Groundwater management and illegality in a differential-evolutionary framework

Abstract

It is estimated that half of all the water extracted, both in developed and developing countries, is unauthorized. This phenomenon makes the management of a groundwater even more difficult to avoid over-exploitation. To study the interaction between farmers, that could be compliant and non-compliant, and a water agency, we built a leader-follower differential game. However, we assumed that the water agency does not know neither ex-ante nor ex-post the number of compliant farmers. After illustrating the results of the dynamic game through numerical simulation using the Western La Mancha (Spain) data, we endogenize the types’ choice in an evolutionary context. Finally, we perform comparative dynamics in the steady state to understand the role of the sanction to counter illegal behaviors.

Mathematical appendix

1.1 Proof of Proposition 1

The first order conditions with respect to water pumped are:

$$\begin{aligned} \begin{aligned}&\frac{\partial \pi }{\partial w_c} =p\left( \alpha -\beta w_c\right) -\tau (t)-c_0+c_1 H(t)=0 \\&\frac{\partial \pi }{\partial w_{nc}} =p\left( \alpha -\beta w_{nc}\right) -c_0+c_1 H(t)-(\sigma +\tau (t))\phi =0 \end{aligned} \end{aligned}$$

Solving, we get:

$$\begin{aligned} \begin{aligned}&{\widetilde{w}}_c= \frac{p\alpha -c_0+c_1H(t)-\tau (t)}{p\beta } \\&{\widetilde{w}}_{nc}= \frac{p\alpha -c_0+c_1H(t)-(\sigma +\tau (t))\phi }{p\beta } \end{aligned} \end{aligned}$$

Notice that ${\widetilde{w}}_c>0$ if and only if:

$$\begin{aligned} \tau (t)<{\widetilde{\tau }}_c:=p\alpha -c_0+c_1H(t) \end{aligned}$$

Moreover, ${\widetilde{\tau }}_c>0$ if and only if:

$$\begin{aligned} H(t)>{\widetilde{H}}_c:=\frac{c_0-p\alpha }{c_1} \end{aligned}$$

Analogously, ${\widetilde{w}}_{nc}>0$ if and only if:

$$\begin{aligned} \tau (t)<{\widetilde{\tau }}_{nc}:=\frac{p\alpha -c_0+c_1H(t)}{\phi }-\sigma \end{aligned}$$

Notice that ${\widetilde{\tau }}_{nc}>0$ if and only if:

$$\begin{aligned} H(t)>{\widetilde{H}}_{nc}:=\frac{c_0-p\alpha +\sigma \phi }{c_1} \end{aligned}$$

This concludes the proof. $\square$

1.2 HJB equation solution

Assuming an interior solution and differentiating the right-side of equation (13) with respect to $\tau$, we lead:

$$\begin{aligned} \tau (t)=\frac{(1-\gamma )[(1-x)\phi +x]V'(H,t)}{x S_a} \end{aligned}$$

(17)

Replacing $\tau$ given by (17) in ${\widetilde{w}}_{nc}$ and ${\widetilde{w}}_c$ given by (10) and (11), we obtain:

$$\begin{aligned} w_{nc}=\frac{p\alpha -c_0+c_1H(t)-\sigma \phi }{p\beta }-\frac{\phi (1-\gamma )[(1-x)\phi +x]V'(H,t)}{px\beta S_a} \end{aligned}$$

(18)

and

$$\begin{aligned} w_c=\frac{p\alpha -c_0+c_1H(t)}{p\beta }-\frac{(1-\gamma )[(1-x)\phi +x]V'(H,t)}{px\beta S_a} \end{aligned}$$

(19)

Sustituting (18) and (19) in HJB, and rearranging the terms, it follows:

$$\begin{aligned} \begin{aligned}&rV(H,t)=\frac{N(1-\gamma )^2[(1-x)\phi +x]^2}{2p\beta x S_a^2}\cdot (V'(H,t))^2+ \\&\quad +\frac{(1-\gamma )N[(1-x)\sigma \phi -p\alpha +c_0-c_1H(t)]+\beta p[R+\Omega -\delta H(t)]}{p\beta S_a}\\&\quad \cdot V'(H,t) \\&\quad +\frac{[\alpha p-c_0+c_1H(t)]^2xN}{2 p \beta }-d_0+d_1H(t) \end{aligned} \end{aligned}$$

(20)

