The uniqueness of meromorphic function shared values with meromorphic solutions of a class of q-difference equations

1.   Introduction

Throughout this paper, a meromorphic function will always mean meromorphic in the whole complex plane. In what follows, we assume that the reader is familiar with the fundamental concepts of Nevanlinna's value distribution theory ^[5,8,10].

Let $f$ and $g$ be meromorphic functions and $a$ be a complex number. If $f-a$ and $g-a$ have the same zeros with the same multiplicities, then we say that $f(z)$ and $g(z)$ share $a$ counting multiplicity (CM). If $f-a$ and $g-a$ have the same zeros (ignoring multiplicity), then $f(z)$ and $g(z)$ share $a$ IM. If $f(z)$ and $g(z)$ have the same poles (CM), then $f(z)$ and $g(z)$ share $\infty$ CM. In this paper, we suppose that $f(z)$ shares the value $a$ partially with $g(z)$, and that $\overline {N}_{(m, n)}(r, a)$ denotes the reduced counting function of those zeros of $f(z)-a$ with multiplicity $k$, and of $g(z)-a$ with multiplicity $l$ in $\{z: |z| < r\}$.

Definition 1. Let $f(z)$ be a meromorphic function in the complex plane. The order $\rho$ and lower order $\mu$ of $f(z)$ are defined respectively by the order of $T(r, f)$, that is,

Definition 2. If the meromorphic function $a(z)(\not\equiv\infty)$ is satisfied, it follows that

where $E\subset[0, \infty)$ is a set of real numbers with finite measures, that is,

then, $a$ is called a small function of $f(z)$.

Definition 3. Let $f(z)$ be meromorphic in the complex plane. If the order and the lower order of $f(z)$ are equal, then $f(z)$ is called a function with normal growth.

In recent years, with the research and development of the theory of difference equations, the value distribution and uniqueness of meromorphic solutions in complex domain difference equations (see ^[2,3,6]) has gradually became a hot research direction in the field of complex analysis. In $2017$, Cui Ning and Chen Zongxuan^[4] considered the problem that meromorphic solutions of a class of linear difference equations share values with arbitrary meromorphic function, and they obtained the following results.

Theorem 1.1. (see ^[4]) Let $a_1(z)$, $a_0(z)$, and $F(z)$ be non-constant polynomials satisfying that $a_1(z)+a_0(z)\not\equiv0$; $f(z)$ is the finite-order transcendental meromorphic solutions of the following difference equation:

If meromorphic functions $g(z)$ and $f(z)$ share $0$, $1$, $\infty$ CM, then one of the following situations must occur:

(i) $f(z)\equiv g(z)$;

(ii) $f(z)+g(z)\equiv f(z)g(z)$;

(iii) There is a polynomial $\beta(z) = az+b_0$ and a constant $a_0$ satisfying that $e^{a_0}\neq e^{b_0}$; then, $f(z) = \frac{1-e^{\beta(z)}}{e^{\beta(z)}(e^{a_0-b_0}-1)}$ and $g(z) = \frac{1-e^{\beta(z)}}{1-e^{b_0-a_0}}$, where $a(\neq0)$, $b_0$ are constants.

Then, Yang Yin and Ye Yasheng^[9] studied the problem for the solutions of $q$-difference equations with shared values for any meromorphic function.

Theorem 1.2. (see ^[9]) Let $a_1(z)$, $a_0(z)$, and $F(z)$ be non-zero meromorphic functions whose order is less than $1$; $f(z)$ is a finite transcendental meromorphic solution of the following difference equation:

where $p$, $q$ are constants, $n\in\mathbb{N^+}$, $q^n\neq\pm1$, and $q\neq0$. If $g(z)$ is any meromorphic function that shares $0$, $1$, $\infty$ CM with $f(z)$, then $f\equiv g$.

Based on the conclusion of the first-order $q$-difference equations of Theorems 1.1 and 1.2, we naturally consider the existence of the meromorphic solutions of the second-order $q$-difference equations with the meromorphic coefficients, as well as the uniqueness of the meromorphic solutions with any non-constant meromorphic function with shared values; it is a difficult and interesting problem. Therefore, we put forward the following question:

Question 1. For the second order homogeneous $q$-difference equation:

with the meromorphic functions as coefficients. Are there uniqueness conclusions when meromorphic solutions share values with a non-constant meromorphic function?

