Ordinary Differential Equation Note

Ordinary Differential Equation

Definition

An equation that contains one or several derivatives of unknown functions

Types

• Linear ODE
  • Homogeneous Linear ODE
  • Non-Homegeneous Linear ODE
• Non-linear ODE

Facts

Initial Value Problem

Definition

ODE $F(x,y,y^{\prime })=0$ or $y^{\prime }=G(x,y)$ with an initial condition $y(x_0)=y_0$

Facts

The general solution of an ODE is a family of infinitely many solution curves

The solution curve of an IVP sould pass through the given point $(x_0,y_0)$

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Transclude of First-Order-ODE

Superposition Principle

Definition

If $y_1,y_2$ are solutions of homogeneous linear ODE $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=r(x)$ on $I=(\alpha ,\beta )$ then, any linear combination $c_1{y}_1+c_2{y}_2(c_1,c_2\in \mathbb{R})$ is also a solution (on $I$)

Facts

Suppose that each of infinitely many functions $u_1,u_2,\dots$ satisfies a linear homogeneous differential equation or boundary condition $L(u)=0$

Then, the infinite series $u={\sum }_{n=1}^{\infty }{c}_n{u}_n$ where $c_n\in \mathbb{R}\text{or}\mathbb{C}$ that satisfies following conditions satisfies $L(u)=0$

• the infinite series converges and required differentiability in $L(u)=0$
• the required boundary condition is satisfied

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Transclude of Reduction-of-Order

Linear Independence

Definition

Let $y_1,y_2,\dots ,y_n$ be $n$ vectors or functions

Linearly independent

If $\sum _{i=1}^n{k}_i{y}_i=0\Rightarrow k_1=k_2=\cdots =k_n=0$,
then $\{y_i|i=1,2,\dots ,n\}$ is linearly independent.

$y_n$ can’t be expressed using the others

Linearly dependent

If $\exists k_1,k_2,\dots ,k_n$, not all zero, s.t. $\sum _{i=1}^n{k}_i{y}_i=0$,
then $\{y_i|i=1,2,\dots ,n\}$ is linearly dependent.

Facts

Basis(solutions of the ODE) = linearly independent solutions

Mutually orthogonal $y_i\perp y_j(i\ne j)$ vectors $y_1,y_2,\dots ,y_n$ are linearly independent

Check of linear independence of a set of vectors

1. make a matrix using the given column vectors
2. apply Gauss Elimination to make an upper triangular or Echelon matrix
3. check the number of non-zero pivots

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Higher-order Linear ODE

Definition

Second-Order Liner ODE

$y^{\prime \prime }+p(x)y^{\prime }+q(x)y=r(x)$

Higher-Order Liner ODE

$L(y)=\underset{=\sum _{k=0}^n{p}_k(x)y^{(k)},p_n(x)=1}{\underbrace{y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots +p_1(x)y^{\prime }+p_0(x)y}}=r(x)$

Solution

Second-Order Homogeneous Linear ODE with Constant Coefficients

Definition

$y^{\prime \prime }=a{y}^{\prime }+by=0,a,b\in \mathbb{R}$

Solution

Assume the solutions will be $y=e^{\lambda x}$ form then ${\lambda }^2{e}^{\lambda x}+a\lambda e^{\lambda X}+b{e}^{\lambda x}=({\lambda }^2+a\lambda +b)\underset{>0}{\underbrace{e^{\lambda x}}}=0$ $\Rightarrow {\lambda }^2+a\lambda +b=0$^[characteristic equation]

${\lambda }_{1,2}=\frac{-a\pm \sqrt{a^2-4b}}{2}\Rightarrow y_1=e^{{\lambda }_{1}x}$, $y_2=e^{{\lambda }_{2}x}$ are solutions

Two Distinct Real Roots ${\lambda }_1\ne {\lambda }_2$

General Solution

$y=c_1{e}^{{\lambda }_{1}x}+c_2{e}^{{\lambda }_{2}x},c_1,c_2\in \mathbb{R}$

Multiple Real Roots $\lambda =-\frac{a}{2}$

$y_1=e^{\lambda x}=e^{-\frac{a}{2}x}$

by using the Reduction of Order, we can get $y_2$.

$U=\frac{1}{y_1^2}{e}^{-\int pdx}=\frac{1}{e^{-ax}}{e}^{-ax}=1$ $\Rightarrow y_2=u{y}_1=y_1\int Udx=e^{-\frac{a}{2}x}x=x{e}^{\lambda x}$

General Solution

$y=c_1{e}^{\lambda x}+c_{2}x{e}^{\lambda x}=(c_1+c_{2}x)e^{\lambda x},c_1,c_2\in \mathbb{R}$

Complex Roots ${\lambda }_1=-\frac{a}{2}+i\omega ,{\lambda }_2=-\frac{a}{2}-i\omega$

