Second Order Equations

Definitions

Second-Order Differential Equation

$y'' = f(t,y,y')$y′′=f(t,y,y′)

Solution

$y''(t) = f(t,y(t),y'(t))$y′′(t)=f(t,y(t),y′(t))

Linear Equations

$y'' + p(t)y' + q(t)y = g(t)$y′′+p(t)y′+q(t)y=g(t)

Where coefficients p, q, and g can be arbitrary functions of independent variable t, but y, y’, and y’’ must all be first order.

g(t) is called forcing term.

If g(t) = 0, the equation is said to be homogeneous:
$y''+p(t)y'+q(t)y=0$y′′+p(t)y′+q(t)y=0

Existence and Uniqueness

Suppose the functions p(t), q(t), and g(t) are continuous on the interval $(\alpha, \beta)$(α,β). Let $t_0$t0​ be any point in $(\alpha, \beta)$(α,β). Then for any real numbers $y_0$y0​ and $y_1$y1​ there is one and only one function $y(t)$y(t) defined on $(\alpha, \beta)$(α,β), which is a solution to
$y''+p(t)y'+q(t)y=g(t) \quad for \ \alpha \lt t \lt \beta$y′′+p(t)y′+q(t)y=g(t)for α<t<β
and $y(t)$y(t) satisfies the initial conditions
$y(t_0)=y_0 \\ y'(t_0)=y_1$y(t0​)=y0​y′(t0​)=y1​

Structure of General Solutions

Suppose that $y_1$y1​ and $y_2$y2​ are both solutions to the homogeneous, linear equation
$y''+p(t)y'+q(t)y=0.$y′′+p(t)y′+q(t)y=0.
Then the function
$y=C_1y_1+C_2y_2$y=C1​y1​+C2​y2​
is also a solution.

Linear Combination

A linear combination of the two functions $u$u and $v$v is any function of the form
$w=Au+Bv,$w=Au+Bv,
where $A$A and $B$B are constants.

Linear Independent

Two functions u and v are said to be linearly independent on the interval $(\alpha, \beta)$(α,β) if neither is a constant multiple of the other on that interval. If one is a constant multiple of the other on $(\alpha, \beta)$(α,β) they are said to be linearly dependent there.

General Solution

Suppose that $y_1$y1​ and $y_2$y2​ are linearly independent solutions to the homogeneous, linear equation
$y''+p(t)y'+q(t)y=0.$y′′+p(t)y′+q(t)y=0.
Then the general solution is
$y=C_1y_1+C_2y_2,$y=C1​y1​+C2​y2​,
where $C_1$C1​ and $C_2$C2​ are arbitrary constants.

$y_1$y1​ and $y_2$y2​ form a fundamental set of solutions.

Wronskian

The Wronskian of two functions $u$u and $v$v is defined to be
KaTeX parse error: Undefined control sequence: \matrix at position 16: W(t)=det\left(\̲m̲a̲t̲r̲i̲x̲{u(t) & v(t)\\ …
Suppose the functions $u$u and $v$v are sulutions to the linear, homogeneous equation
$y''+p(t)y'+q(t)y=0$y′′+p(t)y′+q(t)y=0
in the interval $(\alpha, \beta)$(α,β). Then the Wronskian of $u$u and $v$v is either identically equal to zero on $(\alpha, \beta)$(α,β) or it is never equal to zero there.

Use Wronskian to Check Linear Dependency

Suppose the functions $u$u and $v$v are solutions to the linear, homogeneous equation
$y''+p(t)y'+q(t)y=0$y′′+p(t)y′+q(t)y=0
in the interval $(\alpha, \beta)$(α,β). Then $u$u and $v$v are linearly dependent if and only if their Wronskian is identically zero in $(\alpha, \beta)$(α,β).

If $W(t_0)\neq 0$W(t0​)​=0 for some $t_0$t0​ in the interval $(\alpha, \beta)$(α,β), then u and v are linearly independent in $(\alpha, \beta)$(α,β). On the other hand, if u and v are linearly independent in $(\alpha, \beta)$(α,β), then $W(t)$W(t) never vanishes in $(\alpha, \beta)$(α,β)

Second-Order Equations and Systems

这一节就是想说明一阶方程和更高阶方程之间的关系

A planner system of first-order equation is a set of two first-order differential equations involving two unknown functions. It might be written as
$x'=f(t,x,y)\\ y'=g(t,x,y),$x′=f(t,x,y)y′=g(t,x,y),
where $f$f and $g$g are functions of the independent variable $t$t and the two unknowns $x$x and $y$y.

二阶方程$y''=F(t,y,y')$y′′=F(t,y,y′)可以写成以下一阶系统：
$y' = v\\ v' = F(t,y,v)$y′=vv′=F(t,y,v)
如果y是二阶方程的一个解，那么y和v就是上面一阶系统的解

The yv-plan is called the phase plane.

Plotting y and v versus t is the composite plot.

General Solutions to Linear, Homogeneous Equations with Constant Coefficients

Characteristic Equation for a differential equation $y''+py'+qy=0$y′′+py′+qy=0 is
$\lambda^2+p\lambda + q=0.$λ2+pλ+q=0.
The polynomial $\lambda^2+p\lambda + q$λ2+pλ+q is called the characteristic polynomial for the equation. The root(s) of the characteristic equation is called characteristic root.

