the original statement in [184, Theorem 1.1], the assumption was that f (z) is of order zero. However, the result remains valid for ρ(f ) < 1 as well: Theorem 10.2.3. Let f (z) be a meromorphic function with ρ(f ) < 1, and let a1 , a2 , a3 ∈ ℂ ∪ {∞} be three distinct values. If f (z) and f (qz) share a1 , a2 CM and a3 IM, then f (z) = f (qz). Remark 10.2.4. (1) The assumption in Theorem 10.2.3 that a3 is shared IM can be replaced by one of the following assumptions: (i) There exists a point z0 such that f (z0 ) = f (qz0 ) = a3 ; (ii) a3 is a Picard exceptional value of f . (2) Theorem 10.2.3 is not valid for meromorphic functions with ρ(f ) ≥ 1. This can be seen by the following two examples: (a) If f (z) = ez and q = −1, then f (z) and f (qz) share 0, 1, ∞ CM, but f (z) ≠ f (qz). 2 (b) If f (z) = ez and q = i, then f (z) and f (qz) share 0, 1, ∞ CM, but f (z) ≠ f (qz).

222 | 10 Complex q-delay-differential equations (3) The condition a1 , a2 CM and a3 IM in Theorem 10.2.3 cannot be reduced to a1 , a2 2z CM. This can be seen by the following example: If f (z) = (z+1) 2 and q = −1, then

f (qz) =

−2z . (1−z)2

Now f (z) and f (qz) share 0, 1 CM, whereas f (z) ≠ f (qz).

Considering entire functions f of order zero, the assumptions in Theorem 10.2.3 can be reduced as follows; see [184, Theorem 1.2]. Theorem 10.2.5. Let f be an entire function of order zero, and let a1 , a2 ∈ ℂ be two distinct values. If f (z) and f (qz) share a1 and a2 IM, then f (z) = f (qz). Note that Theorem 10.2.1 is related to the value sharing problem on f (z) and f 󸀠 (z), whereas Theorems 10.2.3 and 10.2.5 are related to the value sharing problem on f (z) and f (qz). Proceeding to the corresponding q-delay-differential cases, we first consider the case where f 󸀠 (z) and f (qz) share common values when q is a nonzero constant. Before treating such a value sharing problem, we should consider properties of meromorphic solutions of the q-delay-differential equation f 󸀠 (z) = f (qz),

(10.14)

where q is a nonzero constant. Clearly, the nontrivial entire solutions of (10.14) are transcendental. We now first show the following: Theorem 10.2.6. Nontrivial entire solutions of (10.14) must be of the form a0 n(n−1) q 2 zn, n! n=0 +∞

f (z) = ∑ where a0 is a free complex parameter.

Proof. Let f (z) = a0 + a1 z + a2 z 2 + ⋅ ⋅ ⋅ + an z n + ⋅ ⋅ ⋅. Thus f (qz) = a0 + a1 qz + a2 (qz)2 + ⋅ ⋅ ⋅ + an (qz)n + ⋅ ⋅ ⋅

(10.15)

f 󸀠 (z) = a1 + 2a2 z + ⋅ ⋅ ⋅ + nan z n−1 + ⋅ ⋅ ⋅ .

(10.16)

and

By comparing the coefficients of (10.15) and (10.16) we get a1 = a0 q0 , a a2 = 0 q1 , 2 a0 1+2 a3 = q , 3! a a4 = 0 q1+2+3 . 4!

10.2 Uniqueness on q-delay-differential polynomials | 223

Using a simple induction, we get that f (z) must be of the form a0 n(n−1) q 2 zn. n! n=0 +∞

f (z) = ∑

n Remark 10.2.7. Recall that the order of an entire function g(z) = ∑∞ n=0 an z equals

ρ(g) = limn→∞

n log n

log |a1 |

.

n

Thus we conclude that ρ(f ) = 0 if |q| ≠ 1 and ρ(f ) = 1 if |q| = 1 in Theorem 10.2.6. Clearly, if q = 1, then a0 n z = a0 ez . n! n=0 +∞

f (z) = ∑

Using Lemma 10.2.2, Liu and Cao [152] obtained the following result. Theorem 10.2.8. Let f be a meromorphic function with ρ(f ) < 1, and let a1 , a2 ∈ ℂ ∪ {∞} and a3 ∈ ℂ be three distinct values. If f (qz) and f 󸀠 (z) share a1 , a2 CM and a3 IM, then f 󸀠 (z) = f (qz). Proof. First, suppose that a1 , a2 , a3 ∈ ℂ. Define F(z) := f (qz)−a1 a3 −a2 . . f (qz)−a2 a3 −a1

f 󸀠 (z)−a1 a3 −a2 . f 󸀠 (z)−a2 a3 −a1

and G(z) :=

Thus we have that F(z) and G(z) share 0, ∞ CM and 1 IM. Since F(z) and G(z) are meromorphic functions with ρ(f ) < 1, from Lemma 10.2.2 it follows that F(z) = kG(z). If the value 1 is not a Picard exceptional value of F and G, then k = 1, and F(z)−1 thus F(z) = G(z). If the value 1 is a Picard exceptional value, then we see that G(z)−1 has no zeros and poles. Hence we have

F(z)−1 G(z)−1

= C. Therefore k = 1, and so F(z) = G(z).

We conclude that f 󸀠 (z) = f (qz). If one of a1 , a2 is ∞, we may assume, without loss of generality, that a1 = ∞. Let F(z) = f 󸀠 (z) − a2 and G(z) = f (qz) − a2 . Then F(z) and G(z) share 0, ∞ CM. From Lemma 10.2.2 we see that F(z) = kG(z). Combining this with the condition that a3 is shared IM, we have k = 1, and thus f 󸀠 (z) = f (qz). Remark 10.2.9. Theorem 10.2.8 is not valid in general for meromorphic functions with ρ(f ) ≥ 1. As an example, look at f (z) = ez and q = −1. Then f (qz) and f 󸀠 (z) share 0, 1, −1 CM, but f 󸀠 (z) ≠ f (qz). Theorem 10.2.10. Let f (z) be a nonconstant entire function, and let q be a nonzero constant. If f (qz) and f 󸀠 (z) share two distinct constants a, b ∈ ℂ CM and one of a, b is a Picard exceptional value, then f 󸀠 (z) = f (qz) or f (z) = e−Az+B , where −Ae2B = b2 and q = −1. For the proof of Theorem 10.2.10, we need the following lemma. Lemma 10.2.11. [90, Theorem 3.7] Let f (z) be an entire function. If f (z) and f (l) (z) (l ≥ 2) have no zeros, then f (z) = eAz+B , where A, B are constants.

224 | 10 Complex q-delay-differential equations Now we are ready to proceed to proving Theorem 10.2.10. Proof. Since one of a, b is a Picard exceptional value, we may assume, without loss of generality, that a is a Picard exceptional value. Then, by the sharing value assumption, f (qz) − a = eα(z)

(10.17)

f 󸀠 (z) − a = eβ(z) ,

(10.18)

and

where α(z) and β(z) are nonconstant entire functions, and T(r, α󸀠 (z)) = S(r, f (qz)). Differentiating f (qz), we have f 󸀠 (qz) =

1 󸀠 α (z)eα(z) = a + eβ(qz) . q

(10.19)

From (10.17) and (10.19) we get T(r, eα(z) ) = T(r, f (qz)) + O(1) ≤ T(r, f 󸀠 (qz)) + S(r, f (qz))

(10.20)

T(r, eβ(qz) ) = T(r, f 󸀠 (qz)) + O(1).

(10.21)

and

If a ≠ 0, from (10.19) we conclude that 1 󸀠 eβ(qz) α (z)eα(z) − = 1. aq a

(10.22)

Using the second main theorem for three small functions (Theorem 1.1.22), we have T(r, eα(z) ) ≤ N (r,

1 eα(z) −

aq α󸀠 (z)

) + S(r, eα(z) )

1 ) + S(r, eα(z) ) eβ(qz) = S(r, eα(z) ),

= N (r,

(10.23)

which is a contradiction. Thus a = 0. From (10.19) we get eβ(qz)−α(z) =

α󸀠 (z) . q

(10.24)

(z)−b Since b ≠ 0 is shared CM by f 󸀠 (z) and f (qz), we get ff (qz)−b = eγ(z) , where γ(z) is an entire function. Thus combining this with (10.17), (10.18), and (10.24), we have 󸀠

eβ(z) − b = f 󸀠 (z) − b = eγ(z) (f (qz) − b) = eγ(z) (eα(z) − b).

10.2 Uniqueness on q-delay-differential polynomials | 225

Since b ≠ 0, we have eβ(z) eγ(z)+α(z) + eγ(z) − = 1. b b If now eγ(z) ≡ 1, then it follows that f 󸀠 (z) = f (qz). By Theorem 1.1.34, if eβ(z) ≡ b, γ(z)+α(z) then clearly f 󸀠 (z) ≡ b and f (qz) ≡ b, which is impossible. If e −b ≡ 1, then we also have eβ(z)−γ(z) ≡ −b. Therefore, as we can easily see, eα(z)+β(z) = b2 . This implies that α(z) + β(z) ≡ d, where d is a constant. So β(qz) ≡ −α(qz) + d. Combining this with (10.24), we obtain e−α(qz)−α(z)+d =

α󸀠 (z) . q

(10.25)

The left-hand side of (10.25) has no zeros, and thus α󸀠 (z) has no zeros, and therefore either α(z) = Az + B, or α(z) is a transcendental entire function. If α(z) = Az + B, then from (10.25) we have q = −1 and ed−2B = −A. From (10.17) we have f (z) = e−Az+B and −Ae2B = b2 . Finally, suppose α(z) is a transcendental entire function. Since f (qz)f 󸀠 (z) = eα(z)+β(z) = b2 , f (qz) and f 󸀠 (z) share 0, ∞, b CM and −b IM. By [69, Theorem 1] sharing of −b is CM as well, and, by a classical result, f (qz) is a Möbius transformation of f 󸀠 (z), say f (qz) = (c1 f 󸀠 (z) + c2 )/(c3 f 󸀠 (z) + c4 ). From this it immediately follows by looking at ±b-values of f (qz) and f 󸀠 (z) that f (qz) = (c1 f 󸀠 (z) + c3 b2 )/(c3 f 󸀠 (z) + c1 ). Recall that f (qz) and f 󸀠 (z) share 0 CM, thus c3 = 0, and thus f (qz) = f 󸀠 (z). This completes the proof. Remark 10.2.12. If a, b are not Picard exceptional values, then Theorem 10.2.10 is not valid in general. This can be seen by looking at the function f (z) = 3a − ea2z and q = −1. Then f 󸀠 (z) = e2a2z and f (qz) = 3a − ae2z share a and 2a CM, whereas f 󸀠 (z) ≠ f (qz). We next proceed to giving a result on the uniqueness of f (qz)n f 󸀠 (z) and g(qz)n g 󸀠 (z), when they share a nonzero polynomial. Theorem 10.2.13. Let f (z) and g(z) be transcendental entire functions of order zero, and let q ∈ ℂ \ {0} and n ≥ 5. If f (qz)n f 󸀠 (z) and g(qz)n g 󸀠 (z) share a nonzero polynomial p(z) CM, then f (qz)n f 󸀠 (z) = g(qz)n g 󸀠 (z). Proof. From the conditions we get n 󸀠

n 󸀠

f (qz)n f 󸀠 (z)−p(z) g(qz)n g 󸀠 (z)−p(z)

f (qz) f (z) = g(qz) g (z). If c ≠ 1, then we have

= c. If c = 1, then it follows that

f (qz)n f 󸀠 (z) − cg(qz)n g 󸀠 (z) = p(z)(1 − c).

(10.26)

Using the second main theorem for three small functions (Theorem 1.1.22), we get T(r, f (qz)n f 󸀠 (z)) ≤ N(r, f (qz)n f 󸀠 (z)) + N (r, + N (r,

1 ) f (qz)n f 󸀠 (z)

1 ) + S(r, f (qz)n f 󸀠 (z)) f (qz)n f 󸀠 (z) − (1 − c)p(z)

226 | 10 Complex q-delay-differential equations 1 1 1 ) + N (r, 󸀠 ) + N (r, ) + S∗ (r, f ) f (qz) f (z) g(qz)n g 󸀠 (z) ≤ 2T(r, f ) + 2T(r, g) + S∗ (r, f ). ≤ N (r,

(10.27)

Similarly, for g, we also obtain T(r, g(qz)n g 󸀠 (z)) ≤ 2T(r, f ) + 2T(r, g) + S∗ (r, g).

(10.28)

Combining (10.27) and (10.28) with (10.12), we get n[T(r, f ) + T(r, g)] ≤ 4[T(r, f ) + T(r, g)] + S∗ (r, f ) + S∗ (r, g), which is a contradiction with the condition n ≥ 5. Remark 10.2.14. It remains open whether f ≡ g follows from f (qz)n f 󸀠 (z) = g(qz)n g 󸀠 (z).

10.3 q-delay-differential equations We continue now by first looking at linear q-delay-differential equations. It seems to us that these linear equations have not yet been much investigated. Recalling [9], we consider equations of the form n

m

p=1

j=1

f 󸀠󸀠 (z) + ∑ ap (z)f 󸀠 (λp z) + ∑ bj (z)f (σj z) = 0,

(10.29)

where the coefficients ap , bj are entire functions, and λp , σj are nonzero complex numbers in the open unit disc. Obviously, all meromorphic solutions to (10.29) are entire. We now prove the following: Theorem 10.3.1. Suppose that in equation (10.29) all coefficients ap , bj are entire functions of order ≤ ρ and that all q-delays λp , σj are nonzero complex numbers in the open unit disc. Then all meromorphic solutions to (10.29) are of order ≤ ρ as well. Proof. Given an arbitrary entire solution f to (10.29), we may assume that f is transcendental. Denote λ := max{λ1 , . . . , λn }, σ := max{σ1 , . . . , σm }, τ := max{λ, σ}, s := n + m, and C(r) := maxp,j {M(r, ap ), M(r, bj )}. Clearly, ρ(C) ≤ ρ. Recall that M(r, f ) ≤ |f (0)| + rM(r, f 󸀠 ),

M(r, f 󸀠 ) ≤ |f 󸀠 (0)| + rM(r, f 󸀠󸀠 ).

(10.30)

Therefore, by the maximum modulus principle and (10.30), n

m

M(r, f 󸀠󸀠 ) ≤ ∑ M(r, ap )M(|λp |r, f 󸀠 ) + ∑ M(r, bj )M(|σj |r, f ) p=1

j=1

≤ sC(r)(M(τr, f ) + M(τr, f )) ≤ sC(r)(|f (0)| + 2τr|f 󸀠 (0)| + 2τ2 r 2 M(τr, f 󸀠󸀠 )). 󸀠

10.3 q-delay-differential equations | 227

Since f is transcendental by assumption, the maximum moduli of f and its derivatives are unbounded. Therefore, letting r be large enough, we may use log+ instead of log, whenever needed, to estimate log M(r, f 󸀠󸀠 ) as follows: log M(r, f 󸀠󸀠 ) ≤ log C(r) + 3 log r + O(1) + log M(τr, f 󸀠󸀠 ).

(10.31)

To continue, using the Hadamard three-circle theorem for the radii 1, τr, r, we obtain log M(τr, f 󸀠󸀠 ) ≤

log τr − log τ log M(1, f 󸀠󸀠 ) + log M(r, f 󸀠󸀠 ) log r log r

and log M(τr, f 󸀠󸀠 ) ≤

− log τ log τ log M(1, f 󸀠󸀠 ) + (1 + ) log M(r, f 󸀠󸀠 ). log r log r

Combining this inequality with (10.31), we obtain −

log τ log τ log M(r, f 󸀠󸀠 ) ≤ log C(r) + 3 log r − M(1, f 󸀠󸀠 ) + O(1), log r log r

and hence log M(r, f 󸀠󸀠 ) ≤ (log τ)−1 ((log r)(log C(r)) + 4 log2 r). Applying log, resp., log+ , once more, we get log log M(r, f 󸀠󸀠 ) ≤ log log r + log log C(r) + 8 log log r + O(1). Since ρ(C) ≤ ρ, we conclude that ρ(f 󸀠󸀠 ) ≤ ρ, and hence ρ(f ) ≤ ρ, completing the proof. Remark 10.3.2. (1) Note that in the preceding theorem the assumption that all q-delays in (10.29) belong to the open unit disc is necessary. As an example [9, p. 45], consider the equation f 󸀠󸀠 (z) + f (z) + f (z/2) = 0, which has a solution (−1)n n−1 ∏(1 + 1/4j )z 2n (2n)! n=0 j=0 ∞

f (z) = ∑ of order one.

