1. Introduction
The very successful theory of General Relativity (GR) [1, 2] can be validly considered as the foundation of many other theories of gravity. One more proof to confirm once again this gigantic theory is the recent appearance of gravitational waves [3]. But perfection is not of this world. Some limitations of GR can be mentioned like the inescapable existence of singularities and the lack of effective explanations of phenomena such as dark matter and dark energy problems. In order to provide answers to some of these problems, several researchers have built alternative theories which deviate from the GR in the distributions of matter. Therefore, apart from quantum gravity, we have several modified gravities. As modified gravity, we have the one inspired by Born-Infeld (BI) electrodynamics [4], where the infiniteness of the electric field at the location of a point charge is regularized. Then, Deser and Gibbons [5] used this BI structure to set up a new theory of gravity in the metric formulation. It should not be forgotten that the pioneer Eddington had proposed a determining form of gravity in this reformulation of GR for a de Sitter spacetime [6], but the coupling of matter remained a problem in his approach. A few years later, Vollick [7] set up the Palatini formulation of BI gravity and worked on several axes, namely the non-trivial couplings of matter [8, 9]. Subsequently, Banados and Ferreira [10] proposed to the scientific world a new formulation, called Eddington-inspired-Born-Infeld (EiBI) theory of gravity, whose material coupling was different and simple than that proposed by Vollick in [7]. The EiBI theory is a theory of bi-gravity [11–14], which reduces to GR theory in a vacuum. The main characteristic of the EiBI bi-gravity theory is the existence of a physical metric g which is coupled to matter and of an auxiliary metric q which is not coupled to matter. Throughout the work, the Universe described by the bi-metric (g, q) will be referred to as a bi-universe. The action of the EiBI theory proposed by [10] is given by1 ) with respect to the fields gαβ and ${{\rm{\Gamma }}}_{\alpha \beta }^{\gamma }$, respectively, where
$\begin{eqnarray}\begin{array}{l}{S}_{{EiBI}}[g,{\rm{\Gamma }},{\psi }_{M}]=\displaystyle \frac{2}{k}{\displaystyle \int }_{}^{}{{\rm{d}}}^{4}x\\ \times \,[\sqrt{\left|{g}_{\alpha \beta }+{{kR}}_{\alpha \beta }({\rm{\Gamma }})\right|}-\lambda \sqrt{\left|g\right|}]+{S}_{M}[g,{\psi }_{M}],\end{array}\end{eqnarray}$
where k is the Eddington parameter linked to the cosmological constant Λ by the dimensionless constant λ = 1 + kΛ; $\left|g\right|$ denotes the absolute value of the determinant of the matrix (gαβ); SM[g, ψM] is the action of field. The equations of motion are obtained by varying the action ( $\begin{eqnarray}{{\rm{\Gamma }}}_{\alpha \beta }^{\gamma }=\displaystyle \frac{1}{2}{q}^{\gamma \delta }({q}_{\beta \delta ,\alpha }+{q}_{\alpha \delta ,\beta }-{q}_{\alpha \beta ,\delta }).\end{eqnarray}$
This yields the following equations: $\begin{eqnarray}\sqrt{q}{q}^{\alpha \beta }=\lambda \sqrt{g}{g}^{\alpha \beta }-k\sqrt{g}{T}^{\alpha \beta },\end{eqnarray}$
$\begin{eqnarray}{q}_{\alpha \beta }={g}_{\alpha \beta }+{{kR}}_{\alpha \beta },\end{eqnarray}$
with qβδ,α standing for $\tfrac{\partial {q}_{\beta \delta }}{\partial {x}^{\alpha }}$. Here Rαβ is the symmetric part of the Ricci tensor built from the connection ${{\rm{\Gamma }}}_{\alpha \beta }^{\gamma }$, given by $\begin{eqnarray}{R}_{\alpha \beta }({\rm{\Gamma }})=\displaystyle \frac{\partial {{\rm{\Gamma }}}_{\alpha \beta }^{\gamma }}{\partial {x}^{\gamma }}-\displaystyle \frac{\partial {{\rm{\Gamma }}}_{\alpha \mu }^{\gamma }}{\partial {x}^{\beta }}+{{\rm{\Gamma }}}_{\alpha \beta }^{\nu }{{\rm{\Gamma }}}_{\gamma \nu }^{\gamma }-{{\rm{\Gamma }}}_{\alpha \gamma }^{\nu }{{\rm{\Gamma }}}_{\nu \beta }^{\gamma }.\end{eqnarray}$
(qαβ) is the inverse of (qαβ), and the energy-momentum tensor is given by $\begin{eqnarray}{T}^{\alpha \beta }=\displaystyle \frac{2}{\sqrt{-g}}\displaystyle \frac{\delta {S}_{M}}{\delta {g}_{\alpha \beta }}.\end{eqnarray}$
In GR, the energy-momentum tensor Tαβ fulfils the following conservation equations, $\begin{eqnarray}{{\rm{\nabla }}}_{\alpha }{T}^{\alpha \beta }=0\quad {\rm{with}}\quad \beta =0,...,3,\end{eqnarray}$
where ∇ is the covariant derivative.Many research activities have been carried out on EiBI theory in recent years. Astrophysical scenarios have been discussed in [15–20, 46, 22–28], spherically symmetric solutions have been obtained in [14, 28–33], and some interesting cosmological solutions have been derived in [9, 13, 14, 46, 29, 31, 33–35]. In the present paper, we assume the following equation of state for the fluid, p = wρ, where p, w and ρ are the pressure of the fluid, a constant which defines the fluid and the energy density of the fluid, respectively [14, 36]. We study the evolution of an uncharged perfect fluid of pure radiation type, i.e. when $w=\tfrac{1}{3}$, the geometry is a Bianchi type I spacetime, which is in fact a generalization of Friedmann–Lemaître–Robertson–Walker (FLRW) spacetime [14, 31, 33]. The motivation for choosing a radiation fluid is that it naturally shows an upper bound on its energy density, which further induces an upper bound for scalar curvature equivalent of the inverse Eddington parameter $\tfrac{1}{k}\gt 0$ [14]. Indeed, our motivation for considering the Bianchi type I geometry is that, in Physics, the Universe as we observe it today is very well described by the cosmological and homogeneous models of FLRW. We assume that the Universe is initially anisotropic and becomes asymptotically isotropic. Here we consider six functions a, b, c, A, B and C, of the time variable t. These functions are determined by the equations (3 )–(4 ) of the EiBI theory, with the material content of spacetime represented by a material stress tensor defined by the density of material ρ. Equations of the EiBI theory, coupled with the laws of conservation, turn out to be a differential system allowing the determination of the functions a, b, c, A, B, C and ρ. By setting suitable initial data, we arrive at the solution of the model considered. We study the asymptotic behaviours of the volume of the scalar factor ${ \mathcal V }$, the Hubble parameter H, the Anisotropic parameter ${ \mathcal A }$ and the deceleration parameter Q. Indeed, the study of these different parameters makes it possible to foresee the cosmological applications of the EiBI bi-gravity theory for homogeneous and isotropic backgrounds. In [14], the authors obtain and discuss the stabilities of different interesting cosmological solutions for early universe, including loitering, quasi de Sitter and bouncing solutions. Therefore, the anomalies observed in the cosmic microwave background (CMB) might be due to the presence of small anisotropies. Moreover, by studying asymptotic behaviours, we show that these solutions tend toward the vacuum at the late time under appropriate conditions.
The present work is organized as follows. In section 2 we formulate the model by using the Bianchi type I geometry and consider a metric tensor g, an auxiliary tensor q together with a relativistic perfect fluid of density ρ in the Eddington-inspired-Born-Infeld theory. In section 3 , we obtain explicit expressions of functions a, b, c, A, B and C, and study the asymptotic behaviour of the bi-universe described by the physical metric g and the auxiliary metric q. The conclusion is elaborated in section 4 .
2. The EiBI equations
We consider a Bianchi type I bi-universe $(g,q,{{\mathbb{R}}}^{4})$ and denote by xα = (x0, xi) the usual coordinates in ${{\mathbb{R}}}^{4};$ g and q stand for physical metric and auxiliary metric of signature ( − , + , + , + ) which can be written respectively as [14, 31]:
$\begin{eqnarray}g=-{\rm{d}}{t}^{2}+{a}^{2}(t){\left({\rm{d}}{x}^{1}\right)}^{2}+{b}^{2}(t){\left({\rm{d}}{x}^{2}\right)}^{2}+{c}^{2}(t){\left({\rm{d}}{x}^{3}\right)}^{2},\end{eqnarray}$
$\begin{eqnarray}q=-{\rm{d}}{t}^{2}+{A}^{2}(t){\left({\rm{d}}{x}^{1}\right)}^{2}+{B}^{2}(t){\left({\rm{d}}{x}^{2}\right)}^{2}+{C}^{2}(t){\left({\rm{d}}{x}^{3}\right)}^{2},\end{eqnarray}$
where a > 0, b > 0, c > 0, A > 0, B > 0, and C > 0 are unknown functions of the single time variable t = x0.The general expression of the stress matter tensor of a relativistic perfect fluid is given by
$\begin{eqnarray}{T}^{\alpha \beta }=\left(\rho +p\right){u}^{\alpha }{u}^{\beta }+{{pg}}^{\alpha \beta },\end{eqnarray}$
where ρ > 0 and p > 0 are unknown functions of t representing the matter density and the pressure, respectively. We consider a perfect fluid of pure radiation type, which means that $p=\tfrac{\rho }{3}$. The matter tensor then can be written as: $\begin{eqnarray}{T}^{\alpha \beta }=\displaystyle \frac{4}{3}\rho {u}^{\alpha }{u}^{\beta }+\displaystyle \frac{1}{3}\rho {g}^{\alpha \beta },\end{eqnarray}$
where $u=\left({u}^{\alpha }\right)$ is a unit vector tangent to the geodesics of g.In order to simplify, we consider a frame in which the fluid is spatially at rest. This implies ui = ui = 0. Recall that indices are raised and lowered using the usual rules: Vα = gαβVβ; Vα = gαβVβ, $\left({g}^{\alpha \beta }\right)$ standing for the inverse matrix of $\left({g}_{\alpha \beta }\right)$.
Solving EiBI equations (3 )–(4 ) consists of determining, on one hand, the gravitational field represented by the physical metric $g=\left({g}_{\alpha \beta }\right)$ and auxiliary metric $q=\left({q}_{\alpha \beta }\right)$, through the six unknown functions a > 0, b > 0, c > 0, A > 0, B > 0, and C > 0, and, on the order hand, its source represented by the matter density ρ.
