# The Oscillation on Solutions of Some Classes of Linear Differential Equations with Meromorphic Coefficients of Finite[p,q]-Order

Authors: Hong-Yan Xu; Jin Tu; Zu-Xing Xuan
Journal: The Scientific World Journal
 (2013)

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## Abstract

This paper considers the oscillation on
meromorphic solutions of the second-order linear differential equations
with the formf′′+A(z)f=0, where A(z) is a meromorphic function
with [p,q]-order.  We obtain some theorems which are the improvement
and generalization of the results given by Bank and Laine, Cao and
Li, Kinnunen, and others.


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## Body

## 1. Introduction and Main Results

The purpose of this paper is to study the oscillation on solutions of linear differential equations in the complex plane. It is well known that Nevanlinna theory has appeared to be a powerful tool in the field of complex differential equations. We assume that readers are familiar with the standard notations and the fundamental results of the Nevanlinna’s value distribution theory of meromorphic functions (see [1, 3, 4]). Throughout the paper, a meromorphic function f means meromorphic in the complex plane ℂ. In addition, we use σ(f) and λ(f) to denote the order and the exponent of convergence of zero sequence of meromorphic function f(z), respectively. For sufficiently large r∈[1,∞), we define logi+1r=logi(logr)(i∈N) and expi+1r=exp(expir)(i∈N) and exp0r=r=log0r, exp-1r=logr.For the second-order linear differential equation,(1)f′′+A(z)f=0,
where A(z) is an entire function or meromorphic function of finite order. In 1982, Bank and Laine [5] mainly studied the distribution of zeros of solutions of (1) when A is an entire function of finite order.Obviously, all solutions of (1) are entire when A(z) is entire. However, there are some immediate difficulties when A is meromorphic; for example, the solutions of (1) may not be entire, and it is possible that no solution of (1) except the zero solution is single-valued on the plane. In 1983, Bank and Laine [6] investigated the exponent of convergence of zero sequence of nontrivial solutions of (1), when A is meromorphic function and obtained the results as follows.Theorem 1 (see [6, Theorem  5]).
LetA be a transcendental meromorphic function of order σ(A), where 0<σ(A)≤∞. Assume that λ¯(A)<σ(A). Then, if f≢0 is a meromorphic solution of (1), one has
(2)σ(A)≤max{λ¯(f),λ¯(1f)}.Theorem 2 (see [6, Theorem  6]).
LetA be a transcendental meromorphic function. Assume that (1) possesses two linearly independent meromorphic solutions f1 and f2 satisfying λ¯(f1)<∞, λ¯(f2)<∞. Then, any solution f≢0 of (1) which is not a constant multiple of either f1 or f2 satisfies
(3)max{λ¯(f),λ¯(1f)}=∞,
unless all solutions of (1) are of finite order. In the special case where λ¯(1/A)<∞, one can conclude that λ¯(f)=∞ unless all solutions of (1) are of finite order.After their work, many authors have investigated the growth and the exponent of convergence of zero sequence of non-trivial solutions of (1) and obtained many classical results (see [5, 7, 8]).In 1998, Kinnunen [9] further investigated the oscillation results of entire solutions of (1) when A(z) is an entire function with finite iterated order and obtained some theorems which improved some theorems given by Bank and Laine [5]. Later, Chen [10] and Wang and Lü [11] studied the oscillation of solutions of (1) when A(z) is a meromorphic function with finite order by using the Wiman-Valiron theory; they obtained some results which extend some theorems of Kinnunen [9]. In 