常微分方程

1. 基础概念

1.1 常用术语

术语	翻译	术语	翻译	术语	翻译
homogeneous	齐次	constant	常数	derivative	导数
superposition	叠加	multiplication	乘法	substitute	代换
partial derivative	偏导数	arbitrary	任意	reducible	可化简
autonomous	自治	algebra	代数	interval	区间
infinite family	无穷集	proportional	比例	explicit	特解
assertion	断言	intersect	相交	parameter	参数
antiderivative	不定积分	branch	分支	domain	范围
independent variable	自变量	verify	验证	coefficient	系数
identity equation	恒等式	curve	曲线	equation	方程
dependent variable	因变量	integration	积分	slope	斜率
necessary condition	必要条件	exponential	指数的	static	静态
asymptote	渐近线	denote	表示	indeed	的确
hypothesis	假设	error function	误差方程	encompass	包含
equilibrium	平衡	vibration	振动	duplication	重复

1.2 n 阶方程形式

The most general form of an nth-order differential equation with independent \(x\) and unknown function or dependent variable \(y=y(x)\) is \(F(x,y,y',y'',\cdots,y^{(n)})=0\). An nth-order differential equation ordinarily has an n-parameter family of solutions—one involving n different arbitrary constants or parameters.

1.3 积分求解

The first-order equation \(\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=f(x,y)\end{aligned}\) takes an especially simple form if the function \(f\) is independent of the dependent variable \(y\), i.e., \(\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=f(x)\end{aligned}\). In this special case, we only need to integrate both sides:

\[\begin{aligned}y(x)=\int f(x)\mathrm{d}x+C\end{aligned}\]

1.4 解的存在唯一性

Suppose that both the function \(f(x,y)\) and its partial derivative \(D_yf(x,y)\) are continuous on some rectangle \(R\) in the xy-plane that contains the point \((a,b)\) in its interior. Then, for some open interval \(I\) containing the point \(a\), the initial value problem \(\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=f(x,y),\quad y(a)=b\end{aligned}\) has one and only one solution that is defined on the interval \(I\).

If the function \(f(x,y)\) and/or its partial derivative \(\begin{aligned}\frac{\partial f}{\partial y}\end{aligned}\) fail to satisfy the continuity hypothesis, then the initial value problem may have either no solution or many (even infinitely many) solutions.

1.5 皮卡迭代法

To prove the existence and uniqueness theorem rigorously, the typical Picard iteration method follows these steps:

Re-write the ODE as an integral equation.
Construct the Picard iterates.
Show that the Picard iterates converge uniformly on \(|x-x_0|\leq h\).
Show that the above limit function is a solution.
Show that the solution is unique.

2. 一阶方程解法

2.1 分离变量法

The first-order differential equation \(\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=H(x,y)\end{aligned}\) is called separable provided that \(H(x,y)\) can be written as the product of a function of \(x\) and a function of \(y\): \(\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=g(x)h(y)=\frac{g(x)}{f(y)}\end{aligned}\). Solution method:

\[\begin{aligned}\int f(y)\mathrm{d}y=\int g(x)\mathrm{d}x+C,\quad F(y)=\int f(y)\mathrm{d}y,\quad G(x)=\int g(x)\mathrm{d}x\end{aligned}\]

2.2 一阶线性方程

Standard form:

\[\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)\end{aligned},\quad \rho(x)=\mathrm{e}^{\int P(x)\mathrm{d}x}\]

Solution steps:

Begin by calculating the integrating factor \(\rho(x)=e^{\int P(x)\mathrm{d}x}\).
Multiply both sides of the differential equation by \(\rho(x)\).
Recognize the left-hand side of the resulting equation as the derivative of a product: \(D_x[\rho(x)y(x)]=\rho(x)Q(x)\).
Integrate this equation to get the solution: \[\begin{aligned}\rho(x)y(x)=\int\rho(x)Q(x)\mathrm{d}x+C \implies y(x)=\mathrm{e}^{-\int P(x)\mathrm{d}x}\left[\int(Q(x)\mathrm{e}^{\int P(x)\mathrm{d}x})\mathrm{d}x+C\right]\end{aligned}\]

Properties:

Guarantee only a solution on a possibly smaller interval.
Linear first-order differential equation has no singular solutions.

