Difference Equations: Modelling and Solution Strategies | 差分方程的建模与求解思路

📚 Difference Equations: Modelling and Solution Strategies | 差分方程的建模与求解思路

Difference equations describe how a quantity changes in discrete time steps. They are essential in IB Mathematics, especially in the Applications and Interpretation (AI) course, where modelling real-world situations with recurrence relations is a key skill.

差分方程描述的是量在离散时间步长中的变化方式。它们在IB数学中至关重要，尤其是在应用与解释（AI）课程中，用递推关系为现实情境建模是一项核心技能。

This article presents a clear, exam-focused guide to setting up difference equations from word problems, solving first-order and second-order linear cases, and analysing their long-run behaviour. Each section pairs English and Chinese explanations to support bilingual learning.

本文将提供一份清晰、紧扣考点的指南，涵盖如何从应用题中建立差分方程、求解一阶和二阶线性差分方程，并分析它们的长期行为。每一节都配有中英双语解释，以支持双语学习。

1. What Is a Difference Equation? | 什么是差分方程？

A difference equation (also called a recurrence relation) expresses the value of a sequence uₙ in terms of previous terms, such as uₙ₋₁ or uₙ₋₂. The order is the number of previous terms needed.

差分方程（也称为递推关系）用序列的前一项或前几项来表示当前项 uₙ，例如 uₙ₋₁ 或 uₙ₋₂。方程的阶数取决于需要使用的前项个数。

For example, the first-order linear difference equation has the form:

例如，一阶线性差分方程的形式为：

uₙ₊₁ = a uₙ + b, n ≥ 0

where a and b are constants, and an initial value u₀ is usually given. The equation is linear because uₙ appears only to the first power.

其中 a 和 b 是常数，通常还给出初值 u₀。该方程是线性的，因为 uₙ 只以一次方出现。

2. Modelling with Difference Equations | 用差分方程建模

To model a situation, identify the change from one time step to the next. Common examples include population growth, bank balances, drug dosage levels, and cooling temperatures.

建立模型时，需要找出从一个时间步到下一个时间步的变化。常见例子包括人口增长、银行存款、药物剂量水平和温度冷却。

Suppose a population grows by 5% per year and 200 individuals are removed each year. If Pₙ is the population after n years, then:

假设某人口每年增长5%，并且每年移除200个个体。若 Pₙ 表示第 n 年的人口数，则：

Pₙ₊₁ = 1.05 Pₙ − 200

This is a first-order linear difference equation. The coefficient 1.05 represents growth, and −200 represents removal.

这是一个一阶线性差分方程。系数1.05表示增长，−200表示移除。

For a loan of initial amount L₀ with monthly interest rate r and repayment M, the balance after n months is:

对于初始金额为 L₀、月利率为 r、每月还款 M 的贷款，第 n 个月后的余额为：

Lₙ₊₁ = (1 + r) Lₙ − M

These models all share the same mathematical structure, so a single solution method applies to all of them.

这些模型都具有相同的数学结构，因此一种求解方法即可适用于所有此类问题。

3. Solving by Iteration | 用迭代法求解

Iteration means applying the recurrence repeatedly from the initial value. For example, if uₙ₊₁ = 2uₙ + 3 with u₀ = 1, then:

迭代就是从初值开始反复应用递推关系。例如，若 uₙ₊₁ = 2uₙ + 3 且 u₀ = 1，则：

u₁ = 2(1) + 3 = 5, u₂ = 2(5) + 3 = 13, u₃ = 2(13) + 3 = 29

This gives a sequence: 1, 5, 13, 29, … Iteration is straightforward and often used with a calculator or spreadsheet, especially when only a few terms are needed.

这样得到序列：1, 5, 13, 29, … 迭代法直观简单，通常配合计算器或电子表格使用，尤其当只需要少数几项时。

However, iteration does not give a formula for uₙ directly. To find a specific term without computing all previous terms, we need an explicit solution.

然而，迭代法不能直接给出 uₙ 的公式。要在不计算所有前项的情况下找到某一项，就需要显式解。

4. Explicit Solution of First-Order Linear Difference Equations | 一阶线性差分方程的显式解

For uₙ₊₁ = a uₙ + b, the explicit solution depends on whether a = 1 or a ≠ 1.

