常微分方程

nth-order Differential Equation

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

Integrals as General and Particular Solutions

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 $y$ , $\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=f(x)\end{aligned}$ , in this special case only need to integrate both sides $\begin{aligned}y(x)=\int f(x)\mathrm{d}x+C\end{aligned}$

Existence and Uniqueness of Solutions

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.

Separable Equations

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 $y$ : $\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=g(x)h(y)=\frac{g(x)}{f(y)}\end{aligned}$

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

Singular Solutions

It is common for a nonlinear first-order differential equation to have both a general solution involving a n 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.

Linear first-order Equation

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:
$\begin{aligned}&\mathrm{e}^{\int P(x)\mathrm{d}x}\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)\mathrm{e}^{\int P(x)\mathrm{d}x}y=Q(x)\mathrm{e}^{\int P(x)\mathrm{d}x}\\&y(x)=\mathrm{e}^{-\int P(x)\mathrm{d}x}\begin{bmatrix}\begin{aligned}\int(Q(x)\mathrm{e}^{\int P(x)\mathrm{d}x})\end{aligned}\mathrm{d}x+C\end{bmatrix}\end{aligned}$
1. Begin by calculating the integrating factor $\rho(x)=e^{\int P(x)\mathrm{d}x}$
2. Then multiply both sides of the differential equation by $\rho(x)$
3. Next , recognize the left-hand side of the resulting equation as the derivative of a product :
$D_x[\rho(x)y(x)]=\rho(x)Q(x)$
4. Finally , integrate this equation
$\begin{aligned}\rho(x)y(x)=\int\rho((x)Q(x)\mathrm{d}x+C\end{aligned}$
Properties:
1. guarantee only a solution on a possibly smaller interval.
2. linear first-order differential equation has no singular solutions.

Exact Differential Equation

A general solution $y(x)$ of a first-order differential equation is often defined implicitly by an equation of the form $F(x,y(x))=C$ ,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 the equation $F(x,y)=C$ implicitly defines a general solution of $M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0$ ,then $M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0$ is called an exact differential equation.

If $M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0$ is exact and $M$ and $N$ have continuous partial derivatives, it then follows that $\begin{aligned}\frac{\partial M}{\partial y}=F_{xy}=F_{yx}=\frac{\partial N}{\partial x}\end{aligned}$ .
Thus the equation $\begin{aligned}\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\end{aligned}$ is a necessary condition that the differential equation $M\mathrm{d}x+N\mathrm{d}y=0$ to be exact.

whether a given differential equation is exact or not is related to the precise form $M\mathrm{d}x+N\mathrm{d}y=0$ in which it is written.
$\begin{aligned}F(x,y)=\int M(x,y)\mathrm{d}x+g(y)\end{aligned}$
$\begin{aligned}N=\frac{\partial F}{\partial y}=\begin{pmatrix}\begin{aligned}\frac{\partial}{\partial y}\int M(x,y)\mathrm{d}x\end{aligned}\end{pmatrix}+g’(y)\end{aligned}$
$\begin{aligned}g’(x)=N-\frac{\partial}{\partial y}\int M(x,y)\mathrm{d}x\end{aligned}$
$\begin{aligned}F(x,y)=\int M(x,y)\mathrm{d}x+\int\begin{pmatrix}\begin{aligned}N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\mathrm{d}x\end{aligned}\end{pmatrix}\mathrm{d}y\end{aligned}$

