Persistence of Hyperbolic-type Degenerate Lower-dimensional Invariant Tori with Prescribed Frequencies in Hamiltonian Systems

Abstract

It is known that under Kolmogorov’s nondegeneracy condition, the nondegenerate hyperbolic invariant torus with Diophantine frequencies will persist under small perturbations, meaning that the perturbed system still has an invariant torus with prescribed frequencies. However, the degenerate torus is sensitive to perturbations. In this paper, we prove the persistence of two classes of hyperbolic-type degenerate lower-dimensional invariant tori, one of them corrects an earlier work [34] by the second author. The proof is based on a modified KAM iteration and analysis of stability of degenerate critical points of analytic functions.

APPENDIX A. SOME BASIC PROPERTY OF FUNCTION

Lemma 8

Suppose that $P(w)=\sum_{\beta\in{\mathbb{Z}}_{+}^{n+2}}P_{\beta}(x)\bar{w}^{\beta}$ is analytic on $D(s,r).$ If $\|P\|_{s,\varepsilon}\leqslant\varepsilon^{2d+1}$ , then

$$\|P_{\beta}\|_{s}\leqslant\varepsilon^{2d+1-|\beta|}\ \ \mbox{for}\ |\beta|\leqslant 2d.$$

Moreover, if $\|P_{L}\|_{s,\varepsilon}\leqslant\varepsilon^{2d+1}$ and $\varepsilon\leqslant\frac{r_{0}}{4}\leqslant\frac{r}{2}\leqslant\frac{r_{0}}{2}$ , then we have $\|P_{L}\|_{s,r}\lessdot\varepsilon.$

Proof

This lemma can be verified easily according to the definition and so we omit the details.     $\square$

Lemma 9

Suppose that $P(w)=\sum_{|\beta|\geqslant 2d+1}P_{\beta}(x)\bar{w}^{\beta}$ is analytic on $D(s,r).$ Let $\|P\|_{s,r}\leqslant T.$ Then $\|P\|_{s,\varepsilon}\lessdot T\varepsilon^{2d+1}$ if $\varepsilon\leqslant\frac{r_{0}}{4}\leqslant\frac{r}{2}\leqslant\frac{r_{0}}{2}$ .

Proof

Let

$$f(\bar{w})=\sum_{|\beta|\geqslant 2d+1}\|P_{\beta}\|_{s}\bar{w}^{\beta}.$$

Then we have

$$\|P\|_{s,\varepsilon}=\sup_{\bar{w}\in D(\varepsilon)}|f(\bar{w})|.$$

The Cauchy estimate gives $\|P_{\beta}\|_{s}\leqslant\frac{T}{r^{|\beta|}},$ then it follows that

$$|f(\bar{w})|\leqslant T\sum_{|\beta|\geqslant 2d+1}\frac{|\bar{w}^{\beta}|}{r^{|\beta|}}.$$

For all $\bar{w}\in D(\varepsilon)$, we have

$$|f(\bar{w})|\leqslant T\sum_{|\beta|\geqslant 2d+1}\frac{\varepsilon^{|\beta|}}{r^{|\beta|}}\leqslant\frac{T}{r^{2d+1}}\sum_{\beta\in{\mathbb{Z}}_{+}^{n+2}}\frac{1}{2^{|\beta|}}\cdot\varepsilon^{2d+1}\lessdot\ T\varepsilon^{2d+1}\ .$$

$\square$

APPENDIX B. IMPLICIT FUNCTION THEOREM

Lemma 10

Let $\Omega_{\rho}$ be a complex neighborhood of $\Omega$ in $\mathbb{C}^{n}$ with neighborhood radius $\rho>0.$ Suppose $f:\Omega_{\rho}\to\mathbb{C}^{n}$ is an analytic mapping and

$$|f-id|_{\Omega_{\rho}}\leqslant\varepsilon<\frac{\rho}{2},$$

where $|\cdot|_{\Omega_{\rho}}$ indicates the super-norm on $\Omega_{\rho}$ . Then $f$ has an analytic inverse $f^{-1}:\Omega_{\rho-2\varepsilon}\to\Omega_{\rho}$ such that $|f^{-1}-id|_{\Omega_{\rho-2\varepsilon}}\leqslant\varepsilon.$

