Research Article Fixed Points of Meromorphic Solutions for Some Difference Equations Zong-Xuan Chen
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Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 496096, 7 pages
http://dx.doi.org/10.1155/2013/496096
Research Article
Fixed Points of Meromorphic Solutions for
Some Difference Equations
Zong-Xuan Chen1 and Kwang Ho Shon2
1
2
School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China
Department of Mathematics, College of Natural Sciences, Pusan National University, Pusan 609-735, Republic of Korea
Correspondence should be addressed to Kwang Ho Shon; khshon@pusan.ac.kr
Received 1 December 2012; Accepted 17 April 2013
Academic Editor: Patricia J. Y. Wong
Copyright © 2013 Z.-X. Chen and K. H. Shon. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
We investigate fixed points of meromorphic solutions π¦(π§) for the Pielou logistic equation and obtain some estimates of exponents
of convergence of fixed points of π¦(π§) and its shifts π¦(π§+π), differences Δπ¦(π§) = π¦(π§+1)−π¦(π§), and divided differences Δπ¦(π§)/π¦(π§).
1. Introduction and Results
In this paper, we assume the reader is familiar with basic
notions of Nevanlinna’s value distribution theory (see [1–3]).
In addition, we use the notation π(π) to denote the order of
growth of a meromorphic function π and π(π) and π(1/π) to
denote, respectively, the exponents of convergence of zeros
and poles of π. We also use the notation π(π) to denote the
exponent of convergence of fixed points of π that is defined
as
log π (π, 1/ (π (π§) − π§))
.
π→∞
log π
π (π) = lim
(1)
Recently, a number of papers (including [4–17]) focus
on complex difference equations and difference analogues of
Nevanlinna’s theory.
The Pielou logistic equation
π¦ (π§ + 1) =
π
(π§) π¦ (π§)
,
π (π§) + π (π§) π¦ (π§)
(2)
where π(π§), π(π§), and π
(π§) are nonzero polynomials, is
an important difference equation, because it is obtained by
transform from the well-known Verhulst Pearl equation (see
[18, page 99])
π₯σΈ (π‘) = π₯ (π‘) [π − ππ₯ (π‘)]
(π, π > 0) ,
(3)
which is the most popular continuous model of growth of a
population.
Chen [7] obtained the following theorem.
Theorem A. Let π(π§), π(π§), and π
(π§) be polynomials with
π(π§)π(π§)π
(π§) ≡ΜΈ 0 and let π¦(π§) be a finite order transcendental meromorphic solution of (2). Then
1
π ( ) = π (π¦) ≥ 1.
π¦
(4)
Example 1. The function π¦(π§) = π§2π§ /(2π§ − 1) satisfies the
Pielou logistic equation
π¦ (π§ + 1) =
2 (π§ + 1) π¦ (π§)
,
π§ + π¦ (π§)
(5)
where π¦(π§) satisfies
(π¦) = 0,
1
π ( ) = π (π¦) = 1.
π¦
(6)
This example shows that the result of Theorem A is sharp.
One of the main purposes in this paper is to study
fixed points of meromorphic solutions of the Pielou logistic
equation (2).
The problem of fixed points of meromorphic functions is
an important one in the theory of meromorphic functions.
2
Abstract and Applied Analysis
Many papers and books (including [18–20]) investigate fixed
points of meromorphic functions.
Now we consider fixed points of meromorphic functions
and their shifts, differences, and divided differences. We see
that there are many examples to show that either π(π§) may
have no fixed point, for example, π1 (π§) = ππ§ + π§ or the shift
π(π§ + π) of π(π§), or the difference Δ π π(π§) = π(π§ + π) − π(π§) of
π(π§) may have only finitely many fixed points; for example,
for the function π2 (π§) = ππ§ + π§ − 1, its shift π2 (π§ + 1) = πππ§ + π§,
and its difference Δ 2ππ π2 (π§) = π2 (π§ + 2ππ) − π2 (π§) = 2ππ have
only finitely many fixed points. Even if for a meromorphic
function of small growth, Chen and Shon show that there
exists a meromorphic function π0 such that π(π0 ) < 1 and
Δ π π0 (π§) = π0 (π§+π)−π0 (π§) has only finitely many fixed points
(see Theorem 6 of [9]).
