Volume 2013, Article ID 253168, 5 pages http://dx.doi.org/10.1155/2013/253168
Research Article
Some Properties of Solutions of Second-Order
Linear Differential Equations
Zinelaâbidine Latreuch and Benharrat Belaïdi
Laboratory of Pure and Applied Mathematics, Department of Mathematics, University of Mostaganem (UMAB), BP 227, 27000 Mostaganem, Algeria
Correspondence should be addressed to Benharrat Belaïdi; belaidi@univ-mosta.dz Received 13 August 2012; Accepted 25 September 2012
Academic Editor: Rabha W. Ibrahim
Copyright © 2013 Z. Latreuch and B. Belaïdi. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We study the growth and oscillation of 𝑔𝑔𝑓𝑓 = 𝑑𝑑1𝑓𝑓1+ 𝑑𝑑2𝑓𝑓2, where 𝑑𝑑1and 𝑑𝑑2are entire functions of �nite order not all vanishing
identically and 𝑓𝑓1and 𝑓𝑓2are two linearly independent solutions of the linear differential equation 𝑓𝑓′′+ 𝐴𝐴𝐴𝐴𝐴𝐴𝑓𝑓 = 𝐴.
1. Introduction and Main Results
roughout 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 𝜆𝜆𝐴𝑓𝑓𝐴 and 𝜆𝜆𝐴𝑓𝑓𝐴 to denote, respectively, the exponents of convergence of the zero sequence and distinct zeros of a meromorphic function 𝑓𝑓, 𝜌𝜌𝐴𝑓𝑓𝐴 to denote the order of growth of 𝑓𝑓.
De�nition � (see [4, 5]). Let 𝑓𝑓 be a meromorphic function.
en the hyperorder of 𝑓𝑓𝐴𝐴𝐴𝐴 is de�ned by 𝜌𝜌2𝑓𝑓 = lim sup
𝑟𝑟 𝑟 +𝑟
log log 𝑇𝑇 𝑟𝑟𝑟 𝑓𝑓
log 𝑟𝑟 . (1)
De�nition 2 (see [4, 5]). Let 𝑓𝑓 be a meromorphic function.
en the hyper-exponent of convergence of zeros sequence of 𝑓𝑓𝐴𝐴𝐴𝐴 is de�ned by
𝜆𝜆2𝑓𝑓 = lim sup
𝑟𝑟 𝑟 +𝑟
log log 𝑁𝑁 𝑟𝑟𝑟 1𝑟𝑓𝑓
log 𝑟𝑟 𝑟 (2)
where 𝑁𝑁𝐴𝑟𝑟𝑟 1𝑟𝑓𝑓𝐴 is the counting function of zeros of 𝑓𝑓𝐴𝐴𝐴𝐴 in {𝐴𝐴 𝑧 𝑧𝐴𝐴𝑧 𝑧 𝑟𝑟𝑧. Similarly, the hyperexponent of convergence of the sequence of distinct zeros of 𝑓𝑓𝐴𝐴𝐴𝐴 is de�ned by
𝜆𝜆2𝑓𝑓 = lim sup
𝑟𝑟 𝑟 +𝑟
log log 𝑁𝑁 𝑟𝑟𝑟 1𝑟𝑓𝑓
log 𝑟𝑟 𝑟 (3)
where 𝑁𝑁𝐴𝑟𝑟𝑟 1𝑟𝑓𝑓𝐴 is the counting function of distinct zeros of 𝑓𝑓𝐴𝐴𝐴𝐴 in {𝐴𝐴 𝑧 𝑧𝐴𝐴𝑧 𝑧 𝑟𝑟𝑧.
Suppose that 𝑓𝑓1 and 𝑓𝑓2 are two linearly independent solutions of the complex linear differential equation
𝑓𝑓′′+ 𝐴𝐴 𝐴𝐴𝐴𝐴 𝑓𝑓 = 𝐴𝑟 (4) and the polynomial of solutions
𝑔𝑔𝑓𝑓= 𝑑𝑑1𝑓𝑓1+ 𝑑𝑑2𝑓𝑓2𝑟 (5)
where 𝐴𝐴 and 𝑑𝑑𝑗𝑗𝐴𝑗𝑗 = 1𝑟 2𝐴 are entire functions of �nite order in the complex plane. It is clear that if 𝑑𝑑𝑗𝑗𝐴𝑗𝑗 = 1𝑟 2𝐴 are
complex numbers or 𝑑𝑑1 = 𝑐𝑐𝑑𝑑2where 𝑐𝑐 is a complex number,
then 𝑔𝑔𝑓𝑓is a solution of (4) or has the same properties of the solutions.
It is natural to ask what can be said about the properties of 𝑔𝑔𝑓𝑓in the case when 𝑑𝑑1≠ 𝑐𝑐𝑑𝑑2where 𝑐𝑐 is a complex number
and under what conditions 𝑔𝑔𝑓𝑓keeps the same properties of the solutions of (4).
