R E S E A R C H
Open Access
On the growth of solutions of certain
higher-order linear differential equations
Zhiqiang Mao
1*and Huifang Liu
2**Correspondence:
[email protected]; [email protected]
1School of Mathematics and
Computer, Jiangxi Science and Technology Normal University, Nanchang, 330038, China
2College of Mathematics and
Information Science, Jiangxi Normal University, Nanchang, 330022, China
Abstract
In this paper, we investigate the growth of meromorphic solutions of the equations
f(k)+Ak–1(z)f(k–1)+· · ·+A1(z)f+A0(z)f= 0,
f(k)+Ak–1(z)f(k–1)+· · ·+A1(z)f+A0(z)f=F(z),
whereA0(z) (≡0),A1(z),. . .,Ak–1(z) andF(z) (≡0) are entire functions of finite order. We
find some conditions on the coefficients to guarantee that every nontrivial meromorphic solution of such equations is of infinite order.
MSC: 30D35; 34M10
Keywords: entire function; differential equation; order
1 Introduction and main results
It is assumed that the reader is familiar with the standard notations and the fundamental results of the Nevanlinna theory [–]. Letf(z) be a nonconstant meromorphic function in the complex plane. We use notationsσ(f) andσ(f) to denote the order of growth and
the hyper-order off, which are defined by
σ(f) = lim r→∞
log+T(r,f)
logr , σ(f) =rlim→∞
log+log+T(r,f)
logr ,
respectively.
Consider the linear differential equation
f(k)+Ak–(z)f(k–)+· · ·+A(z)f+A(z)f = , (.)
where A(z) (≡ ), A(z), . . . ,Ak–(z) are entire functions, if the coefficients Aj(z) (j=
, . . . ,k– ) are polynomials, then all nontrivial solutions of (.) are of finite order (see [, ]). Ifsis the largest integer such thatAs(z) is transcendental, it is well known that (.)
has at mostslinearly independent finite-order solutions. Thus when at least one of the coefficientsAj(z) is transcendental, most of the solutions of (.) are of infinite order. In
the case when
max
≤j≤k–
σ(Aj)
<σ(A) < +∞,
Chen and Yang [] proved that all nontrivial solutions of (.) are of infinite order. In the case when
max j=d
σ(Aj)
<σ(Ad)≤
,
Hellersteinet al.[] proved that all transcendental solutions of (.) are of infinite order. While in the case whenσ(A) =σ(A) =· · ·=σ(Ak–), (.) may have a solution of finite
order.
Example . The equation
f+ez– f–ezf+ez–ezf =
has a solutionf(z) =ezofσ(f) = . Hereσ(ez– ) =σ(ez) =σ(ez–ez) = .
Thus a natural question is what conditions onAj(z) whenσ(A) =σ(A) =· · ·=σ(Ak–)
will guarantee that all nontrivial solutions of (.) are of infinite order. Concerning this question, we recall the following results as for the special case ofk= .
Theorem A(see []) Let P(z) =anzn+· · ·,Q(z) =bnzn+· · · be polynomials with degree n
(≥),h(z) (≡),h(z)be entire functions of order less than n.Ifargan=argbnor an=cbn
( <c< ),then every nontrivial solution f of the equation
f+h(z)eP(z)f+h(z)eQ(z)f=
satisfiesσ(f) =∞andσ(f)≥n.
Theorem B(see []) Let hj(z) (≡) (j= , )be entire functions of order less than,a,b
be nonzero complex numbers such thatarga=argb or a=cb(c> ).Then every nontrivial
solution f of the equation
f+h(z)eazf+h(z)ebzf = (.)
satisfiesσ(f) =∞.
From Theorems A and B, we obtain that if a=b, then every nontrivial solutionf of Eq. (.) is of infinite order. When the coefficient off is of the formh(z)eaz+h(z)eaz,
wherea=a, Peng and Chen obtained the following result.
Theorem C(see []) Let hj(z) (≡) (j= , )be entire functions of order less than,a,a
be nonzero complex numbers and a=a(suppose that|a| ≤ |a|).Ifarga=πor a< –,
then every nontrivial solution f of the equation
f+e–zf+h(z)eaz+h(z)eaz
f= (.)
In this paper, we continue to investigate the growth of solutions of higher linear differ-ential equations, which has the same form as Eq. (.), and obtain the following results which generalize Theorem C.
