AEM2.DifferentialEquations#
Last Edit: 10/6/25
Ordinary Differential Equations 常微分方程#
- 在 ODE 中，未知函数只依赖于一个 independent variable
$$ \frac{d^2 y}{dx^2} = -k^2 y $$
使用的是 Leibniz notation（莱布尼茨记号）
其中：
- y：dependent variable（因变量）或函数
- x：independent variable（自变量）
与上式等价的拉格朗日记号写法为 Lagrange Notation
$$ y’’ = -k^2 y $$
通用记法：
$$ \frac{d^n y}{d x^n} \equiv y^{(n)} $$
Classifying DE’s#
Order of a DE#
DE 的 order 是方程中出现的最高阶导数的阶数
示例：
$$ y^{(4)} + 2x^2 + y’’ = \sin x $$
是一个四阶常微分方程（4th order ODE）
General form of DE#
$$ F(x, y, y’, y’’, …, y^{(n)}) = 0 $$
Linear and Non-linear ODEs#
- 一个阶数为 n 的微分方程是 linear 的，当它可以写成：
$$ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_0(x)y = f(x) $$
- 系数 \(a_i(x)\) 只依赖于 \(x\)，不能含有 \(y\)
- 示例：
First-order linear DE
$$ y’ + p(x)y = f(x) $$
Second-order linear DE
$$ y’’ + p(x)y’ + q(x)y = f(x) $$
Homogeneous and Non-homogeneous Linear ODE#
- 若 \(f(x) = 0\)，则为 Homogeneous
- 若 \(f(x) \ne 0\)，则为 Non-homogeneous
Constant vs Variable Coefficient#
- 若 \(a_0(x), …, a_n(x)\) 是 constant，则为 constant coefficient DE
- 否则为 Variable coefficient DE
Autonomous ODE#
- 自治微分方程（Autonomous ODE），又叫 Time-Invariant
- 指的是 Independent variable 不会显式出现在方程中
Ex. Autonomous ODE#
- Population growth equation
$$ \frac{dP}{dt} = (\beta - \alpha)P $$
- t 没有显式出现
- \(\beta\)：出生率
- \(\alpha\)：死亡率
General Form#
$$ y^{(n)} = F(y^{(n-1)}, y^{(n-2)}, …, y’, y) $$
Ex. Identify the ODEs#
$$ y’’ + 3(y’)^2 + \cos x = 0 $$
- non-linear（因为有 \((y’)^2\)）
- non-homogeneous（有常数项 \(\cos x\)）
- variable coefficient（\(3(y’)^2\) 中的系数依赖 \(y’\)，不是常数）
这里是 \(y’\) 的平方
$$ 5\sin\left(\frac{dy}{dt}\right) + \tan^2 y = 1 $$
- non-linear
- non-homogeneous
$$ \frac{d^2 y}{dx^2} + x^2 \frac{dy}{dx} + (\sin x) y = e^x $$
- linear（\(y, y’, y’’\) 都是一次）
- non-homogeneous（右边不是 0）
- variable coefficient（系数 \(x^2, \sin x\) 都依赖于 \(x\)）
$$ x y’’ + 4x^2 y’ - \frac{2}{1 + x^2} y = 0 $$
- linear
- homogeneous（右边是 0）
- variable coefficient（所有系数依赖 x）
$$ y’’ + \sin(y’) + x y = x^2 $$
- non-linear（因为 \(\sin(y’)\) 是非线性项）
- non-homogeneous（右边为 \(x^2\)）
- variable coefficient（项 \(x y\) 中系数为 \(x\)）
Solution of ODE#
- Given an ODE of form
$$ y^{(n)} = F(y^{(n-1)}, …, y’, y, x) $$
- 如果一个函数 \(y = \varphi(x)\) 被代入后，使得方程变为 identity 恒等式，那么它就是这个微分方程的一个 solution
- 代入后要满足：
$$ \varphi^{(n)} \equiv F\left(\varphi^{(n-1)}, …, \varphi’, \varphi, x\right) $$
Ex. Verifying Solution#
For DE \(y’ = 6x(y - 1)^{2/3}\)，verify by substitution that is a solution on interval \((-\infty, +\infty)\)，where C is an arbitrary constant
$$ y’ = 2x \cdot 3(x^2 + C)^2 = 6x(x^2 + C)^2 $$
$$ y - 1 = (x^2 + C)^3 \Rightarrow (y - 1)^{2/3} = [(x^2 + C)^3]^{2/3} = (x^2 + C)^2 $$
$$ 6x(y - 1)^{2/3} = 6x(x^2 + C)^2 $$
Existence of Solutions, General, Particular and Singular#
- General Solution 包含一个 C，代表了一个 Solution Set
- Particular Solution 解出了 C 的具体值
- Singular Solution 是不属于 General Solution Set 中的一个解
