# Onq-Difference Riccati Equations and Second-Order Linear q-Difference Equations

Authors: Zhi-Bo Huang
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

We considerq-difference Riccati equations and second-order linear q-difference equations in the complex plane. We present some basic properties, such as the transformations between these two equations, the representations and the value distribution of meromorphic solutions of q-difference Riccati equations, and the q-Casorati determinant of meromorphic solutions of second-order linear q-difference equations. In particular, we find that the meromorphic solutions of these two equations are concerned with the q-Gamma function when q∈ℂ such that 0<|q|<1. Some examples are also listed to illustrate our results.


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## Body

## 1. Introduction and Main Results

In this paper, a meromorphic function means meromorphic in the whole complex planeℂ, unless stated otherwise. We also assume that the reader is familiar with the standard symbols and fundamental results such as m(r,f),N(r,f), and T(r,f), of Nevanlinna theory, see, for example, [1, 2], for a given meromorphic function f(z). A meromorphic function a(z) is said to be a small function relative to f(z) if T(r,a)=S(r,f), where S(r,f) is used to denote any quantity satisfying S(r,f)=o({T(r,f)} as r→∞, possibly outside of a set of finite logarithmic measure, furthermore, possibly outside of a set E of logarithmic density logdens(E)=limr→∞∫[1,r]∩E‍(dt/t)/logr=0. For a small function a(z) relative to f(z), we define
(1)δ(a,f)=lim_r→∞m(r,1/(f-a))T(r,f)=1-lim¯r→∞N(r,1/(f-a))T(r,f).Recently, Ishizaki [3] considered difference Riccati equation
(2)Δf(z)+f(z)2+A(z)f(z)-1=0,
and second-order linear difference equation
(3)Δ2y(z)+A(z)y(z)=0,
where A(z) is meromorphic function, and gave surveys of basic properties of (2) and (3), which are analogues in the differential cases.Now, we are concerned withq-difference Riccati equation
(4)g(qz)=-a1(z)g(z)+a0(z)g(z),
and second-order linear q-difference equation
(5)f(q2z)+a1(z)f(qz)+a0(z)f(z)=0,
where q∈ℂ∖{0}, |q|≠1, a1(z) and a0(z)≢0 are rational functions and will obtain some parallel results for q-difference case. For a meromorphic function h(z), the q-difference operator Δq is defined by Δqh(z)=h(qz)-h(z).This paper is organized as follows. In Section2, we describe the transformation between q-difference Riccati equation (4) and second-order linear q-difference equation (5). In Section 3, we present some properties of q-difference Riccati equation (4), such as q-difference analogue on the property of a cross ratio for four distinct meromorphic solutions of a differential Riccati equation, the meromorphic solutions concerning with q-Gamma function. In Section 4, we study the value distribution of transcendental meromorphic solutions of q-difference Riccati equation (4) and the form of meromorphic solutions of second-order linear q-difference equation (5). In Section 5, we discuss the properties on the q-Casorati determinant of meromorphic solutions of second-order linear q-difference equation (5).


## 2. Transformations betweenq-Difference Riccati Equations and Linear q-Difference Equations of Second-Order

It is well known that a differential Riccati equation(6)w′(z)+w(z)2+A(z)=0
and second-order linear differential equation
(7)u′′(z)+A(z)u(z)=0
are closely related by the transformation
(8)w(z)=-u′(z)u(z),
where A(z) is a meromorphic function, see, for example, [4, pages 103–106].Ishizaki [3] considered a difference analogue of (6) and (7) and obtained that difference Riccati equation (2) and second-order linear difference equation (3) are closely linked by the transformation
(9)f(z)=-Δy(z)y(z),
where A(z) is a meromorphic function.Here, we are concerned with a transformation between (4) and (5), see [5]. For a nontrivial meromorphic solution f(z) of (5), we take
(10)g(z)=f(qz)f(z).
Then g(z) satisfies q-difference Riccati equation (4). In fact, we deduce from (5) that
(11)f(q2z)f(qz)+a1(z)+a0(z)f(z)f(qz)=0,
which implies the desired form of (4).Conversely, if (4) admits a nontrivial meromorphic solution g(z), then meromorphic function f(z) of first-order q-difference equation (10) satisfies (5). In fact, we conclude from (4) and (10) that
(12)f(q2z)=g(qz)f(qz)=(-a1(z)g(z)+a0(z)g(z))f(qz)=-a1(z)f(qz)-a0(z)f(z),
which implies (5).Example 1.
