On the Environmental Efficiency of Water Storage: the Case of a Conjunctive use of Ground and Rainwater

Abstract

Rainwater harvesting, consisting in collecting runoff from precipitation, has been widely developed to stop groundwater declines and even raise water tables. However, this expected environmental effect is not self-evident. We show in a simple setting that the success of this conjunctive use depends on whether the runoff rate is above a threshold value. Moreover, the bigger the storage capacity, the higher the runoff rate must be to obtain an environmentally efficient system. We also extend the model to include other hydrological parameters and ecological damages, which respectively increase and decrease the environmental efficiency of rainwater harvesting.

Appendices

Appendix

1.1 Appendix A: The Solution to the General Problem

Let us introduce some notations used below:

$$\begin{array}{lll} \rho\in[\varepsilon,1-\beta(1-\varepsilon)] & \beta\in(0,1) &{}\varepsilon\in(0,1)\\ R_{\rho}\equiv(1-\rho)R & {} a\equiv1-\beta>0 &{} b\equiv\!1-\!\rho-\!\beta(1-\varepsilon)>0\\ K_{\varepsilon}\equiv\frac{K}{1-\varepsilon} & {\kern-1.7pt} w(t)\equiv w_{g}(t)& {}{\Gamma}\equiv G^{\prime}\left( 0\right)\\&{\kern25pt}+(1-\varepsilon)w_{r}(t) \\ \mu_{r}(t)\equiv\frac{\mu_{r}^{\prime}(t)}{1-\varepsilon}\geq0 & {\kern-1.5pt} \mu_{W}(t)\equiv\frac{\mu_{W}^{\prime}(t)}{1-\varepsilon}\geq0 & \end{array} $$

1.1.1 A.1 The Problem

Let us consider the following optimal control problem:

$$ \begin{array}{l} \underset{w_{g}(t),w_{r}(t)}{\max}\!{\int}_{0}^{+\infty}exp^{-\delta t}\!\left[{\int}_{0}^{w(t)}\!P(x)dx\,-\,C(h(t)) w_{g}(t)\right.\\\hspace*{7.5pc}\left.-Kw_{r}(t)\,-\,G\left( -\dot{h}(t)\right)\right] dt\\ \!\qquad\text{w.r.t }\!\!\left\{\! \begin{array}{l} (sA)\dot{h}(t)\,=\,(1-\rho)\left( R\,-\,w_{r}(t)\right) \,-\,w_{g}(t)\\ \hspace*{4pc}+\beta\left( w_{g}(t)\,+\,(1\,-\,\varepsilon)w_{r}(t)\right)\\ w_{g}(t)\geq0, 0\leq w_{r}(t)\!\leq\! W\\ h_{0}\text{ given and }h(\infty)\text{ free}. \end{array} \right. \end{array} $$

(A.1)

The current value of the Hamiltonian for this problem is:

$$\begin{array}{@{}rcl@{}} \mathcal{H}&=&{\int}_{0}^{w(t)}P(x)dx-C(h(t))w_{g}(t)-Kw_{r}(t)\\ &&-G\left( -\dot {h}(t)\right) +\lambda(t)\frac{1}{sA}\left[ R_{\rho}-aw_{g}(t)-bw_{r} (t)\right] \end{array} $$

where λ(t)≥0 denotes the shadow value of the groundwater resource. Moreover, if we introduce the Lagrange multipliers μ _g (t) and $\mu _{r}^{\prime }(t)$ associated with the non-negativity constraints on w _g (t) and w _r (t), respectively, and the implicit price $\mu _{W}^{\prime }(t)$ of W, the capacity constraint, the Lagrange function associated with this program is:

$$ \mathcal{L}\,=\,\mathcal{H}+\mu_{g}(t)\cdot w_{g}(t)+\mu_{r}^{\prime}(t)\cdot w_{r}(t)+\mu_{W}^{\prime}(t)\cdot\left( W-w_{r}(t)\right). $$

(A.2)

The Hamiltonian conditions are given by systems (CTR1) and (DYN1):

$$(CTR1)\left\{ \begin{array}{ll} P\left( w(t)\right) -C\left( h(t)\right) \\\hspace*{3.13pc}-\frac{a}{sA}\left( \lambda(t)+G^{\prime}\left( -\dot{h}(t)\right) \right) \\\hspace*{3.13pc}+\mu_{g}(t)=0, & (GW_{1})\\ P\left( w(t)\right) -K_{\varepsilon}\\\hspace*{3.13pc}-\frac{b}{\left( 1-\varepsilon\right) sA}\left( \lambda(t)+G^{\prime}\left( -\dot{h}(t)\right) \right) \\\hspace*{3.13pc}+\mu_{r}(t)-\mu_{W}(t)=0, & (RW_{1})\\ \mu_{g}(t)w_{g}(t)=0,\\\quad\mu_{r}(t)w_{r}(t)=0,\\\quad\mu_{W}(t)\left( W-w_{r}(t)\right) =0, & (S1_{1})\\ \mu_{g}(t),\mu_{r}(t),\mu_{W}(t)\geq0, \\\quad w_{g}(t)\geq0,\text{ and }w_{r}(t)\in\left[ 0,W\right], & (S2_{1}) \end{array} \right. $$

and

$$(DYN1)\quad\left\{ \begin{array}{l} \left( sA\right) \dot{h}(t)=R_{\rho}-aw_{g}(t)-bw_{r}(t),\\ \dot{\lambda}(t)=\delta\lambda(t)+C^{\prime}\left( h(t)\right) w_{g}(t). \end{array} \right. $$

1.1.2 A.2 The Construction of the Steady State

Let us consider a steady state satisfying (CTR1) and (DYN1) with $\dot {h}(t)=\dot {\lambda }(t)=0$. There are eight possible cases, since each of the three Lagrange multipliers may or may not be strictly positive. We nevertheless immediately observe that:

Remark 1

At a steady state: (i) both constrains on w _g cannot be simultaneously binding since W > 0, (ii) there must be some GW consumption (i.e., μ _g = 0) otherwise $\dot {h}(t)=0$ implies that $w_{r}=\frac {R_{\rho }}{b}=\frac {\left (1-\rho \right ) }{\left (1-\rho \right ) -\beta (1-\varepsilon )}R>R$, which is impossible since w _r ≤ W < R.

