【word】 ON DIFFERENCE EQUATIONS RELATING TO GAMMA FUNCTION

【word】 ON DIFFERENCE EQUATIONS RELATING TO GAMMA FUNCTION ON DIFFERENCE EQUATIONS RELATING TO GAMMA FUNCTION 1282ACTAMATHEMATICASCIENTIAVo1.31Ser.B whereR(z)isarationalfunction,P(z)isapolynomia1. BankandKaufman…gavethefollowingresultconcerningtheexistenceandthegrowth restrictionofsolutionsofdifferenceequations. TheoremAForanyrationalfunctionR(),equation(1.1)alwayshasameromorphic solutiony(z)suchthatT(r,Y)=O(r)asr__?... Inthispaper,weassumethereaderisfamiliarwiththebasicnotionsofNevanlinna’s valuedistributiontheory(seee.g.[25]).Inaddition,weusethenotations(.厂)todenotethe orderofgrowthofthemeromorphicfunction.厂(z),(. 厂)and(?)todenote,respectively,the exponentsofconvergenceofzerosandpolesof.厂 ().Wealsousethenotation7_(,)todenote theexponentofconvergenceoffixedpointsoffdefinedby 7-(_厂)1imS .. Hp— logN(r,~_z) Recently,anumberofpapers(including[62O])focusedoncomplexdifferenceequations anddifferenceanaloguesofNevanlinna’stheory.AsthedifferenceanaloguesofNevanlinna’s theoryareinvestigated,manyresultsonthecomplexdifferenceequationsweregotrapidly. Manypapers(including[6,11,12,1520])mainlydealwiththegrowthofmeromorphic solutionsofdifferenceequations. Thegoalofourresearchistoconsidertheexistence,thegrowth,poles,zeros,fixedpoints andtheBorelexceptionalvalueofsolutionsof(1.1)and(1.2).wlewillprovethefollowingfour theorems. Theorem1ForanyrationalfunctionR(z)?0,considerequation(1.1),thefollowing statementsho】ds. (1)Equation(1.1)musthaveameromorhpicsolutiony(z)satisfies auitstranscendentalmeromorphicsolutionsoffiniteordersatisf)r()= ()=()=l,and ()1. (2)Everytranscendentalmeromorphicsolutiony(z)offiniteorderofithasatmostone Borelexceptionalvalue. (3)Ifitssolution()hasinfinitelymanypoles,then()1. (4)Ifn(z)?1andy(z)isitstranscendentalmeromorphicsolutionoffiniteorder,then theexponentofconvergenceoffixedpointsofy(z)satisfiesr()=(). The.rem2LetR(z)= polynomials,degP(z)=P,degQ( statementshold. ?0bearationalfunctionwhereP(),Q(z)areirreducible z)=q,P—q=8.Considerequation(1.1),thefollowing polynomialswithdegm(z):m,degn(z)=礼,thenm—n:s+1. (3)Ifs一2andy(z)isdefinedasin(2),then仃一n=s+1orm—n=0. (4)Ifq=0(i.e.Q(z)isanonzeroconstant),then(1.1)musthasapolynomialsolution y(z)=as+lz++-??