Existence of equilibria in exhaustible resource markets with economies of scale and inventories

Abstract

The paper proves the existence of equilibrium in non-renewable resource markets when extraction costs are non-convex and resource storage is possible. Inventories flatten the consumption path and eliminate price jumps at the end of the extraction period, so that market equilibrium becomes possible. We distinguish between two types of solutions, one with immediate and one with delayed buildup of inventories. For both cases, we do not only characterize potential optimal paths but also show that equilibria actually exist under fairly general conditions. It is found that optimum resource extraction involves increasing quantities over a period of time. What is generally interpreted as an indicator of increasing resource abundance is thus perfectly compatible with constant resource stocks.

Appendix: Proof of Proposition 4

In this proof, we check that the necessary conditions of Eqs. (12)–(26) define an equilibrium. To do so, we verify that no deviation is profitable. Deviations can be of three types: (i) the firm may choose to extract but to leave resources below the ground, (ii) the firm may choose not to start the extraction, and (iii) the firm may choose to stop extracting at another date than the optimal date $t_{2}^{*}$. These possible deviations are discussed in “First deviation: extracting but less than $R_{0}$” to “Third deviation: extracting $R_{0}$ but opting for other extraction and sale paths” sections of “Appendix.”

1.1 First deviation: extracting but less than $R_{0}$

We prove that given the price path $P^{*}$, once the extraction has started, the firm will extract everything and leaves no resources below the ground.

We consider a deviation (q, s) corresponding to the extraction of the amount $R\le R_{0}$, while the firm faces the optimal price path $P^{*}$. We denote $t_{2}$ the end of the extraction and $t_{f}$ the end of the sale flow. Remark that the optimal price path $P^{*}$ depends on $R_{0}$ but is independent of R. The plan (q, s) is characterized using a program similar to (3) and involves distinguishing two cases depending on $R_{0}\ge R_{1}^{*}$ or not.

First case: $R_{0}\le R_{1}^{*}$. The intertemporal profit associated with this deviation is:

$$\begin{aligned} J_{R}[q,s]=\int _{0}^{t_{f}}e^{-\delta t}P_{t}^{*}s_{t}\hbox {d}t-\int _{0}^{t_{2}}e^{-\delta t}(f(q_{t})+c_{0})\hbox {d}t, \end{aligned}$$

(39)

where we use the subscript R to highlight the dependence in R and where the controls (q, s) and the optimal price $P^{*}$ can be (indirectly for controls—directly for the price) derived from Eqs. (12)–(18) and can be expressed as follows:

$$\begin{aligned} e^{-\delta t}g(s_{t})&=g(s_{0}),\\ e^{-\delta t}f^{\prime }(q_{t})&=f^{\prime }(q_{0}),\\ \int _{0}^{t_{f}}s_{t}\hbox {d}t&=\int _{0}^{t_{2}}q_{t}\hbox {d}t=R\le R_{0}, \\ e^{-\delta t}P_{t}^{*}&=g(s_{0}^{*}). \end{aligned}$$

Using the above definitions, Eq. (39) simplifies into $J_{R}[q,s]=g(s_{0}^{*})R-\int _{0}^{t_{2}}e^{-\delta t}(f(q_{t})+c_{0})\hbox {d}t$, whose derivation with respect to R yields:

$$\begin{aligned} \frac{\partial J_{R}[q,s]}{\partial R}&=\,g(s_{0}^{*})-e^{-\delta t_{2}}(f(\underline{q})+c_{0})\frac{\partial t_{2}}{\partial R}-\int _{0}^{t_{2}}e^{-\delta t}f^{\prime }(q_{t})\frac{\partial q_{t}}{\partial R}\hbox {d}t\\&=\,g(s_{0}^{*})-e^{-\delta t_{2}}\underline{q}f^{\prime }(\underline{q})\frac{\partial t_{2}}{\partial R}-e^{-\delta t_{2}}f^{\prime }(\underline{q})\int _{0}^{t_{2}}\frac{\partial q_{t}}{\partial R}\hbox {d}t\\&=\,g(s_{0}^{*})-e^{-\delta t_{2}}f^{\prime }(\underline{q}) \end{aligned}$$

since the resource constraint $\int _{0}^{t_{f}}q_{t}\hbox {d}t=R$ implies after derivation $\underline{q}\frac{\partial t_{f}}{\partial R}+\int _{0}^{t_{f}}\frac{\partial q_{t}}{\partial R}\hbox {d}t$.

Since $t_{2}$ is an increasing function of R, $R\mapsto \frac{\partial J_{R}[q,s]}{\partial R}$ admits at most one zero on $[0,R_{0}]$ and $R\mapsto J_{R}[q,s]$ is either maximal in 0 or in $R_{0}$. Once the extraction has started, the firm will extract $R_{0}$. The firm will therefore extract either everything or nothing.

Second case: $R_{0}\ge R_{1}^{*}$. The extraction q is assumed to stop at a date $t_{2}\le t_{2}^{*}$. Controls (q, s) and the optimal price $P^{*}$ can be derived from Eqs. (19)–(26). We distinguish again two cases, depending on $t_{2}\ge t_{1}^{*}$ or not.

Case $t_{2}\le t_{1}^{*}$.

