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KAM theory for the reversible perturbations of 2D linear beam equations

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Abstract

In the present paper, we prove an infinite dimensional reversible Kolmogorov-Arnold-Moser (KAM) theorem. As an application, we study the existence of KAM tori for a class of two dimensional (2D) non-Hamiltonian completely resonant beam equations with derivative nonlinearities. The Birkhoff normal form theory is also used since there are no external parameters in the equations.

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Notes

  1. For simplicity, in what follows, we drop the subscript on domain in the norm

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Acknowledgements

The authors are very grateful to the referee for his/her invaluable suggestions.

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Correspondence to Zhaowei Lou.

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The research was supported by the National Natural Science Foundation of China (NSFC) (Grant No. 11971012). Z. Lou was supported by NSFC (Grant No. 11901291) and the Natural Science Foundation of Jiangsu Province, China (Grant No. BK20190395).

Appendix

Appendix

1.1 Proof of Lemma 4.5

First of all, we give some results which will be used in this section.

Lemma 6.1

For \(k\ne 0\), \(\langle k,\alpha \rangle +\varepsilon ^{-4}(|n|^2-|n'|^2)+\langle Ak+\beta -\beta ',\xi \rangle \not \equiv 0.\)

Proof

If \(Ak+\beta -\beta '=0\), the vector \(k=(k _1,k_2,\cdots ,k_b)\) could be solved directly by Cramer’s rule:

$$\begin{aligned} k_l=\frac{2}{2b+1}\left( \frac{1}{|n'|^2}-\frac{1}{|n|^2}\right) |i_l|^2,\quad 1\le l\le b. \end{aligned}$$

Denote \(k\in {\mathbb Z}^b\) and \(k\ne 0\), so either just one element of the vector k not in \({\mathbb Z}\) or \(|n|^2-|n'|^2=0\) could lead to a contradiction. Now we consider the case \(k_l\) is an integer for \(1\le l\le b\) . At this time

$$\begin{aligned}&\varepsilon ^{4}\langle k,\alpha \rangle +(|n|^2-|n'|^2)\\&\quad =\frac{|n|^2-|n'|^2}{(2b+1)|n|^2|n'|^2} \left[ 2\sum \limits _{k=1}^b |i_k|^4+(2b+1)|n|^2|n'|^2\right] \ne 0 \end{aligned}$$

\(\square \)

Lemma 6.2

For \(k\ne 0\), \(n,n'\in \mathcal L_1\), \(i=i'\), \(j=j'\) and \(Q_nQ_m=Q_{n'}Q_{m'}\) then all the eigenvalues of

$$\begin{aligned}&\langle k,\omega \rangle I + \left( \begin{array}{cc}\Omega _n+\omega _i&{}\frac{1}{(2\pi )^2}Q_m\sqrt{\xi _i\xi _j}\\ \frac{1}{(2\pi )^2}Q_n\sqrt{\xi _i\xi _j}&{}\Omega _m+\omega _j\end{array}\right) \otimes I_2\\&\quad + I_2\otimes \left( \begin{array}{cc}\Omega _{n'}+\omega _{i}&{}\frac{1}{(2\pi )^2}Q_{n'}\sqrt{\xi _{i}\xi _{j}}\\ \frac{1}{(2\pi )^2}Q_{m'}\sqrt{\xi _{i}\xi _{j}}&{}\Omega _{m'}+\omega _{j}\end{array}\right) ^\mathrm {T}, \end{aligned}$$

are not identically zero.

