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Complex linear differential equations with certain analytic coefficients of [p,q]-order in the unit disc

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Abstract

In this paper, the authors study some growth properties of analytic functions of [p,q]-order in the disc and apply them to investigating the growth and zeros of solutions of complex linear differential equations with analytic coefficients of [p,q]-order satisfying certain growth conditions in the unit disc, and they obtain some results which are generalizations and improvements of some previous results.

MSC:30D35, 34M10.

1 Notations and results

We assume that readers are familiar with the fundamental results and the standard notations of the Nevanlinna value distribution theory of meromorphic functions in the unit disc (see [14]). Firstly, we introduce some notations. Let us define inductively, for r(0,+), exp 1 r= e r and exp i + 1 r=exp( exp i r), iN. For all r sufficiently large in (0,+), we define log 1 r=logr and log i + 1 r=log( log i r), iN. We also denote exp 0 r=r= log 0 r and exp 1 r= log 1 r. Moreover, we denote the linear measure of a set E[0,1) by mE= E dt, and the upper and lower density of E[0,1) are defined, respectively, by

dens ¯ Δ E= lim ¯ r 1 m ( E [ r , 1 ) ) 1 r , dens ̲ Δ E= lim ̲ r 1 m ( E [ r , 1 ) ) 1 r .

The complex oscillation theory of linear differential equations

f ( k ) + A k 1 (z) f ( k 1 ) ++ A 0 (z)f=0
(1.1)

and

f ( k ) + A k 1 (z) f ( k 1 ) ++ A 0 (z)f=F(z)
(1.2)

in the unit disc has been developed since the 1980s (see [5]). After that, many important results have been obtained (see [611]). After the work of Liu et al. in [12], there has been an increasing interest in studying the interaction between the analytic coefficients of [p,q]-order and the solutions of (1.1) and (1.2) (see [1316]). In this paper, the authors continue to focus on studying the growth and zeros of solutions of (1.1), (1.2) with analytic coefficients of [p,q]-order which satisfy certain growth conditions in the unit disc.

We use p, q to denote positive integers, and we use Δ={z:|z|<1} to denote the unit disc. In the following, we recall some notations of meromorphic functions and analytic functions in Δ.

Definition 1.1 (see [3, 10])

Let f(z) be a meromorphic function in Δ, and set

D(f)= lim ¯ r 1 T ( r , f ) log ( 1 r ) .

If f(z) is an analytic function in Δ,

D M (f)= lim ¯ r 1 log M ( r , f ) log ( 1 r ) .

If D(f)=, we say that f is admissible, if D(f)<, we say that f is non-admissible. If D M (f)=, we say that f is of infinite degree, if D M (f)<, we say that f is of finite degree.

Definition 1.2 (see [7, 11])

The iterated p-order of a meromorphic function f(z) in Δ is defined by

σ p (f)= lim ¯ r 1 log p T ( r , f ) log ( 1 r ) .

For an analytic function f(z) in Δ, we also define

σ M , p (f)= lim ¯ r 1 log p + 1 M ( r , f ) log ( 1 r ) .

Remark 1.1 If p=1, we denote σ 1 (f)=σ(f) and σ M , 1 (f)= σ M (f), and we have σ(f) σ M (f)σ(f)+1 (see [3]) and σ M , p (f)= σ p (f) for p>1 (see [7, 11]).

Definition 1.3 (see [1315])

Let 1qp or 2q=p+1, and f(z) be a meromorphic function in Δ, then the [p,q]-order of f(z) is defined by

σ [ p , q ] (f)= lim ¯ r 1 log p T ( r , f ) log q ( 1 1 r ) .

For an analytic function f(z) in Δ, we also define

σ M , [ p , q ] (f)= lim ¯ r 1 log p + 1 M ( r , f ) log q ( 1 1 r ) .

Definition 1.4 (see [13])

Let 1qp or 2q=p+1, we use N(r, 1 f ) ( N ¯ (r, 1 f )) to denote the integrated counting function for the (distinct) zero-sequence of a meromorphic function f(z) in Δ. Then the [p,q]-exponents of convergence of (distinct) zero-sequence of f(z) about N(r, 1 f ) ( N ¯ (r, 1 f )) are defined, respectively, by

λ [ p , q ] N (f)= lim ¯ r 1 log p N ( r , 1 f ) log q ( 1 1 r ) , λ ¯ [ p , q ] N ¯ (f)= lim ¯ r 1 log p N ¯ ( r , 1 f ) log q ( 1 1 r ) .

