110 This work is licensed under a Creative Commons Attribution 4.0 International License. On the Growth of Solutions of Nonhomogeneous Higher order Complex Linear Differential Equations Ayad W. Ali Mustansiriyah University/College of Science/Department of Mathematics/Baghdad/Iraq ayad.w.a@uomustansiriyah.edu.iq Abstract The nonhomogeneous higher order linear complex differential equation (HOLCDE) with meromorphic (or entire) functions is considered in this paper. The results are obtained by putting some conditions on the coefficients to prove that the hyper order of any nonzero solution of this equation equals the order of one of its coefficients in case the coefficients are meromorphic functions. In this case, the conditions were put are that the lower order of one of the coefficients dominates the maximum of the convergence exponent of the zeros sequence of it, the lower order of both of the other coefficients and the nonhomogeneous part and that the solution has infinite order. Whiles in case the coefficients are entire functions, any nonzero solution with finite order has hyper order equals to the lower order of one of its coefficients is proved. In this case, the condition that the lower order of one of the coefficients is greater than the maximum of the lower order of the other coefficients and the lower order of the nonhomogeneous part is assumed. Keywords: Complex linear differential equations; meromorphic functions; entire functions; order of growth; lower order of growth. 1. Introduction The theory of meromorphic functions due to Nevanlinna is a good tool in the complex differential equations field. At the forefront of the application of Nevanlinna's theory of meromorphic functions to the complex differential equations was Yoseida in 1932, and since then complex differential equations have become an active field of study by researchers. We refer the reader to, for instance [1], for more information about the differential