Lecture 3: Melnikov method through the Jamilton-Jacobi ... · Jamilton-Jacobi equation for the perturbed pendulum Multiscale Phenomena in Geometry and Dynamics Technical University

Lecture 3: Melnikov method through theJamilton-Jacobi equation for the perturbed

pendulumMultiscale Phenomena in Geometry and Dynamics

Technical University Munich (TUM)

Tere M-Seara

Universitat Politecnica de Catalunya

22-29 July 2019

T. M-Seara (UPC) Lecture 3 22-29 July 2019 1 / 30

The problemWe want to find a way to study the splitting of separatrices of the Hamiltonian:

H(

y , x ,tε

)=

y2

2+ (cos x − 1) + ε(cos x − 1) sin

tε

As we cannot apply the Melnikov formula directly, we will consider the“regular problem”

Hamiltonian

H (y , x , ωt) =y2

2+ (cos x − 1) + ε(cos x − 1) sinωt

Equations

x = yy = sin x + ε sin x sinωt

And we will study the splitting of separatrices of the pendulum “directly”to understand how we obtain the bound of the error.

later we will go to the exponentially small case that will be ω = 1ε and see

if we can improve the bound of the error.


The perturbed pendulum

From now on:

H (y , x , ωt) =y2

2+ (cos x − 1) + ε(cos x − 1) sinωt

Equations

x = yy = sin x + ε sin x sinωt

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

We want to compute the splitting between its invariant manifolds forε 6= 0 small.


Plan

Look for suitable analytic parameterizations of the invariantmanifolds.

Analyze the difference of the parameterizations.

Deduce from this analysis the first order of the difference.

As in our example the unperturbed separatrix is a graph we lookfor the perturbed invariant manifolds as graphs.


Parameterizations of the invariant manifolds

We use that the manifolds are Lagrangian graphs and then they aregiven by the gradient of a function.

y = ∂xSu,s(x , ωt)

The generating functions Su,s(x , τ) satisfy the Hamilton-Jacobi equation(PDE).

ω∂τS (x , τ) + H (x , ∂xS (x , τ) , τ) = 0

Another possibility is to look for

y = f (x , τ)

and using that is invariant by the flow we obtain the PDE:

ω∂τ f (x , τ) + ∂xH(x , f (x , τ), τ) + ∂y H(x , f (x , τ), τ)∂x f (x , τ) = 0

then f = ∂xS.


Parameterizations of the invariant manifolds

Pendulum example, remember:

H (y , x , ωt) =y2

2+ cos x − 1 + ε(cos x − 1) sinωt

Equation for Ss,u:

ω∂τS +(∂xS)2

2+ cos x − 1 + ε(cos x − 1) sin τ = 0

Asymptotic conditions: limx→0

∂xSu (x , τ) = 0

limx→2π

∂xSs (x , τ) = 0


Unperturbed system; Parameterizations of theinvariant manifolds

Unperturbed pendulum:

H0 (y , x) =y2

2+ cos x − 1

In this case we know that the manifolds cincide:

y = ∂xS0(x , ωt)

p

2pipi

q

0


Unperturbed system; Parameterizations of theinvariant manifolds

Unperturbed pendulum: H0 (y , x) = y2

2 + cos x − 1

The Hamilton-Jacobi equation becomes.

ω∂τS (x , τ) + H0 (x , ∂xS (x , τ)) = 0

Equation for S0:

ω∂τS +(∂xS)2

2+ cos x − 1 = 0

It has a solution S0 = S0(x) where:

(∂xS0)2

2+ cos x − 1 = 0

which gives: S0(x) = ± sin(x/2).

