KAM Tori for a Two Dimensional Beam Equation with a Quintic Nonlinear Term and Quasi-periodic Forcing

Abstract

This work studies a two-dimensional beam equation with a quintic nonlinear term and quasi-periodic forcing

$$\begin{aligned} u_{tt}+\Delta ^2 u+ \varepsilon \phi (t)h(u)=0,\quad x\in {\mathbb {T}}^2,\quad t\in {\mathbb {R}} \end{aligned}$$

with periodic boundary conditions, where $\varepsilon$ is a small positive parameter; $\phi (t)$ is a real analytic quasi-periodic function in t with frequency vector $\eta =(\eta _1 ,\eta _2 \ldots ,\eta _{n^*})\subset [\varrho , 2\varrho ]^{n^*}$ for a given positive integer $n^*$ and some constant $\varrho >0$; and h is a real analytic function of the form

$$\begin{aligned} h(u)=c_1u+c_5u^5+\sum _{{\hat{i}}\ge 6}c_{{\hat{i}}}u^{{\hat{i}}},\quad c_1,c_5\ne 0. \end{aligned}$$

Firstly, the linear part of Hamiltonian system corresponding to the equation is transformed to constant coefficients by a linear quasi-periodic change of variables. Then, a symplectic transformation is used to convert the Hamiltonian system into an angle-dependent block-diagonal normal form, which can be achieved by selecting the appropriate tangential sites. Finally, it is obtained that a Whitney smooth family of small-amplitude quasi-periodic solutions for the equation by developing an abstract KAM (Kolmogorov–Arnold–Moser) theorem for infinite dimensional Hamiltonian systems.

Appendix

Lemma 6.1

${\mathcal {I}}$ is the given set with n points in Remark 1.1. For $f,r,s,t,{\widetilde{f}},{\widetilde{r}},{\widetilde{s}},{\widetilde{t}}\in {\mathcal {I}}$, the following results true

• 1. $f-r=0$, $f=r$, $|f|=|r|$ are equivalent.

• 2. $f-r+s-t=0$, $\{f,s\}=\{r,t\}$, $\{|f|,|s|\}=\{|r|,|t|\}$ are equivalent.

• 3. $f-r+s-t+{\widetilde{f}}-{\widetilde{r}}=0$, $\{f,s,{\widetilde{f}}\}=\{r,t,{\widetilde{r}}\}$, $\{|f|,|s|,|{\widetilde{f}}|\}=\{|r|,|t|,|{\widetilde{r}}|\}$ are equivalent.

• 4. $f-r+s-t+{\widetilde{f}}-{\widetilde{r}}+{\widetilde{s}}-{\widetilde{t}}=0$, $\{f,s,{\widetilde{f}},{\widetilde{s}}\}=\{r,t,{\widetilde{r}},{\widetilde{t}}\}$, $\{|f|,|s|,|{\widetilde{f}}|,|{\widetilde{s}}|\}=\{|r|,|t|,|{\widetilde{r}}|,|{\widetilde{t}}|\}$ are equivalent.

• 5. $<f,s>=<r,t>$, $\{f,s\}=\{r,t\}$, $\{|f|,|s|\}=\{|r|,|t|\}$ are equivalent.

• 6. $<f-s,r-t>=0$ is equivalent to $f=s$ or $r=t$.

These results are obvious and omit the proof.

Lemma 6.2

The set ${\mathcal {I}}$ given in Remark 1.1 is admissible.

Proof

By $i+r-j-s=0$ and ${|i|}^2+{|r|}^2-{|j|}^2-{|s|}^2=0$, then $<j-r,i-j>=0$. By $i+d+r-j-l-s=0$ and ${|i|}^2+{|d|}^2+{|r|}^2-{|j|}^2-{|l|}^2-{|s|}^2=0$, then $<r-l,i-j+d-l>=<i-j,j-d>$. To prove that ${\mathcal {I}}$ has the properties (3)–(7) in Definition 1.1, we need to prove that

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<j-r, i-j>=0,\\ <{\widetilde{j}}-{\widetilde{r}},{\widetilde{i}}-{\widetilde{j}}>=0.\\ \end{array} \right. , \end{aligned}$$

(6.1)

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<i-r,j-r>=0,\\ <{\widetilde{i}}-{\widetilde{r}},{\widetilde{j}}-{\widetilde{r}}>=0.\\ \end{array} \right. , \end{aligned}$$

