doi:10.1155/2009/134749
Research Article
Results and Conjectures about Order
q
Lyness’
Difference Equation
un
qun
a
un
q
−
1
· · ·
un
1
in
R
∗
, with a Particular Study of the Case
q
3
G. Bastien
1, 2and M. Rogalski
2, 31Institut de Math´ematiques de Jussieu, UPMC-Paris 06, UMR-CNRS 7586, 75251 Paris, France 2Equipe d’Analyse fonctionnelle, IMJ, 16 rue Clisson, 75013 Paris, France
3Laboratoire Paul Painlev´e, USTL Lille 1, UMR-CNRS 8524, 59655 Villeneuve, France
Correspondence should be addressed to G. Bastien,[email protected]
Received 4 March 2009; Accepted 14 July 2009
Recommended by Istvan Gyori
We study orderqLyness’ difference equation inR∗ :unqunaunq−1· · ·un1, witha >0 and the associated dynamical systemFainR
q
∗ . We study its solutionsdivergence, permanency, local
stability of the equilibrium. We prove some results, about the first three invariant functions and the topological nature of the corresponding invariant sets, about the differential at the equilibrium, about the role of 2-periodic points whenqis odd, about the nonexistence of some minimal periods, and so forth and discuss some problems, related to the search of common period to all solutions, or to the second and third invariants. We look at the caseq 3 with new methods using new invariants for the mapF2
aand state some conjectures on the associated dynamical system inR
q
∗ in
more general cases.
Copyrightq2009 G. Bastien and M. Rogalski. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
We will study the global behavior of solutions of the orderqLyness’ difference equation for
q≥3 anda >0:
unq
aunq−1· · ·un1
un , u0, u1, . . . , uq−1>0. 1.1
The associated dynamical system inR∗qis given by the mapFa:
Fa:xx1, . . . , xq
−→Fax
ax1· · ·xq−1
xq
, x1, x2, . . . , xq−1
The orbit ofM0 uq−1, . . . , u1, u0underFais{M0, M1, . . . , Mn , . . .}where the pointMnis
Mn unq−1, . . . , un1, un.
For q 2, we have so the ordinary order 2 Lyness’ difference equation, which is studied in some previous papers; see, for example,1,7. This equation was introduced in 1942 by Lyness, for the casea 1, while he was working on a problem in number theory. This equation is also used in geometry and frieze patterns see3for references. When
q3, it is “Todd’s equation,” whose deep study fora /1 is made in4.
InSection 2we study the elementary properties of the solutions of1.1: divergence, permanent character, and local stability of the equilibrium. Then we look at the first invariant function for orderqLyness’ equation, and prove that the invariant manifolds associated to it are homeomorphic to the sphereSq−1.
InSection 3we look at the second invariant for orderqLyness’ equationq≥3, prove results about it, and make some conjectures, that we prove forq3,q4, andq5.
InSection 4we study, for oddq ≥ 5, the third invariant of orderqLyness’ equation found in5and prove about it results analogous to these ones of Sections2and3.
In Sections5and6we study the eigenvalues and the global nature of the differential
dFaat the equilibrium, give some elementary results about periods of solutions of1.1, and study the possible common periods to all solutions of1.1.
InSection 7we study the particular caseq 3. We use another approach than in4
which solves almost completely the caseq3, using new invariants of the mapF2
a.
InSection 8we make some general conjectures for the behaviour of the solutions of orderqLyness’ difference equation.
We have recently received the nice preprint6, which studies also orderqLyness’ equation, with new and very interesting tools. Some points are common to this paper and to our one.
is invariant under the action ofFa. So, the sequence
GaMn
Lemma 2.2. (1) The quantityGaxtends to infinity whenxtends to the point at infinity ofR∗q; that is, the sets{x|Gax≤K}are compact.
(2)Gahas an unique critical point, at its absolute minimum, attained at the equilibrium of
1.1L: , , . . . , , whereis the unique positive solution of the equation
X2−q−1X−a0. 2.3
One has > q−1, and the minimum ofGais
Ka:
1 1
q
aq 1
q1
q−1 >
q√a2. 2.4
Proof. 1On the closed sets{x|Gax≤K}ofR∗qwe haveixi ≤K−aand∀i11/xi ≤ K/a, and these estimates define a compact set inR∗q.
2Substracting the two equations∂Ga/∂xi0 and∂Ga/∂xj0 we havexi−xjxi xj1 0, and so all thexi are equal at a critical point. Hence, the equation∂Ga/∂x1 0
givesx2
1−q−1x1−a 0, and then for alli xi . So pointLis the unique critical point
ofGa, and thenGaattains only its absolute minimumKaat this point. Now, evaluation of
this minimum is obvious.
So we can apply 17, Proposition 2.1, to give a synthesis of new and older results about the dynamical systemFa,R∗q see also8.
Proposition 2.3. (1) ForK > Ka, the sets
ΣaK:{x|Gax K}, ΣaK:{x|Gax≤K} 2.5
are compact inR∗q and invariant byFa. The sets ΣaK are the boundaries of the open sets {x | Gax< K}, which are connected and constitute a fundamental system of neighborhoods ofL.
IfGaM0 K, then the orbit ofM0remains onΣaKand so is bounded.
(2) IfM0/L, then the sequenceundiverges, but for alln a/K−a≤ un ≤ K−aK
GaM0,unis permanent.
(3) The equilibriumLofFais locally stable.
Forq 2, one knows that ΣaKis homeomorphic to a circle (see [1]). Forq ≥ 3 one can
generalize this fact. First one proves a local result.
Proposition 2.4. IfK > Kais sufficiently near toKa, then the setΣaKis starlike with respect
to the pointL, and a ray fromLcutsΣaKat exactly one point. HenceΣaKis homeomorphic to
Proof. Put, for ρ > 0, ρ < , and→−u an unitary vector of Rq,δ→−
3. The Second Invariant of Order
q
Lyness’ Equation
3.1. General Results
In5the authors found new invariants for orderqLyness’ equation.
Proposition 3.15. Ifq≥3, the following functionHaonR∗q,
is invariant under the action ofFa.
It is proved in5thatHaandGaare independent.