How to solve first-order ODE#
Separable First-order ODE#
- 对于形如 \(y’ = g(x)h(y)\) 的 ODE
- 解法是将变量 x 和 y 分离，积分两边：
$$ \int \frac{1}{h(y)} dy = \int g(x) dx $$
- 这是标准的 Separation of Variables
Integrating Factor Method#
适用于 linear first-order ODE
标准形式为：
$$ y’ + P(x)y = f(x) $$
有 Integrating Factor
$$ \mu(x) = e^{\int P(x),dx} $$
- 将方程两边都乘以 \(\mu(x)\)
$$ \mu(x) \cdot y’ + \mu(x) \cdot P(x) \cdot y = \mu(x) \cdot f(x) $$
- 左边可以写成一个积的导数：
$$ \frac{d}{dx}[\mu(x) \cdot y] = \mu(x) \cdot f(x) $$
- 两边同时积分
$$ \int \frac{d}{dx}[\mu(x) \cdot y], dx = \int \mu(x) \cdot f(x) , dx $$
- 得到：
$$ \mu(x) \cdot y = \int \mu(x) \cdot f(x), dx + C $$
- 解出 y(x)
$$ y(x) = \frac{1}{\mu(x)} \left[ \int \mu(x) \cdot f(x) , dx + C \right] $$
Ex.#
$$ xy’ - y = 2x^2 \quad \text{with} \quad y(x_0) = y_0 $$
- 将 DE 转换成标准形式，有 \(y’ - \frac{1}{x} y = 2x\)
- 所以 \(P(x) = -\frac{1}{x}，f(x) = 2x\)
计算积分因子（Integrating Factor）：
$$ \mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} $$
两边乘以积分因子：
$$ \frac{d}{dx} \left( \frac{1}{x} y \right) = 2 $$
积分并解出 y：
$$ \frac{1}{x} y = \int 2 dx + C = 2x + C \quad \Rightarrow \quad y = x(2x + C) = 2x^2 + Cx $$
若 IC 为 \(y(1) = 1\)：
- 解代入得：
$$ 1 = 2(1)^2 + C(1) \Rightarrow C = -1 $$
若 IC 为 y(1) = 0
- 解代入得：
$$ 0 = 2 + C \Rightarrow C = -2 $$
若 IC 为 \(y’(1) = 2, y(1) = 2\)
$$ y = 2x^2 + Cx \Rightarrow y’ = 4x + C $$
- \(y’(1) = 4 + C = 2 \Rightarrow C = -2\)
- 代入 y(1) 得：
$$ y(1) = 2 + (-2) = 0 \ne 2 $$
- ❌ 无解（No solution）
Existence and Uniqueness of Solutions for Linear 1st Order IVPs#
Existence of a Unique Solution for First-Order IVPs#
Initial-value problem (IVP, 初值问题)：
$$ y’ = f(x,y), \quad y(x_0) = y_0 $$
给到一个 1st Order Differential Equation，并给了某个点 \((x_0, y_0)\) 的初始条件 去研究解到底存不存在？如果存在，是不是唯一的？
- 如果 \(f(x,y)\) 和 \(\frac{\partial f}{\partial y}\)（对 y 的偏导）在矩形区域 R (矩形区域) 内连续
- 也就是说如果一个微分方程 \(y’ = f(x,y)\) 里面的函数 \(f(x,y)\) 和它对 y 的偏导数 \(\frac{\partial f}{\partial y}\) 都是“平滑的”（数学上 = 连续）
- 那么在区间 \(I_0 (a < x_0 < b)\) 内，Solution Exist 并且是 Unique 的，存在唯一解 \(y(x)\)，并且这个区间 \(I_0\) 包含在更大的区间 \(I (a \leq x \leq b)\) 中
这种情况下，只保证在 \(x_0\) 附近一个小区间 \(I_0\) 内，存在唯一解
Existence of a Unique Solution for First-Order Linear IVPs#
Linear IVP (线性初值问题)：
$$ y’ + P(x)y = f(x), \quad y(x_0) = y_0 $$
多了一个 Linear 的条件
- 如果函数 P(x) 和 f(x) 在包含 \(x_0\) 的开区间 I (开区间) 上连续，
- 那么这个初值问题在整个区间 I 上都有唯一解
解不仅唯一，而且在整个区间 I 上都成立
Why Linear Follow From Non-Linear#
- 两个 Theorem，第一个 Non-Linear 下要满足的条件是
- f(x,y) 连续；\(\dfrac{\partial f}{\partial y}\) 在该区域也连续
- 而在 Linear 的特殊情况下
$$ y’ + P(x)y = f(x) $$
- 可以改写成为
$$ y’ = f(x) - P(x)y = \hat{f}(x,y) $$
也就来到了 Non-Linear 的第一个条件，想要让 \(\hat{f}(x,y)\) Cont. 在 Linear Function 中只需要 \(f(x)、P(x)\) 在区间 II 上连续就行
同时 \(f(x)、P(x)\) 连续的情况下，也满足了 Partial Derivative 连续的条件，这是因为
$$ \frac{\partial \hat{f}(x,y)}{\partial y} = -P(x) $$
所以只要 P(x) 连续，所以这个偏导也连续
Ex. Applying Non-Linear Case#
“What is the interval R in this example?”
