1 Introduction

It can be said with certainly that solving soliton equations is a crucial subject of theoretical research of solitons and integrable system, and that ever-growing number of methods for solving the soliton equations are proposed with the efforts of researchers. Algebro-geometric solution is an important category of solutions to nonlinear equations, which can not only reveal the intrinsic structure of solutions, but also characterize the quasi-periodic behavior of the nonlinear phenomena.

During the past five decades, numbers of researchers made significant contributions to solving soliton equations with the algebro-geometric approach. The modern origin of this method is that the researchers Burchnall and Chanundy found that there is a connection between linear differential operators and algebraic curves with the aid of the study of the linear differential operators [1, 2]. In the mid-twentieth century, Novikov and Krichever proposed a systematic algebraic geometric approach to solve nonlinear equations with the aim of obtaining quasi-periodic solutions of the Riemann theta function [3,4,5]. Moreover thrilling results appeared later. In 1988, Cao managed to use the nonlinear features of Lax pairs to study algebro-geometric solutions of finite-dimensional integrable Hamiltonian systems and it was applied to many soliton equations successfully [6, 7]. Then Geng generalized a new method called finite-order expansion of the Lax matrix, which has obtained great achievements and gave an alternative way of solving the soliton equations [8, 9]. In the 1990s, F. Gestesy and H. Holden developed a polynomial recursion method that successfully extended algebro-geometric solutions from a single equation to a hierarchy [10, 11]. In view of the current situation, the study of solving soliton equations using algebraic geometric methods attracted increasing interest and it has been applied successfully to soliton equations connected with \(2\times 2\) and \(3\times 3\) discrete matrix spectral problems [12,13,14,15,16]. From the perspective of development, there is hope that we can apply algebro-geometric methods to solving the soliton equations linked to the \(4\times 4\) discrete matrix spectral problems [17].

It is universally acknowledged that solving the soliton equations connected with \(2\times 2\) matrix spectral problem is a prerequisite and basis for solving the soliton equations linked to higher-order matrix spectral problem. The whole solving process can be divided into four main steps as follows.

Step 1. Generating an integrable hierarchy with bi-Hamiltonian structure We first introduce a discrete spectral parameter and then the hierarchy can be derived with the aid of the discrete zero curvature equation. Subsequently, the bi-Hamiltonian structure of the hierarchy can be obtained through the trace identity.

Step 2. Separating the above hierarchy into soluble ordinary differential equations (ODEs) The Lenard recursion gradients and a Lax matrix are introduced, from which the relationship between the elliptic variables and the potentials can be established, so we can separate the corresponding hierarchy into soluble ODEs.

Step 3. Straightening out the of continuous flow For the sake of completing this step, we need to introduce the theory of hyperelliptic Riemann surface, then the flow can be straightened successfully by the Abel–Jacobi coordinates.

Step 4. Getting the algebro-geometric solutions In order to solve the soliton equation, the main tool is to straighten out the discrete flow to linearize the equations and then express the linearized equations as the Riemann theta function. There are two ways to straighten out it. One way is to define a fundamental solution matrix of soliton equations, using the expression of Dubrovin–Novikov’s type to straighten out the discrete flow to linearize the soliton equations [18]. The other way is to establish the meromorphic function \(\Phi \), then analyze the asymptotic properties of them, the solutions can be obtained if we can get the Riemann theta function representations of the function \(\Phi \) in a concise form [19], if not, we need to define the Baker–Akhiezer function \(\Psi \) on the basis of the function \(\Phi \) and analyze its asymptotic behavior, the explicit Riemann theta function representations of the functions \(\Phi \) and \(\Psi \) can be generated after that [20].

Though using algebro-geometric method to solve soliton equations is systematically provided in the case of \(2\times 2\) matrix spectral problem [21,22,23,24], inferring the solutions is challenging because of concerning the knowledge of the discrete variables and algebraic curve. For this paper, we will discuss the following integrable hierarchy of semi-discrete equations

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle {u_{n_t}=u_n(u_nv_{n-1}-u_{n+1}v_n),}\\ \displaystyle {v_{n_t}=v_n(u_n-u_{n+1})+v_n(u_nv_{n-1}-u_{n+1}v_n),} \end{array} \right. \end{aligned}$$
(1)

which has been studied explicit solutions by Darboux transformation [25], where \(u_n\) and \(v_n\) are two potentials and t is the time variable. As an application, the main aim of this paper is to use algebraic geometric methods to solve the hierarchy (1).

The outline of the present paper is as follows. In Sect. 2, we derive a hierarchy of semi-discrete equations (1) by use of the discrete zero curvature equation and its bi-Hamiltonian system is constructed through the trace identity. In Sect. 3, the Lax matrix and elliptic variables are introduced, through which we can decompose the hierarchy (1) into soluble ODEs. In Sect. 4, we straighten out the continuous flow on the basis of the hyperelliptic Riemann surface and the Abel-Jacobi coordinates. In Sect. 5, we define the stationary meromorphic function and analyze its asymptotic behavior, then in the time-dependent case, the algebro-geometric constructions of the hierarchy (1) are proposed owing to the Riemann theta function and the asymptotic behavior of the functions. In Sect. 6, the conclusions and remarks of this paper are summarized.

2 The Integrable Hierarchy and Its Bi-Hamiltonian Structure

To get the corresponding hierarchy (1), we design the following isospectral problem

$$\begin{aligned} E\varphi _n=U_n(u_n,\lambda )\varphi _n,\quad U_n(u_n,\lambda )=\left( \begin{array}{cc}\frac{\displaystyle u_n}{\displaystyle \lambda }&{}\frac{\displaystyle 1+v_n}{\displaystyle \lambda }\\ u_n&{}1\end{array}\right) , \end{aligned}$$
(2)

where \(\varphi _n=(\varphi _n^1,\varphi _n^2)^T\), E is a shift operator \((Eu_n=u_{n+1})\) and \(\lambda \) is a constant spectral parameter \((\lambda _t=0)\). Then, we introduce the stationary discrete zero curvature equation

$$\begin{aligned} \displaystyle {(EV_{n}^{[m]})U_n-U_nV_{n}^{[m]}=0.} \end{aligned}$$
(3)

Let us suppose that a solution \(V_{n}^{[m]}\) of (3) is given by

$$\begin{aligned} V_{n}^{[m]}=\left( \begin{array}{cc}a_n&{}b_n\\ \lambda c_n&{}-a_n\end{array}\right) , \end{aligned}$$

then (3) is equivalent to

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle {\frac{\displaystyle u_n}{\displaystyle \lambda }(a_{n+1}-a_n)+u_nb_{n+1}-(1+v_n)c_n=0,}\\ \displaystyle {\frac{\displaystyle 1+v_n}{\displaystyle \lambda }(a_{n+1}+a_n)+b_{n+1}- \frac{\displaystyle u_n}{\displaystyle \lambda }b_n=0,}\\ \displaystyle {-u_n(a_{n+1}+a_n)+u_nc_{n+1}-\lambda c_{n}=0,}\\ \displaystyle {(1+v_n)c_{n+1}-u_nb_n-(a_{n+1}-a_n)=0,} \end{array} \right. \end{aligned}$$
(4)

where \(a_n, b_n, c_n\) are polynomials of the spectral \(\lambda \) with

$$\begin{aligned} \begin{array}{ll} \displaystyle {a_n=\sum _{m\ge 0}a_{n,{+}}^{(m)}\lambda ^{-m}+\sum _{m\ge 0}a_{n,{-}}^{(m)}\lambda ^{m}},\\ \displaystyle {b_n=\sum _{m\ge 0}b_{n,{+}}^{(m)}\lambda ^{-m}+\sum _{m\ge 0}b_{n,{-}}^{(m)}\lambda ^{m}},\\ \displaystyle {c_n=\sum _{m\ge 0}c_{n,{+}}^{(m)}\lambda ^{-m}+\sum _{m\ge 0}c_{n,{-}}^{(m)}\lambda ^{m}}. \end{array} \end{aligned}$$
(5)

Taking

$$\begin{aligned} \begin{array}{ll} a_{n,{+}}^{(0)}=-\frac{1}{2},~b_{n,{+}}^{(0)}=0,~c_{n,{+}}^{(0)}=0,~ a_{n,{-}}^{(0)}=\frac{1}{2},~b_{n,{-}}^{(0)}=\frac{1+v_n}{u_n},~c_{n,{-}}^{(0)}=1, \end{array} \end{aligned}$$

and substituting the expanding expressions (5) into (4), the recurrence relation can be derived as follows

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle {u_n(a_{n+1,+}^{(m)}-a_{n,+}^{(m)})+u_nb_{n+1,+}^{(m+1)}-(1+v_n)c_{n,+}^{(m+1)}=0,}\\ \displaystyle {(1+v_n)(a_{n+1,+}^{(m)}+a_{n,+}^{(m)})-u_nb_{n,+}^{(m)}+b_{n+1,+}^{(m+1)}=0,}\\ \displaystyle {-u_n(a_{n+1,+}^{(m)}+a_{n,+}^{(m)})+u_nc_{n+1,+}^{(m)}-c_{n,+}^{(m+1)}=0,}\\ \displaystyle {(1+v_n)c_{n+1,+}^{(m)}-u_nb_{n,+}^{(m)}-(a_{n+1,+}^{(m)}-a_{n,+}^{(m)})=0,} \end{array} \right. \end{aligned}$$
(6)

and

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle {u_n(a_{n+1,-}^{(m+1)}-a_{n,-}^{(m+1)})+u_nb_{n+1,-}^{(m)}-(1+v_n)c_{n,-}^{(m)}=0,}\\ \displaystyle {(1+v_n)(a_{n+1,-}^{(m+1)}+a_{n,-}^{(m+1)})-u_nb_{n,-}^{(m+1)}+b_{n+1,-}^{(m)}=0,}\\ \displaystyle {-u_n(a_{n+1,-}^{(m+1)}+a_{n,-}^{(m+1)})+u_nc_{n+1,-}^{(m+1)}-c_{n,-}^{(m)}=0,}\\ \displaystyle {(1+v_n)c_{n+1,-}^{(m+1)}-u_nb_{n,-}^{(m+1)}-(a_{n+1,-}^{(m+1)}-a_{n,-}^{(m+1)})=0.} \end{array} \right. \end{aligned}$$
(7)

Under the initial-value conditions of \(a_{n,\pm }^{(0)}\), \(b_{n,\pm }^{(0)}\) and \(c_{n,\pm }^{(0)}\), the recursion relations (5) uniquely decide \(a_{n,\pm }^{(m)},\) \(b_{n,\pm }^{(m)},\) \(c_{n,\pm }^{(m)},\) \(m\ge 1\) and the first few quantities are given by

$$\begin{aligned} a_{n,+}^{(1)}= & {} u_n(1+v_{n-1}),\quad b_{n+1,+}^{(1)}=1+v_n,\quad c_{n,+}^{(1)}=u_n,\\ a_{n,+}^{(2)}= & {} -{u_n}^2(1+v_{n-1})^2-u_nu_{n-1}v_{n-1}(1+v_{n-2})-u_nu_{n+1}v_{n}(1+v_{n-1}),\\ b_{n+1,+}^{(2)}= & {} -u_{n+1}(1+v_n)^2-u_nv_n(1+v_{n-1}),\\ c_{n,+}^{(2)}= & {} -u_nu_{n+1}v_{n}-{u_n}^2(1+v_{n-1}),\\ a_{n,-}^{(1)}= & {} -\frac{1+v_n}{u_n},\quad b_{n,-}^{(1)}=-\left( \frac{v_n(1+v_{n+1})}{u_{n+1}}+\frac{(1+v_n)^2}{u_n}\right) ,\quad c_{n+1,-}^{(1)}\\= & {} -\left( \frac{1+v_n}{u_{n+1}}+\frac{v_n}{u_n}\right) ,\ldots . \end{aligned}$$

Since the formula (6) and (7) imply that \(V_{n}^{[m]}\) unsatisfied the compatibility condition, so we must take an addictive constant to adjust the expression of \(a_n\). For \(m\ge 0\) \((m\in Z)\), we select

$$\begin{aligned} V_{n}^{[m]}=\left( \begin{array}{cc}\sum \limits _{i=0}^m a_i\lambda ^{m-i}&{}\sum \limits _{i=0}^m b_i\lambda ^{m-i}\\ \sum \limits _{i=0}^m c_i\lambda ^{m+1-i}&{}-\sum \limits _{i=0}^m a_i\lambda ^{m-i}\end{array}\right) , \end{aligned}$$
(8)

and make a modification

$$\begin{aligned} \Delta _n^{(m)}=\left( \begin{array}{cc}-2a_n^{(m)}+c_n^{(m)}&{}0\\ 0&{}0\end{array}\right) . \end{aligned}$$
(9)

Then, we set

$$\begin{aligned} V_n^{(m)}=V_{n}^{[m]}+\Delta _n^{(m)}, \end{aligned}$$

and the following equation holds

$$\begin{aligned}&E(V_n^{(m)})U_n-U_nV_n^{(m)}\nonumber \\&\quad = \left( \begin{array}{cc}\frac{\displaystyle u_n}{\displaystyle \lambda }(a_{n}^{(m)}-a_{n+1}^{(m)})+\frac{\displaystyle u_n}{\displaystyle \lambda }(c_{n+1}^{(m)}-c_{n}^{(m)})&{}\frac{\displaystyle v_n}{\displaystyle \lambda }(a_{n}^{(m)}-a_{n+1}^{(m)})\\ u_n(a_{n}^{(m)}-a_{n+1}^{(m)})+u_n(c_{n+1}^{(m)}-c_{n}^{(m)})&{}0\end{array}\right) . \end{aligned}$$
(10)

Let the time evolution of the eigenfunction of the spectral problem (2) obey the differential equation

$$\begin{aligned} \varphi _{n,{t_m}}=V_n^{(m)}\varphi _n, \end{aligned}$$
(11)

and the compatible conditions of equations (2) and (11) are

$$\begin{aligned} U_{n,{t_m}}=E(V_n^{(m)})U_n-U_nV_n^{(m)},\quad m\ge 0. \end{aligned}$$
(12)

According to (12), the semi-discrete equations are implied as follows

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle {u_{n,{t_m}}=u_n(a_{n}^{(m)}-a_{n+1}^{(m)})+u_n(c_{n+1}^{(m)}-c_{n}^{(m)}),}\\ \displaystyle {v_{n,{t_m}}=v_n(a_{n}^{(m)}-a_{n+1}^{(m)}),} \end{array} \right. \end{aligned}$$
(13)

so the discrete zero curvature representation of (13) is (12). It is easy to verify that the first nonlinear lattice equation in (13) is (1), namely if and only if \(m=1\) and \(t_1=t\).

