# Value Distribution of Meromorphic Solutions and Their Derivatives of Complex Differential Equations

Authors: Abdallah El Farissi
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

We deal with the relationship between the small functions and the derivatives of solutions of higher-order linear differential equationsf(k)+Ak-1f(k-1)+⋯+A0f=0,   k≥2, where Aj(z)  (j=0,1,…,k-1) are meromorphic functions. The theorems of this paper improve the previous results given by El Farissi, Belaïdi, Wang, Lu, Liu, and Zhang.


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## Body

## 1. Introduction and Statement of Result

Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna's value distribution theory (see [1, 2]). In addition, we will use λ(f) and λ(1/f) to denote, respectively, the exponents of convergence of the zero sequence and the pole sequence of a meromorphic function f, ρ(f) to denote the order of growth of f, σ(f) to denote the type of the entire function with 0<ρ(f)<∞, and λ¯(f) and λ¯(1/f) to denote, respectively, the exponents of convergence of the sequence of distinct zeros and distinct poles of f. A meromorphic function φ(z) is called a small function of a meromorphic function f(z) if T(r,φ)=o(T(r,f)) as r→+∞, where T(r,f) is the Nevanlinna characteristic function of f. In order to express the rate of growth of meromorphic solutions of infinite order, we recall the following definitions.Definition 1 (see [2–4]).
Letf be a meromorphic function, and let z1,z2,…, such that (|zj|=rj,0<r1≤r2≤⋯), be the sequence of the fixed points of f, with each point being repeated only once. The exponent of convergence of the sequence of distinct fixed points of f is defined by the following:
(1)τ¯(f)=inf{τ>0:∑j=1+∞|zj|-τ<+∞}.
Clearly,
(2)τ¯(f)=limr→+∞¯logN¯(r,1/(f-z))logr,
where N¯(r,1/(f-z)) is the counting function of distinct fixed points of f(z) in {|z|<r}.Definition 2 (see [4–6]).
Letf be a meromorphic function. Then the hyperorder ρ2(f)of f(z) is defined by the following:
(3)ρ2(f)=limr→+∞¯loglogT(r,f)logr.Definition 3 (see [4, 5]).
Letf be a meromorphic function. Then the hyperexponent of convergence of the sequence of distinct zeros of f(z) is defined by the following:
(4)λ¯2(f)=limr→+∞¯loglogN¯(r,1/f)logr,
where N¯(r,1/f) is the counting function of distinct zeros of f(z) in {|z|<r}.Fork≥2, we consider the following linear differential equation:
(5)f(k)+Af=0,
where A(z) is a transcendental meromorphic function of finite order ρ(A)=ρ>0. Many important results have been obtained on the fixed points of general transcendental meromorphic functions for almost four decades (see [7]). However, there are a few studies on the fixed points of solutions of differential equations. In [8], Wang and Lü have investigated the fixed points and hyperorder of solutions of second-order linear differential equations with meromorphic coefficients and their derivatives, and they have obtained the following result.Theorem A (see [8]).
Suppose thatA(z) is a transcendental meromorphic function satisfying δ(∞,A)=limr→+∞¯(m(r,A)/T(r,A))=δ>0, ρ(A)=ρ<+∞. Then, every meromorphic solution  f(z)≢0 of the equation
(6)f′′+A(z)f=0
satisfies that f and f′, f′′ all have infinitely many fixed points and
(7)τ¯(f)=τ¯(f′)=τ¯(f′′)=ρ(f)=+∞,τ¯2(f)=τ¯2(f′)=τ¯2(f′′)=ρ2(f)=ρ.Theorem A has been generalized to higher-order differential equations by Liu Ming-Sheng and Zhang Xiao-Mei as follows.Theorem B (see [4]).
Suppose thatk≥2 and A(z) are transcendental meromorphic functions satisfying δ(∞,A)=δ>0, ρ(A)=ρ<+∞. Then every meromorphic solution f(z)≢0 of (5) satisfies that f and f′,f′′,…,f(k) all have infinitely many fixed points and
