Ecosystems under stress: A-Level Edexcel mathematical modelling | 压力下的生态系统：A-Level Edexcel 数学建模

📚 Ecosystems under stress: A-Level Edexcel mathematical modelling | 压力下的生态系统：A-Level Edexcel 数学建模

Ecosystems rarely sit in a steady state: populations grow, resources run short, predators hunt, and human activity adds extra pressure. In A-Level Edexcel Mathematics, differential equations give us a precise language to describe how an ecosystem responds to stress, from a fishery’s collapse to the recovery of a forest.

生态系统很少处于静止状态：种群会增长、资源会短缺、捕食者会捕猎，而人类活动还会增加额外压力。在 A-Level Edexcel 数学中，微分方程为我们提供了一种精确的语言，用来描述生态系统如何对压力做出反应，从渔业的崩溃到森林的恢复。

1. Modelling ecosystems with differential equations | 用微分方程建立生态系统模型

Suppose N(t) is the size of a population at time t. The rate of change dN/dt equals the birth rate minus the death rate, possibly plus migration. This is the starting point for all the models in this article.

假设 N(t) 表示某一时刻 t 的种群数量。变化率 dN/dt 等于出生率减去死亡率，可能再加上迁移量。这是本文所有模型的起点。

dN/dt = B(N) − D(N)

Here B(N) and D(N) are functions of the current population size, so the model captures density-dependent feedback. Ecosystems under stress usually show strong feedback, because overcrowding, disease or resource depletion tends to reduce growth.

这里的 B(N) 和 D(N) 都是当前种群数量的函数，因此模型捕捉了密度依赖的反馈。受压的生态系统通常表现出强烈的反馈，因为过度拥挤、疾病或资源枯竭往往会抑制增长。

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2. Exponential growth and its limits | 指数增长及其局限

The simplest model is Malthusian growth, where each individual reproduces at a constant rate r and deaths are also proportional to N. This gives dN/dt = rN, with solution N(t) = N₀eʳᵗ.

最简单的模型是马尔萨斯增长，即每个个体以恒定速率 r 繁殖，死亡也与 N 成正比。这给出了 dN/dt = rN，其解为 N(t) = N₀eʳᵗ。

Exponential growth is a useful baseline, but it cannot describe an ecosystem under stress for long. If r > 0, the population grows without bound, which is impossible when food, space and shelter are finite.

指数增长是一个有用的基准，但它无法长期描述受压的生态系统。如果 r > 0，种群将无限增长，这在食物、空间和庇护所都有限的情况下是不可能的。

dN/dt = rN ⇒ N(t) = N₀eʳᵗ

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3. Logistic growth model | 逻辑斯蒂增长模型

To include resource limits, Edexcel exam questions often use the logistic equation. The term (1 − N/K) reduces growth as N approaches the carrying capacity K.

为了纳入资源限制，Edexcel 考试题经常使用逻辑斯蒂方程。因子 (1 − N/K) 会在 N 接近环境容纳量 K 时降低增长率。

dN/dt = rN(1 − N/K)

The solution is an S-shaped curve: slow growth at first, rapid growth in the middle, then levelling off at K. In an exam you may be asked to find the time when the population is growing fastest; this occurs when N = K/2.

其解是一条 S 形曲线：起初增长缓慢，中期快速增长，然后稳定在 K。考试中可能会要求你找出种群增长最快的时间，这发生在 N = K/2 时。

N(t) = K / [1 + ((K − N₀) / N₀)e⁻ʳᵗ]

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4. Introducing stress: harvesting and pollution | 引入压力：捕捞与污染

An ecosystem under stress can be modelled by subtracting a constant harvesting rate h, or by adding a mortality term due to pollution. For a fishery, h represents the number of fish removed per unit time.

受压的生态系统可以通过减去一个恒定的捕捞率 h，或加入因污染造成的死亡项来建模。对渔业来说，h 代表单位时间内被捕获的鱼的数量。

dN/dt = rN(1 − N/K) − h

When h is small, the population can still recover; when h is too large, the population may collapse to zero. This threshold behaviour is a key idea in sustainability and appears in modelling questions.

当 h 较小时，种群仍然可以恢复；当 h 过大时，种群可能会崩溃到零。这种阈值行为是可持续性的关键思想，也出现在建模题中。

The stressed logistic model combines three parameters: r (intrinsic growth rate), K (carrying capacity) and h (stress). Changing h is an easy way to investigate how much pressure an ecosystem can absorb.

受压的逻辑斯蒂模型结合了三个参数：r（内禀增长率）、K（环境容纳量）和 h（压力）。改变 h 是研究生态系统能承受多大压力的一种简单方法。

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5. Equilibrium points and stability | 平衡点与稳定性

At equilibrium, dN/dt = 0. For the stressed logistic model, this means solving rN(1 − N/K) − h = 0, which rearranges to a quadratic.

在平衡状态，dN/dt = 0。对于受压的逻辑斯蒂模型，这意味着求解 rN(1 − N/K) − h = 0，整理后得到一个二次方程。

rN(1 − N/K) − h = 0 ⇒ rN − rN²/K − h = 0

The roots are the equilibrium population sizes. Using the quadratic formula gives the following, where the square root must be real for biological equilibria to exist.

这些根就是平衡种群数量。使用二次公式可以得出以下结果，其中平方根必须为实数，生物平衡才存在。

N* = (K/2) × [1 ± √(1 − 4h/(rK))]

If h > rK/4, the expression under the square root becomes negative, so there is no real equilibrium above zero. The ecosystem cannot support any positive population; extinction is the only steady state. If h < rK/4, there are two equilibria: a smaller, unstable one and a larger, stable one.

如果 h > rK/4，根号内的表达式变为负数，因此不存在正的实数平衡。生态系统无法支持任何正种群数量，灭绝是唯一的稳定状态。如果 h < rK/4，则有两个平衡点：一个较小的不稳定平衡点和一个较大的稳定平衡点。

Stability is determined by the sign of the derivative of f(N) = rN(1 − N/K) − h at the equilibrium. If f'(N*) < 0, the equilibrium is stable; if f'(N*) > 0, it is unstable.

稳定性由 f(N) = rN(1 − N/K) − h 在平衡点处的导数符号决定。如果 f'(N*) < 0，平衡是稳定的；如果 f'(N*) > 0，则不稳定。

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6. Predator-prey dynamics under stress | 压力下的捕食者-猎物动态

The Lotka-Volterra model links two populations: prey P and predator Q. The prey grows logistically or exponentially, while the predator relies on eating prey to survive.

Lotka-Volterra 模型联系了两个种群：猎物 P 和捕食者 Q。猎物以逻辑斯蒂或指数方式增长，而捕食者依靠捕食猎物生存。

dP/dt = aP − bPQ
dQ/dt = cPQ − dQ

Here a is the prey’s natural growth rate, b is the attack rate, c is the conversion efficiency of eaten prey into new predators, and d is the predator death rate. Under stress, one or more of these parameters may change, for example if hunting reduces the predator death rate or pollution lowers the prey’s growth rate.

这里 a 是猎物的自然增长率，b 是捕食率，c 是将吃掉的猎物转化为新捕食者的效率，d 是捕食者死亡率。在压力下，这些参数中的一个或多个可能发生变化，例如捕猎降低了捕食

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