The Hyperorder of Solutions of Second-Order Linear Differential Equations

Abstract

We prove that the hyperorder of every nontrivial solution of homogenous linear differential equations of type 𝑓′′+𝐴₁(𝑧)𝑒^𝑎𝑧𝑓′+𝐴₀(𝑧)𝑒^𝑏𝑧𝑓=0 and nonhomogeneous equation of type 𝑓′′+𝐴₁(𝑧)𝑒^𝑎𝑧𝑓′+𝐴₀(𝑧)𝑒^𝑏𝑧𝑓=𝐻(𝑧) is one, where 𝐴₀,𝐴₁,𝐻(𝑧) are entire functions of order less than one, improving the previous results of Chen, Wang, and Laine.

1. Introduction

We assume that the reader is familiar with the usual notations and the basic results of the Nevanlinna theory (see [1–4]). We also use basic notions and the results of the Wiman-Valiron theory; see [5]. Let 𝑓(𝑧) be a nonconstant meromorphic function in the complex plane. We remark that 𝜎(𝑓), respectively, 𝜎₂(𝑓) will be used to denote the order, respectively, the hyperorder, of 𝑓. In particular, the hyperorder 𝜎₂(𝑓) is defined as 𝜎₂𝑓=limsup_𝑟→∞loglog𝑇𝑟,𝑓;log𝑟 (1) see [1, 2, 4]. For a set 𝐸⊂𝑅⁺, let 𝑚(𝐸), respectively, 𝜆(𝐸), denote the linear measure, respectively, the logarithmic measure of 𝐸. Moreover, the upper logarithmic density and the lower logarithmic density of 𝐸 are defined by logdens(𝐸)=limsup_𝑟→∞𝜆𝐸⋂[1,𝑟],log𝑟logdens(𝐸)=liminf_𝑟→∞𝜆𝐸⋂[1,𝑟].log𝑟 (2) Observe that 𝐸 may have a different meaning at different occurrences in what follows.

We now recall some previous results concerning linear differential equations of type 𝑓^′′+𝐴₁(𝑧)𝑒^𝑎𝑧𝑓^′+𝐴₀(𝑧)𝑒^𝑏𝑧𝑓𝑓=0,_′′+𝐴₁(𝑧)𝑒^𝑎𝑧𝑓^′+𝐴₀(𝑧)𝑒^𝑏𝑧𝑓=𝐻(𝑧), (3)(4) where 𝐴₀,𝐴₁,𝐻(𝑧) are entire functions of order less than one, and 𝑎,𝑏 are complex constants.

Chen proved the following theorem; see [6].

Theorem A. Let 𝐴₀≢0, 𝐴₁≢0 be entire functions of order less than one, and the complex constants 𝑎,𝑏 satisfy 𝑎𝑏≠0 and 𝑎=𝑐𝑏(𝑐>1). Then every nontrivial solution 𝑓 of (3) is of infinite order.

Wang and Laine investigated the nonhomogeneous equation (4) and got the following; see [7].

Theorem B. Suppose that 𝐴₀≢0, 𝐴₁≢0, 𝐻 are entire functions of order less than one, and the complex constants 𝑎,𝑏 satisfy 𝑎𝑏≠0 and 𝑏≠𝑎. Then every nontrivial solution 𝑓 of (4) is of infinite order.

Theorem C. Suppose that 𝐴₀≢0, 𝐴₁≢0, 𝐷₀, 𝐷₁, 𝐻 are entire functions of order less than one, and the complex constants 𝑎,𝑏 satisfy 𝑎𝑏≠0 and 𝑏/𝑎<0. Then every nontrivial solution 𝑓 of equation 𝑓^′′+𝐴₁(𝑧)𝑒^𝑎𝑧+𝐷₁(𝑧)𝑓^′+𝐴₀(𝑧)𝑒^𝑏𝑧+𝐷₀(𝑧)𝑓=𝐻(𝑧) (5) is of infinite order.

In this paper, we investigate the hyperorder of the nontrivial solutions of (3), (4), and (5) and obtain the following theorems.

Theorem 1. Suppose that 𝐴₀≢0, 𝐴₁≢0, 𝐻 are entire functions of order less than one, and the complex constants 𝑎,𝑏 satisfy 𝑎𝑏≠0 and 𝑏≠𝑎. Then the hyperorder of every nontrivial solution 𝑓 of (4) is one.

Corollary 2. Let 𝐴₀≢0, 𝐴₁≢0 be entire functions of order less than one, and the complex constants 𝑎,𝑏 satisfy 𝑎𝑏≠0 and 𝑎=𝑐𝑏(𝑐>1). Then the hyperorder of every nontrivial solution 𝑓 of (3) is one.

Theorem 3. Suppose that 𝐴₀≢0, 𝐴₁≢0, 𝐷₀, 𝐷₁, 𝐻 are entire functions of order less than one, and the complex constants 𝑎,𝑏 satisfy 𝑎𝑏≠0 and 𝑏/𝑎<0. Then the hyperorder of every nontrivial solution 𝑓 of (5) is one.

