On Subnormal Solutions of Periodic Differential Equations

Abstract

We investigate the existence and the form of subnormal solutions of higher-order linear periodic differential equations, and precisely estimate the growth of all solutions.

1. Introduction and Results

In this paper we use standard notations from the value distribution theory (see [1–3]). In addition, we denote the order of growth of 𝑓(𝑧) by 𝜎(𝑓) and also use the notation 𝜎₂(𝑓) to denote the hyperorder of 𝑓(𝑧), which is defined as 𝜎₂𝑓=lim_𝑟→∞loglog𝑇𝑟,𝑓.log𝑟 (1.1)

Consider the second-order homogeneous linear periodic differential equation 𝑓^′′+𝑃𝑒^𝑧𝑓^′+𝑄𝑒^𝑧𝑓=0, (1.2) where 𝑃(𝑧) and 𝑄(𝑧) are polynomials in 𝑧, but both are not constants. It is well known that every solution 𝑓 of (1.2) is an entire function.

Suppose 𝑓(≢0) is a solution of (1.2), and if 𝑓 satisfies the condition lim_𝑟→∞log𝑇𝑟,𝑓𝑟=0, (1.3) then we say that 𝑓 is a nontrivial subnormal solution of (1.2). For convenience, we also say that 𝑓≡0 is a subnormal solution of (1.2) (see [4, 5]).

It clearly follows that (i)if 0<lim_𝑟→∞log𝑇𝑟,𝑓𝑟<∞, (1.4) then 𝜎₂(𝑓)=1; (ii) if 𝜎₂(𝑓)<1, then (1.3) holds.

Wittich [5] investigated the subnormal solution of (1.2), which gives the form of all subnormal solutions of (1.2) in the following theorem.

Theorem A (see [5]). If 𝑓≢0 is a subnormal solution of (1.2), then 𝑓 must have the form 𝑓(𝑧)=𝑒^𝑐𝑧ℎ₀+ℎ₁𝑒^𝑧+⋯+ℎ_𝑚𝑒^𝑚𝑧, (1.5) where 𝑚≥0 is an integer and 𝑐,ℎ₀,…,ℎ_𝑚 are constants with ℎ₀≠0 and ℎ_𝑚≠0.

Gundersen and Steinbart [4] refined Theorem A and got the following theorem.

Theorem B (see [4]). Under the assumption of Theorem A, the following statements hold.
(i) If deg𝑃>𝑑𝑒𝑔𝑄 and 𝑄≢0, then any subnormal solution 𝑓(≢0) of (1.2) must have the form 𝑓(𝑧)=^𝑚_𝑘=0ℎ_𝑘𝑒^−𝑘𝑧, (1.6) where 𝑚≥1 is an integer and ℎ₀,ℎ₁,…,ℎ_𝑚 are constants with ℎ₀≠0 and ℎ_𝑚≠0.(ii)If 𝑄≡0 and deg𝑃≥1, then any subnormal solution of (1.2) must be a constant. (iii) If deg𝑃<𝑑𝑒𝑔𝑄, then the only subnormal solution of (1.2) is 𝑓≡0.

In [6], Chen and Shon proved that supposing that 𝑃𝑒^𝑧=𝑎_𝑛(𝑧)𝑒^𝑛𝑧+⋯+𝑎₁(𝑧)𝑒^𝑧,𝑄𝑒^𝑧=𝑏_𝑠(𝑧)𝑒^𝑠𝑧+⋯+𝑏₁(𝑧)𝑒^𝑧, (1.7) where 𝑎_𝑛(𝑧),…,𝑎₁(𝑧),𝑏_𝑠(𝑧),…,𝑏₁(𝑧) are polynomials satisfying 𝑎_𝑛(𝑧)𝑏_𝑠(𝑧)≢0, if 𝑛≠𝑠, then every solution 𝑓(≢0) of (1.2) satisfies 𝜎₂(𝑓)=1.

In [6], the condition “all constant terms of 𝑃(𝑒^𝑧) and 𝑄(𝑒^𝑧) are equal to zero” plays an important role in the growth of solutions of (1.2). This makes us consider that the condition may be applied to higher-order differential equations.

Gundersen and Steinbart [4] consider a subnormal solution of higher-order linear nonhomogeneous differential equation 𝑓^(𝑘)+𝑃_𝑘−1𝑒^𝑧𝑓^(𝑘−1)+⋯+𝑃₀𝑒^𝑧𝑓=𝑄₁𝑒^𝑧+𝑄₂𝑒^−𝑧, (1.8) where 𝑄_𝑑(𝑧)(𝑑=1,2), 𝑃_𝑗(𝑧)(𝑗=0,…,𝑘−1) are polynomials in 𝑧 and obtain the following theorem.

Theorem C. Suppose that, in (1.8), one has 𝑘≥2 and deg𝑃₀>𝑑𝑒𝑔𝑃_𝑗 (1.9) for all 1≤𝑗≤𝑘−1. Then any subnormal solution 𝑓 of (1.8) must have the form 𝑓(𝑧)=𝑆₁𝑒^𝑧+𝑆₂𝑒^−𝑧, (1.10) where 𝑆₁(𝑧) and 𝑆₂(𝑧) are polynomials in 𝑧.

From the proof of Theorem C, we see that the condition (1.9) of Theorem C guarantees that the corresponding homogeneous differential equation of (1.8) 𝑓^(𝑘)+𝑃_𝑘−1𝑒^𝑧𝑓^(𝑘−1)+⋯+𝑃₀𝑒^𝑧𝑓=0 (1.11) has no nontrivial subnormal solution.

Thus, a natural question is whether or not (1.11) has a nontrivial subnormal solution if the condition (1.9) is replaced by the condition “there exists some 𝑠 satisfying deg𝑃_𝑠>𝑑𝑒𝑔𝑃_𝑗(𝑗≠𝑠)”.

Examples 1.1 and 1.2 show that if deg𝑃_𝑠>𝑑𝑒𝑔𝑃_𝑗(𝑗≠𝑠), (1.11) may have a nontrivial subnormal solution.

Example 1.1. The equation 𝑓^′′′−14𝑒^3𝑧+2𝑒^2𝑧𝑓^′′+12𝑒^𝑧𝑓^′+𝑒^𝑧+8𝑓=0 (1.12) has a subnormal solution 𝑓=𝑒^−2𝑧+1.

Example 1.2. The equation 𝑓^′′′+𝑒^3𝑧+4𝑓^′′+𝑒^4𝑧+𝑒^𝑧𝑓^′+3𝑒^𝑧−9𝑓=0 (1.13) has a subnormal solution 𝑓=𝑒^−3𝑧+1.

Thus, a natural question is what conditions will guarantee that (1.11) has no nontrivial subnormal solution under the condition deg𝑃_𝑠>𝑑𝑒𝑔𝑃_𝑗(𝑗≠𝑠).

