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# Sections and Chapters

Medine<EMAIL_ADDRESS>and Metin
<EMAIL_ADDRESS>1Department of Physics, Bogazici
University, Bebek, Istanbul, Turkiye

# Analytic Solutions of Brans-Dicke Cosmology: Early Inflation and Late Time
Accelerated Expansion

Medine<EMAIL_ADDRESS>and Metin
<EMAIL_ADDRESS>1Department of Physics, Bogazici
University, Bebek, Istanbul, Turkiye

###### Abstract

We investigate the most general exact solutions of Brans-Dicke cosmology by
choosing the scale factor $"a"$ as the new independent variable. It is shown
that a set of three field equations can be reduced to a constraint equation
and a first order linear differential equation. Thus this new set of equations
is solvable when one supplies one of the following pairs of functions:
($\phi(a)$, $\rho(a)$), ($\phi(a)$, $V(a)$) or ($\phi(a)$, $H(a)$). A universe
with a single component energy-matter density is studied. It is seen that when
there is no constant energy density, the Hubble function still contains a
constant term which causes exponential expansion. This constant is expressed
in terms of the initial values of the universe. An early universe and the
present universe with dark energy are studied. In addition late time
accelerated expansion is also explained with cosmic domain walls. If we take
Brans-Dicke parameter $w>4\times 10^{4}$ formulas of the Hubble function
reduce to solutions of $\Lambda CDM$ cosmology. Therefore comparison of our
results with recent observations of type Ia supernovae indicates that eighty-
nine percent of present universe may consist of domain walls while rest is
matter.

##  1 Introduction

Since Einstein’s General Theory of Relativity, many new modified theories have
been introduced to extend it. Dirac proposed that Newton’s constant may be
varying with time [1], after influential ideas were developed by Weyl [2, 3]
and Eddington [4]. Spacetime dependent gravitational constant can be described
according to Jordan’s study in [5]. Thus his work has been accepted as the
origin of the scalar-tensor theory.
A new approach to general relativity which is based on Mach’s principle was
developed by Brans and Dicke in 1961 [6]. They took effective gravitational
constant to be proportional to the reciprocal of a scalar field. This is
called as nonminimal coupling of scalar field. More complicated nonminimal
coupling terms were porposed in [7, 8, 9, 10, 11, 12, 13, 14, 15].
Two different revolutionary steps have been taken in cosmology since 1980.
First one was the construction of inflationary cosmology by A.H. Guth and
A.D.Linde. It had been proposed as a solution for problems of the standard
model of cosmology which are the flatness problem, the horizon problem and the
monopole problem [16, 17]. Second one was the observational evidence of
accelerated expansion of the present universe [18, 19, 20]. Standard cosmology
explains this behaviour by contribution of dark energy ($\sim 68\%$), dark
matter ($\sim 27\%$) and baryonic matter ($\sim 5\%$) to the total density
parameter. Nature of dark energy and dark matter are still unknown.
In the last decades cosmologists established several modified theories of
gravity to investigate these two concepts [21, 22, 23, 24, 25, 26, 27, 28, 29,
30]. All of these theories have different field equations which govern the
dynamics of the universe. Thus all these theories have different physical
conclusions.
In recent years searching for exact solutions of the field equations of scalar
tensor theories has been taken more attention of scientists [31, 32, 33, 34,
35, 36]. In addition in 1990 it has been shown that there are potentials
$V(\phi)$ leading to desired behaviour for the scale factor without the use of
’slow-roll approximation’ usually assumed in inflationary models [37].
In this text we pick up Brans-Dicke cosmology and we focus on it. Up to now
some exact solutions for specific cases has been presented. While for a
radiation filled universe exact solution was given in [38], parametric
solution for the case $p=\epsilon\rho$ was given in [39]. For Bianchi type
universes with arbitrary barotropic perfect fluid exact solution were found
[40]. Parametric exact solutions for potential free universe filled with
ordinary matter were given in [41]. On the other hand Noether’s theorem was
used to determine conservation laws in Brans-Dicke cosmology which includes
two scalar fields and exact solutions were found [42]. An exact solutions of
modified Brans-Dicke theory was studied in [43].
The major purpose of this article is to solve field equations of Brans-Dicke
cosmology analytically. First, we present field equations in section 2. Then
we introduce our mathematical techniques in section 3 and we rewrite the field
equations with the new independent variable $"a"$. This set of equations can
be reduced to a constraint equation and a first order differential equation.
Exact solutions are found in section 4 for given $\phi(a)$ and $\rho(a)$, in
section 5 for given $\phi(a)$ and $V(a)$, in section 6 for given $\phi(a)$ and
$H(a)$. A universe consisting of single component energy-matter density is
investigated in section 7. The early epoch of the universe is studied in
section 8. We present a solution for a universe which contains dark energy and
matter in section 9.1 and we present a solution for a universe which consists
of cosmic domain walls and matter in section 9.2. Comparison of both models
with observation is illustrated by a plot. Discussion is given in section 10
and we summarize our results in conclusion. In this study we choose $w>4\times
10^{4}$ to be compatible with results of Einstein telescope [44] and time
delay experiments [45].

##  2 Field Equations

In this study we will use FLWR metric given by,

$\displaystyle
ds^{2}=dt^{2}-a^{2}(t)[\frac{dr^{2}}{1-kr^{2}}+r^{2}(d\theta^{2}+\sin^{2}\theta
d\varphi^{2})].$ (1)

where $k$ is the curvature parameter, $a(t)=\dfrac{R(t)}{R(t_{0})}$ is the
normalized scale factor of the universe. Here $r$ has dimension of $lenght$,
$a(t)$ is dimensionless and $k$ has diemsion of $lenght^{-2}$. In BDJT the
Lagrange density can be written as [46, 47, 48, 49, 6]

$\displaystyle\mathcal{L}$ $\displaystyle=\left[-\Phi
R+\omega\dfrac{1}{\Phi}g^{\mu\nu}\partial_{\mu}{\Phi}\partial_{\nu}{\Phi}-V(\Phi)+L_{M}\right]\sqrt{-g},$
(2)
$\displaystyle=\left[-\frac{1}{8\omega}\phi^{2}R+\frac{1}{2}g^{\mu\nu}\partial_{\mu}{\phi}\partial_{\nu}{\phi}-V(\phi)+L_{M}\right]\sqrt{-g},$
(3)

where we will call $\phi$ as the Jordan scalar field and $\Phi$ as the Brans-
Dicke scalar field, the two being related by

$\displaystyle\Phi=\frac{1}{8\omega}\phi^{2},$ (4)

where $\omega$ is the dimensionless Brans-Dicke parameter, $R$ is the Ricci
scalar and $L_{M}$ represents the contribution due to matter fields. We use
the metric signature $(+,-,-,-)$ and units with $\hbar=1$, $c=1$. We prefer to
use the Jordan scalar for the scalar field since in flat spacetime the
Lagrangian then becomes the standard Lagrangian for a scalar field

$\displaystyle\mathcal{L}=\frac{1}{2}\partial_{\mu}{\phi}\partial^{\mu}{\phi}-V(\phi).$
(5)

The homogeneous and isotropic cosmological field equations obtained from this
Lagrangian density have already been calculated for a potential
$V(\phi)=\dfrac{1}{2}m^{2}\phi^{2}$ by Arik, Calik and Sheftel in [50, 51,
52]. Thus their general form for a potential $V(\phi)$ are given by,

$\displaystyle\frac{3}{4\omega}\phi^{2}\left(\frac{\dot{a}^{2}}{a^{2}}+\frac{k}{a^{2}}\right)-\frac{1}{2}\dot{\phi}^{2}-V(\phi)+\frac{3}{2\omega}\frac{\dot{a}}{a}\dot{\phi}\phi=\rho_{m},$
(6)
$\displaystyle\frac{-1}{4\omega}\phi^{2}\left(2\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}+\frac{k}{a^{2}}\right)-\frac{1}{\omega}\frac{\dot{a}}{a}\dot{\phi}\phi-\frac{1}{2\omega}\ddot{\phi}\phi-\left(\frac{1}{2}+\frac{1}{2\omega}\right)\dot{\phi}^{2}+V(\phi)=p_{m},$
(7)
$\displaystyle\ddot{\phi}+3\frac{\dot{a}}{a}\dot{\phi}+\derivative{V(\phi)}{\phi}-\frac{3}{2\omega}\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}+\frac{k}{a^{2}}\right)\phi=0.$
(8)

##  3 Mathematical Techniques

It is apparent that the independent variable time $t$ is not seen explicitly
in our set of differential equations given by (6-8). For this reason one can
choose a new variable in terms of old dependent variables. Hence the order of
differential equation can be reduced by one [53]. Our choice is
$\dot{a}=aH(a)$. Then $a$ becomes new independent variable and field equations
transform into the following form

$\displaystyle\frac{3}{4\omega}\phi^{2}\left(H^{2}+\dfrac{k}{a^{2}}\right)-\dfrac{1}{2}\left(\phi^{{}^{\prime}}Ha\right)^{2}-V(a)+\frac{3}{2\omega}\phi\phi^{{}^{\prime}}H^{2}a=\rho_{m}$
(9)
$\displaystyle-\frac{1}{4\omega}\phi^{2}\left(3H^{2}+2H^{{}^{\prime}}Ha+\dfrac{k}{a^{2}}\right)-\dfrac{3}{2\omega}\phi^{{}^{\prime}}\phi
H^{2}a-\dfrac{1}{2\omega}\left(\phi^{{}^{\prime\prime}}H^{2}a^{2}+\phi^{{}^{\prime}}H^{{}^{\prime}}Ha^{2}\right)\phi$
$\displaystyle-\left(\dfrac{1}{2}+\dfrac{1}{2\omega}\right)\left(\phi^{{}^{\prime}}Ha\right)^{2}+V(a)=p_{m},$
(10)
$\displaystyle\phi^{{}^{\prime\prime}}H^{2}a^{2}+\phi^{{}^{\prime}}H^{{}^{\prime}}Ha^{2}+4\phi^{{}^{\prime}}H^{2}a+\dfrac{V^{{}^{\prime}}}{\phi^{{}^{\prime}}}-\dfrac{3}{2\omega}\left(2H^{2}+H^{{}^{\prime}}Ha+\frac{k}{a^{2}}\right)\phi=0,$
(11)

where the prime denotes $\dfrac{d}{da}$. We will name (9) as the energy
density equation, (10) as the pressure equation and (11) as the $\phi$
equation.
These three equations are connected by the perfect fluid continuity equation,

$\displaystyle\dot{\rho}(a)+3\dfrac{\dot{a}}{a}(\rho(a)+p(a))=0,$ (12)
$\displaystyle[\rho^{{}^{\prime}}(a)a+3(\rho(a)+p(a))]H(a)=0.$ (13)

In other words when we plug energy density and pressure which are given by
(9-10) in (13) we end up with $-a\phi^{{}^{\prime}}(a)H(a)\times eq.(11)$.
This shows that we have two independent equations and four unknown functions:
$H(a)$, $V(a)$, $\phi(a)$ and $\rho(a)$ (one can choose $p(a)$ instead of
$\rho(a)$).
In our calculations we will use the energy density equation and the $\phi$
equation. When we plug solutions in to the pressure equation, $\rho(a)$ and
the resulting $p(a)$ will satisfy the perfect fluid equation of state.
It is seen that energy density equation is a first order variable coefficient
and nonlinear differential equation for $\phi(a)$. In addition this is not a
type of Bernoulli equation which is analytically solvable. In addition (11) is
second order variable coefficient nonlinear differential equation for
$\phi(a)$. Analytic solutions for both of them have not been established yet.
To be able to solve them, $\phi(a)$ should be given. Then we will be left with
$\rho(a)$, $V(a)$ and $H(a)$. Since (9) does not contain any derivative of
last three functions it is just a constraint. On the other hand first
derivatives of $H(a)$ and $V(a)$ are seen in (11). Thus our system consist of
a constraint equation and a first order differential equation. One of the
following functions; $H(a)$, $V(a)$ and $\phi(a)$ is still free. One of them
should be accompanied with $\phi(a)$ as a given function. Three combinations
exist; $\rho(a)$ and $\phi(a)$, $V(a)$ and $\phi(a)$, $H(a)$ and $\phi(a)$. We
will solve our system for these three sets separately.

