# Some Oscillation Results of Higher-Order Linear Differential Equations  with Meromorphic Coefficients

Authors: Zhigang Huang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the growth of solutions of higher-order nonhomogeneous linear differential equations with meromorphic coefficients. We also discuss the relationship between small functions and solutions of such equations.


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## Body

## 1. Introduction and Main Results

Letf(z) be a meromorphic function in the whole complex plane. Throughout this paper, we assume that the reader is familiar with the fundamental results of the Nevanlinna’s value distribution theory of meromorphic functions and the standard notations such as the order σ(f), the exponent of convergence of zero-sequence λ(f), and the exponent of convergence of the sequence of distinct zeros λ¯(f). Moreover, a meromorphic function ψ(z) is called a small function with respect to f(z) if T(r,ψ)=o(T(r,f)) as r→∞, possible outside of a set of r with finite measure, where T(r,f) is the Nevanlinna characteristic function of f(z). The study of oscillation of solutions of linear differential equations has attracted many interests since the work of Bank and Laine; for more details see [1].  One of the main subject of this research is the growth and zero distribution of solutions of linear differential equations. In this paper, we first discuss the growth of solutions of higher-order linear differential equation(1.1)f(k)+Hk-1f(k-1)+⋯+H0f=F(z),k≥2,
where Hj(j=1,…,k-1), H0≢0 and F≢0 are entire functions of finite order. Some results on the growth of solutions of (1.1) have been obtained by several researchers, see [2–4]. In the case that the coefficients Hj(j=0,…,k-1) are polynomials, the growth of solutions of (1.1) has been extensively studied; see [2]. In 1992, Hellerstein et al. [3] proved that every transcendental solution of (1.1) is of infinite order, if there exists some d∈{0,1,…,k-1} such that maxj≠d{σ(Hj),σ(F)}<σ(Hd)≤1/2. As for sectorial growth conditions on the coefficients of (1.1) that imply that all solutions are of infinite order, see, for example, [5]. In addition, a special case for this was investigated by Wang and Laine [6]. Recently, Wang and Laine [4] obtain the following.Theorem A.
Suppose thatHj(z)=Dj(z)ePj(z)(j=0,…,k-1) where Pj(z)=ajnzn+⋯+aj0(j=0,…,k-1) are polynomials with degree n≥1, Dj(z) are entire functions of order less than n, not all vanishing, and F(z)≢0 is an entire function of order less than n. If ajn(j=0,…,k-1) are distinct complex numbers, then every solution of (1.1) is of infinite order.Thus a natural question is whether every meromorphic solution of (1.1) has infinite order, if the coefficients of (1.1) are meromorphic? The following theorem partially answers this question.Theorem 1.1.
LetAj(z)(j=1,…,k-1), A0≢0 and F≢0 be meromorphic functions with max{σ(Aj),j=0,1,…,k-1),σ(F)}<1, and let a0,a1,…,ak-1 be complex constants such that (i) argaj=arga0 and aj=cja0(0<cj<1) or (ii) argaj≠arga0(j=1,…,k-1). Then every meromorphic solution f≢0 of the equation
(1.2)f(k)+Ak-1eak-1zf(k-1)+⋯+A0ea0zf=F,k≥2,
satisfies λ(f)=λ¯(f)=σ(f)=∞.Remark 1.2.
In Theorem1.1, if ajz(j=0,…,k-1) are replaced by Pj(z)=ajnzn+⋯+aj0(j=0,…,k-1), with additional hypothesis on ajn similar to aj, and furthermore σ(F) is replaced by σ(F)<n, then we have the same conclusion with Theorem 1.1.In 2000, Chen [7] first studied the fixed points of solutions of second-order linear differential equations and obtained some precise estimation on the number of fixed points. After that, a number of results on fixed points of solutions of differential equations with entire coefficients were obtained; see [8–10]. In 2006, Chen and Shon [11] further studied the relation between small functions and solutions of differential equations and obtain the following.Theorem B.
LetAj(z)≢0(j=0,1) be entire functions with σ(Aj)<1, and let a,b be complex constants such that ab≠0 and arga≠argb or a=cb(0<c<1). If ψ(z)≢0 is an entire function with finite order, then every solution f≢0 of equation
(1.3)f′′+A1(z)eazf′+A0(z)ebzf=0
satisfies λ¯(f-ψ)=λ¯(f′-ψ)=λ¯(f′′-ψ)=∞.Motivated by Theorem B, we try to consider the relation between small functions with meromorphic solutions of (1.2). Indeed, such relationship on higher order differential equations is more difficult than that of second order differential equations. Moreover, the method used in the proof of Theorem B cannot deal with the case of higher-order linear differential equations.Theorem 1.3.
Under the assumption of Theorem1.1, if ϕ≢0 is a meromorphic function of finite order and satisfies max{λ(ϕ),λ(1/ϕ)}<1, then every non-trivial meromorphic solution of (1.2) satisfies λ¯(f-ϕ)=λ¯(f′-ϕ)=∞.Remark 1.4.
As the remark on Theorem1.1 in Remark 1.2, the conclusion of Theorem 1.3 can also be extended to the case that ajz(j=0,…,k-1) are replaced by Pj(z)=ajnzn+⋯+aj0(j=0,…,k-1), with some similar additional hypothesis on ajn, F and ϕ in Theorem 1.3.


