# Some Properties of Solutions of Second-Order Linear Differential Equations

Authors: Zinelaâbidine Latreuch; Benharrat Belaïdi
Journal: Journal of Complex Analysis
 (2013)

---

## Abstract

We study the growth and oscillation ofgf=d1f1+d2f2, where d1 and
d2 are entire functions of finite order not all vanishing identically and f1 and
f2 are two linearly independent solutions of the linear differential equation f′′+A(z)f=0.


---

## Body

## 1. Introduction and Main Results

Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna value distribution theory (see [1–4]). In addition, we will use λ(f) and λ¯(f) to denote, respectively, the exponents of convergence of the zero sequence and distinct zeros of a meromorphic function f, ρ(f) to denote the order of growth of f.Definition 1 (see [4, 5]).
Letf be a meromorphic function. Then the hyperorder of f(z) is defined by
(1)ρ2(f)=limsupr→+∞loglogT(r,f)logr.Definition 2 (see [4, 5]).
Letf be a meromorphic function. Then the hyper-exponent of convergence of zeros sequence of f(z) is defined by
(2)λ2(f)=limsupr→+∞loglogN(r,1/f)logr,
where N(r,1/f) is the counting function of zeros of f(z) in {z:|z|<r}. Similarly, the hyperexponent of convergence of the sequence of distinct zeros of f(z) is defined by
(3)λ¯2(f)=limsupr→+∞loglogN¯(r,1/f)logr,
where N¯(r,1/f) is the counting function of distinct zeros of f(z) in {z:|z|<r}.Suppose thatf1 and f2 are two linearly independent solutions of the complex linear differential equation
(4)f′′+A(z)f=0,
and the polynomial of solutions
(5)gf=d1f1+d2f2,
where A and dj(j=1,2) are entire functions of finite order in the complex plane. It is clear that if dj(j=1,2) are complex numbers or d1=cd2 where c is a complex number, then gf is a solution of (4) or has the same properties of the solutions.It is natural to ask what can be said about the properties ofgf in the case when d1≠cd2 where c is a complex number and under what conditions gf keeps the same properties of the solutions of (4).In [6], Chen studied the fixed points and hyper-order of solutions of second-order linear differential equations with entire coefficients and obtained the following results.Theorem A (see [6]).
For all nontrivial solutionsf of (4) the following hold.(i)
IfA is a polynomial with degA=n≥1, then one has
(6)λ(f-z)=ρ(f)=n+22.(ii)
IfA is transcendental and ρ(A)<∞, then one has
(7)λ(f-z)=ρ(f)=∞,λ2(f-z)=ρ2(f)=ρ(A).Before we state our results we defineh and ψ by
(8)h=|d10d20d1′d1d2′d2d1′′-d1A2d1′d2′′-d2A2d2′d1′′′-3d1′A-d1A′d1′′-d1A+2d1′′d2′′′-3d2′A-d2A′d2′′-d2A+2d2′′|,ψ(z)=2(d1d2d2′-d22d1′)hφ(3)+ϕ2φ′′+ϕ1φ′+ϕ0φ,
where φ≢0 is entire function of finite order and
(9)ϕ2=3d22d1′′-3d1d2d2′′h,ϕ1=2d1d2d2′A+6d2d1′d2′′-6d2d2′d1′′-2d22d1′Ah,ϕ0=2d2d1′d2′′′-2d1d2′d2′′′-3d1d2d2′′A-3d2d1′′d2′′h+2d1d2d2′A′-4d2d1′d2′A-6d1′d2′d2′′+3d1(d2′′)2h+4d1(d2′)2A+3d22d1′′A+6(d2′)2d1′′-2d22d1′A′h.
The subject of this paper is to study the controllability of solutions of the differential equation (4). In fact, we study the growth and oscillation of gf=d1f1+d2f2 where f1 and f2 are two linearly independent solutions of (4) and d1  and d2 are entire functions of finite order not all vanishing identically and satisfying d1≠cd2 where c is a complex number, and we obtain the following results.Theorem 3.
LetA(z) be a transcendental entire function of finite order. Let dj(z)(j=1,2) be finite-order entire functions that are not all vanishing identically such that max{ρ(d1),ρ(d2)}<ρ(A). If f1 and  f2 are two linearly independent solutions of (4), then the polynomial of solutions (5) satisfies
(10)ρ(gf)=ρ(fj)=∞(j=1,2),ρ2(gf)=ρ2(fj)=ρ(A)(j=1,2).Theorem 4.
Under the hypotheses of Theorem3, let φ(z)≢0 be an entire function with finite order such that ψ(z)≢0. If f1 and f2 are two linearly independent solutions of (4), then the polynomial of solutions (5) satisfies
(11)λ¯(gf-φ)=λ(gf-φ)=ρ(fj)=∞(j=1,2),λ¯2(gf-φ)=λ2(gf-φ)=ρ2(fj)=ρ(A)(j=1,2).Theorem 5.
LetA(z) be a polynomial of degA=n. Let dj(z)  (j=1,2) be finite-order entire functions that are not all vanishing identically such that h≢0 and max{ρ(d1),ρ(d2)}<(n+2)/2. If f1f2 are two linearly independent solutions of (4), then the polynomial of solutions (5) satisfies
(12)ρ(gf)=ρ(fj)=n+22(j=1,2).Theorem 6.
Under the hypotheses of Theorem5, let φ(z)≢0 be an entire function with ρ(φ)<(n+2)/2 such that ψ(z)≢0. If f1 and f2 are two linearly independent solutions of (4), then the polynomial of solutions (5) satisfies
(13)λ¯(gf-φ)=λ(gf-φ)=n+22.


