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On the distance between separatrices for the discretized pendulum equation
Hocine Sellama
To cite this version:
Hocine Sellama. On the distance between separatrices for the discretized pendulum equation. 2008.
�hal-00287701�
disretized pendulum equation
IRMA -UMR7501CNRS/ULP
7rueRenéDesartes-67084StrasbourgCedex, Frane
email: sellamamath.u-strasbg.fr
abstrat
Weonsiderthedisretization
q(t+ε) +q(t−ε)−2q(t) =ε2sin q(t) ,
ε > 0 asmall parameter, of thependulum equation q′′ = sin(q); in system
form,wehavethedisretization
q(t+ε)−q(t) =εp(t+ε), p(t+ε)−p(t) =εsin q(t) .
ofthesystem
q′=p, p′= sin(q).
Thelattersystemofordinarydierentialequationshastwosaddlepoints
atA= (0,0),B = (2π,0)andnearboth,thereexiststableandunstableman-
ifolds. Italsoadmits aheteroliniorbitonnetingthestationary pointsB
andA parametrisedbyq0(t) = 4 arctan e−t
andwhihontainsthestable
manifoldofthissystematAaswellasitsunstablemanifoldatB. Weprove
thatthestablemanifoldofthepointAandtheunstablemanifoldofthepoint B do notoinide for the disretization. More preisely, we show that the vertialdistanebetweenthesetwomanifoldsisexponentiallysmallbut not
zeroand in partiular wegiveanasymptoti estimateof this distane. For
this purpose we usea method adapted from the artileof Shäfke-Volkmer
[10℄usingformalseriesandaurateestimatesoftheoeients. Ourresult
issimilartothatofLazutkinet.al.[9℄;ourmethodofproof,however,isquite
dierent.
Keywords: Dierene equation; Manifolds; Linear operator; Formal solu-
tion;Gevreyasymptoti;Quasi-solution
1 Introdution
We onsiderthe following dierene equation
q(t + ε) + q(t − ε) − 2q(t) = ε
2sin q(t)
.
(1.1)This seond order equation is a disretization of the pendulum equation
q
′′= sin(q)
. It isequivalent to the following systemof rstorder diereneequations(
q(t + ε) = q(t) + εp(t + ε), p(t + ε) = p(t) + ε sin q(t)
.
(1.2)
whih an beonsidered asadisretization ofthesystem
(
q
′= p, p
′= sin(q).
(1.3)
Thelattersystemhastwo saddlepointsat
A = (0, 0)
,B = (2π, 0)
and thereexiststableand unstable manifolds. For the disretized equation (1.2) and suiently
small
ε > 0
,these manifoldsstill exist.The system (1.3) has
q
0(t), q
0′(t)
, where
q
0(t) = 4 arctan e
−t,as a hetero-
lini orbit onneting the stationary points
B
andA
; it is a parametrisation of the urvep = − 2 sin(q/2)
and ontains thestable manifold of (1.3) at the pointA
aswellasits unstablemanifold atB
. Thisurve,together withp = 2 sin(q/2)
,separates regions with periodi orbits from regions with non-periodi orbits and
is therefore often alled a separatrix. Our purpose is study the behavior of this
separatrixunderdisretizationoftheequationitturnsoutthatthereisnolonger
aheterolini orbit for system (1.2) and its thestable manifoldat
A
and the un-stablemanifold at
B
no longer oinide. More preisely, we want to estimate thedistanebetween thestablemanifold
W
s,ε− of(1.2)atA
andtheunstablemanifoldW
u,ε+ of(1.2) atB
asa funtionof the parameterε
.Lazutkinet.al.[9 ℄,Gelfreih[4℄,(seealsoLazutkin[7℄[8℄) hadgivenanasymp-
totiestimateofthesplittinganglebetween themanifolds. Startingfromahetero-
linisolution of the dierential equation, they study thebehaviorof analyti so-
lutionsofthediereneequationintheneighbourhoodofitssingularities
t = ±
π2i
.We showthatthe distanebetween these twomanifolds isexponentiallysmall
but not zero and we give an asymptoti estimate of this distane. This result
is similar to that of Lazutkin et. al. [9℄; our method of proof, however, is quite
dierent.
Weuseamethodadaptedfromtheartile ofShäfke-Volkmer [10 ℄usingafor-
malpowerseries solutionand aurate estimates of theoeients. Thismethod
wasadapted for thelogisti equation in Sellama[11℄. It turns out that theadap-
tationof thismethodfor thependulum equationismorediult thaninthease
ofthe logistiequation.
