Electronic Journal of Differential Equations, Vol. 2023 (2023), No. 27, pp. 1–14. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu or https://ejde.math.unt.edu GROWTH PROPERTIES OF SOLUTIONS OF COMPLEX DIFFERENTIAL EQUATIONS WITH ENTIRE COEFFICIENTS OF FINITE (α, β, γ)-ORDER BENHARRAT BELAÏDI, TANMAY BISWAS Abstract. In this article, we investigate the complex higher order linear dif- ferential equations in which the coefficients are entire functions of (α, β, γ)- order and obtain some results which improve and generalize some previous results of Tu et al. [29] as well as Beläıdi [1, 2, 3]. 1. Introduction Throughout this article, we assume that the reader is familiar with the fundamen- tal results and the standard notations of the Nevanlinna value distribution theory of entire and meromorphic functions and the theory of complex linear differential equations which are available in [12, 21, 34] and therefore we do not explain those in details. To study the generalized growth properties of entire and meromorphic functions, the concepts of different growth indicators such as the iterated p-order (see [20, 26] ), the (p, q)-th order (see [17, 18]), (p, q)-ϕ order (see [27]) etc. are very useful and during the past decades, several authors made close investigations on the generalized growth properties of entire and meromorphic functions related to the above growth indicators in some different directions. The theory of complex linear equations has been developed since 1960s. Many authors have investigated the complex linear differential equations f (k)(z) +Ak−1(z)f (k−1)(z) + · · ·+A0(z)f(z) = 0, (1.1) f (k)(z) +Ak−1(z)f (k−1)(z) + · · ·+A0(z)f(z) = F (z) (1.2) and achieved many valuable results when the coefficients A0(z), . . . , Ak−1(z), F (z) (k ≥ 2) in (1.1) or (1.2) are entire functions of finite order or finite iterated p-order or (p, q)-th order or (p, q)-ϕ order; see [1, 2, 3, 7, 8, 10, 15, 21, 22, 23, 24, 26, 27, 29, 30, 31, 33]. In [9], Chyzhykov and Semochko showed that both definitions of iterated p- order and the (p, q)-th order have the disadvantage that they do not cover arbitrary growth (see [9, Example 1.4]). They used more general scale, called the ϕ-order (see [9]). In recent times, the concept of ϕ-order is used to study the growth of solutions of complex differential equations which extend and improve many previous results (see [4, 5, 9, 19]). 2020 Mathematics Subject Classification. 30D35, 34M10. Key words and phrases. Complex differential equations; (α, β, γ)-order; growth of solutions. ©2023. This work is licensed under a CC BY 4.0 license. Submitted July 22, 2023. Published March 11, 2023. 1 2 B. BELAÏDI. T. BISWAS EJDE-2023/27 In [25], Mulyava et al. have used the concept of (α, β)-order or generalized order of an entire function in order to investigate the properties of solutions of a hetero- geneous differential equation of the second order and obtained several interesting results. For details about (α, β)-order one may see [25, 28]. In this paper, we investigate the complex higher order linear differential equations in which the coefficients are entire functions of (α, β, γ)-order and obtain some results which improve and generalize some previous results of Tu et al. [29] as well as Beläıdi [1, 2, 3]. 