By the above definitions, the following propositions about the analytic function of [p,q]-order in the unit disc can easily be deduced.

Proposition 1.1 Let f(z) be an analytic function of [p,q]-order in Δ. Then the following five statements hold:

  1. (i)

    If p=q=1, then σ(f) σ M (f)1+σ(f).

  2. (ii)

    If p=q2 and σ [ p , q ] (f)<1, then σ [ p , q ] (f) σ M , [ p , q ] (f)1.

  3. (iii)

    If p=q2 and σ [ p , q ] (f)1, or p>q1, then σ [ p , q ] (f)= σ M , [ p , q ] (f).

  4. (iv)

    If p1 and σ [ p , p + 1 ] (f)>1, then D(f)=; if σ [ p , p + 1 ] (f)<1, then D(f)=0.

  5. (v)

    If p1 and σ M , [ p , p + 1 ] (f)>1, then D M (f)=; if σ M , [ p , p + 1 ] (f)<1, then D M (f)=0.

Proof (i), (iv), (v) hold obviously, we prove (ii) and (iii).

  1. (ii)

    By the standard inequality T(r,f) log + M(r,f) 1 + 3 r 1 r T( 1 + r 2 ,f) (0<r<1) (see [1, 2, 4]), we get

    log p T(r,f) log p + 1 + M(r,f)max { log p ( 4 1 r ) , log p T ( 1 + r 2 , f ) } .
    (1.3)

If p=q2 and σ [ p , q ] (f)<1, from (1.3) we have σ [ p , q ] (f) σ M , [ p , q ] (f)1.

  1. (iii)

    If p=q2 and σ [ p , q ] (f)1, or p>q1, from (1.3), we have σ [ p , q ] (f)= σ M , [ p , q ] (f). □

Proposition 1.2 Let f(z) be a meromorphic function of [p,q]-order in Δ. Then the following statements hold:

  1. (i)

    If p>q1, then λ ¯ [ p , q ] N ¯ (f)= λ ¯ [ p , q ] n ¯ (f).

  2. (ii)

    If p=q=1, then λ ¯ N ¯ (f) λ ¯ n ¯ (f) λ ¯ N ¯ (f)+1.

  3. (iii)

    If p=q2, then λ ¯ [ p , p ] N ¯ (f) λ ¯ [ p , p ] n ¯ (f)max{ λ ¯ [ p , p ] N ¯ (f),1}. Furthermore, we have λ ¯ [ p , p ] N ¯ (f)= λ ¯ [ p , p ] n (f) if λ ¯ [ p , p ] N ¯ (f)1, and if λ ¯ [ p , p ] N ¯ (f)<1 then λ ¯ [ p , p ] N ¯ (f) λ ¯ [ p , p ] n ¯ (f)1.

Proof Without loss of generality, assuming that f(0)0, by N ¯ (r, 1 f )= 0 r n ¯ ( t , 1 f ) t dt, we have

n ¯ ( r , 1 f ) 1 log ( 1 + 1 r 2 r ) r r + 1 r 2 n ¯ ( t , 1 f ) t dt 1 log ( 1 + 1 r 2 r ) N ¯ ( 1 + r 2 , 1 f ) (0<r<1),

since log(1+ 1 r 2 r ) 1 r 2 r (r 1 ), we obtain

lim ¯ r 1 log p n ¯ ( r , 1 f ) log q ( 1 1 r ) max { lim ¯ r 1 log p N ¯ ( 1 + r 2 , 1 f ) log q ( 1 1 r ) , lim ¯ r 1 log p ( 2 r 1 r ) log q ( 1 1 r ) } .

By the above inequality, we obtain:

  1. (i)

    if p>q1, then λ ¯ [ p , q ] n ¯ (f) λ ¯ [ p , q ] N ¯ (f);

  2. (ii)

    if p=q=1, then λ ¯ n ¯ (f) λ ¯ N ¯ (f)+1;

  3. (iii)

    if p=q2, then λ ¯ [ p , p ] n ¯ (f)max{ λ ¯ [ p , p ] N ¯ (f),1}.