Integrating Factor

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

• Assume the integrating factor $\mu$ depends only on $x$ , $\mu=\mu(x)$ , Then:
  $\begin{aligned}\frac{\partial(\mu M)}{\partial y}=\frac{\partial(\mu N)}{\partial x}\end{aligned}$
  $\begin{aligned}\Rightarrow\mu\frac{\partial M}{\partial y}=\frac{\partial\mu}{\partial x}N+\mu\frac{\partial N}{\partial x}\end{aligned}$
  $\begin{aligned}\Rightarrow\mu(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})=\frac{\partial\mu}{\partial x}N\end{aligned}$
  $\begin{aligned}\Rightarrow\frac{\mathrm{d}\mu}{\mu}=\frac{1}{N}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})\mathrm{d}x\end{aligned}$
  $\begin{aligned}\Rightarrow\ln|\mu|=\int\frac{1}{N}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})\mathrm{d}x+C\end{aligned}$
  $\Rightarrow\begin{aligned}\mu=\exp[\int\frac{1}{N}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})\mathrm{d}x+C]\end{aligned}$
• Assume the integrating factor $\mu$ depends only on $y$ , $\mu=\mu(y)$ , Then:
  $\begin{aligned}\frac{\partial(\mu M)}{\partial y}=\frac{\partial(\mu N)}{\partial x}\end{aligned}$
  $\Rightarrow\begin{aligned}\frac{\partial\mu}{\partial y}M+\mu\frac{\partial M}{\partial y}=\mu\frac{\partial N}{\partial x}\end{aligned}$
  $\Rightarrow\begin{aligned}\mu(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})=\frac{\partial\mu}{\partial y}M\end{aligned}$
  $\begin{aligned}\Rightarrow\frac{\mathrm{d}\mu}{\mu}=\frac{1}{M}(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\mathrm{d}y\end{aligned}$
  $\begin{aligned}\Rightarrow\ln|\mu|=\int\frac{1}{M}(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\mathrm{d}y+C\end{aligned}$
  $\Rightarrow\begin{aligned}\mu=\exp[\int\frac{1}{M}(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\mathrm{d}y+C]\end{aligned}$

Substitution Methods

e.g.1:
$\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=f(x,y)\end{aligned}$
Solution:
Substitute $v=\alpha(x,y),\quad y=\beta(x,v)$
$\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\partial\beta}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}x}+\frac{\partial\beta}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}x}=\beta_x+\beta_v\frac{\mathrm{d}v}{\mathrm{d}x}\end{aligned}$
$\begin{aligned}\frac{\mathrm{d}v}{\mathrm{d}x}=g(x,v)\end{aligned}$ with new dependent variable v.
Then solve this either separable or linear new equation.

e.g.2: Bernoulli Equation
$\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)y^n,\quad n\neq0,1\end{aligned}$
Solution:
Substitute $v=y^{1-n}$ ，$\begin{aligned}\frac{\mathrm{d}v}{\mathrm{d}x}=(1-n)y^{-n}\frac{\mathrm{d}y}{\mathrm{d}x}\end{aligned}$
$\begin{aligned}y^{-n}\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y^{1-n}=Q(x)\end{aligned}\quad\Rightarrow\quad\begin{aligned}\frac{1}{1-n}\frac{\mathrm{d}v}{\mathrm{d}x}+P(x)v=Q(x)\end{aligned}$
Then solve this Linear first-order equation.

e.g.3: Homogeneous Equation
$\begin{aligned}\frac{\mathrm{d}y}{\mathrm{d}x}=F(\frac{y}{x})\end{aligned}$
Solution:
Substitute $\begin{aligned}v=\frac{y}{x},\quad y=vx,\quad\frac{\mathrm{d}y}{\mathrm{d}x}=v+x\frac{\mathrm{d}v}{\mathrm{d}x}\end{aligned}$ Then $\begin{aligned}x\frac{\mathrm{d}v}{\mathrm{d}x}=F(v)-v\end{aligned}$
Transformed into the separable equation.

