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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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 2 / 30
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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 3 / 30
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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 4 / 30
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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 5 / 30
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
T. M-Seara (UPC) Lecture 3 22-29 July 2019 6 / 30
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
T. M-Seara (UPC) Lecture 3 22-29 July 2019 7 / 30
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
T. M-Seara (UPC) Lecture 3 22-29 July 2019 8 / 30
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)
T. M-Seara (UPC) Lecture 3 22-29 July 2019 10 / 30
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
T. M-Seara (UPC) Lecture 3 22-29 July 2019 11 / 30
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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 12 / 30
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].
T. M-Seara (UPC) Lecture 3 22-29 July 2019 13 / 30
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 .
T. M-Seara (UPC) Lecture 3 22-29 July 2019 14 / 30
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 .
T. M-Seara (UPC) Lecture 3 22-29 July 2019 15 / 30
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 .
T. M-Seara (UPC) Lecture 3 22-29 July 2019 16 / 30
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)):
T. M-Seara (UPC) Lecture 3 22-29 July 2019 17 / 30
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. M-Seara (UPC) Lecture 3 22-29 July 2019 18 / 30
The fixed point argument
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).
T. M-Seara (UPC) Lecture 3 22-29 July 2019 19 / 30
The fixed point argument
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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 20 / 30
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 |
T. M-Seara (UPC) Lecture 3 22-29 July 2019 21 / 30
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
T. M-Seara (UPC) Lecture 3 22-29 July 2019 22 / 30
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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 23 / 30
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|.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 24 / 30
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
T. M-Seara (UPC) Lecture 3 22-29 July 2019 25 / 30
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‖.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 26 / 30
The fixed point argument
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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 27 / 30
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)
T. M-Seara (UPC) Lecture 3 22-29 July 2019 28 / 30
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.
T. M-Seara (UPC) Lecture 3 22-29 July 2019 29 / 30
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?
T. M-Seara (UPC) Lecture 3 22-29 July 2019 30 / 30