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Existence of entire solutions of some nonlinear differentialdifference equations
Journal of Inequalities and Applications volume 2017, Article number: 90 (2017)
Abstract
In this paper, we investigate the admissible entire solutions of finite order of the differentialdifference 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 nonzero polynomials, \(\alpha(z)\) is a polynomial and \(c\in\mathbb{C}\backslash\{0\}\). In addition, we investigate the nonexistence of entire solutions of finite order of the differentialdifference equation \((f'(z))^{n}+P(z)f^{m}(z+c)=Q(z)\), where \(P(z)\), \(Q(z)\) are two nonconstant polynomials, \(c\in\mathbb{C}\backslash\{0\}\), m, n are positive integers and satisfy \(\frac{1}{m}+\frac{1}{n}<2\) except for \(m=1\), \(n=2\).
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
Let \(f(z)\), \(g(z)\) be two transcendental entire functions. Then if m and n are integers ≥3, the functional equation
cannot hold.
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 differentialdifference equation and obtained the following results.
Theorem B
The transcendental entire solutions with finite order of
must satisfy \(f(z)=\sin(z\pm iB)\), where B is a constant and \(c=2k\pi\) or \(c=2k\pi+\pi\), k is an integer.
Theorem C
The transcendental entire solutions with finite order of
must satisfy \(f(z)=\frac{1}{2}\sin(2z+iB)\), where \(c=k\pi+\frac{\pi}{2}\), k is an integer, and B is a constant.
In [7], Chen and Gao improved Theorem B and obtained the following result.
Theorem D
Let \(P(z)\), \(Q(z)\) be two nonzero polynomials. If the differentialdifference equation
admits a transcendental entire solution of finite order, then \(P(z)\), \(Q(z)\) reduce to constants, and
where \(a=\pm iA\), \(A=\frac{(1)^{k}k\pi}{c}\), k is an integer, b is a constant and p, q, c are nonzero constants.
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 nonconstant 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
Let \(P(z)\), \(Q(z)\) be two nonzero polynomials, \(c\in\mathbb{C}\backslash\{0\}\) and \(\alpha(z)\) be a polynomial. If the differentialdifference equation
admits a transcendental entire solution of finite order, then \(f(z)\) must satisfy one of the following cases:
(i) \(P(z)\) and \(Q(z)\) reduce to constants, and
where \(A_{1}=ie^{A_{1}c}p\), \(A_{2}=ie^{A_{2}c}p\), \(B_{1}\), \(B_{2}\) are constants and \(A_{1}\), \(A_{2}\), \(q_{1}\), \(q_{2}\), p, c are nonzero constants;
(ii) \(P(z)\) reduces to a constant, \(Q(z)\) is a polynomial with degree 1, and
or
where \(A_{1}=ie^{A_{1}c}p\), \(A_{2}=ie^{A_{2}c}p\), \(B_{1}\), \(B_{2}\), \(a_{0}\), \(b_{0}\) are constants and \(A_{1}\), \(A_{2}\), \(q_{1}\), \(q_{2}\), \(a_{1}\), \(b_{1}\), p, c are nonzero constants;
(iii)
where \(B(z)\) satisfies \([B'(z)+AB(z)]^{2}+P^{2}(z)B^{2}(z+c)e^{2Ac}=Q(z)e^{D}\), A, c are nonzero constants, D is a constant.
