# Superposed periodic kink and pulse solutions of coupled nonlinear equations

Avinash Khare<EMAIL_ADDRESS>Physics Department, Savitribai Phule
Pune University, Pune 411007, India  Saikat Banerjee<EMAIL_ADDRESS>Theoretical Division, T-4, Los Alamos National Laboratory, Los Alamos, New
Mexico 87545, USA  Avadh Saxena<EMAIL_ADDRESS>Theoretical Division and
Center for Nonlinear Studies, Los Alamos National Laboratory, Los Alamos, New
Mexico 87545, USA

###### Abstract

We present novel previously unexplored periodic solutions, expressed in terms
of Jacobi elliptic functions, for both a coupled $\phi^{4}$ model and a
coupled nonlinear Schrödinger equation (NLS) model. Remarkably, these
solutions can be elegantly reformulated as a linear combination of periodic
kinks and antikinks, or as a combination of two periodic kinks or two periodic
pulse solutions. However, we also find that for $m=0$ and a specific value of
the periodicity (or at a nonzero value of the elliptic modulus $m$) this
superposition does not hold. These results demonstrate that the notion of
superposed solutions extends to the coupled nonlinear equations as well.

##  I Introduction

A striking characteristic of linear theories is the presence of the
superposition principle. In linear $n$th order differential equations, there
exist $n$ independent solutions, and any other solution can be expressed as a
linear combination of these independent solutions. However, the highly
nontrivial attribute of nonlinear theories arises, in contrast to the linear
theories, from the fact that they do not obey the superposition principle, and
it is not apparant how many independent solutions exist for a nonlinear
equation. Nevertheless, several nonlinear equations, such as $\phi^{4}$, NLS,
Korteweg-de Vries (KdV), and modified KdV (mKdV) etc., have found extensive
applications in various areas of physics, ranging from the dynamics of coupled
Bose-Einstein condensates [1] to nonlinear photonic integrations [2].
Interestingly, these equations have been shown to exhibit a form of
superposition principle [3]. Particularly, it has been demonstrated that if a
nonlinear differential equation admits a periodic pulse solution expressed as
Jacobi elliptic functions [4] ${\rm dn}(x,m)$ and ${\rm cn}(x,m)$, then a
superposed periodic solution ${\rm dn}(x,m)\pm\sqrt{m}\ cn(x,m)$ will also be
allowed with $m$ being the modulus of the Jacobi elliptic function. In a
similar fashion, if a nonlinear equation admits a solution expressed as ${\rm
dn}^{2}(x,m)$, then another solution expressed as ${\rm
dn}^{2}(x,m)\pm\sqrt{m}{\rm dn}(x,m){\rm cn}(x,m)$ will also be allowed.
Notably, most of the nonlinear differential equations mentioned above also
exhibit a different kind of superposition principle. Specifically, it has been
demonstrated that if a nonlinear differential equation admits ${\rm dn}(x,m)$
and/or ${\rm cn}(x,m)$ as solutions, then it will also allow for complex
parity-time (PT)-invariant solutions expressed as ${\rm dn}(x,m)\pm
i\sqrt{m}{\rm sn}(x,m)$ and/or ${\rm cn}(x,m)\pm i{\rm sn}(x,m)$ [5, 6, 7]. In
a similar way, one finds that if there is a periodic kink solution in terms of
${\rm sn}(x,m)$ for a given nonlinear differential equation, then it also
allows for a complex PT-invariant solution expressed as $\sqrt{m}{\rm
sn}(x,m)\pm i{\rm dn}(x,m)$ in addition to the form ${\rm sn}(x,m)\pm i{\rm
cn}(x,m)$. Moreover, if a nonlinear equation hosts a solution expressed as
${\rm dn}^{2}(x,m)$, it will also allow for complex PT-invariant periodic
solutions, represented as ${\rm dn}^{2}(x,m)\pm i\sqrt{m}{\rm dn}(x,m){\rm
sn}(x,m)$ or ${\rm dn}^{2}(x,m)\pm im{\rm sn}(x,m){\rm cn}(x,m)$ [5, 6, 7].

Inspired by a previous theoretical work [8], Tankeyev, Smagin, Borich, and
Zhuravlev [9, 10] obtained a novel solution that can be expressed as a
combination of a kink and an antikink solution. Analogous solutions have been
previously obtained in the realms of condensed matter physics and field theory
[11, 12, 13, 14]. The key step in the work of Tankeyev et al. [9] lies in
their utilization of a hyperbolic identity. Building upon their findings, we
decided to explore similar identities related to the Jacobi elliptic functions
${\rm sn}(x,m)$, ${\rm cn}(x,m)$, and ${\rm dn}(x,m)$, which guided us in
discerning the potential structure of superposed periodic solutions. In our
earlier investigation [15], we showed that a wide class of nonlinear
differential equations, such as the the symmetric and asymmetric $\phi^{4}$
equation, the NLS, the quadratic-cubic NLS, mKdV and mKdV-KdV equations admit
superposed periodic kink and pulse solutions, with some cases also admitting
superposed hyperbolic kink solutions. Subsequently, in another notable study
[16], we presented novel superposed hyperbolic solutions for few coupled
nonlinear equations, in terms of kink-antikink, or two kinks or two pulse
solutions.

Hence, a natural question arises: Can the aforementioned coupled equations
also admit periodic solutions which can be reexpressed as superposition of
periodic kink-antikink or two periodic kinks or two periodic pulse solutions?
In this paper we show that this is indeed the case. Specifically, we extend
our investigation to the same coupled $\phi^{4}$ model and the coupled NLS
model explored in [16] and show that both these models admit periodic
solutions, which can be reexpressed as either the superposition of a periodic
kink and an antikink, or two distinct periodic kinks, or two periodic pulse
solutions.

Throughout this paper a periodic solution refers to a solution that exhibits
spatial periodicity, unless stated otherwise. Furthermore, we employ the term
“periodic kink” (or “periodic pulse”) solution to denote a kink (or pulse)
lattice solution that goes over to the hyperbolic kink (or pulse) solution in
the limit of $m=1$. Finally, we note that so far we have been unable to obtain
periodic solutions of the coupled mKdV equation, which can be expressed as a
superposition of a periodic kink and an antikink, or two periodic kinks, or
two periodic pulse solutions.

The paper is structured as follows. In Section II, we focus on the same
coupled $\phi^{4}$ model examined in [16] and obtain a large number of new
solutions that can be reexpressed as a superposition of either a periodic kink
and an antikink, or two periodic kinks, or two periodic pulse solutions. In
Subsection III.1 and III.2 respectively, we explore the same solutions but in
the special cases of $B=\frac{m}{1-m}>0$ as well as $m=0$. In these cases, the
solutions cannot be reexpressed as a superposition of two periodic kinks, or a
periodic kink-antikink pair, or two periodic pulse solutions. Notably, we find
that some of these solutions in case $B=\frac{m}{1-m}$ exhibit nonreciprocal
behavior. Moving on to Section IV, we delve into the coupled nonlinear
Schrödinger equation (NLS) model discussed in [16]. Interestingly, we
establish a mapping between this model and the coupled $\phi^{4}$ model
discussed in Section II, allowing us to readily extract the superposed
solutions for the NLS model. Lastly, in Section V, we summarize our main
findings and discuss a few open problems. In Appendix A, we discuss the
superposed solutions of the coupled $\phi^{4}$ model under the assumption that
the two fields are proportional to each other and show that these solutions
can be reexpressed as a superposition of either a periodic kink and an
antikink, two periodic kinks, or two periodic pulse solutions (specifically,
of ${\rm dn}(x,m)$ type). In Appendix B we obtain six solutions involving Lamé
polynomials of order two, three solutions with other Lamé polynomials of order
two, and three nonreciprocal solutions involving Lamé polynomials of order
one.

##  II A Coupled $\phi^{4}$ Model

Let us consider the following coupled $\phi^{4}$ model characterized by the
equations [16]

$\displaystyle\phi_{1xx}$
$\displaystyle=a_{1}\phi_{1}+(b_{1}\phi_{1}^{2}+d_{1}\phi_{2}^{2})\phi_{1},$
(1a) $\displaystyle\phi_{2xx}$
$\displaystyle=a_{2}\phi_{2}+(b_{2}\phi_{1}^{2}+d_{2}\phi_{2}^{2})\phi_{2}.$
(1b)

Note that these coupled equations can be derived from an interaction potential
$V(\phi_{1},\phi_{2})$ only if $d_{1}=b_{2}$. Before we discuss the coupled
superposed periodic kink and pulse solutions, let us note that these coupled
equations also admit several periodic kink and pulse solutions most of which
are valid even when $b_{2}=d_{1}$.

Next, we show that there are 19 periodic solutions (fifteen in this section
and four in Appendix A) for these coupled equations. All of them can be
expressed as distinct superposed solutions either of the form ${\rm cn}(\beta
x+\Delta)\pm{\rm cn}(\beta x-\Delta)$ , or ${\rm dn}(\beta x+\Delta)\pm{\rm
dn}(\beta x-\Delta)$ , or ${\rm sn}(\beta x+\Delta)\pm{\rm sn}(\beta
x-\Delta)$ . To this end, we will make use of certain identities satisfied by
these superposed combinations. These identities are readily derived by using
the following addition theorems for ${\rm sn}(x,m)$, ${\rm cn}(x,m)$, and
${\rm dn}(x,m)$ [4]

$\displaystyle{\rm sn}(a+b,m)$ $\displaystyle=\frac{{\rm sn}(a,m){\rm
cn}(b,m){\rm dn}(b,m)+{\rm sn}(b,m){\rm cn}(a,m){\rm dn}(a,m)}{1-m{\rm
sn}^{2}(a,m){\rm sn}^{2}(b,m)},$ (2a) $\displaystyle{\rm cn}(a+b,m)$
$\displaystyle=\frac{{\rm cn}(a,m){\rm cn}(b,m)-{\rm sn}(a,m){\rm dn}(a,m){\rm
sn}(b,m){\rm dn}(b,m)}{1-m{\rm sn}^{2}(a,m){\rm sn}^{2}(b,m)},$ (2b)
$\displaystyle{\rm dn}(a+b,m)$ $\displaystyle=\frac{{\rm dn}(a,m){\rm
dn}(b,m)-m{\rm sn}(a,m){\rm cn}(a,m){\rm sn}(b,m){\rm cn}(b,m)}{1-m{\rm
sn}^{2}(a,m){\rm sn}^{2}(b,m)},$ (2c)

where $m$ is the modulus of the Jacobi elliptic functions [4]. In particular,
on using Eq. (2a) we obtain the identities

$\displaystyle{\rm sn}(y+\Delta,m)+{\rm sn}(y-\Delta,m)$
$\displaystyle=\frac{2{\rm sn}(y,m){\rm cn}(\Delta,m)}{{\rm
dn}(\Delta,m)[1+B{\rm cn}^{2}(y,m)]},$ (3a) $\displaystyle{\rm
sn}(y+\Delta,m)-{\rm sn}(y-\Delta,m)$ $\displaystyle=\frac{2{\rm cn}(y,m){\rm
dn}(y,m){\rm sn}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)[1+B{\rm cn}^{2}(y,m)]},$
(3b)

where

$B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}>0.$ (4)

On the other hand, on using Eq. (2b) we obtain the identities

$\displaystyle{\rm cn}(y+\Delta,m)+{\rm cn}(y-\Delta,m)$
$\displaystyle=\frac{2{\rm cn}(y,m){\rm cn}(\Delta,m)}{{\rm
dn}^{2}(\Delta,m)[1+B{\rm cn}^{2}(y,m)]},$ (5a) $\displaystyle{\rm
cn}(y-\Delta,m)-{\rm cn}(y+\Delta,m)$ $\displaystyle=\frac{2{\rm sn}(y,m){\rm
dn}(y,m){\rm sn}(\Delta,m)}{{\rm dn}(\Delta,m)[1+B{\rm cn}^{2}(y,m)]},$ (5b)

where $B$ is again given by Eq. (4). Finally, on using Eq. (2c) we obtain the
identities

$\displaystyle{\rm dn}(y+\Delta,m)+{\rm dn}(y-\Delta,m)$
$\displaystyle=\frac{2{\rm dn}(y,m)}{{\rm dn}(\Delta,m)[1+B{\rm
cn}^{2}(y,m)]},$ (6a) $\displaystyle{\rm dn}(y-\Delta,m)-{\rm dn}(y+\Delta,m)$
$\displaystyle=\frac{2m{\rm sn}(y,m){\rm cn}(y,m){\rm sn}(\Delta,m){\rm
cn}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)[1+B{\rm cn}^{2}(y,m)]},$ (6b)

where $B$ is again as given by Eq. (4).