2007, Liang and Liu [12] considered the complex oscillation on (1) when A(z) is a meromorphic function with many finite poles. They extended the above oscillation results by using the method of the Wiman-Valiron theory. Although the Wiman-Valiron theory is a powerful tool to investigate entire solutions, it is only useful for the meromorphic function A(z) with the exponent of the convergence sequence of poles which is less than the order of A(z) if we consider (1). In 2010, Cao and Li [13] made use of a result due to Chiang and Hayman [14] instead of the Wiman-Valiron theory and obtained four oscillation theorems and three corollaries which extended the above results due to Bank-Laine, Kinnunen, and Liang and Liu.Theorem 3 (see [13, Theorem  1.6]).
LetA(z) be a meromorphic function with 0<i(A)=n<∞, and assume that λ¯n(A)<σn(A)≠0. Then, if f is a nonzero meromorphic solution of (1), one has σn(A)≤max{λ¯n(f),λ¯n(1/f)}. In the special case where either δ(∞,f)>0 or the poles of f are of uniformly bounded multiplicities, one can conclude that max{λn+1(f),λn+1(1/f)}≤σn+1(A)≤max{λ¯n(f),λ¯n(1/f)}.Theorem 4 (see [13, Theorem  1.7]).
LetA be a meromorphic function with 0<i(A)=n<∞. Assume that (1) possesses two linearly independent meromorphic solutions f1 and f2. Denote E∶=f1f2. If λ¯n(E)<∞, then any nonzero solution f of (1) which is not a constant multiple of either f1 or f2 satisfies λ¯n(f)=∞, unless all solutions of (1) are of finite iterated n-order. In the special case where δ(∞,A)>0  iλ(1/A)<n, or λn(1/A)<σn(A) (e.g., A is an entire function), one can conclude that λ¯n=∞.Theorem 5 (see [13, Theorem  1.9]).
LetA be a meromorphic function with 0<i(A)=n<∞. Assume that f1 and f2 are two linearly independent meromorphic solutions of (1) such that max{λn(f1),λn(f2)}<σn(A). Let Π(z)≢0 be any meromorphic function for which σn(Π)<σn(A). Let g1 and g2 be two linearly independent solutions of the differential equation g′′+(A(z)+Π(z))g=0. Denote E∶=f1f2 and F∶=g1g2. If
(4)max{iλ(1E),iλ(1F)}<normax{λn(1E),λn(1F)}<λn(A),
then max{λn(g1),λn(g2)}≥σn(A).In 1976, Juneja and his coauthors [15, 16] firstly introduced the concept of [p,q]-order of entire functions. Recently, Belaïdi [17] and Liu et al. [18] investigated the growth of solutions of complex differential equations
(5)f(n)+An-1(z)f(n-1)+⋯+A1(z)f′+A0(z)f=0,
where Aj(z), (j=0,1,…,n-1) are entire or meromorphic functions with finite [p,q]-order, by using the idea of [p,q]-order, and obtained some interest results which improved and extended some previous theorems given by [5–7, 9, 19].Thus, it is interesting to consider the complex oscillation on the meromorphic solutions of (1) for the case when A is entire or meromorphic functions in the terms of the idea of [p,q]-order.In this paper, we further investigated the complex oscillation of meromorphic solutions of (1) when A(z) is meromorphic by using the idea of [p,q]-order. To state our theorems, we first introduce the concepts of entire functions of [p,q]-order (see [15, 16, 18]). Throughout this paper, we always assume that p, q are positive integers satisfying p≥q≥1.Definition 6.
Iff(z) is a transcendental entire function, the [p,q]-order of f(z) is defined by
(6)σ[p,q](f)=limsupr→∞logp+1M(r,f)logqr=limsupr→∞logpT(r,f)logqr.Remark 7.
Iff(z) is a polynomial, then σ[p,q](f)=0 for any p≥q≥1. By Definition 6, we have that σ[1,1](f)=σ1(f)=σ(f), σ[p+1,1](f)=σp(f).Remark 8.
Iff(z) is an entire function satisfying 0<σ[p,q](f)<∞, then(i)
σ[p-n,q](f)=∞(n<p), σ[p,q-n](f)=0(n<q), σ[p+n,q+n](f)=1(n<p) for n=1,2,….(ii)
