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Research Article

Growth of Logarithmic Derivatives and Their Applications in

Complex Differential Equations

Zinelâabidine Latreuch and Benharrat Bela

\di

Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem (UMAB), B. P. 227, Mostaganem, Algeria

Correspondence should be addressed to Benharrat Bela¨ıdi; belaidi@univ-mosta.dz Received 29 November 2013; Accepted 21 January 2014; Published 4 March 2014 Academic Editor: Chuanxi Qian

Copyright © 2014 Z. Latreuch and B. Bela¨ıdi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We continue the study of the behavior of the growth of logarithmic derivatives. In fact, we prove some relations between the value distribution of solutions of linear differential equations and growth of their logarithmic derivatives. We also give an estimate of the growth of the quotient of two differential polynomials generated by solutions of the equation𝑓󸀠󸀠+ 𝐴(𝑧)𝑓󸀠+ 𝐵(𝑧)𝑓 = 0, where 𝐴(𝑧) and𝐵(𝑧) are entire functions.

1. Introduction and Main Results

Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna’s value distribution theory (see [1–4]). In addi-tion, we will denote by𝜆(𝑓) (resp., 𝜆(𝑓)) and 𝜆(1/𝑓) (resp., 𝜆(1/𝑓)) the exponent of convergence of zeros (resp., distinct zeros) and poles (resp., distinct poles) of a meromorphic function 𝑓, 𝜌(𝑓) to denote the order of growth of 𝑓. A meromorphic function𝜑(𝑧) is called a small function with respect to𝑓(𝑧) if 𝑇(𝑟, 𝜑) = 𝑜(𝑇(𝑟, 𝑓)) as 𝑟 → +∞ except possibly a set of𝑟 of finite linear measure, where 𝑇(𝑟, 𝑓) is the Nevanlinna characteristic function of𝑓.

Definition 1 (see [4]). Let𝑓 be a meromorphic function. Then the hyperorder of𝑓(𝑧) is defined as

𝜌2(𝑓) = lim sup

𝑟 → +∞

log log𝑇 (𝑟, 𝑓)

log𝑟 . (1)

Definition 2 (see [1,3]). The type of a meromorphic function 𝑓 of order 𝜌(0 < 𝜌 < ∞) is defined as

𝜏 (𝑓) = lim sup

𝑟 → +∞

𝑇 (𝑟, 𝑓)

𝑟𝜌 . (2)

If𝑓 is an entire function, then the type of 𝑓 of order 𝜌(0 < 𝜌 < ∞) is defined as

𝜏𝑀(𝑓) = lim sup

𝑟 → +∞

log𝑀 (𝑟, 𝑓)

𝑟𝜌 , (3)

where𝑀(𝑟, 𝑓) = max|𝑧|=𝑟|𝑓(𝑧)| is the maximum modulus function.

Remark 3. For entire function, we can have𝜏𝑀(𝑓) ̸= 𝜏(𝑓). For

example, if𝑓(𝑧) = 𝑒𝑧, then we have𝜏𝑀(𝑓) = 1 and 𝜏(𝑓) = 1/𝜋.

Growth of logarithmic derivative of meromorphic func-tions has been generously considered during the last decades, among others, by Gol’dberg and Grinˇste˘ın [5], Benbourenane and Korhonen [6], Hinkkanen [7], and Heittokangas et al. [8]. All these considerations have been devoted to getting more detailed estimates to the proximity function𝑚(𝑟, 𝑓󸀠/𝑓) than what is the original Nevanlinna estimate that essentially can be written as 𝑚(𝑟, 𝑓󸀠/𝑓) = 𝑆(𝑟, 𝑓). The same applies for the case of higher logarithmic derivatives 𝑚(𝑟, 𝑓(𝑘)/𝑓) as well. The estimation of logarithmic derivatives plays the key role in theory of differential equations. In his paper Gundersen [9] proved some interesting inequalities on the module of logarithmic derivatives of meromorphic functions. Volume 2014, Article ID 342974, 6 pages

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Recently [10], the authors have studied some properties of the behavior of growth of logarithmic derivatives of entire and meromorphic functions and have obtained some relations between the zeros of entire functions and the growth of their logarithmic derivatives. In fact, they have proved the following.

