Existence of entire solutions of some non-linear differential-difference equations
- Minfeng Chen^{1}Email author,
- Zongsheng Gao^{1} and
- Yunfei Du^{1}
https://doi.org/10.1186/s13660-017-1368-1
© The Author(s) 2017
Received: 15 November 2016
Accepted: 18 April 2017
Published: 27 April 2017
Abstract
In this paper, we investigate the admissible entire solutions of finite order of the differential-difference equations \((f'(z))^{2}+P^{2}(z)f^{2}(z+c)=Q(z)e^{\alpha(z)}\) and \((f'(z))^{2}+[f(z+c)-f(z)]^{2}=Q(z)e^{\alpha(z)}\), where \(P(z)\), \(Q(z)\) are two non-zero polynomials, \(\alpha(z)\) is a polynomial and \(c\in\mathbb{C}\backslash\{0\}\). In addition, we investigate the non-existence of entire solutions of finite order of the differential-difference equation \((f'(z))^{n}+P(z)f^{m}(z+c)=Q(z)\), where \(P(z)\), \(Q(z)\) are two non-constant polynomials, \(c\in\mathbb{C}\backslash\{0\}\), m, n are positive integers and satisfy \(\frac{1}{m}+\frac{1}{n}<2\) except for \(m=1\), \(n=2\).
Keywords
MSC
1 Introduction and main results
In this paper, we assume that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory [1, 2]. In addition, we denote by \(S(r,f)\) any quantify satisfying \(S(r,f)=o(T(r,f))\), as \(r\rightarrow\infty\), outside of a possible exceptional set of finite logarithmic measure. We define the logarithmic measure of E to be \(\operatorname{lm}(E)=\int_{E\cap(1,\infty)}\frac{dr}{r}\). A set \(E\subset(1,\infty)\) is said to have finite logarithmic measure if \(\operatorname{lm}(E)<\infty\). Throughout this paper, all constants are complex constants unless otherwise specified.
Nevanlinna’s value theory of meromorphic functions has been used to study the properties of entire or meromorphic solutions of differential equations and difference equations in complex plane, such as [3–5]. In [6], Montel stated the following theorem.
Theorem A
However, when \(n=2\) and \(g(z)\) has a specific relationship with \(f(z)\) in (1.1), the problem that whether we can obtain the accurate expressions of entire solutions or not is worth to be considered. Recently, many results focused on this problem that were obtained by using the Nevanlinna theory, such as [7–14].
In [10], Liu et al. considered Fermat type differential-difference equation and obtained the following results.
Theorem B
Theorem C
In [7], Chen and Gao improved Theorem B and obtained the following result.
Theorem D
Remark 1.1
Equation (1.2) is a special case of (1.4). Theorem D generalized Theorem B. From Theorem D, we see that if \(P(z)\) and \(Q(z)\) are non-constant polynomials, then equation (1.4) has no transcendental entire solution of finite order.
In this paper, we generalize equations (1.2)-(1.4) and obtain the following results.
Theorem 1.1
Remark 1.2
In equation (1.5), when \(\alpha(z)\) is a constant, then equation (1.5) reduces to equation (1.4). That is, Theorem 1.1 generalizes Theorems B and D.
Theorem 1.2
Remark 1.3
In equation (1.6), when \(Q(z)\) is non-zero constant and \(\alpha(z)\) is a constant, then equation (1.6) reduces to equation (1.3). That is, Theorem 1.2 generalizes Theorem C.
Theorem E
Let m, n be positive integers satisfying \(\frac{1}{m}+\frac{1}{n}<1\). Then there are no non-constant entire solutions \(f(z)\) and \(g(z)\) that satisfy (1.7).
Theorem F
Equation (1.8) has no transcendental entire solutions with finite order, provided that \(m\neq n\), where m, n are positive integers.
Now, we generalize (1.8) and obtain the following result.
Theorem 1.3
2 Some lemmas
In order to prove our conclusions, we need some lemmas.
Lemma 2.1
Lemma 2.2
see [19]
- (1)
\(\sum_{j=1}^{n}f_{j}(z)e^{g_{j}(z)}\equiv0\).
- (2)
\(g_{j}(z)-g_{k}(z)\) are not constants for \(1\leq j< k\leq n\).
- (3)
For \(1\leq j\leq n\), \(1\leq h< k\leq n\), \(T(r,f_{j}(z))=o(T(r,e^{g_{h}(z)-g_{k}(z)}))\) (\(r\rightarrow\infty\), \(r\notin E\)).
Lemma 2.3
see [19]
Lemma 2.4
- (i)
if \(b=0\) and \(a\neq c\), then \(Q(z)\) reduces to a non-zero constant;
- (ii)
if \(b=0\) and \(a=c\), then \(Q(z)\) reduces to a non-zero constant or \(Q(z)=a_{1}z+a_{0}\), where \(a_{1}\) is a non-zero constant, \(a_{0}\) is a constant;
- (iii)
if \(b\neq0\) and \(a\neq c\), then \(Q(z)=a_{1}z+a_{0}\) and \(b=a_{1}(c-a)\), where \(a_{1}\) is a non-zero constant, \(a_{0}\) is a constant;
- (iv)
if \(b\neq0\) and \(a=c\), then \(Q(z)=a_{2}z^{2}+a_{1}z+a_{0}\) and \(b=a_{2}c^{2}\), where \(a_{2}\) is a non-zero constant, \(a_{1}\), \(a_{0}\) are constants.