$$ y’ = 3y^{2/3}, \quad y(2)=0 $$
- Non-Linear Case，检查 \(f(x,y) = 3y^{2/3}\) 是连续的
- 计算偏导：
$$ \frac{\partial f}{\partial y} = 2y^{-1/3} = \frac{2}{y^{1/3}} $$
- 观察可以发现，当 y=0 的时候，\(\dfrac{2}{y^{1/3}}\) 会发散
So, are we guaranteed to have a unique solution?
Theorem 1.2.1 does not guarantee that. But we don’t know if the solution is unique or not.
定理 1.2.1 不能保证。但是我们也不能直接下结论说解一定不唯一，只是“没有保证”。
Solution Curves without Solutions: Direction Field#
- 对于一个微分方程：
$$ y’ = f(x,y), $$
- 它告诉你：在平面上每一个点 \((x,y)\)，解曲线的斜率是多少
- 方向场 就是：在很多点 \((x,y)\) 上画一条小的斜线，表示解曲线在这里的切线方向
- 这样即使你没解出方程，也能大致看到解的走势
Autonomous DE and its Critical Points#
Autonomous DE（自治微分方程） 指的是方程里面 Independent Variable x 不显式出现，形式为：
$$ \frac{dy}{dx} = f(y) $$
或者更一般写成 \(F(y,y’)=0\)
👉 换句话说：方程的变化只依赖于 y，不依赖于 x
Critical Point#
- 给定：\(y’ = f(y)\)，如果存在某个 \(y=c\)，使得：\(f(c)=0\)
- 那么 y=c 就是 临界点 (critical point)
也就是导数 \(y’=0\) 的点
Ex. Critical Point of a Function#
$$ x’ = x(4-x) $$
- Critical Point 出现在 x = 0 和 4 的位置
Stability of Critical Points#
Stable Point#
Unstable Point#
Semi-Stable Point#
General Linear ODE’s and Linear Differential Operator#
- 回顾前面的 Linear ODE，其 Form 为
$$ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_1(x)y’ + a_0(x)y = g(x) $$
- 这是 n-th order linear ordinary differential equation, ODE 的一般
- 其中 \(a_k(x)\) 是系数函数，可能随 \(x\) 变化
- 右边的 \(g(x)\) 是非齐次项
- \(y^{(n)}\) 表示对函数 y 的 n 阶导数
- 其是 Linear 的，当它的所有 Coefficient 都是一次方的
- 其是 Homogeneous 的，当 g(x) = 0 时
- 其是 Constant coefficient differential equation 常系数微分方程，当 \(a_k(x)\) 都是 Constant 时
Linear Differential Operator L#
- 线性微分的算子 L 可以简化我们写高阶 linear ODE 的方式
- 上面提到了，传统的写法是
$$ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_1(x)y’ + a_0(x)y = g(x) $$
- 这整个东西可以简写为：
$$ Ly = g(x) $$
- 其中 L 是一个“作用在 y”上的线性微分算子
- 可以发现 Equation 的左边只有 y，这就说明了 L 不仅包含 Coefficient，还包含对 y 的 Derivative
$$ L = a_n(x) \frac{d^n}{dx^n} + a_{n-1}(x) \frac{d^{n-1}}{dx^{n-1}} + \cdots + a_1(x) \frac{d}{dx} + a_0(x) $$
Ex1. Find Linear Differential Operator L#
$$ 3\ddot{y} - 4 \dot{y} + 2xy = e^{x} \Rightarrow L = 3 \frac{d^2}{dx^2} - 4 \frac{d}{dx} + 2x $$
Ex2. Find Linear Differential Operator L#
$$ y’’ + x \sin(y’) - xy = x^2 $$
- 这并不是一个 Linear 的 DE，所以没有整个 Linear Differential Operator L
