📚 Case Study in Action: Analysing Predator-Prey Dynamics with Differential Equations and Matrix Methods | 案例分析实战演练：用微分方程和矩阵方法分析捕食者-猎物动态
In CIE Year 13 Further Mathematics, applying matrix algebra and differential equations to real-world scenarios is a major skill. This case study demonstrates how to model the interaction between two species — rabbits (prey) and foxes (predators) — using a system of first-order ODEs, then analyse stability through linearisation, Jacobian matrices, and eigenvalues. The step-by-step approach mirrors exam-style questions where you are expected to interpret results in context.
在 CIE 13 年级进阶数学中，将矩阵代数和微分方程应用于现实场景是一项重要技能。本案例分析演示如何用一阶常微分方程组对两个物种——兔子（猎物）和狐狸（捕食者）——的相互作用进行建模，然后通过线性化、雅可比矩阵和特征值分析稳定性。逐步解题方法反映了考卷中要求结合背景解释结果的典型题目风格。
1. The Problem Scenario | 问题情景
A conservation area is initially home to 800 rabbits and 50 foxes. Ecologists observe that the rabbit population grows naturally at a per-capita rate of 2 per year but is reduced by encounters with foxes. Foxes rely on rabbits for food; without rabbits, their death rate would be 1 per year. The interaction strength is measured by the product of the two populations, with coefficients β = 0.02 for predation impact on rabbits and δ = 0.001 for the benefit to foxes. We are tasked with predicting the long-term dynamics of the ecosystem.
一个保护区最初有 800 只兔子和 50 只狐狸。生态学家观察到兔子种群的自然人均年增长率为 2，但会因与狐狸相遇而减少。狐狸依赖兔子为食；如果没有兔子，它们每年的死亡率将是 1。相互作用强度用两个种群数量的乘积来衡量，其中对兔子捕食影响的系数 β = 0.02，对狐狸收益的系数 δ = 0.001。我们的任务是预测生态系统的长期动态。
2. Constructing the Mathematical Model | 建立数学模型
The classic Lotka-Volterra equations are used. Let x(t) be the number of rabbits and y(t) the number of foxes at time t years. The system is:
采用经典的 Lotka-Volterra 方程。设 x(t) 为兔子数量，y(t) 为狐狸数量，时间 t 以年为单位。方程组为：
dx/dt = αx − βxy
dy/dt = δxy − γy
where α = 2 (rabbit growth rate), β = 0.02 (predation rate), δ = 0.001 (conversion efficiency), γ = 1 (fox death rate). These are first-order nonlinear ODEs because of the xy terms. No analytical solution exists in closed form, so we use qualitative methods.
其中 α = 2（兔子增长率），β = 0.02（捕食率），δ = 0.001（转化效率），γ = 1（狐狸死亡率）。由于存在 xy 项，这是一阶非线性常微分方程组。不存在闭合形式的解析解，因此我们使用定性方法。
3. Assigning Numerical Parameters | 赋予数值参数
Substituting the given numbers, the system becomes:
代入给定数值，方程组变为：
dx/dt = 2x − 0.02xy
dy/dt = 0.001xy − y
We denote the right-hand functions as f(x, y) = 2x − 0.02xy and g(x, y) = 0.001xy − y. The initial condition is x(0) = 800, y(0) = 50, but stability analysis focuses on the general behaviour near equilibria rather than solving the IVP directly.
我们记右侧函数为 f(x, y) = 2x − 0.02xy 和 g(x, y) = 0.001xy − y。初始条件为 x(0) = 800, y(0) = 50，但稳定性分析侧重于平衡点附近的总体行为，而不是直接求解初值问题。
4. Finding Equilibrium Points | 寻找平衡点
Equilibria occur when both derivatives are zero. Set dx/dt = 0 and dy/dt = 0:
平衡点出现在两个导数均为零时。令 dx/dt = 0 和 dy/dt = 0：
2x − 0.02xy = 0 ⇒ x(2 − 0.02y) = 0
0.001xy − y = 0 ⇒ y(0.001x − 1) = 0
Solving simultaneously gives two equilibrium points: (0, 0) and the coexistence point where 2 − 0.02y = 0 and 0.001x − 1 = 0, yielding y = 100, x = 1000. Thus we have E₁ = (0, 0) and E₂ = (1000, 100).