On the other hand, by

N ¯ ( r , 1 f ) = r 0 r n ¯ ( t , 1 f ) t d t + N ¯ ( r 0 , 1 f ) n ¯ ( r , 1 f ) log ( r r 0 ) + N ¯ ( r 0 , 1 f ) ( 0 < r 0 < r < 1 ) ,

we can easily get λ ¯ [ p , q ] N ¯ (f) λ ¯ [ p , q ] n ¯ (f) (pq1). Therefore, the conclusions of Proposition 1.2 hold. □

In recent years, Belaïdi has investigated the growth of solutions of (1.1), (1.2) with analytic coefficients of [p,q]-order in the unit disc and obtained the following results.

Theorem A (see [13])

Let pq1 be integers and H 1 be a set of complex numbers satisfying dens ¯ Δ {|z|:z H 1 Δ}>0, and let A 0 , A 1 ,, A k 1 be analytic functions in Δ satisfying max{ σ M , [ p , q ] ( A j ):j=1,,k1} σ M , [ p , q ] ( A 0 )= σ 1 . Suppose that there exists a real number α 1 satisfying 0 α 1 < σ 1 such that, for any given ε (0<ε< σ 1 α 1 ), we have

| A 0 (z)| exp p + 1 { ( σ 1 ε ) log q ( 1 1 | z | ) }

and

| A j (z)| exp p + 1 { α 1 log q ( 1 1 | z | ) } (j=1,2,,k1),

as |z| 1 for z H 1 . Then every solution f0 of (1.1) satisfies σ [ p , q ] (f)= σ M , [ p , q ] (f)= and σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f)= σ M , [ p , q ] ( A 0 )= σ 1 .

Theorem B (see [15])

Let pq1 be integers and H 2 be a set of complex numbers satisfying dens ¯ Δ {|z|:z H 2 Δ}>0, and let A 0 , A 1 ,, A k 1 be analytic functions in Δ satisfying max{ σ [ p , q ] ( A j ):j=1,,k1} σ [ p , q ] ( A 0 )= σ 2 . Suppose that there exists a real number β 1 satisfying 0 β 1 < σ 2 such that, for any given ε (0<ε< σ 2 β 1 ), we have

T(r, A 0 ) exp p { ( σ 2 ε ) log q ( 1 1 | z | ) }

and

T(r, A j ) exp p { β 1 log q ( 1 1 | z | ) } (j=1,2,,k1),

as |z| 1 for z H 2 . Then every solution f0 of (1.1) satisfies σ [ p , q ] (f)= σ M , [ p , q ] (f)= and σ [ p , q ] ( A 0 ) σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f)max{ σ M , [ p , q ] ( A j ):j=0,1,,k1}. Furthermore, if p>q, then σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f)= σ [ p , q ] ( A 0 ).

Theorem C (see [13])

Suppose that the assumptions of Theorem  A are satisfied, and let F0 be an analytic function in Δ of [p,q]-order. Then the following two statements hold:

  1. (i)

    If σ [ p + 1 , q ] (F)< σ M , [ p , q ] ( A 0 ), then every solution f of (1.2) satisfies λ ¯ [ p + 1 , q ] (f)= λ [ p + 1 , q ] (f)= σ [ p + 1 , q ] (f)= σ M , [ p , q ] ( A 0 ) with at most one exceptional solution f 0 satisfying σ [ p + 1 , q ] ( f 0 )< σ M , [ p , q ] ( A 0 ).

  2. (ii)

    If σ [ p + 1 , q ] (F)> σ M , [ p , q ] ( A 0 ), then every solution f of (1.2) satisfies σ [ p + 1 , q ] (f)= σ [ p + 1 , q ] (F).

From Theorems A-C, we obtain the following results.

Theorem 1.1 Let H 3 be a complex set satisfying dens ¯ Δ {|z|:z H 3 Δ}>0. If A j (z) (j=0,1,,k1) are analytic functions in Δ satisfying max{ σ M , [ p , q ] ( A j )|j=0,1,,k1} σ 3 (0< σ 3 <), and if there exist two positive constants α 2 , β 2 (0< β 2 < α 2 ) such that, for all z H 3 and |z| 1 , we have

| A 0 (z)| exp p { α 2 [ log q 1 ( 1 1 r ) ] σ 3 }

and

| A j (z)| exp p { β 2 [ log q 1 ( 1 1 r ) ] σ 3 } (j=1,,k1).

Then the following statements hold:

  1. (i)

    If pq1, then every solution f(z)0 of (1.1) satisfies σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f)= σ 3 .

  2. (ii)

    If 2q=p+1 and σ 3 >1, then every solution f(z)0 of (1.1) satisfies σ [ p + 1 , p + 1 ] (f)= σ M , [ p + 1 , p + 1 ] (f)= σ 3 .

