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On the growth of solutions of certain higher-order linear differential equations

Advances in Difference Equations20142014:33

https://doi.org/10.1186/1687-1847-2014-33

Received: 6 November 2013

Accepted: 8 January 2014

Published: 27 January 2014

Abstract

In this paper, we investigate the growth of meromorphic solutions of the equations

f ( k ) + A k 1 ( z ) f ( k 1 ) + + A 1 ( z ) f + A 0 ( z ) f = 0 , f ( k ) + A k 1 ( z ) f ( k 1 ) + + A 1 ( z ) f + A 0 ( z ) f = F ( z ) ,

where A 0 ( z ) (0), A 1 ( z ) , , A k 1 ( z ) and F ( z ) (0) are entire functions of finite order. We find some conditions on the coefficients to guarantee that every nontrivial meromorphic solution of such equations is of infinite order.

MSC:30D35, 34M10.

Keywords

  • entire function
  • differential equation
  • order

1 Introduction and main results

It is assumed that the reader is familiar with the standard notations and the fundamental results of the Nevanlinna theory [13]. Let f ( z ) be a nonconstant meromorphic function in the complex plane. We use notations σ ( f ) and σ 2 ( f ) to denote the order of growth and the hyper-order of f, which are defined by
σ ( f ) = lim r log + T ( r , f ) log r , σ 2 ( f ) = lim r log + log + T ( r , f ) log r ,

respectively.

Consider the linear differential equation
f ( k ) + A k 1 ( z ) f ( k 1 ) + + A 1 ( z ) f + A 0 ( z ) f = 0 ,
(1.1)
where A 0 ( z ) (0), A 1 ( z ) , , A k 1 ( z ) are entire functions, if the coefficients A j ( z ) ( j = 0 , , k 1 ) are polynomials, then all nontrivial solutions of (1.1) are of finite order (see [2, 4]). If s is the largest integer such that A s ( z ) is transcendental, it is well known that (1.1) has at most s linearly independent finite-order solutions. Thus when at least one of the coefficients A j ( z ) is transcendental, most of the solutions of (1.1) are of infinite order. In the case when
max 1 j k 1 { σ ( A j ) } < σ ( A 0 ) < + ,
Chen and Yang [5] proved that all nontrivial solutions of (1.1) are of infinite order. In the case when
max j d { σ ( A j ) } < σ ( A d ) 1 2 ,

Hellerstein et al. [6] proved that all transcendental solutions of (1.1) are of infinite order. While in the case when σ ( A 0 ) = σ ( A 1 ) = = σ ( A k 1 ) , (1.1) may have a solution of finite order.

Example 1.1 The equation
f + ( e z 1 ) f e 2 z f + ( e 2 z e z ) f = 0

has a solution f ( z ) = e z of σ ( f ) = 1 . Here σ ( e z 1 ) = σ ( e 2 z ) = σ ( e 2 z e z ) = 1 .

Thus a natural question is what conditions on A j ( z ) when σ ( A 0 ) = σ ( A 1 ) = = σ ( A k 1 ) will guarantee that all nontrivial solutions of (1.1) are of infinite order. Concerning this question, we recall the following results as for the special case of k = 2 .

Theorem A (see [7])

Let P ( z ) = a n z n +  , Q ( z ) = b n z n + be polynomials with degree n (≥1), h 0 ( z ) (0), h 1 ( z ) be entire functions of order less than n. If arg a n arg b n or a n = c b n ( 0 < c < 1 ), then every nontrivial solution f of the equation
f + h 1 ( z ) e P ( z ) f + h 0 ( z ) e Q ( z ) f = 0

satisfies σ ( f ) = and σ 2 ( f ) n .

Theorem B (see [8])

Let h j ( z ) (0) ( j = 0 , 1 ) be entire functions of order less than 1, a, b be nonzero complex numbers such that arg a arg b or a = c b ( c > 1 ). Then every nontrivial solution f of the equation
f + h 1 ( z ) e a z f + h 0 ( z ) e b z f = 0
(1.2)

satisfies σ ( f ) = .

From Theorems A and B, we obtain that if a b , then every nontrivial solution f of Eq. (1.2) is of infinite order. When the coefficient of f is of the form h 01 ( z ) e a 1 z + h 02 ( z ) e a 2 z , where a 1 a 2 , Peng and Chen obtained the following result.

Theorem C (see [9])

Let h 0 j ( z ) (0) ( j = 1 , 2 ) be entire functions of order less than 1, a 1 , a 2 be nonzero complex numbers and a 1 a 2 (suppose that | a 1 | | a 2 | ). If arg a 1 π or a 1 < 1 , then every nontrivial solution f of the equation
f + e z f + ( h 01 ( z ) e a 1 z + h 02 ( z ) e a 2 z ) f = 0
(1.3)

satisfies σ ( f ) = and σ 2 ( f ) = 1 .

In this paper, we continue to investigate the growth of solutions of higher linear differential equations, which has the same form as Eq. (1.3), and obtain the following results which generalize Theorem C.

