Periodic solutions to a coupled two-dimensional lattice presented by Blaszak and Szum with Riemann–theta function

Abstract

A coupled two-dimensional lattice presented by Blaszak and Szum is studied with the aid of Riemann–theta function and the bilinear method. By utilizing a bilinear form of the equation, we have obtained one-periodic and two-periodic solutions. In order to analyze the solution, we study asymptotic behavior and draw the solution plots.

1 Introduction

The subject of the discrete system has been attracting interest [1–3]. Especially, Toda lattice equations have been discussed by many researchers. In [4], Dai was dedicated to the study of integrable variable-coefficient Toda lattice by using the dressing method. Nakamura in [5] discussed the $3+1$-dimensional Toda equation and derived the solutions by using the Bessel functions. The authors in [6] obtained the solutions of $2+1$-dimensional Toda lattice by using Darboux transformation. Tian and Hu discussed semi-discrete KP and BKP equations by utilizing nonlocal symmetries in [7]. By using the Hirota bilinear method with the help of Riemann–theta function, Nakamura [8, 9] studied some famous equations such as KdV, Boussinesq, Toda, etc., and Dai et al. demonstrated for KP equation and Toda lattice [10, 11]. Recently, a lot of researchers have been concerned with the method [12–15]. However, the coupled discrete system and high-dimensional equations have less been studied in the previous literature.

In this paper, we consider the two-dimensional lattice presented by Blaszak and Szum in [16]:

$$ \begin{gathered} u_{t}(n)=u(n) \bigl(w(n)-w(n-1)\bigr), \\ v_{t}(n)=u(n+1)-u(n)+w_{y}(n), \\ w_{t}(n+1)+w_{t}(n)=v(n+1)-v(n)-w^{2}(n+1)+w^{2}(n), \end{gathered} $$

(1.1)

which is a coupled discrete system. Tam and Hu in [17] discussed its bilinear forms and its solutions. Yu et al. [18] derived its pfaffianization and molecule solutions. We will obtain one-periodic solution and two-periodic solution by utilizing the bilinear method and the Riemann–theta function presented in [8–10].

The paper is organized as follows. In Sect. 2, we obtain one-periodic wave solution and study its asymptotic behavior. The solution is also studied graphically. In Sect. 3, we obtain two-periodic wave solutions whose asymptotic behaviors are studied and the plots are given.

2 One-periodic solution and its asymptotic behavior

In the section, we study one-periodic solution of (1.1). Through the transformation [17],

$$ \begin{gathered} u(n)=\frac{f(n+1)f(n-1)}{f^{2}(n)},\qquad v(n)=\frac{D_{t}^{2}f(n)\cdot f(n+1)}{f(n) f(n+1)},\\ w(n)= \biggl(\ln\frac{f(n+1)}{f(n)}\biggr)_{t}, \end{gathered} $$

(2.1)

then (1.1) can be written as

$$\begin{aligned}& \bigl(D_{z}e^{\frac{1}{2}D_{n}}-D_{t}^{2}e^{\frac {1}{2}D_{n}}+c_{1} \bigr)f(n)\cdot f(n)=0, \end{aligned}$$

(2.2)

$$\begin{aligned}& (D_{t}D_{z}-D_{t}D_{y}-2\cosh D_{n}+2+c_{2})f(n)\cdot f(n)=0, \end{aligned}$$

(2.3)

where z is an auxiliary variable, $c_{1}$ and $c_{2}$ are integration constants. The Hirota bilinear differential operator is defined as [19]

$$ D^{m}_{x}D^{n}_{y}a(x,y)\cdot b(x,y) \equiv(\partial_{x}-\partial_{x'})^{m}( \partial_{y}-\partial _{y'})^{n}a(x,y)\times b \bigl(x',y'\bigr)\big|x'=x,y'=y $$

and the difference operator is defined as

$$\begin{gathered} e^{D_{n}}a_{n}\cdot b_{n}=a_{n+1}b_{n-1}, \qquad e^{-D_{n}}a_{n}\cdot b_{n}=a_{n-1}b_{n+1}, \\ \cosh D_{n} a_{n}\cdot b_{n}=\frac{1}{2}(a_{n+1}b_{n-1}+a_{n-1}b_{n+1}).\end{gathered} $$

From the definition of Hirota bilinear operator, we have

$$ D^{m}_{x}D^{l}_{y} e^{\zeta_{1}} \cdot e^{\zeta_{2}}=(l_{1}-l_{2})^{m}( \rho_{1}-\rho_{2})^{l} e^{\zeta_{1}+\zeta_{2}}, $$

where $\zeta_{j}=l_{j}x+\rho_{j}y+\eta_{j}n+\zeta_{j0}$ ($j=1,2$). Moreover, it is easy to deduce

$$\begin{aligned}& \cosh D_{n} e^{\zeta_{1}}\cdot e^{\zeta_{2}}=\cosh( \eta_{1}-\eta_{2})e^{\zeta _{1}+\zeta_{2}}, \end{aligned}$$

(2.4)

$$\begin{aligned}& G(D_{x},D_{y},\cosh D_{n})e^{\zeta_{1}} \cdot e^{\zeta_{2}}=G(l_{1}-l_{2},\rho_{1}- \rho_{2},\eta_{1}-\eta_{2})e^{\zeta_{1}+\zeta_{2}}. \end{aligned}$$

(2.5)

2.1 One-periodic wave solution

In view of [8, 9], we consider the Riemann–theta function solution of the bilinear form (2.2) and (2.3)

$$ f=\sum_{k\in Z^{N}}e^{\pi i\langle\tau k,k\rangle+2\pi i\langle \zeta,k\rangle}, $$

(2.6)

where $\langle\cdot, \cdot\rangle$ is the inner product, $k=(k_{1},\ldots,k_{N})^{T}$, $\zeta=(\zeta_{1},\ldots,\zeta_{N})^{T}$ and τ is a symmetric matrix, $\zeta_{j}=p_{j}t+l_{j}y+\mu_{j}z+\eta_{j}n+\zeta_{0j}$ ($j=1,\ldots,N$). In order to obtain one-periodic wave solution, we consider the case for $N=1$, and we denote $k=k_{1}$, $\zeta=\zeta_{1}$, $\zeta_{0}=\zeta _{01}$. The direct calculations show that $\pi i\langle\tau k,k\rangle=\pi i k^{2}\tau$, $2\pi i\langle\zeta,k\rangle=2\pi i k\zeta$. Thus (2.6) becomes

$$ f=\sum_{k=-\infty}^{\infty}e^{2\pi i k\zeta+\pi ik^{2}\tau}. $$

(2.7)

Inserting (2.7) into (2.2) and using the bilinear properties, we have

$$ \begin{gathered} F_{1}\bigl(D_{z},D_{t},e^{\frac{1}{2}D_{n}} \bigr)f(n)\cdot f(n) \\ \quad\equiv\bigl(D_{z}e^{\frac{1}{2}D_{n}}-D_{t}^{2}e^{\frac {1}{2}D_{n}}+c_{1} \bigr)f(n)\cdot f(n) \\ \quad=\sum_{k,k'=-\infty}^{\infty}F_{1} \bigl(D_{z},D_{t},e^{\frac {1}{2}D_{n}}\bigr)\exp\bigl(2\pi i k \zeta+\pi ik^{2}\tau\bigr)\cdot\exp\bigl(2\pi i k'\zeta + \pi ik^{\prime2}\tau\bigr) \\ \quad=\sum_{k,m=-\infty}^{\infty}F_{1} \bigl(D_{z},D_{t},e^{\frac{1}{2}D_{n}}\bigr)\exp \bigl(2\pi i k \zeta+\pi ik^{2}\tau\bigr)\cdot\exp\bigl(2\pi i (m-k)\zeta+\pi i(m-k)^{2}\tau\bigr) \\ \quad=\sum_{k,m=-\infty}^{\infty}F_{1}{ \bigl(2\pi i(2k-m)\mu,2\pi i(2k-m)p,e^{\pi i(2k-m)\eta}\bigr)}\\ \qquad{}\times\exp{\bigl(2\pi im \zeta+\pi i\bigl[k^{2}+(k-m)^{2}\bigr]\tau\bigr)} \\ \quad=\sum_{m=-\infty}^{\infty}\tilde{F_{1}}(m)\exp(2\pi im\zeta)=0, \end{gathered} $$

where the new summation index $m=k+k'$ has been introduced and $\tilde{F_{1}}(m)$ is defined by

$$ \tilde{F_{1}}(m)=\sum_{k=-\infty}^{\infty}F_{1}{\bigl(2\pi i(2k-m)\mu,2\pi i(2k-m)p,e^{\pi i(2k-m)\eta}} \bigr)e^{\pi i[k^{2}+(k-m)^{2}]\tau}. $$