+alz+ao(as+l?0),coefficientsas+1,…,alofwhichcanbedecided bycoefficientsofR(名),andthecoefficienta0maytakeanyconstant. ByTheorem1,wecanobtainthefollowingcorollary. NO.4z?x?Chenetal:ONDIFFERENCEEQUATIONSRELATINGTOGAMMAFUNCTION1283 Corollary1FortheGausspsifunction():(see[21,Chapter7]),wehave ()=A()=r()=()= ‘IIheorem3LetP(z)beapolynomialwithdegP(z)=P1. Considerthedifference equation(1.2),then (1)Equation(1.2)hasnononzerorationalsolution: (2)EVerYtranscendentalmeromorphicsolutionof(1.2)satisfies()l, andhasatmost oneBorelexceptionalvalue. ThefollowingExamplesl,6showtheexistencesandtheform sofrationalsolutionsof Theorem2?Examples2-4showthatthereassuredlyexistt wocasesm—n::s+1and m一礼 0undercondition(3)ofTheorem2.Example6showspropertiesontranscendent a1 Example7showsthatthec.nditi.nR(z)?1inTheorelTl1(4)cannotbeomitted. Example1Thedifferenceequation (z+1)一()=z2丽+3z+l hasas.luti.n():z2,wheres=2—2:0andm, n:2一l:1:s+1 Example2Thedifferenceequation y(z+1)一y(z) hasasolutiony(z)=1,wheres=一2andm,礼 Example3Thedifferenceequation v(z+1)一y(z) , 1 z(z+1) —— 1=8J,1 3U (Z+3)+2) hasasolution(z)=z-1,wheres=一2andm—n:0 Example4Thedifferenceequation y(z十1)一y(z)=-4(z1)—11 — 2)(z(z+ hasas.luti.n(z)=‘(z- z( 1 . ) 一 (z 3 -- ) 2) , wheres:一3and一他:0 Example5Thedifferenceequation (+1)一(): Z hasaranscendenta1s.luti.n(z):()andhasn.rationals.lutj.n. Equati.ns y(z+1)一y(z)(Z—1)(z一2)andv(z+1)一()=z(z一1)— (—(—z———-—’——2——)—.(—z———-’———3—’)——(—z———-————4——) havenorationalsolutions. xampe6Thedi舶renceequatjon(+1)一(z)=2z+lhasthef0llOwingthree solutions: 1284ACTAMATHEMATICASCIENTIAVb1. 31Ser.B (1) andm一 (2) -a0(a0maybeanyconstant),wheres=1—0= Apolynomialsolutionyl(z)=-4 n=2—0=2=s-4-1: Afiniteordersolutiony2(z)=e-4-zwhichsatisfies(2)=7_(2)=o(y2)=1 (3)AinfiniteordersolutionY3(z)=ee+2. Example7Thedifferenceequationy(z-4-1)一 y(z)=1hasasolutiony(z)=e+Z, whichhasnofixedpoint. 2ProofofTheorem1 WeneedthefollowinglemmasfortheproofofTheorem1. Lemma2.1[7】Letfbeatranscendentalmeromorphicfunctionsatisfying limsupT(r,f)= 0(2.1) Thenf(z-4-1)一f(z)istranscendenta1. Lemma2.2ForanyrationalfunctionR(z)?0,thedifferenceequation(1.1)musthave atranscendentalmeromorphicsolutiony(z)satisfying盯()=1. ProofByTheoremA,weseethatthedifferenceequation(1.1)alwayshasameromorphic solutionYl()suchthatT(r,Y1)=o(r)asr_?.IfYlistranscendental,thenwesety(z)= 1().NowsupposethatYlisarationalfunction.Sinceanyperiodicfunctionwith period1 satisfiesthecorrespondinghomogeneousdifferenceequation y(z-4-1)一y(z)=0 of(1.1),wetakeameromorphicperiodicfunction y(z)yl(z)-4-yo(z).Thenfortwocasesabove,y(z) Yistranscendentaland()1. yo(z)with satisfiesf1. period1anda(yo1=1.Set 