Entire solutions of two certain types of quadratic trinomial q-difference differential equations

1.   Introduction and main results

The classical Fermat's last theorem that equation $x^n+y^n = 1$ has no non-trivial rational solutions, when $n\ge3$, had been proved by Wiles in ^[1]. Considering $x, y$ in $x^n+y^n = 1$ as elements in function fields, we arrive at looking equations that may be called Fermat type functional equations

In 1966, Gross ^[2] proved the Fermat type functional equation (1.1) has no transcendental meromorphic solutions when $n\ge4$. If $n = 2$, then Eq (1.1) has the entire solutions $f(z) = \sin(h(z))$ and $g(z) = \cos(h(z))$, where $h(z)$ is any entire function, and no other solutions exist ^[3]. Baker ^[4] and Yang ^[5] also obtained some related results on Fermat type functional equation.

In recent years, the analogue of Fermat type equations inspired numerous investigations. Particularly, some authors have gotten a number of interesting results by considering that $g(z)$ has a special relationship with $f(z)$ ^[6,7]. For example, Liu et al. ^[6] considered the difference equation

and obtained the following result:

Theorem 1.1. (see ^[6], Theorem 1.1) The transcendental entire solutions with finite order of Eq (1.2) must satisfy $f(z) = \sin(Az+B)$, where $B$ is a constant and $A = \dfrac{(4k+1)\pi}{2c}$, $k$ is an integer.

Later on, considering a generalization of Eq (1.2) as

where $P(z), Q(z)$ are non-zero polynomials, Liu and Yang obtained a result (see ^[8], Theorem 2.1), which is an improvement of Theorem A. Closely related to difference expressions are q-difference expressions, where the usual shift $f(z+c)$ of a meromorphic function will be replaced by the q-shift $f(qz)$. Liu and Cao ^[9] considered the entire solutions of Fermat type q-difference equations

where $P(z), Q(z)$ are non-zero polynomials and $|q| = 1$. They showed the following theorem:

Theorem 1.2. (see ^[9], Theorem 2.6) If Eq (1.4) admits a transcendental entire solution of finite order, then $P(z)$ must be a constant $P$. This solution can be written as

satisfying one of the following conditions:

(1) $q$ satisfies $p(qz) = p(z)$ and $Q_1(z)-iP Q_1(qz)\equiv0, Q_2(z)+iPQ_2(qz)\equiv0$, $P^4Q(q^2z) = Q(z)$;

(2) $q$ satisfies $p(qz)+p(z) = 2a_0$, and $iP Q_1(qz)e^{2a_0}\equiv-Q_2(z)$, $iPQ_2(qz)\equiv Q_1(z)e^{2a_0}$, $P^4Q(q^2z) = Q(z)$, $e^{8a_0} = 1$, where $Q(z) = Q_1(z)Q_2(z)$ and $p(z)$ is a non-constant polynomial.

Liu and Yang ^[7] in 2016 studied the existence and the forms of solutions of some quadratic trinomial functional equations and obtained some precise properties on the meromorphic solutions of the following equations

and

If $\alpha\ne\pm1, 0$, then Eq (1.5) has no transcendental meromorphic solutions (see ^[7], Theorem 1.3) and the finite order transcendental entire functions of Eq (1.6) must be of order equal to one (see ^[7], Theorem 1.4).

Recently, Luo et al. ^[10] investigated the transcendental entire solutions with finite order of the quadratic trinomial difference equation

and differential difference equation

where $\alpha^2(\ne0, 1), c$ are constants and $g(z)$ is a polynomial.

Theorem 1.3. (see ^[10], Theorem 2.1) Let $\alpha^2\ne0, 1$, $c(\ne0)\in\mathbb{C}$ and $g$ be a polynomial. If the difference equation (1.7) admits a transcendental entire solution $f(z)$ of finite order, then $g(z)$ must be of the form $g(z) = az+b$, where $a, b\in\mathbb{C}$.

