Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 92, pp. 1–20. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu LOWER ORDER FOR MEROMORPHIC SOLUTIONS TO LINEAR DELAY-DIFFERENTIAL EQUATIONS RACHID BELLAAMA, BENHARRAT BELAÏDI Abstract. In this article, we study the order of growth for solutions of the non-homogeneous linear delay-differential equation n∑ i=0 m∑ j=0 Aijf (j)(z + ci) = F (z), where Aij(z) (i = 0, . . . , n; j = 0, . . . ,m), F (z) are entire or meromorphic functions and ci (0, 1, . . . , n) are non-zero distinct complex numbers. Under the condition that there exists one coefficient having the maximal lower order, or having the maximal lower type, strictly greater than the order, or the type, of the other coefficients, we obtain estimates of the lower bound of the order of meromorphic solutions of the above equation. 1. Introduction and statement of main results Throughout this article, a meromorphic function means a function that is mero- morphic in the whole complex plane C. We use the basic notations such as m(r, f), N(r, f), T (r, f) and fundamental results of Nevanlinna’s value distribution theory [6, 10, 12, 23]. Further, we denote respectively by ρ(f), µ(f), τ(f), τ(f), the order, the lower order, the type, and the lower type of a meromorphic function f . Also when f is an entire function, we use τM (f), τM (f) respectively for the type and lower type of f . Recently, a lot of results have been obtained for complex difference and complex difference equations [3, 5, 8, 9, 15]. The back-ground for these studies lies in the recent difference counterparts of Nevanlinna theory. The key result here is the difference analogue of the lemma on the logarithmic derivative obtained by Halburd- Korhonen [8, 9] and Chiang-Feng [5], independently. Properties of meromorphic solutions of complex linear difference equations of type Anf(z + cn) +An−1f(z + cn−1) + · · ·+A1f(z + c1) +A0f(z) = An+1, (1.1) where Aj(z) (j = 0, . . . , n+1) are entire or meromorphic functions and ci (1, . . . , n) are non-zero distinct complex numbers, have been made where one of the coeffi- cients is dominating in comparison with the other coefficients, see e.g. [5, 13]. The following two theorems have been obtained in [1]. 2010 Mathematics Subject Classification. 30D35, 39B32, 39A10. Key words and phrases. Linear difference equation; linear delay-differential equation; meromorphic solution; order; type; lower order, lower type. ©2021. This work is licensed under a CC BY 4.0 license. Submitted June 25, 2021. Published November 18, 2021. 1 2 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 Theorem 1.1 ([1]). Let Aj(z) (j = 0, . . . , n + 1) be entire functions, and let k, l ∈ {0, 1, . . . , n+ 1}. If the following three assumptions hold simultaneously: (1) max{µ(Ak), ρ(Aj), j 6= k, l} = ρ ≤ µ(Al) <∞, µ(Al) > 0; (2) τM (Al) > τM (Ak), when µ(Al) = µ(Ak); (3) max{τM (Aj) : ρ(Aj) = µ(Al), j 6= k, l} = τ1 < τM (Al), when µ(Al) = max{ρ(Aj), j 6= k, l}. Then every meromorphic solution f of (1.1) satisfies ρ(f) ≥ µ(Al) if An+1 6≡ 0. Furthermore, if An+1(z) ≡ 0, then every meromorphic solution f 6≡ 0 of (1.1) satisfies ρ(f) ≥ µ(Al) + 1. Theorem 1.2 ([1]). Let Aj(z) (j = 0, . . . , n + 1) be meromorphic functions, and let k, l ∈ {0, 1, . . . , n+ 1}. If the following five assumptions hold simultaneously. (1) max{µ(Ak), ρ(Aj), j 6= k, l} = ρ ≤ µ(Al) <∞; (2) τ(Al) > τ(Ak), when µ(Al) = µ(Ak); (3) τ1 = ∑ ρ(Aj)=µ(Al), j 6=l,k τ(Aj) < τ(Al) < +∞ when µ(Al) = max{ρ(Aj), j 6= l, k}; (4) τ1 + τ(Ak) < τ(Al) < +∞ when µ(Al) = µ(Ak) = max{ρ(Aj), j 6= k, l}; (5) λ( 1 Al ) < µ(Al) <∞. Then every meromorphic solution f of (1.1) satisfies ρ(f) ≥ µ(Al) if An+1 6≡ 0. Furthermore, if An+1(z) ≡ 0, then every meromorphic solution f 6≡ 0 of (1.1) satisfies ρ(f) ≥ µ(Al) + 1. Historically, the study of complex delay-differential equations can be traced back to Naftalevich’s research. By using operator theory and iteration method, Naf- talevich [18] considered the meromorphic solutions on complex delay-differential equations. Also there are few investigations on complex delay-differential equation field using Nevanlinna theory. Recently Liu, Laine and Yang [15] presented de- velopments and new results on complex delay-differential equations, an area with important and interesting applications, which also gathers increasing attention (see, [14, 17, 19, 20, 21]). Chen and Zheng [4] investigated the growth of solutions of the homogeneous linear delay-differential equation n∑ i=0 m∑ j=0 Aijf (j)(z + ci) = 0 (1.2) and have obtained the following results. Theorem 1.3 ([4]). Let Aij(z) (i = 0, . . . , n; j = 0, . . . ,m) be entire