Kantowski–Sachs Spherically Symmetric Solutions in Teleparallel F(T) Gravity

Abstract

In this paper, we investigate time-dependent Kantowski–Sachs spherically symmetric teleparallel $F\left(T\right)$ gravity in vacuum and in a perfect isotropic fluid. We begin by finding the field equations and solve for new teleparallel $F\left(T\right)$ solutions. With a power-law ansatz for the co-frame functions, we find new non-trivial teleparallel $F\left(T\right)$ vacuum solutions. We then proceed to find new non-trivial teleparallel $F\left(T\right)$ solutions in a perfect isotropic fluid with both linear and non-linear equations of state. We find a great number of new exact and approximated teleparallel $F\left(T\right)$ solutions. These classes of new solutions are relevant for future cosmological applications.

1. Introduction

The teleparallel $F\left(T\right)$-type theory of gravity is a promising alternative theory to General Relativity (GR) [1,2,3]. This is a theory where the geometry is described by the spacetime torsion, a function of the coframe ${\mathbf{h}}^a$ (and its derivatives) and the spin connection ${\omega }_{bc}^a$. The teleparallel gravity mainly works with a frame basis instead of a metric tensor. The role of symmetry is not as clear as in pseudo-Riemannian geometry, where symmetry is defined in terms of Killing Vectors (KVs). Riemannian geometry in GR is defined by the Levi-Civita curvature and is fully dependent on the metric. However, it is very different for the case of teleparallel $F\left(T\right)$-type gravity.

The frame-based approach for determining spacetime symmetries has been explored [4,5,6]. A possible complication arises from the possibility of a non-trivial linear isotropy group: a Lie group of Lorentz frame transformations keeping the associated tensors of the geometry invariant. A new approach was introduced for symmetry determination of any geometry based on an independent frame and spin connection admitting the torsion and curvature tensors as geometric objects [7]. In this case, the spin connection is an independent object, and any geometry with a zero curvature and a zero non-metricity tensor is defined as teleparallel geometry. The approach is based on the existence of a particular class of invariantly defined symmetry frames. For a non-trivial linear isotropy group, the symmetry determination needs to define a set of inhomogeneous differential equations (DEs) for coframes and spin connection in an orthonormal gauge $g_{ab}=diag[-1,1,1,1]$. In addition, the spin connection ${\omega }_{bc}^a$ will be defined in terms of an arbitrary Lorentz transformation ${\Lambda }_b^a$ and obtained from the zero curvature requirement (see Ref. [7] and references within). All these considerations can be summarized as follows [8]:

$$\begin{array}{cc}\hfill {\mathcal{L}}_{\mathbf{X}}{\mathbf{h}}^a=& {\lambda }_b^a{\mathbf{h}}^b,\hfill \end{array}$$

(1a)

$$\begin{array}{cc}\hfill {\mathcal{L}}_{\mathbf{X}}{\omega }_{bc}^a=& 0,\hfill \end{array}$$

(1b)

$$\begin{array}{cc}\hfill {\omega }_{bc}^a=& {\Lambda }_d^a{\mathbf{h}}_c\left({\left({\Lambda }^{-1}\right)}_b^d\right),\hfill \end{array}$$

(1c)

where ${\mathbf{h}}^a$ is the orthonormal coframe basis, ${\mathcal{L}}_{\mathbf{X}}$ is the Lie derivative with respect to the KV $\mathbf{X}$, and ${\lambda }_b^a$ is an invariant Lie algebra generator of Lorentz transformations ${\Lambda }_b^a$.

A well-known subclass of teleparallel gravitational theories equivalent to GR is the Teleparallel Equivalent to General Relativity (TEGR), based on a torsion scalar T constructed from the torsion tensor [1]. The most common generalization of TEGR is the teleparallel $F\left(T\right)$-type gravity, where F is a function of the torsion scalar T [9,10,11]. In the covariant approach to $F\left(T\right)$-type gravity, teleparallel geometry is gauge-invariant with a zero curvature and a zero non-metricity spin connection. This last quantity is zero in all proper frames and non-zero in all other frames [1,3,12]. Therefore, the resulting teleparallel gravity theory is locally invariant with covariant FEs under the Lorentz definition [13]. A proper frame is not invariantly defined in terms of the connection (a non-tensorial quantity), and some problems arise when using such a frame to determine symmetries. However, some previous considerations have also been developed, used, and/or adapted for theories close to teleparallel $F\left(T\right)$ theories. We can think of the New General Relativity (NGR) described essentially by the use of the three irreducible components of the torsion tensor (see Refs. [14,15,16] and references within). Some of the previous elements were also considered for the symmetric teleparallel $F\left(Q\right)$-type gravity, a theory in development with some potential (see refs. [17,18,19,20] and references within). Or again, some of these elements are also useful for improving the study of the geometrical trinity of gravity and some intermediate theories like $F(T,Q)$-type, $F(R,Q)$-type, $F(R,T)$-type and others (see Refs. [21,22,23,24,25,26,27] and references within). These examples show a some number of possible approaches, but the best is to go with the teleparallel $F\left(T\right)$-type approach first.

There are important papers in the literature on spherically symmetric solutions in teleparallel $F\left(T\right)$ gravity [28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44]. The main features can be summarized by power-law $F\left(T\right)$ solutions obtained from power-law frame components (see [28,29,30,31] and references within). Most of these papers use the Weitzenback gauge (proper frames trivially satisfying the antisymmetric FEs), but extra degrees of freedom (DoFs) arise by imposing the zero spin connection. This specificity leads to only symmetric parts of FEs, and the presented $F\left(T\right)$ solutions are essentially limited to the power-law for a big coframe expression. These symmetric FEs and the solutions are similar between the different gauges, but performing a frame changing is necessary for solving this extra DoF potential issue. When a non-proper frame is used, the non-zero spin connections are solutions to the non-trivial antisymmetric parts of FEs, and then all DoFs are covered by the FEs. This method is used in a paper on teleparallel spherically symmetric geometries focusing on vacuum solutions and additional symmetry structures [32]. The general FEs are defined in an orthonormal gauge assuming a diagonal frame and a non-trivial spin connection, leading to non-trivial antisymmetric and symmetric parts of FEs without extra DoFs. This paper also studied the Kantowski–Sachs (KS) geometry case and found the vacuum $F\left(T\right)$ solutions by using a special power-law ansatz. Therefore, we need to find more solutions and to go further for the KS spacetime, as done recently for static spacetime $F\left(T\right)$ solutions for perfect fluids [45]. The KS spacetime’s fourth symmetry is defined by the radial coordinate derivative ${\partial }_r$, leading to time coordinate dependence for coframes, spin connections, and FEs.

In the literature, there are some works on KS spacetimes and solutions in GR and some specific $f\left(R\right)$-type theories of gravity [46,47,48]. In these papers, for some $F\left(R\right)$ solutions, a detailed study is carried out concerning critical points, limits on physical quantities, asymptotes, and also the evolution of curvature, to name a few. There are several other papers on more elaborated KS spacetime models, but they are not made in terms of teleparallel gravity, and they are essentially focused on $f\left(R\right)$-type gravity. For KS teleparallel $F\left(T\right)$ theory, there is a small number of recent works [49,50]. There have recently been published some generalizations of KS models and solutions for $F(T,B)$ gravity, namely, the LRS Bianchi III Universe and even the $F(T,R)$-type and teleparallel $F\left(Q\right)$-type gravities. However, these papers essentially focus on cosmological solutions and possible equilibrium points of these cosmological systems [51,52,53,54,55,56]. Therefore, their approaches do not really provide a new class of teleparallel $F\left(T\right)$-type solutions. All these works suggest that KS spacetime geometry solutions are relevant and can be used for more refined cosmological solutions. Recently, there are some works on teleparallel KS solutions from Paliathanasis that may also lead to scalar field and quantized cosmological solutions [57,58]. However, no quantized solutions will be considered in the current paper because the aim of the current paper is only to find $F\left(T\right)$ solutions for KS spacetimes without scalar field and quantization.

For this paper, we assume a time coordinate-dependent, spherically symmetric teleparallel geometry (a Kantowski–Sachs teleparallel geometry) in an orthonormal gauge, as defined in Ref. [32]. We will first find vacuum $F\left(T\right)$ solutions, then we will focus on finding perfect fluid Kantowski–Sachs teleparallel $F\left(T\right)$ solutions. After a summary of the teleparallel FEs and Kantowski–Sachs class of geometries in Section 2, we will find additional $F\left(T\right)$ solutions in vacuum in Section 3. We then repeat the exercise in Section 4 with the linear equation of state (EoS). In Section 5, we will solve FEs and find several approximated $F\left(T\right)$ solutions for a perfect fluid with a non-linear EoS. This paper has some common features, aims, and structure in common with the paper on static perfect fluid teleparallel $F\left(T\right)$ solutions studied in Ref. [45].

The notation is defined as: coordinate indices $\mu ,\nu ,\dots$, tangent space indices $a,b,\dots$ (see Ref. [7]), spacetime coordinates $x^{\mu }$, frame fields ${\mathbf{h}}_a$ and its dual one-forms ${\mathbf{h}}^a$, vierbein $h_a^{\mu }$ or $h_{\mu }^a$, spacetime metric $g_{\mu \nu }$, Minkowski tangent space metric ${\eta }_{ab}$, spin connection one-form ${\omega }_b^a={\omega }_{bc}^a{\mathbf{h}}^c$, curvature tensor $R_{bcd}^a$, and torsion tensor $T_{bc}^a$. The derivatives with respect to t$F_t=F^{\prime }$ (with a prime).

2. Teleparallel Spherically Symmetric Spacetimes and Field Equations

2.1. Summary of Teleparallel Field Equations

The teleparallel $F\left(T\right)$-type gravity action integral is as follows [1,2,3,32,45]:

$$S_{F\left(T\right)}=\int d^{4}x\left[{\displaystyle \frac{h}{2\kappa }}F\left(T\right)+{\mathcal{L}}_{Matter}\right],$$

(2)

where h is the coframe determinant, and $\kappa$ is the coupling constant. We apply the least-action principle to Equation (2), and the symmetric and antisymmetric parts of the FEs are as follows [32,45]:

$$\begin{array}{ccc}\hfill \kappa {\Theta }_{\left(ab\right)}& =& F_T\left(T\right){\stackrel{\circ }{G}}_{ab}+F_{TT}\left(T\right)S_{\left(ab\right)}^{\mu }{\partial }_{\mu }T+{\displaystyle \frac{g_{ab}}{2}}\left[F\left(T\right)-T{F}_T\left(T\right)\right],\hfill \end{array}$$

(3a)

$$\begin{array}{ccc}\hfill 0& =& F_{TT}\left(T\right)S_{\left[ab\right]}^{\mu }{\partial }_{\mu }T,\hfill \end{array}$$

(3b)

with ${\stackrel{\circ }{G}}_{ab}$ being the Einstein tensor, ${\Theta }_{\left(ab\right)}$ the energy momentum, T the torsion scalar, $g_{ab}$ the gauge metric, $S_{ab}^{\mu }$ the superpotential (torsion dependent), and $\kappa$ the coupling constant. The canonical energy momentum is obtained from the ${\mathcal{L}}_{Matter}$ term of Equation (2) as follows:

$$\begin{array}{c}\hfill {\Theta }_a^{\mu }={\displaystyle \frac{1}{h}}{\displaystyle \frac{\delta {\mathcal{L}}_{Matter}}{\delta h_{\mu }^a}}.\end{array}$$

(4)

The antisymmetric and symmetric parts of Equation (4) are as follows [32]:

$${\Theta }_{\left[ab\right]}=0,{\Theta }_{\left(ab\right)}=T_{ab},$$

(5)

where $T_{ab}$ is the symmetric part of the energy–momentum tensor. Equation (5) is valid only when the matter field interacts with the metric $g_{\mu \nu }$ defined by the coframe $h_{\mu }^a$ and the gauge $g_{ab}$, and when it is not directly coupled to $F\left(T\right)$ gravity. This consideration is valid in the situation of zero hypermomentum (i.e., ${\mathfrak{T}}^{\mu \nu }=0$) as discussed in Refs. [30,45]. The hypermomentum is defined from the components of Equation (3a,b) as follows [30]:

$$\begin{array}{c}\hfill {\mathfrak{T}}_{ab}=\kappa {\Theta }_{ab}-F_T\left(T\right){\stackrel{\circ }{G}}_{ab}-F_{TT}\left(T\right)S_{ab}^{\mu }{\partial }_{\mu }T-{\displaystyle \frac{g_{ab}}{2}}\left[F\left(T\right)-T{F}_T\left(T\right)\right].\end{array}$$

(6)

The conservation of energy–momentum in teleparallel gravity for the ${\mathfrak{T}}^{\mu \nu }=0$ case states that ${\Theta }_a^{\mu }$ satisfies the relation [1,2]:

$$\begin{array}{c}\hfill {\stackrel{\circ }{\nabla }}_{\nu }\left({\Theta }^{\mu \nu }\right)=0,\end{array}$$

(7)

with ${\stackrel{\circ }{\nabla }}_{\nu }$ being the covariant derivative and ${\Theta }^{\mu \nu }$ being the conserved energy–momentum tensor. Equation (7) is also the GR conservation of the energy–momentum expression. Satisfaction of Equation (7) is automatically required because of the ${\mathfrak{T}}^{\mu \nu }=0$ condition (null hypermomentum). For non-zero hypermomentum situations (i.e., ${\mathfrak{T}}^{\mu \nu }\ne 0$), we need to satisfy more complex conservation equations than Equation (7) (see Ref. [30] for details).

For a perfect and isotropic fluid with any EoS, the matter tensor $T_{ab}$ is [45,59,60]:

$$\begin{array}{c}\hfill T_{ab}=\left(P\left(\rho \right)+\rho \right)u_a{u}_b+g_{ab}P\left(\rho \right),\end{array}$$

(8)

where $P\left(\rho \right)=P\left(\rho \right(t\left)\right)$ is the EoS in terms of the time-dependent fluid density $\rho =\rho \left(t\right)$ and $u_a=(-1,0,0,0)$.

2.2. Spherically Symmetric Teleparallel Kantowski–Sachs Geometry

The orthonormal time-dependent Kantowski–Sachs resulting vierbein is [32]:

$$\begin{array}{c}\hfill h_{\mu }^a=Diag\left[1,A_2\left(t\right),A_3\left(t\right),A_3\left(t\right)sin\left(\theta \right)\right],\end{array}$$

(9)

where we are able to choose new coordinates such that $A_1\left(t\right)=1$ without any loss of generality. This will allow us to find cosmological-like solutions.

The spin connection ${\omega }_{abc}$ components for time-dependent spacetimes is [32]:

$$\begin{array}{c}\hfill \begin{array}{cccc}& {\omega }_{341}=W_1\left(t\right),\hfill & \hfill {\omega }_{342}=W_2\left(t\right),& {\omega }_{233}={\omega }_{244}=W_3\left(t\right),\hfill \\ & {\omega }_{234}=-{\omega }_{243}=W_4\left(t\right),\hfill & \hfill {\omega }_{121}=W_5\left(t\right),& {\omega }_{122}=W_6\left(t\right),\hfill \\ & {\omega }_{133}={\omega }_{144}=W_7\left(t\right),\hfill & \hfill {\omega }_{134}=-{\omega }_{143}=W_8\left(t\right),& {\omega }_{344}=-{\displaystyle \frac{cos\left(\theta \right)}{A_{3}sin\left(\theta \right)}}.\hfill \end{array}\end{array}$$

(10)

For Equations (9) and (10), the curvature vanishing requirement implies that the functions $W_i\left(t\right)$ must take the form:

$$\begin{array}{c}\hfill \begin{array}{ccccc}\hfill W_1& =-{\chi }^{\prime },\hfill & \hfill W_2=0,& W_3={\displaystyle \frac{cosh\left(\psi \right)cos\left(\chi \right)}{A_3}},\hfill & \hfill W_4={\displaystyle \frac{cosh\left(\psi \right)sin\left(\chi \right)}{A_3}},\\ \hfill W_5& =-{\psi }^{\prime },\hfill & \hfill W_6=0,& W_7={\displaystyle \frac{sinh\left(\psi \right)cos\left(\chi \right)}{A_3}},\hfill & \hfill W_8={\displaystyle \frac{sinh\left(\psi \right)sin\left(\chi \right)}{A_3}},\end{array}\end{array}$$

(11)

where $\chi$ and $\psi$ are arbitrary functions of the coordinate t (${\chi }^{\prime }={\chi }_t$ and ${\psi }^{\prime }={\psi }_t$).

