Theory and Modern Applications

# Some results of ϖ-Painlevé difference equation

## Abstract

In this article, we mainly investigate some properties of two types of difference equations

$$Y(\varpi z)+Y(z)+Y\biggl(\frac{z}{\varpi}\biggr)=\frac{\xi z+o}{Y(z)}+v$$

and

$$Y(\varpi z)+Y\biggl(\frac{z}{\varpi}\biggr)=\frac{\xi z+o}{Y(z)}+ \frac{v}{Y^{2}(z)}.$$

## 1 Introduction

Halburd and Korhonen [4] used Nevanlinna theory to single out difference equations in this form

$$Y(z+1) + Y(z-1) = R(z,Y),$$
(1.1)

where $$R(z,Y)$$ is rational in O and meromorphic in z, has an admissible meromorphic solution of finite order, then either O satisfies a difference Riccati equation

$$Y(z+1) = \frac{p(z+1)Y(z)+q(z)}{Y(z)+p(z)},$$
(1.2)

where $$p(z), q(z) \in S(Y)$$, where $$S(Y)$$ denotes the field of small functions with respect to Y, or Eq. (1.1) can be transformed to one of the following equations:

\begin{aligned}& Y(z+1)+Y(z)+Y(z-1)=\frac{\varsigma_{1}z+\varsigma_{2}}{Y(z)}+\kappa_{1}, \end{aligned}
(1.3)
\begin{aligned}& Y(z+1)-Y(z)+Y(z-1)=\frac{\varsigma_{1}z+\varsigma _{2}}{Y(z)}+(-1)^{z}\kappa_{1}, \end{aligned}
(1.4)
\begin{aligned}& Y(z+1)+Y(z-1)=\frac{\varsigma_{1}z+\varsigma_{3}}{Y(z)}+\varsigma_{2}, \end{aligned}
(1.5)
\begin{aligned}& Y(z+1)+Y(z-1)=\frac{\varsigma_{1}z+\kappa_{1}}{Y(z)}+\frac{\varsigma _{2}}{Y^{2}(z)}, \end{aligned}
(1.6)
\begin{aligned}& Y(z+1)+Y(z-1)=\frac{(\varsigma_{1}z+\kappa_{1})Y(z)+\varsigma _{2}}{(-1)^{-z}-Y^{2}(z)}, \end{aligned}
(1.7)
\begin{aligned}& Y(z+1)+Y(z-1)=\frac{(\varsigma_{1}z+\kappa_{1})Y(z)+\varsigma _{2}}{1-Y^{2}(z)}, \end{aligned}
(1.8)
\begin{aligned}& Y(z+1)Y(z)+Y(z)Y(z-1)=p, \end{aligned}
(1.9)
\begin{aligned}& Y(z+1)+Y(z-1)=pY(z)+q, \end{aligned}
(1.10)

where $$\varsigma_{k}, \kappa_{k} \in S(Y)$$ are arbitrary finite order periodic functions with period k.

Eqs. (1.3), (1.5), and (1.6) are known alternative forms of difference Painlevé I equation, Eq. (1.8) is a difference Painlevé II, and (1.9) and (1.10) are linear difference equations. Chen and Shon [2, 3] considered some value distribution problems of finite order meromorphic solutions of Eqs. (1.2), (1.5), (1.6), and (1.8). A natural question is: What is the result if we give q-difference analogues of (1.3) and (1.6)? For this question, we consider the following equations:

\begin{aligned}& Y(\varpi z)+Y(z)+Y\biggl(\frac{z}{\varpi}\biggr)=\frac{\xi z+o}{Y(z)}+v, \end{aligned}
(1.11)
\begin{aligned}& Y(\varpi z)+Y\biggl(\frac{z}{\varpi}\biggr)=\frac{\xi z+o}{Y(z)}+ \frac {v}{Y^{2}(z)}. \end{aligned}
(1.12)

### Theorem 1.1

Let $$Y(z)$$ be a transcendental meromorphic solution with zero order of Eq. (1.11) and ξ, o, v be three constants such that ξ, o cannot vanish simultaneously. Then

1. (i)

$$Y(z)$$ has infinitely many poles.

2. (ii)

For any finite value B, if $$\xi\neq0$$, then $$Y(z)-B$$ has infinitely many zeros.

3. (iii)

If $$\xi=0$$ and $$Y(z)-A$$ has finite zeros, then A is a solution of $$3z^{2}-o-vz=0$$.