(2.8)

Thus,

$$\begin{aligned}& \tilde{F_{1}}(0) =\sum_{k=-\infty}^{\infty}\bigl(4\pi ik\mu e^{2\pi ik\eta}-16\pi^{2}k^{2}p^{2}e^{2\pi ik\eta} +c_{1}\bigr)e^{2\pi ik^{2}\tau}=0, \end{aligned}$$

(2.9)

$$\begin{aligned}& \begin{aligned}[b]\tilde{F_{1}}(1) &=\sum_{k=-\infty}^{\infty}\bigl[2\pi i(2k-1)\mu e^{\pi i(2k-1)\eta}-4\pi^{2} (2k-1)^{2}p^{2} e^{\pi i(2k-1)\eta}+c_{1}\bigr]e^{\pi i[k^{2}+(k-1)^{2}]\tau}\hspace{-20pt}\\ &=0.\end{aligned} \end{aligned}$$

(2.10)

We denote

$$ \begin{gathered} d_{1}=\exp2\pi ik^{2}\tau, \qquad d_{2}=\exp\pi i\bigl[k^{2}+(k-1)^{2}\bigr]\tau ,\\\Delta_{1}=\sum_{k=-\infty}^{\infty}d_{1},\qquad \Delta_{2}=\sum_{k=-\infty}^{\infty}d_{2}, \\ a_{11}=\sum_{k=-\infty}^{\infty}4\pi i k d_{1}\exp2\pi ik\eta, \qquad a_{21}=\sum _{k=-\infty}^{\infty}2\pi i (2k-1) d_{2}\exp\pi i(2k-1)\eta. \end{gathered} $$

Then (2.9) and (2.10) are written as

$$ \mu a_{11}+2a_{11,\eta}p^{2}+c_{1} \Delta_{1}=0,\qquad \mu a_{21}+2a_{21,\eta}p^{2}+c_{1} \Delta_{2}=0, $$

from which we have

$$ \mu=2p^{2}\frac{a_{21,\eta}\Delta_{1}-a_{11,\eta}\Delta _{2}}{a_{11}\Delta_{2}-a_{21}\Delta_{1}},\qquad c_{1}=2p^{2} \frac {a_{11,\eta}a_{21}-a_{21,\eta}a_{11}}{a_{11}\Delta_{2}-a_{21}\Delta_{1}}. $$

(2.11)

Similarly, substituting (2.7) into (2.3), we derive

$$ \begin{gathered} F_{2}(D_{z},D_{t},D_{y}, \cosh D_{n})f(n)\cdot f(n) \\ \quad\equiv(D_{t}D_{z}-D_{t}D_{y}-2 \cosh D_{n}+2+c_{2})f(n)\cdot f(n) \\ \quad=\sum_{k,k'=-\infty}^{\infty}F_{2}(D_{z},D_{t},D_{y},\cosh D_{n})\exp \bigl(2\pi i k\zeta+\pi ik^{2}\tau\bigr)\cdot\exp \bigl(2\pi i k'\zeta+\pi ik^{\prime2}\tau\bigr) \\ \quad=\sum_{k,m=-\infty}^{\infty}F_{2}{ \bigl(2\pi i(2k-m)\mu,2\pi i(2k-m)p,2\pi i(2k-m)l,\cosh2\pi i(2k-m)\eta\bigr)} \\ \qquad{}\times\exp{\bigl(2\pi im\zeta+\pi i\bigl[k^{2}+(k-m)^{2} \bigr]\tau\bigr)} \\ \quad=\sum_{m=-\infty}^{\infty}\tilde{F_{2}}(m)\exp(2\pi im\zeta)=0, \end{gathered} $$

where

$$ \begin{aligned} \tilde{F_{2}}(m)={}&\sum _{k=-\infty}^{\infty}F_{2}{\bigl[2\pi i(2k-m)\mu,2 \pi i(2k-m)p,2\pi i(2k-m)l,\cosh2\pi i(2k-m)\eta\bigr]} \\ &\times\exp{\pi i\bigl[k^{2}+(k-m)^{2}\bigr]\tau}. \end{aligned} $$

It is easy to know that if $\tilde{F_{2}}(0)=0$, $\tilde{F_{2}}(1)=0$, then all $\tilde{F_{2}}(m)=0$ are proved.

$$ \begin{gathered} \tilde{F_{2}}(0)=\sum _{k=-\infty}^{\infty}\bigl(-16\pi^{2}k^{2}p \mu+16\pi ^{2}k^{2}pl-2\cosh4\pi ik\eta+2+c_{2} \bigr)e^{2\pi ik^{2}\tau}=0, \\ \tilde{F_{2}}(1)=\sum_{k=-\infty}^{\infty}\bigl(4\pi^{2}(2k-1)^{2}(-p\mu +pl)-2\cosh2\pi i(2k-1) \eta+2+c_{2}\bigr)\\ \phantom{\tilde{F_{2}}(1)=}{}\times e^{\pi i[k^{2}+(k-1)^{2}]\tau}\\ \phantom{\tilde{F_{2}}(1)}=0. \end{gathered} $$

(2.12)

Letting $b_{11}=\sum_{k=-\infty}^{\infty}16\pi^{2}k^{2}d_{1}$, $b_{12}=\sum_{k=-\infty}^{\infty}\cosh(4\pi i k\eta) d_{1}$, $b_{21}=\sum_{k=-\infty}^{\infty}4\pi^{2}(2k-1)^{2}d_{2}$, $b_{22}=\sum_{k=-\infty}^{\infty}\cosh2\pi i (2k-1)\eta d_{2}$, thus, (2.12) can be written as

$$\begin{gathered} p(l-\mu)b_{11}-2b_{12}+(2+c_{2}) \Delta_{1}=0, \\ p(l-\mu)b_{21}-2b_{22}+(2+c_{2}) \Delta_{2}=0.\end{gathered} $$

Solving the above system, we have

$$ l=\mu+\frac{2}{p}\frac{b_{22}\Delta_{1}-b_{12}\Delta_{2}}{b_{21}\Delta _{1}-b_{11}\Delta_{2}},\qquad 2+c_{2}=2 \frac{b_{12}b_{21}-b_{22}b_{11}}{b_{21}\Delta_{1}-b_{11}\Delta_{2}}, $$

(2.13)

from which we find that parameter l is dependent on μ, η, and p. In view of (2.11), we can see that μ is dependent on η and p.

Then we have derived the Riemann–theta function solution $f(n)$ of (2.2) and (2.3). Furthermore, the Riemann–theta function periodic solutions of (1.1) are obtained by using transformation (2.1).

