Exact Solutions
In a row of the problems below [following the paper Marco A. Reyes, On exact solutions to the scalar field equations in standard cosmology, arXiv: 0806.2292] we present a simple algebraic method to find exact solutions for a wide variety of scalar field potentials. Let us consider the function $V_a(\varphi)$ defined as \begin{equation}\label{es_1} V_a(\varphi)\equiv V(\varphi)+\frac12\dot\varphi^2. \end{equation} Derivative of this function reads \[\frac{dV_a}{d\varphi}=\frac{dV}{d\varphi}+\ddot\varphi.\] Hence, equations \[H^2=\frac12\left(\frac12\dot\varphi^2+V(\varphi)\right),\] \[\ddot\varphi+3H\dot\varphi+\frac{dV}{d\varphi}=0\] can now be rewritten as \begin{align} \label{es_4}3H^2 & =V_a,\\ \label{es_5}3H\dot\varphi & =-\frac{dV_a}{d\varphi}. \end{align} To solve them, note that eq.(\ref{es_4}) defines $H$ as a function of $\varphi$, which when inserted into eq.(\ref{es_5}), gives the scalar field $\varphi(t)$ as a function of $t$, at least in quadratures \[-3H(\varphi)\left(\frac{dV_a}{d\varphi}\right)^{-1}d\varphi=dt.\] Finally, inserting $\varphi(t)$ into eqs.(\ref{es_1}) and (\ref{es_4}) gives $V(\varphi)$ and $a(t)$, respectively, and the solution is completed.
One could also use $H(t)$ to determine $\varphi(t)$, since \[\dot H=-\frac12\dot\varphi^2.\] implies that \[\Delta\varphi(t)=\pm\int\sqrt{-2\dot H}dt.\] Since $V_a(t)=3H^2(t)$, a complete knowledge of $H(t)$ fully determines the solution to the problem.
Problem 1
problem id:
For $H(t)=\alpha/t$ find $V(\varphi)$ and $\Delta\varphi(t)$.
\[V(\varphi)=(3\alpha^2-\alpha)\exp[-2\Delta\varphi/\sqrt{2\alpha}];\] \[\Delta\varphi(t)=\sqrt{2\alpha}\ln t.\]
Problem 2
problem id:
Reconstruct $V(\varphi)$ and find $\varphi(t)$ and $a(t)$ for $V_a=\lambda\varphi^2$.
In this case \[H^2=\frac13\lambda\varphi^2,\quad \frac{dV_a}{d\varphi}=2\lambda\varphi.\] Therefore, from \[-3H(\varphi)\left(\frac{dV_a}{d\varphi}\right)^{-1}d\varphi=dt.\] one finds that \[\Delta\varphi(t)=\pm2\sqrt{\frac\lambda3}\Delta t.\] Hence, $\dot\varphi$ is constant. Letting $\varphi(t_0=0)=0$, and using eqs. (\ref{es_1}) and (\ref{es_4}) we get \begin{align} \nonumber V(\varphi) & = \lambda\varphi^2-\frac23\lambda,\\ \nonumber a(t) & =a_0e^{-\frac\lambda3t^2}. \end{align} Obviously, one would be tempted to pick $\lambda<0$ in order to make $a(t)$ a growing function of $t$, but that would make $\varphi(t)$ an imaginary function of $t$.
Problem 3
problem id:
Reconstruct $V(\varphi)$ and find $\varphi(t)$ and $a(t)$ for $V_a=\lambda\varphi^4$.
Proceeding the same way as in the previous problem one finds \begin{align} \nonumber \varphi(t) & =\varphi_0e^{\pm4\sqrt{\frac\lambda3}t},\\ \nonumber V(\varphi) & = \lambda\varphi^4-\frac83\lambda\varphi^2,\\ \nonumber a(t) & =a_0\exp\left[-\frac{\varphi_0^2}8e^{\pm8\sqrt{\frac\lambda3}(t-t_0)}\right]. \end{align}
Problem 4
problem id:
Reconstruct $V(\varphi)$ and find $\varphi(t)$ and $a(t)$ for $V_a=\lambda\varphi^n$, $n>2$.
Proceeding the same way as above one finds \begin{align} \nonumber \varphi(t) & =\left[\varphi_0^{2-n}\pm2n(n-2)\sqrt{\frac\lambda3}(t-t_0)\right]^{-\frac1{n-2}};\\ \nonumber V(\varphi) & = \lambda\varphi^{2n}-\frac23\lambda n^2\varphi^{2(n-1)};\\ \nonumber a(t) & =a_0\exp\left\{-\frac1{4n}\left[\varphi_0^{2-n}\pm2n(n-2)\sqrt{\frac\lambda3}(t-t_0)\right]^{-\frac2{n-2}}\right\}. \end{align}
