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Some Properties on The [p,q]-Order of Meromorphic Solutions of Homogeneous and

Non-homogeneous Linear Differential Equations With Meromorphic Coefficients

Mansouria Saidani, Benharrat Belaïdi∗
Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem

(UMAB), B. P. 227 Mostaganem, Algeria
saidaniman@yahoo.fr, benharrat.belaidi@univ-mosta.dz
∗Correspondence: benharrat.belaidi@univ-mosta.dz

Abstract. In the present paper, we investigate the [p,q]-order of solutions of higher order lineardifferential equations
Ak (z) f

(k)
+Ak−1 (z) f

(k−1)
+ · · ·+A1 (z) f ′ +A0 (z) f =0

and
Ak (z) f

(k)
+Ak−1 (z) f

(k−1)
+ · · ·+A1 (z) f ′ +A0 (z) f =F (z) ,where A0 (z) , A1 (z) , ...,Ak (z) 6≡ 0 and F (z) 6≡ 0 are meromorphic functions of finite [p,q]-order.We improve and extend some results of the authors by using the concept [p,q]-order.

1. Introduction and main results
In this paper, we assume that the reader is familiar with the fundamental results and the standardnotations of the Nevanlinna’s value distribution theory of meromorphic functions (see [7] , [9] , [14] ,
[24]) . In addition, for any integers p ≥ q ≥ 1 and a meromorphic function f in the whole complexplane, we will use ρ[p,q] (f ) , µ[p,q] (f ) to denote respectively the [p,q]-order and the lower [p,q]-order, λ[p,q] (f −a) (or λ[p,q] (f −a)) to denote the [p,q]-convergence exponent of the sequence ofdistinct a-points (or of a-points) and λ[p,q](1f ) to denote the [p,q]-exponent of convergence of thepoles, we refer the reader to see [12] , [15] , [16] and [25] . In particular for q = 1, ρ[p,1] (f ) = ρp (f )is the iterated p-order, µ[p,1] (f ) = µp (f ) is the iterated lower p-order, λ[p,1] (f −a) = λp (f ,a)(or λ[p,1] (f −a) = λp (f ,a)) is the iterated convergence exponent of the sequence of distinct a-points (or of a-points), λ[p,1](1f ) = λp (1f ) is the iterated exponent of convergence of the poles, see
[7] , [11] , [13] , [14] and [24] for notations and definitions.

Received: 6 Sep 2021.
Key words and phrases. linear differential equations; meromorphic functions; [p,q]-order; [p,q]-exponent of conver-gence of zeros. 86

https://adac.ee
https://doi.org/10.28924/ada/ma.1.86
https://orcid.org/0000-0002-6635-2514


Eur. J. Math. Anal. 1 (2021) 87
Several authors have investigated the growth of solutions of second order and higher orderhomogeneous and non-homogeneous linear differential equations with analytic, entire or meromor-phic coefficients, see ([1−3], [6], [8], [11], [13−16], [18] , [20−21], [23], [25]). In the recent years,many authors have studied the complex linear differential equations

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

f (k) + Ak−1 (z) f
(k−1) + · · · + A1 (z) f ′ + A0 (z) f = F (z) , (1.2)where A0 (z) 6≡ 0, A1 (z) , ...,Ak−1 (z) and F (z) 6≡ 0 are meromorphic functions of finite iterated

p-order. In [2] , Belaïdi considered the growth of meromorphic solutions of equations (1.1) and
(1.2) with meromorphic coefficients of finite iterated p−order and obtained some results whichimprove and generalize some previous results.
Theorem A ([2]) Let H ⊂ [0, +∞) be a set with a positive upper density, and let Aj (z) (j = 0, 1, ...,
k − 1) be meromorphic functions with finite iterated p-order. If there exist positive constants
σ > 0,α > 0 such that ρ = max

{
ρp
(
Aj
)

: j = 1, ...,k − 1
}
< σ and |A0 (z) | ≥ expp (αrσ) as

|z| = r ∈ H, r → +∞, then every meromorphic solution f 6≡ 0 of equation (1.1) satisfies

µp (f ) = ρp(f ) = +∞, ρp+1(f ) ≥ σ.

Furthermore, if λp
(
1
f

)
< ∞, then i (f ) = p + 1 and

σ ≤ ρp+1 (f ) ≤ ρp (A0) .

Theorem B ([2]) Let H ⊂ [0, +∞) be a set with a positive upper density, and let Aj (z) (j =
0, 1, ...,k − 1) and F (z) 6≡ 0 be meromorphic functions with finite iterated p-order. If there exist
positive constants σ > 0,α > 0 such that |A0 (z) | ≥ expp (αrσ) as |z| = r ∈ H, r → +∞, and
ρ = max

{
ρp
(
Aj
)

(j = 1, ...,k − 1),ρp (F )
}
< σ, then every meromorphic solution of equation

(1.2) with λp
(
1
f

)
< σ satisfies

λp (f ) = λp(f ) = ρp(f ) = ∞, λp+1 (f ) = λp+1(f ) = ρp+1(f ).

Furthermore, if λp
(
1
f

)
< min{µp (f ) ,σ} , then i (f ) = p + 1 and

λp+1 (f ) = λp+1(f ) = ρp+1 (f ) ≤ ρp (A0) .

Recently, in [18] the authors have studied the growth of solutions of the equations (1.1) and
(1.2) when As(z) to dominate all other coefficients and they got some results about ρp+1 (f ) asfollows.
Theorem C ([18]) Let H ⊂ (1, +∞) be a set with a positive upper logarithmic density (or ml (H) =
+∞), and let Aj (z) (j = 0, 1, ...,k − 1) be meromorphic functions with finite iterated p-order.



Eur. J. Math. Anal. 1 (2021) 88
If there exist positive constants σ > 0,α > 0 and an integer s, 0 ≤ s ≤ k − 1, such that
|As (z) | ≥ expp (αrσ) as |z| = r ∈ H, r → +∞, and ρ = max

{
ρp
(
Aj
)

(j 6= s)
}
< σ, then every

non-transcendental meromorphic solution f 6≡ 0 of (1.1) is a polynomial with deg f ≤ s − 1 and
every transcendental meromorphic solution f of (1.1) with λp

(
1
f

)
< µp (f ) satisfies i (f ) = p + 1

µp (f ) = ρp(f ) = +∞

and
σ ≤ ρp+1 (f ) ≤ ρp (As) .