It satisfies both asymptotic conditions(is an homoclinic):

limx→0

∂xS0(x) = limx→2π

∂xS0(x) = 0


Reparameterization using the time over the separatrix

exercice: Parameterization of the pendulum separatrix:

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

cosh v,

(x = yy = sin x

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

Equation for S(x , τ):

ω∂τS +(∂xS)2

2+ cos x − 1 + ε(cos x − 1) sin τ = 0

exercice:Equation for T (v , τ):

ω∂τT +cosh2 v

8(∂v T )2 − 2

cosh2 v− ε 2

cosh2 vsin τ = 0

Asymptotic conditionslim

Re v→−∞cosh v ∂v T u (v , τ) = 0

limRe v→+∞

cosh v ∂v T s (v , τ) = 0.T. M-Seara (UPC) Lecture 3 22-29 July 2019 9 / 30

New parameterizations

If we have a solution of:

ω∂τT +cosh2 v

8(∂v T )2 − 2

cosh2 v− ε 2

cosh2 vsin τ = 0

Will give us 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.

(x = 4 arctan(ev ), y =2

cosh v)


splitting of separatrices

We want to study the two solutions T u, T s of this equation which givethe invariant manifolds and see that they are different

ω∂τT +cosh2 v

8(∂v T )2 − 2

cosh2 v− ε 2

cosh2 vsin τ = 0

q

p

2pi

t0


New parameterizations

ω∂τT +cosh2 v

8(∂v T )2 − 2

cosh2 v− ε 2

cosh2 vsin τ = 0

Unperturbed pendulum: ∂v T0(v) =4

cosh2 v.

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

We expect T u,s = O(ε).

Parameterizations of the invariant manifolds:x = 4 arctan(ev )

y =2

cosh v+

cosh v2

∂vT u,s (v , τ)

To study the difference between manifolds we have to analyze∂vT u − ∂vT s.


We will prove the following Theorem (Melnikov)Fix V > 0, then there exists ε0 such that for 0 < ε < ε0 we have:

T u(v , τ) is defined for −∞ < v ≤ V , τ ∈ [0, 2π].T s(v , τ) is defined for −V ≤ v < +∞, τ ∈ [0, 2π].∂vT u(v , τ)− ∂vT u(v , τ) = ε∂v L(v , τ) +O(ε2) for −V ≤ v < +V , τ ∈ [0, 2π].L(v , τ) is the Melnikov potential:

L(v , τ) = −∫ +∞

−∞(H1(xh(v + s), τ + ωs)− H1(0, τ + ωs))ds

= −∫ +∞

−∞

2cosh2(v + s)

sin(τ + ωs)ds = −2πω

sinh πω2

sin(τ − ωv).

Remember the parameterizations of the invariant manifolds:x = 4 arctan(ev )

y =2

cosh v+

cosh v2

∂vT u,s (v , τ)

Therefore the distance between the stable and unstable manifolds is given by:

d(v , τ) = εcosh v

2∂v L(v , τ) +O(ε2) = ε

πω2

sinh πω2

cosh v cos(τ − ωv) +O(ε2)

for −V ≤ v < +V , τ ∈ [0, 2π].


ω∂τT +cosh2 v

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 ∂v T0(v) =4

cosh2 v.

New equation (exercice):

ω∂τT + ∂vT +cosh2 v

8(∂vT )2 − ε 2

cosh2 vsin τ = 0

T u,s are expected to be small (order ε).

We can solve this equation by inverting the linear differentialoperator L0 = ω∂τ + ∂v .


The integral operators

Inverses of L0 = ω∂τ + ∂v . (L0g = h, g = G(h).)

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

Gu(h)(v , τ) =

∫ 0

−∞h(v + t , τ + ωt) dt .

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

Gs(h)(v , τ) =

∫ 0

+∞h(v + t , τ + ωt) dt .


The fixed point equation

We look for T u as a solution of

T u = Gu

(−cosh2 v

8(∂vT u)2 + ε

2cosh2 v

sin τ

)

with

Gu(h)(v , τ) =

∫ 0

−∞h(v + t , τ + ωt) dt .