There are two different forms of solutions to the coupled equations Eq. (1a)
and Eq. (1b) depending on if $\phi_{2}(x)\propto\phi_{1}(x)$ or otherwise.
Here we elaborate on those solutions where $\phi_{2}(x)$ and $\phi_{1}(x)$ are
distinct (and not proportional to each other) while those solutions where
$\phi_{2}(x)\propto\phi_{1}(x)$ are discussed in Appendix A.

###  II.1 Solutions where $\phi_{2}(x)$ and $\phi_{1}(x)$ are distinct

We now demonstrate that in case $\phi_{1}$ and $\phi_{2}$ are distinct (i.e.
not proportional to each other) then the coupled equations Eq. (1a) and Eq.
(1b) admit not only superposed periodic kink, i.e. ${\rm sn}(x,m)$, and
periodic ${\rm dn}(x,m)$ pulse solutions but also superposed periodic ${\rm
cn}(x,m)$ pulse solutions except when either $m=0$ or $B=\frac{m}{(1-m)}$. In
both these cases even though there are exact solutions but these are not
superposed solutions. In this section we therefore restrict ourselves to
$0<m\leq 1$ and further assume that $B\neq\frac{m}{(1-m)}$. In the next
section we will discuss the solutions when $m=0$ as well as when
$B=\frac{m}{(1-m)}$; we obtain 15 such superposed solutions. As we show in
Appendix A, in case the two fields $\phi_{1}$ and $\phi_{2}$ are proportional
to each other, in that case one is only able to obtain superposed periodic
kink and pulse solutions of type ${\rm sn}(x,m)$ and ${\rm dn}(x,m)$,
respectively (and not the pulse solutions of the ${\rm cn}(x,m)$ type).

Solution I

Figure 1: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (7) and Eq. (8), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{1}=d_{2}=-1$.

It is not difficult to check that

$\phi_{1}(x)=\frac{A{\rm cn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm sn}(\beta x,m){\rm dn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad B>0,$ (7)

is an exact solution of the coupled equations Eq. (1a), and Eq. (1b), provided

$\displaystyle a_{1}=a_{2}=(2m-1)\beta^{2},\quad
d_{1}D^{2}=3d_{2}D^{2}=-6B\beta^{2},$ $\displaystyle
b_{2}A^{2}=3b_{1}A^{2}=-6(B+1)[m-(1-m)B]\beta^{2}.$ (8)

Since $B>0$, it then follows that for this solution $d_{1},d_{2}<0$. On the
other hand, $b_{1},b_{2}>(<)$ 0 depending on if $m<(>)$ $(1-m)B$. Finally
$a_{1},a_{2}\geq(<)$ 0 depending on if $m\geq(<)$ 1/2. Note that in this
section we are considering the solutions with $m\neq(1-m)B$ and $m\neq 0$.
Further, in case $b_{2}=d_{1}$, then from Eq. (8) it follows that
$d_{2}=b_{1}$. We might add that in all the solutions given below, $B>0$.
Further, in all the solutions given below only $A^{2}$ and $D^{2}$ appear and
without any loss of generality we choose $A,D>0$.

On using the identities Eq. (5a) and Eq. (5b), the coupled solution Eq. (7)
can be re-written as

$\displaystyle\phi_{1}(x)$ $\displaystyle=\sqrt{\frac{m}{2|b_{1}|}}\beta[{\rm
cn}(\beta x+\Delta,m)+{\rm cn}(\beta x-\Delta,m)],$ (9a)
$\displaystyle\phi_{2}(x)$ $\displaystyle=\sqrt{\frac{m}{2|d_{2}|}}\beta[{\rm
cn}(\beta x-\Delta,m)-{\rm cn}(\beta x+\Delta,m)],$ (9b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$. The
structure of these solutions is shown in Fig. 1 for distinct choices of the
shift parameter $\Delta$, and the modulus $m$. We have independently checked
this solution and all subsequent solutions using MATHEMATICA V.13.1 as well.

Solution II

Figure 2: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (10) and Eq. (11), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{1}=d_{2}=-1$.

We note that

$\phi_{1}(x)=\frac{A{\rm dn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm sn}(\beta x,m){\rm cn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad$ (10)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle a_{1}=a_{2}=(2-m)\beta^{2},\quad
d_{1}D^{2}=3d_{2}D^{2}=-6B[m-(1-m)B]\beta^{2},$ $\displaystyle
b_{2}A^{2}=3b_{1}A^{2}=-6(1+B)\beta^{2}.$ (11)

Since $B>0$, it follows that for this solution while $a_{1},a_{2}>0$,
$b_{1},b_{2}<0$. On the other hand $d_{1},d_{2}>(<)$ 0 depending on if $m<(>)$
$(1-m)B$. Further, in case $b_{2}=d_{1}$, then from Eq. (11) it follows that
$d_{2}=b_{1}$.

On using the identities Eq. (6a) and Eq. (6b), the coupled solution Eq. (11)
can be rewritten as a superposition of the Jacobi elliptic functions

$\displaystyle\phi_{1}(x)$ $\displaystyle=\frac{\beta}{\sqrt{2|b_{1}|}}[{\rm
dn}(\beta x+\Delta,m)+{\rm dn}(\beta x-\Delta,m)],$ (12a)
$\displaystyle\phi_{2}(x)$ $\displaystyle=\frac{\beta}{\sqrt{2|d_{2}|}}[{\rm
dn}(\beta x-\Delta,m)-{\rm dn}(\beta x+\Delta,m)],$ (12b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$. The
structure of these solutions is shown in Fig. 2 for three distinct choices of
$m$ and a fixed value of $\Delta$. In this and representative subsequent
figures we will not show the variation with $\Delta$.

Solution III

It is easy to check that

$\phi_{1}(x)=\frac{A{\rm cn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm cn}(\beta x,m){\rm sn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad$ (13)

is an exact solution of the coupled equations Eq. (1a), and Eq. (1b), provided

$\displaystyle a_{1}=[(2m-1)-6(1-m)B]\beta^{2},\quad
b_{1}A^{2}=2(B+1)[3(1-m)B^{2}+4(1-m)B-m]\beta^{2},$ $\displaystyle
d_{1}D^{2}=-6[m+(1-m)B^{2}]B\beta^{2},$ $\displaystyle
a_{2}=[(2m-1)-6(1-m)B]\beta^{2},\quad
b_{2}A^{2}=6(B+1)[(1-m)B^{2}+2(1-m)B-m]\beta^{2},$ $\displaystyle
d_{2}D^{2}=-2[3(1-m)B^{2}+2(1-m)B+m]B\beta^{2}\,.$ (14)

Notice that for this solution $d_{1},d_{2}<0$ while $a_{1}\geq(<)$ 0 depending
on if $\frac{(2m-1)}{6(1-m)}\geq(<)$ $B$. On the other hand $a_{2}\geq(<)$ 0
depending on whether $\frac{5m-4}{6(1-m)}\geq(<)$ $B$. Finally, $b_{1}\geq(<)$
0 depending on whether $(1-m)(B+1)(3B+1)\geq(<)$ 1 while $b_{2}\geq(<)$ 1
depending on if $(1-m)(B+1)^{2}\geq(<)$ 1.

Figure 3: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (13) and Eq. (14), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{2}=d_{1}=-1$.
Since the $\phi_{1}$ amplitude is very large for $m=0.98$ we have instead used
$m=0.9$. Similarly, for $\phi_{2}$ we have kept the elliptic modulus range to
be $0.3<m<0.6$.

On using the identities Eq. (5a) and Eq. (6b), the coupled solution Eq. (13)
can be rewritten as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}^{2}(\Delta,m)}{2{\rm
cn}(\Delta,m)}[{\rm cn}(\beta x+\Delta,)+{\rm cn}(\beta x-\Delta,m)],$ (15a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}^{2}(\Delta,m)}{2m{\rm
cn}(\Delta,m){\rm sn}(\Delta,m)}[{\rm dn}(\beta x-\Delta,m)-{\rm dn}(\beta
x+\Delta,m)],$ (15b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (14). The structure of these solutions is shown in
Fig. 3 for three distinct choices of $m$ and a fixed value of $\Delta$.

Solution IV

It is easy to check that

$\phi_{1}(x)=\frac{A{\rm dn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm sn}(\beta x,m){\rm dn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)}m,\quad$ (16)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle[m-(1-m)B]a_{1}=[(1-m)(4+m)B+m(2-m)]\beta^{2},$
$\displaystyle[m-(1-m)B]b_{1}A^{2}=-2(B+1)[3(1-m)B^{2}+2(1-m)B+m]\beta^{2},$
$\displaystyle m[m-(1-m)B]d_{1}D^{2}=-6B[m-2(1-m)B-(1-m)B^{2}]\beta^{2},$
$\displaystyle[m-(1-m)B]a_{2}=[(1-m)(4m+1)B+m(5-4m)]\beta^{2},$ $\displaystyle
m[m-(1-m)B]d_{2}D^{2}=-B[2m^{2}-5m(1-m)B-3(1-m^{2})B^{2}]\beta^{2},$
$\displaystyle[m-(1-m)B]b_{2}A^{2}=-[6(1-m)B^{3}+6(1-m)B^{2}+(m^{2}+4m+1)B+6m]\beta^{2}\,.$
(17)

Notice that for this solution $a_{1},a_{2}>(<)$ 0 while $b_{1},b_{2}<(>)$ 0
depending on if $m>(<)$ $(1-m)B$. Finally, depending on the values of $B$ and
$m$, $d_{1}$ and $d_{2}$ can be positive or negative.

On using the identities Eq. (5b) and Eq. (6a), the coupled solution Eq. (16)
can be rewritten as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}(\Delta,m)}{2}[{\rm dn}(\beta
x+\Delta,m)+{\rm dn}(\beta x-\Delta,m)],$ (18a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}(\Delta,m)}{2{\rm sn}(\Delta,m)}[{\rm
cn}(\beta x-\Delta,m)-{\rm cn}(\beta x+\Delta,m)],$ (18b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (17).

Hyperbolic Limit

In the limit $m=1$, all four solutions, i.e. the solutions I to IV smoothly
cross over to the hyperbolic limit [15]

$\phi_{1}(x)=\frac{A\cosh(\beta x)}{B+\cosh^{2}(\beta
x)},\quad\phi_{2}(x)=\frac{D\sinh(\beta x)}{B+\cosh^{2}(\beta x)},\quad$ (19)

provided

$\displaystyle
a_{1}=a_{2}=\beta^{2}\,,~{}~{}d_{1}D^{2}=3d_{2}D^{2}=-6B\beta^{2},$
$\displaystyle b_{2}A^{2}=3b_{1}A^{2}=-6(1+B)\beta^{2}.$ (20)

On using Eq. (4), it follows that for this solution
$b_{1},b_{2},d_{1},d_{2}<0$. Further, in case $b_{2}=d_{1}$, then from Eq. (8)
it follows that $d_{2}=b_{1}$.

