Growth of Solutions of Nonhomogeneous Linear Differential Equations

Abstract

This paper is devoted to studying growth of solutions of linear differential equations of type 𝑓^(𝑘)+𝐴_𝑘−1(𝑧)𝑓^(𝑘−1)+⋯+𝐴₁(𝑧)𝑓^′+𝐴₀(𝑧)𝑓=𝐻(𝑧) where 𝐴_𝑗(𝑗=0,…,𝑘−1) and 𝐻 are entire functions of finite order.

1. Introduction and Main Results

We assume that the reader is familiar with the usual notations and basic results of the Nevanlinna theory [1–3]. Let now 𝑓(𝑧) be a nonconstant meromorphic function in the complex plane. We remark that 𝜌(𝑓) will be used to denote the order of 𝑓, and𝜌(𝑓)=limsup_𝑟→∞log𝑇(𝑟,𝑓)log𝑟. (1.1)

We now recall some previous results concerning nonhomogeneous linear differential equations of type𝑓^(𝑘)+𝐴_𝑘−1(𝑧)𝑓^(𝑘−1)+⋯+𝐴₁(𝑧)𝑓^′+𝐴₀(𝑧)𝑓=𝐻(𝑧), (1.2)where 𝐴_𝑗(𝑗=0,1,…,𝑘−1) and 𝐴₀≢0, 𝐻≢0 are entire functions of finite-order, 𝑘≥2. In the case that the coefficients 𝐴_𝑗(𝑗=0,1,…,𝑘−1) are polynomials, growth properties of solutions of (1.2) have been extensively studied, see, for example, [4]. In (1.2), if 𝑝 is the largest integer such that 𝐴_𝑝 is transcendental, it is well known that there exist at most 𝑝 linearly independent finite-order solutions of the corresponding homogeneous equation𝑓^(𝑘)+𝐴_𝑘−1(𝑧)𝑓^(𝑘−1)+⋯+𝐴₁(𝑧)𝑓^′+𝐴₀(𝑧)𝑓=0. (1.3)Thus, when at least one of the coefficients 𝐴_𝑗 is transcendental, most of the solutions of (1.2) and (1.3) are of infinite-order. In the case whenmax_𝑗≠𝑑{𝜌(𝐴_𝑗),𝜌(𝐻)}<𝜌(𝐴_𝑑1)≤2, (1.4) Hellerstein et al. [5] proved that every transcendental solution of (1.2) is of infinite-order. As for sectorial growth conditions on the coefficients of (1.2) that imply that all solutions are of infinite-order, see, for example, [6]. As for the special case of 𝑘=2, Wang and Laine studied equations of type𝑓^′′+𝐴₁(𝑧)𝑒^𝑎𝑧𝑓^′+𝐴₀(𝑧)𝑒^𝑏𝑧𝑓=𝐻(𝑧), (1.5)where 𝐴₁≢0,𝐴₀≢0,𝐻 are entire functions of order less than one, and the complex numbers 𝑎,𝑏 satisfy 𝑎𝑏≢0. They proved that every nontrivial solution of (1.5) is of infinite-order if 𝑎≢𝑏, see [7]. We remark that (1.2) may indeed have solutions of finite-order as soon as 𝜌(𝐻)≥max{𝜌(𝐴_𝑗)(𝑗=0,…,𝑘)}, as shown by the next examples.

Example 1.1. The exponential function 𝑓(𝑧)=𝑒^𝑧 satisfies the equation𝑓^(𝑘)+𝑓^(𝑘−1)+⋯+𝑓^′′+𝑒^−𝑧𝑓^′+𝑄(𝑧)𝑓=(𝑘−1+𝑄(𝑧))𝑒^𝑧+1, (1.6)where 𝑄(𝑧) can be any entire function. Choosing 𝑄(𝑧)=1−𝑘 shows that (1.2) may admit a solution of finite-order even if 𝜌(𝐻)<max{𝜌(𝐴_𝑗)(𝑗=0,…,𝑘)}. On the other hand, taking 𝑄(𝑧)=𝑒^𝑧, we have the case that 𝜌(𝐻)=max{𝜌(𝐴_𝑗)(𝑗=0,…,𝑘)} in (1.2).

Example 1.2. The function 𝑓(𝑧)=𝑒^𝑧2 satisfies the equation𝑓^′′′+𝑒^−𝑧𝑓^′′+𝑒^𝑧𝑓^′+𝑒^2𝑧𝑓=(8𝑧³+12𝑧+4𝑧²𝑒^−𝑧+2𝑒^−𝑧+2𝑧𝑒^𝑧+𝑒^2𝑧)𝑒^𝑧2. (1.7)
In this paper, we continue to consider (1.2) in the case when 𝜌(𝐻)<max{𝜌(𝐴_𝑗)(𝑗=0,…,𝑘)}. Recently, Tu and Yi investigated the growth of solutions of (1.3) when most coefficients have the same order, see [8]. We next prove two results of (1.2), which generalize Theorems 2 and 4 in [8] and Theorem 1.1 in [7].