##  4 Solution for given $\rho(a)$ and $\phi(a)$

One can solve the constraint equation (9) and the first order differential
equation (11) by different methods:

  * •

Method 1) First, find $V(a)$ from the constraint equation then substitute it
in the differential equation and solve it for $H(a)$.

  * •

Method 2) First, solve the differential equation for $V(a)$ then substitute it
in the constraint equation and solve it for $H(a)$.

  * •

Method 3) First, find $H(a)$ from the constraint equation then substitute it
in the differential equation and solve it for $V(a)$.

  * •

Mehod 4) First, solve the differential equation for $H(a)$ then substitute it
in the constraint equation and solve it for $V(a)$.

Results are different representations of the same solution. However they may
have different singular cases which require special attention. To reach
complete set of solutions we perform all four methods. Moreover, most of the
solutions include integrals of different functions and sometimes integration
techniques can be inadequate . For this reason one of the form of solutions
can be superior to others. We would like to remind that the $\rho$ equation is
equivalent to the constraint equation and the first order differential
equation is equivalent to the $\phi$ equation. We will use both names.

###  4.1 Method 1

####  4.1.1 Nonsingular case

Firstly we find $V(a)$ from the constraint equation

$\displaystyle
V(a)=\dfrac{3}{4w}\phi^{2}(a)(H^{2}(a)+\dfrac{k}{a^{2}})-\dfrac{1}{2}(\phi^{{}^{\prime}}(a)H(a)a)^{2}+\dfrac{3}{2w}\phi\phi^{{}^{\prime}}H^{2}a+\rho(a).$
(14)

Then the $\phi$ equation becomes

$\displaystyle
H^{{}^{\prime}}(a)+\dfrac{a[(1+2w)\phi^{{}^{\prime}2}(a)+\phi(a)\phi^{{}^{\prime\prime}}(a)]}{\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))}H(a)=\dfrac{2w\rho^{{}^{\prime}}a^{3}+3k\phi^{2}(a)}{3a^{3}\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))H(a)},$
(15)

which is the Bernoulli equation. Thus we apply change of variable
$\gamma(a)=H^{2}(a)$ and we obtain

$\displaystyle\gamma^{{}^{\prime}}(a)+\dfrac{2a[(1+2w)\phi^{{}^{\prime}2}(a)+\phi(a)\phi^{{}^{\prime\prime}}(a)]}{\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))}\gamma(a)=\dfrac{2(2w\rho^{{}^{\prime}}a^{3}+3k\phi^{2}(a))}{3a^{3}\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))}.$
(16)

Its solution is found as

$\displaystyle\gamma(a)=\dfrac{exp\Big{[}-\int_{a_{in}}^{a}P(a^{{}^{\prime}})da^{{}^{\prime}}\Big{]}}{\Big{[}\phi(\phi+a\phi^{{}^{\prime}})\Big{]}^{2}}\Big{\\{}\int_{a_{in}}^{a}exp\Big{[}\int_{a_{in}}^{a^{{}^{\prime}}}P(a^{{}^{\prime\prime}})da^{{}^{\prime\prime}}\Big{]}Q(a^{{}^{\prime}})da^{{}^{\prime}}+\tilde{\gamma}(a_{in})\Big{\\}},$
(17) $\displaystyle
P(a)=\dfrac{2[2wa\phi^{{}^{\prime}2}(a)-3\phi(a)\phi^{{}^{\prime}}(a)]}{\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))},\hskip
20.0ptQ(a)=\dfrac{2(2w\rho^{{}^{\prime}}a^{3}+3k\phi^{2}(a))\Big{[}\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))\Big{]}}{3a^{3}},$
(18)
$\displaystyle\tilde{\gamma}(a_{in})=\gamma(a_{in})\Big{[}\phi(a_{in})\Big{(}\phi(a_{in})+a_{in}\phi^{{}^{\prime}}(a_{in})\Big{)}\Big{]}^{2}.$
(19)

Hence

$\displaystyle H(a)=\sqrt{\gamma(a)},$ (20)

where $\gamma(a)$ is given by (17). One should choose $\phi(a)$ such that
$\gamma(a)$ is positive. Thus $H(a)$ will be a real function. However if one
goes further by assigning specific form to the Hubble function and tries to
solve (17) for $\phi(a)$, she will end up with a second order, variable
coefficient, nonlinear differential equation for whose solution there is no
known method. For this reason we will continue with trial $\phi(a)$ functions
in our examples.

####  4.1.2 Singular case

The denominator of the function $P(a)$ which is used to formulate the Hubble
function becomes singular when $\phi(a)=F/a$. Hence this case needs special
attention. $V(a)$ is obtained form the $\rho$ equation,

$\displaystyle
V(a)=-\rho(a)+\dfrac{F^{2}\Big{[}3k-(3+2w)a^{2}H^{2}(a)\Big{]}}{4wa^{4}}.$
(21)

Then the $\phi$ equation turns into the following form

$\displaystyle\dfrac{3F^{2}\Big{[}k-(3+2w)a^{2}H^{2}(a)\Big{]}+2wa^{5}\rho^{{}^{\prime}}(a)}{2Fwa^{3}}=0.$
(22)

The Hubble function is found as

$\displaystyle
H(a)=\sqrt{\dfrac{3F^{2}k+2wa^{5}\rho^{{}^{\prime}}(a)}{3(3+2w)F^{2}a^{2}}}.$
(23)

Hence the final form of potential is given by

$\displaystyle
V(a)=-\rho(a)-\dfrac{a}{6}\rho^{{}^{\prime}}(a)+\dfrac{kF^{2}}{2wa^{4}}.$ (24)

If one inserts the energy density of an ordinary matters which is proportional
to $1/a^{n}$ results are

$\displaystyle
H=\sqrt{\dfrac{3F^{2}k-2nw\rho_{n}a^{4-n}}{3(3+2w)F^{2}a^{2}}},$ (25)
$\displaystyle V(a)=\dfrac{F^{2}k}{2wa^{4}}+\dfrac{(n-6)\rho_{n}}{a^{n}},$
(26) $\displaystyle V(\phi)=\dfrac{F^{2}k}{2w}\phi^{4}+(n-6)\rho_{n}\phi^{n}.$
(27)

Except the radiation dominated universe and dark energy dominated universe all
these matters results in an imaginary form of the Hubble function and negative
potential values as easily recognized from (25-26). In addition radiation
dominated universe corresponds to $V=\lambda\phi^{4}/4$ which has been studied
in [54].

###  4.2 Method 2

####  4.2.1 Non-singular Case

In the second method we start with the $\phi$ equation. By using it $V(a)$ is
found as

$\displaystyle
V(a)=-\int_{a_{in}}^{a}\dfrac{\phi^{{}^{\prime}}\Big{\\{}-3\phi\Big{[}k+a^{{}^{\prime}2}H(2H+a^{{}^{\prime}}H^{{}^{\prime}})\Big{]}+2a^{{}^{\prime}3}wH\Big{[}(4H+a^{{}^{\prime}}H^{{}^{\prime}})\phi^{{}^{\prime}}+a^{{}^{\prime}}H\phi^{{}^{\prime\prime}}\Big{]}\Big{\\}}}{2a^{{}^{\prime}2}w}da^{{}^{\prime}}+V(a_{in}).$
(28)

Then we insert $V(a)$ in $\rho$ equation and we take the derivative of both
sides with respect to $a$. We obtain the following equation

$\displaystyle
H^{{}^{\prime}}(a)+\dfrac{a[(1+2w)\phi^{{}^{\prime}2}(a)+\phi(a)\phi^{{}^{\prime\prime}}(a)]}{\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))}H(a)=\dfrac{2w\rho^{{}^{\prime}}(a)a^{3}+3k\phi^{2}(a)}{3a^{3}\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))H(a)}.$
(29)

This equation is equivalent to (15). Its solution has been already presented
in (17-20). Although forms of potentials which are given by (24) and (28) are
different, one can show that they are equal to each other up to a constant.
We have just shown that results of method 1 and method 2 are the same, thus
there is no need to repeat calculations for singular case. In this method we
end up with two integration constants. One of them should be chosen such that
when final forms of formulas are inserted in the constraint equation it must
be satisfied.

###  4.3 Method 3

####  4.3.1 Non Singular Case

First we find $H(a)$ from the $\rho$ equation

$\displaystyle
H(a)=\sqrt{\dfrac{-3k\phi^{2}(a)+4wa^{2}\Big{[}\rho(a)+V(a)\Big{]}}{a^{2}\Big{[}3\phi^{2}(a)+6a\phi(a)\phi^{{}^{\prime}}(a)-2wa^{2}\phi^{{}^{\prime}2}\Big{]}}}.$
(30)

Then the $\phi$ equation turns into the following form

$\displaystyle V^{{}^{\prime}}(a)+P(a)V(a)=Q(a),$ (31)

where

$\displaystyle
P(a)=\Big{\\{}2\phi^{{}^{\prime}}(a)\Big{[}-6\phi^{3}(a)+(3+14w)a^{2}\phi(a)\phi^{{}^{\prime}2}(a)-2w(1+2w)a^{3}\phi^{{}^{\prime}3}(a)+a\phi^{2}(a)((-6+8w)\phi^{{}^{\prime}}(a)$
$\displaystyle+(3+2w)a\phi^{{}^{\prime\prime}}(a))\Big{]}\Big{\\}}/\Big{\\{}\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))\Big{[}3\phi^{2}(a)+6a\phi(a)\phi^{{}^{\prime}}(a)-2wa^{2}\phi^{{}^{\prime}2}(a)\Big{]}\Big{\\}},$
(32)

and

$\displaystyle
Q(a)=-\Big{\\{}\phi^{{}^{\prime}}(a)\Big{[}-36w^{2}a^{4}\phi(a)\rho^{{}^{\prime}}(a)\phi^{{}^{\prime}2}(a)+8w^{3}a^{5}\rho^{{}^{\prime}}(a)\phi^{{}^{\prime}3}(a)-12wa^{2}\phi^{2}(a)\phi^{{}^{\prime}}(a)((-3+w)a\rho^{{}^{\prime}}(a)$
$\displaystyle+k(3+2w)\phi^{{}^{\prime}2}(a))+9a\phi^{3}(a)(2wa\rho^{{}^{\prime}}(a)+3k(3+2w)\phi^{{}^{\prime}2}(a))+9k(3+2w)\phi^{4}(a)(3\phi^{{}^{\prime}}(a)+a\phi^{{}^{\prime\prime}})$
$\displaystyle+12wa\rho(a)(6\phi^{3}(a)-(3+14w)a^{2}\phi(a)\phi^{{}^{\prime}2}(a)+2w(1+2w)a^{3}\phi^{{}^{\prime}3}(a)-a\phi^{2}(a)((-6+8w)\phi^{{}^{\prime}}(a)$
$\displaystyle+(3+2w)a\phi^{{}^{\prime\prime}}(a)))\Big{]}\Big{\\}}/\Big{\\{}6wa\phi(a)(\phi(a)+a\phi^{{}^{\prime}}(a))\Big{[}-3\phi^{2}(a)-6a\phi(a)\phi^{{}^{\prime}}(a)+2wa^{2}\phi^{{}^{\prime}2}(a)\Big{]}\Big{\\}}.$
(33)

Hence potential is

$\displaystyle
V(a)=exp\Big{(}-\int_{a_{in}}^{a}P(a^{{}^{\prime}})da^{{}^{\prime}}\Big{)}\Big{\\{}\int_{a_{in}}^{a}exp\Big{(}\int_{a_{in}}^{a^{{}^{\prime}}}P(a^{{}^{\prime\prime}})da^{{}^{\prime\prime}}\Big{)}Q(a^{{}^{\prime}})da^{{}^{\prime}}+V(a_{in})\Big{\\}}.$
(34)

####  4.3.2 Singular Cases

The solution for method 3 has three singular cases;

I) $\phi(a)=0$,

II)$\phi(a)+a\phi^{{}^{\prime}}(a)=0$,

III)$-3\phi^{2}(a)-6a\phi(a)\phi^{{}^{\prime}}(a)+2wa^{2}\phi^{{}^{\prime}2}(a)=0$.

