# Entire Solutions of the Second-Order Fermat-Type Differential-Difference Equation

Authors: Guoqiang Dang; Jinhua Cai
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, the entire solutions of finite order of the Fermat-type differential-difference equationf″z2+△ckfz2=1 and the system of equations f1″z2+△ckf2z2=1andf2″z2+△ckf1z2=1 have been studied. We give the necessary and sufficient conditions of existence of the entire solutions of finite order.


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## Body

## 1. Introduction

In this work, we assume that the readers are familiar with general definitions and fundamental theories of Nevanlinna theory [1–3]. f is a meromorphic function which means f is meromorphic in the finite complex plane ℂ. If f has no poles, we call f is an entire function. We denote by Sr,f, any function satisfyingSr,f=oTr,f, r⟶∞, outside of a possible exceptional set of finite logarithmic measure. For a meromorphic function fz, we define its shift by fcz=fz+c and its difference operators by(1)△cfz=fz+c−fz,△cnfz=△cn−1△cfz.The classic Fermat-type functional equation(2)fzn+gzn=1,has been intensively studied in recent years. If n≥4, equation (2) has no transcendental meromorphic solutions [4]. If n≥3, equation (2) has no transcendental entire solutions [4]. If n=2, all entire solutions are the forms of fz=sinhz and gz=coshz, where hz is any entire function [5].Yang [6] studied the following generalized Fermat-type functional equation:(3)fzn+gzm=1,where n and m are positive integers and obtained the following theorem.Theorem 1.
If1/n+1/m<1, then equation (3) has no nonconstant entire solutions fz and gz.For further research solutions of the Fermat-type functional equation, Yang and Li [7] considered the following special Fermat-type functional equation:(4)fz2+f′z2=1,and they obtained the following theorem.Theorem 2.
The transcendental meromorphic solutions of (4) must satisfy fz=1/2Peλz+1/2Pe−λz, where P and λ are nonzero constants.In the following,c is a nonzero constant, unless otherwise specified.Liu [8] investigated the following special Fermat-type functional equation:(5)fz2+fz+c2=1.He obtained that each transcendental entire solution of (5) with the finite order must satisfy fz=h1z+h2z/2, where h1z+c/h1z=i,h2z+c/h2z=−i, and h1zh2z=1. Liu et al. [9] proved that the nonconstant finite-order entire solutions of (5) must have order one. Then, Liu et al. [10] considered the entire solutions of the following difference equations:(6)fzn+fz+cm=1,(7)f′zn+fz+cm=1,and obtained the following results.Theorem 3.
Equation (6) has no nonconstant entire solution if n>m>1 or n=m>2; has no transcendental entire solution with finite order if n>m; and has the general solutions fz=1/2+eπzi/chz, where hz is any entire function periodic with period c.Theorem 4.
Equation (7) has no transcendental entire solution with finite order, provided that m≠n, where n and m are positive integers.Liu et al. [10] obtained that fz=ez+1 satisfies the equation f′z+fz+c=1 when m=n=1; fz=ez+1 satisfies the equation f′z+fz+c=1 when m=n=1, where ec=−1; and fz=sinz±Bi satisfies the equation f′z2+fz+c2=1 when m=n=2, where c=2kπ or c=2kπ+π and k is an integer.Liu et al. [10] also investigated the following special Fermat-type functional equation:(8)f′zn+fz+c−fzm=1,and obtained the following theorem.Theorem 5.
