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
has the form of \(f(z_{1}, z_{2})=\sin \left( Az_{1}+Bz_{2}+H(z_{2})\right) ,\) where A, B 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
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
and
where p is a nonconstant entire function on \(\mathbb {C}^{2},\) which gives
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
that is
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
can not be a constant. Furthermore, note that
Hence, we can get from [1, Lemma 3.1] that
Rewrite it to be
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
where A is a nonzero constant in \(\mathbb {C}.\) Submitting (3) into (2) gives
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
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
We may write \(g(z_{2})=Bz_{2}+h(z_{2})\) such that \(A^2=1\) and
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,
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
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
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
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
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
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Work supported by the National Natural Science Foundation of China (No. 11871260, No. 11461042), the Natural Science Foundation of Jiangxi Province (No. 20161BAB201007) and the outstanding young talent assistance program of Jiangxi Province (No. 20171BCB23002) in China.
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Xu, L., Cao, T. Correction to: Solutions of Complex Fermat-Type Partial Difference and Differential-Difference Equations. Mediterr. J. Math. 17, 8 (2020). https://doi.org/10.1007/s00009-019-1438-3
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DOI: https://doi.org/10.1007/s00009-019-1438-3