A Diophantine duality applied to the KAM and Nekhoroshev theorems

Abstract

In this paper, we use geometry of numbers to relate two dual Diophantine problems. This allows us to focus on simultaneous approximations rather than small linear forms. As a consequence, we develop a new approach to the perturbation theory for quasi-periodic solutions dealing only with periodic approximations and avoiding classical small divisors estimates. We obtain two results of stability, in the spirit of the KAM and Nekhoroshev theorems, in the model case of a perturbation of a constant vector field on the $n$-dimensional torus. Our first result, which is a Nekhoroshev type theorem, is the construction of a “partial” normal form, that is a normal form with a small remainder whose size depends on the Diophantine properties of the vector. Then, assuming our vector satisfies the Bruno–Rüssmann condition, we construct an “inverted” normal form, recovering the classical KAM theorem of Kolmogorov, Arnold and Moser for constant vector fields on torus.

Appendices

Appendix A: Bruno–Rüssmann condition

Recall that the set of Bruno–Rüssmann vectors $\mathcal B _d$ was defined in Sect. 2.2, and in Sect. 3 we introduced another condition (A). The aim of this short appendix is to prove the following proposition.

Lemma 4.1

A vector $\alpha$ satisfies condition (A) if and only if it belongs to $\mathcal B _d$.

In the proof, we shall make use of integration by parts and change of variables formulas for Stieltjes integral.

Proof

First let us prove that there exists a continuous, non-decreasing and unbounded function $\Phi : [1,+\infty ) \rightarrow [1,+\infty )$ such that

$$\begin{aligned} \Psi _\alpha (Q) \le \Phi (Q) \le \Psi _\alpha (Q+1), \quad Q\ge 1. \end{aligned}$$

(4.1)

Recall from Proposition 2.2 that $\Psi _\alpha =\Psi _d$ is left-continuous and constant on each interval $(Q_l,Q_{l+1}]$, $l\in \mathbb{N }^*$ and on $[1,Q_1]$ with $Q_1 \ge 1$. For any $l\in \mathbb{N }^*$, let us choose a point $b_l\in (Q_l,Q_{l+1}) \cap [Q_{l+1}-1,Q_{l+1})$ and define

$$\begin{aligned} \Phi _l(Q)\!=\! {\left\{ \begin{array}{ll} \Psi _\alpha (Q_{l+1}), \quad Q\in (Q_l,b_l],\\ \Psi _\alpha (Q_{l+1})+(Q\!-\!b_l)(Q_{l+1}\!-\!b_l)^{-1} \left( \Psi _\alpha (Q_{l+2})\!-\!\Psi _\alpha (Q_{l+1})\right) , \quad Q\!\in \! (b_l,Q_{l+1}]. \end{array}\right. } \end{aligned}$$

If $Q_1>1$, we choose a point $b_0\in [1,Q_{1}) \cap [Q_{1}-1,Q_{1})$ and define $\Phi _0$ on $[1,Q_1]$ similarly, otherwise if $Q_1=1$ we simply define $\Phi _0(1)=\Psi _{\alpha }(1)$, so that finally the function $\Phi$ defined by

$$\begin{aligned} \Phi =\Phi _0 \mathbf 1 _{[1,Q_{1}]}+\sum _{l\in \mathbb{N }^*}\Phi _l \mathbf 1 _{(Q_l,Q_{l+1}]} \end{aligned}$$

has all the wanted properties. The existence of such a function $\Phi$ shows that in Corollary 2.5, the integral condition may be written equivalently in terms of $\Psi _{\alpha }$, or also in terms of this function $\Phi$.

Now let $\Delta _\Phi (Q)=Q\Phi (Q)$ and $\Delta _\Phi ^{-1}$ denotes the functional inverse of $\Delta _\Phi$. From (4.1), we deduce the following inequalities:

$$\begin{aligned} \Delta _\alpha (Q) \le \Delta _\Phi (Q) \le \Delta _\alpha (Q+1), \quad Q\ge 1. \end{aligned}$$

$$\begin{aligned} \Delta ^*_\alpha (x)-1 \le \Delta ^{-1}_\Phi (x) \le \Delta ^*_\alpha (x), \quad x\ge \Delta _{\alpha }(1). \end{aligned}$$

$$\begin{aligned} \frac{1}{\Delta ^*_\alpha (x)} \le \frac{1}{\Delta ^{-1}_\Phi (x)} \le \frac{1}{\Delta ^*_\alpha (x)-1}, \quad x> \Delta _{\alpha }(1). \end{aligned}$$