equations theory in the complex plane. One of the aims of studying this type of equation is the order of growth of its solutions of it. In our work, we shall study the hyper order of the solutions of Eq. (1) below considering the lower orders of coefficients and exponent of convergence of the zeros Ibn Al-Haitham Journal for Pure and Applied Sciences http://jih.uobaghdad.edu.iq/index.php/j/index: Journal homepage Doi: 10.30526\35.3.2777 Article history: Received 17 Janeury 2022, Accepted 14 March 2022, Published in July 2022. https://creativecommons.org/licenses/by/4.0/ mailto:ayad.w.a@uomustansiriyah.edu.iq file:///F:/العدد%20الثاني%202022/:%20http:/jih.uobaghdad.edu.iq/index.php/j/index IHJPAS. 53 (3)2022 111 sequence of one of them. In this paper, we assume that the reader is familiar with the basic concepts and the results regarding the Nevanlinna value distribution theory of meromorphic functions such as 𝑀(𝑟, 𝑓), 𝑇(𝑟, 𝑓) and 𝑁(𝑟, 𝑓) etc., see [2]. A nonhomogeneous HOLCDE is given by 𝑓 (𝑛) + 𝐴𝑛−1(𝑧)𝑓 (𝑛−1) + ⋯ + 𝐴2(𝑧)𝑓 ′′ + 𝐴1(𝑧)𝑓 ′ + 𝐴0(𝑧)𝑓 = 𝐹(𝑧) (1) where 𝑓 = 𝑓 (0) is unknown and 𝐴𝑗 (𝑧), 0 ≤ 𝑗 ≤ 𝑛 − 1, 𝐹(𝑧) are given functions. Many authors studied Eq. (1) and obtained some results. Here we shall mention some of them. The following two results study the hyper order of 𝑓 when the order of one coefficient dominates the order of 𝐹 and exponent of convergence of the zeros sequence of that coefficient. Theorem 1 [3] Let 𝐸 ⊆ ℂ satisfy 𝑚𝑙 ({|𝑧|: 𝑧 ∈ 𝐸}) = ∞ and 𝐴𝑗 (𝑧), 𝑗 = 0,1, … , 𝑛 − 1, 𝐹(𝑧) be meromorphic functions. Suppose that it is 𝑠, 0 ≤ 𝑠 ≤ 𝑛– 1, satisfies max 0≤𝑗≤𝑛−1 𝑗≠𝑠 {𝜌(𝐴𝑗 ), 𝜆 ( 1 𝐴𝑠 ) , 𝜌(𝐹)} ≤ 𝜌(𝐴𝑠) = 𝜌 < ∞ (2) and for some constants 0 ≤ 𝛽 < 𝛼, we have |𝐴𝑗 (𝑧)| ≤ exp(𝛽|𝑧| 𝜌− ) , 𝑗 ≠ 𝑠 (3) |𝐴𝑠(𝑧)| ≥ exp(𝛼|𝑧| 𝜌− ) (4) hold for > 0 and as |𝑧| → ∞, 𝑧 ∈ 𝐸. Then any meromorphic solution 𝑓 ≠ 0 of Eq. (1) with poles of uniformly bounded multiplicities satisfies 𝜌2(𝑓) ≥ 𝜌(𝐴𝑠). Theorem 2 [3] Let 𝐴𝑗 (𝑧), 𝑗 = 0,1, … , 𝑛 − 