functions, and a, l ∈ {0, 1, . . . , n}, b ∈ {0, 1, . . . ,m} such that (a, b) 6= (l, 0). If the following three assumptions hold simultaneously: (1) max{µ(Aab), ρ(Aij), (i, j) 6= (a, b), (l, 0)} = ρ ≤ µ(Al0) <∞, µ(Al0) > 0; (2) τM (Al0) > τM (Aab), when µ(Al0) = µ(Aab); (3) τM (Al0) > max{τM (Aij) : ρ(Aij) = µ(Al0), (i, j) 6= (a, b), (l, 0)}, when µ(Al0) = max{ρ(Aij) : (i, j) 6= (a, b), (l, 0)}. Then any non zero meromorphic solution f of (1.2) satisfies ρ(f) ≥ µ(Al0) + 1. EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 3 Theorem 1.4 ([4]). Let Aij(z) (i = 0, . . . , n; j = 0, . . . ,m) be meromorphic func- tions, and a, l ∈ {0, 1, . . . , n}, b ∈ {0, 1, . . . ,m} such that (a, b) 6= (l, 0). If the following four assumptions hold simultaneously: (1) δ(∞, Al0) = lim infr→+∞ m(r,Al0) T (r,Al0) = δ > 0; (2) max{µ(Aab), ρ(Aij), (i, j) 6= (a, b), (l, 0)} = ρ ≤ µ(Al0) <∞, µ(Al0) > 0; (3) δτ(Al0) > τ(Aab), when µ(Al0) = µ(Aab); (4) δτ(Al0) > max{τ(Aij) : ρ(Aij) = µ(Al0), (i, j) 6= (a, b), (l, 0)} when µ(Al0) = max{ρ(Aij) : (i, j) 6= (a, b), (l, 0)}. Then any non zero meromorphic solution f of (1.2) satisfies ρ(f) ≥ µ(Al0) + 1. In this article, by combining complex differential and difference equations, we extend the results of Theorems 1.3 and 1.4 for the complex non-homogeneous linear delay-differential equation n∑ i=0 m∑ j=0 Aijf (j)(z + ci) = F. (1.3) Let us define S := {F,Aij : (i, j) 6= (l, 0), (k, p)}, ρ(S) := max{ρ(g) : g ∈ S}. The main results of this paper reads as follows. Theorem 1.5. Consider a delay-differential equation (1.3) with entire coefficients. Suppose that one of the coefficients, say Al0 with µ(Al0) > 0, is dominante in the sense that: (1) ρ := max{µ(Akp), ρ(S)} ≤ µ(Al0) <∞; (2) τM (Al0) > τM (Akp), whenever µ(Al0) = µ(Akp); (3) τ1 := max{τM (g) : ρ(g) = µ(Al0), g ∈ S} < τM (Al0), whenever µ(Al0) = ρ(S). Then every meromorphic solution f of (1.3) satisfies ρ(f) ≥ µ(Al0) if F (z) 6≡ 0. Further, if F (z) ≡ 0, then every meromorphic solution f 6≡ 0 of (1.2) satisfies ρ(f) ≥ µ(Al0) + 1. Theorem 1.6. Consider a delay-differential equation of type (1.3) with meromor- phic coefficients. Suppose that one of the coefficients, say Al0, is dominate in the sense that (1) ρ := max{µ(Akp), ρ(S)} ≤ µ(Al0) <∞; (2) τ(Al0) > τ(Akp), whenever µ(Al0) = µ(Akp); (3) τ1 = ∑ ρ(Aij)=µ(Al0), (i,j)6=(l,0),(k,p) τ(Aij) + τ(F ) < τ(Al0) < +∞ whenever µ(Al0) = ρ(S); (4) τ1 + τ(Akp) < τ(Al0) < +∞ whenever µ(Al0) = µ(Akp) = ρ(S); (5) λ( 1 Al0 ) < µ(Al0) <∞. Then every meromorphic solution f of (1.3) satisfies ρ(f) ≥ µ(Al0) if F (z) 6≡ 0. Further, if F (z) ≡ 0, then every meromorphic solution f 6≡ 0 of (1.2) satisfies ρ(f) ≥ µ(Al0) + 1. 4 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 2. Some preliminary lemmas Lemma 2.1 ([7]). Let f be a transcendental meromorphic function of finite order ρ(f), and let k and j be integers satisfying k > j ≥ 0. Then for every ε(> 0), there exists a subset E1 ⊂ (1,+∞) which has finite logarithmic measure such that for all z satisfying |z| = r /∈ [0, 1] ∪ E1, we have∣∣f (k)(z) f (j)(z) ∣∣ ≤ |z|(k−j)(ρ(f)−1+ε). Lemma 2.2 ([5]). Let f be a meromorphic function of finite order ρ, and let c1, c2(c1 6= c2) be two arbitrary complex numbers. Let ε > 0 be given, then there exists a subset E2 ⊂ (1,+∞) with finite logarithmic measure such that for all z satisfying |z| = r /∈ [0, 1] ∪ E2, we have exp{−rρ−1+ε} ≤ ∣∣f(z + c1) f(z + c2) ∣∣ ≤ exp{rρ−1+ε}. Lemma 2.3 ([6]). Let f be a meromorphic function, c be a non-zero complex constant. Then we have that as r → +∞ (1 + o(1))T (r − |c|, f(z)) ≤ T (r, f(z + c)) ≤ (1 + o(1))T (r + |c|, f(z)). Consequently ρ(f(z + c)) = ρ(f), µ(f(z + c)) = µ(f). Lemma 2.4 ([2]). Let f be a meromorphic function of finite order ρ. Then for any given ε > 0, there exists a set E3 ⊂ (1,+∞) having finite linear measure and finite logarithmic measure such that for all z satisfying |z| = r /∈ [0, 1] ∪ E3 and sufficiently large r, we have exp{−rρ+ε} ≤ ∣∣f(z)| ≤ exp{rρ+ε}. Lemma 2.5 ([11]). Let f be an entire function with µ(f) <∞. Then for any given ε(> 0), there exists a subset E4 ⊂ (1,+∞) with infinite logarithmic measure such that for all r ∈ E4, we have µ(f) = lim r→+∞, r∈E4 log logM(r, f) log r , M(r, f) < exp{rµ(f)+ε}. Lemma 2.6 ([22]). Let f be an entire function with 0 < µ(f) <∞. Then for any given ε(> 0), there exists a subset E5 ⊂ (1,+∞) with infinite logarithmic measure such that for all r ∈ E5, we have τM (f) = lim r→+∞, r∈E5 logM(r, f) log r , M(r, f) < exp{(τM (f) + ε)rµ(f)}. Lemma 2.7 ([5]). Let f be a meromorphic function of finite order ρ(f) <∞, and let c1, c2 be two distinct complex numbers. Then for each ε > 0, we have m ( r, f(z + c1) f(z + c2) ) = O(rρ(f)−1+ε). EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 5 Lemma 2.8 ([24]). Let f be a meromorphic function with µ(f) <∞. Then for any given ε(> 0), there exists a subset E6 ⊂ (1,+∞) with infinite logarithmic measure such that for all r ∈ E6, we have T (r, f) < rµ(f)+ε. Lemma 2.9 ([16]). Let f be a meromorphic function with 0 < µ(f) < ∞. Then for any given ε(> 0), there exists a subset E7 ⊂ (1,+∞) with infinite logarithmic measure such that for all r ∈ E7, we have T (r, f) < (τ(f) + ε)rµ(f). 