DOI: 10.2478/auom-2013-0022 An. Şt. Univ. Ovidius Constanţa Vol. 21(2),2013, 35–52 On the Growth of Solutions of Some Second Order Linear Differential Equations With Entire Coefficients Benharrat BELAÏDI and Habib HABIB Abstract In this paper, we investigate the order and the hyper-order of growth of solutions of the linear differential equation
f ′′ + Q e−z f ′ + (A1 ea1 z + A2 ea2 z )n f = 0, where n > 2 is an integer, Aj (z) (6≡ 0) (j = 1, 2) are entire functions with max {σ (Aj ) : j = 1, 2} < 1, Q (z) = qm z m + · · · + q1 z + q0 is a nonconstant polynomial and a1 , a2 are complex numbers. Under some conditions, we prove that every solution f (z) 6≡ 0 of the above equation is of infinite order and hyper-order 1. 1 Introduction and statement of results Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna’s value distribution theory (see [8], [13]). Let σ (f ) denote the order of growth of an entire function f and the hyper-order σ2 (f ) of f is defined by (see [9], [13]) σ2 (f ) = lim sup r→+∞ log log log M (r, f ) log log T (r, f ) = lim sup , log r log r r→+∞ Key Words: Linear differential equations, Entire solutions, Order of growth, Hyperorder, Fixed points. 2010 Mathematics Subject Classification: Primary 34M10; Secondary 30D35. Received: February 2013 Accepted: June 2013 35 Unauthenticated Download Date | 3/16/16 1:53 PM 36 Benharrat BELAÏDI and Habib HABIB where T (r, f ) is the Nevanlinna characteristic function of f and M (r, f ) = max|z|=r |f (z)|. In order to give some estimates of fixed points, we recall the following definition. Definition 1.1 ([3], [10]) Let f be a meromorphic function. Then the exponent of convergence of the sequence of distinct fixed points of f (z) is defined by   1 log N r, f −z τ (f ) = λ (f − z) = lim sup , log r r→+∞   where N r, f1 is the counting function of distinct zeros of f (z) in {z : |z| < r}. We also define λ (f − ϕ) = lim sup 1 log N (r, f −ϕ ) r→+∞ log r for any meromorphic function ϕ (z). In [11], Peng and Chen have investigated the order and hyper-order of solutions of some second order linear differential equations and have proved the following result. Theorem A ([11]) Let Aj (z) (6≡ 0) (j = 1, 2) be entire functions with σ (Aj ) < 1, a1 , a2 be complex numbers such that a1 a2 6= 0, a1 6= a2 (suppose that |a1 | 6 |a2 |). If arg a1 6= π or a1 < −1, then every solution f 6≡ 0 of the equation f ′′ + e−z f ′ + (A1 ea1 z + A2 ea2 z ) f = 0 has infinite order and σ2 (f ) = 1. The main purpose of this paper is to extend and improve the results of Theorem A to some second order linear differential equations. In fact we will prove the following results. Theorem 1.1 Let n > 2 be an integer, Aj (z) (6≡ 0) (j = 1, 2) be entire functions with max {σ (Aj ) : j = 1, 2} < 1, Q (z) = qm z m +· · ·+q1 z+q0 be nonconstant polynomial and a1 , a2 be complex numbers such that a1 a2 6= 0, a1 6= a2 . If (1) arg a1 6= π and arg a1 6= arg a2 or (2) arg a1 6= π, arg a1 = arg a2 and Unauthenticated Download Date | 3/16/16 1:53 PM ON THE GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS 37 |a2 | > n |a1 | or (3) a1 < 0 and arg a1 6= arg a2 or (4) − n1 (|a2 | − m) < a1 < 0, |a2 | > m and arg a1 = arg a2 , then every solution f 6≡ 0 of the equation
n f ′′ + Q e−z f ′ + (A1 ea1 z + A2 ea2 z ) f = 0 (1.1) satisfies σ (f ) = +∞ and σ2 (f ) = 1. Theorem 1.2 Let Aj (z) (j = 1, 2), Q (z), a1 , a2 , n satisfy the additional hypotheses of Theorem 1.1. If ϕ 6≡ 0 is an entire function of order σ (ϕ) < +∞, then every solution f 6≡ 0 of equation (1.1) satisfies λ (f − ϕ) = λ (f − ϕ) = σ (f ) = +∞, λ2 (f − ϕ) = λ2 (f − ϕ) = σ2 (f ) = 1. Theorem 1.3 Let Aj (z) (j = 1, 2), Q (z), a1 , a2 , n satisfy the additional hypotheses of Theorem 1.1. If ϕ 6≡ 0 is an entire function of order σ (ϕ) < 1, then every solution f 6≡ 0 of equation (1.1) satisfies λ (f − ϕ) = λ (f ′ − ϕ) = +∞. Furthermore, if (i) (2n + 2) a1 6= (2 − p) a1 + pa2 − k ( p = 0, 1, 2; k = 0, 1, · · ·, 2m), (n + 2 − p) a1 + pa2 − k ( p = 0, 1, · · ·, n + 2; k = 0, 1, · · ·, m) or (ii) (2n + 2) a2 6= (2 − p) a1 + pa2 − k ( p = 0, 1, 2; k = 0, 1, · · ·, 2m), (n + 2 − p) a1 + pa2 − k ( p = 0, 1, · · ·, n + 2; k = 0, 1, · · ·, m), then λ (f ′′ − ϕ) = +∞. Corollary 1.1 Let Aj (z) (j = 1, 2), Q (z), a1 , a2 , n satisfy the additional hypotheses of Theorem 1.1. If f 6≡ 0 is any solution of equation (1.1), then f , f ′ all have infinitely many fixed points and satisfy τ (f ) = τ (f ′ ) = ∞. Furthermore, if (i) (2n + 2) a1 6= (2 − p) a1 + pa2 − k ( p = 0, 1, 2; k = 0, 1, · · ·, 2m), (n + 2 − p) a1 + pa2 − k ( p = 0, 1, · · ·, n + 2; k = 0, 1, · · ·, m) or (ii) (2n + 2) a2 6= (2 − p) a1 + pa2 − k ( p = 0, 1, 2; k = 0, 1, · · ·, 2m), (n + 2 − p) a1 + pa2 − k ( p = 0, 1, · · ·, n + 2; k = 0, 1, · · ·, m), then f ′′ has infinitely many fixed points and satisfies τ (f ′′ ) = ∞. Unauthenticated Download Date | 3/16/16 1:53 PM 38 2 Benharrat BELAÏDI and Habib HABIB Preliminary lemmas To prove our theorems, we need the following lemmas. Lemma 2.1 ([7]) Let f be a transcendental meromorphic function with σ (f ) = σ < +∞, H = {(k1 , j1 ) , (k2 , j2 ) , · · ·, (kq , jq )} be a finite set of distinct pairs of integers satisfying ki > ji > 0 (i = 1, · · ·, q) and let ε > 0 be a given constant. Then,
 with linear measure zero, such that, if (i) there a set E1 ⊂ − π2 , 3π 2  π exists
ψ ∈ − 2 , 3π \ E , then there is a constant R0 = R0 (ψ) > 1, such that for 1 2 all z satisfying arg z = ψ and |z| > R0 and for all (k, j) ∈ H, we have f (k) (z) (k−j)(σ−1+ε) 6 |z| , f (j) (z) (2.1) (ii) there exists a set E2 ⊂ (1, +∞) with finite logarithmic measure, such that for all z satisfying |z| ∈ / E2 ∪ [0, 1] and for all (k, j) ∈ H, we have f (k) (z) (k−j)(σ−1+ε) 6 |z| , f (j) (z) (2.2) (iii) there exists a set E3 ⊂ (0, +∞) with finite linear measure, such that for all z satisfying |z| ∈ / E3 and for all (k, j) ∈ H, we have f (k) (z) (k−j)(σ+ε) 6 |z| . f (j) (z) (2.3) Lemma 2.2 ([4]) Suppose that P (z) = (α + iβ) z n + · · · (α, β are real numbers, |α| + |β| 6= 0) is a polynomial with degree n > 1, that A (z) (6≡ 0) is an entire function with σ (A) < n. Set g (z) = A (z) eP (z) , z = reiθ , δ (P, θ) = α cos nθ − β sin nθ. Then for any given ε > 0, there is a set E4 ⊂ [0, 2π) that has linear measure zero, such that for any θ ∈ [0, 2π)  (E4 ∪ E5 ), there is R > 0, such that for |z| = r > R, we have (i) if δ (P, θ) > 0, then
exp {(1 − ε) δ (P, θ) rn } 6 g reiθ 6 exp {(1 + ε) δ (P, θ) rn } ; (2.4) (ii) if δ (P, θ) < 0, then exp {(1 + ε) δ (P, θ) rn } 6 g reiθ
6 exp {(1 − ε) δ (P, θ) rn } , where E5 = {θ ∈ [0, 2π) : δ (P, θ) = 0} is a finite set. Unauthenticated Download Date | 3/16/16 1:53 PM (2.5) ON THE GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS 39 Lemma 2.3 ([11]) Suppose that n > 1 is a positive entire number. Let Pj (z) = ajn z n + · · · (j = 1, 2) be nonconstant polynomials, where ajq (q = 1, · · ·, n) are complex numbers and a1n a2n 6= 0. Set z = reiθ , ajn = |ajn | eiθj ,
π 3π θj ∈ − 2 , 2
, δ (Pj , θ) = |ajn | cos (θj + nθ), then there is a set π 3π measure zero. If θ1 6= θ2 , then there exists a , 2n that has linear E6 ⊂ − 2n
π π ray arg z = θ, θ ∈ − 2n , 2n \ (E6 ∪ E7 ), such that δ (P1 , θ) > 0, δ (P2 , θ) < 0 (2.6) or  where E7 = θ ∈ measure zero.  δ (P1 , θ) < 0, δ (P2 , θ) > 0, (2.7)
: δ (Pj , θ) = 0 is a finite set, which has linear π 3π − 2n , 2n
π π Remark 2.1 ([11]) In Lemma 2.3, if θ ∈ − 2n \ (E6 ∪ E7 ) is replaced by , 2n
π 3π , 2n \ (E6 ∪ E7 ), then we obtain the same result. θ ∈ 2n Lemma 2.4([5]) Suppose that k > 2 and B0 , B1 , · · ·, Bk−1 are entire functions of finite order and let σ = max {σ (Bj ) : j = 0, · · ·, k − 1}. Then every solution f of the equation f (k) + Bk−1 f (k−1) + · · · + B1 f ′ + B0 f = 0 (2.8) satisfies σ2 (f ) 6 σ. Lemma 2.5 ([7]) Let f (z) be a transcendental meromorphic function, and let α > 1 be a given constant. Then there exist a set E8 ⊂ (1, ∞) with finite logarithmic measure and a constant B > 0 that depends only on α and i, j (0 6 i < j 6 k), such that for all z satisfying |z| = r ∈ / [0, 1] ∪ E8 , we have  j−i T (αr, f ) f (j) (z) α 6 B . (2.9) (log r) log T (αr, f ) r f (i) (z) Lemma 2.6([2]) Let A0 , A1 , · · ·, Ak−1 , F 6≡ 0 be finite order meromorphic functions. If f is a meromorphic solution with σ (f ) = +∞ of the equation f (k) + Ak−1 f (k−1) + · · · + A1 f ′ + A0 f = F, (2.10) then f satisfies λ (f ) = λ (f ) = σ(f ) = +∞. Lemma 2.7 ([1]) Let A0 , A1 , · · ·, Ak−1 , F 6≡ 0 be finite order meromorphic functions. If f is a meromorphic solution of equation (2.10) with σ (f ) = +∞ and σ2 (f ) = σ, then f satisfies λ2 (f ) = λ2 (f ) = σ2 (f ) = σ. Unauthenticated Download Date | 3/16/16 1:53 PM (2.11) 40 Benharrat BELAÏDI and Habib HABIB Lemma 2.8([6], [13]) Suppose that f1 (z) , f2 (z) , · · ·, fn (z) (n > 2) are meromorphic functions and g1 (z) , g2 (z) , · · ·, gn (z) are entire functions satisfying the following conditions: n P (i) fj (z) egj (z) ≡ 0; j=1 (ii) gj (z) − gk (z) are not constants for 1 6 j < k 6 n;
(iii) For 1 6 j 6 n, 1 6 h < k 6 n, T (r, fj ) = o T r, egh (z)−gk (z) (r → ∞, r∈ / E9 ), where E9 is a set with finite linear measure. Then fj (z) ≡ 0 (j = 1, · · ·, n). Lemma 2.9 ([12]) Suppose that f1 (z) , f2 (z) , · · ·, fn (z) (n > 2) are meromorphic functions and g1 (z) , g2 (z) , · · ·, gn (z) are entire functions satisfying the following conditions: n P (i) fj (z) egj (z) ≡ fn+1 ; j=1 (ii) If 1 6 j 6 n + 1, 1 6 k 6 n, the order of fj is less than the order of egk (z) . If n > 2, 1 6 j 6 n + 1, 1 6 h < k 6 n, and the order of fj is less than the order of egh −gk . Then fj (z) ≡ 0 (j = 1, 2, · · ·, n + 1). 3 Proof of Theorem 1.1 Assume that f (6≡ 0) is a solution of equation (1.1). First step: We prove that σ (f ) = +∞. Suppose that σ (f ) = σ < +∞. We rewrite (1.1) as n−1 X
f′ f ′′ Cnp A1n−p e(n−p)a1 z Ap2 epa2 z = 0. + Q e−z + An1 ena1 z + An2 ena2 z + f f p=1 (3.1) By Lemma 2.1, for any given ε,   |a2 | − n |a1 | 1 0 < ε < min , , 2 [(2n − 1) |a2 | + n |a1 |] 2 (2n − 1)
 of linear measure zero, such that if θ ∈ there exists a set E1 ⊂ − π2 , 3π 2  π 3π
− 2 , 2 \ E1 , then there is a constant R0 = R0 (θ) > 1, such that for all z satisfying arg z = θ and |z| = r > R0 , we have f (j) (z) 6 rj(σ−1+ε) (j = 1, 2) . (3.2) f (z)
 Let z = reiθ , a1 = |a1 | eiθ1 , a2 = |a2 | eiθ2 , θ1 , θ2 ∈ − π2 , 3π 2 . We know that δ (pa1 z, θ) = pδ (a1 z, θ) and δ (pa2 z, θ) = pδ (a2 z, θ), where p > 0. Unauthenticated Download Date | 3/16/16 1:53 PM ON THE GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS 41 Case 1: Assume that arg a1 6= π and arg a1 6= arg a2 , which is θ1 6= π and θ1 6= θ2 . By Lemma 2.2 and
Lemma 2.3, for the above ε, there is a ray arg z = θ such that θ ∈ − π2 , π2 \ (E1 ∪ E6 ∪ E7 ) (where E6 and E7 are defined as in Lemma 2.3, E1 ∪ E6 ∪ E7 is of the linear measure zero), and satisfying δ (a1 z, θ) > 0, δ (a2 z, θ) < 0 or δ (a1 z, θ) < 0, δ (a2 z, θ) > 0. a) When δ (a1 z, θ) > 0, δ (a2 z, θ) < 0, for sufficiently large r, we get by Lemma 2.2 |An1 ena1 z | > exp {(1 − ε) nδ (a1 z, θ) r} , (3.3) |An2 ena2 z | 6 exp {(1 − ε) nδ (a2 z, θ) r} < 1, (3.4) A1n−p e(n−p)a1 z 6 exp {(1 + ε) (n − p) δ (a1 z, θ) r} 6 exp {(1 + ε) (n − 1) δ (a1 z, θ) r} , p = 1, · · ·, n − 1, |Ap2 epa2 z | For θ ∈ − π2 , π2
(3.5) 6 exp {(1 − ε) pδ (a2 z, θ) r} < 1, p = 1, · · ·, n − 1. (3.6) we have Q e−z
= qm e−mz + · · · + q1 e−z + q0 6 |qm | e−mz + · · · + |q1 | e−z + |q0 | 6 |qm | e−mr cos θ + · · · + |q1 | e−r cos θ + |q0 | 6 M, (3.7) where M > 0 is a some constant. By (3.1) − (3.7), we get exp {(1 − ε) nδ (a1 z, θ) r} 6 |An1 ena1 z | 6 n−1 X
f′ f ′′ Cnp A1n−p e(n−p)a1 z |Ap2 epa2 z | + Q e−z + |An2 ena2 z | + f f p=1 6 r2(σ−1+ε) + M rσ−1+ε + 2n exp {(1 + ε) (n − 1) δ (a1 z, θ) r} 6 M1 rM2 exp {(1 + ε) (n − 1) δ (a1 z, θ) r} , where M1 > 0 and M2 > 0 are some constants. By 0 < ε < we have   1 exp δ (a1 z, θ) r 6 M1 rM2 . 2 Unauthenticated Download Date | 3/16/16 1:53 PM (3.8) 1 2(2n−1) and (3.8), (3.9) 42 Benharrat BELAÏDI and Habib HABIB By δ (a1 z, θ) > 0 we know that (3.9) is a contradiction. b) When δ (a1 z, θ) < 0, δ (a2 z, θ) > 0, using a proof similar to the above, we can also get a contradiction. Case 2: Assume that arg a1 6= π, arg a1 = arg a2 and |a2 | > n |a1 |, which is θ1 6= π and θ1 = θ2 and |a2 | > n |a1 |.
