# Growth of Logarithmic Derivatives and Their Applications in Complex Differential Equations

Authors: Zinelâabidine Latreuch; Benharrat Belaïdi
Journal: International Journal of Analysis
 (2014)

---

## Abstract

We continue the study of the behavior of the growth of logarithmic derivatives. In fact, we prove some relations between the value distribution of solutions of linear differential equations and growth
of their logarithmic derivatives. We also give an estimate of the growth of the quotient of two differential polynomials generated by solutions of the equationf
″
+
A
(
z
)
f
′
+
B
(
z
)
f
=
0, where A
(
z
) and B
(
z
) are entire functions.


---

## Body

## 1. Introduction and Main Results

Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna's value distribution theory (see [1–4]). In addition, we will denote by λ
(
f
) (resp., λ
¯
(
f
)) and λ
(
1
/
f
) (resp., λ
¯
(
1
/
f
)) the exponent of convergence of zeros (resp., distinct zeros) and poles (resp., distinct poles) of a meromorphic function f, ρ
(
f
) to denote the order of growth of f. A meromorphic function φ
(
z
) is called a small function with respect to f
(
z
) if T
(
r
,
φ
)
=
o
(
T
(
r
,
f
)
) as r
→
+
∞ except possibly a set of r of finite linear measure, where T
(
r
,
f
) is the Nevanlinna characteristic function of f.Definition 1 (see [4]).
Letf be a meromorphic function. Then the hyperorder of f
(
z
) is defined as
(1)
ρ
2
(
f
)
=
limsup
⁡
r
→
+
∞
log
⁡
log
⁡
T
(
r
,
f
)
log
⁡
r
.Definition 2 (see [1, 3]).
The type of a meromorphic functionf of order ρ
(
0
<
ρ
<
∞
) is defined as
(2)
τ
(
f
)
=
limsup
⁡
r
→
+
∞
T
(
r
,
f
)
r
ρ
.
If f is an entire function, then the type of f of order ρ
(
0
<
ρ
<
∞
) is defined as
(3)
τ
M
(
f
)
=
limsup
⁡
r
→
+
∞
log
⁡
M
(
r
,
f
)
r
ρ
,
where M
(
r
,
f
)
=
max
⁡
|
z
|
=
r
|
f
(
z
)
| is the maximum modulus function.Remark 3.
For entire function, we can haveτ
M
(
f
)
≠
τ
(
f
). For example, if f
(
z
)
=
e
z, then we have τ
M
(
f
)
=
1 and τ
(
f
)
=
1
/
π.Growth of logarithmic derivative of meromorphic functions has been generously considered during the last decades, among others, by Gol'dberg and Grinšteĭn [5], Benbourenane and Korhonen [6], Hinkkanen [7], and Heittokangas et al. [8]. All these considerations have been devoted to getting more detailed estimates to the proximity function m
(
r
,
f
′
/
f
) than what is the original Nevanlinna estimate that essentially can be written as m
(
r
,
f
′
/
f
)
=
S
(
r
,
f
). The same applies for the case of higher logarithmic derivatives m
(
r
,
f
(
k
)
/
f
