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Deterministic Asymmetric-cost Differential Games for Energy Production with Production Bounds


We study a continuous optimal control problem which models competition in the energy market. Competing agents maximize profits from selling crude oil by determining optimal production rates by solving Hamilton–Jacobi–Bellman (HJB) equations. The HJB equations arise from a differential game between two types of players: a single finite-reserve producer and multiple high-cost infinite-reserve producers. We extend an earlier similar model, deterministic unbounded-production to a bounded-production game, in which we show that the upper (lower) bound decreases (increases) the profit of finite-reserve player and the low-cost opponents and increases (decreases) the profit of high-cost opponents, due to the effects on the finite-reserve player’s exit time and the market price.


Differential Game in Energy Market

This paper presents a stylized model of a world oil market in which various production blocs are characterized by having their own production costs and reserve levels. We consider a world having a number of producers with large reserves which we model as infinite, as well as single producer with a finite reserve. This finite-reserve producer must carefully balance future and current profit levels in its production decisions. The setting is very similar to that of Ledvina and Sircar (2012) [1], which, however, does not allow for constraints on production, which may be physically or economically quite meaningful to be implemented. Therefore, we extend this model by adding new stylized constraints. In our model, producers can, within upper and lower bounds, choose their own production levels. These bounds may arise for geological, technical or political reasons.

Our goal is to study the outcome of the interaction of these oil producing blocs, each optimizing its own production to maximize discounted profit. We use a simple linear price model for the supply–price relationship. Oil-producing blocs must consider both their own traits and those of their opponents to make optimal profits. This constitutes a differential game among the blocs. The mathematical solution to this game yields optimal production strategies and maximum profits for those blocs.

Literature Review

To optimize the profit from the producing energy resources, optimal control methods have been applied to obtain the optimal production strategies. The production strategy for exhaustible resources has long been a popular topic of study. Early research comes from Hotelling (1931) [2], where price and energy quantity follows a deterministic differential equation. Hotelling (1931) [2] derives the dynamic of the price of exhaustible energy, demonstrating this price grows exponentially with the discount rate for a monopolist. Even the earlier work Gray (1914) [3] provides a similar result. Hubbert (1956) [23] provides a bell-shape “Hubbert Curve” for production rate of exhaustible energy. Deshmukh and Pliska (1983) [21] and Kamien and Schwartz (1978) [24] developed the monopolistic optimal production in response to external changes. 

A classical problem is that of how a duopoly or oligopoly manages a renewable common resource. The so-called productive asset problem is to maximize the accumulated profit of each player by deciding the production strategy for each as a control in continuous time. The resulting model is an example of a differential game. Differential game models have been widely applied to the productive asset duopoly problem (see Benchekroun (2003) [4]) and to the productive asset oligopoly problems (see Benchekroun (2008) [5], Benchekroun et al. (2009) [6] and Colombo and Labrecciosa (2013) [7]).

A differential game model is a particular type of continuous-time control problem, where multiple players each must find their own strategies (optimal controls). In our case, the optimal control is the production rate \(q_i(t)\) for player i. Mathematically, the player i aims to compute the objective function in integral form,

$$\begin{aligned} J_i = \sup _{q_i} \int _0^T \pi _i(q_1^*(t), q_2^*(t), \dots , q_i(t), \dots , q_N^*(t))\mathrm {d}t, \end{aligned}$$

where \(\pi _i(\,\cdot \,)\) is player i’s profit. An N-player differential game leads to simultaneous Hamilton–Jacobi–Bellman (HJB) equations. Solving the HJB equations analytically or numerically supplies the optimal production strategy.

Harris et al. (2010) [8] use a “relative prudence” bound of \(\rho = -\frac{QP''(Q)}{P'(Q)}\), where P(Q) is the price as a function of production Q, in static Cournot games to ensure the existence of the Nash equilibrium, where all producers held infinite reserves, and obtain the approximate solution to dynamic games where all producers held low-cost exhaustible resources and high-cost inexhaustible resources. Then, Ludkovski and Sircar (2016) [25]  incorporate this work with R&D (research and development) effect. Ledvina and Sircar (2011) [9] investigate the stochastic Bertrand oligopolistic game, where demand function was individual for each player, and obtain an approximate solution, which measures the substitutability of energy produced by different players.

Ledvina and Sircar (2012) [1] find the solution to oligopolistic games with asymmetric costs in energy market, where only one producer had a exhaustible (finite) reserve and others held renewable (infinite) reserve. They assumed a linear and deterministic price model and computed so-called blockading points quantifying entry or exit prices for high-cost, infinite-reserve, producers. Based on the solution to asymmetric oligopolistic game, Dasarathy and Sircar (2015) [10] added effects of R&D, varying costs, resource discovery and energy policy and computed the numerical solutions to the problem. Furthermore, differential games in energy markets can incorporate stochastic factors (see Brown and Sircar (2016) [15], Ludkovski and Yang (2015) [26]), and be extended to mean-field games (see Campi and Fischer (2018) [16], Chan and Sircar (2015) [17] [18], Chan and Sircar (2017) [19]). 

The Novelty of Our Results

We set constraints on production of the finite-reserve producer based on the asymmetric-cost model from Ledvina and Sircar (2012) [1]. The current paper computes the lower bound of the production for the finite-reserve producer to keep the profit higher than specified value, but we also work on the upper bounds to consider the capacity of production. This concept can be traced back to Simaan and Takayama (1978) [11], which considered a constant maximum capacity of production in a duopoly game and analyzed each firm’s production with this capacity. Benchekroun (2003) [4], for a renewable resource, assumes a production restriction of one player, depending on current asset’s stock, and argues that this production restriction would increase the long-run profit, an intuition-challenging result replicated in this paper. Moreover, Geras’kin and Chkhartishvili (2017) [1222] also incorporated constraints on capacity and competitiveness into a static oligopolistic game with nonlinear cost functions and observe that the combination of the two constraints allows the total production of players to exceed the free market niche.

In this paper, we utilize a linear price model for profits per time unit and explain its rationale in Sec. 2. Section 3 gives the completed results of the N-player asymmetric-cost static and dynamic games from Ledvina and Sircar (2012) [1]. Depending on the solutions to profits with, respectively, unconstrained, upper and lower, production bounds, our research makes the following contributions, as discussed in Sec. 4:

  • Taking the setting of the deterministic problem based on Ledvina and Sircar (2012) [1], we add production constraints to the production/price of the finite-reserve players, assuming that the behavior of the finite-reserve player is controlled by external circumstances.

  • We set a lower bound on production to protect a minimum oil price. An unexpected result shows an increased trend of the total accumulated profit with the minimal-production constraint compared to the unconstrained case.

  • We analyze why the upper and lower bounds on the production of the finite-reserve producer have different directions of impacts on the opponents, depending on their costs of production. The profits of higher-cost opponents decrease with the upper bound, while the lower-cost opponents show an inverse trend. The opposite trends hold for the lower bound.

  • We present the impacts of minimum-profit and maximum-production constraints on oil prices.

Our study will be of interest to those studying competition between energy producing blocs and the related price formation. It also illustrates a somewhat counterintuitive result about the impact of adding constraints to a Nash equilibrium. Although this insight is already present in even simple game theory models, it nonetheless surprises many who see it in this context.

Price Model

Typically, price is a decreasing function of demand and, for a commodity that is difficult to store in large quantities like crude oil, production must approximately meet demand over annual timescales. As such Eq. 5 of Ledvina and Sircar (2012) [1] presents a linear price model widely used by economists and statisticians because it is easy to apply linear regression between price \(P(Q_{\text {total}})\) and total oil production \(Q_{\text {total}}\) to fit data to the model. Mathematically, the price model is

$$\begin{aligned} P(Q_{\text {total}}) = M - \alpha Q_{\text {total}}, \end{aligned}$$

where M and \(\alpha\) are coefficients. It helps to work in nondimensional units,

$$\begin{aligned} \frac{P(Q_{\text {total}})}{M} = 1-\frac{\alpha Q_{\text {total}}}{M}. \end{aligned}$$

We define nondimensional price \(P_0(Q_{\text {total}}) := \frac{P(Q_{\text {total}})}{M}\). Since \(\frac{\alpha }{M}\) is a constant measured by dollar/barrel, we can absorb the constant and control our production through \(q:=\frac{\alpha Q_{\text {total}}}{M}\), instead of \(Q_{\text {total}}\). Therefore, the linear nondimensionalized price model which we will use in all subsequent sections is

$$\begin{aligned} P_0(q)=1-q. \end{aligned}$$

Similarly, the cost of production S, measured in dollars per barrel, may be nondimensionalized by \(s := \frac{S}{M}\) with s taken as the corresponding nondimensional production cost.

Deterministic Game of Finite-reserve producer versus Infinite-reserve producers

We assume a differential game with N players in total, including one player with finite reserve and \(N-1\) players with infinite reserves. We follow the setting of Tsur and Zemel (2003) [13], Lafforgue (2008) [14] and Ledvina and Sircar (2012) [1] and assume that the finite-reserve producer has initial reserve \(x(0) = x\), which depletes until exhausted thus,

$$\begin{aligned} \mathrm {d}x(t) = -q_0(t) \mathbbm {1}_{\left\{ {x(t)>0}\right\} }\mathrm {d}t. \end{aligned}$$

Here, \(q_0(t)\ge 0\) is the rate at which the finite-reserve producer extracts the resource. It is both the rate of depletion and the rate of production of the finite-reserve player.

We let \(q_i(t)\) be the production rates of the infinite-reserve players, where \(i = 1,2,\dots , N-1\). We need not track the reserve for infinite-reserve producers.

The differential game is based on the Nash equilibrium, in which players aim to select their optimal strategy, given that their opponents have selected their own optimal strategies. A mathematically precise definition is provided in Def. 1.

Definition 1

Assume there are N players. Player i has a control \(q_i\in \mathcal {Q}_i\) and a profit function \(J_i(q_i,\mathbf {Q}_{-i})\), where \(\mathbf {Q}_{-i}=(q_1,\dots ,q_{i-1},q_{i+1},\dots ,q_N)\) are the controls of the opponents. The Nash equilibrium is the set of controls \(\mathbf {q}=(q_1^*, \dots , q_n^*)\) such that

$$\begin{aligned} J_i(q_i^*, \mathbf {Q}_{-i}^*)\ge J_i(q_i, \mathbf {Q}_{-i}^*) \end{aligned}$$

for \(i = 1,\dots ,N\) and \(\forall q_i\in \mathcal {Q}_i\).

Remark 1

The Nash equilibrium of Def. 1 means that each player will optimize their own profit function, under the assumption that the opponents have obtained their own optimal strategies. Hence, each cannot profit more by unilaterally changing their own strategy. However, the Nash equilibrium may not result in the largest-profit strategy for any given player, or even the total profit for all players, as illustrated in Sec. 4.2.

From the perspective of control \(q_i\), Def. 2 provides different types of controls for player i.