2.2 Asymptotic behavior of the one-periodic wave solution

In what follows, we will prove that the soliton solution can be regarded as the limit of the following periodic solution. Therefore, we write $q=\exp\pi i \tau$ and take a limit $q\rightarrow0$ (or $\operatorname{Im}\tau\rightarrow\infty$).

Theorem 1

Under the condition $q\rightarrow0$ (or $\operatorname{Im}\tau\rightarrow\infty$), the Riemann–theta function periodic solution (2.7) of (2.2) and (2.3) tends to the one-soliton solutions of (1.1) via (2.1).

$$ \begin{gathered} u(n)=\frac{(1+e^{\widetilde{\zeta}+\widetilde{\eta}})(1+e^{\widetilde {\zeta}-\widetilde{\eta}})}{(1+e^{\widetilde{\zeta}})^{2}}, \\ v(n)=-2\pi^{2}p^{2}e^{\frac{\widetilde{\zeta}}{2}} \operatorname{sech} \frac{\widetilde{\zeta}}{2} \frac{1+e^{\widetilde{\eta }}}{1+e^{\widetilde{\zeta}+\widetilde{\eta}}}, \\ w(n)=2\pi ip \frac{e^{\widetilde{\zeta}}(e^{\widetilde{\eta }}-1)}{(1+e^{\widetilde{\zeta}})(1+e^{\widetilde{\zeta}+\widetilde{\eta}})}, \end{gathered} $$

(2.14)

where

$$ \begin{gathered} \widetilde{\zeta}=2\pi i(pt+ly+\mu z+\eta n)+ \widetilde{\zeta _{0}},\qquad\widetilde{\eta}=2\pi i \eta,\qquad \widetilde{\zeta _{0}}=\zeta_{0}+\frac{1}{2}\tau, \\ \mu\rightarrow-2\pi p^{2}\cot\pi\eta,\qquad l\rightarrow\mu+ \frac{\cos ^{2} \pi\eta}{p\pi^{2}},\qquad c_{1}\rightarrow0,\qquad c_{2}\rightarrow0. \end{gathered} $$

Proof

Utilizing $q=\exp\pi i \tau$, the quantities defined above are then expanded in powers of q

$$\begin{aligned}& \Delta_{1}=\sum _{k=-\infty}^{\infty}e^{2\pi ik^{2}\tau }=1+2q^{2}+o \bigl(q^{2}\bigr), \qquad \Delta_{2}=\sum _{k=-\infty}^{\infty}e^{2\pi i[k^{2}+(k-1)^{2}]\tau }=2q^{2}+o \bigl(q^{2}\bigr), \\& a_{11}=\sum_{k=-\infty}^{\infty}4\pi ik e^{2\pi ik\eta} e^{2\pi ik^{2}\tau}=-8\pi\sin(2\pi\eta)q^{2}+o \bigl(q^{2}\bigr), \\& a_{11,\eta}=\sum_{k=-\infty}^{\infty}-8 \pi^{2} k^{2} e^{2\pi ik\eta} e^{2\pi ik^{2}\tau}=-16 \pi^{2} \cos(2\pi\eta)q^{2}+o\bigl(q^{2}\bigr), \\& a_{21}=\sum_{k=-\infty}^{\infty}2\pi i(2k-1) e^{\pi i(2k-1)\eta} e^{\pi i[k^{2}+(k-1)^{2}]\tau}=-4\pi\sin(\pi\eta) q+o \bigl(q^{5}\bigr), \\& a_{21,\eta}=\sum_{k=-\infty}^{\infty}-2 \pi^{2}(2k-1)^{2} e^{\pi i(2k-1)\eta} e^{\pi i[k^{2}+(k-1)^{2}]\tau}=-4 \pi^{2} \cos(\pi\eta) q+o\bigl(q^{5}\bigr), \\& b_{11}=\sum_{k=-\infty}^{\infty}16 \pi^{2}k^{2}e^{2\pi ik^{2}\tau}=32\pi ^{2}q^{2}+o \bigl(q^{8}\bigr), \\& b_{12}=\sum_{k=-\infty}^{\infty}\cosh(4 \pi i k\eta)e^{2\pi ik^{2}\tau }=1+2\cos(4\pi\eta)q^{2}+o \bigl(q^{8}\bigr), \\& b_{21}=\sum_{k=-\infty}^{\infty}4 \pi^{2}(2k-1)^{2}e^{\pi i[k^{2}+(k-1)^{2}]\tau}=8\pi^{2}q+o \bigl(q^{5}\bigr), \\& b_{22}=\sum_{k=-\infty}^{\infty}\cosh2\pi i(2k-1)\eta e^{\pi i[k^{2}+(k-1)^{2}]\tau}=2\cos(2\pi\eta) q+o\bigl(q^{5}\bigr). \end{aligned}$$

Using (2.11) and (2.13), we have $\mu\rightarrow-2\pi p^{2}\cot\pi\eta$, $l\rightarrow\mu+\frac{\cos2\pi\eta}{2p\pi^{2}}$, $c_{1}\rightarrow0$, $c_{2}\rightarrow0$ for $q\rightarrow0$.

In order to consider the convergence to the one-periodic wave solution (2.7) in the limit of $q\rightarrow0$, under the transformation $\zeta _{0}=\widetilde{\zeta_{0}}-\frac{1}{2}\tau$, we can get the following convergent forms:

$$ \begin{gathered} f(n)=1+\exp{\widetilde{\zeta}}+o \bigl(q^{2}\bigr), \\ f(n-1)=1+\exp{(\widetilde{\zeta}-\widetilde{\eta})}+o\bigl(q^{2} \bigr), \\ f(n+1)=1+\exp{(\widetilde{\zeta}+\widetilde{\eta})}+o\bigl(q^{2} \bigr), \\ f_{t}(n)=2\pi ip\exp{\widetilde{\zeta}}+o\bigl(q^{2} \bigr), \\ f_{tt}(n)=-4\pi^{2}p^{2}\exp{\widetilde{\zeta}}+o \bigl(q^{2}\bigr), \\ f_{t}(n+1)=2\pi ip\exp(\widetilde{\zeta}+\widetilde{\eta})+o \bigl(q^{2}\bigr), \\ f_{tt}(n+1)=-4\pi^{2}p^{2}\exp(\widetilde{\zeta}+ \widetilde{\eta})+o\bigl(q^{2}\bigr). \end{gathered} $$

(2.15)

After some tedious calculations, we derive (2.14). □

In what follows, Fig. 1, Fig. 2, and Fig. 3 describe the plots of $u(t,y,n)$, $v(t,y,n)$, and $w(t,y,n)$, respectively. From these figures, we find that the plots of $v(t,y,n)$ and $w(t,y,n)$ have similar forms.

Figure 1

One-periodic solution plot of $u(t,y,n)$ for $z=0.6$, $n=2$, $\eta=0.1$, $\tilde{\zeta}_{0}=2$, $l=4$, $p=0.8$, $\tau=0.8i$, $t\in[-15,15]$, $y\in[-20,20]$

Figure 2

One-periodic solution plot of $v(t,y,n)$ for $z=0.6$, $n=2$, $\eta=0.1$, $\tilde{\zeta}_{0}=2$, $l=4$, $p=0.8$, $\tau=0.8i$, $t\in[-15,15]$, $y\in[-20,20]$

Figure 3

One-periodic solution plot of $w(t,y,n)$ for $z=0.6$, $n=2$, $\eta=0.1$, $\tilde{\zeta}_{0}=2$, $l=4$, $p=0.8$, $\tau=0.8i$, $t\in[-15,15]$, $y\in[-20,20]$

3 Two-periodic wave solution and its asymptotic behavior

In what follows, similar to the one-periodic wave solution, we consider a two-periodic wave solution of the coupled two-dimensional lattice (1.1).