Next, we will build up the bi-Hamiltonian structure and prove its Liouville integrability. Depends on the discrete trace identity

$$\begin{aligned} \frac{\delta }{\delta u_n}\sum _{n\in Z}\left\langle {R_n},{\frac{\partial {U_n}}{\partial {\lambda }}} \right\rangle =\left( \lambda ^{-\varepsilon } \left( \frac{\partial }{\partial {\lambda }} \right) \lambda ^\varepsilon \right) \left\langle {R_n},{\frac{\partial {U_n}}{\partial {u_n^i}}}\right\rangle ,\quad i=1,2, \end{aligned}$$
(14)

we have

$$\begin{aligned} U_{n,{t_m}}=\left( \begin{array}{cc} u_n\\ v_n \end{array} \right) _{t_m} =\displaystyle {J\frac{\displaystyle \delta H_n^{(m)}}{\displaystyle \delta u_n} =K\frac{\displaystyle \delta H_n^{(m-1)}}{\displaystyle \delta u_n} ,} \end{aligned}$$
(15)

here

$$\begin{aligned} J=\left( \begin{array}{cc} 0&{}u_n(E^{-1}-1)v_n\\ v_n(1-E)u_n&{}0\end{array}\right) , \end{aligned}$$
(16)

and

$$\begin{aligned} K=\left( \begin{array}{cc}u_nv_nE{u_n}^2-{u_n}^2E^{-1}u_nv_n&{} u_nv_nEu_nv_n-{u_n}^2E^{-1}(v_n+{v_n}^2)+{u_n}^2v_n\\ (v_n+{v_n}^2)E{u_n}^2-u_nv_nE^{-1}u_nv_n-{u_n}^2v_n&{}(v_n+{v_n}^2)Eu_nv_n-u_nv_nE^{-1}(v_n+{v_n}^2)\end{array}\right) . \end{aligned}$$
(17)

Using (5), we derive the recursion constructions

$$\begin{aligned} \frac{\displaystyle \delta H_n^{(m)}}{\displaystyle \delta u_n}=\Upsilon \frac{\displaystyle \delta H_n^{(m-1)}}{\displaystyle \delta u_n}, \end{aligned}$$
(18)

the recursion operator is

$$\begin{aligned} \Upsilon =\left( \begin{array}{cc} \Upsilon _{11}&{}\Upsilon _{12}\\ \Upsilon _{21}&{}\Upsilon _{22}\end{array}\right) , \end{aligned}$$
(19)

with

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle {\Upsilon _{11}=-\frac{1}{u_n}(E-1)^{-1}(1+v_n)E{u_n}^2+\frac{1}{u_n} (E-1)^{-1}u_nE^{-1}u_nv_n+\frac{1}{u_n}(E-1)^{-1}{u_n}^2,}\\ \displaystyle {\Upsilon _{12}=-\frac{1}{u_n}(E-1)^{-1}(1+v_n)Eu_nv_n+\frac{1}{u_n}(E-1)^{-1}u_nE^{-1}(v_n+{v_n}^2),}\\ \displaystyle {\Upsilon _{21}=\frac{1}{v_n}(1-E^{-1})^{-1}u_nE^{-1}u_nv_n-\frac{1}{v_n}(1-E^{-1})^{-1}v_nE{u_n}^2,}\\ \displaystyle {\Upsilon _{22}=\frac{1}{v_n}(1-E^{-1})^{-1}u_nE^{-1}(v_n+{v_n}^2)- \frac{1}{v_n}(1-E^{-1})^{-1}v_nEu_nv_n-\frac{1}{v_n}(1-E^{-1})^{-1}u_nv_n},\\ \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} H_n^{(0)}=-\frac{1}{2}\sum _{n\in z}(\ln (u_nv_n)),H_n^{(m)}=-\sum _{n\in z}\frac{a_n^{(m+1)}}{m},m>0. \end{aligned}$$

Therefore, it is not difficult to prove the Liouville integrability of the hierarchy (1).

3 Elliptic Variables and Soluble Ordinary Differential Equations

Firstly, the hierarchy (1) should be converted to soluble ordinary differential equations. Suppose that (3) has two essential solutions \(\psi (n)=(\psi ^{(1)}(n),\psi ^{(2)}(n))\) and \(\chi (n)=(\chi ^{(1)}(n),\chi ^{(2)}(n))\), we then give a Lax matrix \(W_n\) with three functions f(n), h(n) and g(n) of \(\lambda \),

$$\begin{aligned} W_n=\frac{1}{2}(\chi (n)\psi (n)^T+\psi (n)\chi (n)^T)\sigma =\left( \begin{array}{cc} f(n) &{} g(n) \\ \lambda h(n) &{} -f(n) \end{array}\right) , ~~~\sigma =\left( \begin{array}{cc} 0 &{} -1 \\ 1 &{} 0 \end{array}\right) , \end{aligned}$$
(20)

and \( W_n \) ought to satisfy

$$\begin{aligned} W_{n+1}U_n-U_nW_n=0,~~~W_{n,t_m}=[V_n^{(m)},W_n]. \end{aligned}$$
(21)

It is easy to see that (21) can be extended as

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{u_n}{\lambda }\Delta f(n)+u_nEg(n)-(1+v_n)h(n)=0,\\ Eg(n)+\frac{1+v_n}{\lambda }(E+1)f(n)-\frac{u_n}{\lambda }g(n)=0,\\ u_nE h(n)-\lambda h_n-u_n(E+1)f(n)=0,\\ (1+v_n)Eh(n)-u_ng(n)-\Delta f(n)=0, \end{array} \right. \end{aligned}$$
(22)

and

$$\begin{aligned} \left\{ \begin{array}{ll} f(n)_{t_m}=\lambda b_n^{(m)}h(n)-\lambda c_n^{(m)}g(n),\\ g(n)_{t_m}=2a_n^{(m)}g(n)-2b_n^{(m)}f(n),\\ h(n)_{t_m}=2c_n^{(m)}f(n)-2a_n^{(m)}h(n). \end{array} \right. \end{aligned}$$
(23)

Suppose the following functions are polynomials of finite order in \(\lambda \):

$$\begin{aligned} \begin{array}{ll} f(n)=\sum \limits _{j=0}^{N'} f_{j,+}(n)\lambda ^{N'-j}+\sum \limits _{j=0}^{N'-1} f_{j,-}(n)\lambda ^{-N'+j},\\ g(n)=\sum \limits _{j=0}^{N'} g_{j,+}(n)\lambda ^{N'-j}+\sum \limits _{j=0}^{N'-1} g_{j,-}(n)\lambda ^{-N'+j},\\ h(n)=\sum \limits _{j=0}^{N'} h_{j,+}(n)\lambda ^{N'-j}+\sum \limits _{j=0}^{N'-1} h_{j,-}(n)\lambda ^{-N'+j},\\ \end{array} \end{aligned}$$
(24)

where \(N=2N'-1\).

Then, the Lenard gradient sequences are listed as follows

$$\begin{aligned} {\tilde{J}}_nS_{{j+1},{+}}= & {} {\tilde{K}}_nS_{j,+},~~~{\tilde{J}}_nS_{0,+}=0,~j\ge 0,\nonumber \\ {\tilde{K}}_nS_{{j+1},{-}}= & {} {\tilde{J}}_nS_{j,-},~~~{\tilde{K}}_nS_{0,-}=0,~j\ge 0,\nonumber \\ \end{aligned}$$
(25)

and \(S_{j,\pm }(n)=({\bar{h}}_{j,\pm }(n),{\bar{g}}_{j,\pm }(n),{\bar{f}}_{j,\pm }(n))^{T}\) with

$$\begin{aligned} S_{0,+}=\left( 0,0,-\frac{1}{2}\right) ^{T},~~~~~~~S_{0,-}=\left( 1,\frac{1+v_n}{u_n},\frac{1}{2}\right) ^{T}. \end{aligned}$$

Taking (24) into (22) and defining \(G_{j,_{\pm }}(n)=(h_{j,{\pm }}(n),g_{j,{\pm }}(n),f_{j,{\pm }}(n))^T\), we have

$$\begin{aligned} {\tilde{J}}_nG_{{j+1},+}(n)= & {} {\tilde{K}}_nG_{j,+}(n),~~{\tilde{J}}_nG_{0,+}(n)=0,j\ge 0,\nonumber \\ {\tilde{K}}_nG_{{j+1},-}(n)= & {} {\tilde{J}}_nG_{j,-}(n),~~{\tilde{K}}_nG_{0,-}(n)=0,j\ge 0,\nonumber \\ {\tilde{K}}_nG_{N,+}(n)= & {} {\tilde{J}}_nG_{{N-1},-}(n),~~~~j\ge 0,\nonumber \\ \end{aligned}$$
(26)

where the matrix differential operators \({\tilde{J}}_n,\ {\tilde{K}}_n \) are given by

$$\begin{aligned} {\tilde{J}}_n=\left( \begin{array}{ccc}E &{} 0&{}0\\ 0&{}1&{}0\\ {}-u_n&{}(1+v_n)E&{}1-E\\ \end{array}\right) , ~~~{\tilde{K}}_{n}=\left( \begin{array}{ccc}u_{n}&{} 0&{}-(1+v_n)(1+E)\\ 0&{}u_{n}E&{}-u_n(1+E)\\ {}-u_n&{}(1+v_n)E&{}1-E \end{array}\right) . \end{aligned}$$

It is evident that

$$\begin{aligned} \begin{array}{ll} G_{0,\pm }=\alpha _{0,\pm }S_{0,\pm }(n),\\ \end{array} \end{aligned}$$
(27)

where \(\alpha _{0,{\pm }}\) are constants.

Taking \({\tilde{J}}_n^{-1}{\tilde{K}}_n\) and \({\tilde{K}}_n^{-1}{\tilde{J}}_n\) into (27), we have

$$\begin{aligned} \begin{array}{ll} G_{1,+}(n)=\alpha _{0,+}S_{1,+}(n)+\alpha _{1,+}S_{0,+}(n),\\ G_{1,-}(n)=\alpha _{0,-}S_{1,-}(n)+\alpha _{1,-}S_{0,-}(n),\\ \end{array} \end{aligned}$$

so

$$\begin{aligned} G_{\kappa ,{\pm }}(n)=\sum _{j=0}^{\kappa }\alpha _{j,{\pm }}S_{{\kappa -j},{\pm }}(n),~~~1\le \kappa \le N'. \end{aligned}$$
(28)

According to (27) and (28), we can obtain the values of \(G_{\kappa ,_{\pm }}(n)\) \((\kappa =0,1,2)\) and \(S_{0,\pm }(n)\) as \(\alpha _{0,{\pm }}=1\),

$$\begin{aligned}&\left\{ \begin{array}{ll} h_{0,+}(n)=0,~~~~g_{0,+}(n)=0,~~~~f_{0,+}(n)=-\frac{1}{2},\\ h_{0,-}(n)=1,~~~~g_{0,-}(n)=\frac{1+v_{n}}{u_{n}},~~~~f_{0,-}(n)=\frac{1}{2},\\ \end{array} \right. \end{aligned}$$
(29)
$$\begin{aligned}&\left\{ \begin{array}{ll} {\bar{h}}_{0,+}(n)=0,~~~~{\bar{f}}_{0,+}(n)=-\frac{1}{2},~~~~{\bar{g}}_{0,+}(n)=0,\\ {\bar{h}}_{0,-}(n)=1,~~~~{\bar{f}}_{0,-}(n)=\frac{1}{2},~~~~{\bar{g}}_{0,-}(n)=\frac{1+v_n}{u_n},\\ \end{array} \right. \end{aligned}$$
(30)
$$\begin{aligned}&\left\{ \begin{array}{ll} g_{1,+}(n)=u_n,~~g_{1,-}(n)=-(\frac{v_n(1+v_{n+1})}{u_{n+1}}+\frac{(1+v_n)^2}{u_n}),\\ h_{1,+}(n)=1+v_{n-1},~~h_{1,-}(n)=-(\frac{1+v_{n-1}}{u_{n}}+\frac{v_{n-1}}{u_{n-1}}),\\ f_{1,+}(n)=u_n+u_nv_{n-1}-\frac{1}{2}\alpha _{1,+},~~f_{1,-}(n)=-\frac{1+v_n}{u_n}+\frac{1}{2}\alpha _{1,+}, \end{array} \right. \end{aligned}$$
(31)
$$\begin{aligned}&\left\{ \begin{array}{ll} g_{2,+}(n)=-u_nu_{n+1}v_{n}-{u_n}^2(1+v_{n-1})+u_n\alpha _{1,+},\\ h_{2,+}(n)=-u_{n}(1+v_{n-1})^2-u_{n-1}v_{n-1}(1+v_{n-2})+\alpha _{1,+}(1+v_{n-1}),\\ f_{2,+}(n)=-{u_n}^2(1+v_{n-1})^2-u_nu_{n-1}v_{n-1}(1+v_{n-2})\\ ~~~~~~~~~~~~~-u_nu_{n+1}v_{n}(1+v_{n-1})+\alpha _{1,+}(u_n+u_nv_{n-1})-\frac{1}{2}\alpha _{2,+}. \end{array} \right. \end{aligned}$$
(32)