This lemma is proved in [23] and here we omit the details.     $\square$

APPENDIX C. SOLVING HOMOLOGICAL LINEAR EQUATION

Let $E(u,v)=\frac{1}{2}(au^{2}-bv^{2}).$ Consider the linear operator defined by

$$L:X(u,v)\to\{E,X\}(u,v),\ \ X\in\mathcal{X}=\mbox{span}\{u^{j}v^{l}\ |\ j+l=m\},$$

where $\{E,X\}$ denotes the Poisson bracket of $E$ and $X$. Obviously, $\mathcal{X}$ is the $(m+1)$-dimensional linear space which consists of all homogeneous polynomials of $(u,v)$ of degree $m$ spanned by $\{v^{m},uv^{m-1},\cdots u^{m-1}v,u^{m}\}$. Let $A$ be the matrix representation of the linear operator $L$ on the basis $\{v^{m},uv^{m-1},\cdots u^{m-1}v,u^{m}\}$, then we have the following lemma.

Lemma 11

The matrix $A$ is given by

$$\begin{pmatrix}0&b&0&0&\cdots&0&0&0&0\\ ma&0&2b&0&\cdots&0&0&0&0\\ 0&(m-1)a&0&3b&\cdots&0&0&0&0\\ \vdots&\vdots&\cdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&0&0&0&\cdots&0&(m-2)b&0&0\\ 0&0&0&0&\cdots&3a&0&(m-1)b&0\\ 0&0&0&0&\cdots&0&2a&0&mb\\ 0&0&0&0&\cdots&0&0&a&0\end{pmatrix}$$

(C.1)

Moreover, $A$ has $m+1$ eigenvalues $\{(m-2l)\mu\ \ |\ l=0,1,\cdots m\}$ , where $\mu=\sqrt{ab}$ . In particular, if $ab\geqslant 0$ and $\lambda\in{\mathbb{C}},\lambda\neq 0,\mbox{Re}(\lambda)=0,$ then

$$|\det(\lambda I+A)|\geqslant|\lambda|^{m+1}\ \mbox{and}\ \|(\lambda I+A)^{-1}\|\leqslant\max\{c|\lambda|^{-m-1},\ 2|\lambda|^{-1}\},$$

where $\|\cdot\|$ denotes the norm of operators and $c$ is a constant depending only on $a,b,m.$

Proof

By the definition of the Poisson bracket we have

$$\{E,X\}=E_{u}X_{v}-E_{v}X_{u}=auX_{v}+bvX_{u}.$$

Let $X(u,v)=\sum_{j+l=m}X_{j,l}u^{j}v^{l},$ where $(X_{0,m},X_{1,m-1},\cdots X_{m,0})$ are the coordinates of $X$ under the basis $\{u^{j}v^{m-j},\ j=0,1,\cdots m\}.$ It follows easily that

$$\{E,X\}=a\sum_{j=1}^{m}(m+1-j)X_{j-1,m+1-j}u^{j}v^{m-j}+b\sum_{j=0}^{m-1}(j+1)X_{j+1,m-j-1}u^{j}v^{m-j}.$$

Then we have (C.1).

Below we are going to find the eigenvalues of $A$. If $a=0$ or $b=0$ , then $\mu=0$ and so $A$ has only zero as eigenvalue. Now suppose $a>0$ and $b>0$. Define a linear symplectic transformation $\phi_{1}$ by

$$(u,v)=\phi_{1}(u_{+},v_{+})=(\kappa u_{+},\kappa^{-1}v_{+}),\ \mbox{where}\ \ \kappa=\left(\frac{b}{a}\right)^{\frac{1}{4}}.$$

Let $\mu=\sqrt{ab}.$ Then under the new symplectic basis $(u_{+},v_{+})$ we have

$$E(u_{+},v_{+})=\frac{\mu}{2}(u_{+}^{2}-v_{+}^{2}).$$

Then we define another symplectic transformation $\phi_{2}$ by

$$(u_{+},v_{+})=\phi_{2}(u,v)=\left(\frac{1}{\sqrt{2}}(u-v),\frac{1}{\sqrt{2}}(u+v)\right).$$

Under the new symplectic basis $(u,v)$, we have $E(u,v)=\mu uv.$ Since the operator $L$ is invariant under symplectic transformations, then, under the new symplectic variables $(u,v),$ it follows that

$$LX(u,v)=\{E,X\}(u,v),\ \ \ E(u,v)=\mu uv.$$

It is easy to see that $\mu(m-2j)$ are eigenvalues of $L$ with eigenvectors $u^{j}v^{m-j}$ for all $\ j=0,1,2\cdots m$. Since $\mu\neq 0,$ they are all $m+1$ different eigenvalues of $A$.

