Growth of Solutions to Higher Order Dif f erential Equation with Meromorphic Coeffi cients

WANG LI-JUN AND LIU HUI-FANG

(College of Mathematics and Information Science,Jiangxi Normal University, Nanchang,330022)

Growth of Solutions to Higher Order Dif f erential Equation with Meromorphic Coeffi cients

WANG LI-JUN AND LIU HUI-FANG*

(College of Mathematics and Information Science,Jiangxi Normal University, Nanchang,330022)

Communicated by Shi Shao-yun

In this paper,we study the growth of solutions of higher order dif f erential equation with meromorphic coeffi cients,and f i nd some conditions which guarantee that its every nontrivial solution is of inf i nite order.

meromorphic function,dif f erential equation,order of growth

1 Introduction and Main Results

In this paper,we adopt the standard notations and the fundamental results of the Nevanlinna’s value distribution theory of meromorphic functions as explained in[1]–[3].In addition,we use notationsσ(f)andσ2(f)to denote the order and the hyper-order of a meromorphic functionf,λ(f)and¯λ(f)to denote the exponent of convergence of the zero-sequence and the distinct zero-sequence offrespectively.

Consider the dif f erential equation

whereB(z)is an entire function.Frei[4]proved that ifBis a constant,then(1.1)has a solution of f i nite order if and only ifB=−n2,wherenis an integer.IfBis a nonconstant polynomial orBis transcendental ofσ(B)1,Ozawa[5],Langely[6]and Gundersen[7]proved that all nontrivial solutions of(1.1)have inf i nite order.Thus there arises the following question:IfBis transcendental ofσ(B)=1,when does(1.1)have a solution of inf i niteorder?Lots of work have been done on this direction(see[8]–[12],etc.).

In[8],Chen proved the following result,which improved results of Frei[4],Ozawa[5], Langley[6]and Gundersen[7].

is of inf i nite order.

Recently,the authors in[9]extended Theorem A to some higher order dif f erential equations

and proved the following result.

Theorem B[9]Let k≥2be an integer,A1,A2,Bj,Dj(j=1,···,k−1)be nonzero entire functions of orders less than1and a1,a2,bj,dj(j=1,···,k−1)be nonzero complex numbers such that a12and bj＜0.Suppose that there exists αj∈(0,1),βj∈(0,1)such that dj=αja1+βja2.Set α=If

(1)arga1/π andarga1/arga2;or

(2)arga1arga1=arga2and

(ii)|a2|＜(1−α)|a1|;or

(3)a1＜0andarga1arga2;or

then every solutionof(1.2)satisf i es σ(f)=+∞and σ2(f)=1.

In Theorems A and B,the authors considered the case that all coeffi cients of the above equations are entire functions.In this paper,we are concerned with the more general case. In fact,we consider the following dif f erential equation

with meromorphic coeffi cients,and obtain the following results.

Theorem 1.1Let Pj(z)=ajnzn+···+aj0,Qj(z)=bjnzn+···+bj0(j=0,···,k−1)be polynomials with degree n(≥1)such that a0n/b0nand ajn＜0,bjn=αja0n+βjb0n(0＜αj＜1,0＜βj＜1,j=1,···,k−1).And let Aj,Bj(j=0,···,k−1)be meromorphic functions with orders less than n and A0B00.If one of the following cases occurs:

(i)arga0nargb0n;

(ii)arga0n=argb0nπ and

(iii)arga0n=argb0nπ and

then every nontrivial meromorphic solution f,whose poles are of uniformly bounded multiplicities,of(1.3)satisf i es σ(f)=+∞and σ2(f)=n.

From Theorem 1.1,we obtain the following result.

Corollary 1.1Under the hypothesis of Theorem1.1,if Aj,Bj(j=0,···,k−1)are entire functions,then every nontrivial solution f of(1.3)satisf i es σ(f)=+∞and σ2(f)=n.

Remark 1.1It is obvious that Corollary 1.1 is an improvement of Theorem B.