3.1 Construction of two-periodic wave solution

By letting $N=2$ in (2.6), we have $f(n)=\sum_{k\in Z^{2}} e^{2\pi i \langle\zeta,k\rangle+\pi i\langle\tau k,k\rangle}$ and substitute it into (2.2). For convenience of calculations, we have introduced different forms of k and $k'$. Thus, we obtain

$$\begin{aligned} F_{1}f_{n} \cdot f_{n}={}&\sum_{k,k'\in Z^{2}} F_{1} \bigl(D_{z},D_{t},e^{\frac{1}{2}D_{n}}\bigr)e^{2\pi i \langle\zeta,k\rangle+\pi i\langle\tau k,k\rangle}\cdot e^{2\pi i \langle\zeta,k'\rangle+\pi i\langle\tau k',k'\rangle} \\ ={}&\sum_{k,k'\in Z^{2}}F_{1}\bigl(2\pi i\bigl\langle k-k',\mu\bigr\rangle ,2\pi i\bigl\langle k-k',p \bigr\rangle ,e^{2\pi i\langle k-k',\eta\rangle }\bigr) \\ &\times\exp\bigl(2\pi i\bigl\langle \zeta,k+k'\bigr\rangle \bigr) \exp\pi i\bigl(\bigl\langle \tau k',k'\bigr\rangle + \langle\tau k,k\rangle\bigr) \\ ={}&\sum_{s'\in Z^{2}}\sum_{k_{1},k_{2}=-\infty}^{\infty}F_{1}{\bigl(2\pi i\bigl\langle 2k-s',\mu\bigr\rangle ,2\pi i\bigl\langle 2k-s',p\bigr\rangle ,e^{2\pi i\langle2k-s',\eta\rangle}\bigr)} \\ &\times\exp{\pi i\bigl(\bigl\langle \tau\bigl(k-s' \bigr),k-s'\bigr\rangle +\langle\tau k,k\rangle\bigr)\exp\bigl(2\pi i \bigl\langle \zeta,s'\bigr\rangle \bigr)} \\ \equiv{}&\sum_{s'\in Z^{2}} \tilde{F_{1}} \bigl(s'_{1},s'_{2}\bigr)\exp{ \bigl(2\pi i\bigl\langle \zeta,s'\bigr\rangle \bigr)}=0. \end{aligned}$$

(3.1)

By introducing the new summation index $k+k'=s'$, $s'=(s'_{1},s'_{2})^{T}$, $k=(k_{1},k_{2})^{T}$, $\tilde{F}_{1}(s'_{1},s'_{2})$ is denoted by

$$ \begin{aligned}[b] \tilde{F}_{1} \bigl(s'_{1},s'_{2}\bigr)={}&\sum _{k_{1},k_{2}=-\infty}^{\infty}F_{1}\bigl[2\pi i \bigl\langle 2k-s',\mu\bigr\rangle ,2\pi i\bigl\langle 2k-s',p \bigr\rangle ,e^{2\pi i\langle2k-s',\eta\rangle}\bigr] \\ &\times\exp\pi i\bigl(\bigl\langle \tau\bigl(k-s' \bigr),k-s'\bigr\rangle +\langle\tau k,k\rangle\bigr) \\ ={}&\sum_{k_{j}=-\infty}^{\infty}F_{1} \Biggl(2\pi i\sum_{j=1}^{2} \bigl(2k_{j}-\bigl(s'_{j}-2 \delta_{jl}\bigr)\bigr)\mu_{j},2\pi i\sum _{j=1}^{2}\bigl(2k_{j}- \bigl(s'_{j}-2\delta_{jl}\bigr) \bigr)p_{j}, \\ &{2\pi i\sum_{j=1}^{2}\bigl(2k_{j}- \bigl(s'_{j}-2\delta_{jl}\bigr)\bigr) \eta_{j}}\Biggr) \\ &\times\exp\pi i\sum_{j,l=1}^{2} \bigl[(k_{j}+\delta_{jl})\tau_{jl}(k_{j}+ \delta _{jl})\\ &+\bigl(\bigl(s_{j}'-2 \delta_{jl}-k_{j}\bigr)+\delta_{jl}\bigr) \tau_{jl}\bigl(\bigl(s_{j}'-2\delta _{jl}-k_{j}\bigr)+\delta_{jl}\bigr)\bigr] \\ ={}&\left \{ \textstyle\begin{array}{l@{\quad}l} \tilde{F}_{1}(s'_{1}-2,s'_{2})e^{2\pi i (s_{1}'-1)\tau_{11}+2\pi i s'_{2}\tau_{12}}, & l \text{ is even},\\ \tilde{F}_{1}(s'_{1},s'_{2}-2)e^{2\pi i (s_{2}'-1)\tau_{22}+2\pi i s'_{1}\tau_{12}}, & l \text{ is odd}. \end{array}\displaystyle \right . \end{aligned} $$

(3.2)

This relation implies that if $\tilde{F}_{1}(0,0)=\tilde{F}_{1}(0,1)=\tilde{F}_{1}(1,0)=\tilde{F}_{1}(1,1)=0$, then $\tilde{F}_{1}(s'_{1},s'_{2})=0$ for $s'_{1}$, $s'_{2}\in Z$.

Denoting

$$ \begin{gathered} \delta_{j}(n)=e^{\pi i\langle\tau(k-m^{(j)}),k-m^{(j)}\rangle+\pi i\langle\tau k, k\rangle}, \\ m^{(1)}=(0,0)^{T},\qquad m^{(2)}=(1,0)^{T}, \qquad m^{(3)}=(0,1)^{T},\qquad m^{(4)}=(1,1)^{T}, \end{gathered} $$

we have

$$ \begin{gathered} \tilde{F_{1}}(0,0)=\sum _{k_{1}, k_{2}=-\infty}^{\infty}\bigl[2\pi i\bigl\langle 2k-(0,0)^{T},\mu\bigr\rangle e^{\pi i\langle2k-(0,0)^{T},\eta\rangle} \\ \phantom{\tilde{F_{1}}(0,0)=}+4\pi^{2}\bigl\langle 2k-(0,0)^{T},p\bigr\rangle ^{2}e^{\pi i\langle2k-(0,0)^{T},\eta\rangle}+c_{1}\bigr]e^{2\pi i\langle \tau k,k\rangle}, \\ \tilde{F_{1}}(0,1)=\sum_{k_{1},k_{2}=-\infty}^{\infty} \bigl[2\pi i\bigl\langle 2k-(0,1)^{T},\mu\bigr\rangle e^{\pi i\langle2k-(0,1)^{T},\eta\rangle} \\ \phantom{\tilde{F_{1}}(0,1)=}+4\pi^{2}\bigl\langle 2k-(0,1)^{T},p\bigr\rangle ^{2}e^{\pi i\langle2k-(0,1)^{T},\eta\rangle}+c_{1}\bigr]e^{\pi i\langle \tau(k-(0,1)^{T}),k-(0,1)^{T}\rangle+\pi i\langle\tau k,k\rangle}, \\ \tilde{F_{1}}(1,0)=\sum_{k_{1},k_{2}=-\infty}^{\infty} \bigl[2\pi i\bigl\langle 2k-(1,0)^{T},\mu\bigr\rangle e^{\pi i\langle2k-(1,0)^{T},\eta\rangle} \\ \phantom{\tilde{F_{1}}(1,0)=}+4\pi^{2}\bigl\langle 2k-(1,0)^{T},p\bigr\rangle ^{2}e^{\pi i\langle2k-(1,0)^{T},\eta\rangle}+c_{1}\bigr]e^{\pi i\langle \tau(k-(1,0)^{T}),k-(1,0)^{T}\rangle+\pi i\langle\tau k,k\rangle}, \\ \tilde{F_{1}}(1,1)=\sum_{k_{1},k_{2}=-\infty}^{\infty} \bigl[2\pi i\bigl\langle 2k-(1,1)^{T},\mu\bigr\rangle e^{\pi i\langle2k-(1,1)^{T},\eta\rangle} \\ \phantom{\tilde{F_{1}}(1,1)=}+4\pi^{2}\bigl\langle 2k-(1,1)^{T},p\bigr\rangle ^{2}e^{\pi i\langle2k-(1,1)^{T},\eta\rangle}+c_{1}\bigr]e^{\pi i\langle \tau(k-(1,1)^{T}),k-(1,1)^{T}\rangle+\pi i\langle\tau k,k\rangle}. \end{gathered} $$

