Optimal Environmental Policy for a Mine Under Polluting Waste Rocks and Stock Pollution

Abstract

This study analyzes socially optimal environmental policy for a mine with a model that takes into account waste or waste rock production and abatement possibilities of the mine. We develop a model, in which the mine produces an externality related to the waste rocks, such as acid mine drainage. We find that the extraction rate tends to be lower in a mine with higher waste rock production, and that the optimal tax on the waste rock production is strictly increasing in time. We extend the model to incorporate an additional externality in the form of a stock pollutant. We analyze the optimal taxes and show that the typical result that the time path of the tax on the stock pollutant is inverted U-shaped may be lost in a mine model with abatement possibility and fixed operation period.

Appendix

1.1 Proof of Proposition 1

(i) Since the control set $U:=\{ (q,z) \in \mathbb {R}^2 \ | \ q\ge 0, z\ge 0\}$ is convex and H is strictly concave in (q, z), the control functions are continuous functions.

(ii) First we show that $q(t)>0$ at some t. Assume not, that is $q(t)\equiv 0$. Then $X(T)=x_0>0$, and therefore by (16) $\lambda (T)=0$. This implies $\lambda (t)\equiv 0$. Equation (13) together with boundary condition $S(T) \ge 0$, implies $z(t)\equiv 0$. Then by assumption ${C^s}'(0)=0$ (recall Assumption 1) and by condition (11), $\mu (t)\ge 0$ for every t. This yields a contradiction since (10), that is, condition

$$\begin{aligned} p-C'(0)+\mu \alpha \le 0, \end{aligned}$$

(A.1)

and assumption $p-C'(0)>0$, imply $\mu (t)< 0$. Therefore $q(t)>0$ at some t. Since q(t) is continuous, it must be that $q(t)>0$ at some interval $[t_0,t_1]$.

Now we show that the rate of extraction is strictly decreasing, when the rate of extraction is positive. We first show that $\mu (0)<0$. Note that $z>0$ somewhere: Suppose not, that is, suppose $z\equiv 0$. Then $S(T)>0$ and $G'(S(T))>0$. By (11) and assumption ${C^s}'(0)=0$, $\mu \ge 0$, which implies by (15) that $\mu (0) \ge 0$. Hence $\mu (T)\ge 0$. Because $\mu (T)\ge 0$ and $G'(S(T))>0$, also $\mu (T)+G'(S(T))>0$, which implies by condition (17) that $S(T)=0$. This contradicts $S(T)>0$. Therefore $z>0$ somewhere. Because ${C^s}'(z)=-\mu (0)e^{\rho t}$ for some t, $\mu (0)<0$. To show that the rate of extraction is strictly decreasing let $t_0^+, t_1^+ \in [t_0,t_1]$ with $t_0^+< t_1^+$. We obtain, after using (10) twice and subtracting, that on the interval where $q(t)>0$,

$$\begin{aligned} p-C'(q(t_0^+))-p+C'(q(t_1^+))&= \end{aligned}$$

(A.2)

$$\begin{aligned} \lambda (0)[e^{\rho t_0^+}-e^{\rho t_1^+}]-\mu (0)\left[ e^{\rho t_0^+}-e^{\rho t_1^+}\right]&<0. \end{aligned}$$

(A.3)

This implies $q(t_1^+)<q(t_0^+)$ by assumption $C''(q)>0$.

Continuity of q(t) implies that we can choose $t_0=0$, because otherwise q(t) would be discontinuous or increasing somewhere.

(iii) Recall that $z(t)>0$ at some interval $[t_2,t_3]$. At this interval, control z(t) is strictly increasing. This is true because on this interval, ${C^s}'(z)=-\mu (0)e^{\rho t}$, which clearly implies that z(t) must increase: Let $t_2^+, t_3^+ \in [t_2,t_3]$ with $t_2^+< t_3^+$. Using (11) twice and subtracting, we get

$$\begin{aligned} {C^s}'(z(t_3^+))-{C^s}'(z(t_2^+))=\mu (0)[e^{\rho t_2^+}-e^{\rho t_3^+}]>0. \end{aligned}$$

(A.4)

This implies $z(t_3^+)>z(t_2^+)$ by assumption ${C^s}''(q)>0$. Note in addition that because $\mu (0)<0$, $\mu (t)<0$ for all t. Because z(t) is strictly increasing and continuous, there can be only one interval, where $z(t)>0$. In addition, since $\mu (0)<0$, $z(0)>0$: Suppose not. By condition (11)

$$\begin{aligned} -{C^s}'(0)-\mu (0)\le 0. \end{aligned}$$

(A.5)

Then by assumption ${C^s}'(0)=0$, $\mu (0)\ge 0$, which contradicts $\mu (0)<0$.

