ON THE ALGEBRAIC DIFFERENCE EQUATIONS un+2 un = ψ(un+1 ) IN R+ , RELATED TO A FAMILY ∗ OF pdf

We give the possible periods of solutions, and studytheir global behavior, such as the density of initial periodic points, the density of trajecto-ries in some curves, and a form of sens

un+2un= ψ(un+1) IN R+

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 type un+2 u n = ψ(u n+1),which started in a previous paper Here we study the case where the algebraic curvesrelated to the equations are quarticsQ(K) of the plane We prove, as in “on some alge-

braic difference equations u n+2 u n = ψ(u n+1) inR +

∗, related to families of conics or cubics:generalization of the Lyness’ sequences” (2004), that the solutionsM n =(u n+1,u n) arepersistent and bounded, move on the positive componentQ0(K) of the quartic Q(K)

which passes throughM0, and diverge ifM0is not the equilibrium, which is locally ble In fact, we study the dynamical systemF(x, y) =((a+ bx + cx2)/ y(c + dx + x2),x),

sta-(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 studytheir 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 dichotomybetween a form of pointwise chaotic behavior and the existence of a common minimalperiod to all nonconstant orbits ofF.

In the present paper, we will study the difference equation

u n+2 u n = a + bu n+1+cu2n+1

c + du n+1+u2

n+1

Copyright©2005 Hindawi Publishing Corporation

Advances in Di fference Equations 2005:3 (2005) 227–261

DOI: 10.1155/ADE.2005.227

The dynamical system inR +

∗2which represents this difference equation is

which satisfiesG ◦ F = G, and thus G(u n+1,u n) is constant on every solution of (1.2)

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

x2y2+dxy(x + y) + c

x2+y2 

+b(x + y) + a − Kxy =0, (1.6)passes throughM0

The quarticsQ(K) are invariant on the action of F, and thus the points M nmove onthe quartic passing throughM0, more precisely on its positive componentQ0(K).

The mapF has a geometrical interpretation If M ∈ R+

∗2, letM be the second point ofthe quarticQ(K) which passes through M whose first coordinate is the same as those of

M (there is only one such point M because the point at the vertical infinity is a doublepoint of the quartic) The imageF(M) is the symmetric point of M with respect to thediagonalx = y.

For all this results, we refer to [4]

InSection 2, we give a general topological result useful for our study, which extends aresult 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 and b + c + d > 0, bounded and persistent inR +

∗2, and diverge if (u1,u0)=

(, ), the fixed point of F, and prove that this point is locally stable.

InSection 4, we show that the case whereu n+2 u nis a homographic function ofu n+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 mapF to each curve Q0(K) is conjugated to a rotation onto the circle

(seeTheorem 6.11) We study also in Sections6and7whether the chaotic behavior 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 thegeneral one (Section 8)

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

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 beuseful for the study of difference equations This assertion extends [4, Proposition 1]

Proposition 2.1 Let X be a topological Hausdor ff space Let F : X → X and G : X → R be two maps Suppose first that the following conditions hold:

(a)F is continuous on X;

(b)G is continuous and has a strict minimum K m at a point L;

(c)∀ x ∈ X, G ◦ F(x) = G(x) (the invariance property);

(d)F has at most one fixed point.

If K ≥ K m , the level sets (if nonempty) of G are defined byᏯK = { x ∈ X | G(x) = K } Then the following three results hold:

(1) every point x ∈ X lies in exactly one setᏯK ;

(2) the point L is the (unique) fixed point of F;

(3) if M0∈ X let M n+1 = F(M n ) be the points of the orbit of M0 under F; then M n ∈

ᏯG(M0) , and if M0= L, then the sequence (M n ) does not converge.

Now suppose additional hypotheses:

(e)X is connected and locally compact;

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

(4) eachᏯK is compact and nonempty for K m ≤ K < K ∞ (withᏯK m = { L } ), and the equilibrium point L is locally stable.

