International Journal of Analysis and Applications
ISSN 2291-8639
Volume 5, Number 2 (2014), 198-211
http://www.etamaths.com

PROPERTIES OF MEROMORPHIC SOLUTIONS OF A CLASS

OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS

BENHARRAT BELAÏDI∗ AND HABIB HABIB

Abstract. This paper deals with the growth of meromorphic solutions of
some second order linear differential equations, where it is assumed that the

coefficients are meromorphic functions. Our results extend the previous results

due to Chen and Shon, Xu and Zhang, Peng and Chen and others.

1. Introduction and statement of result

In this paper, we shall assume that the reader is familiar with the fundamental
results and the standard notations of the Nevanlinna value distribution theory of
meromorphic functions (see [13] , [20]). In addition, we will use notations ρ (f) ,
ρ2 (f) to denote respectively the order and the hyper-order of growth of a mero-
morphic function f (z).

For the second order linear differential equation

(1.1) f′′ + e−zf′ + B (z) f = 0,

where B (z) is an entire function, it is well-known that each solution f of equation
(1.1) is an entire function, and that if f1 and f2 are two linearly independent
solutions of (1.1) , then by [7], there is at least one of f1, f2 of infinite order.
Hence, ”most” solutions of (1.1) will have infinite order. But equation (1.1) with
B(z) = −(1 + e−z) possesses a solution f (z) = ez of finite order.

A natural question arises: What conditions on B(z) will guarantee that every
solution f 6≡ 0 of (1.1) has infinite order? Many authors, Frei [8], Ozawa [16],
Amemiya-Ozawa [1] and Gundersen [10], Langley [14] have studied this problem.
They proved that when B(z) is a nonconstant polynomial or B(z) is a transcen-
dental entire function with order ρ(B) 6= 1, then every solution f 6≡ 0 of (1.1) has
infinite order.

In 2002, Chen [3] considered the question: What conditions on B(z) when ρ(B)
= 1 will guarantee that every nontrivial solution of (1.1) has infinite order? He
proved the following result, which improved results of Frei, Amemiya-Ozawa, Oza-
wa, Langley and Gundersen.

2010 Mathematics Subject Classification. 34M10, 30D35.
Key words and phrases. Linear differential equations, meromorphic solutions, order of growth,

hyper-order.

c©2014 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

198



PROPERTIES OF MEROMORPHIC SOLUTIONS 199

Theorem A [3] Let Aj (z) (6≡ 0) (j = 0, 1) be entire functions with max{ρ (Aj)
(j = 0, 1)} < 1, and let a,b be complex constants that satisfy ab 6= 0 and a 6= b.
Then every solution f 6≡ 0 of the differential equation
(1.2) f′′ + A1 (z) e

azf′ + A0 (z) e
bzf = 0

is of infinite order.

In [4], Chen and Shon have considered equation (1.2) when Aj (z) (j = 0, 1)
are meromorphic functions and have proved the following result.

Theorem B ([4]) Let Aj (z) (6≡ 0) (j = 0, 1) be meromorphic functions with ρ (Aj) <
1 (j = 0, 1) , and let a, b be complex numbers such that ab 6= 0 and arg a 6= arg b or
a = cb (0 < c < 1) . Then every meromorphic solution f (z) 6≡ 0 of equation (1.2)
has infinite order.

In [17], Peng and Chen have investigated the order and hyper-order of solutions
of some second order linear differential equations and have proved the following
result.

Theorem C [17] Let Aj (z) ( 6≡ 0) (j = 1, 2) be entire functions with ρ (Aj) < 1,
a1, a2 be complex numbers such that a1a2 6= 0, a1 6= a2 (suppose that |a1| ≤ |a2|).
If arg a1 6= π or a1 < −1, then every solution f ( 6≡ 0) of the differential equation

f′′ + e−zf′ + (A1e
a1z + A2e

a2z) f = 0

has infinite order and ρ2 (f) = 1.

Recently in [2] , the authors extend and improve the results of Theorem C to
some second order linear differential equations as follows.

Theorem D [2] Let n ≥ 2 be an integer, Aj (z) ( 6≡ 0) (j = 1, 2) be entire functions
with max{ρ (Aj) : j = 1, 2} < 1, Q (z) = qmzm + · · · + q1z + q0 be nonconstant
polynomial and a1, a2 be complex numbers such that a1a2 6= 0, a1 6= a2. If (1)
arg a1 6= π and arg a1 6= arg a2 or (2) arg a1 6= π, arg a1 = arg a2 and |a2| > n |a1|
or (3) a1 < 0 and arg a1 6= arg a2 or (4) −1n (|a2|−m) < a1 < 0, |a2| > m and
arg a1 = arg a2, then every solution f 6≡ 0 of the differential equation

f′′ + Q
(
e−z
)
f′ + (A1e

a1z + A2e
a2z)

n
f = 0

satisfies ρ (f) = +∞ and ρ2 (f) = 1.

Recently Xu and Zhang have investigated the order and the hyper-order of
meromorphic solutions of some second order linear differential equations and have
proved the following result.

Theorem E [19] Suppose that Aj (z) (6≡ 0) (j = 0, 1, 2) are meromorphic functions
and ρ (Aj) < 1, and a1, a2 are two complex numbers such that a1a2 6= 0, a1 6= a2
(suppose that |a1| ≤ |a2|). Let a0 be a constant satisfying a0 < 0. If arg a1 6= π
or a1 < a0, then every meromorphic solution f ( 6≡ 0) whose poles are of uniformly
bounded multiplicities of the equation

f′′ + A0e
a0zf′ + (A1e

a1z + A2e
a2z) f = 0



200 BELAÏDI AND HABIB

has infinite order and ρ2 (f) = 1.