(iv) When $q(t)>0$,

$$\begin{aligned} p-C'(q)=(\lambda (0)-\mu (0)\alpha ) e^{\rho t}. \end{aligned}$$

(A.6)

Because $\lambda (0)>0$ and $\mu (0)<0$, the instantaneous marginal profit rises at the rate of interest.

1.2 Proof of Proposition 3

Note first that $\mu (T)<0$ and (17) imply $G'(S(T))>0$. This implies $S(T)>0$. For the second part of the result, we show first that $\alpha q(0)\le z(0)$ is generally impossible at optimum. Suppose that $\alpha q(0)\le z(0)$ holds. Then ${\dot{S}}(0)\le 0$, and because ${\dot{q}}<0$ for every $t\in [0,{\overline{t}})$ and ${\dot{z}}>0$ for every $t\in [0,T]$, ${\dot{S}}(t)<0$ for all $t\in (0,T]$. Then $S(T) > 0$ cannot be satisfied. Therefore, inequality $\beta q(0)>z(0)$ holds implying that ${\dot{S}}(0)>0$. Note that the time path of S cannot have a minimum: Suppose it has a minimum. At such a point extraction is positive and ${\ddot{S}}\ge 0$, which is equivalent with $\alpha {\dot{q}} \ge {\dot{z}}$ by Eq. (13). But since ${\dot{z}}>0$, we obtain ${\dot{q}}>0$, which contradicts Part (ii) Proposition 1.

Hence the time path is either strictly increasing or has a maximum value on (0, T). If the path is strictly increasing, the maximum value is at $t=T$. We show next, that if the path has a maximum value on (0, T), the maximum value is unique. To this end, suppose that the stock has a maximum value at $t_0<T$, which is denoted by $S(t_0)$. The stock cannot have another local maximum: Suppose it does. Two possibilities exists in this case: either there exists $t_1$ with a local maximum satisfying $S(t)<\max \{S(t_0),S(t_1) \}$ for all $t \in (t_0,t_1)$, or there exists a local maximum $t_2$ satisfying $S(t)=S(t_0)$ for all $t\in (t_0,t_2)$. The first case is ruled out, because it implies that S has a local minimum on $(t_0,t_1)$, which contradicts the above reasoning. To rule out the second possibility, we note that ${\dot{S}}(t_0)=0$ and ${\dot{S}}(t)>0$ for every $t \in (0,t_0)$. Equation ${\dot{S}}(t_0)=0$ is equivalent to $\beta q(t_0)=z(t_0)$. Since ${\dot{q}}\le 0$ and ${\dot{z}}>0$, it must hold that after $t_0$ we have ${\dot{S}}(t)<0$. This contradicts $S(t)=S(t_0)$ for all $t\in (t_0,t_2)$.

1.3 Proof of Proposition 4

Let $t^{L}$ be the time instant when the mine is exhausted with the waste rock production rate $\alpha ^L$ and let $t^{H}$ be the respective time instant with coefficient $\alpha ^H$. We show first that $t^{L}>t^{H}$ is impossible, so assume that $t^{L}>t^{H}$ holds. In both cases ${\dot{q}}<0$ whenever $q>0$. Because of this and because extraction is zero after exhaustion, $q^{H}$ hits zero before $t^{L}$. Since the mine is exhausted in both cases,

$$\begin{aligned} \int _{0}^{t^{H}}q^{H}(\tau )\ \mathrm {d}\tau =x_0=\int _{0}^{t^{L}}q^{L}(\tau )\ \mathrm {d}\tau . \end{aligned}$$

(A.7)

It follows from these, that there exists a time instant ${\hat{t}}<t^{H}$ for which

$$\begin{aligned} q^{L}({\hat{t}})=q^{H}({\hat{t}})>0. \end{aligned}$$

(A.8)

From this and from the necessary conditions of the models’ problems, we obtain the following equations

$$\begin{aligned} p-C'(q^{H}({\hat{t}}))&=\lambda ^{H}({\hat{t}})-\alpha ^{H}\mu ^{H}({\hat{t}}), \end{aligned}$$

(A.9)

$$\begin{aligned} p-C'(q^{L}({\hat{t}}))&=\lambda ^{L}({\hat{t}})-\alpha ^{L}\mu ^{L}({\hat{t}}). \end{aligned}$$

(A.10)