Theorem A (see [10]). Let𝑓 be an entire function with finite

number of zeros. Then, for any integer𝑘 ≥ 1,

𝜌 (𝑓(𝑘)

𝑓 ) = 𝜌2(𝑓) . (4)

Theorem B (see [10]). Suppose that𝑘 ≥ 2 is an integer and let

𝑓 be a meromorphic function. Then 𝜌 (𝑓𝑓󸀠) = max {𝜌 (𝑓𝑓(𝑘)) , 𝑘 ≥ 2}

= max {𝜌 (𝑓𝑓(𝑘)) , 𝜌 (𝑓(𝑘+1)𝑓 ) , 𝑘 ≥ 2} . (5)

In [11, pages 457–458], Bank and Langley have obtained the following result.

Theorem C (see [11]). All nontrivial solutions of

𝑓󸀠󸀠+ 𝑒𝑃(𝑧)𝑓 = 0, (6)

where𝑃(𝑧) is a nonconstant polynomial, satisfy 𝜆(𝑓) = ∞.

Theorem D (see [4] (E. Picard)). Any transcendental entire

function𝑓 must take every finite complex value infinitely many times, with at most one exception.

Example 4. The entire function𝑒𝑧 omits the value zero and every other finite value is assumed infinitely often and zero is the exceptional value of Picard.

Question 1. Under what conditions can we ensure that𝑓 has

no Picard’s exceptional value?

Question 2. Under what conditions can we obtain the same

result as Theorem C for higher order linear differential equations with entire coefficients?

The subject of this paper is to give answers to the above questions. In fact, we prove some relations between the value distribution of solutions of linear differential equations and growth of their logarithmic derivatives. Firstly, we study the relationship between the growth of logarithmic derivatives of solutions of complex differential equation

𝑓(𝑘)+ 𝐴𝑘−1𝑓(𝑘−1)+ ⋅ ⋅ ⋅ + 𝐴0𝑓 = 0 (7)

and their exponents of convergence. We will also prove the nonexistence of any Picard exceptional value of solutions of (7), and we obtain the following.

Theorem 5. Let 𝐴0, 𝐴1, . . . , 𝐴𝑘−1be entire functions

satisfy-ing𝜌(𝐴0) = 𝜌 (0 < 𝜌 < ∞), 𝜏𝑀(𝐴0) = 𝜏 (0 < 𝜏 < ∞), and

let𝜌(𝐴𝑗) ≤ 𝜌, 𝜏𝑀(𝐴𝑗) < 𝜏 if 𝜌(𝐴𝑗) = 𝜌 (𝑗 = 1, . . . , 𝑘 − 1). If

𝑓 is a nontrivial solution of (7), such that𝜌(𝑓󸀠/𝑓) > 𝜌(𝐴0),

then𝑓 assume every finite complex value infinitely often. Furthermore, if𝜌(𝑓󸀠/𝑓) = ∞, then

𝜆 (𝑓) = 𝜆 (𝑓) = ∞. (8)

Corollary 6. Under the hypotheses ofTheorem 5, let𝜑(𝑧) be

an entire function with finite order. If𝑓 is a nontrivial solution of (7), such that𝜌(𝑓(𝑗+1)/𝑓(𝑗)) = ∞ for some 𝑗 ∈ N, then

𝜆 (𝑓(𝑗)− 𝜑) = 𝜆 (𝑓(𝑗)− 𝜑) = ∞. (9)

Corollary 7. Under the hypotheses ofTheorem 5, every

non-trivial solution of (7) does not have any Picard exceptional

values.