From the expressions of g and q given respectively by equations (8 ) and (9 ) we deduce that2 ) and (13 ) we obtain by a direct calculation the non-vanishing Christofell coefficients, which are defined by ${{\rm{\Gamma }}}_{i0}^{i}$ and ${{\rm{\Gamma }}}_{{ii}}^{0}$, i = 1, 2, 3. So, we have5 ), we obtain the non-vanishing components of the Ricci tensor Rαβ, given by11 ) and (12 ) we obtain the following non-vanishing contravariant components of the energy-momentum tensor:
$\begin{eqnarray}\left\{\begin{array}{l}{g}_{00}=-1,\quad {g}_{11}={a}^{2},\quad {g}_{22}={b}^{2},\quad {g}_{33}={c}^{2},\\ {g}^{00}=-1,\quad {g}^{11}={a}^{-2},\quad {g}^{22}={b}^{-2},\quad {g}^{33}={c}^{-2},\\ {g}_{\alpha \beta }={g}^{\alpha \beta }=0\quad \mathrm{if}\quad \alpha \ne \beta ,\end{array}\right.\end{eqnarray}$
$\begin{eqnarray}\left\{\begin{array}{l}{q}_{00}=-1,\quad {q}_{11}={A}^{2},\quad {q}_{22}={B}^{2},\quad {q}_{33}={C}^{2},\\ {q}^{00}=-1,\quad {q}^{11}={A}^{-2},\quad {q}^{22}={B}^{-2},\quad {q}^{33}={C}^{-2},\\ {q}_{\alpha \beta }={q}^{\alpha \beta }=0\quad \mathrm{if}\quad \alpha \ne \beta .\end{array}\right.\end{eqnarray}$
Using equations ( $\begin{eqnarray}\begin{array}{rcl}{{\rm{\Gamma }}}_{10}^{1} & = & \displaystyle \frac{\dot{A}}{A},\quad {{\rm{\Gamma }}}_{20}^{2}=\displaystyle \frac{\dot{B}}{B},\quad {{\rm{\Gamma }}}_{30}^{3}=\displaystyle \frac{\dot{C}}{C},\\ {{\rm{\Gamma }}}_{11}^{0} & = & \dot{A}A,\quad {{\rm{\Gamma }}}_{22}^{0}=\dot{B}B,\quad {{\rm{\Gamma }}}_{33}^{0}=\dot{C}C.\end{array}\end{eqnarray}$
Using the equation ( $\begin{eqnarray}\begin{array}{rcl}{R}_{00} & = & -\left(\displaystyle \frac{\ddot{A}}{A}+\displaystyle \frac{\ddot{B}}{B}+\displaystyle \frac{\ddot{B}}{B}\right),\quad {R}_{11}=A\left(\ddot{A}+\dot{A}\displaystyle \frac{\dot{B}}{B}+\dot{A}\displaystyle \frac{\dot{C}}{C}\right),\\ {R}_{22} & = & B\left(\ddot{B}+\dot{B}\displaystyle \frac{\dot{A}}{A}+\dot{B}\displaystyle \frac{\dot{C}}{C}\right),\,{R}_{33}=C\left(\ddot{C}+\dot{C}\displaystyle \frac{\dot{A}}{A}+\dot{C}\displaystyle \frac{\dot{B}}{B}\right).\end{array}\end{eqnarray}$
Using equations ( $\begin{eqnarray}\begin{array}{rcl}{T}^{00} & = & \rho ,\quad {T}^{11}=\displaystyle \frac{1}{3}\rho {a}^{-2},\\ {T}^{22} & = & \displaystyle \frac{1}{3}\rho {b}^{-2},\quad {T}^{33}=\displaystyle \frac{1}{3}\rho {c}^{-2}.\end{array}\end{eqnarray}$
Note that we will have to consider both the evolution and the constraints systems. Therefore, we have the following two Propositions.Let $k\ne 0$. The Eddington-inspired-Born-Infeld evolution system in $\left(g,q,\rho \right)$ can be written as the following nonlinear second order system
$\begin{eqnarray}\displaystyle \frac{\ddot{A}}{A}+\displaystyle \frac{\ddot{B}}{B}+\displaystyle \frac{\ddot{C}}{C}=0,\end{eqnarray}$
$\begin{eqnarray}\ddot{A}+\dot{A}\displaystyle \frac{\dot{B}}{B}+\dot{A}\displaystyle \frac{\dot{C}}{C}=\displaystyle \frac{1}{k}\left(A-\displaystyle \frac{{a}^{2}}{A}\right),\end{eqnarray}$
$\begin{eqnarray}\ddot{B}+\dot{B}\displaystyle \frac{\dot{A}}{A}+\dot{B}\displaystyle \frac{\dot{C}}{C}=\displaystyle \frac{1}{k}\left(B-\displaystyle \frac{{b}^{2}}{B}\right),\end{eqnarray}$
$\begin{eqnarray}\ddot{C}+\dot{C}\displaystyle \frac{\dot{A}}{A}+\dot{C}\displaystyle \frac{\dot{B}}{B}=\displaystyle \frac{1}{k}\left(C-\displaystyle \frac{{c}^{2}}{C}\right),\end{eqnarray}$
$\begin{eqnarray}\dot{\rho }+\displaystyle \frac{4}{3}\left(\displaystyle \frac{\dot{A}}{A}+\displaystyle \frac{\dot{B}}{B}+\displaystyle \frac{\dot{C}}{C}\right)\rho =0,\end{eqnarray}$
where a, b and c are given by (From equations (
$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\nabla }}}_{\alpha }{T}^{\alpha \beta } & = & {{\rm{\nabla }}}_{\alpha }\left(\displaystyle \frac{4}{3}\rho {u}^{\alpha }{u}^{\beta }+\displaystyle \frac{1}{3}\rho {g}^{\alpha \beta }\right)\\ & = & \displaystyle \frac{4}{3}{{\rm{\nabla }}}_{\alpha }\left(\rho {u}^{\alpha }{u}^{\beta }\right)+\displaystyle \frac{1}{3}{g}^{\alpha \beta }{{\rm{\nabla }}}_{\alpha }\rho \\ & = & \displaystyle \frac{4}{3}\left[{u}^{\beta }{{\rm{\nabla }}}_{\alpha }\left(\rho {u}^{\alpha }\right)+\rho {u}^{\alpha }{{\rm{\nabla }}}_{\alpha }{u}^{\beta }\right]+\displaystyle \frac{1}{3}{g}^{\alpha \beta }{{\rm{\nabla }}}_{\alpha }\rho .\end{array}\end{eqnarray*}$
Using the conservation law ( $\begin{eqnarray}4\left[{u}^{\beta }{{\rm{\nabla }}}_{\alpha }\left(\rho {u}^{\alpha }\right)+\rho {u}^{\alpha }{{\rm{\nabla }}}_{\alpha }{u}^{\beta }\right]+{g}^{\alpha \beta }{{\rm{\nabla }}}_{\alpha }\rho =0.\end{eqnarray}$
Differentiating ${u}_{\beta }{u}^{\beta }=-1$ yields ${u}_{\beta }{{\rm{\nabla }}}_{\alpha }{u}^{\beta }=0$. The contracted multiplication of equation ( $\begin{eqnarray*}4\left(-{{\rm{\nabla }}}_{\alpha }\left(\rho {u}^{\alpha }\right)\right)+{u}^{\alpha }{{\rm{\nabla }}}_{\alpha }\rho =0.\end{eqnarray*}$
This is equivalent to $\begin{eqnarray*}-4\left[{\partial }_{0}\left(\rho {u}^{0}\right)+{{\rm{\Gamma }}}_{i0}^{i}\rho {u}^{0}\right]+{u}^{0}{\partial }_{0}\rho =0.\end{eqnarray*}$
Using equation ( $\begin{eqnarray}-4{\partial }_{0}\left(\rho {u}^{0}\right)-4\left(\displaystyle \frac{\dot{A}}{A}+\displaystyle \frac{\dot{B}}{B}+\displaystyle \frac{\dot{C}}{C}\right)\rho {u}^{0}+{u}^{0}{\partial }_{0}\rho =0.\end{eqnarray}$
Since ${u}_{\alpha }{u}^{\alpha }=-1$ and ${u}^{i}={u}_{i}=0$, we have ${\left({u}^{0}\right)}^{2}={\left({u}_{0}\right)}^{2}=1$. So u0 is different from 0. From equation (Assuming that $\lambda +k\rho \gt 0$ and $\lambda -\tfrac{k}{3}\rho \gt 0$. The Eddington-inspired-Born-Infeld constraints system can be written as
$\begin{eqnarray}\displaystyle \frac{{ABC}}{{abc}}=\lambda +k\rho ,\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{{aBC}}{{Abc}}=\lambda -\displaystyle \frac{k}{3}\rho ,\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{{AbC}}{{aBc}}=\lambda -\displaystyle \frac{k}{3}\rho ,\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{{ABc}}{{abC}}=\lambda -\displaystyle \frac{k}{3}\rho ,\end{eqnarray}$
where a, b and c are given by (From equations (
Let $k\ne 0$, $\lambda +k\rho \gt 0$ and $\lambda -\tfrac{k}{3}\rho \gt 0$. The Eddington-inspired-Born-Infeld evolution system (
$\begin{eqnarray}\displaystyle \frac{\ddot{A}}{A}+\displaystyle \frac{\ddot{B}}{B}+\displaystyle \frac{\ddot{C}}{C}=0,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\ddot{A}}{A}+\displaystyle \frac{\dot{A}}{A}\displaystyle \frac{\dot{B}}{B}+\displaystyle \frac{\dot{A}}{A}\displaystyle \frac{\dot{C}}{C}=\displaystyle \frac{1}{k}\\ \quad \times \left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\ddot{B}}{B}+\displaystyle \frac{\dot{B}}{B}\displaystyle \frac{\dot{A}}{A}+\displaystyle \frac{\dot{B}}{B}\displaystyle \frac{\dot{C}}{C}=\displaystyle \frac{1}{k}\\ \quad \times \left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\ddot{C}}{C}+\displaystyle \frac{\dot{C}}{C}\displaystyle \frac{\dot{A}}{A}+\displaystyle \frac{\dot{C}}{C}\displaystyle \frac{\dot{B}}{B}=\displaystyle \frac{1}{k}\\ \quad \times \left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{array}\end{eqnarray}$
where a, b and c are given by (Using equations (
$\begin{eqnarray}\displaystyle \frac{A}{a}=\displaystyle \frac{B}{b}=\displaystyle \frac{C}{c}={\left(\lambda +k\rho \right)}^{\tfrac{1}{2}}{\left(\lambda -\displaystyle \frac{k}{3}\rho \right)}^{-\tfrac{1}{2}},\end{eqnarray}$
where $\lambda +k\rho \gt 0$ and $\lambda -\tfrac{k}{3}\rho \gt 0$. Moreover, we have $\begin{eqnarray}\displaystyle \frac{{AB}}{{ab}}=\displaystyle \frac{{BC}}{{bc}}=\displaystyle \frac{{AC}}{{ac}}={\left(\lambda +k\rho \right)}^{\tfrac{1}{2}}{\left(\lambda -\displaystyle \frac{k}{3}\rho \right)}^{\tfrac{1}{2}}.\end{eqnarray}$
Therefore, by dividing equation (We assume that $\lambda +k\rho \gt 0$ and $\lambda -\tfrac{k}{3}\rho \gt 0$. Then the Eddington-inspired-Born-Infeld constraints system can be written as an equation of degree 3 for ρ, as follows:
$\begin{eqnarray}{\rho }^{3}-\displaystyle \frac{9\lambda }{k}{\rho }^{2}+\displaystyle \frac{27\left({\lambda }^{2}+1\right)}{{k}^{2}}\rho -\displaystyle \frac{27\left({\lambda }^{3}-\lambda \right)}{{k}^{3}}=0.\end{eqnarray}$
From equations (
We are looking for solutions a, b, c, A, B, C, ρ, of the EiBI equations satisfying the initial conditions21 ) shows that the matter density ρ will be determined by34 ), is satisfied over the entire domain of the solutions a, b, c, A, B, C, ρ if and only if (34 ) is satisfied for t = t0, i.e. the initial data a0, b0, c0, A0, B0, B0, ρ0 in (35 ) satisfy the initial constraint [37]28 )–(31 ) can be written in terms of U and V.
