Optimal Allocation of Groundwater Resources: Managing Water Quantity and Quality

Abstract

Despite the importance of groundwater in the economy of the Hai River Basin (HRB), falling water tables and salinization of aquifers are both occurring in the region. Hydrological and hydrogeological studies have shown that increases in the salinization of parts of the freshwater aquifers are closely related to the extraction of groundwater. This study uses a framework that considers the interaction between water quantity and quality to examine how the presence of the prehistoric saline water layer affects groundwater management. Simulation results show that in a region where there is a salinization problem like in the HRB, it is optimal to pump at high rates in the early stage of extraction when the quality of groundwater is high. It is then optimal to reduce the pumping rate rapidly as the quality of groundwater deteriorates. Given this characteristic of the optimal pumping path, the heavy extraction currently observed in the HRB does not necessarily indicate that groundwater resources are being overused. However, unregulated extraction by non-cooperative users would eventually cause both the depletion of the water resource and the deterioration of water quality. Hence, joint quantity–quality management is required in the HRB. The study also shows that benefits to groundwater management are higher and costs are lower in regions with salinization problems.

Appendices

Appendix 1. Derivative of the Equation of Motion for the Groundwater Salinity Level (E_t)

In this study, we simplify our analysis by only focusing on the case when the saline water moves into the deep aquifer. In Fig. 6, the extraction of deep aquifer water, Mw_t, leads to an increase in the depth-to-water in the deep aquifer, h_t. The hydraulic head in the deep aquifer, H₁, keeps declining as a result. When the pressure difference between the head in the saline water layer, H₂, and that in the deep aquifer is large enough, the saline water can move into the deep aquifer through the aquitard. The movement of saline water in response to the change in the head difference, Q₁₂, accounts for the phenomenon of increasing salinity level in the deep aquifer.

Fig. 6

The process of saline water intrusion due to groundwater extraction

In the language of hydrology, Darcy’s Law can be employed to formalize the movement of saline water.⁶ Suppose the hydraulic head of the deep aquifer is linear in the depth-to-water: $H_{1t} = - c_{1} h_{t} + d_{1}$ and $H_{2t} = c_{2} \frac{{Q_{t} }}{As} + d_{2}$, where Q_t is the stock of the saline water, A is the area of saline water layer, and s is the specific yield, the volume of saline water that moves into the deep aquifer at time t can be expressed as

$$Q_{t} - Q_{t + 1} = Q_{12} = - KA\frac{{H_{1t} - H_{2t} }}{b} = - \frac{KA}{b}\left[ {( - c_{1} h_{t} + d_{1} ) - \left( {c_{2} \frac{{Q_{t} }}{As} + d_{2} } \right)} \right]$$

(7)

where K is the hydraulic conductivity of the aquitard (unit: volume per unit of time), and b is the thickness of the aquitard. From (7), we have Q_t − Q_t+1 ∝ h_t+1 − h_t.⁷ We assume that the change in the level of salinity, E_t+1 − E_t, is proportional to the total amount of intruded saline water at time t, Q₁₂. Hence, we have E_t+1 − E_t ∝ h_t+1 − h_t. Suppose E_t+1 − E_t= δ(h_t+1 − h_t), we obtain the equation of motion for E_t:

E_t+1 = E_t + (h_t+1 − h_t)

Appendix 2. Derivation of the Euler Equation for the Cooperative Extraction Model

We rewrite (2) as

$$\begin{aligned} \mathop {\text{Max}}\limits_{{\{ w[t]\} }} L &= M\sum\limits_{t = 0}^{\infty } {\beta^{t} f(w_{t} ,E_{t} ,h_{t} )} \\ & \quad - \sum\limits_{t = 0}^{\infty } {\beta^{t + 1} \lambda_{t + 1} [h_{t + 1} - (h_{t} + \phi_{1} Mw_{t} - \phi_{2} R)} ] \\ & \quad - \sum\limits_{t = 0}^{\infty } {\beta^{t + 1} \mu_{t + 1} [} E_{t + 1} - (E_{t} + \delta (\phi_{1} Mw_{t} - \phi_{2} R)]\\ \end{aligned}$$