2. Lemmas

Lemma 4 (see [5]). Let 𝑓 be an entire function of infinite order and let 𝜈_𝑓(𝑟) be the central index of 𝑓(𝑧), then the hyperorder 𝜎₂𝑓=limsup_𝑟→∞loglog𝜈_𝑓(𝑟).log𝑟 (6)

Lemma 5 (see [8]). Let 𝑓 be an entire function of infinite order with 𝜎₂(𝑓)=𝛼(0≤𝛼<∞), and there exists a set 𝐸₂⊂[1,∞) which have a finite logarithmic measure. Then there exists a sequence {𝑧_𝑛=𝑟_𝑛𝑒^𝑖𝜃𝑛} such that |𝑓(𝑧_𝑛)|=𝑀(𝑟_𝑛,𝑓), 𝜃_𝑛∈[0,2𝜋), lim_𝑛→∞𝜃_𝑛=𝜃₀∈[0,2𝜋), and 𝑟_𝑛∉𝐸₂, 𝑟_𝑛→∞ and such that(1)if 𝜎₂(𝑓)=𝛼(0<𝛼<∞), then, for any given 𝜀₁(0<𝜀₁<𝛼), exp𝑟^𝛼−𝜀1_𝑛<𝜈𝑟_𝑛<exp𝑟^𝛼+𝜀1_𝑛; (7)(2)if 𝜎₂(𝑓)=0, then, for any given 𝜀₂(0<𝜀₂<1/2) and for any large 𝑀₁(>0), 𝑟^𝑀1_𝑛<𝜈𝑟_𝑛<exp𝑟^𝜀2_𝑛. (8)

Lemma 6 (see [7]). Suppose that 𝑃(𝑧)=(𝛼+𝑖𝛽)𝑧, where 𝛼,𝛽 are real numbers, |𝛼|+|𝛽|≠0, and that 𝐴(𝑧)(≢0) is a meromorphic function with 𝜎(𝐴)<1. Set 𝑔(𝑧)=𝐴(𝑧)𝑒^𝑃(𝑧), 𝑧=𝑟𝑒^𝑖𝜃, 𝛿(𝑃,𝜃)=𝛼cos𝜃−𝛽sin𝜃. Then, for any given 𝜀>0, there exists a set 𝐸₃⊂(1,∞) of finite linear measure such that, for any 𝜃∈[0,2𝜋)⧵𝐻, there exists 𝑅>0 such that, for |𝑧|=𝑟>𝑅 and 𝑟∉𝐸₃, we have(1)if 𝛿(𝑃,𝜃)>0, then )𝛿()𝑟}<𝑔exp{(1−𝜀𝑃,𝜃𝑟𝑒^𝑖𝜃<exp{(1+𝜀)𝛿(𝑃,𝜃)𝑟}; (9)(2)if 𝛿(𝑃,𝜃)<0, then )𝛿()𝑟}<𝑔exp{(1+𝜀𝑃,𝜃𝑟𝑒^𝑖𝜃<exp{(1−𝜀)𝛿(𝑃,𝜃)𝑟}, (10)where 𝐻={𝜃∈[0,2𝜋)∣𝛿(𝑃,𝜃)=0}.

Lemma 7 (see [6]). Let 𝐴,𝐵 be entire functions of finite order and if 𝑓 is a solution of equation 𝑓^′′+𝐴𝑓^′+𝐵𝑓=0, (11) then the hyperorder 𝜎₂(𝑓)≤max{𝜎(𝐴),𝜎(𝐵)}.

The proof of the lemma below follows the idea of Bergweiler et al.; see [9, Theorem 3.1].

Lemma 8. Let 𝑓(𝑧) be an entire function, and 𝑀(𝑟,𝑓)=𝑓(𝑟𝑒^𝑖𝜃𝑟) for every 𝑟. Set 𝜃_𝑟→𝜃₀, and there exists a constant 𝑙₀>0 and a set 𝐸 with positive lower logarithmic density such that 𝑀𝑟,𝑓^1/5≤𝑓𝑟𝑒^𝑖𝜃 (12) for all 𝑟∈𝐸 large enough and all 𝜃 such that |𝜃−𝜃₀|<𝑙₀.