From (A.8)–(A.10) we obtain that

$$\begin{aligned} \lambda ^{L}({\hat{t}})-\alpha ^{L}\mu ^{L}({\hat{t}})=\lambda ^{H}({\hat{t}})-\alpha ^{H} \mu ^{H}({\hat{t}}). \end{aligned}$$

(A.11)

Furthermore, by the necessary conditions,

$$\begin{aligned} \lambda ^{L}(t)=\lambda ^{L}(0)e^{\rho t}, \quad \lambda ^{H}(t)=\lambda ^{H}(0)e^{\rho t}, \quad \mu ^{H}(t)=\mu ^{H}(0)e^{\rho t}\quad \mathrm { and } \quad \mu ^{L}(t)=\mu ^{L}(0)e^{\rho t}. \end{aligned}$$

(A.12)

Define now function f with $f(t):=\lambda ^{L}(t)-\alpha ^{L} \mu ^{L}(t)-\lambda ^{H}(t)+\alpha ^{H} \mu ^{H}(t)$, and note that $f(t)=f(0)e^{\rho t}$ and $f({\hat{t}})=f(0)e^{\rho {\hat{t}}}$. Since by (A.11) $f({\hat{t}})=0$, we obtain that $f(0)=0$. Hence $f(t)\equiv 0$. Since (A.9) and (A.10) hold also for other time instants with positive extraction, we obtain equation

$$\begin{aligned} p-C'(q^{L}(t))-p+C'\left( q^{H}(t)\right) =f(t)=0. \end{aligned}$$

(A.13)

Hence $q^{L}(t)=q^{H}(t)$ for all t with positive $q^{L}$ and $q^{H}$. But then

$$\begin{aligned} \int _{0}^{t^{L}}q^{L}(\tau )\ \mathrm {d}\tau&=\int _{0}^{t^{H}}q^{L}(\tau )\ \mathrm {d}\tau +\int _{t^{H}}^{t^{L}}q^{L}(\tau )\ \mathrm {d}\tau \end{aligned}$$

(A.14)

$$\begin{aligned}&=\int _{0}^{t^{H}}q^{H}(\tau )\ \mathrm {d}\tau +\int _{t^{H}}^{t^{L}}q^{L}(\tau )\ \mathrm {d}\tau \end{aligned}$$

(A.15)

$$\begin{aligned}&=x_0+\int _{t^{H}}^{t^{L}}q^{L}(\tau )\ \mathrm {d}\tau > x_0, \end{aligned}$$

(A.16)

which contradicts (A.7).

By the same principles and after minor changes, $t^{L}<t^{H}$ is also impossible. Hence $t^{L}=t^{H}$. Similar arguments as used above show that the extraction rates are the same.

1.4 Proof of Proposition 5

(i) Since the mine with coefficient $\alpha ^{H}$ is not exhausted, $\lambda ^H(T)=0$, and therefore $\lambda ^H(t)\equiv 0$. The rate of extraction with coefficient $\alpha ^{H}$, $q^H$, cannot be anywhere greater than the rate of extraction with coefficient $\alpha ^{L}$, $q^L$: Suppose otherwise, that is, suppose that $q^{H}(t)>q^{L}(t)$ for some $t \in [0,T]$. Then there exists a time instant ${\hat{t}}<T$ for which

$$\begin{aligned} q^{L}({\hat{t}})=q^{H}({\hat{t}}), \end{aligned}$$

(A.17)

and we obtain using similar arguments as in the proof of Proposition 4 that $\lambda ^{L}(t)-\alpha ^{L} \mu ^{L}(t)=-\alpha ^{H} \mu ^{H}(t)$ for every t with positive extraction levels. This means that $q^{H}(t)= q^{L}(t)$ (when extraction is positive), which yields a contradiction with $q^{H}(t)>q^{L}(t)$.

Since the mine with $\alpha ^{L}$ is exhausted and the mine with $\alpha ^{H}$ is not, the extraction rates cannot be equal at any time instant, because similar arguments as above show that the extraction rates are equal for all time instants in such a case.

(ii) We will show that $q^H<q^L$ for all $t \in [0,T ]$ by contradiction. To this end, we assume $q^H\ge q^L$ for some $t \in [0,T ]$. Note then that $q^H\ge q^L$ for all $t \in [0,T ]$: Otherwise (that is, if $q^H < q^L$ for some t) there would exist a time instant ${\hat{t}}$ such that $q^H=q^L$, which implies using similar arguments as in the proof of Proposition 4 that $q^H\equiv q^L$, which contradicts inequality $q^H < q^L$ for some t.