We will show
Theorem1.1. Givenany positive
t
0,it isknownthat for suientlysmalε
0> 0
andall
t ∈ ] − ∞ , t
0]
there is exatly oneone pointw
u,ε+(t) = (q
0(t), p ˜
+u,ε(t))
on thestableunstable manifoldhavingrst oordinate
q
0(t)
. There existonstantsα 6 = 0
,suhthat for any positive
t
0distv
w
s,ε+(t), W
s,ε−= 4πα
ε
2cosh(t) sin 2πt ε
e
−π2 ε
+ O
1 ε e
−π2 ε
,
asε ց 0,
uniformly for
− t
0< t < t
0,where distvP, W
u,ε−denotesthe vertial distane of a
point
P
from the unstable manifoldsW
u,ε−.ThisresultorrespondstotheresultofLazutkinet.al.[9℄astheanglebetween
themanifolds at anintersetionpoint isasymptotially equivalent to
1 q′0(t)
d
dt distv
w
+s,ε(t), W
u,ε−,but we do not want to give anydetail here.
Our proof uses the following steps. First, we onstrut a formal solution for
the dierene equation (1.1) in the form of a power series in
d = 2
arsinh(ε/2)
,whose oeients are polynomials in
u = tanh(dt/ε)
. This is done in setion 2;theintrodution of
d
is neessarybeause polynomials aredesired asoeients.Then, we give asymptoti approximations of these oeients using appropriate
norms onspaesof polynomials. To thatpurposeweintrodueoperatorsonpoly-
nomials series. In setion 6 we use the trunated Laplae transform to onstrut
a funtion whih satises(1.1) exeptfor an exponentially small error. The next
andlaststepistogiveanasymptotiestimateforthedistaneofsomepointofthe
stablemanifoldfromthe unstablemanifold. Aalulation showsthat
α = 89.0334
and therefore
4πα = 1118.8267
(See Remark5.4); theorresponding onstantsof Lazutkin have already been alulated with high preision (See Lazutkin et. al.[9℄). A proof that
α 6 = 0
asin [10℄ or [11 ℄ would be possible. Y.B. Suris [12 ℄ hadshownthat
α 6 = 0
.2 Formal solutions
The purpose of this setion is to nd a onvenient formal solution for equation
(1.1). First,we need some preparations. Weput
u : = tanh d
ε t
, q
0d(t) : = 4 arctan exp
− d ε t
!
, q
d(t) =
p1 − u
2A
d(u) + q
0d(t), A
d(u) =
∞
X
n=1
A
n(u)d
nfor a formalsolution of (1.1), where
d = ε +
P∞n=3
d
nε
n is aformal powers seriesin
ε
to be determined.Remark. Thelinearization ofequation(1.1)atthepoint
A
givesthefollowingequation
Z(t + ε) + Z(t − ε) − 2Z (t) = ε
2Z(t).
The parameter
d
issuh thatZ(t) = e
−dt is a solution ofthis equation, thereforeε
andd
are oupledbytherelationd = 2
arsinh(ε/2).
q
0d(t + ε) + q
0d(t − ε) − 2q
0d(t) = 2
+∞
X
n=1
1
(2n) ! q
0d(2n)(t) ε
2n,
(2.1)where
2
(2n)! q
(2n)0d(t)ε
2n/d
2n is anoddpolynomialI
2n−1(u)
multiplied by√ 1 − u
2;we nd
I
2n−1(1) = 4/(2n)!
.Using
cos(q
0d) = 2u
2− 1, sin(q
0d) = 2u √
1 − u
2,we an express our equation(1.1)in theform
A
d(T
+)
r
1 − (T
+)
21 − u
2+ A
d(T
−)
r
1 − (T
−)
21 − u
2− 2A
d(u) = f ε, u, A
d(u)
(2.2)
or equivalently
A
d(T
+)
cosh(d) + u sinh(d) + A
d(T
−)
cosh(d) − u sinh(d) − 2A
d(u) = f ε, u, A
d(u)
(2.3)
where
f ε, u, A
d(u)
= ε
22u cos
A
d(u)
p1 − u
2+ 2u
2− 1
√ 1 − u
2sin
A
d(u)
p1 − u
2−
+∞
X
n=1
I
2n−1(u) d
2n,
T
+= T
+(d, u) = u + tanh(d)
1 + u tanh(d) = tanh d
ε (t + ε) , T
−= T
−(d, u) = u − tanh(d)
1 − u tanh(d) = tanh d
ε (t − ε) .
