lecture 3: exponentially small splitting of separatricesm. guardia (upc) lecture 3 february 8, 2017...

Lecture 3: Exponentially small splitting of

separatrices

Marcel Guardia

Universitat Politecnica de Catalunya

February 8, 2017

M. Guardia (UPC) Lecture 3 February 8, 2017 1 / 26

Exponentially small splitting of separatrices

For the RPC3BP we want to prove transversality of the invariant

manifolds for fixed µ ∈ (0,1/2] and G0 ≫ 1.

The distance between the invariant manifolds is exponentially

small in G0.

Melnikov Theory does not apply unless µ≪ e−G3

0

3 .

To prove splitting for any µ ∈ (0,1/2] is a difficult problem since

classical perturbative techniques do not apply.

We have to deal with exponentially small splitting of separatrices.

We show how to deal with this problem in a simpler model.


The perturbed pendulum

Hamiltonian

H

(

y , x ,t

ε

)

=y2

2+ (cos x − 1) + δ(cos x − 1) sin

t

ε

Equations

x = y

y = sin x + δ sin x sint

ε

This is a model of the resonances of nearly completely integrable

Hamiltonian systems.

Tipically in this problems δ is a fixed non small constant.

In the RPC3BP one has: ε ∼ G−30 and δ ∼ µG−2

0 .

Luckily in the RPC3BP ressembles this model with δ ∼ ε2/3.

This makes a big difference for analyzing the splitting.


The perturbed pendulum

From now on:

H

(

y , x ,t

ε

)

=y2

2+ (cos x − 1) + ε(cos x − 1) sin

t

ε

Equations

x = y

y = sin x + ε sin x sint

ε

t = 1

(x , y) = (0,0) is a hyperbolic periodic orbit for this system.

We want to compute the splitting between its invariant manifolds.


First rigorous result dealing with this problem:

Theorem (Neishtadt 84)

Let us fix ε0 > 0. Then, for ε ∈ (0, ε0), there exists an ε-close to the

identity canonical transformation that transforms system

H

(

y , x ,t

ε

)

=y2

2+ (cos x − 1) + H1

(

x , y ,t

ε

)

with H1 analytic into

H

(

y , x ,t

ε

)

=y2

2+ (cos x − 1) + εH0(y , x , ε) + P

(

y , x ,t

ε

)

and P satisfies

|P| ≤ c2e−c1ε

for certain constants c1, c2 > 0.


This theorem implies that:

The system is exponentially small close to integrable.

The invariant manifolds are exponentially close.

It does not give any information about the splitting of the separatrix

We do not know whether they coincide or not.

We need to compute the first order of the distance between the

invariant manifolds.


Plan

Look for suitable parameterizations of the invariant manifolds.

Extend these parameterizations to certain regions of the complex

plane.

Analyze the difference of the parameterizations in these regions.

Deduce from this analysis the first order of the difference in the

reals.

Simplifications of our example: the unperturbed separatrix is a

graph over the base


Parameterizations of the invariant manifolds

We look for parameterizations of the invariant manifolds as graphs

y = ∂xSu,s(x , t/ε)

The generating functions Su,s satisfies the Hamilton-Jacobi

equation.

ε−1∂τS (x , τ) + H (x , ∂x S (x , τ) , τ) = 0

Pendulum example

ε−1∂τS +(∂xS)2

2+ cos x − 1 + ε(cos x − 1) sin τ = 0

Asymptotic conditions:

limx→0

∂xSu (x , t/ε) = 0

limx→2π

∂xSs (x , t/ε) = 0


Reparameterization using the time over the separatrix

Parameterization of the pendulum separatrix:

x0(v) = 4 arctan(ev ), y0(v) =2

cosh v.

Reparametrize x = x0(v), T u,s(v , τ) = Su,s(x0(v), τ)

Equation

ε−1∂τS +(∂xS)2

2+ cos x − 1 + ε(cos x − 1) sin τ = 0

becomes

ε−1∂τT +cosh2 v

8(∂v T )2 −

2

cosh2 v− ε

2

cosh2 vsin τ = 0

Asymptotic conditions

limRe v→−∞

cosh v ∂v T u (v , τ) = 0

limRe v→+∞

cosh v ∂vT s (v , τ) = 0.