In the above results, Nevanlinna theory of meromorphic functions ^[11,12] and its difference counterparts ^[13,14] play a critical role. For related results, we refer the reader to ^{[15,16,17,18,19,20,21,22,23]} and the references therein.

Motivated by the above equations and results, we investigate the existence and forms of entire solutions of the following two quadratic trinomial q-difference differential equations

where $\alpha^2\ne0, 1$ and $q\ne0, \pm1$ are complex numbers, and $g(z)$ is a polynomial.

where $\alpha^2\ne0, 1$ and $q\ne0, 1$ are complex numbers, and $g(z)$ is a polynomial.

Below, for convenience, let

Theorem 1.4. If Eq (1.9) admits a transcendental entire solution $f(z)$ with finite order, then $g(z)$ must satisfy $\deg(g(z)) > 2$ and $q^{\deg(g(z))} = 1$. Furthermore,

We give an example to show that the result of Theorem 1.4 is precise as follows:

Example 1.1. $f(z) = \pm\frac{\sqrt{6}}{6}e^{\frac{z^3}{2}}$ is a transcendental entire solution of

Here, $g(z) = z^3$, $q = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$, $\alpha = 2$, $A_1 = \frac{\sqrt{3}-3}{6} \mbox{ and } A_2 = \frac{\sqrt{3}+3}{6}.$

Corollary 1.1. If $\deg(g(z))\le2$, then Eq (1.9) has no transcendental entire solution of $f(z)$ with finite order.

Corollary 1.2. If $|q|\ne1$, then Eq (1.9) has no transcendental entire solution of $f(z)$ with finite order.

Theorem 1.5. If Eq (1.10) admits a transcendental entire solution $f(z)$ with finite order, then $g(z)\equiv\beta$, $q = -1$ and

where $t$, $y_1, y_2, \beta\in\mathbb{C}$ satisfying $\beta = y_1+y_2$ and $t = \pm i$.

We give an example to show that the result of Theorem 1.5 is precise as follows:

Example 1.2. $f(z) = \frac{\sqrt{2}}{2i}(\frac{\sqrt{3}-3}{6}e^{iz+\ln{i}}-\frac{\sqrt{3}+3}{6}e^{-iz})$ is a transcendental entire solution of

Here, $g(z)\equiv\ln{i}$, $q = -1$, $\alpha = 2$, $A_1 = \frac{\sqrt{3}-3}{6} \mbox{ and } A_2 = \frac{\sqrt{3}+3}{6}.$

Corollary 1.3. If $\deg(g(z))\ge1$, then Eq (1.10) has no transcendental entire solution of $f(z)$ with finite order.

Corollary 1.4. If $q\ne0, \pm1$, then Eq (1.10) has no transcendental entire solution of $f(z)$ with finite order.

2.   Some lemmas

Lemma 2.1. ^[12] Let $f_j(z)$, $j = 1, 2, 3$ be meromorphic functions and $f_1(z)$ is not a constant. If

and

where $\lambda < 1$, $T(r) = \max_{1\le j\le 3}\{T(r, f_j)\}$ and $I$ represents a set of $r\in(0, \infty)$ with infinite linear measure. Then, $f_2\equiv1$ or $f_3\equiv1$.

Lemma 2.2. ^[12] If $f_j(z)$, $g_j(z)$$(1\le j\le n, n\ge2)$ are entire functions satisfying

(1) $\sum_{j = 1}^n f_j(z)e^{g_{j}(z)}\equiv 0$;

(2) The orders of $f_j$ are less than that of $e^{g_h(z)-g_k(z)}$ for $1\le j\le n$, $1\le h < k\le n$.

Then $f_j(z)\equiv0$ for $1\le j\le n$.

Lemma 2.3. ^[12] Let $p(z)$ be a nonzero polynomial with degree $n$. If $p(qz)-p(z)$ is a constant, then $q^n = 1$ and $p(qz)\equiv p(z)$. If $p(qz)+p(z)$ is a constant, then $q^n = -1$ and $p(qz)+p(z) = 2a_0$, where $a_0$ is the constant term of $p(z)$.