2. Definitions and notation First of all, let L be a class of continuous non-negative on (−∞,+∞) function α such that α(x) = α(x0) ≥ 0 for x ≤ x0 and α(x) ↑ +∞ as x0 ≤ x → +∞. We say that α ∈ L1, if α ∈ L and α(a+ b) ≤ α(a) +α(b) + c for all a, b ≥ R0 and fixed c ∈ (0,+∞). Further we say that α ∈ L2, if α ∈ L and α(x+O(1)) = (1+o(1))α(x) as x→ +∞. Finally, α ∈ L3, if α ∈ L and α(a+ b) ≤ α(a) + α(b) for all a, b ≥ R0, i.e., α is subadditive. Clearly L3 ⊂ L1. Particularly, when α ∈ L3, then one can easily verify that α(mr) ≤ mα(r), m ≥ 2 is an integer. Up to a normalization, subadditivity is implied by concavity. Indeed, if α(r) is concave on [0,+∞) and satisfies α(0) ≥ 0, then for t ∈ [0, 1], α(tx) = α(tx+ (1− t) · 0) ≥ tα(x) + (1− t)α(0) ≥ tα(x), so that by choosing t = a a+b or t = b a+b , we obtain α(a+ b) = a a+ b α(a+ b) + b a+ b α(a+ b) ≤ α ( a a+ b (a+ b) ) + α ( b a+ b (a+ b) ) = α(a) + α(b), a, b ≥ 0. As a non-decreasing, subadditive and unbounded function, α(r) satisfies α(r) ≤ α(r +R0) ≤ α(r) + α(R0) for any R0 ≥ 0. This yields that α(r) ∼ α(r +R0) as r → +∞. Now we add two conditions on α, β and γ: (i) Always α ∈ L1, β ∈ L2 and γ ∈ L3; and (ii) α(log[p] x) = o(β(log γ(x))), p ≥ 2, α(log x) = o(α(x)) and α−1(kx) = o(α−1(x)) (k < 1) as x→ +∞. Throughout this paper, we assume that α, β and γ always satisfy the above two conditions unless otherwise specifically stated. Heittokangas et al. [16] introduced a new concept of ϕ-order of entire and mero- morphic function considering ϕ as subadditive function. For details one may see [16]. Extending this notion, recently Beläıdi and Biswas [6] introduce the definition of the (α, β, γ)-order of a meromorphic function in the following way: Definition 2.1 ([6]). The (α, β, γ)-order denoted by σ(α,β,γ)[f ] of an entire function f(z) is defined by σ(α,β,γ)[f ] = lim sup r→+∞ α(log[2]M(r, f)) β(log γ(r)) . EJDE-2023/27 GROWTH OF SOLUTIONS OF COMPLEX DIFFERENTIAL EQUATIONS 3 By the inequality T (r, f) ≤ log+M(r, f) ≤ R+r R−rT (R, f) (0 < r < R) [12] for an entire function f(z), one can easily verify that [6] σ(α,β,γ)[f ] = lim sup r→+∞ α(log T (r, f)) β(log γ(r)) = lim sup r→+∞ α(log[2]M(r, f)) β(log γ(r)) . Proposition 2.2 ([6]). If f(z) is an entire function, then σ(α(log),β,γ)[f ] = lim sup r→+∞ α(log[2] T (r, f)) β(log γ(r)) = lim sup r→+∞ α(log[3]M(r, f)) β(log γ(r)) . Similar to Definition 2.1, one can define the (α, β, γ)-exponent convergence of the zero-sequence of a meromorphic the following way: Definition 2.3 ([6]). The (α, β, γ)-exponent convergence of the zero-sequence de- noted by λ(α,β,γ)[f ] of a meromorphic function f(z) is defined by λ(α,β,γ)[f ] = lim sup r→+∞ α(log n(r, 1/f)) β(log γ(r)) . Analogously, the (α, β, γ)-exponent convergence of the distinct zero-sequence de- noted by λ(α,β,γ)[f ] of f(z) is defined by λ(α,β,γ)[f ] = lim sup r→+∞ α(log n(r, 1/f)) β(log γ(r)) . Accordingly, the values λ(α(log),β,γ)[f ] = lim sup r→+∞ α(log[2] n(r, 1/f)) β(log γ(r)) , λ(α(log),β,γ)[f ] = lim sup r→+∞ α(log[2] n(r, 1/f)) β(log γ(r)) are respectively called as (α(log), β, γ)-exponent convergence of the zero-sequence and (α(log), β, γ) -exponent convergence of the distinct zero-sequence of a mero- morphic function f(z). The linear measure of a set E ⊂ [0,+∞) is defined as m(E) = ∫ +∞ 0 χE(t)dt. The logarithmic measure of a set E ⊂ [1,+∞) is defined by