$\begin{array}{rl}\Rightarrow & {\hat{y}}_1=e^{-\frac{a}{2}x}{e}^{+i\omega x}=e^{-\frac{a}{2}x}(\cos \omega x+i\sin \omega x)\\ & {\hat{y}}_2=e^{-\frac{a}{2}x}{e}^{-i\omega x}=e^{-\frac{a}{2}x}(\cos \omega x-i\sin \omega x)\\ \Rightarrow & y_1=\frac{{\hat{y}}_1+{\hat{y}}_2}{2}=e^{-\frac{a}{2}x}\cos \omega x\\ & y_2=\frac{{\hat{y}}_1-{\hat{y}}_2}{2i}=e^{-\frac{a}{2}x}\sin \omega x\end{array}$

General Solution

$y=A{e}^{-\frac{a}{2}x}\cos \omega x+B{e}^{-\frac{a}{2}x}\sin \omega x,A,B\in \mathbb{R}$

Examples

$$\begin{array}{rl}& y^{\prime \prime }+5y^{\prime }+6y=0\\ & \text{assume}y=e^{\lambda x}\\ & {\lambda }^2+5\lambda +6=0\Rightarrow (\lambda +3)(\lambda +2)=0\Rightarrow \lambda =-3\text{or}-2\\ & y=c_1{e}^{-3x}+c_2{e}^{-2x}\end{array}$$Link to original

Transclude of Higher-Order-Homogeneous-Linear-ODE-with-Constant-Coefficients

Euler-Cauchy Equations

Definition

$x^2{y}^{\prime \prime }+ax{y}^{\prime }+by=0(x\ne 0)$

Solution

Assume the solutions $y=x^m$ form
then $x^{2}m(m-1)x^{m-2}+axm{x}^{m-1}+b{x}^m=m(m-1)+am+b=0$

$\Rightarrow m^2+(a-1)m+b=0$^[auxiliary equation]

$m=\frac{(1-a)\pm \sqrt{(1-a{)}^2-4b}}{2}=\frac{1-a}{2}\pm \sqrt{\frac{(1-a{)}^2}{4}-b}$

Two Distinct Real Roots $m_1,m_2$

$y_1=x^{m_1},y_2=x^{m_2}$

General Solution

$y=c_1{x}^{m_1}+c_2{x}^{m_2},c_1,c_2\in \mathbb{R}$

Multiple Real Roots $m=\frac{1-a}{2}$

$y_1=x^m=x^{\frac{1-a}{2}}$

By using the Reduction of Order, we can get $y_2$.

$U=\frac{1}{y_1^2}{e}^{-\int pdx}=\frac{1}{y_1^2}{e}^{-\int \frac{a}{x}dx}=\frac{1}{x^{1-a}}{e}^{-a\ln x}=\frac{1}{x^{1-a}}{x}^{-a}=\frac{1}{x},p=\frac{ax}{x^2}=\frac{a}{x}$ $\Rightarrow y_2=u{y}_1=y_1\int Udx=x^m\ln x(x>0)$

General Solution

$y=c_1{x}^m+c_2{x}^m\ln x,c_1,c_2\in \mathbb{R}$

Complex Roots $m=\mu \pm i\nu ,\mu ,\nu \in \mathbb{R}$

$\Rightarrow {\hat{y}}_1=x^{\mu +i\nu }=x^{\mu }{x}^{i\nu }=x^{\mu }(e^{\ln x}{)}^{i\nu }=x^{\mu }{e}^{i\nu \ln x}=x^{\mu }(\cos (\nu \ln x)+i\sin (\nu \ln x))$ ${\hat{y}}_2=x^{\mu -i\nu }=x^{\mu }{x}^{-i\nu }=x^{\mu }(e^{\ln x}{)}^{-i\nu }=x^{\mu }{e}^{-i\nu \ln x}=x^{\mu }(\cos (\nu \ln x)-i\sin (\nu \ln x))$ $\Rightarrow y_1=\frac{{\hat{y}}_1+{\hat{y}}_2}{2}=x^{\mu }\cos (\nu \ln x)$ $y_2=\frac{{\hat{y}}_1-{\hat{y}}_2}{2i}=x^{\mu }\sin (\nu \ln x)$

General Solution

$y=A{x}^{\mu }\cos (\nu \ln x)+B{x}^{\mu }\sin (\nu \ln x),A,B\in \mathbb{R}$

Examples

$$\begin{array}{rl}& x^2{y}^{\prime \prime }-3x{y}^{\prime }-12y=0\\ & \text{substitute}{x}^m:=y\\ & x^2(x^m{)}^{\prime \prime }-3x(x^m{)}^{\prime }-12(x^m)=0\\ & m(m-1)x^m-3m{x}^m-12x^m=0\\ & m^2-4m-12=0\Rightarrow (m-6)(m+2)=0\Rightarrow m=6\text{or}-2\\ & y=C_1{x}^6+C_2{x}^{-2}\end{array}$$Link to original

Higher-Order Non-Homegeneous Linear ODE

Definition

Second-order

$y^{\prime \prime }+p(x)y^{\prime }+q(x)y=r(x)\ne 0⑴$^[Non-homogeneous ODE]

$y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0⑵$^[Corresponding homogeneous ODE]

Higher-order

$L(y)=y^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_0{y}^{(0)}=r(x)\ne 0$^[Non-homogeneous ODE]

