# Complex Oscillation of Higher-Order Linear Differential Equations with Coefficients Being Lacunary Series of Finite Iterated Order

Authors: Jin Tu; Hong-Yan Xu; Hua-ming Liu; Yong Liu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The authors introduce the lacunary series of finite iterated order and use them to investigate the growth of solutions of higher-order linear differential equations with entire coefficients of finite iterated order and obtain some results which improve and extend some previous results of Belaidi, 2006, Cao and Yi, 2007, Kinnunen, 1998, Laine and Wu, 2000, Tu and Chen, 2009, Tu and Deng, 2008, Tu and Deng, 2010, Tu and Liu, 2009, and Tu and Long, 2009.


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## Body

## 1. Definitions and Notations

In this paper, we assume that readers are familiar with the fundamental results and standard notations of the Nevanlinna theory of meromorphic functions (see [1–3]). In order to describe the growth of order of entire functions or meromorphic functions more precisely, we first introduce some notations about finite iterated order. Let us define inductively, for r∈(0,+∞),exp1r=er and expi+1r=exp(expir),i∈ℕ. For all sufficiently large r, we define log1r=logr and logi+1r=log(logir),i∈ℕ. We also denote exp0r=r=log0r and exp-1r=log1r. Moreover, we denote the logarithmic measure of a set E⊂(0,+∞) by mlE=∫E‍dt/t, and the upper logarithmic density of E⊂(0,+∞) is defined by
(1)logdens¯E=limr→∞¯ml(E⋂‍[1,r])logr.
Throughout this paper, we use p to denote a positive integer. In the following, we recall some definitions of entire functions or meromorphic functions of finite iterated order (see [4–10]).Definition 1.
Thep-iterated order of a meromorphic function f(z) is defined by
(2)σp(f)=limr→∞¯logpT(r,f)logr.Remark 2.
Iff(z) is an entire function, then the p-iterated order of f(z) is defined by
(3)σp(f)=limr→∞¯logpT(r,f)logr=limr→∞¯logp+1M(r,f)logr.
If p=1, the classical growth of order of f(z) is defined by
(4)σ(f)=limr→∞¯logT(r,f)logr=limr→∞¯log2M(r,f)logr.
If p=2, the hyperorder of f(z) is defined by
(5)σ2(f)=limr→∞¯log2T(r,f)logr=limr→∞¯log3M(r,f)logr.Definition 3.
Iff(z) is an entire function with 0<σp(f)=σ<∞, then the p-iterated type of f(z) is defined by
(6)τp(f)=limr→∞¯logpM(r,f)rσ.Definition 4.
Thep-iterated lower order of an entire function f(z) is defined by
(7)μp(f)=limr→∞¯logpT(r,f)logr=limr→∞¯logp+1M(r,f)logr.Definition 5.
The finiteness degree of the iterated order of a meromorphic functionf(z) is defined by
(8)i(f)={0nnnnforf(z)rational,min{p∈ℕ:σp(f)<∞}p∈ℕforftranscendentalforwhichsomep∈ℕp∈ℕwithσp(f)<∞exists,∞nnnorf(z)withσp(f)=∞∀p∈ℕ.Definition 6.
Thep-iterated exponent of convergence of a-point of a meromorphic function f(z) is defined by
(9)λp(f,a)=limr→∞¯logpn(r,a)logr=limr→∞¯logpN(r,a)logr.
If a=0, the p-iterated exponent of convergence of zero-sequence and distinct zero-sequence of a meromorphic function f(z) are defined, respectively, by
(10)λp(f)=limr→∞¯logpn(r,1/f)logr=limr→∞¯logpN(r,1/f)logr,λ¯p(f)=limr→∞¯logpn¯(r,1/f)logr=limr→∞¯logpN¯(r,1/f)logr.
The p-iterated lower exponent of convergence of zero-sequence and distinct zero-sequence of a meromorphic function f(z) are defined, respectively, by
(11)λ_p(f)=limr→∞¯logpn¯(r,1/f)logr=limr→∞¯logpN¯(r,1/f)logr,λ_¯p(f)=limr→∞¯logpn¯(r,1/f)logr=limr→∞¯logpN¯(r,1/f)logr.