We look for T s as a solution of

T s = Gs

(−cosh2 v

8(∂vT s)2 + ε

2cosh2 v

sin τ

)

with

Gs(h)(v , τ) =

∫ 0

+∞h(v + t , τ + ωt) dt .


The fixed point equation

T u is a fixed point of

T u = Fu(∂vT ) = Gu

(−cosh2 v

8(∂vT )2 + ε

2cosh2 v

sin τ

)

T s is a fixed point of

T s = Fs(∂vT ) = Gs

(−cosh2 v

8(∂vT )2 + ε

2cosh2 v

sin τ

)

Therefore we need to solve these two fixed point equations and thencompute T u(v , τ)− T s(v , τ) (and ∂vT u(v , τ)− ∂vT s(v , τ)):


The fixed point argument

T = Fu(∂vT ) = Gu

(−cosh2 v

8(∂vT )2 + ε

2cosh2 v

sin τ

)We solve the equation for the unstable manifold (The stable one isanalogous)

As we want to deal with analytic functions we will take:Re v ≤ V , |Im v | ≤ ρ, Re τ ∈ [0,2π], |Im τ | ≤ σ, where ρ and σ are small.

We will consider χ the (Banach) space of functions T (v , τ), 2π- periodicin τ and analytic in the previous domain.

Then Fu is a function that acts in this (Banach) space.

We will need a good norm in χ which controls the exponential decay at−∞ of the function T u.

As we expect the fix point to be small a good approximation to begin isuse 0.



T = Fu(∂vT ) = Gu

(−cosh2 v

8(∂vT )2 + ε

2cosh2 v

sin τ

)

The equation has a problem, the operator Fu acts in the derivatives of T

We differenciate both sides respect to v :

∂vT = ∂vFu(∂vT ) = ∂vGu

(−cosh2 v

8(∂vT )2 + ε

2cosh2 v

sin τ

)

We will solve an equation whose unknown will be hu = ∂vT u:

h = ∂vFu(h) = ∂vGu

(−cosh2 v

8h2 + ε

2cosh2 v

sin τ

)

We can recover T u(v , τ) = Fu(hu).



We look for hu such that is a fixed point of h = ∂vFu(h).

∂vFu(h)(v , τ) = ∂vGu

(−

cosh2 v8

(h(v , τ))2 + ε2

cosh2 vsin τ

)

∂vFu(0)(v , τ) = ∂vGu(ε

2cosh2 v

sin τ)

= ε∂v

∫ 0

−∞

2cosh2(v + t)

sin (τ + ωt) dt

∂vFu(0)(v , τ) = ε

∫ 0

−∞

−4 sinh(v + t)cosh3(v + t)

sin (τ + ωt) dt

As in our domain (exercice): |4 sinh vcosh3 v

| ≤ Ke2Re v , we have that:

|∂vFu(0)(v , τ)| ≤ Kεe2Re v

The constant K does not depend on ε.


The fixed point argument: weighted norms

Now we choose the norm in χ to control the function and its decreasingat −∞:

||h|| = sup |h(v , τ)e−2Re v |where the supremum is taken in our domain.

Then is equivalent that:

|‖h‖ ≤ M to |h(v , τ)| ≤ Me2Re v . (as Re v → −∞)

Exercice There exist constants K1, K2 such that for |Im v | ≤ ρ andRe v ≤ V , we have

k1e−Re v ≤ | cosh v | ≤ K2e−Re v (1K2

eRe v ≤ 1| cosh v |

≤ 1K1

eRe v )

Therefore is equivalent to use the norm:

||h|| = sup |h(v , τ)cosh2v |


The fixed point argument: weighted normsWe look for hu such that is a fixed point of h = ∂vFu(h).

∂vFu(h)(v , τ) = ∂vGu

(−

cosh2 v8

(h(v , τ))2 + ε2

cosh2 vsin τ

)

We had that: |∂vFu(0)(v , τ)| ≤ K εe2Re v

Equivalently: |∂vFu(0)(v , τ)| ≤ ε Kcosh2 v

Clearly we have seen that ||∂vFu(0)|| ≤ K ε

Take BR(0) the ball of radious R in the space χ, where R = 2K ε.