A divided difference (π(π§ + π) − π(π§))/π(π§) may also
have only finitely many fixed points; for example, the function
π(π§) = π§ππ§ satisfies that its divided difference (π(π§ + 1) −
π(π§))/π(π§) = ((π§ + 1)π − π§)/π§ has only finitely many fixed
points. Chen and Shon obtained Theorem B.
Theorem B (see [9]). Let π ∈ C \ {0} be a constant and let π
be a transcendental meromorphic function of order of growth
π(π) = π < 1 or of the form π(π§) = β(π§)πππ§ , where π =ΜΈ 0 is
a constant and β(π§) is a transcendental meromorphic function
with π(β) < 1. Suppose that π(π§) is a nonconstant polynomial.
Then
πΊ (π§) =
π (π§ + π) − π (π§)
− π (π§)
π (π§)
(7)
From Theorem B, we easily see that under conditions
of Theorem B, the divided difference πΊ1 (π§) = (π(π§ + π) −
π(π§))/π(π§) has infinitely many fixed points. The previous
example π(π§) = π§ππ§ shows that result of Theorem B is sharp.
However, we discover that the properties on fixed points
of meromorphic solutions of (2) are very good. We prove the
following theorem.
Theorem 2. Let π(π§), π(π§), and π
(π§) be nonzero polynomials such that
deg π (π§) ≥ 1.
(8)
Set Δπ¦(π§) = π¦(π§ + 1) − π¦(π§). Then every finite order
transcendental meromorphic solution π¦(π§) of (2) satisfies the
following:
(i) π(π¦(π§ + π)) = π(π¦(π§)) ≥ 1 (π = 0, 1, . . .);
(ii) if π
(π§) − (π§ + 1)π(π§) ≡ΜΈ 0, then π(Δπ¦(π§)/π¦(π§)) =
π(π¦(π§));
(iii) if there is a polynomial β(π§) satisfying
2
2
(−π
(π§) + π (π§) + π§π (π§)) − 4π§π (π§) π (π§) = β(π§) ,
then π(Δπ¦(π§)) = π(π¦(π§)).
π (π1 (π§)) = 0 =ΜΈ π (π1 (π§ + 1)) = 1.
(10)
2. Proof of Theorem 2
We need the following lemmas for the proof of Theorem 2.
Lemma 4 (see [12, 17]). Let π€(π§) be a nonconstant finite order
meromorphic solution of
π (π§, π€) = 0,
(11)
where π(π§, π€) is a difference polynomial in π€(π§). If π(π§, π) ≡ΜΈ 0
for a meromorphic function π(π§) satisfying π(π, π) = π(π, π€),
then
π (π,
1
) = π (π, π€)
π€−π
(12)
holds for all π outside of a possible exceptional set with finite
logarithmic measure.
Remark 5. Using the same method as in the proof of
Lemma 4 (see [12]), we can prove that in Lemma 4, if all
coefficients ππ (π§) of π(π§, π€) satisfy π(ππ (π§)) = π1 < π(π€(π§)) =
π and if π(π§, π) ≡ΜΈ 0 for a meromorphic function π(π§)
satisfying π(π, π) = π(π, π€), then for a given π (0 < π < π−π1 ),
π (π,
has infinitely many zeros.
deg π (π§) ≥ max {deg π
(π§) , deg π (π§)} ,
Remark 3. Generally, π(π(π§)) =ΜΈ π(π(π§+π)) for a meromorphic
function π(π§) of finite order. For example, the function
π1 (π§) = ππ§ + π§ satisfies
1
) = π (π, π€ (π§)) + π (ππ1 +π )
π€ (π§) − π (π§)
(13)
holds for all π outside of a possible exceptional set with finite
logarithmic measure.
