The properties of solutions for several types of Painlevé equations concerning fixed-points, zeros and poles

Hong Yan Xu; Xiu Min Zheng

doi:10.1515/math-2019-0079

Article Open Access

The properties of solutions for several types of Painlevé equations concerning fixed-points, zeros and poles

Hong Yan Xu and Xiu Min Zheng

Published/Copyright: August 28, 2019

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$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

The purpose of this manuscript is to study some properties on meromorphic solutions for several types of q-difference equations. Some exponents of convergence of zeros, poles and fixed points related to meromorphic solutions for some q-difference equations are obtained. Our theorems are some extension and improvements to those results given by Qi, Peng, Chen, and Zhang.

Keywords: fixed point; q-difference equation; meromorphic solution

MSC 2010: 39A13; 30D35

1 Introduction and main results

Around 2006, Halburd-Korhonen [1] and Chiang-Feng [2] established independently some important fundamental results of Nevanlinna theory about the complex difference and difference operators. After their wonderful work, considerable attention has been paid in studying complex difference equations, and a lot of important and interesting results (see [2, 3, 4]) focusing on complex difference equations and difference analogues of Nevanlinna theory were obtained. Halburd-Korhonen [1, 5, 6] studied the equation

f(z+1)+f(z−1)=R(z,f), (1.1)

where R(z, f) is rational in f and meromorphic in z, and they singled out the difference Painlevé I equation

f(z+1)+f(z−1)=az+bf(z)+c, (1.2)

and the difference Painlevé II equation

f(z+1)+f(z−1)=(az+b)f(z)+c1−f(z)2. (1.3)

Later, Ronkainen [7] in 2010 further discussed the equation

f(z+1)f(z−1)=R(z,f) (1.4)

where R(z, f) is rational and irreducible in f and meromorphic in z. He pointed out that either f satisfies the difference Riccati equation

f(z+1)=A(z)f(z)+B(z)f(z)+C(z),

or equation (1.4) can be transformed to one of the following equations

f(z+1)f(z−1)=η(z)f(z)2−λ(z)f(z)+μ(z)(f(z)−1)(f(z)−υ(z)),f(z+1)f(z−1)=η(z)f(z)2−λ(z)f(z)f(z)−1,f(z+1)f(z−1)=η(z)(f(z)−λ(z))f(z)−1,f(z+1)f(z−1)=h(z)f(z)m,

where η(z), λ(z), υ(z) satisfy some conditions. The above four equations can be called as the difference Painlevé III equations.

In what follows, we should assume that the readers are familiar with the fundamental theorems and the standard notations in the theory of Nevanlinna value distribution (see Hayman [8], Yang [9] and Yi-Yang [10]). Let f be a meromorphic function, we denote σ(f), λ(f) and λ(1f) to be the order, the exponent of convergence of zeros and the exponent of convergence of poles of f(z), respectively, and denote τ(f) to be the exponent of convergence of fixed points of f(z), which is defined by

τ(f)=lim supr→+∞log⁡N(r,1f(z)−z)log⁡r.

In 2010, Chen-Shon [11] considered the difference Painlevé I,II equation (1.2),(1.3) and obtained the following theorems.

Theorem 1.1

(see [11, Theorem 4]). Let a, b, c be constants, where a, b are not both equal to zero. Then

if a ≠ 0, then (1.2) has no rational solution;
if a = 0, and b ≠ 0, then (1.2) has a nonzero constant solution w(z) = A, where A satisfies 2A²−cA−b = 0.

The other rational solution w(z) satisfies w(z) = P(z)Q(z) + A, where P(z) and Q(z) are relatively prime polynomials and satisfy deg P < deg Q.

Theorem 1.2

(see [11, Theorem 1]). Let a, b, c be constants with ac ≠ 0. If f(z) is a finite order transcendental meromorphic solution of equation (1.3), then

f has at most one nonzero finite Borel exceptional value for σ(f) > 0;
λ(1/f) = λ(f) = σ(f).

In 2013 and 2014, Zhang-Yi [12], Zhang-Yang [13] studied the difference Painlevé III equations with the constant coefficients, and obtained

Theorem 1.3

(see [13]). If f(z) is a transcendental finite order meromorphic solution of

f(z+1)f(z−1)(f(z)−1)=ηf(z)orf(z+1)f(z−1)(f(z)−1)=f(z)2−λf(z),

where η(≠ 0), λ(≠ 0, 1) are constants, then

λ(f) = σ(f);
f has at most one nonzero Borel exceptional value for σ(f) > 0.

