An infinite dimensional KAM theorem with application to two dimensional completely resonant beam equation

TL;DR

This paper develops an infinite-dimensional KAM theorem and applies it to prove the existence of small amplitude quasi-periodic solutions for a resonant beam equation on a two-dimensional torus.

Contribution

It introduces a new abstract KAM theorem for infinite dimensions and demonstrates its application to a specific resonant PDE, expanding the understanding of quasi-periodic solutions.

Findings

Existence of Whitney smooth small amplitude quasi-periodic solutions.

Application of the KAM theorem to a 2D resonant beam equation.

Construction of solutions on finite-dimensional tori.

Abstract

In this paper we consider the completely resonant beam equation on \T^2 with cubic nonlinearity on a subspace of L^2 (\T^2) which will be explained later. We establish an abstract infinite dimensional KAM theorem and apply it to the completely resonant beam equation. We prove the existence of a class of Whitney smooth small amplitude quasi-periodic solutions corresponding to finite dimensional tori.

Equations

$$u_{tt}+{\Delta }_x^{2}u+u^3=0u=u(t,x),t\in R,x\in T^2$$

$$\mathcal{U}=\{u=\sum _{n\in Z_{odd}^2}{u}_n{\varphi }_n,{\varphi }_n(x)=e^{i⟨n,x⟩}\}$$

$$Z_{odd}^2=\{n=(n_1,n_2):n_1\in 2Z-1,n_2\in 2Z\}$$

$$\displaystyle\left\{\begin{array}[]{rcl}&&n-m+i-j=0\\ &&|n|^{2}-|m|^{2}+|i|^{2}-|j|^{2}=0\end{array}\right.$$

$$\displaystyle\left\{\begin{array}[]{rcl}&&n+m-i-j=0\\ &&|n|^{2}+|m|^{2}-|i|^{2}-|j|^{2}=0\end{array}\right.$$

$$u(t,x)=\sum _{j=1}^b\sqrt{{\xi }_j}(e^{i{\omega }_{j}t}{\varphi }_{i_j}+e^{-i{\omega }_{j}t}{\bar{\varphi }}_{i_j})+O(|\xi {|}^{\frac{3}{2}})$$

$$\Vert z{\Vert }_{a,\rho }=\sum _{n\in Z_1^2}|z_n||n{|}^a{e}^{\rho |n|}a,\rho >0$$

$$D(r,s)=\left\{(\theta ,I,z,\bar{z}):|\mathrm{Im}\theta |<r,|I|<s^2,\Vert z{\Vert }_{a,\rho }<s,\Vert \bar{z}{\Vert }_{a,\rho }<s\right\}$$

$$F(\theta ,I,z,\bar{z})=\sum _{k,l,\alpha ,\beta }{F}_{kl\alpha \beta }(\xi )e^{i⟨k,\theta ⟩}{I}^l{z}^{\alpha }{\bar{z}}^{\beta }$$

$$\Vert F{\Vert }_{D(r,s),\mathcal{O}}$$

$$|F_{kl\alpha \beta }{|}_{\mathcal{O}}$$

$$X_F=(F_I,-F_{\theta },i\{F_{z_n}{\}}_{n\in Z_1^2},-i\{F_{{\bar{z}}_n}{\}}_{n\in Z_1^2})$$

$$\Vert X_F{\Vert }_{D(r,s),\mathcal{O}}$$

$$H_0=N+\mathcal{A}+\mathcal{B}+\bar{\mathcal{B}}$$

$$N=⟨\omega (\xi ),I⟩+\sum _{n\in Z_1^2}{\Omega }_n{z}_n{\bar{z}}_n$$

$$\mathcal{A}=\sum _{n\in {\mathcal{L}}_1}{a}_n(\xi )e^{i({\theta }_i-{\theta }_j)}{z}_n{\bar{z}}_m$$

$$\mathcal{B}=\sum _{n\in {\mathcal{L}}_2}{a}_n(\xi )e^{-i({\theta }_i+{\theta }_j)}{z}_n{z}_m$$

$$\bar{\mathcal{B}}=\sum _{n\in {\mathcal{L}}_2}{a}_n(\xi )e^{i({\theta }_i+{\theta }_j)}{\bar{z}}_n{\bar{z}}_m$$

$${\Omega }_n={\epsilon }^{-p}|n{|}^2+{\tilde{\Omega }}_{n}p>0$$

$$A_n={\Omega }_{n}n\in Z_1^2\setminus ({\mathcal{L}}_1\cup {\mathcal{L}}_2)$$

$$A_{n}=\left(\begin{array}[]{cc}\Omega_{n}+\omega_{i}&a_{n}\\ a_{m}&\Omega_{m}+\omega_{j}\end{array}\right)\qquad n\in{\mathcal{L}}_{1}$$

$$A_{n}=\left(\begin{array}[]{cc}\Omega_{n}-\omega_{i}&a_{n}\\ \bar{a}_{m}&\Omega_{m}-\omega_{j}\end{array}\right)\qquad n\in{\mathcal{L}}_{2}$$

$$|⟨k,\omega ⟩|\ge \frac{\gamma }{|k{|}^{\tau }}k\ne 0$$

$$|\det (⟨k,\omega ⟩+A_n)|\ge \frac{\gamma }{|k{|}^{\tau }}$$

$$|\det (⟨k,\omega ⟩+A_n\otimes I_2\pm I_2\otimes A_{n^{{}^{\prime }}})|\ge \frac{\gamma }{|k{|}^{\tau }}k\ne 0$$

$$\Vert X_{\mathcal{A}}{\Vert }_{D_{a,\rho }(r,s),\mathcal{O}}+\Vert X_{\mathcal{B}}{\Vert }_{D_{a,\rho }(r,s),\mathcal{O}}+\Vert X_{\bar{\mathcal{B}}}{\Vert }_{D_{a,\rho }(r,s),\mathcal{O}}<1,\Vert X_P{\Vert }_{D_{a,\rho }(r,s),\mathcal{O}}<\epsilon$$

$$\mathcal{A}+\mathcal{B}+\bar{\mathcal{B}}+P=\sum _{k\in Z^b,l\in N^b,\alpha ,\beta }(\mathcal{A}+\mathcal{B}+\bar{\mathcal{B}}+P{)}_{kl\alpha \beta }(\xi )e^{i⟨k,\theta ⟩}{I}^l{z}^{\alpha }{\bar{z}}^{\beta }$$