Theorem 1.3. Suppose that 𝐴_𝑗(𝑧)=ℎ_𝑗(𝑧)𝑒^{𝑃𝑗(𝑧)}(𝑗=0,…,𝑘−1) where 𝑃_𝑗(𝑧)=𝑎_𝑗𝑛𝑧^𝑛+⋯+𝑎_𝑗0 are polynomials with degree 𝑛≥1, ℎ_𝑗(𝑧) are entire functions of order less than 𝑛, not all vanishing, and 𝐻(𝑧)≢0 is an entire function of order less than 𝑛. If 𝑎_𝑗𝑛(𝑗=0,…,𝑘−1) are distinct complex numbers, then every solution of (1.2) is of infinite-order.

Theorem 1.4. Suppose that 𝐴_𝑗(𝑧)=ℎ_𝑗(𝑧)𝑒^{𝑃𝑗(𝑧)}(𝑗=0,…,𝑘−1) where 𝑃_𝑗(𝑧)=𝑎_𝑗𝑛𝑧^𝑛+⋯+𝑎_𝑗0 are polynomials with degree 𝑛≥1, ℎ_𝑗(𝑧) and 𝐻(𝑧)≢0 are entire functions of order less than 𝑛. Moreover, suppose that there are two coefficients 𝐴_𝑠,𝐴_𝑙 so that for 𝑎_𝑠𝑛=|𝑎_𝑠𝑛|𝑒^𝑖𝜃𝑠 and 𝑎_𝑙𝑛=|𝑎_𝑙𝑛|𝑒^𝑖𝜃𝑙, where 0≤𝑠<𝑙≤𝑘−1, 𝜃_𝑠,𝜃_𝑙∈[0,2𝜋), 𝜃_𝑠≢𝜃_𝑙, ℎ_𝑠ℎ_𝑙≢0, and for all 𝑗≢𝑠,𝑙, 𝑎_𝑗𝑛 satisfies either 𝑎_𝑗𝑛=𝑑_𝑗𝑎_𝑠𝑛(0<𝑑_𝑗<1) or 𝑎_𝑗𝑛=𝑑_𝑗𝑎_𝑙𝑛(0<𝑑_𝑗<1). Then every transcendental solution of (1.2) is of infinite-order.

In the case when 𝐴_𝑗(𝑧)=ℎ_𝑗𝑒^𝑎𝑗𝑧+𝑔_𝑗 where ℎ_𝑗,𝑔_𝑗(𝑗=0,…,𝑘−1) are polynomials, Chen considered the growth of solutions of (1.3) with some additional conditions imposed upon on 𝑎_𝑗, see [9]. Our last results generalizes his result and [7, Theorem 1.3].

Theorem 1.5. Suppose that 𝐴_𝑗(𝑧)=ℎ_𝑗(𝑧)𝑒^{𝑃𝑗(𝑧)}+𝑔_𝑗(𝑧)(𝑗=0,…,𝑘−1) where 𝑃_𝑗(𝑧)=𝑎_𝑗𝑛𝑧^𝑛+⋯+𝑎_𝑗0 are polynomials with degree 𝑛≥1, ℎ_𝑗(𝑧),𝑔_𝑗(𝑧) and 𝐻(𝑧)≢0 are entire functions of order less than 𝑛. Moreover, suppose that there exist 𝑎_𝑠𝑛=𝑑_𝑠𝑒^𝑖𝜑 and 𝑎_𝑙𝑛=−𝑑_𝑙𝑒^𝑖𝜑 with 𝑑_𝑠>0,𝑑_𝑙>0 and 0≤𝑠<𝑙≤𝑘−1 such that for 𝑗≠𝑠,𝑙, 𝑎_𝑗𝑛=𝑑_𝑗𝑒^𝑖𝜑(𝑑_𝑗≥0) or 𝑎_𝑗𝑛=−𝑑_𝑗𝑒^𝑖𝜑(𝑑_𝑗≥0), and max{𝑑_𝑗,𝑗≢𝑠,𝑙}=𝑑<min{𝑑_𝑠,𝑑_𝑙}. If ℎ_𝑠ℎ_𝑙≢0, then every transcendental solution of (1.2) is of infinite-order.

Remark 1.6. Under the assumptions of Theorem 1.4, respectively, of Theorem 1.5, polynomial solutions may exist. However, such possible polynomial solutions must be of degree less than 𝑠. If not, a contradiction immediately follows by combining (5.1) with Lemma 2.1, if 𝐹≡0, respectively, with Lemma 2.2, if 𝐹≢0.