Theorem 1.2 Let H 4 be a complex set satisfying dens ¯ Δ {|z|:z H 4 Δ}>0. If A j (z) (j=0,1,,k1) are analytic functions in Δ satisfying max{ σ M , [ p , q ] ( A j )|j=0,1,,k1} σ 4 (0< σ 4 <), and there exist two positive constants α 3 , β 3 such that, for all z H 4 and |z| 1 , we have

T ( r , A 0 ( z ) ) exp p 1 { α 3 [ log q 1 ( 1 1 r ) ] σ 4 }

and

T ( r , A j ( z ) ) exp p 1 { β 3 [ log q 1 ( 1 1 r ) ] σ 4 } (j=1,,k1).

Then the following statements hold:

  1. (i)

    If pq2 and 0< β 3 < α 3 , then every solution f(z)0 of (1.1) satisfies σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f)= σ 4 .

  2. (ii)

    If 3q=p+1, 0< β 3 < α 3 and σ 4 >1, then every solution f(z)0 of (1.1) satisfies σ [ p + 1 , p + 1 ] (f)= σ M , [ p + 1 , p + 1 ] (f)= σ 4 .

  3. (iii)

    If p=1, q=2, 0<k β 3 < α 3 and σ 4 >1, then every solution f(z)0 of (1.1) satisfies σ [ 2 , 2 ] (f)= σ M , [ 2 , 2 ] (f)= σ 4 .

Theorem 1.3 Let F(z)0, A j (z) (j=0,1,,k1) be analytic functions in Δ. Suppose that H 3 , A j (z) (j=0,1,,k1) satisfy the hypotheses in Theorem  1.1, then we have the following statements:

  1. (i)

    Let 1qp, if σ [ p + 1 , q ] (F)> σ 3 , then all solutions of (1.2) satisfy σ [ p + 1 , q ] (f)= σ [ p + 1 , q ] (F); if σ [ p + 1 , q ] (F) σ 3 , then all solutions of (1.2) satisfy λ ¯ [ p + 1 , q ] N ¯ (f)= λ [ p + 1 , q ] N (f)= σ [ p + 1 , q ] (f)= σ 3 with at most one exceptional solution f 0 satisfying σ [ p + 1 , q ] ( f 0 )< σ 3 .

  2. (ii)

    Let 2q=p+1, σ 3 >1, if σ [ p + 1 , p + 1 ] (F)> σ 3 , then all solutions of (1.2) satisfy σ [ p + 1 , p + 1 ] (f)= σ [ p + 1 , p + 1 ] (F); if σ [ p + 1 , p + 1 ] (F) σ 3 , then all solutions of (1.2) satisfy λ ¯ [ p + 1 , p + 1 ] N ¯ (f)= λ [ p + 1 , p + 1 ] N (f)= σ [ p + 1 , p + 1 ] (f)= σ 3 , with at most one exceptional solution f 0 satisfying σ [ p + 1 , p + 1 ] ( f 0 )< σ 3 .

Corollary 1.4 Let F(z)0, A j (z) (j=0,1,,k1) be analytic functions in Δ. Suppose that H 4 , A j (z) (j=0,1,,k1) satisfy the hypotheses in Theorem  1.2, then we have the following statements:

  1. (i)

    Let 2qp, 0< β 3 < α 3 , if σ [ p + 1 , q ] (F)> σ 4 , then all solutions of (1.2) satisfy σ [ p + 1 , q ] (f)= σ [ p + 1 , q ] (F); if σ [ p + 1 , q ] (F) σ 4 , then all solutions of (1.2) satisfy λ ¯ [ p + 1 , q ] N ¯ (f)= λ [ p + 1 , q ] N (f)= σ [ p + 1 , q ] (f)= σ 4 with at most one exceptional solution f 0 satisfying σ [ p + 1 , q ] ( f 0 )< σ 4 .

  2. (ii)

    Let 3q=p+1, 0< β 3 < α 3 , σ 4 >1, if σ [ p + 1 , p + 1 ] (F)> σ 4 , then all solutions of (1.2) satisfy σ [ p + 1 , p + 1 ] (f)= σ [ p + 1 , p + 1 ] (F); if σ [ p + 1 , p + 1 ] (F) σ 4 , then all solutions of (1.2) satisfy λ ¯ [ p + 1 , p + 1 ] N ¯ (f)= λ [ p + 1 , p + 1 ] N (f)= σ [ p + 1 , p + 1 ] (f)= σ 4 with at most one exceptional solution f 0 satisfying σ [ p + 1 , p + 1 ] ( f 0 )< σ 4 .