On using the identities Eq. (6a) and Eq. (6b), the coupled solution Eq. (19)
can be rewritten as [15]

$\displaystyle\phi_{1}(x)=\frac{\beta}{\sqrt{2|b_{1}|}}[{\rm sech}(\beta
x+\Delta)+{\rm sech}(\beta x-\Delta)],$ (21a)
$\displaystyle\phi_{2}(x)=\frac{\beta}{\sqrt{2|d_{2}|}}[{\rm sech}(\beta
x-\Delta)-{\rm sech}(\beta x+\Delta)],$ (21b)

where $B=\sinh^{2}(\Delta)$.

Solution V

It is readily checked that

$\phi_{1}(x)=\frac{A{\rm cn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm sn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad$ (22)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle Ba_{1}=[(2m-1)B+6m]\beta^{2},\quad
Bd_{1}D^{2}=-6[m+(1-m)B^{2}]\beta^{2},$ $\displaystyle
Bb_{1}A^{2}=-2(B+1)[3m+4mB-(1-m)B^{2}]\beta^{2},$ $\displaystyle
Ba_{2}=[(5m-1)B+6m]\beta^{2},\quad
Bd_{2}D^{2}=-2[3m+2mB+(1-m)B^{2}]\beta^{2},$ $\displaystyle
Bb_{2}A^{2}=-6(B+1)[m+2mB-(1-m)B^{2}]\beta^{2}.$ (23)

Notice that for this solution $d_{1},d_{2}<0$ while $b_{1}\geq(<)$ 0 provided
$B^{2}\geq(<)$ $m(B+1)(B+3)$. On the other hand, $b_{2}\geq(<)$ 0 if
$B^{2}\geq(<)$ $m(B+1)^{2}$. Further, $a_{1}\geq(<)$ 0 if $(2m-1)B+6m\geq(<)$
0 while $a_{2}\geq(<)$ 0 provided $(5m-1)B+6m\geq(<)$ 0.

On using the identities Eq. (3a) and Eq. (5a), the coupled solution Eq. (22)
can be rewritten as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}^{2}(\Delta,m)}{2{\rm
cn}(\Delta,m)}[{\rm cn}(\beta x+\Delta,m)+{\rm cn}(\beta x-\Delta,m)],$ (24a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}(\Delta,m)}{2{\rm cn}(\Delta,m)}[{\rm
sn}(\beta x+\Delta,m)+{\rm sn}(\beta x-\Delta,m)],$ (24b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (23).

Solution VI

It is readily checked that

$\phi_{1}(x)=\frac{A{\rm dn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm sn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad$ (25)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle
Ba_{1}=[(5m-4)B+6]\beta^{2}\,,~{}~{}Ba_{2}=[((5m-1)B+6m]\beta^{2},$
$\displaystyle Bd_{1}D^{2}=-6[m-(1-m)B][m-2(1-m)B-(1-m)B^{2}]\beta^{2},$
$\displaystyle Bb_{1}A^{2}=-2(B+1)[3m+2(3m-1)B-3(1-m)B^{2}]\beta^{2},$
$\displaystyle
Bd_{2}D^{2}=-2[m-(1-m)B][3m+2(3m-2)B-3(1-m)^{2}B^{2}]\beta^{2},$
$\displaystyle Bb_{2}A^{2}=-6(B+1)[m+2mB-(1-m)B^{2}]\beta^{2}.$ (26)

Notice that for this solution $a_{1}\geq(<)$ 0 depending on if
$(5m-4)B+6\geq(<)$ 0 while $a_{2}\geq(<)$ 0 provided $(5m-1)B+6m\geq(<)$ 0\.
Similarly depending on the values of $B$ and $m$, $d_{1},d_{2},b_{1},b_{2}$
can be positive or negative.

On using the identities Eq. (3a) and Eq. (6a), the coupled solution Eq. (25)
can be rewritten as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}(\Delta)}{2}[{\rm dn}(\beta
x+\Delta,m)+{\rm dn}(\beta x-\Delta,m)],$ (27a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}(\Delta,m)}{2{\rm cn}(\Delta,m)}[{\rm
sn}(\beta x+\Delta,m)+{\rm sn}(\beta x-\Delta,m)],$ (27b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ and $A$ and
$D$ are as given by Eq. (26).

Hyperbolic Limit

In the limit $m=1$, the solutions V and VI reduce to the hyperbolic superposed
solution [15]

$\phi_{1}(x)=\frac{A\cosh(\beta x)}{B+\cosh^{2}(\beta
x)},\quad\phi_{2}(x)=\frac{D\sinh(\beta x)\cosh(\beta x)}{B+\cosh^{2}(\beta
x)},\quad$ (28)

provided

$\displaystyle Ba_{1}=(B+6)\beta^{2},\quad Bd_{1}D^{2}=-6\beta^{2},$
$\displaystyle Bb_{1}A^{2}=-2(B+1)(4B+3)\beta^{2},$ $\displaystyle
Ba_{2}=2(2B+3)\beta^{2},\quad Bb_{2}A^{2}=-6(B+1)(1+2B)\beta^{2},$
$\displaystyle Bd_{2}D^{2}=-2(3+2B)\beta^{2}.$ (29)

Notice that for this solution $a_{1},a_{2}>0$ while
$b_{1},b_{2},d_{1},d_{2}<0$. Further, in case $b_{2}=d_{1}$, then from Eq.
(29) it follows that $\frac{d_{2}}{b_{1}}=\frac{(2B+1)(2B+3)}{(4B+3)}$.

On using the identities Eq. (3a) and Eq. (5a), the coupled solution Eq. (28)
can be re-expressed as [15]

$\displaystyle\phi_{1}(x)=\frac{\sqrt{3\cosh(2\Delta)}\beta}{2\sqrt{|b_{2}|}}[{\rm
sech}(\beta x+\Delta)+{\rm sech}(\beta x-\Delta)],$ (30a)
$\displaystyle\phi_{2}(x)=\frac{\sqrt{3}\beta}{\sqrt{2|d_{1}|}\sinh(\Delta)}[\tanh(\beta
x+\Delta)+\tanh(\beta x-\Delta)],$ (30b)

where $B=\sinh^{2}(\Delta)$.

Figure 4: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (31) and Eq. (32), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{2}=d_{1}=-1$.

Solution VII

It is readily checked that

$\phi_{1}(x)=\frac{A{\rm cn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm cn}(\beta x,m){\rm dn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad$ (31)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle a_{1}=[(2m-1)-6(1-m)B]\beta^{2},\quad
md_{1}D^{2}=6B[m+(1-m)B^{2}]\beta^{2},$ $\displaystyle
mb_{1}A^{2}=-2[m-(1-m)B][m+4mB-3(1-m)B^{2}]\beta^{2},$ $\displaystyle
a_{2}=[(5m-1)-6(1-m)B]\beta^{2},\quad
md_{2}D^{2}=2B[m-2mB+3(1-m)B^{2}]\beta^{2},$ $\displaystyle
mb_{2}A^{2}=-6[m-(1-m)B][m+2mB-(1-m)B^{2}]\beta^{2}.$ (32)

Notice that for this solution $d_{1}>0$ while $d_{2}\geq(<)$ 0 if
$3B^{2}\geq(<)$ $(B+1)(3B-1)m$. On the other hand, $a_{1}\geq(<)$ 0 provided
$(2m-1)\geq(<)$ $6(1-m)B$ while $a_{2}\geq(<)$ 0 if $(5m-1)\geq(<)$ $6(1-m)B$.
Finally, depending on the values of $B$ and $m$, both $b_{1}$ and $b_{2}$ can
be positive or negative.

On using the identities Eq. (3b) and Eq. (5a), the coupled solution Eq. (31)
can be expressed differently as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}^{2}(\Delta)}{2{\rm
cn}(\Delta,m)}[{\rm cn}(\beta x+\Delta,m)+{\rm cn}(\beta x-\Delta,m)],$ (33a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}^{2}(\Delta,m)}{2{\rm
sn}(\Delta,m)}[{\rm sn}(\beta x+\Delta,m)-{\rm sn}(\beta x-\Delta,m)],$ (33b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (32). The structure of these solutions is shown in
Fig. 4 for three distinct choices of $m$ and a fixed value of $\Delta$.

Solution VIII

It is straightforward to check that

$\phi_{1}(x)=\frac{A{\rm dn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm cn}(\beta x,m){\rm dn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad$ (34)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle[m-(1-m)B]a_{1}=[m(2-m)+(1-m)(4+m)B]\beta^{2},$
$\displaystyle[m-(1-m)B]b_{1}A^{2}=-2[m+2(1+m)B-(1-m)B^{2}]\beta^{2},$
$\displaystyle[m-(1-m)B]d_{1}D^{2}=6B[m-2(1-m)B-(1-m)B^{2}]B\beta^{2},$
$\displaystyle[m-(1-m)B]a_{2}=[(5-m)m+(1-m^{2})B]\beta^{2},$
$\displaystyle[m-(1-m)B]b_{2}A^{2}=-6[m+2mB-(1-m)B^{2}]\beta^{2},$
$\displaystyle[m-(1-m)B]d_{2}D^{2}=2B[m-2(2-m)B-(1-m)B^{2}]\beta^{2}.$ (35)

Notice that for this solution while $a_{1},a_{2}>(<)$ 0 depending on if
$m>(<)$ $(1-m)B$, depending on the values of $B$ and $m$,
$b_{1},b_{2},d_{1},d_{2}$ can take positive or negative values.

On using the identities Eq. (3b) and Eq. (6a), the coupled solution Eq. (34)
can be expressed differently as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}(\Delta,m)}{2}[{\rm dn}(\beta
x+\Delta,m)+{\rm dn}(\beta x-\Delta,m)],$ (36a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}^{2}(\Delta,m)}{2{\rm
sn}(\Delta,m)}[{\rm sn}(\beta x+\Delta,m)-{\rm sn}(\beta x-\Delta,m)],$ (36b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (35).

Hyperbolic Limit

In the limit $m=1$, the solutions VII and VIII become the well-known
hyperbolic solution [15]

$\phi_{1}(x)=\frac{A\cosh(\beta x)}{B+\cosh^{2}(\beta
x)},\quad\phi_{2}(x)=\frac{D}{B+\cosh^{2}(\beta x)},\quad$ (37)

provided

$\displaystyle a_{1}=\beta^{2}\,,~{}~{}d_{1}D^{2}=6B\beta^{2},\quad
b_{1}A^{2}=-2(1+4B)\beta^{2},$ $\displaystyle
a_{2}=4\beta^{2}\,,~{}~{}d_{2}D^{2}=2B(1-2B)\beta^{2},\quad
b_{2}A^{2}=-6(1+2B)\beta^{2}.$ (38)

Notice that for this solution while $a_{1},a_{2}>0$, $b_{1},b_{2}<0$, and
$d_{1}>0$, while $d_{2}>(<)$ 0 depending on if $B<(>)$ 1/2.

On using the identities Eq. (3b) and Eq. (5a), the coupled solution Eq. (37)
can be rewritten as [15]

$\displaystyle\phi_{1}(x)=\frac{\sqrt{3\cosh(2\Delta)}\beta}{2\sqrt{|b_{2}|}\cosh(\Delta)}[{\rm
sech}(\beta x+\Delta)+{\rm sech}(\beta x-\Delta)],$ (39a)
$\displaystyle\phi_{2}(x)=\frac{\sqrt{3}\beta}{\sqrt{2|d_{1}|}{\rm
sech}(\Delta)}[\tanh(\beta x+\Delta)-\tanh(\beta x-\Delta)],$ (39b)

where $B=\sinh^{2}(\Delta)$.

