# Robustness of the data-driven approach in limited angle tomography

Yiran Wang and Yimin Zhong  Yiran Wang
Department of Mathematics, Emory University<EMAIL_ADDRESS>Yimin Zhong
Department of Mathematics and Statistics, Auburn University
<EMAIL_ADDRESS>
(Date: . YW is partly supported by NSF under grant DMS-2205266. YZ is partly
supported by NSF under grant DMS-2309530.)

###### Abstract.

The limited angle Radon transform is notoriously difficult to invert due to
the ill-posedness. In this work, we give a mathematical explanation that the
data-driven approach based on deep neural networks can reconstruct more
information in a stable way compared to traditional methods.

##  1\. Introduction

Consider the limited angle Radon transform on ${\mathbb{R}}^{2}$. Let
$(x_{1},x_{2})$ be the coordinate for ${\mathbb{R}}^{2}$. For
$\delta\in(0,\pi/2),$ let
${\mathcal{S}}_{\delta}=\\{(\cos\theta,\sin\theta)\in{\mathbb{S}}^{1}:|\theta|\leq\delta\\}.$
We use the parallel beam parametrization of the Radon transform. Let
$v\in{\mathcal{S}}_{\delta}$ and $v^{\perp}$ be the unit vector orthogonal to
$v$ so that $\\{v,v^{\perp}\\}$ has the same orientation as the coordinate
frame. The limited angle Radon transform of a function $f$ on
${\mathbb{R}}^{2}$ is

(1)
$R_{\delta}f(p,v)=\int_{{\mathbb{R}}}\chi_{\delta}(v)f(pv+tv^{\perp})dt,\quad
v\in{\mathbb{S}}^{1},p\in{\mathbb{R}}$

where $\chi_{\delta}$ is the characteristic function of
${\mathcal{S}}_{\delta}$ in ${\mathbb{S}}^{1}.$ For simplicity, we assume that
$f$ is supported in the unit disk
${\mathcal{B}}=\\{x\in{\mathbb{R}}^{2}:|x|<1\\}$ where $|\cdot|$ stands for
the Euclidean norm. It is a classical result that $R_{\delta}$ is injective on
compactly supported distributions, see for instance [9, 14]. However, the
inverse problem of recovering $f$ from $R_{\delta}f$ is severely ill-posed.
This was demonstrated from the behavior of the singular values in [4]. On the
microlocal level, it’s clear that some information of $f$ is lost. In
particular, Quinto in [10] gave a microlocal description of the singularities
of $f$ that can be recovered, called “visible singularities”, and those that
cannot, called “invisible singularities”, see Section 2. Due to the ill-
posedness, traditional methods such as filtered back-projection or iterative
methods often produce artifacts in the reconstruction, and it is very
challenging to recover the invisible information.

Recently, there are increasing efforts in applying the data-driven approach
such as deep neural networks to the limited angle tomography, see for example
[1, 2, 3, 6, 7, 12] and references therein. Multiple numerical experiments
reported that the data-driven approach can significantly reduce the artifact
phenomena and even learn the invisible singularities. The success seems
difficult to understand from the existing mathematical theory. Also, the
robustness of the data-driven approach is a concern, see for example [8]. It
is suspected that the data-driven approach should be tailored to specific
tasks, not for general image reconstruction. But a precise understanding is
still missing.

The goal of this paper is to provide a mathematical explanation. It is useful
to interpret the instability of the limited angle Radon transform from the
following view point: there is no stability estimate of the form

(2) $\|f\|_{H^{s}({\mathbb{R}}^{2})}\leq
C(s,t)\|R_{\delta}f\|_{H^{s+t}({\mathbb{R}}\times{\mathbb{S}}^{1})}$

for any $s,t\in{\mathbb{R}}$, see Section VI.2 of [9]. This is because,
roughly speaking, after taking the limited angle Radon transform, the Fourier
transform $\hat{f}$ of $f$ is lost in a conic region (corresponding to the
missing angles), see Figure 1. Recovery of $\hat{f}$ in the rest of the
Fourier space (${\mathcal{K}}_{\delta}$ in Figure 1) however is stable, see
Theorem 3.1. This result is sharp. For the data-driven approach, we consider
$f$ in a training data set ${\mathcal{D}}$ and show that $\hat{f}$ can be
stably recovered in a larger cone (${\mathcal{K}}_{\delta+\epsilon}$ in Figure
1), see Theorem 3.2. The size of the cone depends on ${\mathcal{D}}$ and it is
possible to cover the whole Fourier space for certain ${\mathcal{D}}$. This
explains the phenomena of “learning the invisible”. We remark that our theory
does not rely on the specific structure of the neural network. In Section 4,
we use a simple U-net to test our theory. The concluding summary is offered in
Section 5.

Figure 1. Illustration of the increased stability in the Fourier space.
Traditional methods can recover $\hat{f}$ in the cone ${\mathcal{K}}_{\delta}$
(defined in Section 3). The data-driven approach can recover $\hat{f}$ in a
larger cone ${\mathcal{K}}_{\delta+\epsilon}$. The size of the cone depends on
the specific training set.

##  2\. The visible and invisible singularities

In [11], Quinto described some principles on recovering features of an object
from limited X-ray CT. Roughly speaking, if a boundary of a feature of the
body is tangent to a line in a limited data set, then that boundary should be
easy to reconstruct (called visible boundaries). Otherwise, the boundary
should be difficult to reconstruct from the limited data (called invisible
boundaries). We use the filtered backprojection reconstruction to demonstrate
the principle.

Let $\theta\in(-\delta,\delta)$ and define the limited angle back-projection
for $R_{\delta}$ as

$R^{*}_{\delta}g(x)=\int_{-\delta}^{\delta}g(v(\theta),x\cdot
v(\theta))d\theta.$

Let $\Lambda$ be the Riesz potential

$\Lambda
g(v,p)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i\tau(p-s)}|\tau|g(v,s)dsd\tau.$

Then we consider the following FBP for limited angle transform

(3) $L_{\delta}f=\frac{1}{4\pi}R_{\delta}^{*}\Lambda R_{\delta}f.$

Note that this is not an inversion formula. We describe what can be
reconstructed from this formula.

We introduce some notions from microlocal analysis. Let $f$ be a function on
${\mathbb{R}}^{2}$ and let
$x_{0}\in{\mathbb{R}}^{2},\xi_{0}\in{\mathbb{R}}^{2}\backslash 0$. By
definition, $f$ is smooth at $x_{0}$ in direction $\xi_{0}$ if there is a
smooth function $\phi\in C_{c}^{\infty}({\mathbb{R}}^{2})$ with
$\phi(x_{0})\neq 0$ and an open cone ${\mathscr{V}}$ containing $\xi_{0}$ such
that for any $N\in{\mathbb{N}}$ there is $C_{N}$ such that

$|\widehat{\phi f}(\xi)|\leq C_{N}(1+|\xi|)^{-N},\text{for all
}\xi\in{\mathscr{V}}.$

where $\hat{}$ denotes the Fourier transform in ${\mathbb{R}}^{2}.$ If $f$ is
not smooth at $x_{0}$ in direction $\xi_{0}$, then
$(x_{0},\xi_{0})\in\text{WF}(f)$ the wavefront set of $f$. Let
$(x_{0},\xi_{0})\in\text{WF}(f).$ We say $(x_{0},\xi_{0})$ is a visible
singularity of $f$ (for $L_{\delta}$) if $\xi_{0}$ is parallel to some
$v\in{\mathcal{S}}_{\delta}$. We say $(x_{0},\xi_{0})$ is an invisible
singularity of $f$ if $\xi_{0}$ is not parallel for any
$v\in{\mathcal{S}}_{\delta}$. We have

######  Theorem 2.1 (Theorem 2 of [11]).

Let $f$ be a function of compact support and $(x_{0},\xi_{0})\in\text{WF}(f)$.

  1. (1)

If $(x_{0},\xi_{0})$ is a visible singularity of $f$ then
$(x_{0},\xi_{0})\in\text{WF}(L_{\delta}f)$.

  2. (2)

If $(x_{0},\xi_{0})$ is an invisible singularity of $f$ then
$(x_{0},\xi_{0})\notin\text{WF}(L_{\delta}f)$.

We remark that it is possible to obtain the more precise microlocal Sobolev
regularity of the visible singularities, see Theorem 3.1 of [11]. Within the
microlocal framework, one can also explain the artifact phenomena. For our
setup, the artifact can occur on lines parallel to
$v\in\partial{\mathcal{S}}_{\delta}$ and the lines are tangent to the boundary
of some feature in the object, see Principle 2 of [10].

##  3\. The stable reconstruction

We establish quantitative estimate for recovering visible singularities. Our
starting point is the simple injectivity proof for $R_{\delta}$, see [9,
Theorem 3.4]. The transform $R_{\delta}f(p,v)$ is a function defined on
${\mathbb{R}}\times{\mathbb{S}}^{1}.$ We take Fourier transform of
$R_{\delta}f$ in $p$ to get

(4) $\displaystyle\widehat{R_{\delta}f}(\tau,v)$
$\displaystyle=\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-\imath
p\tau}\chi_{\delta}(v)f(pv+tv^{\perp})dtdp$
$\displaystyle=\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-\imath(pv+tv^{\perp})\cdot\tau
v}\chi_{\delta}(v)f(pv+tv^{\perp})dtdp=\hat{f}(\tau v)\chi_{\delta}(v).$

Here, $\imath^{2}=-1$. Because $R_{\delta}f(p,v)=0$, we get $\hat{f}(\xi)=0$
for $\xi\in{\mathbb{R}}^{2}$ parallel to $v$. Thus $\hat{f}(\xi)=0$ in a cone
${\mathcal{K}}_{\delta}=\\{\xi\in{\mathbb{R}}^{2}:\xi/|\xi|\in{\mathcal{S}}_{\delta}\\}$
with non-empty interior for $\delta>0$. Because $\hat{f}$ is analytic in
$\xi$, we conclude that $\hat{f}(\xi)=0$ for all $\xi\in{\mathbb{R}}^{2}.$ So
$f=0$ after taking the inverse Fourier transform.

We can further derive some stability estimate. Below we also use
$\chi_{\delta}$ for the characteristic function of ${\mathcal{K}}_{\delta}$ in
${\mathbb{R}}^{2}.$ It should be clear from the context which one we are
referring to.

######  Theorem 3.1.

For $\delta\in(0,\pi/2]$ and $f\in H^{-1/2}({\mathbb{R}}^{2}),$ we have

(5)
$\|\chi_{\delta}(D)f\|_{H^{-1/2}({\mathbb{R}}^{2})}\leq\|R_{\delta}f\|_{L^{2}({\mathbb{R}}\times{\mathbb{S}}^{1})}.$

###### Proof.

Applying Plancherel’s theorem and using (4), we get

(6)
$\displaystyle\|R_{\delta}f\|^{2}_{L^{2}({\mathbb{R}}\times{\mathbb{S}}^{1})}=$
$\displaystyle\|\widehat{R_{\delta}f}\|^{2}_{L^{2}({\mathbb{R}}\times{\mathbb{S}}^{1})}=\int_{{\mathbb{S}}^{1}}\int_{{\mathbb{R}}}\chi_{\delta}(v)|\hat{f}(\tau
v)|^{2}d\tau dv$ $\displaystyle=$
$\displaystyle\int_{{\mathcal{K}}_{\delta}}|\xi|^{-1}|\hat{f}(\xi)|^{2}d\xi\geq\|\chi_{\delta}(D)f\|_{H^{-1/2}({\mathbb{R}}^{2})}^{2}.$

This finishes the proof. ∎

The theorem essentially is a quantitative version of Theorem 2.1 on stably
recovering visible singularities of $f$. It is also clear that the stability
estimate (2) fails because of $(1-\chi_{\delta}(D))f$ which sits in the kernel
of $R_{\delta}.$

We seek improvement of the estimate to include invisible singularities. For
$\epsilon>0$ small we define $\chi_{\delta+\epsilon}(D)$ as a Fourier
multiplier of $\chi_{\delta+\epsilon}(\xi),\xi\in{\mathbb{R}}^{2}$. For $N\geq
0$, we define

(7) $\begin{gathered}{\mathcal{D}}_{N,\epsilon}=\\{f\in
L^{2}({\mathcal{B}}):\|f\|_{L^{2}}\leq
N\|\chi_{\delta+\epsilon}(D)f\|_{H^{-1/2}}\\}.\end{gathered}$

Observe that $L^{2}({\mathcal{B}})=\cup_{N\geq 0}{\mathcal{D}}_{N,\epsilon}.$
The following theorem says that for functions in ${\mathcal{D}}_{N,\epsilon}$,
the stability estimate (5) can be improved.

