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Research article

Meromorphic solutions of three certain types of non-linear difference equations

  • In this paper, the representations of meromorphic solutions for three types of non-linear difference equations of form

    fn(z)+Pd(z,f)=u(z)ev(z),

    fn(z)+Pd(z,f)=p1eλz+p2eλz

    and

    fn(z)+Pd(z,f)=p1eα1z+p2eα2z

    are investigated, where n2 is an integer, Pd(z,f) is a difference polynomial in f of degree dn1 with small coefficients, u(z) is a non-zero polynomial, v(z) is a non-constant polynomial, λ,pj,αj(j=1,2) are non-zero constants. Some examples are also presented to show our results are best in certain sense.

    Citation: Min Feng Chen, Zhi Bo Huang, Zong Sheng Gao. Meromorphic solutions of three certain types of non-linear difference equations[J]. AIMS Mathematics, 2021, 6(11): 11708-11722. doi: 10.3934/math.2021680

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  • In this paper, the representations of meromorphic solutions for three types of non-linear difference equations of form

    fn(z)+Pd(z,f)=u(z)ev(z),

    fn(z)+Pd(z,f)=p1eλz+p2eλz

    and

    fn(z)+Pd(z,f)=p1eα1z+p2eα2z

    are investigated, where n2 is an integer, Pd(z,f) is a difference polynomial in f of degree dn1 with small coefficients, u(z) is a non-zero polynomial, v(z) is a non-constant polynomial, λ,pj,αj(j=1,2) are non-zero constants. Some examples are also presented to show our results are best in certain sense.



    In 1964, Hayman [6,P69] extended and improved the results of Tumura [16] and Clunie [1], and obtained the following theorem.

    Theorem 1.1. (See [6]) Suppose that f(z),g(z) are non-constant meromorphic function and satisfy the equation

    fn(z)+Qd(z,f)=g(z), (1.1)

    where Qd(z,f) is a differential polynomial in f of degree dn1. If

    N(r,f)+N(r,1g)=S(r,f), (1.2)

    then g(z)=(γ0(z)+f(z))n, where γ0(z) is a small function of f(z).

    With the establishment of difference analogue of the lemma of the logarithmic derivative, the Clunie and Mohon'ko lemmas, see Halburd and Korhonen [8], Chiang and Feng [5], more and more researchers [2,3,4,13,15,17,20] come to study the properties of complex difference equations or complex differential-difference equations. In particular, there has been a renewed interest [11,12,14,17] in solvability and existence for entire or meromorphic solutions of Eq (1.1). Replacing the differential polynomial Qd(z,f) in Eq (1.1) by a differential-difference polynomial, and satisfying the condition (1.2), Chen et al. [4] proved the following result.

    Theorem 1.2. (See [4]) Let n2 be an integer. Suppose that f(z) is a non-constant finite order meromorphic solution of

    fn(z)+Qd(z,f)=u(z)ev(z), (1.3)

    where Qd(z,f) is a differential-difference polynomial in f with degree dn2, u(z) is a non-zero rational function, v(z) is a non-constant polynomial. If N(r,f)=S(r,f), then

    u(z)ev(z)=fn(z)andQd(z,f)=0.

    Remark 1.1. With other conditions fixed, if Qd(z,f) is a difference polynomial in f, the Theorem 1.2 is still valid. If the condition dn2 is omitted, then the Theorem 1.2 is impossible. For example, f(z)=1+ez is an entire solution of equation f2(z)f(z+ln2)=e2z, where n=2, d=n1=1. So it's natural to ask what will happen to the solutions of Eq (1.3) when d=n1? In this paper, we study this problem and obtain the following result.

    Theorem 1.3. Let n2 be an integer, Pd(z,f) be a difference polynomial in f of degree dn1 with polynomial coefficients, u(z) be a non-zero polynomial, v(z) be a non-constant polynomial. If f(z) is a transcendental meromorphic solution of equation

    fn(z)+Pd(z,f)=u(z)ev(z) (1.4)

    and N(r,f)=S(r,f), then σ(f)=degv(z), and one of the following holds:

    (1) fn(z)=u(z)ev(z) and Pd(z,f)=0;

    (2) If φ(z)=(u(z)+u(z)v(z))f(z)nu(z)f(z)0, then T(r,f)=N1)(r,1f)+S(r,f). Furthermore, if φ(z) is a non-zero polynomial, then f(z)=γ0(z)+p(z)eq(z), where γ0(z),p(z),q(z) are non-zero polynomials satisfying pn(z)=u(z),nq(z)=v(z).