(2) A related paper treating certain second-order linear delay-differential equations is due to Bèlair and Giroux [10]. Proceeding to nonlinear q-delay-differential equations, we first consider Malmquist-type situations. Malmquist-type theorems for delay-differential equations were treated in Chapter 8. In this section, we shortly look at some results on

228 | 10 Complex q-delay-differential equations q-difference or q-delay-differential Malmquist-type theorems. Recalling standard notations, denote R(z, f ) =

∑ni=0 ai (z)f (z)i , ∑nj=0 bj (z)f (z)j

where the coefficients ai (z) (i = 0, 1, . . . , n) and bj (z) (j = 0, 1, . . . , m) are small functions with respect to f (z). Moreover, denote d = degf R(z, f ) = max{m, n}. The nonautonomous Schröder q-difference equation f (qz) = R(z, f (z))

(10.32)

has been recently studied, and considerations of the autonomous Schröder equation f (qz) = R(f (z)) go back to Ritt [193]. Gundersen et al. [78] considered the order of growth of meromorphic solutions to (10.32), from which a q-difference analogue of the classical Malmquist theorem is given: If equation (10.32) admits a meromorphic solution of order zero, then (10.32) reduces to a q-difference Riccati equation, that is, d = 1. Using Theorem 10.1.4, Zhang and Korhonen [245, Theorem 3.1] extended this result as follows. Theorem 10.3.3. Let q1 , . . . , qs be nonzero constants, and let s be a positive integer. If a q-difference equation of the form ⋁ f (qi z) = R(z, f ),

(10.33)

where the symbol ⋁ stands for either ∑si=1 or ∏si=1 , admits a transcendental meromorphic solution of order zero, then d ≤ s. Furthermore, Gundersen et al. [78, Theorem 3.2] proved that if f (z) is a transcend . Zheng and Chen dental meromorphic solution of (10.32) and |q| > 1, then ρ(f ) = log log q [251, Theorem 1] also considered the growth for meromorphic solutions to more general complex q-difference equations: Theorem 10.3.4. Suppose that f is a transcendental meromorphic solution to n

d

j=1

i=1

∑ aj (z)f (qj z) = ∑ bi (z)f (z)i ,

(10.34)

where |q| > 1, d ≥ 2, and the coefficients aj (z) and bi (z) are rational functions. (i) If f is entire or has finitely many poles only, then there exist constants K > 0 and log r

r0 > 0 such that log M(r, f ) ≥ Kd n log |q| for all r ≥ r0 . (ii) If f has infinitely many poles, then there exist constants K > 0 and r0 > 0 such that log r

n(r, f ) ≥ Kd n log |q| for all r ≥ r0 .

Thus the lower order of f always satisfies μ(f ) ≥

log d . n log |q|

10.3 q-delay-differential equations | 229

For more results and examples on q-difference equations, see Zheng and Chen [251], Wang [207], and Chen [29, Chapter 10]. In the q-delay-differential case, Chen, Jiang, and Chen [34, Theorem 1.1] obtained the following result. Theorem 10.3.5. Let f (z) be a meromorphic solution to (f 󸀠 (qz))s = R(z, f ),

(10.35)

where s is a positive integer. Then one of the following possibilities holds: (i) If |q| > 1, f (z) is a transcendental meromorphic solution to (10.35), and d > 2s, then log d − log 2s ≤ μ(f ) ≤ ρ(f ). log |q| Moreover, if d > s, then log d − log s ≤ μ(f ) ≤ ρ(f ). log |q| (ii) If |q| < 1, f (z) is a transcendental meromorphic solution to (10.35), and d ≤ 2s, then ρ(f ) ≤

log 2s − log d . − log |q|

ρ(f ) ≤

log s − log d . − log |q|

Moreover, if d ≤ s, then

(iii) If |q| = 1 and f (z) is a transcendental meromorphic solution to (10.35), then d ≤ 2s. If f (z) is a transcendental entire solution to (10.35), then d ≤ s. We now proceed to considering q-delay-differential equations of the form f (qz) = qf (z)f 󸀠 (z).

(10.36)

Note that z, sin z, and sinh z are entire solutions to (10.36), provided that q = 2 and f (0) = 0. All formal power series solutions to (10.36) were considered by Beardon [8]. To state his results, we first give some notations. We define the formal series 𝒪 and ℐ as 2

𝒪 := 0 + 0z + 0z + ⋅ ⋅ ⋅

and 2

3

ℐ := 0 + z + 0z + 0z + ⋅ ⋅ ⋅ .

230 | 10 Complex q-delay-differential equations Define the sets p

𝒦p := {z : z = p + 2},

p = 1, 2, . . . ,

and 𝒦 := 𝒦1 ∪ 𝒦2 ∪ ⋅ ⋅ ⋅ .

Obviously, 𝒦1 = {3}, 𝒦2 = {−2, 2}, and 𝒦p contains exactly p points equally spaced 1

around the circle |z| = Rp , where Rp = (p + 2) p and Rp ∈ 𝒦p . We also note that Rp is decreasing with p since log(x+2) is decreasing and x lim Rp = 1.

p→+∞

Beardon [8] obtained the following existence result for formal solutions. Theorem 10.3.6. Any transcendental solution to (10.36) is of the from f (z) = z + z(bz p + ⋅ ⋅ ⋅), where p is a positive integer, b ≠ 0, and q ∈ 𝒦p . In particular, if q ∈ 𝒦, then the only formal solutions to (10.36) are 𝒪 and ℐ . Before stating another result due to Beardon [8], we recall that by definition f and g are linear conjugates if there is a nonzero constant c such that g(z) = cf ( zc ). Thus if f satisfies (10.36) and g is a linear conjugate to f , then g also satisfies (10.36). Indeed, z z g 󸀠 (z) = [cf ( )] = f 󸀠 ( ), c c 󸀠

so that g(qz) = cf (

qz z z ) = cqf ( ) [f ( )]󸀠 = qg(z)g 󸀠 (z). c c c

Theorem 10.3.7. For each positive integer p, there is a unique real entire solution Fp = z(1 + z p + b2 z 2p + b3 z 3p + ⋅ ⋅ ⋅) to (10.36) for each q ∈ 𝒦p . Further, if q ∈ 𝒦p , then the only transcendental solutions to (10.36) are the linear conjugates of functions in Fp . Proof. See [8, p. 275–277]. When estimating the order of solutions to q-difference or to q-delay-differential equations, the following lemma plays an important part; see [78], Lemma 3.1. We also note that it remains true for Φ ≡ 0.

10.3 q-delay-differential equations | 231

Lemma 10.3.8. Let Φ : (1, ∞) → (0, ∞) be an increasing function, and let f be a nonconstant meromorphic function. If for some real constant α ∈ (0, 1), there exist real constants K1 > 0 and K2 ≥ 1 such that T(r, f ) ≤ K1 Φ(αr) + K2 T(αr, f ) + S(αr, f ), then the order of growth of f satisfies ρ(f ) ≤

log K2 log Φ(r) + lim sup . − log α log r r→∞

For the upper bound of the order of solutions to (10.36), see Zhang [241, Theorems 1 and 2]: Theorem 10.3.9. If f (z) is a transcendental entire solution to (10.36) for q ∈ 𝒦, then log 2 ρ(f ) ≤ log . Furthermore, if q ≥ 3, then (10.36) has no periodic solutions. |q| Proof. By the definition of 𝒦 we know that |q| > 1. We have T(r, f (qz)) = T(r, qf (z)f 󸀠 (z)) ≤ 2T(r, f ) + S(r, f ).

(10.37)

Thus (10.1) implies that

Setting α =

1 , |q|

T(|q|r, f (z)) ≤ 2T(r, f ) + S(r, f ).

(10.38)

T(r, f ) ≤ 2T(αr, f ) + S(αr, f ).

(10.39)

we have

log 2 Now applying Lemma 10.3.8, we deduce that ρ(f ) ≤ log . Obviously, if q ≥ 3, then we |q| see that ρ(f ) < 1, which is impossible for periodic functions f (z).

A slightly more general result was obtained by Xu, Yang, and Wang [223]. They considered the order of solutions to f (qz)n = qf (z)[f 󸀠 (z)]s , which can also be written as [f 󸀠 (z)]s =

f (qz)n . qf (z)

n Their result reads as ρ(f ) ≤ log(s+1)−log . log |q| Looking at a more general situation, Chen, Jiang, and Chen [34, Theorem 1.5] obtained the following result.

232 | 10 Complex q-delay-differential equations Theorem 10.3.10. Let |q| > 1. Suppose that f (z) is a transcendental meromorphic solution to n

∑ αs (z)f (λs ) (z) =

s=1

P(qz, f (qz)) , Q(z, f (z))

(10.40)

where P(z, f ), Q(z, f ) are rational functions irreducible in f with meromorphic coefficients of small growth S(r, f ). Moreover, assume that each αs is small in the sense of T(r, αs ) = S(r, f ). Denote and assume that 1 ≤ pP := degf P ≤ degf Q =: qQ and λ := ∑ns=1 λs . Then ρ(f ) ≤

log(qQ + λ + n) − log pP . log |q|

Moreover, if qQ > pP + λ + n, then log(qQ − λ − n) − log pP log(qQ + λ + n) − log pP ≤ μ(f ) ≤ ρ(f ) ≤ . log |q| log |q| Proof. Obviously, f is a transcendental meromorphic function. By standard Nevanlinna theory estimates we see that T(r, P(qz, f (qz))) = pP T(r, f (qz)) + S(r, f (qz)) = pP T(|q|r, f ) + S(|q|r, f ) n

= T (r, Q(z, f (z)) ∑ αs (z)f (λs ) (z)) s=1

n

≤ T(r, Q(z, f (z))) + T (r, ∑ f (λs ) ) + S(r, f ) n

s=1

≤ qQ T(r, f ) + ∑(T(r, f ) + λs N(r, f ) + S(r, f )) + S(r, f ) s=1 n

≤ qQ T(r, f ) + ∑(1 + λs )T(r, f ) + S(r, f ) s=1

= (qQ + λ + n)T(r, f ) + S(r, f ), and hence pP T(|q|r, f ) + S(|q|r, f ) ≤ (qQ + λ + n)T(r, f ) + S(r, f ), which is equal to pP T(r, f ) + S(r, f ) ≤ (qQ + λ + n)T(r/|q|, f ) + S(r/|q|, f ) for sufficiently large r outside a possible exceptional set of finite logarithmic measure. By a standard reasoning, see [72, Lemma 5], we may remove the exceptional set to obtain, with β > 1, β/|q| < 1,

pP (1 − ε)T(r, f ) ≤ (qQ + λ + n)(1 + ε)T(βr/|q|, f )

10.3 q-delay-differential equations | 233

for all r sufficiently large. Since 1 ≤ pP ≤ qQ , we have Lemma 10.3.8, we conclude that ρ(f ) ≤

(qQ +λ+n)(1+ε) pP (1−ε)

≥ 1. Recalling

log(qQ + λ + n)(1 + ε) − log(pP (1 − ε)) , − log(β/|q|)

and hence ρ(f ) ≤

log(qQ + λ + n) − log pP log |q|

by letting ε → 0 and β → 1. To prove the second statement, let us first note that T(r, Q(z, f (z))) = qQ T(r, f ) + S(r, f ) = T (r, n

P(qz, f (qz)) ) ∑ns=1 αs f (λs )

≤ T(r, P(qz, f (qz))) + T(r, ∑ αs f (λs ) ) + O(1) n

s=1

≤ pP T(|q|r, f ) + ∑(T(r, f ) + λs N(r, f ) + S(r, f )) + S(|q|r, f ) s=1

≤ pP T(|q|r, f ) + (λ + n)T(r, f ) + S(r, f ) + S(|q|r, f ),

(10.41)

and hence qQ T(r, f ) ≤ pP T(|q|r, f ) + (λ + n)T(r, f ) + S(|q|r, f ). Given ε > 0, we obtain (qQ − λ − n)(1 − ε)T(r, f ) ≤ pP (1 + ε)T(|q|r, f ) outside a possible exceptional set of finite logarithmic measure. Similarly as before, we remove the exceptional set by taking β > 1 to obtain (qQ − λ − n)(1 − ε)T(r, f ) ≤ pP (1 + ε)T(β|q|r, f ) for all r large enough. Since βqα > 1, we have (10.42) with Lemma 2.2 in [191], we obtain μ(f ) = lim inf r→∞

(qQ −λ−n)(1−ε) pP (1+ε)

(10.42)

> 1. Combining inequality

log+ T(r, f ) log((qQ − λ − n)(1 − ε)) − log(pP (1 + ε)) ≥ , log r log β|q|

and hence ρ(f ) ≥ μ(f ) ≥

log((qQ − λ − n)) − log pP log |q|

by proceeding to limit with ε, β. Note that this final inequality is nontrivial, since qQ > pP + λ + n.

234 | 10 Complex q-delay-differential equations Corollary 10.3.11. Let |q| > 1. Suppose that f (z) is a transcendental meromorphic solution to (f 󸀠 (z))n =

P(qz, f (qz)) , Q(z, f (z))

(10.43)

where P(z, f ), Q(z, f ) are rational functions irreducible in f with meromorphic coefficients of small growth S(r, f ). Assume again that 1 ≤ pP := degf P ≤ degf Q =: qQ . Then ρ(f ) ≤

log(qQ + 2n) − log pP . log |q|

Moreover, if qQ > pP + 2n, then log(qQ − 2n) − log pP log(qQ + 2n) − log pP ≤ μ(f ) ≤ ρ(f ) ≤ . log |q| log |q| Remark 10.3.12. We remark that there are no many papers treating q-delay-differential equations by using the Nevanlinna theory. It seems to us that more considerations in this topic are needed. In Chapter 6, Fermat-type difference and delay-differential equations were treated. To close this chapter, we shortly consider the corresponding results for q-difference and q-delay-differential Fermat-type equations. First, consider the q-difference Fermat-type equations f (z)2 + P(z)2 f (qz)2 = Q(z),

(10.44)

where P(z) and Q(z) are nonzero polynomials. Liu and Cao [146, Theorem 2.6] obtained the following result. Theorem 10.3.13. Let P(z) and Q(z) be nonzero polynomials, and let |q| = 1. If the qdifference equation (10.44) admits a transcendental entire solution of finite order, then P(z) must be a constant P. These solutions can be written in the form f (z) =

Q1 (z)ep(z) + Q2 (z)e−p(z) 2

satisfying one of the following conditions: (i) q satisfies p(qz) = p(z), Q1 (z) − iPQ1 (qz) ≡ 0, Q2 (z) + iPQ2 (qz) ≡ 0, and P 4 Q(q2 z) = Q(z); (ii) q satisfies p(qz) + p(z) = 2a0 , iPQ1 (qz)e2a0 ≡ −Q2 (z), iPQ2 (qz) ≡ Q1 (z)e2a0 , P 4 Q(q2 z) = Q(z), e8a0 = 1, and Q1 (z)Q2 (z) = Q(z), where p(z) is a nonconstant polynomial. Remark 10.3.14. (1) We omit the proof of Theorem 10.3.13. However, we can explain why the condition |q| = 1 is needed. First of all, it is easy to see that the entire solutions of order zero to (10.44) must be polynomials by the factorization of (10.44). To

10.3 q-delay-differential equations | 235

continue, recall that Bergweiler, Ishizaki, and Yanagihara [13] considered the linear q-difference equation n

∑ aj (z)f (qj z) = Q(z),

j=0

(10.45)

where Q(z) and aj (z) (j = 0, 1, . . . , n) are polynomials without common zeros, an (z)a0 (z) ≠ 0, and 0 < |q| < 1. They showed that meromorphic solutions f (z) to (10.45) satisfy T(r, f ) = O((log r)2 ), and hence ρ(f ) = 0. Assume now that f (z) is an entire solution to (10.44) and denote F(z) := f (z)2 . If |q| < 1, then (10.44) takes the form F(z) + P(z)2 F(qz) = Q(z). Then ρ(F) = 0 and ρ(f ) = 0, and thus f (z) must be a polynomial. If |q| > 1, then denote G(z) := f (qz)2 . Thus (10.44) takes the form 1 G ( z) + P(z)2 G(z) = Q(z). q From (10.1) we see that 1 T(r, f (z)) ≤ T(|q|r, f (z)) = T(r, f (qz)) + O(1) = T(r, G(z)) + O(1) 2 = O((log r)2 ) + O(1). Therefore ρ(f ) = 0, and thus f (z) must be a polynomial again. If P(z)2 is replaced by a more general polynomial in (10.44), then the existence of entire solutions and meromorphic solutions to (10.44) remains largely open. As a completion to the preceding theorem, we give some examples. (1) The equation f (z)2 − f (−z)2 = z 3 has a transcendental entire solution f (z) =

zep1 (z) + z 2 e−p1 (z) , 2

where p1 (z) is a polynomial that satisfies p1 (−z) = p1 (z). Write, as we may, p1 (z) = a2n z 2n + a2n−2 z 2(n−1) + ⋅ ⋅ ⋅ + a2 z 2 + a0 .

236 | 10 Complex q-delay-differential equations Then, of course, ρ(f ) = 2n. (2) The equation f (z)2 + f (−iz)2 = z 4 has a transcendental entire solution f (z) =

zep2 (z) + z 3 e−p2 (z) , 2

where p2 (z) is a polynomial that satisfies p2 (−iz) = p2 (z). Now write p2 (z) = a4n z 4n + a4n−4 z 4(n−1) + ⋅ ⋅ ⋅ + a4 z 4 + a0 . This time, we have ρ(f ) = 4n. (3) The equation f (z)2 + f (−z)2 = z 2 has a transcendental entire solution f (z) =

zep(z) + ze−p(z) , 2

where p(z) is a polynomial that satisfies p(−z) + p(z) = 4kiπ +

iπ . 2

This implies that we may write p(z) = a2n+1 z 2n+1 + a2n−1 z 2n−1 + ⋅ ⋅ ⋅ + a1 z + 2kiπ +

iπ , 4

where k ∈ ℤ. Now we have ρ(f ) = 2n + 1. For the q-delay-differential Fermat-type equations, Liu and Cao [146, Theorem 3.2] considered f 󸀠 (z)2 + f (qz)2 = 1.