Theorem 1.4

(see [12, Theorem 4.3]). If f(z) is a transcendental finite order meromorphic solution of

f(z+1)f(z−1)(f(z)−1)2=f(z)2−λf(z)+μ,

where λ, μ are constants and λμ ≠ 0, then λ(f) = σ(f).

In 2007, Barnett, Halburd, Korhonen and Morgan [14] first established the Logarithmic Derivative Lemma on complex q-difference operators. Then by applying those fundamental results, many mathematicians have done a lot of work about the value distribution of complex q-difference operators, solutions for complex q-difference equations, by replacing the difference f(z + c) with the q-difference f(qz), q ∈ ℂ ∖ {0, 1} for the meromorphic function f(z) in some expression concerning complex difference equations and complex difference operators (see [15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28]).

In 2015, Qi and Yang [29] considered the following equation

f(qz)+fzq=az+bf(z)+c, (1.5)

which can be seen as q-difference analogues of (1.2), and obtained the result as follows.

Theorem 1.5

[29, Theorem 1.1]. Let f(z) be a transcendental meromorphic solution with zero order of equation (1.5), and a, b, c be three constants such that a, b cannot vanish simultaneously. Then,

f(z) has infinitely many poles.
if a ≠ 0, then f(z) − d has infinitely many zeros for any d ∈ ℂ;
if a = 0 and f(z) takes a finite value A finitely often, then A is a solution of 2z² − cz − b = 0.

Motivated by the idea from [29] and [13], our main aim of this aritcle is further to investigate some properties of meromorphic solutions for some q-difference equations, which can be called as q-difference Painlevé III equations. We obtain the following four results.

Theorem 1.6

Let q ∈ ℂ∖{0} and ∣q∣ ≠ 1, and let f(z) be a zero order transcendental meromorphic solution of the following equation

f(qz)f(zq)(f(z)−1)2=f(z)2−λf(z)+μ, (1.6)

where λ, μ are constants satisfying λμ ≠ 0. Then

for any η ∈ ℂ − {0, 1}, f(η z) has infinitely many fixed-points and τ(f(η z)) = σ(f), especially, f(q^jz) has infinitely many fixed-points and τ(f(q^jz)) = σ(f), where j is a positive integer;
Δqf,Δqff have infinitely many poles, and

λ(1Δqf)=λ1Δqff;
f(z) has infinitely many zeros and poles, and the Nevanlinna exceptional value of f(z) can only come from a set E = {z ∣z⁴ − 2z³ + λ z − μ = 0}.

Theorem 1.7

For q(≠ 0) ∈ ℂ and ∣q∣ ≠ 1, and let f(z) be a zero order transcendental meromorphic solution of the following equation

f(qz)f(zq)(f(z)−1)2=f(z)2. (1.7)

Then

for any η ∈ ℂ − {0, 1}, f(η z) has infinitely many fixed-points and τ(f(η z)) = σ(f);
f(z), Δqf,Δqff have infinitely many poles, and

λ1f=λ(1Δqf)=λ1Δqff.

Theorem 1.8

Let q ∈ ℂ∖{0} and ∣q∣ ≠ 1, and let f(z) be a zero order transcendental meromorphic solution of the following equation

f(qz)f(zq)(f(z)−1)=λ(z)f(z), (1.8)

where λ(z) is a nonconstant polynomial. Then

for any η ∈ ℂ∖{0, 1}, f(η z) has infinitely many fixed-points and τ(f(η z)) = σ(f);
f(z), Δ_qf have infinitely many zeros and poles, and Δqff has infinitely many poles, and

λ(f)=λ(Δqf)=λ(1f)=λ(1Δqf)=λ1Δqff;
f(z) has no Nevanlinna exceptional value.