Theorem D ([18]) Let H ⊂ (1, +∞) be a set with a positive upper logarithmic density (or ml (H) =
+∞), and let Aj (z) (j = 0, 1, ...,k−1) and F (z) 6≡ 0 be meromorphic functions with finite iterated
p-order. If there exist positive constants σ > 0,α > 0 and an integer s, 0 ≤ s ≤ k − 1, such
that |As (z) | ≥ expp (αrσ) as |z| = r ∈ H, r → +∞, and max

{
ρp
(
Aj
)

(j 6= s), ρp (F )
}
< σ,

then every non-transcendental meromorphic solution f of (1.2) is a polynomial with deg f ≤ s−1
and every transcendental meromorphic solution f of (1.2) with λp

(
1
f

)
< min{σ,µp(f )} satisfies

i (f ) = p + 1

λp (f ) = λp(f ) = ρp(f ) = µp (f ) = +∞

and
σ ≤ λp+1 (f ) = λp+1(f ) = ρp+1 (f ) ≤ ρp (As) .Thus, the following question arises: can we have the same properties as in Theorems C andD for the solutions of equations

Ak (z) f
(k) + Ak−1 (z) f

(k−1) + · · · + A1 (z) f ′ + A0 (z) f = 0 (1.3)
and

Ak (z) f
(k) + Ak−1 (z) f

(k−1) + · · · + A1 (z) f ′ + A0 (z) f = F (z) , (1.4)when the coefficients Aj (j = 0, 1, ...,k) are of [p,q]−order? In this paper, we proceed this wayand we obtain the following results.
Theorem 1.1 Let H ⊂ (1, +∞) be a set with a positive upper logarithmic density (or ml (H) =
+∞) and let Aj (z) (j = 0, 1, ...,k) with Ak (z) 6≡ 0 be meromorphic functions with finite [p,q]-
order. If there exist a positive constant σ > 0 and an integer s, 0 ≤ s ≤ k, such that for
sufficiently small ε > 0, we have |As (z) | ≥ expp+1

{
(σ −ε) logq r

}
as |z| = r ∈ H, r → +∞ and

ρ = max
{
ρ[p,q]

(
Aj
)

(j 6= s)
}
< σ, then every non-transcendental meromorphic solution f 6≡ 0 of

(1.3) is a polynomial with deg f ≤ s−1 and every transcendental meromorphic solution f of (1.3)
with λ[p,q]

(
1
f

)
< µ[p,q] (f ) satisfies

ρ[p,q](f ) = µ[p,q] (f ) = +∞, σ ≤ ρ[p+1,q] (f ) ≤ ρ[p,q] (As) .

Remark 1.1 Putting Ak (z) ≡ 1 and q = 1 in Theorem 1.1, we obtain Theorem C.



Eur. J. Math. Anal. 1 (2021) 89
Corollary 1.1 Under the hypotheses of Theorem 1.1, suppose further that ϕ is a transcendental
meromorphic function satisfying ρ[p+1,q] (ϕ) < σ. Then, every transcendental meromorphic solution
f of equation (1.3) with λ[p,q]

(
1
f

)
< µ[p,q] (f ) satisfies

σ ≤ λ[p+1,q] (f −ϕ) = λ[p+1,q] (f −ϕ)

= ρ[p+1,q] (f −ϕ) = ρ[p+1,q] (f ) ≤ ρ[p,q] (As) .

Considering the non-homogeneous linear differential equation (1.4), we obtain the followingresults.
Theorem 1.2 Let H ⊂ (1, +∞) be a set with a positive upper logarithmic density (or ml (H) =
+∞), and let Aj (z) (j = 0, 1, ...,k) with Ak (z) 6≡ 0 and F (z) 6≡ 0 be meromorphic functions with
finite [p,q]-order. If there exist a positive constant σ > 0 and an integer s, 0 ≤ s ≤ k, such that for
sufficiently small ε > 0, we have |As (z) | ≥ expp+1

{
(σ −ε) logq r

}
as |z| = r ∈ H, r → +∞ and

max
{
ρ[p,q]

(
Aj
)

(j 6= s), ρ[p,q] (F )
}
< σ, then every non-transcendental meromorphic solution f

of (1.4) is a polynomial with deg f ≤ s − 1 and every transcendental meromorphic solution f of
(1.4) with λ[p,q]

(
1
f

)
< min

{
σ,µ[p,q](f )

}
satisfies

λ[p,q] (f ) = λ[p,q](f ) = ρ[p,q](f ) = µ[p,q] (f ) = +∞

and

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

Remark 1.2 Putting Ak (z) ≡ 1 and q = 1 in Theorem 1.2, we obtain Theorem D.
Corollary 1.2 Let Aj (z) (j = 0, 1, ...,k) , F (z) , H satisfy all the hypotheses of Theorem 1.2, and
let ϕ be a transcendental meromorphic function satisfying ρ[p+1,q] (ϕ) < σ. Then, every tran-
scendental meromorphic solution f with λ[p,q]

(
1
f

)
< min{σ,µ[p,q] (f )} of equation (1.4) satisfies

σ ≤ λ[p+1,q] (f −ϕ) = λ[p+1,q] (f −ϕ) = ρ[p+1,q] (f −ϕ) ≤ ρ[p,q] (As) .

Remark 1.3 In [17, 19] , the authors have studied the growth and the oscillation of solutionsof equations (1.3) and (1.4) when the coefficients Aj (z) (j = 0, 1, ...,k) and F (z) are entirefunctions of iterated p-order or of [p,q]-order. However, in the present paper the coefficients Aj (z)
(j = 0, 1, ...,k) and F (z) are meromorphic functions with reduction of the hypotheses in Theorems1.1 and 1.2. So, this article may be understood as an extension and an improvement of [17, 19] .



Eur. J. Math. Anal. 1 (2021) 90
2. Some auxiliary lemmas

In order to prove our theorems, we need the following definition, proposition and lemmas. TheLebesgue linear measure of a set E ⊂ [0, +∞) is m (E) = ∫
E

dt, and the logarithmic measure of a
set F ⊂ [1, +∞) is ml (F ) = ∫

F

dt
t
. The upper density of E ⊂ [0, +∞) is given by
dens (E) = lim sup

r→∞

m (E ∩ [0, r])
r

and the upper logarithmic density of the set F ⊂ [1, +∞) is defined by
log dens (F ) = lim sup

r−→+∞

ml (F ∩ [1, r])
log r

.

Proposition 2.1 ([2]) For all H ⊂ (1, +∞) the following statements hold:
(i) If ml (H) = +∞, then m (H) = +∞;
(ii) If dens (H) > 0, then m (H) = +∞;
(iii) If log dens (H) > 0, then ml (H) = +∞.
Lemma 2.1 ([5]) Let f be a transcendental meromorphic function in the plane, and let α > 1 be
a given constant. Then, there exist a set E1 ⊂ (1, +∞) that has a finite logarithmic measure, and
a constant B > 0 depending only on α and (i, j) ((i, j) positive integers with i > j) such that for
all z with |z| = r 6∈ [0, 1] ∪E1, we have∣∣∣∣∣f (i)(z)f (j)(z)

∣∣∣∣∣ ≤ B
(
T (αr,f )

r
(logα r) log T (αr,f )

)i−j
.

Lemma 2.2 ([4]) Let p ≥ q ≥ 1 be integers and g be an entire function such that ρ[p,q] (g) < +∞.
Then, there exist entire functions u(z) and v(z) such that

g (z) = u(z)ev(z),

ρ[p,q] (g) = max
{
ρ[p,q] (u) ,ρ[p,q]

(
ev(z)

)}
and

ρ[p,q] (u) = lim sup
r→+∞

logp N
(
r, 1
g

)
logq r

.

