for simplicity. This is exactly realized when the potential is exponential-type: $$\begin{aligned}
\frac{\kappa}{3}V(\phi)
=H_*^2\exp\Bigl[\sqrt{2\kappa\epsilon_{*}}(\phi -\phi_{*})\Bigr]\,.\end{aligned}$$ To see that an exact tracking solution exists in this case, we solve Eqs. , and under the assumption $$\begin{aligned}
\phi (t)\propto \ln t\,. \end{aligned}$$ Only if $\epsilon_{*}>1$, can one solve these equations consistently to find $$\begin{aligned}
&&a_{{\rm R},{\rm track}}(t)
=\frac{t}{\sqrt{1-1/\epsilon_{*}}}
\,,\label{eq:exponential sol1}\\
&&\phi_{\rm track}(t)
=\phi_{*}
+\sqrt{\frac{2}{\kappa\epsilon_{*}}}
\ln\biggl[\sqrt{\frac{3}{2}\epsilon_{*}}H_*t\biggr]\,.
\label{eq:exponential sol2}\end{aligned}$$ It is obvious that the above solution is an exactly tracking solution. Furthermore, one can show that it is an attractor solution.

We should note that the tracking occurs even for $\epsilon_{*}={\cal O}(1)$. The difference from the case with $\epsilon_{*}\gg 1$ is that the curvature term is subdominant for the expansion of the universe in this case during the tracking phase. This means that the curvature length is well outside the horizon scale for $\epsilon_{*}={\cal O}(1)$, while it remains comparable to the horizon scale for $\epsilon_{*}\gg 1$.

![The evolution of the energy densities in the region R with the exponential-type potential defined by Eq.  with $\epsilon_{*}=0.1$, $0.5$, $1$, $10^2$ and $10^4$. The dash-dotted line in red is the contribution from the spatial curvature, the solid curves in blue are the potential energy and the dashed 