(8)τ¯(f)=τ¯(f′)=τ¯(f′′)=⋯=τ¯(f(k))=ρ(f)=+∞,τ¯2(f)=τ¯2(f′)=τ¯2(f′′)=⋯=τ¯2(f(k))=ρ2(f)=ρ.In [9], Belaïdi and El Farissi extended the result of Theorem B, and they gave the following theorem.Theorem C (see [9]).
Suppose thatk≥2 and A(z) are transcendental meromorphic functions satisfying δ(∞,A)=δ>0 and ρ(A)=ρ  (0<ρ<+∞). If φ(z)≢0 is a meromorphic function with finite order ρ(φ)<+∞, then every meromorphic solution f(z)≢0 of (5) satisfies
(9)λ¯(f-φ)=λ¯(f′-φ)=⋯=λ¯(f(k)-φ)λ¯(f-φ)=ρ(f)=+∞,λ¯2(f-φ)=λ¯2(f′-φ)=⋯=λ¯2(f(k)-φ)λ¯2(f-φ)=ρ2(f)=ρ.Recently, Xu et al. [10] investigated the relationship between small functions and the derivatives of solutions of the following equation:
(10)f(k)+Ak-1f(k-1)+⋯+A0f=0,k≥2,
and obtained the following theorems which improve the results given by Chen, Wang, Lu, Liu, Zhang, Belaïdi, and El Farissi.Theorem D (see [10]).
Letk≥2 and Aj(j=0,1,…,k-1) be entire functions with finite order and satisfy one of the following conditions:(i)
max{ρ(Aj),(j=1,…,k-1)}<ρ(A0)=ρ<+∞,(ii)
0<ρ(Ak-1)=ρ(Ak-2)=⋯=ρ(A0)<∞ and max{σ(Aj),(j=1,…,k-1)} = σ1<σ(A0) = σ;Then, for every solution f≢0 of (10) and for any entire function φ≢0 satisfying ρ2(φ)<ρ(A0), one has the following:
(11)λ¯2(f(i)-φ)=λ2(f(i)-φ)=ρ2(f)=ρ,i∈ℕ.Theorem E (see [10]).
Letk≥2 and Aj(j=0,1,…,k-1) be meromorphic functions satisfying max{ρ(Aj)(j=1,…,k-1)}<ρ(A0)=ρ<+∞and δ(∞,A0)>0. Then for every meromorphic solution f≢0 of (10) and for any meromorphic function φ≢0 satisfying ρ2(φ)<ρ(A0), one has λ¯2(f(i)-φ)=λ2(f(i)-φ)⩾ρ(i=0,1,…).In this paper, we will deal with the above equation, investigate the relationship between small functions and derivatives of solutions of (10), obtain some results, which improve the previous results given by Xu, Tu, and Zheng, and prove the following theorems.Theorem 4.
Letk≥2 and Aj(j=0,1,…,k-1) be meromorphic functions of finite order such that all solutions of (10) satisfy ρ(f)=+∞ and ρ2(f)=ρ. Then if φ(z)≢0 is a meromorphic function with ρ2(φ)<ρ, then every solution f(z)≢0 of (10) satisfies
(12)λ¯(f(i)-φ)=λ(f(i)-φ)=ρ(f)=+∞,i∈ℕ,λ¯2(f(i)-φ)=λ2(f(i)-φ)=ρ2(f)=ρ,i∈ℕ.Theorem 5.
Letk≥2 and Aj(j=0,1,…,k-1) be meromorphic functions of finite order such that all solutions of (10) satisfy ρ(f)=+∞. Then if φ(z)≢0 is a meromorphic function with ρ(φ)<∞, then every solution f(z)≢0 of (10) satisfies (12). Furthermore, if ρ2(f)⩾ρ, and ρ2(φ)<ρ, then
(13)λ¯2(f(i)-φ)=λ2(f(i)-φ)=ρ2(f)⩾ρ,i∈ℕ.Remark 6.
The following example shows that Theorem4 is not valid when (10) has a finite order solution.Example 7.
For the following equation*f′′+e2z+ez-11-ezf′+-e2z1-ezf=0
we can easily get that (*) has solutions f1(z)=eez+αez, α∈ℂ-{0}, and f2(z)=ez. Solution f1 satisfies ρ(f1)=+∞ and ρ2(f)=1. Take φ(z)=αez; then ρ2(φ)<1. Thus, we can get that λ¯(f′-φ)=0≠∞=ρ(f) and λ¯2(f′-φ)=0≠1=ρ2(f).Remark 8.
The proof of Theorem4 is quite different from that of Theorem D (see [10]). The main ingredient in the proof is Lemma 17.Corollary 9.
Under the assumptions of Theorem4, if φ(z)=z, then for every solution f of (10), one has the following:
(14)τ¯(f(i))=τ(f(i))=ρ(f)=+∞,i∈ℕ,τ¯2(f(i))=τ2(f(i))=ρ2(f)=ρ(A0)=ρ,i∈ℕ.Remark 10.
Theorem4 is the improvement of Theorems A, B, C, and D, and Theorem 5 is the improvement of Theorem E.Corollary 11.
Letk≥2 and hj(z)(j=0,1,…,k-1)(hj≢0) be entire functions that have finite order such that ρ=max{ρ(hj):j=0,1,…,k-1}<n. Let Pj(z)=an,jzn+an-1,jzn-1+⋯+a1,jz+a0,jbe polynomials, where an,j≠0(j=0,1,…,k-1), ai,j(j=0,1,…,k-1); i=(0,1,…,n-1) are complex numbers, and suppose that argan,j≠argan,0 or an,j=cn,jan,0, where 0<cn,j<1, j∈{1,2,…,k-1}. If φ(z)≢0 is an entire function with ρ2(φ)<n, then every solution f(z)≢0 of the equation
(15)f(k)+hk-1ePk-1(z)f(k-1)+⋯+hsePs(z)f(s)f(k)+⋯+h1eP1(z)f′+h0eP0(z)f=0satisfies(12) with ρ=n.