Suppose at last the additional hypothesis:

(g)G has only one local minimum (its global one at L); then

(5) for K > K m the set ᏯK is the boundary of the open set U K = { G < K } which is a connected relatively compact set.

Proof Assertions (1) and (2) are obvious If M n+1 = F(M n), thenM n ∈ᏯG(M0) SupposethatM nconverges to a pointN Then G(N) = G(M0) andF(N) = N, so by (d) and (1)

N = L But G(M n)= G(N) = G(L) = K m, and by (b)M n = N for all n Thus, if M0= L,

thenM ndoes not converge

If (e) and (f) hold, it is easy to see thatᏯKis nonempty and compact for everyK ≥ K m;

in particular, sequences (M n) are bounded (i.e., relatively compact)

We prove now that the setsU K = { G < K }form a basis of neighborhoods ofL Let V

be an open neighborhood ofL The sets { G ≤ K }, forK > K m, are compact, and theirintersection is{ L }; so there is aK > K msuch that{ G ≤ K } ⊂ V , and thus U K ⊂ V

We can now prove easily thatL is locally stable: if V is a neighborhood of L, there exists

K > K msuch thatU K ⊂ V If M0∈ U K, then, for everyn, M n ∈ U K by (c), andM n ∈ V

We prove now assertion (5), if (g) also holds We haveU K ⊂ { G ≤ K }, andU K \ U K ⊂ { G ≤ K } \ { G < K } =ᏯK Thus,∂U K ⊂ᏯK Now, ifᏯK ⊂ ∂U K, there existsx ∈ᏯK,x / ∈

∂U K, thus there exists a neighborhoodV of x such that V ∩ U K = ∅ Thus,G ≥ K on

V , and G(x) = K; thus x is a local minimum of G, and x = L because K > K m: this isimpossible, and∂U K =ᏯK

Finally, we prove thatU K is connected IfU K is the union of two disjoint nonemptyopen setsA and B (which are relatively compact), put α =infA G and β =infB G; we have

α, β < K If α = G(u) with u ∈ A and β = G(v) with v ∈ B, then u and v are two distinct

local minima ofG, which is impossible Thus we can suppose that α = G(u) with u ∈

A \ A But A ⊂ X \ B (because U K is open), thusu / ∈ B, and u ∈ U K \ U K = ∂U K =ᏯK So

We will useProposition 2.1whenX is an open subset ofR 2; in this context,F and G

are given by (2) and (3) In the general case, we can ask the question whether a form ofchaotic behavior may be described for the mapF (we will study in this paper if it is the

case withF and G given by (2) and (3)) Precisely, one may ask the question whether F

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 thatX is a metric space, with distance d We say that the

dynam-ical system (X, F) with invariant G has IPCB if we have the following three properties.

(a) There exists a partition ofX \{ L }into two dense subsetsA and B which both are

union of “curves”ᏯK, and then invariant underF : A is the set of initial periodic

pointsM0,B is the set of initial points M0whose orbit is dense in the curveᏯK

which passes throughM0(that isᏯG(M0))

(b) Every pointM0∈ X \{ L }has sensitivity to initial conditions, that is, there exists

δ(M0)> 0 (this dependance on M0explains the term “pointwise”) such that everyneighborhood ofM0contains a pointM0whose iteratesM n  satisfyd(M n,M n )≥ δ(M0) for infinitely many integersn.

(c) There exists an integerN such that every integer n ≥ N is the minimal period of

some periodic orbit ofF.