(6.2)

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<r-l,i-j+d-l>=<i-j,j-d>,\\<r-{\widetilde{l}},{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}>=<{\widetilde{i}}-{\widetilde{j}},{\widetilde{j}}-{\widetilde{d}}>.\\ \end{array} \right. , \end{aligned}$$

(6.3)

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<l-r,i-j+d-r>=<i-j,j-d>,\\<{\widetilde{l}}-r,{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-r>=<{\widetilde{i}}-{\widetilde{j}},{\widetilde{j}}-{\widetilde{d}}>.\\ \end{array} \right. , \end{aligned}$$

(6.4)

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<j-r,i-j>=0,\\ <{\widetilde{i}}-{\widetilde{r}},{\widetilde{j}}-{\widetilde{r}}>=0.\\ \end{array} \right. , \end{aligned}$$

(6.5)

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<j-r,i-j>=0,\\<r-{\widetilde{l}},{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}>=<{\widetilde{i}}-{\widetilde{j}},{\widetilde{j}}-{\widetilde{d}}>.\\ \end{array} \right. , \end{aligned}$$

(6.6)

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<j-r,i-j>=0,\\<{\widetilde{l}}-r,{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-r>=<{\widetilde{i}}-{\widetilde{j}},{\widetilde{j}}-{\widetilde{d}}>.\\ \end{array} \right. , \end{aligned}$$

(6.7)

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<i-r,j-r>=0,\\<r-{\widetilde{l}},{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}>=<{\widetilde{i}}-{\widetilde{j}},{\widetilde{j}}-{\widetilde{d}}>.\\ \end{array} \right. , \end{aligned}$$

(6.8)

$$\begin{aligned}{} & {} \left\{ \begin{array}{l}<i-r,j-r>=0,\\<{\widetilde{l}}-r,{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-r>=<{\widetilde{i}}-{\widetilde{j}},{\widetilde{j}}-{\widetilde{d}}>.\\ \end{array} \right. , \end{aligned}$$

(6.9)

and

$$\begin{aligned} \left\{ \begin{array}{l}<r-l,i-j+d-l>=<i-j,j-d>,\\<{\widetilde{l}}-r,{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-r>=<{\widetilde{i}}-{\widetilde{j}},{\widetilde{j}}-{\widetilde{d}}>.\\ \end{array} \right. \end{aligned}$$

(6.10)

have no solution in $r=(r_1, r_2)\in {{\mathbb {Z}}}^{2,*}$ for ${\widetilde{i}},{\widetilde{j}},{\widetilde{d}},{\widetilde{l}},i,j,d,l\in {\mathcal {I}}$ and $\{{\widetilde{i}},{\widetilde{j}},{\widetilde{d}},{\widetilde{l}}\}\ne \{i,j,d,l\}$.

(I) The property (2) is first proved by contradiction. The proof for the property (1) is similar and simpler. Suppose $i,j,d,l,r\in {\mathcal {I}}$ satisfies $<r-l, i-j+d-l>=<i-j,j-d>$.

Case 1.1. Only one element of $\{|i|,|j|,|d|,|l|,|r|\}$ is at its maximum.

Case 1.1.1. Assuming $|r|=\mathrm{{max}}\{|i|,|j|,|d|,|l|,|r|\}$. Write $<r-l,i-j+d-l>=<i-j,j-d>$ in terms of the following components

$$\begin{aligned}{} & {} (r_1-l_1)(i_1-j_1+d_1-l_1)+(r_2-l_2)(i_2-j_2+d_2-l_2)\\{} & {} \qquad =(i_1-j_1)(j_1-d_1)+(i_2-j_2)(j_2-d_2). \end{aligned}$$

According to the calculation, we have

$$\begin{aligned} \begin{aligned} r_2&=l_2+\displaystyle \frac{(l_1-r_1)(i_1-j_1+d_1-l_1)+(i_1-j_1)(j_1-d_1)+(i_2-j_2)(j_2-d_2)}{i_2-j_2+d_2-l_2}\\&\le \displaystyle l_2 +\big |(l_1-r_1)(i_1-j_1+d_1-l_1)\big |+ \big |(i_1-j_1)(j_1-d_1)\big |\\&\quad +\big |(i_2-j_2)(j_2-d_2)\big |\\ {}&\le \displaystyle l_2+2r_1^2+ r_1^2+r_1^2\le l_2+ 4r_1^2< 5r_1^2. \end{aligned} \end{aligned}$$

This is contradictory to $r_2=r_1^5$.

