# Exponential Polynomials and Nonlinear Differential-Difference Equations

Authors: Junfeng Xu; Jianxun Rong
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we study finite-order entire solutions of nonlinear differential-difference equations and solve a conjecture proposed by Chen, Gao, and Zhang when the solution is an exponential polynomial. We also find that any exponential polynomial solution of a nonlinear difference equation should have special forms.


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## Body

## 1. Introduction and Main Result

Extensive application of Nevanlinna theory has prompted scholars to acquire a number of results on differential equations, difference equations, and differential-difference equations. In this paper, we assume readers are familiar with the standard notations and fundamental results, see [1–4].Given a meromorphic functionf and a constant c. We take c=1 for simplicity. Δfz=fz+1−fz and Δnfz=ΔΔn−1fz are the first-order difference operator and n-th order difference operator of f, respectively. We adopt the notations ρf and λf to denote the order and the exponent of convergence of zeros of f, respectively.Recall the definition of exponential polynomial of the form(1)fz=P1zeQ1z+P2zeQ2z+⋯+PkzeQkz,where Pj′s and Qj′s are polynomials in z. Denote(2)Γ0=eαzαzis a nonconstant polynomial,Γ1=eαz+dαzis a  nonconstant  polynomial andd∈ℂ,Γ0′=pzeαzpzis  a polynomial  andαzis a  nonconstant  polynomial.Many papers recently have focused on solvability and existence of solutions of nonlinear differential-difference equations, see [5–16].In 2012, Wen et al. [17] classified finite-order entire solutions of the following nonlinear difference equation:(3)fnz+qzeQzfz+c=Pz,where n≥2 is an integer and qz, Qz, and Pz are polynomials such that qz is not identically zero and Qz is not a constant. They obtained the following result.Theorem 1 (see [17]).
Letn≥2 be an integer, c∈ℂ, and qz, Qz, and Pz be polynomials such that qz is not identically zero and Qz is not a constant. Then, the finite-order entire solutions f of equation (3) should satisfy the following:(a)
Every solutionf satisfies ρf=degQ and is of mean type.(b)
Every solutionf satisfies λf=ρf if and only if Pz≡0.(c)
A solutionf belongs to Γ0 if and only if Pz≡0. In particular, this is the case n≥3.(d)
If a solutionf belongs to Γ0 and g is any other finite-order entire solution to equation (3), then f=ηg, where ηn−1=1.(e)
Iff is an exponential polynomial solution of form (1), then f∈Γ1. Moreover, if f∈Γ1/Γ0, then ρf=1.In 2016, Liu [9] investigated finite-order transcendental entire solutions of the following nonlinear differential-difference equation:(4)fnz+qzeQzfkz+c=Pz,where n≥2 and k≥1 are integers and qz, Qz, and Pz are polynomials such that qz is not identically zero and Qz is not a constant. He obtained a result which is similar to Theorem 1.In 2019, Chen et al. [6] considered solutions of equation (3), where Pz is replaced by p1eλz+p2e−λz. They obtained the following result.Theorem 2 (see [6]).
Letn≥3 be an integer and c, λ, p1, and p2 be nonzero constants. Suppose qz and Qz are polynomials such that qz is nonvanishing and Qz is not a constant. If f is an entire solution of finite order of(5)fnz+qzeQzfz+c=p1eλz+p2e−λz,then the following conclusions hold:(1)
Every solutionf satisfies ρf=degQ=1.(2)
If a solutionf belongs to Γ0, then qz must be a constant and one of the following two relation groups holds:(a)
fz=eλ/nz+B and Qz=−n+1/nλz+b(b)
fz=e−λ/n+B and Qz=n+1/nλz+b, where both b and B are constants.Remark 1.
Chen et al. [6] gave an example: fz=ez is an entire solution of finite order of the following difference equation:(6)f2z+2e−3zfz−log2=e2z+e−2z.
From the example, they conjectured that the conclusions of Theorem2 are still valid if n=2.
We consider the conjecture and prove a more generalized result. Moreover, we solve Chen, Gao, and Zhang’s conjecture whenfz is an exponential polynomial of form (1).Theorem 3.
Letk≥0 be an integer and c, α1, α2, p1, and p2 be nonzero constants such that α1≠α2. Suppose qz is a nonvanishing polynomial and Qz is a nonconstant polynomial. If the differential-difference equation(7)f2z+qzeQzfkz+c=p1eα1z+p2eα2z,has a transcendental entire solution f, then(1)
