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# Analysis of long-term strategies of riparian countries in transboundary river basins

Assume n countries ((nge 2)) are located in a transboundary river basin and they are the players of an evolutionary game in which the countries’ strategies concerning water sharing in the basin evolve over time. Each country can choose between a cooperative strategy or a non-cooperative strategy. The game’s interactions and players’ payoffs vary with the number n and the location of the countries within the river basin, specifically, in relation to whether they are upstream-located or downstream-located countries within the river basin. The probability of country i choosing a cooperative strategy is herein denoted by ({x}_{1}^{(i)}), (i=1, 2, dots , n), and there are ({2}^{n}) payoff sets for all the combinations of the countries’ strategies. This paper assesses the interactions between three countries sharing a transboundary river basin.

### Problem description

Let 1, 2, and 3 denote three countries sharing a transboundary river basin. Country 1 is upstream and countries 2 and 3 are located downstream. Country 1 can use maximum amount of the water of the river and choose not to share it with the downstream countries. This strategy, however, may trigger conflict with the two other countries of political, social, economic, security, and environmental natures. Instead, Country 1 can release excess water to be shared by Countries 2 and 3. Countries 2 and 3 are inclined to cooperate with Country 1 unless other benefits emerge by being non-cooperative with Country 1.

There are two types of benefits and one type of cost in the payoff matrix of the assumed problem that are economic in nature. The first is a water benefit earned by a country from receiving the water from the transboundary river. The set of benefits related to water use includes economic benefits earned from agricultural, urban, and industrial development benefits. It should be noted that the water benefit for Country 1 means the economic benefit of consuming more water than its water right from the river. So, water benefits of Country 2 and 3 are the economic benefit of consuming excess water of upstream which is released by Country 1.

The second is a potential benefit earned from the cooperative strategy of a country. Cooperation benefits stem from sustainability conditions like social interests, environmental benefits and political conjunctures such as international alliances and harmony from amicable interactions with neighboring countries. The parameters F and E (water benefit and potential benefit, respectively) encompass a number of benefit parameters; nevertheless, parameters were simplified to two benefit parameters to simplify the complexity of the water-sharing problem. Costs forced on other countries from non-cooperation by a country involves commercial, security, political, diplomatic, military, and environmental costs. Figure 1 displays the locations of three countries and their shifting interactions in a transboundary river basin.

### Basic assumptions

The evolutionary game model of interactions between riparian countries in the transboundary river basin rests on the following assumptions:

### Assumption 1

There are three countries (i.e., players) in the game of transboundary water sharing, each seeking to maximize its payoff from the game.

### Assumption 2

Country 1 has two possible strategies. One is for Country 1 to release a specified amount of water to the downstream countries (this would be Country 1’s cooperative strategy). The cooperative strategy by Country 1 would produce benefits F2 and F3 to Countries 2 and 3, respectively. By being cooperative Country 1 would attain a benefit E1 called the potential benefit from cooperative responses from the downstream countries. The other strategy is for Country 1 to deny water to the downstream countries (this would be Country 1’s non-cooperative strategy), in which case Country 1 would earn the water benefit F1 from using water that would otherwise be released, but would forego the potential benefit E1. Moreover, by pursuing a non-cooperative strategy Country 1 would inflict a cost C1m to the downstream countries.

### Assumption 3

There are two possible strategies for Country 2. One is for Country 2 to accept the behavior of Country 1 (this would be Country 2’s cooperative strategy), which would cause earning a potential benefit E2 to Country 2. Recall that if Country 2 acquiesces to Country 1’s cooperative behavior it would receive a benefit F2. Or, Country 2 may disagree with Country 1 (this would be Country 2’s non-cooperative strategy), in which case, Country 2 would lose benefit E2, and it would inflict a cost C2m to the other countries.

### Assumption 4

Similar to Country 2, Country 3 has two possible strategies. One is for Country 3 to agree Country 1’s behavior (this would be Country 3’s cooperative strategy) attaining a potential benefit E3. Recall that if Country 3 agrees with Country 1’s cooperative behavior it would gain a benefit F3. Another strategy for Country 3 is to oppose Country 1 (this would be Country 3’s non-cooperative strategy) missing the benefit E3 and forcing a cost C3m to the other countries.