The main purpose of this paper is to extend and improve the results of Theo-
rems A-E to some second order linear differential equations. In fact we will prove
the following result.

Theorem 1.1 Let Aj (z) ( 6≡ 0) (j = 1, · · · , l1) (l1 ≥ 3) and Bj (z) ( 6≡ 0) (j = 1, · · · , l2)
(l2 ≥ 1) be meromorphic functions with

max{ρ (Aj) (j = 1, · · · , l1) ,ρ (Bj) (j = 1, · · · , l2)} < 1

and aj 6= 0 (j = 1, · · · , l1) be distinct complex numbers and bj (j = 1, · · · , l2) be dis-
tinct real numbers such that bj < 0. Suppose that there exist αj, βj (j = 3, · · · , l1)
where 0 < αj < 1, 0 < βj < 1 and aj = αja1 + βja2. Set α = max{αj : j =
3, · · · , l1}, β = max{βj : j = 3, · · · , l1} and b = min{bj : j = 1, · · · , l2}. If
(1) arg a1 6= π and arg a1 6= arg a2

or
(2) arg a1 6= π, arg a1 = arg a2 and (i) |a2| >

|a1|
1−β or (ii) |a2| < (1 −α) |a1|

or
(3) a1 < 0 and arg a1 6= arg a2

or
(4) (i) (1 −β) a2 − b < a1 < 0, a2 < b1−β or (ii) a1 <

a2+b
1−α and a2 < 0, then every

meromorphic solution f (6≡ 0) whose poles are of uniformly bounded multiplicities
of the differential equation

(1.3) f′′ +


 l2∑
j=1

Bje
bjz


f′ +


 l1∑
j=1

Aje
ajz


f = 0

satisfies ρ (f) = +∞ and ρ2 (f) = 1.

2. Preliminary lemmas

We define the linear measure of a set E ⊂ [0, +∞) by m(E) =
∫ +∞

0
χE(t)dt and

the logarithmic measure of a set F ⊂ (1, +∞) by lm(F) =
∫ +∞

1
χF (t)
t
dt, where χH

is the characteristic function of a set H.

Lemma 2.1 [11] Let f be a transcendental meromorphic function with ρ (f) = ρ <
+∞. Let ε > 0 be a given constant, and let k, j be integers satisfying k > j ≥ 0.
Then, there exists a set E1 ⊂

[
−π

2
, 3π

2

)
with linear measure zero, such that, if

ψ ∈
[
−π

2
, 3π

2

)
\E1, then there is a constant R0 = R0 (ψ) > 1, such that for all z

satisfying arg z = ψ and |z| ≥ R0, we have

(2.1)

∣∣∣∣f(k) (z)f(j) (z)
∣∣∣∣ ≤ |z|(k−j)(ρ−1+ε) .

Lemma 2.2 ([4] , [15]) Consider g (z) = A (z) eaz, where A (z) 6≡ 0 is a mero-
morphic function with order ρ (A) = α < 1, a is a complex constant, a = |a|eiϕ
(ϕ ∈ [0, 2π)). Set E2 = {θ ∈ [0, 2π) : cos (ϕ + θ) = 0}, then E2 is a finite set. Then
for any given ε (0 < ε < 1 −α) there is a set E3 ⊂ [0, 2π) that has linear measure
zero such that if z = reiθ, θ ∈ [0, 2π) � (E2 ∪E3), then we have when r is suffi-
ciently large:



PROPERTIES OF MEROMORPHIC SOLUTIONS 201

(i) If cos (ϕ + θ) > 0, then

(2.2) exp{(1 −ε) rδ (az,θ)}≤ |g (z)| ≤ exp{(1 + ε) rδ (az,θ)} .
(ii) If cos (ϕ + θ) < 0, then

(2.3) exp{(1 + ε) rδ (az,θ)}≤ |g (z)| ≤ exp{(1 −ε) rδ (az,θ)} ,
where δ (az,θ) = |a|cos (ϕ + θ) .

Lemma 2.3 [17] Suppose that n ≥ 1 is a natural number. Let Pj (z) = ajnzn +
· · · (j = 1, 2) be nonconstant polynomials, where ajq ( q = 1, · · · ,n) are complex
numbers and a1na2n 6= 0. Set z = reiθ, ajn = |ajn|eiθj , θj ∈

[
−π

2
, 3π

2

)
, δ (Pj,θ) =

|ajn|cos (θj + nθ), then there is a set E4 ⊂
[
− π

2n
, 3π

2n

)
that has linear measure zero

such that if θ1 6= θ2, then there exists a ray arg z = θ with θ ∈
(
− π

2n
, π

2n

)
\(E4 ∪E5)

satisfying either

(2.4) δ (P1,θ) > 0, δ (P2,θ) < 0

or

(2.5) δ (P1,θ) < 0, δ (P2,θ) > 0,

where E5 =
{
θ ∈

[
− π

2n
, 3π

2n

)
: δ (Pj,θ) = 0

}
is a finite set, which has linear measure

zero.

Remark 2.1 [17] We can obtain, in Lemma 2.3, if θ ∈
(
− π

2n
, π

2n

)
\ (E4 ∪E5) is

replaced by θ ∈
(
π
2n
, 3π

2n

)
\ (E4 ∪E5), then it has the same result.

Lemma 2.4 [4] Let f (z) be a transcendental meromorphic function of order
ρ (f) = α < +∞. Then for any given ε > 0, there is a set E6 ⊂

[
−π

2
, 3π

2

)
that

has linear measure zero such that if θ ∈
[
−π

2
, 3π

2

)
�E6, then there is a constant

R1 = R1 (θ) > 1, such that for all z satisfying arg z = θ and |z| ≥ R1, we have
(2.6) exp

{
−rα+ε

}
≤ |f (z)| ≤ exp

{
rα+ε

}
.