e.g.4: Reducible 2nd-Order Equation
$\begin{aligned}\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=f(x,y,\frac{\mathrm{d}y}{\mathrm{d}x}),\quad F(x,y,y’,y’’)=0\end{aligned}$
Solution:
1. If $y$ is missing, then $F(x,y’,y’’)=0$ , substitute $\begin{aligned}p=y’=\frac{\mathrm{d}y}{\mathrm{d}x},\quad y’’=\frac{\mathrm{d}p}{\mathrm{d}x}\end{aligned}$ ,results in the first-order equation $F(x,p,p’)=0$ , if we can solve this equation for a general solution $p(x,C_1)$ involving an arbitrary constant $C_1$ , then $\begin{aligned}y(x)=\int y’(x)\mathrm{d}x=\int p(x,C_1)\mathrm{d}x+C_2\end{aligned}$
2. If $x$ is missing, then $F(y,y’,y’’)=0$ , substitute $\begin{aligned}p=y’=\frac{\mathrm{d}y}{\mathrm{d}x},\quad y’’=\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{\mathrm{d}p}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}=p\frac{\mathrm{d}p}{\mathrm{d}y}\end{aligned}$ , results in the first order equation $\begin{aligned}F(y,p,p\frac{\mathrm{d}p}{\mathrm{d}y})=0\end{aligned}$ , if we can solve this equation for a general solution $p(y,C_1)$ involving an arbitrary constant $C_1$ , then $\begin{aligned}x(y)=\int\frac{\mathrm{d}x}{\mathrm{d}y}\mathrm{d}y=\int\frac{1}{\mathrm{d}y/\mathrm{d}x}\mathrm{d}y=\int\frac{1}{p}\mathrm{d}y=\int\frac{\mathrm{d}y}{p(y,C_1)}+C_2\end{aligned}$

Singular solutions

There are special solutions thar cannot be obtained from the general solution $y=\varphi(x;C)$ for some value of $C$ ,but are always tangent to one of the $y=\varphi(x,C)$ solution at some point.
If there is a singular solution, at each point on the curve has more than 1 solution.
if $y=\varphi(x)$ is a singular solution $\Rightarrow$ It’s a p-discriminant

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

Principle of the p-Discriminant Method

Consider a first-order OED 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 $\begin{aligned}p:\frac{\partial F}{\partial p}=0\end{aligned}$
  By eliminating $p$ from these equations, we obtain the p-discriminant curve, which may represent the singular solution.

Proving the Correctness of the p-Discriminant Method

Singular Solution: A solution of a differential equation that is not tangent to any solution in the family of general solutions and cannot be obtained by choosing specific values for the constants in the general solution.
Envelop: 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 general solution family.
p-Discriminant Method: By solving the system $F(x,y,p)=0$ and $\begin{aligned}\frac{\partial F}{\partial p}=0\end{aligned}$ , the curve obtained after eliminating $p$ may represent a singular solution.

Assume the family of general solutions is $\varphi(x,y,C)=0$ where $C$ is a constant.
The envelope must satisfy: $\varphi(x,y,C)=0\quad\begin{aligned}\frac{\partial\varphi}{\partial C}=0\end{aligned}$
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 $C$ vanishes, leading to $\begin{aligned}\frac{\partial F}{\partial p}=0\end{aligned}$

Homogeneous Second-Order Linear Equations

Let $y_1$ and $y_2$ be two solutions of the homogeneous linear equation in $y’’+p(x)y’+q(x)y=0$ on the interval $I$.If $c_1$ and $c_2$ are constants, then the linear combination $y=c_1y_1+c_2y_2$ is also a solution on $I$.

The second-order equation $y’’+p(x)y’+q(x)y=f(x)$ has infinitely many solution curves passing through the point $(a,b_0)$ , namely one for each (real number) value of the initial slope $y’(a)=b_1$ . That is, instead of there being only one line through $(a,b_0)$ tangent to a solution curve, every nonvertical straight line through $(a,b_0)$ is tangent to some solution curve of $y’’+p(x)y’+q(x)y=f(x)$.

Linear Independence of Two Functions

Two functions defined on an open interval $I$ are said to be linearly independent on $I$ provided that neither is a constant multiple of the other.
Given two functions $f$ and $g$ , the Wronskian of $f$ and $g$ is the determinant

$$W=\begin{vmatrix}f&g\\f'&g'\end{vmatrix}=fg'-f'g$$

Suppose that $y_1$ and $y_2$ are solutions of the homogeneous second-order linear equation $y’’+p(x)y’+q(x)y=0$ on an open interval $I$ on which $p$ and $q$ are continuous