Remark 1.7. In the preceding three theorems, if 𝜌(𝑓)=∞, then we also have 𝜆(𝑓)=∞ for the exponent of convergence of the zero-sequence of 𝑓. Indeed, rewriting (1.2) in the form1𝑓=1𝐻(𝑓^(𝑘)𝑓+𝐴_𝑘−1𝑓^(𝑘−1)𝑓+⋯+𝐴₀), (1.8)we have1𝑚(𝑟,𝑓1)≤𝑚(𝑟,𝐻)+^𝑘−1_𝑗=0𝑚(𝑟,𝐴_𝑗)+^𝑘−1_𝑗=0𝑓𝑚(𝑟,^(𝑗)𝑓)=𝑂(𝑟^𝛽)+𝑆(𝑟,𝑓), (1.9)for some finite 𝛽. Therefore, 𝑁(𝑟,1/𝑓) must be of infinite-order.

2. Preliminary Lemmas

Lemma 2.1 (see [10]). Suppose that 𝑓₁(𝑧),𝑓₂(𝑧),…,𝑓_𝑛(𝑧)(𝑛≥2) are meromorphic functions and 𝑔₁(𝑧),𝑔₂(𝑧),…,𝑔_𝑛(𝑧) are entire functions satisfying the following conditions:
(i)∑^𝑛_𝑗=1𝑓_𝑗(𝑧)𝑒^{𝑔𝑗(𝑧)}≡0,(ii)𝑔_𝑗(𝑧)−𝑔_𝑘(𝑧) are not constants for 1≤𝑗<𝑘≤𝑛,(iii)for 1≤𝑗≤𝑛,1≤ℎ<𝑘≤𝑛,𝑇(𝑟,𝑓_𝑗)=𝑜{𝑇(𝑟,𝑒^{𝑔ℎ−𝑔𝑘})},(𝑟→∞,𝑟∉𝐸), (2.1)where 𝐸 is a set with finite linear measure. Then 𝑓_𝑗≡0(𝑗=1,2,…,𝑛).

Lemma 2.2 (see [10]). Suppose that 𝑓₁(𝑧),𝑓₂(𝑧),…,𝑓_𝑛(𝑧)(𝑛≥2) are linearly independent meromorphic functions satisfying the following identity:^𝑛_𝑗=1𝑓_𝑗≡1. (2.2)Then for 1≤𝑗≤𝑛, one has 𝑇(𝑟,𝑓_𝑗)≤^𝑘_𝑗=11𝑁(𝑟,𝑓_𝑘)+𝑁(𝑟,𝑓_𝑗)+𝑁(𝑟,𝐷)−^𝑛_𝑘=1𝑁(𝑟,𝑓_𝑘1)−𝑁(𝑟,𝐷)+𝑆(𝑟), (2.3)where 𝐷 is the Wronskian determinant 𝑊(𝑓₁,𝑓₂,…,𝑓_𝑛),𝑆(𝑟)=𝑜(max_{1≤𝑘≤𝑛}{𝑇(𝑟,𝑓_𝑘)}),(𝑟→∞,𝑟∉𝐸), (2.4)𝐸 is a set with finite linear measure.

Lemma 2.3 (see [11, 12]). Suppose that 𝑃(𝑧)=(𝛼+𝑖𝛽)𝑧^𝑛+⋯ (𝛼,𝛽 are real numbers, |𝛼|+|𝛽|≠0) is a polynomial with degree 𝑛≥1, and that 𝐴(𝑧)(≢0) is an entire function with 𝜌(𝐴)<𝑛. Set 𝑔(𝑧)=𝐴(𝑧)𝑒^𝑃(𝑧),𝑧=re^𝑖𝜃,𝛿(𝑃,𝜃)=𝛼cos(𝑛𝜃)−𝛽sin(𝑛𝜃). Then for any given 𝜀>0, there exists a set 𝐻₁⊂[0,2𝜋) of finite linear measure such that for any 𝜃∈[0,2𝜋)⧵(𝐻₁∪𝐻₂), there is 𝑅>0 such that for |𝑧|=𝑟>𝑅, one has
(i)if 𝛿(𝑃,𝜃)>0, thenexp{(1−𝜀)𝛿(𝑃,𝜃)𝑟^𝑛}<|𝑔(re^𝑖𝜃)|<exp{(1+𝜀)𝛿(𝑃,𝜃)𝑟^𝑛}; (2.5)(ii) if 𝛿(𝑃,𝜃)<0, thenexp{(1+𝜀)𝛿(𝑃,𝜃)𝑟^𝑛}<|𝑔(re^𝑖𝜃)|<exp{(1−𝜀)𝛿(𝑃,𝜃)𝑟^𝑛}, (2.6)where 𝐻₂={𝜃∈[0,2𝜋);𝛿(𝑃,𝜃)=0}.