2.3 恰当方程

A general solution \(y(x)\) of a first-order differential equation is often defined implicitly by an equation of the form \(F(x,y)=C\). Differentiate each side with respect to \(x\):

\[\begin{aligned}&\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}=0 \Rightarrow M(x,y)+N(x,y)\frac{\mathrm{d}y}{\mathrm{d}x}=0 \Rightarrow M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0\end{aligned}\]

If there exists a function \(F(x,y)\) such that \(\begin{aligned}\frac{\partial F}{\partial x}=M, \quad \frac{\partial F}{\partial y}=N\end{aligned}\), then \(M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0\) is called an exact differential equation.

Thus the equation \(\begin{aligned}\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\end{aligned}\) is a necessary condition for the differential equation to be exact.

Solution:

\[\begin{aligned}F(x,y)=\int M(x,y)\mathrm{d}x+g(y)\end{aligned}\]

\[\begin{aligned}N=\frac{\partial F}{\partial y}=\begin{pmatrix}\begin{aligned}\frac{\partial}{\partial y}\int M(x,y)\mathrm{d}x\end{aligned}\end{pmatrix}+g'(y) \implies g'(y)=N-\frac{\partial}{\partial y}\int M(x,y)\mathrm{d}x\end{aligned}\]

\[\begin{aligned}F(x,y)=\int M(x,y)\mathrm{d}x+\int\begin{pmatrix}\begin{aligned}N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\mathrm{d}x\end{aligned}\end{pmatrix}\mathrm{d}y = C\end{aligned}\]

2.4 积分因子

If \(M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0\) is not exact, we can multiply it by an integrating factor \(\mu\) to make it exact.

If \(\mu\) depends only on \(x\): \[ \begin{aligned} \frac{\mathrm{d}\mu}{\mu} = \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right)\mathrm{d}x \Rightarrow \mu(x) = \exp\left[\int \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right)\mathrm{d}x\right] \end{aligned} \]
If \(\mu\) depends only on \(y\): \[ \begin{aligned} \frac{\mathrm{d}\mu}{\mu} = \frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)\mathrm{d}y \Rightarrow \mu(y) = \exp\left[\int \frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)\mathrm{d}y\right] \end{aligned} \]

2.5 变量代换法

一阶普遍代换:

Consider \(\frac{\mathrm{d}y}{\mathrm{d}x} = f(x,y)\). Introduce a new dependent variable \(v = \alpha(x,y)\), so that \(y = \beta(x,v)\).

Substituting \(\frac{\mathrm{d}y}{\mathrm{d}x} = \beta_x + \beta_v \frac{\mathrm{d}v}{\mathrm{d}x}\) into the original equation yields a new first‑order

equation for \(v\): \(\frac{\mathrm{d}v}{\mathrm{d}x} = g(x,v)\).

伯努利方程:

Form: \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)\,y = Q(x)\,y^{\,n}, \quad n \neq 0,1\).

Solution: Set \(v = y^{1-n}\). Multiply the original equation by \((1-n)y^{-n}\) to obtain a linear first‑order equation for \(v\):

\[\frac{\mathrm{d}v}{\mathrm{d}x} + (1-n)P(x)\,v = (1-n)Q(x)\]

Solve for \(v(x)\), then recover \(y = v^{1/(1-n)}\).

一阶齐次方程:

Form: \(\frac{\mathrm{d}y}{\mathrm{d}x} = F\!\left(\frac{y}{x}\right)\).

Solution: Introduce the substitution \(v = \dfrac{y}{x}\), so \(y = vx\) and \(\frac{\mathrm{d}y}{\mathrm{d}x} = v + x\frac{\mathrm{d}v}{\mathrm{d}x}\).

Substituting gives \(x\frac{\mathrm{d}v}{\mathrm{d}x} = F(v) - v\), which is a separable equation in \(x\) and \(v\).

3. 奇解与p-判别式法

3.1 奇解与包络

It is common for a nonlinear first-order differential equation to have both a general solution involving an arbitrary constant \(C\) and one or several particular solutions that cannot be obtained by selecting a value for \(C\). These exceptional solutions are frequently called singular solutions.

Singular Solution: A solution of a differential equation that is not tangent to any solution in the family of general solutions (wait, actually it is tangent to them) and cannot be obtained by choosing specific values for the constants in the general solution.
Envelope: A curve that is tangent at each point to some solution in the family of general solutions. A singular solution is typically the envelope of the general solution family.

If there is a singular solution, at each point on the curve the ODE has more than 1 solution (uniqueness fails).