In Theorem 1.3, we answer this question. We conclude that if all constant terms of 𝑃_𝑗 are equal to zero under the conditions deg𝑃_𝑠>𝑑𝑒𝑔𝑃_𝑗(𝑗≠𝑠) and 𝑃₀≢0, then (1.11) has no nontrivial subnormal solution, and we also prove that all solutions of (1.11) satisfy 𝜎₂(𝑓)=1.

Examples 1.1 and 1.2 show that the condition “all constant terms of 𝑃_𝑗 are equal to zero” cannot be deleted in Theorem 1.3.

In this paper, we firstly investigate the existence of subnormal solutions. It is an important problem in theory of periodic differential equations.

Theorem 1.4 generalizes the result of Theorem C, shows that (1.8) has at most one nontrivial subnormal solution, gives the form of subnormal solution of (1.8), and proves that all other solutions 𝑓 of (1.8) satisfy 𝜎₂(𝑓)=1.

Theorem 1.6 refines Theorem C.

Our method for obtaining the proof is totally different from the method applied in [4, 5].

Theorem 1.3. Let 𝑃_𝑗(𝑧)(𝑗=0,…,𝑘−1) be polynomials in 𝑧 such that all constant terms of 𝑃_𝑗 are equal to zero and deg𝑃_𝑗=𝑚_𝑗, that is, 𝑃_𝑗𝑒^𝑧=𝑎_𝑗𝑚𝑗𝑒^𝑚𝑗𝑧+𝑎_{𝑗(𝑚𝑗−1)}𝑒^{(𝑚𝑗−1)𝑧}+⋯+𝑎_𝑗1𝑒^𝑧, (1.14) where 𝑎_𝑗𝑚𝑗,𝑎_{𝑗(𝑚𝑗−1)},…,𝑎_𝑗1 are constants and 𝑎_𝑗𝑚𝑗≠0;𝑚_𝑗≥1 are integers. Suppose that there exists 𝑚_𝑠(𝑠∈{0,…,𝑘−1}) satisfying 𝑚_𝑠>𝑚𝑎𝑥𝑚_𝑗∶𝑗=0,…,𝑠−1,𝑠+1,…,𝑘−1=𝑚. (1.15) Then, one has the following properties.(i) If 𝑃₀≢0, then (1.11) has no nontrivial subnormal solution and every solution of (1.11) is of hyper order 𝜎₂(𝑓)=1.(ii) If 𝑃₀≡⋯≡𝑃_𝑑−1≡0 and 𝑃_𝑑≢0(𝑑<𝑠), then any polynomials with degree ≤𝑑−1 are subnormal solutions of (1.11) and all other solutions 𝑓 of (1.11) satisfy 𝜎₂(𝑓)=1.

Considering proof of theorems, if the set 𝑒^𝑧=𝜁, then (1.8) (or (1.11)) becomes an equation with rational coefficients, but the equation with rational coefficients may have nonmeromorphic solution. For example, the equation 𝑧𝑓^′′+𝑧𝑓^′−2𝑓=0 (1.16) has a solution 𝑓=exp{1/𝑧}. This shows that we cannot use the transformation 𝑒^𝑧=𝜁 to prove that every solution of (1.11) is of 𝜎₂(𝑓)=1.

Theorem 1.4. Let 𝑃_𝑗(𝑒^𝑧)(𝑗=0,…,𝑘−1) satisfy (1.14) and (1.15). Let 𝑄₁(𝑧) and 𝑄₂(𝑧) be polynomials in 𝑧. If 𝑃₀≢0, then(i) (1.8) possesses at most one nontrivial subnormal solution 𝑓₀, and 𝑓₀ is of the form (1.10), where 𝑆₁(𝑧) and 𝑆₂(𝑧) are polynomials in 𝑧;(ii) all other solutions 𝑓 of (1.8) satisfy 𝜎₂(𝑓)=1 except the possible subnormal solution in (i).

Example 1.5 shows the existence of subnormal solution in Theorem 1.4.

Example 1.5. The equation 𝑓^′′′−2𝑒^2𝑧𝑓^′′−𝑒^𝑧𝑓^′+𝑒^𝑧𝑓=−2𝑒^3𝑧−𝑒^−𝑧+2 (1.17) has a subnormal solution 𝑓=𝑒^𝑧+𝑒^−𝑧+1.

Theorem 1.6. Under the assumption of Theorem C, the following statements hold.(i) Equation (1.11) has no nontrivial subnormal solution, and all solutions of (1.11) satisfy 𝜎₂(𝑓)=1.(ii) Equation (1.8) has at most one nontrivial subnormal solution 𝑓₀, and 𝑓₀ is of the form (1.10); all other solutions 𝑓 of (1.8) satisfy 𝜎₂(𝑓)=1.

2. Lemmas for the Proofs of Theorems

Lemma 2.1 (see [7, 8]). Let 𝑓_𝑗(𝑧)(𝑗=1,…,𝑛)(𝑛≥2) be meromorphic functions, and let 𝑔_𝑗(𝑧)(𝑗=1,…,𝑛) be entire functions and satisfy
(i)∑^𝑛_𝑗=1𝑓_𝑗(𝑧)𝑒^{𝑔𝑗(𝑧)}≡0;(ii) when 1≤𝑗<𝑘≤𝑛, then 𝑔_𝑗(𝑧)−𝑔_𝑘(𝑧) is not a constant; (iii) when 1≤𝑗≤𝑛, 1≤ℎ<𝑘≤𝑛, then 𝑇𝑟,𝑓_𝑗=𝑜𝑇𝑟,𝑒^{𝑔ℎ−𝑔𝑘}(𝑟→∞,𝑟∉𝐸), (2.1) where 𝐸⊂(1,∞) is of finite linear measure or logarithmic measure.
Then 𝑓_𝑗(𝑧)≡0(𝑗=1,…,𝑛).

Lemma 2.2. Let 𝑃_𝑗, 𝑚_𝑗, 𝑚_𝑠, and 𝑚 satisfy the hypotheses of Theorem 1.3.(i)If 𝑃₀≢0, then (1.11) has no nonzero polynomial solution.(ii) If 𝑃₀≡⋯≡𝑃_𝑑−1≡0 and 𝑃_𝑑≢0(𝑑<𝑠), then all polynomials with degree ≤𝑑−1 are solutions of (1.11), and any polynomial with degree ≥𝑑 is not solution of (1.11).