The following result is obtained.

Theorem 1.3. Let $a_1(z)$ and $a_0(z)$ be non-zero meromorphic functions whose order is less than $1$; $f$ is a transcendental meromorphic solution of finite order of (1.1), where $q\not\in E$ and the set $E$ satisfies that $E = \{q|A_4q^{4n}+A_3q^{3n}+A_2q^{2n}+A_1q^n+A_0 = 0$, $n\in\mathbf{N^+}$, $A_j(j = 0, 1, 2, 3, 4)\in\mathbb{Z}$ and $|A_4|\leq2$, $|A_3|\leq4$, $|A_2|\leq6$, $|A_1|\leq4$, $|A_0|\leq2\}$. If $g$ is any non-constant meromorphic function that shares $0$, $1$, $\infty$ CM with $f$, then $f\equiv g$.

Furthermore, based on the above study of meromorphic solutions and meromorphic functions with shared values, we naturally have the following question:

Question 2. Do meromorphic solutions and meromorphic functions of higher-order $q$-difference equations have the same uniqueness conclusions with shared values?

In $1998$, Bergweiler et al. (see ^[1]) studied the meromorphic solution existence of a class of non-homogeneous $q$-difference equations described by

where $0 < |q| < 1$ is a complex number, $a_j(z)$, $j = 0, 1, \cdots, n$; $Q(z)$ denotes rational functions, and $a_0(z)\not\equiv0$, $a_n (z)\equiv1$.

For the study of Question 2, we obtain the following results.

Theorem 1.4. Let $f$ be a meromorphic solution of (1.2). If $g$ is any non-constant meromorphic function that shares $0$, $1$, $c \; (c\neq0, 1)$ IM and $\infty$ CM with $f$, then $f\equiv g$.

Remark 1.5. The number of "3IM+1CM" shared values in Theorem 1.4 is accurate.

Example 1.6. For the meromorphic solution of $q$-difference equation (1.2) to $f(z) = z$, for ${g(z) = \frac{2z}{1+z^2}}$, $f$ and $g$ share $1$, $-1$ IM and $0$ CM, but $f(z)\not\equiv g(z)$.

2.   Some lemmas

In order to prove our main results, we shall recall some lemmas as follows.

Lemma 2.1. (see ^[10], p. 65) Let $h(z)$ be a non-constant entire function and $f(z) = e^{h(z)}$. Let $\rho$ and $\mu$ be the order and the lower order of $f(z)$, respectively. We have the following:

(i) If $h(z)$ is a polynomial of degree $p$, then $\rho = \mu = p$.

(ii) If $h(z)$ is a transcendental entire function, then $\rho = \mu = \infty$.

Lemma 2.2. (see ^[10], p. 75) Suppose that $f_1(z)$, $f_2(z), \cdots, f_n(z) \; (n\geq 2)$ are meromorphic functions and $g_1(z)$, $g_2(z), \cdots, g_n(z)$ are entire functions satisfying the following conditions:

(i) ${\sum\limits_{j = 1}^n f_j(z)e^{g_j(z)}\equiv0}$.

(ii) $g_j(z)-g_k(z)$ are not constants for $1\leq j < k\leq n$.

(iii) For $1\leq j\leq n$, $1\leq h < k\leq n$,

Then $f_j(z)\equiv0(j = 1, 2, \cdots, n)$.

Lemma 2.3. (see ^[11], Theorem A) Suppose that $f_i \; (i = 1, 2, \cdots, n, n+1, n+2, \cdots, n+m)$ represents meromorphic functions, where $f_j \; (j = 1, 2, \cdots, n)$ is not constant, $f_k\not\equiv0 \; (k = n+1, n+2, \cdots, n+m)$, and

If

where $\lambda < 1$ and $m$ is an arbitrary positive integer, then there is $t_i\in\{0, 1\} \; (i = 1, 2, \cdots, m)$ such that