3.2 p-判别式原理

Consider a first-order ODE in implicit form:

\[\begin{aligned}F(x,y,p)=0,\quad p=\frac{\mathrm{d}y}{\mathrm{d}x}\end{aligned}\]

A singular solution must satisfy two conditions simultaneously:

The original equation: \(F(x,y,p)=0\)
The partial derivative of \(F\) with respect to \(p\): \(\begin{aligned}\frac{\partial F}{\partial p}=0\end{aligned}\) By eliminating \(p\) from these equations, we obtain the p-discriminant curve, which may represent the singular solution.

3.3 判别式法证明

Assume the family of general solutions is \(\varphi(x,y,C)=0\) where \(C\) is a constant. The envelope must satisfy: \(\varphi(x,y,C)=0\) and \(\begin{aligned}\frac{\partial\varphi}{\partial C}=0\end{aligned}\). Eliminating \(C\) gives the envelope’s equation. For the differential equation \(F(x,y,p)=0\), the envelope (singular solution) of its general solution family must satisfy:

Every point \((x,y)\) on the envelope satisfies \(F(x,y,p)=0\) (since it is a solution).
The slope \(p\) of the envelope at that point matches the slope of the corresponding solution in the general family.
The envelope represents a limiting case where distinct solutions “coalesce”, implying the dependence on \(p\) in relation to \(C\) vanishes, leading to \(\begin{aligned}\frac{\partial F}{\partial p}=0\end{aligned}\).

4. 高阶线性方程

4.1 二阶方程降阶

Consider \(F(x,y,y',y'')=0\). Two common cases allow reduction to a first‑order problem:

Dependent variable \(y\) missing (\(F(x,y',y'')=0\)):
Set \(p = y'\), so \(y'' = \frac{\mathrm{d}p}{\mathrm{d}x}\). The equation becomes \(F(x,p,p')=0\). Solve for \(p(x,C_1)\), then \(y(x) = \int p(x,C_1)\,\mathrm{d}x + C_2\).
Independent variable \(x\) missing (\(F(y,y',y'')=0\)):
Set \(p = y'\). Use the chain rule: \(y'' = \frac{\mathrm{d}p}{\mathrm{d}x} = \frac{\mathrm{d}p}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x} = p\frac{\mathrm{d}p}{\mathrm{d}y}\). The equation reduces to \(F\!\left(y,p,p\frac{\mathrm{d}p}{\mathrm{d}y}\right)=0\). Solve for \(p(y,C_1)\), then \(x(y) = \int \frac{1}{p(y,C_1)}\,\mathrm{d}y + C_2\).

4.2 线性无关与朗斯基

Two functions defined on an open interval \(I\) are linearly independent provided neither is a constant multiple of the other. The Wronskian of \(n\) functions \(f_1, \cdots, f_n\) is:

\[W=\begin{vmatrix}f_1&f_2&\cdots&f_n\\f_1'&f_2'&\cdots&f_n'\\\vdots&\vdots&&\vdots\\f^{(n-1)}_1&f^{(n-1)}_2&\cdots&f^{(n-1)}_n\end{vmatrix}\]

For solutions \(y_1, y_2, \cdots, y_n\) of a homogeneous nth-order linear ODE:

If they are linearly dependent, then \(W=0\) on \(I\).
If they are linearly independent, then \(W\neq0\) at each point of \(I\).

4.3 通解结构

Let \(y_1, \cdots, y_n\) be \(n\) linearly independent solutions of the homogeneous equation \(y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_n(x)y=0\) on \(I\). Then the linear combination \(y=c_1y_1+\cdots+c_ny_n\) is also a solution, giving the general solution.

4.4 常系数齐次方程

For the differential equation \(a_ny^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_1y'+a_0y=0\), its characteristic equation is \(a_nr^n+a_{n-1}r^{n-1}+\cdots+a_1r+a_0=0\).

Real and distinct roots: \(y(x)=c_1e^{r_1x}+c_2e^{r_2x}+\cdots+c_ne^{r_nx}\).
Repeated roots: If \(r_1\) has multiplicity \(k\), the solution includes \((c_1+c_2x+\cdots+c_kx^{k-1})e^{r_1x}\).
Complex roots (Euler’s Formula): \(e^{(a\pm bi)x}=e^{ax}(\cos{bx}\pm i\sin{bx})\). Complex conjugate roots yield solutions in the form \(e^{ax}(d_1\cos{bx}+d_2\sin{bx})\).