联立求解得到两个平衡点：(0, 0) 以及共存点，由 2 − 0.02y = 0 和 0.001x − 1 = 0 解得 y = 100, x = 1000。因此我们有 E₁ = (0, 0) 和 E₂ = (1000, 100)。
5. Linearisation and the Jacobian Matrix | 线性化与雅可比矩阵
For a nonlinear system, stability is determined by the linearised system near each equilibrium. The Jacobian matrix J consists of partial derivatives:
对于非线性系统，稳定性由各平衡点附近的线性化系统决定。雅可比矩阵 J 由偏导数构成：
| ∂f/∂x = 2 − 0.02y | ∂f/∂y = −0.02x |
| ∂g/∂x = 0.001y | ∂g/∂y = 0.001x − 1 |
At any point (x, y), the matrix J(x, y) characterises the local behaviour. We then evaluate J at each equilibrium and compute its eigenvalues.
在任意点 (x, y)，矩阵 J(x, y) 刻画了局部行为。然后我们在每个平衡点处计算 J 并求其特征值。
6. Evaluating the Jacobian at Each Equilibrium | 在各平衡点计算雅可比
First, at E₁ = (0, 0): substituting x = 0, y = 0 gives
首先，在 E₁ = (0, 0)：代入 x = 0, y = 0 得到
| 2 | 0 |
| 0 | −1 |
Next, at E₂ = (1000, 100): ∂f/∂x = 2 − 0.02×100 = 0; ∂f/∂y = −0.02×1000 = −20; ∂g/∂x = 0.001×100 = 0.1; ∂g/∂y = 0.001×1000 − 1 = 0. Hence
接下来，在 E₂ = (1000, 100)：∂f/∂x = 2 − 0.02×100 = 0；∂f/∂y = −0.02×1000 = −20；∂g/∂x = 0.001×100 = 0.1；∂g/∂y = 0.001×1000 − 1 = 0。因此
| 0 | −20 |
| 0.1 | 0 |
These matrices form the basis of the linearised system u’ = Ju at each equilibrium, where u represents small perturbations.
这些矩阵构成了各平衡点处线性化系统 u’ = Ju 的基础，其中 u 表示微小扰动。
7. Eigenvalue Analysis at the Origin | 原点处的特征值分析
For E₁, the Jacobian is diagonal, so eigenvalues are simply the diagonal entries: λ₁ = 2, λ₂ = −1. Since one eigenvalue is positive and the other negative, the origin is a saddle point. This means small perturbations grow along the eigenvector associated with λ = 2 (rabbit growth when foxes are absent), while the system is attracted along the direction of λ = −1 (fox decline without prey). The saddle is unstable, so both species cannot simultaneously vanish unless the initial condition lies on the stable manifold.
对于 E₁，雅可比矩阵是对角阵，因此特征值即为对角线元素：λ₁ = 2，λ₂ = −1。由于一个特征值为正，另一个为负，原点是鞍点。这意味着微小扰动会沿着与 λ = 2 相关的特征向量增长（无狐狸时兔子增长），同时系统沿着 λ = −1 方向被吸引（无猎物时狐狸减少）。鞍点是不稳定的，因此两个物种不可能同时消失，除非初始条件恰好位于稳定流形上。
8. Eigenvalue Analysis at the Coexistence Equilibrium | 共存平衡点的特征值分析
At E₂, the Jacobian matrix is J₂ = [[0, −20], [0.1, 0]]. The characteristic equation is det(J₂ − λI) = (−λ)(−λ) − (−20)(0.1) = λ² + 2 = 0. Thus λ = ± i√2. Since the real part is zero, linearisation yields a centre. In linear terms, the trajectories are closed ellipses, suggesting periodic oscillations around (1000, 100). However, because the system is nonlinear, the true behaviour may be a centre or a stable/unstable spiral depending on higher-order terms. A centre in a conservative Lotka-Volterra model indicates neutrally stable cycles.