Every solutionf satisfies ρf=degQ≥1.(2)
Iff is an exponential polynomial of form (1), then ρf=degQ=1.(3)
Iff belongs to Γ0′, then one of the following two relation groups holds:(a)
fz=gzeα2/2z+B, Qz=α1−α2/2z+b, gz2e2B=p2, and qz∑s=0kksα2/2sgz+ck−seα2c/2+b+B=p1(b)
fz=gzeα1/2z+B, Qz=α2−α1/2z+b, gz2e2B=p1, and qz∑s=0kksα1/2sgz+ck−seα1c/2+b+B=p2, where both b and B are constants and gz is a polynomial.Corollary 1.
Letk≥0 be an integer and c, α1, α2, p1, and p2 be nonzero constants such that α1≠α2. Suppose qz is a nonvanishing polynomial and Qz is a nonconstant polynomial. If the differential-difference equation (7) has solutions f satisfying f∈Γ0, then ρf=degQ=1 and qz must be a constant and one of the following two relation groups holds:(1)
fz=eα2/2z+B, Qz=α1−α2/2z+b, e2B=p2, and qα2/2keα2c/2+b+B=p1(2)
fz=eα1/2z+B, Qz=α2−α1/2z+b, e2B=p1, and qα1/2keα2c/2+b+B=p2, where both b and B are constants.
Next, we give two examples to illustrate equation (7).Example 1.
f
z
=
e
z
+
π
i is an entire solution of finite order of the following differential-difference equation:(8)f2z+2e−3zf′z−πi=2e−2z+e2z,where k=1, α1=−2, α2=2, and gz2e2B=1=p2. Thus, case (1) occurs.Example 2.
f
z
=
2
e
2
z is an entire solution of finite order of the following difference equation:(9)f2z+3ez−πifz+2πi=4e4z−6e3z,where k=0, α1=4, α2=3, and gz2e2B=4=p1. Thus, case (2) occurs.
In 2015, Zhang et al. [18] studied the existence of entire solutions of the following nonlinear difference equation:(10)f3z+q3zΔ3f+q2zΔ2f+q1zΔf=λ1eαz+λ2e−αz+pz.
They obtained the following result.Theorem 4 (see [18]).
Letλ1, λ2, and α be nonzero constants. Suppose qjzj=1,2,3 and pz are polynomials. Then, the nonlinear difference equation (10) possesses solutions of finite order of the form fz=c1eα/3z+c2e−α/3z with c13=λ1, c23=λ2, and pz≡0. v=eα/3 and q1, q2, and q3 satisfy the following condition:(11)v−13v3+1q3+v−12v3−vq2+v−1v3+v2q1=0.
Moreover, one of the following conclusions holds:(1)
Ifα=6nπ±3πi, then 8q3−4q2+2q13=27λ1λ2(2)
Ifα=6nπ±πi, then q1+q2=0 and ±3iq1−q33=27λ1λ2, where n is an integer(3)
Ifv≠−1 and v≠1±3i/2, then q1, q2, and q3 satisfy the following equation:(12)v−12v2−v+1q3+vv−12q2+v2q1=0.In the following, we consider a difference equation which is similar to (10) and obtain the following result.Theorem 5.
Suppose thatp1, p2, and λ are nonzero constants and that a1z and a2z are nonzero polynomials. If f is a nontrivial exponential polynomial solution of(13)f2z+a2zΔ2f+a1zΔf=p1eλz+p2e−λz,then f has solutions of finite order of the following form:(14)fz=c0z+c1eλ/2z+c2e−λ/2z,where eλ/2=±i, a22=−2c1c2, c12=p1, and c22=p2. c0z is a nonzero polynomial, v=eλ/2, and a1z and a2z satisfy(15)v2−1a1v+a2v−12=0.
Moreover, one of the following conclusions holds:(1)
Ifv=−1, then λ=4kπ+2πi, c0z=a1z−2a2z≠0, and a1z and a2z satisfy(16)2c1c2+a2za1z+2−2a2z+2−a1z+2a2z+a1z−2a2za1z+1−2a2z+1=0.Especially, ifa1 and a2 are constants, then c02=−2c1c2.(2)
Ifv≠−1 and v is the solution of a1v+a2v−12=0, then c0z=a1z/2 and a1z and a2z satisfy(17)a1z2−2a1za1z+1−2a2zΔ2a1z=8c1c2.Especially, ifa1 is a constant, then c02=−2c1c2.
The following examples show the existences of solution of equation (13).Example 3.
An entire solutionfz=−2+eπiz−2e−πiz solves the following difference equation:(18)f2z+3Δ2f+4Δf=e2πiz+4e−2πiz,where λ=2πi, a1=4,a2=3, c0=−2. The case v=expλ/2=−1 occurs.Example 4.
An entire solutionfz=2+eπi/2z−2e−πi/2z solves the following difference equation:(19)f2z+2Δ2f+4Δf=eπiz+4e−πiz,where λ=πi, a1=4,a2=2,c0=a1/2=2. The case v=eπ/2=i satisfies a1v+a2v−12=0.Example 5.
An entire solutionfz=−2+2eπi/2z−e−πi/2z solves the following difference equation:(20)f2z−2Δ2f−4Δf=4e−πiz+eπiz,where λ=−πi, a1=−4,a2=−2,c0=a1/2=−2. The case v=expλ/2=−1 satisfies a1v+a2v−12=0.
This paper is organized as follows. In Section2, we introduce the background of exponential polynomials and some indispensable lemmas. Sections 3 and 4 contain the detailed proofs on Theorems 3 and 5. In Section 5, we will discuss the methods of the main results obtained in the paper.