if[p′,q′] is any pair of integers satisfying q′=p′+q-p and p′<p, then σ[p′,q′](f)=0 if 0<σ[p,q](f)<1 and σ[p′,q′](f)=∞ if 1<σ[p,q](f)<∞;(iii)
ρ[p′,q′](f)=∞ for q′-p′>q-p and ρ[p′,q′](f)=0 for q′-p′<q-p.Definition 9.
A transcendental meromorphic functionf(z) is said to have index-pair [p,q] if 0<σ[p,q](f)<∞ and σ[p-1,q-1](f) is not a nonzero finite number.Definition 10.
Letf1,f2 be two entire functions such that σ[p1,q1](f1)=σ1, σ[p2,q2](f2)=σ2 and p1≤p2. Then the following results about their comparative growth can be easily deduced.(i)
Ifp2-p1>q2-q1, then the growth of f1 is slower than the growth of f2;(ii)
Ifp2-p1<q2-q1, then f1 grows faster than f2.(iii)
Ifp2-p1=q2-q1>0, then the growth of f1 is slower than the growth of f2 if σ2≥1 while the growth of f1 is faster than the growth of f2 if ρ2<1.(iv)
Letp2-p1=q2-q1=0; then f1, f2 are of the same index-pair [p1,q1]. If σ1>σ2, then f1 grows faster than f2, and if σ1<σ2, then f1 grows slower than f2. If σ1=σ2, Definition 6 does not give any precise estimate about the relative growth of f1 and f2.Definition 11 (see [15, 16, 18]).
The[p,q] exponent of convergence of the zero sequence and the [p,q] exponent of convergence of the distinct zero sequence of f(z) are defined respectively, by
(7)λ[p,q](f)=limr→∞¯logpn(r,1/f)logqr=limr→∞¯logpN(r,1/f)logqr,λ¯[p,q](f)=limr→∞¯logpn¯(r,1/f)logqr=limr→∞¯logpN¯(r,1/f)logqr.Remark 12.
It is easy to know that(8)λ¯[p,q](f)≤λ[p,q](f)≤σ[p,q](f).Now, we will show our main results on the complex oscillation on meromorphic solutions of (1) when A(z) is meromorphic with finite [p,q]-order as follows.Theorem 13.
LetA(z) be a transcendental meromorphic function with σ[p,q](A)>0. Assume that λ¯[p,q](A)<σ[p,q](A). Then, if f is a nonzero meromorphic solution of (1), one has
(9)σ[p,q](A)≤max{λ¯[p,q](f),λ¯[p,q](1f)}.
In the special case where either δ(∞,f)>0 or the poles of f are of uniformly bounded multiplicities, one can get that
(10)max{λ[p+1,q](f),λ[p+1,q](1f)}≤σ[p+1,q](A)≤max{λ¯[p,q](f),λ¯[p,q](1f)}.Theorem 14.
LetA(z) be a transcendental meromorphic function with σ[p,q](A)∶=σ(0<σ<+∞). Assume that (1) possesses two linearly independent meromorphic solutions f1 and f2. Denote E∶=f1f2. If λ¯[p,q](E)<∞, then any nonzero solution f of (1) which is not a constant multiple of either f1 or f2 satisfies λ¯[p,q](f)=∞, unless all solutions of (1) are of finite [p,q]-order. In the special case where δ(∞,A)>0 or λ[p,q](1/A)<σ[p,q](A), one can conclude that λ¯[p,q]=∞.Theorem 15.
LetA(z) be a transcendental meromorphic function with σ[p,q](A)>0. Assume that (1) possesses two linearly independent meromorphic solutions f1 and f2. Denote E∶=f1f2. If δ(∞,A)>0 or λ¯[p,q](1/A)<σ[p,q](A), and if either δ(∞,f)>0 or the poles of f are of uniformly bounded multiplicities, then one has
(11)λ¯[p+1,q](E)=λ[p+1,q](E)=σ[p+1,q](E)=max{λ¯[p+1,q](f1),λ¯[p+1,q](f2)}≤λ[p+1,q](f1)=λ[p+1,q](f2)=λ[p,q](A).Theorem 16.
LetA be a meromorphic function with σ[p,q](A)>0. Assume that f1 and f2 are two linearly independent meromorphic solutions of (1) such that
(12)max{λ[p,q](f1),λ[p,q](f2)}<σ[p,q](A).
Let Π(z)≢0 be any meromorphic function for which σ[p,q](Π)<σ[p,q](A). Let g1 and g2 be two linearly independent solutions of the differential equation
(13)g′′+(A(z)+Π(z))g=0.
Denote E∶=f1f2 and F∶=g1g2. If
(14)max{λ[p,q](1E),λ[p,q](1F)}<λ[p,q](A),
then max{λ[p,q](g1),λ[p,q](g2)}≥σ[p,q](A).Remark 17 (Following Hayman [20]).
we will use the abbreviation “n.e.” (nearly everywhere) to mean “everywhere in(0,+∞) except in a set of finite measure” in the proofs of our main results of this paper.Remark 18.
Obviously, Theorems13–16 are the improvement of Theorems 3–5 given by Cao and Li [13].