######  Theorem 3.2.

For $N\geq 0$, there is $\epsilon>0$ small such that

(8) $\|\chi_{\delta+\epsilon}(D)f\|_{H^{-1/2}({\mathbb{R}}^{2})}\leq
2\|R_{\delta}f\|_{L^{2}({\mathbb{R}}\times{\mathbb{S}}^{1})}$

for all $f\in{\mathcal{D}}_{N,\epsilon}.$

###### Proof.

We decompose

(9)
$R_{\delta}f=R_{\delta}\chi_{\delta+\epsilon}(D)f+R_{\delta}(1-\chi_{\delta+\epsilon}(D))f.$

Note that the second term on the right vanishes because
$(1-\chi_{\delta+\epsilon}(D))f$ is in the kernel of $R_{\delta}.$ So we
estimate the first term. Using (4), we have

(10) $\displaystyle
R_{\delta}\chi_{\delta+\epsilon}(D)f(p,v)-R_{\delta+\epsilon}\chi_{\delta+\epsilon}(D)f(p,v)$
$\displaystyle=$ $\displaystyle(2\pi)^{-1}\int_{\mathbb{R}}e^{\imath
p\tau}(\chi_{\delta}(v)-\chi_{\delta+\epsilon}(v))\chi_{\delta+\epsilon}(\tau
v)\hat{f}(\tau v)d\tau.$

Note that $\chi_{\delta}(v)-\chi_{\delta+\epsilon}(v)$ is supported in a set
of ${\mathbb{S}}^{1}$ with measure $<C\epsilon.$ Also, the left hand side of
(10) is compactly supported in $p$. Using Plancherel’s theorem again, we get

(11)
$\displaystyle\|R_{\delta}\chi_{\delta+\epsilon}(D)f-R_{\delta+\epsilon}\chi_{\delta+\epsilon}(D)f\|_{L^{2}}^{2}$
$\displaystyle\leq$ $\displaystyle
C\int_{{\mathbb{R}}}\int_{{\mathbb{S}}^{1}}|(\chi_{\delta}(v)-\chi_{\delta+\epsilon}(v))\chi_{\delta+\epsilon}(\tau
v)\hat{f}(\tau v)|^{2}dvd\tau$ $\displaystyle\leq$ $\displaystyle
C\epsilon^{2}\left(\int_{{\mathbb{R}}}\int_{{\mathbb{S}}^{1}}|\chi_{\delta+\epsilon}(\tau
v)\hat{f}(\tau v)|^{4}d\tau dv\right)^{\frac{1}{2}}$ $\displaystyle\leq$
$\displaystyle
C\epsilon^{2}\pi\|f\|_{L^{2}}\left(\int_{{\mathbb{R}}}\int_{{\mathbb{S}}^{1}}|\hat{f}(\tau
v)|^{2}d\tau dv\right)^{\frac{1}{2}}$ $\displaystyle\leq$ $\displaystyle
C\epsilon^{2}\pi\sqrt{4\pi^{2}+1}\|f\|_{L^{2}}^{2}\leq
20C\epsilon^{2}N^{2}\|\chi_{\delta+\epsilon}(D)f\|^{2}_{H^{-1/2}}.$

where $C$ is an absolute constant independent of $\epsilon,N,f$. Also in the
derivation, we used Hölder’s inequality and the fact that
$\|\hat{h}\|_{L^{\infty}}\leq\sqrt{\pi}\|h\|_{L^{2}}$ for $h$ compactly
supported in $\mathcal{B}$ and

$\displaystyle\int_{{\mathbb{R}}}\int_{{\mathbb{S}}^{1}}|\hat{f}(\tau
v)|^{2}d\tau dv$ $\displaystyle=\int_{|\tau|\geq
1}\int_{{\mathbb{S}}^{1}}|\hat{f}(\tau v)|^{2}d\tau
dv+\int_{|\tau|<1}\int_{{\mathbb{S}}^{1}}|\hat{f}(\tau v)|^{2}d\tau dv$
$\displaystyle\leq\int_{|\tau|\geq
1}\int_{{\mathbb{S}}^{1}}|\tau||\hat{f}(\tau v)|^{2}d\tau
dv+\pi\|f\|_{L^{2}}^{2}\int_{|\tau|<1}\int_{{\mathbb{S}}^{1}}1d\tau dv$
$\displaystyle\leq(1+4\pi^{2})\|f\|_{L^{2}}^{2}.$

We can apply Theorem 3.1 to derive that

$\displaystyle\|\chi_{\delta+\epsilon}(D)f\|_{H^{-1/2}}$
$\displaystyle\leq\|R_{\delta+\epsilon}\chi_{\delta+\epsilon}(D)f\|_{L^{2}}$
$\displaystyle\leq\|R_{\delta}\chi_{\delta+\epsilon}(D)f\|_{L^{2}}+\|R_{\delta}\chi_{\delta+\epsilon}(D)f-R_{\delta+\epsilon}\chi_{\delta+\epsilon}(D)f\|_{L^{2}}$
$\displaystyle\leq\|R_{\delta}\chi_{\delta+\epsilon}(D)f\|_{L^{2}}+\sqrt{20C}\epsilon
N\|\chi_{\delta+\epsilon}(D)f\|_{H^{-1/2}}.$

For $\epsilon>0$ sufficiently small so that $\sqrt{20C}\epsilon N<1/2$, we
obtain

(12) $\|\chi_{\delta+\epsilon}(D)f\|_{H^{-1/2}({\mathbb{R}}^{2})}\leq
2\|R_{\delta}\chi_{\delta+\epsilon}(D)f\|_{L^{2}({\mathbb{R}}\times{\mathbb{S}}^{1})},$

which finishes the proof. ∎

We remark that we chose the Sobolev norms in the condition in (7) for
simplicity. The analysis can be adapted to other Sobolev functions by slightly
modifying the proof of Theorem 3.1 and 3.2. For neural networks, suppose the
training is done on a finite data set ${\mathcal{D}}\subset H^{-1/2}$. Then
one can always find $N>0$ and $\epsilon>0$ so that
${\mathcal{D}}\subset{\mathcal{D}}_{N,\epsilon}$. Suppose the neural network
algorithm is capable of learning an integral operator (e.g. universal
approximation property is achieved). Then the learned inversion operator
$R^{-1}_{\delta}$ would potentially satisfy the improved stability estimate
(8) which can explain the stable reconstruction of some invisible
singularities.

##  4\. The experiment

###  4.1. The neural network

We use the same data as in [3] which can be downloaded from [15]. The data set
consists of synthetic images of ellipses, where the number, locations, sizes
and the intensity gradients of the ellipses are chosen at random. The
measurements are computed using the Matlab function radon with added Gaussian
noise. Also, the measurements are simulated at a higher resolution and then
downsampled to an image of resolution $128\times 128$ to avoid the inverse
crime. For the simulation below, we use measurements for a missing wedge of
$80^{\circ}$ (out of $180^{\circ}$). See Figure 3.

For the network structure, we first use Matlab function iradon to compute the
FBP inversion. Then we use a basic U-net [13] with network structure plotted
in Figure 2. We use $500$ images for training the network among which $80\%$
are used for training and $20\%$ for validation. We show two examples of the
reconstruction on test images in Figure 4. We observe that in both examples
artifacts are almost completely removed, and some of the invisible
singularities are also constructed well. But we do notice that the
reconstructions are not perfect. Several thin ellipses with mainly invisible
singularities are not reconstructed at all in both examples.

Figure 2. Structure of the U-net. The input is a $128\times 128$ image and the
output is also a $128\times 128$ image. The number of channels is indicated in
the boxes. The arrows indicate the flow of the data.

###  4.2. The robustness

We consider violation of the condition in (7). For $a>0$, we consider the
characteristic function for the disk centered at $0$ with radius $a$

$f_{a}(x)=1\text{ for }|x|<a\text{ and }f_{a}(x)=0\text{ for }|x|\geq a.$

Taking Fourier transform in ${\mathbb{R}}^{2}$ and using polar coordinate, we
get

$\begin{gathered}\hat{f}_{a}(\xi)=\int_{{\mathbb{S}}^{1}}\int_{0}^{a}e^{-ir\omega\cdot\xi}rdrd\omega=\int_{{\mathbb{S}}^{1}}\int_{0}^{1}e^{-ir\omega\cdot
a\xi}a^{2}rdrd\omega=a^{2}\hat{f}_{1}(a\xi).\end{gathered}$

Below, we use homogeneous Sobolev spaces to compute

$\begin{gathered}\|\chi_{\delta+\epsilon}(D)f_{a}\|_{\dot{H}^{-1/2}}^{2}=\int_{{\mathbb{R}}^{2}}|\xi|^{-1}\chi_{\delta+\epsilon}(\xi)^{2}|\hat{f}_{a}(\xi)|^{2}d\xi\\\
=\int_{{\mathbb{R}}^{2}}|\xi|^{-1}\chi_{\delta+\epsilon}(\xi)^{2}a^{4}|\hat{f}_{1}(a\xi)|^{2}d\xi=\int_{{\mathbb{R}}^{2}}|\xi/a|^{-1}\chi_{\delta+\epsilon}(\xi/a)^{2}a^{2}|\hat{f}_{1}(\xi)|^{2}d\xi.\end{gathered}$

We see that for $a$ large,

$\|\chi_{\delta+\epsilon}(D)f_{a}\|_{\dot{H}^{-1/2}}\simeq
a^{3/2}\|\chi_{\delta+\epsilon}(D)f_{1}\|_{\dot{H}^{-1/2}}.$

Similarly, we have $\|f_{a}\|_{L^{2}}\simeq a\|f_{1}\|_{L^{2}}$. If
$\|f_{1}\|_{L^{2}}$ $\leq C_{0}\|\chi_{\delta+\epsilon}(D)f_{1}\|_{H^{-1/2}}$
is true for some $C_{0}>0$, then for $f_{a}$, the condition in (7) will be
violated when $a$ is large. We show the test result for this example in Figure
5. As $a$ increases, we start to see instabilities in the network
reconstruction and eventually artifacts start to appear, which agrees with our
theory.

We also test the trained network on other types of images, see the two squares
in Figure 6. We know that the corners could produce strong artifacts as shown
in the FBP reconstruction. Because the two squares are not too large,
condition (7) is likely to be satisfied via similar Fourier analysis as for
$f_{a}$. The numerical result agrees with our observation.

###  4.3. Further discussions

The examples presented in Sections 4.1 and 4.2 indicate that the trained
neural network is a good approximation of $R_{\delta}^{-1}$. However, we do
not claim that every trained network shares the same property. In fact, we
tested training the network for a longer time and observed that the training
error may still decrease, but the test results on general images in Section
4.2 usually become worse. This might be a sign of overfitting. The discussion
of the training mechanism is beyond the scope of this work.

##  5\. Conclusion

In this note, we provided a mathematical explanation of the robustness of the
data-driven approach based on deep neural networks for limited angle
tomography. We showed that prior condition on the training data set can lead
to stable reconstruction of some invisible singularities. We also tested the
theory using a simple U-net and the numerical results agree well with our
analysis.