Example 8. It is clear that the function𝑓(𝑧) = sin 𝑒𝑧satisfies the differential equation

𝑓󸀠󸀠− 𝑓󸀠+ 𝑒2𝑧𝑓 = 0 (10) and 𝜌(𝑓󸀠/𝑓) = 𝜌(𝑓) = ∞. Obviously, the conditions of

Theorem 5are satisfied. Then𝜆(𝑓) = 𝜆(𝑓) = ∞. Secondly, we consider the differential equation

𝑓󸀠󸀠+ 𝐴 (𝑧) 𝑓󸀠+ 𝐵 (𝑧) 𝑓 = 0, (11) where𝐴(𝑧) and 𝐵(𝑧) are entire functions, and let 𝑃𝑛[𝑓] and 𝑄𝑚[𝑓] be two differential polynomials defined by

𝑃𝑛[𝑓] = 𝑑𝑛𝑓(𝑛)+ 𝑑𝑛−1𝑓(𝑛−1)+ ⋅ ⋅ ⋅ + 𝑑0𝑓 (𝑛 ≥ 1) , 𝑄𝑚[𝑓] = 𝑏𝑚𝑓(𝑚)+ 𝑏𝑚−1𝑓(𝑚−1)+ ⋅ ⋅ ⋅ + 𝑏0𝑓 (𝑚 ≥ 1) , (12) where 𝑑𝑖(𝑖 = 0, . . . , 𝑛) and 𝑏𝑗 (𝑗 = 0, . . . , 𝑚) are entire functions not all vanishing identically. In recent years, there is a great interest to investigate the growth and oscillation of differential polynomials generated by solutions of differential equations in the complex plane (see [12–14]). In this part, we give under some conditions an estimate on the growth of the quotient of two differential polynomials generated by solutions of (11), and we obtain the following results.

Theorem 9. Let 𝑓 be a nontrivial solution of (11), and let

𝑑𝑖 (𝑖 = 0, . . . , 𝑛) and 𝑏𝑗 (𝑗 = 0, . . . , 𝑚) be entire functions

not all vanishing identically such that𝑔𝑓 = 𝑃𝑛[𝑓]/𝑄𝑚[𝑓] is irreducible function in𝑓󸀠/𝑓. If max{𝜌 (𝐴) , 𝜌 (𝐵) , 𝜌 (𝑑𝑖) (𝑖 = 0, . . . , 𝑛) , 𝜌 (𝑏𝑗) (𝑗 = 0, . . . , 𝑚)} < 𝜌 (𝑓 󸀠 𝑓) , (13) then 𝜌 (𝑔𝑓) = 𝜌 (𝑓󸀠 𝑓) . (14)

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Remark 10. In Theorem 9, the condition (13) is necessary. For example,𝑓(𝑧) = exp(𝑒𝑧) is a solution of the differential equation 𝑓󸀠󸀠− (𝑒𝑧+ 𝑒2𝑧) 𝑓 = 0 (15) with 𝜌 (𝑓󸀠 𝑓) = 1 (16) and let 𝑔𝑓=𝑑2𝑓 󸀠󸀠+ 𝑑 1𝑓󸀠+ 𝑑0𝑓 𝑏2𝑓󸀠󸀠+ 𝑏 1𝑓󸀠+ 𝑏0𝑓 = 𝑑2(𝑒𝑧+ 𝑒2𝑧) + 𝑑 1𝑒𝑧+ 𝑑0 𝑏2(𝑒𝑧+ 𝑒2𝑧) + 𝑏 1𝑒𝑧+ 𝑏0 . (17) If we take𝑑2= 𝑒𝑧2,𝑑1= 𝑑0= 1, and 𝑏2= 𝑏1= 𝑏0= 1, then

𝜌 (𝑔𝑓) = 2 > 𝜌 (𝑓 󸀠

𝑓) = 1. (18)

Remark 11. InTheorem 9, the condition that𝑔𝑓is irreducible function in𝑓󸀠/𝑓 is necessary. If 𝑑𝑖= 𝑏𝑗 (𝑛 = 𝑚), then 𝑔𝑓≡ 1 and so

𝜌 (𝑔𝑓) = 0 ̸= 𝜌 (𝑓󸀠

𝑓) . (19)

By setting 𝑛 = 𝑚 = 2 in Theorem 9, we can obtain the following special case.

