Symmetric Markovian Games of Commons with Potentially Sustainable Endogenous Growth

Abstract

Differential games of common resources that are governed by linear accumulation constraints have several applications. Examples include political rent-seeking groups expropriating public infrastructure, oligopolies expropriating common resources, industries using specific common infrastructure or equipment, capital flight problems, pollution, etc. Most of the theoretical literature employs specific parametric examples of utility functions. For symmetric differential games with linear constraints and a general time-separable utility function depending only on the player’s control variable, we provide an exact formula for interior symmetric Markovian strategies. This exact solution (a) serves as a guide for obtaining some new closed-form solutions and for characterizing multiple equilibria and (b) implies that if the utility function is an analytic function, then the Markovian strategies are analytic functions, too. This analyticity property facilitates the numerical computation of interior solutions of such games using polynomial projection methods and gives potential for computing modified game versions with corner solutions by employing a homotopy approach.

Appendix

1.1 Derivation of Function $\phi \left( \lambda \right)$ in Eq. (59)

Substituting (58) into (19) leads to

$$\begin{aligned} \phi \left( \lambda \right) =\frac{1}{A-\rho }\left\{ \frac{4\left( N-1\right) }{9\kappa ^{2}}\left( 1-\lambda \right) ^{2}+\left[ N+\xi \left( N-1\right) \right] \frac{4\lambda ^{\xi }}{9\kappa ^{2}}\int \lambda ^{-\xi -1}\left( 1-\lambda \right) ^{2}\mathrm{d}\lambda \right\} \text { .} \nonumber \\ \end{aligned}$$

(97)

To calculate the integral in (97) we expand the quadratic form, namely

$$\begin{aligned} \int \lambda ^{-\xi -1}\left( 1-\lambda \right) ^{2}\mathrm{d}\lambda =\int \left( \lambda ^{-\xi -1}-2\lambda ^{-\xi }+\lambda ^{-\xi +1}\right) \mathrm{d}\lambda \text { ,} \end{aligned}$$

which leads to

$$\begin{aligned} \int \lambda ^{-\xi -1}\left( 1-\lambda \right) ^{2}\mathrm{d}\lambda =\frac{1}{-\xi +2}\lambda ^{-\xi }\left( \lambda ^{2}-2\frac{-\xi +2}{-\xi +1}\lambda + \frac{-\xi +2}{-\xi }\right) \text { .} \end{aligned}$$

(98)

Combining (98) with (97) gives

$$\begin{aligned} \phi \left( \lambda \right) =\alpha \cdot \left( \lambda ^{2}-2\lambda +1\right) +\beta \cdot \left( \lambda ^{2}-2\frac{-\xi +2}{-\xi +1}\lambda + \frac{-\xi +2}{-\xi }\right) \text { ,} \end{aligned}$$

(99)

where

$$\begin{aligned} \alpha \equiv \frac{4\left( N-1\right) }{9\kappa ^{2}\left( A-\rho \right) } \text { and }\beta \equiv \frac{4\left[ N+\xi \left( N-1\right) \right] }{ 9\kappa ^{2}\left( A-\rho \right) \left( -\xi +2\right) }\text { .} \end{aligned}$$

(100)

Collecting terms in (99) leads to

$$\begin{aligned} \phi \left( \lambda \right)&=\left( \alpha +\beta \right) \left[ \lambda ^{2}-2\frac{\alpha +\zeta \beta }{\alpha +\beta }\lambda +\left( \frac{ \alpha +\zeta \beta }{\alpha +\beta }\right) ^{2}\right] \\&\quad +\,\alpha +\beta \frac{-\xi +2}{-\xi }-\left( \alpha +\beta \right) \left( \frac{\alpha +\zeta \beta }{\alpha +\beta }\right) ^{2}\text { ,} \end{aligned}$$

or

$$\begin{aligned} \phi \left( \lambda \right) =\left( \alpha +\beta \right) \left( \lambda - \frac{\alpha +\zeta \beta }{\alpha +\beta }\right) ^{2}+\alpha +\beta \frac{ -\xi +2}{-\xi }-\left( \alpha +\beta \right) \left( \frac{\alpha +\zeta \beta }{\alpha +\beta }\right) ^{2}\text { ,} \end{aligned}$$

(101)

where

$$\begin{aligned} \zeta \equiv \frac{-\xi +2}{-\xi +1}\text { .} \end{aligned}$$

(102)

Substituting the expressions for $\alpha$, $\beta$ and $\zeta$ given by (100) and (102) into (101) gives Eq. (59), together with the expressions given by (60), (61) and (62). $\square$