Proposition 3.2. (1) The quantityHaxtends to infinity whenxtends to the point at infinity of
R
andSaMis bounded. One see that every product ofq−1 variablesxi appears in the left-hand member of the inequalityhax≤Mx1· · ·xq, with coefficienta. So we havexi≥a/M,
and these inequalities imply the compactness ofSaM.
2 We have not succeed in proving that the only critical point of Ha is L see
Conjecture 1, which would give an easy proof of the second point of Proposition 3.2 as forGa, and so we give here a direct proof of this point, with inequalities.
iiiBut for eachj∈ {1,2, . . . q−1}, we havexjxj1≥2xj xj11/2with equality only
with equality only if all thexjare equal, and
ax1xqx1· · ·xq≥ax1xq
with equality only if all thexjare equal.
ivNow we put√x1xquandx2x3· · ·xq−1tq−1. From3.5and3.6we obtain
with equality only if allxjare equal.
vWe have to find the minimum ofψt, uonR∗2. We putst u1/q−1and have has a minimum, which is a critical point. At such a point, we have
u2aq−1s, s2s1−u2uaq−1s0. 3.9
As we noticed before, the proof ofProposition 3.2will be dramatically simpler if the following conjecture would be true.
Conjecture 1. The pointLis the only critical point forHa.
Definition 3.3. A differentiable numerical functionRonR∗q, invariant under the action ofFa, is said “diagonal” if it has the two properties:
ievery critical point ofRwith formt, t, t, . . . , tis the pointL;
iiifqis odd, every critical point ofRwith formt, u, t, u, t, . . . , u, tis the pointL.
Proposition 3.4. LetRbe a differentiable diagonal invariant, and supposeq≥3.
(a) If it exists somej ∈ {1, . . . , q−1}such that every critical point ofR satisfiesxj xj1,
thenRhas a unique critical point which is the pointL.
(b) If the equalitiesx1xqandx2xq−1are true for every critical point ofR, thenRhas only
one critical point which is the pointL.
Proof. We will use the following easy lemma.
Lemma 3.5. If Mis a critical point of a differentiable invariantR, all the pointsFapMare also
critical points ofRforp∈Z.
This lemma results from the relations RFaM RM and from the fact that
dFaManddFa−1Mare invertible, which is easy to see from1.2.
Now, the mapsFaandFa−1act on last or first coordinates ofMas right or left shiftj−1
time forFaandq−j−1 times forF−1
a , and the equalityxj xj1spreads right and left. Then,
such a critical point ist, t, . . . , tfor somet > 0. BecauseRis diagonal, this point isL; this gives assertionaof the proposition.
In order to prove assertionb, we suppose that for every critical pointM x1, . . . , xq
we havex1xqandx2xq−1.
We writeFaM M x1, . . . , xq ax1· · ·xq−1/xq, x1, x2, . . . , xq−2, xq−1,
Fa−1M M x 1, . . . x q
x2, x3, . . . , xq−1, xq,
ax2· · ·xq
x1
. 3.10
So, byLemma 3.5we have the implications, ifq≥4,{x2 xq−1} ⇒ {x1 xq−2},{x2
x q−1} ⇒ {x3 xq}, then,{xq−2 xq} ⇒ {xq−3 xq−1}. And then,{x 1 xq −2} ⇒ {x2 xq−1};
also{x 1x3 } ⇒ {x2x4};{xq−1xq−3} ⇒ {xq−2xq−4}.
At this stage, we havex1 x3 xq−2 xqandx2 x4 xq−3 xq−1. Ifq 4, all the
xi have the same value. Ifq > 4, we can continue in this way and obtain by induction the following:
∗ifqis even, all thexihave the same values;
∗∗ifqis odd, for alli x2i1tand for allj x2jufor some positivetandu.
Ifq3, the hypothesis said already that the critical point has formt, u, t. In all cases, the diagonal character ofRgives the result.
Of course, for proving Conjecture 1 from Proposition 3.4 it suffices to see that Ha
is diagonal, and to prove the hypothesis xj xj1 or {x1 xq andx2 xq−1} for one
Lemma 3.6. InvariantsGa,Ha(forq≥3), andJa(for oddq≥5) are diagonal.
The third invariantJais defined inSection 4.
Proof. It is very easy forGa, and we will not use this result. ForHanotice that for the critical pointt, t, . . . , tthe relation∂Ha/∂x1 0 gives the equationtt12t1 at2qtt1,
that is,t: this is propertyi. If the critical point has the formt, u, t, u, . . . , u, t, the same relation gives1u/t1tu 1t/at2 m1tmu, whereq2m1. But the
relation∂Ha/∂x2 0 gives 2/1tu−1/u1/at2 m1tmu 0. One deduce
from these two relations the equalityu1u/t1t 1t−u, so we havet uand we apply propertyi.
In order to prove i for Ja, the relation ∂Ja/∂x1 0 is t/1 ta q −1t
1/t1t, which givest . Now suppose that a pointt, u, t, u, . . . , u, tis critical. When we write that the two first partial derivatives are zero, we obtaint/1tamtu
u1um/t1tm1and1u/ua1 m1tu u1um/t1tm1. So we haveamtu tu, and by putting this equality in the first previous relation, we get 1umum1 1tmtm1, that is,tu: we can apply propertyi.
Conjecture 1would have many consequences.
Definition 3.7. We put, forM > Ma,
SaM:{x|Hax M}, SaM:{x|Hax≤M}. 3.11
Obviously, these sets are compact and invariant by the action of Fa. From 9,
Proposition 2.1, Conjecture 1 would imply other properties, which are assertions of
Proposition 2.3for the first invariantGaand are still conjectures forHa.
Conjecture 2. The setsSaMare the boundaries of the open sets{x|Hax< M}, which are
connected.
Conjecture 1would imply an important property ofSaM, but we have no proof of it.
Conjecture 3. The setsSaMare, forM > Ma, smooth manifolds ofRq.
In fine , we think thatTheorem 2.5is valid forSa.
Conjecture 4. The invariant sets SaMare, forM > Ma, homeomorphic to the q−1
-dimensional sphereSq−1.
In fact,Conjecture 4will be a consequence ofConjecture 1and of aConjecture 6which will be introduced later. First, we prove a result similar toProposition 2.4.