(4.2)

From (4.2) it follows that $\alpha$ satisfies condition (A) if and only if

$$\begin{aligned} \int \limits _{\Delta _{\Phi }(1)}^{+\infty } \frac{dx}{x\Delta _{\Phi }^{-1}(x)}<\infty . \end{aligned}$$

(4.3)

By a change of variables, letting $Q=\Delta _{\Phi }^{-1}(x)$, (4.3) is equivalent to

$$\begin{aligned} \int \limits _{1}^{+\infty }\frac{d\Delta _{\Phi }(Q)}{Q\Delta _{\Phi }(Q)}<\infty . \end{aligned}$$

(4.4)

Now for $t=\Delta _{\Phi }^{-1}(1)$ and $T>t$, an integration by parts gives

$$\begin{aligned} T^{-1}\ln \Delta _{\Phi }(T)+\int \limits _{t}^{T}Q^{-2} \ln (\Delta _{\Phi }(Q))dQ=\int \limits _{t}^T \frac{d\Delta _{\Phi }(Q)}{Q\Delta _{\Phi }(Q)} \end{aligned}$$

(4.5)

and as $T^{-1}\ln \Delta _{\Phi }(T)>0$, letting $T$ goes to infinity in (4.5), condition (4.4) implies

$$\begin{aligned} \int _{1}^{+\infty }Q^{-2}\ln (\Delta _{\Phi }(Q))dQ<\infty . \end{aligned}$$

(4.6)

Now, using the fact that $\Delta _{\Phi }$ is increasing and assuming that (4.6) holds true, we also have

$$\begin{aligned} T^{-1}\ln \Delta _{\Phi }(T)=\int _{T}^{+\infty }Q^{-2} \ln (\Delta _{\Phi }(T))dQ \le \int _{T}^{+\infty }Q^{-2} \ln (\Delta _{\Phi }(Q))dQ \end{aligned}$$

and therefore letting $T$ goes to infinity in (4.5), condition (4.6) implies condition (4.4). So (4.4) and (4.6) are in fact equivalent. Since $\ln \Delta _{\Phi }(Q)=\ln Q+\ln \Phi (Q)$, (4.6) is clearly equivalent to

$$\begin{aligned} \int _{1}^{+\infty }Q^{-2}\ln (\Phi (Q))dQ<\infty . \end{aligned}$$

(4.7)

Hence conditions (A), (4.3), (4.4), (4.6), (4.7) and $\alpha \in \mathcal B _d$ are all equivalent, and this ends the proof. $\square$

Appendix B: Technical estimates

In this second appendix, we derive technical estimates concerning the Lie series method for vector fields. These are well-known (see [11] for instance, where this formalism was first used as far as we aware of), but for completeness we prove the estimates adapted to our need.

Lemma 5.1

Let $V$ be a bounded real-analytic vector field on $\mathbb{T }^n_s$, $0<\varsigma <s$ and $\tau =\varsigma |V|_{s}^{-1}$. For $t\in \mathbb{C }$ such that $|t|< \tau$, the map $V^t : \mathbb{T }^n_{s-\varsigma } \rightarrow \mathbb{T }^n_s$ is a well-defined real-analytic embedding, and we have

$$\begin{aligned} |V^t-\mathrm{Id }|_{s-\varsigma }\le |V|_s, \quad |t|<\tau . \end{aligned}$$

Moreover, $V^t$ depends analytically on $t$, for $|t|<\tau$.

Proof

This is a direct consequence of the existence theorem for analytic differential equations and the analytic dependence on the initial condition: on the domain $\mathbb{T }^n_{s-\varsigma }$, for $|t|<\tau$, $V^t$ is well-defined, depends analytically on $t$ and satisfies the equality

$$\begin{aligned} V^t=\mathrm{Id }+\int _{0}^{t}V\circ V^u du. \end{aligned}$$

The statement follows. $\square$

Lemma 5.2

Let $X$ and $V$ be two bounded real-analytic vector fields on $\mathbb{T }^n_s$, and $0<\varsigma <s$. Then

$$\begin{aligned} |[X,V]|_{s-\varsigma }\le 2\varsigma ^{-1}|X|_s|V|_s. \end{aligned}$$

