# On the Growth of Solutions of a Class of Higher Order Linear Differential Equations with Extremal Coefficients

Authors: Jianren Long; Chunhui Qiu; Pengcheng Wu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We consider that the linear differential equationsf(k)+Ak-1(z)f(k-1)+⋯+A1(z)f′+A0(z)f=0, where Aj(j=0,1,…,k-1), are entire functions. Assume that there exists l∈{1,2,…,k-1}, such that Al is extremal for Yang's inequality; then we will give some conditions on other coefficients which can guarantee that every solution f(≢0) of the equation is of infinite order. More specifically, we estimate the lower bound of hyperorder of f if every solution f(≢0) of the equation is of infinite order.


---

## Body

## 1. Introduction and Main Results

We will assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna theory of meromorphic functions (see [1, 2] or [3]). In addition, for a meromorphic function f in the complex plane ℂ, we will use the notations ρ(f) and μ(f) to denote its order and lower order, respectively.In order to estimate the rate of growth of meromorphic function of infinite order more precisely, we recall the following definition.Definition 1 (see [4]).
Letf be a meromorphic function in the complex plane ℂ. Then one defines the hyperorder ρ2(f) of f by
(1)ρ2(f)=limsup⁡r→∞log⁡+log⁡+T(r,f)log⁡r.
Consider the second order linear differential equation(2)f′′+A(z)f′+B(z)f=0,
where A and B(≢0) are entire functions. It is well known that if A is an entire function, B(≢0) is a transcendental entire function, and f1, f2 are two linearly independent solutions of (2), then at least one of f1, f2 must have infinite order. On the other hand, there are some equations of form (2) that possess a solution f(≢0) of finite order; for example, f(z)=ez satisfies f′′+e-zf′-(e-z+1)f=0. Therefore, one may ask, what assumptions on A(z) and B(z) will guarantee that every solution f≢0 of (2) is of infinite order? From the works of Gundersen (see [5]) and Hellerstein et al. (see [6]), we know that if A(z) and B(z) are entire functions with ρ(A)<ρ(B), or A(z) is a polynomial, and B(z) is transcendental, or ρ(B)<ρ(A)≤1/2, then every solution f(≢0) of (2) is of infinite order. More results can be found in [7–12]. For entire solutions of infinite order more precise estimates for their rate of growth would be an important achievement. There are many authors investigating the hyperorder ρ2(f) of solutions of (2), such as Chen and Yang (see [8, 13]) and Kwon (see [14, 15]).In this paper, we will introduce the deficient value and Borel direction into the studies of the complex differential equations. In order to give the definition of the Borel direction, we need the following notation. Letα<β such that β-α<2π and r>0; set Ω(α,β)={z:α<argz<β}, Ω(α,β,r)={z:α<argz<β}⋂{z:|z|<r}.Definition 2.
Letf be a meromorphic function in the complex plane ℂ with 0<μ(f)<∞. Let μ(f)≤λ≤ρ(f) be a finite constant. A ray argz=θ(0≤θ<2π) from the origin is called a Borel direction of order ≥λ of f, if for any positive number ε>0 and for any complex number a∈ℂ⋃{∞}, possibly with two exceptions, the following inequality holds:
(3)limsup⁡r→∞log⁡n(Ω(θ-ε,θ+ε,r),f=a)log⁡r≥λ,