## 2. Introductions and Main Results

In the past ten years, many authors have investigated the complex oscillation properties of the higher-order linear differential equations(12)f(k)+Ak-1(z)f(k-1)+⋯+A0(z)f=0,(13)f(k)+Ak-1(z)f(k-1)+⋯+A0(z)f=F(z)
with Aj(z)(j=0,…,k-1),F(z) being entire functions or meromorphic functions of fast growing (e.g., see [4–12]), and obtained the following results.Theorem A (see [8]).
LetA0(z),…,Ak-1(z) be entire functions, if i(Aj)≤p(j=0,…,k-1), then i(f)≤p+1 and σp+1(f)≤max{σp(Aj),j=0,…,k-1} hold for all solutions of (12).Theorem B (see [8]).
LetA0(z),…,Ak-1(z) be entire functions and let i(A0)=p. If i(Aj)<p or σp(Aj)<σp(A0) for all j=1,…,k-1, then i(f)=p+1 and σp+1(f)=σp(A0) hold for all nontrivial solutions of (12).Theorem C (see [4, 12]).
LetA0(z),…,Ak-1(z) be entire functions and let i(A0)=p. Assume that max{σp(Aj):j=1,…,k-1}≤σp(A0)(>0) and max{τp(Aj):σp(Aj)=σp(A0)}<τp(A0). Then, every solution f(z)≢0 of (12) satisfies i(f)=p+1 and σp+1(f)=σp(A0).Theorem D (see [10]).
LetA0(z),…,Ak-1(z) be entire functions of finite iterated order satisfying i(A0)=p, σp(A0)=σ, and limr→∞¯(∑j=1k-1‍m(r,Aj)/m(r,A0))<1. Then, every nontrivial solution f(z) of (12) satisfies σp+1(f)=σp(A0)=σ.Theorem E (see [10]).
LetA0(z),…,Ak-1(z) be entire functions of finite iterated order satisfying max{σp(Aj),j≠0}<μp(A0)=σp(A0) and limr→∞¯(∑j=1k-1‍m(r,Aj)/m(r,A0))<1(r∉E), where E is a set of r of finite linear measure, then every nontrivial solution f(z) of (12) satisfies σp+1(f)=μp(A0)=σp(A0).Theorem F(see [5]).
LetAj(z)(j=0,…,k-1) be entire functions of finite iterated order such that there exists one transcendental Ad(0≤d≤k-1) satisfying σp(Aj)≤σp(Ad)<∞ for all j≠s, then (12) has at least one solution f(z) that satisfies i(f)=p+1 and σp+1(f)=σp(Ad).Remark 7.
Theorems B–E are investigating the growth of solutions of (12) when the coefficients are of finite iterated order and A0(z) grows faster than other coefficients in (12). What can we have if there exists one middle coefficient Ad(z)(1≤d≤k-1) such that Ad(z) grows faster than other coefficients in (12) or (13)? Many authors have investigated this question when Ad(z) is of finite order and obtained many results (e.g., see [13–15]). Here, our question is that under what conditions can we obtain similar results with Theorems B-C if Ad(z)(1≤d≤k-1) is of finite iterated order and grows faster than other coefficients in (12) or (13).
In 2009, Tu and Liu make use of the proposition of lacunary power series to investigate the above question in the casep=1 and obtain the following result.Theorem G (see [15]).
LetAj(z)(j=0,…,k-1),F(z) be entire functions satisfying max{σ(Aj),j≠d,σ(F)}<σ(Ad)<∞(1≤d≤k-1). Suppose that Ad(z)=∑n=0∞‍cλnzλn is an entire function of regular growth such that the sequence of exponents {λn} satisfies Fabry gap condition
(14)λnn⟶∞(n⟶∞),
then one has (i)
ifF(z)≡ 0, then every transcendental solution f(z) of (13) satisfies σ2(f)=σ(Ad);(ii)
ifF(z)≢0, then every transcendental solution f(z) of (13) satisfies λ¯2(f)=λ2(f)=σ2(f)=σ(Ad).In this paper, we continue our research in this area and obtain the following results.Theorem 8.