Solution IX

It is easy to check that

$\phi_{1}(x)=\frac{A{\rm cn}(\beta x,m){\rm sn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad\phi_{2}(x)=\frac{D{\rm cn}(\beta x,m){\rm dn}(\beta
x,m)}{1+B{\rm cn}^{2}(\beta x,m)},\quad$ (40)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided
[15]

$\displaystyle a_{1}=[(5m-4)-6(1-m)B]\beta^{2},\quad
a_{2}=[((5m-1)-6(1-m)B]\beta^{2},$ $\displaystyle
d_{1}D^{2}=-6(B+1)[m-2(1-m)B-(1-m)B^{2}]\beta^{2},$ $\displaystyle
b_{1}A^{2}=-2[m-(1-m)B][3m+2(3m-1)B-3(1-m)B^{2}]\beta^{2},$ $\displaystyle
d_{2}D^{2}=-2m(B+1)[3+2(3m-2)B-3(1-m)B^{2}]\beta^{2},$ $\displaystyle
b_{2}A^{2}=-6[m-(1-m)B][m+2mB-(1-m)B^{2}]\beta^{2}.$ (41)

Notice that for this solution while $a_{1}>(<)$ 0 depending on if
$5m-4\geq(<)$ $6(1-m)B$, $a_{2}\geq(<)$ 0 depending on whether $5m-1\geq(<)$
$6(1-m)B$. Further, depending on the values of $B$ and $m$,
$b_{1},b_{2},d_{1},d_{2}$ can take positive or negative values.

On using the identities Eq. (3b) and Eq. (6b), the coupled solution Eq. (40)
can be rewritten as [15]

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}^{2}(\Delta,m)}{2m{\rm
sn}(\Delta,m){\rm cn}(\Delta,m)}[{\rm dn}(\beta x-\Delta,m)-{\rm dn}(\beta
x+\Delta,m)],$ (42a) $\displaystyle\phi_{2}(x)=\frac{D{\rm
dn}^{2}(\Delta,m)}{2{\rm sn}(\Delta,m)}[{\rm sn}(\beta x+\Delta,m)-{\rm
sn}(\beta x-\Delta,m)],$ (42b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (41). The structure of these solutions is shown in
Fig. 5 for three distinct choices of $m$ and a fixed value of $\Delta$.

Figure 5: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (40) and Eq. (41), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{1}=d_{2}=-1$.

Solution X

It is readily checked that

$\phi_{1}(x)=\frac{A{\rm sn}(\beta x,m){\rm dn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad\phi_{2}(x)=\frac{D{\rm cn}(\beta x,m){\rm dn}(\beta
x,m)}{1+B{\rm cn}^{2}(\beta x,m)},\quad$ (43)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle[m-(1-m)B]a_{1}=[(1-m)(1+4m)B+m(5-4m)]\beta^{2},$
$\displaystyle[m-(1-m)B]b_{1}A^{2}=-2[3m+4mB-(1-m)B^{2}]\beta^{2},$
$\displaystyle[m-(1-m)B]d_{1}D^{2}=-[6m+2m(4-m)B+2(1-m)(2+3m)B^{2}+2(1-m)(2m+1)B^{3}]\beta^{2},$
$\displaystyle[m-(1-m)B]a_{2}=[(5-m)m+(1-m^{2})B]\beta^{2},$
$\displaystyle[m-(1-m)B]b_{2}A^{2}=-6[m+2mB-(1-m)B^{2}]\beta^{2},\quad 0<m<1,$
$\displaystyle[m-(1-m)B]d_{2}D^{2}=-[6m+10mB+2(1+m)B^{2}+(1-m)B^{3}]\beta^{2}.$
(44)

Notice that for this solution while $a_{1},a_{2}>(<)$ 0, $d_{1},d_{2}<(>)$ 0
depending on if $m>(<)$ $(1-m)B$. Further, depending on the values of $B$ and
$m$, both $b_{1},b_{2}$ can take positive or negative values.

On using the identities Eq. (3b) and Eq. (5b), the coupled solution Eq. (43)
can be rewritten as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}(\Delta,m)}{2{\rm sn}(\Delta,m)}[{\rm
cn}(\beta x-\Delta,m)-{\rm cn}(\beta x+\Delta,m)],$ (45a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}^{2}(\Delta,m)}{2{\rm
sn}(\Delta,m)}[{\rm sn}(\beta x+\Delta,m)-{\rm sn}(\beta x-\Delta,m)],$ (45b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (44).

Hyperbolic Limit

In the limit $m=1$, the solutions IX and X go over to the hyperbolic solution
[15]

$\phi_{1}(x)=\frac{A\sinh(\beta x)}{B+\cosh^{2}(\beta
x)},\quad\phi_{2}(x)=\frac{D}{B+\cosh^{2}(\beta x)},\quad A,B,D>0,$ (46)

provided

$\displaystyle a_{1}=\beta^{2}\,,~{}~{}d_{1}D^{2}=-6(B+1)\beta^{2},\quad
b_{1}A^{2}=-2(3+4B)\beta^{2},$ $\displaystyle
a_{2}=4\beta^{2}\,,~{}~{}d_{2}D^{2}=-2(B+1)(3+2B)\beta^{2},\quad
b_{2}A^{2}=-6(1+2B)\beta^{2}.$ (47)

Notice that for this solution while $a_{1},a_{2}>0$,
$b_{1},b_{2},d_{1},d_{2}<0$. Further, in case $b_{2}=d_{1}$, then it follows
from Eq. (47) that $\frac{d_{2}}{b_{1}}=\frac{(3+2B)(1+2B)}{3+4B}$.

On using the identities Eq. (3b) and Eq. (6b), the coupled solution Eq. (46)
can be rewritten as [15]

$\displaystyle\phi_{1}(x)=\frac{\sqrt{3\cosh(2\Delta)}\beta}{2\sqrt{|b_{2}|}{\rm
sech}(\Delta)}[{\rm sech}(\beta x-\Delta)-{\rm sech}(\beta x+\Delta)],$ (48a)
$\displaystyle\phi_{2}(x)=\frac{\sqrt{3}\beta}{\sqrt{2|d_{1}|}\sinh(\Delta)}[\tanh(\beta
x+\Delta)-\tanh(\beta x-\Delta)],$ (48b)

where $B=\sinh^{2}(\Delta)$.

Solution XI

Figure 6: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (49) and Eq. (50), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{2}=1,d_{1}=-1$.

It is straightforward to check that

$\phi_{1}(x)=\frac{A{\rm sn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm sn}(\beta x,m){\rm dn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad$ (49)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle(1+B)a_{1}=[(5-m)B-(1+m)]\beta^{2},\quad(1+B)a_{2}=[(5-4m)B-(4m+1)]\beta^{2},$
$\displaystyle m(1+B)d_{1}D^{2}=-6B[m+2mB-(1-m)B^{2}]\beta^{2},$
$\displaystyle m(1+B)b_{1}A^{2}=2[m-(1-m)B][m-2mB+3(1-m)B^{2}]\beta^{2},$
$\displaystyle m(1+B)d_{2}D^{2}=-2[m+4mB-3(1-m)B^{2}]B\beta^{2},$
$\displaystyle m(1+B)b_{2}A^{2}=6[m-(1-m)B][m+(1-m)B^{2}]\beta^{2}.$ (50)

Notice that for this solution while $b_{2}>(<)$ 0 depending on if $m>(<)$
$(1-m)B$, $a_{1}>(<)$ 0 depending on if $(5-m)B\geq(<)$ $(1+m)$,
$a_{2}\geq(<)$ 0 depending on whether $(5-4m)B\geq(<)$ $4m+1$. Further,
depending on the values of $B$ and $m$, $b_{1},d_{1},d_{2}$ can take positive
or negative values.

On using the identities Eq.(3a) and Eq. (5b), the coupled solution Eq.(49) can
be rewritten as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}(\Delta,m)}{2{\rm cn}(\Delta,m)}[{\rm
sn}(\beta x+\Delta,m)+{\rm sn}(\beta x-\Delta,m)],$ (51a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}(\Delta,m)}{2{\rm sn}(\Delta,m)}[{\rm
cn}(\beta x-\Delta,m)-{\rm cn}(\beta x+\Delta,m)],$ (51b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (50). The structure of these solutions is shown in
Fig. 6 for three distinct choices of $m$ and a fixed value of $\Delta$.

Solution XII

Figure 7: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (52) and Eq. (53), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{2}=d_{1}=-1$.

It is easy to check that

$\phi_{1}(x)=\frac{A{\rm sn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm sn}(\beta x,m){\rm cn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad$ (52)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle(1+B)a_{1}=[(5-m)B-(1+m)]\beta^{2},\quad(1+B)a_{2}=[(2-m)B-(4+m)]\beta^{2},$
$\displaystyle(1+B)d_{1}D^{2}=-6B[m+2mB-(1-m)B^{2}]\beta^{2},$
$\displaystyle(1+B)b_{1}A^{2}=2[m-2(2-m)B-(1-m)B^{2}]\beta^{2},$
$\displaystyle(1+B)d_{2}D^{2}=-2B[m+2(1+m)B-(1-m)B^{2}]\beta^{2},$
$\displaystyle(1+B)b_{2}A^{2}=6[m-2(1-m)B-(1-m)B^{2}]\beta^{2}.$ (53)

Notice that for this solution while $a_{1}>(<)$ 0 depending on if
$(5-m)B\geq(<)$ $(1+m)$, $a_{2}\geq(<)$ 0 depending on whether $(2-m)B\geq(<)$
$4+m$. Further, depending on the values of $B$ and $m$,
$b_{1},b_{2},d_{1},d_{2}$ can take positive or negative values.

On using the identities Eq. (3a) and Eq. (6b), the coupled solution Eq. (52)
can be rewritten as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}(\Delta,m)}{2{\rm cn}(\Delta,m)}[{\rm
sn}(\beta x+\Delta,m)+{\rm sn}(\beta x-\Delta,m)],$ (54a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}^{2}(\Delta,m)}{2{\rm
cn}(\Delta,m){\rm sn}(\Delta,m)}[{\rm dn}(\beta x-\Delta,m)-{\rm dn}(\beta
x+\Delta,m)],$ (54b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (53). The structure of these solutions is shown in
Fig. 7 for three distinct choices of $m$ and a fixed value of $\Delta$.

Hyperbolic Limit

In the limit $m=1$, the solutions XI and XII go over to the hyperbolic
solution [15]

$\phi_{1}(x)=\frac{A\sinh(\beta x)\cosh(\beta x)}{B+\cosh^{2}(\beta
x)},\quad\phi_{2}(x)=\frac{D\sinh(\beta x)}{B+\cosh^{2}(\beta x)},\quad$ (55)

provided

$\displaystyle(1+B)a_{1}=2(2B-1)\beta^{2},\quad(1+B)d_{1}D^{2}=-6B(1+2B)\beta^{2},$
$\displaystyle(1+B)b_{1}A^{2}=2(1-2B)\beta^{2},\quad(1+B)a_{2}=(B-5)\beta^{2},$
$\displaystyle(1+B)d_{2}D^{2}=-2B(1+4B)\beta^{2},\quad(1+B)b_{2}A^{2}=6\beta^{2}.$
(56)

Notice that for this solution whereas $b_{2}>0$, $d_{1},d_{2}<0$, while other
three parameters, i.e. $a_{1},a_{2},b_{1}$ depend on the value of $B$. In
particular, in case $B>(<)$ 1/2 then $a_{1}>(<)$ 0 while $b_{1}<(>)$ 0\. On
the other hand $a_{2}>(<)$ 0 depending on if $B>(<)$ 5\. Thus such a solution
does not exist in case $d_{1}=b_{2}$.

On using the identities Eq. (3a) and Eq. (6b), the coupled solution Eq. (55)
can be rewritten as [15]

$\displaystyle\phi_{1}(x)=\frac{\sqrt{3}\beta}{\sqrt{2b_{2}}\cosh(\Delta)}[\tanh(\beta
x+\Delta)+\tanh(\beta x-\Delta)],$ (57a)
$\displaystyle\phi_{2}(x)=\frac{\sqrt{3\cosh(2\Delta)}\beta}{\sqrt{2|d_{1}|}\cosh(\Delta)}[{\rm
sech}(\beta x-\Delta)-{\rm sech}(\beta x+\Delta)],$ (57b)

where $B=\sinh^{2}(\Delta)$.