Proof. (i) Firstly, by 𝑃₀≢0, we see that all nonzero constants cannot be a solution of (1.11). Now suppose that 𝑓₀=𝑏_𝑛𝑧^𝑛+⋯+𝑏₁𝑧+𝑏₀(𝑛≥1,𝑏_𝑛,…,𝑏₀ are constants, 𝑏_𝑛≠0) is a solution of (1.11). If 𝑛≥𝑠, then 𝑓₀^(𝑠)≢0. Substituting 𝑓₀ into (1.11) and taking 𝑧=𝑟, we conclude that 𝑎_𝑠𝑚𝑠𝑒^𝑚𝑠𝑟𝑏_𝑛𝑛(𝑛−1)⋯(𝑛−𝑠+1)𝑟^𝑛−𝑠((1≤1−𝑜))−𝑃_𝑠𝑒^𝑧𝑓₀^(𝑠)(𝑧≤𝑓_0(𝑘))(𝑧)+𝑃_𝑘−1𝑒^𝑧𝑓₀^(𝑘−1)(𝑧)+⋯+𝑃_𝑠+1𝑒^𝑧𝑓₀^(𝑠+1)(𝑧+𝑃_𝑠−1)𝑒^𝑧𝑓₀^(𝑠−1)(𝑧)+⋯+𝑃₀𝑒^𝑧𝑓₀(𝑧)≤𝑀𝑟^𝑛𝑒^𝑚𝑟(1+𝑜(1)), (2.2) where 𝑀(>0) is some constant. Since 𝑚_𝑠>𝑚, we see that (2.2) is a contradiction. If 𝑛<𝑠, then 𝑃_𝑛𝑒^𝑧𝑓₀^(𝑛)(𝑧)+⋯+𝑃₀𝑒^𝑧𝑓₀(𝑧)=0. (2.3) Set max{deg𝑃_𝑗∶𝑗=0,…,𝑛}=ℎ. If deg𝑃_𝑗=𝑚_𝑗<ℎ, then we can rewrite 𝑃_𝑗𝑒^𝑧=𝑎_𝑗ℎ𝑒^ℎ𝑧+⋯+𝑎_{𝑗(𝑚𝑗+1)}𝑒^{(𝑚𝑗+1)𝑧}+𝑎_𝑗𝑚𝑗𝑒^𝑚𝑗𝑧+⋯+𝑎_𝑗1𝑒^𝑧𝑗=0,…,𝑛, (2.4) where 𝑎_𝑗ℎ=⋯=𝑎_{𝑗(𝑚𝑗+1)}=0. Thus, we conclude by (2.3) and (2.4) that 𝑎_𝑛ℎ𝑓₀^(𝑛)+𝑎_{(𝑛−1)ℎ}𝑓₀^(𝑛−1)+⋯+𝑎_0ℎ𝑓₀𝑒^ℎ𝑧+𝑎+⋯_𝑛𝑗𝑓₀^(𝑛)+𝑎_{(𝑛−1)𝑗}𝑓₀^(𝑛−1)+⋯+𝑎_0𝑗𝑓₀𝑒^𝑗𝑧+𝑎+⋯_𝑛1𝑓₀^(𝑛)+𝑎_(𝑛−1)1𝑓₀^(𝑛−1)+⋯+𝑎₀₁𝑓₀𝑒^𝑧=0. (2.5) Set 𝑄_𝑗(𝑧)=𝑎_𝑛𝑗𝑓₀^(𝑛)+𝑎_{(𝑛−1)𝑗}𝑓₀^(𝑛−1)+⋯+𝑎_0𝑗𝑓₀𝑗=1,…,ℎ. (2.6) Since 𝑓₀ is the polynomial, we see that 𝑚𝑟,𝑄_𝑗=𝑜𝑚𝑟,𝑒^{(𝛼−𝛽)𝑧}1≤𝛽<𝛼≤ℎ. (2.7) By Lemma 2.1, (2.5)–(2.7), we conclude that 𝑄₁(𝑧)≡𝑄₂(𝑧)≡⋯≡𝑄_ℎ(𝑧)≡0. (2.8) Since deg𝑓₀>𝑑𝑒𝑔𝑓^′₀>⋯>deg𝑓₀^(𝑛), by (2.6) and (2.8), we see that 𝑎₀₀=𝑎₀₁=⋯=𝑎_0ℎ=0. (2.9) Since ℎ≥𝑚₀=deg𝑃₀, we have 𝑃₀≡0. This contradicts our assumption that 𝑃₀≢0.
(ii) Since 𝑃₀≡⋯≡𝑃_𝑑−1≡0 and 𝑃_𝑑≢0(𝑑<𝑠), clearly all polynomials with degree ≤𝑑−1 are solutions of (1.11). By 𝑃_𝑑≢0 and (i), we see that 𝑓^(𝑑) cannot be a nonzero polynomial, and hence 𝑓 cannot be a polynomial with degree (≥𝑑).

Lemma 2.3 (see [9]). Let 𝑓 be a transcendental meromorphic function with 𝜎(𝑓)=𝜎<∞, and let 𝐻={(𝑘₁,𝑗₁),(𝑘₂,𝑗₂),…,(𝑘_𝑞,𝑗_𝑞)} be a finite set of distinct pairs of integers that satisfy 𝑘_𝑖>𝑗_𝑖≥0, for 𝑖=1,…,𝑞, and let 𝜀>0 be a given constant. Then, there exists a set 𝐸⊂[−𝜋/2,3𝜋/2) that has linear measure zero, such that if 𝜓∈[−𝜋/2,3𝜋/2)⧵𝐸, then there is a constant 𝑅₀=𝑅₀(𝜓)>1 such that for all 𝑧 satisfying arg𝑧=𝜓 and |𝑧|≥𝑅₀ and for all (𝑘,𝑗)∈𝐻, one has 𝑓^(𝑘)(𝑧)𝑓^(𝑗)(𝑧)≤|𝑧|^{(𝑘−𝑗)(𝜎−1+𝜀)}. (2.10)

Lemma 2.4 (see [10]). Let 𝑓(𝑧) be an entire function and suppose that |𝑓^(𝑘)(𝑧)| is unbounded on some ray arg𝑧=𝜃. Then there exists an infinite sequence of points 𝑧_𝑛=𝑟_𝑛𝑒^𝑖𝜃(𝑛=1,2,…), where 𝑟_𝑛→∞, such that 𝑓^(𝑘)(𝑧_𝑛)→∞ and 𝑓^(𝑗)𝑧_𝑛𝑓^(𝑘)𝑧_𝑛≤𝑧_𝑛^𝑘−𝑗(1+𝑜(1))𝑗=0,…,𝑘−1. (2.11)

Lemma 2.5 (see [11]). Let 𝑓(𝑧) be an entire function with 𝜎(𝑓)=𝜎<∞. Let there exists a set 𝐸⊂[0,2𝜋) that has linear measure zero, such that for any ray arg𝑧=𝜃₀∈[0,2𝜋)⧵𝐸,|𝑓(𝑟𝑒^𝑖𝜃0)|≤𝑀𝑟^𝑘(𝑀=𝑀(𝜃₀)>0 is a constant, and 𝑘(>0) is a constant independent of 𝜃₀). Then 𝑓(𝑧) is a polynomial with deg𝑓≤𝑘.