(10.46)

Chen and Gao [33] treated a generalization of (10.46) and obtained the following result. Theorem 10.3.15. If f (z) a transcendental entire solution of finite order to f (k) (z)2 + f (qz)2 = 1, then q = ±1 and f (qz) = and a2k = ±1. Moreover,

eh(z) +e−h(z) , 2

(10.47)

where h(z) = az + b is a polynomial of degree one,

10.3 q-delay-differential equations | 237

(1) (2) (3) (4) (5)

if q = 1, then f (z) = cos(iaz + ib), ak = −i, and k is odd; if q = −1 and ak = i, then k is odd, and f (z) = ± sin(iaz); if q = −1 and ak = −i, then k is odd, and f (z) = ± cos(iaz); if q = −1 and ak = 1, then k is even, and f (z) = ± sin(iaz − π4 ); if q = −1 and ak = −1, then k is even, and f (z) = ± sin(iaz + π4 ).

Proof. Writing (10.47) in the form (f (qz) + if (k) (z))(f (qz) − if (k) (z)) = 1, we may use the Hadamard factorization theorem to obtain f (qz) + if (k) (z) = eh(z) ,

f (qz) − if (k) (z)) = e−h(z) .

Solving for f (qz) and f (k) (z), we get f (qz) =

eh(z) + e−h(z) , 2

f (k) (z) =

eh(z) − e−h(z) , 2i

where h is a nonconstant polynomial. Differentiating f (qz) k times and using these representations, we obtain eh(qz) − e−h(qz) h1 (z)eh(z) + h2 (z)e−h(z) = , 2i 2qk

(10.48)

where h1 (z) = (h󸀠 (z))k +Pk−1 , h2 (z) = (−h󸀠 (z))k +Qk−1 , and Pk−1 and Qk−1 are polynomials in h󸀠 , . . . , h(k) of degree k − 1. Using (10.48), we obtain −

ih1 (z) h(qz)+h(z) ih2 (z) h(qz)−h(z) e e − + e2h(qz) = 1. qk qk

(10.49)

We may now apply the Borel theorem; see Theorem 1.1.32. Since f is transcendental, we obtain a contradiction unless either h(qz) + h(z) or h(qz) − h(z) is a constant. Suppose now for a while that h(qz) + h(z) = A is constant. Then (10.49) takes the form −

ih1 (z) A ih2 (z) −A 2h(qz) e − e e + e2h(qz) = 1. qk qk

A This is a contradiction, unless we have − ihq1 (z) ≡ 1 and k e

ih2 (z) −A e qk

≡ 1. Therefore

q2k ≡ h1 (z)h2 (z) = (−1)k (h󸀠 (z))2k + ⋅ ⋅ ⋅ . This means that h󸀠 is a constant, and hence h(z) = az +b, a ≠ 0. Then A = a(q+1)z +2b, k k iak A a −A = 1 and i(−1) e = 1, we get (−1)k a2k = 1. If now k is and so q = −1. Since − (−1) ke (−1)k

238 | 10 Complex q-delay-differential equations odd, then we have ak = ±i. If ak = i, then eA = e−A = −1, and b = A/2 = (m + 1/2)πi for m ∈ ℤ. Then 1 f (qz) = f (−z) = (eaz+b + e−(az+b) ), 2 and hence 1 f (z) = (eaz−b + e−(az−b) ) = cos (iaz − b) = cos(iaz + (m + 1/2)π) = ± sin(iaz). 2 If ak = −i, then a similar computation results in f (z) = ± cos(iaz). Then if k is even and ak = ±1, a completely similar reasoning applies to prove claims (4) and (5). It remains to consider the case where h(qz) − h(z) = B is a constant. Again, we conclude that h(z) = az + b, and thus B = a(q − 1)z, meaning that q = 1. Then a simple computation shows that ak = −i and that k is odd. Similarly, we get f (z) = cos(iaz +ib), completing the proof. Corollary 10.3.16. There do not exist transcendental entire solutions to equation (10.47) if q ≠ ±1. We next add a related theorem; see [33], Theorem 1.3. We omit the proof. The reader may consult [33]. Theorem 10.3.17. All transcendental entire solutions of finite order to the q-delaydifferential equation (f (qz) − f (z))2 + (f (k) (z))2 = 1 are of the form f (z) = ± 21 sin(iaz) + c, where c is a constant. We close this chapter by adding a result where the q-shift, the translation shift, and the differentiation appear in the equation to be considered. Theorem 10.3.18. All transcendental entire solutions of finite order to f 󸀠 (z + c)2 + f (qz)2 = 1

(10.50)

satisfy one of the following conditions: (1) If q = −1, then f (z) = sin(z−Ci−c), where C = 21 ic+kπi, k ∈ ℤ, or f (z) = sin(z−Ai+c), where C = − 21 ic + (k + 21 )πi, k ∈ ℤ. (2) If q = 1, then f (z) = sin(−z − Ai), where c = 2πi, or f (z) = sin(−z − Ci + π), where c = (2k + 1)π, k ∈ ℤ. Proof. Starting as in the proof of Theorem 10.3.15, we have f 󸀠 (z + c) =

eh(z) + e−h(z) 2

(10.51)

10.3 q-delay-differential equations | 239

and f (qz) =

eh(z) − e−h(z) , 2i

(10.52)

where h is a polynomial. Differentiating (10.52) and combining with f 󸀠 (qz + c) = 1 h(qz) (e + e−h(qz) ), from (10.51) we get 2 1 󸀠 1 h (z + c/q)eh(z+c/q)+h(qz) + h󸀠 (z + c/q)eh(qz)−h(z+c/q) − e2h(qz) = 1. iq iq

(10.53)

Again applying the Borel theorem (Theorem 1.1.32), as the proof of Theorem 10.3.15, we conclude than either h(z + c/q) + h(qz) = A or h(qz) − h(z + c/q) = B is a constant. Now first suppose that we have h(z + c/q) + h(qz) = A. Repeating the reasoning as in the proof of Theorem 10.3.15, we obtain that iq1 h󸀠 (z + c/q)eA = 1 and iq1 h󸀠 (z + c/q)e−A = 1,

implying that e2A = 1. Suppose first that eA = 1. Then h󸀠 (z + c/q) = iq, meaning that h󸀠 (z) ≡ iq, and so h(z) = iqz + C0 . Changing the constant C0 , as we may, we write h(z) = iqz + C − ic. We now conclude that h(z + c/q) + h(qz) = 2kiπ, and further q = −1, 2C − ic = 2kiπ, where k is an integer. Hence h(z) = −iz + C − ic, and so f (z) =

eiz+C−ic − e−iz−C+ic = sin(z − Ci + c). 2i

Secondly, if eA = −1, then a completely similar reasoning implies that q = −1, 2C + ic = 2kiπ + iπ, k ∈ ℤ, and f (z) =

eiz+C+ic − e−iz−C−ic = sin(z − Ci − c). 2i

It remains to treat the case where h(qz) − h(z + c/q) = B is a constant. As before, we have either eB = 1 or eB = −1. In the first case, we get q = 1, c = 2kπ, k ∈ ℤ, and h(z) = iz + C − 2kiπ. Then f (z) =

1 iz+C−2kiπ (e − e−iz−C+2kiπ ) = sin(−z − Ci). 2i

In the final case eB = −1, we obtain q = 1, c = 2kπ + π, h(z) = −iz + C − 2kiπ − iπ, and f (z) =

1 −iz+C−2kiπ−iπ (e − eiz−C+2kiπ+iπ ) = sin(−z − Ci + π). 2i

Corollary 10.3.19. If q ≠ ±1, then equation (10.50) cannot admit a transcendental entire solution of finite order.

11 Systems of complex delay-differential equations 11.1 Background The systems of complex differential, difference, and delay-differential equations are an interesting research field in complex analysis, although not many investigations yet appeared from the point of view of value distribution theory. A simple example in this field is the elliptic functions that satisfy the following system of difference equations: f (z + α) − f (z) = 0,

{

f (z + β) − f (z) = 0,

Im( αβ )

where ≠ 0. As a background to this chapter, we refer the reader to Naftalevich and Gylys [177] and references therein. We also refer to [200], pointing out some works on systems of difference and delay-differential equations. In particular, there are some works on systems of difference equations by Löwig [163, 164]. In fact, Löwig succeeded in generalizing certain Picard’s results to systems of the form fk (z + h) = Qk (z, f1 (z), . . . , fn (z)),

k = 1, 2, . . . , n,

where Qk (z, f1 (z), . . . , fn (z)) depends not only on f1 , . . . , fn , but also on the variable z. The main results due to Löwig are concerned with the existence and uniqueness of ω-periodic solutions when the coefficients of the equation are ω-periodic functions. This chapter is devoted to certain types of systems of complex delay-differential equations, using Nevanlinna theory as an important tool.

11.2 Malmquist-type delay-differential systems Malmquist-type differential, difference, and delay-differential equations were previously treated in Chapter 8. For the corresponding results on Malmquist-type systems of complex differential equations, see, e. g., [56, 205]. For Malmquist-type systems of complex difference equations, not to be considered here, see, e. g., [57, 206]. For Malmquist-type delay-differential systems, we refer to Li and Gao [130]. The next theorem is a modification of [130, Theorem 2.1]. Denote { { R1 (z, f1 (z)) = { { { { { { { {R2 (z, f2 (z)) = {

a10 (z) + a11 (z)f1 (z) + ⋅ ⋅ ⋅ + a1p1 (z)f1 (z)p1 b10 (z) + b11 (z)f1 (z) + ⋅ ⋅ ⋅ + b1q1 (z)f1 (z)q1

=

a20 (z) + a21 (z)f2 (z) + ⋅ ⋅ ⋅ + a2p2 (z)f2 (z)p2

P1 (z, f1 (z)) , Q1 (z, f1 (z))

b20 (z) + b21 (z)f2 (z) + ⋅ ⋅ ⋅ + b2q2 (z)f2 (z)q2

=

P2 (z, f2 (z)) , Q2 (z, f2 (z))

where aik (z) and bit (z) are small functions with respect to fi (z), and Pi (z, fi (z)) and Qi (z, fi (z)) are irreducible polynomials in fi , i = 1, 2, k = 0, 1, 2, . . . , pi , t = 0, 1, 2, . . . , qi . Moreover, denote di := degfi Ri (z, fi (z)) = max{pi , qi }. https://doi.org/10.1515/9783110560565-011

242 | 11 Systems of complex delay-differential equations Theorem 11.2.1. Let c1 , c2 , . . . , cn be nonzero distinct complex numbers. Suppose that (f1 (z), f2 (z)) is a transcendental meromorphic solution to n

(λ ) { { ∑ α1j (z)f1 1j (z + cj ) = R2 (z, f2 (z)), { { { j=1 { { n (λ2j ) { { { ∑ α2j (z)f2 (z + cj ) = R1 (z, f1 (z)), { j=1

(11.1)

where the coefficients αij (z) are small functions with respect to fi (z), and λij are nonnegative integers, i = 1, 2, j = 1, 2, . . . , n. Denote n

λi := ∑(λij + 1), j=1

i = 1, 2.

If max{ρ2 (f1 ), ρ2 (f2 )} < 1, then d1 d2 ≤ λ1 λ2 . Proof. Since the coefficients in (11.1) are small functions with respect to fi (z) (i = 1, 2), we have that T(r, αij ) = S(r, fi ), i = 1, 2, j = 1, 2, . . . , n, { { { T(r, aik ) = S(r, fi ), i = 1, 2, k = 1, 2, . . . , pi , { { { {T(r, bit ) = S(r, fi ), i = 1, 2, t = 1, 2, . . . , qi , for all r outside a possible exceptional set E1 of finite logarithmic measure. Let (f1 (z), f2 (z)) be a meromorphic solution to (11.1) with max{ρ2 (f1 ), ρ2 (f2 )} < 1. Taking the Nevanlinna characteristic function on both sides of each equation of (11.1) and applying Theorem 1.1.29, we obtain d2 T(r, f2 (z)) = T(r, R2 (z, f2 (z))) + S(r, f2 (z)) n

= T(r, ∑ α1j (z)f1 n

(λ1j )

j=1

≤ ∑ T(r, f1 j=1

(λ1j )

(z + cj )) + S(r, f2 (z))

(z + cj )) + S(r, f1 (z)) + S(r, f2 (z))

n

≤ ∑(λ1j + 1)T(r, f1 (z + cj )) + S(r, f1 (z)) + S(r, f2 (z)) j=1 n

= ∑(λ1j + 1)T(r, f1 (z)) + S(r, f1 (z)) + S(r, f2 (z)) j=1

= λ1 T(r, f1 (z)) + S(r, f1 (z)) + S(r, f2 (z)). Hence d2 T(r, f2 (z)) ≤ λ1 T(r, f1 (z)) + S(r, f1 (z)) + S(r, f2 (z)).

(11.2)

11.2 Malmquist-type delay-differential systems | 243

Similarly, we have d1 T(r, f1 (z)) ≤ λ2 T(r, f2 (z)) + S(r, f1 (z)) + S(r, f2 (z)).

(11.3)

Therefore (d1 + o(1))T(r, f1 ) ≤ (λ2 + o(1))T(r, f2 ), and (d2 + o(1))T(r, f2 ) ≤ (λ1 + o(1))T(r, f1 ). By multiplying these estimates we have (d1 + o(1))(d2 + o(1)) ≤ (λ1 + o(1))(λ2 + o(1)), and hence d1 d2 + o(1)(d1 + d2 + o(1)) ≤ λ1 λ2 + o(1)(λ1 + λ2 + o(1)). Clearly, we must have d1 d2 ≤ λ1 λ2 to avoid a contradiction. An example following the next corollary points out that the equality d1 d2 = λ1 λ2 indeed may appear. Corollary 11.2.2. Let (f1 (z), f2 (z)) be a transcendental meromorphic solution with max{ρ2 (f1 ), ρ2 (f2 )} < 1 to f1󸀠 (z + 1) = R2 (z, f2 (z)),

{

f2󸀠 (z + 1) = R1 (z, f1 (z)).

Then d1 d2 ≤ 4 = λ1 λ2 . Example 11.2.3. The system π π 2 󸀠 {f1 (z + 1) = + f2 (z) , { 2 2 { {f 󸀠 (z + 1) = − π − π f (z)2 , {2 2 2 1

(11.4)

admits a transcendental meromorphic solution (f1 (z), f2 (z)) = (tan

π π z, cot z) . 2 2

Remark 11.2.4. It is not difficult to see that d1 = d2 = 1 for transcendental entire solutions with max{ρ2 (f1 ), ρ2 (f2 )} < 1 to (11.4). Using the notations of Theorem 11.2.1, we add the following result from [130, Theorem 2.3].

244 | 11 Systems of complex delay-differential equations Theorem 11.2.5. Suppose (f1 (z), f2 (z)) is a transcendental meromorphic solution to a delay-differential system n

(λ ) { { ∑ α1j (z)f1 1j (z + cj ) = f2 (p(z)), { { { j=1 { { n (λ2j ) { { { ∑ α2j (z)f2 (z + cj ) = f1 (p(z)), { j=1

(11.5)

where p(z) is a polynomial of degree k ≥ 2. Moreover, assume that λi = ∑nj=1 (λij + 1) ≥ k for i = 1, 2. Then T(r, f1 ) + T(r, f2 ) = O((log r)α ), where α=

log max{λ1 , λ2 } . log k

Proof. Let p(z) := ak z k + ⋅ ⋅ ⋅ + a0 , ak ≠ 0. Given 0 < δ < ak , denote α := |ak | + δ and β := |ak | − δ. Moreover, denote C := max{|c1 |, . . . , |cn |}. Given ε > ε0 , using [63, Lemma 2], we see that n

(1 − ε) T (βr k , f1 ) ≤ T(r, f1 (p(z))) = T (r, ∑ α2j (z)f2 n

j=1

≤ ∑ T(r, f2 j=1

(λ2j )

(λ2j )

(z + cj ))

(z + cj )) + S(r, f2 )

(11.6)

n

≤ ∑(λ2j + 1)T(r, f2 (z + cj )) + S(r, f2 ) j=1

≤ λ2 T(r + C, f2 ) + S(r, f2 ). Given γ > 1, we clearly have T(r + C, f2 ) ≤ T(γr, f2 ) for all r large enough. Therefore we obtain (1 − ε)T(βr k , f1 ) ≤ λ2 T(γr, f2 ) + S(r, f2 ) outside a possible exceptional set of finite linear measure. By a standard reasoning to remove the exceptional set, given μ > 1, we get (1 − ε)T(βr k , f1 ) ≤ λ2 T(μγr, f2 ). Obviously, the same inequality follows by changing λ1 , f1 and λ2 , f2 . Adding the results, we conclude that, with λ := max{λ1 , λ2 }, (1 − ε)(T(βr k , f1 ) + T(βr k , f2 )) ≤ λ(T(μγr, f1 ) + T(μγr, f2 )).