3. Proof of main results Proof of Theorem 1.5. If f has infinite order, then the result holds. Now, we sup- pose that ρ(f) <∞. We divide (1.3) by f(z + cl) to obtain −Al0(z) = n∑ i=0,i6=l,k m∑ j=0 Aij f (j)(z + ci) f(z + ci) f(z + ci) f(z + cl) + m∑ j=0,j 6=p Akj f (j)(z + ck) f(z + ck) f(z + ck) f(z + cl) +Akp f (p)(z + ck) f(z + ck) f(z + ck) f(z + cl) + m∑ j=1 Alj f (j)(z + cl) f(z + cl) − F (z) f(z + cl) . (3.1) Therefore |Al0(z)| ≤ n∑ i=0,i6=l,k m∑ j=0 |Aij | ∣∣f (j)(z + ci) f(z + ci) ∣∣ ∣∣f(z + ci) f(z + cl) ∣∣ + m∑ j=0,j 6=p |Akj | ∣∣f (j)(z + ck) f(z + ck) ∣∣ ∣∣f(z + ck) f(z + cl) ∣∣ + |Akp| ∣∣f (p)(z + ck) f(z + ck) ∣∣ ∣∣f(z + ck) f(z + cl) ∣∣ + m∑ j=1 |Alj | ∣∣f (j)(z + cl) f(z + cl) ∣∣+ ∣∣ F (z) f(z + cl) ∣∣. (3.2) From Lemmas 2.1 and 2.3, for any given ε(> 0), there exists a subset E1 ⊂ (1,+∞) which has finite logarithmic measure such that for all z satisfying |z| = r /∈ [0, 1] ∪ E1, we have∣∣f (j)(z + ci) f(z + ci) ∣∣ ≤ |z|j(ρ(f+ci)−1+ε) = |z|j(ρ(f)−1+ε), (i, j) 6= (l, 0). (3.3) It follows by Lemma 2.2 that for any ε(> 0), there exists a subset E2 ⊂ (1,+∞) with finite logarithmic measure such that for all z satisfying |z| = r /∈ [0, 1] ∪ E2, we have ∣∣f(z + ci) f(z + cl) ∣∣ ≤ exp{rρ(f)−1+ε}, i 6= l. (3.4) From Lemma 2.3, we obtain ρ(f(z + cl)) = ρ ( 1 f(z + cl) ) = ρ(f). 6 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 So, by Lemma 2.4, for any given ε > 0, there exists a subset E3 ⊂ (1,+∞) having finite linear measure and finite logarithmic measure such that for all z satisfying |z| = r /∈ [0, 1] ∪ E3 sufficiently large, we have∣∣ 1 f(z + cl) ∣∣ ≤ exp{rρ(f)+ε}. (3.5) We divide the rest of the proof into four cases. Case 1: ρ < µ(Al0). For g ∈ S, by the definition of ρ(S), for any given ε > 0 and sufficiently large r, we have |g(z)| ≤ exp{rρ(S)+ε} ≤ exp{rρ+ε}. (3.6) From the definition of µ(Al0), for sufficiently small ε > 0 and sufficiently large r, we have |Al0(z)| ≥ exp{rµ(Al0)−ε}. (3.7) It also follows by the definition of µ(Akp) and Lemma 2.5, that for any ε(> 0), there exists a subset E4 ⊂ (1,+∞) with infinite logarithmic measure such that for all r ∈ E4, we have |Akp(z)| ≤ exp{rµ(Akp)+ε}. (3.8) By substituting (3.3)–(3.8) into (3.2), for all z satisfying |z| = r ∈ E4 \ ([0, 1]∪E1∪ E2 ∪ E3), we obtain exp{rµ(Al0)−ε} ≤ n∑ i=0,i6=l,k m∑ j=0 exp{rρ+ε}|z|j(ρ(f)−1+ε) exp{rρ(f)−1+ε} + m∑ j=0,j 6=p exp{rρ+ε}|z|j(ρ(f)−1+ε) exp{rρ(f)−1+ε} + |z|p(ρ(f)−1+ε) exp{rµ(Akp)+ε} exp{rρ(f)−1+ε} + m∑ j=1 exp{rρ+ε}|z|j(ρ(f)−1+ε) + exp{rρ+ε} exp{rρ(f)+ε} ≤ ((n− 1)(m+ 1) + 2m)rm(ρ(f)−1+ε) exp{rρ+ε} exp{rρ(f)−1+ε} + rp(ρ(f)−1+ε) exp{rµ(Akp)+ε} exp{rρ(f)−1+ε} + exp{rρ+ε} exp{rρ(f)+ε}. (3.9) Now, we choose ε sufficiently small to satisfy 0 < 3ε < µ(Al0)−ρ. We deduce from (3.9) that for |z| = r ∈ E4 \ ([0, 1] ∪ E1 ∪ E2 ∪ E3), r → +∞, exp{rµ(Al0)−2ε} ≤ exp{rρ(f)+ε}. Therefore, µ(Al0) ≤ ρ(f) + 3ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0). Further, if F ≡ 0, then by substituting (3.3), (3.4) and (3.6)–(3.8) into (3.2), for all z satisfying |z| = r ∈ E4 \ ([0, 1] ∪ E1 ∪ E2), we obtain exp{rµ(Al0)−ε} ≤ (nm+ n+m− 1)rm(ρ(f)−1+ε) exp{rρ+ε} exp{rρ(f)−1+ε} + rp(ρ(f)−1+ε) exp{rµ(Akp)+ε} exp{rρ(f)−1+ε}. (3.10) By choosing sufficiently small ε satisfying 0 < 3ε < µ(Al0) − ρ, we deduce from (3.10) that for |z| = r ∈ E4 \ ([0, 1] ∪ E1 ∪ E2), r → +∞, exp{rµ(Al0)−2ε} ≤ exp{rρ(f)−1+ε}, EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 7 that is, µ(Al0) ≤ ρ(f)− 1 + 3ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0) + 1. Case 2: β = ρ(S) < µ(Al0) = µ(Akp) and τM (Al0) > τM (Akp). For g ∈ S, by the definition of ρ(S), for any given ε(> 0), and sufficiently large r, we have |g(z)| ≤ exp{rρ(S)+ε} ≤ exp{rβ+ε}. (3.11) From the definition of τM (Al0), for sufficiently small ε > 0 and sufficiently large r, we have |Al0(z)| ≥ exp{(τM (Al0)− ε)rµ(Al0)}. (3.12) Also, from the definition of τM (Akp) and Lemma 2.6, for any given ε(> 0), there exists a subset E5 ⊂ (1,+∞) with infinite logarithmic measure such that for all r ∈ E5, we have |Akp(z)| ≤ exp{(τM (Akp) + ε)rµ(Akp)} = exp{(τM (Akp) + ε)rµ(Al0)}. (3.13) By substituting (3.3)–(3.5) and (3.11)–(3.13) into (3.2), for