Uniqueness and Multiple Trajectories for the Case of Lucas Model

Abstract

The main aim of this paper is to prove that the model introduced by Lucas and further analyzed by Caballe and Santos, Mulligand and Sala-I-Martin, Benhabib and Perli and finally by Boucekkine and Ruiz-Tamarit, has two interesting properties. If the externality parameter in the production of human capital is greater than the elasticity of output with respect to physical capital, then the system is characterized by multiple transitional paths, indexed by the starting value of the fraction of labor allocated to the production of physical capital, leading to different steady-states equilibrium. Alternatively, if the externality parameter in the production of human capital is smaller than the elasticity of output with respect to physical capital, then the system is characterized by an unique transitional path, convergent to the unique steady-state equilibrium. For the special case where the inverse of the elasticity of intertemporal substitution equals the elasticity of output with respect to physical capital, we obtain closed-form solutions and thus, our approach contains as particular cases some other results, as those obtained by Boucekkine and Ruiz-Tamarit. More than this, differently to Boucekkine and Ruiz-Tamarit we obtain closed-form solutions for all variables of the model and thus we are able to verify the two transitional conditions.

Appendix

In order to prove the Theorem 1, we need the following preliminary result, which use the definition of the finite limit of a function at infinity.

Lemma 1

If the function $f = f(x)$ has a finite limit l, when $x\rightarrow \infty$, then there exists a finite $x_{*} > 0$ such that for all $x > x_{*}$, $\left| f(x) - l\right| < \varepsilon$¹. Applying this result to our function $\frac{c(t)}{k(t)}$, under the hypotheses of Proposition 1, we can claim that, for every real $\varepsilon > 0$, there exists a finite $t_{*} > 0$ such that for all $t > t_{*}$, $\left| \frac{c(t)}{k(t)} - \xi \right| < \varepsilon$.

Proof of Theorem 1

According to the relations (11), the Eq. (10) can be written as

$$\begin{aligned} \frac{\dot{u}}{u} = - \dot{D} -\theta u \end{aligned}$$

whose solution is given by (12). We observe that the solution (12) can also be written $u(t) = \frac{\dot{F}(t)}{\theta F(t)}$ and substituting this result into the second equation of the system (3) we obtain $l(t) = l_0 e^{\phi t}F^{-\frac{\phi }{\theta }}$ from where we get (13). Substituting these two results into the first equation of the system (3) we obtain

$$\begin{aligned} \frac{\dot{k}}{k} = \frac{\left( l_0u_0\right) ^{1-\beta }e^{(1-\beta )(\phi t - D)}}{F^{\frac{\gamma }{ \gamma -\beta }}}\frac{1}{k^{1-\beta }}-\eta -\dot{D}, \end{aligned}$$

whose solution is given by (14). Substituting these results into the third equation of the system (3) we find

$$\begin{aligned} \frac{\dot{c}}{c} = -\frac{\rho }{\sigma } + \frac{\beta }{\sigma }\left( \frac{l_0u_0}{k_0}\right) ^{1-\beta }\frac{\dot{\Omega }}{\Omega }, \end{aligned}$$

where

$$\begin{aligned} \Omega (t) = k_0^{1-\beta } + Q_0\Phi (t),\;\Omega (0) = k_0^{1-\beta }, \end{aligned}$$

and the solution for c will be given by (15). Knowing that $\lambda = c^{-\sigma }$, the solution for $\lambda$ is

$$\begin{aligned} \lambda (t) = c_0^{-\sigma }k_0^{\beta }\left[ k_0^{1-\beta }+Q_0 \Phi (t)\right] ^{-\frac{\beta }{1-\beta }}e^{\rho t}. \end{aligned}$$

(27)

