# Admissible Solutions of the Schwarzian Type Difference Equation

Authors: Baoqin Chen; Sheng Li
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

This paper is to investigate the Schwarzian type difference equationΔ3f/Δf-3/2Δ2f/Δf2k=Rz,f=P(z,f)/Q(z,f), where R(z,f) is a rational function in f with polynomial coefficients, P(z,f), respectively Q(z,f) are two irreducible polynomials in f of degree p, respectively q. Relationship between p and q is studied for some special case. Denote d=max⁡p,q. Let f(z) be an admissible solution of (*) such that ρ2(f)<1; then for s (≥2) distinct complex constants α1,…,αs, q+2k∑j=1sδ(αj,f)≤8k. In particular, if N(r,f)=S(r,f), then d+2k∑j=1sδ(αj,f)≤4k.


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## Body

## 1. Introduction and Results

Throughout this paper, a meromorphic function always means being meromorphic in the whole complex plane, andc always means a nonzero constant. For a meromorphic function f(z), we define its shift by f(z+c) and define its difference operators by
(1)Δcf(z)=f(z+c)-f(z),Δcnf(z)=Δcn-1(Δcf(z)),mmmmmmmmmmmmmmmmmmmmmn∈ℕ,n≥2.
In particular, Δcnf(z)=Δnf(z) for the case c=1. We use standard notations of the Nevanlinna theory of meromorphic functions such as T(r,f), m(r,f), and N(r,f) and as stated in [1–3]. For a constant a, we define the Nevanlinna deficiency by
(2)δ(a,f)=lim⁡inf⁡r→∞m(r,1/(f-a))T(r,f)=1-lim⁡sup⁡r→∞N(r,1/(f-a))T(r,f).Recently, numbers of papers (see, e.g., [4–12]) are devoted to considering the complex difference equations and difference analogues of Nevanlinna theory. Due to some idea of [13], we consider the admissible solution of the Schwarzian type difference equation:
(3)S~k(f):=[Δ3fΔf-32(Δ2fΔf)2]k=R(z,f)=P(z,f)Q(z,f),
where R(z,f) is a rational function in f with polynomial coefficients,  P(z,f), respectively Q(z,f), are two irreducible polynomials in f of degree p, respectively, q. Here and in the following, “admissible” always means “transcendental.” And we denote d=max⁡{p,q} from now on. For the existence of solutions of (3), we give some examples below.Examples.
(1)f(z)=sinπz+z is an admissible solution of the Schwarzian type difference equation:
(4)Δ3fΔf-32(Δ2fΔf)2=-8[f2+(1-2z)f+z(z-1)]4f2-4(2z+1)f+(2z+1)2.
(2)f(z)=(ezln⁡2/sin2πz)+z is an admissible solution of the Schwarzian type difference equation
(5)Δ3fΔf-32(Δ2fΔf)2=-f2+2(z+1)f-z2-2z2f2-4(z-1)f+2(z-1)2.
(3) Letf(z)=z2+z, then f(z) solves the Schwarzian type difference equation:
(6)Δ3fΔf-32(Δ2fΔf)2=-32[f2-2(z2-1)f+(z2-1)2].
This example shows that (3) may admit polynomial solutions.Considering the relationship betweenp and q in those examples above, we prove the following result.Theorem 1.
For the Schwarzian type difference equation (3) with polynomial coefficients, note the following.(i)
If it admits an admissible solutionf(z) such that ρ2(f)<1, then
(7)pm(r,f)≤qm(r,f)+S(r,f).In particular, ifm(r,f)≠S(r,f), then p≤q.(ii)
If its coefficients are all constants and it admits a polynomial solutionf(z) with degree s, then s≥2 and qs=ps+2k.Remark 2.
From examples (1) and (2), we conjecture thatp=q in Theorem 1(i). However, we cannot prove it currently. From example (3) given before, we see that the restriction on the coefficients in Theorem 1(ii) cannot be omitted.
For the Schwarzian differential equation,(8)Sk(f)=[f′′′f′-32(f′′f′)2]k=R(z,f)=P(z,f)Q(z,f),
where R(z,f),  P(z,f), and Q(z,f) are as stated before; Ishizaki [13] proved the following result (see also Theorem 9.3.2 in [2]).Theorem A (see [2, 13]).
Letf(z) be an admissible solution of (8) with polynomial coefficients, and let α1,…,αs be s(≥2) distinct complex constants. Then
(9)d+2k∑j=1s‍δ(αj,f)≤4k.For the Schwarzian type difference equation (3), we prove the following result.Theorem 3.
Letf(z) be an admissible solution of (3) with polynomial coefficients such that ρ2(f)<1, and let α1,…,αs be s(≥2) distinct complex constants. Then
(10)q+2k∑j=1s‍δ(αj,f)≤8k.
In particular, if N(r,f)=S(r,f), then
(11)d+2k∑j=1s‍δ(αj,f)≤4k.Remark 4.
From Theorem1, under the condition N(r,f)=S(r,f) in Theorem 3, we have d=q in (11). The behavior of the zeros and the poles of f(z) in S~k(f) is essentially different from that in the Sk(f). We wonder whether the restriction N(r,f)=S(r,f) can be omitted or not.