Moreover, for any given ε > 0, we have

|u(z)| ≥ exp
{
−expp

{(
ρ[p,q] (u) + ε

)
logq r

}}
(r /∈ E2) ,

where E2 ⊂ (1, +∞) is a set of r of finite linear measure.



Eur. J. Math. Anal. 1 (2021) 91
Lemma 2.3 Let p ≥ q ≥ 1 be integers. Suppose that f is a meromorphic function such that
ρ[p,q] (f ) < +∞. Then, there exist entire functions u1 (z) , u2 (z) and v (z) such that

f (z) =
u1 (z) e

v(z)

u2 (z)
(2.1)

and
ρ[p,q](f ) = max

{
ρ[p,q](u1),ρ[p,q](u2), ρ[p,q](e

v(z))
}
. (2.2)

Moreover, for any given ε > 0, we have

exp
{
−expp

{
(ρ(p,q) (f ) + ε) logq r

}}
≤ |f (z)|

≤ expp+1
{

(ρ(p,q) (f ) + ε) logq r
}

(r /∈ E3) , (2.3)
where E3 ⊂ (1, +∞) is a set of r of finite linear measure.

Proof. When p ≥ q = 1, the lemma is due to Tu and Long [21]. Thus, we assume that p > q > 1 or
p = q > 1. By Hadamard factorization theorem, we can write f as f (z) = g(z)

d(z)
, where g (z) and

d (z) are entire functions satisfying
µ[p,q] (g) = µ[p,q] (f ) = µ ≤ ρ[p,q] (f ) = ρ[p,q] (g) < +∞

and
λ[p,q] (d) = ρ[p,q] (d) = λ[p,q]

(
1

f

)
< µ.

By Lemma 2.2, there exist entire functions u(z) and v(z) such that
g (z) = u(z)ev(z), ρ[p,q] (g) = max

{
ρ[p,q] (u) ,ρ[p,q]

(
ev(z)

)}
.

So, there exist entire functions u(z), v(z) and d (z) such that
f (z) =

u(z)ev(z)

d (z)and
ρ[p,q](f ) = max

{
ρ[p,q] (u) ,ρ[p,q](d),ρ[p,q]

(
ev(z)

)}
.

Thus (2.1) and (2.2) hold. Set f (z) = u1(z)ev(z)
u2(z)

, where u1 (z) , u2 (z) are the canonical productsformed with the zeros and poles of f respectively. By the definition of [p,q]-order, for sufficientlylarge r and any given ε > 0, we have
|u1 (z)| ≤ expp+1

{(
ρ[p,q] (u1) +

ε
3

)
logq r

}
,

|u2 (z)| ≤ expp+1
{(
ρ[p,q] (u2) +

ε
3

)
logq r

}
.

(2.4)
Since max {ρ[p,q](u1),ρ[p,q](u2), ρ[p,q](ev(z))} = ρ[p,q](f ), then we obtain

|u1 (z)| ≤ expp+1
{(
ρ[p,q] (f ) +

ε

3

)
logq r

}
, (2.5)

|u2 (z)| ≤ expp+1
{(
ρ[p,q] (f ) +

ε

3

)
logq r

}
, (2.6)



Eur. J. Math. Anal. 1 (2021) 92∣∣∣ev(z)∣∣∣ ≤ expp+1{(ρ[p,q] (f ) + ε
3

)
logq r

}
. (2.7)

By Lemma 2.2, there exists a set E3 ⊂ (1, +∞) of r with a finite linear measure such that for anygiven ε > 0, we have
|u1 (z)| ≥ exp

{
−expp

{(
ρ[p,q] (u1) +

ε

3

)
logq r

}}
≥ exp

{
−expp

{(
ρ[p,q] (f ) +

ε

3

)
logq r

}}
, (r /∈ E3) , (2.8)

|u2 (z)| ≥ exp
{
−expp

{(
ρ[p,q] (u2) +

ε

3

)
logq r

}}
≥ exp

{
−expp

{(
ρ[p,q] (f ) +

ε

3

)
logq r

}}
, (r /∈ E3) . (2.9)

Then, by using (2.5) , (2.7) and (2.9), we obtain for sufficiently large r /∈ E3 and any given ε > 0
|f (z)| =

|u1 (z)|
∣∣ev(z)∣∣

|u2 (z)|

≤
expp+1

{(
ρ[p,q] (f ) +

ε
3

)
logq r

}
expp+1

{(
ρ[p,q] (f ) +

ε
3

)
logq r

}
exp

{
−expp

{(
ρ[p,q] (f ) +

ε
3

)
logq r

}}
≤ expp+1

{(
ρ[p,q] (f ) + ε

)
logq r

}
. (2.10)

On the other hand, we have ρ[p−1,q] (v) = ρ[p,q](ev(z)) ≤ ρ[p,q] (f ) and ∣∣ev(z)∣∣ ≥ e−|v(z)|. Makinguse of the definition of [p,q]-order, we obtain
|v (z)| ≤ M(r,v) ≤ expp

{(
ρ(p−1,q) (v) +

ε

3

)
logq r

}
≤ expp

{(
ρ[p,q] (f ) +

ε

3

)
logq r

}
.

Then, for sufficiently large r and any given ε > 0, we have∣∣∣ev(z)∣∣∣ ≥ e−|v(z)| ≥ exp {−expp {(ρ[p,q] (f ) + ε
3

)
logq r

}}
. (2.11)

By (2.6) , (2.8) and (2.11), we can easily obtain
|f (z)| =

|u1 (z)|
∣∣ev(z)∣∣

|u2 (z)|

≥
exp

{
−expp

{(
ρ[p,q] (f ) +

ε
3

)
logq r

}}
exp

{
−expp

{(
ρ[p,q] (f ) +

ε
3

)
logq r

}}
expp+1

{(
ρ[p,q] (f ) +

ε
3

)
logq r

} .
= exp

{
−3 expp

{(
ρ[p,q] (f ) +

ε

3

)
logq r

}}
≥ exp

{
−expp

{(
ρ[p,q] (f ) + ε

)
logq r

}}
.

Thus, we complete the proof of Lemma 2.3.
Lemma 2.4 Under the assumptions of Theorem 1.1 or Theorem 1.2, we have ρ[p,q] (As) = β ≥ σ.