## 2. Auxiliary Lemmas

Lemma 7 (see [7, 8]).
LetA0,A1,…,Ak-1,F≢0 be finite-order meromorphic functions. If f is a meromorphic solution of the equation
(14)f(k)+Ak-1f(k-1)+⋯+A1f′+A0f=F
with ρ(f)=+∞ and ρ2(f)=ρ, then f satisfies
(15)λ¯(f)=λ(f)=ρ(f)=+∞,λ¯2(f)=λ2(f)=ρ2(f)=ρ.Here, we give a special case of the result due to Cao et al. in [9].Lemma 8.
LetA0,A1,…,Ak-1,F≢0 be finite-order meromorphic functions. If f is a meromorphic solution of (14)  with
(16)max{ρ(Aj)(j=0,1,…,k-1),ρ(F)}<ρ(f)<+∞,
then
(17)λ¯(f)=λ(f)=ρ(f).


## 3. Proofs of the Theorems

Proof of Theorem3.
Suppose thatf1 and f2 are two linearly independent solutions of (4). Then by Theorem A, we have
(18)ρ(f1)=ρ(f2)=∞,ρ2(f1)=ρ2(f2)=ρ(A).
Suppose that d1=cd2, where c is a complex number. Then, by (5) we obtain
(19)gf=cd2f1+d2f2=(cf1+f2)d2.
Since f=cf1+f2 is a solution of (4) and ρ(d2)<ρ(A), then we have
(20)ρ(gf)=ρ(cf1+f2)=∞,ρ2(gf)=ρ2(cf1+f2)=ρ(A).
Suppose now that d1≢cd2 where c is a complex number. Differentiating both sides of (5), we obtain
(21)gf′=d1′f1+d1f1′+d2′f2+d2f2′.
Differentiating both sides of (21), we obtain
(22)gf′′=d1′′f1+2d1′f1′+d1f1′′+d2′′f2+2d2′f2′+d2f2′′.
Substituting fj′′=-Afj(j=1,2) into (22), we obtain
(23)gf′′=(d1′′-d1A)f1+2d1′f1′+(d2′′-d2A)f2+2d2′f2′.
Differentiating both sides of (23) and by substituting fj′′=-Afj(j=1,2), we obtain
(24)gf′′′=(d1′′′-3d1′A-d1A′)f1+(d1′′-d1A+2d1′′)f1′+(d2′′′-3d2′A-d2A′)f2+(d2′′-d2A+2d2′′)f2′.
By (5), (21), (23), and (24) we have
(25)gf=d1f1+d2f2,gf′=d1′f1+d1f1′+d2′f2+d2f2′,gf′′=(d1′′-d1A)f1+2d1′f1′+(d2′′-d2A)f2+2d2′f2′,gf′′′=(d1′′′-3d1′A-d1A′)f1+(d1′′-d1A+2d1′′)f1′+(d2′′′-3d2′A-d2A′)f2+(d2′′-d2A+2d2′′)f2′.
To solve this system of equations, we need first to prove that h≢0. By simple calculations we obtain
(26)h=|d10d20d1′d1d2′d2d1′′-d1A2d1′d2′′-d2A2d2′d1′′′-3d1′A-d1A′d1′′-d1A+2d1′′d2′′′-3d2′A-d2A′d2′′-d2A+2d2′′|=(4d12(d2′)2+4d22(d1′)2-8d1d2d1′d2′)A+2d1d2d1′d2′′′+2d1d2d2′d1′′′-6d1d2d1′′d2′′-6d1d1′d2′d2′′-6d2d1′d2′d1′′+6d1(d2′)2d1′′+6d2(d1′)2d2′′-2d22d1′d1′′′-2d12d2′d2′′′+3d12(d2′′)2+3d22(d1′′)2.
To show that 4d12(d2′)2+4d22(d1′)2-8d1d2d1′d2′≢0, we suppose that
(27)d12(d2′)2+d22(d1′)2-2d1d2d1′d2′=0.
Dividing both sides of (27) by (d1d2)2, we obtain
(28)(d2′d2)2+(d1′d1)2-2d1′d1d2′d2=0
equivalent to
(29)(d1′d1-d2′d2)2=0,
which implies that d1=cd2 where c is a complex number and this is a contradiction. Since max{ρ(d1),ρ(d2)}<ρ(A) and 4d12(d2′)2+4d22(d1′)2-8d1d2d1′d2′≢0, we can deduce from (26) that
(30)ρ(h)=ρ(A)>0.
Hence h≢0. By Cramer's method we have
