The Poverty-Environment Trap in a Modified Romer Model with Dirtiness

Abstract

The model presented in this paper describes an economy with endogenous technical change and polluting production techniques. The main question we want to address is whether the adoption of “dirty” production processes might lead to a sustainable unique steady state, or guarantee the emergence of multiple equilibria. The application of the original Bogdanov–Takens theorem allows us to characterize the regions of the parametric space where the model exhibits either a global indeterminate equilibrium or a poverty-environment trap.

Appendix

• 1. The arrow sufficiency theorem holds since the maximized Hamiltonian, evaluated along the optimal control variables, is concave in all the state variables, as we can simply check through the sign of the minors of the Hessian matrix, whose determinant implies the following:

  $$\left\vert H\right\vert =\left\vert \begin{array}{ccc} - & 0 & + \\ 0 & - & 0 \\ + & 0 & - \end{array} \right\vert $$

  and thus, |H ₁| < 0, |H ₂| > 0, and |H ₃| < 0 iff A > 1, that is, the number of designs must necessarily be greater than one.

• 2. Transversality conditions for a free terminal state hold for all shadow prices and are given by

  $$\begin{array}{@{}rcl@{}} \lim\limits_{t\rightarrow \infty }\lambda Ke^{-\rho t} &=&\tilde{\lambda}e^{(1-2\sigma )gt}\tilde{K}e^{gt}e^{-\rho t}\\ &=&\tilde{\lambda}\tilde{K}e^{-(2\sigma g+\rho )t}=0 \\ \lim\limits_{t\rightarrow \infty }\mu Ee^{-\rho t} &=&\tilde{\mu}e^{(1-2\sigma )gt}\tilde{E}e^{gt}e^{-\rho t}\\ &=&\tilde{\mu}\tilde{E}e^{-(2\sigma g+\rho )t}=0 \\ \lim\limits_{t\rightarrow \infty }\vartheta Ae^{-\rho t} &=&\tilde{\vartheta}e^{(\rho -\varphi -\gamma )t}\tilde{A}e^{gt}e^{-\rho t}\\ &=&\tilde{\vartheta}\tilde{A}e^{(g-\varphi -\gamma )t}=0 \end{array} $$

  where $\tilde{\lambda}, \tilde{\mu}, \tilde{\vartheta}$, and $\tilde{K}, \tilde{E}, \tilde{A}$, are the shadow prices and the state values on the balanced growth path;

• 3. Moreover, for free time t, we need to show that $\underset{ t\rightarrow \infty }{\lim }H=0$, which is always verified due to convergence towards zero of both the discounted utility function, $\underset{t\rightarrow \infty}{\lim }U(\cdot )e^{-\rho t}=0$, and all the multipliers, as proved above.

• 4. We need first to translate the fixed point associated to ℛ to the origin, by substituting $\tilde{x}\equiv x-\bar{x}^{\ast }, \tilde{m}\equiv m-\bar{m}^{\ast }, \tilde{q}=q-\bar{q}^{\ast }$, and introduce the new trivial variables $\mu =\gamma -\bar{\gamma}$ and $\nu=\rho -\bar{\rho}$, such that

  $$\begin{array}{@{}rcl@{}} \overset{\cdot }{\tilde{x}} &=&-(\bar{\rho}+\nu )(\tilde{x}+\bar{x}^{\ast })\\ &&+\frac{(1-\alpha )(\bar{\gamma}+\mu )-\sigma (1+\bar{\gamma}+\mu )}{\sigma (1+\bar{\gamma}+\mu )}\\ &&\times(\tilde{m}+\bar{m}^{\ast })(\tilde{x}+\bar{x}^{\ast })+(\tilde{x}+\bar{x}^{\ast })^{2}\smallskip \\ \overset{\cdot }{\tilde{m}} &=&\frac{\theta +\alpha (\bar{\gamma}+\mu)(\varphi +\bar{\gamma}+\mu )}{(\bar{\gamma}+\mu )(1-\alpha )}(\tilde{m}+\bar{m}^{\ast})\\ &&-\frac{1+\alpha (\bar{\gamma}+\mu )}{1+\bar{\gamma}+\mu}(\tilde{m}+\bar{m}^{\ast })^{2}\\ &&+\frac{1+\bar{\gamma}+\mu }{(\bar{\gamma}+\mu )(1-\alpha )}(\tilde{m}+\bar{ m}^{\ast })(\tilde{q}+\bar{q}^{\ast })^{\omega}\\ \overset{\cdot }{\tilde{q}} &=&-\frac{(\bar{\rho}+\nu )}{\omega }(\tilde{q}+\bar{q}^{\ast })+\frac{(\tilde{m}+\bar{m}^{\ast })(\tilde{q}+\bar{q}^{\ast})^{1+\omega }}{\omega (\tilde{x}+\bar{x}^{\ast })}\\ &&+\frac{(1+\bar{\gamma}+\mu )}{\omega }(\tilde{q}+\bar{q}^{\ast })^{1+\omega} \end{array} $$