  3. (iii)

    Let p=1, q=2, 0<k β 3 < α 3 , σ 4 >1, if σ [ 2 , 2 ] (F)> σ 4 , then all solutions of (1.2) satisfy σ [ 2 , 2 ] (f)= σ [ 2 , 2 ] (F); if σ [ 2 , 2 ] (F) σ 4 , then all solutions of (1.2) satisfy λ ¯ [ 2 , 2 ] N ¯ (f)= λ [ 2 , 2 ] N (f)= σ [ 2 , 2 ] (f)= σ 4 with at most one exceptional solution f 0 satisfying σ [ 2 , 2 ] ( f 0 )< σ 4 .

Remark 1.2 If a set E[0,1) satisfies dens ¯ E>0, then E d t 1 t =+.

2 Preliminary lemmas

Lemma 2.1 (see [10])

Let f(z) be a meromorphic function in Δ, and let k1 be an integer. Then

m ( r , f ( k ) f ) =S(r,f),

where S(r,f)=O{ log + T(r,f)+log( 1 1 r )}, possibly outside a set E 1 [0,1) with E 1 d t 1 t <.

Remark 2.1 Throughout this paper, we use E 1 [0,1) to denote a set satisfying E 1 d t 1 t <, not always the same at each occurrence.

Lemma 2.2 (see [9])

Let k and j be integers satisfying k>j0, and let ε>0 and d(0,1). If f is a meromorphic function in Δ such that f ( j ) does not vanish identically, then

| f ( k ) ( z ) f ( j ) ( z ) | ( ( 1 1 | z | ) 2 + ε max { log ( 1 1 | z | ) , T ( s ( | z | ) , f ) } ) k j ( | z | E 1 ) ,

where s(|z|)=1d(1|z|).

Lemma 2.3 (see [14])

Let 1qp be integers. If A 0 (z),, A k 1 (z) are analytic functions of [p,q]-order in the unit disc. Then every solution f of (1.1) satisfies σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f)max{ σ M , [ p , q ] ( A j )|j=0,1,,k1}.

By a similar proof to Lemma 2.3, we have the following lemma.

Lemma 2.4 If A 0 (z),, A k 1 (z) are analytic functions of [p,p+1]-order in the unit disc with max{ σ M , [ p , p + 1 ] ( A j )|j=0,1,,k1}<. Then every solution f of (1.1) satisfies σ [ p + 1 , p + 1 ] (f) σ M , [ p + 1 , p + 1 ] (f)max{ σ M , [ p , p + 1 ] ( A j )|j=0,1,,k1}.

Lemma 2.5 Let 1qp or 2q=p+1 and f(z) be an analytic function in Δ satisfying 0 σ [ p , q ] (f)= σ 5 (or 0 σ M , [ p , q ] (f)= σ 5 ), then there exists a set E 2 [0,1) satisfying E 2 d t 1 t =+ such that, for all r E 2 , we have

lim r 1 log p T ( r , f ) log q ( 1 1 r ) = σ 5 ( lim r 1 log p + 1 M ( r , f ) log q ( 1 1 r ) = σ 5 ) .

Proof If 1qp, by Definition 1.3, there exists a sequence { r n } n = 1 1 satisfying 1d(1 r n )< r n + 1 (0<d<1) and

lim n log p T ( r n , f ) log q ( 1 1 r n ) = σ [ p , q ] (f)= σ 5 .

Therefore there exists an n 1 (N) such that, for n n 1 and for any r E 2 = n = n 1 [ r n ,1d(1 r n )], we have

log p T ( r , f ) log q ( 1 1 r ) log p T ( r n , f ) log q [ 1 1 [ 1 d ( 1 r n ) ] ] = log p T ( r n , f ) log q [ 1 d ( 1 r n ) ] .

Hence

lim ̲ r 1 log p T ( r , f ) log q ( 1 1 r ) σ 5 (r E 2 ).

Since σ [ p , q ] (f)= σ 5 , for any r E 2 , we have

lim r 1 log p T ( r , f ) log q ( 1 1 r ) = σ 5 ,

where

m l E 2 = n = n 1 r n 1 d ( 1 r n ) d t 1 t = n = n 1 log 1 d =+.