Definition 2

A closed-loop strategy is a decision rule \(q_i(x_0, x(t), t)\), continuous in t and uniformly Lipschitz in x for each t. This strategy depends both current state x(t) and time t.

By Def. 1, the goal of the finite-reserve producer (player 0) is to find \(q_0(t)\) by maximizing the total profit given \(q_1^*(t),\dots ,q_n^*(t)\) in Eq. 6:

$$\begin{aligned} v(x) = \sup _{q_0\ge 0} \int _0^{\tau } \mathrm {e}^{-rt} q_0(t)\left( 1-q_0(t)-\sum _{i=1}^{N-1}q_i^*(t)-s_0\right) \mathrm {d}t, \end{aligned}$$

given the general objective function in Eq. 1 and price function in Eq. 3. Similarly, the infinite-reserve producer (player \(n\ge 1\)) wants to find \(q_n(t)\) by Eq. 7

$$\begin{aligned} v_n(x) = \sup _{q_n\ge 0} \int _0^{\infty } \mathrm {e}^{-rt} q_n(t)\left( 1-q_n(t)-\sum _{i=0,i\ne n}^{N-1}q_i^*(t)-s_n\right) \mathrm {d}t, \end{aligned}$$

where \(s_0, \dots , s_{N-1}\) are the cost of production, r is the interest rate, and the stopping time of the finite-reserve player is \(\tau := \inf \{ t:x(t) = 0 \}\). We maintain the condition \(q_n(t) \ge 0\) because oil cannot be put back into the ground. These production strategy of \(q_i\) depends on the level of x. As Definition 2 indicates, the strategies of players should be closed loop. With the production, the nondimensionalized price given those value functions is

$$\begin{aligned} P_0(q_0+Q) = 1-q_0-Q \end{aligned}$$

where \(q_0, Q = \sum _{i=1}^{N-1}q_i\) denote the production of finite-reserve player, total production of infinite-reserve players. In the next sections, we would use the nondimensionalized price model in our value functions.

Now, turn to the asymmetric-cost oligopolistic game of Ledvina and Sircar (2012) [1] in which there are \(N>1\) players in this game. In that case, assume that production from opponents with asymmetric costs \(s_1<s_2<\dots<s_{N-1}<1\). Assume \(s_0\) as the lowest cost results in the most complicated calculations. The other cases in which \(s_0\) is not the lowest are simpler to handle.

Since \(\tau\) is the stopping time for the exhaustible energy producer, when \(t>\tau\), the finite-reserve producer will exit the game. So we can divide the profit function of the infinite-reserve producers into two parts:

$$\begin{aligned} \begin{aligned} v_n(x)&= \sup _{q_n} \int _0^{\infty } \mathrm {e}^{-rt} q_n(t)\left( 1-q_n(t)-\sum _{i=0,i\ne n}^{N-1}q_i^*(t)-s_n\right) \mathrm {d}t\\&= \sup _{q_n} \int _0^{\tau } \mathrm {e}^{-rt} q_n(t)\left( 1-q_n(t)-\sum _{i=0,i\ne n}^{N-1}q_i^*(t)-s_n\right) \mathrm {d}t +\frac{1}{r}\mathrm {e}^{-r\tau } G_n, \end{aligned} \end{aligned}$$

where \(G_n\) is the constant equilibrium profit of player n in a static game among infinite resource producers.

Static Game

We first discuss the simple static game with no finite-reserve player and \(N-1\) infinite-reserve players. This case will occur after the finite-reserve player exits the game at \(t>\tau\). This static game involves players with a invariant price function and a constant optimal strategy over time for each player. This is the static Cournot game, created in Cournot (1883) [20]. We must solve this to determine the residual values of the infinite-reserve players.

The profit of player n is

$$\begin{aligned} G_n = q_n^*\left( 1-q_n^*-\sum _{i=1;i\ne n}^{N-1} q_i^* - s_n\right) = \sup _{q_n} q_n\left( 1-q_n-\sum _{i=1;i\ne n}^{N-1} q_i^* - s_n\right) . \end{aligned}$$

By maximizing value \(G_n\) with \(q_n\), the production rate for player n is

$$\begin{aligned} q^*_n = \max \left( \frac{1-\sum _{i=1;i\ne n}^{N-1} q_i^*-s_n}{2},0 \right) , \end{aligned}$$

where \(n = 1,2,\dots ,N-1\).

We present a proposition illustrating the exact number of players.

Proposition 1

Define \(\rho _n := \frac{1+\sum _{i=1}^{n-1}s_i}{n}\) and \(\bar{\rho } := \min \{\rho _i|i = 2, \dots , N\}\). In this \(N-1\)-player game, just \(n-1\) players are active where n is given by

$$\begin{aligned} n = \min \{i|\rho _i = \bar{\rho }, i = 2, \dots , N\}. \end{aligned}$$

Then, players \(1, 2, \dots , n-1\) are active, and players \(n, n+1, \dots , N-1\) are not.


See Proposition 2.1 of Dasarathy and Sircar (2015) [10] in the absence of player 0. \(\square\)

Dasarathy and Sircar (2015) [10] make the assumption \(\rho _N>s_N\), to ensure that all infinite-reserve producer is active in the market. With the assumption, there are \(N-1\) producers in the game, and \(q_i^*\) solves simultaneous linear equations with production

$$\begin{aligned} q_n^* = \frac{1+\sum _{i=1}^{N-1}s_i-Ns_n}{N}, \end{aligned}$$

where we must assume that \(\frac{1+\sum _{i=1}^{N-1}s_i}{N}>s_n\) for all n, in order to ensure that all producers will be active.

The time-unit profit \(G_n = \left( \frac{1+\sum _{i=1}^{N-1}s_i-Ns_n}{N}\right) ^2 = (q_n^*)^2\). Moreover, the total static production is

$$\begin{aligned} Q^* = \sum _{i=1}^{N-1}q^*_i = \frac{N-1-\sum _{i=1}^{N-1}s_i}{N}. \end{aligned}$$

Example 1

We consider an simple example of static game. If a two-player static game takes \(s_1 = 0.05\), \(s_2=0.2\), the Nash equilibrium production strategy in the static game \((q_1^*, q_2^*) = (0.3667, 0.2167)\) with computation given in Appendix 1. We also manually increase the production of player A to be \(q_1 = 0.4\) and obtain the corresponding optimal production for player B, \(q_2 = 0.2\).

In fact, the profit of player i (\(i=1,2\)) as an objective function is,

$$\begin{aligned} \sup _{q_i}q_i(1-q_1-q_2-s_i), \end{aligned}$$

which lead to a simultaneous equation

$$\begin{aligned} \begin{aligned} q_2&= s_1+2q_1-1 \text { for player 1}\\ q_1&= s_2+2q_2-1 \text { for player 2}. \end{aligned} \end{aligned}$$

Performing the similar computation of profits with the setting of \(q_1^* = 0.3667, q_2^* = 0.2167\), and \(q_1 = 0.4, q_2 = 0.2\), we obtain the profits as in Table 1:

Table 1 Profit of this simple game

We plot the movement from the unconstrained case to the constrained case in Fig. 1, which shows that the constraint of \(q_1=0.4\) will also shift the production of player 2. The orange and green contours are the profits of players 1 and 2, respectively. If the production rate of player 2 is not changed, changing player 1’s production rate naturally decreases its profit, according to Def. 1 of the Nash Equilibrium. But, Fig. 1 shows that an increase in player 1’s production sacrifices player 2’s production and profit. Therefore, this production constraint benefits player 1.

Fig. 1

Moving from the unconstrained to the constrained case of the static game. Horizontal axis—player 1 action; yellow contours—player 1 profits. Vertical axis—player 2 action, green contours—player 2 profits. All numbers nondimensional, parameters for the model as in \(s_1 = 0.05\), \(s_2=0.2\)

Therefore, although the Nash equilibrium is \((q_1^*, q_2^*) = (0.3667, 0.2167)\), setting \(q_1 = 0.4\) returns a larger profit of (0.1400, 0.0400) for player 1 (\(0.1400>0.1344\)). Hence, a seemingly counterintuitive result occurs because a constraint on the strategy of player 1 benefits that player’s profit. This counterintuitive conclusion is due to the property of the Nash equilibrium. Unlike the single-player optimization problem, the Nash equilibrium may not give the optimal strategy for any given player but an equilibrium between/among players.

Dynamic Game

To understand the novel result which follows, we quickly revisit and review the work of Ledvina and Sircar (2012) [1]. We discuss the case of dynamic games at \(0<t<\tau\). The term “dynamic” denotes that the strategy for each player may change over time. This is because the finite-reserve player producer at each \(t<\tau\) balances varying current and residual values. We need to consider the production of the finite-reserve producer. The HJB equations for players \(n = 0, 1, \dots , N-1\) are

$$\begin{aligned} \begin{aligned} \sup _{q_0\ge 0} q_0\left( 1-q_0-\sum _{i=1}^{N-1}q_i^*-s_0 -v'\right)&= rv \\ \sup _{q_n\ge 0} q_n\left( 1-q_n-\sum _{i=0, i\ne n}^{N-1}q_i^*-s_0 \right)&= rv + q_0^*v_n'. \end{aligned} \end{aligned}$$

However, not all producers will produce energy at all times during the game. The level of reserve x(t) affects the production of the finite-reserve player, and this production of the finite-reserve player will affect the production of these opponents. Therefore, we define the so-called blockading points for the infinite-reserve producers n.

Definition 3

The blockading point of player n is

$$\begin{aligned} x_b^n = \inf \{x>0:q_n^*(x) = 0\}. \end{aligned}$$

The blockading time for player n is defined to be

$$\begin{aligned} \tau _b^n = \sup \{t>0:q_n^*(x(t)) = 0\}. \end{aligned}$$

We use Fig. 2 to explain the blockading points and blockading times. As the reserve x decreases past \(x_b^{n}\), player n starts production and enters the market. Equation 5 demonstrates that the reserve is monotonically decreasing with regard to time t. So the number of players increases as x(t) passes the blockading points \(x_b^n\).