Theorem . Suppose that h(z) (≡),h(z),hj(z) (j= , . . . ,k– )are entire functions of order less than n,Pi(z) =ainzn+· · ·+ai,Qj(z) =bjnzn+· · ·+bj(i= , ;j= , . . . ,k– )
are polynomials with degree n(≥),where ail,bjl(i= , ;j= , . . . ,k– ;l= , . . . ,n)are complex constants.If an=an,bjn(j= , . . . ,k– )satisfies the following conditions:
(i) there exists somes∈ {, . . . ,k– }such thatargbsn=argan;
(ii) bjn=cjan( <cj< )forj∈Iandbjn=djbsn( <dj< )forj∈I,where
I∪I={, . . . ,k– } \ {s},I∩I=∅,
then every solution f (≡)of the equation
f(k)+hk–(z)eQk–(z)f(k–)+· · ·+h(z)eQ(z)f+
h(z)eP(z)+h(z)eP(z)
f= (.)
satisfiesσ(f) =∞andσ(f) =n.
From the proof of Theorem ., we also obtain the following results.
Corollary . Suppose that h(z) (≡),h(z),hj(z) (j= , . . . ,k– )are entire functions of order less than n,Pi(z) =ainzn+· · ·+ai,Qj(z) =bjnzn+· · ·+bj(i= , ;j= , . . . ,k– )
are polynomials with degree n(≥).If an=anand there exists some s∈ {, . . . ,k– }such thatargbsn=argan,bjn=djbsn( <dj< , ≤j≤k– ,j=s),then every solution f (≡) of (.)satisfiesσ(f) =∞andσ(f) =n.
Corollary . Suppose that h(z) (≡),h(z),hj(z) (j= , . . . ,k– )are entire functions of order less than n,Pi(z) =ainzn+· · ·+ai,Qj(z) =bjnzn+· · ·+bj(i= , ;j= , . . . ,k– )
are polynomials with degree n(≥).If an=anand bjn=cjan( <cj< , ≤j≤k– ), then every solution f (≡)of (.)satisfiesσ(f) =∞andσ(f) =n.
Remark . Corollaries . and . are extensions of Theorem C, because Theorem C is just the case fork= ,n= ,bn= – in Corollaries . and ..
Remark . From Theorem . and Corollaries ., ., we obtain that every solutionf
(≡) of the equation
f(k)+hk–(z)eQk–(z)f(k–)+· · ·+h(z)eQ(z)f+B(z)cos
P(z)f =
or
f(k)+hk–(z)eQk–(z)f(k–)+· · ·+h(z)eQ(z)f+B(z)sin
P(z)f=
satisfiesσ(f) =∞andσ(f) =n, whereP(z) = –iP,B(z) is an entire function of order less
Recently, Wang and Laine investigated the growth of solutions of the non-homogeneous linear differential equation
f+h(z)eazf+h(z)ebzf =F(z) (.)
corresponding to (.) and obtained the following result.
Theorem D(see []) Suppose that h(≡),h(≡),F(≡)are entire functions of order
less than,and the complex constants a,b satisfy ab= and a=b.Then every nontrivial
solution f of Eq. (.)is of infinite order.
Thus the other purpose of this paper is to investigate the growth of solutions of the non-homogeneous linear differential equation
f(k)+hk–(z)eQk–(z)f(k–)+· · ·+h(z)eQ(z)f+
h(z)eP(z)+h(z)eP(z)
f=F(z) (.)
corresponding to (.). We obtain the following results.
Theorem . Under the hypotheses of Theorem.,if bsn=an, F(z) (≡)is an entire function of order less than n,then every solution f of Eq. (.)satisfiesσ(f) =∞.
From the proof of Theorem ., we also obtain the following results.
Corollary . Suppose that h(z) (≡),h(z),hj(z) (j= , . . . ,k– ),F(z) (≡)are entire functions of order less than n,Pi(z) =ainzn+· · ·+ai,Qj(z) =bjnzn+· · ·+bj(i= , ;j=
, . . . ,k– )are polynomials with degree n(≥). If an=an,and there exists some s∈
{, . . . ,k– }such thatargbsn=argan,bsn=an,bjn=djbsn( <dj< , ≤j≤k– ,j=s), then every solution f of Eq. (.)satisfiesσ(f) =∞.
Corollary . Suppose that h(z) (≡),h(z),hj(z) (j= , . . . ,k– ),F(z) (≡)are entire functions of order less than n,Pi(z) =ainzn+· · ·+ai,Qj(z) =bjnzn+· · ·+bj(i= , ;j=
, . . . ,k– )are polynomials with degree n(≥). If an=an and bjn=cjan( <cj< ,
≤j≤k– ),then every solution f of Eq. (.)satisfiesσ(f) =∞.