$$(\mathcal{A}+\mathcal{B}+\bar{\mathcal{B}}+P{)}_{kl\alpha \beta }\ne 0⟹\sum _{j=1}^b{k}_j{i}_j+\sum _{n\in Z_1^2}({\alpha }_n-{\beta }_n)n=0$$

$$\Phi :T^b\times {\mathcal{O}}_{\gamma }\to D_{a,\rho }(r,s),\tilde{\omega }:{\mathcal{O}}_{\gamma }\to R^b$$

$$u_{tt}+{\Delta }^{2}u+\epsilon u^3=0$$

$$u_t=v$$

$$v_t=-{\Delta }^{2}u-\epsilon u^3$$

$$-i{q}_t=-\Delta q+\epsilon \frac{1}{\sqrt{2}}(-\Delta {)}^{-\frac{1}{2}}{\left((-\Delta {)}^{-\frac{1}{2}}(\frac{q+\bar{q}}{\sqrt{2}})\right)}^3$$

$$H=\frac{1}{2}⟨-\Delta q,q⟩+\frac{1}{4}\epsilon {\int }_{T^2}{\left((-\Delta {)}^{-\frac{1}{2}}(q+\bar{q})\right)}^{4}dx$$

Full text

Mathematics Subject Classification: Primary 37K55; 35B10

Keywords: KAM theory; Hamiltonian systems; Beam equation; Birkhoff normal form

1 Introduction

In this paper we consider the two dimensional completely resonant beam equation with cubic nonlinearity on a subspace $\mathcal{U}$ of $L^{2}({{T}}^{2})$:

[TABLE]

Here $t$ is time and $x$ is the spatial variable. The subspace $\mathcal{U}$ is defined by

[TABLE]

where the integer set ${{Z}}^{2}_{odd}$ is defined as

[TABLE]

This idea comes from the work by M.Procesi [29] and we will explain it later in section 2. The solution of ”real” completely resonant beam equation (not on ${{Z}}^{2}_{odd}$, just on ${{Z}}^{2}$) will be handled in our forthcoming paper.

The infinite dimensional KAM theory with applications to Hamiltonian PDEs has attracted great interests since $1980$s. Starting from the remarkable work [6,19,31], a lot of achievements have been made in 1-dimensional Hamiltonian PDEs about the existence of quasi-periodic solutions by the methods of KAM theory. For these work, just refer to [5,12,13,17,18,20-25,32]. But when people turn to the higher dimensional case, the multiplicity of eigenvalues became a great obstacle because it leads to much more complicated small divisor conditions and measure estimates. The first breakthrough comes from Bourgain’s work [3] in $1998$. In this work, the cumbersome second Melnikov condition is avoided due to the application of the method of multiscale analysis, which essentially is a Nash-Moser iterative procedure instead of Newtonian iteration being widely used in KAM theory. Following this idea, a lot of important work has been made in higher dimensional case (refer to [1,2,4,30]).

However, despite the advantage of avoiding the difficulty of the second Melnikov conditions, there are also drawbacks of multiscale analysis methods. For example, we couldn’t see the linear stability of the small-amplitude solutions and it couldn’t show us a description of the normal form, which is fundamental in knowing the dynamical structure of an equation. For these reasons, KAM approach is also expected in dealing with higher dimensional equations. The first work comes from Geng and You [14] in 2006, which established the KAM theorem solving higher dimensional beam equations and nonlocal smooth Schrödinger equations with Fourier multiplier. They used the “zero-momentum condition” to avoid the multiplicity of eigenvalues and the regularity property to do the measure estimate. Later in 2010, a remarkable work [8] by Eliasson and Kuksin dealt with quite general case: higher dimensional Schrödinger equations with convolutional type potential and without “zero-momentum condition”. To overcome the multiple eigenvalues they studied the distribution of integer points on a sphere and got a normal form with block-diagonal structure, and conducted the measure estimates by developing the technique named “Lipschitz domain”. Motivated by their method, the quasi-periodic solutions of completely resonant Schrödinger equation on 2-dimension torus was developed by Geng, Xu and You [11] in 2011, with a very elaborate construction of tangential sites. In this paper, they defines the conception of “Töplitz-Lipschitz” condition and proved that the perturbation satisfies “Töplitz-Lipschitz” condition. Later, in [26,27] C.Procesi and M.Procesi extended this result to higher dimensional case. For other work about higher dimensional equation, just refer to [7,9,10,15,16,28,29].

Let us turn to beam equation now. In [15] Geng and You got the quasi-periodic solutions of beam equation in high dimension with typical constant potential and the nonlinearity is independent on the spatial variable $x$. Recently, in [7] Eliasson, Grebert and Kuksin got the quasi-periodic solutions of beam equation having typical constant potential in higher dimensional case, and with an elaborate but quite general choice of tangential sites in the sense of probability. they allow that their normal form contain hyperbolic terms which is cumbersome in solving homological equations. Motivated by their work, we want to consider the completely resonant beam equation (1.1). In our case, there are no outer parameters and only the amplitude provides parameters. Compared with the case of typical constant potential, although we have “zero-momentum condition” here, but when doing the normal form before KAM procedure, some terms still couldn’t be eliminated because of the loss of outer parameters. We could only get a block-diagonal normal form with finite dimensional block. As a consequence, our normal form is always related to the angle variable $\theta$, so here the linear stability is not available. Compared with [11], our convenience is that we have regularity property here and needn’t verify the complicated ”Töeplitz-Lipschitz condition” at each step. But except for this, our normal form structure and KAM iteration is similar to that in [11].

Now we state the choice of tangential sites. Let $S=\{i_{j}\in{{Z}}^{2}_{odd}:1\leq j\leq b\}$ here $b\geq 2$. We say $S$ is admissible if it satisfies the following conditions.

Proposition 1

*(Structure of $S$)

① Any three of them are not vertices of a rectangle.*

② For any $n\in{{Z}}^{2}_{odd}\setminus S$, there exists at most one triple $\{i,j,m\}$ with $i,j\in S,m\in{{Z}}^{2}_{odd}\setminus S$ such that

[TABLE]

and if it exists, we say $(n,m)$ are resonant in the first type and denote all such $n$ by ${\mathcal{L}}_{1}$.

③ For any $n\in{{Z}}^{2}_{odd}\setminus S$, there exists at most one triple $\{i,j,m\}$ with $i,j\in S,m\in{{Z}}^{2}_{odd}\setminus S$ such that

[TABLE]

and if it exists, we say $(n,m)$ are resonant in the second type and denote all such $n$ by ${\mathcal{L}}_{2}$.

