We begin by discussing the case of differential equations, and then move on to difference equations. Concerning the case of first-order differential equations, Malmquist [15] showed a century ago that the only equation of the form

η'=R(z,η),

where R is rational in both arguments, that can have transcendental meromorphic solutions, is the Riccati equation:

η'=a0(z)+a1(z)η+a2(z)η2.

In 1954, Wittich [17] obtained the result that if the coefficients aj(z) are rational functions, then all meromorphic solutions of the Riccati equation are of finite order.

We consider a more general case of the following first-order algebraic differential equation

(1.1)C(z,η)(η')2+B(z,η)η'+A(z,η)=0,

where C(z,η)≡0,B(z,η) and A(z,η) are polynomials in z and η. In 1980, Steinmetz [16] showed that if (1.1) has a transcendental meromorphic solution, then the equation (1.1) can be reduced to the form

(1.2)a(z)η'2+(b2(z)η2+b1(z)η+b0(z))η'=d4(z)η4(z)+d3(z)η3+d2(z)η2+d1(z)η+d0(z).

where a(z),bi(z) for i=0,1,2 and dj(z) for j=0,...,4 are polynomials.

In this paper, we adopt the standard notation of Nevanlinna theory, as found in [9, 18]. Moreover, the forward difference △cη is defined as △cη=η(z+c)-η(z). In recent years, there has been tremendous interest in developing the value distribution of meromorphic functions with respect to a difference analogue, see [5, 6]. In 2018, Ishizaki and Korhonen [10] investigated meromorphic solutions of a difference equation of the form

△η(z)2=A(z)(η(z)η(z+1)-B(z)).

They proved that the above difference equation possesses a continuous limit to the difference equation

(η')2=A(z)(η2-1),

which extends to solutions in certain cases.

For a more general case, next let us consider the difference analogue of (1.2). It is interesting to consider the nature of a meromorphic solution η of

(1.3)a(z)△cηη2+(b2(z)η2(z)+b1(z)η(z)+b0(z))△cηη=d4(z)η4(z)+d3(z)η3(z)+d2(z)η2(z)+d1(z)η(z)+d0(z).

Our first theorem is about the growth of meromorphic solutions of (1.3).

Theorem 1.1. Let c∈C∖{0}, let T(r,a(z))=S(r,η), let T(r,bi(z))=S(r,η) for i=0,1,2, let T(r,dj(z))=S(r,η) for j=0,⋯,4 and let d4(z)≡0. If η(z) is a transcendental meromorphic solution of (1.3), then ρ(η)=∞.

Here ρ(η) denotes the order of growth of the meromorphic function η(z). In what follows λ(η) and λ(1η) denote the exponents of convergence of the zeros and poles of η(z), respectively. While the above result was about the case d4(z)≡0, the following is about the case d4(z)≡0. Indeed, taking d4(z)≡0, (1.3) becomes

(1.4)a(z)η'2+(b2(z)η2+b1(z)η+b0(z))η'=d4(z)η4(z)+d3(z)η3+d2(z)η2+d1(z)η+d0(z)

Using the method from Liao and Yang [13], we obtain

Theorem 1.2. Let c∈C∖{0}, and let a(z),bi(z) for i=0,1,2), and dj(z) for j=0,1,2,3 be polynomials. If η(z) is a finite order transcendental entire solution of (1.4), and either dega(z)≠degd0(z)+1, or, dega(z)=degd0(z)+1 and ρ(η)≠12, then

ρ(η)≥1.

Let cj,j=1,⋯,n, be a finite collection of complex numbers. Then a difference polynomial in η(z) is a function which is polynomial in η(z+cj) for j=1,⋯,n, with meromorphic coefficients aλ(z) such that T(r,aλ)=S(r,η) for all λ. As for difference counterparts of the Clunie Lemma [3], see [7, Corollary 3.3]. The following lemma is a more general version. The following lemma due to Laine and Yang [11] is an analogue of a result due to A. Z. Mohon'ko and V. D. Mohon'ko [14] on differential equations. We start by recalling some lemmas.

Lemma 2.1 [9] Let η be a transcendental meromorphic solution of finite order of a difference equation of the form

(2.1)U(z,η)P(z,η)=Q(z,η),

where U(z,η),P(z,η), and Q(z,η) are difference polynomials such that the total degree degU(z,η)=n in η(z) and its shifts, and degQ(z,η)≤n. Moreover, we assume that U(z,η) contains just one term of maximal total degree in η(z) and its shifts. Then

m(r,P(z,η))=S(r,η).

We need one more lemma from [8]. We say that η has more than S(r,η) poles of a certain type, if the integrated counting function of these poles is not of type S(r,η). We use the notation D(z0,r) to denote an open disc of radius r centered at z0∈C. Also, ∞k denotes a pole of η with multiplicity k. Similarly, 0k and a+0k denote a zero and a-point of η, respectively, with the multiplicity k.