$\begin{eqnarray}\left\{\begin{array}{l}A({t}_{0})={A}_{0},\quad B({t}_{0})={B}_{0},\quad C({t}_{0})={C}_{0}\\ \dot{A}({t}_{0})={C}_{1},\quad \dot{B}({t}_{0})={C}_{2},\quad \dot{C}({t}_{0})={C}_{3}\\ a({t}_{0})={a}_{0},\quad b({t}_{0})={b}_{0},\quad c({t}_{0})={c}_{0}\\ \dot{a}({t}_{0})={C}_{4},\quad \dot{b}({t}_{0})={C}_{5},\quad \dot{c}({t}_{0})={C}_{6}\\ \rho ({t}_{0})={\rho }_{0},\end{array}\right.\end{eqnarray}$
where a0 > 0, b0 > 0, c0 > 0 A0 > 0, B0 > 0, C0 > 0 ρ0 > 0 and Ci, i = 1, 2, 3, 4, 5, 6, are given real constants. Our aim is to prove the existence of solutions on [0, ∞) of the above Cauchy problem. Equation ( $\begin{eqnarray}\rho (t)={\rho }_{0}{{\rm{e}}}^{\left[-\tfrac{4}{3}{\int }_{{t}_{0}}^{t}\left(\displaystyle \frac{\dot{A}}{A}+\displaystyle \frac{\dot{B}}{B}+\displaystyle \frac{\dot{C}}{C}\right)(s){\rm{d}}s\right]},\end{eqnarray}$
which shows that ρ is known once A, B and C are known. It is well known that the EiBI constraint equation ( $\begin{eqnarray}\left\{\begin{array}{l}\lambda -\displaystyle \frac{k}{3}{\rho }_{0}\gt 0,\quad \lambda +k{\rho }_{0}\gt 0,\\ {\rho }_{0}^{3}-\displaystyle \frac{9\lambda }{k}{\rho }_{0}^{2}+\displaystyle \frac{27\left({\lambda }^{2}+1\right)}{{k}^{2}}{\rho }_{0}-\displaystyle \frac{27\left({\lambda }^{3}-\lambda \right)}{{k}^{3}}=0.\end{array}\right.\end{eqnarray}$
By setting the following change of variables $\begin{eqnarray}\left\{\begin{array}{l}U=\displaystyle \frac{\dot{A}}{A},\\ V=\displaystyle \frac{\dot{B}}{B},\\ W=\displaystyle \frac{\dot{C}}{C},\end{array}\right.\end{eqnarray}$
we get $\begin{eqnarray}\left\{\begin{array}{l}\dot{U}=\displaystyle \frac{\ddot{A}}{A}-{U}^{2},\\ \dot{V}=\displaystyle \frac{\ddot{B}}{B}-{V}^{2},\\ \dot{W}=\displaystyle \frac{\ddot{C}}{C}-{W}^{2}.\end{array}\right.\end{eqnarray}$
Hence, the evolution equations (Let $k\ne 0$. The Eddington-inspired-Born-Infeld evolution system (
$\begin{eqnarray}\dot{U}+\dot{V}+\dot{W}+{U}^{2}+{V}^{2}+{W}^{2}=0,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\dot{U}+{U}^{2}+{UV}+{UW}=\displaystyle \frac{1}{k}\\ \quad \times \left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\dot{V}+{V}^{2}+{UV}+{VW}=\displaystyle \frac{1}{k}\\ \quad \times \left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\dot{W}+{W}^{2}+{UW}+{VW}=\displaystyle \frac{1}{k}\\ \quad \times \left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{array}\end{eqnarray}$
where U, V and W are given by (By substituting equations (
We look for solutions $\left(U,V,W\right)$ to system (40 )–(43 ) satisfying at t = t0, the following initial conditions U(t0) = U0, V(t0) = V0, W(t0) = W0 with ${U}_{0}=\tfrac{{C}_{1}}{{A}_{0}}$, ${V}_{0}=\tfrac{{C}_{2}}{{B}_{0}}$, ${W}_{0}=\tfrac{{C}_{3}}{{C}_{0}}$.
Notice that system (28 )–(31 ) imply (40 )–(43 ), and, conversely, the knowledge of $U=\tfrac{\dot{A}}{A}$, $V=\tfrac{\dot{B}}{B}$, $W=\tfrac{\dot{C}}{C}$ gives A, B and C by direct integration.
Assume that $k\gt 0$ and λ satisfy the inequality $\sqrt[3]{2-\sqrt{4{\lambda }^{2}+1}}+\sqrt[3]{2+\sqrt{4{\lambda }^{2}+1}}\leqslant \lambda $ or $k\lt 0$ and λ satisfy the inequality $\sqrt[3]{2-\sqrt{4{\lambda }^{2}+1}}+\sqrt[3]{2+\sqrt{4{\lambda }^{2}+1}}\geqslant \lambda $. Then the initial constraint equation (
$\begin{eqnarray}\begin{array}{rcl}{\rho }_{0} & = & 3\left[\displaystyle \frac{\lambda }{k}+\sqrt[3]{-\displaystyle \frac{2}{{k}^{3}}+\displaystyle \frac{1}{{k}^{2}| k| }\sqrt{4{\lambda }^{2}+1}}\right.\\ & & \left.-\sqrt[3]{\displaystyle \frac{2}{{k}^{3}}+\displaystyle \frac{1}{{k}^{2}| k| }\sqrt{4{\lambda }^{2}+1}}\right].\end{array}\end{eqnarray}$
By setting ${\gamma }_{0}={\rho }_{0}-\tfrac{3\lambda }{k}$ i.e. ${\rho }_{0}={\gamma }_{0}+\tfrac{3\lambda }{k}$, equation (
$\begin{eqnarray}{\gamma }_{0}^{3}+\displaystyle \frac{27}{{k}^{2}}{\gamma }_{0}+\displaystyle \frac{108\lambda }{{k}^{3}}=0.\end{eqnarray}$
According to the Cardan method, the discriminant of this polynomial of degree 3 is given by ${\bigtriangleup }_{1}=\tfrac{54}{{k}^{2}| k| }\sqrt{4{\lambda }^{2}+1}$. Therefore, we get the value of ${\gamma }_{0}$, given by $\begin{eqnarray}\begin{array}{l}{\gamma }_{0}=3\left[\sqrt[3]{-\displaystyle \frac{2}{{k}^{3}}+\displaystyle \frac{1}{{k}^{2}| k| }\sqrt{4{\lambda }^{2}+1}}\right.\\ \,\left.-\,\sqrt[3]{\displaystyle \frac{2}{{k}^{3}}+\displaystyle \frac{1}{{k}^{2}| k| }\sqrt{4{\lambda }^{2}+1}}\right].\end{array}\end{eqnarray}$
Then, we obtain the value of ${\rho }_{0}\gt 0$ for values of k and λ fixed under the prescribed assumptions.□According to equation (37 ), we have $\lambda -\tfrac{k}{3}{\rho }_{0}\gt 0$ and λ + kρ0 > 0, this makes it possible to determine the domains of the Eddington parameter k and the dimensionless constant λ, given by the following equation
$\begin{eqnarray}\left\{\begin{array}{l}\mathrm{if}\quad k\gt 0,\quad \mathrm{then}\quad \lambda \in \left(\displaystyle \frac{k}{3}{\rho }_{0},\infty \right),\\ \mathrm{if}\quad k\lt 0,\quad \mathrm{then}\quad \lambda \in \left(-k{\rho }_{0},\infty \right).\end{array}\right.\end{eqnarray}$
Let $k\gt 0$ and $\lambda \in \left(\tfrac{k}{3}{\rho }_{0},\infty \right)$ or $k\lt 0$ and $\lambda \in (-k{\rho }_{0},\infty )$. The Eddington-inspired-Born-Infeld evolution system (
$\begin{eqnarray}\begin{array}{l}\dot{\chi }=-{\chi }^{2}+\displaystyle \frac{3}{k}\\ \times \,\left\{1-\left[\lambda -\displaystyle \frac{k}{3}{\rho }_{0}{{\rm{e}}}^{\left(-\tfrac{4}{3}{\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)}\right]{\left[\lambda +k{\rho }_{0}{{\rm{e}}}^{\left(-\tfrac{4}{3}{\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)}\right]}^{-1}\right\},\end{array}\end{eqnarray}$
where $\chi =U+V+W$ and ${\rho }_{0}$ is given by (By summing up equations (
$\begin{eqnarray}{UV}+{UW}+{VW}=\displaystyle \frac{3}{2k}\left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right].\end{eqnarray}$
Moreover ( $\begin{eqnarray}\dot{U}+\dot{V}+\dot{W}=-{\left(U+V+W\right)}^{2}+2\left({UV}+{UW}+{VW}\right).\end{eqnarray}$
By setting $\chi =U+V+W$ and using (To advance with our analysis, we will first establish an existence theorem of the solution χ for the nonlinear ODE (48 ).
There is a variety of methods that can be used to solve equation (48 ). We can cite: (1) the solution-tube method, (2) the structure-preserving method, (3) the Lie group method, (4) the symmetry method, (5) the multiplication method, (6) the generalized multi-symplectic method, to name but the few. The solution-tube method has been used in [38] for solving systems of ordinary differential and time-scale equations. In [39], Hu and his collaborators used the preserving structure method to design a strategy for removing space debris by a tether tow. The Lie group method has been developed in [40] to efficiently solve a nonlinear modified Gardner type partial differential equation (PDE) and its time-fractional form. In [41], the authors used the symmetry method to solve generalized KdV-Burgers-Kuramoto equations and its fractional version. The multiplication method has been developed in [42] to find analytical solutions of a (3+1)-dimensional sine-Gordon and a sinh-Gordon equations. In [43, 44], the authors combined the Runge-Kutta numerical method to a generalized multi-symplectic method for solving some PDEs arising from physical phenomena such as wave propagation and signal processes.
In this work, we will mainly use the solution-tube method.
For this purpose, we need the following notations.
