Jump Equilibria in Public-Good Differential Games with a Single State Variable

Abstract

A simple sufficient condition is proved for symmetric Markov subgame perfect Nash equilibria in public-good differential games with a single state variable. The condition admits equilibria in feedback strategies that have discontinuous dependence on the state variable. The application of the condition is demonstrated in the Dockner–Long model for international pollution control. The existence is shown of equilibria that are arbitrarily close to Pareto dominance for all initial conditions. In the limit as the discount rate tends to 0, the equilibrium strategies differ from the optimal strategies under full coordination, but nevertheless the agents’ payoffs do converge to those obtained from the coordinated (first-best) solution. For positive values of the discount rate, the supremal value function associated with the globally Pareto dominant equilibrium is a continuously differentiable function that is not a solution of the Hamilton–Jacobi–Bellman equation.

Appendices

Appendix

Critical Parameter Values

This appendix provides derivations of conditions used in the main text for the location of various specific levels of the state variables with respect to each other. In particular, the following levels are of interest: the right and left tangent points $x_R$ and $x_L$; the asymptotic level $x_C$ under the coordinated policy; the x-coordinate of the point in the (x, p) plane where the unstable manifold of the auxiliary system intersects the line $\mathrm{d}x/\mathrm{d}t=0$; and the x-coordinate of the point where the unstable manifold intersects the reflection of the line $\mathrm{d}x/\mathrm{d}t=0$ with respect to the line $\mathrm{d}x/\mathrm{d}s=0$. The latter two points will be denoted by $x_U$ and $x_V$, respectively; they are the x-coordinates of the corner points of the middle shaded triangles shown in Fig. 5. The relevance of these two points is that they relate to the minimal number of jumps needed to construct trajectories with particular properties; for instance, if $x_R > x_V$, then at least five jumps are needed in a trajectory that connects the left tangent orbit to the right tangent orbit. In this appendix, it is assumed throughout that $N>1$. The relation $x_L< x_C < x_R$ has already been shown in the main text; the discussion below focuses on the location of these three points with respect to $x_U$ and $x_V$.

Consider first the right tangent point $(x_R,p_R)$. By definition, this point lies on a southern orbit, and hence, the inequality $x_R > x_U$ holds. Note that the reflection of $(x_R,p_R)$ with respect to the line $\mathrm{d}x/\mathrm{d}s=0$, denoted by $(x_R,p_R')$, lies either in the northern or in the eastern part of the (x, p) plane and that the inequality $x_R > x_V$ holds if and only if the latter condition is true. Let A denote the matrix appearing in the auxiliary system (4.11). Generally speaking, a point (x, p) is in the eastern or in the western region if and only if the determinant of the $2\times 2$ matrix $[v \;\; Av]$ is positive, where v is defined by $v = [ x-x_E \;\; p-p_E]'$ and $(x_E,p_E)$ is the stationary point of the auxiliary system as identified in (4.14). Therefore, the criterion based on the sign of the determinant of $[v \;\; Av]$, where $v = [ x_R-x_E \;\; p_R'-p_E ]'$, can be used to determine whether or not the relation $x_R > x_V$ holds. One can now calculate as follows.

From (4.18) and from the fact the reflection of a point (x, p) on the line $\mathrm{d}x/\mathrm{d}t=0$ is given by $(x,p/(2N-1))$, we have

$$\begin{aligned} x_R = \frac{\rho + k/N}{\alpha +(k/N)(\rho + k/N)}\,, \qquad p_R' = \frac{-\alpha /(2N-1)}{\alpha +(k/N)(\rho + k/N)}\,. \end{aligned}$$

Using these relations as well as (4.14), one finds

$$\begin{aligned} x_R-x_E&\propto \Big (\rho +\frac{k}{N}\Big ) \big ( (2N-1)\alpha + k(\rho +k) \big ) - N(\rho +k) \Big ( \alpha + \frac{k}{N} \Big (\rho +\frac{k}{N} \Big ) \Big ) \\&= (N-1)\alpha \Big ( \rho - \frac{N-1}{N} \,k \Big ) \end{aligned}$$

and

$$\begin{aligned} p_R' - p_E&\propto \frac{-\alpha }{2N-1}\big ( (2N-1)\alpha + k(\rho +k) \big ) + N\alpha \Big ( \alpha + \frac{k}{N} \Big (\rho +\frac{k}{N} \Big ) \Big ) \\&= (N-1)\alpha \Big [ \alpha + \frac{k}{2N-1}\,\Big ( 2\rho + \frac{k}{N} \,\Big )\Big ] \end{aligned}$$

where the proportionality constant is positive and the same in both cases. Further computation shows that

$$\begin{aligned} \begin{bmatrix} -k &{} 2N-1 \\ \alpha &{} \rho +k \end{bmatrix} \begin{bmatrix} \rho - \frac{N-1}{N} \,k \\ \alpha + \frac{k}{2N-1}\,\Big ( 2\rho + \frac{k}{N} \Big ) \end{bmatrix} = \Big ( \alpha + \frac{k(\rho +k)}{2N-1}\, \Big ) \begin{bmatrix} 2N-1 \\ 2\rho +k/N \end{bmatrix}. \end{aligned}$$