IPCB is the essential result of [3] about the behavior of Lyness’ difference equation

u n+2 u n = k + u n+1if 0< k =1 (ifk =1, 5 is a common minimal period to all nonconstantsolutions)

In [4], we prove also that IPCB holds for the solutions of difference equations inR +

tual property supposes that each setᏯK is homeomorphic to a circle Then the study ofthe properties of functionθ would be essential: continuity (analyticity if X is an open set

ofR 2), limits whenK → K mandK → K ∞

3 First general results of divergence and stability

We begin by identifying the fixed point

Lemma 3.1 If a = b = 0, then sequence ( 1.2 ) tends to 0 If a + b > 0, then the fixed point of the dynamical system ( 1.3 ) is the unique positive root  of the equation

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

Figure 3.1

Proof It is obvious that a fixed point of F has the form (Y , Y ) where Y satisfies (3.1), andthat (3.1) has a unique positive root such that (, ) is invariant by F.

For the limit of the sequence (u n) 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 point M n =(u n+1,u n), whichlies onQ(K) ∩ R+

d ≥0 Thus ρ n = u n+1 /u nis decreasing and tends to a limitλ If λ < 1, then u n →0 If

λ would satisfy λ ≥1, thenu nwould be increasing, and would tend to infinity But then

c/(c + du n+1+u2

n+1)≤1/2 for big n, and thus we would have λ =0, which is a

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

Thus we havexy + a/xy ≤ K, d(x + y) ≤ K, c(x/ y + y/x) ≤ K, b/x ≤ K, and b/ y ≤ K.

Ifb > 0, then x ≥ b/K and y ≥ b/K, and thus, with the condition xy ≤ K, the set A K iscompact Ifb =0, we will supposea > 0 (the case a = b =0 is trivial byLemma 3.1), andthe conditionxy + a/xy ≤ K implies that 0 < r1≤ xy ≤ r2: the point (x, y) is between two

hyperbolas But then ifc or d is positive, we have x/ y + y/x ≤ K/c and thus 0 < s1≤ y/x ≤

s2, orx + y ≤ K/b In the two cases, A K is compact; seeFigure 3.1

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

Indeed, consider the difference equation un+2 u n = a/u2

n+1, witha > 0 Its solutions are

the sequencesu n = a1/4exp[(−1)n(A + Bn)], with A =ln(u0a −1/4) andB =−ln(u0u1a −1/2),which are neither bounded nor persistent ifu0u1= √ a.

Then, we must identify the minimum ofG The equations of critical points are x2y2+

dx2y + c(x2− y2)− by − a =0 andx2y2+d y2x + c(y2− x2)− bx − a =0 The difference

of these two equations gives (x − y)(dxy + 2c(x + y) + b) =0 But if (x, y) ∈ R+

∗2, the onlysolution isx = y, so the previous equations give x4+dx3− bx − a =0, and thus we have

x = y = : G has a unique critical point at the equilibrium L =(, ), the minimum of G

is achieved only at this point, and the value of the minimum is

com-∗2passes a unique curveQ0(K).

We can thus applyProposition 2.1and obtain the following theorem

Theorem 3.2 If a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0, a + b > 0, and b + c + d > 0, every solution of the difference equation ( 1.2 )

is bounded and persistent inR +

∗2 If (u1,u0)=(, ), then (u n ) diverges, the point M n =

(u n+1,u n ) moves on the curve Q0(K) which passes through M0, and K > K m Moreover the equilibrium L is locally stable.

4 The homographic case

In [5], the authors study the difference equation

u n+2 = αu n+1+β

u n

γu n+1+δ, withα, β, γ, δ ≥0,α + β > 0, γ + δ > 0. (4.1)

Ifγ = α =0, we find the classical sequenceu n+2 =(β/δ)/u nwhich is always 4-periodic

Ifγ =0,α =0, the sequencev n =(δ/α)u nsatisfiesv n+2 v n = v n+1+βδ/α2: it is a Lynesssequence, and its behavior is known and given in [3]

So, we supposeγ > 0, and thus we can suppose γ =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, weexamine some cases

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

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

se-(iv) Ifδ =0,α > 0, β > 0, we put u n = β/v n, and obtainv n+2 v n = β2/(v n+1+α), which

has the form (α  v n+1+β )/(v n+1+δ ) Thus, (1.2) for (v n) with the valuesa = c =

