The main purpose of this article is to study Fermat-type complex differential-difference equations f(k)(z)2+[αf(z+c)−βf(z)]2=R(z). Our results improve some results due to Wang–Xu–Tu [AIMS. Mathematics, 2020], Zhang [Bull. Korean. Math. Soc, 2018], and Long–Qin [Applied Mathematics-A Journal of Chinese Universities, 2024]. Moreover, we provide some examples to show the existence of the solutions.
Citation: Zhiyong Xu, Junfeng Xu. Solutions to some generalized Fermat-type differential-difference equations[J]. AIMS Mathematics, 2024, 9(12): 34488-34503. doi: 10.3934/math.20241643
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The main purpose of this article is to study Fermat-type complex differential-difference equations f(k)(z)2+[αf(z+c)−βf(z)]2=R(z). Our results improve some results due to Wang–Xu–Tu [AIMS. Mathematics, 2020], Zhang [Bull. Korean. Math. Soc, 2018], and Long–Qin [Applied Mathematics-A Journal of Chinese Universities, 2024]. Moreover, we provide some examples to show the existence of the solutions.
We first assume that the reader is familiar with the basic results and notations of the Nevanlinna theory and difference Nevanlinna theory with one complex variable, which can be found in [3,7,10,22]. In the past thirty years, there were lots of research focusing on the solutions of Fermat-type differential-difference equations; readers can refer to [1,4,5,6,9,11,12,15,16,17,18,19,20,21].
A. Wiles [13] in 1995 pointed out: The Fermat equation xm+ym=1 does not admit nontrivial solutions in rational numbers as m≥3, and this equation possesses nontrivial rational solutions as m=2. About sixty years ago, Gross [2] investigated the existence of solutions for the Fermat-type functional equation fm+gm=1 and obtained: For m=2, the entire solutions are f=cosα(z), g=sinα(z), where α(z) is an entire function; for m>2, there are no nonconstant entire solutions. In 2009, Liu [7] proved that the Fermat-type equation f(z)2+[f(z+c)−f(z)]2=a2 has no nonconstant entire solutions of finite order, where a is a nonzero constant. In 2012, Liu [8] studied that the Fermat-type equation f′(z)2+[f(z+c)−f(z)]2=1 has the transcendental entire solutions with finite order. In 2018, Zhang [23] generalized Liu's [7,8] theorem, and obtained
Theorem 1.1. Let f be a transcendental meromorphic function with finitely many poles and σ(f)<∞. Then f can not be a solution of the difference equation
f(z)2+[f(z+c)−f(z)]2=R(z), |
where R(z) is a nonzero rational function and c is a nonzero constant.
Theorem 1.2. Let f be a transcendental meromorphic function with finitely many poles and σ(f)<∞. If f is a solution of the differential-difference equation
f′(z)2+[f(z+c)−f(z)]2=R(z), |
where R(z) is a nonzero rational function and c is a nonzero constant, then R(z) is a nonzero constant, and f is of the form
f(z)=c1e2iz+c2e−2iz+b,c=kπ+π/2, |
where c1, c2 are two nonzero constants such that 16c1c2=R(z), b is a constant, and k is an integer.
In 2020, Wang et al. [14] promoted Zhang's [23] form and obtained
Theorem 1.3. Let α(≠0), β∈C, k be an integer. Let f be a transcendental meromorphic solution of difference-differential equation of Fermat type
f′(z)2+[αf(z+c)−βf(z)]2=R(z), |
where R(z) is a nonzero rational function and c is a nonzero constant. If f is of finite order and has finitely many poles, then iαc=±1, and R(z) is a nonconstant polynomial with degzR≤2, or R(z) is a nonzero constant. Furthermore
(I) If R(z) is a nonconstant polynomial and degzR≤2, then f is of the form
f(z)=s1(z)eaz+b+s2(z)e−(az+b)2, |
where R(z)=(as1(z)+m1)(−as2(z)+m2), a≠0, b∈C, and a,b,c,α,β satisfy α≠±β, a=−i(α+β), c=(2k+1)πai, iαc=−1 or a=i(α−β), c=2kπai, iαc=1, where sj=mjz+nj, mj, nj∈C(j=1,2);
(II) If R(z) is a nonzero constant, then f is of the form
f(z)=n1eaz+b+n2e−(az+b)2+d, |
R(z)=−a2n1n2, a≠0, b∈C, and a,b,c,α,β satisfy the following cases:
(II1) if α=β, then a=−2αi, c=(2k+1)πai and d∈C;
(II2) if α=−β, then a=2αi, c=2kπai and d=0;
(II3) if α≠±β, then d=0 and a=−i(α+β), c=(2k+1)πai or a=i(α−β),c=2kπai.
Theorem 1.4. Let α(≠0), β∈C, k be an integer. Let f be a transcendental meromorphic solution of difference-differential equation of Fermat type
f″(z)2+[αf(z+c)−βf(z)]2=R(z), | (1.1) |
where R(z) is a nonzero rational function and c is a nonzero constant.
(I) If α=±β, then Eq (1.1) has no finite-order transcendental meromorphic solution with finitely many poles;
(II) If α≠±β, and Eq (1.1) has a finite-order transcendental meromorphic solution f with finitely many poles, then R(z) must be a nonzero polynomial with degzR≤1. Furthermore,
(II1) if R(z) is a nonzero polynomial of degree one, then f(z) is of the form
f(z)=s1(z)eaz+b+n2e−(az+b)2, |
where a4=α2−β2, b∈C, c=loga2+iβiα+2kπia, eac=2aiαc≠±1 and R(z)=a3n2[as1(z)+2m1], s1(z)=m1z+n1, or f(z) is of the form
f(z)=n1eaz+b+s2(z)e−(az+b)2, |
where a4=α2−β2, b∈C, c=log−a2+iβiα+(2k+1)πia, eac=iαc2a≠±1 and R(z)=a3n1[as2(z)−2m2], s2(z)=m2z+n2;
(II2) if R(z) is a nonzero constant, then f(z) is of the form
f(z)=c1eaz+b+c2e−(az+b)2, |
where a,b,c,α,β,c1,c2,R(z) satisfy a4=α2−β2,b∈C, c=loga2+iβiα+2kπia and R(z)=a4c1c2.