Suppose thatq∈ℂ∖{0} and |q|≠1. Let a0(z)=(q2z2-(q2-2q-1)z+1)/(1-z2) and a1(z)=2/(z-1). Then g(z)=(qz+1)/(z+1) and f(z)=z+1 satisfy q-difference Riccati equation (4) and second-order linear q-difference equation (5), respectively, which both satisfy the transformation (10).


## 3. Representations of Solutions ofq-Difference Riccati Equations

The representations on meromorphic solutions of Riccati equations are interesting. Bank et al. [6, pages 371–373] obtained that differential Riccati equation (6) possesses a one parameter family of meromorphic solutions (fc)c∈ℂ if (6) has three distinct meromorphic solutions α1(z),α2(z), and α3(z). Ishizaki extended this property to difference Riccati equation (2) and obtained a difference analogue of this property, see [3, Proposition 2.1]. Now, we present this property for q-difference case below, which can also be seen as a q-difference analogue of the fact that a cross ratio for four distinct meromorphic solutions of a differential Riccati equation is a constant, see, for example, [4, pages 108-109]. Furthermore, we find that meromorphic solutions of q-difference Riccati equations (4) are concerned with q- Gamma function if q∈ℂ such that 0<|q|<1.Theorem 2.
Suppose that (4) possesses three distinct meromorphic solutions g1(z),g2(z), and g3(z). Then any meromorphic solution g(z) of (4) can be represented by
(13)g(z)=(g1(z)g2(z)-g2(z)g3(z)-g1(z)g2(z)ϕ(z)+g1(z)g3(z)ϕ(z))×(g1(z)-g3(z)-g2(z)ϕ(z)+g3(z)ϕ(z))-1,
where ϕ(z) is a meromorphic function satisfying ϕ(qz)=ϕ(z). Conversely, if for any meromorphic function ϕ(z) satisfying ϕ(qz)=ϕ(z), we define a function g(z) by (13), then g(z) is a meromorphic solution of (4).Proof of Theorem2.
Lethj(z),j=1,2,3,4 be distinct meromorphic functions. We denote a cross ratio of hj(z),j=1,2,3,4 by
(14)R(h1,h2,h3,h4;z)=h1(z)-h3(z)h1(z)-h4(z):h2(z)-h3(z)h2(z)-h4(z).
Suppose that g(z) is meromorphic solution of (4) and is also distinct from g1(z),g2(z), and g3(z). We first show that g(z) is a meromorphic solution of q-difference Riccati equation (4) if and only if R(qz)=R(z), where R(z)=R(g1,g2,g3,g;z). In fact, we conclude from (4) that
(15)R(qz)=g1(qz)-g3(qz)g1(qz)-g(qz):g2(qz)-g3(qz)g2(qz)-g(qz)=a0(z)(g1(z)-g3(z))/g1(z)g3(z)a0(z)(g1(z)-g(z))/g1(z)g(z):a0(z)(g2(z)-g3(z))/g2(z)g3(z)a0(z)(g2(z)-g(z))/g2(z)g(z)=g1(z)-g3(z)g1(z)-g(z):g2(z)-g3(z)g2(z)-g(z)=R(z).
Conversely, if R(qz)=R(z), then
(16)a0(z)(g1(z)-g3(z))/g1(z)g3(z)-(a1(z)g(z)+a0(z)/g(z))-g(qz):a0(z)(g2(z)-g3(z))/g2(z)g3(z)-(a2(z)g(z)+a0(z)/g(z))-g(qz)=g1(z)-g3(z)g1(z)-g(z):g2(z)-g3(z)g2(z)-g(z).
We conclude from (16) that g(qz)=-(a1(z)g(z)+a0(z))/g(z), which shows that g(z) satisfies (4).
Thus, for any meromorphic functionϕ(z) satisfying ϕ(qz)=ϕ(z), we define g(z) by
(17)R(g1,g2,g3,g;z)=ϕ(z).
