📚 Ecosystems and Sustainability: Differential Equation Modelling | 生态系统与可持续性：微分方程建模
In A-Level Mathematics, the study of ecosystems and sustainability provides a rich context for applying differential equations, exponential and logistic models, and equilibrium analysis. This article connects Edexcel Pure Mathematics techniques to real-world ecological questions, including population growth, carrying capacity, harvesting limits and the maximum sustainable yield.
在 A-Level 数学中，生态系统与可持续性为微分方程、指数模型、逻辑斯蒂模型以及平衡分析提供了丰富的应用背景。本文将 Edexcel 纯数学技巧与真实的生态问题联系起来，包括种群增长、环境容纳量、收获限制以及最大可持续产量。
1. Modelling Population Growth | 种群增长建模
Ecology often begins with a simple question: how does a population change over time? In mathematics, we let P(t) denote the population size at time t. The rate of change dP/dt represents the instantaneous growth rate. If the growth rate is proportional to the current population, we obtain the exponential growth model.
生态学通常从一个简单的问题开始：种群随时间如何变化？在数学中，我们用 P(t) 表示时刻 t 的种群大小。变化率 dP/dt 表示瞬时增长率。如果增长率与当前种群成正比，我们就得到指数增长模型。
dP/dt = kP
Here k is a constant called the per-capita growth rate. The model assumes unlimited resources, so it is most suitable for describing the early stage of population growth or the growth of microorganisms in a laboratory culture.
这里 k 是一个常数，称为人均增长率。该模型假设资源无限，因此最适合描述种群增长的早期阶段或实验室培养中微生物的生长。
In Edexcel Pure Mathematics, this is a first-order separable differential equation. You are expected to solve it by separating the variables and applying an initial condition. The general solution is found by writing ∫ (1/P) dP = ∫ k dt, which gives ln|P| = kt + C.
在 Edexcel 纯数学中，这是一个一阶可分离变量微分方程。你需要通过分离变量并代入初始条件来求解。通解的求法为 ∫ (1/P) dP = ∫ k dt，得到 ln|P| = kt + C。
2. Exponential Growth in an Ideal Ecosystem | 理想生态系统中的指数增长
Solving the exponential model gives P = P₀ e^(kt), where P₀ is the initial population at t = 0. For example, if P₀ = 100 and k = 0.05 per day, then after 10 days the predicted population is 100 e^0.5 ≈ 164.9.
求解指数模型得到 P = P₀ e^(kt)，其中 P₀ 是 t = 0 时的初始种群。例如，若 P₀ = 100 且 k = 0.05/天，则 10 天后预测的种群为 100 e^0.5 ≈ 164.9。
P = P₀ e^(kt)
This exponential curve shows explosive growth. In an ideal ecosystem with no resource limits, the population doubles at a constant rate. The doubling time is given by T = ln 2 / k, which is approximately 0.693/k. However, this cannot continue indefinitely in real ecosystems because resources such as food, water and space become limiting.
这条指数曲线显示出爆发式增长。在一个没有资源限制的理想生态系统中，种群以恒定速率翻倍。倍增时间由 T = ln 2 / k 给出，约为 0.693/k。然而，在真实生态系统中，由于食物、水和空间等资源变得有限，这种增长不可能无限持续。
Understanding exponential growth is essential before introducing the logistic model. It also appears in compound interest and radioactive decay, so mastery of the technique has wide applications in the Edexcel syllabus.
在引入逻辑斯蒂模型之前，理解指数增长至关重要。它还出现在复利和放射性衰变中，因此掌握这一技巧在 Edexcel 大纲中具有广泛的应用。
3. Logistic Growth and Carrying Capacity | 逻辑斯蒂增长与环境容纳量
A more realistic ecosystem model introduces a carrying capacity K, which is the maximum population size that the environment can support sustainably. The logistic model modifies the growth rate by the factor (1 − P/K), so growth slows as P approaches K.
更现实的生态系统模型引入了环境容纳量 K，即环境能够持续支撑的最大种群数量。逻辑斯蒂模型通过因子 (1 − P/K) 修正增长率，因此当 P 接近 K 时增长放缓。
dP/dt = rP(1 − P/K)
In this equation, r is the intrinsic growth rate. When P is very small compared with K, the factor (1 − P/K) is close to 1, so the model behaves like exponential growth. When P approaches K, the factor approaches 0, so the growth rate tends to zero. If P exceeds K, the factor becomes negative and the population declines.
在这个方程中，r 是内禀增长率。当 P 相对于 K 非常小时，因子 (1 − P/K) 接近 1，因此模型表现为指数增长。当 P 接近 K 时，因子接近 0，因此增长率趋于零。如果 P 超过 K，因子变为负值，种群下降。
This S-shaped curve is widely observed in populations such as yeast cells, sheep in Tasmania, and fish in limited lakes. It provides a bridge between pure calculus and applied ecology.