Figure 3. Illustration of the experimental setup and the corresponding
visible/invisible directions according to the theory in Section 2. The red
lines indicate the possible artifact directions.

Figure 4. Two numerical reconstructions using FBP and U-net. Each row shows
one example. Left: the original image. Middle: the FBP reconstruction. Right:
the U-net reconstruction

Figure 5. Numerical test of the stability condition in (7). Each row shows one
example. Left: the original image. Middle: the FBP reconstruction. Right: the
U-net reconstruction

Figure 6. Left: the original image. Middle: the FBP reconstruction. Right: the
U-net reconstruction

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Explicit Kink Solutions in Several One-Parameter Family of Higher Order Field
Theory Models

Avinash Khare

Physics Department, Savitribai Phule Pune University

Pune 411007, India

Ayhan Duzgun and Avadh Saxena

Theoretical Division and Center for Nonlinear Studies, Los Alamos National
Laboratory, Los Alamos, New Mexico 87545, USA

Abstract: We present several one-parameter family of higher order field theory
models some of which admit explicit kink solutions with an exponential tail
while others admit explicit kink solutions with a power-law tail. Various
properties of these families of kink solutions are examined in detail. In
particular for models harboring kink solutions with a power-law tail, we show
that there is no gap between the zero mode and the beginning of the continuum
in the kink stability potential. Further, by applying the recent Manton
formalism, we provide estimates for the kink-kink and antikink-kink
acceleration and hence the ratio of the corresponding antikink-kink and kink-
kink forces.

##  1 Introduction

During the last few years there has been a resurgence in studying kink
solutions with exponential as well as power-law tails in several higher (than
6th) order field theory models including $\phi^{8}$, $\phi^{10}$, and
$\phi^{12}$ [1, 2]. The study of higher order field theories, their attendant
kink excitations as well as the associated kink interactions and scattering
[3, 4] are important in a variety of physical contexts ranging from successive
phase transitions [1, 2, 5, 6] to isostructural phase transitions [7] to
models involving long-range interaction between massless mesons [8], as well
as from protein crystallization [9] to successive phase transitions presumably
driving the late time expansion of the Universe [10]. Thus, understanding kink
behavior in these models provides useful insight into the properties of domain
walls in materials, condensed matter, high energy physics, biology and
cosmology, which also serves as one of our main motivations here.

So far implicit kink solutions have been obtained in these higher order models
using which one has studied in detail [11] several aspects of these kink
solutions such as the nature of kink tail, kink-kink and kink-antikink forces
[12, 13, 14, 15], stability analysis, etc. However, one important missing
piece has been the absence of analytically solvable explicit kink solutions in
these higher order field theory models. Few exceptions include explicit kink
solutions in a one-parameter family of $\phi^{4n+2}-\phi^{2n+2}-\phi^{2}$
models ($n=1,2,3$) [1, 9], and one example of $\phi^{8}$ [16] as well as one
example of $\phi^{10}$ [11] field theory models. The main objective of this
paper is to partially fill this gap.

In this context it is worth pointing out that during the last four decades
explicit kink solutions have been obtained in a number of lower order field
theory models like $\phi^{4}$, $\phi^{6}$ [17] and also several non-polynomial
hyperbolic [18] and periodic potentials. Beyond sine-Gordon [17], the latter
include double sine-Gordon [19] and Lamé [20] solitons as examples. Further,
recently explicit kink solutions with super-exponential [21, 22] and super-
super-exponential [23], power-tower [24] as well as power-law tail [25, 26,
27, 28, 29] have also been obtained. Thus, it is of immense interest to obtain
explicit kink solutions in as many higher order field theory models as
possible.

The purpose of this paper is to partially construct some of the exact
solutions. In particular, we present four one-parameter families of higher
order field theory models for which we are able to obtain explicit kink
solutions. Two of these models have a kink solution with an exponential tail
while the other two admit kink solutions with a power-law tail. To the best of
our knowledge, these are the first known examples of a one-parameter family of
higher order field theory models admitting kink solutions with a power-law
tail. We study in detail various properties of these kink solutions including
the kink mass, nature of the kink tail, the kink linear stability analysis and
offer predictions for the kink-kink (KK) and the kink-antikink (K-AK) forces
by using the recent Manton formalism [12] and its extension to higher order
field theories [15]. For the one-parameter family with an exponential tail,
the calculation of the KK and K-AK forces is straightforward using the
celebrated Manton formalism [30].

The plan of the paper is as follows. In Sec. II we obtain explicit kink
solutions in one-parameter families of potentials of the type
$\phi^{2}(a^{2n}-\phi^{2n})^{2}(b^{2n}-\phi^{2n})^{2}$ in case $b$ and $a$ are
related to each other. We study in detail various properties of these kink
solutions all of which have an exponential tail. In Secs. III and IV we
further extend these results. In Sec. III we obtain explicit kink solutions in
a one-parameter family of potentials of the type
$\phi^{2}(a^{2n}-\phi^{2n})^{2}(b^{2n}-\phi^{2n})^{2}(c^{2n}-\phi^{2n})^{2}$
in case $c$, $b$ and $a$ are related to each other. We study in detail the
properties of these kink solutions all of which have an exponential kink tail.
Section IV contains yet another family of solutions with the potential
$V(\phi)=\lambda^{2}\phi^{2}(a^{2n}-\phi^{2n})^{2}(b^{2n}+\phi^{2n})^{2}$. In
the next two sections we study one-parameter families of higher order field
theory models and obtain explicit kink solutions with a power-law tail. In
particular, in Sec. V we obtain explicit kink solutions in a one-parameter
family of potentials of the type $\phi^{2n+2}|a^{2n}-\phi^{2n}|^{3/2}$ all of
which admit kink solutions with a power law-tail. We study the properties of
these solutions including the stability analysis and show that in this case,
as expected, there is no gap between the zero mode and the beginning of the
continuum. By using the recent Manton formalism [12] we provide predictions
for the K-K, AK-K and K-AK acceleration as well as the ratio of the AK-K and
K-K forces. In Sec. VI we present another one-parameter family of potentials
of the form $|a^{2n}-\phi^{2n}|^{(2n+1)/2n}$ and obtain explicit kink
solutions with a power-law tail. In this case too, we study in detail the
various properties of the kink solutions and make predictions for the K-K and
AK-K acceleration as well as the ratio of the corresponding forces. Finally,
in Sec. VII we summarize the results obtained in this paper and point out some
of the open problems. In Appendices A and B we provide few more explicit kink
solutions which are obtained by solving quartic equations. Further, for
completeness, we also briefly discuss known explicit kink solutions in
Appendix C.

##  2 Explicit Kink Solutions with Exponential Tail for a One-Parameter Family
of Potentials

Let us consider the one-parameter family of potentials as given by

$V(\phi)=\lambda^{2}\phi^{2}(a^{2n}-\phi^{2n})^{2}(b^{2n}-\phi^{2n})^{2}\,,~{}~{}n=1,2,3,...,~{}~{}b>a>0\,.$
(1)

Note that this potential has 5 degenerate minima at $\phi=0,\pm a,\pm b$ (see
Fig. 1) and hence two kink and two mirror kink solutions and corresponding
four antikink solutions. We now obtain these kink solutions explicitly.

Figure 1: Potential $V(\phi)$ with five degenerate minima for three different
values of $n=1,2,3$. Parameters are $\lambda=1$, $a=1$ and $b=2^{1/(2n)}$. See
Eq. (1).

###  2.1 Kink Solution From $0$ to $a$

In this case, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\lambda\phi(a^{2n}-\phi^{2n})(b^{2n}-\phi^{2n})\,,$ (2)

which can be achieved by using the partial fraction

$\int
d\phi\,\bigg{[}\frac{A}{\phi}+\frac{B\phi^{2n-1}}{(a^{2n}-\phi^{2n})}+\frac{D\phi^{2n-1}}{(b^{2n}-\phi^{2n})}\bigg{]}\,.$
(3)

It is straightforward to check that

$A=\frac{1}{a^{2n}b^{2n}}\,,~{}~{}B=\frac{1}{a^{2n}(b^{2n}-a^{2n})}\,,~{}~{}D=-\frac{1}{b^{2n}(b^{2n}-a^{2n})}\,.$
(4)

Eq. (3) is then easily integrated giving an implicit kink solution

$e^{\mu
x}=\frac{(\phi^{2n})^{(b^{2n}-a^{2n})}(b^{2n}-\phi^{2n})^{a^{2n}}}{(a^{2n}-\phi^{2n})^{b^{2n}}}\,,$
(5)

where

$\mu=2\sqrt{2}n\lambda a^{2n}b^{2n}(b^{2n}-a^{2n})\,.$ (6)

Note that the implicit kink solution is valid for arbitrary values of $a,b$
with the constraint $b>a>0$.

It is straightforward to calculate the kink mass in this case. We find

$M_{K}=\int_{0}^{a}d\phi\,\sqrt{2V(\phi)}=\frac{\lambda
na^{2(n+1)}[(2n+1)b^{2n}-a^{2n}]}{\sqrt{2}(2n+1)(n+1)}\,.$ (7)

We now consider three special cases in which Eq. (5) can be easily inverted
yielding an explicit kink solution. Two of these cases we discuss below while
the third case is discussed in Appendix A.

####  2.1.1 Case I: $b^{2n}=2a^{2n}$

In this case the implicit kink solution takes the form

$e^{\mu px}=\frac{\phi^{2n}(2a^{2n}-\phi^{2n})}{(a^{2n}-\phi^{2n})^{2}}\,,$
(8)

where $p$ is an arbitrary positive number with $pa^{2n}=1$, while $\mu$ now
has a simple form

$\mu=\frac{4\sqrt{2}n\lambda}{p^{3}}\,.$ (9)

Eq. (8) is easily inverted yielding an explicit kink solution from $0$ to $a$
(see Fig. 2)

$\phi_{K}(x)=a\left[1-\frac{1}{\sqrt{1+e^{\mu px}}}\right]^{1/2n}\,.$ (10)

Thus while the implicit kink solution is known for arbitrary values of $a,b$
satisfying $b>a>0$, the explicit kink solution while valid for arbitrary
values of $a$, the corresponding $b$ is however fixed, i.e. $b^{2n}=2a^{2n}$.

Figure 2: Kink solution $\phi(x)$ given by Eq. (10) for $a=1$, $p=1$ and
$\mu=4\sqrt{2}n$.

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=a2^{-1/2n}e^{\mu
px/2n}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=a-(2n)^{-1}ae^{-\mu px/2}\,.$
(11)

Further, the zero mode in this case is easily calculated

$\psi^{0,K}\propto\frac{e^{\mu px}}{(1+e^{\mu px})^{3/2}}\frac{1}{[1-(1+e^{\mu
px})^{-1/2}]^{1-1/2n}}\,.$ (12)

Finally, it is straightforward to carry out kink stability analysis. The kink
stability potential $V_{K}(x)$ which appears in the Schrödinger-like equation

$-\frac{d^{2}\psi}{dx^{2}}+V_{K}(x)\psi=\omega^{2}\psi\,,$ (13)

is easily calculated

$\displaystyle V(x)=\frac{d^{2}V(\phi)}{d\phi^{2}}\Bigg{|}_{\phi=\phi_{K}(x)}$
$\displaystyle=2\lambda^{2}\bigg{[}(a^{2n}-\phi_{K}^{2n})^{2}(2a^{2n}-\phi_{K}^{2n})^{2}+16n^{2}\phi_{K}^{4n}(a^{2n}-\phi_{K}^{2n})(2a^{2n}-\phi_{K}^{2n})$
$\displaystyle+4n^{2}\phi_{K}^{4n}[(2a^{2n}-\phi_{K}^{2n})^{2}+(a^{2n}-\phi_{K}^{2n})^{2}]$
$\displaystyle-2n(2n+3)\phi_{K}^{2n}(a^{2n}-\phi_{K}^{2})(2a^{2n}-\phi_{K}^{2})(3a^{2n}-2\phi_{K}^{2n})\bigg{]}\,,$
(14)

and substituting the kink solution (10) in it. In particular, since the kink
solution goes from $0$ to $a$ as $x$ goes from $-\infty$ to $\infty$, it is
easy to see that

$\displaystyle V(x=\infty)=\frac{8\lambda
n^{2}}{p^{4}}\,,~{}~{}V(x=-\infty)=\frac{8\lambda}{p^{4}}\,,$ (15)
$\displaystyle
V(x=0)=\frac{[(3+2\sqrt{2})/4-(\sqrt{2}-1)n^{2}-(3+\sqrt{2})n]}{p^{4}}\,.$

It will be interesting to study if this potential admits other discrete modes
than the zero mode.