where n(Ω(θ-ε,θ+ε,r),f=a) denotes the number of zeros, counting the multiplicities, of f-a in the region Ω(θ-ε,θ+ε,r).The fundamental result in angular distribution, due to Valiron, says that a meromorphic function of orderρ>0 must have at least one Borel direction of order ρ; for example, see [3].It is well known that deficient values and Borel directions are very important concepts in Nevanlinna theory of meromorphic functions. These two concepts are extensively studied. There is a striking relationship between them which was found by Yang and Zhang and says that, for a meromorphic functionf of order ρ, the number of deficient values is less or equal to the number of Borel directions of order ρ of f. In 1988, Yang extended the above Y-Z inequality to the case of entire function of finite lower order. In order to useYang’s result to study the complex differential equations, we will use the following Theorem which can be easily derived from [16].Theorem 3 (see [16]).
Suppose thatf is an entire function of finite lower order μ>0. Let q (<∞) denote the number of Borel directions of order ≥μ and p denote the number of finite deficient values of f; then p≤q/2.Note that Theorem3 is explicitly stated in [17]. To see the valid of the conclusion of the theorem, we note that, in [17, Corollary 1], Wu has proved that if f(z) is of finite lower order μ and the number of Borel directions of order ≥μ is finite, then the order ρ of f(z) is also finite. As each Borel direction of order ρ is also a Borel direction of order ≥μ, this implies that, for f(z), the number of the Borel directions of order ρ is fewer or equal to the number of the Borel directions of order ≥μ. Therefore Theorem 3 follows from Theorem 6.7 in [3].In the sequel, we will say that an entire functionf is extremal forYang’s inequality if f satisfies the assumptions of Theorem 3 with p=q/2.The simplest entire function extremal forYang’s inequality is ez. A little bit complicated example is A(z)=∫0z‍e-tndt, (n≥2). We know that (see [3]) A(z) has n deficient values
(4)al=ei(2πl/n)∫0∞‍e-tndt,(l=1,2,…,n)
and q=2n Borel direction argz=((2k-1)/2n)π(k=1,2,…,2n). So p=q/2.Furthermore, we state the following result due to present authors (see, [18]).Theorem 4 (see [18]).
LetA be an entire function extremal forYang’s inequality, and let B be a transcendental entire function such that ρ(A)≠ρ(B). Then every solution f(≢0) of (2) is of infinite order.In this paper, we will consider the higher order linear differential equation(5)f(k)+Ak-1(z)f(k-1)+⋯+A1(z)f′+A0(z)f=0,
where Aj(j=0,1,…,k-1) are entire functions. Many authors have also investigated the growth of solutions of (5) and obtained lots of results on order and hyperorder of the solutions of (5) (see [19–23]). We will introduce the deficient value and Borel directions into the studies of (5). The main result in the paper is as follows.Theorem 5.
LetAj(j=0,1,…,k-1) be entire functions. Suppose that there exists an integer l∈{1,2,…,k-1}, such that Al(z) is extremal forYang’s inequality. Suppose that A0 is a transcendental entire function with ρ(A0)≠ρ(Al) and ρ(Ai)<ρ(A0) for i≠l(1≤i≤k-1). Then every solution f(≢0) of (5) satisfies ρ(f)=∞ and ρ2(f)≥ρ(A0).The paper is organized as follows. In Section2, we will give some lemmas. In Section 3, we will prove Theorem 5. In Section 4, we will discuss some further results related to the two entire coefficients in (5) which are extremal forYang’s inequality.