For all of them both functions, $P(a)$ and $Q(a)$ which are used to formulate
the potential become undefined.
First case is not acceptable because Brans-Dicke theory collapses for
$\phi(a)=0$. Second case has been already investigated in Method 1. We are
left with the last one. We write $\phi(a)=Fexp(\int\alpha(a)da)$. Then
singularity III becomes

$\displaystyle 2wa^{2}\alpha^{2}-6a\alpha-3=0.$ (35)

Thus

$\displaystyle\alpha(a)=\dfrac{3\pm\sqrt{9+6w}}{2wa},\hskip
20.0pt\phi(a)=Fa^{(3\pm\sqrt{9+6w})/(2w)}.$ (36)

Then the potential is found from the $\rho$ equation

$\displaystyle V(a)=\dfrac{3F^{2}k}{4w}a^{(3-2w\pm\sqrt{9+6w})/w}-\rho(a).$
(37)

After writing $H(a)=\sqrt{\gamma(a)}$, $\phi$ equation becomes

$\displaystyle\gamma^{{}^{\prime}}(a)+P(a)\gamma(a)=Q(a).$ (38)

For $\alpha(a)=\dfrac{3+\sqrt{9+6w}}{2wa}$,

$\displaystyle P(a)=\dfrac{2(3+3w+\sqrt{9+6w})}{wa},$ (39) $\displaystyle
Q(a)=\dfrac{4wa^{-(3+3w+\sqrt{9+6w})/w}}{3F^{2}(3+2w+\sqrt{9+6w})}\Big{[}3F^{2}ka^{(3+\sqrt{9+6w})/w}+2wa^{3}\rho^{{}^{\prime}}(a)\Big{]}.$
(40)

Then

$\displaystyle\gamma(a)=\dfrac{1}{a^{2(3+3w+\sqrt{9+6w})/w}}\Big{\\{}b\int_{a_{in}}^{a}a^{{}^{\prime}(3+3w+\sqrt{9+6w})/w}\Big{[}3F^{2}ka^{{}^{\prime}(3+\sqrt{9+6w})/w}+2wa^{{}^{\prime}3}\rho^{{}^{\prime}}\Big{]}da^{{}^{\prime}}+\tilde{\gamma}(a_{in})\Big{\\}},$
(41) $\displaystyle b=\dfrac{4w}{3F^{2}(3+2w+\sqrt{9+6w})},\hskip
25.0pt\tilde{\gamma}(a_{in})=a_{in}^{2(3+3w+\sqrt{9+6w})/w}\gamma(a_{in})$
(42)

For $\alpha(a)=\dfrac{3-\sqrt{9+6w}}{2wa}$,

$\displaystyle P(a)=\dfrac{2(3+3w-\sqrt{9+6w})}{wa},$ (43) $\displaystyle
Q(a)=\dfrac{4wa^{-3(1+w)/w}}{3F^{2}(3+2w-\sqrt{9+6w})}\Big{[}3F^{2}ka^{3/w}+2wa^{(3w+\sqrt{9+6w})/w}\rho^{{}^{\prime}}(a)\Big{]}.$
(44)

Then

$\displaystyle\gamma(a)=\dfrac{1}{a^{2(3+3w-\sqrt{9+6w})/w}}\Big{\\{}b\int_{a_{in}}^{a}a^{{}^{\prime}(3+3w-2\sqrt{9+6w})/w}\Big{[}3F^{2}ka^{{}^{\prime}3/w}+2wa^{{}^{\prime}(3w+\sqrt{9+6w})/w}\rho^{{}^{\prime}}\Big{]}da^{{}^{\prime}}+\tilde{\gamma}(a_{in})\Big{\\}},$
(45) $\displaystyle b=\dfrac{4w}{3F^{2}(3+2w-\sqrt{9+6w})},\hskip
25.0pt\tilde{\gamma}(a_{in})=a_{in}^{2(3+3w-\sqrt{9+6w})/w}\gamma(a_{in})$
(46)

At the next step we examine nature of the Hubble function. First, we take

$\displaystyle\phi(a)=Fa^{(3+\sqrt{9+6w})/(2w)},\hskip
20.0pt\rho=\dfrac{\rho_{n}}{a^{n}}.$ (47)

Hubble function and the potential are found as

$\displaystyle
H(a)=\dfrac{1}{a^{(3+3w+\sqrt{9+6w})/w}}\Big{\\{}\dfrac{2w^{2}a^{(3-(n-4)w+\sqrt{9+6w})/w}}{3F^{2}(3+2w+\sqrt{9+6w})^{2}[3+(6-n)w+\sqrt{9+6w}]}$
$\displaystyle\Big{[}-4(3+2w+\sqrt{9+6w})nw\rho_{n}a^{2}+3F^{2}k\Big{(}3+(6-n)w+\sqrt{9+6w}\Big{)}a^{(3+nw+\sqrt{9+6w})/w}\Big{]}+\tilde{\gamma}(ain)\Big{\\}}^{1/2},$
(48)
$\displaystyle\tilde{\gamma}(a_{in})=a_{in}^{2(3+3w+\sqrt{9+6w})/w}\gamma(a_{in})-\Big{\\{}\dfrac{2w^{2}a^{(3-(n-4)w+\sqrt{9+6w})/w}_{in}}{3F^{2}(3+2w+\sqrt{9+6w})^{2}[3+(6-n)w+\sqrt{9+6w}]}$
$\displaystyle\Big{[}-4(3+2w+\sqrt{9+6w})nw\rho_{n}a^{2}_{in}+3F^{2}k\Big{(}3+(6-n)w+\sqrt{9+6w}\Big{)}a^{(3+nw+\sqrt{9+6w})/w}_{in}\Big{]}\Big{\\}},$
(49) $\displaystyle
V(a)=-\dfrac{\rho_{n}}{a^{n}}+\dfrac{3F^{2}k}{4wa^{(2w-3-\sqrt{9+6w})/w}}.$
(50)

The term containing energy density in $H(a)$ has a negative coefficient for
$0<n<6$ and $w\gg 1$.
Then we take

$\displaystyle\phi(a)=Fa^{(3-\sqrt{9+6w})/(2w)},\hskip
20.0pt\rho=\dfrac{\rho_{n}}{a^{n}}.$ (51)

Hubble function and the potential are found as

$\displaystyle
H(a)=\dfrac{1}{a^{(3+3w-\sqrt{9+6w})/w}}\Big{\\{}\dfrac{2w^{2}a^{(3-(n-4)w-2\sqrt{9+6w})/w}}{3F^{2}(-3-2w+\sqrt{9+6w})^{2}[-3+(n-6)w+\sqrt{9+6w}]}$
$\displaystyle\Big{[}4(3+2w-\sqrt{9+6w})nw\rho_{n}a^{(2w+\sqrt{9+6w})/w}+3F^{2}k\Big{(}-3+(n-6)w+\sqrt{9+6w}\Big{)}a^{3/w+n}\Big{]}+\tilde{\gamma}(ain)\Big{\\}}^{1/2},$
(52)
$\displaystyle\tilde{\gamma}(a_{in})=a_{in}^{2(3+3w-\sqrt{9+6w})/w}\gamma(a_{in})-\Big{\\{}\dfrac{2w^{2}a^{(3-(n-4)w-2\sqrt{9+6w})/w}_{in}}{3F^{2}(-3-2w+\sqrt{9+6w})^{2}[-3+(n-6)w+\sqrt{9+6w}]}$
$\displaystyle\Big{[}4(3+2w-\sqrt{9+6w})nw\rho_{n}a^{(2w+\sqrt{9+6w})/w}_{in}+3F^{2}k\Big{(}-3+(n-6)w+\sqrt{9+6w}\Big{)}a^{3/w+n}_{in}\Big{]}\Big{\\}},$
(53) $\displaystyle
V(a)=-\dfrac{\rho_{n}}{a^{n}}+\dfrac{3F^{2}k}{4wa^{(2w-3+\sqrt{9+6w})/w}}.$
(54)

The term containing energy density in $H(a)$ has a negative coefficient for
$0<n<6$ and $w\gg 1$.

###  4.4 Method 4

Substitution of $H(a)=\sqrt{\gamma(a)}$ into $\phi$ equation results in the
following form,

$\displaystyle\gamma^{{}^{\prime}}(a)+\tilde{P}(a)\gamma=\tilde{Q}(a),$ (55)
$\displaystyle\tilde{P}(a)=\dfrac{4\Big{[}-3\phi(a)+wa(4\phi^{{}^{\prime}}(a)+a\phi^{{}^{\prime\prime}}(a))\Big{]}}{a(-3\phi(a)+2wa\phi^{{}^{\prime}}(a))},\hskip
20.0pt\tilde{Q}(a)=\dfrac{-4wa^{2}V^{{}^{\prime}}(a)+6k\phi(a)\phi^{{}^{\prime}}(a)}{a^{3}\phi^{{}^{\prime}}(a)(-3\phi(a)+2wa\phi^{{}^{\prime}}(a))}.$
(56)

Hence

$\displaystyle\gamma(a)=exp\Big{(}-\int_{a_{in}}^{a}\tilde{P}(a^{{}^{\prime}})da^{{}^{\prime}}\Big{)}\Big{\\{}\int_{a_{in}^{a}}exp\Big{(}\int_{a_{in}}^{a^{{}^{\prime}}}\tilde{P}(a^{{}^{\prime\prime}})da^{{}^{\prime\prime}}\Big{)}\tilde{Q}(a^{{}^{\prime}})da^{{}^{\prime}}+\gamma(a_{in})\Big{\\}},$
(57) $\displaystyle H(a)=\sqrt{\gamma(a)}.$ (58)

After plugging $H(a)$ into $\rho$ equation we multiply both sides by

$\displaystyle\dfrac{exp\Big{(}\int_{a_{in}}^{a}\tilde{P}(a^{{}^{\prime}})da^{{}^{\prime}}\Big{)}}{3\phi^{2}(a)+6a\phi(a)\phi^{{}^{\prime}}(a)-2wa^{2}\phi^{{}^{\prime}2}}.$
(59)

Then we take derivative of both sides with respect $a$. The resulting equation
is equivalent to

$\displaystyle V^{{}^{\prime}}(a)+P(a)V(a)=Q(a)$ (60)

where coefficients $P(a)$ and $Q(a)$ are given by (32) and (33) respectively.
Thus solution is given by (34). In this method we end up with two integration
constants. One of them should be chosen such that when final forms of formulas
are inserted in the constraint equation it must be satisfied.