(N-1) ${C}^{1}({\mathbb{R}},{\mathbb{R}})$, the set of differentiable functions on ${\mathbb{R}}$ such that $\dot{\chi }$ is continuous on ${\mathbb{R}}$. This space is endowed with the norm $\parallel \chi {\parallel }_{{C}^{1}({\mathbb{R}},{\mathbb{R}})}=\max \{\parallel \chi {\parallel }_{0},\parallel \dot{\chi }{\parallel }_{0}\}$, where $\parallel {\chi }^{(i)}{\parallel }_{0}={\sup }_{t\in {\mathbb{R}}}\parallel \chi (t)\parallel $. (N-2) ${L}^{1}({\mathbb{R}},{\mathbb{R}})$, is the space of measurable functions χ such that ∣χ∣ is integrable.
We now solve the following problem51 ) in ${C}^{1}({\mathbb{R}},{\mathbb{R}})$. We now give the necessary mathematical tools, for the resolution of problem (51 ).
$\begin{eqnarray}\left\{\begin{array}{l}\dot{\chi }(t)=F\left(t,\chi (t)\right),\quad t\in {\mathbb{R}},\\ \chi ({t}_{0})=\chi (T),\end{array}\right.\end{eqnarray}$
where T > t0 and F is the function defined by $\begin{eqnarray}\begin{array}{l}F\left(t,\chi (t)\right)=-{\chi }^{2}+\displaystyle \frac{3}{k}\left\{1-\left[\lambda -\displaystyle \frac{k}{3}{\rho }_{0}{{\rm{e}}}^{\left(-\tfrac{4}{3}{\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)}\right]\right.\\ \quad \left.\times {\left[\lambda +k{\rho }_{0}{{\rm{e}}}^{\left(-\tfrac{4}{3}{\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)}\right]}^{-1}\right\}.\end{array}\end{eqnarray}$
We will find the solution to problem (Let's introduce a notion of solution-tube for this problem.
Let $(v,M)\in {C}^{1}({\mathbb{R}},{\mathbb{R}})\times {C}^{1}({\mathbb{R}},[0,\infty ))$. $(v,M)$ is called a solution-tube of (
| i | (i)$\left(x-v(t)\right)\left(F(t,x)-\dot{v}(t)\right)\geqslant M(t)\dot{M}(t)$ for all $t\in {\mathbb{R}}$, $x\in {\mathbb{R}}$ such that $| x-v(t)| $=$M(t);$ |
| ii | (ii)$\dot{v}(t)=F(t,v(t))$ and $\dot{M}(t)=0$ for all $t\in {\mathbb{R}}$ such that $M(t)=0;$ |
| iii | (iii)$| v(T)-v({t}_{0})| \leqslant M(T)-M({t}_{0})$. |
The results below will help to establish the existence theorem for problem (51 ).
[38] Let W be an open set of ${\mathbb{R}}$ and t be a real number. If $g:{\mathbb{R}}\longrightarrow {\mathbb{R}}$ is differentiable in t and if $f:W\longrightarrow {\mathbb{R}}$ is differentiable in $g(t)\in W$, then $f\circ g$ is differentiable in t and $\dot{\overbrace{\left(f\circ g\right)}}(t)=\langle \dot{f}(g(t)),\dot{g}(t)\rangle $.
Theorem 2 gives the following example.
Let a function $\chi :{\mathbb{R}}\longrightarrow {\mathbb{R}}$ differentiable in $t\in {\mathbb{R}}$. We know that the function $| \bullet | :{\mathbb{R}}-\{0\}\longrightarrow (0,\infty )$ is differentiable. If $| \chi (t)| \gt 0$, then by continuity, there exists $\delta \gt 0$ such that $| \chi (s)| \gt 0$ for $s\in \left(t-\delta ,t+\delta \right)$. Moreover, by virtue of theorem
The remark below allows us to define explicitly the exponential function which is of capital utility for the proof of the existence theorem of the Problem (51 ).
For $\epsilon \gt 0$, the exponential function ${{\rm{e}}}_{\epsilon }(\bullet ,{t}_{1}):{\mathbb{R}}\longrightarrow {\mathbb{R}}$ can be defined as being the unique solution to the problem with the initial value
$\begin{eqnarray}\left\{\begin{array}{l}\dot{\chi }(t)=\epsilon \chi (t),\quad t\in {\mathbb{R}},\\ \chi ({t}_{1})=1.\end{array}\right.\end{eqnarray}$
More explicitly, the exponential function ${{\rm{e}}}_{\epsilon }(\bullet ,{t}_{1})$ can be given by the formula $\begin{eqnarray}{{\rm{e}}}_{\epsilon }(t,{t}_{1})={{\rm{e}}}^{\left({\int }_{{t}_{1}}^{t}{\zeta }_{\epsilon }(s){\rm{d}}s\right)},\end{eqnarray}$
where for $h\geqslant 0$, we define ${\zeta }_{\epsilon }(h)$ by $\begin{eqnarray}{\zeta }_{\epsilon }(h)=\left\{\begin{array}{l}\epsilon ,\quad \mathrm{if}\quad h=0,\\ \displaystyle \frac{\mathrm{log}(1+h\epsilon )}{h},\quad \mathrm{otherwise}.\end{array}\right.\end{eqnarray}$
This remark also allows us to state the following Theorem.
[38]If $g\in {L}^{1}({\mathbb{R}},{\mathbb{R}})$, the function $\chi :{\mathbb{R}}\longrightarrow {\mathbb{R}}$ defined by
$\begin{eqnarray}\begin{array}{l}\chi (t)={{\rm{e}}}_{1}(t,{t}_{0})\left[\displaystyle \frac{{{\rm{e}}}_{1}(T,{t}_{0})}{1-{{\rm{e}}}_{1}(T,{t}_{0})}\right.\\ \quad \times {\displaystyle \int }_{[{t}_{0},T)\cap {\mathbb{R}}}^{}\displaystyle \frac{g(s)}{{{\rm{e}}}_{1}(s,{t}_{0})}{\rm{d}}s+{\displaystyle \int }_{[{t}_{0},t)\cap {\mathbb{R}}}^{}\displaystyle \frac{g(s)}{{{\rm{e}}}_{1}(s,{t}_{0})}{\rm{d}}s],\end{array}\end{eqnarray}$
is a solution to the problem $\begin{eqnarray}\left\{\begin{array}{l}\dot{\chi }(t)-\chi (t)=g(t),\quad t\in {\mathbb{R}},\\ \chi ({t}_{0})=\chi (T).\end{array}\right.\end{eqnarray}$
Let a function $L\in {C}^{1}({\mathbb{R}},{\mathbb{R}})$ such that $\dot{L}(t)\gt 0$ for all $t\in \{s\in {\mathbb{R}}:L(s)\gt 0\}$. If $L({t}_{0})\geqslant L(T)$, then $L(t)\leqslant 0$, for all $t\in {\mathbb{R}}$.
Suppose that there exists $t\in {\mathbb{R}}$ such that $L(t)\gt 0$, then there exists ${t}_{1}\in {\mathbb{R}}$ such that $L({t}_{1})={\max }_{t\in {\mathbb{R}}}L(t)\gt 0$.
| • | – If ${t}_{1}\lt T$, then there exists an interval $[{t}_{1},{t}_{2}]$ such that $L(t)\gt 0$ for all $t\in [{t}_{1},{t}_{2}]$. So, $0\lt {\int }_{{t}_{1}}^{{t}_{2}}\dot{L}(s){\rm{d}}{s}=L({t}_{2})-L({t}_{1})$, which contradicts the maximality of $L({t}_{1})$. |
| • | – If ${t}_{1}=T$, then by hypothesis of lemma |
| • | – By taking ${t}_{1}={t}_{0}$, by what precedes, we would find that $L({t}_{0})\leqslant 0$. |
Let us impose the following hypothesis:
(H) There exists (v, M) a solution-tube of (51 ).
In order to prove the existence Theorem, we will have recourse to the following modified problem54 ).
$\begin{eqnarray}\left\{\begin{array}{l}\dot{\chi }(t)-\chi (t)=F(t,\overline{\chi }(t))-\overline{\chi }(t)\quad \mathrm{for}\quad \mathrm{all}\quad t\in {\mathbb{R}},\\ \chi ({t}_{0})=\chi (T),\end{array}\right.\end{eqnarray}$
where $\begin{eqnarray}\overline{\chi }(t)=\left\{\begin{array}{ll}\displaystyle \frac{M(t)}{| \chi (t)-v(t)| }(\chi (t)-v(t))+v(t) & \mathrm{if}\,| \chi (t)-v(t)| \gt M(t),\\ \chi (t) & \mathrm{otherwise}.\end{array}\right.\end{eqnarray}$
We define the operator ${T}_{{P}^{* }}:C\left({\mathbb{R}},{\mathbb{R}}\right)\longrightarrow C\left({\mathbb{R}},{\mathbb{R}}\right)$ by $\begin{eqnarray}\begin{array}{l}{T}_{{P}^{* }}(\chi )(t)={{\rm{e}}}_{1}(t,{t}_{0})\left[\displaystyle \frac{{{\rm{e}}}_{1}(T,{t}_{0})}{1-{{\rm{e}}}_{1}(T,{t}_{0})}\right.\\ \quad \times {\displaystyle \int }_{[{t}_{0},T)\cap {\mathbb{R}}}^{}\displaystyle \frac{\left(F(s,\overline{\chi }(s))-\overline{\chi }(s)\right)}{{{\rm{e}}}_{1}(s,{t}_{0})}{\rm{d}}s\\ \quad +\,{\displaystyle \int }_{[{t}_{0},t)\cap {\mathbb{R}}}^{}\displaystyle \frac{\left(F(s,\overline{\chi }(s))-\overline{\chi }(s)\right)}{{{\rm{e}}}_{1}(s,{t}_{0})}{\rm{d}}s],\end{array}\end{eqnarray}$
where the function e1(•, t0) is defined by (We can easily see that for k ≠ 0, $\lambda -\tfrac{k}{3}\rho \gt 0$ and λ + kρ > 0, the function $F:{\mathbb{R}}\times {\mathbb{R}}\longrightarrow {\mathbb{R}}$ defined in (52 ) is continuous. Therefore we state the following Proposition.
Let $F:{\mathbb{R}}\times {\mathbb{R}}\longrightarrow {\mathbb{R}}$ be a continuous function. If the hypothesis $(H)$ is satisfied, then the operator ${T}_{{P}^{* }}$ is compact.
This result is established in two main steps.
Step 1. Let us first show the continuity of the operator ${T}_{{P}^{* }}$. Let ${\{{\chi }_{n}\}}_{n\in {\mathbb{N}}}$ be a sequence in $C({\mathbb{R}},{\mathbb{R}})$ converging towards an element $\chi \in C({\mathbb{R}},{\mathbb{R}})$.