Lemma 2.5 [11] Let f(z) be a transcendental meromorphic function, and let α > 1
be a given constant. Then there exist a set E7 ⊂ (1,∞) with finite logarithmic
measure and a constant B > 0 that depends only on α and i,j (0 ≤ i < j ≤ k),
such that for all z satisfying |z| = r /∈ [0, 1] ∪E7, we have

(2.7)

∣∣∣∣f(j)(z)f(i)(z)
∣∣∣∣ ≤ B

{
T(αr,f)

r
(logα r) log T(αr,f)

}j−i
.

Lemma 2.6 [12] Let ϕ : [0, +∞) → R and ψ : [0, +∞) → R be monotone non-
decreasing functions such that ϕ (r) ≤ ψ (r) for all r /∈ E8 ∪ [0, 1], where E8 ⊂
(1, +∞) is a set of finite logarithmic measure. Let γ > 1 be a given constant. Then
there exists an r1 = r1 (γ) > 0 such that ϕ (r) ≤ ψ (γr) for all r > r1.

Lemma 2.7 [5] Let k ≥ 2 and A0,A1, · · · ,Ak−1 be meromorphic functions. Let
ρ = max{ρ (Aj) : j = 0, · · · ,k − 1} and all poles of f are of uniformly bounded
multiplicities. Then every transcendental meromorphic solution f of the differential
equation

f(k) + Ak−1f
(k−1) + · · · + A1f′ + A0f = 0

satisfies ρ2 (f) ≤ ρ.



202 BELAÏDI AND HABIB

Lemma 2.8 ([9] , [20]) Suppose that f1 (z) ,f2 (z) , · · · ,fn (z) (n ≥ 2) are meromor-
phic functions and g1 (z) ,g2 (z) , · · · ,gn (z) are entire functions satisfying the fol-
lowing conditions:

(i)
n∑
j=1

fj (z) e
gj (z) ≡ 0;

(ii) gj (z) −gk (z) are not constants for 1 ≤ j < k ≤ n;
(iii) For 1 ≤ j ≤ n, 1 ≤ h < k ≤ n, T (r,fj) = o

{
T
(
r,egh(z)−gk(z)

)}
(r → ∞,

r /∈ E9), where E9 is a set with finite linear measure.
Then fj (z) ≡ 0 (j = 1, · · · ,n).

Lemma 2.9 [18] Suppose that f1 (z) ,f2 (z) , · · · ,fn (z) (n ≥ 2) are meromorphic
functions and g1 (z) ,g2 (z) , · · · ,gn (z) are entire functions satisfying the following
conditions:

(i)
n∑
j=1

fj (z) e
gj (z) ≡ fn+1;

(ii) If 1 ≤ j ≤ n + 1, 1 ≤ k ≤ n, the order of fj is less than the order of egk(z). If
n ≥ 2, 1 ≤ j ≤ n + 1, 1 ≤ h < k ≤ n, and the order of fj is less than the order of
egh−gk .
Then fj (z) ≡ 0 (j = 1, 2, · · · ,n + 1).

3. Proof of Theorem 1.1

First of all we prove that equation (1.3) can’t have a meromorphic solution f 6≡ 0
with ρ (f) < 1. Assume a meromorphic solution f 6≡ 0 with ρ (f) < 1. We can
rewrite (1.3) in the following form

(3.1)

l2∑
j=1

Bjf
′ebjz +

l1∑
j=1

Ajfe
ajz = −f′′.

Obviously, ρ (Bjf
′) < 1 (j = 1, · · · , l2) and ρ (Ajf) < 1 (j = 1, · · · , l1). Set I =

{aj (j = 1, · · · , l1) , bj (j = 1, · · · , l2)}.
1) By the conditions (1) or (2) or (4) (ii) of Theorem 1.1, we can see that a1 6=
a2,a3, · · · ,al1,b1, · · · ,bl2 . Then, we can rewrite (3.1) in the following form

(3.2) A1fe
a1z +

∑
λ∈Γ1

fλe
λz = −f′′,

where Γ1 ⊆ I \ {a1} and fλ (λ ∈ Γ1) are meromorphic functions with order less
than 1 and a1, λ (λ ∈ Γ1) are distinct numbers. By Lemma 2.8 and Lemma 2.9,
we get A1 ≡ 0, which is a contradiction.
2) By the conditions (3) or (4) (i) of Theorem 1.1, we can see that a2 6= a1,a3, · · · ,al1 ,
b1, · · · ,bl2 . Then, we can rewrite (3.1) in the following form

(3.3) A2fe
a2z +

∑
λ∈Γ2

fλe
λz = −f′′,

where Γ2 ⊆ I \ {a2} and fλ (λ ∈ Γ2) are meromorphic functions with order less
than 1 and a2, λ (λ ∈ Γ2) are distinct numbers. By Lemma 2.8 and Lemma 2.9,
we get A2 ≡ 0, which is a contradiction. Therefore ρ (f) ≥ 1.



PROPERTIES OF MEROMORPHIC SOLUTIONS 203

First step. We prove that ρ (f) = +∞. Assume that f 6≡ 0 is a meromorphic
solution whose poles are of uniformly bounded multiplicities of equation (1.3) with
1 ≤ ρ (f) = σ1 < +∞. From equation (1.3), we know that the poles of f (z) can
occur only at the poles of Aj (j = 1, · · · ., l1) and Bj (j = 1, · · · , l2). Note that the
multiplicities of poles of f are uniformly bounded, and thus we have [6]

N (r,f) ≤ M1N (r,f) ≤ M1


 l1∑
j=1

N (r,Aj) +

l2∑
j=1

N (r,Bj)




≤ M max{N (r,Aj) (j = 1, · · · , l1) ,N (r,Bj) (j = 1, · · · , l2)} ,
where M1 and M are some suitable positive constants. This gives

λ

(
1

f

)
≤ γ = max{ρ (Aj) (j = 1, · · · , l1) ,ρ (Bj) (j = 1, · · · , l2)} < 1.