Corollary 12. Let 𝐴(𝑧) and 𝐵(𝑧) be entire functions of finite

order such that𝜌(𝐴) < 𝜌(𝐵) and 0 < 𝜏(𝐴) < 𝜏(𝐵) < ∞ if

𝜌(𝐵) = 𝜌(𝐴) > 0. Let 𝑑𝑖(𝑧)(𝑖 = 0, 1, 2), 𝑏𝑗(𝑧)(𝑗= 0, 1, 2) be

entire functions that are not all vanishing identically such that

max{𝜌 (𝑑𝑖) , 𝜌 (𝑏𝑗) , 0 ≤ 𝑖, 𝑗 ≤ 2} < 𝜌 (𝐵) (20)

and𝑑2(𝑧)b1(𝑧)−𝑑1(𝑧)𝑏2(𝑧) ̸≡ 0. If 𝑓 ̸≡ 0 is a solution of (11),

such that𝜌(𝐵) < 𝜌(𝑓󸀠/𝑓), then

𝜌 (𝑑𝑏2𝑓󸀠󸀠+ 𝑑1𝑓󸀠+ 𝑑0𝑓

2𝑓󸀠󸀠+ 𝑏1𝑓󸀠+ 𝑏0𝑓) = 𝜌 (

𝑓󸀠

𝑓) . (21)

Corollary 13. Under the hypotheses ofCorollary 12, if𝑓 is a

nontrivial solution of (11), such that𝜌(𝑓󸀠/𝑓) = ∞, then 𝜆 (𝑓) = 𝜆 (𝑓󸀠) = ∞. (22)

2. Auxiliary Lemmas

Lemma 14 (see [1]). Let𝑓 be a meromorphic function and let

𝑘 ∈ N. Then

𝑚 (𝑟,𝑓(𝑘)

𝑓 ) = 𝑆 (𝑟, 𝑓) , (23)

where𝑆(𝑟, 𝑓) = 𝑂(log 𝑇(𝑟, 𝑓) + log 𝑟), possibly outside a set

𝐸 ⊂ (0, ∞) with a finite linear measure. If 𝑓 is a finite order of

growth, then

𝑚 (𝑟,𝑓𝑓(𝑘)) = 𝑂 (log 𝑟) . (24)

Lemma 15 (see [15]). Let𝐴0(𝑧), . . . , 𝐴𝑘−1(𝑧) be entire

func-tions of finite order such that

max{𝜌 (𝐴𝑗) : 𝑗 = 1, . . . , 𝑘 − 1} < 𝜌 (𝐴0) . (25)

Then every solution𝑓 ̸≡ 0 of (7) satisfies𝜌(𝑓) = +∞ and 𝜌2(𝑓) = 𝜌(𝐴0).

Lemma 16 (see [16]). Let𝐴0, 𝐴1, . . . , 𝐴𝑘−1be entire functions

satisfying𝜌(𝐴0) = 𝜌 (0 < 𝜌 < ∞), 𝜏𝑀(𝐴0) = 𝜏 (0 < 𝜏 < ∞), and let𝜌(𝐴𝑗) ≤ 𝜌, 𝜏𝑀(𝐴𝑗) < 𝜏 if 𝜌(𝐴𝑗) = 𝜌 (𝑗 = 1, . . . , 𝑘 − 1). Then every solution𝑓 ̸≡ 0 of (7) satisfies𝜌(𝑓) = +∞ and 𝜌2(𝑓) = 𝜌(𝐴0).

Here, we give a special case of the result given by Xu et al. in [17].

Lemma 17. Let 𝐴0, 𝐴1, . . . , 𝐴𝑘−1 be entire functions with

finite order and let them satisfy one of the following conditions:

(i) max{𝜌(𝐴𝑗) : 𝑗 = 1, . . . , 𝑘 − 1} < 𝜌(𝐴0) < ∞;

(ii)0 < 𝜌(𝐴𝑘−1) = ⋅ ⋅ ⋅ = 𝜌(𝐴0) < ∞ and max{𝜏𝑀(𝐴𝑗) : 𝑗 = 1, . . . , 𝑘 − 1} < 𝜏𝑀(𝐴0) < ∞.