Theorem 1.1 Suppose that h 01 ( z ) (0), h 02 ( z ) , h j ( z ) ( j = 1 , , k 1 ) are entire functions of order less than n, P i ( z ) = a i n z n + + a i 0 , Q j ( z ) = b j n z n + + b j 0 ( i = 1 , 2 ; j = 1 , , k 1 ) are polynomials with degree n (≥1), where a i l , b j l ( i = 1 , 2 ; j = 1 , , k 1 ; l = 0 , , n ) are complex constants. If a 1 n a 2 n , b j n ( j = 1 , , k 1 ) satisfies the following conditions:
  1. (i)

    there exists some s { 1 , , k 1 } such that arg b s n arg a 1 n ;

     
  2. (ii)

    b j n = c j a 1 n ( 0 < c j < 1 ) for j I 1 and b j n = d j b s n ( 0 < d j < 1 ) for j I 2 , where I 1 I 2 = { 1 , , k 1 } { s } , I 1 I 2 = ,

     
then every solution f (0) of the equation
f ( k ) + h k 1 ( z ) e Q k 1 ( z ) f ( k 1 ) + + h 1 ( z ) e Q 1 ( z ) f + ( h 01 ( z ) e P 1 ( z ) + h 02 ( z ) e P 2 ( z ) ) f = 0
(1.4)

satisfies σ ( f ) = and σ 2 ( f ) = n .

From the proof of Theorem 1.1, we also obtain the following results.

Corollary 1.1 Suppose that h 01 ( z ) (0), h 02 ( z ) , h j ( z ) ( j = 1 , , k 1 ) are entire functions of order less than n, P i ( z ) = a i n z n + + a i 0 , Q j ( z ) = b j n z n + + b j 0 ( i = 1 , 2 ; j = 1 , , k 1 ) are polynomials with degree n (≥1). If a 1 n a 2 n and there exists some s { 1 , , k 1 } such that arg b s n arg a 1 n , b j n = d j b s n ( 0 < d j < 1 , 1 j k 1 , j s ), then every solution f (0) of (1.4) satisfies σ ( f ) = and σ 2 ( f ) = n .

Corollary 1.2 Suppose that h 01 ( z ) (0), h 02 ( z ) , h j ( z ) ( j = 1 , , k 1 ) are entire functions of order less than n, P i ( z ) = a i n z n + + a i 0 , Q j ( z ) = b j n z n + + b j 0 ( i = 1 , 2 ; j = 1 , , k 1 ) are polynomials with degree n (≥1). If a 1 n a 2 n and b j n = c j a 1 n ( 0 < c j < 1 , 1 j k 1 ), then every solution f (0) of (1.4) satisfies σ ( f ) = and σ 2 ( f ) = n .

Remark 1.1 Corollaries 1.1 and 1.2 are extensions of Theorem C, because Theorem C is just the case for k = 2 , n = 1 , b 1 n = 1 in Corollaries 1.1 and 1.2.

Remark 1.2 From Theorem 1.1 and Corollaries 1.1, 1.2, we obtain that every solution f (0) of the equation
f ( k ) + h k 1 ( z ) e Q k 1 ( z ) f ( k 1 ) + + h 1 ( z ) e Q 1 ( z ) f + B ( z ) cos ( P ( z ) ) f = 0
or
f ( k ) + h k 1 ( z ) e Q k 1 ( z ) f ( k 1 ) + + h 1 ( z ) e Q 1 ( z ) f + B ( z ) sin ( P ( z ) ) f = 0

satisfies σ ( f ) = and σ 2 ( f ) = n , where P ( z ) = i P 1 , B ( z ) is an entire function of order less than n.

Recently, Wang and Laine investigated the growth of solutions of the non-homogeneous linear differential equation
f + h 1 ( z ) e a z f + h 0 ( z ) e b z f = F ( z )
(1.5)

corresponding to (1.2) and obtained the following result.

Theorem D (see [10])

Suppose that h 0 (0), h 1 (0), F (0) are entire functions of order less than 1, and the complex constants a, b satisfy a b 0 and a b . Then every nontrivial solution f of Eq. (1.5) is of infinite order.

Thus the other purpose of this paper is to investigate the growth of solutions of the non-homogeneous linear differential equation
f ( k ) + h k 1 ( z ) e Q k 1 ( z ) f ( k 1 ) + + h 1 ( z ) e Q 1 ( z ) f + ( h 01 ( z ) e P 1 ( z ) + h 02 ( z ) e P 2 ( z ) ) f = F ( z )
(1.6)

corresponding to (1.4). We obtain the following results.

Theorem 1.2 Under the hypotheses of Theorem  1.1, if b s n a 2 n , F ( z ) (0) is an entire function of order less than n, then every solution f of Eq. (1.6) satisfies σ ( f ) = .

From the proof of Theorem 1.2, we also obtain the following results.

Corollary 1.3 Suppose that h 01 ( z ) (0), h 02 ( z ) , h j ( z ) ( j = 1 , , k 1 ), F ( z ) (0) are entire functions of order less than n, P i ( z ) = a i n z n + + a i 0 , Q j ( z ) = b j n z n + + b j 0 ( i = 1 , 2 ; j = 1 , , k 1 ) are polynomials with degree n (≥1). If a 1 n a 2 n , and there exists some s { 1 , , k 1 } such that arg b s n arg a 1 n , b s n a 2 n , b j n = d j b s n ( 0 < d j < 1 , 1 j k 1 , j s ), then every solution f of Eq. (1.6) satisfies σ ( f ) = .