Proposition 3.8. Supposeq 3. IfM > Mais sufficiently near toMa, then the setSaM
is starlike with respect to the pointL, and a ray fromLcutsSaKat exactly one point. Moreover, SaMis the boundary ofSaMand of the set{Ha < M}. HenceSaMis homeomorphic to an euclidean ball, andSaMis homeomorphic to a sphere, ifMis sufficiently near toMa.
SaM, and the fact that a ray fromLcutsSaMat exactly a point. So it is easy to see that
SaMis the boundary of the two sets given in the proposition, and then the existence of the
announced homeomorphisms is easy to prove.
Now we suppose that q3. We have to find an upper bound on the unit sphere of the quantity in the sqare brackets. It is a quadratic formB, and the eigenvalues of its matrix are
form ofλ3<0< λ2< λ1 23−1/2 1/4−251
2−5−1232212
So the quadratic formBis majorized byλ3. Hence the proposition results from the following
easy lemma.
Corollary 3.10. Conjectures1and5implyConjecture 4(and then all the four conjectures).
Proof. It will be an easy consequence of the classical following result in differential geometry see10, Theorem 50, pages 109-110.
Fact 1. LetVbe a differential manifold, andf:V →Ra smooth function, such that ifhas a unique critical pointα, at its absolute minimummfα;
iifor everyλ, μsuch thatm≤λ < μ, the subsetVλμ :{x|λ≤fx≤μ}is compact. Then there is a diffeomorphism ofV which mapsVmλontoVmμ.
We apply this fact withV R∗q,f Ha,α L, andm Ma. IfMis sufficiently
near toMa,SaMis homeomorphic to the ball, and forM > Mthe set{M≤Ha≤M}is
compact. So for everyM > Mthe setSaMis homeomorphic to a ball, and the setSaM
is homeomorphic to the unit sphereSq−1: this isConjecture 4.
Remark 3.11. Of course the previous corollary is true for the invariantGa, and then gives an
other proof ofTheorem 2.5which does not use Reeb’s theorem. The analogue ofConjecture 6
is in the proof ofProposition 2.4.
3.3. The Truth of the Four Conjectures for
q
3
Theorem 3.12. Ifq3, Conjectures1,2,3, and4are true.
Proof. As we saw before, it is sufficient to proveConjecture 1Conjecture 6is true forq3. We have
Ha
x, y, z
1xy1yzaxzxyz
xyz . 3.15
We write the two equalitiesHax 0,Haz 0. Easy calculations give first the relations
xz1xy1 y1axzxzyandzx1zy1 y1axzxzy. So we getxz. Now the third equationHay0 is written as
2yax22xyxy1ax22x, 3.16
and the two first equations reduced to the only one
xx1xy1y1ax22xy. 3.17
We write3.16in the formy−xy2xa1−x1x2−2x−a 0 and3.17in the
formy−xx24x2ya x2−2x−a 0. These two relations imply the equalities
xyandx2−2x−a0. So we obtainxyz.
Corollary 3.13. Letq3. IfGaM0 KandHaM0 M, then the orbit ofM0remains on the
invariant compact curve (part inR∗3of the intersection of two regular surfaces)
C
From Theorems2.5and3.12, the two surfaces of previous corollary are homeomorphic to the 2-dimensional sphereS2. In4, other and more complicated proofs of these facts are
given.
3.4. The Truth of the First Three Conjectures for
q
4
Theorem 3.14. Ifq4, then Conjectures1,2, and3are true.
Proof. It is sufficient to proveConjecture 1. We have now
Hax, y, z, t
1xy1yz1ztaxtxyzt
xyzt . 3.19
We will useProposition 3.4; so it suffices to see that a critical pointx, y, z, tsatisfiesy z
andx t. We putD :axtxyzt. The logarithmic derivation gives for a critical point ofHathe four equationsthey exchange if we exchangexandt, andyandz:
∂Ha
∂x 0 or
1
D
1y
x1xy1t , 3.20
∂Ha
∂t 0 or
1
D
1z
t1tz1x , 3.21
∂Ha
∂y 0 or
1
D
1x1z−y2
y1xy1yz , 3.22
∂Ha
∂z 0 or
1
D
1t1y−z2
z1tz1yz. 3.23
From these equations, we will prove two fondamental relations:
xzz1 tyy1, 3.24
t2x1 z1axyz. 3.25
In order to prove3.24we make the ratio of relations3.20and3.21, subtract3.23from 3.22, and make the ratio of the two results. The second one is a consequence ofLemma 3.5: the pointFax, y, z, t axyz/t, x, y, zis a critical point, so its coordinates satisfy
relation3.24:axyz/tyy1 zxx1. We use once more3.24forx, y, z, t
and obtain3.25. Now we putA:ayz. Relation3.25becomest2x1 z1Ax.
By exchange ofyandzand ofxandtwe obtain alsox2t1 y1At. We substract these two relations and obtain
t−xx1t1 Az−yxz−ty. 3.26
3.26and obtain the factorizationz−y/zz1tx1t1yz1tyz−Azz1 0.
The development of the first term in square brackets in formula3.27contains the last term
zt2x1, and so the factor ofz−yis strictly positive. So we obtainyz, but the relation
which follows formula3.26givesxt, and this proves the proposition.
3.5. The Truth of the First Three Conjectures for
q
5
In this section we prove the following result.
Theorem 3.15. Ifq5, then Conjectures1,2, and3are true.
By writingx1−x5x11−x5/x1and using3.33forx5/x1, we have also
x1−x5 x2−x4x11x2x3x4
x21x2x3
. 3.35
Finally, we writex2x1 −x4x5 x2x11−x4/x2x5/x1and put in this relation the ratio
x5/x1from relation3.33; we obtain after easy calculations
x2x1−x4x5 x2−x4
x1
x2
1x3x2x4 x22x2x4x24
1x2x3
. 3.36
The essential fact in formulas3.34,3.35, and3.36is that the three factors ofx2−x4are
positive. Now3.28givesD x11x51x1x2/1x2, and from invariance of the
left-hand member by exchange ofx1withx5and ofx2withx4we get
x11x41x51x1x2−x51x21x11x5x4 0. 3.37
The goal is to make appear the three first member of3.34,3.35and3.36in3.37. We obtain1x11x5x1−x5 x1x4−x2x5 x1x21x41x5−x5x41x11x2 0,
or1x11x5x1−x5 x1x4−x2x5 x2x1−x4x5 x2x4x1−x5 x1x5x2−x4 0.