1, 𝐹(𝑧) are satisfied (2). Then every meromorphic solution 𝑓 ≠ 0 with 𝜌(𝑓) = ∞ has poles are of uniformly bounded multiplicities of Eq. (1) satisfies 𝜌2(𝑓) ≤ 𝜌(𝐴𝑠). The following result studies the property of 𝑓 when the order of one coefficient dominates the maximum orders of 𝐹 and the other coefficients. Theorem 3 [4] Let 𝐴𝑗 (𝑧), 𝑗 = 0,1, … , 𝑛 − 1, 𝐹(𝑧) entire functions. Suppose there is 0 ≤ 𝑠 ≤ 𝑛– 1, such that max { max 0≤𝑗≤𝑛−1 𝑗≠𝑠 𝜌(𝐴𝑗 ), 𝜌(𝐹)} < 𝜌(𝐴𝑠) ≤ 1 2 Then every solution of Eq. (1) is either a polynomial or infinite order entire function. Theorem 4 [5] Let 𝐴𝑗 (𝑧), 𝑗 = 0,1, … , 𝑛 − 1, 𝐹(𝑧) be defined as in Theorem 3 such that max{𝜌(𝐴𝑗 ), 𝜌(𝐹)} < 𝜌(𝐴𝑠) < 1 2 Then every transcendental solution of Eq. (1) satisfies 𝜌2(𝑓) = 𝜌(𝐴𝑠). Furthermore, if 𝐹 ≠ 0 then 𝜆2(𝑓) = 𝜌(𝐴𝑠). 2. Material We recall the following definitions. Definition 1 [6] Let 𝐸 ⊆ [0, ∞). Then the linear measure of 𝐸 is 𝑚(𝐸) = ∫ 𝑑𝑡 𝐸 IHJPAS. 53 (3)2022 112 Definition 2 [7, 8] Let 𝐸 ⊆ [1, ∞), then the logarithmic measure of 𝐸 is 𝑚𝑙 (𝐸) = ∫ 𝑑𝑡 𝑡 𝐸 Definition 3 [9, 10] Let 𝑓 be a meromorphic function. We define the order of growth 𝜌(𝑓) (respectively) and lower order of growth 𝜇(𝑓), by 𝜌(𝑓) = lim 𝑟→∞ 𝑠𝑢𝑝 𝑙𝑜𝑔+𝑇(𝑟, 𝑓) 𝑙𝑜𝑔 𝑟 and 𝜇(𝑓) = lim 𝑟→∞ 𝑖𝑛𝑓 𝑙𝑜𝑔+𝑇(𝑟, 𝑓) 𝑙𝑜𝑔 𝑟 If 𝑓 is entire function, then 𝑇(𝑟, 𝑓) is replaced with 𝑙𝑜𝑔+𝑀(𝑟, 𝑓), where 𝑀(𝑟, 𝑓) = max |𝑧|=𝑟 |𝑓(𝑧)| Definition 4 [11, 12] We define the hyper-order 𝜌2(𝑓) of meromorphic function 𝑓 by 𝜌2(𝑓) = lim 𝑟→∞ 𝑠𝑢𝑝 𝑙𝑜𝑔+𝑙𝑜𝑔+𝑇(𝑟, 𝑓) 𝑙𝑜𝑔 𝑟 If 𝑓 is entire function, then 𝜌2(𝑓) = lim 𝑟→∞ 𝑠𝑢𝑝 𝑙𝑜𝑔+𝑙𝑜𝑔+𝑙𝑜𝑔+𝑀(𝑟, 𝑓) 𝑙𝑜𝑔 𝑟 Definition 5 [3] Let 𝑓 be meromorphic function. By 𝜆(𝑓) = lim 𝑟→∞ sup 𝑙𝑜𝑔+𝑁(𝑟, 1 𝑓 ) 𝑙𝑜𝑔𝑟 we meant the convergence exponent of the zeros sequence of 𝑓, while 𝜆 ( 1 𝑓 ) = lim 𝑟→∞ sup 𝑙𝑜𝑔+𝑁(𝑟, 𝑓) 𝑙𝑜𝑔𝑟 is called the convergence exponent of the poles sequence of 𝑓. Definition 6 [3] The lower and upper logarithmic densities of 𝐸 ⊆ [1, ∞) are as follows 𝑙𝑜𝑔𝑑𝑒𝑛𝑠𝐸 = 𝑙𝑖𝑚 𝑟→∞ 𝑖𝑛𝑓 𝑚𝑙 (𝐸⋂[1, 𝑟]) 𝑙𝑜𝑔𝑟 and 