## 2. Preliminary Lemmas

To prove our theorems, we need some lemmas.Lemma 2.1 (see [12]).
Letw(z) be a transcendental meromorphic function with σ(f)=σ<∞. Let Γ={(k1,j1),…,(km,jm)} be a finite set of distinct pairs of integers satisfying ki>ji≥0 for i=1,2,…,m. Also let ϵ>0 be a given constant. Then, there exists a set E1⊂(1,∞) that has finite logarithmic measure, such that for all z satisfying z∉E∪[0,1] and for all (k,j)∈Γ, One has
(2.1)|w(k)(z)||w(j)(z)|≤|z|(k-j)(σ-1+ɛ).LetP(z)=(α+βi)zn+⋯ is a nonconstant polynomial and α,β are real constants. For θ∈[0,2π), set δ(P(z),θ)=αcosnθ-βsinnθ.Lemma 2.2 (see [13]).
LetP(z) be a non-constant polynomial of degree n. Let w(z) be a meromorphic function, not identically zero, of order less than n, and set g(z)=w(z)eP(z). Then for any given ɛ>0 there exists a zero measure set H1⊂[0,2π) such that if θ∈θ∈[0,2π)∖(H1∪H2), then for |z|>r(θ).(1)
Ifδ(P,θ)<0, then exp((1+ϵ)δ(P,θ)rn)≤|g(reiθ)|≤exp((1-ϵ)δ(P,θ)rn);(2)
Ifδ(P,θ)>0, then exp((1-ϵ)δ(P,θ)rn)≤|g(reiθ)|≤exp((1+ϵ)δ(P,θ)rn), where H2={θ:δ(P,θ)=0,0≤θ<2π} is a finite set.Lemma 2.3 (see [14]).
LetA0,…,Ak-1,F≢0 be finite-order meromorphic functions. If f is an infinite-order meromorphic solution of the equation
(2.2)f(k)+Ak-1f(k-1)+⋯+A0f=F,
then f satisfies λ(f)=λ¯(f)=σ(f)=∞.Lemma 2.4 (see [15], page 79).
Suppose thatf1(z),f2(z),…,fn(z)(n≥2) are meromorphic functions and g1(z),g2(z),…,gn(z) are entire functions satisfying the following conditions:(1)
∑j=1nfj(z)egj(z)≡0;(2)
gj(z)-gk(z) are not constants for 1≤j<k≤n;(3)
For1≤j≤n,1≤h<k≤n, T(r,fj)=o{T(r,egh-gk)}(r→∞,r∉E), where E has a finite measure.Thenfj(z)≡0(j=1,2,…,n).Lemma 2.5 (see [4], Lemma 2.6).
Letf(z) be a an entire function of order σ(f)=σ<∞. Suppose that there exists a set E⊂[0,2π) which has linear measure zero, such that log+|f(reiθ)|≤Mrρ for any ray argz=θ∈[0,2π)∖E, where M is a positive constant depending on θ, while ρ is a positive constant independent of θ. Then σ(f)≤ρ.Lemma 2.6 (see [16]).
Letf(z) be a transcendental meromorphic function of order σ(f)=σ<∞. Then for any given ɛ>0, there exists a set E3⊂[0,2π) which has linear measure zero, such that if ψ1∈[0,2π)∖E3, then there is a constant R1=R1(ψ1)>1 such that for all z satisfying argz=ψ1 and |z|=r≥R2, One has
(2.3)exp⁡{-rσ+ɛ}≤|f(z)|≤exp⁡{rσ+ɛ}.