Proof

First consider a real-analytic function $f$ defined on $\mathbb{T }^n_s$, and let $\mathcal L _V f$ the Lie derivative of $f$ along $V$, that is

$$\begin{aligned} \mathcal L _V f=\frac{d}{dt}(f\circ V^t)_{|t=0}=F^{\prime }(0), \quad F(t)=f\circ V^t. \end{aligned}$$

Now for $|t|<\tau =\varsigma |V|_{s}^{-1}$, by Lemma 5.1, $F(t)$ is a well-defined real-analytic function on $\mathbb{T }^n_{s-\varsigma }$ and moreover the map $F$ is analytic in $t$, hence by the classical Cauchy estimate

$$\begin{aligned} |\mathcal L _V f|_{s-\varsigma }=|F^{\prime }(0)|\le \tau ^{-1}\sup _{|t|<\tau }|f\circ V^t|_{s-\varsigma }\le \tau ^{-1} |f|_s =\varsigma ^{-1} |V|_{s}|f|_s. \end{aligned}$$

Similarly, we have

$$\begin{aligned} |\mathcal L _X f|_{s-\varsigma } \le \varsigma ^{-1} |X|_{s}|f|_s. \end{aligned}$$

Now, for $1\le i \le n$, if $X^i$ and $V^i$ are the components of $X$ and $V$, then $\mathcal L _X V_i-\mathcal L _V X_i$ are the components of $[X,V]$, so each component of $[X,V]$ is bounded, on the domain $\mathbb{T }^n_{s-\varsigma }$, by $2\varsigma ^{-1}|X|_s|V|_s$ and therefore

$$\begin{aligned} |[X,V]|_{s-\varsigma }\le 2\varsigma ^{-1}|X|_s|V|_s \end{aligned}$$

which is the desired estimate. $\square$

Lemma 5.3

Let $X$ and $V$ be two bounded real-analytic vector fields on $\mathbb{T }^n_s$, and $0<\varsigma <s$. Assume that $|V|_s \le (4e)^{-1}\varsigma$. Then for all $|t|\le 1$,

$$\begin{aligned} |\big (V^t\big )^*X|_{s-\varsigma }\le 2|X|_s. \end{aligned}$$

Proof

From the general identity

$$\begin{aligned} \frac{d}{dt}\big (V^t\big )^*X=\big (V^t\big )^*[X,V] \end{aligned}$$

we have the formal Lie series expansion

$$\begin{aligned} \big (V^t\big )^*X=\sum _{n\in \mathbb{N }}(n!)^{-1}X_nt^n, \quad X_0=X, \quad X_{n}=[X_{n-1},V], \; n\ge 1. \end{aligned}$$

Let $s_i=s-in^{-1}\varsigma$ for $1\le i \le n$, so that $s_0=s$ and $s_n=s-\varsigma$. Using Lemma 5.2, we have

$$\begin{aligned} |X_n|_{s-\varsigma }=|X_n|_{s_n}=|[X_{n-1},V]|_{s_n}\le 2n\varsigma ^{-1}|X_{n-1}|_{s_{n-1}}|V|_{s_{n-1}} \end{aligned}$$

and by induction

$$\begin{aligned} |X_n|_{s-\varsigma }\le (2n\varsigma ^{-1})^n|X_0|_{s_0}|V|_{s_0}^n= (2n\varsigma ^{-1})^n|X|_{s}|V|_{s}^n. \end{aligned}$$

By assumption, $|V|_s \le (4e)^{-1}\varsigma$ so

$$\begin{aligned} |X_n|_{s-\varsigma }\le (2e)^{-n}n^n |X|_s \end{aligned}$$

and therefore

$$\begin{aligned} |\big (V^t\big )^*X|_{s-\varsigma }\le |X|_s \sum _{n\ge 0}(n!)^{-1}(2e)^{-n}(n|t|)^n. \end{aligned}$$

Now for any $n\in \mathbb{N }^*$, $n!\ge n^ne^{-n}$, hence $(n!)^{-1}(2e)^{-n}(n|t|)^n \le (2^{-1}|t|)^n$ and since $|t|\le 1$, the above series is bounded by $2$. This proves the lemma. $\square$

The above estimate can be easily improved, but for $t\ne 0$, the constant $2$ cannot be replaced by $1$.