Now let us observe:

Lemma 1

If the capacity constraint on RW harvesting is not binding (i.e., μ _W =0) at the steady state, then K _ε >C(h).

Proof Lemma 1

Since $\mu _{g}=\mu _{W}=\dot {h}=0,$ the difference between Eqs. (G W ₁) and (R W ₁) is:

$$ K_{\varepsilon}-C\left( h\right) +\frac{1}{sA}\left( \frac{b} {1-\varepsilon}-a\right) \left[ \lambda+G^{\prime}\left( 0\right) \right] -\mu_{r}=0. $$

(A.3)

Using (i) μ _r ≥ 0 and (ii) $\lambda =-\frac {1}{\delta }C^{\prime }\left (h\right ) w_{g}$ since $\dot {\lambda }=0$, it follows that:

$$ K_{\varepsilon}-C\left( h\right) \geq\frac{1}{sA}\left( \frac {b}{1-\varepsilon}-a\right) \left( \frac{1}{\delta}C^{\prime}\left( h\right) w_{g}-G^{\prime}\left( 0\right) \right). $$

(A.4)

(iii) Notice that $\frac {b}{1-\varepsilon }-a=-\frac {\rho -\varepsilon }{1-\varepsilon }<0$ since ρ > ε, and (iv) by assumption $C^{\prime }<0,G^{\prime }>0.$ The lhs of Eq. A.4 is therefore positive and we can conclude that K _ε >C(h). □

From Lemma 1, we assert :

Lemma 2

If we assume that

$$ P\left( \frac{R}{a}\right) >K_{\varepsilon}-\frac{1}{\delta sA}C^{\prime }\left( C^{-1}\left( K_{\varepsilon}\right) \right) \left( R-aW\right) +\frac{a}{sA}G^{\prime}\left( 0\right) $$

(A.5)

the RW storage capacity is always binding at a steady state.

Proof

Assume the contrary, w _r < W, so that μ _W = 0. Since $\lambda =-\frac {1}{\delta }C^{\prime }\left (h\right ) w_{g}$, (R W ₁) at the steady state can be written as:

$$ \mu_{r}=-P\left( w\right) +K_{\varepsilon}-\frac{b}{\left( 1-\varepsilon \right) sA}\left[ \frac{1}{\delta}C^{\prime}\left( h\right) w_{g}-G^{\prime}\left( 0\right) \right]. $$

(A.6)

Using Lemma 1, i.e., $K_{\varepsilon }>C(h)\Leftrightarrow C^{-1}\left (K_{\varepsilon }\right ) <h$, and since $C^{\prime \prime }\geq 0$, we deduce that:

$$\begin{array}{@{}rcl@{}} \mu_{r}&\leq&-P\left( w\right) +K_{\varepsilon}-\frac{b}{\delta (1-\varepsilon)sA}C^{\prime}\left( C^{-1}\left( K_{\varepsilon}\right) \right) w_{g}\\ &&+\frac{b}{\left( 1-\varepsilon\right) sA}G^{\prime}\left( 0\right). \end{array} $$

(A.7)

Moreover, from system (DYN1) at steady state, we know that:

• (i) $R_{\rho }-aw_{g}-bw_{r}=0\Leftrightarrow aw_{g}=R_{\rho } -bw_{r}<R_{\rho }-bW$ since we have assumed that w _r < W. Moreover, b = 1−ρ−β(1−ε) and 1>ρ > ε, we can therefore say:

  $$\begin{array}{@{}rcl@{}} aw_{g}&<&\left( 1-\rho\right) \left( R-W\right) +\beta(1-\varepsilon )W<R\\ &&-(1-\beta)W=R-aW. \end{array} $$

  (A.8)

• (ii) The total consumption of water is $w=\frac {R_{\rho }-bw_{r}} {a}+(1-\varepsilon )w_{r}=\frac {1}{a}\left (R_{\rho }-\left (b-a(1-\varepsilon )\right ) w_{r}\right )$. Since b−a(1 − ε) = ε−ρ, we have $w=\frac {1}{a}\left [ R_{\rho }+\left (\rho -\varepsilon \right ) w_{r}\right ]$. If we now remember that w _r < W < R and ε < ρ < 1, we observe that:

  $$ w<\frac{1}{a}\left( (1-\rho)+\left( \rho-\varepsilon\right) \right) R<\frac{R}{1-\beta}. $$

  (A.9)