## 2. Lemmas

The following lemma plays a very important role in the theory of complex differential equations and difference equations. It can be found in Mohon’ko [14] and Valiron [15] (see also Theorem 2.2.5 in the book of Laine and Yang [2]).Lemma 5 (see [14, 15]).
Letf(z) be a meromorphic function. Then, for all irreducible rational functions in f,
(12)R(z,f)=P(z,f)Q(z,f)=∑i=0p‍ai(z)fi∑j=0q‍bj(z)fj,
with meromorphic coefficients ai(z),bj(z) such that
(13)T(r,ai)=S(r,f),i=0,…,p,T(r,bj)=S(r,f),j=0,…,q,
and the characteristic function of R(z,f) satisfies
(14)T(r,R(z,f))=dT(r,f)+S(r,f),
where d=max⁡{p,q}.The following two results can be found in [10]. In fact, Lemma 6 is a special case of Lemma 8.3 in [10].Lemma 6 (see [10]).
Letf(z) be a meromorphic function of hyper order ρ2(f)=ς<1, c∈ℂ and ε>0. Then
(15)T(r,f(z+c))=T(r,f)+S(r,f),
possibly outside of a set of r with finite logarithmic measure.Lemma 7 (see [10]).
Letf(z) be a meromorphic function of hyper order ρ2(f)=ς<1, c∈ℂ and ε>0. Then
(16)m(r,f(z+c)f(z))=o(T(r,f)r1-ς-ε)=S(r,f),
possibly outside of a set of r with finite logarithmic measure.From Lemma7, we can easily get the following conclusion.Lemma 8.
Letf(z) be a meromorphic function of hyper order ρ2(f)=ς<1, c∈ℂ and ε>0. Then
(17)m(r,Δcnf(z)f(z))=S(r,f),m(r,Δckf(z)Δcjf(z))=S(r,f),k>j,
possibly outside of a set of r with finite logarithmic measure.Lemma 9.
Letf be an admissible solution of (3) with coefficients. Then, using the notation Q(z):=Q(z,f(z)),
(18)qT(r,f)+S(r,f)≤N(r,1Q).
In particular, if N(r,f)=S(r,f), then
(19)dT(r,f)+S(r,f)≤N(r,1Q).Proof.
We use the idea by Ishizaki [13] (see also [2]) to prove Lemma 9. It follows from Lemma 8 that
(20)m(r,R)=m(r,[Δ3fΔf-32(Δ2fΔf)2]k)≤km(r,Δ3fΔf)+2km(r,Δ2fΔf)+S(r,f)=S(r,f).
From this and Lemma 5, we get
(21)dT(r,f)+S(r,f)=T(r,R)=N(r,R)+S(r,f),
and hence
(22)dT(r,f)=N(r,R)+S(r,f).
Ifd=p>q, since all coefficients of P(z,f) and Q(z,f) are polynomials, there are at the most finitely many poles of R(z,f), neither the poles of f(z) nor the zeros of Q(z,f). Therefore, we see that
(23)N(r,R)≤(p-q)N(r,f)+N(r,1Q)+S(r,f)≤(p-q)T(r,f)+N(r,1Q)+S(r,f).
We obtain (18) from this and (22) immediately.
Ifd=q≥p, there are at most finitely many poles of R(z,f), not the zeros of Q(z,f), then
(24)N(r,R)≤N(r,1Q)+S(r,f).
Now (18) follows from (22) and (24).
Notice that ifN(r,f)=S(r,f), then (24) always holds. This finishes the proof of Lemma 9.