Case 1.1.2. Assuming $|j|=\mathrm{{max}}\{|i|,|j|,|d|,|l|,|r|\}$, then

$$\begin{aligned} \left|<i-j,j-d>\right|>\left|<\displaystyle \frac{j}{2},\displaystyle \frac{j}{2}>\right| =\displaystyle \frac{1}{4}{|j|}^2>\left| <r-l,i-j+d-l>\right| . \end{aligned}$$

This is contradictory to the hypothesis $<r-l,i-j+d-l>=<i-j,j-d>$.

Case 1.2. Two elements of $\{|i|,|j|,|d|,|l|,|r|\}$ is at its maximum.

Case 1.2.1. Assuming $|l|=|j|=\mathrm{{max}}\{|i|,|j|,|d|,|l|,|r|\}$, then

$$\begin{aligned} \begin{aligned}&<r-l,i-j+d-l>=<i-j,j-d>\\&\Rightarrow \quad<r-j,i+d-2j>=<i-j,j-d>\\&\Rightarrow \quad 3<j,j>-2<r,j>-2<i,j>-2<d,j>+<i,r>+<d,r>\\&\qquad +<i,d>=0. \end{aligned} \end{aligned}$$

But we have

$$\begin{aligned} \begin{aligned}&3<j,j>-2<r,j>-2<i,j>-2<d,j>+<i,r>\\&\qquad +<d,r>+<i,d> \\ =&3<j,j>-2<r+i+d,j>+<i,r>+<d,r>+<i,d> \\ >&0. \end{aligned} \end{aligned}$$

This is a contradiction between.

Other situations are similar to the three above. So the hypothesis $<r-l,i-j+d-l>=<i-j,j-d>$ is not true. That is ${\mathcal {I}}$ meets the property (2) in Definition 1.1. Similarly, ${\mathcal {I}}$ meets the property (1) in Definition 1.1.

(II) Let’s prove the equation (6.3) has no solution in ${{\mathbb {Z}}}^{2,*}$. The proof for (6.1) and (6.6) is similar and simpler to the proof for (6.3).

Case 2.1. Only one of $\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}},||i|,|j|,|d|,|l|\}$ is at its maximum.

Case 2.1.1. Assuming $|i|=\mathrm{{max}}\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}},||i|,|j|,|d|,|l|\}$. According to the calculation, then

$$\begin{aligned} r_2=l_2+j_2-d_2+\displaystyle \frac{\alpha _{211}}{\beta _{211}}, \end{aligned}$$

where

$$\begin{aligned} \alpha _{211}= & {} i_1\left[<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}> +<j-d+l-{\widetilde{l}},{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}> \right] \\{} & {} +(d_1-j_1-l_1)\left[<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>+<j-d+l-{\widetilde{l}},{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}> \right] \\{} & {} +({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)<d-j,d-l>,\\ \beta _{211}= & {} ({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)(i_2-j_2+d_2-l_2) +(j_1-i_1+l_1-d_1)\\{} & {} \times ({\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2-{\widetilde{l}}_2). \end{aligned}$$

To prove $\alpha _{211}\ne 0$ by contradiction. Assuming $\alpha _{211}= 0$, then

$$\begin{aligned}\left\{ \begin{array}{l}<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}> +<j-d+l-{\widetilde{l}},{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}>=0,\\ (d_1-j_1-l_1)\left[<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>+<j-d+l-{\widetilde{l}},{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}> \right] \\ +({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)<d-j,d-l>=0. \end{array} \right. \end{aligned}$$

From the system above we get $({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)<d-j,d-l>=0$. And in view of ${\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1\ne 0$, then $<d-j,d-l>=0$ is obtained. According to Lemma 6.1, then $d=j$ or $d=l$. It’s contradictory to $r\in {\mathcal {N}}_3$. That is $\alpha _{211}\ne 0$. Due to the order of the numerator $\alpha _{211}$ doesn’t exceed $i_1$ and the order of the divisor $\beta _{211}$ is $i_2$, we have $r_2\not \in {{\mathbb {Z}}}$.