Under assumptions $P^{\prime }<0$ and $C^{\prime }<0$ and items (i) and (ii), we can write condition (A.7) as:

$$\begin{array}{@{}rcl@{}} \mu_{r}&<&-P\left( \frac{R}{1-\beta}\right) +K_{\varepsilon}\\ &&-\frac {b}{a\delta\left( 1-\varepsilon\right) sA}C^{\prime}\left( C^{-1}\left( K_{\varepsilon}\right) \right) \left( R-aW\right) \\ &&+\frac{b}{\left( 1-\varepsilon\right) sA}G^{\prime}\left( 0\right). \end{array} $$

(A.10)

Finally, since $0<\frac {b}{\left (1-\varepsilon \right ) a}<1$ (see (iii) of the Proof of Lemma 1) $C^{\prime }<0$ and $G^{\prime }\left (0\right ) >0$, we can state that:

$$\begin{array}{@{}rcl@{}} \mu_{r}&<&-P\left( \frac{R}{a}\right) +K_{\varepsilon}-\frac{1}{\delta sA}C^{\prime}\left( C^{-1}\left( K_{\varepsilon}\right) \right) \left( R-aW\right) \\ &&+\frac{a}{sA}G^{\prime}\left( 0\right). \end{array} $$

(A.11)

Given the assumption introduced in Eq. A.5, this implies that μ _r < 0 which is a contradiction. □

If we now combine Remark 1 and Lemmas 1 and 2, we can say that a steady state, if it exists, verifies:

$$ \left\{ \begin{array}{ll} P\left( w\right) \,-\,C\left( h\right) \,-\,\frac{a}{sA}\left( \lambda \,+\,G^{\prime}\left( 0\right) \right) \,=\,0, & w_{g}\,=\,\frac{1}{a}\left( R_{\rho }\,-\,bW\right), \\ \mu_{W}\,=\,P\left( w\right) \,-\,K_{\varepsilon}\,-\,\frac{b}{\left( 1-\varepsilon \right) sA}\left( \lambda\,+\,G^{\prime}\left( 0\right) \right) , & \lambda\,=\,-\frac{1}{\delta a}C^{\prime}\left( h\right) \left( R_{\rho }\,-\,bW\right), \\ \mu_{W}\!\geq\!0\text{ and }w_{r}\,=\,W, & w\,=\,\frac{1}{a}\left( R_{\rho} \,+\,(\rho\,-\,\varepsilon)W\right). \end{array} \right. $$

(A.12)

In other words, a unique steady state exists, if there exists a unique solution $h^{\ast }\in \left [ 0,\bar {h}\right ]$ to

$$\begin{array}{@{}rcl@{}} \phi_{1}(h^{\ast})&=&P\left( \frac{1}{a}\left( R_{\rho}+(\rho-\varepsilon )W\right) \right) -C\left( h^{\ast}\right)\\ &&+\frac{1}{\delta sA}C^{\prime }\left( h^{\ast}\right) \left( R_{\rho}\,-\,bW\right) \,-\,\frac{a}{sA}G^{\prime }\left( 0\right)\! =\!0 \end{array} $$

(A.13)

which verifies:

$$\begin{array}{@{}rcl@{}} \mu_{W}&=&P\left( \frac{1}{a}\left( R_{\rho}+(\rho-\varepsilon)W\right) \right)\\ &&-K_{\varepsilon}+\frac{b}{a\delta\left( 1-\varepsilon\right) sA}C^{\prime}\left( h^{\ast}\right) \left( R_{\rho}-bW\right)\\ &&-\frac {b}{\left( 1-\varepsilon\right) sA}G^{\prime}\left( 0\right) \geq0. \end{array} $$

(A.14)

So, let us first observe:

Lemma 3

If condition (A.5) is satisfied, any solution (it is exists) to ϕ ₁ (h ^∗ )=0 verifies μ _W >0.

Proof

Using Eq. A.12 and W < R, we get $w<\frac {1}{a}\left (R_{\rho }+\rho W\right ) <\frac {R}{a}$ and by Eq. A.8 we know that R _ρ −b W < R−a W. If we now keep in mind that $P^{\prime }<0$ and $C^{^{\prime }}<0$, we can deduce from condition (A.5) that:

$$\begin{array}{@{}rcl@{}} 0 & \!<\!&P\left( \frac{R}{a}\right) -K_{\varepsilon}+\frac{1}{\delta sA}C^{\prime}\left( C^{-1}\left( K_{\varepsilon}\right) \right) \left( R-aW\right) -\frac{a}{sA}G^{\prime}\left( 0\right)\\ &\!<\!&\underbrace{P(w)\,-\,K_{\varepsilon}\,+\,\frac{1}{\delta sA}C^{\prime}\left( C^{-1}\left( K_{\varepsilon}\right) \right) \left( R_{\rho}\,-\,bW\right) \,-\,\frac{a} {sA}G^{\prime}\left( 0\right)}_{\equiv\phi_{1}(C^{-1}\left( K_{\varepsilon }\right) )} . \end{array} $$

(A.15)

But ϕ ₁(h) is increasing in h since $C^{\prime }<0$ and C ” ≥ 0. We deduce that the steady state $h^{\ast }<C^{-1}\left (K_{\varepsilon }\right )$ or C(h ^∗)>K _ε . We can therefore say from Eq. (A.13) that:

$$\begin{array}{@{}rcl@{}} &&P\left( \frac{1}{a}\left( R_{\rho}\,+\,(\rho\,-\,\varepsilon)W\right) \right) \,-\,K_{\varepsilon}\!+\frac{1}{\delta sA}C^{\prime}\left( h^{\ast}\right) \left( R_{\rho}-bW\right)\\ &&-\frac{a}{sA}G^{\prime}\left( 0\right) >0. \end{array} $$

(A.16)