The intertemporal profit $J_{R}[q,s]$ can be expressed as follows:

$$\begin{aligned} J_{R}[q,s]=\int _{0}^{t_{f}}e^{-\delta t}(P_{t}^{*}q_{t}-f(q_{t})-c_{0})\hbox {d}t, \end{aligned}$$

(40)

where q follows:

$$\begin{aligned} \forall t \in [0,t_{f}],\ e^{-\delta t}(P_{t}^{*}-f^{\prime }(q_{t}))&=e^{-\delta t_{f}}(P_{t_{f}}^{*}-f^{\prime }(\underline{q})), \int _{0}^{t_{f}}q_{t}\hbox {d}t&=R. \end{aligned}$$

(41)

The derivation of (40) with respect to R yields:

$$\begin{aligned} \frac{\partial J_{R}[q,s]}{\partial R}&=e^{-\delta t_{f}}(P_{t_{f}}^{*}\underline{q}-f(\underline{q})-c_{0})\frac{\partial t_{f}}{\partial R}\nonumber \\&\quad +\int _{0}^{t_{f}}e^{-\delta t}(P_{t}^{*}-f^{\prime }(q_{t}))\frac{\partial q_{t}}{\partial R}\hbox {d}t\nonumber \\&=e^{-\delta t_{f}}(P_{t_{f}}^{*}-f^{\prime }(\underline{q}))\left( \underline{q}\frac{\partial t_{f}}{\partial R}+\int _{0}^{t_{f}}\frac{\partial q_{t}}{\partial R}\hbox {d}t\right) \nonumber \\&=e^{-\delta t_{f}}(P_{t_{f}}^{*}-f^{\prime }(\underline{q})), \end{aligned}$$

(42)

since the derivation of (41) wrt R implies $\underline{q}\frac{\partial t_{f}}{\partial R}+\int _{0}^{t_{f}}\frac{\partial q_{t}}{\partial R}\hbox {d}t=1$.

Case $t_{2}\ge t_{1}^{*}$.

We have:

$$\begin{aligned} J_{R}[q,s]=\int _{0}^{t_{1}^{*}}e^{-\delta t}(P_{t}^{*}q_{t}-f(q_{t})-c_{0})\hbox {d}t+\int _{t_{1}^{*}}^{t_{f}}e^{-\delta t}P_{t}^{*}s_{t}\hbox {d}t-\int _{t_{1}^{*}}^{t_{2}}e^{-\delta t}(f(q_{t})+c_{0})\hbox {d}t, \end{aligned}$$

(43)

where controls (q,s) verify:

$$\begin{aligned} e^{-\delta t}(P_{t}^{*}-f^{\prime }(q_{t}))&=e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*}-f^{\prime }(q_{1}))\nonumber \\ e^{-\delta t}g(s_{t})&=e^{-\delta t_{1}^{*}}g(q_{1}),\nonumber \\ e^{-\delta t}f^{\prime }(q_{t})&=e^{-\delta t_{1}^{*}}f^{\prime }(q_{1}),\nonumber \\ \int _{t_{1}^{*}}^{t_{f}}s_{t}\hbox {d}t&=\int _{t_{1}^{*}}^{t_{2}}q_{t}\hbox {d}t=R-\int _{0}^{t_{1}^{*}}q_{t}\hbox {d}t. \end{aligned}$$

(44)

The derivation of (43) yields:

$$\begin{aligned} \frac{\partial J_{R}[q,s]}{\partial R}&=\int _{0}^{t_{1}^{*}}e^{-\delta t}(P_{t}^{*}-f^{\prime }(q_{t}))\frac{\partial q_{t}}{\partial R}\hbox {d}t+e^{-\delta t_{1}^{*}}P_{t_{1}^{*}}^{*}\left( 1-\int _{0}^{t_{1}^{*}}\frac{\partial q_{t}}{\partial R}\hbox {d}t\right) \\&\quad -\,e^{-\delta t_{2}}(f(\underline{q})+c_{0})\hbox {d}t\frac{\partial t_{2}}{\partial R}-\int _{t_{1}^{*}}^{t_{2}}e^{-\delta t}f^{\prime }(q_{t})\frac{\partial q_{t}}{\partial R}\hbox {d}t\\&=e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*}-f^{\prime }(q_{t_{1}^{*}}))\int _{0}^{t_{1}^{*}}\frac{\partial q_{t}}{\partial R}\hbox {d}t+e^{-\delta t_{1}^{*}}P_{t_{1}^{*}}^{*}\left( 1-\int _{0}^{t_{1}^{*}}\frac{\partial q_{t}}{\partial R}\hbox {d}t\right) \\&\quad -\,e^{-\delta t_{1}^{*}}f^{\prime }(q_{t_{1}^{*}})\left( \underline{q}\frac{\partial t_{2}}{\partial R}+\int _{t_{1}^{*}}^{t_{2}}\frac{\partial q_{t}}{\partial R}\hbox {d}t\right) \end{aligned}$$

Moreover, we have by derivation of (44), $\underline{q}\frac{\partial t_{2}}{\partial R}+\int _{t_{1}^{*}}^{t_{2}}\frac{\partial q_{t}}{\partial R}\hbox {d}t=1-\int _{0}^{t_{1}^{*}}\frac{\partial q_{t}}{\partial R}\hbox {d}t$, which implies:

$$\begin{aligned} \frac{\partial J_{R}[q,s]}{\partial R}=e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*}-f^{\prime }(q_{t_{1}^{*}}))=e^{-\delta t_{f}}(P_{t_{f}}^{*}-f^{\prime }(\underline{q})) \end{aligned}$$

(45)

Conclusion of the second case $R_{0}\ge R_{1}^{*}$.