(3.3)

Denote

$$\begin{gathered} a_{j1}=\sum_{k_{1}, k_{2}=-\infty}^{\infty}2 \pi i\bigl(2k_{1}-m_{1}^{(j)} \bigr)e^{\pi i\langle k-m^{(j)},\eta\rangle}\delta_{1}, \\ a_{j2}=\sum_{k_{1},k_{2} =-\infty}^{\infty}2\pi i \bigl(2k_{2}-m_{2}^{(j)}\bigr)e^{\pi i\langle k-m^{(j)},\eta\rangle} \delta_{2}, \\ a_{j3}=\sum_{k_{1},k_{2}=-\infty}^{\infty}4\pi ^{2}\bigl(2k_{1}-m_{1}^{(j)} \bigr)^{2}e^{\pi i\langle k-m^{(j)},\eta\rangle}\delta_{3}, \\ a_{j4}=\sum_{k_{1},k_{2}=-\infty}^{\infty}4\pi ^{2}\bigl(2k_{2}-m_{2}^{(j)} \bigr)^{2}e^{\pi i\langle k-m^{(j)},\eta\rangle}\delta_{4}, \\ b_{j}=-c_{1}\sum_{k_{1},k_{2}=-\infty}^{\infty} \delta_{j}-8\pi ^{2}\bigl(2k_{1}-m_{1}^{(j)} \bigr) \bigl(2k_{2}-m_{2}^{(j)}\bigr)p_{1}p_{2} e^{\pi i\langle k-m_{1}^{(j)},\eta\rangle}\delta_{j}, \end{gathered} $$

then (3.3) can be written as

$$ A\left ( \textstyle\begin{array}{c}\mu_{1}\\ \mu_{2}\\ p^{2}_{1}\\ p^{2}_{2} \end{array}\displaystyle \right )=\vec{b}, $$

from which we have $\mu_{1}=\frac{\triangle_{1}}{\triangle}$, $\mu_{2}=\frac{\triangle _{2}}{\triangle}$, $p^{2}_{1}=\frac{\triangle_{3}}{\triangle}$, $p^{2}_{2}=\frac{\triangle _{4}}{\triangle}$, where $\triangle=\det A$ and $\triangle_{1}$, $\triangle_{2}$, $\triangle _{3}$, $\triangle_{4}$ are given △ by replacing 1st, 2nd, 3rd, 4th columns with b⃗, respectively.

Similarly, by letting $N=2$ in (2.6), then we have $f(n)=\sum_{k\in Z^{2}} e^{2\pi i \langle\zeta,k\rangle+\pi i\langle\tau k,k\rangle}$ and substitute it into (2.3). For convenience of calculations, we have introduced k and $k'$ of different form. We have derived

$$ \begin{aligned}[b] F_{2}(f_{n}\cdot f_{n})={}&\sum_{k,k'\in Z^{2}} F_{2}(D_{z},D_{t},D_{y}, \cosh D_{n})e^{2\pi i \langle\zeta,k\rangle+\pi i\langle\tau k,k\rangle}\cdot e^{2\pi i \langle\zeta,k'\rangle+\pi i\langle\tau k',k'\rangle} \\ ={}&\sum_{k,k'\in Z^{2}}F_{2}\bigl(2\pi i\bigl\langle k-k',\mu\bigr\rangle ,2\pi i\bigl\langle k-k',p \bigr\rangle ,2\pi i\bigl\langle k-k',l\bigr\rangle ,\cosh2\pi i\bigl\langle k-k',\eta\bigr\rangle \bigr) \\ &\times\exp\bigl(2\pi i\bigl\langle \zeta,k+k'\bigr\rangle \bigr) \exp(\pi i\bigl(\bigl\langle \tau k',k'\bigr\rangle + \langle\tau k,k\rangle\bigr) \\ ={}&\sum_{s'\in Z^{2}}\sum_{k_{1},k_{2}=-\infty}^{\infty}F_{2}\bigl(2\pi i\bigl\langle 2k-s',\mu\bigr\rangle ,2\pi i \bigl\langle 2k-s',p\bigr\rangle ,2\pi i\bigl\langle 2k-s',l \bigr\rangle ,\\ &\cosh2\pi i\bigl\langle 2k-s',\eta\bigr\rangle \bigr) \\ &\times\exp{\pi i\bigl(\bigl\langle \eta\bigl(k-s' \bigr),k-s'\bigr\rangle +\langle\tau k,k\rangle\bigr)\exp\bigl(2\pi i \bigl\langle \zeta,s'\bigr\rangle \bigr)} \\ \equiv{}&\sum_{s'\in Z^{2}} \tilde{F_{2}} \bigl(s'_{1},s'_{2}\bigr)\exp{ \bigl(2\pi i\bigl\langle \zeta,s'\bigr\rangle \bigr)}=0. \end{aligned} $$

(3.4)

By introducing the new summation index $k+k'=s'$, $k=(k_{1},k_{2})^{T}$, $\tilde{F}_{2}(s'_{1},s'_{2})$ is denoted by

$$ \begin{aligned}[b] \tilde{F}_{2} \bigl(s'_{1},s'_{2}\bigr)={}&\sum _{k_{1},k_{2}=-\infty}^{\infty}F_{2}\bigl[2\pi i \bigl\langle 2k-s',\mu\bigr\rangle ,2\pi i\bigl\langle 2k-s',p \bigr\rangle ,2\pi i\bigl\langle 2k-s',l\bigr\rangle ,\cosh2\pi i\bigl\langle 2k-s',\eta\bigr\rangle \bigr] \\ & \times\exp\pi i\bigl(\bigl\langle \tau\bigl(k-s'\bigr),\bigl(k-s' \bigr)\bigr\rangle +\langle\tau k,k\rangle\bigr) \\ ={}&\left \{ \textstyle\begin{array}{l@{\quad}l} \tilde{F}_{2}(s'_{1}-2,s'_{2})e^{2\pi i (s_{1}'-1)\tau_{11}+2\pi i s'_{2}\tau_{12}}, & l \text{ is even},\\ \tilde{F}_{2}(s'_{1},s'_{2}-2)e^{2\pi i (s_{2}'-1)\tau_{22}+2\pi i s'_{1}\tau_{12}}, & l \text{ is odd}, \end{array}\displaystyle \right . \end{aligned} $$

(3.5)

which means that if $\tilde{F}_{2}(m^{(j)})=0$, thus all $\tilde {F}_{2}(s'_{1},s'_{2})=0$. Through direct calculations, we have derived

$$ \begin{aligned}[b] \tilde{F}_{2} \bigl(m^{(j)}\bigr)={}&\sum_{k_{1},k_{2}=-\infty}^{\infty}\bigl[-4\pi ^{2}\bigl\langle 2k-m^{(j)},p\bigr\rangle \bigl\langle 2k-m^{(j)},\mu\bigr\rangle +4\pi ^{2}\bigl\langle 2k-m^{(j)},p\bigr\rangle \bigl\langle 2k-m^{(j)},l\bigr\rangle \\ & -2\cosh2\pi i\bigl\langle 2k-m^{(j)},\eta\bigr\rangle +2+c_{2}\bigr]\\ &\times e^{\pi i\langle \tau k,k\rangle+\pi i\langle\tau(k-m^{(j)}),k-m^{(j)}\rangle }. \end{aligned} $$