Proof. Since 𝑓 is entire function, we know that 𝑀(𝑟,𝑓) is nondecrease, 𝑀(𝑟,𝑓)→∞ as 𝑟→∞, and |𝑓(𝑟𝑒^𝑖𝜃)| is continuous on the circle |𝑧|=𝑟 for 𝑧=𝑟𝑒^𝑖𝜃. Set 𝜃_𝑟→𝜃₀∈[0,2𝜋) as 𝑟→∞. 𝐴(𝑢^𝑎,𝑢^𝑏) denotes an annulus for 0<𝑎<𝑏 and sufficiently large 𝑢. Then, there exists a constant 4𝑙₀(<𝜋) such that |𝑓(𝑟𝑒^𝑖𝜃)|>1 for 𝑧=𝑟𝑒^𝑖𝜃∈𝐷, where 𝐷∶={(𝑟,𝜃)∣𝑟∈𝐴(𝑢^𝑎,𝑢^𝑏),|𝜃−𝜃₀|<4𝑙₀} for 𝑢 sufficiently large. Then the function ℎ(𝑧)∶=log|𝑓(𝑧)| is a positive harmonic in 𝐷. So 𝐻(𝑡)=ℎ(𝑒^𝑡) is a positive harmonic in the domain 𝑆∶={𝑡∣𝑎log𝑢<ℜ(𝑡)<𝑏log𝑢,𝜃₀−4𝑙₀<ℑ(𝑡)<𝜃₀+4𝑙₀}. Thus, if 𝑡₁ and 𝑡₂ satisfy 𝑎log𝑢+3𝑙₀<ℜ(𝑡₁)=ℜ(𝑡₂)<𝑏log𝑢−3𝑙₀ and |ℑ(𝑡₁)−ℑ(𝑡₂)|<2𝑙₀, where ℑ(𝑡_𝑗)∈(𝜃₀−𝑙₀,𝜃₀+𝑙₀), 𝑗=1,2, then {𝑡∣|𝑡−𝑡₁|<3𝑙₀} and |𝑡₁−𝑡₂|<2𝑙₀<3𝑙₀. So 15=3𝑙₀−2𝑙₀3𝑙₀+2𝑙₀≤𝐻𝑡₂𝐻𝑡₁≤3𝑙₀+2𝑙₀3𝑙₀−2𝑙₀=5 (13) by Harnack's inequality; see [10, Theorem 1.3.1]. Therefore, if 𝑧₁ and 𝑧₂ are in the domain 𝐷₁∶={(𝑟,𝜃)∣𝑟∈𝐴(𝑢^𝑎,𝑢^𝑏),|𝜃−𝜃₀|<𝑙₀}, where 𝑢 is sufficiently large, then 15≤ℎ𝑧₂ℎ𝑧₁≤5. (14) Set 𝑚^∗(𝑟,𝑓)=min_{|𝑧|=𝑟,|𝜃−𝜃0|<𝑙0}|𝑓(𝑟𝑒^𝑖𝜃)|. Then, we have 1/5log𝑀(𝑟,𝑓)≤log𝑚^∗(𝑟,𝑓) for 𝑧∈𝐷₁. If let 𝑢→∞, and then the set 𝐸 of 𝑟∈(𝑢^𝑎,𝑢^𝑏) is of positive lower logarithmic density. Thus, the conclusion of this lemma holds.

Lemma 9. Let 𝑓(𝑧) be an entire function with infinite order and let hyperorder 𝜎₂(𝑓)≤1, 𝑔(𝑧) be an entire function with finite order 𝜎(𝑔)<∞. For 𝑟∈𝐸, where 𝐸 is the infinite logarithmic measure set which is given in Lemma 8. Then, for any given 𝜀₀, 𝑔(𝑧)𝑓(𝑧)<𝜀₀ (15) for all 𝑧 such that |𝑧|=𝑟∈𝐸 is sufficiently large and that |𝑓(𝑧)|=𝑀(𝑟,𝑓).

Proof. Since, for the entire function 𝑔(𝑧), 𝜎𝑔=limsup_𝑟→∞loglog𝑀𝑟,𝑔log𝑟 (16) for any given 𝜀, we have 𝑔(𝑧)≤exp𝑟^{𝜎(𝑔)+𝜀} (17) for all 𝑟 sufficiently large. Since the order of 𝑓 is infinite, for 𝑟∈𝐸, there exists a sufficiently large real number 𝐴 such that 𝜎𝑓=limsup_𝑟→∞loglog𝑀𝑟,𝑓log𝑟>𝐴. (18) Thus, for 𝑟∈𝐸 is sufficiently large, 𝑀𝑟,𝑓≥exp𝑟^𝐴−𝜀. (19) By (17) and (19), we conclude that 𝑔(𝑧)𝑓(𝑧)≤exp𝑟^{𝜎(𝑔)+𝜀}𝑀≤𝑟𝑟,𝑓exp^{𝜎(𝑔)+𝜀}exp𝑟^𝐴−𝜀⟶0 (20) for all 𝑧 satisfying |𝑓(𝑧)|=𝑀(𝑟,𝑓) such that 𝑟∈𝐸 is sufficiently large. Thus, the conclusion holds.