Inspired by [14], can f′ and f″ in Theorems 1.3 and 1.4 be replaced by f(k)? In this paper, we consider this question. Our results are listed as follows.
Theorem 1.5. Let α(≠0), β∈C, k∈Z+, and k be an odd number. Let f be a transcendental meromorphic solution of difference-differential equation of Fermat type
f(k)(z)2+[αf(z+c)−βf(z)]2=R(z), | (1.2) |
where R(z) is a nonzero rational function and c is a nonzero constant.
If f is of finite order and has finitely many poles, then iαc=±1, and R(z) is a polynomial with degzR=1, or R(z) is a nonzero constant. Let sj(z)=mjz+nj, mj, nj∈C(j=1,2).
Case I: If R(z) is a polynomial with degzR=1, then f is of the form
f(z)=s1(z)eaz+b+s2(z)e−(az+b)2, |
where R(z)=−n2a2k−1[as1(z)+km1], m1≠0, m2=0, a≠0, b∈C, and a,b,c,α,β satisfy α≠±β, ak=−i(α+β), c=(2l+1)πai, l∈Z, iαc=−kak−1=−1 or ak=i(α−β), c=2lπai, l∈Z, iαc=kak−1=1; or R(z)=n1a2k−1[−as2(z)+km2], m1=0, m2≠0, a≠0, b∈C, and a,b,c,α,β satisfy α≠±β, ak=−i(α+β), c=(2l+1)πai, l∈Z, iαc=kak−1=−1 or ak=i(α−β), c=2lπai, l∈Z, iαc=−kak−1=1.
Case II: If R(z) is a nonzero constant, then mj=0(j=1,2), f is of the form
f(z)=n1eaz+b+n2e−(az+b)2+c0, |
where R(z)=−a2kn1n2,a≠0,b∈C, and a,b,c,α,β satisfy the following cases:
(II1) if α=β, then ak=−2αi, c=(2l+1)πai and c0∈C, l∈Z;
(II2) if α=−β, then ak=2αi, c=2lπai and c0=0, l∈Z;
(II3) if α≠±β, then c0=0 and ak=−i(α+β), c=(2l+1)πai, l∈Z or ak=i(α−β), c=2lπai, l∈Z.
Remark 1. When k=1, Theorem 1.5 becomes Theorem 1.3.
Remark 2. In [14], we find that Case (I) of Theorem 1.3 is not accurate, including the corresponding examples. On page 695, line 11 of [14], substituting (5.10) into the first equation in (5.3), we have R2(z)=−as2(z)+m2. Meanwhile, substituting (5.10) into the second equation in (5.3), we have R2(z)=−as2(z)−m2. From this, we have m2=0, which is a contradiction. Therefore, if f′(z)2+[αf(z+c)−βf(z)]2=R(z) has a solution, then degzR≤1.
Next, we will provide some examples to explain the existence of solutions to Eq (1.2) in the above situation.
Example 1.1. For Case I, the transcendental entire function
f(z)=(z+1)e√33z+b+e−(√33z+b)2 |
satisfies
f(3)(z)2+[1√3πf(z+c)−(√39i−1√3π)f(z)]2=−√39[√39(z+1)+1], |
where k=3, s1(z)=z+1, s2(z)=1, c=√3πi, a=√33, b∈C, α=1√3π, β=√39i−1√3π and R(z)=−√39[√39(z+1)+1].
The transcendental entire function
f(z)=e√33iz+b+(z+1)e−(√33iz+b)2 |
satisfies
f(3)(z)2+[i√3πf(z+c)−(√39−i√3π)f(z)]2=−√39i[√39i(z+1)−1], |
where k=3, s1(z)=1, s2(z)=z+1, c=√3π, a=√33i, b∈C, α=i√3π, β=√39−i√3π and R(z)=−√39i[√39i(z+1)−1].
Example 1.2. For Case II1, the transcendental entire function
f(z)=ez+b+2e−(z+b)2+c0 |
satisfies
f(3)(z)2+[i2f(z+c)−i2f(z)]2=−2, |
where k=3, n1=1, n2=2, a=1, b,c0∈C, c=πi, α=i2, β=i2 and R(z)=−2.
Example 1.3. For Case II2, the transcendental entire function
f(z)=e2z+b+e−(2z+b)2, |
satisfies
f(3)(z)2+[−4if(z+c)−4if(z)]2=−64, |
where k=3, n1=1, n2=1, a=2, b∈C, c=πi, α=−4i, β=4i and R(z)=−64.
Example 1.4. For Case II3, the transcendental entire function
f(z)=2ez+b+e−(z+b)2 |
satisfies
f(3)(z)2+[f(z+c)−(1+i)f(z)]2=−64, |
where k=3, n1=2, n2=1, a=1, b∈C, c=2πi, α=1, β=1+i and R(z)=−2.
Theorem 1.6. Let α(≠0), β∈C, k∈Z+ and k be an even number. Let f be a transcendental meromorphic solution of Eq (1.2).
If f is of finite order and has finitely many poles, then R(z) is a polynomial with degzR=1, or R(z) is a nonzero constant.