Then g(z) is represented by (13), and also satisfies q-difference Riccati equation (4). The proof of Theorem 2 is completed.Now, we recall some results of transcendental meromorphic solutions concerned withq-difference Riccati equation (4). Bergweiler et al. [7, 8] pointed out that all transcendental meromorphic solutions of (5) satisfy T(r,f)=O((logr)2) if q∈ℂ and 0<|q|<1. Since (10) is a transformation between (4) and (5), we obtain that all transcendental meromorphic solutions of (4) are of order zero if q∈ℂ and 0<|q|<1. On the other hand, if g(z) is a transcendental meromorphic solution of
(18)g(qz)=R(z,g(z)),
where q∈ℂ,|q|>1 and the coefficients of R(z,g(z)) are small functions relative to g(z), Gundersen et al. [9] showed that the order of growth of (18) is equal to logdegg(R)/log|q|, where degg(R) is the degree of irreducible rational function R(z,g(z)) in g(z). Thus, from the above two cases, we obtain that all transcendental meromorphic solutions of (4) are of order zero for all q∈ℂ∖{0} and |q|≠1.We also illustrate some of the results onq-difference equations, which are explicitly solvable in terms of known zero-order meromorphic functions (see [5]). Let q∈ℂ be such that 0<|q|<1. Then q-Gamma function Γq(x) is defined by
(19)Γq(x):=(q;q)∞(qx;q)∞(1-q)1-x,
where (a;q)∞=∏k=0∞‍(1-aqk). It is a meromorphic function with poles at x=-n±2πik/logq, where k and n are nonnegative integers, see [10]. By defining
(20)γq(z):=(1-q)x-1Γq(x),z=qx,
and γq(0):=(q;q)∞, we see that γq(z) is a meromorphic function of zero-order with no zeros, having its poles at {qk}k=0∞.Therefore, the first-order linearq-difference equation
(21)h(qz)=(1-z)h(z)
is solved by the function γq(z). Moreover, for general first-order linear q-difference equation,
(22)h(qz)=a(z)h(z),
where a(z) is a rational function. If a(z)≡a is a constant, (22) is solvable in terms of rational functions if and only if logqa is an integer. If a(z) is nonconstant, let αi,i=1,2,…,n and βj,j=1,2,…,m be the zeros and poles of a(z), respectively, repeated according to their multiplicities. Then a(z) can be written in the form
(23)a(z)=c(1-z/α1)(1-z/α2)⋯(1-z/αn)(1-z/β1)(1-z/β2)⋯(1-z/βm),
where c≠0 is a complex number depending on a(z). So, (22) is solved by
(24)h(z)=zlogqcγq(z/α1)γq(z/α2)⋯γq(z/αn)γq(z/β1)γq(z/β2)⋯γq(z/βm),
which is meromorphic if and only if logqc is an integer.Now, letc1(z) and c2(z) be two distinct rational solutions of the differential Riccati equation (6). If there exists a rational solution c3(z) distinct from cj(z),j=1,2, then all meromorphic solutions of (6) are rational solutions. If there exists a transcendental meromorphic solution w(z), then there is no rational solution other than cj(z),j=1,2, see, for example, [6, pages 393-394]. For difference Riccati equation (2), Ishizaki obtained a difference analogue, see [3, Proposition 2.2]. In the following, we give a q-difference case for q-difference Riccati equation (4).Theorem 3.
Letq∈ℂ be such that 0<|q|<1. Suppose that q-difference Riccati equation (4) possesses two distinct rational solutions g1(z) and g2(z). Then there exists a meromorphic solution g3(z) distinct from g1(z) and g2(z) so that any meromorphic solution g(z) of (4) is represented in the form (13).Proof of Theorem3.
Sinceg1(z) and g2(z) are two distinct rational solutions of (4), we define a translation
(25)g(z)=g1(z)h(z)+g2(z)h(z)+1.
Then σ(h)=σ(g)=0. Substituting (25) into (4), we conclude that
(26)h(qz)=g1(z)g2(z)h(z),
which is type of (22). So, h(z) is a meromorphic solution of (26) as in the form (24). Therefore, we conclude from (25) that g3(z) is a meromorphic solution of (4), which is distinct from g1(z) and g2(z). So, we now deduce from Theorem 2 that any meromorphic solution of (4) is represented in the form (13). The proof of Theorem 3 is completed.Example 4.
Letq=-(1/2), a1(z)=(5z+4)/2(z+2) and a0(z)=(z-4)/(z+2) in (4) and (5). Then functions
(27)g1(z)=-2,g2(z)=-z-22(z+1)
satisfy q-difference Riccati equation (4), and (26) turns into
(28)h(-12z)=4(z+1)2(z-2)h(z)=-2[1-z/(-1)]1-z/2h(z).
We note that
(29)γq(z)=(1-q)x-1Γq(x)=(q;q)∞(qx;q)∞=(q;q)∞(z;q)∞.
