A Formal KAM Theorem for Hamiltonian Systems and Its Application to Hyperbolic Lower Dimensional Invariant Tori

Abstract

In this paper we reformulate a formal KAM theorem for Hamiltonian systems with parameters under Bruno-Rüssmann condition. The proof is based on KAM iteration and the key is to adjust the parameters for small divisors after KAM iteration instead of in each KAM step. By this formal KAM theorem we can follow some well known KAM-type results for hyperbolic tori. Moreover, it can also be applied to the persistence of invariant tori with prescribed frequencies.

Appendix

Lemma 5.1

Let $\lambda _1,\lambda _2,\cdots \lambda _{2m}$ be the eigenvalues of matrix $Q_0$ with $|\textrm{Re} \lambda _i|\ge \delta _0>0$, for any $i=1,2,\cdots 2m$. Set $g(x)\in {\mathcal {A}}$, where ${\mathcal {A}}$ denotes the analytic function space defined on the strip $T_s$, among which $T_s=\{x\in {\mathbb {C}}^n/2\pi {\mathbb {Z}}^n: |\textrm{Im}x|_{\infty }\le s \}$. There exists a sufficiently small $\epsilon _0>0$, such that for ${\hat{Q}}(x)\in {\mathcal {A}}$, if $||{\hat{Q}}||_{T_s}\le \epsilon _0$, then the equation

$$\begin{aligned} \big \langle \omega ,\partial _{x}f(x)\big \rangle -(Q_0+{\hat{Q}}(x))f(x)=g(x) \end{aligned}$$

has a unique solution $f(x)\in {\mathcal {A}}$, with

$$\begin{aligned} ||f||_{ T_{s}}\le c||g||_{ T_{s}}. \end{aligned}$$

Proof

Let ${\mathcal {L}}{{f}}=\big \langle \omega ,\partial _{x}f(x)\big \rangle -Q_0f(x),\ f \in {\mathcal {A}}$. Then ${\mathcal {L}}:{\mathcal {A}}\rightarrow {\mathcal {A}}$ is a linear operator. By assumption, the matrix $Q_0$ is hyperbolic, the operator ${\mathcal {L}}$ has a bounded inverse with $\Vert {\mathcal {L}}^{-1}\Vert \le \frac{1}{\delta _0}$. By Banach fixed point theorem, it is easy to follow this lemma and we omit the details. $\square$

Let ${\bar{\Xi }}$ be an approximation function satisfying (2.2) and (2.3). Let

$$\begin{aligned} {\bar{\Gamma }}(\sigma )=\sup \limits _{t\ge 0}{\bar{\Xi }}(t)e^{-\sigma t}. \end{aligned}$$

Define

$$\begin{aligned} {\bar{\Delta }}(\sigma )={ \inf }\mathop {\Pi }\limits _{j=0}^{\infty }{\bar{\Gamma }}(\sigma _j)^{\kappa _j}, \ \ \kappa _j=\frac{\kappa -1}{\kappa ^{j+1}} \ \text{ with } \ \kappa =\frac{4}{3}, \end{aligned}$$

(5.1)

where the infimum is taken for sequences $\{\sigma _j\}$ satisfying

$$\begin{aligned} \sigma _0\ge \sigma _1\ge \sigma _2\ge \cdots >0 \ \ \text{ and } \ \ \sigma _0+\sigma _1+\sigma _2+\cdots \le \sigma . \end{aligned}$$

We first state a lemma which is proved in [17]. We also refer it to [18].

Lemma 5.2

(Lemma A.1 [17]) For all $\sigma >0$, the function ${\bar{\Delta }}(\sigma )$ is finite. More precisely, if

$$\begin{aligned} \frac{1}{\textrm{log}\kappa }\int _T^{\infty }\frac{\textrm{log}\bigl ({\bar{\Xi }}(t)\bigr )}{t^2}\ dt\le \sigma , \end{aligned}$$

then

$$\begin{aligned} {\bar{\Delta }}(\sigma )\le e^{(\kappa -1)\sigma T}. \end{aligned}$$

Let $\Xi$ be an approximation function satisfying (2.2) and (2.3). Let $\ell \ge 0$ and

$$\begin{aligned} \Gamma (\sigma )=\sup \limits _{t\ge 0}(1+t)^{\ell +2}\Xi ^{\ell +2}(t)e^{-\sigma t}. \end{aligned}$$

Define

$$\begin{aligned} \Delta (\sigma )={ \inf }\mathop {\Pi }\limits _{j=0}^{\infty }\Gamma (\sigma _j)^{\kappa _j}, \ \ \kappa _j=\frac{\kappa -1}{\kappa ^{j+1}} \ \text{ with } \ \kappa =\frac{4}{3}, \end{aligned}$$

