# Connection between the symmetric discrete AKP system and bilinear ABS
lattice equations

Jing Wang1,2,3111Email<EMAIL_ADDRESS><EMAIL_ADDRESS>Da-jun Zhang1,2222Email<EMAIL_ADDRESS>Ken-ichi Maruno3333Email:
<EMAIL_ADDRESS>
1Department of Mathematics, Shanghai University, Shanghai 200444, P.R. China

2Newtouch Center for Mathematics of Shanghai University, Shanghai 200444, P.R.
China

3Department of Applied Mathematics, Faculty of Science and Engineering, Waseda
University,

Shinjuku-ku, Tokyo 169-8555, Japan

###### Abstract

In this paper, we show that all the bilinear Adler-Bobenko-Suris (ABS)
equations (except Q2 and Q4) can be obtained from symmetric discrete AKP
system by taking proper reductions and continuum limits. Among the bilinear
ABS equations, a simpler bilinear form of the ABS H2 equation is given. In
addition, an 8-point 3-dimensional lattice equation and an 8-point
4-dimensional lattice equation are obtained as by-products. Both of them can
be considered as extensions of the symmetric discrete AKP equation.

Keywords:

Adler-Bobenko-Suris equations, symmetric discrete AKP system, bilinear
equation, reduction, continuum limit

##  1 Introduction

There are intrinsic relations between integrable systems in different
dimensions. In discrete case, the discrete AKP (A-type Kadomtsev-Petviashvili)
equation [1, 2], also known as the Hirota equation or the Hirota-Miwa
equation, is fundamental in the sense that many lower dimensional (bilinear)
discrete equations, such as the discrete Korteweg-de Vries (dKdV), discrete
modified KdV (dmKdV), discrete sine-Gordon, discrete Boussinesq and Toda
lattice equations, can be reduced from it [3, 4, 5]. Note that the discrete
AKP (dAKP) equation is multi-dimensionally consistent [6]. This means the high
order dAKP equations of $d$-dimensions (see [7]) can be essentially considered
as combinations of the dAKP equations defined on any three directions in the
$d$-dimensional space. Another notable progress was made by [8] in which it
was shown that the dAKP $\tau$-function with reflection symmetry can be shared
by the discrete BKP (dBKP, or Miwa’s equation) equation as well, in alignment
with which the result is the so-called “KdV solves BKP” in continuous case
[9]. Note also that when the dAKP $\tau$-function allows reflection symmetry,
it is composed by the plane wave factors (PWFs) of the dKdV-type. All these
facts indicate that when implementing reductions of the dAKP equation, not
only the dAKP equation itself, but also its reflected forms and high order
equations (collectively referred to as the symmetric dAKP system) and also the
symmetric dBKP system can be utilized for reductions in getting the bilinear
dKdV-type equations.

The Adler-Bobenko-Suris (ABS) equations (see (2.4)) provide a complete
classification of multi-dimensionally consistent affine linear quadrilateral
models with certain additional conditions [10], which are known as H1, H2, H3,
A1, A2, Q1, Q2, Q3 and Q4. These equations have been solved in various ways
[11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22]. In fact, the PWFs of
solutions (except Q4 which is the elliptic case) and continuum limits [23, 24]
indicate that the ABS equations are of the dKdV-type. In this paper, we aim to
derive the ABS equations (except Q2 and Q4) in their bilinear forms from the
symmetric dAKP system. Since A1 is connected to Q1 by
$u\rightarrow(-1)^{n_{1}+n_{2}}u$ and A2 to $\rm{Q3}_{\delta=0}$ by
$u\rightarrow{u^{(-1)}}^{n_{1}+n_{2}}$, these two equations will not be
considered in this paper. Some bilinear forms for H1, H2, H3, Q1 and Q3 in the
ABS list involve more than one dependent variables (see [14, 22], while the
symmetric dAKP equations exclusively involve a single $\tau$-function. In this
paper, we aim to bridge the gap between the bilinear ABS equations and the
symmetric dAKP system.

The paper is organized as follows. Firstly, as preliminary we list necessary
notations and the ABS equations in section 2. Then, in section 3 we endeavor
to reconfigure the bilinear ABS equations using a single $\tau$-function by
introducing suitable auxiliary continuous variables, which means some
functions will be explained as derivatives of the $\tau$-function with respect
to the auxiliary variables. Correspondingly, the $\tau$-function will be
presented in both Casoratian and Wronskian forms. In section 4, we recall the
the symmetric dAKP system and their $\tau$-function. Then in section 5 we show
what continuum limits and reductions to apply on the symmetric dAKP system so
that all the bilinear ABS equations can be obtained from reductions. Besides,
as by-products, an 8-point 3-dimensional lattice equation and an 8-point
4-dimensional lattice equation associated with the symmetric dAKP system are
found. Appendix A contains more details about these two equations. Finally,
conclusions are given in section 6.

##  2 Mathematical preliminary

The functions of our interest in the fully discrete case are defined on
$\mathbb{Z}^{r}$, for example,

$\tau:=\tau((n_{1},a_{1}),(n_{2},a_{2}),\cdots,(n_{r},a_{r})),$ (2.1)

where $(n_{1},n_{2},\cdots,n_{r})\in\mathbb{Z}^{r}$ stand for the discrete
coordinates and $(a_{1},a_{2},\cdots,a_{r})$ serve as spacing parameters for
the corresponding directions. By $E_{n_{i}}$ we denote the forward shift
operator in $n_{i}$ direction, e.g.,

$E_{n_{i}}^{\pm j}\tau=\tau((n_{1},a_{1}),\cdots,(n_{i-1},a_{i-1}),(n_{i}\pm
j,a_{i}),(n_{i+1},a_{i+1}),\cdots,(n_{r},a_{r})).$

We also introduce numerical superscripts and subscripts to denote shifts in
different directions, say,

$\displaystyle\tau_{i}=E_{n_{i}}\tau,~{}~{}\tau^{i}=E_{n_{i}}^{-1}\tau,~{}~{}\tau_{ij}=E_{n_{j}}E_{n_{i}}\tau,~{}~{}\tau^{ij}=E_{n_{j}}^{-1}E_{n_{i}}^{-1}\tau,~{}~{}\tau_{i}^{j}=E_{n_{j}}^{-1}E_{n_{i}}\tau,$
(2.2a) $\displaystyle\tau_{\,\bar{i}}=E_{n_{1}}E_{n_{2}}\cdots
E_{n_{i-1}}E_{n_{i+1}}\cdots E_{n_{r}}\tau,~{}~{}~{}\cdots\cdots.$ (2.2b)

######  Remark 1.

To avoid any confusion, throughout the paper we only use such notations for
the functions denoted by $\tau$, $u$, $f,g$, $h$, $s$ and $v$.

With these settings, the (bilinear) dAKP equation [2] takes the form

$A\equiv(a_{1}-a_{2})\tau_{3}\tau_{12}+(a_{2}-a_{3})\tau_{1}\tau_{23}+(a_{3}-a_{1})\tau_{2}\tau_{13}=0,$
(2.3)

and the ABS equations are given as [10]

$\displaystyle\rm{H}1:~{}~{}$ $\displaystyle(u-u_{12})(u_{1}-u_{2})-p+q=0,$
(2.4a) $\displaystyle\mathrm{H}2:~{}~{}$
$\displaystyle(u-u_{12})(u_{1}-u_{2})-(p-q)(u+u_{1}+u_{2}+u_{12}+p+q)=0,$
(2.4b) $\displaystyle\mathrm{H}3:~{}~{}$ $\displaystyle
p(uu_{1}+u_{2}u_{12})-q(uu_{2}+u_{1}u_{12})+\delta(p^{2}-q^{2})=0,$ (2.4c)
$\displaystyle\rm{A1}:~{}~{}$ $\displaystyle
p(u+u_{2})(u_{1}+u_{12})-q(u+u_{1})(u_{2}+u_{12})-\delta^{2}pq(p-q)=0,$ (2.4d)
$\displaystyle\rm{A2}:~{}~{}$ $\displaystyle
p(1\\!-\\!q^{2})(uu_{1}+u_{2}u_{12})\\!-\\!q(1-p^{2})(uu_{2}+u_{1}u_{12})-(p^{2}-q^{2})(1+uu_{1}u_{2}u_{12})=0,$
(2.4e) $\displaystyle\mathrm{Q}1:~{}~{}$ $\displaystyle
p(u-u_{2})(u_{1}-u_{12})-q(u-u_{1})(u_{2}-u_{12})+\delta^{2}pq(p-q)=0,$ (2.4f)
$\displaystyle\mathrm{Q}2:~{}~{}$ $\displaystyle
p(u-u_{2})(u_{1}-u_{12})-q(u-u_{1})(u_{2}-u_{12})$
$\displaystyle+pq(p-q)(u+u_{1}+u_{2}+u_{12}-p^{2}+pq-q^{2})=0,$ (2.4g)
$\displaystyle\mathrm{Q}3:~{}~{}$ $\displaystyle
p(1-q^{2})(uu_{2}+u_{1}u_{12})-q(1-p^{2})(uu_{1}+u_{2}u_{12})$
$\displaystyle-\left(p^{2}-q^{2}\right)\left(u_{1}u_{2}+uu_{12}+\frac{\delta^{2}\left(1-p^{2}\right)\left(1-q^{2}\right)}{4pq}\right)=0,$
(2.4h) $\displaystyle\mathrm{Q}4:~{}~{}$
$\displaystyle\mathrm{sn}(p)(uu_{1}+u_{2}u_{12})-\mathrm{sn}(q)(uu_{2}+u_{1}u_{12})-\mathrm{sn}(p-q)(u_{1}u_{2}+uu_{12})$
$\displaystyle+\mathrm{sn}(p)\mathrm{sn}(q)\mathrm{sn}(p-q)(1+k^{2}uu_{1}u_{2}u_{12})=0,$
(2.4i)

where $u=u((n_{1},p),(n_{2},q))$ and the present form of Q4 is due to [25].

Next, we introduce the Wronskian and Casoratian. Define a $N$-th order column
vector

$\displaystyle\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T},$ (2.5)

where the entry

$\psi_{j}:=\psi_{j}(n_{1},n_{2},\cdots,n_{r},x_{1},x_{2},\cdots,x_{l})$ (2.6)

is a function defined on $\mathbb{Z}^{r}\times\mathbb{R}^{l}$, $\\{n_{j}\\}$
and $\\{x_{i}\\}$ serve as discrete and continuous independent variables
respectively. A Wronskian composed by the vector (2.5) and its derivatives
with respect to a certain variable $x_{i}$ is written as

$\displaystyle W(\psi_{1},\psi_{2},\cdots,\psi_{N})$
$\displaystyle:=\left\lvert\psi,\partial_{x_{i}}\psi,\cdots,\partial_{x_{i}}^{N-1}\psi\right\rvert\equiv\left\lvert
0,1,\cdots,N-1\right\rvert_{[x_{i}]}\equiv\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[x_{i}]},$
(2.7)

where by the subscript $[x_{i}]$ we specially denote that the Wronskian is
defined in terms of $x_{i}$. Note that we follow [26] to use the compact
notation $|\widehat{N-1}|$. Similarly, we also have
$\bigl{\lvert}\widehat{N-2},N\bigr{\rvert}_{[x_{i}]}=\left\lvert
0,1,\cdots,N-2,N\right\rvert_{[x_{i}]}$. Casoratians composed by the vector
(2.5) with respect to the shift in $n_{i}$-direction are given as

$\displaystyle C_{1}(\psi_{1},\psi_{2},\cdots,\psi_{N})$
$\displaystyle\\!:=\left\lvert\psi,E_{n_{i}}\psi,\cdots\\!,E_{n_{i}}^{N-1}\psi\right\rvert\equiv\left\lvert
0,1,\cdots,N-1\right\rvert_{[n_{i}]}\equiv\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[n_{i}]},$
$\displaystyle C_{2}(\psi_{1},\psi_{2},\cdots,\psi_{N})$
$\displaystyle:=\left\lvert\psi,E_{n_{i}}\psi,\cdots,E_{n_{i}}^{N-2}\psi,E_{n_{i}}^{N}\psi\right\rvert$
$\displaystyle\equiv\left\lvert
0,1,\cdots,N-2,N\right\rvert_{[n_{i}]}\equiv\bigl{\lvert}\widehat{N-2},N\bigr{\rvert}_{[n_{i}]},$
$\displaystyle C_{3}(\psi_{1},\psi_{2},\cdots,\psi_{N})$
$\displaystyle:=\left\lvert
E_{n_{i}}^{-1}\psi,E_{n_{i}}\psi,\cdots,E_{n_{i}}^{N-2}\psi,E_{n_{i}}^{N}\psi\right\rvert$
$\displaystyle\equiv\left\lvert-1,1,\cdots,N-2,N\right\rvert_{[n_{i}]}\equiv\bigl{\lvert}-1,\widetilde{N-2}\bigr{\rvert}_{[n_{i}]}.$

Note that for a generic notation
$\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[\lambda]}$, it can be either a
Wronskian or a Casoratian, depending on $\lambda=x_{i}$ or $\lambda=n_{i}$.

Finally, we look at bilinear derivatives and bilinear equations. Hirota’s
bilinear $D$-operator is defined as

$D_{x}^{n}D_{y}^{m}f(x,y)\cdot
g(x,y)=(\partial_{x}-\partial_{x^{\prime}})^{n}(\partial_{y}-\partial_{y^{\prime}})^{m}f(x,y)g(x^{\prime},y^{\prime})|_{x^{\prime}=x,y^{\prime}=y},$
(2.8)

where $f$ and $g$ are $\mathrm{C}^{\infty}$ functions with respect to $(x,y)$.
This definition can be easily extended to multidimensional cases, while (2.8)
serves as an illustration. One can introduce Hirota’s bilinear equation of the
form

$P(D_{x},D_{y})f(x,y)\cdot g(x,y)=0$ (2.9)

where $P(x,y)$ is a binary polynomial. Note that such an equation has the so-
called gauge property, say,

$P(D_{x},D_{y})e^{ax+by}f(x,y)\cdot
e^{ax+by}g(x,y)=e^{2(ax+by)}P(D_{x},D_{y})f(x,y)\cdot g(x,y)=0,$ (2.10)

where $a,b$ are constants. A discrete bilinear equation takes a form [14]

$\sum_{j=1}^{\sigma}c_{j}f^{(j)}(n_{1}+\nu^{(j)}_{1},\cdots,n_{r}+\nu^{(j)}_{r})g^{(j)}(n_{1}+\mu^{(j)}_{1},\cdots,n_{r}+\mu^{(j)}_{r})=0,$
(2.11)

where $\\{\nu_{i}^{(j)},\mu_{i}^{(j)}\\}$ are integers, and for each $i$,
subject to

$t_{i}=\nu_{i}^{(j)}+\mu_{i}^{(j)}=\nu_{i}^{(k)}+\mu_{i}^{(k)},~{}~{}\forall
j,k\in\\{1,2,\cdots,\sigma\\},$

which guarantees the gauge property in discrete cases [14]. In this paper, we
call $(t_{1},t_{2},\cdots,t_{\sigma})$ to be the _total index_ of the bilinear
equation (2.11).