Let f = g/d, d be the canonical product formed with the nonzero poles of f (z),

with ρ (d) = λ (d) = λ
(

1
f

)
= σ2 ≤ γ < 1, g is an entire function and 1 ≤ ρ (g) =

ρ (f) = σ1 < ∞. Substituting f = g/d into (1.3), we can get

g′′

g
+




 l2∑
j=1

Bje
bjz


− 2d′

d


 g′
g

+ 2

(
d′

d

)2
−
d′′

d
−


 l2∑
j=1

Bje
bjz


 d′
d

(3.4) +A1e
a1z + A2e

a2z +

l1∑
j=3

Aje
(αja1+βja2)z = 0.

By Lemma 2.4, for any given ε (0 < ε < 1 −γ), there is a set E6 ⊂
[
−π

2
, 3π

2

)
that

has linear measure zero such that if θ ∈
[
−π

2
, 3π

2

)
�E6, then there is a constant

R1 = R1 (θ) > 1, such that for all z satisfying arg z = θ and |z| ≥ R1, we have

(3.5) |Bj (z)| ≤ exp
{
rγ+ε

}
(j = 1, · · · , l2) .

By Lemma 2.1, for any given ε (0 < ε < 1 −γ), there exists a set E1 ⊂
[
−π

2
, 3π

2

)
of linear measure zero, such that if θ ∈

[
−π

2
, 3π

2

)
\ E1, then there is a constant

R0 = R0 (θ) > 1, such that for all z satisfying arg z = θ and |z| = r ≥ R0, we have

(3.6)

∣∣∣∣g(j) (z)g (z)
∣∣∣∣ ≤ rj(σ1−1+ε) (j = 1, 2) ,

(3.7)

∣∣∣∣d(j) (z)d (z)
∣∣∣∣ ≤ rj(σ2−1+ε) (j = 1, 2) .

Let z = reiθ, a1 = |a1|eiθ1 , a2 = |a2|eiθ2 , θ1,θ2 ∈
[
−π

2
, 3π

2

)
. We know that

δ (αja1z,θ) = αjδ (a1z,θ), δ (βja2z,θ) = βjδ (a2z,θ) (j = 3, · · · , l1) and α < 1,
β < 1.

Case 1. Assume that arg a1 6= π and arg a1 6= arg a2, which is θ1 6= π and θ1 6= θ2.
By Lemma 2.2 and Lemma 2.3, for any given ε

0 < ε < min

{
1 −γ,

1 −α
2 (1 + α)

,
1 −β

2 (1 + β)

}
,



204 BELAÏDI AND HABIB

there is a ray arg z = θ such that θ ∈
(
−π

2
, π

2

)
\(E1 ∪E4 ∪E5 ∪E6) (where E4 and

E5 are defined as in Lemma 2.3, E1 ∪E4 ∪E5 ∪E6 is of linear measure zero), and
satisfying

δ (a1z,θ) > 0, δ (a2z,θ) < 0

or

δ (a1z,θ) < 0, δ (a2z,θ) > 0.

(a) When δ (a1z,θ) > 0, δ (a2z,θ) < 0, for sufficiently large r, we get by Lemma
2.2

(3.8) |A1ea1z| ≥ exp{(1 −ε) δ (a1z,θ) r} ,

(3.9) |A2ea2z| ≤ exp{(1 −ε) δ (a2z,θ) r} < 1,

|Ajeαja1z| ≤ exp{(1 + ε) αjδ (a1z,θ) r}

(3.10) ≤ exp{(1 + ε) αδ (a1z,θ) r} (j = 3, · · · , l1) ,

(3.11)
∣∣eβja2z∣∣ ≤ exp{(1 −ε) βjδ (a2z,θ) r} < 1 (j = 3, · · · , l1) .

By (3.10) and (3.11), we get∣∣∣∣∣∣
l1∑
j=3

Aje
(αja1+βja2)z

∣∣∣∣∣∣ ≤
l1∑
j=3

|Ajeαja1z|
∣∣eβja2z∣∣

(3.12) ≤ (l1 − 2) exp{(1 + ε) αδ (a1z,θ) r} .

For θ ∈
(
−π

2
, π

2

)
by (3.5) we have∣∣∣∣∣∣
l2∑
j=1

Bje
bjz

∣∣∣∣∣∣ ≤
l2∑
j=1

|Bj|
∣∣ebjz∣∣ ≤ exp {rγ+ε} l2∑

j=1

∣∣ebjz∣∣

(3.13) = exp
{
rγ+ε

} l2∑
j=1

ebjr cos θ ≤ l2 exp
{
rγ+ε

}
because bj < 0 and cos θ > 0. By (3.4), we obtain

|A1ea1z| ≤
∣∣∣∣g′′g

∣∣∣∣ +


∣∣∣∣∣∣
l2∑
j=1

Bje
bjz

∣∣∣∣∣∣ + 2
∣∣∣∣d′d
∣∣∣∣

∣∣∣∣g′g

∣∣∣∣ + 2
∣∣∣∣d′d
∣∣∣∣2 +

∣∣∣∣d′′d
∣∣∣∣

(3.14) +

∣∣∣∣∣∣
l2∑
j=1

Bje
bjz

∣∣∣∣∣∣
∣∣∣∣d′d
∣∣∣∣ + |A2ea2z| +

∣∣∣∣∣∣
l1∑
j=3

Aje
(αja1+βja2)z

∣∣∣∣∣∣ .
Substituting (3.6) − (3.9) , (3.12) and (3.13) into (3.14), we have

exp{(1 −ε) δ (a1z,θ) r}≤ |A1ea1z|

≤ r2(σ1−1+ε) +
[
l2 exp

{
rγ+ε

}
+ 2rσ2−1+ε

]
rσ1−1+ε + 3r2(σ2−1+ε)

+l2 exp
{
rγ+ε

}
rσ2−1+ε + 1 + (l1 − 2) exp{(1 + ε) αδ (a1z,θ) r}

(3.15) ≤ M1rM2 exp
{
rγ+ε

}
exp{(1 + ε) αδ (a1z,θ) r} ,



PROPERTIES OF MEROMORPHIC SOLUTIONS 205

where M1 > 0 and M2 > 0 are some constants. By 0 < ε <
1−α

2(1+α)
and (3.15), we

have

(3.16) exp

{
1 −α

2
δ (a1z,θ) r

}
≤ M1rM2 exp

{
rγ+ε

}
.