• If $y_1$ and $y_2$ are linearly dependent, then $W(y_1,y_2)=0$ on $I$
• If $y_1$ and $y_2$ are linearly independent, then $W(y_1,y_2)\neq0$ at each point of $I$

Linear Second-Order Equations with Constant Coefficients

$ar^2+br+c=0$ this quadratic equation is called the characteristic equation of the homogeneous linear differential equation $ay’’+by’+cy=0$

• If the roots $r_1$ and $r_2$ of the characteristic equation are real and distinct, then $y(x)=c_1e^{r_1x}+c_2e^{r_2x}$ is a general solution.
• If the characteristic equation has equal roots $r_1=r_2$ , then $y(x)=(c_1+c_2x)e^{r_1x}$ is a general solution.

General Solutions of Linear Equations

Let $y_1,y_2,\cdots,y_n$ be $n$ solutions of the homogeneous linear equation in $y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_{n-1}(x)y’+p_n(x)y=0$ on the interval $I$ . If $c_1,c_2,\cdots,c_n$ are constants, then the linear combination $y=c_1y_2+c_2y_c+\cdots+c_ny_n$ is also a solution.

The $n$ functions $f_1,f_2,\cdots,f_n$ are said to be linearly dependent on the interval $I$ provided that there exist constants $c_1,c_2,\cdots,c_n$ not all zero such that $c_1f_1+c_2f_2+\cdots+c_nf_n=0$
Their Wronskian is the $n\times n$ determinant $W=\begin{vmatrix}f_1&f_2&\cdots&f_n\\f_1’&f_2’&\cdots&f_n’\\\vdots&\vdots&&\vdots\\f^{(n-1)}_1&f^{(n-1)}_2&\cdots&f^{(n-1)}_n\end{vmatrix}$
Suppose that $y_1,y_2,\cdots,y_n$ are $n$ solutions of the homogeneous nth-order linear equation $y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_{n-1}(x)y’+p_n(x)y=0$ on an open interval $I$ where each $p_i$ is continuous. Let $W=W(y_1,y_2,\cdots,y_n)$

• If $y_1,y_2,\cdots,y_n$ are linearly dependent, then $W=0$ on $I$
• If $y_1,y_2,\cdots,y_n$ are linearly independent, then $W\neq0$ at each point of $I$

Nonhomogeneous Equations

Let $y_p$ be a particular solution of the nonhomogeneous equation $y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_{n-1}(x)y’+p_n(x)y=f(x)$ on an open interval $I$ where the functions $p_i$ and $f$ are continuous. Let $y_1,y_2,\cdots,y_n$ be linearly independent solutions of the associated homogeneous equation $y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_{n-1}(x)y’+p_n(x)y=0$ . If $Y$ is any solution whatsoever of $y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_{n-1}(x)y’+p_n(x)y=f(x)$ on $I$ ,then there exist number $c_1,c_2,\cdots,c_n$ such that

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

for all $x$ in $I$

Polynomial Operators

$\begin{aligned}L&=a_n\frac{d^n}{dx^n}+a_{n-1}\frac{d^{n-1}}{dx^{n-1}}+\cdots+a_2\frac{d^2}{dx^2}+a_1\frac{d}{dx}+a_0\\&=a_nD^n+a_{n-1}D^{n-1}+\cdots+a_nD^2+a_1D+a_0\end{aligned}$

Homogeneous Equations with Constant Coefficients

$a_nr^n+a_{n-1}r^{n-1}+\cdots+a_2r^2+a_1r+a_0=0$ is called the characteristic equation or auxiliary equation of the differential equation of $a_ny^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_1y’+a_0y=0$

• If the roots $r_1,r_2,\cdots,r_n$ of the characteristic equation are real and distinct, then $y(x)=c_1e^{r_1x}+c_2e^{r_2x}+\cdots+c_ne^{r_nx}$ is a general solution.
• If the characteristic equation has repeated roots $r_1$ , the desired solution of $(D-r_1)^ky=0$ is $y(x)=ue^{r_1x}=(c_1+c_2x+c_3x^2+\cdots+c_kx^{k-1})e^{r_1x}$