1)andT(r,Y)=0(r),namely, If盯()<1,thenYsatisfies(2.1).ByLemma2.1,weseey(z+1)一 y(z)istranscendental, whichcontradictsourassumptionthatR(z)isarationalfunction.Soo(y)=1. Lemma2.3[11,13】LetfbeameromorphicfunctionoffiniteorderandC? C.Then m (r,)=s(r,,), wheres(r,f)denotess(r,f)=o{T(r,,)). Lemma2.4[13,20】 Letw(z)beanonconstantfiniteordermeromorphicsolutionof P(z,W)=0 wheref(z,W)isadifferencepolynomialin(z).IfP(z,a)?0forameromorphicfu nction a(z)satisfyingT(r,a)=s(r,),then W一0 ,:(,)J Lemma2.5(see[22])Letg(z)beanentirefunctionoforder()=<...Then foranygiven,>0,thereisasetE? (1,?)thathasfinitelinearmeasuremEandfinite logarithmicmeasureImE,suchthatforallZsatisfying=r[0,1]UE, exp{一r+)?l9()lexp{r+)(2.2) No.4Z.X.Cheneta1:ONDIFFERENCEEQUATIONSRELATINGTOGAMMAFUNCTION1285 Remark2.6Letgbeameromorphicfunctionoforder(g)=<(3o.Thenby Lemma2.5,weeasilyobtainthatforanygiven,>0,thereisasetE? (1,..)thathas finitelinearmeasuremEandfinitelogarithmicmeasurelmE,suchthatforallzsatisfying 1zI=r【0,1]UE,(2.2)holds. ProofofTheorem1f1)ByLemma2.2,weseethat(1.1)musthaveatranscendental meromorphicsolutiony(z)whichisoforderofgrowth()=1.ByLemma2.3and (1.1),we obtainthat m ()=m(一)=m(r.)砌 ? ()=?(1)+.(1ogr)砌 HenceA(y)=()=1. Ify(z)isatranscendentalsolutionwith()<1,theny(z+1)一 y(z)istranscendentalby Lemma2.1.ThiscontradictsoursuppositionthatR(z)isarationalfunction.He nce()?1. Usingthesamemethodasabove,wehave()=(). (2)Now,weprovethatafiniteordermeromorphicsolutiony(z)hasatmostone Borel exceptionalvalue. Supposethaty(z)hastwoBorelexceptionalvalueaand6(?0,口).Wewillresultina contradiction. First,wesupposethata,barefinitevalue.Weset Then(,)=()and f(z)= A(f):(一a)<() Hence,f(z)mayberewrittenas (z)一a (z)一b’ ()=(y-b)<c f(z)=7r(z)e, (2.3) (2.4) where(?0)isaconstant,nisapositiveintegerand(,)=n,7r()(?0)isameromorp hic functionwith By(2.3),wehave (7r)<(.厂)=n )= Substituting(2.6)into(1.1),weobtainthat R(z)f(z+1)f(z)+(b—a—R()),(+1)+(a—b—R())I厂()+R(z)=0 By(2.4),wehave .,(+1):7r(+1)e5(z十)”=7r(+1)ez+n”+…十=71”1(z)e” (2.5) (2.6) (2.7) (2.8) 1286ACTAMATHEMATICASCIENTIAVo1.31Ser.B where7r1()=7r(+1)enz”一十…+(?0)and 0”(71”1)max{(7r),n Substituting(2.4)and(2.8)into(2.7),weobtainthat R(z)7r(z)7r1()e.+[(6—0一R())7r1z)+(0一b—R(z))7r()]e”+R(z)=0 By(2.5)and(2.9),weseethat max{G(Tr),(不1))=d<(,)=n (2.9) (2.10) ByLemma2.5andRemark2.6,weseethat,foranygiven,(0<2,<n—d),thereisaset E(1,..)withfinitelinearmeasure,suchthatforallZsatisfyingIZI=r[0,1】UE, exp{一r+)lR()7r1(z)7r(z)lexp{rd+) (b一0一R(z))7r1z)+(0一b—R())7r(z)lexp{r+)(2.12) ForsufficientlylargelZl=r,wehavelR(z)1rwherek(>0)issomeconstant. Nowwecanchoosepointszj=e,J=1,2,…,satisfyingrj..,rj[0,1]UE,and exp{25z~)=exp{2151n) Thus,by(2.11)一(2.13)andd+,<礼,weobtainthat IR(zj)7r()7r1(zj)e2~z+[(6一.