Existence and Uniqueness of nth order Linear ODE#
- 之前提到过 1st Linear ODE 的存在与唯一性定理
- 那么对于 n 阶线性 ODE + 初始条件（initial value problem, IVP），是否也存在类似定理？
- 要求 nth ODE 形如
$$ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_1(x)y’ + a_0(x)y = g(x) $$
- initial conditions 为
$$ y(x_0) = y_0,\quad y’(x_0) = y_1,\quad \cdots,\quad y^{(n-1)}(x_0) = y_{n-1} $$
Theorem: Existence of a Unique Solution#
- 若函数 \(a_1(x), a_2(x), …, a_n(x), g(x)\) 在区间 I 上连续（continuous），且 \(a_n(x) \ne 0\)，那么对任意 \(x_0 \in I\)，初值问题：
$$ a_n(x)y^{(n)} + \cdots + a_0(x)y = g(x), \quad y(x_0) = y_0, \quad y’(x_0) = y_1, \quad \cdots, \quad y^{(n-1)}(x_0) = y_{n-1} $$
- 在区间 I 上有唯一解
这里的 \(a_n(x)\) 指的是最高项的那个 Coefficient
Ex1. Existence of a Unique Solution#
$$ y^{(3)} + 3y’’ + 4y’ + 12y = 0 $$
- 初始条件为：
$$ y(1) = y’(1) = y’’(1) = 0 $$
- 系数全是常数，连续
- \(a_3 = 1 \ne 0\)，最高阶导数项系数非零
- 所以 Exist Unique Solution
Ex2. Existence of a Unique Solution#
$$ (1 - t^2)y’’ - 2ty’ + 2y = 0 $$
- 初始条件是：
$$ y(0) = 1,\quad y’(0) = 2 $$
- 问：在哪个区间上存在唯一解？
- 总的来说，除了最高次项系数不是常数，其他都是 Constant，连续，
- \(1 - t^2 = 0\) 当且仅当 \(t = \pm 1\)，这些点会使 \(a_2(t) = 0\)，不满足定理条件，所以不能包含 \(t = \pm 1\)
Homogenous Linear DE’s: General Solutions#
Theorem： General Solutions for Homogeneous Linear DEs#
- 对于一个 nth Order 的 Homogenous Linear DEs
$$ L(y) = 0 $$
- 一定有 n 个 linearly independent 的解，这组解被称为：fundamental set of solutions（基本解组）
- 当知道了完整的 Solution Set：
$$ y_1(x), y_2(x), …, y_n(x) $$
- 那么 Particular Solution 就是它们的 linear combination
- 如果 IVP 给了超过 n 个初值条件，会 over-determined，可能无解
- 这是因为 L 是线性算子（linear operator）
$$ L(\alpha y_1 + \beta y_2) = \alpha L(y_1) + \beta L(y_2) $$
- 也就是说假设 \(y_1, y_2, …, y_n\) 是解，那么任意线性组合：
$$ c_1 y_1 + c_2 y_2 + \cdots + c_n y_n $$
- 也一定是解，这就是为什么我们把所有解写成“线性组合”
Definition: Linearly Independence of Functions#
- 回忆线性代数里的定义：向量 \(v_1, …, v_n\) 是线性无关的意思是：
$$ c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0 $$
- 仅当所有 \(c_k = 0\) 时才成
- 在函数空间中也一样：如果只有 \(c_1 = c_2 = \cdots = c_n = 0\) 才能让 \(c_1 f_1(x) + c_2 f_2(x) + \cdots + c_n f_n(x) = 0\) 成立，则函数 \(f_1, …, f_n\) 是 linearly independent
Ex1. Finding the Linear Independency of function#
Are functions \({1 - 3x + 2x², 1 - x + x², 1 + x}\) linearly independent?