####  2.1.2 Case II: $b^{2n}=3a^{2n}$

In this case the implicit kink solution takes the form

$e^{\mu qx}=\frac{\phi^{4n}(3a^{2n}-\phi^{2n})}{(a^{2n}-\phi^{2n})^{3}}\,,$
(16)

where $q$ is any positive number satisfying $a^{2n}q=1$ and $\mu$ now has a
simple form

$\mu=\frac{12\sqrt{2}n\lambda}{q^{3}}\,.$ (17)

Eq. (16) is easily inverted yielding a cubic equation

$u^{3}+\frac{3u}{(e^{\mu qx}-1)}-\frac{2}{(e^{q\mu x}-1)}=0\,,$ (18)

where $u=1-(\frac{\phi}{a})^{2n}$. Thus the explicit kink solution is given by

$\displaystyle
u=1-\left(\frac{\phi}{a}\right)^{2n}=\bigg{[}\frac{(\sqrt{e^{\mu
qx}-1)}+e^{\mu qx/2}}{(e^{\mu
qx}-1)^{3/2}}\bigg{]}^{1/3}+\bigg{[}\frac{\sqrt{(e^{\mu qx}-1)}-e^{\mu
qx/2}}{(e^{\mu qx}-1)^{3/2}}\bigg{]}^{1/3}\,.$ (19)

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=3^{-1/4n}ae^{\mu
qx/4n}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=a-\frac{a}{2^{2/3}n}e^{-\mu
qx/3}\,.$ (20)

Thus while the implicit kink solution is known for arbitrary values of $a,b$
satisfying $b>a>0$, the explicit kink solution (19) while valid for arbitrary
values of $a$, the corresponding $b$ is however fixed, i.e. $b^{2n}=3a^{2n}$.

Proceeding in the same way, by choosing $b^{2n}=ma^{2n}$ one can in principle
obtain an explicit kink solution by solving an $m$’th order equation.

###  2.2 Kink Solution From $a$ to $b$

In this case, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\lambda\phi(a^{2n}-\phi^{2n})(b^{2n}-\phi^{2n})\,,$ (21)

which is easily achieved by using the partial fraction

$\int
d\phi\,\bigg{[}\frac{A}{\phi}+\frac{B\phi^{2n-1}}{(\phi^{2n}a^{2n})}+\frac{D\phi^{2n-1}}{(b^{2n}-\phi^{2n})}\bigg{]}\,.$
(22)

It is straightforward to check that

$A=-\frac{1}{a^{2n}b^{2n}}\,,~{}~{}B=\frac{1}{a^{2n}(b^{2n}-a^{2n})}\,,~{}~{}D=-\frac{1}{b^{2n}(b^{2n}-a^{2n})}\,.$
(23)

Eq. (22) is then easily integrated giving an implicit kink solution

$e^{-\mu
x}=\frac{(\phi^{2n})^{(b^{2n}-a^{2n})}(b^{2n}-\phi^{2n})^{a^{2n}}}{(a^{2n}-\phi^{2n})^{b^{2n}}}\,,$
(24)

where

$\mu=2\sqrt{2}n\lambda a^{2n}b^{2n}(b^{2n}-a^{2n})\,.$ (25)

It is straightforward to calculate the kink mass in this case. We find

$\displaystyle M_{K}=\int_{a}^{b}d\phi\,\sqrt{2V(\phi)}=\frac{\lambda
na^{2}}{\sqrt{2}}\bigg{[}\frac{([b/a]^{4n+2}-1)}{(n+1)(n+2)}-\frac{(b/a)^{2n}([b/a]^{2}-1)}{(n+1)}\bigg{]}\,.$
(26)

We now consider three special cases in which Eq. (24) can be easily inverted
yielding an explicit kink solution, two given below and the third in Appendix
A.

####  2.2.1 Case I: $b^{2n}=2a^{2n}$

In this case the implicit kink solution takes the form

$e^{-\mu px}=\frac{\phi^{2n}(2a^{2n}-\phi^{2n})}{(\phi^{2n}-a^{n})^{2}}\,,$
(27)

where $p$ is an arbitrary positive number satisfying $a^{2n}p=1$ while $\mu$
now has a simple form given by Eq. (9). Eq. (27) is easily inverted yielding
an explicit kink solution (see Fig. 3)

$\phi(x)=a\left[1+\frac{1}{\sqrt{1+e^{-\mu px}}}\right]^{1/2n}\,.$ (28)

Figure 3: Kink solution $\phi(x)$ given by Eq. (28) with $a=1$, $p=1$ and
$\mu=4\sqrt{2}n$.

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=a+(2n)^{-1}ae^{\mu
px/2}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=2^{1/2n}a-(4n)^{-1}ae^{-\mu
px}\,.$ (29)

Further, the zero mode in this case is easily calculated

$\psi^{0,K}\propto\frac{e^{-\mu px}}{(1+e^{-\mu
px})^{3/2}}\frac{1}{[1+(1+e^{-\mu px})^{-1/2}]^{1-1/2n}}\,.$ (30)

Finally, it is straightforward to do the kink stability analysis. The kink
stability potential that appears in the Schrödinger like equation (13) can be
easily calculated by using

$\displaystyle V(x)=\frac{d^{2}V(\phi)}{d\phi^{2}}\Bigg{|}_{\phi=\phi_{K}(x)}$
$\displaystyle=2\lambda^{2}\bigg{[}(\phi_{K}^{2n}-a^{2n})^{2}(2a^{2n}-\phi_{K}^{2n})^{2}-16n^{2}\phi_{K}^{4n}(\phi_{K}^{2n}-a^{2n})(2a^{2n}-\phi_{K}^{2n})+4n^{2}\phi_{K}^{4n}[(2a^{2n}-\phi_{K}^{2n})^{2}+(\phi_{K}^{2n}-a^{2n})^{2}]$
$\displaystyle-2n(2n+3)\phi_{K}^{2n}(\phi_{K}^{2}-1)(2-\phi_{K}^{2})(2\phi_{K}^{2n}-3a^{2n})\bigg{]}\,,$
(31)

and substituting the kink solution (28) in it. In particular, since the kink
solution goes from $a$ to $2^{1/2n}a$ as $x$ goes from $-\infty$ to $\infty$,
it is easy to observe that

$\displaystyle V(x=\infty)={32\lambda
n^{2}}{p^{4}}\,,~{}~{}V(x=-\infty)=\frac{8\lambda n^{2}}{p^{4}}\,,$
$\displaystyle
V(x=0)=\frac{[(3-2\sqrt{2})/4-(3\sqrt{2}+2)n^{2}-(3+\sqrt{2})n/2]}{p^{4}}\,.$
(32)

For large $n$, thus $V(x=\pm\infty)$ is large and it is certain that this
stability potential has several discrete modes apart from the zero mode. In
fact in the $n\rightarrow\infty$ limit, this potential has only discrete
modes. This is rather unusual.

####  2.2.2 Case II: $b^{2n}=3a^{2n}$

In this case the implicit kink solution takes the form

$e^{\mu qx}=\frac{\phi^{4n}(3-\phi^{2n})}{(1-\phi^{2n})^{3}}\,,$ (33)

where $q$ is any positive number satisfying $qa^{2n}=1$ while $\mu$ now has a
simple form as given by Eq. (16). Eq. (33) is easily inverted yielding a cubic
equation

$u^{3}-\frac{3u}{(e^{-\mu qx}+1)}-\frac{2}{(e^{-\mu qx}+1)}=0\,,$ (34)

where $u=(\frac{\phi}{a})^{2n}-1$. Thus the explicit kink solution is given by

$\displaystyle
u=\left(\frac{\phi}{a}\right)^{2n}-1=\bigg{[}\frac{(\sqrt{e^{-\mu
qx}+1)}+e^{-\mu qx/2}}{(e^{-\mu
qx}+1)^{3/2}}\bigg{]}^{1/3}+\bigg{[}\frac{\sqrt{(e^{-\mu qx}+1)}-e^{-\mu
px/2}}{(e^{-\mu qx}+1)^{3/2}}\bigg{]}^{1/3}\,.$ (35)

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=a+a\frac{\sqrt{3}}{2n}e^{\mu
qx/2}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=3^{1/2n}a-\frac{8a}{9}e^{-\mu
qx}\,.$ (36)

Another exact solution in this family of potentials is provided in Appendix A.

##  3 Explicit Kink Solutions with Exponential Tail for Another One-Parameter
Family of Potentials

We now obtain exact kink solutions in yet another one-parameter family of
higher order field theories characterized by the potential (see Fig. 4)

$V(\phi)=\lambda^{2}\phi^{2}(a^{2n}-\phi^{2n})^{2}(b^{2n}-\phi^{2n})^{2}(c^{2n}-\phi^{2n})^{2}\,,~{}~{}c>b>a>0\,.$
(37)

Note that this potential has 7 degenerate minima at $\phi=0,\pm a,\pm b,\pm c$
and hence three kink and three mirror kink solutions and corresponding six
antikink solutions. We now discuss the three kink solutions one by one.

Figure 4: Potential $V(\psi)$ with seven degenerate minima for three different
values of $n=1,2,3$. Parameters are $\lambda=1$, $a=1$, $b=3^{1/(2n)}$ and
$c=4^{1/(2n)}$. See Eq. (37).

###  3.1 Kink Solution From $0$ to $a$

In this case, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\lambda\phi(a^{2n}-\phi^{2n})(b^{2n}-\phi^{2n})(c^{2n}-\phi^{2n})\,,$
(38)

which is easily achieved by using the partial fraction

$\int
d\phi\,\bigg{[}\frac{A}{\phi}+\frac{B\phi^{2n-1}}{(a^{2n}-\phi^{2n})}+\frac{D\phi^{2n-1}}{(b^{2n}-\phi^{2n})}+\frac{E\phi^{2n-1}}{(c^{2n}-\phi^{2n})}\bigg{]}\,.$
(39)

We find that

$\displaystyle
A=\frac{1}{a^{2n}b^{2n}c^{2n}}\,,~{}~{}~{}B=\frac{1}{a^{2n}(b^{2n}-a^{2n})(c^{2n}-a^{2n})}\,,$
$\displaystyle
D=-\frac{1}{b^{2n}(b^{2n}-a^{2n})(c^{2n}-b^{2n})}\,,~{}~{}~{}E=\frac{1}{c^{2n}(c^{2n}-a^{2n})(c^{2n}-b^{2n})}\,.$
(40)

Eq. (39) is then easily integrated giving an implicit kink solution

$e^{\mu
x}=\frac{(\phi^{2n})^{[(b^{2n}-a^{2n})(c^{2n}-a^{2n})(c^{2n}-b^{2n})]}(b^{2n}-\phi^{2n})^{a^{2n}c^{2n}(c^{2n}-a^{2n})}}{(a^{2n}-\phi^{2n})^{b^{2n}c^{2n}(c^{2n}-b^{2n})}(c^{2n}-\phi^{2n})^{b^{2n}a^{2n}(b^{2n}-a^{2n})}}\,,$
(41)

where

$\mu=2\sqrt{2}n\lambda
a^{2n}b^{2n}c^{2n}(b^{2n}-a^{2n})(c^{2n}-a^{2n})(c^{2n}-b^{2n})\,.$ (42)

Note that the implicit kink solution is valid for arbitrary values of $a,b,c$
with the constraint $c>b>a>0$.