-n(zj))7r1()+(.一6 exp{一d+}exp{2lI吁)一exp{rjd+)exp{l1哆)一k exp{215lr~(1一o(1))}(1一o(1)). (2.13) R(乃))7r()】e+R()1 (2.14) Thus,(2.14)contradicts(2.10). Second,wesupposethataisafinitevalueandb=?.Wesetf(z)=y(z)一a.Then (,)=()and c.厂=c一n,<c,入()=()<c Hence,f(z)mayberewrittenas f(z)=不()e, wheren,and7r()aredefinedasabove.And/(z+1)mayberewritten嬲 /(z+1)=71”1()e where71”1(Z)isdefinedasabove.Since/(z+1)一/(z)y(z+1)一(z),wehave f(z+1)一f(z)=(不1z)一7r())e=R(z)(2.15) ByR(z)?0,weseethat不1(z)一不 (z)?0.Thusby(2.5),(2.9)andthefactthatR()isa rationalfunction,weseethat(2.15)isacontradiction. No.4Z.X.Chenetal:ONDIFFERENCEEQUATIONSRELATINGTOGAMMAFUNCTION1287 (3)Now,wesupposethaty(z)hasinfinitelymanypoles.Thenwewillprove(1,) 1.SupposethatasetA={+iyjlJ=1,…,s)consistsofallpolesofR().SetM= max{Ixj『+ll+1:lJs}.Thenthereisnopoleofn(z)inregionsD1=z:Re>M}, D2={:Rez<一M),D3=z:Imz>M)andD4=z:Imz<一M}. Sincey(z)hasinfinitelymanypoles,weseethatthereisatleastoneDjsuchthat Yhas infinitelymanypolesinp.IfYhasinfinitelymanypolesinD1,thenthereexists apoint Zl?Dlsuchthaty(z1)=co.Thusforanyn?N,Zl+佗?DIandR(zl+ 佗)?..,andby (1.1),weseey(zl+n)=?.Hence()1. IfYhasinfinitelymanypolesinD3orD4,wecanusethesamemethodtoprove( 1)1. IfYhasinfinitelymanypolesinD2,thenthereexistsapointz2? D2suchthaty(z2)=... Wemayrewrite(1.1)as y(z)一y(z一1)=R(z一1) InD2,n(z一 1)hasnopole,usingthesamemethodasabove,weobtainthat,foranyn? N,y(z2,n)=?.So,()1. (4)SupposeR(z)?1.Weprovethattheexponentofconvergenceoffixedpoints ofY satisfies丁()=().Set f(z)=y(z)一z. Thenf(z)isatranscendentalmeromorphicfunctionand Substitutingy(z) (.厂)=(),丁()=(,),s(r,f)=s(r,y) f(z)+zinto(1.1),weobtainthat P(z,f):,(+1)一,()+l—n(z)=0 SinceP(z,0)=1一n(z)?0,byLemma2.4,weseethat ? (,71)=(卅+,,) Hencer()A(f):(,)=() 3ProofofTheorem2 WeneedthefollowinglemmasfortheproofofTheorem2. Lemma3.1Letm(z)and佗 (z)beirreduciblepolynomialswithdegm(z)=mand deg礼(z)=凡,m+n1.Then (1)Ifm?礼,thendeg[m(z+1)n(z)一m(z)n(z+1)】=m+n一1; (2)Ifm=礼,thendeg[m(z十1)n(z)一m(z)凡(z十1)]m+礼一2. ProofSupposethat m()0mz+am-lZ一+…+ao,n(z)=bnz+bn-1”一+...