## 3. Proof of Theorem1

Case 1.
Equation (3) admits an admissible solution f(z) such that ρ2(f)<1. Since all coefficients of P(z,f) and Q(z,f) are polynomials, there are at the most finitely many poles of f(z) that are not the poles of P(z,f) and Q(z,f).  This implies that
(25)N(r,P)=pN(r,f)+S(r,f),N(r,Q)=qN(r,f)+S(r,f).From Lemma5, we get
(26)T(r,P)=pT(r,f)+S(r,f),T(r,Q)=qT(r,f)+S(r,f).We can deduce from (3), (25), (26), and Lemma 8 that
(27)pT(r,f)+S(r,f)=T(r,P)=m(r,P)+N(r,P)≤pN(r,f)+m(r,S~k(f)Q)+S(r,f)≤pN(r,f)+m(r,S~k(f))+m(r,Q)+S(r,f)=pN(r,f)+T(r,Q)-N(r,Q)+S(r,f)=pN(r,f)+qT(r,f)-qN(r,f)+S(r,f)=pN(r,f)+qm(r,f)+S(r,f).
It follows from this that
(28)pm(r,f)≤qm(r,f)+S(r,f).
What is more is that if m(r,f)≠S(r,f), then we obtain from (28) that p≤qCase 2.
The coefficients of (3) are all constants and it admits a polynomial solution f(z) with degree s. Set
(29)f(z)=aszs+as-1zs-1+⋯+a1z+a0,
then
(30)f(z+1)=aszs+bs-1zs-1+⋯+b1z+b0,
where
(31)bs-j=asCsj+as-1Cs-1j-1+⋯+as-j+1Cs-j+11+as-j.
From (29) and (30), we obtain that
(32)Δf=saszs-1+(bs-2-as-2)zs-2+⋯+(b1-a1)z+(b0-a0).Ifs=1, then Δ2f=Δ3f≡0, which yields that P(z,f)≡0. That is a contradiction to our assumption. Thus, s≥2.Ifs=2, then Δf=2a2z+a2+a1, Δ2f=2a2, and Δ3f≡0. Now from (3), we get
(33)(-3)kQ(z,f)(Δ2f)2k=2kP(z,f)(Δf)2k.
Considering degrees of both sides of the equation above, we can see that q=p+k.Ifs≥3, we can deduce similarly that
(34)Δ2f=s(s-1)aszs-2+P1(z),Δ3f=s(s-1)(s-2)aszs-3+P2(z),
where P1(z),P2(z) are polynomials such that deg⁡P1≤s-3,deg⁡P2≤s-4.Rewrite (3) as follows:
(35)Q(z,f)[2Δ3f·Δf-3(Δ2f)2]k=2kP(z,f)(Δf)2k.From (34), we find that the leading coefficient of 2Δ3f·Δf-3(Δ2f)2 is
(36)-as2s2(s-1)(s+1)≠0.Considering degrees of both sides of (35), we prove that qs=ps+2k.


## 4. Proof of Theorem3

Firstly, we consider the general case. As mentioned in Remark 1 in [13], due to Jank and Volkmann [16], if (3) admits an admissible solution, then there are at most S(r,f) common zeros of P(z,f) and Q(z,f). Since all coefficients of Q(z,f) are polynomials, there are at the most finitely many poles of f that are the zeros of Q(z,f). Therefore, from (3), we have
(37)12kN(r,1Q)≤N(r,1Δf)+S(r,f)≤T(r,Δf)+S(r,f)=T(r,f(z+1)-f(z))+S(r,f)≤2T(r,f)+S(r,f).
Combining this and Lemma 9, applying the second main theorem, we get
(38)q2kT(r,f)+∑j=1s‍m(r,1f-αj)≤q2kT(r,f)+m(r,f)+∑j=1s‍m(r,1f-αj)≤12kN(r,1Q)+m(r,f)+∑j=1s‍m(r,1f-αj)+S(r,f)≤2T(r,f)+m(r,f)+∑j=1s‍m(r,1f-αj)+S(r,f)≤4T(r,f)+S(r,f).
Thus, we prove that (10) holds.Secondly, we consider the case thatN(r,f)=S(r,f). From (3) and Lemma 8, we similarly get that
(39)12kN(r,1Q)≤N(r,1Δf)+S(r,f)≤T(r,Δf)+S(r,f)=m(r,Δf)+N(r,Δf)+S(r,f)≤m(r,Δff)+m(r,f)+S(r,f)≤m(r,f)+S(r,f).
From this and applying Lemma 9 with (19), as arguing before, we can prove that (11) holds.



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