1.2 Proof of Inequality (66)

Fix any value of $\rho$ and observe that

$$\begin{aligned} \theta =F\left( A\right) G\left( N\right) \text { ,} \end{aligned}$$

(103)

where

$$\begin{aligned} F\left( A\right) =\frac{3A-2\rho }{2A-\rho }\text { } \end{aligned}$$

(104)

and

$$\begin{aligned} G\left( N\right) =\frac{2N-1}{3N-2}\text { .} \end{aligned}$$

(105)

Notice that, according to (65), $A>2/3\rho >1/2\rho$, and therefore, $F\left( A\right) >0$. Moreover,

$$\begin{aligned} F^{\prime }\left( A\right) =\frac{\rho }{\left( 2A-\rho \right) ^{2}}>0\text { ,} \end{aligned}$$

(106)

and

$$\begin{aligned} \underset{A\downarrow \frac{2}{3}\rho }{\lim }F\left( A\right) =0\text { , and }\underset{A\rightarrow \infty }{\lim }F\left( A\right) =\frac{3}{2} \text { .} \end{aligned}$$

(107)

Combining (107) with (106) gives

$$\begin{aligned} 0<F\left( A\right) <\frac{3}{2}\text { for all }A\text {, given any }\rho ~ \text {complying with }(65). \end{aligned}$$

(108)

Similarly, notice that

$$\begin{aligned} G^{\prime }\left( N\right) =\frac{-1}{\left( 3N-2\right) ^{2}}<0\text { ,} \end{aligned}$$

(109)

while

$$\begin{aligned} G\left( 1\right) =1\text { and }\underset{N\rightarrow \infty }{\lim } G\left( N\right) =\frac{2}{3}\text { . } \end{aligned}$$

(110)

Therefore, (103), (109) and (110) imply

$$\begin{aligned} \frac{2}{3}<G\left( N\right) \le 1\text { , for all }N\in \left\{ 1,2,\ldots \right\} \text { .} \end{aligned}$$

(111)

Combining (103), (109) and (111) proves inequality (66). $\square$

1.3 Why the Case of $\lambda -\theta \ge 0$ in Eq. (63) is not Admissible

Substituting $\lambda -\theta \ge 0$ into (63) gives

$$\begin{aligned} \lambda =\phi ^{-1}\left( k\right) =\frac{1}{\eta ^{\frac{1}{2}}}\left( k-\psi \right) ^{\frac{1}{2}}+\theta \text { , if }\lambda \ge \theta \text { .} \end{aligned}$$

(112)

Recall that $\lambda =J^{\prime }\left( k\right)$. Differentiating the right-hand side of (112), we can see that $J^{\prime \prime }\left( k\right) =1/\left( 2\eta ^{1/2}\right) \left( k-\psi \right) ^{-1/2}>0$ for all $k\ge \max \left\{ 0,\psi \right\}$. Yet, $J^{\prime \prime }\left( k\right) >0$ is not a property of the value function that complies with the transversality condition. To see this, consider the first-order condition given by (5), which implies

$$\begin{aligned} u^{\prime }\left( C\left( k\right) \right) =J^{\prime }\left( k\right) \text { .} \end{aligned}$$

(113)

Differentiating both sides of (113) implies

$$\begin{aligned} J^{\prime \prime }\left( k\right) =u^{\prime \prime }\left( C\left( k\right) \right) C^{\prime }\left( k\right) \text { .} \end{aligned}$$

(114)

Because $u^{\prime \prime }\left( c\right) <0$ for all $c\$complying with (57),

$$\begin{aligned} J^{\prime \prime }\left( k\right) >0~\ \text {combined with } (114)\text { imply that }C^{\prime }\left( k\right) <0\text { .} \end{aligned}$$

(115)

Yet, remember that the budget constraint given by (1) implies

$$\begin{aligned} {\dot{k}}\left( t\right) =Ak\left( t\right) -NC\left( k\left( t\right) \right) \text { .} \end{aligned}$$

(116)

Combining (116) with $C^{\prime }\left( k\right) <0$ means that the right-hand side of Eq. (116) is upward sloping in $k\left( t\right)$. Based on (21), we can combine (112) with (58) to obtain the explicit formula for $C\left( k\right)$, namely

$$\begin{aligned} C\left( k\right) =\frac{4}{9\kappa ^{2}}\left[ 1-\theta -\frac{1}{\eta ^{ \frac{1}{2}}}\left( k-\psi \right) ^{\frac{1}{2}}\right] ^{2}\text { .} \end{aligned}$$

(117)

