Biodiversity, Infectious Diseases, and the Dilution Effect

Abstract

Biologists point out that biodiversity loss contributes to promote the transmission of diseases. In epidemiology, this phenomenon is known as dilution effect. Our paper aims to introduce this effect in an economic model where the spread of an infectious disease is considered. More precisely, we embed a SIS model into a Ramsey model (1928) where a pollution externality coming from production affects the evolution of biodiversity. Biodiversity is assimilated to a renewable resource and affects the infectivity of the disease (dilution effect). A green tax is levied on production at the firm level to finance depollution according to a balanced budget rule. In the long run, a disease-free and an endemic regime are possible. We focus only on the second case and we find that the magnitude of the dilution effect determines the number of steady states. When the dilution effect remains low, there are two cases depending on the environmental impact of production: (1) a low impact implies two steady states with high and low biodiversity respectively; (2) a large impact rules out any steady state. Conversely, when the dilution effect becomes high, a (unique) steady state always exists: a strong dilution effect works as a buffer and prevents the human pressure from being lethal for biodiversity in the long run. Moreover, under a low dilution effect, a higher green tax rate always impairs biodiversity at the low steady state, while this green paradox is over under a high dilution effect. In the short run, we show that a limit cycle can arise around the high biodiversity steady state when the dilution effect is low. Surprisingly, the limit cycle is preserved under a high dilution effect. In other words, even if a strong dilution effect preserves the biodiversity in the long run and prevents the economy from the green paradox, it does not shelter the economy from the occurrence of biodiversity fluctuations.

Appendix

Proof

of Proposition 2

The agent maximizes the intertemporal utility function (9) under the budget constraint (7). Setting the Hamiltonian $H_{t}=e^{-\theta t}u\left (c_{t},B_{t}\right ) +\lambda _{t}\left [ \left (r_{t} -\delta \right ) h_{t}+\bar {\omega }_{t}-c_{t}\right ]$, deriving the first-order conditions ∂H_t/∂c_t = 0, $\partial H_{t}/\partial h_{t}=-\dot {\lambda }_{t}$ and $\partial H_{t}/\partial \mu _{t}=\dot {h}_{t}$, and defining μ_t ≡ λ_te^𝜃t, we get Eqs. 10, 11, and 12. □

Proof

of Proposition 3

Consider Eqs. 4, 14, 18, and Proposition 2. □

Proof

of Proposition 7

We differentiate system (19)–(22) to find the following:

$$ \begin{array}{l} \left[ \begin{array}{c} \frac{\tau}{\mu^{\ast}}\frac{d\mu^{\ast}}{d\tau}\\ \frac{\tau}{k^{\ast}}\frac{\text{dk}^{\ast}}{d\tau}\\ \frac{\tau}{l^{\ast}}\frac{\text{dl}^{\ast}}{d\tau}\\ \frac{\tau}{B}\frac{\text{dB}}{d\tau} \end{array} \right] \mathbf{=}\left[ \begin{array}{cccc} 0 & \frac{1-\alpha}{\sigma} & 0 & 0\\ -\frac{\phi}{\varepsilon_{\text{cc}}} & \theta & \phi+\beta\left( 1-l\right) & \phi\frac{\varepsilon_{\text{cB}}}{\varepsilon_{\text{cc}}}-\gamma d\frac{1-l}{l}\\ 0 & 0 & 1 & -d\\ 0 & \alpha & 1 & -\frac{1-2B}{1-B} \end{array}\right]^{-1}\left[ \begin{array}{c} -\frac{\tau}{1-\tau}\\ \frac{\theta+\delta}{\alpha}\frac{\tau}{1-\tau}\\ 0\\ \frac{b\tau}{a-b\tau} \end{array} \right],\\\\ \text{where}\\ \phi\equiv\frac{\theta+\left( 1-\alpha\right) \delta}{\alpha}=\frac{c^{\ast }}{k^{\ast}l^{\ast}},\\ \text{That is,} \\ \left[ \begin{array}{c} \frac{\tau}{\mu^{\ast}}\frac{d\mu^{\ast}}{d\tau}\\ \frac{\tau}{k^{\ast}}\frac{\text{dk}^{\ast}}{d\tau}\\ \frac{\tau}{l^{\ast}}\frac{\text{dl}^{\ast}}{d\tau}\\ \frac{\tau}{B}\frac{\text{dB}}{d\tau} \end{array} \right] =\left[ \begin{array}{c} \varepsilon_{\text{cc}}\left[ \left( \frac{b\tau}{a-b\tau}+\frac{\alpha\sigma }{1-\alpha}\frac{\tau}{1-\tau}\right) \frac{1-B}{\bar{B}-B}\dfrac {d+\frac{\varepsilon_{\text{cB}}}{\varepsilon_{\text{cc}}}}{d-2}-\left[ 1+\frac{\alpha }{1-\alpha}\frac{\sigma\theta+\left( 1-\alpha\right) \delta}{\theta+\left( 1-\alpha\right) \delta}\right] \frac{\tau}{1-\tau}\right] \\ -\frac{\sigma}{1-\alpha}\frac{\tau}{1-\tau}\\ \left( \frac{\alpha\sigma}{1-\alpha}\frac{\tau}{1-\tau}+\frac{b\tau}{a-b\tau }\right) \frac{1-B}{\bar{B}-B}\frac{d}{d-2}\\ \left( \frac{\alpha\sigma}{1-\alpha}\frac{\tau}{1-\tau}+\frac{b\tau}{a-b\tau }\right) \frac{1-B}{\bar{B}-B}\frac{1}{d-2} \end{array} \right] \text{.} \\\\\\\\\\\\\\\\\\\\\\\\\\ \end{array} $$