3. Proofs of Theorems

Proof of Theorem 1. Suppose that 𝑓 is a solution of (4), and then 𝑓 is an entire function.
Step  1. We prove that 𝜎₂(𝑓)≤1. Since 𝜎(𝐴₀;𝐴₁;𝐻)<1, set 𝜎(𝐻)=𝜆<1. Then for any given 𝜀 satisfying 𝜀<1−𝜆, when 𝑟 is sufficiently large, we have 𝐴₁𝑒^𝑎𝑧≤exp𝑟^1+𝜀𝐴₀,𝑒^𝑏𝑧≤exp𝑟^1+𝜀,|𝐻(𝑧)|≤exp𝑟^𝜆+𝜀. (21)(22)(23) From the Wiman-Valiron theory, there is a set 𝐸₁ having finite logarithmic measure, such that 𝑓^(𝑗)(𝑧)𝑓(𝑧)=𝜈_𝑓(𝑟)𝑧^𝑗(1+𝑂(1))𝑗=1,2 (24) whenever |𝑓(𝑧)|=𝑀(𝑟,𝑓),𝑟∉𝐸₁, where the 𝜈_𝑓(𝑟) is the central index of 𝑓(𝑧), and we know that 𝜈_𝑓(𝑟)→∞ as 𝑟→∞. When 𝑟 sufficiently large, we have |𝑓(𝑧)|=𝑀(𝑟,𝑓)>1. From (4) we have 𝑓^′′𝑓≤𝐴₁𝑒^𝑎𝑧𝑓^′𝑓+𝐴₀𝑒^𝑏𝑧+𝐻𝑓. (25) Substituting (21), (22), (23), and (24) into (25), we obtain 𝜈_𝑓(𝑟)|𝑧|²((1𝑟1+𝑂))≤exp^1+𝜀𝜈_𝑓(𝑟)|𝑧|((1𝑟1+𝑂))+exp^1+𝜀+exp𝑟^𝜆+𝜀, (26) where 𝑧 satisfies |𝑧|=𝑟∉𝐸₁ and 𝑟 sufficiently large. By (26) we get limsup_𝑟→∞loglog𝜈_𝑓(𝑟)log𝑟≤1+𝜀. (27) Since 𝜀 is arbitrary, by (27) and Lemma 4, we have 𝜎₂(𝑓)≤1.
Step  2. By Theorem B, we know that the order of 𝑓 is infinite, and, by the first step, we clear that the hyperorder of 𝑓 is less than one. Thus, by Lemma 9 and (23), we have 𝐻𝑓<𝜀₀ (28) for all 𝑧 satisfying |𝑓(𝑧)|=𝑀(𝑟,𝑓) such that 𝑟∈𝐸 is sufficiently large, where 𝐸 is of infinite logarithmic measure. Set 𝜎₂(𝑓)=𝛼₀, and we assert that 𝛼₀=1. Now we assume that 𝛼₀<1, and prove that 𝜎₂(𝑓)=𝛼₀<1 results in contradictions. 𝐸₂,𝐸₃ are the sets in Lemmas 5 and 6, respectively.
Since 𝜆(𝐸₁∪𝐸₂∪𝐸₃)<∞, we have that 𝜆(𝐸⧵(𝐸₁∪𝐸₂∪𝐸₃)) is infinite. Thus, by Lemma 5, we see that there exists a sequence of points {𝑧_𝑛=𝑟_𝑛𝑒^𝑖𝜃𝑛} such that |𝑓(𝑧_𝑛)|=𝑀(𝑟_𝑛,𝑓), 𝜃_𝑛∈[0,2𝜋), lim_𝑛→∞𝜃_𝑛=𝜃₀∈[0,2𝜋), 𝑟_𝑛∈𝐸⧵(𝐸₁∪𝐸₂∪𝐸₃), 𝑟_𝑛→∞, and if 𝜎₂(𝑓)=𝛼₀(0<𝛼₀<∞), then, for any given 𝜀₁(0<𝜀₁<min{𝛼₀,1−𝛼₀}), exp𝑟^{𝛼0−𝜀1}_𝑛<𝜈_𝑓𝑟_𝑛<exp𝑟^𝛼0+𝜀1_𝑛; (29) if 𝜎₂(𝑓)=0, then, for any given 𝜀₂(0<𝜀₂<1/2) and for any large 𝑀₁(>0), 𝑟^𝑀1_𝑛<𝜈_𝑓𝑟_𝑛<exp𝑟^𝜀2_𝑛. (30) Firstly, we prove the case when 𝜎₂(𝑓)=𝛼₀(>0). It can separate into three cases to discuss.