Theorem 1.9

Let q ∈ ℂ∖{0} and ∣q∣ ≠ 1, and let f(z) be a zero order transcendental meromorphic solution of the following equation

f(qz)f(zq)f(z)2=h(z), (1.9)

where h(z) is a nonconstant rational function. Then

for any η ∈ ℂ − {0, 1}, f(η z) has infinitely many fixed-points and τ(f(η z)) = σ(f);
f(z), Δqf,Δqff have infinitely many zeros and poles, and

λ(f)=λ(1f)=λ(Δqf)=λ(1Δqf)=λ(Δqff)=λ1Δqff;
f(z) has no Nevanlinna exceptional value.

2 Proofs of Theorems 1.6 and 1.7

2.1 The proof of Theorem 1.6

Suppose that f(z) is a zero order transcendental meromorphic solution of equation (1.6).

For any η ∈ ℂ∖{0, 1}, substituting η z into (1.6), we have

f(qηz)f(ηzq)(f(ηz)−1)2=f(ηz)2−λf(ηz)+μ. (2.1)

Denote g(z) = f(η z), then (2.1) can be represented as

g(qz)g(zq)(g(z)−1)2=g(z)2−λg(z)+μ.

Let

P1(z,g):=g(qz)g(zq)(g(z)−1)2−g(z)2+λg(z)−μ=0.

Thus, it follows

P1(z,z)=z2(z−1)2−z2+λz−μ≢0.

In view of P₁(z, z) ≢ 0 and by Theorem 2.5 in [14], it follows that

m(r,1g(z)−z)=S(r,g).

Thus, since f is of zero order, from Theorems 1.1 and 1.3 in [25], it yields

N(r,1f(ηz)−z)=N(r,1g(z)−z)=T(r,g)+S(r,g)=T(r,f(ηz))+S(r,f(ηz))=T(r,f)+S(r,f).

Therefore, for any η ∈ ℂ∖{0, 1}, τ(f(η z)) = σ(f).
Next, we divide the proof into three cases: ii₁). λ − μ ≠ 1; ii₂). λ − μ = 1,μ = 1; ii₃). λ − μ = 1,μ ≠ 1.
1. λ − μ ≠ 1. We can rewrite (1.6) as the following form
  
  f(qz)f(zq)f(z)2=f(z)2−λf(z)+μf(z)2(f(z)−1)2. (2.2)
  
  Thus, in view of Theorems 1.1 and 1.3 in [25], Theorem 1.1 in [14] and Valiron-Mohon’ko Lemma [30], and from (2.2) and λ − μ ≠ 1, it follows that
  
  4T(r,f)=Tr,f(z)2−λf(z)+μf(z)2(f(z)−1)2+O(1)=Tr,f(qz)f(zq)f(z)2+O(1)≤2Tr,f(qz)f(z)+S(r,f)+O(1)≤2Tr,Δqff+S(r,f),
  
  that is,
  
  2T(r,f)≤T(r,Δqff)+S(r,f). (2.3)
  
  Then, we have
  
  N(r,Δqff)=T(r,Δqff)−m(r,Δqff)≥2T(r,f)+S(r,f). (2.4)
  
  Hence, we can conclude that Δqff has infinitely many poles and λ1Δqff=σ(f) by combining with
  
  N(r,Δqff)≤T(r,Δqff)≤2T(r,f)+S(r,f). (2.5)
  
  Since Δ_qf(z) = f(qz) − f(z) and Δ_q⁻¹f(z) = f(zq) − f(z), then it follow from (1.6) that
  
  f(qz)f(zq)=(Δqf+f)(Δq−1f+f)=f(z)2−λf(z)+μ(f(z)−1)2,
  
  that is,
  
  (Δqf+Δq−1f)f+ΔqfΔq−1f=−f4+2f3−λf+μ(f−1)2. (2.6)
  
  Since λμ ≠ 0 and λ − μ ≠ 1, then in view of Theorems 1.1 and 1.3 in [25],Theorem 1.1 in [14] and Valiron-Mohon’ko Lemma [30], it yields
  
  4T(r,f)=T(r,−f4+2f3−λf+μ(f−1)2)+O(1)=Tr,(Δqf+Δq−1f)f+ΔqfΔq−1f+O(1)≤T(r,f)+2T(r,Δqf)+2T(r,Δq−1f)+O(1)≤T(r,f)+4T(r,Δqf)+S(r,f),
  
  that is,
  
  34T(r,f)≤T(r,Δqf)+S(r,f). (2.7)
  
  On the other hand, we can rewrite (1.6) as
  
  f(qz)f(zq)f(z)2=2f(qz)f(zq)f(z)−f(qz)f(zq)+f(z)2−λf(z)+μ. (2.8)
  
  By applying Theorem 2.5 in [17] for (2.8), it yields
  
  m(r,f)=S(r,f). (2.9)
  
  Thus, in view of (2.7) and (2.9), and Theorem 1.1 in [14], it follows
  
  N(r,Δqf)=T(r,Δqf)−,m(r,Δqf)≥T(r,Δqf)−(m(r,Δqff)+m(r,f))≥34T(r,f)+S(r,f).
  
