The 'Global Commons' from a Mathematical Perspective | “全球公地”的数学视角

📚 The ‘Global Commons’ from a Mathematical Perspective | “全球公地”的数学视角

The ‘global commons’ refers to resources that are shared by all of humanity, such as the atmosphere, the oceans, outer space, and Antarctica. These resources are finite yet open to all, creating a classic challenge for sustainable management. Mathematics provides powerful tools to analyse how individuals, companies, and nations make decisions about these shared resources, and to design strategies that avoid depletion or collapse.

“全球公地”指的是全人类共享的资源，例如大气、海洋、外层空间和南极。这些资源是有限的，却又向所有人开放，因此构成了可持续管理的经典挑战。数学提供了强大的工具，用以分析个人、企业和国家如何就这些共享资源作出决策，并设计能够避免资源耗竭或系统崩溃的策略。

1. The Tragedy of the Commons | 公地悲剧

In 1968, ecologist Garrett Hardin described the ‘tragedy of the commons’: when individuals act independently and rationally according to their own self-interest, they deplete a shared resource even though it is not in anyone’s long-term interest. Mathematically, this can be modelled as a non-cooperative game where the dominant strategy for each player leads to a suboptimal outcome for all.

1968年，生态学家加勒特·哈丁描述了“公地悲剧”：当每个人根据自己的利益独立而理性地行动时，他们最终会耗尽共享资源，即使这种耗尽并不符合任何人的长期利益。在数学上，这可以建模为非合作博弈，其中每个参与者的占优策略导致了对于所有人来说都非最优的结果。

2. Game Theory: The Prisoner’s Dilemma | 博弈论：囚徒困境

Most shared-resource problems reduce to a Prisoner’s Dilemma. Each player can cooperate (restrain use) or defect (overuse). The payoffs are arranged so that defecting always gives a higher individual payoff, regardless of what the other does. Yet if both defect, both are worse off than if both had cooperated.

大多数共享资源问题都可归结为囚徒困境。每个参与者可以选择合作（限制使用）或背叛（过度使用）。收益的排列使得无论对方怎么做，背叛总能带来更高的个人收益。然而，如果双方都背叛，其结果比双方都合作时要差。

As a matrix, with two players A and B, where C = cooperate and D = defect:

用矩阵表示，假设两个参与者A和B，C代表合作，D代表背叛：

	B: C	B: D
A: C	3, 3	1, 4
A: D	4, 1	2, 2

Payoff (A, B): higher numbers are better.

收益 (A, B)：数值越高越好。

Here, defect is the strictly dominant strategy for both players; the unique Nash equilibrium is (D, D) with payoff (2, 2), which is worse than the cooperative outcome (3, 3).

在矩阵中，背叛是双方的严格占优策略；唯一的纳什均衡是（D, D），收益为（2, 2），比合作结果（3, 3）要差。

3. Nash Equilibrium Formally | 纳什均衡的形式定义

In a game with n players, a strategy profile (s₁*, s₂*, …, sₙ*) is a Nash equilibrium if, for every player i, the strategy sᵢ* is a best response to the others’ strategies. That is, no player can gain by unilaterally changing their strategy. The mathematical condition is:

在一个有 n 个参与者的博弈中，策略组合 (s₁*, s₂*, …, sₙ*) 是纳什均衡，当且仅当对于每个参与者 i，策略 sᵢ* 是对其他参与者策略的最佳反应。也就是说，没有任何参与者能够通过单方面改变策略而获益。其数学条件为：

uᵢ(s₁*, …, sᵢ*, …, sₙ*) ≥ uᵢ(s₁*, …, sᵢ, …, sₙ*) ∀ sᵢ ∈ Sᵢ

where uᵢ is the payoff function and Sᵢ is the set of possible strategies for player i. For public-goods and commons problems, the Nash equilibrium is usually not Pareto-efficient, which is why international cooperation is needed.

其中 uᵢ 是收益函数，Sᵢ 是参与者 i 的策略集合。对于公共物品和公地问题，纳什均衡通常不是帕累托有效的，这就是需要国际合作的原因。

4. A Two-Fishermen Model | 双渔民模型

Consider two fisherman who share a lake. Each can choose to fish with either 1 boat or 2 boats. If they both use 1 boat, the annual catch per boat is 10 tons. If one uses 1 boat and the other uses 2 boats, the sustainable harvest per boat is 8 tons for the 1-boat fisher and 6 tons for the 2-boat fisher. If both use 2 boats, each boat harvests only 4 tons.