Case I: If α=±β, then Eq (1.2) has no-finite order transcendental meromorphic solution with finitely many poles;
Case II: If α≠±β, and Eq (1.2) has a finite-order transcendental meromorphic solution f with finitely many poles, then R(z) must be a nonzero polynomial with degzR≤1. Let sj=mjz+nj, mj, nj∈C(j=1,2),
(II1) if R(z) is a polynomial with degzR=1, then f(z) is of the form
f(z)=s1(z)eaz+b+n2e−(az+b)2, |
where a2k=α2−β2,b∈C, c=logak+iβiα+2lπia, l∈Z, eac=kak−1iαc≠±1 and R(z)=n2a2k−1[as1(z)+km1], or f(z) is of the form
f(z)=n1eaz+b+s2(z)e−(az+b)2, |
where a2k=α2−β2, b∈C, c=logak+iβiα+2lπia, l∈Z, eac=iαckak−1≠±1 and R(z)=n1a2k−1[as2(z)−km2].
(II2) if R(z) is a nonzero constant, then f(z) is of the form
f(z)=n1eaz+b+n2e−(az+b)2, |
where a,b,c,α,β,c1,c2, R(z) satisfy a2k=α2−β2, b∈C, l∈Z, c=logak+iβiα+2lπia and R(z)=a2kn1n2.
Remark 3. When k=2, Theorem 1.6 becomes Theorem 1.4.
Next, we will provide some examples to explain the existence of solutions to Eq (1.2) in the above situation.
Example 1.5. For Case II1, let c0 be a solution of equation e2c(2−c)=2, the transcendental entire function
f(z)=zez+b+e−(z+b)2 |
satisfies
f(4)(z)2+[4ic0ec0f(z+c)−4−c0ic0f(z)]2=−64, |
where k=4, s1(z)=z, n2=1, a=1, b∈C, α=4ic0ec0, β=4−c0ic0, α2−β2=1, c0=log1+iβiα, eac0=4iαc0 and R(z)=z+4.
Example 1.6. For Case II2, the transcendental entire function
f(z)=ez+b+e−(z+b)2 |
satisfies
f(4)(z)2+[i2f(z+c)−√52if(z)]2=1, |
where k=4, n1=1, n2=1, ec=√5−2, a=1, b∈C, α=i2, β=√52i, α2−β2=1, c=log(√5−2) and R(z)=1.
In 2024, Long and Qin [9] studied this equation
f(k)(z)2+P(z)2f(z+c)2=Q(z), |
and obtained
Theorem 1.7. There is no transcendental entire solution with finite order of the equation
f(k)(z)2+P(z)2f(z+c)2=Q(z), | (1.3) |
where P(z) is a non-constant polynomial, and Q(z) is a non-zero polynomial.
Motivating from Theorems 1.5 and 1.6, we replace f(z+c) by the αf(z+c)−βf(z) in Theorem 1.7 and obtain
Theorem 1.8. There is no transcendental entire solution with finite order of the equation
f(k)(z)2+P(z)2(αf(z+c)−βf(z))2=Q(z), | (1.4) |
where k∈Z+, α(≠0), β∈C, P(z) is a non-constant polynomial and Q(z) is a non-zero polynomial.
We can use these lemmas to prove our theorems.
Lemma 2.1. Let c, a, α be three nonzero constants, k∈Z+ be an odd number, and ac≠±(k−1). If R1,R2 are two nonzero rational functions satisfying the following differential-difference equations
{iα[R1(z+c)−R1(z)]=k−1∑i=0CikR(k−i)1ai,iα[R2(z+c)−R2(z)]=k−1∑i=0(−1)i+1CikR(k−i)2ai. | (2.1) |
Then iαc=kak−1 and Ri are nonzero polynomials with degzRi≤1 (i=1,2).
Wang et al. [14] proved the case of k=1, below we prove the case of k≥3.
Proof. First, we prove that R1(z) has no poles. On the contrary, suppose that z0 is a pole of R1(z). We can write (2.1) in a new form
{R1(z+c)=k−1∑i=0CikR(k−i)1aiiα+R1(z),R2(z+c)=k−1∑i=0(−1)i+1CikR(k−i)2aiiα+R2(z). | (2.2) |
It is easy to see that z0+c is also a pole of R1(z) by comparing the order of pole z0 on both sides of (2.2). By recycling this operation, we obtain a sequence of poles of R1(z) that are z0+2c, z0+3c, ⋯, z0+tc, this is impossible since R1(z) is a nonzero rational function. Then, R1(z) is a polynomial. Similarly, it can be inferred that R2(z) also is a polynomial.
Therefore, R1(z) and R2(z) are two nonzero polynomials. Let
R1(z)=apzp+ap−1zp−1+⋯+a1z+a0,R2(z)=btzt+bt−1zt−1+⋯+b1z+b0, |
where aj∈C, j∈{0,1,⋯,p}, bj∈C, j∈{0,1,⋯,t}, ap≠0, bt≠0. Substituting R1(z) and R2(z) into (2.1) and comparing the coefficients of zp−1, zp−2, zt−1, and zt−2 on both sides of these two equations, we have
{iαappc=C1kak−1app,iαbttc=C1kak−1btt,iα[apC2pc2+ap−1(p−1)c]=C1kak−1(p−1)ap−1+C2kak−2p(p−1)ap,iα[btC2tc2+bt−1(t−1)c]=C1kak−1(t−1)bt−1−C2kak−2t(t−1)bt, | (2.3) |
which means
{iαc=kak−1,(C1kac2−C2k)p(p−1)=0,(C1kac2+C2k)t(t−1)=0. | (2.4) |
It follows that iαc=kak−1, p=0 or 1 and t=0 or 1. Therefore, this completes the proof of Lemma 2.1.