## 2. Lemmas

In this section, we need some auxiliary results. The following lemma is by Gundersen.Lemma 6 (see [24]).
Let(f,Γ) denote a pair that consists of a transcendental meromorphic function f and a finite set
(6)Γ={(k1,j1),(k2,j2),…,(kq,jq)}
of distinct pairs of integers that satisfy ki>ji≥0 for i=1,…,q. Let α>1 and ε>0 be given real constants. Then the following three statements hold.
(i) There exists a setE1⊂[0,2π) that has linear measure zero, and there exists a constant c>0 that depends only on α and Γ, such that if ψ0∈[0,2π)-E1, then there is a constant R0=R0(ψ0)>1 such that, for all z satisfying argz=ψ0 and |z|=r≥R0 and for all (k,j)∈Γ, we have
(7)|f(k)(z)f(j)(z)|≤c(T(αr,f)rlog⁡αrlog⁡T(αr,f))k-j.
In particular, iff has finite order ρ(f), then (7) is replaced by
(8)|f(k)(z)f(j)(z)|≤|z|(k-j)(ρ(f)-1+ε).
(ii) There exists a setE2⊂(1,∞) that has finite logarithmic measure, and there exists a constant c>0 that depends only on α and Γ, such that, for all z satisfying |z|=r∉E2⋃[0,1] and for all (k,j)∈Γ, inequality (7) holds.
In particular, iff has finite order ρ(f), then inequality (8) holds.
(iii) There exists a setE3⊂[0,∞) that has finite linear measure, and there exists a constant c>0 that depends only on α and Γ, such that, for all z satisfying |z|=r∉E3 and for all (k,j)∈Γ, we have
(9)|f(k)(z)f(j)(z)|≤c(T(αr,f)rεlog⁡T(αr,f))k-j.
In particular, iff has finite order ρ(f), then (9) is replaced by
(10)|f(k)(z)f(j)(z)|≤|z|(k-j)(ρ(f)+ε).Letf be an entire function extremal forYang’s inequality p=q/2. Suppose that the rays argz=θk (k=1,2,…,q) (0≤θ1<θ2<⋯<θq<θq+1=θ1+2π) are the q distinct Borel directions of order ≥μ of f. In [17], Wu studied the entire functions which are extremal forYang’s inequality systematically. The following results play an important role in the proof of our results.Lemma 7 (see [17]).
Suppose thatf is extremal forYang’s inequality. Then μ(f)=ρ(f). Moreover, for every deficient value ai (i=1,2,…,p) there exists a corresponding angular domain Ω(θki,θki+1)={z:θki<argz<θki+1} such that for every ε>0 inequality
(11)log⁡1|f(z)-ai|>C(θki,θki+1,ε,δ(ai,f))T(|z|,f)
holds for z∈Ω(θki+ε,θki+1-ε,r,+∞)={z:θki+ε<argz<θki+1-ε}⋂{z:r<|z|<∞}, where C(θki,θki+1,ε,δ(ai,f)) is a positive constant depending only on θki, θki+1, ε and δ(ai,f).In the sequel, we will say thatf decays exponentially to ai in Ω(θki,θki+1), if (11) holds in Ω(θki,θki+1). Note that if f is extremal forYang’s inequality, then μ(f)=ρ(f). Thus, for these functions, we need only to consider the Borel direction of order ρ(f).Lemma 8 (see [18]).
Letf be extremal forYang’s inequality. Suppose that there exists θ∈Ω(θj,θj+1)(1≤j≤q) such that
(12)limsup⁡r→∞log⁡+log⁡+|f(reiθ)|log⁡r=ρ(f),
where argz=θj (j=1,2,…,q) are Borel directions of f. Then θj+1-θj=π/ρ(f).Before stating the following lemmas, forE⊂[0,∞), we define the Lebesgue measure of E by mes(E) and the logarithmic measure of E⊂[1,∞) by ml(E)=∫E‍(dt/t) and define the upper and lower logarithmic density of E⊂[1,∞), respectively, by
(13)log⁡dens¯E=lim⁡r→∞¯ml(E⋂‍[1,r])log⁡r,log⁡dens_E=lim⁡_r→∞ml(E⋂‍[1,r])log⁡r.Lemma 9 (see [25]).
Letf be an entire function with ρ(f)=ρ<1/2 and suppose that m(r) is defined as
(14)m(r)=inf⁡|z|=rlog⁡|f(z)|.
If σ<ρ, then the set {r:m(r)>rσ} has a positive upper logarithmic density.Lemma 10 (see [26]).
Letg(r) and h(r) be monotone nondecreasing functions on (0,∞) such that g(r)≤h(r) for all r outside some set of finite logarithmic measure. Let λ>1 be a given real constant. Then there exists a constant r0>0 such that g(r)≤h(λr) for all r≥r0.Lemma 11.
Letf(z) be an entire function with order ρ(0<ρ<∞), let and Ω(ϕ1,ϕ2)={z:ϕ1<argz<ϕ2} be a sector with ϕ2-ϕ1<π/ρ. If there is Borel direction of f(z) in Ω(ϕ1,ϕ2), then there exists at least one of the two rays Lj:argz=ϕj(j=1,2), without lose of generality, says, L2, such that
(15)limsup⁡r→∞log⁡+log⁡+|f(reiϕ2)|log⁡r=ρ.Lemma11 can be founded in [27, Lemma 1], which can be proved by using a result in [28, Page 119-120].