We assume that the reader is familiar with the basic notions of Nevanlinna theory (see, e.g., [5, 6]).

### Theorem 1.2

Let $$c\in\mathbb{C}\setminus\{0\}$$, $$|\varpi|\neq1$$, and $$V(z)=\frac {X(z)}{B(z)}$$ be an irreducible rational function, where $$X(z)$$ and $$B(z)$$ are polynomials with $$\deg X(z)=x$$ and $$\deg B(z)=b$$.

1. (i)

Suppose that $$x\geq b$$ and $$x-b$$ is zero or an even number. If the equation

$$Y(\varpi z)+Y(z)+Y\biggl(\frac{z}{\varpi}\biggr)=\frac{V(z)}{Y(z)}+c$$
(1.13)

has an irreducible rational solution $$Y(z)=\frac{I(z)}{J(z)}$$, where $$i(z)$$ and $$J(z)$$ are polynomials with $$\deg i(z)=i$$ and $$\deg J(z)=j$$, then

$$i-j=\frac{x-b}{2}.$$
2. (ii)

Suppose that $$x< b$$. If Eq. (1.13) has an irreducible rational solution $$Y(z)=\frac{i(z)}{J(z)}$$, then $$Y(z)$$ satisfies one of the following two cases:

1. (1)

$$Y(z)=\frac{i(z)}{J(z)}=\frac{c}{3}+\frac{T(z)}{D(z)}$$, where $$T(z)$$ and $$D(z)$$ are polynomials with $$\deg T(z)=t$$ and $$\deg D(z)=d$$, and $$b-x=d-t$$.

2. (2)

$$i-j=x-b$$.

### Theorem 1.3

Let $$Y(z)$$ be a transcendental meromorphic solution with zero order of Eq. (1.12) and ξ, o, v be three constants such that ξ, o cannot vanish simultaneously. Then

1. (i)

$$Y(z)$$ has infinitely many poles.

2. (ii)

For any finite value B, if $$\xi\neq0$$ and $$v\neq0$$, then $$Y(z)-B$$ has infinitely many zeros.

3. (iii)

If $$\xi=0$$ and $$Y(z)-A$$ has finite zeros, then A is a solution of $$2z^{2}-oz-v=0$$.

### Theorem 1.4

Let ξ, o, π be constants with $$\xi\pi\neq0$$ and $$|\varpi|\neq 1$$. Suppose that a rational function

$$Y(z)=\frac{F(z)}{U(z)}=\frac{\mu_{0}z^{m}+\mu_{1}z^{m-1}+\cdots+\mu _{m}}{\lambda_{0}z^{n}+\lambda_{1}z^{n-1}+\cdots+\lambda_{n}}$$

is a solution of (1.12), where $$F(z)$$ and $$U(z)$$ are relatively prime polynomials, $$\mu_{0}\neq0$$, $$\mu_{1},\ldots, \mu_{m}$$, and $$\lambda _{0}\neq0$$, $$\lambda_{1},\ldots, \lambda_{n}$$ are constants. Then $$n=m+1$$ and $$\mu_{0}=-\frac{\pi}{\xi}\lambda_{0}$$.

## 2 Some lemmas

### Lemma 2.1

([1])

Let $$Y(z)$$ be a non-constant zero order meromorphic solution of

$$Y(z)^{n}P(z, Y)=Q(z, Y),$$

where $$P(z, Y)$$ and $$Q(z, Y)$$ are ϖ-difference polynomials in $$Y(z)$$. If the degree of $$Q(z, Y)$$ as a polynomial in $$Y(z)$$ and its ϖ-shifts is at most n, then

$$m\bigl(r, P(z, Y)\bigr)=o\bigl(T(r, Y)\bigr)$$

on a set of logarithmic density 1.

### Lemma 2.2

([1])

Let $$Y(z)$$ be a non-constant zero order meromorphic solution of

$$H(z, Y)=0,$$

where $$H(z,O)$$ is a ϖ-difference polynomial in $$Y(z)$$. If $$H(z, Y)\not\equiv0$$ for a slowly moving target $$a(z)$$, then

$$m\biggl(r, \frac{1}{Y-a}\biggr)=o\bigl(T(r, Y)\bigr)$$

on a set of logarithmic density 1.