## 2. Some Lemmas

For the proof of our results we need the following lemmas.Lemma 19 (see [20, Theorem  4]).
Letf(z) be a transcendental meromorphic function not of the form eαz+β. Then
(15)T(r,ff′)≤3N¯(r,f)+7N¯(r,1f)+4N¯(r,1f′′)+S(r,ff′).
Using the same proof of Remark  1.3 in [9], one can easily prove the following lemma.Lemma 20.
Letf(z) be a meromorphic function of [p,q]-order. Then
(16)σ[p,q](f)=σ[p,q](f′).Lemma 21.
Letf(z) be a meromorphic function with [p,q]-order and σ[p,q](f)=σ, and let k≥1 be an integer. Then for any ε>0,
(17)m(r,f(k)f)=O{expp-1{(σ+ε)logqr}}
holds outside of an exceptional set E1 of finite linear measure.Proof.
Letk≥1. Since σ=σ[p,q](f)<∞, we have for all sufficiently large r(18)T(r,f)<expp{(σ+ε)logqr}.
By the lemma of the logarithmic derivative, we have
(19)m(r,f(k)f)=O{logT(r,f)+logr},(r∉E1),
where E1⊂(1,∞) is a set of finite linear measure, not necessarily the same at each occurrence. Hence we have
(20)m(r,f′f)=O{expp-1{(σ+ε)logqr}},(r∉E1).
Next, assume that we have
(21)m(r,f(k)f)=O{expp-1{(σ+ε)logqr}},(r∉E1),
for some k∈N. Since N(r,f(k))≤(k+1)N(r,f), it holds that
(22)T(r,f(k))≤m(r,f(k))+N(r,f(k))≤m(r,f(k)f)+m(r,f)+(k+1)N(r,f)≤(k+1)T(r,f)+O{expp-1{(σ+ε)logqr}},iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii(r∉E1).
By (20), we again obtain
(23)m(r,f(k+1)f(k))=O{expp-1{(σ+ε)logqr}},(r∉E1),
and hence, for sufficiently large r∉E1,
(24)m(r,f(k+1)f)≤m(r,f(k+1)f(k))+m(r,f(k)f)=O{expp-1{(σ+ε)logqr}}.Using the same proof of Lemma  3.6 in [19], we can easily prove the following lemma.Lemma 22.
LetΦ(r) be a continuous and positive increasing function, defined for r on (0,+∞), with [p,q]-order σ[p,q](Φ)=limsupr→+∞(logpΦ(r)/logqr). Then for any subset E of [0,+∞) that has finite linear measure, there exists a sequence {rm}, rm∉E such that
(25)σ[p,q](Φ)=limrm→+∞logpΦ(rm)logqrm.Lemma 23.
Letg1(z) and g2(z) be two entire functions of [p,q]-order, and denote F=g1g2. Then
(26)λ[p,q](F)=max{λ[p,q](g1),λ[p,q](g2)}.Proof.
Letn(r,F) denote the number of the zeros of E(z) in disk={z:|z|≤r} and so on for g1 and g2. Since, for any given r>0, we have n(r,F)≥n(r,g1) and n(r,F)≥n(r,g2), thus by Definition 11, we have
(27)λ[p,q](F)≥max{λ[p,q](g1),λ[p,q](g2)}.
On the other hand, since the zero of E(z) must be the zero of g1 or g2, then for any given r>0, we have
(28)n(r,F)≤n(r,g1)+n(r,g2)≤2max{n(r,g1),n(r,g2)}.
Therefore, by Definition 11, we have
(29)λ[p,q](F)≤max{λ[p,q](g1),λ[p,q](g2)}.
Thus we complete the proof of Lemma22.Lemma 24.
A meromorphic functionf(z) with [p,q] index can be represented by the form
(30)f(z)=U(z)eg(z)V(z),
where U(z), V(z), and g(z) are entire functions such that
(31)λ[p,q](f)=λ[p,q](U)=σ[p,q](U),λ[p,q](1f)=λ[p,q](V)=σ[p,q](V),σ[p,q](f)=max{σ[p,q](U),σ[p,q](V),σ[p,q](eg)}.Proof.
By using the essential part of the factorization theorem for meromorphic function of finite[p,q]-order and similar to the proof of Lemma  1.8 in [9], we can get the conclusions of this lemma easily.Lemma 25 (see [14, Theorem  6.2]).
Letf be a meromorphic solution of
(32)f(k)+Ak-1(z)f(k-1)+⋯+A1(z)f′+A0(z)f=0,
where A0,…,Ak-1 are meromorphic functions in the plane ℂ. Assume that not all coefficients Aj are constants. Given a real constant γ>1, and denoting T(r)∶=∑j-0k-1‍T(r,Aj), one has