Since $\frac {b}{a\left (1-\varepsilon \right ) }\in \left (0,1\right )$, $C^{\prime }<0$ and $G^{\prime }\left (0\right ) >0$, we can even state:

$$\begin{array}{@{}rcl@{}} &&P\left( \frac{1}{a}\left( R_{\rho}+(\rho-\varepsilon)W\right) \right) \\&&-K_{\varepsilon}+\frac{b}{a\delta\left( 1-\varepsilon\right) sA}C^{\prime }\left( h^{\ast}\right) \left( R_{\rho}-bW\right) \\&&-\frac{b}{\left( 1-\varepsilon\right) sA}G^{\prime}\left( 0\right) >0 \end{array} $$

(A.17)

or, in other words, that μ _W >0. □

We can finally affirm that:

Proposition 10

Under condition (A.5), the path solution to optimization problem (A.1) admits a unique steady state given by Eq. A.12.

Proof

It simply remains to show that ϕ ₁(h) = 0 admits a unique solution $h^{\ast }\in \left [ 0,\bar {h}\right ]$. This is immediate since (i) $\phi _{1}^{\prime }>0$ because $C^{\prime }<0$ and C” ≥ 0, (ii) $\lim _{h\rightarrow 0}\phi _{1}(h)=-\infty$ because $\lim _{h\rightarrow 0}C(h)=\infty$ and (iii) $\lim _{h\rightarrow \bar {h}}\phi _{1}(h)>0$. However, this last point is less obvious. So let us first observe that $\lim _{h\rightarrow \bar {h}}\phi _{1}(h)=P\left (\frac {1} {a}\left (R_{\rho }+(\rho -\varepsilon )W\right ) \right ) -aG^{\prime }\left (0\right )$ since we have assumed that $C(\bar {h})=C^{\prime }(\bar {h})=0$. Now remark that $\frac {1}{a}\left (R_{\rho }+(\rho -\varepsilon )W\right ) <\frac {R}{1-\beta }$ since W < R and use condition (A.5) in order to obtain:

$$\begin{array}{@{}rcl@{}} \lim_{h\rightarrow0}\phi_{1}(\bar{h})\!&>&\!P\left( \frac{R}{a}\right) -\frac {a}{sA}G^{\prime}\left( 0\right)\\ &>&\!K_{\varepsilon}\,-\,\frac{1}{\delta sA}C^{\prime}\left( \!C^{-1}\left( K_{\varepsilon}\right)\! \right) \left( R\,-\,aW\right)\! >\!0 \end{array} $$

(A.18)

(remember $C^{\prime }<0$ and R−a W > 0). □

1.1.3 A.3 The Local Saddle Point Property

Let us return to the systems (CTR1) and (DYN1) and set μ _g = 0, and μ _r = 0. Since μ _W >0 at the steady state, we can choose a neighborhood around the steady state such that μ _W (t)>0 and therefore set w _r (t) = W. In this case, equation (G W ₁) becomes:

$$\begin{array}{@{}rcl@{}} &&{}\phi_{2}\left( w_{g}(t),h(t),\lambda(t)\right) =0, \end{array} $$

(A.19)

$$\begin{array}{@{}rcl@{}} &&{}P\left( w_{g}(t)+\left( 1-\varepsilon\right) W\right) -C\left( h(t)\right)\\ &&{}-\!\frac{a}{sA}\left( \lambda(t)\,+\,G^{\prime}\left( \frac{1} {sA}\left( \,-\,R_{\rho}\,+\,aw_{g}(t)\,+\,bW\right) \right) \right) \!\,=\,0. \end{array} $$

(A.20)

Moreover, $\partial _{w_{g}}\phi _{2}=P^{\prime }-\left (\frac {a}{sA}\right )^{2}G"<0$ by our assumptions. We can therefore apply the implicit function theorem around the steady state and build $w_{g}(t)=\phi _{3}\left (h(t),\lambda (t)\right ) >0$. We even know that:

$$ \partial_{h}\phi_{3}\,=\,\frac{C^{\prime}}{P^{\prime}-\left( \frac{a} {sA}\right)^{2}G\text{"}}\!>\!0\text{ and }\partial_{\lambda}\phi_{3}\,=\,\frac{\frac {a}{sA}}{P^{\prime}-\left( \frac{a}{sA}\right)^{2}G\text{"}}\!<\!0. $$

(A.21)

It follows that the dynamics of the system is locally given by:

$$ \left\{ \begin{array}{l} \dot{h}(t)=\frac{1}{sA}\left( R_{\rho}-a\phi_{3}\left( h(t),\lambda (t)\right) -bW\right), \\ \dot{\lambda}(t)=\delta\lambda(t)+C^{\prime}\left( h(t)\right) \phi_{3}\left( h(t),\lambda(t)\right). \end{array} \right. $$

(A.22)

It remains to verify that the Jacobian J of this two dimensional system has a positive trace T r(J) and a negative determinant $\det (J)$. By computation, we obtain:

$$ J=\left[ \begin{array}{cc} -\frac{\frac{a}{sA}C^{\prime}}{P^{\prime}-\left( \frac{a}{sA}\right)^{2}G\text{"}} & -\frac{\left( \frac{a}{sA}\right)^{2}}{P^{\prime}-\left( \frac{a}{sA}\right)^{2}G\text{"}}\\ C\text{"}\cdot\phi_{3}+\frac{\left( C^{\prime}\right)^{2}}{P^{\prime}-\left( \frac{a}{sA}\right)^{2}G\text{"}} & \delta+\frac{\frac{a}{sA}C^{\prime} }{P^{\prime}-\left( \frac{a}{sA}\right)^{2}G\text{"}} \end{array} \right]. $$