## 3. Proof of Theorem5

Now we prove our main result.Sinceρ(Ai)<ρ(A0)(i≠l,1≤i≤k-1), we know that for any given constant η with max⁡{ρ(Ai),i≠l,1≤i≤k-1}<η<ρ(A0), there exists a constant R1>0 such that
(16)|Ai(z)|≤exp⁡(rη)
holds for all |z|=r>R1.We consider the following two cases.Case 1.
We suppose thatρ(Al)<ρ(A0). Now to the contrary assume that there is a solution f(≢0) of (5) with ρ(f)<+∞. We will seek a contradiction. By Lemma 6(ii), there exists a set E1⊂[1,+∞) that has finite logarithmic measure, such that the following inequality
(17)|f(m)(z)f(z)|≤|z|kρ(f),m=1,2,…,k
holds for all z with |z|=r∉E1⋃[0,1].
We deduce from (17) and (5) that
(18)|A0(z)|≤|f(k)f|+|Ak-1(z)||f(k-1)f|+⋯+|Al(z)||f(l)f|+⋯+|A1(z)||f′f|≤|z|kρ(f)(1+|Ak-1(z)|aaaaaaaaa+⋯+|Al(z)|+⋯+|A1(z)|)
holds for all z with |z|=r∉E1⋃[0,1].
Thus,(19)T(r,A0)≤kρ(f)log⁡r+(k-1)T(r,Ai)
holds for all z with |z|=r∉E1⋃[0,1], where T(r,Ai)=max⁡1≤h≤k-1T(r,Ah). By Lemma 10, we have ρ(A0)≤ρ(Ai); this is a contradiction. Therefore, every solution f(≢0) of (5) is of infinite order.
By using similar methods as [14], we can easily prove that ρ2(f)≥ρ(A0) in this case. We omit the details here.Case 2.
We suppose thatρ(Al)>ρ(A0). Now to the contrary assume that there is a solution f(≢0) of (5) with ρ(f)<+∞. We will seek a contradiction.
Suppose thatai(i=1,2,…,p) are all the finite deficient values of Al(z). Thus we have 2p angular domains Sj={z∣θj<argz<θj+1} (j=1,2,…,2p). For any ε>0, by using Lemmas 7 and 11, we can easily obtain that Al(z) has the following properties: in each sector Sj, either there exists some ai such that
(20)log⁡1|Al(z)-ai|>C(θj,θj+1,ε,δ(ai,Al))T(|z|,Al)
holds for z∈Ω(θj+ε,θj+1-ε,r,+∞), where C(θj,θj+1,ε,δ(ai,Al)) is a positive constant depending only on θj, θj+1, ε and δ(ai,Al), or there exists θ∈Sj such that
(21)limsup⁡r→∞log⁡+log⁡+|Al(reiθ)|log⁡r=ρ(Al)
holds. For the sake of simplicity, in the sequel we use C to represent C(θj,θj+1,ε,δ(ai,Al)). Note that if there exists some ai such that (20) holds in Sj, then there exists θ such that (21) holds in Sj-1 and Sj+1. And if there exists θ∈Sj such that (21) holds, then there are ai (ai′) such that (20) holds in Sj-1 and Sj+1, respectively.Without loss of generality, we assume that there is a rayargz=θ in S1 such that (21) holds. Therefore, there exists a ray in each sector S3,S5,…,S2p-1, such that (21) holds. By using Lemma 8, we know that all the sectors have the same magnitude π/ρ(Al).Firstly, suppose thatρ(A0)≥1/2. Since A0(z) must have a Borel direction of order ρ(A0), by using Lemma 11, we can see that there exists a sector Ω(α,β)(α<β) such that β-α≥π/ρ(A0) and such that for all the rays argz=θ(α<θ<β) we have