Fig. 2

Picture explaining blockading points and blockading times

Intuitively, if all opponents have blockading points, it should be that \(x_b^1>x_b^2>\dots >x_b^{N-1}\) because \(s_1<s_2<\dots <s_{N-1}\) by assumption. In other words, producers with higher costs are more easily excluded. Then, in the interval \([x_b^n, x_b^{n-1})\), there are \(n-1\) active infinite-reserve producers. Moreover, the production rate of producers would satisfy

$$\begin{aligned} \begin{aligned} q_0^*&= \frac{1-\sum _{i=1}^{n-1}q^*_i-s_0-v'(x)}{2}\\ q_k^*&= \frac{1-\sum _{i=0, i\ne k}^{n-1}q^*_i-s_k}{2},\\ \end{aligned} \end{aligned}$$

where \(k = 1,2,\dots , n-1\), while \(q_k^* = 0\) for \(k = n,n+1,\dots ,N-1\). So by taking \(s_0+v'(x)\) as the total “cost” for finite-reserve producer and \(v'(x)\) as the “shadow cost” of depleting the reserve, we can easily obtain the production rate for all acting producers in Eq. 21

$$\begin{aligned} \begin{aligned} q_0^*(x)&= \frac{1+\sum _{i=1}^{n-1}s_i-n(s_0+v'(x))}{n+1}\\ q_k^*(x)&= \frac{1+s_0+v'(x)+\sum _{i=1,i\ne k}^{n-1}s_i-n s_k}{n+1}. \end{aligned} \end{aligned}$$

However, if the costs of infinite-reserve producers are low enough, they will not have a blockading point, as they are profitable in all price conditions. Assume that there are \(N-K\) players having blockading points \(\{x_b^{ \, N-1},x_b^{ \, N-2},\dots ,x_b^{ \, K}\}\). We will provide an exact formula for K later. By plugging in the \(q^*_n\)’s, the HJB equation for the finite-reserve producer will be

$$\begin{aligned} rv = \sum _{n=K}^{N} \frac{1}{(n+1)^2}\left( 1+\sum _{i=1}^{n-1}s_i-n(s_0+v')\right) ^2 \mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} } \end{aligned}$$

where \(v(0)=0\), \(x_b^{k-1}:=\infty\) and \(x_b^{N}:=0\). Indeed, Eq. 22 is a piecewise differential equation between two consecutive blockading points \([x_b^{n},x_b^{n-1}]\), w.r.t. the reserve of the finite-reserve producers x.

Lemma 1

Consider the ODE

$$\begin{aligned} {\left\{ \begin{array}{ll} (a-v')^2 = bv\\ v(x_0) = v_0 \end{array}\right. } \end{aligned}$$

where \(v(0)\ge 0\) and \(a,b>0\). The solution to the ODE is

$$\begin{aligned} v(x) = \frac{a^2}{b}(1+W(\theta (x-x_0)))^2 \end{aligned}$$

where \(W(\,\cdot \,)\) is the Lambert-W function, \(\theta (x)=\beta \mathrm {e}^{\beta -\frac{bx}{2a}}\) and \(\beta = -1+\frac{\sqrt{bv_0}}{a}\). The Lambert-W function satisfies \(z=W(z)\mathrm {e}^{W(z)}\) given \(z\ge -e^{-1}\).


See Appendix 2. \(\square\)

Using Lemma 1, assuming \(v_b^{N}=0\) leads to Proposition 2.

Proposition 2

The solution to Eq. 22 is

$$\begin{aligned} v(x) = \sum _{n=K}^{N} \frac{a_n^2}{b_n}(1+W(\theta _n(x-x_b^n)))^2 \mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} } \end{aligned}$$

where coefficients \(a_n = \frac{1+\sum _{i=1}^{n-1}s_i-ns_0}{n}\), \(b_n = r\left( \frac{n+1}{n}\right) ^2\), \(\theta _n(x) = \beta _n\mathrm {e}^{\beta _n-\frac{b_n x}{2a_n}}\) and \(\beta _n = -1+\frac{\sqrt{b_nv(x_b^n)}}{a_n}\). The notation \(a_n, b_n, \theta _n(x), \beta _n\) are used over this whole paper. Moreover, since continuity of v(x) is required at \(x_b^{n-1}\), we must ensure that the terminal condition over \([x_b^{n}, x_b^{n-1})\) satisfies

$$\begin{aligned} v(x_b^{n-1}) = \frac{a_{n}^2}{b_{n}}(1+W(\theta _n(x_b^{n-1}-x_b^{n})))^2 \end{aligned}$$

for \(n = N, N,\dots , K+1\). We can give an explicit solution to v(x) recursively with initial condition \(v(x_b^{N})=0\).

Moreover, by Lemma 1, we can compute the piecewise derivative for v(x) for every interval, which is

$$\begin{aligned} v'(x) = -\sum _{n=K}^{N} a_nW(\theta _n(x-x_b^n))\mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} }. \end{aligned}$$


Over the interval \([x_b^n, x_b^{n-1})\), Eq. 22 can be transformed into

$$\begin{aligned} \frac{r(n+1)^2}{n^2} v = \left( \frac{1+\sum _{i=1}^{n-1}s_i}{n}-s_0-v'\right) ^2 . \end{aligned}$$

Therefore, taking \(a = a_n, b = b_n, v_0 = v(x_b^{n})\) as parametric setting in 1 piecewisely leads to 25. \(\square\)

Blockading Points

Now, we present the explicit formula for the blockading point \(x_b^n\) for \(n=K,K+1, \dots ,N-1\). By the definition of blockading points in Eq. 18 and the formula of \(q_{n-1}^*(x)\) in Eq. 21, we obtain the formula of blockading point,

$$\begin{aligned} \begin{aligned} q_{n-1}^*(x_b^{n-1})&= \frac{1+s_0+v'(x_b^{n-1})+\sum _{i=1}^{n-2}s_i-n s_{n-1}}{n} = 0\\ v'(x_b^{n-1})&= n s_{n-1} - \sum _{i=1}^{n-2}s_i - s_0 -1. \end{aligned} \end{aligned}$$

Define \(\delta _n = (n+1)s_n-(1+s_0+\sum _{i=1}^{n-1}s_i)\). By taking \(v'(x_b^{n-1})\) from Eq. 29, the condition of existing a blockading point for n is \(v'(x_b^{n-1}) = \delta _{n-1}>0\), because v(x) must be an increasing function. In other words, the profit function should increase with the reserve.

As Ledvina and Sircar [1] indicate, intuitively, \(q_k^*(x)\) is continuous w.r.t. x, because \(q_k^*\) adjusts itself with the change of reserve x. Therefore, we require the continuity of derivative \(v'(x) = \sum _{n=1}^N a_nW(\theta _n(x-x_b^n)) \mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} }\) in Eq. 27. This gives

$$\begin{aligned} \lim _{x\rightarrow x_b^{n-1}-0}v'(x) = v'(x_b^{n-1}) = -a_nW(\theta _{n}(x_b^{n-1}-x_b^n)). \end{aligned}$$

Equating Eq. 30 and the \(v'(x)\) in Eq. 29 gives the formula of blockading points in Proposition 3.

Proposition 3

Let \(K = \min \{n:\delta _n>0\}\) and define \(x_b^{K-1}:=\infty\) and \(x_b^{N}:=0\). Then, the infinite-reserve producers \(K,K+1,\dots , N-1\) have blockading points, given by

$$\begin{aligned} x_b^{N-1} = \frac{1}{\mu _N}\left( -1+\frac{\delta _{N-1}}{a_N}-\log \left( \frac{\delta _{N-1}}{a_N} \right) \right) \end{aligned}$$

and for \(n = N-1, N-2, \dots , K+1\),

$$\begin{aligned} x_b^{n-1} = x_b^{n} + \frac{1}{\mu _n}\left( \log \left( \frac{\delta _n}{\delta _{n-1}}\right) -\frac{(n+1)(s_n-s_{n-1})}{a_n} \right) . \end{aligned}$$

where we define \(\mu _n = \frac{b_n}{2a_n} = \frac{r}{2a_n}\left( \frac{n+1}{n}\right) ^2\).

Moreover, the value function at the blockading points \(v(x_b^{n}), n=N-1, N-2, \dots , K+1\) is

$$\begin{aligned} v(x_b^n) = \frac{1}{r}(s_n - s_0 - \delta _{n})^2. \end{aligned}$$

which ensure the continuity of \(v'(x)\) at \(x_b^n\). But the second derivative \(v''(x)\) is not necessarily continuous at \(x_b^{n}\).


See Appendix 3. \(\square\)

This proposition provides an exact formulae for the blockading points. At blockading points \(x_b^n,\) the second derivative \(v''(x)\) is not continuous, intuitively because the production \(q_k^*(x)\), which contains the item \(v'(x),\) decreases with x and suddenly stops at zero after x passes \(x_b^{k}\).

Optimal Production of Each Player

From the definition of Lambert-W function, \(v'(x)\) is a positive and decreasing function. So the production rate function is

$$\begin{aligned} q_0^*(x) = \sum _{n=K}^{N}\frac{na_n\left( 1+W(\theta _n(x-x_b^n))\right) }{n+1}\mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} }. \end{aligned}$$

From the property of \(W(\theta (\,\cdot \,))\ge -1\), we can easily prove that \(q_0^*(x)\ge 0\). Therefore, we confirm that the production of the finite-reserve player \(q_0^*(x)\) must be positive over \(x\in (0,\infty )\) and \(q_0^*(x)\) always depends on x and is a closed-loop strategy. The limit

$$\begin{aligned} \lim _{x\rightarrow 0} q_0^*(x) = \frac{Na_N\left( 1+W(\theta _N(0))\right) }{N+1} = 0. \end{aligned}$$

This addresses that production of the finite-reserve player approaches zero as the reserve \(x\rightarrow 0\). Moreover, \(q^*_0(x)\) is a increasing function. By taking \(x\rightarrow \infty\), the supremum of production is \(q_0^*(\infty )=\frac{1+\sum _{i=1}^{k-1}s_i-ks_0}{k+1}\). In this case, this game becomes a static game with N players with the finite-reverse player becoming infinite reserve.

Remark 2

In fact, the condition on the cost \(s_0\) to keep the production \(q_0^*>0\) in Eq. 34 can be relaxed to \(a_n>0\). By assumption on \(\rho _n\) where \(n=1,2,\dots ,N\) in Sec. 3.1, we can easily derive that \(a_1>a_2>\dots>a_{N}>0\). Therefore, the condition for \(s_0\) is equivalent to

$$\begin{aligned} \rho _N=\frac{1+\sum _{i=1}^{N-1}s_i}{N}>s_0. \end{aligned}$$

Moreover, this condition on \(s_0\) is equivalent to the case in which the finite-reserve producer must be in the game, which becomes a N-player static game as the reserve \(x\rightarrow \infty\).

Over the interval \([x_b^{n}, x_b^{n-1})\), we can also derive the closed-loop strategy of production of opponents by inserting this \(v'(x)\) into Eq. 21,

$$\begin{aligned} q_k^*(x) = \max \left( \frac{1+s_0-a_nW(\theta _n(x-x_b^n))+\sum _{i=1,i\ne k}^{n-1}s_i-ns_k}{n+1},0\right) . \end{aligned}$$

where x is in the interval \([x_b^n,x_b^{n-1})\). Over the interval, the relative value of k and n decides whether player k is blockaded. In particular, when \(k<n\), this production takes positive part and player k is not blockaded. Otherwise, player k is blockaded and \(q_k^*(x) = 0\).