Proof

The eigenvalues we are considering have the form as following:

$$\begin{aligned}&\langle k,\omega \rangle +\frac{1}{2}(\Omega _n+\omega _i)+\frac{1}{2}(\Omega _m+\omega _j)\pm \frac{1}{2} \sqrt{root 1} +\frac{1}{2}(\Omega _{n'} +\omega _i)\\&\quad +\frac{1}{2}(\Omega _{m'}+\omega _j)\pm \frac{1}{2} \sqrt{root 2} \end{aligned}$$

with

$$\begin{aligned} root 1= & {} (\Omega _n+\omega _i)^2+(\Omega _m+\omega _j)^2-2(\Omega _n+\omega _i)(\Omega _m+\omega _j) +\frac{4}{(2\pi )^4}Q_nQ_m\xi _i\xi _j\\ root 2= & {} (\Omega _{n'}+\omega _i)^2+(\Omega _{m'}+\omega _j)^2-2(\Omega _{n'}+\omega _i)(\Omega _{m'}+\omega _j) +\frac{4}{(2\pi )^4}Q_{n'}Q_{m'}\xi _i\xi _j. \end{aligned}$$

Obviously root1 and root2 are not square number in the sense of measure. When \(root 1\ne root 2\), all the eigenvalues are not identically zero due to the presence of the square root terms. When \(root 1=root 2\), at this time we consider the eigenvalue

$$\begin{aligned}&\langle k,\omega \rangle +\frac{1}{2}(\Omega _n+\omega _i)+\frac{1}{2}(\Omega _m+\omega _j) +\frac{1}{2}(\Omega _{n'}+\omega _i)+\frac{1}{2}(\Omega _{m'}+\omega _j)\\&\quad =\langle k+e_i+e_j,\omega \rangle +\frac{1}{2}(\Omega _n+\Omega _m+\Omega _{n'}+\Omega _{m'})\\&\quad =\langle k',\alpha \rangle +\frac{\varepsilon ^{-4}}{2}(|n|^2+|n'|^2+|m|^2+|m'|^2)+\langle Ak'+\zeta ,\xi \rangle \end{aligned}$$

where

$$\begin{aligned} k'=k+e_i+e_j,\quad \zeta =\frac{1}{4(2\pi )^2}\left( \frac{1}{|n|^2}+\frac{1}{|m'|^2}+\frac{1}{|m|^2}+\frac{1}{|n'|^2}\right) (1,1,\cdots ,1). \end{aligned}$$

When \(k=-e_i-e_j\) there is \(k'=0\) which implies \(\frac{1}{2}(\Omega _n+\Omega _m+\Omega _{n'}+\Omega _{m'})\ne 0.\) If \(Ak'+\zeta =0\) the vector \(k'=(k' _1,k'_2,\cdots ,k'_b)\) could be solved directly with Cramer’s rule:

$$\begin{aligned} k'_l=\frac{-1}{2b+1}\left( \frac{1}{|n|^2}+\frac{1}{|m'|^2}+\frac{1}{|m|^2}+\frac{1}{|n'|^2}\right) |i_l|^2,\quad 1\le l\le b. \end{aligned}$$

Obviously for \( 1\le l\le b\) it is true that \(k'_l\ne 0\). Denote \(k'\in {\mathbb Z}^b\) , so just one element of the vector \(k'\) not in \({\mathbb Z}\) could lead to a contradiction. Now we consider the case \(k'_l\) is an integer for \(1\le l\le b\) . At this time

$$\begin{aligned}&\varepsilon ^{4}\langle k',\alpha \rangle +\frac{1}{2}\left( |n|^2+|n'|^2+|m|^2+|m'|^2\right) \\&\quad =\frac{|n|^2+|m'|^2}{2(2b+1)|n|^2|m'|^2} \left[ (2b+1)|n|^2|m'|^2-2\sum \limits _{k=1}^b |i_k|^4\right] \\&\quad + \frac{|m|^2+|n'|^2}{2(2b+1)|m|^2|n'|^2} \left[ (2b+1)|m|^2|n'|^2-2\sum \limits _{k=1}^b |i_k|^4\right] .\\ \end{aligned}$$

Since \(n,m'\in {\mathbb Z}_{odd}^2\), \(|n|^2|m'|^2\) is an odd number which guarantees \((2b+1)|n|^2|m'|^2-2\sum \limits _{k=1}^b |i_k|^4\) is also an odd number. \((2b+1)|m|^2|n'|^2-2\sum \limits _{k=1}^b |i_k|^4\) has the same result.