## 2. Preliminaries

We recollect a basic result on exponential polynomials. LetPz=bqzq+bq−1zq−1+⋯+b0, where bq≠0. We know ([1], p.7) that(21)Tr,ePz=bqrqπ+orq.For exponential polynomialsfz of form (1), Wen et al. [17] followed the reasoning in [19] and acquired some instrumental tools.Suppose the polynomialsQjz in (1) are pairwise different and normalized by Qj0=0. Then, representation (1) is uniquely determined and the functions PjzeQjz are linearly independent. Let(22)q=maxdegQjQjz≡0,and let w1,w2,…,wm be pairwise different leading coefficients of the polynomials Qjz of maximum degree q. Thus, (1) can be written in the following normalized form:(23)fz=H0z+H1zew1zq+H2zew2zq+⋯+Hmzewmzq,where Hiz0≤i≤m are either exponential polynomials of degree less than q or ordinary polynomials in z. Hjz≡0 hold for 1≤j≤m.A convex hull of a setW⊂ℂ, denoted by coW, is the intersection of all convex sets containing W. If W contains only finitely many elements, then coW is obtained as an intersection of finitely many closed half-planes. Hence, coW is either a compact polygon (with a nonempty interior) or a line segment. We denote the perimeter of coW by CcoW. If coW is a line segment, then CcoW equals to twice the length of this line segment. We fix the notation for W=w1¯,w2¯,…,wm¯, W0=0,w1¯,…,wm¯, and Qz=bqzq+⋯+b0.Theorem 6 (see [19], Satz 1]).
Letf be given by (23). Then,(24)Tr,f=CcoW0rq2π+orq.Next, we can find the following consequence from the result of Steinmetz ([20], Satz 1), i.e.,(25)mr,fkz+cfz=orq,holds for an exponential polynomial fz in form (23) (also see [21], Section 3).Some auxiliary results are necessary. The first one is a difference analogue of logarithmic derivative lemma given by Chiang and Feng.Lemma 1 (see [22], Corollary 2.5).
Letfz be a meromorphic function with finite order ρf. Suppose c is a fixed nonzero complex constant. Then, for each ε>0, we have(26)mr,fz+cfz+mr,fzfz+c=Orρf−1+ε+Ologr.
The following lemma is a useful tool to solve differential-difference equations and difference equations.Lemma 2 (see [3]).
Suppose thatf1z,f2z,…,fnzn≥2 are meromorphic functions and that g1z,g2z,…,gnzn≥2 are entire functions. They satisfy the following conditions:(1)
f1zeg1z+f2zeg2z+⋯+fnzegnz≡0(2)
gjz−gkz are not constants for 1≤j<k≤n(3)
Tr,fjz=oTr,eghz−gkzr⟶∞,r∉E holds, for 1≤j≤n and 1≤h<k≤n, where E⊂1,∞ is finite linear measure or finite logarithmic measure.
Then,fjz≡0 j=1,2,…,n.Halburd and Korhonen proved a difference analogue of Clunie lemma under the condition finite order.Lemma 3 (see [23]).
Letfz be a nonconstant finite-order meromorphic solution of(27)fnzPz,f=Qz,f,where Pz,f and Qz,f are difference polynomials in f with small meromorphic coefficients. Suppose c∈ℂ and δ<1. If the total degree of Qz,f is a polynomial in f and its shifts are less than or equal to n, then(28)mr,Pz,f=oTr+c,frδ+oTr,f,for all r outside of a possible exceptional set with finite logarithmic measure.Remark 2.
Similar to Lemma3, if f is a transcendental exponential polynomial in form (23), Pz,f and Qz,f are differential-difference polynomials in f and the coefficients of Pz,f and Qz,f are polynomials aizi=1,2,…,n, for each ε>0, then an obtained result is(29)mr,Pz,f=orq,where r is sufficiently large.
Chen and Yang proved the following lemma.Lemma 4 (see [24]).
Letλ be a nonzero constant and Hz be a nonvanishing polynomial. Then, the differential equation(30)4f″−λ2f=Hz,has a special solution c0z which is a nonzero polynomial.In addition, the following lemma is similar to Lemma 5.3 of [17] and Lemma 2.7 of [9]. The proof can be given word by word.Lemma 5.
Letf be given by (23), where q≥2. If f is a solution of equation (7), then m=1.