Eur. J. Math. Anal. 1 (2021) 93
Proof. Assume that ρ[p,q] (As) = β < σ. According to the hypotheses of Theorems 1.1 or 1.2, thereexists a positive constant σ > 0 such that for sufficiently small ε > 0, we have

|As (z) | ≥ expp+1
{

(σ −ε) logq r
} (2.12)

as |z| = r ∈ H, r → +∞, where H ⊂ (1, +∞) is a set with a positive upper logarithmic density
(by Proposition 2.1, we have ml (H) = +∞). By Lemma 2.3, we can find a set E3 ⊂ (1, +∞) thathas finite linear measure (and so of finite logarithmic measure) such that when |z| = r /∈ E3, wehave for any given ε (0 < 2ε < σ−β)

|As (z) | ≤ expp+1
{

(β + ε) logq r
}
. (2.13)

By (2.12) and (2.13) , we obtain for |z| = r ∈ H rE3, r → +∞
expp+1

{
(σ −ε) logq r

}
≤ |As (z) | ≤ expp+1

{
(β + ε) logq r

}
and by ε (0 < 2ε < σ−β) this is a contradiction. Hence ρ[p,q] (As) = β ≥ σ.
Lemma 2.5 (Wiman-Valiron, [10] , [22]) Let f be a transcendental entire function, and let z be
a point with |z| = r at which |f (z)| = M (r, f ). Then the estimation

f (j) (z)

f (z)
=

(
νf (r)

z

)j
(1 + o (1)) (j ≥ 1 is an integer )

holds for all |z| outside a set E4 of r of finite logarithmic measure, where νf (r) is the central
index of f .

Lemma 2.6 ([12]) Let f be an entire function of [p,q]-order and let νf (r) be the central index of
f . Then

ρ[p,q] (f ) = lim sup
r→+∞

logp νf (r)

logq r
, µ[p,q] (f ) = lim inf

r→+∞

logp νf (r)

logq r
.

The following two lemmas were given in [4] without proof, so for the convenience of the reader,we prove them.
Lemma 2.7 Let f (z) = g(z)

d(z)
be a meromorphic function, where g (z) , d (z) are entire functions

satisfying µ[p,q] (g) = µ[p,q] (f ) = µ ≤ ρ[p,q] (f ) = ρ[p,q] (g) ≤ +∞ and λ[p,q] (d) = ρ[p,q] (d) =
β = λ[p,q]

(
1
f

)
< µ. Then, there exists a set E5 ⊂ (1, +∞) of finite logarithmic measure such that

for all |z| = r /∈ [0, 1] ∪E5 and |g (z) | = M (r,g) , we have

f (n) (z)

f (z)
=

(
νg (r)

z

)n
(1 + o (1)) , n ∈N,

where νg (r) denote the central index of g.



Eur. J. Math. Anal. 1 (2021) 94
Proof. By mathematical induction, we obtain

f (n) =
g(n)

d
+

n−1∑
j=0

g(j)

d

∑
(j1...jn)

Cjj1...jn

(
d′

d

)j1
×···×

(
d(n)

d

)jn
, (2.14)

where Cjj1...jn are constants and j + j1 + 2j2 + · · · + njn = n. Hence
f (n)

f
=
g(n)

g
+

n−1∑
j=0

g(j)

g

∑
(j1...jn)

Cjj1...jn

(
d′

d

)j1
×···×

(
d(n)

d

)jn
. (2.15)

From Lemma 2.5, there exists a set E4 ⊂ (1, +∞) with finite logarithmic measure such that for apoint z satisfying |z| = r /∈ E4 and |g (z)| = M (r,g), we have
g(j)(z)

g(z)
=

(
νg (r)

z

)j
(1 + o (1)) (j = 1, 2, ...,n) , (2.16)

where νg (r) is the central index of g. Substituting (2.16) into (2.15) yields
f (n) (z)

f (z)
=

(
νg (r)

z

)n
[(1 + o (1))

+

n−1∑
j=0

(
νg (r)

z

)j−n
(1 + o (1))

∑
(j1...jn)

Cjj1...jn

(
d′

d

)j1
×···×

(
d(n)

d

)jn . (2.17)
Since ρ[p,q] (d) = β < µ, then for any given ε (0 < 2ε < µ−β) and sufficiently large r , we have

T (r,d) ≤ expp
{(
β +

ε

2

)
logq r

}
By using Lemma 2.1, for α = 2, there exist a set E1 ⊂ (1, +∞) with ml(E1) < ∞ and a constant
B > 0, such that for all z satisfying |z| = r /∈ [0, 1] ∪E1, we have∣∣∣∣∣d(m) (z)d (z)

∣∣∣∣∣ ≤ B [T (2r,d)]m+1 ≤ B[expp {(β + ε2) logq (2r)}]m+1
≤ expp

{
(β + ε) logq r

}m
, m = 1, 2, ...,n. (2.18)

By Lemma 2.6 and µ[p,q] (g) = µ[p,q] (f ) = µ, it follows that
νg (r) > expp

{
(µ−ε) logq r

}
for sufficiently large r . Thus, by using j1 + 2j2 + · · · + njn = n− j, we obtain∣∣∣∣∣∣

(
νg (r)

z

)j−n (
d′

d

)j1
×···×

(
d(n)

d

)jn∣∣∣∣∣∣ ≤
[

expp
{

(µ−ε) logq r
}

r

]j−n
×
[
expp

{
(β + ε) logq r

}]n−j
=

[
r expp

{
(β + ε) logq r

}
expp

{
(µ−ε) logq r

} ]n−j → 0 (2.19)
as r → +∞, where |z| = r /∈ [0, 1] ∪E5, E5 = E1 ∪E4 and |g (z)| = M (r,g) . From (2.17) and
(2.19), we obtain our assertion.



Eur. J. Math. Anal. 1 (2021) 95
Lemma 2.8 Let f (z) = g(z)

d(z)
be a meromorphic function, where g (z), d (z) are entire functions

satisfying µ[p,q] (g) = µ[p,q] (f ) = µ ≤ ρ[p,q] (f ) = ρ[p,q] (g) ≤ +∞ and λ[p,q] (d) = ρ[p,q] (d) =
λ[p,q]

(
1
f

)
< µ. Then, there exists a set E6 ⊂ (1, +∞) of finite logarithmic measure such that for

all |z| = r /∈ [0, 1] ∪E6 and |g (z) | = M (r,g), we have∣∣∣∣ f (z)f (s) (z)
∣∣∣∣ ≤ r2s, (s ∈N) .

Proof. By Lemma 2.7, there exists a set E5 of finite logarithmic measure such that the estimation
f (s)(z)

f (z)
=

(
νg (r)

z

)s
(1 + o (1)) (s ≥ 1 is an integer) (2.20)

holds for all |z| = r /∈ [0, 1] ∪E5 and |g (z)| = M (r,g), where νg (r) is the central index of g. Onthe other hand, by Lemma 2.6, for any given ε (0 < ε < 1), there exists R > 1 such that for all
r > R, we have

νg (r) > expp
{

(µ−ε) logq (r)
}
. (2.21)

If µ = +∞, then µ−ε can be replaced by a large enough real number M. Set E6 = [1,R] ∪E5,
lm (E6) < +∞. Hence from (2.20) and (2.21), we obtain∣∣∣∣ f (z)f (s) (z)

∣∣∣∣ = ∣∣∣∣ zνg (r)
∣∣∣∣s 1|1 + o (1)| ≤ rs(expp {(µ−ε) logq (r)})s ≤ r2s,where |z| = r /∈ [0, 1] ∪E6, r → +∞ and |g (z)| = M (r,g) .