In [6], Chen studied the �xed points and hyper-order of solutions of second-order linear differential equations with entire coefficients and obtained the following results. eorem A (see [6]). For all nontrivial solutions 𝑓𝑓 of (4) the
(i) If 𝐴𝐴 is a polynomial with deg 𝐴𝐴 𝐴 𝐴𝐴 𝐴 𝐴, then one has 𝜆𝜆 𝑓𝑓 𝑓 𝑓𝑓 𝐴 𝜌𝜌 𝑓𝑓 𝐴 𝐴𝐴 𝑛 𝑛
𝑛 . (6)
(ii) If 𝐴𝐴 is transcendental and 𝜌𝜌𝜌𝐴𝐴𝜌 𝜌 𝜌, then one has 𝜆𝜆 𝑓𝑓 𝑓 𝑓𝑓 𝐴 𝜌𝜌 𝑓𝑓 𝐴 𝜌,
𝜆𝜆𝑛𝑓𝑓 𝑓 𝑓𝑓 𝐴 𝜌𝜌𝑛𝑓𝑓 𝐴 𝜌𝜌 𝜌𝐴𝐴𝜌 .
(7) �efore we state our results we de�ne ℎ and 𝜓𝜓 by
ℎ 𝐴 𝑑𝑑𝐴 0 𝑑𝑑𝑛 0 𝑑𝑑′ 𝐴 𝑑𝑑𝐴 𝑑𝑑′𝑛 𝑑𝑑𝑛 𝑑𝑑′′ 𝐴𝑓 𝑑𝑑𝐴𝐴𝐴 𝑛𝑑𝑑′𝐴 𝑑𝑑′′𝑛𝑓 𝑑𝑑𝑛𝐴𝐴 𝑛𝑑𝑑′𝑛 𝑑𝑑′′′ 𝐴 𝑓 3𝑑𝑑′𝐴𝐴𝐴 𝑓 𝑑𝑑𝐴𝐴𝐴′ 𝑑𝑑′′𝐴𝑓 𝑑𝑑𝐴𝐴𝐴 𝑛 𝑛𝑑𝑑′′𝐴 𝑑𝑑′′′𝑛 𝑓 3𝑑𝑑′𝑛𝐴𝐴 𝑓 𝑑𝑑𝑛𝐴𝐴′ 𝑑𝑑′′𝑛𝑓 𝑑𝑑𝑛𝐴𝐴 𝑛 𝑛𝑑𝑑′′𝑛 , 𝜓𝜓 𝜌𝑓𝑓𝜌 𝐴 𝑛 𝑑𝑑𝐴𝑑𝑑𝑛𝑑𝑑 ′ 𝑛𝑓 𝑑𝑑𝑛𝑛𝑑𝑑′𝐴 ℎ 𝜑𝜑𝜌3𝜌𝑛 𝜙𝜙𝑛𝜑𝜑′′𝑛 𝜙𝜙𝐴𝜑𝜑′𝑛 𝜙𝜙0𝜑𝜑, (8) where 𝜑𝜑 ̸≡ 0 is entire function of �nite order and
𝜙𝜙𝑛𝐴 3𝑑𝑑𝑛𝑛𝑑𝑑′′𝐴 𝑓 3𝑑𝑑𝐴𝑑𝑑𝑛𝑑𝑑′′𝑛 ℎ , 𝜙𝜙𝐴𝐴 𝑛𝑑𝑑𝐴𝑑𝑑𝑛𝑑𝑑′𝑛𝐴𝐴 𝑛 𝐴𝑑𝑑𝑛𝑑𝑑′𝐴𝑑𝑑𝑛′′𝑓 𝐴𝑑𝑑𝑛𝑑𝑑′𝑛𝑑𝑑′′𝐴𝑓 𝑛𝑑𝑑𝑛𝑛𝑑𝑑′𝐴𝐴𝐴 ℎ , 𝜙𝜙0𝐴 𝑛𝑑𝑑𝑛𝑑𝑑′𝐴𝑑𝑑′′′𝑛 𝑓 𝑛𝑑𝑑𝐴𝑑𝑑′𝑛𝑑𝑑𝑛′′′𝑓 3𝑑𝑑ℎ 𝐴𝑑𝑑𝑛𝑑𝑑′′𝑛𝐴𝐴 𝑓 3𝑑𝑑𝑛𝑑𝑑′′𝐴𝑑𝑑′′𝑛 𝑛𝑛𝑑𝑑𝐴𝑑𝑑𝑛𝑑𝑑 ′ 𝑛𝐴𝐴′𝑓 4𝑑𝑑𝑛𝑑𝑑′𝐴𝑑𝑑′𝑛𝐴𝐴 𝑓 𝐴𝑑𝑑𝐴′𝑑𝑑′𝑛𝑑𝑑′′𝑛 𝑛 3𝑑𝑑𝐴𝑑𝑑′′𝑛 𝑛 ℎ 𝑛4𝑑𝑑𝐴𝑑𝑑 ′ 𝑛 𝑛 𝐴𝐴 𝑛 3𝑑𝑑𝑛𝑛𝑑𝑑′′𝐴𝐴𝐴 𝑛 𝐴𝑑𝑑′𝑛 𝑛 𝑑𝑑′′𝐴𝑓 𝑛𝑑𝑑𝑛𝑛𝑑𝑑′𝐴𝐴𝐴′ ℎ . (9) e 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 𝑔𝑔𝑓𝑓𝐴 𝑑𝑑𝐴𝑓𝑓𝐴𝑛 𝑑𝑑𝑛𝑓𝑓𝑛where 𝑓𝑓𝐴 and 𝑓𝑓𝑛 are two linearly independent solutions of (4) and 𝑑𝑑𝐴 and 𝑑𝑑𝑛
are entire functions of �nite order not all vanishing identically and satisfying 𝑑𝑑𝐴≠ 𝑐𝑐𝑑𝑑𝑛where 𝑐𝑐 is a complex number, and we
obtain the following results.