对于 uₙ₊₁ = a uₙ + b，显式解取决于 a = 1 还是 a ≠ 1。

If a = 1, the equation becomes uₙ₊₁ = uₙ + b, an arithmetic sequence:

若 a = 1，方程变为 uₙ₊₁ = uₙ + b，即等差数列：

uₙ = u₀ + n b

If a ≠ 1, the solution can be found by shifting to a fixed point (equilibrium). Let the fixed point be L satisfying L = a L + b, so L = b / (1 − a).

若 a ≠ 1，可以通过平移至不动点（平衡点）来求解。设不动点 L 满足 L = a L + b，因此 L = b / (1 − a)。

Then the sequence vₙ = uₙ − L satisfies vₙ₊₁ = a vₙ, which is geometric. Hence:

于是序列 vₙ = uₙ − L 满足 vₙ₊₁ = a vₙ，即等比序列。因此：

uₙ = L + (u₀ − L) aⁿ

This formula is central to IB exams. For the population example Pₙ₊₁ = 1.05 Pₙ − 200 with P₀ = 10000, we have L = (−200) / (1 − 1.05) = 4000, so:

这个公式是IB考试的核心。对于人口例子 Pₙ₊₁ = 1.05 Pₙ − 200 且 P₀ = 10000，有 L = (−200) / (1 − 1.05) = 4000，因此：

Pₙ = 4000 + (10000 − 4000)(1.05)ⁿ = 4000 + 6000(1.05)ⁿ

5. Equilibrium Points and Stability | 平衡点与稳定性

The fixed point L is also called an equilibrium value. Its stability determines the long-term behaviour of the sequence.

不动点 L 也称为平衡值。其稳定性决定了序列的长期行为。

If |a| < 1, then aⁿ → 0 as n → ∞, so uₙ → L. The equilibrium is stable, and the sequence converges to L.

若 |a| < 1，则当 n → ∞ 时 aⁿ → 0，因此 uₙ → L。平衡点是稳定的，序列收敛于 L。

If |a| > 1, then aⁿ grows without bound, so uₙ moves away from L unless u₀ = L. The equilibrium is unstable.

若 |a| > 1，则 aⁿ 无界增长，因此除 u₀ = L 外，uₙ 会远离 L。平衡点是不稳定的。

If a = −1, the sequence oscillates between two values and is neither convergent nor divergent.

若 a = −1，序列在两个值之间振荡，既不收敛也不发散。

|a| < 1 ⇒ stable; |a| > 1 ⇒ unstable; a = −1 ⇒ periodic oscillation

In the loan model Lₙ₊₁ = (1 + r) Lₙ − M, the coefficient 1 + r > 1, so the balance is unstable unless it exactly equals the equilibrium. This means that if the monthly payment is too small, the debt grows forever.

在贷款模型 Lₙ₊₁ = (1 + r) Lₙ − M 中，系数 1 + r > 1，因此除非余额恰好等于平衡点，否则它是不稳定的。这意味着如果月还款额太小，债务将永远增长。

6. Second-Order Linear Difference Equations | 二阶线性差分方程

A second-order linear difference equation relates uₙ₊₂ to uₙ₊₁ and uₙ. Its standard form is:

二阶线性差分方程将 uₙ₊₂ 与 uₙ₊₁ 和 uₙ 联系起来。其标准形式为：

uₙ₊₂ + p uₙ₊₁ + q uₙ = r

This type appears in models of oscillating systems, such as predator-prey populations or vibrating strings in discrete form. It requires two initial values, usually u₀ and u₁.

这类方程出现在振荡系统的模型中，例如捕食者-猎物种群或离散形式的振动弦。它需要两个初值，通常是 u₀ 和 u₁。

When r = 0, the equation is homogeneous; when r ≠ 0, it is non-homogeneous. The solution combines the complementary solution and a particular solution, similar to solving differential equations.