  Hence, we conclude that Δ_qf has infinitely many poles and λ(1Δqf)=σ(f) by combining with
  
  Nr,Δqf≤2T(r,f)+S(r,f). (2.10)
2. λ − μ = 1 and μ = 1. Thus, λ = 2. Then equation (1.6) becomes
  
  f(qz)f(zq)=1,
  
  and
  
  1f(z)2=f(qz)f(zq)f(z)2.
  
  Thus, in view of Theorems 1.1 and 1.3 in [25],Theorem 1.1 in [14] and Valiron-Mohon’ko Lemma [30], we deduce
  
  2T(r,f)=T(r,1f2)+O(1)=Tr,f(qz)f(zq)f(z)2+O(1)≤2T(r,Δqff)+S(r,f),
  
  that is,
  
  T(r,f)≤T(r,Δqff)+S(r,f). (2.11)
  
  So, in view of Theorem 1.1 in [14] and (2.11), it yields
  
  N(r,Δqff)=T(r,Δqff)−m(r,Δqff)≥T(r,f)+S(r,f),
  
  which implies that Δqff has infinitely many poles and λ1Δqff=σ(f) by combining with (2.5).
  
  For λ = 2 and μ = 1, we can rewrite (1.6) as
  
  (Δqf+f)(Δq−1f+f)=1,
  
  that is,
  
  (Δqf+Δq−1f)f+ΔqfΔq−1f=1−f(z)2. (2.12)
  
  Thus, by applying Theorems 1.1 and 1.3 in [25] for (2.12), it follows
  
  2T(r,f)=T(r,1−f2)+O(1)=T(Δqf+Δq−1f)f+ΔqfΔq−1f)+O(1)≤T(r,f)+4T(r,Δqf)+S(r,f),
  
  that is,
  
  14T(r,f)≤T(r,Δqf)+S(r,f). (2.13)
  
  In view of (1.6) and Theorem 1.1 in [14], it follows
  
  2m(r,f)=m(r,f2)=mr,f(z)2f(qz)f(zq)≤m(r,f(z)f(qz))+mr,f(z)f(zq)=S(r,f),
  
  that is,
  
  m(r,f)=S(r,f). (2.14)
  
  In view of (2.13) and (2.14) and Theorem 1.1 in [14], we can deduce
  
  N(r,Δqf)=T(r,Δqf)−m(r,Δqf)≥T(r,Δqf)−(m(r,Δqff)+m(r,f))≥14T(r,f)+S(r,f).
  
  Hence, we conclude that Δ_qf has infinitely many poles and λ1Δqff=σ(f) by combining with (2.10).
3. λ − μ = 1 and μ ≠ 1. Thus, λ = μ+1 and (1.6) becomes
  
  f(qz)f(zq)(f(z)−1)=f(z)−μ, (2.15)
  
  and
  
  f(qz)f(zq)f(z)2=f(z)−μf(z)2(f(z)−1). (2.16)
  
  By applying Valiron-Mohon’ko Lemma [30], and in view of Theorems 1.1 and 1.3 in [25] and μ ≠ 1, it follows
  
  3T(r,f)=Tr,f−μf2(f−1)+O(1)=Tr,f(qz)f(zq)f(z)2+O(1)≤2T(r,Δqff)+S(r,f),
  
  that is,
  
  32T(r,f)≤T(r,Δqff)+S(r,f). (2.17)
  
  Thus, we can conclude from (2.17) and Theorem 1.1 in [14] that
  
  N(r,Δqff)≥32T(r,f)+S(r,f).
  
  By combining with (2.5), it means that Δqff has infinitely many poles and λ1Δqff=σ(f) .
  
