Correction to: Solutions of Complex Fermat-Type Partial Difference and Differential-Difference Equations

• Ling Xu
• Tingbin Cao
Correction

1 Correction to: Mediterr. J. Math. (2018) 15:227  https://doi.org/10.1007/s00009-018-1274-x

Abstract. We give a correction to Theorem 1.2 in a previous paper [Mediterr. J. Math. (2018) 15:227]. Two examples are given to explain the corrected conclusion.

Mathematics Subject Classification. Primary 39A45 Secondary 32H30 39A14 35A20

Keywords. Several complex variables meromorphic functions fermat-type equations Nevanlinna theory partial differential-difference equations

2 Introduction and main result

Recently, the present authors originally considered solutions of complex partial differential-difference equations of the Fermat type by making use of Nevanlinna theory. Unfortunately, there was an error in the proof of [1, Theorem 1.2] (that is lines -1 to -3 on the Page 11), and thus its conclusion was stated wrong. Here we correct it as follows.s

Theorem 1.1

Let $$c=(c_{1}, c_{2})$$ be a constant in $$\mathbb {C}^{2}.$$ Then any transcendental entire solution with finite order of the partial difference-differential equation of the Fermat type
\begin{aligned} \left( \frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}\right) ^{2}+f^{2}(z_{1}+c_{1}, z_{2}+c_{2})=1 \end{aligned}
(1)
has the form of $$f(z_{1}, z_{2})=\sin \left( Az_{1}+Bz_{2}+H(z_{2})\right) ,$$ where AB are constants on $$\mathbb {C}$$ satisfying $$A^2=1$$ and $$Ae^{i(A c_{1}+Bc_{2})}=1,$$ and $$H(z_{2})$$ is a polynomial in one variable $$z_{2}$$ such that $$H(z_{2})\equiv H(z_{2}+c_{2}).$$ In the special case whenever $$c_{2}\ne 0,$$ we have $$f(z_{1}, z_{2})=\sin \left( Az_{1}+Bz_{2}+\text{ Constant }\right) .$$

We show the details of the proof as follows.