From (42) and (45), we deduce that for any $R\in [0,R_{0}]$ and thus for any $t_{2}\in [0,t_{2}^{*}]$, we have $\frac{\partial J_{R}[q,s]}{\partial R}=e^{-\delta t_{f}}(g(q_{t_{f}}^{*})-f^{\prime }(\underline{q}))$. Since $\pi (q_{1}^{*})>0$, we know from Proposition 3 that $t\mapsto P_{t}^{*}$ is increasing. Thus, $R\mapsto \frac{\partial J_{R}[q,s]}{\partial R}$ admits at most one zero for $R\in [0,R_{0}]$ and $R\mapsto J_{R}[q,s]$ is maximal either for $R=0$ ($t_{2}=0$, i.e., no extraction) or $R=R_{0}$ ($t_{2}=t_{2}^{*}$, i.e., full extraction). Once the extraction has started, the firm will extract everything and leaves no resources below the ground.

1.2 Second deviation: not starting the extraction

We now check that not starting the extraction cannot be a possible deviation. We denote $J_{R_{0}}[q^{*},s^{*}]$ the intertemporal profit associated with the extraction of $R_{0}$, when the firm faces the price path $P^{*}$ and follows the plan $(q^{*},s^{*})$. Note that in this case (compared to $R\mapsto J_{R}$), both the optimal plan $(q^{*},s^{*})$ and the price $P^{*}$ depend on $R_{0}$. These quantities are defined in Eqs. (12)–(18).

Since no extraction corresponds to a zero intertemporal profit: $J_{R_{0}=0}[q^{*},s^{*}]=0$, we need to prove that $J_{R_{0}}[q^{*},s^{*}]>0$ for any $0<R_{0}\le R_{1}^{*}$.

The intertemporal profit can also be expressed as follows:

$$\begin{aligned} J_{R_{0}}[q^{*},s^{*}]=g(s_{0}^{*})R_{0}-\int _{0}^{t_{2}^{*}}e^{-\delta t}(f(q_{t}^{*})+c_{0})\hbox {d}t \end{aligned}$$

Since $e^{-\delta t}f^{\prime }(q_{t}^{*})=e^{-\delta t_{2}^{*}}f^{\prime }(\underline{q})=f^{\prime }(q_{0}^{*})$ for any $0\le t\le t_{2}^{*}$, we obtain:

$$\begin{aligned} J_{R_{0}}[q^{*},s^{*}]&=\frac{1}{\delta }(g(s_{0}^{*})\delta R_{0}+\varphi (q_{0}^{*})-\varphi (\underline{q})),\\ \text {where: }\varphi (q)&=qf^{\prime }(q)-f(q). \end{aligned}$$

Since $\delta R_{0}=\int _{q_{0}^{*}}^{\underline{q}}u\frac{f^{\prime \prime }(u)}{f^{\prime }(u)}du$, we have:

$$\begin{aligned} J_{R_{0}}[q^{*},s^{*}]=\frac{1}{\delta }(g(s_{0}^{*})\int _{q_{0}^{*}}^{\underline{q}}u\frac{f^{\prime \prime }(u)}{f^{\prime }(u)}du+\varphi (q_{0}^{*})-\varphi (\underline{q})). \end{aligned}$$

(46)

Using that f is convex, we obtain:

$$\begin{aligned} \int _{q_{0}^{*}}^{\underline{q}}u\frac{f^{\prime \prime }(u)}{f^{\prime }(u)}du&\ge \frac{1}{f^{\prime }(\underline{q})}\int _{q_{0}^{*}}^{\underline{q}}uf^{\prime \prime }(u)du\\&\ge \frac{\varphi (\underline{q})-\varphi (q_{0}^{*})}{f^{\prime }(\underline{q})}\ge \frac{\varphi (\underline{q})-\varphi (q_{0}^{*})}{g(\underline{q})}, \end{aligned}$$

where the last equality comes from the fact that $\pi (\underline{q})\ge 0$ and $\underline{q}\ge q_{0}^{*}$. We deduce:

$$\begin{aligned} J_{R_{0}}[q^{*},s^{*}]\ge \frac{1}{\delta }\frac{g(s_{0}^{*})-g(\underline{q})}{g(\underline{q})}(\varphi (\underline{q})-\varphi (q_{0}^{*}))\ge 0, \end{aligned}$$

where the inequality sign is strict for any $R_{0}>0$.

To conclude, we define $\overline{R}_{a}=\inf \{R_{0}\ge R_{1}^{*},\,J_{R_{0}}[q^{*},s^{*}]>0\}$. Using our above result, we deduce that $\overline{R}_{a}>R_{1}^{*}$ by continuity, which proves the result.

1.3 Third deviation: extracting $R_{0}$ but opting for other extraction and sale paths

We prove that there exists a resource level $\overline{R}_{b}>R_{1}^{*}$ (possibly equal to infinity), such that if $R_{0}<\overline{R}_{b}$, a firm extracting $R_{0}$ always follows the optimal extraction plan $(q^{*},s^{*})$ and no deviation is profitable. The proof involves two steps: (i) we start with proving that there is no deviation when $R_{0}\le R_{1}^{*}$ and (ii) we find a threshold $\overline{R}_{b}>R_{1}^{*}$ such that no deviation holds when $R_{0}<\overline{R}_{b}$.

No deviation when $R_{0}\le R_{1}^{*}$.

We denote (q, s) the optimal plan of the firm. We denote $\tau _{2}$ the date at which the extraction q stops. The plan (q, s) is characterized by equation similar to (12)–(17). We distinguish two cases depending on whether $\tau _{2}$ is smaller than $t_{f}^{*}$ or not.

