Optimal Pollution, Optimal Population, and Sustainability

Abstract

This paper develops a long-run consumer optimization model with endogenous pollution and endogenous population. The positive check increases mortality if pollution increases. The optimal path is sustainable if it provides non-decreasing consumption for a non-decreasing population. As usually, optimality and sustainability may conflict; with population endogenous to pollution, this conflict may ultimately lead the human species toward self-imposed extinction. Not even technical progress can warrant sustainability.

Appendix: Local Stability of the Steady States

Lets write $\dot{S}=\varphi ( S,E )$ and $\dot{E}=\phi ( S,E )$. In a steady state it hods $\dot{S}=\dot{E}=0$ implying

$$ \delta+\delta^{\prime}S+\rho-\theta n=\frac{n^{\prime}}{(\rho -n) ( \theta-1 ) }\delta S. $$

(18)

The Jacobian of the model is

As evaluated around a steady state, its elements become

in which the last row is derived by using (12) and (11b). Because ϕ _E contains the undefined second derivative n(S), we write

The expression in the square brackets is the difference in the slopes of the phase lines $\dot{S}=0$ and $\dot{E}=0$ and $( -\varphi_{E} ) \cdot\phi_{E}=-\frac{n^{\prime}E}{ ( \rho-n ) ( \theta -1 ) }$ is positive for all E>0. In steady states 1 and 3 the slope of the $\dot{E}=0$-line is steeper than that of the $\dot{S}=0$-line (see Fig. 2) making the square brackets negative. Thus, $\operatorname{DET} J<0$ and these steady states are saddles. In steady state 2 the slope of the $\dot {E}=0$-line is smaller (possibly negative) than the slope of the $\dot{S}=0$-line and the value of the square brackets is positive. The trace of the Jacobian is

Therefore, this steady state is an unstable node or focus.