(A.23)

We can therefore say, under our assumptions, that :

$$ Tr(J)\,=\,\delta>0\text{ and }\det(J)=\frac{-\delta\frac{a}{sA}C^{\prime}+\left( \frac{a}{sA}\right)^{2}C\text{"}\phi_{3}}{P^{\prime}-\left( \frac{a}{sA}\right)^{2}G\text{"}}<0. $$

(A.24)

Appendix B: The Study of the Threshold

Let us introduce the system:

$$\begin{array}{@{}rcl@{}} && \left\{ \begin{array}{l} \phi_{4}(\rho,h;R,\beta,{\Gamma},s,A)=0\\ \phi_{5}(\rho,h;W,R,\varepsilon,\beta,{\Gamma},s,A)=0 \end{array} \right. \end{array} $$

(B.1)

$$\begin{array}{@{}rcl@{}} && \Leftrightarrow\left\{ \begin{array}{l} P\left( \frac{1}{a}R_{\rho}\right) \,-\,C\left( h\right) \,+\,\frac{1}{\delta sA}C^{\prime}\left( h\right) R_{\rho}\,-\,\frac{a}{sA}{\Gamma}\,=\,0\\ P\left( \frac{1}{a}\left( R_{\rho}\,+\,(\rho\,-\,\varepsilon)W\right) \right) \,-\,C\left( h\right) \\{\kern18pt} \,+\,\frac{1}{\delta sA}C^{\prime}\left( h\right)\left( R_{\rho}\,-\,bW\right) \,-\,\frac{a}{sA}{\Gamma}\,=\,0 \end{array} \right. \end{array} $$

(B.2)

with ${\Gamma }=G^{\prime }\left (0\right )$. The first equation gives us the stationary water table without RWH while the second includes this option. So if we solves this system in h and ρ, we obtain the threshold value, $\bar {\rho }$ for which the stationary water table h is the same without or with rainwater harvesting. Moreover, this threshold can then be linked to the different parameters $\left (W,R,\varepsilon ,\beta ,{\Gamma }\right )$.

1.1 B.1 Existence of a Threshold

Let us observe that for any admissible $\left (W,R,\varepsilon ,\beta ,{\Gamma },s,A\right )$, ∀{i = 4,5}, (i) ∂ _h ϕ _i >0 since $C^{\prime }<0$ and C”≥ 0, (ii) $\lim _{h\rightarrow 0}\phi _{i} =-\infty$ since $\lim _{h\rightarrow 0}C(h)=-\infty$ and (iii) $\lim _{h\rightarrow \bar {h}}\phi _{i}>0$ (apply the same argument as in the proof of Proposition 10). We can therefore, by the implicit function theorem, define two continuous functions $h_{g}\left (\rho \right )$ and $h_{r}\left (\rho \right )$ which respectively solve in h, ϕ ₄(ρ, h;R, β,Γ)=0 and ϕ ₅(ρ, h;W, R, ε, β,Γ)=0. It remains now to study $\left (h_{g}\left (\rho \right ) -h_{r}\left (\rho \right ) \right )$ since the threshold solves $h_{g}\left (\bar {\rho }\right ) -h_{r}\left (\bar {\rho }\right ) =0$.

Since $\rho \in \left [ \varepsilon ,1-\beta \left (1-\varepsilon \right ) \right ]$, let us first observe that $\left (h_{g}\left (\rho \right ) -h_{r}\left (\rho \right ) \right )_{\rho =\varepsilon }>0$. To verify this point, let us observe that, for ρ = ε, h _g and h _r respectively solve:

$$ \left\{ \begin{array}{l} P\left( \frac{1-\varepsilon}{1-\beta}R\right) \,-\,C\left( h_{g}\right) \,+\,\frac{1-\varepsilon}{\delta sA}C^{\prime}\left( h_{g}\right) R\,-\,\frac {(1-\beta)}{sA}{\Gamma}\,=\,0,\\ P\left( \frac{1-\varepsilon}{1-\beta}R\right) \,-\,C\left( h_{r}\right) \,+\,\frac{1-\varepsilon}{\delta sA}C^{\prime}\left( h_{r}\right) \left( R\,-\,\left( 1\,-\,\beta\right) W\right) \,-\,\frac{(1-\beta)}{sA}{\Gamma}\,=\,0. \end{array} \right. $$

(B.3)

This implies:

$$\begin{array}{@{}rcl@{}} && -C\left( h_{g}\right) \,+\,\frac{(1\,-\,\varepsilon)R}{\delta sA}C^{\prime }\left( h_{g}\right) \,=\,-C\left( h_{r}\right) \,+\,\frac{1\,-\,\varepsilon}{\delta sA}C^{\prime}\left( h_{r}\right) \left( R\,-\,\left( 1\,-\,\beta\right) W\right) \\ && \Leftrightarrow\left[ \!-C\left( h_{g}\right) \,+\,\frac{(1\,-\,\varepsilon )R}{\delta sA}C^{\prime}\left( h_{g}\right) \right] \,-\,\left[ -C\left( h_{r}\right) \,+\,\frac{(1\,-\,\varepsilon)R}{\delta sA}C^{\prime}\left( h_{r}\right) \right]\\ &&=-\frac{(1\,-\,\varepsilon)(1-\beta)W}{\delta sA} C^{\prime}\left( h_{r}\right). \end{array} $$

(B.4)

It is straightforward that the right-hand side of Eq. (B.4) is positive, so that:

$$ \left[ -C\left( h_{g}\right) +\frac{(1-\varepsilon)R}{\delta sA}C^{\prime }\left( h_{g}\right) \right] \!>\!\left[ -C\left( h_{r}\right) +\frac{(1-\varepsilon)R}{\delta sA}C^{\prime}\left( h_{r}\right) \right]. $$

(B.5)

Finally, since $C^{\prime }<0$ and C ” ≥ 0, we deduce that $h_{g}\left (\varepsilon \right ) >h_{r}\left (\varepsilon \right )$.