If $ab<0$, instead of the above $\phi_{2}$, we use a symplectic transformation with multiplier ${\rm i}=\sqrt{-1}$, which is defined by

$$\phi_{2}^{\prime}:(u_{+},v_{+})\to(u,v)\ \mbox{with}\ \ u=\frac{1}{\sqrt{2}}(u_{+}-{\rm i}v_{+}),v=\frac{1}{\sqrt{2}}(u_{+}+{\rm i}v_{+}).$$

In the same way the result follows easily and we omit the details.

Below we estimate $(\lambda I+A)^{-1}$. If $ab\geqslant 0$, obviously, we have

$$|\det(\lambda I+A)|=\Big{|}\prod_{j=0}^{m}(\lambda+\lambda_{j})\Big{|},$$

where $\lambda_{j}=\sqrt{ab}(m-2j),\ j=0,1,2,\cdots m.$ Obviously, it follows that $|\det(\lambda I+A)|\geqslant|\lambda|^{m+1}.$

Let $\lambda_{0}=\max\{1,2\|A\|\}$, where $\|A\|$ is the operator norm of $A$. If $|\lambda|\geqslant\lambda_{0}$, we have $\|\lambda^{-1}A\|\leqslant\frac{1}{2}$ and so $\|(I+\lambda^{-1}A)^{-1}\|\leqslant 2.$ Thus,

$$\|(\lambda I+A)^{-1}\|=|\lambda|^{-1}\|(I+\lambda^{-1}A)^{-1}\|\leqslant 2|\lambda|^{-1}.$$

If $|\lambda|\leqslant\lambda_{0},$ since $(\lambda I+A)^{-1}=(\det(\lambda I+A))^{-1}(\lambda I+A)^{*}$, where $(\lambda I+A)^{*}$ is the adjoint matrix. It is easy to see that $\|(\lambda I+A)^{*}\|\leqslant c,$ where $c$ depends on $a,b,m,\lambda_{0}.$ So

$$\ \|(\lambda I+A)^{-1}\|\leqslant c|\lambda|^{-m-1}.$$

Hence,

$$\ \|(\lambda I+A)^{-1}\|\leqslant\max\{c|\lambda|^{-m-1},\ 2|\lambda|^{-1}\}.$$

$\square$

Lemma 12

Suppose that $\omega$ is a Diophantine vector, i. e., (1.2) holds. Let

$$E(u,v)=\frac{1}{2}(au^{2}-bv^{2}),\ \ P=\sum_{j+l=m}P_{jl}(x)u^{j}v^{l}\ \ \mbox{ and}\ \ [P]=0,$$

where $[P]$ denotes the average of $P$ with respect to $x$ on ${\mathbb{T}}^{n}.$ Suppose $a\geqslant 0$ and $b\geqslant 0.$ Let

$$P_{jl}(x)=\sum_{k\neq 0}P_{ijk}e^{\rm{i}\langle k,x\rangle}.$$

Then there exists a unique homogeneous function $W=\sum_{j+l=m}W_{jl}(x)u^{j}v^{l}$ satisfying

$$\tilde{L}W=\langle\omega,W_{x}\rangle+\{E,W\}=P.$$

(C.2)

Moreover, we have

$$\|W\|_{s-\sigma,r}\lessdot\ \alpha^{-m-1}\sigma^{-\tau(m+1)-n}\|P\|_{s,r},\ \forall\ 0<\sigma<s,r>0.$$