Using equation (
$\begin{eqnarray*}\begin{array}{l}\parallel {T}_{{P}^{* }}({\chi }_{n})(t)-{T}_{{P}^{* }}(\chi )(t)\parallel \leqslant (1+C)\parallel {{\rm{e}}}_{1}(t,{t}_{0})\parallel \\ \quad \times \parallel {\displaystyle \int }_{[{t}_{0},T)\cap {\mathbb{R}}}^{}\displaystyle \frac{F(s,{\overline{\chi }}_{n}(s))-F(s,\overline{\chi }(s))-({\overline{\chi }}_{n}(s)-\overline{\chi }(s))}{{{\rm{e}}}_{1}(s,{t}_{0})}{\rm{d}}s\parallel \\ \quad \leqslant \displaystyle \frac{\left(1+C\right)K}{M}{\displaystyle \int }_{[{t}_{0},T)\cap {\mathbb{R}}}^{}\left(| F(s,{\overline{\chi }}_{n}(s))-F(s,\overline{\chi }(s))| \right.\\ \quad \left.+| {\overline{\chi }}_{n}(s)-\overline{\chi }(s)| \right){\rm{d}}s,\end{array}\end{eqnarray*}$
where $K:={\max }_{t\in {\mathbb{R}}}| {{\rm{e}}}_{1}(t,{t}_{0})| $, $M:={\min }_{{t}_{1}\in {\mathbb{R}}}| {{\rm{e}}}_{1}({t}_{1},{t}_{0})| $ and $C:=\left|\tfrac{{{\rm{e}}}_{1}(T,{t}_{0})}{1-{{\rm{e}}}_{1}(T,{t}_{0})}\right|$. Since there exists a constant $R\gt 0$ such that $| \overline{\chi }| \lt R$, there exits an index N such that $| {\overline{\chi }}_{n}| \lt R$ for all $n\gt N$. Thus, F is uniformly continuous over ${\mathbb{R}}\times {B}_{R}(0)$. So, for $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for all $x,y\in {\mathbb{R}}$, we have $| x-y| \lt \delta \lt \tfrac{\epsilon M}{2K(1+C)(T-{t}_{0})}$, $| F(s,x)-F(s,y)| \lt \delta \lt \tfrac{\epsilon M}{2K(1+C)(T-{t}_{0})}$ for all $s\in {\mathbb{R}}$. Thus, it is possible to find an index $\overline{N}\gt N$ such that $| {\hat{\chi }}_{n}-\overline{\chi }| \lt \delta $ for $n\gt \overline{N}$ such that $\parallel {T}_{{P}^{* }}({\chi }_{n})(t)-{T}_{{P}^{* }}(\chi )(t)\parallel \lt 2\tfrac{K(1+C)}{M}{\int }_{[{t}_{0},T)\cap {\mathbb{R}}}^{}\tfrac{\epsilon M}{2K(1+C)(T-{t}_{0})}{\rm{d}}s\leqslant \epsilon $. Thus ${T}_{{P}^{* }}$ is continuous.
Step 2. Let us now show that the set ${T}_{{P}^{* }}\left(C({\mathbb{R}},{\mathbb{R}})\right)$ is relatively compact. Consider a sequence ${\{{y}_{n}\}}_{n\in {\mathbb{N}}}$ in ${T}_{{P}^{* }}\left(C({\mathbb{R}},{\mathbb{R}})\right)$. For all $n\in {\mathbb{N}}$, there exits ${\chi }_{n}\in C({\mathbb{R}},{\mathbb{R}})$ such that ${y}_{n}={T}_{{P}^{* }}({\chi }_{n})$. Using equation (
$\begin{eqnarray*}\begin{array}{l}\parallel {T}_{{P}^{* }}({\chi }_{n})(t)\parallel \leqslant \tfrac{K(1+C)}{M}\\ \quad \times \left({\displaystyle \int }_{[{t}_{0},T)\cap {\mathbb{R}}}^{}| F(s,{\overline{\chi }}_{n}(s)| {\rm{d}}s+{\displaystyle \int }_{[{t}_{0},T){\mathbb{R}}}^{}| {\overline{\chi }}_{n}(s)| {\rm{d}}s\right).\end{array}\end{eqnarray*}$
By definition, there exists $R\gt 0$ such that $| {\overline{\chi }}_{n}(s)| \leqslant R$ for all $s\in {\mathbb{R}}$ and all $n\in {\mathbb{N}}$. The function F being uniformly continuous over ${\mathbb{R}}\times {B}_{R}(0)$, we can deduce the existence of a constant $A\gt 0$ such that $| F(s,{\overline{\chi }}_{n}(s))| \leqslant A$ for all $s\in {\mathbb{R}}$ and $n\in {\mathbb{N}}$. Thus, the sequence ${\{{y}_{n}\}}_{n\in {\mathbb{N}}}$ is uniformly bounded. Note also from the above fact that for ${t}_{1},{t}_{2}\in {\mathbb{R}}$, we have $\begin{eqnarray*}\begin{array}{l}\parallel {T}_{{P}^{* }}({\chi }_{n})({t}_{2})-{T}_{{P}^{* }}({\chi }_{n})({t}_{1})\parallel \\ \quad \leqslant B\parallel {{\rm{e}}}_{1}({t}_{2},{t}_{0})-{{\rm{e}}}_{1}({t}_{1},{t}_{0})\parallel \\ \quad +\,K\parallel {\displaystyle \int }_{[{t}_{1},{t}_{2})\cap {\mathbb{R}}}^{}\displaystyle \frac{F(s,{\overline{\chi }}_{n}(s))-{\overline{\chi }}_{n}(s)}{{{\rm{e}}}_{1}(s,{t}_{0})}{\rm{d}}s\parallel \\ \quad \lt B\parallel {{\rm{e}}}_{1}({t}_{2},{t}_{0})-{{\rm{e}}}_{1}({t}_{1},{t}_{0})\parallel \\ \quad +\displaystyle \frac{K(A+R)}{M}| {t}_{2}-{t}_{1}| ,\end{array}\end{eqnarray*}$
where B is a constant which can be chosen such that it is greater than $\begin{eqnarray*}\begin{array}{l}\mathop{\sup }\limits_{n\in {\mathbb{N}}}\parallel \tfrac{{{\rm{e}}}_{1}(T,{t}_{0})}{1-{{\rm{e}}}_{1}(T,{t}_{0})}{\displaystyle \int }_{[{t}_{0},T)\cap {\mathbb{R}}}^{}\tfrac{\left(F(s,{\overline{\chi }}_{n}(s))-{\overline{\chi }}_{n}(s)\right)}{{{\rm{e}}}_{1}(s,{t}_{0})}{\rm{d}}s\\ \quad +{\displaystyle \int }_{[{t}_{0},{t}_{1})\cap {\mathbb{R}}}^{}\tfrac{\left(F(s,{\overline{\chi }}_{n}(s))-{\overline{\chi }}_{n}(s)\right)}{{{\rm{e}}}_{1}(s,{t}_{0})}{\rm{d}}s\parallel .\end{array}\end{eqnarray*}$
Thus the sequence ${\{{y}_{n}\}}_{n\in {\mathbb{N}}}$ is also equicontinuous and by virtue of the Arzel$\grave{a}$–Ascoli Theorem, ${\{{y}_{n}\}}_{n\in {\mathbb{N}}}$ has a convergent subsequence.
Steps 1 and 2 show that ${T}_{{P}^{* }}$ is compact.□
We can now state and prove the existence Theorem.
If the hypothesis $(H)$ is satisfied, then problem (
where
$\begin{eqnarray*}\begin{array}{l}T(v,M)=\left\{\chi \in {C}^{1}\left({\mathbb{R}},{\mathbb{R}}\right):| \chi (t)\right.\\ \quad \left.-v(t)| \leqslant M(t)\quad \forall t\in {\mathbb{R}}\right\}.\end{array}\end{eqnarray*}$
Step 1. We need to show that a.e. on the set $A=\{t\in {\mathbb{R}}:| \chi (t)-v(t)| \gt M(t)\}$, we have
$\begin{eqnarray}\begin{array}{l}\dot{\overbrace{\left(| \chi (t)-v(t)| -M(t)\right)}}\\ \quad \geqslant \displaystyle \frac{\left(\chi (t)-v(t)\right)\left(\dot{\chi }(t)-\dot{v}(t)\right)}{| \chi (t)-v(t)| }-\dot{M}(t).\end{array}\end{eqnarray}$
This is done by using Example 1. Step 2. We will now show that almost everywhere on A, we have $\dot{\overbrace{\left(| \chi (t)-v(t)| -M(t)\right)}}\gt 0$.
*If $M(t)\gt 0$, then by the hypothesis of solution-tube, and by (
$\begin{eqnarray*}\begin{array}{l}\dot{\overbrace{\left(| \chi (t)-v(t)| -M(t)\right)}}\\ \,\geqslant \displaystyle \frac{\left(\chi (t)-v(t)\right)\left(F(t,\overline{\chi }(t))-\left(\overline{\chi }(t)-\chi (t)\right)-\dot{v}(t)\right)}{| \chi (t)-v(t)| }\\ \,-\,\dot{M}(t)\\ \quad =\displaystyle \frac{\left(\overline{\chi }(t)-v(t)\right)\left(F(t,\overline{\chi }(t))-\dot{v}(t)\right)}{M(t)}\\ \quad -\dot{M}(t)-\left(M(t)-| \chi (t)-v(t)| \right)\\ \quad \gt \displaystyle \frac{M(t)\dot{M}(t)}{M(t)}-\dot{M}(t)=0.\end{array}\end{eqnarray*}$
Thus, we have shown that $\dot{\overbrace{\left(| \chi (t)-v(t)| -M(t)\right)}}\gt 0$, almost everywhere on A. *If $M(t)=0$, by the hypothesis of the solution-tube and by (
$\begin{eqnarray*}\begin{array}{l}\dot{\overbrace{\left(| \chi (t)-v(t)| -M(t)\right)}}\\ \quad \geqslant \displaystyle \frac{\left(\chi (t)-v(t)\right)\left(F(t,\overline{\chi }(t))-\left(\overline{\chi }(t)\,-\,\chi (t)\right)-\dot{v}(t)\right)}{| \chi (t)-v(t)| }\\ -\dot{M}(t)\\ \quad =\displaystyle \frac{\left(\chi (t)-v(t)\right)\left(F(t,\chi (t))-\dot{v}(t)\right)}{| \chi (t)-v(t)| }\\ \,-\,\dot{M}(t)+| \chi (t)-v(t)| \\ \quad \gt 0.\end{array}\end{eqnarray*}$
Hence, almost everywhere on A, we have $\dot{\overbrace{\left(| \chi (t)-v(t)| -M(t)\right)}}\gt 0$. Step 3. By setting $L(t)=| \chi (t)-v(t)| -M(t)$, we have for all $t\in \{t\in {\mathbb{R}}:L(t)\gt 0\}$, $\dot{L}(t)\gt 0$. Moreover, by the hypothesis of the solution-tube, notice that
$\begin{eqnarray*}\begin{array}{l}L(T)-L({t}_{0})\leqslant | v(T)-v({t}_{0})| \\ \quad -\,\left(M(T)-M({t}_{0})\right)\leqslant 0.\end{array}\end{eqnarray*}$
Thus, the assumptions of Lemma As F is continuous, we also have the following result.
