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Complex Oscillation of Solutions and Their Arbitrary-Order Derivatives of Linear Differential

Equations With Analytic Coefficients of [p,q]-Order in the Unit Disc

Benharrat Belaïdi∗, Meriem Belmiloud
Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem

(UMAB), B. P. 227 Mostaganem, Algeria
benharrat.belaidi@univ-mosta.dz, meriem.belmiloud27@gmail.com

∗Correspondence: benharrat.belaidi@univ-mosta.dz

Abstract. Throughout this article, we investigate the growth and fixed points of solutions of complexhigher order linear differential equations in which the coefficients are analytic functions of [p,q]−orderin the unit disc. This work improves some results of Belaïdi [3–5], which is a generalization of recentresults from Chen et al. [9].

1. Introduction and main results
Consider for k ≥ 2 the following complex linear differential equations

f (k) + Ak−1 (z) f
(k−1) + · · · + A1 (z) f ′ + A0 (z) f = 0, (1.1)

Ak (z) f
(k) + Ak−1 (z) f

(k−1) + · · · + A1 (z) f ′ + A0 (z) f = 0, (1.2)where Ai 6≡ 0 (i = 0, 1, ...,k) are analytic functions in the unit disc D = {z ∈ C : |z| < 1}. It iswell-known that the solutions of (1.1) are analytic in D too and that there are exactly k linearlyindependent solutions of (1.1), see [13]. Bernal [6] was the first to use the concept of iteratedorder to study the growth of fast growing solutions of equation (1.1) . After that, the iterated orderof solutions of higher order equations was investigated by Cao in [8], he extended the results ofChen and Yang [10], Belaïdi [2] on C. In addition, Cao [8] obtained some results concerning thefixed points of homogeneous linear differential equations (1.1) and (1.2) . In [15, 16], Juneja and hisco-authors have investigated some properties of entire functions of [p,q]-order, and obtained someresults of their growth. In [20], by using the concept of [p,q]-order Liu, Tu and Shi have consideredthe equation (1.1) with entire coefficients and obtained different results concerning the growth of itssolutions in the complex plane. In [3], the [p,q]−order was introduced in the unit disc D, and manyresults on [p,q]−order of solutions of (1.1) have been found by different researchers [3–5,14,18,22]in D. Recently, Chen et al. in [9] gave some results about the growth and fixed points of solutions
Received: 30 Sep 2022.
Key words and phrases. linear differential equations; analytic function; [p,q]−order; fixed points.1

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 2
of higher-order linear differential equations in the unit disc, they studied and estimated the fixedpoints of solutions of (1.1) and (1.2), and also extended the coefficient conditions to a type ofone-constant-control coefficient comparison and obtained the same estimates of iterated order ofsolutions. The aim of this paper is to contrast coefficients by producing better estimates of thegrowth of solutions by using the concept of [p,q]−order, and optimizing the coefficients’s conditionswith less control constants of the coefficients’s modulus or characteristic functions and we will obtainresults which improve and generalize those of Chen et al., Belaïdi, Cao, Tu and Xuan.

Throughout this paper, we shall assume that the reader is familiar with the fundamental resultsand the standard notations of the Nevanlinna’s theory in the unit disc D = {z ∈ C : |z| < 1}(see, [12, 13, 17, 21]).
Now, we give the definitions of iterated order and growth index to classify generally thefunctions of fast growth in D as those in C (see, [6]). Let us define inductively, for r ∈R, exp1 r := erand expp+1 r := exp (expp r) , p ∈ N. We also define for all r sufficiently large in (0, +∞) ,

log1 r := log r and logp+1 r := log (logp r) , p ∈N. Moreover, we denote by exp0 r := r, log0 r := r,
log−1 r := exp1 r and exp−1 r := log1 r.
Definition 1.1 (see [7]) Let f be a meromorphic function in D. Then the iterated n−order of f is
defined by

σn (f ) = lim sup
r→1−

log+n T (r, f )

log
(
1
1−r
) (n ≥ 1 is an integer ) ,

where log+1 x = log
+x = max{log x, 0} , log+n+1x = log

+
(

log+n x
)
. For n = 1, this notation is

called order (σ1 (f ) = σ (f )) and for n = 2 hyper-order ([19]). If f is an analytic in D, then the
iterated n−order of f is defined by

σM,n (f ) = lim sup
r→1−

log+n+1M (r, f )

log
(
1
1−r
) (n ≥ 1 is an integer ) .

For n = 1, σM,1 (f ) = σM (f ) .

Now, we introduce the concept of [p,q]-order of meromorphic and analytic functions in theunit disc.
Definition 1.2 ([3]) Let p ≥ q ≥ 1 be integers and f be a meromorphic function in D. Then, the
[p,q]-order of f is defined by

σ[p,q] (f ) = lim sup
r→1−

log+p T (r, f )

logq
(
1
1−r
) .

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 3
For an analytic function f in D, we also define

σM,[p,q] (f ) = lim sup
r→1−

log+p+1M (r, f )

logq
(
1
1−r
) .

Remark 1.1 It is easy to see that 0 ≤ σ[p,q] (f ) ≤ ∞ (0 ≤ σM,[p,q] (f ) ≤ ∞), for any p ≥
q ≥ 1. By Definition 1.2, we have that σ[1,1] = σ (f ) (σM,[1,1] = σM (f )) and σ[2,1] = σ2 (f )(
σM,[2,1] = σM,2 (f )

).
Proposition 1.1 ([3]) Let p ≥ q ≥ 1 be integers, and let f be an analytic function in D of
[p,q]-order. The following two statements hold:
(i) If p = q, then

σ[p,q] (f ) ≤ σM,[p,q] (f ) ≤ σ[p,q] (f ) + 1.

(ii) If p > q, then
σ[p,q] (f ) = σM,[p,q] (f ) .

Definition 1.3 ([4]) Let p ≥ q ≥ 1 be integers and f be a meromorphic function in D. Then, the
[p,q]-exponent of convergence of the sequence of zeros of f is defined by

λ[p,q] (f ) = lim sup
r→1−

log+p N
(
r, 1
f

)
logq

(
1
1−r
) ,

where N
(
r, 1
f

)
is the integrated counting function of zeros of f in {z : |z| ≤ r}. Similarly, the

[p,q]-exponent of convergence of the sequence of distinct zeros of f is defined by

λ[p,q] (f ) = lim sup
r→1−

log+p N
(
r, 1
f

)
logq

1
1−r

,

where N
(
r, 1
f

)
is the integrated counting function of distinct zeros of f in {z : |z| ≤ r}.

Definition 1.4 Let p ≥ q ≥ 1 be integers and f be a meromorphic function in D. Then, the
[p,q]-exponent of convergence of the sequence of fixed points of f is defined by

λ[p,q] (f −z) = lim sup
r→1−

log+p N
(
r, 1
f−z
)

logq
(
1
1−r
) .

Similarly, the [p,q]-exponent of convergence of the sequence of distinct fixed points of f is defined
by

λ̄[p,q] (f −z) = lim sup
r→1−

log+p N̄
(
r, 1
f−z
)

logq
(
1
1−r
) .

Recall that for a measurable set E ⊂ [0, 1) , the upper and lower densities of E are defined by
densDE = lim sup

r→1−

m (E ∩ [0, r))
m ([0, r))

and densDE = lim inf
r→1−

m (E ∩ [0, r))
m ([0, r))

,

respectively, where m (F ) = ∫
F

dt
1−t for F ⊂ [0, 1). It is clear that 0 ≤ densDE ≤ densDE ≤ 1for any measurable set E ⊂ [0, 1) .

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 4
Proposition 1.2 If a set E satisfies densDE > 0, then m (E) =

∫
E

dt
1−t = +∞.

Proof. Suppose that m (E) = ∫
E

dt
1−t = δ < ∞. We have
m ([0, r)) = − log (1 − r) .

Since m (E ∩ [0, r)) ≤ m (E) , then
densDE = lim sup

r→1−

m (E ∩ [0, r))
m ([0, r))

≤ lim sup
r→1−

δ

− log (1 − r)
= 0.