Proof

Let $X=(W_{0,m},W_{1,m-1},\cdots W_{m,0}),$ $Y=(P_{0,m},P_{1,m-1},\cdots P_{m,0}).$ Then Eq. (C.2) is equivalent to

$$\langle\omega,X_{x}\rangle+AX=Y.$$

Let $\{X_{k}\}$ and $\{Y_{k}\}$ be the Fourier coefficients of $X$ and $Y$, respectively, then

$$(\lambda I+A)X_{k}=Y_{k},\ \ \lambda={\rm i}\langle\omega,k\rangle.$$

It follows that

$$X_{k}=(\lambda I+A)^{-1}Y_{k}.$$

Note that $|\lambda|\geqslant\frac{\alpha}{|k|^{\tau}},\ \ \forall|k|\neq 0$ and $0<\alpha\leqslant 1.$ By Lemma 11, it follows that

$$\|(\lambda I+A)^{-1}\|\leqslant\max\{c|\lambda|^{-m-1},\ 2|\lambda|^{-1}\}\leqslant\max\{c,2\}\left(\frac{\alpha}{|k|^{\tau}}\right)^{-m-1}.$$

Then we have

$$\ \ \|X_{k}\|\lessdot\ \left(\frac{\alpha}{|k|^{\tau}}\right)^{-m-1}\|Y_{k}\|.$$

Then we have

$$\|X\|_{s-\sigma}\lessdot\ \alpha^{-m-1}\sigma^{-(m+1)\tau-n}\|Y\|_{s}$$

and so

$$\|W\|_{s-\sigma,r}\lessdot\ \alpha^{-m-1}\sigma^{-\tau(m+1)-n}\|P\|_{s,r},\ \ \forall\ 0<\sigma<s,r>0.$$

$\square$

Lemma 13 (Solving Homological Equation)

Let

$$N=\langle\omega,y\rangle+\frac{1}{2}\langle My,y\rangle+\langle y,E_{1}(u,v)\rangle+E_{2}(u,v)+\tilde{N}(\bar{w}),$$

where

$$E_{1}(u,v)=\tilde{a}u+\tilde{b}v,\ \ E_{2}(u,v)=\frac{1}{2}(au^{2}-bv^{2}),\ \ \tilde{N}(\bar{w})=\sum_{3\leqslant|\beta|\leqslant 2d}N_{\beta}\bar{w}^{\beta},$$

where $\tilde{a},\tilde{b}\in{\mathbb{R}}^{n}.$ Suppose $a\geqslant 0,b\geqslant 0$ and $\omega$ is a Diophantine vector with (1.2) holding.

Write as

$$\mathcal{L}W=\{N,W\}=\tilde{L}W+H^{0}(W)+H^{1}(W)+H^{2}(W),$$

where

$$\tilde{L}W=\langle\omega,W_{x}\rangle+\{E_{2},W\},$$

$$H^{0}(W)=\langle y,\{E_{1},W\}\rangle=W_{u}\langle\tilde{a},y\rangle-W_{v}\langle\tilde{b},y\rangle,$$

(C.3)

and

$$H^{1}(W)=H^{1}_{0}(W)+H^{1}_{1}(W),\ \ H^{2}(W)=\{\tilde{N},W\},$$

(C.4)

where

$$H^{1}_{0}(W)=\langle E_{1},W_{x}\rangle=\langle\tilde{a},W_{x}\rangle u+\langle\tilde{b},W_{x}\rangle v,\ \ H^{1}_{1}(W)=-\langle MW_{x},y\rangle.$$

(C.5)

Recall that $w=(x,y,u,v)$ and $\bar{w}=(y,u,v)$ , $\beta=(i,j,l)\in{\mathbb{Z}}_{+}^{n+2},\ \frac{r_{0}}{2}\leqslant r\leqslant r_{0}.$ Let

$$P(w)=\sum_{|\beta|\leqslant 2d}P_{\beta}(x)\bar{w}^{\beta},\ \ P_{\beta}(x)=\sum_{|k|\leqslant K}P_{\beta,k}e^{{\rm i}\langle k,x\rangle}.$$