[38] Let $F:{\mathbb{R}}\times {\mathbb{R}}\longrightarrow {\mathbb{R}}$ be a continuous function. If there exists non-negative constants α and K such that
$\begin{eqnarray*}\left|\tfrac{F(t,p)-p}{h(t)}\right|\leqslant 2\alpha {pF}(t,p)+K,\end{eqnarray*}$
for all $(t,p)\in {\mathbb{R}}\times {\mathbb{R}}$ where $h:{\mathbb{R}}\longrightarrow {\mathbb{R}}$ is defined by $h(t)={{\rm{e}}}^{\left({\int }_{{t}_{0}}^{t}{\varepsilon }_{-1}(s){\rm{d}}s\right)}$ with $\begin{eqnarray}{\zeta }_{\epsilon }(s)=\left\{\begin{array}{l}-1,\quad \mathrm{if}\quad s=0,\\ \displaystyle \frac{\mathrm{log}(1-s)}{s},\quad \mathrm{otherwise}.\end{array}\right.\end{eqnarray}$
Then the problem (In particular $F:\{0,1,...,N,N+1\}\times {\mathbb{R}}\longrightarrow {\mathbb{R}}$ (N being a positive integer) is continuous, theorem 5 yields another important result.
[38] Let $F:\{0,1,...,N,N+1\}\times {\mathbb{R}}\longrightarrow {\mathbb{R}}$ be a continuous function. If there exists non-negative constants α and K such that
$\begin{eqnarray*}\tfrac{| F(t,p)-p| }{{2}^{t+1}}\leqslant 2\alpha {pF}(t,p)+K,\end{eqnarray*}$
for all $(t,p)\in \{0,1,...,N,N\,+\,1\}\times {\mathbb{R}}$. Then the following system of finite difference equations $\begin{eqnarray}\left\{\begin{array}{l}\chi (t+1)-\chi (t)=F(t,\chi (t)),\quad t\in \{0,1,...,N,N+1\},\\ \chi (0)=\chi (N+1),\end{array}\right.\end{eqnarray}$
has a solution.Since we have just shown that χ exists, then equations (41 )–(43 ) of lemma 3 admit solutions given by the following Lemma.
Let $k\gt 0$ and $\lambda \in (\tfrac{k}{3}{\rho }_{0},\infty )$ or $k\lt 0$ and $\lambda \in (-k{\rho }_{0},\infty )$. The Eddington-inspired-Born-Infeld evolution system (
$\begin{eqnarray}\begin{array}{rcl}U(t) & = & \left[{\displaystyle \int }_{{t}_{0}}^{t}\left(1-\displaystyle \frac{\lambda -\tfrac{k}{3}\rho (s)}{\lambda +k\rho (s)}\right){{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}{s}\right)}{\rm{d}}s\right.\\ & & \left.+\displaystyle \frac{{C}_{1}}{{A}_{0}}\right]{{\rm{e}}}^{\left(-{\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}V(t) & = & \left[{\displaystyle \int }_{{t}_{0}}^{t}\left(1-\displaystyle \frac{\lambda -\tfrac{k}{3}\rho (s)}{\lambda +k\rho (s)}\right){{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}{s}\right)}{\rm{d}}s\right.\\ & & \left.+\displaystyle \frac{{C}_{2}}{{B}_{0}}\right]{{\rm{e}}}^{\left(-{\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}W(t) & = & \left[{\displaystyle \int }_{{t}_{0}}^{t}\left(1-\displaystyle \frac{\lambda -\tfrac{k}{3}\rho (s)}{\lambda +k\rho (s)}\right){{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}{s}\right)}{\rm{d}}s\right.\\ & & \left.+\displaystyle \frac{{C}_{3}}{{C}_{0}}\right]{{\rm{e}}}^{\left(-{\displaystyle \int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)},\end{array}\end{eqnarray}$
where ${C}_{1}={A}_{0}{U}_{0}$, ${C}_{2}={B}_{0}{V}_{0}$ and ${C}_{3}={C}_{0}{W}_{0}$ are given real constants, $\rho (t)={\rho }_{0}{{\rm{e}}}^{\left(-\tfrac{4}{3}{\int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)}$, $\chi =U+V+W$, ${\rho }_{0}$ is given by (Since χ exists, equations (
$\begin{eqnarray}\dot{U}+U\chi =\displaystyle \frac{1}{k}\left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{eqnarray}$
$\begin{eqnarray}\dot{V}+V\chi =\displaystyle \frac{1}{k}\left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{eqnarray}$
$\begin{eqnarray}\dot{W}+W\chi =\displaystyle \frac{1}{k}\left[1-\left(\lambda -\displaystyle \frac{k}{3}\rho \right){\left(\lambda +k\rho \right)}^{-1}\right],\end{eqnarray}$
which is a linear system of ODEs with non-constant coefficients. Consequently, by using the constant variation method one arrives at the solution given by (We have just proved that the auxiliary metric q exists. Indeed, according to the above discussion, we obtain the component functions A, B and C which are given respectively by64 ), (65 ) and (66 ).
$\begin{eqnarray}\begin{array}{rcl}A(t) & = & {A}_{0}{{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}U(s){\rm{d}}s\right)},\\ B(t) & = & {B}_{0}{{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}V(s){\rm{d}}s\right)},\\ C(t) & = & {C}_{0}{{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}W(s){\rm{d}}s\right)},\end{array}\end{eqnarray}$
where U, V and W are defined respectively by (Now, we are going to prove the existence of the physical metric g. Using equation (32 ), the expressions of a, b and c are given by
$\begin{eqnarray}\begin{array}{rcl}a(t) & = & {A}_{0}{\left(\lambda +k\rho \right)}^{-\tfrac{1}{2}}{\left(\lambda -\displaystyle \frac{k}{3}\rho \right)}^{\tfrac{1}{2}}{{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}U(s){\rm{d}}s\right)},\\ b(t) & = & {B}_{0}{\left(\lambda +k\rho \right)}^{-\tfrac{1}{2}}{\left(\lambda -\displaystyle \frac{k}{3}\rho \right)}^{\tfrac{1}{2}}{{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}V(s){\rm{d}}s\right)},\\ c(t) & = & {C}_{0}{\left(\lambda +k\rho \right)}^{-\tfrac{1}{2}}{\left(\lambda -\displaystyle \frac{k}{3}\rho \right)}^{\tfrac{1}{2}}{{\rm{e}}}^{\left({\displaystyle \int }_{{t}_{0}}^{t}W(s){\rm{d}}s\right)}.\end{array}\end{eqnarray}$
The following theorem gives a general summary of the existence of the physical metric g, the auxiliary metric q and the matter density ρ.Let $k\gt 0$ and $\lambda \in \left(\tfrac{k}{3}{\rho }_{0},\infty \right)$ or $k\lt 0$ and $\lambda \in (-k{\rho }_{0},\infty )$. Assume that ${a}_{0}\gt 0$, ${b}_{0}\gt 0$ and ${c}_{0}\gt 0$. Then, there exists a solution $(g,q,\rho )$ given by
$\begin{eqnarray}\begin{array}{rcl}g & = & -{\rm{d}}{t}^{2}+\left(\displaystyle \frac{\lambda -\tfrac{k}{3}\rho }{\lambda +k\rho }\right)\left\{{A}_{0}^{2}{{\rm{e}}}^{\left(2{\displaystyle \int }_{{t}_{0}}^{t}U(s){\rm{d}}s\right)}{\left({\rm{d}}{x}^{1}\right)}^{2}\right.\\ & & +{B}_{0}^{2}{{\rm{e}}}^{\left(2{\displaystyle \int }_{{t}_{0}}^{t}V(s){\rm{d}}s\right)}{\left({\rm{d}}{x}^{2}\right)}^{2}\\ & & \left.{\left.+{C}_{0}^{2}{{\rm{e}}}^{\left(2{\displaystyle \int }_{{t}_{0}}^{t}W(s){\rm{d}}s\right)}({\rm{d}}{x}^{3}\right)}^{2}\right\},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}q=-{\rm{d}}{t}^{2}+{A}_{0}^{2}{{\rm{e}}}^{\left(2{\displaystyle \int }_{{t}_{0}}^{t}U(s){\rm{d}}s\right)}{\left({\rm{d}}{x}^{1}\right)}^{2}\\ \quad +{B}_{0}^{2}{{\rm{e}}}^{\left(2{\displaystyle \int }_{{t}_{0}}^{t}V(s){\rm{d}}s\right)}{\left({\rm{d}}{x}^{2}\right)}^{2}\\ \quad +{C}_{0}^{2}{{\rm{e}}}^{\left(2{\displaystyle \int }_{{t}_{0}}^{t}W(s){\rm{d}}s\right)}{\left({\rm{d}}{x}^{3}\right)}^{2},\end{array}\end{eqnarray}$
$\begin{eqnarray}\rho (t)={\rho }_{0}{\left(\displaystyle \frac{{A}_{0}{B}_{0}{C}_{0}}{{ABC}}\right)}^{\tfrac{4}{3}},\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{l}{\rho }_{0}=3\left[\displaystyle \frac{\lambda }{k}+\sqrt[3]{-\displaystyle \frac{2}{{k}^{3}}+\displaystyle \frac{1}{{k}^{2}| k| }\sqrt{4{\lambda }^{2}+1}}\right.\\ \quad \left.-\sqrt[3]{\displaystyle \frac{2}{{k}^{3}}+\displaystyle \frac{1}{{k}^{2}| k| }\sqrt{4{\lambda }^{2}+1}}\right],\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{A}_{0} & = & {a}_{0}\sqrt{\displaystyle \frac{\lambda +k{\rho }_{0}}{\lambda -\tfrac{k}{3}{\rho }_{0}}},\quad {B}_{0}={b}_{0}\sqrt{\displaystyle \frac{\lambda +k{\rho }_{0}}{\lambda -\tfrac{k}{3}{\rho }_{0}}}\quad \mathrm{and}\\ \quad {C}_{0} & = & {c}_{0}\sqrt{\displaystyle \frac{\lambda +k{\rho }_{0}}{\lambda -\tfrac{k}{3}{\rho }_{0}}},\end{array}\end{eqnarray}$
and U, V and W are defined respectively by (We now study the asymptotic behaviour of the solutions obtained in Theorem 7.
3. Asymptotic behaviour
From values of the functions a, b, c, A, B, C and ρ, we now define important cosmological indicators in the directions of the physical metric g and the auxiliary metric q.