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
Let \(Q(z)\) be a nonzero polynomial, \(\alpha(z)\) be a polynomial and \(c\in\mathbb{C}\backslash\{0\}\). If the differentialdifference equation
admits a transcendental entire solution of finite order, then \(f(z)\) must satisfy one of the following cases:
(i)
where \(B(z)\) satisfies \([B'(z)+AB(z)]^{2}+[B(z+c)e^{Ac}B(z)]^{2}=Q(z)e^{D}\), A, c are nonzero constants, \(c_{0}\), D are constants; In particular, if \(A=\pm i\), then
or
or
or
where \(c_{1}\), \(c_{2}\), \(c_{3}\), \(c_{4}\), \(B_{1}\), \(B_{2}\), \(a_{0}\), \(b_{0}\) are constants and \(a_{1}\), \(b_{1}\), \(q_{1}\), \(q_{2}\), c are nonzero constants;
(ii)
or
where \(B_{1}\), \(B_{2}\), \(c_{5}\), \(c_{6}\) are constants and \(A_{1}\), \(A_{2}\), \(q_{1}\), \(q_{2}\) are nonzero constants;
(iii)
or
or
or
where \(e^{A_{1}c}1=iA_{1}\), \(e^{A_{2}c}1=iA_{2}\), \(B_{1}\), \(B_{2}\), \(a_{0}\), \(b_{0}\) are constants and \(A_{1}\), \(A_{2}\), \(q_{1}\), \(q_{2}\), \(a_{1}\), \(b_{1}\), c are nonzero constants.
Remark 1.3
In equation (1.6), when \(Q(z)\) is nonzero constant and \(\alpha(z)\) is a constant, then equation (1.6) reduces to equation (1.3). That is, Theorem 1.2 generalizes Theorem C.
Fermat type functional equations were investigated by Gross [15, 16] and many others. In [17], Yang studied the Fermat type functional equation
where \(a(z)\), \(b(z)\) are small functions with respect to \(f(z)\) and obtained the following result.
Theorem E
Let m, n be positive integers satisfying \(\frac{1}{m}+\frac{1}{n}<1\). Then there are no nonconstant entire solutions \(f(z)\) and \(g(z)\) that satisfy (1.7).
When \(g(z)\) has a specific relationship with \(f(z)\) in (1.7), Liu et al. [10] studied the differentialdifference equation
and obtained the following result.
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
Let \(P(z)\), \(Q(z)\) be two nonconstant polynomials and \(c\in\mathbb{C}\backslash\{0\}\), then the equation
has no transcendental entire solutions with finite order, provided that \(\frac{1}{m}+\frac{1}{n}<2\) except for \(m=1\), \(n=2\), where m, n are positive integers.
Some lemmas
In order to prove our conclusions, we need some lemmas.
Lemma 2.1
Let \(f(z)\) be a transcendental meromorphic solution of
where \(P(z,f)\) and \(Q(z,f)\) are polynomials in \(f(z)\) and its derivatives with meromorphic coefficients, say \(\{a_{\lambda}\lambda\in I\}\), such that \(m(r,a_{\lambda})=S(r,f)\) for all \(\lambda\in I\). If the total degree of \(Q(z,f)\) as a polynomial in \(f(z)\) and its derivatives is ≤n, then \(m(r,P(z,f))=S(r,f)\).
Lemma 2.2
see [19]
Suppose that \(f_{1}(z), f_{2}(z),\ldots,f_{n}(z)\) (\(n\geq2\)) are meromorphic functions and \(g_{1}(z), g_{2}(z),\ldots,g_{n}(z)\) are entire functions satisfying the following conditions:

(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 \(1\leq j\leq n\), \(1\leq h< k\leq n\), \(T(r,f_{j}(z))=o(T(r,e^{g_{h}(z)g_{k}(z)}))\) (\(r\rightarrow\infty\), \(r\notin E\)).
Then \(f_{j}(z)\equiv0\) (\(j=1,\ldots, n\)).
Lemma 2.3
see [19]
Suppose that \(f_{1}(z), f_{2}(z),\ldots,f_{n}(z)\) (\(n\geq3\)) are meromorphic functions which are not constants except for \(f_{n}(z)\). Furthermore, let \(\sum_{j=1}^{n} f_{j}(z)\equiv1\). If \(f_{n}(z)\not\equiv0\) and
where \(r\in I\), \(k=1,2,\ldots, n1\) and \(\lambda<1\), then \(f_{n}(z)\equiv1\).