Lemma 2.6 can be obtained from [12, Theorem 4] or [2, Theorem 7.3].

Lemma 2.6. Let 𝐴₀,…,𝐴_𝑘−1 be entire functions of finite order. If 𝑓(𝑧)  is a solution of equation 𝑓^(𝑘)+𝐴_𝑘−1𝑓^(𝑘−1)+⋯+𝐴₀𝑓=0, (2.12) then 𝜎₂(𝑓)≤max{𝜎(𝐴_𝑗)∶𝑗=0,…,𝑘−1}.

Lemma 2.7 (see [13]). Let 𝑔(𝑧) be an entire function of infinite order with the hyperorder 𝜎₂(𝑔)=𝜎, and let 𝜈(𝑟) be the central index of 𝑔. Then lim_𝑟→∞loglog𝜈(𝑟)log𝑟=𝜎₂𝑔=𝜎. (2.13)

Lemma 2.8 (see [6]). Let 𝑓(𝑧) be an entire function of infinite order with 𝜎₂(𝑓)=𝛼(0≤𝛼<∞), and let a set 𝐸⊂[1,∞) have finite logarithmic measure. Then there exists {𝑧_𝑘=𝑟_𝑘𝑒^𝑖𝜃𝑘} such that |𝑓(𝑧_𝑘)|=𝑀(𝑟_𝑘,𝑓), 𝜃_𝑘∈[−𝜋/2,3𝜋/2), lim_𝑘→∞𝜃_𝑘=𝜃₀∈[−𝜋/2,3𝜋/2),𝑟_𝑘∉𝐸,𝑟_𝑘→∞ such that(i)if 𝜎₂(𝑓)=𝛼(0<𝛼<∞), then for any given 𝜀₁(0<𝜀₁<𝛼), exp𝑟^𝛼−𝜀1_𝑘<𝜈𝑟_𝑘<exp𝑟^𝛼+𝜀1_𝑘; (2.14)(ii)if 𝜎(𝑓)=∞ and 𝜎₂(𝑓)=0, then for any given 𝜀₂(0<𝜀₂<1/2) and for any large 𝑀(>0), one has as 𝑟_𝑘 sufficiently large 𝑟^𝑀_𝑘<𝜈𝑟_𝑘<exp𝑟^𝜀2_𝑘. (2.15)

Lemma 2.9 (see [9]). Let 𝑓 be a transcendental meromorphic function, and let 𝛼>1 be a given constant. Then there exist a set 𝐸⊂(1,∞) with finite logarithmic measure and a constant 𝐵>0 that depends only on 𝛼 and 𝑖,𝑗(𝑖<𝑗(𝑖,𝑗∈ℕ)), such that for all 𝑧 satisfying |𝑧|=𝑟∉[0,1]∪𝐸, one has 𝑓^(𝑗)(𝑧)𝑓^(𝑖)(𝑧)𝑇≤𝐵𝛼𝑟,𝑓𝑟log^𝛼𝑟log𝑇𝛼𝑟,𝑓^𝑗−𝑖. (2.16)

Remark 2.10. From the proof of Lemma 2.9 (i.e., Theorem 3 in [9]), we can see that exceptional set 𝐸 satisfies that if 𝑎_𝑛 and 𝑏_𝑚(𝑛,𝑚=1,2,…) denote all zeros and poles of 𝑓, respectively, 𝑂(𝑎_𝑛) and 𝑂(𝑏_𝑚) denote sufficiently small neighborhoods of 𝑎_𝑛 and 𝑏_𝑚, respectively, then 𝐸=|𝑧|∶𝑧∈^+∞_𝑛=1𝑂𝑎_𝑛^+∞_𝑚=1𝑂𝑏_𝑚. (2.17) Hence, if 𝑓(𝑧) is a transcendental entire function and 𝑧 is a point such that it satisfies that |𝑓(𝑧)| is sufficiently large, then (2.16) holds.

Lemma 2.11 (see [4]). Consider an 𝑛th-order linear differential equation of the form 𝑃₀𝑒^𝑧,𝑒^−𝑧𝑓^(𝑛)+𝑃₁𝑒^𝑧,𝑒^−𝑧𝑓^(𝑛−1)+⋯+𝑃_𝑛𝑒^𝑧,𝑒^−𝑧𝑓=𝑃_𝑛+1𝑒^𝑧,𝑒^−𝑧, (2.18) where each 𝑃_𝑗(𝑧,𝑤) is a polynomial in 𝑧 and 𝑤 with 𝑃₀(𝑧,𝑤)≢0. Suppose that 𝑓=𝜙(𝑧) is an entire subnormal solution of (2.18), that is, an entire solution of (2.18) that also satisfies (1.3). If 𝜙 is periodic with period 2𝜋𝑖, then 𝜙(𝑧)=𝑆₁𝑒^𝑧+𝑆₂𝑒^−𝑧, (2.19) where 𝑆₁(𝑧) and 𝑆₂(𝑧) are polynomials in 𝑧.

3. Proof of Theorem 1.3

(i) Suppose that 𝑃₀≢0 and 𝑓(≢0) are the solution of (1.11). Then 𝑓 is an entire function. By Lemma 2.2 (i), we see that 𝑓 is transcendental.