lm(E) = ∫ +∞ 1 χE(t) t dt, where χE(t) is the characteristic function of E. The upper and lower densities of E are densE = lim sup r→+∞ m(E ∩ [0, r]) r , densE = lim inf r→+∞ m(E ∩ [0, r]) r . Proposition 2.4 ([6]). If f(z) is a meromorphic function, then λ(α,β,γ)[f ] = lim sup r→+∞ α(log n(r, 1/f)) β(log γ(r)) = lim sup r→+∞ α(logN(r, 1/f)) β(log γ(r)) and λ(α,β,γ)[f ] = lim sup r→+∞ α(log n(r, 1/f)) β(log γ(r)) = lim sup r→+∞ α(logN(r, 1/f)) β(log γ(r)) . Proposition 2.5 ([6]). If f(z) is a meromorphic function, then λ(α(log),β,γ)[f ] = lim sup r→+∞ α(log[2] n(r, 1/f)) β(log γ(r)) = lim sup r→+∞ α(log[2]N(r, 1/f)) β(log γ(r)) 4 B. BELAÏDI. T. BISWAS EJDE-2023/27 and λ(α(log),β,γ)[f ] = lim sup r→+∞ α(log[2] n(r, 1/f)) β(log γ(r)) = lim sup r→+∞ α(log[2]N(r, 1/f)) β(log γ(r)) . Proposition 2.6 ([6]). Let f1(z), f2(z) be non-constant meromorphic functions with σ(α(log),β,γ)[f1] and σ(α(log),β,γ)[f2] as their (α(log), β, γ)-order. Then (i) σ(α(log),β,γ)[f1 ± f2] ≤ max{σ(α(log),β,γ)[f1], σ(α(log),β,γ)[f2]}; (ii) σ(α(log),β,γ)[f2 · f2] ≤ max{σ(α(log),β,γ)[f1], σ(α(log),β,γ)[f2]}; (iii) If σ(α(log),β,γ)[f1] 6= σ(α(log),β,γ)[f2], then σ(α(log),β,γ)[f1 ± f2] = max{σ(α(log),β,γ)[f1], σ(α(log),β,γ)[f2]}; (iv) If σ(α(log),β,γ)[f1] 6= σ(α(log),β,γ)[f2], then σ(α(log),β,γ)[f2 · f2] = max{σ(α(log),β,γ)[f1], σ(α(log),β,γ)[f2]}. 3. Main Results In this section we present our main results which considerably extend the results by Tu et al. [29] as well as those by Beläıdi [1, 2, 3]. Theorem 3.1. Let A0(z), A1(z), . . . , Ak−1(z) be entire functions with A0(z) 6≡ 0 such that for real constants a, b, µ, θ1, θ2 with 0 ≤ b < a, µ > 0, θ1 < θ2, we have |A0(z)| ≥ exp{a exp(α−1(µβ(log γ(|z|))))} (3.1) and |Aj(z)| ≤ exp{b exp(α−1(µβ(log γ(|z|))))}, j = 1, . . . , k − 1, (3.2) as z → ∞ with θ1 ≤ arg z ≤ θ2. Then σ(α(log),β,γ)[f ] ≥ µ holds for all non-trivial solutions of (1.1). Theorem 3.2. Let H be a set of complex numbers satisfying dens{|z| : z ∈ H} > 0, and let A0(z), A1(z), . . . , Ak−1(z) be entire functions and satisfy (3.1) and (3.2) as z → ∞ for z ∈ H, where 0 ≤ b < a, µ > 0. Then every solution f(z) 6≡ 0 of (1.1) satisfies σ(α(log),β,γ)[f ] ≥ µ. Theorem 3.3. Let H be a set of complex numbers satisfying dens{|z| : z ∈ H} > 0, and let A0(z), A1(z), . . . , Ak−1(z) be entire functions of (α, β, γ)-order with max{σ(α,β,γ)[Aj ] : j = 1, . . . , k − 1} ≤ σ(α,β,γ)[A0] = σ < +∞ such that for some constants 0 ≤ b < a and for any given ε > 0, we have |A0(z)| ≥ exp{a exp(α−1((σ − ε)β(log γ(|z|))))} (3.3) and |Aj(z)| ≤ exp{b exp(α−1((σ − ε)β(log γ(|z|))))}, j = 1, . . . , k − 1, (3.4) as z →∞ for z ∈ H. Then every solution f(z) 6≡ 0 of (1.1) satisfies σ(α(log),β,γ)[f ] = σ(α,β,γ)[A0] = σ. Theorem 3.4. Let H, A0(z), A1(z), . . . , Ak−1(z) satisfy the hypotheses of Theorem 3.3, and let F (z) 6≡ 0 be an entire function of (α, β, γ)-order. (i) If σ(α(log),β,γ)[F ] < σ(α,β,γ)[A0], then every solution f(z) of (1.2) satisfies λ(α(log),β,γ)[f ] = λ(α(log),β,γ)[f ] = σ(α(log),β,γ)[f ] = σ, with at most one exceptional solution f0(z) satisfying σ(α(log),β,γ)[f0] < σ. (ii) If σ(α,β,γ)[A0] ≤ σ(α(log),β,γ)[F ] < +∞, then every solution f(z) of (1.2) satisfies σ(α(log),β,γ)[f ] = σ(α(log),β,γ)[F ]. EJDE-2023/27 GROWTH OF SOLUTIONS OF COMPLEX DIFFERENTIAL EQUATIONS 5 4. Some Lemmas In this section we present some