(33)logm(r,f)<T(r){(logr)logT(r)}γ,ift=0,logm(r,f)<r2t+γ-1T(r){logT(r)}γ,ift=0,
outside of an exceptional set Et with ∫Et‍lt-1dl<+∞.Remark 26.
From the above lemma, we can see thatt=1 corresponds to Euclidean measure and t=0 to logarithmic measure.Using the above lemma, we can get the following lemma.Lemma 27.
LetA0,A1,…,Ak-1 be meromorphic functions with [p,q] index and 0≤σ[p,q](Aj)<∞(j=0,1,…,k-1). If f is a meromorphic solution of (20) whose poles are of uniformly bounded multiplicities or δ(∞,f)>0, then σ[p+1,q](f)≤max{σ[p,q](Aj):j=0,1,…,k-1}.Proof.
Firstly, suppose thatσ[p,q](f)<∞; then we have σ[p+1,q](f)=0. Since 0≤σ[p,q](Aj)<∞(j=0,1,…,k-1), then we have σ[p+1,q](f)≤max{σ[p,q](Aj):j=0,1,…,k-1}. Second, suppose that σ[p,q](f)=∞. From (32), we know that the poles of f(z) can only occur at the poles of A0,A1,…,Ak-1. Since the multiplicities of poles of f are uniformly bounded, we have
(34)N(r,f)≤K1N¯(r,f)≤K1∑j=0k-1‍N¯(r,Aj)≤K0max{N(r,Aj):j=0,1,…,k-1},
where K0, K1 are some suitable positive constants. Thus, we can get
(35)T(r,f)=m(r,f)+O(max{N(r,Aj):j=0,1,…,k-1}).
Set δ(∞,f)∶=α1>0; for sufficiently large r, we have
(36)m(r,f)≥α12T(r,f).
From Lemma 24 and (35) or (36), we obtain
(37)logT(r,f)≤logm(r,f)+O(logT(r))≤O{T(r){(logr)logT(r)}γ}
or
(38)logT(r,f)≤(2α1m(r,f))≤O{T(r){(logr)logT(r)}γ}
outside of an exceptional set E0 with finite logarithmic measure.
Using a standard method to deal with the finite logarithmic measure set, one immediately gets from the previous inequalities thatσ[p+1,q](f)≤max{σ[p,q](Aj):j=0,1,…,k-1}.
Thus, we complete the proof of this lemma.Lemma 28.
LetA be a meromorphic function with [p,q] index, and let f be a nonzero meromorphic solution of (1). Then(i)
if eitherδ(∞,f)>0 or the poles of f are of uniformly bounded multiplicities, then σ[p+1,q](f)≤σ[p,q](A);(ii)
ifδ(∞,A)>0 or λ[p,q](1/A)<σ[p,q](A), then σ[p+1,q](f)≥σ[p,q](A).Proof.
Suppose thatf is a nonzero meromorphic solution of (1). It is easy to see that (i) holds since (i) is just a special case of Lemma 25.
Next, we assume thatA satisfies δ(∞,A)>0 or λ[p,q](1/A)<σ[p,q](A). By (1), we have
(39)-A(z)=f′′f.
By (39) and Lemma 21, we can get that
(40)m(r,A)≤m(r,f′′f)=O{log(rT(r,f))}
holds for all sufficiently large r∉E, where E⊂(0,+∞) has finite linear measure. Hence
(41)T(r,A)=m(r,A)+N(r,A)≤N(r,A)+O{log(rT(r,f))}
holds for all sufficiently large |z|=r∉E.
Ifσ[p,q](A)=0, since A is a meromorphic function with [p,q] index, by Definition 6 and Lemma 22, there exists a sequence {rm} such that, for all rm∉E3,
(42)T(rm,A)≥expp-1{Mlogqrm}
holds for any sufficiently large constant M>0. If σ[p,q](A)=σ>0, by Lemma 22 there exists a sequence {rm} such that, for all rm∉E3,
(43)T(rm,A)≥expp{(σ-ε)logqrm}
holds for any given ε(0<ε<σ).
We will consider two cases as follows.
Case  1. Suppose that δ(∞,A)∶=α2>0. Then for sufficiently large r, we have
(44)α22T(r,A)≤m(r,A)=O{log(rT(r,f))}.
If σ[p,q](A)=0 and A has [p,q] index, from (42) and (44), we have σ[p+1,q](f)≥0=σ[p,q](A). If σ[p,q](A)>0, then from (43) and (44), we can get that σ[p+1,q](f)≥σ[p,q](A).
Case  2. Suppose that λ[p,q](1/A)<σ[p,q](A)=σ. Then the inequality
(45)N(r,A)≤expp{(λ[p,q]1A+ε)logqr}
holds for any given ε(0<2ε<σ-λ[p,q](1/A)). Then from (41), (43), and (45), we can get that σ[p+1,q](f)≥σ[p,q](A).
Thus, we complete the proof of Lemma28.