Proof
(i) If \(b=0\) and \(a\neq c\), comparing the coefficients of \(z^{s-1}\) on both sides of (2.1), we see that \(sa_{s}c=asa_{s}\), it contradicts with \(a\neq c\) and \(a_{s}\neq0\).
(ii) If \(b=0\), \(a=c\) and \(s\geq2\), comparing the coefficients of \(z^{s-2}\) on both sides of (2.1), we see that \(a_{s}C_{s}^{2}c^{2}+a_{s-1}C_{s-1}^{1}c=a(s-1)a_{s-1}\), then \(a_{s}C_{s}^{2}c^{2}=0\), a contradiction. Thus \(s\leq1\), that is, \(Q(z)\) reduces to a non-zero constant or \(Q(z)\) is a non-constant polynomial with degree 1.
(iii) If \(b\neq0\) and \(a\neq c\), \(Q(z)\) is a non-zero constant, note that \(b\neq0\), clearly (2.1) is a contradiction. If \(s\geq2\), comparing the coefficients of \(z^{s-1}\) on both sides of (2.1), we see that \(sa_{s}c=asa_{s}\), a contradiction. If \(s=1\), by (2.1), we see that \(b=a_{1}(c-a)\neq0\).
(iv) If \(b\neq0\) and \(a=c\), \(Q(z)\) is a non-zero constant, note that \(b\neq0\), clearly (2.1) is a contradiction. If \(s=1\), by (2.1), we see that \(b=a_{1}(c-a)=0\), a contradiction. If \(s\geq3\), comparing the coefficients of \(z^{s-2}\) on both sides of (2.1), we see that \(a_{s}C_{s}^{2}c^{2}+a_{s-1}C_{s-1}^{1}c=a(s-1)a_{s-1}\), then \(a_{s}C_{s}^{2}c^{2}=0\), a contradiction. If \(s=2\), by (2.1), we see that \(b=a_{2}c^{2}\neq0\). □
Lemma 2.5
see [3]
3 Proof of Theorem 1.1
Similarly, we can prove that \([Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z)]P(z)-Q_{2}(z)P'(z)\not\equiv0\).
Thus, \([Q'_{1}(z)+Q_{1}(z)\alpha_{1}'(z)]P(z)-Q_{1}(z)P'(z)\not \equiv0\) and \([Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z)]P(z)-Q_{2}(z)P'(z)\not\equiv0\), by (3.4) and Lemma 2.3, we see that if any two of \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are not constants, then the third term must be constant. If any two of them are constants, then the third term also must be constant. In what follows, we discuss four cases: Case 1, \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha _{2}(z)-\alpha_{1}(z+c)}\) are not constants; Case 2, \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are not constants; Case 3, \(e^{\alpha_{2}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are not constants; Case 4, \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z)-\alpha _{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are all constants.
By \(\alpha_{1}(z)-\alpha_{1}(z+2c)=[\alpha_{1}(z)-\alpha _{2}(z+c)]+[\alpha_{2}(z+c)-\alpha_{1}(z+2c)]\), we see that \(\alpha_{1}(z)-\alpha_{1}(z+2c)\) is a constant, then \(\alpha_{1}(z)\) is a constant or a polynomial with degree 1, which implies that \(\alpha_{1}(z)-\alpha_{1}(z+c)\) is also a constant, a contradiction.
Note that \(A_{1}=ie^{A_{1}c}p\) and \(A_{2}=-ie^{A_{2}c}p\), we see that \(A_{1}\neq A_{2}\), that is, \(\frac{1}{A_{1}}\neq\frac{1}{A_{2}}\). In what follows, we discuss three subcases: Subcase 3.1, \(\frac{1}{A_{1}}\neq c\) and \(\frac{1}{A_{2}}\neq c\); Subcase 3.2, \(\frac{1}{A_{1}}=c\) and \(\frac{1}{A_{2}}\neq c\); Subcase 3.3, \(\frac{1}{A_{1}}\neq c\) and \(\frac{1}{A_{2}}=c\).
Case 4, \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are all constants, that is, \(\alpha_{1}(z)-\alpha_{1}(z+c)\), \(\alpha_{2}(z)-\alpha_{1}(z+c)\) and \(\alpha_{2}(z+c)-\alpha_{1}(z+c)\) are all constants. Note that \(\alpha_{1}(z)\) and \(\alpha_{2}(z)\) are not constants simultaneously, then \(\alpha_{1}(z)=Az+B_{1}\), \(\alpha_{2}(z)=Az+B_{2}\) and \(\alpha(z)=2Az+D\), where A is non-zero constant and \(B_{1}\), \(B_{2}\), D (\(=B_{1}+B_{2}\)) are constants. Therefore, by (1.5), (3.2) and (3.3), we have \(f(z)=B(z)e^{Az}\), where \(B(z)\) satisfies \([B'(z)+AB(z)]^{2}+P^{2}(z)B^{2}(z+c)e^{2Ac}=Q(z)e^{D}\).