lemmas which will be needed in the sequel. Lemma 4.1 ([11]). Let f(z) be a nontrivial entire function, and let κ > 1 and ε > 0 be given constants. Then there exist a constant c > 0 and a set E1 ⊂ [0,+∞) having finite linear measure such that for all z satisfying |z| = r /∈ E1, we have∣∣f (k)(z) f(z) ∣∣ ≤ c[T (κr, f)rε log T (κr, f)]k (k ∈ N). (4.1) Lemma 4.2 ([13, 14, 21, 32]). Let f(z) be a transcendental entire function, and let z be a point with |z| = r at which |f(z)| = M(r, f). Then, for all |z| outside a set E2 of r of finite logarithmic measure, we have f (k)(z) f(z) = (ν(r, f) z )k (1 + o(1)) (k ∈ N, r /∈ E2), (4.2) where ν(r, f) is the central index of f(z). Lemma 4.3 ([6]). Let f(z) be an entire function satisfying σ(α(log),β,γ)[f ] = σ1, and let ν(r, f) be the central index of f(z). Then lim sup r→+∞ α(log[2] ν(r, f)) β(log γ(r)) = σ1. Lemma 4.4. Let f(z) be a transcendental entire function. Then σ(α(log),β,γ)[f ] = σ(α(log),β,γ)[f ′]. Proof. By Cauchy’s integral formula, we have f ′(z) = 1 2πi ∮ Γ f(ζ) (ζ − z)2 dζ, where Γ = {ζ : |ζ − z| = R− r}, |z| = r < R. Set ζ − z = (R− r)eiθ (0 ≤ θ ≤ 2π), dζ = (R− r)ieiθdθ. Since max{|f(ζ)| : ζ ∈ Γ} ≤M(R, f), then we obtain M(r, f ′) = |f ′(z)| ≤ 1 2π ∫ 2π 0 |f(ζ)| |ζ − z|2 (R− r)dθ ≤ M(R, f) R− r . Setting R = r + 1, it follows that M(r, f ′) ≤M(r + 1, f). Since γ(r +R0) ∼ γ(r) as r → +∞, it follows that σ(α(log),β,γ)[f ′] = lim sup r→+∞ α(log[3]M(r, f ′)) β(log γ(r)) ≤ lim sup r→+∞ (α(log[3]M(r + 1, f)) β(log γ(r + 1)) · β(log γ(r + 1)) β(log γ(r)) ) = lim sup r→+∞ (α(log[3]M(r + 1, f)) β(log γ(r + 1)) · β(log γ(r)) β(log γ(r)) ) = lim sup r→+∞ α(log[3]M(r + 1, f)) β(log γ(r + 1)) . Thus, from the above we obtain σ(α(log),β,γ)[f ′] ≤ σ(α(log),β,γ)[f ]. (4.3) 6 B. BELAÏDI. T. BISWAS EJDE-2023/27 On the other hand, for an entire function f(z), we have f(z) − f(0) = ∫ z 0 f ′(t)dt, where the integral being taken along the straight line from 0 to z, so we obtain that M(r, f) ≤ ∣∣ ∫ z 0 f ′(t)dt|+ |f(0)| ≤ rM(r, f ′) + |f(0)|. Therefore from above we have log[3]M(r, f) ≤ log[3]M(r, f ′) + log[3] r + log[3] |f(0)|+O(1). Since α(a + b) ≤ α(a) + α(b) + c, c > 0 and α(log[3] x) = o(β(log γ(x))), so from above we get that σ(α(log),β,γ)[f ] ≤ σ(α(log),β,γ)[f ′]. (4.4) Hence the lemma follows from (4.3) and (4.4). � Remark 4.5. In the line of Lemma 4.4 one can easily deduce that σ(α,β,γ)[f ] = σ(α,β,γ)[f ′], where f(z) is an entire transcendental function. Lemma 4.6. Let f(z) be an entire function of (α, β, γ)-order that satisfies σ(α,β,γ)[f ] = σ. Then there exists a set E3 ⊂ (1,+∞) having infinite logarithmic measure such that for all r ∈ E3, we have lim r→+∞ α(log T (r, f)) β(log γ(r)) = σ (r ∈ E3). Proof. By Definition 2.1, there exists an increasing sequence {rn}+∞n=1 tending to +∞ that satisfying (1 + 1 n )rn < rn+1 and lim rn→+∞ α(log T (rn, f)) β(log γ(rn)) = σ(α,β,γ)[f ] = σ. So, there exists an n1 ∈ N such that for n ≥ n1 and for any r ∈ E3 = ∪+∞ n=n1 [rn, (1+ 1 n )rn], we have α(log T (rn, f)) β(log γ((1 + 1 n )rn)) ≤ α(log T (r, f)) β(log γ(r)) ≤ α(log T ((1 + 1 n )rn, f)) β(log γ(rn)) . (4.5) From this inequality and γ((1 + 1 n )rn) ≤ γ(2rn) ≤ 2γ(rn), we have lim r→+∞, r∈E3 α(log T (r, f)) β(log γ(r)) ≥ lim rn→+∞ (α(log T (rn, f)) β(log γ(rn)) β ( log γ(rn) ) β ( log γ((1 + 1 n )rn)) ) ≥ lim rn→+∞ (α(log T (rn, f)) β ( log γ(rn) ) β(log γ(rn)) β ( log(2γ(rn)) )) = lim rn→+∞ (α(log T (rn, f)) β(log γ(rn)) β(log γ(rn)) β((1 + log 2 log γ(rn) ) log γ(rn)) ) = lim rn→+∞ (α(log T (rn, f)) β ( log γ(rn) ) β(log γ(rn)) β ( (1 + o(1)) log γ(rn) )). (4.6) EJDE-2023/27 GROWTH OF SOLUTIONS OF COMPLEX DIFFERENTIAL EQUATIONS 7 From this inequality and β(x+ o(1)) = (1 + o(1))β(x) as x→ +∞, we obtain that lim r→+∞, r∈E3 α(log T (r, f)) β(log γ(r)) ≥ lim rn→+∞ (α(log T (rn, f)) β(log γ(rn)) β(log γ(rn)) (1 + o(1))β(log γ(rn)) ) = lim rn→+∞ α(log T (rn, f)) β(log γ(rn)) = σ. (4.7) On the other hand, by (4.5), γ((1 + 1 n )rn) ≤ γ(2rn) ≤ 2γ(rn) and β(x + o(1)) = (1 + o(1))β(x) as x→ +∞, we have lim r→+∞, r∈E3 α(log T (r, f)) β(log γ(r)) ≤ lim rn→+∞ (α(log T ((1 + 1 n )rn, f)) β(log γ(1 + 1 n )rn) β(log γ((1 + 1 n )rn)) β(log γ(rn)) ) ≤ lim rn→+∞ (α(log T ((1 + 1 n )rn, f)) β(log γ(1 + 1 n )rn) β(log(2γ(rn))) β(log γ(rn)) ) = lim rn→+∞ (α(log T ((1 + 1 n )rn, f)) β(log γ(1 + 1 n )rn) β((1 + o(1)) log γ(rn)) β(log γ(rn)) ) = lim rn→+∞ (α(log T ((1 + 1 n )rn, f)) β(log γ(1 + 1 n )rn) (1 + o(1))β(log γ(rn)) β(log γ(rn)) ) = σ. (4.8) Therefore, by (4.7) and (4.8), we obtain lim r→+∞, r∈E3 α(log T (r, f)) β(log γ(r)) = σ, where lm(E3) = ∑+∞ n=n1 ∫ (1+ 1 n ) rn 1 t dt = ∑+∞ n=n1 log(1 + 1 n ) = +∞. This completes the proof. � Lemma 4.7. Let f(z) be an entire function of (α, β, γ)-order with σ(α,β,γ)[f ] = σ > 0, and let f1(z) be an entire function of (α1, β1, γ1)-order with σ(α1,β1,γ1)[f1] = σ1 < +∞. If σ(α,β,γ)[f ] and σ(α1,β1,γ1)[f1] satisfy one of the following conditions: (i) α(r) = α1(r), β(r) = β1(r), γ(r) = γ1(r). and σ(α1,β1,γ1)[f1] < σ(α,β,γ)[f ]; (ii) limr→+∞ α−1 1 (r) α−1(r) = 0, β(r) = β1(r), γ(r) = γ1(r) and σ(α1,β1,γ1)[f1] < σ(α,β,γ)[f ]; then there exists a set E4 ⊂ (1,+∞) having infinite logarithmic measure such that for all r ∈ E4, we have lim r→+∞ T (r, f1) T (r, f) = 0 (r ∈ E4). Proof. (i) By definition, for all sufficiently large values of r, we obtain T (r, f1) ≤ exp{α−1((σ1 + ε)β(log γ(r)))}. (4.9) From σ(α,β,γ)[f ] = σ and Lemma 4.6, there exists a set E4 of infinite logarithmic measure satisfying lim r→+∞ α(log T (r, f)) β(log γ(r)) = σ (r ∈ E4). 8 B. BELAÏDI. T. BISWAS EJDE-2023/27 Then T (r, f) ≥ exp{α−1((σ − ε)β(log γ(r)))} (r ∈ E4), (4.10) where 0 < 2ε < σ − σ1. Now by (4.9) and (4.10), we obtain that T (r, f1) T (r, f) ≤ exp{α−1((σ1 + ε)β(log γ(r)))} exp{α−1((σ − ε)β(log γ(r)))} = exp { α−1((σ1 + ε)β(log γ(r)))− α−1((σ − ε)β(log γ(r))) } = exp { α−1((σ − ε)β(log γ(r))) (α−1((σ1 + ε)β(log γ(r))) α−1((σ − ε)β(log γ(r))) − 1 )} = exp { α−1((σ − ε)β(log γ(r))) (α−1(σ1+ε σ−ε (σ − ε)β(log γ(r))) α−1((σ − ε)β(log γ(r))) − 1 )} = exp { α−1((σ − ε)β(log γ(r))) (α−1(k(σ − ε)β(log γ(r))) α−1((σ − ε)β(log γ(r))) − 1 )} → 0, r → +∞ (r ∈ E4), k = σ1 + ε σ − ε < 1. From the above inequality we obtain lim r→+∞ T (r, f1) T (r, f) = 0 (r ∈ E4). (ii) By definition, we obtain for all sufficiently large values of r that T (r, f1) ≤ exp{α−1 1 ((σ1 + ε)β(log γ(r)))}. (4.11) Now by (4.10) and (4.11), for any given ε with 0 < 2ε < σ − σ1. we obtain that T (r, f1) T (r, f) ≤ exp{α−1 1 ((σ1 + ε)β(log γ(r)))} exp{α−1((σ − ε)β(log γ(r)))} = exp{α−1 1 ((σ1 + ε)β(log γ(r)))} exp{α−1((σ1 + ε)β(log γ(r)))} exp{α−1((σ1 + ε)β(log γ(r)))} exp{α−1((σ − ε)β(log γ(r)))} = exp { α−1((σ1 + ε)β(log γ(r))) (α−1 1 ((σ1 + ε)β(log γ(r))) α−1((σ1 + ε)β(log γ(r))) − 1 )} × exp{α−1((σ1 + ε)β(log γ(r)))} exp{α−1((σ − ε)β(log γ(r)))} . Since limr→+∞ α−1 1 (r) α−1(r) = 0 and limr→+∞ α−1(kr) α−1(r) = 0 (k < 1), then by the inequality obove, we obtain lim r→+∞ T (r, f1) T (r, f) = 0 (r ∈ E4). � Lemma 4.8. Let F (z) 6≡ 0, Aj(z) (j = 0, . . . , k − 1) be entire functions. Also let f(z) be a solution of (1.2) satisfying max{σ(α(log),β,γ)[Aj ], σ(α(log),β,γ)[F ] : j = 0, 1, . . . , k − 1} < σ(α(log),β,γ)[f ]. Then λ(α(log),β,γ)[f ] = λ(α(log),β,γ)[f ] = σ(α(log),β,γ)[f ]. EJDE-2023/27 GROWTH OF SOLUTIONS OF COMPLEX DIFFERENTIAL EQUATIONS 9 Proof. By (1.2) we have 1 f = 1 F (f (k) f +Ak−1(z) f (k−1) f + · · ·+A1(z) f ′ f +A0 ) . (4.12) Now it is easy to see that if f(z) has a zero at z0 of order a (a > k), and A0, . . . , Ak−1 are analytic at z0, then F (z) must have a zero at z0 of order a− k, hence n ( r, 1 f ) ≤ kn ( r, 1 f ) + n ( r, 1 F ) (4.13) and N ( r, 1 f ) ≤ kN ( r, 1 f ) +N ( r, 1 F ) . (4.14) By the lemma on logarithmic derivative and (4.12), we have m ( r, 1 f ) ≤ m ( r, 1 F ) + k−1∑ j=0 m(r,Aj) +O(log T (r, f) + log r) (r /∈ E5), (4.15) where E5 is a set of r of finite linear measure. By (4.14) and ( 4.15), we obtain that T (r, f) = T ( r, 1 f ) +O(1) ≤ kN ( r, 1 f ) + T (r, F ) + k−1∑ j=0 T (r,Aj) +O(log(rT (r, f))) (4.16) for r /∈ E5. Since max{σ(α(log),β,γ)[Aj ], σ(α(log),β,γ)[F ] : j = 0, 1, . . . , k − 1} < σ(α(log),β,γ)[f ], by Lemma 4.7, there exists a set E4 having infinite logarithmic measure such that max {T (r, F ) T (r, f) , T (r,Aj) T (r, f) } → 0, r → +∞ (r ∈ E4, j = 0, . . . , k − 1). (4.17) Since f(z) is transcendental, we have O(log(rT (r, f))) = o(T (r, f)) as r → +∞. (4.18) Therefore, by substituting (4.17) and (4.18) into (4.16), for all |z| = r ∈ E4\E5, we obtain T (r, f) ≤ O ( N ( r, 1 f )) . Hence from above we have σ(α(log),β,γ)[f ] ≤ λ(α(log),β,γ)[f ]. Therefore, λ(α(log),β,γ)[f ] = λ(α(log),β,γ)[f ] = σ(α(log),β,γ)[f ]. Hence the lemma follows. � Lemma 4.9. Let f be a meromorphic function. If σ(α,β,γ)[f ] = σ < +∞, then σ(α(log),β,γ)[f ] = 0. Proof. Suppose that σ(α,β,γ)[f ] = σ < +∞. Then, for any given ε > 0 and suffi- ciently large r, we have T (r, f) ≤ exp{α−1((σ + ε)β(log γ(r)))}. 10 B. BELAÏDI. T. BISWAS EJDE-2023/27 Then, we immediately obtain σ(α(log),β,γ)[f ] = lim sup r→+∞ α(log[2] T (r, f)) β(log γ(r)) ≤ lim sup r→+∞ α(log[2](exp{α−1((σ + ε)β(log γ(r)))})) β(log γ(r)) = lim sup r→+∞ α(logα−1((σ + ε)β(log γ(r)))) β(log γ(r)) = lim sup x→+∞ α(logα−1((σ + ε)x)) x = (σ + ε)lim sup x→+∞ α(log x) α(x) = 0. � 5. Proof of main results Proof of Theorem 3.1. Let f(z) 6≡ 0 be a solution of (1.1) and rewrite (1.1) as A0(z) = − (f (k)(z) f(z) +Ak−1(z) f (k−1)(z) f(z) + · · ·+A1(z) f ′(z) f(z) ) . Therefore, |A0(z)| ≤ ∣∣f (k)(z) f(z) ∣∣+ |Ak−1(z)| ∣∣f (k−1)(z) f(z) ∣∣+ · · ·+ |A1(z)| ∣∣f ′(z) f(z) ∣∣. (5.1) By Lemma 4.1, there exist a constant c > 0 and a set E1 ⊂ [0,+∞) having finite linear measure such that |z| = r /∈ E1 for all z = reiθ, we have |f (j)(z) f(z) | ≤ c[rT (2r, f)]2k, j = 1, . . . , k. (5.2) By (5.1), (5.2), and the hypotheses of Theorem 3.1, we have exp{a exp(α−1(µβ(log γ(|z|))))} ≤ |A0(z)| ≤ k exp{b exp(α−1(µβ(log γ(|z|))))}c[rT (2r, f)]2k (5.3) as z →∞ with |z| = r /∈ E1, θ1 ≤ arg z = θ ≤ θ2. Now from (5.3) we have exp{(a− b) exp(α−1(µβ(log