④ Any $n\in{{Z}}^{2}_{odd}\setminus S$ shouldn’t be in ${\mathcal{L}}_{1}$ and ${\mathcal{L}}_{2}$ at the same time. It means that ${\mathcal{L}}_{1}\cap{\mathcal{L}}_{2}=\emptyset$.

(Here $|\cdot|$ means $l^{2}$ norm)

The proof of the existence of admissible sets is postponed in the Appendix, which is a modification of [11].

Now we could state the main theorem.

Theorem 1

Let $S=(i_{1},i_{2},\cdots,i_{b})\subseteq{{Z}}^{2}_{odd}$ be an admissible set. There exists a Cantor set $\mathcal{C}$ of positive measure, s.t. $\forall\xi=(\xi_{1},\xi_{2},\cdots,\xi_{b})\in\mathcal{C}$, equation (1.1) admits a small-amplitude real-valued quasi-periodic solution

[TABLE]

The outline of this paper is as follows: In section 2 we state some preliminaries and the abstract KAM theorem. In section 3 we deal with the normal form before KAM iteration. In section 4 we conduct one step of KAM iteration: solving homological equation and verifing the new normal form and perturbation. In section 5 we prove uniform convergence and get the invariant torus. In section 6 we complete the measure estimate. The choice of tangential sites is put into the appendix.

2 Preliminaries and statement of the abstract KAM theorem

In this section we introduce some notations and state the abstract KAM theorem which allows the existence of some terms dependent on $\theta$ in the normal form part.

To simplify, we only consider the subspace ${{Z}}^{2}_{odd}$ (defined in (1.3)) instead of ${{Z}}^{2}$. Given $b$ points $\{i_{1},i_{2},\cdots,i_{b}\}$ ($b\geq 2$) in ${{Z}}^{2}_{odd}$, denoted by $S$, which should be an admissible set(defined in Propositon 1), and let ${{Z}}^{2}_{1}$ be the complementary set of $S$ in ${{Z}}^{2}_{odd}$. Denote $z=(z_{n})_{n\in{{Z}}^{2}_{1}}$ with its conjugate $\bar{z}=(\bar{z}_{n})_{n\in{{Z}}^{2}_{1}}$. We introduce the weighted norm as follows:

[TABLE]

Here $|n|=\sqrt{|n_{1}|^{2}+|n_{2}|^{2}}$,$n=(n_{1},n_{2})\in{{Z}}^{2}_{1}$. Denote a neighborhood of ${{T}}^{b}\times\{I=0\}\times\{z=0\}\times\{\bar{z}=0\}$ by

[TABLE]

Here $|\cdotp|$ means the sup-norm of complex vectors.

Let $\alpha=\{\alpha_{n}\}_{n\in{{Z}}^{2}_{1}},\beta=\{\beta_{n}\}_{n\in{{Z}}^{2}_{1}}$, $\alpha_{n},\beta_{n}\in{{N}}$ with only finitely many non-vanishing components. Denote $z^{\alpha}\bar{z}^{\beta}=\prod_{n\in{{Z}}^{2}_{1}}z^{\alpha_{n}}_{n}\bar{z}^{\beta_{n}}_{n}$ and let

[TABLE]

where $\xi\in{\mathcal{O}}\subseteq{{R}}^{b}$ is the parameter set. $k=(k_{1},\cdots,k_{b})\in{{Z}}^{b}$ and $l=(l_{1},\cdots,l_{b})\in{{N}}^{b}$, $I^{l}=I^{l_{1}}_{1}\cdots I^{l_{b}}_{b}$. Denote the weighted norm of $F$ by

[TABLE]

where the derivatives with respect to $\xi$ are in the sense of Whitney.

To a function $F$ we define its Hamiltonian vector field by

[TABLE]

and the associated weighted norm is

[TABLE]

where $\bar{a}>0$ is a constant and we need $\bar{a}>a$ to measure the regularity property of the perturbation at each iterative step.

The normal form has the following form:

[TABLE]

where $\xi\in\mathcal{O}$ is the parameter. For each $n\in{\mathcal{L}}_{1}$ or $n\in{\mathcal{L}}_{2}$, the 3-triple $(m,i,j)$ is uniquely determined.

For this unperturbed system, it’s easy to see that it admits a special solution $(\theta,0,0,0)\rightarrow(\theta+\omega t,0,0,0)$ corresponding to an invariant torus in the phase space. Our goal is to prove that, after removing some parameters, the perturbed system $H=H_{0}+P$ still admits invariant torus provided that $\|X_{P}\|_{D_{a,\rho}(r,s),{\mathcal{O}}}$ is sufficiently small. To achieve this goal, we require that Hamiltonian $H$ satisfies some conditions:

$(\bf A1)$ Nondegeneracy: The map $\xi\rightarrow\omega(\xi)$ is a $C^{4}_{W}$ diffeomorphism between $\mathcal{O}$ and its image ($C^{4}_{W}$ means $C^{4}$ in the sense of Whitney).

$(\bf A2)$ Asymptotics of normal frequencies:

[TABLE]

here $\tilde{\Omega}_{n}$ is a $C^{4}_{W}$ function of $\xi$, and $\tilde{\Omega}_{n}=O(|n|^{-\iota})\quad\iota>0$

$(\bf A3)$ Melnikov conditions: Let

[TABLE]

and

[TABLE]

[TABLE]

Then we assume that there exists $\gamma,\tau>0$, such that

[TABLE]

$(\bf A4)$ Boundedness: ${\mathcal{A}}+{\mathcal{B}}+\bar{\mathcal{B}}+P$ is real analytic in each variable $\theta,I,z,\bar{z}$ and Whitney smooth in $\xi$. And we have

[TABLE]

$(\bf A5)$ Zero-momentum condition:

The normal form part ${\mathcal{A}}+{\mathcal{B}}+\bar{\mathcal{B}}+P$ satisfy the following condition:

[TABLE]

we have

[TABLE]

Now we state our abstract KAM theorem, and as a corollary, we get Theorem $\bf 1$.