### Lemma 2.3

([7])

Let $$Y(z)$$ be a zero order meromorphic function, and $$\varpi\in\mathbb {C}\setminus\{0\}$$. Then

\begin{aligned}& T\bigl(r, Y(\varpi z)\bigr)=\bigl(1+o(1)\bigr)T\bigl(r, Y(z)\bigr); \\& N\bigl(r, Y(\varpi z)\bigr)=\bigl(1+o(1)\bigr)N\bigl(r, Y(z)\bigr). \end{aligned}

## 3 Proof of Theorem 1.1

(i): Suppose that $$Y(z)$$ is a zero order transcendental meromorphic solution of (1.11). By (1.11), we have

$$Y(z)P(z, Y)=Q(z, Y),$$
(3.1)

where $$P(z, Y)=Y(\varpi z)+Y(z)+Y(\frac{z}{\varpi})$$, $$Q(z, Y)=\xi z+o+vY(z)$$. Lemma 2.1 implies that

$$m\bigl(r, P(z, Y)\bigr)=o\bigl(T(r, Y)\bigr)$$
(3.2)

on a set of logarithmic density 1. By the Valiron–Mohon’ko theorem, we obtain that

$$T\biggl(r, Y(\varpi z)+Y(z)+Y\biggl(\frac{z}{\varpi}\biggr)\biggr)=T(r, Y)+S(r, Y).$$
(3.3)

By Lemma 2.3, we obtain

\begin{aligned}[b] N\biggl(r, Y(\varpi z)+Y(z)+Y\biggl(\frac{z}{\varpi}\biggr)\biggr)&\leq N \bigl(r, Y(\varpi z)\bigr)+N(r, Y)+N\biggl(r, Y\biggl(\frac{z}{\varpi}\biggr) \biggr)\\ &=3\bigl(1+o(1)\bigr)N(r, Y). \end{aligned}
(3.4)

(3.2), (3.3), and (3.4) imply that

$$T(r, Y)\leq3\bigl(1+o(1)\bigr)N(r, Y)+S(r, Y)$$
(3.5)

on a set of logarithmic density 1. Hence, $$Y(z)$$ has infinitely many poles.

(ii): For any finite value B, and let

$$Y_{1}(z)=Y(z)-B.$$

Substituting $$Y_{1}(z)=Y(z)-B$$ into (3.1), we obtain

$$\bigl(Y_{1}(z)+B\bigr) \biggl(Y_{1}(\varpi z)+Y_{1}(z)+Y_{1}\biggl(\frac{z}{\varpi}\biggr)+3B\biggr)= \xi z+o+v\bigl(Y_{1}(z)+B\bigr).$$

Let

$$P\bigl(z, Y_{1}(z)\bigr)=\bigl(Y_{1}(z)+B\bigr) \biggl(Y_{1}(\varpi z)+Y_{1}(z)+Y_{1}\biggl( \frac{z}{\varpi }\biggr)+3B\biggr)-\xi z-o-v\bigl(Y_{1}(z)+B \bigr).$$
(3.6)

If $$\xi\neq0$$, by (3.6), we have $$P(z, 0)=3B^{2}-\xi z-o-vB\not\equiv0$$. Lemma 2.2 implies that

$$m\biggl(r, \frac{1}{Y_{1}}\biggr)=o\bigl(T(r, Y_{1})\bigr)$$

on a set of logarithmic density 1. Hence

$$N\biggl(r, \frac{1}{Y-B}\biggr)=N\biggl(r, \frac{1}{Y_{1}}\biggr)=T(r, Y_{1}) \bigl(1+o(1)\bigr)=T(r, Y) \bigl(1+o(1)\bigr)$$

on a set of logarithmic density 1. Hence, $$Y(z)$$ has infinitely many finite values.

(iii): If $$\xi=0$$ and B is not a solution of $$3z^{2}-o-vz=0$$, then $$P(z, 0)=3B^{2}-o-vB\not\equiv0$$. Using a similar method as above, we can obtain that

$$N\biggl(r, \frac{1}{Y-B}\biggr)=T(r, Y) \bigl(1+o(1)\bigr),$$

which contradicts the assumption of Theorem 1.1, hence the conclusion holds.