Equation (8) has no transcendental entire solutions with finite order, provided that m≠n>1, where n and m are positive integers.Liu et al. built exact solutions for equation (8); for m=n=1, the equation f′z+fz+c−fz=1 has a transcendental entire solution fz=e−z+Az/ln2, where A=ln2/1+c and e−c=2; for m=2,n=1, the equation f′z+fz+c−fz2=1 admits a transcendental entire solution fz=BeEz+eE/2z+Az, where B=−E/4,A=−E/4kiπ,c=2kiπ/E, and E2=64kiπ/kiπ+4, where k is odd; and for m=n=2, the equation f′z2+fz+c−fz2=1 admits a transcendental entire solution fz=1/2sin2z+Bi, where c=kπ+π/2,k is an integer, and B is a constant.In 2019, Liu et al. [11] researched the entire solutions with finite order of the Fermat-type differential-difference equation(9)fkz2+△cfz2=1,and the following system of differential-difference equations:(10)f1kz2+△cf2z2=1,f2kz2+△cf1z2=1.In 2019, Dang and Chen [12] obtained the meromorphic solutions of the following special Fermat-type functional equation:(11)afn+bf′m≡1.In the following, we will consider the entire solutions with finite order of the Fermat-type differential-difference equation(12)f″z2+△ckfz2=1,and obtain the following result.Theorem 6.
fz be entire solutions of finite order of differential-difference equation (12) if and only if fz be the following forms:(13)fz=1ikeac−1kcosi1+k/2eac−1k/2z+ib+ez+d,where a,b,d, and e are constants, eac≠1, eack+−1k=0, and a2=ieac−1k; fz=az2+bz+d, where a=0,b=±1/c for k=1, 4a2c4+4a2=1 for k=2, and a=±1/2 for k≥3.Obviously, from Theorem6, we immediately get the following example.Example 1.
The transcendental entire function solutions with finite order of the differential-difference equation(14)f″z2+△c2fz2=1,must satisfy(15)fz=1−eac−12coseac−1z−ib+ez+d,where a,b,d, and e are constants, eac≠1, e2ac+1=0, and a2=ieac−12; fz=az2+bz+d, where 4a2c4+4a2=1.
Then, we will study the following system of differential-difference equations:(16)f1″z2+△ckf2z2=1,f2″z2+△ckf1z2=1,and obtain the next result.Theorem 7.
f1z,f2z be the transcendental entire solutions of finite order of the system of differential-difference equation (16) if and only if f1z,f2z be the following forms:(17)f1z=cosiieac−1k1/2z+ib1ieac−1k+e1z+d1,f2z=f1z+ez+d,where ekac=−1k, eac≠1, and eb1−b2=1;(18)f1z=cosi−ieac−1k1/2z+ib1−ieac−1k+e1z+d1,f2z=−f1z+ez+d,where ekac+−1k=0, eac≠1, and eb1−b2=−1; fjz=±1/2z2+bjz+dj for k≥3; f1z=a1z2+b1z+d1,f2z=±a1z2+b2z+d2, where a22=a12 for k=2; f1z=a1z2+b1z+d1,f2z=a2z2+b2z+d2, where c4=1 for k=2; and fjz=±1/cz+d for k=1, where aj,bj,dj, and d are constants, j=1,2.Obviously, from Theorem7, we immediately get the following example.Example 2.
Letk=2. Then, the transcendental entire function solutions with finite order of the system of differential-difference equations(19)f1″z2+△c2f2z2=1,f2″z2+△c2f1z2=1,must be in the following forms:(20)f1z=cosiieac−121/2z+ib1ieac−12+e1z+d1,f2z=f1z+ez+d,where ekac=−1k, eac≠1, and eb1−b2=1;(21)f1z=cosi−ieac−12+−121/2z+ib1−ieac−12+e1z+d1,f2z=−f1z+ez+d,where ekac+−1k=0, eac≠1, and eb1−b2=−1.