From formulas3.34,3.35, and 3.36we obtainin finea relationx2−x4A 0, where
A >0. So ifxis a critical point we havex2x4, and formula3.35givesx1x5.
Then it results fromProposition 3.4bthat the only critical point isL.
4. The Third Invariant of Order
q
Lyness’ Equation When
q
2
m
1
≥
5
is Odd
In5, the authors give a simple third invariant of orderqLyness’ equation for oddq≥5.
Proposition 4.111. Whenq2m1≥5is odd, the function
Jax1, x2, . . . , xq
:
aqi1xi
m
j1x2j
1x2j
m
j0x2j1
1x2j1
q
i1xi
4.1
is invariant under the action ofFa.
SeeRemark 7.2 ofSection 7.1for an easy proof ofProposition 4.1. It is to be noticed thatJais invariant under a permutation of variables of odd ranks, and the same fact holds
for even ranks. Then the analogue of Propositions2.3and3.2is the following.
Proposition 4.2. (1) The quantityJaxtends to infinity whenxtends to the point at infinity ofR∗q
(q2m1≥5).
(2)Jahas a strict minimum at the pointL , . . . , , whose value is
(3)Jahas only one critical point, its minimumL , . . . , . (4) ForN > Nathe sets
ΔaN:{x|Jax N} 4.3
are compactq−1-dimensional smooth manifolds inR∗qwhich are invariant under the action ofFa.
As for invariantHawe make the following conjecture.
Conjecture 6. For N > Na, the invariant compact q−1-dimensional manifoldΔaN
{x|Jax N}is homeomorphic to the sphereSq−1.
We can reduce this conjecture to the following, which can be tested on a computer.
Conjecture 7. For integerm≥2, real numbersa >0 andN >0 the two curves in the positive planeR∗2
u1t t1u Num11t−m, 4.7
a m1tmutm21tmu−m11u1−m 4.8
have at most two common pointst1, u1andt2, u2.
Tests with a graphic computer give a great evidence to Conjecture 7, but its proof seems not easy.
Proof ofConjecture 7⇒Conjecture 6.As for invariantGawe will use a smooth function which
will have, byConjecture 7, exactly two critical points on manifoldΔaN; so the theorem of
Reeb will give us the result similarly to the proof ofTheorem 2.5.
Letfx qi1xi. We search critical pointxoff onΔaN. The relationdf λdJa
means that all the partial derivatives of Ja are equal. We write with evident denotations
Jax a Σ
even coordinates of a critical point have the same valueu.
Then relation4.9becomesa m1tmu1um/tm1um/1tm1 t1 because a compactq−1-manifold inRq cannot be included in an hyperplane. So we can
5. Nature of the Differential
dF
aL
at the Equilibrium
and, for each such eigenvalueλ, a 1-dimensional eigenspace generated by the eigenvector
Vλ λq−1, λq−2, . . . , λ,1.
So we have to determine the nature of solutions of5.3. The results are not the same ifqis odd or if it is even. Recall that fromLemma 2.2we have 1/ <1/q−1.
5.1. The Eigenvalues When
q
Is Even
First we study the case whereq2mis even. We use the following result.
Lemma 5.1. If0< z≤1/2m−1, the polynomial
Proof. First it is obvious that 0, 1,−1 are not roots ofP2m. Then the proof has three steps.
First step.P2mhas no real roots.
BecauseP2mis reciprocal, it suffices to show that there is no real rootλwith 0<|λ|<1.
If such a root exists, then one has 1< λ2m1λz1λλ2· · ·λ2m−2< z2m−1, and so
Second step. The roots of P2mhave modulus 1.
The numberz λ2m1λ−1/λλ2m−1−1is real, andz1 also. Hence we have
λ2m1−1λ2m−λ λ2m1−1λ2m−λ, or
λ−λλλ−1&λλ2m−1λλ2m−2· · ·1−
λ2m−1λ2m−2λ· · ·λ2m−1
' 0.
5.5
Butλ /λ; ifλλ1, we have|λ|1. If not, putr :λλ >0. BecauseP2mis reciprocal, we can
suppose that|λ|<1. From the equality
λλ2m−1λλ2m−2· · ·1−
λ2m−1λ2m−2λ· · ·λ2m−1
0, 5.6
we getr2m−1r2m−2· · ·1≤2mrm−1/2, orrm−1/21/rm−1/2· · ·r1/21/r1/2≤2m. But if
r /1, the first member is greater than 2m, which gives a contradiction.
Third step. Roots ofP2mare all distinct.
The proof of this step will be given in common with the one of the third step in the proof ofLemma 5.3.
5.2. The Eigenvalues When
q
Is Odd
Now we study the case whereq2m1 is odd. First we have the following result.
Lemma 5.2. If q 2m 1 is odd, then −1 is an eigenvalue of dFaL, for the eigen vector
V−1 1,−1,1, . . . ,−1,1, which is a vector tangent at the pointLto the hyperbolaHa defined in
Theorem 6.1inSection 6.1.
Proof. A first proof is to see that the vectorV−1satisfies the relationAIV−1 0, which is
obvious. One can also see that−1 is a root of5.3, which is also obvious. A deeper proof is in relation with the 2-periodic points of the hyperbolaHaseeTheorem 6.1inSection 6.1. We
have, byTheorem 6.1, the relationsFaht−Fah/t− hφamt−hφma/t−
see the definition ofhandφm
a inTheorem 6.1and formula6.4. But the first term tends to
the vectordFaL◦hwhent → , and the second tends to the vectorφm
a◦h
−h. Sohis an eigenvector ofdFaLfor the eigenvalue−1. And it is obvious to see that h 1,−1, . . . ,−1,1 V−1.
So5.3can be writtenλ1λ2m−λ2m−1· · ·−λ1 λ/λ11λ2λ4· · ·λ2m−2.
Then we cancel the factorλ1 and add to the two members the quantityλλ3· · ·λ2m−1.
