Novel Deformation Function Creating or Destroying any Number of Even Kink
Solutions

Avinash Khare

Physics Department, Savitribai Phule Pune University

Pune 411007, India

Avadh Saxena

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

Abstract: We present a one-parameter family of deformation functions $f(\phi)$
which have novel properties. Firstly, the deformation function is its own
inverse. We show that a class of potentials remains invariant under this
deformation. Further, when applied to a certain class of kink bearing
potentials, one obtains potentials which are not bounded from below. Besides,
we show that there is a wide class of potentials having power law kink tails
which are connected by this deformation but the corresponding kink tails of
the two potentials have different asymptotic behavior. Finally, we show that
when this deformation function is applied to an appropriate one-parameter
family of potentials having two kink solutions, it creates new potentials with
an arbitrary even number ($2n$) of kink solutions. Conversely, by starting
from this potential with $2n$ kink solutions the deformation function
annihilates $2n-2$ kinks and we get back the potential which we started with
having only two kink solutions.

##  1 Introduction

During the last few years there has been a resurgence in studying kink
solutions with power-law tails in several higher (than 6th) order field theory
models including $\phi^{8}$, $\phi^{10}$, and $\phi^{12}$ [1, 2] and even
$\phi^{14}$, $\phi^{16}$, and $\phi^{18}$ [3]. The study of higher order field
theories, their attendant kink excitations as well as the associated kink
interactions and scattering are important in a variety of physical contexts
ranging from successive phase transitions [1, 2, 4, 5] to isostructural phase
transitions [6] to models involving long-range interaction between massless
mesons [7], as well as from protein crystallization [8] to successive phase
transitions presumably driving the late time expansion of the Universe [9].
Thus, it is of interest to study kink solutions in additional higher order
field theories and study the various properties of these solutions including
the nature of the kink-kink and the kink-antikink forces and scattering in
these theories.

In this context we mention that almost two decades ago, in an interesting
paper, Bazeia, Losano and Malbonisson [10] introduced the idea of deformation
function which takes one from a kink bearing potential to another kink bearing
(or in some cases pulse bearing) potential. In a subsequent series of papers,
these and other authors [11, 12, 13, 14] studied a variety of deformation
functions and also discussed possible applications of some of these
deformation functions.

We mention in particular the deformation function $f(\phi)=(1-\phi^{2})^{1/2}$
introduced in [12] where the authors briefly discussed some of its properties.
The purpose of this paper is to discuss a one-parameter family of deformation
functions

$f(\phi)=(1-\phi^{2n})^{1/2n}\,,~{}~{}n=1,2,3,...\,,$ (1)

where $n$ is any positive integer. In particular we point out its several
remarkable properties.

The plan of the paper is the following. In Sec. II we set up the formalism for
the kink solutions in a neutral scalar field theory in $1+1$ dimensions and
then spell out the basic steps of the deformation prescription. We then
discuss some of the interesting properties of the deformation function (1). In
particular, if under the deformation (1), a scalar potential $V_{0}(\phi)$
goes to another potential $V_{1}(\phi)$, then the same deformation function
acting on $V_{1}(\phi)$ gives back the original potential $V_{0}(\phi)$ i.e.
$V_{0}$ and $V_{1}$ are deformation-dual potentials. In other words, the
deformation (1) is inverse of itself. We then exhibit a wide class of
potentials $V_{0}(\phi)$ which are invariant under this deformation; we call
such potentials as self-deformed potentials. Further, we exhibit a wide class
of kink bearing potentials $V_{0}(\phi)$ which under the deformation (1) go to
potentials $V_{1}(\phi)$ which are not bounded from below. In Sec. III we then
discuss a wide class of deformation-dual potentials both of which admit 2 kink
solutions and show that in case at least one of the kink tails of $V_{0}$ and
$V_{1}$ is a power law tail, then the power law tail behavior of $V_{0}$ and
the corresponding tail of $V_{1}$ are different. This is unlike most of the
examples discussed so far in the literature. We illustrate this feature by
discussing one example in detail. In Sec. IV we exhibit perhaps the most
remarkable property of the deformation (1). In particular, we show that by
appropriately choosing the potential $V_{0}(\phi)$ with two kink solutions and
by applying the deformation (1), one can obtain the potential $V_{1}$ with as
many even number $2m$ of kink solutions as desired. We discuss one example in
detail to illustrate this aspect. Needless to say that, if instead we apply
the deformation (1) on $V_{1}$ having $2m$ kink solutions, we obtain the
potential $V_{0}$ with two kink solutions. In other words, by an appropriate
choice of the potential, one can create or destroy as many even number of kink
solutions as desired. Finally, in Sec. V we summarize the main results
obtained in this paper and point out a few open questions.

##  2 Formalism and Important Properties of the Deformation Function

Consider a relativistic neutral scalar field theory in $1+1$ dimensions with
the Lagrangian density being [15]

${\rm{\cal{L}}}=\frac{1}{2}\left(\frac{d\phi}{dt}\right)^{2}-\frac{1}{2}\left(\frac{d\phi}{dx}\right)^{2}-V(\phi)\,,$
(2)

which leads to the equation of motion

$\frac{d^{2}\phi}{dt^{2}}-\frac{d^{2}\phi}{dx^{2}}=\frac{dV}{d\phi}\,.$ (3)

We assume that the potential $V(\phi)$ is smooth and nonnegative and attains
its global minimum value of $V=0$ for two or more values of $\phi$ which are
the global minima of the theory. Thus one has static kink and antikink
solutions interpolating between the two adjoining global minima as $x$ goes
from $-\infty$ to +$\infty$. While the field equation for a static kink is a
second order ODE, it can be reduced to a first order ODE using the so called
Bogomolnyi technique [16]. The first order ODE is given by

$\frac{d\phi}{dx}=\pm\sqrt{2V(\phi)}\,.$ (4)

Let us now point out a few basic properties of the deformation formalism [10,
14]. In particular, if we start with a potential $V_{0}(\phi)$ and apply a
deformation function $f(\phi)$, then the new potential $V_{1}(\phi)$ is
related to the original potential $V_{0}(\phi)$ by

$V_{1}(\phi)=\frac{V_{0}(\phi)}{\left[\frac{df(\phi)}{d\phi}\right]^{2}}\,.$
(5)

Further, if the explicit kink solutions corresponding to the two potentials
$V_{0}(\phi)$ and $V_{1}(\phi)$ are $\phi^{(0)}_{K}(x)$ and
$\phi^{(1)}_{K}(x)$ respectively, then they are related by

$\phi^{(1)}_{K}(x)=f^{-1}[\phi^{(0)}_{K}(x)]\,.$ (6)

On the other hand, if instead of the explicit kink solution, only the implicit
kink solutions are known and if corresponding to the potentials $V_{0}(\phi)$
and $V_{1}(\phi)$ they are given by $x^{(0)}_{K}(\phi)$ and
$x^{(1)}_{K}(\phi)$ respectively, then

$x^{(1)}_{K}(\phi)=x^{(0)}_{K}(f[\phi])\,.$ (7)

Let us now discuss the various properties of the deformation function given by
Eq. (1).

  1. 1.