This completes the proof of Theorem 1.1.
4 Proof of Theorem 1.2
According to the above proof, we can see that \(Q'_{1}(z)+Q_{1}(z)\alpha_{1}'(z)+iQ_{1}(z)\equiv0\) and \(Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z)-iQ_{2}(z)\equiv0\) cannot be valid simultaneously. In what follows, we assume that \(Q'_{1}(z)+Q_{1}(z)\alpha_{1}'(z)+iQ_{1}(z)\not\equiv0\) and \(Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z)-iQ_{2}(z)\not\equiv0\). By (4.3) and Lemma 2.3, we see that if any two of \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are not constants, then the third term must be constant. If any two of them are constants, then the third term also must be constant. In the following, we discuss four cases: Case 1, \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha _{2}(z)-\alpha_{1}(z+c)}\) are not constants; Case 2, \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are not constants; Case 3, \(e^{\alpha_{2}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are not constants; Case 4, \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z)-\alpha _{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are all constants.
Case 1 and Case 2, similarly to the proof of Case 1 and Case 2 of Theorem 1.1, we can obtain a contradiction.
By Lemma 2.4, we see that if \(\frac{1}{A_{2}-i}\neq c\), then \(Q_{2}(z)=q_{2}\) (constant); If \(\frac{1}{A_{2}-i}=c\), then \(Q_{2}(z)\equiv q_{2}\) (constant) or \(Q_{2}(z)=b_{1}z+b_{0}\), where \(b_{1}\) is a non-zero constant and \(b_{0}\) is a constant.
If \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\), by (1.6), (4.1) and (4.2), we get \(c=-i\). Note that \(A_{2}-i=-ie^{A_{2}c}\), then \(\frac{1}{A_{2}-i}\neq-i\). Therefore, we only need to consider three subcases: Subcase 3.1.1, \(\frac{1}{A_{2}-i}\neq c\) (\(=-i\)), \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\); Subcase 3.1.2, \(\frac{1}{A_{2}-i}\neq c\) (\(=-i\)), \(Q_{1}(z)=a_{1}z+a_{0}\) and \(Q_{2}(z)\equiv q_{2}\); Subcase 3.1.3, \(\frac{1}{A_{2}-i}=c\) (\(\neq-i\)), \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)=b_{1}z+b_{0}\).
Case 4, \(e^{\alpha_{1}(z)-\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z)-\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)-\alpha_{1}(z+c)}\) are all constants, that is, \(\alpha_{1}(z)-\alpha_{1}(z+c)\), \(\alpha_{2}(z)-\alpha_{1}(z+c)\) and \(\alpha_{2}(z+c)-\alpha_{1}(z+c)\) are all constants. Note that \(\alpha_{1}(z)\) and \(\alpha_{2}(z)\) are not constants simultaneously, then \(\alpha_{1}(z)=Az+B_{1}\), \(\alpha_{2}(z)=Az+B_{2}\) and \(\alpha(z)=2Az+D\), where A is non-zero constant and \(B_{1}\), \(B_{2}\), D (\(=B_{1}+B_{2}\)) are constants. Therefore, by (1.6), (4.1) and (4.2), we have \(f(z)=B(z)e^{Az}+c_{0}\), where \(B(z)\) satisfies \([B'(z)+AB(z)]^{2}+[B(z+c)e^{Ac}-B(z)]^{2}=Q(z)e^{D}\).
This completes the proof of Theorem 1.2.
5 Proof of Theorem 1.3
Suppose that \(f(z)\) is a transcendental entire function of finite order satisfying (1.9). In what follows, we will discuss four cases: Case 1, \(m=n\geq2\); Case 2, \(m>n\); Case 3, \(n>m\geq2\); Case 4, \(n\geq3\), \(m=1\).
Case 1, \(m=n\geq2\). If \(m=n=2\), note that \(P(z)\) and \(Q(z)\) are non-constant polynomials, by Theorem D, we see that (1.9) has no transcendental entire solutions of finite order. If \(m=n\geq3\), rewriting (1.9) as \(\frac{1}{Q(z)}(f'(z))^{n}+\frac{P(z)}{Q(z)}f^{m}(z+c)=1\), by Theorem E, we see that (1.9) has no transcendental entire solutions of finite order.
Case 2 and Case 3, similarly to the proof of the Case 1 and Case 2 of [10], Theorem 1.2, we can also obtain (1.9) has no transcendental entire solutions of finite order.
This completes the proof of Theorem 1.3.
Declarations
Acknowledgements
The authors would like to thank the referee for his/her reading of the original version of the manuscript with valuable suggestions and comments. This work was supported by the National Natural Science Foundation of China (11371225).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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