Post on 23-Jun-2020
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 2 / 33
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
y = sin x + µ sin x sint
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?
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 3 / 33
The problem
One can see here the different role of the parameters µ and ε.Equations:
x = y
y = sin x + µ sin x sint
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
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 4 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 5 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 6 / 33
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
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 7 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 8 / 33
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
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 9 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 10 / 33
The problem
New parameterizations
ε−1∂τT +cosh2 v
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
.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 11 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 12 / 33
The problem
New parameterizations
ε−1∂τT +cosh2 v
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?).
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 13 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 14 / 33
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
)
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 15 / 33
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
cosh2(v + t)sin(τ + ε−1t
)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
cosh2(v + t)sin(τ + ε−1t
)dt.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 16 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 17 / 33
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)
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 18 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 19 / 33
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
cosh2(v + t)sin(τ + ε−1t
)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 .
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 20 / 33
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:
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 21 / 33
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?
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 22 / 33
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
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 23 / 33
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 .
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 24 / 33
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
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 25 / 33
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 .
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 26 / 33
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
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 27 / 33
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
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 28 / 33
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
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 29 / 33
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)
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 30 / 33
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.
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 31 / 33
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?
Tere M-Seara (UPC) Lecture 4 22-29 July 2019 32 / 33