这种 S 形曲线在酵母细胞、塔斯马尼亚绵羊以及有限湖泊中的鱼类等种群中被广泛观察到。它架起了纯微积分与应用生态学之间的桥梁。
4. Formulating the Logistic Differential Equation | 构建逻辑斯蒂微分方程
To analyse the logistic equation, expand the right-hand side to see the density-dependent feedback clearly:
为了分析逻辑斯蒂方程，展开右边以清楚地看到密度制约的反馈：
dP/dt = rP − (r/K)P²
The first term rP drives exponential growth at low densities, while the second term (r/K)P² represents the negative effect of crowding, competition for resources, or disease transmission. This is a separable first-order differential equation, so Edexcel students can solve it by the standard method.
第一项 rP 在低密度下驱动指数增长，而第二项 (r/K)P² 代表拥挤、资源竞争或疾病传播的负面效应。这是一个可分离变量的一阶微分方程，因此 Edexcel 学生可以用标准方法求解。
Separating variables gives:
分离变量得到：
∫ dP / [P(1 − P/K)] = ∫ r dt
The left-hand side requires partial fractions. Write 1/[P(1 − P/K)] = A/P + B/(1 − P/K). Solving for A and B gives A = 1 and B = 1/K. Therefore the integral becomes:
左边需要部分分式。设 1/[P(1 − P/K)] = A/P + B/(1 − P/K)。求解 A 和 B 得 A = 1, B = 1/K。因此积分变为：
∫ (1/P + 1/(K − P)) dP = ∫ r dt
This careful algebraic set-up is a common exam task, so practise forming the partial fractions before integrating.
这种细致的代数设置是常见的考试任务，因此要在积分前练习构造部分分式。
5. Solving the Logistic Equation | 求解逻辑斯蒂方程
Integrating both sides gives:
两边积分得到：
ln|P| − ln|K − P| = rt + C
Using logarithm laws, this becomes ln|P/(K − P)| = rt + C. Exponentiating both sides gives P/(K − P) = A e^(rt), where A = e^C. Applying the initial condition P(0) = P₀ allows us to find A = P₀/(K − P₀). After algebraic rearrangement, the explicit solution is:
利用对数法则，得到 ln|P/(K − P)| = rt + C。两边取指数得到 P/(K − P) = A e^(rt)，其中 A = e^C。代入初始条件 P(0) = P₀ 可求得 A = P₀/(K − P₀)。经过代数整理，显式解为：
P(t) = K P₀ / [P₀ + (K − P₀) e^(−rt)]
As t → ∞, the exponential term e^(−rt) approaches 0, so the whole expression tends to K P₀ / P₀ = K. This confirms that the population approaches the carrying capacity in the long run, as long as P₀ is positive.
当 t → ∞ 时，指数项 e^(−rt) 趋于 0，因此整个表达式趋于 K P₀ / P₀ = K。这证实了只要 P₀ 为正，种群最终会趋近于环境容纳量。
In an exam, you may be asked to verify a given solution by differentiation or to substitute t = 0 to check the initial condition. Always write the final answer with the initial condition explicitly shown.
在考试中，你可能需要求导验证给定的解，或代入 t = 0 检查初始条件。务必写出明确显示初始条件的最终答案。
6. Equilibrium Points and Stability | 平衡点与稳定性
Equilibrium solutions occur when dP/dt = 0. For the logistic model, setting rP(1 − P/K) = 0 gives two equilibria: P = 0 and P = K. These are the constant solutions of the differential equation.
平衡解出现在 dP/dt = 0 时。对于逻辑斯蒂模型，令 rP(1 − P/K) = 0 得到两个平衡点：P = 0 和 P = K。这些是微分方程的常数解。
To determine stability, examine the sign of dP/dt near each equilibrium. For P slightly above 0, the term rP is positive and the factor (1 − P/K) is close to 1, so dP/dt is positive. This means P = 0 is unstable: any small positive population will grow away from it.
为了判断稳定性，考察每个平衡点附近 dP/dt 的符号。当 P 略大于 0 时，rP 为正，因子 (1 − P/K) 接近 1，因此 dP/dt 为正。这意味着 P = 0 不稳定：任何小的正种群都会远离它。
For P slightly below K, the factor (1 − P/K) is positive but small, so dP/dt is positive and the population rises toward K. For P slightly above K, the factor is negative, so dP/dt is negative and the population falls back toward K. Therefore P = K is stable.
当 P 略低于 K 时，因子 (1 − P/K) 为正但很小，因此 dP/dt 为正，种群向 K 上升。当 P 略高于 K 时，因子为负，因此 dP/dt 为负，种群回落到 K。因此 P = K 是稳定的。
| Equilibrium | Sign of dP/dt around equilibrium | Stability |
|---|---|---|
| P = 0 |
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