New parameterizations


8(∂v T )2 −

2

cosh2 v− ε

2

cosh2 vsin τ = 0

Parameterizations of the form

x = 4 arctan(ev ), y =cosh v

2∂v T u,s (v , τ) .

For the unperturbed pendulum: ∂v T0(v) =4

cosh2 v.

We can consider a formal power series expansion

T u,s(v , τ) = T0(v) +∑

k≥1

εk Tu,sk (v , τ)


Formal power series

We have: T uk (v , τ) = T s

k (v , τ) ∀k ∈ N

Namely: T u(v , τ) − T s(v , τ) = O(εk ) ∀k ∈ N

Proceeding formally, their difference is beyond all orders.

What is happening?

1 The power series is convergent: both manifolds coincide in theperturbed case

2 The power series in ε is divergent: the manifolds do not coincide

and their difference is flat with respect ε.

We will see that is happening the second option


New parameterizations


8(∂v T )2 −

2

cosh2 v− ε

2

cosh2 vsin τ = 0

Unperturbed pendulum: ∂vT0(v) =4

cosh2 v.

Perturbed pendulum: T u,s = T0 + T u,s.

Parameterizations of the invariant manifolds:

x = 4 arctan(ev )

y =2

cosh v+

cosh v

2∂vT

u,s (v , τ)

To study the difference between manifolds: analyze T u − T s.



8(∂v T )2 −

2

cosh2 v− ε

2

cosh2 vsin τ = 0

Plug T u,s = T0 + T u,s into the Hamilton-Jacobi equation.

Recall ∂vT0(v) =4

cosh2 v.

New equation:

ε−1∂τT + ∂vT +cosh2 v

8(∂vT )2 − ε

2

cosh2 vsin τ = 0

T u,s are expected to be small.

We can solve this equation by inverting the linear differential

operator.


The integral operators

Inverses of L0 = ε−1∂τ + ∂v .

For functions which decay to zero as Re v −→ −∞,

Gu(h)(v , τ) =

∫ 0

−∞

h(v + t , τ + ε−1t)dt .

For functions which decay to zero as Re v −→ +∞,

Gs(h)(v , τ) =

∫ 0

+∞

h(v + t , τ + ε−1t)dt .


The fixed point equation

T u is a solution of

T u = Gu

(

−cosh2 v

8(∂vT

u)2 + ε2

cosh2 vsin τ

)

with

Gu(h)(v , τ) =

∫ 0

−∞

h(v + t , τ + ε−1t)dt .

Look for fixed points of

Fu(T ) = Gu

(

−cosh2 v

8(∂vT )2 + ε

2

cosh2 vsin τ

)

T s is a fixed point of

Fs(T ) = Gs

(

−cosh2 v

8(∂vT )2 + ε

2

cosh2 vsin τ

)


The fixed point argument in the reals

T u,s = Fu,s(T u,s) = Fu,s(0) +O(ε2).

Fu(0) = ε

∫ 0

−∞

2

cosh2(v + t)sin(

τ + ε−1t)

dt

Fs(0) = ε

∫ 0

+∞

2

cosh2(v + t)sin(

τ + ε−1t)

dt

The difference: T u − T s = Fu(0)−Fs(0) +O(ε2).

Difference between first iteration is the Melnikov potential

Fu(0)−Fs(0) = ε

∫ +∞

−∞

2

cosh2(v + t)sin(

τ + ε−1t)

dt .

We have recovered Melnikov theory through the Hamilton-Jacobi

equation.


First order (Melnikov potential):

Fu(0)−Fs(0) = 4πe−π2ε sin(τ − ε−1v)

Melnikov is exponentially small and the error is O(ε2).

We need better estimates for the error.

We consider the fixed point equation for complex values of v ∈ C.

To compute the Melnikov function, we have analyzed the

singularities of the Melnikov integrand at v = ±iπ

2.

We extend the parameterizations of the invariant manifolds up to a

distance O(ε) of these singularities.


Complex domains:

In τ : Tσ = {τ ∈ C/(2πZ) : |Im τ | ≤ σ}.

In v :


Close to the singularities of the unperturbed separatrix

By the fixed point argument T u,s are defined in these complex

domains and satisfy:

|T u,s(v , τ)| .ε2

(v ± iπ/2)2.