In fine, we see that the eigenvalues ofdFaLdistinct of−1 are the roots of the polynomial P2m1λ:λ2mλ2m−2· · ·λ21−1zλλ2λ3· · ·λ2m−1, forz1/. We have
now the following result, analogous toLemma 5.1.
Lemma 5.3. If0< z≤1/2m, the polynomial
P2m1λ λ2mλ2m−2· · ·λ21−1z
has2mdistinct roots, with modulus 1, different from 1 and−1, hence of the forme±i θj,j 1,2, . . . , m,
withθj/θkifj /k, and0< θj< πfor allj.
Proof. First, it is obvious that 0,1,and −1 are not roots of P2m1. Then, the proof has three
steps, as forLemma 5.1.
First step.P2m1has no real roots.
A real rootλis necessarily positive; becauseP2m1is reciprocal, we can supposeλ <1.
So we have 11/2m≥1z 1λ2· · ·λ2m/λ1λ2· · ·λ2m−2 1/λλ2m−1/1
λ2 · · ·λ2m−2 > 1/λλ2m−1/m. But the function of λin the right-hand member has for minimum on0,∞the positive quantity2m/2m−12m−1/m1/2m >11/2m, and this is a contradiction.
Second step. Roots of P2m1have modulus1.
Remark that 1z λ2m2−1/λλ2m−1is real, hence equals to its conjugated. As in
proof ofLemma 5.1, this gives the relation
λ−λλλ−1&λλ2mλλ2m−1· · ·λλ1−
λ2mλ2m−1λ· · ·λ2m
' 0.
5.8
The same method as in Lemma 5.1, with r : λλ, supposed to satisfy r < 1, gives the inequalityrm1/rmrm−11/rm−1· · ·r1/r1≤2m1. But ifr /1, the first member
is greater than 2m1, and this is a contradiction.
Third step. roots of P2m1are all distinct.
The proof of this step is the same of that one of third step of the proof ofLemma 5.1
and is given below.
Common proof of third steps of Lemmas5.1and5.3.Roots of equationλq−zλq−1· · ·λ
10 are distinct if and only if it is the case for equationλq1−u λqu λ−10, whereuz1,
obtained by multiplication byλ−1. But a multiple root is a common root to this equation and to the derivative one q1λq −q u λq−1u 0. We eliminate ubetween these two
equations, and obtainq 1−λ2q/λq−11−λ2. So we haveλq−1 1λ2λ4· · ·λ2q−2/q.
By steps 1 and 2 of the proofs of Lemmas5.1and5.3we know thatλq−1is an extremal point of
the unit disk, and so the previous barycentric relation implies thatλ21, andλ±1, which
is impossible.
5.3. The Global Nature of
dF
aL
From Lemmas5.1,5.2, and5.3, we get the description ofdFaL.
Theorem 5.4. (1) Theqeigenvalues ofdFaLare all distinct and have modulus 1.
(2) Whenq2mis even,dFaLis linearly conjugated to a product ofmrotations in mutually orthogonal planes, with anglesθ1, θ2, . . . , θm∈0, πall distinct.
(3) Whenq 2m1is odd, dFaLis linearly conjugated to a product ofmrotations in
mutually orthogonal planes in the hyperplanexq 0, of anglesθ1, θ2, . . . , θm ∈0, πall distinct,
Proof. Eigenvalues ofdFaLare distinct in the two cases, and so the matrix is diagonalizable in a basis of eigenvectors inCq. But it is classical that, with conjugated eigenvalues of modulus
1 and a real matrix fordFaL, one can obtain the forms given in the theorem.
6. Results about Possible Periods of Solutions of
1.1
6.1. Role of 2-Periodic Points When
q
2
m
1
Is Odd
In12, the authors found the existence of 2-periodic points whenq2m1 is oddsee also 4,13. There is a theoretical reason to the existence of these points, fromTheorem 2.5and the following result.
Fact 2. Every continuous mapf :S2m → S2mhas a 2-periodic point.
Proof. It is sufficient to prove the result for smooth maps. Putg f◦fandidx xonS2m. Ifg has no fixed point, then for allx∈S2m, gx/ −−idx; sogand−idare homotopic,
and then have the same degree. But degg degf2and deg−id −12m1, and this is impossiblefor elementary properties of the degree, see10, pages 237–239.
These 2-periodic points have particular properties. The relations6.1of the following theorem are already in 12. See also 9 for applications to lacunary Lyness’ difference equations.
Theorem 6.1. Letq 2m1,m≥ 1, anda > 0. The locus of 2-periodic points for mapFais the
branch of hyperbolaHa⊂R∗qwhose equations are
x1 x3 · · ·xq, x2x4· · ·xq−1
mx1a
x1−m
, x1> m. 6.1
This branch of hyperbolaHa passes through the equilibriumLand is invariant by the mapFa. The
map
h:t−→
t,mta t−m , t,
mta
t−m , . . . ,
mta
t−m , t
:m,∞−→m,∞q ⊂R∗q 6.2
is a parametrization ofHa, which has the property that
Faht h
mta
t−m
, Fa
h
mta
t−m
ht. 6.3
Moreover, ifK > Ka,Hacuts the manifoldΣaKat exactly two pointsPa1KandPa2Kwhich
are exchanged byFa, their parameters aret1andt2 mt1a/t1−mwithm < t1< < t2.
Proof. Theqequations satisfied by a pointM0 uq−1, uq−2, . . . , u0for being 2-periodic are
u0 u2 · · · uq−1 , u1 u3 · · ·uq andauq· · ·u2 u1uq−1. If we putu0 xq t
andu1xq−1s, we obtaint amts/s, ors mta/t−m, that is,6.1and the
Now we put, fort > m,
φmat mta
t−m , 6.4
which is an involution with fixed point. We can writeht t, φmat, t, φmat, . . . , t, and hence we have hφm
at φamt, t, φmat, . . . , φamt Faht, for ht is 2-periodic. We
have alsoFahφm
at Fa◦Faht ht.