By δ (a1z,θ) > 0 and γ + ε < 1 we know that (3.16) is a contradiction.

(b) When δ (a1z,θ) < 0, δ (a2z,θ) > 0, for sufficiently large r, we get

(3.17) |A2ea2z| ≥ exp{(1 −ε) δ (a2z,θ) r} ,

(3.18) |A1ea1z| ≤ exp{(1 −ε) δ (a1z,θ) r} < 1,

(3.19) |Ajeαja1z| ≤ exp{(1 −ε) αjδ (a1z,θ) r} < 1 (j = 3, · · · , l1) ,∣∣eβja2z∣∣ ≤ exp{(1 + ε) βjδ (a2z,θ) r}
(3.20) ≤ exp{(1 + ε) βδ (a2z,θ) r} (j = 3, · · · , l1) .
By (3.19) and (3.20), we get∣∣∣∣∣∣

l1∑
j=3

Aje
(αja1+βja2)z

∣∣∣∣∣∣ ≤
l1∑
j=3

|Ajeαja1z|
∣∣eβja2z∣∣

(3.21) ≤ (l1 − 2) exp{(1 + ε) βδ (a2z,θ) r} .
By (3.4), we obtain

|A2ea2z| ≤
∣∣∣∣g′′g

∣∣∣∣ +


∣∣∣∣∣∣
l2∑
j=1

Bje
bjz

∣∣∣∣∣∣ + 2
∣∣∣∣d′d
∣∣∣∣

∣∣∣∣g′g

∣∣∣∣ + 2
∣∣∣∣d′d
∣∣∣∣2 +

∣∣∣∣d′′d
∣∣∣∣

(3.22) +

∣∣∣∣∣∣
l2∑
j=1

Bje
bjz

∣∣∣∣∣∣
∣∣∣∣d′d
∣∣∣∣ + |A1ea1z| +

∣∣∣∣∣∣
l1∑
j=3

Aje
(αja1+βja2)z

∣∣∣∣∣∣ .
Substituting (3.6) , (3.7) , (3.13) , (3.17) , (3.18) and (3.21) into (3.22), we have

exp{(1 −ε) δ (a2z,θ) r}≤ |A2ea2z|

≤ r2(σ1−1+ε) +
[
l2 exp

{
rγ+ε

}
+ 2rσ2−1+ε

]
rσ1−1+ε + 3r2(σ2−1+ε)

+l2 exp
{
rγ+ε

}
rσ2−1+ε + 1 + (l1 − 2) exp{(1 + ε) βδ (a2z,θ) r}

(3.23) ≤ M1rM2 exp
{
rγ+ε

}
exp{(1 + ε) βδ (a2z,θ) r} .

By 0 < ε < 1−β
2(1+β)

and (3.23), we obtain

(3.24) exp

{
1 −β

2
δ (a2z,θ) r

}
≤ M1rM2 exp

{
rγ+ε

}
.

By δ (a2z,θ) > 0 and γ + ε < 1 we know that (3.24) is a contradiction.

Case 2. Assume that arg a1 6= π, arg a1 = arg a2, which is θ1 6= π, θ1 = θ2. By
Lemma 2.3, for any given ε

0 < ε < min

{
1 −γ,

(1 −α) |a1|− |a2|
2 [(1 + α) |a1| + |a2|]

,
(1 −β) |a2|− |a1|

2 [(1 + β) |a2| + |a1|]

}
,



206 BELAÏDI AND HABIB

there is a ray arg z = θ such that θ ∈
(
−π

2
, π

2

)
\(E1 ∪E4 ∪E5 ∪E6) and δ (a1z,θ) >

0. Since θ1 = θ2, then δ (a2z,θ) > 0.

(i) |a2| >
|a1|
1−β . For sufficiently large r, we have (3.10) , (3.17) , (3.20) hold and we

get

(3.25) |A1ea1z| ≤ exp{(1 + ε) δ (a1z,θ) r} .
By (3.10) and (3.20), we obtain∣∣∣∣∣∣

l1∑
j=3

Aje
(αja1+βja2)z

∣∣∣∣∣∣ ≤
l1∑
j=3

|Ajeαja1z|
∣∣eβja2z∣∣

(3.26) ≤ (l1 − 2) exp{(1 + ε) αδ (a1z,θ) r}exp{(1 + ε) βδ (a2z,θ) r} .
Substituting (3.6) , (3.7) , (3.13) , (3.17) , (3.25) and (3.26) into (3.22), we have

exp{(1 −ε) δ (a2z,θ) r}≤ |A2ea2z|

≤ r2(σ1−1+ε) +
[
l2 exp

{
rγ+ε

}
+ 2rσ2−1+ε

]
rσ1−1+ε + 3r2(σ2−1+ε)

+l2 exp
{
rγ+ε

}
rσ2−1+ε + exp{(1 + ε) δ (a1z,θ) r}

+ (l1 − 2) exp{(1 + ε) αδ (a1z,θ) r}exp{(1 + ε) βδ (a2z,θ) r}

(3.27) ≤ M1rM2 exp
{
rγ+ε

}
exp{(1 + ε) δ (a1z,θ) r}exp{(1 + ε) βδ (a2z,θ) r} .