𝑙𝑜𝑔𝑑𝑒𝑛𝑠̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅𝐸 = 𝑙𝑖𝑚 𝑟→∞ 𝑠𝑢𝑝 𝑚𝑙 (𝐸⋂[1, 𝑟]) 𝑙𝑜𝑔𝑟 respectively. We say that 𝐸 has logarithmic density if 𝑙𝑜𝑔𝑑𝑒𝑛𝑠𝐸 = 𝑙𝑜𝑔𝑑𝑒𝑛𝑠̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅𝐸 3. Methods of Work This section has introduced some results that will help us to prove our results. Lemma 1 [13] Let (𝑓, 𝛤) denote a pair that consists of transcendental meromorphic function 𝑓(𝑧) and a finite set Γ = {(𝑘1, 𝑗1), (𝑘2, 𝑗2), … , (𝑘𝑞 , 𝑗𝑞 )} of distinct pairs in ℤ+ 𝑘𝑖 > 𝑗𝑖 ≥ 0 for 𝑖 = 1,2, . . . , 𝑞, let 𝛼 > 0, > 0. Then the following hold: IHJPAS. 53 (3)2022 113 i) There is 𝐸1 ⊆ [0, 2𝜋) with 𝑚(𝐸1) = 0, and there is 𝑐 > 0 that depends only on 𝛼 and Γ such that if 𝜑0 ∈ [0, 2𝜋)\E1, there is 𝑅0 = 𝑅0(𝜑0) > 1 such that for 𝑎𝑟𝑔 𝑧 = 𝜑0 and |𝑧| = 𝑟 ≥ 𝑅0, & for (𝑘, 𝑗) ∈ Γ, we have | 𝑓 (𝑘)(𝑧) 𝑓 (𝑗)(𝑧) | ≤ 𝑐 ( 𝑇(𝛼𝑟, 𝑓) 𝑟 𝑙𝑜𝑔𝛼𝑟 log 𝑇(𝛼𝑟, 𝑓)) 𝑘−𝑗 (5) In particular, if 𝑓(𝑧) with 𝜌(𝑓 ) < ∞, then (5) is replaced by: | 𝑓 (𝑘)(𝑧) 𝑓 (𝑗)(𝑧) | ≤ 𝑐|𝑧|(𝑘−𝑗)(𝜌(𝑓)−1+ ) (6) ii) There is 𝐸2 ⊆ [1, ∞) with 𝑚𝑙 (𝐸2) < ∞, and 𝑐 > 0 that depends only on 𝛼 and Γ such that for |𝑧| = 𝑟 ∉ 𝐸2 ∪ [0, 1] and for (𝑘, 𝑗) ∈ Γ, (5) holds. In particular, if 𝑓(𝑧) is with 𝜌(𝑓) < ∞, then (6) holds. iii) There is 𝐸3 ⊂ [0, ∞) with 𝑚(𝐸3) < ∞, and 𝑐 > 0 that depends only on 𝛼 and Γ such that for |𝑧| = 𝑟 ∉ 𝐸3 and (𝑘, 𝑗) ∈ Γ, we have | 𝑓 (𝑘)(𝑧) 𝑓 (𝑗)(𝑧) | ≤ 𝑐 (𝑇(𝛼𝑟, 𝑓)𝑟 𝑙𝑜𝑔 𝑇(𝛼𝑟, 𝑓))𝑘−𝑗 (7) In particular, if 𝑓(𝑧) with 𝜌(𝑓) < ∞, then (7) is replaced by | 𝑓 (𝑘)(𝑧) 𝑓 (𝑗)(𝑧) | ≤ 𝑐|𝑧|(𝑘−𝑗)(𝜌(𝑓)+ ) (8) Lemma 2 [14] Let 𝑓 be meromorphic with order 𝜌 = 𝜌(𝑓 ) < ∞. Then, for > 0, there is 𝐸 ⊆ (1, ∞) with 𝑚𝑙 (𝐸) < ∞, 𝑚(𝐸) < ∞, s. t. |𝑓(𝑧)| ≤ exp(𝑟𝜌+ ) for |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸. Lemma 3 [3] Let 𝑓(𝑧) = 𝑔(𝑧) 𝑑(𝑧) be meromorphic, where 𝑔(𝑧) and 𝑑(𝑧) are entire that satisfy 𝜇(𝑔) = 𝜇(𝑓) = 𝜇 ≤ 𝜌(𝑔) = 𝜌(𝑓) ≤ ∞ 𝜆(𝑑) = 𝜌(𝑑) = 𝜆 ( 1 𝑓 ) = 𝛽 < 𝜇 Then there is 𝐸 ⊆ (1, ∞) with 𝑚𝑙 (𝐸) < ∞, s. t. 