####  4.4.1 Singularities

This method has five singularities in its formulation. Three of them are the
same singularities given in Method 3. Other two of them are seen in functions
$\tilde{P}(a)$ and $\tilde{Q}(a)$

I)$\phi^{{}^{\prime}}(a)=0$,

II)$-3\phi(a)+2wa\phi^{{}^{\prime}}(a)=0$.

First one makes Brans-Dicke cosmology equivalent to Einstein cosmology which
have been studied in [55]. The second one implies

$\displaystyle\phi(a)=Fa^{3/(2w)}.$ (61)

Then $\phi$ equation becomes

$\displaystyle\dfrac{a^{-3/w-2}}{12Fw^{2}}\Big{\\{}-9F^{2}a^{3/w}\Big{[}2kw-(3+2w)a^{2}H^{2}(a)\Big{]}+8w^{3}a^{3}V^{{}^{\prime}}(a)\Big{\\}}=0.$
(62)

The Hubble function is found easily

$\displaystyle
H(a)=\dfrac{\sqrt{18F^{2}kwa^{3/w}-8w^{3}a^{3}V^{{}^{\prime}}(a)}}{3\sqrt{(3+2w)F^{2}a^{3/w+2}}}.$
(63)

Hence the $\rho$ equation turns into the following form

$\displaystyle V^{{}^{\prime}}(a)+P(a)V(a)=Q(a),$ (64) $\displaystyle
P(a)=\dfrac{3}{wa},\hskip
20.0ptQ(a)=\dfrac{3(3F^{2}ka^{3/w}-2wa^{2}\rho(a))}{2w^{2}a^{3}}.$ (65)

Its solution is

$\displaystyle
V(a)=\dfrac{1}{a^{3/w}}\Big{\\{}\int_{a_{in}}^{a}a^{{}^{\prime}3/w}Q(a^{{}^{\prime}})da^{{}^{\prime}}+\tilde{V}(a_{in})\Big{\\}},$
(66) $\displaystyle\tilde{V}(a_{in})=a_{in}^{3/w}V(a_{in}).$

To investigate the nature of the Hubble function and the potential for usual
forms of energy-matters we plug $\rho=\rho_{n}/a^{n}$. Thus we have

$\displaystyle
H(a)=w\Big{\\{}\frac{2}{3(3+2w)}\Big{[}\dfrac{4a_{in}^{3/w}V(a_{in})}{F^{2}a^{6/w}}-\dfrac{3k}{(w-3)a^{2}}+\dfrac{4nw\rho_{n}}{(nw-3)F^{2}a^{3/w+n}}\Big{]}\Big{\\}}^{1/2},$
(67) $\displaystyle
V(a)=\dfrac{\tilde{V}(a_{in})}{a^{3/w}}+\dfrac{3\rho_{n}}{(nw-3)a^{n}}+\dfrac{9F^{2}k}{4w(3-w)a^{-3/w+2}},$
(68)
$\displaystyle\tilde{V}(a_{in})=a_{in}^{3/w}V(a_{in})-\dfrac{3\rho_{n}}{(nw-3)a^{n}_{in}}-\dfrac{9F^{2}k}{4w(3-w)a^{-3/w+2}_{in}}.$
(69)

Energy density related term in $H(a)$ has a positive coefficient therefore the
Hubble function is always real.

##  5 Solution for given $V(a)$ and $\phi(a)$

Special case of potential in the scalar-tensor theory is important. For this
reason formulation of this case is significant. We establish the solution for
our set of the constraint equation and the first order differential equation
by two following different methods:

  * •

Method 1) First, solve the $\phi$ equation for $H(a)$ then plug it into the
constraint equation to obtain $\rho(a)$.

  * •

Method 2) First, find $H(a)$ from the constraint equation then plug it in to
the $\phi$ equation and solve it for $\rho(a)$.

###  5.1 Method1

####  5.1.1 Non-singular Case

We start our calculations by changing the dependent variable as
$H(a)=\sqrt{\gamma(a)}$ in $\phi$ equation. Hence it turns into the following
form

$\displaystyle\gamma^{{}^{\prime}}(a)+P(a)\gamma(a)=Q(a),$ (70) $\displaystyle
P(a)=\dfrac{4\Big{[}-3\phi(a)+wa(4\phi^{{}^{\prime}}(a)+a\phi^{{}^{\prime\prime}}(a))\Big{]}}{a(-3\phi(a)+2wa\phi^{{}^{\prime}}(a))},$
(71) $\displaystyle
Q(a)=\dfrac{-4wa^{2}V^{{}^{\prime}}(a)+6k\phi(a)\phi^{{}^{\prime}}(a)}{a^{3}\phi^{{}^{\prime}}(a)(-3\phi(a)+2wa\phi^{{}^{\prime}}(a))}.$
(72)

Its solution is given by

$\displaystyle\gamma(a)=exp\Big{(}-\int_{a_{in}}^{a}P(a^{{}^{\prime}})da^{{}^{\prime}}\Big{)}\Big{\\{}\int_{a_{in}}^{a}exp\Big{(}\int_{a_{in}}^{a^{{}^{\prime}}}P(a^{{}^{\prime\prime}})da^{{}^{\prime\prime}}\Big{)}Q(a^{{}^{\prime}})da^{{}^{\prime}}+\gamma(a_{in})\Big{\\}}.$
(73)

Energy density is obtained by inserting $H(a)=\sqrt{\gamma(a)}$ and $V(a)$ in
the $\rho$ equation.

####  5.1.2 Singular Case

When $-3\phi(a)+2wa\phi^{{}^{\prime}}(a)=0$, previous solution becomes
singular. This is the singularity of section 4.4. Thus $\phi(a)=Fa^{3/(2w)}$
and $\phi$ equation is equal to (62) so $H(a)$ is found in (63). Then energy
density is found as

$\displaystyle\rho(a)=\dfrac{3F^{2}ka^{3/w-2}}{2w}-V(a)-\dfrac{waV^{{}^{\prime}}(a)}{3}.$
(74)

When we inspect the results for $V=\dfrac{V_{n}}{a^{n}}$ we obtain

$\displaystyle
H(a)=\dfrac{1}{3\sqrt{3+2w}}\sqrt{\dfrac{18kw}{a^{2}}+\dfrac{8nw^{3}V_{n}}{F^{2}a^{3/w+n}}},$
(75)
$\displaystyle\rho(a)=\dfrac{3F^{2}ka^{3/w-2}}{2w}+\dfrac{(nw-3)V_{n}}{3a^{n}}.$
(76)

###  5.2 Method 2

####  5.2.1 Non-singular Case

First we find $H(a)$ from the $\rho$ equation

$\displaystyle
H(a)=\sqrt{\dfrac{-3k\phi^{2}(a)+4wa^{2}\Big{[}\rho(a)+V(a)\Big{]}}{a^{2}\Big{[}3\phi^{2}(a)+6a\phi(a)\phi^{{}^{\prime}}(a)-2wa^{2}\phi^{{}^{\prime}2}\Big{]}}}.$
(77)

Then $\phi$ equation turns into the following form

$\displaystyle\rho^{{}^{\prime}}(a)+P(a)\rho(a)=Q(a),$ (78)

where

$\displaystyle
P(a)=\Big{\\{}6\Big{[}6\phi^{3}(a)-(3+14w)a^{2}\phi(a)\phi^{{}^{\prime}2}(a)+2(1+2w)wa^{3}\phi^{{}^{\prime}3}(a)-a\phi^{2}(a)((-6+8w)\phi^{{}^{\prime}}(a)$
$\displaystyle+(3+2w)a\phi^{{}^{\prime\prime}}(a))\Big{]}\Big{\\}}/\Big{\\{}a(-3\phi(a)+2wa\phi^{{}^{\prime}}(a))\Big{[}-3\phi^{2}(a)-6a\phi(a)\phi^{{}^{\prime}}(a)+2wa^{2}\phi^{{}^{\prime}2}(a)\Big{]}\Big{\\}},$
(79) $\displaystyle
Q(a)=-\Big{\\{}3\Big{[}4wa^{3}\phi(a)(-(3+14w)V(a)+waV^{{}^{\prime}}(a))\phi^{{}^{\prime}3}(a)+8w^{2}(1+2w)a^{4}V(a)\phi^{{}^{\prime}4}(a)$
$\displaystyle+3a\phi^{3}(a)\phi^{{}^{\prime}}(a)(8wV(a)-6waV^{{}^{\prime}}(a)+3k(3+2w)\phi^{{}^{\prime}2}(a))+4wa^{2}\phi^{2}(a)\phi^{{}^{\prime}}(a)$
$\displaystyle\Big{(}\phi^{{}^{\prime}}(a)((6-8w)V(a)+(-3+w)aV^{{}^{\prime}}(a)-k(3+2w)\phi^{{}^{\prime}2}(a))-(3+2w)aV(a)\phi^{{}^{\prime\prime}}(a)\Big{)}$
$\displaystyle+3\phi^{4}(a)\Big{(}-2waV^{{}^{\prime}}(a)+k(3+2w)\phi^{{}^{\prime}}(a)(3\phi^{{}^{\prime}}(a)+a\phi^{{}^{\prime\prime}}(a))\Big{)}\Big{]}\Big{\\}}/$
$\displaystyle\Big{\\{}2wa^{2}\phi^{{}^{\prime}}(a)(-3\phi(a)+2wa\phi^{{}^{\prime}}(a))\Big{[}-3\phi^{2}(a)-6a\phi(a)\phi^{{}^{\prime}}(a)+2wa^{2}\phi^{{}^{\prime}2}(a)\Big{]}\Big{\\}}.$
(80)

Hence energy density is found as

$\displaystyle\rho(a)=exp\Big{(}-\int_{a_{in}}^{a}P(a^{{}^{\prime}})da^{{}^{\prime}}\Big{)}\Big{\\{}\int_{a_{in}}^{a}exp\Big{(}\int_{a_{in}}^{a^{{}^{\prime}}}P(a^{{}^{\prime\prime}})da^{{}^{\prime\prime}}\Big{)}Q(a^{{}^{\prime}})da^{{}^{\prime}}+\rho(a_{in})\Big{\\}}.$
(81)

####  5.2.2 Singular Cases

This method has two singular cases for non-constant scalar field.
I)$-3\phi(a)+2wa\phi^{{}^{\prime}}(a)=0$,

II)$-3\phi^{2}(a)-6a\phi(a)\phi^{{}^{\prime}}(a)+2wa^{2}\phi^{{}^{\prime}2}(a)=0$.
First case implies $\phi(a)=Fa^{3/(2w)}$. Thus the $\rho$ equation becomes

$\displaystyle\dfrac{3F^{2}a^{3/w-2}\Big{[}2kw+(3+2w)a^{2}H^{2}(a)\Big{]}}{8w^{2}}-V(a)=\rho(a).$
(82)