If \(\varepsilon ^{4}\langle k',\alpha \rangle +\frac{1}{2}(|n|^2+|n'|^2+|m|^2+|m'|^2)=0\), denote \(n,n'\in \mathcal L_1\) then \(|i|^2-|j|^2=|m|^2-|n|^2=|m'|^2-|n'|^2\) which implies \(|m|^2+|n'|^2=|m'|^2+|n|^2\). Accordingly we conclude

$$\begin{aligned} \frac{1}{|n|^2|m'|^2} \left[ (2b+1)|n|^2|m'|^2-2\sum \limits _{k=1}^b |i_k|^4\right] + \frac{1}{|m|^2|n'|^2} \left[ (2b+1)|m|^2|n'|^2-2\sum \limits _{k=1}^b |i_k|^4\right] =0.\end{aligned}$$

Hence

$$\begin{aligned} (2b+1)=\sum \limits _{k=1}^b |i_k|^4\left( \frac{1}{|n|^2|m'|^2}+\frac{1}{|m|^2|n'|^2}\right) , \end{aligned}$$

then

$$\begin{aligned} (2b+1)|n|^2|m'|^2|m|^2|n'|^2=\sum \limits _{k=1}^b |i_k|^4(|n|^2|m'|^2+|m|^2|n'|^2). \end{aligned}$$

However in \({\mathbb Z}_{odd}\), \((2b+1)|n|^2|m'|^2|m|^2|n'|^2\) is an odd number and \(\sum \limits _{k=1}^b |i_k|^4(|n|^2|m'|^2+|m|^2|n'|^2)\) is an even number. It is a contradiction. \(\square \)

Then we give the proof of Lemma 4.5.

At this time the form of \(\langle k,\omega \rangle I \pm A_{n}\otimes I_2\pm I_2\otimes A_{n'}\) is

$$\begin{aligned}&\mathrm{Diag}\left( \begin{array}{c}\langle k,\omega \rangle \pm (\Omega _n+\omega _i)\pm (\Omega _{n'}+\omega _{i'})\\ \langle k,\omega \rangle \pm (\Omega _n+\omega _i)\pm (\Omega _{m'}+\omega _{j'})\\ \langle k,\omega \rangle \pm (\Omega _m+\omega _j)\pm (\Omega _{n'}+\omega _{i'})\\ \langle k,\omega \rangle \pm (\Omega _m+\omega _j)\pm (\Omega _{m'}+\omega _{j'})\end{array}\right) \\&\quad \pm \left( \begin{array}{cc}0&{}\frac{1}{(2\pi )^2}Q_m\sqrt{\xi _i\xi _j}\\ \frac{1}{(2\pi )^2}Q_n\sqrt{\xi _i\xi _j}&{}0\end{array}\right) \otimes I_2\\&\quad \pm I_2\otimes \left( \begin{array}{cc}0&{}\frac{1}{(2\pi )^2}Q_{n'}\sqrt{\xi _{i'}\xi _{j'}}\\ \frac{1}{(2\pi )^2}Q_{m'}\sqrt{\xi _{i'}\xi _{j'}}&{}0\end{array}\right) ^\mathrm {T}, \end{aligned}$$

Set \(\alpha =\varepsilon ^{-4}(|i_1|^2, |i_2|^2, \cdots , |i_b|^2)\), \(\xi =(\xi _{i_1},\xi _{i_2}, \cdots , \xi _{i_b})\),

$$\begin{aligned} \beta= & {} \frac{1}{2(2\pi )^2}\left( \frac{1}{|n|^2}+\frac{1}{|i_i|^2}\right) (1,1,\cdots , 1),\\ \beta '= & {} \frac{1}{2(2\pi )^2}\left( \frac{1}{|n'|^2}+\frac{1}{|i_{i'}|^2}\right) (1,1,\cdots , 1), \end{aligned}$$

and notice that \(|n|^2+|i|^2=|m|^2+|j|^2, |n'|^2+|i'|^2=|m'|^2+|j'|^2.\) We have