Corollary 1.4 Suppose that h 01 ( z ) (0), h 02 ( z ) , h j ( z ) ( j = 1 , , k 1 ), F ( z ) (0) are entire functions of order less than n, P i ( z ) = a i n z n + + a i 0 , Q j ( z ) = b j n z n + + b j 0 ( i = 1 , 2 ; j = 1 , , k 1 ) are polynomials with degree n (≥1). If a 1 n a 2 n and b j n = c j a 1 n ( 0 < c j < 1 , 1 j k 1 ), then every solution f of Eq. (1.6) satisfies σ ( f ) = .

Remark 1.3 When σ ( F ) n , (1.6) may have a solution of finite order.

Example 1.2 The equation
f + e z f + e z f + ( e 2 z 2 e z ) f = ( e 2 z + 2 z e z + 4 z 2 e z + 8 z 3 + 12 z ) e z 2

has a solution f ( z ) = e z 2 of σ ( f ) = 2 . Here P 1 ( z ) = 2 z , P 2 ( z ) = z , Q 1 ( z ) = z , Q 2 ( z ) = z = Q s ( z ) satisfy the hypotheses of Theorem 1.1, and σ ( F ) = 2 > n = 1 .

Example 1.3 The equation
f ( 4 ) + e 2 z f e z f + e z f + ( e 2 z + e z ) f = e 3 z + e 2 z + e z + e z

has a solution f ( z ) = e z of σ ( f ) = 1 . Here P 1 ( z ) = 2 z , P 2 ( z ) = z , Q 1 ( z ) = z , Q 2 ( z ) = z , Q 3 ( z ) = 2 z = Q s ( z ) satisfy the hypotheses of Theorem 1.1, and σ ( F ) = 1 = n .

2 Lemmas

Lemma 2.1 ([11])

Let f j ( j = 1 , , n ) and g j ( j = 1 , , n ) be entire functions such that
  1. (i)

    j = 1 n f j ( z ) e g j ( z ) 0 ,

     
  2. (ii)

    the order of f j is less than the order of e g h g k for n 2 and 1 j n , 1 h < k n .

     

Then f j ( z ) 0 ( j = 1 , , n ).

Lemma 2.2 ([11])

Let f j ( j = 1 , , n + 1 ) and g j ( j = 1 , , n ) be entire functions such that
  1. (i)

    j = 1 n f j ( z ) e g j ( z ) f n + 1 ( z ) ,

     
  2. (ii)

    the order of f j is less than the order of e g k for 1 j n + 1 , 1 k n ; and, furthermore, the order of f j is less than the order of e g h g k for n 2 and 1 j n + 1 , 1 h < k n .

     

Then f j ( z ) 0 ( j = 1 , , n + 1 ).

Lemma 2.3 ([12])

Let f ( z ) be a transcendental meromorphic function of σ ( f ) < , k, j ( k > j 0 ) be integers. Then, for any given ε > 0 , there exists a set E [ 0 , 2 π ) of linear measure zero such that for all z = r e i θ with | z | sufficiently large and θ [ 0 , 2 π ) E , we have
| f ( k ) ( z ) f ( j ) ( z ) | | z | ( k j ) ( σ ( f ) 1 + ε ) .

Lemma 2.4 ([8, 13])

Let P ( z ) = ( α + i β ) z n + (α, β are real numbers, | α | + | β | 0 ) be a polynomial with degree n (≥1), A ( z ) (0) be an entire function with σ ( A ) < n . Set g ( z ) = A ( z ) e P ( z ) , z = r e i θ , δ ( P , θ ) = α cos n θ β sin n θ . Then, for any given ε > 0 , there is a set E 0 [ 0 , 2 π ) that has linear measure zero such that for any θ [ 0 , 2 π ) ( E 0 E 1 ) and sufficiently large r, we have
  1. (i)
    if δ ( P , θ ) > 0 , then
    exp { ( 1 ε ) δ ( P , θ ) r n } | g ( r e i θ ) | exp { ( 1 + ε ) δ ( P , θ ) r n } ;
    (2.1)
     
  2. (ii)
    if δ ( P , θ ) < 0 , then
    exp { ( 1 + ε ) δ ( P , θ ) r n } | g ( r e i θ ) | exp { ( 1 ε ) δ ( P , θ ) r n } ,
    (2.2)
     

where E 1 = { θ [ 0 , 2 π ) : δ ( P , θ ) = 0 } is a finite set.

Lemma 2.5 ([12])

Let f ( z ) be a transcendental meromorphic function, and let α > 1 be a given constant. Then there exist a set H ( 1 , ) that has a finite logarithmic measure and a constant B > 0 depending only on α and k, j ( k > j 0 ) such that for all z with | z | = r [ 0 , 1 ] H , we have
| f ( k ) ( z ) f ( j ) ( z ) | B ( T ( α r , f ) r ( log α r ) log T ( α r , f ) ) k j .
Lemma 2.6 Let Q j ( z ) = b j n z n + + b j 0 ( j = 1 , 2 , 3 ) be polynomials with degree n (≥1), where b j l ( j = 1 , 2 , 3 ; l = 0 , , n ) are complex constants. Set z = r e i θ , arg b j n = φ j [ 0 , 2 π ) , δ ( Q j , θ ) = | b j n | cos ( φ j + n θ ) ( j = 1 , 2 , 3 ). If φ j ( j = 1 , 2 , 3 ) are distinct, then there exist two real numbers θ 1 , θ 2 ( θ 1 < θ 2 ) such that for each θ ( θ 1 , θ 2 ) , we have
δ ( Q 1 , θ ) δ ( Q 2 , θ ) < 0 and δ ( Q 3 , θ ) < 0 .
Proof Set
Ω j k = { z = r e i θ : θ j k < θ < θ j ( k + 1 ) } ( k = 0 , 1 , , 2 n 1 , j = 1 , 2 , 3 ) ,
where θ j k = ( 2 k 1 ) π 2 n φ j n . Then the following results hold.
  1. (i)

    δ ( Q j , θ j k ) = 0 ( k = 0 , 1 , , 2 n 1 , j = 1 , 2 , 3 ).