在 E₂ 处，雅可比矩阵为 J₂ = [[0, −20], [0.1, 0]]。特征方程为 det(J₂ − λI) = (−λ)(−λ) − (−20)(0.1) = λ² + 2 = 0。因此 λ = ± i√2。由于实部为零，线性化给出一个中心。在线性层面，轨线是封闭的椭圆，表明围绕 (1000, 100) 的周期性振荡。然而，由于系统是非线性的，真实行为可能是中心，也可能是稳定或不稳定的螺旋，取决于高阶项。守恒的 Lotka-Volterra 模型中的中心表示中性稳定的循环。
9. Stability Interpretation and Phase Portrait | 稳定性解释与相图
The linearised analysis suggests that E₁ is an unstable saddle, while E₂ exhibits oscillatory motion. In the full nonlinear system, it is known that the Lotka-Volterra equations possess a conserved quantity, meaning the orbits are indeed closed curves for most initial conditions. Therefore E₂ is a nonlinear centre, neither attracting nor repelling. The population sizes oscillate out of phase: when rabbits are abundant, foxes increase with a lag, then rabbit numbers crash, followed by a fox decline, and the cycle repeats.
线性化分析表明 E₁ 是不稳定鞍点，而 E₂ 呈现振荡运动。在完整的非线性系统中，已知 Lotka-Volterra 方程具有一个守恒量，这意味着对大多数初始条件，轨线确实是闭合曲线。因此 E₂ 是非线性中心，既不吸引也不排斥。种群规模异相振荡：当兔子丰富时，狐狸数量滞后增加，然后兔子数量锐减，接着狐狸减少，循环往复。
10. Numerical Simulation Insights | 数值模拟洞察
If we solve the system numerically (e.g., using Euler or Runge-Kutta methods) from x(0)=800, y(0)=50, we obtain closed trajectories in the phase plane. This confirms the centre behaviour. For example, the maximum rabbit population might approach 1200 while foxes peak around 150 before declining. The simulation also reveals that the period of oscillation depends on the amplitude; near the equilibrium, the orbit is almost elliptical, but farther away it becomes distorted.
如果我们从 x(0)=800, y(0)=50 出发对系统进行数值求解（例如用欧拉法或龙格-库塔法），在相平面上将得到闭合轨线。这证实了中心行为。例如，兔子种群的最大值可能接近 1200，而狐狸的峰值约为 150，然后下降。模拟还揭示振荡周期依赖于振幅；在平衡点附近，轨道几乎呈椭圆形，但远离时则变得扭曲。
11. Model Limitations and Refinements | 模型局限与改进
The classic Lotka-Volterra model assumes unlimited growth for prey in the absence of predators, no carrying capacity, and a linear functional response. In reality, rabbit resources are finite, so a logistic term (e.g., dx/dt = αx(1 − x/K) − βxy) would be more realistic. This often changes the coexistence equilibrium to a stable spiral, dampening oscillations towards a steady state. Additionally, stochastic effects and spatial distributions are ignored. Nonetheless, the simple model remains a powerful tool for introducing qualitative analysis in Further Maths.
经典的 Lotka-Volterra 模型假设猎物在没有捕食者时无限增长，没有环境容纳量，且功能反应为线性。现实中兔子的资源是有限的，因此加入逻辑斯蒂项（例如 dx/dt = αx(1 − x/K) − βxy）会更真实。这通常会使共存平衡点变为稳定螺旋，将振荡衰减至稳态。此外，随机效应和空间分布被忽略。尽管如此，这个简单模型仍是进阶数学中引入定性分析的强大工具。
12. Conclusion and Exam Tips | 结论与考试技巧
This case study has walked through the complete pipeline: translating a real-world scenario into a nonlinear ODE system, locating equilibria, linearising via the Jacobian, calculating eigenvalues, and interpreting stability. In CIE examinations, you may be given a system with parameters, asked to classify equilibrium types, and sketch phase portraits. Always present the Jacobian clearly, compute eigenvalues accurately, and link mathematical results back to the biological context. Practice with varying coefficients to see how the nature of equilibria changes.
本案例分析走完了完整流程：将真实场景转化为非线性常微分方程组，寻找平衡点，通过雅可比线性化，计算特征值，并解释稳定性。在 CIE 考试中，你可能会遇到带参数的系统，要求分类平衡点类型并绘制相图。务必清晰地给出雅可比矩阵，准确计算特征值，并将数学结果与生物学背景联系起来。尝试改变系数进行练习，观察平衡点性质如何变化。
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