Lemma 6. Suppose that π
(π§), π(π§), and π(π§) satisfy the
condition (8) in Theorem 2 and that π¦(π§) is a nonconstant
meromorphic function. Then
π1 (π§) = (π
(π§) − π§π (π§)) π¦ (π§) − π§π (π§) ,
π2 (π§) = π (π§) π¦ (π§) + π (π§)
(14)
have at most finitely many common zeros.
Proof. Suppose that π§0 is a common zero of π1 (π§) and π2 (π§).
Then π2 (π§0 ) = π(π§0 )π¦(π§0 ) + π(π§0 ) = 0. Thus, π¦(π§0 ) =
−π(π§0 )/π(π§0 ). Substituting π¦(π§0 ) = −π(π§0 )/π(π§0 ) into π1 (π§),
we obtain
π1 (π§0 ) = −
=−
π (π§0 )
(π
(π§0 ) − π§0 π (π§0 )) − π§0 π (π§0 )
π (π§0 )
π
(π§0 ) π (π§0 )
π (π§0 )
(15)
= 0.
(9)
Since π
(π§)π(π§)/π(π§) has only finitely many zeros, we see that
π1 (π§) and π2 (π§) have at most finitely many common zeros.
Abstract and Applied Analysis
3
Lemma 7 (see [14]). Let π(π§) be a nonconstant finite order
meromorphic function. Then
π (π + 1, π (π§)) = π (π, π (π§)) + π (π, π (π§)) .
(16)
Gol’dberg and Ostrovskii [21, page 66] give that for any
constant π,
(1 + π (1)) π (π − |π| , π (π§))
≤ π (π, π (π§ + π)) ≤ (1 + π (1)) π (π + |π| , π (π§)) .
(17)
for all π outside of a possible exceptional set with finite
logarithmic measure. Thus,
π (π,
for all π outside of a possible exceptional set with finite
logarithmic measure. So, by Theorem A and (24), we obtain
π(π¦(π§)) = π(π¦(π§)) ≥ 1.
Now suppose that π = 1. By (2), we obtain
Lemma 8. Let π(π§) be a nonconstant finite order meromorphic function. Then
=
π (π, π (π§ + 1)) = π (π, π (π§)) + π (π, π (π§)) .
=
(18)
Using the same method as in the proof of Lemma 6, we
can prove Lemmas 9 and 10.
Lemma 9. Suppose that π
(π§), π(π§), and π(π§) satisfy the
condition (8) in Theorem 2 and that π¦(π§) is a nonconstant
meromorphic function. Then
π2 (π§) = π (π§) π¦ (π§) + π (π§)
(19)
(π
(π§) − π§π (π§)) π¦ (π§) − π§π (π§)
π (π§) + π (π§) π¦ (π§)
(π
(π§) − π§π (π§)) [π¦ (π§) − π§π (π§) / (π
(π§) − π§π (π§))]
.
π (π§) + π (π§) π¦ (π§)
(25)
By (8), we see that π
(π§) − π§π(π§) ≡ΜΈ 0. Since π(π§), π(π§), and
π
(π§) are polynomials, by (25), we see that π¦(π§)−π§π(π§)/(π
(π§)−
π§π(π§)) and π(π§) + π(π§)π¦(π§) have the same poles, except
possibly finitely many poles. By Lemma 6, we see that (π
(π§) −
π§π(π§))π¦(π§) − π§π(π§) and π(π§) + π(π§)π¦(π§) have at most finitely
many common zeros. Hence, by (25), we have that
π (π¦ (π§ + 1)) = π (π¦ (π§ + 1) − π§)
= π (π¦ (π§) −
have at most finitely many common zeros.
Lemma 10. Suppose that π
(π§), π(π§), and π(π§) satisfy the
condition (8) in Theorem 2 and π¦(π§) is a nonconstant meromorphic function. Then
π1 (π§) = π (π§) π¦(π§)2 + [−π
(π§) + π (π§) + π§π (π§)] π¦ (π§)
+ π§π (π§) ,
(20)
π2 (π§) = π (π§) π¦ (π§) + π (π§)
have at most finitely many common zeros.