## 3. Proof of Theorem1.1

Assume thatf(z) is a meromorphic solution of (1.2) of finite order. We will deduce a contradiction later. First, we have σ(f)≥1. Otherwise, it follows from Lemma 2.4 that A0≡0 which is a contradiction.Sincef is a meromorphic solution of (1.2), we know that the poles of f can occur only at the poles of Aj(j=0,…,k-1) and F. Let f=g(z)/d(z), where d(z) is the canonical product formed with the nonzero poles of f(z), with σ(d)≤max{σ(F),σ(Aj),j=0,…,k-1}<1, and g is an entire function with 1≤σ(g)=σ(f)=σ<∞. Substituting f=g/d into (1.2), by some calculation we can get(3.1)dF=g(k)+g(k-1)[Ak-1eak-1z+Bk,k-1]+⋯+g′[A1ea1z+∑i=2k-1AieaizBi,1+Bk,1]+g[A0ea0z+∑i=1k-1AieaizBi,1+Bk,0],
where Bi,j are defined as a sum of a finite number of terms of the type(3.2)∑(j1⋯ji)Cjj1⋯ji(d′d)j1⋯(d(i)d)ji.Cjj1⋯ji are constants, and j+j1+2j2+⋯+iji=k. Now, we rewrite (3.1) into(3.3)dFg=g(k)g+⋯+[A1ea1z+∑i=2k-1AieaizBi,1+Bk,1]g′g+[A0ea0z+∑i=1k-1AieaizBi,1+Bk,0].Setmax{σ(Aj),σ(F),σ(d),j=0,…,k-1}=β. By Lemma 2.1, for any given ɛ>0 there exists a set E1∈[0,2π) which has linear measure zero, such that if θ∈[0,2π)∖E1, then there is a constant R1=r1(θ)>1 such that for all z satisfying argz=θ and |z|≥R1, we have(3.4)|g(i)(z)||g(z)|≤|z|k(σ-1+ɛ),|d(i)(z)||d(z)|≤|z|k(β-1+ɛ),i=1,…,k.
Meanwhile, by Lemma 2.6, for the above ɛ there exists a set E2 of zero linear measure, such that if θ∈[0,2π)∖E2, we have for sufficiently large r=|z|,(3.5)|Aj(z)|≤exp⁡{rβ+ɛ},j=0,…,k-1.
Next we divide our proof into two cases.Case 1.
Suppose thatargaj=arga0 and aj=cja0(0<cj<1),j=1,…,k-1. Then for any ray argz=θ, we have
(3.6)δ(ajz,θ)=cjδ(a0z,θ).
By Lemma2.2, for any given ɛ we can find a ray argz=θ∈[0,2π)∖E1∪E2∪H1∪H2, where H1 and H2 are defined in Lemma 2.2, E1∪E2∪H1∪H2 has zero linear measure, such that δ(ajz,θ)=cjδ(a0z,θ)>0, and for sufficiently large r,
(3.7)|A0(reiθ)ea0reiθ|≥exp⁡{(1-ɛ)δ(a0z,θ)r},|Aj(reiθ)eajreiθ|≤exp⁡{(1+ɛ)cjδ(a0z,θ)r},j=1,…,k-1.
Thus, by (3.4) and (3.7), we have for sufficiently large r,
(3.8)|A0ea0z+∑i=1k-1AieaizBi,1+Bk,0|≥M0exp⁡{(1-ɛ)δ(a0z,θ)r},|Ak-1eak-1z+Bk,k-1|≤Mk-1exp⁡{(1-ɛ)δ(ajz,θ)r}=Mk-1exp⁡{(1-ɛ)ck-1δ(a0z,θ)r},⋯|A1ea1z+∑i=2k-1AieaizBi,1+Bk,1|≤M1exp⁡{(1+ɛ)δ(a1z,θ)r}=M1exp⁡{(1+ɛ)c1δ(a0z,θ)r},