Lemma 2.2. Let c, a, α be three nonzero constants and k∈Z+ is an even number. If R1,R2 are two nonzero rational functions satisfying the following differential-difference equations:
{iαeac[R1(z+c)−R1(z)]=k−1∑i=0CikR(k−i)1ai,iαe−ac[R2(z+c)−R2(z)]=k−1∑i=0(−1)i+1CikR(k−i)2ai. | (2.5) |
Then eac=±1 and Ri are nonzero polynomials with degzRi≤1 (i=1,2).
Proof. First, we prove that R1(z) has no poles. On the contrary, suppose that z0 is a pole of R1(z). We can write (2.5) in a new form
{R1(z+c)=k−1∑i=0CikR(k−i)1aiiαeac+R1(z),R2(z+c)=k−1∑i=0(−1)i+1CikR(k−i)2aiiαeac+R2(z). | (2.6) |
Similar to the proof of Lemma 2.1, we can get R1(z) and R2(z) are two nonzero polynomials. Substituting R1(z) and R2(z) into (2.5) and comparing the coefficients of zp−1, zp−2, zt−1 and zt−2 on both sides of such two equations, it yields
{iαeacappc=C1kak−1app,iαe−acbttc=C1kak−1btt,iαeac[apC2pc2+ap−1(p−1)c]=C1kak−1(p−1)ap−1+C2kak−2p(p−1)ap,iαe−ac[btC2tc2+bt−1(t−1)c]=C1kak−1(t−1)bt−1−C2kak−2t(t−1)bt, | (2.7) |
which means
{eac=±1,(C1kac2−C2k)p(p−1)=0,(C1kac2+C2k)t(t−1)=0. | (2.8) |
It follows that eac=±1, p=0 or 1, and t=0 or 1. Therefore, this completes the proof of Lemma 2.2.
Lemma 2.3. [14] Let R be a nonconstant rational function and p(z)=az+b(a≠0). Denote A1=R′+Rp′, An=A′n−1+An−1p′, B1=R′−Rp′, Bn=B′n−1+Bn−1(−p)′. Then
lim|z|→∞A′nR=0,lim|z|→∞AnR=an,lim|z|→∞B′nR=0,lim|z|→∞BnR=(−a)n. |
Lemma 2.4. [22] Suppose that f1,f2,⋯,fn(n≥2) are meromorphic functions and g1,g2,⋯,gn are entire functions satisfying the following conditions:
(i) n∑j=1fjegi≡0;
(ii) gj−gk are not constants for 1≤j<k≤n;
(iii) For 1≤j≤n,1≤h<k≤n,T(r,fj)=o{T(r,egh−gk)}(r→∞,r∉E), where E is a set of (0, ∞) with finite linear measure.
Then fj≡0 (j=1,2,⋯,n).
Lemma 2.5. [22] Let f be a meromorphic function of finite order ρ(f). Write
f(z)=ckzk+ck+1zk+1+⋯,(ck≠0) |
near z=0 and let {a1,a2,⋯} and {b1,b2,⋯} be the zeros and poles of f in C\{0}, respectively. Then
f(z)=zkeQ(z)P1(z)P2(z), |
where P1(z) and P2(z) are the canonical products of f formed with the non-null zeros and poles of f, respectively, and Q(z) is a polynomial of the degree ≤ρ(f).
Lemma 2.6. [22] Suppose that f1(z),f2(z),⋯,fn(z), (n≥3) are meromorphic functions that are not constants except for fn(z). Furthermore, let
n∑j=1fj(z)=1. |
If fn(z)≢ and
\begin{equation*} \sum\limits_{j = 1}^{n}N(r, \frac{1}{f_{j}})+(n-1)\sum\limits_{j = 1}^{n} \overline{N}(r, f_{j}) < (\lambda+o(1))T(r, f_{k}), \end{equation*} |
where \lambda < 1 and k = 1, 2, \cdots, n-1 , then f_{n}(z)\equiv 1 .