So densDE = 0. Hence
densDE > 0 =⇒ m (E) =

∫
E

dt

1 − t
= +∞.

In 2012, Belaïdi in [4] and [5] treated the growth of solutions of homogeneous linear differentialequations in which the coefficients are analytic functions of [p,q]−order in D. As for the equation
(1.1), he got the following results.
Theorem A (see [4]) Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying
densD {|z| : z ∈ H ⊆ D} > 0, and let A0 (z) , ...,Ak−1 (z) be analytic functions in the unit disc
D such that for real constants α,β, where 0 ≤ β < α, we have

|A0 (z)| ≥ expp+1
{
α logq

(
1

1 −|z|

)}
and

|Ai (z)| ≤ expp+1
{
β logq

(
1

1 −|z|

)}
(i = 1, ...,k − 1)

as |z| → 1− for z ∈ H. Then every solution f 6≡ 0 of equation (1.1) satisfies σ[p,q] (f ) =
σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) ≥ α.

Theorem B (see [5]) Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying
densD {|z| : z ∈ H ⊆ D} > 0, and let A0 (z) , ...,Ak−1 (z) be analytic functions in the unit disc
D such that for real constants α,β, where 0 ≤ β < α, we have

T (r,A0) ≥ expp
{
α logq

(
1

1 −|z|

)}
and

T (r,Ai ) ≤ expp
{
β logq

(
1

1 −|z|

)}
(i = 1, ...,k − 1)

as |z| = r → 1− for z ∈ H. Then every solution f 6≡ 0 of equation (1.1) satisfies σ[p,q] (f ) =
σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) ≥ α.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 5
After that in 2021, Chen et al. [9] investigated the growth of solutions of equations (1.1) and

(1.2) in D by using the iterated order, and they got the following results.
Theorem C (see [9]) Let n ≥ 1 be an integer. Let H be a set of complex numbers satisfying
densD {|z| : z ∈ H ⊆ D} > 0, and let A0,A1, ...,Ak−1 be analytic functions in the unit disc D
such that

max{σM,n (Ai ) : i = 1, 2, ...,k − 1}≤ σM,n (A0) = µ (0 < µ < ∞) ,

and for a constant α ≥ 0, we have

lim inf
|z|→1−,z∈H

(
(1 −|z|)µ logn |A0 (z)|

)
> α

and
|Ai (z)| ≤ expn

{
α

(
1

1 −|z|

)µ}
, (i = 1, 2, ...,k − 1)

as |z|→ 1− for z ∈ H. Then every solution f 6≡ 0 of equation (1.1) satisfies σn (f ) = σM,n (f ) =
∞ and σn+1 (f ) = σM,n (A0) = µ.

Theorem D (see [9]) Let n ≥ 1 be an integer. Let H be a set of complex numbers satisfying
densD {|z| : z ∈ H ⊆ D} > 0, and let A0,A1, ...,Ak be analytic functions in the unit disc D, and
for some constants α ≥ 0 and µ > 0, we have

lim inf
|z|→1−,z∈H

(
(1 −|z|)µ logn−1T (r,A0)

)
> α

and
T (r,Ai ) ≤ expn−1

{
α

(
1

1 −|z|

)µ}
, (i = 1, 2, ...,k)

as |z| = r → 1− for z ∈ H. Then every meromorphic (or analytic) solution f 6≡ 0 of equation
(1.2) satisfies σn (f ) = ∞ and σn+1 (f ) ≥ µ.

Theorem E (see [9]) Assume that the assumptions of Theorem C hold. Then every solution f 6≡ 0
of equation (1.1) satisfies

λ̄n

(
f (j) −z

)
= λ̄n (f −z) = σn (f ) = ∞,

λ̄n+1

(
f (j) −z

)
= λ̄n+1 (f −z) = σn+1 (f ) = µ, (j = 1, 2, ...) .In this paper, we improve and generalize the recent results of Chen et al. [9] by using theconcept of [p,q]−order instead of the iterated order with less control constant. At the same time,our work improve some results of Belaïdi in [4] and [5]. To be specific, we will decrease the controlconstants of the coefficients’ modulus or characteristic functions and obtain the same results ofBelaïdi, Tu and Xuan. Here, we study the problem and get the following results.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 6
Theorem 1.1 Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying

densD {|z| : z ∈ H ⊆ D} > 0,

and let A0, ...,Ak−1 be analytic functions in the unit disc D such that

max
{
σM,[p,q] (Ai ) : i = 1, 2, ...,k − 1

}
≤ σM,[p,q] (A0) = µ (0 < µ < +∞)

and for a constant α ≥ 0, we have

lim inf
|z|→1−,z∈H

logp |A0 (z)|(
logq−1

(
1
1−|z|

))µ > α (1.3)
and

|Ai (z)| ≤ expp
{
α
(

logq−1

(
1
1−|z|

))µ}
(i = 1, ...,k − 1) (1.4)

as |z| → 1− for z ∈ H. Then every solution f 6≡ 0 of equation (1.1) satisfies σ[p,q] (f ) =
σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) = µ.

By Theorem 1.1, we easily obtain the following corollary.
Corollary 1.1 ([22]) Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying
densD {|z| : z ∈ H ⊆ D} > 0, and let A0, ...,Ak−1 be analytic functions in the unit disc D such
that

max
{
σM,[p,q] (Ai ) : i = 1, 2, ...,k − 1

}
≤ σM,[p,q] (A0) = µ (0 < µ < +∞)

and for some real constants α,β where 0 ≤ β < α, we have

|A0 (z)| ≥ expp
{
α

(
logq−1

(
1

1 −|z|

))µ}
and

|Ai (z)| ≤ expp
{
β

(
logq−1

(
1

1 −|z|

))µ}
, i = 1, ...,k − 1

as |z| → 1− for z ∈ H. Then every solution f 6≡ 0 of equation (1.1) satisfies σ[p,q] (f ) =
σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) = µ.

Theorem 1.2 Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying

densD {|z| : z ∈ H ⊆ D} > 0,

and let A0, ...,Ak−1 be analytic functions in the unit disc D such that

max
{
σM,[p,q] (Ai ) : i = 1, 2, ...,k − 1

}
≤ σM,[p,q] (A0) = µ (0 < µ < +∞)

and
lim sup
|z|→1−,z∈H

logp |Ai (z)|(
logq−1

(
1
1−|z|

))µ < lim inf
|z|→1−,z∈H

logp |A0 (z)|(
logq−1

(
1
1−|z|

))µ (i = 1, ...,k − 1) (1.5)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 7
as |z| → 1− for z ∈ H.Then every solution f 6≡ 0 of equation (1.1) satisfies σ[p,q] (f ) =
σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) = µ.

Theorem 1.3 Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying

densD {|z| : z ∈ H ⊆ D} > 0,

and let A0, ...,Ak−1 be analytic functions in the unit disc D such that

max
{
σM,[p,q] (Ai ) : i = 1, 2, ...,k − 1

}
≤ σM,[p,q] (A0) = µ (0 < µ < +∞)

and for a constant α ≥ 0, if p ≥ q ≥ 2 we have
lim inf

|z|→1−,z∈H

logp−1T (r,A0)(
logq−1

(
1
1−|z|

))µ > α (1.6)
and

T (r,Ai ) ≤ expp−1
{
α
(

logq−1

(
1
1−|z|

))µ}
, (i = 1, ...,k − 1) (1.7)

as |z| = r → 1− for z ∈ H, then every solution f 6≡ 0 of equation (1.1) satisfies σ[p,q] (f ) =
σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) = µ. If p = q = 1, we have

lim inf
|z|→1−,z∈H

T (r,A0)(
1
1−|z|

)µ > (k − 1) α (1.8)
and

T (r,Ai ) ≤ α
(
1
1−|z|

)µ
, (i = 1, ...,k − 1) (1.9)

as |z| = r → 1− for z ∈ H, then every nontrivial solution f of equation (1.1) satisfies σ(f ) =
σM(f ) = ∞ and σ2 (f ) = σM,2 (f ) = µ.

