Convergence to Normal Forms of Integrable PDEs

Abstract

In an infinite dimensional Hilbert space we consider a family of commuting analytic vector fields vanishing at the origin and which are nonlinear perturbations of some fundamental linear vector fields. We prove that one can construct by the method of Poincaré normal form a local analytic coordinate transformation near the origin transforming the family into a normal form. The result applies to the KdV and NLS equations and to the Toda lattice with periodic boundary conditions. One gets existence of Birkhoff coordinates in a neighborhood of the origin. The proof is obtained by directly estimating, in an iterative way, the terms of the Poincaré normal form and of the transformation to it, through a rapid convergence algorithm.

Appendix A. A Technical Lemma

Lemma A.1

Equation (5.63) holds.

Proof

Denote by $d_k$ the l.h.s. of (5.63), one has

$$\begin{aligned}&d_k=\exp \left( \sum _{l=0}^{k-1}\ln m^{-\frac{b}{m}}\right) =\exp \left( -\sum _{l=0}^{k-1} \frac{b}{2^l}\ln 2^l\right) =\exp \left( -\frac{b\ln 2}{2}\sum _{l=0}^{k-1} \frac{l}{2^{l-1}}\right) \\&\quad =\exp \left( -\frac{b\ln 2}{2}4\left( 1-\frac{k+1}{2^k}\right) \right) , \end{aligned}$$

where we used the formula

$$\begin{aligned} \sum _{l=0}^{k-1} \frac{l}{2^{l-1}}=4\left( 1-\frac{k+1}{2^k}\right) . \end{aligned}$$

Now, the result immediately follows. $\quad \square$

Lemma A.2

Equation (5.64) holds.

Proof

We use the discrete analogue of the formula of the Duhamel formula, namely we make the substitution $r_k=d_ks_k$, where $d_k$ was defined in the proof of Lemma A.1. One gets

$$\begin{aligned} r_{k+1}=d_{k+1}s_{k+1}=q_{2^k}d_ks_{k+1}=q_{2^k}(d_ks_k-\delta _k) \end{aligned}$$

and thus

$$\begin{aligned} s_{k+1}=s_k-\frac{\delta _k}{d_k} ,\quad s_1=\frac{r_1}{d_1}=r_1 , \end{aligned}$$

from which

$$\begin{aligned} s_k=s_1-\sum _{l=1}^{k-1}\frac{\delta _l}{d_l}. \end{aligned}$$

Now, one has

$$\begin{aligned} \sum _{l=1}^{k-1}\frac{\delta _l}{d_l}=\sum _{l=1}^{k-1}\frac{\delta }{4^l} \frac{4^b}{4^{b\frac{l+1}{2^l}}}\le \sum _{l=1}^{k-1}\frac{\delta }{4^l} {4^b}=\frac{4^b}{3}\delta . \end{aligned}$$

Thus,

$$\begin{aligned} r_k\ge d_k\left( \frac{r_1}{d_1}- \frac{4^b}{3}\delta \right) . \end{aligned}$$

$\square$

Lemma A.3

Equation (5.66) holds.

Proof

The first inequality is trivial. We discuss the second one. Using the definition of $r_{k+1}$, we have

$$\begin{aligned} \frac{\epsilon _k}{r_k-r_{k+1}}=\frac{\epsilon _k}{r_k(1-q_{2^k})+ q_{2^k}\delta _k} \le \frac{\epsilon _k}{r_k(1-q_{2^k})} \ ; \end{aligned}$$

(A.86)

now, one has

$$\begin{aligned} 1-q_m=1-\exp \left( -\frac{b}{m}\ln m\right) , \end{aligned}$$

which is of the form $1-e^{-x}$ with x varying from 0 to $\frac{b}{2}\ln 2$. Remarking that in an interval $[0,x_0]$ one has

$$\begin{aligned} 1-e^{-x}\ge e^{-x_0}x , \end{aligned}$$

we get

$$\begin{aligned} 1-q_m\ge 2^{-b/2}\left( \frac{b}{m}\ln m\right) =\frac{b}{2^{k+b/2}}\ln 2^k=\frac{k}{2^k} \frac{b}{2^{b/2}}\ln 2 , \end{aligned}$$

and thus, for $k\ge 1$,

$$\begin{aligned} \frac{\epsilon _k}{r_k(1-q_{2^k})}\le \frac{\epsilon _0}{r_{\infty }} \frac{2^{b/2}}{b\ln 2}\frac{2^k}{k}\frac{1}{4^k}=\frac{\epsilon _0}{r_{\infty }} \frac{2^{b/2}}{b\ln 2}\frac{1}{k2^k}. \end{aligned}$$

Now one has

$$\begin{aligned}&\sum _{k\ge 1}\frac{x^k}{k}=\sum _{k\ge 1}\int _0^xy^{k-1}dy=\sum _{k\ge 0 }\int _0^xy^{k}dy=\int _0^x\frac{1}{1-y}dy\\&\quad =\left[ -\ln \left| 1-y\right| \right] _0^x =-\ln \left| 1-x\right| , \end{aligned}$$

which, for $x=1/2$, gives

$$\begin{aligned} \sum _{k\ge 1}\frac{1}{k2^k}=\ln 2 , \end{aligned}$$

and thus

$$\begin{aligned} \sum _{l\ge {2}}\frac{\epsilon _{l-1}}{ {r_{l-1}-r_l}}\le \frac{\epsilon _0}{r_{\infty }} 2^{b/2}. \end{aligned}$$

adding the first term, namely $\frac{8 C}{8C-1}\frac{\epsilon _0}{r_0}$, one gets the thesis immediately follows. $\quad \square$

Lemma A.4

Let $\mathbf{F}$ be a summable normally analytic vector fields with $F^i$ having a zero of order m at the origin for all i. Let $0<\alpha \le 1$, then

$$\begin{aligned} \left\| \underline{\mathbf{F}} \right\| _{\alpha r}\le \alpha ^m\left\| \underline{\mathbf{F}} \right\| _{r}. \end{aligned}$$

(A.87)

Proof

Consider the function ${\underline{\mathbf{F}}}(z)=\sum _{Q,i}F_{Q,i}z^Q\vec {e}_i$; since all the coefficients are positive one has, for any i,

$$\begin{aligned} \sum _{Q}F_{Q,i}(\alpha z)^Q=\alpha ^m\sum _{Q,i}\alpha ^{|Q|-m}F_{Q,i}z^Q\le \alpha ^m\sum _{Q}F_{Q,i}z^Q \end{aligned}$$

Thus one gets

$$\begin{aligned} \sup _{\left\| z \right\| \le \alpha r}\left\| {\underline{\mathbf{F}}}(z) \right\| _+=\sup _{\left\| z \right\| \le r}\left\| {\underline{\mathbf{F}}}(\alpha z) \right\| _+ \le \alpha ^m \sup _{\left\| z \right\| \le r}\left\| {\underline{\mathbf{F}}}(z) \right\| _+\ . \end{aligned}$$

$\square$