Remark . Whenσ(F)≥n, (.) may have a solution of finite order.
Example . The equation
f+e–zf+ezf+ez– e–zf =ez+ zez+ ze–z+ z+ zez
has a solutionf(z) =ezofσ(f) = . HereP
(z) = z,P(z) = –z,Q(z) =z,Q(z) = –z=Qs(z)
satisfy the hypotheses of Theorem ., andσ(F) = >n= .
Example . The equation
f()+e–zf–e–zf+ezf+ez+e–zf =ez+ez+ez+e–z
has a solutionf(z) =ezofσ(f) = . HereP
(z) = z,P(z) = –z,Q(z) =z,Q(z) = –z,Q(z) =
2 Lemmas
Lemma .([]) Let fj(j= , . . . ,n)and gj(j= , . . . ,n)be entire functions such that
(i) nj=fj(z)egj(z)≡,
(ii) the order offjis less than the order ofegh–gk forn≥and≤j≤n,≤h<k≤n. Then fj(z)≡ (j= , . . . ,n).
Lemma .([]) Let fj(j= , . . . ,n+ )and gj(j= , . . . ,n)be entire functions such that
(i) nj=fj(z)egj(z)≡fn+(z),
(ii) the order offjis less than the order ofegkfor≤j≤n+ ,≤k≤n;and,furthermore, the order offjis less than the order ofegh–gk forn≥and≤j≤n+ ,≤h<k≤n. Then fj(z)≡ (j= , . . . ,n+ ).
Lemma .([]) Let f(z)be a transcendental meromorphic function ofσ(f) <∞, k, j
(k>j≥)be integers.Then,for any givenε> ,there exists a set E⊂[, π)of linear measure zero such that for all z=reiθwith|z|sufficiently large andθ∈[, π)\E,we have
ff((kj))((zz))
≤ |z|(k–j)(σ(f)–+ε).
Lemma .([, ]) Let P(z) = (α+iβ)zn+· · · (α,βare real numbers,|α|+|β| = )be a polynomial with degree n(≥),A(z) (≡)be an entire function withσ(A) <n.Set g(z) =
A(z)eP(z),z=reiθ,δ(P,θ) =αcosnθ–βsinnθ.Then,for any givenε> ,there is a set E
⊂
[, π)that has linear measure zero such that for anyθ∈[, π)\(E∪E)and sufficiently
large r,we have
(i) ifδ(P,θ) > ,then
exp( –ε)δ(P,θ)rn≤greiθ≤exp( +ε)δ(P,θ)rn; (.)
(ii) ifδ(P,θ) < ,then
exp( +ε)δ(P,θ)rn≤greiθ≤exp( –ε)δ(P,θ)rn, (.)
where E={θ∈[, π) :δ(P,θ) = }is a finite set.
Lemma .([]) Let f(z)be a transcendental meromorphic function,and letα> be a
given constant.Then there exist a set H⊂(,∞)that has a finite logarithmic measure
and a constant B> depending only onα and k, j(k>j≥)such that for all z with
|z|=r∈/[, ]∪H,we have
ff((kj))((zz))
≤B
T(αr,f)
r
logαrlogT(αr,f)
k–j
.
Lemma . Let Qj(z) =bjnzn+· · ·+bj(j= , , )be polynomials with degree n(≥),
where bjl(j= , , ;l= , . . . ,n)are complex constants.Set z=reiθ,argbjn=ϕj∈[, π), δ(Qj,θ) =|bjn|cos(ϕj+nθ) (j= , , ).Ifϕj(j= , , )are distinct,then there exist two real numbersθ,θ(θ<θ)such that for eachθ∈(θ,θ),we have
Proof Set
jk=
z=reiθ:θjk<θ<θj(k+)
(k= , , . . . , n– ,j= , , ),
whereθjk= (k– )πn– ϕj
n. Then the following results hold.
(i) δ(Qj,θjk) = (k= , , . . . , n– ,j= , , ).
(ii) δ(Qj,θ) > holds forz∈jkandkis even,δ(Qj,θ) < holds forz∈jkandkis
odd.
(iii) θj(k+)–θjk=πn (k= , , . . . , n– ,j= , , ).
Next we discuss the following four cases.
Case .|ϕi–ϕj| =π(≤i<j≤). By (ii), among the angular domainsk, we may take
an angular domain k such that forz∈k,δ(Q,θ) > . Hence whenz∈(k–)∪
(k+), we haveδ(Q,θ) < . For the sake of convenience, we write
k=
z=reiθ:α<θ<β.