Notice that the inverse of this deformation is equal to itself, i.e.

$f^{-1}(\phi)=f(\phi)\,.$ (8)

Stated differently, if we start with a potential $V_{0}(\phi)$ and after
applying the deformation function $f(\phi)$ as given by Eq. (1) if we get the
potential $V_{1}(\phi)$, then we know that $V_{0}$ and $V_{1}$ are related to
each other as given by Eq. (5). Similarly, if we apply the same deformation
function to $V_{1}(\phi)$, then we get back the potential $V_{0}(\phi)$, i.e.

$V_{0}(\phi)=\frac{V_{1}(\phi)}{\left[\frac{df(\phi)}{d\phi}\right]^{2}}\,,$
(9)

where $f(\phi)$ is given by Eq. (1). In that case, we call $V_{0}$ and $V_{1}$
as deformation-dual or deformation pair. As a result, the relation between the
explicit kink solutions of the potentials $V_{0}(\phi)$ and $V_{1}(\phi)$
given by Eq. (6) can also be expressed as

$\phi^{(1)}_{K}(x)=f[\phi^{(0)}_{K}(x)]\,.$ (10)

  2. 2.

For every integer $n$, a wide class of potentials are invariant under the
deformation (1). We call such potentials as self-dual potentials. Using Eq.
(1) it is easy to check that for an arbitrary integer $n$, the potential

$V_{0}(\phi)=\frac{1}{2}\phi^{4nm+2}[1-\phi^{2n}]^{2m+2}=V_{1}(\phi)\,,~{}~{}m=0,1,2,...\,.$
(11)

We term these wide classes of potentials as self-deformed potentials. As an
illustration, in the case of $n=1$, the self-deformed one parameter family of
potentials is

$V_{0}(\phi)=\frac{1}{2}\phi^{4m+2}[1-\phi^{2}]^{2m+2}=V_{1}(\phi)\,,~{}~{}m=0,1,2,...\,.$
(12)

  3. 3.

For every integer $n$, there is a two-parameter family of kink bearing
potentials $V_{0}$ for which when we apply the kink deformation function (1)
to them, the resulting deformed-dual potential $V_{1}$ is not bounded from
below. In particular, for an arbitrary integer $n$, consider the two-parameter
family of potentials

$V_{0}(\phi)=\frac{1}{2}\phi^{4nm+2(2n-1)(n-1)}(1-\phi^{2n})^{2p+2}\,,~{}~{}~{}m,p=0,1,2,...\,,$
(13)

and if we apply the deformation function (1) to this $V_{0}$, then we find
that the corresponding potential $V_{1}$ is given by

$V_{1}(\phi)=\frac{1}{2}\phi^{4np+2}(1-\phi^{2n})^{2m+2n-1}\,.$ (14)

Notice that $V_{1}(\phi)$ is not bounded from below. As an illustration,
consider the special case when $n=m=1,p=0$, in which case the deformed-dual
potentials are

$V_{0}(\phi)=\frac{1}{2}\phi^{4}(1-\phi^{2})^{2}\,,~{}~{}~{}V_{1}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{2})^{3}\,.$
(15)

It is worth pointing out that the potential $V_{0}$ in Eq. (15) has recently
received a lot of attention [7, 17, 18, 19, 20, 21] in the context of kink
solutions with a power law tail around $\phi=0$ while an exponential tail
around $\phi=\pm 1$. As is clear from Eq. (15), the corresponding dual
potential $V_{1}(\phi)$ is not bounded from below.

##  3 Comparison of Kink Tails of the Deformed-Dual Potentials

We now consider for any arbitrary integer $n$, a two-parameter family of
deformed-dual potentials given by

$V_{0}(\phi)=\frac{1}{2}\phi^{4np+2}(1-\phi^{2n})^{2m+2}\,,~{}~{}p,m=0,1,2,...\,,$
(16)

and

$V_{1}(\phi)=\frac{1}{2}\phi^{4nm+2}(1-\phi^{2n})^{2p+2}\,.$ (17)

Note that here $p\neq m$ since in that case the two potentials $V_{0},V_{1}$
are one and the same, i.e. it is the self-deformed potential. We note that for
any set of integers $p,m,n$, both the deformed-dual potentials (16) and (17)
have two kinks and two antikink solutions. We now show that (for $p\neq m$) at
least one of the kink (and hence the antikink) tails of $V_{0}$ and the
corresponding kink tails of $V_{1}$ have different asymptotic behavior. In
this context it is worth reminding that for most of the examples discussed in
the literature, the kink tails of $V_{0}$ and the corresponding kink tails of
$V_{1}$ are the same up to an overall factor.

In this context let us note the well known result that as $x$ goes from
$-\infty$ to +$\infty$, if a kink solution for a potential $V(\phi)$ goes from
$0$ to $b$ and if $V(\phi)$ behaves as $\phi^{q}$ around $\phi=0$, then the
kink tail around $\phi=0$ has the asymptotic behavior

$\lim_{x\rightarrow-\infty}\phi_{K}(x)=x^{-(q-2)/2}\,,$ (18)

where $q$ is a real number. Thus if $q=2$ then the kink tail is exponential
while for $q>2$ the kink has a power law tail. Let us now use this result and
look at the behavior of the kink solution around $\phi=0$ and $\phi=1$ for the
dual potentials given above. Using Eq. (18), it is straightforward to show
that for the potential $V_{0}(\phi)$ as given by Eq. (16), for the kink
solution between $0$ and $1$, the behavior of the kink tail is given by

$\lim_{x\rightarrow-\infty}\phi^{(0)}_{K}(x)=\frac{A_{0}}{x^{4p}}\,,~{}~{}~{}\lim_{x\rightarrow\infty}\phi^{(0)}_{K}(x)=1-\frac{A_{1}}{x^{m}}\,.$
(19)

On the other hand, for the potential $V_{1}(\phi)$ as given by Eq. (17), for
the kink solution between $0$ and $1$, the behavior of the kink tail is given
by

$\lim_{x\rightarrow-\infty}\phi^{(0)}_{K}(x)=\frac{B_{0}}{x^{4m}}\,,~{}~{}~{}\lim_{x\rightarrow\infty}\phi^{(0)}_{K}(x)=1-\frac{B_{1}}{x^{p}}\,.$
(20)

Here $A_{0},A_{1},B_{0},B_{1}$ are constants. Since under the deformation (1),
while $\phi=0$ and $\phi=1$ of $V_{0}(\phi)$ get mapped to $\phi=1$ and
$\phi=0$ respectively of $V_{1}(\phi)$, one would expect that the behavior of
the kink tail around $\phi=0(1)$ for the potential $V_{0}(\phi)$ should be the
same as that of the kink tail around $\phi=1(0)$ for the deformation-dual
potential $V_{1}(\phi)$. However it is clear from Eqs. (19) and (20) that this
is not so and the asymptotic behavior of the kink tails in the two dual
potentials (16) and (17) is very different unless $p=m$ but in that case the
two potentials are self-deformed (i.e. $V_{1}(\phi)=V_{0}(\phi)$). It is worth
pointing out that in case, say $p=0$ but $m\neq 0$, then the kink tail of
$V_{0}(\phi)$ around $\phi=0$ and the kink tail of $V_{1}(\phi)$ around
$\phi=1$ are both exponential while the kink tail of $V_{0}(\phi)$ around
$\phi=1$ and the kink tail of $V_{1}(\phi)$ around $\phi=0$ are both power law
tails having different asymptotic behavior. On the other hand when both
$p,m>0$ but $p\neq m$, then both the kink tails of $V_{0}(\phi)$ are different
from the corresponding kink tails of $V_{1}(\phi)$,