At last it is to find the points whereHa cutsΣaK. First, we change the parameter
for Ha; we have m
√
m2a and we put b : √m2a − m; so we
take the new parameter v : t − m/b. Hence, v 1 ⇔ t , and if ψ : φm
a − m/b, then ψv 1/v. So we put the values t bv m for x1, x3, . . . , xq
and φm
at m b/v for x2, x4, . . . , xq−1 in the equation Gax K of ΣaK. Some
easy calculations using the relation a m2 b2 for simplifying give the equation
1/vbm1v2 m12b2vbm1m1/bmv2 m2b2vbmm K. But
the two quadratic polynomials in vare reciprocal, and so we put X v1/v ≥ 2 and
X 2⇔v1 ⇔t. Then we definef :2,∞→ RbyfX: m1lnbm1X
m12b2−mlnbmXm2b2. We havefX bm12/bm1X m12b2−
bm2/bmXm2b2 γm1−γm, where the functionm→γm bm2/bmXm2b2
is obviously increasing. So we havefX > 0, andf is increasing, withf∞ ∞. But f2 lnbm12m2/bm2mln1q1/q−1 lnKa, and so there is a unique
solutionX >2 to the equationfX lnKifK > Ka. This solution gives two solutionsv
to the equationv1/vX, 0< v1<1< v2. And this gives two solutionsm < t1< < t2, and
from the invariance ofHaandΣaKbyFawe havet2 φmat1.
6.2. The General Case
q
≥
2
We start with a formula easy to prove.
Lemma 6.2. Ifunnis a solution of1.1andKGauq−1, . . . , u1, u0, one has the relation
∀n≥0
q
i0
unia1K q−1
i1
uni
1uni
. 6.5
We deduce from6.5the possibility forq1to be a period of some solutions of 1.1.
Proposition 6.3. The onlyq1-periodic solutions of1.1are the following:
iifqis even, the equilibriumL;
iiifqis odd, the 2-periodic solutions ofTheorem 5.4.
Proof. Ifunnisq1-periodic, the left-hand member of6.5is constant whennvaries, and sovn :
q−1
i1uni/1uniis also constant; so we havevn1/vn1 unq/1unq1
un1/un1, and then for alln un1 unq, that is,unnisq−1-periodic. Ifpis the minimal
period of the sequenceun,pdividesq1 andq−1, and thenp1 orp2. Ifpis odd, then
2, and thenqis odd; in this case,un is a 2-periodic solution ofTheorem 5.4which may be constant.
As a consequence, we see that order 2 Lyness’ equation has no solution with minimal period 3, which was already known. For order 3 equation, consequences are in the following.
6.3. The Case
q
3
Proposition 6.4. (1) The only 4-periodic solutions of order 3 Lyness’ equation are 2-periodic, and
given byTheorem 5.4.
(2) Order 3 Lyness’ equation has no solution with minimal period 3. (3) Order 3 Lyness’ equation has no solution with minimal period 5.
Remark that in13authors find an infinity of starting points which are 5-periodic for order 3 Lyness’ equation, but in13this equation is studied inRand not, as we make here, inR∗: their 5-periodic solutions take negative values. InSection 7, we will find again points 2and3ofProposition 6.4.
Proof. 1This point is an immediate consequence ofProposition 6.3.
2We putu0 x, u1 y, u2z, and for finding 3-periodic solutions we must write
the relationsu3x, u4y, u5z, that is,x2ayz, y2azx,and z2axy. By
substracting we gety−xxy1 0, and thenxy. In the same way we obtainyz, and so the solution is the equilibrium, whose minimal period is 1.
3We put alsou3s, u4 t. We have to write three conditions:u5u0, u6 u1, u7
u2, and the definition ofsandt. We obtainixzast, iiysatx, iiiztaxy,
ivsxayz, andvtyasz. Fromi,ii, andivwe obtainvizx2−y−1
y1ay−x2. Relationiiwheresandtare replaced by their value obtained fromiv
andvgivesviizy2−x−1 xyax axay−y2ay. In the same way, relation
igivesviiiz y1axya/x1xy−y−1, for the denominator cannot be zero because the numerator is positive. The difference of equationsviandviigives, after factorization,x−yzxy1 y1axy 0. So we havex−y0.
In this caseviiibecomesix z axxa/x2−x−1. The difference ofiand
vgivesxz−t t−z, that is,zt. Soiiibecomesz2a2x, orx z2−a/2. We put this
value ofxinixand obtain, after factorization, the relationx z2−2z−az32z2−a−
2z−2a−1 0. Butz2> a, so the second factor is greater thanaz2−a−2z−2a−1
2z2 >2, and we havez2−2z−a0, that is,z. But the two relationsz2 2xaand
z2 2zaimply the relationxz, and soxyz: the only 5-periodic solution is the
pointL.
6.4. Eigenvalues of
dF
aL
and the Possible Common Periods of
Solutions of
1.1
The question is for which values ofa > 0 andq ≥ 2 have all solutions of1.1a common periodp? What are these common periods? In the case where such a common period exists, one say that1.1is “globally periodic” or “globallyp-periodic”.
Proposition 6.5. There is no periodpcommon to all solutions of orderqLyness’ equation1.1with parametera >0, except the casesq2,a1(withp5), andq3,a1(withp8).
The proof of14is difficult. For the casea1, there is in15a nice short proof, which uses the characteristic polynomial ofdF1L∗, withL∗ ∗, . . . , ∗,∗being the negative root
ofX2−q−1X−10. Unfortunately this proof does not work fora /1.
For every value ofa >0, the existence of a common periodpto all positive solutions of orderqLyness’ equation would imply that the characteristic polynomialBqλ:λq1−
1/λq−1· · ·λ2λwould be afactorof the polynomialAp:λp−1λp−2· · ·λ1. Indeed, the relationFapIdonR∗q; gives obviously the equalitydFaLpIdonRq, so eigenvalues
ofdFaLarepth-roots of unity or equal to 1, and by Lemmas5.1,5.2, and5.3, roots ofBqare all distinct. SoProposition 6.5is related to the following questionwhere in factz1/.
Question 1. Let integersq≥2,p≥2, and a numberz∈0,1/q−1. Can it happen that
Bqλ λq−zλq−1· · ·λ1 6.6
were a factor of the polynomial
Apλ λp−1λp−2· · ·λ1? 6.7
Of course a negative answer toQuestion 1whenq≥4, orq3 anda /1, orq2 and
a /1, would proveProposition 6.5, because it is known that fora1 andq2 orq31.1
is globallyp-periodic, withp5 or 8.
