The Water Cycle: Mathematical Modelling with Differential Equations | 水循环：微分方程建模

📚 The Water Cycle: Mathematical Modelling with Differential Equations | 水循环：微分方程建模

The water cycle is a fundamental concept in geography and environmental science, but it also provides a rich context for mathematical modelling, especially within A-Level Mathematics. By analysing inflows, outflows, evaporation and precipitation, we can construct differential equations that describe how water volume in a reservoir or lake changes over time. This article explores the key mathematical techniques required to tackle such problems, focusing on setting up differential equations, solving them using separation of variables and integrating factors, and interpreting the solutions in real-world terms. These skills are directly relevant to Edexcel A-Level Mathematics, particularly in the contexts of connected rates of change and differential equations.

水循环是地理和环境科学中的基本概念，但它也为数学建模提供了丰富的背景，尤其是在 A-Level 数学中。通过分析流入、流出、蒸发和降水，我们可以构建微分方程来描述水库或湖泊的水量如何随时间变化。本文探讨了解决此类问题所需的关键数学技巧，重点包括建立微分方程、使用分离变量法和积分因子法求解，以及从实际角度解读解的意义。这些技能与 Edexcel A-Level 数学直接相关，特别是在相关联变化率和微分方程的应用中。

1. Understanding the Water Cycle in Context | 理解水循环的背景

In nature, the water cycle involves evaporation from oceans and lakes, condensation into clouds, precipitation as rain or snow, and surface runoff that returns water to bodies of water. For a mathematician, a lake or reservoir can be treated as a stock of water with inputs (rainfall, incoming streams) and outputs (evaporation, outflow, seepage). The rate of change of the volume is simply the net flow rate. This simplification allows us to apply calculus to predict future volumes, times to fill or empty, and equilibrium levels.

在自然界中，水循环涉及海洋和湖泊的蒸发、凝结成云、以雨或雪形式降水，以及使水返回水体的地表径流。对数学家来说，一个湖泊或水库可以被视为具有一定输入（降雨、流入溪流）和输出（蒸发、流出、渗漏）的水量储存。体积的变化率就是净流量。这种简化使我们能够运用微积分来预测未来的水量、注满或排空的时间以及平衡水位。

2. Rate of Change Fundamentals | 变化率基础

At its core, a water cycle model relies on the idea that the derivative of volume V with respect to time t, dV/dt, equals the total rate of water entering minus the total rate of water leaving. If we denote the inflow rate as R_in(t) and the outflow rate as R_out(t), we can write the basic balance equation:

dV/dt = R_in(t) – R_out(t)

Both rates may be constant, depend on time, or depend on the volume itself. For instance, outflow through a pipe often depends on the depth, which is related to volume.

水循环模型的核心在于：体积 V 对时间 t 的导数 dV/dt 等于进水的总速率减去出水的总速率。如果我们用 R_in(t) 表示流入速率，R_out(t) 表示流出速率，那么基本的平衡方程可以写为：

dV/dt = R_in(t) – R_out(t)

这两个速率可能是常数，也可能依赖于时间，或依赖于体积本身。例如，通过管道的流出通常依赖于水深，而水深又与体积相关。

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3. Building a Differential Equation for a Lake | 为湖泊建立微分方程

Consider a lake that receives water from a river at a constant rate of a m³/s and loses water through evaporation at a rate proportional to the surface area. If the lake has a uniform cross-section, the volume V is proportional to depth, and evaporation rate can be modelled as kV, where k is a positive constant. The differential equation becomes:

dV/dt = a – kV

If we also allow water to be extracted for irrigation at a constant rate b, the equation modifies to dV/dt = a – b – kV = (a – b) – kV. This is a first-order linear differential equation that can be solved by standard methods.

考虑一个从河流获得恒定速率 a m³/s 进水的湖泊，并以与表面积成比例的速率蒸发损失水分。若湖泊截面均匀，体积 V 与水深成正比，蒸发速率可建模为 kV，其中 k 为正常数。微分方程即为：

dV/dt = a – kV

如果我们还考虑以恒定速率 b 抽取灌溉用水，方程修正为 dV/dt = a – b – kV = (a – b) – kV。这是一阶线性微分方程，可以用标准方法求解。

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4. Solving the Linear Model | 求解线性模型

For the equation dV/dt = P – kV, where P = a – b is the net constant input, we can separate variables or use an integrating factor. Separating variables gives ∫ dV/(P – kV) = ∫ dt. Integrating yields –(1/k) ln|P – kV| = t + C. After rearranging and applying the initial condition V(0) = V₀, we obtain:

V(t) = P/k + (V₀ – P/k)e^–kt

This solution shows that the volume approaches the equilibrium value P/k as t → ∞, provided k > 0. If initially V₀ is less than P/k, the lake fills towards equilibrium; if greater, it drains down.