Lemma 2.4
Let \(Q(z)\) be a nonzero polynomial and satisfy
where a, c are nonzero constants, b is a constant, then one of the following cases holds:

(i)
if \(b=0\) and \(a\neq c\), then \(Q(z)\) reduces to a nonzero constant;

(ii)
if \(b=0\) and \(a=c\), then \(Q(z)\) reduces to a nonzero constant or \(Q(z)=a_{1}z+a_{0}\), where \(a_{1}\) is a nonzero constant, \(a_{0}\) is a constant;

(iii)
if \(b\neq0\) and \(a\neq c\), then \(Q(z)=a_{1}z+a_{0}\) and \(b=a_{1}(ca)\), where \(a_{1}\) is a nonzero 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 nonzero constant, \(a_{1}\), \(a_{0}\) are constants.
Proof
Denote
Then
(i) If \(b=0\) and \(a\neq c\), comparing the coefficients of \(z^{s1}\) 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 \(z^{s2}\) on both sides of (2.1), we see that \(a_{s}C_{s}^{2}c^{2}+a_{s1}C_{s1}^{1}c=a(s1)a_{s1}\), then \(a_{s}C_{s}^{2}c^{2}=0\), a contradiction. Thus \(s\leq1\), that is, \(Q(z)\) reduces to a nonzero constant or \(Q(z)\) is a nonconstant polynomial with degree 1.
(iii) If \(b\neq0\) and \(a\neq c\), \(Q(z)\) is a nonzero constant, note that \(b\neq0\), clearly (2.1) is a contradiction. If \(s\geq2\), comparing the coefficients of \(z^{s1}\) 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 \(b=a_{1}(ca)\neq0\).
(iv) If \(b\neq0\) and \(a=c\), \(Q(z)\) is a nonzero constant, note that \(b\neq0\), clearly (2.1) is a contradiction. If \(s=1\), by (2.1), we see that \(b=a_{1}(ca)=0\), a contradiction. If \(s\geq3\), comparing the coefficients of \(z^{s2}\) on both sides of (2.1), we see that \(a_{s}C_{s}^{2}c^{2}+a_{s1}C_{s1}^{1}c=a(s1)a_{s1}\), 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]
Let \(\eta_{1}\), \(\eta_{2}\) be two complex numbers such that \(\eta_{1}\neq\eta_{2}\) and let \(f(z)\) be a finite order meromorphic function. Then we have
Proof of Theorem 1.1
Suppose that \(f(z)\) is a transcendental entire solution of finite order of (1.5), then
Thus, both \(f'(z)+iP(z)f(z+c)\) and \(f'(z)iP(z)f(z+c)\) are entire functions with finitely many zeros. Combining (3.1) with the Hadamard factorization theorem [19], Theorem 2.5, we assume that
and
where \(Q_{1}(z)\), \(Q_{2}(z)\) are two nonzero polynomials, \(\alpha_{1}(z)\), \(\alpha_{2}(z)\) are two polynomials and cannot be constants simultaneously, otherwise \(f(z)\) is a polynomial. Thus, we have
and
Differentiating (3.3) and shifting (3.2) by replacing z with \(z+c\), we have
We deduce that \([Q'_{1}(z)+Q_{1}(z)\alpha _{1}'(z)]P(z)Q_{1}(z)P'(z)\not\equiv0\) and \([Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z)]P(z)Q_{2}(z)P'(z)\not\equiv0\). If \([Q'_{1}(z)+Q_{1}(z)\alpha_{1}'(z)]P(z)Q_{1}(z)P'(z)\equiv0\), then \(P(z)\equiv AQ_{1}(z)e^{\alpha_{1}(z)}\), where A is a nonzero constant. Note that \(P(z)\), \(Q_{1}(z)\) are nonzero polynomials, then \(\alpha_{1}(z)\) must be a constant. Let \(\alpha_{1}(z)\equiv A_{1}\). Since \(\alpha_{1}(z)\) and \(\alpha_{2}(z)\) cannot be constants simultaneously, thus \(\alpha_{2}(z)\) cannot be a constant, then \([Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z)]P(z)Q_{2}(z)P'(z)\not\equiv0\). Then (3.4) can be rewritten as