Case 2.1.2. Assuming $|l|=\mathrm{{max}}\{|i|,|j|,|d|,|l|,|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|\}$. According to the calculation, then

$$\begin{aligned}r_2=l_2+\displaystyle \frac{\alpha _{212}l_2+\beta _{212}}{\gamma _{212}l_2+\delta _{212}}=l_2+\displaystyle \frac{\alpha _{212}}{\gamma _{212}} +\displaystyle \frac{\beta _{212}\gamma _{212}-\alpha _{212}\delta _{212}}{\gamma _{212}(\gamma _{212}l_2+\delta _{212})},\end{aligned}$$

where

$$\begin{aligned}{} & {} \alpha _{212}=(i_1-j_1+d_1-l_1)({\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2-{\widetilde{l}}_2), \\{} & {} \begin{aligned} \beta _{212}&=l_1({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)(i_1-j_1+d_1-l_1) -({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)<j\\&\quad -i,j-d> +(i_1-j_1+d_1-l_1)\big (<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>\\&\quad -<{\widetilde{l}},{\widetilde{i}} -{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}>\big ), \end{aligned} \\{} & {} \gamma _{212}=-\big ({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1\big ), \\{} & {} \delta _{212}=({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)(i_2-j_2+d_2) -(i_1-j_1+d_1-l_1)({\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2-{\widetilde{l}}_2). \end{aligned}$$

Set

$$\begin{aligned} \sigma _{212}:=\beta _{212}\gamma _{212}-\alpha _{212}\delta _{212}, \quad \mu _{212}:=\gamma _{212}(\gamma _{212}l_2+\delta _{212}), \end{aligned}$$

then according to the calculation,

$$\begin{aligned} \sigma _{212}=\left| {\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}\right| ^2l_1^2+\widetilde{\sigma }_{212}, \quad \mu _{212}=({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)^2l_2+\widetilde{\mu }_{212}, \end{aligned}$$

where the order of $\widetilde{\sigma }_{212}$ and $\widetilde{\mu }_{212}$ don’t exceed $l_1$. According to Lemma 6.1, then ${\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}\ne 0$. So we have that $\displaystyle \frac{\alpha _{212}}{\gamma _{212}}$ is an integer or $\left| \displaystyle \frac{\alpha _{212}}{\gamma _{212}}-\left[ \displaystyle \frac{\alpha _{212}}{\gamma _{212}}\right] \right| > \displaystyle \frac{1}{\root 25 \of {l_{1}}}$ , and $0<\left| \displaystyle \frac{\sigma _{212}}{\mu _{212}} -\left[ \displaystyle \frac{\sigma _{212}}{\mu _{212}}\right] \right| <\displaystyle \frac{1}{\root 5 \of {l_{1}^2}},$ where $\left[ \bullet \right]$ is the integer operation, and then

$$\begin{aligned}\left| \displaystyle \frac{\alpha _{212}}{\gamma _{212}}+\displaystyle \frac{\sigma _{212}}{\mu _{212}} -\left[ \displaystyle \frac{\alpha _{212}}{\gamma _{212}}+\displaystyle \frac{\sigma _{212}}{\mu _{212}}\right] \right| \in (0,1).\end{aligned}$$

Hence $r_2\not \in {{\mathbb {Z}}}$.

Other situations are similar to the above cases.

Case 2.2. Two elements of $\{|i|,|j|,|d|,|l|,|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|\}$ are at its maximum.

Case 2.2.1. Assuming $|i|=|d|=\mathrm{{max}}\{|i|,|j|,|d|,|l|,|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|\}$, then $i=d$. According to the calculation, then

$$\begin{aligned}r_2= & {} l_2+\displaystyle \frac{\alpha _{221}i_2^2+\beta _{221}i_2+\gamma _{221}}{\delta _{221}i_2+\sigma _{221}} \\ {}= & {} l_2+\displaystyle \frac{\alpha _{221}}{\delta _{221}}i_2 +\displaystyle \frac{\beta _{221}\delta _{221}-\alpha _{221}\sigma _{221}}{\delta _{221}^2} +\displaystyle \frac{\gamma _{221}\delta _{221}^2-\beta _{221}\delta _{221}\sigma _{221} +\alpha _{221}\sigma _{221}^2}{\delta _{221}^2(\delta _{221}i_2+\sigma _{221})}, \end{aligned}$$

where

$$\begin{aligned} \alpha _{221}= & {} -\big ({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1\big ), \quad \beta _{221}=2({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)j_2,\\ \gamma _{221}= & {} -\left( {\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1\right) \left( i_1^2-2i_1j_1+|j|^2\right) \\ {}{} & {} -\left( j_1-2i_1+l_1\right) \left(<l-{\widetilde{l}},{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}> +<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>\right) ,\\ \delta _{221}= & {} 2({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1), \\ \sigma _{221}= & {} -({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)(j_2+l_2) +(j_1-2i_1+l_1)({\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2-{\widetilde{l}}_2). \end{aligned}$$