We first assume that $\gamma - \beta > 0$ and prove that this restriction generates indeterminacy because there are a continuum of transitional paths leading the system to the unique steady state equilibrium, indexed by the values the control variable $u_0$. In order to do this we need to establish some properties of the functions D(t) and B(t). If $\gamma - \beta > 0$ or equivalently $\eta -\xi > 0$, then, according to the Lemma 1, for all $t > t_{*}$

$$\begin{aligned} D(t)= & {} \int \limits _{0}^t\frac{c(s)}{k(s)}ds -\eta t = \int \limits _{0}^{t_{*}}\frac{c(s)}{k(s)}ds - \eta t + \xi \int \limits _{t_{*}}^t ds \\= & {} \tilde{D}\left( t_{*}\right) - \eta t + \xi \left( t-t_{*}\right) = \tilde{D}\left( t_{*}\right) - \xi t_{*}- \left( \eta -\xi \right) t, \end{aligned}$$

and therefore $\lim \limits _{t\rightarrow \infty }D(t) = -\infty$, $\lim \limits _{t\rightarrow \infty } B(t) = \lim \limits _{t\rightarrow \infty } F(t) =+\infty$. Applying l’Hôpital rule in Eq. (12) we get

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }u(t) = \lim \limits _{t\rightarrow \infty }\frac{u_0e^{-D(t)}}{1 + \theta u_0 B(t)} = -\frac{1}{\theta }\lim \limits _{t\rightarrow \infty }\dot{D}(t) = -\frac{1}{\theta } \left( \xi -\eta \right) = u_{*}. \end{aligned}$$

(28)

It remains to prove that $u(t) < 1$ for all $t > 0$.

$$\begin{aligned} 1 - u(t) = \frac{1 + u_0 \theta B(t) - u_0e^{-D(t)}}{1 + \theta u_0 B(t)}. \end{aligned}$$

Denoting by $G(t) = 1 + \theta u_0 B(t) - u_0e^{-D(t)}$, we have $G(0) = 1 - u_0 > 0$ and $\lim \limits _{t\rightarrow \infty }G(t) = +\infty .$

$$\begin{aligned} \dot{G}(t) = u_0 e^{-D(t)}\left[ \frac{c(t)}{k(t)} + \theta - \eta \right] . \end{aligned}$$

$\frac{c}{k} + \theta - \eta = \theta (1 - u_{*}) > 0$ for all $t > t_{*}$. If there exists $t_1 > 0$ such that $\dot{G}(t_1) = 0$, then $t_1 < t_{*}$ and therefore G(t) is a decreasing function for all $t \in (0, t_1)$ and an increasing function for all $t > t_1$. Since $G(t_1) > 0$ we conclude that $G(t) > 0$ for all $t > 0$ and hence $1 - u(t) > 0$. If $\frac{c}{k} + \theta - \eta > 0$ for all $t > 0$, then the conclusion doesn’t change and thus $u(t) < 1$.

We need now to examine the properties of positive function $\Phi (t)$. Also we have $\lim \limits _{t\rightarrow \infty }\tilde{D}(t) = \infty$ and therefore $\lim \limits _{t\rightarrow \infty }e^{-\tilde{D}(t)} = 0.$

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }F(t)^{-\frac{\gamma }{\gamma -\beta }}e^{\eta t} = \lim \limits _{t\rightarrow \infty }\left[ \frac{e^{\frac{\eta (\gamma -\beta )}{\gamma }t}}{F(t)}\right] ^{\frac{\gamma }{\gamma -\beta }} = \left[ \frac{\eta (\gamma -\beta )}{\gamma \theta u_0}\lim \limits _{t\rightarrow \infty } e^{\frac{\eta (\gamma -\beta )}{\gamma }t+D(t)}\right] ^{\frac{\gamma }{\gamma -\beta }}. \end{aligned}$$

Similarly we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\frac{\eta (\gamma -\beta )}{\gamma }t\left[ 1+\lim \limits _{t\rightarrow \infty }\frac{D(t)}{\frac{\eta (\gamma -\beta )}{\gamma }t}\right] = \lim \limits _{t\rightarrow \infty }\frac{\theta \left[ 1-\beta +\gamma (1-u_{*})\right] }{\gamma }t = \infty \end{aligned}$$

and therefore we get

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\Phi (t)= +\infty . \end{aligned}$$

(29)

The transversality condition for k is given by

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }k(t)\lambda (t)e^{-\rho t} = c_0^{-\sigma }k_0^{\beta }\lim \limits _{t\rightarrow \infty } \left[ k_0^{1-\beta }+Q_0 \Phi (t)\right] e^{-\tilde{D}(t)}. \end{aligned}$$

To evaluate the above limit we apply the l’Hôpital rule.