考虑两名渔民共享一个湖泊。每人可以选择使用1艘船或2艘船捕鱼。若双方都用1艘船，每艘船的年渔获量为10吨。若一人使用1艘船、另一人使用2艘船，则1艘船者可捕8吨，2艘船者每艘船可捕6吨。若双方都用2艘船，每艘船仅能捕4吨。

Find the Nash equilibrium for the number of boats chosen. Let the payoff be total annual catch for each player. We compute:

为了找出关于船只数量的纳什均衡，设收益为每个玩家的年总渔获量。计算得：

Both 1 boat: (10, 10)

双方1艘船：(10, 10)
A:1, B:2: A gets 8, B gets 12 (since 2 boats × 6 tons)

A:1, B:2：A得8吨，B得12吨（2艘船×6吨）
A:2, B:1: A gets 12, B gets 8

A:2, B:1：A得12吨，B得8吨
Both 2 boats: each gets 8 (2 boats × 4 tons)

双方2艘船：每方得8吨（2艘船×4吨）

For each player, choosing 2 boats is always better: if the other chooses 1, 12 > 10; if the other chooses 2, 8 > (10? wait, if other chooses 2, then if you choose 1 you get 8, if you choose 2 you get 8, so 2 is still at least as good). Thus (2, 2) is the Nash equilibrium, giving total catch 16, whereas the cooperative optimum (1, 1) gives total catch 20.

对每个玩家而言，选择2艘船总是更好：如果对方选择1艘，则12 > 10；如果对方选择2艘，则你选1艘得8，选2艘也得8，因此2至少不差。因此（2, 2）是纳什均衡，总渔获量为16，而合作最优（1, 1）的总渔获量为20。

5. Logistic Growth and Harvesting | 逻辑斯蒂增长与捕捞

For renewable resources, a common model is the logistic equation. Let N(t) be the population size. In the absence of harvesting, the growth rate is described by:

对于可再生资源，常见的模型是逻辑斯蒂方程。设 N(t) 为种群数量。在没有捕捞的情况下，增长率可表示为：

dN/dt = r N (1 − N/K)

where r is the intrinsic growth rate and K is the carrying capacity. When a harvest rate H is applied, the equation becomes:

其中 r 为内禀增长率，K 为环境容纳量。当引入捕捞速率 H 后，方程为：

dN/dt = r N (1 − N/K) − H

A sustainable harvest occurs when dN/dt = 0, meaning that H = r N (1 − N/K). To keep the population stable at size N, the harvest must exactly equal the natural growth at that population.

可持续捕捞发生在 dN/dt = 0 时，即 H = r N (1 − N/K)。要使种群稳定在规模 N，捕捞量必须恰好等于该种群规模下的自然增长量。

6. Maximum Sustainable Yield | 最大可持续产量

The maximum sustainable yield (MSY) is the largest harvest that can be taken indefinitely without depleting the resource. It occurs at the population size that maximises the growth function G(N) = r N (1 − N/K). We find the critical point by differentiating:

最大可持续产量（MSY）是指在资源不枯竭的前提下可以永久捕获的最大渔获量。它发生在使增长函数 G(N) = r N (1 − N/K) 最大的种群规模处。通过求导找到临界点：

G'(N) = r − (2r N)/K = 0 ⟹ N = K/2

At N = K/2, the sustainable harvest is:

当 N = K/2 时，可持续捕捞量为：

H_max = r (K/2)(1 − 1/2) = rK/4

This result shows that the biologically optimal population for maximum harvest is half the carrying capacity, and the MSY is rK/4.

这一结果表明，达到最大渔获量的生物最优种群规模为环境容纳量的一半，而最大可持续产量为 rK/4。

7. Stability Analysis | 稳定性分析

If the harvest is constant H, the equilibria satisfy H = r N (1 − N/K), a quadratic equation. Its roots are:

若捕捞量恒定 H，则平衡点满足 H = r N (1 − N/K)，这是一个二次方程。其根为：

N₁ = (K/2) (1 + √(1 − 4H/(rK))) and N₂ = (K/2) (1 − √(1 − 4H/(rK)))

For H < rK/4, there are two equilibria. The larger one N₁ is stable, because if N falls slightly, dN/dt becomes positive (growth exceeds harvest) and N recovers. The smaller one N₂ is unstable: a small decrease sends the population to extinction. This threshold behaviour is critical for management.

当 H < rK/4 时，存在两个平衡点。较大的 N₁ 是稳定的，因为如果 N 略微下降，dN/dt 变为正（增长超过捕捞）从而恢复。较小的 N₂ 是不稳定的：稍微减少就会导致种群灭绝。这种阈值行为对于管理至关重要。

8. Discounting the Future | 未来贴现

In intertemporal optimisation, future benefits are discounted. The present value of a stream of harvest benefits H(t) with discount rate δ is:

在跨期优化中，未来收益会被贴现。收益流 H(t) 以贴现率 δ 计算的现值为：

PV = ∫₀^∞ H(t) e^(−δt) dt

If δ is high, future resource stocks are valued much less than current harvest, which encourages overexploitation. A high discount rate effectively tilts the Nash equilibrium even further away from sustainability.