(5.2)

where the infimum is taken for sequences $\{\sigma _j\}$ satisfying

$$\begin{aligned} \sigma _0\ge \sigma _1\ge \sigma _2\ge \cdots >0 \ \ \text{ and } \ \ \sigma _0+\sigma _1+\sigma _2+\cdots \le \sigma . \end{aligned}$$

Since $\Xi$ is an approximation function, it is easy to check that $(1+t)^{\ell +2}\Xi ^{\ell +2}(t)$ is also an approximation function. By Lemma 5.2 with ${\bar{\Xi }}(t)=(1+t)^{\ell +2}\Xi ^{\ell +2}(t),$ it follows that for all $\sigma >0$, $\Delta (\sigma )$ is finite.

Lemma 5.3

The supremum in the definition of $\Gamma (\sigma )$ can be attained. Moreover, the infimum in the definition of $\Delta (\sigma )$ can also be attained. More precisely, for any $\sigma >0$, there exists a sequence $\sigma _0^*\ge \sigma _1^*\ge \sigma _2^*\ge \cdots >0$ such that $\sum \nolimits _{i=0}^{\infty }\sigma _i^*=\sigma$ and

$$\begin{aligned} \Delta (\sigma )=\mathop {\Pi }\limits _{j=0}^{\infty }\Gamma (\sigma _j^*)^{\kappa _j}. \end{aligned}$$

Proof

This lemma is actually proved in [17, 18]. However, because of small divisor conditions, our definitions of $\Gamma$ and $\Delta$ are different. For the convenience of readers, we give the proof, but the idea is the same as in [17, 18].

At first, by assumption (2.2) we have $\frac{{\textrm{log}}(\Xi (t))}{t}\rightarrow 0, \ 0\le t\rightarrow \infty$. Then it is easy to see that the supremum $\Gamma (\sigma )=\sup \nolimits _{t\ge 0}(1+t)^{\ell +2}\Xi ^{\ell +2}(t)e^{-\sigma t}$ is attained and finite.

Note that

$$\begin{aligned} \Delta (\sigma )={ \inf }\mathop {\Pi }\limits _{j=0}^{\infty }\Gamma (\sigma _j)^{\kappa _j}, \ \ \kappa _j=\frac{\kappa -1}{\kappa ^{j+1}} \ \text{ with } \ \kappa =\frac{4}{3}. \end{aligned}$$

Let

$$\begin{aligned} f({{\tilde{\sigma }}})=\mathop {\Pi }\limits _{j=0}^{\infty }\Gamma (\sigma _j)^{\kappa _j}, \end{aligned}$$

where

$$\begin{aligned} {{\tilde{\sigma }}}=(\sigma _0, \sigma _1, \sigma _2, \cdots \sigma _n, \cdots )\in l^1. \end{aligned}$$

We consider $f({{\tilde{\sigma }}})$ as a functional on $l^1$.

Note that the weakly convergence in $l^1$ implies the pointwise convergence. Then $f({{\tilde{\sigma }}})$ is weakly lower semi-continuous on the set:

$$\begin{aligned} A= \left\{ (\sigma _0, \sigma _1, \sigma _2, \cdots \sigma _n, \cdots )\ | \ \sum _{j\ge 0}\sigma _j\le \sigma , \sigma _j>0, \forall j\ge 0 \right\} . \end{aligned}$$

In fact, let ${{\tilde{\sigma }}}_k\rightharpoonup {{\tilde{\sigma }}}$, that is,

$$\begin{aligned} \sigma _{kj}\rightarrow \sigma _j,\ k\rightarrow \infty , \ \forall j=0, 1,2 \cdots . \end{aligned}$$

Moreover,

$$\begin{aligned} \varliminf _{k\rightarrow \infty }\Gamma (\sigma _{kj})\ge \varliminf _{k\rightarrow \infty } (1+t)^{\ell +2}\Xi ^{\ell +2}(t)e^{-\sigma _{kj} t}=(1+t)^{\ell +2}\Xi ^{\ell +2}(t)e^{-\sigma _j t}, \ \forall t\ge 0, \end{aligned}$$

thus,

$$\begin{aligned} \varliminf _{k\rightarrow \infty }\Gamma (\sigma _{kj})\ge \sup \limits _{t\ge 0}(1+t)^{\ell +2}\Xi ^{\ell +2}(t)e^{-\sigma _j t}=\Gamma (\sigma _{j}). \end{aligned}$$