2.3. Teleparallel Kantowski–Sachs Field Equations

The antisymmetric part of the $F\left(T\right)$ FEs are [32]:

$$\begin{array}{cc}\hfill {\displaystyle \frac{F_{TT}\left(T\right)T^{\prime }cosh\left(\psi \right)cos\left(\chi \right)}{A_3}}& =0,{\displaystyle \frac{F_{TT}\left(T\right)T^{\prime }sinh\left(\psi \right)sin\left(\chi \right)}{A_3}}=0.\hfill \end{array}$$

(12)

Equation (12) leads to $\psi =0$ and $\chi ={\displaystyle \frac{\pi }{2}}$ (and also ${\displaystyle \frac{3\pi }{2}}$) for $T\ne$ constant as the solution. By substituting Equation (12) solution into Equation (11), we find that $W_4=\delta =\pm 1$ ($W_j=0$ for $j\ne 4$), and then Equation (10) becomes:

$$\begin{array}{c}\hfill {\omega }_{234}=-{\omega }_{243}=\delta ,{\omega }_{344}=-{\displaystyle \frac{cos\left(\theta \right)}{A_{3}sin\left(\theta \right)}}.\end{array}$$

(13)

The torsion scalar and the symmetric FE components are, for $\chi ={\displaystyle \frac{\pi }{2}}$ ($\delta =+1$) [32], exactly:

$$\begin{array}{cc}\hfill T=& 2{\left(ln\left(A_3\right)\right)}^{\prime }\left({\left(ln\left(A_3\right)\right)}^{\prime }+2{\left(ln\left(A_2\right)\right)}^{\prime }\right)-{\displaystyle \frac{2}{A_3^2}},\hfill \end{array}$$

(14a)

$$\begin{array}{cc}\hfill B^{\prime }=& -{\left(ln\left(A_2{A}_3^2\right)\right)}^{\prime }+{\displaystyle \frac{\frac{1}{A_3^2}-{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{″}}{{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{\prime }}},\hfill \end{array}$$

(14b)

$$\begin{array}{cc}\hfill \kappa \rho +{\displaystyle \frac{F\left(T\right)}{2}}=& \left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right),\hfill \end{array}$$

(14c)

$$\begin{array}{cc}\hfill -\kappa (\rho +P)=& \left[{\left(ln\left(A_3\right)\right)}^{\prime }\left(B^{\prime }+{\left(ln\left(A_3\right)\right)}^{\prime }-{\left(ln\left(A_2\right)\right)}^{\prime }\right)+{\left(ln\left(A_3\right)\right)}^{″}\right]F_T\left(T\right),\hfill \end{array}$$

(14d)

where $F_T\left(T\right)\ne$ constant and $B^{\prime }={\partial }_t\left(ln{F}_T\left(T\right)\right)$. Comparing with the version of the Kantowski–Sachs $F\left(T\right)$-gravity FEs in the literature [49,50], the FEs are different. For the $\delta =-1$ FE set, there are some small minor differences for some terms in Equation (14b–d), mainly some different signs at very specific terms. For the rest, the general form of Equation (14a–d) remains identical, regardless of $\delta$.

Then, the conservation law for non-null $\rho$ and P for time-dependent spacetimes is [32]:

$$\begin{array}{c}\hfill \left(P+\rho \right){\left(ln\left(A_2{A}_3^2\right)\right)}^{\prime }+{\rho }^{\prime }=0,\end{array}$$

(15)

where ${\rho }^{\prime }={\rho }_t$. For the forthcoming steps, we will solve Equations (14a) and (15), and the solutions will also depend on the EoS and the $P\left(\rho \right)$ relationship.

3. Vacuum Solutions

By setting $P=\rho =0$ in Equation (14a–d), we obtain the symmetric FEs:

$$\begin{array}{cc}\hfill T=& 2{\left(ln\left(A_3\right)\right)}^{\prime }\left({\left(ln\left(A_3\right)\right)}^{\prime }+2{\left(ln\left(A_2\right)\right)}^{\prime }\right)-{\displaystyle \frac{2}{A_3^2}},\hfill \end{array}$$

(16a)

$$\begin{array}{cc}\hfill B^{\prime }=& -{\left(ln\left(A_2{A}_3^2\right)\right)}^{\prime }+{\displaystyle \frac{\frac{1}{A_3^2}-{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{\prime \prime }}{{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{\prime }}},\hfill \end{array}$$

(16b)

$$\begin{array}{cc}\hfill F\left(T\right)=& 2\left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right),\hfill \end{array}$$

(16c)

$$\begin{array}{cc}\hfill 0=& \left[{\left(ln\left(A_3\right)\right)}^{\prime }\left(B^{\prime }+{\left(ln\left(A_3\right)\right)}^{\prime }-{\left(ln\left(A_2\right)\right)}^{\prime }\right)+{\left(ln\left(A_3\right)\right)}^{″}\right].\hfill \end{array}$$

(16d)

In this case, conservation laws are trivially satisfied because of null fluid density and pressure. In Ref. [32], we solved the FEs described by Equation (16a–d) by using the special ansatz $A_2=A_3^n$, where n is a real number. We found a linear $A_3$ in time coordinate t and a pure power-law for $F\left(T\right)$ as solutions within this specific ansatz. Some additional solutions are possible, and we first use the power-law ansatz for finding some of them for $A_2$, $A_3$ and $F\left(T\right)$. This type of coframe solution has been frequently used in recent literature and is easy to apply [28,29,30,31,32,33]. But above all, this approach has the advantage of easily distinguishing $F\left(T\right)$ teleparallel solutions from other types of solutions (such as GR solutions). In addition, this approach subsequently makes it possible to generalize with the exponential ansatz (an infinite sum of power-law terms: an appropriate limit case) as well as to refine the coframe and $F\left(T\right)$ solutions. For these purposes, we will then focus on more specific possible solutions, such as the exponential ansatz, to name only this one.

3.1. Power-Law Solutions

We will set the following power-law ansatz:

$$\begin{array}{c}\hfill A_2=t^b\mathrm{and}{A}_3=c_0{t}^c,\end{array}$$

(17)

where $b_0=1$ because we can perform a coordinate transformation $d\tilde{r}=b_{0}dr$ for the $A_2\left(t\right)$ frame component. Equation (16a–d) become:

$$\begin{array}{cc}\hfill T& ={\displaystyle \frac{2c\left(c+2b\right)}{t^2}}-{\displaystyle \frac{2}{c_0^2{t}^{2c}}},\hfill \end{array}$$

(18a)

$$\begin{array}{cc}\hfill t{B}^{\prime }& =(1-b-2c)+{\displaystyle \frac{t^{2-2c}}{c_0^2(b-c)}},\hfill \end{array}$$

(18b)

$$\begin{array}{cc}\hfill F\left(T\right)& =2\left(T+{\displaystyle \frac{2}{c_0^2{t}^{2c}}}\right)F_T\left(T\right),\hfill \end{array}$$

(18c)

$$\begin{array}{cc}\hfill 0& =c\left(t{B}^{\prime }+(c-b-1)\right).\hfill \end{array}$$

(18d)

If $c=0$, Equation (18a) leads to constant torsion scalar: a GR solution. For $c\ne 0$, we can put together Equation (18b,d), leading to the simplified relation:

$$\begin{array}{cc}\hfill 0& =-c_0^2(b-c)(2b+c)+t^{2(1-c)}.\hfill \end{array}$$

(19)

The only possible t independent solution is $c=1$, leading to $A_3=c_{0}t$. Then, Equation (19) becomes:

$$\begin{array}{cc}\hfill 0& =b^2-{\displaystyle \frac{b}{2}}-\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2c_0^2}}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \Rightarrow b={\displaystyle \frac{1+k}{4}},\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill & \Rightarrow A_2=t^{{\displaystyle \frac{1+k}{4}}},\hfill \end{array}$$

(21)

where $k={\delta }_1\sqrt{9+{\displaystyle \frac{8}{c_0^2}}}$ and ${\delta }_1=\pm 1$ ($k<-3$ and $k>3$ for real values). By substituting $c=1$ and Equation (20), we find from Equation (18a) the relation $t\left(T\right)$:

$$\begin{array}{c}\hfill t^{-2}\left(T\right)={\displaystyle \frac{4T}{(3+k)(7-k)}}.\end{array}$$

(22)

By substituting Equation (22) into Equation (18c), we find and solve the DE for $F\left(T\right)$:

$$\begin{array}{cc}\hfill {\displaystyle \frac{F_T\left(T\right)}{F\left(T\right)}}=& \left({\displaystyle \frac{7-k}{8}}\right){\displaystyle \frac{1}{T}},\hfill \\ \hfill & \Rightarrow F\left(T\right)=F_0{T}^{{\displaystyle \frac{7-k}{8}}},\hfill \end{array}$$

(23)

where $F_0$ is an integration constant, and $k\ne 7$. By comparison with the solution in Ref. [32], we can set $A_2=A_3^{{\displaystyle \frac{1+k}{4}}}$, leading to $c_0=1$ for all k and $c_0=-1$ for $k=8k^{\prime }-1$ where $k^{\prime }$ is an integer. Equation (23) is a power-law $F\left(T\right)$ solution as in Ref. [32], but the $A_2=A_3^n$ ansatz also leads to a similar power-law solution as shown in [32].

3.2. $A_3=c_{0}t$ Solutions

In Section 3.1, we find that $A_3=c_{0}t$ is the only possible solution for $A_3$. From this point, we may find some possible $A_2$ exact expressions and $F\left(T\right)$ solutions that are not necessarily a power-law solution. The Equation (16a–d) become:

$$\begin{array}{cc}\hfill T& ={\displaystyle \frac{4}{t}}{\left(ln\left(A_2\right)\right)}^{\prime }+{\displaystyle \frac{2}{t^2}}-{\displaystyle \frac{2}{c_0^2{t}^2}},\hfill \end{array}$$

(24a)

$$\begin{array}{cc}\hfill B^{\prime }& ={\displaystyle \frac{\left(-{\left(ln\left(A_2\right)\right)}^{\prime }{t}^{-1}-{\left(ln\left(A_2\right)\right)}^{″}-{\left(ln\left(A_2\right)\right)}^{\prime 2}+\frac{\left(c_0^{-2}+1\right)}{t^2}\right)}{{\left(ln\left(A_2\right)\right)}^{\prime }-t^{-1}}},\hfill \end{array}$$

(24b)

$$\begin{array}{cc}\hfill F\left(T\right)& =2\left(T+{\displaystyle \frac{2}{c_0^2{t}^2}}\right)F_T\left(T\right),\hfill \end{array}$$

(24c)

$$\begin{array}{cc}\hfill B^{\prime }& ={\left(ln\left(A_2\right)\right)}^{\prime }.\hfill \end{array}$$

(24d)

By substituting Equation (24d) into (24b), we obtain the DE for $A_2$:

$$\begin{array}{c}\hfill 0={\left(ln\left(A_2\right)\right)}^{\prime \prime }+2{\left(ln\left(A_2\right)\right)}^{\prime 2}-{\displaystyle \frac{1+c_0^{-2}}{t^2}}.\end{array}$$

(25)

By setting $y=ln\left(A_2\right)$ and $c_0^{-2}={\displaystyle \frac{1}{8}}(k-3)(k+3)$, Equation (25) becomes $y^{″}\left(t\right)+2y^{\prime }{\left(t\right)}^2+\left({\displaystyle \frac{1-k^2}{8}}\right)t^{-2}=0$. The general solution is exactly:

$$\begin{array}{c}\hfill y=ln\left(A_2\right)={\displaystyle \frac{1}{2}}ln\left[t^k+y_1\right]+{\displaystyle \frac{1}{4}}\left(1-k\right)ln\left(t\right),\end{array}$$

(26)

which leads to:

$$\begin{array}{c}\hfill A_2={\left[t^k+y_1\right]}^{{\displaystyle \frac{1}{2}}}{t}^{{\displaystyle \frac{1}{4}}\left(1-k\right)},\end{array}$$

(27)

where $y_1$ is an arbitrary constant. Then, by using Equations (26) and (27), Equation (24a) becomes:

$$\begin{array}{cc}\hfill T& ={\displaystyle \frac{2k{t}^{-2}}{\left[1+y_1{t}^{-k}\right]}}-{\displaystyle \frac{(k-3)(k+7)}{4}}{t}^{-2}.\hfill \end{array}$$

(28)

For the ${\mathbf{y}}_{\mathbf{1}}=\mathbf{0}$ case: We obtain that Equation (28) is the same as Equation (22). Then, by substituting Equation (22) into Equation (24c), we find the Equation (23) exactly.

For the ${\mathbf{y}}_{\mathbf{1}}\ne \mathbf{0}$ case: Equation (28) will be a characteristic equation for the $t\left(T\right)$ relationship before solving Equation (24c) for the $F\left(T\right)$ specific solution. For this, we set in Equation (28) the parameter k, and then Equation (28) becomes:

$$\begin{array}{cc}\hfill T& ={\displaystyle \frac{2k{t}^{-2}}{\left[1+y_1{t}^{-k}\right]}}-\left({\displaystyle \frac{(k-3)(k+7)}{4}}\right)t^{-2},\hfill \end{array}$$

(29)

where $k>3$ and $k<-3$ for a real value of $c_0$. Then, Equation (24c) will be simplified as follows:

$$\begin{array}{cc}\hfill F\left(T\right)& =2\left(T+{\displaystyle \frac{(k-3)(k+3)}{4}}{t}^{-2}\left(T\right)\right)F_T\left(T\right)\hfill \\ \hfill \Rightarrow & F\left(T\right)=F\left(0\right)exp\left[{\displaystyle \frac{1}{2}}\int {\displaystyle \frac{dT}{\left(T+\frac{(k-3)(k+3)}{4}{t}^{-2}\left(T\right)\right)}}\right].\hfill \end{array}$$

(30)

There are some specific values of k (we set integer values of k for Equation (29) for possible exact solutions) leading to analytic $t\left(T\right)$ solutions for Equation (29):

• Limit of $\mathbf{k}=\pm \mathbf{3}$: If $c_0\to \pm \infty$, we obtain that Equation (30) simplifies as follows:

  $$\begin{array}{cc}\hfill F\left(T\right)& =2T{F}_T\left(T\right)\hfill \\ \hfill \Rightarrow & F\left(T\right)=F_0\sqrt{T}.\hfill \end{array}$$

  (31)
  We find as the limit an usual power-law solution for $k\to \pm 3$. Equation (31) is also obtained in the case of dark energy perfect fluids, where the EoS is $P=-\rho$ in Ref. [50].
• $\mathbf{k}=\pm \mathbf{4}$ and $\pm \mathbf{6}$ cases: We obtain that $c_0^2={\displaystyle \frac{8}{7}}$ and ${\displaystyle \frac{8}{27}}$, respectively. We find, respectively, degree 3 and 4 characteristic equations from Equation (29), and then Equation (30) will not lead in either case to an analytical and closed form for $F\left(T\right)$.

All these subcases lead to new teleparallel $F\left(T\right)$ solutions and confirm some recent solutions, such as those in Ref. [50].

3.3. Exponential Ansatz Solutions

We can also introduce another approach for solving by using an exponential ansatz. We can see this ansatz as an infinite superposition of power-law terms as follows:

$$\begin{array}{c}\hfill A_2\left(t\right)=exp\left(bt\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(bt\right)}^n}{n!}},A_3\left(t\right)=c_{0}exp\left(ct\right)=c_0\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(ct\right)}^n}{n!}}.\end{array}$$

(32)

By substituting Equations (32), Equation (16a–d) become:

$$\begin{array}{cc}\hfill T& =2c(c+2b)-{\displaystyle \frac{2}{c_0^2}}exp(-2ct),\hfill \end{array}$$

(33a)

$$\begin{array}{cc}\hfill B^{\prime }& =-(2c+b)+{\displaystyle \frac{1}{c_0^2(b-c)}}exp(-2ct),\hfill \end{array}$$

(33b)

$$\begin{array}{cc}\hfill F\left(T\right)& =2\left(T+{\displaystyle \frac{2}{c_0^2}}exp(-2ct)\right)F_T\left(T\right),\hfill \end{array}$$

(33c)

$$\begin{array}{cc}\hfill 0& =c\left[B^{\prime }+c-b\right].\hfill \end{array}$$

(33d)

From Equation (33d), we find two GR solutions:

• $c=0$ case: We obtain from Equation (33a–c) that $T=-{\displaystyle \frac{2}{c_0^2}}=$ constant, $A_3=c_0=$ constant, $B^{\prime }=-b+{\displaystyle \frac{1}{c_0^{2}b}}$, and $F\left(T\right)=0$.
• $B^{\prime }=b-c$: Equation (33b) leads to $c=0$, then $A_3=c_0$ constant and $b={\displaystyle \frac{{\delta }_1}{\sqrt{2}{c}_0}}$ (i.e., ${\delta }_1=\pm 1$), leading to $A_2=exp\left({\displaystyle \frac{{\delta }_{1}t}{\sqrt{2}{c}_0}}\right)$ for a t-independent solution. Then, Equation (33a) leads to $T=-{\displaystyle \frac{2}{c_0^2}}=$ constant, and finally Equation (33c) leads to $F\left(T\right)=0$.