Case  1. First assume that 𝛿(𝑎𝑧,𝜃₀)>0. From the continuity of 𝛿(𝑎𝑧,𝜃₀), we have 12𝛿𝑎𝑧,𝜃₀<𝛿𝑎𝑧,𝜃_𝑛<32𝛿𝑎𝑧,𝜃₀ (31) for sufficiently large 𝑛. From (9), we deduce that exp1−𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛≤𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛3()≤exp1+𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛 (32) for all 𝑛 sufficiently large.
From (4), we have 𝑓^′𝑧_𝑛𝑓𝑧_𝑛+𝐴₀𝑧_𝑛𝐴₁𝑧_𝑛𝑒^{(𝑏−𝑎)𝑧𝑛}≤1𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛𝑓^′′𝑧_𝑛𝑓𝑧_𝑛+𝐻𝑧_𝑛𝑓𝑧_𝑛. (33)
Subcase  1.1. We first assume that 𝜃₀ satisfies 𝜂∶=𝛿((𝑏−𝑎)𝑧,𝜃₀)>0. From the continuity of 𝛿((𝑏−𝑎)𝑧,𝜃), we also have exp1−𝜀2𝜂𝑟_𝑛≤𝐴₀𝑧_𝑛𝐴₁𝑧_𝑛𝑒^{(𝑏−𝑎)𝑧𝑛}3()≤exp1+𝜀2𝜂𝑟_𝑛 (34) for all 𝑛 sufficiently large. From (33), we get 𝐴₀𝑧_𝑛𝐴₁𝑧_𝑛𝑒^{(𝑏−𝑎)𝑧𝑛}≤𝑓^′𝑧_𝑛𝑓𝑧_𝑛+1𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛𝑓^′′𝑧_𝑛𝑓𝑧_𝑛+𝐻𝑧_𝑛𝑓𝑧_𝑛. (35) Substituting  (24), (28), (29), (32), and (34) into (35), we obtain exp1−𝜀2𝜂𝑟_𝑛≤exp𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻¹((1−1+𝑂))+exp1−𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛×exp2𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻²(1+𝑂(1))+𝜀₀. (36) Since 𝛼₀+𝜀₁<1, we see that (36) are contradictory as 𝑛→∞.
Subcase  1.2. Next assume that 𝜂∶=𝛿((𝑏−𝑎)𝑧,𝜃₀)<0. Then, from (10), for 𝑛 large enough, we deduce that 3()exp1+𝜀2𝜂𝑟_𝑛≤𝐴₀𝑧_𝑛𝐴₁𝑧_𝑛𝑒^{(𝑏−𝑎)𝑧𝑛}≤exp1−𝜀2𝜂𝑟_𝑛. (37) From (33), we get 𝑓^′𝑧_𝑛𝑓𝑧_𝑛≤𝐴₀𝑧_𝑛𝐴₁𝑧_𝑛𝑒^{(𝑏−𝑎)𝑧𝑛}+1𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛𝑓^′′𝑧_𝑛𝑓𝑧_𝑛+𝐻𝑧_𝑛𝑓𝑧_𝑛. (38) Substituting  (24), (28), (29), (32), and (37) into (38), we obtain 𝜈_𝑓𝑟_𝑛𝑟_𝑛(1+𝑂(1))≤exp1−𝜀2𝜂𝑟_𝑛+exp−1−𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛×exp2𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻²(1+𝑂(1))+𝜀₀ (39) as 𝑛→∞. Since 𝛼₀+𝜀₁<1, this implies that 𝜈_𝑓(𝑟)→0, 𝑛→∞, which is impossible.
Subcase  1.3. Assume finally that 𝜂∶=𝛿((𝑏−𝑎)𝑧,𝜃₀)=0. Here, (12) may be used to construct another sequence of points {𝑧^∗_𝑛=𝑟_𝑛𝑒^{𝑖𝜃∗𝑛}} with lim_𝑛→∞𝜃^∗_𝑛=𝜃^∗₀, such that 𝜂₁∶=𝛿((𝑏−𝑎)𝑧,𝜃^∗₀)>0. Indeed, we may suppose, without less of generality, that ))𝜃𝜂∶=𝛿((𝑏−𝑎𝑧,𝜃>0,𝜃∈₀+2𝑘𝜋,𝜃₀+()𝜋,))𝜃2𝑘+1𝜂∶=𝛿((𝑏−𝑎𝑧,𝜃<0,𝜃∈₀+(2𝑘−1)𝜋,𝜃₀+2𝑘𝜋 (40) with 𝑘∈ℤ. When 