The evolution of a Universe filled with interacting components can be effectively analyzed in terms of dynamical systems theory. Let us consider the following coupled differential equations for two variables
\begin{equation}
\label{IDE_s6_1}
\begin{array}{l}
\dot x=f(x,y,t),\\
\dot y=g(x,y,t).
\end{array}
\end{equation}
We will be interested in the so-called autonomous systems, for which the functions $f$ and $g$ do not contain explicit time-dependent terms.
A point $(x_c,y_c)$ is said to be a fixed (a.k.a. critical) point of the autonomous system if
\[f(x_c,y_c)=g(x_c,y_c)=0.\]
A critical point $(x_c,y_c)$ is called an attractor when it satisfies the condition \(\left(x(t),y(t)\right)\to(x_c,y_c)\) for $t\to\infty$.
Let's look at the behavior of the dynamical system (\ref{IDE_s6_1}) near the critical point. For this purpose, let us consider small perturbations around the critical point
\[x=x_c+\delta x,\quad y=y_c+\delta y.\]
Substituting it into (\ref{IDE_s6_1}) leads to the first-order differential equations:
\[\frac{d}{dN}\left(\begin{array}{c}\delta x\\ \delta y\end{array}\right) = \hat M \left(\begin{array}{c}\delta x\\ \delta y\end{array}\right).\]
Taking into account the specifics of the problem that we are solving, we made the change \[\frac{d}{dt}\to\frac{d}{dN},\]
where $N=\ln a$. The matrix $\hat M$ is given by
\[\hat M =
\left(
\begin{array}{lr}
\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\
{} & {}\\
\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}
\end{array}
\right)
\]
The general solution for the linear perturbations reads
\[\delta x=C_1e^{\lambda_1 N} + C_2e^{\lambda_2 N},\]
\[\delta y=C_3e^{\lambda_1 N} + C_4e^{\lambda_2 N},\]
The stability around the fixed points depends on the nature of the eigenvalues.
Let us treat the interacting dark components as a dynamical system described by the equations
\[\rho'_{de}+3(1+w_{de})\rho_{de}=-Q\]
\[\rho'_{dm}+3(1+w_{dm})\rho_{dm}=Q\]
Here, the prime denotes the derivative with respect to $N=\ln a$. Note that although the interaction can significantly change the cosmological evolution, the system is still autonomous. We consider the following specific interaction forms, which were already analyzed above:
\[Q_1=3\gamma_{dm}\rho_{dm},\quad Q_1=3\gamma_{de}\rho_{de},\quad Q_1=3\gamma_{tot}\rho_{tot}\]
11.6.1
Problem 11.6.1
Find effective EoS parameters $w_{(dm)eff}$ and $w_{(de)eff}$ for the interactions $Q_1$, $Q_2$ and $Q_3$.
solution
\begin{align}
\nonumber Q & = Q_1, & w_{(de)eff}& = w_{de}(\Omega_{de}) + \gamma_{dm}\frac{1-\Omega_{de}}{\Omega_{de}}, & w_{(dm)eff} & = w_{dm} - \gamma_{dm},\\
\nonumber Q & = Q_2, & w_{(de)eff}& = w_{de}(\Omega_{de}) + \gamma_{de}, & w_{(dm)eff} & = w_{dm} - \gamma_{de}\frac{1-\Omega_{de}}{\Omega_{de}},\\
\nonumber Q & = Q_3, & w_{(de)eff}& = w_{de}(\Omega_{de}) + \gamma_{tot}\frac{1}{\Omega_{de}}, & w_{(dm)eff} & = w_{dm} - \frac{\gamma_{tot}}{\Omega_{de}}.
\end{align}
11.6.2
Problem 11.6.2
Find the critical points of equation for ratio $r=\rho_{dm}/\rho_{de}$ if $Q=3\alpha H(\rho_{dm}+\rho_{de})$, where the phenomenological parameter $\alpha$ is a dimensionless, positive constant, $w_{dm}=0$, $w_{de}=const$.
solution
In the considered case
\[\dot r=3Hr\left[w_{de}+ \frac{Q}{9H^3}\frac{(r+1)^2}{r}\right]\to\dot r=3Hr\left[w_{de}+ \alpha\frac{(r+1)^2}{r}\right].\]
The critical points are determined by setting $\dot r=0$. Consequently, stationary solutions of the
resulting quadratic equation
\[r^2+2(1-2b)r+1=0,\quad b\equiv-\frac{w_{de}}{4\alpha}\]
are
\[r_s^\pm=-1+2b\pm b\sqrt{b(b-1)}.\]
11.6.3
Problem 11.6.3
Show, that the remarkable property of the model, considered in the previous problem, is that for the interaction parameter $\alpha$, consistent with the current observations $\alpha<2.3\times10^{-3}$ the ratio $r$ tends to a stationary but unstable value at early times, $r_s^+$, and to a stationary and stable value, $r_s^-$ (an attractor), at late times. Consequently, as the Universe expands, $r(a)$ smoothly evolves from $r_s^+$ to the attractor solution $r_s^-$.