As
u → 1,
the expressionsT
+ andT
− redue to 1, the denominators in (2.3) simplifytoe
±dand heneequation (2.3)redues to(e
−d+ e
d− 2)A
d(1) = ε
2(2 + A
d(1)) − 4(cosh(d) − 1).
This is equivalent to
(2 cosh(d) − 2 − ε
2)(2 + A
d(1)) = 0
and hene we haveneessarily
ε = 2 sinh(d/2)
ifwe want a formalsolution suh that theoeientshave limitsas
u → 1
.Theorem 2.1. (On the formal solution) If
ε = 2 sinh(d/2)
, then equation(2.2)has a unique formalsolution of the form
A
d(u) =
+∞
X
n=1
A
2n−1(u)d
2n,
(2.4)where
A
2n−1(u)
are oddpolynomials of degree≤ 2n − 1
.Proof. We will use theIndution Priniple to show that there exist unique odd
polynomials
A
1, A
3, A
5...A
2n−1 suhthatZ
n(d, u) =
n
X
k=1
A
2k−1(u)d
2k (2.5)satisfy
R
n(d, u) = O(d
2n+4)
(2.6)where
R
n(d, u) = Z
n,dT
+r
1 − (T
+)
21 − u
2+ Z
n,dT
−r
1 − (T
−)
21 − u
2− 2Z
n,d(u) − f ε, u, Z
n,d(u)
(2.7)
For
n = 1
,a short alulation shows that we must haveA
1(u) = −
14u
and heneZ
1,d(u) = −
14ud
2. We obtainR
1(d, u) = ( − 91
48 u
5+ 137
48 u
3− 23
24 )d
6+ O(d
8).
Suppose now thatthereexists
A
1, A
3, A
5...A
2n−1 suhthatZ
n(d, u) =
n
X
k=1
A
2k−1(u)d
2k (2.8)satises (2.6), (2.7). We show that there is a unique polynomial
A
2n+1(u)
suhthat
Z
n+1(d, u) = Z
n(d, u) + A
2n+1(u)d
2n+2 (2.9)satises(2.6). We put
R
n(d, u) = R
2n+3(u)d
2n+4+ O d
2n+6(2.10)
where
R
2n+3(u)
isoddand deg(R
2n+3(u)) ≤ 2n + 3
.We substitute
Z
n+1(d, u)
in equation (2.7). Using Taylor expansion, (2.9),(2.10) and
ε = 2 sinh(d/2)
, we obtainZ
n+1,dT
+r
1 − (T
+)
21 − u
2− Z
n+1,dT
−r
1 − (T
−)
21 − u
2− 2Z
n+1,d(u) − f ε, u, Z
n+1,d=
(u
4− 2u
2+ 1)A
′′2n+1(u) + (4u
3− 4u)A
′2n+1(u) + R
2n+3(u)
d
2n+4+ O d
2n+6(1 − u
2)
2A
′2n+1(u)
′+ R
2n+3(u) = 0
(2.11)This dierential equation has a unique solution vanishing at
u = 0
withoutsingularityat
u = 1
,namelyA
2n+1(u) = −
Z u0
Rt
1
R
2n+3(s)ds
(1 − t
2)
2dt.
(2.12)We now show that this solution is an odd polynomial of
u
. It is lear that Rt1
R
2n+3(s)ds
vanishes fort = 1
and asR
2n+3(s)
is odd, it also vanishes fort = − 1
. It suesto show thatR
2n+3(s)
also vanishesatt = ± 1
. Indeed,takingthe limit of(2.7) as
u → 1
aswe didfor (2.3) and usingu→1
lim f ε, u, Z(d, u)
= ε
2Z(d, 1)
we obtain
R
2n+3(1)d
2n+4= e
d+ e
−d− 2 − ε
2!
Z(d, 1) + O(d
2n+6).
By our hoie of
ε = 2 sinh(d/2)
, we obtainR
2n+3(1)d
2n+4= O(d
2n+6)
. Conse-quently
R
2n+3(1) = 0
. AsR
2n+3(u)
isodd,wealsohaveR
2n+3( − 1) = − R
2n+3(1) = 0
. ThisprovesthatA
2n+1(u)
isan oddpolynomialof degreeA
2n+1(u)
≤ 2n + 1
and
A
2n+1(0) = 0.