  (8)

A second-order Taylor expansion of the translated vector field (8) with respect to $(\tilde{x},\tilde{m},\tilde{q},\mu ,\nu)$ is therefore made, which leads to

$$\begin{array}{@{}rcl@{}} \left( \begin{array}{c} \overset{\cdot }{\tilde{x}} \\ \overset{\cdot }{\tilde{m}} \\ \overset{\cdot }{\tilde{q}} \end{array} \right) &=&\left( \begin{array}{ccc} \bar{x}^{\ast } & \frac{(1-\alpha )\bar{\gamma}-\sigma (1+\bar{\gamma})}{ \sigma (1+\bar{\gamma})}\bar{x}^{\ast } & 0 \\ 0 & \frac{\left( 1+\alpha \bar{\gamma}\right) \bar{m}^{\ast }}{1+\bar{\gamma} } & \frac{(1+\bar{\gamma})\bar{m}^{\ast }}{\bar{\gamma}(1-\alpha )} \\ -\frac{\bar{m}^{\ast }\bar{q}^{\ast }{}^{2}}{\bar{x}^{\ast }{}^{2}} & \frac{ \bar{q}^{\ast }{}^{2}}{\bar{x}^{\ast }} & \rho \end{array} \right) \left( \begin{array}{c} \tilde{x} \\ \tilde{m} \\ \tilde{q} \end{array} \right) \\ &&+\left( \begin{array}{ccc} \frac{(1-\alpha )\bar{m}^{\ast }}{(1+\bar{\gamma})^{2}}\mu -v & \frac{ (1-\alpha )\bar{x}^{\ast }}{(1+\bar{\gamma})^{2}}\mu & 0 \\ 0 & \left[ \frac{\alpha \bar{\gamma}^{2}-\theta }{(1-\alpha )\bar{\gamma}^{2} }+\frac{2(1-\alpha )}{(1+\bar{\gamma})^{2}}\bar{m}^{\ast }-\frac{1}{ (1-\alpha )(\bar{\gamma}+\mu )^{2}}\bar{q}^{\ast }\right] \mu & -\frac{\bar{m }^{\ast }}{(1-\alpha )\bar{\gamma}^{2}}\mu \\ 0 & 0 & 2\bar{q}^{\ast }\mu -v \end{array} \right) \left( \begin{array}{c} \tilde{x} \\ \tilde{m} \\ \tilde{q} \end{array} \right) \\ &&+\left( \begin{array}{c} \tilde{f}_{1}^{nl}(\tilde{x},\tilde{m},\tilde{q},\mu ,\nu ) \\ \tilde{f}_{2}^{nl}(\tilde{x},\tilde{m},\tilde{q},\mu ,\nu ) \\ \tilde{f}_{3}^{nl}(\tilde{x},\tilde{m},\tilde{q},\mu ,\nu ) \end{array} \right) \end{array} $$

(9)

where

$$\begin{array}{@{}rcl@{}} \tilde{f}_{1}^{nl}(\tilde{x},\tilde{m},\tilde{q},\mu ,\nu )&=&\tilde{x}^{2}- \frac{(1-\alpha )\bar{m}^{\ast }\bar{x}^{\ast }}{(1+\bar{\gamma})^{3}}\mu ^{2}-\frac{1+\alpha \bar{\gamma}}{1+\bar{\gamma}}\tilde{m}\tilde{x}\\ \tilde{f}_{2}^{nl}(\tilde{x},\tilde{m},\tilde{q},\mu ,\nu )&=&-\frac{1+\alpha \bar{\gamma}}{1+\bar{\gamma}}\tilde{m}^{2}+\frac{1+\bar{\gamma}}{\bar{\gamma} (1-\alpha )}\tilde{m}\tilde{q}+\left[ \frac{\theta \bar{m}^{\ast }}{ (1-\alpha )\bar{\gamma}^{3}}-\frac{(1-\alpha )\bar{m}^{\ast }{}^{2}}{(1+\bar{ \gamma})^{3}}-\frac{\bar{m}^{\ast }\bar{q}^{\ast }}{(1-\alpha )\bar{\gamma} ^{3}}\right] \mu ^{2}\\ \tilde{f}_{3}^{nl}(\tilde{x},\tilde{m},\tilde{q},\mu ,\nu )&=&\frac{\bar{m} ^{\ast }\bar{q}^{\ast }{}^{2}}{\bar{x}^{\ast }{}^{3}}\tilde{x}^{2}+\left[ \frac{\bar{m}^{\ast }}{\bar{x}^{\ast }}+(1+\bar{\gamma})\right] \tilde{q} ^{2}-\frac{\bar{q}^{\ast }{}^{2}}{\bar{x}^{\ast }{}^{2}}\tilde{x}\tilde{m}- \frac{2\bar{m}^{\ast }\bar{q}^{\ast }}{\bar{x}^{\ast }{}^{2}}\tilde{x}\tilde{ q}+\frac{2\bar{q}^{\ast }}{\bar{x}^{\ast }}\tilde{m}\tilde{q}\\ \end{array} $$