No purely teleparallel $F\left(T\right)$ solution is possible for an exponential ansatz in vacuum.

4. Linear Perfect Fluid Solutions

A perfect isotropic fluid with a linear EoS $P=\alpha \rho$, where $-1<\alpha \le 1$, is now the matter source. We solve Equation (15) for a specific $\rho \left(t\right)$ expression in terms of $A_2$ and $A_3$:

$$\begin{array}{cc}\hfill & \left(1+\alpha \right){\left(ln\left(A_2{A}_3^2\right)\right)}^{\prime }+{\left(ln\rho \right)}^{\prime }=0,\hfill \\ \hfill & \Rightarrow \rho \left(t\right)={\displaystyle \frac{{\rho }_0}{{\left(A_2\left(t\right)A_3^2\left(t\right)\right)}^{\left(1+\alpha \right)}}}.\hfill \end{array}$$

(34)

Equation (14a,d) become:

$$\begin{array}{cc}\hfill T=& 2{\left(ln\left(A_3\right)\right)}^{\prime }{\left(ln\left(A_2^2{A}_3\right)\right)}^{\prime }-{\displaystyle \frac{2}{A_3^2}},\hfill \end{array}$$

(35a)

$$\begin{array}{cc}\hfill B^{\prime }=& -{\left(ln\left(A_2{A}_3^2\right)\right)}^{\prime }+{\displaystyle \frac{\frac{1}{A_3^2}-{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{″}}{{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{\prime }}},\hfill \end{array}$$

(35b)

$$\begin{array}{cc}\hfill \kappa \rho +{\displaystyle \frac{F\left(T\right)}{2}}=& \left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right),\hfill \end{array}$$

(35c)

$$\begin{array}{cc}\hfill -\kappa \rho =& {\displaystyle \frac{F_T\left(T\right)}{\left(1+\alpha \right)}}\left[{\left(ln\left(A_3\right)\right)}^{\prime }\left(B^{\prime }+{\left(ln\left(A_3\right)\right)}^{\prime }-{\left(ln\left(A_2\right)\right)}^{\prime }\right)+{\left(ln\left(A_3\right)\right)}^{″}\right].\hfill \end{array}$$

(35d)

By adding Equation (35c,d) and then substituting Equation (35b), we obtain a linear DE in $F\left(T\right)$:

$$\begin{array}{cc}\hfill F\left(T\right)& ={\displaystyle \frac{2F_T\left(T\right)}{\left(1+\alpha \right)}}\left[\left(\alpha +{\displaystyle \frac{1}{2}}\right)\left(T+{\displaystyle \frac{2}{A_3^2}}\right)+{\displaystyle \frac{{\left(ln\left(A_3\right)\right)}^{\prime }\left(\frac{1}{A_3^2}-{\left(ln\left(A_2\right)\right)}^{″}\right)+{\left(ln\left(A_2\right)\right)}^{\prime }{\left(ln\left(A_3\right)\right)}^{″}}{{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{\prime }}}\right].\hfill \end{array}$$

(36)

4.1. Power-Law Solutions

As in Section 3.1, we use the Equation (17) ansatz. Then, Equations (34) and (35d) become:

$$\begin{array}{cc}\hfill T& ={\displaystyle \frac{2c(c+2b)}{t^2}}-{\displaystyle \frac{2}{c_0^2{t}^{2c}}},\hfill \end{array}$$

(37a)

$$\begin{array}{cc}\hfill B^{\prime }& =\left[{\displaystyle \frac{(1-b-2c)}{t}}+{\displaystyle \frac{t^{1-2c}}{c_0^2(b-c)}}\right],\hfill \end{array}$$

(37b)

$$\begin{array}{cc}\hfill \kappa \rho +{\displaystyle \frac{F\left(T\right)}{2}}& =\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)F_T\left(T\right),\hfill \end{array}$$

(37c)

$$\begin{array}{cc}\hfill -\kappa \rho & ={\displaystyle \frac{F_T\left(T\right)c}{\left(1+\alpha \right)}}\left[{\displaystyle \frac{1}{t}}{B}^{\prime }+{\displaystyle \frac{(c-b-1)}{t^2}}\right],\hfill \end{array}$$

(37d)

$$\begin{array}{cc}\hfill \rho & ={\rho }_0{c}_0^{-2\left(1+\alpha \right)}{t}^{-\left(1+\alpha \right)(b+2c)}.\hfill \end{array}$$

(37e)

With Equation (37a), we find the characteristic equation to solve for each specific value of c:

$$\begin{array}{c}\hfill 0=2c(c+2b)t^{-2}-{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}-T.\end{array}$$

(38)

By putting Equation (37c,d) together and then substituting Equation (37a,b), we obtain as the equation:

$$\begin{array}{cc}\hfill F\left(T\right)& =F_T\left(T\right)\left[{\displaystyle \frac{\left(1+2\alpha \right)}{\left(1+\alpha \right)}}\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\left(T\right)\right)+{\displaystyle \frac{2c{t}^{-2c}\left(T\right)}{c_0^2\left(1+\alpha \right)(b-c)}}\right],\hfill \\ \hfill \Rightarrow & F\left(T\right)=F\left(0\right)exp\left[(1+\alpha )\int dT{\left[(1+2\alpha )T+{\displaystyle \frac{2}{c_0^2}}\left(1+2\alpha +{\displaystyle \frac{c}{(b-c)}}\right)t^{-2c}\left(T\right)\right]}^{-1}\right].\hfill \end{array}$$

(39)

For the $\alpha =-{\displaystyle \frac{1}{2}}$ case, Equation (39) becomes simpler, as follows:

$$\begin{array}{c}\hfill F\left(T\right)=F\left(0\right)exp\left[{\displaystyle \frac{c_0^2(b-c)}{4c}}\int dT{t}^{2c}\left(T\right)\right],\end{array}$$

(40)

where $b\ne c$.

For the particular $\mathbf{c}=-\mathbf{2}\mathbf{b}$ case, Equation (38) simplifies as follows:

$$\begin{array}{c}\hfill t^{4b}\left(T\right)={\displaystyle \frac{c_0^2}{2}}(-T).\end{array}$$

(41)

Then, Equation (39) becomes:

$$\begin{array}{cc}\hfill F\left(T\right)=& F\left(0\right)T^{{\displaystyle \frac{3\left(1+\alpha \right)}{2}}}.\hfill \end{array}$$

(42)

Equation (42) is a pure power-law solution for $\alpha \ne -1$.

There are in principle solutions to Equation (38) in the cases $\mathbf{c}={\displaystyle \frac{\mathbf{1}}{\mathbf{2}}},-{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}},\mathbf{1},-\mathbf{1},{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}},\mathbf{2},$$-\mathbf{2},\mathbf{3},-\mathbf{3},\mathbf{4}$ ($c\ne -2b$ and $c\ne 0$). For analytically solvable solutions, we will solve for the following cases:

• $\mathbf{c}={\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}$: Equation (38) becomes:

  $$\begin{array}{cc}\hfill & 0=\left({\displaystyle \frac{1}{2}}+2b\right)t^{-2}-{\displaystyle \frac{2}{c_0^2}}{t}^{-1}-T,\hfill \\ \hfill & \Rightarrow t^{-1}\left(T\right)={\displaystyle \frac{\left[1+{\delta }_1\sqrt{1+\left(\frac{1}{2}+2b\right)c_0^{4}T}\right]}{\left(\frac{1}{2}+2b\right)c_0^2}}.\hfill \end{array}$$

  (43)
  Equation (39) becomes, by substituting Equation (43) (i.e., $b\ne -{\displaystyle \frac{1}{4}}$ and $\alpha \ne -{\displaystyle \frac{1}{2}}$):

  $$\begin{array}{cc}\hfill F\left(T\right)=& F\left(0\right)[-\left({\displaystyle \frac{1}{2}}+2b\right)c_0^4((1+2\alpha )T+{\displaystyle \frac{4}{c_0^4}}\left(1+2\alpha +{\displaystyle \frac{1}{(2b-1)}}\right)\hfill \\ \hfill & \times {\displaystyle \frac{\left[1+{\delta }_1\sqrt{1+\left(\frac{1}{2}+2b\right)c_0^{4}T}\right]}{\left(1+4b\right)}}){]}^{{\displaystyle \frac{\left(1+\alpha \right)}{\left(1+2\alpha \right)}}}exp[2(1+\alpha )\left(2b-1+{\displaystyle \frac{1}{1+2\alpha }}\right)\hfill \\ \hfill & \times {tanh}^{-1}\left[\left(1+2\alpha \right)(2b-1)\left(1+{\delta }_1\sqrt{1+2c_0^4\left({\displaystyle \frac{1}{4}}+b\right)T}\right)+1\right]],\hfill \end{array}$$

  (44)

  where $b\ne \left\{-{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{2}}\right\}$, $\alpha \ne \left\{-1,-{\displaystyle \frac{1}{2}}\right\}$, and ${\delta }_1=\pm 1$. For $\alpha =-{\displaystyle \frac{1}{2}}$, Equation (40) becomes, by using Equation (43):

  $$\begin{array}{c}\hfill F\left(T\right)=F\left(0\right){\left[1+{\delta }_1\sqrt{1+\left({\displaystyle \frac{1}{2}}+2b\right)c_0^{4}T}\right]}^{{\displaystyle \frac{1}{2}}-b}exp\left[{\delta }_1\left(b-{\displaystyle \frac{1}{2}}\right)\sqrt{1+\left({\displaystyle \frac{1}{2}}+2b\right)c_0^{4}T}\right],\end{array}$$

  (45)

  where $b\ne {\displaystyle \frac{1}{2}}$.
• $\mathbf{c}=\mathbf{1}$: Equation (38) becomes:

  $$\begin{array}{cc}\hfill & 0=\left(2(1+2b)-{\displaystyle \frac{2}{c_0^2}}\right)t^{-2}-T,\hfill \\ \hfill & \Rightarrow t^{-2}\left(T\right)={\displaystyle \frac{T}{2\left(1+2b-\frac{1}{c_0^2}\right)}}.\hfill \end{array}$$

  (46)
  Then, Equation (39) becomes, with Equation (46) (i.e., $\alpha \ne -{\displaystyle \frac{1}{2}}$):

  $$\begin{array}{c}\hfill F\left(T\right)=F\left(0\right)T^{{\displaystyle \frac{(1+\alpha )(b-1)\left(c_0^2(1+2b)-1\right)}{c_0^2(1+2\alpha )(2b+1)(b-1)+1}}},\end{array}$$

  (47)

  where $b\ne 1$. We obtain a power-law $F\left(T\right)$ solution, and Equation (47) is the most general solution for the linear $A_3=c_{0}t$ subcase. For $\alpha =-{\displaystyle \frac{1}{2}}$, Equation (40) becomes, with Equation (46):

  $$\begin{array}{c}\hfill F\left(T\right)=F\left(0\right)T^{{\displaystyle \frac{(b-1)}{2}}\left[(1+2b)c_0^2-1\right]},\end{array}$$

  (48)

  where $b\ne 1$.
• $\mathbf{c}=-\mathbf{1}$: Equation (38) becomes:

  $$\begin{array}{ccc}\hfill & 0=t^4+{\displaystyle \frac{c_0^{2}T}{2}}{t}^2-c_0^2(1-2b),\hfill & \hfill \\ \hfill & \Rightarrow t^2\left(T\right)={\displaystyle \frac{c_0^2}{4}}\left[-T+{\delta }_1\sqrt{T^2+16(1-2b)c_0^{-2}}\right].\hfill \end{array}$$

  (49)
  Then, Equation (39) becomes, with Equation (49) (i.e., $\alpha \ne -{\displaystyle \frac{1}{2}}$):

  $$\begin{array}{cc}\hfill F\left(T\right)=& F\left(0\right){\left[{\displaystyle \frac{{\left[\frac{4(1+2\alpha )}{(b+1)}{T}^2-16{\left(1+2\alpha -\frac{1}{(b+1)}\right)}^2\frac{(1-2b)}{c_0^2}\right]}^{\left(b+\frac{2(1+\alpha )}{(1+2\alpha )}\right)}}{{\left[T+\sqrt{T^2+16(1-2b)c_0^{-2}}\right]}^{2{\delta }_1\left(b+\frac{2\alpha }{(1+2\alpha )}\right)}}}\right]}^{{\displaystyle \frac{(1+\alpha )}{4}}}\hfill \\ \hfill & \times exp[{\displaystyle \frac{(1+\alpha )}{2}}\left(b+{\displaystyle \frac{2(1+\alpha )}{(1+2\alpha )}}\right)\hfill \\ \hfill & \times {tanh}^{-1}\left[{\displaystyle \frac{{\delta }_1\left(1+2\alpha +\frac{1}{(b+1)}\right)T}{\left(1+2\alpha -\frac{1}{(b+1)}\right)\sqrt{T^2+16(1-2b)c_0^{-2}}}}\right]],\hfill \end{array}$$

  (50)

  where $b\ne -1$, $\alpha \ne \left\{-1,-{\displaystyle \frac{1}{2}}\right\}$ and ${\delta }_1=\pm 1$. For $\alpha =-{\displaystyle \frac{1}{2}}$, Equation (40) becomes, with Equation (49):

  $$\begin{array}{cc}\hfill F\left(T\right)=& F\left(0\right){\left[T+\sqrt{T^2+16(1-2b)c_0^{-2}}\right]}^{-{\displaystyle \frac{{\delta }_1}{2}}(b+1)}\hfill \\ \hfill & \times exp\left[-{\displaystyle \frac{c_0^2(b+1)}{32(1-2b)}}T\left(T+{\delta }_1\sqrt{T^2+16(1-2b)c_0^{-2}}\right)\right],\hfill \end{array}$$

  (51)

  where $b\ne \left\{-1,{\displaystyle \frac{1}{2}}\right\}$.
• $\mathbf{c}=\mathbf{2}$: Equation (38) becomes:

  $$\begin{array}{cc}\hfill & 0=t^{-4}-4c_0^2(1+b)t^{-2}+{\displaystyle \frac{c_0^{2}T}{2}},\hfill \\ \hfill & \Rightarrow t^{-2}\left(T\right)=2c_0^2\left[(1+b)+{\delta }_1\sqrt{{(1+b)}^2-{\displaystyle \frac{T}{8c_0^2}}}\right],\hfill \end{array}$$

  (52)

  where ${\delta }_1=\pm 1$. Then, Equation (39) becomes, with Equation (52) (i.e., $\alpha \ne -{\displaystyle \frac{1}{2}}$):

  $$\begin{array}{cc}\hfill & F\left(T\right)=F\left(0\right)exp[(1+\alpha )\left(2-b-{\displaystyle \frac{2}{1+2\alpha }}\right)\hfill \\ \hfill & \times {tanh}^{-1}\left[1+{\displaystyle \frac{2}{(1+2\alpha )(b-2)}}\left(1+{\delta }_1\sqrt{1-{\displaystyle \frac{T}{8c_0^2{(1+b)}^2}}}\right)\right]]\hfill \\ \hfill & \times {\left[{\displaystyle \frac{2T}{2-b}}+16c_0^2{(1+b)}^2\left(1+2\alpha +{\displaystyle \frac{2}{(b-2)}}\right)\left(1+{\delta }_1\sqrt{1-{\displaystyle \frac{T}{8c_0^2{(1+b)}^2}}}\right)\right]}^{{\displaystyle \frac{(1+\alpha )(2-b)}{2}}},\hfill \end{array}$$

  (53)

  where $b\ne 2$, $\alpha \ne \left\{-1,-{\displaystyle \frac{1}{2}}\right\}$, and ${\delta }_1=\pm 1$. For $\alpha =-{\displaystyle \frac{1}{2}}$, Equation (40) becomes, with Equation (52):

  $$\begin{array}{c}\hfill F\left(T\right)=F\left(0\right){\left[4(1+b)+4{\delta }_1\sqrt{{(1+b)}^2-{\displaystyle \frac{T}{8c_0^2}}}\right]}^{{\displaystyle \frac{(2-b)}{2}}}exp\left[{\displaystyle \frac{(2-b)(1+b)}{2\left((1+b)+{\delta }_1\sqrt{{(1+b)}^2-\frac{T}{8c_0^2}}\right)}}\right],\end{array}$$

  (54)

  where $b\ne 2$.

For the other values of c, their Equation (39) integral will not lead to analytic and closed $F\left(T\right)$ solutions. All these previous teleparallel $F\left(T\right)$ solutions are new results. However, Equations (47) and (48) for the case $c=1$ are a bit similar to the solutions of Ref. [50]. In this last paper, the solutions obtained are only for $\alpha =-1$, 0, ${\displaystyle \frac{1}{3}}$, and combinations of these solutions.

