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
figure 1

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

Full size image
Figure 2
figure 2

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

Full size image
Figure 3
figure 3

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

Full size image

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. □