On the Growth of Solutions of a Class of Higher Order Linear Differential Equations with Extremal Coefficients

and Applied Analysis 3 (iii) There exists a set E3 ⊂ [0,∞) that has finite linear measure, and there exists a constant c > 0 that depends only on α and Γ, such that, for all z satisfying |z| = r ∉ E3 and for all (k, j) ∈ Γ, we have 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 f (z) f (z) 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ≤ c(T (αr, f) r logT (αr, f)). (9) In particular, if f has finite order ρ(f), then (9) is replaced by 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 f (z) f (z) 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ≤ |z| . (10) Let f be an entire function extremal for Yang’s inequality p = q/2. Suppose that the rays arg z = θk (k = 1, 2, . . . , q) (0 ≤ θ1 < θ2 < ⋅ ⋅ ⋅ < θq < θq+1 = θ1 + 2π) are the q distinct Borel directions of order≥ μ off. In [17],Wu studied the entire functions which are extremal for Yang’s inequality systematically.The following results play an important role in the proof of our results. Lemma 7 (see [17]). Suppose that f is extremal for Yang’s inequality. Then μ(f) = ρ(f). Moreover, for every deficient value ai (i = 1, 2, . . . , p) there exists a corresponding angular domain Ω(θk i , θk i +1) = {z : θk i < arg z < θk i +1} such that for every ε > 0 inequality log 1 󵄨󵄨󵄨f (z) − ai 󵄨󵄨󵄨 > C (θk i , θk i +1, ε, δ (ai, f)) T (|z| , f) (11) holds for z ∈ Ω(θk i +ε, θk i +1−ε, r, +∞) = {z : θk i +ε < arg z < θk i +1 − ε}⋂{z : r < |z| < ∞}, where C(θk i , θk i +1, ε, δ(ai, f)) is a positive constant depending only on θk i , θk i +1, ε and δ(ai, f). In the sequel, we will say that f decays exponentially to ai in Ω(θk i , θk i +1), if (11) holds in Ω(θk i , θk i +1). Note that if f is extremal for Yang’s inequality, then μ(f) = ρ(f). Thus, for these functions, we need only to consider the Borel direction of order ρ(f). Lemma 8 (see [18]). Let f be extremal for Yang’s inequality. Suppose that there exists θ ∈ Ω (θj, θj+1) (1 ≤ j ≤ q) such that lim sup r→∞ loglog 󵄨󵄨󵄨󵄨f (re ) 󵄨󵄨󵄨󵄨 log r = ρ (f) , (12) where arg z = θj (j = 1, 2, . . . , q) are Borel directions of f. Then θj+1 − θj = π/ρ(f). Before stating the following lemmas, for E ⊂ [0,∞), we define the Lebesgue measure of E by mes(E) and the logarithmic measure of E ⊂ [1,∞) by ml(E) = ∫E(dt/t) and define the upper and lower logarithmic density ofE ⊂ [1,∞), respectively, by log densE = lim r→∞ ml (E⋂ [1, r]) log r , log densE = lim r→∞ ml (E⋂ [1, r]) log r . (13) Lemma 9 (see [25]). Let f be an entire function with ρ(f) = ρ < 1/2 and suppose thatm(r) is defined as m(r) = inf |z|=r log 󵄨󵄨󵄨f (z) 󵄨󵄨󵄨 . (14) If σ < ρ, then the set {r : m(r) > r} has a positive upper logarithmic density. Lemma 10 (see [26]). Let g(r) and h(r) be monotone nondecreasing functions on (0,∞) such that g(r) ≤ h(r) for all r outside some set of finite logarithmic measure. Let λ > 1 be a given real constant. Then there exists a constant r0 > 0 such that g(r) ≤ h(λr) for all r ≥ r0. Lemma 11. Let f(z) be an entire function with order ρ(0 < ρ < ∞), let and Ω(φ1, φ2) = {z : φ1 < arg z < φ2} be a sector with φ2 − φ1 < π/ρ. If there is Borel direction of f(z) in Ω(φ1, φ2), then there exists at least one of the two rays Lj : arg z = φj (j = 1, 2), without lose of generality, says, L2, such that lim sup r→∞ loglog 󵄨󵄨󵄨󵄨f (re iφ ) 󵄨󵄨󵄨󵄨 log r = ρ. (15) Lemma 11 can be founded in [27, Lemma 1], which can be proved by using a result in [28, Page 119-120]. 