On the growth of solutions of a class of second-order complex differential equations
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RESEARCH
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On the growth of solutions of a class of second-order complex differential equations Cai Feng Yi1* , Xu-Qiang Liu1 and Hong Yan Xu2 *
Correspondence: [email protected] 1 Department of Mathematics and Informatics, Jiangxi Normal University, Nanchang, Jiangxi 330022, P.R. China Full list of author information is available at the end of the article
Abstract In this paper, we consider the differential equation f + h(z)eP(z) f + Q(z)f = 0, where h(z) and Q(z) ≡ 0 are meromorphic functions, P(z) is a non-constant polynomial. Assume that Q(z) has an infinite deficient value and finitely many Borel directions. We give some conditions on P(z) which guarantee that every solution f ≡ 0 of the equation has infinite order. MSC: 34AD20; 30D35 Keywords: complex differential equation; meromorphic function; Borel direction; deficient value; hyper-order
1 Introduction and main results In this paper, we shall involve the deficient value and the Borel direction in investigating the growth of solutions of the second-order linear differential equation f + h(z)eP(z) f + Q(z)f = ,
()
where h(z) and Q(z) ≡ are meromorphic functions, P(z) is a non-constant polynomial. We assume that the reader is familiar with the Nevanlinna theory of meromorphic functions and the basic notions such as N(r, f ), m(r, f ), T(r, f ) and δ(r, f ). For the details, see [] or []. The order σ and the hyper-order σ are defined as follows: σ (f ) = lim sup r−→∞
log+ T(r, f ) , log r
σ (f ) = lim sup r−→∞
log+ log+ T(r, f ) . log r
It is well known that if A(z) = h(z)eP(z) and B(z) = Q(z) are transcendental entire functions in equation () and f , f are two linearly independent solutions of equation (), then at least one of f , f must have infinite order. Hence, ‘most’ solutions of equation () will have infinite order. On the other hand, there are some equations of the form () that possess a solution f ≡ which has finite order; for example, f (z) = ez satisfies the equation f + e–z f – (e–z + )f = . Thus the main problem is what condition on A(z) and B(z) can guarantee that every solution f ≡ of equation () has infinite order? There has been much work on this subject. For example, it follows from the work by Gunderson [], Hellerstein et al. [] that if A(z) and B(z) are entire functions with σ (A) < σ (B) or A(z) is a polynomial and B(z) is transcendental; or if σ (B) < σ (A) ≤ , then every solution f ≡ of equation © 2013 Yi et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Yi et al. Advances in Difference Equations 2013, 2013:188 http://www.advancesindifferenceequations.com/content/2013/1/188
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() has infinite order. Furthermore, if A is an entire function with finite order having a finite deficient value and B(z) is a transcendental entire func
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