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

Abstract

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.

Appendices

Nash Equilibrium Computation in Ex. 1

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

(70)

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

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

(71)

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

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

(72)

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

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

(73)

Therefore, the derivative for v(x) is

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

(74)

Then, it is easy to see that

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

(75)

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

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

(76)

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

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

(77)

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

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

(78)

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

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

(79)

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 $n \in \{K,K+1,\dots ,N-1\}$

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

(80)

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

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

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

We already know that the left limit

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

(82)

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

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

(83)

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),

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

(84)

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

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

(85)

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

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

(86)

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

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

(87)

We have demonstrated that the derivative

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

(88)

which simplifies this ODE into

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

(89)

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,

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

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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

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

(91)

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

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

(92)

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

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

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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

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

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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

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

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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

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

(96)

which indicates the solution to this inequality,

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

(97)

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

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

(98)

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.,

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

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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

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

(100)

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

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

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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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