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\frac{k_0^{1-\beta }+Q_0 \Phi (t)}{e^{\tilde{D}(t)}}= & {} \frac{Q_0}{\xi } \lim \limits _{t\rightarrow \infty }\frac{\left[ 1 + \theta u_0 B(t)\right] ^{-\frac{\gamma }{\gamma -\beta }}}{ e^{ D(t)}} \\= & {} \lim \limits _{t\rightarrow \infty }\left[ \frac{1+\theta u_0 B(t)}{e^{-\frac{\gamma -\beta }{\gamma }D(t)}}\right] ^{- \frac{\gamma }{\gamma -\beta }} \\= & {} \left[ \frac{\gamma \delta u_0}{\eta -\xi }\right] ^{-\frac{\gamma }{\gamma -\beta }} \lim \limits _{t\rightarrow \infty }e^{ \frac{\beta }{\gamma -\beta }D(t)} = 0. \end{aligned}$$

The transversality condition for h gives

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }h(t)\mu (t)e^{-\rho t} = h_0\mu _0\lim \limits _{t\rightarrow \infty } e^{\delta t}F^{-\frac{\delta }{\theta }}e^{(\rho -\delta ) t}e^{-\rho t}= h_0\mu _0\lim \limits _{t\rightarrow \infty }\left[ F(t)\right] ^{-\frac{\delta }{\theta }} = 0. \end{aligned}$$

From here we deduce that the transversality conditions hold for any choice of $u_0$, but $u_0$ cannot be determined and we can conclude that the restriction $\gamma > \beta$ generates indeterminacy. As it is well-known, equilibrium indeterminacy arises because there are a continuum of transitional paths indexed by the values of (one of) the control variable leading to the steady state equilibrium. In our case, this variable is $u_0$.

We now assume that $\gamma < \beta$ and prove that this restriction generates a unique transitional paths, leading the system to the unique steady state equilibrium. To prove this statement we need to establish some properties of the functions D(t) and B(t). If $\gamma - \beta < 0$ or equivalently $\eta -\xi < 0$, then for all $t > t_{*}$

$$\begin{aligned} D(t)= & {} \int \limits _{0}^t \frac{c(s)}{k(s)}ds -\eta t = \int \limits _{0}^{t_{*}}\frac{c(s)}{k(s)}ds -\eta t + \xi \int \limits _{t_{*}}^t ds \\= & {} \tilde{D}\left( t_{*}\right) - \xi t_{*} + (\xi -\eta )t, \end{aligned}$$

and therefore $\lim \limits _{t\rightarrow \infty }D(t) = +\infty$ and $\lim \limits _{t\rightarrow \infty }e^{-D(t)} = 0$. First observe that B(t) is an increasing function of time. Since $\lim \limits _{t\rightarrow \infty }e^{-D(t)} = 0$ it immediately follows that $B_{*} = \lim \limits _{t\rightarrow \infty }B(t)$ must be finite because otherwise $\lim \limits _{t\rightarrow \infty }u(t) = 0$. In fact it is just a simply exercise to prove that

$$\begin{aligned} \int \limits _{0}^{\infty } e^{-D(t)}dt \le \frac{1}{\xi -\eta }\;\;\Leftrightarrow \;\; B_{*} \le \frac{1}{\xi -\eta }. \end{aligned}$$

If $B_{*} \ne -\left( \theta u_0\right) ^{-1}$ then $\lim \limits _{t\rightarrow \infty }u(t) = 0$. For u to be an admissible solution we need $B_{*} = -\left( \theta u_0\right) ^{-1}$ and therefore

$$\begin{aligned} u_0 = \frac{\beta }{ \delta (\beta -\gamma )B_{*}} \end{aligned}$$

(30)

and hence we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }u(t) = \frac{\beta }{\delta (\beta -\gamma )}\lim \limits _{t\rightarrow \infty } \frac{ e^{-D(t)}}{B_{*} - B(t)} = \frac{\beta }{\delta (\beta -\gamma )}\lim \limits _{t\rightarrow \infty }\left[ \frac{c(t)}{k(t)}-\eta \right] = u_{*}. \end{aligned}$$

(31)

It remains to prove that $u(t) < 1$ for all $t > 0$.