By deduction, (24) can also be represented as by \((j\in {\mathbb {N}}_0)\),

$$\begin{aligned} \begin{array}{ll} h_{j,{\pm }}(n)=\sum \limits _{l=0}^{j}\alpha _{{j-l},{\pm }}{\bar{h}}_{l,{\pm }}(n),\\ g_{j,{\pm }}(n)=\sum \limits _{l=0}^{j}\alpha _{{j-l},{\pm }}{\bar{g}}_{l,{\pm }}(n),\\ f_{j,{\pm }}(n)=\sum \limits _{l=0}^{j}\alpha _{{j-l},{\pm }}{\bar{f}}_{l,{\pm }}(n). \end{array} \end{aligned}$$
(33)

The corresponding homogeneous polynomials can also be defined as by \((s\in {\mathbb {N}})\),

$$\begin{aligned} \begin{array}{ll} {\bar{H}}_{s,+}(\lambda )=\sum \limits _{\kappa =0}^{s-1}{\bar{h}}_{{s-\kappa },+}\lambda ^\kappa ,~~{\bar{G}}_{s,+}(\lambda )=\sum \limits _{\kappa =0}^{s}{\bar{g}}_{{s-\kappa },+}\lambda ^\kappa ,~~{\bar{F}}_{s,+}(\lambda )=\sum \limits _{\kappa =0}^{s}{\bar{f}}_{{s-\kappa },+}\lambda ^\kappa ,\\ {\bar{H}}_{s,-}(\lambda )=\sum \limits _{\kappa =1}^{s}{\bar{h}}_{{s-\kappa },-}\lambda ^{-\kappa },~~{\bar{G}}_{s,-}(\lambda )=\sum \limits _{\kappa =0}^{s}{\bar{g}}_{{s-\kappa },-}\lambda ^{-\kappa },~~{\bar{F}}_{s,-}(\lambda )=\sum \limits _{\kappa =0}^{s}{\bar{f}}_{{s-\kappa },-}\lambda ^{-\kappa },\\ \end{array} \end{aligned}$$
(34)

and

$$\begin{aligned} \left\{ \begin{array}{ll} a_n=\sum \limits _{\kappa =0}^{n}\gamma _{{n-\kappa },+}{{\bar{F}}}_{\kappa ,+}(\lambda )+\sum \limits _{\kappa =0}^{n}\gamma _{{n-\kappa },-}{{\bar{F}}}_{\kappa ,-}(\lambda ),\\ b_n=\sum \limits _{\kappa =0}^{n}\gamma _{{n-\kappa },+}{{\bar{G}}}_{\kappa ,+}(\lambda )+\sum \limits _{\kappa =0}^{n}\gamma _{{n-\kappa },-}{{\bar{G}}}_{\kappa ,-}(\lambda ),\\ c_n=\sum \limits _{\kappa =0}^{n}\gamma _{{n-\kappa },+}{{\bar{H}}}_{\kappa ,+}(\lambda )+\sum \limits _{\kappa =0}^{n}\gamma _{{n-\kappa },-}{{\bar{H}}}_{\kappa ,-}(\lambda ),\\ \end{array} \right. \end{aligned}$$
(35)

with

$$\begin{aligned} \begin{array}{ll} a_{n,\pm }^{(m)}=\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },\pm }{{\bar{f}}}_{m,\pm },~~b_{n,\pm }^{(m)}=\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },\pm }{{\bar{g}}}_{m,\pm },~~c_{n,\pm }^{(m)}=\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },\pm }{{\bar{h}}}_{m,\pm }. \end{array} \end{aligned}$$

We write g(n) and h(n) in the following expression

$$\begin{aligned} \begin{array}{ll} h(n)=(1+v_{n-1})\lambda ^{-N'}\prod \limits _{j=1}^{N}(\lambda -\mu _j(n)),\\ g(n)=u_n\lambda ^{-N'}\prod \limits _{j=1}^{N}(\lambda -\nu _j(n)), \end{array} \end{aligned}$$
(36)

where the elements \(\mu _j(n)\) and \(\nu _j(n)\) named elliptic variables. Owing to (36), contrasting the coefficients of \(\lambda \) in g(n) and h(n), we can get

$$\begin{aligned} \begin{array}{ll} \sum \limits _{j=1}^{N}{\mu _j}(n)=u_n(1+v_{n-1})+u_{n-1}v_{n-1}\frac{1+v_{n-2}}{1+v_{n-1}}-\alpha _{1,+},\\ \sum \limits _{j=1}^{N}{\nu _j}(n)=u_{n+1}v_n+u_n(1+v_{n-1})-\alpha _{1,+},\\ \prod \limits _{j=1}^{N}{\mu _j(n)}=(-1)^N\frac{1}{1+v_{n-1}},~~\prod \limits _{j=1}^{N}{\nu _j(n)}=(-1)^{N}(1+v_n)(u_n^{-2}). \end{array} \end{aligned}$$
(37)

As \(\det W_n\) is a \((2N+2)\) th-order polynomial of \(\lambda \) and its coefficients are constants of the n-flow and \(t_m\)-flow, we obtain

$$\begin{aligned} -\det W_n=f^2(n)+\lambda g(n)h(n)=(\frac{1}{2{\lambda }^{N'}})^2 \prod \limits _{j=1}^{2N+2}(\lambda -\lambda _j)=R(\lambda ). \end{aligned}$$
(38)

According to (38), we have

$$\begin{aligned} f(n)|_{\lambda =\mu _\kappa (n)}=\sqrt{R(\mu _\kappa (n))},~ f(n)|_{\lambda =\nu _\kappa (n)}=\sqrt{R(\nu _\kappa (n))}. \end{aligned}$$
(39)

Based on (24) and (39), we obtain

$$\begin{aligned} \begin{array}{ll} h(n)_{t_m}|_{\lambda =\mu _\kappa (n)}=-(1+v_{n-1})(\partial _{t_1}\mu _\kappa (n))\prod \limits _{j=1,j\ne \kappa }^{N}(\mu _\kappa (n)-\mu _j(n))\\ \quad =-2\sqrt{R(\mu _\kappa (n))}c_n|_{\lambda =\mu _\kappa (n)},\\ g(n)_{t_m}|_{\lambda =\nu _\kappa (n)}=-u_n(\partial _{t_1}\nu _\kappa (n))\prod \limits _{j=1,j\ne \kappa }^{N}(\nu _\kappa (n)-\nu _j(n))=2\sqrt{R(\nu _k(n))}b_n|_{\lambda =\nu _\kappa (n)}, \end{array} \end{aligned}$$
(40)

which means

$$\begin{aligned} \partial _{t_m} \mu _\kappa (n)= & {} \frac{2\sqrt{R(\mu _\kappa (n))}c_n|_{\lambda =\mu _\kappa (n)}}{(1+v_{n-1})\prod \limits _{j=1,j\ne \kappa }^{N}(\mu _\kappa (n)-\mu _j(n))},\nonumber \\ ~~\partial _{t_m} \nu _\kappa (n)= & {} -\frac{2\sqrt{R(\nu _\kappa (n))}b_n|_{\lambda =\nu _\kappa (n)}}{u_n\prod \limits _{j=1,j\ne \kappa }^{N}(\nu _k(n)-\nu _j(n))}. \end{aligned}$$
(41)

To sum up, we have separated the hierarchy of semi-discrete equations into soluble ordinary differential equations.

4 Straightening Out of the Continuous Flow

In the chapter, we introduce the Riemann surface \({\mathcal {K}}_N\) and then give a list of independent canoncial basis cycles: \(r_1,\ldots , r_N; z_1, \ldots , z_N\), which have the following intersection numbers

$$\begin{aligned} r_i\circ r_j=z_i\circ z_j=0,~~r_i\circ z_j=\delta _{ij},~~1\le i,~~j\le N. \end{aligned}$$
(42)

For \({\mathcal {K}}_N\), we select the differentials of holomorphic

$$\begin{aligned} {\widetilde{\omega }}_s=\frac{\lambda ^{s-1}d\lambda }{\sqrt{4\lambda ^{2N^{'}}R(\lambda )}},~~1\le s\le N, \end{aligned}$$
(43)

and define

$$\begin{aligned} {\mathcal {L}}_{ij}=\int _{r_j}{\widetilde{\omega }}_i,~~{\mathcal {M}}_{ij}=\int _{z_j}{\widetilde{\omega }}_i, \end{aligned}$$
(44)

where \({\mathcal {L}}\) and \({\mathcal {M}}\) are period matrices.

Now, on the basis of matrices \({\mathcal {L}}\) and \({\mathcal {M}}\), we define two new matrices \({\mathcal {N}}\) and \(\tau \) to be of the following form

$$\begin{aligned} {\mathcal {N}}={\mathcal {L}}^{-1},~~\tau ={\mathcal {L}}^{-1}{\mathcal {M}}, \end{aligned}$$

and the \(\tau \) can be proved to be symmetric \((\tau _{ij}=\tau _{ji})\) and its imaginary part \(Im \tau >0\).

Substituting \({\widetilde{\omega }}_s\) into the new basis \(\omega _j\):

$$\begin{aligned} \omega _j=\sum _{s=1}^{N}{\mathcal {N}}_{js}{\widetilde{\omega }}_s,~~j=1,2,...,N, \end{aligned}$$
(45)

which meets

$$\begin{aligned} \begin{array}{ll} \int _{r_i}\omega _j=\sum \limits _{s=1}^{N}{\mathcal {N}}_{js}\int _{r_i}{\widetilde{\omega }}_s=\sum \limits _{s=1}^{N}{\mathcal {N}}_{js}{\mathcal {M}}_{si}=\delta _{ji},\\ \int _{z_i}\omega _j=\sum \limits _{s=1}^{N}{\mathcal {N}}_{js}\int _{z_i}{\widetilde{\omega }}_s=\sum \limits _{s=1}^{N}{\mathcal {N}}_{js}{\mathcal {M}}_{si}=\tau _{ji}. \end{array} \end{aligned}$$
(46)

Let us propose the Abel map \({\mathcal {A}}(P):Div({\mathcal {K}}_N)\rightarrow {{\mathcal {J}}} \) \( (x, \kappa , Q_0\in {\mathcal {K}}_N)\),

$$\begin{aligned} {\mathcal {A}}(x)=\int _{Q_0}^x {\underline{\omega }},~~~ {\mathcal {A}}\left( \sum n_\kappa x_\kappa \right) =\sum n_\kappa {\mathcal {A}}(x_\kappa ), \end{aligned}$$
(47)

and we can define the Riemann theta function as follows

$$\begin{aligned} \theta ({\underline{\lambda }}(x,{\hat{\mu }}(n,t_m)))=\theta ({\underline{\Lambda }}-{\mathcal {A}}(x)+{\underline{\rho }}^{(i)}(n,t_{m})),\ i=1,2, \end{aligned}$$
(48)

where \({\underline{\Lambda }}=(\Lambda _1,\cdots \Lambda _n)\) is interpreted through

$$\begin{aligned} \Lambda _j=\frac{1}{2}(1+\tau _{jj})-\sum _{i=1,i\ne j}\int _{r_i}w_i\int _{Q_0}^x w_j. \end{aligned}$$