## 3. Proof of Theorem 3

Proof of Conclusion 1.
Suppose thatfz is a finite-order entire solution of equation (7). Applying the lemma on the logarithmic derivative and Lemma 1 to equation (7), we obtain(31)2Tr,fz=2mr,fz=mr,p1eα1z+p2eα2z−qzeQzfkz+c≤mr,fzfz+cfzfkz+cfz+c+mr,eQz+Or≤mr,fz+mr,fz+cfz+mr,fkz+cfz+c+mr,eQz+Or≤Tr,fz+Tr,eQz+Or+Sr,f.
Thus,(32)Tr,f≤Tr,eQ+Or+Sr,f,which implies that ρf≤degQ.
Ifρf≤degQ, then the order of left side of equation (7) is equal to degQ. Since the order of right side of equation (7) is equal to 1, we have degQ=1 and ρf<1. Let Qz=a˜z+b˜, where a˜≠0. Equation (7) can be written as(33)p1eα1z+p2eα2z+p3ea˜z+b˜+p4=0,where p3=−qzfkz+c and p4=−fz2 satisfy ρp3<1 and ρp4<1, respectively. Next, we consider the following three cases:Case 1. a˜≠α1 and a˜≠α2.By equation (33) and Lemma 2, we have p1=p2=0, which is a contradiction.Case 2. a˜=α1 and a˜≠α2.Equation (33) can be rewritten as(34)p1+p3eb˜eα1z+p2eα2z+p4=0.Using Lemma2, we have p2=0, which is a contradiction.Case 3. a˜≠α1 and a˜=α2. Similar to the proof of Case 2, we can get a contradiction.Thus, we haveρf=degQ. Noting degQ≥1, we obtain ρf=degQ≥1.Proof of Conclusion 2.
Sincef is an exponential polynomial in form (1), we can consider its equivalent form (23). Suppose q≥2, by Lemma 5 we know m=1. That is, we have fz=H0z+H1zew1zq. Substituting the expression of fz into equation (7) yields(35)p1eα1z+p2eα2z−H0z2=2H1zH0zew1zq+qzeQ0zH0kz+cebqzq+H1z2e2w1zq+qzeQ0z+P1zH1Tz+cebq+w1zq,where Q0z=Qz−bqzq and P1z=w1z+cq−w1zq. In addition, H1Tz+c is a differential polynomial in H1z+c, w1z+cq, and their derivatives. We see that Q0z and P1z are two polynomials with degree less than or equal to q−1. We discuss two cases bq≠w1 and bq=w1:Case 1. bq≠w1.Takingbq=−w1,bq=2w1 and bq∉±w1,2w1, respectively, we apply Lemma 2 to equation (35) to obtain H1z≡0, which is a contradiction.Case 2. bq=w1.Equation (35) can be rewritten as(36)p1eα1z+p2eα2z−H0z2=2H1zH0z+qzeQ0zH0kz+cew1zq+H1z2+qzeQ0z+P1zH1Tz+ce2w1zq.
We utilize Lemma2 again to obtain(37)H0z2=p1eα1z+p2eα2z.