If \(\deg\alpha_{2}(z)\geq2\), then \(\deg\alpha_{2}(z)=\deg\alpha_{2}(z+c)\geq2\), \(\deg(\alpha_{2}(z+c)\alpha_{2}(z))\geq1\), and \(e^{\alpha_{2}(z)}\), \(e^{\alpha_{2}(z+c)}\), \(e^{\alpha_{2}(z+c)\alpha_{2}(z)}\) are of regular growth, by Lemma 2.2, we have
a contradiction. Thus \(\deg\alpha_{2}(z)\leq1\), note that \(\alpha_{2}(z)\) cannot be a constant, then \(\alpha_{2}(z)=A_{2}z+B_{2}\), where \(A_{2}\) is a nonzero constant. Rewriting (3.5) as
where \(H(z)=i P^{2}(z)Q_{2}(z+c)e^{A_{2}c+B_{2}}+ [(Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z))P(z)Q_{2}(z)P'(z)]e^{B_{2}}\). If \(H(z)\equiv0\), since \(i P^{2}(z)Q_{1}(z+c)e^{A_{1}}\not\equiv0\), clearly (3.6) is a contradiction. If \(H(z)\not\equiv0\), we can see that the left side of (3.6) is a transcendental entire function, and the right side of (3.6) is a nonzero polynomial, a contradiction.
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.
Case 1, \(e^{\alpha_{1}(z)\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z)\alpha_{1}(z+c)}\) are not constants, by (3.4) and Lemma 2.3, we have
which implies that \(\alpha_{2}(z+c)\alpha_{1}(z+c)\) is a constant, and
which implies that \(\alpha_{1}(z)\alpha_{2}(z)\) is a constant.
Denote \(e^{\alpha_{2}(z+c)\alpha_{1}(z+c)}=e^{\alpha _{2}(z)\alpha_{1}(z)}=k\) (≠0), by (3.7), we get \(Q_{1}(z)=kQ_{2}(z)\), substituting it into (3.8) yields
Since \(P(z)\) and \(Q_{2}(z)\) are nonzero polynomials, \(\alpha_{1}(z)\) and \(\alpha_{2}(z)\) are polynomials, from the above identity, we get \(\alpha_{1}'(z)+\alpha_{2}'(z)\equiv0\), that is, \(\alpha_{1}(z)+\alpha_{2}(z)\) is a constant. Note that \(\alpha_{1}(z)\alpha_{2}(z)\) is a constant, then both \(\alpha_{1}(z)\) and \(\alpha_{2}(z)\) are constants, a contradiction.
Case 2, \(e^{\alpha_{1}(z)\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)\alpha_{1}(z+c)}\) are not constants, by (3.4) and Lemma 2.3, we have
which implies that \(\alpha_{2}(z)\alpha_{1}(z+c)\) is a constant, then \(\alpha_{2}(z+c)\alpha_{1}(z+2c)\) is also a constant, and
which implies that \(\alpha_{1}(z)\alpha_{2}(z+c)\) is a constant.
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.
Case 3, \(e^{\alpha_{2}(z)\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)\alpha_{1}(z+c)}\) are not constants, by (3.4) and Lemma 2.3, we have
which implies that \(\alpha_{1}(z)\alpha_{1}(z+c)\) is a constant, note that \([Q'_{1}(z)+Q_{1}(z)\alpha_{1}'(z)]P(z)Q_{1}(z)P'(z)\not \equiv0\), then \(\alpha_{1}(z)\) cannot be a constant, therefore, \(\alpha_{1}(z)\) can only be a polynomial with degree 1. Denote \(\alpha_{1}(z)=A_{1}z+B_{1}\), where \(A_{1}\) is a nonzero constant, and \(B_{1}\) is a constant. Rewriting (3.9) as
\(P(z)\) must be a constant, denoted \(P(z)\equiv p\) (≠0). Then (3.10) can be rewritten as
By Lemma 2.4, we have (i) if \(\frac{1}{A_{1}}\neq c\), then \(Q_{1}(z)\equiv q_{1}\) (constant); (ii) if \(\frac{1}{A_{1}}=c\), then \(Q_{1}(z)\equiv q_{1}\) (constant) or \(Q_{1}(z)=a_{1}z+a_{0}\), where \(a_{1}\) is a nonzero constant and \(a_{0}\) is a constant.