    Example 1.1. f(z)=z+ez is a solution of the following difference equation

    f3(z)zf(z)f(z+ln3)+z(z+ln3)f(z)z3=e3z.

    Here, u(z)=1, v(z)=3z, n=3, φ=(u+uv)f3uf=3(z1). It implies that the solution satisfying Theorem 1.3 does exist.

    If g(z)=p1eλz+p2eλz and the condition (1.2) is not met, what will happen to the solutions of Eq (1.1)? Li [11] obtained the following result.

    Theorem 1.4. (See [11]) Let n2 be an integer, Qd(z,f) be a differential polynomial in f of degree dn1, and let λ,p1,p2 be non-zero constants. If f(z) is a transcendental meromorphic solution of the equation

    fn(z)+Qd(z,f)=p1eλz+p2eλz (1.5)

    and N(r,f)=S(r,f), then f(z)=c0(z)+c1eλnz+c2eλnz, where c0(z) is small function of f(z), cj are constants satisfying cnj=pj,j=1,2.

    It's natural to ask: what will happen to the solutions of Eq (1.5) when Qd(z,f) is a difference polynomial in f of degree dn1? In this paper, we consider this problem and obtain the following result.

    Theorem 1.5. Let n2 be an integer, Pd(z,f) be a difference polynomial in f of degree dn1 with polynomial coefficients, λ,p1,p2 be non-zero constants. If f(z) is a transcendental meromorphic solution of equation

    fn(z)+Pd(z,f)=p1eλz+p2eλz (1.6)

    and N(r,f)=S(r,f), then σ(f)=1, and one of the following holds:

    (1) f(z)=c1eλnz+c2eλnz;

    (2) If φ(z)=λ2fn2f0, then T(r,f)=N2)(r,1f)+S(r,f). Furthermore, if φ(z) is a non-zero polynomial, then f(z)=γ0(z)+c1eλnz+c2eλnz, where γ0(z) is a non-zero polynomial, cj are constants satisfying cnj=pj,j=1,2.

    Example 1.2. f(z)=z+ez+ez is a solution of the following difference equation

    f2(z)+2zf(z+iπ)(3z2+2iπz+2)=e2z+e2z.

    Here, λ=2, n=2, φ=λ2fn2f=4z. It implies that the solution satisfying Theorem 1.5 does exist.

    If g(z)=p1eα1z+p2eα2z and the condition (1.2) is not met, what will happen to the solutions of Eq (1.1)? Liu et al. [14] obtained the following result.

    Theorem 1.6. (See [14]) Let n2 be an integer, Qd(z,f) be a differential polynomial in f of degree dn1 with polynomial coefficients, and let pj,αj(j=1,2) be non-zero constants satisfying

    α1α2{tn,nt:1tn}. (1.7)

    If equation

    fn(z)+Qd(z,f)=p1eα1z+p2eα2z (1.8)

    admits a meromorphic solution f(z) satisfying N(r,f)=S(r,f), then one of the following holds:

    (1) f(z)=γ1(z)+c1eα1z, α1α2=nt,

    (2) f(z)=γ2(z)+c2eα2z, α1α2=tn, where γj(z) are small functions of f(z), cj are constants satisfying cnj=pj,j=1,2.

    What will happen to the solutions of Eq (1.8) when Qd(z,f) is a difference polynomial in f of degree dn1? In this paper, we study this problem and obtain the following result.

