Chapter 3 Inner Problem
3.4 Analytic Invariant Manifolds
In this section we prove the existence of an unstable (resp. stable) analytic manifold immersed in C4. We also provide an asymptotic expansion for both manifolds. More
concretely, following (3.4) we look for parametrisations as solutions of the following PDE,
Dx=XH(x). (3.48)
Now, given a formal solution Γˆ in the class τ−1T4C[[τ−1]] of equation (3.48), which exists due to Theorem 3.2.2, we prove the existence of an unique solution Γ− (resp.
Γ+) of equation (3.48) belonging to the space X1(Sh×Dr−) (resp. X1(Sh×D+r)) such thatΓ± ≍Γˆ, i.e.
∀n∈N, ∃C >0, Γ±(ϕ, τ)−Γn(ϕ, τ)
≤Cτ−n−1, in Sh×D±r, (3.49)
where Γn denotes a truncation of Γˆ as defined in Remark 3.2.2.4. We will prove the existence of such solution for the−case only as the+case is completely analogous mod- ulus minor modifications in the definitions of the sets where the functions are analytic. Then we have the following,
Theorem 3.4.1 (Analytic unstable parametrisation). Given a formal solution Γˆ ∈
τ−1T4C[[τ−1]] of equation (3.48) there is an r0 > 0 sufficiently large such that for
every r > r0 the equation (3.48) has an unique analytic solution Γ− ∈ X1(Sh×Dr−) such thatΓ−−Γn∈Xn+1(Sh×Dr−) for alln≥6.
Proof. Letn≥6andr >0(to be chosen later in the proof). Let us look for a solution of equation (3.48) of the form,
Γ−=Γn+ξ, (3.50)
whereξ ∈Xn(Sh×D−r) and Γn is defined as in Remark 3.2.2.4. Substituting (3.50) into equation (3.48) we obtain,
Now we rewrite the previous equation as follows,
L(ξ) =Q(ξ) +Rn, (3.51) whereL is a linear operator acting according to the formula L(ξ) =Dξ−DXH(Γn)ξ and
Q(ξ) =XH(Γn+ξ)−XH(Γn)−DXH(Γn)ξ, Rn=XH(Γn)− DΓn. Note that it follows from Remark 3.2.2.4 that Rn ∈Xn+1(Sh×D−r). We focus our attention in solving equation (3.51) with respect to ξ. For that purpose we want to invert the linear operatorLand obtain a new equation from which we can apply a fixed point argument to get the desired solution.
According to Theorem 2.4.1 we can invert the linear operatorLas long as it has a fundamental matrix U andQ(ξ) ∈Xn+1(Sh×Dr−) given ξ ∈Xn(Sh×D−r). Due to Theorem 3.3.1 there exist anr0>0 such that for everyr > r0 the linear operatorL has a fundamental matrixU such that U−Un−3 ∈X4n−2(Sh×D−r).
Now let us show thatQ(ξ)∈Xn+1(Sh×D−r). Denote the components of the vector fieldXH by (v1, v2, v3, v4) and consider the following auxiliary functions,
γi(t) =vi(Γn+tξ)−vi(Γn)−t∇vi(Γn)ξ, i= 1, . . . ,4.
Note thatγi(0) = 0for i= 1, . . . ,4 andQ(ξ) = (γ1(1), γ2(1), γ3(1), γ4(1))T. Now we can integrate by parts each functionγi to obtain,
γi(1) =
Z 1 0
(1−s)γi′′(s)ds, i= 1, . . . ,4.
Then by the intermediate value theorem there existti∈[0,1]fori= 1, . . . ,4such that γi(1) = (1−ti)γi′′(ti) for i = 1, . . . ,4 where the second derivative ofγi can be easily computed
γi′′(s) =ξT Hess (vi)|Γn+sξξ. (3.52)
Now taking into account that ξ ∈ Xn(Sh×D−r) and the analyticity of XH it is not difficult to get the following estimate,
where k·kC3 is the usual C3 norm of a smooth function. Using this upper bound and the fact that givenr1 >max
n
r0,sin1θ0
o
then for every r > r1 we have |τ|−2 ≤ |τ|−1 forτ ∈D−r, then we can estimatekQ(ξ)kn+1 in the following way,
kQ(ξ)kn+1 ≤8kHkC3kξk2n sup τ∈Dr−
|τ|−n+2 ≤ 8kHkC3kξk2n (r1sinθ0)n−2
, (3.53)
where this last estimate holds sincen≥6. Thus Q(ξ)∈ Xn+1(Sh×D−r).
