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

  • Ling Xu
  • Tingbin CaoEmail author
Correction
  • 69 Downloads

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 \(z_{2}.\)

Notes

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).  https://doi.org/10.1007/s00009-018-1274-x MathSciNetCrossRefzbMATHGoogle Scholar

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© Springer Nature Switzerland AG 2019

Authors and Affiliations

  1. 1.Department of MathematicsNanchang UniversityNanchangChina

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