$$\begin{aligned}&\langle k,\omega \rangle I \pm A_{n}\otimes I_2\pm I_2\otimes A_{n'}\\&\quad =\left( \langle k,\alpha \rangle \pm \varepsilon ^{-4}(|n|^2+|i|^2)\pm \varepsilon ^{-4}(|n'|^2+|i'|^2)+\langle Ak,\xi \rangle \right) I\\&\quad \pm \mathrm{Diag}\left( \begin{array}{l}\langle \pm \left( \frac{1}{|n|^2}+\frac{1}{|i|^2}\right) v \pm \left( \frac{1}{|n'|^2}+\frac{1}{|i'|^2}\right) v,\xi \rangle \\ \langle \pm \left( \frac{1}{|n|^2}+\frac{1}{|i|^2}\right) v\pm \left( \frac{1}{|m'|^2}+\frac{1}{|j'|^2}\right) v,\xi \rangle \\ \langle \pm \left( \frac{1}{|m|^2}+\frac{1}{|j|^2}\right) v \pm \left( \frac{1}{|n'|^2}+\frac{1}{|i'|^2}\right) v,\xi \rangle \\ \langle \pm \left( \frac{1}{|m|^2}+\frac{1}{|j|^2}\right) v \pm \left( \frac{1}{|m'|^2}+\frac{1}{|j'|^2}\right) v,\xi \rangle \end{array}\right) \\&\quad \pm \left( \begin{array}{ll}\frac{1}{4(2\pi )^2|i|^2}\xi _i&{}\frac{1}{(2\pi )^2}Q_m\sqrt{\xi _i\xi _j}\\ \frac{1}{(2\pi )^2}Q_n\sqrt{\xi _i\xi _j}&{}\frac{1}{4(2\pi )^2|j|^2}\xi _j \end{array}\right) \otimes I_2\\&\quad \pm I_2\otimes \left( \begin{array}{ll}\frac{1}{4(2\pi )^2|i'|^2}\xi _{i'} &{}\frac{1}{(2\pi )^2}Q_{n'}\sqrt{\xi _{i'}\xi _{j'}}\\ \frac{1}{(2\pi )^2}Q_{m'}\sqrt{\xi _{i'}\xi _{j'}}&{}\frac{1}{4(2\pi )^2|j'|^2}\xi _{j'}\end{array}\right) ^\mathrm {T} \end{aligned}$$

with \(v=\frac{1}{(2\pi )^2}(1,1,\cdots , 1).\) Its eigenvalues has the form as follows:

$$\begin{aligned}&\langle k,\alpha \rangle \pm \varepsilon ^{-4}(|n|^2+|i|^2)\pm \varepsilon ^{-4}(|n'|^2+|i'|^2)+\langle Ak\pm \beta \pm \beta ',\xi \rangle \\&\quad \pm \left( \frac{1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j\pm \frac{\sqrt{\Delta }}{2}\,\right) \\&\quad \pm \left( \frac{1}{8(2\pi )^2|i'|^2}\xi _{i'}+ \frac{1}{8(2\pi )^2|j'|^2}\xi _{j'}\pm \frac{\sqrt{\Delta '}}{2}\,\right) , \end{aligned}$$

where \(\Delta =H_i^2\xi _i^2+H_j^2\xi _j^2+(-2H_iH_j+4\frac{1}{(2\pi )^4}Q_nQ_m)\xi _i\xi _j\) with

$$\begin{aligned} H_i=\frac{1}{4(2\pi )^2|i|^2},\quad \quad H_j=\frac{1}{4(2\pi )^2|j|^2}, \end{aligned}$$
(6.1)

and \(\Delta '\) has the same form. According to the form of \(Q_nQ_m\), if \(Q_nQ_m=\frac{1}{16|i|^2|j|^2}\) there is