Proof

Assume the f is a transcendental entire solution with finite order of equation (1), then
\begin{aligned} \left[ \frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}+if(z_{1}+c_{1}, z_{2}+c_{2})\right] \left[ \frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}-if(z_{1}+c_{1}, z_{2}+c_{2})\right] =1. \end{aligned}
From the equation we see that both $$\frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}+if(z_{1}+c_{1}, z_{2}+c_{2})$$ and $$\frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}-if(z_{1}+c_{1}, z_{2}+c_{2})$$ have no zeros in $$\mathbb {C}^{2}.$$ Hence similarly as the proof of [1, Theorem 1.4], we may also assume that
\begin{aligned} \frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}+if(z_{1}+c_{1}, z_{2}+c_{2})=e^{ip(z)} \end{aligned}
and
\begin{aligned} \frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}-if(z_{1}+c_{1}, z_{2}+c_{2})=e^{-ip(z)} \end{aligned}
where p is a nonconstant entire function on $$\mathbb {C}^{2},$$ which gives
\begin{aligned} f(z_{1}+c_{1}, z_{2}+c_{2})=\frac{e^{ip(z)}-e^{-ip(z)}}{2i}=\sin p(z). \end{aligned}
Furthermore, it follows immediately from [1, Lemma 3.3] for any variable $$z_{j}$$ $$(j\in 1, 2)$$ that p should be a polynomial function on $$\mathbb {C}^{2}.$$ Hence, p is a nonconstant polynomial on $$\mathbb {C}^{2}.$$ From these equations above, we get from [1, Lemma 3.2] that
\begin{aligned} \frac{\partial f(z_{1}+c_{1}, z_{2}+c_{2})}{\partial z_{1}}= & {} \frac{e^{ip(z_{1}+c_{1}, z_{2}+c_{2})}+e^{-ip(z_{1}+c_{1}, z_{2}+c_{2})}}{2}\\= & {} \frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}\cdot \frac{e^{ip(z_{1},z_{2})}+e^{-ip(z_{1},z_{2})}}{2}, \end{aligned}
that is
\begin{aligned}&\frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}e^{ip(z_{1}, z_{2})+ip(z_{1}+c_{1}, z_{2}+c_{2})}+\frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}e^{ip(z_{1}+c_{1},z_{2}+c_{2})-ip(z_{1}, z_{2})}\nonumber \\&\quad -e^{i2p(z_{1}+c_{1}, z_{2}+c_{2})}=1. \end{aligned}
(2)
From the assertion that p is a nonconstant polynomial, we see that $$ip(z_{1}, z_{2})+ip(z_{1}+c_{1}, z_{2}+c_{2})$$ can not be a constant. This implies that both $$e^{i2p(z_{1}+c_{1}, z_{2}+c_{2})}$$ and $$e^{ip(z_{1}, z_{2})+ip(z_{1}+c_{1}, z_{2}+c_{2})}$$ must be nonconstant and transcendental on $$\mathbb {C}^{2},$$ and that
\begin{aligned} \frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}e^{ip(z_{1}, z_{2})+ip(z_{1}+c_{1}, z_{2}+c_{2})} \end{aligned}
can not be a constant. Furthermore, note that
\begin{aligned}&N(r, e^{i2p(z_{1}+c_{1}, z_{2}+c_{2})})=N(r, \frac{1}{e^{i2p(z_{1}+c_{1}, z_{2}+c_{2})}})=0\\&N(r, \frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}e^{ip(z_{1}, z_{2})+ip(z_{1}+c_{1}, z_{2}+c_{2})})=0\\&N(r, \frac{1}{\frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}e^{ip(z_{1}, z_{2})+ip(z_{1}+c_{1}, z_{2}+c_{2})}})=o(T(r,f)). \end{aligned}
Hence, we can get from [1, Lemma 3.1] that
\begin{aligned} \frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}e^{ip(z_{1}+c_{1},z_{2}+c_{2})-ip(z_{1}, z_{2})}\equiv 1. \end{aligned}
(3)
Rewrite it to be
\begin{aligned} \frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}\equiv e^{ip(z_{1},z_{2})-ip(z_{1}+c_{1}, z_{2}+c_{2})}, \end{aligned}
which implies $$ip(z_{1},z_{2})-ip(z_{1}+c_{1}, z_{2}+c_{2}),$$ and thus $$e^{ip(z_{1},z_{2})-ip(z_{1}+c_{1}, z_{2}+c_{2})},$$ must be a constant. Otherwise, we obtain a contradiction from the fact that the left of the above equation is nontranscendental but the right is transcendental. Assume that
\begin{aligned} \frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}\equiv e^{ip(z_{1},z_{2})-ip(z_{1}+c_{1}, z_{2}+c_{2})}\equiv A, \end{aligned}
where A is a nonzero constant in $$\mathbb {C}.$$ Submitting (3) into (2) gives
\begin{aligned}&\frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}\equiv e^{ip(z_{1}+c_{1}, z_{2}+c_{2})-ip(z_{1}, z_{2})}\equiv \frac{1}{A}. \end{aligned}
Hence $$A^{2}=1.$$ Further, by $$\frac{\partial p(z_{1}, z_{2})}{\partial z_{1}}\equiv A,$$ we know that $$p(z_{1}, z_{2})$$ is only a nonconstant polynomial of the form
\begin{aligned} p(z_{1}, z_{2})=Az_{1}+g(z_{2}), \end{aligned}
where $$g(z_{2})$$ should be a polynomial function in one variable $$z_{2}$$ (Note that the present authors made a mistake of $$p(z_1, z_2) = Az_{1}+ B$$ with a constant B in the original proof in [1], and thus the following is different from the original proof). Since $$p(z_{1}, z_{2})-p(z_{1}+c_{1}, z_{2}+c_{2})=-i Ln A,$$ we get that
\begin{aligned} g(z_{2})-g(z_{2}+c_{2})=Ac_{1}-iLn A. \end{aligned}
We may write $$g(z_{2})=Bz_{2}+h(z_{2})$$ such that $$A^2=1$$ and
\begin{aligned} Ae^{i(Ac_{1}+Bc_{2})}=1, \end{aligned}
where $$h(z_{2})$$ is a polynomial in one variable $$z_{2}.$$ This implies $$Ac_{1}+Bc_{2}=-k\pi (k\in \mathbb {N})$$ and $$h(z_{2})\equiv h(z_{2}+c_{2}).$$ Hence,
\begin{aligned} f(z_{1}, z_{2})= & {} \sin (Az_{1}-Ac_{1}+Bz_{2}-Bc_{2}+h(z_{2}-c_{2}))\\= & {} \sin (Az_{1}+Bz_{2}-(Ac_{1}+Bc_{2})+h(z_{2}))\\= & {} \sin (Az_{1}+Bz_{2}+k\pi +h(z_{2}))\\:= & {} \sin (Az_{1}+Bz_{2}+H(z_{2})), \end{aligned}
where $$H(z_{2})$$ is a polynomial in one variable $$z_{2}$$ satisfying $$H(z_{2})\equiv H(z_{2}+c_{2}).$$ It is clear that $$H(z_{2})$$ should be a constant whenever $$c_{2}\ne 0.$$ $$\square$$