all z satisfying |z| = r ∈ E5 \ ([0, 1] ∪ E1 ∪ E2 ∪ E3), we obtain exp{(τM (Al0)− ε)rµ(Al0)} ≤ (nm+ n+m− 1)rm(ρ(f)−1+ε) exp{rβ+ε} exp{rρ(f)−1+ε} + rp(ρ(f)−1+ε) exp{(τM (Akp) + ε)rµ(Al0)} exp{rρ(f)−1+ε} + exp{rβ+ε} exp{rρ(f)+ε}. (3.14) Therefore, we may choose ε sufficiently small, 0 < 2ε < min{µ(Al0)−β, τM (Al0)− τM (Akp)}, then from (3.14) for r ∈ E5 \ ([0, 1]∪E1 ∪E2 ∪E3) sufficiently large, we obtain exp{(τM (Al0)− τM (Akp)− 2ε)rµ(Al0)−ε} ≤ exp{rρ(f)+ε}. Then, µ(Al0) ≤ ρ(f) + 2ε, since ε > 0 is arbitrary, so ρ(f) ≥ µ(Al0). Further, if F ≡ 0, then by substituting (3.3), (3.4) and (3.11)–(3.13) into (3.2), for all z satisfying |z| = r ∈ E5 \ ([0, 1] ∪ E1 ∪ E2), we have exp{(τM (Al0)− ε)rµ(Al0)} ≤ (nm+ n+m− 1)rm(ρ(f)−1+ε) exp{rβ+ε} exp{rρ(f)−1+ε} + rp(ρ(f)−1+ε) exp{(τM (Akp) + ε)rµ(Al0)} exp{rρ(f)−1+ε}. (3.15) Now, we choose ε sufficiently small, 0 < 2ε < min{µ(Al0)−β, τM (Al0)−τM (Akp)}. Then from (3.15) for r ∈ E5 \ ([0, 1] ∪ E1 ∪ E2) sufficiently large, we obtain exp{(τM (Al0)− τM (Akp)− 2ε)rµ(Al0)−ε} ≤ exp{rρ(f)−1+ε}, that is, µ(Al0) ≤ ρ(f)− 1 + 2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0) + 1. Case 3:µ(Al0) = ρ(S) > µ(Akp) and max{τM (g) : ρ(g) = µ(Al0), g ∈ S} = τ1 < τM (Al0). For g ∈ S, by the definitions of ρ(S) and τM (g), for any given ε > 0 and sufficiently large r, we have |g(z)| ≤ { exp{rρ(S)+ε} ≤ exp{rµ(Al0)−ε}, if ρ(S) < µ(Al0), exp{(τ1 + ε)rµ(Al0)}, if ρ(S) = µ(Al0). (3.16) 8 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 Then, by substituting (3.3)–(3.5), (3.8), (3.12) and (3.16) into (3.2), for all z satis- fying |z| = r ∈ E4 \ ([0, 1] ∪ E1 ∪ E2 ∪ E3) sufficiently large, we obtain exp{(τM (Al0)− ε)rµ(Al0)} ≤ O ( rm(ρ(f)−1+ε) exp{(τ1 + ε)rµ(Al0)} exp{rρ(f)−1+ε} ) +O ( rm(ρ(f)−1+ε) exp{rµ(Al0)−ε} exp{rρ(f)−1+ε} ) + rp(ρ(f)−1+ε) exp{rµ(Akp)+ε} exp{rρ(f)−1+ε} +O ( rm(ρ(f)−1+ε) exp{rµ(Al0)−ε} ) +O ( rm(ρ(f)−1+ε) exp { (τ1 + ε)rµ(Al0) }) + exp{(τ1 + ε)rµ(Al0)} exp{rρ(f)+ε}. (3.17) Now, we choose ε sufficiently small satisfying 0 < 2ε < min{µ(Al0)− µ(Akp), τM (Al0)− τ1}, then from (3.17) for sufficiently large r ∈ E4 \ ([0, 1] ∪ E1 ∪ E2 ∪ E3), we obtain exp{(τM (Al0)− τ1 − 2ε)rµ(Al0)−ε} ≤ exp{rρ(f)+ε}. That means, µ(Al0) ≤ ρ(f) + 2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0). Further, if F ≡ 0, then by substituting (3.3), (3.4), (3.8), (3.12) and (3.16) into (3.2), for all z satisfying |z| = r ∈ E4 \ ([0, 1] ∪ E1 ∪ E2) sufficiently large, we have exp{(τM (Al0)− ε)rµ(Al0)} ≤ O ( rm(ρ(f)−1+ε) exp{(τ1 + ε)rµ(Al0)} exp{rρ(f)−1+ε} ) +O ( rm(ρ(f)−1+ε) exp{rµ(Al0)−ε} exp{rρ(f)−1+ε} ) + rp(ρ(f)−1+ε) exp{rµ(Akp)+ε} exp{rρ(f)−1+ε} +O(rm(ρ(f)−1+ε) exp{rµ(Al0)−ε}) +O(rm(ρ(f)−1+ε) exp{(τ1 + ε)rµ(Al0)}). (3.18) Now, we choose ε sufficiently small satisfying 0 < 2ε < min{µ(Al0)− µ(Akp), τM (Al0)− τ1}. Then from (3.18) for sufficiently large r ∈ E4 \ ([0, 1] ∪ E1 ∪ E2), we obtain exp{(τM (Al0)− τ1 − 2ε)rµ(Al0)−ε} ≤ exp{rρ(f)−1+ε}. That means, µ(Al0) ≤ ρ(f)−1+2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0)+1. Case 4: ρ(S) = µ(Akp) = µ(Al0) and max{τM (Akp), τM (g) : ρ(g) = µ(Al0), g ∈ S} = τ2 < τM (Al0). It follows by substituting (3.3)–(3.5), (3.12), (3.13) and (3.16) into (3.2), for all z satisfying |z| = r ∈ E5 \ ([0, 1]∪E1 ∪E2 ∪E3) sufficiently large, EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 9 we have exp{(τM (Al0)− ε)rµ(Al0)} ≤ O ( rm(ρ(f)−1+ε) exp{(τ2 + ε)rµ(Al0)} exp{rρ(f)−1+ε} ) +O ( rm(ρ(f)−1+ε) exp{rµ(Al0)−ε} exp { rρ(f)−1+ε }) + rp(ρ(f)−1+ε) exp{(τM (Akp) + ε)rµ(Al0)} exp { rρ(f)−1+ε } +O ( rm(ρ(f)−1+ε) exp{rµ(Al0)−ε} ) +O ( rm(ρ(f)−1+ε) exp { (τ2 + ε)rµ(Al0) }) + exp{(τ2 + ε)rµ(Al0)} exp { rρ(f)+ε } . (3.19) Now, we choose ε sufficiently small satisfying 0 < 2ε < τM (Al0)− τ2, from (3.19) for sufficiently large r ∈ E5 \ ([0, 1] ∪ E1 ∪ E2 ∪ E3), we obtain exp{(τM (Al0)− τ2 − 2ε)rµ(Al0)−ε} ≤ exp { rρ(f)+ε } , this means, µ(Al0) ≤ ρ(f)+2ε, since ε > 0 is arbitrary, it follows that ρ(f) ≥ µ(Al0). Further, if F ≡ 0, by substituting (3.3), (3.4), (3.12), (3.13) and (3.16) into (3.2), for all z satisfying |z| = r ∈ E5 \ ([0, 1] ∪ E1 ∪ E2) sufficiently large, we have exp{(τM (Al0)− ε)rµ(Al0)} ≤ O ( rm(ρ(f)−1+ε) exp{(τ2 + ε)rµ(Al0)} exp { rρ(f)−1+ε }) +O ( rm(ρ(f)−1+ε) exp{rµ(Al0)−ε} exp { rρ(f)−1+ε }) + rp(ρ(f)−1+ε) exp{(τM (Akp) + ε)rµ(Al0)} exp { rρ(f)−1+ε } +O ( rm(ρ(f)−1+ε) exp{rµ(Al0)−ε} ) +O ( rm(ρ(f)−1+ε) exp { (τ2 + ε)rµ(Al0) }) . (3.20) Now, we choose ε sufficiently small satisfying 0 < 2ε < τM (Al0)− τ2, from (3.20) for sufficiently large r ∈ E5 \ ([0, 1] ∪ E1 ∪ E2), we obtain exp{(τM (Al0)− τ2 − 2ε)rµ(Al0)−ε} ≤ exp { rρ(f)−1+ε } . That means, µ(Al0) ≤ ρ(f)−1+2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0)+1. The