Set

$$\begin{aligned}{} & {} \mu _{221}:=\displaystyle \frac{\beta _{221}\delta _{221}-\alpha _{221}\sigma _{221}}{\delta _{221}^2}, \\{} & {} \omega _{221}:=\gamma _{221}\delta _{221}^2-\beta _{221}\delta _{221}\sigma _{221}+\alpha _{221}\sigma _{221}^2, \quad \nu _{221}:=\delta _{221}^2(\delta _{221}i_2+\sigma _{221}),\end{aligned}$$

then according to the calculation,

$$\begin{aligned} \omega _{221}=4\left| {\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}-{\widetilde{l}}\right| ^2i_1^2+\widetilde{\omega }_{221}, \quad \nu _{221}=-8({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-{\widetilde{l}}_1)^2i_2+\widetilde{\nu }_{221}, \end{aligned}$$

where the order of $\widetilde{\omega }_{221}$ and $\widetilde{\nu }_{221}$ don’t exceed $i_1$. According to Lemma 6.1, then $-{\widetilde{i}}+{\widetilde{j}}-{\widetilde{d}}+{\widetilde{l}}\ne 0$. So we know that $\displaystyle \frac{\alpha _{221}}{\delta _{221}}=-\displaystyle \frac{1}{2}$, $\mu _{221}$ is an integer or $\left| \mu _{221}- \left[ \mu _{221}\right] \right| >\displaystyle \frac{1}{\root 25 \of {i_{1}}},$ and $0<\left| \displaystyle \frac{\omega _{221}}{\nu _{221}} -\left[ \displaystyle \frac{\omega _{221}}{\nu _{221}}\right] \right| <\displaystyle \frac{1}{i_{1}^2},$ and then

$$\begin{aligned}\left| \displaystyle \frac{\alpha _{221}}{\delta _{221}}i_2 +\mu _{221} +\displaystyle \frac{\omega _{221}}{\nu _{221}}-\left[ \displaystyle \frac{\alpha _{221}}{\delta _{221}}i_2 +\mu _{221} +\displaystyle \frac{\omega _{221}}{\nu _{221}}\right] \right| \in (0,1).\end{aligned}$$

Hence $r_2\not \in {{\mathbb {Z}}}$.

Case 2.2.2. Assuming $|l|=|{\widetilde{l}}|=\mathrm{{max}}\{|i|,|j|,|d|,|l|,|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|\}$, then $l={\widetilde{l}}$. According to the calculation, then

$$\begin{aligned} r_2=l_2+\displaystyle \frac{\alpha _{222}}{\beta _{222}}, \end{aligned}$$

where

$$\begin{aligned} \alpha _{222}= & {} (<j-i,j-d>-<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>)l_1\\{} & {} -({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1)<j-i,j-d>+(i_1-j_1+d_1)<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>,\\ \beta _{222}= & {} (i_1-j_1+d_1-{\widetilde{i}}_1+{\widetilde{j}}_1-{\widetilde{d}}_1)l_2 +({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-l_1)(i_2-j_2+d_2)\\{} & {} -(i_1-j_1+d_1-l_1)({\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2). \end{aligned}$$

To prove $\alpha _{222}\ne 0$ by contradiction. Assumeing $\alpha _{222}= 0$, then

$$\begin{aligned}\left\{ \begin{array}{l}<j-i,j-d>-<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>=0,\\ -({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1)<j-i,j-d>+(i_1-j_1+d_1)<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>=0. \end{array} \right. \end{aligned}$$

From the system above we get $({\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-i_1+j_1-d_1)<j-i,j-d>=0$. And in view of ${\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1-i_1+j_1-d_1\ne 0$, then $<j-i,j-d>=0$ is obtained. According to Lemma 6.1, we have $j=i$ or $j=d$. It’s contradictory to $r\in {\mathcal {N}}_3$. That is $\alpha _{222}\ne 0$. Due to the order of the numerator $\alpha _{222}$ doesn’t exceed $l_1$ and the order of the divisor $\beta _{222}$ is $l_2$, we have $r_2\not \in {{\mathbb {Z}}}$.

