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u

n+2

u

n=

ψ(u

n+1

) IN

R+

, RELATED TO A FAMILY OF ELLIPTIC QUARTICS IN THE PLANE

G. BASTIEN AND M. ROGALSKI

Received 20 October 2004 and in revised form 27 January 2005

We continue the study of algebraic difference equations of the typeun+2un=ψ(un+1), which started in a previous paper. Here we study the case where the algebraic curves related to the equations are quarticsQ(K) of the plane. We prove, as in “on some alge- braic difference equationsun+2un=ψ(un+1) inR+, related to families of conics or cubics:

generalization of the Lyness’ sequences” (2004), that the solutionsMn=(un+1,un) are persistent and bounded, move on the positive componentQ0(K) of the quarticQ(K) which passes throughM0, and diverge ifM0is not the equilibrium, which is locally sta- ble. In fact, we study the dynamical systemF(x,y)=((a+ bx+cx2)/ y(c+dx+x2),x), (a,b,c,d)R+4,a+b >0,b+c+d >0, inR+2, and show that its restriction toQ0(K) is conjugated to a rotation on the circle. We give the possible periods of solutions, and study their global behavior, such as the density of initial periodic points, the density of trajecto- ries in some curves, and a form of sensitivity to initial conditions. We prove a dichotomy between a form of pointwise chaotic behavior and the existence of a common minimal period to all nonconstant orbits ofF.

1. Introduction

In [4], we study the difference equations

un+2un=a+bun+1+u2n+1, un+2un=a+bun+1+cu2n+1

c+un+1 (1.1)

which generalize the Lyness’ difference equationsun+2un=a+un+1(see [2,7,8,9]). The first of these equations is related to a family of conics, and the second to a family of cubics (whose Lyness’ cubics are particular cases). The results of [4] in the two cases are analogous to the results obtained in [3] about the global behavior of the solutions of Lyness’ difference equation.

In the present paper, we will study the difference equation un+2un=a+bun+1+cu2n+1

c+dun+1+u2n+1 . (1.2)

Copyright©2005 Hindawi Publishing Corporation Advances in Dierence Equations 2005:3 (2005) 227–261 DOI:10.1155/ADE.2005.227

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The dynamical system inR+2which represents this difference equation is F(x,y)=

a+bx+cx2 yc+dx+x2,x

. (1.3)

It is well defined as a homeomorphism ofR+2whena,b,c,d0 anda+b+c >0, as we always assume. We have

Mn+1=

un+2,un+1

=FMn

=Fun+1,un

. (1.4)

There is an invariant function

G(x,y)=xy+d(x+y) +c x

y+ y x

+b 1

x+1 y

+ a

xy, (1.5)

which satisfiesGF=G, and thusG(un+1,un) is constant on every solution of (1.2).

IfK=G(u1,u0), the quarticQ(K) with equationG(x,y)=K, or

x2y2+dxy(x+y) +cx2+y2+b(x+y) +aKxy=0, (1.6) passes throughM0.

The quarticsQ(K) are invariant on the action ofF, and thus the pointsMnmove on the quartic passing throughM0, more precisely on its positive componentQ0(K).

The mapFhas a geometrical interpretation. IfMR+2, letMbe the second point of the quarticQ(K) which passes throughMwhose first coordinate is the same as those of M(there is only one such pointMbecause the point at the vertical infinity is a double point of the quartic). The imageF(M) is the symmetric point ofMwith respect to the diagonalx=y.

For all this results, we refer to [4].

InSection 2, we give a general topological result useful for our study, which extends a result of [4], and we define a general property of weak chaotic behavior, whose proof for (1.2) is the goal of this paper.

InSection 3, we use this result to show that the solutions of difference equation (1.2) are, ifa+b >0 andb+c+d >0, bounded and persistent inR+2, and diverge if (u1,u0)= (,), the fixed point ofF, and prove that this point is locally stable.

InSection 4, we show that the case whereun+2unis a homographic function ofun+1, studied in [5], comes down to our general model (1.2). This gives again, in a simpler way, results of [5], and improvements of them.

InSection 5, we study the casea=0, where the quartic passes through the point (0, 0).

This case is easy, because a simple birational map transforms every quarticQ(K) into a cubic curve studied in [4]. So we can apply the results of [4] without more work.