Then, we introduce the Abel-Jacobi coordinates

$$\begin{aligned}{\underline{\rho }}^{(i)}={\mathcal {A}}\left( \sum _{\kappa =1}^N x_\kappa ^{(i)}\right) =\sum _{\kappa =1}^N {\mathcal {A}}(x_\kappa ^{(i)})=\sum _{\kappa =1}^N \int _{Q_0}^{x_\kappa ^{(i)}} {\underline{w}}, ~~ i=1,2,\end{aligned}$$

with \(x_\kappa ^{(1)}={\hat{\mu }}_\kappa (n,t_{m})\), \(x_\kappa ^{(2)}={\hat{\nu }}_\kappa (n,t_{m})\) and \({\underline{\omega }}=(\omega _1,\cdots \omega _N)\), the elements are

$$\begin{aligned} \begin{array}{ll} \rho _j^{(1)}(n,t_m)&{}=\sum \limits _{\kappa =1}^{N}\displaystyle \int _{Q_0}^{x{({\hat{\mu }}_\kappa (n, t_{m}))}}\omega _j\\ &{}=\sum \limits _{\kappa =1}^{N}\sum \limits _{s=1}^{N}\int _{{\lambda }{(Q_0)}}^{{\hat{\mu }}_\kappa (n)}{\mathcal {N}}_{js}\frac{{\lambda }^{s-1}d{\lambda }}{\sqrt{4\lambda ^{2N'}R({\lambda })}},~~~1\le \kappa \le N,\\ \rho _j^{(2)}(n,t_m)&{}=\sum \limits _{\kappa =1}^{N}\displaystyle \int _{Q_0}^{x{({\hat{\nu }}_\kappa (n, t_{m}))}}\omega _j\\ &{}=\sum \limits _{\kappa =1}^{N}\sum \limits _{s=1}^{N}\int _{{\lambda }{(Q_0)}}^{{\hat{\nu }}_\kappa (n)}{\mathcal {N}}_{js}\frac{{\lambda }^{s-1}d{\lambda }}{\sqrt{4\lambda ^{2N'}R({\lambda })}},~~~~1\le \kappa \le N, \end{array} \end{aligned}$$
(49)

where

$$\begin{aligned} \begin{array}{ll} {\hat{\mu }}_j(n,t_m)=(\mu _j(n,t_m),y({\hat{\mu }}_j(n,t_m)))\\ \quad =(\mu _j(n,t_m),2\mu _j(n,t_m)^{N^{'}}f(\mu _j(n,t_m),n,t_m)),\ 1\le j\le N,\\ {\hat{\nu }}_j(n,t_m)=(\nu _j(n,t_m),y({\hat{\nu }}_j(n,t_m)))\\ \quad =(\nu _j(n,t_m),-2\nu _j(n,t_m)^{N^{'}}f(\nu _j(n,t_m),n,t_m)), \ 1\le j\le N. \end{array} \end{aligned}$$
(50)

According to (41), we can obtain

$$\begin{aligned} \begin{array}{ll} \partial _{t_m}\rho _{j}^{(1)}(n,t_m)=\sum \limits _{s=1}^{N}\sum \limits _{\kappa =1}^{N}{\mathcal {N}}_{js}\frac{\mu _\kappa ^{s-1}c_n\mid _{\lambda =\mu _\kappa (n)}}{(1+v_{n-1})\prod \limits _{j=1,j\ne \kappa }^{N}(\mu _\kappa (n)-\mu _j(n))},\\ \\ \partial _{t_m}\rho _{j}^{(2)}(n,t_m)=-\sum \limits _{s=1}^{N}\sum \limits _{\kappa =1}^{N}{\mathcal {N}}_{js}\frac{\nu _\kappa ^{s-1}b_n\mid _{\lambda =\nu _\kappa (n)}}{u_n\prod \limits _{j=1,j\ne \kappa }^{N}(\nu _\kappa (n)-\nu _j(n))}. \end{array} \end{aligned}$$
(51)

In order to finalize the straightening out of the continuous flow, we need to first introduce the following equations and lemmas.

We start with a few elementary results, let \({\underline{\lambda }}=(\lambda _1,\ldots ,\lambda _{2N+2}), ~{\bar{\gamma }}_0({\underline{\lambda }})=\alpha _0({\underline{\lambda }})=1\), then one can get

$$\begin{aligned} \Bigg (\prod \limits _{j=1}^{2N+2}(1-\lambda _j\varsigma )\Bigg )^{-\frac{1}{2}}=\sum \limits _{\kappa =0}^{\infty }{{{\bar{\gamma }}}}_\kappa ({\underline{\lambda }})\varsigma ^\kappa , \end{aligned}$$
(52)

and

$$\begin{aligned} \Bigg (\prod \limits _{j=1}^{2N+2}(1-\lambda _j\varsigma )\Bigg )^{\frac{1}{2}}=\sum \limits _{\kappa =0}^{\infty }{\alpha }_\kappa ({\underline{\lambda }})\varsigma ^\kappa . \end{aligned}$$
(53)

Next, we can deduce that \(\lambda _N\)-dependent summation constants \(\alpha _{0,+},\cdots ,\alpha _{{N'},+}\) and \(\alpha _{0,-},\cdots ,\alpha _{{N'-1},-}\) in f(n), g(n), and h(n)

$$\begin{aligned} \begin{array}{ll} \alpha _{\kappa ,+}=\alpha _{0,+}\alpha _\kappa (\lambda ),~~~\kappa =0,...,N',\\ \alpha _{\kappa ,-}=\alpha _{0,-}\alpha _\kappa ({\underline{\lambda }}^{-1}),~~~\kappa =0,...,N'-1, \end{array} \end{aligned}$$
(54)

with \(\alpha _0({\underline{\lambda }})^{\pm 1}=1.\) Hence, the relationships between the homogeneous and nonhomogeneous can be derived as follows,

$$\begin{aligned}&G_{s,\pm }=\alpha _{0,\pm }\sum \limits _{\kappa =0}^{s}\alpha _{s-\kappa }({\underline{\lambda }}^{\pm 1})S_{\kappa ,\pm }, \end{aligned}$$
(55)
$$\begin{aligned}&\alpha _{0,\pm }S_{s,\pm }=\sum \limits _{\kappa =0}^{s}{{\bar{\gamma }}}_{s-\kappa }{({\underline{\lambda }}^{\pm 1})}G_{\kappa ,\pm }, \end{aligned}$$
(56)

where \(s=0,...,N'-\delta _{\pm }\), and the \(\delta _{\pm }\) follows the following convention,

$$\begin{aligned}\begin{array}{ll} \delta _\pm =\left\{ \begin{array}{ll} 0,~~~~~+,\\ 1,~~~~~-.\\ \end{array}\right. \end{array}\end{aligned}$$

In addition, we introduce the symmetric functions according to the relevant formula of the Lagrange interpolation,

$$\begin{aligned}\begin{array}{ll} {\mathbf {I}}_0({\underline{\mu }})&{}=1,~~\mathbf {I}_\kappa ({\underline{\mu }})=(-1)^{\kappa }\sum \limits _{{\underline{s}}\in {{\mathcal {I}}_\kappa }}{\mu _s}_1\cdots {\mu _s}_\kappa ,~~{{\mathcal {I}}_\kappa }\\ &{}=\{{\underline{s}}\in {N}^{\kappa }\mid {1}\le {s_1}<\cdots <{s_\kappa }\le N\},~~~\kappa =1,\ldots N,\\ {\mathbf {H}}_{\kappa }^{j}({\underline{\mu }})&{}=(-1)^\kappa \sum \limits _{s\in {\mathcal {H}}_{\kappa }^{(j)}}{\mu _s}_1,\cdots {\mu _s}_\kappa ,~~\mathbf {H}_{0}^{j}({\underline{\mu }})=1,~~\mathbf {H}_{N}^{j}({\underline{\mu }})=0,\\ {\mathcal {H}}_{\kappa }^{(j)}&{}=\{{\underline{s}}\in {\mathcal {I}}\mid s_i\ne j,i=1,\cdots ,\kappa \},~~{\underline{\mu }}=(\mu _1,\cdots ,\mu _N),~~\kappa \\ &{}=1,\ldots ,N-1,~~j=1,\ldots ,N, \end{array}\end{aligned}$$

with

$$\begin{aligned}\prod \limits _{j=1}^{N}(\lambda -\mu _j)=\sum \limits _{s=0}^{N}\mathbf {I}_s({\underline{\mu }})\lambda ^{N-s}.\end{aligned}$$

Lemma 1

Assuming that \({\underline{\mu }}\) are N distinct complex numbers. Then,

$$\begin{aligned} \sum \limits _{s=0}^{\kappa }\mathbf {I}_{\kappa -s}({\underline{\mu }})\mu _{j}^{s}=\mathbf {H}_{\kappa }^{(j)}({\underline{\mu }}),~~~j=1,\ldots ,N,~~~\kappa =0,\cdots ,N. \end{aligned}$$
(57)

Lemma 2

Suppose that \(\mu _j\ne \mu _{j'} (j\ne {j'})\), \(\mathbf {T}_{N}({\underline{\mu }})\) which is a \(N\times N\) matrix can be introduced on this basis,

$$\begin{aligned} \mathbf {T}_1({\underline{\mu }})=1,~~~\mathbf {T}_N({\underline{\mu }})=\left( \frac{\mu _{\kappa }^{j-1}}{\prod \limits _{j=1,j\ne \kappa }^{N}(\mu _\kappa -\mu _j)}\right) _{j,\kappa =1}^{N}, \end{aligned}$$
(58)

then, one have \(\mathbf {T}_N({\underline{\mu }})^{-1}=\Bigl (\mathbf {H}_{N-\kappa }^{(j)}({\underline{\mu }})\Bigl )_{j,\kappa =1}^{N}\).

Theorem 1

Straightening out of the continuous flows.

$$\begin{aligned}&\partial _{t_m}{\underline{\rho }}^{(1)}(n,t_m)=\mathbf {{\underline{Z}}}^{(m)}, \end{aligned}$$
(59)
$$\begin{aligned}&\partial _{t_m}{\underline{\rho }}^{(2)}(n,t_m)=-\mathbf {{\underline{Z}}}^{(m)}, \end{aligned}$$
(60)

where

$$\begin{aligned}\begin{array}{ll} \mathbf {{\underline{Z}}}^{(m)}&{}=\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },+}\sum \limits _{q=0}^{\ell -1}\underline{{\mathcal {N}}}_{N-q}{\bar{\gamma }}_{\ell -q-1}({\underline{\lambda }}) -\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },-}\sum \limits _{q=0}^{\ell -1}\underline{{\mathcal {N}}}_{\ell -q}{\bar{\gamma }}_{q}({\underline{\lambda }}^{-1}),~~1\le m\le {N},\\ {\underline{\rho }}^{(i)}(n,t_m)&{}=(\rho _1^{(i)}(n,t_m),\ldots ,\rho _N^{(i)}(n,t_m)),~~~i=1,2,\\ \underline{{\mathcal {N}}}_{\kappa }&{}=({\mathcal {N}}_{1\kappa },\ldots ,{\mathcal {N}}_{N\kappa }),~~~1\le \kappa \le N. \end{array}\end{aligned}$$

Proof

From (24) and (36), we obtain

$$\begin{aligned} \begin{array}{ll} \lambda ^{N'}h(n)&{}=\sum \limits _{j=0}^{N'}h_{j,+}(n)\lambda ^{N+1-j}+\sum \limits _{j=0}^{N'-1}h_{j,-}(n)\lambda ^j=(1+v_{n-1})\prod \limits _{j=1}^{N}(\lambda -\mu _j(n))\\ &{}=(1+v_{n-1})\sum \limits _{s=0}^{N}\mathbf {I}_s({\underline{\mu }})\lambda ^{N-s}. \end{array} \end{aligned}$$
(61)

Using (56) and noticing that \(h_{0,+}={\bar{h}}_{0,+}=0\) and \(\alpha _{0,\pm }=1\), we arrive

$$\begin{aligned}&\begin{array}{ll} h_{s,+}(n)=(1+v_{n-1})\mathbf {I}_{s-1}({\underline{\mu }}),\\ {{\bar{h}}}_{s,+}(n)=(1+v_{n-1})\sum \limits _{\kappa =0}^{s-1}{\bar{\gamma }}_{s-\kappa -1}({\underline{\lambda }})\mathbf {I}_{\kappa }({\underline{\mu }}),~~1\le s\le N',\\ \end{array} \end{aligned}$$
(62)
$$\begin{aligned}&\begin{array}{ll} h_{s,-}(n)=(1+v_{n-1})\mathbf {I}_{N-s}({\underline{\mu }}),\\ {{\bar{h}}}_{s,-}(n)=(1+v_{n-1})\sum \limits _{\kappa =0}^{s}{\bar{\gamma }}_{s-\kappa }({\underline{\lambda }})^{-1}\mathbf {I}_{N-\kappa }(\mu ),~1\le s\le {N'}-1.\\ \end{array} \end{aligned}$$
(63)

According to (34), we obtain

$$\begin{aligned} \begin{array}{ll} {\bar{H}}_{s,+}(\lambda )&{}=\sum \limits _{\kappa =1}^{s}{\bar{h}}_{\kappa ,+}\lambda ^{s-\kappa }=(1+v_{n-1})\sum \limits _{\kappa =1}^{s}\sum \limits _{\ell =0}^{\kappa -1}{\bar{\gamma }}_{{\kappa -\ell -1}}({\underline{\lambda }})\mathbf {I}_\ell ({\underline{\mu }})\lambda ^{s-\kappa }\\ &{}=(1+v_{n-1})\sum \limits _{\kappa =0}^{s-1}{\bar{\gamma }}_{s-\kappa -1}({\underline{\lambda }})\sum \limits _{\ell =0}^{\kappa }\mathbf {I}_\ell ({\underline{\mu }})\lambda ^{\kappa -\ell },~~~~~~~1\le s\le N',\\ {\bar{H}}_{s,-}(\lambda )&{}=\sum \limits _{\kappa =0}^{s-1}{\bar{h}}_{\kappa ,-}\lambda ^{\kappa -s}=(1+v_{n-1})\sum \limits _{\kappa =0}^{s-1}\sum \limits _{\ell =0}^{\kappa }{\bar{\gamma }}_{{\kappa -\ell }}({\underline{\lambda }}^{-1})\mathbf {I}_{N-\ell }({\underline{\mu }})\lambda ^{\kappa -s}\\ &{}=(1+v_{n-1})\sum \limits _{\kappa =0}^{s-1}{\bar{\gamma }}_{\kappa }({\underline{\lambda }}^{-1})\sum \limits _{\ell =N-s+\kappa +1}^{N}\mathbf {I}_\ell ({\underline{\mu }})\lambda ^{N-s+\kappa -\ell },~~1\le s\le {N'}. \end{array} \end{aligned}$$
(64)