Proof. Suppose that Eq (1.2) admits a finite order transcendental meromorphic solution f(z) with finitely many poles. We can rewrite Eq (1.2) in the following form:
\begin{equation} [f^{(k)}(z)+i(\alpha f(z+c)-\beta f(z))][f^{(k)}(z)-i(\alpha f(z+c)-\beta f(z))] = R(z). \end{equation} | (3.1) |
Since f(z) has finitely many poles and R(z) is a nonzero rational function, then f^{(k)}(z)+i(\alpha f(z+c)-\beta f(z)) and f^{(k)}(z)-i(\alpha f(z+c)-\beta f(z)) both have finitely many poles and zeros. Thus, in view of Lemma 2.5, (3.1) can be written as
\begin{equation} \left\{ \begin{array}{ll} f^{(k)}(z)+i(\alpha f(z+c)-\beta f(z)) = R_{1}e^{p(z)}, \\ f^{(k)}(z)-i(\alpha f(z+c)-\beta f(z)) = R_{2}e^{-p(z)}, \end{array} \right. \end{equation} | (3.2) |
where R_{1}, R_{2} are two nonzero rational functions such that R_{1}R_{2} = R and p(z) is a nonzero polynomial. By solving the above equations system, we have
\begin{equation} \left\{ \begin{array}{ll} f^{(k)}(z) = \dfrac{R_{1}e^{p(z)}+R_{2}e^{-p(z)}}{2}, \\ \alpha f(z+c)-\beta f(z) = \dfrac{R_{1}e^{p(z)}-R_{2}e^{-p(z)}}{2i}, \end{array} \right. \end{equation} | (3.3) |
In view of the second equation of (3.3), it follows that
\begin{equation} \alpha f^{(k)}(z+c)-\beta f^{(k)}(z) = \dfrac{A_{k}e^{p(z)}-B_{k}e^{-p(z)}}{2i}, \end{equation} | (3.4) |
where A_{1} = R'_{1}+R_{1}p' , B_{1} = R'_{2}-R_{2}p' , A_{k} = A'_{k-1}+A_{k-1}p' and B_{k} = B'_{k-1}-B_{k-1}p' . Substituting the first equation of system (3.3) into (3.4), it yields that
\begin{eqnarray} e^{p(z)}[i\alpha R_{1}(z+c)e^{p(z+c)-p(z)}-i\beta R_{1}(z)-A_{k}(z)] \quad \quad \quad \quad \quad \\ +e^{-p(z)}[i\alpha R_{2}(z+c)e^{-p(z+c)+p(z)}-i\beta R_{2}(z)+B_{k}(z)] = 0. \end{eqnarray} | (3.5) |
By Lemma 2.4, it follows from (3.5) that
\begin{equation} \left\{ \begin{array}{ll} i\alpha R_{1}(z+c)e^{p(z+c)-p(z)}-i\beta R_{1}(z)-A_{k}(z) = 0, \\ i\alpha R_{2}(z+c)e^{-p(z+c)+p(z)}-i\beta R_{2}(z)+B_{k}(z) = 0. \end{array} \right. \end{equation} | (3.6) |
Since R_{1}, R_{2} are two nonzero rational functions, which implies that p(z) is a polynomial of degree one. Let p(z) = az+b , a\ne 0 , b\in \mathbb{C} . Substituting p(z) , A_{k} , and B_{k} into (3.6), and letting |z| \to \infty , thus, we can conclude from Lemma 2.3 that
\begin{equation} \left\{ \begin{array}{ll} \lim\limits_{|z|\to \infty}i(\alpha \dfrac{R_{1}(z+c)}{R_{1}(z)}e^{p(z+c)-p(z)} -\beta) = i(\alpha e^{ac}-\beta) = \lim\limits_{|z|\to \infty}\dfrac{A_{k}(z)}{R_{1}(z)} = a^{k}, \\ \lim\limits_{|z|\to \infty}i(\alpha \dfrac{R_{2}(z+c)}{R_{2}(z)}e^{-p(z+c)+p(z)}-\beta) = i(\alpha e^{-ac}-\beta) = \lim\limits_{|z|\to \infty}\dfrac{-B_{k}(z)}{R_{2}(z)} = -(-a)^{k}. \end{array} \right. \end{equation} | (3.7) |
Two equations of (3.7), which mean that
\begin{equation} \left\{ \begin{array}{ll} i(\alpha e^{ac}-\beta) = a^{k}, \\ i(\alpha e^{-ac}-\beta) = a^{k}. \end{array} \right. \end{equation} | (3.8) |
Hence, it yields e^{ac} = \pm 1 .
If e^{ac} = 1 , then a^{k} = i \alpha -i\beta . Thus, we can rewrite (3.6) in the following form:
\begin{equation} \left\{ \begin{array}{ll} i\alpha [R_{1}(z+c)-R_{1}(z)] = \sum\limits_{i = 0}^{k-1}C^{i}_{k}R_{1}^{(k-i)}a^{i}, \\ i\alpha [R_{2}(z+c)-R_{2}(z)] = \sum\limits_{i = 0}^{k-1}(-1)^{i+1}C^{i}_{k}R_{2}^{(k-i)}a^{i}. \end{array} \right. \end{equation} | (3.9) |
If R_{j}(j = 1, 2) are two nonzero rational functions, then in view of Lemma 2.1, it follows that i\alpha c = ka^{k-1} and R_{i} are nonzero polynomials with \deg_{z} R_{i} \le 1 (i = 1, 2) . In view of R = R_{1}R_{2} , thus R is a nonzero polynomial with \deg_{z} R \le 2 .
If e^{ac} = -1 , then a^{k} = -i\alpha -i\beta . Thus, we can rewrite (3.6) in the following form
\begin{equation} \left\{ \begin{array}{ll} -i\alpha [R_{1}(z+c)-R_{1}(z)] = \sum\limits_{i = 0}^{k-1}C^{i}_{k}R_{1}^{(k-i)}a^{i}, \\ -i\alpha [R_{2}(z+c)-R_{2}(z)] = \sum\limits_{i = 0}^{k-1}(-1)^{i+1}C^{i}_{k}R_{2}^{(k-i)}a^{i}. \end{array} \right. \end{equation} | (3.10) |
Similar to the discussion above, we can obtain that i\alpha c = -ka^{k-1} and R_{i} are nonzero polynomials with \deg_{z} R_{i} \le 1 (i = 1, 2) . In view of R = R_{1}R_{2} , thus R is a nonzero polynomial with \deg_{z}R \le 2 .
Hence, we can obtain that i\alpha c = \pm ka^{k-1} , R is a nonzero polynomial with \deg_{z} R \le 2 .
Suppose that R(z) is a nonzero polynomial with \deg_{z}R\le 2 , then in view of the first equation of (3.3), it follows that f(z) is of the form
\begin{equation} f(z) = \frac{s_{1}(z)e^{az+b}+s_{2}(z)e^{-(az+b)}}{2}+c_{k-1}z^{k-1}+\cdots+c_{0}, \end{equation} | (3.11) |
where s_{j}(z) = m_{j}z+n_{j} , m_{j} , n_{j}\in \mathbb{C}(j = 1, 2) and c_{0}, \cdots, c_{k-1} are constants.