Being the game a linear-quadratic variety, we postulate a quadratic function of the form:

$$\begin{aligned} V(H,t)=AH^2(t)+BH(t)+C \end{aligned}$$

with the first derivative:

$$\begin{aligned} V'(H,t)=2AH(t)+B \end{aligned}$$

where A, B and C are constant parameters of the unknown value function which are to be determined. Substituting the equations V(H, t) and $V'(H,t)$ in the HJB, we obtain a system of three Riccati equations for the coefficients of the value function:

$$\begin{aligned}{} & {} rA=\frac{2N(1-\gamma )^2[(1-x)\phi +x]^2}{px\beta S_a^2}A^2 - \frac{2[(1-\gamma )c_1N+\delta \beta p]}{p\beta S_a}A+\frac{xNc_1^2}{2\beta p} \end{aligned}$$

(21)

$$\begin{aligned}{} & {} rB=\frac{2N(1-\gamma )^2[(1-x)\phi +x]^2}{p x\beta S_a^2}AB \nonumber \\{} & {} \quad + \frac{2\left\{ (1-\gamma )[(1-x)\phi \sigma -p\alpha +c_0]N+\beta p (R+\Omega )\right\} }{p \beta S_a}A \nonumber \\{} & {} \quad -\frac{(1-\gamma )c_1N+\delta \beta p}{p\beta S_a}B+\frac{(\alpha p-c_0)c_1xN}{\beta }+d_1 \end{aligned}$$

(22)

$$\begin{aligned}{} & {} \quad rC= \frac{N(1-\gamma )^2[\phi (1-x)+x]^2 }{2 p \beta x S_a^2}B^2+ \frac{(1-\gamma )[(1-x)\phi \sigma -p\alpha +c_0]N+\beta p(R+\Omega ) }{p\beta S_a}B \nonumber \\{} & {} \quad +\frac{(\alpha p-c_0)^2xN}{p\beta S_a}-d_0 \end{aligned}$$

(23)

Equation (21) admits two real and distinct solutions $A_1$ and $A_2$:

$$\begin{aligned} A_{1,2}=\frac{x S_a \left\{ \beta p(r S_a+2\delta )+2(1-\gamma )c_1N\pm \sqrt{D}\right\} }{4N(1-\gamma )^2[(1-x)\phi +x]^2} \end{aligned}$$

where

$$\begin{aligned}{} & {} D=4\left\{ N(1-\gamma )c_1[(1-x)\phi +1+x]+\frac{\beta p(r S_a+2\delta )}{2} \right\} \cdot \\{} & {} \quad \left\{ N(1-\gamma )(1-x)(1-\phi )c_1 + \frac{\beta p(r S_a+2\delta )}{2} \right\} \end{aligned}$$

is always positive. The solution has to satisfy the stability condition $\frac{d\dot{H}}{dH}<0$. Substituting (18) and (19) in the dynamics of the water table (4), and considering that $V'(H,t)=2AH(t)+B$, the stability condition becomes:

$$\begin{aligned} \frac{d\dot{H}}{dH}<0 \iff \frac{1}{S_a}\left\{ \frac{N(1-\gamma )[2(1-\gamma )((1-x)\phi +x)^2 A-c_1 S_a x]}{px\beta S_a}-\delta \right\} <0 \end{aligned}$$

satisfied for $A=A_2$. Moreover, from equation (22), we determine the value of $B=B_2$ as:

$$\begin{aligned} B_2=-\frac{x S_a\left\{ [2(1-\gamma )[(1-x)\phi \sigma -p\alpha +c_0]A_2 +x S_a c_1(\alpha p-c_0)]N+p\beta [2(R+\Omega )A_2+d_1 S_a] \right\} }{N(1-\gamma )\{ 2[(1-x)\phi +x]^2(1-\gamma )A_2-xc_1 S_a\} -p\beta x S_a (r S_a + \delta ) } \end{aligned}$$

Finally,

$$\begin{aligned} w_c^*=\frac{\alpha p-c_0+c_1H(t)}{\beta p}-\frac{(1-\gamma )[(1-x)\phi +x](2A_2H(t)+B_2)}{x\beta S_a p} \end{aligned}$$

(24)

$$\begin{aligned} w_{nc}^*=\frac{\alpha p-c_0+c_1H(t)-\sigma \phi }{\beta p}-\frac{\phi (1-\gamma )[(1-x)\phi +x](2A_2H(t)+B_2)}{x\beta S_a p} \end{aligned}$$

(25)

and

$$\begin{aligned} \tau ^*=\frac{(1-\gamma )[(1-x)\phi +x]\left( B_2+2A_2H(t)\right) }{x S_a} \end{aligned}$$

(26)

1.3 Proof of Proposition 3

Substituting the values of $w_c^*$ and $w_{nc}^*$, given by (24) and (24), respectively, in the water table dynamics (4) we get:

$$\begin{aligned} \dot{H}={\widehat{Y}} H + Y \end{aligned}$$

(27)

Solving (27) we obtain the optimal trajectory (15). $\square$