Other situations are similar to the above cases.

Case 2.3. Three elements of $\{|i|,|j|,|d|,|l|,|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|\}$ are at its maximum. It can be seen by [25] that such situation is similar to those mentioned above, therefore omit the proof.

Case 2.4. Four elements of $\{|i|,|j|,|d|,|l|,|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|\}$ are at its maximum. It can be seen by [25] that such situation is similar to those mentioned above, therefore omit the proof. Thus the equation (6.3) has no solution in ${{\mathbb {Z}}}^{2,*}$. That is ${\mathcal {I}}$ meets the property (5) in Definition 1.1. Similarly, ${\mathcal {I}}$ meets the property (3) in Definition 1.1 and ${\mathcal {N}}_2\bigcap {\mathcal {N}}_4=\emptyset .$

(III) Let’s prove the eq (6.4) has no solution in ${{\mathbb {Z}}}^{2,*}$. The proof for (6.2), (6.5), (6.7)–(6.10), is similar and simpler to the proof for (6.4).

According to the calculation, the eq (6.4) is equivalent to

$$\begin{aligned} \left\{ \begin{array}{l}<l-r,i-j+d-r>=<i-j,j-d>,\\<r,{\widetilde{i}}+{\widetilde{d}}+{\widetilde{l}}-{\widetilde{j}}-i-d-l+j> \\ =<i-j,j-d>-<{\widetilde{i}}-{\widetilde{j}},{\widetilde{j}}-{\widetilde{d}}>+<{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}},{\widetilde{l}}>\\ \quad -<i-j+d,l>. \end{array} \right. \nonumber \\ \end{aligned}$$

(6.11)

We assert |j| and $|{\widetilde{j}}|$ is not at its maximum. Assuming $|j|=\mathrm{{max}}\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|, |j|,|d|,|l|\}$. From ${|l|}^2+{|d|}^2+{|i|}^2={|r|}^2+{|s|}^2+{|j|}^2\ge {|j|}^2$ and the definition of ${\mathcal {I}}$, then there exists one element of the set $\{l,d,i\}$ that is identical to j. If $i=j$, then $r+s-d-l=0$ and ${|r|}^2+{|s|}^2-{|d|}^2-{|l|}^2=0$. That is $r\in {\mathcal {N}}_2$. This is contradictory to $r\in {\mathcal {N}}_4$. That is $|j|\ne \mathrm{{max}}\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$. Similarly, we have $|{\widetilde{j}}|\ne \mathrm{{max}}\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$.

Case 3.1. Only one element of $\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$ is at its maximum. Assuming $|d|=\mathrm{{max}}\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$, then

$$\begin{aligned} {|r|}^2={|{\widetilde{i}}|}^2-{|{\widetilde{j}}|}^2+{|{\widetilde{d}}|}^2+{|{\widetilde{l}}|}^2-{|{\widetilde{s}}|}^2\ll d_1. \end{aligned}$$

According to the calculation to (6.11),

$$\begin{aligned} r_2=i_2-j_2+l_2+\displaystyle \frac{\alpha _{31}}{\beta _{31}}, \end{aligned}$$

where

$$\begin{aligned} \alpha _{31}= & {} \left( i_1-j_1+l_1-r_1\right) d_1+<j-i,j-l>-<{\widetilde{j}}-{\widetilde{d}},{\widetilde{j}}-{\widetilde{l}}>\\{} & {} +<{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}-{\widetilde{l}}>- r_1\left( i_1-j_1+l_1-{\widetilde{i}}_1+{\widetilde{j}}_1-{\widetilde{d}}_1-{\widetilde{l}}_1\right) \\{} & {} +(i_2+l_2-j_2)(-i_2+j_2-l_2+{\widetilde{i}}_2+{\widetilde{d}}_2+{\widetilde{l}}_2-{\widetilde{j}}_2),\\ \beta _{31}= & {} d_2+(i_2+l_2-j_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2-{\widetilde{l}}_2). \end{aligned}$$

To prove $\alpha _{31}\ne 0$ by contradiction. Assuming $\alpha _{31}= 0$, then

$$\begin{aligned} i_1-j_1+l_1-r_1=0,\quad r_2=i_2-j_2+l_2. \end{aligned}$$

That is $i-j+l-r=0.$ Thus we have $d=s$ by $i-j+d+l-r-s=0.$ This is contradictory to $d\in {\mathcal {I}}$ and $s\in {{\mathbb {Z}}}^{2,s}$. That is $\alpha _{31}\ne 0$. Due to the order of the numerator $\alpha _{31}$ does not exceed $d_1$ and the order of the divisor $\beta _{31}$ is $d_2$, we have $r_2\not \in {{\mathbb {Z}}}$.