which implies that \(\alpha_{2}(z)\alpha_{2}(z+c)\) is a constant, note that \([Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z)]P(z)Q_{2}(z)P'(z)\not \equiv0\), so \(\alpha_{2}(z)\) cannot be a constant, then \(\alpha_{2}(z)\) can only be a polynomial with degree 1, denote \(\alpha_{2}(z)=A_{2}z+B_{2}\), where \(A_{2}\) is a nonzero constant and \(B_{2}\) is a constant. Note that \(P(z)\equiv p\) (≠0), then (3.11) can be rewritten as
By Lemma 2.4, we have (i) if \(\frac{1}{A_{2}}\neq c\), then \(Q_{2}(z)\equiv q_{2}\) (constant); (ii) if \(\frac{1}{A_{2}}=c\), then \(Q_{2}(z)\equiv q_{2}\) (constant) or \(Q_{2}(z)=b_{1}z+b_{0}\), where \(b_{1}\) is a nonzero constant and \(b_{0}\) is a constant.
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 \(\frac{1}{A_{2}}=c\).
Subcase 3.1, \(\frac{1}{A_{1}}\neq c\) and \(\frac{1}{A_{2}}\neq c\), then \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\), by (1.5), (3.2) and (3.3), we obtain
and
Subcase 3.2, \(\frac{1}{A_{1}}=c\) and \(\frac{1}{A_{2}}\neq c\), then \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\) or \(Q_{1}(z)=a_{1}z+a_{0}\) and \(Q_{2}(z)\equiv q_{2}\). If \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\), the same as Subcase 3.1. If \(Q_{1}(z)=a_{1}z+a_{0}\) and \(Q_{2}(z)\equiv q_{2}\), by (1.5), (3.2) and (3.3), we obtain
and
Subcase 3.3, \(\frac{1}{A_{1}}\neq c\) and \(\frac{1}{A_{2}}=c\), then \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\) or \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)=b_{1}z+b_{0}\). If \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\), the same as Subcase 3.1. If \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)=b_{1}z+b_{0}\), by (1.5), (3.2) and (3.3), we obtain
and
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 nonzero 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.
Proof of Theorem 1.2
As in the beginning of the proof of Theorem 1.1, from (1.6), we have
and
where \(Q_{1}(z)\), \(Q_{2}(z)\) are two nonzero polynomials, \(\alpha_{1}(z)\), \(\alpha_{2}(z)\) are two polynomials and cannot be constants simultaneously, otherwise \(f(z)\) is a polynomial. Differentiating (4.2), shifting (4.1) by replacing z with \(z+c\) and combining (4.1), we have
If \(Q'_{1}(z)+Q_{1}(z)\alpha_{1}'(z)+iQ_{1}(z)\equiv0\), that is, \(Q'_{1}(z)=(\alpha_{1}'(z)+i)Q_{1}(z)\), then we have \(\alpha_{1}'(z)+i\equiv0\) and \(Q_{1}(z)\equiv q_{1}\) (constant). Thus \(\alpha_{1}(z)=iz+B_{1}\), where \(B_{1}\) is a constant, substitute it into (4.3) yields