Because neither of the mines are exhausted, $\lambda ^H=\lambda ^L\equiv 0$. Since $q^H\ge q^L$, $p-C'(q^H)\le p-C'(q^L)$. Combining this with (10) gives us $-\alpha ^H \mu ^H\le -\alpha ^L \mu ^L$ when extraction is positive. Then

$$\begin{aligned} -\mu ^H \le -\frac{\alpha ^L}{\alpha ^H} \mu ^L < -\mu ^L, \end{aligned}$$

(A.18)

since $\alpha ^H>\alpha ^L$. This implies together with equation ${C^s}'(z)=-\mu$ from (11), that $z^L>z^H$ for all $t \in [0,T ]$. This and $q^H\ge q^L$ for all $t \in [0,T ]$ imply by (13) that ${\dot{S}}^H>{\dot{S}}^L$ for all $t \in [0,T ]$. Combining this with $S^H(0)=S^L(0)=0$ gives

$$\begin{aligned} S^H(T)>S^L(T). \end{aligned}$$

(A.19)

Hence $G'(S^H(T))> G'(S^L(T))$. We obtain using these and (17) that

$$\begin{aligned} -\,\mu ^H(T)=G'(S^H(T))>G'(S^L(T))= -\mu ^L(T), \end{aligned}$$

(A.20)

but $-\mu ^H(T)>-\mu ^L(T)$ contradicts (A.18). Hence $q^H<q^L$ for all $t \in [0,T ]$.

1.5 Proof of Proposition 6

Since the mine with a high waste rock production rate is exhausted and the one with a low rate is not, we have

$$\begin{aligned} x_0=\int _{0}^{t^{H}}q^{H}(\tau )\ \mathrm {d}\tau >\int _{0}^{T}q^{L}(\tau )\ \mathrm {d}\tau . \end{aligned}$$

(A.21)

This clearly implies that there exists a time instant at which $q^H>q^L$. Then $q^H>q^L$ for all $t \in [0,T ]$ (otherwise we would again obtain a contradiction). Note that $\lambda ^L\equiv 0$. When $q^L>0$, condition (10) gives us after subtraction that

$$\begin{aligned} p-C'\left( q^H \right) -p+C'\left( q^L \right) =\lambda ^H-\alpha ^H \mu ^H+\alpha ^L \mu ^L. \end{aligned}$$

(A.22)

Since the left-side of this is strictly negative, we obtain

$$\begin{aligned} \lambda ^H-\alpha ^H \mu ^H+\alpha ^L \mu ^L<0. \end{aligned}$$

(A.23)

This implies

$$\begin{aligned} - \mu ^H<-\frac{\lambda ^H}{\alpha ^H}-\frac{\alpha ^L}{\alpha ^H} \mu ^L<-\frac{\alpha ^L}{\alpha ^H} \mu ^L <- \mu ^L. \end{aligned}$$

(A.24)

Because $q^H>q^L$ for all $t \in [0,T ]$ and $-\mu ^H<-\,\mu ^L$ for all $t \in [0,T ]$, we obtain a contradiction using similar lines of reasoning as in the proof of Part (ii) of Proposition 5. Hence it is impossible that a mine with a high waste rock production rate is exhausted, but the mine with a lower rate is not.

1.6 Proof of Proposition 7

(i) Similar to the proof of Part (i) of Proposition 1. (ii) Assume that q is always zero. With similar reasoning as in the proof of Part (ii) of Proposition 1, it follows from this that $\mu (t)\ge 0$ for all t. (Note: Because of $\gamma (t)$, we cannot argue directly for a contradiction using condition $H_q\le 0$ in (27).) Since $q(t)\equiv 0$, (32) becomes

$$\begin{aligned} {\dot{N}}=-a-\delta N, \end{aligned}$$

(A.25)

whose solution using the given initial condition is

$$\begin{aligned} N(t)=\int _{0}^{t}-\,a(\tau )e^{-\delta (t-\tau )} \ \mathrm {d}\tau . \end{aligned}$$

(A.26)

Then, because $a(t)\ge 0$,

$$\begin{aligned} N(T)=\int _{0}^{T}-\,a(\tau )e^{-\delta (T-\tau )} \ \mathrm {d}\tau \le 0, \end{aligned}$$