$$\begin{aligned} 1 - u(t) = \frac{\delta (\beta -\gamma )\left[ B_{*} - B(t)\right] - \beta e^{-D(t)}}{\delta (\beta -\gamma ) \left[ B_{*} - B(t)\right] }. \end{aligned}$$

Denoting by $G(t) = \delta (\beta -\gamma )\left[ B_{*} - B(t)\right] - \beta e^{-D(t)}$ we observe that

$$\begin{aligned} G(0)= & {} \delta (\beta -\gamma )\left[ B_{*} - \frac{\beta }{\delta (\beta -\gamma )}\right] > 0\;\; \text{ and } \;\; \lim \limits _{t\rightarrow \infty }G(t) = 0. \\ \dot{G}(t)= & {} \frac{e^{-D(t)}}{\beta }\left[ \frac{c(t)}{k(t)} - \eta - \frac{\delta (\beta -\gamma )}{\beta } \right] = \frac{e^{-D(t)}}{\beta }\left[ \frac{c(t)}{k(t)} - \eta + \theta \right] . \end{aligned}$$

$\frac{c}{k} + \theta - \eta = \theta (1 - u_{*}) < 0$ for all $t > t_{*}$. If there exists $t_1 > 0$ such that $\dot{G}(t_1) = 0$, then $t_1 < t_{*}$ and therefore G(t) is an increasing function for all $t \in (0, t_1)$ and a decreasing function for all $t > t_1$. Since $G(t_1) > 0$ we conclude that $G(t) > 0$ for all $t > 0$ and hence $1 - u(t) > 0$. If $\frac{c}{k} + \theta - \eta > 0$ for all $t > 0$, then the conclusion doesn’t change and thus $u(t) < 1$. We need now to examine the properties of positive function $\Phi (t)$ for the case $\gamma - \beta < 0$ or equivalently $\eta -\xi < 0$. Observe that $\lim \limits _{t\rightarrow \infty }\left[ \eta t + D(t)\right] = \infty$ and therefore $e^{-[\eta t + D(t)]}\rightarrow 0.$ Since $\lim \limits _{t\rightarrow \infty }F(t) = 0$ we need to apply l’Hôpital rule to evaluate the following limit

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }F(t)^{-\frac{\gamma }{\gamma -\beta }}e^{\eta t} = \lim \limits _{t\rightarrow \infty }\left[ \frac{F(t)}{e^{-\frac{\eta (\beta -\gamma )}{\gamma }t}}\right] ^{\frac{\gamma }{\beta -\gamma }} = \left[ \frac{\gamma u_0}{1-\beta +\gamma }\lim \limits _{t\rightarrow \infty } e^{\frac{\beta \eta }{\gamma }t-\tilde{D}(t)}\right] ^{\frac{\gamma }{\beta -\gamma }}. \end{aligned}$$

Applying now the special case of l’Hôpital rule we obtain

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\left[ 1-\frac{\tilde{D}(t)}{\frac{\beta \eta }{\gamma }t}\right] = \frac{\delta (1-\gamma )(1-\beta +\gamma )+\gamma \delta (\beta -\gamma )u_{*}}{\beta \delta (1-\beta +\gamma )} > 0 \end{aligned}$$

and therefore

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\Phi (t) = +\infty . \end{aligned}$$

(32)

The transversality condition for k is given by

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }k(t)\lambda (t)e^{-\rho t} = c_0^{-\sigma }k_0^{\beta }\lim \limits _{t\rightarrow \infty } \left[ k_0^{1-\beta }+Q_0 \Phi (t)\right] e^{-\tilde{D}(t)}. \end{aligned}$$

(33)

To evaluate the above limit we apply l’Hôpital rule.

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\frac{k_0^{1-\beta }+Q_0 \Phi (t)}{e^{\tilde{D}(t)}} = \frac{Q_0}{\xi } \lim \limits _{t\rightarrow \infty }\frac{\left[ 1 + \theta u_0 B(t)\right] ^{\frac{\gamma }{\beta -\gamma }}}{ e^{ D(t)}} = 0. \end{aligned}$$

The transversality condition for h gives

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }h(t)\mu (t)e^{-\rho t} = h_0\mu _0\lim \limits _{t\rightarrow \infty }\left[ F(t)\right] ^{-\frac{\delta }{\theta }} = 0, \end{aligned}$$

(34)

and thus the proof is completed. $\square$