(3.6)

Let

$$\begin{aligned}& c_{j1}=\sum _{k_{1},k_{2}=-\infty}^{\infty}4\pi ^{2}\bigl(2k_{1}-m_{1}^{(j)} \bigr)^{2}\delta_{j}(n), \\ & c_{j2}=\sum_{k_{1},k_{2}=-\infty}^{\infty}4\pi ^{2}\bigl(2k_{2}-m_{2}^{(j)} \bigr)^{2}\delta_{j}(n), \\ & c_{j3}=\sum_{k_{1},k_{2}=-\infty}^{\infty}4\pi ^{2}\bigl(2k_{1}-m_{1}^{(j)}\bigr) \bigl(2k_{2}-m_{2}^{(j)}\bigr)\delta_{j}(n), \\& c_{j4}=\sum_{k_{1},k_{2}=-\infty}^{\infty}\delta_{j}(n), \\& d_{j}=\sum_{k_{1},k_{2}=-\infty}^{\infty}4\cosh2 \pi i\bigl\langle 2k-m^{(j)},\eta\bigr\rangle \delta_{j}(n), \end{aligned}$$

(3.7)

then $\tilde{F}_{2}(m^{(j)})=0$ can be rewritten as

$$ C\left ( \textstyle\begin{array}{c}p_{1}(l_{1}-\mu_{1})\\ p_{2}(l_{2}-\mu_{2})\\ p_{1}(l_{2}-\mu_{2})+p_{2}(l_{1}-\mu_{1})\\ 2+c_{2} \end{array}\displaystyle \right )=\vec{d}, $$

from which we have $p_{1}(l_{1}-\mu_{1})=\frac{\triangle_{1}}{\triangle}$, $p_{2}(l_{2}-\mu _{2})=\frac{\triangle_{2}}{\triangle}$, $p_{1}(l_{2}-\mu_{2})+p_{2}(l_{1}-\mu_{1})=\frac{\triangle_{3}}{\triangle }$, $2+c_{2}=\frac{\triangle_{4}}{\triangle}$, where $\triangle=\det C$ and $\triangle_{1}$, $\triangle_{2}$, $\triangle _{3}$, $\triangle_{4}$ are given △ by replacing 1st, 2nd, 3rd, 4th columns with d⃗.

3.2 Asymptotic behavior of the two-periodic wave solution

In what follows, we can verify the asymptotic behavior of the two-periodic wave solution to be the well-known two-soliton solution given by the Hirota method.

Theorem 2

Let $\lambda_{1}=\exp\tau_{11}\rightarrow0$, $\lambda_{2}=\exp\tau _{22}\rightarrow0$, the periodic solution (2.1) of (1.1) tends to the two-soliton solution

$$ \begin{gathered} u(n)=\frac{(1+e^{\tilde{\zeta}_{1}+\tilde{\eta}_{1}}+e^{\tilde{\zeta }_{2}+\tilde{\eta}_{2}} +e^{\tilde{\zeta}_{1}+\tilde{\eta}_{1}+\tilde{\zeta}_{2}+\tilde{\eta }_{2}+2\pi i \tau_{12}})(1+e^{\tilde{\zeta}_{1}-\tilde{\eta }_{1}}+e^{\tilde{\zeta}_{2}-\tilde{\eta}_{2}} +e^{\tilde{\zeta}_{1}-\tilde{\eta}_{1}+\tilde{\zeta}_{2}-\tilde{\eta }_{2}+2\pi i \tau_{12}})}{(1+\exp\tilde{\zeta}_{1}+\exp\tilde{\zeta }_{2}+\exp(\tilde{\zeta}_{1}+\tilde{\zeta}_{2}+2\pi i\tau_{12}))^{2}}, \\ v(n)=\frac {f_{tt}(n)f(n+1)-2f_{t}(n)f_{t}(n+1)+f(n)f_{tt}(n+1)}{f(n)f(n+1)}, \\ w(n)=\frac{f_{t}(n+1)f(n)-f_{t}(n)f(n+1)}{f(n)f(n+1)}, \end{gathered} $$

(3.8)

with

$$\begin{aligned}& p_{1}=\frac{1-\cosh\tilde{\eta}_{1}}{2\pi^{2}(l_{1}-\mu_{1})},\qquad p_{2}=\frac{1-\cosh\tilde{\eta}_{2}}{2\pi^{2}(l_{2}-\mu_{2})},\\& \tilde{\zeta}_{i}=2\pi i(p_{i}t+l_{i}y+\mu_{i}z+ \eta_{i}n)+\tilde{\zeta _{0i}},\qquad\tilde{ \zeta_{0i}}=\zeta_{0i}+\frac{1}{2} \tau_{ii}, \qquad \tilde{\eta_{i}}=2\pi i \eta_{i},\quad i=1,2, \\& e^{2\pi i\tau_{12}}=-\frac{2\pi^{2}(p_{1}-p_{2})(l_{1}-\mu_{1}+\mu _{2}-l_{2})-\cosh (\tilde{\eta}_{1}-\tilde{\eta}_{2})+1}{2\pi ^{2}(p_{1}+p_{2})(l_{1}-\mu_{1}+l_{2}-\mu_{2})-\cosh(\tilde{\eta }_{1}+\tilde{\eta}_{2})+1}, \\& 2\pi p_{2}^{2}\cos\pi\eta_{2}-\mu_{2} \sin\pi\eta_{2}=0,\qquad 2\pi p_{1}^{2}\cos\pi \eta_{1}-\mu_{1}\sin\pi\eta_{1}=0, \\& \frac{2\pi(p_{1}-p_{2})^{2}\cos\pi(\eta_{1}-\eta_{2})-(\mu_{1}-\mu _{2})\sin\pi(\eta_{1}-\eta_{2})}{ 2\pi(p_{1}+p_{2})^{2}\cos\pi(\eta_{1}+\eta_{2})-(\mu_{1}+\mu_{2})\sin \pi(\eta_{1}+\eta_{2})} \\& \quad=\frac{2\pi^{2}(p_{1}-p_{2})(l_{1}-l_{2}-\mu_{1}+\mu_{2})-\cos2\pi(\eta _{1}-\eta_{2})+1}{ 2\pi^{2}(p_{1}+p_{2})(l_{1}+l_{2}-\mu_{1}-\mu_{2})-\cos2\pi(\eta _{1}+\eta_{2})+1}, \\& f(n)\rightarrow1+\exp\tilde{\zeta}_{1}+ \exp\tilde{\zeta}_{2}+\exp(\tilde {\zeta}_{1}+\tilde{ \zeta}_{2}+2\pi i\tau_{12}), \\& f_{t}(n)\rightarrow2\pi ip_{1}e^{\tilde{\zeta_{1}}}+2\pi ip_{2}e^{\tilde {\zeta_{2}}}+2\pi i(p_{1}+p_{2})e^{\tilde{\zeta_{1}}+\tilde{\zeta _{2}}+2\pi i\tau_{12}}, \\& f_{tt}(n)\rightarrow-4\pi^{2} p^{2}_{1}e^{\tilde{\zeta_{1}}}-4 \pi^{2} p^{2}_{2}e^{\tilde{\zeta_{2}}}-4\pi^{2} (p_{1}+p_{2})^{2}e^{\tilde{\zeta _{1}}+\tilde{\zeta_{2}}+2\pi i\tau_{12}}, \\& f_{t}(n+1)\rightarrow2\pi ip_{1}e^{\tilde{\zeta_{1}}+\tilde{\eta _{1}}}+2\pi ip_{2}e^{\tilde{\zeta_{2}}+\tilde{\eta_{1}}}+2\pi i(p_{1}+p_{2})e^{\tilde{\zeta_{1}}+\tilde{\zeta_{2}}+\tilde{\eta _{1}}+\tilde{\eta_{2}}+2\pi i\tau_{12}}, \\& f_{tt}(n+1)\rightarrow-4\pi^{2} p^{2}_{1}e^{\tilde{\zeta_{1}}+\tilde {\eta_{1}}}-4 \pi^{2} p^{2}_{2}e^{\tilde{\zeta_{2}}+\tilde{\eta _{2}}}-4\pi^{2} (p_{1}+p_{2})^{2}e^{\tilde{\zeta_{1}}+\tilde{\zeta _{2}}+\tilde{\eta_{1}}+\tilde{\eta_{2}}+2\pi i\tau_{12}}. \end{aligned}$$