Suppose \(\deg\alpha_{2}(z)\geq2\). Clearly, \(Q'_{2}(z)+Q_{2}(z)(\alpha_{2}'(z)i)\not\equiv0\). According to \(\deg\alpha_{2}(z)=\deg\alpha_{2}(z+c)\geq2\), \(\deg\alpha_{1}(z+c)=1\), \(\deg(\alpha_{2}(z+c)\alpha_{1}(z+c))=\deg(\alpha_{2}(z)\alpha _{1}(z+c))\geq2\) and \(\deg(\alpha_{2}(z+c)\alpha_{2}(z))\geq1\), and \(e^{\alpha_{2}(z)}\), \(e^{\alpha_{2}(z+c)}\), \(e^{\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z+c)\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z)\alpha_{1}(z+c)}\), \(e^{\alpha_{2}(z+c)\alpha_{2}(z)}\) are of regular growth, by Lemma 2.2, we have
a contradiction. Thus \(\deg\alpha_{2}(z)\leq1\), that is, \(\alpha_{2}(z)=B_{2}\) (constant) or \(\alpha_{2}(z)=A_{2}z+B_{2}\), where \(A_{2}\) is a nonzero constant. If \(\alpha_{2}(z)\equiv B_{2}\), by (4.4), we have \(e^{\alpha_{1}(z+c)}\equiv\frac {e^{B_{2}}[Q'_{2}(z)iQ_{2}(z)+iQ_{2}(z+c)]}{iQ_{1}(z+c)}\), the left side of this identity is a transcendental entire function, and the right side of this identity is a rational function, a contradiction. Hence, \(\alpha_{2}(z)=A_{2}z+B_{2}\). Rewriting (4.4) as
If \(iA_{2}\neq0\), clearly the above identity is a contradiction. Then \(A_{2}=i\). The above identity can be rewritten as
If \(Q_{2}(z)\equiv q_{2}\) (constant), by (4.5), we get \(q_{1}e^{B_{1}}=q_{2}e^{B_{2}}(2e^{ic}1)\). By (1.6), (4.1) and (4.2), we have
and
If \(Q_{2}(z)\) is a nonconstant polynomial, by (4.5), we obtain
Note that \(iq_{1}e^{ic+B_{1}B_{2}}\neq0\) and \(\frac{1}{2i}\neq c\), by Lemma 2.4, we see that \(Q_{2}(z)=b_{1}z+b_{0}\), where \(b_{1}\) is a nonzero constant and \(b_{0}\) is a constant. From (4.5), we get \(b_{1}(2ic+1)=2iq_{1}e^{B_{1}B_{2}}\), by (1.6), (4.1) and (4.2), we have
and
Similarly, if \(Q'_{2}(z)+Q_{2}(z)\alpha_{2}'(z)iQ_{2}(z)\equiv0\), then we have
or
where \(c_{3}\), \(c_{4}\), \(B_{1}\), \(B_{2}\), \(a_{0}\) are constants and \(a_{1}\), \(q_{1}\), \(q_{2}\) are nonzero constants.
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.
Case 3, \(e^{\alpha_{2}(z)\alpha_{1}(z+c)}\) and \(e^{\alpha_{2}(z+c)\alpha_{1}(z+c)}\) are not constants, by (4.3) and Lemma 2.3, we have
which implies that \(\alpha_{1}(z)\alpha_{1}(z+c)\) is a constant, then \(\alpha_{1}(z)=A_{1}z+B_{1}\) or \(\alpha_{1}(z)\equiv B_{1}\), where \(A_{1}\) is a nonzero constant and \(B_{1}\) is a constant. By (4.3) and (4.7), we also have
which implies that \(\alpha_{2}(z)\alpha_{2}(z+c)\) is a constant, then \(\alpha_{2}(z)=A_{2}z+B_{2}\) or \(\alpha_{2}(z)\equiv B_{2}\), where \(A_{2}\) is a nonzero constant and \(B_{2}\) is a constant. Note that \(\alpha_{1}(z)\) and \(\alpha_{2}(z)\) cannot be constants simultaneously, In what follows, we discuss three subcases: Subcase 3.1, \(\alpha_{1}(z)\equiv B_{1}\) and \(\alpha_{2}(z)=A_{2}z+B_{2}\); Subcase 3.2, \(\alpha_{1}(z)=A_{1}z+B_{1}\) and \(\alpha_{2}(z)\equiv B_{2}\); Subcase 3.3, \(\alpha_{1}(z)=A_{1}z+B_{1}\) and \(\alpha_{2}(z)=A_{2}z+B_{2}\).