## 2. Lemmas

Lemma 1.
(see [1]). Let fjz be a meromorphic function, fjz:j=1,2,…,n−1, being nonconstant, satisfying ∑j=1nfjz=1 and n≥3. If fnz≡0 and(22)∑j=1nNr,1fj+n−1∑j=1nN¯r,fj<λ+o1Tr,fk,where λ<1 and k=1,2,…,n−1, then fn≡1.Lemma 2.
(see [1]). Suppose that f1z,f2z,…,fnzn≥2 are meromorphic functions and g1z,gz,…,gnz are entire functions satisfying the following conditions:(1)
∑j=1nfjzegjz≡0(2)
The orders offj are less than those of egk−gl for 1≤j≤n,1≤k<l≤n
Then,fjz≡0i=1,2,…,n.Lemma 3.
(Hadamard’s factorization theorem; see [1]). Let f be an entire function of finite order ρf with zeros z1,z2,…⊂ℂ\0 and k-fold zero at the origin. Then,(23)fz=zkPzeQz,where Pz is the canonical product of f formed with nonnull of f and Qz is a polynomial of degree less than ρf.


## 3. Proof of Theorem6

Suppose thatfz is an entire solution with finite order which satisfies (12). We rewrite (12) as follows:(24)f″z+i△ckfzf″z−i△ckfz=1.It follows thatf″z+i△ckfz and f″z−i△ckfz have no zeros. By Lemma 3, we have(25)f″z+i△ckfz=epz,f″z−i△ckfz=e−pz,where pz is a polynomial.From (25), we get(26)f″z=epz+e−pz2,(27)△ckfz=epz−e−pz2i.Case 1.
Suppose thatf is a transcendental entire function. Then, it follows from (26) that pz is a nonconstant polynomial. Let degpz=m; then, m≥1. It follows from (26) that(28)f″z+jc=epz+jc+e−pz+jc2.
It follows from (27) that(29)△ckfz′=∑j=0k−1k−jCkjf′z+jc=p′z2iepz+e−pz.
Then,(30)△ckfz″=∑j=0k−1k−jCkjf″z+jc=p″z+p′z22iepz+p″z−p′z22ie−pz.
Combining (28) and (30), we get(31)∑j=0k−1k−jCkjepz+jc+e−pz+jc2=p″z+p′z22iepz+p″z−p′z22ie−pz.
This implies(32)∑j=1k−1k−jCkjepz+jc+e−pz+jcepz+ip″z+p′z2e2pz+ip″z−p′z2+−1kepz+e−pzepz=0.
Therefore,(33)∑j=1k−1k−jCkjepz+jc+e−pz+jc+−1k+ip″z+ip′z2epz+−1k+ip″z−ip′z2e−pz=0.
Ifm≥2, then for 0≤j<l≤k, we have(34)ρepz+jc−pz+lc=m−1≥1,ρepz+jc+pz+lc=m≥2,and by (33) and Lemma 2, we obtain −1k−kCkk=0, which is contradicting.
Hence,m=1. Let pz=az+b, where a≠0; then, pz+kc=az+b+akc=pz+akc, pz+jc=az+b+ajc=pz+ajc, pz+kc−pz+jc=k−jac, and pz+kc+pz+jc=2pz+k+jac. Then, by (33), we have(35)∑j=1k−1k−jCkjepz+kc+pz+jc+epz+kc−pz+jc+−1k+ip″z+ip′z2epz+kc+pz+−1k+ip″z−ip′z2epz+kc−pz=0.
This implies(36)∑j=1k−1k−jCkje2pz+k+jac+ek−jac+−1k+ip″z+ip′z2e2pz+kac+−1k+ip″z−ip′z2ekac=0.
Then,(37)∑j=1k−1k−jCkje2pz+k+jac+−1ke2pz+kac+∑j=1k−1k−jCkjek−jac+−1kekac+ip″z+ip′z2e2pz+kac+ip″z−ip′z2ekac=0.
This implies(38)∑j=0k−1k−jCkje2pz+k+jac+∑j=0k−1k−jCkjek−jac+ip″z+ip′z2e2pz+kac+ip″z−ip′z2ekac=0.
Then,(39)∑j=0k−1k−jCkje2pz+k+jac+∑j=0k−1k−jCkjek−jac+ip′z2e2pz+kac−ip′z2ekac=0.
This implies(40)∑j=0k−1k−jCkje2pz+k+jac+∑j=0k−1k−jCkjek−jac+ia2e2pz+kac−ia2ekac=0.
Then,(41)∑j=0k−1k−jCkjejac+ia2e2pz+kac+∑j=0k−1k−jCkje−jac−ia2ekac=0.
Obviously,∑j=0k−1k−jCkjejac=eac−1k, and ∑j=0k−1k−jCkje−jac=e−ac−1k. From (41) and Lemma 2, we obtain that(42)eac−1k+ia2=0,e−ac−1k−ia2=0.
By (42), it is easy to get eac≠1 and eack+−1k=0. It follows from (42) that a2=ieac−1k. Thus, it follows from (26) that(43)fz=eieac−ik/2z+b+e−ieac−ik/2z−b2ieac−ik+ez+d=ei−iieac−ik/2z+b+e−i−iieac−ik/2z+b2ieac−ik+ez+d=1ieac−ikcos−iieac−ik/2z+b+ez+d=1ieac−ikcosieac−ik/2iz+ib+ez+d=1ieac−ikcosi1+k/2eac−1k/2z+ib+ez+d.Case 2.
Suppose thatf is a polynomial. Then, it follows from (26) that pz is a constant, and fz=az2+bz+d.Ifk≥3, then △ckfz=0. It follows from (12) that a=±1/2.Ifk=2, then △ckfz=2ac2. It follows from (12) that 4a2c4+4a2=1.Ifk=1, then △ckfz=2acz+ac2+bc. It follows from (12) that fz=±1/cz+d.
Thus, Theorem6 is proved.