Table 1 defines the benefits and costs that enter in the transboundary water-sharing game described in this work. The payoff to country (i=mathrm{1,2},3) depends on its own strategy and on the strategies of the other countries, and each country may choose to be cooperative or non-cooperative. The strategies of country (i) are denoted by 1 (cooperation) and 2 (non-cooperation). The probabilities of country (i)’s strategies are denoted by ({x}_{1}^{(i)}) and by ({x}_{2}^{(i)}), in which the former represents cooperation and the latter represents non-cooperation. Clearly, ({x}_{1}^{(i)})+ ({x}_{2}^{(i)}) = 1. The payoff to country (i=1, 2, 3) when the strategies of Countries 1, 2, 3 are (j, k,l), respectively, where (j, k,l) may take the value 1 (cooperation) or 2 (non-cooperation) is denoted by ({U}_{jkl}^{left(iright)}). Thus, for instance, the payoff to country (i=2) is represented by ({U}_{212}^{(2)}) when Countries 1 and 3 are non-cooperative and Country 2’s strategy is cooperative. Evidently, there are 23 payoffs to each country given there are three countries involved and each can be cooperative or non-cooperative. Table 2 shows the symbols for the payoffs that accrue to each country under the probable strategies.

### Formulation of the transboundary water-sharing strategies as an evolutionary game

The expected payoff to country (i) is expressed by the following equation:

$${U}^{(i)}=sumlimits_{j = 1}^2 {sumlimits_{k = 1}^2 {sumlimits_{l = 1}^2} } {x}_{j}^{(1)}{x}_{k}^{(2)}{x}_{l}^{(3)} {U}_{jkl}^{(i)} quad i=1, 2, 3$$

(1)

The following describe the expected payoffs of Country 1 when it acts cooperatively (({U}_{1}^{(1)})) or non-cooperatively (({U}_{2}^{(1)})):

$${U}_{1}^{(1)}={x}_{1}^{(2)}{x}_{1}^{(3)}{U}_{111}^{left(1right)}+{x}_{1}^{(2)}{x}_{2}^{(3)}{U}_{112}^{left(1right)}+{x}_{2}^{(2)}{x}_{1}^{(3)}{U}_{121}^{(1)}+{x}_{2}^{(2)}{x}_{2}^{(3)}{U}_{122}^{(1)}$$

(2)

$${U}_{2}^{(1)}={x}_{1}^{(2)}{x}_{1}^{(3)}{U}_{211}^{left(1right)}+{x}_{1}^{(2)}{x}_{2}^{(3)}{U}_{212}^{left(1right)}+{x}_{2}^{(2)}{x}_{1}^{(3)}{U}_{221}^{left(1right)}+{x}_{2}^{(2)}{x}_{2}^{(3)}{U}_{222}^{left(1right)}$$

(3)

Therefore, the expected payoff of Country 1 is ({U}^{(1)}) which is equal to:

$${U}^{(1)}={x}_{1}^{(1)}{U}_{1}^{(1)}+{x}_{2}^{(1)}{U}_{2}^{(1)}= sumlimits_{j = 1}^2 {sumlimits_{k = 1}^2 {sumlimits_{l = 1}^2 } }{x}_{j}^{(1)}{x}_{k}^{(2)}{x}_{l}^{(3)} {U}_{jkl}^{(1)}$$

(4)

The expected payoffs of Countries 2 and 3 can be similarly obtained as done for Country 1. The cooperative and non-cooperative expected payoffs of all countries can be expressed in terms of the payoffs listed in Table 1. The results are found in Appendix A.

### Replication dynamics equations

The replication dynamics equations describe the time change of the probabilities of a player’s strategies. The replication dynamics equation of Countries (i) is denoted by ({G}^{(i)}left({x}_{1}^{(i)}right)) which is as follow22:

$${G}^{(i)}left({x}_{1}^{(i)}right)=frac{d{x}_{1}^{(i)}}{dt}={x}_{1}^{(i)}left({U}_{1}^{(i)}-{U}^{(i)}right)$$

(5)

The replication dynamics equations of Countries 1, 2 and 3 are presented in Appendix B according to the benefits and costs showed in Table 1.