(A.27)

implying that $a(t)\equiv 0$. Condition (29) and assumption $A'(0)=0$, imply $-\,\gamma \le 0$ or equivalently $\gamma \ge 0$. Condition $H_q\le 0$ becomes now

$$\begin{aligned} p-C'(0)+\mu \alpha +\gamma \beta \le 0, \end{aligned}$$

(A.28)

implying that $\mu < 0$. This contradicts $\mu \ge 0$. Therefore there exists a time instant where $q(t)>0$. Again, by continuity of q, there is an interval, where $q(t)>0$. (iii) Similar to the proof of Part (iii) of Proposition 1. (iv) Let $t_1,t_2\in I \subset [0,T]$ with $t_1<t_2$, where I is an interval with $a(t)>0$. Using (29) twice, we obtain equation

$$\begin{aligned} A'(a(t_2))-A'(a(t_1))=\gamma (t_1)-\gamma (t_2). \end{aligned}$$

(A.29)

The conclusion follows from this, since $A''>0$.

1.7 Proof of Lemma 1

(i) Suppose on the contrary to the claim that $N(t)=0$ on [0, T]. Recall that by Proposition 7 there exists an interval $I \in [0,T]$, where $q>0$. Then on that interval $\beta q=a>0$ (by (32)) and ${\dot{\gamma }}=(\delta +\rho )\gamma$ (by (35) and by $D'(0)=0$). By (29), $A'(a)=-\gamma >0$ on I, implying that ${\dot{\gamma }}<0$. Then a grows on I by Proposition 7. But by (27), $p-C'(q)=\lambda -\mu \alpha -\gamma \beta$ on I, which implies that q is decreasing on I. This contradicts equation $\beta q=a$ for all $t\in I$. (ii) Suppose on the contrary to the claim that there exists an interval $I \subset [0,T]$ such that ${\dot{\gamma }}=0$ and $\gamma \ne 0$ for every $t \in I$. Then it holds on the interval I that

$$\begin{aligned} \gamma&=-\frac{D'(N)}{\delta +\rho }, \end{aligned}$$

(A.30)

$$\begin{aligned} {\dot{a}}&=0, \end{aligned}$$

(A.31)

$$\begin{aligned} {\dot{q}}&=\frac{-\rho \lambda +\rho \alpha \mu }{C''(q)}<0. \end{aligned}$$

(A.32)

Since ${\dot{\gamma }}=0$ and (A.30) holds, ${\dot{N}}=0$. This implies that $\beta q-a=\delta N$ for every $t \in I$. Differentiating this with respect to time yields $\beta {\dot{q}}-{\dot{a}}=0$. This is equivalent with $\beta {\dot{q}}=0$, since (A.31) holds. But this contradicts (A.32).

1.8 Proof of Lemma 2

(i) Suppose on the contrary to the claim that such a time instant does not exists, that is, suppose that $a(t)=0$ for every $t \in (t^0,{\hat{t}})$, where $|{\hat{t}}-t^0| < \epsilon$ for any small $\epsilon > 0$. By the definition of ${\hat{t}}$, $N(t)>0$ on the interval $(t^0,{\hat{t}})$ for sufficiently small $\epsilon$. Let $t^1 \in (t^0,{\hat{t}})$. Then we obtain that

$$\begin{aligned} N({\hat{t}})&=\int _{0}^{{\hat{t}}}q(\tau )e^{-\delta ({\hat{t}}-\tau )} \ \mathrm {d}\tau -\int _{0}^{{\hat{t}}}a(\tau )e^{-\delta ({\hat{t}}-\tau )} \ \mathrm {d}\tau \end{aligned}$$

(A.33)

$$\begin{aligned}&=\int _{0}^{{\hat{t}}}q(\tau )e^{-\delta ({\hat{t}}-\tau )} \ \mathrm {d}\tau -\int _{0}^{t^1}a(\tau )e^{-\delta (t^1-\tau )} \ \mathrm {d}\tau , \end{aligned}$$

(A.34)

by the above assumption about abatement. Because $q(t)\ge 0$, we obtain that $N({\hat{t}})\ge N(t^1)$. However, the definition of ${\hat{t}}$ implies that $N(t^1)>N({\hat{t}})$. These yield a contradiction $N({\hat{t}})>N({\hat{t}})$. (ii) Since $\gamma (t)<0$ for every $t\in (t^+,{\hat{t}})$, $\gamma ({\hat{t}})\le 0$ by the continuity of the costate. The solution to (35) is

$$\begin{aligned} \gamma (t)=\int _{0}^{t}D'(N(\tau ))e^{-(\delta +\rho )(\tau -t)} \ \mathrm {d}\tau +e^{(\delta +\rho )t}\gamma (0). \end{aligned}$$

(A.35)