The extraction stops before $t_{f}^{*}$: $\tau _{2}\le t_{f}^{*}$. The extraction plan is characterized as follows:

$$\begin{aligned} \forall t\in [0,\tau _{2}],\ f^{\prime }(q_{t})=f^{\prime }(\underline{q})e^{\delta (t-\tau _{2})}. \end{aligned}$$

We know that if $\tau _{2}=t_{2}^{*}$ or equivalently $q_{0}=q_{0}^{*}$, we have $q_{t}=q_{t}^{*}$. Let us consider $\tau _{2}\mapsto \int _{0}^{\tau _{2}}q_{t}\hbox {d}t$. After a change of variable and using resource constraint in Eq. (17), we have:

$$\begin{aligned} \int _{0}^{\tau _{2}}q_{t}\hbox {d}t-R_{0}=\frac{1}{\delta }\int _{q_{0}}^{q_{0}^{*}}u\frac{f^{\prime \prime }(u)}{f^{\prime }(u)}du. \end{aligned}$$

The resource constraint therefore imposes $q_{0}=q_{0}^{*}$ and therefore $q_{t}=q_{t}^{*}$.

The extraction stops after $t_{f}^{*}$: $\tau _{2}>t_{f}^{*}$. The extraction plan q is characterized as follows:

$$\begin{aligned} \forall t\in [0,t_{f}^{*}],\ f^{\prime }(q_{t})&=f^{\prime }(q_{t_{f}^{*}})e^{\delta (t-t_{f}^{*})},\\ \forall t\in [t_{f}^{*},\tau _{2}],\ f^{\prime }(q_{t})&=g(0)+(f^{\prime }(\underline{q})-g(0))e^{\delta (t-\tau _{2})}. \end{aligned}$$

Since $f^{\prime }(\underline{q})-g(0)\le -\pi (\underline{q})<0$, $q_{t}$ is decreasing over $[t_{f}^{*},\tau _{2}]$. It implies that if $\tau _{2}>t_{f}^{*}$, then $q_{t_{f}^{*}}>\underline{q}$ and for any $t\in [0,t_{f}^{*}]$, $f^{\prime }(q_{t})>f^{\prime }(q_{t}^{*})$: The resource constraint cannot hold when $\tau _{2}>t_{f}^{*}$.

Conclusion for the extraction. The extraction must therefore stop at date $\tau _{2}=t_{2}^{*}$, which implies that the firm must follow the optimal extraction: $\forall t\in [0,t_{2}^{*}],\ q_{t}=q_{t}^{*}$.

The inventory. We assume that the inventory is exhausted at date $\tau _{f}$ (remark: we do not prevent $q=s$ and $\tau _{f}=t_{2}^{*}$). Since $e^{-\delta t}P_{t}^{*}$ is decreasing for $t\ge t_{f}^{*}$, it is optimal for the firm to exhaust inventories before $t_{f}^{*}$: $\tau _{f}\le t_{f}^{*}$.

We denote $J_{R_{0}}[q,s]$ the intertemporal profit associated with the plan (q, s).

$$\begin{aligned} J_{R_{0}}[q,s]=\int _{0}^{\tau _{f}}e^{-\delta t}P_{t}^{*}s_{t}\hbox {d}t-\int _{0}^{t_{2}^{*}}(f(q_{t})+c_{0})\hbox {d}t. \end{aligned}$$

Using that $q=q^{*}$ and $e^{-\delta t}P_{t}^{*}=e^{-\delta t_{f}^{*}}g(0)$ for $0\le t\le t_{f}^{*}$, we deduce that the difference with the intertemporal profit derived from the optimal extraction plan $J_{R_{0}}[q,s]-J_{R_{0}}[q^{*},s^{*}]$ can be expressed as follows:

$$\begin{aligned} J_{R_{0}}[q,s]-J_{R_{0}}[q^{*},s^{*}]=e^{-\delta t_{f}^{*}}g(0)\left( \int _{0}^{\tau _{f}}s_{t}\hbox {d}t-\int _{t_{1}^{*}}^{t_{f}^{*}}s_{t}^{*}\hbox {d}t\right) =0, \end{aligned}$$

because of resource constraints. Therefore, the deviation never strictly dominates the initial allocation when $R_{0}\le R_{1}^{*}$.

Ruling out deviations when $R_{0}\ge R_{1}^{*}$

Dealing with the case where $R_{0}\ge R_{1}^{*}$ is more complex and involves three steps: (i) we show that no storage occurs before date $t_{1}^{*}$; (ii) we show that for a deviation to hold, the extraction must stop after date $t_{1}^{*}$; and (iii) for deviations whose extraction stops after date $t_{1}^{*}$, we show that they cannot occur provided that $R_{0}\le \overline{R}_{b}$, where $\overline{R}_{b}>R_{1}^{*}$ is a threshold that we define.

We consider that a firm facing the optimal price path $P^{*}$ defined in (26) chooses the plan (q, s). We assume that the extraction stops at date $\tau _{2}$.

No storage before date $t_{1}^{*}$.

We start with the following lemma.

Lemma 1

(No storage before) $t_1^*$ Given a price path $P^{*}$, it is never optimal for the firm to store before date $t_{1}^{*}$.