Other situations are similar to the above cases.

Case 3.2. Two of $\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$ are at its maximum.

Case 3.2.1. Assuming $|l|=|{\widetilde{l}}|=\mathrm{{max}}\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$, then $l={\widetilde{l}}$. According to the eq (6.11), then

$$\begin{aligned} \alpha _{321}r_1^2+\beta _{321}r_1+\gamma _{321}=0, \end{aligned}$$

where

$$\begin{aligned} \alpha _{321}= & {} \left| i+d-j+{\widetilde{j}}-{\widetilde{i}}-{\widetilde{d}}\right| ^2, \\ \beta _{321}= & {} 2\left( i_1+d_1-j_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) \left( -<j-i,j-d>+<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}> \right) \\ {}{} & {} -2\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right)<l,i-j+d+{\widetilde{j}}-{\widetilde{i}}-{\widetilde{d}}> \\ {}{} & {} +\left( i_1-j_1+d_1+l_1\right) \left( i_2-j_2+d_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \left( {\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2+{l}_2\right) \\ {}{} & {} -\left( {\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1+{l}_1\right) \left( i_2-j_2+d_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \left( i_2-j_2+d_2+l_2\right) , \\ \gamma _{321}= & {} \left( -<j-i,j-d>+<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>-<l,i-j+d+{\widetilde{j}}-{\widetilde{i}}\right. \\ {}{} & {} \left. -{\widetilde{d}}> \right) ^2 +\left( i_2-j_2+d_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \left( i_2-j_2+d_2+l_2\right) \left(<{\widetilde{j}}\right. \\ {}{} & {} \left. -{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>+<l,{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}>\right) -\left( i_2-j_2+d_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \\ {}{} & {} \left( {\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2+l_2\right) \left(<j-i,j-d>+<l,{i}-{j}+{d}>\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \Delta= & {} \beta _{321}^2-4\alpha _{321}\gamma _{321} \\ {}= & {} \delta _{321}^2\cdot \left[ l_2+\displaystyle \frac{\sigma _{321}}{\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) ^2} \right] ^2 \\ {}{} & {} +4\left( i_2-j_2+d_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) ^2\cdot \displaystyle \frac{\left| i-j+d+{\widetilde{j}}-{\widetilde{i}}-{\widetilde{d}}\right| ^2}{\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) }\cdot \mu _{321}, \end{aligned}$$

where

$$\begin{aligned} \delta _{321}= & {} \left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) \left( i_2-j_2+d_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) ,\\{\sigma }_{321}= & {} -\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) \left( i_2-j_2+d_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) l_1 \\ {}{} & {} +2\left( i_2-j_2+d_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \left(<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>-<j-i,j-d>\right) \\ {}{} & {} +\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) \left( {\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1\right) \left( i_2-j_2+d_2\right) \\ {}{} & {} -\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) \left( i_1-j_1+d_1\right) \left( {\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2\right) , \\{\mu }_{321}= & {} \left(<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>-<j-i,j-d>\right) l_1 \\ {}{} & {} +<j-i,j-d>\left( {\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1\right) -<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>\left( {i}_1-{j}_1+{d}_1\right) \\ {}{} & {} -\displaystyle \frac{\left(<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>-<j-i,j-d>\right) ^2}{\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) }. \end{aligned}$$

Due to $\mu _{321}$ is of order $l_1$ which is far less than $l_2$, we have

$$\begin{aligned} \Delta= & {} \delta _{321}^2\cdot \left[ l_2+\displaystyle \frac{\sigma _{321}}{\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) ^2} -\omega _{321}\right] ^2, \end{aligned}$$

where

$$\begin{aligned}\omega _{321}\sim \displaystyle \frac{l_1}{l_2}\ll \displaystyle \frac{1}{\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) ^2}.\end{aligned}$$