$y^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_0{y}^{(0)}=0$^[Corresponding homogeneous ODE]

Facts

Thm 1

The sum of solution $y$ of ⑴ on some open interval $I$ and a solution $\tilde{y}$ of ⑵ on $I$ $y+\tilde{y}$ is a solution of ⑴ on $I$

The difference of two solutions of ⑴ on $I$ is a solution of ⑵ on $I$

Thm 2

If $p(x),q(x),r(x)$ in Non-homogeneous ODE ⑴ are continuous on $I$, then every solution of Non-homogeneous ODE ⑴ on $I$ can be obtained from ⑶

Solution

General Solutions of Non-homogeneous Linear ODE

$y(x)=y_h(x)+y_p(x)⑶$, where $y_h(x)$ is a general solution of the corresponding homogeneous ODE ⑵ $y_p(x)$ is any solution (particular solution) of the Non-homogeneous ODE ⑴

By Thm 1 $y(x)$ is a solution of Non-homogeneous ODE ⑴

Variation of parameters

Second-order

for $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=r(x)\Rightarrow y=y_h+y_p$

If $p,q,r$ are continuous on $I$ and if $y_1,y_2$ form a basis of solutions of the corresponding homogeneous ODE, then $y_p(x)=-y_1\int \frac{y_{2}r}{W}dx+y_2\int \frac{y_{1}r}{W}dx$, $W$ is Wronskian Determinant

Higher-order

If $y_1,\dots ,y_n$ are linearly independent solutions of the corresponding homogeneous equation of $\sum _{k=0}^n{p}_k(x)y^{(k)}=r(x),p_n(x)=1$

Then, $y_p$ becomes $y_p(x)=\sum _{k=1}^n{y}_k(x)\int \frac{W_k(x)}{W(x)}r(x)dx$, where Wronskian Determinant $W(x)=\left|\begin{array}{ccc}{y}_1& \dots & y_n\\ y_1^{\prime }& \dots & y_n^{\prime }\\ ⋮& \ddots & ⋮\\ y_1^{(n-1)}& \dots & y_n^{(n-1)}\end{array}\right|$, $W_k(x)=\left|\begin{array}{ccccc}{y}_1& \dots & 0& \dots & y_n\\ y_1^{\prime }& \dots & 0& \dots & y_n^{\prime }\\ ⋮& \ddots & ⋮& \ddots & ⋮\\ y_1^{(n-1)}& \dots & \underset{\text{k-th column}}{\underbrace{1}}& \dots & y_n^{(n-1)}\end{array}\right|$

Method of Undetermined Coefficients

Definition

To solve $y^{\prime \prime }+a{y}^{\prime }+by=r(x)$, we have to solve corresponding homogeneous ODE $y^{\prime \prime }+a{y}^{\prime }+by=0$ and find a particular solution $y_p$

Table

Terms in $r(x)$	Choice for $y_p(x)$
$k{e}^{\gamma x}$	$C{e}^{\gamma x}$
k x^n,\quad _{n=0, 1, 2}	$K_n{x}^n+K_{n-1}{x}^{n-1}+\cdots +K_{1}x+K_0$
$k\cos \omega x$ $k\sin \omega x$	$K\cos \omega x+M\sin \omega x$
$k{e}^{\alpha x}\cos \omega x$ $k{e}^{\alpha x}\sin \omega x$	$e^{\alpha x}(K\cos \omega x+M\sin \omega x)$

Choice Rules

1. Basic Rule: If $r(x)$ belongs a column in the left, choose corresponding right $y_p$
2. Modification Rule: If $y_p$ is a solution of a corresponding homogeneous ODE, multiply the selected $y_p$ by $x$ or $x^2$
3. Sum Rule: If $r(x)$ is a sum of several functions, $y_p$ is selected as the sum of functions corresponding to each.

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Examples

&y'' + 4y = x^{2}+ 3e^{x}\\ &\text{find the solution of the corresponding homogene ODE:}\\ &y'' + 4y = 0\\ &\text{by the solution of separable ODE}\\ &y_{h} = C_{1}\cos(2x) + C_{2}\sin(2x)\\ \\ &\text{by method of undetermined coefficients, assume}\\ &y_{p} = ax^{2}+bx+c+de^{x}\\ &y_{p}'' + 4y_{p} = (2a+de^{x}) + 4(ax^{2}+bx+c+de^{x}) = 5de^{x}+4ax^{2}+4bx+(2a+4c) = x^{2}+ 3e^{x}\\ &\Rightarrow a=\frac{1}{4}, b=0, c=-\frac{1}{8},d=\frac{3}{5}\\ &y_{p}= \frac{1}{4}x^{2} + \frac{3}{5}e^{x} -\frac{1}{8}\\ \\ &y = y_{h}+y_{p} = C_{1}\cos(2x) + C_{2}\sin(2x) +\frac{1}{4}x^{2} + \frac{3}{5}e^{x} -\frac{1}{8} \end{aligned}$$ $$\begin{aligned} &y''+4y=3\sin(2x)\\ &\text{find the solution of the corresponding homogene ODE:}\\ &y''+4y=0\\ &\text{by the solution of separable ODE}\\ &y_{h} = C_{1}\cos(2x)+C_{2}\sin(2x)\\ \\ &\text{by method of undetermined coefficients (modification rule), assume}\\ &y_{p}=ax\cos(2x)+bx\sin(2x)\\ &y_{p}''=-4a\sin(2x)+4b\cos(2x)-4ax\cos(2x)-4bx\sin(2x)\\ &y_{p}''+4y_{p}=-4a\sin(2x)+4b\cos(2x)=3\sin(2x)\\ &\Rightarrow a=-\frac{3}{4},b=0\\ &y_{p}= -\frac{3}{4}x\cos(2x)\\ \\ &y = y_{h}+y_{p} = C_{1}\cos(2x)+C_{2}\sin(2x)-\frac{3}{4}x\cos(2x) \end{aligned}$$