eorem 3. Let 𝐴𝐴𝜌𝑓𝑓𝜌 be a transcendental entire function
of �nite order. Let 𝑑𝑑𝑗𝑗𝜌𝑓𝑓𝜌 𝜌𝑗𝑗 𝐴 𝐴, 𝑛𝜌 be �nite�order entire
functions that are not all vanishing identically such that
max{𝜌𝜌𝜌𝑑𝑑𝐴𝜌, 𝜌𝜌𝜌𝑑𝑑𝑛𝜌} 𝜌 𝜌𝜌𝜌𝐴𝐴𝜌. If 𝑓𝑓𝐴 and 𝑓𝑓𝑛 are two linearly independent solutions of (4), then the polynomial of solutions
(5) satis�es
𝜌𝜌 𝑔𝑔𝑓𝑓 𝐴 𝜌𝜌 𝑓𝑓𝑗𝑗 𝐴 𝜌 𝑗𝑗 𝐴 𝐴, 𝑛 ,
𝜌𝜌𝑛𝑔𝑔𝑓𝑓 𝐴 𝜌𝜌𝑛𝑓𝑓𝑗𝑗 𝐴 𝜌𝜌 𝜌𝐴𝐴𝜌 𝑗𝑗 𝐴 𝐴, 𝑛 . (10)
eorem 4. Under the hypotheses of eorem 3, let 𝜑𝜑𝜌𝑓𝑓𝜌 ̸≡ 0
be an entire function with �nite order such that 𝜓𝜓𝜌𝑓𝑓𝜌 ̸≡ 0. If 𝑓𝑓𝐴
and 𝑓𝑓𝑛are two linearly independent solutions of (4), then the
polynomial of solutions (5) satis�es
𝜆𝜆 𝑔𝑔𝑓𝑓𝑓 𝜑𝜑 𝐴 𝜆𝜆 𝑔𝑔𝑓𝑓𝑓 𝜑𝜑 𝐴 𝜌𝜌 𝑓𝑓𝑗𝑗 𝐴 𝜌 𝑗𝑗 𝐴 𝐴, 𝑛 , 𝜆𝜆𝑛𝑔𝑔𝑓𝑓𝑓 𝜑𝜑 𝐴 𝜆𝜆𝑛𝑔𝑔𝑓𝑓𝑓 𝜑𝜑 𝐴 𝜌𝜌𝑛𝑓𝑓𝑗𝑗 𝐴 𝜌𝜌 𝜌𝐴𝐴𝜌 𝑗𝑗 𝐴 𝐴, 𝑛 .
(11) eorem 5. Let 𝐴𝐴𝜌𝑓𝑓𝜌 be a polynomial of deg 𝐴𝐴 𝐴 𝐴𝐴. Let 𝑑𝑑𝑗𝑗𝜌𝑓𝑓𝜌 𝜌𝑗𝑗 𝐴 𝐴, 𝑛𝜌 be �nite�order entire functions that are not all
vanishing identically such that ℎ ̸≡ 0 and max{𝜌𝜌𝜌𝑑𝑑𝐴𝜌, 𝜌𝜌𝜌𝑑𝑑𝑛𝜌} 𝜌
𝜌𝐴𝐴𝑛𝑛𝜌𝑛𝑛. If 𝑓𝑓𝐴𝑓𝑓𝑛are two linearly independent solutions of (4), then the polynomial of solutions (5) satis�es
𝜌𝜌 𝑔𝑔𝑓𝑓 𝐴 𝜌𝜌 𝑓𝑓𝑗𝑗 𝐴 𝐴𝐴 𝑛 𝑛𝑛 𝑗𝑗 𝐴 𝐴, 𝑛 . (12) eorem 6. Under the hypotheses of eorem 5, let 𝜑𝜑𝜌𝑓𝑓𝜌 ̸≡ 0
be an entire function with 𝜌𝜌𝜌𝜑𝜑𝜌 𝜌 𝜌𝐴𝐴 𝑛 𝑛𝜌𝑛𝑛 such that 𝜓𝜓𝜌𝑓𝑓𝜌 ̸≡ 0.