(22)limsup⁡r→∞log⁡+log⁡+|A0(reiθ)|log⁡r=ρ(A0).Note thatρ(Al)>ρ(A0). It is not hard to see that there exists a sector Ω(α′,β′)(α<α′<β′<β) such that there is an aj0 such that
(23)log⁡1|Al(reiθ)-aj0|>CT(r,Al)
holds for all α′≤θ≤β′.By Lemma6(i), there exist θ0 (α′≤θ0≤β′) and R2>0 such that
(24)|f(m)(reiθ0)f(reiθ0)|≤rkρ(f),m=1,2,…,k
holds for all r>R2.Note that(25)limsup⁡r→∞log⁡+log⁡+|A0(reiθ0)|log⁡r=ρ(A0).
Thus there is a sequence of {rn} with lim⁡n→∞rn=∞ such that
(26)|A0(rneiθ0)|≥exp⁡(rnρ(A0)-ε)
holds for every 0<ε<(ρ(A0)-η)/2. Therefore, we deduce from (16), (23), and (24) that
(27)|A0(rneiθ0)|≤rnkρ(f)(|A1(rneiθ0)|1+|Ak-1(rneiθ0)|+⋯+|Al(rneiθ0)-aj0|aaaaaaaaaaa+|aj0|+⋯+|A1(rneiθ0)|)≤rnkρ(f)((rnη)rnkρ(f)1+exp⁡(rnη)+⋯+exp⁡{-CT(rn,Al)}aaaaaaaaaaa+|aj0|+⋯+exp⁡(rnη))
holds for all sufficiently large n. Therefore, combining (26) with (27), we have that
(28)exp⁡{rnρ(A0)-ε}<rnkρ(f)(1+(k-2)exp⁡(rnη)+exp⁡{-CT(rn,Al)}+|aj0|)
holds for all sufficiently large n. This is a contradiction, so every solution f(≢0) of (5) is infinite order in this case.Secondly, suppose that0<ρ(A0)<1/2. By Lemma 9, there is a sequence of {rn} with lim⁡n→∞rn=∞ such that for any θ∈[0.2π), we must have
(29)|A0(rneiθ)|>exp⁡(rnρ(A0)-ε).Thus we can get a contradiction by using similar argument for the proof of caseρ(A0)≥1/2. So every solution f(≢0) of (5) is infinite order in this case.Lastly, suppose thatρ(A0)=0. Note that A0(z) is a transcendental entire function. By using the results of [29] or [8], for any θ∈[0.2π), we have
(30)limsup⁡r→∞log⁡|A0(reiθ)|log⁡r=∞.
Thus we can get a contradiction by using similar argument for the proof of case ρ(A0)≥1/2. Therefore, every solution f(≢0) of (5) is infinite order in this case.Next we prove thatρ2(f)≥ρ(A0). Firstly, suppose that ρ(A0)>0. By using similar argument as above, there exists θ0(α′≤θ0≤β′) and there is a sequence of {rn} with lim⁡n→∞rn=∞ such that (23) and (26) (or (29)) hold for all sufficiently large n.By Lemma6(i) that there exist θ0(α′≤θ0≤β′) and constants R3>0, d>0, such that the following inequality
(31)|f(m)(z)f(z)|≤dT(2r,f)2k,m=1,…,k
holds for |z|=r>R3.Hence, calculating at the pointszn=rneiθ0 with rn∉[0,R1]⋃[0,R3], from (16), (23), (26) (or (29)), and (31), we get
(32)exp⁡(rnρ(A0)-ε)≤dT(2rn,f)2k(|aj0|1+(k-2)exp⁡(rnη)aaaaaaaaaaaaa+exp⁡{-CT(rn,Al)}+|aj0|).
Thus
(33)lim⁡r→∞¯log⁡+log⁡+T(r,f)log⁡r≥ρ(A0).
This gives ρ2(f)≥ρ(A0).Ifρ(A0)=0, obviously, ρ2(f)≥0. The proof of Theorem 5 is completed.