##  3 Bilinear ABS equations in terms of single $\tau$-function

Some bilinear forms for H1, H2, H3, Q1 and Q3 in the ABS list involve more
than one dependent variables (see [14, 22], while the symmetric dAKP equations
exclusively involve a single $\tau$-function. In this section, we will revisit
known bilinear ABS equations and try to reconfigure them using a unified
$\tau$-function and its derivatives with respect to some auxiliary independent
variables.

###  3.1 H1

For the H1 equation (2.4a), after parameterizing $(p,q)$ by

$p=-a_{1}^{2},~{}~{}q=-a_{2}^{2}~{},$ (3.1)

we consider

$(u-u_{12})(u_{1}-u_{2})-(a_{2}^{2}-a_{1}^{2})=0.$ (3.2)

Introducing the transformation [14]

$u=Z-\frac{g}{f},~{}~{}Z=a_{1}n_{1}+a_{2}n_{2}+\gamma_{0},$ (3.3)

where $\gamma_{0}$ is an arbitrary constant, it then follows from (3.2) that
[14]

$\displaystyle g_{2}f_{1}-g_{1}f_{2}+(a_{1}-a_{2})(f_{1}f_{2}-ff_{12})=0,$
(3.4a) $\displaystyle gf_{12}-g_{12}f+(a_{1}+a_{2})(ff_{12}-f_{1}f_{2})=0.$
(3.4b)

The Casoratian solutions to the above bilinear H1 equation (3.4) are [14]

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[l_{1}]},~{}~{}g=\bigl{\lvert}\widehat{N-2},N\bigr{\rvert}_{[l_{1}]},$
(3.5)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with

$\displaystyle\psi_{j}(n_{1},n_{2},l_{1};k_{j})=\rho_{j}^{+}k_{j}^{l_{1}}(a_{1}+k_{j})^{n_{1}}(a_{2}+k_{j})^{n_{2}}+\rho_{j}^{-}(-k_{j})^{l_{1}}(a_{1}-k_{j})^{n_{1}}(a_{2}-k_{j})^{n_{2}}.$
(3.6)

To write the bilinear H1 equation (3.4) using a single $\tau$-function, due to
the arbitrariness of $\rho_{j}^{\pm}$, we can replace $\rho_{j}^{\pm}$ in
(3.6) with $\rho_{j}^{\pm}e^{\pm k_{j}x}$ and consider the following entry in
the vector $\psi$:

$\displaystyle\psi_{j}(n_{1},n_{2},x;k_{j})=\rho_{j}^{+}e^{k_{j}x}k_{j}^{l_{1}}(a_{1}+k_{j})^{n_{1}}(a_{2}+k_{j})^{n_{2}}+\rho_{j}^{-}e^{-k_{j}x}(-k_{j})^{l_{1}}(a_{1}-k_{j})^{n_{1}}(a_{2}-k_{j})^{n_{2}}.$
(3.7)

Then we have the following.

######  Lemma 1.

For the Casoratians $f$ and $g$ defined in (3.5) composed by $\psi$ with
entries (3.7), the following relations hold:

$\displaystyle
f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[l_{1}]}=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[x]}=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[n_{\sigma}]},~{}~{}(\sigma=1,2),$
(3.8a) $\displaystyle
g=\bigl{\lvert}\widehat{N-2},N\bigr{\rvert}_{[l_{1}]}=\bigl{\lvert}\widehat{N-2},N\bigr{\rvert}_{[x]}=f_{x}=\bigl{\lvert}\widehat{N-2},N\bigr{\rvert}_{[n_{\sigma}]}-a_{\sigma}Nf,~{}~{}(\sigma=1,2).$
(3.8b)

###### Proof.

From (3.7) we can find that

$\displaystyle
E_{l_{1}}\psi_{j}=\partial_{x}\psi_{j},~{}~{}E_{l_{1}}^{i}\psi_{j}=\partial_{x}^{i}\psi_{j},$
(3.9a) $\displaystyle
E_{l_{1}}\psi_{j}=(E_{n_{\sigma}}-a_{\sigma})\psi_{j},~{}~{}E_{l_{1}}^{i}\psi_{j}=(E_{n_{\sigma}}-a_{\sigma})^{i}\psi_{j},~{}~{}\sigma=1,2.$
(3.9b)

The relation (3.9a) immediately gives rise to
$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[x]}$. Using (3.9b) we have

$\displaystyle f$
$\displaystyle=\bigl{\lvert}\widehat{N-1}\bigr{|}_{[l_{1}]}=\bigl{\lvert}\psi,E_{l_{1}}\psi,\cdots,E_{l_{1}}^{N-1}\psi\bigr{\rvert}$
$\displaystyle=\left\lvert\psi,(E_{n_{\sigma}}-a_{\sigma})\psi,\cdots,(E_{n_{\sigma}}-a_{\sigma})^{N-1}\psi\right\rvert$
$\displaystyle=\bigl{\lvert}\psi,E_{n_{\sigma}}\psi,\cdots,E_{n_{\sigma}}^{N-1}\psi\bigr{\rvert}$
$\displaystyle=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[n_{\sigma}]},~{}~{}~{}(\sigma=1,2).$

Similarly, one can obtain (3.8b).

∎

This Lemma immediately indicates that the Casoratians

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[\lambda]},~{}~{}g=\bigl{\lvert}\widehat{N-2},N\bigr{\rvert}_{[\lambda]},~{}~{}(\lambda=n_{1},n_{2},x),$
(3.10)

composed by $\psi$ with entries (3.7) are solutions to the bilinear H1
equation (3.4). In addition, noting that $g=f_{x}$, we are led to the
following.

######  Theorem 1.

Through the transformation

$u=Z-\frac{f_{x}}{f},~{}~{}~{}Z=a_{1}n_{1}+a_{2}n_{2}+\gamma_{0},$ (3.11)

the H1 equation (3.2) allows the following bilinear form with a single
$\tau$-function $f$:

$\displaystyle D_{x}f_{2}\cdot f_{1}+(a_{1}-a_{2})(f_{1}f_{2}-ff_{12})=0,$
(3.12a) $\displaystyle D_{x}f\cdot
f_{12}+(a_{1}+a_{2})(ff_{12}-f_{1}f_{2})=0,$ (3.12b)

and its solutions are given by

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[\lambda]},~{}~{}(\lambda=n_{1},n_{2},x),$
(3.13)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with the entries
$\psi_{j}$ given in (3.7). Note that the terms $(\pm k_{j})^{l_{1}}$ in (3.7)
can be absorbed into $\rho_{j}^{\pm}$.

###  3.2 H2

H2 equation (2.4b) with the parametrization (3.1) is

$(u-u_{12})(u_{1}-u_{2})+(a_{1}^{2}-a_{2}^{2})(u+u_{1}+u_{2}+u_{12}-a_{1}^{2}-a_{2}^{2})=0.$
(3.14)

This equation allows a bilinearization with 5 equations (see equation (4.21)
in [14]), while in the following we provide a simpler form. By the
transformation

$u=Z^{2}-2Z\frac{g}{f}+\frac{h}{f},~{}~{}Z=a_{1}n_{1}+a_{2}n_{2}+\gamma_{0},$
(3.15)

where $\gamma_{0}$ is a constant, H2 equation (3.14) is bilinearized as444Note
that this bilinear form has been obtained in [24] using the connection between
H1 and H2. However, in this paper we find a direct formulation of the
bilinearization, see (3.17).

$\displaystyle\mathcal{H}_{1}\equiv
2a_{1}(f_{1}g-fg_{1})-2gg_{1}+h_{1}f+hf_{1}=0,$ (3.16a)
$\displaystyle\mathcal{H}_{2}\equiv
2a_{2}(f_{2}g-fg_{2})-2gg_{2}+h_{2}f+hf_{2}=0,$ (3.16b)

which are connected with (3.14) as

$\mathrm{H}_{2}=P_{1}\mathcal{H}_{1}+P_{2}\mathcal{H}_{2}+P_{3}\mathcal{\widehat{H}}_{1}+P_{4}\mathcal{\widetilde{H}}_{2}+P_{5}\mathcal{H}_{1}\mathcal{\widehat{H}}_{1}+P_{6}\mathcal{H}_{2}\mathcal{\widetilde{H}}_{2}$
(3.17)

with

$\displaystyle
P_{1}=\frac{2}{ff_{1}}\left(Z+a_{2}-\frac{g_{2}}{f_{2}}\right)\left(Z+a_{1}+a_{2}-\frac{g_{12}}{f_{12}}\right),$
$\displaystyle
P_{2}=\frac{-2}{ff_{2}}\left(Z+a_{1}-\frac{g_{1}}{f_{1}}\right)\left(Z+a_{1}+a_{2}-\frac{g_{12}}{f_{12}}\right),$
$\displaystyle
P_{3}=\frac{2}{ff_{1}}\left(Z-\frac{g}{f}\right)\left(Z+a_{1}-\frac{g_{1}}{f_{1}}\right),$
$\displaystyle
P_{4}=\frac{-2}{ff_{2}}\left(Z-\frac{g}{f}\right)\left(Z+a_{2}-\frac{g_{2}}{f_{2}}\right),$
$\displaystyle
P_{5}=\frac{1}{ff_{1}f_{2}f_{12}},~{}~{}~{}P_{6}=\frac{-1}{ff_{1}f_{2}f_{12}},$

where the $\mathcal{\widehat{H}}_{1}$ (or $\mathcal{\widetilde{H}}_{2}$)
denotes that a $n_{2}$-direction (or $n_{1}$-direction) forward shift on the
whole expression $\mathcal{H}_{1}$ (or $\mathcal{H}_{2}$).

As for solutions to (3.16), we have the following (see [24]).

######  Lemma 2.

The bilinear H2 equation (3.16) admits Casoratian solutions

$\displaystyle
f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[l_{1}]},~{}~{}g=\bigl{\lvert}\widehat{N-2},N\bigr{\rvert}_{[l_{1}]},~{}~{}h=\bigl{\lvert}\widehat{N-3},N-1,N\bigr{\rvert}_{[l_{1}]}+\bigl{\lvert}\widehat{N-2},N+1\bigr{\rvert}_{[l_{1}]},$
(3.18)

composed by $\psi=(\psi_{1},\psi_{2}\cdots,\psi_{N})^{T}$ with entries
$\psi_{j}$ given in (3.6).

Similar to the treatment to H1, we introduce $\psi_{j}$ defined by (3.7).
Then, it follows that the relations $g=f_{x}$ and $h=f_{xx}$ hold and we
obtain the following Theorem.

######  Theorem 2.

By the transformation

$u=Z^{2}-2Z\frac{f_{x}}{f}+\frac{f_{xx}}{f},~{}~{}Z=a_{1}n_{1}+a_{2}n_{2}+\gamma_{0},$
(3.19)

H2 equation (2.4b) allows the following bilinear form

$\displaystyle(D_{x}^{2}+2a_{1}D_{x})f\cdot f_{1}=0,$ (3.20a)
$\displaystyle(D_{x}^{2}+2a_{2}D_{x})f\cdot f_{2}=0,$ (3.20b)

which has a solution

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[\lambda]},~{}~{}(\lambda=n_{1},n_{2},x)$
(3.21)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with the entries
$\psi_{j}$ given in (3.7).

###  3.3 H3

For H3 equation (2.4c), i.e.,

$p(uu_{1}+u_{2}u_{12})-q(uu_{2}+u_{1}u_{12})-\delta(q^{2}-p^{2})=0,$ (3.22)

it allows two bilinear forms, both of which consist of a single
$\tau$-function [14]. With the parametrisation [14]

$p^{2}=\frac{r^{2}a_{3}^{2}}{a_{3}^{2}-a_{1}^{2}},~{}~{}q^{2}=\frac{r^{2}a_{3}^{2}}{a_{3}^{2}-a_{2}^{2}},~{}~{}\alpha^{2}=-\frac{a_{1}-a_{3}}{a_{1}+a_{3}},~{}~{}\beta^{2}=-\frac{a_{2}-a_{3}}{a_{2}+a_{3}},~{}~{}c_{1}c_{2}=-\frac{1}{4}r\delta$
(3.23)

and the transformation [14]

$u=c_{1}\alpha^{n_{1}}\beta^{n_{2}}\frac{f_{3}}{f}+c_{2}\alpha^{-n_{1}}\beta^{-n_{2}}\frac{f^{3}}{f}~{},$
(3.24)

H3 can be bilinearized as either [14]

$\displaystyle(a_{2}+a_{3})f_{12}f_{3}+(a_{1}-a_{3})ff_{123}-(a_{1}+a_{2})f_{1}f_{23}=0,$
(3.25a)
$\displaystyle(a_{3}-a_{2})f_{12}f^{3}-(a_{1}+a_{3})ff_{12}^{3}+(a_{1}+a_{2})f_{1}f_{2}^{3}=0,$
(3.25b) $\displaystyle
2a_{3}(a_{1}-a_{2})ff_{12}+(a_{3}-a_{1})(a_{2}+a_{3})f_{13}f_{2}^{3}+(a_{1}+a_{3})(a_{2}-a_{3})f_{23}f_{1}^{3}=0,$
(3.25c)

or

$\displaystyle
2a_{3}ff_{1}+(a_{1}-a_{3})f_{13}f^{3}-(a_{1}+a_{3})f_{3}f_{1}^{3}=0,$ (3.26a)
$\displaystyle
2a_{3}ff_{2}+(a_{2}-a_{3})f_{23}f^{3}-(a_{2}+a_{3})f_{3}f_{2}^{3}=0.$ (3.26b)

Both forms have a solution

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[\lambda]},~{}~{}(\lambda=n_{1},n_{2},n_{3})$
(3.27)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with entries

$\psi_{j}(n_{1},n_{2},n_{3};k_{j})=\rho_{j}^{+}\prod^{3}_{i=1}(a_{i}+k_{j})^{n_{i}}+\rho_{j}^{-}\prod^{3}_{i=1}(a_{i}-k_{j})^{n_{i}}.$
(3.28)

###  3.4 Q1

####  3.4.1 Q1 with power background

For Q1 equation (2.4f), i.e.,

$p(u-u_{2})(u_{1}-u_{12})-q(u-u_{12})(u_{2}-u_{12})-\delta^{2}pq(q-p)=0,$
(3.29)

after the parametrization [14]

$p=\frac{ra_{3}^{2}}{a_{1}^{2}-a_{3}^{2}},~{}~{}q=\frac{ra_{3}^{2}}{a_{2}^{2}-a_{3}^{2}},~{}~{}\alpha=\frac{a_{1}-a_{3}}{a_{1}+a_{3}},~{}~{}\beta=\frac{a_{2}-a_{3}}{a_{2}+a_{3}},~{}~{}c_{1}c_{2}=\frac{1}{16}r^{2}\delta^{2}$
(3.30)

and through the transformation

$u=c_{1}\alpha^{n_{1}}\beta^{n_{2}}\frac{f_{33}}{f}+c_{2}\alpha^{-n_{1}}\beta^{-n_{2}}\frac{f^{33}}{f}~{},$
(3.31)

it can be bilinearized as [14]

$\displaystyle
2a_{3}ff_{1}+(a_{1}-a_{3})f_{13}f^{3}-(a_{1}+a_{3})f_{3}f_{1}^{3}=0,$ (3.32a)
$\displaystyle
2a_{3}ff_{2}+(a_{2}-a_{3})f_{23}f^{3}-(a_{2}+a_{3})f_{3}f_{2}^{3}=0,$ (3.32b)

which is nothing but (3.26) and has a solution $f$ given in (3.27) composed by
$\psi$ with entries (3.28).