11.3 Fermat-type delay-differential systems | 245

Therefore, using the notation t := μγr, we may write this as T(

β k β k λ (T(t, f1 ) + T(t, f2 )). t , f1 ) + T ( t , f2 ) ≤ 1−ε (μγ)k (μγ)k

(11.7)

By (11.7) and [63, Lemma 3(i)] we conclude that T(r, f1 ) + T(r, f2 ) = O((log r)τ ), where τ =

log(λ/(1−ε)) . log k

Letting ε → 0, the statement follows.

11.3 Fermat-type delay-differential systems The existence of meromorphic solutions of Fermat-type delay-differential equations was treated in Chapter 6.3.3. We now proceed to introduce Fermat-type systems of complex delay-differential equations. Gao [59] considered systems of the form {

f1󸀠 (z)n1 + [f2 (z + c)]m1 = Q1 (z),

f2󸀠 (z)n2 + [f1 (z + c)]m2 = Q2 (z),

(11.8)

where Q1 (z) and Q2 (z) are nonzero polynomials, and n1 , n2 , m1 , m2 are positive integers. The following result is a modification of the case of finite order in [59, Theorem 1] to hyperorder less than one. Theorem 11.3.1. System (11.8) does not admit a transcendental entire solution (f1 (z), f2 (z)) with max{ρ2 (f1 ), ρ2 (f2 )} < 1 if one of the following conditions is satisfied: (i) m1 m2 > n1 n2 , ni (ii) mi > n −1 , i = 1, 2. i

Proof. Assume that (f1 (z), f2 (z)) is a transcendental entire solution to (11.8) with max{ρ2 (f1 ), ρ2 (f2 )} < 1. (i) If m1 m2 > n1 n2 , then we may apply Theorem 1.1.29 and Lemma 1.2.10 to the first equation of (11.8): m1 T(r, f2 ) + S(r, f2 ) = m1 T(r, f2 (z + c)) = T(r, f1󸀠 (z)n1 − Q1 (z)) = n1 T(r, f1󸀠 (z)) + S(r, f1 ) ≤ n1 T(r, f1 (z)) + S(r, f1 ),

that is, (m1 + o(1))T(r, f2 ) ≤ (n1 + o(1))T(r, f1 (z)).

(11.9)

Completely similarly with the second equation of (11.8) we get (m2 + o(1))T(r, f1 ) ≤ (n2 + o(1))T(r, f2 (z)).

(11.10)

246 | 11 Systems of complex delay-differential equations It follows from (11.9) and (11.10) that m1 m2 ≤ n1 n2 , which is a contradiction to m1 m2 > n1 n2 . n

i , i = 1, 2, then applying the distinct-root variant of the second main (ii) If mi > n −1 i theorem (Theorem 1.1.21), we first obtain

n1 T(r, f1󸀠 ) = T(r, (f1󸀠 )n1 ) ≤ N(r, 1/(f1󸀠 )) + N (

1 ) + S(r, f1󸀠 ), (f1󸀠 )n1 − Q1

and hence 1 ) + S(r, f1󸀠 (z)) f1󸀠 (z)n1 − Q1 (z) 1 ) + S(r, f1 ) ≤ N (r, f2 (z + c) ≤ T(r, f2 (z + c)) + S(r, f1 )

(n1 − 1)T(r, f1󸀠 (z)) ≤ N (r,

≤ T(r, f2 (z)) + S(r, f1 ) + S(r, f2 ).

(11.11)

From the first equation of system (11.8) and from (11.11) we get m1 T(r, f2 (z + c)) = n1 T(r, f1󸀠 (z)) + S(r, f1 ) n1 ≤ T(r, f2 (z)) + S(r, f1 ) + S(r, f2 ). n1 − 1

(11.12)

Similarly, m2 T(r, f1 (z + c)) = n2 T(r, f2󸀠 (z)) + S(r, f2 ) n2 ≤ T(r, f1 (z)) + S(r, f1 ) + S(r, f2 ). n2 − 1

(11.13)

It follows from (11.12) and (11.13) that, respectively, (m1 −

n1 + o(1)) T(r, f2 ) ≤ S(r, f1 ) n1 − 1

(11.14)

(m2 −

n2 + o(1)) T(r, f1 ) ≤ S(r, f2 ). n2 − 1

(11.15)

and

Combining (11.14) and (11.15), we obtain (m1 −

n1 n2 S(r, f1 ) S(r, f2 ) + o(1)) (m2 − + o(1)) ≤ . n1 − 1 n2 − 1 T(r, f1 ) T(r, f2 )

(11.16)

n1 n2 ) (m2 − ) ≤ 0, n1 − 1 n2 − 1

(11.17)

Thus (m1 − which is a contradiction to mi >

ni , ni −1

i = 1, 2.

11.3 Fermat-type delay-differential systems | 247

The case n1 = n2 = m1 = m2 in (11.8) can be called a Fermat-type system of complex delay-differential equations. We start looking at the system {

f1󸀠 (z)2 + [f2 (z + c)]2 = 1, f2󸀠 (z)2 + [f1 (z + c)]2 = 1;

(11.18)

see Gao [59], who also considered the slightly more general case where the constant had been replaced by nonzero polynomials on the right-hand side of (11.18). Unfortunately, we do not know examples showing the existence of nonentire meromorphic solutions or transcendental entire solutions of infinite order to (11.18). Considering the transcendental entire solutions of finite order, Gao [58, Theorem 1.1] obtained the following result. Theorem 11.3.2. Let (f1 (z), f2 (z)) be a transcendental entire solution of finite order to (11.18). Then (f1 (z), f2 (z)) = (sin(z−bi), sin(z−b1 i)) and c = kπ, where b, b1 are constants. We omit the proof of this theorem. Indeed, we may consider slightly more general systems, where the shift has been replaced by a nonconstant polynomial g(z): {

f1󸀠 (z)2 + [f2 (g(z))]2 = 1, f2󸀠 (z)2 + [f1 (g(z))]2 = 1.

(11.19)

However, by [62], Lemma 4 we immediately conclude that g must be a polynomial of degree one, and hence g(z) = Az + c. Therefore we may proceed to considering transcendental entire solutions to {

f1󸀠 (z)2 + [f2 (Az + c)]2 = 1, f2󸀠 (z)2 + [f1 (Az + c)]2 = 1;

(11.20)

see Liu and Ma [158, Theorem 3.1]. We now proceed to prove the following: Theorem 11.3.3. Let (f1 (z), f2 (z)) be a transcendental entire solution of finite order to (11.20). Then (f1 (z), f2 (z)) = (sin(z + b󸀠 ), sin(z + b󸀠󸀠 )), where b󸀠 , b󸀠󸀠 are constants. In this case, we do not give precise expressions of the solutions of (11.20). However, the semi-precise expressions show that the solutions are of order one. For the proof of Theorem 11.3.3, we need the following simple lemma: Lemma 11.3.4. Suppose h1 (z), h2 (z) are nonconstant polynomials. If now sin(h1 (z)) = p(z) sin(h2 (z)), then p(z) should be a constant p and p2 = 1. Proof. From sin(h1 (z)) = p(z) sin(h2 (z)) we have eih1 (z) − e−ih1 (z) = p(z)eih2 (z) − p(z)e−ih2 (z) , and thus eih1 (z)+ih2 (z) eih2 (z)−ih1 (z) + + e2ih2 (z) = 1. −p(z) p(z)

(11.21)

248 | 11 Systems of complex delay-differential equations Obviously, e2ih2 (z) ≢ 1. By Theorem 1.1.34 we have either eih1 (z)+ih2 (z) ≡1 −p(z)

(11.22)

eih2 (z)−ih1 (z) ≡ 1. p(z)

(11.23)

or

Therefore p(z) should be a constant p, since the left-hand side of (11.22), resp., of (11.23), has no poles. ih2 (z)−ih1 (z) If (11.22) occurs, then from (11.21) it follows that e p + e2ih2 (z) = 0, and thus

p2 = 1. If (11.23) occurs, then from (11.21) it follows that

p2 = 1.

eih1 (z)+ih2 (z) −p

+ e2ih2 (z) = 0, and thus

Proof of Theorem 11.3.3. From a classical result in Chapter 6.2 due to Gross we obtain f1󸀠 (z) = sin h1 (z), { f2 (Az + c) = cos h1 (z),

(11.24)

f2󸀠 (z) = sin h2 (z), { f1 (Az + c) = cos h2 (z),

(11.25)

and

where h1 (z) and h2 (z) are nonconstant polynomials, since f1 (z) and f2 (z) are transcendental entire functions of finite order. Using these two systems, we have f 󸀠 (Az + c) = sin h1 (Az + c), { {1 󸀠 { {f 󸀠 (Az + c) = −h2 (z) sin h (z), 2 {1 A and f 󸀠 (Az + c) = sin h2 (Az + c), { {2 󸀠 { {f 󸀠 (Az + c) = −h1 (z) sin h (z). 1 {2 A Thus we have sin h1 (Az + c) =

−h󸀠2 (z) sin h2 (z) A

(11.26)

sin h2 (Az + c) =

−h󸀠1 (z) sin h1 (z). A

(11.27)

and

11.3 Fermat-type delay-differential systems | 249

By Lemma 11.3.4, h󸀠1 (z), h󸀠2 (z) are constants ±A. So, we may assume that h1 (z) = a1 z + b1 and h2 (z) = a2 z + b2 , where a1 = ±A and a2 = ±A. By (11.26) and (11.27) we see that sin h1 (A2 z + Ac + c) =

h󸀠2 (Az + c)h󸀠1 (z) sin h1 (z). A2

(11.28)

By Lemma 11.3.4 we must have that h󸀠2 (Az +c)h󸀠1 (z) = ±A2 . We separately consider these two possibilities. Case 1. h󸀠2 (Az + c)h󸀠1 (z) = A2 = a1 a2 . Then (11.28) implies that sin h1 (A2 z + Ac + c) = sin h1 (z), so h1 (A2 z + Ac + c) − h1 (z) = 2kπ, where k is an integer. By h1 (z) = a1 z + b1 we must have A2 = 1. Case 1.1. If A = 1, then we have to consider two subcases. Subcase 1.1.1. If a1 = 1, then a2 = 1 as well. Then h1 (z) = z + b1 , h2 (z) = z + b2 , and c = kπ. The second equation of (11.25) implies that f1 (z + c) = cos(z + b2 ) = sin (z + b2 +

π + 2k1 π) , 2

and hence f1 (z) = sin(z + b󸀠 ), where b󸀠 = b2 + π2 + 2k1 π − kπ, and k1 is an integer. Moreover, the first equation of (11.24) implies that f2 (z + c) = cos(z + b1 ) = sin (z + b1 + and hence f2 (z) = sin(z + b󸀠󸀠 ), where b󸀠 = b1 +

π 2

π + 2k2 π) , 2

+ 2k2 π − kπ, and k2 is an integer.

Subcase 1.1.2. If a1 = −1, then h1 (z) = −z + b1 and h2 (z) = −z + b2 . We also get f1 (z) = sin(z + b󸀠 ), f2 (z) = sin(z + b󸀠󸀠 ) with suitable modifications of b󸀠 , b󸀠󸀠 . Case 1.2. If A = −1, we again need two subcases to be discussed. Subcase 1.2.1. A = −1, and c is a nonzero constant. Thus a1 a2 = 1. We also get f1 (z) = sin(z + b󸀠 ), f2 (z) = sin(z + b󸀠󸀠 ) by using similar discussions as above with even cos z. Subcase 1.2.2. If A = −1 and c = 0, then from (11.27) we get that a1 and a2 are either 1 or −1. Then we see, similarly as before, that f1 (z) = sin(z + b󸀠 ), f2 (z) = sin(z + b󸀠󸀠 ). Case 2. If h󸀠2 (Az + c)h󸀠1 (z) = −A2 = a1 a2 , then from (11.28) we have sin h1 (A2 z + Ac + c) = − sin h1 (z),

250 | 11 Systems of complex delay-differential equations and hence h1 (A2 z + Ac + c) + h1 (z) = 2kπ. This implies that A2 = −1 by h1 (z) = a1 z + b1 . Case 2.1. If A = i, then c =

2kπ−2b1 . a1 (i+1)

Then (11.27) reduces to

sin h2 (z + c) =

−a1 sin h1 (z), A

and so a21 = −1 follows by Lemma 11.3.4. Subcase 2.1.1. If a1 = i, then h1 (z) = iz + b1 and h2 (z) = −iz + b2 . From the second equation of (11.25) we get f1 (iz + c) = cos(−iz + b2 ) = sin (iz − b2 +

π + 2k1 π) , 2

and hence f1 (z) = sin(z + b󸀠 ), where b󸀠 is a constant, and k1 is an integer. From the second equation of (11.24) we also have f2 (iz + c) = cos(iz + b1 ) = sin (iz + b1 +

π + 2k2 π) , 2

and hence f2 (z) = sin(z + b󸀠󸀠 ), where b󸀠󸀠 is a constant, and k2 is an integer. Subcase 2.1.2. If a1 = −i, then h1 (z) = −iz + b1 and h2 (z) = iz + b2 . Similarly as before, we now obtain f1 (z) = sin(z +b󸀠 ), f2 (z) = sin(z +b󸀠󸀠 ) with suitable modifications of b󸀠 , b󸀠󸀠 . Case 2.2. If A = −i, then more or less the same reasoning results in f1 (z) = sin(z+b󸀠 ) and f2 (z) = sin(z + b󸀠󸀠 ) with suitable modifications of b󸀠 , b󸀠󸀠 , completing the proof. To close this section, we shortly remind the readers that Fermat-type partial difference equations were considered by Xu and Cao [227]. Fermat-type systems of partial difference equations f1 (z1 , z2 )2 + f2 (z1 + c1 , z2 + c2 )2 = 1,

{

f2 (z1 , z2 )2 + f1 (z1 + c1 , z2 + c2 )2 = 1,

and Fermat-type systems of partial delay-differential equations n1

𝜕f (z , z ) { ( 1 1 2 ) + f2 (z1 + c1 , z2 + c2 )m1 = 1, { { { 𝜕z1 n2 { { { { ( 𝜕f2 (z1 , z2 ) ) + f1 (z1 + c1 , z2 + c2 )m2 = 1, 𝜕z1 {

11.4 Systems of q-delay-differential equations | 251

and 𝜕f1 (z1 , z2 ) 2 { ( ) + [f2 (z1 + c1 , z2 + c2 ) − f1 (z1 , z2 )]2 = 1, { { { 𝜕z1 { { { 𝜕f2 (z1 , z2 ) 2 { ( ) + [f1 (z1 + c1 , z2 + c2 ) − f2 (z1 , z2 )]2 = 1, 𝜕z1 { were treated by Xu, Liu, and Li [225]. In these systems, c1 , c2 are constants, and n1 , n2 , m1 , m2 are positive integers.

11.4 Systems of q-delay-differential equations Inspired by the q-delay-differential equations that we previously considered in Chapter 10.3, Xu and Tu [224] investigated the growth of solutions of some systems of q-delay-differential equations. Some of their results are as follows. Theorem 11.4.1. Suppose that (f1 , f2 ) is a solution to a delay-differential system {

f1 (q1 z) = c1 f2 (z)f2󸀠 (z),

(11.29)

f2 (q2 z) = c2 f1 (z)f1󸀠 (z),

where |q1 | > 1, |q2 | > 1, and c1 , c2 are nonzero constants. If f1 , f2 are transcendental entire functions, then ρ(fi ) ≤

2 log 2 , log |q1 | + log |q2 |

i = 1, 2.

Proof. From (11.29) we have T(r, f1 (q1 z)) ≤ T(r, f2 ) + T(r, f2󸀠 (z)) + O(1), { T(r, f2 (q2 z)) ≤ T(r, f1 ) + T(r, f1󸀠 (z)) + O(1). Since f1 , f2 are transcendental entire functions, it follows by Theorem 10.1.1 that T(|q1 |r, f1 (z)) ≤ 2T(r, f2 ) + S(r, f2 ),

(11.30)

{

T(|q2 |r, f2 (z)) ≤ 2T(r, f1 ) + S(r, f1 ),

and {

T(|q1 q2 |r, f1 (z)) ≤ 4T(r, f1 ) + S(r, f1 ),

(11.31)

T(|q1 q2 |r, f2 (z)) ≤ 4T(r, f2 ) + S(r, f2 ).

Since |q1 | > 1, |q2 | > 1, and f1 , f2 are transcendental, it follows from (11.31) that T(r, fi (z)) ≤ 4T(αr, fi ) + S(αr, fi ), where α = i = 1, 2.

1 . |q1 q2 |

i = 1, 2,

Since 0 < α < 1, it follows by Lemma 10.3.8 that ρ(fi ) ≤

2 log 2 log |q1 q2 |

for

252 | 11 Systems of complex delay-differential equations The following example shows that system (11.29) may admit a transcendental entire solution. Example 11.4.2. Let q1 = q2 = 2, c1 = 2, and c2 = −2. Then (f1 (z), f2 (z)) = (sin z, − sin z) satisfies the system f1 (2z) = 2f2 (z)f2󸀠 (z), { f2 (2z) = −2f1 (z)f1󸀠 (z), and ρ(fi ) = 1 =

2 log 2 , 2 log 2

i = 1, 2.