Lemma 2.4 (see [13]). Let 𝑓(𝑧) be a transcendental meromorphic function of finite-order 𝜌, and let 𝜀>0 be a given constant. Then there exists a set 𝐻⊂(1,∞) that has finite logarithmic measure, such that for all 𝑧 satisfying |𝑧|∉𝐻∪[0,1] and for all 𝑘,𝑗,0≤𝑗<𝑘, one has |𝑓^(𝑘)(𝑧)𝑓^(𝑗)(𝑧)|≤|𝑧|^{(𝑘−𝑗)(𝜌−1+𝜀)}. (2.7)Similarly, there exists a set 𝐸⊂[0,2𝜋) of linear measure zero such that for all 𝑧=re^𝑖𝜃 with |𝑧| sufficiently large and 𝜃∈[0,2𝜋)⧵𝐸, and for all 𝑘,𝑗,0≤𝑗<𝑘, the inequality (2.7) holds.

Lemma 2.5. Let 𝑓(𝑧) be an entire function and suppose that𝐺(𝑧)∶=log⁺|𝑓^(𝑘)(𝑧)||𝑧|^𝜌 (2.8)is unbounded on some ray arg𝑧=𝜃 with constant 𝜌>0. Then there exists an infinite sequence of points 𝑧_𝑛=𝑟_𝑛𝑒^𝑖𝜃(𝑛=1,2,…), where 𝑟_𝑛→∞, such that 𝐺(𝑧_𝑛)→∞ and|𝑓^(𝑗)(𝑧_𝑛)𝑓^(𝑘)(𝑧_𝑛)1|≤(𝑘−𝑗)!(1+𝑜(1))𝑟_𝑛^𝑘−𝑗,𝑗=0,…,𝑘−1, (2.9)as 𝑛→∞.

Proof. The first assertion is trivial. Denoting𝑀(𝑟,𝐺,𝜃)=max{𝐺(𝑧)∶0≤|𝑧|≤𝑟,arg𝑧=𝜃}, (2.10)we may take the sequence {𝑧_𝑛} in the first assertion so that 𝐺(𝑧_𝑛)=𝑀(𝑟_𝑛,𝐺,𝜃). Since𝐺(𝑧_𝑛)→∞ (2.11)as 𝑛→∞, we immediately see that|𝑓^(𝑘)(𝑧_𝑛)|=𝑀(𝑟_𝑛,𝑓^(𝑘),𝜃)→∞ (2.12)as 𝑛→∞. Using now the same reasoning as in the proof of [14, Lemma 4], see also [15, Lemma 3.1], the second assertion (2.9) follows.

Lemma 2.6. Let 𝑓(𝑧) be an entire function with 𝜌(𝑓)=𝜌<∞. Suppose that there exists a set 𝐸⊂[0,2𝜋) which has linear measure zero, such that log⁺|𝑓(re^𝑖𝜃)|≤𝑀𝑟^𝜎 for any ray arg𝑧=𝜃∈[0,2𝜋)⧵𝐸, where 𝑀 is a positive constant depending on 𝜃, while 𝜎 is a positive constant independent of 𝜃. Then 𝜌(𝑓)≤𝜎.

Proof. Clearly, we may assume that 𝜎<𝜌. Since 𝐸 has linear measure zero, we may choose 𝜃_𝑗∈[0,2𝜋)⧵𝐸 such that 0≤𝜃₁<𝜃₂<⋯<𝜃_𝑛+1=2𝜋, andmax{𝜃_𝑗+1−𝜃_𝑗𝜋,1≤𝑗≤𝑛}≤𝜌+1. (2.13)We first treat the sector𝐻₁∶={𝑧∣𝜃₁≤arg𝑧≤𝜃₂}, (2.14)defining𝜙(𝑧)=𝑓(𝑧)exp{−𝑏𝑒^−𝑖𝜃0𝑧^𝜎}, (2.15)where 𝜃₀=𝜎(𝜃₁+𝜃₂)/2 and 𝑏 is a positive constant, to be determined in what follows. Then 𝜙(𝑧) is a holomorphic inside the sector 𝐻₁. By (2.13), we have 𝜌≤𝜋/(𝜃₂−𝜃₁)−1. Therefore,0>arg(𝑒^−𝑖𝜃0𝑧^𝜎)=arg(𝑒^−𝑖𝜃0𝑟^𝜎𝑒^{𝑖𝜎𝜃1})=𝜎(𝜃₁−𝜃₂)2≥−𝜋2+(𝜃₂−𝜃₁)2 (2.16)on the ray arg𝑧=𝜃₁, and, respectively,0<arg(𝑒^−𝑖𝜃0𝑧^𝜎)=arg(𝑒^−𝑖𝜃0𝑟^𝜎𝑒^{𝑖𝜎𝜃2})=𝜎(𝜃₂−𝜃₁)2≤𝜋2−(𝜃₂−𝜃₁)2 (2.17)on the ray arg𝑧=𝜃₂. Hence, we may now fix 𝑏>0 so that𝜋𝑏cos(2−(𝜃₂−𝜃₁)2)>𝑀. (2.18)By elementary computation, |𝜙(𝑧)|≤𝑀 on the boundary of 𝐻₁, where 𝑀>0 is a bounded constant, not the same at each occurrence. By the definition of 𝜙 in (2.15), it is immediate to see that 𝜙 is of order at most 𝜌. By the Phragmén-Lindelöf theorem, we conclude that |𝜙(𝑧)|≤𝑀 holds on the whole sector 𝐻₁. Hence|𝑓(𝑧)|≤|exp{𝑏𝑒^−𝑖𝜃0𝑧^𝜎}|≤exp{𝑏𝑟^𝜎} (2.19)on 𝐻₁. Repeating the same reasoning for all the sectors 𝐻_𝑗={𝑧∣𝜃_𝑗≤arg𝑧≤𝜃_𝑗+1} where 𝜃_𝑗 are determined in (2.13), the assertion immediately follows.