For simplicity we now explicitly discuss the case of $n=m=1,p=0$ and
demonstrate that indeed the appropriate kink tails of the two deformed-dual
potentials are different. In case $n=m=1$, $p=0$, $V_{0}(\phi)$ and
$V_{1}(\phi)$ are given by

$V_{0}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{2n})^{4}\,,$ (21)
$V_{1}(\phi)=\frac{1}{2}\phi^{6}(1-\phi^{2n})^{2}\,.$ (22)

These two potentials are depicted in Fig. 1 with three degenerate minima each.
The kink solutions in both the cases can be implicitly obtained.

Figure 1: Potential $V_{0}(\phi)$ in Eq. (21) and its dual potential
$V_{1}(\phi)$ in Eq. (22) for $n=1$.

For example, in order to obtain the kink solution of $V_{0}(\phi)$ as given by
Eq. (21) from $0$ to $1$, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\phi(1-\phi^{2})^{2}\,.$ (23)

This equation is easily integrated using partial fractions and we obtain

$x^{(0)}=\frac{1}{2(1-\phi^{2})}+\frac{1}{2}\ln\left[\frac{\phi^{2}}{(1-\phi^{2})}\right]\,.$
(24)

It then immediately follows that for the potential $V_{0}(\phi)$ as given by
Eq. (21), the asymptotic behavior of the kink solution from $0$ to $1$ is
given by

$\lim_{x\rightarrow-\infty}\phi^{(0)}_{K}(x)=e^{(x-1/2)}\,,~{}~{}\lim_{x\rightarrow\infty}\phi^{(0)}_{K}(x)=1-\frac{1}{4x}\,.$
(25)

The kink solution given by Eq. (24) is shown in Fig. 2 with the exponential
and power law tails clearly visible at the two ends.

Figure 2: Kink solution obtained by inverting implicit Eq. (24).

In order to obtain the kink solution of $V_{1}(\phi)$ as given by Eq. (22)
from $0$ to $1$, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\phi^{3}(1-\phi^{2})\,.$ (26)

This equation is easily integrated using partial fractions and we obtain

$x^{(1)}=-\frac{1}{2\phi^{2}}+\frac{1}{2}\ln\left[\frac{\phi^{2}}{(1-\phi^{2})}\right]\,.$
(27)

It then immediately follows that for the potential $V_{1}(\phi)$ as given by
Eq. (22), the asymptotic behavior of the kink solution from $0$ to $1$ is
given by

$\lim_{x\rightarrow-\infty}\phi^{(1)}_{K}(x)=\frac{1}{\sqrt{-2x}}\,,~{}~{}~{}\lim_{x\rightarrow\infty}\phi^{(1)}_{K}(x)=1-\frac{1}{2}e^{-2(x+1/2)}\,,$
(28)

so that while the kink tail of $V_{0}$ goes like $x^{-1}$ around $\phi=1$, the
kink tail of $V_{1}$ around $\phi=0$ goes like $(-x)^{1/2}$. The kink solution
given by Eq. (27) is shown in Fig. 3 with the power law and exponential tails
clearly visible at the two ends. Needless to say that alternatively we can
also calculate the kink tails of the potential (22) by using Eqs. (7) and (24)
and we would get the same kink tails as given by Eq. (28) up to an overall
factor.

Figure 3: Kink solution obtained by inverting implicit Eq. (27).

##  4 Kink Creating and Kink Annihilating Deformation

Finally, we discuss perhaps the most interesting property of the deformation
function (1). In particular, by a judicious choice of the potential
$V_{0}(\phi)$ having two kink solutions, this deformation can create a new
potential $V_{1}(\phi)$ having as many even number of kink solutions as
desired. We give below three wide classes of $V_{0}$ and $V_{1}$ satisfying
this property.

Let us start from the potential

$V_{0}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{2n})^{2}(b_{1}^{2n}+\phi^{2n})^{2}(b_{2}^{2n}+\phi^{2n})^{2}(b_{3}^{{2n}}+\phi^{2n})^{2}...(b_{m}^{2n}+\phi^{2n})^{2}\,,$
(29)

which has two kink and two antikink solutions. On applying the deformation
function (1) on this $V_{0}(\phi)$ we obtain the deformed potential

$V_{1}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{2n})^{2}(b_{1}^{2n}+1-\phi^{2n})^{2}(b_{2}^{2n}+1-\phi^{2n})^{2}...(b_{m}^{2n}+1-\phi^{2n})^{2}\,,$
(30)

which clearly has $2m+2$ kink solutions and $2m+2$ antikink solutions. Thus in
this case, for $V_{0}(\phi)$ as given by Eq. (29), the deformation function
(1) acts like a kink (and antikink) creating operator, creating extra $2m$
kink solutions. From here it immediately follows that, as far as the potential
$V_{1}(\phi)$ as given by Eq. (30) is concerned, the deformation function (1)
acts like a kink (and antikink) annihilating operator destroying $2m$ kinks
out of its $2m+2$ kinks.

One can generalize the potential $V_{0}$ given by Eq. (29) and choose more
general potentials of the form

$V_{0}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{2n})^{2}(b_{1}^{2n}+\phi^{2n})^{2p_{1}}(b_{2}^{2n}+\phi^{2n})^{2p_{2}}...(b_{m}^{2n}+\phi^{2n})^{2p_{m}}\,,$
(31)

where $p_{1},p_{2},...,p_{m}$ are arbitrary positive integers. This $V_{0}$
obviously has has two kink and two antikink solutions. On applying the
deformation function (1) on this $V_{0}(\phi)$ we obtain the deformed
potential

$V_{1}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{2n})^{2}(b_{1}^{2n}+1-\phi^{2n})^{2p_{1}}(b_{2}^{2n}+1-\phi^{2n})^{2p_{2}}...(b_{m}^{2n}+1-\phi^{2n})^{2p_{m}}\,,$
(32)

which clearly has $2m+2$ kinks and $2m+2$ antikink solutions.

We can further generalize the potential $V_{0}(\phi)$ of Eq. (31) and choose
more general potentials of the form

$V_{0}(\phi)=\frac{1}{2}\phi^{4np+2}(1-\phi^{2n})^{2k+2}(b_{1}^{2n}+\phi^{2n})^{2p_{1}}(b_{2}^{2n}+\phi^{2n})^{2p_{2}}...(b_{m}^{2n}+\phi^{2n})^{2p_{m}}\,,$
(33)

where $p,k=0,1,2,...$ while $p_{1},p_{2},...,p_{m}$ are arbitrary positive
integers. Clearly this $V_{0}$ has two kinks and two antikink solutions. On
applying the deformation function (1) on this $V_{0}(\phi)$, we obtain the
deformed potential

$V_{1}(\phi)=\frac{1}{2}\phi^{4nk+2}(1-\phi^{2n})^{2p+2}(b_{1}^{2n}+1-\phi^{2n})^{2p_{1}}(b_{2}^{2n}+1-\phi^{2n})^{2p_{2}}...(b_{m}^{2n}+1-\phi^{2n})^{2p_{m}}\,,$
(34)

which clearly has $2m+2$ kink solutions.