Xu and Tu [224, Theorem 1.11] considered a more general situation. Theorem 11.4.3. Let (f1 , f2 ) be a transcendental meromorphic solution to a system of type f1 (q1 z)n1 = R1 (z)f2 (z)[f2 (z)]s1 , (j)

{

(11.32)

f2 (q2 z)n2 = R2 (z)f1 (z)[f1 (z)]s2 , (j)

where q1 , q2 , ∈ ℂ, |q1 | > 1, |q2 | > 1, n1 , n2 , j, s1 , s2 are positive integers, and R1 (z), R2 (z) are rational functions in z. If n1 n2 ≤ (s1 + 1)(s2 + 1), then ρ(fi ) ≤

log[(s1 j + s1 + 1)(s2 j + s2 + 1)] − log(n1 n2 ) , log |q1 | + log |q2 |

i = 1, 2.

Moreover, if n1 = n2 = 1 and f1 , f2 are meromorphic functions with infinitely many poles, then, for i = 1, 2, log[(s1 j + s1 + 1)(s2 j + s2 + 1)] log[(s1 + 1)(s2 + 1)] ≤ μ(fi ) ≤ ρ(fi ) ≤ . log |q1 | + log |q2 | log |q1 | + log |q2 | Proof. Since Ri (z) (i = 1, 2) are rational functions, we have T(r, Ri (z)) = O(log r), (i = 1, 2). By the Valiron–Mohon’ko theorem (Theorem 1.1.29) and the well-known fact that N(r, f (j) ) = N(r, f ) + jN(r, f ), it follows from (11.32) that { T(r, f1 (q1 z)) ≤ { { { { { { { { { ≤ { { { { { { { T(r, f2 (q2 z)) ≤ { { { { { { { { { ≤ {

s 1 (j) T(r, f2 ) + 1 T(r, f2 (z)) + O(log r) n1 n1 s1 + 1 js T(r, f2 ) + 1 N(r, f2 ) + S(r, f2 ), n1 n1 s2 1 (j) T(r, f1 ) + T(r, f1 (z)) + O(log r) n2 n2 s2 + 1 js T(r, f1 ) + 2 N(r, f1 ) + S(r, f1 ). n2 n2

11.4 Systems of q-delay-differential equations | 253

By Theorem 10.1.1 we get s j + s1 + 1 T(|q1 |r, f1 (z)) ≤ 1 T(r, f2 ) + S(r, f2 ), { { { n1 { { {T(|q |r, f (z)) ≤ s2 j + s2 + 1 T(r, f ) + S(r, f ). 2 2 1 1 { n2 Then we have T(|q1 q2 |r, fi ) ≤

s1 j + s1 + 1 s2 j + s2 + 1 T(r, fi ) + S(r, fi ), n1 n2

Since |q1 | > 1 and |q2 | > 1, we have 0 < α < 1 for α = T(r, fi ) ≤

1 . |q1 q2 |

i = 1, 2.

(11.33)

From (11.33) we obtain

s1 j + s1 + 1 s2 j + s2 + 1 T(αr, fi ) + S(αr, fi ), n1 n2

i = 1, 2.

Since n1 n2 ≤ (s1 j + s1 + 1)(s2 j + s2 + 1), again applying Lemma 10.3.8, we get ρ(fi ) ≤

log[(s1 j + s1 + 1)(s2 j + s2 + 1)] − log(n1 n2 ) , log |q1 | + log |q2 |

i = 1, 2.

Proceeding to the second claim, since R1 (z) and R2 (z) are rational functions, we can choose a sufficiently large constant R > 0 such that R1 (z) and R2 (z) have no zeros or poles in {z ∈ ℂ : |z| > R}. Since f1 has infinitely many poles, we can choose a pole z0 of f1 of multiplicity τ ≥ 1 satisfying |z0 | > R. Since n2 = 1, the right-hand side of the second equation in system (11.32) has a pole of multiplicity τ1󸀠 = (s2 + 1)τ + s2 j at z0 . Then f2 has a pole of multiplicity τ1󸀠 at q2 z0 . Replacing z by q2 z0 in the first equation in system (11.32), we have that f1 has a pole of multiplicity τ1 = (s1 +1)(s2 +1)τ+s2 (s1 +1)j+s1 j at q1 q2 z0 . We proceed to follow the step above. Since R1 (z) and R2 (z) have no zeros or poles in {z ∈ ℂ : |z| > R} and f1 , f2 have infinitely many poles, we may construct the poles ζk = |q1 q2 |k z0 (k ∈ N+ ) of f of multiplicity τk satisfying τk = (s1 + 1)(s2 + 1)τk−1 + s2 (s1 + 1)j + s1 j

= [(s1 + 1)(s2 + 1)]k τ + j[s2 (s1 + 1) + s1 ] {[(s1 + 1)(s2 + 1)]k−1 + ⋅ ⋅ ⋅ + 1} = [(s1 + 1)(s2 + 1)]k τ + j[s2 (s1 + 1) + s1 ]

[(s1 + 1)(s2 + 1)]k − 1 (s1 + 1)(s2 + 1) − 1

as k → ∞, k ∈ ℕ. Since |q| > 1, |ζk | → ∞ as k → ∞, so for k sufficiently large, we have [(s1 + 1)(s2 + 1)]k τ ≤ τk ≤ τ + τ1 + ⋅ ⋅ ⋅ + τk ≤ n(|ζk |, f1 )

(11.34)

k

≤ n(|q1 q2 | |z0 |, f1 ). Thus, for each r sufficiently large, there exists k ∈ ℕ such that r ∈ (|q1 q2 |k |z0 |, |q1 q2 |(k+1) |z0 |),

i.e. k >

log r − log r0 − log |q1 q2 | . log |q1 q2 |

(11.35)

254 | 11 Systems of complex delay-differential equations Thus it follows from (11.34) and (11.35) that n(r, f1 ) ≥ [(s1 + 1)(s2 + 1)]k τ ≥ τ[(s1 + 1)(s2 + 1)]

log r−log r0 −log |q1 q2 | log |q1 q2 |

log r

≥ K1 [(s1 + 1)(s2 + 1)] log |q1 |+log |q2 | , where K1 = τ[(s1 + 1)(s2 + 1)]

− log r0 −log |q1 |−log |q2 | log |q1 |+log |q2 |

.

Since for all r ≥ r0 , log r

K1 [(s1 + 1)(s2 + 1)] log |q1 |+log |q2 ≤ n(r, f1 ) ≤

1 1 N(2r, f1 ) ≤ T(2r, f1 ), log 2 log 2

we see that ρ(f1 ) ≥ μ(f1 ) ≥

log[(s1 + 1)(s2 + 1)] . log |q1 | + log |q2 |

By a similar argument we also get ρ(f2 ) ≥ μ(f2 ) ≥

log[(s1 + 1)(s2 + 1)] . log |q1 | + log |q2 |

This completes the proof of Theorem 11.4.3.

12 Periodicity of entire functions with delay-differential polynomials In this chapter, we first recall some basic results on the periodicity of transcendental entire functions. Then we consider the periodicity of transcendental entire functions f with their differential, difference or delay-differential polynomials.

12.1 Basic results on periodic functions If a transcendental meromorphic function f (z) satisfies f (z+c) = f (z) for all z ∈ ℂ, then f (z) is called a periodic function with period c. Of course, nonconstant polynomials cannot be periodic. For a periodic entire function f (z) with period c and two integers k and l, kc and lc are periods of f (z) as well, and hence their ratio is a rational number k . Of course, there exist nonconstant meromorphic periodic functions possessing two l periods, the ratio of them being imaginary. Such functions are called doubly periodic, resp., elliptic functions. However,we do not consider elliptic functions in the section. For these key observations, see [171, p. 135–141] for periodic functions in general and [171, p. 146–151] for elliptic functions. We can immediately construct periodic entire functions by using of exponential functions. As an example, n

f (z) = ∑ aj e

2iπ jz c

j=1

is a periodic function with period c. Such functions are called trigonometric polynomials with constant coefficients aj . An arbitrary periodic entire function f (z) with period c that is not a trigonometric polynomial, may be written as an everywhere convergent series of the form +∞

f (z) = ∑ an e −∞

2iπ nz c

;

see [171, p. 141–146]. In what follows in this introduction to periodic functions, we restrict ourselves to looking at periodicity with respect to composite entire and meromorphic functions. To start with, we recall Lemma 5.1 from [235]. Theorem 12.1.1. If f is a nonconstant periodic meromorphic function, then ρ(f ) ≥ μ(f ) ≥ 1, where ρ(f ) and μ(f ) are the order and lower order of f , respectively. Ozawa [180, Theorems 1, 2] further proved that the existence of prime periodic functions. Here the primeness of f (z) means that one of h and s must be linear for every factorization of f (z) = h(s(z)), where h, s are entire functions. https://doi.org/10.1515/9783110560565-012

256 | 12 Periodicity of entire functions with delay-differential polynomials Theorem 12.1.2. There exist prime periodic entire functions f of order ρ(f ) ∈ [1, +∞) or of the hyperorder ρ2 (f ) ∈ [1, +∞). Proceeding to periodicity of composite functions, we first recall a result due to Rényi, see [190, Theorem 2]. n Lemma 12.1.3. If F(z) = ∑+∞ n=0 an z is a periodic entire function and Z(N) denotes the number of 0s among the numbers a0 , a1 , . . . , aN−1 , then

lim inf N→+∞

Z(N) 1 ≤ . N 2

A direct consequence of Lemma 12.1.3 is that f (z s ) cannot be periodic if s ≥ 3, where s is a positive integer. Indeed, for F(z) = f (z s ) and s ≥ 3, we easily see that lim inf N→+∞

1 1 Z(N) ≥1− > . N s 2

For the more general case where P(z) is an arbitrary polynomial, we state the following result by Rényi and Rényi [189, Theorem 1]. Theorem 12.1.4. If f is an arbitrary nonconstant entire function and P(z) is an arbitrary polynomial of deg(P(z)) ≥ 3, then the entire function f (P(z)) cannot be periodic. Remark 12.1.5. If deg(P(z)) = 2, then there exists a transcendental entire function f (z) such that f (P(z)) is periodic. As an example, take P(z) := Az 2 + Bz + c, where A ≠ 0, and ∞

f (z) := cos √4A(z − c) + B2 = ∑ (−1)k k=0

(4A(z − c) + B2 )k . (2k)!

Then f (P(z)) = cos(2Az + B) is a periodic function with period Aπ . In particular, we may take f (z) = cos √z and P(z) = z 2 . Halász [81] improved Theorem 12.1.4 as follows. Theorem 12.1.6. If g(z) is a transcendental entire function of order ρ(g) < 1 and f (z) is a nonconstant entire function, then f (g(z)) cannot be periodic. In Theorem 12.1.6 the possibility that g(z) is itself periodic had to be ruled out. Therefore ρ(g) < 1 is supposed. Gross and Yang [68], on the other hand, gave a variant to Theorem 12.1.6 by looking at the problem when an entire function F of the form F(z) = μ1 (z)f (P(z)) + μ2 (z)

(12.1)

is periodic, assuming that f and μi (z) (i = 1, 2) are entire functions and P(z) is a polynomial; see [68, Theorem 2].

12.1 Basic results on periodic functions | 257

Theorem 12.1.7. Let f (z) be a nonconstant entire function, let P(z) be a polynomial of deg(P(z)) = n ≥ 3, and let μ1 (z) ≢ 0 and μ2 (z) be two entire functions. Suppose that for any positive constants k and ε, for r ≥ r0 (k, ε), and for i = 1, 2, k

M((1 + ε)r, μi ) ≤ M(r, f (P(z))) log r

(12.2)

and, moreover, that for some constant A > 0, m(r, μ1 (z)) > M(r, μ1 (z))−A on a set 𝔼 of r of positive lower logarithmic density. Then μ1 (z)f (P(z)) + μ2 (z) cannot be a periodic function. The statement of Theorem 12.1.7 remains valid if condition (12.2) is replaced by the requirement that the orders of μ1 (z) and μ2 (z) are less than n; see [68, Corollary 1]. If P(z) is a polynomial with deg(P(z)) ≥ 2, Gross and Yang [68, Theorem 3] also proved the following: Theorem 12.1.8. Let P(z) be a polynomial with deg(P(z)) ≥ 2, let μ2 (z) be a nonperiodic entire function of order ρ(μ2 ) < 2, and let μ1 (z) be a periodic entire function. Then μ1 (z)eP(z) + μ2 (z) cannot be a periodic function. Looking at the periodicity of P(f (z)), where P(z) is a polynomial and f (z) is an entire function, Rényi and Rényi [189, Theorem 2] obtained the following result. Theorem 12.1.9. If P(z) is a nonconstant polynomial and f (z) is not a entire periodic function, then P(f (z)) cannot be a periodic function. Returning to the situations related to Theorem 12.1.6, Gross [67, Theorem 2] proved the following: Theorem 12.1.10. Let f and g be entire functions such that f 󸀠 has no zeros and g 󸀠 has at most finitely many zeros. If f (g) is a periodic function, then g is either periodic or linear. A direct corollary of Theorem 12.1.10 is that if f (z) is entire, f 󸀠 has no zeros, and f (f ) is a periodic function, then f (z) is a periodic function. More generally, Gross [67, Theorem 3] obtained the following: Theorem 12.1.11. Let f and g be entire functions such that each of the functions f 󸀠 , g, g 󸀠 has at most finitely many zeros. If f (g) is periodic function, then g is a periodic function without zeros. It is natural to ask what we can say about a periodic function f (g) when f 󸀠 and g 󸀠 each has at most finitely many zeros? For simplicity, assume that f (g) has period 1. For any complex a and any integer t, let St (g) = {z : g(z + t) − g(t) = 0},

Ta (g) = {z : g(z) = a}.

258 | 12 Periodicity of entire functions with delay-differential polynomials Let F := {g : Sln0 (g) ∩ Ta (g) be finite for all complex numbers a for some integer n0 and l = 1, 2}. Gross [67, Theorem 4] obtained the following result. Theorem 12.1.12. Let f and g be entire functions such that each of f 󸀠 , g 󸀠 has at most finitely many zeros and g ∈ F. If f (g) is a periodic function with period 1, then g has is of the form g(z) = (az + b)P2 (z) + P1 (z), where Pi (z) is periodic with common integral period for i = 1, 2, P2 (z) has no zeros, and a, b are constants. A sequence S = {sn } is said to be a periodic set of period c(≠ 0) if S∗ = {sn + c} can be rearranged into a sequence coinciding with S. Let Zf (z) denote the zero set of f (z). Obviously, if f (z) is a periodic function with period c, then all the sets Zf (n) (z) , n = 0, 1, 2, . . . , are periodic sets with the same period c. The converse of this is not true. We can see this by looking at f (z) = eibz sin z, where b is not a rational number. Then all the zero sets of Zf (n) (z) (n = 0, 1, 2, . . . ) are periodic sets with the same period, but f (z) is not a periodic function. More generally, for any constant b and any periodic entire function g(z), f (z) = ebz g(z) implies that all the zero sets Zf (n) (z) (n = 0, 1, 2, . . . ) are periodic sets with the same period. Yang [233] obtained a result that answers the following problem: If all the zero sets Zf (n) (z) (n = 0, 1, 2, . . . ) are periodic sets with the same period, is it true that f (z) = ebz g(z)?

Theorem 12.1.13. Let f be an entire function of order ρ2 (f ) < 1, that is, lim sup r→∞

log log log M(r, f ) < 1. log r

Then f (z) is of the form f (z) = ebz g(z) if and only if the zero sets Zf (z) and Zf 󸀠 (z) are periodic sets with the same period c, where b is a constant, and g(z) is a periodic entire function. Remark 12.1.14. Theorem 12.1.13 may be false if ρ2 (f ) < 1 is removed. This can be seen

by taking f (z) = ee

z 2

z

−e √2

.

12.2 Periodicity of differential, difference, and delay-differential polynomials Looking at transcendental entire solutions to differential equations of the form f (z)f 󸀠󸀠 (z) = p(z) sin2 z,

(12.3)

where p(z) is a nonzero polynomial, Li, Lü, and Yang [135, Theorem 1] proved that whenever p(z) is a nonvanishing polynomial with real coefficients and real zeros, then p(z) must be a nonzero constant p, and f (z) = a sin z, where a is a constant satisfying a2 = −p.

12.2 Periodicity of differential, difference, and delay-differential polynomials | 259

They also raised an interesting question on the periodicity of transcendental entire functions, also mentioned in [208, Conjecture 1.1]. We formulate this question as follows. Yang’s conjecture. Let f (z) be a transcendental entire function, and let k be a positive integer. Is it true that if f (z)f (k) (z) is a periodic function, then f (z) is a periodic function as well. Obviously, Yang’s conjecture is related to the properties of transcendental entire solutions to complex delay-differential equations f (z)f (k) (z) = f (z + c)f (k) (z + c), provided that f (z)f (k) (z) is a periodic function with period c, where c is a nonzero constant. Yang’s conjecture is true for k = 1, see Wang and Hu [208, Theorem 1.1], since (f 2 )(k) in this result may be written as 2f (z)f 󸀠 (z) for k = 1. However, we give a more general result; see Liu and Yu [156, Corollary 1.3]. Theorem 12.2.1. Let f be a transcendental entire function, let k be a positive integer, and let n ≥ 2 be an integer. If (f n )(k) is a periodic function, then f is a periodic function as well. Proof. Assume first that n = 2. Since (f 2 )(k) is a periodic function with period c, we have (f (z + c)2 )(k) = (f (z)2 )(k) .