Proof of Theorem 2. (i) We prove that π(π¦(π§+π)) = π(π¦(π§)) ≥
1 (π = 0, 1, . . .). Suppose that π = 0. Set π¦(π§) − π§ = π(π§). So,
π(π§) is transcendental, π(π, π(π§)) = π(π, π¦(π§)) + π(log π), and
π(π, π) = π(π, π¦). Substituting π¦(π§) = π(π§) + π§ into (2), we
obtain
(26)
where β(π§) is a polynomial with deg β(π§) ≤ π(π¦(π§)), π0 (π§)
and π»0 (π§) are canonical products (π0 (π§) may be a polynomial) formed by nonzero zeros and poles of π¦(π§) −
π§π(π§)/(π
(π§)−π§π(π§)), respectively, and π is an integer; if π ≥ 0,
then π(π§) = π§π π0 (π§), π»(π§) = π»0 (π§)π−β(π§) ; if π < 0, then
π(π§) = π0 (π§), π»(π§) = π§−π π»0 (π§)π−β(π§) . Combining Theorem A
with properties of canonical product, we see that
π (π (π§)) = π (π (π§))
= π (π¦ (π§) −
(21)
− π
(π§) [π (π§) + π§] = 0.
π§π (π§)
).
π
(π§) − π§π (π§)
Suppose that π(π¦(π§)−π§π(π§)/(π
(π§)−π§π(π§))) < π(π¦(π§)). Thus,
π¦(π§) − π§π(π§)/(π
(π§) − π§π(π§)) can be rewritten as the following
form:
π (π§) β(π§)
π (π§)
π§π (π§)
= π§π 0
,
π
=
π¦ (π§) −
(27)
π
(π§) − π§π (π§)
π»0 (π§)
π» (π§)
πΎ0 (π§, π) := π (π§) [π (π§ + 1) + π§ + 1] [π (π§) + π§]
+ π (π§) [π (π§ + 1) + π§ + 1]
(24)
π¦ (π§ + 1) − π§
This and Lemma 7 give the following lemma.
π1 (π§) = π
(π§) − (π§ + 1) π (π§) − (π§ + 1) π (π§) π¦ (π§) ,
1
) = π (π, π¦ (π§)) + π (π, π¦ (π§))
π¦ (π§) − π§
π§π (π§)
)
π
(π§) − π§π (π§)
(28)
< π (π¦ (π§)) ,
π (π» (π§)) = π (π» (π§)) = π (π¦ (π§)) .
Thus,
By (27), we obtain
πΎ0 (π§, 0) = π§ (π§ + 1) π (π§) + (π§ + 1) π (π§) − π§π
(π§) .
(22)
By (8) and (22), we see that πΎ0 (π§, 0) ≡ΜΈ 0. Thus, by Lemma 4
and πΎ0 (π§, 0) ≡ΜΈ 0, we obtain
π (π,
1
) = π (π, π (π§)) + π (π, π (π§))
π (π§)
(23)
π¦ (π§) =
π§π (π§)
+ π (π§) π (π§) ,
π
(π§) − π§π (π§)
π¦ (π§ + 1) =
(π§ + 1) π (π§ + 1)
π
(π§ + 1) − (π§ + 1) π (π§ + 1)
+ π (π§ + 1) π (π§ + 1) ,
(29)
4
Abstract and Applied Analysis
where π(π§) = 1/π»(π§). Thus, by (28) and Lemma 8, we have
that
π (π (π§)) = π (π» (π§)) = π (π¦ (π§)) ,
π (π (π§ + 1)) = π (π (π§)) < π (π (π§)) .
(30)
(π§ + 1) π (π§ + 1)
+ π (π§ + 1) π (π§ + 1)}
π
(π§ + 1) − (π§ + 1) π (π§ + 1)
π (π¦ (π§ + 2)) = π (π (π§ + 1)) = π (π (π§)) = π (π¦ (π§)) . (39)
Continuing to use the same method as above, we can obtain
π§π (π§)
+ π (π§) π (π§)]}
π
(π§) − π§π (π§)
π§π (π§)
+ π (π§) π (π§)] = 0.