Complex-Valued Functions and Euler’s Formula

$\begin{cases}e^{ix}=\cos{x}+i\sin{x}\\e^{(a+bi)x}=e^{ax}(\cos{bx}+i\sin{bx})\\e^{(a-bi)x}=e^{ax}(\cos{bx}-i\sin{bx})\end{cases}$

$\begin{aligned}y(x)&=c_1e^{r_1x}+c_2e^{r_2x}=c_1e^{(a+bi)x}+c_2e^{(a-bi)x}\\&=c_1e^{ax}(\cos{bx}+i\sin{bx})+c_2e^{ax}(\cos{bx}-i\sin{bx})\\&=(c_1+c_2)e^{ax}\cos{bx}+i(c_1-c_2)e^{ax}\sin{bx}\\&=e^{ax}(d_1\cos{bx}+d_2\sin{bx})\end{aligned}$

Reduction of Order

Suppose that one solution $y_1(x)$ of the homogeneous second-order linear differential equation $y’’+p(x)y’+q(x)y=0$ is known. The method of reduction of order consists of substituting $y_2(x)=v(x)y_1(x)$ and attempting to determine the function $v(x)$ so that $y_2(x)$ is a second linearly independent solution.

Method of Undetermined Coefficients

$f(x)$ can be written as a sum of terms each of the form $P_m(x)e^{rx}\cos{kx}$ or $P_m(x)e^{rx}\sin{kx}$ take the trail solution $y_p(x)=x^s[(A_0+A_1x+\cdots+A_mx^m)e^{rx}\cos{kx}+(B_0+B_1x+\cdots+B_mx^m)e^{rx}\sin{kx}]$
where $s$ is the smallest nonnegative integer such that no term in $y_p$ duplicates a term in the complementary function $y_c$

Variation of Parameters

If the nonhomogeneous equation $y’’+P(x)y’+Q(x)y=f(x)$ has complementary function $y_c(x)=c_1y_1(x)+c_2y_2(x)$ , then a particular solution is given by

$$\begin{cases}u_1'y_1+u_2'y_2=0\\u_1'y_1'+u_2'y_2'=f(x)\end{cases}$$ $$\begin{aligned}y_p(x)=-y_1(x)\int\frac{y_2(x)f(x)}{W(x)}dx+y_2\int\frac{y_1(x)f(x)}{W(x)}dx\end{aligned}$$

where $W=W(y_1,y_2)$ is the Wronskian of the two independent solutions $y_1$ and $y_2$ of the associated homogeneous equation.

Endpoint Problems and Eigenvalues

$$y''+p(x)y'+q(x)y=0,y(a)=0,y(b)=0$$

to find a solution of the differential equation on the interval $(a,b)$ that satisfies the condition $y(a)=0$ and $y(b)=0$ at the endpoints of the interval. Such a problem is called an endpoint or boundary value problem.

$$y''+p(x)y'+\lambda q(x)y=0,y(a)=0,y(b)=0$$

An endpoint problem containing a parameter $\lambda$ is called an eigenvalue problem. If there are values of the parameter $\lambda$ exist a nontrivial solution of the endpoint problem, such a value of $\lambda$ is called an eigenvalue of characteristic value of the problem.
Suppose that $\lambda_{\ast}$ is an eigenvalue of the problem and that $y_{\ast}$ is a nontrivial solution of the problem with this value of $\lambda$ inserted

$$y_*''+p(x)y_*'+\lambda_*q(x)y_*=0,y_*(a)=0,y_*(b)=0$$

Then we call $y_\ast$ an eigenfunction associated with the eigenvalue $\lambda_{\ast}$ .