0,b = β2,d = α, is exactly (4.1) foru n = β/v n

In any cases, (4.1) comes down to (1.2) or to a known sequence (Lyness’ one) or toelementary 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), butalmost without calculation, and we can improve it

Proposition 4.1 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 erty of solutions of (1.2) that we will prove in the following parts, see corollaries ofTheorems5.1and7.1, where examples of (4.1) which have IPCB are given

prop-5 The casea =0

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

Theorem 5.1 Let the di fference equation inR +

(a) There exists a well-defined number θ b,c,d(K) ∈]0, 1/2[ such that the restriction of F

to Q0b,c,d(K) is conjugated to a rotation on the circle, of angle 2πθ b,c,d(K) ∈]0,π[.

(b) For every b, c, d satisfying the conditions of ( 5.1 ) and b2= c3or bd =2c2, the ence equation ( 5.1 ) has IPCB.

differ-(c) Every integer n ≥ 4 is the minimal period of some solution of ( 5.1 ) for some b, c, d, and some initial point M0.

One has b2= c3and bd =2c2if and only if every solution of ( 5.1 ) is 5-periodic.

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



b d

Point (3) of [4, Theorem 6] has to be modified, because here we have onlyα > 0 and

β ≥0 (arbitrary), but in [4] we haveα ≥0 andβ > −2√

α So, in [4, Lemma 10], we mustreplace the domainD by D= R+2and the functionf () =(1/π) cos −1(1/2)(1 −1/ √

 + 1)

by the functionf () =(1/π) cos −1(1/2)

1 + 3/( + 1) Then it is easy to show that we have

only

o

ψ( D) =]0, 1/3[ Thus every integer n ≥4 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) If c > 0, then solutions (u n) of (5.1) are rational if and only if thev n’s arerational, whenb, c, d are 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 = q2withq ∈ Q+

∗ In this case, and withb rational, 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:

v n = √ c/u nsatisfiesv n+2 v n =1 +v n+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

Lemma 5.4 (case a = c = d =0,b > 0) The positive solutions of the difference equation

u n+2 u n+1 u n = b are 3-periodic.

6 General results in the casea > 0

It is easy to see that ifa > 0 we can suppose that a =1 (putu n = v n √4

a) 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 rection, which are ordinary ifd2−4c =0, the asymptotes being then the linesx = r1,

di-x = r2, y = r1, and y = r2, where the r i are the roots (real or complex) of the tions2+ds + c =0 Moreover, if K > K m, the quarticsQ(K) have no singular point in

equa-R +

∗2 Indeed, if the equation ofQ(K) is p(x, y) − Kxy =0, singular points are given by

p  x − K y =0, p  y − Kx =0, andp − Kxy =0 These relations givex p  x = p and y p  y = p,

and these last relations are the equations whose solutions are the critical points of thefunctionG(x, y) = p(x, y)/xy But we have seen that G has no critical point inR +

∗2exceptforL =(, ), and so the only finite singular point of Q(K) inR +

∗2would beL, but this

point is not onQ(K) if K > K m

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 atransformation, 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

f1 f0 − λ0− λ f2  f3

t

K m

K G(t, t)

Figure 6.1

Lemma 6.1 For K > K m , the coordinates of the intersection points of Q(K) with the diagonal

δ are solutions of the equation

h K(0)=1, and soh K has two roots f2and f3which satisfy 0< f2<  < f3 We have also

h K(− ) = h K() −4d3−4b < 0, and thus we have two other roots f1 and− λ which

satisfyf1< −  < − λ < 0 At last, h K(λ) = h K(− λ) + 4dλ3+ 4bλ =4dλ3+ 4bλ ≥0 Ifb + d >

