\[\ddot\varphi+3H\dot\varphi+\frac{dV}{d\varphi}=-\Gamma,\quad \Gamma\equiv\frac{Q}{\dot\varphi}.\] Here $Q$ is the constant of interaction between dark energy and dark matter.

In the presence of this dark sector interaction, the energy density in the dark matter no longer redshifts as $a^{-3}$ but instead scales as \[\rho_{dm}\propto\frac{\rho_{dm0}}{a^3}f(\varphi/M_{Pl}).\] The conservation equation in this case transforms into \[\dot\rho_{dm}+3H\rho_{dm}=\frac{\rho_{dm0}}{a^3}\frac{df/d\varphi}{f_0},\] where $f_0=f(\varphi/M_{Pl})$ with $\varphi_0$ is the field value today. Consequently the usual Klein-Gordon equation for quintessence models \[\ddot\varphi+3H\dot\varphi+\frac{dV}{d\varphi}=0\] is transformed into \[\ddot\varphi+3H\dot\varphi+\frac{dV}{d\varphi}=-\frac{\rho_{dm0}}{a^3}\frac{df/d\varphi}{f_0}.\]

Evolution of the ratio $r\equiv\rho_{dm}/\rho_\varphi$ is described by the equation \[\dot r=r\left(\frac{\dot\rho_{dm}}{\rho_{dm}}-\frac{\dot\rho_\varphi}{\rho\varphi}\right).\] Introducing the barotropic index $\gamma_i=w_i+1$, $(i=dm,\varphi)$, one obtains \[\dot r =3Hr\left(\gamma_{dm}-\gamma_\varphi+\frac{1+r}{\rho_\varphi r}\Pi_{dm}\right).\] The existence of a stationary solution $\dot r=0$ is guaranteed by the condition \[\Pi_{dm}=(\gamma_\varphi-\gamma_{dm})\frac{r}{1+r}\rho_\varphi.\]

For cold dark matter, $\gamma_{dm}\approx1$, and for dark energy as quintessence, $\gamma_\varphi=\dot\varphi^2/\rho_\varphi$. The coupling term $Q$ in this case is \[Q=-3H(\gamma\varphi-1)\frac{r}{1+r}\rho_\varphi.\] In a spatially flat Universe \[H^2=\frac13(\rho_\varphi+\rho_{dm}),\] and consequently \[Q=-\sqrt3(\gamma_\varphi-1)\frac{\rho_\varphi\rho_{dm}}{\sqrt{\rho_\varphi+\rho_{dm}}}.\]

For the interaction under consideration, the conservation equation for $\rho_{dm}$ and $\rho_\varphi$ transform into \[\dot\rho_{dm}+3H\rho_{dm} =-3H(\gamma_\varphi-1)\frac{r}{1+r}\rho_{dm},\] \[\dot\rho_\varphi+3H\gamma_\varphi\rho_\varphi= -3H(\gamma_\varphi-1)\frac{r}{1+r}\rho_\varphi.\] Assuming $\gamma_\varphi$ to be constant, one obtains \[\rho_\varphi\propto a^{-\nu},\quad \rho_{dm}\propto a^{-\nu},\quad \nu=3\frac{\gamma_\varphi+r}{1+r}.\] Both energy densities happen to redshift at the same rate because we have chosen $Q$ to correspond to the $r=const$.

With $\rho\propto a^{-\nu}$ (see the previous problem) one can solve the first Friedmann equation to find $a\propto t^{2/\nu}$. Using \[\Omega_\varphi=\frac{1}{1+r}=\frac{8\pi G}{3}\frac{\nu^2}{4}t^2\rho_\varphi,\quad \nu=3\frac{\gamma_\varphi+r}{1+r},\] one finds \[\rho_\varphi=\frac{1}{6\pi G}\frac{1+r}{(\gamma_\varphi+r)^2}\frac{1}{t^2}.\] Taking into account that \[\gamma_\varphi=\frac{\dot\varphi^2}{\rho_\varphi},\] one finds \[\dot\varphi=\sqrt{\frac{\gamma_\varphi(1+r)}{6\pi G}}\frac{1}{\gamma_\varphi+r}\frac1t,\] i.e., $\varphi$ evolves logarithmically with time.

With the help of \[\rho_\varphi= \frac{\dot\varphi^2}2+V(\varphi)\] and \[\rho_\varphi= \frac{\dot\varphi^2}{\gamma_\varphi},\] one finds \[V(\varphi)=\frac{\dot\varphi^2}{\gamma_\varphi}\left(1-\frac{\gamma_\varphi}2\right).\] Using the expression for $\dot\varphi$, obtained in the previous problem, one obtains \[V(\varphi)=\frac{1}{6\pi G}\left(1-\frac{\gamma_\varphi}2\right)\frac{1+r}{(\gamma_\varphi+r)^2}\frac1{t^2}.\] Since \[\frac{dV}{d\varphi}\dot\varphi=\frac{dV}{dt}=-2\frac V t,\] then one obtains the following equation for the scalar field potential \[\frac{dV(\varphi)}{d\varphi}=-\lambda V(\varphi),\quad \lambda\equiv\sqrt{\frac{24\pi G}{\gamma_\varphi(1+r)}}(\gamma_\varphi+r),\] and, consequently, \[V(\varphi)=V_0\exp\left[-\lambda(\varphi-\varphi_0)\right].\]