Step 1. We prove that 𝜎(𝑓)=∞. Suppose to the contrary that 𝜎(𝑓)=𝜎<∞. By Lemma 2.3, we know that for any given 𝜀>0, there exists a set 𝐸⊂[−𝜋/2,3𝜋/2) of linear measure zero, such that if 𝜓∈[−𝜋/2,3𝜋/2)⧵𝐸, then there is a constant 𝑅₀=𝑅₀(𝜓)>1 such that for all 𝑧 satisfying arg𝑧=𝜓 and |𝑧|=𝑟>𝑅₀, we have 𝑓^(𝑗)(𝑧)𝑓^(𝑠)(𝑧)≤𝑟^{(𝜎−1+𝜀)(𝑗−𝑠)},𝑗=𝑠+1,…,𝑘. (3.1)
Now we take a ray arg𝑧=𝜃∈(−𝜋/2,𝜋/2)⧵𝐸, then cos𝜃>0. We assert that |𝑓^(𝑠)(𝑟𝑒^𝑖𝜃)| is bounded on the ray arg𝑧=𝜃. If |𝑓^(𝑠)(𝑟𝑒^𝑖𝜃)| is unbounded on the ray arg𝑧=𝜃, then by Lemma 2.4, there exists an infinite sequence of points 𝑧_𝑡=𝑟_𝑡𝑒^𝑖𝜃(𝑡=1,2,…) such that, as 𝑟_𝑡→∞, 𝑓^′(𝑧_𝑡)→∞ and 𝑓^(𝑖)𝑧_𝑡𝑓^(𝑠)𝑧_𝑡≤𝑟_𝑡^𝑠−𝑖(1+𝑜(1))(𝑖=0,…,𝑠−1). (3.2) By (1.11), we get that −𝑃_𝑠𝑒^𝑧𝑡=𝑓^(𝑘)𝑧_𝑡𝑓^(𝑠)𝑧_𝑡+𝑃_𝑘−1𝑒^𝑧𝑡𝑓^(𝑘−1)𝑧_𝑡𝑓^(𝑠)𝑧_𝑡+⋯+𝑃_𝑠+1𝑒^𝑧𝑡𝑓^(𝑠+1)𝑧_𝑡𝑓^(𝑠)𝑧_𝑡+𝑃_𝑠−1𝑒^𝑧𝑡𝑓^(𝑠−1)𝑧_𝑡𝑓^(𝑠)𝑧_𝑡+⋯+𝑃₀𝑒^𝑧𝑡𝑓𝑧_𝑡𝑓^(𝑠)𝑧_𝑡. (3.3) Since cos𝜃>0 and (1.14), we have 𝑃_𝑠𝑒^𝑧𝑡=𝑎_𝑠𝑚𝑠𝑒^{𝑚𝑠𝑟𝑡cos𝜃}((1,𝑃1+𝑜))_𝑗𝑒^𝑧𝑡≤𝑀𝑒^{𝑚𝑟𝑡cos𝜃}(1+𝑜(1))𝑗=0,…,𝑠−1,𝑠+1,…,𝑘−1, (3.4) where 𝑀(>0) is some constant. Substituting (3.1), (3.2), and (3.4) into (3.3), we get that 𝑎_𝑠𝑚𝑠𝑒^{𝑚𝑠𝑟𝑡cos𝜃}(1+𝑜(1))≤𝑘𝑀𝑟_𝑡^(𝜎+1)𝑘𝑒^{𝑚𝑟𝑡cos𝜃}(1+𝑜(1)). (3.5) By 𝑚_𝑠>𝑚, we know that when 𝑟_𝑡→∞, (3.5) is a contradiction. So, 𝑓𝑟𝑒^𝑖𝜃≤𝑀₁𝑟^𝑠 (3.6) on the ray arg𝑧=𝜃∈(−𝜋/2,𝜋/2)⧵𝐸, where 𝑀₁(>0) is some constant.
Now we take a ray arg𝑧=𝜃∈(𝜋/2,3𝜋/2)⧵𝐸. Then, cos𝜃<0. If |𝑓^(𝑘)(𝑟𝑒^𝑖𝜃)| is unbounded on the ray arg𝑧=𝜃, then by Lemma 2.4, there exists an infinite sequence of points 𝑧^′_𝑡=𝑟^′_𝑡𝑒^𝑖𝜃(𝑡=1,2,…) such that, as 𝑟^′_𝑡→∞, 𝑓^(𝑘)(𝑧^′_𝑡)→∞ and 𝑓^(𝑖)𝑧^′_𝑡𝑓^(𝑘)𝑧^′_𝑡≤𝑟^′_𝑡^𝑘−𝑖(1+𝑜(1))(𝑖=0,…,𝑘−1). (3.7) By (1.11), we get that |−1|≤𝑃_𝑘−1𝑒^𝑧′𝑡𝑓^(𝑘−1)𝑧^′_𝑡𝑓^(𝑘)𝑧^′_𝑡+⋯+𝑃₀𝑒^𝑧′𝑡𝑓𝑧^′_𝑡𝑓^(𝑘)𝑧^′_𝑡. (3.8) Since cos𝜃<0 and (3.7), for 𝑗=0,…,𝑘−1 as 𝑟^′_𝑡→∞, 𝑃_𝑗𝑒^𝑧′𝑡𝑓^(𝑗)𝑧^′_𝑡𝑓^(𝑘)𝑧^′_𝑡≤𝑎_𝑗𝑚𝑗𝑒^{𝑚𝑗𝑟′𝑡cos𝜃}+⋯+𝑎_𝑗1𝑒^{𝑟′𝑡cos𝜃}𝑟^′_𝑡^𝑘−𝑗(1+𝑜(1))⟶0. (3.9) By (3.8) and (3.9), we get that 1≤0; this is a contradiction. So, 𝑓𝑟𝑒^𝑖𝜃≤𝑀₁𝑟^𝑘 (3.10) on the ray arg𝑧=𝜃∈(𝜋/2,3𝜋/2)⧵𝐸.

By Lemma 2.5, (3.6), and (3.10), we know that 𝑓(𝑧) is a polynomial, which contradicts the above assertion that 𝑓(𝑧) is transcendental. Therefore 𝜎(𝑓)=∞.

Step 2. We prove that (1.11) has no nontrivial subnormal solution. Now suppose that (1.11) has a nontrivial subnormal solution 𝑓₀, and we will deduce a contradiction. By the conclusion in Step 1, 𝑓₀ satisfies (1.3) and and 𝜎(𝑓₀)=∞. By Lemma 2.6, we see that 𝜎₂(𝑓)≤1. Set 𝜎₂(𝑓)=𝛼≤1. By Lemma 2.9, we see that there exist a subset 𝐸₁⊂(1,∞) having finite logarithmic measure and a constant 𝐵>0 such that for all 𝑧 satisfying |𝑧|=𝑟∉[0,1]∪𝐸₁, we have 𝑓₀^(𝑗)(𝑧)𝑓₀(𝑧)≤𝐵𝑇2𝑟,𝑓₀^𝑘+1𝑗=1,…,𝑘. (3.11)
From the Wiman-Valiron theory (see [2, page 51]), there is a set 𝐸₂⊂(1,∞) having finite logarithmic measure, so we can choose 𝑧 satisfying |𝑧|=𝑟∉[0,1]∪𝐸₂ and |𝑓₀(𝑧)|=𝑀(𝑟,𝑓₀). Thus, we get 𝑓₀^(𝑗)(𝑧)𝑓₀(𝑧)=𝜈(𝑟)𝑧^𝑗(1+𝑜(1)),𝑗=1,…,𝑘, (3.12) where 𝜈(𝑟) is the central index of 𝑓₀(𝑧).
By Lemma 2.8, we see that there exists a sequence {𝑧_𝑛=𝑟_𝑛𝑒^𝑖𝜃𝑛} such that |𝑓(𝑧_𝑛)|=𝑀(𝑟_𝑛,𝑓), 𝜃_𝑛∈[−𝜋/2,3𝜋/2), lim𝜃_𝑛=𝜃₀∈[−𝜋/2,3𝜋/2), 𝑟_𝑛∉[0,1]∪𝐸₁∪𝐸₂, 𝑟_𝑛→∞, and if 𝛼>0, then by (2.14), we see that for any given 𝜀₁(0<𝜀₁<𝛼), and for sufficiently large 𝑟_𝑛, exp𝑟^𝛼−𝜀1_𝑛<𝜈𝑟_𝑛<exp𝑟^𝛼+𝜀1_𝑛, (3.13) and if 𝛼=0, then by 𝜎(𝑓)=∞ and (2.15), we see that for any given 𝜀₂(0<𝜀₂<1/2) and for any sufficiently large 𝑀₂>2𝑘+3, as 𝑟_𝑛 is sufficiently large, 𝑟^𝑀2_𝑛<𝜈𝑟_𝑛<exp𝑟^𝜀2_𝑛. (3.14)

Since 𝜃₀ may belong to (−𝜋/2,𝜋/2), (𝜋/2,3𝜋/2), or {−𝜋/2,𝜋/2}, we divide this proof into three cases to prove.