We have seen that ∂vFu(0) ∈ BR(0)

take h1,h2 ∈ BR(0), then, as Gu is linear:

∂vFu(h1)− ∂vFu(h2) = ∂vGu

(−cosh2 v

8(h2

1 − h22)

)We need to study a little the operator ∂vGu


the operator ∂vGu

lemmaTake g(v , τ) ∈ χ, then, ∂vGu(g) ∈ χ and there exists a constant M > 0 such that:

||∂vGu(g)|| ≤ M||g||

To see this we need to bound the derivative: ∂vGu(g) = ∂v∫ 0−∞ g(v + t , τ + ωt)dt :

Exercice: There exists a constant M > 0 such that, for |Im v | ≤ ρ and Re v ≤ V

| cosh2 v |∫ 0

−∞

1| cosh2(v + t)|

dt ≤ M

It is easy to bound Gu(g)

|Gu(g) cosh2 v | ≤ | cosh2 v∫ 0

−∞

1cosh2(v + t)

cosh2(v + t)g(v + t , τ + ωt)dt |

≤ sup | cosh2 v g(v , τ)|| cosh2 v |∫ 0

−∞

1| cosh2(v + t)|

dt ≤ M||g||

This bound is not good because it does not give bounds of ∂vGu(g) in terms of gWe need to use that g is periodic to obtain the bounds and change a little the domain towork in sectors instead of strips.


The operator ∂vGu

Exercice: If we write g in Fourier series: g(v , τ) =∑

g[k ](v)eikτ , one hasthat:G(v , τ) := Gu(g)(v , τ) is given by

G(v , τ) = Gu(g)(v , τ) =∑k∈Z

G[k ](v)eikτ

where

G[k ](v) =

∫ 0

−∞g[k ](v + t)eikωtdt

which is integrable for g ∈ χ.

In order to prove the lemma we want bounds of dG[k ]

dv in terms of g[k ].First we give “good” bounds of G[k ]. These bounds are computed indifferent ways depending on the domain and whether k 6= 0 or k = 0.

We change the domain of v . Take V > 0 and 0 < β0 <π4 .

New domain: Re v ≤ V and |Imv | ≤ tanβ0|Re v |.

We will use the Fourier norm: ‖g(v , τ)‖ =∑

k∈Z

∥∥∥g[k ]∥∥∥ek|σ|.


The operator ∂vGu

First we bound G[k ]:

cosh2 v ·G[k ](v) = cosh2 v∫ 0

−∞g[k ](v + t) · eikωtdt .

Since the integrand has exponential decay for Re t → −∞, we change theintegration path:

to the line from 0 to −∞ with angle +β0/2, when k > 0

to the line from 0 to −∞ with angle −β0/2, when k < 0

Writting t = ξe∓iβ0/2:

| cosh2 u ·G[k ](v)| ≤ M||g[k ]||∫ 0−∞ |e

ikωξe∓iβ0/2 · e∓iβ0/2|dξ¯ ≤ M||g[k ]||

∫ 0−∞ |e

−|k|ωξsin(β0/2|dξ≤ M ||g[k ]||

|kω| sin(β0/2) ≤ M 1|k|ω‖g

[k ].

and we obtain: ‖G[k ]‖ ≤ M 1|k|ω‖g

[k ]‖, if k 6= 0


The operator ∂vGu

Now we can bound ∂vGu(g) = ∂v G =∑

k∈Zddv G[k ](v)eikτ .

ddv

G[k ](v) =

∫ 0

−∞

ddv

g[k ](v + t)eikωtdt

=

∫ 0

−∞

ddt

(g[k ](v + t)eikωt )dt − ikω∫ 0

−∞g[k ](v + t)eikωtdt

= g[k ](v)− limt→−∞

(g[k ](v + t)eikωt )− ikωG[k ](v)

= g[k ](u)− ikωG[k ](v).