    Theorem 1.7. Let n2 be an integer, Pd(z,f) be a differential polynomial in f of degree dn1 with polynomial coefficients, and let pj,αj(j=1,2) be non-zero constants satisfying the condition (1.7). If f(z) is a meromorphic solution of equation

    fn(z)+Pd(z,f)=p1eα1z+p2eα2z (1.9)

    and N(r,f)=S(r,f), then σ(f)=1 and T(r,f)=N2)(r,1f)+S(r,f). If φ(z)=α1α2f(z)n(α1+α2)f(z)+n2f(z) is a non-zero polynomial and then one of the following holds:

    (1) f(z)=γ0(z)+c1eα1nz, α1α2=nt;

    (2) f(z)=γ0(z)+c2eα2nz, α1α2=tn, where γ0(z) is a non-zero polynomial, cj are constants satisfying cnj=pj,j=1,2.

    Remark 1.2. The condition (1.7) in Theorem 1.7 is necessary, see example 1.3.

    Example 1.3. f(z)=ez+e2z is a solution of the following difference equation

    f2(z)12f(z+ln2)+f(z)=e4z+2e3z.

    Here, α1=4, α2=3, n=2, α1α2=2nn+1.

    Remark 1.3. The following examples 1.4 and 1.5 show that the solution satisfying Theorem 1.7 does exist.

    Example 1.4. f(z)=1+ez is a solution of the following difference equation

    f3(z)+2f(z)f(z+iπ)3f(z)=e3z+e2z.

    Here, α1=3, α2=2, n=3, α1α2=nn1, φ=α1α2fn(α1+α2)f+n2f=6.

    Example 1.5. f(z)=1ez is a solution of the following difference equation

    f3(z)f(z+ln3)f(z)+f(z+iπ)1=2eze3z.

    Here, α1=1, α2=3, n=3, α1α2=n2n, φ=α1α2fn(α1+α2)f+n2f=3.

    In this paper, we assume familiarity with the basic results and standard notations of Nevanlinna theory [6,9,19]. f is meromorphic function in the whole complex plane C. In addition, we use σ(f) to denote the order of growth of f. For simplicity, we denote by S(r,f) any quantify satisfying S(r,f)=o(T(r,f)), as r, outside of a possible exceptional set of finite logarithmic measure. A meromorphic function φ defined in C is called a small function of f if T(r,φ)=S(r,f). Let Nk)(r,1f) denote the counting function of those zeros of f (counting multiplicity) whose multiplicities are not greater than k, and let N(k(r,1f) denote the counting function of those zeros of f (counting multiplicity) whose multiplicities are not less than k.

    Lemma 2.1. (See [5,Corollary 2.6]) Let η1,η2 be two complex numbers such that η1η2 and let f(z) be a finite order meromorphic function. Let σ be the order of f(z), then for each ε>0, we have

    m(r,f(z+η1)f(z+η2))=O(rσ1+ε).

    Lemma 2.2. (See [7,Corollary 3.3]) Let f be a non-constant finite order meromorphic solution of

    fn(z)P(z,f)=Q(z,f),

    where P(z,f) and Q(z,f) are difference polynomials in f with small meromorphic coefficients, and let cC, δ<1. If the total degree of Q(z,f) as a polynomial in f and its shifts is n, then

    m(r,P(z,f))=o(T(r+|c|,f)rδ)+o(T(r,f))

    for all r outside of a possible exceptional set E with finite logarithmic measure Edrr<.

    Using the same methods as in proof of [9,Lemma 2.4.2] and Lemma 2.1, we have a similar conclusion as follows.

    Lemma 2.3. Let f be a non-constant finite order meromorphic solution of

    fn(z)P(z,f)=Q(z,f),

    where P(z,f) and Q(z,f) are differential-difference polynomials in f with small meromorphic coefficients. If the total degree of Q(z,f) as a polynomial in f and its derivatives and its shifts is n, then

    m(r,P(z,f))=S(r,f)

    for all r outside of a possible exceptional set with finite logarithmic measure.

    Lemma 2.4. Suppose that α(z),β(z),φ(z) are non-zero polynomials and satisfy the differential equation

    α(z)f(z)+β(z)f(z)=φ(z), (2.1)

    where degφ(z)degα(z)degβ(z). Then Eq (2.1) has a special solution γ0(z) which is a non-zero polynomial.

    Proof. Now we distinguish two cases below. Case 1: degα(z)=degβ(z); Case 2: degα(z)>degβ(z).