Therst polynomials
A
2n−1(u)
withn > 0
an be alulated usingMaple.n 1 2 3
A
2n−1(u) −
14u
91
864
u
3−
57647u
−
2880319u
5+
1152185u
3−
691203703u
Now, wentrodue theoperators
C
2, C , S
2, S
dened byC (Z )(d, u) =
12Z (d, T
+12) + Z(d, T
−12) S (Z)(d, u) =
12Z(d, T
+12) − Z (d, T
−12) C
1(Z)(d, u) =
12Z(d, T
+) + Z(d, T
−) S
1(Z)(d, u) =
12Z(d, T
+) − Z (d, T
−)
(2.13)
where
T
+12= T
+(
d2, u)
,T
−12= T
−(
d2, u)
andZ(d, u)
is aformalpower series ind
whose oeientsarepolynomials. We an showthat
C
1= 2 S
2+ Id
S
1= 2 SC
(2.14)C
1(Q · G) = C
1(Q) C
1(G) + S
1(Q) S
1(G) S
1(Q · G) = C
1(Q) S
1(G) + S
1(Q) C
1(G)) C (Q · G) = C (Q) C (G) + S (Q) S (G) S (Q · G) = C (Q) S (G) + S (Q) C (G)
(2.15)
if
Q, G
areformal powerseries ind
whoseoeients arepolynomials ofu
.3 Norms for polynomials and basis
In this setion we reall some denitions and results of [10℄. Using a ertain
suquene of polynomials. we dene onvenient norms on spaes of polynomials
whihsatisessome useful proprieties. We denoteby
• P
the setof allpolynomial whose oeentsare omplex• P
n the spaes ofall polynomials ofdegree lessthan or equalto nProposition 3.1. [10℄. We dene the sequene of polynomials
τ
n(u)
byτ
0(u) = 1, τ
1(u) = u, τ
n+1(u) = 1
n Dτ
n(u)
forn ≥ 1,
where the operator
D
is dened byD := (1 − u
2) ∂
∂u .
Thenwe have
1.
T
+(d, u) =
P∞n=0
τ
n+1(u)d
n,2.
τ
n(u)
has exatly degreen
and heneτ
0(u), ..., τ
n(u)
form a basisofP
n,3.
τ
n(tanh(z)) =
(n−1) !1 dzdn−1tanh(z)
Denition 3.2. Let
p ∈ P
n. Asτ
0(u), ..., τ
n(u)
form a basisofP
n, we an writep ∈ P
n asp =
n
X
k=0
a
kτ
k(u).
Thenwe dene the norms
k p k
n=
n
X
i=0
| a
i| π 2
n−i
.
(3.1)Theorem3.3. [10℄. Letn,mbepositiveintegers and
p ∈ P
n,q ∈ P
m.
The norms(3.1)have the followingproperties:
1.
k Dp k
n+1≤ n k p k
n.2. If the onstant term of
p
in the basis{ τ
0, τ
1.., τ
n}
iszero, we havek p k
n≤ k Dp k
n+1.
3. There exists a onstant
M
2 suh thatk pq k
n+m≤ M
2k p k
nk q k
m.4. There isa onstant
M
3 suh that that for alln > 1
,| p(u) | ≤ M
3 π2nk p k
n( − 1 ≤ u ≤ 1)
.5. There is a onstant
M
4 suh that for alln > 1
withp(1) = p( − 1) = 0
p τ
2
n−2
≤ M
4k p k
n4 Operators
Inthis setionwewillusesomedenitionsofShäfke-Volkmer[10℄andadapt their
resultson operatorsonpolynomialseries to our ontext. Let
Q :=
(
Q(d, u) =
∞
X
n=0
Q
n(u)d
n,
whereQ
n(u) ∈ P
n,
for alln ∈
N ).
Byabuseof notation, let
k Q k
n= k Q
nk
n for a polynomial seriesQ(d, u) =
∞
X
n=0
Q
n(u)d
n.
