Competition and cooperation in the exploitation of the groundwater resource

Abstract

We study the exploitation of a common groundwater resource, first as a static and then as a differential game, in order to take into account the strategic and dynamic interactions among the users of the resource. We suppose that firms can form coalitions or can decide not to cooperate. The non-cooperation regime is characterized by pumping that lead to depletion of the aquifer; the cooperation preserves the resource. Open-loop and feedback equilibria have been computed and compared in order to characterize the existence of cooperators and defectors in water resources management.

Appendix

Proof of Proposition 2

In order to solve the problem proposed, we use the maximum principle. Let us define the current value of the Hamiltonian $\mathcal {H}$ in the standard way.

$$\begin{aligned} \mathcal {H} _i= & {} \displaystyle \sum _{i=1}^{m}\left\{ \displaystyle \frac{N}{2k}{w_i}^2-\displaystyle \frac{g}{k} w_i-(c_0+c_1H)w_i-\beta \left[ -(\alpha -1)w_i-\frac{R}{N} \right] \right\} \\&+\,\displaystyle \frac{\lambda _i}{ AS }\left[ R+(\alpha -1)\displaystyle \sum _{i=1}^{m}w_i+(\alpha -1)\displaystyle \sum _{j=m+1}^{N}w_j\right] \quad i=1,\ldots ,m\\ \mathcal {H}_j= & {} \displaystyle \frac{N}{2k}{w_j}^2-\displaystyle \frac{g}{k} w_j-(c_0+c_1H)w_j-\beta \left[ -(\alpha -1)w_j-\frac{R}{N} \right] \\&+\,\displaystyle \frac{\lambda _j}{ AS }\left[ R+(\alpha -1)\displaystyle \sum _{i=1}^{m}w_i+(\alpha -1)\displaystyle \sum _{j=m+1}^{N}w_j\right] \quad j=m+1,\ldots ,N \end{aligned}$$

where $\lambda _i$ and $\lambda _j$ are the adjoint variables. We obtain the following set of necessary conditions for an interior open-loop equilibrium:

$$\begin{aligned} \displaystyle \frac{\partial \mathcal {H}_i}{\partial w_i}= & {} 0\Longleftrightarrow \displaystyle \frac{N}{k}{w_i}-\displaystyle \frac{g}{k} -c_0-c_1H+\beta (\alpha -1)+\displaystyle \frac{\lambda _i}{ AS }(\alpha -1)=0 \quad i=1,\ldots ,m\\ \displaystyle \frac{\partial \mathcal {H}_j}{\partial w_j}= & {} 0\Longleftrightarrow \displaystyle \frac{N}{k}{w_j}-\displaystyle \frac{g}{k} -c_0-c_1H+\beta (\alpha -1)+\displaystyle \frac{\lambda _j}{ AS }(\alpha -1)=0 \\&\qquad j=m+1,\ldots ,N \end{aligned}$$

which give us:

$$\begin{aligned} w_i= & {} \displaystyle \frac{k}{N} \left[ \displaystyle \frac{g}{k}+c_0+c_1H-\beta (\alpha -1)- \displaystyle \frac{\lambda _i}{ AS }(\alpha -1) \right] \quad i=1,\ldots ,m \end{aligned}$$

(28)

$$\begin{aligned} w_j= & {} \displaystyle \frac{k}{N} \left[ \displaystyle \frac{g}{k}+c_0+c_1H-\beta (\alpha -1)- \displaystyle \frac{\lambda _j}{ AS }(\alpha -1) \right] \quad j=m+1,\ldots ,N\nonumber \\ \end{aligned}$$

(29)