Proof

Using the expression (26) of $P^{*}$ together with (19), we obtain for any $t\in [0,t_{1}^{*}]$:

$$\begin{aligned} e^{-\delta t}P_{t}^{*}&=e^{-\delta t}g(\pi ^{-1}(e^{\delta (t-t_{1}^{*})}\pi (q_{1}^{*})))\\&=e^{-\delta t}f^{\prime }(\pi ^{-1}(e^{\delta (t-t_{1}^{*})}\pi (q_{1}^{*})))+e^{-\delta t_{1}^{*}}\pi (q_{1}^{*}) \end{aligned}$$

Since $t\mapsto e^{-\delta t}f^{\prime }(\pi ^{-1}(e^{\delta (t-t_{1}^{*})}\pi (q_{1}^{*})))$ is a decreasing function (product of two positive decreasing functions), the function $t\mapsto e^{-\delta t}P_{t}^{*}$ is decreasing over $[0,t_{1}^{*}]$. It implies that the firm prefers to sell as much as possible, rather than to save. There is therefore no storage before $t_{1}^{*}$. $\square$

No deviation when the extraction stops before date $t_{1}^{*}$: $\tau _{2}\le t_{1}^{*}$.

Since there is no storage before date $t_{1}^{*}$ (Lemma 1), using a similar technique as for the problem (4) of optimal extraction without storage implies that the extraction path q is such that $e^{-\delta t}(f^{\prime }(q_{t})-P_{t}^{*})$ is constant for any $t\in [0,\tau _{2}]$:

$$\begin{aligned} f^{\prime }(q_{t})-P_{t}^{*}&=e^{\delta (t-\tau _{2})}(f^{\prime }(\underline{q})-P_{\tau _{2}}^{*}) \end{aligned}$$

Since $e^{-\delta t}(f^{\prime }(q_{t}^{*})-P_{t}^{*})$ is also constant for any $t\in [0,\tau _{2}]$ (Eq. 19), we have:

$$\begin{aligned} f^{\prime }(q_{t})= f^{\prime }(q_{t}^{*}) +e^{\delta (t-\tau _{2})}(f^{\prime }(\underline{q}) -f^{\prime }(q_{\tau _{2}}^{*})). \end{aligned}$$

(47)

Using definition (47) of the deviation $q_{t}$ for $t\in [0,\tau _{2}]\subset [0,t_{1}^{*}]$, we obtain:

$$\begin{aligned} \int _{0}^{\tau _{2}}q_{t}\hbox {d}t =\int _{0}^{\tau _{2}}f^{\prime -1}(f^{\prime }(q_{t}^{*}) +e^{\delta (t-\tau _{2})}(f^{\prime }(\underline{q}) -f^{\prime }(q_{\tau _{2}}^{*}))\hbox {d}t. \end{aligned}$$

(48)

We know that for any $t\in [0,t_{1}^{*}]$, $e^{-\delta t}(P_{t}^{*}-f^{\prime }(q_{t}^{*}))$ is constant. We have already proved (see proof of Lemma 1) that $t\mapsto e^{-\delta t}P_{t}^{*}$ is decreasing on $[0,t_{1}^{*}]$. Therefore, $t\mapsto -e^{-\delta t}f^{\prime }(q_{t}^{*})$ is increasing. This implies that $\tau _{2}\mapsto f^{\prime -1}(f^{\prime }(q_{t}^{*}) +e^{\delta (t-\tau _{2})}(f^{\prime }(\underline{q})-f^{\prime }(q_{\tau _{2}}^{*}))$ is increasing and strictly positive on $[0,t_{1}^{*}]$. From (48), we deduce that $\tau _{2}\mapsto \int _{0}^{\tau _{2}}q_{t}\hbox {d}t$ is strictly increasing on $[0,t_{1}^{*}]$. It implies that if we guarantee that $\int _{0}^{t_{1}^{*}}q_{t}\hbox {d}t<R_{0}$, no deviation is possible if $\tau _{2}\le t_{1}^{*}$.

Deviations whose extraction stops after $t_{1}^{*}$: $\tau _{2}>t_{1}^{*}$.

Using a similar technique as in the optimal control problem (3), we obtain that the extraction q is defined as follows:

$$\begin{aligned} {\left\{ \begin{array}{ll} \forall t\in [0,t_{1}^{*}],&{} f^{\prime }(q_{t}) =f^{\prime }(q_{t}^{*})+e^{\delta (t-t_{1}^{*})}(f^{\prime }(q_{1})-f^{\prime }(q_{1}^{*})),\\ \forall t\in [t_{1}^{*},\tau _{2}],&{} f^{\prime }(q_{t}) =e^{\delta (t-t_{1}^{*})}f^{\prime }(q_{1})=e^{\delta (t-\tau _{2})}f^{\prime }(\underline{q}). \end{array}\right. } \end{aligned}$$

(49)

We conclude the proof in two steps. First, we define $\overline{R}_{b}$ and shows that it exists and is strictly larger than $R_{1}^{*}$. Second, we show that no deviation exists when the resource level is below $\overline{R}_{b}$.

Definition and existence of $\overline{R}_{b}$.

We define $\overline{R}_{b}\in {\mathbb {R}}^{+}\cup \{ \infty \}$ as follows:

$$\begin{aligned} \overline{R}_{b}&=\sup \left\{ R_{0}\ge R_{1}^{*}|\exists t_{2}^{**}\ge t_{2}^{*},\ \left\{ \tau _{2}\ge t_{1}^{*}|\int _{0}^{\tau _{2}}q_{t}\hbox {d}t=R_{0}\right\} =\{t_{2}^{*},t_{2}^{**}\} \right\} . \end{aligned}$$

(50)

Notice that we allow for $\overline{R}_{b}$ to be infinite. Note that definition (50) hides several implicit dependencies. First, as defined in (49), the plan q depends on $\tau _{2}$. Second, q also depends on $R_{0}$ through $t_{1}^{*}$ and $q_{t}^{*}$. Finally, both dates $t_{1}^{*}$ and $t_{2}^{*}$ depends on $R_{0}$.