If 𝑓𝑓𝐴and 𝑓𝑓𝑛are two linearly independent solutions of (4), then
the polynomial of solutions (5) satis�es
𝜆𝜆 𝑔𝑔𝑓𝑓𝑓 𝜑𝜑 𝐴 𝜆𝜆 𝑔𝑔𝑓𝑓𝑓 𝜑𝜑 𝐴 𝐴𝐴 𝑛 𝑛𝑛 . (13)
2. Auxiliary Lemmas
Lemma 7 (see [7, 8]). Let 𝐴𝐴0, 𝐴𝐴𝐴, … , 𝐴𝐴𝑘𝑘𝑓𝐴, 𝐹𝐹 ̸≡ 0 be �nite�
order meromorphic functions. If 𝑓𝑓 is a meromorphic solution of the equation
𝑓𝑓𝜌𝑘𝑘𝜌𝑛 𝐴𝐴𝑘𝑘𝑓𝐴𝑓𝑓𝜌𝑘𝑘𝑓𝐴𝜌𝑛 ⋯ 𝑛 𝐴𝐴𝐴𝑓𝑓′𝑛 𝐴𝐴0𝑓𝑓 𝐴 𝐹𝐹 (14)
with 𝜌𝜌𝜌𝑓𝑓𝜌 𝐴 𝑛𝜌 and 𝜌𝜌𝑛𝜌𝑓𝑓𝜌 𝐴 𝜌𝜌, then 𝑓𝑓 satis�es
𝜆𝜆 𝑓𝑓 𝐴 𝜆𝜆 𝑓𝑓 𝐴 𝜌𝜌 𝑓𝑓 𝐴 𝑛𝜌, 𝜆𝜆𝑛𝑓𝑓 𝐴 𝜆𝜆𝑛𝑓𝑓 𝐴 𝜌𝜌𝑛𝑓𝑓 𝐴 𝜌𝜌.
(15) Here, we give a special case of the result due to Cao et al. in [9].
Lemma 8. Let 𝐴𝐴0, 𝐴𝐴𝐴, … , 𝐴𝐴𝑘𝑘𝑓𝐴, 𝐹𝐹 ̸≡ 0 be �nite�order mero�
morphic functions. If 𝑓𝑓 is a meromorphic solution of (14) with
max 𝜌𝜌 𝐴𝐴𝑗𝑗 𝑗𝑗 𝐴 0, 𝐴, … , 𝑘𝑘 𝑓 𝐴 , 𝜌𝜌 𝜌𝐹𝐹𝜌 𝜌 𝜌𝜌 𝑓𝑓 𝜌 𝑛𝜌, (16)
then
𝜆𝜆 𝑓𝑓 𝐴 𝜆𝜆 𝑓𝑓 𝐴 𝜌𝜌 𝑓𝑓 . (17)
3. Proofs of the Theorems
Proof of eorem 3. Suppose that 𝑓𝑓𝐴and 𝑓𝑓𝑛are two linearly
independent solutions of (4). en by eorem A, we have 𝜌𝜌 𝑓𝑓𝐴 𝐴 𝜌𝜌 𝑓𝑓𝑛 𝐴 𝜌,
Suppose that 𝑑𝑑1 = 𝑐𝑐𝑑𝑑2, where 𝑐𝑐 is a complex number. en,
by (5) we obtain
𝑔𝑔𝑓𝑓= 𝑐𝑐𝑑𝑑2𝑓𝑓1+ 𝑑𝑑2𝑓𝑓2= 𝑐𝑐𝑓𝑓1+ 𝑓𝑓2 𝑑𝑑2. (19)
Since 𝑓𝑓 = 𝑐𝑐𝑓𝑓1+𝑓𝑓2is a solution of (4) and 𝜌𝜌𝜌𝑑𝑑2) < 𝜌𝜌𝜌𝜌𝜌), then
we have
𝜌𝜌 𝑔𝑔𝑓𝑓 = 𝜌𝜌 𝑐𝑐𝑓𝑓1+ 𝑓𝑓2 = ∞,
𝜌𝜌2𝑔𝑔𝑓𝑓 = 𝜌𝜌2c𝑓𝑓1+ 𝑓𝑓2 = 𝜌𝜌 𝜌𝜌𝜌) . (20) Suppose now that 𝑑𝑑1 ̸≡ 𝑐𝑐𝑑𝑑2 where 𝑐𝑐 is a complex number.
