Rainwater Harvesting and Groundwater Conservation: When Endogenous Heterogeneity Matters

Abstract

In this paper, we focus on resource conservation in a model of decentralized management of groundwater and rainwater. We show that a conservation policy may have opposite effects on the level of the resource, depending on the outcome of the decentralized management. More precisely, we consider identical farmers who can use two water resources (groundwater and/or rainwater) and we study the symmetric and asymmetric feedback stationary Nash equilibria of the dynamic game. We show that a subsidy on the use of rainwater may increase the level of the aquifer at the symmetric equilibrium, whereas it decreases the level of the aquifer at the asymmetric equilibrium. This suggests that the usual focus on (interior) symmetric equilibria in dynamic games may provide misleading policy implications.

Appendix

1.1 Proof of Proposition 4

We first show a preliminary result. Since we focus on asymmetric equilibria with specialized farmers, we have $n_{g}\ge 1,n_{r}\ge 1$ and $n_{b}=0$.

Lemma 1

Assume that $w_{r}^{A}$, the solution of $\max _{w_{r}}\left[ F(\theta w_{r})-C_{r}(w_{r})\right]$, is positive and let $\bar{R}\equiv R-n_{r}w_{r}^{A}>0$. Also, assume that

$$\begin{aligned} \bar{V}_{gi}(h_{0})&= \underset{w_{gi}\ge 0}{max}\int _{0}^{\infty }e^{-\delta t}\left[ F(\mu w_{gi})-C_{g}(h,w_{gi})\right] dt\text { for all } i\in G, \\ \dot{h}&= \rho \bar{R}-\sum _{i\in G}w_{gi},\quad h(0)=h_{0}\text { given,} \end{aligned}$$

has a solution such that $w_{gi}^{A}(t)>0$ for all $i\in G$ and $h(t)>0$ for all $t$. Then, $w_{ri}=w_{r}^{A}$ for all $i\in R$ and $w_{gi}=w_{gi}^{A}$ for all $i\in G$ is a feedback equilibrium of our differential game.

Proof of Lemma 1

Suppose that a feedback equilibrium with $n_{g}\ge 1,n_{r}\ge 1,\,n_{b}=0$ and $h(t)>0$ exists. Let $V_{gi}$ be the value function of player $i$ in group $G$ and $V_{ri}$ the value function of player $i$ in group $R$. Let $\phi _{gi}$ be the feedback equilibrium strategy of groundwater use for player $i$ in group $G$ and $\phi _{ri}$ be the feedback equilibrium strategy $i$ in group $R$. These value functions are a solution of the following Hamilton-Jacobi-Bellman equations:

$$\begin{aligned} \delta V_{gi}(h)&= \max _{w_{g}\ge 0}\left[ F(\mu w_{g})-C_{g}(h,w_{g})\right. \nonumber \\&\left. +\frac{ \partial V_{g}(h)}{\partial h}\left( \rho (R-\sum _{j\in R}\phi _{gj}(h))-w_{g}-\sum _{j\ne i}\phi _{gj}(h)\right) \right] , \end{aligned}$$

(38)

and,

$$\begin{aligned} \delta V_{ri}(h)=\max _{w_{r}\ge 0}\left[ F(\theta w_{r})\!-\!C_{r}(w_{r})\!+\!\frac{ \partial V_{r}^{i}(h)}{\partial h}\left( \rho (R-w_{r}\!-\!\sum _{j\ne i}\phi _{r}^{j}(h))-\sum _{i}\phi _{g}^{i}(h)\right) \right] . \end{aligned}$$

(39)

Now, let us define the following strategies. Let $\phi _{r}=arg\max _{w_{g}}\left[ F(\theta w_{g})-C_{r}(w_{g})\right]$. We have

$$\begin{aligned} \max _{w_{g}}\left[ F(\theta w_{g})-C_{r}(w_{g})\right] =F(\theta \phi _{r})-C_{r}(\phi _{r}), \end{aligned}$$

and

$$\begin{aligned} \bar{V}_{r}=\int _{0}^{\infty }e^{-\delta t}\left[ F(\theta \phi _{r})-C_{r}(\phi _{r})\right] dt=\frac{F(\theta \phi _{r})-C_{r}(\phi _{r})}{\delta }, \end{aligned}$$

(40)

and $\bar{R}=R-n_{r}\phi _{r}$. Notice that $\phi _{r}$ is a constant and it does not depend on $h$. Also, $\bar{V}_{r}$ does not depend on $h$.

