Multiscale Phenomena in Geometry and Dynamics Technical University Munich … · 2019-07-24 ·...

Lecture 4: Exponentially small splitting of separatricesMultiscale Phenomena in Geometry and Dynamics

Technical University Munich (TUM)

Tere M. Seara

Universitat Politecnica de Catalunya

22-29 July 2019

Tere M-Seara (UPC) Lecture 4 22-29 July 2019 1 / 33

The problem

The rapidly perturbed pendulum

The Hamiltonian system we want to understand:

H(

y , x ,t

ε

)=

y 2

2+ (cos x − 1) + ε(cos x − 1) sin

t

ε

we will study a slightly more general problem:

Hamiltonian

H(

y , x ,t

ε

)=

y 2

2+ (cos x − 1) + µ(cos x − 1) sin

t

ε

Equations

x = y

y = sin x + µ sin x sint

ε

Tipically in near integrable Hamiltonian systems µ = 1, therefore µ is a fixednon small constant.

We will deal with this simplified problem taking µ small.

Therefore we have two perturvative small parameters µ and ε.

This will have already the difficulties to see that classical perturbativemethods do not apply.


The problem

The rapidly perturbed pendulum

H

(y , x ,

t

ε

)=

y2

2+ (cos x − 1) + µ(cos x − 1) sin

t

ε

Equations

x = y


ε

Λε = {(0, 0)} is a hyperbolic periodic orbit for this system.

We want to compute the splitting between its invariant manifolds.

The direct application of the Melnikov method (using µ as the small perturvativeparameter) gives, if we measure the distance at x = π (v = 0):

d(0, τ) = πµ

ε2

cos τ

sinh π2ε

+O(µ2) = 2πµ

ε2e−

π2ε (1 +O(e−

πε )) cos τ +O(µ2)

If µ = ε:

d(0, τ) = 2π1

εe−

π2ε cos τ +O(ε2)

This means that the method is not valid? or that we can prove that the error is in factsmaller?


The problem

One can see here the different role of the parameters µ and ε.Equations:

x = y


ε

Change time to τ = tε :

x ′ = εy

y ′ = ε sin x + µε sin x sin τ

is Hamiltonian and the new Hamiltonian is:

K (x , y , τ ; ε, µ) = ε

(y 2

2+ (cos x − 1)(1 + µ sin τ)

)One can see here the singular character of the parameter ε


The problem

Remember Neishtadt theorem:

Theorem (Neishtadt 84)

Take any value of µ, and fix ε0 > 0. Then, for ε ∈ (0, ε0), there exists anε-close to the identity canonical transformation that transforms system

H(

y , x ,t

ε;µ)

=y 2

2+ (cos x − 1) + µH1

(x , y ,

t

ε

)into

H(

y , x ,t

ε;µ)

=y 2

2+ (cos x − 1) + H0(y , x ;µ, ε) + P

(y , x ,

t

ε;µ, ε

)and P satisfies

|P| ≤ c2e−c1ε

for certain constants c1, c2 > 0.


The problem

We know by Neishtadt theorem that:

For any value of µ, the system is exponentially small close tointegrable in ε.

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 know that the distance is exponentially small and also theMelnikov formula is exponentially small. But does this mean thatMelnikov function gives the dominant term of the difference???.

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


The problem

The result for µ small

We will prove the following Theorem:

There exists ε0 > 0 and µ0 such that for |µ| < µ0, ε ∈ (0, ε0),

The invariant manifolds split and their distance at the section x = π(v = 0) is given by

d(0, τ) = 2πµ

ε2e−

π2ε (cos τ +O (µ)) .

Consequently:

The error is of order O(µ2

ε2 e−π2ε )

The Melnikov function gives de correct first order of the splittingwhen µ is small


The problem

Plan to prove the theorem

Look for suitable parameterizations of the invariant manifolds.

Extend these parameterizations to certain regions of the complexplane.

