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

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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.

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  1. In the theory of discontinuous dynamical systems [10], points that are subject to pressure at a positive rate from both sides, as expressed in (2.9), are sometimes referred to as “chattering points”. The terminology is avoided here in the one-dimensional case, because the modeling in this paper does not presume that actual chattering takes place.

  2. Generally speaking, for a continuous function x(t) defined on an interval [0, T] and a given point \({\hat{x}}\), it is possible that the set \(\{ t \in [0,T] \mid x(t) = {\hat{x}}\}\) has measure arbitrarily close to the length T of the interval [0, T], while there is no interval \([\tau ,\tau '] \subset [0,T]\) with \(\tau ' > \tau \) such that \(x(t) = {\hat{x}}\) for almost all \(t \in [\tau ,\tau ']\). An example can be constructed by making use of the “\(\varepsilon \)-Cantor set” [2, p. 140]. Episodicity of the solution of \({\dot{x}}=g(x,u)\) therefore implies in particular that the control function u(t) does not induce this type of singular behavior.

  3. When used with a subscript indicating a region of the state space, the symbol \(\mathbbm {1}\) denotes the function that takes the value 1 in the indicated region, and the value 0 elsewhere.

  4. In the terminology of physics, this means that pollution is modeled here as an extensive quantity (such as mass), rather than as an intensive quantity (such as temperature). Such modeling can be reasonable, for instance, in the case of chemical pollution.

  5. The symbol p is used both to denote a function of s in (4.10) and to denote a function of x in (4.9). This is an abuse of notation.

  6. The prime mark, when applied to vectors, denotes transposition.

  7. An analogous observation is in [11]; see their comments following Def. 4.13 in the cited paper.


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We are grateful to the Lead Guest Editor of the special issue (Florian Wagener) and to two anonymous reviewers for their stimulating and helpful comments.

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Correspondence to Johannes M. Schumacher.

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This article is part of the topical collection “Dynamic Games in Environmental Economics and Management” edited by Florian Wagener and Ngo Van Long.

Jacob C. Engwerda-Retired, Tilburg School of Economics and Management, Tilburg University, Tilburg, The Netherlands.



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 (xp) 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 (xp) 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 (xp) 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 (xp) 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

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


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

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 (xp) 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 (xp) 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 (xp) 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:

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

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:

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

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

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


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

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)

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

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:

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

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

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

From (B.7), one obtains

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

Also, we have

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


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

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

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

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(xC) 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

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


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

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:

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

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:

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

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

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Schumacher, J.M., Reddy, P.V. & Engwerda, J.C. Jump Equilibria in Public-Good Differential Games with a Single State Variable. Dyn Games Appl 12, 784–812 (2022).

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