Then, for every solution 𝑓 ̸≡ 0 of (7) and for any entire

function𝜑(𝑧) ̸≡ 0 satisfying 𝜌(𝜑) < ∞, we have

𝜆 (𝑓(𝑗)− 𝜑) = 𝜆 (𝑓(𝑗)− 𝜑) = 𝜌 (𝑓) = ∞ (𝑗 ∈ N) , 𝜆2(𝑓(𝑗)− 𝜑) = 𝜆2(𝑓(𝑗)− 𝜑) = 𝜌 (𝐴0) (𝑗 ∈ N) . (26)

Lemma 18 (see [10]). Let𝑓 be a meromorphic function. If

there exists an integer𝑘 ≥ 1, such that 𝜌(𝑓(𝑘)/𝑓) = 𝜌(𝑓) and

𝜌(𝑓) > 𝜌2(𝑓), then max{𝜆 (𝑓) , 𝜆 (1 𝑓)} = max {𝜆 (𝑓) , 𝜆 ( 1 𝑓)} = 𝜌 (𝑓) . (27)

Furthermore, if𝑓 is entire function, then

𝜆 (𝑓) = 𝜆 (𝑓) = 𝜌 (𝑓) . (28) By the principle of mathematical induction we can obtain easily the following lemma.

Lemma 19. Suppose that 𝑓󸀠󸀠 = −𝐴(𝑧)𝑓󸀠− 𝐵(𝑧)𝑓, where 𝐴(𝑧)

and𝐵(𝑧) are entire functions. Then, for an integer 𝑗 ≥ 2, there exist entire functions𝛼𝑗and𝛽𝑗such that

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Lemma 20 (see [18]). Let𝑓 be a solution of (7) where the

coefficients𝐴𝑗(𝑧) (𝑗 = 0, . . . , 𝑘−1) are analytic functions in the discΔ𝑅 = {𝑧 ∈ C : |𝑧| < 𝑅}, 0 < 𝑅 ≤ ∞. Let 𝑛𝑐 ∈ {1, . . . , 𝑘} be the number of nonzero coefficients𝐴𝑗(𝑧)(𝑗= 0, . . . , 𝑘 − 1), and let𝜃 ∈ [0, 2𝜋[ and 𝜀 > 0. If 𝑧𝜃 = ]𝑒𝑖𝜃 ∈ Δ𝑅such that

𝐴𝑗(𝑧𝜃) ̸= 0 for some 𝑗 = 0, . . . , 𝑘 − 1, then, for all ] < 𝑟 < 𝑅, 󵄨󵄨󵄨󵄨 󵄨𝑓 (𝑟𝑒𝑖𝜃)󵄨󵄨󵄨󵄨󵄨 ≤ 𝐶exp(𝑛𝑐∫ 𝑟 ] 𝑗=0,...,𝑘−1max 󵄨󵄨󵄨󵄨󵄨𝐴𝑗(𝑡𝑒 𝑖𝜃 )󵄨󵄨󵄨󵄨󵄨1/(𝑘−𝑗)𝑑𝑡) , (30)

where𝐶 > 0 is a constant satisfying

𝐶 ≤ (1 + 𝜀) max 𝑗=0,...,𝑘−1( 󵄨󵄨󵄨󵄨 󵄨𝑓(𝑗)(𝑧𝜃)󵄨󵄨󵄨󵄨󵄨 (𝑛𝑐)𝑗max𝑛=0,...,𝑘−1󵄨󵄨󵄨󵄨𝐴𝑛(𝑧𝜃)󵄨󵄨󵄨󵄨𝑗/(𝑘−𝑛) ) . (31) Here, we give a special case of the result given by Cao et al. in [19].

Lemma 21. Let 𝑓 be a meromorphic function with finite order

0 < 𝜌(𝑓) < ∞ and type 0 < 𝜏(𝑓) < ∞. Then for any given 𝛽 < 𝜏(𝑓) there exists a subset 𝐼 of [1, +∞) that has infinite

logarithmic measure such that𝑇(𝑟, 𝑓) > 𝛽𝑟𝜌(𝑓)holds for all

𝑟 ∈ 𝐼.