Analyze the difference of the parameterizations in these regions.

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


The problem

Parameterizations of the invariant manifolds

As before, we look for

y = ∂xSu,s(x , t/ε)

The generating functions Su,s satisfy the Hamilton-Jacobi equation.

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

Atention: ω = 1ε !!!!

Pendulum example

ε−1∂τS +(∂xS)2

2+ cos x − 1 + µ(cos x − 1) sin τ = 0

Asymptotic conditions: limx→0

∂xSu (x , τ) = 0

limx→2π

∂xS s (x , τ) = 0


The problem

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)

New equation

ε−1∂τT +cosh2 v

8(∂vT )2 − 2

cosh2 v− µ 2

cosh2 vsin τ = 0

Asymptotic conditionslim

<v→−∞cosh v ∂vT u (v , τ) = 0

lim<v→+∞

cosh v ∂vT s (v , τ) = 0.


The problem

New parameterizations


8(∂vT )2 − 2

cosh2 v− µ 2

cosh2 vsin τ = 0

Parameterizations of the form

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

2∂vT u,s (v , τ) .

For the unperturbed pendulum: ∂vT0(v) = 4cosh2 v

.


The problem

Formal power series in the singular parameter ε

Fix µ. We can try to look for the solutions using a power series expansion inε

T u,s(v , τ) = T u,s(v , τ ;µ, ε) = T0(v) +∑k≥1

εkT u,sk (v , τ ;µ)

It turns out that: T uk (v , τ ;µ) = T s

k (v , τ ;µ) ∀k ∈ N

This implies that: T u(v , τ ;µ, ε)− T s(v , τ ;µ, ε) = O(εk) ∀k ∈ N

Proceeding formally, their difference is beyond all orders in the singularparameter ε .

What is happening?

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

2 The power series is divergent: the manifolds do not coincide and theirdifference is flat with respect ε.

We will see that is happening the second option.


The problem

New parameterizations


8(∂vT )2 − 2

cosh2 v− µ 2

cosh2 vsin τ = 0

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)

T u,s satisfy the equation:

ε−1∂τT + ∂vT +cosh2 v

8(∂vT )2 − µ 2

cosh2 vsin τ = 0

T u,s are expected to be small (size ε? size µ?).


The problem

The integral operators

Inverses of L0 = ε−1∂τ + ∂v are the same as before but now L0 andtherefore its inverses Gu,s will depend on ε.

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

Gu(h)(v , τ) =

∫ 0

−∞h(v + t, τ + ε−1t) dt.

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

Gs(h)(v , τ) =

∫ 0

+∞h(v + t, τ + ε−1t) dt.


The problem

The fixed point equation

T u is a solution of

T u = Gu(

cosh2 v

8(∂vT u)2 − µ 2

cosh2 vsin τ

)

h = ∂vT u is solution of:

h = ∂vGu(

cosh2 v

8h2 − µ 2

cosh2 vsin τ

)


The problem

The “old” fixed point argument

WE know that if we proceed as before we will get:

∂vT u,s = ∂vFu,s(∂vT u,s) = ∂vFu,s(0) +O(µ2).

∂vFu(0) = µ∂v

∫ 0

−∞

2

cosh2(v + t)sin(τ + ε−1t

)dt

∂vF s(0) = µ∂v

∫ 0

+∞

2


)dt

The difference: ∂vT u − ∂vT s = ∂vFu(0)− ∂vF s(0) +O(µ2).

Difference between first iteration is the derivative of the Melnikovpotential

∂vFu(0)− ∂vF s(0) = µ∂v

∫ +∞

−∞

2


)dt.


The problem

First order:∂vFu(0)− ∂vF s(0) = 4π

µ

ε2e−

π2ε (1 +O(e−

πε )) cosh v cos(τ − ε−1v)

Melnikov is exponentially small in ε and the error is O(µ2).