Lemma 2.9 ([6]) Let ϕ : [0, +∞) →R and ψ : [0, +∞) →R be monotone nondecreasing functions
such that ϕ(r) ≤ ψ(r) for all r /∈ (E7 ∪ [0, 1]) , where E7 is a set of finite logarithmic measure.
Let α > 1 be a given constant. Then, there exists an r1 = r1(α) > 0 such that ϕ(r) ≤ ψ(αr) for
all r > r1.

Lemma 2.10 ([19]) Let f (z) = g(z)
d(z)

be a meromorphic function, where g (z), d (z) are entire
functions. If 0 ≤ ρ[p,q] (d) < µ[p,q] (f ) , then µ[p,q] (g) = µ[p,q] (f ) and ρ[p,q] (g) = ρ[p,q] (f ) .
Moreover, if ρ[p,q] (f ) = +∞, then ρ[p+1,q] (g) = ρ[p+1,q] (f ) .

Lemma 2.11 Assume that k ≥ 2 and A0, A1, ...,Ak 6≡ 0, F are meromorphic functions. Let
ρ = max

{
ρ[p,q]

(
Aj
)

(j = 0, 1, ...,k),ρ[p,q] (F )
}
< ∞ and let f be a meromorphic solution of

infinite [p,q]-order of equation (1.4) with λ[p,q]
(
1
f

)
< µ[p,q] (f ) . Then, ρ[p+1,q](f ) ≤ ρ.

Proof. Let f be a meromorphic solution of infinite [p,q]-order of equation (1.4) with λ[p,q](1f ) <
µ[p,q] (f ) . So, we can use Hadamard factorization theorem and write f as f (z) = g(z)d(z) , where g(z)and d(z) are entire functions satisfying µ[p,q] (g) = µ[p,q] (f ) = µ ≤ ρ[p,q] (f ) = ρ[p,q] (g) ≤ +∞



Eur. J. Math. Anal. 1 (2021) 96
and λ[p,q] (d) = ρ[p,q] (d) = λ[p,q](1f ) < µ. By Lemma 2.3, there exists a set E3 ⊂ (1, +∞)of r with a finite linear measure such that for all |z| = r /∈ E3 and any given ε (0 < 2ε <
µ[p,q] (f ) −ρ[p,q] (d)), we have

|Aj (z) | ≤ expp+1
{

(ρ(p,q)
(
Aj
)

+ ε) logq r
}

≤ expp+1
{

(ρ + ε) logq r
}
, j = 0, 1, ...,k − 1, (2.22)

|Ak (z) | ≥ exp
{
−expp

{
(ρ(p,q) (Ak) + ε) logq r

}}
≥ exp

{
−expp

{
(ρ + ε) logq r

}} (2.23)
and

|F (z)| ≤ expp+1
{

(ρ(p,q) (F ) + ε) logq r
}
≤ expp+1

{
(ρ + ε) logq r

} . (2.24)
By (2.24), for all z satisfying |z| = r /∈ E3 at which |g(z)| = M(r,g) and any given
ε
(

0 < 2ε < µ[p,q] (f ) −ρ[p,q] (d)
)
, we obtain∣∣∣∣F (z)f (z)

∣∣∣∣ = |F (z)||g(z)| |d (z)|
≤

expp+1
{

(ρ[p,q] (d) + ε) logq r
}

expp+1
{

(ρ + ε) logq r
}

expp+1
{

(µ[p,q] (f ) −ε) logq r
}

≤ expp+1
{

(ρ + ε) logq r
}
. (2.25)

By Lemma 2.7, there exists a set E5 ⊂ (1, +∞) of finite logarithmic measure such that for all
|z| = r /∈ [0, 1] ∪E5 and |g (z) | = M (r,g) , we have

f (j) (z)

f (z)
=

(
νg (r)

z

)j
(1 + o (1)) , j = 1, ...,k. (2.26)

We can rewrite (1.4) as∣∣∣∣∣f (k) (z)f (z)
∣∣∣∣∣ ≤ 1|Ak (z) |

|A0 (z) | + ∣∣∣∣F (z)f (z)
∣∣∣∣ + k−1∑

j=1

|Aj (z) |

∣∣∣∣∣f (j) (z)f (z)
∣∣∣∣∣
 . (2.27)

By substituting (2.22) , (2.23) , (2.25) and (2.26) into (2.27), we obtain∣∣∣∣νg (r)z
∣∣∣∣k |1 + o (1)| ≤ 1exp {−expp {(ρ + ε) logq r}}×

1 +
k−1∑
j=1

∣∣∣∣νg (r)z
∣∣∣∣j |1 + o (1)|

 expp+1{(ρ + ε) logq r}
+ expp+1

{
(ρ + ε) logq r

})
=

2 +
k−1∑
j=1

∣∣∣∣νg (r)z
∣∣∣∣j |1 + o (1)|

 exp {2 expp {(ρ + ε) logq r}} .
Hence

|νg (r)| |1 + o (1)| ≤ (k + 1) r |1 + o (1)|exp
{

2 expp
{

(ρ + ε) logq r
}} (2.28)



Eur. J. Math. Anal. 1 (2021) 97
holds for all z satisfying |z| = r /∈ [0, 1] ∪E3 ∪E5 and |g (z) | = M (r,g) , r → +∞. By (2.28),we get

lim sup
r→+∞

logp+1νg (r)

logq r
≤ ρ + ε. (2.29)

Since ε > 0 is arbitrary, by (2.29) and Lemma 2.6, we obtain ρ[p+1,q] (g) ≤ ρ. Since ρ[p,q] (d) <
µ[p,q] (f ) , so by Lemma 2.10, we have ρ[p+1,q] (g) = ρ[p+1,q] (f ) . Thus, ρ[p+1,q] (f ) ≤ ρ. Therefore,Lemma 2.11 is proved.
Lemma 2.12 ([19]) Let Aj (z) (j = 0, 1, ...,k) , Ak (z) (6≡ 0) ,F (z) (6≡ 0) be meromorphic functions
and let f be a meromorphic solution of (1.4) of infinite [p,q]-order satisfying the following condition

b = max
{
ρ[p+1,q] (F ) , ρ[p+1,q]

(
Aj
)

(j = 0, 1, ...,k)
}
< ρ[p+1,q] (f ) .

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

Lemma 2.13 Let H ⊂ (1, +∞) be a set with a positive upper logarithmic density (or infinite
logarithmic measure), and let Aj (z) (j = 0, 1, ...,k) with Ak (z) 6≡ 0 and F (z) 6≡ 0 be meromorphic
functions with finite [p,q]-order. If there exist a positive constant σ > 0 and an integer s,
0 ≤ s ≤ k, such that for sufficiently small ε > 0, we have |As (z) | ≥ expp+1

{
(σ −ε) logq r

}
as

|z| = r ∈ H, r → +∞ and

max
{
ρ[p,q]

(
Aj
)

(j 6= s), ρ[p,q] (F )
}
< σ,

then every transcendental meromorphic solution f of equation (1.4) satisfies ρ[p,q](f ) ≥ σ.