Lemma 2.4. Let

where $l_i$, $m_i$, $u_i$, $v_i\in\{-1, 0, 1\}(i = 1, 2, 3)$, $q$ is a non-zero constant, and $\alpha(z)$, $\beta(z)$ are polynomials with the degree $n(\geq1)$. If $f_1(z)$, $f_2(z)$ are constants, then $q\in E$, where $E = \{q|A_4q^{4n}+A_3q^{3n}+A_2q^{2n}+A_1q^n+A_0 = 0$, $n\in\mathbf{N^+}$, $A_j(j = 0, 1, 2, 3, 4)\in\mathbb{Z}$ and $|A_4|\leq2$, $|A_3|\leq4$, $|A_2|\leq6$, $|A_1|\leq4$, $|A_0|\leq2\}$.

Proof. If $f_1(z)$, $f_2(z)$ are constant functions, then the polynomials of $l_1\alpha(q^2z)+l_2\alpha(qz)+l_3\alpha(z)+m_1\beta(q^2z)+m_2\beta(qz) +m_3\beta(z)$, $u_1\alpha(q^2z)+u_2\alpha(qz)+u_3\alpha(z)+v_1\beta(q^2z) +v_2\beta(qz)+v_3\beta(z)$ have only constant terms; then,

where $l_i$, $m_i$, $u_i$, $v_i\in\{-1, 0, 1\}$.

As a result of $\alpha(z)$, $\beta(z)$ being polynomials, that is, $a_{n}\neq0$ and $b_{n}\neq0$, it holds that

so we have

Namely

where $A_j \; (j = 0, 1, 2, 3, 4)\in\mathbb{Z}$ and $|A_4|\leq2$, $|A_3|\leq4$, $|A_2|\leq6$, $|A_1|\leq4$, $|A_0|\leq2$.

This completes the proof of Lemma 2.4. □

Lemma 2.5. (see ^[10], p. 220) Let $f$ and $g$ be non-constant meromorphic functions that share four distinct values $a_j \; (j = 1, 2, 3, 4)$ IM. If $f(z)\not\equiv g(z)$, then the following holds:

(i) $T(r, f) = T(r, g)+O(\log r)$, $T(r, g) = T(r, f)+O(\log r)$;

(ii) ${\sum\limits_{j = 1}^4 \overline {N}(r, \frac{1}{f-a_j}) = 2T(r, f)+O(\log r)}$;

(iii) ${\overline {N}(r, \frac{1}{f-b}) = T(r, f)+O(\log r)}$, ${\overline {N}(r, \frac{1}{g-b}) = T(r, g)+O(\log r)}$, where $b\neq a_j \; (j = 1, 2, 3, 4)$;

(iv) ${N_0(r, \frac{1}{f'}) = O(\log r)}$, ${N_0(r, \frac{1}{g'}) = O(\log r)}$, where ${N_0(r, \frac{1}{f'})}$ is the counting function of the zeros of $f'$ but not the zeros of the $f-a_j \; (j = 1, 2, 3, 4)$; the notation ${N_0(r, \frac{1}{g'})}$ can be similarly defined;

(v) ${\sum\limits_{j = 1}^4{N}^{*}(r, a_j) = O(\log r)}$, where ${N}^{*}(r, a_j)$ is the counting function of the multiple common zeros of $f-a_j$ and $g-a_j$, which counts multiplicities according to minor one.

Lemma 2.6. (see ^[1], Theorems 1.1, 1.2) Let $f$ be a meromorphic solution of the following $q$-difference equation:

where $0 < |q| < 1$ is a complex number and $a_j(z)$, $j = 0, 1, \cdots, n$ and $Q(z)$ are rational functions with $a_0(z)\not\equiv0$, $a_n(z)\equiv1$. Then, the following holds:

(i) All meromorphic solutions of the equation satisfy that $T(r, f) = O(\log^2r)$;

(ii) All transcendental meromorphic solutions of the equation satisfy that $\log^2r = O(T(r, f))$.

Lemma 2.7. (see ^[10], p. 110) Let $f$ and $g$ be non-constant rational functions. If $f$ and $g$ share distinct $a_1$, $a_2$, $a_3$ and $a_4$ IM, then $f\equiv g$.