Transit to derivatives of the scalar field with respect to $u=\ln a$ in the Klein-Gordon equation and take into account that $Q=-\lambda\dot\varphi\rho_{dm}$ in the considered case to obtain \[H^2\varphi''+(\dot H+3H^2)\varphi'=\lambda\rho_{dm}-\eta V.\] Substitute \[H^2=\frac13\frac{\rho_{dm}+V}{1-\frac16\varphi''}\] and \[\dot H=-\frac12(\rho_{tot}+p_{tot})\] to find \[\frac13\frac{\rho_{dm}+V}{1-\frac16\varphi''}\varphi''+\frac12(\rho_{dm}+2V)\varphi'=\lambda\rho_{dm}-\eta V.\]

Like in the previous problem, transit to derivatives of the scalar field with respect to $u=\ln a$ in the Klein-Gordon equation and take into account that \[Q=-\alpha\frac{\dot\varphi}{\varphi}\rho_{dm}\] in the considered case to obtain \[\frac13\frac{\rho_{dm}+V}{1-\frac16\varphi''}\varphi''+\frac12(\rho_{dm}+2V)\varphi'=\frac{\alpha}{\varphi} \rho_{dm}-\frac{\beta}{\varphi} V.\]

If only $\delta$ depends on the scale factor, then \[\rho_{dm}=\rho_{dm0}a^{-3}e^{\int\delta(a)d\ln a}.\] One more time use the assumption \[r\equiv\frac{\rho_{dm}}{\rho_{de}}=\frac{\rho_{dm0}}{\rho_{de0}}a^{-\xi}=A^{-1}a^{-\xi},\quad A\equiv\frac{\rho_{dm0}}{\rho_{de0}}=\frac{\Omega_{dm0}}{\Omega_{de0}}.\] Consequently, \[\rho_{de}=\frac{Aa^\xi}{1+Aa^\xi}\rho_{tot},\quad \rho_{dm}=\frac{1}{1+Aa^\xi}\rho_{tot}.\] Then the total energy density satisfies the equation \[\frac{d\rho_{tot}}{da}+\frac3a\frac{1+(1+w_{de})Aa^\xi}{1+Aa^\xi}\rho_{tot}=0.\] Integrate this expression to obtain \[\rho_{tot}=\rho_{tot0}a^{-3}\left[1-\Omega_{de0}(1-a^\xi)\right]^{-3w_{de}/\xi},\quad \rho_{tot0}=\rho_{de0}+\rho_{dm0}.\] Consequently the first Friedmann equation can be written as \[H^2=H_0^2 a^{-3}\left[1-\Omega_{de0}(1-a^\xi)\right]^{-3w_{de}/\xi}.\] Using the conservation equation for DM one can get the coupling function \[\delta=3+\frac1H\frac{\dot\rho_{dm}}{\rho_{dm}}=-\frac{(3+w_{de})Aa^\xi}{1+Aa^\xi}= -(3+w_{de}) \frac{\rho_{de}}{\rho_{tot}}.\] This relation can be expressed as \[\delta=\frac{\delta_0}{\Omega_{de0}+(1-\Omega_{de0})a^{-\xi}}, \quad \delta_0\equiv-\Omega_{de0}(\xi+3w_{de}).\] Let us analyze the obtained expression. Asymptote of the interaction $\delta$ represents a constant, $\delta(a\to\infty)=\delta_0/\Omega_{de0}$. Therefore, if the expansion dynamics is such that $\xi>-3w_{de}$, then $\delta<0$, which implies that the energy flow is from the dark matter to dark energy. On the contrary, when $0<\xi<-3w_{de}$, the energy flow is from the phantom dark energy to dark matter. Further, we can see that there is no coupling between the dark energy and dark matter at $\xi=-3w_{de}$. Specifically, there is no coupling in SCM, for which $\xi=3$, $w_{de}=-1$.

\[q=-\frac{\ddot a}{aH^2}=-1+\frac{\dot H}{H^2}=-1+\frac32\frac{1-\Omega_{de0}+(1+w_{de})\Omega_{de0}a^\xi}{1-\Omega_{de0}(1-a^\xi)}.\] Note that the deceleration parameter values $q(a\to1)$ and $q(a\to\infty)$ are negative, as expected in the dark energy dominated epoch.

Use the result of problem #IDE_63 \[Q=-3H(\gamma\varphi-1)\frac{r}{1+r}\rho_\varphi\] to find that the conservation equation for the fantom field takes on the form \[\dot\rho_\varphi+3\frac{\gamma_\varphi+r}{1+r}H\rho_\varphi=0.\] It then follows that \[\rho_\varphi\propto a^{-3\nu},\quad \nu\equiv\gamma_\varphi\frac{r}{1+r}.\] If the expansion is matter dominated until the time $t_m$, then we can write the scale factor as \[a(t)=a(t_m)\left(1-\nu+\nu\frac t {t_m}\right)^{\frac2{3\nu}}.\] Using the definitions of $r$ and $\gamma_varphi$, the Friedmann equation gives \[a\propto\exp\left(\frac{\sqrt{1+r}}{\sqrt{-3(1+w_\varphi)}}\kappa\varphi\right),\quad \kappa^2\equiv 8\pi G.\] In terms of $\varphi$, $\rho_\varphi$ becomes \[\rho_\varphi\propto\exp\left(\frac{\sqrt{3(1+r)}\nu}{\sqrt{-(1+w_\varphi)}}\kappa\varphi\right).\]

For the scalar field $V(\varphi)\propto\rho_\varphi-p_\varphi$, and for $w_\varphi=const$ one has $V(\varphi)\propto\rho_\varphi$. It follows that \[V(\varphi)=V_0 \exp\left(\frac{\sqrt{3(1+r)}\nu}{\sqrt{-(1+w_\varphi)}}\kappa\varphi\right).\]