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Theory and Modern Applications

Growth and fixed points of solutions to second-order LDE with certain analytic coefficients in the unit disc

Abstract

In this article, the authors investigate the growth and fixed points of solutions of certain second-order linear differential equations with analytic coefficients in the unit disc and obtain some results which improve and generalize previous results.

MSC:30D35, 34M10.

1 Introduction and results

In this paper, we shall assume that the readers are familiar with the fundamental results and standard notations of the Nevanlinna value distribution theory in the complex plane and in the unit disc Δ={z:|z|<1} (see [16]). Before we state our main results, we need to recall some definitions and notations.

Definition 1.1 ([1, 2, 5])

For a meromorphic function f(z) in Δ, the order of f(z) is defined by

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

where T(r,f) is the characteristic function of f(z). And for an analytic function f(z) in Δ, we define σ M (f) by

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

where M(r,f)= max | z | = r |f(z)| is the maximum modulus function of f(z).

Remark 1.2 ([5])

If f(z) is an analytic function in Δ, then

σ(f) σ M (f)σ(f)+1.

Definition 1.3 ([7, 8])

Let f(z) be a meromorphic function in Δ, the hyper-order of f(z) is defined by

σ 2 (f)= lim r 1 ¯ log + log + T ( r , f ) log 1 1 r = lim r 1 ¯ log 2 + T ( r , f ) log 1 1 r .

If f(z) is an analytic function in Δ, then the hyper-order about maximum modulus of f(z) is also defined by

σ M , 2 (f)= lim r 1 ¯ log 3 + M ( r , f ) log 1 1 r .

Definition 1.4 Let f(z) be a meromorphic function in Δ, the hyper-lower-order of f(z) is defined by

μ 2 (f)= lim r 1 ¯ log 2 + T ( r , f ) log 1 1 r .

If f(z) is an analytic function in Δ, we define μ M , 2 (f) by

μ M , 2 (f)= lim r 1 ¯ log 3 + M ( r , f ) log 1 1 r .

Remark 1.5 ([7])

If f(z) is an analytic function in Δ, then

(i) σ 2 (f)= σ M , 2 (f),(ii) μ 2 (f)= μ M , 2 (f).

Definition 1.6 The hyper convergence exponent and the hyper-lower convergence exponent of fixed points of a meromorphic function f in Δ are defined by

λ 2 (fz)= lim r 1 ¯ log 2 + N ( r , 1 f z ) log 1 1 r , λ ̲ 2 (fz)= lim r 1 ¯ log 2 + N ( r , 1 f z ) log 1 1 r .

And we also define λ ¯ 2 (fz) and λ ̲ ¯ 2 (fz), respectively, by

λ ¯ 2 (fz)= lim r 1 ¯ log 2 + N ¯ ( r , 1 f z ) log 1 1 r and λ ̲ ¯ 2 (fz)= lim r 1 ¯ log 2 + N ¯ ( r , 1 f z ) log 1 1 r .

Many authors investigate the linear differential equation

f +A(z) e a z f +B(z) e b z f=0,
(1.1)

where A(z),B(z)0 are entire functions (e.g., see [912]). In [10], Chen proved that if ab0 and ab, then every solution f(z)0 of (1.1) is of infinite order; furthermore, if ab0, ab, A(z)1, B(z) is a polynomial, then every solution f(z)0 of (1.1) satisfies σ 2 (f)=1. In 2012, Hamouda investigated the equation

f +A(z) e a ( z 0 z ) μ f +B(z) e b ( z 0 z ) μ f=0,
(1.2)

where A(z) and B(z) are analytic functions in Δ, and he obtained the following results.

Theorem 1.7 ([13])

Let A(z) and B(z)0 be analytic functions in the unit disc. Suppose that μ>1 is a real constant, a, b and z 0 are complex numbers such that ab0, argaargb, | z 0 |=1. If A(z) and B(z) are analytic on z 0 , then every solution f(z)0 of (1.2) is of infinite order.

Theorem 1.8 ([13])

Let A(z) and B(z)0 be analytic functions in the unit disc. Suppose that μ>1 is a real constant, a, b and z 0 are complex numbers such that ab0, a=cb (0<c<1), | z 0 |=1. If A(z) and B(z) are analytic on z 0 , then every solution f(z)0 of (1.2) is of infinite order.

Remark 1.9 Throughout this paper, we choose the principal branch of logarithm of the function e λ ( z 0 z ) μ if μ is not an integer (λC0).

In this paper, we focus on studying the hyper-order and fixed points of the solutions of (1.2) and obtain the following results.