) as well. The estimation of logarithmic derivatives plays the key role in theory of differential equations. In his paper Gundersen [9] proved some interesting inequalities on the module of logarithmic derivatives of meromorphic functions. Recently [10], the authors have studied some properties of the behavior of growth of logarithmic derivatives of entire and meromorphic functions and have obtained some relations between the zeros of entire functions and the growth of their logarithmic derivatives. In fact, they have proved the following.Theorem A (see [10]).
Letf be an entire function with finite number of zeros. Then, for any integer k
≥
1,
(4)
ρ
(
f
(
k
)
f
)
=
ρ
2
(
f
)
.Theorem B (see [10]).
Suppose thatk
≥
2 is an integer and let f be a meromorphic function. Then
(5)
ρ
(
f
′
f
)
=
max
⁡
{
ρ
(
f
(
k
)
f
)
,
k
≥
2
}
=
max
⁡
{
ρ
(
f
(
k
)
f
)
,
ρ
(
f
(
k
+
1
)
f
)
,
k
≥
2
}
.In [11, pages 457–458], Bank and Langley have obtained the following result.Theorem C (see [11]).
All nontrivial solutions of(6)
f
′′
+
e
P
(
z
)
f
=
0
,
where P
(
z
) is a nonconstant polynomial, satisfy λ
(
f
)
=
∞.Theorem D (see [4] (E. Picard)).
Any transcendental entire functionf must take every finite complex value infinitely many times, with at most one exception.Example 4.
The entire functione
z omits the value zero and every other finite value is assumed infinitely often and zero is the exceptional value of Picard.Question 1.
Under what conditions can we ensure thatf has no Picard's exceptional value?Question 2.
Under what conditions can we obtain the same result as Theorem C for higher order linear differential equations with entire coefficients?The subject of this paper is to give answers to the above questions. In fact, we prove some relations between the value distribution of solutions of linear differential equations and growth of their logarithmic derivatives. Firstly, we study the relationship between the growth of logarithmic derivatives of solutions of complex differential equation(7)
f
(
k
)
+
A
k
-
1
f
(
k
-
1
)
+
⋯
+
A
0
f
=
0
and their exponents of convergence. We will also prove the nonexistence of any Picard exceptional value of solutions of (7), and we obtain the following.Theorem 5.
LetA
0
,
A
1
,
…
,
A
k
-
1 be entire functions satisfying (
A
0
)
=
ρ
(
0
<
ρ
<
∞
), τ
M
(
A
0
)
=
τ
(
0
<
τ
<
∞
), and let ρ
(
A
j
)
≤
ρ, τ
M
(
A
j
)
<
τ if ρ
(
A
j
)
=
ρ
(
j
=
1
,
…
,
k
-
1
). If f is a nontrivial solution of (7), such that ρ
(
f
′
/
f
)
>
ρ
(
A
0
), then f assume every finite complex value infinitely often.
Furthermore, ifρ
(
f
′
/
f
)
=
∞, then
(8)
λ
¯
(
f
)
=
λ
(
f
)
=
∞
.Corollary 6.
Under the hypotheses of Theorem5, let φ
(
z
) be an entire function with finite order. If f is a nontrivial solution of (7), such that ρ
(
f
(
j
+
1
)
/
f
(
j
)
)
=
∞ for some j
∈
ℕ, then
(9)
λ
¯
(
f
(
j
)
-
φ
)
=
λ
(
f
(
j
)
-
φ
)
=
∞
.Corollary 7.
Under the hypotheses of Theorem5, every nontrivial solution of (7) does not have any Picard exceptional values.Example 8.
It is clear that the functionf
(
z
)
=
sin
e
z satisfies the differential equation