Then

$$\begin{aligned} \varliminf _{k\rightarrow \infty }f({{\tilde{\sigma }}}_{k})\ge \prod _{j=0}^{\infty }\varliminf _{k\rightarrow \infty }\Gamma (\sigma _{kj})\ge \prod _{j=0}^{\infty }\Gamma (\sigma _{j})=f({{\tilde{\sigma }}}). \end{aligned}$$

Also note that if $\sigma _j\rightarrow 0$, we have $f({{\tilde{\sigma }}})\rightarrow +\infty .$ And A is a bounded set of $l^1$. Then the infimum $\inf _{{{\tilde{\sigma }}}\in A}f({{\tilde{\sigma }}})$ can be attained. Thus, there exists a sequence $\sigma _0^*\ge \sigma _1^*\ge \sigma _2^*\ge \cdots >0$ such that $\sum \nolimits _{i=0}^{\infty }\sigma _i^*\le \sigma$ and

$$\begin{aligned} \Delta (\sigma )=f({{\tilde{\sigma }}}_*) =\mathop {\Pi }\limits _{j=0}^{\infty }\Gamma \left( \sigma _j^* \right) ^{\kappa _j}, \end{aligned}$$

where ${{\tilde{\sigma }}}_*=(\sigma _0^*, \sigma _1^*, \sigma _2^*, \cdots ).$ Now we can prove $\sum \nolimits _{i=0}^{\infty }\sigma _i^*=\sigma$ by contradiction. If not so, that is, $\sum \nolimits _{i=0}^{\infty }\sigma _i^*<\sigma ,$ then there exists a sequence $\{\sigma _i'\}$ such that $\sigma _i^*<\sigma _i', \ \forall i\ge 0$, and $\sum \nolimits _{i=0}^{\infty }\sigma _i'=\sigma .$ Obviously, $f({{\tilde{\sigma }}}')<f({{\tilde{\sigma }}}_*)$, which is a contradiction. Thus, $\sum \nolimits _{i=0}^{\infty }\sigma _i^*=\sigma$.

In addition, the infimum is not only attainable, but also finite. This conclusion can follow from Lemma 5.2. $\square$

If the approximation function $\Xi (t)$ is absolutely continuous and for almost every $t\ge 0$, assume that

$$\begin{aligned} \frac{d}{dt}\textrm{log}\bigl ( \Xi (t)\bigr )\ge \frac{1}{1+t}, \end{aligned}$$

(5.3)

then $\Xi (t)$ is called a sufficiently increasing function. Without saying it directly, we assume that all approximation functions in this paper are sufficiently increasing.

Lemma 5.4

If the approximation function $\Xi (t)$ is sufficiently increasing, then it follows

$$\begin{aligned} \Gamma _{\ell +1}(\sigma )\le \frac{\sigma }{2({\ell +1})} \Gamma _{{\ell +2}}(\sigma ), \ \ \ell \ge 0, \end{aligned}$$

where $\Gamma _{\ell +1}(\sigma )=\sup \limits _{t\ge 0}(1+t)^{\ell +1}\Xi ^{\ell +1}(t)e^{-\sigma t}$.

Proof

Assume that the approximation function $\Xi (t)$ is sufficiently increasing, if $\sigma (1+t)\le 2({\ell +1})$, we can get

$$\begin{aligned} \frac{d}{dt}\textrm{log}\bigl ((1+t)^{\ell +1}\Xi ^{\ell +1}(t)-\sigma t\bigr )\ge \frac{d}{dt}\textrm{log}\bigl ((1+t)^{\ell +1}\Xi ^{\ell +1}(t)\bigr )-\frac{2({\ell +1})}{1+t}\ge 0, \end{aligned}$$

then $(1+t)^{\ell +1}\Xi ^{\ell +1}(t) e^{-\sigma t}$ can get the supremum at some point $t_*$ and the inequality $\sigma (1+t_*)\ge 2({\ell +1})$ holds true. Thus, it follows

$$\begin{aligned} \Gamma _{\ell +1}(\sigma )= & {} (1+t_*)^{\ell +1}\Xi ^{\ell +1}(t_*)e^{-\sigma t_*}\le \frac{\sigma }{2({\ell +1})} (1+t_*)^{{\ell +2}}\Xi ^{\ell +1}(t_*)e^{-\sigma t_*}\\{} & {} \quad \le \frac{\sigma }{2({\ell +1})} \Gamma _{{\ell +2}}(\sigma ). \end{aligned}$$

The conclusion is proven. $\square$