From (3.27), we obtain

(3.28) exp{η1r}≤ M1rM2 exp
{
rγ+ε

}
,

where
η1 = (1 −ε) δ (a2z,θ) − (1 + ε) δ (a1z,θ) − (1 + ε) βδ (a2z,θ) .

Since 0 < ε <
(1−β)|a2|−|a1|

2[(1+β)|a2|+|a1|]
,θ1 = θ2 and cos (θ1 + θ) > 0, then

η1 = [1 −β −ε (1 + β)] δ (a2z,θ) − (1 + ε) δ (a1z,θ)
= [1 −β −ε (1 + β)] |a2|cos (θ1 + θ) − (1 + ε) |a1|cos (θ1 + θ)

= {[1 −β −ε (1 + β)] |a2|− (1 + ε) |a1|}cos (θ1 + θ)
= {(1 −β) |a2|− |a1|−ε [(1 + β) |a2| + |a1|]}cos (θ1 + θ)

>
(1 −β) |a2|− |a1|

2
cos (θ1 + θ) > 0.

By η1 > 0 and γ + ε < 1 we know that (3.28) is a contradiction.

(ii) |a2| < (1 −α) |a1|. For sufficiently large r, we have (3.8) , (3.10) , (3.20) , (3.26)
hold and we obtain

(3.29) |A2ea2z| ≤ exp{(1 + ε) δ (a2z,θ) r} .
Substituting (3.6) , (3.7) , (3.8) , (3.13) , (3.26) and (3.29) into (3.14), we have

exp{(1 −ε) δ (a1z,θ) r}≤ |A1ea1z|

≤ r2(σ1−1+ε) +
[
l2 exp

{
rγ+ε

}
+ 2rσ2−1+ε

]
rσ1−1+ε + 3r2(σ2−1+ε)

+l2 exp
{
rγ+ε

}
rσ2−1+ε + exp{(1 + ε) δ (a2z,θ) r}

+ (l1 − 2) exp{(1 + ε) αδ (a1z,θ) r}exp{(1 + ε) βδ (a2z,θ) r}

(3.30) ≤ M1rM2 exp
{
rγ+ε

}
exp{(1 + ε) αδ (a1z,θ) r}exp{(1 + ε) δ (a2z,θ) r} .



PROPERTIES OF MEROMORPHIC SOLUTIONS 207

From (3.30), we obtain

(3.31) exp{η2r}≤ M1rM2 exp
{
rγ+ε

}
,

where

η2 = (1 −ε) δ (a1z,θ) − (1 + ε) αδ (a1z,θ) − (1 + ε) δ (a2z,θ) .
Since 0 < ε <

(1−α)|a1|−|a2|
2[(1+α)|a1|+|a2|]

,θ1 = θ2 and cos (θ1 + θ) > 0, then we get

η2 = {(1 −α) |a1|− |a2|−ε [(1 + α) |a1| + |a2|]}cos (θ1 + θ)

>
(1 −α) |a1|− |a2|

2
cos (θ1 + θ) > 0.

By η2 > 0 and γ + ε < 1 we know that (3.31) is a contradiction.

Case 3. Assume that a1 < 0 and arg a1 6= arg a2, which is θ1 = π and θ2 6= π.
By Lemma 2.2, for the above ε, there is a ray arg z = θ such that θ ∈

(
−π

2
, π

2

)
\

(E1 ∪E4 ∪E5 ∪E6) and δ (a2z,θ) > 0. Because cos θ > 0, we have δ (a1z,θ) =
|a1|cos (θ1 + θ) = −|a1|cos θ < 0. Using the same reasoning as in Case 1(b), we
can get a contradiction.

Case 4. Assume that (i) (1 −β) a2 − b < a1 < 0 and a2 < b1−β or (ii) a1 <
a2+b
1−α

and a2 < 0, which is θ1 = θ2 = π. By Lemma 2.2, for any given ε

0 < ε < min

{
1 −γ,

(1 −α) |a1|− |a2| + b
2 [(1 + α) |a1| + |a2|]

,
(1 −β) |a2|− |a1| + b
2 [(1 + β) |a2| + |a1|]

}
,

there is a ray arg z = θ such that θ ∈
(
π
2
, 3π

2

)
\ (E1 ∪E4 ∪E5 ∪E6), then cos θ <

0, δ (a1z,θ) = |a1|cos (θ1 + θ) = −|a1|cos θ > 0, δ (a2z,θ) = |a2|cos (θ2 + θ) =
−|a2|cos θ > 0.

(i) (1 −β) a2 − b < a1 < 0 and a2 < b1−β . For sufficiently large r, we get
(3.10) , (3.17) , (3.20) , (3.25) and (3.26) hold. For θ ∈

(
π
2
, 3π

2

)
by (3.5) we have∣∣∣∣∣∣

l2∑
j=1

Bje
bjz

∣∣∣∣∣∣ ≤
l2∑
j=1

|Bj|
∣∣ebjz∣∣ ≤ exp {rγ+ε} l2∑

j=1

∣∣ebjz∣∣

(3.32) = exp
{
rγ+ε

} l2∑
j=1

ebjr cos θ ≤ l2 exp
{
rγ+ε

}
ebr cos θ

because b ≤ bj < 0 and cos θ < 0. Substituting (3.6) , (3.7) , (3.17) , (3.25) , (3.26)
and (3.32) into (3.22), we obtain

exp{(1 −ε) δ (a2z,θ) r}≤ |A2ea2z|

(3.33)
≤ M1rM2ebr cos θ exp

{
rγ+ε

}
exp{(1 + ε) δ (a1z,θ) r}exp{(1 + ε) βδ (a2z,θ) r} .