− π
(π§) [
π
(π§) − π§π (π§)
π (π¦ (π§ + π)) = π (π¦ (π§))
(31)
(π = 1, 2, . . .) .
(40)
(ii) Suppose that π
(π§) − (π§ + 1)π(π§) ≡ΜΈ 0. We prove that
π(Δπ¦(π§)/π¦(π§)) = π(π¦(π§)). By (2), we obtain
Δπ¦ (π§)
−π§
π¦ (π§)
By (31), we obtain
πΎ1 (π§, 0)
= [(π§ + 1) π (π§ + 1) π (π§) π
(π§)
−π§π (π§) π
(π§) [π
(π§ + 1) − (π§ + 1) π (π§ + 1)]]
By (8), we see that in the numerator of the right side of (32),
there exists only one term π§(π§ + 1)π(π§)π
(π§)π(π§ + 1) being of
the highest degree. So,
πΎ1 (π§, 0) ≡ΜΈ 0.
1
) = π (π, π (π§)) + π (π, π (π§))
π (π§)
+ π (π
(34)
1
) = 0.
π (π§)
(35)
Thus, by this and (28), we see that (34) is a contradiction.
Hence, π(π¦(π§) − π§π(π§)/(π
(π§) − π§π(π§))) = π(π¦(π§)). By (26),
we obtain
π (π¦ (π§ + 1)) = π (π¦ (π§)) .
(36)
Now suppose that π = 2. By (2), we obtain
π (π§ + 1) =
=
π
(π§) − (π§ + 1) π (π§) − (π§ + 1) π (π§) π¦ (π§)
π (π§) + π (π§) π¦ (π§)
π
(π§ + 1) π (π§)
,
π (π§ + 1) + π (π§ + 1) π (π§)
× (π¦ (π§) − (π
(π§) − (π§ + 1) π (π§)) / (π§ + 1) π (π§))
−1
× (π(π§) + π(π§)π¦(π§)) .
Since π¦(π§) − (π
(π§) − (π§ + 1)π(π§))/(π§ + 1)π(π§) and π(π§) +
π(π§)π¦(π§) have the same poles, except possibly finitely many
poles, by Lemma 9 and (41), we only need to prove that
π (π¦ (π§) −
π
(π§) − (π§ + 1) π (π§)
) = π (π¦ (π§)) .
(π§ + 1) π (π§)
(42)
Set
β1 (π§) = π
(π§) − (π§ + 1) π (π§) ,
β2 (π§) = (π§ + 1) π (π§) .
(43)
Suppose that π(π¦(π§) − β1 (π§)/β2 (π§)) < π(π¦(π§)). Using the
same method as in the proof of (i), π¦(π§) − β1 (π§)/β2 (π§) can
be rewritten as the following form:
π¦ (π§) =
β1 (π§)
+ π2 (π§) π2 (π§) ,
β2 (π§)
(44)
where π2 (π§) = 1/π»2 (π§) and π2 (π§) and π»2 (π§) are nonzero
entire functions, such that
π (π2 (π§)) = π (π2 (π§)) < π (π¦ (π§)) ,
(37)
(41)
= − (π§ + 1) π (π§)
)
holds for all π outside of a possible exceptional set with finite
logarithmic measure.
On the other hand, by π(π§) = 1/π»(π§) and the fact that
π»(π§) is an entire function, we see that
π (π,
π¦ (π§ + 1) − π¦ (π§)
−π§
π¦ (π§)
(33)
Thus, by (28), (33), Lemma 4, and its Remark 5, we obtain that
for any given π (0 < π < π(π¦(π§)) − π(π(π§)))
π(π(π§))+π
=
(32)
× ([π
(π§ + 1) − (π§ + 1) π (π§ + 1)] [π
(π§) − π§π(π§)])−1 .
π (π,
(38)
Thus, for (37), applying the conclusion of π = 1 above, we
obtain
πΎ1 (π§, π)
⋅ {π (π§) + π (π§) [
deg π (π§ + 1) ≥ max {deg π
(π§ + 1) , deg π (π§ + 1)} ,
deg π (π§ + 1) ≥ 1.