𝑛 is large enough, we have |𝜃_𝑛−𝜃₀|≤𝑙₀, where 𝑙₀ is a small constant. Choose now 𝜃^∗_𝑛 such that 𝑙₀/2≤𝜃^∗_𝑛−𝜃_𝑛≤𝑙₀. Then 𝜃₀+𝑙₀/2≤𝜃^∗₀≤𝜃₀+𝑙₀. For sufficiently large 𝑛, we have (12) for 𝑧^∗_𝑛 and 𝜂₁∶=𝛿((𝑏−𝑎)𝑧,𝜃^∗₀)>0. Therefore 𝐻𝑧^∗_𝑛𝑓𝑧^∗_𝑛≤𝑀𝑟_𝑛,𝐻𝑀𝑟_𝑛,𝑓^1/5⟶0,exp1−𝜀2𝜂₁𝑟_𝑛≤𝐴₀𝑧^∗_𝑛𝐴₁𝑧^∗_𝑛𝑒^{(𝑏−𝑎)𝑧∗𝑛}3()≤exp1+𝜀2𝜂₁𝑟_𝑛 (41)(42) for sufficiently large 𝑛. Taking now 𝑙₀ small enough, we have 𝛿(𝑎𝑧,𝜃^∗₀)>0, by the continuity of 𝛿(𝑎𝑧,𝜃). This yields exp1−𝜀2𝛿𝑎𝑧,𝜃^∗₀𝑟_𝑛≤𝐴₁𝑧^∗_𝑛𝑒^{𝑎𝑧∗𝑛}3()≤exp1+𝜀2𝛿𝑎𝑧,𝜃^∗₀𝑟_𝑛. (43) Similarly as (36), a contradiction easily follows.
Case  2. Suppose now that 𝛿(𝑎𝑧,𝜃₀)<0. Then, from the continuity of 𝛿(𝑎𝑧,𝜃) and (10), we have 3()exp1+𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛≤𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛≤exp1−𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛 (44) for 𝑛 large enough.
Subcase  2.1. Assume first that 𝛿(𝑏𝑧,𝜃₀)>0. From the continuity of 𝛿(𝑏𝑧,𝜃) and (9), we deduce that exp1−𝜀2𝛿𝑏𝑧,𝜃₀𝑟_𝑛≤𝐴₀𝑧_𝑛𝑒^𝑏𝑧𝑛3()≤exp1+𝜀2𝛿𝑏𝑧,𝜃₀𝑟_𝑛 (45) for 𝑛 large enough. From (4), we have 𝐴₀𝑒^𝑏𝑧≤𝑓^′′𝑓+𝐴₁𝑒^𝑎𝑧𝑓^′𝑓+𝐻𝑓. (46) Substituting  (24), (28), (29), (44), and (45) into (46), we obtain exp1−𝜀2𝛿𝑏𝑧,𝜃₀𝑟_𝑛≤exp2𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻²(1+𝑂(1))+exp1−𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛×exp𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻¹(1+𝑂(1))+𝜀₀. (47) Since 𝛼₀+𝜀₁<1, we see that (47) is contradictory as 𝑛→∞.
Subcase  2.2. Assume that 𝛿(𝑏𝑧,𝜃₀)<0. From the continuity of 𝛿(𝑏𝑧,𝜃) and (9), we deduce that 3()exp1+𝜀2𝛿𝑏𝑧,𝜃₀𝑟_𝑛≤𝐴₀𝑧_𝑛𝑒^𝑏𝑧𝑛≤exp1−𝜀2𝛿𝑏𝑧,𝜃₀𝑟_𝑛 (48) for 𝑛 large enough. From (4), we have 𝑓^′′𝑓≤𝐴₀𝑒^𝑏𝑧+𝐴₁𝑒^𝑎𝑧𝑓^′𝑓+𝐻𝑓. (49) Substituting  (24), (28), (29), (44), and (48) into (49), we obtain 𝜈_𝑓(𝑟_𝑛)𝑟_𝑛²(1+𝑂(1))≤exp1−𝜀2𝛿𝑏𝑧,𝜃₀𝑟_𝑛+exp1−𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛×exp𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻¹(1+𝑂(1))+𝜀₀ (50) as 𝑛→∞. Since 𝛼₀+𝜀₁<1, this implies that 𝜈_𝑓(𝑟)→0, 𝑛→∞, which is impossible.