∂vT u(v , τ)−∂vT s(v , τ) = 4πµ

ε2e−

π2ε cosh v cos(τ−ε−1v)+O(µ2,

µ

ε2e−

πε )

This formula is Asymptotic only if µ = o(e−π2ε )!!

If we want results for better values of µ (not so small) we need betterestimates for the error.

Let’s see first how we compute the Melnikov potential.


The problem

Computation of the Melnikov potential (or function)

L(v , τ) =−∫ +∞

−∞

2

cosh2(v + t)sin

(τ +

t

ε

)dt =

∫ +∞

−∞

2

cosh2(s)sin(τ −

v

ε+

s

ε

)dt

=−e i(τ−

vε

) − e−i(τ− vε

)

iI = −2 I sin(τ −

v

ε)

where

I =

∫ +∞

−∞

1

cosh2(s)e

sε dt

This integral is computed using residums Theorem.The term 2

cosh(v+t)comes from the unperturbed homoclinic orbit!

cosh s has zeros at s = ±iπ/2.

The function 1cosh2(s)

esε has a pole at s = ±iπ/2.

Exercice:I =

π

ε sinh π2ε

Then

L(v , τ) = −2π

ε sinh π2ε

sin(τ −v

ε), M(v , τ) = ∂vL(v , τ) =

2π

ε2 sinh π2ε

cos(τ −v

ε)


The problem

Computation of the Melnikov potential (or function)

To compute the Melnikov function (in fact, the Melnikov potential), we have

analyzed the singularities of the Melnikov integrand at v = ±iπ

2.

Let’s now see if we can bound the Melnikov function (or potential) withoutthe “exact” computation, just knowing bounds of the integrand.

This will give us an heuristic idea of the need to extend the manifolds to thecomplex plane up to a distance O(ε) of the singularities of the unperturbedhomoclinic orbit ±i π2 to improve the bounds of the error.


The problem

Bound for the Melnikov function

Recall the Melnikov function (one can do the same with the Melnikov potential):

M(v , τ) = −∂v∫ +∞

−∞

2


)dt =

∫ +∞

−∞

4 sinh(v + t)

cosh3(v + t)cos(τ + ε−1t

)dt

How can I bound it without computing it?

M(v , τ) is 2π-periodic in τ , M(v , τ) = M [−1](v)e−iτ + M [1](v)e iτ .

for −π2

+ ε ≤ =v ≤ π2− ε we have |M(v , τ) ≤ K µ

ε2 , therefore M [k](v) ≤ K µε2 .

M(v , τ) = M(0, τ − vε

), and M(0, θ) is a periodic function: M(0, θ) = Λ[−1]e−iθ + Λ[1]e iθ.

M [k](v) = Λ[k]e−ikv/ε and this equality is true for any value of v !!!

Λ[1] = M [1](v)e iv/ε, take v = i(π2− ε), and get |Λ[1]| ≤ K µ

ε2 e−π2−εε ≤ 3K µ

ε2 e− π

2ε

Analogously, using v = −i(π2− ε) we get: |Λ[−1]| ≤ 3K µ

ε2 e− π

2ε

Now, for real values of v we have |M(v , τ)| ≤ 6K µε2 e− π

2ε .


The problem

Complex domains:

To bound the error of the Melnikov function we need to “control” the manifoldsclose to the singularities of the unperturbed homoclinic. In fact we need to get toa neighborhood of order ε of theses singularities. Therefore, we will work in thefollowing domains:

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


The problem

The fixed point argument

Remember the equation: h = ∂vFu(h) = ∂vGu(− cosh2 v

8 h2 + µ 2cosh2 v

sin τ)

as we

will work in the domain close to the singularities we note that1

cosh v ∼1

(v−i π2 )(v+i π2 )

|∂vFu(0)| ≤ K µε2

One can prove that there exist ε0 > 0 and µ0 > 0 small enough such thatfor ε ∈ (0, ε0), |µ| ≤ µ0, hu(v , τ) = ∂vT u(v , τ) is defined in the complexdomain and satisfies:

|hu(v , τ)| ≤ b1|µ|ε2

|hu(v , τ)− ∂vFu(0)| ≤ b1|µ|2

ε2

|∂vhu(v , τ)| ≤ b1|µ|ε3

Observe that hu is very big if we don’t assume more restrictions on µ. Howcan we do a fix point argument if the solution is big?