####  3.4.2 Q1 with linear background

Q1 equation (3.29) can have a linear background solution $u=\alpha n_{1}+\beta
n_{2}+\gamma_{0}$ under the parametrization [14]

$p=\frac{c^{2}/r-\delta^{2}r}{a_{1}^{2}-\delta^{2}},~{}~{}q=\frac{c^{2}/r-\delta^{2}r}{a_{2}^{2}-\delta^{2}},~{}~{}\alpha=pa_{1},~{}~{}\beta=qa_{2},$
(3.33)

where $c,r$ and $\gamma_{0}$ are all constants. Through the transformation

$u=\alpha n_{1}+\beta n_{2}+\gamma_{0}-(c^{2}/r-\delta^{2}r)\frac{g}{f},$
(3.34)

there exists the following bilinearization [14]

$\displaystyle\mathcal{Q}_{1}^{(1)}\equiv(a_{2}-\delta)f_{123}f+(a_{1}+\delta)f_{12}f_{3}-(a_{1}+a_{2})f_{13}f_{2}=0,$
(3.35a)
$\displaystyle\mathcal{Q}_{2}^{(1)}\equiv(a_{1}-a_{2})f_{123}f+(a_{2}+\delta)f_{13}f_{2}-(a_{1}+\delta)f_{23}f_{1}=0,$
(3.35b) $\displaystyle\mathcal{Q}_{3}^{(1)}\equiv
f_{1}f_{23}-f_{13}f_{2}+(\delta-
a_{1})f_{13}g_{2}+(a_{2}-\delta)f_{23}g_{1}+(a_{1}-a_{2})f_{3}g_{12}=0,$
(3.35c) $\displaystyle\mathcal{Q}_{4}^{(1)}\equiv
f_{23}f_{1}-f_{13}f_{2}-(a_{1}+\delta)f_{23}g_{1}+(a_{2}+\delta)f_{13}g_{2}+(a_{1}-a_{2})f_{123}g=0,$
(3.35d)

with solutions

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[n_{3}]},~{}~{}g=\bigl{\lvert}-1,\widetilde{N-1}\bigr{\rvert}_{[n_{3}]},$
(3.36)

composed by $\psi$ with entries

$\psi_{j}(n_{1},n_{2},n_{3};k_{j})=\rho_{j}^{+}(a_{1}+k_{j})^{n_{1}}(a_{2}+k_{j})^{n_{2}}(\delta+k_{j})^{n_{3}}+\rho_{j}^{-}(a_{1}-k_{j})^{n_{1}}(a_{2}-k_{j})^{n_{2}}(\delta-
k_{j})^{n_{3}}.$ (3.37)

In order to express the bilinear form using a single $\tau$-function, we first
introduce

$\psi_{j}(n_{1},n_{2},n_{3},y;k_{j})=\,\rho_{j}^{+}e^{\frac{y}{a_{3}+k_{j}}}\prod^{3}_{i=1}(a_{i}+k_{j})^{n_{i}}+\rho_{j}^{-}e^{\frac{y}{a_{3}-k_{j}}}\prod^{3}_{i=1}(a_{i}-k_{j})^{n_{i}}$
(3.38)

with $a_{3}=\delta$. Thus we have $g=f_{y}$ and then we can rewrite
$\mathcal{Q}_{3}^{(1)}$ and $\mathcal{Q}_{4}^{(1)}$ as

$\displaystyle\mathcal{Q}_{3}^{(2)}\equiv-
f_{13}f_{2}+f_{1}f_{23}+(a_{3}-a_{1})f_{13}f_{2,y}+(a_{2}-a_{3})f_{23}f_{1,y}+(a_{1}-a_{2})f_{3}f_{12,y}=0,$
(3.39a) $\displaystyle\mathcal{Q}_{4}^{(2)}\equiv
f_{23}f_{1}-f_{13}f_{2}-(a_{1}+a_{3})f_{23}f_{1,y}+(a_{2}+a_{3})f_{13}f_{2,y}+(a_{1}-a_{2})f_{123}f_{y}=0.$
(3.39b)

Note that these two equations are not ready to be written in terms of Hirota’s
$D$-operator. However, using the known bilinear equations (see (2.3))

$\mathcal{Q}_{0}\equiv
A\equiv(a_{1}-a_{2})f_{3}f_{12}+(a_{2}-a_{3})f_{1}f_{23}+(a_{3}-a_{1})f_{2}f_{13}=0,$
(3.40)

we have the following relations

$\mathcal{Q}_{3}=2\mathcal{Q}_{3}^{(2)}-\mathcal{Q}_{0,y}=0,~{}~{}\mathcal{Q}_{4}=2\mathcal{Q}_{4}^{(2)}-\mathcal{Q}^{(1)}_{2,y}=0,$
(3.41)

which give rise to the expected form

$\displaystyle\\!2f_{1}f_{23}-\\!2f_{13}f_{2}\\!+\\!(a_{3}-a_{1})D_{y}f_{2}\cdot
f_{13}\\!+\\!(a_{2}-a_{3})D_{y}f_{1}\cdot
f_{23}\\!-\\!(a_{1}-a_{2})D_{y}f_{3}\cdot f_{12}\\!=0,$ (3.42a) $\displaystyle
2f_{23}f_{1}-2f_{13}f_{2}\\!-\\!(a_{1}+a_{3})D_{y}f_{1}\cdot
f_{23}\\!+\\!(a_{2}+a_{3})D_{y}f_{2}\cdot
f_{13}\\!+\\!(a_{1}-a_{2})D_{y}f\cdot f_{123}=0.$ (3.42b)

Thus we arrive at the following.

######  Theorem 3.

For Q1 equation (3.29) with parametrization (3.33), through the transformation

$u=\alpha n_{1}+\beta n_{2}+\gamma_{0}-(c^{2}/r-\delta^{2}r)\frac{f_{y}}{f},$
(3.43)

it has bilinear equations (with $a_{3}=\delta$)

$\displaystyle\mathcal{Q}_{0}\equiv(a_{1}-a_{2})f_{3}f_{12}+(a_{2}-a_{3})f_{1}f_{23}+(a_{3}-a_{1})f_{2}f_{13}=0,$
(3.44a)
$\displaystyle\mathcal{Q}_{1}\equiv(a_{2}-a_{3})f_{123}f+(a_{1}+a_{3})f_{12}f_{3}-(a_{1}+a_{2})f_{13}f_{2}=0,$
(3.44b)
$\displaystyle\mathcal{Q}_{2}\equiv(a_{1}-a_{2})f_{123}f+(a_{2}+a_{3})f_{13}f_{2}-(a_{1}+a_{3})f_{23}f_{1}=0,$
(3.44c) $\displaystyle\mathcal{Q}_{3}\equiv
2f_{1}f_{23}\\!-\\!2f_{13}f_{2}\\!+\\!(a_{3}\\!-\\!a_{1})D_{y}f_{2}\\!\cdot\\!f_{13}\\!+\\!(a_{2}\\!-\\!a_{3})D_{y}f_{1}\\!\cdot\\!f_{23}\\!-\\!(a_{1}\\!-\\!a_{2})D_{y}f_{3}\\!\cdot\\!f_{12}=0,$
(3.44d) $\displaystyle\mathcal{Q}_{4}\equiv
2f_{23}f_{1}\\!-\\!2f_{13}f_{2}\\!-\\!(a_{1}\\!+\\!a_{3})D_{y}f_{1}\\!\cdot\\!f_{23}\\!+\\!(a_{2}\\!+\\!a_{3})D_{y}f_{2}\\!\cdot\\!f_{13}\\!+\\!(a_{1}\\!-\\!a_{2})D_{y}f\\!\cdot\\!f_{123}=0,$
(3.44e)

which allow a solution

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[\lambda]},~{}~{}(\lambda=n_{1},n_{2},n_{3}),$
(3.45)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi^{N})^{T}$ with entries
(3.38).

###  3.5 Q3

For Q3 equation (2.4h), via the parametrization

$p^{2}=\frac{a_{1}^{2}-a_{4}^{2}}{a_{1}^{2}-a_{3}^{2}},\quad
q^{2}=\frac{a_{2}^{2}-a_{4}^{2}}{a_{2}^{2}-a_{3}^{2}},\quad
P=\frac{\left(a_{4}^{2}-a_{3}^{2}\right)p}{1-p^{2}},\quad
Q=\frac{\left(a_{4}^{2}-a_{3}^{2}\right)q}{1-q^{2}},$ (3.46)

and replacing $u$ with $(a_{3}^{2}-a_{4}^{4})u$, the Q3 in (2.4h) is written
as [12, 13]

$P(uu_{2}+u_{1}u_{12})-Q(uu_{1}+u_{2}u_{12})-\left(a_{1}^{2}-a_{2}^{2}\right)(u_{2}u_{1}+uu_{12})=\delta^{2}\frac{\left(a_{1}^{2}-a_{2}^{2}\right)}{4PQ}.$
(3.47)

Solutions to this equation can be given through those of the Nijhoff-Quispel-
Capel (NQC) equation, which reads [28]

$\\!\\!\\!\\!\\!\\!\frac{1\\!+\\!(a_{1}\\!-\\!a_{3})s(a_{3},a_{4})\\!-\\!(a_{1}\\!+\\!a_{4})s_{1}(a_{3},a_{4})}{1\\!+\\!(a_{2}\\!-\\!a_{3})s(a_{3},a_{4})\\!-\\!(a_{2}\\!+\\!a_{4})s_{2}(a_{3},a_{4})}\\!=\\!\frac{1\\!+\\!(a_{2}\\!-\\!a_{4})s_{1}(a_{3},a_{4})\\!-\\!(a_{2}\\!+\\!a_{3})s_{12}(a_{3},a_{4})}{1\\!+\\!(a_{1}\\!-\\!a_{4})s_{2}(a_{3},a_{4})\\!-\\!(a_{1}\\!+\\!a_{3})s_{12}(a_{3},a_{4})},$
(3.48)

where $a_{3}$ and $a_{4}$ are two extra parameters, $s(a_{3},a_{4})$ is a
function of $(n_{1},n_{2})$ and it holds specially that

$s(a_{3},a_{4})=s(a_{4},a_{3}).$ (3.49)

Note that here and below in (3.53) $s_{1},s_{2},s_{12},v_{1},v_{2}$, etc. are
defined as in (2.2), see Remark 1. One can also consider that $s(a_{3},a_{4})$
is a function of $(n_{1},n_{2},n_{3},n_{4})$ with $(a_{1},a_{2},a_{3},a_{4})$
as spacing parameters, while $n_{3}$ and $n_{4}$ are dummy variables that do
not appear in the NQC equation (3.48). Solution of the Q3 equation (3.47) can
be expressed as follows [13]

$\displaystyle u=$ $\displaystyle
c_{1}F(a_{3},a_{4})[1-(a_{3}+a_{4})s(a_{3},a_{4})]+c_{2}F(a_{3},-a_{4})[1-(a_{3}-a_{4})s(a_{3},-a_{4})]$
(3.50)
$\displaystyle+c_{3}F(-a_{3},a_{4})[1\\!+\\!(a_{3}\\!-\\!a_{4})s(-a_{3},a_{4})]+c_{4}F(-a_{3},-a_{4})[1+(a_{3}+a_{4})s(-a_{3},-a_{4})],$

where the function $F(a_{3},a_{4})$ is defined as

$F(a_{3},a_{4})=\left(\frac{(a_{1}+a_{3})(a_{1}+a_{4})}{(a_{1}-a_{3})(a_{1}-a_{4})}\right)^{\frac{n_{1}}{2}}\left(\frac{(a_{2}+a_{3})(a_{2}+a_{4})}{(a_{2}-a_{3})(a_{2}-a_{4})}\right)^{\frac{n_{2}}{2}}$
(3.51)

and the constants $c_{i},~{}i=1\cdots 4$, subject to

$c_{1}c_{4}(a_{3}+a_{4})^{2}-c_{2}c_{3}(a_{3}-a_{4})^{2}=-\frac{\delta^{2}}{16a_{3}a_{4}}.$
(3.52)

Thus, the bilinear form of Q3 can be obtained from the bilinearization of the
NQC equation (3.48) [22].

Note that the NQC equation can be derived from the following equation set [28,
13, 19, 29]

$\displaystyle
1+(a_{1}-a_{3})s(a_{3},a_{4})-(a_{1}+a_{4})s_{1}(a_{3},a_{4})=v_{1}(a_{3})v(a_{4}),$
(3.53a) $\displaystyle
1+(a_{2}-a_{3})s(a_{3},a_{4})-(a_{2}+a_{4})s_{2}(a_{3},a_{4})=v_{2}(a_{3})v(a_{4}),$
(3.53b) $\displaystyle
1+(a_{1}-a_{4})s(a_{3},a_{4})-(a_{1}+a_{3})s_{1}(a_{3},a_{4})=v_{1}(a_{4})v(a_{3}),$
(3.53c) $\displaystyle
1+(a_{2}-a_{4})s(a_{3},a_{4})-(a_{2}+a_{3})s_{2}(a_{3},a_{4})=v_{2}(a_{4})v(a_{3}),$
(3.53d)

together with symmetry (3.49) and assumption that $v(a_{3})$ is a function
related to $a_{3}$ but independent of $a_{4}$, i.e.,
$v(a_{4})=v(a_{3})|_{a_{3}\rightarrow a_{4}}$. One can bilinearize the above
system via the transformation

$s(a_{3},a_{4})=\frac{1}{a_{3}+a_{4}}\biggl{(}1-\frac{f^{34}}{f}\biggr{)}\biggr{|}_{n_{3}=n_{4}=0},~{}~{}v(a_{3})=\frac{f^{3}}{f}\biggr{|}_{n_{3}=n_{4}=0},~{}~{}v(a_{4})=\frac{f^{4}}{f}\biggr{|}_{n_{3}=n_{4}=0},$
(3.54)

and get the bilinear form of the NQC equation as well as of the Q3 equation:
(cf.[22])

$\displaystyle(a_{1}-a_{3})ff_{134}-(a_{1}+a_{4})f_{1}f_{34}+(a_{3}+a_{4})f_{14}f_{3}=0,$
(3.55a)
$\displaystyle(a_{2}-a_{3})ff_{234}-(a_{2}+a_{4})f_{2}f_{34}+(a_{3}+a_{4})f_{24}f_{3}=0,$
(3.55b)
$\displaystyle(a_{2}-a_{4})ff_{234}-(a_{2}+a_{3})f_{2}f_{34}+(a_{3}+a_{4})f_{4}f_{23}=0,$
(3.55c)
$\displaystyle(a_{1}-a_{4})ff_{134}-(a_{1}+a_{3})f_{1}f_{34}+(a_{3}+a_{4})f_{4}f_{13}=0.$
(3.55d)

This bilinear system involves only the function $f$ and admits the solution in
Casoratian form [22]:

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[\lambda]},~{}~{}(\lambda=n_{1},n_{2},n_{3},n_{4}),$
(3.56)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with entries

$\psi_{j}(n_{1},n_{2},n_{3},n_{4};k_{j})=\rho_{j}^{+}\prod^{4}_{i=1}(a_{i}+k_{j})^{n_{i}}+\rho_{j}^{-}\prod^{4}_{i=1}(a_{i}-k_{j})^{n_{i}}.$
(3.57)

One can check that the above defined $s$ and $v$ satisfy (3.49) and
$v(a_{4})=v(a_{3})|_{a_{3}\rightarrow a_{4}}$.