3. Proof of Theorem 1.3

Suppose, contrary to the assertion, that 𝑓 is a solution of (1.2) with 𝜌(𝑓)=𝜌<∞, then 𝑛≤𝜌. Indeed, if 𝑓^(𝑘)=𝐻, we may apply Lemma 2.1 to conclude that ℎ_𝑠𝑓^(𝑠)≡0 for some 𝑠, 0≤𝑠≤𝑘−1 such that ℎ_𝑠≠0. Then 𝑓 has to be a polynomial of degree less than 𝑠, so 𝐻(𝑧)≡0, a contradiction. Therefore, we may assume that 𝑓^(𝑘)≠𝐻. By Lemma 2.2, it is easy to see that 𝑛≤𝜌 since the exponential functions 𝑒^𝑃𝑗(𝑗=0,1,…,𝑘−1) are linearly independent.

By Lemma 2.3, there is a set 𝐸⊂[0,2𝜋) of linear measure such that whenever 𝜃∈[0,2𝜋)⧵𝐸, then 𝛿(𝑃_𝑗,𝜃)≠0 for all 0≤𝑗≤𝑘−1 and 𝛿(𝑃_𝑗−𝑃_𝑖,𝜃)≠0 for all 𝑖,𝑗 with 0≤𝑖<𝑗≤𝑘−1. If, moreover, 𝑧=re^𝑖𝜃 has 𝑟 large enough, then each 𝐴_𝑗(𝑧) satisfies either (2.5) or (2.6). By Lemma 2.4, we may assume, at the same time, that|𝑓^(𝑗)(𝑧)𝑓^(𝑖)(𝑧)|≤|𝑧|^𝑘𝜌,0≤𝑖<𝑗≤𝑘. (3.1)Since 𝑎_𝑗𝑛 are distinct complex numbers, then for any fixed 𝜃∈[0,2𝜋)⧵𝐸, there exists exactly one 𝑠∈{0,…,𝑘−1} such that𝛿(𝑃_𝑠,𝜃)=𝛿∶=max{𝛿(𝑃_𝑗,𝜃)∣𝑗=0,…,𝑘−1}. (3.2)Denoting 𝛿₁=max{𝛿(𝑃_𝑗,𝜃)∣𝑗≠𝑠}, then 𝛿₁<𝛿 and 𝛿≠0. We now discuss two cases separately.