We ought to put the translated system (9) in the Jordan normal form characterized by a double-zero eigenvalue and a third eigenvalue given by the trace trJ ^∗. This computation is straightforward and needs

$$ \left( \begin{array}{c} \tilde{x} \\ \tilde{m} \\ \tilde{q} \end{array} \right) =\text{\textbf{T}}\left( \begin{array}{c} \tau _{1} \\ \tau _{2} \\ \tau _{3} \end{array} \right) $$

(10)

where T is a matrix of the eigenvectors

$$\text{\textbf{T}}=\left[ \begin{array}{ccc} \frac{j_{12}^{\ast }j_{23}^{\ast }}{j_{11}^{\ast }j_{22}^{\ast }} & \frac{ (j_{22}^{\ast })^{2}+j_{23}^{\ast }j_{32}^{\ast }}{j_{31}^{\ast }(j_{22}^{\ast })^{2}} & \frac{j_{12}^{\ast }j_{23}^{\ast }}{j_{23}^{\ast }(j_{22}^{\ast }+j_{33}^{\ast })} \\ -\frac{j_{23}^{\ast }}{j_{22}^{\ast }} & -\frac{j_{23}^{\ast }}{ (j_{22}^{\ast })^{2}} & 1 \\ 1 & 0 & \frac{j_{11}^{\ast }+j_{33}^{\ast }}{j_{23}^{\ast }} \end{array} \right] =\left[\begin{array}{ccc} u_{1} & v_{1} & z_{1} \\ u_{2} & v_{2} & 1 \\ 1 & 0 & z_{3} \end{array} \right] $$

which allows us to put the vector field (10) in the new coordinate change and the following normal form:

$$\begin{array}{@{}rcl@{}} \left( \begin{array}{c} \dot{\tau}_{1} \\ \dot{\tau}_{2} \\ \dot{\tau}_{3} \end{array} \right) &=&\left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \text{Tr}(\text{J}^{\mathbf{\ast }}(0)) \end{array} \right] \left( \begin{array}{c} \tau _{1} \\ \tau _{2} \\ \tau _{3} \end{array} \right)\\ &&+\mathbf{B}(\mu ,v)\left(\begin{array}{c} \tau _{1} \\ \tau _{2} \\ \tau _{3} \end{array} \right) +\left( \begin{array}{c} \bar{F}_{1}\left( \mathbf{\tau }\right) \\ \bar{F}_{2}\left( \mathbf{\tau }\right) \\ \bar{F}_{3}\left( \mathbf{\tau }\right) \end{array} \right) \end{array} $$

(11)

where ${\bf{\dot{\tau }}} = {{\rm{T}}^{{{\rm{ - 1}}}}}{J^{ * }}(0){\bf{T\tau }} + {{\bf{T}}^{{{\rm{ - 1}}}}}\tilde{f}_{i}^{{nl}}\left( {{\bf{T\tau }}} \right),{\bar{F}_{i}}\left( {\bf{\tau }} \right) = {{\bf{T}}^{{{\rm{ - 1}}}}}\tilde{f}_{i}^{{nl}}\left( {{\bf{T\tau }}} \right) = {{\bf{T}}^{{{\rm{ - 1}}}}}\tilde{f}_{i}^{{nl}}\left( {{\bf{Tw}}} \right)$ , and B(μ, v) is a matrix that vanishes at μ = 0, and v = 0:

$$ \text{\textbf{B}}(\mu ,v)=\text{\textbf{T}}^{\text{-1}}\left( \begin{array}{ccc} \frac{(1-\alpha )\bar{m}^{\ast }}{(1+\bar{\gamma})^{2}}\mu -v & \frac{ (1-\alpha )\bar{x}^{\ast }}{(1+\bar{\gamma})^{2}}\mu & 0 \\ 0 & \left[ \frac{\alpha \bar{\gamma}^{2}-\theta }{(1-\alpha )\bar{\gamma}^{2} }+\frac{2(1-\alpha )}{(1+\bar{\gamma})^{2}}\bar{m}^{\ast }-\frac{1}{ (1-\alpha )(\bar{\gamma}+\mu )^{2}}\bar{q}^{\ast }\right] \mu & -\frac{\bar{m }^{\ast }}{(1-\alpha )\bar{\gamma}^{2}}\mu \\ 0 & 0 & 2\bar{q}^{\ast }\mu -v \end{array} \right) \text{\textbf{T}} $$

(12)

The vector field (11) can be then suspended via a second-order center manifold reduction, which implies $\dot{\tau}_{3}=0$, (see the mathematical procedure in [21]) and therefore becomes

$$\left( \begin{array}{c} \dot{\tau}_{1} \\ \dot{\tau}_{2} \end{array} \right) =\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right) \left( \begin{array}{c} \tau _{1} \\ \tau _{2} \end{array} \right) +\boldsymbol{A}(\mu ,v)\left( \begin{array}{c} \tau _{1} \\ \tau _{2} \end{array} \right) +\left( \begin{array}{c} F_{1} \\ F_{2} \end{array} \right) $$

(13)

where $F_{1}+\mathcal{O}(w_{i}^{3})=\bar{F}_{1}\left( \boldsymbol{\tau }\right)$, and $F_{2}+\mathcal{O}(w_{i}^{3})=\bar{F}_{2}\left( \boldsymbol{\tau }\right)$, and

$$\boldsymbol{A}(\mu ,\nu )=\frac{1}{D}\left( \begin{array}{cc} -v_{2}z_{3}A_{1}+v_{1}z_{3}B_{1}+(-v_{1}+v_{2}z_{1})C_{1} & -v_{2}z_{3}A_{2}+v_{1}z_{3}B_{2} \\ (u_{2}z_{3}-1)A_{1}+(z_{1}-u_{1}z_{3})B_{1}+(u_{1}-u_{2}z_{1})C_{1} & (u_{2}z_{3}-1)A_{2}+(z_{1}-u_{1}z_{3})B_{2} \end{array} \right) $$

with

$$\begin{array}{l}A_{1}=\left[ \frac{(1-\alpha )\bar{m}^{\ast }}{(1+\bar{\gamma})^{2}}\mu -v \right] u_{1}+\frac{(1-\alpha )\bar{x}^{\ast }}{(1+\bar{\gamma})^{2}}\mu u_{2} \\ A_{2}=\left[ \frac{(1-\alpha )\bar{m}^{\ast }}{(1+\bar{\gamma})^{2}}\mu -v\right] v_{1}+\frac{(1-\alpha )\bar{x}^{\ast }}{(1+\bar{\gamma})^{2}}\mu v_{2} \\ A_{3}=\left[ \frac{(1-\alpha )\bar{m}^{\ast }}{(1+\bar{\gamma})^{2}}\mu -v\right] z_{1}+\frac{(1-\alpha )\bar{x}^{\ast }}{(1+\bar{\gamma})^{2}}\mu \\ B_{1}=\left[ \frac{\alpha \bar{\gamma}^{2}-\theta }{(1-\alpha )\bar{\gamma}^{2}}+\frac{2(1-\alpha )}{(1+\bar{\gamma})^{2}}\bar{m}^{\ast }-\frac{1}{(1-\alpha )(\bar{\gamma}+\mu )^{2}}\bar{q}^{\ast }\right] \mu u_{2}-\frac{\bar{m}^{\ast }}{(1-\alpha )\bar{\gamma}^{2}}\mu \\ B_{2}=\left[ \frac{\alpha \bar{\gamma}^{2}-\theta }{(1-\alpha )\bar{\gamma}^{2}}+\frac{2(1-\alpha )}{(1+\bar{\gamma})^{2}}\bar{m}^{\ast }-\frac{1}{(1-\alpha )(\bar{\gamma}+\mu )^{2}}\bar{q}^{\ast }\right] \mu v_{2} \\ B_{3}=\left[ \frac{\alpha \bar{\gamma}^{2}-\theta }{(1-\alpha )\bar{\gamma}^{2}}+\frac{2(1-\alpha )}{(1+\bar{\gamma})^{2}}\bar{m}^{\ast }-\frac{1}{(1-\alpha )(\bar{\gamma}+\mu )^{2}}\bar{q}^{\ast }\right] \mu -\frac{\bar{m}^{\ast }}{(1-\alpha )\bar{\gamma}^{2}}\mu z_{3} \\C_{1}=2\bar{q}^{\ast }\mu -v \\C_{2}=0 \\ C_{3}=(2\bar{q}^{\ast }\mu -v)z_{3} \end{array} $$