Subcase  2.3. Assume that 𝛿(𝑏𝑧,𝜃₀)=0. Arguing similarly as in Subcase  1.3, we may again construct another sequence of points {𝑧^∗_𝑛=𝑟_𝑛𝑒^{𝑖𝜃∗𝑛}} with lim_𝑛→∞𝜃^∗_𝑛=𝜃^∗₀, such that 𝛿(𝑎𝑧,𝜃^∗₀)<0<𝛿(𝑏𝑧,𝜃^∗₀). Replace 𝛿(𝑎𝑧,𝜃₀) with 𝛿(𝑎𝑧,𝜃^∗₀) in (44) and 𝛿(𝑏𝑧,𝜃₀) with 𝛿(𝑏𝑧,𝜃^∗₀) in (45), respectively. We obtain (44) and (45) for the sequence of {𝑧^∗_𝑛=𝑟_𝑛𝑒^{𝑖𝜃∗𝑛}}. Similarly as (47), we get a contradiction as 𝑛→∞.
Case  3. In this final case, we suppose that 𝛿(𝑎𝑧,𝜃₀)=0. We discuss three subcases according to 𝛿(𝑏𝑧,𝜃₀) as follows.
Subcase  3.1. Suppose that 𝛿(𝑏𝑧,𝜃₀)>0. By an argument similar to that in Subcase  1.3, we can choose another sequence of points {𝑧^∗_𝑛=𝑟_𝑛𝑒^{𝑖𝜃∗𝑛}} with lim_𝑛→∞𝜃^∗_𝑛=𝜃^∗₀, and 𝑙₀/2≤𝜃^∗₀−𝜃₀≤𝑙₀, such that 𝛿(𝑎𝑧,𝜃^∗₀)<0<𝛿(𝑏𝑧,𝜃^∗₀). Similarly as in Subcase  2.3, a contradiction follows as 𝑛→∞.
Subcase  3.2. Suppose that 𝛿(𝑏𝑧,𝜃₀)<0. By an argument similar to the Subcase  3.2 of the proof of Theorem 1.1 in [7], we can choose another sequence of points {𝑧^∗_𝑛=𝑟_𝑛𝑒^{𝑖𝜃∗𝑛}}, with lim_𝑛→∞𝜃^∗_𝑛=𝜃^∗₀, such that 𝛿(𝑏𝑧,𝜃^∗₀)<0<𝛿(𝑎𝑧,𝜃^∗₀). From (4), for {𝑧^∗_𝑛=𝑟_𝑛𝑒^{𝑖𝜃∗𝑛}}, we get 𝑓^′𝑧^∗_𝑛𝑓𝑧^∗_𝑛≤1𝐴₁𝑧^∗_𝑛𝑒^{𝑎𝑧∗𝑛}×𝐴₀𝑧^∗_𝑛𝑒^{𝑏𝑧∗𝑛}+𝑓^′′𝑧^∗_𝑛𝑓𝑧^∗_𝑛+𝐻𝑧^∗_𝑛𝑓𝑧^∗_𝑛. (51) Replace 𝛿(𝑎𝑧,𝜃₀) with 𝛿(𝑎𝑧,𝜃^∗₀) in (32) and 𝛿(𝑏𝑧,𝜃₀) with 𝛿(𝑏𝑧,𝜃^∗₀) in (48), respectively. We obtain (32) and (48) for the sequence of {𝑧^∗_𝑛=𝑟_𝑛𝑒^{𝑖𝜃∗𝑛}}. Substituting them into (51), this implies that 𝜈_𝑓(𝑟)→0, 𝑛→∞, which is impossible.
Subcase  3.3. Finally, suppose that 𝛿(𝑏𝑧,𝜃₀)=0. We now have 𝑎/𝑏=𝑐∈ℝ, 𝑐≠0,1, and so 𝑎𝑧=𝑐𝑏𝑧, (𝑏−𝑎)𝑧=(1−𝑐)𝑏𝑧.
If 𝑐<0, we may choose another sequence such that 𝛿(𝑏𝑧,𝜃)<0<𝛿(𝑎𝑧,𝜃). By an argument similar to that in Subcase 3.2, we can get 𝜈_𝑓(𝑟)→0, 𝑛→∞, a contradiction.
If 0<𝑐<1, we similarly obtain 𝛿((𝑏−𝑎)𝑧,𝜃)>0 and 𝛿(𝑎𝑧,𝜃)>0 for another sequence. By an argument similar to that in Subcase  1.3, a contradiction follows.
Finally, if 𝑐>1, we obtain 𝛿((𝑏−𝑎)𝑧,𝜃)<0<𝛿(𝑎𝑧,𝜃) for another sequence. Similarly as in Subcase  1.2, a contradiction again follows.
Thus, we complete the proof when 0<𝛼₀<1. When 𝛼₀=0, we have (30). Similarly as the case when 0<𝛼₀<1, it results in contradiction. Hence, we get 𝜎₂(𝑓)=𝛼₀=1.