The volumes of scalar factors ${{ \mathcal V }}_{q}$ and ${{ \mathcal V }}_{g}$ are defined respectively by
$\begin{eqnarray}{{ \mathcal V }}_{q}(t)=A(t)B(t)C(t)={A}_{0}{B}_{0}{C}_{0}{{\rm{e}}}^{\left({\int }_{{t}_{0}}^{t}\chi (s){\rm{d}}s\right)},\end{eqnarray}$
$\begin{eqnarray}{{ \mathcal V }}_{g}(t)=a(t)b(t)c(t)={\left(\displaystyle \frac{\lambda -\tfrac{k}{3}\rho (t)}{\lambda +k\rho (t)}\right)}^{\tfrac{3}{2}}{{ \mathcal V }}_{q}(t).\end{eqnarray}$
The Hubble's functions Hq and Hg are defined respectively by $\begin{eqnarray}{H}_{q}(t)=\displaystyle \frac{1}{3}\left(\displaystyle \frac{\dot{A}(t)}{A(t)}+\displaystyle \frac{\dot{B}(t)}{B(t)}+\displaystyle \frac{\dot{C}(t)}{C(t)}\right)=\displaystyle \frac{1}{3}\chi (t),\end{eqnarray}$
$\begin{eqnarray}{H}_{g}(t)=\displaystyle \frac{1}{3}\left(\displaystyle \frac{\dot{a}(t)}{a(t)}+\displaystyle \frac{\dot{b}(t)}{b(t)}+\displaystyle \frac{\dot{c}(t)}{c(t)}\right)=\displaystyle \frac{1}{3}\left(\chi (t)-{\chi }_{0}\right),\end{eqnarray}$
where χ0 = U0 + V0 + W0 = χ(t0).It is of great cosmological importance to evaluate the anisotropy parameters ${{ \mathcal A }}_{g}$ related to the physical metric g and ${{ \mathcal A }}_{q}$ related to the auxiliary metric q. Indeed, these different parameters are useful indicators to study the behaviour of anisotropic cosmological models. In GR, these parameters are finite for singular states [31].
We have
$\begin{eqnarray}\begin{array}{rcl}{{ \mathcal A }}_{q}(t) & = & \tfrac{1}{3{H}_{q}^{2}(t)}\left[{\left({H}_{q}(t)-\tfrac{\dot{A}(t)}{A(t)}\right)}^{2}\right.\\ & & \left.+{\left({H}_{q}(t)-\tfrac{\dot{B}(t)}{B(t)}\right)}^{2}+{\left({H}_{q}(t)-\tfrac{\dot{C}(t)}{C(t)}\right)}^{2}\right]\\ & = & -1+\tfrac{3\left[{\left(U(t)\right)}^{2}+{\left(V(t)\right)}^{2}+{\left(W(t)\right)}^{2}\right]}{{\chi }^{2}(t)}.\end{array}\end{eqnarray}$
Similarly $\begin{eqnarray}\begin{array}{l}{{ \mathcal A }}_{g}(t)=-1\\ \quad +\,\displaystyle \frac{3\left[{\left(U(t)-{U}_{0}\right)}^{2}+{\left(V(t)-{V}_{0}\right)}^{2}+{\left(W(t)-{W}_{0}\right)}^{2}\right]}{{\left(\chi (t)-{\chi }_{0}\right)}^{2}}.\end{array}\end{eqnarray}$
We presently define the decelerations parameters Qg and Qq in the directions of g and q, respectively. The role of these parameters in Cosmology is to know when the bi-universe described by metrics g and q inflate or not. We have $\begin{eqnarray}{Q}_{q}(t)=\displaystyle \frac{{{\rm{d}}{H}_{q}}^{-1}(t)}{{\rm{d}}t}=-\displaystyle \frac{3\dot{\chi }(t)}{{\chi }^{2}(t)},\end{eqnarray}$
$\begin{eqnarray}{Q}_{g}(t)=\displaystyle \frac{{{\rm{d}}{H}_{g}}^{-1}(t)}{{\rm{d}}t}=-\displaystyle \frac{3\dot{\chi }(t)}{{\left(\chi (t)-{\chi }_{0}\right)}^{2}}.\end{eqnarray}$
| i | (i)Assuming that $U(t){U}_{0}+V(t){V}_{0}+W(t){W}_{0}\lt \tfrac{{U}_{0}^{2}+{V}_{0}^{2}+{W}_{0}^{2}}{2}$, we can easily show that $\dot{\chi }(t)\lt 0$. Then ${Q}_{q}(t)\geqslant 0$ and ${Q}_{g}(t)\geqslant 0$. So, in this case, there is standard deceleration of the bi-universe in the direction of the auxiliary metric q and in the direction of the physical metric g. |
| ii | (ii)Assuming that $U(t){U}_{0}+V(t){V}_{0}+W(t){W}_{0}\geqslant \tfrac{{U}_{0}^{2}+{V}_{0}^{2}+{W}_{0}^{2}}{2}$, we can easily show that $\dot{\chi }(t)\geqslant 0$. Then ${Q}_{q}(t)\lt 0$ and ${Q}_{g}(t)\lt 0$. So the bi-universe inflate in the direction of the auxiliary metric q and in the direction of the physical metric g. |
| iii | (iii)By choosing $\chi (t)=-\tfrac{3}{t-{t}_{0}}$ with ${\chi }_{0}=0$, we readily get ${Q}_{q}(t)={Q}_{g}(t)=-1$, for all $t\in {\mathbb{R}}$. Hence, one obtains de Sitter phase in both directions at early and at late time bi-universe. |
| iv | (iv)For $t={t}_{0}$, notice that when ${\chi }_{0}^{2}={U}_{0}^{2}+{V}_{0}^{2}+{W}_{0}^{2}={\left({U}_{0}+{V}_{0}+{W}_{0}\right)}^{2}$, we have ${V}_{0}{W}_{0}+{U}_{0}{V}_{0}+{U}_{0}{W}_{0}=0$ and ${{ \mathcal A }}_{q}(t)=2$. Therefore the auxiliary metric q describes the Kasner geometry (see [31]). |
| v | (v)Assuming that ${A}_{0}={B}_{0}={C}_{0}$, we have $A(t)=B(t)=C(t)={A}_{0}{{\rm{e}}}^{\left({\int }_{{t}_{0}}^{t}{H}_{q}(s){\rm{d}}s\right)}$. Then the physical metric g and the auxiliary metric q are given by $\begin{eqnarray}\begin{array}{rcl}g & = & -{\rm{d}}{t}^{2}+{A}_{0}^{2}\left(\displaystyle \frac{\lambda -\tfrac{k}{3}\rho }{\lambda +k\rho }\right){{\rm{e}}}^{\left(2{\displaystyle \int }_{{t}_{0}}^{t}{H}_{q}(s){\rm{d}}s\right)}\\ & & \left.\times {\left[{\left({\rm{d}}{x}^{1}\right)}^{2}+{\left({\rm{d}}{x}^{2}\right)}^{2}+({\rm{d}}{x}^{3}\right)}^{2}\right],\end{array}\end{eqnarray}$ $\begin{eqnarray}\begin{array}{rcl}q & = & -{\rm{d}}{t}^{2}+{A}_{0}^{2}{{\rm{e}}}^{\left(2{\displaystyle \int }_{{t}_{0}}^{t}{H}_{q}(s){\rm{d}}s\right)}\\ & & \times \left[{\left({\rm{d}}{x}^{1}\right)}^{2}+{\left({\rm{d}}{x}^{2}\right)}^{2}+{\left({\rm{d}}{x}^{3}\right)}^{2}\right].\end{array}\end{eqnarray}$ So in both directions of the bi-universe, the physical metric g and auxiliary metric q describe the FLRW metrics with cosmological expansion factors ${A}_{0}^{2}\left(\tfrac{\lambda -\tfrac{k}{3}\rho }{\lambda +k\rho }\right){{\rm{e}}}^{\left(2{\int }_{{t}_{0}}^{t}{H}_{q}(s){\rm{d}}s\right)}$ and ${A}_{0}^{2}{{\rm{e}}}^{\left(2{\int }_{{t}_{0}}^{t}{H}_{q}(s){\rm{d}}s\right)}$, respectively. |
| vi | (vi)Assuming that at $t={t}_{0}$, we have ${A}_{0}={B}_{0}={C}_{0}=1$. Then the physical metric g and the auxiliary metric q are given by $\begin{eqnarray}\begin{array}{l}\tilde{g}=\left(\displaystyle \frac{\lambda +k{\rho }_{0}}{\lambda -\tfrac{k}{3}{\rho }_{0}}\right)g=-\left(\displaystyle \frac{\lambda +k{\rho }_{0}}{\lambda -\tfrac{k}{3}{\rho }_{0}}\right){\rm{d}}{t}^{2}\\ \quad +\,{\left({\rm{d}}{x}^{1}\right)}^{2}+{\left({\rm{d}}{x}^{2}\right)}^{2}+{\left({\rm{d}}{x}^{3}\right)}^{2},\end{array}\end{eqnarray}$ $\begin{eqnarray}q=-{\rm{d}}{t}^{2}+{\left({\rm{d}}{x}^{1}\right)}^{2}+{\left({\rm{d}}{x}^{2}\right)}^{2}+{\left({\rm{d}}{x}^{3}\right)}^{2},\end{eqnarray}$ which describe Minkowski spacetimes. |
| vii | (vii)For $t={t}_{0}$, the volumes of scalar factors satisfy the initial conditions ${V}_{q}({t}_{0})={A}_{0}{B}_{0}{C}_{0}\geqslant 8\pi k{\rho }_{0}$ and ${V}_{g}({t}_{0})={\left(\tfrac{\lambda -\tfrac{k}{3}{\rho }_{0}}{\lambda +k{\rho }_{0}}\right)}^{\tfrac{3}{2}}{V}_{q}({t}_{0})\geqslant 8\pi k{\rho }_{0}$. Therefore, in EiBI bi-gravity the stiff causal bi-universe starts its evolution from a non-singular state (see [31]). |
| viii | (viii)The asymptotic behaviours of Hg2, Hq2 and the time derivative $\tfrac{{\rm{d}}{H}_{g}}{{\rm{d}}t}$, $\tfrac{{\rm{d}}{H}_{q}}{{\rm{d}}t}$ of the Hubble rate in the EiBI regime are $\begin{eqnarray}\left\{\begin{array}{l}{H}_{g}^{2}(t)\mathop{\longrightarrow }\limits_{t\to {t}_{0}}0\\ \mathrm{and}\\ \displaystyle \frac{{\rm{d}}{H}_{g}}{{\rm{d}}t}(t)\mathop{\longrightarrow }\limits_{t\to {t}_{0}}0.\end{array}\right.\end{eqnarray}$ Similarly, we have $\begin{eqnarray}\left\{\begin{array}{l}{H}_{q}^{2}(t)\mathop{\longrightarrow }\limits_{t\to {t}_{0}}0\\ \mathrm{and}\\ \displaystyle \frac{{\rm{d}}{H}_{q}}{{\rm{d}}t}(t)\mathop{\longrightarrow }\limits_{t\to {t}_{0}}0.\end{array}\right.\end{eqnarray}$ So, the solutions obtained are unstable when cosmic time t tends to t0 (see figure 37 in [14]). |
| ix | (ix)It is worth noting that when $\lambda =1$, both physical and auxiliary metrics g and q coincide, and the bi-universe described by them is asymptotically flat [14]. |
In theorem 8 below, we summarize the asymptotic behaviour of the physical metric g and the auxiliary metric q.
Let $\left(k\gt 0\quad \mathrm{and}\quad \lambda \in \left(\tfrac{k}{3}{\rho }_{0},\infty \right)\right)$ or $\left(k\lt 0\quad \mathrm{and}\quad \lambda \in \left(-k{\rho }_{0},\infty \right)\right)$, ${A}_{0}\gt 0$, ${B}_{0}\gt 0$ and ${C}_{0}\gt 0$. Assume that $\rho (t)\mathop{\longrightarrow }\limits_{t\to \infty }0$. Then the bi-universe $\left({{\mathbb{R}}}^{4},{g}_{\alpha \beta },{q}_{\alpha \beta },{T}^{\alpha \beta }\right)$ tends towards the vacuum at late time.