3.   Proof of Theorem 1.4

Let $f(z)$ be a transcendental entire solution with finite order of Eq (1.9). Denote

where $\mu$, $\nu$ are entire functions. It can be deduced from Eq (1.9) that

From Eq (3.1), we have

The above equation leads to

We observe that both $\frac{\sqrt{1+\alpha}\mu}{e^{\frac{g(z)}{2}}}+i\frac{\sqrt{1-\alpha}\nu}{e^{\frac{g(z)}{2}}}$ and $\frac{\sqrt{1+\alpha}\mu}{e^{\frac{g(z)}{2}}}-i\frac{\sqrt{1-\alpha}\nu}{e^{\frac{g(z)}{2}}}$ have no zeros. Combining Eq (3.2) with the Hadamard factorization theorem, there exists a polynomial $p(z)$ such that

Set

It follows from Eq (3.3) that

This leads to

and

where $A_1$ and $A_2$ are defined as Eq (1.11).

It follows from Eq (3.5) that

Since $\alpha^2\ne0, 1$, we have that both $A_1$ and $A_2$ are nonzero constants. Combining with Eqs (3.6) and (3.7), we have

Case 1.$\gamma_1(z)-\gamma_2(qz)$ is a non-constant polynomial. Using Lemma 2.1 in Eq (3.8), we have

If $\frac{A_1}{A_2}e^{\gamma_2(z)-\gamma_2(qz)}\equiv1$, then $\gamma_2(z)-\gamma_2(qz)$ is a constant. By Lemma 2.3, $\gamma_2(z)-\gamma_2(qz)\equiv0$. Thus, we have $\frac{A_1}{A_2} = 1$, which contradicts with $\alpha\ne0, 1$.

If $-\frac{A_1}{A_2}e^{\gamma_1(qz)-\gamma_2(qz)}\equiv1$, then it follows from Eq (3.8) that $e^{\gamma_1(z)-\gamma_2(z)} = -\frac{A_1}{A_2}.$ In view of Eq (3.4), we get that

It is easy to get that $p(z)$ is a constant and $\frac{A_2}{A_1} = \frac{A_1}{A_2}$. This leads to $A_1^2 = A_2^2$, which contradicts with $\alpha^2\ne 0, 1$.

Case 2. $\gamma_1(z)-\gamma_2(qz)$ is a constant. Let $\kappa = \gamma_1(z)-\gamma_2(qz), \kappa\in\mathbb{C}$. Then, $\gamma_2(qz) = \gamma_1(z)-\kappa$. In view of Eq (3.4), $2p(z) = \gamma_1(z)-\gamma_2(z)$. Equation (3.8) reduces to

Case 2.1. $\kappa = \gamma_1(z)-\gamma_2(qz)\equiv0$. From Eq (3.9) we have $e^{2(p(z)+p(qz))} = 1$, which gives that $p(z)+p(qz)\equiv0$. It follows from Eq (3.4) that

Further, we have $\gamma_1(z)\equiv\gamma_1(q^2z)$ and $\gamma_2(z)\equiv\gamma_2(q^2z)$. Recall that $f(z)$ is transcendental, then from Eq (3.5) we have that $\gamma_1(z)$ and $\gamma_2(z)$ cannot be constant at the same time. By the assumption that $q\ne0, \pm1$, we get a contradiction.

Case 2.2. $\kappa = \gamma_1(z)-\gamma_2(qz)\not\equiv0$. Using the Nevanlinna second fundamental theorem for $e^{2p(qz)}$, we have

which shows that $p(qz)$ is a constant.

We claim that $g(z)$ is a polynomial. If $g(z)$ is a constant, then by combining with $p(qz)$ as a constant and Eq (3.4), we have both $\gamma_1(z)$ and $\gamma_2(z)$ are constants. From Eq (3.5), we have $f(z)$ is a constant, which contradicts with $f(z)$ is transcendental.

