Dynamic Population Models and Differential Equations | 动态人口模型与微分方程

📚 Dynamic Population Models and Differential Equations | 动态人口模型与微分方程

Population dynamics is one of the most compelling applications of differential equations in A-Level Mathematics. By translating real-world biological and demographic changes into mathematical language, we can model how populations grow, stabilise, or collapse over time.

人口动态是 A-Level 数学中微分方程最引人入胜的应用之一。通过将现实世界中的生物学和人口统计学变化转化为数学语言，我们能够模拟人口数量如何随时间增长、稳定或崩溃。

1. Why Differential Equations for Population? | 为什么用微分方程研究人口？

A differential equation describes a rate of change. For population models, the key quantity is dP/dt — the rate at which a population P changes with respect to time t. Observing how this rate depends on the current population allows us to build predictive models that align with observed data.

微分方程描述的是变化率。对于人口模型，核心量是 dP/dt —— 即人口 P 相对于时间 t 的变化速率。观察这个速率如何依赖当前人口数量，我们可以构建与观测数据相符的预测模型。

In essence, we are asking: “If we know how the growth rate behaves, can we determine the population at any future time?”

本质上，我们在问：”如果我们知道增长率如何变化，能否确定未来任意时刻的人口数量？”

2. The Malthusian Model: Exponential Growth | 马尔萨斯模型：指数增长

The simplest population model, proposed by Thomas Malthus, assumes that the growth rate is proportional to the current population:

最简单的种群模型由托马斯·马尔萨斯提出，它假设增长率与当前人口数量成正比：

dP/dt = kP

where k is a positive constant called the intrinsic growth rate. This is a first-order separable differential equation. Rearranging and integrating both sides gives:

其中 k 是正常数，称为内在增长率。这是一个一阶可分离变量的微分方程。整理并两边积分可得：

∫ (1/P) dP = ∫ k dt

ln|P| = kt + C

Exponentiating both sides and applying the initial condition P(0) = P₀ yields the familiar exponential growth solution:

两边取指数并应用初始条件 P(0) = P₀，得到熟悉的指数增长解：

P(t) = P₀ e^(kt)

This model predicts unbounded growth: as t → ∞, P(t) → ∞. It is appropriate for short-term growth of small populations, such as bacteria in a fresh culture.

该模型预测无界增长：当 t → ∞ 时，P(t) → ∞。它适用于小种群的短期增长，例如新鲜培养基中的细菌。

3. Solving Step by Step | 分步求解

Let us solve dP/dt = kP rigorously using separation of variables. The method is standard and examinable in A-Level Mathematics.

让我们用分离变量法严格求解 dP/dt = kP。这一方法是 A-Level 数学中的标准考点。

Step 1: Separate the variables.

第一步：分离变量。

(1/P) dP = k dt

Step 2: Integrate both sides.

第二步：两边积分。

ln|P| = kt + C

Step 3: Solve for P by exponentiating.

第三步：通过取指数解出 P。

P = e^(kt+C) = A e^(kt), where A = e^C

Step 4: Use the initial condition. If P(0) = P₀, then A = P₀, hence:

第四步：使用初始条件。若 P(0) = P₀，则 A = P₀，因此：

P(t) = P₀ e^(kt)

Note that we dropped the absolute value because populations are positive for all t in practical contexts.

注意，我们省略了绝对值符号，因为在实际情境中，种群数量在所有 t 时刻均为正。

4. Limitations of the Malthusian Model | 马尔萨斯模型的局限性

While the exponential model is mathematically elegant, it fails in the long run. No population can grow indefinitely because resources — food, space, water — are finite. When a population becomes too large, competition increases and the growth rate declines.

尽管指数模型在数学上十分优雅，但长期来看它并不成立。没有任何种群能够无限增长，因为资源——食物、空间、水——是有限的。当种群数量过大时，竞争加剧，增长率就会下降。

A more realistic model must incorporate a carrying capacity K — the maximum population that the environment can sustain. This leads to the logistic model.

更现实的模型必须引入环境承载力 K —— 即环境能够维持的最大种群数量。这就引出了逻辑斯蒂模型。

5. The Logistic Model | 逻辑斯蒂模型

The logistic differential equation modifies the exponential model by introducing a limiting factor:

逻辑斯蒂微分方程通过引入一个限制因子来修正指数模型：

dP/dt = rP(1 − P/K)

Here, r is the intrinsic growth rate and K is the carrying capacity. When P is very small compared to K, the factor (1 − P/K) is close to 1, and growth is nearly exponential. As P approaches K, the factor approaches 0, slowing growth to a halt.