     
  2. (ii)

    δ ( Q j , θ ) > 0 holds for z Ω j k and k is even, δ ( Q j , θ ) < 0 holds for z Ω j k and k is odd.

     
  3. (iii)

    θ j ( k + 1 ) θ j k = π n ( k = 0 , 1 , , 2 n 1 , j = 1 , 2 , 3 ).

     

Next we discuss the following four cases.

Case 1. | φ i φ j | π ( 1 i < j 3 ). By (ii), among the angular domains Ω 1 k , we may take an angular domain Ω 1 k 0 such that for z Ω 1 k 0 , δ ( Q 1 , θ ) > 0 . Hence when z Ω 1 ( k 0 1 ) Ω 1 ( k 0 + 1 ) , we have δ ( Q 1 , θ ) < 0 . For the sake of convenience, we write
Ω 1 k 0 = { z = r e i θ : α < θ < β } .
Then by (iii) we write
Ω 1 ( k 0 1 ) = { z = r e i θ : α π n < θ < α } , Ω 1 ( k 0 + 1 ) = { z = r e i θ : β < θ < β + π n } .
Since | φ i φ j | π , by (i) and (iii) we know that there exist two distinct rays
arg z = γ 2 Ω 1 k 0 , arg z = γ 3 Ω 1 k 0
such that
δ ( Q 2 , γ 2 ) = 0 , δ ( Q 3 , γ 3 ) = 0 ,

respectively. Hence by (ii) we obtain that δ ( Q i , θ ) δ ( Q i , θ ) < 0 for any θ ( α , γ i ) , θ ( γ i , β ) and i = 2 , 3 . Without loss of generality, we assume that γ 2 < γ 3 . Now we discuss the following four subcases.

Subcase 1. If δ ( Q 2 , θ ) < 0 when θ ( α , γ 2 ) and δ ( Q 3 , θ ) < 0 when θ ( α , γ 3 ) , then we take θ 1 = α , θ 2 = γ 2 . Hence when θ ( θ 1 , θ 2 ) , we have δ ( Q 1 , θ ) > 0 , δ ( Q 2 , θ ) < 0 and δ ( Q 3 , θ ) < 0 .

Subcase 2. If δ ( Q 2 , θ ) < 0 when θ ( α , γ 2 ) and δ ( Q 3 , θ ) > 0 when θ ( α , γ 3 ) , then we take θ 1 = β , θ 2 = γ 2 + π n . Hence when θ ( θ 1 , θ 2 ) , we have δ ( Q 1 , θ ) < 0 , δ ( Q 2 , θ ) > 0 and δ ( Q 3 , θ ) < 0 .

Subcase 3. If δ ( Q 2 , θ ) > 0 when θ ( α , γ 2 ) and δ ( Q 3 , θ ) < 0 when θ ( α , γ 3 ) , then we take θ 1 = γ 3 π n , θ 2 = α . Hence when θ ( θ 1 , θ 2 ) , we have δ ( Q 1 , θ ) < 0 , δ ( Q 2 , θ ) > 0 and δ ( Q 3 , θ ) < 0 .

Subcase 4. If δ ( Q 2 , θ ) > 0 when θ ( α , γ 2 ) and δ ( Q 3 , θ ) > 0 when θ ( α , γ 3 ) , then we take θ 1 = γ 3 , θ 2 = β . Hence when θ ( θ 1 , θ 2 ) , we have δ ( Q 1 , θ ) > 0 , δ ( Q 2 , θ ) < 0 and δ ( Q 3 , θ ) < 0 .

Case 2. | φ 1 φ 2 | = π . Since φ j [ 0 , 2 π ) and φ j ( j = 1 , 2 , 3 ) are distinct, we obtain that
| φ 3 φ 1 | π , | φ 3 φ 2 | π .
(2.3)
By (ii), among the angular domains Ω 1 k , we may take an angular domain Ω 1 k 0 = { z = r e i θ : θ 1 k 0 < θ < θ 1 ( k 0 + 1 ) } such that for z Ω 1 k 0 , δ ( Q 1 , θ ) > 0 . Note that
δ ( Q 2 , θ ) = | b 2 n | cos ( φ 2 + n θ ) = | b 2 n | cos ( φ 1 + n θ ) ,

so when z Ω 1 k 0 , we have δ ( Q 2 , θ ) < 0 . By (2.3), (i) and (iii), we know that there exists a ray arg z = γ 3 Ω 1 k 0 such that δ ( Q 3 , γ 3 ) = 0 . Hence by (ii) we obtain that δ ( Q 3 , θ ) < 0 for θ ( θ 1 k 0 , γ 3 ) or δ ( Q 3 , θ ) < 0 for θ ( γ 3 , θ 1 ( k 0 + 1 ) ) . Hence we may take ( θ 1 , θ 2 ) = ( θ 1 k 0 , γ 3 ) or ( θ 1 , θ 2 ) = ( γ 3 , θ 1 ( k 0 + 1 ) ) such that for θ ( θ 1 , θ 2 ) , δ ( Q 1 , θ ) > 0 , δ ( Q 2 , θ ) < 0 and δ ( Q 3 , θ ) < 0 .