Given the production of each player, we can easily obtain the total optimal production of opponents and all players over the interval \([x_b^{n}, x_b^{n-1})\),

$$\begin{aligned} \begin{aligned}&Q^*(x) = \sum _{i=1}^{N-1}q_i^* = \frac{(n-1)(1+s_0-a_nW(\theta _n(x-x_b^n))-2\sum _{i=1}^{n-1}s_i)}{n+1}\\&q_0^*(x) + Q^*(x) = \frac{n-s_0+a_nW(\theta _n(x-x_b^n))-\sum _{i=1}^{n-1}s_i}{n+1}. \end{aligned} \end{aligned}$$

where we define \(Q^*(x)\) as the total production of all active opponents. Since \(W(\theta _n(x-x_b^n))\) increases with x, the production of opponents, \(q_k^*(x), Q^*(x),\) is decreasing with x, due to increasing production \(q_0^*(x)\). In contrast, the total production \(q_0^*(x)+Q^*(x)\) is overall increasing. Example 2 gives the visual effects of the production, profit and blockading points.

Example 2

Table 2 Parametric setting of Ex. 2 – 10

In all examples which follow in this paper, the set of parameters are listed in Table 2. Given the blockading points \(x_b^n\) in Eq. 31 and 32, as well as the profit function v(x), the production rate of each producer \(q_i^*(x)\) for \(i=0,1,\dots , 4\) is given in the upper-left plot. The vertical dotted lines are the two blockading points \(x_{b}^{4}\) and \(x_{b}^{3}\). In that plot, players 0, 1, 2 hold their production, while players 3, 4 stop their production at \(x > x_b^3 \approx 0.76\) and \(x>x_b^4 \approx 0.07,\) respectively. In other words, the increase of production \(q_0^*(x)\) with x expels players 3, 4 from the market.

The profit function is the upper-right plot of Fig. 3, where we observe that v(x) increases with x, including the marginal effect. The total production of opponents \(Q^*\) and all players \(q_0^*+Q^*\) in Eq. 25 is also displayed. \(Q^*\) is decreasing with x, because increases of x squeeze the market share of opponents. But the total production \(q_0^*+Q^*\) increases with x overall. As a result, the lower-right plot addresses the decreasing trend of energy price. Those results naturally hold because larger x allows the finite-reserve player to account for a greater market share and increases the accumulated production.

Fig. 3

Production rate of each player (upper-left), profit (upper-right), total production (lower-left) and price (lower-right) in the game with parameters as in Table 2

Blockading Time and Optimal Stopping Time

Given initial reserve \(x(0) = x_0\), we compute the remaining reserve x(t) and production rate \(q^*(x(t))\) at time t. By Eq. 34, we obtain the ODE

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathrm {d}x(t) = -\sum _{n=K}^{N}\frac{na_n\left( 1+W(\theta _n(x(t)-x_b^n))\right) }{n+1}\mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} }\mathrm {d}t \\ x(0) = x_0. \end{array}\right. } \end{aligned}$$

Solving the ODE leads to the relationship between time t and reserve x as presented in Proposition 4.

Proposition 4

Assume that \(x_b^l\le x_0<x_b^{l-1}\) and \(x_b^m\le x<x_b^{m-1}\) where \(m \le l\). Then, the relationship between t and x is

$$\begin{aligned} t(x;x_0) = \frac{2(l-l)}{rl}\ln \left( \frac{v'(x_b^{l-1})}{v'(x_0)}\right) +\sum _{n=m+1}^{l-1}\frac{2n}{r(n+1)}\ln \left( \frac{v'(x_b^{n-1})}{v'(x_b^{n})}\right) +\frac{2m}{r(m+1)}\ln \left( \frac{v'(x)}{v'(x_b^{m})}\right) . \end{aligned}$$


See Appendix 4. \(\square\)

Equation 19 defines the blockading time \(\tau _{b}^l\) as the time at which player l starts oil production, while initial reserve satisfies \(x_b^m\le x<x_b^{m-1}\), over which players \(m, m+1, \dots , N-1\) do not yet participate in the market. The blockading times follow from Eq. 40,

$$\begin{aligned} \tau _b^m(x_0) = \frac{2(l-l)}{rl}\ln \left( \frac{v'(x_b^{l-1})}{v'(x_0)}\right) +\sum _{n=m+1}^{l-1}\frac{2n}{r(n+1)}\ln \left( \frac{v'(x_b^{n-1})}{v'(x_b^{n})}\right) \end{aligned}$$

where the initial number of players \(l=m,m+1,\dots ,N-1\). Without loss of generality, define \(\tau _b^0(x_0):=0\). If we set \(m = N\), we can obtain the stopping time \(\tau _N^b(x_0)\).

Moreover, we can compute the inverse function of t(x), the remaining reserve at time t,

$$\begin{aligned} x(t;x_0) = x_b^m+\frac{2a_m}{b_m}\left[ \beta _m+\frac{v'(x_b^{m-1})}{a_m}\mathrm {e}^{\frac{r(m+1)}{2m}(t-\tau _b^{m-1}(x_0))}-\frac{r(m+1)}{2m}(t-\tau _b^{m-1}(x_0))-\ln \left( \frac{-v'(x_b^{m-1})}{a_m\beta _m}\right) \right] \end{aligned}$$

where \(t\in [\tau _{b}^{m-1},\tau _{b}^m)\) and \(t-\tau _b^{m-1}\) is the time remaining after the last blockading point before \(x(t;x_0)\). Therefore, based on this result, the production rate at time t will be

$$\begin{aligned} \begin{aligned} q_0^*(x(t;x_0))&= \frac{ma_m-v'(x_b^{m-1})\mathrm {e}^{\frac{r(m+1)}{2m}(t-\tau _b^{m-1})}}{m+1} \\ q_k^*(x(t;x_0))&= \max \left( \frac{1+s_0+v'(x_b^{m-1})\mathrm {e}^{\frac{r(m+1)}{2m}(t-\tau _b^{m-1})}/m+\sum _{i=1,i\ne k}^{m-1}s_i-ms_k}{m+1},0\right) . \end{aligned} \end{aligned}$$

Example 3

The upper-left plot of Fig.4 gives the stopping time \(\tau _N^b(x)\). The green vertical line in the right plot represents the stopping time for \(\tau _b^3(5)\) and \(\tau _b^4(5)\). As is shown, the stopping time \(\tau (x)\) increases with the initial reserve x, containing a marginal effect. When x becomes sufficiently large, the slope becomes almost linear because \(q_0^*(\infty )\) is a constant.

The production rate \(q_0^*(x(t))\) given initial reserve \(x_0=5\) is given in the upper-right plot. The stopping time is approximately \(t = 34.0\). As the reserve x(t) is depleted, the production rate is decreasing with higher rate. On the other hand, the opponents’ total production rate \(Q^*(x(t))\) shows an opposite increasing trend. Overall, the total production \(q_0^*(x(t))+Q^*(x(t))\) is decreasing. Given the connection between production and price, it is not surprising that the energy price increases with t, as x(t) depletes.

Fig. 4

Stopping time (upper-left), production rate (upper-right) and energy price (lower) in dynamic game, parameters as in Table 2

Profits of Opponents

We finished the discussion on the finite-reserve producer. Now, consider the value function of player k, \(v_k(x)\). Recall Eq. 37 for the closed-loop production rate of player k,

$$\begin{aligned} q_k^*(x) = {\left\{ \begin{array}{ll} \frac{1+s_0-a_nW(\theta _n(x-x_b^n))+\sum _{i=1,i\ne k}^{n-1}s_i-ns_k}{n+1} \text { when } n \ge k\\ 0 \text { Otherwise}. \end{array}\right. } \end{aligned}$$

Inserting this \(q_k^*\) into the HJB Eq. 17 leads to Proposition 5:

Proposition 5

The profit function of player k is a piecewise function over interval \([x_b^n, x_b^{n-1})\)

$$\begin{aligned} v_k(x) = {\left\{ \begin{array}{ll} A_n(x) v_k(x_b^n) \text { when } n<k\\ A_n(x)v_k(x_b^n)+\frac{c_{k,n}^2}{r}(1-A_n(x))-\frac{4a_nc_{k,n}n}{r(n-1)(n+1)}(W(\theta _n(x-x_b^n))-\beta _nA_n(x))\\ \quad \quad \quad -\frac{na_n^2}{r(n+1)^2}(W^2(\theta _n(x-x_b^n))-\beta _n^2A_n(x)) \text { when } n\ge k, \end{array}\right. } \end{aligned}$$

where \(A_n(x):=\left( \frac{W(\theta _n(x-x_b^n))}{\beta _n}\right) ^{\frac{2n}{n+1}}\), \(c_{k,n} := \frac{1+\sum _{i=0}^{n-1}s_i}{n+1}-s_k\) and the initial condition is \(v_k(0)=\frac{1}{r}G_k\), given \(n= N, N-1, \dots , K\).


See Appendix 5. \(\square\)

Example 4

For each player, the plots of profits given \(x\in [0,3]\) are presented in Fig. 5. All of the accumulated profits decreases with the initial reserve x of the finite-reserve player. But the decreasing trends of players 1 and 2 with lower costs are almost linear but not exponential and have higher profits (from 1.04 to 0.88 for player 1, and from 0.87 to 0.73 for player 2), because they are active in the game for arbitrary value of x. In comparison, players 3 and 4 with higher costs have an exponentially decreasing trend when they are expelled from the market (\(x>0.76\) for player 3, and \(x>0.07\) for player 4).

Fig. 5

Profit function for opponents, parameters as in Table 2

Oligopolistic Game with Constrained Production on the Finite-reserve Producer

In this section, we extend the work of Ledvina and Sircar (2012) [1] to include upper and lower production constraints. As indicated in Sec. 1.2, Simaan and Takayama (1978) [11] consider a duopoly model in which each firm has a maximum capacity. They compute some structural properties of the solution without computing it exactly. In this sense similarly, Benchekroun (2003) [4], in a duopoly sharing the same renewable resource, finds production restrictions result in apparently counterintuitive behavior.

We introduce similar constant production constraints and find a similar counterintuitive result in what follows. Additional computations from these results are also included to build intuition of how our production bounds work.

Limited Production for Finite-reserve producer

In this section, we consider the maximal production limit. The finite-reserve producer may encounter a case in which oil production amount is restricted. This case is possible because organizations such as OPEC may make an agreement on production reduction, or extractors have physical limitations. There may also be geological production constraints.