### Stability analysis of a country’s strategies

Under the assumption of bounded rationality each country does not know which strategies may lead to the optimal solution in the game. Therefore, the countries’ strategies change over time until a stable (i.e., time-independent) solution named evolutionary stable strategy (ESS) is attained. The evolutionary stable theorem for replication dynamics equation states that a stable probability of cooperation ({x}_{1}^{(i)}) for country (i) occurs if the following conditions hold25: (1) ({G}^{(i)}left({x}_{1}^{(i)}right)=0), and (2) (d{G}^{(i)}left({x}_{1}^{(i)}right)/d{x}_{1}^{(i)}<0) when evaluated at ({x}_{1}^{(i)}), that is, when ({G}^{(i){^{prime}}}left({x}_{1}^{(i)}right)<0). The probability of cooperation ({x}_{1}^{(i)}) represents an ESS point if the former two conditions apply. Evidently, a stable probability of non-cooperation equals (1-{x}_{1}^{(i)}) when ({x}_{1}^{(i)}) exists. The method to obtain the stability probabilities, when they exist, is herein described in full for Country 1. It can be concluded from Equation (B1) of Appendix B that ({G}^{left(1right)}left({x}_{1}^{left(1right)}right)={x}_{1}^{left(1right)}{x}_{2}^{left(1right)}{A}_{0}left({x}_{1}^{left(2right)},{x}_{1}^{left(3right)}right)=0) if any of the following occurs: (i) ({x}_{1}^{(1)}=0), (ii) ({x}_{1}^{(1)}=1), (iii) ({A}_{0}left({x}_{1}^{(2)},{x}_{1}^{(3)}right)=0). Condition (iii) holds if the following is true (the constants ({a}_{1}), ({a}_{2}), ({a}_{3}), ({a}_{4}) are defined in Equation (B1) of Appendix B):

$${x}_{1}^{(2)}=-frac{{a}_{3 }{x}_{1}^{(3)}+{a}_{4}}{{a}_{1 }{x}_{1}^{(3)}+{a}_{2}}$$

(6)

Furthermore, ({G}^{(1){^{prime}}}=d{G}^{(1)}/d{x}_{1}^{(1)}=left(1-2{x}_{1}^{(1)}right){A}_{0}left({x}_{1}^{(2)},{x}_{1}^{(3)}right)). Therefore, the ESSs of Country 1 are evaluated as follows:

1. 1.

If ({x}_{1}^{(2)}=-frac{{a}_{3}{x}_{1}^{(3)}+{a}_{4}}{{a}_{1}{x}_{1}^{(3)}+{a}_{2}}), then ({A}_{0}=0), ({G}^{left(1right)}left({x}_{1}^{left(1right)}right)=0), and ({G}^{{left(1right)}^{{prime}}}left({x}_{1}^{left(1right)}right)=0), which means Country 1’ probabilities remain invariant over time;

2. 2.

If (0<{x}_{1}^{(2)}<-frac{{a}_{3}{x}_{1}^{(3)}+{a}_{4}}{{a}_{1}{x}_{1}^{(3)}+{a}_{2}}), then ({A}_{0}<0) and ({G}^{left(1right){^{prime}}}left({x}_{1}^{left(1right)}=0right)<0). So ({x}_{1}^{left(1right)}=0) is the ESS in which Country 1’s ESS is non-cooperative;

3. 3.

If (-frac{{a}_{3}{x}_{1}^{left(3right)}+{a}_{4}}{{a}_{1}{x}_{1}^{left(3right)}+{a}_{2}} <{x}_{1}^{left(2right)}<1), then ({A}_{0}>0) and ({G}^{left(1right){^{prime}}}left({x}_{1}^{left(1right)}=1right)<0). Thus ({x}_{1}^{left(1right)}=1) is the ESS in which Country 1’s ESS is cooperative.

The procedure for determining Country 1’s ESS was applied to Countries 2 and 3 to determine their ESSs. The results are as follows:

Country 2’s ESSs (the constants ({b}_{1}), ({b}_{2}), ({b}_{3}), ({b}_{4}) are defined in Equation (B6) Appendix B):

1. 1.

If ({x}_{1}^{(1)}=-frac{{b}_{3}{x}_{1}^{(3)}+{b}_{4}}{{b}_{1}{x}_{1}^{(3)}+{b}_{2}}), then ({B}_{0}=0), ({G}^{left(2right)}left({x}_{1}^{left(2right)}right)=0), and ({G}^{{left(2right)}^{{prime}}}left({x}_{1}^{left(2right)}right)=0), which means Country 2’ probabilities remain invariant over time;

2. 2.