Remarkably, there are several potentials $V_{0}$ and $V_{1}$ in which kink
solutions are explicitly or implicitly known and they fall in the above
category. As an illustration we discuss two such examples, in one of which
while $V_{0}$ has two kink solutions which are explicitly known, the
corresponding $V_{1}$ has four kink solutions which are all explicitly known
as well. In another case, while $V_{0}$ has two kink solutions which are
implicitly known, the corresponding $V_{1}$ has six kink solutions all of
which are implicitly known as well.

Example I:

Few years ago, we along with Christov [2] had explicitly obtained two kink and
two antikink solutions for the one-parameter family of potentials

$V_{0}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{4n})^{2}\,,~{}~{}n=1,2,3,...\,.$ (35)

On the other hand, very recently, two of us [22] have obtained explicit four
kink and four antikink solutions of the model

$V_{1}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{2n})^{2}(2-\phi^{2n})^{2}\,,~{}~{}n=1,2,3,...\,.$
(36)

Note that these $V_{0}$ and $V_{1}$ are in fact special cases of the
potentials given in Eqs. (29) and (30), respectively, in case $b_{1}=1$ while
$b_{2}=...=b_{m}=0$ and hence $V_{0}(\phi)$ and $V_{1}(\phi)$ as given by Eqs.
(35) and (36) are deformation-dual. The potentials given by Eqs. (35) and (36)
for $n=1$ are depicted in Fig. 4 with three and five degenerate minima,
respectively.

Figure 4: Potential $V_{0}(\phi)$ in Eq. (35) and its dual potential
$V_{1}(\phi)$ in Eq. (36) for $n=1$.

As has been shown in [1], for the potential (35) with arbitrary $n$, the kink
from $0$ to $1$, the corresponding mirror kink from $-1$ to $0$ and the
corresponding two antikink solutions are given by

$\phi_{0K,0aK}(x)=\pm\left[\frac{1\pm\tanh(2nx)}{2}\right]^{1/4n}\,.$ (37)

For $n=1$ this kink solution is depicted in Fig. 5. Having obtained all the
kink and the antikink solutions of the potential (35) from $0$ to $1$, the
corresponding kink, mirror kink and the two anti-kink solutions of the dual
potential (36) are then immediately obtained from the deformation formalism.
In particular using Eq. (10) we obtain

$\phi_{1K,1aK}(x)=\pm[1-\phi^{2}_{0K,0aK}(x)]^{1/4n}=\pm\left[1-\sqrt{\frac{1\pm\tanh(2nx)}{2}}\right]^{1/4n}\,.$
(38)

For $n=1$ his kink solution is depicted in Fig. 6. It is worth pointing out
that these are precisely the explicit kink solutions that we have recently
obtained [22] for the potential (36). Thus, one has found a novel relationship
between the two seemingly very different potentials, one admitting two while
the other admitting four kink solutions.

Figure 5: Explicit kink solution given by Eq. (37) for $n=1$.  Figure 6:
Explicit kink solution given by Eq. (38) for $n=1$.

Unfortunately, at present we do not know how to obtain the kink solution from
$1$ to $(2)^{1/2n}$, the corresponding mirror kink and the corresponding two
antikink solutions of potential (36) from the kink solution of the deformed-
dual potential (35) since it does not have a kink solution from $1$ to
$(2)^{1/2n}$.

Example II:

We have recently obtained [22] implicit six kink and six antikink solutions of
the model characterized by the potential

$V_{1}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{2n})^{2}(2-\phi^{2n})^{2}(3-\phi^{2n})^{2}\,,~{}~{}n=1,2,3,...\,,$
(39)

without realizing that this potential is deformation-dual to the potential

$V_{0}(\phi)=\frac{1}{2}\phi^{2}(1-\phi^{4n})^{2}(2+\phi^{2n})^{2}\,,~{}~{}n=1,2,3,...\,,$
(40)

i.e. the above $V_{1}(\phi)$ can be obtained from the $V_{0}(\phi)$ as given
by Eq. (40) by applying the deformation function (1) on it. Potentials given
by Eqs. (39) and (40) are shown in Fig. 7 for $n=1$ with seven and three
degenerate minima, respectively.

Figure 7: Potential $V_{1}(\phi)$ in Eq. (39) and its dual potential
$V_{0}(\phi)$ in Eq. (40) for $n=1$.

In order to obtain the kink solution of $V_{0}(\phi)$ as given by Eq. (40)
from $0$ to $1$, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\phi(1-\phi^{4n})(2+\phi^{2n})\,.$ (41)

This equation is easily integrated using partial fractions and we obtain

$x^{(0)}=\frac{1}{12}\ln\bigg{[}\frac{\phi^{6n}(2+\phi^{2n})}{(1-\phi^{2n})(1+\phi^{2n})^{3}}\bigg{]}\,.$
(42)

It then immediately follows that for the potential $V_{0}(\phi)$ as given by
Eq. (40), the asymptotic behavior of the kink solution from $0$ to $1$ is
given by

$\lim_{x\rightarrow-\infty}\phi^{(0)}_{K}(x)=(2)^{-1/6}e^{2x/n}\,,~{}~{}~{}\lim_{x\rightarrow\infty}\phi^{(0)}_{K}(x)=1-\frac{3}{16n}e^{-12x}\,.$
(43)

The kink solution given by Eq. (42) for $n=1$ is depicted in Fig. 8 with both
exponential tails clearly visible at the two ends.

Figure 8: Kink solution obtained by inverting implicit Eq. (45) for $n=1$.
Figure 9: Kink solution obtained by inverting implicit Eq. (42) for $n=1$.

In order to obtain the kink solution of $V_{1}(\phi)$ as given by Eq. (39)
from $0$ to $1$, we need to solve the self-dual equation

$\frac{d\phi}{dx}=\phi(1-\phi^{2n})(2-\phi^{2n})(3-\phi^{2n})\,.$ (44)

This equation is easily integrated using partial fractions and as we have
shown in our recent paper [22] we obtain

$x^{(1)}=\frac{1}{12}\ln\bigg{[}\frac{\phi^{2n}(2-\phi^{2n})^{3}}{(1-\phi^{2n})^{3}(3-\phi^{2n})}\bigg{]}\,.$
(45)

It then immediately follows that for the potential $V_{1}(\phi)$ as given by
Eq. (39), the asymptotic behavior of the kink solution from $0$ to $1$ is
given by

$\lim_{x\rightarrow-\infty}\phi^{(0)}_{K}(x)=(3/8)^{1/2n}e^{6x/n}\,,~{}~{}~{}\lim_{x\rightarrow\infty}\phi^{(0)}_{K}(x)=1-\frac{1}{2^{4/3}n}e^{-4x}\,.$
(46)

The kink solution given by Eq. (45) for $n=1$ is depicted in Fig. 9 with both
exponential tails clearly visible at the two ends.