Denition 4.1. Let
f
be formalpower series ofz
whose oeients are omplex.We dene a linear operator
f (dD)
onQ
byf (dD)Q(d, u) =
∞
X
n=0
Xn
i=0
f
iD
iQ
n−i(u)
d
n (4.1)where
f (z) =
P∞i=0
f
iz
i andQ ∈ Q
.BytheaboveDenition and (1)ofProposition 3.1we an showthat
Q(d, T
+θd, u)
= exp(θdD)Q
(d, u)
forQ ∈ Q
and allθ ∈
CC (Q) = cosh(
d2D)Q, S (Q) = sinh(
d2D)Q
C
1(Q) = cosh(dD)Q, S
1(Q) = sinh(dD)Q
(4.2)for polynomial series
Q
inQ
.Remark. Aording to the denitionof norms in(3.1),we have
If
Q ∈ Q ,
thendQ ∈ Q
andk dQ k
n= π
2 k Q k
n−1 for alln ≥ 1.
(4.3)Theorem4.2. [10℄Let
f (z)
be formalpowerserieshavingaradiusofonvergenegreater than
2π
and letk be apositve integer. There isa onstantK
suhthat: IfQ
isa polynomialseries havingthe following propertyk Q k
n≤
0
forn < k
M(n − k) !(2π)
−n forn ≥ k
where
M
is independent ofn
andQ ∈ Q
then the polynomial seriesf (dD)Q
satises
k f (dD)Q k
n≤
0
forn < k
M K(n − k) !(2π)
−n forn ≥ k
Nowwedene on
Q
the following operatorJ = S
dD .
(4.4)where the notation
S
dD means simply
F (dD)
withF(z) = 1
z sinh(
z2)
.Lemma 4.3. For eah integer
k
there exist a positive onstantK
suh that: IfQ
is a polynomial series with odd
Q
n of degree at mostn
,k Q k
n= 0
forn < k
inase of positive
k
andk dDQ k
n≤ M(n − k) !(2π)
−n forn ≥ max(0, k),
where
M
is independent ofn
,then the polynomialseriesJ
−1(Q)
satiseskJ
−1(Q) k
n≤ M K(2π)
−n
(n − k + 1) !
fork ≤ 1
(n − 1) ! log(n)
fork = 2
(n − 1) !
fork ≥ 3
Proof. We an see easily that
J
−1= π C ˜
−1+ g(dD)
, whereC ˜ = cosh(
14dD)
andg(z)
isanalyti for| z | < 4π
,and usethe proofof [10 ℄.We have
S = dD J = J dD
, but using this relation for the inversion ofS
wouldgive aninsuient result. Usingof theformula
1 = 2 z sinh( z
2 ) + F (z)z,
whereF(z) = z
−2(z − 2 sinh( z 2 ))
isan entire funtion,we obtaintherelation
Q = 2 J Q + F(dD) dDQ
(4.5)forpolynomialseries
Q ∈ Q
. Thiswill beessential intheproof ofTheorem 4.4. For eah integer
k
there exist a positive onstantK
suh that: IfQ
is a polynomialseries with oddQ
n of degree atmostn
,k Q k
n= 0
forn < k
inase of positive
k
, andkS (Q) k
n≤ M (n − k) !(2π)
−n forn ≥ max(0, k),
where
M
isindependentofn
, then the polynomial seriesQ
satisesk Q k
n≤ M K(2π)
−n
(n − k + 1) !
fork ≤ 1 (n − 1) ! log(n)
fork = 2 (n − 1) !
fork ≥ 3
Proof. Bythepreedingtheorem,wehavethewantedinequalitiesfor
dDQ = J
−1S Q
in theplae ofQ
. Herewe used againk dDZ k
n≤ (n − 1) k Z k
n−1 for anypolynomialseries
Z ∈ Q
. Usingtheorem4.2impliesthesameforF (dD)dDQ
withthe entire funtion
F
of (4.5) Ask Z k
n≤ k dDZ k
n+1 bytheorem 3.3, we nd thewanted inequalities (and even something better in the ases
k ≥ 2
) also forJ Q
beause
dD J = S
. Thus formula(4.5) yields theresultInordertoobtainanasymptotiapproximationfortheoeientsoftheformal
solution, we will need to reverse some operators. This is not possible for the
operators
S
anddD
on the setQ
,but we an dene a subsetQ
∗ ofQ
on whihtheseoperatorshave aright inverses.
Ifwe dene
Q
∗:=
(
Q(d, u) =
∞
X
n=1
P
n(u)d
n,
whereP
n(u) ∈ P
n∗,
for alln ≥ 1
).
where
P
n∗ isthe subspae ofP
n dened byP
n∗:=
n
X
i=0