Lemma 2.8. (see ^[10], p. 30) Suppose that $f(z)$ and $g(z)$ are two non-constant meromorphic functions in the complex plane with $\rho(f)$ and $\rho(g)$ as their orders, respectively. Then

and

Lemma 2.9. (see ^[10], p. 28) Let $f(z)$ be a non-constant meromorphic function in the complex plane and ${R(f) = \frac{P(f)}{Q(f)}}$, where

are two mutually prime polynomials in $f$. If the coefficients $\{a_k(z)\}$, $\{b_j(z)\}$ are small functions of $f$ and $a_p(z)\not\equiv 0$, $b_q(z)\not\equiv 0$, then

3.   Proof of Theorem 1.3

The idea of proving this theorem is mainly derived from literature^[4].

Since $f(z)$ and $g(z)$ share $0$, $1$, $\infty$ CM, by using the Nevanlinna second fundamental theorem, we have

Similarly, we can get that $T(r, f)\leq3T(r, g)+S(r, f)$.

By the definition of order, we arrive at

Similarly, we have that $\rho\left(f\right)\leq\rho(g)$, that is $\rho(f) = \rho(g)$.

Since $f(z)$ is a finite-order meromorphic function, $g(z)$ is also a finite-order meromorphic function.

Suppose that

where $\alpha(z)$ and $\beta(z)$ are two polynomials.

If $e^{\alpha(z)}\equiv e^{\beta(z)}$, from (3.1), we have that $f(z)\equiv g(z)$.

Suppose, on the contrary, that $e^{\alpha(z)}\not\equiv e^{\beta(z)}$; then from (3.1) we have

If $\alpha(z)$ and $\beta(z)$ are both constants, by (3.2), we have that $f(z)$ is a constant, which is a contradiction with $f(z)$ being a transcendental function.

Now, assume that at least one of the functions among $\alpha(z)$ and $\beta(z)$ is not a constant, and discuss it in three cases:

Case 1. Let $\alpha(z)$ be a constant and $\beta(z)$ be a non-constant polynomial. Sign $e^{\alpha} = c_1(\neq0)$ is a constant; thus (3.2) can be rewritten as

If $c_1 = 1$, from (3.1), we have that $f(z)\equiv g(z)$.

If $c_1\neq1$, substituting (3.3) into (1.1), we have

therefore,

hence,

That is,

where $h_0(z) = 0$ and

Since we have $e^{\beta(z)}$ with normal growth, and by Lemma 2.1 we have that $\rho(e^{\beta(z)})$ is the degree of $\beta(z)$. Because $\beta(z)$ is a non-constant polynomial; hence, $\rho(e^{\beta(z)}) = \deg\beta(z)\geq 1$, which holds for other exponential function terms.

On the other hand, we have that $\rho(a_1(z)) < 1$, $\rho(a_0(z)) < 1$ and $c_1$ is a constant. So for $j = 1, 2, \cdots, 8$, we have

Thus, applying Lemma 2.2 to (3.4), we get that $B_{1j}(z)\equiv 0 \; (j = 1, 2, \cdots, 8)$. So,

we have that $c_1 = 1$, which is a contradiction.

Case 2. $\beta(z)$ is a constant and $\alpha(z)$ is a non-constant polynomial. Let $e^{\beta} = c_2(\neq0)$ be a constant; thus (3.2) can be rewritten as

If $c_2 = 1$, from (3.1), we have that $f(z)\equiv g(z)$.

If $c_2\neq 1$, substituting (3.5) into (1.1), we have

therefore,

hence,

that is,

where $h_0(z) = 0$ and

Since we have $e^{\alpha(z)}$ with normal growth, and by Lemma 2.1, we have that $\rho(e^{\alpha(z)})$ is the degree of $\alpha(z)$. Because $\alpha(z)$ is a non-constant polynomial, $\rho(e^{\alpha(z)}) = \deg\alpha(z)\geq 1$, which holds for other exponential function terms.

On the other hand, we have that $\rho(a_1(z)) < 1$, $\rho(a_0(z)) < 1$, and $c_2$ is a constant. So, for $j = 1, 2, \cdots, 7$, we have

Thus, applying Lemma 2.2 to (3.6), we get that $B_{2j}(z)\equiv 0 \; (j = 1, 2, \cdots, 7)$. Clearly, for $B_{25}(z) = 1$, this is a contradiction.