where Mj(j=0,…,k-1) are positive constants.
Forθ∈[0,2π)∖E1∪E2∪H1∪H2, we claim that
(3.9)log⁡+|g(z)||z|β+ɛ
is bounded on the ray argz=θ. Otherwise, there exists a sequence of points zm=rmeiθ, such that rm→∞, and that
(3.10)log⁡+|g(zm)|rmβ+ɛ⟶∞.
From (3.10) and the definition of order, we see that
(3.11)|d(zm)F(zm)g(zm)|⟶0
for m is large enough.
By (3.3),(3.4), and (3.8)–(3.11), we get for sufficiently large m,
(3.12)M0exp⁡{(1-ɛ)δ(a0z,θ)rm}≤|A0ea0rmeiθ+∑i=1k-1AieairmeiθBi,1+Bk,0|≤|-d(rmeiθ)F(rmeiθ)g|+|g(k)g|+⋯+|[A1ea1rmeiθ+∑i=2k-1AieairmeiθBi,1+Bk,1]g′g|≤∑j=1k-1rk(σ+ɛ)Mjexp⁡{(1+ɛ)cjδ(a0z,θ)rmeiθ}+rk(σ-1+ɛ).
Clearly, we can choose sufficiently small ɛ such that 0<ɛ<min{(1-cj)/(1+cj):j=1,…,k-1}. Then by (3.12), we can obtain a contradiction provided that m is sufficiently large.
Therefore,(3.13)log⁡+|g(z)||z|β+ɛ
is bounded, and we have |g(z)|≤Mexp{rβ+ɛ} on the ray argz=θ.Case 2.
Suppose thatargaj≠arga0(j=1,…,k-1). By Lemma 2.2, there exists a ray argz=θ∈[0,2π)∖E1∪E2∪H1∪H2, where E1,E2,H1, and H2 are defined, respectively, as in Case 1, such that
(3.14)δ(a0z,θ)>0,δ(ajz,θ)<0,j=1,…,k-1.
Further, for any given ε>0, we have for sufficiently large r,
(3.15)|A0(reiθ)ea0reiθ|≥exp⁡{(1-ɛ)δ(a0z,θ)r},|Aj(reiθ)eajreiθ|≤exp⁡{(1-ɛ)δ(ajz,θ)r},j=1,…,k-1.
Thus, by (3.4) and (3.15), we have for sufficiently large r,
(3.16)|A0ea0z+∑i=1k-1AieaizBi,1+Bk,0|≥M0exp⁡{(1-ɛ)δ(a0z,θ)r},|Ak-1eak-1z+Bk,k-1|≤Mk-1rk(σ-1+ɛ),⋯|A1ea1z+∑i=2k-1AieaizBi,1+Bk,1|≤M1rk(σ-1+ɛ),
where Mj(j=0,…,k-1) are positive constants.
Forθ∈[0,2π)∖E1∪E2∪H1∪H2, we claim that
(3.17)log⁡+|g(z)||z|β+ɛ
is bounded on the ray argz=θ. Otherwise, there exists a sequence of points zm=rmeiθ, such that rm→∞, and that
(3.18)log⁡+|g(zm)|rmβ+ɛ⟶∞.
From (3.18) and the definition of order, we see that
(3.19)|d(zm)F(zm)g(zm)|⟶0
for m is large enough.
Then by similar reasoning as in Case1, we can also obtain a contradiction. So
(3.20)log⁡+|g(z)||z|β+ɛ
is bounded, and we have |g(z)|≤Mexp{rβ+ɛ} on the ray argz=θ.
Combining Case1 and Case 2, for any given ray argz=θ∈[0,2π)∖E, E of linear measure zero, we have |g(z)|≤Mexp{rβ+ɛ} on the ray argz=θ, provided that r is sufficiently large. Thus by Lemma 2.5, we get σ(g)≤β+ɛ<1, which is a contradiction. Again by Lemma 2.3, we complete the proof of Theorem 1.1.