This can be used to rewrite $\gamma ({\hat{t}})\le 0$ as

$$\begin{aligned} \gamma ({\hat{t}})&=\int _{0}^{{\hat{t}}}D'(N(\tau ))e^{-(\delta +\rho )(\tau -{\hat{t}})} \ \mathrm {d}\tau +e^{(\delta +\rho ){\hat{t}}}\gamma (0) \end{aligned}$$

(A.36)

$$\begin{aligned}&=e^{(\delta +\rho ){\hat{t}}}\bigg [\int _{0}^{{\hat{t}}}D'(N(\tau ))e^{-(\delta +\rho )\tau } \ \mathrm {d}\tau +\gamma (0)\bigg ] \end{aligned}$$

(A.37)

$$\begin{aligned}&\le 0. \end{aligned}$$

(A.38)

Let $t\ge {\hat{t}}$. Then

$$\begin{aligned} \gamma (t)&=\int _{0}^{t}D'(N(\tau ))e^{-(\delta +\rho )(\tau -t)} \ \mathrm {d}\tau +e^{(\delta +\rho )t}\gamma (0) \end{aligned}$$

(A.39)

$$\begin{aligned}&=e^{(\delta +\rho )t} \bigg [\int _{0}^{t}D'(N(\tau ))e^{-(\delta +\rho )\tau } \ \mathrm {d}\tau +\gamma (0)\bigg ] \end{aligned}$$

(A.40)

$$\begin{aligned}&=e^{(\delta +\rho )t} \bigg [\int _{0}^{{\hat{t}}}D'(N(\tau ))e^{-(\delta +\rho )\tau } \ \mathrm {d}\tau +\gamma (0)\bigg ] \end{aligned}$$

(A.41)

$$\begin{aligned}&\le 0, \end{aligned}$$

(A.42)

where the last equality follows from assumption $D'(0)=0$, since after ${\hat{t}}$ we know that $N(t)=0$ by definition, and the inequality follows from (A.37)–(A.38).

1.9 Proof of Lemma 3

(i) Since $N(t) = 0$ for all $t\in [{\hat{t}},T]$, it must be that $\beta q(t)=a(t)$ there. Assume contrary to the claim that there exists $I \subset [{\hat{t}}, T]$ such that $a(t)>0$ on I. Then $\gamma (t)<0$ on I and therefore ${\dot{a}}>0$ on I, because ${\dot{\gamma }}=(\delta +\rho )\gamma <0$ on I. But by $p-C'(q)=\lambda -\mu \alpha -\gamma \beta$, ${\dot{q}}<0$ on I. This contradicts condition

$$\begin{aligned} \beta q(t)=a(t) \quad {\text {for all }} \quad t\in [{\hat{t}},T]. \end{aligned}$$

(A.43)

(ii) Let $t\ge {\hat{t}}$. Part (ii) of Lemma 2 says that $\gamma (t) \le 0$. Part (i) of the current Lemma says that $a(t)=0$. Condition (29) and assumption $A'(0)=0$ give then that $-\gamma (t) \le 0$. These imply $\gamma (t)=0$ for all $t \in [{\hat{t}},T]$.

1.10 Proof of Proposition 9

The proof has similarities with the proof of Proposition 1 in Tahvonen (1997).

Note first that ${\dot{\gamma }}(t) > 0$ for all $t \in [0,{\hat{t}})$ is not true, since ${\dot{\gamma }}(0)=D'(N(0))+(\delta +\rho )\gamma (0)=(\delta +\rho )\gamma (0)<0$.

Recall that $\gamma (T)=0$ and suppose that the time path has an inverted U-shape somewhere. There must exist a closed interval $I=[t_1,t_2]$, where ${\dot{\gamma }}<0$ on the interior of I and ${\dot{\gamma }}(t_1)={\dot{\gamma }}(t_2)=0$. This implies that ${\dot{q}}<0$ and ${\dot{a}}\ge 0$ on I, so that

$$\begin{aligned} \beta {\dot{q}}-{\dot{a}}<0 \quad {\text { for all}}~ t\in I. \end{aligned}$$

(A.44)

Note that $\ddot{\gamma }(t_1)\le 0$ and $\ddot{\gamma }(t_2)\ge 0$. By Eq. (35), $\ddot{\gamma }(t)=D''(N(t)){\dot{N}}(t)$ at $t=t_1$ and at $t=t_2$, implying that

$$\begin{aligned} {\dot{N}}(t_1)\le 0 \quad {\text { and }} \quad {\dot{N}}(t_2)\ge 0. \end{aligned}$$

(A.45)

The derivative $\ddot{\gamma }=D''(N(t)){\dot{N}}(t)+(\delta +\rho ){\dot{\gamma }}$ is positive somewhere on the interval $[t_1,t_2]$, implying that ${\dot{N}}(t)>0$ there, since ${\dot{\gamma }}<0$ on $(t_1,t_2)$. Then there exists $t_3 \in [t_1,t_2)$ such that ${\dot{N}}(t_3)=0$. But then

$$\begin{aligned} {\ddot{N}}(t_3)=\beta {\dot{q}}(t_3)-{\dot{a}}(t_3)\ge 0, \end{aligned}$$

(A.46)

which is a contradiction with (A.44). Hence the time path is U-shaped.