proof of Theorem 1.5 is complete. � Proof of the Theorem 1.6. If f has infinite order, then the result holds. Now, we suppose that ρ(f) <∞. By (3.1), we have T (r,Al0(z)) = m(r,Al0(z)) +N(r,Al0(z)) ≤ n∑ i=0,i6=l,k m∑ j=0 m(r,Aij(z)) +m(r,Akp(z)) 10 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 + m∑ j=0,j 6=p m(r,Akj(z)) + m∑ j=1 m(r,Alj(z)) + n∑ i=0,i6=l,k m∑ j=0 m ( r, f (j)(z + ci) f(z + ci) ) + n∑ i=0,i6=l,k m ( r, f(z + ci) f(z + cl) ) + m∑ j=1 m ( r, f (j)(z + ck) f(z + ck) ) + 2m ( r, f(z + ck) f(z + cl) ) + m∑ j=1 m ( r, f (j)(z + cl) f(z + cl) ) +m(r, F (z)) +m ( r, 1 f(z + cl) ) +N(r,Al0(z)) +O(1) ≤ n∑ i=0,i6=l,k m∑ j=0 T (r,Aij(z)) + T (r,Akp(z)) + m∑ j=0,j 6=p T (r,Akj(z)) + m∑ j=1 T (r,Alj(z)) + n∑ i=0,i6=l,k m∑ j=1 m ( r, f (j)(z + ci) f(z + ci) ) + n∑ i=0,i6=l,k m(r, f(z + ci) f(z + cl) ) + m∑ j=1 m(r, f (j)(z + ck) f(z + ck) ) + 2m(r, f(z + ck) f(z + cl) ) + m∑ j=1 m ( r, f (j)(z + cl) f(z + cl) ) + T (r, F (z)) + T (r, 1 f(z + cl) ) +N(r,Al0(z)) +O(1). By Lemma 2.3 and the first main theorem of Nevanlinna, when r sufficiently large, we have T ( r, 1 f(z + cl) ) = T (r, f(z + cl)) +O(1) ≤ (1 + o(1))T (r + |cl|, f) ≤ 2T (2r, f). So, for r sufficiently large, we obtain T (r,Al0(z)) ≤ n∑ i=0,i6=l,k m∑ j=0 T (r,Aij(z)) + T (r,Akp(z)) + m∑ j=0,j 6=p T (r,Akj(z)) + m∑ j=1 T (r,Alj(z)) + n∑ i=0,i6=l,k m∑ j=1 m(r, f (j)(z + ci) f(z + ci) ) + n∑ i=0,i6=l,k m ( r, f(z + ci) f(z + cl) ) + T (r, F (z)) + 2T (2r, f) + m∑ j=1 m ( r, f (j)(z + cl) f(z + cl) ) + m∑ j=1 m(r, f (j)(z + ck) f(z + ck) ) + 2m ( r, f(z + ck) f(z + cl) ) +N(r,Al0(z)) +O(1). (3.21) By Lemma 2.7, for any positive ε, we have m ( r, f(z) f(z + cl) ) = O(rρ(f)−1+ε), m ( r, f(z + cj) f(z + cl) ) = O(rρ(f)−1+ε), (3.22) EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 11 for j 6= l. By the lemma of logarithmic derivative [10], there exists a subset E8 ⊂ [0,+∞[ of a finite linear measure such that for all r /∈ E8 sufficiently large, we have m ( r, f (j)(z + ci) f(z + ci) ) = O(log r) (i = 0, . . . , n; j = 1, . . . ,m). (3.23) From the definition of λ( 1 Al0 ), for any ε > 0 and sufficiently large r, we have N(r,Al0) ≤ rλ( 1 Al0 )+ε . (3.24) We divide the rest of the proof into four cases. Case 1: ρ < µ(Al0). For g ∈ S, from the definition of ρ(S) and ρ(f) for any given ε > 0 and sufficiently large r, we have T (r, g) ≤ rρ(S)+ε ≤ rρ+ε, (3.25) T (r, f) ≤ rρ(f)+ε. (3.26) It follows from the definition of µ(Al0), for sufficiently small ε > 0 and sufficiently large r, we have T (r,Al0) ≥ rµ(Al0)−ε. (3.27) It follows from the definition of µ(Akp) and Lemma 2.8, for any ε(> 0), there exists a subset E6 ⊂ (1,+∞) with infinite logarithmic measure such that for all r ∈ E6, we have T (r,Akp) ≤ rµ(Akp)+ε. (3.28) By substituting (3.22)–(3.28) into (3.21) for sufficiently large r ∈ E6\E8, we obtain rµ(Al0)−ε ≤ ((n− 1)(m+ 1) + 2m)rρ+ε + rµ(Akp)+ε +O(rρ(f)−1+ε) + 2(2r)ρ(f)+ε + rρ+ε + r λ( 1 Al0 )+ε +O(log r). (3.29) We may choose ε sufficiently small satisfying 0 < 3ε < min { µ(Al0)− ρ, µ(Al0)− λ ( 1 Al0 )} , it follows from (3.29) that for r ∈ E6 \ E8, r → +∞, rµ(Al0)−2ε ≤ rρ(f)+ε, this means, µ(Al0) ≤ ρ(f) + 3ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0). Further, if F ≡ 0, then by substituting (3.22)–(3.25), (3.27) and (3.28) into (3.21) for sufficiently large r ∈ E6 \ E8, we obtain rµ(Al0)−ε ≤ ((n− 1)(m+ 1) + 2m)rρ+ε + rµ(Akp)+ε +O(rρ(f)−1+ε) + r λ( 1 Al0 )+ε +O(log r). (3.30) We choose ε sufficiently small satisfying 0 < 3ε < min { µ(Al0)− ρ, µ(Al0)− λ( 1 Al0 ) } , from (3.30) that for r ∈ E6 \ E8, r → +∞, rµ(Al0)−2ε ≤ rρ(f)−1+ε, this means, µ(Al0) ≤ ρ(f)−1 + 3ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0) + 1. 12 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 Case 2: β = ρ(S) < µ(Al0) = µ(Akp), and τ(Al0) > τ(Akp). For g ∈ S, by the definition of ρ(S), for any given ε(> 0) and sufficiently large r, we obtain T (r, g) ≤ rρ(S)+ε ≤ rβ+ε. (3.31) From the definition of τ(Al0), for sufficiently small ε > 0 and sufficiently large r, we have T (r,Al0) ≥ (τ(Al0)− ε)rµ(Al0). (3.32) It follows from the definition of τ(Akp) and Lemma 2.9, that for any positive ε, there exists a subset E7 ⊂ (1,+∞) with infinite logarithmic measure such that for all r ∈ E7, we have T (r,Akp) ≤ (τ(Akp) + ε)rµ(Akp) ≤ (τ(Akp) + ε)rµ(Al0). (3.33) By substituting (3.22)–(3.24), (3.26) and (3.31)–(3.33) into (3.21), for sufficiently large r ∈ E7 \ E8, we obtain (τ(Al0)− ε)rµ(Al0) ≤ ((n− 1)(m+ 1) + 2m)rβ+ε + (τ(Akp) + ε)rµ(Al0) +O(rρ(f)−1+ε) + 2(2r)ρ(f)+ε + rβ+ε + r λ( 1 Al0 )+ε +O(log r). (3.34) Now, we choose ε sufficiently small satisfying 0 < 2ε < min { µ(Al0)− β, τ(Al0)− τ(Akp), µ(Al0)− λ ( 1 Al0 )} , so from (3.34) for