By Lemma 2.3, for the above ε, there is a ray arg z = θ such that θ ∈ − π2 , π2 \ (E1 ∪ E6 ∪ E7 ) and δ (a1 z, θ) > 0. Since |a2 | > n |a1 | and n > 2, then |a2 | > |a1 |, thus δ (a2 z, θ) > δ (a1 z, θ) > 0. For sufficiently large r, we have by using Lemma 2.2 |An2 ena2 z | > exp {(1 − ε) nδ (a2 z, θ) r} , (3.10) |An1 ena1 z | 6 exp {(1 + ε) nδ (a1 z, θ) r} , (3.11) A1n−p e(n−p)a1 z 6 exp {(1 + ε) (n − 1) δ (a1 z, θ) r} , p = 1, · · ·, n − 1, (3.12) |Ap2 epa2 z | 6 exp {(1 + ε) (n − 1) δ (a2 z, θ) r} , p = 1, · · ·, n − 1. (3.13) By (3.1) , (3.2) , (3.7) and (3.10) − (3.13) we get exp {(1 − ε) nδ (a2 z, θ) r} 6 |An2 ena2 z | n−1 X
f′ f ′′ Cnp A1n−p e(n−p)a1 z |Ap2 epa2 z | + Q e−z + |An1 ena1 z | + f f p=1 6 6 r2(σ−1+ε) + M rσ−1+ε + exp {(1 + ε) nδ (a1 z, θ) r} +2n exp {(1 + ε) (n − 1) δ (a1 z, θ) r} exp {(1 + ε) (n − 1) δ (a2 z, θ) r} 6 M1 rM2 exp {(1 + ε) nδ (a1 z, θ) r} exp {(1 + ε) (n − 1) δ (a2 z, θ) r} . (3.14) Therefore, by (3.14), we obtain exp {αr} 6 M1 rM2 , where α = [1 − ε (2n − 1)] δ (a2 z, θ) − (1 + ε) nδ (a1 z, θ) . Since 0 < ε < |a2 |−n|a1 | 2[(2n−1)|a2 |+n|a1 |] , θ1 = θ2 and cos (θ1 + θ) > 0, then α = [1 − ε (2n − 1)] |a2 | cos (θ2 + θ) − (1 + ε) n |a1 | cos (θ1 + θ) = {|a2 | − n |a1 | − ε [(2n − 1) |a2 | + n |a1 |]} cos (θ1 + θ) Unauthenticated Download Date | 3/16/16 1:53 PM (3.15) ON THE GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS > 43 |a2 | − n |a1 | cos (θ1 + θ) > 0. 2 Hence (3.15) is a contradiction. Case 3: Assume that a1 < 0 and arg a1 6= arg a2 , which is θ1 = π and θ2 6= π.
By Lemma 2.3, for the above ε, there is a ray arg z = θ such that θ ∈ − π2 , π2 \ (E1 ∪ E6 ∪ E7 ) and δ (a2 z, θ) > 0. Because cos θ > 0, we have δ (a1 z, θ) = |a1 | cos (θ1 + θ) = − |a1 | cos θ < 0. For sufficiently large r, we obtain by Lemma 2.2 |An2 ena2 z | > exp {(1 − ε) nδ (a2 z, θ) r} , (3.16) |An1 ena1 z | 6 exp {(1 − ε) nδ (a1 z, θ) r} < 1, (3.17) A1n−p e(n−p)a1 z 6 exp {(1 − ε) (n − p) δ (a1 z, θ) r} < 1, p = 1, · · ·, n − 1, (3.18) p pa2 z | 6 exp {(1 + ε) (n − 1) δ (a2 z, θ) r} , p = 1, · · ·, n − 1. (3.19) |A2 e Using the same reasoning as in Case 1(a), we can get a contradiction. Case 4. Assume that − n1 (|a2 | − m) < a1 < 0, |a2 | > m and arg a1 = arg a2 , which is θ1 = θ2 = π and |a1 | < n1 (|a2 | − m), then |a2 | > n |a1 | + m, hence |a2 | > n |a1 |.