Let $\phi _{g}^{i}$ be the feedback Nash equilibrium strategy of player $i$ of the following differential game:

$$\begin{aligned}&\bar{V}_{gi}(h_{0})=\underset{w_{gi}\ge 0}{max}\int _{0}^{\infty }e^{-\delta t}\left[ F(\mu w_{gi})-C_{g}(h,w_{gi})\right] dt,\\&\quad \dot{h}=\rho \bar{R} -\sum _{j\in G}w_{gj},\quad h(0)=h_{0}\text { given.} \end{aligned}$$

The associated HJB equation is

$$\begin{aligned} \delta \bar{V}_{gi}(h)=\max _{w_{gi}\ge 0}\left[ F(\mu w_{gi})-C_{g}(h,w_{gi})+\frac{\partial \bar{V}_{g}^{i}(h)}{\partial h}\left( \rho \bar{R}-w_{gi}-\sum _{j\ne i}\phi _{g}^{j}(h)\right) \right] . \end{aligned}$$

(41)

Suppose that the feedback Nash equilibrium is such that $h(t)>0$. Hence $\bar{V}_{gi}(h)$ and $\bar{V}_{g}$ (with the respective equilibrium strategies) are solutions of (38) and (39). $\square$

Now, we can establish the following result:

Lemma 2

There exists an asymmetric SNE with specialized farmers where an increase in the cost of rainwater (through an increase in $\sigma$) has the following effect:

$$\begin{aligned} \frac{\partial w_{r}^{A}}{\partial \sigma }<0<\frac{\partial \left( \sum _{i\in G}w_{gi}^{A}\right) }{\partial \sigma } \end{aligned}$$

Moreover, we have $\frac{\partial h^{A}}{\partial \sigma }>0$ when $n_{g}=1$.

Proof of Lemma 2

Let $C_{r}\equiv \left( 1+\sigma \right) \widetilde{C}_{r}$. Using Lemma 1, we have $w_{r}^{A}=\max _{w_{r}}\left[ F(\theta w_{r})\right. \left. -\left( 1+\sigma \right) \widetilde{C}_{r}(w_{r})\right]$ and $\bar{R}\equiv R-n_{r}w_{r}^{A}$. Hence, when $\sigma$ increases, $w_{r}^{A}$ decreases and $\bar{R}$ increases. At the steady state, $\sum _{i\in G}\phi _{Gi}=\sum _{i\in G}w_{gi}^{A}=\rho \bar{R}$, and then $\sum _{i\in G}w_{gi}^{A}$ increases with $\sigma$. If all the players in group $G$ use the same level of groundwater, then $w_{gi}^{A}$ increases with $\sigma$.

Now, let us focus on the case in which $n_{g}=1$. Since $w_{r}^{A}$ does not depend on $h$, we have $\delta _{i}\left( h\right) =\delta$ for $i\in G$. Let us show that $\frac{\partial h^{A}}{\partial \sigma }>0$. We first show that $\frac{\mu }{\theta }>\frac{1}{\rho }$. Let $RC_{i}\equiv MC_{gi}(h,w_{gi})/MC_{ri}\left( h,w_{gi},w_{ri}\right)$ denote the ratio of marginal costs, where

$$\begin{aligned} MC_{gi}(h,w_{gi})\equiv \frac{\partial C_{g}}{\partial w_{g}}(h,w_{gi})- \frac{1}{\delta _{i}\left( h\right) }\frac{\partial C_{g}}{\partial h} (h,w_{gi}), \end{aligned}$$

(42)

and,

$$\begin{aligned} MC_{ri}\left( h,w_{gi},w_{ri}\right) \equiv C_{r}^{\prime }\left( w_{ri}\right) -\rho \frac{1}{\delta _{i}\left( h\right) }\frac{\partial C_{g} }{\partial h}(h,w_{gi}). \end{aligned}$$

(43)

Let $w_{g}^{A}\equiv w_{gi}^{A}$. Using condition (30), condition (33) and because the ratio of

the marginal costs is a decreasing function of $w_{ri}$, we have

$$\begin{aligned} \frac{MC_{g}(h,w_{g}^{A})}{MC_{r}(h,w_{g}^{A},0)}\le \frac{\mu }{\theta }< \frac{MC_{g}(h,0)}{MC_{r}(h,0,0)}. \end{aligned}$$

(44)