If \deg_{z}R = 2 , then it follows that m_{j}\ne 0(j = 1, 2) .
If i\alpha c = ka^{k-1} and a^{k} = i(\alpha -\beta) , then e^{ac} = 1 , i.e., c = \frac{2l\pi i}{a} , l\in \mathbb{Z} . According to (3.8), if \alpha = \beta , we have a = 0 , it is a contradiction. If \alpha = -\beta , then a^{k} = 2i\alpha . Combining i\alpha c = ka^{k-1} , a^{k} = 2i\alpha and e^{ac} = 1 , we have 1 = e^{ac} = e^{2k} , it is a contradiction. Hence, \alpha \ne \pm \beta . Substituting (3.11) into the second equation of (3.3), it follows that c_{0} = \cdots = c_{k-1} = 0 , we have
\begin{equation} f(z) = \frac{s_{1}(z)e^{az+b}+s_{2}(z)e^{-(az+b)}}{2}. \end{equation} | (3.12) |
Substituting (3.12) into the first equation of (3.3), it yields
\begin{equation} R_{1}(z) = a^{k}s_{1}(z)+ka^{k-1}m_{1} \quad and \quad R_{2}(z) = (-a)^{k}s_{2}(z)+ka^{k-1}m_{2}. \end{equation} | (3.13) |
Substituting (3.12) into the second equation of (3.3), it yields
\begin{equation} R_{1}(z) = a^{k}s_{1}(z)+m_{1} \quad and \quad R_{2}(z) = (-a)^{k}s_{2}(z)-m_{2}. \end{equation} | (3.14) |
Comparing (3.13) and (3.14), we have ka^{k-1} = 1 and ka^{k-1} = -1 , it is a contradiction.
If i\alpha c = -ka^{k-1} and a^{k} = -i(\alpha +\beta) , then e^{ac} = -1 , i.e., c = \frac{(2l+1)\pi i}{a} , l\in \mathbb{Z} . Similar to the discussion above, we can obtain a contradiction. Therefore, there are two categories below:
Case I: If \deg_{z}R = 1 , one of m_{1} and m_{2} is zero, without loss of generality, assume m_{2} = 0 . Substituting (3.12) into (3.3), it follows that R_{1} is a polynomial of degree one and R_{2} is a constant, where i\alpha c = ka^{k-1} = 1 and a^{k} = i(\alpha -\beta) or i\alpha c = -ka^{k-1} = -1 and a^{k} = -i(\alpha +\beta) . Similar to the discussion above, it is easy to prove that \alpha \ne \pm \beta and c_{0} = \cdots = c_{k-1} = 0 .
Therefore, f(z) is of the form
\begin{equation*} f(z) = \frac{s_{1}(z)e^{az+b}+n_{2}e^{-(az+b)}}{2}, \end{equation*} |
where R(z) = -n_{2}a^{2k-1}[as_{1}(z)+km_{1}] , m_{1}\ne 0 , a\ne 0, b\in \mathbb{C} , and a, b, c, \alpha, \beta satisfy \alpha \ne \pm \beta , a^{k} = -i(\alpha +\beta) , c = \frac{(2l+1)\pi i}{a} , l\in \mathbb{Z} , i\alpha c = -ka^{k-1} = -1 , or a^{k} = i(\alpha -\beta) , c = \frac{2l\pi i}{a} , l\in \mathbb{Z} , i\alpha c = ka^{k-1} = 1 .
If m_{1} = 0 , similar to the discussion above, it is easy to prove that f(z) is of the form
\begin{equation*} f(z) = \frac{n_{1}e^{az+b}+s_{2}e^{-(az+b)}}{2}, \end{equation*} |
where R(z) = n_{1}a^{2k-1}[-as_{2}(z)+km_{2}] , m_{2}\ne 0 , a\ne 0, b\in \mathbb{C} , and a, b, c, \alpha, \beta satisfy \alpha \ne \pm \beta , a^{k} = -i(\alpha +\beta) , c = \frac{(2l+1)\pi i}{a} , l\in \mathbb{Z} , i\alpha c = ka^{k-1} = -1 , or a^{k} = i(\alpha -\beta) , c = \frac{2l\pi i}{a} , l\in \mathbb{Z} , i\alpha c = -ka^{k-1} = 1 .
Case II: If R(z) is a nonzero constant, by using the first equation of (3.3), it follows that f(z) is of the form
\begin{equation} f(z) = \frac{n_{1}e^{az+b}+n_{2}e^{-(az+b)}}{2}+c_{k-1}z^{k-1}+\cdots+c_{0}, \end{equation} | (3.15) |
where n_{1}, n_{2}\in \mathbb{C} and c_{0}, \cdots, c_{k-1}\in \mathbb{C} . Substituting (3.15) into the second equation of (3.3), it yields R = -a^{2k}n_{1}n_{2} .
( II_{1} ) If \alpha = \beta , in view of (3.8), it follows that e^{ac} = \pm 1 . If e^{ac} = 1 , then a = 0 , as i\alpha (e^{ac}-1) = a^{k} , a contradiction. Thus, e^{ac} = -1 . Hence, it follows that c = \frac{(2l+1) \pi i}{a} , l\in \mathbb{Z} , a^{k} = -2i\alpha , and c_{0}\in \mathbb{C} , c_{1} = \cdots = c_{k-1} = 0 .