Lemma 22. Let 𝐴(𝑧) and 𝐵(𝑧) be entire functions such that

𝜌(𝐵) = 𝜌 (0 < 𝜌 < ∞), 𝜏(𝐵) = 𝜏 (0 < 𝜏 < ∞), and let 𝜌(𝐴) < 𝜌(𝐵) and 𝜏(𝐴) < 𝜏(𝐵) if 𝜌(𝐴) = 𝜌(𝐵). If 𝑓 ̸≡ 0 is a solution of 𝑓󸀠󸀠+ 𝐴 (𝑧) 𝑓󸀠+ 𝐵 (𝑧) 𝑓 = 0, (32)

then𝜌(𝑓) = ∞ and 𝜌2(𝑓) = 𝜌(𝐵).

Proof. If 𝜌(𝐴) < 𝜌(𝐵), then, byLemma 15, we obtain the result. We prove only the case when𝜌(𝐴) = 𝜌(𝐵) = 𝜌 and 𝜏(𝐴) < 𝜏(𝐵). Since 𝑓 ̸≡ 0, then, by (32), we have

𝐵 = − (𝑓󸀠󸀠 𝑓 + 𝐴

𝑓󸀠

𝑓) . (33)

Suppose that𝑓 is of finite order; then by (33) andLemma 14

we obtain

𝑇 (𝑟, 𝐵) ≤ 𝑇 (𝑟, 𝐴) + 𝑂 (log 𝑟) (34) which implies the contradiction

𝜏 (𝐵) ≤ 𝜏 (𝐴) . (35)

Hence𝜌(𝑓) = ∞. By using inequality (30) for𝑅 = ∞, we have

𝜌2(𝑓) ≤ max {𝜌 (𝐴) , 𝜌 (𝐵)} = 𝜌 (𝐵) . (36) On the other hand, since 𝜌(𝑓) = ∞, then, by (33) and

Lemma 14,

𝑇 (𝑟, 𝐵) ≤ 𝑇 (𝑟, 𝐴) + 𝑆 (𝑟, 𝑓) , (37)

where𝑆(𝑟, 𝑓) = 𝑂(log 𝑇(𝑟, 𝑓) + log 𝑟), possibly outside a set 𝐸 ⊂ (0, +∞) with a finite linear measure. By 𝜏(𝐴) < 𝜏(𝐵) we choose𝛼0,𝛼1satisfying𝜏(𝐴) < 𝛼1< 𝛼0< 𝜏(𝐵) such that, for 𝑟 → +∞, we have

𝑇 (𝑟, 𝐴) ≤ 𝛼1𝑟𝜌. (38)

ByLemma 21, there exists a subset𝐸1 ⊂ [1, +∞) of infinite logarithmic measure such that

𝑇 (𝑟, 𝐵) > 𝛼0𝑟𝜌. (39)

By (37)–(39) we obtain, for all𝑟 ∈ 𝐸1\ 𝐸,

(𝛼0− 𝛼1) 𝑟𝜌≤ 𝑂 (log 𝑇 (𝑟, 𝑓)) + 𝑂 (log 𝑟) (40) which implies

𝜌 (𝐵) = 𝜌 ≤ 𝜌2(𝑓) . (41)

By using (36) and (41), we obtain𝜌2(𝑓) = 𝜌(𝐵).

Remark 23. Lemma 22was obtained by Tu and Yi in [16] for higher order linear differential equations by using the type 𝜏𝑀(Lemma 16in the present paper).

Lemma 24 (see [20]). Let𝑓 and 𝑔 be meromorphic functions

such that0 < 𝜌(𝑓), 𝜌(𝑔) < ∞ and 0 < 𝜏(𝑓), 𝜏(𝑔) < ∞. Then we have the following.

(i) If𝜌(𝑓) > 𝜌(𝑔), then we obtain

𝜏 (𝑓 + 𝑔) = 𝜏 (𝑓𝑔) = 𝜏 (𝑓) . (42) (ii) If𝜌(𝑓) = 𝜌(𝑔) and 𝜏(𝑓) ̸= 𝜏(𝑔), then we get

𝜌 (𝑓 + 𝑔) = 𝜌 (𝑓𝑔) = 𝜌 (𝑓) = 𝜌 (𝑔) . (43)

3. Proof of Theorems and Corollaries

Proof ofTheorem 5. By the conditions of Theorem 5 and

Lemma 16, every nontrivial solution𝑓 of (7) is of infinite order and 𝜌2(𝑓) = 𝜌(𝐴0). We need to prove that every solution𝑓 attains every complex value without exception; that is, for all𝑎 ∈ C, the equation 𝑓(𝑧) = 𝑎 has infinitely many solutions. We have two cases

(i) If 𝑎 ̸= 0, then, by Lemma 17, we have 𝜆(𝑓 − 𝑎) = +∞ and the equation 𝑓(𝑧) = 𝑎 has infinitely many solutions.