But we can have a positive answer toQuestion 1: somea, p, andqfor whichBqdivides
Ap, that is, all eigenvalues ofdFaLarepth root of unity, but without1.1being globallyp
-periodic, such cases are given by the following proposition.
Proposition 6.6. (1) Ifq 2, for everyp ≥ 13there is a number (perhaps not unique) zp,0 <
zp<1(and then anap), such that the polynomialB2 λ2−zpλ1is a factor ofAp; that is,
eigenvalues ofdFapLarepthroot of unity.
(2) Ifq3, for every evenp≥20there is a number (perhaps not unique)zp,0< zp<1/2
(and so anap), such that the polynomialB3λ3−zpλ2λ 1 λ1λ2−zp 1λ1
is a factor ofAp; that is, eigenvalues ofdFapLarepthroot of unity.
Proof ofProposition 6.6. 1If polynomialλ2−zλ1 dividesAp, then it isλ2−2 cos2kπ/pλ1
for some integerksatisfying 1≤k < p/4. So we havez2 cos2kπ/panda 1−z/z2,
with only the constraint that 0< z < 1 anda > 0, that is, 1 ≤k < p/4 andk > p/6. This is possible ifp5,9,10,11, and ifp≥13. This last case gives the first point ofProposition 6.5.
2If polynomialλ1λ2−z1λ1dividesA
p, thenp 2mis even, and the
roots ofλ2−z1λ1 arepthroots of unity, and then this polynomial has the formλ2−
2 coskπ/mλ1, for some integerksatisfying 1≤k≤m−1. So we havez2 coskπ/m−1 anda 1−2z/z2. The only constraints area >0 and 11/2> z1>1, that is, 1≤k < m/3
and k > mθ0 whereθ0 1/πcos−13/4 0.230. But if m > 1/1/3−θ0 9.68, there
is an integerk ≥ 1 satisfying the constraints. So ifp 2m ≥ 20, we have the point2 of
Example 6.7. 1Ifq2 andp24, we find that for the equationun2 2
√
3−2√3
un1/un, the eigenvalues ofdFaLare 24th roots of unity, but 24 is not a common period to
all solutions.
2 For p 30, we find the same fact for 1.1 with q 3 and a30 3 − 4 cos4π/15/2 cos4π/15−12.
We make the following conjecture.
Conjecture 8. If q ≥ 4, the polynomial Bq does not divide the polynomial Ap for 0 < z <
1/q−1 and then for all a >01.1is not globally periodic ifq≥4.
7. Order 3 Lyness’ Difference Equation
Before to study this case, we give general results for oddq.
7.1. Invariants for
F
2a
When
q
2
m
1
Is Odd
Ifqis odd, we have nice invariants forF2
a.
Proposition 7.1. Whenq2m1is odd, the quantities
R0x1, . . . , xq
: 1x11x3· · ·1x2m1
x2x4· · ·x2m ,
7.1
R1ax1, . . . , xq
: 1x21x4· · ·1x2max1x2· · ·x2mx2m1
x1x3· · ·x2m1
7.2
are exchanged byFa:
R0◦FaR1a, R1a◦FaR0 7.3
and then are invariant under the action ofF2
a.
Moreover one has the relations
R0R1aGa, if q≥5 R0R1aJa. 7.4
Proof. Relations7.3come from and easy calculation, and relations7.4are obvious from the definitions ofGaandJa.
Remark 7.2. Relations7.4give simple proof of the invariance ofGaandJaunder the action
ofFaand also of every symmetric function ofR0andR1a.
Proposition 7.1has significant consequences in the caseq3.
7.2. The Case
q
3
This case was intensively studied in the nice paper 4, where the authors proved, for the first time, that there are two curves inR∗3invariant by the mapF2
circles, such that the restriction ofF2
ato these curves is conjugated to rotations on the circles,
and that these curves are exchanged by the mapFa. The authors gave also some interessant
results about possible periods forF2
a of some starting points for some values of parametera
and about initial points whose orbits are dense in the invariant curves.
Authors of16have found a remarkable property of even termsu2n and odd terms u2n1of the order 3 Lyness’ sequence: they are solutions of two-order 2 difference equations
related to two elliptic quartics in the plane. In4, the authors make the conjecture, in a note added in proof, that theirdifficultpaper would be simpler with the aid of16. We will see that in fact this is right, at least partially, with the aid of invariantsR0 andR1
aofFa2 these
invariants are implicit in16for the caseq3. In the caseq3 these invariants become
R0x, y, z 1x1z
y , R
1
a
x, y, z
1yaxyz
xz . 7.5
If we denote, as in16,
c0:R0M0, c1:R1aM0, 7.6
the orbit of the starting pointM0under the action ofFa2is on the curveDM0defined by the
two equations
1x1z−c0y0,
1yaxyz−c1xz0. 7.7
These two surfaces are just quadratic surfaces“quadrics”Q0 andQ1, and thenDM0is
merely a quartic curve. In fact, this curve is included in the curveCM0given in3.18.
Theorem 7.3. (1) In addition to the first relation7.5one has an expression ofHain terms ofR0
andR1
a, and so one has:
GaR0R1a, HaR0R1aR0R1a2−a. 7.8
(2) The polynomial equationsgx, y, z 0andhx, y, z 0ofΣaKand ofSaMgiven
by formulas2.5and3.11(whereKGaM0andMHaM0) are polynomial combinations
of the two equations of the curveDM0given in7.7:
2gx, y, zAx, y, z(1x1z−c0y
)
Bx, y, z(1yaxyz−c1xz
)
,
7.9
whereAx, y, z c1xz 1yaxyzandBx, y, z 1x1z c0y, and
hx, y, zCx, y, z(1x1z−c0y
)
Dx, y, z(1yaxyz−c1xz
)
,
7.10
(3) One has the inclusion
DM0∪FaDM0⊂ C, 7.11
and the partDM0ofDM0which is inR
3
∗ is compact.
Proof. 1The first relation 7.8is in7.4, and an easy but tedious calculation proves the second relation.
2The proofs of the two announced expressions for 2gandhcome from simple but also tedious calculations using the two relationsKc0c1andMc0c1c0c12−a, which
come from relations7.8applied to the pointM0.