Case 3. | φ 1 φ 3 | = π . Then, by a similar reasoning to that of Case 2, we can prove the result.

Case 4. | φ 2 φ 3 | = π . Since φ j [ 0 , 2 π ) and φ j ( j = 1 , 2 , 3 ) are distinct, we obtain that
| φ 1 φ 2 | π , | φ 1 φ 3 | π .
(2.4)
By (ii), among the angular domains Ω 2 k , we may take an angular domain Ω 2 k 0 = { z = r e i θ : θ 2 k 0 < θ < θ 2 ( k 0 + 1 ) } such that for z Ω 2 k 0 , δ ( Q 2 , θ ) > 0 . Note that
δ ( Q 3 , θ ) = | b 3 n | cos ( φ 3 + n θ ) = | b 3 n | cos ( φ 2 + n θ ) ,

so when z Ω 2 k 0 , we have δ ( Q 3 , θ ) < 0 . By (2.4), (i) and (iii), we know that there exists a ray arg z = γ 1 Ω 2 k 0 such that δ ( Q 1 , γ 1 ) = 0 . Hence by (ii) we obtain that δ ( Q 1 , θ ) < 0 for θ ( θ 2 k 0 , γ 1 ) or δ ( Q 1 , θ ) < 0 for θ ( γ 1 , θ 2 ( k 0 + 1 ) ) . Hence we may take ( θ 1 , θ 2 ) = ( θ 2 k 0 , γ 1 ) or ( θ 1 , θ 2 ) = ( γ 1 , θ 2 ( k 0 + 1 ) ) such that for θ ( θ 1 , θ 2 ) , δ ( Q 1 , θ ) < 0 , δ ( Q 2 , θ ) > 0 and δ ( Q 3 , θ ) < 0 . □

Remark 2.1 From the proof of Lemma 2.6, we also obtain the following result.

Let Q j ( z ) = b j n z n + + b j 0 ( j = 1 , 2 ) be polynomials with degree n (≥1), where b j l ( j = 1 , 2 ; l = 0 , , n ) are complex constants. Set z = r e i θ , arg b j n = φ j [ 0 , 2 π ) , δ ( Q j , θ ) = | b j n | cos ( φ j + n θ ) . If φ 1 φ 2 , then there exist two real numbers θ 1 , θ 2 ( θ 1 < θ 2 ) such that for each θ ( θ 1 , θ 2 ) , we have
δ ( Q 1 , θ ) > 0 and δ ( Q 2 , θ ) < 0 .

Lemma 2.7 ([2])

Let g : ( 0 , ) R and h : ( 0 , ) R be monotone nondecreasing functions such that g ( r ) h ( r ) outside of an exceptional set H of finite logarithmic measure. Then, for any α > 1 , there exists r 0 > 0 such that g ( r ) h ( r α ) holds for all r > r 0 .

Lemma 2.8 ([14])

Let A j ( z ) ( j = 0 , , k 1 ) be entire functions of finite order. If f ( z ) (0) is a solution of the equation
f ( k ) + A k 1 f ( k 1 ) + + A 0 f = 0 ,

then σ 2 ( f ) max 0 j k 1 { σ ( A j ) } .

Lemma 2.9 ([15])

Let f ( z ) be an entire function and β > 0 . If G ( z ) = log + | f ( k ) ( z ) | | z | β is unbounded on some ray arg z = θ , then there exists an infinite sequence of points z m = r m e i θ ( m = 1 , 2 , ) with r m such that
G ( z m )
and
| f ( j ) ( z m ) f ( k ) ( z m ) | 1 ( k j ) ! ( 1 + o ( 1 ) ) r m k j ( j = 0 , , k 1 )

as m .

Lemma 2.10 ([15])

Let f ( z ) be an entire function of σ ( f ) < . If there exists a set E [ 0 , 2 π ) which has linear measure zero such that log + | f ( r e i θ ) | M r α for any ray arg z = θ [ 0 , 2 π ) E , where M is a positive constant depending on θ, while α is a positive constant independent of θ, then σ ( f ) α .

3 Proofs of the results

Proof of Theorem 1.1 Let f (0) be a solution of (1.4), then f is an entire function.

Step 1. We prove that σ ( f ) n . Suppose that σ ( f ) < n , rewrite (1.4) in the form
l B l e d j l Q s + h s f ( s ) e Q s + q B q e c j q P 1 + h 01 f e P 1 + h 02 f e P 2 = f ( k ) ,
(3.1)

where c j q { c j : j I 1 } and c j q are distinct, d j l { d j : j I 2 } and d j l are distinct, B l , B q are entire functions of order less than n. Now we discuss the relations between the coefficients of the term z n of polynomials d j l Q s , Q s , c j q P 1 , P 1 and P 2 .

Case 1. There exists some l such that deg ( d j l Q s P 2 ) < n . Then merging the terms B l e d j l Q s and h 02 f e P 2 , by (3.1) we get
l B l e d j l Q s + h s f ( s ) e Q s + q B q e c j q P 1 + h 01 f e P 1 = f ( k ) ,

where B l are entire functions of order less than n. By the hypothesis of Theorem 1.1, we know that the coefficients of the term z n of polynomials d j l Q s , Q s , c j q P 1 , P 1 are distinct. Then, by Lemma 2.1 or Lemma 2.2, we get h 01 f 0 . This is absurd.