LetAj(z)(j=0,…,k-1),F(z) be entire functions of finite iterated order and satisfying 0<max{σp(Aj),j≠d}≤σp(Ad)<∞ and max{τp(Aj):σp(Aj)=σp(Ad)}<τp(Ad)(0≤d≤k-1). Suppose that Ad(z)=∑n=0∞‍cλnzλn is an entire function such that the sequence of exponents {λn} satisfies
(15)λnn>(logn)2+η(η>0,n∈ℕ),
for some η>0, then one has (i)
ifσp(F)<σp(Ad) or σp(F)=σp(Ad) and τp(F)<τp(Ad), then every transcendental solution f(z) of (13) satisfies σp+1(f)=σp(Ad); furthermore if F(z)≢0, then every transcendental solution f(z) of (13) satisfies λ¯p+1(f)=λp+1(f)=σp+1(f)=σp(Ad);(ii)
ifσp(F)>σp(Ad) and σp+1(F)≤σp(Ad), then all solutions of (13) satisfy σp(f)≥σp(F) and σp+1(f)≤σp(Ad);(iii)
ifσp+1(F)>σp(Ad), then all solutions of (13) satisfy σp+1(f)=σp+1(F), and λ¯p+1(f)=λp+1(f)=σp+1(F) holds for all solutions of (13) with at most one exceptional solution f0 satisfying λp+1(f0)<σp+1(F).Remark 9.
Iff(z)=∑n=0∞‍cλnzλn is an entire function of finite order and the sequence of exponents {λn} satisfies (14), then (18) in Lemma 15 holds for f(z), but for entire functions of infinite order, (14) certainly does not imply (18) in Lemma 15 (see [8]); therefore, we need more stringent gap condition (15) which is sufficient and unnecessary for Theorem 8.Theorem 10.
LetAj(z)(j=0,…,k-1),F(z) be entire functions of finite iterated order satisfying max{σp(Aj),j≠d,σp(F)}<μp(Ad)=σp(Ad)=σ<∞(0≤d≤k-1). Suppose that Ad(z)=∑n=0∞‍cλnzλn is an entire function such that the sequence of exponents {λn} satisfies gap condition (15), then every transcendental solution f(z) of (13) satisfies μp+1(f)=σp+1(f)=σ. Furthermore if F(z)≢0, then every transcendental solution f(z) of (13) satisfies λ¯_p+1(f)=λ_p+1(f)=λ¯p+1(f)=λp+1(f)=σ.Theorem 11.
LetAj(z)(j=0,…,k-1),F(z) be entire functions satisfying max{σp(Aj),j≠d}<σp(Ad)<∞(0≤d≤k-1). Suppose that T(r,Ad)~logM(r,Ad) as r→∞ outside a set of r of finite logarithmic measure, then one has(i)
ifσp(F)<σp(Ad), then every transcendental solution f(z) of (13) satisfies σp+1(f)=σp(Ad); furthermore, if F(z)≢0, then every transcendental solution f(z) of (13) satisfies λ¯p+1(f)=λp+1(f)=σp+1(f)=σp(Ad);(ii)
ifσp(F)>σp(Ad) and σp+1(F)≤σp(Ad), then all solutions of (13) satisfy σp(f)≥σp(F) and σp+1(f)≤σp(Ad);(iii)
ifσp+1(F)>σp(Ad), then all solutions of (13) satisfy σp+1(f)=σp+1(F), and λ¯p+1(f)=λp+1(f)=σp+1(F) holds for all solutions of (13) with at most one exceptional solution f0 satisfying λp+1(f0)<σp+1(F);(iv)
ifμp(Ad)=σp(Ad)=σ and F(z)≢0 and σp(F)<σ, then every transcendental solution f(z) of (13) satisfies λ¯_p+1(f)=λ_p+1(f)=λ¯p+1(f)=λp+1(f)=σ.Remark 12.
Theorem10 implies that all the solutions of (13) are of regular growth if Ad is of regular growth under some conditions; Theorem 11 is an improvement of the Theorem in [14, page 2694] and Theorems 1-2 in [16, page 624]. In fact, by Lemma 15, the gap condition (15) in Theorem 8 implies that T(r,Ad)~logM(r,Ad) as r→∞ outside a set of r of finite logarithmic measure; therefore, Theorem 11 is a generalization of Theorem 8 in a sense, but the condition on Ad in Theorem 8 is more stringent than that in Theorem 11. In addition, Theorems 8–11 may have polynomial solutions of degree <d if d>1.


## 3. Preliminary Lemmas

Lemma 13 (see [17]).