## 4. Proof of Theorem7

Suppose thatf1z,f2z be an entire solution with finite order which satisfies equation (16). We rewrite (16) as follows:(44)f1″z+i△ckf2zf1″z−i△ckf2z=1,f2″z+i△ckf1zf2″z−i△ckf1z=1.It follows that(45)f1″z+i△ckf2z≠0,f1″z−i△ckf2z≠0,f2″z+i△ckf1z≠0,f2″z−i△ckf1z≠0.By Lemma3, we have(46)f1″z+i△ckf2z=epz,f1″z−i△ckf2z=e−pz,f2″z+i△ckf1z=eqz,f2″z−i△ckf1z=e−qz,where pz and qz are polynomials.From (46), we get(47)f1″z=epz+e−pz2,(48)△ckf2z=epz−e−pz2i,(49)f2″z=eqz+e−qz2,(50)△ckf1z=eqz−e−qz2i.Case 3.
Suppose thatf1 is a transcendental entire function with finite order. Then, it follows from (47)–(50) that pz and qz are two nonconstant polynomials. Let degpz=m1anddegqz=m2; then, m1≥1andm2≥1.
It follows from (47)–(50) that(51)f1″z+jc=epz+jc+e−pz+jc2,(52)f2″z+jc=eqz+jc+e−qz+jc2,(53)△ckf1z′=∑j=0k−1k−jCkjf1′z+jc=q′z2ieqz+e−qz,(54)△ckf2z′=∑j=0k−1k−jCkjf2′z+jc=p′z2iepz+e−pz.
Then,(55)△ckf1z″=∑j=0k−1k−jCkjf1″z+jc=q″z+q′z22ieqz+q″z−q′z22ie−qz,(56)△ckf2z″=∑j=0k−1k−jCkjf2″z+jc=p″z+p′z22iepz+p″z−p′z22ie−pz.
From (51) and (55), we get(57)∑j=0k−1k−jCkjepz+jc+e−pz+jc2=q″z+q′z22ieqz+q″z−q′z22ie−qz.
From (52) and (56), we get(58)∑j=0k−1k−jCkjeqz+jc+e−qz+jc2=p″z+p′z22iepz+p″z−p′z22ie−pz.
It follows that(59)∑j=0k−1k−jCkjepz+jc+e−pz+jc+iq″z+q′z2eqz+iq″z−q′z2e−qz=0,(60)∑j=0k−1k−jCkjeqz+jc+e−qz+jc+ip″z+p′z2epz+ip″z−p′z2e−pz=0.
Then,(61)∑j=0k−1k−jCkjepz+jc+qz+e−pz+jc+qz+iq″z+iq′z2e2qz+iq″z−iq′z2=0,(62)∑j=0k−1k−jCkjeqz+jc+pz+e−qz+jc+pz+ip″z+ip′z2e2pz+ip″z−ip′z2=0.
Ifpz+jc+qzand−pz+jc+qz are nonconstant polynomials for any j0≤j≤k, then we obtain that eqz±pz+jc are nonconstants for any j0≤j≤k. Assume m2≥1; then, −1k+iq″z−iq′z2 does not vanish; then, we multiply −1/iq″z−iq′z2 on both sides of (61), and we have(63)−1iq″z−iq′z2∑j=0k−1k−jCkjepz+jc+qz+e−pz+jc+qz−iq″z+iq′z2iq″z−iq′z2e2qz=1.
Then, by Lemma1, we have −ip″z+ip′z2/iq″z−iq′z2e2qz≡1; this contradicts with m2≥1.
Hence, there existj00≤j0≤k such that pz+j0c+qz=A or −pz+j0c+qz=A, where A is a constant.
Suppose that−pz+j0c+qz=A. Obviously, m1=m2. We assert that m1=m2=1. Otherwise, we assume that m1=m2≥2; then, deg−pz+jc+qz=deg−pz+jc+pz+j0c−pz+j0c+qz=degpz−1≥1 for any j0≤j≠j0≤k and degpz+jc+qz=m1≥2 for any j0≤j≠j0≤k.
From (63), we have(64)−1iq″z−iq′z2∑j=0k−1k−jCkjepz+jc+qz−1iq″z−iq′z2∑j=0,j≠j0k−1k−jCkje−pz+jc+qz−iq″z+iq′z2iq″z−iq′z2e2qz−−1k−j0Ckj0e−pz+j0c+qziq″z−iq′z2=1.
Then, it follows from (64) and Lemma 2 that −−1k−j0Ckj0eA/iq″z−iq′z2=1. This implies iq″z−iq′z2 is a constant, and m2=m1=1, which is contradicting; therefore, the assertion is proved.
Accordingly,−pz+j0c+qz=A and m1=m2. Thus, we can assume that pz=az+b1,qz=az+b2; then,(65)pz+jc=az+ajc+b1,qz+jc=az+ajc+b2,for any 0≤j≤k.