The Hubble function is found as

$\displaystyle
H(a)=\sqrt{\dfrac{w}{3+2w}}\sqrt{-\dfrac{2k}{a^{2}}+\dfrac{8w}{3F^{2}a^{3/w}}(\rho(a)+V(a))}.$
(83)

Then the $\phi$ equation becomes

$\displaystyle-\dfrac{3Fk}{w}a^{3/(2w)-2}+\dfrac{2a^{-3/(2w)}}{3F}\Big{[}3(\rho(a)+V(a))+waV^{{}^{\prime}}(a)\Big{]},$
(84)

and energy density is found as

$\displaystyle\rho(a)=\dfrac{3F^{2}k}{2w}a^{3/w-2}-V(a)-\dfrac{waV^{{}^{\prime}}(a)}{3}.$
(85)

Examination of this singularity with $V(a)=\dfrac{V_{n}}{a^{n}}$ gives us

$\displaystyle
H(a)=\sqrt{\dfrac{w}{3+2w}}\sqrt{\dfrac{2k}{a^{2}}+\dfrac{8nw^{2}V_{n}}{9F^{2}a^{3/w+n}}},$
(86)
$\displaystyle\rho(a)=\dfrac{3F^{2}k}{2w}a^{3/w-2}+\dfrac{(nw-3)V_{n}}{3a^{n}}.$
(87)

Thus we have the energy density in usual form.
Second singularity has been already seen in section 4.3.2. It implies
$\phi(a)=Fa^{(3\pm\sqrt{9+6w})/(2w)}$.
Then the $\rho$ equation has the following form

$\displaystyle\dfrac{3F^{2}k}{4w}a^{(3-2w\pm\sqrt{9+6w})/w}-V(a)=\rho(a).$
(88)

After writing $H(a)=\sqrt{\gamma(a)}$, $\phi$ equation becomes

$\displaystyle\gamma^{{}^{\prime}}(a)+P(a)\gamma(a)=Q(a).$ (89)

For $\phi(a)=a^{(3+\sqrt{9+6w})/(2w)}$,

$\displaystyle P(a)=\dfrac{2(3+3w+\sqrt{9+6w})}{wa},$ (90) $\displaystyle
Q(a)=\dfrac{a^{-(3+3w+\sqrt{9+6w})/w}}{3\sqrt{3+2w}F^{2}}\Big{[}6\sqrt{3}F^{2}ka^{(3+\sqrt{9+6w})/w}+4w(\sqrt{3}-\sqrt{3+2w})a^{3}V^{{}^{\prime}}(a)\Big{]}.$
(91)

Then

$\displaystyle\gamma(a)=\dfrac{1}{a^{2(3+3w+\sqrt{9+6w})/w}}\Big{\\{}b\int_{a_{in}}^{a}a^{{}^{\prime}(3+3w+\sqrt{9+6w})/w}\Big{[}6\sqrt{3}F^{2}ka^{{}^{\prime}(3+\sqrt{9+6w})/w}$
$\displaystyle+4w(\sqrt{3}-\sqrt{3+2w})a^{{}^{\prime}3}V^{{}^{\prime}}(a^{{}^{\prime}})\Big{]}da^{{}^{\prime}}+\tilde{\gamma}(a_{in})\Big{\\}},$
(92) $\displaystyle b=\dfrac{1}{3\sqrt{3+2w}F^{2}},\hskip
25.0pt\tilde{\gamma}(a_{in})=a_{in}^{2(3+3w+\sqrt{9+6w})/w}\gamma(a_{in}).$
(93)

For $\phi(a)=a^{(3-\sqrt{9+6w})/(2w)}$,

$\displaystyle P(a)=\dfrac{2(3+3w-\sqrt{9+6w})}{wa},$ (94) $\displaystyle
Q(a)=\dfrac{a^{-3(1+w)/w}}{3F^{2}(3+2w-\sqrt{9+6w})}\Big{[}6F^{2}k(3-\sqrt{9+6w})a^{3/w}-8w^{2}a^{(3w+\sqrt{9+6w})/w}V^{{}^{\prime}}(a)\Big{]}.$
(95)

Then

$\displaystyle\gamma(a)=\dfrac{1}{a^{2(3+3w-\sqrt{9+6w})/w}}\Big{\\{}b\int_{a_{in}}^{a}a^{{}^{\prime}(3+3w-2\sqrt{9+6w})/w}\Big{[}6F^{2}k(3-\sqrt{9+6w})a^{{}^{\prime}3/w}$
$\displaystyle+8w^{2}a^{{}^{\prime}(3w+\sqrt{9+6w})/w}V^{{}^{\prime}}(a^{{}^{\prime}})\Big{]}da^{{}^{\prime}}+\tilde{\gamma}(a_{in})\Big{\\}},$
(96) $\displaystyle b=\dfrac{1}{3(3+2w-\sqrt{9+6w})F^{2}},\hskip
25.0pt\tilde{\gamma}(a_{in})=a_{in}^{2(3+3w-\sqrt{9+6w})/w}\gamma(a_{in}).$
(97)

At the next step we examine nature of the Hubble function. First, we take

$\displaystyle\phi(a)=Fa^{(3+\sqrt{9+6w})/(2w)},\hskip
20.0ptV(a)=\dfrac{V_{n}}{a^{n}}.$ (98)

The Hubble function and the energy density are found as

$\displaystyle
H(a)=\Big{\\{}\dfrac{\sqrt{3}k(\sqrt{3+2w}-\sqrt{3})}{2(3+2w)a^{2}}+\dfrac{8nw^{3}V_{n}a^{-(3+nw+\sqrt{9+6w})/w}}{3(3+2w+\sqrt{9+6w})\Big{[}3+(6-n)w+\sqrt{9+6w}\Big{]}F^{2}}$
$\displaystyle+\tilde{\gamma}(a_{in})a^{-2(3+3w+\sqrt{9+6w})/w}\Big{\\}}^{1/2},$
(99)
$\displaystyle\tilde{\gamma}(a_{in})=\Big{\\{}\gamma(a_{in})-\dfrac{\sqrt{3}k(\sqrt{3+2w}-\sqrt{3})}{2(3+2w)a^{2}_{in}}$
$\displaystyle-\dfrac{8nw^{3}V_{n}a^{-(3+nw+\sqrt{9+6w})/w}_{in}}{3(3+2w+\sqrt{9+6w})\Big{[}3+(6-n)w+\sqrt{9+6w}\Big{]}F^{2}}\Big{\\}}a^{2(3+3w+\sqrt{9+6w})/w}_{in},$
(100)
$\displaystyle\rho(a)=\dfrac{3F^{2}k}{4wa^{(2w-3-\sqrt{9+6w})/w}}-\dfrac{V_{n}}{a^{n}}.$
(101)

Although the Hubble function is real for $0<n<6$ and $w\gg 1$, the energy
density always has a negative component.
Then we take

$\displaystyle\phi(a)=Fa^{(3-\sqrt{9+6w})/(2w)},\hskip
20.0ptV(a)=\dfrac{V_{n}}{a^{n}}.$ (102)

The Hubble function and the potential are found as

$\displaystyle
H(a)=\Big{\\{}-\dfrac{\sqrt{3}k(\sqrt{3+2w}+\sqrt{3})}{2(3+2w)a^{2}}+\dfrac{8nw^{3}V_{n}a^{-(3+nw-\sqrt{9+6w})/w}}{3(3+2w-\sqrt{9+6w})\Big{[}3+(6-n)w-\sqrt{9+6w}\Big{]}F^{2}}$
$\displaystyle+\tilde{\gamma}(a_{in})a^{-2(3+3w-\sqrt{9+6w})/w}\Big{\\}}^{1/2},$
(103)
$\displaystyle\tilde{\gamma}(a_{in})=\Big{\\{}\gamma(a_{in})+\dfrac{\sqrt{3}k(\sqrt{3+2w}+\sqrt{3})}{2(3+2w)a^{2}_{in}}$
$\displaystyle-\dfrac{8nw^{3}V_{n}a^{-(3+nw-\sqrt{9+6w})/w}_{in}}{3(3+2w-\sqrt{9+6w})\Big{[}3+(6-n)w-\sqrt{9+6w}\Big{]}F^{2}}\Big{\\}}a^{2(3+3w-\sqrt{9+6w})/w}_{in},$
(104)
$\displaystyle\rho(a)=\dfrac{3F^{2}k}{4wa^{(2w-3+\sqrt{9+6w})/w}}-\dfrac{V_{n}}{a^{n}}.$
(105)

Although the Hubble function is real for $0<n<6$ and $w\gg 1$, the energy
density always has a negative component.

##  6 Solution for given $H(a)$ and $\phi(a)$

The solution for specific form of the Hubble function can be crucial. To
construct the solution one can follow two different paths:

  * •

Method 1) First, find $V(a)$ from the constraint equation then insert it into
the $\phi$ equation and solve it for $\rho(a)$.

  * •

Method 2) First, solve the $\phi$ equation for the $V(a)$ then insert it into
the constraint equation and solve it for $\rho(a)$.

###  6.1 Method1

We find $V(a)$ from the $\rho$ equation

$\displaystyle
V(a)=-\rho(a)+\dfrac{3k\phi^{2}(a)}{4wa^{2}}+\Big{[}3\phi^{2}(a)+6a\phi(a)\phi^{{}^{\prime}}(a)-2wa^{2}\phi^{{}^{\prime}2}(a)\Big{]}\dfrac{H^{2}(a)}{4w}.$
(106)

Then the $\phi$ equation turns into the following form

$\displaystyle\rho^{{}^{\prime}}(a)=Q(a),$ (107) $\displaystyle
Q(a)=\dfrac{-3\phi^{2}(k-a^{3}HH^{{}^{\prime}})+a^{3}[3(1+2w)aH^{2}\phi^{{}^{\prime}2}]+3a^{4}H\phi(H^{{}^{\prime}}\phi^{{}^{\prime}}+H\phi^{{}^{\prime\prime}})}{2wa^{3}}.$
(108)

Thus

$\displaystyle\rho(a)=\int_{a_{in}}^{a}Q(a^{{}^{\prime}})da^{{}^{\prime}}+\rho(a_{in}),$
(109) $\displaystyle
V(a)=-\Big{[}\int_{a_{in}}^{a}Q(a^{{}^{\prime}})da^{{}^{\prime}}+\rho(a_{in})\Big{]}+\dfrac{3k\phi^{2}(a)}{4wa^{2}}+\Big{[}3\phi^{2}(a)+6a\phi(a)\phi^{{}^{\prime}}(a)-2wa^{2}\phi^{{}^{\prime}2}(a)\Big{]}\dfrac{H^{2}(a)}{4w}.$
(110)

###  6.2 Method2

We solve the $\phi$ equation for the potential,

$\displaystyle
V(a)=-\int_{a_{in}}^{a}\phi^{{}^{\prime}}\Big{\\{}-3\phi\Big{[}k+a^{{}^{\prime}2}H(2H+a^{{}^{\prime}}H^{{}^{\prime}})\Big{]}+2wa^{{}^{\prime}3}H\Big{[}(4H+a^{{}^{\prime}}H^{{}^{\prime}})\phi^{{}^{\prime}}+a^{{}^{\prime}}H\phi^{{}^{\prime\prime}}\Big{]}\Big{\\}}da^{{}^{\prime}}+V(a_{in}).$
(111)