Case 1. Assume first that 𝛿>0. By Lemma 2.3, for any given 𝜀 with 0<3𝜀<min{(𝛿−𝛿₁)/𝛿,𝑛−𝜌(𝐻)}, we have|𝐴_𝑠(re^𝑖𝜃)|≥exp{(1−𝜀)𝛿𝑟^𝑛},|𝐴_𝑗(re^𝑖𝜃)|≤exp{(1+𝜀)𝛿₁𝑟^𝑛}, (3.3)for 𝑗≠𝑠, provided that 𝑟 is sufficiently large. We now proceed to show thatlog⁺|𝑓^(𝑠)(𝑧)||𝑧|^{𝜌(𝐻)+𝜀} (3.4)is bounded on the ray arg𝑧=𝜃. Supposing that this is not the case, then by Lemma 2.5, there is a sequence of points 𝑧_𝑚=𝑟_𝑚𝑒^𝑖𝜃, such that 𝑟_𝑚→∞, and thatlog⁺|𝑓^(𝑠)(𝑧_𝑚)|𝑟_𝑚^{𝜌(𝐻)+𝜀}|𝑓→∞,_(𝑗)(𝑧_𝑚)𝑓^(𝑠)(𝑧_𝑚)|≤(1+𝑜(1))𝑟_𝑚^𝑠−𝑗,(𝑗=0,…,𝑠−1). (3.5)(3.6)From (3.5) and the definition of order, it is easy to see that|𝐻(𝑧_𝑚)𝑓^(𝑠)(𝑧_𝑚)|→0, (3.7)for 𝑚 is large enough. From (1.2), we obtain|𝐴_𝑠(𝑧_𝑚𝑓)|≤|^(𝑘)(𝑧_𝑚)𝑓^(𝑠)(𝑧_𝑚)|+⋯+|𝐴_𝑠+1(𝑧_𝑚𝑓)||^(𝑠+1)(𝑧_𝑚)𝑓^(𝑠)(𝑧_𝑚)|+|𝐴_𝑠−1(𝑧_𝑚𝑓)||^(𝑠−1)(𝑧_𝑚)𝑓^(𝑠)(𝑧_𝑚)|+⋯+|𝐴₀(𝑧_𝑚)||𝑓(𝑧_𝑚)𝑓^(𝑠)(𝑧_𝑚)|+|𝐻(𝑧_𝑚)𝑓^(𝑠)(𝑧_𝑚)|. (3.8)Using inequalities (3.1), (3.3), (3.6), and the limit (3.7), we conclude from the preceding inequality thatexp{(1−𝜀₁)𝛿𝑟^𝑛_𝑚}≤(𝑘+1)exp{(1+𝜀₁)𝛿₁𝑟^𝑛_𝑚}𝑟^𝑀_𝑚, (3.9)where 𝑀>0 is a bounded constant, which is a contradiction. Therefore, log⁺|𝑓^(𝑠)(𝑧)|/|𝑧|^{𝜌(𝐻)+𝜀} is bounded, and we have |𝑓^(𝑠)(𝑧)|≤𝑀exp{𝑟^{𝜌(𝐻)+𝜀}} on the ray arg𝑧=𝜃. By the same reasoning as in the proof of [15, Lemma 3.1], we immediately conclude that|𝑓(𝑧)|≤(1+𝑜(1))𝑟^𝑠|𝑓^(𝑠)(𝑧)|≤(1+𝑜(1))𝑀𝑟^𝑠𝑒^{𝑟𝜌(𝐻)+𝜀}≤𝑀𝑒^{𝑟𝜌(𝐻)+2𝜀} (3.10)on the ray arg𝑧=𝜃.

Case 2. Suppose now that 𝛿<0. From (1.2), we get−1=𝐴_𝑘−1𝑓^(𝑘−1)𝑓^(𝑘)+⋯+𝐴_𝑗𝑓^(𝑗)𝑓^(𝑘)+⋯+𝐴₀𝑓𝑓^(𝑘)−𝐻𝑓^(𝑘). (3.11)Again by Lemma 2.3, for any given 𝜀 with 0<3𝜀<min{1,𝑛−𝜌(𝐻)}, we have|𝐴_𝑗(re^𝑖𝜃)|≤exp{(1−𝜀)𝛿𝑟^𝑛},(𝑗=0,1,…,𝑘−1), (3.12)for 𝑟 sufficiently large. As in Case 1, we prove thatlog⁺|𝑓^(𝑘)(𝑧)||𝑧|^{𝜌(𝐻)+𝜀} (3.13)is bounded on the ray arg𝑧=𝜃. If not, similarly as in Case 1, it follows from Lemma 2.5 that there is a sequence of points 𝑧_𝑚=𝑟_𝑚𝑒^𝑖𝜃, such that|𝑓^(𝑗)(𝑧_𝑚)𝑓^(𝑘)(𝑧_𝑚)|≤𝑟_𝑚^𝑘−𝑗|(1+𝑜(1)),(𝑗=0,…,𝑘−1),𝐻(𝑧_𝑚)𝑓^(𝑘)(𝑧_𝑚)|→0, (3.14)for all 𝑚 large enough. Substituting the inequalities (3.12) and (3.14) into (3.11), a contradiction immediately follows. Hence, we have |𝑓^(𝑘)(𝑧)|≤𝑀exp{𝑟^{𝜌(𝐻)+𝜀}} on the ray arg𝑧=𝜃. This implies, as in Case 1, that|𝑓(𝑧)|≤𝑀exp{𝑟^{𝜌(𝐻)+2𝜀}}. (3.15) Therefore, for any given 𝜃∈[0,2𝜋)⧵𝐸, 𝐸 of linear measure zero, we have got (3.15) on the ray arg𝑧=𝜃, provided that 𝑟 is large enough. Then by Lemma 2.6, 𝜌(𝑓)≤𝜌(𝐻)+2𝜀<𝑛, a contradiction. Hence, every transcendental solution of (1.2) must be of infinite-order.