Unfortunately, at present we do not know how to obtain the kink solutions from
$1$ to $(2)^{1/2n}$ and from $(2)^{1/2n}$ to $(3)^{1/2n}$, the corresponding
mirror kinks and the corresponding four antikink solutions of the potential
(39) from the kink solutions of the deformed-dual potential (40), since it
only has a kink solution from $0$ to $1$ and the corresponding mirror kink
solution.

##  5 Conclusions and Open Questions

In this paper we have discussed several novel properties of the deformation
function as given by Eq. (1). We have shown that this deformation function is
its own inverse. As a result, as shown by us, apart from a wide class of self-
deformed potentials (which are invariant under this deformation), all other
potentials break up into deformed pairs. As also shown by us, for a wide class
of kink bearing potentials, the corresponding deformed-dual potential is not
bounded from below. On the other hand, there is a wide class of deformed pair
of potentials both of which admit two kink and two antikink solutions and for
which at least one of the kink tails has a power law tail. In such cases, as
shown by us, the appropriate kink tails of the two deformed pairs with power
law tail have different asymptotic behavior. As a result the kink-kink and the
kink-antikink forces in the two deformed pair of potentials are very
different. Finally, we have pointed out what we consider to be the most
remarkable feature of the deformation function (1), i.e. it can act as an
arbitrary even number $2m$ of kink creating or kink annihilating function.
That is, starting from a wide class of potentials with two kinks, this
deformation can take one to the corresponding deformed-dual potential having
arbitrary even $2m+2$ number of kink solutions. Conversely, by starting from
the deformed-dual potential with $2m+2$ kink solutions and applying the
deformed function (1), one goes to its partner potential having only two kink
solutions.

This paper raises several interesting questions, some of which are as follows.

  1. 1.

In this paper we have considered a deformation function which is its own
inverse, i.e. naively it is a deformation with period two. Generalizing, are
there deformation functions of period $2^{n}$, where $n$ is any positive
integer? If yes, what are its key properties? Can it also act as a kink
creating and/or kink annihilating deformation? How does this deformation
function behave for large $n$?

  2. 2.

Are there deformation functions of prime number (like $3,5,7,11,...$) period
as well as an arbitrary multiple of them? If yes, what are the properties of
such deformation functions?

  3. 3.

Are there deformations which can act as a kink creating function with
creating, say $n$, $2n$, $3n$ … kink solutions or conversely can it act as a
kink annihilating function with reducing $n$, $2n$, $3n$, … number of kink
solutions?

  4. 4.

So far the various deformation functions suggested in the literature are such
that the qualitative nature of the kink tail in a given potential and the
corresponding appropriate kink tail of the potential obtained after the
deformation are similar, i.e. both could be exponential or both could be power
law tail or other tails, such as super-exponential [23], etc. Is there a
deformation for which the qualitative nature of the appropriate kink tails are
qualitatively different?

It would be quite insightful if one could answer some of the questions raised
here.

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

## References

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# Reconfigurable Intelligent Surface MIMO Simulation using Quasi Deterministic
Radio Channel Model ††thanks: The authors are with Huawei Technologies Co.,
Ltd. Their e-mails are {semyon.dorokhin, aderkina.anastasia,
<EMAIL_ADDRESS>and<EMAIL_ADDRESS>
Semyon Dorokhin, Pavel Lysov, Anastasia Aderkina, Vladimir Lyashev

###### Abstract

Reconfigurable Intelligent Surface (RIS) is a planar array that can control
reflection and thus can implement the concept of partially controllable
propagation environment. RIS received a lot of attention from industry and
academia, but the majority of the researchers who study RIS-assisted systems
use simple Rician model. Though it is suitable for theoretical analysis,
stochastic Non Line-of-Sight (NLoS) component in Rician model does not account
for the geometry of deployment. Furthermore, Rician model is not eligible to
evaluate 3GPP standardization proposals. In this article we adapt the popular
Quasi Deterministic Radio channel Generator (QuaDRiGa) for RIS-assisted
systems and compare it against Rician model. The comparison shows that
geometry-inconsistent NLoS Rician modeling results in higher estimated
achievable rate. Our method, in contrast, inherits the advantages of QuaDRiGa:
spatial consistency of Large Scale Fading, User Equipment mobility support as
well as consistency between Large Scale and Small Scale Fading. Moreover,
QuaDRiGa comes with calibrated scenario parameters that ensure 3GPP
compatibility. Finally, the proposed method can be applied to any model or
software originally designed for conventional MIMO, so every researcher can
use it to build a simulation platform for RIS-assisted systems.

###### Index Terms:

Reconfigurable Intelligent Surface, Intelligent Reflective Surface, MIMO,
QuaDRiGa, Channel model, Capacity

##  I Introduction

Reconfigurable intelligent surface (RIS) is a passive planar array that can
control the reflection and either perform beam-steering and beam-focusing or
split the incident beam into multiple reflected beams [1]. Such surfaces have
a potential for capacity improvement and coverage extension [2] as a result of
partial control of propagation environment.

The RIS concept quickly gained popularity, there are plenty of papers on
channel estimation, joint beamforming and interference mitigation in RIS-
assisted systems [3]. However, the majority of the researchers use Rician
model for the analysis and this can greatly delay the standardization of RIS-
assisted systems for several reasons. Firstly, the Non Line-Of-Sight (NLoS)
component in Rician model is modeled as random, which contradicts with the
deterministic nature of RIS deployment. NLoS component in Rician model does
not account for the scenario geometry, which may become crucial in distributed
RIS-assisted systems. Secondly, this model does not comply with 3GPP standard
[4] and it is unclear how to calibrate it to match field tests.

One of the possible techniques for geometry-consistent simulations is Ray
Tracing (RT). Though 3GPP standard [4] includes RT-based model as an
alternative, such simulations need detailed maps of the environment to yield
accurate results. For Macro-cell simulations it will require large number of
digital maps, which is impractically difficult.

Geometry-Based Stochastic Models (GBSMs) offer a tradeoff between Ray-tracing
and Rician models. Instead of using digital map, GBSMs generate clusters of
scatterers according to calibrated distribution and only then use Ray Tracing
approach. Large Scale and Small Scale fading parameters can also be included.
Several authors suggested GBSMs for RIS-assisted systems [5, 6, 7, 8], but
these models have serious limitations on frequency range, number of elements,
mobility of terminals, scenarios etc., as we demonstrate in section II-A.
Moreover, most of them are not 3GPP-compatible and only one of them is freely
accessible online. This motivated us to build a new GBSM simulation method for
RIS-assisted MIMO systems using one of the state-of-the-art conventional MIMO
models. Based on the comparison that we present in section II-B, we decided to
use QuaDRiGa simulation platform [9, 10]. QuaDRiGa is free software that
supports user mobility, features spatial consistency of fading parameters and
offers a wide range of 3GPP-compliant scenarios.

The main contributions of our paper are as follows:

  * •

in section III we propose a new simulation method for RIS-assisted MIMO
systems based on QuaDRiGa;

  * •

in section IV we compare the achievable rate of a system simulated with Rician
and the proposed models and explain the huge difference.