Case 1. Suppose that 𝜃₀∈(−𝜋/2,𝜋/2), then cos𝜃₀>0. If we take 𝛿=(1/4)(𝜋/2−|𝜃₀|), then [𝜃₀−𝛿,𝜃₀+𝛿]⊂(−𝜋/2,𝜋/2). By 𝜃_𝑛→𝜃₀, we see that there is a constant 𝑁(>0) such that, as 𝑛>𝑁, 𝜃_𝑛∈[𝜃₀−𝛿,𝜃₀+𝛿], and 0<cos(|𝜃₀|+𝛿)≤cos𝜃_𝑛. By (3.11), we see that for any given 𝜀₃ satisfying 0<𝜀₃<(1/(4(𝑘+1)))cos(|𝜃₀|+𝛿), 𝑇2𝑟_𝑛,𝑓₀^𝑘+1≤𝑒^{𝜀3(𝑘+1)2𝑟𝑛}≤𝑒^{(1/2)cos(|𝜃0|+𝛿)𝑟𝑛}≤𝑒^{(1/2)cos𝜃𝑛𝑟𝑛} (3.15) holds for 𝑛>𝑁. By (3.11), (3.12), and (3.15), we see that 𝑓₀^{(𝑘−𝑠)}𝑧_𝑛𝑓₀𝑧_𝑛=𝜈𝑟_𝑛𝑟_𝑛^𝑘−𝑠(1+𝑜(1))≤𝐵𝑇2𝑟_𝑛,𝑓₀^𝑘+1≤𝐵𝑒^{(1/2)cos𝜃𝑛𝑟𝑛}. (3.16) By (1.11), we get −𝑓₀^(𝑠)𝑧_𝑛𝑓₀𝑧_𝑛𝑎_𝑠𝑚𝑠𝑒^{𝑚𝑠𝑧𝑛}+⋯+𝑎_𝑠1𝑒^𝑧𝑛=𝑓₀^(𝑘)𝑧_𝑛𝑓₀𝑧_𝑛+𝑃_𝑘−1𝑒^𝑧𝑛𝑓₀^(𝑘−1)𝑧_𝑛𝑓₀𝑧_𝑛+⋯+𝑃_𝑠+1𝑒^𝑧𝑛𝑓₀^(𝑠+1)𝑧_𝑛𝑓₀𝑧_𝑛𝑃_𝑠−1𝑒^𝑧𝑛𝑓₀^(𝑠−1)𝑧_𝑛𝑓₀𝑧_𝑛+⋯+𝑃₁𝑒^𝑧𝑛𝑓^′₀𝑧_𝑛𝑓₀𝑧_𝑛+𝑃₀𝑒^𝑧𝑛. (3.17) Since cos𝜃_𝑛>0 and (1.14), we get that 𝑎_𝑠𝑚𝑠𝑒^{𝑚𝑠𝑧𝑛}+⋯+𝑎_𝑠1𝑒^𝑧𝑛=𝑎_𝑠𝑚𝑠𝑒^{𝑚𝑠𝑟𝑛cos𝜃𝑛}((1,𝑃1+𝑜))_𝑗𝑒^𝑧𝑛≤𝑀₃𝑒^{𝑚𝑟𝑛cos𝜃𝑛}𝑗=0,…,𝑠−1,𝑠+1,…𝑘−1, (3.18) where 𝑀₃(>0) is some constant. Substituting (3.12) and (3.18) into (3.17), we deduce that for sufficiently large 𝑟_𝑛, 𝜈𝑟_𝑛𝑟_𝑛^𝑠𝑎_𝑠𝑚𝑠𝑒^{𝑚𝑠𝑟𝑛cos𝜃𝑛}((1≤𝜈𝑟1+𝑜))_𝑛𝑟_𝑛^𝑘(1+𝑜(1))+𝑀₃𝑒^{𝑚𝑟𝑛cos𝜃𝑛𝑘−1}_{𝑗=0,𝑗≠𝑠}𝜈𝑟_𝑛𝑟_𝑛^𝑗(1+𝑜(1)). (3.19) From (3.13) or (3.14), we have 𝜈𝑟_𝑛>𝑟^𝑀2_𝑛>𝑟_𝑛^2𝑘+3>𝑟_𝑛. (3.20) By (3.16), (3.19), and (3.20), we get 𝜈𝑟_𝑛𝑟_𝑛^𝑠𝑎_𝑠𝑚𝑠𝑒^{𝑚𝑠𝑟𝑛cos𝜃𝑛}(1+𝑜(1))≤𝑘𝑀₃𝑒^{𝑚𝑟𝑛cos𝜃𝑛}𝜈𝑟_𝑛𝑟_𝑛^𝑘((1,𝑎1+𝑜))_𝑠𝑚𝑠𝑒^{(𝑚𝑠−𝑚)𝑟𝑛cos𝜃𝑛}(1+𝑜(1))≤𝑘𝑀₃𝐵𝑒^{(1/2)𝑟𝑛cos𝜃𝑛}. (3.21)(3.22) Since 𝑚_𝑠−𝑚≥1>1/2, we see that (3.22) is a contradiction.