Power Series

$\begin{aligned}\sum_{n=0}^{\infty}c_n(x-a)^n=c_0+c_1(x-a)+\cdots+c_n(x-a)^n\end{aligned}+\cdots$
$\begin{aligned}f(x)=\sum_{n=0}^\infty c_nx^n=c_0+c_1x+c_2x^2+\cdots\end{aligned}$
$\begin{aligned}f’(x)=\sum_{n=1}^\infty nc_nx^{n-1}=c_1+2c_2x+3c_3x^2+\cdots\end{aligned}$

Radius of Convergence

Given the power series $\sum c_nx^n$ ,suppose that the limit $\begin{aligned}\rho=\lim_{n\rightarrow\infty}|\frac{c_n}{c_{n+1}}|\end{aligned}$
exists ($\rho$ is finite) or infinite, Then

• if $\rho=0$ ,then the series diverges for all $x\neq 0$
• if $0<\rho<\infty$ ,then $\sum c_nx^n$ converges if $|x|<\rho$ and diverges if $|x|>\rho$
• if $\rho=\infty$ ,then the series converges for all $x$
  $A(x)y’’+B(x)y’+C(x)y=0$ the point $x=a$ is called an ordinary point of $y’’+P(x)y’+Q(x)y=0$ provided that the function $P(x)$ and $Q(x)$ are both analytic at $x=a$ ,Otherwise $x=a$ is a singular point

Solutions Near an Ordinary Point

Suppose thar $a$ is an ordinary point of the equation $A(x)y’’+B(x)y’+C(x)y=0$ that is the functions $P=B/A$ and $Q=C/A$ are analytic at $x=a$ ,then the equation has two linearly independent solutions , each of the form $\begin{aligned}y(x)=\sum_{n=0}^\infty c_n(x-a)^n\end{aligned}$
The radius of convergence of any such series solution is at least as large as the distance from $a$ to the nearest (real or complex) singular point of the equation, the coefficients in the series can be determined by its substitution.

Regular Singular Points

A differential equation having a singular point at 0 ordinarily will not have power series solutions of the form $y(x)=\sum c_nx^n$ . To investigate the form that a solution of such an equation might take, assume that analytic coefficient functions and rewrite it in the standard form $\begin{aligned}&y''+P(x)y'+Q(x)y=0\\&y''+\frac{p(x)}{x}y'+\frac{q(x)}{x^2}y=0,p(x)=xP(x),q(x)=x^2Q(x)\end{aligned}$
The singular point $x=0$ is a regular singular point if the functions $p(x)=xP(x)$ and $q(x)=x^2Q(x)$ are both analytic at $x=0$. Otherwise it is an irregular singular point.
$\begin{aligned}p_0=p(0)=\lim_{x\rightarrow0}p(x)=\lim_{x\rightarrow0}xP(x)\end{aligned}$
$\begin{aligned}q_0=q(0)=\lim_{x\rightarrow0}q(x)=\lim_{x\rightarrow0}x^2Q(x)\end{aligned}$

• if $p_0=q_0=0$ then $x=0$ may be an ordinary point
• if the limits exist and are finite, then $x=0$ is a regular singular point
• if either limit fails to exist or is infinite, then $x=0$ is an irregular singular point

The Method of Frobenius

In general case, $p(x)$ and $q(x)$ are power series rather than constants, it is a reasonable conjecture that out differential equation might have a solution of the form

$$\begin{aligned}y(x)=x^r\sum_{n=0}^\infty c_nx^n=\sum_{n=0}^\infty c_nx^{n+r}=c_0x^r+c_1x^{r+1}+\cdots\end{aligned}$$

such series is called a Frobenius series.
To investigate the possible existence of Frobenius series solutions, begin with the equation

$$x^2y''+xp(x)y'+q(x)y=0$$

If $x=0$ is a regular singular point, then $p(x)$ and $q(x)$ are analytic at $x=0$ ,so

$$\begin{align} p(x)=p_0+p_1x+p_2x^2+p_3x^3+\cdots \\ q(x)=q_0+q_1x+q_2x^2+q_3x^3+\cdots \end{align}$$