Note that this implies that it is not possible $\frac{MC_{g}(h,w_{g})}{MC_{r}(h,w_{g},0)}\ge 0$ for all $w_{g}\in [0,w_{g}^{A}]$ (because if it is the case $w_{g}^{A}<0$). Then there must exist $\widetilde{w}_{g}\in [0,w_{g}^{A}]$ such that the derivative of the ratio of marginal cost, $\frac{MC_{g}(h,\widetilde{w}_{g})}{MC_{r}(h,\widetilde{w}_{g},0)}$ is negative. Notice that this derivative is:

$$\begin{aligned}&\frac{\left( \begin{array}{c} \left[ \frac{\partial ^{2}C_{g}}{\partial w_{g}^{2}}(h,\widetilde{w}_{g})- \frac{1}{\delta }\frac{\partial ^{2}C_{g}}{\partial w_{g}\partial h}(h, \widetilde{w}_{g})\right] \left[ C_{r}^{\prime }\left( 0\right) -\rho \frac{1 }{\delta }\frac{\partial C_{g}}{\partial h}(h,\widetilde{w}_{g})\right] \\ +\rho \frac{1}{\delta }\frac{\partial ^{2}C_{g}}{\partial w_{g}\partial h}(h, \tilde{g})\left[ \frac{\partial C_{g}}{\partial w_{g}}(h,\widetilde{w}_{g})- \frac{1}{\delta }\frac{\partial C_{g}}{\partial h}(h,\widetilde{w}_{g}) \right] \end{array} \right) }{\left[ C_{r}^{\prime }\left( 0\right) -\rho \frac{1}{\delta } \partial _{h}C_{g}(h,\widetilde{w}_{g})\right] ^{2}} \\&= \frac{\frac{\partial ^{2}C_{g}}{\partial w_{g}^{2}}(h,\widetilde{w}_{g}) \left[ C_{r}^{\prime }\left( 0\right) -\rho \frac{1}{\delta }\frac{\partial C_{g}}{\partial h}(h,\widetilde{w}_{g})\right] -\frac{1}{\delta }\frac{ \partial ^{2}C_{g}}{\partial w_{g}\partial h}(h,\widetilde{w}_{g})\left[ C_{r}^{\prime }\left( 0\right) -\rho \frac{\partial C_{g}}{\partial w_{g}}(h, \widetilde{w}_{g})\right] }{\left[ C_{r}^{\prime }\left( 0\right) -\rho \frac{1}{\delta }\frac{\partial C_{g}}{\partial h}(h,\widetilde{w}_{g}) \right] ^{2}}. \end{aligned}$$

Hence, we must have

$$\begin{aligned} C_{r}^{\prime }\left( 0\right) -\rho \frac{\partial C_{g}}{\partial w_{g}}(h, \widetilde{w}_{g})<-\frac{\frac{\partial ^{2}C_{g}}{\partial w_{g}^{2}}(h, \widetilde{w}_{g})}{-\frac{1}{\delta }\frac{\partial ^{2}C_{g}}{\partial w_{g}\partial h}(h,\widetilde{w}_{g})}\left[ C_{r}^{\prime }\left( 0\right) -\rho \frac{1}{\delta }\frac{\partial C_{g}}{\partial h}(h,\widetilde{w}_{g}) \right] <0. \end{aligned}$$

(45)

Using (45) and $\frac{\partial ^{2}C_{g}}{\partial w_{g}^{2}}\ge 0$, we have

$$\begin{aligned} \frac{1}{\rho }C_{r}^{\prime }(0)-\frac{\partial C_{g}}{\partial w_{g}} (h,w_{g}^{A})<\frac{1}{\rho }C_{r}^{\prime }(0)-\frac{\partial C_{g}}{ \partial w_{g}}(h,\widetilde{w}_{g})<0. \end{aligned}$$

This implies

$$\begin{aligned} \frac{MC_{g}(h,w_{g}^{A})}{MC_{r}(h,w_{g}^{A},0)}=\frac{1}{\rho }\;\frac{ C_{g}(h,w_{g}^{A})-\left( 1/\delta \right) \frac{\partial C_{g}}{\partial h} (h,w_{g}^{A})}{C_{r}^{\prime }(0)/\rho -\left( 1/\delta \right) \frac{ \partial C_{g}}{\partial h}(h,w_{g}^{A})}>\frac{1}{\rho }, \end{aligned}$$

Using (44), we have $\frac{\mu }{\theta }>\frac{1}{\rho }$.