Substituting (29) into (2), we obtain
:= {
where π(π§) = π¦(π§ + 1). By Lemma 8, we have that π(π(π§)) =
π(π¦(π§)). By (8), we have
π (π»2 (π§)) = π (π»2 (π§)) = π (π¦ (π§)) .
(45)
Abstract and Applied Analysis
5
Since π(π§), π(π§), and π
(π§) are polynomials, we see that poles
of π(π§) + π(π§)π¦(π§) must be poles of π(π§)π¦(π§)2 + [−π
(π§) +
π(π§) + π§π(π§)]π¦(π§) + π§π(π§). Thus, poles of π(π§) + π(π§)π¦(π§)
are not zeros of π¦(π§ + 1) − π¦(π§) − π§. By Lemma 10, we see that
the numerator and the denominator of the right side of (51)
have at most finitely many common zeros. Thus, in order to
prove π(Δπ¦(π§)) = π(π¦(π§)), by (52), we only need to prove that
Substituting (44) into (2), we obtain
π·2 (π§, π2 (π§))
:= {
β1 (π§ + 1)
+ π2 (π§ + 1) π2 (π§ + 1)}
β2 (π§ + 1)
× {π (π§) + π (π§)
− π
(π§) {
β1 (π§)
+ π (π§) π2 (π§) π2 (π§)}
β2 (π§)
β1 (π§)
+ π2 (π§) π2 (π§)} = 0,
β2 (π§)
π·2 (π§, 0)
π (π¦ (π§) −
(46)
β (π§)
β (π§)
β1 (π§ + 1)
{π (π§) + π (π§) 1
} − π
(π§) 1
β2 (π§ + 1)
β2 (π§)
β2 (π§)
=
− π
(π§)
= π
(π§)
or
π (π¦ (π§) −
β1 (π§ + 1)
π
(π§) − (π§ + 1) π (π§)
{π (π§) + π (π§)
}
β2 (π§ + 1)
(π§ + 1) π (π§)
=
β1 (π§)
β2 (π§)
− (π§π (π§) + π (π§) − π
(π§)) + β (π§) ,
Since β1 (π§) and β2 (π§) are polynomials, we see that
(48)
Using the same method as in the proof of (i), we obtain a
contradiction. Hence, (42) holds; that is, π(Δπ¦(π§)/π¦(π§)) =
π(π¦(π§)).
(iii) Suppose that there is a polynomial β(π§) satisfying
(−π
(π§) + π (π§) + π§π (π§))2 − 4π§π (π§) π (π§) = β(π§)2 .
(49)
Thus, by (8) and (49), we see that
deg β (π§) = deg π (π§) + 1.
(50)
Now we prove that π(Δπ¦(π§)) = π(π¦(π§)). By (2), we obtain
π¦ (π§ + 1) − π¦ (π§) − π§
=−
π (π§) π¦(π§)2 + [−π
(π§) + π (π§) + π§π (π§)] π¦ (π§) + π§π (π§)
.
π (π§) + π (π§) π¦ (π§)
(51)
By (49) and (51), we obtain
(55)
such that its degree is equal to deg π(π§) + 1. Without loss of
generality, we may suppose that
deg (− (π§π (π§) + π (π§) − π
(π§)) + β (π§)) = deg π (π§) + 1.
(56)
Now we prove that (53) holds. Suppose that
π (π¦ (π§) −
− (π§π (π§) + π (π§) − π
(π§)) + β (π§)
) < π (π¦ (π§)) .
2π (π§)
(57)
Using a similar method as in the proof of (i), we see that π¦(π§)−
(−(π§π(π§) + π(π§) − π
(π§)) + β(π§))/2π(π§) can be rewritten as the
following form:
π¦ (π§) =
− (π§π (π§) + π (π§) − π
(π§)) + β (π§)
2π (π§)
(58)
+ π3 (π§) π3 (π§) ,
π¦ (π§ + 1) − π¦ (π§) − π§
= −π (π§) {π¦ (π§) −
− (π§π (π§) + π (π§) − π
(π§)) − β (π§) ,
(47)
that is,
π·2 (π§, 0) ≡ΜΈ 0.