If (0<{x}_{1}^{(1)}<-frac{{b}_{3}{x}_{1}^{(3)}+{b}_{4}}{{b}_{1}{x}_{1}^{(3)}+{b}_{2}}), then ({B}_{0}<0) and ({G}^{left(2right){^{prime}}}left({x}_{1}^{left(2right)}=0right)<0). Therefore, ({x}_{1}^{left(2right)}=0) is the ESS in which Country 2’s ESS is non-cooperative;

3. 3.

If (-frac{{b}_{3}{x}_{1}^{left(3right)}+{b}_{4}}{{b}_{1}{x}_{1}^{left(3right)}+{b}_{2}} <{x}_{1}^{left(1right)}<1), then ({B}_{0}>0) and ({G}^{left(2right){^{prime}}}left({x}_{1}^{left(2right)}=1right)<0). Thus, ({x}_{1}^{left(2right)}=1) is the ESS in which Country 2’s ESS is cooperative.

Country 3’s ESSs (the constants ({c}_{1}), ({c}_{2}), ({c}_{3}), ({c}_{4}) are defined in Equation (B11) of Appendix B):

1. 1.

If ({x}_{1}^{(1)}=-frac{{c}_{3}{x}_{1}^{(2)}+{c}_{4}}{{c}_{1}{x}_{1}^{(2)}+{c}_{2}}), then ({C}_{0}=0), ({G}^{left(3right)}left({x}_{1}^{left(3right)}right)=0), and ({G}^{{left(2right)}^{{prime}}}left({x}_{1}^{left(3right)}right)=0), which means Country 3’ probabilities remain invariant over time;

2. 2.

If (0<{x}_{1}^{(1)}<-frac{{c}_{3}{x}_{1}^{(2)}+{c}_{4}}{{c}_{1}{x}_{1}^{(2)}+{c}_{2}}), then ({C}_{0}<0) and ({G}^{left(3right){^{prime}}}left({x}_{1}^{left(3right)}=0right)<0). Thus, ({x}_{1}^{left(3right)}=0) is the ESS in which Country 3’s ESS is non-cooperative;

3. 3.

If (-frac{{c}_{3}{x}_{1}^{left(3right)}+{c}_{4}}{{c}_{1}{x}_{1}^{left(3right)}+{c}_{2}} <{x}_{1}^{left(1right)}<1), then ({C}_{0}>0) and ({G}^{left(3right){^{prime}}}left({x}_{1}^{left(3right)}=1right)<0). So, ({x}_{1}^{left(3right)}=1) is the ESS in which Country 3’s ESS is cooperative.

It is evident that the evolutionary strategy of each country is dependent on the other countries’ strategies. The ESS replication dynamic equations were obtained for each country under the specified conditions.

### Stability analysis of multi-country strategies

The stable evolutionary strategies of each country were analyzed individually in the previous sections. Yet, the three countries interact with each other and their strategies may change simultaneously. Table 1 provides the possible payoffs that can arise from this paper’s game. This section evaluates the stability of the equilibrium strategies for the countries’ strategies listed in Table 2. Each country’s strategy becomes stable when the replication dynamics equations are equal to 0 (i.e., ({G}^{left(iright)}({x}_{1}^{left(iright)})) = 0, for (i=1, 2, 3)). The solution of the set of replication dynamics equations being equal to 0 yields the equilibrium probabilities governing the countries’ ESSs (evolutionary stable strategies). Björnerstedt and Jörgen26 demonstrated that the stable solution to tripartite problems of the type herein considered must be a strict Nash equilibrium point, which means the probable stable solutions to the problem herein entertained are (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1) and (1,1,1).

According to Friedman’s5 proposed method the Jacobian matrix of the replication dynamics system and an analysis of eigenvalues of the matrix are needed to investigate the stability of equilibrium points. The Jacobian matrix J for the replication dynamics system of interactions of n = 3 countries is as follows:

$$J=left[begin{array}{ccc}frac{partial {G}^{(1)}({x}_{1}^{(1)})}{partial {x}_{1}^{(1)}}& frac{partial {G}^{(1)}({x}_{1}^{(1)})}{partial {x}_{1}^{(2)}}& frac{partial {G}^{(1)}({x}_{1}^{(1)})}{partial {x}_{1}^{(3)}} frac{partial {G}^{(2)}({x}_{1}^{(2)})}{partial {x}_{1}^{(1)}}& frac{partial {G}^{(2)}({x}_{1}^{(2)})}{partial {x}_{1}^{(2)}}& frac{partial {G}_{2}(y)}{partial {x}_{1}^{(3)}} frac{partial {G}^{left(3right)}({x}_{1}^{left(3right)})}{partial {x}_{1}^{(1)}}& frac{partial {G}^{left(3right)}({x}_{1}^{left(3right)})}{partial {x}_{1}^{(2)}}& frac{partial {G}^{left(3right)}({x}_{1}^{left(3right)})}{partial {x}_{1}^{(3)}}end{array}right]=left[begin{array}{ccc}{J}_{11}& {J}_{12}& {J}_{13} {J}_{21}& {J}_{22}& {J}_{23} {J}_{32}& {J}_{32}& {J}_{33}end{array}right]$$