By using the similar argument to that of Theorem 1.1,we also obtain the following result:

Theorem 1.2Let B1,B2,Aj(j=1,···,k−1)be entire functions with orders less than n and B1B20,b1,b2be nonzero complex numbers such that b12.If

(i)at least one ofargb1,argb2is not equal to π;or

(ii)argb1=argb2=π,b1＜−1,

then every solutionof the dif f erential equation

satisf i es σ(f)=+∞and σ2(f)=n.

Remark 1.2It is obvious that Theorem 1.2 improves results of Frei[4],Ozawa[5], Langley[6]and Gundersen[7].

2 Lemmas

Lemma 2.1[13]Let f1(z),···,fn(z)(n≥2)be meromorphic functions and g1(z),···, gn(z)be entire functions satisfying the following conditions:

(ii)For1≤j＜k≤n,gj(z)−gk(z)are not constants;

(iii)For1≤j≤n,1≤t＜k≤n,T(r,fj)=o{T(r,egt−gk)}(),where E⊂(1,∞)is a set of f i nite linear measure.Then fj(z)≡0(j=1,···,n).

Lemma 2.2[14]Let f(z)be a transcendental meromorphic function,and α＞1be a given constant.Then there exists a set H⊂(1,∞)that has f i nite logarithmic measure,and a constant B＞0depending only on α and k,j(k＞j≥0),such that for all z with |z|=[0,1]∪H,we have

Lemma 2.3[15]−[16]Let f be a meromorphic function of f i nite order.Then for any given ε＞0,there exists a set H⊂(1,+∞)that has f i nite logarithmic measure,such that for all z with|z|=[0,1]∪H,we have|f(z)|≤exp{rσ(f)+ε}.

Lemma 2.4[17]Let Qj(z)=bjnzn+···+bj0(j=1,2,3)be polynomials with degree n(≥1),where bjl(j=1,2,3;l=0,···,n)are complex constants.Set z=reiθ,argbjn=φj∈[0,2π),δ(Qj,θ)=|bjn|cos(φj+nθ)(j=1,2,3).If φj(j=1,2,3)are distinct,then there exist two real numbers θ1,θ2(θ1＜θ2)such that for each θ∈(θ1,θ2),we have

Remark 2.1[17]For polynomialsQj(z)=bjnzn+···+bj0(j=1,2),if argb1nargb2n, then there exist two real numbersθ1,θ2(θ1＜θ2)such that for eachθ∈(θ1,θ2),we have

Lemma 2.5[8]Let P(z)=(α+iβ)zn+···(α,β are real numbers,|α|+0)be a polynomial with degree n(≥1),A(z)be a meromorphic function with σ(A)＜n.Set g(z)=A(z)eP(z),z=reiθ,δ(P,θ)=αcosnθ−βsinnθ.Then for any given ε＞0,there is a set E0⊂[0,2π)that has linear measure zero,such that for any θ∈[0,2π)(E0∪E1)and sufficiently large r,we have

(i)if δ(P,θ)＞0,then

(ii)if δ(P,θ)＜0,then

where E1={θ∈[0,2π):δ(P,θ)=0}is a f i nite set.

Lemma 2.6[18]Let A0,A1,···,Ak−1be meromorphic functions of f i nite order.Then any meromorphic solution f,whose poles are of uniformly bounded multiplicities,of the equation

satisf i es σ2(f)≤max{σ(Aj),j=0,···,k−1}.

3 Proof of Theorem 1.1

In the sequel,we useEto denote a set of linear measure zero and useHto denote a set of f i nite logarithmic measure,not necessarily the same at each occurrence.We also useM(＞0),θ1,θ2to denote any real numbers,not necessarily the same at each occurrence.

We divide two steps to complete the proof of Theorem 1.1.

Step 1.We prove that(1.3)has no nonzero rational solutionf.