Theorem 1.10 Let A(z) and B(z)0 be analytic functions in the unit disc, and let μ>1 be a real constant, a, b and z 0 be complex numbers such that ab0, argaargb, | z 0 |=1. If A(z) and B(z) satisfy one of the following conditions:

  1. (1)

    max{ σ M (A), σ M (B)}μ, A(z) and B(z) are analytic on z 0 ;

  2. (2)

    σ M (A)<μ, σ M (B)μ and B(z) is analytic on z 0 ;

then every solution f(z)0 of (1.2) satisfies

  1. (i)

    μ M , 2 (f)= μ 2 (f)= σ 2 (f)= σ M , 2 (f)=μ;

  2. (ii)

    λ ̲ ¯ 2 (fz)= λ ̲ 2 (fz)= λ ¯ 2 (fz)= λ 2 (fz)=μ.

Theorem 1.11 Under the assumptions of Theorem  1.10, with the exception that ab0, a=cb (0<c<1), | z 0 |=1, every solution f(z)0 of (1.2) satisfies

  1. (i)

    μ M , 2 (f)= μ 2 (f)= σ 2 (f)= σ M , 2 (f)=μ;

  2. (ii)

    λ ̲ ¯ 2 (fz)= λ ̲ 2 (fz)= λ ¯ 2 (fz)= λ 2 (fz)=μ.

Corollary 1.12 Let A(z),B(z)0, C(z) and D(z) be analytic functions in the unit disc, and let a, b and z 0 be complex numbers such that ab0, argaargb or a=cb (0<c<1), | z 0 |=1. If one of the following conditions holds,

  1. (1)

    max{ σ M (A), σ M (B), σ M (C), σ M (D)}μ and A(z), B(z), C(z), D(z) are analytic on z 0 ;

  2. (2)

    max{ σ M (A), σ M (C), σ M (D)}<μ, σ M (B)μ and B(z) is analytic on z 0 ;

then every solution f(z)0 of

f + ( A ( z ) e a ( z 0 z ) μ + C ( z ) ) f + ( B ( z ) e b ( z 0 z ) μ + D ( z ) ) f=0
(1.3)

satisfies

  1. (i)

    μ M , 2 (f)= μ 2 (f)= σ 2 (f)= σ M , 2 (f)=μ;

  2. (ii)

    λ ̲ ¯ 2 (fz)= λ ̲ 2 (fz)= λ ¯ 2 (fz)= λ 2 (fz)=μ.

Theorem 1.13 Let A(z) and B(z)0 be analytic functions in the unit disc. Suppose that μ>1 and ν are real constants, a, b, z 0 and z 1 are complex numbers such that b0, z 0 z 1 and | z 0 |=| z 1 |=1. If A(z) and B(z) satisfy one of the following conditions:

  1. (1)

    A(z) and B(z) are analytic on z 0 ;

  2. (2)

    σ M (A)<μ and B(z) is analytic on z 0 ;

then every solution f(z)0 of

f +A(z) e a ( z 1 z ) ν f +B(z) e b ( z 0 z ) μ f=0
(1.4)

satisfies μ M , 2 (f) μ 2 (f)μ.

Theorem 1.14 Let A(z) and B(z)0 be analytic functions in the unit disc. Suppose that μ, ν (μ>1, μν) are real constants, a, b, z 0 and z 1 are complex numbers such that b0, z 0 z 1 and | z 0 |=| z 1 |=1. If A(z) and B(z) satisfy one of the following conditions:

  1. (1)

    max{ σ M (A), σ M (B)}μ, A(z) and B(z) are analytic on z 0 ;

  2. (2)

    σ M (A)<μ, σ M (B)μ and B(z) is analytic on z 0 ;

then every solution f(z)0 of (1.4) satisfies

  1. (i)

    μ 2 (f)= μ M , 2 (f)= σ 2 (f)= σ M , 2 (f)=μ;

  2. (ii)

    λ ̲ ¯ 2 (fz)= λ ̲ 2 (fz)= λ ¯ 2 (fz)= λ 2 (fz)=μ.

2 Lemmas

Lemma 2.1 ([14])

Let k and j be integers satisfying k>j0, and let ε>0. If f(z) is meromorphic 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 E 1 [0,1) is a set with E 1 1 1 r dr< and s(|z|)=1d(1|z|).

