📚 The Revolution in Prussia | 普鲁士的革命
In 1848, the Prussian state was shaken by a wave of revolutionary fervour that demanded constitutional reform and national unity. While historians debate the causes and consequences, mathematics offers a powerful toolkit for modelling the dynamics of such social upheavals. In this article, we will apply differential equations, population dynamics and numerical methods covered in the Edexcel A-Level Mathematics course to build a simplified model of the Prussian Revolution, exploring how support for revolution and government repression interact over time.
1848 年，普鲁士国家被一波要求宪政改革和国家统一的革命热潮所撼动。当历史学家争论其原因和后果时，数学提供了一套强大的工具来建模此类社会动荡的动态。在本文中，我们将运用 Edexcel A-Level 数学课程中涉及的微分方程、种群动力学和数值方法，构建一个关于普鲁士革命的简化模型，探索革命支持度与政府镇压如何随时间相互作用。
1. The Prussian Revolution of 1848: A Brief Overview | 1848年普鲁士革命：简要概述
The Revolutions of 1848 erupted across much of Europe, and Prussia was no exception. Middle-class liberals, workers and nationalists took to the streets of Berlin, demanding a free press, a unified Germany and a constitution limiting royal power. King Frederick William IV initially conceded, but by late 1848 the revolutionary momentum faded as the monarchy regained control. Understanding this ebb and flow of revolutionary support can be reframed as a dynamical system, where the ‘population’ of revolutionary supporters and the strength of government repression evolve according to mathematical rules.
1848 年革命在欧洲大部分地区爆发，普鲁士也不例外。中产阶级自由派、工人和民族主义者走上柏林街头，要求新闻自由、德国统一以及一部限制王权的宪法。国王腓特烈·威廉四世起初作出了让步，但到 1848 年末，随着君主制重新夺回控制权，革命势头逐渐消退。理解这种革命支持度的涨落可以被重新定义为一个动力系统，其中革命支持者的“种群”和政府镇压力量按照数学规则演化。
By treating revolutionary sentiment as a variable amenable to calculus, we can explore questions like: Under what conditions does a revolution grow uncontrollably? How does government repression dampen revolutionary fervour? What factors determine whether a revolution succeeds or fails?
通过将革命情绪视为可用微积分处理的变量，我们可以探索如下问题：在什么条件下革命会不受控制地增长？政府镇压如何削弱革命热情？哪些因素决定了一场革命的成败？
2. Why Model a Revolution? | 为什么对革命建模？
Mathematical modelling provides a structured way to abstract complex social phenomena into tractable equations. In Edexcel A-Level Mathematics, you have already encountered models for population growth, chemical reactions and mechanics. The same principles can be applied to social dynamics. A revolution involves multiple interacting ‘populations’: supporters of change, loyalists, and the repressive capacity of the state. By translating qualitative historical narratives into differential equations, we can test hypotheses, identify tipping points and even make qualitative predictions.
数学建模提供了一种结构化的方法，将复杂的社会现象抽象为可处理的方程。在 Edexcel A-Level 数学中，你已经接触过种群增长、化学反应和力学模型。同样的原理可以应用于社会动态。一场革命涉及多个相互作用的“种群”：变革支持者、效忠者以及国家的镇压能力。通过将定性的历史叙述转化为微分方程，我们可以检验假说、识别临界点，甚至做出定性预测。
This approach does not reduce history to mere numbers; rather, it supplements historical understanding by revealing the underlying mathematical skeleton of revolutionary waves. The Prussian revolution of 1848 serves as an excellent case study because it exhibited rapid growth, interaction with state forces and eventual collapse—behaviours reminiscent of predator-prey systems in biology.