(10)
f
′′
-
f
′
+
e
2
z
f
=
0
and ρ
(
f
′
/
f
)
=
ρ
(
f
)
=
∞. Obviously, the conditions of Theorem 5 are satisfied. Then λ
¯
(
f
)
=
λ
(
f
)
=
∞.Secondly, we consider the differential equation(11)
f
′′
+
A
(
z
)
f
′
+
B
(
z
)
f
=
0
,
where A
(
z
) and B
(
z
) are entire functions, and let P
n
[
f
] and Q
m
[
f
] be two differential polynomials defined by
(12)
P
n
[
f
]
=
d
n
f
(
n
)
+
d
n
-
1
f
(
n
-
1
)
+
⋯
+
d
0
f
(
n
≥
1
)
,
Q
m
[
f
]
=
b
m
f
(
m
)
+
b
m
-
1
f
(
m
-
1
)
+
⋯
+
b
0
f
(
m
≥
1
)
,
where d
i
(
i
=
0
,
…
,
n
) and b
j
(
j
=
0
,
…
,
m
) are entire functions not all vanishing identically. In recent years, there is a great interest to investigate the growth and oscillation of differential polynomials generated by solutions of differential equations in the complex plane (see [12–14]). In this part, we give under some conditions an estimate on the growth of the quotient of two differential polynomials generated by solutions of (11), and we obtain the following results.Theorem 9.
Letf be a nontrivial solution of (11), and let d
i
(
i
=
0
,
…
,
n
) and b
j
(
j
=
0
,
…
,
m
) be entire functions not all vanishing identically such that g
f
=
P
n
[
f
]
/
Q
m
[
f
] is irreducible function in f
′
/
f. If
(13)
max
⁡
{
(
b
j
)
(
j
=
0
,
…
,
m
)
ρ
(
A
)
,
ρ
(
B
)
,
ρ
(
d
i
)
(
i
=
0
,
…
,
n
)
,
d
d
d
d
ρ
(
b
j
)
(
j
=
0
,
…
,
m
)
}
<
ρ
(
f
′
f
)
,
then
(14)
ρ
(
g
f
)
=
ρ
(
f
′
f
)
.Remark 10.
In Theorem9, the condition (13) is necessary. For example, f
(
z
)
=
exp
⁡
(
e
z
) is a solution of the differential equation
(15)
f
′′
-
(
e
z
+
e
2
z
)
f
=
0with (16)
ρ
(
f
′
f
)
=
1
and let
(17)
g
f
=
d
2
f
′′
+
d
1
f
′
+
d
0
f
b
2
f
′′
+
b
1
f
′
+
b
0
f
=
d
2
(
e
z
+
e
2
z
)
+
d
1
e
z
+
d
0
b
2
(
e
z
+
e
2
z
)
+
b
1
e
z
+
b
0
.
If we take d
2
=
e
z
2, d
1
=
d
0
=
1, and b
2
=
b
1
=
b
0
=
1, then
(18)
ρ
(
g
f
)
=
2
>
ρ
(
f
′
f
)
=
1
.Remark 11.
In Theorem9, the condition that g
f is irreducible function in f
′
/
f is necessary. If d
i
=
b
j
(
n
=
m
), then g
f
≡
1 and so
(19)
ρ
(
g
f
)
=
0
≠
ρ
(
f
′
f
)
.
By setting n
=
m
=
2 in Theorem 9, we can obtain the following special case.Corollary 12.
LetA
(
z
) and B
(
z
) be entire functions of finite order such that ρ
(
A
)
<
ρ
(
B
) and 0
<
τ
(
A
)
<
τ
(
B
)
<
∞ if ρ
(
B
)
=
ρ
(
A
)
>
0. Let d
i
(
z
)
(
i
=
0,1
,
2
)
,
b
j
(
z
)
(
j
=
0,1
,
2
) be entire functions that are not all vanishing identically such that
(20)
max
⁡
{
ρ
(
d
i
)
,
ρ
(
b
j
)
,
0
≤
i
,
j
≤
2
}
<
ρ
(
B
)
and d
2
(
z
)
b
1
(
z
)
-
d
1
(
z
)
b
2
(
z
)
≢
0. If f
≢
0 is a solution of (11), such that ρ
(
B
)
<
ρ
(
f
′
/
f
), then
(21)
ρ
(
d
2
f
′′
+
d
1
f
′
+
d
0
f
b
2
f
′′
+
b
1
f
′
+
b
0
f
)
=
ρ
(
f
′
f
)
.Corollary 13.
Under the hypotheses of Corollary12, if f is a nontrivial solution of (11), such that ρ
(
f
′
/
f
)
=
∞, then
(22)
λ
¯
(
f
)
=
λ
¯
(
f
′
)
=
∞
.