Letf(z) be a transcendental meromorphic function, and let α>1 be a given constant, for any given constant and for any given ε>0,(i)
there exist a constantB>0 and a set E1⊂(0,+∞) having finite logarithmic measure such that for all z satisfying |z|=r∉E1, one has
(16)|f(j)(z)f(i)(z)|≤B[T(αr,f)r(logr)αlogT(αr,f)]j-i(0≤i<j).(ii)
There exists a setH1⊂[0,2π) that has linear measure zero a constant B>0 that depends only on α, for any θ∈[0,2π)∖H1, there exists a constant R0=R0(θ)>1 such that for all z satisfying argz=θ and |z|=r>R0, one has
(17)|f(j)(z)f(i)(z)|≤B[T(αr,f)logT(αr,f)]j-i(0≤i<j).Remark 14.
Throughout this paper, we useE1⊂(0,∞) to denote a set having finite logarithmic measure or finite linear measure, not always the same at each occurrence.Lemma 15 (see [18]).
Letf(z)=∑n=0∞‍cλnzλn be an entire function and the sequence of exponents {λn} satisfies the gap condition (15). Then for any given ε>0,
(18)logL(r,f)>(1-ε)logM(r,f)
holds outside a set of finite logarithmic measure, where M(r,f)=sup|z|=r|f(z)|, L(r,f)=inf|z|=r|f(z)|.Lemma 16 (see [4]).
Letf(z) be an entire function of finite iterated order satisfying 0<σp(f)=σ<∞ and τp(f)=τ>0, then for any given β<τ, there exists a set E2⊂(0,+∞) having infinite logarithmic measure such that for all r∈E2, one has
(19)M(r,f)>expp{βrσ}.Lemma 17.
Letf(z)=∑n=0∞cλnzλn be an entire function of finite iterated order satisfying 0<σp(f)=σ<∞ and τp(f)=τ>0 such that the sequence of exponents {λn} satisfies the gap condition (15). Then, for any given β>τ, there exists a set E3⊂(0,+∞) having infinite logarithmic measure such that for all |z|=r∈E3, one has
(20)|f(z)|>expp{βrσ}.Proof.
By Lemma15, for any given ε>0, we have
(21)|f(z)|≥L(r,f)>M(r,f)(1-ε),(r∉E1).
For any given β<τ, we can choose β1 and sufficiently small ε>0 such that β<β1<τ and β<(1-ε)β1<τ; by Lemma 16, there exists a set  E2⊂(0,+∞) having infinite logarithmic measure such that for all |z|=r∈E2∖E1, we have
(22)|f(z)|>M(r,f)(1-ε)>(expp{β1rσ})(1-ε)>expp{βrσ},
where E3=E2∖E1 is a set having infinite logarithmic measure.Lemma 18 (see [13]).
Letf(z) be a transcendental entire function. Then, there is a set E1⊂(0,+∞) having finite logarithmic measure such that for all z satisfying |z|=r∉E1 and |f(z)|=M(r,f), one has
(23)|f(z)f(j)(z)|≤2rj,(j∈ℕ).Lemma 19 (see [7, 9, 10]).
LetA0(z),…,Ak-1(z),F(z)≢0 be meromorphic functions, and let f(z) be a meromorphic solution of (13) satisfying one of the following conditions:(i)
max{i(F)=q,i(Aj)(j=0,…,k-1)}<i(f)=p+1(p,q∈ℕ);(ii)
b=max{σp+1(F),σp+1(Aj)(j=0,…,k-1)}<σp+1(f),
thenλ¯p+1(f)=λp+1(f)=σp+1(f).Lemma 20 (see [8]).
LetA0(z),…,Ak-1(z),F(z) be entire functions of finite iterated order, if i(Aj)≤p,i(F)≤p(j=0,…,k-1). Then σp+1(f)≤max{σp(Aj),σp(F),j=0,…,k-1} holds for all solutions of (13).Lemma 21 (see [2]).
Letg:(0,+∞)→R,h:(0,+∞)→R be monotone increasing functions such that(i)
g(r)≤h(r) outside of an exceptional set of finite linear measure. Then, for any α>1, there exists r0>0 such that g(r)≤h(αr) for all r>r0.(ii)
g(r)≤h(r) outside of an exceptional set of finite logarithmic measure. Then, for any α>1, there exists r0>0 such that g(r)≤h(rα) for all r>r0.Lemma 22 (see [19]).