0, thus we have 0< λ < f2 Ifb = d =0, the roots of h K are f1,− λ, λ, and − f1, whoseproduct 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 functiont → G(t, t) is easy to determine, seeFigure 6.1 It is immediatefrom this graph that the roots are continuous functions ofK Their limits when K →+∞

are obvious and given by (6.3) If K → K m, then f2 and f3 tend to, and f1 and − λ

have limits f0and− λ0which are the two other roots of equationh K m(t) =0, which hasalready the double root Thus these two other roots are easy to obtain, they are given by

x  = T (X +λT ), y  = T (Y +λT ), t  = X  Y  − λ2T 2. (6.9)

OnQ(K) this transformation φ K is not defined only at the point (− λ, − λ) if d + b > 0,

and at the points (− λ, − λ) and (λ, λ) if b = d =0

Now we determine the image ofQ(K) under φ K 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, y the product of D by the following factor

α(K) = dλ2−4cλ + b, β(K) = K + 2c + 2dλ −2 2. (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 φ K Identification of the images of Q(K) \ {(− λ, − λ) }andQ0(K) is given in the following results.

Lemma 6.2 (1) If b + d > 0, then XY > λ2on Q0(K), and if b = d = 0, then XY ≥ λ2on

Q0(K), with equality only at the point (λ, λ).

(2) If b + d > 0, then the positive componentΓ0(K) of the cubic Γ(K) is compact, and φ K

is a homeomorphism of Q0(K) ontoΓ0(K) If b = d = 0, thenΓ0(K) is unbounded and has

a point at infinity in direction x = y, which is the image by φ K of the point (λ, λ) of Q0(K), and φ K is a homeomorphism of Q0(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 { G ≤ K }, lying on the hyperbolaXY = λ2 Then we have

Lemma 6.1, and thusQ0(K) ⊂ { XY > λ2}

Now ifb = d =0, we choose (X, Y ) =(λ, λ) Then X + Y > 2λ if XY = λ2, and the samecalculation gives, on ({ G ≤ K } \ {(λ, λ) })∩ { XY = λ2}, the impossible inequality 0> 0.

But this calculation proves also that{ G < K } ∩ { XY = λ2} = ∅ So{ G ≤ K }is contained

in{ XY ≥ λ2}or in{ XY ≤ λ2}, and we conclude, with the aid of the point (f3, 3), that

{ G ≤ K } \ {(λ, λ) } ⊂ { XY > λ2}

(2) Ifb + d > 0, we have XY > λ2 on Q0(K), and formulas (6.7) show thatφ K is ahomeomorphism ofQ0(K) onto the positive componentΓ0(K) of Γ(K) (note that Γ(K)

does not intersect the axis{ y =0} ∩ { x ≥0}nor{ x =0} ∩ { y ≥0}, by formula (6.11))

So the setΓ0(K) is compact inR +

∗2

Ifb = d =0, (6.7) shows thatφ Kis a homeomorphism ofQ0(K) \ {(λ, λ) }ontoΓ0(K),

and soΓ0(K) cannot be bounded Equation (6.11) becomes

λc(x − y)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) on Q0(K), then (x, y) tends to infinity in

6.2 The algebraic-geometric interpretation of the transformed sequence, and the liptic nature of the quartic and cubic curves We begin with the transformation of the

el-sequenceM n =(u n+1,u n) byφ K: what is the behavior of the pointsm n = φ K(M n)?

Lemma 6.3 Let m n be the image inΓ0(K) of the sequence M n =(u n+1,u n ) on Q0(K) by the birational transformation φ K Then the symmetric point of m n+1 with respect to the diagonal x = y lies onΓ0(K) and on the straight line (P, m n ), where P =(−1/λ, 0) ∈ Γ(K)

(see Figure 6.2 ).

need the regularity of the curveΓ(K), that is its elliptic nature With this objective, we

make a new transformationκ, independant from K, and defined it by



+ (c + dλ)U2−(d + λ)U − β(K) U

2− V2

4 + 1=0, (6.16)or

V2 

α(K)U + β(K)

−b + dλ2 

U3+γ(K)U2−4(d + λ)U + 4 =0, (6.17)where

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

Lemma 6.4 For every K > K m ,

β(K) > 4c + 3d + b

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 thus b > 0

andd > 0), then α(K) ≥(bd −4c2)/d > 0 In the general case of condition (3.3) only, we

u  = U , v  = V , w  = W +p(K)U  (6.24)

We obtain a new cubic curveE(K).