So, we can get the following recursion formula by (57),

$$\begin{aligned} \begin{array}{ll} {\bar{H}}_{j,+}(\mu _j)&{}=(1+v_{n-1})\sum \limits _{l=0}^{s-1}{\bar{\gamma }}_{s-\kappa -1}({\underline{\lambda }})\sum \limits _{\ell =0}^{\kappa }\mathbf {I}_\ell ({\underline{\mu }}){\mu _j}^{\kappa -\ell }\\ &{}=(1+v_{n-1})\sum \limits _{\kappa =0}^{s-1}{\bar{\gamma }}_{s-\kappa -1}({\underline{\lambda }})\mathbf {H}_{\kappa }^{j}({\underline{\mu }}),~~~~~~~~~~~~1\le s\le N',\\ {\bar{H}}_{j,-}(\mu _j)&{}=-(1+v_{n-1})\sum \limits _{l=0}^{s-1}{\bar{\gamma }}_{\kappa }({\underline{\lambda }}^{-1})\sum \limits _{\ell =0}^{N-s+\kappa }\mathbf {T}_\ell ({\underline{\mu }}){\mu }_j^{N-s+\kappa -\ell }\\ &{}=-(1+v_{n-1})\sum \limits _{\kappa =0}^{s-1}{\bar{\gamma }}_{\kappa }({\underline{\lambda }}^{-1})\mathbf {H}_{N-s+\kappa }^{j}({\underline{\mu }}),~~~~~~1\le s\le {N'}.\\ \end{array} \end{aligned}$$
(65)

Considering the case \(1\le m\le N'\), using (35), (58) and \({\bar{H}}_{0,\pm }=0\), we can continue the transformation of the above equations

$$\begin{aligned} \begin{array}{ll} c_n(\mu _j)&{}=\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },+}{\bar{H}}_{\ell ,+}(\mu _j)+\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },-}{\bar{H}}_{\ell ,-}(\mu _j)\\ &{}=(1+v_{n-1})\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },+}\sum \limits _{q=0}^{\ell -1}{\bar{\gamma }}_{\ell -q-1}({\underline{\lambda }})\mathbf {H}_{q}^{(j)}({\underline{\mu }})\\ &{}\quad -(1+v_{n-1})\sum \limits _{\ell =1}^{m}{\gamma }_{{m-\ell },-}\sum \limits _{q=0}^{\ell -1}{\bar{\gamma }}_{q}({\underline{\lambda }}^{-1})\mathbf {H}_{N-\ell +q}^{(j)}({\underline{\mu }})\\ &{}=(1+v_{n-1})\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },+}\sum \limits _{q=0}^{\ell -1}{\bar{\gamma }}_{\ell -q-1}({\underline{\lambda }})\mathbf {T}_N({\underline{\mu }})_{j,N-q}^{-1}\\ &{}\quad -(1+v_{n-1})\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },-}\sum \limits _{q=0}^{\ell -1}{\bar{\gamma }}_{q}({\underline{\lambda }}^{-1})\mathbf {T}_N({\underline{\mu }})_{j,\ell -q}^{-1}. \end{array} \end{aligned}$$
(66)

Using (49) and (58), we obtain

$$\begin{aligned}\begin{array}{ll} \partial _{t_m}\rho _{j}^{(1)}(n,t_m)=\sum \limits _{s=1}^{N}\sum \limits _{\kappa =1}^{N}{\mathcal {N}}_{js}\frac{\mu _\kappa ^{s-1}c_n|_{\lambda =\mu _\kappa (n)}}{(1+v_{n-1})\prod \limits _{j=1,j\ne k}^{N}(\mu _\kappa (n)-\mu _j(n))}\\ \quad =\sum \limits _{j=1}^{N}\sum \limits _{s=1}^{N}{\mathcal {N}}_{js}\mathbf {T}_N({\underline{\mu }})_{s,j}\frac{c_n|_{\lambda =\mu _\kappa (n)}}{1+v_{n-1}}\\ \quad =\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },+}\sum \limits _{q=0}^{\ell -1}{\mathcal {N}}_{j,N-q}{\bar{\gamma }}_{\ell -q-1}({\underline{\lambda }})-\sum \limits _{\ell =1}^{m}\gamma _{{m-\ell },-}\sum \limits _{q=0}^{\ell -1}{\mathcal {N}}_{j,\ell -q}{\bar{\gamma }}_q({\underline{\lambda }})^{-1}\\ \quad =\mathbf {Z}_j^{(m)},~~~~~1\le m\le {N'},~1\le j \le N.\\ \end{array}\end{aligned}$$

Therefore, the equation (59) can be proven and we can get (60) in a similar way. In summary, we have straightened out of the continuous flow. \(\square \)

5 Algebro-geometric Solutions

For this paper, we will choose the second way of the Step 4 mentioned above to get the solutions. During this section, we will analyze the asymptotic behavior of the function \(\Phi (x,n)\), and then we will introduce the function \(\Psi (x,n)\) to analyze its asymptotic behavior since we can not get the simple Riemann theta function representations of the function \(\Phi (x,n)\). Eventually, the algebro-geometric solutions of the time-dependent case can be deduced.

5.1 The Stationary Meromorphic Function

In what follows, we will give the stationary meromorphic function of the integrable hierarchy (1). According to (38), the hyperelliptic curve \({\mathcal {K}}_N\) of arithmetic genus N can be denoted by

$$\begin{aligned} {\mathcal {K}}_N: {\mathcal {F}}(\lambda ,y)=y^2-(2\lambda ^{N^{'}})^2R_{2N+2}(\lambda )=0. \end{aligned}$$
(67)

From (38) and (67), we know that

$$\begin{aligned}\left( \frac{y}{2\lambda ^{N'}}\right) ^2=f^2(n)+\lambda g(n)h(n),\end{aligned}$$

and

$$\begin{aligned}\left( \frac{y}{2\lambda ^{N'}}+f(n)\right) \left( \frac{y}{2\lambda ^{N'}}-f(n)\right) =\lambda g(n)h(n).\end{aligned}$$

Then, the stationary function \(\Phi (x,n)\) on \({\mathcal {K}}_N\) can be defined as the following two forms

$$\begin{aligned} \begin{array}{ll} \Phi (x,n)=\frac{(1/2)\lambda ^{-N'}y+f(n)}{g(n)}=\frac{\lambda h(n)}{(1/2)\lambda ^{-N'}y-f(n)}, \end{array} \end{aligned}$$
(68)

and

$$\begin{aligned} \begin{array}{ll} \Phi (x,n)=\frac{(1/2)\lambda ^{-N'}y+f(n)}{h(n)}=\frac{\lambda g(n)}{(1/2)\lambda ^{-N'}y-f(n)}. \end{array} \end{aligned}$$
(69)

The curve \({\mathcal {K}}_N\) can be compacted by adding two points at \(x_{\infty +}\) and \(x_{\infty -}\). Based on the definition of (67), we can define the points \(\mu _j(n)\) and \(\nu _j(n)\) of \({\mathcal {K}}_N\)

$$\begin{aligned} \begin{array}{ll} {\hat{\mu }}_j(n)&{}=(\mu _j(n),2\mu _j(n)^{N^{'}}f(\mu _j(n),n)),\\ {\hat{\nu }}_j(n)&{}=(\nu _j(n),-2\nu _j(n)^{N^{'}}f(\nu _j(n),n)). \end{array} \end{aligned}$$
(70)

Furthermore, the points \(x_{0,+}\) and \(x_{0,-}\) can be defined through

$$\begin{aligned}x_0,_\pm =\left( 0,\pm \frac{\alpha _0,_-}{\alpha _0,_+}\right) \in {\mathcal {K}}_N,\end{aligned}$$

with

$$\begin{aligned}\prod \limits _{j=1}^{2N+2}\lambda _j=\frac{\alpha _0,_-^2}{\alpha _0,_+^2}=1,\end{aligned}$$

and we introduce the holomorphic sheet exchange map \(*\),

$$\begin{aligned} *: x=(\lambda , y_j(\lambda ))\rightarrow x*=(\lambda , -y_j(\lambda )),\ x^{**}=(x^*)^*, j=0,1, \end{aligned}$$
(71)

where \(y_j(\lambda ), j=0,1\).

Next, we define the corresponding Baker-Akhiezer function \(\Psi (x,n,n_0)\) by

$$\begin{aligned} \begin{array}{ll} \Psi (x,n,n_0)= \left( \begin{array}{ll} \Psi _1(x,n,n_0)\\ \Psi _2(x,n,n_0) \end{array} \right) ,\\ E \Psi (x,n,n_0)=U(u(n),v(n),\lambda (x))\Psi (x,n,n_0),\\ V^{(m)}(u(n),v(n),\lambda (x))\Psi (x,n,n_0)=y(x)\Psi (x,n,n_0),\\ \Psi _2(x,n_0,n_0)=1, \ x\in {\mathcal {K}}_N \setminus \{x_{\infty \pm }, x_{0\pm }\},\ n, n_0 \in Z, \end{array} \end{aligned}$$
(72)

and

$$\begin{aligned} \begin{array}{ll} \Psi _2(x,n,n_0)=\left\{ \begin{array}{ll} \prod \limits _{n'=n_0}^{n-1}(1+u(n')\Phi (x,n')),~~~~n\ge n_0+1,\\ ~~1,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ n=n_0,\\ \prod \limits _{n'=n}^{n_0-1}(1+u(n')\Phi (x,n'))^{-1}),~n\le n_0-1, \end{array}\right. \\ \Psi _1(x,n,n_0)=\Phi (x,n,n_0)\left\{ \begin{array}{ll} \prod \limits _{n'=n_0}^{n-1}(\frac{u(n')}{\lambda }+\frac{1+v(n')}{\lambda }\Phi (x,n')^{-1}),~n\ge n_0+1,\\ ~~1,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~n=n_0,\\ \prod \limits _{n'=n}^{n_0-1}(\frac{u(n')}{\lambda }+\frac{1+v(n')}{\lambda }\Phi (x,n')^{-1})^{-1}, n\le n_0-1. \end{array}\right. \end{array} \end{aligned}$$
(73)

The stationary meromorphic function \(\Phi (x,n)\) on \({\mathcal {K}}_N\) is expressed by

$$\begin{aligned} \Psi _1(x,n)=\Phi (x,n)\Psi _2(x,n). \end{aligned}$$
(74)

According to (3) and (73), the function \(\Phi (x,n)\) meets the following Riccati-type

$$\begin{aligned} 1+v(n)+u(n)\Phi (x,n)-\lambda u(n)\Phi (x,n)\Phi (x,n+1)=\lambda \Phi (x,n+1), \end{aligned}$$
(75)

which also satisfies the system through (69)

$$\begin{aligned}\begin{array}{ll} \left\{ \begin{array}{ll} \Phi (x,n)+\Phi (x^*,n)=\frac{ 2f(n)}{h(n)},\\ \Phi (x,n)\Phi (x^*,n)=-\frac{4\lambda ^2g(n)^2}{\lambda ^{-2N'}y^2-4f(n)^2},\\ \Phi (x,n)-\Phi (x^*,n)=-\frac{ \lambda ^{-N'}y}{h(n)}=\frac{4\lambda ^{-N'+1}yg(n)}{\lambda ^{-2N'}y^2-4f(n)^2}.\\ \end{array}\right. \end{array}\end{aligned}$$

In addition, we introduce the expressions of \({\widetilde{F}},{\widetilde{G}},{\widetilde{H}}\) and according to (69), we can get

$$\begin{aligned} \left\{ \begin{array}{ll} {\widetilde{F}}=\frac{\Phi (x,n)+\Phi (x^*,n)}{\Phi (x,n)-\Phi (x^*,n)}=\frac{2\lambda ^{N'}f}{y},\\ {\widetilde{G}}=\frac{-2\Phi (x,n)\Phi (x^*,n)\lambda ^{-1}}{\Phi (x,n)-\Phi (x^*,n)}=\frac{2\lambda ^{N'}g}{y},\\ {\widetilde{H}}=\frac{2}{\Phi (x,n)-\Phi (x^*,n)}=\frac{2\lambda ^{N'}h}{y}. \end{array}\right. \end{aligned}$$
(76)

Assuming that (76) have the following asymptotic behavior. As \(x\rightarrow x_{\infty \pm }\), we have

$$\begin{aligned} \frac{\lambda ^{N'}f}{y}=\mp \sum \limits _{s=0}^{\infty }{{\widetilde{f}}}_{s,+}\zeta ^s,~~~\frac{\lambda ^{N'}g}{y}=\mp \sum \limits _{s=0}^{\infty }{{\widetilde{g}}}_{s,+}\zeta ^s,~~~\frac{\lambda ^{N'}h}{y}=\mp \sum \limits _{s=0}^{\infty }{{\widetilde{h}}}_{s,+}\zeta ^s, \end{aligned}$$
(77)

and as \(x\rightarrow x_{0_\pm }\), we have

$$\begin{aligned} \frac{\lambda ^{N'}f}{y}=\pm \sum \limits _{s=0}^{\infty }{{\widetilde{f}}}_{s,-}\zeta ^s,~~~\frac{\lambda ^{N'}g}{y}=\pm \sum \limits _{s=0}^{\infty }{{\widetilde{g}}}_{s,-}\zeta ^s,~~~\frac{\lambda ^{N'}h}{y}=\pm \sum \limits _{s=0}^{\infty }{{\widetilde{h}}}_{s,-}\zeta ^s, \end{aligned}$$
(78)

of which \(\zeta \) is the local coordinate and \(\zeta =\lambda ^{-1}/\lambda \) beside \( x_{\infty \pm }/ x_{0\pm }\).