We obtain energy density by substituting $V(a)$ into the $\rho$ equation

$\displaystyle\rho(a)=\int_{a_{in}}^{a}\phi^{{}^{\prime}}\Big{\\{}-3\phi\Big{[}k+a^{{}^{\prime}2}H(2H+a^{{}^{\prime}}H^{{}^{\prime}})\Big{]}+2wa^{{}^{\prime}3}H\Big{[}(4H+a^{{}^{\prime}}H^{{}^{\prime}})\phi^{{}^{\prime}}+a^{{}^{\prime}}H\phi^{{}^{\prime\prime}}\Big{]}\Big{\\}}da^{{}^{\prime}}-V(a_{in})$
$\displaystyle+\dfrac{3k\phi^{2}(a)}{4wa^{2}}+\Big{[}3\phi^{2}(a)+6a\phi(a)\phi^{{}^{\prime}}(a)-2wa^{2}\phi^{{}^{\prime}2}(a)\Big{]}\dfrac{H^{2}(a)}{4w}.$
(112)

##  7 Single Component Universe

We investigate the universe with given energy density and the scalar field
which are in the following form

$\displaystyle\rho(a)=\dfrac{\rho_{n}}{a^{n}},\hskip
30.0pt\phi(a)=\dfrac{F}{a^{d}}.$ (113)

We have obtained the Hubble function and the potential function by applying
the procedure explained in section 4.1, we obtain the following formulas

$\displaystyle
H(a)=\sqrt{\dfrac{k}{\mathcal{B}a^{2}}-\dfrac{4n\rho_{n}wa^{2d-n}}{3\mathcal{C}F^{2}}+\tilde{\gamma}(a_{in})a^{\mu}},$
(114) $\displaystyle
V(a)=\dfrac{d^{2}(3+2w)F^{2}k}{2\mathcal{B}wa^{2+2d}}+\dfrac{\mathcal{A}\rho_{n}a^{-n}}{3\mathcal{C}}+\dfrac{\mathcal{F}\tilde{\gamma}(a_{in})F^{2}a^{-2d+\mu}}{4w},$
(115)
$\displaystyle\tilde{\gamma}(a_{in})=a_{in}^{-\mu}H^{2}(a_{in})-\dfrac{ka^{-2-\mu}_{in}}{\mathcal{B}}+\dfrac{4n\rho_{n}wa^{2d-n-\mu}_{in}}{3\mathcal{C}F^{2}},$
(116) $\displaystyle\mu=\dfrac{2d[1+2d(1+w)]}{-1+d},$ (117)
$\displaystyle\mathcal{A}=d\\{3(-4+n)+2d[-3+(-6+n)w]\\},$ (118)
$\displaystyle\mathcal{B}=[-1+2d(1+d+dw)],$ (119)
$\displaystyle\mathcal{C}=\\{-n+d[4+n+d(2+4w)]\\},$ (120)
$\displaystyle\mathcal{F}=[-3+2d(3+dw)],$ (121) $\displaystyle
p(a)=\dfrac{(n-3)\rho_{n}}{3a^{n}}.$ (122)

We should choose $d$ such that $\mathcal{B}<0$ and $\mathcal{C}<0$. Then we
have real Hubble function. In addition if we also satisfy $\mu=0$ at the same
time this will have important physical implications which will be discussed
soon. Thus

$\displaystyle\mu=0\hskip 20.0pt\Rightarrow\hskip 20.0ptd=-\dfrac{1}{2(1+w)},$
(123) $\displaystyle\mathcal{B}=-1-\dfrac{1}{2(1+w)},\hskip
20.0pt\mathcal{C}=-\dfrac{(3+2w)(1+n+nw)}{2(1+w)^{2}}.$ (124)

Then the Hubble function and potential are simplified to the following forms

$\displaystyle
H(a)=\sqrt{-\dfrac{2(1+w)k}{(3+2w)a^{2}}+\dfrac{8n(1+w)^{2}w\rho_{n}a^{-1/(1+w)-n}}{3(3+2w)[1+n(1+w)]F^{2}}+\tilde{\gamma}(ain)},$
(125) $\displaystyle
V(a)=\dfrac{(-3+n)\rho_{n}a^{-n}}{3(1+n(1+w))}-\dfrac{kF^{2}a^{1/(1+w)-2}}{4w(1+w)}+\dfrac{(3+2w)(4+3w)F^{2}}{8w(1+w)^{2}}\tilde{\gamma}(a_{in})a^{1/(1+w)},$
(126)
$\displaystyle\tilde{\gamma}(a_{in})=H^{2}(a_{in})+\dfrac{2(1+w)k}{(3+2w)a^{2}_{in}}-\dfrac{8n(1+w)^{2}w\rho_{n}a^{-1/(1+w)-n}_{in}}{3(3+2w)[1+n(1+w)]F^{2}}.$
(127)

The deceleration parameter is found by applying the chain rule

$\displaystyle q(t)=\dfrac{d}{dt}(\dfrac{1}{H(t)})-1,$ (128) $\displaystyle
q(a)=\dfrac{d}{da}(\dfrac{1}{H(a)})aH(a)-1,$ (129) $\displaystyle
q(a)=-1+\dfrac{2(1+w)[1+n(1+w)][-3F^{2}ka^{1/(1+w)+n}+2nw\rho_{n}a^{2}]}{8nw(1+w)^{2}\rho_{n}a^{2}+3[1+n(1+w)]F^{2}[-2k(1+w)+(3+2w)\tilde{\gamma}(a_{in})a^{2}]a^{1/(1+w)+n}}.$
(130)

When finding $q(a_{in})$ one should plug $\tilde{\gamma}(a_{in})$ which is
given by (127)

$\displaystyle
q(a_{in})=-1+\dfrac{2(1+w)[-3F^{2}ka^{1/(1+w)+n}_{in}+2nw\rho_{n}a^{2}_{in}]}{3(3+2w)F^{2}H^{2}(a_{in})a^{2}_{in}}.$
(131)

Since $a_{in}\ll 1$, $q(a_{in})$ is simplified further

$\displaystyle\lim_{a\rightarrow
a_{in}}q(a)=\begin{cases}-1+\dfrac{4(1+w)nw\rho_{n}}{3(3+2w)F^{2}H^{2}(a_{in})},\hskip
20.0ptif\hskip 20.0ptn+\dfrac{1}{1+w}>2,\\\ \\\
-1+\dfrac{2(1+w)[-3F^{2}k+2nw\rho_{n}]}{3(3+2w)F^{2}H^{2}(a_{in})}\hskip
20.0ptif\hskip 20.0ptn+\dfrac{1}{1+w}=2,\\\ \\\
-1-\dfrac{2(1+w)k}{(3+2w)H^{2}(a_{in})a^{-1/(1+w)-n+2}_{in}},\hskip
20.0ptif\hskip 20.0pt0<n+\dfrac{1}{1+w}<2.\end{cases}$

When we set today values of the scale factor $a=1$, $\rho_{n}$ and $F$ become
today’s values of the energy density and the scalar field respectively. Thus
dynamics of the early universe depends on not only initial value of the Hubble
function and the scale factor but also today’s values of the energy density
and scalar field. In addition behaviour of the universe also depends on the
curvature parameter $k$ for $0<n+\dfrac{1}{1+w}\leq 2$.
The fate of the universe can be described by the following number

$\displaystyle\lim_{a\rightarrow\infty}q(a)=-1.$ (132)

Now, we formulate $a(t)$ for a spatially flat universe as

$\displaystyle
t=\int_{a_{in}}^{a}\dfrac{da^{{}^{\prime}}}{a^{{}^{\prime}}H(a^{{}^{\prime}})},$
(133) $\displaystyle H(a)=\sqrt{\delta a^{-1/(1+w)-n}+\alpha},\hskip
20.0pt\delta=\dfrac{8n(1+w)^{2}w\rho_{n}}{3(3+2w)[1+n(1+w)]F^{2}},\hskip
20.0pt\alpha=\tilde{\gamma}(a_{in}).$ (134)

Thus

$\displaystyle
t=\dfrac{1}{\kappa\sqrt{\alpha}}ln\Big{[}\dfrac{a^{\kappa}+\sqrt{a^{2\kappa}+\dfrac{\delta}{\alpha}}}{a^{\kappa}_{in}+\sqrt{a^{2\kappa}_{in}+\dfrac{\delta}{\alpha}}}\Big{]},\hskip
25.0pt\kappa=\dfrac{1}{2}\Big{[}\dfrac{1+n(1+w)}{1+w}\Big{]}.$ (135)

Hence

$\displaystyle
a(t)=\Big{[}\dfrac{(a^{\kappa}_{in}+\sqrt{a^{2\kappa}_{in}+\dfrac{\delta}{\alpha}})e^{\kappa\sqrt{\alpha}t}+(a^{\kappa}_{in}-\sqrt{a^{2\kappa}_{in}+\dfrac{\delta}{\alpha}})e^{-\kappa\sqrt{\alpha}t}}{2}\Big{]}^{1/\kappa},$
(136)

where $\delta$, $\alpha$ and $\kappa$ are given in (134-135). This formula
indicates a very significant interpretation. When there is no constant energy
density, the universe still expands exponentially because of nonzero initial
conditions. We will investigate this phenomena in depth by adding constant
term to the energy density in sections 8 and 9
As we said at the beginning of section 2 we have obtained perfect fluid
equation of state

$\displaystyle\nu=\dfrac{p(a)}{\rho(a)}=\dfrac{(n-3)}{3}.$ (137)

Potential formula as a function of the scalar field are written as

$\displaystyle
V(\phi)=\dfrac{(-3+n)\rho_{n}}{3(1+n(1+w))}(\dfrac{\phi}{F})^{-2n(1+w)}-\dfrac{kF^{2}}{4w(1+w)}(\dfrac{\phi}{F})^{-2-4w}+\dfrac{(3+2w)(4+3w)}{8w(1+w)^{2}}\tilde{\gamma}(a_{in})\phi^{2}.$
(138)

Most probably $\phi^{2}$ term in the potential is responsible for late time
accelerated expansion. This formula is simpler than the corresponding formula
$V(\phi)$ in our previous study [55].