3The inclusion7.11and then the announced compacity are obvious consequences of point2and of the compacity ofCM0 R
3
∗ ∩ CM0.
Remark 7.4. 1The second formula7.8gives, forq 3, an easy proof of the invariance of
Haunder the action ofFa.
2It is not the case forq≥5, because the second relation7.8has no corresponding one: we know from5thatGa,Ha, andJaare independent and so cannot be all three regular
functions ofR0andR1
a.
We will see that significant properties of mapsF2
aandFacan be deduce only with the
use of invariantsR0andR1
a. We start with the identification ofFaDM0.
Lemma 7.5. (1) If DM0 has equations R0x, y, z c0 and R1ax, y, z c1, then the curve
FaDM0is a quartic curve whose equations are
R0x, y, zc1, R1a
x, y, zc0. 7.12
These two curves are disjoint if and only ifM0does not belong to the surfaceWaofR∗3with equation
ωx, y, z:R0x, y, z−R1ax, y, z0, 7.13
that is,
ωx, y, z:xz1x1z−y3−y21axz−yaxz 0, 7.14
whereωx, y, z xyz ωx, y, z.
(2) The two parts ofR∗3\ Wa, denoted by
W
a
R0> R1a, W−aR0< R1a, 7.15
are two open sets whose union is dense in R∗3, which are exchanged byFa, and so both globally invariant under the action ofF2
a. IfM0/∈ Wa, one has
DM
0⊂ Wa⇐⇒c0> c1⇐⇒FaDM0⊂ W−a,
DM
0⊂ W−a⇐⇒c0< c1⇐⇒FaDM0⊂ Wa.
(3) The pointLlies onWa, andWacuts the diagonal inR∗3only atL.
(4) One hasWa∩ Ha∩R∗3{L}, whereHais the hyperbola of 2-periodic points.
(5) In fine ,Wais the graph of a continuous function φ : x, z → φx, zonR∗2, whose epigraph isW−aand hypograph isWa, which so are connected sets.
Proof. One has M ∈ FaDM0 ⇐⇒ R0Fa−1M c0 and R1aFa−1M c1. With the
expression F−1
a x, y, z y, z,a y z/x this last condition is easily translated as R1
ax, y, z c0 and R0x, y, z c1. Now conditions7.13or 7.14are obvious, and the
conditionsDM0⊂ WaandDM0⊂ W−aalso.
For the point3, the equationωt, t, t 0 ist2−2t−a0, whose only positive solution
is.
Fort >1 the equationωt,ta/t−1, t 0 ist2−2t−at4t3at2 3a−2t
a−12 0. The polynome of degree 4 intis a degree 2 polynomial ina, which is positive fort >1 because its discriminant is−tt−123t4. So the only solution ist.
In finethe last assertion is obvious by the study of the functiony→ωx, y, z.
Remark 7.6. 1 In4 the surfaceWa with 7.14is obtained as part of the locus of points
where the gradients ofGaandHaare parallelthe other points are these ones of the hyperbola Ha.
2 Points 1, 2, and 3 of Lemma 7.5 are obviously true for every oddq, from
Proposition 7.10. Some easy calculations show that the point 4 is also true for an odd dimensionq.
3The curvesDM0andFaDM0are not disjoint inP3; they have in common four
points at infinity:0,0,1,0,0,1,−1,0,1,0,0,0, and1,−1,0,0.
From the fact that ifM0/∈ Wa,DM0∩FaDM0 ∅, the following result is obvious
and has an obvious generalization to orderqLyness’ equation whenqis odd, which is in 4and gives again points2and3ofProposition 6.4.
Corollary 7.7. No point ofR∗3\ Wais periodic under the action ofFawith an odd period.
Now we can assert the principal qualitative result on the mapF2
a. ViaProposition 7.13
below we find again a result of4, but it comes in this last paper from new and interessant, but sophisticated, arguments, whereas we use here only previous results of17.
Theorem 7.8. Let M0/∈ Wa ∪ Ha. Then the two quartics DM0 and FaDM0 are disjoint,
exchanged by Fa, both globally invariant under the action of F2
a. The positive and compact parts
DM
0and FaDM0 are both homeomorphic to the circle, and the restrictions ofFa2 to each
of these two curves are conjugated to rotations on the circle, with the same angle.
Proof. We can suppose thatM0∈ Wa, that is,c0 > c1, the proof is the same in the other case.
1Letπbe the projection on thex, z-plane, and put
E0M0:πDM0. 7.17
First we remark thatπis injective on the surfaceR0x, y, z c
0, because it is the graph of the
functionx, z→y 1x1z/c0. So the projectionπis ahomeomorphismof the compact
curveDM0onto its projectionE0M0 E0M0∩R∗2.
2We search the conjugated of the restriction ofF2
atoDM0by the projectionπ. We
have easilyF2
we havey 1x1z/c0, and so we obtainafter simplification by the factor 1zthat
ifx, z∈E0M0, then
π F2aπ−1x, z!
1c0a 1c0x
z1x , x
. 7.18
So we conclude that the restriction ofF2
atoDM0is conjugated byπto the restriction
toE0M0of the map
Tac0:R∗2−→R∗2:x, z−→
1c0a 1c0x
z1x , x
. 7.19
3 But this mapTac0in R∗2 is studied in17. We know from this papersee its
introduction and part 4: “The homographic case”that there is an invariant of this map and of the associated difference equation xn2xn 1c0a 1c0xn1/1xn1. More
precisely, put
p01c0a, q01c0; 7.20
then one has
p0q0x
1x
p0q0x
q0x
1xq0x
a bxcx2
c dxx2 , 7.21
a p0q0, b p0q02, c q0, d q01. 7.22
Then we see in17that the quantity
V0x, z:
1
xz
x2z2dxzxz cx2z2bxz a 7.23
is invariant under the action of Tac0. So the curves Γk with equations V0x, z k are
invariant byTac0, and it is proved in17that these curves are, ifk > km, elliptic quartics,
which are homeomorphic to circles, and that the restriction ofTac0toΓkis conjugated to a
rotation. So the theorem will be proved if we show that the curveE0M0is a curveΓk, for a ksatisfyingk > km, that is, ifΓkdoes not reduce to the fixed point ofTac0.
So we will determine the equation of