## 2. Auxiliary Lemmas

The following lemmas will be used in the proofs of Theorem4 and Corollary 11.Lemma 12 (see [11]).
Letk≥2 and hj(z)(j=0,1,…,k-1)(hj≢0) be entire functions that have finite order such that ρ=max{ρ(hj):j=0,1,…,k-1}<n. Let Pj(z)=an,jzn + an-1,jzn-1 + ⋯ + a1,jz + a0,jbe polynomials, where an,j≠0(j=0,1,…,k-1), ai,j(j=0,1,…,k-1;i=0,1,…,n-1) are complex numbers, and suppose that argan,j≠argan,0 or an,j=cn,jan,0, where 0<cn,j<1, j∈{1,2,…,k-1}. Then every solution of (15) of infinite order and ρ2(f)=n.Lemma 13 (see [3]).
LetA0,A1,…,Ak-1,  F(≢0) be finite order meromorphic functions. If f is a meromorphic solution with ρ(f)=+∞ of the following equation:
(16)f(k)+Ak-1f(k-1)+⋯+A1f′+A0f=F,
then λ¯(f)=λ(f)=ρ(f)=+∞.Lemma 14 (see [12]).
LetA0,A1,…,Ak-1,  F(≢0) be finite order meromorphic functions. If f is a meromorphic solution of (16), then we have the following statements:(i)
ifmax{σ(F),σ(Aj);j=0,1,…,k-1}<σ(f)=σ≤∞, then σ(f)=λ(f)=λ¯(f);(ii)
ifmax{σ2(F),σ(Aj);j=0,1,…,k-1}<σ2(f)=σ≤∞, then σ2(f)=λ2(f)=λ¯2(f).LetAj(j=0,1,…,k-1) be meromorphic functions. We define the sequences of functions by the following:
(17)Aj0=Aj,j=0,1,…,k-1,Ak-1i=Ak-1i-1+(A0i-1)′A0i-1,i∈ℕ,Aji=Aji-1+Aj+1i-1(Ψj+1i-1)′Ψj+1i-1,j=0,1,…,k-2,i∈ℕ,
where Ψj+1i-1=Aj+1i-1/A0i-1.Remark 15.
In the case where one of the functions ofAji  (j=0,1,…,k-1) is equal to zero, then Aji+1=Aj-1i(j=0,1,…,k-1).Lemma 16.
Suppose thatf is a solution of (10). Then gi=f(i) is a solution of the following equation:
(18)g(k)+Ak-1ig(k-1)+⋯+A0ig=0,
where Aji(j=0,1,…,k-1) are given by (17).Proof.
Assume thatf is a solution of (10), and let gi=f(i). We prove that gi is an entire solution of (18). Our proof is by induction. For i=1, differentiating both sides of (10), we obtain
(19)f(k+1)+Ak-1f(k)+(Ak-1′+Ak-2)f(k-1)+⋯f(k+1)+(A1′+A0)f′+A0′f=0,
and replacing f by
(20)f=-(f(k)+Ak-1f(k-1)+⋯+A1f′)A0,
we get
(21)f(k+1)+(Ak-1-A0′A0)f(k)f(k+1)+(Ak-1′+Ak-2-Ak-1A0′A0)f(k-1)⋯f(k+1)+(A1′+A0-A1A0′A0)f′=0.
That is,
(22)g1(k)+Ak-11g1(k-1)+Ak-21g1(k-2)⋯+A01g1=0.
Suppose that the assertion is true for the values which are strictly smaller than a certain i. We suppose gi-1 is a solution of the following equation:
(23)gi-1(k)+Ak-1i-1gi-1(k-1)+Ak-2i-1gi-1(k-2)+⋯+A0i-1gi-1=0.
Differentiating (23), we can write the following:
(24)gi-1(k+1)+Ak-1i-1gi-1(k)+((Ak-1i-1)′+Ak-2)gi-1(k-1)+⋯gi-1(k+1)+((A1i-1)′+A0i-1)gi-1′+A0′gi-1=0.
In (24), replacing gi-1 by
(25)gi-1=-(gi-1(k)+Ak-1i-1gi-1(k-1)+Ak-2i-1gi-1(k-2)+⋯+A(gi-1)′)A0i-1
yields the following:
(26)gi-1(k+1)+(Ak-1i-1-(A0i-1)′A0i-1)gi-1(k)gi-1(k+1)+((Ak-1i-1)′+Ak-2-Ak-1i-1(A0i-1)′A0i-1)gi-1(k-1)+⋯gi-1(k+1)+((A1i-1)′+A0i-1-A1i-1(A0i-1)′A0i-1)gi-1′=0.
That is,
(27)gi(k)+Ak-1i-1gi(k-1)+Ak-2i-1gi(k-2)+⋯+A0i-1gi=0.
Lemma 16 is thus proved.Lemma 17.
LetAj(j=0,1,…,k-1) be meromorphic functions of finite order such that all solutions of (10) have infinite order and ρ2(f)=ρ, and let Aji(j=0,1,…,k-1) be defined as in (17). Then the nontrivial meromorphic solution g of the equation
(28)g(k)+Ak-1ig(k-1)+⋯+A0ig=0,k≥2,
satisfies ρ(g)=+∞ and ρ2(g)=ρ.Proof.
Let{f1,f2,…,fk} be a fundamental system of solutions of (10). We show that {f1(i),f2(i),…,fk(i)} is a fundamental system of solutions of (28). By Lemma 16, it follows that f1(i),f2(i),…,fk(i) are solutions of (28). Let α1,α2,…,αk be constants such that
(29)α1f1(i)+α2f2(i)+⋯+αkfk(i)=0.
Then, we have
(30)α1f1+α2f2+⋯+αkfk=P(z),
where P(z) is a polynomial of a degree less than i. Since α1f1+α2f2+⋯+αkfk is a solution of (10), then P is a solution of (10), and by the conditions of the lemma, we conclude that P is an infinite order solution of (10); this leads to a contradiction. Therefore, P is a trivial solution. We deduce that α1f1+α2f2+⋯+αkfk=0. Using the fact that {f1,f2,…,fk} is a fundamental solution of (10), we get α1=α2=⋯=αk=0. Now, let g be a nontrivial solution of (28). Then, using the fact that {f1(i),f2(i),…,fk(i)} is a fundamental solution of (28), we claim that there exist constants α1,α2,…,αk not all equal to zero, such that g=α1f1(i)+α2f2(i)+⋯+αkfk(i). Let h=α1f1+α2f2+⋯+αkfk, h be a solution of (10), and h(i)=g. Hence, by conditions of the lemma, we get ρ(h)=ρ(g)=+∞ and ρ2(h)=ρ2(g)=+∞.Proof of Theorem4.
Assume thatf is a solution of (10). By the conditions of the theorem, we get ρ(f)=∞ and ρ2(f)=ρ. Taking gi=f(i), then ρ(gi)=∞ and ρ2(gi)=ρ. Now, let w(z)=gi(z)-φ(z), where φ is a meromorphic function with ρ2(φ)<ρ.
Thenρ(w)=∞ and ρ2(w)=ρ.
In order to proveλ¯(gi-φ)=λ(gi-φ)=∞ and λ¯2(gi-φ)=λ2(gi-φ)=ρ, we need to prove only that λ¯(w)=λ(w)=∞ and λ¯2(w)=λ2(w)=ρ. Using the fact that gi=w+φ and by Lemma 16 we get the following:
(31)w(k)+Ak-1iw(k-1)+⋯+A0iw=-(φ(k)+Ak-1iφ(k-1)+⋯+A0iφ)=F.
By ρ(Aji)<∞, ρ2(φ)<ρ, φ≢0 and Lemma 17, we get F≢0 and ρ2(F)<ρ. By Lemmas 13 and 14 we get λ¯(w)=λ2(w)=ρ(w)=∞ and λ¯2(w)=λ2(w)=ρ2(w)=ρ. The proof of Theorem 4 is complete.Proof of Theorem5.
By the same reasoning as before we can prove Theorem5.Proof of Corollary11.
By Lemma12, all solutions of (15) satisfy ρ(f)=+∞ and ρ2(f)=n. Applying Theorem 4 we get λ¯(f(i)-φ)=ρ(f)=+∞, i∈ℕ and λ¯2(f(i)-φ)=n, i∈ℕ.



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