Proof of Theorem 3. Suppose that 𝑓 is a nontrivial solution of (5), and then 𝑓 is an entire function. Since 𝜚=max{𝜎(𝐷₀),𝜎(𝐷₁)}<1, we have 𝐷_𝑗(𝑧)≤exp𝑟^𝜚+𝜀𝑗=0,1 (52) for any such that 0<3𝜀<1−𝜚. Similarly as in the proof of Theorem 1, we may choose a sequence of points {𝑧_𝑛=𝑟_𝑛𝑒^𝑖𝜃𝑛} that satisfy |𝑓(𝑧_𝑛)|=𝑀(𝑟_𝑛,𝑓), with lim_𝑛→∞𝜃_𝑛=𝜃₀, 𝑟_𝑛∈𝐸⧵(𝐸₁∪𝐸₂∪𝐸₃), 𝑟_𝑛→∞.
Step  1. We will prove that 𝜎₂(𝑓)≤1. Since 𝜎(𝐴₀;𝐴₁;𝐻)<1, set 𝜎(𝐻)=𝜆<1. Then for any given 𝜀 satisfying 𝜀<min{1−𝜆,(1−𝜚)/3}, when 𝑟 is sufficiently large, we have 𝐴₁𝑒^𝑎𝑧+𝐷₁(𝑧)≤exp𝑟^1+𝜀,𝐴₀𝑒^𝑏𝑧+𝐷₀(𝑧)≤exp𝑟^1+𝜀,|𝐻(𝑧)|≤exp𝑟^𝜆+𝜀. (53)(54) From the Wiman-Valiron theory, we have (24). By Theorem C, we know that 𝜎(𝑓)=∞. So we have (28). From (5) we have 𝑓^′′𝑓+𝐴₁𝑒^𝑎𝑧+𝐷₁𝑓^′𝑓+𝐴₀𝑒^𝑏𝑧+𝐷₀=𝐻𝑓. (55) Substituting  (24), (28), and (53) into (55), we obtain (26) and (27); thus, we have 𝜎₂(𝑓)≤1.
Step  2. Set 𝜎₂(𝑓)=𝛼₀, and we assert that 𝛼₀=1. Now we assume that 𝛼₀<1, and prove that 𝜎₂(𝑓)=𝛼₀<1 results in contradiction. By Lemma 5, we have (29) and (30). Next we only prove the case 0<𝜎₂(𝑓)=𝛼₀≤1 by using (29). The case 𝜎₂(𝑓)=𝛼₀=0 also can be proved by the same method, a little different is that we use (30) instead of (29). Since 𝑎=𝑏𝑐,𝑐(<0) is a real number, there are three cases to be discussed, according to the signs of 𝛿(𝑎𝑧,𝜃₀) and 𝛿(𝑏𝑧,𝜃₀).
Case  1. First assume that 𝛿(𝑏𝑧,𝜃₀)<0<𝛿(𝑎𝑧,𝜃₀), so we have (32) and (48). Combining (52), (32), and (48), we deduce 𝐴₀𝑧_𝑛𝑒^𝑏𝑧𝑛+𝐷₀𝑧_𝑛≤exp𝑟^𝜚+2𝜀,𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛+𝐷₁𝑧_𝑛≥exp1−2𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛 (56)(57)
provided that 𝑛 is large enough. From (5), we have 𝑓^′𝑧_𝑛𝑓𝑧_𝑛≤1𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛+𝐷₁𝑧_𝑛×𝑓^′′𝑧_𝑛𝑓𝑧_𝑛+𝐴₀𝑧_𝑛𝑒^𝑏𝑧𝑛+𝐷₀𝑧_𝑛+𝐻𝑧_𝑛𝑓𝑧_𝑛. (58) Substituting  (24), (28), (56), and (57) into (58), we obtain 𝜈_𝑓𝑟_𝑛𝑟_𝑛(1+𝑂(1))≤exp−1−2𝜀2𝛿𝑎𝑧,𝜃₀𝑟_𝑛×exp2𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻²((1𝑟1+𝑂))+exp_𝑛^𝜚+2𝜀+𝜀₀ (59) for 𝑛 large enough. Since 𝛼₀+𝜀₁<1, this implies that 𝜈_𝑓(𝑟_𝑛)→0, 𝑛→∞, which is impossible.
Case  2. Next, assume that 𝛿(𝑎𝑧,𝜃₀)<0<𝛿(𝑏𝑧,𝜃₀), so we have (44) and (45). Combining (52), (44), and (45), we deduce 𝐴₀𝑧_𝑛𝑒^𝑏𝑧𝑛+𝐷₀𝑧_𝑛≥exp1−2𝜀2𝛿𝑏𝑧,𝜃₀𝑟_𝑛,𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛+𝐷₁𝑧_𝑛≤exp𝑟_𝑛^𝜚+2𝜀 (60)(61)
for 𝑛 is large enough. From (5), we have 𝐴₀𝑧_𝑛𝑒^𝑏𝑧𝑛+𝐷₀𝑧_𝑛≤𝑓^′′𝑧_𝑛𝑓𝑧_𝑛+𝐴₁𝑧_𝑛𝑒^𝑎𝑧𝑛+𝐷₁𝑧_𝑛×𝑓^′𝑧_𝑛𝑓𝑧_𝑛+𝐻𝑧_𝑛𝑓𝑧_𝑛. (62) Substituting  (24), (28), (60), and (61) into (62), we obtain exp1−2𝜀2𝛿𝑏𝑧,𝜃₀𝑟_𝑛≤exp2𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻²((1𝑟1+𝑂))+exp_𝑛^𝜚+2𝜀exp𝑟^𝛼0+𝜀1_𝑛𝑟_𝑛⁻¹(1+𝑂(1))+𝜀₀ (63) for 𝑛 large enough. Since 𝛼₀+𝜀₁<1, 𝜚+2𝜀<1, this leads to a contradiction.
Case  3. Finally, we have to assume that 𝛿(𝑎𝑧,𝜃₀)=𝛿(𝑏𝑧,𝜃₀)=0. Similarly as in Subcase  1.3 of the proof of Theorem 1, we may again construct a sequence of points {𝑧^∗_𝑛=𝑟_𝑛𝑒^{𝑖𝜃∗𝑛}}, with lim_𝑛→∞𝜃^∗_𝑛=𝜃^∗₀, such that 𝛿(𝑎𝑧,𝜃^∗₀)<0. Indeed, without loss of generality, 𝛿(𝑎𝑧,𝜃)>0,𝜃∈𝜃₀+2𝑘𝜋,𝜃₀+()𝜋,𝛿()𝜃2𝑘+1𝑎𝑧,𝜃<0,𝜃∈₀+(2𝑘−1)𝜋,𝜃₀+2𝑘𝜋 (64)
for all 𝑘∈ℤ. Provided that 𝑛 is large enough, we have |𝜃_𝑛−𝜃₀|≤𝑙₀. Choosing now 𝜃^∗₀ such that 𝑙₀/2≤𝜃_𝑛−𝜃^∗_𝑛≤𝑙₀, then 𝑙₀/2≤𝜃₀−𝜃^∗₀≤𝑙₀, thus, 𝜃₀−𝑙₀≤𝜃^∗₀≤𝜃₀−𝑙₀/2, and 𝛿(𝑎𝑧,𝜃^∗₀)<0. Since now 𝛿(𝑏𝑧,𝜃^∗₀)>0, a contradiction follows as in Case  2 above.

Acknowledgments

The author wishes to express his thanks to the referee for his/her valuable suggestions and comments. The present investigation was supported by the National Natural Science Foundation under Grant no. 11226088 and the Key Project of Natural Science Foundation of Educational Committee of Henan Province under Grant no. 12A110002 of the People's Republic of China.