InSection 6, we prove general results in the casea >0, which lead to the fact that the restriction of the mapFto each curveQ0(K) is conjugated to a rotation onto the circle (seeTheorem 6.11). We study also in Sections6and7whether the chaotic behavior de- fined inSection 2holds in the general case of (1.2), with a general property of dichotomy (seeTheorem 6.18), and what happens in some particular cases (Section 7) and in the general one (Section 8).

InSection 9, we determine the possible periods of solutions of (1.2).

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2. A topological tool for difference equations with an invariant

In this section, we give an abstract and more or less classical general result which will be useful for the study of difference equations. This assertion extends [4, Proposition 1].

Proposition2.1. LetXbe a topological Hausdorffspace. LetF:XXandG:XRbe two maps. Suppose first that the following conditions hold:

(a)Fis continuous onX;

(b)Gis continuous and has a strict minimumKmat a pointL;

(c)xX,GF(x)=G(x)(the invariance property);

(d)Fhas at most one fixed point.

IfKKm, the level sets (if nonempty) ofG are defined byK= {xX|G(x)=K}. Then the following three results hold:

(1)every pointxXlies in exactly one setK; (2)the pointLis the (unique) fixed point ofF;

(3)ifM0X let Mn+1=F(Mn)be the points of the orbit ofM0 underF; thenMnG(M0), and ifM0=L, then the sequence(Mn)does not converge.

Now suppose additional hypotheses:

(e)Xis connected and locally compact;

(f)K:=limx→∞G(x)+exists, andG < K; then

(4)eachK is compact and nonempty for KmK < K (withKm= {L}), and the equilibrium pointLis locally stable.

Suppose at last the additional hypothesis:

(g)Ghas only one local minimum (its global one atL); then

(5)for K > Km the setK is the boundary of the open set UK = {G < K}which is a connected relatively compact set.

Proof. Assertions (1) and (2) are obvious. IfMn+1=F(Mn), thenMnG(M0). Suppose thatMnconverges to a pointN. ThenG(N)=G(M0) andF(N)=N, so by (d) and (1) N=L. ButG(Mn)=G(N)=G(L)=Km, and by (b)Mn=N for alln. Thus, ifM0=L, thenMndoes not converge.

If (e) and (f) hold, it is easy to see thatᏯKis nonempty and compact for everyKKm; in particular, sequences (Mn) are bounded (i.e., relatively compact).

We prove now that the setsUK= {G < K}form a basis of neighborhoods ofL. LetV be an open neighborhood ofL. The sets{GK}, forK > Km, are compact, and their intersection is{L}; so there is aK > Kmsuch that{GK} ⊂V, and thusUKV.

We can now prove easily thatLis locally stable: ifVis a neighborhood ofL, there exists K > Kmsuch thatUKV. IfM0UK, then, for everyn,MnUK by (c), andMnV.

We prove now assertion (5), if (g) also holds. We haveUK⊂ {GK}, andUK\UK {GK} \ {G < K} =K. Thus,∂UKK. Now, ifᏯK∂UK, there existsxK,x /

∂UK, thus there exists a neighborhoodV ofxsuch thatVUK = ∅. Thus,GK on V, andG(x)=K; thus x is a local minimum ofG, andx=LbecauseK > Km: this is impossible, and∂UK=K.

Finally, we prove thatUK is connected. IfUK is the union of two disjoint nonempty open setsAandB(which are relatively compact), putα=infAGandβ=infBG; we have α,β < K. Ifα=G(u) withuAandβ=G(v) withvB, thenuandvare two distinct

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local minima ofG, which is impossible. Thus we can suppose thatα=G(u) withu A\A. ButAX\B(becauseUK is open), thusu /B, anduUK\UK=∂UK=K. So

we haveG(u)=K > α, which is a contradiction.

We will useProposition 2.1whenXis an open subset ofR2; in this context,FandG are given by (2) and (3). In the general case, we can ask the question whether a form of chaotic behavior may be described for the mapF(we will study in this paper if it is the case withFandGgiven by (2) and (3)). Precisely, one may ask the question whetherF has an “invariant pointwise chaotic behavior,” denoted IPCB.