Theorem 2

Assume that the Hamiltonian $H=N+{\mathcal{A}}+{\mathcal{B}}+\bar{\mathcal{B}}+P$ satisfies condition $\bf(A1)-(A5)$. Let $\gamma>0$ be sufficiently small, then there exists $\varepsilon>0$ and $a,\rho>0$ such that if $\|X_{P}\|_{D_{a,\rho}(r,s),\mathcal{O}}<\varepsilon$, the following holds: There exists a Cantor subset ${\mathcal{O}}_{\gamma}\subseteq{\mathcal{O}}$ with $meas({\mathcal{O}}\setminus{\mathcal{O}}_{\gamma})=O(\gamma^{\varsigma})$ ($\varsigma$ is a positive constant) and two maps which are analytic in $\theta$ and $C_{W}^{4}$ in $\xi$.

[TABLE]

where $\Phi$ is $\frac{\varepsilon}{\gamma^{16}}$-close to the trivial embedding $\Phi_{0}:{{T}}^{b}\times{\mathcal{O}}\rightarrow{{T}}^{b}\times\{0,0,0\}$ and $\tilde{\omega}$ is $\varepsilon$-close to the unperturbed frequency $\omega$, such that $\forall\xi\in{\mathcal{O}}_{\gamma}$ and $\theta\in{{T}}^{b}$, the curve $t\rightarrow\Psi(\theta+\tilde{\omega}t,\xi)$ is a quasi-periodic solution of the Hamiltonian equation governed by $H=N+{\mathcal{A}}+{\mathcal{B}}+\bar{\mathcal{B}}+P$.

3 Normal Form

Consider the equation (1.1). The linear operator $-\Delta$ has eigenvalues $\lambda_{n}=|n|^{2}$and corresponding eigenfunctions $\phi_{n}=\frac{1}{2\pi}e^{i\left\langle n,x\right\rangle}$. By scaling $u\rightarrow\varepsilon^{\frac{1}{2}}u$, (1.1)becomes

[TABLE]

Now introduce $v=u_{t}$ and (3.1) is turned into

[TABLE]

Let $q=\frac{1}{\sqrt{2}}((-\Delta)^{\frac{1}{2}}u-i(-\Delta)^{-\frac{1}{2}}v)$ and (3.2) becomes

[TABLE]

Write it in the form of Hamiltonian equation $q_{t}=i\frac{\partial H}{\partial\bar{q}}$ and we get the Hamiltonian

[TABLE]

where $\left\langle\cdot,\cdot\right\rangle$ is the inner product in $L^{2}({{T}}^{2})$. Notice that in ${{Z}}^{2}_{odd}$ the origin is avoided so $(-\Delta)^{-\frac{1}{2}}$ is well defined. (That is why we use it instead of the whole ${{Z}}^{2}$) Now expand $q$ into Fourier series

[TABLE]

so the Hamiltonian becomes (justify $\varepsilon$ if necessary)

[TABLE]

Now we state the normal form theorem of $H$.

Propsition 3.1

Let $S$ be admissible. For Hamiltonian function (3.6), there exists a symplectic transformation $\Phi$ satisfying

[TABLE]

where

[TABLE]

and

[TABLE]

[TABLE]

$\bf Proof:$ We construct a Hamiltonian function $F$ to induce $\Phi=X^{1}_{F}$ which is the time-1 map of $F$. For convenience, we define three sets as below:

[TABLE]

and similarly

[TABLE]

[TABLE]

we define $F$ as

[TABLE]

(3.6) is put into (set $z_{n}=q_{n},\bar{z}_{n}=\bar{q}_{n},n\notin S$)

[TABLE]

Here we need to state a fact: For four points $n,m,i,j\in{{Z}}^{2}_{odd}$ , it could never satisfy $|n|^{2}+|m|^{2}+|i|^{2}-|j|^{2}=0$. If not, we assume $n=(n_{1},n_{2}),m=(m_{1},m_{2}),i=(i_{1},i_{2}),j=(j_{1},j_{2})$ and in each one the first component is odd and the second component is even. Then we have

[TABLE]

The right one can be divided by 4 but the left one couldn’t, which is a contradiction. By this fact we know that the set

[TABLE]

is empty.

Introduce the action-angle variable in the tangential sites:

[TABLE]

so we have

[TABLE]

By scaling in variables:

[TABLE]

and scale time $t\rightarrow\varepsilon^{9}t$ we get the Hamiltonian function as follows:

[TABLE]

where

[TABLE]

Now we verify that the normal form $(3.7)-(3.13)$ satisfy condition $\bf(A1)-(A5)$.

$Verifying\bf(A1)$: By (3.8) we get

[TABLE]

where $a_{ij}=\frac{2}{\lambda_{i}^{2}}$ if $i=j$ and $a_{ij}=\frac{4}{\lambda_{i}\lambda_{j}}$ if $i\neq j$. It’s easy to see that this matrix is non–degenerate.

$Verifying\bf(A2)$: By $(3.9)$, just take $p=4,\iota=2$.

$Verifying\bf(A3)$: Recall the definition in condition $(A3)$, we only verify the most complicated case:

[TABLE]

where $n,n^{{}^{\prime}}\in{\mathcal{L}}_{1}\cup{\mathcal{L}}_{2}$. We verify two facts: (3.28) is a polynomial of parameter $\xi$ with degree 4 and it couldn’t be equivalently zero. For the former one, notice that $\lambda I+A\otimes I_{2}-I_{2}\otimes B=(\lambda I+A)\otimes I-I\otimes B$ (here $|\cdot|$ means determinant) and using the formula

[TABLE]

then we get it. For the latter one, it’s the same as that in [11]. By this, we could get

[TABLE]

So by excluding parameters with measure $O(\gamma^{\frac{1}{4}})$, we have

[TABLE]

For the verification of $\bf(A4)$ and $\bf(A5)$, just refer to [14].

4 KAM Iteration

We prove Theorem $\bf 2$ by a KAM iteration which involves an infinite sequence of change of variables. Each step of KAM iteration makes the perturbation smaller than the previous step at the cost of excluding a small set of parameters. We have to prove the convergence of the iteration and estimate the measure of the excluded set after infinite KAM steps.

At the $\nu$-step of the KAM iteration, we consider a Hamiltonian vector field with

[TABLE]

where $A_{\nu}+B_{\nu}+\bar{B}_{\nu}+P_{\nu}$ is defined in $D(r_{\nu},s_{\nu})\times{\mathcal{O}}_{\nu-1}$

We construct a map

[TABLE]

so that the vector field $X_{H_{\nu}\circ\Phi_{\nu}}$ defined on $D(r_{\nu+1},s_{\nu+1})$satisfies

[TABLE]

and the new Hamiltonian still satisfies $\bf(A1)-(A5)$.