Combining (61) and (62), we get(66)∑j=0k−1k−jCkjeaz+ajc+b1+e−az−ajc−b1eaz+b2+ia2e2az+2b2−ia2=0,(67)∑j=0k−1k−jCkjeaz+ajc+b2+e−az−ajc−b2eaz+b1+ia2e2az+2b1−ia2=0.
Furthermore,(68)∑j=0k−1k−jCkjeajc+ia2eb2−b1e2az+b1+b2+∑j=0k−1k−jCkje−ajc−b1+b2−ia2=0,(69)∑j=0k−1k−jCkjeajc+ia2eb1−b2e2az+b1+b2+∑j=0k−1k−jCkje−ajc+b1−b2−ia2=0.
It follows from (68) and (69) that(70)∑j=0k−1k−jCkjeajc+ia2eb2−b1=0,(71)∑j=0k−1k−jCkje−ajc−b1+b2−ia2=0,(72)∑j=0k−1k−jCkjeajc+ia2eb1−b2=0,(73)∑j=0k−1k−jCkje−ajc+b1−b2−ia2=0.
By the binomial formula and (70)–(73), we have(74)eac−1k+ia2eb2−b1=0,(75)eb2−b1e−ac−1k−ia2=0,(76)eac−1k+ia2eb1−b2=0,(77)eb1−b2e−ac−1k−ia2=0.
From (75) and (77), we get eb1−b2=eb2−b1; hence, eb1−b2=±1. This implies b2=b1+inπ, where n is an integer. From (74) and (75) and eb1−b2=±1, we obtain that ekac=−1k and eac≠1, where eb1−b2=1; ekac+−1k=0 and eac≠1, where eb1−b2=−1.
It follows from (47) and (49) that(78)f1″z=epz+e−pz2=eaz+b1+e−az−b12,f2″z=eqz+e−qz2=eaz+b2+e−az−b22.
Therefore,(79)f1z=eaz+b1+e−az−b12a2+e1z+d1,f2z=eaz+b2+e−az−b22a2+e2z+d2,where ei and di are constants.
Combining (79) and eb1−b2=±1, we get that f2z=±f1z+ez+d, where e and d are constants.
Ifb2=b1+2niπ, then it follows from (74) and (79) that a=ieac−1k1/2 and(80)f1z=eieac−1k1/2z+b1+e−ieac−1k1/2z−b12ieac−1k+e1z+d1=ei−iieac−1k1/2z−iib1+eiiieac−1k1/2z+iib12ieac−1k+e1z+d1=cos−iieac−1k1/2z−ib1ieac−1k+e1z+d1=cosiieac−1k1/2z+ib1ieac−1k+e1z+d1,f2z=f1z+ez+d.
Ifb2=b1+2n+1iπ, then it follows from (74) and (79) that a=−ieac−1k1/2 and(81)f1z=e−ieac−1k1/2z+b1+e−ieac−1k1/2z−b12−ieac−1k+e1z+d1=ei−i−ieac−1k1/2z−iib1+eii−ieac−1k1/2z+iib12−ieac−1k+e1z+d1=cosi−ieac−1k1/2z+ib1−ieac−1k+e1z+d1,f2z=−f1z+ez+d.Case 4.
Suppose thatf is a polynomial. Then, it follows from (47)–(50) that pz,qz are two constants. Hence, f1z=a1z2+b1z+d1 and f2z=a2z2+b2z+d2, where aj,bj, and dj are constants.Ifk≥3, then △ckfjz=0,j=1,2. It follows from (16) that aj=±1/2, and fjz=±1/2z2+bjz+dj.Ifk=2, then △ckfjz=2ajc2,j=1,2. It follows from (16) that(82)a12c4+a22=14,a22c4+a12=14.
Therefore,a22=a12, or c4=1. Therefore,(83)f1z=a1z2+b1z+d1,f2z=±a1z2+b2z+d2,where a22=a12, and(84)f1z=a1z2+b1z+d1,f2z=a2z2+b2z+d2,where c4=1.
Ifk=1, we have △ckfjz=2ajcz+ajc2+bjc,j=1,2. It follows from (16) that fjz=±1/cz+d, where d is a constant.
Thus, Theorem7 is proved.


## 5. Discussion

Consider the differential-difference equation(85)f″zn+△ckfzm=1,where n and m are positive integers. If n>m>1 or m>n>1 or n=m>2, it follows easily that (85) has no nonconstant entire solution by Theorem 1. For the case n=m=2, the entire solutions with finite order have been illustrated in Theorem 6.



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