Case 3. $\alpha{(z)}$ and $\beta{(z)}$ are non-constant polynomials. Substituting (3.2) into (1.2), we have

therefore,

Thus, by the degree relationship between $\alpha{(z)}$ and $\beta{(z)}$, there are three subcases:

Case 3.1. $\deg\alpha{(z)} > \deg\beta{(z)}\geq 1$; (3.8) can be rewritten as

where $h_0(z) = 0$ and

We have $e^{\alpha(z)}$, $e^{\beta(z)}$ with normal growth and $\deg\alpha{(z)} > \deg\beta{(z)}\geq 1$. Then, $\rho(a_1(z)) < 1$ and $\rho(a_0(z)) < 1$; hence, $T(r, e^{\beta(z)}) = o\{T(r, e^{\alpha(z)})\}$ and

Similarly, the above formulas have similar expressions for the other exponential terms of $B_{3j}(z) \; (j = 1, 2, \cdots, 7)$ of (3.9); so, for $j = 1, 2, \cdots, 7$, we have

Applying Lemma 2.2 to (3.9), we get that $B_{3j}(z)\equiv 0 \; (j = 1, 2, \cdots, 7)$. Therefore,

we have that $e^{\beta(q^{2}z)} = 1$, where ${\beta(q^{2}z)} = 2k\pi i$, $k\in\mathbb{Z}$.

Since ${\beta(z)}$ is a polynomial, which is a contradiction.

Case 3.2. $\deg\beta{(z)} > \deg\alpha{(z)}\geq 1$; (3.8) can be rewritten as

where $h_0(z) = 0$ and

We have $e^{\alpha(z)}$, $e^{\beta(z)}$ with normal growth and $\deg\beta{(z)} > \deg\alpha{(z)}\geq 1$. Then, $\rho(a_1(z)) < 1$ and $\rho(a_0(z)) < 1$; thus, $T(r, e^{\alpha (z)}) = o\{T(r, e^{\beta(z)})\}$ and

Similarly, for the other exponential terms of $B_{4j}(z) \; (j = 1, 2, \cdots, 8)$, we have similar expressions; so, for $j = 1, 2, \cdots, 8$, we have

Applying Lemma 2.2 to (3.10) we get that $B_{4j}(z)\equiv 0 \; (j = 1, 2, \cdots, 8)$, we have

hence,

That is, $1-e^{\alpha(q^{2}z)} = 0$, where ${\alpha(z)}$ is a non-constant polynomial, which is a contradiction.

Case 3.3. $\deg\beta{(z)} = \deg\alpha{(z)} = n\geq 1$. Set

where $a_{n}(\neq 0)$, $a_{n-1}, \cdots, a_0$, $b_{n}(\neq 0)$, $b_{n-1}, \cdots, b_0$ are constants.

Let us consider the relationship between the highest coefficients of the polynomial number of each exponential function term in (3.8), and the items with the same coefficients are merged. Therefore, we discuss the following two situations according to whether $a_{n}$ is equal to $b_{n}$:

Subcase 3.3.1. If $a_{n} = b_{n}$, (3.8) can be rewritten as

where

For $a_{n} = b_{n}$, we have

hence $\deg(\alpha(z)-\beta(z))\leq n-1$.

Similarly, we can get the following formulas:

By Lemma 2.4, we have that $q\not\in E$; then,

We have $e^{\alpha(z)}$, $e^{\beta(z)}$ with normal growth. Then, $\rho(a_1(z)) < 1$ and $\rho(a_0(z)) < 1$; so, for $B_{5j}(z) \; (j = 1, 2, 3, 4)$, we have

Thus, applying Lemma 2.2 to (3.11), we get that $B_{5j}(z)\equiv 0 \; (j = 1, 2, 3, 4)$. By

that is,

it follows that

since $e^{\alpha(z)}\not\equiv e^{\beta(z)}$, $1-e^{\beta(z)-\alpha(z)}\neq0$, and $e^{\alpha(qz)-\beta(qz)}-1\neq0$, thus, the above formula is obviously not equal to $0$, which is a contradiction.