Proof

Let $\tilde{\zeta_{i}}=2\pi i\zeta_{i}+\pi i \frac{\tau_{ii}}{2}$, $\tilde{\eta_{i}}=2\pi i \eta_{i}$ for $i=1,2$. We expand the two-periodic wave solution (2.6) ($N=2$) of (2.2) and (2.3):

$$ \begin{aligned}[b] f(n)={}&1+\exp(2\pi i \zeta_{1}+\pi i\tau_{11})+\exp(-2\pi i\zeta_{1}+ \pi i\tau_{11})+\exp(2\pi i\zeta_{2}+\pi i \tau_{22}) \\ &+\exp(-2\pi i\zeta_{2}+\pi i\tau_{22})+\exp\bigl(2\pi i( \zeta_{1}+\zeta_{2})+\pi i(\tau_{11}+2 \tau_{12}+\tau_{22})\bigr) \\ &+\exp\bigl(-2\pi i(\zeta_{1}+\zeta_{2})+\pi i( \tau_{11}+2\tau_{12}+\tau_{22})\bigr)+\cdots \\ \rightarrow{}& 1+\exp\tilde{\zeta}_{1}+\exp\tilde{ \zeta}_{2}+\exp(\tilde {\zeta}_{1}+\tilde{ \zeta}_{2}+2\pi i\tau_{12}), \end{aligned} $$

(3.9)

then we have

$$ \begin{gathered} f_{t}(n)\rightarrow2\pi ip_{1}e^{\tilde{\zeta_{1}}}+2\pi ip_{2}e^{\tilde{\zeta_{2}}}+2\pi i(p_{1}+p_{2})e^{\tilde{\zeta _{1}}+\tilde{\zeta_{2}}+2\pi i\tau_{12}}, \\ f_{tt}(n)\rightarrow-4\pi^{2} p^{2}_{1}e^{\tilde{\zeta_{1}}}-4 \pi^{2} p^{2}_{2}e^{\tilde{\zeta_{2}}}-4\pi^{2} (p_{1}+p_{2})^{2}e^{\tilde{\zeta _{1}}+\tilde{\zeta_{2}}+2\pi i\tau_{12}}, \\ f_{t}(n+1)\rightarrow2\pi ip_{1}e^{\tilde{\zeta_{1}}+\tilde{\eta _{1}}}+2\pi ip_{2}e^{\tilde{\zeta_{2}}+\tilde{\eta_{1}}}+2\pi i(p_{1}+p_{2})e^{\tilde{\zeta_{1}}+\tilde{\zeta_{2}}+\tilde{\eta _{1}}+\tilde{\eta_{2}}+2\pi i\tau_{12}}, \\ f_{tt}(n+1)\rightarrow-4\pi^{2} p^{2}_{1}e^{\tilde{\zeta_{1}}+\tilde {\eta_{1}}}-4 \pi^{2} p^{2}_{2}e^{\tilde{\zeta_{2}}+\tilde{\eta _{2}}}-4\pi^{2} (p_{1}+p_{2})^{2}e^{\tilde{\zeta_{1}}+\tilde{\zeta _{2}}+\tilde{\eta_{1}}+\tilde{\eta_{2}}+2\pi i\tau_{12}}. \end{gathered} $$

(3.10)

For convenience, we denote $\lambda_{1}=e^{\pi i\tau_{11}}$, $\lambda _{2}=e^{\pi i\tau_{22}}$. In what follows, we expand each function in $\tilde{F_{1}}(0,0)=\tilde {F_{1}}(0,1)=\tilde{F_{1}}(1,0)=\tilde{F_{1}}(1,1)=0$, $\tilde {F_{2}}(0,0)=\tilde{F_{2}}(0,1)=\tilde{F_{2}}(1,0)=\tilde{F_{2}}(1,1)=0$ into series of $\lambda_{1}$, $\lambda_{2}$,

$$ \begin{aligned}[b] \tilde{F_{1}}(0,0)={}&c_{1}+ \bigl[4\pi i\mu_{1}\bigl(e^{2\pi i\eta_{1}}-e^{-2\pi i\eta_{1}}\bigr)+16 \pi^{2}p^{2}_{1}\bigl(e^{2\pi i\eta_{1}}+e^{-2\pi i\eta _{1}} \bigr)+c_{1}\bigr]e^{2\pi i\tau_{11}} \\ &+\bigl[4\pi i\mu_{2}\bigl(e^{2\pi i\eta_{2}}-e^{-2\pi i\eta_{2}}\bigr)+16 \pi ^{2}p^{2}_{2}\bigl(e^{2\pi i\eta_{2}}+e^{-2\pi i\eta_{2}} \bigr)+c_{1}\bigr]e^{2\pi i\tau_{22}}\\ &+o\bigl(\lambda_{1}^{r_{1}} \lambda_{2}^{r_{2}}\bigr), \end{aligned} $$

(3.11)

when $r_{1}+r_{2}\geq4$, it is easy to see that $c_{1}\rightarrow0$.

$$ \begin{aligned}[b] \tilde{F_{1}}(0,1)={}& \bigl[\bigl(2\pi i\mu_{2}\bigl(e^{\pi i\eta_{2}}-e^{-\pi i\eta _{2}} \bigr)+4\pi^{2}p^{2}_{2}\bigl(e^{\pi i\eta_{2}}+e^{-\pi i\eta_{2}} \bigr)+c_{1}\bigr) \\ &+\bigl(2\pi i(2\mu_{1}-\mu_{2})+4\pi^{2}(2p_{1}-p_{2})^{2}+c_{1} \bigr)e^{2\pi i(\tau_{11}-\tau_{12})} \\ &+\bigl(2\pi i(2\mu_{1}+\mu_{2})+4\pi^{2}(2p_{1}+p_{2})^{2}+c_{1} \bigr)e^{2\pi i(\tau_{11}+\tau_{12})}\bigr]e^{\pi i\tau_{22}}\\&+o\bigl(\lambda_{1}^{r_{1}} \lambda _{2}^{r_{2}}\bigr) \end{aligned} $$

(3.12)

in view of $c_{1}\rightarrow0$, from which we have $2\pi p^{2}_{2}\cos \pi\eta_{2}-\mu_{2}\sin\pi\eta_{2}=0$.

$$ \begin{aligned}[b] \tilde{F_{1}}(1,0)={}& \bigl[\bigl(2\pi i\mu_{1}\bigl(e^{\pi i\eta_{1}}-e^{-\pi i\eta _{1}} \bigr)+4\pi^{2}p^{2}_{1}\bigl(e^{\pi i\eta_{1}}+e^{-\pi i\eta_{1}} \bigr)+c_{1}\bigr) \\ &+\bigl(2\pi i(2\mu_{2}-\mu_{1})+4\pi^{2}(2p_{2}-p_{1})^{2}+c_{1} \bigr)e^{2\pi i(\tau_{22}-\tau_{12})} \\ &+\bigl(2\pi i(2\mu_{2}+\mu_{1})+4\pi^{2}(2p_{2}+p_{1})^{2}+c_{1} \bigr)e^{2\pi i(\tau_{22}+\tau_{12})}\bigr]e^{\pi i\tau_{11}}\\ &+o\bigl(\lambda_{1}^{r_{1}} \lambda _{2}^{r_{2}}\bigr). \end{aligned} $$