sufficiently large r ∈ E7 \ E8, we have (τ(Al0)− τ(Akp)− 2ε)rµ(Al0)−ε ≤ rρ(f)+ε, this means, µ(Al0) ≤ ρ(f) + 2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0). Further, if F ≡ 0, then by substituting (3.22)–(3.24) and (3.31)–(3.33) into (3.21), for sufficiently large r ∈ E7 \ E8, we obtain (τ(Al0)− ε)rµ(Al0) ≤ ((n− 1)(m+ 1) + 2m)rβ+ε + (τ(Akp) + ε)rµ(Al0) +O(rρ(f)−1+ε) + r λ( 1 Al0 )+ε +O(log r). (3.35) Now, we choose ε sufficiently small satisfying 0 < 2ε < min { µ(Al0)− β, τ(Al0)− τ(Akp), µ(Al0)− λ ( 1 Al0 )} . From (3.35) for sufficiently large r ∈ E7 \ E8, we obtain (τ(Al0)− τ(Akp)− 2ε)rµ(Al0)−ε ≤ rρ(f)−1+ε, this means, µ(Al0) ≤ ρ(f)−1 + 2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0) + 1. Case 3: µ(Al0) = ρ(S) > µ(Akp) and τ1 = ∑ ρ(Aij)=µ(Al0), (i,j) 6=(l,0),(k,p) τ(Aij) + τ(F ) < τ(Al0). Then there exists a subset J ⊆ {0, 1, . . . , n} × {0, 1, . . . ,m} \ {(l, 0), (k, p)} such that for all (i, j) ∈ J , when ρ(Aij) = µ(Al0), we have ∑ (i,j)∈J τ(Aij) < τ(Al0)−τ(F ), EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 13 and for (i, j) ∈ Π = {0, 1, . . . , n} × {0, 1, . . . ,m} \ (J ∪ {(l, 0), (k, p)}) we have ρ(Aij) < µ(Al0). Hence, for any ε > 0 and sufficiently large r, we obtain T (r,Aij) ≤ { (τ(Aij) + ε)rµ(Al0), if (i, j) ∈ J, rρ(Aij)+ε ≤ rµ(Al0)−ε, if (i, j) ∈ Π (3.36) and T (r, F ) ≤ { (τ(F ) + ε)rµ(Al0), if ρ(F ) = µ(Al0), rρ(F )+ε ≤ rµ(Al0)−ε, if ρ(F ) < µ(Al0). (3.37) Then, by substituting (3.22)–(3.24), (3.26), (3.28), (3.32), (3.36) and (3.37) into (3.21), for all z satisfying |z| = r ∈ E6 \ E8 sufficiently large r, we obtain (τ(Al0)− ε)rµ(Al0) ≤ ∑ (i,j)∈J (τ(Aij) + ε)rµ(Al0) + ∑ (i,j)∈Π rµ(Al0)−ε + rµ(Akp)+ε + (τ(F ) + ε)rµ(Al0) + r λ( 1 Al0 )+ε + 2(2r)ρ(f)+ε +O(rρ(f)−1+ε) +O(ln r) ≤ (τ1 + (nm+ n+m)ε)rµ(Al0) +O(rµ(Al0)−ε) + rµ(Akp)+ε + r λ( 1 Al0 )+ε + 2(2r)ρ(f)+ε +O(rρ(f)−1+ε) +O(log r). (3.38) Now, we choose ε sufficiently small satisfying 0 < ε < min {µ(Al0)− µ(Akp) 2 , τ(Al0)− τ1 nm+ n+m+ 1 , µ(Al0)− λ( 1 Al0 ) 2 } , then from (3.38) for sufficiently large r ∈ E6 \ E8, we obtain (τ(Al0)− τ1 − (nm+ n+m+ 1)ε)rµ(Al0)−ε ≤ rρ(f)+ε, this means, µ(Al0) ≤ ρ(f) + 2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0). Further, if F ≡ 0, then by substituting (3.22)–(3.24), (3.28), (3.32) and (3.36) into (3.21), for all z satisfying |z| = r ∈ E6 \ E8 sufficiently large r, we obtain (τ(Al0)− ε)rµ(Al0) ≤ (τ1 + (nm+ n+m− 1)ε)rµ(Al0) +O ( rµ(Al0)−ε) + rµ(Akp)+ε + r λ( 1 Al0 )+ε +O ( rρ(f)−1+ε ) +O(log r). (3.39) Now, we choose ε sufficiently small satisfying 0 < ε < min {µ(Al0)− µ(Akp) 2 , τ(Al0)− τ1 nm+ n+m , µ(Al0)− λ( 1 Al0 ) 2 } , then from (3.39) for sufficiently large r ∈ E6 \ E8, we obtain (τ(Al0)− τ1 − (nm+ n+m)ε)rµ(Al0)−ε ≤ rρ(f)−1+ε, this means, µ(Al0) ≤ ρ(f)−1 + 2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0) + 1. Case 4: ρ(S) = µ(Al0) = µ(Akp) with τ1 + τ(Akp) < τ(Al0). It follows by substituting (3.22)–(3.24), (3.26), (3.32), (3.33), (3.36) and (3.37) into (3.21), for all sufficiently large r ∈ E7 \ E8, we have (τ(Al0)− ε)rµ(Al0) ≤ (τ1 + (nm+ n+m)ε)rµ(Al0) +O ( rµ(Al0)−ε) + (τ(Akp) + ε)rµ(Al0) + r λ( 1 Al0 )+ε + 2(2r)ρ(f)+ε +O(rρ(f)−1+ε) +O(log r). (3.40) 14 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 Now, we choose ε sufficiently small satisfying 0 < ε < min {τ(Al0)− τ1 − τ(Akp) nm+ n+m+ 2 , µ(Al0)− λ( 1 Al0 ) 2 } , then from (3.40) for sufficiently large r ∈ E7 \ E8, we obtain (τ(Al0)− τ1 − τ(Akp)− (nm+ n+m+ 2)ε)rµ(Al0)−ε ≤ rρ(f)+ε, this means, µ(Al0) ≤ ρ(f) + 2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0). Further, if F ≡ 0, then by substituting (3.22)–(3.24), (3.32), (3.33) and (3.36) into (3.21), for all sufficiently large r ∈ E7 \ E8, we have (τ(Al0)− ε)rµ(Al0) ≤ (τ1 + (nm+ n+m− 1)ε)rµ(Al0) +O ( rµ(Al0)−ε) + (τ(Akp) + ε)rµ(Al0) + r λ( 1 Al0 )+ε +O(rρ(f)−1+ε) +O(log r). (3.41) Now, we choose ε sufficiently small satisfying 0 < ε < min {τ(Al0)− τ1 − τ(Akp) nm+ n+m+ 1 , µ(Al0)− λ( 1 Al0 ) 2 } , then from (3.41) for sufficiently large r ∈ E7 \ E8, we obtain (τ(Al0)− τ1 − τ(Akp)− (nm+ n+m+ 1)ε)rµ(Al0)−ε ≤ rρ(f)−1+ε, so this means, µ(Al0) ≤ ρ(f)−1+2ε, since ε > 0 is arbitrary, then ρ(f) ≥ µ(Al0)+1. The proof of Theorem 1.6 is complete. � 4. Examples Example 4.1. We consider the non-homogeneous linear delay-differential equation with entire coefficients A02(z)f ′′(z) +A11(z)f ′(z + 1) +A01(z)f ′(z) +A10(z)f(z + 1) +A00(z)f(z) = F (z). (4.1) Case 1: ρ(S) < µ(Al0). In (4.1), for A00(z) = π2 + 2π4z2, A10(z) = e−π 2z2−π2 , A01(z) = 2π2(z + 1)e2π2z+π2 , A11(z) = −2π2z, A02(z) = −1 2 , F (z) = e2π2z, we have max{µ(A11), ρ(F ), ρ(Aij) : (i, j) 6= (1, 0), (1, 1)} = 1 < µ(A10) = 2. We see that the conditions of Theorem 1.5 