By Lemma 2.3, for the above ε, there is a ray arg z = θ such that θ ∈ π2 , 3π 2 \ (E1 ∪ E6 ∪ E7 ), then cos θ < 0, δ (a1 z, θ) = |a1 | cos (θ1 + θ) = − |a1 | cos θ > 0, δ (a2 z, θ) = |a2 | cos (θ2 + θ) = − |a2 | cos θ > 0. Since |a2 | > n |a1 | and n > 2, then |a2 | > |a1 |, thus δ (a2 z, θ) > δ (a
1 z, θ) > 0, for sufficiently large r, we get we have (3.10) − (3.13) hold. For θ ∈ π2 , 3π 2 Q e−z
6 M e−mr cos θ . (3.20) By (3.1) , (3.2) , (3.10) − (3.13) and (3.20), we get exp {(1 − ε) nδ (a2 z, θ) r} 6 |An2 ena2 z | 6 n−1 X
f′ f ′′ + Q e−z + |An1 ena1 z | + Cnp A1n−p e(n−p)a1 z |Ap2 epa2 z | f f p=1 6 r2(σ−1+ε) + M rσ−1+ε e−mr cos θ + exp {(1 + ε) nδ (a1 z, θ) r} +2n exp {(1 + ε) (n − 1) δ (a1 z, θ) r} exp {(1 + ε) (n − 1) δ (a2 z, θ) r} Unauthenticated Download Date | 3/16/16 1:53 PM 44 Benharrat BELAÏDI and Habib HABIB 6 M1 rM2 e−mr cos θ exp {(1 + ε) nδ (a1 z, θ) r} exp {(1 + ε) (n − 1) δ (a2 z, θ) r} . (3.21) Therefore, by (3.21), we obtain exp {βr} 6 M1 rM2 , (3.22) where β = [1 − ε (2n − 1)] δ (a2 z, θ) − (1 + ε) nδ (a1 z, θ) + m cos θ. Since |a2 | − n |a1 | − m > 0, then 2 [(2n − 1) |a2 | + n |a1 |] > |a2 | − n |a1 | − m > 0. Therefore, |a2 | − n |a1 | − m < 1. 2 [(2n − 1) |a2 | + n |a1 |] |a2 |−n|a1 |−m Then, we can take 0 < ε < 2[(2n−1)|a . Since 0 < ε < 2 |+n|a1 |] θ1 = θ2 = π and cos θ < 0, then |a2 |−n|a1 |−m 2[(2n−1)|a2 |+n|a1 |] , β = − cos θ {|a2 | − n |a1 | − m − ε [(2n − 1) |a2 | + n |a1 |]} 1 > − (|a2 | − n |a1 | − m) cos θ > 0. 2 Hence, (3.22) is a contradiction. σ (f ) = +∞. Concluding the above proof, we obtain Second step: We prove that σ2 (f ) = 1. By
n max{σ(Q e−z ), σ((A1 ea1 z + A2 ea2 z ) )} = 1 and the Lemma 2.4, we get σ2 (f ) 6 1. By Lemma 2.5, we know that there exists a set E8 ⊂ (1, +∞) with finite logarithmic measure and a constant B > 0, such that for all z satisfying |z| = r ∈ / [0, 1] ∪ E8 , we get f (j) (z) j+1 6 B [T (2r, f )] (j = 1, 2) . f (z) (3.23) Case 1: θ1 6= π and θ1 6= θ2
. In first step, we have proved that there is a ray arg z = θ where θ ∈ − π2 , π2 \ (E1 ∪ E6 ∪ E7 ), satisfying δ (a1 z, θ) > 0, δ (a2 z, θ) < 0 or δ (a1 z, θ) < 0, δ (a2 z, θ) > 0. Unauthenticated Download Date | 3/16/16 1:53 PM ON THE GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS 45 a) When δ (a1 z, θ) > 0, δ (a2 z, θ) < 0, for sufficiently large r, we get (3.3) − (3.7) holds. By (3.1) , (3.3) − (3.7) and (3.23), we obtain exp {(1 − ε) nδ (a1 z, θ) r} 6 |An1 ena1 z | n−1 X
f′ f ′′ −z n na2 z + Q e + |A2 e |+ Cnp A1n−p e(n−p)a1 z |Ap2 epa2 z | 6 f f p=1 3 2 6 B [T (2r, f )] + M B [T (2r, f )] + 2n exp {(1 + ε) (n − 1) δ (a1 z, θ) r} 3 6 M1 exp {(1 + ε) (n − 1) δ (a1 z, θ) r} [T (2r, f )] . By 0 < ε < 1 2(2n−1) (3.24) and (3.24), we have exp  1 δ (a1 z, θ) r 2  3 6 M1 [T (2r, f )] . (3.25) By δ (a1 z, θ) > 0 and (3.25), we have σ2 (f ) > 1, then σ2 (f ) = 1. b) When δ (a1 z, θ) < 0, δ (a2 z, θ) > 0, using a proof similar to the above, we can also get σ2 (f ) = 1. Case 2: θ1 6= π, θ1 = θ2 and |a2 | > n |a1 |.
In first step, we have proved that there is a ray arg z = θ where θ ∈ − π2 , π2 \ (E1 ∪ E6 ∪ E7 ), satisfying δ (a2 z, θ) > δ (a1 z, θ) > 0 and for sufficiently large r, we get (3.7) and (3.10) − (3.13) hold. By (3.1) , (3.7) , (3.10) − (3.13) and (3.23) , we get 3 exp {αr} 6 M1 [T (2r, f )] , (3.26) where α = [1 − ε (2n − 1)] δ (a2 z, θ) − (1 + ε) nδ (a1 z, θ) > 0. By α > 0 and (3.26), we have σ2 (f ) > 1, then σ2 (f ) = 1. Case 3: a1 < 0 and θ1 6= θ2
. In first step, we have proved that there is a ray arg z = θ where θ ∈ − π2 , π2 \ (E1 ∪ E6 ∪ E7 ), satisfying δ (a2 z, θ) > 0 and δ (a1 z, θ) < 0 and for sufficiently large r, we get (3.16) − (3.19) hold. Using the same reasoning as in second step ( Case 1 (a)), we can get σ2 (f ) = 1. Unauthenticated Download Date | 3/16/16 1:53 PM 46 Benharrat BELAÏDI and Habib HABIB Case 4: − n1 (|a2 | − m) < a1 < 0, |a2 | > m and θ1 = θ2