Assume thatz0 is a zero of the above equation. Obviously, z0 is a simple zero of p1eα1z+p2eα2z, but z0 is the multiple zero of H0z2. This is a contradiction. We have q=1. fz is reduced to H0z+H1zew1z, which implies ρf=degQ=1.Proof of Conclusion 3.
Sincefz belongs to Γ0′, from Conclusion 2, we know that ρf=degQ=1.
Let(38)fz=gzeAz+B,(39)Qz=az+b,where a and A are nonzero constants, b and B are constants, and gz is a nonvanishing polynomial. It follows from formula (38) that(40)fkz+c=eAc+B∑s=0kksAsgz+ck−seAz.
Substituting formulas (38)–(40) into equation (7), we have(41)gzeB2e2A−α2z−p1eα1−α2z−p2+eAc+b+Bqz∑s=0kksAsgz+ck−sea+A−α2z=0.
We consider the following four cases:Case 1. 2A−α2=0 and A+a−α2=0.Using Lemma2, it follows from equation (41) that p1=0. It is a contradiction.Case 2. 2A−α2≠0 and A+a−α2≠0.If2A−α2≠A+a−α2≠α1−α2, then we obtain p1=p2≡qz≡0 by equation (41) and Lemma 2. A contradiction occurs.Now, we consider that only two of2A−α2, A+a−α2, and α1−α2 coincide. Without loss of generality, assuming 2A−α2=A+a−α2≠α1−α2, we see that equation (41) is represented as(42)gzeB2+eAc+b+Bqz∑s=0kksAsgz+ck−se2A−α2z−p1eα1−α2z−p2=0.From the above equation, using Lemma2, we have p1=p2=0, which implies a contradiction.Thus,2A−α2=A+a−α2=α1−α2. We write equation (41) as(43)gzeB2+eAc+b+Bqz∑s=0kksAsgz+ck−s−p1eα1−α2z−p2=0.We use Lemma2 again to lead to p2=0. It is a contradiction.Case 3. 2A−α2=0 and A+a−α2≠0.IfA+a−α2≠α1−α2, then we have p1=qz≡0 by equation (41) and Lemma 2. A contradiction occurs.Thus,A+a−α2=α1−α2. We deduce A=α2/2 and a=α1−α2/2. Equation (41) can be represented as(44)qz∑s=0kksα22sgz+ck−seα2c/2+b+B−p1eα1−α2z−p2−gzeB2=0.Because of Lemma2, we have(45)gz2e2B=p2,qz∑s=0kksα22sgz+ck−seα2c/2+b+B=p1.We proceed to obtainfz=gzeα2/2z+B and Qz=α1−α2/2z+b.Case 4. 2A−α2≠0 and A+a−α2=0.If2A−α2≠α1−α2, then we obtain p1=0 by equation (41) and Lemma 2. A contradiction occurs.Thus,2A−α2=α1−α2. We derive A=α1/2 and a=α2−α1/2. Equation (41) is equivalent to(46)gzeB2−p1eα1−α2z−p2−qz∑s=0kksα22sgz+ck−seα1c/2+b+B=0.
By Lemma2, we have(47)gz2e2B=p1qz∑s=0kksα22sgz+ck−seα1c/2+b+B=p2.
Consequently, we obtainfz=gzeα1/2z+B and Qz=α2−α1/2z+b.