The criterion $\det [ v \;\; Av ] > 0$ for the inequality $x_R > x_V$ to hold, with $v = [ x_R-x_E \;\; p_R'-p_E ]'$, can therefore be written as:

$$\begin{aligned} \Big (\rho - \frac{N-1}{N} \,k\Big )(2\rho + k/N) - (2N-1) \Big ( \alpha + \frac{k}{2N-1}\,\Big ( 2\rho + \frac{k}{N} \,\Big ) \Big ) > 0. \end{aligned}$$

From this, one finds

$$\begin{aligned} x_R > x_V \quad \text {if and only if} \quad \alpha < \Big ( 2\rho + \frac{k}{N} \Big ) \Big ( \frac{\rho }{2N-1} - \frac{k}{N} \Big ). \end{aligned}$$

The analysis of the location of the points $x_L$ and $x_C$ with respect to the points $x_U$ and $x_V$ proceeds in a similar way. We have $x_L < x_U$ if and only if the point $(x_L,p_L')$ lies in the western part of the (x, p) plane. Since this point must lie either in the southern or in the western part, the determinant criterion can again be used. Calculations along the same lines as above lead to the conclusion that

$$\begin{aligned} x_L< x_U \quad \text {if and only if} \quad \alpha < \Big ( \frac{2\rho }{2N-1} \,+\, \frac{k}{N} \Big ) \Big ( \rho -\frac{k}{N} \Big ). \end{aligned}$$

The point $(x_C,p_C)$ is located either in the southern or in the western part of the (x, p) plane, and the inequality $x_C < x_U$ holds if and only if the latter condition is true. Applying the determinant condition as before, one finds

$$\begin{aligned} x_C< x_U \quad \text {if and only if} \quad \alpha < \frac{\rho +k}{N^2} \big ( (N-1)\rho - k \big ). \end{aligned}$$

Finally, the inequality $x_C > x_V$ holds if and only if the point $(x_C,p_C')$ (the reflection of $(x_C,p_C)$) is located in the eastern part of the (x, p) plane, rather than the northern part. Going through the calculations as above, one finds that the corresponding condition is

$$\begin{aligned} \alpha < -\frac{1}{N^2} \, (\rho +k) \Big ( \frac{N-1}{2N-1} \, \rho +k \Big ) \end{aligned}$$

which cannot be satisfied under the standing assumption that all three of $\alpha$, $\rho$, and k are positive.

Solution of a Differential Equation

The evolution of the state under the canonical single-jump strategy is given by:

$$\begin{aligned} {\dot{x}}(t) = -kx(t) + Nu(x(t)) \end{aligned}$$

where $u(x) = p(x)+1$, and where the function p(x) satisfies $G(x,p(x)) = G(x_C,p_C)$. Computation shows that

$$\begin{aligned} G(x_C,p_C) = \frac{\frac{1}{2}k^2}{N^2 \alpha +k^2} \end{aligned}$$

so that the condition $G(x,p(x)) = G(x_C,p_C)$ can be written more explicitly as:

$$\begin{aligned} \tfrac{1}{2}\, (2N-1) p^2 + (N-kx)p = \tfrac{1}{2}\,\alpha x^2 - \frac{\frac{1}{2}N^2 \alpha }{N^2 \alpha + k^2}\,. \end{aligned}$$

(B.1)

For given x, this equation has two solutions for p, corresponding to the two different branches of the canonical single-jump solution. Concentrate first on the behavior for $x > x_C$. Introduce a new variable v by

$$\begin{aligned} v = -kx + N(p+1) \end{aligned}$$

so that $p = (v+kx-N)/N$. It follows from (B.1) that the variables x and v are related by:

$$\begin{aligned} (2N-1)v^2 -2N(N-1) \Big ({ -\frac{k}{N}\,(x-x_C) + \frac{N^2\alpha }{N^2\alpha +k^2} }\Big ) v = (N^2\alpha +k^2) (x-x_C)^2. \end{aligned}$$

(B.2)

Dividing through by the coefficient of $v^2$ and replacing $x-x_C$ by a new variable also called x (which should cause no confusion), one can rewrite the relationship above as

$$\begin{aligned} v^2 - 2(ax+b)v = cx^2 \end{aligned}$$

(B.3)

with

$$\begin{aligned} a = - \frac{N-1}{2N-1}\,k, \quad b = \frac{N(N-1)}{2N-1} \, \frac{N^2\alpha }{N^2\alpha +k^2}, \quad c = \frac{1}{2N-1}\,(N^2\alpha +k^2). \end{aligned}$$

(B.4)

Note that both b and c are positive. For initial conditions $x_0>x_C$ in the original coordinates, the evolution of the state is therefore given by the differential equation (omitting the time variable)

$$\begin{aligned} {\dot{x}} = ax+b - \sqrt{(ax+b)^2+cx^2} \end{aligned}$$

(B.5)

where now x represents $x-x_C$. This is a scalar differential equation that can be solved by separation of variables:

$$\begin{aligned} \int \frac{1}{ax+b - \sqrt{(ax+b)^2+cx^2}} \, \mathrm{d}x = \int \mathrm{d}t. \end{aligned}$$