We give two examples to explain the conclusion of the theorem.

Example 1.2

Let $$A=1,$$ $$B=2,$$ and let two constants $$c_{1}$$ and $$c_{2}$$ satisfy $$e^{ic_{1}}=1$$ and $$c_{2}=0.$$ Then $$Ae^{i(Ac_{1}+Bc_{2})}=1.$$ The entire function $$f(z)=\sin (z_{1}+2z_{2}+z_{2}^{3}+1)$$ satisfies the Fermat type partial differential difference equation
\begin{aligned} \left( \frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}\right) ^{2}+f^{2}(z_{1}+c_{1}, z_{2}+c_{2})=1 \end{aligned}
in $$\mathbb {C}^{2},$$ where $$c=(c_{1},c_{2}).$$ This shows that the function $$H(z_{2})$$ in the conclusion of Theorem 1.1 may be a nonconstant polynomial whenever $$c_{2}=0.$$

Example 1.3

Let $$A=1,$$ $$B=2i,$$ and let two constants $$c_{1}$$ and $$c_{2}$$ satisfy $$c_{1}+2ic_{2}=0.$$ Then $$Ae^{i(Ac_{1}+Bc_{2})}=1.$$ The entire funcion $$f(z)=\sin (z_{1}+2iz_{2})$$ satisfies the Fermat type partial differential difference equation
\begin{aligned} \left( \frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}\right) ^{2}+f^{2}(z_{1}+c_{1}, z_{2}+c_{2})=1 \end{aligned}
in $$\mathbb {C}^{2},$$ where $$c=(c_{1},c_{2}).$$ This shows that the function $$H(z_{2})$$ in the conclusion of Theorem 1.1 may be a constant whenever $$c_{2}\ne 0$$ or not.

If there are no differences, that is $$c=(0,0),$$ then Theorem 1.1 implies the following corollary.

Corollary 1.4

Any transcendental entire solution with finite order of the partial differential equation of the Fermat type
\begin{aligned} \left( \frac{\partial f(z_{1}, z_{2})}{\partial z_{1}}\right) ^{2}+f^{2}(z_{1}, z_{2})=1 \end{aligned}
(4)
has the form of $$f(z_{1}, z_{2})=\sin \left( z_{1}+g(z_{2})\right) ,$$ where $$g(z_{2})$$ is a polynomial in one variable $$z_{2}.$$

Reference

1. 1.
Xu, L., Cao, T.B.: Solutions of Complex Fermat-Type Partial Difference and Differential-Difference Equations. Mediterr. J. Math. 15, 227 (2018).