Assume that the maximal production for the finite-reserve producer is a constant c. In this case, the profit function in Eq. 6 changes to

$$\begin{aligned} v_{c,0}(x) = \sup _{0\le q_{c,0}\le c} \int _0^{\tau } \mathrm {e}^{-rt} q_{c,0}(t)\left( 1-q_{c,0}(t)-\sum _{i=1}^{N-1}q_{c,i}^*(t)-s_0\right) \mathrm {d}t. \end{aligned}$$

And the HJB Eq. 17 will become

$$\begin{aligned} \sup _{0\le q_{c,0}\le c} q_{c,0}\left( 1-q_{c,0}-\sum _{i=1}^{N-1}q_{c,i}^*-s_0 -v_{c,0}'\right) = rv_{c,0}. \end{aligned}$$

We discuss Eq. 47 in detail. In the last part of Sec. 3.2.2, we have derived the maximal production \(q^*_0(\infty )\) of finite-reserve producer. So if \(q^*_0(\infty )\le c\), the constraint will not bind. But for \(q^*_0(\infty )> c\), the production will be limited at least some of the time. We define the constraint-touching point \(x_c\) to be the point at \(q^*_0(x_c)=c\). Then, the bounded closed-loop strategy of production rate will be

$$\begin{aligned} q_{c,0}^*(x)={\left\{ \begin{array}{ll} q_0^*(x) \quad \text {if }x\le x_c\\ c \quad \text {if }x>x_c \end{array}\right. }. \end{aligned}$$

Assume that \(x_c\) is on the interval \([x_b^{n_c},x_b^{n_c-1})\), i.e., \(q^*_0(x_b^{n_c})\le c \le q^*_0(x_b^{n_c-1})\). Over this interval, players \(n_c, \dots , N-1\) are blockaded. By Eq. 34, the expression of \(x_c\) is

$$\begin{aligned} x_c = x_b^{n_c} -\frac{2a_{n_c}}{b_{n_c}}\left[ \ln \left( \frac{1}{\beta _{n_c}}\left( -1+\frac{(n_c+1)c}{n_ca_{n_c}} \right) \right) -1+\frac{(n_c+1)c}{n_ca_{n_c}}-\beta _{n_c} \right] \end{aligned}$$

The production of opponents is already computed in Eq. 37. Therefore, with the upper bound of player 0, the closed-loop production of player k becomes

$$\begin{aligned} q_{c,k}^*(x)= {\left\{ \begin{array}{ll} q_k^*(x) \text { when }x \le x_c\\ \max \left( \frac{1-c+\sum _{i=1,i\ne k}^{n_c-1}s_i-(n_c-1)s_k}{n_c},0\right) \text { when }x > x_c \end{array}\right. }. \end{aligned}$$

The lower part (i.e., \(x>x_c\)) indicates that the number of players is \(n_c\), where players \(n_c, \dots , N-1\) are blockaded out.

Summing production of opponents results in the HJB equation for the finite-reserve player,

$$\begin{aligned} rv_{c,0}={\left\{ \begin{array}{ll} \sum _{n=n_c}^{N} \frac{1}{(n+1)^2}\left( 1+\sum _{i=1}^{n-1}s_i-n(s_0+v_{c,0}')\right) ^2 \mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} } \quad \text {when } x\le x_c \\ c\left( \frac{1-c+\sum _{i=1}^{n_c-1}s_i-n_cs_0}{n_c}-v_{c,0}'\right) \quad \text {when } x>x_c. \end{array}\right. } \end{aligned}$$

Therefore, if \(x>x_c\), the HJB equation for the producer is a first-order linear ODE. To simplify, we define \(a_c=\frac{1-c+\sum _{i=1}^{n_c-1}s_i-n_cs_0}{n_c}\). So the solution to Eq. 51 is

$$\begin{aligned} v_{c,0}(x) = {\left\{ \begin{array}{ll} v(x) \quad \text {when }x\le x_c \\ \mathrm {e}^{-\frac{r(x-x_c)}{c}}v(x_c)+\frac{c}{r}a_c\left( 1-\mathrm {e}^{-\frac{r(x-x_c)}{c}} \right) \quad \text {when }x> x_c \end{array}\right. } \end{aligned}$$

Figure 6 explains the effect of the limited-production bound. When \(x\le x_c\), the constraint is inactive and production is identical to unconstrained case. And the blockading points \(x_b^n\) for \(n=N, N-1, \dots , n_c\) are effective. But when \(x>x_c\), the number of players stays at \(n_c\), and blockading points larger than \(x_c\) no longer play a role.

Fig. 6

Picture explaining blockading points and times in the case of limited production

Example 5

Fig. 7

Production (upper-left), energy price (upper-right), profit (lower-left) and profit difference (lower-right) for the limited-production constraint, parameters as in Table 2

The upper-left plot of Fig. 7 presents production rates of the finite-reserve player \(q_c^*(x)\) and the opponents \(Q_c^*(x)\), where we set the limit \(c=0.15\). Corresponding to the upper limit, the red vertical line is the constraint-touching point \(x_c\approx 1.3\). When \(x\le x_c\), the production is identical to the unconstrained case. In contrast, when \(x> x_c\), this constraint is active; hence, the productions \(q_c^*\) and \(Q_c^*\) are constants. Moreover, this maximal-production constraint sets a lower bound of the price at \(P_0(q_c^*+Q_c^*) = 0.490\).

The lower-left plot gives the profit function \(v_{c,0}(x)\), and the lower-right plot compares profits between constrained and unconstrained cases by taking the difference \(v(x)-v_c(x)\). When \(x>x_c\), the curve is bent upward and indicates \(v_0(x)>v_{c,0}(x)\). Therefore, the upper constraint has the effect of reducing profits. This result appears trivial because the production bound directly reduces the profit. However, the stopping time is extended due to the maximal-production bound. Therefore, strictly speaking, the effect of the maximal-production bound on lowering profit per time unit is larger than the effect of extending the production time.

Remaining Reserve and Stopping Time

Assume that the initial reserve is \(x_0\). From the production rate in Eq. 48, we can easily conclude that the remaining reserve \(x_c(t)\) is

$$\begin{aligned} x_c(t;x_0) = {\left\{ \begin{array}{ll} x(t;x_0)\quad \text {when }x_0\le x_c\\ x_0-ct\quad \text {when }x_0>x_c \text { and }0 \le t< \frac{x_0-x_c}{c}\\ x\left( t-\frac{x_0-x_c}{c};x_c\right) \quad \text {when }x_0>x_c \text { and }t\ge \frac{x_0-x_c}{c} \end{array}\right. } \end{aligned}$$

where \(x(t;x_0)\) is given in Eq. 42. Moreover, we define the time \(\frac{x_0-x_c}{c}\) to be the constraint-touching time. As t passes this constraint-touching time, the constraint stops binding and production of each player is identical to that from the unconstrained case.

Inserting \(x_c(t;x_0)\) into \(q_c^*(x)\), we obtain the production rate at time t,

$$\begin{aligned} q^*_c(x_c(t;x_0)) = {\left\{ \begin{array}{ll} q^*(t;x_0)\quad \text {when }x_0\le x_c\\ c \quad \text {when }x_0>x_c \text { and }0 \le t< \frac{x_0-x_c}{c}\\ q^*\left( t-\frac{x_0-x_c}{c}\right) \quad \text {when }x_0>x_c \text { and }t\ge \frac{x_0-x_c}{c} \end{array}\right. }. \end{aligned}$$

with the associated stopping time,

$$\begin{aligned} \tau _c(x_0) = {\left\{ \begin{array}{ll} \tau _N^b(x_0)\quad \text {when }x_0\le x_c\\ \frac{x_0-x_c}{c}+\tau _N^b(x_c) \quad \text {when }x_0>x_c \end{array}\right. } \end{aligned}$$

where \(\tau _N^b\) is given in Sec. 3.2.3.

Example 6

The upper-left plot of Fig. 8 gives stopping time \(\tau _N^b(x)\) versus initial reserve x. We set the limited production rate to be \(c = 0.15\). The constraint-touching point \(x_c\), as the red vertical line indicates, addresses that production rate when \(x>x_c\) is a constant; hence, the slope is strictly constant. Moreover, compared to the stopping time in Fig. 4, the constraint naturally extends the stopping due to lower production when \(x>x_c\).

The production rate \(q_0^*(x(t))\) given initial reserve \(x_0=5\) versus t is given in the upper-right plot. The production rate of all players is constant initially, because the constraint is active. After the time touches \(\frac{x_0-x_c}{c}\) and the reserve falls below the constraint-touching point \(x_c\), the constraint stops binding. So when \(t>\frac{x_0-x_c}{c}\), the production will be complete at the same time as unconstrained case. At the stopping time \(t=40.6\), the finite-reserve player stops. For total production of opponents, \(Q_c^*\), this constraint allows them to have the minimal production rate. But the total production of all players, \(q_c^*+Q_c^*\) is constrained. As a result, as the energy price in the lower plot indicates, the constraint sets the lower price bound until \(\frac{x_0-x_c}{c}\).

Fig. 8

Stopping time (upper-left), production rate (upper-right) and energy price (lower) in the limited-production case

Comparison of Profits of Opponents

In order to compute the profits of opponents of player k, we list the ODEs for opponents depending on the constraint-touching point \(x_c\), in a similar fashion as done in Sec. 4.1.

$$\begin{aligned} {\left\{ \begin{array}{ll} rv_{c,k}(x)+q_0^*(x)v_{c,k}'(x)=\left( q_k^*(x)\right) ^2 \quad \text {when } 0<x\le x_c\\ v_{c,k}(x)+cv_{c,k}'(x) = u_{c,k}^2 \quad \text {when }x> x_c. \end{array}\right. } \end{aligned}$$

where \(u_{c,k} = \max \left( \frac{1-c+\sum _{i=1}^{n_c-1}s_i}{n_c}-s_k,0\right)\) is a constant. When \(x\le x_c\), the ODE is completely equal to the unconstrained case in Sec. 3.2.4. The constraint is active, so \(q_c^* = c\) when \(x>x_c\). To be more specific, Proposition 6 gives the exact solution.

Proposition 6

The profit function of player k with the limited-production boundary is

$$\begin{aligned} v_k(x) = {\left\{ \begin{array}{ll} A_n(x) v_k(x_b^n) \text { when } n<k, x \le x_c\\ A_n(x)v_k(x_b^n)+\frac{c_{k,n}^2}{r}(1-A_n(x))-\frac{4a_nc_{k,n}n}{r(n-1)(n+1)}(W(\theta _n(x-x_b^n))-\beta _nA_n(x))\\ \quad \quad \quad -\frac{na_n^2}{r(n+1)^2}(W^2(\theta _n(x-x_b^n))-\beta _n^2A_n(x)) \text { when } n\ge k , x \le x_c\\ \\ \exp \left( {\frac{r(x-x_c)}{c}}\right) v(x_c)+\frac{u_{c,k}^2}{r}\left( 1- \exp \left( {\frac{r(x-x_c)}{c}}\right) \right) \text { when } x>x_c. \end{array}\right. } \end{aligned}$$

where parametric setting follows Proposition 5.


Over the interval \(x\le x_c\), the solution is identical to Proposition 5. When \(x>x_c\), the ODE is a constant-coefficient first-order ODE. \(\square\)

Example 7

To make a explicit comparison of profits to Fig. 5, the difference of profits is shown in Fig. 9. The three plots shows that the upper bound on the finite-reserve producer affects the profits of opponents in different ways. The profits of lower-cost players 1 and 2 decrease. In contrast, players 3 and 4 benefit from player 0’s upper production bound. But overall, the lower plot indicates that this production bound does not benefit the whole market, as total profit \(v+\sum _{k=1}^4 v_k > v_c+\sum _{k=1}^4 v_{c,k}\).