4.2. $A_2=A_3^n$ Ansatz Solutions

There are other possible solutions to Equations (34) and (35d) by using the ansatz $A_2=A_3^n$, as in Ref. [32]. In this case, Equations (34) and (35d) become:

$$\begin{array}{cc}\hfill T=& 2(1+2n){\left(ln\left(A_3\right)\right)}^{\prime 2}-{\displaystyle \frac{2}{A_3^2}},\hfill \end{array}$$

(55a)

$$\begin{array}{cc}\hfill B^{\prime }=& -{\displaystyle \frac{\left((n+2){\left(ln\left(A_3\right)\right)}^{\prime 2}+{\left(ln\left(A_3\right)\right)}^{″}+\frac{1}{(n-1)A_3^2}\right)}{{\left(ln\left(A_3\right)\right)}^{\prime }}},\hfill \end{array}$$

(55b)

$$\begin{array}{cc}\hfill \kappa \rho +{\displaystyle \frac{F\left(T\right)}{2}}=& \left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right),\hfill \end{array}$$

(55c)

$$\begin{array}{cc}\hfill -\kappa \rho =& {\displaystyle \frac{F_T\left(T\right)}{\left(1+\alpha \right)}}\left[{\left(ln\left(A_3\right)\right)}^{\prime }{B}^{\prime }+{\left(ln\left(A_3\right)\right)}^{″}-(n-1){\left(ln\left(A_3\right)\right)}^{\prime 2}\right],\hfill \end{array}$$

(55d)

$$\begin{array}{cc}\hfill \rho =& {\rho }_0{A}_3^{-(2+n)\left(1+\alpha \right)},\hfill \end{array}$$

(55e)

where $n\in \mathbb{R}$. We find from Equation (55a) that:

$$\begin{array}{c}\hfill (2n+1){\left(ln\left(A_3\right)\right)}^{\prime 2}={\displaystyle \frac{1}{2}}\left(T+{\displaystyle \frac{2}{A_3^2}}\right).\end{array}$$

(56)

By putting Equation (55b–d) together, and then by substituting Equation (56), we find that:

$$\begin{array}{cc}\hfill F\left(T\right)& =F_T\left(T\right)\left[\left({\displaystyle \frac{1+2\alpha }{1+\alpha }}\right)T+{\displaystyle \frac{2\left(n\left(1+2\alpha \right)-2(1+\alpha )\right)}{A_3^2(n-1)\left(1+\alpha \right)}}\right].\hfill \end{array}$$

(57)

The Equation (57) solutions depend on the $A_3$ components. A constant $A_3$ leads to constant torsion scalar according to Equation (55a) and then to a GR solution. So, we need $A_3\ne$ constant as well as an $F\left(T\right)$ solution not depending on $A_3$. For this requirement, we need to satisfy the relation $n\left(1+2\alpha \right)-2(1+\alpha )=0$, leading to the solution $n={\displaystyle \frac{2(1+\alpha )}{1+2\alpha }}$, where $\alpha \ne -{\displaystyle \frac{1}{2}}$ for a power-law solution. Equation (57) leads to the following $F\left(T\right)$ solutions:

• $\alpha \ne -{\displaystyle \frac{1}{2}}$: Equation (57) becomes a simple DE:

  $$\begin{array}{cc}\hfill F\left(T\right)& =\left({\displaystyle \frac{1+2\alpha }{1+\alpha }}\right)T{F}_T\left(T\right),\hfill \\ \hfill \Rightarrow & F\left(T\right)=F_0{T}^{{\displaystyle \frac{1+\alpha }{1+2\alpha }}}\equiv F_0{T}^{{\displaystyle \frac{n}{2}}},\hfill \end{array}$$

  (58)

  where $F_0$ is an integration constant, $\alpha \ne \left\{-{\displaystyle \frac{1}{2}},-1,0\right\}$, and $A_2=A_3^{{\displaystyle \frac{2(1+\alpha )}{1+2\alpha }}}$. We also refind the Equation (58) by the general way. The dust matter case $\alpha =0$ leads to GR solutions.
• $\alpha =-{\displaystyle \frac{1}{2}}$: We can approximate Equation (57) by setting $\alpha =-{\displaystyle \frac{1}{2}}+\Delta \alpha$ for very small $\Delta \alpha$ as:

  $$\begin{array}{cc}\hfill F\left(T\right)& \approx 4F_T\left(T\right)\left[\Delta \alpha T-{\displaystyle \frac{\left(1-2n\Delta \alpha \right)}{A_3^2(n-1)}}\right].\hfill \end{array}$$

  (59)
  If $n\to \infty$ and $\Delta \alpha \to 0$ for all $A_3$, we obtain the GR solutions. For an $A_3$ independent $F\left(T\right)$ solution, we need to satisfy the condition $\Delta \alpha \approx {\displaystyle \frac{1}{2n}}$ for a large n. In this case, Equation (59) becomes, for $n\gg 1$:

  $$\begin{array}{cc}\hfill F\left(T\right)& \approx {\displaystyle \frac{2}{n}}T{F}_T\left(T\right).\hfill \end{array}$$

  (60)
  The solution is exactly Equation (58), and then Equation (60) proves that the function $F\left(T\right)$ is defined around $\alpha =-{\displaystyle \frac{1}{2}}$ and that n is very large.

We need to find a possible $A_3$ satisfying Equation (55a–e) for the Equation (58) solution for all $\alpha \ne -{\displaystyle \frac{1}{2}}$ cases. By using Equation (55c,e) and then by substituting Equation (58), we obtain as the characteristic equation for T:

$$\begin{array}{cc}\hfill 0& ={\displaystyle \frac{2\kappa {\rho }_0}{F_0}}{A}_3^{-{\displaystyle \frac{2(1+\alpha )(2+3\alpha )}{1+2\alpha }}}-\left({\displaystyle \frac{1}{1+2\alpha }}\right)T^{{\displaystyle \frac{1+\alpha }{1+2\alpha }}}-{\displaystyle \frac{4}{A_3^2}}\left({\displaystyle \frac{1+\alpha }{1+2\alpha }}\right)T^{-{\displaystyle \frac{\alpha }{1+2\alpha }}}.\hfill \end{array}$$

(61)

Then, Equation (55a) in terms of $\alpha$ is:

$$\begin{array}{cc}\hfill T=& 2\left({\displaystyle \frac{5+6\alpha }{1+2\alpha }}\right){\displaystyle \frac{A_3^{\prime 2}}{A_3^2}}-{\displaystyle \frac{2}{A_3^2}}.\hfill \end{array}$$

(62)

From Equation (61), we can isolate T in terms of $A_3$ and then equate to Equation (62), leading to the DE for $A_3\left(t\right)$, depending on $\alpha$ and power of T inside Equation (61). There are possible simple analytical solutions:

• $\alpha =-{\displaystyle \frac{1}{3}}$ case: Equation (58) becomes as simple as $F\left(T\right)=F_0{T}^2$ and $n=4$, and Equation (55e) leads to $\rho ={\rho }_0{A}_3^{-4}$. Equations (61) and (62) become:

  $$\begin{array}{cc}\hfill 0=& T^2+{\displaystyle \frac{8}{3A_3^2}}T-{\displaystyle \frac{2\kappa {\rho }_0}{3F_0}}{A}_3^{-4},\hfill \end{array}$$

  (63a)

  $$\begin{array}{cc}\hfill T=& 18{\displaystyle \frac{A_3^{\prime 2}}{A_3^2}}-{\displaystyle \frac{2}{A_3^2}}.\hfill \end{array}$$

  (63b)
  By putting Equation (63a,b) together, we find as DE and as the solution:

  $$\begin{array}{cc}\hfill A_3^{\prime }=& {\delta }_2\sqrt{{\displaystyle \frac{2}{27}}\left({\displaystyle \frac{1}{2}}+{\delta }_1\sqrt{1+{\displaystyle \frac{3\kappa {\rho }_0}{8F_0}}}\right)}=\mathrm{constant},\hfill \\ \hfill & \Rightarrow A_3={\delta }_2\sqrt{{\displaystyle \frac{2}{27}}\left({\displaystyle \frac{1}{2}}+{\delta }_1\sqrt{1+{\displaystyle \frac{3\kappa {\rho }_0}{8F_0}}}\right)}t=c_{0}t,\hfill \end{array}$$

  (64)

  where $\left({\delta }_1,{\delta }_2\right)=\left(\pm 1,\pm 1\right)$. Equation (64) leads to a linear $A_3$ and this confirms the power-law ansatz result.
• $\alpha =-{\displaystyle \frac{2}{3}}$ case: Equation (58) becomes $F\left(T\right)=F_0{T}^{-1}$, $n=-2$ and Equation (55e) leads to $\rho \left(t\right)={\rho }_0=$ constant. Then Equations (61) and (62) become:

  $$\begin{array}{cc}\hfill 0& ={\displaystyle \frac{2\kappa {\rho }_0}{F_0}}+3T^{-1}+{\displaystyle \frac{4}{A_3^2}}{T}^{-2},\hfill \end{array}$$

  (65a)

  $$\begin{array}{cc}\hfill T=& -6{\displaystyle \frac{A_3^{\prime 2}}{A_3^2}}-{\displaystyle \frac{2}{A_3^2}}.\hfill \end{array}$$

  (65b)

  By putting Equation (65a,b) together, we obtain as DE:

  $$\begin{array}{cc}\hfill 3A_3^{\prime 2}=& {\displaystyle \frac{3F_0}{8\kappa {\rho }_0}}{A}_3\left[A_3+{\delta }_1\sqrt{A_3^2-{\displaystyle \frac{32\kappa {\rho }_0}{9F_0}}}\right]-1,\hfill \\ \hfill \Rightarrow {\displaystyle \frac{{\delta }_2(t-t_0)}{\sqrt{3}}}=& \int d{A}_3{\left[{\displaystyle \frac{3F_0}{8\kappa {\rho }_0}}{A}_3\left[A_3+{\delta }_1\sqrt{A_3^2-{\displaystyle \frac{32\kappa {\rho }_0}{9F_0}}}\right]-1\right]}^{-1},\hfill \end{array}$$

  (66)

  where ${\delta }_2=\pm 1$. This last integral is complex to solve. However, there are two possible limits:

  (a)

  Low fluid density limit ${\displaystyle \frac{\kappa {\rho }_0}{F_0}}\ll 1$: Equation (66) simplifies for ${\delta }_1=+1$:

  $$\begin{array}{cc}\hfill {\delta }_2(t-t_0)\approx & -{\displaystyle \frac{4\kappa {\rho }_0}{\sqrt{3}{F}_0}}{A}_3^{-1},\hfill \\ \hfill \Rightarrow A_3\approx & {\displaystyle \frac{4{\delta }_2\kappa {\rho }_0}{\sqrt{3}{F}_0(t_0-t)}}.\hfill \end{array}$$

  (67)

  We find at Equation (67) an $A_3$ component for a contracting universe.

  (b)

  High fluid density limit ${\displaystyle \frac{\kappa {\rho }_0}{F_0}}\gg 1$: Equation (66) simplifies for ${\delta }_1=+1$:

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_2(t-t_0)}{\sqrt{3}}}\approx & -A_3+{\delta }_1\sqrt{-{\displaystyle \frac{F_0}{8\kappa {\rho }_0}}}{A}_3^2,\hfill \\ \hfill \Rightarrow A_3\approx & {\left(-{\displaystyle \frac{2\kappa {\rho }_0}{F_0}}\right)}^{1/2}\left[1+{\delta }_3\sqrt{1+{\displaystyle \frac{2{\delta }_1{\delta }_2}{\sqrt{3}}}{\left(-{\displaystyle \frac{F_0}{2\kappa {\rho }_0}}\right)}^{1/2}(t-t_0)}\right],\hfill \\ \hfill \approx & {\delta }_1{\left(-{\displaystyle \frac{2\kappa {\rho }_0}{F_0}}\right)}^{1/2}\left(1+{\delta }_3\right)+{\displaystyle \frac{{\delta }_2{\delta }_3}{\sqrt{3}}}(t-t_0),\hfill \end{array}$$

  (68)

  where ${\delta }_3=\pm 1$. We find from Equation (68) a linear $A_3$ as the highest fluid density limit as for vacuum solutions and also confirm the power-law ansatz result.

• $\alpha =-{\displaystyle \frac{1}{4}}$ case: Equation (58) becomes $F\left(T\right)=F_0{T}^{{\displaystyle \frac{3}{2}}}$ and $n=3$, and Equation (55e) leads to $\rho \left(t\right)={\rho }_0{A}_3^{-15/4}$. Then, Equations (61) and (62) become:

  $$\begin{array}{cc}\hfill 0& =T^{{\displaystyle \frac{3}{2}}}+{\displaystyle \frac{3}{A_3^2}}{T}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{\kappa {\rho }_0}{F_0}}{A}_3^{-{\displaystyle \frac{15}{4}}},\hfill \end{array}$$

  (69a)

  $$\begin{array}{cc}\hfill T=& 14{\left(ln\left(A_3\right)\right)}^{\prime 2}-{\displaystyle \frac{2}{A_3^2}}.\hfill \end{array}$$

  (69b)

  By putting together Equation (69a,b), we find the DE:

  $$\begin{array}{cc}\hfill A_3^{\prime }=& {\displaystyle \frac{{\delta }_1}{\sqrt{14}{2}^{1/3}{A}_3^{1/4}}}[{\left(\sqrt{{\displaystyle \frac{{\left(\kappa {\rho }_0\right)}^2}{F_0^2}}+4A_3^{3/2}}+{\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^{2/3}\hfill \\ \hfill & +2^{4/3}{A}_3{\left(\sqrt{{\displaystyle \frac{\left(\kappa {\rho }_0\right)}{F_0^2}}+4A_3^{3/2}}+{\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^{-2/3}{]}^{1/2}\hfill \\ \hfill \Rightarrow & {\displaystyle \frac{{\delta }_{1}t}{\sqrt{14}}}=\int d{A}_3[{\left(\sqrt{1+{\left({\displaystyle \frac{\kappa {\rho }_0}{2F_0{A}_3^{3/4}}}\right)}^2}+{\displaystyle \frac{\kappa {\rho }_0}{2F_0{A}_3^{3/4}}}\right)}^{2/3}\hfill \\ \hfill & +{\left(\sqrt{1+{\left({\displaystyle \frac{\kappa {\rho }_0}{2F_0{A}_3^{3/4}}}\right)}^2}+{\displaystyle \frac{\kappa {\rho }_0}{2F_0{A}_3^{3/4}}}\right)}^{-2/3}{]}^{-1/2}.\hfill \end{array}$$

  (70)

  Equation (70) is complex to solve. However, there are two limit cases where we can solve this equation:

  (a)

  Low fluid density limit ${\displaystyle \frac{\kappa {\rho }_0}{2F_0{A}_3^{3/4}}}\ll 1$: In this situation, Equation (70) will be approximated at the 1st order level:

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_{1}t}{\sqrt{7}}}\approx & \int d{A}_3\left[1-{\displaystyle \frac{1}{36}}{\left({\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^2{A}_3^{-3/2}\right],\hfill \\ \hfill \approx & A_3+{\displaystyle \frac{1}{18\sqrt{A_3}}}{\left({\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^2+C_1,\hfill \\ \hfill \Rightarrow & A_3^{3/2}-{\displaystyle \frac{{\delta }_1\left(t-t_0\right)}{\sqrt{7}}}{A}_3^{1/2}+{\displaystyle \frac{1}{18}}{\left({\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^2\approx 0,\hfill \end{array}$$

  (71)

  where $C_1$ is an integration constant, and $t_0$ depends on this constant. For weak ${\displaystyle \frac{\kappa {\rho }_0}{F_0}}$, the only relevant solution to Equation (71) leading to a real function for $A_3$ is with the ${\delta }_1=-1$ subcase:

  $$\begin{array}{cc}\hfill A_3\left(t\right)\approx & {\displaystyle \frac{1}{6^{4/3}·7^{2/3}}}{\left(\sqrt{48\sqrt{7}{\left(t-t_0\right)}^3+49{\left({\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^4}-7{\left({\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^2\right)}^{2/3}\hfill \\ \hfill & +{\displaystyle \frac{4{\left(t-t_0\right)}^2}{6^{2/3}\sqrt[3]{7}}}{\left(\sqrt{48\sqrt{7}{\left(t-t_0\right)}^3+49{\left({\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^4}-7{\left({\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^2\right)}^{-2/3}-{\displaystyle \frac{2\left(t-t_0\right)}{3\sqrt{7}}}\hfill \end{array}$$

  (72)

  For the ${\displaystyle \frac{\kappa {\rho }_0}{F_0}}$ very weak limit, Equation (72) will simplify as:

  $$\begin{array}{cc}\hfill A_3\left(t\right)\approx & {\displaystyle \frac{7}{324}}{\left({\displaystyle \frac{\kappa {\rho }_0}{F_0}}\right)}^4{\left(t-t_0\right)}^{-2},\hfill \\ \hfill \approx & c_0{t}^{-2}\mathrm{for}{t}_0=0.\hfill \end{array}$$

  (73)

  (b)

  High fluid density limit ${\displaystyle \frac{\kappa {\rho }_0}{2F_0{A}_3^{3/4}}}\gg 1$ (or ${\displaystyle \frac{F_0}{\kappa {\rho }_0}}\ll 1$): In this last case, Equation (70) will be approximated as:

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_{1}t}{\sqrt{14}}}\approx & {\left({\displaystyle \frac{F_0}{\kappa {\rho }_0}}\right)}^{1/3}\int d{A}_3{A}_3^{1/4},\hfill \\ \hfill \approx & {\displaystyle \frac{4}{5}}{\left({\displaystyle \frac{F_0}{\kappa {\rho }_0}}\right)}^{1/3}{A}_3^{5/4}+C_2,\hfill \\ \hfill \to A_3\left(t\right)\approx & {\left({\displaystyle \frac{5}{4\sqrt{14}}}\right)}^{4/5}{\left({\displaystyle \frac{F_0}{\kappa {\rho }_0}}\right)}^{4/15}{\left(t-t_0\right)}^{4/5},\hfill \\ \hfill \approx & c_0{t}^{4/5}\mathrm{for}{t}_0=0,\hfill \end{array}$$

  (74)

  where $C_2$ is an integration constant, and $t_0$ depends on this constant.