Then by (iii) we write
(k–)=
z=reiθ:α–π
n <θ<α
,
(k+)=
z=reiθ:β<θ<β+π
n
.
Since|ϕi–ϕj| =π, by (i) and (iii) we know that there exist two distinct rays
argz=γ∈k, argz=γ∈k such that
δ(Q,γ) = , δ(Q,γ) = ,
respectively. Hence by (ii) we obtain thatδ(Qi,θ)δ(Qi,θ) < for anyθ∈(α,γi),θ∈
(γi,β) andi= , . Without loss of generality, we assume thatγ<γ. Now we discuss the
following four subcases.
Subcase . Ifδ(Q,θ) < whenθ∈(α,γ) andδ(Q,θ) < whenθ∈(α,γ), then we take
θ=α,θ=γ. Hence whenθ∈(θ,θ), we haveδ(Q,θ) > ,δ(Q,θ) < andδ(Q,θ) < .
Subcase . Ifδ(Q,θ) < whenθ ∈(α,γ) and δ(Q,θ) > when θ ∈(α,γ), then we
take θ=β, θ=γ+πn. Hence whenθ ∈(θ,θ), we haveδ(Q,θ) < , δ(Q,θ) > and
δ(Q,θ) < .
Subcase . Ifδ(Q,θ) > whenθ ∈(α,γ) and δ(Q,θ) < when θ ∈(α,γ), then we
take θ=γ– πn,θ=α. Hence whenθ ∈(θ,θ), we haveδ(Q,θ) < , δ(Q,θ) > and
δ(Q,θ) < .
Subcase . Ifδ(Q,θ) > whenθ∈(α,γ) andδ(Q,θ) > whenθ∈(α,γ), then we take
θ=γ,θ=β. Hence whenθ∈(θ,θ), we haveδ(Q,θ) > ,δ(Q,θ) < andδ(Q,θ) < .
Case .|ϕ–ϕ|=π. Sinceϕj∈[, π) andϕj(j= , , ) are distinct, we obtain that
By (ii), among the angular domainsk, we may take an angular domaink={z=reiθ:
θk<θ<θ(k+)}such that forz∈k,δ(Q,θ) > . Note that
δ(Q,θ) =|bn|cos(ϕ+nθ) = –|bn|cos(ϕ+nθ),
so whenz∈k, we haveδ(Q,θ) < . By (.), (i) and (iii), we know that there exists a rayargz=γ∈k such thatδ(Q,γ) = . Hence by (ii) we obtain thatδ(Q,θ) < for
θ ∈(θk,γ) orδ(Q,θ) < forθ ∈(γ,θ(k+)). Hence we may take (θ,θ) = (θk,γ) or (θ,θ) = (γ,θ(k+)) such that forθ∈(θ,θ),δ(Q,θ) > ,δ(Q,θ) < andδ(Q,θ) < .
Case .|ϕ–ϕ|=π. Then, by a similar reasoning to that of Case , we can prove the
result.
Case .|ϕ–ϕ|=π. Sinceϕj∈[, π) andϕj(j= , , ) are distinct, we obtain that
|ϕ–ϕ| =π, |ϕ–ϕ| =π. (.)
By (ii), among the angular domainsk, we may take an angular domaink={z=reiθ:
θk<θ<θ(k+)}such that forz∈k,δ(Q,θ) > . Note that
δ(Q,θ) =|bn|cos(ϕ+nθ) = –|bn|cos(ϕ+nθ),
so whenz∈k, we haveδ(Q,θ) < . By (.), (i) and (iii), we know that there exists a rayargz=γ∈k such thatδ(Q,γ) = . Hence by (ii) we obtain thatδ(Q,θ) < for
θ ∈(θk,γ) orδ(Q,θ) < forθ∈(γ,θ(k+)). Hence we may take (θ,θ) = (θk,γ) or (θ,θ) = (γ,θ(k+)) such that forθ ∈(θ,θ),δ(Q,θ) < ,δ(Q,θ) > andδ(Q,θ) < .
Remark . From the proof of Lemma ., we also obtain the following result.
LetQj(z) =bjnzn+· · ·+bj(j= , ) be polynomials with degreen(≥), wherebjl(j= , ; l= , . . . ,n) are complex constants. Setz=reiθ,argb
jn=ϕj∈[, π),δ(Qj,θ) =|bjn|cos(ϕj+ nθ). Ifϕ=ϕ, then there exist two real numbers θ,θ (θ<θ) such that for eachθ ∈
(θ,θ), we have
δ(Q,θ) > and δ(Q,θ) < .