Suppose the equation has the Frobenius series solution

$$\begin{aligned}y=\sum_{n=0}^\infty c_nx^{n+r}\quad y'=\sum_{n=0}^{\infty}c_n(n+r)x^{n+r-1}\quad y''=\sum_{n=0}^\infty c_n(n+r)(n+r-1)x^{n+r-2}\end{aligned}$$

the coefficient $r(r-1)c_0+p_0rc_0+q_0c_0$ of $x^r$ must vanish, because $c_0\neq0$ ,it follows that $r$ must satisfy the quadratic equation $r(r-1)+p_0r+q_0=0$ , which is called the indicial equation, and its two roots are the exponents of the differential equation at $x=0$
If $r_1\neq r_2$ ,it follows that there are two possible Frobenius series solutions
If $r_1=r_2$ ,there is only one possible Frobenius series solution

Frobenius Series Solutions

Suppose that $x=0$ is a regular singular point of the equation $x^2y’’+xp(x)y’+q(x)y=0$
Let $\rho>0$ denote the minimum of the radii of convergence of the power series

$$\begin{aligned}p(x)=\sum_{n=0}^\infty p_nx^n\quad\mathrm{and}\quad q(x)=\sum_{n=0}^\infty q_nx^n\end{aligned}$$

Let $r_1$ and $r_2$ be the roots, with $r_1\geq r_2$

• For $x>0$ , there exists a solution of the form $\begin{aligned}y_1(x)=x^{r_1}\sum_{n=0}^\infty a_nx^n\quad(a_0\neq 0)\end{aligned}$
• If $r_1-r_2$ is neither zero nor a positive integer, there exists a second linearly independent solution for $x>0$ of the form $\begin{aligned}y_2(x)=x^{r_2}\sum_{n=0}^\infty b_nx^n\quad(b_0\neq 0)\end{aligned}$
• If $r_1-r_2$ is a positive integer there may or may not exist a second Frobenius series solution of the equation

The Nonlogarithmic Case with $r_1=r_2+N$

Derived the indicial equation by substituting the power series $p(x)=\sum p_nx^n$ and $\sum q_nx^n$ and the Frobenius series $\begin{aligned}y(x)=x\sum_{n=0}^\infty c_nx^{n+r},c_0\neq0\end{aligned}$
in the differential equation in the form $x^2y’’+xp(x)y’+q(x)y=0$
The result of this substitution, after collection of the coefficients of like powers of $x$ , is an equation of the form $\begin{aligned}\sum_{n=0}^\infty F_n(r)x^{n+r}=0\end{aligned}$
$F_0(r)=[r(r-1)+p_0r+q_0]c_0=\phi(r)c_0$
$F_n(r)=\phi(r+n)c_n+L_n(r:c_0,c_1,\cdots,c_{n-1})$
$\begin{aligned}L_n=\sum_{k=0}^{n-1}[(r+k)p_{n-k}+q_{n-k}]c_k\end{aligned}$
$\begin{aligned}y_2=y_1\int\frac{\exp(-\int P(x)dx)}{y_1^2}dx\end{aligned}$

• $N=0$ , $\begin{aligned}y_2(x)=y_1(x)\ln{x}+x^{1+r_1}\sum_{n=0}^{\infty}b_nx^n\end{aligned}$
• $N\neq0$ , $\begin{aligned}y_2(x)=C_Ny_1(x)\ln{x}+x^{r_2}\sum_{n=0}^\infty b_nx^n\end{aligned}$

First-Order Systems and Applications

Consider the system consisting of the single nth-order equation

$$x^{(n)}=f(t,x,x',\cdots,x^{(n-1)})$$

Introduce the dependent variables $x_1,x_2,\cdots,x_n$ defined as follows:

$$x_1=x,\quad x_2=x',\quad x_3=x'',\quad\cdots,\quad x_n=x^{(n-1)}$$

The substitution yields the system

$$x_n'=f(t,x_1,x_2,\cdots,x_n)$$

Linear systems

Consider a linear first-order system of the form

$$\begin{align} x_1'&=p_{11}(t)x_1+p_{12}(t)x_2+\cdots+p_{1n}(t)x_n+f_1(t)\\ x_2'&=p_{21}(t)x_1+p_{22}(t)x_2+\cdots+p_{2n}(t)x_n+f_2(t)\\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\vdots\\ x_n'&=p_{n1}(t)x_1+p_{n2}(t)x_2+\cdots+p_{nn}(t)x_n+f_n(t) \end{align}$$