Case 2. deg ( Q s P 2 ) < n . Then merging the terms h s f ( s ) e Q s and h 02 f e P 2 , by (3.1) we get
l B l e d j l Q s + D s e Q s + q B q e c j q P 1 + h 01 f e P 1 = f ( k ) ,

where D s is an entire function of order less than n. By the hypothesis of Theorem 1.1, we know that the coefficients of the term z n of polynomials d j l Q s , Q s , c j q P 1 , P 1 are distinct. Then, by Lemma 2.1 or Lemma 2.2, we get h 01 f 0 . This is absurd.

Case 3. There exists some q such that deg ( c j q P 1 P 2 ) < n . Then merging the terms B q e c j q P 1 and h 02 f e P 2 , by (3.1) we get
l B l e d j l Q s + h s f ( s ) e Q s + q B q e c j q P 1 + h 01 f e P 1 = f ( k ) ,

where B q are entire functions of order less than n. By the hypothesis of Theorem 1.1, we know that the coefficients of the term z n of polynomials d j l Q s , Q s , c j q P 1 , P 1 are distinct. Then by Lemma 2.1 or Lemma 2.2, we get h 01 f 0 . This is absurd.

Step 2. We prove that σ ( f ) = and σ 2 ( f ) n . By Step 1 we know that f is transcendental. Then by Lemma 2.5 there exists a set H ( 1 , ) that has a finite logarithmic measure, and a constant B > 0 such that for all z with | z | = r [ 0 , 1 ] H , we have
| f ( j ) ( z ) f ( z ) | B ( T ( 2 r , f ) ) k + 1 ( j = 1 , , k 1 ) .
(3.2)
Set
z = r e i θ , δ ( P i , θ ) = | a i n | cos ( arg a i n + n θ ) ( i = 1 , 2 ) , δ ( Q j , θ ) = | b j n | cos ( arg b j n + n θ ) ( j = 1 , , k 1 ) .

Next we discuss the following three cases.

Case 1. arg a 1 n arg a 2 n and arg a 2 n arg b s n . By the hypothesis of Theorem 1.1, we know that arg a 1 n , arg a 2 n , arg b s n are distinct. Then by Lemma 2.6 there exist two real numbers θ 1 , θ 2 ( θ 1 < θ 2 ) such that for each θ ( θ 1 , θ 2 ) , we have
δ ( P 1 , θ ) δ ( P 2 , θ ) < 0 and δ ( Q s , θ ) < 0 .
Without loss of generality, we assume that δ ( P 1 , θ ) > 0 , δ ( P 2 , θ ) < 0 . Let c = max j I 1 { c j } , then c < 1 . By Lemma 2.4, for ε ( 0 < 2 ε < 1 c 1 + c ), there exists a set E [ 0 , 2 π ) of linear measure zero such that for z = r e i θ with θ ( θ 1 , θ 2 ) E and sufficiently large r, we have
| h 01 ( z ) e P 1 ( z ) | exp { ( 1 ε ) δ ( P 1 , θ ) r n } ,
(3.3)
| h 02 ( z ) e P 2 ( z ) | exp { ( 1 ε ) δ ( P 2 , θ ) r n } ,
(3.4)
| h j ( z ) e Q j ( z ) | exp { ( 1 + ε ) c j δ ( P 1 , θ ) r n } ( j I 1 ) ,
(3.5)
| h j ( z ) e Q j ( z ) | exp { ( 1 ε ) d j δ ( Q s , θ ) r n } ( j I 2 { s } , d s = 1 ) .
(3.6)
Then combining (3.2)-(3.6) and (1.4), for z = r e i θ with θ ( θ 1 , θ 2 ) E and sufficiently large r H , we have
exp { ( 1 ε ) δ ( P 1 , θ ) r n } | h 01 ( z ) e P 1 ( z ) | | f ( k ) ( z ) f ( z ) | + j = 1 k 1 | h j ( z ) e Q j ( z ) | | f ( j ) ( z ) f ( z ) | + | h 02 ( z ) e P 2 ( z ) | ( k + 1 ) B ( T ( 2 r , f ) ) k + 1 exp { ( 1 + ε ) c δ ( P 1 , θ ) r n } .
(3.7)

Then by Lemma 2.7 and (3.7) we get σ ( f ) = and σ 2 ( f ) n .

Case 2. arg a 1 n arg a 2 n and arg a 2 n = arg b s n . Since arg a 1 n arg b s n , by Remark 2.1 there exist two real numbers θ 1 , θ 2 ( θ 1 < θ 2 ) such that for each θ ( θ 1 , θ 2 ) , we have
δ ( P 1 , θ ) > 0 and δ ( Q s , θ ) < 0 .

Then, using a similar proof to that of Case 1, we get σ ( f ) = and σ 2 ( f ) n .