However, \(2\not \mid E_1\) and \(2\mid E_2\) lead to a contradiction. So it is true that \(-2H_iH_j+4\frac{1}{(2\pi )^4}Q_nQ_m\ne 2H_iH_j\). So \(\Delta \) is not a square number. If \(i\ne i'\) or \(Q_nQ_m\ne Q_{n'}Q_{m'}\), all the eigenvalues are not identically zero due to the presence of the square root terms. If \(Q_nQ_m=Q_{n'}Q_{m'}\) and \(i=i'\), consequently \(j= j'\), hence if the eigenvalue is

$$\begin{aligned}&\langle k,\alpha \rangle + \varepsilon ^{-4}(|n|^2+|i|^2)- \varepsilon ^{-4}(|n'|^2+|i|^2)+\langle Ak+\beta -\beta ',\xi \rangle \\&\quad + \left( \frac{1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j+ \frac{\sqrt{\Delta }}{2}\,\right) \\&\quad - \left( \frac{1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j+ \frac{\sqrt{\Delta }}{2}\,\right) , \\&= \langle k,\alpha \rangle +\varepsilon ^{-4}(|n|^2-|n'|^2)+\langle Ak+\beta -\beta ',\xi \rangle . \end{aligned}$$

By the result of Lemma 6.1 this eigenvalue is not identically zero; if the eigenvalue is

$$\begin{aligned}&\langle k,\alpha \rangle + \varepsilon ^{-4}(|n|^2+|i|^2)+ \varepsilon ^{-4}(|n'|^2+|i|^2)+\langle Ak+\beta +\beta ',\xi \rangle \\&\quad + \left( \frac{1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j+ \frac{\sqrt{\Delta }}{2}\,\right) \\&\quad + \left( \frac{1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j- \frac{\sqrt{\Delta }}{2}\,\right) . \end{aligned}$$

Since the result of Lemma 6.2 this eigenvalue is not identically zero. Thus all eigenvalues are not identically zero.

1.2 Proof of Lemma 4.6

Set

$$\begin{aligned} {\bar{\beta }}= & {} \frac{1}{2(2\pi )^2}\left( \frac{1}{|n|^2}-\frac{1}{|i_i|^2}\right) (1,1,\cdots , 1),\\ {\bar{\beta }}'= & {} \frac{1}{2(2\pi )^2}\left( \frac{1}{|n'|^2}-\frac{1}{|i_{i'}|^2}\right) (1,1,\cdots , 1). \end{aligned}$$

For \(n\in \mathcal L_1, n'\in \mathcal L_2,\) the form of the eigenvalues of \(\langle k,\omega \rangle I \pm A_{n}\otimes I_2\pm I_2\otimes A_{n'}\) is

$$\begin{aligned}&\langle k,\alpha \rangle \pm \varepsilon ^{-4}(|n|^2+|i|^2)\pm \varepsilon ^{-4}(|n'|^2-|i'|^2)+\langle Ak\pm \beta \pm {\bar{\beta }}',\xi \rangle \\&\quad \pm \left( \frac{1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j\pm \frac{\sqrt{\Delta }}{2}\,\right) \\&\quad \pm \left( -\frac{1}{8(2\pi )^2|i'|^2}\xi _{i'}+ \frac{1}{8(2\pi )^2|j'|^2}\xi _{j'}\pm \frac{\sqrt{{{\bar{\Delta }}}'}}{2}\,\right) , \end{aligned}$$

where \({\bar{\Delta }}=H_i^2\xi _i^2+H_j^2\xi _j^2+(2H_iH_j-4\frac{1}{(2\pi )^4}{\bar{Q}}_n{\bar{Q}}_m)\xi _i\xi _j\) with

$$\begin{aligned} H_i=\frac{1}{4(2\pi )^2|i|^2},\quad \quad H_j=\frac{1}{4(2\pi )^2|j|^2}, \end{aligned}$$
(6.2)

and \({{\bar{\Delta }}}'\) has the same form. According to the form of \({\bar{Q}}_n{\bar{Q}}_m\), if \({\bar{Q}}_n{\bar{Q}}_m=\frac{1}{16|i|^2|j|^2}\) there is