## 4 Proof of Theorem 1.2

By (1.13) and $$Y(z)=\frac{I(z)}{J(z)}$$, we have

$$\begin{gathered}[b] B(z)I(z)I(\varpi z)J(z)J\biggl(\frac{z}{\varpi} \biggr)+B(z)I^{2}(z)J\biggl(\frac {z}{\varpi}\biggr)J(\varpi z) +B(z)I(z)I\biggl(\frac{z}{\varpi}\biggr)J(\varpi z)J(z) \\ \quad {}-c B(z)I(z)J(\varpi z)J\biggl(\frac{z}{\varpi}\biggr)J(z)=X(z)J(\varpi z)J\biggl(\frac {z}{\varpi}\biggr)J^{2}(z). \end{gathered}$$
(4.1)

Obviously, we have

\begin{aligned}& \begin{aligned}[b] &\deg\biggl(B(z)I(z)I(\varpi z)J(z)J\biggl(\frac{z}{\varpi} \biggr)+B(z)I^{2}(z)J\biggl(\frac {z}{\varpi}\biggr)J(\varpi z) \\ &\quad{}+B(z)I(z)I\biggl(\frac{z}{\varpi}\biggr)J(\varpi z)J(z)\biggr) =b+2i+2j; \end{aligned} \end{aligned}
(4.2)
\begin{aligned}& \deg\biggl(c B(z)I(z)J(\varpi z)J\biggl(\frac{z}{\varpi}\biggr)J(z) \biggr)=b+i+3j; \end{aligned}
(4.3)
\begin{aligned}& \deg\biggl(X(z)J(\varpi z)J\biggl(\frac{z}{\varpi}\biggr)J^{2}(z) \biggr)=x+4j. \end{aligned}
(4.4)

(i): Suppose first that $$x>b$$ and $$x-b$$ is an even number. If $$\deg i(z)=i< j=\deg J(z)$$, then (4.1)–(4.4) imply that $$x+4j=b+i+3j$$, that is, $$0>i-j=x-b>0$$. This is impossible.

If $$i=j$$, then we use a similar method as above, we can obtain $$0< x-b=i-j=0$$, this is impossible. So, $$i>j$$. By (4.1), we have $$x+4j=b+2i+2j$$, that is, $$i-j=\frac {x-b}{2}$$.

(ii): Suppose that $$x< b$$. If $$i>j$$, then (4.1)–(4.4) yield that $$x+4j=b+2i+2j$$, that is, $$0>x-b=2(i-j)>0$$, which is a contradiction.

If $$i=j$$, then we can assume

$$Y(z)=\iota_{0}+\frac{T(z)}{D(z)},$$
(4.5)

where $$\iota_{0} \neq0$$, $$T(z)$$ and $$D(z)$$ are polynomials, and $$\deg T(z)=t<\deg D(z)=d$$. Thus, as $$z\rightarrow\infty$$, (1.13) and (4.5) imply that

$$3\iota_{0}\bigl(1+o(1)\bigr)=\frac{o(1)}{\iota_{0}(1+o(1))}+c,$$
(4.6)

which implies $$\iota_{0}=\frac{c}{3}$$. Hence,

$$Y(z)=\frac{c}{3}+\frac{T(z)}{D(z)}.$$
(4.7)

Substituting (4.7) into (1.13), we get

$$\begin{gathered} B(z) \biggl(\frac{c}{3} D(z)+T(z)\biggr) \biggl(T(z)D( \varpi z)D\biggl(\frac{z}{\varpi }\biggr)+T(\varpi z)D(z)D\biggl( \frac{z}{\varpi}\biggr) \\ \quad{}+T\biggl(\frac{z}{\varpi}\biggr)D(z)D(\varpi z)\biggr) =X(z)D^{2}(z)D(\varpi z)D\biggl(\frac{z}{\varpi}\biggr). \end{gathered}$$

Obviously,

$$\begin{gathered} \deg\biggl[B(z) \biggl(\frac{c}{3} D(z)+T(z)\biggr) \biggl(T(z)D(\varpi z)D\biggl(\frac{z}{\varpi }\biggr)+T(\varpi z)D(z)D\biggl( \frac{z}{\varpi}\biggr)\\ \quad {}+T\biggl(\frac{z}{\varpi}\biggr)D(z)D(\varpi z)\biggr) \biggr]=3d+b+t; \\ \deg X(z)D^{2}(z)D(\varpi z)D\biggl(\frac{z}{\varpi}\biggr)=x+4d. \end{gathered}$$

Hence, $$b-x=d-t$$.