The first-order condition for this problem gives:

$$\frac{\partial L}{{\partial w_{t} }} = f_{{w_{t} }} (w_{t} ,E_{t} ,h_{t} ) + \beta \phi_{1} (\lambda_{t + 1} + \delta \mu_{t + 1} ) = 0$$

(8)

$$\frac{\partial L}{{\partial h_{t} }} = Mf_{{h_{t} }} (w_{t} ,E_{t} ,h_{t} ) + \beta \lambda_{t + 1} - \lambda_{t} = 0$$

(9)

$$\frac{\partial L}{{\partial E_{t} }} = Mf_{{E_{t} }} (w_{t} ,E_{t} ,h_{t} ) + \beta \mu_{t + 1} - \mu_{t} = 0$$

(10)

$$= > \lambda_{t + 1} + \delta \mu_{t + 1} = - f_{{w_{t} }} (w_{t} ,E_{t} ,h_{t} )/(\beta \phi_{1} )$$

(8)

Lagging (8) by one period gives:

$$\lambda_{t} + \delta \mu_{t} = - f_{{w_{t - 1} }} (w_{t - 1} ,E_{t - 1} ,h_{t - 1} )/(\beta \phi_{1} )$$

(11)

(9) +δ^∗ (10)⇒

$$\beta (\lambda_{t + 1} + \delta \mu_{t + 1} ) - (\lambda_{t} + \delta \mu_{t} ) = - M[f_{{h_{t} }} (w_{t} ,E_{t} ,h_{t} ) + \delta f_{{E_{t} }} (w_{t} ,E_{t} ,h_{t} )]$$

(12)

Plugging (8) and (11) into (12) gives:

$$\begin{aligned} f_{{w_{{t - 1}} }} (w_{{t - 1}} ,E_{{t - 1}} ,h_{{t - 1}} ) & = \beta f_{{w_{t} }} (w_{t} ,E_{t} ,h_{t} ) - \beta \phi _{1} M[f_{{h_{t} }} (w_{t} ,E_{t} ,h_{t} ) & \\ & \quad + \delta f_{{E_{t} }} (w_{t} ,E_{t} ,h_{t} )] \\ \end{aligned}$$

(11)

Rolling equation (11) forward one period gives:

$$f_{{w_{t} }} \begin{array}{*{20}c} { = \beta f_{{w_{t + 1} }} - \beta \phi_{1} M(f_{{h_{t + 1} }} + \delta f_{{E_{t + 1} }} )} & {} \\ \end{array}$$

$$\begin{aligned} & = \beta[\beta f_{{w_{t + 2} }} - \beta \phi_{1} M\left( {f_{{h_{t + 1} }} + \delta f_{{E_{t + 1} }} } \right)] - \beta \phi_{1} M\left( {f_{{h_{t + 2} }} + \delta f_{{E_{t + 2} }} } \right) \\ & = \cdot \cdot \cdot \\ & = \beta^{s} f_{{w_{t + s} }} - \sum\limits_{\ell = 1}^{s} {\beta^{\ell } \phi_{1} M\left( {f_{{h_{t + \ell } }} + \delta f_{{E_{t + \ell } }} } \right)} \\ \end{aligned}$$

(12)

Leading it forward into infinite future, we obtain

$$f_{{w_{t} }} = \mathop {\lim }\limits_{s \to \infty } \left\{ {\beta^{s} f_{{w_{t + s} }} - \sum\limits_{\ell = 1}^{s} {\beta^{\ell } \phi_{1} M(f_{{h_{t + \ell } }} + \delta f_{{E_{t + \ell } }} )} } \right\}$$

(13)

Since β is the discount factor that is well within the range of 0 and 1, (13) gives

$$f_{{w_{t} }} = - M\sum\limits_{\ell = 1}^{\infty } {\beta^{\ell } \phi_{1} f_{{h_{t + \ell } }} } - M\sum\limits_{\ell = 1}^{\infty } {\beta^{\ell } \phi_{1} \delta f_{{E_{t + \ell } }} }$$

(14)