From (3.33) we have

(3.34) exp{η3r}≤ M1rM2 exp
{
rγ+ε

}
,

where

η3 = (1 −ε) δ (a2z,θ) − (1 + ε) δ (a1z,θ) − (1 + ε) βδ (a2z,θ) − b cos θ.



208 BELAÏDI AND HABIB

Since (1 −β) a2 − b < a1, a2 = −|a2| and a1 = −|a1|, then we get (1 −β) |a2|−
|a1| + b > 0. We can see that 0 < (1 −β) |a2| − |a1| + b < (1 −β) |a2| − |a1| <
2 [(1 + β) |a2| + |a1|]. Therefore

0 <
(1 −β) |a2|− |a1| + b
2 [(1 + β) |a2| + |a1|]

< 1.

By 0 < ε <
(1−β)|a2|−|a1|+b
2[(1+β)|a2|+|a1|]

,θ1 = θ2 = π and cos θ < 0, we obtain

η3 = [1 −β −ε (1 + β)] δ (a2z,θ) − (1 + ε) δ (a1z,θ) − b cos θ

= − [1 −β −ε (1 + β)] |a2|cos θ + (1 + ε) |a1|cos θ − b cos θ
= (−cos θ){[1 −β −ε (1 + β)] |a2|− (1 + ε) |a1| + b}

= (−cos θ){(1 −β) |a2|− |a1| + b−ε [(1 + β) |a2| + |a1|]}

>
−1
2

[(1 −β) |a2|− |a1| + b] cos θ > 0.

By η3 > 0 and γ + ε < 1 we know that (3.34) is a contradiction.

(ii) a1 <
a2+b
1−α and a2 < 0. For sufficiently large r, we get (3.8) , (3.10) , (3.20) ,

(3.26) and (3.29) hold. Substituting (3.6) , (3.7) , (3.8) , (3.26) , (3.29) and (3.32)
into (3.14), we obtain

exp{(1 −ε) δ (a1z,θ) r}≤ |A1ea1z|

(3.35)
≤ M1rM2ebr cos θ exp

{
rγ+ε

}
exp{(1 + ε) αδ (a1z,θ) r}exp{(1 + ε) δ (a2z,θ) r} .

From (3.35) we have

(3.36) exp{η4r}≤ M1rM2 exp
{
rγ+ε

}
,

where

η4 = (1 −ε) δ (a1z,θ) − (1 + ε) αδ (a1z,θ) − (1 + ε) δ (a2z,θ) − b cos θ.

Since a1 <
a2+b
1−α ,a2 = −|a2| and a1 = −|a1|, then we get (1 −α) |a1| − |a2| +

b > 0. We can see that 0 < (1 −α) |a1| − |a2| + b < (1 −α) |a1| − |a2| <
2 [(1 + α) |a1| + |a2|]. Therefore

0 <
(1 −α) |a1|− |a2| + b
2 [(1 + α) |a1| + |a2|]

< 1.

By 0 < ε <
(1−α)|a1|−|a2|+b
2[(1+α)|a1|+|a2|]

,θ1 = θ2 = π and cos θ < 0, we get

η4 = (−cos θ){(1 −α) |a1|− |a2| + b−ε [(1 + α) |a1| + |a2|]}

>
−1
2

[(1 −α) |a1|− |a2| + b] cos θ > 0.

By η4 > 0 and γ + ε < 1 we know that (3.36) is a contradiction. Concluding the
above proof, we obtain ρ (f) = ρ (g) = +∞.

Second step. We prove that ρ2 (f) = 1. By

max


ρ

 l2∑
j=1

Bje
bjz


 ,ρ


 l1∑
j=1

Aje
ajz




 = 1



PROPERTIES OF MEROMORPHIC SOLUTIONS 209

and Lemma 2.7, we obtain ρ2 (f) ≤ 1. By Lemma 2.5, we know that there exists a
set E7 ⊂ (1, +∞) with finite logarithmic measure and a constant C > 0, such that
for all z satisfying |z| = r /∈ [0, 1] ∪E7, we get

(3.37)

∣∣∣∣f(j)(z)f(z)
∣∣∣∣ ≤ C [T(2r,f)]j+1 (j = 1, 2) .

By (1.3), we have

(3.38) |A1ea1z| ≤
∣∣∣∣f′′f

∣∣∣∣ +
∣∣∣∣∣∣
l2∑
j=1

Bje
bjz

∣∣∣∣∣∣
∣∣∣∣f′f
∣∣∣∣ + |A2ea2z| +

∣∣∣∣∣∣
l1∑
j=3

Aje
(αja1+βja2)z

∣∣∣∣∣∣ ,
(3.39) |A2ea2z| ≤

∣∣∣∣f′′f
∣∣∣∣ +
∣∣∣∣∣∣
l2∑
j=1

Bje
bjz

∣∣∣∣∣∣
∣∣∣∣f′f
∣∣∣∣ + |A1ea1z| +

∣∣∣∣∣∣
l1∑
j=3

Aje
(αja1+βja2)z

∣∣∣∣∣∣ .
Case 1. arg a1 6= π and arg a1 6= arg a2. In the first step, we have proved that
there is a ray arg z = θ where θ ∈

(
−π

2
, π

2

)
\ (E1 ∪E4 ∪E5 ∪E6), satisfying

δ (a1z,θ) > 0, δ (a2z,θ) < 0 or δ (a1z,θ) < 0, δ (a2z,θ) > 0.