Polynomial Operators:

\[\begin{aligned}L=a_nD^n+a_{n-1}D^{n-1}+\cdots+a_2D^2+a_1D+a_0\end{aligned}\]

4.5 降阶法

Suppose one solution \(y_1(x)\) of \(y''+p(x)y'+q(x)y=0\) is known. Substitute \(y_2(x)=v(x)y_1(x)\) to find a second linearly independent solution \(v(x)\). (Also yields the integral formula: \(\begin{aligned}y_2=y_1\int\frac{\exp(-\int P(x)dx)}{y_1^2}dx\end{aligned}\))

4.6 非齐次方程通解

If \(y_p\) is a particular solution of the nonhomogeneous equation and \(y_1,\cdots,y_n\) are linearly independent solutions of the associated homogeneous equation, then the general solution is:

\[Y(x)=c_1y_1(x)+c_2y_2(x)+\cdots+c_ny_n(x)+y_p(x)\]

4.7 待定系数法

Take the trial solution \(y_p(x)=x^s[(A_0+\cdots+A_mx^m)e^{rx}\cos{kx}+(B_0+\cdots+B_mx^m)e^{rx}\sin{kx}]\) where \(s\) is the smallest nonnegative integer such that no term in \(y_p\) duplicates a term in \(y_c\).

\(f(x)\)	\(y_p\) 设定形式
\(P_m=b_0+b_1x+\cdots+b_mx^m\)	\(x^s(A_0+A_1x+\cdots+A_mx^m)\)
\(a\cos{kx}+b\sin{kx}\)	\(x^s(A\cos{kx}+B\sin{kx})\)
\(e^{rx}(a\cos{kx}+b\sin{kx})\)	\(x^se^{rx}(A\cos{kx}+B\sin{kx})\)
\(P_m(x)e^{rx}\)	\(x^s(A_0+A_1x+\cdots+A_mx^m)e^{rx}\)
\(P_m(x)(a\cos{kx}+b\sin{kx})\)	\(\begin{align}&x^s[(A_0+A_1x+\cdots+A_mx^m)\cos{kx}\\&+(B_0+B_1x+\cdots+B_mx^m)\sin{kx}]\end{align}\)

4.8 参数变易法

For nonhomogeneous equation \(y''+P(x)y'+Q(x)y=f(x)\), if \(y_c(x)=c_1y_1(x)+c_2y_2(x)\), set \(y_p=u_1y_1+u_2y_2\). We solve the system:

\[\begin{cases}u_1'y_1+u_2'y_2=0\\u_1'y_1'+u_2'y_2'=f(x)\end{cases}\]

Which yields:

\[\begin{aligned}y_p(x)=-y_1(x)\int\frac{y_2(x)f(x)}{W(x)}dx+y_2(x)\int\frac{y_1(x)f(x)}{W(x)}dx\end{aligned}\]

5. 级数解法

5.1 幂级数与收敛半径

Given \(\sum c_nx^n\), suppose the limit \(\begin{aligned}\rho=\lim_{n\rightarrow\infty}\left|\frac{c_n}{c_{n+1}}\right|\end{aligned}\) exists.

If \(\rho=0\), the series diverges for all \(x\neq 0\).
If \(0<\rho<\infty\), the series converges if \(|x|<\rho\) and diverges if \(|x|>\rho\).
If \(\rho=\infty\), the series converges for all \(x\).

5.2 常点级数解

For \(y''+P(x)y'+Q(x)y=0\), \(x=a\) is an ordinary point if \(P(x)\) and \(Q(x)\) are both analytic at \(x=a\). It will have two linearly independent power series solutions of the form \(\begin{aligned}y(x)=\sum_{n=0}^\infty c_n(x-a)^n\end{aligned}\).

5.3 正则奇点

Rewrite as \(y''+\frac{p(x)}{x}y'+\frac{q(x)}{x^2}y=0\). The singular point \(x=0\) is a regular singular point if \(p(x)=xP(x)\) and \(q(x)=x^2Q(x)\) are both analytic at \(x=0\). Calculate limits: \(p_0=\lim_{x\rightarrow0}xP(x)\) and \(q_0=\lim_{x\rightarrow0}x^2Q(x)\).

If limits are finite, \(x=0\) is a regular singular point.
If either limit fails to exist or is infinite, \(x=0\) is an irregular singular point.