We say that this system is homogeneous if the functions $f_1,f_2,\cdots,f_n$ are all identically zero, Otherwise, it is nonhomogeneous

Any system of two linear differential equations with constant coefficients can be written in the form $L_1x+L_2y=f_1(t),L_3x+L_4y=f_2(t)$
$\begin{cases}L_1x+L_2y=f_1(t)\Rightarrow L_3L_1x+L_3L_2y=L_3f_1(t)\\L_3x+L_4y=f_2(t)\Rightarrow L_1L_3x+L_1L_4y=L_1f_2(t)\end{cases}\Rightarrow(L_1L_4-L_2L_3)y=L_1f_2(t)-L_3f_1(t)$
define the operational determinant

$$\begin{vmatrix}L_1&L_2\\L_3&L_4\end{vmatrix}=L_1L_4-L_2L_3$$

Then the equation related to $x$ and $y$ can be rewritten as

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

Similarly, for the system of three linear equations, $x$ satisfies the single linear equation

$$\begin{vmatrix}L_{11}&L_{12}&L_{13}\\L_{21}&L_{22}&L_{23}\\L_{31}&L_{32}&L_{33}\end{vmatrix}x=\begin{vmatrix}f_1(t)&L_{12}&L_{13}\\f_2(t)&L_{22}&L_{23}\\f_3(t)&L_{32}&L_{33}\end{vmatrix}$$

Suppose that $x_1,x_2,\cdots x_n$ are $n$ solutions of the homogeneous linear equation $x’=P(t)x$ on an open interval $I$ ,suppose that $P(t)$ is continuous on $I$ ,let $W=W(x_1,x_2,\cdots,x_n)$

• If $x_1,x_2,\cdots,x_n$ are linearly dependent on $I$ ,then $W=0$ at every point of $I$
• If $x_1,x_2,\cdots,x_n$ are linearly independent on $I$ ,then $W\neq0$ at every point of $I$

The Eigenvalue Method for Homogeneous Systems

Write the system in the matrix form $x’=Ax$ where $A=[a_{ij}]$ , When substitute the trial solution $x=ve^{\lambda t}$ with derivative $x’=\lambda ve^{\lambda t}$ , the result is $\lambda ve^{\lambda t}=Ave^{\lambda t}\Rightarrow Av=\lambda v$
The number $\lambda$ is called an eigenvalue of the $n\times n$ matrix $A$ provided that $|A-\lambda I|=0$
The vector $v$ is called an eigenvector of $\lambda$

Chains of Generalized Eigenvalues

Begin with a nonzero solution of $(A-\lambda I)^{d+1}u=0$ and successively multiply by the matrix $A-\lambda I$ until the zero vector is obtained, If

$$\begin{aligned}&(A-\lambda I)u_1=u_2\neq 0\\&\quad\quad\quad\quad\vdots\\&(A-\lambda I)u_{k-1}=u_k\neq0\end{aligned}$$

but $(A-\lambda I)u_k=0$ ,then the vectors $\{v_1,v_2,\cdots,v_k\}=\{u_k,u_{k_1}\cdots,u_2,u_1\}$ form a length $k$ chain of generalized eigenvectors based on the eigenvector $v_1$

Fundamental Matrix Solutions

Let $\Phi(t)$ be a fundamental matrix for the homogeneous linear system $x’=Ax$
Then the unique solution of the initial value problem $x’=Ax,x(0)=x_0$
is given by $x(t)=\Phi(t)\Phi(0)^{-1}x_0=e^{At}x_0$

$$\begin{aligned}x(t)=e^{At}x_0+e^{At}\int_0^te^{-As}f(s)ds\end{aligned}$$ $$\begin{aligned}x(t)=\Phi(t)\Phi(a)^{-1}x_a+\Phi(t)\int_a^t\Phi(s)^{-1}f(s)ds\end{aligned}$$

Some words