(3.13)

In view of $c_{1}\rightarrow0$, from (3.13), we have $2\pi p^{2}_{1}\cos\pi\eta_{1}-\mu_{1}\sin\pi\eta_{1}=0$.

$$ \begin{aligned}[b] \tilde{F_{1}}(1,1)={}& \bigl[\bigl[2\pi i(\mu_{1}+\mu_{2}) \bigl(e^{\pi i(\eta_{1}+\eta _{2})}-e^{-\pi i(\eta_{1}+\eta_{2})} \bigr)\\ &+4\pi^{2}(p_{1}+p_{2})^{2} \bigl(e^{\pi i(\eta_{1}+\eta_{2})}+e^{-\pi i(\eta_{1}+\eta_{2})}\bigr)\bigr]e^{2\pi i\tau _{12}} \\ &+2\pi i(\mu_{1}-\mu_{2}) \bigl(e^{\pi i(\eta_{1}-\eta_{2})}-e^{\pi i(\eta _{1}-\eta_{2})} \bigr)\\ &+4\pi^{2}(p_{1}-p_{2})^{2} \bigl(e^{\pi i(\eta_{1}-\eta _{2})}+e^{-\pi i(\eta_{1}+\eta_{2})}\bigr)\bigr]e^{\pi i(\tau_{11}+\tau _{22})}+o\bigl( \lambda_{1}^{r_{1}}\lambda_{2}^{r_{2}}\bigr). \end{aligned} $$

(3.14)

From (3.14), we have

$$ e^{2\pi i\tau_{12}}=-\frac{2\pi i(\mu_{1}-\mu_{2})(e^{\pi i(\eta _{1}-\eta_{2})}-e^{-\pi i(\eta_{1}-\eta_{2})})+4\pi ^{2}(p_{1}-p_{2})^{2}(e^{\pi i(\eta_{1}-\eta_{2})}+e^{-\pi i(\eta _{1}-\eta_{2})})}{2\pi i(\mu_{1}+\mu_{2})(e^{\pi i(\eta_{1}+\eta _{2})}-e^{-\pi i(\eta_{1}+\eta_{2})})+4\pi^{2}(p_{1}+p_{2})^{2}(e^{\pi i(\eta_{1}+\eta_{2})}+e^{-\pi i(\eta_{1}+\eta_{2})})}. $$

(3.15)

In the following, we will consider

$$ \begin{aligned}[b] \tilde{F_{2}}(0,0)={}&c_{2}+ \bigl[16\pi^{2}p_{2}(l_{2}-\mu_{2})-2 \cosh4\pi i\eta_{2}+2+c_{2}\bigr]e^{2\pi i\tau_{22}} \\ &+\bigl[16\pi^{2}p_{1}(l_{1}- \mu_{1})-2\cosh4\pi i\eta_{1}+2+c_{2} \bigr]e^{2\pi i\tau_{11}}+o\bigl(\lambda_{1}^{r_{1}} \lambda_{2}^{r_{2}}\bigr). \end{aligned} $$

(3.16)

From $\tilde{F_{2}}(0,0)\rightarrow0$, we have $c_{2}\rightarrow0$.

$$ \begin{aligned}[b] \tilde{F_{2}}(0,1)={}& \bigl(8\pi^{2}p_{2}(l_{2}-\mu_{2})-4\cosh2 \pi i\eta _{2}+4+2c_{2}\bigr)e^{\pi i\tau_{22}} \\ &+\bigl[4\pi^{2}(2p_{1}-p_{2}) (2l_{1}-2\mu_{1}+l_{2}-\mu_{2})-2 \cosh2\pi i(2\eta_{1}-\eta_{2})+2+c_{2} \bigr]\\ &\times e^{\pi i(2\tau_{11}-2\tau_{12}+\tau _{22})} \\ &+\bigl[4\pi^{2}(2p_{1}+p_{2}) (2l_{1}+2\mu_{1}-l_{2}-\mu_{2})-2 \cosh2\pi i(2\eta_{1}+\eta_{2})+2+c_{2} \bigr]\\ &\times e^{\pi i(2\tau_{11}+2\tau_{12}+\tau _{22})}+o\bigl(\lambda_{1}^{r_{1}} \lambda_{2}^{r_{2}}\bigr). \end{aligned} $$

(3.17)

From $\tilde{F_{2}}(0,1)\rightarrow0$, in view of $c_{2}\rightarrow0$, we have $2\pi^{2}p_{2}(\mu_{2}-l_{2})+\cosh2\pi i\eta_{2}-1=0$.

$$ \begin{aligned}[b] \tilde{F_{2}}(1,0)={}&\bigl(8 \pi^{2}p_{1}(l_{1}-\mu_{1})-4\cosh2 \pi i\eta _{1}+4+2c_{2}\bigr)e^{\pi i\tau_{11}} \\ &+\bigl[4\pi^{2}(p_{2}-p_{1}) (2l_{2}-2\mu_{2}-l_{1}+\mu_{1})-2 \cosh2\pi i(2\eta_{2}-\eta_{1})+2+c_{2} \bigr]\\ &\times e^{\pi i(2\tau_{22}-2\tau_{12}+\tau _{11})} \\ &+\bigl[4\pi^{2}(p_{1}+2p_{2}) (2l_{2}-2\mu_{2}+l_{1}-\mu_{1})-2 \cosh2\pi i(2\eta_{2}+\eta_{1})+2+c_{2} \bigr]\\ &\times e^{\pi i(\tau_{11}+2\tau_{12}+2\tau _{22})}+o\bigl(\lambda_{1}^{r_{1}} \lambda_{2}^{r_{2}}\bigr). \end{aligned} $$

(3.18)

For $\tilde{F_{2}}(1,0)\rightarrow0$, we have $2\pi^{2}p_{1}(\mu _{1}-l_{1})+\cosh2\pi i\eta_{1}-1=0$.

$$ \begin{aligned}[b] \tilde{F_{2}}(1,1)={}&2 \bigl[4\pi^{2}(p_{1}+p_{2}) (l_{1}+l_{2}- \mu_{1}-\mu_{2}) -2\cosh2\pi i(\eta_{1}+ \eta_{2})+2+c_{2}\bigr]e^{2\pi i\tau_{12}} \\ &+2\bigl[4\pi^{2}(p_{1}-p_{2}) (l_{1}-l_{2}-\mu_{1}+\mu_{2}) -2 \cosh2\pi i(\eta_{1}-\eta_{2})+2+c_{2} \bigr]e^{\pi i(\tau_{11}+\tau _{22})}\hspace{-24pt}\\ &+o\bigl(\lambda_{1}^{r_{1}} \lambda_{2}^{r_{2}}\bigr). \end{aligned} $$

(3.19)

In view of $\tilde{F_{2}}(1,1)\rightarrow0$, we have

$$ e^{2\pi i\tau_{12}}=-\frac{2\pi^{2}(p_{1}-p_{2})(l_{1}-\mu_{1}+\mu _{2}-l_{2})-2\cosh2\pi i(\eta_{1}-\eta_{2})+1}{2\pi ^{2}(p_{1}+p_{2})(l_{1}-\mu_{1}-\mu_{2}+l_{2})-2\cosh2\pi i(\eta _{1}+\eta_{2})+1}. $$

(3.20)

This completes the proof of Theorem 2. □