Solution XIII

It is straightforward to check that

$\phi_{1}(x)=\frac{A{\rm sn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm cn}(\beta x,m){\rm dn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad$ (58)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle a_{1}=a_{2}=-(1+m)\beta^{2},\quad
d_{1}D^{2}=3d_{2}D^{2}=6B(B+1)\beta^{2},$ $\displaystyle
b_{2}A^{2}=3b_{1}A^{2}=6[m-(1-m)B]\beta^{2}.$ (59)

Thus for such a solution while $a_{1},a_{2}<0$, $d_{1},d_{2}>0$. On the other
hand, $b_{1},b_{2}>(<)$ 0 depending on whether $m>(<)$ $(1-m)B$.

On using the identities Eq.(3a) and Eq. (3b), the coupled solution Eq. (58)
can be rewritten as

$\displaystyle\phi_{1}(x)={\rm
dn}(\Delta,m)\frac{\sqrt{m}\beta}{\sqrt{2b_{1}}}[{\rm sn}(\beta
x+\Delta,m)+{\rm sn}(\beta x-\Delta,m)],$ (60a)
$\displaystyle\phi_{2}(x)=\frac{\sqrt{m}\beta}{\sqrt{2d_{2}}}[{\rm sn}(\beta
x+\Delta,m)-{\rm sn}(\beta x-\Delta,m)],$ (60b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$. The
structure of these solutions is shown in Fig. 8 for three distinct choices of
$m$ and a fixed value of $\Delta$.

Figure 8: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (58) and Eq. (59), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{2}=d_{1}=-1$.

Hyperbolic Limit

In the limit $m=1$, the solution XIII goes over to the hyperbolic solution
[15]

$\phi_{1}(x)=\frac{A\sinh(\beta x)\cosh(\beta x)}{B+\cosh^{2}(\beta
x)},\quad\phi_{2}(x)=\frac{D}{B+\cosh^{2}(\beta x)},\quad A,B,D>0,$ (61)

provided

$\displaystyle
a_{1}=a_{2}=-2\beta^{2}\,,~{}~{}d_{1}D^{2}=3d_{2}D^{2}=6B(B+1)\beta^{2},$
$\displaystyle b_{2}A^{2}=3b_{1}A^{2}=6\beta^{2}.$ (62)

Thus for such a solution while $a_{1},a_{2}<0$, $b_{1},b_{2},d_{1},d_{2}>0$.
Further, if $d_{1}=b_{2}$, Eq. (62) implies that $d_{2}=b_{1}$.

On using the identities Eq. (3a) and Eq. (3b), the coupled solution Eq. (61)
can be rewritten as [15]

$\displaystyle\phi_{1}(x)=\frac{\beta}{\sqrt{2b_{1}}}[\tanh(\beta
x+\Delta)+\tanh(\beta x-\Delta)],$ (63a)
$\displaystyle\phi_{2}(x)=\frac{\beta}{\sqrt{2d_{2}}}[\tanh(\beta
x+\Delta)-\tanh(\beta x-\Delta)],$ (63b)

where $B=\sinh^{2}(\Delta)$.

Solution XIV

It is easy to check that

$\phi_{1}(x)=\frac{A{\rm cn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad\phi_{2}(x)=\frac{D{\rm dn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad$ (64)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle
Ba_{1}=[(2m-1)B+6m]\beta^{2},\quad(1-m)Bd_{1}D^{2}=-6[(1-m)B^{2}+m]\beta^{2},$
$\displaystyle(1-m)Bb_{1}A^{2}=2[m-(1-m)B][3m-4(1-m)B-(1-m)B^{2}]\beta^{2},$
$\displaystyle
Ba_{2}=[(5m-4)B+6m]\beta^{2},\quad(1-m)Bd_{2}D^{2}=2[2(1-m)B-(1-m)B^{2}-3m]\beta^{2},$
$\displaystyle(1-m)Bb_{2}A^{2}=6[2m^{2}-3m(1-m)B+(1-m)(2-3m)B^{2}+(1-m)^{2}B^{3}]\beta^{2}.$
(65)

Notice that for this solution $0<m<1$ while $d_{1}<0$. Further, $a_{1}\geq(<)$
0 depending on if $2m(B+3)\geq(<)$ $B$, $a_{2}\geq(<)$ 0 depending on whether
$m(5B+6)\geq(<)$ $4B$. Finally, depending on the values of $B$ and $m$,
$b_{1},b_{2},d_{2}$ can take positive or negative values.

On using the identities Eq. (5a) and Eq. (6a), the coupled solution Eq. (64)
can be expressed differently as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}^{2}(\Delta,m)}{2{\rm
cn}(\Delta,m)}[{\rm cn}(\beta x+\Delta,m)+{\rm cn}(\beta x-\Delta,m)],$ (66a)
$\displaystyle\phi_{2}(x)=\frac{D{\rm dn}(\Delta,m)}{2}[{\rm dn}(\beta
x+\Delta,m)+{\rm dn}(\beta x-\Delta,m)],$ (66b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (65).

Solution XV

It is easy to check that

$\phi_{1}(x)=\frac{A{\rm cn}(\beta x,m){\rm sn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad\phi_{2}(x)=\frac{D{\rm sn}(\beta x,m){\rm dn}(\beta
x,m)}{1+B{\rm cn}^{2}(\beta x,m)},\quad$ (67)

is an exact solution of the coupled equations Eq. (1a) and Eq. (1b), provided

$\displaystyle(B+1)a_{1}=[(2-m)B-(4+m)]\beta^{2},\quad(B+1)a_{2}=[(5-4m)B-(4m+1)]\beta^{2},$
$\displaystyle(1-m)(B+1)d_{1}D^{2}=6[1-(1-m)(1+B)^{2}]\beta^{2},$
$\displaystyle(1-m)(B+1)b_{1}A^{2}=-2[m-(1-m)^{2}B][3m-2(1-m)B+(1-m)^{2}B^{2}]\beta^{2},$
$\displaystyle(1-m)(B+1)b_{2}A^{2}=-6[m-(1-m)B][m+(1-m)B^{2}]\beta^{2},$
$\displaystyle(1-m)(B+1)d_{2}D^{2}=2[3m-4(1-m)B-(1-m)B^{2}]\beta^{2}.$ (68)

Notice that for this solution $0<m<1$ while $b_{2}<(>)$ 0 depending on if
$m>(<)$ $(1-m)B$. Further, $a_{1}\geq(<)$ 0 depending on if $(2-m)B)\geq(<)$
$4+m$, $a_{2}\geq(<)$ 0 depending on whether $(5-4m)B\geq(<)$ $4m+1$. Finally,
depending on the values of $B$ and $m$, $b_{1},d_{1},d_{2}$ can take positive
or negative values.

On using the identities Eq. (5b) and Eq. (6b), the coupled solution Eq. (67)
can be expressed differently as

$\displaystyle\phi_{1}(x)=\frac{A{\rm dn}^{2}(\Delta,m)}{2m{\rm
sn}(\Delta,m){\rm cn}(\Delta,m)}[{\rm dn}(\beta x-\Delta,m)-{\rm dn}(\beta
x+\Delta,m)],$ (69a) $\displaystyle\phi_{2}(x)=\frac{D{\rm
dn}(\Delta,m)}{2{\rm sn}(\Delta,m)}[{\rm cn}(\beta x-\Delta,m)-{\rm cn}(\beta
x+\Delta,m)],$ (69b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$ while $A$ and
$D$ are as given by Eq. (68). The structure of these solutions is shown in
Fig. 9 for three distinct choices of $m$ and a fixed value of $\Delta$.

Figure 9: Variations of coupled fields $\phi_{1}(x,m)$ and $\phi_{2}(x,m)$ as
a function of the position following Eq. (67) and Eq. (68), respectively. Note
that the scaling parameter $\beta$ is assumed to be 1, and $b_{2}=d_{1}=-1$.

##  III The Periodic But Not Superposed Solutions When $B=m/(1-m)$ and when
$m=0$

We now discuss the fifteen solutions obtained in the previous section in case
$m=(1-m)B$ as well as when $m=0$. We will show that in both these cases while
we have exact solutions but there are no superposed solutions. Let us discuss
the two cases one by one.

###  III.1 The Periodic Solutions When $B=m/(1-m)$

If we look at the 15 solutions obtained in the previous section, one common
thing in all these solutions is that $\phi_{1},\phi_{2}$ in these solutions
always have $1+B{\rm cn}^{2}(\beta x,m)$ in the denominator. But in case
$B=\frac{m}{(1-m)}$, then

$1+B{\rm cn}^{2}(\beta x,m)=\frac{{\rm dn}^{2}(\beta x,m)}{(1-m)}\,.$ (70)

Further, since in all the solutions in the previous section $B=\frac{m{\rm
sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$, hence we find that for the
special case of $B=\frac{m}{(1-m)}$ one has ${\rm sn}^{2}(\Delta,m)=1$ so that
$\Delta=\pm K(m)$. Now using the well known relations [4]

${\rm dn}(y-\Delta)={\rm dn}(y+\Delta),\quad{\rm sn}(y-\Delta)=-{\rm
sn}(y+\Delta),\quad{\rm cn}(y-\Delta)=-{\rm cn}(y+\Delta),$ (71)

it is clear that in this special case, while we have exact solutions, these
solutions cannot be reexpressed as a nontrivial superposition of a periodic
kink-antikink or superposition of two periodic kinks or two periodic pulse
solutions.

Let us now analyze the 15 exact solutions in this special case. We find that
in this special case, $\phi_{1},\phi_{2}$ in the three solutions IV, VIII and
X are all in terms of (inverse) Lamé polynomials of order one, i.e.
$\frac{cn(\beta x,m)}{{\rm dn}(\beta x,m)},\frac{{\rm sn}(\beta x,m)}{{\rm
dn}(\beta x,m)},\frac{1}{{\rm dn}(\beta x,m)}$. On the other hand, we find
that $\phi_{1}$ and $\phi_{2}$ in the three solutions III, V and XII are all
in terms of the (inverse) Lamé polynomials of order two, i.e. $\frac{cn(\beta
x,m)}{{\rm dn}^{2}(\beta x,m)},\frac{{\rm sn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\frac{{\rm cn}(\beta x,m){\rm sn}(\beta x,m)}{{\rm dn}^{2}(\beta x,m)}$.
Remarkably, the remaining 9 solutions have an unusual form with one of
$\phi_{1}$ and $\phi_{2}$ is (inverse) Lamé polynomial of order one as given
above, while the other one is the (inverse) Lamé polynomial of order two as
given above. We now present exact solutions in all the fifteen cases.

Case I: Three Solutions in Terms of (inverse) Lamé Polynomials of Order One

Solution I

It is easy to check that

$\phi_{1}=\frac{A}{{\rm dn}(\beta x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm
sn}(\beta x,m)}{{\rm dn}(\beta x,m)},$ (72)

provided

$\displaystyle b_{1}A^{2}+d_{1}D^{2}$
$\displaystyle=b_{2}A^{2}+d_{2}D^{2}=-2(1-m)\beta^{2},$ (73a) $\displaystyle
a_{1}-d_{1}D^{2}$
$\displaystyle=(2-m)\beta^{2}\,,~{}~{}a_{2}-d_{2}D^{2}=\beta^{2}.$ (73b)

Note that this solution is only valid for $0<m<1$. In fact this is true for
all the fifteen solutions presented below as well as for the six solutions
presented in Appendix B.