− (π§π (π§) + π (π§) − π
(π§)) − β (π§)
) = π (π¦ (π§)) .
2π (π§)
(54)
By (50), we have deg(−π§π(π§)) = deg β(π§) = deg π(π§) + 1.
Combining this with (8), we see that there exists at least one
of
β1 (π§ + 1) β2 (π§) − (π§ + 1) β1 (π§) β2 (π§ + 1)
.
(π§ + 1) β2 (π§ + 1) β2 (π§)
β1 (π§ + 1) β2 (π§) − (π§ + 1) β1 (π§) β2 (π§ + 1) ≡ΜΈ 0,
− (π§π (π§) + π (π§) − π
(π§)) + β (π§)
) = π (π¦ (π§))
2π (π§)
(53)
− (π§π (π§) + π (π§) − π
(π§)) + β (π§)
}
2π (π§)
− (π§π (π§) + π (π§) − π
(π§)) − β (π§)
}
× {π¦ (π§) −
2π (π§)
−1
× (π(π§) + π (π§) π¦ (π§)) .
(52)
where π3 (π§) = 1/π»3 (π§) and π3 (π§) and π»3 (π§) are nonzero
entire functions, such that
π (π3 (π§)) = π (π3 (π§)) < π (π¦ (π§)) ,
π (π»3 (π§)) = π (π»3 (π§)) = π (π¦ (π§)) .
(59)
6
Abstract and Applied Analysis
Substituting (58) into (2), we obtain
Acknowledgments
π·3 (π§, π3 (π§))
The first author was supported by the National Natural
Science Foundation of China (no. 11171119). The second
author was supported by Basic Science Research Program
through the National Research Foundation of Korea (NRF)
funded by the Ministry of Education, Science and Technology
(2010-0009646).
:= {
− ((π§ + 1) π (π§ + 1) + π (π§ + 1) − π
(π§ + 1)) + β (π§ + 1)
2π (π§ + 1)
+π3 (π§ + 1) π3 (π§ + 1) }
⋅ {π (π§) + π (π§)
− (π§π (π§) + π (π§) − π
(π§)) + β (π§)
2π (π§)
References
+π (π§) π3 (π§) π3 (π§) }
− π
(π§)
− (π§π (π§) + π (π§) − π
(π§)) + β (π§)
2π (π§)
− π
(π§) π3 (π§) π3 (π§) = 0,
π·3 (π§, 0)
:= {
− (π§ + 1) π (π§ + 1) − π (π§ + 1) + π
(π§ + 1) + β (π§ + 1)
}
2π (π§ + 1)
⋅{
−π§π (π§) + π (π§) + π
(π§) + β (π§)
}
2
− π
(π§)
=
−π§π (π§) − π (π§) + π
(π§) + β (π§)
2π (π§)
π1 (π§) − π2 (π§)
,
4π (π§ + 1) π (π§)
(60)
where
π1 (π§) = [− (π§ + 1) π (π§ + 1) − π (π§ + 1)
+π
(π§ + 1) + β (π§ + 1)]
× [−π§π (π§) + π (π§) + π
(π§) + β (π§)] π (π§) ,
π2 (π§) = 2 (−π§π (π§) − π (π§) + π
(π§) + β (π§)) π
(π§) π (π§ + 1) .
(61)
By (8), (50), and (56), we see that
deg (− (π§ + 1) π (π§ + 1) − π (π§ + 1) + π
(π§ + 1) + β (π§ + 1))
= deg π (π§) + 1,
deg (−π§π (π§) + π (π§) + π
(π§) + β (π§)) = deg π (π§) + 1.
(62)
Thus, we obtain
deg π1 (π§) = 3 deg π (π§) + 2,
deg π2 (π§) ≤ 3 deg π (π§) + 1.
(63)
So, by (60) and (63), we see that π·3 (π§, 0) ≡ΜΈ 0.
Using the same method as in the proof of (i), we see that
(53) holds.
Thus, Theorem 2 is proved.
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7
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