Lemma 2.2 ([13])

Let A(z) be an analytic function on a point z 0 C, set g(z)=A(z) e a ( z z 0 ) μ (μ>0 is a real constant), a=α+iβ0, z 0 z= Re i φ , δ a (φ)=αcos(μφ)+βsin(μφ) and H={φ[0,2π): δ a (φ)=0} (obviously, H is of linear measure zero). Then, for any given ε>0 and for any φ[0,2π)H, there exists R 0 >0 such that for 0<R< R 0 , we have

  1. (i)

    if δ a (φ)>0, then

    exp { ( 1 ε ) δ a ( φ ) 1 R μ } | g ( z ) | exp { ( 1 + ε ) δ a ( φ ) 1 R μ } ;
  2. (ii)

    if δ a (φ)<0, then

    exp { ( 1 + ε ) δ a ( φ ) 1 R μ } | g ( z ) | exp { ( 1 ε ) δ a ( φ ) 1 R μ } .

Remark 2.3 ([13])

Set δ a (φ)=γcos(μφ+ φ 0 ), where γ= α 2 + β 2 , μ>1, φ[0,2π). It is easy to know that δ a (φ) changes its sign on each interval I[0,2π) satisfying μmI>π, where mI denotes the linear measure of the interval I.

Lemma 2.4 ([15])

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 2 [0,1), for which E 2 1 1 r dr<. 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 2.5 ([7])

If A 0 (z), A 1 (z),, A k 1 (z) are analytic functions of finite order in the unit disc, then every solution f0 of

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

satisfies

σ 2 (f)= σ M , 2 (f)max { σ M ( A j ) : j = 0 , 1 , , k 1 } .

Remark 2.6 Lemma 2.5 is a special case of Theorem 2.1 in [7].

Lemma 2.7 ([4])

Let f be a meromorphic function in the unit disc, and let kN. Then

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

where S(r,f)=O( log + T(r,f))+O(log( 1 1 r )), possibly outside a set E 3 [0,1) with E 3 1 1 r dr<.

Lemma 2.8 ([8])

Suppose that A 0 , A 1 ,, A k 1 ,F0 are meromorphic functions in Δ, and let f(z) be a meromorphic solution of the equation

f ( k ) + A k 1 f ( k 1 ) ++ A 0 f=F(z),
(2.2)

such that max{ σ i (F), σ i ( A j ):j=0,1,,k1}< σ i (f), where i=1,2, then

λ ¯ i (f)= λ i (f)= σ i (f).

Lemma 2.9 Suppose that A 0 , A 1 ,, A k 1 ,F0 are meromorphic functions in Δ, and let f(z) be a meromorphic solution of equation (2.2) such that max{ σ i (F), σ i ( A j ):j=0,1,,k1}< μ i (f), where i=1,2, then

λ ̲ ¯ i (f)= λ ̲ i (f)= μ i (f).

Proof Suppose that f(z)0 is a solution of (2.2), by (2.2), we get

1 f = 1 F ( f ( k ) f + A k 1 f ( k 1 ) f + + A 0 ) ,
(2.3)

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 ) .
(2.4)

By Lemma 2.7 and (2.3), 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 3 ),
(2.5)

where E 3 1 1 r dr<. By (2.4)-(2.5), we get

T ( r , 1 f ) 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 3 ) .
(2.6)

Since max{ σ i (F), σ i ( A j ):j=0,1,,k1}< μ i (f), then we have

max { T ( r , F ) T ( r , f ) , T ( r , A j ) T ( r , f ) } 0 ( r 1 , j = 0 , , k 1 ) .
(2.7)

By (2.6)-(2.7) and by Lemma 2.4, for all |z|=r[0,1), we have

( 1 o ( 1 ) ) T(r,f)k N ¯ ( s ( r ) , 1 f ) +O { log + T ( s ( r ) , f ) + log ( 1 1 s ( r ) ) } ,
(2.8)

where s(r)=1d(1r), d(0,1). By (2.8), we have

λ ̲ ¯ i (f)= λ ̲ i (f)= μ i (f)(i=1,2).

 □

3 Proofs of theorems

Proof of Theorem 1.10 (i) Suppose that f(z)0 is a solution of (1.2), we obtain

|B(z) e b ( z 0 z ) μ || f ( z ) f ( z ) |+|A(z) e a ( z 0 z ) μ || f ( z ) f ( z ) |.
(3.1)