$$\begin{aligned} |m|^2(63|n|^2-32|j|^2)=16|i|^2(2|n|^2-|j|^2). \end{aligned}$$

Obviously the right one could be divided by 2 but the left could not which is a contradiction. So it is true that \(2H_iH_j-\frac{4}{(2\pi )^4}{\bar{Q}}_n{\bar{Q}}_m\ne -2H_iH_j\). So \({\bar{\Delta }}\) is not a square number. Meanwhile if \({\bar{Q}}_{n'}{\bar{Q}}_{m'}+ Q_n Q_m=\frac{1}{16|i|^2|j|^2}\) there is

$$\begin{aligned} E_1:= & {} |m|^2|j|^2|m'|^2\left( 125|i|^2|n|^2|n'|^2-64|i|^2|j|^2|n|-128|j|^2|n|^2|n'|-128|i|^2|j|^2|n'|\right) \\= & {} 32|i|^4|j|^2|m|^2|m'|^2(2|n'|^2-|j|^2)\\&-128|i|^2|n'|^2|m'|^2(|j|^2+|m|^2)(-|i|^2|n|^2+|j|^2|n|^2+|j|^2|i|^2):=E_2. \end{aligned}$$

However, \(2\not \mid E_1\) and \(2\mid E_2\) lead to a contradiction. So it is true that \(-2H_iH_j+\frac{4 }{(2\pi )^4} Q_n Q_m\ne 2H_iH_j-\frac{4 }{(2\pi )^4}{\bar{Q}}_{n'}{\bar{Q}}_{m'}\). Hence all the eigenvalues are not identically zero due to the presence of the square root terms.

1.3 Proof of Lemma 4.7

Because \(n, n'\in \mathcal L_2,\) the form of the eigenvalues of \(\langle k,\omega \rangle I \pm A_{n}\otimes I_2\pm I_2\otimes A_{n'}\) is

$$\begin{aligned}&\langle k,\alpha \rangle \pm \varepsilon ^{-4}(|n|^2-|i|^2)\pm \varepsilon ^{-4}(|n'|^2-|i'|^2)+\langle Ak\pm {\bar{\beta }}\pm {\bar{\beta }}',\xi \rangle \\&\quad \pm \left( \frac{-1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j\pm \frac{\sqrt{{\bar{\Delta }}}}{2}\,\right) \\&\quad \pm \left( \frac{-1}{8(2\pi )^2|i'|^2}\xi _{i'}+ \frac{1}{8(2\pi )^2|j'|^2}\xi _{j'}\pm \frac{\sqrt{{{\bar{\Delta }}}'}}{2}\,\right) . \end{aligned}$$

If \(i\ne i'\) or \({\bar{Q}}_n{\bar{Q}}_m\ne {\bar{Q}}_{n'}{\bar{Q}}_{m'}\), all the eigenvalues are not identically zero due to the presence of the square root terms. If \({\bar{Q}}_n{\bar{Q}}_m={\bar{Q}}_{n'}{\bar{Q}}_{m'}\) and \(i=i'\), consequently \(j= j'\), hence if the eigenvalue is

$$\begin{aligned}&\langle k,\alpha \rangle + \varepsilon ^{-4}(|n|^2-|i|^2)- \varepsilon ^{-4}(|n'|^2-|i|^2)+\langle Ak+{\bar{\beta }}-{\bar{\beta }}',\xi \rangle \\&\quad + \left( \frac{-1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j+ \frac{\sqrt{{\bar{\Delta }}}}{2}\,\right) \\&\quad - \left( \frac{-1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j+ \frac{\sqrt{{\bar{\Delta }}}}{2}\,\right) , \\&= \langle k,\alpha \rangle +\varepsilon ^{-4}(|n|^2-|n'|^2)+\langle Ak+\beta -\beta ',\xi \rangle . \end{aligned}$$