  In view of λ − μ = 1 and μ ≠ 1, we can rewrite (1.6) as the following
  
  f(qz)f(zq)=(Δqf+f)(Δq−1f+f)=f−μf−1,
  
  that is,
  
  (Δqf+Δq−1f)f+ΔqfΔq−1f=−f3+f2+f−μf−1. (2.18)
  
  Thus, by applying Valiron-Mohon’ko Lemma [30] and Theorem 1.1 in [14] for (2.18), we have
  
  3T(r,f)=Tr,−f3+f2+f−μf−1+O(1)=T(r,(Δqf+Δq−1f)f+ΔqfΔq−1f)+O(1)≤T(r,f)+4T(r,Δqf)+S(r,f),
  
  that is,
  
  12T(r,f)≤T(r,Δqf)+S(r,f). (2.19)
  
  And in view of (1.6), we have
  
  f(qz)f(zq)f(z)=f(qz)f(zq)+f(z)−μ. (2.20)
  
  By Theorem 2.5 in [14], it yields
  
  m(r,f)=S(r,f).
  
  Thus, it follows from Theorem 1.1 in [14] that
  
  N(r,Δqf)=T(r,Δqf)−m(r,Δqf)≥T(r,Δqf)−(m(r,Δqff)+m(r,f))≥12T(r,f)+S(r,f).
  
  Hence, by combining with (2.10), we conclude that Δ_qf has infinitely many poles and λ(1Δqf)=σ(f) .
  
  Therefore, this completes the proof of Theorem 1.6 (ii).
By the process of the proof of Theorem 1.6 (ii), we have m(r, f) = S(r, f), this means N(r, f) = T(r, f) + S(r, f), that is, ∞ is not a Nevanlinna exceptional value of f(z).

Besides, set

P2(z,f(z)):=f(qz)f(zq)(f(z)−1)2−f(z)2+λf(z)−μ≡0.

Since μ ≠ 0, then it follows P₂(z, 0) = μ ≠ 0. Thus, in view of Theorem 2.5 in [14], we have

mr,1f=S(r,f),

which implies that 0 is not a Nevanlinna exceptional value of f.

Now, let β ∉ E, then it follow that

P2(z,β)=β4−2β3+λβ−μ≠0.

From Theorem 2.5 in [14], it yields mr,1f−β = S(r, f), which implies that β is not a Nevanlinna exceptional value of f(z).

Hence, the conclusion of Theorem 1.6 (iii) is true.

Therefore, this completes the proof of Theorem 1.6.

2.2 The proof of Theorem 1.7

By using the similar argument as in the proof of Theorem 1.6, it is easy to get the conclusions of Theorem 1.7.

3 Proofs of Theorems 1.8 and 1.9

3.1 The proof of Theorem 1.8

Similar to the argument as in the proof of Theorem 1.6, we can prove that τ(f(η z)) = σ(f) and Δqf,Δqff have infinitely many poles and λ(1Δqf)=λ(1Δqff). Now, we only need to prove that Δ_qf has infinitely many zeros and λ(Δ_qf) = σ(f). We rewrite (1.8) as the form

f(qz)f(zq)f(z)=f(qz)f(zq)+λ(z)f(z). (3.1)

So, it yields from Lemma Theorem 2.5 in [17] that

m(r,f)=S(r,f), (3.2)

which leads to N(r, f) = T(r, f) + S(r, f). This means that f(z) has infinitely many poles and λ(1f) = σ(f).

Besides, we can rewrite (1.8) again as the form

λ(z)f(z)=f(qz)f(z)f(zq)f(z)(f(z)−1). (3.3)

Thus, in view of (3.2) and (3.3), and by Theorem 1.1 in [14], we can deduce

mr,1f≤mr,f(qz)f(z)+mr,f(z)f(zq)+m(r,f)+m(r,1λ)=S(r,f), (3.4)

that is, N(r,1f) = T(r, f) + S(r, f), which implies that f(z) has infinitely many zeros and λ(f) = σ(f).

