\begin{figure}[h,t,b,p]
\renewcommand{\figurename}{Fig.}
\centering
\includegraphics
[width=0.8\textwidth,viewport=125 160 490 640,clip]
{figs808_phase}
\caption{ \textbf{The Unique Balanced Growth Path} \hspace{10pt} {\footnotesize The phase diagram of the transformed system ($x$, $\lambda$) shows the two branches of $\dot{x}$ $=$ $0$ separated by an asymptote at $x$ $=$ $\hat{x}$ $\simeq$ $0.7$. The branch located on the left of $\hat{x}$ lies in the negative quadrant for $x$ $\geq$ $0$, with the exception of a contact with the origin (this branch is associated with a negative shadow price of capital, $\lambda_{k}$ $<$ $0$). The phase diagram also shows that the origin is a globally unstable stationary state -- the transversality condition (\ref{foc2}) is violated. The stationary state must therefore lie to the right of $\hat{x}$. The branch located on the right of $\hat{x}$ crosses zero at $x$ $=$ $\tilde{x}$ $\simeq$ $1.3$. The portion of the branch located on the right of $\tilde{x}$ lies in the negative quadrant (this branch is associated with a positive shadow price of (harmful) habit, $-\lambda_{S}$ $>$ $0$). An optimum stationary state must lie between $\hat{x}$ and $\tilde{x}$, and is unique. Arrows of motion show the saddle-point stability of the positive stationary state. Benchmark simulation~: $\rho$ $=$ $0.05$, $\gamma$ $=$ $0.5$, $\sigma$ $=$ $4$, $\mu$ $=$ $0.5$, $\delta_{S}$ $=$ $1$, $r$ $=$ $A-\delta_{k}$ $=$ $0.1$, $B$ $=$ $1$. The stationary state is approximately ($1.0$, $1.6$). Arrows are not shown for negative values of $\lambda$.}}
\label{Returns}
\end{figure}