From Lemma 2.1, for any given ε>0, there exists a set E 1 [0,1) with E 1 1 1 r dr< such that for all zΔ satisfying |z|=r E 1 , we have

| f ( k ) ( z ) f ( z ) | ( 1 1 r ) M [ T ( s ( r ) , f ) ] k (k=1,2),
(3.2)

where s(r)=1d(1r), d(0,1), M>0 is a constant, not necessarily the same at each occurrence. Set I={θ:z z 0 =R e i θ ,zΔ}[0,2π), we have mIπ as R0. Since μ>1, there exists R 0 ( R 0 is sufficiently small and not necessarily the same at each occurrence) such that for 0<R< R 0 , we have μmI>π. By Remark 2.3 and argaargb, for all 0<R< R 0 , there exists some φI such that δ b (φ)>0 and δ a (φ)<0. From Lemma 2.2, for any given ε (0<ε<1) and for all z{z:z z 0 =R e i φ ,zΔ}, there exists R 0 such that for 0<R< R 0 , we have

|B(z) e b ( z 0 z ) μ |exp { ( 1 ε ) δ b ( φ ) 1 R μ } .
(3.3)

By condition (1) that A(z) is analytic on z 0 and by Lemma 2.2, for all z E 4 ={z:z z 0 =R e i φ ,0<R< R 0 ,zΔ}, we have

|A(z) e a ( z 0 z ) μ |exp { ( 1 ε ) δ a ( φ ) 1 R μ } <M.
(3.4)

By the metric relations in the triangle o z 0 z, we have

| z | 2 =1+ R 2 2Rcos φ ( φ = o z 0 z [ 0 , π 2 ) ).
(3.5)

By (3.5), for all z{z:z z 0 =R e i θ ,0<R< R 0 ,θI}, there exists certain ε 0 (0< ε 0 <1) such that

1 1 | z | 1 R = 1 1 | z | 2 cos φ R 1 + | z | > ε 0 1 | z | .
(3.6)

Equation (3.6) implies R0r 1 . By condition (2), σ M (A)<μ and δ a (φ)<0, for all zΔ, |z|=r 1 and 0<R< R 0 , we have

|A(z) e a ( z 0 z ) μ |= | A ( z ) | exp { δ a ( φ ) R μ } Mexp { 1 ( 1 r ) μ 0 } ,
(3.7)

where μ 0 satisfies σ M (A)< μ 0 <μ. By (3.1)-(3.4), (3.6) and (3.7), for all z E 4 and |z|=r E 1 1 , we have

exp { ( 1 ε ) δ b ( φ ) ε 0 μ ( 1 r ) μ } M ( 1 1 r ) M exp { 1 ( 1 r ) μ 0 } [ T ( s ( r ) , f ) ] 2 ,
(3.8)

where s(r)=1d(1r), d(0,1). By (3.8) and Lemma 2.4, we have

μ μ 2 (f)= μ M , 2 (f).
(3.9)

On the other hand, by Lemma 2.5, we have

σ 2 (f)= σ M , 2 (f)μ.
(3.10)

By (3.9) and (3.10), we have

μ M , 2 (f)= μ 2 (f)= σ 2 (f)= σ M , 2 (f)=μ.

(ii) Set g(z)=f(z)z, zΔ, where f(z)0 is a solution of (1.2). It is obvious that λ ¯ 2 (g)= λ ¯ 2 (fz), λ 2 (g)= λ 2 (fz), σ 2 (g)= σ 2 (fz)= σ 2 (f)=μ. Then equation (1.2) becomes

g +A(z) e a ( z 0 z ) μ g +B(z) e b ( z 0 z ) μ g= ( A ( z ) e a ( z 0 z ) μ + z B ( z ) e b ( z 0 z ) μ ) .
(3.11)

By (3.3), (3.4) and (3.7), it is easy to see A(z) e a ( z 0 z ) μ +zB(z) e b ( z 0 z ) μ 0 by modulus estimation. By Lemma 2.8 and (3.11), we have

λ ¯ 2 (g)= λ 2 (g)= σ 2 (g)= σ 2 (f)=μ.

Also, by Lemma 2.9 and (3.11), we deduce λ ¯ ̲ 2 (g)= λ ̲ 2 (g)= μ 2 (g)=μ. Therefore, we obtain

λ ̲ ¯ 2 (fz)= λ ̲ 2 (fz)= λ ¯ 2 (fz)= λ 2 (fz)=μ.

 □

Proof of Theorem 1.11 (i) Similar to the proof of Theorem 1.10, we can obtain (3.1)-(3.3). Since a=cb (0<c<1) and δ b (φ)>0, we have δ a (φ)=c δ b (φ)>0. From conditions (1)-(2) and by (3.7), Lemma 2.2, for any given ε>0 and for all z E 4 ={z:z z 0 =R e i φ ,0<R< R 0 ,zΔ}, we have

|A(z) e a ( z 0 z ) μ |exp { ( 1 + ε ) δ a ( φ ) 1 R μ } .
(3.12)