The adjoint equations are:

$$\begin{aligned} \dot{\lambda _i}= & {} r\lambda _i+c_1mw_i \quad i=1,\ldots ,m\\ \dot{\lambda _j}= & {} r\lambda _j+c_1w_j \quad j=m+1,\ldots ,N \end{aligned}$$

and the transversality conditions being:

$$\begin{aligned}&\lim _{t\longrightarrow +\infty }e^{-rt} \lambda _i(t)\ge 0 \quad \lim _{t\longrightarrow +\infty }e^{-rt} \lambda _i(t)H(t)= 0 \quad i=1,\ldots ,m\\&\lim _{t\longrightarrow +\infty }e^{-rt} \lambda _j(t)\ge 0 \quad \lim _{t\longrightarrow +\infty }e^{-rt} \lambda _j(t)H(t)= 0 \quad j=m+1,\ldots ,N \end{aligned}$$

Taking into account that, at the steady state, $\dot{H}=\dot{\lambda } =0$, the first-order conditions and the adjoint equations are used to determine the stationary equilibrium

$$\begin{aligned} \dot{\lambda _i}&=0\Longleftrightarrow \lambda _i=-c_1 m[g+k(c_0+c_1H-\beta (\alpha -1))]\left[ \displaystyle \frac{ AS }{{ rASN }-(\alpha -1)c_1mk}\right] \\ i&=1,\ldots ,m \\ \dot{\lambda _j}&=0\Longleftrightarrow \lambda _j=-c_1 [g+k(c_0+c_1H-\beta (\alpha -1))]\left[ \displaystyle \frac{ AS }{{ rASN }-(\alpha -1)c_1k}\right] \\ j&=m+1,\ldots ,N \end{aligned}$$

and moreover

$$\begin{aligned} \dot{\lambda _i}=0\Longleftrightarrow w_i=-\displaystyle \frac{r\lambda _i}{mc_1} \quad \text{ and } \quad \dot{\lambda _j}=0\Longleftrightarrow w_j=-\displaystyle \frac{r\lambda _j}{c_1} \end{aligned}$$

From differential equation, we have

$$\begin{aligned} \dot{H}=0\Longleftrightarrow R+m(\alpha -1)w_i+(N-m)(\alpha -1)w_j=0 \end{aligned}$$

By substitution of $\lambda _i$ and $\lambda _j$, we obtain the steady-state value of water table

$$\begin{aligned} H^*=-\displaystyle \frac{R}{rc_1k(\alpha -1)[m\mu +(N-m)\xi ]}-\displaystyle \frac{1}{c_1}\left[ \displaystyle \frac{g}{k}+c_0-\beta (\alpha -1)\right] \end{aligned}$$

where

$$\begin{aligned} \mu = \left[ \displaystyle \frac{ AS }{{ rASN }-(\alpha -1)c_1mk}\right] \quad \text{ and } \quad \xi =\left[ \displaystyle \frac{ AS }{{ rASN }-(\alpha -1)c_1k}\right] \end{aligned}$$

and so

$$\begin{aligned} \lambda _i^*= & {} \displaystyle \frac{\mu c_1mR}{r(\alpha -1)[m\mu +(N-m)\xi ]}\\ \lambda _j^*= & {} \displaystyle \frac{\xi c_1R}{r(\alpha -1)[m\mu +(N-m)\xi ]} \end{aligned}$$

and

$$\begin{aligned} w _i^*= & {} -\displaystyle \frac{\mu R}{(\alpha -1)[m\mu +(N-m)\xi ]}\\ w _j^*= & {} -\displaystyle \frac{\xi R}{(\alpha -1)[m\mu +(N-m)\xi ]} \end{aligned}$$

$\square$

Proof of Proposition 3

We have

$$\begin{aligned} \displaystyle \frac{\partial H^*}{\partial m}= & {} \displaystyle \frac{R[c_1k(\alpha -1)-{ NrAS }][m^2c_1k(\alpha -1)-{ NrAS }(2m-1)]}{rAS[c_1km(m-N-1)(\alpha -1)+N^2rAS]^2}>0 \\ \displaystyle \frac{\partial H^*_{PO}}{\partial m}= & {} \displaystyle \frac{R}{{ rNAS }}>0 \end{aligned}$$

$\square$