Subcase 3.3.2. If $a_{n}\neq b_{n}$, dividing (3.8) by $e^{\alpha(qz)+\beta(z)}$; we have

For the convenience of the following description, we use $f_j(z)$ and $g_j(z) \; (j = 1, 2, \ldots, 18)$ to represent the coefficient functions and exponential functions, that is,

By Lemma 2.4, we deduce that $-e^{\beta(qz)+\alpha(z)-\alpha(qz)-\beta(z)}$, $e^{\alpha(z)-\beta(z)}$, $-a_1(z)e^{\alpha(q^2z)-\alpha(qz)}$, $-a_0(z)e^{\alpha(q^2z)}$, $-a_0(z)e^{\beta(q^2z)-\beta(z)}$, $(1+a_0(z))e^{\beta(q^2z)}$ and $e^{\beta(qz)-\alpha(qz)}$, these are not constant functions in (3.12) for $q\not\in E$; then applying Lemma 2.3 to (3.12), there exists $t_i\in\{0, 1\} \; (i = 1, 2, \cdots, 11)$ for the following equation:

we assume that there are at least two values of $t_i \; (i = 1, 2, \cdots, 11)$ that are equal to $1$; without loss of generality, set $t_1 = t_7 = 1$ and the rest are equal to $0$; then,

from (3.13), we see that

If $e^{g_4(z)} = c_1$ is a constant, then, considering (3.14), we have

If $g_{12}(z)$ is a non-constant polynomial, we get contradiction from the following equation:

If $g_{12}(z)$ is a constant, since $g_4(z)$, $g_{12}(z)$ are constants, it follows that

we get that $q\in E$, which is a contradiction.

So, $e^{g_4}$ and $e^{g_{12}}$ are not constants; considering (3.14) we have

if $a_1^{'}(z)+a_1g_4^{'} = 0$, we obtain that $\frac{a_1^{'}(z)}{a_1(z)} = -g_4^{'}(z)$, that is

where $C$ is an arbitrary constant.

Since $g_4(z)$ is a non-constant polynomial, we have that $\rho(a_1(z)) < 1$, which is a contradiction.

So, $a'_1(z)+a_1g'_4\neq0$; then, there is

if $g_4-g_{12}$ is not a constant, since $\rho(a_1) < 1$, we have

which is a contradiction.

If $g_4-g_{12} = c_2$ is a constant function, $g_4 = g_{12}+c_2$; then, also applying (3.14) we have

since $g_{12}$ is a non-constant polynomial, by

which is a contradiction.

Therefore, for (3.13), there is only one $t_i \; (i = 1, 2, \cdots, 11)$ that is equal to $1$, and the rest are all zeros. Without loss of generality, we assume that $t_1 = 1$; then,

if $g_4(z) = \alpha({q^{2}z)}+\alpha{(z)}-\alpha(qz)-\beta(z)$ is a non-constant polynomial, then

also consider the order of the following equation

which is a contradiction.

Therefore, we deduce that $g_4(z)$ is a constant; thus, we have

On the other hand, together with (3.12) and (3.15), we have

that is,

We assert that $g_k$, $g_k-g_j \; (1\leq k < j\leq18, k, j\neq 4)$ in (3.18) are not constants. If $g_j(z) \; (1\leq j\leq18, j\neq 4)$ is a constant, we take $j = 6$; then $\beta({q^{2}z)}+\alpha{(z)}-\alpha(qz)-\beta(z)$ is a constant; taking into considering with $(3.16)$, we have the following:

we get that $q\in E$, which is a contradiction.

Similarly, if $g_k-g_j \; (1\leq k < j\leq18, k, j\neq 4)$ is a constant, we can take $k = 1$ and $j = 6$; then, $\alpha(qz)-\beta({q^2z)}$ is a constant; incorporating (3.16), we have the following:

we have that $q\in E$, which is contradictory.

Above all, for (3.18), we have $g_k$, $g_k-g_j \; (1\leq k < j\leq18, k, j\neq 4)$ are polynomials. Given that $\rho(f_j) < 1 \; (1\leq k < j\leq18)$, we can deduce that when $1\leq k < j\leq18, k, j\neq 4$, we have

where $E$ is a set of finite logarithmic measures.