By the result of Lemma 6.1 this eigenvalue is not identically zero; if the eigenvalue is

$$\begin{aligned}&\langle k,\alpha \rangle + \varepsilon ^{-4}(|n|^2-|i|^2)+ \varepsilon ^{-4}(|n'|^2-|i|^2)+\langle Ak+{\bar{\beta }}+{\bar{\beta }}',\xi \rangle \\&\quad + \left( \frac{-1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j+ \frac{\sqrt{{\bar{\Delta }}}}{2}\,\right) \\&\quad + \left( \frac{-1}{8(2\pi )^2|i|^2}\xi _i+ \frac{1}{8(2\pi )^2|j|^2}\xi _j- \frac{\sqrt{{\bar{\Delta }}}}{2}.\right) \end{aligned}$$

Let us give the following lemma for the above eigenvalue.

Lemma 6.3

For \(k\ne 0\), \(n,n'\in \mathcal L_2\), \(i=i'\), \(j=j'\) and \({\bar{Q}}_n{\bar{Q}}_m={\bar{Q}}_{n'}{\bar{Q}}_{m'}\) then all the eigenvalues of

$$\begin{aligned}&\langle k,\omega \rangle I + \left( \begin{array}{cc}\Omega _n-\omega _i&{}\frac{-1}{(2\pi )^2}{\bar{Q}}_m\sqrt{\xi _i\xi _j}\\ \frac{1}{(2\pi )^2}{\bar{Q}}_n\sqrt{\xi _i\xi _j}&{}-\Omega _m+\omega _j\end{array}\right) \otimes I_2\\&\quad + I_2\otimes \left( \begin{array}{cc}\Omega _{n'}-\omega _{i}&{}\frac{1}{(2\pi )^2}{\bar{Q}}_{n'}\sqrt{\xi _{i}\xi _{j}}\\ \frac{-1}{(2\pi )^2}{\bar{Q}}_{m'}\sqrt{\xi _{i}\xi _{j}}&{}-\Omega _{m'}+\omega _{j}\end{array}\right) ^\mathrm {T}, \end{aligned}$$

are not identically zero.

Proof

The eigenvalues we are considering have the form as following:

$$\begin{aligned}&\langle k,\omega \rangle +\frac{1}{2}(\Omega _n-\omega _i)-\frac{1}{2}(\Omega _m-\omega _j)\pm \frac{1}{2} \sqrt{root 1'} +\frac{1}{2}(\Omega _{n'}-\omega _i)\\&\quad -\frac{1}{2}(\Omega _{m'}-\omega _j)\pm \frac{1}{2} \sqrt{root 2'} \end{aligned}$$

with

$$\begin{aligned} root 1'= & {} (\Omega _n-\omega _i)^2+(\Omega _m-\omega _j)^2+2(\Omega _n-\omega _i)(\Omega _m-\omega _j) -\frac{4}{(2\pi )^2}{\bar{Q}}_n{\bar{Q}}_m\xi _i\xi _j\\ root 2'= & {} (\Omega _{n'}-\omega _i)^2+(\Omega _{m'}-\omega _j)^2+2(\Omega _{n'}-\omega _i)(\Omega _{m'}-\omega _j) -\frac{4}{(2\pi )^2}{\bar{Q}}_{n'}{\bar{Q}}_{m'}\xi _i\xi _j. \end{aligned}$$

Obviously \(root 1'\) and \(root 2'\) are not square number in the sense of measure. When \(root 1'\ne root 2'\), all the eigenvalues are not identically zero due to the presence of the square root terms. When \(root 1'=root 2'\), at this time we consider the eigenvalue

$$\begin{aligned}&\langle k,\omega \rangle +\frac{1}{2}(\Omega _n-\omega _i)-\frac{1}{2}(\Omega _m-\omega _j) +\frac{1}{2}(\Omega _{n'}-\omega _i)-\frac{1}{2}(\Omega _{m'}-\omega _j)\\&\quad =\langle k-e_i+e_j,\omega \rangle +\frac{1}{2}(\Omega _n-\Omega _m+\Omega _{n'}-\Omega _{m'})\\&\quad =\langle k'',\alpha \rangle +\frac{\varepsilon ^{-4}}{2}(|n|^2+|n'|^2-|m|^2-|m'|^2)+\langle Ak''+\zeta ' ,\xi \rangle \end{aligned}$$

where

$$\begin{aligned} k''=k-e_i+e_j,\quad \zeta '=\frac{1}{4(2\pi )^2}(\frac{1}{|n|^2}-\frac{1}{|m'|^2}-\frac{1}{|m|^2}+\frac{1}{|n'|^2})(1,1,\cdots ,1). \end{aligned}$$