Ex2. Finding the Linear Independency of function#
Are functions \({e^x, xe^x, x²e^x}\) linearly independent?\
How to check if n functions are independent?#
- 如果我们有 n 个多项式函数，这种情况是容易判断的
- 找到它们的线性组合形式：
$$ \sum_{k=1}^{n} c_k y_k $$
- 也就是形式如 \(c_1 y_1 + c_2 y_2 + \cdots + c_n y_n\)
- 观察每一项（例如 \(x^k\)）的系数，如果所有系数都为零，说明这些函数线性无关
- 这就是上面的两个例子
Case in non Polynomial#
- 当函数不是 Polynomial，可以通过 Wronskian 行列式
The Wronskian#
对于一组函数 \(f_1(x), f_2(x), \dots, f_n(x)\)，如果它们在区间 \([a, b]\) 上有 \(n-1\) 阶导数，则可以构造一个如下的行列式：
$$ W(f_1, f_2, \dots, f_n) = \begin{vmatrix} f_1 & f_2 & \cdots & f_n \ f_1’ & f_2’ & \cdots & f_n’ \ \vdots & \vdots & \ddots & \vdots \ f_1^{(n-1)} & f_2^{(n-1)} & \cdots & f_n^{(n-1)} \end{vmatrix} $$
假设这组函数在区间 \(I = [a, b]\) 上都有 \(n-1\) 阶导数
如果它们的 Wronskian 行列式 在整个区间 I 上不恒等为零（not identically zero），那么这些函数是 Linearly independent 的
The beauty of Math (or how seemingly unrelated mathematical topics connect to each other)#
- 考虑一个 Function Set
$$ \left{ e^{m_1 x}, e^{m_2 x} \right}, \quad m_1, m_2 \in \mathbb{C} $$
- 想要找这些函数是否 linearly independent
- 如果通过之前的通过 Polynomial 构造的方式，可以列出 Linear Combination
$$ C_1 e^{m_1 x} + C_2 e^{m_2 x} = 0, \quad \forall x \in \mathbb{R} \tag{I} $$
- 目前这是一个方程、两个未知数，方程不够
- 于是可以尝试用 Wronskain，先对 Functions 求导，得到
$$ C_1 m_1 e^{m_1 x} + C_2 m_2 e^{m_2 x} = 0, \quad \forall x \in \mathbb{R} $$
- 将他们写入 Wronskain 可以得到
$$ \begin{bmatrix} e^{m_1 x} & e^{m_2 x} \ m_1 e^{m_1 x} & m_2 e^{m_2 x} \end{bmatrix} \begin{bmatrix} C_1 \ C_2 \end{bmatrix}#
\begin{bmatrix} 0 \ 0 \end{bmatrix} \tag{II} $$
- 令左边的 Matrix 为 \(A(x)\)
- 向量为 \(\vec{C} = [C_1, C_2]^T\)
- 也就是寻找一个向量 \(\vec{C}\) 使得：
$$ A(x)\vec{C} = \vec{0}, \quad \forall x \in \mathbb{R} $$
- 如果唯一满足这个条件的是 \(\vec{C} = \vec{0}\)，那说明函数是线性无关的
- 而要保证这一结论，可以通过证明如果某个点 \(x \in \mathbb{R}\) 使得 \(A(x)\) Invertible，那么系统只有唯一解 \(\vec{C} = \vec{0}\)
- 而想要证明 \(A(x_0)\) 可逆，就可以通过 Determinant，即
$$ \vec{C} = A(x_0)^{-1} \vec{0} = \vec{0} $$
- 而这个 \(\det A(x)\) 就是 Wronskian，有
$$ W(f_1, f_2) = \begin{vmatrix} e^{m_1 x} & e^{m_2 x} \ m_1 e^{m_1 x} & m_2 e^{m_2 x} \end{vmatrix} = e^{(m_1 + m_2)x}(m_2 - m_1) $$
- 如果 \(m_1 \ne m_2\)，那么 \(W(f_1, f_2) \ne 0\)，Wronskian 不为零
- 因此函数集合 \({e^{m_1 x}, e^{m_2 x}}\) Linearly Independent
Case when W = 0 for all x in I#
- 已知一组函数 \({f_1(x), f_2(x), \dots, f_n(x)}\)
- 计算得 \(W(f_1, f_2, \dots, f_n) = 0\) 对所有 \(x \in I\) 成立
- 不能否断定这些函数是 **linearly dependent，**这不是定理所说的内容
$$ W \not\equiv 0 \quad \Longrightarrow \quad \text{线性无关} $$
- 但是它的逆命题 不成立：
$$ W \equiv 0 \quad NOT\Rightarrow \quad \text{线性相关} $$
Ex. Case when W = 0 for all x in I#
$$ f_1(x) = \begin{cases} x^2, & x \geq 0 \ 0, & x < 0 \end{cases}, \quad f_2(x) = \begin{cases} 0, & x \geq 0 \ x^2, & x < 0 \end{cases} $$
- 这两个函数在各自区域里互补，显然是线性无关的
- 但它们的 Wronskian 在整个区间恒为零，所以 Wronskian=0 并不能保证线性相关
Special Case of Wronskian#
- 如果一组函数 \({y_1(x), \dots, y_n(x)}\) 是某个n阶 homogeneous linear ODE 的解
$$ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_1(x)y’ + a_0(x)y = 0 $$
- 会有一个更强的结论，也就是
- 这些函数在区间 II 上线性无关 当且仅当 iff：
$$ W(y_1, y_2, \dots, y_n)(x) \neq 0, \quad \forall x \in I $$
- 也就是说
- \(W \equiv 0\) ⇒ 必定线性相关
- \(W \neq 0\) ⇒ 必定线性无关
Ex. Determine Fundamental Set of Solution#
Is the set \({e^x,e^{-x},\cosh x}\) the Fundamental Set Solution of \(y-y’’’=0\) ?