Proof. Assume that f is a transcendental meromorphic solution of equation (1.4) with ρ[p,q](f ) < σ.From (1.4) , we have
As =

F

f (s)
−

k∑
j=0
j 6=s

Aj
f (j)

f (s)
. (2.30)

Since max {ρ[p,q](Aj) (j 6= s), ρ[p,q] (F )} < σ and ρ[p,q] (f ) < σ, then from (2.30) we obtainthat
ρ1 = ρ[p,q] (As) ≤ max

{
ρ[p,q]

(
Aj
)

(j 6= s), ρ[p,q] (F ) , ρ[p,q] (f )
}
< σ.

By Lemma 2.3, for any ε (0 < 2ε < σ−ρ1) , there exists a set E3 ⊂ (1, +∞) with a finite linearmeasure such that
|As (z)| ≤ expp+1

{
(ρ(p,q) (As) + ε) logq r

}
= expp+1

{
(ρ1 + ε) logq r

} (2.31)
holds for all z satisfying |z| = r /∈ E3. From the hypotheses of Lemma 2.13, there exists a set Hwith log densH > 0 (or ml (H) = +∞) such that

|As (z)| ≥ expp+1
{

(σ −ε) logq r
} (2.32)



Eur. J. Math. Anal. 1 (2021) 98
holds for all z satisfying |z| = r ∈ H, r → +∞. By (2.31) and (2.32), we conclude that for all zsatisfying |z| = r ∈ H rE3, r → +∞, we have

expp+1
{

(σ −ε) logq r
}
≤ expp+1

{
(ρ1 + ε) logq r

}
and by ε (0 < 2ε < σ−ρ1) this is a contradiction as r → +∞. Consequently, any transcendentalmeromorphic solution f of equation (1.4) satisfies ρ[p,q] (f ) ≥ σ.
Lemma 2.14 ([23]) Let p ≥ q ≥ 1 be integers. Let f be a meromorphic function for which
ρ[p,q] (f ) = β < +∞, and let k ≥ 1 be an integer. Then for any ε > 0,

m

(
r,
f (k)

f

)
= O

(
expp−1

{
(β + ε) logq r

})
,

holds outside of a possible exceptional set E8 of finite linear measure.

Lemma 2.15 Let A0,A1, ...,Ak 6≡ 0,F 6≡ 0 be finite [p,q]-order meromorphic functions. If f is a
meromorphic solution with ρ[p,q] (f ) = +∞ and ρ[p+1,q] (f ) = ρ < +∞ of equation (1.4) , then
λ[p,q] (f ) = λ[p,q](f ) = ρ[p,q](f ) = +∞ and λ[p+1,q] (f ) = λ[p+1,q](f ) = ρ[p+1,q](f ) = ρ.

Proof Let f be a meromorphic solution of (1.4) with infinite [p,q]-order and ρ[p+1,q] (f ) =
ρ < +∞. Note first that by definition, we have λ[p+1,q] (f ) ≤ λ[p+1,q] (f ) ≤ ρ[p+1,q] (f ) . Then, itremains to show that

ρ[p+1,q] (f ) ≤ λ[p+1,q] (f ) ≤ λ[p+1,q] (f ) .

We rewrite (1.4) as
1

f
=

1

F

(
Ak (z)

f (k)

f
+ Ak−1 (z)

f (k−1)

f
+ · · · + A1 (z)

f ′

f
+ A0 (z)

)
. (2.33)

By using Lemma 2.14 and (2.33), for |z| = r outside a set E8 of a finite linear measure and anygiven ε > 0, we get
m

(
r,

1

f

)
≤ m

(
r,

1

F

)
+

k∑
j=1

m

(
r,
f (j)

f

)
+

k∑
j=0

m
(
r,Aj

)
+ O (1)

≤ m
(
r,

1

F

)
+

k∑
j=0

m
(
r,Aj

)
+ O

(
expp

{
(ρ + ε) logq r

})
. (2.34)

On the other hand, by (1.4), if f has a zero at z0 of order α (α > k), and A0, A1, ..., Ak are allanalytic at z0, then F must have a zero at z0 of order at least α−k. Hence,
n

(
r,

1

f

)
≤ kn

(
r,

1

f

)
+ n

(
r,

1

F

)
+

k∑
j=0

n
(
r,Aj

)



Eur. J. Math. Anal. 1 (2021) 99
and

N

(
r,

1

f

)
≤ kN

(
r,

1

f

)
+ N

(
r,

1

F

)
+

k∑
j=0

N
(
r,Aj

)
. (2.35)

Therefore, by (2.34) and (2.35), for all sufficiently large r /∈ E8 and any given ε > 0, we have
T (r, f ) = T (r,

1

f
) + O (1) ≤ T (r,F ) +

k∑
j=0

T
(
r,Aj

)
+ kN

(
r,

1

f

)
+ O

(
expp

{
(ρ + ε) logq r

})
. (2.36)

Noting c = max {ρ[p,q](Aj) (j = 0, 1, ...,k),ρ[p,q] (F )} . Then, by using the definition of the [p,q]-order, for the above ε and sufficiently large r , we have
T (r,F ) ≤ expp

{
(c + ε) logq r

}
, (2.37)

T
(
r,Aj

)
≤ expp

{
(c + ε) logq r

}
, j = 0, 1, ...,k. (2.38)

Replacing (2.37) and (2.38) into (2.36) , for r /∈ E8 sufficiently large and any given ε > 0, weobtain
T (r, f ) ≤ kN

(
r,

1

f

)
+ (k + 2) expp

{
(c + ε) logq r

}
+ O

(
expp

{
(ρ + ε) logq r

})
. (2.39)

Hence, for any f with ρ[p,q] (f ) = +∞ and ρ[p+1,q](f ) = ρ, by (2.39) , we have
λ[p,q] (f ) ≥ ρ[p,q] (f ) = +∞, λ[p+1,q] (f ) ≥ ρ[p+1,q] (f ) ,

so
ρ[p+1,q] (f ) ≤ λ[p+1,q] (f ) ≤ λ[p+1,q] (f ) .

And the fact that λ[p+1,q] (f ) ≤ λ[p+1,q] (f ) ≤ ρ[p+1,q] (f ) , we obtain
λ[p+1,q] (f ) = λ[p+1,q] (f ) = ρ[p+1,q] (f ) = ρ.

3. Proof of Theorem 1.1
Assume that f 6≡ 0 is a rational solution of (1.3). First, we will prove that f must be a polynomialwith deg f ≤ s − 1. For this, if f is a rational function, which has a pole at z0 of degree m ≥ 1,or f is a polynomial with deg f ≥ s, then f (s)(z) 6≡ 0. By (1.3) and Lemma 2.4, we obtain

σ ≤ ρ[p,q](As) = ρ[p,q](Asf
(s)) = ρ[p,q]

−
 k∑
j=0, j 6=s

Ajf
(j)


≤ max
j=0,1,...,k, j 6=s

{
ρ[p,q]

(
Aj
)}

which is a contradiction. Therefore, f must be a polynomial with deg f ≤ s − 1.