( II_{2} ) If \alpha = -\beta , in view of (3.8), it follows that e^{ac} = \pm 1 . If e^{ac} = -1 , then a = 0 , as i\alpha (e^{ac}+1) = a^{k} , a contradiction. Thus, e^{ac} = 1 . Hence, it follows that c = \frac{2l \pi i}{a} , l\in \mathbb{Z} , a^{k} = 2i\alpha , and c_{0} = \cdots = c_{k-1} = 0 .
( II_{3} ) If \alpha \ne \pm \beta , substituting (3.15) into the second equation of (3.3), it yields c_{0} = \cdots = c_{k-1} = 0 . In view of (3.8), it follows that e^{ac} = \pm 1 . If e^{ac} = 1 , it follows that c = \frac{2l\pi i}{a} and a^{k} = i(\alpha -\beta) , l\in \mathbb{Z} . If e^{ac} = -1 , it follows that c = \frac{(2l+1) \pi}{a} and a^{k} = -i(\alpha +\beta) , l\in \mathbb{Z} . Therefore, this completes the proof of Theorem 1.5.
Proof. Similar to the method of proving Theorem 1.5, we can obtain the expression (3.8).
When k is an even number, two equations of (3.7), which mean that
\begin{equation} \left\{ \begin{array}{ll} i(\alpha e^{ac}-\beta) = a^{k}, \\ i(\alpha e^{-ac}-\beta) = -a^{k}. \end{array} \right. \end{equation} | (4.1) |
Hence, it follows a^{2k} = \alpha^{2}-\beta^{2} .
Case I: If \alpha = \pm \beta , this is a contradiction with a\ne 0 .
Case II: If \alpha \ne \pm \beta . Substituting p(z) = az+b and (4.1) into (3.6), it yields
\begin{equation} \left\{ \begin{array}{ll} i\alpha e^{ac}[R_{1}(z+c)-R_{1}(z)] = \sum\limits_{i = 0}^{k-1}C^{i}_{k}R_{1}^{(k-i)}a^{i}, \\ i\alpha e^{-ac}[R_{2}(z+c)-R_{2}(z)] = \sum\limits_{i = 0}^{k-1}(-1)^{i+1}C^{i}_{k}R_{2}^{(k-i)}a^{i}. \end{array} \right. \end{equation} | (4.2) |
Suppose that R_{1}, R_{2} are nonzero rational functions; in view of Lemma 2.2, we can conclude that e^{ac} = \pm 1 and R_{i} are nonzero polynomials with \deg_{z} R_{i} \le 1 (i = 1, 2) . In view of R = R_{1}R_{2} , thus R is a nonzero polynomial with \deg_{z} R \le 2 . Set \deg_{z}R_{1} = p and \deg_{z}R_{2} = t .
When p = 1 and t = 1 , if e^{ac} = 1 , then from (4.1), it follows that i\alpha -i\beta = a^{k} and i\alpha -i\beta = -a^{k} , a contradiction. If e^{ac} = -1 , then from (4.1), it follows that -i\alpha -i\beta = a^{k} and -i\alpha -i\beta = -a^{k} , a contradiction. Hence, there is at most a polynomial of degree one in R_{1} and R_{2} .
( II_{1} ) Suppose that p = 1 , t = 0 . In view of (3.3), it follows that f is of the form
\begin{equation} f(z) = \frac{s_{1}(z)e^{az+b}+n_{2}e^{-(az+b)}}{2}+P(z), \end{equation} | (4.3) |
where a\ne 0 , b\in \mathbb{C} , s_{1}(z) = m_{1}z+n_{1} , m_{1}(\ne 0) , n_{1} , n_{2}\in \mathbb{C} , and P(z) is a polynomial of degree k-1 . Since \alpha \ne \beta , then it yields from the second equation of (3.3) that P(z) \equiv 0 . And by using the first equation in (4.2), it follows that i\alpha e^{ac}c = ka^{k-1} . Hence, f(z) is of the form
\begin{equation*} f(z) = \frac{s_{1}(z)e^{az+b}+n_{2}e^{-(az+b)}}{2}, \end{equation*} |
where a^{2k} = \alpha^{2}-\beta^{2} , b\in \mathbb{C} , c = \dfrac{\log \frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a} , l\in \mathbb{Z} , e^{ac} = \frac{ka^{k-1}}{i\alpha c}\ne \pm 1 and R = n_{2}a^{2k-1}[as_{1}(z)+km_{1}] .
Suppose that p = 0 , t = 1 . Similar to the above argument as in ( II_{1} ), we obtain
\begin{equation*} f(z) = \frac{n_{1}e^{az+b}+s_{2}(z)e^{-(az+b)}}{2}, \end{equation*} |
where a^{2k} = \alpha^{2}-\beta^{2}, b\in \mathbb{C} , c = \dfrac{\log \frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a} , l\in \mathbb{Z} , e^{ac} = \frac{i\alpha c }{ka^{k-1}}\ne \pm 1 and R = n_{1}a^{2k-1}[as_{2}(z)-km_{2}] .
( II_{2} ) Suppose that p = 0 , t = 0 . By using (3.3), it follows that f is of the form
\begin{equation} f(z) = \frac{n_{1}e^{az+b}+n_{2}e^{-(az+b)}}{2}+P(z), \end{equation} | (4.4) |
where a\ne 0 , b\in \mathbb{C} , n_{1} , n_{2}\in \mathbb{C} \verb|\| \{0\} , and P(z) is a polynomial of degree k-1 . Since \alpha \ne \beta , then it yields from the second equation of (3.3) that P(z) \equiv 0 . Hence, f(z) is of the form
\begin{equation*} f(z) = \frac{n_{1}e^{az+b}+n_{2}e^{-(az+b)}}{2}, \end{equation*} |
where a^{2k} = \alpha^{2}-\beta^{2} , b\in \mathbb{C} , c = \dfrac{\log \frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a} , l\in \mathbb{Z} and R = a^{2k}n_{1}n_{2} . Therefore, this completes the proof of Theorem 1.6.