By Theorem 1.3, we easily obtain the following corollary.
Corollary 1.2 ([22]) Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying
densD {|z| : z ∈ H ⊆ D} > 0, and let A0, ...,Ak−1 be analytic functions in the unit disc D such
that

max
{
σM,[p,q] (Ai ) : i = 1, 2, ...,k − 1

}
≤ σM,[p,q] (A0) = µ (0 < µ < +∞)

and for some real constants α,β, where 0 ≤ β < α, we have

T (r,A0) ≥ expp−1
{
α

(
logq−1

(
1

1 −|z|

))µ}
and

T (r,Ai ) ≤ expp−1
{
β
(

logq−1

(
1
1−|z|

))µ}
(i = 1, ...,k − 1)

as |z| = r → 1− for z ∈ H. Then the following statements hold:

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 8
(i) If p = q = 1 and 0 ≤ (k − 1) β < α, then every nontrivial solution f of equation (1.1) satisfies
σ(f ) = σM(f ) = ∞ and σ2 (f ) = σM,2 (f ) = µ.
(ii) If p ≥ q ≥ 2 and 0 ≤ β < α, then every nontrivial solution f of equation (1.1) satisfies
σ[p,q] (f ) = σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) = µ.

Theorem 1.4 Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying

densD {|z| : z ∈ H ⊆ D} > 0,

and let A0, ...,Ak−1 be analytic functions in the unit disc D such that

max
{
σM,[p,q] (Ai ) : i = 1, 2, ...,k − 1

}
≤ σM,[p,q] (A0) = µ (0 < µ < +∞)

and if p ≥ q ≥ 2, we have

lim sup
|z|→1−,z∈H

logp−1T (r,Ai )(
logq−1

(
1
1−|z|

))µ < lim inf
|z|→1−,z∈H

logp−1T (r,A0)(
logq−1

(
1
1−|z|

))µ , (i = 1, ...,k − 1) (1.10)
as |z| = r → 1− for z ∈ H, then every nontrivial solution f of equation (1.1) satisfies σ[p,q] (f ) =
σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) = µ. If p = q = 1, we have

lim sup
|z|→1−,z∈H

(k − 1) T (r,Ai )(
1
1−|z|

)µ < lim inf
|z|→1−,z∈H

T (r,A0)(
1
1−|z|

)µ , (i = 1, ...,k − 1) (1.11)
as |z| = r → 1− for z ∈ H, then every nontrivial solution f of equation (1.1) satisfies σ(f ) =
σM(f ) = ∞ and σ2 (f ) = σM,2 (f ) = µ.

Theorem 1.5 Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying

densD {|z| : z ∈ H ⊆ D} > 0,

and let A0, ...,Ak be analytic functions in the unit disc D such that for some constants α ≥ 0 and
µ > 0, we have (1.3) and

|Ai (z)| ≤ expp
{
α
(

logq−1

(
1
1−|z|

))µ}
, (i = 1, ...,k)

as |z| → 1− for z ∈ H. Then every meromorphic (or analytic) solution f 6≡ 0 of equation
(1.2) satisfies σ[p,q] (f ) = ∞ and σ[p+1,q] (f ) ≥ µ.

Theorem 1.6 Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying

densD {|z| : z ∈ H ⊆ D} > 0,

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 9
and let A0, ...,Ak be analytic functions in the unit disc D such that for a constant µ > 0, we
have

lim sup
|z|→1−,z∈H

logp |Ai (z)|(
logq−1

(
1
1−|z|

))µ < lim inf
|z|→1−,z∈H

logp |A0 (z)|(
logq−1

(
1
1−|z|

))µ , (i = 1, ...,k)
as |z| → 1− for z ∈ H.Then every meromorphic (or analytic) solution f 6≡ 0 of equation
(1.2) satisfies σ[p,q] (f ) = ∞ and σ[p+1,q] (f ) ≥ µ.

Theorem 1.7 Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying

densD {|z| : z ∈ H ⊆ D} > 0,

and let A0, ...,Ak be analytic functions in the unit disc D such that for some constants α ≥ 0 and
µ > 0, if p ≥ q ≥ 2 we have

lim inf
|z|→1−,z∈H

logp−1T (r,A0)(
logq−1

(
1
1−|z|

))µ > α (1.12)
and

T (r,Ai ) ≤ expp−1
{
α
(

logq−1

(
1
1−|z|

))µ}
, (i = 1, ...,k) (1.13)

as |z| = r → 1− for z ∈ H, then every meromorphic (or analytic) solution f 6≡ 0 of equation
(1.2) satisfies σ[p,q] (f ) = ∞ and σ[p+1,q] (f ) ≥ µ. If p = q = 1, we have

lim inf
|z|→1−,z∈H

T (r,A0)(
1
1−|z|

)µ > kα (1.14)
and

T (r,Ai ) ≤ α
(
1
1−|z|

)µ
, (i = 1, ...,k) (1.15)

as |z| = r → 1− for z ∈ H, then every meromorphic (or analytic) solution f 6≡ 0 of equation
(1.2) satisfies σ (f ) = ∞ and σ2 (f ) ≥ µ.

Theorem 1.8 Let p ≥ q ≥ 1 be integers. Let H be a set of complex numbers satisfying

densD {|z| : z ∈ H ⊆ D} > 0,

and let A0 (z) , ...,Ak (z) be analytic functions in the unit disc D such that for a constant µ > 0,
if p ≥ q ≥ 2 we have

lim sup
|z|→1−,z∈H

logp−1T (r,Ai )(
logq−1

(
1
1−|z|

))µ < lim inf
|z|→1−,z∈H

logp−1T (r,A0)(
logq−1

(
1
1−|z|

))µ , (i = 1, ...,k) (1.16)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 10
as |z| = r → 1− for z ∈ H, then every meromorphic (or analytic) solution f 6≡ 0 of equation
(1.2) satisfies σ[p,q] (f ) = ∞ and σ[p+1,q] (f ) ≥ µ. If p = q = 1, we have

lim sup
|z|→1−,z∈H

kT (|z| ,Ai )(
1
1−|z|

)µ < lim inf
|z|→1−,z∈H

T (|z| ,A0)(
1
1−|z|

)µ , (i = 1, ...,k) (1.17)
as |z| = r → 1− for z ∈ H, then every meromorphic (or analytic) solution f 6≡ 0 of equation (1.2)
satisfies σ(f ) = ∞ and σ2 (f ) ≥ µ.

Remark 1.2 For equation (1.1) , we can easily conclude that Theorems A-C are generalized toTheorems 1.1-1.4.
In the same paper, Chen et al. [9] obtained some results of the fixed points of solutions andtheir arbitrary order derivatives of equations (1.1) and (1.2). Here, we generalize these results,and we obtain our theorems as following.

Theorem 1.9 Assume that the assumptions of Theorem 1.1 or Theorem 1.2 hold. Then every solution
f 6≡ 0 of equation (1.1) satisfies

λ̄[p,q]

(
f (j) −z

)
= λ[p,q] (f −z) = σ[p,q] (f ) = ∞,

λ̄[p+1,q]

(
f (j) −z

)
= λ̄[p+1,q] (f −z) = σ[p+1,q] (f ) = µ, (j = 1, 2, ...) .

Theorem 1.10 Assume that the assumptions of Theorem 1.3 or Theorem 1.4 hold. Then every
solution f 6≡ 0 of equation (1.1) satisfies

λ̄[p,q]

(
f (j) −z

)
= λ[p,q] (f −z) = σ[p,q] (f ) = ∞,

λ̄[p+1,q]

(
f (j) −z

)
= λ̄[p+1,q] (f −z) = σ[p+1,q] (f ) = µ, (j = 1, 2, ...) .

Theorem 1.11 Assume that the assumptions of one of Theorem 1.5 to Theorem 1.8 hold. Then every
meromorphic (or analytic) solution f 6≡ 0 of equation (1.2) satisfies

λ̄[p,q]

(
f (j) −z

)
= λ[p,q] (f −z) = σ[p,q] (f ) = ∞,

λ̄[p+1,q]

(
f (j) −z

)
= λ̄[p+1,q] (f −z) = σ[p+1,q] (f ) ≥ µ, (j = 1, 2, ...) .