Lemma .([]) Let g: (,∞)→R and h: (,∞)→R be monotone nondecreasing
func-tions such that g(r)≤h(r)outside of an exceptional set H of finite logarithmic measure.
Then,for anyα> ,there exists r> such that g(r)≤h(rα)holds for all r>r.
Lemma .([]) Let Aj(z) (j= , . . . ,k– )be entire functions of finite order.If f(z) (≡) is a solution of the equation
f(k)+Ak–f(k–)+· · ·+Af= ,
Lemma .([]) Let f(z)be an entire function andβ> .If G(z) = log+||zf|(βk)(z)|is unbounded entire functions of order less thann. Now we discuss the relations between the coefficients of the termznof polynomialsd
jlQs,Qs,cjqP,PandP.
whereBlare entire functions of order less thann. By the hypothesis of Theorem ., we know that the coefficients of the termznof polynomialsd
jlQs,Qs,cjqP,P are distinct.
whereDsis an entire function of order less thann. By the hypothesis of Theorem ., we
know that the coefficients of the termznof polynomialsd
whereBqare entire functions of order less thann. By the hypothesis of Theorem ., we know that the coefficients of the termznof polynomialsdjlQs,Qs,cjqP,P are distinct.
Then by Lemma . or Lemma ., we gethf ≡. This is absurd.
Step . We prove thatσ(f) =∞andσ(f)≥n. By Step we know thatfis transcendental.
Case .argan=argan. Sincean=an, without loss of generality, we assume that|an|>
|an|. Sinceargan=argbsn, by Remark . there exist two real numbersθ,θ(θ<θ) such
that for eachθ∈(θ,θ), we have
δ(P,θ) > and δ(Qs,θ) < .
Letc=maxj∈I{cj, |an|
|an|}, thenc< . By Lemma ., for∀ε( < ε<
–c
+c), there exists a set E⊂[, π) of linear measure zero such that forz=reiθwithθ∈(θ
,θ)\Eand sufficiently
larger, we have (.), (.), (.) and h(z)eP(z)≤exp
( +ε)δ(P,θ)rn
. (.)
By (.), (.), (.), (.), (.) and (.), forz=reiθ withθ∈(θ
,θ)\E and sufficiently
larger∈/H, we have
exp( –ε)δ(P,θ)rn
≤h(z)eP(z)
≤f(k)(z)
f(z) +
k–
j=
hj(z)eQj(z)
ff(j()(zz))+h(z)eP(z)
≤(k+ )BT(r,f)k+exp( +ε)cδ(P,θ)rn
. (.)
Then, by Lemma . and (.), we getσ(f) =∞andσ(f)≥n.
Step . We prove that σ(f) =n. By Step , we getσ(f)≥n. On the other hand, by
Lemma . we getσ(f)≤n. Hence we obtain thatσ(f) =n.
Proof of Theorem. Letfbe a solution of (.), thenfis a nonzero entire function. Using
a similar proof to Step in the proof of Theorem ., we obtain thatσ(f)≥n. Now we prove thatσ(f) =∞. Suppose thatσ(f) <∞, by Lemma ., for any givenε> , there exists a set
E⊂[, π) of linear measure zero such that for allz=reiθwith|z|sufficiently large and
θ∈[, π)\E, we have
ff((ji))((zz))
≤ |z|(j–i)(σ(f)–+ε) (≤i<j≤k). (.)
Letα,βbe two real numbers such thatσ(F) <α<β<n≤σ(f), then
F(z)≤exprα (.)
holds for sufficiently large|z|=r. Set
δ(Pi,θ) =|ain|cos(argain+nθ) (i= , ),
δ(Qj,θ) =|bjn|cos(argbjn+nθ) (j= , . . . ,k– ).
By Lemma ., there exists a setE⊂[, π) of linear measure zero such that whenever
θ∈[, π)\E, we have
δ(Qs,θ)= , δ(P,θ)= , δ(P,θ)= ,
and forz=reiθ withθ ∈[, π)\E
Then by (.), (.), (.) and (.), for sufficiently largerm, we get
The authors declare that they have no competing interests.
Authors’ contributions
Both authors drafted the manuscript, read and approved the final manuscript.
Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 11201195, 11171119), the Natural Science Foundation of Jiangxi, China (No. 20132BAB201008, 20122BAB201012). The authors thank the referee for his/her valuable suggestions to improve the present article.
Received: 6 November 2013 Accepted: 8 January 2014 Published:27 Jan 2014
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Cite this article as:Mao and Liu:On the growth of solutions of certain higher-order linear differential equations.