Thus, we conclude from (24) and (29) that
(30)h(z)=z{log-1/2-2}·γ-1/2(-z)γ-1/2(z/2)=z-1(z/2;-1/2)∞(-z;-1/2)∞=z-1∏k=0∞(1-(z/2)(-1/2)k)∏k=0∞(1+z(-1/2)k)=z-1∏k=0∞(1+z(-1/2)k+1)∏k=0∞(1+z(-1/2)k)=z-1(([1+(-12)z][1+(-12)2z]⋯[1+(-12)kz]⋯)×((1+z)[1+(-12)z][1+(-12)2z]⋯[1+(-12)kz]⋯)-1)=1z(z+1),g3(z)=g1(z)h(z)+g2(z)h(z)+1=-(z-2)22(z2+z+1)
is a meromorphic solution of (4), which is distinct from g1(z) and g2(z). Moreover, we also conclude from (10), (27), and (5) that
(31)f1(-12z)=-2f1(z),f2(-12z)=1-z/21-z/(-1)f2(z),
which are corresponding to g1(z) and g2(z), respectively, and are also the types of (22). Thus, we deduce from (24) that
(32)f1(z)=1z,f2(z)=γ-1/2(z/2)γ-1/2(-z)=z+1
satisfy second-order linear q-difference equation (5).


## 4. Value Distribution of Solutions ofq-Difference Riccati Equations and Form of Solutions of Second-Order Linear q-Difference Equations

We first consider the value distribution of transcendental meromorphic solution ofq-difference Riccati equation (4).Theorem 5.
Leta1(z) and a0(z) be nonconstant rational functions. If g(z) is a zero-order transcendental meromorphic solution of q-difference Riccati equation
(33)g(qz)=-a1(z)g(z)+a0(z)g(z)
with q∈ℂ∖{0} and |q|≠1, then (i)
if(34)|q|>1,N¯(r,g)+N¯(r,1g)=S(r,g),
then g(z) has at most one Borel exceptional value;(ii)
if|q|≠1, then Nevanlinna deficiencies δ(0,g)=δ(∞,g)=0;(iii)
if|q|≠1 and qz2+za1(z)+a0(z)≢0, then g(z) has infinitely many fixed points.In particular, we obtain the following theorem.Theorem 6.
Ifa1(z)=a1 and a0(z)=a0(≠0) are constants, and if q∈ℂ∖{0} and |q|≠1, then q-difference Riccati equation (4) has only rational solutions. Furthermore, if a1(z)≡0 and a0(z)=a0 is nonzero constant, then (4) has only a nonzero constant solution g(z)=d, which satisfies d2+a0=0.We need some preliminaries to prove Theorems5 and 6.The theorem of Tumura and Clunie is an important result in Nevanlinna theory, see [11, 12]. Weissenborn extended it and obtained the following lemma.Lemma 7 (see [13, Theorem]).
Leth(z) be a meromorphic function and let ϕ be given by
(35)ϕ(z)=cn(z)h(z)n+cn-1(z)h(z)n-1+⋯+c0(z),T(r,cj)=S(r,h),j=0,1,…,n-1,n.
Then either
(36)ϕ≡(h+cn-1(z)ncn(z))n,
or
(37)T(r,h)≤N¯(r,1ϕ)+N¯(r,h)+S(r,h).Lemma 8.
Suppose thath(z) is a nonconstant meromorphic function satisfying
(38)N¯(r,h)+N¯(r,1h)=S(r,h).
Let
(39)ϕ(z)=cn(z)h(z)n+cn-1(z)h(z)n-1+⋯+c0(z)
be a polynomial in h(z) with n∈ℕ, and coefficients satisfying
(40)T(r,cj)=S(r,h),j=0,1,…,n-1,n,cn(z)c0(z)≢0.
Then
(41)N(r,1ϕ)=nT(r,h)+S(r,h)
or
(42)N¯(r,1ϕ)≥T(r,h)+S(r,h).
Thus, ϕ(z)≢0.Proof of Lemma8.
By differentiating both sides of (39), we conclude that
(43)ϕ′(z)=∑j=1n‍(cj′(z)+jcj(z)h′(z)h(z))h(z)j.
Thus, we deduce from (39) and (43) that
(44)(cn′(z)+ncn(z)h′(z)h(z))ϕ(z)-cnϕ′(z)=∑j=1n-1‍[cj(z)(cn′(z)+ncn(z)h′(z)h(z))-cn(z)(cj′(z)+jcj(z)h′(z)h(z))]h(z)j+c0(z)(cn′(z)+ncn(z)h′(z)h(z)).
Therefore, (cn′(z)+ncn(z)(h′(z)/h(z)))ϕ(z)-cnϕ′(z) is a polynomial in h(z) with degree no greater than n-1 and the term of degree zero is c0(z)(cn′(z)+ncn(z)(h′(z)/h(z)))≢0. Then
(45)cn′(z)+ncn(z)h′(z)h(z)≢0.