To simplify notations, in the following text, the quantities without subscripts refer to quantities at the $\nu$ th step, while the quantities with subscripts $+$ denote the corresponding quantities at the $(\nu+1)$ th step. Let’s consider the Hamiltonian defined in $D(r,s)\times\mathcal{O}$:

[TABLE]

We assume that for $\xi\in\mathcal{O}$, one has

[TABLE]

where $A_{n}=\Omega_{n}$ for $n\in{{{Z}}^{2}_{1}}\setminus({\mathcal{L}}_{1}\cup{\mathcal{L}}_{2})$ and

[TABLE]

[TABLE]

where $(n,m)$ are resonant pairs and $(i,j)$ is uniquely determined by $(n,m)$.

Expand $P$ into Fourier-Taylor series $P=\sum_{k,l,\alpha,\beta}P_{kl\alpha\beta}I^{l}e^{i\left\langle k,\theta\right\rangle}z^{\alpha}\bar{z}^{\beta}$ and by $(A5)$ we get that

[TABLE]

We now let $0<r_{+}<r$ and define

[TABLE]

Here and later, the letter $c$ denotes suitable(possible different) constants independent of the iteration steps.

Now we describe how to construct a set $\mathcal{O}_{+}\subseteq\mathcal{O}$ and a change of variables $\Phi:D_{+}\times{\mathcal{O}}_{+}=D(r_{+},s_{+})\times{\mathcal{O}}_{+}\rightarrow D(r,s)\times{\mathcal{O}}$ such that the transformed Hamiltonian $H_{+}=H\circ\Phi=N+{\mathcal{A}}_{+}+{\mathcal{B}}_{+}+\bar{\mathcal{B}}_{+}+P_{+}$ satisfies assumptions $\bf(A1)-(A5)$ with new parameters $\varepsilon_{+},r_{+},s_{+}$ and with $\xi\in\mathcal{O}_{+}$.

4.1 Homological Equation

Expand $P$ into Fourier-Taylor series

[TABLE]

where $k\in{{Z}}^{b},l\in{{N}}^{b}$and the multi-indices $\alpha$,$\beta$ run over the set of all infinite dimensional vectors $\alpha=(\cdots,\alpha_{n},\cdots)_{n\in{{Z}}^{2}_{1}}$, $\beta=(\cdots,\beta_{n},\cdots)_{n\in{{Z}}^{2}_{1}}$ with finitely many nonzero components of positive integers. And by $(A5)$ we get that

[TABLE]

Consider its quadratic truncation $R$:

[TABLE]

where $P^{k10}_{n}=P_{kl\alpha\beta}$ with $\alpha=e_{n}$,$\beta=0$, $P^{k01}_{n}=P_{kl\alpha\beta}$ with $\alpha=0$,$\beta=e_{n}$, here $e_{n}$ denotes the vector with the $n$ th component being $1$ and the other components being zero. Similarly, $P^{k20}_{nm}=P_{kl\alpha\beta}$ with $\alpha=e_{n}+e_{m}$, $\beta=0$; $P^{k11}_{nm}=P_{kl\alpha\beta}$ with $\alpha=e_{n}$,$\beta=e_{m}$; $P^{k02}_{nm}=P_{kl\alpha\beta}$ with $\alpha=0$, $\beta=e_{n}+e_{m}$.

Rewrite $H$ as $H=N+{\mathcal{A}}+{\mathcal{B}}+\bar{\mathcal{B}}+R+(P-R)$. Due to the choice of $s_{+}\ll s$ and the definition of the norm, it follows immediately

[TABLE]

and in $D(r,s_{+})$

[TABLE]

In the following, we will construct a Hamiltonian function $F$ satisfying $(A5)$ and with the same form of $R$ defined in $D_{+}=D(r_{+},s_{+})$ such that the time one map $X^{1}_{F}$ of the Hamiltonian vector field $X_{F}$ defines a map from $D_{+}$ to $D$ and puts $H$ into $H_{+}$. Precisely, one has

[TABLE]

So we get the linearized homological equation:

[TABLE]

where

[TABLE]

We define $N_{+}=N+\hat{N},{\mathcal{A}}_{+}={\mathcal{A}}+\hat{\mathcal{A}},{\mathcal{B}}_{+}={\mathcal{B}}+\hat{\mathcal{B}},\bar{\mathcal{B}_{+}}=\bar{\mathcal{B}}+\hat{\bar{\mathcal{B}}}$ and

[TABLE]

We construct the Hamiltonian function $F$ as below:

[TABLE]

Now $(4.13)$ turns to

[TABLE]

and the most complicated

[TABLE]

$\bf Solving(4.20)$: $F^{j}(\theta)=\sum\limits_{k\neq 0}F^{j}_{k}e^{i\left\langle k,\theta\right\rangle},\quad j=0,1.$ By comparing the Fourier coefficients we get

[TABLE]

and according to assumption $(4.2)$ we get

[TABLE]

so one has the estimate

[TABLE]

$\bf Solving(4.21)$: We decompose this part into three cases:

$(1)$ If $n\in{{Z}}^{2}_{1}\setminus\{{\mathcal{L}}_{1}\cup{\mathcal{L}}_{2}\}$, one has

[TABLE]

$(2)$ If $n\in{\mathcal{L}}_{1}$ and the corresponding resonant group $m,i,j$, one has

[TABLE]

$(3)$ If $n\in{\mathcal{L}}_{2}$, one has

[TABLE]

The above three equations have the coefficient matrix of the form $\left\langle k,\omega\right\rangle I+A_{n}$ and by the assumption $(4.2)$ we know that $|det(\left\langle k,\omega\right\rangle I+A_{n})|\geq\frac{\gamma}{|k|^{\tau}}$ So we get the estimate

$(1)$: If $n\in{{Z}}^{2}_{1}\setminus\{{\mathcal{L}}_{1}\cup{\mathcal{L}}_{2}\}$, one has

[TABLE]

$(2)$: If $n\in{\mathcal{L}}_{1}$ and the corresponding resonant group $m,i,j$, one has

[TABLE]

$(3)$ If $n\in{\mathcal{L}}_{2}$, one has

[TABLE]

$\bf Solving(4.22)$: Similarly, we also decompose them into three parts. In this case, the coefficient matrix has the form of

[TABLE]

By the assumption:

[TABLE]

$(1)$: If $n,n^{\prime}\in{{Z}}^{2}_{1}\setminus({\mathcal{L}}_{1}\cup{{\mathcal{L}}_{2}})$, one has

[TABLE]

and we get the estimate

[TABLE]

$(2)$: If $n\in{{Z}}^{2}_{1}\setminus\{{\mathcal{L}}_{1}\cup{\mathcal{L}}_{2}\}$, $n^{\prime}\in{\mathcal{L}}_{1}$, one has

[TABLE]

we have the estimate

[TABLE]

when $n^{\prime}\in{\mathcal{L}}_{2}$ is similar, the estimate is similar.