Case 2. Suppose that 𝜃₀∈(𝜋/2,3𝜋/2). By cos𝜃₀<0 and 𝜃_𝑛→𝜃₀, we see that for sufficiently large 𝑛, cos𝜃_𝑛<0. By (1.11), we get −𝑓₀^(𝑘)𝑧_𝑛𝑓₀𝑧_𝑛=𝑃_𝑘−1𝑒^𝑧𝑛𝑓₀^(𝑘−1)𝑧_𝑛𝑓₀𝑧_𝑛+⋯+𝑃₁𝑒^𝑧𝑛𝑓^′₀𝑧_𝑛𝑓₀𝑧_𝑛+𝑃₀𝑒^𝑧𝑛. (3.23) Since cos𝜃_𝑛<0 and (1.14), we get that for 𝑗=0,…,𝑘−1𝑃_𝑗𝑒^𝑧𝑛≤𝑎_𝑗𝑚𝑗𝑒^{𝑚𝑗𝑟𝑛cos𝜃𝑛}+⋯+𝑎_𝑗1𝑒^{𝑟𝑛cos𝜃𝑛}≤𝐶_𝑗, (3.24) where 𝐶_𝑗(𝑗=0,…,𝑘−1) are constants. By (3.12), (3.20), (3.23), and (3.24), we get that 𝜈𝑟_𝑛𝑟_𝑛^𝑘((1=𝑓1+𝑜))^0(𝑘)𝑧_𝑛𝑓₀𝑧_𝑛≤𝐶_𝑘−1+⋯+𝐶₀𝜈𝑟_𝑛𝑟_𝑛^𝑘−1(1+𝑜(1)). (3.25) Thus, we have 𝜈𝑟_𝑛(1+𝑜(1))≤𝐶_𝑘−1+⋯+𝐶₀𝑟_𝑛(1+𝑜(1)). (3.26) By (3.20), we see that (3.26) is also a contradiction.

Case 3. Suppose that 𝜃₀=𝜋/2 or 𝜃₀=−𝜋/2. Since the proof of 𝜃₀=−𝜋/2 is the same as the proof of 𝜃₀=𝜋/2, we only prove the case that 𝜃₀=𝜋/2. Since 𝜃_𝑛→𝜃₀, for any given 𝜀₄(0<𝜀₄<1/10), we see that there is an integer 𝑁(>0), as 𝑛>𝑁, 𝜃_𝑛∈[𝜋/2−𝜀₄,𝜋/2+𝜀₄], and 𝑧_𝑛=𝑟_𝑛𝑒^𝑖𝜃𝑛∈𝜋Ω=𝑧∶2−𝜀₄𝜋≤arg𝑧≤2+𝜀₄. (3.27)
By Lemma 2.9, there exist a subset 𝐸₃⊂(1,∞) having finite logarithmic measure and a constant 𝐵>0 such that for all 𝑧 satisfying |𝑧|=𝑟∉[0,1]∪𝐸₃, we have 𝑓₀^(𝑑)(𝑧)𝑓₀^(𝑠)(𝑧)≤𝐵𝑇2𝑟,𝑓₀^(𝑠)𝑘+1(𝑑=𝑠+1,…,𝑘). (3.28)

Now we consider the growth of 𝑓₀(𝑟𝑒^𝑖𝜃) on a ray arg𝑧=𝜃∈Ω⧵{𝜋/2}. If 𝜃∈[𝜋/2−𝜀₄,𝜋/2), then cos𝜃>0. By (1.3), for any given 𝜀₅ satisfying 0<𝜀₅<(1/(4(𝑘+1)))cos𝜃, 𝑇2𝑟,𝑓₀^(𝑠)𝑘+1≤𝑒^{𝜀5(𝑘+1)2𝑟}≤𝑒^{(1/2)cos𝜃𝑟}. (3.29) If |𝑓₀^(𝑠)(𝑟𝑒^𝑖𝜃)| is unbounded on the ray arg𝑧=𝜃, then by Lemma 2.4, there exists a sequence {𝑦_𝑗=𝑅_𝑗𝑒^𝑖𝜃} such that, as 𝑅_𝑗→∞, 𝑓₀^(𝑠)(𝑦_𝑗)→∞ and 𝑓₀^(𝑖)𝑦_𝑗𝑓₀^(𝑠)𝑦_𝑗≤𝑅_𝑗^𝑠−𝑖(1+𝑜(1))(𝑖=0,…,𝑠−1). (3.30) By Remark 2.10 and 𝑓₀^(𝑠)(𝑦_𝑗)→∞, we know that 𝑦_𝑗 satisfies (3.28). By (3.28) and (3.29), we see that for sufficiently large 𝑗, 𝑓₀^(𝑑)𝑦_𝑗𝑓₀^(𝑠)𝑦_𝑗≤𝐵𝑇2𝑅_𝑗,𝑓₀^(𝑠)𝑘+1≤𝑒^{(1/2)cos𝜃𝑅𝑗}(𝑑=𝑠+1,…,𝑘). (3.31) By (1.11), (3.18), (3.30), and (3.31), we deduce that 𝑎_𝑠𝑚𝑠𝑒^{𝑚𝑠𝑅𝑗cos𝜃}(1+𝑜(1))=−𝑃_𝑠𝑒^𝑦𝑗≤𝑘𝑀₃𝐵𝑒^{(𝑚+(1/2))𝑅𝑗cos𝜃}𝑅^𝑠_𝑗(1+𝑜(1)). (3.32) Since 𝑚_𝑠>𝑚+1/2, we know that when 𝑅_𝑗→∞, (3.32) is a contradiction. Hence 𝑓₀𝑟𝑒^𝑖𝜃≤𝑀𝑟^𝑠 (3.33) on the ray arg𝑧=𝜃∈[𝜋/2−𝜀₄,𝜋/2).

If 𝜃∈(𝜋/2,𝜋/2+𝜀₄], then cos𝜃<0. We assert that |𝑓₀^(𝑘)(𝑟𝑒^𝑖𝜃)| is bounded on the ray arg𝑧=𝜃. If |𝑓₀^(𝑘)(𝑟𝑒^𝑖𝜃)| is unbounded on the ray arg𝑧=𝜃, then by Lemma 2.4, there exists a sequence {𝑦^∗_𝑗=𝑅^∗_𝑗𝑒^𝑖𝜃} such that, as 𝑅^∗_𝑗→∞, 𝑓₀^(𝑘)(𝑦^∗_𝑗)→∞ and 𝑓₀^(𝑖)𝑦^∗_𝑗𝑓₀^(𝑘)𝑦^∗_𝑗≤𝑅^∗_𝑗^𝑘−𝑖(1+𝑜(1))(𝑖=0,…,𝑘−1). (3.34) Since cos𝜃<0, for fixed 𝑡∈{0,1,…,𝑘−1}, we deduce that as 𝑅^∗_𝑗→∞𝑃_𝑡𝑒^𝑦∗𝑗𝑅^∗_𝑗^𝑘≤𝑎_𝑡𝑚𝑡𝑒^{𝑚𝑡𝑅∗𝑗cos𝜃}𝑅^∗_𝑗^𝑘+⋯+𝑎_𝑡1𝑒^{𝑅∗𝑗cos𝜃}𝑅^∗_𝑗^𝑘⟶0. (3.35) By (1.11), (3.34), and (3.35), we deduce that as 𝑅^∗_𝑗→∞1≤𝑃_𝑘−1𝑒^𝑦∗𝑗𝑓₀^(𝑘−1)𝑦^∗_𝑗𝑓₀^(𝑘)𝑦^∗_𝑗+⋯+𝑃₀𝑒^𝑦∗𝑗𝑓₀𝑦^∗_𝑗𝑓₀^(𝑘)𝑦^∗_𝑗≤𝑃_𝑘−1𝑒^𝑦∗𝑗𝑅^∗_𝑗^𝑘(1+𝑜(1))+⋯+𝑃₀𝑒^𝑦∗𝑗𝑅^∗_𝑗^𝑘(1+𝑜(1))⟶0. (3.36) This, (3.36) is a contradiction. Hence 𝑓₀𝑟𝑒^𝑖𝜃≤𝑀𝑟^𝑘 (3.37) on the ray arg𝑧=𝜃∈(𝜋/2,𝜋/2+𝜀₄].