##  8 Early Universe, Dark Energy and Radiation

In this section the scalar field and the energy density is taken as

$\displaystyle\phi(a)=\dfrac{F}{a^{d}},\hskip
20.0ptd=-\dfrac{1}{2(1+w)},\hskip
20.0pt\rho(a)=\dfrac{\rho_{r}}{a^{4}}+\Lambda.$ (139)

To see whether $\Lambda$ contributes to $H(a)$ we have performed calculations
in all four different methods. Results are found in the following form

$\displaystyle
H(a)=\sqrt{-\dfrac{2(1+w)k}{(3+2w)a^{2}}+\dfrac{32(1+w)^{2}w\rho_{r}a^{-1/(1+w)-4}}{3(3+2w)(5+4w)F^{2}}+\alpha},$
(140) $\displaystyle V(a)=-\dfrac{F^{2}ka^{1/(1+w)-2}}{4w(1+w)}+\beta
a^{1/(1+w)}+\dfrac{\rho_{r}}{3(5+4w)a^{4}}-\Lambda.$ (141)

In method-1 we find

$\displaystyle\alpha_{1}=H^{2}(a_{in})+\dfrac{2(1+w)k}{(3+2w)a_{in}^{2}}-\dfrac{32(1+w)^{2}w\rho_{r}a^{-1/(1+w)-4}_{in}}{3(3+2w)(5+4w)F^{2}},$
(142)
$\displaystyle\beta_{1}=\dfrac{(3+2w)(4+3w)F^{2}}{8w(1+w)^{2}}\alpha_{1}.$
(143)

In method-2 we find

$\displaystyle\alpha_{2}=\alpha_{1},\hskip 20.0pt\beta_{2}=\beta_{1}.$ (144)

In this method to be able to satisfy the constraint equation with final
formulas of $V(a)$ and $H(a)$ initial conditions must satisfy the following
equation

$\displaystyle
V(a_{in})=-\dfrac{F^{2}ka^{1/(1+w)-2}_{in}}{4w(1+w)}+\beta_{2}a^{1/(1+w)}_{in}+\dfrac{\rho_{r}}{3(5+4w)a^{4}_{in}}-\Lambda,$
(145)

which is consistent with the formula of $V(a)$.
In method-3 we find

$\displaystyle\alpha_{3}=\dfrac{8w(1+w)^{2}\tilde{V}(a_{in})}{(3+2w)(4+3w)F^{2}},$
(146)
$\displaystyle\tilde{V}(a_{in})=V(a_{in})a^{-1/(1+w)}_{in}-\dfrac{\rho_{r}a^{-1/(1+w)-4}_{in}}{3(5+4w)}+\dfrac{F^{2}k}{4w(1+w)a^{2}_{in}}+\Lambda
a^{-1/(1+w)}_{in},$ (147)
$\displaystyle\beta_{3}=\dfrac{(3+2w)(4+3w)F^{2}}{8w(1+w)^{2}}\alpha_{3}.$
(148)

Although at first sight it seems that $\alpha_{1}\neq\alpha_{3}$, one can show
that they are equal to each other by substituting $a_{in}$ in $H(a)$ and
solving the equation for $V(a_{in})$.
In method-4 we find

$\displaystyle\alpha_{4}=\alpha_{3},\hskip 20.0pt\beta_{4}=\beta_{3}.$ (149)

By the same reasoning in method-2, constraint equation implies the following
condition on initial values

$\displaystyle
a^{1/(1+w)+3}H^{2}(a_{in})-\dfrac{32w(1+w)^{2}\rho_{r}}{3(3+2w)(5+4w)F^{2}a_{in}}-\dfrac{8w(1+w)^{2}\tilde{V}(a_{in})a^{1/(1+w)+3}_{in}}{(3+2w)(4+3w)F^{2}}+\dfrac{2(1+w)ka^{1/(1+w)+1}_{in}}{3+2w}=0.$
(150)

$V(a_{in})$ which is obtained by solving this equation is consistent with the
formula of $V(a)$.
The deceleration parameter is found as

$\displaystyle
q(a)=-1+\dfrac{2(1+w)(5+4w)[-3F^{2}ka^{1/(1+w)+2}+8w\rho_{r}]}{32w(1+w)^{2}\rho_{r}+3(5+4w)F^{2}[-2k(1+w)+(3+2w)\alpha_{1}a^{2}]a^{1/(1+w)+2}},$
(151)
$\displaystyle\alpha_{1}=H^{2}(a_{in})+\dfrac{2(1+w)k}{(3+2w)a^{2}_{in}}-\dfrac{32(1+w)^{2}w\rho_{r}a^{-1/(1+w)-4}_{in}}{3(3+2w)(5+4w)F^{2}}.$
(152)

Thus initial value of the deceleration parameter is found

$\displaystyle
q(a_{in})=-1+\dfrac{2(1+w)[-3F^{2}ka^{1/(1+w)+2}_{in}+8w\rho_{r}]}{3(3+2w)F^{2}H^{2}(a_{in})}.$
(153)

The deceleration parameter which is given in (151) is same as the deceleration
parameter found in single component universe without constant term, since
$\alpha_{1}$ given in (152) is equal to $\tilde{\gamma}(a_{in})$ given in
(127) with $n=4$.
The scale factor as a function of time is written as

$\displaystyle
a(t)=\Big{[}\dfrac{(a^{\kappa}_{in}+\sqrt{a^{2\kappa}_{in}+\dfrac{\delta}{\alpha}})e^{\kappa\sqrt{\alpha}t}+(a^{\kappa}_{in}-\sqrt{a^{2\kappa}_{in}+\dfrac{\delta}{\alpha}})e^{-\kappa\sqrt{\alpha}t}}{2}\Big{]}^{1/\kappa},$
(154) $\displaystyle\kappa=\dfrac{1}{2}\Big{(}\dfrac{5+4w}{1+w}\Big{)},\hskip
20.0pt\delta=\dfrac{32(1+w)^{2}w\rho_{r}}{3(3+2w)(5+4w)F^{2}},$ (155)

where $\alpha=\alpha_{1}$ and $\alpha_{1}$ is given by (142).
Formula of the potential as function of the scalar field is found as

$\displaystyle
V(\phi)=-\dfrac{F^{2}k}{4w(1+w)}(\dfrac{\phi}{F})^{-2-4w}+\beta(\dfrac{\phi}{F})^{2}+\dfrac{\rho_{r}}{3(5+4w)}(\dfrac{\phi}{F})^{-8(1+4w)}-\Lambda,$
(156)

where $\beta=\beta_{1}$ which is given by (143). Pressure is found as in
expected form

$\displaystyle p(a)=\dfrac{\rho_{r}}{3a^{4}}-\Lambda.$ (157)

##  9 Late Time expansion of the universe

###  9.1 Dark Energy(!) dominated universe

We study in spatially flat universe where

$\displaystyle\phi(a)=\dfrac{F}{a^{d}},\hskip 20.0ptd=-\frac{1}{2(1+w)},\hskip
20.0pt\rho(a)=\dfrac{\rho_{m}}{a^{3}}+\Lambda.$ (158)

Then related function is formulated as

$\displaystyle
H(a)=\sqrt{\dfrac{8w(1+w)^{2}\rho_{m}}{(3+2w)(4+3w)F^{2}a^{1/(1+w)+3}}+\tilde{\gamma}(a_{in})},$
(159) $\displaystyle
V(a)=\dfrac{(3+2w)(4+3w)}{8w(1+w)^{2}}F^{2}\tilde{\gamma}(a_{in})a^{1/(1+w)}-\Lambda,$
(160) $\displaystyle
q(a)=-1+\dfrac{4w(1+w)(4+3w)\rho_{m}}{8w(1+w)^{2}\rho_{m}+(3+2w)(4+3w)F^{2}\tilde{\gamma}(a_{in})a^{1/(1+w)+3}},$
(161)
$\displaystyle\tilde{\gamma}(a_{in})=H^{2}(a_{in})-\dfrac{8(1+w)^{2}w\rho_{m}a^{-1/(1+w)-3}_{in}}{(3+2w)(4+3w)F^{2}},$
(162) $\displaystyle p(a)=-\Lambda.$ (163)

The scale factor as function of time is found as

$\displaystyle
a(t)=\Big{[}\dfrac{(a^{\kappa}_{in}+\sqrt{a^{2\kappa}_{in}+\dfrac{\delta}{\alpha}})e^{\kappa\sqrt{\alpha}t}+(a^{\kappa}_{in}-\sqrt{a^{2\kappa}_{in}+\dfrac{\delta}{\alpha}})e^{-\kappa\sqrt{\alpha}t}}{2}\Big{]}^{1/\kappa},$
(164) $\displaystyle\kappa=\dfrac{1}{2}\Big{(}\dfrac{4+3w}{1+w}\Big{)},\hskip
20.0pt\delta=\dfrac{8(1+w)^{2}w\rho_{m}}{(3+2w)(4+3w)F^{2}}.$ (165)

where $\alpha=\tilde{\gamma}(a_{in})$ and $\tilde{\gamma}(a_{in})$ is given by
(162).
In Brans-Dicke theory effective gravitational constant is defined as
$G_{eff}=\dfrac{w}{2\pi\phi^{2}}$. Thus it’s present value becomes
$G_{0}=\dfrac{w}{2\pi F^{2}}$ where $F=\phi(t_{0})$. In addition when we take
observational value of Brans-Dicke parameter $w\gg 1$, the Hubble function
becomes

$\displaystyle H(a)=\sqrt{\dfrac{8\pi
G_{0}\rho_{m}}{3a^{3}}+\tilde{\gamma}(a_{in})}.$ (166)

By using following definitions of cosmological parameters

$\displaystyle\Omega_{m,0}=\dfrac{8\pi G}{3H^{2}_{0}}\rho_{m},\hskip
20.0pt\Omega_{\tilde{\gamma},0}=\dfrac{8\pi
G}{3H^{2}_{0}}\rho_{\tilde{\gamma}},\hskip
20.0pt\rho_{\tilde{\gamma}}=\dfrac{3\tilde{\gamma}(a_{in})}{8\pi G},$ (167)

the Hubble function is written as

$\displaystyle
H(a)=H_{0}\sqrt{\dfrac{\Omega_{m,0}}{a^{3}}+\Omega_{\tilde{\gamma},0}}.$ (168)

When one replace $\Omega_{\tilde{\gamma},0}$ with $\Omega_{\Lambda,0}$ in
(166) we will have the Hubble function for the late epoch of the universe in
standard cosmology where dark energy dominates.
We have already compared this model with observation of type Ia supernovae
data in [55]. In our previous study we have obtained $H_{0}=71.80\pm 0.22$,
$\Omega_{\Lambda,0}=0.715\pm 0.012$, $\Omega_{m}=0.285\pm 0.012$ and
$\chi^{2}/\nu=0.990$ for absolute magnitude $M=-19.30$. Present value of
deceleration parameter has been found as $q_{0}=-0.572$ with this cosmological
parameters.
On the other hand potential can be rewritten as

$\displaystyle
V(\phi)=\dfrac{(3+2w)(4+3w)}{8w(1+w)^{2}}\tilde{\gamma}(a_{in})\phi^{2}-\Lambda.$
(169)

Most probably $\phi^{2}$ potential is responsible from the accelerated
expansion.