    Case 1. In this case, we consider three subcases. Subcase 1.1. If α,β,φ are non-zero constants, then γ0=φα. Subcase 1.2. If α,β are non-zero constants, and φ is non-constant polynomial. Let

    φ(z)=anzn+an1zn1++a0,n1,

    where an(0),an1,,a0 are constants. Assuming

    γ0(z)=bnzn+bn1zn1++b0,

    and using the method of undetermined coefficients, it follows from Eq (2.1) that

    {bn=anα;bj=ajnβαaj+1α,j=n1,n2,,0.

    So, we get the expression of the special solution γ0(z) and degγ0(z)=degφ(z). Subcase 1.3. If α,β are non-constant polynomials, then φ(z) must be non-constant polynomial. Using the method of undetermined coefficients and Eq (2.1), similar to the proof of subcase 1.2, we can also obtain the expression of the special solution γ0(z) and degγ0(z)=degφ(z)degα(z).

    Case 2. If degα(z)>degβ(z), then φ(z) must be are non-constant polynomials. By using the method of undetermined coefficients and equation (2.1), similar to the proof of subcase 1.2, we can also obtain the expression of the special solution γ0(z) and degγ0(z)=degφ(z)degα(z).

    Lemma 2.5. (See [3,Lemma 2.6]) Let n1 be an integer, λ be a non-zero constant and φ(z) be a non-zero polynomial. Then the differential equation

    λ2f(z)n2f(z)=φ(z)

    has a special solution γ0(z) which is a non-zero polynomial.

    Lemma 2.6. Let n2 be an integer, Pd(z,f) be a difference polynomial in f of degree dn1 with small meromorphic coefficients, and let pj,αj(j=1,2) be non-zero constants satisfying α1α2. If f(z) is a meromorphic solution of equation

    fn(z)+Pd(z,f)=p1eα1z+p2eα2z (2.2)

    and N(r,f)=S(r,f), then σ(f)=1.

    Proof. Set

    Pd(z,f)=μIaμ(z)tμj=1f(z+δμj)lμj, (2.3)

    where I is a finite set of the index μ, tμ, lμj (μI,j=1,,tμ) are natural numbers, δμj (μI,j=1,,tμ) are distinct complex constants. Denote gμj(z):=f(z+δμj)f(z) and substitute this equality into (2.3) yields

    Pd(z,f)=μI(aμ(z)tμj=1glμjμj(z))flμ(z):=dk=0bk(z)fk(z), (2.4)

    where lμ=tμj=1lμj, d=maxμI{lμ}, bk(z)=lμ=k(aμ(z)tμj=1glμjμj(z)) (k=0,,d). By applying Lemma 2.1, we get m(r,bk(z))=S(r,f) (k=0,,d). Differentiating (2.4) yields

    Pd(z,f)=μI((aμ(z)+tμj=1aμ(z)lμjf(z+δμj)f(z+δμj))tμj=1glμjμj(z))flμ(z):=dk=0ck(z)fk(z). (2.5)

    By using Lemma 2.1 and the lemma of the logarithmic derivative, we get m(r,ck(z))=S(r,f) (k=0,,d). Noting that m(r,bk(z))=S(r,f) (k=0,,d) and by induction, we obtain

    m(r,Pd(z,f))dm(r,f)+S(r,f). (2.6)

    Combining (2.2) and (2.6), and noting that N(r,f)=S(r,f), we get

    T(r,p1eα1z+p2eα2z)=T(r,fn+Pd(z,f))=m(r,fn+Pd(z,f))+S(r,f)nT(r,f)+S(r,f),

    and

    T(r,p1eα1z+p2eα2z)=T(r,fn+Pd(z,f))=m(r,fn+Pd(z,f))+S(r,f)m(r,fn)m(r,Pd(z,f))+S(r,f)(nd)T(r,f)+S(r,f).

    From the above two inequalities, we have

    (nd)T(r,f)+S(r,f)T(r,p1eα1z+p2eα2z)nT(r,f)+S(r,f).

    Then σ(f)=1.

    Lemma 2.7. (See [14,Lemma 2.5]) Let n1 be an integer, α1,α2 be non-zero constants satisfying α1α2, and let φ(z) be a non-vanishing polynomial. Then the differential equation

    α1α2f(z)n(α1+α2)f(z)+n2f(z)=φ(z)

    has a special solution γ0(z) which is a non-vanishing polynomial.