We just have to show that ${T}^{\alpha \beta }(t)\mathop{\longrightarrow }\limits_{t\to \infty }0$.
According to the hypothesis of theorem
$\begin{eqnarray}\left\{\begin{array}{l}{T}^{00}(t)=\rho (t)\mathop{\longrightarrow }\limits_{t\to \infty }0,\\ {T}^{11}(t)=\displaystyle \frac{\rho (t)}{3{a}^{2}(t)}\mathop{\longrightarrow }\limits_{t\to \infty }0,\\ {T}^{22}(t)=\displaystyle \frac{\rho (t)}{3{b}^{2}(t)}\mathop{\longrightarrow }\limits_{t\to \infty }0,\\ {T}^{33}(t)=\displaystyle \frac{\rho (t)}{3{c}^{2}(t)}\mathop{\longrightarrow }\limits_{t\to \infty }0.\end{array}\right.\end{eqnarray}$
We notice that (4. Conclusion
In this work, we were in the quest of finding new exact solutions of Bianchi type I, comprising a relativistic perfect fluid, in Eddington-inspired-Born-Infeld theory of gravity. For this purpose, we first established a system of second order ordinary differential equations (Proposition 1), supplemented by a suitable constraints system (Proposition 2). Then, after an appropriate change of variables, combined with a thorough analysis of the systems obtained, we arrived at a nonlinear first order ordinary differential equation (see equation (48 ) of Lemma 4 ). To establish a solution to equation (48 ), we consider problem (51 ). By using the mathematical tools of nonlinear analysis and the notion of solution-tube for nonlinear first order differential equations, we succeed in establishing an existence result (Theorem 4 ) for problem (51 ). From this solution, we retrieve the physical metric g, the auxiliary metric q and the matter density ρ (Theorem 7 ). Finally, by dint of many meticulous calculations, we investigated the asymptotic behaviours of the physical metric g and the auxiliary metric q obtained. As a matter of fact, we provided the expressions of the volumes of scalar factors, Hubble's functions, anisotropy parameters and deceleration parameters related to both metrics. These aforementioned functions are very important in Cosmology. We saw from Remark 4 that the auxiliary metric q is equivalent to the Kasner metric when the anisotropy parameter ${{ \mathcal A }}_{q}(t)$ equals 2 for t = t0 so that ${U}_{0}^{2}+{V}_{0}^{2}+{W}_{0}^{2}={\left({U}_{0}+{V}_{0}+{W}_{0}\right)}^{2}$. We also found from equations (89 ) and (90 ) that in both directions the solutions obtained are unstable when the cosmic time t tends to t0. We also noticed that when λ = 1 both physical metric g and auxiliary metric q coincide. From Theorem 8, we concluded that, under appropriate conditions $\left(k\gt 0\quad \mathrm{and}\quad \lambda \in \left(\tfrac{k}{3}{\rho }_{0},\infty \right)\right)$ or $\left(k\lt 0\quad \mathrm{and}\quad \lambda \in \left(-k{\rho }_{0},\infty \right)\right)$, A0 > 0, B0 > 0 and C0 > 0, when t tends towards a late time the solution obtained in the EiBI theory tends towards a vacuum.
It is worth stressing the fact that, in the quest for searching exact solutions to the model studied in this work, we have set the lapse function to 1 in (8 ) and (9 ). However, it would be interesting to consider the cases where one of the lapse function is 1 and the other one is unknown.
Firstly, let's consider the physical metric g given by equation (8 ) and the auxiliary metric q given by the following equation17 )–(20 )94 ) is the counterpart of equation (17 ) in this case. So there is no way to avoid this. Also, equations (94 )–(97 ) are quite involved and not easy to handle with the approach used for solving equations (17 )–(20 ) in the present work. The nonlinear second order ODEs system (94 )–(97 ) is difficult to manipulate for the search of exact solutions.
$\begin{eqnarray}\begin{array}{l}q=-{N}^{2}(t){\rm{d}}{t}^{2}+{A}^{2}(t){\left({\rm{d}}{x}^{1}\right)}^{2}\\ \quad +\,{B}^{2}(t){\left({\rm{d}}{x}^{2}\right)}^{2}+{C}^{2}(t){\left({\rm{d}}{x}^{3}\right)}^{2},\end{array}\end{eqnarray}$
where N > 0 represents the lapse function. In this case the non-vanishing components of the Ricci tensor are now given by $\begin{eqnarray}\begin{array}{rcl}{R}_{00} & = & \displaystyle \frac{\ddot{N}}{N}+\displaystyle \frac{\dot{N}}{N}\left(\displaystyle \frac{\dot{A}}{A}+\displaystyle \frac{\dot{B}}{B}+\displaystyle \frac{\dot{C}}{C}\right)\\ & & -\left(\displaystyle \frac{\ddot{A}}{A}+\displaystyle \frac{\ddot{B}}{B}+\displaystyle \frac{\ddot{C}}{C}\right),\\ {R}_{11} & = & A\left[\ddot{A}-2\displaystyle \frac{\dot{N}}{N}+\dot{A}\displaystyle \frac{\dot{N}}{{N}^{3}}+\displaystyle \frac{1}{{N}^{2}}\left(\dot{A}\displaystyle \frac{\dot{B}}{B}+\dot{A}\displaystyle \frac{\dot{C}}{C}\right)\right]\\ & & +{\left(\dot{A}\right)}^{2}\left[1-\displaystyle \frac{1}{{N}^{2}}\right],\\ {R}_{22} & = & B\left[\ddot{B}-2\displaystyle \frac{\dot{N}}{N}+\dot{B}\displaystyle \frac{\dot{N}}{{N}^{3}}+\displaystyle \frac{1}{{N}^{2}}\left(\dot{B}\displaystyle \frac{\dot{A}}{A}+\dot{B}\displaystyle \frac{\dot{C}}{C}\right)\right]\\ & & +{\left(\dot{B}\right)}^{2}\left[1-\displaystyle \frac{1}{{N}^{2}}\right],\\ {R}_{33} & = & C\left[\ddot{C}-2\displaystyle \frac{\dot{N}}{N}+\dot{C}\displaystyle \frac{\dot{N}}{{N}^{3}}+\displaystyle \frac{1}{{N}^{2}}\left(\dot{C}\displaystyle \frac{\dot{A}}{A}+\dot{C}\displaystyle \frac{\dot{B}}{B}\right)\right]\\ & & +{\left(\dot{C}\right)}^{2}\left[1-\displaystyle \frac{1}{{N}^{2}}\right].\end{array}\end{eqnarray}$
After lengthy and tedious calculations, we arrive at the following equations which are counterparts of equations ( $\begin{eqnarray}\begin{array}{l}\displaystyle \frac{\ddot{N}}{N}+\displaystyle \frac{\dot{N}}{N}\left(\displaystyle \frac{\dot{A}}{A}+\displaystyle \frac{\dot{B}}{B}+\displaystyle \frac{\dot{C}}{C}\right)\\ \quad -\,\left(\displaystyle \frac{\ddot{A}}{A}+\displaystyle \frac{\ddot{B}}{B}+\displaystyle \frac{\ddot{C}}{C}\right)=\displaystyle \frac{1}{k}\left(1-{N}^{2}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}A\left[\ddot{A}-2\displaystyle \frac{\dot{N}}{N}+\dot{A}\displaystyle \frac{\dot{N}}{{N}^{3}}+\displaystyle \frac{1}{{N}^{2}}\left(\dot{A}\displaystyle \frac{\dot{B}}{B}+\dot{A}\displaystyle \frac{\dot{C}}{C}\right)\right]\\ \quad +\,{\left(\dot{A}\right)}^{2}\left[1-\displaystyle \frac{1}{{N}^{2}}\right]=\displaystyle \frac{1}{k}\left({A}^{2}-{a}^{2}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}B\left[\ddot{B}-2\displaystyle \frac{\dot{N}}{N}+\dot{B}\displaystyle \frac{\dot{N}}{{N}^{3}}+\displaystyle \frac{1}{{N}^{2}}\left(\dot{B}\displaystyle \frac{\dot{A}}{A}\right.\right.\\ \quad \left.\left.+\,\dot{B}\displaystyle \frac{\dot{C}}{C}\right)\right]+{\left(\dot{B}\right)}^{2}\left[1-\displaystyle \frac{1}{{N}^{2}}\right]=\displaystyle \frac{1}{k}\left({B}^{2}-{b}^{2}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}C\left[\ddot{C}-2\displaystyle \frac{\dot{N}}{N}+\dot{C}\displaystyle \frac{\dot{N}}{{N}^{3}}+\displaystyle \frac{1}{{N}^{2}}\left(\dot{C}\displaystyle \frac{\dot{A}}{A}\right.\right.\\ \quad \left.\left.+\,\dot{C}\displaystyle \frac{\dot{B}}{B}\right)\right]+{\left(\dot{C}\right)}^{2}\left[1-\displaystyle \frac{1}{{N}^{2}}\right]=\displaystyle \frac{1}{k}\left({C}^{2}-{c}^{2}\right).\end{array}\end{eqnarray}$
It should be noted that equation (Secondly, we now consider the auxiliary metric defined by (9 ) and the physical metric defined as follows:17 ) of Proposition 1 is changed to its following analogue:17 ) and its analogues cannot be avoided. We postpone the complete study of these two cases to a forthcoming paper. Another perspective pertaining to the present work is to take into consideration the spatial dependence of the physical and the auxiliary metrics. This would yield a more intricate system of partial differential equations to analyze mathematically and interpret physically. One could begin with the spherical symmetry assumption on g and q.
$\begin{eqnarray}\begin{array}{l}g=-{M}^{2}(t){\rm{d}}{t}^{2}+{a}^{2}(t){\left({\rm{d}}{x}^{1}\right)}^{2}\\ \quad +{b}^{2}(t){\left({\rm{d}}{x}^{2}\right)}^{2}+{c}^{2}(t){\left({\rm{d}}{x}^{3}\right)}^{2},\end{array}\end{eqnarray}$
where M > 0 represents the lapse function. In this case, direct calculations show that only equation ( $\begin{eqnarray}\displaystyle \frac{\ddot{A}}{A}+\displaystyle \frac{\ddot{B}}{B}+\displaystyle \frac{\ddot{C}}{C}=\displaystyle \frac{1}{k}\left(1-{M}^{2}\right),\end{eqnarray}$
where M verifies the following equation $\begin{eqnarray}{M}^{4}-\displaystyle \frac{3{A}^{2}{\left(\lambda -\tfrac{k}{3}\rho \right)}^{2}}{4{k}^{2}{\rho }^{2}{a}^{2}}{M}^{2}+\displaystyle \frac{\lambda -\tfrac{k}{3}\rho }{4k\rho }=0.\end{eqnarray}$
So, in all cases, equation (