So, the formula (3.18) satisfies all conditions of Lemma 2.2; then, we have that $f_j(z)\equiv 0 \; (j = 1, 2, \cdots, 18)$. It follows that $f_1(z) = 1$, $f_2(z) = -1$, $f_3(z) = 1$ and $f_{12}(z) = -1$ are non-zero constants, which is a contradiction.

Hence, $f(z)\equiv g(z)$. This completes the proof of Theorem 1.3.

4.   Proof of Theorem 1.4

The idea of proving this theorem is mainly derived from literature^[7].

Since $f(z)$ and $g(z)$ are non-constant meromorphic functions, assume that $f(z)\not\equiv g(z)$; we have

By (4.1), we deduce the following.

Assertion 1: $N(r, H) = O(\log r)$ and $N(r, \frac{1}{H}) = O(\log r)$.

The possible poles of $H(z)$ are derived from the zeros of $g'(z)$, $f(z)$, $f(z)-1$, $f(z)-c$ and the poles of $f(z)$ and $g(z)$.

We first prove that the poles of $f(z)$ and $g(z)$ are not zeros or poles of $H(z)$. Since $f(z)$ and $g(z)$ share $\infty$ CM, suppose that $z_0$ represents $k(\geq1)$ multiplicities of $f(z)$ and $g(z)$; then,

where $a_{-k}$, $b_{-k}$ are non-zero constants.

So, we have

then,

Similarly,

substituting (4.2) and (4.3) into (4.1), we have

Since $a_{-k}$, $b_{-k}$ is not $0$, $z = z_0$ is not the zero or pole of $H(z)$; that is, the poles of $f(z)$ and $g(z)$ are not zeros or poles of $H(z)$.

Next, we prove that the same zeros of the zeros, $1-$value points, $c-$value points of $f(z)$ and the zeros, $1-$value points, $c-$value points of $g(z)$ are not the zeros or poles of $H(z)$. Assume that $z_1$ is the public zero of $f(z)-a$ and $g(z)-a$, and that the multiplicities of $f(z)$ and $g(z)$ are $s(>0)$, $t(>0)$ respectively, where $a\in\{0, 1, c\}$. For the two sides of the above formula with the following Laurent expansion in the neighborhood of $z_1$, assume that

Hence,

substituting (4.4) and $(4.5)$ into (4.1), we have

Since $f_1(z_1)$ and $g_1(z_1)$ are not zero, then

then, the public zeros of $f(z)$ and $g(z)$ are not zeros or poles of $H(z)$.

Similarly, the $1-$value points of $f(z)$ and the $1-$value points of $g(z)$ are not zeros or poles of $H(z)$. Moreover, the $c-$value points of $f(z)$ and the $c-$value points of $g(z)$ are not zeros or poles of $H(z)$. So the poles of $H(z)$ can only come from the zero points of $g'(z)$ but not the zero points of $g(z)$, $g(z)-1$ and $g(z)-c$; by Lemma 2.5(iv), we have

where ${N_0(r, \frac{1}{g'})}$ denotes the zero points of $g'$ but not the counting function of the zero points of $g-a_i \; (i = 1, 2, 3)$.

Above all, we can draw a conclusion that assertion $1$ is established. Therefore

where $R(z)$ is a rational function and $\alpha(z)$ is a polynomial.

By Lemma 2.6 for the meromorphic solution $f$ of (1.2), we have

where $k_{1}$, $k_{2}$ are non-zero constants.

Since

we have that $\mu(f) = \rho(f) = 0$.

By Lemma 2.5(i),

hence,

thus, for $H(z)$, we have that $\rho(H(z)) = 0$. Therefore, $\alpha(z)$ is a constant. Otherwise,

which is a contradiction. So, $H(z)$ is a rational function.