Eur. J. Math. Anal. 1 (2021) 100
Now, we assume that f is a transcendental meromorphic solution of (1.3) such that λ[p,q](1f ) <

µ[p,q](f ). By Lemma 2.3, for any given ε (0 < 2ε < σ−ρ) , there exists a set E3 ⊂ (1, +∞) witha finite linear measure (and so of finite logarithmic measure) such that
|Aj (z) | ≤ expp+1

{
(ρ + ε) logq r

}
, j = 0, 1, ...,k, j 6= s (3.1)

holds for all z satisfying |z| = r /∈ E3. In view of Lemma 2.8, there exists a set E6 ⊂ (1, +∞) offinite logarithmic measure such that |z| = r /∈ [0, 1] ∪E6, |g (z) | = M (r,g) and for r sufficientlylarge, we have ∣∣∣∣ f (z)f (s) (z)
∣∣∣∣ ≤ r2s (s ≥ 1 is an integer) . (3.2)

According to Lemma 2.1, there exist a set E1 ⊂ (1, +∞) with ml(E1) < ∞ and a constant B > 0,such that for all z satisfying |z| = r /∈ [0, 1] ∪E1, we have∣∣∣∣∣f (j) (z)f (z)
∣∣∣∣∣ ≤ B [T (2r, f )]k+1 , j = 1, 2, ...,k, j 6= s. (3.3)

From the hypotheses of Theorem 1.1, there exists a set H ⊂ (1, +∞) with ml (H) = +∞, suchthat for all z satisfying |z| = r ∈ H, r → +∞ and sufficiently small ε > 0, we have
|As (z) | ≥ expp+1

{
(σ −ε) logq r

}
. (3.4)

Now, by rewriting equation (1.3) in the form
|As| ≤

∣∣∣∣ ff (s)
∣∣∣∣
|A0| + k∑

j=1
j 6=s

∣∣Aj∣∣
∣∣∣∣∣f (j)f

∣∣∣∣∣
 (3.5)

and substituting (3.1) , (3.2) , (3.3) and (3.4) into (3.5), for all z satisfying |z| = r ∈ Hr([0, 1]∪
E1 ∪E3 ∪E6), r → +∞, we obtain

expp+1
{

(σ −ε) logq r
}
≤ Bkr2s expp+1

{
(ρ + ε) logq r

}
[T (2r, f )]

k+1
.

Since 0 < 2ε < σ−ρ, then we have
exp

{
(1 −o (1)) expp

{
(σ −ε) logq r

}}
≤ Bkr2s [T (2r, f )]k+1 . (3.6)

From (3.6) and Lemma 2.9, for any given γ > 1 and sufficiently large r > R, we get
exp

{
(1 −o (1)) expp

{
(σ −ε) logq r

}}
≤ Bk (γr)2s [T (2γr,f )]k+1

which gives
ρ[p,q](f ) = µ[p,q] (f ) = +∞, σ ≤ ρ[p+1,q] (f ) . (3.7)

By using Lemma 2.4, we have
max

{
ρ[p,q]

(
Aj
)

: j = 0, 1, ...,k
}

= ρ[p,q] (As) = β < +∞.



Eur. J. Math. Anal. 1 (2021) 101
Since f is of infinite [p,q]-order meromorphic solution of equation (1.3) satisfying λ[p,q](1f ) <
µ[p,q] (f ), then by Lemma 2.11, we obtain

ρ[p+1,q] (f ) ≤ max
{
ρ[p,q]

(
Aj
)

: j = 0, 1, ...,k
}

= ρ[p,q] (As) . (3.8)
By (3.7) and (3.8) , we conclude that µ[p,q] (f ) = ρ[p,q] (f ) = +∞ and σ ≤ ρ[p+1,q] (f ) ≤
ρ[p,q] (As) .

4. Proof of Corollary 1.1
Assume that ϕ is a transcendental meromorphic function such that ρ[p+1,q] (ϕ) < σ. Noting g =
f − ϕ, then ρ[p+1,q] (g) = ρ[p+1,q] (f ), so by Theorem 1.1, σ ≤ ρ[p+1,q] (g) ≤ ρ[p,q] (As) . Bysubstituting f = g + ϕ into (1.3), we obtain

Ak (z) g
(k) + Ak−1 (z) g

(k−1) + · · · + A1 (z) g′ + A0 (z) g

= −
(
Ak (z) ϕ

(k) + Ak−1 (z) ϕ
(k−1) + · · · + A1 (z) ϕ′ + A0 (z) ϕ

)
= G (z) . (4.1)

It is clear that the right side G of equation (4.1) is non-zero, because by Theorem 1.1, ϕ is not asolution of equation (1.3). Moreover, the [p + 1,q]-order of G satisfies
ρ[p+1,q] (G) ≤ max

{
ρ[p+1,q] (ϕ) , ρ[p+1,q]

(
Aj
)

(j = 0, 1, ...,k)
}
< σ,

which implies
max

{
ρ[p+1,q] (G) , ρ[p+1,q]

(
Aj
)

(j = 0, 1, ...,k)
}
< σ ≤ ρ[p+1,q] (g) .

Then by Lemma 2.12, we obtain
σ ≤ λ[p+1,q] (g) = λ[p+1,q] (g)

= ρ[p+1,q] (g) = ρ[p+1,q] (f ) ≤ ρ[p,q] (As) ,that is
σ ≤ λ[p+1,q] (f −ϕ) = λ[p+1,q] (f −ϕ)

= ρ[p+1,q] (f −ϕ) = ρ[p+1,q] (f ) ≤ ρ[p,q] (As) .

5. Proof of Theorem 1.2
Assume that f is a rational solution of (1.4). First, we will prove that f must be a polynomial with
deg f ≤ s − 1. For this, if f is a rational function, which has a pole at z0 of degree m ≥ 1, or f isa polynomial with deg f ≥ s, then f (s)(z) 6≡ 0. By (1.4) and Lemma 2.4, we obtain

σ ≤ ρ[p,q](As) = ρ[p,q](Asf
(s)) = ρ[p,q]

F − k∑
j=0
j 6=s

Aj (z) f
(j)


≤ max
j=0,1,...,k, j 6=s

{
ρ[p,q]

(
Aj
)
, ρ[p,q] (F )

}
,



Eur. J. Math. Anal. 1 (2021) 102
which is a contradiction. Therefore, f must be a polynomial with deg f ≤ s − 1.
Now, we assume that f is a transcendental meromorphic solution of (1.4) such that λ[p,q](1f ) <
µ[p,q](f ). From Lemma 2.13, we know that f satisfies ρ[p,q] (f ) ≥ σ. By the hypothesis λ[p,q](1f ) <
min{µ[p,q](f ),σ} and Hadamard factorization theorem, we can write f as f (z) = g(z)d(z), where g (z)and d (z) are entire functions satisfying

µ[p,q](g) = µ[p,q](f ) = µ ≤ ρ[p,q](g) = ρ[p,q](f ),

ρ[p,q](d) = λ[p,q]

(
1

f

)
= β < min{µ[p,q](f ),σ}.