\begin{aligned} &y”-2y’+y=xe^{x}+4\ &\text{find the solution of the corresponding homogene ODE:}\ &y”-2y’+y=0\ &\text{by the solution of separable ODE}\ &y_{h}=C_{1}e^{x}+C_{2}xe^{x}\ \ &\text{by method of undetermined coefficients (modification rule), assume}\ &y_{p}=ax^{3}e^{x}+b\ &y_{p}‘=3ax^{2}e^{x}+ax^{3}e^{x}\ &y_{p}”=6axe^{x} + 6ax^{2}e^{x} + ax^{3}e^{x}\ &y_{p}”-2y_{p}‘+y_{p} = 6axe^{x} + b = xe^{x}+4\ &\Rightarrow a=\frac{1}{6}, b=4\ &y_{p} = \frac{1}{6}x^{3}e^{x}+4\ \ &y=y_{h}+y_{p} = C_{1}e^{x}+C_{2}xe^{x} + \frac{1}{6}x^{3}e^{x}+4 \end{aligned}

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Existence and Uniqueness Theorem

Definition

First-order

Let $f:(x,y)\to \mathbb{R}$ be a continuous as a function of $x,y$ and lipschitz continuous in $y$. Then, $y^{\prime }=f(x,y),y(x_0)=y_0$ has a unique solution $y$

Thm 2.4.1

$p(t),g(t)$ are continuous on an open interval $I=(\alpha ,\beta ),t_0\in I,y_0\in \mathbb{R}$

$\Rightarrow$ There exist a unique function $y=\varphi (t)$ satisfying a linear ODE $y^{\prime }=p(t)y=g(t)$
for each $t\in I$ and satisfying the initial condition $y(t_0)=y_0$

Thm 2.4.2

Consider an I.V.P $y^{\prime }=f(t,y),y(t_0)=y_0$

$f$ and $\frac{\partial f}{\partial d}$ are continuous in $R=\{(x,y)|\alpha <t<\beta ,\gamma <y<\delta \}$ and $(t_0,y_0)\in \mathbb{R}$

$\Rightarrow$ There exists $h>0$ s.t. $(t_0-h,t_0+R)\subset (\alpha ,\beta )$
and a unique solution $y=\varphi (t)$ defined on $(t_0-h,t_0+h)$ of the I.V.P

${\varphi }^{\prime }(t)=f(y,\varphi (t))\forall t\in (y_0-h,t_0+h)$ and $\varphi (t_0)=y_0$

If ODE is linear, $f(t,y)=-p(t)y+g(t)$, so $f=-py+g,\frac{\partial f}{\partial y}=-p$ both are continuous.
(condition of Thm 2.4.2 $\iff$ condition of Thm 2.4.1)

Thm 2.4.2 guarantee the existence of a unique solution of I.V.P in $(t_0-h,t_0+h)$ not in $(\alpha ,\beta )$

Second-order

Thm 1 (Existence and Uniqueness Theorem for I.V.P)

Let $p(t),g(t),r(t)$ be continuous on some open interval $I=(\alpha ,\beta )$

If $L(y)=y^{\prime \prime }+p{y}^{\prime }+qy=r,y(t_0)=k_0,y^{\prime }(t_0)=k_1$ for some $t_0\in I$ then, this I.V.P has a unique solution $y=\varphi (t)$ on $I$

Thm 2 (Linear Independence / Dependence of Solutions)

Let $p,q$ be continuous on $I$ and $y_1,y_2$ be solutions of $L(y)=y^{\prime \prime }+p{y}^{\prime }+qy=0$. Then,

$y_1,y_2$ are linearly independent $\iff W(y_1,y_2)(t)\ne 0\text{for all}t\in I$ $\iff W(y_1,y_2)(t)\ne 0\text{for some}t\in I$

$y_1,y_2$ are linearly dependent $\iff W(y_1,y_2)(t)=0\text{for all}t\in I$ $\iff W(y_1,y_2)(t)=0\text{for some}t\in I$

where $W$ is Wronskian Determinant

Thm 3 (Existence of a general solution)

Let $p,q$ be continuous on an open interval $I=(\alpha ,\beta )$

Then, the ODE $y^{\prime \prime }+p{y}^{\prime }+qy=0$ has a general solution

Thm 4 (General solutions includes all solutions)