. In first step, we have proved that there is a ray arg z = θ where θ ∈ π2 , 3π \ (E1 ∪ E6 ∪ E7 ), 2 satisfying δ (a2 z, θ) > δ (a1 z, θ) > 0 and for sufficiently large r, we get (3.10)−(3.13) hold. By (3.1) , (3.10)−(3.13) , (3.20) and (3.23) we obtain 3 exp {βr} 6 M1 [T (2r, f )] , (3.27) where β = [1 − ε (2n − 1)] δ (a2 z, θ) − (1 + ε) nδ (a1 z, θ) + m cos θ > 0. By β > 0 and (3.27), we have σ2 (f ) > 1, then σ2 (f ) = 1. Concluding the above proof, we obtain σ2 (f ) = 1. The proof of Theorem 1.1 is complete. Example 1.1 Consider the differential equation

2 f ′′ + −4e−3z − 4ie−z − 1 f ′ + iez + 2e−z f = 0, (3.28) where Q (z) = −4z 3 − 4iz − 1, a1 = 1, a2 = −1, A1 (z) = i and A2 (z) = 2. Obviously, the conditions of Theorem 1.1 (1) are satisfied. The entire function z f (z) = ee , with σ (f ) = +∞ and σ2 (f ) = 1, is a solution of (3.28). Example 1.2 Consider the differential equation   π 2 3  2π 1 π f ′′ + −8e−2z − 12ei 3 e−z − 1 − 6ei 3 f ′ + ei 3 e 3 z + 2e− 3 z f = 0, (3.29) π π 2π where Q (z) = −8z 2 − 12ei 3 z − 1 − 6ei 3 , a1 = 32 , a2 = − 31 , A1 (z) = ei 3 and A2 (z) = 2. Obviously, the conditions of Theorem 1.1 (1) are satisfied. The z entire function f (z) = ee , with σ (f ) = +∞ and σ2 (f ) = 1, is a solution of (3.29). Example 1.3 Consider the differential equation   1 4  3π π 1 π f ′′ + −e−3z − 4ei 4 e−2z − 6ie−z − 1 − 4ei 4 f ′ + e− 2 z + ei 4 e 2 z f = 0, (3.30) a2 = A1 (z) = 1 where Q (z) = −z − 4e z − 6iz − 1 − 4e , a1 = π and A2 (z) = ei 4 . Obviously, the conditions of Theorem 1.1 (3) are satisfied. z The entire function f (z) = ee , with σ (f ) = +∞ and σ2 (f ) = 1, is a solution of (3.30). 3 iπ 4 2 i 3π 4 Unauthenticated Download Date | 3/16/16 1:53 PM − 12 , 1 2, ON THE GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS 4 47 Proof of Theorem 1.2 We prove that λ (f − ϕ) = λ (f − ϕ) = σ (f ) = +∞ and λ2 (f − ϕ) = λ2 (f − ϕ) = σ2 (f ) = 1. First, setting ω = f − ϕ. Since σ (ϕ) < ∞, then we have σ (ω) = σ (f ) = +∞. From (1.1), we have
n ω ′′ + Q e−z ω ′ + (A1 ea1 z + A2 ea2 z ) ω = H, (4.1) n where H = − [ϕ′′ + Q (e−z ) ϕ′ + (A1 ea1 z + A2 ea2 z ) ϕ] . Now we prove that H 6≡ 0. In fact if H ≡ 0, then
n ϕ′′ + Q e−z ϕ′ + (A1 ea1 z + A2 ea2 z ) ϕ = 0. (4.2) Hence ϕ is a solution of equation (1.1) with σ (ϕ) = ∞ and by Theorem 1.1, it is a contradiction. Since σ (f ) = ∞, σ (ϕ) < ∞ and σ2 (f ) = 1, we get σ2 (ω) = σ2 (f − ϕ) = σ2 (f ) = 1. By the Lemma 2.6 and Lemma 2.7, we have λ (ω) = λ (ω) = σ(ω) = σ (f ) = +∞ and λ2 (ω) = λ2 (ω) = σ2 (ω) = σ2 (f ) = 1, i.e., λ (f − ϕ) = λ (f − ϕ) = σ (f ) = +∞ and λ2 (f − ϕ) = λ2 (f − ϕ) = σ2 (f ) = 1. 5 Proof of Theorem 1.3 Suppose that f 6≡ 0 is a solution of equation (1.1), then σ (f ) = +∞ by Theorem 1.1. Since σ (ϕ) < 1, then by Theorem 1.2, we have λ (f − ϕ) = +∞. Now we prove that λ (f ′ − ϕ) = ∞. Set g1 (z) = f ′ (z) − ϕ (z), then σ (g1 ) = σ (f ′ ) = σ (f ) = ∞. Set B (z) = Q (e−z ) and R (z) = A1 ea1 z + A2 ea2 z , then B ′ (z) = −e−z Q′ (e−z ) and R′ = (A′1 + a1 A1 ) ea1 z + (A′2 + a2 A2 ) ea2 z . Differentiating both sides of equation (1.1), we have f ′′′ + Bf ′′ + (B ′ + Rn ) f ′ + nR′ Rn−1 f = 0. (5.1) By (1.1), we have f =− 1 (f ′′ + Bf ′ ) . Rn Substituting (5.2) into (5.1), we have     R′ R′ ′′′ ′′ ′ n f + B−n f + B + R − nB f ′ = 0. R R (5.2) (5.3) Substituting f ′ = g1 + ϕ, f ′′ = g1′ + ϕ′ , f ′′′ = g1′′ + ϕ′′ into (5.3), we get g1′′ + E1 g1′ + E0 g1 = E, Unauthenticated Download Date | 3/16/16 1:53 PM (5.4) 48 Benharrat BELAÏDI and Habib HABIB where R′ R′ E1 = B − n , E0 = B ′ + Rn − nB , R R       ′ R′ R ′′ ′ ′ n ϕ + B + R − nB ϕ . E =− ϕ + B−n R R Now we prove that E 6≡ 0. In fact, if E ≡ 0, then we get ϕ′′ ϕ′ R+ (BR − nR′ ) + B ′ R − nBR′ + Rn+1 = 0. (5.5) ϕ ϕ  ′′   ′ ′ ′′ Obviously ϕϕ , ϕϕ are meromorphic