Property of IPCB. We suppose that X, F, and G satisfy properties (a), (b),. . ., (g) of Proposition 2.1, and thatXis a metric space, with distanced. We say that the dynam- ical system (X,F) with invariantGhas IPCB if we have the following three properties.

(a) There exists a partition ofX\{L}into two dense subsetsAandBwhich both are union of “curves”ᏯK, and then invariant underF:Ais the set of initial periodic pointsM0,Bis the set of initial pointsM0whose orbit is dense in the curveᏯK

which passes throughM0(that isᏯG(M0)).

(b) Every pointM0X\{L}has sensitivity to initial conditions, that is, there exists δ(M0)>0 (this dependance onM0explains the term “pointwise”) such that every neighborhood ofM0contains a pointM0whose iteratesMn satisfyd(Mn,Mn) δ(M0) for infinitely many integersn.

(c) There exists an integerNsuch that every integernNis the minimal period of some periodic orbit ofF.

IPCB is the essential result of [3] about the behavior of Lyness’ difference equation un+2un=k+un+1if 0< k=1 (ifk=1, 5 is a common minimal period to all nonconstant solutions).

In [4], we prove also that IPCB holds for the solutions of difference equations inR+2 un+2un=a+bun+1+u2n+1, un+2un=a+bun+1+cu2n+1

c+un+1 . (2.1)

An important tool to study the dynamical system linked to (1.2) may be an eventual property in the abstract case ofProposition 2.1: for everyK]Km,K[, is the dynamical systemF|K conjugated to a rotation on the circle with angle 2πθ(K)]0,π[? This even- tual property supposes that each setᏯK is homeomorphic to a circle. Then the study of the properties of functionθwould be essential: continuity (analyticity ifXis an open set ofR2), limits whenKKmandKK.

3. First general results of divergence and stability We begin by identifying the fixed point.

Lemma3.1. Ifa=b=0, then sequence (1.2) tends to0. Ifa+b >0, then the fixed point of the dynamical system (1.3) is the unique positive rootof the equation

Y4+dY3bYa=0, (3.1)

and it is the unique possible limit for sequence (1.2).

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Figure 3.1

Proof. It is obvious that a fixed point ofFhas the form (Y,Y) whereYsatisfies (3.1), and that (3.1) has a unique positive rootsuch that (,) is invariant byF.

For the limit of the sequence (un) of (1.2), we must be more careful ifa=0, because in this caseY=0 is solution of (3.1). But thenQ(K) passes through (0, 0) and has as tangent at this point the linex+y=0 ifb >0, and so the pointMn=(un+1,un), which lies onQ(K)R+2, cannot tend to (0, 0).

Ifa=b=0, the fixed point is solution of Y4+dY3=0, which has no solution in R+. In this case, we have un+2/un+1=(un+1/un)(c/(c+dun+1+u2n+1)), with c >0 and d0. Thus ρn=un+1/unis decreasing and tends to a limitλ. Ifλ <1, thenun0. If λwould satisfyλ1, thenunwould be increasing, and would tend to infinity. But then c/(c+dun+1+u2n+1)1/2 for bign, and thus we would haveλ=0, which is a contradic-

tion.

With the objective of usingProposition 2.1, it is necessary to study the functionG. The first question is to know ifG(x,y)+if (x,y) tends to the infinite point of the locally compact spaceR+2. It appears that this condition fails in the general case. Indeed, we look for a condition for the setsAK:= {GK} ∩R+2to be compact. The hypothesis is

xy+d(x+y) +c x

y+ y x

+b 1

x+1 y

+ a

xy K. (3.2)

Thus we havexy+a/xyK,d(x+y)K,c(x/ y+y/x)K,b/xK, andb/ yK. Ifb >0, thenxb/Kandyb/K, and thus, with the conditionxyK, the setAK is compact. Ifb=0, we will supposea >0 (the casea=b=0 is trivial byLemma 3.1), and the conditionxy+a/xyKimplies that 0< r1xyr2: the point (x,y) is between two hyperbolas. But then ifcordis positive, we havex/ y+y/xK/cand thus 0< s1y/x s2, orx+yK/b. In the two cases,AK is compact; seeFigure 3.1.