Let

Then we have

Rewrite(1.3)in the form

(i)Suppose that arga0nargb0n,then at least one of arga0n,argb0nis not equalπ.Without loss of generality,we assume that arga0nπ.For eachj∈{1,···,k−1}, since argajn=π,we geta0najn.On the other hand,by arga0nargb0n,we also geta0nbjnfor eachj∈{0,1,···,k−1}.Hence,by(3.1),(3.2)and Lemma 2.1,we getφ0(z)≡0,which impliesA0(z)≡0.This is absurd.

(ii)Suppose that arga0n=argb0nπandthen we geta0najnfor eachj∈{1,···,k−1}.Since

for eachj∈{1,···,k−1},we geta0nbjn.Hence,by(3.1),(3.2)and Lemma 2.1,we getφ0(z)≡0,which impliesA0(z)≡0.This is absurd.

Step 2.We prove that every transcendental solutionf,whose poles are of uniformly bounded multiplicities,of(1.3)satisf i esσ(f)=+∞andσ2(f)=n.

By Lemma 2.6,we get

Next we proveσ2(f)≥n.By Lemma 2.2,there exists a setH⊂(1,∞)that has f i nite logarithmic measure,and a constantB＞0,such that for allzwith|z|=r/∈[0,1]∪H,we have

Set

Then we have

where Re(bjnzn)denotes the real part ofbjnzn.

(i)Suppose that arga0nargb0n.Now we discuss the following two cases.

Case 2.1 arga0nπand argb0nπ.

By argajn=π(1≤j≤k−1)and Lemma 2.4,there exist two real numbersθ1,θ2(θ1＜θ2)such that for eachθ∈(θ1,θ2)we have

wherej=1,···,k−1.Hence,by(1.3),(3.4),(3.8)–(3.11),for anyθ∈(θ1,θ2)Eand suffi ciently large,we have

which deducesσ2(f)≥n.

Hence,by(1.3),(3.4),(3.12)–(3.14),for anyθ∈(θ1,θ2)Eand suffi ciently large, we have

which deducesσ2(f)≥n.

Case 2.2 arga0n=πor argb0n=π.

Without loss of generality,we assume arga0n=π.Then by argajn=π(1≤j≤k−1) and Remark 2.1,there exist two real numbersθ1,θ2(θ1＜θ2)such that for eachθ∈(θ1,θ2), we have

Hence,by(3.5),(3.6),(3.15)and Lemma 2.5,for any givenε＞0 and anyθ∈(θ1,θ2)E, asrsuffi ciently large,we still have(3.12)–(3.14).Then by the similar argument as above, we getσ2(f)≥n.

(ii)Suppose arga0n=argb0nπandThen by argajn=π(1≤j≤k−1)and Remark 2.1,there exist two real numbersθ1,θ2(θ1＜θ2)such that for eachθ∈(θ1,θ2),we have

Note that when arga0n=argb0n,we have

Hence by(3.5),(3.6),(3.16),(3.17)and Lemma 2.5,for any givenε＞0 and anyθ∈(θ1,θ2)E,asrsuffi ciently large,we have(3.8),(3.9)and

Hence by(1.3),(3.4),(3.8),(3.9),(3.17)–(3.19),for anyθ∈(θ1,θ2)Eand suffi ciently large,we have

which deducesσ2(f)≥n.

Hence,there exist two real numbersθ1,θ2(θ1＜θ2)such that for eachθ∈(θ1,θ2),we haveδ(P0,θ)＞0 and

Then by Lemma 2.5,for any givenε＞0 and anyθ∈(θ1,θ2)E,asrsuffi ciently large,we have(3.8)and

forj=0,···,k−1.Hence,by(1.3),(3.4),(3.8),(3.22)and(3.23),for anyθ∈(θ1,θ2)Eand suffi ciently large,we have

Then by(3.24),we getσ2(f)≥n.

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tion:30D35,34M10

A

1674-5647(2017)02-0135-08

10.13447/j.1674-5647.2017.02.05

Received date:Sept.9,2015.

Foundation item:The NSF(11201195)of China,the NSF(20132BAB201008)of Jiangxi Province.

*Corresponding author.

E-mail address:925268196@qq.com(Wang L J),liuhuifang73@sina.com(Liu H F).