(31)f1=|gf0d20gf′d1d2′d2gf′′2d1′d2′′-d2A2d2′gf′′′d1′′-d1A+2d1′′d2′′′-3d2′A-d2A′d2′′-d2A+2d2′′|h=2(d1d2d2′-d22d1′)hgf(3)+ϕ2gf′′+ϕ1gf′+ϕ0gf,
where ϕj(j=0,1,2) are meromorphic functions of finite order defined in (9). Suppose now that ρ(gf)<∞, then by (31) we obtain ρ(f1)<∞, which is a contradiction, hence ρ(gf)=∞. By (5) we have ρ2(gf)≤ρ(A). Suppose that ρ2(gf)<ρ(A), then by (31) we obtain ρ2(f1)<ρ(A), which is a contradiction. Hence ρ2(gf)=ρ(A).Proof of Theorem4.
By Theorem3 we have ρ(gf)=∞ and ρ2(gf)=ρ(A). Set w(z)=d1f1+d2f2-φ. Since ρ(φ)<∞, then we have ρ(w)=ρ(gf)=∞ and ρ2(w)=ρ2(gf)=ρ(A). In order to prove that λ¯(gf-φ)=λ(gf-φ)=∞,  λ¯2(gf-φ)=λ2(gf-φ)=ρ(A) we need to prove only that λ¯(w)=λ(w)=∞ and λ¯2(w)=λ2(w)=ρ(A). By gf=w+φ we get from (31)
(32)f1=2(d1d2d2′-d22d1′)hw(3)+ϕ2w′′+ϕ1w′+ϕ0w+ψ,
where
(33)ψ=2(d1d2d2′-d22d1′)hφ(3)+ϕ2φ′′+ϕ1φ′+ϕ0φ.
Substituting (32) into (4), we obtain
(34)2(d1d2d2′-d22d1′)hw(5)+∑j=04βjw(j)=-(ψ′′+Aψ)=B,
where βj(j=0,…,4) are meromorphic functions of finite order. Since ψ≢0 and ρ(ψ)<∞, it follows that ψ is not a solution of (4), which implies that B≢0. Then by applying Lemma 7 we obtain (11).Proof of Theorem5.
Suppose thatf1 and f2 are two linearly independent solutions of (4). Then by Theorem A
(35)ρ(f1)=ρ(f2)=n+22.
By the same reasoning as in Theorem 3, we have
(36)h=|d10d20d1′d1d2′d2d1′′-d1A2d1′d2′′-d2A2d2′d1′′′-3d1′A-d1A′d1′′-d1A+2d1′′d2′′′-3d2′A-d2A′d2′′-d2A+2d2′′|.
Since h≢0 and by Cramer's method we have
(37)f1=|gf0d20gf′d1d2′d2gf′′2d1′d2′′-d2A2d2′gf′′′d1′′-d1A+2d1′′d2′′′-3d2′A-d2A′d2′′-d2A+2d2′′|h=2(d1d2d2′-d22d1′)hgf(3)+ϕ2gf′′+ϕ1gf′+ϕ0gf,
where ϕj(j=0,1,2) are meromorphic functions with ρ(ϕj)<(n+2)/2(j=0,1,2) defined in (9). By (5) we have ρ(gf)≤(n+2)/2. Suppose that ρ(gf)<(n+2)/2, then by (37) we obtain ρ(f1)<(n+2)/2, which is a contradiction. Hence, ρ(gf)=(n+2)/2.Proof of Theorem6.
By Theorem5 we have ρ(gf)=(n+2)/2. Set w(z)=d1f1+d2f2-φ. Since ρ(φ)<(n+2)/2, then we have ρ(w)=ρ(gf)=(n+2)/2. In order to prove that λ¯(gf-φ)=(n+2)/2, we need to prove only that λ¯(w)=(n+2)/2. By gf=w+φ we get from (37)
(38)f1=2(d1d2d2′-d22d1′)hw(3)+ϕ2w′′+ϕ1w′+ϕ0w+ψ,
where
(39)ψ=2(d1d2d2′-d22d1′)hφ(3)+ϕ2φ′′+ϕ1φ′+ϕ0φ.
Substituting (38) into (4), we obtain
(40)2(d1d2d2′-d22d1′)hw(5)+∑j=04βjw(j)=-(ψ′′+Aψ)=B,
where βj(j=0,…,4) are meromorphic functions with ρ(βj)<(n+2)/2. Since ψ≢0 and ρ(ψ)<(n+2)/2, it follows that ψ is not a solution of (4), which implies that B≢0. Then by applying Lemma 8 we obtain (13).



---

*Source: 253168-2013-01-30.xml*