Let $p,q$ be continuous on an open interval $I=(\alpha ,\beta )$

Let $y_1,y_2$ be a basis of the solutions of the ODE $y^{\prime \prime }+p{y}^{\prime }+qy=0$ on $I$

Then every solution $y=Y(t)$ of $y^{\prime \prime }+p{y}^{\prime }+qy=0$ on $I$ is of the form $Y(x)=c_1{y}_1+c_2+y_2$ for some $c_1,c_2\in \mathbb{R}$

Higher-order ODE

Thm 4.1.1 (Existence and uniqueness)

Let $L(y)=\underset{=\sum _{k=0}^n{p}_k(x)y^{(k)},p_n(x)=1}{\underbrace{y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots +p_1(x)y^{\prime }+p_0(x)y}}=r(x)⊛$ be an n-th order linear ODE

Then, I.V.P $(⊛,y(x_0)=k_0,y^{\prime }(x_0)=k_1,\dots ,y^{(n-1)}(x_0)=k_{n-1})$ has a unique solution on $I$

Thm 4.1.2 & 4.1.3

When $p_0,p_1,\dots ,p_{n-1}$ continuous on $I$, let $y_1,\dots y_n$ be solution of $⊛$. Then,

$W(y_1,\dots y_n)(x_0)\ne 0$ for some $x_0\in I$ $\iff (y_1,\dots y_n)(x)\ne 0$ for all $x\in I$ $\iff y_1,\dots y_n$ are linearly independent

Also, equivalently,

$W(y_1,\dots y_n)(x_0)=0$ on $I$ $\iff (y_1,\dots y_n)(x)=0$ for some $I$ $\iff y_1,\dots y_n$ are linearly dependent

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Ordinary and Singular Points

Definition

Let $y^{\prime \prime }+p(x)+y^{\prime }+q(x)y=0$ be a 2nd-order ODE

Ordinary point

$p(x),q(x)$ are analytic at $x=x_0$ $\iff p,q$ is rational and no $(x-x_0)$ in the denominators

Singular point

Regular singular point

$(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are analytic at $x=x_0$ $\iff p,q$ is rational and no $(x-x_0)$ in the denominators of $(x-x_0)p(x),(x-x_0{)}^{2}q(x)$ $\iff \underset{x\to x_0}{\lim }(x-x_0)p(x),\underset{x\to x_0}{\lim }(x-x_0{)}^{2}q(x)$ are exist

Irregular Singular point

otherwise, an irregular singular point

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Series Solutions

Facts

Existence of Power Series Solutions

Let $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0⊛$ be a 2nd order ODE.

If $x_0$ is an ordinary point of the ODE, then Every solution of the ODE is analytic at $x_0\iff R>0$ s.t. $y(x)=\sum _{n=0}^{\infty }{a}_n(x-x_0{)}^n$ on $|x|<R$

$y(x)=a_0{y}_1(x)+a_1{y}_2(x)$, where $y_1,y_2$ are solutions with $y_1(x_0)=1,y_1^{\prime }(x_0)=0,$ such $\{y_1,y_2\}$ is basis

The radius of convergence $R$ of $y(x)$ is at least the minimum of those of $p(x),q(x)$

Solution

Transclude of Frobenius-Method

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Laplace Transform

Definition

$\mathcal{L}:\{f(t)\}\to \{F(s)\}$

for $f(t)$, function defined on $t\ge 0$ ( on $[0,\infty )$)

Laplace transform of $f$

$F(s)=\mathcal{L}(f(t)):={\int }_0^{\infty }{e}^{-st}f(t)dt$

Inverse transform of $F$

${\mathcal{L}}^{-1}(F(s))=f(t)$

Solution

For given equation $y^{\prime \prime }+a{y}^{\prime }+by=r(t),y(0),y^{\prime }(0)$

1. make a subsidiary equation Let $Y(s)=\mathcal{L}(y(t)),R(s)=\mathcal{L}(r(t))$ $\Rightarrow (s^{2}Y-Sy(0)-y^{\prime }(0))+\alpha (sY-y(0))+bY=R$ $\Rightarrow (s^2+as+b)Y=(s+a)y(0)+y^{\prime }(0)+R$^[subsidiary equation]

2. find a solution of a subsidiary equation by algebra Let $Q(s):=\frac{1}{s^2+as+b}=(\frac{1}{(s+\frac{1}{2}a{)}^2+b^2-\frac{1}{4}{a}^2})$^[Transfer function] $\Rightarrow Y(s)=((s+a)y(0)+y^{\prime }(0))Q(s)+R(s)Q(s)$

3. find Inversion of $Y(s)$ $y(t)={\mathcal{L}}^{-1}(Y(s))$

Facts

Linearity of the Laplace Transform

$\mathcal{L}(af(t)+bg(t))=a\mathcal{L}(f(t))+b\mathcal{L}(g(t)),\forall a,b\in \mathbb{R}$