Since $f(z)$, $g(z)$ share $0$, $1$, $c(\neq0, 1)$ IM and $\infty$ CM and $f(z)\not\equiv g(z)$, by Lemma 2.7 for the transcendental functions $f(z)$, we have

Thus, given (4.1), we have

For the convenience of the proof, suppose that $f(z)$ shares the value $a$ partially with $g(z)$, and that $\overline {N}_{(m, n)}(r, a)$ denotes the reduced counting function of those zeros of $f(z)-a$ with multiplicity $m$, and of $g(z)-a$ with multiplicity $n$ in $\{z: |z| < r\}$. We make the following assumption:

Assertion 2. For any positive integer pair $(m, n)$, we have

Suppose, on the contrary, that $\overline{N}_{(m, n)}(r, a)\neq S(r, f)$, $a\in\{0, 1, c\}$. Next, we consider the following two cases.

Case 1. Suppose that $m = n$.

Let $z_2\in\{z: |z| < r\}$ be the public zeros of $f(z)-a$ and $g(z)-a$ for $m$ multiplicities; applying this in consideration with (4.6), we have that $H(z_2) = 1$.

If $H(z)\not\equiv1$, combining (4.7) with the first fundamental Nevanlinna theorem, we have

which is a contradiction.

If $H(z)\equiv1$, by (4.9), we have

for all $z\in\mathbb{C}$.

Since $f(z)$ and $g(z)$ share $0$, $1$, $c(\neq0, 1)$ IM, taking into consideration (4.10), we arrive at $f(z)$ and $g(z)$ sharing $0$, $1$, $c$, $\infty$ CM. Then, there is

given that $\rho(f) = \rho(g) = 0$, we have that $\alpha(z)$ and $\beta(z)$ are constants. Thus, $f(z)$ is a constant, which is a contradiction.

Case 2. Suppose that $m\neq n$.

Let $z_3$ be the public zeros of $f(z)-a$ and $g(z)-a$ for $m$ and $n$ multiplicities respectively, where $a\in\{0, 1, c\}$. Also, by (4.6), we have that $H(z_3) = \frac{m}{n}$.

If $H(z_3)\not\equiv\frac{m}{n}$, combining (4.7) with the first fundamental Nevanlinna theorem, we have

which is a contradiction.

If $H(z_3)\equiv\frac{m}{n}$, then

for all $z\in\mathbb{C}$.

Let $z_3$ be the public zeros of $f(z)-a$ and $g(z)-a$ and the corresponding multiplicities be $p$ and $q$, where $a\in\{0, 1, c\}$. Combining (4.6) with $(4.11)$ on both sides with Laurent expansion of $z_3$, then, for all $a\in\{0, 1, c\}$ we have

Set

from the above we get that $L(z)$ is analytic in $z_3$. Thus,

Given that $m\neq n$ and $m (r, \frac{f'}{f}) = O(\log r)$, we deduce that

that is, $L(z)$ is a polynomial.

Then for, (4.13), we have

therefore,

since $L(z)$ is a polynomial, by Lemma 2.9, we get

then, by $m\neq n$ and Lemma 2.5(i), we have

which is a contradiction.

In summary, for any positive integer pair $(m, n)$, we have

this completes the proof of Assertion $2$.

Above all, we give the rest of the proof of the theorem.

the first item on the right sides of (4.14) is equivalent to $S(r, f)$; the second item, the third item and the fourth item can be estimated by applying the upper bound as $\frac{1}{10}N(r, \frac{1}{f})$, $\frac{1}{10}N(r, \frac{1}{g})$ and $\frac{1}{10}N(r, \frac{1}{f})$ (or $\frac{1}{10}N(r, \frac{1}{g}))$. Combining Lemma 2.5(i) with (4.14), we have

Similarly,

Combining this with the above estimates and Lemma 2.5(ii), we have

we get $T(r, f) = S(r, f)$, which is a contradiction.

This completes the proof of Theorem 1.4.

5.   Conclusions

In this paper, we have investigated the meromorphic solutions of a class of homogeneous second-order q-difference equations and the uniqueness problem of a meromorphic function with three shared values. We have also discussed the uniqueness problem of the meromorphic solutions of a class of nonhomogeneous q-difference equations and a meromorphic function with four shared values.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

We would like to thank the referee for his or her valuable comments and helpful suggestions. They have led to an improvement of the presentation of this paper.

Conflict of interest

The authors declare no conflict of interest in this paper.