functions with σ ϕϕ < 1, σ ϕϕ < 1. We can rewrite (5.5) in the form m X k=0 fk e(a1 −k)z + m X l=0 hl e(a2 −l)z + n X p Cn+1 A1n+1−p Ap2 e[(n+1−p)a1 +pa2 ]z p=1 +An+1 e(n+1)a1 z + An+1 e(n+1)a2 z = 0, 1 2 (5.6) where fk (k = 0, 1, · · ·, m) and hl (l = 0, 1, · · ·, m) are meromorphic functions with σ (fk ) < 1 and σ (fl ) < 1. Set I ={a1 − k (k = 0, 1, · · ·, m), a2 − l (l = 0, 1, · · ·, m), (n + 1 − p) a1 + pa2 (p = 1, 2, · · ·, n), (n + 1) a1 , (n + 1) a2 }. By the conditions of the Theorem 1.1, it is clear that (n + 1) a1 6= a1 , (n + 1) a2 , (n + 1 − p) a1 + pa2 (p = 1, 2, · · ·, n). (i) If (n + 1) a1 6= a1 − k (k = 1, · · ·, m), a2 − l (l = 0, 1, · · ·, m), then we write (5.6) in the form X An+1 e(n+1)a1 z + αβ eβz = 0, 1 β∈Γ1 where Γ1 ⊆ I \ {(n + 1) a1 }. By Lemma 2.8 and Lemma 2.9, we get A1 ≡ 0, it is a contradiction. (ii) If (n + 1) a1 = γ such that γ ∈{a1 − k (k = 1, · · ·, m), a2 − l (l = 0, 1, · · ·, m)}, then (n + 1) a2 6= β for all β ∈ I \ {(n + 1) a2 }. Hence, we write (5.6) in the form X αβ eβz = 0, An+1 e(n+1)a2 z + 2 β∈Γ2 where Γ2 ⊆ I \ {(n + 1) a2 }. By Lemma 2.8 and Lemma 2.9, we get A2 ≡ 0, it is a contradiction. Hence, E 6≡ 0 is proved. We know that the functions E1 , E0 and E are of finite order. By Lemma 2.6 and (5.4), we have λ (g1 ) = λ (f ′ − ϕ) = ∞. Unauthenticated Download Date | 3/16/16 1:53 PM ON THE GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS 49 Now we prove that λ (f ′′ − ϕ) = ∞. Set g2 (z) = f ′′ (z) − ϕ (z), then σ (g2 ) = σ (f ′′ ) = σ (f ) = ∞. Differentiating both sides of equation (1.1), we have
f (4) + Bf ′′′ + (2B ′ + Rn ) f ′′ + B ′′ + 2nR′ Rn−1 f ′   +n R′′ Rn−1 + (n − 1) R′2 Rn−2 f = 0. (5.7) Combining (5.2) with (5.7), we get   R′2 R′′ (4) ′′′ ′ n − n (n − 1) 2 f ′′ f + Bf + 2B + R − n R R   R′′ R′2 + B ′′ + 2nR′ Rn−1 − nB − n (n − 1) B 2 f ′ = 0. R R ′ (5.8) ′ Now we prove that B ′ + Rn − nB RR 6≡ 0. Suppose that B ′ + Rn − nB RR ≡ 0, then we have B ′ R + Rn+1 − nBR′ = 0. (5.9) We can write (5.9) in the form (5.6), then by the same reasoning as in the ′ proof of λ (f ′ − ϕ) = ∞ we get a contradiction. Hence B ′ + Rn − nB RR 6≡ 0 is proved. Set ψ (z) = B ′ R + Rn+1 − nBR′ , (5.10) S1 = 2B ′ R2 + Rn+2 − nR′′ R − n (n − 1) R′2 , ′′ 2 ′ S2 = B R + 2nR R n+1 ′′ (5.11) ′2 − nBR R − n (n − 1) BR , S3 = BR − nR′ . (5.12) (5.13) By (5.3) , (5.10) and (5.13), we get f′ = − R ψ (z)  f ′′′ +  S3 ′′ f . R By (5.14) , (5.11) , (5.12) and (5.8), we obtain     S1 S2 S3 S2 ′′′ (4) f + f ′′ = 0. − 2 f + B− Rψ (z) R2 R ψ (z) (5.14) (5.15) Substituting f ′′ = g2 + ϕ, f ′′′ = g2′ + ϕ′ , f (4) = g2′′ + ϕ′′ into (5.15) we get g2′′ + H1 g2′ + H0 g2 = H, where H1 = B − S2 S1 S2 S3 , H0 = 2 − 2 , Rψ (z) R R ψ (z) Unauthenticated Download Date | 3/16/16 1:53 PM (5.16) 50 Benharrat BELAÏDI and Habib HABIB −H = ϕ′′ + ϕ′ H1 + ϕH0 . We can get L0 (z) L1 (z) , H0 = , Rψ (z) Rψ (z) H1 = (5.17) where L1 (z) = B ′ BR2 + BRn+2 − nB 2 R′ R − B ′′ R2 − 2nR′ Rn+1 +nBR′′ R + n (n − 1) BR′2 , ′2 2 ′ L0 (z) = 2B R + 3B R n+2 ′ ′ − 2nB BR R + R (5.18) 2n+2 ′ − 3nBR R n+1 −nB ′ R′′ R − nR′′ Rn+1 − n (n − 1) B ′ R′2 + n2 + n R′2 Rn − B ′′ BR2
+nB 2 R′′ R + n (n − 1) B 2 R′2 + nB ′′ R′ R. (5.19) Therefore −H 1 = ϕ Rψ (z)   ϕ′′ ϕ′ Rψ (z) + L1 (z) + L0 (z) , ϕ ϕ (5.20) Rψ (z) = B ′ R2 + Rn+2 − nBR′ R. (5.21) Now we prove that −H 6≡ 0. In fact, if −H ≡ 0, then by (5.20) we have ϕ′ ϕ′′ Rψ (z) + L1 (z) + L0 (z) = 0. ϕ ϕ Obviously, ϕ′′ ϕ and ϕ′ ϕ are meromorphic functions with σ (5.22)  ϕ′′ ϕ  < 1, σ 1. By (5.18) , (5.19) and (5.21), we can rewrite (5.22) in the form A2n+2 e(2n+2)a1 z + A2n+2 e(2n+2)a2 z + 1 2 2n+1 X  ϕ′ ϕ  < p C2n+2 A12n+2−p Ap2 e[(2n+2−p)a1 +pa2 ]z p=1 + X fp,k e[(2−p)a1 +pa2 −k]z + X 06p62 06p6n+2 06k62m 06k6m hp,k e[(n+2−p)a1 +pa2 −k]z = 0, (5.23) where fp,k (0 6 p 6 2, 0 6 k 6 2m) and hp,k (0 6 p 6 n + 2, 0 6 k 6 m) are