(a) When δ (a1z,θ) > 0, δ (a2z,θ) < 0, for sufficiently large r, we get (3.8) − (3.12)
hold. Substituting (3.8) , (3.9) , (3.12) , (3.13) and (3.37) into (3.38), we obtain for
all z = reiθ satisfying |z| = r /∈ [0, 1] ∪E7, θ ∈

(
−π

2
, π

2

)
\ (E1 ∪E4 ∪E5 ∪E6)

exp{(1 −ε) δ (a1z,θ) r}≤ |A1ea1z|

(3.40) ≤ M exp
{
rγ+ε

}
exp{(1 + ε) αδ (a1z,θ) r} [T (2r,f)]

3
,

where M > 0 is a some constant. From (3.40) and 0 < ε < 1−α
2(1+α)

, we get

(3.41) exp

{
1 −α

2
δ (a1z,θ) r

}
≤ M exp

{
rγ+ε

}
[T (2r,f)]

3
.

Since δ (a1z,θ) > 0 and γ + ε < 1, then by using Lemma 2.6 and (3.41), we obtain
ρ2 (f) ≥ 1. Hence ρ2 (f) = 1.

(b) When δ (a1z,θ) < 0, δ (a2z,θ) > 0, for sufficiently large r, we get (3.17)−(3.21)
hold. By using the same reasoning as above, we can get ρ2 (f) = 1.

Case 2. arg a1 6= π, arg a1 = arg a2. In the first step, we have proved that there is
a ray arg z = θ where θ ∈

(
−π

2
, π

2

)
\ (E1 ∪E4 ∪E5 ∪E6), satisfying δ (a1z,θ) > 0

and δ (a2z,θ) > 0.

(i) |a2| >
|a1|
1−β . For sufficiently large r, we have (3.10) , (3.17) , (3.20) , (3.25) and

(3.26) hold. Substituting (3.13) , (3.17) , (3.25) , (3.26) and (3.37) into (3.39), we ob-
tain for all z = reiθ satisfying |z| = r /∈ [0, 1]∪E7, θ ∈

(
−π

2
, π

2

)
\(E1 ∪E4 ∪E5 ∪E6)

exp{(1 −ε) δ (a2z,θ) r}≤ |A2ea2z|
(3.42)

≤ M exp
{
rγ+ε

}
exp{(1 + ε) δ (a1z,θ) r}exp{(1 + ε) βδ (a2z,θ) r} [T (2r,f)]

3
.

From (3.42), we obtain

(3.43) exp{η1r}≤ M exp
{
rγ+ε

}
[T (2r,f)]

3
,



210 BELAÏDI AND HABIB

where

η1 = (1 −ε) δ (a2z,θ) − (1 + ε) δ (a1z,θ) − (1 + ε) βδ (a2z,θ) .
Since η1 > 0 and γ + ε < 1, then by using Lemma 2.6 and (3.43), we obtain
ρ2 (f) ≥ 1. Hence ρ2 (f) = 1.

(ii) |a2| < (1 −α) |a1|. For sufficiently large r, we have (3.8) , (3.10) , (3.20) , (3.26)
and (3.29) hold. By using the same reasoning as above, we can get ρ2 (f) = 1.

Case 3. a1 < 0 and arg a1 6= arg a2. In the first step, we have proved that there is
a ray arg z = θ where θ ∈

(
−π

2
, π

2

)
\ (E1 ∪E4 ∪E5 ∪E6), satisfying δ (a2z,θ) > 0

and δ (a1z,θ) < 0. Using the same reasoning as in the second step ( Case 1 (b)),
we can get ρ2 (f) = 1.

Case 4. (i) (1 −β) a2 − b < a1 < 0 and a2 < b1−β or (ii) a1 <
a2+b
1−α and a2 < 0.

In the first step, we have proved that there is a ray arg z = θ where θ ∈
(
π
2
, 3π

2

)
\

(E1 ∪E4 ∪E5 ∪E6), satisfying δ (a2z,θ) > 0 and δ (a1z,θ) > 0.

(i) (1 −β) a2 − b < a1 < 0 and a2 < b1−β . For sufficiently large r, we get (3.10) ,
(3.17) , (3.20) , (3.25) and (3.26) hold. Substituting (3.17) , (3.25) , (3.26) , (3.32)
and (3.37) into (3.39), we obtain for all z = reiθ satisfying |z| = r /∈ [0, 1] ∪ E7,
θ ∈

(
π
2
, 3π

2

)
\ (E1 ∪E4 ∪E5 ∪E6)

exp{(1 −ε) δ (a2z,θ) r}≤ |A2ea2z|

≤ Mebr cos θ exp
{
rγ+ε

}
exp{(1 + ε) δ (a1z,θ) r}

(3.44) ×exp{(1 + ε) βδ (a2z,θ) r} [T (2r,f)]
3
.

From (3.44) we obtain

(3.45) exp{η3r}≤ M exp
{
rγ+ε

}
[T (2r,f)]

3
,

where

η3 = (1 −ε) δ (a2z,θ) − (1 + ε) δ (a1z,θ) − (1 + ε) βδ (a2z,θ) − b cos θ.
Since η3 > 0 and γ + ε < 1, then by using Lemma 2.6 and (3.45), we obtain
ρ2 (f) ≥ 1. Hence ρ2 (f) = 1.

(ii) a1 <
a2+b
1−α and a2 < 0. For sufficiently large r, we get (3.8) , (3.10) , (3.20) ,

(3.26) and (3.29) hold. By using the same reasoning as above, we can get ρ2 (f) = 1.
Concluding the above proof, we obtain that every meromorphic solution f ( 6≡ 0)
whose poles are of uniformly bounded multiplicities of (1.3) satisfies ρ (f) = ∞ and
ρ2 (f) = 1. The proof of Theorem 1.1 is complete.

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Department of Mathematics, Laboratory of Pure and Applied Mathematics, Univer-

sity of Mostaganem (UMAB), B. P. 227 Mostaganem-(Algeria)

∗Corresponding author