Letf(z) be an entire function of finite iterated order satisfying μp(f)=μ<∞. Then, for any given ε>0, there exists a set E4⊂(0,+∞) having infinite logarithmic measure such that for all r∈E4, one has
(24)M(r,f)<expp{rμ+ε}.Lemma 23 (see [2, 20]).
Letf(z) be a transcendental entire function, let 0<η1<1/4 and zr a point such that |zr|=r and that |f(zr)|>M(r,f)νf(r)-(1/4)+η1 holds. Then, there exists a set E1⊂(0,+∞) of finite logarithmic measure such that
(25)f(j)(zr)f(zr)=(νf(r)zr)j(1+o(1))
holds for all j∈ℕ and all r∉E1, where νf(r) is the central index of f(z).Lemma 24 (see [7, 9]).
Letf(z) be an entire function of finite iterated order satisfying σp(f)=σ,μq(f)=μ,p,q∈ℕ. Then, one has
(26)limr→∞¯logpνf(r)logr=σ,limr→∞¯logqνf(r)logr=μ.Lemma 25.
LetAj(z)(j=0,…,k-1),F(z) be entire functions of finite iterated order satisfying max{σp(Aj),j≠d,σp(F)}≤μp(Ad)<∞. Then, every solution f(z) of (13) satisfies μp+1(f)≤μp(Ad).Proof.
By (13), we have
(27)|f(k)f|≤|Ak-1f(k-1)f|+⋯+|A1f′f|+|A0|+|F(z)f(z)|.
By Lemma 23, there exists a set E1 having finite logarithmic measure such that for all |z|=r∉E1 and |f(z)|=M(r,f), we have
(28)f(j)(z)f(z)=(νf(r)z)j(1+o(1)),(j∈ℕ).
By Lemma 22, for any given ε>0, there exists a set E4 having infinite logarithmic measure such that for all |z|=r∈E4 and |f(z)|=M(r,f)>1, we have
(29)|Ad(z)|≤expp{rμp(Ad)+ε},|Aj(z)|≤expp{rμp(Ad)+ε},|F(z)f(z)|≤|F(z)|<expp{rμp(Ad)+ε}.
Hence from (27)–(29), for any given ε>0 and for all z satisfying |z|=r∈E4∖E1 and |f(z)|=M(r,f), we have
(30)(νf(r)r)k(1+o(1))≤(k+1)(νf(r)r)k-1(1+o(1))expp{rμp(Ad)+ε}.
By (30) and Lemma 24, we have μp+1(f)≤μp(Ad). Therefore, we complete the proof of Lemma 25.Lemma 26.
LetA0,A1,…,Ak-1,F≢0 be meromorphic functions of finite iterated order; if f is a meromorphic solution of the (13) and satisfies b=max{σp+1(F),σp+1(Aj),j=0,…,k-1}<μp+1(f), then λ¯_p+1(f)=λ_p+1(f)=μp+1(f).Proof.
By (13), we have
(31)1f=1F(f(k)f+Ak-1f(k-1)f+⋯+A0).
By (31), we get
(32)N(r,1f)≤kN¯(r,1f)+N(r,1F)+∑j=0k-1‍N(r,Aj).
By the lemma of logarithmic derivative and (31), we have
(33)m(r,1f)≤m(r,1F)+∑j=0k-1‍m(r,Aj)+O{log(rT(r,f))}(r∉E1).
By (32) and (33), we have
(34)T(r,f)=T(r,1f)+O(1)≤kN¯(r,1f)+T(r,1F)+∑j=0k-1‍T(r,Aj)+O{log(rT(r,f))}=kN¯(r,1f)+T(r,F)+∑j=0k-1‍T(r,Aj)+O{log(rT(r,f))},(r∉E1).
Since max{σp+1(F),σp+1(Aj),  j=0,1,…,k-1}<μp+1(f), for sufficiently large r, we have
(35)T(r,F)=o{T(r,f)},T(r,Aj)=o{T(r,f)},j=0,…,k-1,log(rT(r,f))=o{T(r,f)}.
By (34)-(35), we have
(36)(1-o(1))T(r,f)≤kN¯(r,1f),(r∉E1).