## 2. Auxiliary Lemmas

Lemma 14 (see [1]).
Letf be a meromorphic function and let k
∈
ℕ. Then
(23)
m
(
r
,
f
(
k
)
f
)
=
S
(
r
,
f
)
,
where S
(
r
,
f
)
=
O
(
log
⁡
T
(
r
,
f
)
+
log
⁡
r
), possibly outside a set E
⊂
(
0
,
∞
) with a finite linear measure. If f is a finite order of growth, then
(24)
m
(
r
,
f
(
k
)
f
)
=
O
(
log
⁡
r
)
.Lemma 15 (see [15]).
LetA
0
(
z
)
,
…
,
A
k
-
1
(
z
) be entire functions of finite order such that
(25)
max
⁡
{
ρ
(
A
j
)
:
j
=
1
,
…
,
k
-
1
}
<
ρ
(
A
0
)
.
Then every solution f
≢
0 of (7) satisfies ρ
(
f
)
=
+
∞ and ρ
2
(
f
)
=
ρ
(
A
0
).Lemma 16 (see [16]).
LetA
0
,
A
1
,
…
,
A
k
-
1 be entire functions satisfying (
A
0
)
=
ρ
(
0
<
ρ
<
∞
), τ
M
(
A
0
)
=
τ
(
0
<
τ
<
∞
), and let ρ
(
A
j
)
≤
ρ, τ
M
(
A
j
)
<
τ if ρ
(
A
j
)
=
ρ
(
j
=
1
,
…
,
k
-
1
). Then every solution f
≢
0 of (7) satisfies ρ
(
f
)
=
+
∞ and ρ
2
(
f
)
=
ρ
(
A
0
).Here, we give a special case of the result given by Xu et al. in [17].Lemma 17.
LetA
0
,
A
1
,
…
,
A
k
-
1 be entire functions with finite order and let them satisfy one of the following conditions:(i)
max
⁡
{
ρ
(
A
j
)
:
j
=
1
,
…
,
k
-
1
}
<
ρ
(
A
0
)
<
∞;
(ii)
0
<
ρ
(
A
k
-
1
)
=
⋯
=
ρ
(
A
0
)
<
∞ and max
⁡
{
τ
M
(
A
j
)
:
j
=
1
,
…
,
k
-
1
}
<
τ
M
(
A
0
)
<
∞
. Then, for every solution f
≢
0 of (7) and for any entire function φ
(
z
)
≢
0 satisfying ρ
(
φ
)
<
∞, we have
(26)
λ
¯
(
f
(
j
)
-
φ
)
=
λ
(
f
(
j
)
-
φ
)
=
ρ
(
f
)
=
∞
(
j
∈
ℕ
)
,
λ
¯
2
(
f
(
j
)
-
φ
)
=
λ
2
(
f
(
j
)
-
φ
)
=
ρ
(
A
0
)
(
j
∈
ℕ
)
.Lemma 18 (see [10]).
Letf be a meromorphic function. If there exists an integer k
≥
1, such that ρ
(
f
(
k
)
/
f
)
=
ρ
(
f
) and ρ
(
f
)
>
ρ
2
(
f
), then
(27)
max
⁡
{
λ
¯
(
f
)
,
λ
¯
(
1
f
)
}
=
max
⁡
{
λ
(
f
)
,
λ
(
1
f
)
}
=
ρ
(
f
)
.
Furthermore, if f is entire function, then
(28)
λ
¯
(
f
)
=
λ
(
f
)
=
ρ
(
f
)
.By the principle of mathematical induction we can obtain easily the following lemma.Lemma 19.
Suppose thatf
′′
=
-
A
(
z
)
f
′
-
B
(
z
)
f, where A
(
z
) and B
(
z
) are entire functions. Then, for an integer j
≥
2, there exist entire functions α
j and β
j such that
(29)
f
(
j
)
=
α
j
f
′
+
β
j
f
.Lemma 20 (see [18]).
Letf be a solution of (7) where the coefficients A
j
(
z
)
(
j
=
0
,
…
,
k
-
1
) are analytic functions in the disc Δ
R
=
{
z
∈
ℂ
:
|
z
|
<
R
}, 0
<
R
≤
∞. Let n
c
∈
{
1
,
…
,
k
} be the number of nonzero coefficients A
j
(
z
)
(
j
=
0
,
…
,
k
-
1
), and let θ
∈
[
0,2
π
[ and ɛ
>
0. If z
θ
=
ν
e
i
θ
∈
Δ
R such that A
j
(
z
θ
)
≠
0 for some j
=
0
,
…
,
k
-
1, then, for all ν
<
r
<
R,
(30)
|
f
(
r
e
i
θ
)
|
≤
C
exp
⁡
(
n
c
∫
ν
r
max
⁡
j
=