Suppose

$$[P]=0,\ \|P\|_{s,\varepsilon}\leqslant\varepsilon^{2d+1}.$$

Then there exist $W$ and $R$ such that $\mathcal{L}W=P+R,$ where

$$W(w)=\sum_{|\beta|\leqslant 2d}W_{\beta}(x)\bar{w}^{\beta},\ \ R=\sum_{2d+1\leqslant|\beta|\leqslant 4d-1}R_{\beta}(x)\bar{w}^{\beta}$$

with $[W]=[R]=0.$ Moreover, if $\eta=\frac{\varepsilon}{\alpha^{\mu}\sigma^{\mu\tau+\nu n+2d}}\leqslant 1$ , we have

$$\|W\|_{s-\sigma,r}\lessdot\eta,\ \ \|W\|_{s-\sigma,\varepsilon}\lessdot\ \varepsilon^{2d}\eta,\ \ \|R\|_{s-\sigma,r}\lessdot\ \eta/\sigma,$$

where

$$\mu=\sum_{j=1}^{2d+1}\sum_{i=1}^{j+1}i,\ \ \ \ \nu=\sum_{j=1}^{2d+1}j,\ \ \ 0<\sigma<s.$$

Proof

It is easy to see that $H^{0}(W)$ has the same order as $W$ in $(y,u,v)$. More precisely, it makes the order of $(u,v)$ one less and the order of $y$ one more. However, $H^{1}(W)$ raises the order of $(y,u,v)$ one more. So we start with $0$-order terms, and then do it in the same way by increasing the order one by one. Inductively, after we normalize all $m$-order terms $\sum_{|i|+j+l=m}P_{ijl}y^{i}u^{j}v^{l}$ , we turn to $(m+1)$-order terms. For $m$-order terms, we begin with $0$-order terms of $y$ with $|i|=0$. When solving $|i|$-order terms of $y$, it can produce some $(|i|+1)$-order terms of $y$, which is put into the next step. When $|i|=m$, the term like $W(\bar{w})=\sum_{|i|=m}P_{i00}y^{i}$ does not depend on $(u,v)$, so $H^{0}(W)=0$ and it does not produce any new term.

When solving all $m$-order terms of $(y,u,v)$, $H^{1}(W)$ may produce some $(m+1)$-order terms of $(y,u,v)$, which are considered in solving $(m+1)$-order terms later. Also note that

$$H^{2}(W)=\{\tilde{N}(\bar{w}),W\}=\sum_{3\leqslant|\beta|\leqslant 2d}\{N_{\beta}\bar{w}^{\beta},W\},$$

which makes the order of $W$ higher than $2$. These terms will be considered in later steps. After $2d$ steps we finish the proof with some higher terms remaining.

1. Solving $0$-order terms: Let $P_{0}=P$, first we consider the $0$-order term in $P_{0}$. For convenience, define the project operators $\Pi^{\bar{w}}_{m}$ and $\Pi^{y}_{k}$ by

$$\Pi^{\bar{w}}_{m}\Big{(}\sum_{\beta}P_{\beta}\bar{w}^{\beta}\Big{)}=\sum_{|\beta|=m}P_{\beta}\bar{w}^{\beta},\ \ \Pi^{y}_{k}\Big{(}\sum_{\beta}P_{\beta}\bar{w}^{\beta}\Big{)}=\sum_{\beta=(i,j,l),|i|=k}P_{\beta}\bar{w}^{\beta}.$$

Let $P^{0}_{0}=P_{000}(x)$ be the $0$-order term of $P_{0}$. By Lemma 12 we find

$$W^{0}(w)=W_{000}(x)=\tilde{L}^{-1}\big{(}P_{0}^{0}(x)\big{)},$$

in which $W_{000,k}=\frac{P_{000,k}}{\rm{i}\langle\omega,k\rangle},k\neq 0.$ Moreover,

$$\|W^{0}\|_{s-\sigma/2,r}\leqslant\frac{\varepsilon}{\alpha\sigma^{\tau+n}},\ \ \|W^{0}\|_{s-\sigma/2,\varepsilon}\lessdot\ \frac{\varepsilon^{2d+1}}{\alpha\sigma^{\tau+n}},\ \ [W^{0}]=0.$$

Noting that $H^{0}(W^{0})=0$, we have

$$\mathcal{L}W^{0}=P_{0}^{0}+H^{1}(W^{0})+H^{2}(W^{0}).$$

(C.6)

2. Solving $1$-order terms: Now we consider $1$-order terms. Let

$$P_{1}=(P_{0}-P^{0}_{0})-H^{1}(W^{0})-H^{2}(W^{0}).$$

Then (C.6) becomes

$$\mathcal{L}W^{0}=P_{0}-P_{1}.$$

(C.7)

