Difference between revisions of "Phase space structure of models with interaction"
From Universe in Problems
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=== Problem 1 === | === Problem 1 === | ||
| − | + | Find effective EoS parameters $w_{(dm)eff}$ and $w_{(de)eff}$ for the interactions $Q_1$, $Q_2$ and $Q_3$. | |
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<div class="NavHead">solution</div> | <div class="NavHead">solution</div> | ||
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| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">\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}</p> | ||
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| − | === Problem | + | === Problem 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, $w_{dm}=0$, $w_{de}=const$. | |
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<div class="NavHead">solution</div> | <div class="NavHead">solution</div> | ||
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| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">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 | ||
| + | \[r_s^\pm=-1+2b\pm b\sqrt{b(b-1)}.\]</p> | ||
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| − | === Problem | + | === Problem 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 $r_s^-$. | |
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<div class="NavHead">solution</div> | <div class="NavHead">solution</div> | ||
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| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">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 $r_s^-$.</p> | ||
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| − | === Problem | + | === Problem 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 [http://arxiv.org/abs/1003.2788]) | ||
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| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">\[\Omega'_{dm}=3f_j \Omega_{dm}\Omega_{de},\] |
| + | \[\Omega'_{de}=-3f_j \Omega_{dm}\Omega_{de},\] | ||
| + | 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}</p> | ||
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| − | === Problem | + | === Problem 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). | ||
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| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">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.</p> |
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| − | === Problem | + | === Problem 6 === |
| − | + | Construct the stability matrix for the dynamical system considered in the [[#IDE_44|problem]] and determine its eigenvalues. | |
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<div class="NavHead">solution</div> | <div class="NavHead">solution</div> | ||
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| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">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.</p> | ||
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| − | === Problem | + | === Problem 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) $f_j = 0$. | |
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<div class="NavHead">solution</div> | <div class="NavHead">solution</div> | ||
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| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">\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}</p> | ||
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| − | === Problem | + | === Problem 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 $Q=3\gamma_{dm}\rho_{dm}$. | |
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| − | <p style="text-align: left;"> | + | <p style="text-align: left;">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 $\gamma_{dm}<\gamma$.</p> | |
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| − | === Problem | + | === Problem 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 [http://arxiv.org/abs/1112.5095]) | |
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<div class="NavFrame collapsed"> | <div class="NavFrame collapsed"> | ||
<div class="NavHead">solution</div> | <div class="NavHead">solution</div> | ||
<div style="width:100%;" class="NavContent"> | <div style="width:100%;" class="NavContent"> | ||
| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">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.</p> | ||
</div> | </div> | ||
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Latest revision as of 11:33, 8 November 2013
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: \[Q_1=3\gamma_{dm}\rho_{dm},\quad Q_1=3\gamma_{de}\rho_{de},\quad Q_1=3\gamma_{tot}\rho_{tot}\]
Problem 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}
Problem 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 \[r_s^\pm=-1+2b\pm b\sqrt{b(b-1)}.\]
Problem 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 $r_s^-$.
Problem 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}
Problem 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.
Problem 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.
Problem 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}
Problem 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 $\gamma_{dm}<\gamma$.
Problem 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.