    Lemma 2.8. (See [7,p.247]) Suppose that f(z) is a transcendental meromorphic function and K>1. Then there exists a set M(K) of upper logarithmic density at most δ(K)=min{(2eK11)1,(1+e(K1))exp(e(1K))} such that for every positive integer k, we have

    lim suprrM(K)T(r,f)T(r,f(k))3eK.

    Remark 2.1. By Lemma 2.8, we see that if f is a transcendental meromorphic function, and if φ satisfying T(r,φ(k))=S1(r,f), then T(r,φ)=S1(r,f), where S1(r,f) is defined to be any quantity such that for any positive number ε there exists a set E(ε) whose upper logarithmic density is less than ε, and S1(r,f)=o(T(r,f)) as r, rE(ε).

    Lemma 2.9. (See [18,Theorem 1.51]) Suppose that f1,f2,,fn(n2) are meromorphic functions and g1,g2,,gn are entire functions satisfying the following conditions:

    (1) nj=1fjegj0.

    (2) gjgk are not constants for 1j<kn.

    (3) For 1jn,1h<kn,

    T(r,fj)=o(T(r,eghgk))(r,rE),

    where E[1,) is finite linear measure Edr< or finite logarithmic measure Edrr<. Then fj0(j=1,,n).

    Lemma 2.10. (See [10,Lemma 3]) Suppose that h is a nonconstant meromorphic function satisfying

    ¯N(r,h)+¯N(r,1h)=S(r,h).

    Let f=a0hp+a1hp1++ap, g=b0hq+b1hq1++bq be polynomials in h with coefficients a0,a1,,ap,b0,b1,,bq being small functions of h and a0b0ap0. If qp, then m(r,g/f)=S(r,f).

    Proof. Combining (1.4), (2.6) and N(r,f)=S(r,f), we get

    T(r,u(z)ev(z))=T(r,fn+Pd(z,f))=m(r,fn+Pd(z,f))+S(r,f)nT(r,f)+S(r,f),

    and

    T(r,u(z)ev(z))=T(r,fn+Pd(z,f))=m(r,fn+Pd(z,f))+S(r,f)m(r,fn)m(r,Pd(z,f))+S(r,f)(nd)T(r,f)+S(r,f).

    From the above two inequalities, we get

    (nd)T(r,f)+S(r,f)T(r,u(z)ev(z))nT(r,f)+S(r,f).

    Then σ(f)=degv(z).

    Denote Pd(z,f)=P. By differentiating (1.4), we have

    nfn1f+P=(u+uv)ev. (3.1)

    Eliminating ev from (1.4) and (3.1), we have

    fn1φ=uP(u+uv)P, (3.2)

    where

    φ=(u+uv)fnuf. (3.3)

    By Lemma 2.3, we have

    m(r,φ)=S(r,f).

    Noting that N(r,f)=S(r,f), then T(r,φ)=S(r,f). We consider two cases as follows.

    Case 1. If φ0, then (u+uv)f=nuf, that is ff=u+uvnu=1n(uu+v). By integration, we get fn=cuev(c0). Substituting this equality into (1.4) yields (11c)fn=P. If c1, by Lemma 2.2, we have m(r,f)=S(r,f). Noting that N(r,f)=S(r,f), then T(r,f)=S(r,f), a contradiction. Therefore c=1, and f(z)=p(z)eq(z), where p(z),q(z) are non-zero polynomials satisfying pn(z)=u(z),nq(z)=v(z).

    Case 2. If φ0, and z0 is a multiple zero of f, it follows from (3.3) that z0 is a zero of φ. Then

    N(2(r,1f)=S(r,f). (3.4)

    We claim that u+uv0, otherwise f=φnu, then T(r,f)=S1(r,f), by Lemma 2.8, we have T(r,f)=S1(r,f), which is impossible. Rewrite (3.3) as 1f=1φ[(u+uv)nuff], by the lemma of the logarithmic derivative and T(r,φ)=S(r,f), we get

    m(r,1f)m(r,1φ)+m(r,u+uv)+m(r,nu)+m(r,ff)+O(1)=S(r,f). (3.5)

    Combining (3.4) and (3.5), we get

    T(r,f)=m(r,1f)+N(r,1f)+O(1)=N1)(r,1f)+S(r,f).