We also can prove lim r 1 log p + 1 M ( r , f ) log q ( 1 1 r ) = σ 5 (r E 2 ) by the above proof.

By the above proof, this lemma also holds for the case 2q=p+1. □

Lemma 2.6 Let A j (z) (j=0,1,,k1), F(z)0 be analytic functions in Δ. Then the following statements hold:

  1. (i)

    If pq1 and f(z) is a solution of (1.2) satisfying max{ σ [ p , q ] ( A j ), σ [ p , q ] (F)|j=0,1,,k1}< σ [ p , q ] (f), then λ ¯ [ p , q ] N ¯ (f)= λ [ p , q ] N (f)= σ [ p , q ] (f).

  2. (ii)

    If f(z) is a solution of (1.2) satisfying max{ σ [ p , p + 1 ] ( A j ), σ [ p , p + 1 ] (F),1|j=0,1,,k1}< σ [ p , p + 1 ] (f), then λ ¯ [ p , p + 1 ] N ¯ (f)= λ [ p , p + 1 ] N (f)= σ [ p , p + 1 ] (f).

Proof (i) Suppose that f(z)0 is a solution of (1.2). By (1.2), we get

1 f = 1 F ( f ( k ) f + A k 1 f ( k 1 ) f + + A 0 ) ,
(2.1)

and it is easy to see that if f has a zero at z 0 of order α (α>k), and A 0 ,, A k 1 are analytic at z 0 , then F must have a zero at z 0 of order αk, hence

n ( r , 1 f ) k n ¯ ( r , 1 f ) +n ( r , 1 F )

and

N ( r , 1 f ) k N ¯ ( r , 1 f ) +N ( r , 1 F ) .
(2.2)

By Lemma 2.1 and (2.1), we have

m ( r , 1 f ) m ( r , 1 F ) + j = 0 k 1 m(r, A j )+O { log + T ( r , f ) + log ( 1 1 r ) } (r E 1 ).
(2.3)

By (2.2)-(2.3), we get

T ( r , f ) = T ( r , 1 f ) + O ( 1 ) k N ¯ ( r , 1 f ) + T ( r , F ) + j = 0 k 1 T ( r , A j ) + O { log + T ( r , f ) + log ( 1 1 r ) } ( r E 1 ) .
(2.4)

Since max{ σ [ p , q ] (F), σ [ p , q ] ( A j )|j=0,1,,k1}< σ [ p , q ] (f), by Lemma 2.5 and Definition 1.3, there exists a set E 3 with E 3 d t 1 t =+ such that

max { T ( r , F ) T ( r , f ) , T ( r , A j ) T ( r , f ) } 0 ( r 1 , r E 3 , j = 0 , , k 1 ) .
(2.5)

By (2.4)-(2.5), for all |z|=r E 3 E 1 , we have

( 1 o ( 1 ) ) T(r,f)k N ¯ ( r , 1 f ) +O { log + T ( r , f ) + log ( 1 1 r ) } ,

then we get σ [ p , q ] (f) λ ¯ [ p , q ] N ¯ (f). Therefore λ ¯ [ p , q ] N ¯ (f)= λ [ p , q ] N (f)= σ [ p , q ] (f).

  1. (ii)

    By a similar proof to case (i), we can easily obtain the conclusion of case (ii). □

Lemma 2.7 (see [17])

Let g:(0,1)R and h:(0,1)R be monotone increasing functions such that g(r)h(r) holds outside of an exceptional set E 1 [0,1) with E 1 d t 1 t <. Then there exists a constant d(0,1) such that if s(r)=1d(1r), then g(r)h(s(r)) for all r[0,1).

Lemma 2.8 (see [18])

Suppose that f(z) is meromorphic in Δ with f(0)=0. Then

m(r,f) [ 1 + φ ( r R ) ] T ( R , f ) +N ( R , f ) ,
(2.6)

where 0<r<R<1, φ(t)= 1 π log 1 + t 1 t .

Lemma 2.9 Let f(z) be an analytic function of [p,q]-order in Δ. Then the following statements hold:

  1. (i)

    If pq1, then σ [ p , q ] (f)= σ [ p , q ] ( f ).

  2. (ii)

    If 3q=p+1, then σ [ p , p + 1 ] ( f )max{ σ [ p , p + 1 ] (f),1} and σ [ p , p + 1 ] (f)max{ σ [ p , p + 1 ] ( f ),1}.