Unperturbed separatrix

T0(v) ∼1

v ± iπ/2

Both become larger close to the singularity.

T0 is always larger than T u,s (otherwise the fixed point argument

would not work!).


Difference between manifolds

Parameterizations of the invariant manifolds:

x = 4 arctan(ev )

y =2

cosh v+

cosh2 v

2∂vT

u (v , τ)

We study ∆ = T u − T s.

Subtract the equation

ε−1∂τT + ∂vT +cosh2 v

8(∂vT )2 − ε

2

cosh2 vsin τ = 0

for T u,s.

∆ satisfies L∆ = 0 for the linear operator

L = ε−1∂τ + ∂v +cosh2 v

8

(

∂vTu(v , τ) + ∂vT

s(v , τ))

∂v

L is close to L0 = ε−1∂τ + ∂v .


Difference between manifolds

Assume that ∆ ∈ KerL0, L0 = ε−1∂τ + ∂v .

Main idea: functions in KerL0 bounded in a complex strip are

exponentially small in the reals.

Proposition

Let ψ(v , τ) ∈ KerL0 be an analytic in τ function in [−ir , ir ] × Tσ. Then,

ψ can be extended analytically to {|Im v | ≤ r} × Tσ

Define M = max(v ,τ)∈[−ir ,ir ]×Tσ

|∂vψ(v , τ)|

Then, for ε≪ 1,

∀(v , τ) ∈ R× T, |∂vψ(v , τ)| ≤ 2Me−ε−1r


Why?

ε−1∂τψ + ∂vψ = 0 implies ψ(v , τ) = Λ(τ − ε−1v) for some Λ.

We can extend analytically ψ to {|Im v | ≤ r} × Tσ.

Let us take and example ψ(v , τ) = A sin(τ − ε−1v) for some

constant A > 0.

We have defined M = max(v ,τ)∈[−ir ,ir ]×Tσ

|∂vψ(v , τ)|.

Take (v , τ) = (ir ,0),

M ≥ Aε−1 cosh(ε−1r).

So A ≤ 2Mεe−rε .

For real values of (v , τ): |ψ(v , τ)| ≤ 2Mεe− rε .


Better estimates for the error

We apply this lemma to ∆− Melnikov = T u − T s − Melnikov:

(Approximately) ∆− Melnikov ∈ KerL0

The fixed point argument gives bounds for ∆− Melnikov for

(v , τ) ∈ [−ir , ir ]× Tσ with r =π

2− ε.

The bounds of ∆− Melnikov are smaller than the size of Melnikov

(in the complex).

We deduce exponentially small bounds for real values of the

variables.


Conclusion

Parameterizations

x = x0(v)

y = y0(v) + y−10 (v)∂vT

u

(

v ,t

ε

)

We obtain

∂v∆(v , τ) = ∂vTu(v , τ) − ∂vT

s(v , τ)

= −4πε−1e−π2ε

(

sin(τ − ε−1v) +O(ε))

.

We have a first order of the difference between the invariant

manifolds.


Splitting of separatrices

We have proved splitting of separatrices for

H

(

y , x ,t

ε

)

=y2

2+ (cos x − 1) + ε(cos x − 1) sin

t

ε

Typically in resonances of nearly integrable systems

H

(

y , x ,t

ε

)

=y2

2+ (cos x − 1) + δ(cos x − 1) sin

t

ε

for fixed δ ∈ R.

This case is much harder: Melnikov does not give the first order.

Analysis of the invariant manifolds close to the singularities gives

a new first order.


Back to the circular problem

One can apply the explained ideas to the circular problem

Then, we have an asymptotic formula for the distance between the

invariant manifolds for any µ ∈ (0,1/2) and G0 ≫ 1

∂vTu(v , ξ)−∂vT

s(v , ξ) = G32

0 e−G3

0

3

(

C(µ) sin(ξ + G30v) +O(G

− 12

0 )

)

where

C(µ) =1

2

√

π

2µ(1 − µ)(1 − 2µ).

This allows us to prove their transversality in this range of

parameters.


lecture 3: exponentially small splitting of separatricesm. guardia (upc) lecture 3 february 8, 2017...

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