###  9.2 Domain wall dominated universe

In this section we study in spatially flat universe with scalar field and the
energy density is given as

$\displaystyle\phi(a)=\dfrac{F}{a^{d}},\hskip
20.0ptd=-\dfrac{1}{2(1+w)},\hskip
20.0pt\rho(a)=\dfrac{\rho_{r}}{a^{3}}+\dfrac{\rho_{w}}{a}.$ (170)

The Hubble function, potential, deceleration parameter and the pressure is
formulated as

$\displaystyle
H(a)=\sqrt{\dfrac{8w(1+w)^{2}}{3(3+2w)F^{2}}\Big{[}\dfrac{3\rho_{m}}{(4+3w)a^{3}}+\dfrac{\rho_{w}}{(2+w)a}\Big{]}a^{-1/(1+w)}+\tilde{\gamma}(a_{in})},$
(171) $\displaystyle
V(a)=\dfrac{(3+2w)(4+3w)}{8w(1+w)^{2}}\tilde{\gamma}(a_{in})F^{2}a^{1/(1+w)}-\dfrac{2\rho_{w}}{3(2+w)a},$
(172) $\displaystyle
q(a)=-1+\dfrac{4w(1+w)(2+w)(4+3w)(3\rho_{m}+\rho_{w}a^{2})}{3(2+w)(3+2w)(4+3w)\tilde{\gamma}(a_{in})F^{2}a^{1/(1+w)+3}+8w(1+w)^{2}[3(2+w)\rho_{m}+(4+3w)\rho_{w}a^{2}]},$
(173)
$\displaystyle\tilde{\gamma}(a_{in})=H^{2}(a_{in})-\dfrac{8w(1+w)^{2}}{3(3+2w)F^{2}}\Big{[}\dfrac{3\rho_{m}}{(4+3w)a^{3}_{in}}+\dfrac{\rho_{w}}{(2+w)a_{in}}\Big{]}a^{-1/(1+w)}_{in},$
(174) $\displaystyle p(a)=-\dfrac{2\rho_{w}}{3a}.$ (175)

Now we take $\tilde{\gamma}(a_{in})=0$, to investigate whether cosmic domain
walls cause accelerated expansion of the universe or not. We again apply the
fundamental idea of the Brans-Dicke theory as we did in section 8. Thus we
take $G_{0}=\dfrac{w}{2\pi F^{2}}$ and we take $w\gg 1$. Then the Hubble
function is simplified to

$\displaystyle H(a)=\sqrt{\dfrac{8\pi
G_{0}}{3}(\dfrac{\rho_{m}}{a^{3}}+\dfrac{\rho_{w}}{a})}.$ (176)

The relation between the scale factor and time is found as

$\displaystyle
t=\int_{a_{in}}^{a}\dfrac{da^{{}^{\prime}}}{a^{{}^{\prime}}H(a^{{}^{\prime}})},$
$\displaystyle
t=\int_{a_{in}}^{a}\dfrac{da^{{}^{\prime}}}{a^{{}^{\prime}}\sqrt{\dfrac{8\pi
G}{3}(\dfrac{\rho_{m}}{a^{3}}+\dfrac{\rho_{w}}{a})}},$ (177) $\displaystyle
t=\sqrt{\dfrac{3}{2\pi
G\rho_{w}}}\Big{[}\sqrt{a^{{}^{\prime}}}{}_{2}F_{1}(\dfrac{1}{2},-\dfrac{1}{4};\dfrac{3}{4};-\dfrac{\rho_{m}}{\rho_{w}a^{{}^{\prime}2}})\Big{]}_{a_{in}}^{a}.$
(178)

${}_{2}F_{1}$ is the hypergeometric function

$\displaystyle{}_{2}F_{1}(1/2,b;b+1;u)=\sum_{n=0}^{\infty}\dfrac{(1/2)_{n}(b)_{n}}{(b+1)_{n}}\dfrac{u^{n}}{n!},$
(179)

where $(b)_{n}$ is the Pochhammer symbol which is defined as

$\displaystyle(b)_{n}=\begin{cases}1\hskip 20.0pt\text{if}\hskip 20.0ptn=0,\\\
b(b+1)(b+2)...(b+n-1),\hskip 20.0pt\text{if}\hskip 20.0ptn>0.\end{cases}$
(180)

Type Ia supernovaes are known as standard candles since their measurements
makes comparison of theory and observations is possible. This is done by the
following relation

$\displaystyle m=5log_{10}(\dfrac{d_{L}}{1Mpc})+25+M,$ (181)

where $m$ and $M$ are the apparent and the absolute magnitudes respectively.
Then the distance modulus is defined as $\mu=m-M$.
Now we borrow one of our plots from our previous work [55]. In our previous
study we have already performed curve fitting with type Ia supernovae data. We
have found $H_{0}=71.03\pm 0.20$, $\Omega_{w,0}=0.889\pm 0.015$,
$\Omega_{m,0}=0.111\pm 0.015$ and $\chi^{2}/\nu=1.008$ for $M=-19.30$ in
domain wall dominated universe. We obtain $q_{0}=-0.389$ with these values of
cosmological density parameters. On the other hand for dark energy dominated
universe we have found $H_{0}=71.80\pm 0.22$, $\Omega_{\Lambda,0}=0.715\pm
0.012$, $\Omega_{m,0}=0.285\pm 0.012$ and $\chi^{2}/\nu=0.990$ for $M=-19.30$.
We obtain $q_{0}=-0.572$. All these results are shown in Figure 1.

Figure 1: Figure I:Distance modulus vs redshift plot for $M=-19.30$. Dots
represent observations of Pantheon data while green line represents domain
wall dominated universe and red line represents dark energy dominated universe

In addition

$\displaystyle\lim_{a\Rightarrow\infty}q(a)=-0.5.$ (182)

$V(\phi)$ is formulated as

$\displaystyle V(\phi)=-\dfrac{2\rho_{w}}{3(2+w)}(\dfrac{\phi}{F})^{-2(1+w)}.$
(183)

Although power of $\phi$ in the potential is different from $2$ it causes
accelerated expansion of the universe.

##  10 Discussion

We have revealed three major points of cosmology in this study. These are the
answers for the following questions:

  * 1

Is there any corresponding energy for the cosmological constant?

  * 2

Is it possible to have accelerated universe without cosmological constant?

  * 3

Is ratio of dark mater to baryonic matter smaller than $\dfrac{0.05}{0.27}$?

It is known that constant energy density contributes to the Hubble function.
We have examined this phenomena in section 8 and in section 9.1. Firstly we
have solved our system of equations for early universe by taking

$\displaystyle\phi(a)=\dfrac{F}{a^{d}},\hskip
20.0ptd=-\dfrac{1}{2(1+w)},\hskip
20.0pt\rho(a)=\dfrac{\rho_{r}}{a^{4}}+\Lambda.$ (184)

Then $H(a)$ is found in the form of (140) where there is a constant term
$\alpha$. At first sight there is no contribution of $\Lambda$ in $\alpha_{1}$
given by (142). However one can also claim that $\Lambda$ contributes in
$\alpha_{3}$ by substituting (147) in (146). This dilemma can be explained as
follows. One can write

$\displaystyle
V(a_{in})=\dfrac{3F^{2}ka^{1/(1+w)-2}_{in}}{4w}+\dfrac{(3+2w)(4+3w)}{8w(1+w)^{2}}F^{2}H^{2}(a_{in})a^{1/(1+w)}(a_{in})-\dfrac{\rho_{r}}{a^{4}_{in}}-\Lambda.$
(185)

Then one can rewrite $H(a_{in})$ in terms of $V(a_{in})$ and $\Lambda$.
We have also examined the spatially flat late universe in section 9.1 by
supplying the following functions

$\displaystyle\phi(a)=\dfrac{F}{a^{d}},\hskip
20.0ptd=-\dfrac{1}{2(1+w)},\hskip
20.0pt\rho(a)=\dfrac{\rho_{m}}{a^{3}}+\Lambda.$ (186)

The Hubble function has been found in the form given in (159) which has a
constant term. At first sight this constant term which is given by (162) does
not contain $\Lambda$. One can apply the same trick and one can write

$\displaystyle
V(a_{in})=\dfrac{(3+2w)(4+3w)}{8w(1+w)^{2}}F^{2}H^{2}(a_{in})a^{1/(1+w)}_{in}-\dfrac{\rho_{m}}{a^{3}_{in}}-\Lambda.$
(187)

Again, one can rewrite $H(a_{in})$ in terms of $V(a_{in})$ and $\Lambda$ and
one can claim that $H(a)$ contains $\Lambda$.
There are two possible interpretations;

  * I

Constant energy density does not contribute to the Hubble function, it only
modifies value of potential at the beginning of the universe.

  * II

Initial value of the Hubble function $H(a_{in})$, can be rewritten in terms of
$V(a_{in})$ and $\Lambda$ so constant energy density contributes to the Hubble
function.

Both of them can be reasonable according to one’s perspective.
Although one chooses second explanation which is in agreement with common
trend, one can still ask the question in another way: _Is it possible to have
a constant term in the Hubble function when there is no constant energy
density in the universe?_ To answer this question we have investigated single
component universe in section 7 by giving following functions

$\displaystyle\phi(a)=\dfrac{F}{a^{d}},\hskip
20.0ptd=-\dfrac{1}{2(1+w)},\hskip 20.0pt\rho(a)=\dfrac{\rho_{n}}{a^{n}}.$
(188)

Then $H(a)$ is found in the form given by (125). The Hubble function has a
constant term although there is no constant energy density. This constant is
given in (127). Now $V(a_{in})$ does not contain $\Lambda$. In addition,
expected exponential expansion of $a(t)$ is given in (136). Thus there is one
possible explanation. When there is no constant energy density, the Hubble
function still has a constant term which causes exponential expansion.
We have shown that universe has an accelerated expansion when one introduces
cosmic domain walls with matter in the energy density in section 9.2. We had
studied this case with Friedmann cosmology in [55], and we have compared our
model with the latest supernovae data. It is seen that both dark energy
dominated and domain wall dominated universe have perfect fit with data. The
differences between these two models most probably will be seen when bigger
redshift data are available.
While comparison of dark energy dominated universe with observation has
resulted in $\Omega_{\Lambda}=0.715\pm 0.012$ and $\Omega_{m}=0.285\pm 0.012$,
comparison of domain wall dominated universe with observation has resulted in
$\Omega_{w}=0.889\pm 0.015$ and $\Omega_{m}=0.111\pm 0.015$. Hence a new
question appears; is the ratio of dark matter to baryonic matter less than
$\dfrac{0.05}{0.27}$?
In 1980s Modified theories of Newtonian Dynamics or MOND, were proposed [56,
57]. It mainly claims that observational aspects of galaxies can be understood
without dark matter. Recent observational evidence for the external field
effect in MOND which was proposed as an alternative to dark matter is
presented in [58]. A new relativistic MOND theory [59] successfully reproduces
cosmic microwave background power spectra. These theories may explain non-
baryonic part of domain wall dominated universe which covers six percent of
the energy-matter content of the universe.

##  11 Conclusion

In this study we have applied change of variable $a=\dot{a}H(a)$ to field
equations of Brans-Dicke theory and have written all equations in terms of
independent variable $"a"$. Then we have ended up with a constraint equation
and a Bernoulli type differential equation which can be linearized. We have
presented analytic solutions for supplied pairs of functions; ($\phi(a)$ and
$\rho(a)$) , ($\phi(a)$ and $V(a)$), or ($\phi(a)$ and $H(a)$).
Investigation of single component universe has shown that one will have a
constant term in the Hubble function although there is no constant energy
density. Early epoch of the universe with dark energy and radiation have been
studied, and exponential expansion is seen in $a(t)$. Late-time acceleration
is obtained for both dark energy dominated universe and domain wall dominated
universe. When Brans-Dicke parameter $w\gg 1$, the Hubble function reduces to
$H(a)$ of the Einstein cosmology. In all cases potential is found as
combination of power law potentials.
We would like to emphasize that so-called dark energy term may be just a
number without corresponding constant energy-matter density. Details of this
subject has been presented in the discussion. Hopefully astronomers and
cosmologists will pay more attention for searches of cosmic domain walls.

## Acknowledgement

We thank Teoman Turgut for reading first draft of this article and for
fruitfull discussion. We thank Bogazici University for the financial support
provided by the Scientific Research Fund (BAP), research project No 16521.

## Appendices

$\displaystyle\dot{a}=Ha$ (189) $\displaystyle\frac{\dot{a}^{2}}{a^{2}}=H^{2}$
(190) $\displaystyle\frac{\ddot{a}}{a}=H^{2}+H^{{}^{\prime}}Ha$ (191)
$\displaystyle\dot{\phi}=\phi^{{}^{\prime}}Ha$ (192)
$\displaystyle\ddot{\phi}=\phi^{{}^{\prime\prime}}H^{2}a^{2}+\phi^{{}^{\prime}}H^{{}^{\prime}}Ha^{2}+\phi^{{}^{\prime}}H^{2}a$
(193)

where the prime denotes $\dfrac{d}{da}$.

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