The problem

Remember the equation: h = ∂vFu(h) = ∂vGu(− cosh2 v

8h2 + µ 2

cosh2 vsin τ

)In fact, what we prove is that for v ∈ Du with −V1 ≤ <v ≤ V , where V1 > 0 is independent ofε, µ: ∣∣cosh3(v)hu(v , τ)

∣∣ ≤ b1|µ|ε∣∣cosh3(hu(v , τ)− ∂vFu(0))∣∣ ≤ b1|µ|2ε∣∣cosh3 ∂vh

u(v , τ)∣∣ ≤ b1|µ|

Observe that:

| cosh v

∫ 0

−∞

1

cosh2(v + t)dt| ≤ K

This is better because this bound gives more information about the size of hu :

|hu(v , τ)| ≤ b1|µ|ε if v ∈ R, v ≤ V

|hu(v , τ)| ≤ b1|µ

ε2| if v ∈ C, near

π

2i

Observe that

|∂vT0(v)| ≤ K if v ∈ R, v ≤ V

|∂vT0(v)| ≤ K |1

ε2| if v ∈ C, near ±

π

2i

Therefore, if µ is small we are still in a perturbative setting near the singularities ±π2i


The problem

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(∂vT u(v , τ) + ∂vT s(v , τ))∂v

The term in red is of size µ, therefore L is close to L0 = ε−1∂τ + ∂v .


The problem

Difference between manifolds

Assume that ∆ ∈ KerL0, L0 = ε−1∂τ + ∂v .Main idea: functions in KerL0 bounded in a complex strip areexponentially 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 {|=v | ≤ r} × TσIts mean value 〈ψ〉 = (2π)−1

∫ 2π0 ψ(v , τ) dτ does not depend on v.

Define M = max(v ,τ)∈[−ir ,ir ]×Tσ

|∂vψ(v , τ)|

Then, for ε� 1,

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


The problem

Proof

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

function Λ.

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

Fourier series ψ(v , τ) =∑k∈Z

ψk(v)e ikτ =∑k∈Z

Λke ik(τ− vε

).

ψk(v) = Λke−ikvε =⇒ ψ0(v) = Λ0 is independent of v .

Fourier coefficients: |∂vψk(v)| ≤ sup |∂vψ|e−|k|σ ≤ Me−|k|σ

Taking v = +ir if k ≥ 0 and v = −ir if k ≤ 0

|kε

Λk | = sup∣∣∣∂vψk(v)e ikε

−1v∣∣∣ ≤ Me−|k|

rε e−|k|σ.

For real values of (v , τ): |∂vψ(v , τ)| ≤∑k 6=0

|ε−1kΛk | ≤ 2Me−rε .


The problem

Non (completely) rigourous exponentially small bounds

As ∆ = T u − T s satisfies L∆ = 0 for the linear operator

L = ε−1∂τ + ∂v +cosh2 v

8(∂vT u(v , τ) + ∂vT s(v , τ))∂v

and the term in red is of size µ,L is µ-close to L0 = ε−1∂τ + ∂v .