Thus, $\deg(g(z))\ge1$. Set $p(z)\equiv\eta$, where $\eta\in\mathbb{C}$. Then, it follows from Eqs (3.4) and (3.8) that

If $g(z)-g(qz)$ is a non-constant polynomial, then by using Lemma 2.2 in Eq (3.10), we have

It gives $A_1^2 = A_2^2$, which contradicts with $\alpha^2\ne 0, 1$. Thus, $g(z)-g(qz)$ is a constant.

Further, by Lemma 2.3, we obtain $g(z)-g(qz)\equiv0$ and $q^{\deg(g(z))} = 1$. Since $q\ne\pm1$, then $\deg(g(z))\ne\; 1, 2$. Combining with $\deg(g(z))\ge1$, we have $\deg(g(z)) > 2$. Moreover, Eq (3.10) reduces to

Thus, we have $\frac{A_1}{A_2}-1 = \left(\frac{A_1}{A_2}-1\right)e^{2\eta}$. Since $A_1\ne A_2$, then $\frac{A_1}{A_2}-1\ne0$. Hence, we have $e^{2\eta} = 1$. It gives $e^\eta = \pm1$, i.e., $e^{p(z)}\equiv\pm1$.

From Eqs (3.4) and (3.5), we have

And together with Eq (1.11), we obtain

We completed the proof of Theorem 1.4.

4.   Proof of Theorem 1.5

Let $f(z)$ be a transcendental entire solution with finite order of Eq (1.10). Using the same argument as in the proof of Theorem 1.4, we have

and

In view of Eqs (4.1) and (4.2), it follows that

This leads to

Case 1.$\gamma_1(qz)-\gamma_2(qz)$ is a constant. From Eq (3.4), we have $\gamma_1(qz)-\gamma_2(qz) = 2p(qz)$. Thus, $p(z)$ is a constant. Let $\iota\equiv e^{p(z)}$, where $\iota\in\mathbb{C}\verb|\|\{0\}$.

Furthermore, we have $\deg(g(z))\ge1$. Otherwise, from Eq (3.4), we have that both $\gamma_1(z)$ and $\gamma_2(z)$ are constants. It follows from Eq (4.1) that $f'(z)$ is a constant, which conflicts with $f(z)$ being transcendental.

Combining with Eqs (3.4) and (4.3), we get that

If $g(z)-g(qz)$ is a non-constant polynomial, then by using Lemma 2.2 in Eq (4.4), we get that

The second equation of (4.5) gives that $\iota^2 = -\frac{A_2}{A_1}$. Substituting this into the first equation of (4.5), we have

Since $\deg(g(z))\ge1$ and $q\ne0, 1$, then we have $\frac{-A_2}{A_1}+\frac{A_1}{A_2} = 0$. It gives that $A_1^2 = A_2^2$, which contradicts with $\alpha^2\ne0, 1$.

If $g(z)-g(qz)$ is a constant, by Lemma 2.3, we have $g(z)-g(qz)\equiv0$ and $q^{\deg(g(z))} = 1$. Since $q\ne1$, then $\deg(g(z))\ne1$. Note that $\deg(g(z))\ge1$, then $\deg(g(z))\ge2$.

Equation (4.4) reduces to

This implies that $\dfrac{\iota^2}{q}+\dfrac{A_1}{qA_2} = 0$ and $1+\dfrac{A_1}{A_2}\iota^2 = 0$. Similar to the above, we also have $A_1^2 = A_2^2$, which is a contradiction.

Case 2.$\gamma_1(qz)-\gamma_2(qz)$ is a non-constant polynomial. Since $\gamma_1(qz)-\gamma_2(qz) = 2p(qz)$, then we have $p(z)$ is a non-constant polynomial.

Next, we show that $\gamma_1'(z)\not\equiv0$ and $\gamma_2'(z)\not\equiv0$. From Eq (4.3), it is easy to get that $\gamma_1'(z)\equiv0$ and $\gamma_2'(z)\equiv0$ cannot hold at the same time.

If $\gamma_1'(z)\equiv0$ and $\gamma_2'(z)\not\equiv0$, then Eq (4.3) reduces to

Using the Nevanlinna second fundamental theorem for $e^{\gamma_1(qz)-\gamma_2(qz)}$, we have that

which is a contradiction.