4. Proof of Theorem 1.4

Suppose that 𝑓 is a transcendental solution of (1.2) with 𝜌(𝑓)=𝜌<∞.

If 𝑓^(𝑘)≡𝐻 and 𝜌<𝑛, it follows from (1.2) that𝑓^(𝑙)ℎ_𝑙𝑒^{𝑃𝑙(𝑧)}+𝑓^(𝑠)ℎ_𝑠𝑒^{𝑃𝑠(𝑧)}+^𝑝_𝑢=1𝐵_𝑢(𝑧)𝑒^{𝑑𝑗𝑢𝑃𝑙(𝑧)}+^𝑞_𝑣=1𝐶_𝑣(𝑧)𝑒^{𝑑𝑗𝑣𝑃𝑠(𝑧)}=0, (4.1)where 𝐵_𝑢(𝑢=1,…,𝑝),𝐶_𝑣(𝑣=1,…,𝑞) are entire functions of order less than 𝑛. Collecting terms of the same type together, if needed, we may assume that the coefficients 𝑑_𝑗𝑢(𝑢=1,…,𝑝), respectively, 𝑑_𝑗𝑣(𝑣=1,…,𝑞), are distinct. Since 𝜃_𝑠≠𝜃_𝑙 and 𝜃_𝑠,𝜃_𝑙∈[0,2𝜋), we conclude that 𝑑_𝑗𝑢𝑃_𝑙(𝑧)−𝑑_𝑗𝑣𝑃_𝑠(𝑧) are polynomials of degree 𝑛. Indeed, if 𝑑_𝑗𝑢𝑎_𝑙𝑛=𝑑_𝑗𝑣𝑎_𝑠𝑛, we have𝑑0<^𝑗𝑢𝑑_𝑗𝑣|𝑎_𝑙𝑛𝑎_𝑠𝑛|=𝑒^{𝑖(𝜃𝑠−𝜃𝑙)} (4.2)which is impossible. Similarly, 𝑃_𝑙(𝑧)−𝑃_𝑠(𝑧),𝑃_𝑙(𝑧)−𝑑_𝑗𝑣𝑃_𝑠(𝑧), and 𝑃_𝑠(𝑧)−𝑑_𝑗𝑢𝑃_𝑙(𝑧) are also polynomials of degree 𝑛. Therefore, applying Lemma 2.1 to (4.1), we infer that 𝑓^(𝑙)ℎ_𝑙≡𝑓^(𝑠)ℎ_𝑠≡0. Since ℎ_𝑠ℎ_𝑙≢0, 𝑓 has to be a polynomial of degree less than 𝑠, then 𝐻≡0, a contradiction.

Therefore, we may proceed under the assumption that 𝑓^(𝑘)≢𝐻. By Lemma 2.2, if 𝑓^(𝑘)≢𝐻, then 𝑛≤𝜌 since the exponential functions 𝑒^𝑃𝑙,𝑒^𝑃𝑠,𝑒^{𝑑𝑗𝑢𝑃𝑙}(𝑢=1,2,…,𝑝) and 𝑒^{𝑑𝑗𝑣𝑃𝑠}(𝑣=1,2,…,𝑞) are linearly independent.

Since 𝜃_𝑠≠𝜃_𝑙, by Lemmas 2.3 and 2.4, there exists a set 𝐸⊂[0,2𝜋) of linear measure zero such that whenever 𝜃∈[0,2𝜋)⧵𝐸 then 𝐴_𝑗(re^𝑖𝜃) satisfies either (2.5) or (2.6), (3.1) holds, and𝛿(𝑃_𝑠,𝜃)≠𝛿(𝑃_𝑙,𝜃),𝛿₂∶=max{𝛿(𝑃_𝑠,𝜃),𝛿(𝑃_𝑙,𝜃)}≠0. (4.3)In what follows, we apply the notations 𝛿,𝛿₁ from the proof of Theorem 1.5 as well.

Case 1. Firstly assume that 𝛿₂>0. Without loss of generality, we may assume that 𝛿₂=𝛿(𝑃_𝑠,𝜃). From the hypothesis of 𝑎_𝑗𝑛, we know that 𝛿₁<𝛿₂=𝛿. Therefore, (3.3) holds by Lemma 2.3. Using the same reasoning as in Case 1 of the proof of Theorem 1.3, we obtain the inequality (3.15) on the ray arg𝑧=𝜃.

Case 2. Finally, assume that 𝛿₂<0. Again by the condition on 𝑎_𝑗𝑛, we see that 𝛿<0. Then the same argument as in Case 2 of the proof of Theorem 1.3 applies, and we again obtain (3.15).

Therefore, by Lemma 2.6, we obtain a contradiction, so 𝜌(𝑓)=∞.