𝑓 (𝑛)(𝑧) 𝑓(𝑧) = ( 𝑣(𝑟, 𝑔) 𝑧 ) 𝑛 (1 + 𝑜(1)), 𝑛 ≥ 1 holds for 𝑧 with |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸, |𝑔(𝑧)| = 𝑀(𝑟, 𝑔), 𝑀(𝑟, 𝑔) = max |𝑧|=𝑟 |𝑔(𝑧)|, where 𝑣(𝑟, 𝑔) is the central index of 𝑔. Lemma 4 [11] Let 𝑔(𝑟) and ℎ(𝑟) be monotone nondecreasing functions on [0, ∞), such that 𝑔(𝑟) ≤ ℎ(𝑟) for 𝑟 ∉ 𝑆 where 𝑆 is a set with 𝑚𝑙 (𝑆) < ∞, let 𝛼 > 1. Then there is 𝑟0 > 1, such that 𝑔(𝑟) ≤ ℎ(𝛼𝑟) for 𝑟 > 𝑟0. Lemma 5 [15] Assume that 𝑔(𝑧) is an entire function with 0 ≤ 𝜇(𝑔) < 1. Then, for 𝛼 ∈ (𝜇(𝑔), 1), there is 𝐸 ⊆ [0, ∞) such that 𝑙𝑜𝑔𝑑𝑒𝑛𝑠̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅(𝐸) ≥ 1 − 𝜇(𝑔) 𝛼 , where 𝐸 = {𝑟 ∈ [0, ∞) ∶ 𝑚(𝑟) > 𝑀(𝑟) 𝑐𝑜𝑠 𝜋𝛼} where 𝑚(𝑟) = 𝑖𝑛𝑓|𝑧|=𝑟 𝑙𝑜𝑔|𝑔(𝑧)| 𝑀(𝑟) = 𝑠𝑢𝑝|𝑧|=𝑟 𝑙𝑜𝑔|𝑔(𝑧)| Lemma 6 [16] Let 𝑔(𝑧) satisfy the hypothesis of Lemma 5 with 𝜌(𝑔) = 𝜌 instead of 𝜇(𝑔). If 𝜌 < 𝛼 < 1 , then IHJPAS. 53 (3)2022 114 𝑙𝑜𝑔𝑑𝑒𝑛𝑠(𝐸) ≥ 1 − 𝜌 𝛼 Lemma 7 [15] Let 𝑓(𝑧) be an entire function with 𝜇(𝑓 ) = 𝜇 < 1 2 and 𝜇 < 𝜌 = 𝜌(𝑓 ). If 𝜇 ≤ 𝛿 < 𝑚𝑖𝑛(𝜌, 1 2 ) and 𝛿 < 𝛼 < 1 2 , then 𝑙𝑜𝑔𝑑𝑒𝑛𝑠{𝑟 ∈ [0, ∞): 𝑚(𝑟) > (𝑐𝑜𝑠𝜋𝛼)𝑀(𝑟) > 𝑟𝛿 } > 𝐶(𝜌, 𝛿, 𝛼) where 𝐶(𝜌, 𝛿, 𝛼) > 0 and 𝑚(𝑟) and 𝑀(𝑟) are given as in Lemma 5. Lemma 8 [5] Let 𝑓(𝑧) be transcendental entire function. Then there is 𝐸 ⊆ (1, ∞) with 𝑚𝑙 (𝐸) < ∞, s. t. for a point 𝑧 with |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸 & |𝑓(𝑧)| = 𝑀(𝑟, 𝑓) , we have | 𝑓(𝑧) 𝑓 (𝑠)(𝑧) | ≤ 2𝑟 𝑠 (𝑠 ∈ ℕ) Lemma 9 [3] Let 𝑓 be an infinite order entire function, with 𝜌2(𝑓 ) < ∞, and let 𝜈(𝑟, 𝑔) be the central index of 𝑓. Then lim 𝑟→∞ sup 𝑙𝑜𝑔+𝑙𝑜𝑔+𝜈(𝑟, 𝑔) 𝑙𝑜𝑔𝑟 = 𝜌2(𝑓 ) Lemma 10 [3] Let 𝑓(𝑧) = 𝑔(𝑧) 𝑑(𝑧) be given as in Lemma 3. If 0 ≤ 𝜌(𝑑) < 𝜇(𝑓), then 𝜇(𝑔) = 𝜇(𝑓), 𝜌(𝑔) = 𝜌(𝑓). Moreover, if 𝜌(𝑓) = ∞, then 𝜌2(𝑓) = 𝜌2(𝑔). 4. Results and Discussion The coefficients in the following two results are meromorphic and one of them has lower order that dominates the others on some subset of ℂ with a finite order solution. Theorem 5 Let 𝐸 ⊆ ℂ and define 𝑆 = {|𝑧|: 𝑧 ∈ 𝐸} and let 𝐸 satisfies 𝑚𝑙 (𝑆) = ∞. Suppose that 𝐴𝑗 (𝑧), 𝑗 = 0,1, … , 𝑛 − 1 and 𝐹(𝑧) are meromorphic