To determine an explicit expression for the left-hand side, apply Euler’s substitution:

$$\begin{aligned} y = \sqrt{(ax+b)^2+cx^2} + mx, \qquad m = \sqrt{a^2+c}\,. \end{aligned}$$

(B.6)

The function $y=y(x)$ satisfies $y(0)=b$ and is monotonically increasing, since

$$\begin{aligned} y'(x)&= \frac{a(ax+b)+cx}{\sqrt{(ax+b)^2+cx^2}} + m \\&= \frac{(a^2+c)x+ab + \sqrt{((a^2+c)x+ab)^2+b^2}}{\sqrt{(ax+b)^2+cx^2}}\, > 0. \end{aligned}$$

By taking squares of both sides of the equation $y-mx=\sqrt{(ax+b)^2+cx^2}$, one finds

$$\begin{aligned} x = \frac{y^2-b^2}{2(ab+my)}\,. \end{aligned}$$

(B.7)

From (B.7), one obtains

$$\begin{aligned} \mathrm{d}x = \frac{my^2+2aby+mb^2}{2(ab+my)^2}\,\mathrm{d}y. \end{aligned}$$

(B.8)

Also, we have

$$\begin{aligned} ax+b-\sqrt{(ax+b)^2+cx^2} = ax+b - y+mx = (a-m) \, \frac{(y-b)^2}{2(ab+my)}\,. \end{aligned}$$

Therefore,

$$\begin{aligned}&\int \frac{1}{ax+b-\sqrt{(ax+b)^2+cx^2}} \, \mathrm{d}x \\&\quad = \frac{1}{a-m} \, \int \frac{my^2+2aby+mb^2}{(ab+my)(y-b)^2} \, \mathrm{d}y \\&\quad = \frac{1}{a-m} \, \int \Big (\, \frac{2b}{(y-b)^2} + \frac{2a}{m+a} \, \frac{1}{y-b} + \frac{m-a}{m+a} \, \frac{m}{ab+my} \, \Big ) \, \mathrm{d}y \\&\quad = \frac{2b}{m-a} \, \frac{1}{y-b} -\frac{2a}{c}\,\log (y-b) -\frac{1}{m+a} \log (ab+my) \, + C \end{aligned}$$

where C is an integration constant. For $x>0$, define t(x, C) by

$$\begin{aligned} t(x,C) = \frac{2b}{(m-a)(y(x)-b)} - \frac{2a}{c}\log (y(x)-1) - \frac{1}{m+a}\log (ab+my(x)) + C \end{aligned}$$

(B.9)

where y(x) is defined in (B.6). Since y(x) tends to b from above as x tends to 0 from above, we have $\lim _{x\downarrow 0} t(x,C) = \infty$. Also, $\lim _{x\rightarrow \infty } t(x,C) = -\infty$ for any value of C. By construction, we have for $x>0$

$$\begin{aligned} \frac{\mathrm{d}t}{\mathrm{d}x}(x,C) = \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}(t(x,C))} < 0 \end{aligned}$$

where $\mathrm{d}x/\mathrm{d}t$ is defined in (B.5); therefore, t(x, C) is monotonically decreasing as a function of x. It follows that for any initial condition $x_0>0$ we can find $C_0 = C_0(x_0)$ such that $t(x_0,C_0)=0$. The inverse of the mapping $t(\cdot ,C_0):(0,x_0] \rightarrow [0,\infty )$ is the solution of (B.5) with $x(0)=x_0$.

The behavior of the solution for large values of t can be described as follows. From expression (B.9), we have

$$\begin{aligned} \lim _{t\rightarrow \infty } tx(t) = \lim _{x \downarrow 0} xt(x,C) = \lim _{x \downarrow 0} \frac{2bx}{(m-a)(y(x)-b)} = \frac{2b}{c} \end{aligned}$$

(B.10)

since

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{y(x)-b}{x} = \lim _{x \rightarrow 0} y'(x) = m+a. \end{aligned}$$

In other words, the solutions of (B.5) with any positive initial condition behave approximately as $x(t) = 2b/(ct)$ as t tends to infinity. In terms of the original closed-loop system under the canonical limit-case equilibrium strategy, relation (4.36) follows from (B.4) and (B.10).

For initial conditions $x_0 < x_C$, the differential equation (B.5) needs to be replaced by:

$$\begin{aligned} {\dot{x}} = ax+b + \sqrt{(ax+b)^2+cx^2} \end{aligned}$$

(B.11)

where the variable x and the parameters a, b, and c have the same meaning as before. In a similar way as above, one derives that the general solution of this equation is given by:

$$\begin{aligned} \frac{2b}{m+a} \, \frac{1}{y+b} -\frac{2a}{c}\,\log (y+b) +\frac{1}{m-a} \log (ab+my) = t+C \end{aligned}$$

(B.12)

where $y=y(x)$ is given by (B.6).