## 4. Further Results

In this section, we will study (5) with coefficients A0 and Al which are both extremal forYang’s inequality.Theorem 12.
LetAj(j=0,1,…,k-1) be entire functions. Suppose that there exists an integer l∈{1,2,…,k-1}, such that Al is extremal forYang’s inequality p1=q1/2. Suppose that A0 is an entire function extremal forYang’s inequality p2=q2/2 and ρ(Ai)<ρ(A0) for i≠l(1≤i≤k-1). Suppose that one of the following conditions holds:(1)
q1≠q2,(2)
q1=q2, and the set of Borel directions of Al is different from that of A0.Then every solution f(≢0) of (5) satisfies ρ(f)=∞ and ρ2(f)≥ρ(A0).Proof.
We first treat the case that the entire functionsAl and A0 satisfy condition (1).
Note that ifρ(Al)≠ρ(A0), then the conclusion of Theorem 12 follows from Theorem 5. Now suppose that ρ(Al)=ρ(A0)=ρ. We divide the proof into two cases: (a) q1<q2 and (b) q1>q2.
Now suppose that (a)q1<q2 holds. It is easy to see from Lemmas 7 and 8 that there are q2/2 sectors with magnitude π/ρ such that
(34)limsup⁡r→∞log⁡+log⁡+|A0(reiθ)|log⁡r=ρ,
while there are q1/2 sectors with magnitude π/ρ such that
(35)limsup⁡r→∞log⁡+log⁡+|Al(reiθ)|log⁡r=ρ.
Note that q1<q2. It is easy to see that there exists a sector Ω(α,β)(0<α<β<2π) such that, for every θ∈(α,β), Al(z) must be bounded in Ω(α,β), while A0(z) satisfies (34). So, by using the same argument in the proof of Theorem 5, we can easily prove the theorem.
We next suppose that (b)q1>q2. It is not hard to see that there must exist a sector Ω(α,β) such that Al(z) is bounded in Ω(α,β), while for any θ∈Ω(α,β)A0(z) satisfies (34). By using similar arguments as we did before, we can prove the theorem under the condition that Al(z) and A0(z) satisfy (1).
We turn to the case thatAl(z) and A0(z) satisfy (2). In this case, it is easy to see that there exists a sector such that in it Al(z) is bounded, while A0(z) satisfies (34). By using similar arguments as we did before, we can prove the theorem in this case. We omit the details here. The proof of Theorem 12 is completed.Finally, in [16], we note that if an entire function f is extremal forYang’s inequality p=q/2, then for any positive integer m, f(m) also has some special properties. In the sectors where, for any θ, f satisfies
(36)limsup⁡r→∞log⁡+log⁡+|f(reiθ)|log⁡r=ρ,
for any θ, f(m) satisfies
(37)limsup⁡r→∞log⁡+log⁡+|f(m)(reiθ)|log⁡r=ρ,f decays to some deficient values exponentially and f(m) decays to 0 exponentially. Therefore, in the same manner as in the proofs of Theorems 5 and 12, we have the following result.Theorem 13.
LetAj(j=0,1,…,k-1) be entire functions. Suppose that there exists an integer l∈{1,2,…,k-1}, such that Al is extremal forYang’s inequality p1=q1/2. Suppose that A0 is a transcendental entire function with ρ(A0)≠ρ(Al) and ρ(Ai)<ρ(A0) for i≠l(1≤i≤k-1). Then every solution f≢0 of
(38)f(k)+Ak-1(z)f(k-1)+⋯+Al(m)(z)f(l)+⋯+A0(z)f=0
satisfies ρ(f)=∞ and ρ2(f)≥ρ(A0).Moreover, suppose thatA0 is an entire function extremal forYang’s inequality p2=q2/2 and that one of the following assumptions holds:(1)
q1≠q2,(2)
q1=q2, and the set of Borel directions of Al is different from that of A0.Then every solution f(≢0) of
(39)f(k)+Ak-1f(k-1)+⋯+Al(m)(z)f(l)+⋯+A0(n)(z)f=0
satisfies ρ(f)=∞ and ρ2(f)≥ρ(A0), where m and n are two nonnegative integers.Finally we give an example satisfying the conditions of Theorem12. Let again An(z)=∫0z‍e-tndt, (n≥2). So if we let Al=An(z), A0=Am(z)=∫0z‍e-tmdt with n≠m and all other coefficients Ai satisfy ρ(Ai)<ρ(A0) for i≠l(1≤i≤k-1), then, by Theorem 12(1), every solution f(≢0) of (5) satisfies ρ(f)=∞ and ρ2(f)≥ρ(A0). Furthermore, if we let A0(z)=An(eiθz) with θ∈(0,π/2n), then, by Theorem 12(2), every solution f(≢0) of (5) satisfies ρ(f)=∞ and ρ2(f)≥ρ(A0).



---

*Source: 305710-2014-04-10.xml*