In conclusion, we have the following for Q3.

######  Theorem 4.

Q3 equation (2.4h) can be bilinearized as the system (3.55) via the
transformation

$\displaystyle
u=\left[c_{1}F(a_{3},a_{4})\frac{f^{34}}{f}+c_{2}F(a_{3},-a_{4})\frac{f_{4}^{3}}{f}+c_{3}F(-a_{3},a_{4})\frac{f_{3}^{4}}{f}+c_{4}F(-a_{3},-a_{4})\frac{f_{34}}{f}\right]\biggr{|}_{n_{3}=n_{4}=0},$
(3.58)

where $F$ is defined as (3.51) and the constants $c_{1},c_{2},c_{3},c_{4}$ are
subject to (3.52). The bilinear form allows the Casoratian solution (3.56)
composed by $\psi$ with entries (3.57).

###  3.6 Universal $\tau$-function for the bilinear ABS

Let us quickly summarize this section about the universal $\tau$-function for
the bilinear ABS equations.

######  Theorem 5.

The bilinear ABS equations, including (3.12), (3.20), (3.25), (3.26), (3.32),
(3.44) and (3.55), share the same single $\tau$-function

$f=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[\lambda]},~{}~{}(\lambda=n_{1},n_{2},n_{3},n_{4},x)$
(3.59)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with universal
entries

$\\!\\!\psi_{j}(n_{1},n_{2},n_{3},n_{4},x,y;k_{j})=\rho_{j}^{+}e^{k_{j}x+\frac{y}{a_{3}+k_{j}}}\prod^{4}_{i=1}(a_{i}+k_{j})^{n_{i}}+\rho_{j}^{-}e^{-k_{j}x+\frac{y}{a_{3}-k_{j}}}\prod^{4}_{i=1}(a_{i}-k_{j})^{n_{i}}.$
(3.60)

##  4 Symmetric dAKP system

In the previous section we have expressed the bilinear forms of H1, H2, H3, Q1
and Q3 using a single $\tau$-function in Wronskian-Casoratian form,
$f=\big{|}\widehat{N-1}\bigr{|}_{[\lambda]}$, composed by a universal vector
$\psi$ (see Theorem 5). To explain how those bilinear ABS equations are
related to the dAKP equation and its deformations, in the following we will
have a close look at the dAKP equations and its reflected forms, which will be
used in reduction.

Consider the following (high order dAKP) compact form [7]

$\left|\begin{matrix}1&a_{1}&a_{1}^{2}&\cdots&a_{1}^{\sigma-2}&\tau_{1}\tau_{\bar{1}}\\\
1&a_{2}&a_{2}^{2}&\cdots&a_{2}^{\sigma-2}&\tau_{2}\tau_{\bar{2}}\\\
\vdots&\vdots&\vdots&&\vdots&\vdots\\\
1&a_{\sigma}&a_{\sigma}^{2}&\cdots&a_{\sigma}^{\sigma-2}&\tau_{\sigma}\tau_{\bar{\sigma}}\\\
\end{matrix}\right|=0,~{}~{}(\mathrm{for}~{}\sigma=3,4,5),$ (4.1)

where the notation $\tau_{\bar{\sigma}}$ can be referred from (2.2) in Sec.2.
When $\sigma=3$, it is the dAKP equation (2.3), i.e.,

$A\equiv(a_{1}-a_{2})\tau_{3}\tau_{12}+(a_{2}-a_{3})\tau_{1}\tau_{23}+(a_{3}-a_{1})\tau_{2}\tau_{13}=0,$
(4.2)

and when $\sigma=4$, it gives the 4D dAKP equation

$\displaystyle\mathfrak{A}\equiv$
$\displaystyle(a_{2}-a_{3})(a_{3}-a_{4})(a_{2}-a_{4})\tau_{1}\tau_{234}-(a_{1}-a_{3})(a_{3}-a_{4})(a_{1}-a_{4})\tau_{2}\tau_{134}$
$\displaystyle+(a_{1}-a_{2})(a_{2}-a_{4})(a_{1}-a_{4})\tau_{3}\tau_{124}-(a_{1}-a_{2})(a_{2}-a_{3})(a_{1}-a_{3})\tau_{4}\tau_{123}=0,$
(4.3)

which can be considered as a result of the 4D consistency of the dAKP equation
(4.2) [8]. The 5D dAKP equation is the case for $\sigma=5$, which we do not
write out here the expanded form. Note that the dAKP equation is 4D consistent
[6], and we believe the 5D dAKP equation is also a result of the
multidimensional consistency of the dAKP equation.

Solutions of (4.1) can be given either in Hirota’s exponential-polynomial form

$\tau=\sum_{\mu=0,1}\exp\left[\sum_{j=1}^{N}\mu_{j}\eta_{j}+\sum_{1\leqslant
i<j}^{N}\mu_{i}\mu_{j}a_{ij}\right]$ (4.4)

with the PWF and phase factor

$\rho=\mathrm{e}^{\eta_{j}}=\prod_{r=1}^{\sigma}\left(\frac{a_{r}-q_{j}}{a_{r}-k_{j}}\right)^{n_{r}}\eta_{j}^{(0)},~{}~{}\quad
A_{ij}=e^{a_{ij}}=\frac{\left(k_{i}-k_{j}\right)\left(q_{i}-q_{j}\right)}{\left(k_{i}-q_{j}\right)\left(q_{i}-k_{j}\right)},~{}~{}~{}\sigma=3,4,5,$
(4.5)

where the summation over $\mu$ means to take all possible
$\mu_{j}=0,1~{}(j=1,2,\cdots,N)$, or in Casoratian form

$\tau=\left\lvert\widehat{N-1}\right\rvert_{[\lambda]},~{}~{}\lambda=n_{1},n_{2},\cdots,n_{\sigma},~{}~{}\sigma=3,4,5$
(4.6)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with entries

$\psi_{j}(n_{1},n_{2},\cdots,n_{\sigma},;k_{j},q_{j})=\rho_{j}^{+}\prod_{r=1}^{\sigma}(a_{r}-q_{j})^{n_{r}}+\rho_{j}^{-}\prod_{r=1}^{\sigma}(a_{r}-k_{j})^{n_{r}},~{}~{}~{}~{}\sigma=3,4,5.$
(4.7)

To approach to the bilinear ABS equations, we impose constraints

$q_{j}=-k_{j}$ (4.8)

for $j=1,2,\cdots,N$, which yield reflection symmetry for the above
$\tau$-function [8]:

$\displaystyle\tau((n_{1},a_{1}),n_{2},\cdots,n_{\sigma})=\tau((-n_{1},-a_{1}),n_{2},\cdots,n_{\sigma})=S_{1}\tau,$
(4.9a)
$\displaystyle\tau(n_{1},(n_{2},a_{2}),\cdots,n_{\sigma})=\tau(n_{1},(-n_{2},-a_{2}),\cdots,n_{\sigma})=S_{2}\tau,$
(4.9b)
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\vdots$
$\displaystyle\tau(n_{1},n_{2},\cdots,(n_{\sigma},a_{\sigma}))=\tau(n_{1},n_{2},\cdots,(-n_{\sigma},-a_{\sigma}))=S_{\sigma}\tau,$
(4.9c)

for $\sigma=3,4,5$, where $S_{i}$ stands for the operator to reflect
$(n_{1},a_{1})$ to $(-n_{1},-a_{1})$. Such symmetries then lead us to the
reflected forms of the original equations555 Note that the symmetries can be
extended to more general cases, see Remark 2 in [8]. For example, when
$\sigma=3$ and $\sigma=4$, the explicit reflected forms of the dAKP eqauation
(4.2) and 4D dAKP equation (4) are respectively

$\displaystyle
A_{1}\equiv(a_{2}-a_{3})\tau\tau_{123}+(a_{3}+a_{1})\tau_{3}\tau_{12}-(a_{1}+a_{2})\tau_{2}\tau_{13}=0,$
(4.10a) $\displaystyle
A_{2}\equiv(a_{3}-a_{1})\tau\tau_{123}+(a_{1}+a_{2})\tau_{1}\tau_{23}-(a_{2}+a_{3})\tau_{3}\tau_{12}=0,$
(4.10b) $\displaystyle
A_{3}\equiv(a_{1}-a_{2})\tau\tau_{123}+(a_{2}+a_{3})\tau_{2}\tau_{13}-(a_{3}+a_{1})\tau_{1}\tau_{23}=0,$
(4.10c)

and

$\displaystyle\mathfrak{A}_{1}\equiv$
$\displaystyle(a_{2}-a_{3})(a_{3}-a_{4})(a_{2}-a_{4})\tau\tau_{1234}-(a_{1}+a_{3})(a_{3}-a_{4})(a_{1}+a_{4})\tau_{12}\tau_{34}$
$\displaystyle+(a_{1}+a_{2})(a_{2}-a_{4})(a_{1}+a_{4})\tau_{13}\tau_{24}-(a_{1}+a_{2})(a_{2}-a_{3})(a_{1}+a_{3})\tau_{14}\tau_{23}=0,$
(4.11a) $\displaystyle\mathfrak{A}_{2}\equiv$
$\displaystyle(a_{2}+a_{3})(a_{3}-a_{4})(a_{2}+a_{4})\tau_{12}\tau_{34}-(a_{1}-a_{3})(a_{3}-a_{4})(a_{1}-a_{4})\tau\tau_{1234}$
$\displaystyle-(a_{1}+a_{2})(a_{2}+a_{4})(a_{1}-a_{4})\tau_{23}\tau_{14}+(a_{1}+a_{2})(a_{2}+a_{3})(a_{1}-a_{3})\tau_{24}\tau_{13}=0,$
(4.11b) $\displaystyle\mathfrak{A}_{3}\equiv$
$\displaystyle-(a_{2}+a_{3})(a_{3}+a_{4})(a_{2}-a_{4})\tau_{13}\tau_{24}+(a_{1}+a_{3})(a_{3}+a_{4})(a_{1}-a_{4})\tau_{23}\tau_{14}$
$\displaystyle+(a_{1}-a_{2})(a_{2}-a_{4})(a_{1}-a_{4})\tau\tau_{1234}-(a_{1}-a_{2})(a_{2}+a_{3})(a_{1}+a_{3})\tau_{34}\tau_{12}=0,$
(4.11c) $\displaystyle\mathfrak{A}_{4}\equiv$
$\displaystyle(a_{2}-a_{3})(a_{3}+a_{4})(a_{2}+a_{4})\tau_{14}\tau_{23}-(a_{1}-a_{3})(a_{3}+a_{4})(a_{1}+a_{4})\tau_{24}\tau_{13}$
$\displaystyle+(a_{1}-a_{2})(a_{2}+a_{4})(a_{1}+a_{4})\tau_{34}\tau_{12}-(a_{1}-a_{2})(a_{2}-a_{3})(a_{1}-a_{3})\tau\tau_{1234}=0.$
(4.11d)

There are also reflected forms for the 5D dAKP equation. Thus, the following
results can be obtained.

######  Theorem 6.

The so-called symmetric dAKP system, consisting of the dAKP equation (4.2), 4D
dAKP equation (4), 5D dAKP equation and their reflected forms, share the same
soliton solutions in Hirota’s exponential-polynomial form

$\tau=\sum_{\mu=0,1}\exp\left[\sum_{j=1}^{N}\mu_{j}\eta_{j}+\sum_{1\leqslant
i<j}^{N}\mu_{i}\mu_{j}a_{ij}\right]$ (4.12)

with PWF and phase factor

$\rho=\mathrm{e}^{\eta_{j}}=\prod_{r=1}^{\sigma}\left(\frac{a_{r}+k_{j}}{a_{r}-k_{j}}\right)^{n_{r}}\eta_{j}^{(0)},~{}~{}\quad
A_{ij}=e^{a_{ij}}=\frac{\left(k_{i}-k_{j}\right)^{2}}{\left(k_{i}+k_{j}\right)^{2}},~{}~{}~{}\sigma=3,4,5,$
(4.13)

or in Casoratian form

$\tau=\big{\lvert}\widehat{N-1}\big{\rvert}_{[\lambda]},~{}~{}\lambda=n_{1},n_{2},\cdots,n_{\sigma}~{},~{}~{}\sigma=3,4,5$
(4.14)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with entries

$\psi_{j}(n_{1},n_{2},\cdots,n_{\sigma},;k_{j})=\rho_{j}^{+}\prod_{r=1}^{\sigma}(a_{r}+k_{j})^{n_{r}}+\rho_{j}^{-}\prod_{r=1}^{\sigma}(a_{r}-k_{j})^{n_{r}},~{}~{}~{}~{}\sigma=3,4,5.$
(4.15)

######  Remark 2.

Under the constraint condition (4.8), besides the symmetries (4.9), the
$\tau$-function allows some other symmetries as well. For example, when
$\sigma=4$, there is a symmetry

$\tau=S_{2}\circ S_{1}\tau,$

which, from the 4D AKP equation (4), leads to a reflected 4D dAKP equation

$\displaystyle\mathfrak{A}_{5}\equiv$
$\displaystyle(a_{2}+a_{3})(a_{3}-a_{4})(a_{2}+a_{4})\tau_{2}\tau_{134}-(a_{1}+a_{3})(a_{3}-a_{4})(a_{1}+a_{4})\tau_{1}\tau_{234}$
$\displaystyle-(a_{1}-a_{2})(a_{2}+a_{4})(a_{1}+a_{4})\tau_{123}\tau_{4}+(a_{1}-a_{2})(a_{2}+a_{3})(a_{1}+a_{3})\tau_{124}\tau_{3}=0.$
(4.16a) There are also symmetries $\tau=S_{3}\circ
S_{1}\tau,~{}~{}\tau=S_{4}\circ S_{1}\tau,$ which give rise to the following
reflected 4D dAKP equations, respectively,
$\displaystyle\mathfrak{A}_{6}\equiv$
$\displaystyle-(a_{2}+a_{3})(a_{3}+a_{4})(a_{2}-a_{4})\tau_{3}\tau_{124}+(a_{1}-a_{3})(a_{3}+a_{4})(a_{1}+a_{4})\tau_{123}\tau_{4}$
$\displaystyle+(a_{1}+a_{2})(a_{2}-a_{4})(a_{1}+a_{4})\tau_{1}\tau_{234}-(a_{1}+a_{2})(a_{2}+a_{3})(a_{1}-a_{3})\tau_{134}\tau_{2}=0,$
(4.16b) $\displaystyle\mathfrak{A}_{7}\equiv$
$\displaystyle(a_{2}-a_{3})(a_{3}+a_{4})(a_{2}+a_{4})\tau_{4}\tau_{123}-(a_{1}+a_{3})(a_{3}+a_{4})(a_{1}-a_{4})\tau_{124}\tau_{3}$
$\displaystyle+(a_{1}+a_{2})(a_{2}+a_{4})(a_{1}-a_{4})\tau_{134}\tau_{2}-(a_{1}+a_{2})(a_{2}-a_{3})(a_{1}+a_{3})\tau_{1}\tau_{234}=0,$
(4.16c)

and these equations allow the same soliton solution as those given in Theorem
6. Equations in (4.16) also belong to the symmetric dAKP system which will
generate bilinear ABS equations in reduction.