functions. Suppose that it is 𝑠, 0 ≤ 𝑠 ≤ 𝑛– 1, such that 𝑝 = max 0≤𝑗≤𝑛−1 𝑗≠𝑠 {𝜇(𝐴𝑗 ), 𝜆 ( 1 𝐴𝑠 ) , 𝜇(𝐹)} ≤ 𝜇(𝐴𝑠) = 𝜇 < ∞, (9) holds and for a constant 0 ≤ 𝛽 < 𝛼, the relations (3) and (4) hold, for > 0 and as |𝑧| → ∞, 𝑧 ∈ 𝐸. Then any solution 𝑓 ≠ 0 of Eq. (1) with 𝜌(𝑓) < ∞ satisfies 𝜌2(𝑓 ) ≥ 𝜌(𝐴𝑠 ). Proof From Eq. (1), we have 𝐴𝑠 = 𝐹 𝑓 𝑓 𝑓 (𝑠) − { 𝑓 (𝑛) 𝑓 (𝑠) + 𝐴𝑛−1 𝑓 (𝑛−1) 𝑓 (𝑠) + ⋯ + 𝐴𝑠+1 𝑓 (𝑠+1) 𝑓 (𝑠) + 𝑓 𝑓 (𝑠) (𝐴𝑠−1 𝑓 (𝑠−1) 𝑓 + ⋯ + 𝐴1 𝑓 ′ 𝑓 + 𝐴0)} (10) Using Lemma 1 (ii), with 𝛼 = 2, there is 𝐸1 ⊆ [1, ∞) with 𝑚𝑙 (𝐸1) < ∞ and 𝐵 > 0, such that | 𝑓 (𝑗)(𝑧) 𝑓 (𝑠)(𝑧) | ≤ 𝐵𝑟(𝑇(2𝑟, 𝑓)) 𝑗−𝑠 , 𝑗 = 𝑠 + 1, 𝑠 + 2, … , 𝑛 (11) and | 𝑓 (𝑗)(𝑧) 𝑓(𝑧) | ≤ 𝐵𝑟(𝑇(2𝑟, 𝑓)) 𝑗 , 𝑗 = 1, 2, … , 𝑠 − 1, (12) hold for |𝑧| = 𝑟 ∉ ([0, 𝑅1] ∪ 𝐸1), 𝑅1 > 1. IHJPAS. 53 (3)2022 115 Put 𝑓(𝑧) = 𝑔(𝑧) 𝑑(𝑧) where 𝑔(𝑧) is entire, 𝑑(𝑧) is a product of poles sequence of 𝑓. Let 𝜂 be such that 𝑝 < 𝜂 < 𝜇(𝐴𝑠). Using Lemma 2, there is 𝐸2 ⊆ (1, ∞) with 𝑚𝑙 (𝐸2) < ∞, such that |𝐹(𝑧)𝑑(𝑧)| ≤ exp(𝑟𝜂) (13) holds for |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸2. Thus, there is 𝑅2 (> 𝑅1), such that, for all 𝑧 satisfying |𝑧| = 𝑟 > 𝑅2, 𝑣 (𝑟, 𝑔) > 1, |1 + 𝑜(1)| > 1 2 and |𝑔(𝑧)| = 𝑀(𝑟, 𝑔) > 1 the following holds: | 𝑓(𝑧) 𝑓 (𝑠)(𝑧) | ≤ 2𝑟 𝑠 (14) Set 𝐻 = {|𝑧| ∶ 𝑧 ∈ 𝐸}\([0, 𝑅2] ∪ 𝐸1 ∪ 𝐸2) Then 𝑚𝑙 (𝐻) = ∞. It follows from (13) that | 𝐹(𝑧) 𝑓(𝑧) | = | 𝐹(𝑧) 𝑔(𝑧) 𝑑(𝑧)| = | 𝐹(𝑧) 𝑀(𝑟, 𝑔) | |𝑑(𝑧)| ≤ exp(𝑟𝜂) (15) for |𝑧| = 𝑟 ∈ 𝐻, 𝑟 > 𝑅2, and |𝑔(𝑧)| = 𝑀(𝑟, 𝑔). It follows from (10), (11), (12), (14), (15) and (3), (4), that exp (𝛼𝑟𝜌− ) ≤ 2(𝑛 + 1)𝐵𝑟 𝑠+1(𝑇(2𝑟, 𝑓)) 𝑛+1 exp (𝛽𝑟𝜌− )exp (𝑟𝜂 ) for 𝑧 with |𝑧| = 𝑟 ∈ 𝐻, ∈ (0, 𝜇(𝐴𝑠)–𝜂 2 ) and |𝑔(𝑧)| = 𝑀(𝑟, 𝑔). Since 𝜂 < 𝜇(𝐴𝑠), we obtain 𝜌2(𝑓 ) ≥ 𝜌(𝐴𝑠) In the following result, the same conditions described in the previous Theorem are given with a solution that has infinite order and obtains the opposite result. Theorem 6 Let 𝐴𝑗 (𝑧), 𝑗 = 0,1, … , 𝑛 − 1, 𝐹(𝑧) be defined as in Theorem 5 and satisfy inequality (9). Then every solution 𝑓 ≠ 0 with 𝜌(𝑓) = ∞ of Eq. (1) satisfies 𝜌2(𝑓 ) ≤ 𝜌(𝐴𝑠). Proof From Eq. (1), we have − 𝑓 (𝑛) 𝑓 = 𝐴𝑛−1 𝑓 (𝑛−1) 𝑓 + ⋯ + 𝐴𝑠 𝑓 (𝑠) 𝑓 + ⋯ + 𝐴1 𝑓 ′ 𝑓 + 𝐴0 − 𝐹 𝑓 (16) Using Lemma 2, for > 0, there is 𝐸3 ⊆ (1, ∞) with 𝑚𝑙 (𝐸3) < ∞, such that |𝐴𝑗 (𝑧)| ≤ exp(𝑟 𝜌(𝐴𝑠)+ ) , 𝑗 = 0,1, … , 𝑛 − 1 (17) and |𝐹(𝑧)| ≤ exp(𝑟𝜌(𝐴𝑠)+ ) , (18) holds for |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸3. Put 𝑓(𝑧) = 𝑔(𝑧) 𝑑(𝑧) as in the proof of Theorem 6. Thus by Lemma 3, there is 𝐸4 ⊆ (1, ∞) with 𝑚𝑙 (𝐸4) < ∞, such that 𝑓 (𝑗)(𝑧) 𝑓(𝑧) = ( 𝑣(𝑟, 𝑔) 𝑧 ) 𝑗 (1 + 𝑜(1)), 𝑗 = 1, … , 𝑛 (19) holds for |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸4, |𝑔(𝑧)| = 𝑀(𝑟, 𝑔) > 1. Hence from (16), (17) and (19) there is 𝑅 > 1, such that IHJPAS. 53 (3)2022 116 |( 𝑣(𝑟, 𝑔) 𝑧 ) 𝑛 (1 + 𝑜(1))| {|( 𝑣(𝑟, 𝑔) 𝑧 ) 𝑛−1 (1 + 𝑜(1))| + ⋯ + |( 𝑣(𝑟, 𝑔) 𝑧 ) 𝑠 (1 + 𝑜(1))| + ⋯ + |( 𝑣(𝑟, 𝑔) 𝑧 ) (1 + 𝑜(1))| + 1} exp(𝑟𝜌(𝐴𝑠)+ ) + | 𝐹(𝑧) 𝑓(𝑧) | (20) holds for |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸3 ∪ 𝐸4. Since 𝜌(𝑑) < 𝜌(𝐴𝑠), we have for 𝑟 > 𝑅, |𝑑(𝑧)| ≤ exp(𝑟𝜌(𝐴𝑠)+ ) (21) Then from (18) and (21) we have | 𝐹 𝑓 | = | 𝐹 𝑔 𝑑| = | 𝐹 𝑀(𝑟, 𝑔) 𝑑| ≤ exp(2𝑟𝜌(𝐴𝑠)+ ) Combining (20) and the above inequality, we get (𝑣(𝑟, 𝑔)) 𝑛 |1 + 𝑜(1)| ≤ (𝑛 + 1) exp(2𝑟𝜌(𝐴𝑠)+ ) |𝑧|𝑛(𝑣(𝑟, 𝑔)) 𝑛−1 |1 + 𝑜(1)| Hence 𝑣𝑔(𝑟) ≤ (𝑛 + 1) exp(2𝑟 𝜌(𝐴𝑠)+ )|𝑧|𝑛 Combining Lemma 9 and Lemma 4 and the above inequality, we get 𝜌2(𝑔 ) ≤ 𝜌(𝐴𝑠 ) Combining Lemma 10 and the above inequality we get 𝜌2(𝑓 ) ≤ 𝜌(𝐴𝑠) This completes the proof. Theorem 7 Under the assumptions of the previous two Theorems we have 𝜌2(𝑓 ) = 𝜌(𝐴𝑠). In what follows we shall consider Eq. (1) when 𝐴𝑗 (𝑧) and 𝐹(𝑧) are entire. Theorem 8 Assume that 𝐴𝑗 (𝑧), 𝑗 = 0,1, … , 𝑛 − 1 and 𝐹(𝑧) are entire functions and it is 𝑠, 0 ≤ 𝑠 ≤ 𝑛– 1, such that 𝑞 = max 0≤𝑗≤𝑛−1 𝑗≠𝑠 {𝜇(𝐴𝑗 ), 𝜇(𝐹)} < 𝜇(𝐴𝑠) < 1 2 Let 𝑓 ≠ 0 be any solution with 𝜌(𝑓) < ∞. Then 𝜌2(𝑓 ) = 𝜇(𝐴𝑠). Proof From Eq. (1) we have Eq. (10). Using Lemma 1 (ii), there is 𝐸1 ⊆ [1, ∞) with 𝑚𝑙 (𝐸1) < ∞ such that | 𝑓 (𝑗)(𝑧) 𝑓 (𝑠)(𝑧) | ≤ 