$(3)$: If $n\in{\mathcal{L}}_{1}$,$n^{\prime}\in{\mathcal{L}}_{2}$, one has

[TABLE]

and we get the estimate

[TABLE]

when $n\in{\mathcal{L}}_{1},n^{\prime}\in{\mathcal{L}}_{1}$ or $n\in{\mathcal{L}}_{2},n^{\prime}\in{\mathcal{L}}_{2}$, the estimate is similar.

Now we could give the small-divisor condition in the next step with new parameters. For simplicity, we only consider the most complicated case: the second Melnikov condition. Assume that

[TABLE]

then we have

[TABLE]

provided $|k|<K$ where $K=c(\frac{\gamma-\gamma_{+}}{\varepsilon})^{\frac{1}{\tau+1}}$. So the small divisor condition in the next step holds automatically for $|k|<K$ and we will deal with other terms in section 6.

4.2 Estimation of coordinate transformation and new perturbation

With the similar methods in [14], we could get the estimates of $X_{F}$ and $\phi^{t}_{F}$, just with different parameters.

Lemma 4.1

Let $D_{i}=D(r_{+}+\frac{i}{4}(r-r_{+}),\frac{i}{4}s),\quad 0\leq i\leq 4$. Then we get

[TABLE]

Lemma 4.2

Let $\eta=\varepsilon^{\frac{1}{3}},D_{i\eta}=D(r_{+}+\frac{1}{4}(r-r_{+}),\frac{i}{4}\eta s),\quad 0\leq i\leq 4$. If $\varepsilon\ll\gamma^{16}(r-r_{+})^{c}$, then we have

[TABLE]

and

[TABLE]

With above estimates, we could give the estimate of new perturbations. We have

[TABLE]

where $R(t)=R+(1-t)\{N,F\}=(1-t)(N_{+}-N)+tR$ and

[TABLE]

By Lemma $\bf 4.1$, we get

[TABLE]

At the same time, by Cauchy estimate, one has

[TABLE]

One the other hand, we have

[TABLE]

To sum up, $P_{+}$ is bounded by

[TABLE]

5 Iterative Lemma and Convergence

For fixed parameters $r,s,\varepsilon,\gamma$, at the $\nu$ th step of the iterative procedure, we define the sequence

[TABLE]

where $c$ is a constant and the parameters $r_{0},s_{0},\varepsilon_{0},\gamma_{0},K_{0}$ are defined as $r,s,\varepsilon,\gamma,1$ respectively.

For later use, we define the resonant sets useful for the part of measure estimate:

[TABLE]

where each part is defined by

[TABLE]

Now we could state the iterative lemma as follows:

Lemma 5.1

Let $\varepsilon$ is small enough and $\nu\geq 0$, assume that we are at the $\nu$ th step.

$(1)$* $N_{\nu}+\mathcal{A}_{\nu}+\mathcal{B}_{\nu}+$ ${\bar{\mathcal{B}}}_{\nu}$ is the normal form depending on the parameter $\xi$, where*

[TABLE]

and satisfying the following small divisor conditions:

[TABLE]

where the matrix is defined as

[TABLE]

[TABLE]

[TABLE]

and the parameter $\xi$ is in a closed set $\mathcal{O}_{\nu}$ of ${{R}}^{b}$.

$(2)$* $\omega_{\nu},\Omega_{\nu}$ and $a_{n}^{\nu}$ are $C_{W}^{4}$ smooth and satisfy the condition $(\delta=\min\{\bar{a}-a,\iota\})$*

[TABLE]

$(3)$* The perturbation $P_{\nu}$ satisfy condition $(A5)$ and $\|X_{P_{\nu}}\|_{D(r_{\nu},s_{\nu}),\mathcal{O}_{\nu}}<\varepsilon_{\nu}$.*

Then there exists a closed subset ${\mathcal{O}}_{\nu+1}\subseteq\mathcal{O}_{\nu}$ defined by

[TABLE]

and a symplectic transformation of variables

[TABLE]

such that on the domain $D(r_{\nu+1},s_{\nu+1})\times{\mathcal{O}}_{\nu}$, the Hamiltonian has the form

[TABLE]

with

[TABLE]

The new perturbation $P_{\nu+1}$ satisfy condition $(A5)$ and

[TABLE]

Now assume that the assumption of $\bf(A1)-(A5)$ is satisfied. We could apply the iterative lemma at the $\nu=0$ step as long as $\varepsilon_{0},\gamma_{0}$ are sufficiently small. By an inductive way, we get the sequence

[TABLE]

Let $\breve{\mathcal{O}}=\bigcap^{\infty}_{\nu=0}{\mathcal{O}}_{\nu}$. It’s easy to conclude that $N_{\nu},{\mathcal{A}}_{\nu},{\mathcal{B}}_{\nu},\bar{\mathcal{B}}_{\nu},\Phi^{\nu},D\Phi^{\nu}$ all converge uniformly on $D(\frac{1}{2}r,0)\times\breve{\mathcal{O}}$ with

[TABLE]

and

[TABLE]

provided that $\varepsilon$ is sufficiently small.