这种方法不会把历史简化为单纯的数字；相反，它通过揭示革命浪潮背后的数学骨架来补充历史理解。1848 年的普鲁士革命是一个极好的案例研究，因为它表现出快速增长、与国家力量的相互作用以及最终崩溃——这些行为让人联想到生物学中的捕食者-猎物系统。
3. Defining the Fundamental Variables and Parameters | 定义基本变量与参数
To build our model, let t denote time measured in weeks after the outbreak of protests in March 1848. Define R(t) as the number of active revolutionary supporters (in thousands) at time t, and G(t) as the government’s effective repressive strength (in arbitrary units). The model will rely on several parameters: α, the intrinsic growth rate of revolutionary support in the absence of repression; β, the per-capita dampening effect of government action on revolutionaries; γ, the natural reinforcement rate of government forces; and δ, the attrition rate of government strength due to revolutionary activity.
为了构建模型，设 t 表示 1848 年 3 月抗议爆发后以周为单位的时间。定义 R(t) 为时刻 t 活跃革命支持者的数量（以千人为单位），G(t) 为政府的有效镇压力量（任意单位）。该模型将依赖于几个参数：α，无镇压时革命支持度的内在增长率；β，政府行动对革命者的人均削弱效应；γ，政府力量的自然增援率；δ，由于革命活动导致的政府力量消耗率。
These parameters encapsulate the political and social climate. A high α might reflect widespread discontent, while a high β indicates efficient policing. By adjusting these values, we can simulate different historical scenarios and examine their outcomes.
这些参数概括了政治和社会气候。高 α 可能反映了广泛的不满情绪，而高 β 则表示高效的警察行动。通过调整这些数值，我们可以模拟不同的历史情境并检查其结果。
4. Isolated Growth: dR/dt = αR | 孤立增长：dR/dt = αR
First, consider a naive scenario where revolutionaries face no opposition. The growth of R would then follow a simple exponential law: the rate of change of revolutionary support is proportional to current support. The differential equation is written as
dR/dt = αR.
分离变量并积分，得到解 R(t) = R₀e^(αt)，其中 R₀ 为初始支持度。这预示革命将无限制地呈指数增长，显然不符合实际，但它为更现实的模型铺平了道路。
Separating variables and integrating gives the solution R(t) = R₀e^(αt), where R₀ is the initial support. This predicts unbounded exponential growth of revolution—clearly unrealistic—but it paves the way for more realistic models.
5. Logistic Growth for Revolutionary Support | 革命支持度的 Logistic 增长
A more plausible model assumes that revolutionary support cannot grow indefinitely due to limited resources, such as the number of politically active individuals or the saturation of sympathisers. This introduces the concept of a carrying capacity K. The logistic differential equation is
dR/dt = rR (1 – R/K),
where r is the intrinsic growth rate. The solution is R(t) = K / (1 + A e^(–rt)), with A determined by initial conditions. In the context of the Prussian Revolution, K could represent the maximum number of people willing to actively support the revolution.
一个更合理的模型假定革命支持度不能无限增长，因为资源有限，例如政治活跃个体的数量或同情者的饱和。这引入了环境容量 K 的概念。Logistic 微分方程为 dR/dt = rR (1 – R/K)，其中 r 是内在增长率。其解为 R(t) = K / (1 + A e^(–rt))，其中 A 由初始条件决定。在普鲁士革命的语境中，K 可以代表愿意积极支持革命的最大人数。
This model exhibits an S-shaped curve: early exponential growth followed by a slowdown as support approaches K. However, it fails to capture the observed decline of the Prussian revolution after spring 1848. We must incorporate the interaction with government forces to replicate that decline.
该模型呈现 S 形曲线：早期指数增长，随后随着支持度接近 K 而减缓。然而，它未能捕捉到的 1848 年春季后普鲁士革命的衰退。我们必须纳入与政府力量的相互作用才能重现这一衰退。
6. Modelling Interaction: Revolutionaries vs Government | 建模相互作用：革命者对政府
We now introduce a two-variable system inspired by predator-prey dynamics. Revolutionary support R and government strength G interact. When revolutionaries and government forces meet, revolutionary support is suppressed (like prey being eaten), while government forces also suffer losses (like predators being injured). The coupled differential equations are:
dR/dt = αR – βRG,
dG/dt = γG – δRG.