So the good condition in (1.2), which we suppose in all the sequel, is

a+b >0, b+c+d >0. (3.3) It is to be noticed that ifb=c=d=0, the function Gdoes not tend to +at the point at infinity ofR+2, and then the solutions of the difference equation (1.2) may be unbounded or not persistent. In fact, it is always the case.

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Indeed, consider the difference equationun+2un=a/u2n+1, witha >0. Its solutions are the sequencesun=a1/4exp[(1)n(A+Bn)], withA=ln(u0a1/4) andB=−ln(u0u1a1/2), which are neither bounded nor persistent ifu0u1=

a.

Then, we must identify the minimum ofG. The equations of critical points arex2y2+ dx2y+c(x2y2)bya=0 andx2y2+d y2x+c(y2x2)bxa=0. The difference of these two equations gives (xy)(dxy+ 2c(x+y) +b)=0. But if (x,y)R+2, the only solution isx=y, so the previous equations givex4+dx3bxa=0, and thus we have x=y=:Ghas a unique critical point at the equilibriumL=(,), the minimum ofG is achieved only at this point, and the value of the minimum is

Km=d+ 2c+3b +2a

2. (3.4)

IfK > Km, thenQ0(K)=Q(K)R+2= {(x,y)R+2|G(x,y)=K}is a nonempty com- pact component ofQ(K), and through every pointMR+2passes a unique curveQ0(K).

We can thus applyProposition 2.1and obtain the following theorem.

Theorem3.2. Ifa0,b0,c0,d0,a+b >0, andb+c+d >0, every solution of the difference equation (1.2)

un+2un=a+bun+1+cu2n+1

c+dun+1+u2n+1 (3.5)

is bounded and persistent inR+2. If (u1,u0)=(,), then(un)diverges, the point Mn= (un+1,un)moves on the curveQ0(K)which passes throughM0, andK > Km. Moreover the equilibriumLis locally stable.

4. The homographic case

In [5], the authors study the difference equation un+2= αun+1+β

unγun+1+δ, withα,β,γ,δ0,α+β >0,γ+δ >0. (4.1) Ifγ=α=0, we find the classical sequenceun+2=(β/δ)/unwhich is always 4-periodic.

Ifγ=0,α=0, the sequencevn=(δ/α)unsatisfiesvn+2vn=vn+1+βδ/α2: it is a Lyness sequence, and its behavior is known and given in [3].

So, we supposeγ >0, and thus we can supposeγ=1.

Under this hypothesis, if we suppose that the two quadratic polynomials of (1.2) have a common rootx= −p <0, then (4.1) is a particular case of (1.2). To see this fact, we examine some cases.

(i) Ifδ=0, easy calculation shows that with a=αβ

δ , b=α2

δ +β, c=α, d=α

δ+δ, (4.2)

(1.2) is exactly (4.1) withγ=1.

(ii) If δ=α=0, (4.1) becomesun+2un=β/un+1, which is a classical 3-periodic se- quence.

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(iii) Ifδ=0,α >0,β=0,un+2=α/unis another case of the previous 4-periodic se- quence.

(iv) Ifδ=0,α >0,β >0, we putun=β/vn, and obtainvn+2vn=β2/(vn+1+α), which has the form (αvn+1+β)/(vn+1+δ). Thus, (1.2) for (vn) with the valuesa=c= 0,b=β2,d=α, is exactly (4.1) forun=β/vn.

In any cases, (4.1) comes down to (1.2) or to a known sequence (Lyness’ one) or to elementary sequences (3- or 4-periodic). Thus, with the aid of elementary results on Ly- ness’ equation (see [3]), we deduce again fromSection 2the result of [5] about (4.1), but almost without calculation, and we can improve it.

Proposition4.1. The solutions of (4.1) are bounded and persistent, and diverge if(u1,u0) is different than the fixed point. Moreover the equilibrium point is locally stable.

Of course, other properties of the solutions of (4.1) will follow from the general prop- erty of solutions of (1.2) that we will prove in the following parts, see corollaries of Theorems5.1and7.1, where examples of (4.1) which have IPCB are given.

5. The casea=0

In this part, we solve the case whena=0, which is simple, because an easy birational map reduces the associated quartic curves to cubic ones which give a previous case already solved (see [4]). So we obtain the following general result.