(46)

Under Assumption 4 and $B\in \left (0,1\right )$, we obtain easily Proposition 7. □

Proof

of Proposition 8

In the Cobb–Douglas case, σ = 1 and, according to expressions (46), (37) yields the following:

$$ \frac{\tau}{c^{\ast}}\frac{\partial c^{\ast}}{\partial\tau}=-\frac{1} {1-\alpha}\frac{\tau}{1-\tau}+\left( \frac{\alpha}{1-\alpha}\frac{\tau }{1-\tau}+\frac{b\tau}{a-b\tau}\right) \frac{1-B}{\bar{B}-B}\frac{d} {d-2}\text{.} $$

In the case a = b, we obtain the following:

$$ \begin{array}{@{}rcl@{}} \frac{\tau}{c^{\ast}}\frac{\partial c^{\ast}}{\partial\tau}&=&\frac{1}{1-\alpha }\frac{\tau}{1-\tau}\left( \frac{1-B}{\bar{B}-B}\frac{d}{d-2}-1\right)\\ &=&\frac{1}{1-\alpha}\frac{\tau}{1-\tau}\frac{1-2B}{\left( d-2\right) \left( \bar{B}-B\right) }\text{.} \end{array} $$

Proposition 8 immediately follows. □

Proof

of Proposition 9

According to Corollary 15 of [7], a Hopf bifurcation arises iff S₂ = S₃/T + DT/S₃ and T and S₃ have the same sign.

Let us rewrite S₂ and S₃ as follows:

$$ \begin{array}{@{}rcl@{}} S_{2} & =&\frac{Z}{m}-\frac{T}{m}\frac{S_{3}}{T}, \end{array} $$

(47)

$$ \begin{array}{@{}rcl@{}} S_{3} & =&z_{3}-m\alpha q\left( 1-B\right) \frac{\varepsilon_{2} }{\varepsilon_{1}}\text{.} \end{array} $$

(48)

Replacing (47), equation

$$ S_{2}=\frac{S_{3}}{T}+D\frac{T}{S_{3}}, $$

becomes the following:

$$ \frac{S_{3}}{T}=\frac{Z\pm\sqrt{Z^{2}-4mD\left( m+T\right) }}{2\left( m+T\right) }\text{.} $$

(49)

We observe that, if 0 < d < 1, D < 0 < m + T iff $\bar {B}<B<\left (1+\theta \right ) /2$. In this case, m + T > 0 and $4mD\left (m+T\right ) <0$. If d > 1, then D < 0 and, thus, D < 0 < m + T iff $B<\left (1+\theta \right ) /2$. Even in this case, m + T > 0 and $4mD\left (m+T\right ) <0$.

Then, in both the cases,

$$ \left( \frac{S_{3}}{T}\right)_{-}<0<\left( \frac{S_{3}}{T}\right)_{+}\text{.} $$

Clearly, the solution ε₂ of

$$ \frac{S_{3}\left( \varepsilon_{2}\right) }{T}=\left( \frac{S_{3}} {T}\right)_{-}<0, $$

is not acceptable as Hopf bifurcation value because T and S₃ have opposite sign. Let ε_H be solution of the following:

$$ \frac{S_{3}\left( \varepsilon_{2}\right) }{T}=\left( \frac{S_{3}} {T}\right)_{+}\text{.} $$

Replacing Eq. 48 in the LHS and Eq. 49 in the RHS, we obtain Eq. 44. □