Fig. 9

Difference of profits of low-cost opponents (upper-left), high-cost opponents (upper-right) and all players (lower) in the limited-production case

In fact, the upper bound on production of the finite-reserve player has two impacts on opponents: prolonged stopping time resulting in longer nonmaximal production, and increased production with large reserve leading to more profits per time unit. Players 1 and 2 already have large enough productions owing to their lower cost of production. Hence, their negative effect of the prolonged stopping time overwhelms the positive effect of increased production. In contrast, this effect is opposite for higher-cost opponents 3 and 4. The high-cost opponents account for only a tiny share of the market. The limited production of \(q_c^*\) allows them to profit more because the positive effect of increased production overwhelms the negative prolonged stopping time.

Minimal Profit for Finite-reserve Producer

In this section, we consider an entity which must maintain sufficient production to keep running. For example, an energy company may need to ensure a minimal profit per time unit for fixed consumption. A country whose economy mostly depends on oil exports may also need to keep a level of oil production rate to maintain its economy.

Assume that p is the required minimal profit per time unit. Then, by Eq. 6, the producer must ensure that the profit

$$\begin{aligned} q_{p,0}\left( 1-q_{p,0}-\sum _{i=1}^{N-1}q_{p,i}^*-s_0\right) \ge p. \end{aligned}$$

Proposition 7

The minimal profit constraint in Eq. 58 is equivalent to setting a minimal closed-loop production strategy as

$$\begin{aligned} q_{p,0}^*(x)= {\left\{ \begin{array}{ll} \frac{n_p a_{n_p}-\sqrt{n_p^2 a_{n_p}^2-4pn_p}}{2} \quad \text {when }x\le x_p\\ q_0^*(x) \quad \text {when }x>x_p \end{array}\right. }, \end{aligned}$$

where we similarly define \(x_p\) be the constraint-touching point with regard to minimal production. The expression of \(x_p\) is

$$\begin{aligned} x_p = x_b^{n_p} -\frac{2a_{n_p}}{b_{n_p}}\left[ \ln \left( \frac{1}{\beta _{n_p}}\left( -1+\frac{(n_p+1)q_{p,0}^*}{n_pa_{n_p}} \right) \right) -1+\frac{(n_p+1)q_{p,0}^*}{n_pa_{n_p}}-\beta _{n_p} \right] , \end{aligned}$$

where \(n_p\) is the number of total players when the constraint is active. In order to find \(n_p\), the number of total players, we can test the condition \(n_p = \#\{k:q_k^*>0\}+1\) by trying \(n_p=1,2,\dots ,N\).


See Appendix 6. \(\square\)

Define the constant production of the finite-reserve player, \(c_p = \frac{n_p a_{n_p}-\sqrt{n_p^2 a_{n_p}^2-4pn_p}}{2}\) when the constraint is active. The value function of the finite-reserve player becomes

$$\begin{aligned} v_{p,0}(x) = \sup _{q_0 > c_p} \int _0^{\tau } \mathrm {e}^{-rt} q_{p,0}(t)\left( 1-q_{p,0}(t)-\sum _{i=1}^{N-1}q_{p,i}^*(t)-s_0\right) \mathrm {d}t. \end{aligned}$$

Given \(x_p\) and \(n_p\) in Proposition 7, we can compute closed-loop production of opponent k,

$$\begin{aligned} q_{p,k}^*(x) = {\left\{ \begin{array}{ll} \max \left( \frac{1-c_p+\sum _{i=1}^{n_p-1}s_i}{n_p}-s_k,0\right) \text { when } x\le x_p\\ q_{k}^*(x) \text { when } x>x_p \end{array}\right. } \end{aligned}$$

Therefore, summing the production of opponents from Eq. 62 and inserting into Eq. 61 result in the following HJB equation:

$$\begin{aligned} rv_{p,0}={\left\{ \begin{array}{ll} q_{p,0}^*\left( \frac{1-q_{p,0}^*+\sum _{i=1}^{n_p-1}s_i-n_ps_0}{n_p}-v_{p,0}'\right) \quad \text {when } 0\le x\le x_p \\ \sum _{n=n_p}^{N} \frac{1}{(n+1)^2}\left( 1+\sum _{i=1}^{n-1}s_i-n(s_0+v_{p,0}')\right) ^2 \mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} } \quad \text {when } x> x_p \end{array}\right. }. \end{aligned}$$

Therefore, if \(x\le x_p\), the HJB equation for the producer is a first-order linear ODE. For simplicity, we define \(a_p=\frac{1-c_p+\sum _{i=1}^{n_p-1}s_i-n_ps_0}{n_p}\). So the solution to Eq. 63 is

$$\begin{aligned} v_{p,0}(x) = {\left\{ \begin{array}{ll} \frac{c_p}{r}a_p\left( 1-\mathrm {e}^{-\frac{rx}{c_p}} \right) \quad \text {when }x\le x_p \\ \frac{a_{n_p}^2}{b_{n_p}}(1+W(\theta _{n_p}(x-x_p)))^2 \quad \text {when }x_p<x\le x_b^{n_p+1}\\ \sum _{n=n_p+1}^{N} \frac{a_n^2}{b_n}(1+W(\theta _n(x-x_b^n)))^2 \mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} } \quad \text {when }x> x_b^{n_p+1} \end{array}\right. } \end{aligned}$$

where \(\theta _{n_p}(x) = \beta _{n_p}\mathrm {e}^{\beta _{n_p}-\frac{b_{n_p} x}{2a_{n_p}}}\) and \(\beta _{n_p} = -1+\frac{\sqrt{b_{n_p}v_{p,0}(x_p)}}{a_{n_p}}\).

Figure 10 can explain the effect of the limited-production bound. In contrast to the case of limited-production boundary, when \(x> x_p\), the constraint is inactive and production is identical to unconstrained cases. And the blockading points \(x_b^n\) for \(n=n_p, n_p+1, \dots , K\) are effective. But when \(x\le x_p\), the number of players stays at \(n_p\), and blockading points less than \(x_p\) are not effective.

Fig. 10

Picture showing blockading points and times in the case of minimal-profit constraint

Example 8

The upper-left plot of Fig. 11 shows the production rates of the finite-reserve player \(q_p^*(x)\) and the opponents \(Q_p^*(x)\), where the minimal profit is \(p=0.05\). The red vertical dotted line is the constraint-touching point \(x_p\approx 0.62\). When \(x<x_p\), the constraint is active and the production rate \(q_p^*(x)\approx 0.12\). When \(x>x_p\), the constraint does not play a role; hence, \(q_p^*(x)\) and \(q_0^*(x)\) become equal. On the contrary, the total production of opponents has a upper bound, because of this constraint. And the total production of all players, \(q_p^*+Q_p^*\), has a similar trend as \(q_p^*\). Therefore, as the upper-right plot indicates, the energy price has an upper-bound of approximately 0.503. Moreover, there is only one green line \(x_b^3\) in each plot of Fig. 11, which demonstrates that the player 4 with the highest cost never participates in the game until the player 0 depletes its own reserves.

Fig. 11

Production (upper-left), energy price (upper-right), profit (lower-left) and difference of profit (lower-right) in the minimal-profit case, parameters as in Table 2

The lower-left plot of Fig. 11 gives the profit function \(v_{p,0}(x)\). And the lower-right plot compares the constrained and unconstrained cases, represented by the difference \(v_0(x)-v_{p,0}(x)\) presented in the upper-right and lower plots. The difference \(v(x) - v_p(x)<0\) over the whole \([0,\infty )\) excludes the initial region, which means \(v(x)<v_{p,0}(x)\). This result indicates that the profit \(v_{p,0}(x)\) in the Nash equilibrium with minimal profit constraint is higher than v(x) without the constraint. The result also appears counterintuitive as discussed in Ex. 1.

For a simple optimization problem with a single decision maker operating outside a game setting, unconstrained control cannot perform worse than the constrained control. Otherwise, the decision maker would change the control accordingly. But a game using the Nash equilibrium does not give a strategy with the highest profit. The effect of this lower bound of production ensures the profit per time unit of the finite-reserve player, but makes an earlier stopping time. As a result, the higher profit per time unit overwhelms the harm of shorter earning time.

Remaining Reserve and Stopping Time

We can also give the remaining reserve x(t) based on the result in Sec. 3.2.3 for this case. For initial reserve \(x_0\), the remaining reserve is

$$\begin{aligned} x_q(t;x_0) = {\left\{ \begin{array}{ll} x(t;x_0)\quad \text {when }0\le t<t(x_p) \text { and }x_0>x_p \\ x(t(x_p);x_0)-c_p(t-t(x_p)) \quad \text {when } t(x_p)\le t<t(x_p)+\frac{x_p}{c_p} \text { and }x_0>x_p\\ x_0-q_{p,0}^*(t-t(x_p)) \quad \text {when } 0\le x_0\le x_p \end{array}\right. }. \end{aligned}$$

We define the time \(t(x_p)\) to be the constraint-touching point of the minimal-profit bound. When \(t< t(x_p)\), the bound is not active and \(q_p^*(x(t)) = q_0^*(x(t))\). When \(t \ge t(x_p)\), the production bound binds and \(q_p^*(x(t))\) reaches the constraint.

Then, we can easily compute the rate of production at time t.

$$\begin{aligned} q_{p,0}^*(x_q(t;x_0)) = {\left\{ \begin{array}{ll} q(x(t;x_0))\quad \text {when }0\le t <t(x_p) \text { and }x_0>x_p \\ c_p \quad \text {Otherwise} \end{array}\right. }. \end{aligned}$$

Moreover, the stopping time \(\tau _p(x_0)\) will be

$$\begin{aligned} \tau _p(x_0) = {\left\{ \begin{array}{ll} \frac{x_0}{c_p}\quad \text {when }0\le x_0<x_p\\ t(x_p;x_0) + \frac{x_p}{q_{p,0}^*}\quad \text {when }x_0\ge x_p \end{array}\right. }. \end{aligned}$$

Example 9

The upper-left plot of Fig. 12 gives stopping time \(\tau _p(x)\). When \(x<x_p\), the constraint of minimal profit is active; hence, the stopping time increases linearly with a constant production rate \(q_{p,0}^*\). And when \(x\ge x_p\), the production rates are the same for constrained and unconstrained cases.

The upper-right plot shows the production rate \(q_0^*(x(t))\) given initial reserve \(x_0=5\). At first, the constraint makes no difference. The red vertical line \(t(x_p;x_0) = 23\) is the constraint-touching time. On the approximate interval \(23<t<29\), the production rate becomes a constant 0.11 to achieve a minimal profit of \(p=0.05\). When \(t\approx 29\), the reserve is completely depleted; hence, the finite-reserve producer exits the market and the production rate falls to 0. On the contrary, the production of opponents \(Q_p^*\) jumps to 0.49.

