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

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

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

• (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 $\deg B(z)=b$.

• (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

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

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

  • (2) $i-j=x-b$.

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

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

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

• (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 $\xi\mu_{0}\lambda_{0}+\pi\lambda_{0}^{2}=0$.