其中 r 是内在增长率，K 是环境承载力。当 P 远小于 K 时，因子 (1 − P/K) 接近 1，增长近似指数；当 P 接近 K 时，该因子趋于 0，增长逐渐停止。

This single equation captures a rich spectrum of behaviour: rapid initial growth, inflection, and saturation at the carrying capacity.

仅这一个方程就捕捉了丰富的行为：初期快速增长、拐点和在环境承载力处饱和。

6. Solving the Logistic Equation | 求解逻辑斯蒂方程

The logistic equation is separable. We rewrite it as:

逻辑斯蒂方程是可分离变量的。我们将其改写为：

∫ dP / [P(1 − P/K)] = ∫ r dt

Using partial fractions, we express the integrand as:

使用部分分式，我们将被积函数表示为：

1/[P(1 − P/K)] = A/P + B/(1 − P/K)

Solving for A and B gives A = 1 and B = 1/K. Thus:

解出 A 和 B 得 A = 1，B = 1/K。因此：

∫ (1/P)dP + (1/K) ∫ 1/(1 − P/K) dP = rt + C

Evaluating the integrals:

求积分得：

ln|P| − ln|1 − P/K| = rt + C

ln|P / (1 − P/K)| = rt + C

Exponentiating:

取指数：

P / (1 − P/K) = Be^(rt)

Solving for P with initial condition P(0) = P₀ gives the logistic growth function:

使用初始条件 P(0) = P₀ 解出 P，得到逻辑斯蒂增长函数：

P(t) = KP₀ / [P₀ + (K − P₀)e^(−rt)]

This solution is a cornerstone of population biology and appears frequently in exam problems.

该解是种群生物学的基石，在考试题目中频繁出现。

7. Equilibrium Solutions and Stability | 平衡解与稳定性

Equilibrium solutions occur when dP/dt = 0. For the logistic equation, this happens at P = 0 and P = K.

平衡解出现在 dP/dt = 0 时。对于逻辑斯蒂方程，这发生在 P = 0 和 P = K 处。

P = 0: Unstable equilibrium. Any small positive perturbation leads to growth away from zero.
P = K: Stable equilibrium. If P is slightly below K, it increases; if slightly above K, it decreases.

P = 0：不稳定平衡。任何微小的正向扰动都会导致种群偏离零而增长。
P = K：稳定平衡。若 P 略低于 K，种群增长；若略高于 K，种群减少。

This stability analysis is often tested conceptually in examinations. Students should be able to classify equilibrium points from the sign of dP/dt.

这种稳定性分析在考试中经常以概念题形式出现。学生应当能够根据 dP/dt 的符号对平衡点进行分类。

8. The Inflection Point: Maximum Growth Rate | 拐点：最大增长率

A key property of the logistic curve is that its growth rate dP/dt is not constant. It increases when P < K/2 and decreases when P > K/2. The maximum growth rate occurs at P = K/2.

逻辑斯蒂曲线的一个关键性质是，其增长率 dP/dt 并不恒定。当 P < K/2 时增长率上升，当 P > K/2 时增长率下降。最大增长率出现在 P = K/2 处。

We can verify this by differentiating dP/dt with respect to P. Setting the derivative equal to zero:

我们可以通过对 dP/dt 关于 P 求导来验证。令导数为零：

d/dP [rP(1 − P/K)] = r(1 − 2P/K) = 0

Hence P = K/2. At this population size, the graph of P(t) has an S-shaped (sigmoid) curve with its steepest slope.

因此 P = K/2。在这个种群规模下，P(t) 的图像呈 S 形（乙状），且斜率最陡。

9. Real-World Applications | 现实世界应用

Dynamic population models extend far beyond biology. They are used in epidemiology to model the spread of infectious diseases, in economics to describe the adoption of new technologies, and in ecology to manage wildlife populations.

动态人口模型的应用远远超出生物学领域。它们在流行病学中用于模拟传染病的传播，在经济学中描述新技术的采用，在生态学中用于管理野生动物种群。

Fisheries management: A fish population subject to harvesting can be modelled by dP/dt = rP(1 − P/K) − h, where h is the harvest rate. Overfishing occurs when h exceeds the maximum sustainable yield.
Epidemiology: The SIR model splits a population into Susceptible, Infected, and Recovered compartments, leading to systems of differential equations.
Conservation biology: Minimum viable population sizes can be estimated using stochastic population models.