Therefore

$$\begin{aligned} r_1= & {} \displaystyle \frac{-\beta _{321}\pm \sqrt{\Delta }}{2\alpha _{321}} \\ {}= & {} \displaystyle \frac{-\beta _{321}}{2\alpha _{321}}\pm \displaystyle \frac{|\delta _{321}|l_2 +\displaystyle \frac{|\delta _{321}|\sigma _{321}}{\left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) ^2}-|\delta _{321}|\omega _{321}}{2\alpha _{321}}. \end{aligned}$$

In view of $0<\left| \displaystyle \frac{|\delta _{321}|\omega _{321}}{2\alpha _{321}}\right| \ll \displaystyle \frac{1}{2\alpha _{321} \left( i_1-j_1+d_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) ^2}$, we have $r_1\not \in {{\mathbb {Z}}}$.

Other situations are similar to the above cases.

Case 3.3. Three elements of $\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$ are at its maximum. It can be seen by [25] that such situation is similar to those mentioned above, So omit the proof.

Case 3.4. Four elements of $\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$ are at its maximum.

Case 3.4.1. Assuming $|i|=|d|=|l|=|{\widetilde{l}}|=\mathrm{{max}}\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$, then $i=d=l={\widetilde{l}}$. According to the calculation to (6.11),

$$\begin{aligned}\alpha _{341}r_1^2+\beta _{341}r_1+\gamma _{341}=0, \end{aligned}$$

where

$$\begin{aligned} \alpha _{341}= & {} \left| 2l-j+{\widetilde{j}}-{\widetilde{i}}-{\widetilde{d}}\right| ^2, \\ \beta _{341}= & {} 2\left( 2l_1-j_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right) \left( -\left| j-l\right| ^2+<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>\right) \\ {}{} & {} +2\left( 2l_1-j_1+{\widetilde{j}}_1-{\widetilde{i}}_1-{\widetilde{d}}_1\right)<l,-2l+j+{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}> \\ {}{} & {} +\left( 3l_1-j_1\right) \left( 2l_2-j_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \left( {l}_2+{\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2\right) \\ {}{} & {} -\left( {l}_1+{\widetilde{i}}_1-{\widetilde{j}}_1+{\widetilde{d}}_1\right) \left( 2l_2-j_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \left( 3l_2-j_2\right) , \\ \gamma _{341}= & {} -\left| j-l\right| ^2+\left(<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}> +<l,-2l+j+{\widetilde{i}}-{\widetilde{j}}+{\widetilde{d}}> \right) ^2 \\ {}{} & {} +\left( 2l_2-j_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \left( 3l_2-j_2\right) \cdot \left(<{\widetilde{j}}-{\widetilde{i}},{\widetilde{j}}-{\widetilde{d}}>+<l,{\widetilde{i}}-{\widetilde{j}}\right. \\ {}{} & {} \left. +{\widetilde{d}}>\right) -\left( 2l_2-j_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) \left( {l}_2+{\widetilde{i}}_2-{\widetilde{j}}_2+{\widetilde{d}}_2\right) \cdot \\ {}{} & {} \times \left( |j-l|^2+<l,2l-j>\right) . \end{aligned}$$

The equation (6.11) has no solution in ${{\mathbb {Z}}}^{2,*}$ because

$$\begin{aligned} \Delta =\beta _{341}^2-4\alpha _{341}\gamma _{341} =\left( 2l_2-j_2+{\widetilde{j}}_2-{\widetilde{i}}_2-{\widetilde{d}}_2\right) ^2\cdot l_2^2\cdot \left( -8l_2^2+\delta _{341}\right) <0, \end{aligned}$$

where the order of ${\delta }_{341}$ does not exceed $l_2$.

Other situations are similar to the above cases.

Case 3.5. Five elements of $\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$ are at its maximum. It can be seen by [25] that such situation is similar to those mentioned above, so omit the proof.

Case 3.6. Six elements of $\{|{\widetilde{i}}|,|{\widetilde{j}}|,|{\widetilde{d}}|,|{\widetilde{l}}|,|i|,|j|,|d|,|l|\}$ are at its maximum. It can be seen by [25] that such situation is similar to those mentioned above, so omit the proof.

So the equation (6.4) has no solution in ${{\mathbb {Z}}}^{2,*}$. That is ${\mathcal {I}}$ meets the property (6) in Definition 1.1. Similarly, ${\mathcal {I}}$ meets the properties (4) and (7) in Definition 1.1. $\square$