Appendix 3. Derivation of Proposition 1

At the steady state, w_t-1= w_t= w∗. We use w^*, E^*, and h^*to denote the value at the steady state. Equation (14) now becomes:

$$(1 - \beta )f_{w} (w^{*} ,E^{*} ,h^{*} ) + \beta \phi_{1} f_{h} (w^{*} ,E^{*} ,h^{*} ) + \beta \phi_{1} \delta f_{E} (w^{*} ,E^{*} ,h^{*} ) = 0$$

Using the implicit function theorem gives:

$$\frac{{\partial h^{{^{*} }} }}{\partial \delta } = - \frac{{\beta \phi_{1} f_{E} (w^{*} ,E^{*} ,h^{*} )}}{{(1 - \beta )f_{wh} (w^{*} ,E^{*} ,h^{*} ) + \beta \phi_{1} f_{hh} (w^{*} ,E^{*} ,h^{*} )}}$$

We have f_E < 0, 1− > 0, f_wh< 0 and f_hh< 0. Therefore,

$$\frac{{\partial h^{*} }}{\partial \delta } = - \frac{( - )}{( + )( - ) + ( - )} < 0$$

Appendix 4. Derivation of Proposition 2

Bellman equation for problem s(2) is:

$$\begin{aligned} &V(h,E) = \mathop {\text{Max}}\limits_{w} \left\{ {f(w,h,E) + \beta V(h^{\prime},E^{\prime})} \right\} \\ &\quad {\text{s}} . {\text{t}} .\quad h^{\prime} = h + \phi_{1} Mw - \phi_{2} R \\& \quad \quad \quad E^{\prime} = E + \delta (h^{\prime} - h) \\ \end{aligned}$$

(15)

Using the Envelope Theorem gives:

$V_{\delta } = \beta V_{{E^{\prime}}} (h^{\prime},\;E^{\prime}) \cdot (h^{\prime} - h)$ ⇒ $V_{\delta } < 0$ since $V_{{E^{\prime}}} < 0$ and $h^{\prime} > h$.

Appendix 5. Derivation of the Euler Equation for Non-cooperative Extraction Model

we rewrite (4) as

$$\begin{aligned} \mathop {{\text{Max}}}\limits_{{\{ w_{{it}} \} }} L & = \sum\limits_{{t = 0}}^{\infty } {\beta ^{t} f^{i} (w_{{it}} ,E_{t} ,h_{t} )} - \sum\limits_{{t = 0}}^{\infty } {\beta ^{{t + 1}} \lambda _{{t + 1}} } (h_{{t + 1}} - h_{t} - \phi _{1} w_{{it}} - \phi _{1} \sum\limits_{{j \ne i}}^{M} {w_{j}^{*} } + \phi _{2} R) \\ & \quad - \sum\limits_{{t = 0}}^{\infty } {\beta ^{{t + 1}} } \mu _{{t + 1}} (E_{{t + 1}} - E_{t} - \phi _{1} \delta w_{{it}} & \\ & \quad - \phi _{1} \delta \sum\limits_{{j \ne i}}^{M} {w_{j}^{*} } + \phi _{2} \delta R) \\ \end{aligned}$$

The first-order condition for this problem gives:

$$\frac{\partial L}{{\partial w_{t} }} = f_{{w_{t} }}^{i} (w_{t} ,E_{t} ,h_{t} ) + \beta \phi (\lambda_{t + 1} + \delta \mu_{t + 1} ) = 0$$

(16)

$$\frac{\partial L}{{\partial h_{t} }} = f_{{h_{t} }} (w_{t} ,E_{t} ,h_{t} ) + \beta \lambda_{t + 1} - \lambda_{t} + \phi_{1} \beta (\lambda_{t + 1} + \delta \mu_{t + 1} )\sum\limits_{j \ne i}^{M} {\frac{{\partial w_{j}^{*} }}{{\partial h_{t} }}} = 0$$