Since $H^{1}(W^{0})$ and $H^{2}(W^{0})$ include $W^{0}_{x}$, by the estimate for $W^{0}$, it is easy to see

$$\|P_{1}\|_{s-\sigma,r}\leqslant\frac{\varepsilon}{\alpha\sigma^{\tau+n+1}}.$$

Moreover, if $\varepsilon$ is sufficiently small such that $\frac{\varepsilon}{\alpha\sigma^{\tau+n+1}}\ll 1,$ then it follows that

$$\ \ \|P_{1}\|_{s-\sigma,\varepsilon}\lessdot\ \bigl{(}\varepsilon^{2d+1}+\frac{\varepsilon^{2d+2}}{\alpha\sigma^{\tau+n+1}}\bigr{)}\lessdot\ \varepsilon^{2d+1}.$$

Let

$$P^{1}_{1}=\Pi_{1}^{\bar{w}}P_{1}=\Pi_{1}^{\bar{w}}(P_{0}-P^{0}_{0})-H^{1}(W^{0}).$$

Let

$$W^{1}_{0}=\tilde{L}^{-1}(\Pi_{0}^{y}P^{1}_{1}),\ \ W^{1}_{1}=\tilde{L}^{-1}\big{(}(\Pi_{1}^{y})P^{1}_{1}-H^{0}(W^{1}_{0})\bigr{)},W^{1}=W^{1}_{0}+W^{1}_{1}.$$

Note that $W^{1}_{1}$ does not depend on $(u,v)$ and $H^{0}(W^{1}_{1})=0$. It follows that

$$\mathcal{L}W^{1}=P^{1}_{1}+H^{1}(W^{1})+H^{2}(W^{1}).$$

Let

$$P_{2}=P_{1}-P^{1}_{1}-H^{1}(W^{1})-H^{2}(W^{1}).$$

Then $\Pi^{\bar{w}}_{m}P_{2}=0,\ \forall\ m\leqslant 1.$ Moreover,

$$\mathcal{L}W^{1}=P_{1}-P_{2},\ \ \ \mathcal{L}(W^{0}+W^{1})=P_{0}-P_{2}.$$

(C.8)

Below we estimate $W^{1}$ and $P_{2}$. Obviously $P^{1}_{1}$ and $\Pi_{0}^{y}P^{1}_{1}$ have the same estimates as $P^{1}$, by Lemma 12 it follows that

$$\|W^{1}_{0}\|_{s-\frac{4}{3}\sigma,r}\lessdot\frac{\varepsilon}{\alpha^{2+1}\sigma^{2\tau+n+\tau+n+1}},\ \ \|W^{1}_{0}\|_{s-\frac{4}{3}\sigma,\varepsilon}\lessdot\ \frac{\varepsilon^{2d+1}}{\alpha^{2}\sigma^{2\tau+n}},\ \ [W^{1}_{0}]=0.$$

Also note that $H^{0}(W^{1}_{0})$ has the same estimate as $W^{1}_{0}$, so we have

$$\|W^{1}_{1}\|_{s-\frac{5}{3}\sigma,r}\lessdot\frac{\varepsilon}{\alpha^{2+1+1}\sigma^{(2+1)\tau+2n+\tau+n+1}},\ \ \|W^{1}_{1}\|_{s-\frac{5}{3}\sigma,\varepsilon}\lessdot\ \frac{\varepsilon^{2d+1}}{\alpha^{2+1}\sigma^{(2+1)\tau+2n}},\ \ [W^{1}_{1}]=0.$$

By the above estimate for $W^{1}_{0},W^{1}_{1}$, it is easy to see

$$\|P_{2}\|_{s-2\sigma,r}\leqslant\frac{\varepsilon}{\alpha^{2+1+1}\sigma^{(2+1)\tau+2n+\tau+n+2}}.$$

Moreover, if $\varepsilon$ is sufficiently small such that $\frac{\varepsilon}{\alpha^{2+1}\sigma^{(2+1)\tau+2n+1}}\ll 1,$ then it follows that

$$\ \ \|P_{2}\|_{s-2\sigma,\varepsilon}\lessdot\ \bigl{(}\varepsilon^{2d+1}+\frac{\varepsilon^{2d+2}}{\alpha^{2+1}\sigma^{(2+1)\tau+2n+1}}\bigl{)}\lessdot\ \varepsilon^{2d+1}.$$