  3. (iii)

    If p=1, q=2, then σ [ 1 , 2 ] ( f )max{ σ [ 1 , 2 ] (f),1} and σ [ 1 , 2 ] (f)1+ σ [ 1 , 2 ] ( f ).

Proof By Lemma 2.1, we have

T ( r , f ) 2T(r,f)+m ( r , f f ) 3T(r,f)+O { log 1 1 r } (0<r<1,r E 1 ).
(2.7)

By (2.7) and Lemma 2.7, it is easy to see σ [ p , q ] ( f ) σ [ p , q ] (f) (pq1) and σ [ p , p + 1 ] ( f )max{ σ [ p , p + 1 ] (f),1}. On the other hand, set R= 1 + r 2 , 0<r<1, by Lemma 2.8, we have

T(r,f)< ( 3 + log 4 ( 1 r ) ) T ( 1 + r 2 , f ) .
(2.8)

By (2.8), we have, if pq1, then σ [ p , q ] (f) σ [ p , q ] ( f ) and if 3q=p+1, then σ [ p , p + 1 ] (f)max{ σ [ p , p + 1 ] ( f ),1}; and we can easily obtain the conclusion (iii) by (2.7) and (2.8). Therefore Lemma 2.9 holds. □

3 Proofs of Theorems 1.1-1.3

Proof of Theorem 1.1 (i) Let H 5 ={|z|:z H 3 Δ}, since dens ¯ Δ {|z|:z H 3 Δ}>0, then by Remark 1.2, H 5 is a set of r with H 5 d t 1 t =+. For any |z|=r H 5 and r 1 , we have

| A 0 ( z ) | exp p { α 2 [ log q 1 ( 1 1 r ) ] σ 3 } , | A j ( z ) | exp p { β 2 [ log q 1 ( 1 1 r ) ] σ 3 } ( j = 1 , , k 1 ) .
(3.1)

If f0, from (1.1), we get

| A 0 || f ( k ) f |+| A k 1 || f ( k 1 ) f |++| A 1 || f f |.
(3.2)

By Lemma 2.2, for |z|=r E 1 , we get

| f ( j ) ( z ) f ( z ) | ( 1 1 r ) M T ( s ( r ) , f ) j (j=1,,k),
(3.3)

where M denotes a positive constant, not always the same at each occurrence. By (3.1)-(3.3), for all z satisfying |z|=r H 5 E 1 and r 1 , we have

exp p { α 2 [ log q 1 ( 1 1 r ) ] σ 3 } k exp p { β 2 [ log q 1 ( 1 1 r ) ] σ 3 } ( 1 1 r ) M T ( s ( r ) , f ) k .
(3.4)

If pq1, by (3.4), then σ 3 σ [ p + 1 , q ] (f). On the other hand, by Lemma 2.3, we have σ [ p + 1 , q ] (f)max{ σ M , [ p , q ] ( A j )|j=0,1,,k1} σ 3 . Therefore every solution f(z)0 of (1.1) satisfies σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f)= σ 3 .

  1. (ii)

    If 2q=p+1 and σ 3 >1, by a similar proof to case (i), we obtain the conclusion. □

Proof of Theorem 1.2 (i) Let H 6 ={|z|:z H 4 Δ}, since dens ¯ Δ {|z|:z H 4 Δ}>0, then by Remark 1.2, H 6 is a set of r with H 6 d t 1 t =+. For any |z|=r H 6 and r 1 , we have

T ( r , A 0 ) exp p 1 { α 3 [ log q 1 ( 1 1 r ) ] σ 4 } , T ( r , A j ) exp p 1 { β 3 [ log q 1 ( 1 1 r ) ] σ 4 } ( j = 1 , , k 1 ) .
(3.5)

If f0, from (1.1), we get

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

then

T(r, A 0 ) i = 1 k 1 T(r, A j )+ j = 1 k m ( r , f ( k ) f ) +O(1).
(3.6)

By Lemma 2.1 and (3.6), there exists a set E 1 [0,1) with E 1 d t 1 t < such that, for all z satisfying |z|=r E 1 , we have

T(r, A 0 ) i = 1 k 1 T(r, A j )+O { log + T ( r , f ) + log ( 1 1 r ) } .
(3.7)

By (3.5), (3.7), for |z|=r H 6 E 1 and r 1 , we have

exp p 1 { α 3 [ log q 1 ( 1 1 r ) ] σ 4 } ( k 1 ) exp p 1 { β 3 [ log q 1 ( 1 1 r ) ] σ 4 } + O { log + T ( r , f ) + log ( 1 1 r ) } .