Differentiating both sides of (5), we obtain
𝑔𝑔′𝑓𝑓= 𝑑𝑑′1𝑓𝑓1+ 𝑑𝑑1𝑓𝑓1′ + 𝑑𝑑′2𝑓𝑓2+ 𝑑𝑑2𝑓𝑓′2. (21) Differentiating both sides of (21), we obtain
𝑔𝑔′′𝑓𝑓= 𝑑𝑑′′1𝑓𝑓1+ 2𝑑𝑑′1𝑓𝑓′1+ 𝑑𝑑1𝑓𝑓1′′+ 𝑑𝑑′′2𝑓𝑓2+ 2𝑑𝑑′2𝑓𝑓′2+ 𝑑𝑑2𝑓𝑓′′2. (22) Substituting 𝑓𝑓′′𝑗𝑗 = −𝜌𝜌𝑓𝑓𝑗𝑗 𝜌𝑗𝑗 = 1, 2) into (22), we obtain
𝑔𝑔′′𝑓𝑓= 𝑑𝑑′′1− 𝑑𝑑1𝜌𝜌 𝑓𝑓1+ 2𝑑𝑑′1𝑓𝑓1′ + 𝑑𝑑′′2− 𝑑𝑑2𝜌𝜌 𝑓𝑓2+ 2𝑑𝑑′2𝑓𝑓′2. (23) Differentiating both sides of (23) and by substituting 𝑓𝑓′′𝑗𝑗 = −𝜌𝜌𝑓𝑓𝑗𝑗𝜌𝑗𝑗 = 1, 2), we obtain 𝑔𝑔′′′𝑓𝑓 = 𝑑𝑑′′′1 − 3𝑑𝑑′1𝜌𝜌 − 𝑑𝑑1𝜌𝜌′ 𝑓𝑓1+ 𝑑𝑑′′1 − 𝑑𝑑1𝜌𝜌 + 2𝑑𝑑′′1 𝑓𝑓′1 + 𝑑𝑑′′′2 − 3𝑑𝑑′2𝜌𝜌 − 𝑑𝑑2𝜌𝜌′ 𝑓𝑓2+ 𝑑𝑑′′2 − 𝑑𝑑2𝜌𝜌 + 2𝑑𝑑′′2 𝑓𝑓′2. (24) By (5), (21), (23), and (24) we have 𝑔𝑔𝑓𝑓= 𝑑𝑑1𝑓𝑓1+ 𝑑𝑑2𝑓𝑓2, 𝑔𝑔′𝑓𝑓= 𝑑𝑑′1𝑓𝑓1+ 𝑑𝑑1𝑓𝑓1′ + 𝑑𝑑′2𝑓𝑓2+ 𝑑𝑑2𝑓𝑓′2, 𝑔𝑔′′𝑓𝑓= 𝑑𝑑′′1− 𝑑𝑑1𝜌𝜌 𝑓𝑓1+ 2𝑑𝑑′1𝑓𝑓1′ + 𝑑𝑑′′2− 𝑑𝑑2𝜌𝜌 𝑓𝑓2+ 2𝑑𝑑′2𝑓𝑓′2, 𝑔𝑔′′′𝑓𝑓 = 𝑑𝑑′′′1 − 3𝑑𝑑′1𝜌𝜌 − 𝑑𝑑1𝜌𝜌′ 𝑓𝑓1+ 𝑑𝑑′′1− 𝑑𝑑1𝜌𝜌 + 2𝑑𝑑′′1 𝑓𝑓′1 + 𝑑𝑑′′′2 − 3𝑑𝑑′2𝜌𝜌 − 𝑑𝑑2𝜌𝜌′ 𝑓𝑓2+ 𝑑𝑑′′2− 𝑑𝑑2𝜌𝜌 + 2𝑑𝑑′′2 𝑓𝑓′2. (25)
To solve this system of equations, we need �rst to prove that ℎ ̸≡ 0. By simple calculations we obtain
ℎ = 𝑑𝑑1 0 𝑑𝑑2 0 𝑑𝑑′ 1 𝑑𝑑1 𝑑𝑑′2 𝑑𝑑2 𝑑𝑑′′ 1− 𝑑𝑑1𝜌𝜌 2𝑑𝑑′1 𝑑𝑑′′2− 𝑑𝑑2𝜌𝜌 2𝑑𝑑′2 𝑑𝑑′′′ 1 − 3𝑑𝑑′1𝜌𝜌 − 𝑑𝑑1𝜌𝜌′ 𝑑𝑑′′1− 𝑑𝑑1𝜌𝜌 + 2𝑑𝑑′′1 𝑑𝑑′′′2 − 3𝑑𝑑′2𝜌𝜌 − 𝑑𝑑2𝜌𝜌′ 𝑑𝑑′′2− 𝑑𝑑2𝜌𝜌 + 2𝑑𝑑′′2 = 4𝑑𝑑21𝑑𝑑′22+ 4𝑑𝑑22𝑑𝑑′12− 8𝑑𝑑1𝑑𝑑2𝑑𝑑′1𝑑𝑑′2 𝜌𝜌 + 2𝑑𝑑1𝑑𝑑2𝑑𝑑′1𝑑𝑑′′′2 + 2𝑑𝑑1𝑑𝑑2𝑑𝑑′2𝑑𝑑′′′1 − 6𝑑𝑑1𝑑𝑑2𝑑𝑑′′1𝑑𝑑′′2 − 