0
,
…
,
k
-
1
|
A
j
(
t
e
i
θ
)
|
1
/
(
k
-
j
)
d
t
)
,
where C
>
0 is a constant satisfying
(31)
C
≤
(
1
+
ɛ
)
max
⁡
j
=
0
,
…
,
k
-
1
(
|
f
(
j
)
(
z
θ
)
|
(
n
c
)
j
max
⁡
n
=
0
,
…
,
k
-
1
|
A
n
(
z
θ
)
|
j
/
(
k
-
n
)
)
.Here, we give a special case of the result given by Cao et al. in [19].Lemma 21.
Letf be a meromorphic function with finite order 0
<
ρ
(
f
)
<
∞ and type 0
<
τ
(
f
)
<
∞. Then for any given β
<
τ
(
f
) there exists a subset I of [
1
,
+
∞
) that has infinite logarithmic measure such that T
(
r
,
f
)
>
β
r
ρ
(
f
) holds for all r
∈
I.Lemma 22.
LetA
(
z
) and B
(
z
) be entire functions such that ρ
(
B
)
=
ρ
(
0
<
ρ
<
∞
), τ
(
B
)
=
τ
(
0
<
τ
<
∞
), and let ρ
(
A
)
<
ρ
(
B
) and τ
(
A
)
<
τ
(
B
) if ρ
(
A
)
=
ρ
(
B
). If f
≢
0 is a solution of
(32)
f
′′
+
A
(
z
)
f
′
+
B
(
z
)
f
=
0
,
then ρ
(
f
)
=
∞ and ρ
2
(
f
)
=
ρ
(
B
).Proof.
Ifρ
(
A
)
<
ρ
(
B
), then, by Lemma 15, we obtain the result. We prove only the case when ρ
(
A
)
=
ρ
(
B
)
=
ρ and τ
(
A
)
<
τ
(
B
). Since f
≢
0, then, by (32), we have
(33)
B
=
-
(
f
′′
f
+
A
f
′
f
)
.
Suppose that f is of finite order; then by (33) and Lemma 14 we obtain
(34)
T
(
r
,
B
)
≤
T
(
r
,
A
)
+
O
(
log
⁡
r
)
which implies the contradiction
(35)
τ
(
B
)
≤
τ
(
A
)
.
Hence ρ
(
f
)
=
∞. By using inequality (30) for R
=
∞, we have
(36)
ρ
2
(
f
)
≤
max
⁡
{
ρ
(
A
)
,
ρ
(
B
)
}
=
ρ
(
B
)
.
On the other hand, since ρ
(
f
)
=
∞, then, by (33) and Lemma 14,
(37)
T
(
r
,
B
)
≤
T
(
r
,
A
)
+
S
(
r
,
f
)
,
where S
(
r
,
f
)
=
O
(
log
⁡
T
(
r
,
f
)
+
log
⁡
r
), possibly outside a set E
⊂
(
0
,
+
∞
) with a finite linear measure. By τ
(
A
)
<
τ
(
B
) we choose α
0, α
1 satisfying τ
(
A
)
<
α
1
<
α
0
<
τ
(
B
) such that, for r
→
+
∞, we have
(38)
T
(
r
,
A
)
≤
α
1
r
ρ
.
By Lemma 21, there exists a subset E
1
⊂
[
1
,
+
∞
) of infinite logarithmic measure such that
(39)
T
(
r
,
B
)
>
α
0
r
ρ
.
By (37)–(39) we obtain, for all r
∈
E
1
∖
E,
(40)
(
α
0
-
α
1
)
r
ρ
≤
O
(
log
⁡
T
(
r
,
f
)
)
+
O
(
log
⁡
r
)
which implies
(41)
ρ
(
B
)
=
ρ
≤
ρ
2
(
f
)
.
By using (36) and (41), we obtain ρ
2
(
f
)
=
ρ
(
B
).Remark 23.
Lemma22 was obtained by Tu and Yi in [16] for higher order linear differential equations by using the type τ
M (Lemma 16 in the present paper).Lemma 24 (see [20]).
Letf and g be meromorphic functions such that 0
<
ρ
(
f
), ρ
(
g
)
<
∞ and 0
<
τ
(
f
), τ
(
g
)
<
∞. Then we have the following.(i)
Ifρ
(
f
)
>
ρ
(
g
), then we obtain
(42)
τ
(
f
+
g
)
=
τ
(
f
g
)
=
τ
(
f
)
.
(ii)
Ifρ
(
f
)
=
ρ
(
g
) and τ
(
f
)
≠
τ
(
g
), then we get
(43)
ρ
(
f
+
g
)
=
ρ
(
f
g
)
=
ρ
(
f
)
=
ρ
(
g
)
.