+b0, wheream,…,a0,bn,…,b0areconstants,ambn?0.Since m(z+1)n(z)一m()礼(z+1)Azm+一J_Bzm+ 1288ACTAMATHEMATICASCIENTIAVo1.31Ser.B whereA=0m6n(仃—n)and B: l a~bn(m(m-1)-n(n-1))+ambn-l(m-n+1)+am-lbn(m-n-1)) weseethat (1)Ifm?n,thenby0mb(m—n)?0,?reha deg【m(+1)礼()一m(z)礼(z+1)】:m十礼一1 (2)Ifm:n,thenbynbn(m一礼)=0,Weh陀 m(+1)n(z)一m(z)札(z+1):(nm6一.m一1bn).+ ;Supposethat’s=--iand…(1.…1)has…ti0na1s0lution , m…哪2’?polyn0degm():deg” n(z)一wherere(z)),n(z)areirredu )(? cible 0)絮一C1ea!ecannotbeaconstantbyn(z)(?0)?So,m+礼’u钉儿 y(z+1)一y(z)m(z+1)— n(z)-— m(z)n(z+1)一 n(z+1)札() ByLemma3.1,wehave m+他一1+q:2几+p, m—n:P—q+1=s+1:0 Thiscontradictsoursuppositionthatm>礼 Ifm:n,thenbyLemma3.1,wehave P(z) Q(z)’ deg[m(+1)礼()一m()礼(+1)】m+n一2? Ifdeg[m(+1)n(z)一m(z)礼(z+1)] 1.e., :盯+钆一2,thenby(3.1),wehave m+n一2+q=P+2n, 0:m一礼=P—q+2=一1+2=1 (3.1) Thisisac.ntradict_l0n.Ifdeg[m(+1)n(z)一m(z)礼(+1)】:m+佗一 (3),thenwe obtainthat Thisisalsoacontradiction 0:m—n:s+k:一1(3) Ifm<礼,thenbyLemma3.1and(3.1),weobtainthat 0>m一亿:P—q+1=s+1=一1+1=0? Thisisalsoacontradiction.Hence(1.1)has (2)Supposethats0and(z) aredefinedasintheproofofLemma norationalsolution 3.1 isarationalsolutionof(1.1)wherere(z)and No.4Z.X.Chenetal:ONDIFFERENCEEQUATIONSRELATINGTOGA MMAFUNCTION1289 First,supposethats>0.Then have()一,(+1)__+abm.So,_?(asz-_?o0).Ifm=佗,then,asZ-_?0(3,we }? aSZ— y(z+1)一u(z)一0 Thiscontradicts(1.1).Ifm< 扎,then,usingthesamemethodasabove,wehavey(z+1)一 ()__+0.Thisisalsoacontradiction.Hencewehavem>n.ByLemma3.1an d(3.1),we obtainthatm—n=8+1. Second,supposethatS:0.Usingthesamemethodasabove,wegetm一礼:s+1 m,n=s+1一2+1=一1. Thisisacontradiction.Hencewehavemn. Ifm<n,thenbyLemma3.1,wehavem一佗=s+1. Hencey(z)satisfiesm,n=s+1 orm=n. (4)Supposethatq=0,ieR(z)兰P(z)isapolynomialofdegP=S.Set Supposethat P(z)=dsz+ds,iz一+…+dl0(ds?0) y(z)=nm+am--1Z一+…+a0(n?0) isasolutionof(1.1).Then (+1),()=ammZ一+【0m—一1am--1z一 +…+[c锄.m+11.一1+…+Clm_ (j_1)am-(J一1)]zm-j +…+【a+am--1+…+01], (3.2) (3.3) where(h=2,…,m;d=1,…,h)areusualnotationsforthebin0mia1coefficients. sut)- stituting(3.2)and(3.3)into(1.1),weobtainthat m=s+1. am~as+l= 1 d+卜l, l 0m一10s+1—1 S上1——1 am—U一1)as+l一(J一1) (ds+1—2一C2+10+1) 8+1,—1)(d~+l-J—+n叶 一 j + -- 1 I — l.s+1—1一…一c1一 .一2)as+l-.一2)), al口8+1—8 8+1一s (d0一as+l—a8一???一02). (3.4) Thus,coefficients0s+l,…,alcanbedecidedbycoefficientsofR(z),andthecoe fficientn0may takeanyconstant. Theorem2isthusproved. 1290ACTAMATHEMATICASCIENTIAVo1.31Ser.B 4ProofofCorollary1 By[21,Chapter7】,weseethat?(z)satisfiesthedifferenceequation q2(z+1)一()= Sincewehaveknownthat(专)=()=1,byTheorem1(1),weseethat()=()=1. ByTheorem1(4)andR(z)=1(?1),weseethat皿 (z)hasinfinitelymanyfixedpoints andtheexponentofconvergenceofitsfixedpointssatisfies7_(皿)=()=1. Thus,Corollary1isproved. 