Hence, it is clear that for k 6= 0,∥∥∥∥ d

dvG[k ]

∥∥∥∥ ≤ (1 + M)‖g[k ]‖.

Moreover, recalling thatddv

G[0](v) = g[0](v), we obtain

‖∂vGu(g)‖ = ‖∂v G‖ =∑k∈Z

∥∥∥ ddv

G[k ]∥∥∥e|k|σi ≤ M‖g‖.



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

8 h2 + ε 2cosh2 v

sin τ)

||∂vFu(0)|| ≤ Kε

Take BR(0) the ball of radious R in this space, where R = 2Kε.

We have seen that ∂vFu(0) ∈ BR(0)

Take h1, h2 ∈ BR(0), then

‖∂vFu(h1)− ∂vFu(h2)‖ = ‖∂vGu

(−

cosh2 v8

(h21 − h2

2)

)‖ ≤ ‖

cosh2 v8

(h21 − h2

2)‖

≤ sup | cosh4 v (h1 + h2)(h1 − h2)| ≤ ‖h1 + h2‖‖h1 − h2‖ ≤ 2R‖h1 − h2‖ ≤ 4Kε‖h1 − h2‖

Therefore, if 0 < ε < ε0 = 18K

the map ∂vFu is a contraction from BR(0) in BR(0) and it

has a unique fixed point hu ∈ BR(0), that is, ||hu || ≤ 2Kε.


Moreover the fix point hu satisfies:

hu = ∂vFu(hu) = ∂vFu(0) + ∂vFu(hu)− ∂vFu(0)

‖hu − ∂vFu(0)‖ = ‖∂vFu(hu)− ∂vFu(0)‖ ≤ 4K ε‖hu‖ ≤ 8K 2ε2

Therefore for −∞ < v ≤ V :

hu(v , τ) = ∂vFu(0)(v , τ) +O(ε2)


difference between the manifolds

We have seen that there is hu solution of:hu(v , τ) = ∂vFu(hu)(v , τ) = ∂vGu

(− cosh2 v

8 (hu(v , τ))2 + ε 2cosh2 v sin τ

)such that: hu(v , τ) = ∂vFu(0)(v , τ) +O(ε2), for v ≤ V

Analogously, there is a solution hs solution of:hs(v , τ) = ∂vFs(hs)(v , τ) = ∂vGs

(− cosh2 v

8 (hs(v , τ))2 + ε 2cosh2 v sin τ

)such that: hs(v , τ) = ∂vFs(0)(v , τ) +O(ε2), for v ≥ −V

Now we can compute the difference at v , such that v ∈ R, −V ≤ v ≤ V :

hu(v , τ)− hs(v , τ) = ∂vFu(0)(v , τ)− ∂vFs(0)(v , τ) +O(ε2)

= ε∂v

∫ +∞

−∞

2cosh2(v + t)

sin (τ + ωt) dt +O(ε2)

= ε∂v L(v , τ) +O(ε2) = εM(v , τ) +O(ε2)

As hu = ∂vT u and hs = ∂vT s, this concludes the proof of the theorem.


The exponentially small case

In the case ω = 1ε all the proof works the same.

The difference at v , such that v ∈ R, −V ≤ v ≤ V :

∂vT u(v , τ)− ∂vT s(v , τ) = ε∂v

∫ +∞

−∞

2cosh2(v + t)

sin(τ +

tε

)dt +O(ε2)

= ε∂v L(v , τ) +O(ε2) = εM(v , τ) +O(ε2)

But now L(v , τ) = −4π 1εe−

π2ε (1 +O(e−

πε )) sin(τ − v

ε ).

How can we improve the results?


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Lecture 3: Melnikov method through the Jamilton-Jacobi ... · Jamilton-Jacobi equation for the perturbed pendulum Multiscale Phenomena in Geometry and Dynamics Technical University