## 3. Proof of Theorems and Corollaries

Proof of Theorem5.
By the conditions of Theorem5 and Lemma 16, every nontrivial solution f of (7) is of infinite order and ρ
2
(
f
)
=
ρ
(
A
0
). We need to prove that every solution f attains every complex value without exception; that is, for all a
∈
ℂ, the equation f
(
z
)
=
a has infinitely many solutions. We have two cases(i)
Ifa
≠
0, then, by Lemma 17, we have λ
(
f
-
a
)
=
+
∞ and the equation f
(
z
)
=
a has infinitely many solutions.
(ii)
Ifa
=
0, then we suppose that f has a finite number of zeros. By Theorem A
(44)
ρ
(
A
0
)
<
ρ
(
f
′
f
)
=
ρ
2
(
f
)
=
ρ
(
A
0
) which is a contradiction, and hence f assume every finite complex value infinitely often. Now, if ρ
(
f
′
/
f
)
=
ρ
(
f
)
=
∞ and since ρ
2
(
f
)
=
ρ
(
A
0
)
<
ρ
(
f
)
=
∞, then by using Lemma 18 we have
(45)
λ
¯
(
f
)
=
λ
(
f
)
=
ρ
(
f
)
=
∞
.Proof of Corollary6.
By using Lemma17, we have for any finite order entire function φ
(
z
)
≢
0
(46)
λ
¯
(
f
(
j
)
-
φ
)
=
λ
(
f
(
j
)
-
φ
)
=
ρ
(
f
)
=
∞
(
j
∈
ℕ
)
.
Now, if φ
(
z
)
≡
0, then we put g
(
z
)
=
f
(
j
)
(
z
)
(
j
∈
ℕ
). It follows that
(47)
ρ
(
f
(
j
+
1
)
f
(
j
)
)
=
ρ
(
g
′
g
)
=
ρ
(
g
)
=
∞
,
and, since ρ
2
(
g
)
=
ρ
2
(
f
)
=
ρ
(
A
0
)
<
ρ
(
g
)
=
∞, then, by applying Lemma 18, we obtain
(48)
λ
¯
(
g
)
=
λ
(
g
)
=
∞
;
that is, λ
¯
(
f
(
j
)
)
=
λ
(
f
(
j
)
)
=
∞
(
j
∈
ℕ
) and the proof of Corollary 6 is completed.Proof of Theorem9.
By using Lemma19, we have
(49)
g
f
=
P
n
[
f
]
Q
m
[
f
]
=
∑
i
=
0
n
d
i
f
(
i
)
∑
j
=
0
m
b
j
f
(
j
)
=
d
0
f
+
d
1
f
′
+
∑
i
=
2
n
d
i
(
α
i
f
′
+
β
i
f
)
b
0
f
+
b
1
f
′
+
∑
j
=
2
m
b
j
(
α
j
f
′
+
β
j
f
)
,
which we can write as
(50)
g
f
=
P
(
z
)
f
′
/
f
+
Q
(
z
)
R
(
z
)
f
′
/
f
+
S
(
z
)
,
where
(51)
P
=
d
1
+
∑
i
=
2
n
d
i
α
i
,
Q
=
d
0
+
∑
i
=
2
n
d
i
β
i
,
R
=
b
1
+
∑
j
=
2
m
b
j
α
j
,
S
=
b
0
+
∑
j
=
2
m
b
j
β
j
.
It is clear that
(52)
max
⁡
{
ρ
(
P
)
,
ρ
(