Proof. Suppose, on the contrary, to the assertion that there exists a transcendental entire solution f of (1.4) with finite order. We aim for a contradiction. By using a similar reason as in the proof of Theorem 1.5, we obtain
\begin{equation} f^{(k)}(z) = \dfrac{Q_{1}(z)e^{h(z)}+Q_{2}(z)e^{-h(z)}}{2} \end{equation} | (5.1) |
and
\begin{equation} \alpha f(z+c)-\beta f(z) = \dfrac{Q_{1}(z)e^{h(z)}-Q_{2}(z)e^{-h(z)}}{2iP(z)}, \end{equation} | (5.2) |
where h(z) is a non-constant polynomial, Q_{1}(z) and Q_{2}(z) are non-zero polynomials such that Q_{1}(z)Q_{2}(z) = Q(z) . Combining (5.1) and (5.2), we obtain
\begin{eqnarray} &&\alpha f^{(k)}(z+c)-\beta f^{(k)}(z) \\ = &&\dfrac{\alpha Q_{1}(z+c)e^{h(z+c)}+\alpha Q_{2}(z+c)e^{-h(z+c)}}{2}-\dfrac{\beta Q_{1}(z)e^{h(z)}+\beta Q_{2}(z)e^{-h(z)}}{2} \\ = &&\dfrac{h_{1}(z)e^{h(z)}-h_{2}(z)e^{-h(z)}}{2iP(z)^{k+1}}, \end{eqnarray} | (5.3) |
where
\begin{eqnarray*} h_{1}(z) = &&\sum\limits_{i = 0}^{k-1}C_{k-1}^{i}\sum\limits_{t = 0}^{k-i}C_{k-i}^{t}Q_{1}^{(k-t-i)}[(h')^{t}+M_{t}(h, h', \cdots, h^{(t)})]P^{(i)}P^{k-1} \nonumber \\ &&-\sum\limits_{i = 0}^{k-1}C_{k-1}^{i}\sum\limits_{t = 0}^{i}C_{i}^{t}Q_{1}^{(h-t)}[(h')^{t}+M_{t}(h, h', \cdots, h^{(t)})]P^{(k-i)}P^{k-1}+o(h_{1}(z)), \end{eqnarray*} |
\begin{eqnarray*} h_{2}(z) = &&\sum\limits_{i = 0}^{k-1}C_{k-1}^{i}\sum\limits_{t = 0}^{k-i}C_{k-i}^{t}Q_{2}^{(k-t-i)}[(h')^{t}+N_{t}(h, h', \cdots, h^{(t)})]P^{(i)}P^{k-1} \nonumber \\ &&-\sum\limits_{i = 0}^{k-1}C_{k-1}^{i}\sum\limits_{t = 0}^{i}C_{i}^{t}Q_{2}^{(h-t)}[(h')^{t}+N_{t}(h, h', \cdots, h^{(t)})]P^{(k-i)}P^{k-1}+o(h_{2}(z)), \end{eqnarray*} |
M_{t} and N_{t} are differential polynomials of (h, h', \cdots, h^{(t)}) . Thus from (5.3), we get
\begin{eqnarray} \dfrac{h_{1}(z)+\beta iP(z)^{k+1}Q_{1}(z)}{\alpha iP(z)^{k+1}Q_{2}(z+c)}e^{h(z)+h(z+c)}-\dfrac{h_{2}(z)-\beta iP(z)^{k+1}Q_{2}(z)}{\alpha iP(z)^{k+1}Q_{2}(z+c)}e^{h(z+c)-h(z)} \\ -\dfrac{Q_{1}(z+c)}{Q_{2}(z+c)}e^{2h(z+c)}\equiv 1. \end{eqnarray} | (5.4) |
It is easy to see that both \dfrac{h_{1}(z)+\beta iP(z)^{k+1}Q_{1}(z)}{\alpha iP(z)^{k+1}Q_{2}(z+c)}e^{h(z)+h(z+c)} and \dfrac{Q_{1}(z+c)}{Q_{2}(z+c)}e^{2h(z+c)} are not constants. Using Lemma 2.6, we obtain -[h_{2}(z)-\beta iP(z)^{k+1}Q_{2}(z)]e^{h(z+c)-h(z)} = \alpha iP(z)^{k+1}Q_{2}(z+c) , so h(z) = Az+B , A is a non-zero constant, and B is a constant. Thus, we obtain
\begin{equation} iP(z)^{k+1}[\beta Q_{2}(z)e^{Ac}-\alpha Q_{2}(z+c)] = h_{2}(z)e^{Ac}. \end{equation} | (5.5) |
Set \deg(P(z)) = p , \deg(Q(z)) = q , \deg(Q_{1}(z)) = q_{1} , \deg(Q_{2}(z)) = q_{2} and \deg(h(z)) = h . By comparing the degree of both sides of (5.5), it is not difficult to find that the degree of the left hand-side is (k + 1)p+q_{2} or (k + 1)p+q_{2}-1 , and the degree of right-hand side is kp+q_{2}-1 ; this is a contradiction. Therefore, this completes the proof of Theorem 1.8.
Zhiyong Xu: Conceptualization, Methodology, Writing-original draft; Junfeng Xu: Supervision, Writing-review and editing. Both of the authors have read and approved the final version of the manuscript for publication.
This research was supported by Fund of Education Department of Guangdong (Nos. 2022ZDZX1034, 2023GXJK517).
The authors declare that none of the authors have any competing interests in the manuscript.
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