6𝑑𝑑1𝑑𝑑′1𝑑𝑑′2𝑑𝑑′′2− 6𝑑𝑑2𝑑𝑑′1𝑑𝑑′2𝑑𝑑′′1 + 6𝑑𝑑1𝑑𝑑′22𝑑𝑑1′′+ 6𝑑𝑑2𝑑𝑑′12𝑑𝑑2′′− 2𝑑𝑑22𝑑𝑑′1𝑑𝑑1′′′− 2𝑑𝑑21𝑑𝑑′2𝑑𝑑′′′2 + 3𝑑𝑑21𝑑𝑑′′22+ 3𝑑𝑑22𝑑𝑑′′12. (26) To show that 4𝑑𝑑21𝜌𝑑𝑑′2) 2 +4𝑑𝑑22𝜌𝑑𝑑′1)2−8𝑑𝑑1𝑑𝑑2𝑑𝑑′1𝑑𝑑′2 ̸≡ 0, we suppose that 𝑑𝑑21𝑑𝑑′2 2 + 𝑑𝑑22𝑑𝑑′1 2 − 2𝑑𝑑1𝑑𝑑2𝑑𝑑′1𝑑𝑑′2= 0. (27)
Dividing both sides of (27) by 𝜌𝑑𝑑1𝑑𝑑2)2, we obtain
𝑑𝑑′2 𝑑𝑑2 2 + 𝑑𝑑′1 𝑑𝑑1 2 − 2𝑑𝑑′1 𝑑𝑑1 𝑑𝑑′2 𝑑𝑑2 = 0 (28) equivalent to 𝑑𝑑𝑑𝑑′1 1 − 𝑑𝑑 ′ 2 𝑑𝑑2 2 = 0, (29)
which implies that 𝑑𝑑1 = 𝑐𝑐𝑑𝑑2 where 𝑐𝑐 is a complex number
and this is a contradiction. Since max{𝜌𝜌𝜌𝑑𝑑1), 𝜌𝜌𝜌𝑑𝑑2)} < 𝜌𝜌𝜌𝜌𝜌)
and 4𝑑𝑑21𝜌𝑑𝑑′2)2+4𝑑𝑑22𝜌𝑑𝑑1′)2−8𝑑𝑑1𝑑𝑑2𝑑𝑑′1𝑑𝑑′2 ̸≡ 0, we can deduce from
(26) that
𝜌𝜌 𝜌ℎ) = 𝜌𝜌 𝜌𝜌𝜌) > 0. (30) Hence ℎ ̸≡ 0. By Cramer’s method we have
𝑓𝑓1= 𝑔𝑔𝑓𝑓 0 𝑑𝑑2 0 𝑔𝑔′𝑓𝑓 𝑑𝑑1 𝑑𝑑′2 𝑑𝑑2 𝑔𝑔′′ 𝑓𝑓 2𝑑𝑑′1 𝑑𝑑′′2−𝑑𝑑2𝜌𝜌 2𝑑𝑑′2 𝑔𝑔′′′ 𝑓𝑓 𝑑𝑑′′1−𝑑𝑑1𝜌𝜌+2𝑑𝑑′′1𝑑𝑑′′′2−3𝑑𝑑′2𝜌𝜌−𝑑𝑑2𝜌𝜌′𝑑𝑑′′2−𝑑𝑑2𝜌𝜌+2𝑑𝑑′′2 ℎ = 2 𝑑𝑑1𝑑𝑑2𝑑𝑑 ′ 2− 𝑑𝑑22𝑑𝑑′1 ℎ 𝑔𝑔𝜌3)𝑓𝑓 + 𝜙𝜙2𝑔𝑔′′𝑓𝑓+ 𝜙𝜙1𝑔𝑔′𝑓𝑓+ 𝜙𝜙0𝑔𝑔𝑓𝑓, (31)
where 𝜙𝜙𝑗𝑗𝜌𝑗𝑗 = 0, 1, 2) are meromorphic functions of �nite order de�ned in (9). Suppose now that 𝜌𝜌𝜌𝑔𝑔𝑓𝑓) < ∞, then by
(31) we obtain 𝜌𝜌𝜌𝑓𝑓1) < ∞, which is a contradiction, hence 𝜌𝜌𝜌𝑔𝑔𝑓𝑓) = ∞. By (5) we have 𝜌𝜌2𝜌𝑔𝑔𝑓𝑓) ≤ 𝜌𝜌𝜌𝜌𝜌). Suppose that 𝜌𝜌2𝜌𝑔𝑔𝑓𝑓) < 𝜌𝜌𝜌𝜌𝜌), then by (31) we obtain 𝜌𝜌2𝜌𝑓𝑓1) < 𝜌𝜌𝜌𝜌𝜌), which is a contradiction. Hence 𝜌𝜌2𝜌𝑔𝑔𝑓𝑓) = 𝜌𝜌𝜌𝜌𝜌).