    If φ is a non-zero polynomial and deg(u+uv)deg(nu), by Lemma 2.4, we have

    f(z)=γ0(z)+p(z)eq(z), (3.6)

    where γ0(z),p(z),q(z) are non-zero polynomials. Substituting (3.6) into (1.4), and using Lemma 2.9, we get pn(z)=u(z), nq(z)=v(z).

    This completes the proof of Theorem 1.3.

    Proof. Suppose that f is a transcendental meromorphic solution of finite order of Eq (1.6) and N(r,f)=S(r,f), by Lemma 2.6, we get σ(f)=1. Denote Pd(z,f)=P. Differentiating (1.6) gives

    nfn1f+P=λ(p1eλzp2eλz). (4.1)

    Differentiating (4.1) yields

    n(n1)fn2(f)2+nfn1f+P=λ2(p1eλz+p2eλz). (4.2)

    Combining (1.6) and (4.1), we get

    λ2f2nn2f2(n1)(f)2+2λ2Pfn2nPfn1f+λ2P2(P)24λ2p1p2=0. (4.3)

    Combining (1.6) and (4.2), we get

    λ2fnn(n1)fn2(f)2nfn1f+λ2PP=0. (4.4)

    Eliminating (f)2 from (4.3) and (4.4), we have

    f2n1φ=Q(z,f), (4.5)

    where

    φ=λ2fn2f (4.6)

    and

    Q(z,f)=[(n2)λ2P+nP]fn2n(n1)Pfn1f+(n1)[λ2P2(P)24λ2p1p2], (4.7)

    Q(z,f) is a differential-difference polynomial in f of of degree 2n1. By Lemma 2.3, we have m(r,φ)=S(r,f). Noting that N(r,f)=S(r,f), then

    T(r,φ)=S(r,f). (4.8)

    We distinguish two cases below:

    Case 1. If φ0, that is

    λ2fn2f=0, (4.9)

    which has two fundamental solutions f1(z)=eλnz, f2(z)=eλnz. Then the general solution of Eq (4.9) can be expressed as

    f(z)=c1eλnz+c2eλnz.

    Substituting the above formula into (1.6), and applying Lemma 2.9, we obtain cnj=pj,j=1,2.

    Case 2. If φ0, and z0 is a multiple zero of f whose multiplicities are not less than 3, it follows from (4.6) that z0 is a zero of φ. Then

    N(3(r,1f)=S(r,f). (4.10)

    Rewriting (4.6) as 1f=1φ(λ2n2ff), by the lemma of the logarithmic derivative and (4.8), we get

    m(r,1f)m(r,1φ)+m(r,λ2)+m(r,n2ff)+O(1)=S(r,f). (4.11)

    It follows from (4.10) and (4.11) that

    T(r,f)=m(r,1f)+N(r,1f)+O(1)=N2)(r,1f)+S(r,f).

    If φ is a non-zero polynomial, by Lemma 2.5, we have

    f(z)=γ0(z)+c1eλnz+c2eλnz, (4.12)

    where γ0(z) is a non-zero polynomial, cj(j=1,2) are constants. Substituting (4.12) into (1.6), and using Lemma 2.9, we get cnj=pj,j=1,2.

    This completes the proof of Theorem 1.5.

    Proof. Suppose that f is a transcendental meromorphic solution of finite order of Eq (1.9) and N(r,f)=S(r,f), by Lemma 2.6, we get σ(f)=1. Denote Pd(z,f)=P. Differentiating (1.9) gives

    nfn1f+P=α1p1eα1z+α2p2eα2z. (5.1)

    Eliminating eα1z and eα2z from (1.4) and (4.1), respectively, we have

    α1fnnfn1f+α1PP=(α1α2)p2eα2z, (5.2)
    α2fnnfn1f+α2PP=(α2α1)p1eα1z. (5.3)

    Differentiating (5.3) yields

    nα2fn1fn(n1)fn2(f)2nfn1f+α2PP=α1(α2α1)p1eα1z. (5.4)