Remark that if for someK one has p(K) =0, thenψ K = Id.

Ifp(K) =0, then the line with equationU = −1/ p(K) is an asymptote of inflexion of

Of course, ifp =0, we find again the cubic curveΓ(K) itself.

Now we can prove the essential result of this part

Proposition 6.5 (1) If K > K m , the cubic curve E(K) is regular: it is an elliptic curve So there is on E(K) an abelian group law whose unit element is the point at infinity in vertical direction, and whose addition is defined by A + B + C = 0 if and only if the three points A,

B, C of E(K) are on the same straight line (and the opposite − A is the symmetric of A with respect to the u-axis).

(2) Let mn be the images of points m n by ψ K ◦ κ, and P the image of P by the same map.

Then, for every n, mn+1 =  m n+P, and mn =  m0+n P, for the addition of the group law on

E(K).

Proof Equation of E(K) has the form βv2= P3(u), with β > 0 and deg(P3)≤3 So weknow (see [1,6]) thatE(K) is regular if and only if

(i) the coefficient of u3inP3is nonzero (deg(P3)=3);

(ii) the three roots ofP are distinct

p(K) =0 So, ifp(K) =0, thenb + dλ2> 0.

Suppose then thatp(K) =0 and that the coefficient of u3inP3is zero ThenE(K) splits

into a conic and the line at infinity, that isΓ(K) splits into its asymptote U = −1/ p(K)

and a conic (besides one see easily that the left hand of (6.17) has the factorU + 1/ p(K)).

So, it is also the case ofΓ(K): it is the union of the real line J with equation x + y =1/ p(K)

and a conicB But Γ(K) is symmetric with respect to the diagonal, contains the points

P =(−1/λ, 0), P  =(0,−1/λ), and the three distinct points F i  =(1/( f i − λ), 1/( f i − λ)),

i =1, 2, 3, except if b = d =0, whenF2 is the point at infinity in directionx = y (see

Lemma 6.2) One has F2 ∈ / J and F3 ∈ / J (because Γ(K) ∩ { x ≥0} ∩ { y =0} = ∅), so

F1 ∈ J.

Ifb + d > 0,Γ0(K) is compact and contains F2andF3(seeLemma 6.2), soΓ0(K) = B is

necessarily an ellipse, but in this case the real pointsP and P cannot lie onΓ(K) (because

they do not belong toJ: F1∈ / ∆λ), and this is a contradiction (seeFigure 6.3)

Ifb = d =0,Γ0(K) is necessarily a parabola with axis x = y, and the same contradiction

holds

Now we prove point (ii) The three roots ofP3are distinct These roots are the firstcoordinates of the images of the (f i, i) by the transformationψ K ◦ κ ◦ φ K So on the coor-dinates we have the transformationst →1/(t − λ), t → −2t, t → t/(1 + p(K)t) From the

numbers f1< 0 < f2< f3, we obtain first f1 < 0 < f3 < f2 ≤+∞, and then−∞ ≤  f2< f3<

0< f1 If p(K) =0, the images of these last numbers byt → t/(1 + p(K)t) are finite (P3

has degree exactly 3), so fi = −1/ p(K) for i =1, 2, 3, and thus we obtain imagese1,e2,e3

which are distinct Ifp(K) =0, thenb + d > 0 and f2> −∞, andei =  f iare distinct So

point (1) of the proposition is proved

Finally, the transformationψ k ◦ κ is projective, so it preserves the alignment of points;

thus point (2) of the proposition is obvious fromLemma 6.3

Corollary 6.6 If K > K m and a > 0, then the quartic curve Q(K) is elliptic.