Solution II

It is easy to check that

$\phi_{1}=\frac{A}{{\rm dn}(\beta x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm
cn}(\beta x,m)}{{\rm dn}(\beta x,m)},$ (74)

is an exact solution of the coupled Eq. (1a) and Eq. (1b) provided

$\displaystyle b_{1}A^{2}-(1-m)d_{1}D^{2}$
$\displaystyle=b_{2}A^{2}-(1-m)d_{2}D^{2}=-2(1-m)\beta^{2},$ (75a)
$\displaystyle a_{1}+d_{1}D^{2}$ $\displaystyle=(2-m)\beta^{2},\quad
a_{2}+d_{2}D^{2}=(1-m)\beta^{2}.$ (75b)

Solution III

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}(\beta x,m)},$
(76)

is an exact solution of the coupled Eq. (1a) and Eq. (1b) provided

$\displaystyle b_{1}A^{2}-(1-m)d_{1}D^{2}$
$\displaystyle=b_{2}A^{2}-(1-m)d_{2}D^{2}=-2(1-m)\beta^{2},$ (77a)
$\displaystyle a_{1}+md_{1}D^{2}$ $\displaystyle=(2m-1)\beta^{2},\quad
a_{2}+md_{2}D^{2}=-(1-m)\beta^{2}.$ (77b)

Case II: Three Solutions in Terms of (inverse) Lamé Polynomials of Order Two

Solution IV

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},$ (78)

is an exact solution of the coupled Eq. (1a) and Eq. (1b) provided

$\displaystyle b_{1}=b_{2},\quad d_{1}=d_{2},\quad
mb_{1}A^{2}=-6(1-m)^{2}\beta^{2},\quad md_{1}D^{2}=-6(1-m)\beta^{2},$ (79a)
$\displaystyle a_{1}=(5-m)\beta^{2},\quad a_{2}=(5-4m)\beta^{2}.$ (79b)

Solution V

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m){\rm sn}(\beta
x,m)}{{\rm dn}^{2}(\beta x,m)},$ (80)

is an exact solution of the coupled Eq. (1a) and Eq. (1b) provided

$\displaystyle b_{1}=b_{2},\quad d_{1}=d_{2},\quad
b_{1}A^{2}=-6(1-m)^{2}\beta^{2},\quad d_{1}D^{2}=-6(1-m)\beta^{2},$ (81a)
$\displaystyle a_{1}=(5m-1)\beta^{2},\quad a_{2}=(5m-4)\beta^{2}.$ (81b)

Solution VI

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m){\rm sn}(\beta
x,m)}{{\rm dn}^{2}(\beta x,m)},$ (82)

is an exact solution of the coupled Eq. (1a) and Eq. (1b) provided

$b_{1}=b_{2}=-d_{1}=-d_{2},\quad D=\pm A,\quad b_{1}A^{2}=6\beta^{2},\quad
a_{1}=-(4m+1)\beta^{2},\quad a_{2}=-(4+m)\beta^{2}.$ (83)

We might add here that there is also a fourth (inverse) Lamé polynomial of
order two, i.e. $A[\frac{1}{{\rm dn}^{2}(\beta x,m)}+p]$, where $p$ is a
constant, which has not occured in any of the 15 solutions that we have
considered. Instead, it would follow if we had considered
$\phi_{1}=\frac{A{\rm cn}^{2}(\beta x,m)}{1+B{\rm cn}^{2}(\beta x,m)}$.
However, we did not consider it as it cannot be re-expressed as a
superposition of a periodic kink-antikink or two periodic kinks or two
periodic pulse solutions. For completeness, we present the three exact
solutions of this (inverse) Lamé polynomial of order two with the other three
(inverse) Lamé polynomials of order two in Appendix B.

Case III: Nine Solutions in Terms of (inverse) Lamé Polynomials of Order One
and Two

If we rearrange so that in the remaining nine solutions $\phi_{1}$ is always
(inverse) Lamé polynomial of order two while $\phi_{2}$ is (inverse) Lamé
polynomial of order one, then one finds that in all the nine solutions
$b_{1}=b_{2}=0$ and hence effectively we need to solve rather unusual simpler
coupled $\phi^{4}$ equations as

$\displaystyle\phi_{1xx}=a_{1}\phi_{1}+d_{1}\phi_{2}^{2}\phi_{1},$ (84a)
$\displaystyle\phi_{2xx}=a_{2}\phi_{2}+d_{2}\phi_{2}^{3}.$ (84b)

So far as we are aware of, such solutions have never been discussed in the
literature. We now present these one by one.

Solution VII

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}(\beta x,m)},$
(85)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=6\beta^{2},\quad a_{1}=-(4m+1)\beta^{2},\quad
d_{2}D^{2}=2\beta^{2},\quad a_{2}=-(1+m)\beta^{2}.$ (86)

This solution is valid for arbitrary nonzero values of $A$ since none of the
relations in Eq. (86) depends on $A$. This is in fact true for all the 9
solutions presented here as well as for similar three solutions in Appendix B.
Note that Eqs. (84a) and (84b) are nonreciprocal equations since while the
solution for $\phi_{1}$ depends on $\phi_{2}$, the latter is independent of
$\phi_{1}$. This is in fact true for all the 9 solutions presented here as
well as for the similar 3 solutions presented in Appendix B.

Solution VIII

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}(\beta x,m)},$
(87)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=-6(1-m)\beta^{2},\quad a_{1}=(2m-1)\beta^{2},\quad
d_{2}D^{2}=-2(1-m)\beta^{2},\quad a_{2}=(2m-1)\beta^{2}.$ (88)

Solution IX

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D}{{\rm dn}(\beta x,m)},$ (89)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=-6(1-m)\beta^{2},\quad a_{1}=(5-4m)\beta^{2},\quad
d_{2}D^{2}=-2(1-m)\beta^{2},\quad a_{2}=(2-m)\beta^{2}.$ (90)

Solution X

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}(\beta x,m)}$
(91)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=6\beta^{2},\quad a_{1}=-(m+1)\beta^{2},\quad
d_{2}D^{2}=2\beta^{2},\quad a_{2}=-(1+m)\beta^{2}.$ (92)

Solution XI

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}(\beta x,m)},$
(93)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=-6(1-m)\beta^{2},\quad a_{1}=(5m-1)\beta^{2},\quad
d_{2}D^{2}=-2(1-m)\beta^{2},\quad a_{2}=(2m-1)\beta^{2}.$ (94)

Solution XII

It is easy to check that

$\phi_{1}=\frac{A\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D}{{\rm dn}(\beta x,m)},$ (95)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=-6(1-m)\beta^{2},\quad a_{1}=(5-m)\beta^{2},\quad
d_{2}D^{2}=-2(1-m)\beta^{2},\quad a_{2}=(2-m)\beta^{2}.$ (96)

Solution XIII

It is easy to check that

$\phi_{1}=\frac{Am{\rm sn}(\beta x,m){\rm cn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m)}{{\rm dn}(\beta x,m)},$
(97)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=6\beta^{2},\quad a_{1}=-(4+m)\beta^{2},\quad
d_{2}D^{2}=2\beta^{2},\quad a_{2}=-(1+m)\beta^{2}.$ (98)

Solution XIV

It is easy to check that

$\phi_{1}=\frac{Am{\rm sn}(\beta x,m){\rm cn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D\sqrt{m}{\rm sn}(\beta x,m)}{{\rm dn}(\beta x,m)},$
(99)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=-6(1-m)\beta^{2},\quad a_{1}=(5m-4)\beta^{2},\quad
d_{2}D^{2}=-2(1-m)\beta^{2},\quad a_{2}=(2m-1)\beta^{2}.$ (100)

Solution XV

It is easy to check that

$\phi_{1}=\frac{Am{\rm sn}(\beta x,m){\rm cn}(\beta x,m)}{{\rm dn}^{2}(\beta
x,m)},\quad\phi_{2}=\frac{D}{{\rm dn}(\beta x,m)},$ (101)

is an exact solution of coupled Eq. (84a) and Eq. (84b) provided

$d_{1}D^{2}=-6(1-m)\beta^{2},\quad a_{1}=(2-m)\beta^{2},\quad
d_{2}D^{2}=-2(1-m)\beta^{2},\quad a_{2}=(2-m)\beta^{2}.$ (102)

For completeness we have presented the three solutions corresponding to the
fourth inverse Lamé polynomial of order two with the three Lamé polynomials of
order one in Appendix B, even though they do not follow from any of the 15
solutions that we have considered. Instead, it would follow if we had
considered $\phi_{1}=\frac{A{\rm cn}^{2}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)}$. However, we did not consider it as it cannot be re-expressed as a
superposition of a periodic kink-antikink or two periodic kinks or two
periodic pulse solutions.

One obvious question (though not related with the present paper) is: are there
similar solutions of coupled Eq. (84a) and Eq. (84b) in terms of Lamé
polynomials of order one and two. The answer is in the affirmative. In fact
there are 12 such periodic solutions which in turn lead to 6 hyperbolic
nonreciprocal solutions. However, we will not present them in this paper.

###  III.2 The Trigonometric Solutions: When $m=0$

In case $m=0$, the addition theorems for $\sin(x)$ and $\cos(x)$ are rather
simple and well known and clearly there are no superposed solutions in this
limit. One way to discern it is that, in this case $\Delta=0$. It turns out
that in the limit $m=0$, 13 out of the 15 superposed solutions discussed in
the last section reduce to 6 nontrivial trigonometric solutions while
solutions VII and XI are not valid in this limit. We now discuss the allowed 6
solutions one by one.

Solution I

It is easy to check that

$\phi_{1}=\frac{A\cos(\beta x)}{1+B\cos^{2}(\beta
x)},\quad\phi_{2}=\frac{D\sin(\beta x)}{1+B\cos^{2}(\beta x)},\quad B>0,$
(103)

is an exact solution of coupled Eq. (1a) and Eq. (1b) provided

$a_{1}=a_{2}=-\beta^{2},\quad d_{1}D^{2}=3d_{2}D^{2}=-6B\beta^{2},\quad
b_{2}A^{2}=3b_{1}A^{2}=6B(B+1)\beta^{2}.$ (104)

For all the solutions given below, $B>0$ and so we avoid mentioning it again.

Solution II

It is easy to check that

$\phi_{1}=\frac{A\cos(\beta x)}{1+B\cos^{2}(\beta
x)},\quad\phi_{2}=\frac{D}{1+B\cos^{2}(\beta x)},$ (105)

is an exact solution of coupled Eq. (1a) and Eq. (1b) provided

$\displaystyle a_{1}=-\beta^{2},\quad d_{1}D^{2}=-6B\beta^{2},\quad
b_{1}A^{2}=2B(B+4)\beta^{2},$ (106a) $\displaystyle a_{2}=-4\beta^{2},\quad
d_{2}D^{2}=2(2-B)\beta^{2},\quad b_{2}A^{2}=6B(B+2)\beta^{2}.$ (106b)

Solution III

It is easy to check that

$\phi_{1}=\frac{A}{1+B\cos^{2}(\beta x)},\quad\phi_{2}=\frac{D\sin(\beta
x)}{1+B\cos^{2}(\beta x)},$ (107)

is an exact solution of coupled Eq. (1a) and Eq. (1b) provided

$\displaystyle a_{1}=-4\beta^{2},\quad d_{1}D^{2}=-6B(B+2)\beta^{2},\quad
b_{1}A^{2}=2(B+1)(3B+2)\beta^{2},$ (108a) $\displaystyle
a_{2}=-\beta^{2},\quad d_{2}D^{2}=-2B(3B+4)\beta^{2},\quad
b_{2}A^{2},=6B(B+1)\beta^{2}\,.$ (108b)

Solution IV

It is easy to check that

$\phi_{1}=\frac{A}{1+B\cos^{2}(\beta x)},\quad\phi_{2}=\frac{D\sin(\beta
x)\cos(\beta x)}{1+B\cos^{2}(\beta x)},$ (109)

is an exact solution of coupled Eq. (1a) and Eq. (1b) provided

$a_{1}=a_{2}=2\beta^{2},\quad d_{1}D^{2}=3d_{2}D^{2}=6B^{2}\beta^{2},\quad
b_{2}A^{2}=3b_{1}A^{2}=-6(B+1)\beta^{2}.$ (110)