    Eliminating eα1z from (5.3) and (5.4), we have

    α1α2fnn(α1+α2)fn1f+n(n1)fn2(f)2+nfn1f=Q(z,f), (5.5)

    where

    Q(z,f)=α1α2P(α1+α2)P+P (5.6)

    is a differential-difference polynomial in f of degree n1. It follows from (5.2) and (5.3) that

    α1α2f2nn(α1+α2)f2n1f+n2f2n2(f)2=(α1α2)2p1p2e(α1+α2)zR(z,f), (5.7)

    where

    R(z,f)=(α1fnnfn1f)(α2PP)+(α2fnnfn1f)(α1PP)+(α1PP)(α2PP) (5.8)

    is a differential-difference polynomial in f of degree 2n1. Eliminating (f)2 from (5.5) and (5.7), we have

    f2n1φ=(n1)(α1α2)2p1p2e(α1+α2)z+T(z,f), (5.9)

    where

    φ=α1α2fn(α1+α2)f+n2f, (5.10)

    and

    T(z,f)=(n1)R(z,f)nfnQ(z,f) (5.11)

    is a differential-difference polynomial in f of degree 2n1. It follows from (1.9), (2.6) and N(r,f)=S(r,f) that

    O(T(r,ez))T(r,p1eα1z+p2eα2z)=T(r,fn+Pd(z,f))=m(r,fn+Pd(z,f))+S(r,f)m(r,fn)m(r,Pd(z,f))+S(r,f)(nd)T(r,f)+S(r,f). (5.12)

    It follows from (2.4), (2.5) and (5.2) that

    T(r,eα2z)=m(r,eα2z)m(r,((α1nff)fn+dk=0(α1bkck)fk))+O(1)nT(r,f)+S(r,f). (5.13)

    Combining (5.12) and (5.13), we have

    S(r,f)=S(r,ez). (5.14)

    Substituting (2.4) into (1.9) yields

    1p1eα1z+p2eα2zb0+dk=1bkp1eα1z+p2eα2zb0(1f)nk=(1f)n. (5.15)

    Without loss of generality, we assume that b0=Pd(z,0)0. Otherwise, we make the transformation ˆf=fc for a suitable constant c satisfying cn+Pd(z,c)0. Then (1.9) is changed to the form (ˆf)n(z)+Pd(z,ˆf)=p1eα1z+p2eα2z, where Pd(z,ˆf) is a difference polynomial in ˆf of degree at most n1 with polynomial coefficients, and Pd(z,0)=cn+Pd(z,c)0. Noting that b0=Pd(z,0)0, by applying Lemma 2.10, it follows from (1.7) that

    m(r,1p1eα1z+p2eα2zb0)=S(r,ez),m(r,eαjzp1eα1z+p2eα2zb0)=S(r,ez),j=1,2. (5.16)

    It follows from (5.14)(5.16) that

    nm(r,1f)(n1)m(r,1f)+S(r,f),m(r,eαjzfn)(n1)m(r,1f)+S(r,f),j=1,2,

    then

    m(r,1f)=S(r,f),m(r,eαjzfn)=S(r,f),j=1,2. (5.17)

    Without loss of generality, if α1α2=tn, for some t{1,2,,n1}, then

    |e(α1+α2)zf2n1|=|eα2z|n+tn|f|nn+tn+n1t=|eα2zfn|n+tn1|f|n1t.

    From (5.17) and the above equality, we have

    m(r,e(α1+α2)zf2n1)=S(r,f). (5.18)

    By applying Lemma 2.1 and the lemma of the logarithmic derivative, from (5.9), (5.17) and (5.18), we get m(r,φ)=S(r,f). Noting that N(r,f)=S(r,f), then T(r,φ)=S(r,f).

    We consider two cases as follows.