Using (117), we derive the first and second derivatives of the strategies $C\left( k\right)$, i.e.,

$$\begin{aligned} C^{\prime }\left( k\right) =\frac{-4}{9\kappa ^{2}\eta ^{\frac{1}{2}}}\left[ \left( 1-\theta \right) \left( k-\psi \right) ^{-\frac{1}{2}}-\frac{1}{\eta ^{\frac{1}{2}}}\right] \text { .} \end{aligned}$$

(118)

Notice that (118) combined with (115) implies that

$$\begin{aligned} C^{\prime }\left( k\right)<0~\ \text {holds if }k<\psi +\eta \left( 1-\theta \right) ^{2}\text { .} \end{aligned}$$

(119)

In addition, (118) implies

$$\begin{aligned} C^{\prime \prime }\left( k\right) =\frac{2}{9\kappa ^{2}\eta ^{\frac{1}{2}}} \left( 1-\theta \right) \left( k-\psi \right) ^{-\frac{3}{2}}>0~\text {.} \end{aligned}$$

(120)

Fig. 3

Properties of the decision rule if $\lambda -\theta \ge 0$

Fig. 4

Resource dynamics in the case where A is such that $\psi >0$

Fig. 5

Resource dynamics in the case where A is such that $\psi <0$

All properties of $C\left( k\right)$ described by (57), (117), (118), (119) and (120) are depicted in Fig. 3, where the shaded areas indicate value regions where the strategies $C\left( k\right)$ are not defined. Without loss of generality, Fig. 3 depicts a case where $\psi >0$. The case of $\psi \le 0$ would simply depict a picture with $C\left( k\right)$ exhibiting the same properties for $k\in \left[ 0,\psi +\eta \left( 1-\theta \right) ^{2}\right]$.

Introducing strategies $C\left( k\right)$ into (1), we obtain

$$\begin{aligned} {\dot{k}}=Ak-NC\left( k\right) \text { .} \end{aligned}$$

(121)

Differentiating (121) with respect to k, we obtain

$$\begin{aligned} \frac{\partial {\dot{k}}}{\partial k}=A-NC^{\prime }\left( k\right) >0\text { , for all }k\in \left[ \max \left\{ 0,\psi \right\} ,\psi +\eta \left( 1-\theta \right) ^{2}\right] \text { .} \end{aligned}$$

(122)

Equation (122) is a consequence of Eq. (119). Eq. (122) implies unstable dynamics of k. These unstable dynamics of k further imply a violation of the feature that the solution is interior. In the absence of an interior solution, Proposition 1 does not apply, and therefore, the closed-form solution of the strategies, $C\left( k\right)$, given by (117), is invalid.

Figures 4 and 5 depict (121) and the dynamics of k, based on all parametric cases. Specifically, we distinguish cases of parametric values of A such that $\psi >0$ and otherwise. Based on Eq. (60), after some algebra, and making use of the parametric constraint given by (65), we can show that

$$\begin{aligned} \psi> & {} 0\Leftrightarrow \left( A-\frac{3N-2}{4N-3}\rho \right) \left( A-\rho N\right) >0\Leftrightarrow A\in \left( \frac{2}{3}\rho ~,~\frac{3N-2}{4N-3} \rho \right) \cup \left( \rho N~,~\infty \right) \text { ,} \nonumber \\\end{aligned}$$

(123)

$$\begin{aligned} \psi= & {} 0\Leftrightarrow A=\rho N\text { or }A=\frac{3N-2}{4N-3}\rho \text { ,} \end{aligned}$$

(124)

and

$$\begin{aligned} \psi <0\Leftrightarrow A\in \left( \frac{3N-2}{4N-3}\rho ~,~\rho N\right) \text { .} \end{aligned}$$

(125)

A common feature between Figs. 4 and 5 is that when $k=\psi +\eta \left( 1-\theta \right) ^{2}$, which is the upper bound of k for which $C\left( k\right)$ is admissible in this case of $\lambda -\theta \ge 0$, ${\dot{k}}>0$. To see this, insert $k=\psi +\eta \left( 1-\theta \right) ^{2}$ into (121) to obtain

$$\begin{aligned} \left. {\dot{k}}\right| _{k=\psi +\eta \left( 1-\theta \right) ^{2}}=A \left[ \psi +\eta \left( 1-\theta \right) ^{2}\right] >0\text { .} \end{aligned}$$

(126)

Inequality (126) justifies why in both Figs. 4 and 5 the curve depicting the law of motion for k is above the 0 line.