All these previous teleparallel $F\left(T\right)$ solutions are new results. However, there are some specific solutions similar to the cases of quintessence processes ($\alpha <-{\displaystyle \frac{1}{3}}$ cases) and some $F\left(T\right)$ solutions similar to Ref. [50], but the approach and aims of this current section are different from recent works [49,50,51,52].

4.3. Exponential Ansatz Solutions

By using the Equation (32) exponential ansatz, we can also find FEs from Equation (35a–d) as follows:

$$\begin{array}{cc}\hfill T& =2c(c+2b)-{\displaystyle \frac{2}{c_0^2}}exp(-2ct),\hfill \end{array}$$

(75a)

$$\begin{array}{cc}\hfill B^{\prime }& =-(2c+b)+{\displaystyle \frac{1}{c_0^2(b-c)}}exp(-2ct),\hfill \end{array}$$

(75b)

$$\begin{array}{cc}\hfill \kappa \rho +{\displaystyle \frac{F\left(T\right)}{2}}& =\left(T+{\displaystyle \frac{2}{c_0^2}}exp(-2ct)\right)F_T\left(T\right),\hfill \end{array}$$

(75c)

$$\begin{array}{cc}\hfill -\kappa \rho & ={\displaystyle \frac{c{F}_T\left(T\right)}{\left(1+\alpha \right)}}\left[B^{\prime }+c-b\right].\hfill \end{array}$$

(75d)

In addition, Equation (34) becomes:

$$\begin{array}{c}\hfill \rho \left(t\right)={\displaystyle \frac{{\rho }_0}{c_0^{2\left(1+\alpha \right)}}}exp\left(-\left(1+\alpha \right)(b+2c)t\right).\end{array}$$

(76)

From Equation (75a), we find that:

$$\begin{array}{c}\hfill {\displaystyle \frac{exp(-2ct)}{c_0^2}}=c(c+2b)-{\displaystyle \frac{T}{2}}.\end{array}$$

(77)

Then, by substituting Equation (77) into Equations (75b) and (76) and then by adding Equation (75c,d), we find that:

$$\begin{array}{cc}\hfill F\left(T\right)& ={\displaystyle \frac{c{F}_T\left(T\right)}{\left(1+\alpha \right)(c-b)}}\left[T-2(c+2b)\left(b(1+2\alpha )-2\alpha c\right)\right],\hfill \end{array}$$

(78a)

$$\begin{array}{cc}\hfill \rho \left(T\right)& ={\rho }_0{c}_0^{\left(1+\alpha \right){\displaystyle \frac{b}{c}}}{\left(c(c+2b)-{\displaystyle \frac{T}{2}}\right)}^{\left(1+\alpha \right)\left(\frac{b}{2c}+1\right)},\hfill \end{array}$$

(78b)

where $c\ne b$, $c\ne 0$, and $\alpha \ne -1$. There are three possible cases:

• $c=-2b$: Equation (78a) becomes:

  $$\begin{array}{cc}\hfill F\left(T\right)& ={\displaystyle \frac{2}{3\left(1+\alpha \right)}}T{F}_T\left(T\right),\hfill \\ \hfill & \Rightarrow F\left(T\right)=F\left(0\right)T^{{\displaystyle \frac{3\left(1+\alpha \right)}{2}}}.\hfill \end{array}$$

  (79)
  We obtain a pure power-law $F\left(T\right)$ solution as expected, and Equation (78b) is:

  $$\begin{array}{cc}\hfill \rho \left(T\right)& ={\displaystyle \frac{{\rho }_0}{{(-2)}^{\frac{3\left(1+\alpha \right)}{4}}{c}_0^{\frac{\left(1+\alpha \right)}{2}}}}{T}^{{\displaystyle \frac{3\left(1+\alpha \right)}{4}}}.\hfill \end{array}$$

  (80)
• $b={\displaystyle \frac{2\alpha }{(1+2\alpha )}}c$: Equation (78a) becomes:

  $$\begin{array}{cc}\hfill F\left(T\right)& ={\displaystyle \frac{(1+2\alpha )}{\left(1+\alpha \right)}}T{F}_T\left(T\right),\hfill \\ \hfill & \Rightarrow F\left(T\right)=F\left(0\right)T^{{\displaystyle \frac{\left(1+\alpha \right)}{(1+2\alpha )}}},\hfill \end{array}$$

  (81)

  where $\alpha \ne \left\{-1,-{\displaystyle \frac{1}{2}}\right\}$. Then, Equation (78b) is:

  $$\begin{array}{cc}\hfill \rho \left(T\right)& ={\rho }_0{c}_0^{{\displaystyle \frac{2\alpha \left(1+\alpha \right)}{(1+2\alpha )}}}{\left({\displaystyle \frac{(1+6\alpha )}{(1+2\alpha )}}{c}^2-{\displaystyle \frac{T}{2}}\right)}^{{\displaystyle \frac{(1+3\alpha )\left(1+\alpha \right)}{(1+2\alpha )}}},\hfill \end{array}$$

  (82)

  where $c\ne 0$ is a free parameter, and $\alpha \ne \left\{-1,-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{3}},0\right\}$ for a teleparallel $F\left(T\right)$ solution.
• $c\ne -2b$: The general solution to Equation (78a) is:

  $$\begin{array}{c}\hfill F\left(T\right)=F\left(0\right){\left[T-2(c+2b)\left(b(1+2\alpha )-2\alpha c\right)\right]}^{{\displaystyle \frac{(c-b)\left(1+\alpha \right)}{c}}},\end{array}$$

  (83)

  where $b\ne c$, $c\ne 0$, and $\rho \left(T\right)$ is Equation (78b).

All these previous teleparallel $F\left(T\right)$ solutions are new results. These new classes of $F\left(T\right)$ solutions generalize the solutions of Ref. [50] and are simple functions.

5. Non-Linear Perfect Fluid Solutions

After solving and finding KS solutions for a perfect isotropic linear fluid, we need to know what is happening if the perfect fluid is not linear. We will use as matter source the perfect fluid with a non-linear EoS as $P\left(t\right)=\alpha \rho \left(t\right)+\beta {\left[\rho \left(t\right)\right]}^w$, valid for all $\alpha \le 1$, where $\alpha \ne -1$ with $w>1$ and $\beta {\rho }^{w-1}\ll \alpha$. We have in this non-linear EoS the linear dominating term plus a small power-law correction to compare with Section 4 solutions by highlighting the new terms. Different from Ref. [45] and Section 4 solutions, we will find several $\alpha <-1$ teleparallel $F\left(T\right)$ solutions, leading to some analytically phantom energy solutions. Equation (15) for the conservation law becomes:

$$\begin{array}{cc}\hfill & \left[\left(1+\alpha \right)\rho +\beta {\rho }^w\right]{\left(ln\left(A_2{A}_3^2\right)\right)}^{\prime }+{\rho }^{\prime }=0,\hfill \\ \hfill & \Rightarrow \rho \left(t\right)={\displaystyle \frac{{\rho }_0}{{\left[1-C{\left(A_2\left(t\right)A_3^2\left(t\right)\right)}^{\left(1+\alpha \right)(w-1)}\right]}^{\frac{1}{w-1}}}},\hfill \end{array}$$

(84)

where ${\rho }_0={\left(-{\displaystyle \frac{(1+\alpha )}{\beta }}\right)}^{{\displaystyle \frac{1}{w-1}}}$ ($\beta <0$ for a positive ${\rho }_0$), and C is an integration constant. There are, in principle, an infinite number of possibilities. With the current non-linear EoS, Equation (14a–d) become:

$$\begin{array}{cc}\hfill T=& 2{\left(ln\left(A_3\right)\right)}^{\prime }\left({\left(ln\left(A_3\right)\right)}^{\prime }+2{\left(ln\left(A_2\right)\right)}^{\prime }\right)-{\displaystyle \frac{2}{A_3^2}},\hfill \end{array}$$

(85a)

$$\begin{array}{cc}\hfill B^{\prime }=& -{\left(ln\left(A_2{A}_3^2\right)\right)}^{\prime }+{\displaystyle \frac{\frac{1}{A_3^2}-{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{″}}{{\left(ln\left(\frac{A_2}{A_3}\right)\right)}^{\prime }}},\hfill \end{array}$$

(85b)

$$\begin{array}{cc}\hfill \kappa \rho =& \left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}},\hfill \end{array}$$

(85c)

$$\begin{array}{cc}\hfill -\kappa \left(1+\alpha \right)\rho -\kappa \beta {\rho }^w=& \left[{\left(ln\left(A_3\right)\right)}^{\prime }\left(B^{\prime }+{\left(ln\left(A_3\right)\right)}^{\prime }-{\left(ln\left(A_2\right)\right)}^{\prime }\right)+{\left(ln\left(A_3\right)\right)}^{″}\right]F_T\left(T\right).\hfill \end{array}$$

(85d)

As in Section 4, we use similar ansatzes to solve Equation (85a–d) for the $A_2$, $A_3$, and $F\left(T\right)$ solutions.

5.1. Power-Law Solutions

By using the Equation (17) ansatz, Equation (84) becomes:

$$\begin{array}{c}\hfill \rho \left(t\right)={\displaystyle \frac{{\rho }_0}{{\left[1-C_1{t}^{(b+2c)(1+\alpha )(w-1)}\right]}^{\frac{1}{w-1}}}},\end{array}$$

(86)

where $b\ne -2c$ and $C_1=C{c}_0^{2\left(1+\alpha \right)(w-1)}$. For $b=-2c$, we obtain that $\rho ={\rho }_0=$ constant from Equation (84). Then, Equation (85a–d) become:

$$\begin{array}{cc}\hfill T& ={\displaystyle \frac{2c(c+2b)}{t^2}}-{\displaystyle \frac{2}{c_0^2}}{t}^{-2c},\hfill \end{array}$$

(87a)

$$\begin{array}{cc}\hfill B^{\prime }& =\left({\displaystyle \frac{(1-b-2c)}{t}}+{\displaystyle \frac{t^{1-2c}}{c_0^2(b-c)}}\right),\hfill \end{array}$$

(87b)

$$\begin{array}{cc}\hfill \kappa \rho & =\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}},\hfill \end{array}$$

(87c)

$$\begin{array}{cc}\hfill -\kappa \left(1+\alpha \right)\rho -\kappa \beta {\rho }^w& =c\left[{\displaystyle \frac{B^{\prime }}{t}}+{\displaystyle \frac{(c-b-1)}{t^2}}\right]F_T\left(T\right).\hfill \end{array}$$

(87d)

By substituting Equation (87b,c) into Equation (87d), we find as DE:

$$\begin{array}{cc}\hfill 0=& {\left(\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right)}^w+\left({\displaystyle \frac{{\kappa }^{w-1}(1+\alpha )}{\beta }}\right)\left(\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right)\hfill \\ \hfill & +{\displaystyle \frac{{\kappa }^{w-1}c}{\beta }}\left[-(c+2b)t^{-2}+{\displaystyle \frac{t^{-2c}}{c_0^2(b-c)}}\right]F_T\left(T\right).\hfill \end{array}$$

(88)

Equation (88) is the general and non-linear DE to solve for $F\left(T\right)$. We need to transform Equation (88) into a solvable DE. By setting $\mathbf{w}=\mathbf{2}$ in Equation (88), we find that:

$$\begin{array}{cc}\hfill & \left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\hfill \\ \hfill & ={\displaystyle \frac{\kappa (1+\alpha )}{2\beta }}\left[-1+{\delta }_1\sqrt{1-{\displaystyle \frac{2\beta }{\kappa {(1+\alpha )}^2}}\left[-2c(c+2b)t^{-2}+{\displaystyle \frac{2c{t}^{-2c}}{c_0^2(b-c)}}\right]F_T\left(T\right)}\right],\hfill \end{array}$$

(89)

where ${\delta }_1=\pm 1$. From Equation (87a), we isolate the relation:

$$\begin{array}{c}\hfill 2c(c+2b)t^{-2}=T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}.\end{array}$$

(90)

Then, Equation (89) becomes:

$$\begin{array}{cc}\hfill & 2\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)F_T\left(T\right)-F\left(T\right)\hfill \\ \hfill & ={\displaystyle \frac{\kappa (1+\alpha )}{\beta }}\left[-1+{\delta }_1\sqrt{1+{\displaystyle \frac{2\beta }{\kappa {(1+\alpha )}^2}}\left[\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)-{\displaystyle \frac{2c{t}^{-2c}}{c_0^2(b-c)}}\right]F_T\left(T\right)}\right].\hfill \end{array}$$

(91)

Equation (91) is a non-linear DE, and we need to simplify this equation. For a small quadratic correction (i.e., $\beta \ll (1+\alpha )$) to the linear perfect fluid EoS, Equation (91) will simplify as:

$$\begin{array}{cc}\hfill & 2\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)F_T\left(T\right)-F\left(T\right)\hfill \\ \hfill & \approx {\displaystyle \frac{\kappa (1+\alpha )}{\beta }}\left[\left({\delta }_1-1\right)+{\displaystyle \frac{{\delta }_1\beta }{\kappa {(1+\alpha )}^2}}\left[\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\right)-{\displaystyle \frac{2c{t}^{-2c}}{c_0^2(b-c)}}\right]F_T\left(T\right)\right].\hfill \end{array}$$

(92)

For ${\delta }_1=+1$, we obtain Equation (39) exactly for a linear perfect fluid. However, for ${\delta }_1=-1$, Equation (92) becomes, for $\alpha \ne -{\displaystyle \frac{3}{2}}$:

$$\begin{array}{cc}\hfill F\left(T\right)-& {\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}\approx {\displaystyle \frac{F_T\left(T\right)}{(1+\alpha )}}\left[(3+2\alpha )\left(T+{\displaystyle \frac{2}{c_0^2}}{t}^{-2c}\left(T\right)\right)-{\displaystyle \frac{2c{t}^{-2c}\left(T\right)}{c_0^2(b-c)}}\right],\hfill \\ \hfill \Rightarrow F\left(T\right)\approx & {\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}+\left(F\left(0\right)-{\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}\right)\hfill \\ \hfill & \times exp\left[(1+\alpha )\int dT{\left[(3+2\alpha )T+{\displaystyle \frac{2}{c_0^2}}\left(3+2\alpha -{\displaystyle \frac{c}{(b-c)}}\right)t^{-2c}\left(T\right)\right]}^{-1}\right],\hfill \end{array}$$

(93)

where $\beta \ne 0$, $b\ne c$, and $c\ne 0$. In Equation (93), we obtained the general solution for possible $F\left(T\right)$. In addition, we need to solve the characteristic Equation (38), such as for perfect fluid cases in Section 4, for each subcase.

For $\alpha =-{\displaystyle \frac{3}{2}}$ cases, Equation (93) simplifies as:

$$\begin{array}{cc}\hfill F\left(T\right)\approx & -{\displaystyle \frac{\kappa }{\beta }}+\left(F\left(0\right)+{\displaystyle \frac{\kappa }{\beta }}\right)exp\left[{\displaystyle \frac{c_0^2(b-c)}{4c}}\int dT{t}^{2c}\left(T\right)\right],\hfill \end{array}$$

(94)

where $b\ne c$ and $c\ne 0$. Equation (94) is very similar to Equation (40) because the same integrals are used to solve. The only changes from the $F\left(T\right)$ solutions of Section 4.1 are for the shifting constants depending on $\beta$.

For the $\mathbf{c}=-\mathbf{2}\mathbf{b}$ case, Equation (38) simplifies as $t^{4b}\left(T\right)={\displaystyle \frac{c_0^2(-T)}{2}}$, and then Equation (93) becomes:

$$\begin{array}{c}\hfill F\left(T\right)\approx {\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}+\left(F\left(0\right)-{\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}\right)T^{-{\displaystyle \frac{3(1+\alpha )}{2}}},\end{array}$$

(95)

where $\beta \ne 0$.