Proposition 6.7 (1) The coe fficient of u3in ( 6.25 ) is positive:

q(K) : = b + dλ2+pγ + 4p2(d + λ) + 4p3> 0. (6.27)

(2) The rootse i of P3, images of the f i by ψ K , satisfy the inequalities



e2< e 3< e 1. (6.28)

Proof We have seen that q(K) is never zero, and that the three numbers eiare distinct

We will use an argument of continuity and connectedness LetΩ be the subset ofR 3defined byb ≥0,c ≥0,d ≥0, and b + c + d > 0; it is a convex set We consider K m =

2c + d + 3b/ + 2/2as a continuous function of (b, c, d) ∈ Ω (it is easy to see that  is a

continuous function onΩ) Let Σ be the subset ofR 4defined by

(b,c,d)∈Ω

(b, c, d) ×K m(b, c, d), + ∞ , (6.29)

that is the strict epigraph ofK m It is easy to see thatΣ is a connected set So q, as

con-tinuous function onΣ, has a constant sign But the function p vanishes for some value of

(b, c, d, K) ∈ Σ, because if c = d =0 andb > 0, then α = b > 0, and if b = d =0 andc > 0,

thenα = −4cλ < 0 : p must vanish at some point by the intermediate value theorem But

in this pointq assumes the value b + dλ2> 0 because if p =0, thenb + d > 0 So we have

q > 0 onΣ

For the same reason, the functionsei −  e jdo not vanish (i = j), and for p =0 we have



e i =  f iand f2< f3< f1: these inequalities hold also fore i.

For a later use ofProposition 6.5, we look at the coordinates of the pointP Easy cal-

culations give its coordinates:

becauseP does not belong to the asymptote of Γ(K), so P is a finite point of Γ(K).

We need the position of the coordinates ofP.

Lemma 6.8 The coordinates of the point P satisfy the inequalities

1

p + λ > e 1> 0, − p + λ1 < 0. (6.31)

Proof First, on the set Σ one has 1/(p + λ) =  e1: otherwise, we would have 1/λ =  f1=

−2f1 =2/(λ − f1), and then f1= − λ, which is false byLemma 6.1 Then, ifp =0,e1=  f1,and the inequality 1/(p + λ) =1/λ >e1is equivalent to 2/(λ − f1)< 1/λ, or f1< − λ, which

is true byLemma 6.1 Thus, the same argument with the connectedness ofΣ gives thefirst inequality

Ifp > 0, it is easy to see that e 1> 0 If p =0,e 1=  f1= −2f1 > 0 If p < 0, we look at the

graph of the functiont → t/(1 + pt) (seeFigure 6.4): if f1> −1/ p, we obtain the

inequal-itiese1< 1/ p <e2< e 3< 0, which is impossible byProposition 6.7; so 0< f1< −1/ p, and

Remark 6.9 If p > 0,Figure 6.4andProposition 6.7prove that we have−1/ p <f2< f3<

0< f1.

Propositions6.5and 6.7and Lemma 6.8give immediately the following importantresult

Proposition 6.10 (1) The cubic curve E(K) has the form given in Figure 6.5

(2) A solution ( u m ) of di fference equation ( 1.2 ) with a > 0 and b + c + d > 0 has minimal period n if and only if the point P is of order exactly n in the group E(K).

(3) If a point M0of Q0(K) has minimal period n, it is also the case for every other point

M0 ∈ Q0(K).

Now we can transform the cubic E(K) (which is Γ(K) if p(K) =0) by a last affinemapτ K, to get the canonical form of a regular cubic curve that we will parametrize by

(b, c, d, K) ∈ ? ?, because if c =< /small> d =< /small>0 andb > 0, then α =< /small> b... coordinates of the pointP Easy cal-

culations give its coordinates:

becauseP does not belong to the asymptote of Γ(K ), so P is a finite point of< /i>... the connectedness of? ? gives thefirst inequality