Obviously $W^{1}$ has the same estimate as $W^{1}_{1}$ and we have

$$\|W^{1}\|_{s-2\sigma,r}\leqslant\frac{\varepsilon}{\alpha^{4}\sigma^{4\tau+3n+1}},\ \ \|W^{1}\|_{s-2\sigma,\varepsilon}\lessdot\ \frac{\varepsilon^{2d+1}}{\alpha^{3}\sigma^{3\tau+2n}},\ \ [W^{1}]=0.$$

3. Solving $2d$-order terms: After $2d$ steps, we have solved all $(2d-1)$-order terms and obtained

$$W^{2d-1}=W^{2d-1}_{0}+W^{2d-1}_{1}+\cdots W^{2d-1}_{2d-1}$$

and $P_{2d}$ such that

$$\mathcal{L}(W^{2d-1})=P^{2d-1}_{2d-1}+H^{1}(W^{2d-1})+H^{2}(W^{2d-1})$$

and

$$\mathcal{L}(W^{1}+\cdots W^{2d-1})=P_{0}-P_{2d}$$

where

$$P_{2d}=P_{2d-1}-P^{2d-1}_{2d-1}-H^{1}(W^{2d-1})-H^{2}(W^{2d-1}).$$

Here also note that $W^{2d-1}_{2d-1}$ does not depend on $(u,v)$ and so $H^{0}(W^{2d-1}_{2d-1})=0.$

Moreover, for $W^{2d-1}$ and $P_{2d}$ we have estimates

$$\|W^{2d-1}\|_{s-(2d+\frac{1}{2})\sigma,r}\leqslant\frac{\varepsilon}{\alpha^{\mu_{2d-1}}\sigma^{\mu_{2d-1}\tau+\nu_{2d-1}n+2d-1}},$$

$$\|W^{2d-1}\|_{s-(2d+\frac{1}{2})\sigma,\varepsilon}\lessdot\ \frac{\varepsilon^{2d+1}}{\alpha^{\nu_{2d}}\sigma^{\nu_{2d}+2dn}},$$

and $[W^{2d-1}]=0,$ where the notations $\mu_{m}$ and $\nu_{k}$ are defined by

$$\mu_{m}=\nu_{0}+\nu_{1}+\nu_{2}\cdots+\nu_{m+1},\ \ \nu_{k}=1+2+\cdots+k+1.$$

Then it follows that

$$\|P_{2d}\|_{s-2d\sigma,r}\leqslant\frac{\varepsilon}{\alpha^{\mu_{2d-1}}\sigma^{\mu_{2d-1}\tau+\nu_{2d-1}n+2d}}.$$

Moreover, if $\varepsilon$ is sufficiently small such that $\frac{\varepsilon}{\alpha^{\nu_{2d}}\sigma^{\nu_{2d}+2dn+1}}\leqslant 1,$ it follows that

$$\ \ \|P_{2d}\|_{s-2\sigma,\varepsilon}\lessdot\ \frac{\varepsilon^{2d+2}}{\alpha^{\nu_{2d}}\sigma^{\nu_{2d}+2dn+1}}\lessdot\ \varepsilon^{2d+1}.$$

Below we consider $2d$-order terms. Let $P_{2d}^{2d}=\Pi_{2d}^{\bar{w}}P_{2d}$. Obviously, $P_{2d}^{2d}=\sum_{j=0}^{2d}\Pi_{j}^{y}P_{2d}^{2d}.$ Let

$$W^{2d}_{0}=\tilde{L}^{-1}(\Pi_{0}^{y}P^{2d}_{2d}),\ \ W^{2d}_{1}=\tilde{L}^{-1}\big{(}\Pi_{1}^{y}P^{2d}_{2d}-H^{0}(W^{2d}_{0})\bigr{)},\ $$

$$W^{2d}_{2}=\tilde{L}^{-1}\big{(}\Pi_{2}^{y}P^{2d}_{2d}-H^{0}(W^{2d}_{1})\bigr{)},\quad\ldots,\quad W^{2d}_{2d}=\tilde{L}^{-1}\big{(}\Pi_{2d}^{y}P^{2d}_{2d}-H^{0}(W^{2d}_{2d-1})\bigr{)}.$$