As the result from the production, the lower plot presents the energy price as a function of t. The curve over \(t<x_p\) is identical to the unconstrained case. But the energy price remains constant when the bound is active and jumps to 0.528 after the finite-reserve player depletes the reserve.

Fig. 12

Stopping time (left) and production rate (right) in the minimal-profit case, parameters as in Table 2

Profits of Opponents

The computation of the profits of opponents in the minimal-profit case is similar to the limited-production case. We need to list the ODEs for opponents depending on the constraint-touching point \(x_p\).

$$\begin{aligned} {\left\{ \begin{array}{ll} v_{p,k}(x)+c_pv_{p,k}'(x) = l_{p,k}^2 \quad \text {when }0<x\le x_p \\ rv_{p,k}(x)+q_0^*(x)v_{p,k}'(x)=\left( q_k^*(x)\right) ^2 \quad \text {when }x>x_p \end{array}\right. }. \end{aligned}$$

where \(l_{p,k} = \max \left( \frac{1-c_p+\sum _{i=1}^{n_p-1}s_i}{n_p}-s_k,0\right)\) is a constant. When \(x\le x_p\), the constraint is active and \(q_p^* = c_p\). When \(x>x_p\), the constraint is not active; hence, the ODE is the same as Sec. 3.2.4, except for the initial conditions. Proposition 5 gives the exact solution to the ODE.

Proposition 8

The profit function of player k with minimal-profit boundary is

$$\begin{aligned} v_k(x) = {\left\{ \begin{array}{ll} \exp \left( -\frac{rx}{c_p}\right) +\frac{l_{p,k}^2}{r}\left( 1-\exp \left( -\frac{rx}{c_p}\right) \right) \text { when } 0\le x \le x_p\\ \\ A_p(x) v_k(x_p) \text { when } n_p<k, x_p<x\le x_{n_p-1} \\ A_p(x)v_k(x_p)+\frac{c_{k,n_p}^2}{r}(1-A_p(x))-\frac{4a_{n_p}c_{k,n_p}n_p}{r(n_p-1)(n_p+1)}(W(\theta _n(x-x_b^{n_p}))-W(\theta _{n_p}(x_p-x_b^n))A_p(x))\\ \quad \quad \quad -\frac{n_pa_{n_p}^2}{r(n_p+1)^2}(W^2(\theta _{n_p}(x-x_b^{n_p}))-W^2(\theta _{n_p}(x_p-x_b^n))A_{n_p}(x)) \text { when } n_p\ge k, x_p<x\le x_{n_p-1} \\ \\ A_n(x) v_k(x_b^n) \text { when } n<k, x>x_{n_p-1} \\ A_n(x)v_k(x_b^n)+\frac{c_{k,n}^2}{r}(1-A_n(x))-\frac{4a_nc_{k,n}n}{r(n-1)(n+1)}(W(\theta _n(x-x_b^n))-\beta _nA_n(x))\\ \quad \quad \quad -\frac{na_n^2}{r(n+1)^2}(W^2(\theta _n(x-x_b^n))-\beta _n^2A_n(x)) \text { when } n\ge k, x>x_{n_p-1}, \end{array}\right. } \end{aligned}$$

where \(A_p(x) =\left( \frac{W(\theta _{n_p}(x-x_b^n))}{W(\theta _{n_p}(x_p-x_b^n))}\right) ^{\frac{2n_p}{n_p+1}}\), and other parametric setting follows Proposition 5.


When \(x\le x_p\), the ODE is a constant-coefficient first-order ODE. When \(x>x_p\), the solving process is identical to the unconstrained case except that the initial condition is \(x_p\) but not the blockading points over the interval \(x_p<x<x_{n_p-1}\). \(\square\)

Example 10

To make a explicit comparison of profits to Fig. 5, the profit difference of low-cost, high-cost opponents and all players is shown in Fig. 13. The three plots show that the upper bound on the finite-reserve producer affects the profits of opponents in different ways. The profits of lower-cost players 1 and 2 decrease. In contrast, players 3 and 4 benefit from the minimal-profit production bound. But the whole market depends on the initial reserve x(t).

Setting this minimal-profit bound, as lower plot of Fig. 13 presents, increases (resp. decreases) when x is high (resp. low).

Fig. 13

Difference of profits of high-cost opponents (upper-left), low-cost opponents (upper-right) and all players (lower) in the minimal-profit case, parameter as in Table 2

In contrast to Sec. 4.1.2, the lower bound on production of the finite-reserve player has two impacts on opponents: reduced stopping time resulting in shorter production period and decreased production given small reserve leading to less profits per time unit. Players 1 and 2 already have sufficiently large productions owing to lower cost of production. Therefore, the decreased amount of production when the constraint is active has limited impact. In contrast, for players 3 and 4, who already account for only a tiny share of the market, the impact of their decreased production outweighs the earlier stopping time.

Remark 3

Similar to Ex. 1, we also obtain a counterintuitive result for the player 0 manually lifting the production and obtaining a higher profit. Via the analysis of the profits of opponents, the manual increase in the production has different direction impact on opponents. But as a whole, since the lower-cost opponents take advantages over those with higher costs, who possess a much larger market share, the overall effect benefits the finite-reserve player.

Price comparison

Fig. 14

Comparison of prices over time in difference cases, parameter as in Table 2

The price for the unconstrained, upper-bound and lower-bound cases is available in Fig. 14. Compared to the unconstrained case (blue curve), the upper bound (orange curve) is initially higher (\(t<18\)), then presenting a parallel trend of price movement when the constraint does not bind (\(t>25\)).

In contrast, the lower bound (green curve) follows identical price path when this profit constraint does not bind (\(t<23\)), and presents a lower price between \(23<t<29\). All of those price curves eventually lead to the same price of 0.53 as t passes the stopping times of \(\tau = 34.0, 40.6, 28.9\), at which x depletes in the unconstrained, upper-bound and lower-bound cases, respectively (see Table 3).


Table 3 Summary of constraints on production to profits of players

In this paper, we construct a type of differential game models between a finite-reserve player and multiple infinite-reserve players. We focus on the relationship between the production rate \(q^*_i\) and the profit \(v_i(x)\) for player i w.r.t. the oil reserve x of the finite-reserve player.

Theoretically, a higher initial oil reserve x lifts the production rate of the finite-reserve player \(q^*_0(x)\) and the profit \(v_0(x)\) up with marginal effect because higher reserve enables the finite-reserve player to produce more crude oil. But the finite-reserve player cannot increase production rate \(q_0^*\) without limit due to potential loss of profit from the effect of price model P(Q) from a higher total production Q. On the other hand, the opponents have to decrease their production rate \(q_i^*\) for \(i\ge 1\) because their market shares are squeezed by the finite-reserve producer.

We also manually set the cases of limited production and minimal profit for the finite-reserve producer, represented by the upper and lower production bounds. The upper and lower production bounds have opposite effects on the profits of different types of players, as Table 3 indicates. The upper (lower) bound increases (decreases) the profits of low-cost players and decreases (increases) the profit of the high costs. The reason behind this result is due to the aggregated effect of profits and losses of changed production and stopping time.

Minimal-profit constraints on the finite-reserve producers help high-cost producers when they are blockaded out at high x because they are hurt by decreased market price when they are blockaded out, but are benefit by earlier exit of the finite firm. It hurts low-cost firms when they are not blockaded out. The reader may consider the impact of this game theoretic insight on the understanding of the world oil market.

Data Availability Statement

We claim that this work is a theoretical result and there is no available data source.


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We acknowledge the support of the Natural Sciences and Engineering Research Council of Canada (NSERC) [funding reference number: 2020-06667].

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Nash Equilibrium Computation in Ex. 1

$$\begin{aligned} \begin{aligned} q_1^*&= \frac{1+s_1+s_2-3s_1}{3} = \frac{1+0.05+0.2}{3}-0.05 = 0.3667 \\ q_2^*&= \frac{1+s_1+s_2-3s_2}{3} = \frac{1+0.05+0.2}{3}-0.2 = 0.2167. \end{aligned} \end{aligned}$$

In the Nash equilibrium, the profits for the two players become

$$\begin{aligned} \begin{aligned} G_1 = \left( \frac{1+s_1+s_2-3s_1}{3}\right) ^2 = 0.3667^2 = 0.1344 \\ G_2 = \left( \frac{1+s_1+s_2-3s_2}{3}\right) ^2 = 0.2167^2 = 0.0469. \end{aligned} \end{aligned}$$

If player 1 were to increase the production to \(q_1 = 0.4\) slightly, then for player 2, the optimal \(q_2=\frac{1-0.4-0.2}{2} = 0.2\). Then, the profit becomes

$$\begin{aligned} \begin{aligned} G_1 = q_1\left( 1 - q_1 - q_2 - s_1\right) = 0.4 (1-0.4-0.2-0.05) = 0.1400 \\ G_2 = q_n\left( 1 - q_2 - q_1 - s_2\right) = 0.2 (1-0.4-0.2-0.2) = 0.0400. \end{aligned} \end{aligned}$$

Proof of Lemma 1

From the definition of Lambert-W function, \(z=W(z)\mathrm {e}^{W(z)}\), we can compute the derivative for W(z) to be

$$\begin{aligned} W'(z) = \frac{W(z)}{z(1+W(z))}. \end{aligned}$$

Therefore, the derivative for v(x) is

$$\begin{aligned} \begin{aligned} v'(x)&= \frac{2a^2}{b}\theta '(x)W'(\theta (x))(1+W(\theta (x)))\\&= -a\theta (x)(1+W(\theta (x)))\frac{W(\theta (x))}{\theta (x)(1+W(\theta (x)))}\\&= -a W(\theta (x)) \end{aligned} \end{aligned}$$

Then, it is easy to see that

$$\begin{aligned} (a-v'(x))^2 = a^2(1+W(\theta (x)))^2 = bv(x). \end{aligned}$$

So the solution satisfies the ODE. Moreover, it also satisfies the initial condition

$$\begin{aligned} v(0) = \frac{a^2}{b}(1+W(\beta \mathrm {e}^{\beta }))^2 = \frac{a^2}{b}(1-\beta )^2 = v(0). \end{aligned}$$

Proof of Proposition 3

By combining Eq. 29 and Eq. 30, and considering the first blockading point \(x_b^{N-1}\), we can compute that

$$\begin{aligned} W(\theta _N(x_b^{ \, N-1}-x_b^{ \, N})) = -\frac{\delta _{N-1}}{a_n}. \end{aligned}$$

Therefore, by the definition of the Lambert-W function and since \(x_b^N=0\), the first blockading point \(x_b^{N-1}\) is

$$\begin{aligned} x_b^{ \, N-1} = \frac{1}{\mu _N}\left( -1+\frac{\delta _{N-1}}{a_N}-\log \left( \frac{\delta _{N-1}}{a_N} \right) \right) \end{aligned}$$

where we define \(\mu _n = \frac{b_n}{2a_n} = \frac{r}{2a_n}\left( \frac{n+1}{n}\right) ^2\). Inserting Eq. 77 into Eq. 26 in the case of \(n = N\) reveals that

$$\begin{aligned} v(x_b^{\, N-1}) = \frac{1}{r}(s_{N-1}-s_0-\delta _{N-1})^2. \end{aligned}$$

Let \(K = \min \{n:\delta _n>0\}\). Then, the infinite-reserve producers \(K,K+1,\dots , N-1\) have blockading points. Now, we find other blockading points \(x_b^{\, K}, x_b^{\, K+1},\dots , x_b^{\, N-2}\).