As for the linear perfect fluid case, we obtain the same cases $\mathbf{c}={\displaystyle \frac{\mathbf{1}}{\mathbf{2}}},-{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}},\mathbf{1},-\mathbf{1},{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}},\mathbf{2}$, $-\mathbf{2},\mathbf{3},$$-\mathbf{3},\mathbf{4}$ for $c\ne -2b$. In this section, we will develop only the $\mathbf{c}={\displaystyle \frac{\mathbf{1}}{\mathbf{2}}},\mathbf{1},-\mathbf{1}$, and $\mathbf{2}$ subcases, as done in Section 4, because all other subcases will not lead to an analytic and closed $F\left(T\right)$ solution. The analytic and closed teleparallel $F\left(T\right)$ solutions are:

• $\mathbf{c}={\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}$: By substituting Equation (43) for $t^{-1}\left(T\right)$, Equation (93) becomes (i.e., $\alpha \ne -{\displaystyle \frac{3}{2}}$):

  $$\begin{array}{cc}\hfill F\left(T\right)\approx & {\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}+\left(F\left(0\right)-{\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}\right)\hfill \\ \hfill & \times [-c_0^4\left({\displaystyle \frac{1}{2}}+2b\right)((3+2\alpha )T+{\displaystyle \frac{4}{c_0^4}}\left(3+2\alpha +{\displaystyle \frac{1}{(1-2b)}}\right)\hfill \\ \hfill & \times {\displaystyle \frac{\left[1+{\delta }_2\sqrt{1+\left(\frac{1}{2}+2b\right)c_0^{4}T}\right]}{\left(1+4b\right)}}){]}^{{\displaystyle \frac{1+\alpha }{3+2\alpha }}}exp[2(1+\alpha )\left(1-2b+{\displaystyle \frac{1}{3+2\alpha }}\right)\hfill \\ \hfill & \times {tanh}^{-1}\left(1+(1-2b)(3+2\alpha )\left(1+{\delta }_2\sqrt{1+\left({\displaystyle \frac{1}{2}}+2b\right)c_0^{4}T}\right)\right)],\hfill \end{array}$$

  (96)

  where $b\ne -{\displaystyle \frac{1}{4}}$, $\alpha \ne -{\displaystyle \frac{3}{2}}$, and ${\delta }_2=\pm 1$. For $\alpha =-{\displaystyle \frac{3}{2}}$, Equation (94) becomes, with Equation (43):

  $$\begin{array}{cc}\hfill F\left(T\right)=& -{\displaystyle \frac{\kappa }{\beta }}+\left(F\left(0\right)+{\displaystyle \frac{\kappa }{\beta }}\right){\left[1+{\delta }_1\sqrt{1+\left({\displaystyle \frac{1}{2}}+2b\right)c_0^{4}T}\right]}^{{\displaystyle \frac{1}{2}}-b}\hfill \\ \hfill & \times exp\left[{\delta }_1\left(b-{\displaystyle \frac{1}{2}}\right)\sqrt{1+\left({\displaystyle \frac{1}{2}}+2b\right)c_0^{4}T}\right],\hfill \end{array}$$

  (97)

  where $b\ne {\displaystyle \frac{1}{2}}$ and $\beta \ne 0$.
• $\mathbf{c}=\mathbf{1}$: By substituting Equation (46) for $t^{-2}\left(T\right)$, Equation (93) becomes (i.e., $\alpha \ne -{\displaystyle \frac{3}{2}}$):

  $$\begin{array}{cc}\hfill F\left(T\right)\approx & {\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}+\left(F\left(0\right)-{\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}\right)T^{\left[{\displaystyle \frac{(1+\alpha )(b-1)\left(c_0^2\left(1+2b\right)-1\right)}{c_0^2(3+2\alpha )(b-1)\left(2b+1\right)-1}}\right]},\hfill \end{array}$$

  (98)

  where $b\ne 1$. For $\alpha =-{\displaystyle \frac{3}{2}}$, Equation (94) becomes, with Equation (46):

  $$\begin{array}{c}\hfill F\left(T\right)=-{\displaystyle \frac{\kappa }{\beta }}+\left(F\left(0\right)+{\displaystyle \frac{\kappa }{\beta }}\right)T^{{\displaystyle \frac{(b-1)}{2}}\left[(1+2b)c_0^2-1\right]},\end{array}$$

  (99)

  where $b\ne 1$ and $\beta \ne 0$.
• $\mathbf{c}=-\mathbf{1}$: By substituting Equation (49) for $t^2\left(T\right)$, Equation (93) becomes (i.e., $\alpha \ne -{\displaystyle \frac{3}{2}}$):

  $$\begin{array}{cc}\hfill F\left(T\right)\approx & {\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}+\left(F\left(0\right)-{\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}\right)\hfill \\ \hfill & \times {\left[-{\displaystyle \frac{4(3+2\alpha )}{b+1}}{T}^2-{\displaystyle \frac{16(1-2b)}{c_0^2}}{\left(3+2\alpha +{\displaystyle \frac{1}{b+1}}\right)}^2\right]}^{{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1+\alpha }{3+2\alpha }}-(b+1)(1+\alpha )\right)}\hfill \\ \hfill & \times {\left[T+\sqrt{T^2+16(1-2b)c_0^{-2}}\right]}^{{\displaystyle \frac{{\delta }_1}{2}}\left({\displaystyle \frac{1+\alpha }{3+2\alpha }}+(b+1)(1+\alpha )\right)}\hfill \\ \hfill & \times exp[{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1+\alpha }{3+2\alpha }}-(b+1)(1+\alpha )\right)\hfill \\ \hfill & \times {tanh}^{-1}\left[{\displaystyle \frac{{\delta }_1\left(3+2\alpha -\frac{1}{b+1}\right)T}{\left(3+2\alpha +\frac{1}{b+1}\right)\sqrt{T^2+16(1-2b)c_0^{-2}}}}\right]],\hfill \end{array}$$

  (100)

  where $b\ne -1$, $\alpha \ne -{\displaystyle \frac{3}{2}}$, and ${\delta }_1=\pm 1$. For $\alpha =-{\displaystyle \frac{3}{2}}$, Equation (94) becomes, with Equation (49):

  $$\begin{array}{cc}\hfill F\left(T\right)=& -{\displaystyle \frac{\kappa }{\beta }}+\left(F\left(0\right)+{\displaystyle \frac{\kappa }{\beta }}\right){\left[T+\sqrt{T^2+16(1-2b)c_0^{-2}}\right]}^{-{\displaystyle \frac{{\delta }_1}{2}}(b+1)}\hfill \\ \hfill & \times exp\left[-{\displaystyle \frac{c_0^2(b+1)}{32(1-2b)}}T\left(T+{\delta }_1\sqrt{T^2+16(1-2b)c_0^{-2}}\right)\right],\hfill \end{array}$$

  (101)

  where $b\ne \left\{-1,{\displaystyle \frac{1}{2}}\right\}$ and $\beta \ne 0$.
• $\mathbf{c}=\mathbf{2}$: By substituting Equation (52) for $t^{-2}\left(T\right)$, Equation (93) becomes (i.e., $\alpha \ne -{\displaystyle \frac{3}{2}}$):

  $$\begin{array}{cc}\hfill F\left(T\right)\approx & {\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}+\left(F\left(0\right)-{\displaystyle \frac{2\kappa (1+\alpha )}{\beta }}\right)\hfill \\ \hfill & \times {\left[{\displaystyle \frac{T}{4c_0^2{(1+b)}^2(b-2)}}+2\left(3+2\alpha -{\displaystyle \frac{2}{b-2}}\right)\left[1+{\delta }_1\sqrt{1-{\displaystyle \frac{T}{8c_0^2{(1+b)}^2}}}\right]\right]}^{{\displaystyle \frac{(1+\alpha )(b-2)}{2}}}\hfill \\ \hfill & \times exp[{\delta }_1(1+\alpha )\left(b-2-{\displaystyle \frac{2}{3+2\alpha }}\right)\hfill \\ \hfill & \times {tanh}^{-1}\left[{\displaystyle \frac{{\delta }_1\left((3+2\alpha )(b-2)-2\right)-2\sqrt{1-\frac{T}{8c_0^2{(1+b)}^2}}}{(b-2)(3+2\alpha )}}\right]],\hfill \end{array}$$

  (102)

  where $b\ne 2$, $\alpha \ne -{\displaystyle \frac{3}{2}}$, and ${\delta }_1=\pm 1$. For $\alpha =-{\displaystyle \frac{3}{2}}$, Equation (94) becomes, with Equation (52):

  $$\begin{array}{cc}\hfill F\left(T\right)=& -{\displaystyle \frac{\kappa }{\beta }}+\left(F\left(0\right)+{\displaystyle \frac{\kappa }{\beta }}\right){\left[4(1+b)+4{\delta }_1\sqrt{{(1+b)}^2-{\displaystyle \frac{T}{8c_0^2}}}\right]}^{{\displaystyle \frac{(2-b)}{2}}}\hfill \\ \hfill & \times exp\left[{\displaystyle \frac{(2-b)(1+b)}{2\left((1+b)+{\delta }_1\sqrt{{(1+b)}^2-\frac{T}{8c_0^2}}\right)}}\right],\hfill \end{array}$$

  (103)

  where $b\ne 2$ and $\beta \ne 0$.

All these previous teleparallel $F\left(T\right)$ solutions are new results. Therefore, by comparing the $F\left(T\right)$ solutions described by Equations (39)–(54) in Section 4.1 with Equations (93)–(103), we see a number of similarities. Aside from a few additional and specific constants and parameters, Equations (93)–(103) of this section are related to the $F\left(T\right)$ solutions of Section 4.1. One of the logical explanation is that the same characteristic equations described by Equation (38) are used for both classes of $F\left(T\right)$ solutions. An additional explanation is that the DEs expressed by Equations (39) and (93), respectively, have very similar terms: this consideration goes more towards the similarity of the solutions. Because there are new non-linear fluid solutions, these new $F\left(T\right)$ solutions automatically further generalize the solutions in Ref. [50].

5.2. $A_2=A_3^n$ Ansatz Solutions

By setting $A_2=A_3^n$, we find as fluid density from Equation (84):

$$\begin{array}{c}\hfill \rho \left(t\right)={\displaystyle \frac{{\rho }_0}{{\left[1-C{\left(A_3\left(t\right)\right)}^{(2+n)\left(1+\alpha \right)(w-1)}\right]}^{\frac{1}{w-1}}}},\end{array}$$

(104)

where $n\ne -2$. For $n=-2$, Equation (104) leads to a constant fluid density for any $A_3\left(t\right)$. Then, Equation (85a–d) become:

$$\begin{array}{cc}\hfill T& =2(1+2n){\left(ln\left(A_3\right)\right)}^{\prime 2}-{\displaystyle \frac{2}{A_3^2}}\hfill \end{array}$$

(105a)

$$\begin{array}{cc}\hfill B^{\prime }& =-{\displaystyle \frac{\left({\left(ln\left(A_3\right)\right)}^{\prime \prime }+(2+n){\left(ln\left(A_3\right)\right)}^{\prime 2}+\frac{1}{(1-n)A_3^2}\right)}{{\left(ln\left(A_3\right)\right)}^{\prime }}}\hfill \end{array}$$

(105b)

$$\begin{array}{cc}\hfill \kappa \rho & =\left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}},\hfill \end{array}$$

(105c)

$$\begin{array}{cc}\hfill -\kappa \left(1+\alpha \right)\rho -\kappa \beta {\rho }^w& =\left[{\left(ln\left(A_3\right)\right)}^{\prime }{B}^{\prime }+{\left(ln\left(A_3\right)\right)}^{″}+(1-n){\left(ln\left(A_3\right)\right)}^{\prime 2}\right]F_T\left(T\right).\hfill \end{array}$$

(105d)

By substituting Equation (105b,c) into Equation (105d), we obtain as DE:

$$\begin{array}{cc}\hfill 0=& {\left[\left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right]}^w+{\displaystyle \frac{\left(1+\alpha \right){\kappa }^{w-1}}{\beta }}\left[\left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right]\hfill \\ \hfill & -{\displaystyle \frac{{\kappa }^{w-1}}{\beta }}\left[(1+2n){\left(ln\left(A_3\right)\right)}^{\prime 2}+{\displaystyle \frac{1}{(1-n)A_3^2}}\right]F_T\left(T\right).\hfill \end{array}$$

(106)

Equation (106) is the general and non-linear DE to solve for $F\left(T\right)$. From Equation (105a), we can isolate the expression:

$$\begin{array}{c}\hfill (1+2n){\left(ln\left(A_3\right)\right)}^{\prime 2}={\displaystyle \frac{1}{2}}\left(T+{\displaystyle \frac{2}{A_3^2}}\right),\end{array}$$

(107)

and then Equation (106) becomes:

$$\begin{array}{cc}\hfill 0=& {\left[\left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right]}^w+{\displaystyle \frac{\left(1+\alpha \right){\kappa }^{w-1}}{\beta }}\left[\left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right]\hfill \\ \hfill & -{\displaystyle \frac{{\kappa }^{w-1}}{2\beta }}\left[\left(T+{\displaystyle \frac{2}{A_3^2}}\right)+{\displaystyle \frac{2}{(1-n)A_3^2}}\right]F_T\left(T\right).\hfill \end{array}$$

(108)

For the $\mathbf{w}=\mathbf{2}$ fluid case and, in the situation where $\beta \ll 1+\alpha$, Equation (108) will become:

$$\begin{array}{cc}\hfill & \left(T+{\displaystyle \frac{2}{A_3^2}}\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\hfill \\ \hfill & ={\displaystyle \frac{\left(1+\alpha \right)\kappa }{2\beta }}\left[-1+{\delta }_1\sqrt{1+{\displaystyle \frac{2\beta }{\kappa {\left(1+\alpha \right)}^2}}\left[\left(T+{\displaystyle \frac{2}{A_3^2}}\right)+{\displaystyle \frac{2}{(1-n)A_3^2}}\right]F_T\left(T\right)}\right],\hfill \\ \hfill & \approx {\displaystyle \frac{\left(1+\alpha \right)\kappa }{2\beta }}\left({\delta }_1-1\right)+{\displaystyle \frac{{\delta }_1}{2\left(1+\alpha \right)}}\left[\left(T+{\displaystyle \frac{2}{A_3^2}}\right)+{\displaystyle \frac{2}{(1-n)A_3^2}}\right]F_T\left(T\right).\hfill \end{array}$$

(109)

For ${\delta }_1=+1$, we obtain Equation (57) exactly for a linear perfect fluid. We will solve only for the ${\delta }_1=-1$ situation, and Equation (109) becomes:

$$\begin{array}{cc}\hfill F\left(T\right)\approx & {\displaystyle \frac{2\left(1+\alpha \right)\kappa }{\beta }}+{\displaystyle \frac{1}{\left(1+\alpha \right)}}\left[(3+2\alpha )T+{\displaystyle \frac{4}{A_3^2(1-n)}}\left((2+\alpha )-n\left({\displaystyle \frac{3}{2}}+\alpha \right)\right)\right]F_T\left(T\right)\hfill \end{array}$$

(110)

For a pure $F\left(T\right)$ solution valid for all $A_3$, we need to set $n={\displaystyle \frac{2+\alpha }{\frac{3}{2}+\alpha }}$, where $n\ne 1$ and $\alpha \ne -{\displaystyle \frac{3}{2}}$. From Equation (110), there are two possible situations:

• $\alpha \ne -{\displaystyle \frac{3}{2}}$ general case: Equation (110) simplifies and the solution is:

  $$\begin{array}{cc}\hfill F\left(T\right)\approx & {\displaystyle \frac{2\left(1+\alpha \right)\kappa }{\beta }}+{\displaystyle \frac{(3+2\alpha )}{\left(1+\alpha \right)}}T{F}_T\left(T\right),\hfill \\ \hfill \Rightarrow F\left(T\right)\approx & {\displaystyle \frac{2\left(1+\alpha \right)\kappa }{\beta }}+\left(F\left(0\right)-{\displaystyle \frac{2\left(1+\alpha \right)\kappa }{\beta }}\right)T^{{\displaystyle \frac{\left(1+\alpha \right)}{2\left(\frac{3}{2}+\alpha \right)}}}.\hfill \end{array}$$

  (111)
• $\alpha =-{\displaystyle \frac{3}{2}}$: We set $\alpha =-{\displaystyle \frac{3}{2}}+\Delta \alpha$ for studying $F\left(T\right)$ solutions around $\alpha =-{\displaystyle \frac{3}{2}}$ (where $\Delta \alpha \to 0$). Then, Equation (110) becomes:

  $$\begin{array}{cc}\hfill F\left(T\right)\approx & {\displaystyle \frac{\left(-1+2\Delta \alpha \right)\kappa }{\beta }}-4\left[\Delta \alpha T+{\displaystyle \frac{\left(1-2(n-2)\Delta \alpha \right)}{A_3^2(1-n)}}\right]F_T\left(T\right).\hfill \end{array}$$

  (112)
  If $n\to \infty$ and $\Delta \alpha \to 0$, we obtain a GR solution. For an $A_3$ independent solution, we need to satisfy $\Delta \alpha \approx {\displaystyle \frac{1}{2(n-2)}}$, where $n\gg 1$, and then Equation (112) becomes:

  $$\begin{array}{cc}\hfill F\left(T\right)\approx & -{\displaystyle \frac{\kappa }{\beta }}-{\displaystyle \frac{2}{(n-2)}}T{F}_T\left(T\right),\hfill \\ \hfill \Rightarrow & F\left(T\right)\approx -{\displaystyle \frac{\kappa }{\beta }}+\left(F\left(0\right)+{\displaystyle \frac{\kappa }{\beta }}\right)T^{1-{\displaystyle \frac{n}{2}}}.\hfill \end{array}$$

  (113)
  We find at Equation (113) a finite limit of $F\left(T\right)$ valid for large n. If there is a singularity at $\alpha =-{\displaystyle \frac{3}{2}}$, the $F\left(T\right)$ solution is well-defined close to this point.