Lemma 2.2 [6] Let η be a meromorphic function having more than S(r,η) poles, and let as,s=1,⋯,n, be small meromorphic functions with respect to η. Denote by mj the maximum order of zeros and poles of the functions as at zj. Then for any ε>0, there are at most S(r,η) points zj such that q

η(zj)=∞kj,

where mj≥εkj.

Proof of Theorem 1.1. Let η be a meromorphic solution of (1.3). We assume that ρ(η)=ρ<∞. (1.3) can be written as follows

(2.1)d4(z)η6(z)=Q(z,η),

where Q(z,η)=a(z)(△cη)2+(b2(z)η2(z)+b1(z)η(z)+b0(z))△cηη-d3(z)η5(z)-d2(z)η4(z)-d1(z)η3(z)-d0(z)η2(z). Since the total degree of Q(z,η) as a polynomial in η(z) and its shifts, degQ(z,η)≤5, by Lemma 2.1 and (2.1), we have

m(r,η)=S(r,η).

So, η has more than S(r,η) poles, counting multiplicities. Using zj to denote points in the pole sequence. By Lemma 2.2, we obtain that there exist more than S(r,η) points such that η(zj)=∞kj, where εkj>mj. Here mj refers to the coefficients a(z),bi(z)(i=0,1,2),dj(z)(j=0,1,2,3). Denoting the sequence of such poles by z1,j, we take this sequence as our starting point. For ε<18, (1.3) implies that η(z1,j+c)=∞k2,j, where k2,j≥(2-ε)k1,j. Lemma 2.2 implies that η has more than S(r,η) such points z2,j such that η(z2,j)=∞k2,j, where εk2,j>m2,j. Then we only pick one of these points and denote it by z2,j. Continuing to the next phase. By (1.3), we deduce that z3,j:=z2,j+c is a pole of η of multiplicity k3,j, where

k3,j≥(2-ε)k2,j≥(2-ε)2k1,j.

Following the steps above, we can find a sequence zn of poles of η, the multiplicity of which is kn, and kn≥(2-ε)n-1k1≥(2-ε)n-1.

By a simple geometric observation, we have

zn∈D(z1,(n-1)|c|)⊂D(0,|z1|+(n-1)|c|)=D(0,rn).

As n→∞, we have rn≤2(n-1)|c|. So,

n(rn,f)≥(2-ε)n-1>(158)n-1.

Hence, we have λ(η)=∞, a contradiction. So ρ(η)=∞.

Lemma 3.1 [2] Let η be a transcendental entire function of order ρ(η)=ρ<1, let 0<ε<18 and z be such that |z|=r, where

|η(z)|>M(r,η)(ν(r,η))-18+ε

holds. Then for each positive integer k, there exists a set E⊂(1,∞) that has finite logarithmic measure, such that for all r∈E∪[0,1],

△cηη=cν(r,η)z(1+o(1)).

Lemma 3.2 [10] Suppose that η(z) is a transcendental entire function of finite order ρ(η)=ρ<∞, and that a set Er⊂R+ has a finite logarithmic measure. Then, there exists a sequence of positive numbers rk satisfying rk∈Er and rk→∞ such that for given ε>0, as rk sufficiently lager, we have rkρ-ε<ν(rk,η)<rkρ+ε and exprkρ-ε<M(rk,η)<exprkρ+ε.

Proof of Theorem 1.2. Suppose that η(z) is a transcendental entire function. Suppose, contrary to the assertion, that ρ(η)=ρ<1. If d3(z)≡0, then we can write (1.4) in this form

(4.1)d3(z)=a(z)η3△cηη2+b2(z)η△cηη+b1(z)η2△cηη +b0(z)η3△cηη-d2(z)η-d1(z)η2-d0(z)η3.

By Lemma 3.1, we know that there exists a set H⊂(1,∞) of finite logarithmic measure, such that

(4.2)△cηη=cν(r,η)z(1+o(1)),  |z|=r∈H,

where z satisfy |z|=r and |η(z)|=M(r,η),  ν(r,η) is the central index of η(z). By Lemma 3.2, we see that there exist some infinite sequence of points zk such that |η(zk)|=M(rk,η), and such that for any given ε(0<ε<1-ρ2), as rk→∞, and |zk|=rk∈H1∪H∪[0,1] ,where H1⊂(1,∞) is a subset with finite logarithmic measure, we have

(4.3)ν(rk,η)rk<rkρ+ε-1→0.