Theorem5.1. Let the difference equation inR+be un+2un= bun+1+cu2n+1

c+dun+1+u2n+1 withb >0,c0,d0,c+d >0, (5.1) whose solutions diverge if(u1,u0)=(,). LetL=(,)be the equilibrium, withpositive solution of the equationY3+dY2b=0. LetF(x,y)=((bx+cx2)/ y(c+dx+x2),x)be the homeomorphism ofR+2 associated to (5.1):Mn:=(un+1,un)=Fn(M0). LetQb,c,d(K) be the quartic curve with equation

x2y2+dxy(x+y) +cx2+y2+b(x+y)Kxy=0 (5.2) which passes throughM0=(u1,u0), andQ0b,c,d(K)its positive component, globally invariant under the action ofF.

(a)There exists a well-defined numberθb,c,d(K)]0, 1/2[such that the restriction ofF toQ0b,c,d(K)is conjugated to a rotation on the circle, of angle2πθb,c,d(K)]0,π[.

(b)For everyb,c,dsatisfying the conditions of (5.1) andb2=c3orbd=2c2, the differ- ence equation (5.1) has IPCB.

(c)Every integern4is the minimal period of some solution of (5.1) for someb,c,d, and some initial pointM0.

One hasb2=c3andbd=2c2if and only if every solution of (5.1) is5-periodic.

Proof. Ifa=0, then the quartic curve (1.6) reduces to (5.2), and then it passes through (0, 0).

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(1) Casec=0 andd >0.

We define the birational map X=

b d 1

x, Y= b

d 1

y, (5.3)

that is the transformation on the solutions (un) of difference equation (5.1) by the for- mula

vn= b

d 1

un. (5.4)

Under map (5.3) the quartic (5.2) becomes the cubic of paper [4]:

Γα(K) withα=

b

d3,K=K

bd, (5.5)

associated to the difference equation

vn+2vn+1vn=α+vn+1 (5.6)

whose solutions are studied in [4]. Then results ofTheorem 5.1are nothing else but [4, Proposition 8 and Theorem 4].

(2) Casec >0.

We define now the birational map X=b

c 1

x, Y=b c 1

y, (5.7)

that is the transformation on the solutions (un) of difference equation (5.1) by the for- mula

vn=b c

1

un. (5.8)

Under map (5.7) the quartic (5.3) becomes the cubic of (see [4]) Γα,β(K) withα=b2

c3,β=bd

c2,K=K

c, (5.9)

associated to the difference equation

vn+2vn=α+βvn+1+v2n+1

vn+1+ 1 (5.10)

whose solutions are studied in [4]. Then the results inTheorem 5.1are nothing but [4, Proposition 11 and Theorem 6]. The case of 5-periodicity in [4] corresponds to the values α=1 andβ=2 (see [4, Lemma 8]), which gives the end of assertion (d) of the theorem.

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Point (3) of [4, Theorem 6] has to be modified, because here we have onlyα >0 and β0 (arbitrary), but in [4] we haveα0 andβ >2α. So, in [4, Lemma 10], we must replace the domainDbyD=R+2and the functionf()=(1/π) cos1(1/2)(11/+ 1) by the functionf()=(1/π) cos1(1/2)1 + 3/(+ 1). Then it is easy to show that we have only

o

ψ(D) =]0, 1/3[. Thus every integern4 is actually a period.

Corollary 5.2. The solutions of (4.1) studied in [5], where αβγ >0, δ =0, satisfy Theorem 5.1.

Remark 5.3. (1) Ifc >0, then solutions (un) of (5.1) are rational if and only if thevn’s are rational, whenb,c,dare rational. Then, in this case, a rational periodic solution of (5.1) may have only periods which belong to the set{3, 4, 5, 6, 7, 8, 9, 10, 12}(see [4]).

But ifc=0, the map (5.3) does not preserve rationality of real numbers, except if b/d=q2withqQ+. In this case, and withbrational, the periodic rational solutions of (5.1) may have only periods 7 or 10 (see [4, corollary of Proposition 7]).

(2) The 5-periodic caseb2=c3andbd=2c2corresponds to initial Lyness’ sequence:

vn=

c/unsatisfiesvn+2vn=1 +vn+1.

We give now two easy cases witha=0, which are not covered byTheorem 5.1.

First, the casea=b=0 is given byLemma 3.1: the sequence tends to 0.