5ProofofTheorem3 WleneedthefollowingremarkandlemmasfortheproofofTheorem3. Remark5.1FollowingHayman 【23,p.75—76】,wedefineanc-settobeacountableunion ofopendiscsnotcontainingtheorigin,andsubtendinganglesattheoriginwhosesumisfinite. IfEiSanE—set,thenthesetofr>1forwhichthecircles(0,r1meetsEhasfinitelogarithmic measure.andforalmostallreal0theintersectionofEwiththerayargZ=0isbounded. Lemma5.217lLetgbeafunctiontranscendentalandmeromorphicinthecomplex planeoforderlessthani.Leth>0.Thenthereexistsan,.setEsuchthat C 一..q(+)一asz一一\E) uniformlyinCforlCIh.Furthermore,EmaybechosenSOthatforlargenotinEthe functionghasnozerosorpolesinl(一zlh. Lemma5.3[11]LetCl,C2betwocomplexnumberssuchthatCl?C2andletf(z)be afiniteordermeromorphicfunction.Let盯betheorderof.厂(z),thenforeach ,>0,wehave O(r一+) wherem(z)andn(z)areirreduciblepolynomialswithdegm(z)=m,degn(z) 1.By(1.2),wehave m(z+1)n(z)一P(z)m(z)n(z+1)=0. nandm一_n> (5.1) Buttheleftsideof(5.1)isapolynomialofdegreep+m+n.So,(5.1)isacontradic tion.Hence y(z)hasnononzerorationalsolution. Nowsupposethaty(z)istranscendentaland()<1.ThenbyLemma5.2,there exists an,一setEsuchthat y(z+1)=()(1+0(1))asz一?inC\E By(1.2)and(5.2),wehavethat (z)(1+o(1)一P())=0asz—..inC\E (5.2) (5.3) No.4z.X.Chenetal:ONDIFFERENCEEQUATIONSRELATINGTOGAMMAFUNCTION1291 SinceforsufficientlylargeJZI=randinC\E 1+o(1)一P()I>1, weobtain()三0by(5.3).Thisisacontradiction.So,()1. Now,weprovethat()hasatmostoneBorelexceptionalvalue.First,supposeth at a(?..)and..aretwoBorelexceptionalvaluesof(z).Setf(z)=y(z)一 a.Thenf(z)has Borelexceptionalvalues0and...andcanberewrittenas ,()=H(z)e() whereH(z)isameromorphicfunction,h(z)isanonconstantpolynomial o(H)<degh(z)=h.By(1.2)and(5.4),weobtainthat H(z+11 H(z)P(ze(外1)一h(z):1) (5.4) ofdegh(z)=h,and (5.5) Ifh=1,seth(z):+(OL?0,areconstants),thenh(z+1)一h(z)=.Thus,by Lemma5.2and(日)<h=1,thereexistsanE—setEsuchthat : 1+.(1)H(z1一asz—}?inCkE Thus,byh(z+1)一h(z)=,degP(z)1and(5.6),weseethat H(z+11 g(z)P(z Thiscontradicts(5.5). Ifh2,thendeg[h(z+ havethat e(舛)一h()0as_..in\E 1)一(z)] (5.6) h一11and(日)=<h.By(1.2)and(5.4),we : P(z)eh?)H( z) ByLemma5.3,weseethat,foranygivenE(0<2s<h一), whereC?0isaconstant. Sinceh一>2s,by(5.8)and Second,supposethata(??) Set = O(r—H) P()eh’’一h外1’1=Crh一 (5.7) (5.8) (5MATHEMATICASCIENTIAVl0】.3lSer.B Substituting(5.11)into(1.2),weobtainthat (1一P(z))bf(z+1),(z)+(0P()一b)f(z+1)+(6P(z)一口),()+n—aP(z):0 Usingthe