对于方程 dV/dt = P – kV，其中 P = a – b 为净恒定输入，我们可以分离变量或使用积分因子。分离变量得 ∫ dV/(P – kV) = ∫ dt。积分后得到 –(1/k) ln|P – kV| = t + C。重新整理并应用初始条件 V(0) = V₀，得到：

V(t) = P/k + (V₀ – P/k)e^–kt

此解表明，只要 k > 0，当 t → ∞ 时体积趋近于平衡值 P/k。如果初始 V₀ 小于 P/k，湖泊将逐渐注满至平衡值；如果大于平衡值，则会排水下降。

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5. Analysing the Steady State | 分析稳态

The equilibrium volume V_eq = P/k is found by setting dV/dt = 0. This steady-state condition occurs when inflow exactly balances outflow. The stability of this equilibrium is clear: if a sudden rainfall increases V above V_eq, then dV/dt becomes negative, driving V back down. Conversely, a drought that reduces V below V_eq causes dV/dt > 0, restoring the volume. Such behaviour is a hallmark of negative feedback in dynamical systems.

平衡体积 V_eq = P/k 通过令 dV/dt = 0 得出。当流入与流出恰好平衡时，就会达到这一稳态条件。该平衡的稳定性显而易见：如果一场突降暴雨使 V 升高到 V_eq 以上，dV/dt 变为负值，驱使 V 回落；反之，干旱使 V 低于 V_eq 会导致 dV/dt > 0，从而恢复体积。这种行为是动力系统中负反馈的标志性特征。

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6. Modelling Evaporation and Precipitation | 模拟蒸发和降水

In reality, precipitation and evaporation vary seasonally. We can incorporate sinusoidal functions to model these periodic variations. For example, suppose net inflow rate is P(t) = P₀ + A sin(2πt/365) (t in days). The differential equation dV/dt + kV = P₀ + A sin(ωt) is still linear. Using an integrating factor e^kt, we solve it to find V(t) consisting of a transient term and a steady periodic term. This models the seasonal filling and draining of a reservoir.

实际上，降水和蒸发随季节变化。我们可以引入正弦函数来模拟这些周期性变化。例如，假设净流入速率为 P(t) = P₀ + A sin(2πt/365)（t 以天计）。微分方程 dV/dt + kV = P₀ + A sin(ωt) 仍然是线性的。利用积分因子 e^kt，我们可以求解出 V(t)，它包含一个瞬态项和一个稳态周期项。这模拟了水库随季节的蓄水和放水过程。

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7. Separable Equations for Leakage | 渗漏的可分离变量方程

If a water tank drains through a small hole, Torricelli’s law states that the outflow velocity is proportional to √(2gh), where h is the height of water above the hole. For a cylindrical tank, V = Ah, so dV/dt = A dh/dt. The outflow rate is then –a√h (constant a). Hence, A dh/dt = –a√h, a separable equation: ∫ h^–½ dh = –(a/A) ∫ dt. Integrating gives 2√h = –(a/A)t + C. If initial height is h₀, we have √h = √h₀ – (a/(2A))t. The tank empties when h = 0, giving the emptying time T = (2A√h₀)/a.

如果水箱通过一个小孔排水，托里拆利定律指出出水速度与 √(2gh) 成正比，其中 h 是孔上方的水头高度。对于圆柱形水箱，V = Ah，所以 dV/dt = A dh/dt。此时出水速率为 –a√h（常数 a）。于是有 A dh/dt = –a√h，这是一个可分离变量方程：∫ h^–½ dh = –(a/A) ∫ dt。积分得 2√h = –(a/A)t + C。若初始高度为 h₀，则 √h = √h₀ – (a/(2A))t。当 h = 0 时水箱排空，由此得排空时间 T = (2A√h₀)/a。

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8. Connected Rates of Change in Water Tanks | 水箱的相关变化率问题

Many exam questions involve connected rates of change: water is poured into a tank of a given shape, and we need to find the rate at which the water level rises. For a conical tank of radius r and height H, the volume of water when the depth is h is V = (π/3)(r²/H²)h³ (if the cone vertex is at the bottom). Differentiating with respect to t gives dV/dt = (π r²/H²) h² dh/dt. If water enters at constant rate Q, we equate Q = (π r²/H²) h² dh/dt and solve for dh/dt. This demonstrates the power of implicit differentiation linking geometry and rates.