Finally, in section V we summarize our research findings.

##  II Models for RIS-assisted systems

Since the RIS concept is relatively young, there is still no standardized
model for RIS-assisted systems. The articles on RIS use different channel
models with different parameters, as we show in section II-A. At the same
time, the models for conventional MIMO systems are well-developed and can be
adapted to RIS-assisted systems. In section II-B we choose the best one for
this task.

###  II-A RIS-assisted system models

Let us consider a downlink RIS MIMO system with $N_{RX}$ antennas at the User
Equipment (UE), $N_{TX}$ antennas at the Base Station (BS) and $N_{RIS}$
elements at the RIS. All RIS-assisted channel models express the total channel
$\mathbf{H}\in\mathbb{C}^{N_{RX}\times N_{TX}}$ using the BS-UE channel
$\mathbf{H}_{0}\in\mathbb{C}^{N_{RX}\times N_{TX}}$, BS-RIS channel
$\mathbf{H}_{A}\in\mathbb{C}^{N_{RIS}\times N_{TX}}$ and RIS-UE channel
$\mathbf{H}_{B}\in\mathbb{C}^{N_{RX}\times N_{RIS}}$:

$\mathbf{H}=\mathbf{H}_{0}+\mathbf{H}_{B}\mathbf{Q}\mathbf{H}_{A},$ (1)

where $\mathbf{Q}=diag(\mathbf{\mu})\in\mathbb{C}^{N_{RIS}\times N_{RIS}}$ is
the RIS control matrix with diagonal elements $|\mu_{i}|=1\ \forall i=1,\dots
N_{RIS}$. Next, some model should be selected for $\mathbf{H}_{0}$,
$\mathbf{H}_{A}$ and $\mathbf{H}_{B}$ channels.

Simple Freespace model can be used for theoretical estimations, e.g. to
compare RIS against Decode-and-Forward relays [11]. However, this model lacks
Non Line-Of-Sight (NLoS) components and therefore it is inappropriate for
realistic scenarios. To account for the multipath components, the majority of
researchers [3] adopt Rician channel model, splitting every channel matrix
$\mathbf{H}_{i}$ in (1) into LoS and NLoS components:

$\mathbf{H}_{i}=\beta\left(\sqrt{\frac{K}{K+1}}\mathbf{H}_{i}^{LoS}+\sqrt{\frac{1}{K+1}}\mathbf{H}_{i}^{NLoS}\right),$
(2)

where $\beta$ is the pathloss, $K$ is the Rician factor,
$\mathbf{H}_{i}^{LoS}$ is calculated according to the freespace model and
$\mathbf{H}_{i}^{NLoS}$ is random. For example, in [12] the authors took into
account spatial correlation by taking columns of $\mathbf{H}_{i}^{NLoS}$ from
$\mathcal{CN}(\mathbf{0},\mathbf{R})$ with correlation matrix $\mathbf{R}$
determined by $sinc$ function.

Nevertheless, such model of NLoS component is unlikely to give realistic
results. Firstly, during the derivation of $\mathbf{R}$ authors in [12]
assumed that the scatterers are distributed uniformly in front of the the RIS,
which hardly happens in real life. Secondly, the RIS position is
deterministic, therefore modeling the channels as random matrices yields an
uncalibrated tool. It can be used to study properties of algorithms, but it is
unclear how to tune it to match real-world measurements.

To bring RIS-assisted systems closer to standardization, researchers should
use geometry-consistent channel models. Good results can be achieved with Ray
Tracing method, especially for indoor scenarios that allow creating highly-
accurate digital maps [13]. However, in outdoor scenarios the environment can
vary a lot, especially between urban and rural environments. Generating a
sufficiently detailed map for every specific cell is impractically difficult.

Figure 1: General structure of GBSM model for RIS-assisted systems.

Geometry-Based Stochastic Models (GBSM) offer a trade-off between the fully-
determenistic Ray Tracing and almost stochastic Rician methods. GBSMs do not
need the exact map of the environment, as the position of scaterrer clusters
is generated randomly according to calibrated distributions with specific
bodily angles $\sigma$, as in Fig. 1. After the cluster generation, Ray
Tracing can be applied. Finally, Large and Small Scale fading parameters [4]
can be introduced. GBSM models offer calibration for every typical scenario
and at the same time are more accurate than stochastic ones.

Several researchers suggested different GBSMs for RIS-assisted communication.
For example, the model introduced in [6] considers only LoS part of
$\mathbf{H}_{A}$ and $\mathbf{H}_{B}$ channels and NLoS part of
$\mathbf{H}_{0}$ channel, while the model in [5] features cluster generation.
However, [5] does not take into account user mobility and [6] considers only
one cluster in the NLoS part. Moreover, they are limited to narrow bands and
support only omnidirectional antennas.

The GBSM model from [7] supports Tx and Rx mobility as well. In addition, it
includes Shadow Factor and cluster evolution with death probability.
Nevertheless, it calculates the cascaded channel
$\mathbf{H}_{B}\mathbf{Q}\mathbf{H}_{A}$ pathloss based on the optimal RIS
coefficients. These coefficients are derived for freespace model and may not
be optimal in general. Thus, the pathloss expression may be incorrect if
$\mathbf{H}_{B}$ or $\mathbf{H}_{A}$ are NLoS.

The GBSM model introduced in [8] features pathloss for Urban Microcellular and
Indoor Hotspot scenarios from 3GPP standard [4] and is available online as
simulation software [14]. However, it supports neither Tx/Rx mobility nor
wideband simulations and is implemented only for single-antenna case.
Moreover, only LoS component is supported in $\mathbf{H}_{B}$ in indoor
scenario and the number of clusters is the same as for mm-Wave band, limiting
simulation frequency.

Although there are several GBSMs for RIS-assisted systems, they all have
rather strict constraints related either to the number of antenna elements,
radiation patterns, pathloss expression, frequency range or channel
conditions. These constraints motivated us to use one of the available MIMO
models and to adapt it to RIS-assisted MIMO systems.

###  II-B MIMO GBSM models

Among all 5G GBSM simulation platforms and models, there are three that
received significant popularity [15], namely, the More General 5G model (MG5G)
[16], the NYUSIM channel simulator [17] and the Quasi Determenistic Radio
Channel Generator (QuaDRiGa) [10].

TABLE I: State of the art MIMO GBSM models Simulator or Model |  Calibration according to 3GPP model |  Massive MIMO support |  LOS/NLOS transition |  Moving clusters |  Custom antenna patterns
---|---|---|---|---|---
QuaDRiGa |  Yes |  Yes |  Yes |  No |  Yes
NYUSIM |  No |  No |  Yes |  No |  No
MG5G |  No |  Yes |  No |  Yes |  Yes

The main advantage of the MG5G model is its detailed cluster evolution. In
addition to ’newborn’ and ’disappearance’ cluster states, MG5G features
’survival’ state. The parameters of survived clusters are regenerated based on
the new geometry and the channel coefficients are recalculated for every time
interval. However, MG5G model does not support transition between LoS and NLoS
scenarios.