3. Proof of Theorem 5 Now we prove our main result. Since ρ(A i) < ρ(A0) (i ̸ = l, 1 ≤ i ≤ k − 1), we know that for any given constant ηwithmax{ρ(A i), i ̸ = l, 1 ≤ i ≤ k−1} < η < ρ(A0), there exists a constant R1 > 0 such that 󵄨󵄨󵄨A i (z) 󵄨󵄨󵄨 ≤ exp (r ) (16) holds for all |z| = r > R1. We consider the following two cases. Case 1. We suppose that ρ(A l) < ρ(A0). Now to the contrary assume that there is a solutionf( ̸ ≡ 0) of (5) with ρ(f) < +∞. We will seek a contradiction. By Lemma 6(ii), there exists a set E1 ⊂ [1, +∞) that has finite logarithmic measure, such that the following inequality 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 f (z) f (z) 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ≤ |z| , m = 1, 2, . . . , k (17) holds for all z with |z| = r ∉ E1⋃[0, 1]. We deduce from (17) and (5) that 󵄨󵄨󵄨A0 (z) 󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 f f 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨Ak−1 (z) 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 f f 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 + ⋅ ⋅ ⋅ + 󵄨󵄨󵄨A l (z) 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 f f 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 + ⋅ ⋅ ⋅ + 󵄨󵄨󵄨A1 (z) 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 f f 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ≤ |z| kρ(f) (1 + 󵄨󵄨󵄨Ak−1 (z) 󵄨󵄨󵄨 + ⋅ ⋅ ⋅ + 󵄨󵄨󵄨A l (z) 󵄨󵄨󵄨 + ⋅ ⋅ ⋅ + 󵄨󵄨󵄨A1 (z) 󵄨󵄨󵄨) (18) holds for all z with |z| = r ∉ E1⋃[0, 1]. 4 Abstract and Applied Analysis


Introduction and Main Results
We will assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna theory of meromorphic functions (see [1,2] or [3]).In addition, for a meromorphic function  in the complex plane C, we will use the notations () and () to denote its order and lower order, respectively.
In order to estimate the rate of growth of meromorphic function of infinite order more precisely, we recall the following definition.
Definition 1 (see [4]).Let  be a meromorphic function in the complex plane C. Then one defines the hyperorder  2 () of  by Consider the second order linear differential equation where  and ( ̸ ≡ 0) are entire functions.It is well known that if  is an entire function, ( ̸ ≡ 0) is a transcendental entire function, and  1 ,  2 are two linearly independent solutions of (2), then at least one of  1 ,  2 must have infinite order.On the other hand, there are some equations of form (2) that possess a solution ( ̸ ≡ 0) of finite order; for example, () =   satisfies   +  −   − ( − + 1) = 0. Therefore, one may ask, what assumptions on () and () will guarantee that every solution  ̸ ≡ 0 of (2) is of infinite order?From the works of Gundersen (see [5]) and Hellerstein et al. (see [6]), we know that if () and () are entire functions with () < (), or () is a polynomial, and () is transcendental, or () < () ≤ 1/2, then every solution ( ̸ ≡ 0) of ( 2) is of infinite order.More results can be found in [7][8][9][10][11][12].For entire solutions of infinite order more precise estimates for their rate of growth would be an important achievement.There are many authors investigating the hyperorder  2 () of solutions of (2), such as Chen and Yang (see [8,13]) and Kwon (see [14,15]).
The fundamental result in angular distribution, due to Valiron, says that a meromorphic function of order  > 0 must have at least one Borel direction of order ; for example, see [3].