5. Proof of Theorem 1.5

Contrary to the assertion, suppose that 𝑓 is a transcendental solution of (1.2) of finite-order. If 𝜌<𝑛, then it follows from (1.2) that𝑓^(𝑙)ℎ_𝑙𝑒^{𝑃𝑙(𝑧)}+𝑓^(𝑠)ℎ_𝑠𝑒^{𝑃𝑠(𝑧)}+^𝑝_𝑢=1𝐵_𝑢(𝑧)𝑒^{𝑑𝑗𝑢𝑃𝑙(𝑧)}+^𝑞_𝑣=1𝐶_𝑣(𝑧)𝑒^{𝑑𝑗𝑣𝑃𝑠(𝑧)}=𝐹(𝑧), (5.1)where 𝐵_𝑢(𝑢=1,…,𝑝),𝐶_𝑣(𝑣=1,…,𝑞), and 𝐹(𝑧) are entire functions of order less than 𝑛, 𝑑_𝑗𝑢≠0(𝑢=1,…,𝑝) are distinct, and 𝑑_𝑗𝑣≠0(𝑣=1,…,𝑞) are also distinct. Similarly as in the proof of Theorem 1.4, we may assume that 𝑛≤𝜌. Since 𝜎=max{𝜌(𝑔_𝑗)(𝑗=0,…,𝑘−1)}<𝑛, we havemax{|𝑔_𝑗(𝑧)|(𝑗=0,…,𝑘−1),|𝐻(𝑧)|}≤exp{𝑟^𝜎+𝜀} (5.2)for any 𝜀 with 0<3𝜀<𝑛−𝜎, and for |𝑧| sufficiently large. Since 𝑑_𝑠 and 𝑑_𝑙 in 𝑎_𝑠𝑛=𝑑_𝑠𝑒^𝑖𝜑 and 𝑎_𝑙𝑛=−𝑑_𝑙𝑒^𝑖𝜑 are strictly positive, the set {𝜃∈[0,2𝜋),𝛿(𝑃_𝑠,𝜃)=𝛿(𝑃_𝑙,𝜃)} is of linear measure zero. Therefore, again by Lemmas 2.3 and 2.4, there exists a set 𝐸⊂[0,2𝜋) of linear measure zero such that for any given 𝜃∈[0,2𝜋)⧵𝐸, ℎ_𝑗𝑒^𝑃𝑗 satisfies either (2.5) or (2.6), and (3.1) holds. Moreover, 𝛿(𝑃_𝑠,𝜃)≠𝛿(𝑃_𝑙,𝜃). Without loss of generality, we may assume that 𝛿₂∶=max{𝛿(𝑃_𝑠,𝜃),𝛿(𝑃_𝑙,𝜃)}=𝛿(𝑃_𝑙,𝜃)=−𝑑_𝑙cos(𝜑+𝑛𝜃), where cos(𝜑+𝑛𝜃)<0. Then from (2.5) and (5.2), for any 𝜀 also satisfying 0<3𝜀<(𝑑_𝑙−𝑑)⧵𝑑_𝑙, we obtain for |𝑧| sufficiently large that|𝐴_𝑙(re^𝑖𝜃)|≥exp{−(1−𝜀)𝑑_𝑙cos(𝜑+𝑛𝜃)𝑟^𝑛}. (5.3)For all other coefficients 𝐴_𝑗(𝑗≠𝑠), considering the hypothesis of 𝑎_𝑗𝑛, we have|𝐴_𝑗(re^𝑖𝜃)|≤exp{−(1+𝜀)𝑑cos(𝜑+𝑛𝜃)𝑟^𝑛}, (5.4)when 𝑟 is large enough. It follows from (1.2) that−𝐴_𝑙=𝑓^(𝑘)𝑓^(𝑠)+⋯+𝐴_𝑙+1𝑓^(𝑙+1)𝑓^(𝑙)+𝐴_𝑙−1𝑓^(𝑙−1)𝑓^(𝑙)+⋯+𝐴₀𝑓𝑓^(𝑙)−𝐻𝑓^(𝑙). (5.5)Similarly as in Case 1 of the proof of Theorem 1.4, and using Lemma 2.5, we may prove thatlog⁺|𝑓^(𝑙)(𝑧)||𝑧|^{𝜌(𝐻)+𝜀} (5.6)is bounded on the ray arg𝑧=𝜃. Therefore, the inequality (3.15) always holds on the ray arg𝑧=𝜃. Then, by Lemma 2.6, a contradiction follows, and so 𝜌(𝑓)=∞.

Acknowledgments

The authors wish to express their thanks to the referee for his valuable suggestions and comments. The authors have been partially supported by the National Natural Science Foundation of China (Grant No. 10871047) and the Academy of Finland (Grant no. 210245 and no. 124954).