Otherwise, \(({{\widetilde{f}}}_s,_{\pm },{{\widetilde{g}}}_s,_{\pm },{{\widetilde{h}}}_s,_{\pm })\) and \(({{\bar{f}}}_s,_{\pm },{{\bar{g}}}_s,_{\pm },{{\bar{h}}}_s,_{\pm })\) have the same recursion relations, so we can obtain

$$\begin{aligned}{{\widetilde{f}}}_{s,{\pm }}={{\bar{f}}}_{s,{\pm }},~~{{\widetilde{g}}}_{s,{\pm }}={{\bar{g}}}_{s,{\pm }},~~{{\widetilde{h}}}_{s,{\pm }}={{\bar{h}}}_{s,{\pm }}.\end{aligned}$$

Lemma 3

Suppose that u and v meet the discrete hierarchy (1) and define \(x=(\lambda , y)\in {\mathcal {K}}_N\setminus \{x_{\infty \pm }, x_{0\pm }\}.\) Based on these, the meromorphic function has the following asymptotic behavior

$$\begin{aligned} \begin{array}{ll} \Phi (x,n)=\left\{ \begin{array}{ll} \frac{1}{u_n}\zeta ^{-1}+(1+{v_{n-1}}^2)(1+u_n)+O(\zeta ),~\ x\rightarrow x_{\infty +},\\ v_nu_{n+1}(1+v_{n-1})+O(\zeta ),~~~~~~~~~~~~~~~\ x\rightarrow x_{\infty -},\\ \end{array}\right. \zeta =\frac{1}{\lambda },\\ \\ \Phi (x,n)=\left\{ \begin{array}{ll} \frac{u_n}{1+v_n}+O(\zeta ),~~~~~~~~~~~~~\ x\rightarrow x_{0+},\\ -\frac{1}{u_{n+1}v_n}\zeta +O(\zeta ^2),~~~~~~\ x\rightarrow x_{0-}, \end{array}\right. ~~~~~~\zeta =\lambda , \end{array} \end{aligned}$$
(79)

where \(u_n, v_n\) satisfy the hierarchy of nonlinear semi-discrete equations (1).

Proof

Giving a local coordinate \(\zeta =\frac{1}{\lambda }\) beside \(x_{\infty \pm }\) and \(\zeta =\lambda \) beside \(x_{0 \pm }\). In terms of (29), (35) and (68), we get

$$\begin{aligned}\begin{array}{ll} f(n)=\left\{ \begin{array}{ll} \zeta ^{-N'}(f_0,_{+}+f_1,_{+}\zeta +f_2,_{+}\zeta ^2+O(\zeta ^3),~~~~~~~~~~~~~~\ x\rightarrow x_{\infty \pm },\\ \zeta ^{-N'}(f_0,_{-}+f_1,_{-}\zeta +f_2,_{-}\zeta ^2+O(\zeta ^3),~~~~~~~~~~~~~~\ x\rightarrow x_{0\pm }, \end{array}\right. \\ \\ y=\left\{ \begin{array}{ll} \mp \frac{1}{2}\zeta ^{-(N+1)}\prod \limits _{j=1}^{2N+2}(1-\lambda _j\zeta )^{\frac{1}{2}}=\mp \frac{1}{2}\zeta ^{-N-1}\sum \limits _{\kappa =0}^{\infty }\alpha _\kappa (\lambda )\zeta ^{\kappa },~\ x\rightarrow x_{\infty \pm },\\ \frac{1}{2}(\prod \limits _{j=1}^{2N+2}\lambda _j)^{\frac{1}{2}}\prod \limits _{j=1}^{2N+2}(1-\lambda ^{-1}\zeta )^{\frac{1}{2}}=\frac{1}{2}\sum \limits _{\kappa =0}^{\infty }\alpha _\kappa (\lambda )^{-1}\zeta ^{\kappa },~~~~~\ x\rightarrow x_{0\pm }, \end{array}\right. \\ \\ g(n)^{-1}=\left\{ \begin{array}{ll} \frac{1}{u_n}\zeta ^{N'-1}\prod \limits _{j=1}^{N}(1-v_j\zeta )^{-1}=\frac{\zeta ^{N'-1}}{u_n}(1+\sum \limits _{j=1}^{N}v_j\zeta +O(\zeta ^{2})),~\ x\rightarrow x_{\infty \pm },\\ \frac{(-1)^{N}\zeta ^{N'}}{u_n\prod \limits _{j=1}^{N}v_j}\prod \limits _{j=1}^{N}(1-{v_j}^{-1}\zeta )^{-1}=\zeta ^{N'}\frac{u_n}{1+v_n}(1+O(\zeta )),~~~~~~~\ x\rightarrow x_{0\pm }. \end{array}\right. \\ \end{array}\end{aligned}$$

Then, according to (68), we can easily obtain the following equations

$$\begin{aligned}\begin{array}{ll} \Phi (x,n)=((2\lambda ^{N'})^{-1}y+f(n))\times g(n)^{-1}\\ \quad =\left\{ \begin{array}{ll} [\mp \frac{1}{2}(1+\alpha _1(\lambda )\zeta +\alpha _2(\lambda )\zeta ^{2}+O(\zeta ^{3}))+f_0,_{+}+f_1,_{+}\zeta +f_2,_{+}\zeta ^{2}+ O(\zeta ^3)]\\ \times (1+\sum \limits _{j=1}^{N}v_j\zeta +O(\zeta ^2))\frac{1}{u_n}\zeta ^{-1},~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\ x\rightarrow x_{\infty \pm },\\ {}[\pm \frac{1}{2}(1+\alpha _1(\lambda ^{-1})\zeta +O(\zeta ^2))+f_0,_-+f_1,_-\zeta +O(\zeta ^2)]\times \frac{u_n}{1+v_n}(1+O(\zeta )),~~\ x\rightarrow x_{0\pm }. \end{array}\right. \\ \\ \quad =\left\{ \begin{array}{ll} \frac{1}{u_n}\zeta ^{-1}+(1+{v_{n-1}}^2)(1+u_n)+O(\zeta ),~~~~~~~~~\ x\rightarrow x_{\infty +},\\ v_nu_{n+1}(1+v_{n-1})+O(\zeta ),~~~~~~~~~~~~~~~~~~~~~~\ x\rightarrow x_{\infty -},\\ \frac{u_n}{1+v_n}+O(\zeta ),~~~~~~~~~~~~~~~\ x\rightarrow x_{0+},\\ {}-\frac{1}{u_{n+1}v_n}\zeta +O(\zeta ^2),~~~~~~~~\ x\rightarrow x_{0-}. \end{array}\right. \\ \end{array}\end{aligned}$$

which prove the Lemma 3. \(\square \)

5.2 Algebro-geometric Solutions of the Time-Dependent Case

For the situation of time-dependent, let us define the Baker-Akhiezer function \(\Psi \) by

$$\begin{aligned} \begin{array}{ll} \Psi (x,n,n_0,t_m,t_{0m})&{}= \left( \begin{array}{ll} \Psi _1(x,n,n_0,t_m,t_{0m})\\ \Psi _2(x,n,n_0,t_m,t_{0m}) \end{array} \right) ,\\ E \Psi (x,n,n_0,t_m,t_{0m})&{}=U(u(n,t_m),u(n,t_m),\lambda (x))\Psi (x,n,n_0,t_m,t_{0m}),\\ \Psi _{t_m}(x,n,n_0,t_m,t_{0m})&{}=V^{(m)}(u(n,t_m),u(n,t_m),\lambda (x))\Psi (x,n,n_0,t_m,t_{0m}),\\ &{}\quad V^{(m)}(u(n,t_m),u(n,t_m),\lambda (x))\Psi (x,n,n_0,t_m,t_{0m})=y(x)\Psi (x,n,n_0,t_m,t_{0m}),\\ \Psi _2(x,n_0,n_0,t_{0m},t_{0m})&{}=1, \ x\in {\mathcal {K}}_N \setminus \{x_{\infty \pm }, x_{0\pm }\},\ n, n_0 \in Z. \end{array} \end{aligned}$$
(80)

The following function \(\Phi (x,n,t_m)\) on \({\mathcal {K}}_N\), which tightly connected with \(\Psi (x,n,n_0,t_m,t_{0m})\), can be defined by

$$\begin{aligned} \Psi _1(x,n,n_0,t_m,t_{0m})=\Phi (x,n,t_m)\Psi _2(x,n,n_0,t_m,t_{0m}), \end{aligned}$$
(81)

such that

$$\begin{aligned} \begin{array}{ll} \Psi _2(x,n,n_0,t_m,t_{0m})=\left\{ \begin{array}{ll} \exp \Bigl (\displaystyle \int _{t_{0m}}^{t_m}d\ell (\lambda c(\lambda ,n_0,\ell )-a(\lambda ,n_0,\ell )\Phi (x,n_0,\ell )\Bigl )\\ ~\times \prod \limits _{n\prime =n_0}^{n-1}(1+u(n\prime )\Phi (x,n\prime )), ~n\ge n_0+1,\\ ~~1, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~n=n_0,\\ \prod \limits _{n\prime =n}^{n_0-1}(1+u(n\prime )\Phi (x,n\prime ))^{-1}),~ ~n\le n_0-1. \end{array}\right. \\ \Psi _1(x,n,n_0,t_m,t_{0m})=\exp \Bigl (\displaystyle \int _{t_{0m}}^{t_m}d\ell (\lambda c(\lambda ,n_0,\ell )-a(\lambda ,n_0,\ell )\Phi (x,n_0,\ell )\Bigl )\\ ~~~~~~~~\times \Phi (x,n_0,t_m)\left\{ \begin{array}{ll} \prod \limits _{n\prime =n_0}^{n-1}(\frac{u(n\prime )}{\lambda }+\frac{1+v(n\prime )}{\lambda }\Phi (x,n\prime )^{-1}) ,~~~~n\ge n_0+1,\\ ~~1,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~n=n_0,\\ \prod \limits _{n\prime =n}^{n_0-1}((\frac{u(n\prime )}{\lambda }+\frac{1+v(n\prime )}{\lambda }\Phi (x,n\prime )^{-1})^{-1},~n\le n_0-1. \end{array}\right. \end{array} \end{aligned}$$
(82)

Naturally, because \(\Phi (x,n,t_m)\) also fulfills (75), we have

$$\begin{aligned}&1+v(n ,t_m)+u(n,t_m)\Phi (x,n ,t_m)-\lambda u(n ,t_m)\Phi (x,n ,t_m)\Phi (x,n+1 ,t_m)\nonumber \\&\quad =\lambda \Phi (x,n+1 ,t_m), \end{aligned}$$
(83)

and

$$\begin{aligned} \begin{array}{ll} \left\{ \begin{array}{ll} \Phi (x,n,t_m)+\Phi (x^*,n,t_m)=\frac{ 2f(n,t_m)}{h(n,t_m)},\\ \Phi (x,n,t_m)\Phi (x^*,n,t_m)=-\frac{4\lambda ^2g(n,t_m)^2}{\lambda ^{-2N'}y^2-4f(n,t_m)^2},\\ \Phi (x,n,t_m)-\Phi (x^*,n,t_m)=-\frac{ \lambda ^{-N'}y}{h(n,t_m)}=\frac{4\lambda ^{-N'+1}yg(n,t_m)}{\lambda ^{-2N'}y^2-4f(n,t_m)^2}.\\ \end{array}\right. \end{array} \end{aligned}$$
(84)