If $$i< j$$, by $$i< j$$, $$x< b$$, (4.1)–(4.4), then we have

$$i-j=x-b.$$

## 5 Proof of Theorem 1.3

(i). Suppose that $$Y(z)$$ is a zero order transcendental meromorphic solution of (1.12). By (1.12), we have

$$Y^{2}(z) \biggl(Y(\varpi z)+Y\biggl(\frac{z}{\varpi}\biggr)\biggr)=( \xi z+o)Y+v.$$
(5.1)

Lemma 2.1 implies that

$$m\biggl(r, Y(\varpi z)+Y\biggl(\frac{z}{\varpi}\biggr)\biggr)=o\bigl(T(r, Y) \bigr)$$
(5.2)

on a set of logarithmic density 1. By the Valiron–Mohon’ko theorem, we get that

$$T\biggl(r, Y(\varpi z)+Y\biggl(\frac{z}{\varpi}\biggr)\biggr)=T(r, Y)+S(r, Y).$$
(5.3)

By Lemma 2.3, we obtain

$$N\biggl(r, Y(\varpi z)+Y\biggl(\frac{z}{\varpi}\biggr)\biggr)\leq N\bigl(r, Y( \varpi z)\bigr)+N\biggl(r, Y\biggl(\frac {z}{\varpi}\biggr)\biggr)=2\bigl(1+o(1) \bigr)N(r, Y).$$
(5.4)

(5.2), (5.3), and (5.4) yield that

$$T(r, Y)\leq2\bigl(1+o(1)\bigr)N(r, Y)+S(r, Y)$$
(5.5)

on a set of logarithmic density 1. Hence, $$Y(z)$$ has infinitely many poles.

(ii). For any finite value B, let

$$Y_{1}(z)=Y(z)-B.$$

Substituting $$Y_{1}(z)=Y(z)-B$$ into (5.1), we obtain

$$\bigl(Y_{1}(z)+B\bigr)^{2}\biggl(Y_{1}(\varpi z)+Y_{1}\biggl(\frac{z}{\varpi}\biggr)+2B\biggr)=(\xi z+o) \bigl(Y_{1}(z)+B\bigr)+v.$$

Let

$$P\bigl(z, Y_{1}(z)\bigr)=\bigl(Y_{1}(z)+B \bigr)^{2}\biggl(Y(\varpi z)+Y\biggl(\frac{z}{\varpi}\biggr)+2B \biggr)-(\xi z+o) \bigl(Y_{1}(z)+B\bigr)-v.$$
(5.6)

By (5.6), we have $$P(z, 0)=2B^{2}-(\xi z+o)B-\pi$$.

If $$B=0$$ and $$\pi\neq0$$, then we obtain that $$P(z, 0)=-v\not\equiv0$$.

If $$B\neq0$$, then we have $$P(z, 0)=2B^{2}-(\xi z+o)B-v\not\equiv0$$ since $$\xi\neq0$$. Using a method similar to Theorem 1.1, we can obtain that

$$N\biggl(r, \frac{1}{Y-A}\biggr)=N\biggl(r, \frac{1}{Y_{1}}\biggr)=T(r, Y_{1}) \bigl(1+o(1)\bigr)=T(r, Y) \bigl(1+o(1)\bigr)$$

on a set of logarithmic density 1. Hence, $$Y(z)$$ has infinitely many finite values.

(iii). If $$\xi=0$$ and A is not a solution of $$2z^{2}-oz-\pi=0$$, then using a method similar to Theorem 1.1, we also obtain that

$$N\biggl(r, \frac{1}{Y-A}\biggr)=T(r, Y) \bigl(1+o(1)\bigr),$$

which contradicts the assumption of Theorem 1.3, hence the conclusion holds.

## 6 Proof of Theorem 1.4

Assume that (1.12) has a rational solution $$Y(z)$$ and has poles $$t_{1}, t_{2},\ldots,t_{k}$$. Next, let

$$\frac{c_{j\lambda_{j}}}{(z-t_{j})^{\lambda_{j}}}+\cdots+\frac {c_{j_{1}}}{(z-t_{j})}\quad (j=1,\ldots,k)$$

be the principal parts of Y at $$t_{j}$$, respectively, where $$c_{j\lambda _{j}},\ldots,c_{j1}$$ are constants, $$c_{j\lambda_{j}}\neq0$$. Hence

$$Y(z)=\frac{F(z)}{U(z)}=\sum^{k}_{j=1} \biggl[\frac{c_{j\lambda _{j}}}{(z-t_{j})^{\lambda_{j}}}+\cdots+\frac{c_{j_{1}}}{(z-t_{j})}\biggr]+\tau_{0}+ \tau _{1}z+\cdots+\tau_{s}z^{s},$$
(6.1)