NYUSIM focuses on mm-Wave range and features a channel model similar to the
3GPP [4] model. However, the NYUSIM platform has restrictions on the number of
antenna elements: no more than 128 Tx antennas and no more than 64 RX
antennas. Such restrictions do not allow modeling a sufficiently large RIS to
provide reasonable gain. Moreover, NYUSIM lacks custom antenna pattern
support, which is crucial for physically-consistent simulations.

Compared to these two simulators, QuaDRiGa platform has no limit on the number
of antenna elements and supports transition between LoS and NLoS scenarios.
Apart from that, the main advantage of QuaDRiGa is that it offers predefined
channel models with parameters calibrated according to 3GPP standard channel
model [4]. Different simulation scenarios can be loaded both for sub-6GHz and
mm-Wave ranges.

We choose QuaDRiGa for our RIS-assisted MIMO simulations based on the
comparison summarized in Table I. Compared to MG5G, it supports LoS/NLoS
transition and in comparison to NYUSIM, it has no limit on antenna array size
and supports custom antenna patterns. Finally, the main advantage of QuaDRiGa
is that it comes with 3GPP-calibrated model configurations unlike the two
other competitors. Until the goal of RIS standardization is accomplished,
using QuaDRiGa for simulations is a natural step towards it.

##  III RIS-assisted MIMO with QuaDRiGa

Figure 2: The geometry of the RIS-assisted system.

In this section we propose the novel RIS-assisted MIMO simulation method that
is based on the well-known QuaDRiGa platform [10] . The main advantage of the
proposed method is that QuaDRiGa is compatible [9] with 3GPP TR 38.901 v16.1.0
and 3GPP TR 36.873 v12.5.0 standards, as well as with the mmMAGIC channel
model. Moreover, QuaDRiGa features spatial consistency of Large Scale Fading
parameters and consistency between Large Scale and Small Scale parameters. In
addition to this, it supports mobility of UE, which means that the proposed
simulation method can be used for RIS-assisted MIMO systems with moving UEs or
RIS.

The QuaDRiGa simulation platform was originally designed for MIMO systems, so
of-the-shelf QuaDRiGa does not support simulations with RIS. To extend it to
RIS-assisted communication we model every channel in (1) as conventional MIMO
channel and use QuaDRiGa to obtain the corresponding channel matrix.

More specifically, we obtain the direct channel $\mathbf{H}_{0}$ modeling MIMO
system BS-UE in a usual way. To get the BS-RIS channel $\mathbf{H}_{A}$, we
represent the RIS as a virtual receiver and simulate the BS-RIS subsystem as
conventional MIMO system. Next, we represent the RIS as a virtual transmitter
and obtain the RIS-UE channel $\mathbf{H}_{B}$. Thus in our QuaDRiGa-based
simulation platform the RIS is represented by a virtual receiver and a virtual
transmitter that have the same coordinates and the same element array. At the
same time, we omit the simulation between the virtual transmitter and receiver
by setting the corresponding pairing parameter in configuration [9].

Apart from 3GPP-compliant calibrated model, another detail that ensures
physically-consistent simulation is the scattering pattern of RIS element. In
this section we demonstrate how it can be included in the proposed simulation
method.

First, let us explain the concept of scattering pattern using a simple single-
antenna LoS example with single-element RIS. In this case, it is possible to
express the received power in the cascaded subchannel using the radar equation
approach:

$P_{RX}=\frac{P_{TX}G_{TX}(\varphi_{d},\theta_{d})G_{RX}(\varphi_{a},\theta_{a})\sigma_{RIS}\lambda^{2}}{{4\pi}^{3}R_{TI}^{2}R_{IR}^{2}},$
(3)

where $P_{TX}$ is the transmitter power, $\sigma_{RIS}$ is the RIS element
radar cross-section, $G_{TX}(\varphi_{d},\theta_{d})$ and
$G_{RX}(\varphi_{a},\theta_{a})$ are the transmitter and receiver antenna
patterns, respectively, $\varphi_{d},\theta_{d}$ are the transmitter’s Angles
of Departure and $\varphi_{a},\theta_{a}$ are the receiver’s Angles of
Arrival. We denote the carrier wavelength as $\lambda$, and the distances from
Tx to RIS and from RIS to Rx as $R_{TI}$ and $R_{IR}$ respectively, as it is
shown in Fig. 2.

(a) CDF of $\mathbf{H}_{A}$ and $\mathbf{H}_{B}$ eigenvalues. Solid lines for
$\mathbf{H}_{A}$, dashed lines for $\mathbf{H}_{B}$.

(b) CDF of $\mathbf{H}_{0}$ channel. Solid lines for first eigenvalues, dashed
for the second.

(c) Achievable Rate CDF with (dashed) and without (solid) RIS, RIS control
from [18].

Figure 3: Comparison between the Rician and the proposed model.

The antenna patterns in (3) can be expressed in normalized form:
$G_{TX}=G_{TX}^{max}F_{TX}(\varphi_{d},\theta_{d})$ and
$G_{RX}=G_{RX}^{max}F(\varphi_{a},\theta_{a})$, as in [19]. Supposing one RIS
element can be represented as a rectangular plate of conductive material of
size $a\times b$, we can obtain similar expression using eq. (11.44) from
[20]:

$\sigma_{RIS}(\varphi_{i},\varphi_{r},\theta_{i},\theta_{r})=4\pi\left(\frac{ab}{\lambda}\right)^{2}F(\varphi_{i},\varphi_{r},\theta_{i},\theta_{r}),$
(4)

where $\varphi_{i},\theta_{i}$ are azimuth and elevation angles of incidence,
$\varphi_{r},\theta_{r}$ \- corresponding angles of reflection, as in Fig. 2.
$F(\varphi_{i},\varphi_{r},\theta_{i},\theta_{r})$ is the isolated element
scattering pattern, a normalized function that takes values between 0 and 1
and describes angular properties of scattered field.

As it follows from macroscopic reradiation model introduced in [21], the
scattering pattern of RIS element can be factorized as:

$F(\varphi_{i},\varphi_{r},\theta_{i},\theta_{r})=F(\varphi_{i},\theta_{i})F(\varphi_{r},\theta_{r}),$
(5)

where $F(\varphi,\theta)$ is the antenna pattern of RIS element. Decomposition
(5) helps us to model the scattering pattern quite easily using QuaDRiGa
antenna pattern options. The antenna patterns of the virtual transmitter and
receiver’s element can be set to
$F(\varphi,\theta)=\left(sin(\theta)\right)^{\alpha}$ as in [21]. Note that
the definition of angle $\theta$ in [21] is different, that is why in our
article the antenna pattern includes $sin(\theta)$ instead of $cos(\theta)$.

In general, due to mutual coupling the center element pattern is different
from the edge element pattern [22]. However, for large arrays this effect
hardly impacts array gain [22] and we can assume that all the elements have
the same pattern. Furthermore, mutual coupling causes the difference between
the isolated and the embedded element pattern. The embedded pattern can be
obtained via EM-simulation or measured in a setup like in [23] and uploaded to
QuaDRiGa as a custom one. Thus, specifying the pattern of single RIS element
is enough, but in this article we used omnidirectional element model for the
RIS, BS and the UE so that it would be easier to compare the proposed
simulation method with the Rician model.

