Problem 11.11.1
Consider a simple model of transient acceleration with decaying cosmological constant \begin{equation} \label{ec} \dot{\rho}_{m} + 3\frac{\dot{a}}{a}\rho_{m} = - \dot{\rho}_\Lambda\;, \end{equation} where $\rho_{m}$ and $\rho_\Lambda$ energy density DE and cosmological constant $\Lambda$. At the early stages of the expansion of the Universe, when $\rho_\Lambda$ is quite small, such a decay does not influence cosmological evolution in any way. At later stages, as the DE contribution increases, its decay has an ever increasing effect on the standard dependence of the DM energy density $\rho_{m} \propto a^{-3}$ on the scale factor $a$. We consider the deviation to be described by a function of the scale factor - $\epsilon(a)$. \begin{equation} \label{dm} \rho_{m} = \rho_{m, 0}a^{-3 + \epsilon(a)}\;, \end{equation} where $a_0 = 1$ in the present epoch. Other fields of matter (radiation, baryons) evolve independently and are conserved. Hence, the DE density has the form \begin{equation}\label{decayv} \rho_{\Lambda} = \rho_{m0} \int\limits_{a}^{1}\frac{\epsilon(\tilde{a}) + \tilde{a}\epsilon' \ln(\tilde{a})}{\tilde{a}^{4 - \epsilon(\tilde a)}} d\tilde{a} + {\rm{X}}\;, %\tilde\rho_{\Lambda 0} + \end{equation} where the prime denotes the derivative with respect to the scale factor, and ${\rm{X}}$ is the integration constant. If radiation is neglected, the first Friedmann equation takes the form \begin{equation} \label{friedmann} {H}= H_0\left[\Omega_{b,0}{a}^{-3} + \Omega_{m0}\varphi(a) + {\Omega}_{\rm{X,0}}\right]^{1/2}, \end{equation} Using the assumption that the function $\epsilon (a)$ has the following simple form \begin{equation} \label{Parametrization_a} \epsilon(a) = \epsilon_0a^\xi\ = \epsilon_0(1+z)^{-\xi}, \end{equation} where $\epsilon_0$ and $\xi$ can take both positive and negative values, find function $\varphi(a)$ and relative energy density $\Omega_b(a)$, $\Omega_{m}(a)$ and $\Omega_{\Lambda}(a)$.
The function $\varphi(a)$ is then written as \begin{equation} \label{f(a)} \varphi(a) = a^{-3 +\epsilon(a)} + \int\limits_{a}^{1}\frac{\epsilon(\tilde{a}) + \tilde{a}\epsilon' \ln(\tilde{a})}{\tilde{a}^{4 - \epsilon(a)}} d\tilde{a}\;, \end{equation} where ${\Omega}_{{\rm{X,0}}}$, is the relative contribution of the constant ${\rm{X}}$ to the common relative density. It follows from the expression (\ref{decayv}) that \begin{equation}\label{decayv2} \rho_{\Lambda} = \rho_{m0}\epsilon_{0} \int\limits_{a}^{1}\frac{[1 + \ln(\tilde{a}^{\xi})]}{\tilde{a}^{4 - \xi - \epsilon_{0}\tilde{a}^{\xi}}} d\tilde{a} + {\rm{X}}\;. \end{equation} We note that the case $\epsilon_0 = 0$ corresponds to the SCM, i.e., ${\rm{X}} \equiv {\rho}_{\Lambda}$. Using the formulas presented above, it is not difficult to obtain the dependences for the relative densities $\Omega_b(a)$, $\Omega_{m}(a)$ and $\Omega_{\Lambda}(a)$: \begin{equation} \label{8a} \Omega_{b}(a) = \frac{a^{-3}}{{\rm{A}} + a^{-3} + {\rm{B^{-1}}}\varphi(a)}\;, \end{equation} \begin{equation} \label{8b} \Omega_{m}(a) = \frac{a^{-3 + \epsilon(a)}}{{\rm{D}} + {\rm{B}}a^{-3} + \varphi(a)}\;, \end{equation} \begin{equation} \label{8c} \Omega_{{\rm{\Lambda}}}(a) = \frac{{\rm{D}} + \varphi(a) - a^{-3 + \epsilon(a)}}{{\rm{D}} + {\rm{B}}a^{-3} + \varphi(a)}\;, \end{equation} where ${\rm{A}} = {\Omega_{{\rm{X}},0}}/{\Omega_{b,0}}$, ${\rm{B}} = {\Omega_{b,0}}/{\Omega_{m0}}$ and ${\rm{D}} = {\Omega_{{\rm{X}},0}}/{\Omega_{m0}}$.