By Lemma 21  (i) and by (36), we have λ¯_p+1(f)=λ_p+1(f)=μp+1(f).Lemma 27.
Letf(z) be a transcendental entire function, for each sufficiently large |z|=r, and let zr=reiθr be a point satisfying |f(zr)|=M(r,f). Then, there exists a constant δr(>0) such that for all z satisfying |z|=r∉E1 and argz=θ∈[θr-δr,θr+δr], one has
(37)f(j)(z)f(z)=(νf(r)z)j(1+o(1))(j∈ℕ).Proof.
Ifzr=reiθr is a point satisfying |f(zr)|=M(r,f), since |f(z)| is continuous in |z|=r, then there exists a constant δr(>0) such that for all z satisfying |z|=r (large enough) and argz=θ∈[θr-δr,θr+δr], we have
(38)||f(z)|-|f(zr)||<ε;thatis,|f(z)|>12|f(zr)|nnnnnnnnnn=12M(r,f)nnnnnnnnnn>M(r,f)νf(r)-(1/4)+η1.
By Lemma 23, we have
(39)f(j)(z)f(z)=(νf(r)z)j(1+o(1)),(j∈ℕ)
holds for all z satisfying |z|=r∉E1 and argz=θ∈[θr-δr,θr+δr].Lemma 28.
Letf(z) be a transcendental entire function, for each sufficiently large |z|=r, and let zr=reiθr be a point satisfying |f(zr)|=M(r,f). Then, there exists a constant δr(>0) such that for all z satisfying |z|=r∉E1 and argz=θ∈[θr-δr,θr+δr], one has
(40)|f(z)f(j)(z)|≤2rj(j∈ℕ).Proof.
Ifzr=reiθr is a point satisfying |f(zr)|=M(r,f), then by Lemma 27 there exists a constant δr(>0) such that for all z satisfying |z|=r∉E1 and argz=θ∈[θr-δr,θr+δr], we have
(41)f(j)(z)f(z)=(νf(r)z)j(1+o(1)).
Since f(z) is transcendental, we have νf(r)→∞(r→∞). Hence by (41), for all z satisfying |z|=r∉E1 and argz=θ∈[θr-δr,θr+δr], we have
(42)|f(j)(z)f(z)|≥12r-j;thatis,|f(z)f(j)(z)|≤2rj(j∈ℕ).
Therefore, we complete the proof of Lemma 28.Lemma 29.
Letf(z) be an entire function of order 0<σp(f)=σ<∞. Then for any given ε>0, there exists a set E5⊂(0,+∞) with positive upper logarithmic density such that for all |z|=r∈E5, one has
(43)M(r,f)>expp{rσ-ε}.Proof.
Since0<σp(f)=σ<∞, then for any given ε>0, there exists an increasing sequence {rn} tending to ∞ such that for n≥n4, we have
(44)M(rn,f)>expp{rnσ-(ε/2)}.
Since ε>0, we can choose α to satisfy 1<α<(σ-(ε/2))/(σ-ε). Then for all r∈[rn,rnα](n≥n4), we have
(45)M(r,f)≥M(rn,f)>expp{rnσ-(ε/2)}≥expp{r(σ-(ε/2))/α}>expp{rσ-ε}.
Setting E5=⋃n=n4∞‍[rn,rnα], we have
(46)logdens¯E5=limr→∞¯ml(E5∩[1,r])logr≥limn→∞¯ml(E5∩[1,rnα])logrnα≥limn→∞ml([rn,rnα])logrnα=α-1α>0.
Thus, Lemma 29 is proved.Lemma 30.
Letf(z) be a transcendental entire function satisfying 0<σp(f)=σ<∞ and T(r,f)~logM(r,f) as r→∞ outside a set r of finite logarithmic measure. Then for any given ε>0, there exists a set E6⊂(0,+∞) with positive upper logarithmic density and a set H2⊂[0,2π) with linear measure zero such that for all z satisfying r∈E6 and argz=θ∈[0,2π)∖H2, one has
(47)|f(reiθ)|>expp{rσ-ε}.Proof.
Sincem(r,f)~logM(r,f) as r→∞(r∉E1), then by the definition of m(r,f), there exists a set H2⊂[0,2π) having linear measure zero such that for all z satisfying argz=θ∈[0,2π)∖H2, we have
(48)|f(reiθ)|>M(r,f)1-ε(r∉E1).