γ(|z|))))} ≤ kc[rT (2r, f)]2k, (a− b) exp(α−1(µβ(log γ(|z|)))) ≤ 2k(log r + log T (2r, f)) + log(kc), exp(α−1(µβ(log γ(|z|)))) ≤ 2k a− b (log r + log T (2r, f)) + log(kc) a− b . By using α(a + b) ≤ α(a) + α(b) + c for all x, y ≥ R0 and fixed c ∈ (0,+∞), from the above we obtain α−1(µβ(log γ(|z|))) ≤ log[2] T (2r, f) + log[2] r +O(1), µβ(log γ(r)) ≤ α((log[2] T (2r, f) + log[2] r +O(1))), µβ(log γ(r)) ≤ α(log[2] T (2r, f)) + α(log[2] r) + c. (5.4) EJDE-2023/27 GROWTH OF SOLUTIONS OF COMPLEX DIFFERENTIAL EQUATIONS 11 By using γ(2r) ≤ 2γ(r), β(r+o(1)) = (1+o(1))β(r) as r → +∞, and α(log[2] r) β(log γ(r)) → 0 as r → +∞, then by (5.4) and Proposition 2.2, we have σ(α(log),β)[f ] ≥ µ. This completes the proof. � Proof of Theorem 3.2. Let f(z) 6≡ 0 be a solution of (1.1). By the hypotheses of Theorem 3.2, there exists a set H with dens{|z| : z ∈ H} > 0 such that for all z satisfying z ∈ H, we have |A0(z)| ≥ exp{a exp(α−1(µβ(log γ(|z|))))}, (5.5) |Aj(z)| ≤ exp{b exp(α−1(µβ(log γ(|z|))))}, j = 1, . . . , k − 1 , (5.6) as z → ∞. We set H1 = {|z| = r : z ∈ H}, since dens{|z| : z ∈ H} > 0, it follows that H1 is a set with ∫ H1 dr = +∞. Therefore from, by substituting (5.2), (5.5) and (5.6) into (5.1), it follows that for all z satisfying |z| = r ∈ H1 \ E1, we have exp{a exp(α−1(µβ(log γ(|z|))))} ≤ k exp{b exp(α−1(µβ(log γ(|z|))))}c[rT (2r, f)]2k as |z| = r → +∞. Thus exp{(a− b) exp(α−1(µβ(log γ(|z|))))} ≤ kc[rT (2r, f)]2k (5.7) as |z| = r ∈ H1 \ E1, r → +∞. Since α(a + b) ≤ α(a) + α(b) + c for all x, y ≥ R0 and fixed c ∈ (0,+∞), γ(2r) ≤ 2γ(r), β(r + o(1)) = (1 + o(1))β(r) as r → +∞, and α(log[2] r) β(log γ(r)) → 0 as r → +∞, then by (5.7) and Proposition 2.2, we obtain σ(α(log),β,γ)[f ] ≥ µ. � Proof of Theorem 3.3. By Theorem 3.2, we have σ(α(log),β,γ)[f ] ≥ σ−ε, since ε > 0 is arbitrary, we obtain σ(α(log),β,γ)[f ] ≥ σ(α,β,γ)[A0] = σ. On the other hand, by Lemma 4.2, there exists a set E2 ⊂ [1,+∞) having finite logarithmic measure such that (4.2) holds for all z satisfying |z| = r /∈ [0, 1] ∪ E2 and |f(z)| = M(r, f). Now for any given ε > 0 and for sufficiently large r, we obtain |Aj(z)| ≤ exp[2]{α−1((σ + ε)β(log γ(r)))}, j = 0, 1, . . . , k − 1. (5.8) Substituting (4.2) and (5.8) into (1.1), for all z satisfying |z| = r /∈ [0, 1] ∪ E2 and |f(z)| = M(r, f), we have(ν(r, f) |z| )k |1 + o(1)| ≤ k (ν(r, f) |z| )k−1 |1 + o(1)| exp[2]{α−1((σ + ε)β(log γ(r)))}. It follows that ν(r, f) ≤ kr|1 + o(1)| exp[2]{α−1((σ + ε)β(log γ(r)))}. (5.9) Therefore in view of (5.9), α(a + b) ≤ α(a) + α(b) + c for all x, y ≥ R0 and fixed c ∈ (0,+∞) and α(log[2] r) β(log γ(r)) → 0 as r → +∞, we obtain lim sup r→+∞ α(log[2] ν(r, f)) β(log γ(r)) ≤ σ + ε. (5.10) Since ε > 0 is arbitrary, by (5.10) and Lemma 4.3, we obtain that σ(α(log),β,γ)[f ] ≤ σ. This and the fact that σ(α(log),β,γ)[f ] ≥ σ yield σ(α(log),β,γ)[f ] = σ. The proof is complete. � 12 B. BELAÏDI. T. BISWAS EJDE-2023/27 Proof of Theorem 3.4. (i) Suppose that σ(α(log),β,γ)[F ] < σ(α,β,γ)[A0]. First, we show that (1.2) can possess at most one solution f0(z) satisfying σ(α(log),β,γ)[f0] < σ. In fact, if f∗(z) is a second solution with σ(α(log),β,γ)[f ∗] < σ, then σ(α(log),β,γ)[f0− f∗] < σ. But f0(z)−f∗(z) is a solution of the corresponding homogeneous equation (1.1) of (1.2), this contradicts Theorem 3.3. We assume that f(z) is a solution with