Q
)
,
ρ
(
R
)
,
ρ
(
S
)
}
<
ρ
(
f
′
f
)
and, since g
f is irreducible function in f
′
/
f, then P
S
-
Q
R
≢
0. So
(53)
ρ
(
g
f
)
≤
max
⁡
{
ρ
(
P
)
,
ρ
(
Q
)
,
ρ
(
R
)
,
ρ
(
S
)
,
ρ
(
f
′
f
)
}
=
ρ
(
f
′
f
)
.
On the other hand, we have
(54)
f
′
f
=
-
S
(
z
)
g
f
+
Q
(
z
)
R
(
z
)
g
f
-
P
(
z
)
which implies
(55)
ρ
(
f
′
f
)
≤
max
⁡
{
ρ
(
P
)
,
ρ
(
Q
)
,
ρ
(
R
)
,
ρ
(
S
)
,
ρ
(
g
f
)
}
=
ρ
(
g
f
)
.
By (53) and (55), we obtain ρ
(
g
f
)
=
ρ
(
f
′
/
f
).Proof of Corollary12.
Under the conditions of Corollary12 and Lemma 22, we have ρ
(
f
)
=
∞ and ρ
2
(
f
)
=
ρ
(
B
). Set
(56)
g
f
=
d
2
f
′′
+
d
1
f
′
+
d
0
f
b
2
f
′′
+
b
1
f
′
+
b
0
f
.
Substituting f
′′
=
-
A
(
z
)
f
′
-
B
(
z
)
f into g
f, we obtain
(57)
g
f
=
d
2
f
′′
+
d
1
f
′
+
d
0
f
b
2
f
′′
+
b
1
f
′
+
b
0
f
=
(
d
1
-
d
2
A
)
f
′
+
(
d
0
-
d
2
B
)
f
(
b
1
-
b
2
A
)
f
′
+
(
b
0
-
b
2
B
)
f
=
(
d
1
-
d
2
A
)
w
+
(
d
0
-
d
2
B
)
(
b
1
-
b
2
A
)
w
+
(
b
0
-
b
2
B
)
,
where w
=
f
′
/
f. In order to prove that g
f is irreducible function in w, we need only to prove that h
≢
0, where
(58)
h
=
(
d
1
-
d
2
A
)
(
b
0
-
b
2
B
)
-
(
d
0
-
d
2
B
)
(
b
1
-
b
2
A
)
=
b
0
d
1
-
b
1
d
0
+
A
(
b
2
d
0
-
b
0
d
2
)
+
B
(
b
1
d
2
-
b
2
d
1
)
.
Since ρ
(
A
)
<
ρ
(
B
) and 0
<
τ
(
A
)
<
τ
(
B
)
<
∞, if ρ
(
B
)
=
ρ
(
A
)
>
0, d
2
(
z
)
b
1
(
z
)
-
d
1
(
z
)
b
2
(
z
)
≢
0, then, by Lemma 24, we have ρ
(
h
)
=
ρ
(
B
)
>
0, so h
≢
0. By applying Theorem 9 we obtain ρ
(
f
′
/
f
)
=
ρ
(
g
f
).Proof of Corollary13.
Suppose thatf is a nontrivial solution of (11) such that ρ
(
f
′
/
f
)
=
∞. Then, by Lemma 22, we have
(59)
ρ
(
f
)
=
+
∞
,
ρ
2
(
f
)
=
ρ
(
B
)
.
By setting d
2
=
b
1
≡
1 and d
1
=
d
0
=
b
2
=
b
0
≡
0 and using Corollary 12 we have
(60)
ρ
(
f
′
f
)
=
ρ
(
f
′′
f
′
)
=
∞
which implies by using Lemma 18 that
(61)
λ
¯
(
f
)
=
λ
¯
(
f
′
)
=
∞
.



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*Source: 342974-2014-03-04.xml*