The meaning of the level $\overline{R}_{b}$ is the following one. For any resource level $R_{0}$ smaller than $\overline{R}_{b}$, there are at most two dates ($t_{2}^{*}$ and $t_{2}^{**}$) at which the resource constraint holds. It implies first that it is not possible to find a deviation which extracts resources faster than the optimal extraction $q^{*}$: The other possible deviation will stop extracting at $t_{2}^{**}$ after the date $t_{2}^{*}$. Because of the time preference and of the fixed cost $c_{0}$, the firm may prefer deviations that extract faster than $q^{*}$ to $q^{*}$. It is therefore important to rule them out.

Lemma 2

(Existence of) $\overline{R}_{b}$ The maximal resources level $\overline{R}_{b}$ defined in (50) exists, is strictly larger than $R_{1}^{*}$ and may be equal to infinity.

Proof

First, it is clear from the case $R_{0}\le R_{1}^{*}$ treated above that the set

$$\begin{aligned} \left\{ R_{0}\ge R_{1}^{*}|\exists t_{2}^{**}\ge t_{2}^{*},\ \left\{ \tau _{2}\ge t_{1}^{*}|\int _{0}^{\tau _{2}}q_{t}\hbox {d}t=R_{0}\right\} =\{t_{2}^{*},t_{2}^{**}\} \right\} \end{aligned}$$

is not empty, since $R_{1}^{*}$ belongs to it (in that case, $t_{2}^{*}$ is the unique date).

Second, we can show than whenever $R_{0}-R_{1}^{*}$ is not too large but positive, $\tau _{2}\mapsto \int _{0}^{\tau _{2}}q_{t}\hbox {d}t$ is increasing for $\tau _{2}\le t_{2}^{*}$. We have from (49):

$$\begin{aligned} \int _{0}^{\tau _{2}}q_{t}\hbox {d}t&=\int _{0}^{t_{1}^{*}}q_{t}\hbox {d}t +\int _{t_{1}^{*}}^{\tau _{2}}q_{t}\hbox {d}t\nonumber \\&=\int _{0}^{t_{1}^{*}}q_{t}\hbox {d}t +\int _{q_{1}}^{\underline{q}}u\frac{f^{\prime \prime }(u)}{f^{\prime }(u)}\hbox {d}u=Q(q_{1}) \end{aligned}$$

(51)

with $e^{\delta \tau _{2}} =\frac{f^{\prime }(\underline{q})}{f^{\prime }(q_{1})}$: proving that $\tau _{2}\mapsto \int _{0}^{\tau _{2}}q_{t}\hbox {d}t$ is increasing for $\tau _{2}\in [t_{1}^{*},t_{2}^{*}]$ is equivalent to prove that Q is decreasing for $q_{1}\in [q_{1}^{*},\underline{q}]$. We have

$$\begin{aligned} Q^{\prime }(q_{1})&=\int _{0}^{t_{1}^{*}}\frac{\partial q_{t}}{\partial q_{1}}\hbox {d}t-q_{1}\frac{f^{\prime \prime }(q_{1})}{f^{\prime }(q_{1})}\le t_{1}^{*}\sup _{t\in [0,t_{1}^{*}]}\frac{\partial q_{t}}{\partial q_{1}}-q_{1}\frac{f^{\prime \prime }(q_{1})}{f^{\prime }(q_{1})}. \end{aligned}$$

(52)

From (49), $f^{\prime \prime }(q_{t})\frac{\partial q_{t}}{\partial q_{1}}=e^{\delta (t-t_{1}^{*})}f^{\prime \prime }(q_{1})$, so $0<\sup _{t\in [0,t_{1}^{*}]}\frac{\partial q_{t}}{\partial q_{1}}\le \frac{f^{\prime \prime }(q_{1})}{\inf _{u\in [q_{1},q_{0}]}f^{\prime \prime }(u)}$.

Moreover, from Eqs. (25) and (22), we have:

$$\begin{aligned} R_{0}-R_{1}^{*}= \int _{q_{1}^{*}}^{q_{0}^{*}}u\frac{-\pi ^{\prime }(u)}{\pi (u)}\hbox {d}u>(q_{0}^{*}-q_{1}^{*})\inf _{u\in [q_{1}^{*},q_{0}^{*}]}u\frac{-\pi ^{\prime }(u)}{\pi (u)} \end{aligned}$$

and

$$\begin{aligned} t_{1}^{*}=\frac{1}{\delta }\ln \frac{\pi (q_{1}^{*})}{\pi (q_{0}^{*})}\le \frac{1}{\delta }\ln \frac{\pi (q_{1}^{*})}{\pi \left( q_{1}^{*} +\frac{R_{0}-R_{1}^{*}}{\inf _{u\in [q_{1}^{*},q_{0}^{*}]}u\frac{-\pi ^{\prime }(u)}{\pi (u)}}\right) }. \end{aligned}$$