Second, observe that $\left (h_{g}\left (\rho \right ) -h_{r}\left (\rho \right ) \right )_{\rho =1-\beta \left (1-\varepsilon \right ) }<0.$ In this case, the quantities h _g and h _r solve:

$$ \left\{ \begin{array}{l} P\left( \frac{\beta\left( 1-\varepsilon\right) }{1-\beta}R\right) \,-\,C\left( h_{g}\right) \,+\,\frac{\beta\left( 1-\varepsilon\right) }{\delta sA}C^{\prime}\left( h_{g}\right) R-\frac{(1-\beta)}{sA}{\Gamma}=0,\\ P\left( \frac{\beta\left( 1-\varepsilon\right) }{1-\beta}R+\left( 1-\varepsilon\right) W\right) -C\left( h_{r}\right) +\frac{\beta\left( 1-\varepsilon\right) }{\delta sA}C^{\prime}\left( h_{r}\right) R-\frac{(1-\beta)}{sA}{\Gamma}=0. \end{array} \right. $$

(B.6)

Since $P\left (\frac {\beta \left (1-\varepsilon \right ) }{1-\beta }R\right ) >P\left (\frac {\beta \left (1-\varepsilon \right ) }{1-\beta }R+\left (1-\varepsilon \right ) W\right )$ (remember that $P^{\prime }<0$), we can write:

$$ C\left( h_{g}\right) -\frac{\beta\left( 1-\varepsilon\right) }{\delta sA}C^{\prime}\left( h_{g}\right) R>C\left( h_{r}\right) -\frac {\beta\left( 1-\varepsilon\right) }{\delta sA}C^{\prime}\left( h_{r}\right) R. $$

(B.7)

Since $C\left (h\right ) -\frac {\beta \left (1-\varepsilon \right ) }{\delta sA}C^{\prime }\left (h\right ) R$ is decreasing in h, we conclude that $h_{g}\left (1-\beta \left (1-\varepsilon \right ) \right ) <h_{r}\left (1-\beta \left (1-\varepsilon \right ) \right )$.

Finally by continuity, we can affirm that there exists at least one $\bar {\rho }\in \left [ \varepsilon ,1-\beta \left (1-\varepsilon \right ) \right ]$ such that $h_{g}\left (\bar {\rho }\right ) -h_{r}\left (\bar {\rho }\right ) =0$.

1.2 B.2 Uniqueness of the Threshold

Since this threshold and the associated water table are also obtained as a solution to system (B.1) for a given set $\left (W,R,\varepsilon ,\beta ,{\Gamma }\right )$ of parameters, we can use the Gale-Nikaido theorem (see Gale-Nikaido [11] or Mas-Colell [22]) which states that if every principal minor of $\partial _{\rho ,h}\left (\phi _{i=4,5}\right )$ is positive and the domain $K=\left [ 0,\bar {h}\right ] \times \left [ 0,1\right ]$ of the function is a rectangle, the solution, if it exists, is unique. Let us verify this point.

$$\begin{array}{@{}rcl@{}} &&\partial_{\rho,h}\left( \phi_{i=4,5}\right) =\left[ \begin{array}{cc} -R\left( \frac{1}{a}P_{g}^{\prime}+\frac{1}{\delta sA}C^{\prime}\right) & -C^{\prime}+\frac{R_{\rho}}{\delta sA}C\text{"}\\ -(R-W)\left( \frac{1}{a}P_{r}^{\prime}+\frac{1}{\delta sA}C^{\prime}\right) & -C^{\prime}+\frac{R_{\rho}-bW}{\delta sA}C\text{"} \end{array} \right]\\ &&\text{ with }\left( \begin{array}{l} P_{g}^{\prime}\\ P_{r}^{\prime} \end{array} \right) =\left( \begin{array} [c]{l} P^{\prime}\left( \frac{1}{a}R_{\rho}\right) \\ P^{\prime}\left( \frac{1}{a}\left( R_{\rho}+(\rho-\varepsilon)W\right) \right) \end{array} \right). \end{array} $$

Concerning the principal minors of order 1, we notice that:

• (i) $-R\left (\frac {1}{a}P_{g}^{\prime }+\frac {1}{\delta sA} C^{\prime }\right ) >0$ since $C^{\prime }<0$ and $P^{\prime }<0$;

• (ii) $-C^{\prime }+\frac {R_{\rho }-bW}{\delta sA}C">0$ since $C^{\prime }<0$, C ” ≥ 0 and R _ρ −b W = (1−ρ)(R−W) + β(1−ε)W > 0;