## 4. Proof of Theorem 5

Assume that the difference equation (13) has a transcendental entire solution f of finite order.Applying Lemma1 to equation (13), we have(48)Tr,p1eλz+p2e−λz=Tr,f2z+a2zΔ2f+a1zΔf≤Tr,f2+Tr,a2Δ2f+a1Δf+O1≤Tr,f2+mr,a2Δ2f+a1Δff+mr,f≤Tr,f2+mr,f+Sr,f≤3Tr,f+Sr,f.On the other hand, we deduce(49)Tr,p1eλz+p2e−λz=Tr,f2+a2zΔ2f+a1zΔf≥Tr,f2−Tr,a2zΔ2f+a1zΔf+O1≥2Tr,f−mr,a2Δ2f+a1Δff−mr,f+O1≥2Tr,f−Tr,f+Sr,f=Tr,f+Sr,f.Combining equations (48) and (49), it follows that(50)Tr,f+Sr,f≤Tr,p1eλz+p2e−λz≤3Tr,f+Sr,f,which implies ρf=1.DenotingP1f≔a2zΔ2f+a1zΔf, we rewrite equation (13) as(51)f2+P1f=p1eλz+p2e−λz.Differentiating equation (51) twice on both sides, we have(52)2ff′+P1′f=λp1eλz−p2e−λz,(53)2f′2+2ff″+P1″f=λ2p1eλz+p2e−λz.By equations (51) and (53), we obtain(54)f′2=12λ2f2−ff″+Q1f,where Q1f=1/2λ2P1f−P1″f. Eliminating eλz and e−λz from equations (51) and (52), we have(55)λ2f2+P1f2−2ff′+P1′f2=4p1p2λ2,which implies(56)λ2f4−4f2f′2=R3f,where(57)R3f=4ff′P1′f+P1′f2+4p1p2λ2−2λ2f2P1f−λ2P1f2.Substituting equation (54) into equation (56) yields(58)f34f″−λ2f=T3f,where T3f=4f2Q1f+R3f is a differential-difference polynomial in f and its total degree is not greater than three.Now, we discuss two cases.Case 1. T3f≡0.It follows from equation (58) that(59)4f″−λ2f≡0.The general entire solutionfz of the above equation is(60)fz=c1eλ/2z+c2e−λ/2z,wherec1 and c2 are constants satisfying c1c2≡0. We obtain(61)Δfz=c1eλ/2−1eλ/2z+c2e−λ/2−1e−λ/2z,(62)Δ2fz=c1eλ/2−12eλ/2z+c2e−λ/2−12e−λ/2z.Substituting formulas (60)–(62) into equation (13) yields(63)2c1c2+c12−p1eλz+c22−p2e−λz+c1a1zeλ/2−1+a2zeλ/2−12eλ/2z+c2a1ze−λ/2−1+a2ze−λ/2−12e−λ/2z=0.By Lemma2 and equation (63), we deduce c1c2≡0, which is a contradiction.Case 2. T3f≡0.Noting thatf is an exponential polynomial in (23) with the order 1, we have(64)fz=H0z+H1zew1z+H2zew2z+⋯+Hmzewmz,whereH0z,H1z,…,Hmz are polynomials. Therefore,(65)4f″z−λ2fz=4H0″z−λ2H0z+∑i=1m4Hi″z+4w1Hi′z+4Hi′z+4w1Hiz−λ2Hiz.Since equation (58) satisfies conditions of Lemma 3 and Remark 2, it follows that(66)mr,4f″−λ2f=or.From this, (65), and Theorem 6, we know that that 4f″−λ2f is a polynomial. By equation (58) and T3f≡0, we have(67)4f″−λ2f=Hz,whereHz is a nonvanishing polynomial. By Lemma 4, the above equation has a nonzero polynomial solution c0z. Then, the general entire solution fz of 4f″−λ2f=Hz can be represented