solution
Using the standard analysis methods of critical points, one can easily to show that the critical points, obtained in the previous problem, in the case $\alpha<2.3\times10^{-3}$ represent an unstable stationary solution $r_s^+>0$ and a stable stationary solution $r_s^->0$ respectively. The general solution of equation
\[\dot r=3Hr\left[w_{de}+ \alpha\frac{(r+1)^2}{r}\right].\]
is
\[r(x)=\frac{r-s^-+xr_s^+}{1=x},\]
interpolates between $r_s^+$ and $r_s^-$. Here, $x\equiv(a/a_*)^{-\mu}$, with $\mu\equiv12\alpha\sqrt{b(b-1)}$, and $a_*$ is the scale factor value at which $r$ takes the arithmetic medium value $(r_s^++r_s^-)/2$. In the range $r_s^-<r<r_s^+$ the function $r(x)$ decreases monotonously. Consequently, as the Universe expands, $r(x)$ smoothly evolves from $r_s^+$ to the attractor solution $r_s^-$.
11.6.4
Problem 11.6.4
Transform the system of equations
\[\rho'_{de}+3(1+w_{de})\rho_{de}=-Q,\]
\[\rho'_{dm}+3(1+w_{dm})\rho_{dm}=Q,\]
into the one for the fractional density energies.(after [1])
solution
\[\Omega'_{dm}=3f_j \Omega_{dm}\Omega_{de},\]
\[\Omega'_{de}=-3f_j \Omega_{dm}\Omega_{de},\]
where $j=0,1,2,3$. Here $j=0$ corresponds to non-interacting case with $f_0=w_{de}-w_{dm}$. For $j=1,2,3$ ($Q_1,Q_2,Q_3$):
\begin{align}
\nonumber f_j & =w_{(de)eff,j}-w_{(dm)eff,j},\\
\nonumber f_1 & =f_0+\frac{\gamma_{dm}}{\Omega_{de}},\\
\nonumber f_1 & =f_0+\frac{\gamma_{de}}{1-\Omega_{de}},\\
\nonumber f_1 & =f_0+\frac{\gamma_{tot}}{\Omega_{de}(1-\Omega_{de})}.
\end{align}
11.6.5
Problem 11.6.5
Analyze the critical points of the autonomous system, obtained in the previous problem
\[\Omega'_{dm}=3f_j \Omega_{dm}\Omega_{de},\]
\[\Omega'_{de}=-3f_j \Omega_{dm}\Omega_{de},\]
by imposing the conditions $\Omega'_{dm}=\Omega'_{de}=0$ and $\Omega_{dm}+\Omega_{de}=1$ (flatness of Universe).
solution
The critical points can be divided into the following categories. The critical point $M$ is the matter dominated phase with $\Omega_{dm}=1$, and the critical point $E$ is the dark energy dominated phase with $\Omega_{de}=1$. If $f_j\propto1/\Omega_{dm}$ or $f_j\propto1/\Omega_{de}$, these two fixed points may not exist. Besides the above two fixed points, there are other solutions with $f_j$. Note that an attractor is one of the stable critical points of the autonomous system.
11.6.6
Problem 11.6.6
Construct the stability matrix for the dynamical system considered in the problem and determine its eigenvalues.
solution
If we substitute linear perturbations near the critical point ($\bar\Omega_{dm}, \bar\Omega_{de}$) into the equations of motion
\begin{align}
\nonumber \Omega'_{dm} & = \ 3f_j \Omega_{dm}\Omega_{de},\\
\nonumber \Omega'_{de} & = -3f_j \Omega_{dm}\Omega_{de},
\end{align}
and linearize them, we can get that
\[\hat M=\left( \begin{array}{rr}
3f_j(\bar\Omega_{de})\bar\Omega_{de} & 3(f_j(\bar\Omega_{de})\bar\Omega_{dm} + f'_j\bar\Omega_{dm}\bar\Omega_{de})\\
{}&{}\\
-3f_j(\bar\Omega_{de})\bar\Omega_{de} & -3(f_j(\bar\Omega_{de})\bar\Omega_{dm} + f'_j\bar\Omega_{dm}\bar\Omega_{de})
\end{array}\right).\]
Here $f'\equiv df/d\bar\Omega_{de}$. The two eigenvalues of the matrix $\hat M$ determining the stability of the corresponding critical point are
\begin{align}
\nonumber \lambda_1 &= 0,\\
\nonumber \lambda_2 &= 3f_j(2\bar\Omega_{de}-1) - 3f'_j\bar\Omega_{de}(\bar\Omega_{de}-1).