Otherwise, if cn′(z)+ncn(z)(h′(z)/h(z))≡0, then cn(z)h(z)n is a nonzero constant, a contradiction. We also note that c0(z)(cn′(z)+ncn(z)(h′(z)/h(z))) is a small function relative to h(z) by (38) and the lemma of logarithmic derivative. Set
(46)μ1(z)=cn′(z)+ncn(z)h′(z)h(z),ν1(z)=cn.
Then μ1(z) and ν1(z) are small functions relative to h(z) and
(47)ϕ1(z)=μ1(z)ϕ(z)-ν1(z)ϕ′(z)
is a polynomial in h(z) with degree no greater than n-1 and the term of degree zero is small function relative to h(z).
If the degree ofϕ1(z) is greater than zero, then by repeating the above process, we can get two small functions μ2(z) and ν2(z) such that
(48)ϕ2(z)=μ2(z)ϕ1(z)-ν1(z)ϕ1′(z)
is a polynomial in h(z) with a degree less than the degree of ϕ1(z) and the term of degree zero is a small function relative to h(z).
We note that such process will be terminated at mostn times. Thus, We can proceed this process to obtain small functions μj(z) and νj(z), where j=1,2,…,s,s+1 and s≤n, such that
(49)ϕj(z)=μj(z)ϕj-1(z)-νj(z)ϕj-1′(z)
are polynomial in h(z) with degϕj(z)>degϕj-1(z)(j=1,2,…,s), where ϕ0(z)≡ϕ(z) and
(50)ϕs+1(z)=μs+1(z)ϕs(z)-νs+1(z)ϕs′(z)
is a small function relative to h(z). Thus, we deduce that the small function ϕs+1(z) can be expressed as a linear differential polynomial in ϕ(z) with coefficients being small functions relative to h(z). So,
(51)m(r,1ϕ)=S(r,h).
On the other hand, we deduce from Lemma 7 that either
(52)ϕ≡(h+cn-1(z)ncn(z))n
or
(53)T(r,h)≤N¯(r,1ϕ)+N¯(r,h)+S(r,h).
Thus, we deduce from Valiron-Mohon’ko Lemma, (51), and (52) that (41) holds and obtain from (38) and (53) that (42) holds. Therefore, ϕ(z)≢0. The proof of Lemma 8 is completed.Lemma 9 (see [9, Theorem 5.2]).
Leth(z) be a transcendental meromorphic solution of
(54)h(qz)=R(z,h(z))=∑i=0pai(z)h(z)i∑j=0qbjbj(z)h(z)j
with meromorphic coefficients ai(z),bj(z) relative to h(z) and q∈ℂ such that |q|>1. If N¯(r,h)+N¯(r,1/h)=S(r,h), then (54) is either of the form
(55)f(qz)=ap(z)f(z)porf(qz)=a0(z)f(z)q.Lemma 10 (see [5, Theorem 2.2]).
Letf(z) be a nonconstant zero-order meromorphic solution of
(56)P(z,f)=0,
where P(z,f) is a q-difference polynomials in f(z). If P(z,α)≢0 for a small function α(z) relative to f(z), then
(57)m(r,1f-α)=o(T(r,f))
on a set of logarithmic density 1.Lemma 11 (see [2, Theorem 2.2.5 and Corollary 2.2.7]).
Letf(z) be a meromorphic function. Then for all irreducible rational functions in f(z),
(58)R(z,f(z))=∑i=0pai(z)f(z)i∑j=0qbj(z)f(z)j,
with meromorphic coefficients ai(z),bj(z), the characteristic function of R(z,f(z)) satisfies
(59)T(R(z,f(z)))=dT(r,f)+O(Ψ(r)),
where d=max{p,q} and Ψ(r)=maxi,j{T(r,ai),T(r,bj)}.
In the particular case when(60)T(r,ai)=S(r,f),i=0,1,…,p,T(r,bj)=S(r,f),j=0,1,…,q,
we have
(61)T(r,R(z,f(z)))=dT(r,f)+S(r,f).
We also use the observation [7, page 2] that, for any meromorphic function f(z) and any constant q∈ℂ∖{0},
(62)T(r,f(qz))=T(|q|r,f)+O(1).Proof of Theorem5.
Suppose thatg(z) is a zero-order transcendental meromorphic solution of q-difference Riccati equation (4).
(i) Suppose thatg(z) has two finite Borel exceptional values a and b(≠0,a). For the case where one of a and b is infinite, we can use a similar method to prove. Set
(63)h(z)=g(z)-ag(z)-b.
Since N¯(r,g)+N¯(r,1/g)=S(r,g), we deduce from (63) that
(64)N¯(r,h)+N¯(r,1h)=S(r,h).