Subcase 3.1, \(\alpha_{1}(z)\equiv B_{1}\) and \(\alpha_{2}(z)=A_{2}z+B_{2}\), by (4.6), we have
By Lemma 2.4, we see that if \(c\neqi\), then \(Q_{1}(z)\equiv q_{1}\) (constant); If \(c=i\), then \(Q_{1}(z)\equiv q_{1}\) (constant) or \(Q_{1}(z)=a_{1}z+a_{0}\), where \(a_{1}\) is a nonzero constant and \(a_{0}\) is a constant.
By (4.7), we have
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 nonzero 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 \(\frac{1}{A_{2}i}\neqi\). 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, \(\frac{1}{A_{2}i}=c\) (\(\neqi\)), \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)=b_{1}z+b_{0}\).
Subcase 3.1.1, \(\frac{1}{A_{2}i}\neq c\) (\(=i\)), \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\), by (1.6), (4.1) and (4.2), we get
and
Subcase 3.1.2, \(\frac{1}{A_{2}i}\neq c\) (\(=i\)), \(Q_{1}(z)=a_{1}z+a_{0}\) and \(Q_{2}(z)\equiv q_{2}\), by (1.6), (4.1) and (4.2), we have
and
then \(a_{1}\) must be zero, a contradiction.
Subcase 3.1.3, \(\frac{1}{A_{2}i}=c\) (\(\neqi\)), \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)=b_{1}z+b_{0}\), by (1.6), (4.1) and (4.2), we have
and
then \(c=i\), note that \(c\neqi\), a contradiction.
Subcase 3.2, \(\alpha_{1}(z)=A_{1}z+B_{1}\) and \(\alpha_{2}(z)\equiv B_{2}\), similarly to the proof of Subcase 3.1, (1.6) admits a transcendental entire solution, if and only if \(\frac{1}{A_{1}+i}\neq c\) (=i), \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\). By (1.6), (4.1) and (4.2), we get
and
where \(B_{1}\), \(B_{2}\), \(c_{6}\) are constants and \(q_{1}\), \(q_{2}\) are nonzero constants.
Subcase 3.3, \(\alpha_{1}(z)=A_{1}z+B_{1}\) and \(\alpha_{2}(z)=A_{2}z+B_{2}\), by (4.6) and (4.7), we obtain, respectively,
and
Clearly, \(A_{1}\neq A_{2}\). In what follows, we discuss four subcases: Subcase 3.3.1, \(\frac{1}{A_{1}+i}\neq c\) and \(\frac{1}{A_{2}i}\neq c\); Subcase 3.3.2, \(\frac{1}{A_{1}+i}=c\) and \(\frac{1}{A_{2}i}\neq c\); Subcase 3.3.3, \(\frac{1}{A_{1}+i}\neq c\) and \(\frac{1}{A_{2}i}=c\); Subcase 3.3.4, \(\frac{1}{A_{1}+i}=\frac{1}{A_{2}i}=c\).