2. Some lemmas
In this section we give some lemmas which are used in the proofs of our theorems.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 11
Lemma 2.1 ([11], Theorem 3.1) Let k and j be integers satisfying k > j ≥ 0, and let ε > 0 and
d ∈ (0, 1). If f is a meromorphic function in D such that f (j) does not vanish identically, then for
|z| /∈ E1 ∣∣∣∣∣f (k) (z)f (j) (z)

∣∣∣∣∣ ≤
[(

1

1 −|z|

)2+ε
max

{
log

(
1

1 −|z|

)
; T (s (|z|) , f )

}]k−j
,

where E1 ⊂ [0, 1) is a set with
∫
E1

dr
1−r < ∞ and s (|z|) = 1 −d (1 −|z|) .

Lemma 2.2 ([13]) Let f be a meromorphic function in the unit disc D, and let k ≥ 1 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 E2 ⊂ [0, 1) with ∫E2 dr1−r < ∞.

Lemma 2.3 ([1]) 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 E3 ⊂ [0, 1) for which

∫
E3

dr
1−r < ∞. Then

there exists a constant d ∈ (0, 1) such that if s (r) = 1 −d (1 − r) , then g (r) ≤ h (s (r)) for all
r ∈ [0, 1).

Lemma 2.4 ([3]) Let p ≥ q ≥ 1 be integers. If A0 (z) , ...,Ak−1 (z) are analytic functions of
[p,q]−order in the unit disc D, then every solution f 6≡ 0 of (1.1) satisfies

σ[p+1,q] (f ) = σM,[p+1,q] (f ) ≤ max
{
σM,[p,q]

(
Aj
)

: j = 0, 1, ...,k − 1
}
.

Lemma 2.5 ([4, 18]) Let p ≥ q ≥ 1 be integers. If f and g are non-constant meromorphic functions
of [p,q]−order in D, then we have
(i) σ[p,q] (f ) = σ[p,q](1f ) , σ[p,q] (af ) = σ[p,q] (f ) and σ[p,q] (f + a) = σ[p,q] (f ) (a ∈C∗) ,
(ii) σ[p,q] (f ′) = σ[p,q] (f ) ,
(iii) σ[p,q] (f + g) ≤ max {σ[p,q] (f ) ,σ[p,q] (g)} ,
(iv) σ[p,q] (f g) ≤ max {σ[p,q] (f ) ,σ[p,q] (g)} , if σ[p,q] (f ) > σ[p,q] (g) , then we obtain
σ[p,q] (f + g) = σ[p,q] (f g) = σ[p,q] (f ) .

Lemma 2.6 ([4]) Let p ≥ q ≥ 1 be integers. Let A0, ...,Ak−1 and F 6≡ 0 be finite [p,q]−order
analytic functions in the unit disc D. If f is a solution with σ[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σ < ∞
of equation

f (k) + Ak−1 (z) f
(k−1) + · · · + A1 (z) f ′ + A0 (z) f = F, (2.1)

then
λ̄[p,q] (f ) = λ[p,q] (f ) = σ[p,q] (f ) = ∞,

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

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 12
By using the same arguments of the proof of Lemma 3.5 in the paper [14, p. 4], we obtain thefollowing lemma in the case when σ[p,q] (f ) = σ = ∞.

Lemma 2.7 Let p ≥ q ≥ 1 be integers. Let Aj (j = 0, ...,k − 1) , F 6≡ 0 be meromorphic functions
in D, and let f be a solution of the differential equation (2.1) satisfying

max
{
σ[p,q]

(
Aj
)

(j = 0, ...,k − 1) ,σ[p,q] (F )
}
< σ[p,q] (f ) = σ ≤∞.

Then we have

λ[p,q] (f ) = λ[p,q] (f ) = σ[p,q] (f )

and

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

3. Proofs of Theorems 1.1 to 1.8
Proof of Theorem 1.1. Suppose that every solution f of equation (1.1) not being identically equalto 0. From the conditions of Theorem 1.1, there exists a set H of complex numbers satisfying
densDH1 > 0, where H1 = {r = |z| : z ∈ H ⊆ D} . Then H1 is a set with ∫H1 dr1−r = +∞, suchthat for z ∈ H we have (1.3) and (1.4) as |z|→ 1−. By Lemma 2.1, there exists a set E1 ⊂ [0, 1)with ∫

E1
dr
1−r < ∞ such that for |z| /∈ E1, we have for j = 1, ...,k∣∣∣∣∣f (j) (z)f (z)

∣∣∣∣∣ ≤
[(

1

1 −|z|

)2+ε
max

{
log

(
1

1 −|z|

)
,T (s (|z|) , f )

}]j
, (3.1)

where s (|z|) = 1 −d (1 −|z|) , d ∈ (0, 1). From (1.1) , we get
|A0 (z)| ≤

∣∣∣∣∣f (k)f
∣∣∣∣∣ + |Ak−1 (z)|

∣∣∣∣∣f (k−1)f
∣∣∣∣∣ + · · · + |A1 (z)|

∣∣∣∣f ′f
∣∣∣∣ . (3.2)

By (1.3), we know that
∃γ ∈R : lim inf

|z|→1−,z∈H

logp |A0 (z)|(
logq−1

(
1
1−|z|

))µ > γ > α.
Obviously

logp |A0 (z)|(
logq−1

(
1
1−|z|

))µ > γ > α ≥ 0 (3.3)
as |z|→ 1− for z ∈ H. By (1.4) and (3.3) , we obtain

|A0 (z)| > expp
{
γ

(
logq−1

(
1

1 −|z|

))µ}
> expp

{
α

(
logq−1

(
1

1 −|z|

))µ}
≥ |Ai (z)| (i = 1, 2, ...,k − 1) (3.4)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 13
as |z|→ 1− for z ∈ H. Applying (3.1) and (3.4) into (3.2) , we have

expp

{
γ

(
logq−1

(
1

1 −|z|

))µ}
≤ |A0 (z)|

≤ k

[(
1

1 −|z|

)2+ε
max

{
log

(
1

1 −|z|

)
,T (s (|z|) , f )

}]k
×expp

{
α

(
logq−1

(
1

1 −|z|

))µ}
holds for all z satisfying |z| ∈ H1\E1 as |z|→ 1−. Noting that γ > α, by the last inequality, weobtain

exp

(
(1 −o (1)) expp−1

{
γ

(
logq−1

(
1

1 −|z|

))µ})
≤ k

(
1

1 −|z|

)k(2+ε)
Tk (s (|z|) , f ) (3.5)

for all z satisfying |z| ∈ H1\E1 as |z| → 1−. Then, by (3.5) and combining with Lemma 2.3, weget for all r = |z| ∈ H1
exp

(
(1 −o(1)) expp−1

{
γ

(
logq−1

(
1

1 − r

))µ})
≤ k

(
1

1 − s (r)

)k(2+ε)
Tk (s1 (r) , f ) , (3.6)

where s1 (r) = 1 − d2 (1 − r) with d ∈ (0, 1). Therefore, from (3.6) we obtain σ[p,q] (f ) =
σM,[p,q] (f ) = ∞ and

σ[p+1,q] (f ) = σM,[p+1,q] (f ) = lim sup
s1(r)→1−

log+p+1T (s1 (r) , f )

logq

(
1

1−s1(r)

) ≥ µ. (3.7)
By Lemma 2.4, we get

σ[p+1,q] (f ) = σM,[p+1,q] (f )

≤ max
{
σM,[p,q] (Ai ) : i = 0, 1, ...,k − 1

}
= σM,[p,q] (A0) = µ. (3.8)Therefore, by (3.7) and (3.8) , we obtain σ[p,q] (f ) = σM,[p,q] (f ) = ∞ and

σ[p+1,q] (f ) = σM,[p+1,q] (f ) = σM,[p,q] (A0) = µ.