Find the Second Solution for a 2nd Degree Linear Homo ODE#
Assume that we already know that \(y_1(x)\) is a solution for this ODE, and consider the ODE
$$ y’’+P(x)y’+Q(x)y=0 $$
- Let the second solution as
$$ y_2(x)=U(x)y_1(x) $$
- Where \(U(x)\) is an undetermined function
- Take the derivative on both side to get
$$ y_2’ = U’ y_1 + U y_1’ $$
$$ y_2’’ = U’’ y_1 + 2U’ y_1’ + U y_1’’ $$
- Recall the original function \(y’’ + P(x) y’ + Q(x) y = 0\)
- Sub in \(y_2\) to get
$$ (U y_1)’’ + P(U y_1)’ + Q(U y_1) = 0 $$
Substitution is \(y_2= Uy_1\), since \(y_2\) is also a solution, so it can be sub in the function
- Expand to get
$$ (Uy_1)’’ = U’‘y_1 + 2U’y_1’ + Uy_1’’ $$
$$ (Uy_1)’ = U’y_1 + Uy_1’ $$
- Sub in to get
$$ (U’‘y_1 + 2U’y_1’ + Uy_1’’) + P(U’y_1 + Uy_1’) + Q(Uy_1) = 0 $$
- Arrange them into order of U
$$ y_1U’’+(2y_1’ + P y_1)U’+(y_1’’ + P y_1’ + Q y_1)U = 0\tag{1} $$
- We have the original ODE \(y_1’’ + P y_1’ + Q y_1=0\)
- Thus (1) now become
$$ U y_1’’ + (2 y_1’ + P y_1) U’ + y_1 U’’ = 0 $$
- Now let \(w=U’\), then
$$ y_1 w’ + (2 y_1’ + P y_1) w = 0 $$
- Arrange to get
$$ w’ + \left( 2\frac{y_1’}{y_1} + P \right) w = 0 $$
- Further Arrange to get
$$ \frac{w’}{w} + 2\frac{y_1’}{y_1} + P = 0\Rightarrow \frac{w’}{w} + 2\frac{y_1’}{y_1} = -P $$
- Can see that LHS is a result of Derivative of Products
- Thus have
$$ \frac{d}{dx}(\ln(w y_1^2)) = -P $$
- Integrate both side to get
$$ \ln(w y_1^2) = -\int P(x),dx $$
$$ w y_1^2 = e^{-\int P(x),dx}\Rightarrow w = \frac{1}{y_1^2} e^{-\int P(x),dx} $$
- Recall that we let \(w=U’\) before, now sub back to get
$$ U’ = \frac{1}{y_1^2} e^{-\int P(x),dx} $$
- Integrate both side to get
$$ U = \int \frac{1}{y_1^2} e^{-\int P(x),dx},dx $$
- Recall we let \(y_2=y1U(x)\), so the Final Result is
$$ y_2(x) = y_1(x) \int \frac{e^{-\int P(x),dx}}{y_1(x)^2},dx $$
Ex. Find the second Solution for 2nd Linear Homo ODE#
The function is \(y’’+4y’+4y=0\), \(y_1(x) = e^{-2x}\)
Find the fundamental set of solutions for Higher Order Homo Linear DE#
Consider the function
$$ a y’’’ + b y’’ + c y’ + d y = 0\quad a,b,c,d \in \mathbb{R} $$
- This is a third-order homogeneous linear ODE with constant coefficients
- Consider \(y(x) = e^{mx}\) as a solution
- Now sub in \(y(x)\) into function
$$ y’ = m e^{mx}, \quad y’’ = m^2 e^{mx}, \quad y’’’ = m^3 e^{mx} $$
$$ a m^3 e^{mx} + b m^2 e^{mx} + c m e^{mx} + d e^{mx} = 0 $$
- Since \(e^{mx} \neq 0\), then get rid of it to get
$$ a m^3 + b m^2 + c m + d = 0 $$
- This is the Characteristic Equation（特征方程）
- For that Equation above, clearly it has three solutions \(m_1, m_2, m_3\), so the funciton can become
$$ a (m - m_1)(m - m_2)(m - m_3) = 0 $$
- Thus the three solutions are \(e^{m_1 x},\ e^{m_2 x},\ e^{m_3 x}\)
- And clearly if \(m_1\neq m_2\neq m_3\), then they are linearly independent