are satisfied. The function f(z) = eπ 2z2 is a solution of (4.1) and satisfies ρ(f) = 2 ≥ µ(A10) = 2. Case 2: ρ(S) < µ(Al0) = µ(Akp) with τM (Al0) > τM (Akp). In (4.1), for A00(z) = 2π2, A10(z) = 2π2(z + 1)ez 2 + e−π 2z2−π2 , A01(z) = 2π2z, A11(z) = −ez 2 , A02(z) = −1, F (z) = e2π2z, we obtain max{ρ(F ), ρ(Aij) : (i, j) 6= (1, 0), (1, 1)} = 1 < µ(A10) = µ(A11) = 2 and τM (A10) = π2 > τM (A11) = 1. Hence, the conditions of Theorem 1.5 are satisfied. The function f(z) = eπ 2z2 is a solution of (4.1) and satisfies ρ(f) = 2 ≥ µ(A10) = 2. EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 15 Case 3: µ(Al0) = ρ(S) > µ(Akp) with τM (Al0) > τ1 = max{τM (g) : ρ(g) = µ(Al0), g ∈ S}. In (4.1), for A00(z) = π2 + 2π4z2, A10(z) = e− 4 5π 2z2−π2 , A01(z) = 2π2(z + 1)e2π2z+π2 , A11(z) = −2π2z, A02(z) = −1 2 , F (z) = e 1 5π 2z2+2π2z, we have µ(A10) = max{ρ(F ), ρ(Aij) : (i, j) 6= (1, 0), (0, 1)} = 2 > µ(A01) = 1 and τM (A10) = 4π2 5 > τ1 = τM (F ) = π2 5 . Obviously, the conditions of Theorem 1.5 are satisfied. The function f(z) = eπ 2z2 is a solution of (4.1) and satisfies ρ(f) = 2 ≥ µ(A10) = 2. Case 4: µ(Al0) = µ(Akp) = ρ(S) and τM (A10) > max{τ1, τM (Akp)}. In (4.1), for A00(z) = 2π2, A10(z) = 2π2(z + 1)ez 2 + e− 4 5π 2z2−π2 , A01(z) = 2π2z, A11(z) = −ez 2 , A02(z) = −1, F (z) = e 1 5π 2z2+2π2z, we obtain µ(A10) = µ(A11) = max{ρ(F ), ρ(Aij) : (i, j) 6= (1, 0), (1, 1)} = 2 and τM (A10) = 4π2 5 > max{τ1, τM (A11)} = max{τM (F ), τM (A11)} = π2 5 . We see that the conditions of Theorem 1.5 are satisfied. The function f(z) = eπ 2z2 is a solution of equation (4.1) and satisfies ρ(f) = 2 ≥ µ(A10) = 2. Example 4.2. We consider the homogeneous linear delay-differential equation with entire coefficients A11(z)g′(z − 1) +A20(z)g(z + 3) +A00(z)g(z) = 0. (4.2) Case 1: max{µ(Akp), ρ(Aij) : (i, j) 6= (l, 0), (k, p)} < µ(Al0). In (4.2), for A00(z) = 1, A20(z) = (4πi(1− z)− e4πiz)e−16πiz, A11(z) = 1, we have max{µ(A11), ρ(Aij) : (i, j) 6= (2, 0), (1, 1)} = 0 < µ(A20) = 1. So, the conditions of Theorem 1.5 are satisfied. The function g(z) = e2πiz2 is a solution of (4.2) and g satisfies ρ(g) = 2 ≥ µ(A20) + 1 = 2. Case 2: max{ρ(Aij) : (i, j) 6= (l, 0), (k, p)} < µ(Al0) = µ(Akp) with τM (Al0) > τM (Akp). In (4.2), for A00(z) = 1, A20(z) = (4πi(1− z)− e2πiz)e−14πiz, A11(z) = e2πiz, we obtain µ(A20) = µ(A11) = 1 > max{ρ(Aij) : (i, j) 6= (2, 0), (1, 1)} = 0 and τM (A20) = 14π > τM (A11) = 2π. Obviously, the conditions of Theorem 1.5 are satisfied. The function g(z) = e2πiz2 is a solution of (4.2) and g satisfies ρ(g) = 2 ≥ µ(A20) + 1 = 2. Case 3: µ(Al0) = max{ρ(Aij) : (i, j) 6= (l, 0), (k, p)} > µ(Akp) with τM (Al0) > τ1 = max{τM (Aij) : ρ(Aij) = µ(Al0), (i, j) 6= (l, 0), (k, p)}. In (4.2), for A00(z) = e2πiz, A20(z) = (4πi(1− z)− e6πiz)e−16πiz, A11(z) = 1, we obtain µ(A20) = max{ρ(Aij) : (i, j) 6= (2, 0), (1, 1)} = 1 > µ(A11) = 0 and τM (A20) = 16π > τ1 = τM (A00) = 2π. Obviously, the conditions of Theorem 1.5 are satisfied. The function g(z) = e2πiz2 is a solution of (4.2) and g satisfies ρ(g) = 2 ≥ µ(A20) + 1 = 2. 16 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 Case 4. µ(Al0) = µ(Akp) = max{ρ(Aij) : (i, j) 6= (l, 0), (k, p)} with τM (Al0) > τ2 = max{τM (Akp), τM (Aij) : ρ(Aij) = µ(Al0), (i, j) 6= (l, 0), (k, p)}. In (4.2), for A00(z) = e−2πiz, A20(z) = (4πi(1− z)− 1)e−14πiz, A11(z) = e2πiz, we have µ(A20) = µ(A11) = max{ρ(Aij) : (i, j) 6= (2, 0), (1, 1)} = 1 and τM (A20) = 14π > max{τM (A00), τM (A11)} = 2π. It is clear that the conditions of Theorem 1.5 are satisfied. The function g(z) = e2πiz2 is a solution of (4.2) and g satisfies ρ(g) = 2 ≥ µ(A20) + 1 = 2. Example 4.3. We consider the non-homogeneous linear delay-differential equation with meromorphic coefficients A11(z)f ′(z − 1) +A01(z)f ′(z) +A20(z)f(z + 1) +A10(z)f(z − 1) = F (z). (4.3) Case 1: max{µ(Akp), ρ(S)} < µ(Al0). In (4.3), for A10(z) = e−π 3z3+3π3z2−3π3z+π3 , A20(z) = 3π3(2z − 1)e−3π3z2−3π3z−π3 , A01(z) = −1, A11(z) = e3π3z2−3π3z+π3 , F (z) = tan(πz), we have max{µ(A11), ρ(F ), ρ(Aij) : (i, j) 6= (1, 0), (1, 1)} = 2 < µ(A10) = 3, λ( 1 A10 ) = 0 < µ(A10) = 3. It is easy to see that the conditions of Theorem 1.6 are satisfied. The meromorphic function f(z) = eπ 3z3 tan(πz) is a solution of (4.3) and satisfies ρ(f) = 3 ≥ µ(A10) = 3. Case 2: ρ(S) < µ(Al0) = µ(Akp) with τ(Al0) > τ(Akp). In (4.3), for A10(z) = e−π 3z3+3π3z2−3π3z+π3 + ( 3π3(2z − 1)− 3z2π3 − π tan(πz) + π tan(πz) ) e−z 3 , A20(z) = 3π3z2 + π tan(πz) + π tan(πz) , A01(z) = −e3π3z2+3π3z+π3 , A11(z) = e−z 3 , F (z) = tan(πz), we obtain max{ρ(F ), ρ(Aij) : (i, j) 6= (1, 0), (1, 1)} = 2 < µ(A10) = µ(A11) = 3, λ( 1 A10 ) = 1 < µ(A10) = 3, τ(A10) = π2 > τ(A11) = 1 π . Hence, the