Similarly, if $\gamma_1'(z)\not\equiv0$ and $\gamma_2'(z)\equiv0$, we also get a contradiction.

Then, by using Lemma 2.1 in Eq (4.3), we have

Case 2.1.If $\frac{A_1}{qA_2}\gamma_2'(z)e^{\gamma_2(z)-\gamma_2(qz)}\equiv1$, it implies that $\gamma_2'(z)$ is a nonzero constant, and $\gamma_2(z)-\gamma_2(qz)$ is a constant.

By Lemma 2.3, we have $\gamma_2(z)-\gamma_2(qz)\equiv0$ and $q^{\deg(\gamma_2(z))} = 1$. Since $q\ne1$, then $\deg(\gamma_2(z))\ne1$, which contradicts with $\gamma_2'(z)$ being a nonzero constant.

Case 2.2.If $\frac{\gamma_1'(z)}{q}e^{\gamma_1(z)-\gamma_2(qz)}\equiv1$, then from Eq (4.3) we have

The above two equations give that $\gamma_1(z)-\gamma_2(qz)$ and $\gamma_2(z)-\gamma_1(qz)$ are constants. Moreover, we also have that $\gamma_i'(z)(i = 1, 2)$ are nonzero constants, i.e., $\deg(\gamma_i(z)) = 1(i = 1, 2)$.

Set

where $\eta_1, \eta_2\in\mathbb{C}$.

In view of Eq (3.4), we have

By Lemma 2.3, we get that $q^{\deg(p(z))} = -1$ and $q^{\deg(g(z))} = 1$. Since $q\ne1$, then $\deg(g(z))\ne1$.

We now show that $\deg(g(z)) = 0$. If $\deg(g(z))\ge2$, by combining with $\deg(\gamma_i(z)) = 1$ and Eq (3.4), then we have $\deg(p(z)) = \deg(g(z))$. Therefore, $q^{\deg(p(z))} = q^{\deg(g(z))} = 1$, which contradicts with $q^{\deg(p(z))} = -1$. Hence, we have $g(z)\equiv\beta$, where $\beta\in\mathbb{C}$.

Recall that $\deg\gamma_i(z) = 1(i = 1, 2)$. It follows from Eq (3.4) that $\gamma_1(z)+\gamma_2(z) = g(z)\equiv\beta$.

Set

where $t\in\mathbb{C}\verb|\|\{0\}, y_1, y_2\in\mathbb{C}$ such that $\beta = y_1+y_2$.

It follows from Eqs (3.4) and (4.8) that $p(z) = tz+\frac{y_1-y_2}{2}$. And together with $q^{\deg(p(z))} = -1$, then we have $q = -1$.

By substituting $q = -1$ and Eq (4.8) into $\frac{\gamma_1'(z)}{q}e^{\gamma_1(z)-\gamma_2(qz)}\equiv1$ and Eq (4.6), we obtain

respectively. It gives that $t = \pm i$.

Furthermore, substituting Eq (4.8) into Eq (4.1), we have

Integration of the above equation gives that

We completed the proof of Theorem 1.5.

5.   Conclusions

In this paper, we showed that the explicit forms for entire solutions of two certain types of Fermat-type q-difference differential equations. In addition, we have given specific examples to illustrate our results.

Use of AI tools declaration

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

Acknowledgments

This article was partially supported by National Natural Science Foundation of China (NSFC)(NO.12171127, NO.12301095) and Topic on Fundamental and application fundamental research of Guangzhou in 2023 (No. 2023A04J0648). The corresponding author was supported by the National Natural Science Foundation of China (NSFC)(No. 12171127). The third author was supported by the National Natural Science Foundation of China (NSFC)(NO.12301095) and Topic on Fundamental and application fundamental research of Guangzhou in 2023 (No. 2023A04J0648). The authors thank the anonymous referee for the careful reading and some helpful comments.

Conflict of interest

The authors state no conflict of interest.