现在，我们引入了一个受捕食者-猎物动力学启发的双变量系统。革命支持度 R 和政府力量 G 相互作用。当革命者与政府力量相遇时，革命支持度受到压制（如同猎物被吃掉），而政府力量也遭受损失（如同捕食者受伤）。耦合微分方程为：dR/dt = αR – βRG，dG/dt = γG – δRG。
Here, the term –βRG represents the rate at which revolutionary support is eroded by government intervention. Similarly, the term –δRG represents the government’s attrition due to revolutionary activities. The linear terms αR and γG model natural growth rates in the absence of interaction.
此处，项 –βRG 表示革命支持度被政府干预侵蚀的速率。类似地，项 –δRG 表示政府因革命活动而遭受的损耗。线性项 αR 和 γG 分别模拟无相互作用时的自然增长率。
7. Equilibrium Points and Their Meaning | 平衡点及其含义
To understand the long-term behaviour, we find equilibrium points by setting the derivatives to zero. Solving αR – βRG = 0 and γG – δRG = 0 yields two equilibria:
E₁ = (0, 0), E₂ = (γ/δ, α/β).
为了理解长期行为，我们通过令导数为零来寻找平衡点。解 αR – βRG = 0 和 γG – δRG = 0 得到两个平衡点：E₁ = (0, 0)，E₂ = (γ/δ, α/β)。
E₁ represents total failure of both revolution and government strength—an unlikely state. E₂ is more interesting: it is a coexistence equilibrium where revolutionary support and government forces persist at constant levels. In the Prussian context, this would correspond to a stalemate where the revolution neither triumphs nor is fully crushed. However, historical evidence shows the revolution was suppressed, so our model must allow for outcomes where R declines toward zero.
E₁ 代表革命和政府力量都完全失败的状态——不太可能发生。E₂ 更有趣：它是一个共存平衡点，革命支持度和政府力量在恒定水平上持续存在。在普鲁士语境中，这将对应于一种僵局，革命既不胜利也不被完全镇压。然而，历史证据表明革命被镇压了，因此我们的模型必须允许 R 向零衰减这种结果。
A simple linear stability analysis or phase portrait examination shows that E₂ is a saddle point or an unstable node depending on parameters, meaning trajectories tend to move away from it. This can lead to either revolution explosion or government dominance.
简单的线性稳定性分析或相图检查表明，根据参数的不同，E₂ 是一个鞍点或一个不稳定结点，这意味着轨线倾向于远离它。这可能导致革命爆发或政府占据主导。
8. Phase Plane Analysis | 相平面分析
By sketching the nullclines (where dR/dt = 0 and dG/dt = 0) on an (R, G) plane, we can visualise the system’s behaviour. The R-nullcline consists of the axes R = 0 and G = α/β. The G-nullcline consists of G = 0 and R = γ/δ. The intersection of G = α/β and R = γ/δ gives E₂. Typical trajectories show that if initial revolutionary support is high and government repression low, R can grow rapidly before eventually declining as government forces mount. This captures the initial surge and later collapse of the 1848 revolution in Prussia.
通过在 (R, G) 平面上勾勒零倾线（dR/dt = 0 和 dG/dt = 0 的曲线），我们可以直观地看到系统的行为。R 零倾线由坐标轴 R = 0 和直线 G = α/β 组成。G 零倾线由 G = 0 和 R = γ/δ 组成。G = α/β 与 R = γ/δ 的交点即为 E₂。典型的轨线显示，如果初始革命支持度高而政府镇压低，R 可以先快速增长，随后随着政府力量的集结而下降。这就捕捉到了 1848 年普鲁士革命的初始高涨和后来的崩溃。
Although A-Level students are not required to compute eigenvalues, they can use vector-field reasoning to deduce stability. The model clearly demonstrates how feedback between state repression and revolutionary enthusiasm can produce oscillatory or decaying dynamics.