By Lemma 29, for any given ε>0, there exists a set E6⊂(0,+∞) with positive upper logarithmic density, we have
(49)M(r,f)>expp{rσ-(ε/2)}.
By (48) and (49), for any given ε>0 and for all z satisfying |z|=r∈E6∖E1 and argz=θ∈[0,2π)∖H2, we have
(50)|f(reiθ)|>M(r,f)1-ε>(expp{rσ-(ε/2)})1-ε>expp{rσ-ε}.
Therefore, we complete the proof of Lemma 30.


## 4.  Proofs of Theorems8–11

Proof of Theorem8.
(
i
) Assume that f(z) is a transcendental solution of (13). By (13), we have
(51)|Ad|≤|f(k)f(d)|+⋯+|Ad+1||f(d+1)f(d)|+|ff(d)|(|Ad-1||f(d-1)f|+⋯+|A0|+|Ff|).
By Lemma 13  (i), there exists a set E1⊂[1,∞) having finite logarithmic measure and a constant B>0 such that
(52)|f(j)(z)f(i)(z)|≤B(T(2r,f))2k,(0≤i<j≤k)
holds for all |z|=r∉E1 and for sufficiently large r. Since max{σp(Aj),j≠d,σp(F)}≤σp(Ad) and max{τp(Aj),τp(F):σp(Aj)=σp(Ad)=σp(F)}<τp(Ad), we choose α1,β1 to satisfy max{τp(Aj),j≠d,τp(F)}<α1<β1<τp(Ad); by Lemma 17, there exists a set E3⊂(0,+∞) having infinite logarithmic measure such that for all z satisfying |z|=r∈E3 and for sufficiently large r, we have
(53)|Ad(z)|>expp{β1rσp(Ad)},|Aj(z)|<expp{α1rσp(Ad)},j≠d.
By Lemma 18, there exists a set E1⊂(0,+∞) having finite logarithmic measure such that for all z satisfying |f(z)|=M(r,f)>1 and |z|=r∉E1, we have
(54)|f(z)f(d)(z)|≤2rd,|F(z)f(z)|≤|F(z)|<expp{α1rσp(Ad)}.
Hence from (51)–(54), for all z satisfying |z|=r∈E3∖E1 and |f(z)|=M(r,f), we have
(55)expp{β1rσp(Ad)}≤2B(k+1)rdexp{α1rσp(Ad)}[T(2r,f)]2k.
By (55), we have
(56)σp+1(f)=limr→∞¯logp+1T(r,f)logr≥σp(Ad).
On the other hand, by Lemma20, we have σp+1(f)≤σp(Ad). Therefore, every transcendental solution f(z) of (13) satisfies σp+1(f)=σp(Ad). Furthermore if F(z)≢0, then by Lemma 19, we have that every transcendental solution f(z) of (13) satisfies λ¯p+1(f)=λp+1(f)=σp+1(f)=σp(Ad).
(
ii
) We assume that f is a solution of (13). By the elementary theory of differential equations, all the solutions of (13) are entire functions and have the form
(57)f=f*+C1f1+C2f2+⋯+Ckfk,
where C1,…,Ck are complex constants, {f1,…,fk} is a solution base of (12), and f* is a solution of (13) and has the form
(58)f*=D1f1+D2f2+⋯+Dkfk,
where D1,…,Dk are certain entire functions satisfying
(59)Dj′=F·Gj(f1,…,fk)·W(f1,…,fk)-1(j=1,…,k),
where Gj(f1,…,fk) are differential polynomials in f1,…,fk and their derivative with constant coefficients, and W(f1,…,fk) is the Wronskian of f1,…,fk. By Theorem A, we have σp+1(fj)≤σp(Ad)(j=1,2,…,k); then by (57)–(59), we get
(60)σp+1(f)≤max{σp+1(fj),σp+1(F),j=1,…,k}≤σp(Ad).
Sinceσp(F)>σp(Ad), it is easy to see that σp(f)≥σp(F) by (13).
(
iii
) Suppose that f is a solution of (13), it is easy to see that σp+1(f)≥σp+1(F) by (13). On the other hand, since σp+1(F)>σp(Ad) and by (57)–(59), we have
(61)σp+1(f)≤max{σp+1(fj),σp+1(F),j=1,…,k}≤σp+1(F).