It is well known that deficient values and Borel directions are very important concepts in Nevanlinna theory of meromorphic functions.These two concepts are extensively studied.There is a striking relationship between them which was found by Yang and Zhang and says that, for a meromorphic function  of order , the number of deficient values is less or equal to the number of Borel directions of order  of .In 1988, Yang extended the above - inequality to the case of entire function of finite lower order.In order to use Yang's result to study the complex differential equations, we will use the following Theorem which can be easily derived from [16].
Theorem 3 (see [16]).Suppose that  is an entire function of finite lower order  > 0. Let  (< ∞) denote the number of Borel directions of order ≥  and  denote the number of finite deficient values of ; then  ≤ /2.
Note that Theorem 3 is explicitly stated in [17].To see the valid of the conclusion of the theorem, we note that, in [17, Corollary 1], Wu has proved that if () is of finite lower order  and the number of Borel directions of order ≥  is finite, then the order  of () is also finite.As each Borel direction of order  is also a Borel direction of order ≥ , this implies that, for (), the number of the Borel directions of order  is fewer or equal to the number of the Borel directions of order ≥ .Therefore Theorem 3 follows from Theorem 6.7 in [3].
In the sequel, we will say that an entire function  is extremal for Yang's inequality if  satisfies the assumptions of Theorem 3 with  = /2.
Theorem 5. Let   ( = 0, 1, . . .,  − 1) be entire functions.Suppose that there exists an integer  ∈ {1, 2, . . .,  − 1}, such that   () is extremal for Yang's inequality.Suppose that  0 is a transcendental entire function with ( 0 ) ̸ = (  ) and The paper is organized as follows.In Section 2, we will give some lemmas.In Section 3, we will prove Theorem 5.In Section 4, we will discuss some further results related to the two entire coefficients in (5) which are extremal for Yang's inequality.

Lemmas
In this section, we need some auxiliary results.The following lemma is by Gundersen.
(iii) There exists a set  3 ⊂ [0, ∞) that has finite linear measure, and there exists a constant  > 0 that depends only on  and Γ, such that, for all  satisfying || =  ∉  3 and for all (, ) ∈ Γ, we have           () ()  () ()          ≤ ( (, )   log  (, )) In particular, if  has finite order (), then (9) is replaced by Let  be an entire function extremal for Yang's inequality  = /2.Suppose that the rays arg  =   ( = 1, 2, . . ., ) In [17], Wu studied the entire functions which are extremal for Yang's inequality systematically.The following results play an important role in the proof of our results.

Proof of Theorem 5
Now we prove our main result.
By using similar methods as [14], we can easily prove that  2 () ≥ ( 0 ) in this case.We omit the details here.
Suppose that   ( = 1, 2, . . ., ) are all the finite deficient values of   ().Without loss of generality, we assume that there is a ray arg  =  in  1 such that (21) holds.Therefore, there exists a ray in each sector  3 ,  5 , . . .,  2−1 , such that (21) holds.By using Lemma 8, we know that all the sectors have the same magnitude /(  ).
Proof.We first treat the case that the entire functions   and  0 satisfy condition (1).
We turn to the case that   () and  0 () satisfy (2).In this case, it is easy to see that there exists a sector such that in it   () is bounded, while  0 () satisfies (34).By using similar arguments as we did before, we can prove the theorem in this case.We omit the details here.The proof of Theorem 12 is completed.
Finally, in [16], we note that if an entire function  is extremal for Yang's inequality  = /2, then for any positive integer ,  () also has some special properties.In the sectors where, for any ,  satisfies lim sup decays to some deficient values exponentially and  ()  decays to 0 exponentially.Therefore, in the same manner as in the proofs of Theorems 5 and 12, we have the following result.Moreover, suppose that  0 is an entire function extremal for Yang's inequality  2 =  2 /2 and that one of the following assumptions holds: (1)  1 ̸ =  2 , (2)  1 =  2 , and the set of Borel directions of   is different from that of  0 .
2 and (b)  1 >  2 .Now suppose that (a)  1 <  2 holds.It is easy to see from Lemmas 7 and 8 that there are  2 /2 sectors with magnitude / such that log + log +        (  )