For finding $A_3$ with $\alpha \ne -{\displaystyle \frac{3}{2}}$, we need to put Equations (104) and (105c) together and then find a DE for the $w=2$ fluid case. Then, by substituting Equation (111) inside, we find that:

$$\begin{array}{c}\hfill 0=\left({\displaystyle \frac{\kappa }{\beta }}\right){\displaystyle \frac{\beta {\rho }_0+\left(1+\alpha \right)\left(1-C{A}_3^{(2+n)\left(1+\alpha \right)}\right)}{\left(F\left(0\right)-\frac{2\left(1+\alpha \right)\kappa }{\beta }\right)\left(1-C{A}_3^{(2+n)\left(1+\alpha \right)}\right)}}-{\displaystyle \frac{2}{A_3^2}}\left({\displaystyle \frac{1+\alpha }{3+2\alpha }}\right)T^{{\displaystyle \frac{\left(1+\alpha \right)}{\left(3+2\alpha \right)}}-1}+\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1+\alpha }{3+2\alpha }}\right)T^{{\displaystyle \frac{\left(1+\alpha \right)}{\left(3+2\alpha \right)}}}.\end{array}$$

(114)

Equation (114) is, in principle, hard to solve. However, there are some specific values of $\alpha$ (linear EoS parameter) where $\alpha <-1$ (phantom energy cases), leading to analytical solutions for $A_3$:

• $\alpha =-{\displaystyle \frac{4}{3}}$: Equation (114) becomes:

  $$\begin{array}{cc}\hfill 0=& \left({\displaystyle \frac{\kappa A_3^2}{2\beta }}\right){\displaystyle \frac{\beta {\rho }_0-\frac{1}{3}\left(1-C{A}_3^{-2}\right)}{\left(F\left(0\right)+\frac{2\kappa }{3\beta }\right)\left(1-C{A}_3^{-2}\right)}}+{\displaystyle \frac{3A_3^2}{4}}{T}^{-1}+T^{-2},\hfill \\ \hfill & \Rightarrow T={\left[-{\displaystyle \frac{3A_3^2}{8}}+{\delta }_1\sqrt{{\displaystyle \frac{9A_3^4}{64}}-\left({\displaystyle \frac{\kappa A_3^2}{2\beta }}\right){\displaystyle \frac{\beta {\rho }_0-\frac{1}{3}\left(1-C{A}_3^{-2}\right)}{\left(F\left(0\right)+\frac{2\kappa }{3\beta }\right)\left(1-C{A}_3^{-2}\right)}}}\right]}^{-1},\hfill \end{array}$$

  (115)

  where ${\delta }_1=\pm 1$. Equation (105a) will be in terms of $\alpha$:

  $$\begin{array}{cc}\hfill T& =\left({\displaystyle \frac{11+6\alpha }{\frac{3}{2}+\alpha }}\right){\displaystyle \frac{A_3^{\prime 2}}{A_3^2}}-{\displaystyle \frac{2}{A_3^2}}.\hfill \end{array}$$

  (116)
  Then, by putting Equations (115) and (116) together for $\alpha =-{\displaystyle \frac{4}{3}}$, we find the DE to solve for $A_3$:

  $$\begin{array}{c}\hfill 3{\delta }_2{A}_3^{\prime }={\left[1-{\displaystyle \frac{4}{3}}{\left[1-{\delta }_1\sqrt{1+{\displaystyle \frac{32}{27A_3^2}}\left({\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\right)\left(1-{\displaystyle \frac{3\beta {\rho }_0}{\left(1-C{A}_3^{-2}\right)}}\right)}\right]}^{-1}\right]}^{1/2},\end{array}$$

  (117)

  where $\tilde{F}\left(0\right)=F\left(0\right)+{\displaystyle \frac{2\kappa }{3\beta }}$ is an effective constant. This integral described by Equation (117) is very complex, and there is no exact solution. Therefore, there are two solvable limits leading to approximated $A_3$ solutions:

  (a)

  $\kappa {\rho }_0\ll \tilde{F}\left(0\right)$ (low density limit): Equation (117) will be approximated as:

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_2(t-t_0)}{3\sqrt{3}}}\approx & \int {\left[{\displaystyle \frac{{\delta }_1\sqrt{\frac{32}{27}\left(\frac{\kappa }{\beta \tilde{F}\left(0\right)}\right)+A_3^2}-A_3}{3{\delta }_1\sqrt{\frac{32}{27}\left(\frac{\kappa }{\beta \tilde{F}\left(0\right)}\right)+A_3^2}+A_3}}\right]}^{1/2}[1+\left({\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\right){\displaystyle \frac{32{\delta }_1{A}_3^3}{9\left(C-A_3^2\right)}}\hfill \\ \hfill & \times {\left({\displaystyle \frac{32}{27}}\left({\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\right)+A_3^2\right)}^{-1/2}{\left({\delta }_1\sqrt{{\displaystyle \frac{32}{27}}\left({\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\right)+A_3^2}-A_3\right)}^{-1}\hfill \\ \hfill & \times {\left(3{\delta }_1\sqrt{{\displaystyle \frac{32}{27}}\left({\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\right)+A_3^2}+A_3\right)}^{-1}]d{A}_3.\hfill \end{array}$$

  (118)

  Equation (118) is very complex to solve and will not give an exact solution. But, there are two limit cases for ${\delta }_1=-1$:

  • ${\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\ll 1$: Equation (118) will be approximated as:

    $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_2(t-t_0)}{3\sqrt{3}}}\approx & A_3+{\displaystyle \frac{8\kappa }{27\beta \tilde{F}\left(0\right)}}\left({\displaystyle \frac{1}{A_3}}-{\displaystyle \frac{3}{\sqrt{C}}}{tanh}^{-1}\left({\displaystyle \frac{A_3}{\sqrt{C}}}\right)\right),\hfill \\ \hfill \approx & A_3+f_1\left(A_3\right),\hfill \end{array}$$

    (119)

    where $f_1\left(A_3\right)\ll A_3$. In the ${\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\to 0$ limit, we find that $A_3={\displaystyle \frac{{\delta }_2(t-t_0)}{3\sqrt{3}}}$ as for a vacuum solution. The function $A_3$ is bounded by $0<A_3<+\sqrt{C}$ (or $-\sqrt{C}<A_3<0$), and the limits on $f_1\left(A_3\right)$ are:

    –

    $A_3\to 0^{+}$ and $A_3\to -\sqrt{C}$: $f_1\left(A_3\right)\to +\infty$.

    –

    $A_3\to +\sqrt{C}$ and $A_3\to 0^{-}$: $f_1\left(A_3\right)\to -\infty$.

    With $f_1\left(A_3\right)$ going to infinity, this solution leads to an unstable universe.
  • ${\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\gg 1$: Equation (118) will be simplified as:

    $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}\approx & A_3+{\displaystyle \frac{1}{8}}\sqrt{{\displaystyle \frac{3\beta \tilde{F}\left(0\right)}{2\kappa }}}\left(3Cln\left(A_3^2-C\right)+5A_3^2\right),\hfill \\ \hfill \approx & A_3+f_2\left(A_3\right),\hfill \end{array}$$

    (120)

    where $f_2\left(A_3\right)\ll A_3$. For the ${\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\to 0$ limit, we still find that $A_3={\displaystyle \frac{{\delta }_2(t-t_0)}{3\sqrt{3}}}$, as previously. The $f_2\left(A_3\right)$ correction function is defined for $A_3<-\sqrt{C}$ and $A_3>+\sqrt{6}$ only, and the limits are:

    –

    $A_3\to \pm \infty$: $f_2\left(A_3\right)\to +\infty$.

    –

    $A_3\to \pm \sqrt{C}$: $f_2\left(A_3\right)\to -\infty$.

    With $f_2\left(A_3\right)$ going to infinity, this solution leads to an unstable universe.

  (b)

  $\kappa {\rho }_0\gg \tilde{F}\left(0\right)$ (high density limit): Equation (117) will be approximated as:

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}\approx & A_3-{\displaystyle \frac{{\delta }_1}{4}}{\left({\displaystyle \frac{\tilde{F}\left(0\right)}{2\kappa {\rho }_0}}\right)}^{1/2}\left(A_3\sqrt{C-A_3^2}+Carctan\left({\displaystyle \frac{A_3}{\sqrt{C-A_3^2}}}\right)\right),\hfill \\ \hfill \approx & A_3+g\left(A_3\right),\hfill \end{array}$$

  (121)

  where $g\left(A_3\right)\ll A_3$. Equation (121) is a bounded equation because there is an inverse trigonometric function ($-\sqrt{C}<A_3<+\sqrt{C}$). In the ${\displaystyle \frac{\tilde{F}\left(0\right)}{2\kappa {\rho }_0}}\to 0$ limit, we find that $A_3\to {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}$, as for a vacuum solution. The $g\left(A_3\right)$ correction to $A_3$ will be for both limits of $A_3$:

  • $A_3\to +\sqrt{C}$: $g\left(A_3\right)=-{\displaystyle \frac{{\delta }_{1}C\pi }{8}}{\left({\displaystyle \frac{\tilde{F}\left(0\right)}{2\kappa {\rho }_0}}\right)}^{1/2}$.
  • $A_3\to -\sqrt{C}$: $g\left(A_3\right)=+{\displaystyle \frac{{\delta }_{1}C\pi }{8}}{\left({\displaystyle \frac{\tilde{F}\left(0\right)}{2\kappa {\rho }_0}}\right)}^{1/2}$.

  The $A_3$ solution is bounded by two linear functions of t as:

  $$\begin{array}{c}\hfill {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}-{\displaystyle \frac{{\delta }_{1}C\pi }{8}}{\left({\displaystyle \frac{\tilde{F}\left(0\right)}{2\kappa {\rho }_0}}\right)}^{1/2}<A_3\left(t\right)<{\displaystyle \frac{{\delta }_2(t-t_0)}{3}}+{\displaystyle \frac{{\delta }_{1}C\pi }{8}}{\left({\displaystyle \frac{\tilde{F}\left(0\right)}{2\kappa {\rho }_0}}\right)}^{1/2}.\end{array}$$

  (122)

• $\alpha =-{\displaystyle \frac{5}{3}}$: Equation (114) becomes, in this case:

  $$\begin{array}{cc}\hfill 0=& -\left({\displaystyle \frac{2\kappa }{3\beta }}\right){\displaystyle \frac{\beta {\rho }_0-\frac{2}{3}\left(1-C\right)}{\left(F\left(0\right)+\frac{4\kappa }{3\beta }\right)\left(1-C\right)}}+{\displaystyle \frac{8}{3A_3^2}}T+T^2,\hfill \\ \hfill & \Rightarrow T=-{\displaystyle \frac{4}{3A_3^2}}+{\delta }_1\sqrt{{\displaystyle \frac{16}{9A_3^4}}+\left({\displaystyle \frac{2\kappa }{3\beta }}\right){\displaystyle \frac{\beta {\rho }_0-\frac{2}{3}\left(1-C\right)}{\left(F\left(0\right)+\frac{4\kappa }{3\beta }\right)\left(1-C\right)}}},\hfill \end{array}$$

  (123)

  where ${\delta }_1=\pm 1$. By using Equation (116) for the $\alpha =-{\displaystyle \frac{5}{3}}$ case, and then by merging to Equation (123), we find the DE to solve for $A_3$:

  $$\begin{array}{c}\hfill 3{\delta }_2{A}_3^{\prime }\approx {\left[-1-2{\delta }_1\sqrt{1-\left({\displaystyle \frac{\kappa }{8\beta {\tilde{F}}_0}}\right)\left(2+{\displaystyle \frac{3\beta {\rho }_0}{\left(C-1\right)}}\right)A_3^4}\right]}^{1/2},\end{array}$$

  (124)

  where $\tilde{F}\left(0\right)=F\left(0\right)+{\displaystyle \frac{4\kappa }{3\beta }}$ here. Equation (124) is a very complex integral without an exact solution. But, there are two solvable limits leading to approximated $A_3$ solutions.

  (a)

  $\kappa {\rho }_0\ll \tilde{F}\left(0\right)$ (low density limit): Equation (124) will be approximated as:

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}\approx & \int [{\displaystyle \frac{1}{\sqrt{-2{\delta }_1\sqrt{1-\frac{\kappa }{4\beta \tilde{F}\left(0\right)}{A}_3^4}-1}}}+{\displaystyle \frac{3{\delta }_1\kappa {\rho }_0}{16\tilde{F}{\left(0\right)}^2(1-C)}}\hfill \\ \hfill & \times {\displaystyle \frac{A_3^4}{\sqrt{1-\frac{\kappa }{4\beta \tilde{F}\left(0\right)}{A}_3^4}{\left(-2{\delta }_1\sqrt{1-\frac{\kappa }{4\beta \tilde{F}\left(0\right)}{A}_3^4}-1\right)}^{3/2}}}]d{A}_3.\hfill \end{array}$$

  (125)

  Equation (125) is complex to solve and needs some specific approximations. There are two possible subcases:

  • ${\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\ll 1$: Equation (125) becomes, for ${\delta }_1=-1$ (real number solution):

    $$\begin{array}{cc}\hfill & {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}\approx A_3+{\displaystyle \frac{\kappa }{40\beta \tilde{F}\left(0\right)}}\left(1+{\displaystyle \frac{3\beta {\rho }_0}{2\tilde{F}\left(0\right)(C-1)}}\right)A_3^5,\hfill \\ \hfill \Rightarrow & A_3\left(t\right)={\displaystyle \frac{{\delta }_2(t-t_0)}{3}}\hfill \\ \hfill & {\times }_4{F}_3\left({\displaystyle \frac{1}{5}},{\displaystyle \frac{2}{5}},{\displaystyle \frac{3}{5}},{\displaystyle \frac{4}{5}};{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{4}},{\displaystyle \frac{5}{4}};-{\left({\displaystyle \frac{5}{12}}\right)}^4{\displaystyle \frac{\kappa }{8\beta \tilde{F}\left(0\right)}}\left(1+{\displaystyle \frac{3\beta {\rho }_0}{2\tilde{F}\left(0\right)(C-1)}}\right){(t-t_0)}^4\right).\hfill \end{array}$$

    (126)
  • ${\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\gg 1$: Equation (125) becomes, for ${\delta }_1=-1$:

    $$\begin{array}{cc}\hfill & {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}\approx \left(1+{\displaystyle \frac{3\beta {\rho }_0}{8\tilde{F}\left(0\right)(1-C)}}\right){\left(-{\displaystyle \frac{\beta \tilde{F}\left(0\right)}{\kappa }}\right)}^{1/4}ln{A}_3,\hfill \\ \hfill & \Rightarrow A_3\left(t\right)=A_3\left(0\right)exp\left[{\left(-{\displaystyle \frac{\kappa }{\beta \tilde{F}\left(0\right)}}\right)}^{1/4}{\displaystyle \frac{{\delta }_2}{3\left(1+\frac{3\beta {\rho }_0}{8\tilde{F}\left(0\right)(1-C)}\right)}}t\right].\hfill \end{array}$$

    (127)

    We find in Equation (127) an exponential function for $A_3$. We must have $\tilde{F}\left(0\right)<0$ for a real $A_3$, and the universe is:

    –

    Expanding: ${\delta }_2=+1$ and $C<1$ or ${\delta }_2=-1$ and $C>1$.

    –

    Contracting: ${\delta }_2=-1$ and $C<1$ or ${\delta }_2=+1$ and $C>1$.