Proof of Theorem 1.2. Set
α0 = lim inf

|z|→1−,z∈H

logp |A0 (z)|(
logq−1

(
1
1−|z|

))µ ,
αi = lim sup

|z|→1−,z∈H

logp |Ai (z)|(
logq−1

(
1
1−|z|

))µ , (i = 1, 2, ...,k − 1) .

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 14
By (1.5), there exist real numbers α,γ such that αi < α < γ < α0, i = 1, 2, ...,k − 1. It yields

logp |Ai (z)|(
logq−1

(
1
1−|z|

))µ < α < γ < logp |A0 (z)|(
logq−1

(
1
1−|z|

))µ
as |z| → 1− for z ∈ H. Hence, we have (3.4) as |z| → 1− for z ∈ H. Then, by using the sameproof of Theorem 1.1, we get

σ[p,q] (f ) = σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) ≥ µ
and by Lemma 2.4 we obtain the conclusion of Theorem 1.2.
Proof of Theorem 1.3. Suppose that every solution f of equation (1.1) not being identically equalto 0. By (1.1) , we can write

−A0 (z) =
f (k)(z)

f (z)
+ Ak−1 (z)

f (k−1)(z)

f (z)
+ · · · + A1 (z)

f ′(z)

f (z)
. (3.9)

From (3.9) , we obtain
T (r,A0) = m(r,A0) ≤

k−1∑
i=1

m(r,Ai ) +

k∑
i=1

m

(
r,
f (i)

f

)
+ O(1)

=

k−1∑
i=1

T (r,Ai ) +

k∑
i=1

m

(
r,
f (i)

f

)
+ O(1). (3.10)

If p ≥ q ≥ 2, then by (1.6), we know that
∃γ ∈R : lim inf

|z|→1−,z∈H

logp−1T (r,A0)(
logq−1

(
1
1−|z|

))µ > γ > α.
Obviously

logp−1T (r,A0)(
logq−1

(
1
1−|z|

))µ > γ > α ≥ 0 (3.11)
as |z|→ 1− for z ∈ H. By (1.7) and (3.11) , we obtain

T (r,A0) > expp−1

{
γ

(
logq−1

(
1

1 −|z|

))µ}
> expp−1

{
α

(
logq−1

(
1

1 −|z|

))µ}
≥ T (r,Ai ) , (i = 1, 2, ...,k − 1) (3.12)

as |z|→ 1− for z ∈ H. By applying Lemma 2.2 and substituting (3.12) into (3.10) , we get
expp−1

{
γ

(
logq−1

(
1

1 − r

))µ}
≤ (k − 1) expp−1

{
α

(
logq−1

(
1

1 − r

))µ}
+O

(
log+T (r, f ) + log

(
1

1 − r

))

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 15
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Noting that γ > α, by the last inequality,we have

exp

{
(1 −o (1)) expp−2

{
γ

(
logq−1

(
1

1 − r

))µ}}
≤ O

(
log+T (r, f ) + log

(
1

1 − r

)) (3.13)
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Therefore, from (3.13) we obtain

σ[p,q] (f ) = σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) ≥ µ. (3.14)
By Lemma 2.4, we get

σ[p+1,q] (f ) = σM,[p+1,q] (f )

≤ max
{
σM,[p,q] (Ai ) : i = 0, 1, ...,k − 1

}
= σM,[p,q] (A0) = µ. (3.15)Therefore, by (3.14) and (3.15) , we obtain

σ[p,q] (f ) = σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) = µ.
If p = q = 1, then by (1.8), we know that

∃γ ∈R : lim inf
|z|→1−,z∈H

T (r,A0)(
1
1−|z|

)µ > γ > (k − 1) α.
Obviously

T (r,A0)(
1
1−|z|

)µ > γ > (k − 1) α ≥ 0 (3.16)
as |z|→ 1− for z ∈ H. By (1.9) and (3.16) , we obtain

T (r,A0) > γ

(
1

1 −|z|

)µ
> (k − 1) α

(
1

1 −|z|

)µ
≥ α

(
1

1 −|z|

)µ
≥ T (r,Ai ) , (i = 1, 2, ...,k − 1) (3.17)

as |z|→ 1− for z ∈ H. By applying Lemma 2.2 and substituting (3.17) into (3.10) , we get
γ

(
1

1 − r

)µ
≤ (k − 1) α

(
1

1 − r

)µ
+O

(
log+T (r, f ) + log

(
1

1 − r

))
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Noting that γ > (k − 1) α, by the lastinequality, we have

(γ − (k − 1) α)
(

1

1 − r

)µ
≤ O

(
log+T (r, f ) + log

(
1

1 − r

)) (3.18)
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Therefore, from (3.18) we obtain

σ (f ) = σM (f ) = ∞ and σ2 (f ) = σM,2 (f ) ≥ µ. (3.19)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 16
By Lemma 2.4, we get

σ2 (f ) = σM,2 (f ) ≤ max{σM (Ai ) : i = 0, 1, ...,k − 1} = σM (A0) = µ. (3.20)
Therefore, by (3.19) and (3.20) , we obtain σ (f ) = σM (f ) = ∞ and σ2 (f ) = σM,2 (f ) = µ.
Proof of Theorem 1.4. If p ≥ q ≥ 2, we set

α0 = lim inf
|z|→1−,z∈H

logp−1T (r,A0)(
logq−1

(
1
1−|z|

))µ ,
αi = lim sup

|z|→1−,z∈H

logp−1T (r,Ai )(
logq−1

(
1
1−|z|

))µ , (i = 1, 2, ...,k − 1) .
By (1.10), there exist real numbers α,γ such that αi < α < γ < α0, i = 1, 2, ...,k − 1. It yields

logp−1T (r,Ai )(
logq−1

(
1
1−|z|

))µ < α < γ < logp−1T (r,A0)(
logq−1

(
1
1−|z|

))µ
as |z|→ 1− for z ∈ H. Hence, we have

T (r,A0) > expp−1

{
γ

(
logq−1

(
1

1 −|z|

))µ}
> expp−1

{
α

(
logq−1

(
1

1 −|z|

))µ}
≥ T (r,Ai ) , (i = 1, 2, ...,k − 1)

as |z|→ 1− for z ∈ H. Then, by using the same proof of Theorem 1.3, we get
σ[p,q] (f ) = σM,[p,q] (f ) = ∞ and σ[p+1,q] (f ) = σM,[p+1,q] (f ) ≥ µ,

and by Lemma 2.4 we obtain the conclusion of Theorem 1.4.If p = q = 1, we set
α0 = lim inf

|z|→1−,z∈H

T (r,A0)(
1
1−|z|

)µ ,
αi = lim sup

|z|→1−,z∈H

(k − 1) T (r,Ai )(
1
1−|z|

)µ , (i = 1, 2, ...,k − 1) .
By (1.11), there exist real numbers α,γ such that αi < α < γ < α0, i = 1, 2, ...,k − 1. It yields

(k − 1) T (r,Ai )(
1
1−|z|

)µ < α < γ < T (r,A0)(
1
1−|z|

)µ (3.21)
as |z|→ 1− for z ∈ H. By (3.21) , we obtain

T (r,A0) > γ

(
1

1 −|z|

)µ
> α

(
1

1 −|z|

)µ
≥ (k − 1) T (r,Ai ) , (i = 1, 2, ...,k − 1) (3.22)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 17
as |z|→ 1− for z ∈ H. By applying Lemma 2.2 and substituting (3.22) into (3.10) , we get

γ

(
1

1 − r

)µ
≤ α

(
1

1 − r

)µ
+ O

(
log+T (r, f ) + log

(
1

1 − r

))
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Noting that γ > α, by the last inequality,we have

(γ −α)
(

1

1 − r

)µ
≤ O

(
log+T (r, f ) + log

(
1

1 − r

)) (3.23)
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Therefore, from (3.23) we obtain