where $$\tau_{0},\ldots,\tau_{s}$$ are constants. Assume that $$\tau_{s}\neq0$$ ($$s\geq1$$). When $$z\rightarrow\infty$$,

\begin{aligned}& Y(z)=\tau_{s} z^{s}\bigl(1+o(1)\bigr), \qquad Y(\varpi z)= \varpi^{s}\tau _{s}z^{s}\bigl(1+o(1) \bigr), \end{aligned}
(6.2)
\begin{aligned}& Y\biggl(\frac{z}{\varpi}\biggr)=\frac{1}{\varpi^{s}}\tau_{s}z^{s} \bigl(1+o(1)\bigr). \end{aligned}
(6.3)

By (1.12), we have

$$Y^{2}(z) \biggl(Y(\varpi z)+Y\biggl(\frac{z}{\varpi}\biggr)\biggr)=( \xi z+o)Y+v.$$
(6.4)

When $$z\rightarrow\infty$$, (6.2), (6.3), and (6.4) imply that

$$\biggl(\varpi^{s}+\frac{1}{\varpi^{s}}\biggr)\tau_{s}^{3} z^{3s}\bigl(1+o(1)\bigr)=(\xi z+o) \bigl(\tau_{s} z^{s}\bigl(1+o(1)\bigr)\bigr)+v,$$

which is a contradiction since $$\tau_{s}\neq0$$ and $$s\geq1$$. Assume that $$\tau_{0}\neq0$$, as $$z\rightarrow\infty$$,

\begin{aligned}& Y(z)=\tau_{0} \bigl(1+o(1)\bigr),\qquad Y(\varpi z)= \tau_{0}\bigl(1+o(1)\bigr), \end{aligned}
(6.5)
\begin{aligned}& Y\biggl(\frac{z}{\varpi}\biggr)=\tau_{0}\bigl(1+o(1) \bigr). \end{aligned}
(6.6)

By (6.4) together with (6.5) and (6.6), we obtain that

$$(\xi z+o) \bigl(\tau_{0}\bigl(1+o(1)\bigr)\bigr)=2 \tau_{s}^{3}-\pi.$$

This is impossible since $$\xi\neq0$$ and $$\tau_{0}\neq0$$. Hence

$$Y(z)=\frac{F(z)}{U(z)},$$
(6.7)

where $$\deg F(z)=m<\deg U(z)=n$$. Equation (6.7) and (1.12) imply that

\begin{aligned} &F^{2}(z)F(\varpi z)U\biggl(\frac{z}{\varpi}\biggr)+F^{2}(z)F \biggl(\frac{z}{\varpi }\biggr)U(\varpi z)\\ &\quad =(\xi z+o)F(z)U(z)U(\varpi z)U\biggl( \frac{z}{\varpi}\biggr) +vU^{2}(z)U(\varpi z)U\biggl(\frac{z}{\varpi} \biggr). \end{aligned}

Hence we have $$n=m+1$$. By (1.12) and $$n=m+1$$, we obtain

$$\frac{F(\varpi z)}{U(\varpi z)}+\frac{F(\frac{z}{\varpi})}{U(\frac {z}{\varpi})}=\frac{(\xi z+o)F(z)U(z)+vU^{2}(z)}{F^{2}(z)}.$$

Since as $$z\rightarrow\infty$$, we have

$$\frac{F(\varpi z)}{U(\varpi z)}+\frac{F(\frac{z}{\varpi})}{U(\frac {z}{\varpi})}\rightarrow0;$$

and

$$\frac{(\xi z+o)F(z)U(z)+vU^{2}(z)}{F^{2}(z)}=\frac{(\xi\mu_{0}\lambda _{0}+v\lambda_{0}^{2})z^{2n}(1+o(1))}{\mu_{0}^{2}z^{2n-2}(1+o(1))},$$

we obtain $$\xi\mu_{0}\lambda_{0}+\pi\lambda_{0}^{2}=0$$.

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## Funding

The work was supported by the NNSF of China (No. 10771121, 11301220, 11401387, 11661052), the NSF of Zhejiang Province, China (No. LQ 14A010007), the NSF of Shandong Province, China (No. ZR2012AQ020), and the Fund of Doctoral Program Research of Shaoxing College of Art and Science (20135018).

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Liu, Y., Zhang, Y. Some results of ϖ-Painlevé difference equation. Adv Differ Equ 2018, 282 (2018). https://doi.org/10.1186/s13662-018-1710-z