It is straightforward to calculate the kink mass in this case. We find

$\displaystyle M_{K}=\int_{0}^{a}d\phi\,\sqrt{2V(\phi)}=\frac{\lambda
na^{2(3n+1)}[6\alpha\beta
n^{2}+(5\alpha\beta+2\alpha+2\beta+1)n+\alpha\beta-\alpha-\beta-1]}{\sqrt{2}(3n+1)(2n+1)(n+1)}\,,$
(43)

where

$\alpha=\frac{b^{2n}}{a^{2n}}\,,~{}~{}\beta=\frac{c^{2n}}{a^{2n}}\,.$ (44)

We now consider a special case in which Eq. (41) can be easily inverted
yielding an explicit kink solution.

####  3.1.1 Special Case: $b^{2n}=3a^{2n},~{}c^{2n}=4a^{2n}$

In this case the implicit kink solution takes the form

$e^{\mu
px}=\frac{\phi^{2n}(3a^{2n}-\phi^{2n})^{2}}{(a^{2n}-\phi^{2n})^{2}(4a^{2n}-\phi^{2n})}\,,$
(45)

where $p$ is an arbitrary positive number with $6pa^{6n}=1$, while $\mu$ now
has a simple form

$\mu=\frac{2\sqrt{2}n\lambda}{p^{2}}\,.$ (46)

Eq. (45) is easily inverted yielding a cubic equation

$u^{3}-3u+2\tanh\left(\frac{\mu px}{2}\right)=0\,,$ (47)

where $u=2-(\frac{\phi}{a})^{2n}$. Thus the explicit kink solution is given by
(see Fig. 5)

$u=2-\left(\frac{\phi}{a}\right)^{2n}=2\cos\big{[}(1/3)\cos^{-1}(-\tanh[p\mu
x/2])\big{]}\,,$ (48)

where we have used the fact that $\cos^{-1}(1)=0\,,~{}~{}\cos^{-1}(-1)=\pi$.
It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=(2/3)^{1/n}ae^{\mu
px/2n}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=a-\frac{a}{\sqrt{3}n}e^{-\mu
px/2}\,.$ (49)

Thus while the implicit kink solution is known for arbitrary values of $a,b,c$
satisfying $c>b>a>0$, the explicit kink solution (48), while valid for
arbitrary values of $a$, the corresponding $b,c$ are however fixed, i.e.
$b^{2n}=3a^{2n},c^{2n}=4a^{2n}$.

Figure 5: Kink solution $\phi(x)$ given by Eq. (48) for $a=1$, $p=1/6$ and
$\mu=72\sqrt{2}n$.

###  3.2 Kink Solution From $a$ to $b$

In this case, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\lambda\phi(\phi^{2n}-a^{2n})(b^{2n}-\phi^{2n})(c^{2n}-\phi^{2n})\,,$
(50)

which is easily achieved by using the partial fraction

$\int
d\phi\,\bigg{[}\frac{A}{\phi}+\frac{B\phi^{2n-1}}{(\phi^{2n}-a^{2n})}+\frac{D\phi^{2n-1}}{(b^{2n}-\phi^{2n})}+\frac{E\phi^{2n-1}}{(c^{2n}-\phi^{2n})}\bigg{]}\,.$
(51)

We find

$\displaystyle
A=-\frac{1}{a^{2n}b^{2n}c^{2n}}\,,~{}~{}~{}B=\frac{1}{a^{2n}(b^{2n}-a^{2n})(c^{2n}-a^{2n})}\,,$
$\displaystyle
D=\frac{1}{b^{2n}(b^{2n}-a^{2n})(c^{2n}-b^{2n})}\,,~{}~{}~{}E=-\frac{1}{c^{2n}(c^{2n}-a^{2n})(c^{2n}-b^{2n})}\,.$
(52)

Eq. (51) is then easily integrated giving an implicit kink solution

$e^{\mu
x}=\frac{(\phi^{2n}-a^{2n})^{b^{2n}c^{2n}(c^{2n}-b^{2n})}(c^{2n}-\phi^{2n})^{b^{2n}a^{2n}(b^{2n}-a^{2n})}}{(\phi^{2n})^{[(b^{2n}-a^{2n})(c^{2n}-a^{2n})(c^{2n}-b^{2n})]}(b^{2n}-\phi^{2n})^{a^{2n}c^{2n}(c^{2n}-a^{2n})}}\,,$
(53)

where $\mu$ is given by Eq. (42). Note that the implicit kink solution is
valid for arbitrary values of $a,b,c$ with the constraint $c>b>a>0$.

It is straightforward to calculate the kink mass in this case. We find

$\displaystyle M_{K}=\int_{a}^{b}d\phi\,\sqrt{2V(\phi)}$ $\displaystyle=$
$\displaystyle\frac{\lambda
na^{2(3n+1)}}{\sqrt{2}(3n+1)(2n+1)(n+1)}\bigg{[}\big{(}6\alpha\beta
n^{2}+[5\alpha\beta+2\alpha+2\beta+1]n+\alpha\beta-\alpha-\beta-1\big{)}$ (54)
$\displaystyle+$ $\displaystyle\alpha^{(n+1)/n}\big{(}-6\beta
n^{2}+[3\alpha\beta+3\alpha-5\beta-\alpha^{2}]n+\alpha\beta+\alpha-\alpha^{2}-\beta\big{)}\bigg{]}\,,$

where $\alpha,\beta$ are as given by Eq. (44).

We now consider a special case in which Eq. (53) can be easily inverted
yielding an explicit kink solution.

####  3.2.1 Special Case: $b^{2n}=3a^{2n},~{}c^{2n}=4a^{2n}$

In this case the implicit kink solution takes the form

$e^{\mu
px}=\frac{(\phi^{2n}-a^{2n})^{2}(4a^{2n}-\phi^{2n})}{\phi^{2n}(3a^{2n}-\phi^{2n})^{2}}\,,$
(55)

where p is any positive number with $6pa^{6n}=1$, while $\mu$ now has a simple
form as given by Eq. (46). Eq. (55) is easily inverted yielding a cubic
equation

$u^{3}-3u+2\tanh\left(\frac{\mu px}{2}\right)=0\,,$ (56)

where $u=(\frac{\phi}{a})^{2n}-2$. Thus the explicit kink solution is given by
(see Fig. 6)

$u=\left(\frac{\phi}{a}\right)^{2n}-2=2\cos\big{[}(1/3)\cos^{-1}(-\tanh[p\mu
x/2])-2\pi/3\big{]}\,,$ (57)

where we have used the fact that $\cos^{-1}(1)=0\,,~{}~{}\cos^{-1}(-1)=\pi$.
It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=a+\frac{a}{\sqrt{3}n}e^{\mu
px/2}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=(3)^{1/2n}a-\frac{3^{1/n}a}{3\sqrt{3}n}e^{-\mu
px/2}\,.$ (58)

Figure 6: Kink solution $\phi(x)$ given by Eq. (57) for $a=1$, $p=1/6$ and
$\mu=72\sqrt{2}n$.

###  3.3 Kink Solution From $b$ to $c$

In this case, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\lambda\phi(\phi^{2n}-a^{2n})(\phi^{2n}-b^{2n})(c^{2n}-\phi^{2n})\,,$
(59)

which is easily achieved by using the partial fraction

$\int
d\phi\,\bigg{[}\frac{A}{\phi}+\frac{B\phi^{2n-1}}{(\phi^{2n}-a^{2n})}+\frac{D\phi^{2n-1}}{(\phi^{2n}-b^{2n})}+\frac{E\phi^{2n-1}}{(c^{2n}-\phi^{2n})}\bigg{]}\,.$
(60)

We find

$\displaystyle
A=\frac{1}{a^{2n}b^{2n}c^{2n}}\,,~{}~{}~{}B=-\frac{1}{a^{2n}(b^{2n}-a^{2n})(c^{2n}-a^{2n})}\,,$
$\displaystyle
D=\frac{1}{b^{2n}(b^{2n}-a^{2n})(c^{2n}-b^{2n})}\,,~{}~{}~{}E=\frac{1}{c^{2n}(c^{2n}-a^{2n})(c^{2n}-b^{2n})}\,.$
(61)

Eq. (62) is then easily integrated giving an implicit kink solution

$e^{\mu
x}=\frac{(\phi^{2n})^{[(b^{2n}-a^{2n})(c^{2n}-a^{2n})(c^{2n}-b^{2n})]}(\phi^{2n}-b^{2n})^{a^{2n}c^{2n}(c^{2n}-a^{2n})}}{(\phi^{2n}-a^{2n})^{b^{2n}c^{2n}(c^{2n}-b^{2n})}(c^{2n}-\phi^{2n})^{b^{2n}a^{2n}(b^{2n}-a^{2n})}}\,,$
(62)

where

$\mu=2\sqrt{2}n\lambda
a^{2n}b^{2n}c^{2n}(b^{2n}-a^{2n})(c^{2n}-a^{2n})(c^{2n}-b^{2n})\,.$ (63)

Note that the implicit kink solution is valid for arbitrary values of $a,b,c$
with the constraint $c>b>a>0$.

It is straightforward to calculate the kink mass in this case. We find

$\displaystyle M_{K}=\int_{b}^{c}d\phi\,\sqrt{2V(\phi)}$ $\displaystyle=$
$\displaystyle\frac{\lambda na^{2(3n+1)}}{\sqrt{2}(3n+1)(2n+1)(n+1)}$ (64)
$\displaystyle+$ $\displaystyle\left[\beta^{(n+1)/n}\left(6\beta
n^{2}+[-3\alpha\beta+5\alpha-3\beta-\beta^{2}]n-\alpha\beta+\alpha+\beta^{2}-\beta\right)\right]\,,$

where $\alpha,\beta$ are as given by Eq. (44).

We now consider one special case in which Eq. (62) can be easily inverted
yielding an explicit kink solution.

####  3.3.1 Special Case: $b^{2n}=3a^{2n},~{}c^{2n}=4a^{2n}$

In this case the implicit kink solution takes the form

$e^{\mu
px}=\frac{\phi^{2n}(3a^{2n}-\phi^{2n})^{2}}{(\phi^{2n}-a^{2n})^{2}(4a^{2n}-\phi^{2n})}\,,$
(65)

where p is any positive number with $6pa^{6n}=1$, while $\mu$ now has a simple
form as given by Eq. (45). Eq. (65) is easily inverted yielding a cubic
equation

$u^{3}-3u-2\tanh\left(\frac{\mu px}{2}\right)=0\,,$ (66)

where $u=(\frac{\phi}{a})^{2n}-2$. Thus the explicit kink solution is given by
(see Fig. 7)

$u=\left(\frac{\phi}{a}\right)^{2n}-2=2\cos\big{[}(1/3)\cos^{-1}(\tanh[p\mu
x/2])\big{]}\,,$ (67)

where we have used the fact that $\cos^{-1}(1)=0\,,~{}~{}\cos^{-1}(-1)=\pi$.
It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=(3)^{1/2n}a+\frac{a3^{1/2n}}{3\sqrt{3}n}e^{\mu
px/2}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=(2)^{1/n}a-\frac{2^{1/n}a}{18n}e^{-\mu
px/2}\,.$ (68)

Figure 7: Kink solution $\phi(x)$ given by Eq. (67) for $a=1$, $p=1/6$ and
$\mu=72\sqrt{2}n$.