如果 δ 很高，未来资源存量的价值远低于当前捕捞，从而鼓励过度开发。高贴现率实际上使纳什均衡进一步偏离可持续性。

9. Public Goods and Free-Riding | 公共物品与搭便车

Global commons often involve producing a public good, such as clean air, which is non-excludable and non-rivalrous. Let n countries each choose a contribution level gᵢ ≥ 0. The total public good provided is G = Σ gᵢ. Each country’s utility might be:

全球公地通常涉及提供一种公共物品，如清洁空气，它既不可排他又非竞争性。假设 n 个国家各自选择贡献水平 gᵢ ≥ 0，总公共物品供给量为 G = Σ gᵢ。每个国家的效用可能是：

uᵢ = α ln G − (gᵢ²)/2

Solving for the Nash equilibrium, each country sets its contribution such that the marginal benefit of a unit of G equals its marginal cost. But the social optimal requires ∑ (∂uᵢ/∂G) = (∂cᵢ/∂gᵢ), which leads to a larger total contribution. Thus the non-cooperative equilibrium always under-provides the public good.

在纳什均衡中，每个国家设定其贡献使得 G 的边际收益等于自身边际成本。但社会最优要求对所有国家的边际收益求和等于边际成本，这会导致更大的总贡献量。因此非合作均衡总是导致公共物品供给不足。

10. The Mathematics of Emission Targets | 减排目标的数学

International climate agreements often use formulas to allocate emission reductions. One simple model is to set total allowed emissions E_total, and distribute among countries in proportion to their population Pᵢ or historical emissions Hᵢ. A mathematical criterion for a fair allocation is that each country’s marginal abatement cost equals a common carbon price λ:

国际气候协议通常使用公式来分配减排责任。一个简单模型是设定总允许排放量 E_total，然后按人口 Pᵢ 或历史排放量 Hᵢ 在国家间分配。公平分配的一个数学标准是每个国家的边际减排成本等于统一的碳价格 λ：

MACᵢ(eᵢ) = λ for all i, and Σ eᵢ = E_total

where eᵢ is the emission level of country i. Solving this system gives an efficient allocation. If a country has a lower MAC, it should reduce more; countries with higher MAC can purchase permits from others, leading to a market equilibrium.

其中 eᵢ 是国家 i 的排放水平。求解该系统可得有效分配。如果某国边际减排成本更低，则应承担更多减排；边际成本较高的国家可以从其他国购买配额，从而形成市场均衡。

11. Tipping Points and Catastrophic Shifts | 临界点与灾难性转移

Some commons systems have thresholds beyond which irreversible damage occurs. A dynamical system with a tipping point can be expressed as:

一些公地系统存在阈值，一旦超过便发生不可逆破坏。具有临界点的动力系统可以表示为：

dx/dt = f(x, β)

where β is a control parameter (e.g., temperature). If there exists a critical value β_c such that for β > β_c, the stable equilibrium disappears, the system jumps to an alternative state. The distance from the current state to the threshold can be estimated from recovery rates: systems near a tipping point recover slowly. Mathematically, the recovery rate λ is the derivative of f at the equilibrium:

其中 β 是控制参数（例如温度）。如果存在临界值 β_c，当 β > β_c 时稳定平衡消失，系统会跳变到另一状态。当前状态与阈值的距离可以从恢复速率估计：接近临界点的系统恢复缓慢。数学上，恢复速率 λ 是 f 在平衡点处的导数：

λ = df/dx |_x=x*

As λ → 0, the system experiences critical slowing down, a warning sign of an approaching tipping point.

当 λ → 0 时，系统经历临界减速，这是临界点临近的预警信号。

12. Designing Institutions: Quotas and Taxes | 制度设计：配额与税收

Mathematics can also help design interventions. A common policy is to impose a tax t per unit of harvest. The effective harvest equation becomes dN/dt = r N (1 − N/K) − (h + t), shifting the stable equilibrium. If t is set equal to the externality cost, the Nash equilibrium moves closer to the social optimum. Individually transferable quotas (ITQs) restrict total allowable catch to a level T; because each quota holder can trade, the market price reflects the shadow value of the resource, leading to an efficient distribution of effort.

数学也可以帮助设计干预措施。一种常见政策是对每单位捕捞量征税 t。修正后的捕捞方程为 dN/dt = r N (1 − N/K) − (h + t)，从而改变稳定平衡点。如果 t 等于外部性成本，纳什均衡将接近社会最优。个体可转让配额（ITQ）将总允许捕捞量限制在 T 水平；由于配额持有者可以交易，市场价格反映资源的影子价值，促使捕捞努力实现高效配置。

Conclusion | 结论

The ‘global commons’ is not merely a biological or political issue; it is deeply mathematical. Game theory exposes the strategic logic of overuse, differential equations capture the dynamics of resource stocks, and optimisation theory helps design sustainable policies. Understanding these mathematical foundations empowers students to engage critically with pressing environmental challenges.

“全球公地”不仅是生物或政治问题，它深深植根于数学。博弈论揭示了过度使用的策略逻辑，微分方程刻画了资源存量的动态变化，优化理论则帮助设计可持续政策。理解这些数学基础，能让学生以批判性的视角应对紧迫的环境挑战。

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The ‘Global Commons’ from a Mathematical Perspective | “全球公地”的数学视角