##  5 Reductions to the bilinear ABS

In this section, we explain how the bilinear ABS equations are connected to
the symmetric dAKP system via reductions or continuum limits.

###  5.1 Bilinear H1

Equation (3.12a) in the bilinear H1 (3.12) is nothing but the bilinear
differential-difference AKP equation with one continuous independent variable
(see case (N-2) in [30]), which is a result of a continuum limit of the dAKP
equation (4.2). In fact, let

$a_{3},n_{3}\rightarrow\infty~{}~{}~{}\text{while
keeping}~{}x=n_{3}/a_{3}~{}~{}\text{finite},$ (5.1)

and still write $\tau(n_{1},n_{2},n_{3})$ as $\tau(n_{1},n_{2},x)$ without
making confusion, i.e.,

$\tau=\tau(n_{1},n_{2},n_{3}):=\tau(n_{1},n_{2},x),$ (5.2)

it then implies that

$\tau_{3}=\tau(n_{1},n_{2},n_{3}+1)=\tau\Bigl{(}n_{1},n_{2},x+\frac{1}{a_{3}}\Bigr{)}=\tau+\frac{1}{a_{3}}\partial_{x}\tau+\frac{1}{2a_{3}^{2}}\partial_{x}^{2}\tau+\cdots.$
(5.3)

Inserting this Taylor expansion into equation (4.2), from the leading term we
get

$(a_{1}-a_{2})(\tau\tau_{12}-\tau_{1}\tau_{2})+D_{x}\tau_{1}\cdot\tau_{2}=0,$
(5.4a) which is (3.12a). Starting from the reflected dAKP equation (4.10b) and
implementing the same continuum limit yields
$(a_{1}+a_{2})(\tau\tau_{12}-\tau_{1}\tau_{2})+D_{x}\tau\cdot\tau_{12}=0,$
(5.4b) i.e., (3.12b).

Thus, the bilinear H1 (3.12) consists of two differential-difference AKP
equations, which are the continuum limits of the dAKP equation (4.2) and its
reflected form (4.10b).

As for the $\tau$-function of (5.4) in the form (4.12) or (4.14), considering
the continuum limit of (4.13) and (4.15) with $\sigma=3$, we have

$\rho=e^{2k_{j}x}\left(\frac{a_{1}+k_{j}}{a_{1}-k_{j}}\right)^{n_{1}}\left(\frac{a_{2}+k_{j}}{a_{2}-k_{j}}\right)^{n_{2}}\eta_{j}^{(0)}$
(5.5)

and

$\displaystyle\psi_{j}(n_{1},n_{2},x;k_{j})$
$\displaystyle=\rho_{j}^{+}e^{k_{j}x}(a_{1}+k_{j})^{n_{1}}(a_{2}+k_{j})^{n_{2}}+\rho_{j}^{-}e^{-k_{j}x}(a_{1}-k_{j})^{n_{1}}(a_{2}-k_{j})^{n_{2}}.$
(5.6)

###  5.2 Bilinear H2

We start from the following two reflected equations in the dAKP system:

$\displaystyle(a_{3}-a_{1})\tau\tau_{134}+(a_{1}+a_{4})\tau_{1}\tau_{34}-(a_{4}+a_{3})\tau_{3}\tau_{14}=0,$
(5.7a)
$\displaystyle(a_{3}-a_{2})\tau\tau_{234}+(a_{2}+a_{4})\tau_{2}\tau_{34}-(a_{4}+a_{3})\tau_{3}\tau_{24}=0,$
(5.7b)

where their common PWF is

$\rho=\eta_{j}^{(0)}\prod^{4}_{i=1}\left(\frac{a_{i}+k_{j}}{a_{i}-k_{j}}\right)^{n_{i}},$
(5.8)

and the common Casoratian entries are

$\displaystyle\psi_{j}(n_{1},n_{2},n_{3},n_{4};k_{j})$
$\displaystyle=\rho_{j}^{+}\prod^{4}_{l=1}(a_{l}+k_{j})^{n_{l}}+\rho_{j}^{-}\prod^{4}_{l=1}(a_{l}-k_{j})^{n_{l}}.$
(5.9)

Impose a constraint on the parameters and introduce a new variable $N_{3}$
as666 The constraint $a_{3}=a_{4}$ corresponds to the constraint on $\tau$:
$\tau(n_{1},n_{2},n_{3}+1,n_{4})=\tau(n_{1},n_{2},n_{3},n_{4}+1)$.

$a_{3}=a_{4},~{}~{}N_{3}=a_{3}+a_{4},$ (5.10)

which turn the PWF (5.8) to a form

$\rho=\left(\frac{a_{1}+k_{j}}{a_{1}-k_{j}}\right)^{n_{1}}\left(\frac{a_{2}+k_{j}}{a_{2}-k_{j}}\right)^{n_{2}}\left(\frac{a_{3}+k_{j}}{a_{3}-k_{j}}\right)^{N_{3}},$
(5.11)

then the system (5.7) is reduced to

$\displaystyle(a_{3}-a_{1})\tau\tau_{133}+(a_{1}+a_{3})\tau_{1}\tau_{33}-2a_{3}\tau_{3}\tau_{13}=0,$
(5.12a)
$\displaystyle(a_{3}-a_{2})\tau\tau_{233}+(a_{2}+a_{3})\tau_{2}\tau_{33}-2a_{3}\tau_{3}\tau_{23}=0.$
(5.12b)

Note that here $\tau_{3}$ means $\tau(N_{3}+1)$. Next, let

$a_{3},N_{3}\rightarrow\infty~{}~{}~{}\text{while
keeping}~{}x=N_{3}/a_{3}~{}~{}\text{finite},$ (5.13)

and rewrite the dependent variable as

$\tau=\tau(n_{1},n_{2},N_{3}):=\tau(n_{1},n_{2},x).$ (5.14)

Then, one can expand (5.12) in terms of $1/a_{3}$ and from the leading terms
we have the bilinear equations (3.20) of H2, i.e.,

$\displaystyle(D_{x}^{2}+2a_{1}D_{x})\tau\cdot\tau_{1}=0,$ (5.15a)
$\displaystyle(D_{x}^{2}+2a_{2}D_{x})\tau\cdot\tau_{2}=0.$ (5.15b)

The corresponding PWF $\rho$ and the Casoratian entries $\psi_{j}$ are given
as

$\rho=\left(\frac{a_{1}+k_{j}}{a_{1}-k_{j}}\right)^{n_{1}}\left(\frac{a_{2}+k_{j}}{a_{2}-k_{j}}\right)^{n_{2}}e^{2k_{j}x}\eta_{j}^{(0)},$
(5.16)

and

$\displaystyle\psi_{j}(n_{1},n_{2},x;k_{j})$
$\displaystyle=\rho_{j}^{+}e^{k_{j}x}\prod^{2}_{l=1}(a_{l}+k_{j})^{n_{l}}+\rho_{j}^{-}e^{-k_{j}x}\prod^{2}_{l=1}(a_{l}-k_{j})^{n_{l}}.$
(5.17)

###  5.3 Bilinear H3

As mentioned in subsection 3.3, there are two bilinear forms of H3 equation,
i.e., equations (3.25) and (3.26). In equation (3.25), the first two equations
can be considered as the reflected dAKP equations, which are all featured by
their total index (1,1,1). However, the third equation (3.25c) has total index
(1,1,0). In the next part we will look at the 4D dAKP equation which has total
index (1,1,1,1) and try obtaining (3.25c) through dimensional reductions.

Consider the reflected 4D dAKP equation (4.11a), which has a 4D PWF (5.8).
Introduce constraint $a_{4}=-a_{3}$, which gives rise to the reduction

$\tau(n_{1},n_{2},n_{3}+1,n_{4})=\tau(n_{1},n_{2},n_{3},n_{4}-1).$ (5.18)

The PWF (5.8) is reduced to

$\rho=\left(\frac{a_{1}+k_{j}}{a_{1}-k_{j}}\right)^{n_{1}}\left(\frac{a_{2}+k_{j}}{a_{2}-k_{j}}\right)^{n_{2}}\left(\frac{a_{3}+k_{j}}{a_{3}-k_{j}}\right)^{N_{3}}\eta_{j}^{(0)},$
(5.19)

where $N_{3}=n_{3}-n_{4}$, and the reflected 4D dAKP equation (4.11a) gives
rise to

$\displaystyle
2a_{3}(a_{1}-a_{2})\tau\tau_{12}+(a_{3}-a_{1})(a_{2}+a_{3})\tau_{13}\tau_{2}^{3}+(a_{1}+a_{3})(a_{2}-a_{3})\tau_{23}\tau_{1}^{3}=0,$
(5.20)

which is exactly the third bilinear equation (3.25c). Here by $\tau_{3}$ and
$\tau^{3}$ we mean $\tau(N_{3}+1)$ and $\tau(N_{3}-1)$, respectively. With the
above PWF (5.19), the reflected dAKP equations (4.10b) and (4.10a) hold, and
so do (3.25a) and (3.25b). For convenience, we replace $N_{3}$ by $n_{3}$ in
(5.19) and then the $\tau$-function in (3.25) will be given by Theorem 6 with
$\sigma=3$.

The second bilinear form (3.26) consists of two 2D equations in 3D space. The
total index patterns of them are $(1,0,0)$ and $(0,1,0)$, respectively. Note
that $n_{2}$ is a dummy variable in (3.26a), and likewise $n_{1}$ for (3.26b).
To get these two equations simultaneously, we consider two 3D dAKP equations
in 4D space:

$\displaystyle(a_{1}-a_{3})\tau_{4}\tau_{13}+(a_{3}-a_{4})\tau_{1}\tau_{34}+(a_{4}-a_{1})\tau_{3}\tau_{14}=0,$
(5.21a)
$\displaystyle(a_{2}-a_{3})\tau_{4}\tau_{23}+(a_{3}-a_{4})\tau_{2}\tau_{34}+(a_{4}-a_{2})\tau_{3}\tau_{24}=0,$
(5.21b)

which allow a $\tau$-function (4.12) with the PWF (5.8)777Note that both
equations in (5.21) are dAKP equations, which allow a more general PWF like
the one in (4.5) or in (4.7) with $\sigma=4$. However, to meet the definition
$N_{3}=n_{3}-n_{4}$, we shall consider the dKdV type PWFs in (4.13) or (4.15)
with $\sigma=4$.. Introducing reduction $a_{4}=-a_{3}$ which corresponds to
the reduction (5.18) on $\tau$, the PWF (5.8) goes to (5.19) with
$N_{3}=n_{3}-n_{4}$ and (5.21) gives rise to (3.26).

###  5.4 Bilinear Q1

For Q1 equation with a power background, its bilinear form (3.32) is the same
as (3.26) of H3 equation, which can be obtained from the dAKP equations (5.21)
through a one-step reduction. In the next part we look at the bilinear form
(3.44) for Q1 equation with a linear background, which is related to skew-
limit and much more complicated than the power background case.

The first three equations in (3.44) are the dAKP and its reflected forms,
while (3.44d) and (3.44e) are (3+1)D bilinear equations which have total index
$(1,1,1)$ in discrete directions. To obtain these two equations, we take a
two-step procedure. We will start from certain 5D dAKP system with total index
$(1,1,1,1,1)$ and take dimensional reductions to obtain some 4D equations with
total index $(1,1,1,0)$, on which we then perform a skew continuum limit to
get some (3+1)D semi-discrete equations with total index $(1,1,1)$ in discrete
directions.

Let us consider the 5D dAKP equation (4.1), i.e.,

$\mathcal{A}\equiv\left|\begin{matrix}1&a_{1}&a_{1}^{2}&a_{1}^{3}&\tau_{1}\tau_{2345}\\\
1&a_{2}&a_{2}^{2}&a_{2}^{3}&\tau_{2}\tau_{1345}\\\
1&a_{3}&a_{3}^{2}&a_{3}^{3}&\tau_{3}\tau_{1245}\\\
1&a_{4}&a_{4}^{2}&a_{4}^{3}&\tau_{4}\tau_{1235}\\\
1&a_{5}&a_{5}^{2}&a_{5}^{3}&\tau_{5}\tau_{1234}\\\ \end{matrix}\right|=0,$
(5.22)

and its 5 reflected forms

$\displaystyle\mathcal{A}_{1}=\mathcal{A}|_{a_{1}\rightarrow-
a_{1},n_{1}\rightarrow-n_{1}},$ (5.23a)
$\displaystyle\mathcal{A}_{2}=\mathcal{A}|_{a_{2}\rightarrow-
a_{2},n_{2}\rightarrow-n_{2}},$ (5.23b)
$\displaystyle\mathcal{A}_{3}=\mathcal{A}|_{a_{3}\rightarrow-
a_{3},n_{3}\rightarrow-n_{3}},$ (5.23c)
$\displaystyle\mathcal{A}_{4}=\mathcal{A}|_{a_{4}\rightarrow-
a_{4},n_{4}\rightarrow-n_{4}},$ (5.23d)
$\displaystyle\mathcal{A}_{5}=\mathcal{A}|_{a_{5}\rightarrow-
a_{5},n_{5}\rightarrow-n_{5}},$ (5.23e)

with the PWF

$\rho=\eta_{j}^{(0)}\prod^{5}_{i=1}\left(\frac{a_{i}+k_{j}}{a_{i}-k_{j}}\right)^{n_{i}}.$
(5.24)

Next, by imposing constraint

$a_{4}=-a_{3},$ (5.25)

the PWF (5.24) is reduced to

$\rho=\left(\frac{a_{1}+k_{j}}{a_{1}-k_{j}}\right)^{n_{1}}\left(\frac{a_{2}+k_{j}}{a_{2}-k_{j}}\right)^{n_{2}}\left(\frac{a_{3}+k_{j}}{a_{3}-k_{j}}\right)^{N_{3}}\left(\frac{a_{5}+k_{j}}{a_{5}-k_{j}}\right)^{n_{5}}\eta_{j}^{(0)}$
(5.26)

where $N_{3}=n_{3}-n_{4}$. The constraint (5.25) indicates
$\tau(n_{1},n_{2},n_{3}+1,n_{4},n_{5})=\tau(n_{1},n_{2},n_{3},n_{4}-1,n_{5}),$
which leads the 5D system (5.22) and (5.23) to the following 4D system with
total index $(1,1,0,1)$:

$\displaystyle\mathcal{A}_{1}^{(1)}\equiv$
$\displaystyle-(a_{1}-a_{2})(a_{1}-a_{3})(a_{1}-a_{5})(a_{2}-a_{3})(a_{2}-a_{5})(a_{3}-a_{5})\tau_{1235}\tau^{3}$
$\displaystyle-(a_{1}-a_{2})(a_{1}+a_{3})(a_{1}-a_{5})(a_{2}+a_{3})(a_{2}-a_{5})(a_{3}+a_{5})\tau_{125}^{3}\tau_{3}$
$\displaystyle+2a_{3}(a_{1}-a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\tau_{12}\tau_{5}+2a_{3}(a_{1}-a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{15}\tau_{2}$
$\displaystyle-2a_{3}(a_{2}-a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{1}\tau_{25}=0,$
(5.27a) $\displaystyle\mathcal{A}_{2}^{(1)}\equiv$
$\displaystyle(a_{1}+a_{2})(a_{2}-a_{3})(a_{1}+a_{3})(a_{2}-a_{5})(a_{3}-a_{5})(a_{1}+a_{5})\tau_{1}^{3}\tau_{235}$
$\displaystyle+(a_{1}+a_{2})(a_{1}-a_{3})(a_{2}+a_{3})(a_{2}-a_{5})(a_{1}+a_{5})(a_{3}+a_{5})\tau_{13}\tau_{25}^{3}$
$\displaystyle-2a_{3}(a_{2}-a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau\tau_{125}-2a_{3}(a_{1}+a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\tau_{15}\tau_{2}$
$\displaystyle-2a_{3}(a_{1}+a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{12}\tau_{5}=0,$
(5.27b) $\displaystyle\mathcal{A}_{3}^{(1)}\equiv$
$\displaystyle-(a_{1}+a_{2})(a_{1}-a_{3})(a_{2}+a_{3})(a_{1}-a_{5})(a_{3}-a_{5})(a_{2}+a_{5})\tau_{135}\tau_{2}^{3}$
$\displaystyle-(a_{1}+a_{2})(a_{2}-a_{3})(a_{1}+a_{3})(a_{1}-a_{5})(a_{2}+a_{5})(a_{3}+a_{5})\tau_{15}^{3}\tau_{23}$
$\displaystyle+2a_{3}(a_{2}+a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{12}\tau_{5}+2a_{3}(a_{1}-a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau\tau_{125}$
$\displaystyle+2a_{3}(a_{1}+a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\tau_{1}\tau_{25}=0,$
(5.27c) $\displaystyle\mathcal{A}_{4}^{(1)}\equiv$
$\displaystyle-(a_{1}-a_{2})(a_{1}+a_{3})(a_{2}+a_{3})(a_{1}+a_{5})(a_{2}+a_{5})(a_{3}-a_{5})\tau_{12}^{3}\tau_{35}$
$\displaystyle-(a_{1}-a_{2})(a_{1}-a_{3})(a_{2}-a_{3})(a_{1}+a_{5})(a_{2}+a_{5})(a_{3}+a_{5})\tau_{123}\tau_{5}^{3}$
$\displaystyle+2a_{3}(a_{1}-a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\tau\tau_{125}-2a_{3}(a_{2}+a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{15}\tau_{2}$
$\displaystyle+2a_{3}(a_{1}+a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{1}\tau_{25}=0.$
(5.27d)

Note that here $\tau_{3}=\tau(N_{3}+1)$. To perform skew continuum limit, we
need to convert the above system from the coordinates
$\\{n_{1},n_{2},N_{3},n_{5}\\}$ into the coordinates
$\\{n_{1},n_{2},N_{3}^{\prime}=N_{3}+n_{5},n_{5}\\}$. After that, we come to a
4D system with total index $(1,1,1,1)$:

$\displaystyle\mathcal{A}_{1}^{(2)}\equiv$
$\displaystyle-(a_{1}-a_{2})(a_{1}-a_{3})(a_{1}-a_{5})(a_{2}-a_{3})(a_{2}-a_{5})(a_{3}-a_{5})\tau_{12335}\tau^{3}$
$\displaystyle-(a_{1}-a_{2})(a_{1}+a_{3})(a_{1}-a_{5})(a_{2}+a_{3})(a_{2}-a_{5})(a_{3}+a_{5})\tau_{125}\tau_{3}$
$\displaystyle+2a_{3}(a_{1}-a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\tau_{12}\tau_{35}+2a_{3}(a_{1}-a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{135}\tau_{2}$
$\displaystyle-2a_{3}(a_{2}-a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{1}\tau_{235}=0,$
(5.28a) $\displaystyle\mathcal{A}_{2}^{(2)}\equiv$
$\displaystyle(a_{1}+a_{2})(a_{2}-a_{3})(a_{1}+a_{3})(a_{2}-a_{5})(a_{3}-a_{5})(a_{1}+a_{5})\tau_{1}^{3}\tau_{2335}$
$\displaystyle+(a_{1}+a_{2})(a_{1}-a_{3})(a_{2}+a_{3})(a_{2}-a_{5})(a_{1}+a_{5})(a_{3}+a_{5})\tau_{13}\tau_{25}$
$\displaystyle-2a_{3}(a_{2}-a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau\tau_{1235}-2a_{3}(a_{1}+a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\tau_{135}\tau_{2}$
$\displaystyle-2a_{3}(a_{1}+a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{12}\tau_{35}=0,$
(5.28b) $\displaystyle\mathcal{A}_{3}^{(2)}\equiv$
$\displaystyle-(a_{1}+a_{2})(a_{1}-a_{3})(a_{2}+a_{3})(a_{1}-a_{5})(a_{3}-a_{5})(a_{2}+a_{5})\tau_{1335}\tau_{2}^{3}$
$\displaystyle-(a_{1}+a_{2})(a_{2}-a_{3})(a_{1}+a_{3})(a_{1}-a_{5})(a_{2}+a_{5})(a_{3}+a_{5})\tau_{15}\tau_{23}$
$\displaystyle+2a_{3}(a_{2}+a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{12}\tau_{35}+2a_{3}(a_{1}-a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau\tau_{1235}$
$\displaystyle+2a_{3}(a_{1}+a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\tau_{1}\tau_{235}=0,$
(5.28c) $\displaystyle\mathcal{A}_{4}^{(2)}\equiv$
$\displaystyle-(a_{1}-a_{2})(a_{1}+a_{3})(a_{2}+a_{3})(a_{1}+a_{5})(a_{2}+a_{5})(a_{3}-a_{5})\tau_{12}^{3}\tau_{335}$
$\displaystyle-(a_{1}-a_{2})(a_{1}-a_{3})(a_{2}-a_{3})(a_{1}+a_{5})(a_{2}+a_{5})(a_{3}+a_{5})\tau_{123}\tau_{5}$
$\displaystyle+2a_{3}(a_{1}-a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\tau\tau_{1235}-2a_{3}(a_{2}+a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{135}\tau_{2}$
$\displaystyle+2a_{3}(a_{1}+a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\tau_{1}\tau_{235}=0,$
(5.28d)

and correspondingly, the PWF takes a form

$\rho=\left(\frac{a_{1}+k_{j}}{a_{1}-k_{j}}\right)^{n_{1}}\left(\frac{a_{2}+k_{j}}{a_{2}-k_{j}}\right)^{n_{2}}\left(\frac{a_{3}+k_{j}}{a_{3}-k_{j}}\right)^{N_{3}^{\prime}}\left(\frac{a_{3}-k_{j}}{a_{3}+k_{j}}\cdot\frac{a_{5}+k_{j}}{a_{5}-k_{j}}\right)^{n_{5}}\eta_{j}^{(0)}.$
(5.29)

Note that in (5.28) by $\tau_{3}$ we mean $\tau(N_{3}^{\prime}+1)$. Now set
$\varepsilon=a_{5}-a_{3}$. Then the limit scheme

$N_{3}\rightarrow-\infty,~{}n_{5}\rightarrow\infty,~{}\varepsilon\rightarrow
0,~{}~{}\text{while}~{}n_{3}~{}\text{and}~{}y:=n_{5}\varepsilon~{}\text{being
finite},$ (5.30)

brings the factor
$\left(\frac{a_{3}-k_{j}}{a_{3}+k_{j}}\cdot\frac{a_{5}+k_{j}}{a_{5}-k_{j}}\right)^{n_{5}}$
in the PWF (5.29) to

$\displaystyle\underset{\varepsilon\to
0}{\underset{n_{5}\to\infty}{\lim}}\left(1+\frac{2k_{j}(a_{3}-a_{5})n_{5}}{(a_{3}+k_{j})(a_{5}-k_{j})n_{5}}\right)^{n_{5}}=\underset{\varepsilon\to
0}{\underset{n_{5}\to\infty}{\lim}}\left(1+\frac{-2k_{j}(y-y_{0})}{(a_{3}+k_{j})(a_{3}-k_{j}+\varepsilon)n_{5}}\right)^{n_{5}}$
$\displaystyle=$ $\displaystyle\underset{\varepsilon\to
0}{\underset{n_{5}\to\infty}{\lim}}\left(1+\frac{1}{n_{5}}\cdot\frac{-2k_{j}(y-y_{0})}{(a_{3}^{2}-k_{j}^{2})+\varepsilon(a_{3}+k_{j})}\right)^{n_{5}}={\underset{n_{5}\to\infty}{\lim}}\left(1+\frac{1}{n_{5}}\cdot\frac{-2k_{j}(y-y_{0})}{(a_{3}^{2}-k_{j}^{2})}\right)^{n_{5}}$
$\displaystyle=$
$\displaystyle\exp\left(\frac{-2k_{j}y}{a_{3}^{2}-k_{j}^{2}}\right).$

For convenience, we write

$\tau=\tau(n_{1},n_{2},N_{3}^{\prime},n_{5}):=\tau(n_{1},n_{2},N_{3}^{\prime},y),$
(5.31)

which indicates

$\tau_{5}=\tau(n_{1},n_{2},N_{3}^{\prime},n_{5}+1)=\tau(n_{1},n_{2},N_{3}^{\prime},y+\varepsilon)=\tau+\varepsilon\partial_{y}\tau+\frac{1}{2}\varepsilon^{2}\partial_{y}^{2}\tau+\cdots.$
(5.32)

In such a scheme, the leading terms of the system (5.28) yield

$\displaystyle\mathcal{A}_{1}^{(3)}\equiv$
$\displaystyle-4a_{3}^{2}(a_{1}+a_{3})(a_{1}-a_{3})^{2}\tau_{13}\tau_{2}+(a_{1}-a_{2})(a_{1}-a_{3})^{2}(a_{2}-a_{3})^{2}\tau_{1233}\tau^{3}$
$\displaystyle-(a_{1}-a_{2})(a_{1}+a_{3})(a_{2}+a_{3})(a_{1}(a_{2}-3a_{3})+a_{3}(-3a_{2}+5a_{3}))\tau_{12}\tau_{3}$
$\displaystyle+4a_{3}^{2}(a_{2}+a_{3})(a_{2}-a_{3})^{2}\tau_{1}\tau_{23}+2a_{3}(a_{1}-a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})D_{y}\tau_{3}\\!\cdot\\!\tau_{12}=0,$
(5.33a) $\displaystyle\mathcal{A}_{2}^{(3)}\equiv$ $\displaystyle
4a_{3}^{2}(a_{2}+a_{3})(a_{2}-a_{3})^{2}\tau\tau_{123}-(a_{1}+a_{2})(a_{2}-a_{3})^{2}(a_{1}+a_{3})^{2}\tau_{1}^{3}\tau_{233}$
$\displaystyle+(a_{1}+a_{2})(a_{1}-a_{3})(a_{2}+a_{3})(a_{1}(a_{2}-3a_{3})+(3a_{2}-5a_{3})a_{3})\tau_{13}\tau_{2}$
$\displaystyle+4a_{3}^{2}(a_{1}-a_{3})(a_{1}+a_{3})^{2}\tau_{12}\tau_{3}+2a_{3}(a_{1}+a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})D_{y}\tau_{2}\\!\cdot\\!\tau_{13}=0,$
(5.33b) $\displaystyle\mathcal{A}_{3}^{(3)}\equiv$
$\displaystyle-4a_{3}^{2}(a_{1}+a_{3})(a_{1}-a_{3})^{2}\tau\tau_{123}+(a_{1}+a_{2})(a_{1}-a_{3})^{2}(a_{2}+a_{3})^{2}\tau_{133}\tau_{2}^{3}$
$\displaystyle-(a_{1}+a_{2})(a_{2}-a_{3})(a_{1}+a_{3})(a_{1}(a_{2}+3a_{3})-a_{3}(3a_{2}+5a_{3}))\tau_{1}\tau_{23}$
$\displaystyle-4a_{3}^{2}(a_{2}-a_{3})(a_{2}+a_{3})^{2}\tau_{12}\tau_{3}-2a_{3}(a_{1}+a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})D_{y}\tau_{1}\\!\cdot\\!\tau_{23}=0,$
(5.33c) $\displaystyle\mathcal{A}_{4}^{(3)}\equiv$ $\displaystyle
4a_{3}^{2}(a_{2}-a_{3})(a_{2}+a_{3})^{2}\tau_{13}\tau_{2}+(a_{1}-a_{2})(a_{1}+a_{3})^{2}(a_{2}+a_{3})^{2}\tau_{12}^{3}\tau_{33}$
$\displaystyle-(a_{1}-a_{3})(a_{2}-a_{3})(a_{1}^{2}(a_{2}+3a_{3})-a_{2}a_{3}(3a_{2}+5a_{3})-a_{1}(a_{2}^{2}-5a_{3}^{2}))\tau\tau_{123}$
$\displaystyle-\\!4a_{3}^{2}(a_{1}-a_{3})(a_{1}+a_{3})^{2}\tau_{1}\tau_{23}\\!-\\!2a_{3}(a_{1}-a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})D_{y}\tau\\!\cdot\\!\tau_{123}=0.$
(5.33d)

Thus, we arrive at a (3+1)D system with total index $(1,1,1)$ in three
discrete directions, and the corresponding PWF is

$\rho=\left(\frac{a_{1}+k_{j}}{a_{1}-k_{j}}\right)^{n_{1}}\left(\frac{a_{2}+k_{j}}{a_{2}-k_{j}}\right)^{n_{2}}\left(\frac{a_{3}+k_{j}}{a_{3}-k_{j}}\right)^{N_{3}^{\prime}}\exp\left(\frac{-2k_{j}y}{a_{3}^{2}-k_{j}^{2}}\right)\eta_{j}^{(0)}.$
(5.34)

To recover (3.44d) and (3.44e), i.e., $\mathcal{Q}_{3}=0$ and
$\mathcal{Q}_{4}=0$, we consider combinations

$\displaystyle(a_{1}+a_{2})\mathcal{A}_{1}^{(3)}+(a_{1}-a_{3})\mathcal{A}_{2}^{(3)}+(a_{2}-a_{3})\mathcal{A}_{3}^{(3)}=0,$
(5.35a)
$\displaystyle(a_{2}+a_{3})\mathcal{A}_{2}^{(3)}+(a_{1}+a_{3})\mathcal{A}_{3}^{(3)}-(a_{1}+a_{2})\mathcal{A}_{4}^{(3)}=0,$
(5.35b)

which give rise to

$\displaystyle\\{2a_{1}a_{3}(a_{2}^{2}-a_{3}^{2})+a_{3}^{2}(7a_{2}^{2}-2a_{2}a_{3}-13a_{3}^{2})+a_{1}^{2}(-a_{2}^{2}+2a_{2}a_{3}+7a_{3}^{2})\\}A$
$\displaystyle-(a_{2}-a_{3})(a_{1}-a_{3})\dot{A}-2a_{3}(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\mathcal{Q}_{3}=0,$
(5.36a)
$\displaystyle\\{2a_{1}a_{3}(a_{2}^{2}-a_{3}^{2})+a_{3}^{2}(-7a_{2}^{2}-2a_{2}a_{3}+13a_{3}^{2})+a_{1}^{2}(a_{2}^{2}+2a_{2}a_{3}-7a_{3}^{2})\\}A_{3}$
$\displaystyle+(a_{1}+a_{3})(a_{2}+a_{3})\dot{A}_{1}+2a_{3}(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\mathcal{Q}_{4}=0,$
(5.36b)

where $A$ and $A_{3}$ are the dAKP (4.2) and reflected form (4.10c),
respectively, and the notation $\dot{A}$ stands for

$\displaystyle\dot{A}:=$
$\displaystyle(a_{2}-a_{3})(a_{1}+a_{3})^{2}\tau_{1}^{3}\tau_{233}-(a_{1}-a_{2})(a_{1}-a_{3})(a_{2}-a_{3})\tau^{3}\tau_{1233}$
$\displaystyle-(a_{1}-a_{3})(a_{2}+a_{3})^{2}\tau_{2}^{3}\tau_{133}+4a_{3}^{2}(a_{1}-a_{2})\tau\tau_{123},$
(5.37)

and $\dot{A}_{1}$ is a reflected form of $\dot{A}$ (5.4) by taking

$\tau(n_{1},n_{2},(N_{3}^{\prime},a_{3}))=\tau(n_{1},n_{2},(-N_{3}^{\prime},-a_{3})).$
(5.38)

Thus, if we can prove $\dot{A}=0$, we may recover equations
$\mathcal{Q}_{3}=0$ and $\mathcal{Q}_{4}=0$ from (5.36). Meanwhile, the
$\tau$-function composed by the PWF (5.34) satisfies the first three bilinear
equations in (3.44).