Second, we have the following classical result.

Lemma5.4 (case a=c=d=0,b >0). The positive solutions of the difference equation un+2un+1un=bare3-periodic.

6. General results in the casea >0

It is easy to see that ifa >0 we can suppose thata=1 (putun=vn4a). We make this hypothesis from now on.

6.1. Points on the diagonal and the birational transformation of the quartic. We know that the quartic curve has two double points at infinity in vertical and horizontal di- rection, which are ordinary ifd24c=0, the asymptotes being then the linesx=r1, x=r2, y=r1, and y=r2, where the ri are the roots (real or complex) of the equa- tions2+ds+c=0. Moreover, if K > Km, the quarticsQ(K) have no singular point in R+2. Indeed, if the equation ofQ(K) is p(x,y)Kxy=0, singular points are given by pxK y=0, pyKx=0, andpKxy=0. These relations givex px=pandy py=p, and these last relations are the equations whose solutions are the critical points of the functionG(x,y)=p(x,y)/xy. But we have seen thatGhas no critical point inR+2except forL=(,), and so the only finite singular point ofQ(K) inR+2would beL, but this point is not onQ(K) ifK > Km.

So we can hope that the quartic curveQ(K) is an elliptic one, and thus that it can be transformed in a regular cubic curve by a birational transformation. To make such a transformation, some point ofQ(K) should disappear, and to preserve the symmetry of the curve with respect to the diagonalδ:x=y, we choose this point on this diagonal.

So the fundamental technical result will be the behavior of the points ofQ(K) on the diagonal.

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f1 f0 λ0λ f2 f3

Km t K

G(t, t)

Figure 6.1

Lemma6.1. ForK > Km, the coordinates of the intersection points ofQ(K)with the diagonal δare solutions of the equation

t4+ 2dt3+ (2cK)t2+ 2bt+ 1=0. (6.1) These coordinates are real numbers f1,λ,f2, f3which satisfy

f1< <λ <0< λ < f2< < f3 ifd+b >0, f1= −1

λ<1<λ <0< λ=f2<1= < f3=1

λ ifd=b=0. (6.2) Moreover, numbers fiandλare continuous functions ofKon]Km, +[, whose limits when K+andKKmare

Klim+λ= lim

K+f2=0, lim

K+f1= − lim

K+f3= −∞, (6.3)

KlimKm

f2= lim

KKm

f3=, λ0:= lim

KKm

λ=(d+)

(d+)2 1 2,

KlimKm

f1:= f0= −(d+)

(d+)2 1 2.

(6.4)

Proof. Formula (6.1) is obvious. LethK(t)=t4+ 2dt3+ (2cK)t2+ 2bt+ 1=0. By (3.1) and (3.4) we have

hK()=d3+ 3b+ 2 + (2cK)2< d3+ 3b+ 22

d+3b + 2

2

=0, (6.5) hK(0)=1, and sohK has two roots f2and f3which satisfy 0< f2< < f3. We have also hK()=hK()4d34b <0, and thus we have two other roots f1 andλ which satisfyf1< <λ <0. At last,hK(λ)=hK(λ) + 4dλ3+ 4bλ=4dλ3+ 4bλ0. Ifb+d >

0, thus we have 0< λ < f2. Ifb=d=0, the roots of hK are f1,λ,λ, andf1, whose product is 1. This gives (6.2).

Then we remark that the equationh(t)=0 is equivalent to the relationG(t,t)=K. But the graph of the functiontG(t,t) is easy to determine, seeFigure 6.1. It is immediate from this graph that the roots are continuous functions ofK. Their limits whenK+

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are obvious and given by (6.3). If KKm, then f2 and f3 tend to, and f1 and λ have limits f0andλ0which are the two other roots of equationhKm(t)=0, which has already the double root. Thus these two other roots are easy to obtain, they are given by

(6.4).

Now we write the equation ofQ(K) in the form:

X2Y2+dXY(X+Y) +cX2+Y2+b(X+Y) + 1KXY=0. (6.6) We make the birational transformationφKdefined in affine coordinates, forXY=λ2, by

x= X+λ

XYλ2, y= Y+λ

XYλ2, or X=1 +λx

y , Y=1 +λy

x , (6.7)

or, in homogeneous coordinates, by

X=xt+λx2, Y=yt+λy2, T=xy, (6.8) or

x=T(X+λT), y=T(Y+λT), t=XYλ2T2. (6.9) OnQ(K) this transformationφK is not defined only at the point (λ,λ) ifd+b >0, and at the points (λ,λ) and (λ,λ) ifb=d=0.

