Finite order solutions of second order linear differential equations

Abstract

ABSTRACT. We consider the differential equation /" + A(z)f + B(z)f = 0 where A(z) and B(z) are entire functions.We will find conditions on A(z) and B(z) which will guarantee that every solution / ^ 0 of the equation will have infinite order. We will also find conditions on A(z) and B(z) which will guarantee that any finite order solution / ^ 0 of the equation will not have zero as a Borel exceptional value. We will also show that if A(z) and B(z) satisfy certain growth conditions, then any finite order solution of the equation will satisfy certain other growth conditions. Related results are also proven. Several examples are given to complement the theory.

On the Nevanlinna characteristic of f(z+η) and difference equations in the complex plane

“…The classical growth comparisons between f (z) and f ′ (z) have important applications to differential equations (see e.g. [14], [16]). In the case of applying Nevanlinna theory to difference equations, one of the most basic questions is the growth comparison between T (r, f (z + 1)) and T (r, f (z)).…”

On the growth of solutions of $f”+gf’+hf=0$

“…Gundersen [5,Theorem 6] has extended Theorem A to obtain the conclusion of our theorem under the more restrictive hypothesis p(h) < p(g) < 5 . Thus our contribution is to treat the case p(g) = \ .…”

Completely regular growth solutions of second order complex linear differential equations

“…As noted in [6], if A(z) and B(z) are entire such that ρ(A) < ρ(B), then (1.2) implies that all solutions f ≡ 0 of (1.1) are of infinite order. More specifically, if ρ(A) = ρ(B) > 0 but the T -types satisfy τ (A) < τ (B), where…”

“…The remaining assertion is a trivial consequence of (2.1). If A(z) and B(z) are entire such that ρ(B) < ρ(A) ≤ 1/2, then all solutions of (1.1) are of infinite order [6], [10]. For any integer n ≥ 1, finite order solutions may occur in the case…”