Therefore, all solutions of (13) satisfy σp+1(f)=σp+1(F).
By the same proof in Theorem 4.2 in [8, page 401], we can obtain that all solutions of (13) satisfying λ¯p+1(f)=λp+1(f)=σp+1(F) with at most one exceptional solution f0 satisfying λp+1(f0)<σp+1(F).Proof of Theorem10.
Suppose thatf(z) is a transcendental solution of (13), by the same proof in Theorem 8, we have σp+1(f)=σp(Ad)=σ. Thus, it remains to show that μp+1(f)=μp(Ad)=σ. We choose α2,β2 to satisfy
(62)max{σp(Aj),j≠d,σp(F)}<α2<β2<σ.
Since the sequence of exponents {λn} of Ad satisfies (15) and μp(Ad)=σ, then by Lemma 15, there exists a set E1 having finite logarithmic measure such that for all sufficiently large r∉E1, we have
(63)|Ad(z)|≥expp{rβ2},|Aj(z)|≤expp{rα2},j≠d.
Hence from (51), (52), (54), and (63), for all z satisfying |z|=r∉E1 and |f(z)|=M(r,f), we have
(64)expp{rβ2}≤2B(k+1)rdexpp{rα2}[T(2r,f)]2k.
Since β2 is arbitrarily close to σ, by (64) and Lemma 21  (ii), we have
(65)μp+1(f)=limr→∞¯logp+1T(r,f)logr≥σ.
On the other hand, by Lemma 26, we have μp+1(f)≤μp(Ad)=σ; therefore, every transcendental solution of (13) satisfies μp+1(f)=σ.Proof of Theorem11.
(
i
) By Lemma 20, we know that every solution of (13) satisfies σp+1(f)≤σp(Ad). In the following, we show that every transcendental solution f(z) of (13) satisfies σp+1(f)≥σp(Ad). Suppose that f(z) is a transcendental solution of (13). For each sufficiently large circle |z|=r, we take a point zr=reiθr satisfying |f(zr)|=M(r,f). By Lemma 28, there exist a constant δr>0 and a set E1 such that for all z satisfying |z|=r∉E1 and argz=θ∈[θr-δr,θr+δr], we have
(66)|f(z)f(d)(z)|≤2rd.
By Lemma 13  (ii), there exist a set H1⊂[0,2π) having linear measure zero and a constant B>0 such that for sufficiently large r and for all z satisfying argz=θ∈[θr-δr,θr+δr]∖H1, we have
(67)|f(j)(z)f(i)(z)|≤B[T(2r,f)]2k(0≤i<j≤k).
Setting max{σp(Aj),σp(F),j≠d}=b<σp(Ad), for all z satisfying |z|=r∉E1 and argz=θ∈[θr-δr,θr+δr] and for any given ε(0<2ε<σp(Ad)-b), we have
(68)|Aj(z)|≤exp{rb+ε}(j≠d),|F(z)f(z)|≤|F(z)|≤exp{rb+ε}.
Since T(r,Ad)~logM(r,Ad) as r→∞(r∉E1), by Lemma 30, for any given ε>0, there exists a set E6⊂(0,∞) with logdens¯E6>0 and a set H2⊂[0,2π) with linear measure zero such that for all z satisfying |z|=r∈E6 and argz=θ∈[θr-δr,θr+δr]∖H2, we have
(69)|Ad(z)|>expp{rσp(Ad)-ε}.
Substituting (66)–(69) into (51), for all z satisfying |z|=r∈E6∖E1 and argz=θ∈[θr-δr,θr+δr]∖(H1∪H2), we have
(70)expp{rσp(Ad)-ε}≤(k+1)B[T(2r,f)]2k·2rd·expp{rb+ε}.
From (70), we have σp+1(f)≥σp(Ad). Therefore, every transcendental solution f(z) of (13) satisfies σp+1(f)=σp(Ad). Furthermore, if F(z)≢0, then every transcendental solution f(z) of (13) satisfies λ¯p+1(f)=λp+1(f)=σp+1(f)=σp(Ad).
(
ii
)
–
(
iv
) By the same proof in Theorems 8 and 10, we can obtain the conclusions (ii)–(iv).



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