##  4 Analytical Kink Solutions in Another Higher Order Field Theory Model

We now show that one can obtain exact kink solutions in a one-parameter family
of higher order field theories characterized by the potential (see Fig. 8)

$V(\phi)=\lambda^{2}\phi^{2}(a^{2n}-\phi^{2n})^{2}(b^{2n}+\phi^{2n})^{2}\,.$
(69)

Note that this potential has 3 degenerate minima at $\phi=0,\pm a$ and hence
one kink and one mirror kink solution and corresponding two antikink
solutions.

Figure 8: Potential $V(\psi)$ with three degenerate minima for $n=1,2,3$.
Parameters are $\lambda=1$, $a=1$ and $b=2^{1/(2n)}$. See Eq. (69)

###  4.1 Kink Solution From $0$ to $a$

In this case, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\lambda\phi(a^{2n}-\phi^{2n})(b^{2n}+\phi^{2n})\,,$ (70)

which is easily achieved by using the partial fraction

$\int
d\phi\,\bigg{[}\frac{A}{\phi}+\frac{B\phi^{2n-1}}{(a^{2n}-\phi^{2n})}+\frac{D\phi^{2n-1}}{(b^{2n}-\phi^{2n})}\bigg{]}\,.$
(71)

It is straightforward to check that

$A=\frac{1}{a^{2n}b^{2n}}\,,~{}~{}B=\frac{1}{a^{2n}(b^{2n}+a^{2n})}\,,~{}~{}D=-\frac{1}{b^{2n}(b^{2n}+a^{2n})}\,.$
(72)

Eq. (70) is then easily integrated giving an implicit kink solution

$e^{\mu
x}=\frac{(\phi^{2n})^{(b^{2n}+a^{2n})}}{(b^{2n}+\phi^{2n})^{a^{2n}}(a^{2n}-\phi^{2n})^{b^{2n}}}\,,$
(73)

where

$\mu=2\sqrt{2}n\lambda a^{2n}b^{2n}(b^{2n}+a^{2n})\,.$ (74)

Note that the implicit kink solution is valid for arbitrary values of $a,b$.

It is straightforward to calculate the kink mass in this case. We find

$M_{K}=\int_{0}^{a}d\phi\,\sqrt{2V(\phi)}=\frac{\lambda
na^{2(n+1)}[(2n+1)b^{2n}+a^{2n}]}{\sqrt{2}(2n+1)(n+1)}\,.$ (75)

We now consider three special cases in which Eq. (73) can be easily inverted
yielding explicit kink solutions.

####  4.1.1 Case I: $b^{2n}=2a^{2n}$

In this case the implicit kink solution (5) takes the form

$e^{\mu px}=\frac{\phi^{6n}}{(2a^{2n}+\phi^{2n})(a^{2n}-\phi^{2n})^{2}}\,,$
(76)

where $pa^{2n}=1$, while $\mu$ now has a simple form

$\mu=\frac{12\sqrt{2}n\lambda}{p^{3}}\,.$ (77)

Eq. (76) is easily inverted yielding a cubic equation

$(e^{p\mu x}-1)u^{3}-3e^{\mu px}u+2e^{p\mu x}=0\,,$ (78)

where $u=(\frac{\phi}{a})^{2n}$. Notice that the solution crucially depends on
whether $x$ is positive or negative or is zero. In particular, at $x=0$,
$u=2/3$, i.e. $\phi_{K}=(2/3)^{1/2n}a$. On the other hand the solution has
different forms in case $x>0$ or $x<0$. In particular, for $x>0$ the explicit
kink solution is given by (see Fig. 9)

$u=\left(\frac{\phi_{K}}{a}\right)^{2n}=2\sqrt{\frac{e^{p\mu x}}{e^{p\mu
x}-1}}\cos\bigg{[}\frac{1}{3}\cos^{-1}\bigg{(}-\sqrt{\frac{e^{p\mu
x}-1}{e^{p\mu x}}}\bigg{)}\bigg{]}\,,$ (79)

while for $x<0$ the solution is given by

$\displaystyle
u=\left(\frac{\phi_{K}}{a}\right)^{2n}=\bigg{[}\frac{1}{1-e^{\mu
px}}+\frac{e^{\mu px}}{(1-e^{\mu
px})^{3/2}}\bigg{]}^{1/3}+\bigg{[}\frac{1}{1-e^{\mu px}}-\frac{e^{\mu
px}}{(1-e^{\mu px})^{3/2}}\bigg{]}^{1/3}\,.$ (80)

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi_{K}(x)=2^{1/6n}ae^{\mu
px}\,,~{}~{}\lim_{x\rightarrow\infty}\phi_{K}(x)=a-\frac{a}{2\sqrt{3}n}e^{-\mu
px/2}\,.$ (81)

Thus while the implicit kink solution is known for arbitrary values of $a,b$
satisfying $b>a>0$, the explicit kink solution while valid for arbitrary
values of $a$, the corresponding $b$ is however fixed, i.e. $b^{2n}=2a^{2n}$.

Figure 9: Kink solution $\phi(x)$ given by Eqs. (79) and (80) for $a=1$, $p=1$
and $\mu=12\sqrt{2}n$.

It is straightforward to carry out the stability analysis by calculating the
kink potential
$V_{K}(x)=\frac{d^{2}V(\phi)}{d\phi^{2}}\Big{|}_{\phi=\phi_{K}}$ which appears
in the Schrödinger-like equation. In particular, we find that

$V_{K}(x\rightarrow-\infty)=8\lambda^{2}a^{8n}\,,~{}~{}~{}V_{K}(x\rightarrow\infty)=72n^{2}\lambda^{2}a^{8n}\,.$
(82)

####  4.1.2 Case II: $b^{2n}=(1/2)a^{2n}$

In this case the implicit kink solution (73) takes the form

$e^{2\mu
px}=\frac{\phi^{6n}}{((1/2)a^{2n}+\phi^{2n})^{2}(a^{2n}-\phi^{2n})}\,,$ (83)

where $pa^{2n}=1$, while $\mu$ now has a simple form

$\mu=\frac{3\sqrt{2}n\lambda}{2p^{3}}\,.$ (84)

Eq. (83) is easily inverted yielding a cubic equation

$u^{3}-\frac{3e^{2\mu px}}{4(1+e^{2p\mu x})}u-\frac{e^{2\mu px}}{4(1+e^{2p\mu
x})}=0\,,$ (85)

where $u=(\frac{\phi}{a})^{2n}$. Hence the explicit kink solution is

$\displaystyle u=\left(\frac{\phi_{K}}{a}\right)^{2n}=\bigg{[}\frac{e^{2\mu
px}}{8(1+e^{2p\mu x})}+\frac{e^{2\mu px}}{16(1+e^{2\mu
px})^{3/2}}\bigg{]}^{1/3}+\bigg{[}\frac{e^{2\mu px}}{8(1+e^{2p\mu
x})}-\frac{e^{2\mu px}}{16(1+e^{2\mu px})^{3/2}}\bigg{]}^{1/3}\,.$ (86)

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi_{K}(x)=2^{-1/3n}ae^{\mu
px}\,,~{}~{}\lim_{x\rightarrow\infty}\phi_{K}(x)=a-\frac{2a}{9n}e^{-2\mu
px}\,.$ (87)

Thus while the implicit kink solution is known for arbitrary values of $a,b$,
the explicit kink solution while valid for arbitrary values of $a$, the
corresponding $b$ is however fixed, i.e. $b^{2n}=(1/2)a^{2n}$.

It is straightforward to carry out the stability analysis by calculating the
kink potential
$V_{K}(x)=\frac{d^{2}V(\phi)}{d\phi^{2}}\Big{|}_{\phi=\phi_{K}}$ which appears
in the Schrödinger-like equation (13). In particular, we find that

$V_{K}(x\rightarrow-\infty)=\lambda^{2}a^{8n}/2\,,~{}~{}~{}V_{K}(x\rightarrow\infty)=18n^{2}\lambda^{2}a^{8n}\,.$
(88)

####  4.1.3 Case III: $b=a$

In this case the implicit kink solution (73) is easily inverted yielding the
explicit kink solution

$\phi=a\bigg{[}\frac{e^{p\mu x}}{1+e^{p\mu x}}\bigg{]}^{1/4n}\,.$ (89)

It may be noted that in this case the solution is valid not only for integer
$n$ but even half-integer $n$ and in fact has already been discussed
previously by us [11]. We discuss it briefly in Appendix A.

##  5 Analytical Kink Solutions with Power Law Tail in a One-Parameter Family
of Higher Order Field Theories

We now show that one can obtain exact kink solutions with power law tail in a
one-parameter family of higher order field theories characterized by the
potential (see Fig. 10)

$V(\phi)=\lambda^{2}\phi^{2(n+1)}\left|(a^{2n}-\phi^{2n})\right|^{3}\,.$ (90)

Note that this potential has degenerate minima at $\phi=0,\pm a$ and hence one
kink solution from $0$ to $a$ and the corresponding mirror kink solution from
$-a$ to $0$ and the corresponding two antikinks. Note that for both the kink
and antikink solutions $-a\leq\phi\leq a$. It is worth pointing out that
whereas the potential (90) is continuous, its derivative is discontinuous at
$\phi=\pm a$. However, since for the kink as well for the antikink solutions
$-a\leq x\leq a$, this discontinuity would not matter as far as the kink and
the antikink solutions are concerned.

Figure 10: Potential $V(\psi)$ with three degenerate minima for $n=1,2,3$.
Parameters are $\lambda=1$ and $a=1$. See Eq. (90).

In order to obtain the kink solution from 0 to $a$, we need to solve the self-
dual equation

$\frac{d\phi}{dx}=\sqrt{2}\lambda\phi^{n+1}\left|\phi^{2n}-a^{2n}\right|^{3/2}\,.$
(91)

This is easily integrated yielding

$\frac{(2\phi^{2n}-a^{2n})}{\phi^{n}(a^{2n}-\phi^{2n})^{1/2}}=\mu
x\,,~{}~{}~{}\mu=\sqrt{2}n\lambda a^{4n}\,.$ (92)

Eq. (92) is easily inverted yielding an explicit kink solution (see Fig. 11)

$\phi_{K}(x)=\frac{a}{2^{1/2n}}\left[1+\frac{\mu
x}{\sqrt{\mu^{2}x^{2}+4}}\right]^{1/2n}\,.$ (93)

It is straightforward to observe that

$\lim_{x\rightarrow-\infty}\phi_{K}(x)=\frac{a}{(-\mu
x)^{1/n}}\,,~{}~{}\lim_{x\rightarrow\infty}\phi_{K}(x)=a-\frac{a}{2n(\mu
x)^{2}}\,.$ (94)

Figure 11: Kink solution $\phi(x)$ given by Eq. (93) for $a=1$ and
$\mu=\sqrt{2}n$.

It is easy to calculate the kink mass in this case. We find

$M_{K}=\int_{0}^{a}d\phi\,\sqrt{2V(\phi)}=\frac{3\lambda\sqrt{\pi}a^{4n+2}}{4\sqrt{2}n}\frac{\Gamma[(n+2)/2n]}{\Gamma[(3n+1)/n]}\,.$
(95)

Using the explicit kink solution (93) it is straightforward to calculate the
zero mode and we find

$\psi_{0,K}\propto\frac{d\phi_{K}(x)}{dx}\propto\frac{1}{\left[1+\frac{\mu
x}{\sqrt{\mu^{2}x^{2}+4}}\right]^{1-1/2n}(1+\mu^{2}x^{2})^{3/2}}\,.$ (96)

As expected, this zero mode vanishes as $x\rightarrow\pm\infty$, i.e. as
$\phi\rightarrow 0,a$.

