📚 Logistic Population Modelling: A Case Study | 逻辑斯蒂人口建模案例分析
In AQA Year 13 Further Mathematics, tying theoretical concepts to realistic scenarios is essential for deep understanding. This case study explores the logistic growth model, a classic differential equation that captures population dynamics under limited resources. We will formulate the model, solve it analytically using separation of variables, and then apply Euler's method and the improved Euler method to obtain numerical approximations. By comparing these approximations with the exact solution, you will gain insight into numerical error and the efficacy of different methods.
在AQA Year 13进阶数学中，将理论概念与现实场景结合是深入理解的关键。本案例研究探讨逻辑斯蒂增长模型——一个经典的微分方程，用以描述有限资源下的种群动态。我们将建立模型，使用分离变量法求解析解，然后应用欧拉法和改进欧拉法进行数值近似。通过比较这些近似与精确解，你将深入理解数值误差以及不同方法的效能。
1. The Scenario and Model Formulation | 场景与模型建立
servations suggest that the growth rate slows as the population approaches the environment' ial population is P(0) = 100. The rate of change of P is governed by the logistic differential equation:
实验室培养的一种微生物种群，其增长率随种群接近环境容纳量而减缓。设P(t)为第t天的种群数量，内禀增长率r = 0.5 天⁻¹，环境容纳量K = 1000。初始种群为P(0) = 100。P的变化率由逻辑斯蒂微分方程决定：
dP/dt = r P (1 - P/K)
This first-order, non-linear ordinary differential equation is widely used in ecology and epidemiology. The term (1 - P/K) reduces the effective growth rate as P approaches K, creating an S-shaped (sigmoid) solution curve.
这个一阶非线性常微分方程广泛应用于生态学和流行病学。因子(1 - P/K)使得当P趋近K时有效增长率降低，从而形成S形（sigmoid）的解曲线。
2. Analytical Solution via Separation of Variables | 分离变量求解解析解
The logistic equation can be solved analytically. Rearrange to separate variables P and t:
逻辑斯蒂方程可解析求解。整理以分离变量P和t：
1 / [P(1 - P/K)] dP = r dt
. Note that 1/[P(1 - P/K)] = K/[P(K - P)] = 1/P + 1/(K - P). Integrating both sides gives:
左边利用部分分式积分。注意1/[P(1 - P/K)] = K/[P(K - P)] = 1/P + 1/(K - P)。两边积分得：
∫ (1/P + 1/(K - P)) dP = ∫ r dt
This yields ln|P| - ln|K - P| = r t + C, which simplifies to ln|P/(K - P)| = r t + C. Exponentiate both sides to obtain P/(K - P) = A eʳᵗ, where A = eᶜ.
得到 ln|P| - ln|K - P| = r t + C，简化为 ln|P/(K - P)| = r t + C。两边取指数得 P/(K - P) = A eʳᵗ，其中A = eᶜ。
3. Determining Constants and the Exact Solution | 确定常数与精确解
- P) = A e⁰ gives 100/(1000 - 100) = A, so A = 100/900 = 1/9. Hence the exact solution is:
代入初始条件P(0) = 100。将t = 0, P = 100代入P/(K - P) = A e⁰，得100/(1000 - 100) = A，即A = 100/900 = 1/9。因此精确解为：
P(t) = 1000 / (1 + 9 e⁻⁰·⁵ᵗ)
le, at t = 5 days, P(5) ≈ 1000/(1 + 9e⁻²·⁵) ≈ 1000/(1 + 9 × 0.0821) ≈ 1000/1.7389 ≈ 575.1.