To understand why there are two curves depicting (121) in Fig. 4, which focuses on parameter values implying $\psi >0$, consider the equivalence given by (123) and focus on the specific value of k, $k=\psi$. By inserting $k=\psi ~$into (121),

$$\begin{aligned} \left. {\dot{k}}\right| _{k=\psi }=A\psi -\frac{4}{9\kappa ^{2}}\left( 1-\theta \right) ^{2}>0\Leftrightarrow \left( N-1\right) \left( A-\rho N\right) >0\text { .} \end{aligned}$$

(127)

In the trivial case of $N=1$, $\left. {\dot{k}}\right| _{k=\psi }=0$. Yet, this does not correspond to an interior solution with free initial conditions. We therefore focus on cases with $N\ge 2$. When $N\ge 2$, the equivalence given by (127) implies that

$$\begin{aligned} \left. {\dot{k}}\right| _{k=\psi } > reqqless 0\Leftrightarrow A > reqqless \rho N\text { .} \end{aligned}$$

(128)

Given that

$$\begin{aligned} \frac{3N-2}{4N-3}\in \left( \frac{3}{4},1\right) \text { for all }N\in \left\{ 2,3,\ldots \right\} \text {, } \end{aligned}$$

the two curves depicting (121) in Fig. 4 are justified. The equivalence implied by (123) implies that, in the case where $A\in \left( 2/3\rho ~,~\left( 3N-2\right) /\left( 4N-3\right) \rho \right)$, there is a value $k^{ss}$ for which $\left. {\dot{k}}\right| _{k=k^{ss}}=0$ . Yet, this unstable zero growth value does not correspond to an interior solution with free initial conditions for the problem.

Figure 5 focuses on the case implied by (125). Because of (128), in Fig. 5 we have once more a value $k^{ss}$ for which $\left. {\dot{k}}\right| _{k=k^{ss}}=0$. Again, this unstable zero growth value does not correspond to an interior solution with free initial conditions for the problem. The same problem arises for the two specific values of A given by (124), for which $\psi =0$.

In summary, the case of $\lambda -\theta \ge 0$ does not correspond to an interior solution and it should, therefore, be discarded. $\square$

1.4 Proof of Equivalence (85)

To prove that $\theta > reqqless 1\Leftrightarrow A > reqqless \rho N$, use (61) to obtain

$$\begin{aligned} \theta > reqqless 1\Leftrightarrow \frac{3A-2\rho }{2A-\rho }\frac{2N-1}{ 3N-2} > reqqless 1\text { .} \end{aligned}$$

(129)

Based on the parametric constraint given by (75), numerators and denominators in the fractions appearing in (129) are strictly positive. This feature leads to verifying that

$$\begin{aligned} \frac{3A-2\rho }{2A-\rho }\frac{2N-1}{3N-2} > reqqless 1\Leftrightarrow A > reqqless \rho N\text { ,} \end{aligned}$$

which confirms the first part of (85) that $\theta > reqqless 1\Leftrightarrow A > reqqless \rho N$.

For proving the second part of (85) that ${\bar{k}} > reqqless \eta \Leftrightarrow A > reqqless \rho N$, observe that (62) and (73) imply

$$\begin{aligned} \frac{{\bar{k}}}{\eta }=\frac{\left( 3A-2\rho \right) N}{3AN-2A}\text {.} \end{aligned}$$

(130)

Using the parametric constraint given by (75), which also implies $\eta >0$, together we can show that

$$\begin{aligned} \frac{{\bar{k}}}{\eta } > reqqless 1\Leftrightarrow A > reqqless \rho N\text { , } \end{aligned}$$

which proves the second part of (85) that ${\bar{k}} > reqqless \eta \Leftrightarrow A > reqqless \rho N$.

Finally, for proving that ${\bar{k}}\left( 2-\theta \right) -\eta \theta > reqqless 0\Leftrightarrow A > reqqless \rho N$, use (61) and (62) to see that

$$\begin{aligned}&{\bar{k}}\left( 2-\theta \right) -\eta \theta > reqqless 0\Leftrightarrow \frac{N}{A}\left( 2-\frac{6AN-3A-4\rho N+2\rho }{6AN-4A-3\rho N+2\rho } \right) > reqqless \frac{2N-1}{2A-\rho }\Leftrightarrow \\&\quad \Leftrightarrow \frac{6AN-5A-2\rho N+2\rho }{2A-\rho } > reqqless \frac{A}{N} \frac{2N-1}{2A-\rho }\Leftrightarrow \\&\quad \Leftrightarrow 2\left( 3A-\rho \right) N^{2}-\left( 5A-2\rho \right) N > reqqless A\left( 6N^{2}-7N+2\right) \Leftrightarrow \\&\quad \Leftrightarrow \left( N-1\right) \left( A-\rho N\right) > reqqless 0\text { ,} \end{aligned}$$

confirming that ${\bar{k}}\left( 2-\theta \right) -\eta \theta > reqqless 0\Leftrightarrow A > reqqless \rho N$ for all $N\ge 2$. $\square$