If pq2 and 0< β 3 < α 3 , then every solution f(z)0 of (1.1) satisfies σ 4 σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f). On the other hand, by Lemma 2.3, all solutions of (1.1) satisfy σ [ p + 1 , q ] (f)= σ M , [ p + 1 , q ] (f)max{ σ M , [ p , q ] ( A j )|j=0,1,,k1} σ 4 . Therefore every solution f(z)0 of (1.1) satisfies σ [ p + 1 , q ] (f)= σ 4 .

(ii)-(iii) By a similar proof to case (i), we obtain the conclusions of (ii)-(iii). □

Proof of Theorem 1.3 (i) For 1qp, assume that f is a solution of (1.2), by the elementary theory of differential equations, thus all the solutions of (1.2) have the form

f= f + C 1 f 1 + C 2 f 2 ++ C k f k ,

where C 1 ,, C k are complex constants, f 1 ,, f k is a solution base of (1.1), f is a solution of (1.2) and has the form

f = D 1 f 1 + D 2 f 2 ++ D k f k ,
(3.8)

where D 1 ,, D k are certain analytic functions in Δ satisfying

D j =F G j ( f 1 ,, f k )W ( f 1 , , f k ) 1 (j=1,,k),
(3.9)

where G j ( f 1 ,, f k ) are differential polynomials in f 1 ,, f k and their derivative with constant coefficients, and W( f 1 ,, f k ) is the Wronskian of f 1 ,, f k .

If σ [ p + 1 , q ] (F)> σ 3 , by Lemma 2.3, Lemma 2.9, and (3.8)-(3.9), we find that all solutions of (1.2) satisfy

σ [ p + 1 , q ] (f)max { σ [ p + 1 , q ] ( f j ) , σ [ p + 1 , q ] ( F ) | j = 1 , , k } =max { σ 3 , σ [ p + 1 , q ] ( F ) } σ [ p + 1 , q ] (F).

On the other hand, by a simple order comparison from (1.2), we see that all solutions of (1.2) satisfy σ [ p + 1 , q ] (f) σ [ p + 1 , q ] (F). Therefore all solutions of (1.2) satisfy

σ [ p + 1 , q ] (f)= σ [ p + 1 , q ] (F).

If σ [ p + 1 , q ] (F) σ 3 , by the above proof in (3.8)-(3.9), we can find that all solutions of (1.2) satisfy σ [ p + 1 , q ] (f) σ 3 . We affirm that (1.2) can only possess at most one exceptional solution f 0 satisfying σ [ p + 1 , q ] ( f 0 )< σ 3 . In fact, if f is another solution satisfying σ [ p + 1 , q ] ( f )< σ 3 , then σ [ p + 1 , q ] ( f 0 f )< σ 3 . But f 0 f is a solution of (1.1) and satisfies σ [ p + 1 , q ] ( f 0 f )= σ 3 by Theorem 1.1(i), this is a contradiction. Then σ [ p + 1 , q ] (f)= σ 3 holds for all solutions of (1.2) with at most one exceptional solution f 0 satisfying σ [ p + 1 , q ] ( f 0 )< σ 3 . By Lemma 2.6(i), we get

λ ¯ [ p + 1 , q ] N ¯ (f)= λ [ p + 1 , q ] N (f)= σ [ p + 1 , q ] (f)

holds for all solutions satisfying σ [ p + 1 , q ] (f)= σ 3 with at most one exceptional solution f 0 satisfying σ [ p + 1 , q ] ( f 0 )< σ 3 .

  1. (ii)

    For 2q=p+1, σ 3 >1, by a similar proof to case (i), we draw the conclusions of case (ii). □

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Acknowledgements

This project is supported by the Natural Science Foundation of Jiangxi Province in China (20132BAB211002) and by the Science and Technology Plan Program of Education Bureau of Jiangxi Province (GJJ14271, GJJ14272). Zu-Xing Xuan is supported in part by Beijing Natural Science Foundation (No. 1132013) and The Project of Construction of Innovative Teams and Teacher Career Development for Universities and Colleges under Beijing Municipality (CIT and TCD20130513).

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Keywords

  • linear differential equations
  • unit disc
  • [p,q]-order
  • [p,q]-exponent of convergence of zero-sequence