By a similar calculation as for \(v(x_b^{\, N-1})\), it can be computed that for \(n \in \{K,K+1,\dots ,N-1\}\)

$$\begin{aligned} v(x_b^{n-1}) = \frac{1}{r}(s_{n-1}-s_0-\delta _{n-1})^2 = \frac{a_{n}^2}{b_{n}}(1+W(\theta _n(x_b^{n-1}-x_b^{n})))^2. \end{aligned}$$

By taking \(x_b^{n-1}-x_b^{n}\) out of the Lambert-W function, Eq. 80 can indicate the relation between \(x_b^{n-1}\) and \(x_b^{n}\) as

$$\begin{aligned} x_b^{n-1} = x_b^{n} + \frac{1}{\mu _n}\left( \log \left( \frac{\delta _n}{\delta _{n-1}}\right) -\frac{(n+1)(s_n-s_{n-1})}{a_n} \right) . \end{aligned}$$

where \(n=K,K+1,\dots ,N-1\).

We already know that the left limit

$$\begin{aligned} \lim _{x\rightarrow x_b^n -0} v'(x) = -a_{n+1}W(\theta _{n+1}(x_b^n - x_b^{n+1})) = \delta _n, \end{aligned}$$

Given these \(v(x_b^{n}) = \frac{1}{r}(s_n - s_0 - \delta _n)^2\), we can compute

$$\begin{aligned} \begin{aligned} \lim _{x\rightarrow x_b^n +0} v'(x)&= -a_n W(\theta _n(0)) = a_n - \sqrt{b_n v(x_b^n)} = a_n - \frac{n+1}{n}(s_n-s_0 - \delta _n) \\&= \frac{1+\sum _{i=1}^{n-1} s_i}{n}-s_0 - \frac{n+1}{n}\left( s_n-s_0+(n+1)s_n + \left( 1+s_0+\sum _{i=1}^{n-1}s_i\right) \right) = \delta _n. \end{aligned} \end{aligned}$$

Therefore, this confirms that \(v'(x)\) is also continuous at \(x_b^n\).

However, using the property to compute the second derivative of v(x),

$$\begin{aligned} v''(x) = -\sum _{n=K}^{N} \frac{b_n}{2}\frac{W(\theta _n(x-x_b^n))}{1+W(\theta _n(x-x_b^n))}\mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} }. \end{aligned}$$

Using the left and right limits of \(v'(x)\) simply leads to

$$\begin{aligned} \begin{aligned} \lim _{x\rightarrow x_b^n -0} v''(x)&= -\frac{b_{n+1}\delta _n}{2(a_{n+1}-\delta _n)}\\ \lim _{x\rightarrow x_b^n +0} v''(x)&= -\frac{b_n\delta _n}{2(a_{n}-\delta _n)}. \end{aligned} \end{aligned}$$

This may not be equal given that different set of \(\{s_i\}_{i=0,1, \dots , N-1}\) can lead to \(a_n \ne a_{n+1}, b_n \ne b_{n+1}\). So the second derivative \(v''(x)\) is not continuous at \(x_b^n\).

Proof of Proposition 4

First, we use the chain rule with Lemma 1 to indicate that

$$\begin{aligned} \mathop {}\!\frac{\mathrm {d}}{\mathrm {d}x}{W(\theta _n(x-x_b^n))} = -\frac{b_nW(\theta _n(x-x_b^n))}{2a_n\left( 1+W(\theta _n(x-x_b^n))\right) }. \end{aligned}$$

Over the interval \([x_b^{n}, x_b^{n-1})\), we transform Eq. 5 given \(q_0^*(x)\) as

$$\begin{aligned} \begin{aligned} \mathrm {d}t&= -\frac{n+1}{na_n\left( 1+W(\theta _n(x-x_b^n))\right) }\mathrm {d}x(t) \\&= \frac{2n}{r(n+1)W(\theta _n(x-x_b^n))}\mathrm {d}(W(\theta _n(x-x_b^n)))\\&= \frac{2n}{r(n+1)}\mathrm {d}(\ln W(\theta _n(x-x_b^n))). \end{aligned} \end{aligned}$$

We have demonstrated that the derivative

$$\begin{aligned} v'(x) = -\sum _{n=K}^{N} a_nW(\theta _n(x-x_b^n))\mathbbm {1}_{\left\{ {x_b^n\le x<x_b^{n-1}}\right\} } \end{aligned}$$

which simplifies this ODE into

$$\begin{aligned} \mathrm {d}t = \frac{2n}{r(n+1)}\mathrm {d}(\ln v'(x)). \end{aligned}$$

Aggregating each interval \([x_b^{n}, x_b^{n-1})\) from \(n = l, l-1, \dots , m\) leads to Eq. 40.

Proof of Proposition 5

Inserting explicit formula of \(q_0^*(x)\) and \(q_k^*(x)\) into the HJB Eq. 17, we can obtain for player k,

$$\begin{aligned} rv_k(x)+q_0^*(x)v_k'(x)=\left( q_k^*(x)\right) ^2 , \end{aligned}$$

which is a first-order ODE with initial condition \(v_k(0)=\frac{1}{r}G_k\) at \(x=0\). When \(n<k\), the equation reduces to the homogeneous ODE

$$\begin{aligned} v_k'(x)+\frac{r(n+1)}{na_n[1+W(\theta _n(x-x_b^n))]}v_k(x) = 0. \end{aligned}$$

The integrating factor for this ODE is \(W^{-\frac{2n}{n+1}}(\theta _n(x-x_b^n))\). Then, we can easily obtain the solution

$$\begin{aligned} v_k(x) = \left( \frac{W(\theta _n(x-x_b^n))}{\beta _n}\right) ^{\frac{2n}{n+1}} v_k(x_b^n). \end{aligned}$$

For \(n\ge k\), player k is not blockaded and \(q_k^*(x)\) takes the positive part. Using the same integrating factor, we need to find the solution to the inhomogeneous ODE

$$\begin{aligned} \mathop {}\!\frac{\mathrm {d}}{\mathrm {d}x} \left[ W^{-\frac{2n}{n+1}}(\theta _n(x-x_b^n))v_k(x)\right] =\frac{(n+1)W^{-\frac{2n}{n+1}}(\theta _n(x-x_b^n))}{na_n[1+W(\theta _n(x-x_b^n))]}\left[ c_{k,n}-\frac{a_nW(\theta _n(x-x_b^n))}{n+1}\right] ^2 \end{aligned}$$

where we define that \(c_{k,n} := \frac{1+\sum _{i=0}^{n-1}s_i}{n+1}-s_k\). Taking integral on both sides, we can obtain an explicit solution

$$\begin{aligned} \begin{aligned} v_k(x) =&A_n(x)v_k(x_b^n)+\frac{c_{k,n}^2}{r}(1-A_n(x))-\frac{4a_nc_{k,n}n}{r(n-1)(n+1)}(W(\theta _n(x-x_b^n))-\beta _nA_n(x))\\&-\frac{na_n^2}{r(n+1)^2}(W^2(\theta _n(x-x_b^n))-\beta _n^2A_n(x)), \end{aligned} \end{aligned}$$

where \(A_n(x):=\left( \frac{W(\theta _n(x-x_b^n))}{\beta _n}\right) ^{\frac{2n}{n+1}}\).

Proof of Proposition 7

Assume n players are active in the game. First, the total production of opponents is

$$\begin{aligned} Q^*=\sum _{i=1}^{N-1}q_i^*=\frac{(1-q_0)(n-1)-\sum _{i=1}^{n-1}s_i}{n} \end{aligned}$$

given \(q_0\) is a predetermined production, which can either be a constant or a function of x. Inserting the total production into the equality Eq. 58 leads to

$$\begin{aligned} q_0\left( -\frac{1}{n} q_0+a_n\right) \ge p, \end{aligned}$$

which indicates the solution to this inequality,

$$\begin{aligned} \frac{n a_{n}-\sqrt{n^2 a_{n}^2-4pn}}{2} \le q_0 \le \frac{n a_{n}-\sqrt{n_p^2 a_{n}^2-4pn}}{2}. \end{aligned}$$

Now, we already demonstrate that \(q_0^*(x) = \frac{n}{n+1}(a_n-v'(x))\) is a increasing function in Eq. 34. Therefore, profit of the finite-reserve producer is

$$\begin{aligned} \frac{n}{(n+1)^2}\left( na_n^2 - (n-1)a_nv'(x) - (v'(x))^2\right) . \end{aligned}$$

over the interval \([x_b^n, x_b^{n-1})\). We have already confirmed that that \(v'(x)>0\) is a decreasing function. So this unit profit increases with x. To achieve the minimal profit, the focus should be on left interval, i.e.,

$$\begin{aligned} q_0\ge \frac{n a_{n}-\sqrt{n^2 a_{n}^2-4pn}}{2}. \end{aligned}$$

The remaining problem is to find the number of players the constraint-touching point \(x_p\) is achieved. To find \(x_p\), we must test each n so that when

$$\begin{aligned} q_0 = \frac{n a_{n}-\sqrt{n^2 a_{n}^2-4pn}}{2}, \end{aligned}$$

the number of active players \(\#\{q_k^*:q_k^*>0\} = n-1\) from \(n = N, N-1, \dots , K\). Denote this value to be \(n_p\). And denote the corresponding constant production as \(q_{p,0}^*\).

Inserting this \(q_{p,0}^*\) into formula of \(q_0^*(x)\) leads to

$$\begin{aligned} x_p = x_b^{n_p} -\frac{2a_{n_p}}{b_{n_p}}\left[ \ln \left( \frac{1}{\beta _{n_p}}\left( -1+\frac{(n_p+1)q_{p,0}^*}{n_pa_{n_p}} \right) \right) -1+\frac{(n_p+1)q_{p,0}^*}{n_pa_{n_p}}-\beta _{n_p} \right] , \end{aligned}$$

Therefore, the production with minimal profit is

$$\begin{aligned} q_{p,0}^*(x)= {\left\{ \begin{array}{ll} \frac{n_p a_{n_p}-\sqrt{n_p^2 a_{n_p}^2-4pn_p}}{2} \quad \text {when }x\le x_p\\ q_0^*(x) \quad \text {when }x>x_p \end{array}\right. }. \end{aligned}$$

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Chen, J., Davison, M. Deterministic Asymmetric-cost Differential Games for Energy Production with Production Bounds. Oper. Res. Forum 2, 50 (2021).

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  • Energy markets
  • HJB Equations
  • Differential games