Subcase 3.3.1, \(\frac{1}{A_{1}+i}\neq c\) and \(\frac{1}{A_{2}i}\neq c\), by (4.8), (4.9) and Lemma 2.4, we have \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\), where \(q_{1}\) and \(q_{2}\) are nonzero constants. By (1.6), (4.1) and (4.2), we get
and
Subcase 3.3.2, \(\frac{1}{A_{1}+i}=c\) and \(\frac{1}{A_{2}i}\neq c\). By \(\frac{1}{A_{1}+i}=c\), (4.8) and Lemma 2.4, we have \(Q_{1}(z)\equiv q_{1}\) or \(Q_{1}(z)=a_{1}z+a_{0}\), where \(a_{1}\), \(q_{1}\) are nonzero constants and \(a_{0}\) is a constant. By \(\frac{1}{A_{2}i}\neq c\), (4.9) and Lemma 2.4, we have \(Q_{2}(z)\equiv q_{2}\), where \(q_{2}\) is a nonzero constant. If \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\), the same as Subcase 3.3.1. If \(Q_{1}(z)=a_{1}z+a_{0}\) and \(Q_{2}(z)\equiv q_{2}\), by (1.6), (4.1) and (4.2), we get
and
Subcase 3.3.3, \(\frac{1}{A_{1}+i}\neq c\) and \(\frac{1}{A_{2}i}=c\), similarly to the proof of Subcase 3.3.2, we can obtain
or
where \(B_{1}\), \(B_{2}\), \(b_{0}\), \(c_{7}\), \(c_{9}\) are constants and \(A_{1}\), \(A_{2}\), \(q_{1}\), \(q_{2}\), \(b_{1}\), c are nonzero constants.
Subcase 3.3.4, \(\frac{1}{A_{1}+i}=\frac{1}{A_{2}i}=c\). By \(\frac{1}{A_{1}+i}=c\), (4.8) and Lemma 2.4, we have \(Q_{1}(z)\equiv q_{1}\) or \(Q_{1}(z)=a_{1}z+a_{0}\), where \(a_{1}\), \(q_{1}\) are nonzero constants and \(a_{0}\) is a constant. By \(\frac{1}{A_{2}i}=c\), (4.9) and Lemma 2.4, we have \(Q_{2}(z)\equiv q_{2}\) or \(Q_{2}(z)=b_{1}z+b_{0}\), where \(b_{1}\), \(q_{2}\) are nonzero constants and \(b_{0}\) is a constant. If \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)\equiv q_{2}\), the same as Subcase 3.3.1. If \(Q_{1}(z)=a_{1}z+a_{0}\) and \(Q_{2}(z)\equiv q_{2}\), the same as Subcase 3.3.2. If \(Q_{1}(z)\equiv q_{1}\) and \(Q_{2}(z)=b_{1}z+b_{0}\), the same as Subcase 3.3.3. If \(Q_{1}(z)=a_{1}z+a_{0}\) and \(Q_{2}(z)=b_{1}z+b_{0}\), by (1.6), (4.1) and (4.2), we get
and
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 nonzero 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.
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 nonconstant 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.
Case 4, \(n\geq3\), \(m=1\). Differentiating (1.9), we get
Substituting (1.9) into the above equation yields
Denote \(F(z)=f'(z)\), \(\varphi(z)=nf''(z)\frac{P'(z)}{P(z)}f'(z)=nF'(z)\frac{P'(z)}{P(z)}F(z)\). Then (5.1) can be rewritten as
By Lemma 2.5, we see that \(m (r,\frac{F(z+c)}{F(z)} )=S(r,F)\) and \(m (r,Q'(z)Q(z)\frac{P'(z)}{P(z)} )=S(r,F)\), note that \(n1\geq2\), by Lemma 2.1, we have
We see that \(\varphi(z)\not\equiv0\), otherwise \((f'(z))^{n}=F^{n}(z)\equiv AP(z)\), where A is a nonzero constant, a contradiction. Note that \(f(z)\) is a transcendental entire function, then \(N(r,\varphi(z))=S(r,F)\) and
that is, \(T(r,f'(z))=T(r,F(z))\leq S(r,F)=S(r,f')\), a contradiction.
This completes the proof of Theorem 1.3.
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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).
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Chen, M., Gao, Z. & Du, Y. Existence of entire solutions of some nonlinear differentialdifference equations. J Inequal Appl 2017, 90 (2017). https://doi.org/10.1186/s1366001713681
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DOI: https://doi.org/10.1186/s1366001713681
MSC
 39B32
 39A10
 34M05
Keywords
 differentialdifference equation
 admissible entire solution
 finite order