(12.4)

f (z + c)2 = f (z)2 + p(z),

(12.5)

Integrating (12.4), we obtain

where p(z) is a polynomial of degree deg(p(z)) ≤ k − 1. Then (f (z + c) + f (z))(f (z + c) − f (z)) = p(z).

(12.6)

Case 1. If p(z) vanishes, then f (z + c) = f (z) or f (z + c) − f (z) = 0. This implies that f (z) is a periodic function of period either c or 2c. Case 2. Suppose next that p(z) does not vanish. Since f is entire, there exist two polynomials p1 (z), p2 (z) satisfying p1 (z)p2 (z) = p(z). Then f (z + c) − f (z) = p1 (z)eP(z) , { f (z + c) + f (z) = p2 (z)e−P(z) .

(12.7)

1 f (z + c) = {p1 (z)eP(z) + p2 (z)e−P(z) } 2

(12.8)

From (12.7) we have

260 | 12 Periodicity of entire functions with delay-differential polynomials and 1 f (z) = {p2 (z)e−P(z) − p1 (z)eP(z) }. 2

(12.9)

Then we may also write 1 f (z + c) = {p2 (z + c)e−P(z+c) − p1 (z + c)eP(z+c) }. 2 Therefore p1 (z)eP(z) + p2 (z)eP(z) = p2 (z + c)e−P(z+c) − p1 (z + c)eP(z+c) . Denote p (z) { f1 (z) := − 2 e−2P(z) , { { p1 (z) { { { { { p2 (z + c) −P(z+c)−P(z) f2 (z) := e , { { p1 (z) { { { { { {f (z) := − p1 (z + c) eP(z+c)−P(z) , 3 p1 (z) {

(12.10)

f1 + f2 + f3 = 1.

(12.11)

implying that

Using Theorem 1.1.34, we must have either f2 ≡ 1 or f3 ≡ 1. Suppose first that f3 (z) = −

p1 (z + c) P(z+c)−P(z) e ≡ 1. p1 (z)

(12.12)

Then we also have −

p2 (z) −2P(z) p2 (z + c) −P(z+c)−P(z) e + e ≡ 0. p1 (z) p1 (z)

So eP(z+c)−P(z) =

p2 (z + c) . p2 (z)

(12.13)

Combining (12.12) and (12.13), we have p1 (z + c) p (z) =− 1 . p2 (z) p2 (z + c) This means that we have p(z + c) = −p(z) for all z. This is impossible as p(z) is a nonvanishing polynomial. Finally, if f2 ≡ 1, then we similarly obtain the same contradiction p(z + c) = −p(z).

12.2 Periodicity of differential, difference, and delay-differential polynomials | 261

Assume now that n ≥ 3. We may now apply the proof of Theorem 1 in [232] in the present special case that f is entire and n = m ≥ 3. First, note that T(r, f (z + c)n ) = nT(r, f (z + c)) = T(r, f (z)n − p(z)) = T(r, f (z)n ) + O(log r). Then, observing that Θ(r, 1/f (z)n ) ≥ 1 − 1/n and Θ(r, 1/f (z + c)n ) ≥ 1 − 1/n, and recalling that for an entire f , Θ(r, 1/f (z)n )+Θ(r, 1/f (z +c)n ) ≤ 1, we readily obtain a contradiction

n ≥ 3, where Θ(r, f1 ) = 1 − lim supr→∞

N(r, f1 ) T(r,f )

.

Remark 12.2.2. The function f 2 cannot be replaced by f in Theorem 12.2.1. For example, f (z) = ez + z is not a periodic function, whereas f (k) is a periodic function. A natural question is proposed in [156, Question 1.4]: is it possible to replace f n in Theorem 12.2.1 by a polynomial (of degree n) with a similar conclusion. Partial results of this type are the following two theorems; see [156, Theorems 1.5 and 1.7]. Theorem 12.2.3. Let f be a transcendental entire function, let k be a positive integer, and let a1 and a2 ≠ 0 be constants. If (a2 f 2 + a1 f )(k) is a periodic function, then f is a periodic function as well. Proof. Assume that (a2 f 2 + a1 f )(k) is a periodic function with period c. Thus (a2 f (z + c)2 + a1 f (z + c))

(k)

= (a2 f (z)2 + a1 f (z))

(k)

.

(12.14)

By integration we get a2 f (z + c)2 + a1 f (z + c) = a2 f (z)2 + a1 f (z) + P(z),

(12.15)

where P(z) is a polynomial with deg(P(z)) ≤ k − 1. Therefore [f (z + c) − f (z)][a2 f (z + c) + a2 f (z) + a1 ] = P(z).

(12.16)

Case 1. If P(z) ≡ 0, then either f (z + c) ≡ f (z) or a2 f (z + c) + a2 f (z) + a1 ≡ 0. In the first case, f (z) is periodic with period c, whereas in the second case, we immediately verify that f (z) is a periodic function with period 2c. Case 2. Suppose now that P(z) ≢ 0. Since f (z) is an entire function, we obtain by (12.16) that f (z + c) − f (z) = P1 (z)eQ(z) ,

{

a2 f (z + c) + a2 f (z) + a1 = P2 (z)e−Q(z) ,

(12.17)

where P1 (z)P2 (z) = P(z), and Q(z) is an entire function. By elementary computation, P2 (z)e−Q(z) − a1 + a2 P1 (z)eQ(z) { f (z + c) = , { { { 2a2 { { P (z)e−Q(z) − a1 − a2 P1 (z)eQ(z) { { f (z) = 2 . 2a2 {

262 | 12 Periodicity of entire functions with delay-differential polynomials We easily verify that P2 (z + c) Q(z)−Q(z+c) a2 P1 (z + c) Q(z+c)+Q(z) a2 P1 (z) 2Q(z) e − e − e ≡ 1. P2 (z) P2 (z) P2 (z) 1 (z) 2Q(z) e − 1 does not vanish identically. Using again Theorem 1.1.34, Clearly, − aP2 P(z) 2 we again need to consider two cases.

Case 2.1. Suppose first that −

P2 (z+c) Q(z)−Q(z+c) e P2 (z)

≡ 1 and

a2 P1 (z + c) Q(z+c)+Q(z) P1 (z) 2Q(z) e − e ≡ 0. P2 (z) P2 (z)

In this case, we see that P1 (z + c)/P2 (z + c) ≡ −P1 (z)/P2 (z), and hence P1 /P2 would be periodic with period 2c. This is a contradiction as P1 /P2 is a rational function. Case 2.2. Suppose next that − a2 PP1 (z+c) eQ(z+c)+Q(z) = 1 and (z) 2

P2 (z + c) Q(z)−Q(z+c) P1 (z) 2Q(z) e − e = 0, P2 (z) P2 (z) and a similar contradiction again follows. Theorem 12.2.4. Suppose f is a transcendental entire function of hyperorder ρ2 (f ) < 1 and N(r, f1 ) = S(r, f ). Moreover, let k be a positive integer, and let n ≥ 2. If (an f n + ⋅ ⋅ ⋅ +

a1 f )(k) , an ≠ 0, is a periodic function, then f is periodic as well.

Proof. If (an f n + ⋅ ⋅ ⋅ + a1 f )(k) is a periodic function with period c, then an f (z + c)n + ⋅ ⋅ ⋅ + a1 f (z + c) = an f (z)n + ⋅ ⋅ ⋅ + a1 f + P(z),

(12.18)

where P(z) is a polynomial with deg(P(z)) ≤ k − 1. Define now h(z) := f (z + c)/f (z). Then m(r, h) = S(r, f ) by (1.24). Since N(r, f1 ) = S(r, f ), it follows that N(r, h) ≤ N(r, f (z + c)) + N(r, 1/f (z)) = S(r, f ), and hence T(r, h) = S(r, f ). From (12.18) we obtain an [h(z)n − 1]f (z)n + an−1 [h(z)n−1 − 1]f (z)n−1 + ⋅ ⋅ ⋅ + a1 [h(z) − 1]f (z) = P(z). If now hn ≠ 1, we may apply the Clunie lemma to conclude that m(r, f ) = S(r, f ), a contradiction. Thus we must have hn = 1. If now an−1 = ⋅ ⋅ ⋅ = a1 = 0, then f is periodic by Theorem 12.2.1. So, let now ak ≠ 0 with k ≤ n − 1 being maximal. Repeating the reasoning above, we get hk = 1, and hence hn−k = 1 as well. This means that f (z + c) = hn−k f (z), from which it easily follows that f (z) is a periodic function with period (n − k)c, completing the proof. Remark 12.2.5. Consider the case where a2 f 2 + a1 f is replaced by an f n + ⋅ ⋅ ⋅ + a1 f in Theorem 12.2.1, where an ≠ 0, and at least one ai ≠ 0 (i = 1, . . . , n − 1). Using a similar method to that in the proof of Theorem 12.2.1, it is difficult to obtain the expressions of f (z) and f (z + c).

12.2 Periodicity of differential, difference, and delay-differential polynomials | 263

Theorem 12.1.9 implies that an entire function f (z) must be periodic if it solves the functional equation an f (z)n + ⋅ ⋅ ⋅ + a1 f (z) = an f (z + c)n + ⋅ ⋅ ⋅ + a1 f (z + c). Thus the conditions ρ2 (f ) < 1 and N(r, f1 ) = S(r, f ) can be removed in Theorem 12.2.4 for k = 0. In addition, the result holds for k > 0, as will be shown by Theorem 12.2.8; see Wei, Liu, and Liu [216]. Before proving this result, we first need to add a key lemma; see [134, Theorem 1]. Lemma 12.2.6. Let P(z) and Q(z) be polynomials of degree p and q, respectively. Then there do not exist nonconstant meromorphic functions f and g satisfying P(f ) = Q(g) if P(z) and Q(z) satisfy one of the following conditions: (i) p ≤ q, and there exist a zero r1 of P 󸀠 (z) and a multiple zero r2 of P 󸀠 (z) such that both Q(z) − P(rj ) = 0 (j = 1, 2) have no multiple roots. (ii) p ≤ q, and there exist three zeros rj (j = 1, 2, 3) of P 󸀠 (z) such that all Q(z) − P(rj ) = 0 (j = 1, 2, 3) have no multiple roots. (iii) p ≤ q, and there exists a zero r0 of P 󸀠 (z) of multiplicity k ≥ 3 such that Q(z)−P(r0 ) = 0 has no multiple roots. We recall [43, Definition 3.4]: an entire function F(z) is called periodic mod g with period c if F(z + c) − F(z) = g. Lemma 12.2.7 ([43, Theorem 3.4]). If P(z) is any nonlinear polynomial and g(z) is an arbitrary transcendental entire function, then P(g) is not periodic mod a nonconstant polynomial. Theorem 12.2.8. Let f be a transcendental entire function, let k be a positive integer, and let a1 , a2 , . . . , an be constants with an ≠ 0, and let n ≥ 2. If (an f n + ⋅ ⋅ ⋅ + a1 f )(k) is a periodic function, then f is also a periodic function. Proof. Firstly, we consider the case n ≥ 4. Assume that (an f n + ⋅ ⋅ ⋅ + a1 f )(k) is a periodic function with period c and denote P(z) = an z n + ⋅ ⋅ ⋅ + a1 z with deg(P(z)) = n ≥ 4. By Lemma 12.2.7 we obtain P(f (z)) = P(f (z + c)) − d

(12.19)

for some d ∈ ℂ. If d = 0, then by Theorem 12.1.9, f (z) must be a periodic function. Suppose now that d ≠ 0. Then (12.19) yields P(f (z)) = P(f (z + kc)) − kd

(12.20)

264 | 12 Periodicity of entire functions with delay-differential polynomials for all k ∈ ℕ. Let rl (l ∈ {1, 2, 3}) be a zero of P 󸀠 (z), and assume that there are infinitely many k ∈ ℕ such that the polynomial P(z) − kd − P(rl ) has multiple roots. For each such k, let βk be any one of the multiple roots of P(z)−kd −P(rl ) (there may in principle be more than one). We state that the roots βk are distinct. To this end, suppose on the contrary that βi = βj for i ≠ j, i, j ∈ ℕ. Then P(βi ) − id − P(rl ) = P(βj ) − jd − P(rl ) = 0, and hence (j − i)d = 0, which is a contradiction. Thus βi ≠ βj for all i ≠ j. But now P 󸀠 (βk ) = 0 for infinitely many distinct βk , which implies that P(z) is a constant, a contradiction. Thus the polynomial P(z) − kd − P(rl ) can have multiple roots for only finitely many k ∈ ℕ for all choices of l ∈ {1, 2, 3}. Hence there exists k0 ∈ ℕ such that P(z) − k0 d − P(rl ) has only simple roots for all l ∈ {1, 2, 3}. By letting Q(z) = P(z) − k0 d it follows by (12.20) and Lemma 12.2.6 that f is a constant, which is a contradiction. We proceed to proving the case n ≤ 3. Suppose that (a3 f 3 +a2 f 2 +a1 f )(k) is a periodic function with period c. Then a3 f (z)3 + a2 f (z)2 + a1 f (z) = a3 f (z + c)3 + a2 f (z + c)2 + a1 f (z + c) + p(z), where p(z) is a polynomial with degree less than k. Applying Lemma 12.2.7, we have p(z) is a constant. Then we obtain (f (z) − f (z + c))(a3 (f (z)2 + f (z)f (z + c) + f (z + c)2 ) + a2 (f (z) + f (z + c)) + a1 ) = p.

(12.21)

Case 1. Assume that p = 0 in (12.21). In fact, Theorem 12.1.9 implies that f (z) is a periodic function in this case. Another basic method will be given to show the possible periods when p = 0. From (12.21) we have f (z) − f (z + c) = 0

(12.22)

or a3 (f (z)2 + f (z)f (z + c) + f (z + c)2 ) + a2 (f (z) + f (z + c)) + a1 = 0.

(12.23)

Equation (12.22) implies that f (z) is a periodic function with period c. Shifting forward equation (12.23), we have a3 (f (z + c)2 + f (z + c)f (z + 2c) + f (z + 2c)2 ) + a2 (f (z + c) + f (z + 2c)) + a1 = 0. (12.24) Combining (12.23) with (12.24), we obtain (f (z) − f (z + 2c))(a3 (f (z) + f (z + c) + f (z + 2c)) + a2 ) = 0.

(12.25)

12.2 Periodicity of differential, difference, and delay-differential polynomials | 265

Thus we have f (z) − f (z + 2c) = 0

(12.26)

a3 (f (z) + f (z + c) + f (z + 2c)) + a2 = 0.

(12.27)

or

Equation (12.26) implies that f (z) is a periodic function with period 2c, and equation (12.27) implies that f (z) is a periodic function with period 3c. Case 2. Assume that p ≠ 0 in (12.21). Since f (z) is an entire function, equation (12.21) implies that both f (z) − f (z + c) and a3 (f (z)2 + f (z)f (z + c) + f (z + c)2 ) + a2 (f (z) + f (z + c)) + a1 have no zeros. Using the Hadamard factorization theorem, we have f (z) − f (z + c) = c1 eP(z) , { { { a3 (f (z)2 + f (z)f (z + c) + f (z + c)2 ) + a2 (f (z) + f (z + c)) + a1 { { { −P(z) , { = c2 e

(12.28)

where c1 , c2 are nonzero constants, c1 c2 = p, and P(z) is an entire function. By shifting forward equation (12.28) we obtain f (z + c) − f (z + 2c) = c1 eP(z+c) , { { { a (f (z + c)2 + f (z + c)f (z + 2c) + f (z + 2c)2 ) + a2 (f (z + c) + f (z + 2c)) { { 3 { −P(z+c) . { + a1 = c2 e

(12.29)

By adding the first equation of (12.28) to the first equation of (12.29) we obtain f (z) − f (z + 2c) = c1 eP(z) + c1 eP(z+c) .

(12.30)

By subtracting the second equation of (12.29) from the second equation of (12.28) we obtain (f (z) − f (z + 2c))(a3 (f (z) + f (z + c) + f (z + 2c)) + a2 ) = c2 e−P(z) − c2 e−P(z+c) .