1.11 Proof of Proposition 10

We argue that $a(T)=0$. Assume on the contrary that $a(T)>0$. Then $\gamma (T)<0$. However, condition (38) together with $N(T)=0$ and $R'(0)=0$ imply $\gamma (T)\ge 0$, which contradicts $\gamma (T)<0$. Hence $a(T)=0$. Since ${\hat{t}}=T$, $N(t)>0$ for all $t \in (T-\epsilon ,T)$ for some (small) $\epsilon >0$. Note that

$$\begin{aligned} {\dot{N}}(T)=\beta q(T)-a(T)-\delta N(T)=\beta q(T)-a(T), \end{aligned}$$

(A.47)

which is non-positive at optimum. Since $\beta q(T)\le a(T)=0$, we have that $q(T)=0$. By Lemma 2, we have $a(t)>0$ for all $t \in (T-\epsilon ,T)$. For these instants of time, $\gamma (t)<0$, implying that $\gamma (T)\le 0$. This and $\gamma (T)\ge 0$ obtained using condition (38) as above, imply that $\gamma (T)=0$. Thus, when ${\hat{t}}=T$ and $N(T)=0$, the shadow value of the pollution stock, $\gamma$, has a U-shaped time path by the same principles as in the proof of Proposition 9.

1.12 Proof of Proposition 11

Note that in this case condition (38) implies $-\,\gamma (T)=R'(N(T))$. Since the right-side is strictly positive, we obtain $\gamma (T)<0$. This implies together with

$$\begin{aligned} \gamma (T)=\int _{0}^{T}D'(N(\tau ))e^{-(\delta +\rho )(\tau -T)} \ \mathrm {d}\tau +e^{(\delta +\rho )T}\gamma (0), \end{aligned}$$

(A.48)

that also $\gamma (0)<0$. By (35) and assumption $D'(0)=0$,

$$\begin{aligned} {\dot{\gamma }}(0)=D'(N(0))+(\delta +\rho ) \gamma (0)=(\delta +\rho ) \gamma (0)<0. \end{aligned}$$

(A.49)

Note that ${\dot{\gamma }}(0)<0$ excludes the possibility that ${\dot{\gamma }}(t)>0$ for all $t \in [0,T]$. This leaves four possibilities:

• (1) ${\dot{\gamma }}(t)<0$ for all $t \in [0,T]$;

• (2) ${\dot{\gamma }}(t)<0$ for all $t \in [0,T]$ except at an isolated time instant at which ${\dot{\gamma }}(t)=0$;

• (3) The time path is U-shaped;

• (4) The time path is first decreasing, then increasing and lastly decreasing again.

There are no other possibilities: Suppose on the contrary that the time path has three points with ${\dot{\gamma }}(t)=0$. Then the path has an inverted U-shape somewhere, and the same arguments as in the proof of Proposition 9 produce a contradiction.

1.13 Proof of Proposition 12

(i) Assume on the contrary to the claim that $t^{L}\ge t^{H}$. Given the time invariant taxes and strictly positive extraction rates, the necessary conditions of the mining firms’ problems include the following conditions:

$$\begin{aligned} p-C'(q^{H}(t))&=\lambda ^{H}(t)+\alpha ^{H}v_S+\beta ^{H}v_N, \end{aligned}$$

(A.50)

$$\begin{aligned} p-C'(q^{L}(t))&=\lambda ^{L}(t)+\alpha ^{L}v_S+\beta ^{L}v_N. \end{aligned}$$

(A.51)

Since both mines are exhausted and $t^{L}\ge t^{H}$, the time paths of the mines’ extraction rates must cross at least once.