(7)

The elements ({J}_{ij}) of the Jacobian matrix (7) may be written in terms of the benefits and costs introduced in Table 1. The equations for the nine elements of the Jacobian matrix (7) are presented in Appendix C. The ESSs probabilities are such that all the eigenvalues of the Jacobian matrix are negative. The eigenvalues of the Jacobian matrix are obtained by solving the equation (left|left(J-lambda Iright)right|=0) where | | denotes the determinant of a matrix, (I) and (lambda ) denote respectively the identity matrix and the vector of eigenvalues. The signs of eigenvalues are determined at each equilibrium point shown in Table 3. It is evident from Table 3 that there is no definitive stable solution for this problem, which is one with all certainly negative eigenvalues. Such a solution would be arrived at eventually if it existed. However, any point with uncertain stability can be an ESS point. The points (0,0,0), (1,0,0), (0,1,0), and (0,0,1) could be ESSs depending on the sign of the constants ({a}_{4}), ({b}_{4}), ({c}_{4}) that appear in the replication dynamics equations. The stable point (0,0,0) occurs when no country cooperates. The constraints to reach this stable point are ({E}_{1}-{F}_{1}-{C2}_{3}-{C3}_{3}+{C2}_{7}+{C3}_{7}<0), ({E}_{2}-{H}_{3}-{C3}_{6}+{C1}_{7}+{C3}_{8}<0), and ({E}_{3}-{C1}_{6}-{C2}_{6}+{C1}_{8}+{C2}_{8}<0). The constraints to achieve cooperation of the upstream country and non-cooperation of the downstream countries, i.e. (1,0,0), are ({E}_{1}-{F}_{1}-{C2}_{3}-{C3}_{3}+{C2}_{7}+C3>0), ({E}_{2}-{C1}_{3}-{C3}_{6}+{C1}_{7}+{C3}_{8}<0), and ({E}_{3}-{C1}_{6}-{C2}_{6}+{C1}_{8}+{C2}_{8}<0). The stable point (0,1,0) means only the downstream Country 2 chooses to be cooperative occurs when the following constraints are met: ({E}_{1}-{F}_{1}-{C2}_{3}-{C3}_{3}+{C2}_{7}+{C3}_{7}<0), ({E}_{2}-{H}_{3}-{C3}_{6}+{C1}_{7}+{C3}_{8}>0), and ({E}_{3}-{C1}_{6}-{C2}_{6}+{C1}_{8}+{C2}_{8}<0). The point (0,0,1) may occur when only the downstream Country 3 chooses to be cooperative, in which case the constraints ({E}_{1}-{F}_{1}-{C2}_{3}-{C3}_{3}+{C2}_{7}+{C3}_{7}<0), ({E}_{2}-{C1}_{3}-{C3}_{6}+{C1}_{7}+{C3}_{8}<0), and ({E}_{3}-{C1}_{6}-{C2}_{6}+{C1}_{8}+{C2}_{8}>0) must be satisfied.

The point (0,1,1) means the two downstream countries choose to cooperate in spite of Country 1 being non-cooperative. For this to occur the following constraints must be met: ({E}_{1}<{F}_{1}), ({E}_{2}+{C1}_{5}-{C1}_{1}>0), and ({E}_{3}+{C1}_{4}-{C1}_{2}>0). The point (1,1,1) signifies cooperation by all countries, which takes place when ({E}_{1}>{F}_{1}). In this case, the potential benefit of Country 1 due to the cooperation with downstream countries exceeds the benefit that would accrue if it retained all the water for itself. The points (1,1,0) and (1,0,1) cannot be stable solutions because of the positive eigenvalues in their Jacobian matrix. Therefore, there are six probable stable scenarios for the interaction of three countries in the transboundary river basin.

Source: Ecology - nature.com