(12.31)

A basic computation from (12.28), (12.30), and (12.31) gives c2 −P(z)−P(z+c) a 2 1 f (z) = e G(z) + c1 eP(z) + c1 eP(z+c) − 2 , { { { 3a3 c1 3 3 3a3 { { {f (z + c) = c2 e−P(z)−P(z+c) G(z) − 1 c eP(z) + 1 c eP(z+c) − a2 , 3a3 c1 3 1 3 1 3a3 {

(12.32)

266 | 12 Periodicity of entire functions with delay-differential polynomials P(z+c)−P(z)

where G(z) = eeP(z+c)−P(z) −1 . +1 Since f (z) is an entire function, the expression of f (z) in (12.32) implies that G(z) must be a transcendental entire function or a constant. Obviously, G(z) is not a nonconstant entire function since eP(z+c)−P(z) − 1 and eP(z+c)−P(z) + 1 must have infinitely many zeros unless P(z + c) − P(z) is a constant, in which case, G(z) reduces to a constant. Let G(z) = D and P(z + c) − P(z) = d, where D and d are constants. If D = 0, then eP(z+c)−P(z) − 1 = ed − 1 = 0, which means that d = 2sπi (s = 0, ±1, ±2, . . . ). Then a 1 2 f (z + c) = c1 eP(z+c) + c1 eP(z+2c) − 2 3 3 3a3 a 2 1 = c1 eP(z)+2sπi + c1 eP(z+c)+2sπi − 2 3 3 3a3 a 2 1 = c1 eP(z) + c1 eP(z+c) − 2 3 3 3a3 a 1 1 = − c1 eP(z) + c1 eP(z+c) − 2 , 3 3 3a3 which means that c1 eP(z) = 0, which is impossible. If D ≠ 0, then d ≠ 2kπi, and substituting P(z + c) = P(z) + d into the formulas of f (z) and f (z + c), we have Dc2 −2P(z)−d 2 P(z) 1 P(z)+d a2 e + c1 e , + c1 e − 3a3 c1 3 3 3a3 Dc2 −2P(z+c)+d 1 P(z+c)−d 1 P(z+c) a2 f (z + c) = e − c1 e . + c1 e − 3a3 c1 3 3 3a3 f (z) =

(12.33) (12.34)

Shifting forward (12.33) and combining with (12.34), we have Ce−3P(z+c) = B, where C =

Dc2 (ed a3 c12

− e−d ) and B = 1 + ed + e−d . Obviously, e−3P(z+c) is not a constant;

otherwise, f (z) would be a constant. If e−3P(z+c) is not a constant, then B = C = 0, and by a basic computation from the expressions of B and C we can get a contradiction. Thus we have proved Theorem 12.2.8. Remark 12.2.9. (1) A nonconstant polynomial P(z) is called a uniqueness polynomial for entire functions (UPE) if f = g whenever P(f ) = P(g) for any two nonconstant entire functions f and g; see Li and Yang [132]. Similarly to the above definition, we also introduce the following definition: A nonconstant polynomial P(z) is called a uniqueness polynomial for periodicity of entire functions (UPPE) if f is a periodic function with period c or mc whenever [P(f (z))](k) = [P(f (z + c))](k) , where k is any nonnegative integer, and m is a nonzero integer. (2) Obviously, UPE must be UPPE. However, UPPE may not UPE, for example, z 2 ∘ (f (z)) = z 2 ∘ (f (z + c)) implies that f (z) = f (z + c) or f (z) = −f (z + c) (f (z) = f (z + 2c)),

12.2 Periodicity of differential, difference, and delay-differential polynomials | 267

and hence z 2 is UPPE but not UPE. Theorems 12.2.8 and 12.1.9 imply that any nonlinear polynomial is UPPE. (3) Theorem 12.2.8 is not valid if the polynomial P(z) := an z n + ⋅ ⋅ ⋅ + a1 z is replaced by any transcendental entire function. For example, take P(z) = ez − z and k ≥ 1. Then [ef (z) − f (z)](k) = [ef (z+c) − f (z + c)](k) admits an entire solution f (z) = ez + z and c = 2kiπ, where f (z) is not a periodic function. If f (z) is a transcendental entire function with a nonzero Picard exceptional value, then Liu and Yu [156, Theorem 1.1] also proved the following: Theorem 12.2.10. Let f be a transcendental entire function with nonzero Picard exceptional value, and let k be a positive integer. If ff (k) is a periodic function, then f is periodic. Proof. Let d be a nonzero Picard exceptional value. Then f (z) = eg(z) + d, where g(z) is an entire function, and T(r, g) = S(r, f ). Since ff (k) is a periodic function with period c, we have f (z)f (k) (z) = f (z + c)f (k) (z + c). Hence H(z, g(z))e2g(z) + dH(z, g(z))eg(z) = H(z, g(z + c))e2g(z+c) + dH(z, g(z + c))eg(z+c) , where H(z, g(z)) is a differential polynomial in g(z). Obviously, T(r, eg(z) ) + S(r, eg(z) ) = T(r, eg(z+c) ) + S(r, eg(z+c) ). Hence we have T(r, g(z + c)) = T(r, g(z)) = S(r, eg(z) ). We may write H(z, g(z)) g(z)−g(z+c) 1 g(z+c) H(z, g(z)) e2g(z)−g(z+c) + e − e = 1. dH(z, g(z + c)) H(z, g(z + c)) d Obviously, − d1 eg(z+c) is not a constant. Using again Theorem 1.1.34, we need to consider the following two cases. Case 1. First, suppose that H(z, g(z)) e2g(z)−g(z+c) ≡ 1 dH(z, g(z + c))

268 | 12 Periodicity of entire functions with delay-differential polynomials and H(z, g(z)) g(z)−g(z+c) 1 g(z+c) e − e ≡ 0. H(z, g(z + c)) d Then eg(z)+g(z+c) = d2 , and so g(z) = B − g(z + c), where B is a constant. Thus we must H(z,g(z)) have dH(z,g(z+c)) e3g(z)−B = 1. This implies that T(r, eg(z) ) = S(r, eg(z) ), a contradiction. Case 2. Next, suppose that H(z, g(z)) g(z)−g(z+c) e ≡1 H(z, g(z + c)) and H(z, g(z)) 1 e2g(z)−g(z+c) − eg(z+c) ≡ 0. dH(z, g(z + c)) d Then we must have eg(z) = eg(z+c) , and hence f (z) = f (z + c). Remark 12.2.11. Observe that if the Picard value d = 0 in the preceding theorem and f (z) = eh(z) , where h(z) is a nonconstant polynomial, then Yang’s conjecture is true in this case. Indeed, assume that f (z)f (k) (z) = f (z + c)f (k) (z + c), where c is a nonzero constant. Substituting f (z) = eh(z) into this equation, we have H(z) e2h(z+c)−2h(z) = H(z+c) , where H(z) is a polynomial in h(z) and its derivatives, and hence H(z) is also a polynomial in z. Note that if the rational function H(z) H(z+c)

2h(z+c)−2h(z)

H(z) H(z+c)

has no zeros

and poles, then ≡ 1. Thus we have e ≡ 1, and so f (z) is a periodic function with period c or 2c. This situation remains open in the case where h(z) is transcendental and d = 0. We now proceed now to considering a generalized Yang’s conjecture, which can be stated as follows. Generalized Yang’s conjecture. Let f (z) be a transcendental entire function, and let n, k be positive integers. If f (z)n f (k) (z) is a periodic function, then f (z) is periodic as well. Theorem 12.2.12. Let f (z) be a transcendental entire function, and let n, k be positive integers. Suppose f (z)n f (k) (z) is a periodic function and one of the following conditions is satisfied: (i) k = 1; (ii) f (z) = eh(z) , where h(z) is a nonconstant polynomial; (iii) f (z) has a nonzero Picard exceptional value and is of finite order. Then f (z) is also a periodic function.

12.2 Periodicity of differential, difference, and delay-differential polynomials | 269

Proof. A similar reasoning as in the proofs of the previous results applies. The reader can consider this as an exercise, or see the proof in [159]. Remark 12.2.13. Observe the assumption that f is of finite order in (iii). It remains an open question whether this assumption may be removed. Theorem 12.2.14. Let f (z) be a transcendental entire function, and let n, k be positive integers. If f (z)n f (k) (z) and f (z)n f (k+1) (z) are periodic functions with the same period c, then f (z) is a periodic function as well with period either c, 2c, or (n + 1)c. Proof. By the assumption that f (z)n f (k) (z) and f (z)n f (k+1) (z) are periodic functions with period c we have {

f (z)n f (k) (z) = f (z + c)n f (k) (z + c),

f (z)n f (k+1) (z) = f (z + c)n f (k+1) (z + c).

(12.35)

Denote F(z) := f (k) (z). Thus F 󸀠 (z) = f (k+1) (z), F(z + c) = f (k) (z + c), and F 󸀠 (z + c) = f (k+1) (z + c). By (12.35) we have F 󸀠 (z + c) F 󸀠 (z) = . F(z + c) F(z) Integrating this identity, we get F(z + c) = eA F(z),

(12.36)

f (k) (z + c) = eA f (k) (z),

(12.37)

that is,

where A is a constant. From the first equation of (12.35) and from (12.37) we have f (z + c)n eA = f (z)n .

(12.38)

Case 1. If n = 1, then (12.38) means that f (z + c)eA = f (z).

(12.39)

Differentiating equation (12.39) k times, we have f (k) (z + c)eA = f (k) (z), that is, F(z + c)eA = F(z).

(12.40)

270 | 12 Periodicity of entire functions with delay-differential polynomials Thus e2A = 1 follows by (12.36) and (12.40). Equation (12.39) implies that f (z) is a periodic function with period either c or 2c. Case 2. In n ≥ 2, by integrating (12.37) k times results in f (z + c) = eA (f (z) + P(z)),

(12.41)

where P(z) is a polynomial with degree ≤ k − 1. Thus f (z + c)n = enA (f (z) + P(z))n .

(12.42)

Therefore P(z) ≡ 0, and e(n+1)A = 1 by comparing (12.38) and (12.42). So by (12.41) f (z) must be a periodic function with period (n + 1)c. Remark 12.2.15. Given positive integers k, l, if f (k) (z) and f (l) (z) are periodic entire functions, then base periods are the same. In fact, if not, assume that f (k) (z) = f (k) (z+c) and f (l) (z) = f (l) (z +b) for all z ∈ ℂ, where c, b are the base periods for f (k) (z) and f (l) (z). Without loss of generality, we may assume that k < l. Then f (l) (z) = f (l) (z + c) = f (l) (z + b) for all z ∈ ℂ. So c = b. However, it remains an open question whether f (z)n f (k) (z) and f (z)n f (l) (z) have the same base periods if they are periodic functions. We next consider the periodicity of differential polynomials f (z)n + f (k) (z) and obtain the following result. Theorem 12.2.16. Let f (z) be a transcendental entire function, and let n ≥ 2. If f (z)n + f 󸀠 (z) is a periodic function, then f (z) is also a periodic function. Proof. Assuming that f (z)n + f 󸀠 (z) is a periodic function with period c, we have f (z + c)n + f 󸀠 (z + c) = f (z)n + f 󸀠 (z) and f (z + c)n − f (z)n

= (f (z + c) − f (z))(f (z + c)n−1 + f (z + c)n−2 f (z) + ⋅ ⋅ ⋅ + f (z)n−1 )

= f 󸀠 (z) − f 󸀠 (z + c).

This equation implies that either f (z+c)−f (z) ≡ 0, and hence f (z) is a periodic function with period c, or f (z + c) − f (z) has no zeros at all. In the latter case, f (z + c) − f (z) = eh(z) by the Hadamard factorization theorem, where h(z) is an entire function. Case 1. If n = 2, then, obviously, f (z+c)+f (z) = −h󸀠 (z). Then, of course, T(r, h󸀠 (z)) = S(r, eh(z) ). Hence 1 h(z) + h󸀠 (z)), { {f (z) = − 2 (e { 1 h(z) { − h󸀠 (z)). {f (z + c) = 2 (e

(12.43)

12.2 Periodicity of differential, difference, and delay-differential polynomials | 271

So, we have h󸀠 (z + c) − h󸀠 (z) = −(eh(z+c) + eh(z) ). Then, clearly, T(r, eh(z+c) ) = T(r, eh(z) ) + S(r, eh(z) ). If now h󸀠 (z + c) − h󸀠 (z) ≢ 0, then by the second main theorem 1 1 ) + N (r, h(z) ) + S(r, eh(z) ) eh(z) e − (h󸀠 (z + c) + h󸀠 (z)) 1 ≤ N (r, − h(z+c) ) + S(r, eh(z) ) = S(r, eh(z) ), e

T(r, eh(z) ) ≤ N (r,

a contradiction. Therefore we must have h󸀠 (z + c) ≡ h󸀠 (z), and it follows that eh(z+c) = −eh(z) . Thus from (12.43) we conclude that f (z) is a periodic function with period 2c. Case 2. If n ≥ 3, then we have f (z + c) − f (z) = eh(z) ,

{

f (z + c)n−1 + f (z + c)n−2 f (z) + ⋅ ⋅ ⋅ + f (z)n−1 = −h󸀠 (z).

(12.44)

Equations in (12.44) may be written as f (z + c) { f (z) [ − 1] = eh(z) , { { { f (z) n−1 { { f (z + c)n−2 f (z + c) { {f (z)n−1 [ f (z + c) + + ⋅⋅⋅ + + 1] = −h󸀠 (z). n−1 n−2 f (z) f (z) f (z) {

(12.45)

. If M = 1, then f (z) is a periodic function with period c. Suppose Define M(z) := f (z+c) f (z) now that M(z) ≢ 1. Then equation (12.45) implies that M n−1

(M − 1)n−1 e(n−1)h . =− n−2 h󸀠 +M + ⋅⋅⋅ + M + 1

(12.46)

By the Valiron–Mohon’ko theorem (Theorem 1.1.29) we see that (n − 1)T(r, M) = (n − 1)T(r, eh ) + S(r, eh ).

(12.47)

Moreover, N (r,

M n−1

+

1 1 ) = N (r, 󸀠 ) = S(r, eh(z) ), h (z) + ⋅⋅⋅ + M + 1

M n−2

and, by (12.45), M − 1 has no zeros. Factorizing M n−1 + M n−2 + ⋅ ⋅ ⋅ + M + 1 and using the second main theorem, we have obtain 1 1 ) + N (r, n−1 ) M−1 M + M n−2 + ⋅ ⋅ ⋅ + M + 1 + S(r, M) ≤ S(r, eh ) + S(r, M).

(n − 2)T(r, M) ≤ N (r,

Combining this with (12.47), we conclude that T(r, M) = S(r, eh ) and T(r, eh ) = S(r, eh ), a contradiction.

272 | 12 Periodicity of entire functions with delay-differential polynomials Remark 12.2.17. (1) We also obtain, under certain conditions on the zeros of f (z + c) − f (z), that f is periodic whenever f n + f (k) , k ≥ 2, is periodic; see [159]. Some considerations on the periodicity of f n + ak f (k) + ⋅ ⋅ ⋅ + a1 f 󸀠 can be found in [166], where at least one of ai (i = 1, 2, . . . , k) is a nonzero constant. (2) Theorem 12.2.16 is not true if n = 1. Indeed, f (z) = ze−z is not a periodic function, whereas f (z) + f (k) (z) = (−1)k−1 e−z is a periodic function. We now continue by giving more results related to Yang’s conjecture in difference or delay-differential situations. For all these results below, see Liu and Korhonen [161]. Theorem 12.2.18. Let f (z) be a transcendental entire function of hyperorder ρ2 (f ) < 1, and let n ≥ 2 be a positive integer. If f (z)n f (z + η) is a periodic function, then f (z) is also a periodic function. Proof. Suppose f (z)n f (z + η) is periodic with period c ≠ 0. Then f (z + c)n f (z + η + c) = f (z)n f (z + η), and thus f (z + η + c) f (z)n = . n f (z + c) f (z + η) Define G(z) :=

f (z) . f (z+c)

(12.48)

From (12.48) we have

nT(r, G) = T (r,

1 ) = T(r, G(z + η)) + O(1) = T(r, G(z)) + S(r, G), G(z + η)

which contradicts with n ≥ 2. So G(z) is a constant A such that An = A1 . Hence An+1 = 1, that is, f (z)n+1 = f (z + c)n+1 . Thus f (z) is a periodic function with period (n + 1)c. Remark 12.2.19. (1) Theorem 12.2.18 is not valid in general for transcendental entire functions of hyperorder ρ2 (f ) ≥ 1. This can be seen by looking at the nonperiodic entire z function f (z) = eze such that eη = −n, where n is a positive integer. In this case, z f (z)n f (z + η) = e−nηe is a periodic function. (2) In the case n = 1, it is easy to see that if f (z)f (z + η) is a periodic function with period c = η, then f (z) is also a periodic function with period 2η. However, it remains an open question what may happen when c ≠ η. Theorem 12.2.20. Let f (z) be a transcendental entire function of hyperorder ρ2 (f ) < 1, and let n ≥ 4 be a positive integer. If [f (z)n f (z + η)](k) is a periodic function, then f (z) is periodic. Proof. If [f (z)n f (z + η)](k) is a periodic function with period c ≠ 0, then f (z + c)n f (z + η + c) = f (z)n f (z + η) + P(z),

12.2 Periodicity of differential, difference, and delay-differential polynomials | 273

where P(z) is a polynomial of degree deg P(z) ≤ k − 1. We first show that P(z) identically vanishes. Suppose this is not the case. Since f (z) is transcendental entire of hyperorder ρ2 (f ) < 1, we may apply the second main theorem for three small functions (Theorem 1.1.22) and Lemma 1.2.10 to obtain (n + 1)T(r, f ) = T(r, f (z)n f (z + η)) + S(r, f ) 1 ) ≤ N(r, f (z)n f (z + η)) + N (r, f (z)n f (z + η) 1 + N (r, ) + S(r, f ) f (z)n f (z + η) + P(z) 1 1 ≤ N (r, ) + N (r, ) + S(r, f ) f (z)n f (z + η) f (z + c1 )n f (z + η + c1 ) ≤ 4T(r, f ) + S(r, f ), which is a contradiction. Thus we have P(z) ≡ 0. We can now apply the same proof as in Theorem 12.2.18 to obtain the periodicity of f . Remark 12.2.21. Theorem 12.2.20 is not true in general if n = 1 and k ≥ 2. Indeed, consider f (z) = ez + z such that ec = −1. Now both [f (z)f (z + c)]󸀠󸀠 = −22 e2z + cez + 2 and [f (z)f (z + c)](k) = −2k e2z + cez (k ≥ 3) are periodic functions with period 2c, whereas f (z) is not periodic. Theorem 12.2.22. Let f (z) be either a transcendental entire function of finite order, and let n ≥ 1, or let f (z) be transcendental entire of infinite order, and let n = k = 1. If [f (z)n f (z + η)](k) is a periodic function with period η, then f (z) is periodic as well. Proof. Since [f (z)n f (z + η)](k) is a periodic function with period η, we have f (z)n f (z + η) = f (z + η)n f (z + 2η) + P(z),

(12.49)

where P(z) is a polynomial with deg P(z) ≤ k − 1. Assume that P(z) ≢ 0. Then we have f (z + η)[f (z)n − f (z + η)n−1 f (z + 2η)] = P(z).