(ii) If𝑎 = 0, then we suppose that 𝑓 has a finite number of zeros. By Theorem A

𝜌 (𝐴0) < 𝜌 (𝑓 󸀠

𝑓) = 𝜌2(𝑓) = 𝜌 (𝐴0) (44) which is a contradiction, and hence𝑓 assume every finite complex value infinitely often. Now, if𝜌(𝑓󸀠/𝑓) = 𝜌(𝑓) = ∞ and since𝜌2(𝑓) = 𝜌(𝐴0) < 𝜌(𝑓) = ∞, then by using

Lemma 18we have

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Proof ofCorollary 6. By using Lemma 17, we have for any finite order entire function𝜑(𝑧) ̸≡ 0

𝜆 (𝑓(𝑗)− 𝜑) = 𝜆 (𝑓(𝑗)− 𝜑) = 𝜌 (𝑓) = ∞ (𝑗 ∈ N) . (46)

Now, if𝜑(𝑧) ≡ 0, then we put 𝑔(𝑧) = 𝑓(𝑗)(𝑧) (𝑗∈ N). It follows that

𝜌 (𝑓(𝑗+1)

𝑓(𝑗) ) = 𝜌 (

𝑔󸀠

𝑔) = 𝜌 (𝑔) = ∞, (47) and, since𝜌2(𝑔) = 𝜌2(𝑓) = 𝜌(𝐴0) < 𝜌(𝑔) = ∞, then, by applyingLemma 18, we obtain

𝜆 (𝑔) = 𝜆 (𝑔) = ∞; (48) that is,𝜆(𝑓(𝑗)) = 𝜆(𝑓(𝑗)) = ∞ (𝑗 ∈ N) and the proof of

Corollary 6is completed.

Proof ofTheorem 9. By usingLemma 19, we have 𝑔𝑓= 𝑄𝑃𝑛[𝑓] 𝑚[𝑓] = ∑𝑛𝑖=0𝑑𝑖𝑓(𝑖) ∑𝑚𝑗=0𝑏𝑗𝑓(𝑗) = 𝑑0𝑓 + 𝑑1𝑓 󸀠+ ∑𝑛 𝑖=2𝑑𝑖(𝛼𝑖𝑓󸀠+ 𝛽𝑖𝑓) 𝑏0𝑓 + 𝑏1𝑓󸀠+ ∑𝑚 𝑗=2𝑏𝑗(𝛼𝑗𝑓󸀠+ 𝛽𝑗𝑓) , (49)

which we can write as

𝑔𝑓= 𝑃 (𝑧) 𝑓󸀠/𝑓 + 𝑄 (𝑧) 𝑅 (𝑧) 𝑓󸀠/𝑓 + 𝑆 (𝑧), (50) where 𝑃 = 𝑑1+ 𝑛 ∑ 𝑖=2𝑑𝑖𝛼𝑖, 𝑄 = 𝑑0+ 𝑛 ∑ 𝑖=2𝑑𝑖𝛽𝑖, 𝑅 = 𝑏1+∑𝑚 𝑗=2 𝑏𝑗𝛼𝑗, 𝑆 = 𝑏0+∑𝑚 𝑗=2 𝑏𝑗𝛽𝑗. (51) It is clear that max{𝜌 (𝑃) , 𝜌 (𝑄) , 𝜌 (𝑅) , 𝜌 (𝑆)} < 𝜌 (𝑓 󸀠 𝑓) (52) and, since𝑔𝑓is irreducible function in𝑓󸀠/𝑓, then 𝑃𝑆−𝑄𝑅 ̸≡ 0. So

𝜌 (𝑔𝑓) ≤ max {𝜌 (𝑃) , 𝜌 (𝑄) , 𝜌 (𝑅) , 𝜌 (𝑆) , 𝜌 (𝑓𝑓󸀠)} = 𝜌 (𝑓󸀠

𝑓) .