σ (f ) = σM (f ) = ∞ and σ2 (f ) = σM,2 (f ) ≥ µ. (3.24)
By Lemma 2.4, we get

σ2 (f ) = σM,2 (f ) ≤ max{σM (Ai ) : i = 0, 1, ...,k − 1} = σM (A0) = µ. (3.25)
Therefore, by (3.24) and (3.25) , we obtain σ (f ) = σM (f ) = ∞ and σ2 (f ) = σM,2 (f ) = µ.
Proof of Theorems 1.5 and 1.6. Suppose that every meromorphic (or analytic) solution f ofequation (1.2) not being identically equal to 0. From (1.2) , we get

A0 (z) ≤ |Ak (z)|

∣∣∣∣∣f (k)f
∣∣∣∣∣ + |Ak−1 (z)|

∣∣∣∣∣f (k−1)f
∣∣∣∣∣ + · · · + |A1 (z)|

∣∣∣∣f ′f
∣∣∣∣ . (3.26)

By using a similar proof as in Theorem 1.1 or Theorem 1.2, we obtain
|A0 (z)| > expp

{
γ

(
logq−1

(
1

1 −|z|

))µ}
> expp

{
α

(
logq−1

(
1

1 −|z|

))µ}
≥ |Ai (z)| (i = 1, 2, ...,k) (3.27)

for |z| ∈ H1\E1 as |z|→ 1−. Applying (3.1) and (3.27) into (3.26) , we get
expp

{
γ

(
logq−1

(
1

1 −|z|

))µ}
≤ |A0 (z)|

≤ k

[(
1

1 −|z|

)2+ε
max

{
log

(
1

1 −|z|

)
,T (s (|z|) , f )

}]k
×expp

{
α

(
logq−1

(
1

1 −|z|

))µ}
for all z satisfying |z| ∈ H1\E1 as |z|→ 1−. Noting that γ > α, by the last inequality, we have

exp

(
(1 −o (1)) expp−1

{
γ

(
logq−1

(
1

1 −|z|

))µ})

≤ k
(

1

1 −|z|

)k(2+ε)
Tk (s (|z|) , f ) (3.28)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 18
for all z satisfying |z| ∈ H1\E1 as |z| → 1−. Then, by (3.28) and combining with Lemma 2.3, weget for all r = |z| ∈ H1

exp

(
(1 −o(1)) expp−1

{
γ

(
logq−1

(
1

1 − r

))µ})
≤ k

(
1

1 − s (r)

)k(2+ε)
Tk (s1 (r) , f ) , (3.29)

where s1 (r) = 1−d2 (1 − r) with d ∈ (0, 1). Therefore, from (3.29) we obtain σ[p,q] (f ) = ∞ and
σ[p+1,q] (f ) = lim sup

s1(r)→1−

log+p+1T (s1 (r) , f )

logq

(
1

1−s1(r)

) ≥ µ.
Proof of Theorems 1.7 and 1.8. Suppose that every meromorphic (or analytic) solution f ofequation (1.2) not being identically equal to 0. By (1.2) , we can write

−A0 (z) = Ak (z)
f (k)(z)

f (z)
+ Ak−1 (z)

f (k−1)(z)

f (z)
+ · · · + A1 (z)

f ′(z)

f (z)
. (3.30)

From (3.30) , we have
T (r,A0) = m(r,A0) ≤

k∑
i=1

m(r,Ai ) +

k∑
i=1

m

(
r,
f (i)

f

)
+ O(1)

=

k∑
i=1

T (r,Ai ) +

k∑
i=1

m

(
r,
f (i)

f

)
+ O(1). (3.31)

If p ≥ q ≥ 2, then by using a similar proof as in Theorem 1.3 or Theorem 1.4, we obtain
T (r,A0) > expp−1

{
γ

(
logq−1

(
1

1 −|z|

))µ}
> expp−1

{
α

(
logq−1

(
1

1 −|z|

))µ}
≥ T (r,Ai ) , (i = 1, 2, ...,k) (3.32)

as |z|→ 1− for z ∈ H. By applying Lemma 2.2 and substituting (3.32) into (3.31) , we get
expp−1

{
γ

(
logq−1

(
1

1 − r

))µ}
≤ k expp−1

{
α

(
logq−1

(
1

1 − r

))µ}
+O

(
log+T (r, f ) + log

(
1

1 − r

))
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Noting that γ > α, by the last inequality,we have

exp

{
(1 −o (1)) expp−2

{
γ

(
logq−1

(
1

1 − r

))µ}}
≤ O

(
log+T (r, f ) + log

(
1

1 − r

)) (3.33)
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Therefore, from (3.33) we obtain

σ[p,q] (f ) = ∞ and σ[p+1,q] (f ) ≥ µ.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 19
If p = q = 1, then by using a similar proof as in Theorem 1.3 or Theorem 1.4, we get

T (r,A0) > γ

(
1

1 −|z|

)µ
> kα

(
1

1 −|z|

)µ
> α

(
1

1 −|z|

)µ
≥ T (r,Ai ) , (i = 1, 2, ...,k) (3.34)

as |z|→ 1− for z ∈ H. By applying Lemma 2.2 and substituting (3.34) into (3.31) , we obtain
γ

(
1

1 − r

)µ
≤ kα

(
1

1 − r

)µ
+ O

(
log+T (r, f ) + log

(
1

1 − r

))
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Noting that γ > kα, by the last inequality,we have

(γ −kα)
(

1

1 − r

)µ
≤ O

(
log+T (r, f ) + log

(
1

1 − r

)) (3.35)
for all z satisfying |z| = r ∈ H1\E2 as |z| = r → 1−. Therefore, from (3.35) we obtain

σ (f ) = σM (f ) = ∞ and σ2 (f ) = σM,2 (f ) ≥ µ.
4. Proof of Theorem 1.9

Suppose that every solution f of equation (1.1) not being identically equal to 0.
First step. We consider the fixed points of f . Define the function g by setting

g (z) := f (z) −z, z ∈ D.

It follows from (1.1) that
g(k) + Ak−1g

(k−1) + · · · + A1g′ + A0g = −A1 −zA0 (4.1)
and by Theorem 1.1 or Theorem 1.2, we get

σ[p,q] (g) = σ[p,q] (f ) = ∞, σ[p+1,q] (g) = σ[p+1,q] (f ) = µ,
λ̄[p+1,q] (g) = λ̄[p+1,q] (f −z) .

(4.2)
Now, we prove that −A1 − zA0 6≡ 0. Assume that −A1 − zA0 ≡ 0. Clearly A0 6≡ 0. Then

lim
|z|→1−,z∈H

∣∣∣A1A0∣∣∣ = 1 and by (3.4), we have
∣∣∣∣A1 (z)A0 (z)

∣∣∣∣ < expp
{
α
(

logq−1

(
1
1−|z|

))µ}
expp

{
γ
(

logq−1

(
1
1−|z|

))µ}
=

1

exp
{

(1 −o (1)) expp−1
{
γ
(

logq−1

(
1
1−|z|

))µ}} → 0

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 20
as |z| → 1− for z ∈ H. Then lim

|z|→1−,z∈H

∣∣∣A1A0∣∣∣ = 0. It is easy to see the contradiction. Hence,
−A1 −zA0 6≡ 0. Next by Lemma 2.5, we get

max
{
σ[p,q] (Ai ) (i = 0, 1, ...,k − 1) ,σ[p,q] (−A1 −zA0)

}
< ∞.

We deduce, by using (4.1) , (4.2) and Lemma 2.6 that
λ̄[p,q] (g) = σ[p,q] (g) = ∞, λ̄[p+1,q] (g) = σ[p+1,q] (g) = µ.

Therefore, we obtain
λ̄[p,q] (f −z) =λ̄[p,q] (g) = σ[p,q] (g) = σ[p,q] (f ) = ∞,

λ̄[p+1,q] (f −z) = λ̄[p+1,q] (g) = σ[p+1,q] (g) = σ[p+1,q] (f ) = µ.

Second step. For the following proof, we use the principle of mathematical induction. Set Ak (z) ≡
1, then

|Ak (z)| ≤ expp
{
α

(
logq−1

(
1

1 −|z|

))µ}
and equation (1.1) becomes (1.2) . We consider the fixed points of f (j) (z) (j = 1, 2, ...). Definethe function g1 by setting

g1 (z) := f
′ (z) −z, z ∈ D.