By (3.1)-(3.3) and (3.12), for any given ε (0<ε< 1 c 1 + c ) and for all z E 4 ={z:z z 0 =R e i φ ,0<R< R 0 ,zΔ}, we have

exp { [ ( 1 c ( 1 + c ) ε ) ] δ b ( φ ) 1 R u } M ( 1 1 r ) M [ T ( s ( r ) , f ) ] 2 (r E 1 ),

where s(r)=1d(1r), d(0,1) and E 1 1 1 r dr<. By (3.6) and Lemma 2.4, we obtain

exp { [ ( 1 c ( 1 + c ) ε ) ] δ b ( φ ) ε 0 μ ( 1 r ) μ } ( 1 1 s ( r ) ) M [ T ( s 1 ( r ) , f ) ] 2 ( r 1 ) ,
(3.13)

where s 1 (r)=1 d 2 (1r), d(0,1). By (3.13) and Lemma 2.4, we have

μ μ M , 2 (f)= μ 2 (f).

On the other hand, by Lemma 2.5, we have σ 2 (f)= σ M , 2 (f)μ. Therefore, we obtain

μ M , 2 (f)= μ 2 (f)= σ 2 (f)= σ M , 2 (f)=μ.

(ii) By the similar proof in case (ii) of Theorem 1.10, we have that

λ ̲ ¯ 2 (fz)= λ ̲ 2 (fz)= λ ¯ 2 (fz)= λ 2 (fz)=μ

holds for every solution f(z)0 of (1.2). □

Proof of Theorem 1.13 Suppose that f0 is a solution of (1.4), from (1.4), we obtain

|B(z) e b ( z 0 z ) μ || f ( z ) f ( z ) |+|A(z) e a ( z 1 z ) ν || f ( z ) f ( z ) |.
(3.14)

Since A(z) is analytic on z 0 or σ M (A)< μ 0 <μ, for z near enough z 0 and zΔ, we have

| A ( z ) | <M,or | A ( z ) | <exp { 1 ( 1 r ) μ 0 } .
(3.15)

Since z 1 z 0 , for all z near enough z 0 and zΔ, we have

| z 1 z|| z 1 z 0 || z 0 z| | z 0 z 1 | 2 ,| e a ( z 1 z ) ν | e 2 ν | a | | z 1 z 0 | ν M.
(3.16)

Using (3.1)-(3.3), (3.6) and (3.15)-(3.16), for all z E 4 and |z|=r E 1 1 , we obtain

exp { ( 1 ε ) δ b ( φ ) ε 0 μ ( 1 r ) μ } M ( 1 1 r ) M exp { 1 ( 1 r ) μ 0 } [ T ( s ( r ) , f ) ] 2 .
(3.17)

By (3.17) and Lemma 2.4, we have

μ μ 2 (f)= μ M , 2 (f).

 □

Proof of Theorem 1.14 (i) From Theorem 1.13 we have that every solution f(z)0 of (1.4) satisfies

μ μ 2 (f)= μ M , 2 (f).

On the other hand, by Lemma 2.5, we have that every solution f(z)0 of (1.4) satisfies

σ 2 (f)= σ M , 2 (f)μ.

Therefore every solution f(z)0 of (1.4) satisfies

μ M , 2 (f)= μ 2 (f)= σ 2 (f)= σ M , 2 (f)=μ.

(ii) Set g(z)=f(z)z, zΔ, equation (1.4) becomes

g +A(z) e a ( z 1 z ) ν g +B(z) e b ( z 0 z ) μ g= ( A ( z ) e a ( z 1 z ) ν + z B ( z ) e b ( z 0 z ) μ ) .

It is easy to see A(z) e a ( z 1 z ) μ +zB(z) e b ( z 0 z ) μ 0 by (3.3), (3.15) and (3.16). By the similar proof in case (ii) of Theorem 1.10, we have that every solution f(z)0 of (1.4) satisfies

λ ̲ ¯ 2 (fz)= λ ̲ 2 (fz)= λ ¯ 2 (fz)= λ 2 (fz)=μ.

 □

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Acknowledgements

The authors thank the referees for their valuable suggestions to improve the present article. This project is supported by the National Natural Science Foundation of China (Grant No. 11171119, 11261024, 11271045, 11301233), the Natural Science Foundation of Jiangxi Province in China (Grant 20122BAB211005, 20132BAB211001, 20132BAB211002).

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JT, TYP, HYX, HZ and GYD completed the main part of this article, JT corrected the main theorems. All authors read and approved the final manuscript.

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Tu, J., Peng, TY., Xu, HY. et al. Growth and fixed points of solutions to second-order LDE with certain analytic coefficients in the unit disc. Adv Differ Equ 2013, 342 (2013). https://doi.org/10.1186/1687-1847-2013-342

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