A New KAM Iteration with Nearly Infinitely Small Steps in Reversible Systems of Polynomial Character

Abstract

In this paper we prove the persistence of invariant tori of analytic reversible systems under Brjuno–Rüssmann’s non-resonant condition by an improved KAM iteration, but the frequency may undergo some drift. Furthermore, if the frequency mapping has nonzero Brouwer’s topological degree, then the invariant torus with prescribed frequency can persist. In the proof we propose a new method of KAM iteration for reversible systems, containing an artificial parameter $q, 0< q <1,$ which makes the steps of KAM iteration infinitely small in the speed of the function $q^{n}\epsilon$ rather than super exponential function $\epsilon ^{\lambda ^{n}}, 1<\lambda <2$.

Appendix

In this section we formulate some lemmas which have been used in the previous section.

Lemma 2.4

Let $s_{+}=s-\sigma , r_{+}=\eta r,$ and

$$\begin{aligned} \partial _{\hat{\omega }}h(x)=\sum _{k}i\langle k, \hat{\omega }\rangle h_{k}e^{i\langle k, x\rangle }. \end{aligned}$$

Suppose that $\Vert h\Vert _{s,r}\le 2\Lambda (\tau )\sigma \alpha ^{-1}\epsilon ,$ $|\hat{\omega }|_{s_{+},r_{+}}\le a\epsilon ,$ where $\epsilon <\frac{b^{2}\alpha ^{2} }{4\Lambda ^{2}(\tau )}.$ Then we have

$$\begin{aligned} \Vert \partial _{\hat{\omega }}h(x)\Vert _{s_{+}, r_{+}}\le ab\epsilon . \end{aligned}$$

Proof

$$\begin{aligned} \Vert \partial _{\hat{\omega }}h(x)\Vert _{s_{+}, r_{+}}= & {} \sum _{k}|\langle k,\hat{\omega }\rangle |\cdot \Vert h_{k}\Vert e^{|k|s_{+}}\\\le & {} \sum _{k}|\langle k,\hat{\omega }\rangle | \cdot \Vert h\Vert _{s,r}e^{-|k|s}e^{|k|s_{+}}\\\le & {} |\hat{\omega }|_{s_{+},r_{+}}\Vert h\Vert _{s,r}\sum _{k}|k|e^{-|k|(s-s_{+})}\\\le & {} 2a \epsilon \Lambda (\tau )\alpha ^{-1}\epsilon ^{\frac{1}{2}}=ab\epsilon . \end{aligned}$$

Lemma 2.5

Let $s_{+}=s-\sigma , r_{+}=\eta r,$ and

$$\begin{aligned} \partial _{\hat{\omega }+f_{+}^{1}}g(x)=\sum _{k}i\langle k, \hat{\omega }+f^{1}_{+}\rangle g_{k}e^{i\langle k,x\rangle }. \end{aligned}$$

Suppose that $\Vert g(x)\Vert _{s,r}\le 2\Lambda (\tau )\sigma \alpha ^{-1}\epsilon ,$ $\Vert f^{1}_{+}\Vert _{s_{+},r_{+}}\le 2((1-a)+b+ab)\epsilon ,$ $|\hat{\omega }|_{s_{+},r_{+}}\le a\epsilon ,$ where $\epsilon <\frac{b^{2}\alpha ^{2} }{4\Lambda ^{2}(\tau )}.$ Then we have

$$\begin{aligned} \Vert \partial _{\hat{\omega }+f_{+}^{1}}g(x)\Vert _{s_{+},r_{+}}\le 3((1-a)+b+ab)\epsilon . \end{aligned}$$

Proof

$$\begin{aligned} \Vert \partial _{\hat{\omega }+f_{+}^{1}}g(x)\Vert _{s_{+},r_{+}}= & {} \sum _{k} |\langle k, \hat{\omega }+f^{1}_{+}\rangle |\cdot \Vert g_{k}\Vert e^{|k|s_{+}}\\\le & {} \sum _{k}|\langle k, \hat{\omega }+f^{1}_{+}\rangle |\cdot \Vert g\Vert _{s,r}e^{-|k|s}e^{|k|s_{+}}\\\le & {} (|\hat{\omega }|_{s_{+},r_{+}}+\Vert f^{1}_{+}\Vert _{s_{+},r_{+}})\Vert g\Vert _{s,r}\sum _{k}|k|e^{-|k|(s-s_{+})} \\\le & {} 3((1-a)+b+ab)\epsilon . \end{aligned}$$