  (b)

  $\kappa {\rho }_0\gg \tilde{F}\left(0\right)$ (high density limit): Equation (124) will be approximated as:

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}\approx & \sqrt{-{\delta }_1}{\left({\displaystyle \frac{2\tilde{F}{\left(0\right)}^2(1-C)}{3\kappa {\rho }_0}}\right)}^{1/4}ln{A}_3.\hfill \end{array}$$

  (128)

  For ${\delta }_1=1$, Equation (128) leads to an oscillating $A_3$ solution, which is not physically relevant. For ${\delta }_1=-1$, we find an exponential $A_3$ solution:

  $$\begin{array}{cc}\hfill & {\displaystyle \frac{{\delta }_2(t-t_0)}{3}}\approx \sqrt{-{\delta }_1}{\left({\displaystyle \frac{2\tilde{F}{\left(0\right)}^2(1-C)}{3\kappa {\rho }_0}}\right)}^{1/4}ln{A}_3,\hfill \\ \hfill & \Rightarrow A_3\left(t\right)=A_3\left(0\right)exp\left[{\displaystyle \frac{{\delta }_2}{3}}\left({\displaystyle \frac{3\kappa {\rho }_0}{2\tilde{F}{\left(0\right)}^2(1-C)}}\right)t\right].\hfill \end{array}$$

  (129)

  We have the following scenarios for universe evolution:

  • Expanding: ${\delta }_2=+1$ and $C<1$ or ${\delta }_2=-1$ and $C>1$.
  • Contracting: ${\delta }_2=-1$ and $C<1$ or ${\delta }_2=+1$ and $C>1$.

All these previous teleparallel $F\left(T\right)$ solutions are really new results. However, the solutions applicable to phantom energy and Big Rip models are not covered in Ref. [50]. To this point, the $F\left(T\right)$ solutions newly obtained in this section constitute a very nice advance specific to non-linear perfect fluids with a positive impact on the teleparallel $F\left(T\right)$ gravity possibilities. These new results highlight the differences between the solutions for linear EoS and those for non-linear EoS in the teleparallel gravity context. The main difference lies in the types of physical processes targeted: non-linear perfect fluids lead more easily to the study of phantom energy and Big Rip processes (where $\alpha <-1$), although linear perfect fluids will not do this in general (because they are usually limited to $-1<\alpha <1$). This difference can be seen by comparing the possible subcases of $F\left(T\right)$ solutions between Section 4.2 and Section 5.2. Indeed, the quadratic correction term $\beta {\rho }^2$ in the non-linear EoS induces this difference.

5.3. Exponential Ansatz Solutions

Also, by using the Equation (32) exponential ansatz, we can find FEs from Equation (85a–d) as:

$$\begin{array}{cc}\hfill T& =2c(c+2b)-{\displaystyle \frac{2}{c_0^2}}exp(-2ct),\hfill \end{array}$$

(130a)

$$\begin{array}{cc}\hfill B^{\prime }& =-(2c+b)+{\displaystyle \frac{1}{c_0^2(b-c)}}exp(-2ct),\hfill \end{array}$$

(130b)

$$\begin{array}{cc}\hfill \kappa \rho & =\left(T+{\displaystyle \frac{2}{c_0^2}}exp(-2ct)\right)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}},\hfill \end{array}$$

(130c)

$$\begin{array}{cc}\hfill -\kappa \left(1+\alpha \right)\rho -\kappa \beta {\rho }^w& =c\left[B^{\prime }+c-b\right]F_T\left(T\right).\hfill \end{array}$$

(130d)

In addition, Equation (84) becomes:

$$\begin{array}{c}\hfill \rho \left(t\right)={\rho }_0{\left[1-C{c}_0^{2\left(1+\alpha \right)(w-1)}exp\left(\left(1+\alpha \right)(w-1)(b+2c)t\right)\right]}^{-{\displaystyle \frac{1}{w-1}}}.\end{array}$$

(131)

From Equation (130a) we find as solution Equation (77); then, by substituting Equations (77) and (130b) into Equations (130c) to (131), and by substituting Equation (130c) into (130d), we obtain that:

$$\begin{array}{cc}\hfill & -\left(1+\alpha \right)\left(2c(c+2b)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right)-{\kappa }^{1-w}\beta {\left(2c(c+2b)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right)}^w\hfill \end{array}$$

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{c}{(b-c)}}\left[(c+2b)(2c-b)-{\displaystyle \frac{T}{2}}\right]F_T\left(T\right),\hfill \end{array}$$

(132a)

$$\begin{array}{cc}\hfill \rho \left(T\right)& ={\rho }_0{\left[1-C{c}_0^{-\left(1+\alpha \right)(w-1){\displaystyle \frac{b}{c}}}{\left(c(c+2b)-{\displaystyle \frac{T}{2}}\right)}^{-\left(1+\alpha \right)(w-1)\left({\displaystyle \frac{b}{2c}}+1\right)}\right]}^{-{\displaystyle \frac{1}{w-1}}}.\hfill \end{array}$$

(132b)

There are two possible cases of solutions for Equation (132a):

• $\mathbf{c}=-\mathbf{2}\mathbf{b}$: Equation (132a) simplifies as a simple DE:

  $$\begin{array}{cc}\hfill & F\left(T\right)\left[1+{\displaystyle \frac{\beta }{{(-2\kappa )}^{w-1}\left(1+\alpha \right)}}{\left(F\left(T\right)\right)}^{w-1}\right]={\displaystyle \frac{2}{3\left(1+\alpha \right)}}T{F}_T\left(T\right).\hfill \end{array}$$

  (133)

  By integration, the solution to Equation (133) is:

  $$\begin{array}{c}\hfill F\left(T\right)={\left[\left({\displaystyle \frac{\beta }{{(-2\kappa )}^{w-1}\left(1+\alpha \right)}}+{\left(F\left(0\right)\right)}^{1-w}\right)T^{-{\displaystyle \frac{3(w-1)\left(1+\alpha \right)}{2}}}-{\displaystyle \frac{\beta }{{(-2\kappa )}^{w-1}\left(1+\alpha \right)}}\right]}^{{\displaystyle \frac{1}{1-w}}},\end{array}$$

  (134)

  where $w>1$, $b\ne c$, and $c\ne 0$.
• $\mathbf{c}\ne -\mathbf{2}\mathbf{b}$ (general case): We can solve Equation (132a) for $w=2,3,$ and 4. However, the $w=3$ and 4 cases are complex to solve, and we will restrict ourselves to $w=2$ subcase. Then, Equation (132a) becomes, in this case:

  $$\begin{array}{cc}\hfill 0=& {\left(2c(c+2b)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right)}^2+{\displaystyle \frac{\kappa \left(1+\alpha \right)}{\beta }}\left(2c(c+2b)F_T\left(T\right)-{\displaystyle \frac{F\left(T\right)}{2}}\right)\hfill \\ \hfill & +{\displaystyle \frac{c\kappa }{(b-c)\beta }}\left[(c+2b)(2c-b)-{\displaystyle \frac{T}{2}}\right]F_T\left(T\right),\hfill \\ \hfill \Rightarrow & 4c(c+2b)F_T\left(T\right)-F\left(T\right)\hfill \\ \hfill & ={\displaystyle \frac{\kappa \left(1+\alpha \right)}{\beta }}\left[-1+{\delta }_1\sqrt{1+{\displaystyle \frac{2\beta c}{(c-b)\kappa {\left(1+\alpha \right)}^2}}\left(2(c+2b)(2c-b)-T\right)F_T\left(T\right)}\right].\hfill \end{array}$$

  (135)

  Equation (135) is a non-linear DE, and we need to approximate this relation for $\beta \ll (1+\alpha )$, the weak quadratic term approximation. In this case, Equation (135) will be approximated as:

  $$\begin{array}{cc}\hfill & 4c(c+2b)F_T\left(T\right)-F\left(T\right)\approx {\displaystyle \frac{\kappa \left(1+\alpha \right)}{\beta }}({\delta }_1-1)+{\displaystyle \frac{{\delta }_{1}c}{(c-b)\left(1+\alpha \right)}}\left(2(c+2b)(2c-b)-T\right)F_T\left(T\right).\hfill \end{array}$$

  (136)

  For ${\delta }_1=+1$, Equation (136) leads exactly to Equation (78a) for linear perfect fluids, with Equation (78b) as $\rho \left(T\right)$. For ${\delta }_1=-1$, Equation (136) becomes the DE:

  $$\begin{array}{cc}\hfill & F\left(T\right)-{\displaystyle \frac{2\kappa \left(1+\alpha \right)}{\beta }}\approx {\displaystyle \frac{c{F}_T\left(T\right)}{(b-c)\left(1+\alpha \right)}}\left[T-2(c+2b)\left(2(\alpha +2)c-(3+2\alpha )b\right)\right].\hfill \end{array}$$

  (137)

  There are two possible subcases for solutions to Equation (137) when $c\ne -2b$:

  (a)

  $b={\displaystyle \frac{2(2+\alpha )}{(3+2\alpha )}}c$ case: Equation (137) simplifies as:

  $$\begin{array}{cc}\hfill & F\left(T\right)-{\displaystyle \frac{2\kappa \left(1+\alpha \right)}{\beta }}\approx {\displaystyle \frac{(3+2\alpha )}{\left(1+\alpha \right)}}T{F}_T\left(T\right),\hfill \\ \hfill & \Rightarrow F\left(T\right)\approx {\displaystyle \frac{2\kappa \left(1+\alpha \right)}{\beta }}+\left(F\left(0\right)-{\displaystyle \frac{2\kappa \left(1+\alpha \right)}{\beta }}\right)T^{{\displaystyle \frac{\left(1+\alpha \right)}{(3+2\alpha )}}}.\hfill \end{array}$$

  (138)

  Then, Equation (132b) for the fluid density is:

  $$\begin{array}{cc}\hfill \rho \left(T\right)& ={\rho }_0{\left[1-C{c}_0^{-{\displaystyle \frac{2\left(1+\alpha \right)(2+\alpha )}{(3+2\alpha )}}}{\left(c^2{\displaystyle \frac{(11+6\alpha )}{(3+2\alpha )}}-{\displaystyle \frac{T}{2}}\right)}^{-{\displaystyle \frac{\left(1+\alpha \right)(5+3\alpha )}{(3+2\alpha )}}}\right]}^{-1}.\hfill \end{array}$$

  (139)

  (b)

  General case ($c\ne -2b$ and $b\ne {\displaystyle \frac{2(2+\alpha )}{(3+2\alpha )}}c$): By integrating Equation (137), we obtain that:

  $$\begin{array}{c}\hfill F\left(T\right)\approx {\displaystyle \frac{2\kappa \left(1+\alpha \right)}{\beta }}+\left[F\left(0\right)-{\displaystyle \frac{2\kappa \left(1+\alpha \right)}{\beta }}\right]{\left[T-T_0\right]}^{{\displaystyle \frac{(b-c)\left(1+\alpha \right)}{c}}},\end{array}$$

  (140)

  where $T_0=2(c+2b)\left(2(\alpha +2)c-(3+2\alpha )b\right)$ is a constant, and $F\left(0\right)$ is the integration constant. The fluid density $\rho \left(T\right)$ is described by Equation (132b).

All these previous teleparallel $F\left(T\right)$ solutions are new results. Once again, this new class of $F\left(T\right)$ solutions generalizes the solutions in Ref. [50] because these are non-linear EoS $F\left(T\right)$ solutions. In addition, the $F\left(T\right)$ solutions obtained in the present section are almost identical to those in Section 4.3, apart from a shift taking into account the quadratic correction in the EoS. Once again, the solutions freshly obtained here are similar to those obtained in Section 4.3.

6. Discussion and Conclusions

The main aim of this paper was to obtain KS teleparallel $F\left(T\right)$ solutions for vacuum, linear. and non-linear perfect fluids. This was achieved through the use of various ansatzes, for example, power-law and exponential (infinite superposition of power-laws). We obtained not only power-law $F\left(T\right)$ solutions, but also more complex $F\left(T\right)$, as we can see in Section 4.1 and Section 5.1 for $c={\displaystyle \frac{1}{2}},1,-1,2$ via power-law ansatz solutions. As mentioned before, these teleparallel $F\left(T\right)$ solutions found in Section 4.1 and Section 5.1 by this ansatz have many similar terms and were also obtained by the same form of characteristic equation described by Equation (38). Obviously, for $c=1$, we find a power-law $F\left(T\right)$ solution generalizing the vacuum solution. For the vacuum case, we found that the $A_3$ component can only be linear in t, which limits the possible analytical solutions. For the exponential ansatz, the teleparallel $F\left(T\right)$ solutions appear only for perfect fluids in Section 4.3 and Section 5.3, and they are power-law solutions (no additional constant term for linear and an additional constant term for non-linear); this situation represents an infinite sum limit of power-law terms.

However, some special solutions are found in Section 4.2 and Section 5.2. We find power-law $F\left(T\right)$ solutions as before; therefore, there are various types of solutions appearing for $A_3$. For linear perfect fluids $P=\alpha \rho$ in Section 4.2, we obtain an exact solution for $A_3$ when $\alpha =-{\displaystyle \frac{1}{3}}$, but we must approximate the solutions for low and high cosmological fluid density situations for the cases $\alpha =-{\displaystyle \frac{1}{4}}$ and $-{\displaystyle \frac{2}{3}}$. For low densities, these are relevant models for predicting the future evolution of an expanding universe. For high densities, these models are especially useful for explaining the universe just after the Big Bang. In some specific cases, we find as the limit a linear $A_3$ in t as for the vacuum situation. Even without taking into account that, for $\alpha <-{\displaystyle \frac{1}{3}}$ cases, we are able to model the famous quintessence process, in the latter case, it will be necessary to replace the perfect fluid by a scalar field to achieve this [61,62,63]. The study of this physical process with the teleparallel $F\left(T\right)$ solutions found in this paper would be necessary for a more complete theory on quintessence.

In the case of a non-linear perfect fluids in Section 5.2, we obtain power-law $F\left(T\right)$ solutions with an additional constant term. However, we extend the perfect fluid linear term $\alpha$ definition to $\alpha <-1$ values for finding the analytically solvable $A_3$ solutions. We find that solutions for $\alpha =-{\displaystyle \frac{4}{3}}$ and $\alpha =-{\displaystyle \frac{5}{3}}$ describe the phantom energy cases (negative kinetic energy) that could lead at the end of evolution to the Big Rip, according to some recent works [64,65]. In these latter cases, we carried out high and low fluid density approximations for the $A_3$ solutions for the possible limits. But, we also studied the stability of correction terms with respect to the dominant term, somewhat like that in Refs. [66,67]. We notice that, for the case $\alpha =-{\displaystyle \frac{4}{3}}$ at high fluid density, we obtain an $A_3$ with linear functions as minimal and maximal limits, which means a stable solution for $A_3$. All other subcases do not offer finite limits for $A_3$, so these solutions may be divergent, and the universe model is therefore unstable. Once again, a more detailed study is really necessary on this phenomenon of phantom energy and the Big Rip process in teleparallel $F\left(T\right)$ gravity. These two phenomena are critical and concern dark energy in the universe. They deserve better answers. Nevertheless, the teleparallel $F\left(T\right)$ gravity approach has proven powerful enough to provide this new class of solutions for achieving them in the future.

Apart from the previous cases concerning universe models, there were two recurring situations in Section 4.2 and Section 5.2 of non-possibility for the $F\left(T\right)$ solution when $\alpha =-{\displaystyle \frac{1}{2}}$ for linear EoS and $\alpha =-{\displaystyle \frac{3}{2}}$ for quadratic EoS, respectively. The problem does not arise by using power-law ansatz solutions in Section 4.1 and Section 5.1. For the $\alpha =-{\displaystyle \frac{1}{2}}$ and $-{\displaystyle \frac{3}{2}}$ subcases, respectively, there are several $F\left(T\right)$ solutions as seen previously. Therefore, in the situations in Section 4.2 and Section 5.2, we have proven by approximations on $\alpha$ close to $-{\displaystyle \frac{1}{2}}$ and $-{\displaystyle \frac{3}{2}}$, respectively, that $F\left(T\right)$ described by Equations (58) and (113) is defined all around these two critical values. There are only ansatz-caused singularities because the value of n in this ansatz goes to infinity at $\alpha =-{\displaystyle \frac{1}{2}}$ and $-{\displaystyle \frac{3}{2}}$. These are only a purely mathematical concern, not physical, in both situations.

After the considerations of dark energy, quintessence, and Big Rip models, there is some possible future work in finding KS teleparallel $F\left(T\right)$ solutions for electromagnetic sources. This possible future work can be useful for studying spacetimes with electrically charged particle sources. Another helpful possible work for more complete quintessence models would be to find KS teleparallel $F\left(T\right)$ solutions from a scalar field source. We may also do this type of work for phantom energy models with a scalar field. We can also extend this specific study for quantized scalar fields, as in Ref. [57], by hoping to find similar $F\left(T\right)$ solutions to those in this paper. This is possibly arduous work but full of hope.