σ(α(log),β,γ)[f ] ≥ σ and f1(z), f2(z),. . . , fk(z) is a solution base of the corresponding homogeneous equation (1.1). Then, f(z) can be expressed in the form f(z) = B1(z)f1(z) +B2(z)f2(z) + · · ·+Bk(z)fk(z), (5.11) where B1(z), B2(z), . . . , Bk(z) are determined by B′1(z)f1(z) +B′2(z)f2(z) + · · ·+B′k(z)fk(z) = 0, B′1(z)f ′1(z) +B′2(z)f ′2(z) + · · ·+B′k(z)f ′k(z) = 0, . . . B′1(z)f (k−1) 1 (z) +B′2(z)f (k−1) 2 (z) + · · ·+B′k(z)f (k−1) k (z) = F (z). (5.12) As the Wronskian W (f1, f2, . . . , fk) is a differential polynomial in f1, f2, . . . , fk with constant coefficients, it is easy to deduce that σ(α(log),β,γ)[W ] ≤ σ(α(log),β,γ)[fj ] = σ(α,β,γ)[A0] = σ. (5.13) From (5.12) we obtain B′j = F ·Gj(f1, f2, . . . , fk) ·W (f1, f2, . . . , fk)−1, j = 1, . . . , k, (5.14) where Gj(f1, f2, . . . , fk) are differential polynomials in f1, f2, . . . , fk with constant coefficients. Therefore, σ(α(log),β)[Gj ] ≤ σ(α(log),β)[fj ] = σ(α,β,γ)[A0] = σ, j = 1, . . . , k. (5.15) Since σ(α(log),β,γ)[F ] < σ(α,β,γ)[A0], by Lemma 4.4, (5.13)-(5.15), for j = 1, . . . , k, we obtain σ(α(log),β,γ)[Bj ] = σ(α(log),β,γ)[B ′ j ] ≤ max{σ(α(log),β,γ)[F ], σ(α,β,γ)[A0]} = σ(α,β,γ)[A0] = σ. (5.16) Now, from (5.11) and (5.16), we have σ(α(log),β,γ)[f ] ≤ max { σ(α(log),β,γ)[fj ], σ(α(log),β,γ)[Bj ] (j = 1, . . . , k) } = σ(α,β,γ)[A0] = σ. (5.17) This and the assumption σ(α(log),β,γ)[f ] ≥ σ yield σ(α(log),β,γ)[f ] = σ. By Lemma 4.9, we have max{σ(α(log),β,γ)[F ], σ(α(log),β,γ)[Aj ] (j = 0, 1, . . . , k − 1)} = σ(α(log),β,γ)(F ) < σ(α,β,γ)[A0] = σ(α(log),β,γ)[f ]. So, if f(z) is a solution of equation (1.2) satisfying σ(α(log),β,γ)[f ] = σ, then by Lemma 4.8, we have λ(α(log),β,γ)[f ] = λ(α(log),β,γ)[f ] = σ(α(log),β,γ)[f ] = σ. EJDE-2023/27 GROWTH OF SOLUTIONS OF COMPLEX DIFFERENTIAL EQUATIONS 13 (ii) Suppose that σ(α,β,γ)[A0] ≤ σ(α(log),β)[F ] < +∞. Then, by (5.16), for j = 1, . . . , k, we obtain σ(α(log),β,γ)[Bj ] = σ(α(log),β,γ)[B ′ j ] ≤ max{σ(α(log),β,γ)[F ], σ(α,β,γ)[A0]} = σ(α(log),β,γ)[F ]. (5.18) Now from (5.11) and (5.18), we obtain σ(α(log),β,γ)[f ] ≤ max{σ(α(log),β,γ)[fj ], σ(α(log),β,γ)[Bj ] (j = 1, . . . , k)} ≤ σ(α(log),β,γ)[F ] . (5.19) From (1.2), a simple consideration of (α(log), β, γ)-order implies that σ(α(log),β,γ)[f ] ≥ σ(α(log),β,γ)[F ]. By the above inequality and (5.19), we have σ(α(log),β,γ)[f ] = σ(α(log),β,γ)[F ] which completes the proof. � Acknowledgements. 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Colling- wood, Chelsea Publishing Company, New York, 1949. [33] H. Y. Xu, J. Tu; Oscillation of meromorphic solutions to linear differential equations with coefficients of [p, q]-order. Electron. J. Differential Equations, 2014 (2014), No. 73, 14 pp. [34] C. C. Yang, H. X. Yi; Uniqueness theory of meromorphic functions. Mathematics and its Applications, 557. Kluwer Academic Publishers Group, Dordrecht, 2003. Benharrat Beläıdi Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem (UMAB), B. P. 227 Mostaganem, Algeria Email address: benharrat.belaidi@univ-mosta.dz Tanmay Biswas Rajbari, Rabindrapally, R. N. Tagore Road, P.O.-Krishnagar, P. S. Kotwali, Dist-Nadia, PIN-741101, West Bengal, India Email address: tanmaybiswas math@rediffmail.com 1. Introduction 2. Definitions and notation 3. Main Results 4. Some Lemmas 5. Proof of main results Acknowledgements References