If \(Ak''+\zeta =0\) the vector \(k''=(k'' _1,k''_2,\cdots ,k''_b)\) could be solved directly with Cramer’s rule:

$$\begin{aligned}k''_l=\frac{1}{2b+1}(\frac{1}{|m|^2}-\frac{1}{|n|^2}+\frac{1}{|m'|^2}-\frac{1}{|n'|^2})|i_l|^2,\quad 1\le l\le b.\end{aligned}$$

Denote \(n,n'\in \mathcal L_2\) then \(|i|^2+|j|^2=|m|^2+|n|^2=|m'|^2+|n'|^2\) which implies \(|n'|^2-|m|^2=|n|^2-|m'|^2\). While at this time when \(|n|\ne |m'|\), since \(k''= 0\) if and only if \(\frac{1}{|m|^2}-\frac{1}{|n|^2}+\frac{1}{|m'|^2}-\frac{1}{|n'|^2}=0\) so \(k=e_i-e_j\), then

$$\begin{aligned}&\langle k,\omega \rangle +\frac{1}{2}(\Omega _n-\omega _i)-\frac{1}{2}(\Omega _m-\omega _j) +\frac{1}{2}(\Omega _{n'}-\omega _i)-\frac{1}{2}(\Omega _{m'}-\omega _j)\\&\quad =\frac{1}{2}(\Omega _n-\Omega _m+\Omega _{n'}-\Omega _{m'})\\&\quad =\frac{\varepsilon ^{-4}}{2}(|n|^2+|n'|^2-|m|^2-|m'|^2)+\langle \zeta ' ,\xi \rangle \\&\quad =\varepsilon ^{-4}(|n|^2-|m'|^2)\ne 0 \end{aligned}$$

If \(\frac{1}{|m|^2}-\frac{1}{|n|^2}+\frac{1}{|m'|^2}-\frac{1}{|n'|^2}\ne 0\) there is \(k''_l\ne 0\) for \(1\le l\le b\). Denote \(k''\in {\mathbb Z}^b\) , so just one element of the vector \(k''\) not in \({\mathbb Z}\) could lead to a contradiction. Now we consider the case \(k''_l\) is an integer for \(1\le l\le b\) . At this time

$$\begin{aligned}&\varepsilon ^{4}\langle k'',\alpha \rangle +\frac{1}{2}\left( |n|^2+|n'|^2-|m|^2-|m'|^2\right) \\&\quad =\frac{|n|^2-|m'|^2}{2(2b+1)|n|^2|m'|^2} \left[ (2b+1)|n|^2|m'|^2+2\sum \limits _{k=1}^b |i_k|^4\right] \\&\qquad + \frac{|n'|^2-|m|^2}{2(2b+1)|m|^2|n'|^2} \left[ (2b+1)|m|^2|n'|^2+2\sum \limits _{k=1}^b |i_k|^4\right] \\&\quad =\frac{|n|^2-|m'|^2}{(2b+1)} \left[ (2b+1)+\left( \frac{1}{|m|^2|n'|^2}+\frac{1}{|n|^2|m'|^2}\right) \sum \limits _{k=1}^b |i_k|^4\right] \ne 0 \end{aligned}$$

. \(\square \)

By the result of Lemma 6.3 this eigenvalue is not identically zero, and thus all eigenvalues are not identically zero.

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Ge, C., Geng, J. & Lou, Z. KAM theory for the reversible perturbations of 2D linear beam equations. Math. Z. 297, 1693–1731 (2021). https://doi.org/10.1007/s00209-020-02575-9

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