We also conclude from (63) that
(65)g(z)=a-bh(z)1-h(z).
Now, substituting (65) into (4), we conclude that
(66)h(qz)=(ba1(z)+a0(z)+ab)h(z)-(aa1(z)+a0(z)+a2)(ba1(z)+a0(z)+b2)h(z)-(aa1(z)+a0(z)+ab).
By the assumptions of Theorem 5, we get
(67)(ba1(z)+a0(z)+ab)≢0,(ba1(z)+a0(z)+b2)≢0.
Thus, we deduce from Lemma 9, (64), and (66) that
(68)h(qz)=c(z)h(z)k,k∈ℤ∖{0},
where T(r,c(z))=S(r,h).
Ifk≥1, we conclude from (66) and (68) that
(69)c(z)(ba1(z)+a0(z)+b2)h(z)k+1-c(z)(aa1(z)+a0(z)+ab)h(z)k-(ba1(z)+a0(z)+ab)h(z)+(aa1(z)+a0(z)+a2)=0.
Thus, we deduce from Lemma 8 and (64) that (69) is a contradiction. If k≤-1, we use the same method as above to get another contradiction. Therefore, g(z) at most one Borel exceptional value.
(ii) We first proveδ(0,g)=0. We obtain from (4) that
(70)P1(z,g)=g(z)g(qz)+a1(z)g(z)+a0(z)=0.
Since P1(z,0)=a0(z)≢0, we deduce from Lemma 10 and (70) that
(71)m(r,1g)=S(r,g)
on a set E of logarithmic density 1. Therefore,
(72)0≤δ(0,g)=lim_r→∞m(r,1/g)T(r,g)≤lim_r→∞,r∈Em(r,1/g)T(r,g)=0.
Thus, δ(0,g)=0.
We second proveδ(∞,g)=0. Set y(z)=1/g(z). Then
(73)T(r,y)=T(r,g)+O(1),S(r,y)=S(r,g).
Now, substituting g(z)=1/y(z) into (4), we conclude that
(74)P2(z,y)=y(qz)(a0(z)y(z)+a1(z))+1=0.
Since P2(z,0)=1≢0, we obtain from Lemma 10 and (74) that
(75)m(r,1y)=S(r,y)
on a set E of logarithmic density 1. Therefore,
(76)N(r,1y)=T(r,y)+S(r,y)
on a set E of logarithmic density 1. Thus, we conclude from y(z)=1/g(z) and (76) that
(77)N(r,g)=N(r,1y)=T(r,y)+o(T(r,y))=T(r,g)+S(r,g)
on a set E of logarithmic density 1, and so,
(78)0≤δ(∞,g)=1-lim¯r→∞N(r,g)T(r,g)≤1-lim¯r→∞,r∈EN(r,g)T(r,g)=0.
Thus, δ(∞,g)=0.
(iii) Sety(z)=g(z)-z. Then
(79)T(r,y)=T(r,g)+S(r,g),S(r,y)=S(r,g).
Substituting g(z)=y(z)+z into (4), we conclude that
(80)P3(z,y)=y(z)y(qz)+zy(qz)+qzy(z)+qz2+za1(z)+a0(z)=0.
Since P3(z,0)=qz2+za1(z)+a0(z)≢0, we deduce from Lemma 10 and (80) that
(81)m(r,1y)=S(r,y)
on a set E of logarithmic density 1. Therefore
(82)N(r,1g-z)=N(r,1y)=T(r,y)+o(T(r,y))=T(r,g)+S(r,g)
on a set E of logarithmic density 1. This shows that g(z) has infinitely many fixed points if qz2+za1(z)+a0(z)≢0.Proof of Theorem6.
Suppose first that0<|q|<1 and (4) with nonzero constant coefficients a1(z) and a0(z) admits a meromorphic solution g(z). We assert that g(z) is rational. In fact, we conclude from Lemma 11, (4), and (62) that
(83)T(r,f)≤T(|q|r,f)+A,r≥R0,
where A>Ψ(r)=max{T(r,a0),T(r,a1)}≥0,R0(>0) is fixed number.
Thus, for anyr≥R0, there exists an n∈ℕ such that
(84)R0|q|n-1≤r<R0|q|n.
By an inductive argument, we deduce from (84) that
(85)T(r,f)≤T(|q|nr,f)+An≤T(R0,f)+A(logrlog(1/|q|)-logR0log(1/|q|)+1)=O(logr).
Suppose now that |q|>1 and (4) with nonzero constant coefficients a1(z) and a1(z) admits a meromorphic solution g(z). Replacing z by z/q in (4), we proceed in a similar method as above to get (85) again. Therefore, g(z) is rational solution of (4).