- Then will get the fundamental solution set of the DE as
$$ {e^{m_1 x}, e^{m_2 x}, e^{m_3 x}} $$
Types of roots (solution for DE)#
In above part, we conclude the fundamental solution set of DE, but there may exist multiple situations for those roots (i.e. May have complex roots)
- Now consider a more general case, for a nth order homo DE
$$ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_1 y’ + a_0 y = 0 $$
- Its Characteristic Equation is
$$ a_n m^n + a_{n-1} m^{n-1} + \dots + a_1 m + a_0 = 0 $$
- The possible roots can be
- Real
- Complex
- Repeated
- Now we are going to analysis them one by one
Type 1：Distinct Real Roots#
The most simple case, when all three are Distinct Real Roots
$$ y = c_1 e^{m_1 x} + c_2 e^{m_2 x} + \dots + c_n e^{m_n x} $$
- Then the solution is simple
Type 2：Conjugate Complex Roots#
As Complex Roots need to appear with its Conjugate, thus have
$$ m_{1,2} = \alpha \pm i\beta $$
- Their roots are \(y = e^{(\alpha + i\beta)x}\), expand and get
$$ e^{\alpha x} \cos(\beta x), \quad e^{\alpha x} \sin(\beta x) $$
- Those two complex solution can be represent as
$$ y = e^{\alpha x}(c_1 \cos\beta x + c_2 \sin\beta x) $$
Prove that a Complex Number Solution’s Conjugate is also a Solution We have the first Complex Number as \(A_n m_1^n + A_{n-1} m_1^{n-1} + \cdots + A_1 m_1 + A_0 = 0\), take Conjugate on both side to get \(\overline{A_n} , \overline{m_1}^n + \overline{A_{n-1}} , \overline{m_1}^{,n-1} + \cdots + \overline{A_1} , \overline{m_1} + \overline{A_0} = 0\), thus also a solution
Type 3：Repeated Real Roots#
If one Real Root has repeated r times \(m_1 = m_2 = \dots = m_r = m\), then the solution is
$$ e^{mx},x e^{mx}, x^2 e^{mx}, \dots, x^{r-1} e^{mx} $$
Ex. Specify the general class of solution of function#
Having the function
$$ y^{(6)} - 8y^{(5)} + 23y^{(4)} - 26y^{(3)} - 2y’’ + 32y’ - 24y = 0 $$
- First get the Characteristic Function \(m^6 - 8m^5 + 23m^4 - 26m^3 - 2m^2 + 32m - 24 = 0\)
- This function has the following roots
$$ (m + 1)(m - 3)(m - 2)^2 (m - (1 + i))(m - (1 - i)) = 0 $$
- The roots we have above are
| Root | Type | Solution |
|---|---|---|
| ( m = -1 ) | Real | ( \(e^{-x}\) ) |
| ( m = 3 ) | Real | ( \(e^{3x}\) ) |
| ( m = 2 )（重根） | Repeated Real | ( \(e^{2x},\ x e^{2x}\) ) |
| ( m = 1 \pm i ) | Conjugate Complex | ( \(e^{x}\cos x,\ e^{x}\sin x\) |
- So the General Solution si
$$ { e^{-x},\ e^{3x},\ e^{2x},\ x e^{2x},\ e^{x}\cos x,\ e^{x}\sin x } $$
$$ y_c(x) = C_1 e^{-x} + C_2 e^{3x} + C_3 e^{2x} + C_4 x e^{2x} + C_5 e^{x}\cos x + C_6 e^{x}\sin x $$
Ex. Can the auxiliary equation have repeated complex roots?#
If we have the Characteristic Function as
$$ (m - (1+i))^2 (m - (1-i))^2 = 0 $$
- Then its a Conjugate Pair of \((1+i)\) and \((1-i)\), their solution is then
$$ e^{x}\cos x,\ e^{x}\sin x,\ x e^{x}\cos x,\ x e^{x}\sin x $$
General Solution of Non-homo Linear DE#