We can obtain the following asymptotic behavior

$$\begin{aligned} \begin{array}{ll} \Phi (x,n,t_m)\\ \quad =\left\{ \begin{array}{ll} \frac{1}{u(n,t_m)}\zeta ^{-1}+(1+v(n-1,t_m)^2)(1+u(n,t_m))+O(\zeta ),~~\ x\rightarrow x_{\infty +},\\ v(n,t_m)u(n+1,t_m)(1+v(n-1,t_m)+O(\zeta ),~~~~~~~~~~~~~~\ x\rightarrow x_{\infty -},\\ \end{array}\right. \zeta =\frac{1}{\lambda },\\ \\ \Phi (x,n,t_m)\\ \quad =\left\{ \begin{array}{ll} \frac{u(n,t_m)}{1+v(n,t_m)}+O(\zeta ),~~~~~~~~~~~~~~~~~~~~\ x\rightarrow x_{0+},\\ -\frac{1}{u(n+1,t_m)v(n,t_m)}\zeta +O(\zeta ^2),~~~~~~~\ x\rightarrow x_{0-}, \end{array}\right. ~~~~\zeta =\lambda , \end{array} \end{aligned}$$
(85)

and its divisor is given through [22],

$$\begin{aligned} (\Phi (x,n ,t_m))={\mathcal {D}}_{{\hat{\nu }}{(n,t_m)}}-{\mathcal {D}}_{{\hat{\mu }}{(n,t_m)}}+x_{0-}-x_{\infty +}. \end{aligned}$$
(86)

One perceive that

$$\begin{aligned} \begin{array}{ll} \Psi _2(x,n,n_0,t_m,t_{0m})=\Psi _2(x,n_0,n_0,t_m,t_{0m})\Psi _2(x,n,n_0,t_m,t_{m}).\\ \end{array} \end{aligned}$$
(87)

Theorem 2

Assuming that (2), (13) and (21) hold, the ingredient \(\Psi _2\) of the Baker–Akhiezer function \(\Psi \) has the asymptotic behavior:

$$\begin{aligned}&\begin{array}{ll} \Psi _2(x,n,n_0,t_m,t_{0m})= \Bigl (1+O(\zeta )\Bigl )\exp \biggl (\pm \frac{1}{2}(t_m-t_{0m})\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },-}\zeta ^{-\ell -1}\biggl )\\ \quad \times \left\{ \begin{array}{ll} -\zeta ^{n_0-n},~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x\rightarrow x_{\infty +},\\ 1+v(n,t_m)u(n,t_m)u(n+1,t_m),~~~~~~~\ x\rightarrow x_{\infty -}, \end{array}\right. ~~~~~~\zeta =\frac{1}{\lambda }, \end{array} \end{aligned}$$
(88)
$$\begin{aligned}&\begin{array}{ll} \Psi _2(x,n,n_0,t_m,t_{0m})= \Bigl (1+O(\zeta )\Bigr )\exp \Bigl (\mp \frac{1}{2}(t_m-t_{0m})\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },-}\zeta ^{-\ell -1}\Bigr )\\ \quad \times \left\{ \begin{array}{ll} \exp \Bigl (\displaystyle \int _{t_{0m}}^{t_m}a_{n,+}^{(m)}(n_0,\ell )-a_{n,-}^{(m)}(n_0,\ell )d\ell \Bigl )\times \Bigl (1+\frac{u(n,t_m)^2}{1+v(n,t_m)}\Bigr ),~~~~\ x\rightarrow x_{0+},\\ \exp \Bigl (\displaystyle \int _{t_{0m}}^{t_m}c_{n,+}^{(m)}(n_0,\ell )-c_{n,-}^{(m)}(n_0,\ell )d\ell \Bigl )\times \Bigl (\frac{u(n,t_m)^2}{u(n+1,t_m)v(n,t_m)}\Bigr ),~\ x\rightarrow x_{0-},\end{array}\right. \zeta =\lambda . \end{array} \end{aligned}$$
(89)

The divisors \((\Psi _1(.,n,n_0,t_m,t_{0m})),\ (\Psi _2(.,n,n_0,t_m,t_{0m}))\) are given by [22],

$$\begin{aligned} \begin{array}{ll} (\Psi _1(.,n,n_0,t_m,t_{0m}))={\mathcal {D}}_{{\hat{\mu }}{(n,t_m)}}-{\mathcal {D}}_{{\hat{\mu }}{(n_0,t_{0m})}}+(n-n_0)(x_{0-}-x_{\infty -}),\\ (\Psi _2(.,n,n_0,t_m,t_{0m}))={\mathcal {D}}_{{\hat{\nu }}{(n,t_m)}}-{\mathcal {D}}_{{\hat{\nu }}{(n_0,t_{0m})}}+(n-n_0)(x_{0-}-x_{\infty -})+x_{0-}-x_{\infty +}. \end{array} \end{aligned}$$
(90)

Proof

Above all, we calculate the asymptotic behavior of \((1+u_n\Phi )\),

$$\begin{aligned}\begin{array}{ll} 1+u_n\Phi =\left\{ \begin{array}{ll} -\zeta ^{-1}+O(\zeta ),\\ 1+v_nu_nu_{n+1}(1+v_{n-1})+O(\zeta ),~~~~~\ x\rightarrow x_{\infty \pm },\\ 1+\frac{{u_n}^2}{1+v_n}+O(\zeta ),\\ -\frac{u_n}{u_{n+1}v_n}\zeta +O(\zeta ^2),~~~~~~~~~~~~~~~~~~~~~~~\ x\rightarrow x_{0\pm }, \end{array}\right. \end{array}\end{aligned}$$

for \(t_{0m}=t_m\), we have

$$\begin{aligned} \begin{array}{ll} \Psi _2(x,n,n_0,t_m,t_m)=\left\{ \begin{array}{ll} -\zeta ^{n_0-n}(1-O(\zeta )),~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\ x\rightarrow x_{\infty +},\\ 1+v(n,t_m)u(n,t_m)u(n+1,t_m)+O(\zeta ),~\ x\rightarrow x_{\infty -}, \end{array}\right. \end{array} \end{aligned}$$
(91)

and

$$\begin{aligned} \begin{array}{ll} \Psi _2(x,n,n_0,t_m,t_m)=\left\{ \begin{array}{ll} 1+\frac{u(n,t_m)^2}{1+v(n,t_m)}+O(\zeta ),~~~~~~~~~~~~~~~~~~~~~~~\ x\rightarrow x_{0+},\\ -\zeta ^{n-n_0}\frac{u(n,t_m)}{u(n+1,t_m)v(n,t_m)}(1+O(\zeta )),~~~~~~~\ x\rightarrow x_{0-}. \end{array}\right. \end{array} \end{aligned}$$
(92)

Secondly, we need to look at the following equation

$$\begin{aligned} \Psi _2(x,n,n_0,t_m,t_{0m})=\exp \biggl (\int _{t_{0m}}^{t_m}\lambda c(\lambda ,n_0,\ell )-a(\lambda ,n_0,\ell )\Phi (x,n_0,\ell )\biggl ), \end{aligned}$$
(93)

and calculate the asymptotic behavior of (\(\lambda c_n-a_n\Phi \)).

Concentrating on the homogeneous coefficients first, for \({x}\rightarrow {x_{\infty \pm }}\), one calculates

$$\begin{aligned}\begin{array}{ll} \lambda {{\bar{H}}}_{\ell ,+}-{{\bar{F}}}_{\ell ,+}\Phi &{}=\lambda {{\bar{H}}}_{\ell ,+}-{{\bar{F}}}_{\ell ,+}\frac{\frac{1}{2}\lambda ^{-N'}y+f}{h}\\ &{}=\lambda {{\bar{H}}}_{\ell ,+}-{{\bar{F}}}_{\ell ,+}(\frac{\lambda ^{N'}f}{y}+\frac{1}{2})(\frac{\lambda ^{N'}h}{y})^{-1}\\ &{}=\pm \frac{1}{2}\zeta ^{-\ell -1}+\frac{{{\bar{f}}}_{0,+}\mp \frac{1}{2}}{{{\bar{h}}}_{1,+}}{{\bar{h}}}_{{\ell +1},+}+O(\zeta ), \end{array}\end{aligned}$$

where

$$\begin{aligned} c_n=\sum \limits _{\ell =0}^{m}\gamma _{m-\ell }+{{\bar{H}}}_{\ell ,+}+O(\zeta ),~~~~a_n=\sum \limits _{\ell =0}^{m}\gamma _{m-\ell }+{{\bar{F}}}_{\ell ,+}+O(\zeta ). \end{aligned}$$
(94)

Substituting (94) into (93), we obtain

$$\begin{aligned} \begin{array}{ll} \lambda c_n-a_n\Phi &{}=\pm \frac{1}{2}\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}+\zeta ^{-\ell -1}+\frac{{{\bar{h}}}_{0,+}\mp \frac{1}{2}}{1+v_{n-1}{{\bar{h}}}_{{\ell +1},+}}+O(\zeta )\\ &{}=-\frac{1}{2}\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}+\zeta ^{-\ell -1},~~~~~~~~~x\rightarrow x_{\infty -}, \end{array} \end{aligned}$$
(95)

if \(x\rightarrow x_{\infty +}\), one other term \(c_n\) need to be analyzed,

$$\begin{aligned}\begin{array}{ll} c_n&{}=\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}{{\bar{H}}}_{\ell ,+}+\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}{{\bar{h}}}_{{\ell -1},-}\zeta +O(\zeta ^2)\\ &{}=\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}{{\bar{H}}}_{\ell ,+}+{{\bar{h}}}_{{m-1},-}\zeta +O(\zeta ^2), \end{array}\end{aligned}$$

so

$$\begin{aligned} \lambda c_n-a_n\Phi =\frac{1}{2}\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}+\zeta ^{-\ell -1}+\frac{{{\bar{h}}}_{{n+1},+}-h_{{n-1},-}}{1+v_{n-1}}+O(\zeta ), ~~~~~~~~~x\rightarrow x_{\infty +}. \end{aligned}$$
(96)

In the same way as above for \({x}\rightarrow {x_{\infty \pm }}\), if \(x\rightarrow x_{0\pm }\), we can get

$$\begin{aligned}\begin{array}{ll} \lambda {{\bar{H}}}_{\ell ,-}-{{\bar{F}}}_{\ell ,-}\Phi &{}=\lambda {{\bar{H}}}_{\ell ,-}-{{\bar{F}}}_{\ell ,-}\frac{\frac{1}{2}\lambda ^{-N'}y+f}{h}\\ &{}=\lambda {{\bar{H}}}_{\ell ,-}-{{\bar{F}}}_{\ell ,-}(\frac{\lambda ^{N'}f}{y}+\frac{1}{2})(\frac{\lambda ^{N'}h}{y})^{-1}\\ &{}=\mp \frac{1}{2}\zeta ^{-\ell -1}+\frac{{{\bar{f}}}_{0,-}\pm \frac{1}{2}}{{{\bar{h}}}_{0,-}}{{\bar{h}}}_{\ell ,-}+O(\zeta ), \end{array}\end{aligned}$$

where

$$\begin{aligned} c_n=\sum \limits _{\ell =0}^{m}\gamma _{m-\ell }+{{\bar{H}}}_{\ell ,-}+c_{n,+}^{(m)}O(\zeta ),~~~~a_n=\sum \limits _{\ell =0}^{m}\gamma _{m-\ell }+{{\bar{F}}}_{\ell ,-}+a_{n,+}^{(m)}O(\zeta ), \end{aligned}$$
(97)

so

$$\begin{aligned} \begin{array}{ll} \lambda c_n-a_n\Phi &{}=\mp \frac{1}{2}\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}+\zeta ^{-\ell -1}+a_{n,+}^{(m)}+a_{n,-}^{(m)}+\frac{{{\bar{f}}}_{0,-}\pm \frac{1}{2}}{{{\bar{h}}}_{0,-}}(c_{n,+}^{(m)}-c_{n,-}^{(m)})+O(\zeta ) \\ &{}=\left\{ \begin{array}{ll} -\frac{1}{2}\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}\zeta ^{-\ell -1}+a_{n,+}^{(m)}-a_{n,-}^{(m)}+O(\zeta ),~~~\ x\rightarrow x_{0+},\\ \frac{1}{2}\sum \limits _{\ell =0}^{m}\gamma _{{m-\ell },+}\zeta ^{-\ell -1}+c_{n,+}^{(m)}-c_{n,-}^{(m)}+O(\zeta ),~~~~~\ x\rightarrow x_{0-}. \end{array}\right. \end{array} \end{aligned}$$
(98)

In summary, according to (95), (96) and (98), equations (88) and (89) have been proven.

Calculating the divisors of \(\Psi _2\) is the next step we need to complete, we only require to calculate separately the divisors of \(\Psi _2(.,n,n_0,t_m,t_m/t_{0m})\) based on (87). We need to calculate the divisors of \((1+u_n\Phi )\) for analyzing the divisors of \(\Psi _2(.,n,n_0,t_m,t_m)\), the following is the calculation process.