Solution V

It is easy to check that

$\phi_{1}=\frac{A\cos(\beta x)}{1+B\cos^{2}(\beta
x)},\quad\phi_{2}=\frac{D\sin(\beta x)\cos(\beta x)}{1+B\cos^{2}(\beta x)},$
(111)

is an exact solution of coupled Eq. (1a) and Eq. (1b) provided

$\displaystyle a_{1}=-6(B+1)\beta^{2},\quad d_{1}D^{2}=-6B^{3}\beta^{2},\quad
b_{1}A^{2}=2B(B+1)(3B+4)\beta^{2},$ (112a) $\displaystyle
a_{2}=-6(B+4)\beta^{2},\quad d_{2}D^{2}=-2(3B+2)\beta^{2},\quad
b_{2}A^{2}=6B(B+1)(B+2)\beta^{2}.$ (112b)

Solution VI

It is easy to check that

$\phi_{1}=\frac{A\sin(\beta x)}{1+B\cos^{2}(\beta
x)},\quad\phi_{2}=\frac{D\sin(\beta x)\cos(\beta x)}{1+B\cos^{2}(\beta x)},$
(113)

is an exact solution of coupled Eq. (1a) and Eq. (1b) provided

$\displaystyle(1+B)a_{1}=(5B-1)\beta^{2},\quad(1+B)d_{1}D^{2}=6B^{3}\beta^{2},\quad(1+B)b_{1}A^{2}=-2B(B+4)\beta^{2},$
(114a)
$\displaystyle(1+B)a_{2}=2(B-2)\beta^{2},\quad(1+B)d_{2}D^{2}=2B^{2}(B-2)\beta^{2},\quad(1+B)b_{2}A^{2}=-6B(B+2)\beta^{2}.$
(114b)

Having obtained the 15 solutions of the coupled $\phi^{4}$ model with six
parameters, $a_{1},a_{2},b_{1},b_{2},d_{1},d_{2}$, it is worthwhile asking: is
there a range of these 6 parameters for which the exact solution has not been
obtained so far? We believe that these six parameters with each of them being
positive and negative span a six parameter space and it is highly unlikely
that one has found solutions in this entire space of parameters. Out of the 15
solutions that we have obtained, we can try to draw some conclusions by first
carefully looking at the superposed solutions at $m=1$ (i.e. hyperbolic
superposed solutions) obtained earlier [16] and mentioned in the previous
section as well as the exact solutions when $m=(1-m)B$ and $m=0$.

From the six hyperbolic superposed solutions mentioned in the previous section
as well as the 9 solutions obtained above when $m=(1-m)B$ and the six
solutions that we have obtained at $m=0$ we note that in none of the 21
solutions either all the six parameters $a_{1},a_{2},b_{1},b_{2},d_{1},d_{2}$
are positive or all negative. We have also looked at the 15 periodic
superposed solutions obtained in the last section and for a few values of $B$
and $m$ that we have seen, we obtain the same conclusion.

It is worthwhile looking at the other known solutions of this coupled model
and see if the above conclusion is correct or not.

##  IV Novel periodic superposed solutions of a coupled NLS model

Let us consider the following coupled NLS model [16]

$\displaystyle iu_{1t}+u_{1xx}+6[g_{11}|u_{1}|^{2}+g_{12}|u_{2}|^{2}]u_{1}$
$\displaystyle=0,$ (115a) $\displaystyle
iu_{2t}+u_{2xx}+6[g_{21}|u_{1}|^{2}+g_{22}|u_{2}|^{2}]u_{2}$
$\displaystyle=0.$ (115b)

It is worth pointing out that in case $g_{11}=g_{12}=g_{21}=g_{22}$, then it
corresponds to the celebrated Manakov system which is known to be integrable
[17]. On the other hand, in case $g_{11}=g_{21}=-g_{12}=-g_{22}$, then it
corresponds to the Manakov-Zakharov-Schulman (MZS) system [18, 19, 20].

Before we discuss the superposed periodic kink and pulse solutions of Eq.
(115a) and Eq. (115b), let us note that these coupled equations also admit
periodic kink and pulse solutions.

We now show that these coupled equations admit 19 periodic solutions which can
be rewritten as superposed solutions either of the form ${\rm cn}(\beta
x+\Delta)\pm{\rm cn}(\beta x-\Delta)$ , or ${\rm dn}(\beta x+\Delta)\pm{\rm
dn}(\beta x-\Delta)$ , or ${\rm sn}(\beta x+\Delta)\pm{\rm sn}(\beta
x-\Delta)$ .

We start from Eqs. (115a) and (115b) and make an ansatz

$u_{1}(x,t)=e^{-i\omega_{1}t}u_{1}(x),\quad
u_{2}(x,t)=e^{-i\omega_{2}t}u_{2}(x),$ (116)

so that the two coupled equations take the form

$\displaystyle
u_{1xx}(x)=\omega_{1}u_{1}-6[g_{11}|u_{1}|^{2}+g_{12}|u_{2}|^{2}]u_{1},$
(117a) $\displaystyle
u_{2xx}(x)=\omega_{2}u_{2}-6[g_{21}|u_{1}|^{2}+g_{22}|u_{2}|^{2}]u_{2}.$
(117b)

Observe that in case $u_{1}(x)$ and $u_{2}(x)$ are real, the coupled equations
[Eq. (117a), and Eq. (117b)] can be directly mapped to the coupled $\phi^{4}$
equations [Eq. (1a) and Eq. (1b)] with the identification of
$\omega_{1},\omega_{2}$ with $a_{1},a_{2}$, respectively. Further, the four
coupling constants $g_{11},g_{12},g_{21},g_{22}$ can be identified with
$-b_{1},-d_{1},-b_{2},-d_{2}$, respectively. One can then simply read off the
19 superposed periodic solutions of the coupled NLS model from the
corresponding 15 superposed periodic solutions of the coupled $\phi^{4}$ model
as obtained in Sec. II and 4 superposed solutions of the coupled $\phi^{4}$
model as obtained in Appendix A. For completeness, we simply mention one
solution of the coupled NLS model and the readers can similarly obtain the
remaining 18 superposed solutions of the coupled NLS model including the 4
given in Appendix A.

###  IV.1 Solutions of coupled NLS when $u_{1}(x,t)$ and $u_{2}(x,t)$ are
distinct

Solution I

It is easy to check that

$u_{1}(x,t)=e^{i\omega_{1}t}\frac{A{\rm cn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)},\quad u_{2}(x,t)=e^{i\omega_{2}t}\frac{D{\rm sn}(\beta x,m){\rm
dn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta x,m)},$ (118)

with $B>0$ is an exact solution of the coupled equations, Eq. (115a) and Eq.
(115b), provided

$\displaystyle\omega_{1}=\omega_{2}=(2m-1)\beta^{2},\quad
g_{12}D^{2}=3g_{22}D^{2}=6B\beta^{2},$ $\displaystyle
g_{21}A^{2}=3g_{11}A^{2}=6(B+1)[m-(1-m)B]\beta^{2}.$ (119)

Notice that for this solution $g_{12},g_{22}>$ 0\. Further,
$\omega_{1},\omega_{2}\geq(<)$ 0 in case $m\geq(<)$ 1/2 while
$g_{12},g_{22}>0$. Finally, $g_{11},g_{21}\geq(<)$ 0 depending on if
$m\geq(<)$ $(1-m)B$.

On using the identities Eq. (5a) and Eq. (5b), the coupled solution Eq. (118)
can be rewritten as

$\displaystyle u_{1}(x,t)=e^{i\omega_{1}t}\sqrt{\frac{m}{2g_{11}}}\beta[{\rm
cn}(\beta x+\Delta,m)+{\rm cn}(\beta x-\Delta,m)],$ (120a) $\displaystyle
u_{2}(x,t)=e^{i\omega_{2}t}\sqrt{\frac{m}{2g_{22}}}\beta[{\rm cn}(\beta
x-\Delta,m)-{\rm cn}(\beta x+\Delta,m)],$ (120b)

where $B=\frac{m{\rm sn}^{2}(\Delta,m)}{{\rm dn}^{2}(\Delta,m)}$.

##  V Conclusion and Open Problems

In this article, we have expanded the concept of superposition to encompass
coupled $\phi^{4}$ and coupled NLS models, unveiling their capacity to exhibit
not only hyperbolic solutions but also periodic solutions. These periodic
solutions can be expressed as a superposition of a periodic kink and an
antikink, two periodic kinks, or two periodic pulse solutions. Remarkably, we
have identified fifteen such superposed periodic solutions when the coupled
fields are distinct (and not proportional to each other), and four periodic
solutions when the fields are proportional.

Our findings raise intriguing questions that warrant further exploration. We
highlight a few of these questions:

  1. 1.

Unlike the coupled $\phi^{4}$ or coupled NLS models, we have been unable to
obtain periodic solutions for the coupled mKdV equation that can be expressed
as a superposition of a periodic kink and an antikink, or two periodic kinks,
or two periodic pulse solutions. While coupled hyperbolic solutions of the
mKdV equation have been obtained in a different form as the superposition of a
kink and an antikink [15], it is worthwhile to investigate if the concept of
superposition can be extended to periodic solutions of the coupled mKdV
equation as well.

  2. 2.

Numerous other coupled equations, such as coupled KdV, coupled asymmetric
$\phi^{4}$, and coupled $\phi^{6}$, merit exploration to determine the extent
to which the notion of superposition can be applied to their periodic
solutions.

  3. 3.

Is it possible to extend the concept of superposition in alternative and
unexplored directions for nonlinear equations? Given the vast richness of
nonlinear theories, this pursuit holds promising potential.

  4. 4.

What are the direct implications of superposed periodic kink-antikink or two
periodic kink/pulse solutions? Do they correspond to bound states or merely
excite the system?

  5. 5.

Can the notion of superposition be extended to nonlocal theories? Our recent
endeavors [21, 22] have partially extended this concept to nonlocal NLS [23,
24], nonlocal mKdV [25], nonlocal Hirota equations [26, 27] , as well as
coupled nonlocal Ablowitz-Musslimani variant of NLS and coupled nonlocal mKdV
models [28]. It is worth exploring the possibility of extending this concept
to other nonlocal models.

We intend to address some of the issues raised above in the near future.

##  VI Acknowledgment

One of us (AK) is grateful to Indian National Science Academy (INSA) for the
award of INSA Honorary Scientist position at Savitribai Phule Pune University.
The work at Los Alamos National Laboratory was carried out under the auspices
of the US DOE and NNSA under contract No. DEAC52-06NA25396.

##  Appendix A Superposed periodic solutions of the coupled $\phi^{4}$ model:
$\phi_{2}(x)\propto\phi_{1}(x)$

In case $\phi_{2}(x)=\alpha\phi_{1}(x)$, with $\alpha$ being a real number, it
is straightforward to verify that Eq. (1a) and Eq. (1b) are consistent with
each other provided

$a_{2}=a_{1}\,,~{}~{}b_{2}+\alpha^{2}d_{2}=b_{1}+\alpha^{2}d_{1}.$ (121)

Thus, the relevant equation to solve in this case is

$\phi_{1xx}=a_{1}\phi_{1}+(b_{1}+\alpha^{2}d_{1})\phi^{3}.$ (122)

Now, in a recent publication [15] we have already solved such an equation and
shown that it admits four superposed solutions which we can immediately read
off from that paper. For completeness we provide these solutions below.