Thus,

$$W^{2d}=W^{2d}_{0}+W^{2d}_{1}+\cdots W^{2d}_{2d}.$$

Here we note that $W^{2d}_{2d}$ does not depend on $(u,v)$ and so $H^{0}(W^{2d}_{2d})=0.$

Then

$$\mathcal{L}(W^{2d})=P^{2d}_{2d}+H^{1}(W^{2d})+H^{2}(W^{2d}).$$

Let $W=W^{0}+W^{1}+\cdots W^{2d}$, then

$$\mathcal{L}(W)=P_{0}-P_{2d+1},$$

where

$$P_{2d+1}=P_{2d}-P^{2d}_{2d}-H^{1}(W^{2d})-H^{2}(W^{2d})$$

only consists of over $2d$-order terms of $\bar{w}$.

Below we estimate $W^{2d}$ and $P_{2d+1}.$ By Lemma 12 it follows that

$$\|W^{2d}_{0}\|_{s-(2d+\frac{1}{2d+2})\sigma,r}\lessdot\frac{\|P_{2d}\|_{s-2d\sigma,r}}{\alpha\sigma^{\tau+n}},$$

$$\|W^{2d}_{0}\|_{s-(2d+\frac{1}{2d+2})\sigma,\varepsilon}\lessdot\ \frac{\|P_{2d}\|_{s-2d\sigma,\varepsilon}}{\alpha\sigma^{\tau+n}},$$

and $[W^{2d}_{0}]=0.$

Similarly, for any $1\leqslant j\leqslant 2d$ we have

$$\|W^{2d}_{j}\|_{s-(2d+\frac{j+1}{2d+2})\sigma,r}\lessdot\frac{\|W^{2d}_{j-1}\|_{s-(2d+\frac{j}{2d+2})\sigma,r}}{\alpha^{2d-j+1}\sigma^{(2d-j+1)\tau+(j+1)n}}\leqslant\cdots\lessdot\frac{\|P_{2d}\|_{s-2d\sigma,r}}{\alpha^{\bar{\nu}_{j+1}}\sigma^{\bar{\nu}_{j+1}\tau+(j+1)n}},$$

$$\ \ \|W^{2d}_{j}\|_{s-(2d+\frac{j+1}{2d+2})\sigma,\varepsilon}\lessdot\frac{\|W^{2d}_{j-1}\|_{s-(2d+\frac{j}{2d+2})\sigma,r}}{\alpha^{2d-j+1}\sigma^{(2d-j+1)\tau+(j+1)n}}\lessdot\cdots\lessdot\ \frac{\|P_{2d}\|_{s-2d\sigma,\varepsilon}}{\alpha^{\bar{\nu}_{j+1}}\sigma^{\bar{\nu}_{j+1}\tau+(j+1)n}},$$

where $\bar{\nu}_{j+1}=2d+1+2d+\cdots+(2d+1-j).$ Moreover, we have $[W^{2d}_{j}]=0.$

Thus, it follows that

$$\|W^{2d}\|_{s-(2d+\frac{2d+1}{2d+2})\sigma,r}\leqslant\frac{\|P_{2d}\|_{s-2d\sigma,r}}{\alpha^{\nu_{2d+1}}\sigma^{\nu_{2d+1}\tau+(2d+1)n}}\leqslant\frac{\varepsilon}{\alpha^{\mu_{2d}}\sigma^{\mu_{2d}\tau+\nu_{2d}n+2d}},$$

$$\ \ \|W^{2d}\|_{s-(2d+\frac{2d+1}{2d+2})\sigma,\varepsilon}\lessdot\ \frac{\varepsilon^{2d+1}}{\alpha^{\nu_{2d+1}}\sigma^{\nu_{2d+1}\tau+(2d+1)n}},$$

and $[W^{2d}]=0.$ Moreover,

$$\|P_{2d+1}\|_{s-(2d+1)\sigma,r}\leqslant\frac{\varepsilon}{\alpha^{\mu_{2d}}\sigma^{\mu_{2d}\tau+\nu_{2d}n+2d+1}}.$$

Let $R=-P_{2d+1}$. Noting that $W$ has the same estimate as $W^{2d}$ and replacing $(2d+1)\sigma$ with $\sigma,$ $\mu$ with $\mu_{2d}$ and $\nu$ with $\nu_{2d},$ we have proved this lemma.     $\square$