Thus, by (4.1)-(4.3), we deduce that as zk satisfy |η(zk)|=M(rk,η),|zk|=rk∈H1∪H∪[0,1],rk→∞

(4.4)|d3(zk)|≤|a(zk)M(rk,η)3(△cηη)2|+|b2(zk)M(rk,η)△cηη|+|b1(zk)(M(rk,η)2)△cηη|+|b0(zk)M(rk,η)3△cηη|+|d2(zk)M(rk,η)|+|d1(zk)M(rk,η)2|+|d0(zk)M(rk,η)3|=|a(zk)M(rk,η)3||(cν(rk,η)rk(1+o(1)))2|+|b2(zk)M(rk,η)||cν(rk,η)rk(1+o(1))|+|b1(zk)(M(rk,η))2||cν(rk,η)rk(1+o(1))|+|b0(zk)M(rk,η)3||cν(rk,η)rk(1+o(1))|+|d2(zk)M(rk,η)|+|d1(zk)M(rk,η)2|+|d0(zk)M(rk,η)3|→0.

This is impossible. Hence d3(z)≡0. Now we may write (1.4) as follows

(4.5)d2(z)-b2(z)△cηη=a(z)η2(△cηη)2+b1(z)η△cηη+b0(z)η2△cηη-d1(z)η-d0(z)η2.

By (4.2), (4.3), and (4.5), we know that

(4.6)||d2(zk)|-|b2(zk)cν(rk,η)rk(1+o(1))||≤|d2(zk)-b2(zk)△cηη|=|a(zk)η2(△cηη)2+b1(zk)η△cηη+b0(zk)η2△cηη-d1(zk)η-d0(zk)η2|=|a(zk)M(rk,η)2||(cν(rk,η)rk(1+o(1)))2|+|b1(zk)M(rk,η)||cν(rk,η)rk(1+o(1))|+|b0(zk)M(rk,η)2||cν(rk,η)rk(1+o(1))|+|d1(zk)M(rk,η)|+|d0(zk)M(rk,η)2|→0.

We divide the proof into the following two cases

Case 1. If b2(z)≡0, then (4.6) implies that d2(z)≡0, hence (1.4) can be written as following

(4.7)a(z)△cηη2+(b1(z)η(z)+b0(z))△cηη=d1(z)η(z)+d0(z).

By computing (4.7), we have

(4.8)||d1(z)|-|b1(z)△cηη||≤|d1(z)-b1(z)△cηη|=|a(z)η(△cηη)2+b0(z)η△cηη-d0(z)η|≤|a(z)η(△cηη)2|+|b0(z)η△cηη|+|d0(z)η|.

By (4.1)-(4.3) and (4.8), we obtain that as zk satisfy |η(zk)|=M(rk,η),|zk|=rk∈H1∪H∪[0,1],rk→∞

(4.9)||d1(zk)|-|b1(zk)cν(rk,η)rk(1+o(1))||≤|a(zk)M(rk,η)(cν(rk,η)rk(1+o(1)))2|+|b0(zk)M(rk,η)cν(rk,η)rk(1+o(1))|+|d0(zk)M(rk,η)|→0.

If b1(z)≡0, then by (4.9),

(4.10)||d1(zk)|-|b1(zk)cν(rk,η)rk(1+o(1))||b1(zk)→0.

(4.3) and (4.10) imply that

(4.11)d1(zk)b1(zk)→0,

as k→∞ . Since d1(z) and b1(z) are polynomials, we obtain that by (4.11)

(4.12)zkd1(zk)b1(zk)→q,

as k→∞, and q is a finite constant. Suppose that d1(z)≡0. Then we deduce that from (4.10) and (4.12)

|q|=limk→∞|zkd1(zk)b1(zk)|=|c|limk→∞ν(rk,η)(1+o(1))=∞.

This is impossible. Hence d1(z)≡0. (1.4) can be reduced into

(4.13)a(z)(△cηη)2+(b1(z)η(z)+b0(z))△cηη=d0(z),

By the above assumption, we know b1(z)≡0, then (4.13) implies that

(4.14)|△cη|=|d0(zk)b1(zk)-a(zk)(△cηη)2b1(zk)-b0(zk)△cηηb1(zk)|≤|d0(zk)b1(zk)|+|a(zk)(△cηη)2b1(zk)|+|b0(zk)△cηηb1(zk)|≤MrkN.

where M and N are some finite constants. On the other hand, we know that

(4.15)|△cηMrkN|=|c|ν(rk,η)(1+o(1))M(rk,η)|M|rkN+1→∞,

as k→∞, a contradiction. Hence b1(z)≡0. By (4.9), we also get d1(z)≡0. Hence, we can write (1.4) as follows

(4.16)a(z)(△cηη)2+b0(z)△cηη=d0(z).

We assume that a(z)≡0, next we consider the following two subcases.