Now, we affirm thatg(z) must be nonzero constant if a1(z)≡0 and a0(z)=a0(≠0) is a constant. Otherwise, if g(z) is nonconstant rational and has a pole z0≠0, we conclude from (4) that g(z) has infinitely many poles of the forms q2(n-1)z0 and infinitely many zeros of the forms q2(n-1)+1z0 for all n∈ℕ. Conversely, If g(z) is nonconstant rational and has a zero z0≠0, we conclude from (4) that g(z) has infinitely many zeros of the forms q2(n-1)z0 and infinitely many poles of the forms q2(n-1)+1z0 for all n∈ℕ. These are both impossible since g(z) is rational. Thus, the only possible pole (resp. zero) of g(z) is at 0. So g(z) may have the form g(z)=dzk(k∈ℤ), where d is a nonzero constant. If k≠0, we get a contradiction from (4). Therefore, k=0 and (4) has only a nonzero constant solution g(z)=d, which satisfies d2+a0=0. The proof of Theorem 6 is completed.We now consider the form of meromorphic solutions of (5), which is according to Theorem 6. In fact, more details about meromorphic solutions of (5) have been studied in [7, 14]. Here, we only consider the case that all coefficients are constants.Theorem 12.
Ifa1(z)≡0 and a0(z)=a0 is constant, and if q∈ℂ∖{0} and |q|≠1, then every meromorphic solution f(z) of second-order linear q-difference equation (5) has the form f(z)=βzk, where β∈ℂ∖{0} and k∈ℤ satisfying q2k+a0=0.We first list a lemma needed below.Lemma 13 (see [14, Theorem 2.1]).
Suppose thatq∈ℂ∖{0} and |q|≠1. Let a0,a1,…,an be complex constants and let Q(z) be of the reduced form Q(z)=p1(z)/zl, where p1(z) is a polynomial of degree d and l∈ℕ∪{0}. Then all meromorphic solutions f(z) of
(86)∑j=0n‍aj(z)f(qjz)=Q(z)
are of the reduced form f(z)=p2(z)/zp, where p2(z) is a polynomial and p≥l.Proof of Theorem12.
We deduce from Lemma13 that all meromorphic solutions f(z) of (5) are of the form f(z)=p2(z)/zp, where p2(z) and p are defined as Lemma 13. Thus, we conclude from Theorem 6 and (10) that
(87)d=g(z)=f(qz)f(z)=1qp·p2(qz)p2(z),
where d is defined as Theorem 6. From (87), we obtain that there exists β∈ℂ∖{0} and m∈ℕ∪{0} such that p2(z)=βzm, and so f(z)=p2(z)/zp=βzm/zp=:βzk, where k=m-p∈ℤ. Now, substituting f(z)=βzk into (5), we conclude that k satisfies q2k+a0=0. The proof of Theorem 12 is completed.Example 14.
Letq∈ℂ∖{0},|q|≠1, a1(z)≡0 and a0(z)=-1/q2. Then second-order q-difference equation (5) is solved by f(z)=1/z. Obviously, f(z)=1/z and k=-1 satisfy the conclusions described by Theorem 12.


## 5.  Linearq-Difference Equations of Second-Order

Lety1(z) and y2(z) be meromorphic solutions of (5). We define the q-Casorati determinant of meromorphic functions y1(z) and y2(z) by
(88)Cq^(z)=Cq^(y1,y2;z)=|y1(z)y2(z)y1(qz)y2(qz)|.
Then the q-Casorati determinant Cq^(z) vanishes identically on ℂ if and only if the functions y1(z) and y2(z) are linearly dependent over the field of functions ϕ(qz)=ϕ(z). On the other hand, g1(z) and g2(z) are linear independent if and only if C^q(g1,g2;z)≢0. From this definition, we have some properties on the q-Casorati determinant C^q(z) as follows.Theorem 15.
Ify1(z) and y2(z) are nontrivial meromorphic solutions of (5), then q-Casorati determinant C^q(y1,y2;z) satisfies a first-order q-difference equation
(89)ΔqCq^(z)=(a0-1)Cq^(z).
Conversely, we assume that y1(z)(≢0) and y2(z) satisfy (89). If y1(z) is a meromorphic solution of (5), then y2(z) is also a meromorphic solution of (5).Proof of Theorem15.