We deduce that

$$\begin{aligned} Q^{\prime }(q_{1})\le \frac{1}{\delta }\ln \left( \frac{\pi (q_{1}^{*})}{\pi \left( q_{1}^{*} +\frac{R_{0}-R_{1}^{*}}{\inf _{u\in [q_{1}^{*},q_{0}^{*}]}u\frac{-\pi ^{\prime }(u)}{\pi (u)}}\right) }\right) \frac{f^{\prime \prime }(q_{1})}{\inf _{u\in [q_{1},q_{0}]}f^{\prime \prime }(u)} -q_{1}\frac{f^{\prime \prime }(q_{1})}{f^{\prime }(q_{1})}, \end{aligned}$$

which can be made negative for any $q_{1}\in [q_{1}^{*},\underline{q}]$ provided that $R_{0}-R_{1}^{*}$ is made sufficiently small. Remark that it is not fully obvious because $q_{0}$ and $q_{0}^{*}$ depend (negatively) on $R_{0}-R_{1}^{*}$. However, since both $q_{0}$ and $q_{0}^{*}$ decrease with $R_{0}-R_{1}^{*}$, it is possible to assume without loss of generality that for not too large $R_{0}-R_{1}^{*}$, $q_{0}$ and $q_{0}^{*}$ are bounded by some $\overline{q}_{0}$ and $\overline{q}_{0}^{*}$ independent of $R_{0}-R_{1}^{*}$, such that $\inf _{u\in [q_{1},q_{0}]}f^{\prime \prime }(u)\ge \inf _{u\in [q_{1},\overline{q}_{0}]}f^{\prime \prime }(u)$. This latter lower bound is independent of $R_{0}-R_{1}^{*}$ and strictly positive since $f^{\prime \prime }$ is continuous and $[q_{1},\overline{q}_{0}]$ compact. We can then obtain an explicit bound on $R_{0}-R_{1}^{*}$.

When $Q^{\prime }(q_{1})$ is negative for any $q_{1}\in [q_{1}^{*},\underline{q}]$, $\tau _{2}=t_{2}^{*}$ is the first date at which $\int _{0}^{\tau _{2}}q_{t}\hbox {d}t=R_{0}$. We further need to prove that if $R_{0}-R_{1}^{*}$ is not too large, there exists at most one other date $t_{2}^{**}\ge t_{2}^{*}$ (or another $q_{1}^{**}<q_{1}^{*}$), such that $Q(q_{1}^{**})=R_{0}$. From (52), the sign of $Q^{\prime }$ is determined by the following expression $Q_{1}(q_{1})=f^{\prime }(q_{1})\int _{0}^{t_{1}^{*}}\frac{e^{\delta (t-t_{1}^{*})}}{f^{\prime \prime }(q_{t})}\hbox {d}t-q_{1}$. Provided that $f^{\prime \prime }$ is continuously derivable, one can use the same technique as above and prove that when $R_{0}-R_{1}^{*}$ is not too large, $Q_{1}(q_{1})$ is decreasing for any $q_{1}\le q_{1}^{*}$, which guarantees that $Q_{1}$ admits at most one zero, smaller than $q_{1}^{*}$. Since $Q(q_{1}^{*})=R_{0}$, there therefore exists at most one other $q_{1}^{**}<q_{1}^{*}$, such that $Q(q_{1}^{**})=R_{0}$.

We therefore deduce that $\overline{R}_{b}$ exists and is such that $\overline{R}_{b}>R_{1}^{*}$. $\square$

No deviation when $R_{0}\le \overline{R}_{b}$.

From definition (50) of $\overline{R}_{b}$, it is sufficient to prove that the (possible) deviation that stops at date $t_{2}^{**}$ is dominated by the optimal extraction stopping at $t_{2}^{*}$. We formalize in the following lemma.

Lemma 3

(Optimality of) $q^*$ If there exists another extraction plan $q^{**}$ stopping at date $t_{2}^{**}>t_{2}^{*}$ such that $\int _{0}^{t_{2}^{**}}q_{t}^{**}\hbox {d}t=R_{0}$, then the plan $q^{**}$ is never preferred to the optimal plan $q^{*}$.

Proof

We still consider a plan (q, s), in which the extraction stops at date $\tau _{2}\ge t_{2}^{*}$ and the sale stops at date $\tau _{f}\ge \tau _{2}$. The intertemporal profit J[q, s] can be expressed as follows:

$$\begin{aligned} J[q,s]&=\int _{0}^{t_{1}^{*}}e^{-\delta t}(P_{t}^{*}q_{t}-f(q_{t})-c_{0})\hbox {d}t+e^{-\delta t_{1}^{*}}P_{t_{1}^{*}}^{*}\int _{t_{1}^{*}}^{\tau _{f}}s_{t}\hbox {d}t\\ {}&\quad -\int _{t_{1}^{*}}^{\tau _{2}}e^{-\delta t}(f(q_{t})+c_{0})\hbox {d}t \end{aligned}$$