Now we are copping with the Non-Homo Case, where there’s non zero on the RHS, so with the function
$$ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \dots + a_0(x)y = g(x) $$
- So we are trying to find a Particular solution that
$$ L(y_p) = g(x) $$
- And a Homo Solution where
$$ L(y_c) = 0 $$
Proving the General Solution is Comp + Part#
So we are trying to prove that \(y = y_c + y_p\)
- So sub it back to get
$$ L(y) = L(y_c + y_p) $$
- Because L is Linear, then
$$ L(y) = L(y_c) + L(y_p) $$
- We know that \(L(y_c) = 0\) and \(L(y_p) = g(x)\), thus
$$ L(y) = 0 + g(x) = g(x) $$
- This will also work in the converse way that
$$ y - y_p $$
- has to have the identity that
$$ L(y - y_p) = L(y) - L(y_p) = g(x) - g(x) = 0 $$
Ex. Finding the General Solution#
Having the function \(y’’ - 3y’ + 2y = e^{x}\)
Variation of Parameters#
Finding the Particular Solution of the DE is hard, and this method can help
- Suppose we have the DE
$$ y’’ + P(x)y’ + Q(x)y = f(x) $$
- First the Complementary Solution is
$$ y_c = C_1 y_1 + C_2 y_2 $$
- Then Start the Variation of Parameters
- Let all Constant become function
$$ y_p = U_1(x) y_1 + U_2(x) y_2 $$
- Before we taking the derivatives, we found that
$$ y_p’ = U_1’y_1 + U_1y_1’ + U_2’y_2 + U_2y_2’ $$
- Is hard to due, so we can smartly setting a restriction that let
$$ U_1’ y_1 + U_2’ y_2 = 0 \quad \text{(Eq.1)}\Rightarrow y_p’ =U_1y’_1+U_2y’_2 $$
This helps when calculating \(y’’\)
- Now, calculate the \(y’’\)
$$ y_p’’ = U_1’y_1’ + U_1y_1’’ + U_2’y_2’ + U_2y_2’’ $$
- Then sub back the \(y_p\) into \(y’’ + P y’ + Q y = f(x)\),
$$ (U_1’y_1’ + U_1y_1’’ + U_2’y_2’ + U_2y_2’’)+P(U_1y_1’ + U_2y_2’)+Q(U_1y_1 + U_2y_2)= f(x) $$
- Rearrange \(U_1,U_2\) to get
$$ (U_1’y_1’ + U_2’y_2’) + U_1(y_1’’ + Py_1’ + Qy_1) + U_2(y_2’’ + Py_2’ + Qy_2) = f(x) $$
- Cause \(y_1, y_2\) are solution of the Homo Function, thus
$$ y_1’’ + Py_1’ + Qy_1 = 0, \quad y_2’’ + Py_2’ + Qy_2 = 0 $$
- Then the part left is
$$ U_1’y_1’ + U_2’y_2’ = f(x) $$
- So now, we got a function set
$$ \begin{cases} U_1’y_1 + U_2’y_2 = 0 & (\text{Eq.1}) \ U_1’y_1’ + U_2’y_2’ = f(x) & (\text{Eq.2}) \end{cases} $$
- Write them in Matrix Form is
$$
\begin{bmatrix} y_1 & y_2 \ y_1’ & y_2’ \end{bmatrix} \begin{bmatrix} U_1’ \ U_2’ \end{bmatrix}
\begin{bmatrix} 0 \ f(x) \end{bmatrix}
$$
- We can find out that this matrix is just Wronskian, thus use Cramer’s Rule
$$ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1’ & y_2’ \end{vmatrix} $$
- Thus we got
$$ U_1’ = -\frac{y_2 f}{W(y_1, y_2)}, \quad U_2’ = \frac{y_1 f}{W(y_1, y_2)} $$
- Take integral
$$ U_1 = \int -\frac{y_2 f}{W(y_1, y_2)},dx, \quad U_2 = \int \frac{y_1 f}{W(y_1, y_2)},dx $$
- Then we will get the Particular Solution
Ex. Finding the Particular Solution#
We have the function \(y’’ - 2y’ + y = \frac{e^t}{1+t^2}\)
5.1