Proof of eorem 4. By eorem 3 we have 𝜌𝜌𝜌𝜌𝜌𝑓𝑓) = ∞ and 𝜌𝜌2𝜌𝜌𝜌𝑓𝑓) = 𝜌𝜌𝜌𝜌𝜌). Set 𝑤𝑤𝜌𝑤𝑤) = 𝑤𝑤1𝑓𝑓1+ 𝑤𝑤2𝑓𝑓2− 𝜑𝜑. Since 𝜌𝜌𝜌𝜑𝜑) 𝜌 ∞,
then we have 𝜌𝜌𝜌𝑤𝑤) = 𝜌𝜌𝜌𝜌𝜌𝑓𝑓) = ∞ and 𝜌𝜌2𝜌𝑤𝑤) = 𝜌𝜌2𝜌𝜌𝜌𝑓𝑓) = 𝜌𝜌𝜌𝜌𝜌). In order to prove that 𝜆𝜆𝜌𝜌𝜌𝑓𝑓−𝜑𝜑) = 𝜆𝜆𝜌𝜌𝜌𝑓𝑓−𝜑𝜑) = ∞, 𝜆𝜆2𝜌𝜌𝜌𝑓𝑓−𝜑𝜑) = 𝜆𝜆2𝜌𝜌𝜌𝑓𝑓− 𝜑𝜑) = 𝜌𝜌𝜌𝜌𝜌) we need to prove only that 𝜆𝜆𝜌𝑤𝑤) = 𝜆𝜆𝜌𝑤𝑤) = ∞ and 𝜆𝜆2𝜌𝑤𝑤) = 𝜆𝜆2𝜌𝑤𝑤) = 𝜌𝜌𝜌𝜌𝜌). By 𝜌𝜌𝑓𝑓 = 𝑤𝑤 + 𝜑𝜑 we get from (31) 𝑓𝑓1=2 𝑤𝑤1𝑤𝑤2𝑤𝑤 ′ 2− 𝑤𝑤22𝑤𝑤′1 ℎ 𝑤𝑤𝜌3)+ 𝜙𝜙2𝑤𝑤′′+ 𝜙𝜙1𝑤𝑤′+ 𝜙𝜙0𝑤𝑤 + 𝑤𝑤𝑤 (32) where 𝑤𝑤 = 2 𝑤𝑤1𝑤𝑤2𝑤𝑤 ′ 2− 𝑤𝑤22𝑤𝑤′1 ℎ 𝜑𝜑𝜌3)+ 𝜙𝜙2𝜑𝜑′′+ 𝜙𝜙1𝜑𝜑′+ 𝜙𝜙0𝜑𝜑𝜑 (33)
Substituting (32) into (4), we obtain 2 𝑤𝑤1𝑤𝑤2𝑤𝑤′2− 𝑤𝑤22𝑤𝑤′1 ℎ 𝑤𝑤𝜌5)+ 4 𝑗𝑗=0𝛽𝛽𝑗𝑗𝑤𝑤 𝜌𝑗𝑗)= − 𝑤𝑤′′+ 𝜌𝜌𝑤𝑤 = 𝐵𝐵𝑤 (34) where 𝛽𝛽𝑗𝑗𝜌𝑗𝑗 = 0𝑤 𝑗 𝑤 4) are meromorphic functions of �nite order. Since 𝑤𝑤 ̸≡ 0 and 𝜌𝜌𝜌𝑤𝑤) 𝜌 ∞, it follows that 𝑤𝑤 is not a solution of (4), which implies that 𝐵𝐵 ̸≡ 0. en by applying Lemma 7 we obtain (11).
Proof of eorem 5. Suppose that 𝑓𝑓1and 𝑓𝑓2are two linearly
independent solutions of (4). en by eorem A
𝜌𝜌 𝑓𝑓1 = 𝜌𝜌 𝑓𝑓2 = 𝑛𝑛 + 22 𝜑 (35) By the same reasoning as in eorem 3, we have
ℎ = 𝑤𝑤1 0 𝑤𝑤2 0 𝑤𝑤′ 1 𝑤𝑤1 𝑤𝑤′2 𝑤𝑤2 𝑤𝑤′′ 1− 𝑤𝑤1𝜌𝜌 2𝑤𝑤′1 𝑤𝑤′′2− 𝑤𝑤2𝜌𝜌 2𝑤𝑤′2 𝑤𝑤′′′ 1 − 3𝑤𝑤′1𝜌𝜌 − 𝑤𝑤1𝜌𝜌′ 𝑤𝑤′′1− 𝑤𝑤1𝜌𝜌 + 2𝑤𝑤′′1 𝑤𝑤′′′2 − 3𝑤𝑤′2𝜌𝜌 − 𝑤𝑤2A′ 𝑤𝑤′′2− 𝑤𝑤2𝜌𝜌 + 2𝑤𝑤′′2 𝜑 (36) Since ℎ ̸≡ 0 and by Cramer’s method we have
𝑓𝑓1= 𝜌𝜌𝑓𝑓 0 𝑤𝑤2 0 𝜌𝜌′ 𝑓𝑓 𝑤𝑤1 𝑤𝑤′2 𝑤𝑤2 𝜌𝜌′′ 𝑓𝑓 2𝑤𝑤′1 𝑤𝑤′′2−𝑤𝑤2𝜌𝜌 2𝑤𝑤′2 𝜌𝜌′′′𝑓𝑓 𝑤𝑤′′1−𝑤𝑤1𝜌𝜌+2𝑤𝑤′′1𝑤𝑤′′′2−3𝑤𝑤′2𝜌𝜌−𝑤𝑤2𝜌𝜌′𝑤𝑤′′2−𝑤𝑤2𝜌𝜌+2𝑤𝑤′′2 ℎ = 2 𝑤𝑤1𝑤𝑤2𝑤𝑤 ′ 2− 𝑤𝑤22𝑤𝑤′1 ℎ 𝜌𝜌𝜌3)𝑓𝑓 + 𝜙𝜙2𝜌𝜌′′𝑓𝑓+ 𝜙𝜙1𝜌𝜌′𝑓𝑓+ 𝜙𝜙0𝜌𝜌𝑓𝑓𝑤 (37)