If ∆ was solution of L0∆ = 0, we apply the lemma to ∆:

|∂v∆| ≤ |∂vT u|+ |∂vT s | ≤ b1|µ|ε2 for (v , τ) ∈ [−ir , ir ]× Tσ with r =

π

2− ε

Using the previous lemma we obtain

|∂v∆(u, τ)| = |∂vT u − ∂vT s | ≤ b1|µ|ε2 e−

π2ε , for real values

(v , τ) ∈ [−V ,V ]× TσWe have deduced exponentially small bounds of the difference between themanifolds for real values of the variables:

d(0, τ) ≤ K|µ|ε2

e−π2ε


The problem

Non (completely) rigourous exponentially smallasymptotics

As ∆ = T u − T s satisfies L∆ = 0 for the linear operator

L = ε−1∂τ + ∂v +cosh2 v

8(∂vT u(v , τ) + ∂vT s(v , τ))∂v

and the term in red is of size µ, L is µ-close to L0 = ε−1∂τ + ∂v .

The Melnikov potential L(v , τ) = L(0, τ − vε ) satisfies L0L = 0

If ∆ was solution of L0∆ = 0, then L0(∆− L) = 0.

for (v , τ) ∈ [−ir , ir ]× Tσ with r =π

2− ε.

|∂v∆− ∂vL| = |∂vT u − ∂vT s − ∂vFu(0) + ∂vF s(0)|

≤ |∂vT u − ∂vFu(0)|+ |∂vT s − ∂vF s(0)| ≤ b1|µ|2

ε2

We deduce exponentially small bounds for real values of the variables:

|(∂v∆− ∂vL)(v , τ)| ≤ b1|µ|2ε2 e−

π2ε , for (v , τ) ∈ [−V ,V ]× Tσ


The problem

Next proposition makes these arguments rigourous: There exists ε0 > 0 and µ0

sufficiently small, such that for ε ∈ (0, ε0) and |µ| ≤ µ0, there exists areal-analytic function C defined in Du ∩ Ds × Tσ such that the change

(u, τ)→ (v , τ) v = u + C(u, τ)

conjugates the operators Lε and L0.Moreover, there exists a constant b2 > 0 such that

|C(u, τ)| ≤ b2|µ|ε|∂uC(u, τ)| ≤ b2|µ|.

Furthermore, (v , τ) = (u + C(u, τ), τ) is invertible and its inverse is of the form(u, τ) = (v + V(v , τ), τ) where V is a function defined for (v , τ) ∈ RK⊥3

× Tσ,

which satisfies|V(v , τ)| ≤ b2|µ|ε


The problem

The function ∆(u, τ) = ∆(u + C(u, τ), τ) satisfies L0∆ = 0.

L0(∆(u, τ)− L(u, τ)) = 0

As C is small one can bound

∆(u, τ)− L(u, τ) = ∆(u + C(u, τ), τ)−∆(u, τ) + ∆(u, τ)−M(u, τ)

using the same auguments we get a exponentially small bound for:

∆(u, τ)− L(u, τ)

Now we recover:

∆(v , τ)− L(v , τ) =∆(v + V(v , τ), τ)− L(v + V(v , τ), τ)

+L(v + V(v , τ), τ)− L(v , τ)


The problem

Conclusion

Parameterizations x = x0(v)

y = y0(v) + y−10 (v)∂vT u

(v ,

t

ε

)We obtain

∂v∆(v , τ) = ∂vT u(v , τ)− ∂vT s(v , τ)

= −4πµ

ε2e−

π2ε(cos(τ − ε−1v) +O(µ)

).

We have a first order of the difference between the invariantmanifolds if µ is small enough.


The problem

Splitting of separatrices

We have proved splitting of separatrices for

H(

y , x ,t

ε

)=

y 2

2+ (cos x − 1) + µ(cos x − 1) sin

t

ε

if µ is a small parameter.

Typically in Arnold diffusion

H(

y , x ,t

ε

)=

y 2

2+ (cos x − 1) + µ(cos x − 1) sin

t

ε

for fixed µ ∈ R.

Can we say something in this case?


Multiscale Phenomena in Geometry and Dynamics Technical University Munich … · 2019-07-24 ·...

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Transcript of Multiscale Phenomena in Geometry and Dynamics Technical University Munich … · 2019-07-24 ·...