Case 3. arg a 1 n = arg a 2 n . Since a 1 n a 2 n , without loss of generality, we assume that | a 1 n | > | a 2 n | . Since arg a 1 n arg b s n , by Remark 2.1 there exist two real numbers θ 1 , θ 2 ( θ 1 < θ 2 ) such that for each θ ( θ 1 , θ 2 ) , we have
δ ( P 1 , θ ) > 0 and δ ( Q s , θ ) < 0 .
Let c = max j I 1 { c j , | a 2 n | | a 1 n | } , then c < 1 . By Lemma 2.4, for ε ( 0 < 2 ε < 1 c 1 + c ), there exists a set E [ 0 , 2 π ) of linear measure zero such that for z = r e i θ with θ ( θ 1 , θ 2 ) E and sufficiently large r, we have (3.3), (3.5), (3.6) and
| h 02 ( z ) e P 2 ( z ) | exp { ( 1 + ε ) δ ( P 2 , θ ) r n } .
(3.8)
By (3.2), (3.3), (3.5), (3.6), (3.8) and (1.4), for z = r e i θ with θ ( θ 1 , θ 2 ) E and sufficiently large r H , we have
exp { ( 1 ε ) δ ( P 1 , θ ) r n } | h 01 ( z ) e P 1 ( z ) | | f ( k ) ( z ) f ( z ) | + j = 1 k 1 | h j ( z ) e Q j ( z ) | | f ( j ) ( z ) f ( z ) | + | h 02 ( z ) e P 2 ( z ) | ( k + 1 ) B ( T ( 2 r , f ) ) k + 1 exp { ( 1 + ε ) c δ ( P 1 , θ ) r n } .
(3.9)

Then, by Lemma 2.7 and (3.9), we get σ ( f ) = and σ 2 ( f ) n .

Step 3. We prove that σ 2 ( f ) = n . By Step 2, we get σ 2 ( f ) n . On the other hand, by Lemma 2.8 we get σ 2 ( f ) n . Hence we obtain that σ 2 ( f ) = n . □

Proof of Theorem 1.2 Let f be a solution of (1.6), then f is a nonzero entire function. Using a similar proof to Step 1 in the proof of Theorem 1.1, we obtain that σ ( f ) n . Now we prove that σ ( f ) = . Suppose that σ ( f ) < , by Lemma 2.3, for any given ε > 0 , there exists a set E 0 [ 0 , 2 π ) of linear measure zero such that for all z = r e i θ with | z | sufficiently large and θ [ 0 , 2 π ) E 0 , we have
| f ( j ) ( z ) f ( i ) ( z ) | | z | ( j i ) ( σ ( f ) 1 + ε ) ( 0 i < j k ) .
(3.10)
Let α, β be two real numbers such that σ ( F ) < α < β < n σ ( f ) , then
| F ( z ) | exp { r α }
(3.11)
holds for sufficiently large | z | = r . Set
δ ( P i , θ ) = | a i n | cos ( arg a i n + n θ ) ( i = 1 , 2 ) , δ ( Q j , θ ) = | b j n | cos ( arg b j n + n θ ) ( j = 1 , , k 1 ) .
By Lemma 2.4, there exists a set E 1 [ 0 , 2 π ) of linear measure zero such that whenever θ [ 0 , 2 π ) E 1 , we have
δ ( Q s , θ ) 0 , δ ( P 1 , θ ) 0 , δ ( P 2 , θ ) 0 , δ ( P 1 P 2 , θ ) 0 , δ ( P 1 Q s , θ ) 0 ,

and for z = r e i θ with θ [ 0 , 2 π ) E 1 and sufficiently large r, h 01 ( z ) e P 1 ( z ) , h 02 ( z ) e P 2 ( z ) and each h j ( z ) e Q j ( z ) satisfy either (2.1) or (2.2). Next we discuss the following two cases.

Case 1. arg a 1 n = arg a 2 n . Since a 1 n a 2 n , without loss of generality, we assume that | a 1 n | > | a 2 n | . For any given θ [ 0 , 2 π ) ( E 0 E 1 ) , we discuss the following three subcases.

Subcase 1. δ ( Q s , θ ) > 0 and δ ( P 1 , θ ) > 0 . If δ ( Q s , θ ) > δ ( P 1 , θ ) , let d = max j I 2 { δ ( P 1 , θ ) δ ( Q s , θ ) , d j } , then d < 1 . Now we prove that
log + | f ( s ) ( z ) | | z | β
is bounded on the ray arg z = θ . Suppose that this is not the case, then by Lemma 2.9 there exists an infinite sequence of points z m = r m e i θ with r m such that
log + | f ( s ) ( z m ) | | z m | β
(3.12)
and
| f ( j ) ( z m ) f ( s ) ( z m ) | 1 ( s j ) ! ( 1 + o ( 1 ) ) r m s j ( j = 0 , , s 1 ) .
(3.13)
Combining with (3.11) and (3.12), we get
| F ( z m ) f ( s ) ( z m ) | 0 .
(3.14)
Then by (1.6), (2.1), (2.2), (3.10), (3.13) and (3.14), for sufficiently large r m , we get
exp { ( 1 ε ) δ ( Q s , θ ) r m n } | h s ( z m ) e Q s ( z m ) | | F ( z m ) f ( s ) ( z m ) | + | f ( k ) ( z m ) f ( s ) ( z m ) | + j = 1 j s k 1 | h j ( z m ) e Q j ( z m ) f ( j ) ( z m ) f ( s ) ( z m ) | + { | h 01 ( z m ) e P 1 ( z m ) | + | h 02 ( z m ) e P 2 ( z m ) | } | f ( z m ) f ( s ) ( z m ) | ( k + 2 ) r m k σ ( f ) exp { ( 1 + ε ) d δ ( Q s , θ ) r m n } .
When 0 < ε < 1 d 1 + d , the above inequality does not hold. Therefore log + | f ( s ) ( z ) | / | z | β is bounded, and we have | f ( s ) ( z ) | exp { M | z | β } on the ray arg z = θ , where M > 0 is a real constant, not the same at each occurrence. By the same reasoning as that of [[16], Lemma 3.1], we get
| f ( z ) | | z | s exp { M | z | β } ( 1 + o ( 1 ) )
(3.15)
on the ray arg z = θ . If δ ( Q s , θ ) < δ ( P 1 , θ ) , let d = max j I 1 { δ ( Q s , θ ) δ ( P 1 , θ ) , | a 2 n | | a 1 n | , c j } , then d < 1 . Now we prove that
log + | f ( z ) | | z | β
is bounded on the ray arg z = θ . Suppose that this is not the case, then by Lemma 2.9 there exists an infinite sequence of points z m = r m e i θ with r m such that
log + | f ( z m ) | | z m | β .
Then combining with (3.11), we get
| F ( z m ) f ( z m ) | 0 .
(3.16)
Then by (1.6), (2.1), (2.2), (3.10) and (3.16), for sufficiently large r m , we get
exp { ( 1 ε ) δ ( P 1 , θ ) r m n } | h 01 ( z m ) e P 1 ( z m ) | | F ( z m ) f ( z m ) | + | f ( k ) ( z m ) f ( z m ) | + j = 1 k 1 | h j ( z m ) e Q j ( z m ) f ( j ) ( z m ) f ( z m ) | + | h 02 ( z m ) e P 2 ( z m ) | ( k + 2 ) r m k σ ( f ) exp { ( 1 + ε ) d δ ( P 1 , θ ) r m n } .
When 0 < ε < 1 d 1 + d , the above inequality does not hold. Therefore log + | f ( z ) | / | z | β is bounded, and we have
| f ( z ) | exp { M | z | β }
(3.17)

