Limit Cycles Under a Negative Effect of Pollution on Consumption Demand: The Role of an Environmental Kuznets Curve

Abstract

Since Heal (Explorations in natural resource economics. The Johns Hopkins University Press for Resources for the Future, Baltimore, 1982), there is a theoretical consensus about the occurrence of limit cycles (through a Hopf bifurcation) under a positive effect of pollution on consumption demand (compensation effect) and about the impossibility under a negative effect (distaste effect). However, recent empirical evidence advocates for the relevance of distaste effects. Our paper challenges the conventional view on the theoretical ground and reconciles theory and evidence. The environmental Kuznets curve (EKC) (pollution first increases in the capital level then decreases) plays the main role. Indeed, the standard case à la Heal (limit cycles only under a compensation effect) only works along the upward-sloping branch of the curve while the opposite (limit cycles only under a distaste effect) holds along the downward-sloping branch. Welfare effects of taxation also change according to the slope of the EKC.

Appendix

Proof of Proposition 1

The consumer’s Hamiltonian function writes

$$\begin{aligned} \tilde{H}\equiv e^{-\rho t}u\left( c,P\right) +\tilde{\lambda }\left[ \left( r-\delta \right) h+w-c\right] \end{aligned}$$

The first-order conditions are given by $\partial \tilde{H}/\partial \tilde{\lambda }=\left( r-\delta \right) h+w-c=\dot{h}$, $\partial \tilde{H}/\partial h=\tilde{\lambda }\left( r-\delta \right) =-\tilde{\lambda }^{\prime }$, $\partial \tilde{H}/\partial c=e^{-\rho t}u_{c}-\tilde{\lambda } =0$. Setting $\lambda \equiv e^{\rho t}\tilde{\lambda }$, we find $\dot{\lambda }-\rho \lambda =e^{\rho t}\tilde{\lambda }^{\prime }$ and, therefore, $\lambda \left( r-\delta -\rho \right) =-\dot{\lambda }$. Finally, the budget constraint $\dot{h}=\left( r-\delta \right) h+w-c$, now binding, writes at equilibrium $\dot{k}=\left( r-\delta \right) k+w-c$.   $\square$

Proof of Proposition 3

Focus on Eq. (12). $\dot{P}=0$ implies

$$\begin{aligned} P=\frac{k^{*}}{a}\left[ b-\gamma \left( \tau k^{*}\right) \tau \right] \equiv P\left( k^{*}\right) \end{aligned}$$

(34)

with

$$\begin{aligned} P^{\prime }\left( k^{*}\right) =\frac{1}{a}\left[ b-\tau \left( 1+\theta \right) \gamma \left( \tau k^{*}\right) \right] \end{aligned}$$

(35)

Under Assumption 3.1, if $k^{*}<\tilde{k}$ then $P^{\prime }\left( k\right) >0$ while if $k^{*}>\tilde{k}$ then $P^{\prime }\left( k^{*}\right) <0$.    $\square$

Proof of Proposition 4

At the steady state, $\dot{\lambda }=\dot{k}=\dot{P}=0$. Equation (10) gives (15).

Assumption 1 implies that there exists a unique $k^{*}>0$ verifying (15). Replacing this value into Eq. (12) gives $P^{*}=\left[ bk^{*}-\gamma \left( \tau k^{*}\right) \tau k^{*}\right] /a$. Since $b>\gamma \left( m^{*}\right) \tau$, there exists a unique $P^{*}>0$. Replacing $\left( k^{*},P^{*}\right)$ into Eq. (11), we obtain:

$$\begin{aligned} c^{*}=c\left( \lambda ^{*},P^{*}\right) =\rho k^{*}+w\left( k^{*}\right) >0 \end{aligned}$$

(36)

Equation (6) becomes

$$\begin{aligned} \lambda ^{*}=u_{c}\left( c^{*},P^{*}\right) =u_{c}\left( \rho k^{*}+w\left( k^{*}\right) ,\left[ bk^{*}-\gamma \left( \tau k^{*}\right) \tau k^{*}\right] /a\right) >0 \end{aligned}$$

   $\square$

Proof of Proposition 6

Let

$$\begin{aligned} W^{*}\equiv \int _{0}^{\infty }e^{-\rho t}u\left( c^{*},P^{*}\right) \,dt=u\left( c^{*},P^{*}\right) \int _{0}^{\infty }e^{-\rho t} \,dt=\frac{1}{\rho }u\left( c^{*},P^{*}\right) \end{aligned}$$

be the welfare function evaluated at the steady state. We find

$$\begin{aligned} \frac{\tau }{W^{*}}\frac{\partial W^{*}}{\partial \tau }=\frac{c^{*} }{u^{*}}\frac{\partial u}{\partial c}\frac{\tau }{c^{*}}\frac{\partial c^{*}}{\partial \tau }+\frac{P^{*}}{u^{*}}\frac{\partial u}{\partial P}\frac{\tau }{P^{*}}\frac{\partial P^{*}}{\partial \tau } \end{aligned}$$