First shifting (s-shifting) theorem

If $\mathcal{L}(f)=F(s)$, then $\mathcal{L}(e^{at}f)=F(s-a)$

Existence and uniqueness of Laplace Transform

If there is $M,k$ with $|f(t)|\le M{e}^{kt}\forall t\ge 0$^[growth restriction, growth of exponential order] $f(t)$ has Laplace Transform

If $\exists t_1,\dots ,t_n$ s.t. $f$ is continuous on $(t_{i-1},t_i)$ and has a finite limits as $t\to t_i^{+}$ or $t\to t_i^{-}$ $f$ is piecewise continuous on $a\le t\le b$

If $f(t):[0,\infty )\to \mathbb{R}$ is piecewise continuous and satisfies growth-restriction,
then there exists $\mathcal{L}(f(t))\forall s>k$

Laplace transform of derivatives

If $\mathcal{L}(f(t))=F(s)$ and $f$ satisfies the growth restriction, $\mathcal{L}(f^{\prime }(t))=s\mathcal{L}(f)-f(0)=sF(s)-f(0)$ $\mathcal{L}(f^{\prime \prime }(t))=s\mathcal{L}(f^{\prime })-f^{\prime }(0)=s^{2}F(s)-sf(0)-f^{\prime }(0)$ $\mathcal{L}(f^{(n)}(t))=s^{n}F(s)-s^{n-1}f(0)-s^{n-2}{f}^{\prime }(0)-\cdots -f^{(n-1)}(0)$

Laplace Transform of Integrals

If $\mathcal{L}(f(t))=F(s)$, then $\mathcal{L}({\int }_0^{t}f(\tau )d\tau )=\frac{1}{S}F(s)$, thus ${\int }_0^{t}f(\tau )d\tau ={\mathcal{L}}^{-1}(\frac{1}{S}F(s))$

Laplace transform of Unit step function

$L(u(t-a))=\frac{e^{-as}}{s},s>0$

Second shifting theorem (Time-shifting / t-shifting)

If $\mathcal{L}(f(t))=F(s)$,
then $\mathcal{L}(f(t-a)u(t-a))=e^{-as}F(s)$ and $\mathcal{L}(f(t)u(t-a))=e^{-as}\mathcal{L}(f(t+a))$ where $u$ is unit step function

Laplace transform of Dirac’s delta function

$\mathcal{L}(\delta (t-a))=e^{-as}$

Laplace transform and Convolution

If $\mathcal{L}(f(t))=F(s),\mathcal{L}(g(t))=G(s)$,
then $H(s)=\mathcal{L}(h(t))=\mathcal{L}((f\ast g)(t))=F(s)G(s)$

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Systems of ODEs

Definition

A system of n (1st-order) OSEs

functions $x_1,\dots ,x_n$ on $t$ with n initial conditions $x_1(t_0)=X_1,\dots ,x_n(t_0)=X_n$ has solutions $x_1={\varphi }_1(t),\dots ,x_n={\varphi }_n(t)$ on $(t_0-h,t_0+h)$ where $x_1^{\prime }=F_1(t,x_1,\dots ,x_n),\dots ,x_n^{\prime }=F_n(t,x_1,\dots ,x_n)$

Solution

Homogeneous linear systems differential equation with constant coefficients

Solution

We want to solve ${\mathbf{x}}^{\prime }=\mathbf{Ax}$. where $\mathbf{x}=\mathbf{x}(t)=\left(\begin{array}{c}{x}_1(t)\\ ⋮\\ x_n(t)\end{array}\right)$, $\mathbf{A}=\left(\begin{array}{ccc}{a}_{11}& \dots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \dots & a_{nn}\end{array}\right)$

Try the form of the solution $\mathbf{x}$ for general n, $\mathbf{x}(t)=\xi e^{\lambda t}$, where $\xi$ is constant vector ${\mathbf{x}}^{\prime }=\lambda \xi e^{\lambda t}=\mathbf{A}\xi e^{\lambda t}\Rightarrow \mathbf{A}\xi =\lambda \xi$^[eigenvalue problem]

Real Eigenvalues ${\lambda }_1,\dots ,{\lambda }_n\to {\xi }^{(1)},\dots ,{\xi }^{(n)}$

Solutions: ${\mathbf{x}}^{(1)}={\xi }^{(1)}{e}^{{\lambda }_{1}t},\dots ,{\mathbf{x}}^{(n)}={\xi }^{(n)}{e}^{{\lambda }_{n}t}$

$W({\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)})=\left|\begin{array}{ccc}{\mathbf{x}}^{(1)}& \dots & {\mathbf{x}}^{(n)}\end{array}\right|=\left|\begin{array}{ccc}{\xi }^{(1)}& \dots & {\xi }^{(n)}\end{array}\right|e^{({\lambda }_1+\cdots +{\lambda }_n)t}\ne 0$ So, ${\xi }^{(1)},\dots ,{\xi }^{(n)}$ are linearly independent

General solution: $c_1{\xi }^{(1)}{e}^{{\lambda }_{1}t}+\cdots +c_n{\xi }^{(n)}{e}^{{\lambda }_{n}t},c_1,\dots ,c_n\in \mathbb{R}$