on the ray arg z = θ .

Subcase 2. δ ( Q s , θ ) < 0 and δ ( P 1 , θ ) < 0 . Now we prove that
log + | f ( k ) ( z ) | | z | β
is bounded on the ray arg z = θ . Suppose that this is not the case, then by Lemma 2.9 there exists an infinite sequence of points z m = r m e i θ with r m such that
log + | f ( k ) ( z m ) | | z m | β
(3.18)
and
| f ( j ) ( z m ) f ( k ) ( z m ) | 1 ( k j ) ! ( 1 + o ( 1 ) ) r m k j ( j = 0 , , k 1 ) .
(3.19)
Combining with (3.11) and (3.18), we get
| F ( z m ) f ( k ) ( z m ) | 0 .
(3.20)
Then by (1.6), (2.2), (3.19) and (3.20), for sufficiently large r m , we get
1 | F ( z m ) f ( k ) ( z m ) | + j = 1 k 1 | h j ( z m ) e Q j ( z m ) f ( j ) ( z m ) f ( k ) ( z m ) | + { | h 01 ( z m ) e P 1 ( z m ) | + | h 02 ( z m ) e P 2 ( z m ) | } | f ( z m ) f ( k ) ( z m ) | 0 .
This is absurd. Therefore log + | f ( k ) ( z ) | / | z | β is bounded, and we have | f ( k ) ( z ) | exp { M | z | β } on the ray arg z = θ . By the same reasoning as that of [[16], Lemma 3.1], we get
| f ( z ) | | z | k exp { M | z | β } ( 1 + o ( 1 ) )
(3.21)

on the ray arg z = θ .

Subcase 3. δ ( Q s , θ ) δ ( P 1 , θ ) < 0 . Using a similar argument to that of Subcase 1, we obtain that (3.15) or (3.17) holds on the ray arg z = θ .

Then by (3.15), (3.17) and (3.21), for any given θ [ 0 , 2 π ) ( E 0 E 1 ) , we have
| f ( r e i θ ) | r k exp { M r β } ( 1 + o ( 1 ) ) ,

where E 0 E 1 is of linear measure zero. Hence by Lemma 2.10 we get σ ( f ) β < n , a contradiction. Hence σ ( f ) = .

Case 2. arg a 1 n arg a 2 n . Since b s n a 2 n , E 2 = { θ [ 0 , 2 π ) : δ ( P 2 Q s , θ ) = 0 } is a finite set. Let δ 1 = max { δ ( P 1 , θ ) , δ ( P 2 , θ ) , δ ( Q s , θ ) } , then for any given θ [ 0 , 2 π ) ( E 0 E 1 E 2 ) , δ ( P 1 , θ ) , δ ( P 2 , θ ) , δ ( Q s , θ ) are distinct, and we have δ 1 > 0 or δ 1 < 0 . If δ 1 > 0 , using the same argument as that of Subcase 1 in Case 1, we obtain that (3.15) or (3.17) holds on the ray arg z = θ . If δ 1 < 0 , using the same argument as that of Subcase 2 in Case 1, we obtain that (3.21) holds on the ray arg z = θ . Hence we get σ ( f ) = . □

Declarations

Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 11201195, 11171119), the Natural Science Foundation of Jiangxi, China (No. 20132BAB201008, 20122BAB201012). The authors thank the referee for his/her valuable suggestions to improve the present article.

Authors’ Affiliations

(1)
School of Mathematics and Computer, Jiangxi Science and Technology Normal University, Nanchang, China
(2)
College of Mathematics and Information Science, Jiangxi Normal University, Nanchang, China

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© Mao and Liu; licensee Springer. 2014

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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