渔业管理：受捕捞影响的鱼类种群可用 dP/dt = rP(1 − P/K) − h 建模，其中 h 为捕捞率。当 h 超过最大可持续产量时，就会发生过度捕捞。
流行病学：SIR 模型将人群分为易感者、感染者和康复者三类，形成微分方程组。
保护生物学：最小可行种群规模可以用随机种群模型进行估计。

Understanding the underlying mathematics allows scientists to make informed decisions with limited data.

理解背后的数学原理，使科学家能够在数据有限的情况下做出明智的决策。

10. Exam Technique: Common Pitfalls | 考试技巧：常见陷阱

Students frequently lose marks on population modelling questions due to avoidable errors. Below are the most common pitfalls and how to avoid them.

学生常在人口建模题中因可避免的错误而失分。以下是最常见的陷阱及其规避方法。

Forgetting the constant of integration: Always include C when integrating, and determine it from the initial condition.
Incorrect partial fractions: When decomposing 1/[P(1 − P/K)], verify that the coefficients A and B are correct before integrating.
Misinterpreting K: K is the carrying capacity, not the initial population. Do not confuse it with P₀.
Sign errors in the exponent: In the logistic solution, the exponent is −rt, not +rt. A positive exponent would incorrectly suggest unbounded growth.
Skipping the stability check: After finding equilibrium solutions, always state whether each is stable or unstable.

忘记积分常数：积分时始终加上 C，并利用初始条件确定其值。
部分分式出错：分解 1/[P(1 − P/K)] 时，务必在积分前验证系数 A 和 B 是否正确。
误解 K：K 是环境承载力，不是初始种群。不要将其与 P₀ 混淆。
指数符号错误：在逻辑斯蒂解中，指数是 −rt 而非 +rt。正指数将错误地暗示无界增长。
跳过稳定性检验：求出平衡解后，务必说明每个解是稳定的还是不稳定的。

11. Worked Example: From Start to Finish | 完整例题演练

Problem: A population of rabbits on an island follows the logistic model with r = 0.5 per year, K = 1000, and an initial population of 100 rabbits. Find the population after 2 years, and determine when the population reaches 900.

题目：某岛屿上的兔子种群遵循逻辑斯蒂模型，r = 0.5/年，K = 1000，初始种群为 100 只。求 2 年后的种群数量，并确定种群何时达到 900 只。

Solution:

解答：

Using the logistic solution:

使用逻辑斯蒂解：

P(t) = 1000 × 100 / [100 + (1000 − 100)e^(−0.5t)]

At t = 2:

当 t = 2 时：

P(2) = 100000 / [100 + 900e^(−1)]

Since e^(−1) ≈ 0.3679, we have:

由于 e^(−1) ≈ 0.3679，我们有：

P(2) ≈ 100000 / (100 + 331.1) ≈ 100000 / 431.1 ≈ 232

For the population to reach 900, we solve:

为求种群达到 900 的时间，我们解方程：

900 = 100000 / [100 + 900e^(−0.5t)]

Rearranging:

整理得：

900 × [100 + 900e^(−0.5t)] = 100000

90000 + 810000e^(−0.5t) = 100000

810000e^(−0.5t) = 10000

e^(−0.5t) = 1/81

Taking natural logarithms:

取自然对数：

−0.5t = ln(1/81) = −ln 81

t = 2ln 81 ≈ 2 × 4.394 = 8.79 years

Thus the population reaches 900 after approximately 8.8 years.

因此，种群约在 8.8 年后达到 900 只。

12. Beyond the Logistic: Further Directions | 超越逻辑斯蒂：进一步拓展

For students aiming at higher marks, it is worth knowing that real populations often deviate from the logistic model due to time delays, environmental stochasticity, or Allee effects — where very small populations grow more slowly because individuals struggle to find mates.

对于争取高分的同学，值得了解的是，真实种群常常因时间延迟、环境随机性或阿利效应而偏离逻辑斯蒂模型——阿利效应指种群极小时个体难以找到配偶，导致增长更慢。

Time-delayed models take the form:

时间延迟模型的形式为：

dP/dt = rP(t)(1 − P(t − τ)/K)

where τ represents the delay in the population’s response to crowding. Such equations can produce oscillations and even chaotic behaviour — a fascinating extension of the core A-Level content.

其中 τ 表示种群对拥挤作出反应的延迟时间。此类方程可以产生振荡甚至混沌行为——这是 A-Level 核心内容的一个迷人延伸。

Mastering the basic models, however, remains the essential foundation. Every complex model builds upon the separable differential equation techniques you learn at this level.

然而，掌握基本模型仍然是必要的基础。每一个复杂模型都建立在你在此阶段学习的可分离变量微分方程技巧之上。

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Dynamic Population Models and Differential Equations | 动态人口模型与微分方程