On the Growth of Solutions of Some Second-Order Linear Differential Equations

In this paper, we willassume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna's value distribution theory of meromorphic functions (e.g., see [1–3]). In addition, we will use the notation to denote the order of growth of meromorphic function , to denote the hyper-order of (see [3]). is defined to be

(1.1)

We consider the second-order linear differential equation

(1.2)

where and are entire functions of finite order. It is well known that each solution of (1.2) is an entire function, and most solutions of (1.2) have infinite order.

Thus, a natural question is what conditions on and will guarantee that every solution of (1.2) has infinite order? Ozawa [4], Gundersen [5], Amemiya and Ozawa [6], and Langley [7] have studied the problem with and is complex number or polynomial. For the case that , and is transcendental entire function, Gundersen proved the following in [5, Theorem A].

Theorem A.

If is a transcendental entire function with order , then every solution of equation

(1.3)

has infinite order.

Theorem A states that when , (1.3) may have finite-order solutions. We go deep into the problem: what condition in when will guarantee every solution of (1.3) has infinite order? And more precise estimation for its rate of growth is a very important aspect. Chen investigated the problem and obtain the following in [8, Theorem B and Theorem C].

Theorem B.

Let be entire functions with be complex numbers such that and . then every solution of the equation

(1.4)

has infinite order.

Theorem C.

Let be nonzero complex numbers and be a nonconstant polynomial or where is nonzero polynomial, then every solution of the equation

(1.5)

has infinite order and .

For Theorems B and C, many authors, Wang and Lü [9],Huang, Chen, and Li [10], and Cheng and Kang [11] have made some improvement. In this paper, we areconcerned with the more general problem, and obtain the following theorem that extend and improve the previous results.

Theorem 1.1.

Let be entire functions with be complex numbers such that (suppose that . If or , then every solution of the equation

(1.6)

has infinite order and .

Lemma 2.1 (see [12]).

Let be a transcendental meromorphic function with be a finite set of distinct pairs of integers satisfying . And let be a given constant. Then,

(i)there exists a set with linear measure zero, such that, if , then there is a constant , such that for all satisfying and and for all , one has

(2.1)

(ii)there exists a set with finite logarithmic measure, such that for all satisfying and for all , we have

(2.2)

(iii)there exists a set with finite linear measure, such that for all satisfying and for all , we have

(2.3)

Lemma 2.2 (see [8]).

Suppose that are real numbers, is a polynomial with degree , that is an entire function with . Set . Then for any given , there exists a set that has the linear measure zero, such that for any , there is , such that for , we have

(i)if , then

(2.4)

(ii)if , then

(2.5)

where is a finite set.

Using Lemma 2.2, we can prove Lemma 2.3.

Lemma 2.3.

Suppose that is a positive entire number. Let be nonconstant polynomials, where are complex numbers and . Set , then there is a set that has linear measure zero. If , then there exists a ray , , such that

(2.6)

or

(2.7)

where is a finite set, which has linear measure zero.

Proof.

According to the values of and , we divide our discussion into three cases.

Case 1 ( ).

1. (a)

   If , let , Then there are three cases: (i) ; (ii) ; (iii) .

    
2. (i)

   . By , we know that .

    

Suppose that , then take is any constant in .

Since has linear measure zero, there exists such that . Thus . By and that is , we have

(2.8)

Therefore,

(2.9)

When , then , we can prove it by using similar argument action as in the above proof.

1. (ii)

   , then . Suppose that , then . Let , and take is any constant in .

    

Since has a linear measure zero, there exists such that ,

(2.10)

Therefore

(2.11)

Suppose that , then . Let , and take is any constant in .

Since has linear measure zero, there exists such that ,

(2.12)

Therefore,

(2.13)

Let , take is any constant in .

Since has a linear measure zero, there exists , such that . Then

(2.14)

When , , thus, .

When , , thus, .

Therefore

(2.15)

Case 2.

When , or and , using a proof similar to Case 1, we can get the conclusion.

Case 3 ( and ).

By , there are only two cases: ; or .

If . Take is any constant in .

Since has linear measure zero, there exists such that . Using a proof similar to Case 1(c), we can prove it.

When , we can prove it by using the same reasoning

Remark 2.4.

Using the similar reasoning of Lemma 2.3, we can obtain that, in Lemma 2.3, if is replaced by , then it has the same result.

Lemma 2.5 (see [8]).

Let be entire functions with finite order. If is a solution of the equation

(2.16)

then .

Lemma 2.6 (see [12]).

Let be a transcendental meromorphic function, and let be a given constant, Then there exists a set with finite logarithmic measure and a constant that depends only on and ( ), such that for all satisfying ,

(2.17)

Remark 2.7.

In Lemma 2.6, when , we have

(2.18)

Lemma 2.8 (see [13]).

Suppose that and are nondecreasing functions, such that , where is a set with at most finite measure, then for any constant , there exists such that for all .

Suppose that is a solution of (1.6), then, is an entire function.

First Step

We prove that . Suppose, to the contrary, that . By Lemma 2.1, for any given , there exists a set of linear measure zero, such that if , then, there is a constant , such that for all satisfying and , we have

(3.1)

Let .

Case 1 ( , which is ).

1. (i)
   Suppose that . By Lemmas 2.2 and 2.3, for the above , there is a ray , such that (where and are defined as in Lemma 2.3, and is of the linear measure zero), and satisfying

   

   (3.2)
    

or

(3.3)

When , for sufficiently large , we have

(3.4)

Hence

(3.5)

By (1.6), we obtain

(3.6)

Since , we know that , then . Substituting (3.1) and (3.5) into (3.6), we get

(3.7)

By , we know that (3.7) is a contradiction.

When , using a proof similar to the above, we can also get a contradiction.

1. (ii)
   Suppose that . By Lemma 2.2, for the above , there is a ray such that and . Since , and , then , thus . For sufficiently large , we have

   

   (3.8)
    

Hence,

(3.9)

where .

Since , we see that , then .

Since , we know that , then . Substituting (3.1) and (3.9) into (3.6), we obtain

(3.10)

Since , we know that (3.10) is a contradiction.

Case 2 ( , which is ).

1. (i)
   Suppose that , then . By Lemma 2.2, for the above , there is a ray such that and . Because . For sufficiently large , we have

   

   (3.11)
    

Hence

(3.12)

Since , then .

Using the same reasoning as in Case 1(ii), we can get a contradiction.

Concluding the above proof, we obtain .

Second Step

We prove that .

By Lemma 2.5 and , then .

By Lemma 2.6 and Remark 2.7, we know that there exists a set with finite logarithmic measure and a constant , such that for all satisfying , we get that (2.18) holds.

For Cases 1 and 2(i) in first step, we have proved that there is a ray satisfying , for sufficiently large , we get that (3.5) or (3.9) or (3.12) hold, that is,

(3.13)

where is a constant.

Since , then . By (2.18), (3.6), and (3.13), we obtain

(3.14)

By , (3.14) and Lemma 2.8, we know that there exists , when , we have , then .

For Case 2(ii) in first step, we have proved that there is a ray satisfying , for sufficiently large , we get (3.9) hold, and we also get that .

By (2.18), (3.6), and (3.9), we obtain

(3.15)

By and (3.15) and Lemma 2.8, we know that there exists , when , we have , then .

Concluding the above proof, we obtain .

Theorem 1.1 is thus proved.