该函数从100开始增长，初期呈指数趋势，然后趋于1000。例如，第5天时，P(5) ≈ 1000/(1 + 9e⁻²·⁵) ≈ 1000/(1 + 9 × 0.0821) ≈ 1000/1.7389 ≈ 575.1。
4. Euler's Method for Numerical Approximation | 欧拉法数值近似
Euler' g the derivative at the current point. With step size h = 1 day, the iterative formula is:
欧拉法通过利用当前点的导数向前推进来近似解。取步长h = 1天，迭代公式为：
Pₙ₊₁ = Pₙ + h · r Pₙ (1 - Pₙ/K)
ute subsequent values. The table below shows hand-calculated steps for the first few days:
从t₀ = 0, P₀ = 100开始，我们计算后续值。下表展示了最初几天的逐步手工计算：
| t (days) | P (Euler) | dP/dt at start of step |
|---|---|---|
| 0 | 100 | 0.5 × 100 × (1 - 0.1) = 45 |
| 1 | 100 + 45 = 145 | 0.5 × 145 × (1 - 0.145) ≈ 62.0 |
| 2 | 145 + 62.0 = 207.0 | 0.5 × 207 × (1 - 0.207) ≈ 82.1 |
| 3 | 207 + 82.1 = 289.1 | 0.5 × 289.1 × (1 - 0.2891) ≈ 102.8 |
| 4 | 289.1 + 102.8 = 391.9 | 0.5 × 391.9 × (1 - 0.3919) ≈ 119.2 |
| 5 | 391.9 + 119.2 = 511.1 | — |
Notice that Euler' oticeably lower than the exact value 575.1. This underestimation is typical because the derivative used at the start of each step does not account for the curvature slowing growth.
注意欧拉法得到P(5) ≈ 511.1，明显低于精确值575.1。这种低估是典型的，因为每步开始时使用的导数未计入因曲率而减缓的增长趋势。
5. Improved Euler Method (Heun's Method) | 改进欧拉法（Heun方法）
The improved Euler method reduces error by averaging the derivatives at the beginning and end of the step. The algorithm uses a predictor-corrector approach:
改进欧拉法通过平均步长起点和终点的导数来减少误差。该算法采用预测-校正方法：
Predictor: P* = Pₙ + h f(tₙ, Pₙ)
Corrector: Pₙ₊₁ = Pₙ + (h/2) [f(tₙ, Pₙ) + f(tₙ₊₁, P*)]
h = 1 to our logistic model yields much closer approximations. Starting again at P₀ = 100:
将此方法以h = 1应用于我们的逻辑斯蒂模型，得到更接近的近似。再次从P₀ = 100开始：
- Step 1 (t=0 to t=1): ted at P*: 0.5×145×(1-0.145)=61.9875. Corrector P₁ = 100 + 0.5×(45 + 61.9875) = 153.49375.
- 第1步：预测值P* = 100 + 45 = 145；计算P*处的导数：0.5×145×(1-0.145)=61.9875。校正值P₁ = 100 + 0.5×(45 + 61.9875) = 153.49375。
- Step 2 (t=1 to t=2): P*: 0.5×218.49×(1-0.21849) ≈ 85.37; Corrector P₂ = 153.49 + 0.5×(65.00 + 85.37) = 228.68.
- 第2步：f(1, 153.49) ≈ 0.5×153.49×(1-0.15349) ≈ 65.00；预测值P* = 153.49 + 65.00 = 218.49；P*处导数：0.5×218.49×(1-0.21849) ≈ 85.37；校正值P₂ = 153.49 + 0.5×(65.00 + 85.37) = 228.68。
- Continuing this pattern yields P₃ ≈ 327.2, P₄ ≈ 444.0, and P₅ ≈ 564.2.
- 继续此模式得到P₃ ≈ 327.2，P₄ ≈ 444.0，P₅ ≈ 564.2。
The improved Euler estimate for P(5) is 564.2, much closer to the exact 575.1 than Euler's 511.1.
改进欧拉法对P(5)的估计为564.2，远较欧拉法的511.1更接近精确值575.1。
6. Comparison of Numerical and Exact Solutions | 数值解与精确解的比较
The table below collates the exact values and the approximations from both numerical methods. Exact values are rounded to one decimal place.
下表汇总了精确值及两种数值方法的近似结果。精确值四舍五入至一位小数。
| t (days) | Exact P(t) | Euler P(t) | Improved Euler P(t) |
|---|---|---|---|
| 0 | 100.0 | 100.0 | 100.0 |
| 1 | 154.6 | 145.0 | 153.5 |
| 2 | 231.1 | 207.0 | 228.7 |
| 3 | 331.9 | 289.1 | 327.2 |
| 4 | 452.8 | 391.9 | 444.0 |
| 5 | 575.1 | 511.1 | 564.2 |
The improved Euler method
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