It therefore remains to compute the determinant of $\partial _{\rho ,h}\left (\phi _{i=4,5}\right )$. This quantity is given by:

$$\begin{array}{@{}rcl@{}} \det\left( \partial_{\rho,h}\left( \phi_{i=4,5}\right) \right) & =&\left( \frac{1}{a}P_{g}^{\prime}+\frac{1}{\delta sA}C^{\prime}\right) \left( RC^{\prime}-\frac{(1-\rho)\left( R-W\right) R}{\delta sA} C"\right.\\&&\left.-\frac{\beta(1-\varepsilon)RW}{\delta sA}C"\right) \\ & +&\left( \frac{1}{a}P_{r}^{\prime}+\frac{1}{\delta sA}C^{\prime}\right) \left( \vphantom{\frac{(1-\rho)\left( R-W\right) R}{\delta sA}}\!-RC^{\prime}\right.\\&&\left.+\frac{(1-\rho)\left( R-W\right) R}{\delta sA}C"+WC^{\prime}\right) \\ & =&\underset{>0}{\underbrace{\frac{R}{a}\left( C^{\prime}-\frac {(1-\rho)\left( R-W\right) }{\delta sA}C"\right) \left( P_{g}^{\prime }-P_{r}^{\prime}\right) }}\\&&-\underset{<0}{\underbrace{\frac{\beta (1-\varepsilon)RW}{\delta sA}C"\left( \frac{1}{a}P_{g}^{\prime}+\frac {1}{\delta sA}C^{\prime}\right) }}\\&&+\underset{>0}{\underbrace{WC^{\prime }\left( \frac{1}{a}P_{r}^{\prime}+\frac{1}{\delta sA}C^{\prime}\right) }} \end{array} $$

$$ {\noindent}\text{hence }\det\left( \partial_{\rho,h}\left( \phi_{i=4,5}\right) \right) >0 $$

(B.8)

since $C^{\prime },P_{g}^{\prime },P_{r}^{\prime }<0,C"\geq 0$ and because P ” > 0, $P^{\prime }\left (\frac {1}{a}R_{\rho }\right ) <P^{\prime }\left (\frac {1}{a}\left (R_{\rho }+(\rho -\varepsilon )W\right ) \right )$, i.e. $P_{g}^{\prime }-P_{r}^{\prime }<0$.

1.3 B.3 Comparative Statics

Let us denote by θ one of the following parameters {W, R, β, ε,Γ, s, A} and let us compute $\frac {\partial \rho } {\partial \theta }$. It we differentiate system (B.1) with respect to $\left (h,\rho ,\theta \right )$ and solve the system, we know by the Cramer rule that:

$$ \frac{\partial\rho}{\partial\theta}=\det\left( \begin{array}{cc} -\partial_{\theta}\phi_{4} & \partial_{h}\phi_{4}\\ -\partial_{\theta}\phi_{5} & \partial_{h}\phi_{5} \end{array} \right) \left/ \det\left( \begin{array}{cc} \partial_{\rho}\phi_{4} & \partial_{h}\phi_{4}\\ \partial_{\rho}\phi_{5} & \partial_{h}\phi_{5} \end{array} \right) \right. =\frac{N}{D}. $$

(B.9)

From Eq. B.8, we know that the denominator D is positive. It follows that the sign of $\frac {\partial \rho }{\partial \theta }$ is simply given by its numerator N. Moreover, a simple exercise of computation shows that :

$$\begin{array}{@{}rcl@{}} \partial_{h}\phi_{4}&=&-C^{\prime}(h)+\frac{1}{\delta sA}C^{\prime\prime}(h)R_{\rho}>0\text{ and }\partial_{h}\phi_{5}=-C^{\prime}(h)\\&&+\frac{1}{\delta sA}C^{\prime\prime}(h)\left( R_{\rho}-bW\right) >0 \end{array} $$

(B.10)

and

(B.11)

we can therefore identify, after computation, the effects on the threshold of the following parameters:

• (i) The storage capacity, $\frac {\partial \rho }{\partial W}$.

  $$\begin{array}{@{}rcl@{}} &&{}sign~\left( \frac{\partial\rho}{\partial W}\right) =sign~\left( \left( \frac{\rho-\varepsilon}{a}P_{r}^{\prime}-\frac{b}{\delta}C^{\prime}\right) \underset{>0\text{ since }C^{\prime}<0\text{, }C">0}{\underbrace{\left( -C^{\prime}+\frac{R_{\rho}}{\delta sA}C"\right) }}\right)\\&& {}=sign~\left( \frac{\rho-\varepsilon}{a}P_{r}^{\prime}-\frac{b}{\delta sA}C^{\prime }\right). \end{array} $$

  (B.12)

  Let us now observe that the computation of the difference of the two equations (see Eq. (B.1)) which define the threshold gives:

  $$ P\left( \frac{1}{a}R_{\rho}\right) -P\left( \frac{1}{a}\left( R_{\rho }+(\rho-\varepsilon)W\right) \right) +\frac{bW}{\delta sA}C^{\prime}=0 $$

  (B.13)

  and the convexity of the inverse demand gives:

  $$ P\left( \frac{1}{a}R_{\rho}\right) -P\left( \frac{1}{a}\left( R_{\rho }+(\rho-\varepsilon)W\right) \right) >-\frac{\left( \rho-\varepsilon \right) W}{a}P_{r}^{\prime}. $$

  (B.14)

  These two last equations immediately show that $\left (\frac {\left (\rho -\varepsilon \right ) }{a}P_{r}^{\prime }-\frac {b}{\delta sA}C^{\prime }\right ) >0$ hence that $\frac {\partial \rho }{\partial W}>0.$

• (ii) The evaporation rate, $\frac {\partial \rho }{\partial \varepsilon }$.

  $$ sign\left( \frac{\partial\rho}{\partial\varepsilon}\right) =sign\left( -W\left( \frac{1}{a}P_{r}^{\prime}+\frac{\beta}{\delta}C^{\prime}\right) \left( -C^{\prime}+\frac{R_{\rho}}{\delta}C"\right) \right) >0 $$

  (B.15)

  since $P_{r}^{\prime }<0$, $C^{\prime }<0$ and C ” > 0.