## 4. Proof of Theorem1.3

Letf(z)≢0 be a meromorphic solution of (1.2). Set Bj(z)=Aj(z)eajz. First, we prove that λ¯(f-ϕ)=∞.Setf=g0+ϕ. By Theorem 1.1, we have σ(g0)=σ(f)=∞. Substituting f=g0+ϕ into (1.2), we get(4.1)g0(k)+Bk-1g0(k-1)+⋯+B0g0=F-(ϕ(k)+Bk-1ϕ(k-1)+⋯+B0ϕ).
We remark that (4.1) may have finite-order solutions, but we only need to discuss the solutions whose order are ∞.Ifϕ(k)+Bk-1ϕ(k-1)+⋯+B0ϕ=F, then by Theorem 1.1, we have σ(ϕ)=∞. It is a contradiction. Hence we have F-(ϕ(k)+Bk-1ϕ(k-1)+⋯+B0ϕ)≢0. It follows from Lemmas 3 and (4.1) that λ¯(g0)=λ(g0)=σ(g0)=∞.Now we prove thatλ¯(f′-ϕ)=∞. Set g1=f′-ϕ. Differentiating both side of (1.2), we get(4.2)f(k+1)+Bk-1f(k)+(Bk-1′+Bk-2)f(k-1)+⋯+(B1′+B0)f′+B0′f=F′.
Rewriting (1.2), we have(4.3)f=F-1B0[f(k)+Bk-1f(k-1)+⋯+B1f′].
Substituting (4.3) into (4.2), we get(4.4)f(k+1)+[Bk-1-B0′B0]f(k)+[Bk-1′+Bk-2-B0′B0Bk-1]f(k-1)+⋯+[B1′+B0-B0′B0B1]f′=F′-B0′F.
We rewrite (4.4) into the following form:(4.5)f(k+1)+Ck-1f(k)+Ck-2f(k-1)+⋯+C0f′=F′-B0′F,
where Ck-1=Bk-1-B0′/B0,Cj=Bj+1′+Bj-(B0′/B0)Bj+1,j=0,…,k-2. Substituting f′=g1+ϕ,…,f(k+1)=g1(k)+ϕ(k) into (4.5), we have(4.6)g1(k)+Ck-1g1(k-1)+Ck-2g1(k-2)+⋯+C0g1=F′-B0′F-[ϕ(k)+Ck-1ϕ(k-1)+Ck-2ϕ(k-2)+⋯+C0ϕ].Next we prove(4.7)Q1(z)=ϕ(k)+Ck-1ϕ(k-1)+Ck-2ϕ(k-2)+⋯+C0ϕ-F′+B0′F≢0.
Conversely, assume that Q1(z)≡0.Dividingϕ into both sides of Q1(z)≡0, we have(4.8)Dk-1eak-1z+⋯+D1ea1z+D0ea0z+D(z)≡0,
where(4.9)D0=A0+A0′F+a0A0F,D1=A1ϕ′ϕ+(A1′+A1a1-A0′A0A1),D2=A2ϕ′′ϕ+(A2′+A2a2-A0′A0A2)ϕ′ϕ,⋯Dj=Ajϕ(j)ϕ+(Aj′+Ajaj-A0′A0Aj)ϕ(j-1)ϕ,⋯Dk-1=Ak-1ϕ(k-1)ϕ+(Ak-1′+Ak-1ak-1-A0′A0Ak-1)ϕ(k-2)ϕ,D=ϕ(k)ϕ-A0′A0ϕ(k-1)ϕ-F′.
Set max{σ(Aj),σ(F),j=0,…,k-1}=β. Clearly from (4.9),(4.10)σ(D0(z))≤β.
Meanwhile, by (4.9), we have(4.11)m(r,Dj)≤3m(r,Aj)+m(r,ϕ(j)ϕ)+m(r,ϕ(j-1)ϕ)+m(r,Aj′)+m(r,A0′A0)+O(1)≤4m(r,Aj)+O(log⁡r),j=1,…,k-1,m(r,D)≤m(r,F′)+O(log⁡r),
outside a possible exceptional set E of finite linear measure. On the other hand, again by (4.9), we have(4.12)N(r,Dj)≤N(r,1ϕ)+N(r,1A0)+k(N(r,A0)+N(r,ϕ)+N(r,Aj)),j=1,…,k-1,N(r,D)≤N(r,1ϕ)+N(r,1A0)+2N(r,A0)+kN(r,ϕ)+N(r,F′).
Combining (4.11) and (4.12), we get(4.13)σ(Dj)≤max⁡{λ(ϕ),λ(1ϕ),σ(Aj),j=0,…,k-1}<1,j=1,…,k-1,σ(D)≤max⁡{λ(ϕ),λ(1ϕ),σ(F′),σ(A0)}<1.
By Lemma 2.6, for any given positive constants ɛ, there exists a ray argz=θ∈[0,2π)∖E1, where E1 has zero linear measure, such that for sufficiently large r,(4.14)|D(z)|≤exp⁡{rσ(D)+ɛ}.
Therefore by (4.13), and Lemma 2.2, we can find a ray argz=θ∈[0,2π)∖E1∪H1∪H2, where H1 and H2 are defined in Lemma 2.2, H1∪H2 has zero linear measure, such that δ(ajz,θ)=cjδ(a0z,θ)>0, and for sufficiently large r,(4.15)|D0(reiθ)ea0reiθ|≥exp⁡{(1-ɛ)δ(a0z,θ)r},|Dj(reiθ)eajreiθ|≤exp⁡{(1+ɛ)cjδ(a0z,θ)r},j=1,…,k-1.
Thus, by (4.8) and (4.14) and (4.15), we have(4.16)exp⁡{(1-ɛ)δ(a0z,θ)r}≤∑j=1k-1exp⁡{(1+ɛ)cjδ(a0z,θ)r}+exp⁡{rσ(D)+ɛ}.
Since ɛ can be chosen arbitrarily small, we can obtain a contradiction. Thus we have Q1(z)≢0. By Lemma 2.3 and (4.6), we get λ¯(f′-ϕ)=∞.



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