meromorphic functions with σ (fp,k ) < 1 and σ (hp,k ) < 1. Set J ={(2n + 2) a1 , (2n + 2) a2 , (2n + 2 − p) a1 + pa2 (p = 1, 2, · · ·, 2n + 1), (2 − p) a1 + pa2 − k (p = 0, 1, 2; k = 0, · · ·, 2m), (n + 2 − p) a1 + pa2 − k (p = 0, 1, · · ·, n + 2; k = 0, 1, · · ·, m)}. By the conditions of Theorem 1.3, it is clear that (2n + 2) a1 6= (2n + 2) a2 , (2n + 2 − p) a1 +pa2 (p = 1, 2, · · ·, 2n+1), Unauthenticated Download Date | 3/16/16 1:53 PM ON THE GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS 51 2a1 , (n + 2) a1 and (2n + 2) a2 6= (2n + 2) a1 , (2n + 2 − p) a1 + pa2 (p = 1, 2, · · ·, 2n + 1), 2a2 , (n + 2) a2 . (1) By the conditions of Theorem 1.3 (i), we have (2n + 2) a1 6= β for all β ∈ J \ {(2n + 2) a1 }, hence we write (5.23) in the form A2n+2 e(2n+2)a1 z + 1 X αβ eβz = 0, β∈Γ1 where Γ1 ⊆ J \ {(2n + 2) a1 }. By Lemma 2.8 and Lemma 2.9, we get A1 ≡ 0, it is a contradiction. (2) By the conditions of Theorem 1.3 (ii), we have (2n + 2) a2 6= β for all β ∈ J \ {(2n + 2) a2 }, hence we write (5.23) in the form A2n+2 e(2n+2)a2 z + 2 X αβ eβz = 0, β∈Γ2 where Γ2 ⊆ J \ {(2n + 2) a2 }. By Lemma 2.8 and Lemma 2.9, we get A2 ≡ 0, it is a contradiction. Hence, H 6≡ 0 is proved. We know that the functions H1 , H0 and H are of finite order. By Lemma 2.6 and (5.16), we have λ (g2 ) = λ (f ′′ − ϕ) = ∞. The proof of Theorem 1.3 is complete. References [1] B. Belaı̈di, Growth and oscillation theory of solutions of some linear differential equations, Mat. Vesnik 60 (2008), no. 4, 233–246. [2] Z. X. Chen, Zeros of meromorphic solutions of higher order linear differential equations, Analysis 14 (1994), no. 4, 425–438. [3] Z. X. Chen, The fixed points and hyper-order of solutions of second order complex differential equations (in Chinese), Acta Math. Sci. Ser. A Chin. Ed. 20 (2000), no. 3, 425–432. [4] Z. X. Chen, The growth of solutions of f ′′ + e−z f ′ + Q (z) f = 0 where the order (Q) = 1, Sci. China Ser. A 45 (2002), no. 3, 290–300. [5] Z. X. Chen and K. H. Shon, On the growth of solutions of a class of higher order differential equations, Acta Math. Sci. Ser. B Engl. Ed. 24 (2004), no. 1, 52–60. [6] F. Gross, On the distribution of values of meromorphic functions, Trans. Amer. Math. Soc. 131(1968), 199–214. Unauthenticated Download Date | 3/16/16 1:53 PM 52 Benharrat BELAÏDI and Habib HABIB [7] G. G. Gundersen, Estimates for the logarithmic derivative of a meromorphic function, plus similar estimates, J. London Math. Soc. (2) 37 (1988), no. 1, 88–104. [8] W. K. Hayman, Meromorphic functions, Oxford Mathematical Monographs Clarendon Press, Oxford 1964. [9] K. H. Kwon, Nonexistence of finite order solutions of certain second order linear differential equations, Kodai Math. J. 19 (1996), no. 3, 378–387. [10] M. S. Liu, X. M. Zhang, Fixed points of meromorphic solutions of higher order Linear differential equations, Ann. Acad. Sci. Fenn. Math. 31 (2006), no. 1, 191–211. [11] F. Peng and Z. X. Chen, On the growth of solutions of some second-order linear differential equations, J. Inequal. Appl. 2011, Art. ID 635604, 1–9. [12] J. F. Xu, H. X. Yi, The relation between solutions of higher order differential equations with functions of small growth, Acta Math. Sci., Chinese Series, 53 (2010), 291–296. [13] C. C. Yang and H. X. Yi, Uniqueness theory of meromorphic functions, Mathematics and its Applications, 557. Kluwer Academic Publishers Group, Dordrecht, 2003. Benharrat BELAÏDI Department of Mathematics, Laboratory of Pure and Applied Mathematics University of Mostaganem (UMAB), B. P. 227 Mostaganem-Algeria. Email: belaidi@univ-mosta.dz Habib HABIB Department of Mathematics, Laboratory of Pure and Applied Mathematics University of Mostaganem (UMAB), B. P. 227 Mostaganem-Algeria. Email: habibhabib2927@yahoo.fr Unauthenticated Download Date | 3/16/16 1:53 PM