Now we determine the image ofQ(K) underφK. We substitute the second formulas of (6.7) in (6.6). PuttingD:=1 +λ(x+y), easy calculation gives for the left hand of the equation in variablesx,ythe product ofDby the following factor

2+bxy(x+y) +λcx3+y3+ (c+dλ)(x+y)2+ (d+λ)(x+y)

+22dλ2cKxy+ 1 (6.10)

(the coefficient ofx2y2isλ42dλ3+ (2cK22bλ+ 1 which is 0 because the point (λ,λ)Q(K)).

So we obtain the straight line∆λwith equation 1 +λ(x+y)=0 and the cubic curve Γ(K) with equation

(x+y)λc(x+y)2+α(K)xy+ (c+dλ)(x+y)2+ (d+λ)(x+y)β(K)xy+ 1=0, (6.11) where

α(K)=24cλ+b, β(K)=K+ 2c+ 2dλ2. (6.12) With second formulas of (6.7), one sees that if (x,y)λ\ {(1/λ, 0), (0,1/λ)}, then (X,Y)=(λ,λ) which has no image by φK. Identification of the images of Q(K)\ {(λ,λ)}andQ0(K) is given in the following results.

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Lemma6.2. (1)Ifb+d >0, thenXY > λ2onQ0(K), and ifb=d=0, thenXYλ2on Q0(K), with equality only at the point(λ,λ).

(2)Ifb+d >0, then the positive componentΓ0(K)of the cubicΓ(K)is compact, andφK

is a homeomorphism ofQ0(K)ontoΓ0(K). Ifb=d=0, thenΓ0(K)is unbounded and has a point at infinity in directionx=y, which is the image byφK of the point(λ,λ)ofQ0(K), andφKis a homeomorphism ofQ0(K)\ {(λ,λ)}onΓ0(K).

Proof. (1) We work here inR+2, and begin in the case whenb+d >0. Suppose that there is a point (X,Y) in the set{GK}, lying on the hyperbolaXY =λ2. Then we have X+Y2λand

0X2Y2+dXY(X+Y) +c(X+Y)2(2c+K)XY+b(X+Y) + 1

λ4+ 2dλ3+ (2cK2+ 2bλ+ 1

=hK(λ) + 4dλ3+ 4bλ

=4dλ3+ 4bλ >0,

(6.13)

and this is impossible. So the set {GK}, which is connected byProposition 2.1, is contained in one of the two connected components ofR+2\ {XY =λ2}. But f2> λby Lemma 6.1, and thusQ0(K)⊂ {XY > λ2}.

Now ifb=d=0, we choose (X,Y)=(λ,λ). ThenX+Y >2λifXY=λ2, and the same calculation gives, on ({GK} \ {(λ,λ)})∩ {XY =λ2}, the impossible inequality 0>0.

But this calculation proves also that{G < K} ∩ {XY=λ2} = ∅. So{GK}is contained in{XYλ2}or in{XYλ2}, and we conclude, with the aid of the point (f3,f3), that {GK} \ {(λ,λ)} ⊂ {XY > λ2}.

(2) Ifb+d >0, we have XY > λ2 on Q0(K), and formulas (6.7) show thatφK is a homeomorphism ofQ0(K) onto the positive componentΓ0(K) ofΓ(K) (note thatΓ(K) does not intersect the axis{y=0} ∩ {x0}nor{x=0} ∩ {y0}, by formula (6.11)).

So the setΓ0(K) is compact inR+2.

Ifb=d=0, (6.7) shows thatφKis a homeomorphism ofQ0(K)\ {(λ,λ)}ontoΓ0(K), and soΓ0(K) cannot be bounded. Equation (6.11) becomes

λc(xy)2(x+y) +c(x+y)2+λ(x+y)βxy+ 1=0, (6.14) and this proves thatΓ(K) has the point at infinity in the direction of the diagonal. More- over, (6.7) shows that when (X,Y)(λ,λ) onQ0(K), then (x,y) tends to infinity in

directionx=yonΓ0(K).