Let $\phi^{t}_{H}$ be the Hamiltonian flow induced by $X_{H}$. By $H_{\nu+1}=H\circ\Phi^{\nu}$ one has

[TABLE]

and by the uniform convergence of all the related parameters, we get

[TABLE]

and

[TABLE]

For parameters $\xi\in\breve{\mathcal{O}}$, one has

[TABLE]

This means that $\Phi^{\infty}({{T}}^{b}\times\{\xi\})$ is an embedded torus which is invariant for the original perturbed Hamiltonian system at $\xi\in\breve{\mathcal{O}}$. The frequencies $\omega_{\infty}(\xi)$ associated to the tori $\Phi^{\infty}({{T}}^{b}\times\{\xi\})$ is slightly different from $\omega(\xi)$. The normal behavior of the invariant torus is governed by normal frequencies $\Omega^{\infty}_{n}$.   $\Box$

6 Measure Estimate

Recall the resonant sets at the $\nu$ th step ${\mathcal{R}}^{\nu}=\bigcup_{|k|\geq K_{\nu-1},nm}\left({\mathcal{R}}^{\nu}_{k}\cup{\mathcal{R}}^{\nu}_{k,n}\cup{\mathcal{R}}^{\nu}_{k,nm}\right)$. To estimate its measure, we need to estimate each single set ${\mathcal{R}}^{\nu}_{k},{\mathcal{R}}^{\nu}_{k,n},{\mathcal{R}}^{\nu}_{k,nm}$ first.

Lemma 6.1

Fix $|k|\geq K_{\nu-1},n,m$, one has

[TABLE]

$\bf Proof$: One has that $\omega_{\nu}(\xi)=\omega(\xi)+\sum^{\nu-1}_{j=0}P^{j}_{0l00}(\xi)$ with $\sum_{0\leq j\leq{\nu-1}}|P^{j}_{0l00}(\xi)|_{{\mathcal{O}}_{\nu-1}}<\varepsilon$, and $\Omega^{\nu}_{n}(\xi)=\Omega_{n}(\xi)+\sum_{0\leq j\leq{\nu-1}}P^{011,j}_{nn}(\xi)$ with $\sum_{0\leq j\leq{\nu-1}}|P^{011,j}_{nn}(\xi)|_{{\mathcal{O}}_{\nu-1}}<\frac{\varepsilon}{|n|^{\delta}}$($\delta=\min\{\bar{a}-a,\iota\}>0$ ) Similar results also hold for $a_{n}$. So it’s easy to conclude that

[TABLE]

Then the result is obvious.   $\Box$

Lemma 6.2

The whole measure we need to exclude during the KAM procedure is

[TABLE]

$\bf Proof$: Fix one $\nu$ and one $k$, and we only estimate the most complicated term:

[TABLE]

Consider its diagonal entry, we only consider one element. If $|n|^{2}-|m|^{2}=l\geq c|k|$, then ${\mathcal{R}}^{\nu}_{k,nm}=\emptyset$ otherwise we assume $|n|\geq|m|$, by the regularity property, we get

[TABLE]

so we have that

[TABLE]

it’s easy to see that ${\mathcal{Q}}^{\nu}_{lm}\subseteq{\mathcal{Q}}^{\nu}_{lm_{0}}$ for $|m|\geq|m_{0}|$. By Lemma $\bf 6.1$ we have

[TABLE]

By choosing appropriate $m_{0}$ to reach $\frac{\gamma^{\frac{1}{4}}|m_{0}|^{2}}{|k|^{\frac{\tau}{4}}}=|m_{0}|^{-\delta}$ (just let $|m_{0}|=\left(\frac{|k|^{\tau}}{\gamma}\right)^{\frac{1}{4\delta+8}}$). Then we get the estimate

[TABLE]

justify the parameter $\tau$ appropriately and we get the result.   $\Box$

7 Appendix

In this part we give a precise method to construct the admissible set $S=\{i_{1}=(x_{1},y_{1}),i_{2}=(x_{2},y_{2}),\cdots,i_{b}=(x_{b},y_{b})\}$. It’s modified from the appendix in [11] and we omit some detailed calculation which has been done in [11]. The points in S will be defined in an inductive way. The first point $(x_{1},y_{1})\in{{Z}}^{2}_{odd}$ is chosen as $x_{1}>b^{2},y_{1}=2x_{1}^{5^{b}}$ and the second $x_{2}=x_{1}^{5},y_{2}=2x_{2}^{5^{b}}$. If we have chosen the first $j$ points $i_{1},i_{2},\cdots,i_{j}$, then we define

[TABLE]

Recall the condition of admissible set (Proposition 1). We verify the conditions one by one. Given three points $c,d,f\in S$, it’s easy to see that

[TABLE]

So any three points in S can’t be three vertices of a rectangle.

To verify condition ②-④, following the appendix in [11], it suffices to prove that each equation set in the following has no integer solution in ${{Z}}^{2}_{1}$ for $c,d,f,g\in S$ and $\{c,d\}\neq\{f,g\}$.

[TABLE]

[TABLE]

[TABLE]

The three equation sets correspond to condition ②,③,④ respectively and we denote them by $\bf\@slowromancap i@,\@slowromancap ii@,\@slowromancap iii@$ respectively.

For $\bf\@slowromancap i@$,

$\bf\@slowromancap i@-i:$ If only one element of $\{|c|,|d|,|f|,|f|\}$ reaches the maximun value of them.

$\bf\@slowromancap i@-i-(1):$ $|d|$ or $|f|$ reaches the maximum. Without lose of generality, just assume $d$. It’s easy to get that

[TABLE]

By the fact $d_{2}\gg$ other ones, the second term in the right side couldn’t be an integer, implying $n_{2}\notin{{Z}}$, which is a contradiction.

$\bf\@slowromancap i@-i-(2):$ $|g|$ or $|c|$ reaches the maximum. Without lose of generality, just assume $|g|$. By calculation, we get

[TABLE]

where $A=(c_{1}^{5^{b}-1}+c_{1}^{5^{b}-2}d_{1}+\cdots+c_{1}d_{1}^{5^{b}-2}+d_{1}^{5^{b}-1})$ By the fact $g_{1}\gg$ other ones, we know that the last term is not an integer, so $n_{2}\notin{{Z}}$, which is a contradiction.

$\bf\@slowromancap i@-ii:$ If two elements of $|c|,|d|,|f|,|g|$ reach the maximum, by the structure of S we know that these two points must be equal, and the case when three points reach the maximum will not happen.

$\bf\@slowromancap i@-ii-(1):$ when $|d|=|g|$ reach the maximum, we have $d=g$. By calculation we get

[TABLE]

where $A=(g_{1}^{5^{b}-1}+g_{1}^{5^{b}-2}f_{1}+\cdots+g_{1}f_{1}^{5^{b}-2}+f_{1}^{5^{b}-1})$ and $B=(g_{1}^{5^{b}-1}+g_{1}^{5^{b}-2}c_{1}+\cdots+g_{1}c_{1}^{5^{b}-2}+c_{1}^{5^{b}-1})$ without losing generality, we assume $|c|<|f|$(the case $|c|=|f|$ will not happen). According to the structure of S, it’s easy to see that in the expression above, the denominator is divisible by $|c_{1}|^{4}$, in the numerator, all terms are divisible by $|c_{1}|^{4}$ except for $c_{1}$. It means that $n_{2}\notin{{Z}}$ which is a contradiction.

