This paper is to investigate the Schwarzian type difference equation Δ3f/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=maxp,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.
1. Introduction and Results
Throughout this paper, a meromorphic function always means being meromorphic in the whole complex plane, and c 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)=liminfr→∞m(r,1/(f-a))T(r,f)=1-limsupr→∞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)=(ezln2/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) Let f(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 between p 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.
If it admits an admissible solution f(z) such that ρ2(f)<1, then
(7)pm(r,f)≤qm(r,f)+S(r,f).
In particular, if m(r,f)≠S(r,f), then p≤q.
If its coefficients are all constants and it admits a polynomial solution f(z) with degree s, then s≥2 and qs=ps+2k.
Remark 2.
From examples (1) and (2), we conjecture that p=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]).
Let f(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.
Let f(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 Theorem 1, 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]).
Let f(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]).
Let f(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]).
Let f(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 Lemma 7, we can easily get the following conclusion.
Lemma 8.
Let f(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.
Let f 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).
If d=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.
If d=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 if N(r,f)=S(r,f), then (24) always holds. This finishes the proof of Lemma 9.
3. Proof of Theorem 1Case 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 Lemma 5, 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≤q
Case 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).
If s=1, then Δ2f=Δ3f≡0, which yields that P(z,f)≡0. That is a contradiction to our assumption. Thus, s≥2.
If s=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.
If s≥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 degP1≤s-3,degP2≤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 Theorem 3
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 that N(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.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors would like to thank the referees for their valuable suggestions. This work is supported by the NNSFC (nos. 11226091 and 11301091), the Guangdong Natural Science Foundation (no. S2013040014347), and the Foundation for Distinguished Young Talents in Higher Education of Guangdong (no. 2013LYM_0037).
HaymanW. K.1964Oxford, UKClarendon PressOxford Mathematical MonographsMR0164038LaineI.1993Berlin, GermanyWalter de Gruyter10.1515/9783110863147MR1207139YangC.-C.YiH.-X.2003557Dordrecht, The NetherlandsKluwer Academic Publishers GroupMathematics and Its ApplicationsMR2105668AblowitzM. J.HalburdR.HerbstB.On the extension of the Painlevé property to difference equations200013388990510.1088/0951-7715/13/3/321MR1759006ZBL0956.39003BergweilerW.LangleyJ. K.Zeros of differences of meromorphic functions2007142113314710.1017/S0305004106009777MR2296397ZBL1114.30028ChiangY.-M.FengS.-J.On the Nevanlinna characteristic of f(z+η) and difference equations in the complex plane200816110512910.1007/s11139-007-9101-1MR2407244ZBL1152.30024ChiangY.-M.FengS.-J.On the growth of logarithmic differences, difference quotients and logarithmic derivatives of meromorphic functions200936173767379110.1090/S0002-9947-09-04663-7MR2491899ZBL1172.30009HalburdR. G.KorhonenR. J.Difference analogue of the lemma on the logarithmic derivative with applications to difference equations2006314247748710.1016/j.jmaa.2005.04.010MR2185244ZBL1085.30026HalburdR. G.KorhonenR. J.Existence of finite-order meromorphic solutions as a detector of integrability in difference equations2006218219120310.1016/j.physd.2006.05.005MR2238752ZBL1105.39019HalburdR. G.KorhonenR. J.TohgeK.Holomorphic curves with shift-invariant hyperplane preimagessubmitted to Transactions of the American Mathematical Society, http://arxiv.org/abs/0903.3236HeittokangasJ.KorhonenR.LaineI.RieppoJ.TohgeK.Complex difference equations of malmquist type2001112739MR193160010.1007/BF03320974ZBL1013.39001LaineI.YangC.-C.Clunie theorems for difference and q-difference polynomials200776355656610.1112/jlms/jdm073MR2377111IshizakiK.Admissible solutions of the Schwarzian differential equation1991502258278MR109492310.1017/S1446788700032742ZBL0733.34009Mohon'koA. Z.The nevanlinna characteristics of certain meromorphic functions1971148387MR0298006ValironG.Sur la dérivée des fonctions algébroides1931591739MR1504970ZBL0002.27102JankG.VolkmannL.1985Basel, SwitzerlandBirkhäuser Verlag