6.2. The algebraic-geometric interpretation of the transformed sequence, and the el- liptic nature of the quartic and cubic curves. We begin with the transformation of the sequenceMn=(un+1,un) byφK: what is the behavior of the pointsmn=φK(Mn)?

Lemma6.3. Let mnbe the image inΓ0(K)of the sequenceMn=(un+1,un)onQ0(K)by the birational transformation φK. Then the symmetric point ofmn+1 with respect to the diagonalx=ylies onΓ0(K)and on the straight line(P,mn), whereP=(1/λ, 0)Γ(K) (seeFigure 6.2).

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Mn Mn+1

Q0(K)

α φK

P

λ

mn

mn+1

Γ0(K)

Figure 6.2

Proof. The symmetry with respect to the diagonal is preserved byφK. We look at the images byφKof the vertical linesX=α. By (6.7), they are the straight lines with equations αy=1 +λx, and all these lines pass through the pointP=(1/λ, 0). So the lemma is

obvious.

Now we will, as in [3,4], translate the property of sequencemngiven byLemma 6.3 into the additionmn+1=mn+Pfor a group law on the cubic curveΓ(K). To do this, we need the regularity of the curveΓ(K), that is its elliptic nature. With this objective, we make a new transformationκ, independant fromK, and defined it by

κ(x,y)=(U,V), wherex+y= −U,xy=V. (6.15) SoΓ(K) becomes a new cubic curveΓ(K) with equation

U

λcU2+α(K)U2V2 4

+ (c+dλ)U2(d+λ)Uβ(K)U2V2

4 + 1=0, (6.16) or

V2α(K)U+β(K)

b+2U3+γ(K)U24(d+λ)U+ 4=0, (6.17) where

γ(K)=4c+ 4dλβ(K)=2+ 2dλ+ 2cK. (6.18) Now we need the following result.

Lemma6.4. For everyK > Km,

β(K)>4c+ 3d+b

>0. (6.19)

Proof. It is obvious with formulas (3.1), (3.4), and (6.12), and condition (3.3).

On the other hand, one can see that the quantityα(K) may be zero for some value of K, and positive or negative (see proof ofProposition 6.7). But if 4c2< bd(and thusb >0 andd >0), thenα(K)(bd4c2)/d >0. In the general case of condition (3.3) only, we

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set

p(K)=α(K)

β(K). (6.20)

Now we define a new transformationψK, in affine coordinates, by

U= u

1p(K)u, V= v

1p(K)u, (6.21)

or

u= U

1 +p(K)U, v= V

1 +p(K)U, (6.22)

or, in homogeneous coordinates, by

U=u, V=v, W=wp(K)u, (6.23) or

u=U, v=V, w=W+p(K)U. (6.24) We obtain a new cubic curveE(K).

Remark that if for someKone hasp(K)=0, thenψK=Id.

Ifp(K)=0, then the line with equationU= −1/ p(K) is an asymptote of inflexion of Γ(K) which is sent to the line at infinity byψK; this line is a tangent of inflexion to the cubicE(K).

The equation ofE(K) is

v2β=u3b+2++ 4p2(d+λ) + 4p3

u212p2+ 8p(d+λ) +γ+u4(d+λ) + 12p4. (6.25) Of course, ifp=0, we find again the cubic curveΓ(K) itself.

Now we can prove the essential result of this part.

Proposition6.5. (1)IfK > Km, the cubic curveE(K)is regular: it is an elliptic curve. So there is onE(K)an abelian group law whose unit element is the point at infinity in vertical direction, and whose addition is defined byA+B+C=0if and only if the three pointsA, B,CofE(K)are on the same straight line (and the oppositeAis the symmetric ofAwith respect to the u-axis).

(2)Letmnbe the images of pointsmnbyψKκ, andPthe image ofPby the same map.

Then, for everyn,mn+1=mn+P, and mn=m0+nP, for the addition of the group law on E(K).

Proof. Equation ofE(K) has the formβv2=P3(u), withβ >0 and deg(P3)3. So we know (see [1,6]) thatE(K) is regular if and only if

(i) the coefficient ofu3inP3is nonzero (deg(P3)=3);

(ii) the three roots ofP3are distinct.

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