## 3. Proof of Theorem13

Sinceσ[p,q](A)∶=σ>0 and f is a solution of (1), it is easy to see that f can not be rational nor be of the form eaz+b for constants a and b. Thus, by Lemma 19 we have
(46)T(r,ff′)=O(N¯(r,f)+N¯(r,1f)+N¯(r,1f′′)),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiin.e.asr⟶∞.
From (1), we have
(47)N¯(r,1f′′)≤N¯(r,1f)+N¯(r,1A).
Suppose that (9) fails to hold that is
(48)σ[p,q](A)>max{λ¯[p,q](f),λ¯[p,q](1f)};
by the assumption λ¯[p,q](A)<σ[p,q](A) and from (46)–(48), we can get σ[p,q](f/f′)<σ[p,q](A).Setψ=f′/f; by the first main theorem, we can get that
(49)σ[p,q](ψ)<σ[p,q](A).
From (1) and ψ=f′/f, we can get -A=ψ′+ψ2. Thus, from (49) and Lemma 20, we have σ[p,q](A)≤σ[p,q](ψ)<σ[p,q](A), a contradiction. Thus, (9) is true.In the special case where eitherδ(∞,f)>0 or the poles of f are of uniformly bounded multiplicities, by Lemma 28 we have
(50)max{λ[p+1,q](1f),λ[p+1,q](f)}≤σ[p+1,q](f)≤σ[p,q](A).
Combining the above discussions, we can get (10).Thus, this completes the proof of Theorem13.