as(68)fz=c0z+c1eλ/2z+c2e−λ/2z,wherec0z is nonzero polynomial and c1 and c2 are constants. It is easy to verify(69)Δfz=c0z+1−c0z+c1eλ/2−1eλ/2z+c2e−λ/2−1e−λ/2z,Δ2fz=c0z+2−2c0z+1+c0z+c1eλ/2−12eλ/2z+c2e−λ/2−12e−λ/2z.Substituting formulas (68) and (70) into equation (13) yields(70)c12−p1eλz+c22−p2e−λz+2c1c0z+a1zc1eλ/2−1+a2zc1eλ/2−12eλ/2z+2c2c0z+a1zc2e−λ/2−1+a2zc2e−λ/2−12e−λ/2z+2c1c2+c0z2+a1zc0z+1−a1zc0z+a2zc0z+2−2a2zc0z+1+a2zc0z=0.By Lemma2 and equation (70), we deduce(71)c12=p1≠0,c22=p2≠0,(72)2c0z+a1zeλ/2−1+a2zeλ/2−12≡0,(73)2c0z+a1ze−λ/2−1+a2ze−λ/2−12≡0,(74)2c1c2+c0z2+a1zc0z+1−a1zc0z+a2zc0z+2−2a2zc0z+1+a2zc0z≡0.From (72) and (73), we have(75)a1eλ/2−e−λ/2+a2eλ/2−e−λ/2eλ/2+e−λ/2−2=0,Setv=eλ/2, it follows that(76)v2−1a1v+a2v−12=0.Ifv=1, then λ=4kπi, and substituting v=1 into (72) or (73), we obtain c0z≡0. It is a contradiction.Ifv=−1, then λ=4kπ+2πi, and substituting v=−1 into (72) or (73), we obtain c0z=a1z−2a2z≠0 and (74) can be reduced to(77)2c1c2+a2za1z+2−2a2z+2−a1z+2a2z+a1z−2a2za1z+1−2a2z+1=0.Especially, ifa1 and a2 are constants, then c02=−2c1c2.Ifv≠±1 and v is the solution of a1v+a2v−12=0. From (72) or (73), we have c0z=a1z/2. Equation (74) can be reduced to(78)a1z2−2a1za1z+1−2a2zΔ2a1z=8c1c2.Especially, ifa1 is a constants, then c02=−2c1c2.


## 5. Conclusions

In this study, we mainly consider the solution of two equations when the solution is an exponential polynomial.First, we consider the nonlinear differential-difference equation (7) proposed by Chen et al. [6]. They conjecture that the conclusions of Theorem 2 are still valid. We consider the conjecture in Theorem 3. In the first step, we proved that ρf=degQ. From this, it seems plausible that f is an exponential polynomial of form (1). In the second step, we confirmed that ρf=degQ=1 when f is an exponential polynomial. In the last step, we give the solution when f belongs to Γ0′ by Conclusion 2.Second, we consider a difference equation which is similar to (10), where f3z is also replaced by f2z. Since we cannot prove that 4f″−λ2f is a polynomial if f has no restriction, a new Clunie Lemma is given in Remark 2 where f is an exponential polynomial. We obtain the expression of the solution of equation (13) if the solution is an exponential polynomial by the special Clunie Lemma.



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*Source: 6901270-2020-02-20.xml*