$\bf\@slowromancap i@-ii-(2):$ If $|d|=|f|$ reach the maximum, we have $d=f$, then we have

[TABLE]

In this case $c,d,g$ are three vertices of a rectangular, which is a contradiction.

$\bf\@slowromancap i@-ii-(3):$ If $|c|=|g|$ reach the maximum, we have $c=g$. From

[TABLE]

we know $d,f,g$ lie on the same line, which is a contradiction.

Now we turn to $\bf\@slowromancap iii@$.

$\bf\@slowromancap iii@-i:$ If only one element of $|c|,|d|,|f|,|g|$ reaches the maximum value of them and each one is different from others.

$\bf\@slowromancap iii@-i-(1):$ $|d|$ reaches the maximum.

We have

[TABLE]

We take $g$ as the origin and from above we get an equation about $n_{1}$.

[TABLE]

The discriminant of the equation is

[TABLE]

where $\alpha\sim\frac{d_{1}}{d_{2}}\ll\frac{1}{|f_{1}|}$ so we get

[TABLE]

while $\sqrt{\Delta}=$ some integer plus $\alpha$, so the numerator in the above expression is not an integer. So we conclude that $n_{1}\notin{{Z}}$, which is a contradiction.

$\bf\@slowromancap iii@-i-(2)$ $|f|$ reaches the maximum.

As before, we take $g$ as the origin and we get

[TABLE]

The discriminant is

[TABLE]

by the fact $c_{2}d_{2}\gg(c_{1}-d_{1})^{2}$ and $f_{2}\gg$ other terms.

$\bf\@slowromancap iii@-i-(3)$ $|g|$ reaches the maximum.

It’s easy to see $|n|^{2}>|g|^{2}-|f|^{2}$ and we get

[TABLE]

$\bf\@slowromancap iii@-ii$ If only one element of $\{|c|,|d|,|f|,|g|\}$ reaches the maximun and two of the others are the same.

$\bf\@slowromancap iii@-ii-(1)$ $|d|$ reaches the maximum, and $c=g$, without losing generality, we just assume $c=g$. We could assume $|f|>|c|$, the case $|f|<|c|$ is similar.

In this case, by calculation we get

[TABLE]

By the structure of S, we could conclude that the term

[TABLE]

must has a form of an integer plus $\frac{2(f_{1}-c_{1})^{2}f_{1}^{\kappa}-(f_{2}-c_{2})(f_{1}-c_{1})f_{1}^{\tilde{\kappa}}}{(f_{1}-c_{1})^{2}+(f_{2}-c_{2})^{2}}$ where $0<\kappa\leq 5^{b}$ and $5\leq\tilde{\kappa}\leq 5^{b-1}$. Then we have $\frac{(f_{1}-c_{1})^{2}f^{\kappa}_{1}}{(f_{1}-c_{1})^{2}+(f_{2}-c_{2})^{2}}\leq\frac{1}{f_{1}^{5^{b}-2}}$ and $\frac{c_{1}f_{1}(f_{1}^{5^{b}-2}+f_{1}^{5^{b}-3}c_{1}+\cdots+f_{1}c_{1}^{5^{b}-3}+c_{1}^{5^{b}-2})}{1+(f_{1}^{5^{b}-1}+f_{1}^{5^{b}-2}c_{1}+\cdots+f_{1}c_{1}^{5^{b}-2}+c_{1}^{5^{b}-1})^{2}}\leq\frac{1}{f_{1}^{5^{b}-2}}$ while $\frac{(f_{2}-c_{2})(f_{1}-c_{1})f_{1}^{\tilde{\kappa}}}{(f_{2}-c_{2})^{2}+(f_{1}-c_{1})^{2}}>f_{1}^{6-5^{b}}$ which is much lager than the former two, and these three terms all $\ll 1$, hence $n_{2}\notin{{Z}}$. The case $c=f$ is similar to $\bf\@slowromancap iii@-i-(1)$.

$\bf\@slowromancap iii@-ii-(2)$ $|f|$ reaches the maximum and $g=c$ or $g=d$. Without losing generality, we just assume $g=c$.

We have

[TABLE]

If we take $c$ as the origin, we get $n_{2}=\frac{f_{1}^{2}d_{2}-f_{1}f_{2}d_{1}}{f_{1}^{2}+f_{2}^{2}}\notin{{Z}}$

$\bf\@slowromancap iii@-ii-(3)$ $|g|$ reaches the maximum. It’s similar to $\bf\@slowromancap iii@-i-(3)$.

$\bf\@slowromancap iii@-iii$ If two elements of $\{|c|,|d|,|f|,|g|\}$ reach the maximum of their values.

$\bf\@slowromancap iii@-iii-(1)$ $|g|=|d|$ reach the maximum. We have $g=d$. Then we get

[TABLE]

which implies that $n_{1}\notin{{Z}}$ due to the fact $d_{2}\gg$ other terms.

$\bf\@slowromancap iii@-iii-(2)$ $|f|=|d|$ reach the maximum. We have $f=d$ and take $g$ as the origin as before. Then we get

[TABLE]

The discriminant is

[TABLE]

due to the fact $d_{2}\gg$ other terms. So $n_{1}$ doesn’t exist, which is a contradiction.

At last we turn to $\bf\@slowromancap ii@$.

We know $\{c,d\}\notin\{f,g\}$. If $\{c,d\}\cap\{f,g\}\neq\emptyset$, without losing generality we just assume $c=f$. The equation becomes

[TABLE]

We have proved that it has no solution in ${{Z}}^{2}$ in $\bf\@slowromancap iii@-(iii)-(2)$.

Now we concentrate on the remaining case when the four elements are different from each other. Without losing generality, we just assume $|d|$ reach the maximum of their values. It’s easy to see $|n|^{2}\ll d_{1}$ and

[TABLE]

If $c_{1}=n_{1}$, then in the above expression the numerator is smaller than $d_{1}$, and if $c_{1}\neq n_{1}$, the numerator is smaller than $\frac{d_{2}}{2}$, we still have $n_{2}\notin{{Z}}$, which is a contradiction.