## 4. Proof of Theorem14

Suppose that (1) possesses two linearly independent meromorphic solutions f1 and f2 such that λ¯[p,q](E)<∞, where E1∶=E=f1f2. Let f=af1+bf2, where a and b are nonzero constants, and set E2∶=ff1. From (1), we can see that any pole of f is a pole of A. Since λ¯[p,q](1/A)≤σ[p,q](A), and by the assumptions of Theorem 14, we have λ¯[p,q](1/f)≤λ¯[p,q](1/A)<∞. If λ¯[p,q](f)<∞, from the above discussion, we can get that λ¯[p,q](E1)<∞ and λ¯[p,q](E2)<∞. By Lemma D(e) in [6], there exists a constant c>0 such that n.e. as r→∞,
(51)T(r,Ej)=O(N¯(r,1Ej)+T(r,A)+logr)=O(expp(clogqr)+T(r,A))
for j=1,2. Since E2=af12+bE1, from (51) we can get that n.e. as r→∞,
(52)T(r,f1)=O(expp(clogqr)+T(r,A)).
From A=-(f′′/f) and Lemma 21, we can get
(53)m(r,A)=O(log(rT(r,f))),n.e.r⟶∞.
And from (1), we can see that any pole of A is at most double and is either a zero or pole of f. Then we get
(54)N(r,A)≤2(N¯(r,1f)+N¯(r,f)).
By assumptions λ¯[p,q](f)<∞, λ¯[p,q](1/f)<∞ and (54), we can get that N(r,A)=O(expp(dlogqr)) as r→∞ for some d>0. Together with (52), (53), and (54), we can get that T(r,f1)=O(expp(dlogqr)) n.e. as r→∞. Thus, it follows that f1 is of finite [p,q]-order.By the identity of Abel, we have(55)(f2f1)′=Wf12,
where W is equal to the Wronskian of f1 and f2. Hence, by Lemma 20 and (55), we get
(56)σ[p,q](f2)=σ[p,q](f1f2f1)≤max{σ[p,q](f2f1),σ[p,q](f1)}=σ[p,q](f1).Reversing the roles off1 and f2, we can get that σ[p,q](f1)=σ[p,q](f2). Thus, we can get that all solutions of (1) are of finite [p,q]-order if λ¯[p,q](f)<∞.In special case whereδ(∞,A)>0 or λ[p,q](1/A)<σ[p,q](A), by Lemma 28, we can get that all meromorphic solutions f≢0 of (1) satisfy σ[p+1,q](f)≥σ=σ[p,q](A). Hence, we can obtain that λ¯[p,q](f)=∞ holds for any solution f≢0 of (1) which is not a constant multiple of either f1 or f2.


## 5. Proof of Theorem15

From [6, Page 664], we can get σ[p+1,q](f1)=σ[p+1,q](f2) easily. Suppose that δ(∞,A)∶=α3>0 or λ[p,q](1/A)<σ[p,q](A) and that either δ(∞,f)>0 or the poles of f are of uniformly bounded multiplicities. Then by Lemma 28 we obtain
(57)σ[p+1,q](E)≤max{σ[p+1,q](f1),σ[p+1,q](f2)}≤σ[p+1,q](f1)=σ[p+1,q](f2)=σ[p,q](A)<∞.
By Lemma D(e) in [6], there a constant c>0 such that n.e. as r→∞,
(58)T(r,E)=O(N¯(r,1E)+T(r,A)+logr)=O(expp(clogqr)+T(r,A)).
By Lemmas 21 and 28, we have
(59)m(r,A)=m(r,f1′′f1)=O(log(rT(r,f1)))=O(expp(β1logqr))
for some β1<∞ outside of a possible exceptional set E4⊂[0,∞) with finite linear measure. If δ(∞,A)∶=α3>0, then for sufficiently large r, we have
(60)α32T(r,A)≤m(r,A)=O(expp(β1logqr)),r∉E4.
If λ[p,q](1/A)<σ[p,q](A)<∞, there exists a constant β2<∞ such that
(61)N(r,A)=O(expp(β2logqr)).
From the above equality and (60), we have
(62)T(r,A)=m(r,A)+N(r,A)=O(expp(βlogqr)),
where β=max{β1,β2}.Therefore, together with (58) and either (60) or (62), we obtain
(63)T(r,E)=O(N¯(r,1E)+expp(βlogqr)),r∉E4.Suppose thatλ¯[p+1,q](E)<σ[p+1,q](E)∶=σ1; then from Definitions 6 and 11 we have N¯(r,1/E)=O(expp+1(blogqr)) for some b<σ1. From (63), T(r,E)=O(expp+1(blogqr)), r∉E4, and then by standard reasoning, we obtain σ[p+1,q](E)≤b<σ1=σ[p+1,q](E). Thus, we get a contradiction. Therefore, we have λ¯[p+1,q](E)≥σ1=σ[p+1,q](E). And since σ[p+1,q](E)≥λ[p+1,q](E)≥λ¯[p+1,q](E), then we have σ[p+1,q](E)=λ[p+1,q](E)=λ¯[p+1,q](E).By Lemma D(a) in [6], f1 and f2 have no common zeros. Let fj=(gj/dj)(j=1,2), where gj and dj have no common zeros. This implies that g1 and g2 have no common zeros, that λ[p,q](fj)=λ[p,q](gj) for j=1,2, and that λ[p,q](E)=λ[p,q](g1g2). Then by Lemma 23, we have λ[p+1,q](E)=max{λ[p+1,q](f1),λ[p+1,q](f2)}.Thus, we can get the following conclusion(64)λ¯[p+1,q](E)=λ[p+1,q](E)=σ[p+1,q](E)=max{λ¯[p+1,q](f1),λ¯[p+1,q](f2)}≤λ[p+1,q](f1)=λ[p+1,q](f2)=λ[p,q](A)<∞.
Therefore, we complete the proof of Theorem 15.