Thus, it follows from Theorem 2.4.1 that there is an unique bounded linear operator L−1 such that LL−1 = Id. Thus, in order to solve equation (3.51), it is sufficient to find a fixed point inXn(Sh×Dr−) of the following non-linear operator,
ξ7→ L−1(Q(ξ)) +L−1(Rn).
Let us denote this non-linear operator byG. So in order to apply the contraction mapping theorem we have to check thatG is contracting in some invariant ball
Bρ=
ξ ∈Xn(Sh×Dr−)| kξkn≤ρ ,
where ρ > 0. First we prove that G(Bρ) ⊆ Bρ for some ρ > 0. Indeed, let ρ = 2L−1
n,n+1kRnkn+1 andξ∈Bρ, then (3.53) implies,
L−1(Q(ξ))− L−1(Rn)n≤L−1n,n+1 8kHkC3kξk2n (r1sinθ0)n−2 +kRnkn+1 ! ≤ρ, providedr1 is sufficiently large,
r1 ≥ (16kHkC3 L−1 n,n+1ρ) 1 n−2 sinθ0 . (3.54)
ThusG leaves invariant a closed ballBρ.
To check the contraction we letξ1, ξ2∈Bρ and consider a line connecting both points, i.e.,θt= (1−t)ξ1+tξ2. Clearlyθt∈Bρ for allt∈[0,1]. Similar as before we define the following auxiliary functions,
Note that,
Q(ξ1) = (ψ1(0), ψ2(0), ψ3(0), ψ4(0))T and Q(ξ2) = (ψ1(1), ψ2(1), ψ3(1), ψ4(1))T. By the mean value theorem there exist ti ∈ [0,1] for i = 1, . . . ,4 such that ψi(1)− ψi(0) =ψ′i(ti). Differentiating the functionsψi we get,
ψi(1)−ψi(0) = (∇vi(Γn+θti)− ∇vi(Γn))·(ξ2−ξ1), i= 1, . . . ,4. (3.55)
Now we can easily get the following upper bounds for the differences (3.55),
|ψi(1)−ψi(0)| ≤2kHkC3ρ|τ|−2nkξ2−ξ1kn. Thus, kQ(ξ2)−Q(ξ1)kn+1≤ 8ρkHkC3 (r0sinθ0)n−2k ξ2−ξ1kn. Applying the linear operator L−1 and taking into account (3.54) we get,
L−1(Q(ξ2)−Q(ξ1)) n≤ L−1 n,n+1 8ρkHkC3 (r0sinθ0)n−2 kξ2−ξ1kn ≤ 1 2kξ2−ξ1kn,
which proves thatkG(ξ2)− G(ξ1)kn≤ 12kξ2−ξ1kn. Thus G is contracting in the ball
Bρ provided r > r1 where, r1 >max r0, 1 sinθ0 ,(16kHkC3 L−1 n,n+1ρ) 1 n−2 sinθ0 .
Now let us check that the unique functionΓ−obtained withn≥6is in fact independent of n. Increasing r, if necessary, the distance kΓ−−Γ6k6 can be made as small as we want in order to apply the contraction mapping theorem for n = 6. Hence it is independent of n. Finally,
Γ−−Γn=Γ−−Γn+1+Γn+1−Γn∈Xn+1(Sh×Dr−). This completes the proof of the Theorem.
θ0 r θ0 −r D1 r D+ r+ D−r−
Figure 3.1: The intersection of the domains D±
r±.
As previously observed we can repeat the same arguments of the previous The- orem but now considering the functions defined on the domains Sh×Dr+. We obtain the following,
Theorem 3.4.2(Analytic stable parametrisation). Given a formal solutionΓˆ ∈τ−1T4C[[τ−1]]
of equation (3.48) there is an r0 > 0 sufficiently large such that for every r > r0 the equation (3.48) has an unique analytic solution Γ+ ∈ X1(Sh×Dr+) such that
Γ+−Γn∈Xn+1(Sh×D+r)for alln≥6.