(12.50)

Hence f (z + η) = P1 (z)eh(z) ,

{

f (z)n − f (z + η)n−1 f (z + 2η) = P2 (z)e−h(z) ,

where P1 (z)P2 (z) = P(z), and P1 (z) and P2 (z) are nonzero polynomials. Hence P1 (z − η)n enh(z−η) − P1 (z)n−1 P1 (z + η)e(n−1)h(z)+h(z+η) = P2 (z)e−h(z) , and we have P1 (z − η)n enh(z−η)+h(z) − P1 (z)n−1 P1 (z + η)enh(z)+h(z+η) = P2 (z).

(12.51)

274 | 12 Periodicity of entire functions with delay-differential polynomials Let f1 := P1 (z − η)n enh(z−η)+h(z) and f2 := −P1 (z)n−1 P1 (z + η)enh(z)+h(z+η) . Then (12.51) implies that f1 (z) + f2 (z) = P2 (z). If f1 and f2 are transcendental, then we may use the second main theorem for three small functions to obtain T(r, f1 ) ≤ N(r, f1 ) + N (r, ≤ S(r, f1 ),

1 1 ) + N (r, ) + S(r, f1 ) f1 f1 − P2 (z)

which is impossible. Thus f1 and f2 are polynomials. Then the exponents in (12.51) must be constants. Therefore, clearly, for a constant B, nh(z − η) + h(z) = nh(z) + h(z + η) = B.

(12.52)

Suppose now that f (z) is of finite order and n ≥ 1. Then h(z) is a polynomial. By (12.52) h must be a constant as well, and we can get a contradiction from (12.52), so P(z) ≡ 0 (in this finite-order case). Thus, similarly as in the proof of Theorem 12.2.18, it follows that f is a periodic function. Finally, let f (z) be of infinite order, and let n = k = 1. Then, by (12.52), h(z + η) − h(z − η) = 0, and hence h(z) is periodic function of period 2η. Since k = 1, P must be a constant, as well as P1 and P2 . Therefore f (z) = P1 eh(z−η) is a periodic function, where P1 is a nonzero constant. Theorem 12.2.23. Let f (z) be a transcendental entire function of hyperorder ρ2 (f ) < 1, and let n ≥ 5 be a positive integer. If [f (z)n Δη f ](k) is a periodic function with period η, then f (z) is a periodic function. Proof. We assume that the period of [f (z)n Δη f ](k) is η, where η is a nonzero complex number. Thus f (z + η)n [f (z + 2η) − f (z + η)] = f (z)n [f (z + η) − f (z)] + Q(z), where Q(z) is a polynomial with deg Q(z) ≤ k − 1. If Q(z) ≢ 0, then using the first and second main theorems for three small functions again, we obtain nT(r, f ) ≤ T(r, f (z)n [f (z + η) − f (z)]) + S(r, f ) ≤ N(r, f (z)n [f (z + η) − f (z)]) + N (r,

1 ) f (z)n [f (z + η) − f (z)]

1 ) + S(r, f ) f (z)n [f (z + η) − f (z)] + Q(z) 1 1 1 ≤ N (r, ) + N (r, ) + N (r, ) f (z) f (z + η) − f (z) f (z + η) 1 + N (r, ) + S(r, f ) f (z + 2η) − f (z + η) 1 1 ≤ N (r, ) + T (r, f (z + η) − f (z)) + N (r, ) f (z) f (z + η) + N (r,

12.2 Periodicity of differential, difference, and delay-differential polynomials | 275

+ T (r, f (z + 2η) − f (z + η)) + S(r, f )

≤ 4T(r, f ) + S(r, f ).

Since n ≥ 5, we get a contradiction. So Q = 0. Therefore we have f (z + η)n [f (z + 2η) − f (z + η)] = f (z)n [f (z + η) − f (z)]. If f (z + 2η) − f (z + η) ≡ 0, then f (z) is a periodic function with period η. If f (z + 2η) − f (z + η) ≢ 0, then f (z + η) − f (z) f (z + η)n = = n f (z) f (z + 2η) − f (z + η) = Let G(z) :=

f (z+η) . f (z)

f (z+η) −1 f (z) f (z+2η) f (z+η) f (z+η) − f (z) f (z+η) f (z)

f (z+η) −1 f (z) f (z+2η) f (z+η) − f (z) f (z)

.

Then we have G(z)n =

1 G(z) − 1 . G(z + η) − 1 G(z)

Hence nT(r, G) ≤ T(r, G(z)) + T(r, G(z + η)) + S(r, G(z)) ≤ 2T(r, G) + S(r, G). Since n ≥ 5, we conclude that G must be a constant A ≠ 1. This means that An = we have An+1 = 1. Writing now n+1

An+1 = 1 = ∏ j=1

1 , A

so

f (z + jη) f (z + (n + 1)η) = , f (z + (j − 1)η) f (z)

we deduce that f (z) is a periodic function with period (n + 1)η. Remark 12.2.24. Observe that f (k) (z) + f (l) (z) (k > l) may be a periodic function even if f (z) is not a periodic function. This can be seen by looking at f (z) = ez + z. Concerning more closely the relation on the periodicity of f (k) (z) + f (l) (z) and f (z), we first prove the next lemma to proceed to the theorem below. Lemma 12.2.25. [161, Lemma 2.4] All entire solutions of f (z + c)2 − f (z)2 = e−z are periodic functions, where c is a nonzero constant.

(12.53)

276 | 12 Periodicity of entire functions with delay-differential polynomials Proof. Using Gross’ result in Chapter 6.2, we have f (z + c) = sin(h(z)), { { { e− z2 { { { if (z) = cos(h(z)), z { e− 2

(12.54)

where h(z) is any entire function such that sin h(z) cos h(z) ≢ 0. A basic computation from (12.54) shows c

e− 2 sin(h(z + c) +

π + 2kπ) = i sin h(z), 2

(12.55)

and hence c

c

e− 2 i(h(z+c)+ π2 +2kπ−h(z)) e− 2 −i(h(z+c)+ π2 +2kπ+h(z)) e − e + e−2ih(z) = 1. i i Case 1. If h(z) is a constant h, then by (12.56) h should satisfy e−2ih = 0, 1, −1). Thus f (z) = −ie

− z2

cos h, a periodic function.

(12.56) c

e 2 −1

c e2

+1

(≠

Case 2. If h(z) is not a constant, then e−2ih(z) is not a constant, and both h(z+c)+h(z) and h(z +c)−h(z) are not constants at the same time. Using Theorem 1.1.34, we discuss the following two subcases. Subcase 2.1. If −c

e 2 i(h(z+c)+ π2 +2kπ−h(z)) { { ≡ 1, { { i e c { − { { { − e 2 e−i(h(z+c)+ π2 +2kπ+h(z)) + e−2ih(z) ≡ 0, { i

(12.57)

then e−c = −1. By shifting equation (12.53) forward we have f (z + 2c)2 − f (z + c)2 = −e−z , and thus f (z + 2c)2 = f (z)2 , so that f (z) is a periodic function with period 2c or 4c. Subcase 2.2. If c

e− 2 −i(h(z+c)+ π2 +2kπ+h(z)) { { ≡ 1, { {− i e { { e− c2 π { { ei(h(z+c)+ 2 +2kπ−h(z)) + e−2ih(z) ≡ 0, { i

(12.58)

then e−c = −1. By the same discussions as in Subcase 2.1, f (z) is a periodic function with period 2c or 4c. Theorem 12.2.26. Let f (z) be a transcendental entire function, and let one of the following two conditions be satisfied:

12.2 Periodicity of differential, difference, and delay-differential polynomials | 277

(1) k = 1 and l = 0; (2) ρ(f ) ≥ 2 and k > l. If (f (z)2 )(k) + (f (z)2 )(l) is a periodic function, then f (z) is also a periodic function. Proof. Assume that (f (z)2 )(k) + (f (z)2 )(l) is a periodic function with period c(≠ 0). Thus (f (z + c)2 )(k) + (f (z + c)2 )(l) = (f (z)2 )(k) + (f (z)2 )(l) .

(12.59)

F(z) := f (z + c)2 − f (z)2 .

(12.60)

F (k) (z) = −F (l) (z).

(12.61)

We set

Then (12.59) can be written as

We discuss two cases. Case 1. If k = 1 and l = 0, then we may integrate (12.61) to get F(z) = Ce−z , where C is a nonzero constant, or F = 0. If F(z) = Ce−z , then Lemma 12.2.25 implies that f (z) is a periodic function. If F = 0, then f is a periodic function of period either c or 2c. Case 2. If ρ(f ) ≥ 2 and k > l, then from (12.61) it follows that F(z) must be an exponential polynomial satisfying F (l) (z) = μ1 eλ1 z + ⋅ ⋅ ⋅ + μk−l eλk−l z , where λik−l = −1, and μi are constants (i = 1, 2, . . . , k − l). Thus ρ(F(z)) ≤ 1. We claim that ρ(f (z + c) − f (z)) = ρ(f (z + c) + f (z)) = ρ(f ) ≥ 2. Indeed, if ρ(f (z + c) − f (z)) < 2 and ρ(f (z + c) + f (z)) < 2, then ρ(f ) < 2, a contradiction. Then if only one of ρ(f (z + c) − f (z)) < 2 and ρ(f (z + c) + f (z)) < 2 holds, then ρ(F(z)) ≥ 2, a contradiction again. Thus ρ(f (z + c) − f (z)) = ρ(f (z + c) + f (z)) ≥ 2 by (12.60). Furthermore, using the equation 2f (z) = f (z + c) + f (z) − (f (z + c) − f (z)), we have ρ(f ) ≤ ρ(f (z+c)−f (z)) = ρ(f (z+c)+f (z)). Combining this with ρ(f (z+c)−f (z)) ≤ ρ(f ), we have the claim. If F(z) ≢ 0, then from (12.60) and the Hadamard factorization theorem we have f (z + c) − f (z) = h1 (z)eH(z) , { f (z + c) + f (z) = h2 (z)e−H(z) ,

(12.62)

where max{ρ(h1 ), ρ(h2 )} ≤ 1 and ρ(eH ) ≥ 2. Thus T(r, hi ) = S(r, eH ), i = 1, 2. Then 1 −H(z) − h1 (z)eH(z) ), { {f (z) = 2 (h2 (z)e { 1 { −H(z) + h1 (z)eH(z) ). {f (z + c) = 2 (h2 (z)e

(12.63)

278 | 12 Periodicity of entire functions with delay-differential polynomials Hence h2 (z)e−H(z) + h1 (z)eH(z) = h2 (z + c)e−H(z+c) − h1 (z + c)eH(z+c) . (z) −2H(z) e , f2 = Dividing (12.64) by h1 (z)eH(z) and denoting f1 = − hh2 (z)

and f3 = − h1h(z+c) eH(z+c)−H(z) , it follows that (z)

1

(12.64)

h2 (z+c) −H(z+c)−H(z) e , h1 (z)

1

f1 + f2 + f3 = 1.

(12.65)

Obviously, at least one of −H(z + c) − H(z) and H(z + c) − H(z) is not constant. In fact, only one of −H(z + c) − H(z) and H(z + c) − H(z) is not a constant by Theorem 1.1.34. If −H(z + c) − H(z) is not a constant, then it follows by the same proposition that f3 ≡ 1. Immediately, h1 (z)eH(z) ≡ −h1 (z + c)eH(z+c) .

(12.66)

From the first equation of (12.62) and (12.66) we have f (z + c) − f (z) ≡ −(f (z + 2c) − f (z + c)),

(12.67)

which means that f (z) ≡ f (z + 2c), and thus f is a periodic function with period 2c. If H(z + c) − H(z) is not a constant, then, using Theorem 1.1.34 again, we have that f2 ≡ 1. Thus h1 (z)eH(z) ≡ h2 (z + c)e−H(z+c) .

(12.68)

Then from (12.62) and (12.68) we get f (z + c) − f (z) ≡ f (z + 2c) + f (z + c),

(12.69)

which means that f (z) ≡ f (z + 4c), and thus f is a periodic function with period 4c. Remark 12.2.27. We see that Theorem 12.2.26 is not true for ρ(f ) = 1, k > l ≥ 2. Taking f (z) = e−z + z + 1, by an elementary computation we obtain that (f (z)2 )󸀠󸀠󸀠 + (f (z)2 )󸀠󸀠 = −4e−2z + 2e−z + 2 is a periodic function, but f (z) is not periodic. For the case k = l, see [208, Theorem 1.1]. To close this chapter, for recent considerations on the periodicity of 1)f

(k)

, we refer to Liu, Korhonen, and Liu [162].

f (k) fn

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Index λ-small function 29 ε-set 14 c-difference 11 c-shift 11 q-delay-differential equations 226 q-delay-differential Fermat type equations 236 q-delay-differential polynomials 213 q-difference Fermat type equations 234 q-difference Riccati equation 228 q-Gamma function 219 admissible solution 108

exponential type 99 exponential type functions 59 factorial series 88 Fermat delay-differential equations 132 Fermat difference equations 122 Fermat equations 117 Fermat type delay-differential systems 245 Fermat type differential equations 119 Fermat type functional equations 117 first order complex difference equation 93

binomial series 88 Borel exceptional values 51 Borel theorem 10 Brück Conjecture 71

gamma function 93 general linear shift polynomials 24 generalized Yang’s Conjecture 268 generic zero 173 Gross’s result 117

Cartan identity 3 central index 87 characteristic function 2, 17, 22 Clunie lemma 8, 9, 18, 20, 215 common poles 26 common zeros 26 complex delay-differential equation 96 complex delay-differential polynomials 24, 56 complex differential polynomials 21 composite functions 256 convex hull 59, 89 counting function 2

Hadamard factorization theorem 10 Hadamard three circles theorem 227 Hayman conjecture 218 Hayman’s result 21 higher order homogeneous linear complex difference equations 101 higher order homogeneous linear complex differential equations 100 higher order non-homogeneous linear complex delay-differential equations 106 hyper-exponent of convergence of poles 192 hyper-order 3

deficiency 4 delay-differential type Riccati equations 156 delay-differential version of Brück conjecture 72 difference counterpart to the Malmquist theorem 163 difference quotients 14 difference Riccati equations 150 difference variants of the second main theorem 17 difference variants of Wiman–Valiron theory 88 difference version of Brück conjecture 71 differential Riccati equation 149

iterated differences 11 Laurent series 26 linear conjugates 230 linear delay-differential polynomials 28, 74 linear measure 1 logarithmic derivatives 7 logarithmic derivatives lemma 5 logarithmic measure 1 logarithmic q-difference 90 lower linear density 2 lower logarithmic density 2

exceptional sets 1 exponential polynomials 59 exponential sums 59

Malmquist theorem 149, 228 Malmquist type delay-differential equations 164 Malmquist type systems 241

290 | Index

matrix 205 maximum modulus 87, 227 maximum term 87 Möbius transformation 225 Mohon’ko and Mohon’ko lemma 9, 216 Nevanlinna exceptional value 4 non-linear complex delay-differential equations 196 order 3, 87 pair counting function 17 periodic functions 151, 231 periodic mod g with period c 263 periodic set 258 periodicity 255 Poisson–Jensen formula 2 Pólya’s theorem 122 prime periodic functions 255 proximity function 2, 12 ramification index 5 Schröder q-difference equation 228 Stirling numbers 89 system of difference equations 241 systems of q-delay-differential equations 251

the difference analogue of the logarithmic derivative 12 the exponent of convergence for a-points 10 the exponent of convergence for the poles 10 the exponent of convergence for the zeros 10 the exponent of convergence of the zeros 100 the first main theorem 3 the second main theorem 7, 215, 217 trigonometric formulas 195 uniqueness 67, 221 uniqueness polynomial 266 uniqueness theory of delay-differential polynomials 78 upper linear density 1 upper logarithmic density 1 Valiron–Mohon’ko theorem 10 value distribution 21 value sharing 67, 222 Weierstrass elliptic function 18 Weierstrass factorization theorem 10 weighted counting functions 4, 6 Wiman–Valiron theory 87, 91 Yang’s Conjecture 259 zeros 21

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