Subcase I. degb0(z)≥dega(z), we have by (4.16), (4.2) and (4.3),

(4.17)|lb0(zk)ν(rk,η)rk(1+o(1))|=|d0(zk)|

where l is a finite nonzero constant. So

(4.18)limk→∞|d0(zk)b0(zk)|=limk→∞|l|ν(rk,η)rk(1+o(1))=0

(4.18) implies that

(4.19)limk→∞zkd0(zk)b0(zk)→l1,

where l1 is some finite constant. By (4.19) and (4.17), as k→∞, we obtain

ν(rk,η)≤|d0(zk)b0(zk)rk|→l1.

We can get a contradiction, since ν(rk,η)→∞, as k→∞.

Subcase 2. degb0(z)<dega(z), we have by (4.16), (4.2) and (4.3),

(4.20)|d0(zk)a(zk)|≤(cν(rk,η)rk)2(1+o(1))|+|b0(zk)a(zk)cν(rk,η)rk(1+o(1))|→0,

as k→∞. We assume that d0(z)≡0. If dega(z)=degb0(z)+1, then as k→∞

(4.21)rkd0(zk)a(zk)=l2,   rkb0(zk)a(zk)→l3

where l2 is a finite nonzero constant, and l2 is a finite constant. By Lemma 3.2, we have

(4.22)rnρ(η)-ε<ν(rn,η)<rnρ(η)+ε.

If ρ(η)<12, then by (4.16), (4.21) and (4.22), for any given ε(0<ε<1-2ρ2), we have

|l2|=|rkd0(zk)a(zk)|≤|ν(rk,η)2rk|+|ν(rk,η)rkrkb0(zk)a(zk)|≤rk2ρ+2ε-1+|l3|rkρ+ε-1→0,

a contradiction. If ρ(η)>12, then by (4.16), (4.21) and (4.22), for any given ε(0<ε<1-2ρ2), we have

rk2ρ-1-ε≤|(ν(rk,η)2)rk|≤|(ν(rk,η))rk||rkb0(zk)a(zk)|+|rkd0(zk)a(zk)|≤l4,

where l4 is some finite constant. This is impossible, since rk2ρ-1-ε→∞, as k→∞. If dega(z)>degd0(z)+1, as k→∞, we have

(4.23)rkb0(zk)a(zk)→l3,   rk2d0(zk)a(zk)→l5,

where l5 are some finite constants. By (4.23) and (4.16), as k→∞, we have

ν(rk,η)≤|rkb0(zk)a(zk)|+|rk2d0(zk)a(zk)1ν(rk,η)|→l6,

where l6 is some finite constant, we can get a contradiction, since ν(rk,η)→∞. So d0(z)≡0. By (4.16), we have

ν(rk,η)≤|rkb0(zk)a(zk)|→l3,

a contradiction.

By Subcase 1 and Subcase 2, we have a(z)≡0. So (1.4) can be reduced into

(4.24)b0(z)△cηη=d0(z)

Together (4.24) and (4.2), we obtain

(4.25)b0(zk)cν(rk,η)rk=d0(zk).

(4.25) implies that either limk→∞ν(rk,η)=l7, where l7 is a finite constant, or ν(rk,η)≥l8rkn, where l8 is a finite nonzero constant, and n is a positive integer. This is a contradiction.

Case 2. If b2(z)≡0, then by (4.6)

(4.26)||d2(zk)|-|b2(zk)cν(rk,η)rk(1+o(1))||b2(zk)→0.

By (4.26), we have

(4.27)d2(zk)b2(zk)→0,

as k→∞. Since d2(z) and b2(z) are polynomials, we get by (4.27)

(4.28)zkd2(zk)b2(zk)→l9,

as k→∞, and l9 is a finite constant. Suppose that d2(z)≡0. Then we deduce that from (4.28)

(4.29)l9=limk→∞|zkd2(zk)b2(zk)|=|c|limk→∞ν(rk,η)(1+o(1))=∞.

This is impossible, since l9 is a finite constant. Hence d2(z)≡0. (1.4) can be reduced into

|△cη|=|d1(zk)b2(zk)+d0(zk)b2(zk)1η(z)-b1(zk)△cηηb2(zk)-a(zk)(△cηη)2b2(zk)1η-b0(zk)△cηηb2(zk)1η|≤|d1(zk)b2(zk)|+|d0(zk)b2(zk)1η(z)|+|b1(zk)(△cηη)2b2(zk)|+|a(zk)(△cηη)2b2(zk)1η|+|b0(zk)△cηηb2(zk)1η|≤l10rkl11

where l10 and l11 are some finite constants. On the other hand, we know that

(4.30)△cηl10rkl11=|c|ν(rk,η)(1+o(1))M(rk,η)l10rkl11+1→∞,

as k→∞. This is a contradiction, △cηl10rkl11<1

By Case 1 and Case 2, we know ρ(η)≥1.