5.4 弗罗贝尼乌斯法

If \(x=0\) is a regular singular point, assume a Frobenius series solution:

\[\begin{aligned}y(x)=x^r\sum_{n=0}^\infty c_nx^n=\sum_{n=0}^\infty c_nx^{n+r}\quad (c_0\neq0)\end{aligned}\]

Substituting into the ODE yields the indicial equation: \(r(r-1)+p_0r+q_0=0\). Its two roots \(r_1, r_2\) are the exponents.

If \(r_1-r_2\) is not an integer, there are two Frobenius series solutions.
If \(r_1=r_2\), there is only one Frobenius series solution.
If \(r_1-r_2\) is a positive integer, there may or may not exist a second Frobenius series solution.

5.5 对数与非对数情形

Let \(r_1 \geq r_2\) and \(r_1-r_2=N\). The first solution is always \(y_1(x)=x^{r_1}\sum a_nx^n\). The second solution \(y_2\) takes the form:

If \(N=0\) (\(r_1=r_2\)): \(\begin{aligned}y_2(x)=y_1(x)\ln{x}+x^{1+r_1}\sum_{n=0}^{\infty}b_nx^n\end{aligned}\)
If \(N>0\) (positive integer): \(\begin{aligned}y_2(x)=C_Ny_1(x)\ln{x}+x^{r_2}\sum_{n=0}^\infty b_nx^n\end{aligned}\)

6. 微分方程组

6.1 高阶转一阶

For \(x^{(n)}=f(t,x,x',\cdots,x^{(n-1)})\), introduce variables \(x_1=x, x_2=x', \cdots, x_n=x^{(n-1)}\). The substitution yields the system: \(x_1' = x_2, \cdots, x_{n-1}' = x_n, x_n'=f(t,x_1,x_2,\cdots,x_n)\).

6.2 方程组与消元法

Any system of two linear differential equations with constant coefficients \(L_1x+L_2y=f_1(t), L_3x+L_4y=f_2(t)\) can be uncoupled by defining the operational determinant \(\Delta = L_1L_4-L_2L_3\):

\[\begin{vmatrix}L_1&L_2\\L_3&L_4\end{vmatrix}x=\begin{vmatrix}f_1(t)&L_2\\f_2(t)&L_4\end{vmatrix},\quad\begin{vmatrix}L_1&L_2\\L_3&L_4\end{vmatrix}y=\begin{vmatrix}L_1&f_1(t)\\L_3&f_2(t)\end{vmatrix}\]

If \(x_1,x_2,\cdots x_n\) are \(n\) solutions of \(x'=P(t)x\), their Wronskian \(W \neq 0\) globally if they are linearly independent, or \(W = 0\) globally if dependent.

6.3 特征值法

Write the system as \(x'=Ax\). Substitute the trial solution \(x=ve^{\lambda t} \Rightarrow Av=\lambda v\). \(\lambda\) is an eigenvalue of \(A\) if \(|A-\lambda I|=0\), and \(v\) is the corresponding eigenvector.

6.4 特征向量链

If there’s a defective eigenvalue, find a vector \(u_{k}\) such that \((A-\lambda I)^{k}u_{k}=0\) but \((A-\lambda I)^{k-1}u_{k} \neq 0\). Multiply successively by \((A-\lambda I)\) to build a chain of generalized eigenvectors \(\{v_1,v_2,\cdots,v_k\} = \{u_k, \cdots, u_1\}\).

6.5 基础解矩阵

Let \(\Phi(t)\) be a fundamental matrix for \(x'=Ax\).

The unique solution of \(x'=Ax, x(0)=x_0\) is \(x(t)=\Phi(t)\Phi(0)^{-1}x_0=e^{At}x_0\).
For nonhomogeneous system \(x'=Ax+f(t)\): \[\begin{aligned}x(t)=\Phi(t)\Phi(a)^{-1}x_a+\Phi(t)\int_a^t\Phi(s)^{-1}f(s)ds\end{aligned}\]

7. 边值问题

For the boundary value problem \(y''+p(x)y'+q(x)y=0\) subject to \(y(a)=0, y(b)=0\), the goal is to find solutions defined on \((a,b)\) satisfying conditions at endpoints. When introducing a parameter \(\lambda\): \(y''+p(x)y'+\lambda q(x)y=0\), it is called an eigenvalue problem. Values of \(\lambda_*\) that yield non-trivial solutions are called eigenvalues, and the corresponding non-trivial solutions \(y_*\) are eigenfunctions.