𝑀𝑟𝑐 (𝑇(2𝑟, 𝑓)) 2𝑛 , 𝑗 = 𝑠 + 1, … , 𝑛 (22) and | 𝑓 (𝑗)(𝑧) 𝑓(𝑧) | ≤ 𝑀𝑟𝑐 (𝑇(2𝑟, 𝑓)) 2𝑛 , 1 ≤ 𝑗 ≤ 𝑠 − 1 (23) holds for |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸1. Choose 𝛼, 𝛽 such that, 𝑞 < 𝛼 < 𝛽 < 𝜇(𝐴𝑠). Then we have |𝐴𝑗 (𝑧)| ≤ exp(𝑟 𝛼 ) , 0 ≤ 𝑗 ≤ 𝑛 − 1, 𝑗 ≠ 𝑠 (24) |𝐹(𝑧)| ≤ exp(𝑟𝛼 ) (25) for 𝑟 → ∞. By Lemma 6 or Lemma 7 there is 𝐻 ⊆ (1, ∞) with 𝑚𝑙 (𝐻) = ∞, such that |𝐴𝑠(𝑧)| > exp(𝑟 𝛽 ) (26) holds for |𝑧 | = 𝑟 ∈ 𝐻. IHJPAS. 53 (3)2022 117 Because 𝑀(𝑟, 𝑓 ) > 1, by (25) we have |𝐹(𝑧)| 𝑀(𝑟, 𝑓 ) ≤ exp(𝑟𝛼 ) (27) for 𝑟 → ∞. By Lemma 8, there is 𝐸2 ⊆ (1, ∞) with 𝑚𝑙 (𝐸2) < ∞, such that | 𝑓(𝑧) 𝑓 (𝑠)(𝑧) | ≤ 2𝑟 𝑠 (28) for a point 𝑧 with |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸2 and |𝑓(𝑧)| = 𝑀(𝑟, 𝑓). From (22)-(24), (26-28) and Eq. (10) we have exp(𝑟𝛽 ) < 𝑀𝑟𝑐 (𝑇(2𝑟, 𝑓)) 2𝑛 (𝑛 exp(𝑟𝛼 ))2𝑟 𝑠 and exp (𝑟𝛽 (1 + 𝑜(1)) < (𝑇(2𝑟, 𝑓)) 2𝑛 (29) for a point z with |𝑧| = 𝑟 ∈ 𝐻 ∖ ([0,1] ∪ 𝐸1 ∪ 𝐸2). Thus from (29) and since 𝛽 is arbitrary, we deduce 𝜇(𝐴𝑠) ≤ 𝜌2(𝑓 ) (30) Now, we prove that 𝜌2(𝑓 ) ≤ 𝜇(𝐴𝑠). By Lemma 3, there is 𝐸3 ⊆ (1, ∞) with 𝑚𝑙 (𝐸3) < ∞, such that 𝑓 (𝑗)(𝑧) 𝑓(𝑧) = ( 𝑣(𝑟, 𝑓) 𝑧 ) 𝑗 (1 + 𝑜(1)), 1 ≤ 𝑗 ≤ 𝑛 (31) holds for |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸3. Hence for > 0, we have |𝐴𝑗 (𝑧)| ≤ exp(𝑟 𝜇(𝐴𝑠)+ ) , 0 ≤ 𝑗 ≤ 𝑛 − 1 (32) |𝐹(𝑧)| ≤ exp(𝑟𝜇(𝐴𝑠)+ ) (33) for 𝑟 → ∞. Because (33) and |𝑓(𝑧)| = 𝑀(𝑟, 𝑓 ) > 1, we get | 𝐹(𝑧) 𝑓(𝑧) | ≤ exp(𝑟𝜇(𝐴𝑠)+ ) (34) for 𝑟 → ∞. Take 𝑧 with |𝑧| = 𝑟 ∉ [0,1] ∪ 𝐸3. From Eq. (1) we have | 𝑓 (𝑛) 𝑓 | ≤ |𝐴𝑛−1(𝑧)| | 𝑓 (𝑛−1) 𝑓 | + ⋯ + |𝐴2(𝑧)| | 𝑓 ′′ 𝑓 | + |𝐴1(𝑧)| | 𝑓 ′ 𝑓 | + 𝐴0(𝑧) + | 𝐹 𝑓 | (35) Substituting (31), (32) and (34) into (35) yields | 𝑣(𝑟, 𝑓) 𝑧 | 𝑛 |1 + 𝑜(1)| ≤ (𝑛 + 1) | 𝑣(𝑟, 𝑓) 𝑧 | 𝑛−1 |1 + 𝑜(1)| exp(𝑟𝜇(𝐴𝑠)+ ), This gives 𝜌2(𝑓) = lim 𝑟→∞ sup 𝑙𝑜𝑔+𝑙𝑜𝑔+𝑣(𝑟, 𝑓) 𝑙𝑜𝑔𝑟 ≤ 𝜇(𝐴𝑠) + (36) Because is arbitrary, by Lemma 9 and Eq. (36), we get 𝜌2(𝑓 ) ≤ 𝜇(𝐴𝑠) Combining this and (30) yields 𝜌2(𝑓 ) = 𝜇(𝐴𝑠). This completes the proof. 5. 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