Finally the kink stability potential $V_{K}(x)$ which appears in the
Schrödinger-like equation (13) is easily calculated

$\displaystyle V_{K}(x)$ $\displaystyle=$
$\displaystyle\frac{d^{2}V(\phi)}{d\phi^{2}}\Bigg{|}_{\phi=\phi_{K}(x)}=\frac{\lambda^{2}a^{8n}}{16}\phi_{K}^{2n}(x)(a^{2n}-\phi_{K}^{2n})\bigg{[}(2n+1)(2n+2)(a^{2n}-\phi_{K}^{2n}(x))^{2}$
(97) $\displaystyle-$ $\displaystyle
18n(2n+1)\phi_{K}^{2n}(a^{2n}-\phi_{K}^{2n}(x))+24n^{2}\phi_{K}^{4n}(x)\bigg{]}\,.$

As expected, this kink potential $V_{K}(x)$ vanishes as
$x\rightarrow\pm\infty$ thereby confirming that indeed in this case there is
no gap between the zero mode and the beginning of the continuum. Further, one
finds that $V_{K}(x=0)=-\frac{(4n^{2}+6n-1)}{8}$.

###  5.1 Kink-Kink and Kink-Antikink Force

Using the recent Manton formalism [12] it is straightforward to predict the
acceleration $a$ and hence force $F$ in both the K-K and AK-K case and hence
the ratio of the two forces. We predict that for the model characterized by
the potential (69) (with $a=1$ and $\sqrt{2}\lambda=1$)

$a_{K-K}=\bigg{[}\frac{\Gamma(\frac{1}{2[2n+1]})\Gamma(\frac{n}{2[n+1]})}{2\sqrt{\pi}(n+1)}\bigg{]}^{2(2n+1)/n}\frac{4n\Gamma(\frac{3n+1}{n})}{3\sqrt{\pi}\Gamma(\frac{n+2}{2n})}\frac{1}{d^{2(2n+1)/n}}\,,$
(98)
$a_{AK-K}=\bigg{[}-\frac{\sqrt{\pi}\Gamma(\frac{n}{2[2n+1]})}{\Gamma(-\frac{1}{2[n+1]})}\bigg{]}^{2(2n+1)/n}\frac{4n\Gamma(\frac{3n+1}{n})}{3\sqrt{\pi}\Gamma(\frac{n+2}{2n})}\frac{1}{d^{2(2n+1)/n}}\,,$
(99)

where $d$ is the K-K or AK-K separation distance, and hence the ratio of the
two forces is

$\frac{F_{AK-K}}{F_{K-K}}=-\left[\sin(\frac{\pi
n}{2n+1})\right]^{2(2n+1)}\,,~{}~{}n=1,2,3,...\,.$ (100)

It may be noted that for the model characterized by Eq. (90), the kink tail
around $\phi=1$ has very different behavior compared to that around $\phi=0$
and as a result the kink-antikink acceleration and hence the corresponding
force has very different behavior. Generalizing the new Manton formalism [12]
for the fractional values of $n$, we predict that in this model the kink-
antikink acceleration is given by

$a_{K-AK}=\bigg{[}-\frac{\sqrt{\pi}\Gamma(\frac{1}{6})}{\Gamma(-\frac{2}{3})}\bigg{]}^{6}\frac{2\sqrt{2}n\Gamma(\frac{3n+1}{n})}{3\sqrt{\pi}\Gamma(\frac{n+2}{2n})}\left(\frac{1}{d^{6}}\right)\,.$
(101)

##  6 Explicit Kink Solutions with Power Law Tail for Another One-Parameter
Family of Potentials

We now show that one can obtain exact kink solutions with power law tail in
yet another one-parameter family of higher order field theories characterized
by the potential

$V(\phi)=\lambda^{2}\left|(a^{2n}-\phi^{2n})\right|^{(2n+1)/n}\,.$ (102)

Note that this potential has degenerate minima at $\pm a$ and hence one kink
solution from $-a$ to $a$ and the corresponding antikink solution from $a$ to
$-a$. Note that for the kink as well as the antikink solutions,
$-a\leq\phi\leq a$.

It is worth pointing out that whereas the potential (102) is continuous, its
derivative is discontinuous at $x=\pm a$. However, since for the kink as well
as for the antikink solutions $-a\leq\phi\leq a$, this discontinuity would not
matter as far as the kink and the antikink solutions are concerned.

In order to obtain the kink solution from $-a$ to $a$, we need to solve the
self-dual equation

$\frac{d\phi}{dx}=\sqrt{2}\lambda\left|\phi^{2n}-a^{2n}\right|^{(2n+1)/2n}\,.$
(103)

This is easily integrated yielding

$\frac{\phi}{(a^{2n}-\phi^{2n})^{1/2n}}=\mu x\,,~{}~{}~{}\mu=\sqrt{2}\lambda
a^{2n}\,.$ (104)

Eq. (104) is easily inverted yielding an explicit kink solution

$\phi_{K}(x)=\frac{a\mu x}{[1+(\mu x)^{2n}]^{1/2n}}\,.$ (105)

It is worth pointing out that for the special case of n = 1, the kink solution
(105) has been obtained before [25]. It is straightforward to find that

$\lim_{x\rightarrow-\infty}\phi_{K}(x)=-a+\frac{a}{2n(\mu
x)^{2n}}\,,~{}~{}\lim_{x\rightarrow\infty}\phi_{K}(x)=a-\frac{a}{2n(\mu
x)^{2n}}\,.$ (106)

It is also straightforward to observe that

$M_{K}=\int_{-a}^{a}d\phi\,\sqrt{2V(\phi)}`=\frac{\sqrt{2}\lambda
a^{2n+2}}{n}\frac{\Gamma[(4n+1)/2n]\Gamma[1/2n]}{\Gamma[(2n+1)/n]}\,.$ (107)

Using the explicit kink solution (105) it is easy to calculate the zero mode
and we find

$\psi_{0,K}\propto\frac{d\phi_{K}(x)}{dx}\propto[1+(\mu
x)^{2n}]^{-(2n+1)/2n}\,.$ (108)

It is easily checked that, as expected the zero mode eigenfunction vanishes as
$x\rightarrow\pm\infty$, i.e. as $\phi\rightarrow\pm a$.

Finally the kink stability potential $V_{K}(x)$ which appears in the
Schrödinger-like Eq. (13) is easily calculated using

$\displaystyle
V_{K}(x)=\frac{d^{2}V(\phi)}{d\phi^{2}}\Bigg{|}_{\phi=\phi_{K}(x)}=2(n+1)\lambda^{2}a^{4n}\frac{(\mu
x)^{4n-2}[4n-\frac{2n-1}{(\mu x)^{2n}}]}{[1+(\mu x)^{2n}]^{2}}\,.$ (109)

As expected, this kink potential vanishes as $x\rightarrow\pm\infty$ thereby
confirming that indeed in this case there is no gap between the zero mode and
the beginning of the continuum.

###  6.1 KK and K-AK Forces

Using the recent Manton formalism it is straight forward to predict the
acceleration a and hence force F in the AK-k case. However, it is not clear if
the recent Manton formalism [12] which was developed for calculating the force
between the kink and the mirror kink also applies for the single kink case.
However since it applies for the AK-K case, we have an hunch that it can also
be extended to the calculation of the single K-K acceleration and hence the
force and hence the ratio of the AK-K vs K-K forces. Assuming this, we predict
that for this model characterized by the potential (81) (with $a=1$ and
$\sqrt{2}\lambda=1$)

$a_{K-K}=\bigg{[}\frac{n\Gamma(\frac{1}{2[2n+1]})\Gamma(\frac{n}{2n+1})}{\sqrt{\pi}(2n+1)}\bigg{]}^{2(2n+1}\frac{n\Gamma(\frac{2n+1}{n})}{2\Gamma(\frac{4n+1}{2n})\Gamma(\frac{1}{2n})}\frac{1}{d^{2(2n+1)}}\,,$
(110)
$a_{K-AK}=\bigg{[}-\frac{\sqrt{\pi}\Gamma(\frac{1}{2[2n+1]})}{\Gamma(-\frac{n}{2n+1})}\bigg{]}^{2(2n+1}\frac{n\Gamma(\frac{2n+1}{n})}{2\Gamma(\frac{4n+1}{2n})\Gamma(\frac{1}{2n})}\frac{1}{d^{2(2n+1)}}\,,$
(111)

where $d$ is the K-K or AK-K separation distance, and hence the ratio of the
two forces is

$\frac{F_{K-AK}}{F_{K-K}}=-\left[\sin(\frac{\pi n}{2n+1})\right]^{2(2n+1)}\,.$
(112)

##  7 Summary and Possible Open Problems

In this paper we have presented five one parameter family of models for which
explicit kink solutions can be obtained. In three of these models, the kink
tail has an exponential fall-off and using the standard Manton formalism [30]
the K-K and K-AK forces and hence their ratio is easily calculated. On the
other hand, in the other two models, the kink tail has a power-law fall off.
To the best of our knowledge, these two models are the first two known higher
order field theory examples for which explicit kink solutions with power-law
tail can be obtained. Using the recent Manton formalism [12] the K-K and AK-K
forces and hence their ratio is easily calculated in the model discussed in
Sec. V. However, it is not obvious if the new Manton formalism is also
applicable to calculate the K-K force in the model of Sec. VI since in this
case there is only a single kink and there is no mirror kink. Note, however,
that the AK-K force can certainly be calculated in this case. We have assumed
in Sec. VI that one can extrapolate the new Manton formalism for this case too
and using it we have provided a prediction for the K-K acceleration and hence
the corresponding K-K force as well as the ratio of the AK-K vs K-K force. It
would be desirable if one could check whether this is indeed true or not.

Another open problem is to enquire if there exist higher order field theory
models which have super-exponential [21, 22], super-super-exponential [23] or
power-tower [24] type of tails. One criticism against the models presented in
Sec. V and VI is that these models are discontinuous at $x=\pm a$. While this
should not matter as far as the calculation of the kink solutions in these
models is concerned since for all of them $-a\leq x\leq a$, it is certainly
desirable to obtain explicit kink solutions with power-law tail in models
where no such criticism can be made. Yet another open problem is to understand
why for kinks with power-law tail the ratio of the AK-K to K-K forces
progressively decreases as we steadily go to higher order field theory models.
This is unlike the kinks with exponential tails for which the magnitude of
these two forces is always equal.

Acknowledgment: A.K. is grateful to Indian National Science Academy (INSA) for
the award of INSA Senior Scientist position at Savitribai Phule Pune
University. The work at Los Alamos National Laboratory was carried out under
the auspices of the U.S. DOE and NNSA under Contract No. DEAC52-06NA25396.

##  8 Appendix A: Additional Explicit Kink Solutions With Exponential Tail in
Higher Order Field Theories

As mentioned at the end of Sec. 2, we now show that one can obtain exact kink
solutions in a one-parameter family of higher order field theories
characterized by the potential

$V(\phi)=\lambda^{2}\phi^{2}(a^{2n}-\phi^{2n})^{2}(b^{2n}-\phi^{2n})^{2}\,,~{}~{}~{}b>a\,,$
(113)

in case $b^{2n}=4a^{2n}$. Note that the potential (113) has 5 degenerate
minima at $\phi=0,\pm a,\pm b$ and hence two kink and two mirror kink
solutions and corresponding four antikink solutions.