(17)

$$\frac{\partial L}{{\partial E_{t} }} = f_{{E_{t} }} (w_{t} ,E_{t} ,h_{t} ) + \beta \mu_{t + 1} - \mu_{t} + \phi_{1} \beta (\lambda_{t + 1} + \delta \mu_{t + 1} )\sum\limits_{j \ne i}^{M} {\frac{{\partial w_{j}^{*} }}{{\partial E_{t} }}} = 0$$

(18)

Using the same manipulations as in the steps to obtain the Euler equation for the cooperative model, we obtain:

$$\begin{aligned} f_{{w_{it - 1} }} (w_{it - 1} ,E_{t - 1} ,h_{t - 1} ) &= \beta f_{{w_{it} }} (w_{it} ,E_{t} ,h_{t} )\left\{ {1 + \sum\limits_{j \ne i}^{M} {\phi_{1} \frac{{\partial w_{j}^{*} }}{{\partial h_{t} }} + \sum\limits_{j \ne i}^{M} {\phi_{1} \delta \frac{{\partial w_{j}^{*} }}{{\partial E_{t} }}} } } \right\} \\ & \quad- \beta \phi_{1} f_{{h_{t} }} (w_{it} ,E_{t} ,h_{t} ) - \beta \phi_{1} \delta f_{{E_{t} }} (w_{it} ,E_{t} ,h_{t} ) \\ \end{aligned}$$

(19)

Rolling equation (19) forward one period and continuing to substitute for terms in t + 1 gives:

$$\begin{aligned} f_{{w_{{it}} }} = & \;\beta ^{s} f_{{w_{{it}} }} + \sum\limits_{{\ell = 1}}^{s} {\beta ^{\ell } \sum\limits_{{j \ne i}}^{M} {\phi _{1} \frac{{\partial w_{j}^{*} }}{{\partial h_{{t + \ell }} }}} } + \sum\limits_{{\ell = 1}}^{s} {\beta ^{\ell } \sum\limits_{{j \ne i}}^{M} {\phi _{1} \delta \frac{{\partial w_{j}^{*} }}{{\partial E_{{t + \ell }} }}} } & \\ & - \sum\limits_{{\ell = 1}}^{s} {\beta ^{\ell } \phi _{1} f_{{h_{{t + \ell }} }} } - \sum\limits_{{\ell = 1}}^{s} {\beta ^{\ell } \phi _{1} \delta f_{{E_{{t + \ell }} }} } \\ \end{aligned}$$

(20)

which as s goes to infinity becomes

$$f_{{w_{it} }} = \sum\limits_{\ell = 1}^{s} {\beta^{\ell } \sum\limits_{j \ne i}^{M} {\phi_{1} \frac{{\partial w_{j}^{*} }}{{\partial h_{t + \ell } }}} } + \sum\limits_{\ell = 1}^{s} {\beta^{\ell } \sum\limits_{j \ne i}^{M} {\phi_{1} \delta \frac{{\partial w_{j}^{*} }}{{\partial E_{t + \ell } }}} } - \sum\limits_{\ell = 1}^{s} {\beta^{\ell } \phi_{1} f_{{h_{t + \ell } }} } - \sum\limits_{\ell = 1}^{s} {\beta^{\ell } \phi_{1} \delta f_{{E_{t + \ell } }} }$$

(21)

Appendix 6. Derivation of Proposition 2: Over-Extraction Under Non-cooperative Extraction

Applying the implicit function theorem to (16) gives

$\frac{{\partial w_{t}^{*} }}{{\partial h_{t} }} = - \frac{{f_{wh} }}{{f_{ww} }} = - \frac{( - )}{( - )} < 0$ and $\frac{{\partial w_{t}^{*} }}{{\partial E_{t} }} = - \frac{{f_{wE} }}{{f_{ww} }} = - \frac{( - )}{( - )} < 0$.

Therefore the term, $\phi_{1} \sum\nolimits_{j \ne i}^{M} {\left( {\frac{{\partial w_{j}^{*} }}{{\partial h_{t} }} + \frac{{\partial w_{j}^{*} }}{{\partial E_{t} }}} \right)}$, is negative.

Given the same depth-to-water and the same salinity level, the right-hand side of (5) is lower than that of (3). Consequently, w_it-1 > w_t-1 since f_ww < 0.