Complex Eigenvalues: ${\lambda }_1=\gamma +i\omega \to {\xi }^1,{\lambda }_2=\gamma -i\omega \to {\xi }^2$

Solutions: ${\xi }^{(1)}=\mathbf{a}+i\mathbf{b},{\xi }^{(2)}=\mathbf{a}-i\mathbf{b},\mathbf{a},\mathbf{b}$ are real vectors

${\mathbf{x}}^{(1)}={\xi }^{(1)}{e}^{{\lambda }_{1}t}=(\mathbf{a}+i\mathbf{b})e^{\gamma t}(\cos \omega t+i\sin \omega t)$ $=\underset{\mathbf{u}(t)}{\underbrace{e^{\gamma t}(\mathbf{a}\cos \omega t+\mathbf{b}\sin \omega t)}}+i\underset{\mathbf{v}(t)}{\underbrace{e^{\gamma t}(\mathbf{a}\sin \omega t+\mathbf{b}\cos \omega t)}}=\mathbf{u}(t)+i\mathbf{v}(t)$

$W(\mathbf{u},\mathbf{v})=\left|\begin{array}{cc}\mathbf{a}\cos \omega t+\mathbf{b}\sin \omega t& \mathbf{a}\sin \omega t+\mathbf{b}\cos \omega t\end{array}\right|e^{2\gamma t}$ $\Rightarrow W(\mathbf{u},\mathbf{v})(0)=\left|\begin{array}{cc}\mathbf{a}& \mathbf{b}\end{array}\right|\ne 0$

Repeated (real) root with less eigenvectors: ${\lambda }_1=\cdots ={\lambda }_n,{\xi }^{(1)},\dots ,{\xi }^{(k)},k<n$

Let ${\lambda }_1={\lambda }_2=\lambda ,\xi$ is only eigen vector, then ${\mathbf{x}}^{(1)}=\xi e^{\lambda t}$ (with $\mathbf{A}\xi =\lambda \xi$)

Try ${\mathbf{x}}^{(2)}=\xi t{e}^{\lambda t}+\eta e^{\lambda t}$, where generalized eigenvector $\eta$ is to be determined Then $({\mathbf{x}}^{(2)}{)}^{\prime }=\xi e^{\lambda t}+\xi \lambda t{e}^{\lambda t}+\eta \lambda e^{\lambda t}=\mathbf{A}{\mathbf{x}}^{(2)}=\mathbf{A}(\xi t{e}^{\lambda t}+\eta e^{\lambda t})$ $\Rightarrow (\mathbf{A}-\lambda I)\eta =\xi \Rightarrow (\mathbf{A}-\lambda I{)}^2\eta =\underset{0}{\underbrace{\mathbf{A}\xi -\lambda I\xi }}\Rightarrow (\mathbf{A}-\lambda I{)}^2=0$

Triple roots: ${\mathbf{x}}^{(3)}=\xi \frac{t^2}{2}{e}^{\lambda t}+\eta t{e}^{\lambda t}+\zeta e^{\lambda t}$, where $\zeta$ is to be determined

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Facts

Solution of systems of ODEs

Let x_i' = F_i(t, x_1, \dots, x_n), \quad _{i=1, \dots, n} be a system of n(1th-order) ODEs.

If ${\varphi }_1(t){,}_{i=1,\dots ,n}$ is defined and differentiable on $I$ and $x_i^{\prime }={\varphi }_i^{\prime }(t)=F_i(t,{\varphi }_1(t),\dots ,{\varphi }_n(t))\forall t\in I$ we say that $\{x_1={\varphi }_1(t),\dots ,x_n={\varphi }_n(t)\}$ is a solution on an open interval $I=(\alpha ,\beta )$

Conversion of an ODE

An n-th order ODE $y^{(n)}=F(t,y,y^{\prime },\dots ,y^{(n-1)})$ can be converted to a system of 1st-order ODEs by setting $x_1=y,x_2=y^{\prime },\dots ,x_n=y^{(n-1)}$

This is a system of the form $x_1^{\prime }=x_2,x_2^{\prime }=x_3,\dots ,x_{n-1}^{\prime }=x_n,x_n^{\prime }=F(t,x_1,\dots ,x_n)$

Linear system

If $F_1,\dots ,F_n$ are linear on $x_1,\dots ,x_n$, a system of ODEs is called linear

If a system of 1st-linear ODEs is linear,
then $\exists p_{ij}(t),g_i(t)$, s.t. $x_i^{\prime }=\sum _{j=1}^n{p}_{ij}(t)x_j+g_i(t)$ or $\mathbf{x}=\mathbf{Px}+\mathbf{g}(t)$
where $\mathbf{x}(t)=\left[\begin{array}{c}{x}_1(t)\\ ⋮\\ x_n(t)\end{array}\right],\mathbf{P}(t)={\left[\begin{array}{c}{P}_{ij}(t)\end{array}\right]}_{1\le i,j\le n},\mathbf{g}(t)=\left[\begin{array}{c}{g}_1(t)\\ ⋮\\ g_n(t)\end{array}\right]$

If $g_1(t),\dots ,g_n(t)=0$, the system is called homogeneous
If $g_i(t)\ne 0$, the system is called non-homogeneous