\end{align}
When $\lambda_2$ is positive, the corresponding critical point is an unstable node. "Unstable" means that the Universe will not stay in the phase for long, and eventually it will evolve to other phases. When $\lambda_2$ is negative, the corresponding critical point is a stable node and the phase will last long.
11.6.7
Problem 11.6.7
Using result of the previous problem, determine eigenvalues of the stability matrix for the following cases: i) $\Omega_{dm} = 1$, $\Omega_{de} = 0$, $f_j \ne 0$; ii) $\Omega_{dm} = 0$, $\Omega_{de} = 1$, $f_j \ne 0$; iii) $f_j = 0$.
solution
\begin{align}
\nonumber (\Omega_{dm},\Omega_{de}) =(1,0), \quad f_j & \ne 0: & (\lambda_1,\lambda_2) = & (0,-3f_j);\\
\nonumber {} & {} & {} & {}\\
\nonumber (\Omega_{dm},\Omega_{de}) =(0,1), \quad f_j & \ne 0: & (\lambda_1,\lambda_2) = & (0,3f_j);\\
\nonumber {} & {} & {} & {}\\
\nonumber f_j & = 0: & (\lambda_1,\lambda_2) = & (0,-3f'_j).
\end{align}
11.6.8
Problem 11.6.8
Obtain position and type of the critical points obtained in the previous problem for the case of cosmological constant interacting with dark matter as $Q=3\gamma_{dm}\rho_{dm}$.
solution
In the considered case
\[f = w_{(de)eff}-w_{dm}=-\gamma+\frac{\gamma_{dm}}{\Omega_{de}}, \quad \gamma\equiv w_{dm}+1.\]
If $f=0$ the critical point is
\[(\bar\Omega_{dm},\bar\Omega_{de})=(1-\gamma_{dm}/\gamma,\gamma_{dm}/\gamma).\]
Eigenvalues of the stability matrix in this point equal to
\[(\lambda_1,\lambda_2)=(0,-3f'\Omega_{dm}\Omega_{de})=(0,3\gamma_{dm}\Omega_{dm}\Omega_{de})\]
Condition of existence for this critical point reads
\[0\le\Omega\le1\Rightarrow0\le\gamma_{dm}\le\gamma.\]
As $\lambda_2>0$, the critical point is unstable.
The second critical point always exist for the considered type of interaction and its coordinates are
\[(\bar\Omega_{dm},\bar\Omega_{de}) =(0,1)\]
Eigenvalues of the stability matrix in this point equal to
\[(\lambda_1,\lambda_2)=(0,-3f)=\left(0,-3(\gamma-\gamma_{dm})\right)\]
The critical point is stable under the condition $\gamma_{dm}<\gamma$.
11.6.9
Problem 11.6.9
Construct the stability matrix for the following dynamical system
\begin{align}
\nonumber \rho' & = - \left(1+\frac{w_{de}}{1+r}\right)\rho,\\
\nonumber r' & = r \left[w_{de} - \frac{(1+r)^2}{r\rho}\Pi\right],
\end{align}
and determine its eigenvalues. (After [2])
solution
For The considered system of equations the eigenvalues of the stability matrix equal to roots of the equation
\[\lambda^2 + \left[2+w_{de}-w_{de}(1+w_{de})\frac{\partial_r\Pi}{\Pi}\right]\lambda + (1+w_{de}+ w_{de}\partial_\rho\Pi)=0.\]
\[\partial_r \Pi \equiv \frac{\partial\Pi}{\partial r}, \partial_\rho\Pi \equiv \frac{\partial\Pi}{\partial\rho}.\]
This equation has the following solutions
\[\lambda_\pm = \frac12\left[w_{de}(1+w_{de})\frac{\partial_r\Pi}{\Pi}-(2+w_{de})\right] \left\{1\pm\sqrt{1 - \frac{4(1+w_{de}+w_{de}\partial_\rho\Pi)}{\left(w_{de}(1+w_{de})\frac{\partial_r\Pi}{\Pi}-(2+w_{de})\right)^2}}\right\}\]
where we have to require $1+w_{de}+w_{de}\partial_\rho\Pi\ne0$. In case these solutions are non-degenerate and real, they describe an stable critical point for $\lambda_\pm<0$, an unstable critical point for $\lambda_\pm>0$ and a saddle if $\lambda_+$ and $\lambda_-$ have different signs. For complex eigenvalues $\lambda_\pm=\alpha\pm\beta$, it is the sign of $\alpha$ that determines the type of the stationary point. For $\alpha=0$ the critical point is a center, for $\alpha<0$ it is a stable focus and for $\alpha>0$ it is an unstable focus.