(53)

On the other hand, we have 𝑓󸀠 𝑓 = −𝑆 (𝑧) 𝑔𝑓+ 𝑄 (𝑧) 𝑅 (𝑧) 𝑔𝑓− 𝑃 (𝑧) (54) which implies 𝜌 (𝑓󸀠 𝑓) ≤ max {𝜌 (𝑃) , 𝜌 (𝑄) , 𝜌 (𝑅) , 𝜌 (𝑆) , 𝜌 (𝑔𝑓)} = 𝜌 (𝑔𝑓) . (55) By (53) and (55), we obtain𝜌(𝑔𝑓) = 𝜌(𝑓󸀠/𝑓).

Proof ofCorollary 12. Under the conditions of Corollary 12

andLemma 22, we have𝜌(𝑓) = ∞ and 𝜌2(𝑓) = 𝜌(𝐵). Set 𝑔𝑓= 𝑑2𝑓 󸀠󸀠+ 𝑑 1𝑓󸀠+ 𝑑0𝑓 𝑏2𝑓󸀠󸀠+ 𝑏 1𝑓󸀠+ 𝑏0𝑓. (56) Substituting𝑓󸀠󸀠= −𝐴(𝑧)𝑓󸀠− 𝐵(𝑧)𝑓 into 𝑔𝑓, we obtain

𝑔𝑓= 𝑑𝑏2𝑓󸀠󸀠+ 𝑑1𝑓󸀠+ 𝑑0𝑓 2𝑓󸀠󸀠+ 𝑏1𝑓󸀠+ 𝑏0𝑓 = (𝑑1− 𝑑2𝐴) 𝑓󸀠+ (𝑑 0− 𝑑2𝐵) 𝑓 (𝑏1− 𝑏2𝐴) 𝑓󸀠+ (𝑏 0− 𝑏2𝐵) 𝑓 = (𝑑(𝑏1− 𝑑2𝐴) 𝑤 + (𝑑0− 𝑑2𝐵) 1− 𝑏2𝐴) 𝑤 + (𝑏0− 𝑏2𝐵) , (57) where 𝑤 = 𝑓󸀠/𝑓. In order to prove that 𝑔𝑓 is irreducible function in𝑤, we need only to prove that ℎ ̸≡ 0, where

ℎ = (𝑑1− 𝑑2𝐴) (𝑏0− 𝑏2𝐵) − (𝑑0− 𝑑2𝐵) (𝑏1− 𝑏2𝐴)

= 𝑏0𝑑1− 𝑏1𝑑0+ 𝐴 (𝑏2𝑑0− 𝑏0𝑑2) + 𝐵 (𝑏1𝑑2− 𝑏2𝑑1) . (58) Since𝜌(𝐴) < 𝜌(𝐵) and 0 < 𝜏(𝐴) < 𝜏(𝐵) < ∞, if 𝜌(𝐵) = 𝜌(𝐴) > 0, 𝑑2(𝑧)𝑏1(𝑧) − 𝑑1(𝑧)𝑏2(𝑧) ̸≡ 0, then, byLemma 24, we have𝜌(ℎ) = 𝜌(𝐵) > 0, so ℎ ̸≡ 0. By applyingTheorem 9

we obtain𝜌(𝑓󸀠/𝑓) = 𝜌(𝑔𝑓).

Proof ofCorollary 13. Suppose that𝑓 is a nontrivial solution

of (11) such that𝜌(𝑓󸀠/𝑓) = ∞. Then, byLemma 22, we have 𝜌 (𝑓) = +∞, 𝜌2(𝑓) = 𝜌 (𝐵) . (59) By setting𝑑2 = 𝑏1 ≡ 1 and 𝑑1 = 𝑑0 = 𝑏2 = 𝑏0 ≡ 0 and using

Corollary 12we have

𝜌 (𝑓𝑓󸀠) = 𝜌 (𝑓𝑓󸀠󸀠󸀠) = ∞ (60) which implies by usingLemma 18that

𝜆 (𝑓) = 𝜆 (𝑓󸀠) = ∞. (61)

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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