As for equation $\dot{A}=0$, which is an 8-point 3D equation, i.e.,

$\displaystyle\dot{A}:=$
$\displaystyle(a_{2}-a_{3})(a_{1}+a_{3})^{2}\tau_{1}^{3}\tau_{233}-(a_{1}-a_{2})(a_{1}-a_{3})(a_{2}-a_{3})\tau^{3}\tau_{1233}$
$\displaystyle-(a_{1}-a_{3})(a_{2}+a_{3})^{2}\tau_{2}^{3}\tau_{133}+4a_{3}^{2}(a_{1}-a_{2})\tau\tau_{123}=0,$
(5.39)

we have the following remarks. More details about this equation can be found
in Appendix A.

######  Remark 3.

Equation (5.4) holds when $\tau$ is defined as in Theorem 6 with $\sigma=4$. A
direct proof can be found in Appendix A.

######  Remark 4.

Performing the straight continuum limit

$a_{3},n_{3}\rightarrow\infty,~{}~{}x=n_{3}/a_{3},~{}~{}x~{}\text{keeps
finite},$

and still writing $\tau(n_{1},n_{2},n_{3})=\tau(n_{1},n_{2},x)$, the leading
term of (5.4) gives rise to the semi-discrete AKP equation (5.4a), i.e.,

$(a_{1}-a_{2})(\tau\tau_{12}-\tau_{1}\tau_{2})+D_{x}\tau_{1}\cdot\tau_{2}=0.$
(5.40)

###  5.5 Bilinear Q3

The four equations in the bilinear Q3 system (3.55) are all reflected dAKP
equations.

##  6 Conclusions

In this paper, we have successfully established the connection between the
bilinear ABS equations (except Q2 and Q4) and the symmetric dAKP system. The
later is a system generated by the principle equation (3D) dAKP equation and
its reflected forms. Note that high order dAKP equations in this system are
considered as the consequence of the multi-dimensional consistency of the 3D
dAKP equation. We reviewed the bilinear ABS equations to express them in terms
of a single $\tau$-function. It is notable that for H2 equation we gave a new
formulation (see (3.17)) of its bilinearization. The present bilinear H2
system (3.20) contains two equations. Note that this bilinear H2 system can
also be viewed as the deformed bilinear Bäcklund transformation of the KdV
equation [24, 31]. For Q1 equation with linear background, we rewrote two
bilinear equations so that they can be presented in terms of Hirota’s
$D$-operator.

The symmetric dAKP system allows more high order bilinear equations which are
available in implementing reductions. As for the connection between the
bilinear ABS equations and the symmetric dAKP system, bilinear H1 system
consists of differential-difference forms of the dAKP equation and its
reflected form. Bilinear H2 system contains two bilinear equations that are
differential-difference forms of two reduced dAKP equations. H3 equation has
two bilinear forms, which can be obtained from dimensional reductions of a
reflected 4D dAKP equation and two 3D dAKP equations in 4D space. Q1 equation
with power background is one of cases of H3 equation. However, Q1 equation
with linear background is more complicated. Two equations in its bilinear form
are related to the reduced 5D dAKP and its reflected forms and further skew-
continuum limits. Finally, bilinear Q3 equations contains four reflected 3D
dAKP equations in 4D space.

As by-products, we obtained a 4D 8-point equation $M_{1}^{\prime}=0$, i.e.,
(A.15) and a 3D 8-point equation $\dot{A}=0$, i.e., (5.4), while $\dot{A}=0$
is a dimensional reduction of $M_{1}^{\prime}=0$ (see Appendix A). They have
solutions composed by the dKdV type PWF. Although at present we can not obtain
$M_{1}^{\prime}=0$ as a reduction from the symmetric dAKP system, we proved it
allows Casoratian solutions (and so does the Hirota’s form (4.12)) in Appendix
A.

As for further research, we would like to understand connections between the
discrete bilinear Boussinesq equations (see [32]) and dAKP equations. In
addition, the ABS equations, discrete Boussinesq equations and discrete KP
equations admit elliptic soliton solutions composed by Weierstrass functions
[15, 33]. It would be interesting to extend the connections we have found to
the level of elliptic solitons.

### Acknowledgements

DJZ is supported by the NSF of China (No.12271334). JW is currently supported
by the China Scholarship Council visiting-PhD-student program. KM is supported
by JSPS KAKENHI Grant (No.22K03441). DJZ are grateful to Prof. Hietarinta and
Prof. Nijhoff for their discussions.

##  Appendix A Equation (5.4)

###  A.1 Connection with 4D equations

Equations in the 4D system (5.27) are featured with total index $(1,1,0,1)$.
Consider a combination

$(a_{1}+a_{2})\mathcal{A}_{1}^{(1)}+(a_{1}-a_{5})\mathcal{A}_{2}^{(1)}+(a_{2}-a_{5})\mathcal{A}_{3}^{(1)}=0$
(A.1)

which is expressed as

$\displaystyle(a_{1}\\!+\\!a_{2})(a_{1}\\!-\\!a_{5})(a_{2}\\!-\\!a_{5})(a_{3}\\!-\\!a_{5})M_{1}\\!+\\!(a_{1}\\!+\\!a_{2})(a_{1}\\!-\\!a_{5})(a_{2}\\!-\\!a_{5})(a_{3}\\!+\\!a_{5})M_{2}\\!+\\!M_{3}\\!=\\!0,$
(A.2)

where

$\displaystyle M_{1}\\!:=\,$
$\displaystyle(a_{1}\\!+\\!a_{3})(a_{1}\\!+\\!a_{5})(a_{2}\\!-\\!a_{3})\tau_{1}^{3}\tau_{235}-\\!(a_{1}\\!-\\!a_{2})(a_{1}\\!-\\!a_{3})(a_{2}\\!-\\!a_{3})\tau^{3}\tau_{1235}$
$\displaystyle-\\!(a_{1}\\!-\\!a_{3})(a_{2}\\!+\\!a_{3})(a_{2}\\!+\\!a_{5})\tau_{2}^{3}\tau_{135},$
$\displaystyle M_{2}\\!:=\,$
$\displaystyle(a_{1}\\!-\\!a_{3})(a_{1}\\!+\\!a_{5})(a_{2}\\!+\\!a_{3})\tau_{13}\tau_{25}^{3}\\!-\\!(a_{1}\\!-\\!a_{2})(a_{1}\\!+\\!a_{3})(a_{2}\\!+\\!a_{3})\tau_{3}\tau_{125}^{3}$
$\displaystyle-\\!(a_{1}\\!+\\!a_{3})(a_{2}\\!-\\!a_{3})(a_{2}\\!+\\!a_{5})\tau_{23}\tau_{15}^{3},$
$\displaystyle M_{3}\\!:=\,$ $\displaystyle
2a_{3}(a_{1}-a_{5})(a_{1}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\\{(a_{1}+a_{2})\tau_{2}\tau_{15}-(a_{1}+a_{5})\tau_{5}\tau_{12}+(a_{2}-a_{5})\tau\tau_{125}\\}$
$\displaystyle-2a_{3}(a_{2}-a_{5})(a_{2}^{2}-a_{3}^{2})(a_{3}^{2}-a_{5}^{2})\\{(a_{1}+a_{2})\tau_{1}\tau_{25}+(a_{1}-a_{5})\tau\tau_{125}-(a_{2}+a_{5})\tau_{5}\tau_{12}\\}$
$\displaystyle+2a_{3}(a_{1}+a_{2})(a_{1}^{2}-a_{3}^{2})(a_{2}^{2}-a_{3}^{2})\\{(a_{1}\\!-\\!a_{2})\tau_{5}\tau_{12}\\!-\\!(a_{1}\\!-\\!a_{5})\tau_{2}\tau_{15}\\!+\\!(a_{2}\\!-\\!a_{5})\tau_{1}\tau_{25}\\}.$

One can use equations (4.2) and (4.10) to simplify $M_{3}$ to the form

$4a_{3}(a_{1}-a_{5})(a_{2}-a_{5})(a_{1}^{2}-a_{2}^{2})(a_{3}^{2}-a_{5}^{2})\tau\tau_{125},$

such that (A.2) can be rewritten as

$\displaystyle(a_{3}\\!-\\!a_{5})M_{1}^{\prime}\\!+(a_{3}\\!+\\!a_{5})M_{2}^{\prime}=0,$
(A.3)

where

$\displaystyle M_{1}^{\prime}\\!:=\,$
$\displaystyle(a_{1}\\!+\\!a_{3})(a_{1}\\!+\\!a_{5})(a_{2}\\!-\\!a_{3})\tau_{1}^{3}\tau_{235}\\!-\\!(a_{1}\\!-\\!a_{2})(a_{1}\\!-\\!a_{3})(a_{2}\\!-\\!a_{3})\tau^{3}\tau_{1235}\\!$
$\displaystyle-\\!(a_{1}\\!-\\!a_{3})(a_{2}\\!+\\!a_{3})(a_{2}\\!+\\!a_{5})\tau_{2}^{3}\tau_{135}+2a_{3}(a_{1}-a_{2})(a_{3}+a_{5})\tau\tau_{125},$
(A.4) $\displaystyle M_{2}^{\prime}\\!:=\,$
$\displaystyle(a_{1}\\!-\\!a_{3})(a_{1}\\!+\\!a_{5})(a_{2}\\!+\\!a_{3})\tau_{13}\tau_{25}^{3}\\!-\\!(a_{1}\\!-\\!a_{2})(a_{1}\\!+\\!a_{3})(a_{2}\\!+\\!a_{3})\tau_{3}\tau_{125}^{3}\\!$
$\displaystyle-\\!(a_{1}\\!+\\!a_{3})(a_{2}\\!-\\!a_{3})(a_{2}\\!+\\!a_{5})\tau_{23}\tau_{15}^{3}+2a_{3}(a_{1}-a_{2})(a_{3}-a_{5})\tau\tau_{125}.$
(A.5)

Noted that $M_{2}^{\prime}$ is a reflected form of $M_{1}^{\prime}$ by taking

$\tau(n_{1},n_{2},(N_{3},a_{3}),n_{5})\\!=\\!\tau(n_{1},n_{2},(-N_{3},-a_{3}),n_{5}).$
(A.6)

By employing constraint

$a_{3}=a_{5},~{}~{}\tau(n_{1},n_{2},N_{3},n_{5})\rightarrow\tau(n_{1},n_{2},n_{3}^{\prime}=N_{3}+n_{5})$
(A.7)

on $M_{1}^{\prime}$ (A.4) and $M_{2}^{\prime}$ (A.5), they will reduce to

$\displaystyle(a_{2}-a_{3})(a_{1}+a_{3})^{2}\tau_{1}^{3}\tau_{233}-(a_{1}-a_{2})(a_{1}-a_{3})(a_{2}-a_{3})\tau^{3}\tau_{1233}$
(A.8)
$\displaystyle-(a_{1}-a_{3})(a_{2}+a_{3})^{2}\tau_{2}^{3}\tau_{133}+4a_{3}^{2}(a_{1}-a_{2})\tau\tau_{123}$

and

$(a_{1}+a_{3})(a_{2}+a_{3})\\{(a_{1}-a_{3})\tau_{13}\tau_{2}-(a_{1}-a_{2})\tau_{12}\tau_{3}-(a_{2}-a_{3})\tau_{23}\tau_{1}\\}$
(A.9)

respectively, while (LABEL:dot-A-1) is nothing but $\dot{A}$ (5.4) and (A.9)
contains the dAKP (4.2).

In the next part we will prove both $M_{1}^{\prime}$ and $M_{2}^{\prime}$ are
zero, and so is $\dot{A}$.

###  A.2 Proof of solution

For convenience, we introduce some notations related to the generic column
vector
$\psi=\psi(n_{1},n_{2},\cdots,n_{r};l_{1})=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$.
We define

$\displaystyle
m_{\begin{subarray}{c}n_{i}+1\end{subarray}}:=E_{l_{1}}^{m}E_{n_{i}}\psi,~{}~{}m_{\begin{subarray}{c}n_{i}+1\\\
n_{j}-1\end{subarray}}:=E_{l_{1}}^{m}E_{n_{j}}^{-1}E_{n_{i}}\psi,$
$\displaystyle\widetilde{m}_{n_{i}+j}^{n_{\xi}}:=\Theta_{n_{\xi}}^{-1}E_{l_{1}}^{m}E_{n_{i}}^{j}\psi,~{}~{}~{}\widetilde{m}_{n_{i}+j}^{n_{\xi},n_{\theta}}\\!:=\Theta_{n_{\xi}}^{-1}\Theta_{n_{\theta}}^{-1}E_{l_{1}}^{m}E_{n_{i}}^{j}\psi,$

where

$\Theta_{n_{\xi}}=\mathrm{Diag}(U_{1}(n_{\xi}),U_{2}(n_{\xi}),\cdots,U_{N}(n_{\xi})),~{}~{}U_{i}(n_{\xi})=k_{i}^{2}-a_{\xi}^{2}.$
(A.10)

In the proof, we also replace $N_{3}^{\prime}$ with $n_{3}$ without making
confusion.