First of all, we notice the equations of (91) and (92), the poles of the function \((1+u_n\Phi )\) in \({\mathcal {K}}_N\) are consistent with the ones of \(\Phi \). According to (22), (36), (69) and \(y({{\hat{\mu }}}_j)=-2\mu _j^{N'}f(\mu _j)\), one computes:

$$\begin{aligned} \begin{array}{ll} 1+u_n\Phi &{}=1+u_n\frac{\lambda ^{-N'}y+f(n)}{h(n)}\\ &{}=(1+\lambda )+u_n\frac{h(n+1)}{h(n)}+u_n\frac{a_n-f(n+1)}{h}\\ &{}=(1+\lambda )+u_n\frac{h(n+1)}{h(n)}+u_n\frac{{a_n}^2-f(n+1)^2}{h(n)(u_na_n-u_nf(n+1))}\\ &{}=(1+\lambda )+u_n\frac{h(n+1)}{h(n)}(1+\frac{\lambda u_ng(n+1)}{a_ny-f(n+1)})\\ &{}=(1+\lambda )+u_n\frac{h(n+1)}{h(n)}O(1),~~~(x\rightarrow \mu _j). \end{array} \end{aligned}$$
(99)

Hence the zeros of \(\Psi _2(.,n,n_0,t_m,t_m)\) are at \({{\hat{\mu }}}_j\) \((j=1,2,...,N)\), so,

$$\begin{aligned}\Psi _2(.,n,n_0,t_m,t_m)={\mathcal {D}}_{{{\hat{\mu }}}(n,t_m)}-{\mathcal {D}}_{{{\hat{\mu }}}(n_0,t_m)}+(n-n_0)(x_{0-}-x_{\infty -}).\end{aligned}$$

Then, we demand to calculate the divisors of \(\Psi _2(.,n_0,n_0,t_m,t_{0m})\) and it has zero and poles. What follows, we provisionally limit \(t_{0m}\) and \(t_m\) to one small enough nonempty zone \(I\in R\), for \(\ell \in I\) and \(\mu _j(n_0,\ell )\ne \mu _\kappa (n_0,\ell )\) \((j,\kappa =1,...,N)\).

$$\begin{aligned} \begin{array}{ll} &{}\lambda c(\lambda ,n_0,\ell )-a(\lambda ,n_0,\ell )\Phi (x,n_0,\ell ) \\ &{}\quad =-\frac{a(\mu _j(n_0,\ell ),n_0,\ell )\mu _j^{-N'}y({{\hat{\mu }}}_j)}{(1+v_{n-1})(\lambda -\mu _j(n_0,\ell ))\prod \limits _{i=1,i\ne j}^{N}(\mu _j(n_0,\ell )-\mu _i(n_0,\ell ))}\\ &{}\quad =\partial _\ell \ln (\mu _j(n_0,\ell )-\lambda )+O(1),~(x\rightarrow {{\hat{\mu }}}_j(n_0,\ell )). \end{array} \end{aligned}$$
(100)

Hampering \(x=(\lambda ,y)\) near an adequately little neighbourhood of \({{{\hat{\mu }}}_j(n_0,\ell )\in {\mathcal {K}}_N}\) and \(\ell \in [t_{0m},t_m]\subseteq I\), so (100) hint

$$\begin{aligned} \begin{array}{ll} \Psi _2(x,n_0,n_0,t_m,t_{0m})\\ \quad =\left\{ \begin{array}{ll} (\mu _j(n_0,\ell )-\lambda )O(1),~~~~~~~~~as~~x\rightarrow {{\hat{\mu }}}_j(n_0,t_m)\ne {{\hat{\mu }}}_j(n_0,t_{0m}),\\ O(1),~~~~~~~~~~~~~~~~~~~~~~~~~~~~as~~x\rightarrow {{\hat{\mu }}}_j(n_0,t_m)= {{\hat{\mu }}}_j(n_0,t_{0m}),\\ (\mu _j(n_0,\ell )-\lambda )^{-1}O(1),~~~~~~as~~x\rightarrow {{\hat{\mu }}}_j(n_0,t_{0m})\ne {{\hat{\mu }}}_j(n_0,t_m), \end{array}\right. \end{array} \end{aligned}$$
(101)

where \(O(1)\ne 0\). Therefore, we derive

$$\begin{aligned}\Psi _2(x,n_0,n_0,t_m,t_{0m})={\mathcal {D}}_{{{\hat{\mu }}}(n_0,t_m)}-{\mathcal {D}}_{{{\hat{\mu }}}(n_0,t_{0m})}.\end{aligned}$$

According to (87) and (90), the divisor of \(\Psi _1\) follows immediately using \(\Psi _1=\Psi _2\Phi \).

In the following, we will express the functions \(\Phi ,\Psi _2,u(n)\) and v(n) in the form of (48). For the third kind holomorphic on \({\mathcal {K}}_N\setminus \{x_+,x_-\}\), we give one regular differential \(w_{x_+,x_-}^{(3)}\) and it has poles at \(x_{\pm }\) and residues \(\pm 1\). Particularly, we can get

$$\begin{aligned}w_{x_{0-},x_{\infty \pm }}^{(3)}=\frac{y+y_{0-}}{2\lambda } d {\lambda } \mp \frac{1}{2y}{\prod _{j=1}^N(\lambda -\beta _{\pm ,j}) d \lambda }, ~~ {x_{0-}}=(0, y_{0-}).\end{aligned}$$

So, we inference the following asymptotic expansions. For \(x \rightarrow x_{0\pm }\), we have

$$\begin{aligned} \begin{array}{ll} \displaystyle \int _{Q_0}^x {w_{x_{0-},x_{\infty -}}^{(3)}}=\left\{ \begin{array}{ll} 0 \\ \ln (\zeta ) \end{array} \right\} +w_0^{0\pm }(x_{0-},x_{\infty -})+O(\zeta ), \\ \displaystyle \int _{Q_0}^x {w_{x_{0-},x_{\infty +}}^{(3)}}=\left\{ \begin{array}{ll} 0 \\ \ln (\zeta ) \end{array} \right\} +w_0^{0\pm }(x_{0-},x_{\infty +})+O(\zeta ), \end{array} \end{aligned}$$
(102)

for \(x \rightarrow x_{\infty \pm }\), we have

$$\begin{aligned} \begin{array}{ll} \displaystyle \int _{Q_0}^x {w_{x_{0-},x_{\infty -}}^{(3)}}=\left\{ \begin{array}{ll} 0 \\ {}-\ln (\zeta ) \end{array} \right\} +w_0^{\infty \pm }(x_{0-},x_{\infty -})+O(\zeta ), \\ \displaystyle \int _{Q_0}^x {w_{x_{0-},x_{\infty +}}^{(3)}}=\left\{ \begin{array}{ll} -\ln (\zeta ) \\ 0 \end{array} \right\} +w_0^{\infty \pm }(x_{0-},x_{\infty +})+O(\zeta ). \end{array} \end{aligned}$$
(103)

Similar to the previous step, for the second kind holomorphic, we select \(w_{x_{\infty \pm },q}^{(2)}\) and \(w_{x_{0\pm },q}^{(2)}\) be the regular differentials with a distinctive pole at \(x_{\infty \pm }\) and \(x_{0\pm }\). Between them, the major part as follow,

$$\begin{aligned} w_{x_{\infty \pm },q}^{(2)}=(\zeta ^{-(2+q)}+O(1))d\zeta ,~~ x \rightarrow x_{\infty \pm },~~\zeta =\frac{1}{\lambda }. \end{aligned}$$
(104)

Moreover, we define

$$\begin{aligned} {\tilde{\Omega }}_m^{(2)}=-\frac{1}{2}\sum _{\ell =1}^m \ell {\gamma }_{{m-\ell },+}(w_{x_{\infty +},\ell -1}^{(2)}-w_{x_{\infty -},\ell -1}^{(2)}), ~~{\tilde{U}}_{mj}^{(2)}=\frac{1}{2\pi i}\int _{z_j}{\tilde{\Omega }}_m^{(2)}, \end{aligned}$$
(105)

where \(\gamma _{\ell ,+}\) are the summation constants in \(b_n\).

In any situation, we single out the branch point \(Q_0=(\lambda _{\ell _0}),~\ell _0 \in \{1,\cdots ,2N+2\}\) be a fundamental point. Based on (48) and (50),we can easily get

$$\begin{aligned} \begin{array}{ll} \theta ({\underline{\lambda }}(x,{\hat{\mu }}(n,t_m)))&{}=\theta ({\underline{\Lambda }}-{\mathcal {A}}(x)+{\underline{\rho }}^{(1)}(n,t_{m})),\\ \theta ({\underline{\lambda }}(x,{\hat{\nu }}(n,t_m)))&{}=\theta ({\underline{\Lambda }}-{\mathcal {A}}(x)+{\underline{\rho }}^{(2)}(n,t_{m})). \end{array} \end{aligned}$$
(106)

Then, assume that \(u_n\) and \(v_n\) satisfy the semi-discrete hierarchy. Moreover, let \(x=(\lambda ,y)\in {\mathcal {K}}\setminus \{x_{\infty ,\pm },x_{0,\pm }\}\), suppose that \({\mathcal {D}}_{{{\hat{\mu }}}(n,t_m)}\) and \({\mathcal {D}}_{{{\hat{\nu }}}(n,t_m)}\) are nonspecial so the algebro-geometric solutions can be derived as follows (see [16] for details)

$$\begin{aligned} \begin{array}{rl} \Phi (x,n,t_m)&{}=C(n,t_m)\frac{\theta ({\underline{\lambda }}(x,{\hat{\nu }}(n,t_m)))}{\theta ({\underline{\lambda }}(x,{\hat{\mu }}(n,t_m)))}\exp \Bigl (\displaystyle \int _{Q_0}^xw_{x_{0-},x_{\infty -}}^{(3)}\Bigl ),\\ \Psi _2(x,n,n_0,t_m,t_{0m})&{}=C(n,n_0,t_m,t_{0m})\frac{\theta ({\underline{\lambda }}(x,{\hat{\mu }}(n,t_m)))}{\theta ({\underline{\lambda }}(x,{\hat{\mu }}(n_0,t_{0m})))}\\ &{}\quad \times \exp \Bigl ((n-n_0)\displaystyle \int _{Q_0}^xw_{x_{0-},x_{\infty +}}^{(3)}-(t_m-t_{0m})\displaystyle \int _{Q_0}^x {\tilde{\Omega }}_m^{(2)}\Bigl ), \end{array} \end{aligned}$$
(107)

where \(C(n,t_m)\) is independent of \(x\in {\mathcal {K}}_N\), and

$$\begin{aligned} \begin{array}{rl} C(n,t_m)&{}=\frac{1}{u(n_0,t_{0m})}\frac{\theta ({\underline{\lambda }}(x_{\infty -},{\hat{\mu }}(n,t_m)))}{\theta ({\underline{\lambda }}(x_{\infty -},{\hat{\nu }}(n_0,t_{0m})))}\\ &{}\quad \times \exp \Bigl ((n-n_0)(\omega _{0}^{\infty +}-\omega _{0}^{\infty -})+(t_m-t_{0m})({\tilde{\Omega }}_m^{\infty -}-{\tilde{\Omega }}_m^{\infty +})-\omega _{1}^{\infty -}\Bigl ),\\ C(n,n_0,t_m,t_{0m})&{}=\frac{\theta ({\underline{\lambda }}(x_{\infty +},{\hat{\mu }}(n_0,t_{0m})))}{\theta ({\underline{\lambda }}(x_{\infty +},{\hat{\nu }}(n,t_m)))}\\ &{}\quad \times \exp \Bigl ((n_0-n)\omega _{0}^{\infty +}+(t_m-t_{0m})({\tilde{\Omega }}_m^{\infty -}-{\tilde{\Omega }}_m^{\infty +})\Bigl ), \end{array} \end{aligned}$$
(108)

and u and v are of the forms

$$\begin{aligned} \begin{array}{ll} u(n,t_m)&{}=u(n_0,t_{0m})(-1)^{n-n_0}\frac{\theta ({\underline{\lambda }}(x_{0+},{\hat{\nu }}(n_0,t_{0m})))\theta ({\underline{\lambda }}(x_{0+},{\hat{\mu }}(n,t_m)))}{\theta ({\underline{\lambda }}(x_{0+},{\hat{\mu }}(n_0,t_{0m})))\theta ({\underline{\lambda }}(x_{0+},{\hat{\nu }}(n,t_m)))}\\ &{}\quad \times \exp \Bigl ((n-n_0)(w_0^{0-}(x_{0-}, x_{\infty -})- w_0^{\infty +}(x_{0-}, x_{\infty -}))\Bigl ),\\ v(n,t_m)&{}=v(n_0,t_{0m})(-1)^{n-n_0}\frac{\theta ({\underline{\lambda }}(x_{\infty +},{\hat{\mu }}(n_0,t_{0m})))\theta ({\underline{\lambda }}(x_{\infty +},{\hat{\nu }}(n,t_m)))}{\theta ({\underline{\lambda }}(x_{\infty +},{\hat{\nu }}(n_0,t_{0m})))\theta ({\underline{\lambda }}(x_{\infty +},{\hat{\mu }}(n,t_m)))}\\ &{}\quad \times \exp \Bigl (-(n-n_0)(w_0^{0-}(x_{0-}, x_{\infty -})- w_0^{\infty +}(x_{0-}, x_{\infty -}))\Bigl ). \end{array} \end{aligned}$$
(109)

Therefore, the algebro-geometric solutions of the hierarchy of semi-discrete equations (1) correspond to the form of formula (109) above. \(\square \)

6 Conclusions and Remarks

Throughout the paper, we have inferred a Lax integrable hierarchy and established its Hamiltonian structure. Nextly, the resulting hierarchy (1) is converted to soluble ordinary differential equations. And the algebro-geometric constructions are proposed with the help of the finite-order expansion of the Lax matrix. With the widespread use of algebro-geometric methods in the field of soliton equations, their achievements are well documented. In addition, a great deal of solutions and methods for the hierarchy can be studied, like lump-soliton solutions, breather solutions [26], Hirota bilinear technology, Darboux transformation [27, 28]. It is worth mentioning that discussing super-integrable evolution equations is also appealing and challenging.