当 r = 0 时，方程为齐次的；当 r ≠ 0 时，方程是非齐次的。解由余函数解和特解组成，这与求解微分方程类似。

7. Solving Second-Order Equations: The Characteristic Equation | 二阶方程求解：特征方程

For the homogeneous equation uₙ₊₂ + p uₙ₊₁ + q uₙ = 0, assume a solution of the form uₙ = λⁿ. Substituting gives the characteristic equation:

对于齐次方程 uₙ₊₂ + p uₙ₊₁ + q uₙ = 0，假设解具有形式 uₙ = λⁿ。代入后得到特征方程：

λ² + p λ + q = 0

Let the two roots be λ₁ and λ₂. If they are distinct and real, the general solution is:

设两个根为 λ₁ 和 λ₂。若它们是不相等的实根，则通解为：

uₙ = A λ₁ⁿ + B λ₂ⁿ

If there is a repeated root λ, the solution becomes:

若存在重根 λ，则通解变为：

uₙ = (A + B n) λⁿ

If the roots are complex, the solution involves trigonometric terms, but the general form is still built from λⁿ using Euler’s formula. The constants A and B are found from the initial conditions.

若根为复数，则解涉及三角函数项，但通解仍由 λⁿ 通过欧拉公式构成。常数 A 和 B 由初始条件确定。

For example, the Fibonacci sequence satisfies Fₙ₊₂ = Fₙ₊₁ + Fₙ, with characteristic equation λ² − λ − 1 = 0. Its solution is a sum of two exponential terms based on the golden ratio.

例如，斐波那契数列满足 Fₙ₊₂ = Fₙ₊₁ + Fₙ，其特征方程为 λ² − λ − 1 = 0。其解是基于黄金比例的两种指数项之和。

8. Non-Homogeneous Second-Order Equations | 非齐次二阶差分方程

For uₙ₊₂ + p uₙ₊₁ + q uₙ = r, where r is a constant, the general solution is:

对于 uₙ₊₂ + p uₙ₊₁ + q uₙ = r（其中 r 为常数），通解为：

uₙ = (complementary solution) + (particular solution)

The complementary solution is the general solution of the homogeneous equation. The particular solution is a constant uₙ = k if 1 + p + q ≠ 0. Substituting gives k(1 + p + q) = r, so k = r / (1 + p + q).

余函数解是齐次方程的通解。若 1 + p + q ≠ 0，特解为常数 uₙ = k。代入得 k(1 + p + q) = r，所以 k = r / (1 + p + q)。

If 1 + p + q = 0, then the constant is already a solution of the homogeneous equation, so try uₙ = k n instead. This mirrors the method of undetermined coefficients for differential equations.

若 1 + p + q = 0，则常数已经是齐次方程的解，因此应尝试 uₙ = k n。这类似于微分方程中的待定系数法。

Once the general solution is found, use the two initial conditions to determine A and B. Always check your answer by computing the first few terms.

找到通解后，使用两个初始条件来确定 A 和 B。务必通过计算前几项来检查你的答案。

9. Modelling Case: Cobweb Model | 建模案例：蛛网模型

The cobweb model describes price fluctuations in a market where supply depends on the previous period’s price. Let supply be Sₙ = c + d Pₙ₋₁ and demand be Dₙ = e − f Pₙ.

蛛网模型描述的是供给取决于上一期价格的市场价格波动。设供给为 Sₙ = c + d Pₙ₋₁，需求为 Dₙ = e − f Pₙ。

At equilibrium Sₙ = Dₙ, so:

在均衡条件下 Sₙ = Dₙ，因此：

c + d Pₙ₋₁ = e − f Pₙ

Rearranging gives a first-order difference equation:

整理后得到一阶差分方程：

Pₙ = (−d / f) Pₙ₋₁ + (e − c) / f

The stability condition is | −d / f | < 1, meaning the demand response f is stronger than the supply response d. In that case prices converge to the equilibrium price.

稳定条件为 | −d / f | < 1，即需求响应 f 强于供给响应 d。此时价格会收敛到均衡价格。

This model shows how a simple recurrence relation can explain real economic behaviour, and it is a favourite exam question for applying the explicit solution formula.