许多考题涉及相关变化率：水以一定速率注入特定形状的水箱，我们需要求水位上升的速率。对于一个半径为 r、高为 H 的圆锥形水箱，当水深为 h 时，水的体积为 V = (π/3)(r²/H²)h³（假设圆锥顶点在底部）。对 t 求导得 dV/dt = (π r²/H²) h² dh/dt。如果水以恒定速率 Q 注入，我们令 Q = (π r²/H²) h² dh/dt 并解出 dh/dt。这展示了几何和速率之间通过隐函数求导建立的密切联系。

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9. Exponential Decay in Pollutant Washout | 污染物冲刷的指数衰减

A related application is the cleansing of a polluted lake. Suppose a lake of volume V contains a mass m of pollutant, and clean water flows in at rate R while the mixed solution flows out at the same rate. Assuming instantaneous mixing, the concentration C = m/V, and the rate of change dm/dt = 0 (incoming clean) – (R × C) = –(R/V)m. This is dm/dt = –λ m with λ = R/V, giving exponential decay m = m₀ e^–λt. The time to halve the pollutant is T_½ = ln2/λ. This model is mathematically identical to the decay of water volume with a proportional outflow.

一个相关的应用是污染湖泊的净化。假设一个体积为 V 的湖泊含有质量为 m 的污染物，清洁水以速率 R 流入，同时混合溶液以相同速率流出。假设瞬时混合，浓度 C = m/V，质量变化率 dm/dt = 0（流入的清洁水） – (R × C) = –(R/V)m。即 dm/dt = –λ m，其中 λ = R/V，得出指数衰减 m = m₀ e^–λt。污染物减半所需的时间为 T_½ = ln2/λ。该模型在数学上与带比例流出的水量衰减完全相同。

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10. Using the Integrating Factor Method | 积分因子法的使用

When the differential equation is of the form dV/dt + P(t)V = Q(t), the integrating factor method is essential. The integrating factor is μ(t) = e^{∫ P(t) dt}. For our seasonal model, dV/dt + kV = P₀ + A sin ωt, P(t) = k (constant). Hence μ(t) = e^kt. Multiply through by e^kt: d/dt(V e^kt) = e^kt(P₀ + A sin ωt). Integrate both sides and divide by e^kt to obtain V(t). This technique is a cornerstone of Edexcel Pure Mathematics Year 2 integration.

当微分方程形如 dV/dt + P(t)V = Q(t) 时，积分因子法必不可少。积分因子为 μ(t) = e^{∫ P(t) dt}。对于我们的季节模型 dV/dt + kV = P₀ + A sin ωt，P(t) = k（常数）。因此 μ(t) = e^kt。两边乘以 e^kt 得：d/dt(V e^kt) = e^kt(P₀ + A sin ωt)。对两边积分并除以 e^kt 即可求得 V(t)。该技巧是 Edexcel 纯数第二年积分部分的核心。

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11. Interpreting Graphs and Solutions | 解读图像与解

Graphical interpretation is crucial. For the model V(t) = P/k + (V₀ – P/k)e^–kt, the graph approaches the horizontal asymptote V = P/k. The rate of approach is determined by k: a larger k leads to faster equilibrium. When parameters change, say an increase in abstraction b, the asymptote drops. Understanding these features helps in answering exam questions that ask for the long-term behaviour or the effect of modifying a parameter. A table can summarise key quantities:

Parameter	Symbol	Effect on equilibrium
Constant inflow	a	Increases Veq
Evaporation coefficient	k	Decreases Veq
Abstraction rate	b	Decreases Veq

图像的解读至关重要。对于模型 V(t) = P/k + (V₀ – P/k)e^–kt，图像趋近于水平渐近线 V = P/k。趋近速率由 k 决定：k 越大，达到平衡越快。当参数变化时，例如抽取量 b 增加，渐近线会下移。理解这些特征有助于回答考试中关于长期行为或修改参数效果的问题。下表总结了关键量：

参数	符号	对平衡的影响
恒定流入	a	增加 Veq
蒸发系数	k	减小 Veq
抽取速率	b	减小 Veq

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12. Exam Tips and Common Mistakes | 考试技巧与常见错误

When tackling water cycle problems in Edexcel A-Level exams, always check that units are consistent (e.g., m³, seconds). Clearly define your variables and state assumptions, such as constant cross-section or perfect mixing. A common mistake is confusing net rate with total inflow; remember dV/dt = inflow – outflow. Also, when using Torricelli’s law, ensure you correctly relate outflow rate to height and then to volume via the chain rule. Finally, always verify whether your solution makes physical sense: volumes cannot be negative, and asymptotes should reflect realistic limits.

在应对 Edexcel A-Level 考试中的水循环问题时，务必检查单位是否一致（如立方米、秒）。明确定义变量并陈述假设，如恒定截面或完美混合。常见错误是将净速率与总流入混淆；记住 dV/dt = 流入 – 流出。此外，在使用托里拆利定律时，要确保通过链式法则正确地将出水速率与高度、体积联系起来。最后，始终验证你的解在物理上是否合理：体积不能为负，渐近线应反映真实的极限。

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