Since Δ_qf(z) = f(qz) − f(z) and Δ_q⁻¹f(z) = f(zq) − f(z), thus, substituting these into (1.8), it follows that

(f(z)+Δqf(z))(f(z)+Δq−1f(z))=λ(z)f(z)f(z)−1. (3.5)

Assume that z₀ is a zero of f(z), in view of (3.5), then z₀ must be a zero of Δ_qf or Δ_q⁻¹f. Thus, in view of (3.4) and Theorem 1.1 in [14], it follows

T(r,f)=Nr,1f+S(r,f)≤N(r,1Δqf)+N(r,1Δq−1f)+S(r,f)≤2N(r,1Δqf)+S(r,f),

which implies that Δ_qf has infinitely many zeros and λ(Δ_qf) ≥ σ(f). By combining with (2.10), we have λ(Δ_qf) = σ(f).

Hence, this completes the proof of Theorem 1.8 (i) and (ii).

(iii) In view of (3.2) and (3.4), we have δ(0, f) = 0 and δ(∞, f) = 0, it means that 0 and ∞ are not Nevanlinna exceptional values of f(z).

Set

P3(z,f):=f(qz)f(zq)(f(z)−1)−λ(z)f(z).

Then for any a ∈ C∖{0}, and since λ(z) is a nonconstant polynomial, we have P₃(z, a) = a²(a − 1) − aλ(z) ≢ 0. Thus, in view of Theorem 2.5 in [14], it yields that mr,1f−a = S(r, f), which means that δ(a, f) = 0. Hence, it follows that f(z) has no Nevanlinna exceptional value.

Therefore, this completes the proof of Theorem 1.8.

3.2 The proof of Theorem 1.9

By using the similar argument as in the proof of Theorem 1.8, we can get the conclusions of Theorem 1.9 (i), (iii), Δ_qf has infinitely many zeros and poles, and Δqff has infinitely many poles, and

λ(Δqf)=λ(1Δqf)=λ1Δqff=σ(f),

and f also has infinitely many poles and zeros, and λ(f) = λ(1f). Thus, we only need to prove that Δqff has infinitely many zeros and λ(Δqff)=σ(f) .

In view of (1.9), we have

f(q2z)f(z)f(qz)2=h(qz). (3.6)

Thus, from (1.9) and (3.6), it follows

f(q2z)f(qz)f(z)f(zq)f(qz)2f(z)2=h(qz)h(z). (3.7)

Denote g(z) = f(qz)f(z). Thus, we can rewrite (3.7) as

g(qz)g(zq)g(z)2=h(qz)h(z).

Set

P4(z,g)=g(qz)g(zq)g(z)2−h(qz)h(z).

Since h(z) is a nonconstant rational function and ∣q∣ ≠ 1, then P₄(z, 1) = 1 − h(qz)h(z) ≢ 0. Thus, in view of Theorem 2.5 in [14] and Theorems 1.1 and 1.3 in [25], it yields

mr,1g−1=S(r,g)=S(r,f(qz)f(z))≤S(r,f). (3.8)

Hence, it follows

mr,1Δqff=mr,1f(qz)f(z)−1=m(r,1g(z)−1)=S(r,f). (3.9)

Thus, by applying Theorem 1.1 in [14] and Valiron-Mohon’ko Lemma [30] for (1.9), it follows

4T(r,f)=Tr,h(z)f(z)4+O(log⁡r)=Tr,f(qz)f(z)f(z)f(zq)+O(1)≤2Tr,f(qz)f(z)+Sr,f(qz)f(z)+O(log⁡r)≤4T(r,f)+S(r,f),

that is,

T(r,Δqff)=T(r,f(qz)f(z))+O(1)=2T(r,f)+S(r,f). (3.10)

Hence, in view of (3.9) and (3.10), it follows

Nr,1Δqff=T(r,Δqff)−mr,1Δqff+O(1)=T(r,Δqff)+S(r,f)=2T(r,f)+S(r,f),

which means that Δqff has infinitely many zeros and λ(Δqff)=σ(f) .

Therefore, this completes the proof of Theorem 1.9.

Acknowledgements

This work was supported by the National Natural Science Foundation of China (11561033,11761035), the Natural Science Foundation of Jiangxi Province in China (20181BAB201001, 20171BAB201002, 20151BAB201008) and the Foundation of Education Department of Jiangxi (GJJ180734) of China.

We thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments which improved the presentation.

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Received: 2018-10-01

Accepted: 2019-07-14

Published Online: 2019-08-28

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Articles in the same Issue

https://doi.org/10.1515/math-2019-0079

Keywords for this article

fixed point; q-difference equation; meromorphic solution

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