######  Theorem 7.

The equations

$M_{1}^{\prime}=0,~{}~{}M_{2}^{\prime}=0,~{}~{}\dot{A}=0$ (A.11)

share the $\tau$-function

$\tau=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[n_{\sigma}]}=\bigl{\lvert}\widehat{N-1}\bigr{\rvert}_{[l_{1}]},~{}~{}(\sigma=1,2,3,5)$
(A.12)

composed by $\psi=(\psi_{1},\psi_{2},\cdots,\psi_{N})^{T}$ with entries

$\displaystyle\psi_{j}(n_{1},n_{2},n_{3},n_{5};l_{1})=$
$\displaystyle\rho_{j}^{+}k_{i}^{l_{1}}(a_{1}+k_{j})^{n_{1}}(a_{2}+k_{j})^{n_{2}}(a_{3}+k_{j})^{n_{3}}(a_{5}+k_{j})^{n_{5}}$
$\displaystyle+\rho_{j}^{-}(-k_{j})^{l_{1}}(a_{1}-k_{j})^{n_{1}}(a_{2}-k_{j})^{n_{2}}(a_{3}-k_{j})^{n_{3}}(a_{5}-k_{j})^{n_{5}}.$
(A.13)

###### Proof.

The entry $\psi_{j}$ satisfies

$\displaystyle
a_{3}\psi_{j}(n_{3}+1;l_{1})-\psi_{j}(n_{3}+1;l_{1}+1)=(a_{3}^{2}-k_{j}^{2})\psi_{j}(n_{3};l_{1}),$
$\displaystyle
a_{5}\psi_{j}(n_{5}+1;l_{1})-\psi_{j}(n_{5}+1;l_{1}+1)=(a_{5}^{2}-k_{j}^{2})\psi_{j}(n_{5};l_{1}),$
$\displaystyle
a_{i}\psi_{j}(n_{i}-1;l_{1})+\psi_{j}(n_{i}-1;l_{1}+1)=\psi_{j}(n_{i};l_{1}),$
$\displaystyle(a_{3}+a_{i})\psi_{j}(n_{3}+1,n_{i}-1;l_{1})=\psi_{j}(n_{3}+1,n_{i};l_{1})+(a_{3}^{2}-k_{j}^{2})\psi_{j}(n_{3},n_{i}-1;l_{1}),$
$\displaystyle(a_{5}+a_{i})\psi_{j}(n_{5}+1,n_{i}-1;l_{1})=\psi_{j}(n_{5}+1,n_{i};l_{1})+(a_{5}^{2}-k_{j}^{2})\psi_{j}(n_{5},n_{i}-1;l_{1}),$
$\displaystyle(a_{i}-a_{r})\psi_{j}(n_{i}-1,n_{r}-1;l_{1})=\psi_{j}(n_{i},n_{r}-1;l_{1})-\psi_{j}(n_{i}-1,n_{r};l_{1}),$
$\displaystyle(a_{5}\\!-\\!a_{3})\psi_{j}(n_{3}\\!+\\!1,n_{5}\\!+\\!1;l_{1})\\!=\\!(a_{5}^{2}\\!-\\!k_{j}^{2})\psi_{j}(n_{3}\\!+\\!1,n_{5};l_{1})\\!-\\!(a_{3}^{2}\\!-\\!k_{j}^{2})\psi_{j}(n_{3},n_{5}\\!+\\!1;l_{1}),$

which give rise to (note that because of (A.12) here the Casoratians are
displayed with respect to column shifting in $l_{1}$-direction)

$\displaystyle\tau^{i}=\left\lvert\psi(n_{i}-1;0)~{}~{}\psi(n_{i}-1;1)~{}~{}\cdots~{}~{}\psi(n_{i}-1;N-1)\right\rvert$
$\displaystyle~{}~{}~{}=\left\lvert\psi(n_{i}-1;0)~{}~{}\psi(n_{i};0)~{}~{}\cdots~{}~{}\psi(n_{i};N-2)\right\rvert$
$\displaystyle~{}~{}~{}=\bigl{\lvert}0_{n_{i}-1}~{}~{}0~{}~{}\cdots~{}~{}N-2\bigr{\rvert}=\bigl{\lvert}0_{n_{i}-1}~{}~{}\widehat{N-2}\bigr{\rvert},$
$\displaystyle\tau^{ij}=\Bigl{\lvert}0_{n_{i}-1\atop
n_{j}-1}~{}~{}0_{n_{j}-1}~{}~{}\widehat{N-3}\Bigr{\rvert}=\Bigl{\lvert}0_{n_{i}-1\atop
n_{j}-1}~{}~{}0_{n_{i}-1}~{}~{}\widehat{N-3}\Bigr{\rvert},$
$\displaystyle\tau^{ijr}=\Biggl{\lvert}0_{\begin{subarray}{c}n_{i}-1\\\
n_{j}-1\\\ n_{r}-1\end{subarray}}~{}~{}0_{n_{i}-1\atop
n_{j}-1}~{}~{}0_{n_{j}-1}~{}~{}\widehat{N-4}\Biggr{\rvert},$
$\displaystyle\tau_{3}=\Gamma(a_{3})\bigl{\lvert}\widetilde{0}_{n_{3}+1}^{n_{3}}~{}~{}\widehat{N-2}\bigr{\rvert},~{}~{}~{}~{}\tau_{33}=(\Gamma(a_{3}))^{2}\bigl{\lvert}\widetilde{0}_{n_{3}+2}^{n_{3},n_{3}}~{}~{}\widetilde{0}_{n_{3}+1}^{n_{3}}~{}~{}\widehat{N-3}\bigr{\rvert},$
$\displaystyle\tau_{35}=\Gamma(a_{3})\Gamma(a_{5})\Bigl{\lvert}\widetilde{0}_{\begin{subarray}{c}n_{3}+1\\\
n_{5}+1\end{subarray}}^{n_{3},n_{5}}~{}~{}\widetilde{0}_{n_{5}+1}^{n_{5}}~{}~{}\widehat{N-3}\Bigr{\rvert},$
$\displaystyle\tau_{35}^{j}=\Gamma(a_{3})\Gamma(a_{5})\Biggl{\lvert}\widetilde{0}_{\begin{subarray}{c}n_{3}+1\\\
n_{5}+1\\\
n_{j}-1\end{subarray}}^{n_{3},n_{5}}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{5}+1\\\
n_{j}-1\end{subarray}}^{n_{5}}~{}~{}0_{n_{j}-1}~{}~{}\widehat{N-4}\Biggr{\rvert},$
$\displaystyle(a_{i}-a_{j})\tau^{ij}=\left\lvert
0_{n_{j}-1}~{}~{}0_{n_{i}-1}~{}~{}\widehat{N-3}\right\rvert,$
$\displaystyle(a_{3}+a_{j})(a_{5}+a_{j})\tau_{35}^{j}=\Gamma(a_{3})\Gamma(a_{5})\left\lvert\widetilde{0}_{\begin{subarray}{c}n_{3}+1\\\
n_{5}+1\end{subarray}}^{n_{3},n_{5}}~{}~{}\widetilde{0}_{n_{5}+1}^{n_{5}}~{}~{}0_{n_{j}-1}~{}~{}\widehat{N-4}\right\rvert,$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\cdots\cdots$

where $\Gamma(a_{j})=|\Theta_{n_{j}}|$. Accordingly,
$E_{n_{2}}^{-1}E_{n_{1}}^{-1}M_{1}^{{}^{\prime}}$ turns to

$\begin{split}&\Gamma(a_{3})\Gamma(a_{5})\left(\bigl{\lvert}0_{n_{3}-1}~{}~{}0_{n_{2}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert
0_{n_{1}-1}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{3}+1\\\
n_{5}+1\end{subarray}}^{n_{3},n_{5}}~{}~{}\widetilde{0}_{n_{5}+1}^{n_{5}}~{}~{}\widehat{N-4}\right\rvert\right.\\\
&-\left\lvert\widetilde{0}_{\begin{subarray}{c}n_{3}+1\\\
n_{5}+1\end{subarray}}^{n_{3},n_{5}}~{}~{}\widetilde{0}_{n_{5}+1}^{n_{5}}~{}~{}\widehat{N-3}\right\rvert\bigl{\lvert}0_{n_{3}-1}~{}~{}0_{n_{2}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-4}\bigr{\rvert}\\\
&-\bigl{\lvert}0_{n_{3}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert
0_{n_{2}-1}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{3}+1\\\
n_{5}+1\end{subarray}}^{n_{3},n_{5}}~{}~{}\widetilde{0}_{n_{5}+1}^{n_{5}}~{}~{}\widehat{N-4}\right\rvert\\\
&\left.+\bigl{\lvert}0_{n_{2}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert
0_{n_{3}-1}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{3}+1\\\
n_{5}+1\end{subarray}}^{n_{3},n_{5}}~{}~{}\widetilde{0}_{n_{5}+1}^{n_{5}}~{}~{}\widehat{N-4}\right\rvert\right),\end{split}$

which is the expansion of following determinant

$\begin{split}&\Gamma(a_{3})\Gamma(a_{5})\,\left|\begin{matrix}\widehat{N-4}&N-3&0_{n_{3}-1}&0_{n_{2}-1}&0_{n_{1}-1}&0&0&0\\\
0&N-3&0_{n_{3}-1}&0_{n_{2}-1}&0_{n_{1}-1}&\widetilde{0}_{\begin{subarray}{c}n_{3}+1\\\
n_{5}+1\end{subarray}}^{n_{3},n_{5}}&\widetilde{0}_{n_{5}+1}^{n_{5}}&\widehat{N-4}\end{matrix}\right|,\end{split}$
(A.14)

which is zero. Hence we have

$\displaystyle M_{1}^{\prime}\\!:={}$
$\displaystyle\\!(a_{1}\\!+\\!a_{3})(a_{1}\\!+\\!a_{5})(a_{2}\\!-\\!a_{3})\tau_{1}^{3}\tau_{235}\\!-\\!(a_{1}\\!-\\!a_{2})(a_{1}\\!-\\!a_{3})(a_{2}\\!-\\!a_{3})\tau^{3}\tau_{1235}\\!$
(A.15)
$\displaystyle-\\!(a_{1}\\!-\\!a_{3})(a_{2}\\!+\\!a_{3})(a_{2}\\!+\\!a_{5})\tau_{2}^{3}\tau_{135}+2a_{3}(a_{1}-a_{2})(a_{3}+a_{5})\tau\tau_{125}=0.$

Similarly, for $E_{n_{2}}^{-1}E_{n_{1}}^{-1}M_{2}^{{}^{\prime}}$, we have

$\begin{split}&\Gamma(a_{3})\Gamma(a_{5})\left(\bigl{\lvert}\widetilde{0}_{n_{3}+1}^{n_{3}}~{}~{}0_{n_{2}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert\widetilde{0}_{\begin{subarray}{c}n_{5}+1\\\
n_{3}-1\end{subarray}}^{n_{5}}~{}~{}0_{n_{3}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-4}\right\rvert\right.\\\
&-\left\lvert\widetilde{0}_{\begin{subarray}{c}n_{5}+1\\\
n_{3}-1\end{subarray}}^{n_{5}}~{}~{}0_{n_{3}-1}~{}~{}\widehat{N-3}\right\rvert\bigl{\lvert}\widetilde{0}_{n_{3}+1}^{n_{3}}~{}~{}0_{n_{2}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-4}\bigr{\rvert}\\\
&-\bigl{\lvert}\widetilde{0}_{n_{3}+1}^{n_{3}}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert\widetilde{0}_{\begin{subarray}{c}n_{5}+1\\\
n_{3}-1\end{subarray}}^{n_{5}}~{}~{}0_{n_{3}-1}~{}~{}0_{n_{2}-1}~{}~{}\widehat{N-4}\right\rvert\\\
&\left.+\bigl{\lvert}0_{n_{2}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert\widetilde{0}_{n_{3}+1}^{n_{3}}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{5}+1\\\
n_{3}-1\end{subarray}}^{n_{5}}~{}~{}0_{n_{3}-1}~{}~{}\widehat{N-4}\right\rvert\right),\end{split}$

which corresponds to following determinant

$\begin{split}&\Gamma(a_{3})\Gamma(a_{5})\,\left|\begin{matrix}\widehat{N-4}&N-3&0&0_{n_{2}-1}&0_{n_{1}-1}&0&\widetilde{0}_{n_{3}+1}^{n_{3}}&0\\\
0&N-3&0_{n_{3}-1}&0_{n_{2}-1}&0_{n_{1}-1}&\widetilde{0}_{\begin{subarray}{c}n_{5}+1\\\
n_{3}-1\end{subarray}}^{n_{5}}&\widetilde{0}_{n_{3}+1}^{n_{3}}&\widehat{N-4}\end{matrix}\right|,\end{split}$
(A.16)

which is zero as well. So we have

$\displaystyle M_{2}^{\prime}\\!:={}$
$\displaystyle\\!(a_{1}\\!-\\!a_{3})(a_{1}\\!+\\!a_{5})(a_{2}\\!+\\!a_{3})\tau_{13}\tau_{25}^{3}\\!-\\!(a_{1}\\!-\\!a_{2})(a_{1}\\!+\\!a_{3})(a_{2}\\!+\\!a_{3})\tau_{3}\tau_{125}^{3}\\!$
(A.17)
$\displaystyle-\\!(a_{1}\\!+\\!a_{3})(a_{2}\\!-\\!a_{3})(a_{2}\\!+\\!a_{5})\tau_{23}\tau_{15}^{3}+2a_{3}(a_{1}-a_{2})(a_{3}-a_{5})\tau\tau_{125}=0.$

Finally, $E_{n_{2}}^{-1}E_{n_{1}}^{-1}\dot{A}$ is converted to

$\begin{split}&(\Gamma(a_{3}))^{2}\left(\bigl{\lvert}0_{n_{3}-1}~{}~{}0_{n_{2}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert
0_{n_{1}-1}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{3}+2\end{subarray}}^{n_{3},n_{3}}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{3}+1\end{subarray}}^{n_{3}}~{}~{}\widehat{N-4}\right\rvert\right.\\\
&-\bigl{\lvert}\widetilde{0}_{n_{3}+2}^{n_{3},n_{3}}~{}~{}\widetilde{0}_{n_{3}+1}^{n_{3}}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert
0_{\begin{subarray}{c}n_{3}-1\end{subarray}}~{}~{}0_{n_{2}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-4}\right\rvert\\\
&-\bigl{\lvert}0_{n_{3}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert
0_{n_{2}-1}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{3}+2\end{subarray}}^{n_{3},n_{3}}~{}~{}\widetilde{0}_{\begin{subarray}{c}n_{3}+1\end{subarray}}^{n_{3}}~{}~{}\widehat{N-4}\right\rvert\\\
&\left.+\bigl{\lvert}0_{n_{2}-1}~{}~{}0_{n_{1}-1}~{}~{}\widehat{N-3}\bigr{\rvert}\left\lvert
0_{n_{3}-1}~{}~{}\widetilde{0}_{n_{3}+2}^{n_{3},n_{3}}~{}~{}\widetilde{0}_{n_{3}+1}^{n_{3}}~{}~{}\widehat{N-4}\right\rvert\right),\end{split}$

which is equivalent to the the zero valued determinant

$\begin{split}&(\Gamma(a_{3}))^{2}\,\left|\begin{matrix}\widehat{N-4}&N-3&0_{n_{3}-1}&0_{n_{2}-1}&0_{n_{1}-1}&0&0&0\\\
0&N-3&0_{n_{3}-1}&0_{n_{2}-1}&0_{n_{1}-1}&\widetilde{0}_{\begin{subarray}{c}n_{3}+2\end{subarray}}^{n_{3},n_{3}}&\widetilde{0}_{n_{3}+1}^{n_{3}}&\widehat{N-4}\end{matrix}\right|.\end{split}$
(A.18)

Thus we arrive at

$\displaystyle\dot{A}:=$
$\displaystyle(a_{2}-a_{3})(a_{1}+a_{3})^{2}\tau_{1}^{3}\tau_{233}-(a_{1}-a_{2})(a_{1}-a_{3})(a_{2}-a_{3})\tau^{3}\tau_{1233}$
(A.19)
$\displaystyle-(a_{1}-a_{3})(a_{2}+a_{3})^{2}\tau_{2}^{3}\tau_{133}+4a_{3}^{2}(a_{1}-a_{2})\tau\tau_{123}=0.$

Note that it is also a consequence of (A.15) together with the reduction (A.7)
that (A.19) holds.

∎

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