###  8.1 Kink Solution From $0$ to $a$ in case $b^{2n}=4a^{2n}$

In this case the implicit kink solution takes the form

$e^{\mu fx}=\frac{\phi^{6n}(4a^{2n}-\phi^{2n})}{(\phi^{2n}-a^{n})^{4}}\,,$
(114)

where $f$ is an arbitrary positive number satisfying $a^{2n}f=1$ while $\mu$
now has a simple form

$\mu=\frac{24\sqrt{2}n\lambda}{f^{3}}\,.$ (115)

Eq. (114) is easily inverted yielding a quartic equation

$u^{4}-\frac{6u^{2}}{(1+e^{\mu fx})}+\frac{8u}{(1+e^{\mu
fx})}-\frac{3}{(1+e^{\mu fx}+1)}=0\,,$ (116)

where $u=(1-\frac{\phi}{a})^{2n}$. Thus the explicit kink solution is given by

$\displaystyle
u=1-\Big{(}\frac{\phi}{a}\Big{)}^{2n}=1\mp\sqrt{2m}\mp\sqrt{\frac{12}{(1+e^{f\mu
x})}-2m\pm\frac{8}{\sqrt{m}(1+e^{\mu fx})}}\,,$ (117)

where $m$ has the form

$\displaystyle m=\bigg{[}\frac{e^{\mu fx}[\sqrt{(1+e^{\mu fx})}+1}{(1+e^{\mu
fx})^{3}}\bigg{]}^{1/3}+\bigg{[}\frac{e^{\mu fx}[\sqrt{(1+e^{\mu
fx})}-1}{(1+e^{\mu fx})^{3}}\bigg{]}^{1/3}+\frac{2}{(1+e^{\mu fx})}\,.$ (118)

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=\frac{a}{2^{1/3n}}e^{\mu
fx/6n}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=a-\frac{a3^{1/4}}{2n}e^{-\mu
fx/4}\,.$ (119)

###  8.2 Kink Solution From $a$ to $b$ in case $b^{2n}=4a^{2n}$

In this case the implicit kink solution takes the form

$e^{-\mu fx}=\frac{\phi^{6n}(4a^{2n}-\phi^{2n})}{(\phi^{2n}-a^{n})^{4}}\,,$
(120)

where $f$ is an arbitrary positive number satisfying $a^{2n}f=1$ while $\mu$
now has a simple form as given by Eq. (115).

Eq. (120) is easily inverted yielding a quartic equation

$u^{4}-\frac{6u^{2}}{(1+e^{-\mu fx})}-\frac{8u}{(1+e^{-\mu
fx})}-\frac{3}{(1+e^{-\mu fx}+1)}=0\,,$ (121)

where $u=(\frac{\phi}{a})^{2n}-1$. Thus the explicit kink solution is given by

$\displaystyle
u=\Big{(}\frac{\phi}{a}\Big{)}^{2n}-1=1\pm\sqrt{2m}\pm\sqrt{\frac{12}{(1+e^{-f\mu
x})}-2m\pm\frac{8}{\sqrt{m}(1+e^{-\mu fx})}}\,,$ (122)

where $m$ has the form

$\displaystyle m=\bigg{[}\frac{e^{-\mu fx}[\sqrt{(1+e^{-\mu
fx})}+1]}{(1+e^{-\mu fx})^{3}}\bigg{]}^{1/3}+\bigg{[}\frac{e^{-\mu
fx}[\sqrt{(1+e^{-\mu fx})}+1]}{(1+e^{-\mu
fx})^{3}}\bigg{]}^{1/3}+\frac{2}{(1+e^{-\mu fx})}\,.$ (123)

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi(x)=a+\frac{a3^{1/4}}{2n}e^{\mu
fx/4}\,,~{}~{}\lim_{x\rightarrow\infty}\phi(x)=4^{1/2n}a-\frac{a3^{4}2^{1/n}}{2^{8}}e^{-\mu
fx}\,.$ (124)

##  9 Appendix B: Analytical Kink Solutions with a Power-Law Tail in Yet
Another One-Parameter Family of Higher Order Field Theories

We show that one can obtain exact kink solutions with a power-law tail in a
one-parameter family of higher order field theories characterized by the
potential

$V(\phi)=\lambda^{2}\phi^{2(3n+1)}\left|(a^{2n}-\phi^{2n})\right|^{3}\,.$
(125)

Note that this potential has degenerate minima at $\phi=0,\pm a$ and hence one
kink solution from $0$ to $a$ and the corresponding mirror kink solution from
$-a$ to $0$ and the corresponding two antikinks. Note that for both the kink
solutions $-a\leq\phi\leq a$ and hence even though the potential is
discontinuous at $\phi=\pm a$ that would not matter as far as the kink and
corresponding antikink solutions are concerned.

###  9.1 Explicit Kink Solution

In order to obtain the kink solution, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\sqrt{2}\lambda\phi^{3n+1}\left|\phi^{2n}-a^{2n}\right|^{3/2}\,.$
(126)

This is easily integrated yielding

$\frac{(8\phi^{4n}-4\phi^{4n}-a^{4n})^{2}}{9\phi^{6n}(a^{2n}-\phi^{2n})}=\mu^{2}x^{2}\,,~{}~{}~{}\mu=\sqrt{2}n\lambda
a^{6n}\,.$ (127)

Eq. (127) is easily inverted yielding a quartic equation

$u^{4}+au^{2}+bu+d=0\,,$ (128)

where

$a=-\frac{3}{8}\,,~{}~{}~{}b=-\frac{(192+99\mu^{2}x^{2})}{64(64+9\mu^{2}x^{2})}\,,~{}~{}~{}d=\frac{(576-27\mu^{2}x^{2})}{256(64+9\mu^{2}x^{2})}\,,$
(129)

while $u=(\frac{\phi}{a})^{2n}-1/4$. Thus the explicit kink solution is given
by

$u=\left(\frac{\phi}{a}\right)^{2n}-1/4=\pm\sqrt{2m}\pm\sqrt{-(2a+2m\pm\frac{2b}{\sqrt{2m}})}-a/3\,,$
(130)

where $m$ satisfies the cubic equation

$8m^{3}+8am^{2}+(2a^{2}-8d)m-b^{2}=0\,.$ (131)

It is not difficult to check that in this case $m$ is given by

$m=\bigg{[}-\frac{b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}\bigg{]}^{1/3}+\bigg{[}-\frac{b}{2}-\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}\bigg{]}^{1/3}\,.$
(132)

It is straightforward to check that

$\lim_{x\rightarrow-\infty}\phi_{K}(x)=\frac{a}{(9\mu^{2}x^{2})^{1/6n}}\,,~{}~{}~{}\lim_{x\rightarrow\infty}\phi_{K}(x)=a-\frac{a}{2n\mu^{2}x^{2}}\,.$
(133)

It is easy to calculate the kink mass in this case. We find

$M_{K}=\int_{0}^{a}d\phi\,\sqrt{2V(\phi)}=\frac{3\lambda\sqrt{\pi}a^{6n+2}}{4\sqrt{2}}\frac{\Gamma[(3n+2)/2]}{\Gamma[(3n+7)/2]}\,.$
(134)

Finally, as expected in this case too there is no gap between the zero mode
and the beginning of the continuum.

###  9.2 Kink-Kink and Kink-Antikink Force

Using the recent Manton formalism it is straightforward to predict the
acceleration $a$ and hence the force $F$ in both the K-K and AK-K cases and
thus the ratio of the two forces. We predict that for the model characterized
by the potential (90) (with $a=1$ and $\sqrt{2}\lambda=1$)

$a_{K-K}=\bigg{[}\frac{\Gamma(\frac{1}{2[3n+1]})\Gamma(\frac{3n}{2[3n+1]})}{2\sqrt{\pi}(3n+1)}\bigg{]}^{2(3n+1)/3n}\frac{4\Gamma(\frac{3n+7}{2})}{3\sqrt{\pi}\Gamma(\frac{3n+2}{2})}\frac{1}{d^{2(3n+1)/3n}}\,,$
(135)
$a_{AK-K}=\bigg{[}-\frac{\sqrt{\pi}\Gamma(\frac{3n}{2[3n+1]})}{\Gamma(-\frac{1}{2[3n+1]})}\bigg{]}^{2(3n+1)/3n}\frac{4\Gamma(\frac{3n+7}{2})}{3\sqrt{\pi}\Gamma(\frac{3n+2}{2})}\frac{1}{d^{2(3n+1)/3n}}\,,$
(136)

where $d$ is the K-K or AK-K separation distance, and hence the ratio of the
two forces is

$\frac{F_{AK-K}}{F_{K-K}}=-\left[\sin\left(\frac{\pi}{2[3n+1]}\right)\right]^{2(3n+1)/3n}\,,~{}~{}n=1,2,3,...\,.$
(137)

It may be noted that for the model characterized by Eq. (125), the kink tail
around $\phi=1$ has very different behavior compared to that around $\phi=0$
and as a result the kink-antikink acceleration and hence the corresponding
force have very different behavior. Generalizing the new Manton formalism
[12], we predict that in this model the kink-antikink acceleration is given by

$a_{K-AK}=\bigg{[}-\frac{\sqrt{\pi}\Gamma(\frac{1}{6})}{\Gamma(-\frac{1}{3})}\bigg{]}^{6}\frac{1}{8n^{3}}\frac{4\Gamma(\frac{3n+7}{2})}{3\sqrt{\pi}\Gamma(\frac{3n+2}{2})}\left(\frac{1}{d^{6}}\right)\,.$
(138)

##  10 Appendix C: Known Explicit Solutions in Higher Order Field Theory
Models

For the sake of completeness, we list here the previously known solutions from
[2, 11, 16]. To the best of our knowledge, so far only three explicit kink
solutions are known for higher order (polynomial) field theories, i.e. for
$\phi^{8},\phi^{10},\phi^{12}$ and even higher order field theories. Two of
these are with exponential tails while one has a power-law tail. In
particular, while explicit kink solution with a power-law tail is known for a
$\phi^{10}$ model [11], explicit kink solutions with an exponential tail are
known for a (i) $\phi^{4n+2}-\phi^{2n+2}-\phi^{2}$ ($n=1,2,3,...$) model [11]
and (ii) for a $\phi^{8}$ model [16].

###  10.1 $\phi^{10}$ Model with a Power-Law Tail

It has been shown [11] that for the kink potential

$V(\phi)=\lambda^{2}\left|(a^{2}-\phi^{2})\right|^{5}\,,$ (139)

the explicit kink solution from $-a$ to +$a$ is given by

$\displaystyle
1-(\frac{\phi_{K}}{a})^{2}=\bigg{[}\frac{\sqrt{4+\mu^{2}x^{2}}+\mu
x}{2(4+\mu^{2}x^{2})^{3/2}}\bigg{]}^{1/3}+\bigg{[}\frac{\sqrt{4+\mu^{2}x^{2}}-\mu
x}{2(4+\mu^{2}x^{2})^{3/2}}\bigg{]}^{1/3}\,.$ (140)

The details can be found in the reference mentioned above.

###  10.2 One-Parameter Family of Potentials with an Exponential Tail

It has been shown [11] that for the one-parameter family of potentials

$V(\phi)=\frac{\lambda^{2}}{2}\phi^{2}(a^{2n}-\phi^{2n})^{2}\,,~{}~{}n=1,2,3,...\,,$
(141)

which has a kink solution from $0$ to $a$ and a mirror kink from $-a$ to $0$
and the corresponding two antikinks, the kink solution from $0$ to $a$ is
given by

$\phi_{K}=a\left[\frac{1+\tanh(\beta x)}{2}\right]^{1/2n}\,,$ (142)

where $\beta=n\lambda a^{2n}$. The details can be found in the reference
mentioned above.

###  10.3 $\phi^{8}$ Model with an Exponential Tail

It has been shown [16] that for the $\phi^{8}$ model

$V(\phi)=(1/2)(a^{2}-\phi^{2})^{2}(b^{2}-\phi^{2})^{2}\,,~{}~{}~{}0<a<b\,,$
(143)

which has kink solutions from $a$ to $b$, mirror kinks from $-b$ to $-a$ and a
kink solution from $-a$ to +$a$. In case $b=1,a=1/2$, the three kink solutions
and the corresponding antikink solutions are given by

$\phi_{K}=\cos\bigg{[}\frac{1}{3}\cos^{-1}\Big{(}\tanh\frac{3x}{4}\Big{)}+\frac{\pi
m}{3}\bigg{]}\,,$ (144)

depending on the value of $m=0,1,2,3,4,5$. On the other hand, in case
$b=1,a=1/4$ the three kink and three antikink solutions are given by the
solution of a quartic equation. The details can be found in the reference
mentioned above.

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