where 𝜙𝜙𝑗𝑗𝜌𝑗𝑗 = 0𝑤 1𝑤 2) are meromorphic functions with 𝜌𝜌𝜌𝜙𝜙𝑗𝑗) 𝜌 𝜌𝑛𝑛 + 2)𝑛2 𝜌𝑗𝑗 = 0𝑤 1𝑤 2) de�ned in (9). By (5) we have 𝜌𝜌𝜌𝜌𝜌𝑓𝑓) ≤ 𝜌𝑛𝑛 + 2)𝑛2. Suppose that 𝜌𝜌𝜌𝜌𝜌𝑓𝑓) 𝜌 𝜌𝑛𝑛 + 2)𝑛2, then by (37) we obtain 𝜌𝜌𝜌𝑓𝑓1) 𝜌 𝜌𝑛𝑛 + 2)𝑛2, which is a contradiction. Hence, 𝜌𝜌𝜌𝜌𝜌𝑓𝑓) = 𝜌𝑛𝑛 + 2)𝑛2.
Proof of eorem 6. By eorem 5 we have 𝜌𝜌𝜌𝜌𝜌𝑓𝑓) = 𝜌𝑛𝑛+2)𝑛2.
Set 𝑤𝑤𝜌𝑤𝑤) = 𝑤𝑤1𝑓𝑓1+ 𝑤𝑤2𝑓𝑓2− 𝜑𝜑. Since 𝜌𝜌𝜌𝜑𝜑) 𝜌 𝜌𝑛𝑛 + 2)𝑛2, then
we have 𝜌𝜌𝜌𝑤𝑤) = 𝜌𝜌𝜌𝜌𝜌𝑓𝑓) = 𝜌𝑛𝑛 + 2)𝑛2. In order to prove that 𝜆𝜆𝜌𝜌𝜌𝑓𝑓− 𝜑𝜑) = 𝜌𝑛𝑛 + 2)𝑛2, we need to prove only that 𝜆𝜆𝜌𝑤𝑤) = 𝜌𝑛𝑛 + 2)𝑛2. By 𝜌𝜌𝑓𝑓= 𝑤𝑤 + 𝜑𝜑 we get from (37) 𝑓𝑓1= 2 𝑤𝑤1𝑤𝑤2𝑤𝑤 ′ 2− 𝑤𝑤22𝑤𝑤′1 ℎ 𝑤𝑤𝜌3) + 𝜙𝜙2𝑤𝑤′′+ 𝜙𝜙1𝑤𝑤′+ 𝜙𝜙0𝑤𝑤 + 𝑤𝑤𝑤 (38) where 𝑤𝑤 = 2 𝑤𝑤1𝑤𝑤2𝑤𝑤 ′ 2− 𝑤𝑤22𝑤𝑤′1 ℎ 𝜑𝜑𝜌3) + 𝜙𝜙2𝜑𝜑′′+ 𝜙𝜙1𝜑𝜑′+ 𝜙𝜙0𝜑𝜑𝜑 (39) Substituting (38) into (4), we obtain
2 𝑤𝑤1𝑤𝑤2𝑤𝑤′2− 𝑤𝑤22𝑤𝑤′1 ℎ 𝑤𝑤𝜌5)+ 4 𝑗𝑗=0𝛽𝛽𝑗𝑗𝑤𝑤 𝜌𝑗𝑗)= − 𝑤𝑤′′+ 𝜌𝜌𝑤𝑤 = 𝐵𝐵𝑤 (40) where 𝛽𝛽𝑗𝑗 𝜌𝑗𝑗 = 0𝑤 𝑗 𝑤 4) are meromorphic functions with 𝜌𝜌𝜌𝛽𝛽𝑗𝑗) 𝜌 𝜌𝑛𝑛 + 2)𝑛2. Since 𝑤𝑤 ̸≡ 0 and 𝜌𝜌𝜌𝑤𝑤) 𝜌 𝜌𝑛𝑛 + 2)𝑛2, it follows that 𝑤𝑤 is not a solution of (4), which implies that 𝐵𝐵 ̸≡ 0. en by applying Lemma 8 we obtain (13).
Acknowledgments
e authors would like to thank the referees for their helpful remarks and suggestions to improve the paper. is research is supported by Agence Nationale pour le Développement de la Recherche Universitaire (ANDRU) and University of Mostaganem (UMAB), (PNR Project Code 8/u27/3144).
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