conditions of Theorem 1.6 are satisfied. The function f(z) = eπ 3z3 is a solution of (4.3) and f satisfies ρ(f) = 3 ≥ µ(A10) = 3. Case 3: µ(Al0) = ρ(S) > µ(Akp) with τ(Al0) > τ1 = ∑ ρ(Aij)=µ(Al0), (i,j)6=(l,0),(k,p) τ(Aij) + τ(F ). EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 17 In (4.3), for A10(z) = e−2π3z3+3π3z2−3π3z+π3 , A20(z) = 3π3(2z − 1)e−3π3z2−3π3z−π3 , A01(z) = −1, A11(z) = e3π3z2−3π3z+π3 , F (z) = tan(πz) eπ3z3 , we have µ(A10) = max{ρ(F ), ρ(Aij) : (i, j) 6= (1, 0), (1, 1)} = 3 > µ(A11) = 2, λ ( 1 A10 ) = 0 < µ(A10) = 3 and τ(A10) = 2π2 > τ1 = τ(F ) = π2. We can see that the conditions of Theorem 1.6 are satisfied. The meromorphic function f(z) = eπ 3z3 tan(πz) is a solution of (4.3) and satisfies ρ(f) = 3 ≥ µ(A10) = 3. Case 4: µ(Al0) = µ(Ak0) = ρ(S) and τ(A10) > τ1 + τ(Akp). In (4.3), for A10(z) = e−2π3z3+3π3z2−3π3z+π3 , A20(z) = 3π3(2z − 1)e(π4 z) 3−3π3z2−3π3z−π3 , A01(z) = −e(π4 z) 3 , A11(z) = e(π4 z) 3+3π3z2−3π3z+π3 , F (z) = tan(πz) eπ3z3 , we obtain µ(A10) = µ(A11) = max{ρ(F ), ρ(Aij) : (i, j) 6= (1, 0), (1, 1)} = 3, λ( 1 A10 ) = 0 < µ(A10) = 3, τ1 + τ(A11) = τ(A01) + τ(A20) + τ(F ) + τ(A11) = ( 2 43 + 1)π2 + π2 43 = 67 64 π2 < τ(A10) = 2π2. Obviously, the conditions of Theorem 1.6 are satisfied. The meromorphic function f(z) = eπ 3z3 tan(πz) is a solution of (4.3) and satisfies ρ(f) = 3 ≥ µ(A10) = 3. Example 4.4. We consider the homogeneous linear delay-differential equation with meromorphic coefficients A11(z)h′(z + iπ) +A20(z)h(z + 2iπ) +A00(z)h(z) = 0. (4.4) Case 1: max{µ(Akp), ρ(Aij) : (i, j) 6= (l, 0), (k, p)} < µ(Al0). In (4.4), for A00(z) = −1, A20(z) = e12πz2+24π2iz−16π3 − e6πz2+18π2iz−14π3 , A11(z) = cos(2iz) 6i(z + iπ)2 cos(2iz) + 2i sin(2iz) , we have max{µ(A11), ρ(Aij) : (i, j) 6= (2, 0), (1, 1)} = 1 < µ(A20) = 2 and λ ( 1 A20 ) = 0 < µ(A20) = 2. Obviously, the conditions of Theorem 1.6 are satisfied. The meromorphic function h(z) = e2iz3 cos(2iz) 18 R. BELLAAMA, B. BELAÏDI EJDE-2021/92 is a solution of (4.4) and satisfies ρ(h) = 3 ≥ µ(A20) + 1 = 3. Case 2: max{ρ(Aij) : (i, j) 6= (l, 0), (k, p)} < µ(Al0) = µ(Akp) with τ(Al0) > τ(Akp). In (4.4), for A00(z) = 1, A20(z) = −2e12πz2+24π2iz−16π3 , A11(z) = e6πz2+6π2iz−2π3 cos(2iz) 6i(z + iπ)2 cos(2iz) + 2i sin(2iz) , we obtain µ(A20) = µ(A11) = 2 > max{ρ(Aij) : (i, j) 6= (2, 0), (1, 1)} = ρ(A00) = 0, λ( 1 A20 ) = 0 < µ(A20) = 2, τ(A20) = 12 > τ(A11) = 6. It is clear that the conditions of Theorem 1.6 are satisfied. The meromorphic function h(z) = e2iz3 cos(2iz) is a solution of equation (4.4) and satisfies ρ(h) = 3 ≥ µ(A20) + 1 = 3. Case 3: µ(Al0) = max{ρ(Aij) : (i, j) 6= (l, 0), (k, p)} > µ(Akp) with τ(Al0) >∑ ρ(Aij)=µ(Al0), (i,j)6=(l,0),(k,p) τ(Aij). In (4.4), for A00(z) = −eπz 2 , A20(z) = e13πz2+24π2iz−16π3 − e6πz2+18π2iz−14π3 , A11(z) = cos(2iz) 6i(z + iπ)2 cos(2iz) + 2i sin(2iz) , we obtain µ(A20) = max{ρ(Aij) : (i, j) 6= (2, 0), (1, 1)} = ρ(A00) = 2 > µ(A11) = 1, λ( 1 A20 ) = 0 < µ(A20) = 2, τ(A20) = 13 > ∑ ρ(Aij)=µ(Al0), (i,j)6=(l,0),(k,p) τ(Aij) = τ(A00) = 1. It is clear that the conditions of Theorem 1.6 are satisfied. The meromorphic function h(z) = e2iz3 cos(2iz) is a solution of (4.4) and satisfies ρ(h) = 3 ≥ µ(A20) + 1 = 3. Case 4: µ(Al0) = µ(Akp) = max{ρ(Aij) : (i, j) 6= (l, 0), (k, p)} with τ(Al0) >∑ ρ(Aij)=µ(Al0), (i,j)6=(l,0),(k,p) τ(Aij) + τ(Akp). In (4.4), for A00(z) = eπz 2 , A20(z) = −2e13πz2+24π2iz−16π3 , A11(z) = e7πz2+6π2iz−2π3 cos(2iz) 6i(z + iπ)2 cos(2iz) + 2i sin(2iz) , we have µ(A20) = µ(A11) = max{ρ(Aij) : (i, j) 6= (2, 0), (1, 1)} = ρ(A00) = 2, EJDE-2021/92 LOWER ORDER FOR MEROMORPHIC SOLUTIONS 19 λ( 1 A20 ) = 0 < µ(A20) = 2, τ(A20) = 13 > τ(A00) + τ(A11) = 1 + 7 = 8. It is easy to see that the conditions of Theorem 1.6 are satisfied. The meromorphic function h(z) = e2iz3 cos(2iz) is a solution of equation (4.4) and satisfies ρ(h) = 3 ≥ µ(A20) + 1 = 3. Acknowledgements. The authors would like to thank the anonymous referees for their insightful comments, which allow us to improve our original version. This work was supported by the Directorate-General for Scientific Research and Technological Development (DGRSDT). References [1] B. Beläıdi and R. Bellaama; Meromorphic solutions of higher order non-homogeneous linear difference equations. 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Rachid Bellaama Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem, Algeria Email address: rachidbellaama10@gmail.com Benharrat Beläıdi Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem, Algeria Email address: benharrat.belaidi@univ-mosta.dz 1. Introduction and statement of main results 2. Some preliminary lemmas 3. Proof of main results 4. Examples Acknowledgements References