虽然 A-Level 学生不需要计算特征值，但他们可以使用向量场推理来推断稳定性。该模型清楚地展示了国家镇压与革命热情之间的反馈如何产生振荡或衰减的动态。
9. Numerical Simulation Using Euler’s Method | 使用欧拉法进行数值模拟
The coupled differential equations do not have a simple analytical solution, but we can approximate the behaviour using Euler’s method, a core technique in Edexcel A-Level Mathematics. Let step size h = 0.1 weeks, initial R₀ = 5.0 (thousands), G₀ = 2.0, and parameters α = 0.8, β = 0.3, γ = 0.5, δ = 0.25. The Euler iteration formulas are:
Rₙ₊₁ = Rₙ + h (αRₙ – βRₙGₙ),
Gₙ₊₁ = Gₙ + h (γGₙ – δRₙGₙ).
耦合微分方程没有简单的解析解，但我们可以使用欧拉法（Edexcel A-Level 数学中的核心技术）来近似行为。取步长 h = 0.1 周，初始 R₀ = 5.0（千人），G₀ = 2.0，参数 α = 0.8, β = 0.3, γ = 0.5, δ = 0.25。欧拉迭代公式为：Rₙ₊₁ = Rₙ + h (αRₙ – βRₙGₙ)，Gₙ₊₁ = Gₙ + h (γGₙ – δRₙGₙ)。
The table below shows the first 10 steps of the simulation:
| t (weeks) | R (supporters, thousands) | G (government strength) |
|---|---|---|
| 0.0 | 5.000 | 2.000 |
| 0.1 | 5.100 | 2.075 |
| 0.2 | 5.191 | 2.154 |
| 0.3 | 5.274 | 2.237 |
| 0.4 | 5.348 | 2.325 |
| 0.5 | 5.414 | 2.418 |
| 0.6 | 5.470 | 2.516 |
| 0.7 | 5.517 | 2.620 |
| 0.8 | 5.554 | 2.731 |
| 0.9 | 5.582 | 2.849 |
| 1.0 | 5.599 | 2.974 |
Notice that revolutionary support initially increases, but its growth rate slows as government strength rises. If we continue the simulation further, R will eventually peak and then begin a steady decline, consistent with the historical fact that the Prussian revolution waned as the state reasserted control.
注意到革命支持度最初上升，但随着政府力量的增长，其增长率减慢。如果我们继续模拟，R 将最终达到峰值，然后开始稳步下降，这与普鲁士革命随着国家重新确立控制而衰落的历史事实一致。
10. What the Model Reveals About Revolutionary Success | 模型揭示了革命成功的条件
By experimenting with parameter values, we can investigate the conditions under which a revolution succeeds (R grows unhindered) or fails (R collapses). For instance, if β is very small (weak government per-capita impact), revolutionary support can overwhelm government forces, leading to a state collapse. On the other hand, a high δ (revolutionaries very effective at eroding government strength) also favours revolution. In the Prussian case, the government’s ability to regroup and call in military reinforcements effectively increased β over time, turning the tide.
通过试验参数值，我们可以探究革命成功（R 不受阻碍地增长）或失败（R 崩溃）的条件。例如，如果 β 非常小（政府的人均影响力弱），革命支持度会压垮政府力量，导致国家崩溃。另一方面，高 δ（革命者在削弱政府力量方面非常有效）也有利于革命。在普鲁士的案例中，政府重组并召集军事增援的能力实际上随时间增大了 β，从而扭转了局势。
This sensitivity to parameters illustrates the concept of threshold behaviour: a small change in government efficiency or revolutionary fervour can flip the system from one outcome to another. Such tipping points are a hallmark of nonlinear
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