    Case 1. If φ0, that is α1α2fn(α1+α2)f+n2f0. The corresponding characteristic equation α1α2n(α1+α2)λ+n2λ2=0 has two roots α1n, α2n, we get

    f(z)=c1eα1nz+c2eα2nz, (5.19)

    where cj(j=1,2) are constants. Noting that f(z) is a transcendental meromorphic solution of equation (1.9) and satisfies m(r,1/f)=S(r,f), then c1c20. Substituting (5.19) into (1.9) yields

    (c_{1}^{n}-p_{1})e^{\alpha_{1}z}+(c_{2}^{n}-p_{2})e^{\alpha_{2}z}+ \sum\limits_{j = 1}^{n-1}\binom{j}{n}c_{1}^{j}c_{2}^{n-j}e^{\left(\frac{j}{n}\alpha_{1}+\frac{n-j}{n}\alpha_{2}\right)z} +P_{d}(z, f) = 0.

    From (1.7) and Lemma 2.9, at least one of \binom{j}{n}c_{1}^{j}c_{2}^{n-j}e^{\left(\frac{j}{n}\alpha_{1}+\frac{n-j}{n}\alpha_{2}\right)z} (j = 1, \ldots, n-1) is zero, then c_{1}c_{2} = 0 , which is impossible.

    Case 2. If \varphi\not\equiv 0 , and z_{0} is a multiple zero of f whose multiplicities are not less than 3 , it follows from (5.10) that z_{0} is a zero of \varphi . Then

    \begin{align} N_{(3}\left(r, \frac{1}{f}\right) = S(r, f). \end{align} (5.20)

    It follows from (5.17) and (5.20) that

    T(r, f) = m\left(r, \frac{1}{f}\right)+N\left(r, \frac{1}{f}\right)+O(1) = N_{2)}\left(r, \frac{1}{f}\right)+S(r, f).

    If \varphi is a non-zero polynomial, by applying Lemma 2.6, we have

    \begin{align} f(z) = \gamma_{0}(z)+c_{1}e^{\frac{\alpha_{1}}{n}z}+c_{2}e^{\frac{\alpha_{2}}{n}z}, \end{align} (5.21)

    where \gamma_{0}(z) is a non-zero polynomial, c_{j}\; (j = 1, 2) are constants. Noting that f(z) is a transcendental meromorphic solution of Eq (1.9) and satisfies m(r, 1/f) = S(r, f) , then c_{1}, c_{2} are not complete zeroes. If c_{1}c_{2}\neq 0 , substituting (5.21) into (1.9) yields

    \begin{eqnarray*} && (c_{1}^{n}-p_{1})e^{\alpha_{1}z}+(c_{2}^{n}-p_{2})e^{\alpha_{2}z}+ \sum\limits_{j = 1}^{n-1}\binom{j}{n}c_{1}^{j}c_{2}^{n-j}e^{\left(\frac{j}{n}\alpha_{1}+\frac{n-j}{n}\alpha_{2}\right)z}+\\ && \sum\limits_{j = 1}^{n}\binom{j}{n}\gamma_{0}^{j}(z)(c_{1}e^{\frac{\alpha_{1}}{n}z}+c_{2}e^{\frac{\alpha_{2}}{n}z})^{n-j} +P_{d}(z, f) = 0. \end{eqnarray*}

    From (1.7) and Lemma 2.9, at least one of \binom{j}{n}c_{1}^{j}c_{2}^{n-j}e^{\left(\frac{j}{n}\alpha_{1}+\frac{n-j}{n}\alpha_{2}\right)z} (j = 1, \ldots, n-1) is zero, then c_{1}c_{2} = 0 , which is impossible. Then we get f(z) = \gamma_{0}(z)+c_{1}e^{\frac{\alpha_{1}}{n}z} or f(z) = \gamma_{0}(z)+c_{2}e^{\frac{\alpha_{2}}{n}z} .

    This completes the proof of Theorem 1.7 .

    Using the Nevalinna theory of meromorphic functions, this paper study the meromorphic solutions of three Clunie-Tumura types of non-linear difference equations and get the exact forms of the meromorphic solutions of these difference equations with some added conditions. Improvements and extensions of some results in the literature are presented.

    The authors would like to thank the referee for his/her thorough reviewing with constructive suggestions and comments to the paper. This research was supported by the National Natural Science Foundation of China (Nos: 12001117, 12001503, 11801093) and by the National Natural Science Foundation of Guangdong Province (No. 2018A030313267).

    The authors declare no conflict of interest.



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