##  IV Numerical results

To compare our QuaDRiGa-based RIS channel model against Rician model from
[18], we perform a simulation for a simple single-user, single-RIS scenario.
The BS, the RIS and the UE have coordinates (0, 0, 25), (200, 50, 25) and
(250, 0, 1.5) respectively. The number of elements at the BS, the UE and the
RIS is $N_{TX}=4\times 4=16$, $N_{RX}=4\times 1$ and $N_{RIS}=45\times
45=2025$ respectively. The elements on the RIS, the BS and the UE are
$d=\lambda/2$ apart, and for simplicity we model all the elements as
isotropic. In QuaDRiGa we perform simulation in $1.4$ MHz band with $15$ kHz
subcarrier spacing and set the BS power to $30$ dBm, so that the power per
subcarrier is $10.3$ dBm. The noise power spectral density is $-174$ dBm/Hz
and the receiver noise factor is $9$ dB. Thus the noise power per subband is
$-123$ dBm. To compare with Rician model, we extract the channels at one
subcarrier and use these values for achievable rate calculation.

In a popular scenario the direct BS-UE path is blocked and the RIS is in Line-
of-Sight conditions with respect to the UE and the BS. That is why we choose
3GPP 38.901 UMa NLoS model for $\mathbf{H}_{0}$ and 3GPP 38.901 UMa LoS model
for $\mathbf{H}_{A}$ and $\mathbf{H}_{B}$. To make the comparison fair, we set
$K=9$ dB in Rician model (2) for $\mathbf{H}_{A}$ and $\mathbf{H}_{B}$, like
it is in UMa LoS model. For $\mathbf{H}_{0}$ we set $K=-100$ dB like in UMa
NLoS QuaDRiGa configuration file. In (2) we use the same pathloss expressions
as in UMa LoS and NLoS models.

Additionally, to determine Large-Scale fading parameters impact, we perform
simulation with two modifications of 3GPP 38.901 UMa models. The first one
removes the Shadow Fading, setting its standard deviation to $-100$ dB both in
LoS and NLoS models. The second one additionally sets the $K$ standard
deviation to $\sigma_{K}=-100$ dB in UMa LoS model. We obtain $500$
realizations for every channel and for every model to plot the Cumulative
Density Functions (CDFs).

Fig. 3(a) shows the CDFs of LoS channels $\mathbf{H}_{A}$ and
$\mathbf{H}_{B}$. Though the average eigenvalue is the same for the Rician
(blue lines) and UMa LoS models (red lines), the range of possible eigenvalues
is different. The Rician LoS channel yields eigenvalues in a range of $0.2$
dB, while the eigenvalues in 3GPP UMa LoS channel vary by approximately $20$
dB. Such a large range is caused mainly by the Shadow factor, since setting it
to $-100$ dB reduces the eigenvalue range to approximately $3$ dB.
Additionally setting $K$ variance to $-100$ dB yields almost the same CDF as
with Rician model (2).

Next, we compare the 3GPP UMa NLoS model against the Rician model from [18].
As Fig. 3(b) demonstrates, the mean values of the eigenvalues differ: with UMa
model the first and second eigenvalues are approximately $3$ dB and $7$ dB
lower, respectively. Moreover, with Rician model the two eigenvalues are only
$1.5$ dB apart, while with UMa model the difference between them is $6$ dB.
The reason is that Rician model assumes that the scatterers are distributed
uniformly in the hemisphere in front of the receiver, while 3GPP model
features specific AoA and AoD ranges with non-uniform distribution that
depends on the geometry of the scenario.

Finally, we compare the achievable rate of the RIS-assisted MIMO system for
the two models. We assume perfect CSI and use an algorithm suggested by Zhou
et al. [18] for joint BS-RIS precoding. As Fig. 3(c) shows, there is a huge
performance difference between the two models. The mean achievable rate
without the RIS is $2.2$ times larger with Rician model. Moreover, with Rician
model RIS provides $31\%$ gain, while with QuaDRiGa RIS gives almost $100\%$
gain. That is because $\mathbf{H}_{0}$ eigenvalues are smaller with QuaDRiGa.

To summarize, there is a significant difference in the results obtained with
Rician and with 3GPP-standardized models caused by different modeling of the
Non-Line-of-Sight part of the channel. Though it is possible to modify Rician
model so that the results match for fixed UE, BS and RIS position, the
universal calibration is not feasible.

##  V Conclusion

Rician channel model, which is extremely popular in RIS analysis, is an
obstacle to standardization of RIS-assisted MIMO since it models NLoS
component in geometry-inconsistent way. Moreover, Rician model is not
compatible with channel model calibration procedure specified in the 3GPP
standard. To fill this gap, we introduce new RIS-assisted MIMO channel model
which is based on QuaDRiGa simulation platform with calibrated parameters. The
proposed method inherits realistic 3GPP-compliant channel model from QuaDRiGa.
We demonstrate that, compared with the proposed method, Rician model yields
overestimated achievable rate. In addition to this, we explain how the
scattering pattern of the RIS element can be included in the proposed method.
The channel model we analyzed is difficult to calibrate because it uses
isolated RIS element pattern that is different from the embedded one. In our
future research, we are going to obtain the embedded RIS element pattern via
measurements and fit the proposed model based on field tests.

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$\frac{\partial}{\partial
t}W_{t}(T_{M}f)(\mathbf{n})=-\int_{0}^{\infty}a(s)\frac{\partial}{\partial
t}\frac{\partial}{\partial
s}W_{s+t}f(\mathbf{n})\,ds=-\int_{0}^{\infty}a(s)\frac{\partial^{2}}{\partial
s^{2}}W_{s+t}f(\mathbf{n})\,ds.$

In this way,

$\displaystyle\left|\frac{\partial}{\partial
t}W_{t}(T_{M}f)(\mathbf{n})\right|$ $\displaystyle\leq
C\int_{t}^{\infty}s\left|\frac{\partial^{2}}{\partial
s^{2}}W_{s}f(\mathbf{n})\right|\,\frac{ds}{s}$ $\displaystyle\leq
Ct^{-1/2}\left(\int_{t}^{\infty}s^{2}\left|\frac{\partial^{2}}{\partial
s^{2}}W_{s}f(\mathbf{n})\right|^{2}\,ds\right)^{1/2}.$

Finally,

$\displaystyle(g_{1}(T_{M}f)(\mathbf{n}))^{2}$
$\displaystyle=\int_{0}^{\infty}t\left|\frac{\partial}{\partial
t}W_{t}(T_{M}f)(\mathbf{n})\right|^{2}\,dt\leq
C\int_{0}^{\infty}\int_{t}^{\infty}s^{2}\left|\frac{\partial^{2}}{\partial
s^{2}}W_{s}f(\mathbf{n})\right|^{2}\,ds\,dt$
$\displaystyle=C\int_{0}^{\infty}s^{3}\left|\frac{\partial^{2}}{\partial
s^{2}}W_{s}f(\mathbf{n})\right|^{2}\,ds=C(g_{2}(f)(\mathbf{n}))^{2}$

and the proof of (88) is completed. ∎

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