Using (17) and (18), we obtain

$$\begin{aligned} \frac{\tau }{W^{*}}\frac{\partial W^{*}}{\partial \tau }=\left[ \varepsilon _{c}\frac{\rho +\left( \rho +\delta +\tau \right) \frac{1-\alpha }{\sigma }}{\rho +\left( \rho +\delta +\tau \right) \frac{1-\alpha }{\alpha } }+\varepsilon _{P}\pi \left( k^{*}\right) \right] \frac{\tau }{k^{*} }\frac{\partial k^{*}}{\partial \tau } \end{aligned}$$

   $\square$

Proof of Proposition 7

Necessity In a three-dimensional dynamic system, we require at the bifurcation value: $\lambda _{1}=ib=-\lambda _{2}$ with no generic restriction on $\lambda _{3}$ (see Bosi and Ragot 2011 or Kuznetsov 1998 among others). The characteristic polynomial of J is given by: $P\left( \lambda \right) =\left( \lambda -\lambda _{1}\right) \left( \lambda -\lambda _{2}\right) \left( \lambda -\lambda _{3}\right) =\lambda ^{3} -T\lambda ^{2}+S\lambda -D$. Using $\lambda _{1}=ib=-\lambda _{2}$, we find $D=b^{2}\lambda _{3}$, $S=b^{2}$, $T=\lambda _{3}$. Thus, $D=ST$ and $S>0$.

Sufficiency In the case of a three-dimensional system, one eigenvalue is always real, the others two are either real or nonreal and conjugated. Let us show that, if $D=ST$ and $S>0$, these eigenvalues are nonreal with zero real part and, hence, a Hopf bifurcation generically occurs.

We observe that $D=ST$ implies

$$\begin{aligned} \lambda _{1}\lambda _{2}\lambda _{3}=\left( \lambda _{1}\lambda _{2}+\lambda _{1}\lambda _{3}+\lambda _{2}\lambda _{3}\right) \left( \lambda _{1}+\lambda _{2}+\lambda _{3}\right) \end{aligned}$$

or, equivalently,

$$\begin{aligned} \left( \lambda _{1}+\lambda _{2}\right) \left[ \lambda _{3} ^{2}+\left( \lambda _{1}+\lambda _{2}\right) \lambda _{3}+\lambda _{1} \lambda _{2}\right] =0 \end{aligned}$$

(37)

This equation holds if and only if $\lambda _{1}+\lambda _{2}=0$ or $\lambda _{3}^{2}+\left( \lambda _{1}+\lambda _{2}\right) \lambda _{3} +\lambda _{1}\lambda _{2}=0$. Solving this second-degree equation for $\lambda _{3}$, we find $\lambda _{3}=-\lambda _{1}$ or $-\lambda _{2}$. Thus, (37) holds if and only if $\lambda _{1}+\lambda _{2}=0$ or $\lambda _{1}+\lambda _{3}=0$ or $\lambda _{2}+\lambda _{3}=0$. Without loss of generality, let $\lambda _{1}+\lambda _{2}=0$ with, generically, $\lambda _{3}\ne 0$ a real eigenvalue. Since $S>0$, we have also $\lambda _{1}=-\lambda _{2}\ne 0$. We obtain $T=\lambda _{3}\ne 0$ and $S=D/T=\lambda _{1}\lambda _{2}=-\lambda _{1}^{2}>0$. This is possible only if $\lambda _{1}$ is nonreal. If $\lambda _{1}$ is nonreal, $\lambda _{2}$ is conjugated, and, since $\lambda _{1}=-\lambda _{2}$, they have a zero real part.    $\square$

Proof of Proposition 8

Necessity In the real case, we obtain $D=\lambda _{1}\lambda _{2}\lambda _{3}<0$, $S=\lambda _{1}\lambda _{2}+\lambda _{1}\lambda _{3} +\lambda _{2}\lambda _{3}>0$ and $T=\lambda _{1}+\lambda _{2}+\lambda _{3}<0$.

Sufficiency We want to prove that, if $D,T<0$ and $S>0$, then $\lambda _{1},\lambda _{2},\lambda _{3}<0$. Notice that $D<0$ implies $\lambda _{1},\lambda _{2},\lambda _{3}\ne 0$.

$D<0$ implies that at least one eigenvalue is negative. Let, without loss of generality, $\lambda _{3}<0$. Since $\lambda _{3}<0$ and $D=\lambda _{1} \lambda _{2}\lambda _{3}<0$, we have $\lambda _{1}\lambda _{2}>0$. Thus, there are two subcases: (1) $\lambda _{1},\lambda _{2}<0$, (2) $\lambda _{1},\lambda _{2} >0$. If $\lambda _{1},\lambda _{2}>0$, $T<0$ implies $\lambda _{3}<-\left( \lambda _{1}+\lambda _{2}\right)$ and, hence,

$$\begin{aligned} S=\lambda _{1}\lambda _{2}+\left( \lambda _{1}+\lambda _{2}\right) \lambda _{3}<\lambda _{1}\lambda _{2}-\left( \lambda _{1}+\lambda _{2}\right) ^{2}=-\lambda _{1}^{2}-\lambda _{2}^{2}-\lambda _{1}\lambda _{2}<0 \end{aligned}$$

a contradiction. Then, $\lambda _{1},\lambda _{2}<0$.    $\square$