Existence and Uniqueness Theorem

Let x_i' = F_i(t, x_1, \dots, x_n),\quad _{i=1, \dots, n} be a system of (1st-order) ODEs, with $x_i(t_0)=x_i^0\in \mathbb{R}$

If $F_i,\frac{\partial F_i}{\partial x_j}{,}_{1\le i,j\le n}$ are continuous in $R\subseteq t,x_1,x_2,\dots ,x_n$ space by $R=\{(t,x_1,\dots ,x_n)|\alpha <t<\beta \}$ and if $(t_0,x_1^0,\dots ,x_n^0)\in R$
then, $\exists h>0$ and a unique solution $x_1={\varphi }_1(t),\dots ,x_n={\varphi }_n(t)$ defined on $(t_0-h,t_0+h)$ satisfying I.V.P

Existence and Uniqueness Theorem for Linear ODE

Let x_i' = F_i(t, x_1, \dots, x_n),\quad _{i=1, \dots, n} be a system of (1st-order) linear ODEs

If P_{i, j}, g_i,\quad _{1 \le i, j \le n} are continuous on $I=\{t|\alpha <t<\beta \}$ and $x_i(t_0)=x_i^0\in \mathbb{R},t_0\in I$
then, $\exists$ a unique solution $x_1={\varphi }_1(t),\dots ,x_n={\varphi }_n(t)$ defined on $I=(\alpha ,\beta )$ satisfying I.V.P

Basic Theory of Systems of ODEs

If ${\mathbf{x}}^{(i)}$ is solution of ${\mathbf{x}}^{\prime }=P(t)\mathbf{x}$, then $\forall c_i\in \mathbb{R},\sum _{i=1}^n{c}_i{\mathbf{x}}^{(i)}$ is a solution

Let ${\mathbf{x}}^{\prime }=P(t)\mathbf{x}$ be a homogeneous linear system of n ODEs on $I$.

A basis (or fundamental solution) of solutions of the system is a linearly independent set of n solutions

If $\{{\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)}\}$ is a basis, $\mathbf{x}=c_1{\mathbf{x}}^{(1)}+\cdots +c_n{\mathbf{x}}^{(n)}$ is called a general solution

Let ${\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)}$ be $n\times 1$ vectors (of functions) They are linearly independent $\iff$ if $c_1,\dots ,c_n\in \mathbb{R},\sum _{i=1}^n{c}_i{\mathbf{x}}^{(i)}=0$ then $c_1=\cdots =c_n=0$ They are linearly dependent $\iff \exists c_1,\dots ,c_n$, not all zero, s.t. $\sum _{i=1}^n{c}_i{\mathbf{x}}^{(i)}=0$ They are linearly independent at a point $t_0\iff {\mathbf{x}}^{(1)}(t_0),\dots ,{\mathbf{x}}^{(n)}(t_0)$ $W({\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)}):=\left|\begin{array}{ccc}{\mathbf{x}}^{(1)}& \dots & {\mathbf{x}}^{(n)}\end{array}\right|=|\mathbf{X}(t)|$, where $|\mathbf{X}(t)|$ is the Wronskian Determinant of ${\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)}$

Wronskian and systems of ODEs

If n-th order linear ODE and the corresponding system of linear ODEs is given,
then they have the same Wronskian Determinant

Let $t_0\in I$. $W({\mathbf{x}}^{(1)}(t_0),\dots ,{\mathbf{x}}^{(n)}(t_0))\ne 0\iff {\mathbf{x}}^{(1)}(t_0),\dots ,{\mathbf{x}}^{(n)}(t_0)$ are linearly independent

If $\{{\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)}\}$ is a basis for the solution of a homogeneous linear system on $I$,
then $\forall$ solution $\varphi$, $\exists c_1,\dots ,c_n\in \mathbb{R}$ s.t. $\varphi (t)=\sum _{i=1}^n{c}_i{\mathbf{x}}^{(i)}$

${\mathbf{x}}^{(1)}(t),\dots ,{\mathbf{x}}^{(n)}(t_0)$ are linearly independent on $I=(\alpha ,\beta )$ $\Rightarrow {\mathbf{x}}^{(1)}(t_0),\dots ,{\mathbf{x}}^{(n)}(t_0)$ are linearly independent

So, ${\mathbf{x}}^{(1)}(t),\dots ,{\mathbf{x}}^{(n)}(t_0)$ are linearly independent at some $t=t_0$, then ${\mathbf{x}}^{(1)}(t),\dots ,{\mathbf{x}}^{(n)}(t_0)$ are linearly independent.

If ${\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)}$ are linear solutions of homogeneous linear ODEs on $x_1,\dots ,x_n$ on $I=(\alpha ,\beta )$, then $W({\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)})\ne 0\forall t\in I\iff W({\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)})(t_0)\ne 0$ for some $t_0\in I$

Given a linear system of ODE on $I=(\alpha ,\beta )$, $\exists$ a basis $\{{\mathbf{x}}^{(1)},\dots ,{\mathbf{x}}^{(n)}\}$

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