Then, by Lemma 2.5 and (4.2) , we have
σ[p,q] (g1) = σ[p,q] (f

′) = ∞, σ[p+1,q] (g1) = σ[p+1,q] (f ′) = µ,
λ̄[p+1,q] (g1) = λ̄[p+1,q] (f

′ −z) .
(4.3)

Dividing both sides of (1.2) by A0, we obtain
Ak
A0
f (k) +

Ak−1
A0

f (k−1) + · · · +
A1
A0
f ′ + f = 0. (4.4)

It follows, by differentiating both sides of equation (4.4) that
Ak
A0
f (k+1) +

((
Ak
A0

)′
+

Ak−1
A0

)
f (k) + · · · +

((
A2
A0

)′
+ A1

A0

)
f ′′

+

((
A1
A0

)′
+ 1

)
f ′ = 0.

(4.5)
Multiplying (4.5) by A0, we have

Ak,1f
(k+1) + Ak−1,1f

(k) + · · · + A1,1f ′′ + A0,1f ′ = 0. (4.6)
Substituing f ′ = g1 + z into (4.6) , we obtain

Ak,1g
(k)
1 + Ak−1,1g

(k−1)
1 + · · · + A1,1g

′
1 + A0,1g1 = F1, (4.7)

where
Ak,1 = Ak = 1, Ai,1 = A0

((
Ai+1
A0

)′
+

Ai
A0

)
(i = 1, 2, ...,k − 1) , (4.8)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 21
A0,1 = A0

((
A1
A0

)′
+ 1

)
, (4.9)

F1 = −(A1,1 + zA0,1) . (4.10)
Next, we prove that A0,1 6≡ 0 and F1 6≡ 0. Assume that A0,1 ≡ 0, then A1A0 = −z + C0, where C0 isan arbitrary constant. Hence, we have A1 + (z −C0) A0 = 0. Then, f0 = z −C0 is a solution of
(1.1) and σ[p,q] (f0) < ∞. This contradicts (4.2) . Now, assume that F1 ≡ 0. By (4.6) and (4.10) ,we know that the function f1 such that f ′1 = z is a solution of equation (4.6) and σ[p,q] (f1) < ∞.This contradicts (4.2) . Therefore, A0,1 6≡ 0 and F1 6≡ 0. It follows by (4.8) − (4.10) and Lemma2.5 that

max
{
σ[p,q] (Ai,1) (i = 0, 1, ...,k) ,σ[p,q] (F1)

}
< ∞.

We deduce by using (4.3) , (4.7) and Lemma 2.6 that
λ̄[p,q] (g1) = σ[p,q] (g1) = ∞, λ̄[p+1,q] (g1) = σ[p+1,q] (g1) = µ.

Therefore, we obtain
λ̄[p,q]

(
f ′ −z

)
= λ̄[p,q] (g1) = σ[p,q] (g1) = σ[p,q] (f ) = ∞,

λ̄[p+1,q]
(
f ′ −z

)
= λ̄[p+1,q] (g1) = σ[p+1,q] (g1) = σ[p+1,q] (f ) = µ.Set g2 (z) = f ′′ (z) − z, z ∈ D. Then, by using a similar discussion as in the case of the function

g1, we can get
Ak,2f

(k+2) + Ak−1,2f
(k+1) + · · · + A1,2f (3) + A0,2f ′′ = 0

and
Ak,2g

(k)
2 + Ak−1,2g

(k−1)
2 + · · · + A1,2g

′
2 + A0,2g2 = F2,where

Ak,2 = 1, Ai,2 = A0,1

((
Ai+1,1
A0,1

)′
+

Ai,1
A0,1

)
(i = 1, 2, ...,k − 1) ,

A0,2 = A0,1

((
A1,1
A0,1

)′
+ 1

)
,

F2 = −(A1,2 + zA0,2) .

Therefore, by the same procedure as for g1, we obtain
λ̄[p,q]

(
f ′′ −z

)
= λ̄[p,q] (g2) = σ[p,q] (g2) = σ[p,q] (f ) = ∞,

λ̄[p+1,q]
(
f ′′ −z

)
= λ̄[p+1,q] (g2) = σ[p+1,q] (g2) = σ[p+1,q] (f ) = µ.Now, assume that

A0,s 6≡ 0,
λ̄[p,q]

(
f (s) −z

)
= σ[p,q] (f ) = ∞,

λ̄[p+1,q]
(
f (s) −z

)
= σ[p+1,q] (f ) = µ

(4.11)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 22
for all s = 0, 1, ..., j−1, and we prove that for s = j we have (4.11) holds. Set gj (z) = f (j) (z)−z,
z ∈ D. Then, by using (4.2) , we obtain

σ[p,q]
(
gj
)

= σ[p,q]
(
f (j)
)

= ∞, σ[p+1,q]
(
gj
)

= σ[p+1,q]
(
f (j)
)

= µ,

λ̄[p+1,q]
(
gj
)

= λ̄[p+1,q]
(
f (j) −z

)
.

(4.12)
By following the same procedure as before, we have

Ak,jf
(k+j) + Ak−1,jf

(k+j−1) + · · · + A1,jf (j+1) + A0,jf (j) = 0 (4.13)
and

Ak,jg
(k)
j

+ Ak−1,jg
(k−1)
j

+ · · · + A1,jg′j + A0,jgj = Fj, (4.14)where
Ak,j = 1, Ai,j = A0,j−1

((
Ai+1,j−1
A0,j−1

)′
+

Ai,j−1
A0,j−1

)
(i = 1, 2, ...,k − 1) ,

A0,j = A0,j−1

((
A1,j−1
A0,j−1

)′
+ 1

)
6≡ 0 (A0,0 = A0, A1,0 = A1) ,

Fj = −
(
A1,j + zA0,j

)
6≡ 0.

We deduce by applying Lemma 2.6 in (4.14) that
λ̄[p,q]

(
f (j) −z

)
= λ̄[p,q]

(
gj
)

= σ[p,q]
(
gj
)

= σ[p,q]

(
f (j)
)

= ∞,

λ̄[p+1,q]

(
f (j) −z

)
= λ̄[p+1,q]

(
gj
)

= σ[p+1,q]
(
gj
)

= σ[p+1,q]

(
f (j)
)

= µ (j = 1, 2, ...) .

Therefore, we obtain
λ̄[p,q]

(
f (j) −z

)
= λ̄[p,q] (f −z) = σ[p,q] (f ) = ∞,

λ̄[p+1,q]

(
f (j) −z

)
= λ̄[p+1,q] (f −z) = σ[p+1,q] (f ) = µ (j = 1, 2, ...) .

5. Proofs of Theorem 1.10 and 1.11
Proof of Theorem 1.10. Suppose that every solution f of equation (1.1) not being identicallyequal to 0. By applying Theorem 1.3 or Theorem 1.4, we get

σ[p,q] (f ) = ∞, σ[p+1,q] (f ) = µ.

Now, we prove that −A1 −zA0 6≡ 0. Assume that −A1 −zA0 ≡ 0, then we can easily obtain
T (r,A1) = T (r,−zA0) ≤ T (r,A0) + T (r,z) ,
T (r,A0) = T

(
r, A1−z

)
≤ T (r,A1) + T (r,z) + O (1) .