Differentiating (28) with respect to $\sigma$, we have

$$\begin{aligned} \left[ \frac{1}{\delta }\frac{\partial ^{2}C_{g}}{\partial h^{2}} (h,w_{g}^{A})-\frac{\partial ^{2}C_{g}}{\partial w_{g}\partial h} (h,w_{g}^{A})\right] \frac{\partial h}{\partial \sigma }&= \left[ \frac{ \partial C_{g}}{\partial w_{g}}(h,w_{g}^{A})-\frac{1}{\delta }\frac{\partial ^{2}C_{g}}{\partial w_{g}\partial h}(h,w_{g}^{A})\right. \\&\left. -\mu ^{2}F^{\prime \prime }(\mu w_{g}^{A})\right] \frac{\partial w_{g}^{A}}{\partial \sigma } \end{aligned}$$

Finally, $\frac{\partial h^{A}}{\partial \sigma }>0$.

Moreover, in the quadratic linear specification with two players, the SNE with specialized farmers is unique.

1.2 Proof of Proposition 5

We focus on the SNE in which the two farmers use rainwater and groundwater. Assuming that the two farmers use linear strategies, i.e. that farmer $i$ considers that player $j$ uses $w_{gj}=a_{g}h+b_{g}$ and $w_{rj}=a_{r}h+b_{r}$ and that the value function is quadratic $V(h)=\frac{A}{2}h^{2}+Bh+C$. The farmer has to solve the following problem:

$$\begin{aligned} \delta V(h)&= \max _{w_{r},w_{g}}\left[ F(\mu w_{g}+\theta w_{r})-(c-h)w_{g}-Kw_{r}\right. \\&\left. +\frac{\partial V(h)}{\partial h}\left( \rho (R-w_{r}-a_{r}h-b_{r})-w_{g}-a_{g}h-b_{g}\right) \right] . \end{aligned}$$

For this case, first order condition gives:

$$\begin{aligned} \mu -(\mu w_{g}+\theta w_{r})\mu -c+h-Ah-B=0,\quad \theta -(\mu w_{g}+\theta w_{r})\theta -K-(Ah+B)\rho =0, \end{aligned}$$

from these two equations we obtain

$$\begin{aligned} \frac{\mu }{\theta }=\frac{c-h+Ah+B}{K+(Ah+B)\rho }. \end{aligned}$$

(46)

This states that the solution is singular solution because $h$ does not depend on time.¹² Hence, in equilibrium, we must have $A=B=0$ and $\dot{h}=0$. The unique steady state is characterized by:

$$\begin{aligned} h^{S}=c-\frac{\mu }{\theta }K,\quad w_{g}^{S}=\frac{\rho \left( \theta -K- \frac{\theta ^{2}R}{2}\right) }{\theta (\rho \mu -\theta )},\quad w_{r}^{S}= \frac{-\theta +K+\frac{\theta \mu \rho R}{2}}{\theta (\rho \mu -\theta )}. \end{aligned}$$

We can easily deduce that

$$\begin{aligned} \frac{\partial {h^{S}}}{\partial {\sigma }}=\frac{\partial {h^{S}}}{\partial {K}}<0, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial w_{r}^{S}}{\partial {\sigma }}=\frac{\partial w_{r}^{S}}{ \partial {K}}=-\frac{1}{\rho }\frac{\partial w_{g}^{S}}{\partial {K}}. \end{aligned}$$

Moreover

$$\begin{aligned} w_{g}^{S}=\rho \frac{\mu +h-c-\frac{\theta \mu R}{2}}{\mu (\rho \mu -\theta ) },\quad w_{r}^{S}=\frac{-\mu +c-h+\frac{\mu ^{2}\rho R}{2}}{\mu (\rho \mu -\theta )}, \end{aligned}$$

with

$$\begin{aligned} \frac{\partial w_{g}^{S}}{\partial {h}}=\frac{1}{\mu (\rho \mu -\theta )} ,\quad \frac{\partial w_{r}^{S}}{\partial {h}}=-\frac{1}{\mu (\rho \mu -\theta )}. \end{aligned}$$

One can check that $w_{g}^{S},w_{r}^{S}>0$ implies $\rho \mu -\theta >0$. Hence $\frac{\partial w_{g}^{S}}{\partial {\sigma }}=\frac{\partial w_{g}^{S} }{\partial {h}}\frac{\partial {h^{S}}}{\partial {\sigma }}<0$ and $\frac{ \partial w_{r}^{S}}{\partial {\sigma }}=\frac{\partial w_{r}^{S}}{\partial {h }}\frac{\partial {h^{S}}}{\partial {\sigma }}>0$.