Solution I

It is readily checked that the $\phi^{4}$ field Eq. (122) and hence the
coupled equations Eq. (1a) and Eq. (1b) admit the periodic solution

$\phi_{1}(x)=\frac{A{\rm dn}(\beta x,m){\rm cn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad A,B,D>0,$ (123)

provided Eq. (121) is satisfied and further

$\displaystyle 0<m<1\,,~{}~{}B=\frac{\sqrt{m}}{1-\sqrt{m}},$ $\displaystyle
a_{1}=-[1+m+6\sqrt{m}]\beta^{2}<0,\quad(b_{1}+\alpha^{2}d_{1})A^{2}=\frac{8\sqrt{m}\beta^{2}}{(1-\sqrt{m})^{2}}.$
(124)

Notice that for this solution $a_{1}<0,b_{1}+\alpha^{2}d_{1}>0$. Furthermore,
this solution is only valid if $0<m<1$, i.e. there is no corresponding
superposed hyperbolic solution. On using the identity Eq. (3b), one can
express differently the periodic solution I given by Eq. (123) as
superposition of a periodic kink and and an antikink solutions, i.e.

$\phi_{1}(x)=\frac{\sqrt{2m}\beta}{\sqrt{b_{1}+\alpha^{2}d_{1}}}\bigg{[}{\rm
sn}(\beta x+\Delta,m)-{\rm sn}(\beta x-\Delta,m)\bigg{]}.$ (125)

Here $\Delta$ is defined by ${\rm sn}(\sqrt{m}\Delta,1/m)=\pm m^{1/4}$, where
the following identity has been used

$\sqrt{m}{\rm sn}(y,m)={\rm sn}(\sqrt{m}y,1/m).$ (126)

Solution II

Remarkably, the $\phi^{4}$ Eq. (122) also admits another periodic solution

$\phi_{1}(x)=\frac{A{\rm sn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta
x,m)}\,,~{}~{}A,B,D>0,$ (127)

provided

$0<m<1,\quad B=\frac{\sqrt{m}}{1-\sqrt{m}},\quad
a_{1}=[6\sqrt{m}-(1+m)]\beta^{2},\quad(b_{1}+\alpha^{2}d_{1})A^{2}=-8\sqrt{m}\beta^{2}.$
(128)

Thus for this solution while $b_{1}+\alpha^{2}d_{1}<0$, $a_{1}>(<)$ 0
depending on if $6\sqrt{m}>(<)$ 0\. On using the identity Eq. (3a), the
solution Eq. (127) can be expressed differently as a superposition of two
periodic kink solutions, i.e.

$\phi_{1}(x)=i\sqrt{\frac{2m}{|b_{1}+\alpha^{2}d_{1}|}}\beta\bigg{[}{\rm
sn}(\beta x+\Delta,m)+{\rm sn}(\beta x-\Delta,m)\bigg{]}.$ (129)

Here $\Delta$ is defined by ${\rm sn}(\sqrt{m}\Delta,1/m)=\pm m^{1/4}$, where
we utilize the identity Eq. (126).

It is worth to note that for both the solutions I and II, the value of $B$ is
the same but while $b_{1}+\alpha^{2}d_{1}>0$ for the first solution,
$b_{1}+\alpha^{2}d_{1}<0$ for the second solution. Further, both the solutions
are valid if $0<m<1$.

Solution III

It is readily checked that the $\phi^{4}$ field Eq. (122) admits another
periodic solution

$\phi_{1}(x)=\frac{A{\rm sn}(\beta x,m){\rm cn}(\beta x,m)}{1+B{\rm
cn}^{2}(\beta x,m)},\quad A,B,D>0,$ (130)

provided

$0<m<1,\quad B=\frac{1-\sqrt{1-m}}{\sqrt{1-m}},\quad
a_{1}=(2-m-6\sqrt{1-m})\beta^{2},\quad(b_{1}+\alpha^{2}d_{1})A^{2}=-\frac{8(1-\sqrt{1-m})^{2}\beta^{2}}{\sqrt{1-m}}.$
(131)

Note that for this solution $b_{1}+\alpha^{2}d_{1}<0$ while $a_{1}$ could be
positive or negative depending on the value of $m$. On using the identity Eq.
(6b), the solution Eq. (130) can be expressed differently as a superposition
of two periodic ${\rm dn}(x,m)$ solutions, i.e.

$\phi_{1}(x)=\beta\sqrt{\frac{2}{|b_{1}+\alpha^{2}d_{1}|}}\bigg{(}{\rm
dn}[\beta x-\frac{K(m)}{2},m]-{\rm dn}[\beta x+\frac{K(m)}{2},m]\bigg{)}.$
(132)

Solution IV

Remarkably, the $\phi^{4}$ Eq. (122) also allows another periodic solution

$\phi_{1}(x)=\frac{A{\rm dn}(\beta x,m)}{1+B{\rm cn}^{2}(\beta x,m)},\quad
A,B,D>0,$ (133)

provided

$0<m<1,\quad B=\frac{1-\sqrt{1-m}}{\sqrt{1-m}},\quad
a_{1}=[2-m+6\sqrt{1-m}]\beta^{2},\quad(b_{1}+\alpha^{2}d_{1})A^{2}=-\frac{8}{\sqrt{1-m}}\beta^{2}.$
(134)

Thus for this solution while $b_{1}+\alpha^{2}d_{1}<0$, $a_{1}>0$. On
employing the identity Eq. (6a), the periodic solution Eq. (133) can be
expressed differently as a superposition of two periodic ${\rm dn}(x,m)$ pulse
solutions, i.e.

$\phi_{1}(x)=\sqrt{\frac{2}{|b_{1}+\alpha^{2}d_{1}|}}\beta\bigg{(}{\rm
dn}[\beta x+K(m)/2,m]+{\rm dn}[\beta x-K(m)/2,m]\bigg{)}\,,$ (135)

where $K(m)$ is the complete elliptic integral of the first kind.

It is worth to note that for both the superposed periodic pulse solutions III
and IV, not only the value of $B$ is the same but also
$b_{1}+\alpha^{2}d_{1}<0$ for both the solutions. Further, both of them are
not valid for $m=1$, i.e. there is no hyperbolic superposed pulse solution.

It is also worth pointing out that while the solutions I to IV admit
superposed periodic ${\rm sn}(x,m)$ kink solutions and superposed periodic
${\rm dn}(x,m)$ pulse solutions, Eq. (122) does not admit superposed ${\rm
cn}(x,m)$ solutions. However, as shown in Sec. II, in case the two fields
$\phi_{1}$ and $\phi_{2}$ are distinct then superposed solutions of ${\rm
cn}(x,m)$ type are also admitted.

##  Appendix B Some Exact Solutions of the (Inverse) Lamé Polynomials of Order
II

We first present three exact solutions of the fourth (inverse) Lamé
polynomial, i.e. $A[\frac{1}{{\rm dn}^{2}(\beta x,m)}+p]$ with the other three
(inverse) Lamé polynomials of order two.

Solution I

It is easy to check that

$\phi_{1}=A\left[\frac{1}{{\rm dn}^{2}(\beta
x,m)}+p\right],\quad\phi_{2}=\frac{D\sqrt{m}{\rm sn}(\beta x,m)}{{\rm
dn}^{2}(\beta x,m)},$ (136)

is an exact solution of the coupled equations [Eq. (1a) and Eq. (1a)] provided

$\displaystyle b_{1}=b_{2},\quad d_{1}=d_{2},\quad
b_{1}A^{2}=-d_{1}D^{2},\quad(2p+1)b_{1}A^{2}=-6(1-m)\beta^{2},$ (137a)
$\displaystyle a_{1}=\beta^{2}[-\frac{2}{p}-p^{2}b_{1}A^{2}],\quad
a_{2}=\beta^{2}[(5-m)-p^{2}b_{1}A^{2}],$ (137b)

where

$p=\frac{-(2-m)\pm\sqrt{1-m+m^{2}}}{3(1-m)}.$ (138)

Solution II

It is easy to check that

$\phi_{1}=A\left[\frac{1}{{\rm dn}^{2}(\beta
x,m)}+p\right],\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m)}{{\rm
dn}^{2}(\beta x,m)},$ (139)

is an exact solution of the coupled equations [Eq. (1a) and Eq. (1a)] provided

$\displaystyle b_{1}=b_{2},\quad d_{1}=d_{2},\quad
b_{1}A^{2}=(1-m)d_{1}D^{2},\quad[2p(1-m)+1]d_{1}D^{2}=-6(1-m)\beta^{2},$
(140a) $\displaystyle a_{1}=\beta^{2}[-\frac{2}{p}+(1-m)p^{2}d_{1}D^{2}],\quad
a_{2}=\beta^{2}[(5-4m)-(1-m)p^{2}d_{1}D^{2}],$ (140b)

where $p$ is again given by Eq. (138).

Solution III

It is easy to check that

$\phi_{1}=A\left[\frac{1}{{\rm dn}^{2}(\beta
x,m)}+p\right],\quad\phi_{2}=\frac{Dm{\rm cn}(\beta x,m){\rm sn}(\beta
x,m)}{{\rm dn}^{2}(\beta x,m)},$ (141)

is an exact solution of the coupled equations [Eq. (1a) and Eq. (1a)] provided

$\displaystyle b_{1}=b_{2},\quad d_{1}=d_{2},\quad
b_{1}A^{2}=(1-m)d_{1}D^{2},\quad[2p(1-m)+2-m]d_{1}D^{2}=-6(1-m)\beta^{2},$
(142a) $\displaystyle
a_{1}=\beta^{2}\bigg{(}-\frac{2}{p}+[(1-m)p^{2}-1]d_{1}D^{2}\bigg{)},\quad
a_{2}=\beta^{2}\bigg{(}(2-m)+[1-(1-m)p^{2}]d_{1}D^{2}\bigg{)},$ (142b)

where $p$ is again given by Eq. (138).

We now present three exact nonreciprocal solutions of the fourth (inverse)
Lamé polynomial, i.e. $A[\frac{1}{{\rm dn}^{2}(\beta x,m)}+p]$ with the three
(inverse) Lamé polynomials of order one.

Solution IV

It is easy to check that

$\phi_{1}=A\left[\frac{1}{{\rm dn}^{2}(\beta
x,m)}+p\right],\quad\phi_{2}=\frac{D\sqrt{m}{\rm cn}(\beta x,m)}{{\rm
dn}(\beta x,m)},$ (143)

where $p$ is a number, is an exact solution of coupled equations [Eq. (1a) and
Eq. (1a)] provided

$d_{1}D^{2}=6\beta^{2},\quad a_{1}=-2(3+\frac{1}{p})\beta^{2},$ (144)

where $p$ is again given by Eq. (138) while

$d_{2}D^{2}=2\beta^{2},\quad a_{2}=-(1+m)\beta^{.}$ (145)

Solution V

It is easy to check that

$\phi_{1}=A\left[\frac{1}{{\rm dn}^{2}(\beta
x,m)}+p\right],\quad\phi_{2}=\frac{D\sqrt{m}{\rm sn}(\beta x,m)}{{\rm
dn}(\beta x,m)},$ (146)

is an exact solution of coupled equations [Eq. (84a) and Eq. (84a)] provided

$d_{1}D^{2}=-6(1-m)\beta^{2},\quad a_{1}=-2[3(1-m)+\frac{1}{p}]\beta^{2},$
(147)

where $p$ is again given by Eq. (138). Further,

$d_{2}D^{2}=-2(1-m)\beta^{2},\quad a_{2}=(2m-1)\beta^{2}.$ (148)

Solution VI

It is easy to check that

$\phi_{1}=A\left[\frac{1}{{\rm dn}^{2}(\beta
x,m)}+y\right],\quad\phi_{2}=\frac{D}{{\rm dn}(\beta x,m)},$ (149)

is an exact solution of coupled equations [Eq. (84a) and Eq. (84a)] provided

$d_{1}D^{2}=-6(1-m)\beta^{2},\quad a_{1}=-\frac{2}{p}\beta^{2},$ (150)

where $p$ is again given by Eq. (138). Further,

$d_{2}D^{2}=-2(1-m)\beta^{2},\quad a_{2}=(2-m)\beta^{2}.$ (151)

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