Moreover, following Freire et al. [20] and Gamero et al. [21], we are able to reduce (13) to a topologically equivalent system:

$$\begin{array}{@{}rcl@{}} \dot{\tau}_{1}&=&\tau_{2}\\ \dot{\tau}_{2}&=&\varepsilon_{1}+\varepsilon_{2}\tau_{2}+\tau_{1}^{2}+s\tau_{1}\tau_{2} \end{array} $$

(14)

where ε ₁ and ε ₂ are the unfolding parameters of the versal deformation matrix, V $(\mu ,\nu )=\left ( \begin {array}{cc} 0 & 1 \\ \varepsilon _{1} & \varepsilon _{2} \end {array} \right )$, that is

$$\begin{array}{rll} \varepsilon _{1}&=&\frac{\left[ -v_{2}z_{3}A_{2}+v_{1}z_{3}B_{2}\right] \left[ (u_{2}z_{3}-1)A_{1}+(z_{1}-u_{1}z_{3})B_{1}+(u_{1}-u_{2}z_{1})C_{1}\right] }{ D}- \\ && -\frac{\left[ -v_{2}z_{3}A_{1}+v_{1}z_{3}B_{1}+\left(-v_{1}+v_{2}z_{1}\right)C_{1} \right] \left[ (u_{2}z_{3}-1)A_{2}+(z_{1}-u_{1}z_{3})B_{2}\right] }{D} \end{array} $$

(15)

$$\begin{array}{rll} \varepsilon _{2}&=&\frac{-v_{2}z_{3}A_{1}+v_{1}z_{3}B_{1}+(-v_{1}+v_{2}z_{1})C_{1}}{D}+ \\ && +\frac{(u_{2}z_{3}-1)A_{2}+(z_{1}-u_{1}z_{3})B_{2}}{D} \end{array} $$

(16)

given that D = −v ₁ + v ₂ z ₁ − u ₁ v ₂ z ₃ + u ₂ v ₁ z ₃.

Moreover, the stability of (15), i.e., ε ₁ = 0, implies

$$ \xi _{1}\mu ^{2}+\xi _{2}v\mu +v_{1}v^{2}=0 $$

(17)

with

$$\begin{array}{l} \xi _{1}=b_{1}v_{2}z_{3}a_{1}-z_{3}\left[ b_{1}u_{2}-\frac{\bar{m}^{\ast }}{(1-\alpha )\bar{\gamma}^{2}}\right] c_{1}-2\bar{q}^{\ast }z_{1}b_{1}v_{2}+2 \bar{q}^{\ast }c_{1} \\ \xi _{2}=z_{3}\left[ b_{1}u_{2}-\frac{\bar{m}^{\ast }}{(1-\alpha )\bar{\gamma}^{2}}\right] v_{1}+z_{1}b_{1}v_{2}-b_{1}v_{2}z_{3}u_{1}-\left( 2\bar{q}^{\ast }v_{1}+c_{1}\right) \end{array}$$

where

$$\begin{array}{l} a_{1}=\left[ \frac{(1-\alpha )\bar{m}^{\ast }}{(1+\bar{\gamma})^{2}}u_{1}+\frac{(1-\alpha )\bar{x}^{\ast }}{(1+\bar{\gamma})^{2}}u_{2}\right] \\ b_{1}=\left[ \frac{\alpha \bar{\gamma}^{2}-\theta }{(1-\alpha )\bar{\gamma}^{2}}+\frac{2(1-\alpha )}{(1+\bar{\gamma})^{2}}\bar{m}^{\ast }-\frac{1}{(1-\alpha )\bar{\gamma}^{2}}\bar{q}^{\ast }\right] \\ c_{1}=\left[ \frac{(1-\alpha )\bar{m}^{\ast }}{(1+\bar{\gamma})^{2}}v_{1}+\frac{(1-\alpha )\bar{x}^{\ast }}{(1+\bar{\gamma})^{2}}v_{2}\right] \end{array} $$

Solution to (17) easily gives µ = Φv, where $\Phi =\frac{\sqrt{\xi _{2}^{2}-4v_{1}\xi _{1}}-\xi _{2}}{2\xi _{1}}.$.