Consider first the possibility that $t^{L}>t^{H}$. Since there exists a smallest time instant ${\hat{t}}$ such that $q^{H}({\hat{t}})=q^{L}({\hat{t}})$, Eqs. (51), (A.50) and (A.51) imply that

$$\begin{aligned} \lambda ^{H}({\hat{t}})-\lambda ^{L}({\hat{t}})=-K<0. \end{aligned}$$

(A.52)

Note that the function $t \mapsto \lambda ^{H}(t)-\lambda ^{L}(t)$ is then an exponentially decreasing function, obtains strictly negative values, and

$$\begin{aligned} \lambda ^{H}(t)-\lambda ^{L}(t) \in (-\,K,0) \quad {\text {for all}} \quad t \in [0,{\hat{t}}). \end{aligned}$$

(A.53)

This and Eqs. (A.50) and (A.51) imply

$$\begin{aligned} -C'(q^{H}(t))+C'(q^{L}(t))=\lambda ^{H}(t)-\lambda ^{L}(t)+K>0 \quad {\text {for all}} \quad t \in [0,{\hat{t}}). \end{aligned}$$

(A.54)

Hence

$$\begin{aligned} q^{L}(t)>q^{H}(t) \quad {\text {for all}} \quad t \in [0,{\hat{t}}). \end{aligned}$$

(A.55)

If ${\hat{t}}$ is unique, then $q^{L}(t)>q^{H}(t)$ for all t except at $t={\hat{t}}$. This clearly implies that the total extraction for mine L is strictly greater than $x_0$, which is impossible. Suppose that ${\hat{t}}$ is not unique. Then Eq. (A.52) holds for at least two time instants, which contradicts the fact that the function $\lambda ^{H}(t)-\lambda ^{L}(t)$ decreases exponentially.

Consider the possibility that $t^{L}=t^{H}$. Note that continuity implies that Eqs. (A.50) and (A.51) hold also at $t^{L}$. Because $t^{L}=t^{H}$ and $t^{H}<T$, there are two possibilities: either the extraction paths cross at some instant of time smaller than $t^{L}$, or the paths don’t cross. If they cross, then Eq. (A.52) holds for two time instants, which again contradicts the fact that the function $\lambda ^{H}(t)-\lambda ^{L}(t)$ decreases exponentially. If they don’t cross, then one of the mines is not exhausted, which is a contradiction.

(ii) Assume on the contrary to the claim that $t^{L}>t^{H}$ or $t^{L}<t^{H}$. First, let $t^{L}>t^{H}$. Following the reasoning of Part (i), Eq. (A.52) implies that $\lambda ^{H}({\hat{t}})=\lambda ^{L}({\hat{t}})$. Hence $q^{L}(t)=q^{H}(t)$ for all t. But $q^{L}(t)$ is strictly positive on $[t^{H},t^{L})$. These imply that the total extraction under profile $q^{L}(t)$ is strictly greater than $x_0$, which is a contradiction. The proof for the other case, $t^{L}<t^{H}$, is similar.

1.14 Proof of Proposition 13

The proof is divided into two cases, one with $t^H<T$ and the other with $t^H=T$. Consider the first possibility. The proof of Proposition 12 shows that $t^L<t^H$ and that there exists ${\hat{t}}$ such that $q^{L}({\hat{t}})=q^{H}({\hat{t}})$. There is only one such time instant, since otherwise $\lambda ^{H}(t)-\lambda ^{L}(t)$ would not decrease exponentially. Since $t^{L}<t^{H}$, $q^{L}(t)<q^{H}(t)$ for all $t \in ({\hat{t}},t^H]$. This implies that $q^{L}(t)>q^{H}(t)$ for all $t \in [0,{\hat{t}})$.

Consider then the second possibility. If $t^H=T$, then either $t^L<T$ or $t^L=T$. If $t^L<T$, the extraction profiles of the mines must cross, since otherwise mine L is not exhausted. This can occur only once, since otherwise the function $\lambda ^{H}(t)-\lambda ^{L}(t)$ does not evolve exponentially. Then it must be that the extraction path $q^L$ crosses path $q^H$ from the above. Next it is shown that if $t^L=T$, then $q^H(T)>q^L(T)$. To see this, suppose $q^H(T)\le q^L(T)$, and note that $q^H$ cannot always equal $q^L$, since $K>0$. Hence the paths are not the same, but they nevertheless must cross. As in the proof of Proposition 12, there exists ${\hat{t}}$ such that $q^L(t)>q^H(t)$ for all $t \in [0,{\hat{t}})$. But since $q^H(T)\le q^L(T)$, the paths cross twice, which again means that the function $\lambda ^{H}(t)-\lambda ^{L}(t)$ does not decrease exponentially. Hence $q^H(T)>q^L(T)$, and the extraction path $q^L$ crosses path $q^H$ from the above.