By (3.33) and (3.37), we see that |𝑓₀(𝑟𝑒^𝑖𝜃)| satisfies 𝑓₀𝑟𝑒^𝑖𝜃≤𝑀𝑟^𝑘 (3.38) on the ray arg𝑧=𝜃∈Ω⧵{𝜋/2}.

However, since 𝑓₀(𝑧) is of infinite order and {𝑧_𝑛=𝑟_𝑛𝑒^𝑖𝜃𝑛} satisfies |𝑓₀(𝑧_𝑛)|=𝑀(𝑟_𝑛,𝑓₀), we see that for any large 𝑁(>𝑘), as 𝑛 is sufficiently large 𝑓₀𝑧_𝑛=𝑓₀𝑟_𝑛𝑒^𝑖𝜃𝑛≥exp𝑟^𝑁_𝑛. (3.39) Since 𝑧_𝑛∈Ω, by (3.38) and (3.39), we see that for sufficiently large 𝑛𝜃_𝑛=𝜋2. (3.40) Thus cos𝜃_𝑛=0 and for sufficiently large 𝑛𝑃_𝑗𝑒^𝑧𝑛=𝑎_𝑗𝑚𝑗𝑒^{𝑚𝑗𝑧𝑛}+⋯+𝑎_𝑗1𝑒^𝑧𝑛≤𝐶𝑗=0,…,𝑘−1, (3.41) where 𝐶(>0) is some constant. By (1.11) and (3.12), we get that −𝜈𝑟_𝑛𝑧_𝑛^𝑘(1+𝑜(1))=𝑃_𝑘−1𝑒^𝑧𝑛𝜈𝑟_𝑛𝑧_𝑛^𝑘−1(1+𝑜(1))+⋯+𝑃₁𝑒^𝑧𝑛𝜈𝑟_𝑛𝑧_𝑛(1+𝑜(1))+𝑃₀𝑒^𝑧𝑛. (3.42) By (3.20), (3.41) and (3.42), we get that 𝜈𝑟_𝑛≤𝑘𝐶𝑟^𝑘_𝑛. (3.43) By (3.13) (or (3.14)), we see that (3.43) is a contradiction. Hence (1.11) has no nontrivial subnormal solution.

Step 3. We prove that all solutions of (1.11) satisfy 𝜎₂(𝑓)=1. If there is a solution 𝑓₁ satisfying 𝜎₂(𝑓₁)<1, then 𝑓₁ satisfies (1.3), that is, 𝑓₁ is subnormal, but this contradicts the conclusion in Step 2. Hence every solution 𝑓 satisfies 𝜎₂(𝑓)≥1. By this and 𝜎₂(𝑓)≤1, we get 𝜎₂(𝑓)=1. Theorem 1.3(i) is thus proved.

(ii) Since 𝑃₀≡⋯≡𝑃_𝑑−1≡0 and 𝑃_𝑑≢0, we clearly see that all polynomials with degree ≤𝑑−1 are subnormal solutions of (1.11). By (i), we see that every 𝑓^(𝑑) satisfies 𝜎₂(𝑓^(𝑑))=1 or 𝑓^(𝑑)≡0. Hence 𝜎₂(𝑓)=1 or 𝑓 is a polynomial with degree ≤𝑑−1.

4. Proof of Theorem 1.4

Suppose that 𝑓₁ and 𝑓₂(≢𝑓₁) are nontrivial subnormal solutions of (1.8), then 𝑓₁−𝑓₂(≢0) is a subnormal solution of the corresponding homogeneous equation (1.11) of (1.8). This contradicts the assertion of Theorem 1.3(i). Hence (1.8) possesses at most one nontrivial subnormal solution.

Now suppose that 𝑓₀ is a nontrivial subnormal solution of (1.8), then 𝑓₀(𝑧+2𝜋𝑖) is also nontrivial subnormal solution, so, 𝑓₀(𝑧)=𝑓₀(𝑧+2𝜋𝑖) by the above assertion. Thus, by Lemma 2.11, we see that 𝑓₀ satisfies (1.10).

By Theorem 1.3(i), we see that all solutions of the corresponding homogeneous equation (1.11) of (1.8) are of 𝜎₂(𝑓)=1. By variation of parameters, we see that all solutions of (1.8) satisfy 𝜎₂(𝑓)≤1. If 𝜎₂(𝑓)<1, then 𝑓 clearly satisfies (1.3); that is, 𝑓 is subnormal. Hence all other solutions 𝑓 of (1.8) satisfy 𝜎₂(𝑓)=1 except at most one nontrivial subnormal solution.

5. Proof of Theorem 1.6

(i) By Lemma 2.6 and 𝜎(𝑃_𝑗)=1(𝑗=0,…,𝑘−1), we see that 𝜎₂(𝑓)≤1. By Lemma 2.9, we see that there exist a subset 𝐸⊂(1,∞) having finite logarithmic measure and a constant 𝐵>0 such that for all 𝑧 satisfying |𝑧|=𝑟∉[0,1]∪𝐸, 𝑓^(𝑗)(𝑧)𝑓(𝑧)≤𝐵𝑇2𝑟,𝑓^𝑘+1𝑗=1,…,𝑘. (5.1)

Taking 𝑧=𝑟, by (1.11) and (5.1), we deduce that 𝑎_0𝑚0𝑒^𝑚0𝑟(1+𝑜(1))=−𝑃₀𝑒^𝑦𝑗≤𝑘𝐵𝑇2𝑟,𝑓^𝑘+1𝑀𝑒^𝑚𝑟(1+𝑜(1)). (5.2) Since 𝑚₀>𝑚, by (5.2), we get that 𝜎₂(𝑓)≥1. Hence 𝜎₂(𝑓)=1.

(ii) Using a similar method as in the proof of Theorem 1.4, we can prove (ii).

Acknowledgments

The authors are grateful to referees for a number of helpful suggestions to improve the paper. Z-X. Chen was supported by the National Natural Science Foundation of China (no.: 10871076). K. H. Shon was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2010-0009646).