• (iii) The marginal ecosystem damage, $\frac {\partial \rho }{\partial {\Gamma }}$.

  $$ sign\left( \frac{\partial\rho}{\partial{\Gamma}}\right) =sign\left( -\frac{abW}{\delta\left( sA\right)^{2}}C"\right) \leq0. $$

  (B.16)

• (iv) The recharge, $\frac {\partial \rho }{\partial R}$. Since $\partial _{h}\phi _{5}=\partial _{h}\phi _{4}+\frac {bW}{\delta sA}C^{\prime \prime }(h)$ and $\left (\frac {1-\rho }{a}\right ) >0$, a routine computation gives:

  $$ sign\left( \frac{\partial\rho}{\partial R}\right) =sign\left( \underset{>0}{\underbrace{\partial_{h}\phi_{4}}}\underset{\geq0\text{ (since }P"\geq0\text{)}}{\underbrace{\left( -P_{g}^{\prime}+P_{r}^{\prime}\right) }}+\underset{\geq0}{\underbrace{\frac{bW}{\delta sA}C^{\prime\prime}} }\underset{<0}{\underbrace{\left( P_{g}^{\prime}+\frac{a}{\delta sA}C^{\prime}\right) }}\right). $$

  (B.17)

  This quantity is clearly ambiguous. For instance if the demand is linear, we have $P_{r}^{\prime }-P_{g}^{\prime }=0$ and $\frac {\partial \rho }{\partial R} <0$. But if the unit pumping cost is linear in h so that C ” = 0, we observe $\frac {\partial \rho }{\partial R}>0$. Finally in the Gisser-Sánchez case (i.e., C ” = 0 and P linear), it is immediate that $\frac {\partial \rho }{\partial R}=0.$

• (v) The return flow, $\frac {\partial \rho }{\partial \beta }$.

  $$\begin{array}{@{}rcl@{}} &&sign\left( \frac{\partial\rho}{\partial\beta}\right) =sign\left( \partial_{h}\phi_{4}\left[ \frac{R_{\rho}}{a^{2}}\left( -P_{g}^{\prime }+P_{r}^{\prime}\right)\right.\right.\\ &&\left.+\frac{(\rho-\varepsilon)W}{a^{2}}P_{r}^{\prime }+\frac{(1-\varepsilon)W}{\delta sA}C^{\prime}\right]\\&&\left. +\frac{bW}{\delta sA}\left( \frac{R_{\rho}}{a^{2}}P_{g}^{\prime}+{\Gamma}\right) C^{\prime \prime}\right). \end{array} $$

  (B.18)

  This quantity is again ambiguous. To illustrate this point let us assume that C ” = 0 so that $C^{\prime }=-c_{0}$ a constant with c ₀>0. If we add that P(w) is linear (the Gisser-Sánchez case) then $sign\left (\frac {\partial \rho }{\partial \beta }\right ) =sign\left (\frac {(\rho -\varepsilon )W}{a^{2}}P_{r}^{\prime }-\frac {(1-\varepsilon )W}{\delta sA}c_{0}\right ) <0$. If we now introduce an iso-elastic demand P(w) = w ^−α with a > 1, then

  $$\begin{array}{@{}rcl@{}} &&sign\left( \frac{\partial\rho}{\partial\beta}\right) =sign\left( \alpha\underset{\equiv d>0}{\underbrace{\left( \left( \frac{R_{\rho}} {a}\right)^{-\alpha}-\left( \frac{R_{\rho}+(\rho-\varepsilon)W}{a}\right)^{-\alpha}\right) }}\right.\\&&{\kern90pt}\left.-\underset{\equiv n>0}{\underbrace{a\frac{(1-\varepsilon )W}{\delta sA}c_{0}}}\vphantom{\underset{\equiv d>0}{\underbrace{\left( \left( \frac{R_{\rho}} {a}\right)^{-\alpha}-\left( \frac{R_{\rho}+(\rho-\varepsilon)W}{a}\right)^{-\alpha}\right) }}}\right) \end{array} $$

  (B.19)

  so if $\alpha >\max \left \{ \frac {n}{d},1\right \}$ then $\frac {\partial \rho }{\partial \beta }>0$.

• (vi) The storativity coefficient, $\frac {\partial \rho }{\partial s}$. A routine computation gives:

  $$ sign\left( \frac{\partial\rho}{\partial s}\right) =sign\left( \frac{\Gamma}{s}C"-(C^{\prime})^{2}\right). $$

  (B.20)

  If the marginal damage of the aquifer is not considered, i.e. Γ=0, or if marginal pumping cost is linear in h, we have$\frac {\partial \rho }{\partial s}<0$. Otherwise, this quantity may be unsigned especially if the marginal environmental damage Γ is high.

• (vii) The area of the aquifer, $\frac {\partial \rho }{\partial A}$. Since A and s work in the same way on the threshold, we can say that $sign\left (\frac {\partial \rho }{\partial A}\right ) =sign\left (\frac {\Gamma }{A}C"-(C^{\prime })^{2}\right )$ and we can apply the same comments as in (vi).