(5.1)
It follows from (5.1) that

1 −
T (r,z) + O (1)

T (r,A0)
≤
T (r,A1)

T (r,A0)
≤ 1 +

T (r,z)

T (r,A0)
. (5.2)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 23
By following the same reasoning as in the proof of Theorem 1.3 or Theorem 1.4, we have

T (r,A0) > expp−1

{
γ

(
logq−1

(
1

1 −|z|

))µ}

> expp−1

{
α

(
logq−1

(
1

1 −|z|

))µ}
≥ T (r,A1) (5.3)

as r = |z|→ 1− for z ∈ H. By using (5.3) , we obtain
T (r,z)

T (r,A0)
≤

T (r,z)

expp−1

{
γ
(

logq−1

(
1
1−|z|

))µ} → 0 (5.4)
as |z|→ 1− for z ∈ H. Then, by (5.2) and (5.4) , we get

lim
|z|→1−,z∈H

T (r,A1)

T (r,A0)
= 1. (5.5)

On the other hand, we have for p = q = 1
T (r,A1)

T (r,A0)
<
α

γ
< 1 (5.6)

and for p ≥ q ≥ 2
T (r,A1)

T (r,A0)
<

expp−1

{
α
(

logq−1

(
1
1−|z|

))µ}
expp−1

{
γ
(

logq−1

(
1
1−|z|

))µ} → 0 (5.7)
as |z|→ 1− for z ∈ H. It follows by (5.6) and (5.7) that

lim
|z|→1−,z∈H

T (r,A1)

T (r,A0)
6= 1. (5.8)

Obviously, (5.5) contradicts with (5.8). Hence, −A1 −zA0 6≡ 0. Set Ak (z) ≡ 1, then T (r,Ak) ≤
expp−1

{
α
(

logq−1

(
1
1−|z|

))µ}
. Clearly, A0 6≡ 0. We can get the conclusion of Theorem 1.10, byreasoning in the same way as we did in the proof of Theorem 1.9.

Proof of Theorem 1.11. Suppose that every meromorphic (or analytic) solution f of equation (1.2)not being identically equal to 0. By applying one of Theorem 1.5 to Theorem 1.8, we get
σ[p,q] (f ) = ∞, σ[p+1,q] (f ) ≥ µ.

Then, we can get the conclusion of Theorem 1.11, by reasoning in the same way as we did in theproof of Theorem 1.9 and Theorem 1.10 by using σ[p+1,q] (f ) ≥ µ instead of σ[p+1,q] (f ) = µ and
σ[p+1,q]

(
f (j)
)
≥ µ instead of σ[p+1,q](f (j)) = µ (j = 1, 2, ...) .

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 24
6. Examples

Example 6.1 Consider the following differential equation
f ′′ + K1 (z) exp4

{(
log2

(
1

1 −z

))5}
f ′

+ K0 (z) exp4

{
3

(
log2

(
1

1 −z

))5}
f = 0, (6.1)

where K0 and K1 are analytic functions in the unit disc D such that |K0| > 1, |K1| < 1 and
max

{
σM,[4,3] (K0) ,σM,[4,3] (K1)

}
< 5.

In the equation (6.1) , we have
A0 (z) = K0 (z) exp4

{
3

(
log2

(
1

1 −z

))5}
,

A1 (z) = K1 (z) exp4

{(
log2

(
1

1 −z

))5}
.

Then
max

{
σM,[4,3] (A0) ,σM,[4,3] (A1)

}
= 5.

Let H = {z ∈C : |z| = r < 1 and arg z = 0}⊂ D be a set of complex numbers satisfying
densD {|z| : z ∈ H} = 1 > 0.

Then
|A0 (z)| = |K0 (z)|

∣∣∣∣∣exp4
{

3

(
log2

(
1

1 −z

))5}∣∣∣∣∣
> exp4

{
3

(
log2

(
1

1 − r

))5}
⇒

log4 |A0 (z)|(
log2

(
1
1−r
))5 > 3 ⇒ lim inf

r→1−,z∈H

log4 |A0 (z)|(
log2

(
1
1−r
))5 ≥ 3 > 1,

and
|A1 (z)| = |K1 (z)|

∣∣∣∣∣exp4
{(

log2

(
1

1 −z

))5}∣∣∣∣∣
≤ exp4

{(
log2

(
1

1 − r

))5}
as r → 1− for z ∈ H. It is clear that the conditions of Theorem 1.1 hold with α = 1,µ = 5, p = 4and q = 3 on the set H. By Theorem 1.1, every solution f 6≡ 0 of equation (6.1) satisfies

σ[4,3] (f ) = σM,[4,3] (f ) = ∞ and σ[5,3] (f ) = σM,[5,3] (f ) = σM,[5,3] (A0) = 5.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 25
Example 6.2 Consider the following differential equation

K2 (z) exp4

{(
log2

(
1
1−z
))7}

f ′′+K1 (z) exp4

{
2
(

log2
(
1
1−z
))7}

f ′

+ K0 (z) exp4

{
5
(

log2
(
1
1−z
))7}

f = 0, (6.2)
where K0,K1 and K2 are analytic functions in the unit disc D such that |K0| > 1, |K1| < 1, |K2| < 1and

max
{
σM,[4,3] (K0) ,σM,[4,3] (K1) ,σM,[4,3] (K2)

}
< 7.

In the equation (6.2) , we have
A0 (z) = K0 (z) exp4

{
5
(

log2
(
1
1−z
))7}

,

A1 (z) = K1 (z) exp4

{
2
(

log2
(
1
1−z
))7}

,

A2 (z) = K2 (z) exp4

{(
log2

(
1
1−z
))7}

.

Then
max

{
σM,[4,3] (A0) ,σM,[4,3] (A1) ,σM,[4,3] (A2)

}
= 7.

Let H = {z ∈C : |z| = r < 1 and arg z = 0}⊂ D be a set of complex numbers satisfying
densD {|z| : z ∈ H} = 1 > 0.

Then
|A0 (z)| = |K0 (z)|

∣∣∣exp4{5 (log2( 11−z))7}∣∣∣
> exp4

{
5
(

log2
(
1
1−r
))7}

⇒
log4 |A0 (z)|(
log2

(
1
1−r
))7 > 5 ⇒ lim inf

r→1−,z∈H

log4 |A0 (z)|(
log2

(
1
1−r
))7 ≥ 5 > 2,

and
|A1 (z)| = |K1 (z)|

∣∣∣∣∣exp4
{

2

(
log2

(
1

1 −z

))7}∣∣∣∣∣
≤ exp4

{
2

(
log2

(
1

1 − r

))7}

|A2 (z)| = |K2 (z)|

∣∣∣∣∣exp4
{(

log2

(
1

1 −z

))7}∣∣∣∣∣
≤ exp4

{
2

(
log2

(
1

1 − r

))7}

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Eur. J. Math. Anal. 10.28924/ada/ma.3.10 26
as r → 1− for z ∈ H. It is clear that the conditions of Theorem 1.5 hold with α = 2, µ = 7, p = 4and q = 3 on the set H. By Theorem 1.11, every meromorphic (or analytic) solution f 6≡ 0 ofequation (6.2) satisfies

λ̄[4,3]

(
f (j) −z

)
= λ̄[4,3] (f −z) = σ[4,3] (f ) = ∞

and
λ̄[5,3]

(
f (j) −z

)
= λ̄[5,3] (f −z) = σ[5,3] (f ) ≥ 7, (j = 1, 2, ...) .

References
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https://doi.org/10.28924/ada/ma.3.10
http://www.numdam.org/item/CM_1972__25_1_61_0
http://real.mtak.hu/23284
http://ftp.gwdg.de/pub/EMIS/journals/EJDE/Volumes/2011/156/abstr.html
https://doi.org/10.1090/S0002-9939-1987-0902549-5
https://doi.org/10.1016/j.jmaa.2005.09.050
https://doi.org/10.1016/j.jmaa.2008.11.033
https://doi.org/10.1186/s13662-021-03579-3
https://doi.org/10.2996/kmj/1138044047
https://doi.org/10.1112/S0024611502013965
https://doi.org/10.1515/crll.1977.290.180
https://doi.org/10.1515/crll.1977.290.180


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https://doi.org/10.28924/ada/ma.3.10
http://doi.org/10.5644/SJM.09.1.06
https://doi.org/10.1016/j.jmaa.2010.05.014
https://doi.org/10.1186/1687-1847-2014-167

	1. Introduction and main results
	2. Some lemmas
	3. Proofs of Theorems 1.1 to 1.8
	4. Proof of Theorem 1.9
	5. Proofs of Theorem 1.10 and 1.11
	6. Examples
	References