该模型展示了简单的递推关系如何解释现实经济行为，也是IB考试中应用显式解公式的热门题目。

10. Using Technology: Calculator and Spreadsheet | 使用技术：计算器与电子表格

In IB exams, a graphing calculator can generate terms of a recurrence relation quickly. The u(n) function or spreadsheet-style lists allow you to verify solutions.

在IB考试中，图形计算器可以快速生成递推关系的各项。使用 u(n) 函数或电子表格类型的列表可以帮助你验证解。

For iteration, enter the recurrence with the initial value and observe the sequence. For explicit formulas, store constants and evaluate directly.

对于迭代，输入递推关系和初值，观察序列。对于显式公式，存储常数后直接求值。

When using a spreadsheet, the first column is n, the second is uₙ, and the formula for each cell refers to the previous cell. This is especially useful for modelling projects and exploration tasks.

使用电子表格时，第一列为 n，第二列为 uₙ，每个单元格的公式引用上一单元格。这对于建模项目和探究任务特别有用。

Always check domain restrictions: the recurrence might only be valid for n ≥ 0 or n ≥ 1, depending on how the sequence is defined.

始终检查定义域限制：递推关系可能仅对 n ≥ 0 或 n ≥ 1 有效，具体取决于序列的定义方式。

11. Common Pitfalls and Exam Tips | 常见易错点与考试技巧

One common error is confusing uₙ₊₁ with uₙ when writing the recurrence from a word problem. Always identify clearly which index represents the current period and which represents the next period.

常见错误之一是从应用题写递推关系时混淆 uₙ₊₁ 和 uₙ。务必清楚哪个下标表示当前时期，哪个表示下一时期。

Another mistake is forgetting the initial condition when solving for constants. Without u₀ or u₁, the solution has arbitrary constants and cannot be evaluated numerically.

另一个错误是求常数时忘记初始条件。没有 u₀ 或 u₁，通解中含有任意常数，无法求出具体数值。

When using the explicit formula uₙ = L + (u₀ − L) aⁿ, check that a is not 1. If a = 1, the formula for L is undefined, and you must use the arithmetic sequence solution instead.

使用显式公式 uₙ = L + (u₀ − L) aⁿ 时，检查 a 不等于1。若 a = 1，则 L 的公式无定义，必须改用等差数列解。

For second-order equations, always write the characteristic equation in the correct form. If the original equation is uₙ₊₂ = 5uₙ₊₁ − 6uₙ, then the characteristic equation is λ² − 5λ + 6 = 0, not λ² + 5λ + 6 = 0.

对于二阶方程，务必正确写出特征方程。若原方程为 uₙ₊₂ = 5uₙ₊₁ − 6uₙ，则特征方程是 λ² − 5λ + 6 = 0，而不是 λ² + 5λ + 6 = 0。

Finally, interpret your result in the context. Does the population make sense? Is a negative loan balance impossible? Relate the mathematical answer back to the original problem.

最后，要结合上下文解释结果。人口数量是否合理？贷款余额为负是否可能？将数学答案联系回原始问题。

12. Summary: A Step-by-Step Strategy | 总结：分步求解策略

Follow these steps for any difference equation problem:

对于任何差分方程问题，可按以下步骤进行：

Identify the unknown sequence and the time step.

确定未知序列和时间步长。
Write the recurrence relation from the given information.

根据给定信息写出递推关系。
If it is first order, classify whether a = 1 or a ≠ 1.

若是一阶，判断 a = 1 还是 a ≠ 1。
Use the appropriate explicit formula or solve the characteristic equation for second order.

使用相应的显式公式，或对二阶方程解特征方程。
Apply initial conditions to find constants.

应用初始条件求出常数。
Check stability and interpret the long-term behaviour.

检查稳定性并解释长期行为。
Verify with the first few terms using iteration.

用迭代法验证前几项。

With practice, difference equations become a reliable source of marks in IB Mathematics. They connect algebra, sequences, and real-world modelling in a single framework.

通过练习，差分方程会成为IB数学中稳定的得分点。它将代数、数列和现实建模连接在同一个框架中。

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Difference Equations: Modelling and Solution Strategies | 差分方程的建模与求解思路