Suppose first thaty1(z) and y2(z) are nontrivial meromorphic solutions of (5), we conclude that
(90)Cq^(qz)=Cq^(y1,y2;qz)=|y1(qz)y2(qz)y1(q2z)y2(q2z)|=|y1(qz)y2(qz)-a1y1(qz)-a0(z)y1(z)-a1y2(qz)-a0(z)y2(z)|=|y1(qz)y2(qz)-a0(z)y1(z)-a0(z)y2(z)|=a0(z)|y1(z)y2(z)y1(qz)y2(qz)|=a0(z)Cq^(z).
Therefore,
(91)ΔqCq^(z)=Cq^(qz)-Cq^(z)=(a0(z)-1)Cq^(z).
Second, if y1(z)(≢0) and y2(z) satisfy (89), then we have
(92)|y1(qz)y2(qz)y1(q2z)y2(q2z)|=a0(z)|y1(z)y2(z)y1(qz)y2(qz)|.
We note that, for any meromorphic function c(z)≢0,
(93)|y1(qz)y2(qz)y1(q2z)y2(q2z)|=|y1(qz)y2(qz)y1(q2z)+c(z)y1(qz)y2(q2z)+c(z)y2(qz)|.
In particular, we take c(z)=a1(z). Thus,
(94)|y1(qz)y2(qz)y1(q2z)+a1(z)y1(qz)y2(q2z)+a1(z)y2(qz)|=a0(z)|y1(z)y2(z)y1(qz)y2(qz)|.
So, we have
(95)|y1(qz)y2(qz)y1(q2z)+a1(z)y1(qz)y2(q2z)+a1(z)y2(qz)|+|y1(qz)y2(qz)a0(z)y1(z)a0(z)y2(z)|=0.
From this, we conclude that
(96)y1(qz)[y2(q2z)+a1(z)y2(qz)+a0(z)y2(z)]=y2(qz)[y1(q2z)+a1(z)y1(qz)+a0(z)y1(z)].
Since y1(z)(≢0) is a meromorphic solution of (5), we have
(97)y1(q2z)+a1(z)y1(qz)+a0(z)y1(z)=0,
and so,
(98)y2(q2z)+a1(z)y2(qz)+a0(z)y2(z)=0.
This shows that y2(z) is a meromorphic solution of (5). The proof of Theorem 15 is completed.Theorem 16.
(i) Lety1(z) and y2(z) be linear independent meromorphic solutions of (5), and let C^q(z) be the q-Casoratian determinant of y1(z) and y2(z). Then y2(z) is represented as y2(z)=g(z)y1(z), where g(z) satisfies
(99)Δqg(z)=Cq^(z)y1(z)y1(qz).
(ii) Lety1(z) be a nontrivial meromorphic solution of (5), and let C^q(z) be a meromorphic solution of (89). If g(z) satisfies (99), then y2(z)=g(z)y1(z) is a meromorphic solution of (5).Proof of Theorem16.
(i) From the definition ofC^q(z), we obtain
(100)C^q(z)=y1(z)y2(qz)-y2(z)y1(qz).
This shows that y2(z) satisfies first-order q-difference equation of type
(101)y2(qz)=y2(z)·y1(qz)y1(z)+C^q(z)y1(z).
By substituting y2(z)=g(z)y1(z) into (101), we conclude that
(102)g(qz)y1(qz)=g(z)y1(z)·y1(qz)y1(z)+C^q(z)y1(z),
and so we obtain the desired form (99).
(ii) Obviously, we conclude from (99) and (89) that
(103)g(qz)=g(z)+C^q(z)y1(z)y1(qz),g(q2z)=g(qz)+C^q(qz)y1(qz)y1(q2z)=g(qz)+a0(z)C^q(z)y1(qz)y1(q2z).
Since y2(z)=g(z)y1(z), C^q(z)=y1(z)y2(qz)-y2(z)y1(qz), and
(104)y1(q2z)+a1(z)y1(qz)+a0(z)y1(z)=0,
we conclude from (103), and (104) that
(105)y2(q2z)=g(q2z)y1(q2z)=(g(z)+C^q(z)y1(z)y1(qz)+a0(z)C^q(z)y1(qz)y1(q2z))y1(q2z)=g(z)y1(q2z)+y1(q2z)+a0(z)y1(z)y1(z)y1(qz)·C^q(z)=g(z)y1(q2z)+-a1(z)y1(qz)y1(z)y1(qz)·C^q(z)=y2(z)y1(z)·(-a1(z)y1(qz)-a0(z)y1(z))-a1(z)y1(z)·(y1(z)y2(qz)-y2(z)y1(qz))=-a1(z)y2(qz)-a0(z)y2(z).
This yields that y2(z)=g(z)y1(z) is a meromorphic solution of (5). The proof of Theorem 16 is completed.



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*Source: 938579-2013-01-13.xml*