## 6. Proof of Theorem16

FromE∶=f1f2 and F∶=g1g2, by using a similar argument as in [9, Lemma  1.7], we can get λ[p,q](F)=max{λ[p,q](g1),λ[p,q](g1)}. Suppose that λ[p,q](F)<λ[p,q](A)∶=σ1 from (12), we have
(65)N¯(r,1E)=O(expp(β1logqr))
for some β1<σ[p,q](A)=σ1. Since λ[p,q](A)∶=σ1>0 and from Definition 6, for any ε(>0), we have
(66)T(r,A)=O{expp((σ1+ε)logqr)}.
By Lemma D(e) in [6], we have T(r,E)=O(N¯(r,1/E)+T(r,A)+logr). Thus, we can get that
(67)T(r,E)=O{expp((σ1+ε)logqr)}.
Hence, from the above equality, we have σ[p,q](E)≤σ1. On the other hand, by Lemma B(iv) in [6], we have
(68)4A=(E′E)2-2E′′E-1E2,
which implies that σ1≤σ[p,q](E). Since σ[p,q](Π)<σ[p,q](A), using the same argument as in the above for the function F, we have σ[p,q](E)=σ[p,q](F)=σ[p,q](A)=σ1.From the assumptions of Theorem16 and Lemma 24, we can write
(69)E=U1eG1V1,F=U2eG2V2,
where σ[p,q](U1)=λ[p,q](E)<σ[p,q](A) and σ[p,q](U2)=λ[p,q](F)<σ[p,q](A). And since max{λ[p,q](1/E),λ[p,q](1/F)}<λ[p,q](A), we have
(70)σ[p,q](eG1)=σ[p,q](eG2)=σ[p,q](A).Substituting (69) into (68), and using the same argument as in the proof of Theorem  3.1 in [7], we can get
(71)ce2(G1-G2)=-V12U22U12V22,
where c≠0. From (68), we have
(72)E2F2=U12V22U22V12e2(G1-G2)=-1c.
For the function F, similar to (68), we have
(73)4(A+Π)=(F′F)2-2F′′F-1F2.
From (68), (72), and (73), we have
(74)4(A+Π+1cA)=(F′F)2-2F′′F+1c(E′E)2-2cE′′E.Sinceσ[p,q](E)=σ[p,q](F)=σ[p,q](A)=σ1, by Lemma 22, for any ε(>0), we can get
(75)T(r,A(1+1c)+Π)=m(r,A(1+1c)+Π)+N(r,A(1+1c)+Π)=O(expp-1{(σ1+ε)logqr})n.e.asr⟶∞.
From (75), we can get σ[p,q](A(1+(1/c))+Π)=0<σ1=σ[p,q](A) easily. Thus, we can get c=-1. Since  E2=F2,  we have
(76)E′E=F′F,E′′E=F′′F.
Then from (68) and (73), we have Π≡0, a contradiction.Therefore, we complete the proof of Theorem16.



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*Source: 243873-2013-12-16.xml*