The definition of the lower [p,q]-order assures us that
|g (z)| = M(r,g) ≥ expp+1

{
(µ[p,q] (g) −ε) logq r

}
. (5.1)

Putting
ρ1 = max

{
ρ[p,q]

(
Aj
)

(j 6= s) ,ρ[p,q] (F )
}
< σ.

Then, by Lemma 2.3 and (5.1), for any given ε satisfying
0 < 2ε < min{σ −ρ1,µ[p,q] (g) −ρ[p,q] (d)},

there exists a set E3 ⊂ (1, +∞) with a finite logarithmic measure such that for all z satisfying
|z| = r /∈ E3 at which |g (z) | = M(r,g), we obtain∣∣∣∣F (z)f (z)

∣∣∣∣ = |F (z)||g(z)| |d (z)|
≤

expp+1
{

(ρ[p,q] (d) + ε) logq r
}

expp+1
{

(ρ1 + ε) logq r
}

expp+1
{

(µ[p,q] (g) −ε) logq r
}

≤ expp+1
{

(ρ1 + ε) logq r
}
. (5.2)

By using the same arguments as in the proof of Theorem 1.1, for any given
ε
(

0 < 2ε < min{σ −ρ1,µ[p,q] (g) −ρ[p,q] (d)}
) and all z satisfying |z| = r ∈ Hr(E1 ∪E3 ∪E6) ,

r → +∞ at which |g (z) | = M(r,g), we have (3.2) , (3.3) , (3.4) hold and
|Aj (z) | ≤ expp+1

{
(ρ1 + ε) logq r

}
, j = 0, 1, ...,k, j 6= s. (5.3)

By (1.4) , we have
|As| ≤

∣∣∣∣ ff (s)
∣∣∣∣
|A0| + k∑

j=1
j 6=s

∣∣Aj∣∣
∣∣∣∣∣f (j)f

∣∣∣∣∣ +
∣∣∣∣Ff
∣∣∣∣
 . (5.4)

Hence, by substituting (3.2) , (3.3) , (3.4) , (5.2) and (5.3) into (5.4) , for all z satisfying |z| = r ∈
H r (E1 ∪E3 ∪E6), r → +∞, at which |g (z) | = M (r,g) and any given
ε
(

0 < 2ε < min{σ −ρ1,µ[p,q] (g) −ρ[p,q] (d)}
)
, we obtain

expp+1
{

(σ −ε) logq r
}
≤ r2s

(
expp+1

{
(ρ1 + ε) logq r

}



Eur. J. Math. Anal. 1 (2021) 103
+

k∑
j=1,j 6=s

expp+1
{

(ρ1 + ε) logq r
}
B [T (2r, f )]

k+1

+ expp+1
{

(ρ1 + ε) logq r
})

≤ B (k + 1) r2s [T (2r, f )]k+1 expp+1
{

(ρ1 + ε) logq r
}
. (5.5)

Since 0 < 2ε < σ −ρ1, then we can use Lemma 2.9 with (5.5) such that for any given γ > 1 andsufficiently large r > R, we obtain
exp

{
(1 −o (1)) expp

{
(σ −ε) logq r

}}
≤ B (k + 1) (γr)2s [T (2γr,f )]k+1

which gives
ρ[p,q](f ) = µ[p,q] (f ) = +∞, ρ[p+1,q](f ) ≥ σ. (5.6)

Making use of Lemma 2.4, we have
max

{
ρ[p,q]

(
Aj
)

(j = 0, 1, ...,k) ,ρ[p,q] (F )
}

= ρ[p,q] (As) = β < +∞.

By Lemma 2.11 and since f is of infinite [p,q]-order meromorphic solution of equation (1.4)satisfying λ[p,q](1f ) < µ[p,q] (f ) , we get
ρ[p+1,q] (f ) ≤ max

{
ρ[p,q]

(
Aj
)

(j = 0, 1, ...,k) ,ρ[p,q] (F )
}

= ρ[p,q] (As) . (5.7)
Since F 6≡ 0, then by Lemma 2.15, we have

λ[p,q] (f ) = λ[p,q](f ) = µ[p,q] (f ) = ρ[p,q](f ) = +∞ (5.8)
and

σ ≤ λ[p+1,q] (f ) = λ[p+1,q](f ) = ρ[p+1,q](f ). (5.9)
By (5.7) , (5.8) and (5.9) , we conclude that

λ[p,q] (f ) = λ[p,q](f ) = µ[p,q] (f ) = ρ[p,q](f ) = +∞

and
σ ≤ λ[p+1,q] (f ) = λ[p+1,q](f ) = ρ[p+1,q](f ) ≤ ρ[p,q] (As) .

6. Proof of Corollary 1.2
Assume that ϕ is a transcendental meromorphic function such that ρ[p+1,q] (ϕ) < σ. Noting h =
f − ϕ, then ρ[p+1,q] (h) = ρ[p+1,q] (f ), so by Theorem 1.2, σ ≤ ρ[p+1,q] (h) ≤ ρ[p,q] (As) . Bysubstituting f = h + ϕ into (1.4), we obtain

Ak (z) h
(k) + Ak−1 (z) h

(k−1) + · · · + A1 (z) h′ + A0 (z) h

= F (z) −
(
Ak (z) ϕ

(k) + Ak−1 (z) ϕ
(k−1) + · · · + A1 (z) ϕ′ + A0 (z) ϕ

)
= Ψ (z) . (6.1)



Eur. J. Math. Anal. 1 (2021) 104
It is clear that the right side Ψ of the equation (6.1) is non-zero, because by Theorem 1.2, ϕ isnot a solution of equation (1.4). Moreover, the [p + 1,q]-order of Ψ verifies

ρ[p+1,q] (Ψ) ≤ max
{
ρ[p+1,q] (ϕ) , ρ[p+1,q]

(
Aj
)

(j = 0, 1, ...,k)
}
< σ,

which leads to
max

{
ρ[p+1,q] (Ψ) , ρ[p+1,q]

(
Aj
)

(j = 0, 1, ...,k)
}
< σ ≤ ρ[p+1,q] (h) .

Therefore, by Lemma 2.12, we obtain
σ ≤ λ[p+1,q] (h) = λ[p+1,q] (h)

= ρ[p+1,q] (h) = ρ[p+1,q] (f ) ≤ ρ[p,q] (As) ,

that is
σ ≤ λ[p+1,q] (f −ϕ) = λ[p+1,q] (f −ϕ)

= ρ[p+1,q] (f −ϕ) = ρ[p+1,q] (f ) ≤ ρ[p,q] (As) .

Acknowledgements. This paper was supported by the Directorate-General for Scientific Researchand Technological Development (DGRSDT).
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	1.  Introduction and main results
	2. Some auxiliary lemmas
	3. Proof of Theorem 1.1
	4. Proof of Corollary 1.1
	5. Proof of Theorem 1.2
	6. Proof of Corollary 1.2
	References