We know from (49) that $e^{-\delta t}P_{t}^{*}=e^{-\delta t}f^{\prime }(q_{t})+e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*}-f^{\prime }(q_{1}))$ and from the resource constraint that $\int _{t_{1}^{*}}^{\tau _{f}}s_{t}\hbox {d}t=\int _{t_{1}^{*}}^{\tau _{2}}q_{t}\hbox {d}t$. We deduce:

$$\begin{aligned} J[q,s]&=e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*}-f^{\prime }(q_{1}))\int _{0}^{t_{1}^{*}}q_{t}\hbox {d}t+\int _{0}^{t_{1}^{*}}e^{-\delta t}(f^{\prime }(q_{t})q_{t}-f(q_{t})-c_{0})\hbox {d}t \\&-\int _{t_{1}^{*}}^{\tau _{2}}e^{-\delta t}(f(q_{t})+c_{0})\hbox {d}t+e^{-\delta t_{1}^{*}}P_{t_{1}^{*}}^{*}\int _{t_{1}^{*}}^{\tau _{2}}q_{t}\hbox {d}t\\&=e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*} -f^{\prime }(q_{1}))\int _{0}^{\tau _{2}}q_{t}\hbox {d}t+\int _{0}^{t_{1}^{*}}e^{-\delta t}(f^{\prime }(q_{t})q_{t}-f(q_{t})-c_{0})\hbox {d}t \\&-\int _{t_{1}^{*}}^{\tau _{2}}e^{-\delta t}(f(q_{t})+c_{0})\hbox {d}t-e^{-\delta t_{1}^{*}}f^{\prime }(q_{1})\int _{t_{1}^{*}}^{\tau _{2}}q_{t}\hbox {d}t \end{aligned}$$

Since (49) implies that $e^{-\delta t}f^{\prime }(q_{t})=e^{-\delta t_{1}^{*}}f^{\prime }(q_{1})$ for $t\ge t_{1}^{*}$, we have using the expression (51) of $Q(q_{1})$:

$$\begin{aligned} J[q,s]&=e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*} -f^{\prime }(q_{1}))Q(q_{1})+\int _{0}^{\tau _{2}}e^{-\delta t}(f^{\prime }(q_{t})q_{t}-f(q_{t})-c_{0})\hbox {d}t, \end{aligned}$$

where this expression is valid no matter the value of $Q(q_{1})$. The derivation with respect to $q_{1}$ yields:

$$\begin{aligned} \frac{\partial J[q,s]}{\partial q_{1}}=&e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*} -f^{\prime }(q_{1}))Q^{\prime }(q_{1})-e^{-\delta t_{1}^{*}}f^{\prime \prime }(q_{1})Q(q_{1})\\&+\int _{0}^{\tau _{2}}e^{-\delta t}f^{\prime \prime }(q_{t})q_{t}\frac{\partial q_{t}}{\partial q_{1}}\hbox {d}t, \end{aligned}$$

since $q_{\tau _{2}}=\underline{q}$ and $f^{\prime }(\underline{q})\underline{q}-f(\underline{q})=c_{0}$. Moreover, from (49), we have for any $t\in [0,\tau _{2}]$, $e^{-\delta t}f^{\prime \prime }(q_{t})\frac{\partial q_{t}}{\partial q_{1}}=e^{-\delta t_{1}^{*}}f^{\prime \prime }(q_{1})$, which implies:

$$\begin{aligned} \frac{\partial J[q,s]}{\partial q_{1}}&=e^{-\delta t_{1}^{*}}(P_{t_{1}^{*}}^{*} -f^{\prime }(q_{1}))Q^{\prime }(q_{1}). \end{aligned}$$

(53)

Assume that there exists another plan $q^{**}$ whose terminal date is $t_{2}^{**}>t_{2}^{*}$ and such that $\int _{0}^{t_{2}^{**}}q_{t}^{**}\hbox {d}t=R_{0}$. Noting $q_{1}^{**}=q_{t_{1}^{*}}^{**}$, we have from (49), $q_{1}^{**}<q_{1}^{*}$, since $t_{2}^{**}>t_{2}^{*}$. We denote $J^{**}$ ($J^{*}$) the profit associated with $q^{**}$ ($q^{*}$). By integration by parts of (53):

$$\begin{aligned} J^{**}-J^{*}&=e^{-\delta t_{1}^{*}}\int _{q_{1}^{**}}^{q_{1}^{*}}f^{\prime }(q_{1})Q^{\prime }(q_{1})\hbox {d}q_{1}\\&=e^{-\delta t_{1}^{*}}R_{0}(f^{\prime }(q_{1}^{*})-f^{\prime }(q_{1}^{**}))-e^{-\delta t_{1}^{*}}\int _{q_{1}^{**}}^{q_{1}^{*}}f^{\prime \prime }(q_{1})Q(q_{1})\hbox {d}q_{1}, \end{aligned}$$

Moreover, $Q(q_{1})\ge R_{0}$ for any $q_{1}\in [q_{1}^{**},q_{1}^{*}]$. Indeed, $R_{0}=Q(q_{1}^{*})=Q(q_{1}^{**})$: Otherwise, there would exists another $q_{1}$ in $[q_{1}^{**},q_{1}^{*}]$, such that $R_{0}=Q(q_{1})$, which is not possible. Therefore, we deduce (the inequality is strict otherwise Q would be constant and equal to $R_{0}$ over $[q_{1}^{**},q_{1}^{*}]$):

$$\begin{aligned} J^{**}-J^{*}&<e^{-\delta t_{1}^{*}}R_{0}(f(q_{1}^{*})-f(q_{1}^{**}))-R_{0}\int _{q_{1}^{**}}^{q_{1}^{*}}f^{\prime \prime }(q_{1})\hbox {d}q_{1} <0. \end{aligned}$$

We conclude that the extraction $q^{*}$ is strictly preferred to $q^{**}$. $\square$

Lemma 2 guarantees that we cannot find any extraction plan that terminates before $t_{2}^{*}$, while Lemma 3 ensures that the extraction plan $q^{*}$ is preferred to any extraction plan that stops after $t_{2}^{*}$: We deduce that there is no possible deviation for the extraction.