There are three cases:

• $p^2-4q \gt 0$p2−4q>0, the characteristic equation has two distinct, real roots $\lambda_1$λ1​ and $\lambda_2$λ2​. A fundamental set of solutions is
  $y_1(t)=e^{\lambda_1 t} \quad and \quad y_2(t)=e^{\lambda_2 t}$y1​(t)=eλ1​tandy2​(t)=eλ2​t

• $p^2-4q = 0$p2−4q=0, one repeated real root $\lambda$λ. Fundamental set of solutions:
  $y_1(t)=e^{\lambda t} \quad and \quad y_2(t)=te^{\lambda t}$y1​(t)=eλtandy2​(t)=teλt

• $p^2-4q \lt 0$p2−4q<0, two complex conjugate roots $a \pm ib$a±ib. Fundamental set of solutions:
  $y_1(t)=e^{at}\cos(bt) \quad and \quad y_2(t)=e^{at}\sin(bt)$y1​(t)=eatcos(bt)andy2​(t)=eatsin(bt)

Inhomogeneous Equations

$y''+py'+qy = f$y′′+py′+qy=f

where $p=p(t)$p=p(t), $q=q(t)$q=q(t), and $f=f(t)$f=f(t) are functions of the independent variable $t$t. $f$f is called the inhomogeneous term, or the forcing term.

General solution to inhomogeneous equation

Suppose that $y_p$yp​ is a particular solution to the inhomogeneous equation, and that $y_1$y1​ and $y_2$y2​ form a fundamental set of solutions to the associated homogeneous equation $y''+py'+qy=0$y′′+py′+qy=0.

Then the general solution to the inhomogeneous equation is given by
$y = y_p + C_1 y_1 + C_2 y_2,$y=yp​+C1​y1​+C2​y2​,
where $C_1$C1​ and $C_2$C2​ are arbitrary constants.

The general solution can also be written as
$y=y_p + y_h$y=yp​+yh​
where $y_h = C_1 y_1 + C_2 y_2$yh​=C1​y1​+C2​y2​

The Method of Undetermined Coefficients

$y''+py'+qy=f$y′′+py′+qy=f

where $p$p and $q$q are constants.

因为通解的形式是$y=y_p + y_h$y=yp​+yh​，而我们之前已经知道了$y_h$yh​的解法，所以现在只要找到一个特解$y_p$yp​就行。怎么找，看$f(t)$f(t)的形式，然后用待定系数法解出p和q。

特殊情况

$f(t)$f(t)本身就是齐次方程(homogeneous equation)的解，直接用上面的trial solution就可能导致方程两边不相等。这时候，就在trial solution基础上再乘上t，不行的就再乘t… 例如尝试的trial solution是$ae^{rt}$aert，带入方程后造成两边不相等，那就尝试$ate^{rt}$atert，如果还不行，再试$at^2e^{rt}$at2ert。

The Method of Variation of Parameters

$y''+py'+qy = f$y′′+py′+qy=f

1. Find a fundamental set of solutions $y_1$y1​, $y_2$y2​ to the associated homogeneous equation $y''+py'+qy=0$y′′+py′+qy=0

2. Form $y_p=v_1 y_1 + v_2 y_2$yp​=v1​y1​+v2​y2​, where $v_1$v1​ and $v_2$v2​ are functions to be determined.

3. Find $v_1$v1​ and $v_2$v2​ by solving the equations and integrating:
   $v_1' y_1 + v_2' y_2 = 0\\ v_1' y_1' + v_2' y_2' = f(t)$v1′​y1​+v2′​y2​=0v1′​y1′​+v2′​y2′​=f(t)

4. Substitute $v_1$v1​ and $v_2$v2​ into $y_p=v_1 y_1 + v_2 y_2$yp​=v1​y1​+v2​y2​

Harmonic Motion

The equation for the motion of a vibrating spring is
$my'' + \mu y' + ky = F(t),$my′′+μy′+ky=F(t),
where $m$m is mass, $\mu$μ is damping constant, and $k$k is spring constant, F(t) is the external force.

Rewrite the equation as
$\frac{d^2 y}{dt^2} + \frac{\mu}{m} \frac{dy}{dt} + \frac{k}{m} y = \frac{1}{m} F(t),$dt2d2y​+mμ​dtdy​+mk​y=m1​F(t),
and make $c = \frac{\mu}{2m}$c=2mμ​, $\omega_0 = \sqrt{\frac{k}{m}}$ω0​=mk​​, $f(t)=\frac{F(t)}{m}$f(t)=mF(t)​, and $x=y$x=y, we get the equation
$x'' + 2cx' + \omega_0^2 x = f(t).$x′′+2cx′+ω02​x=f(t).
We refer to this equation as the equation for harmonic motion. $c$c is called damping constant, and $f$f is the forcing term.

Unforced Harmonic Motion

$x'' + 2cx' + \omega_0^2 x = 0$x′′+2cx′+ω02​x=0

where $c \ge 0$c≥0 and $\omega_0 \gt 0$ω0​>0 are constants.

Simple harmonic motion

When there is no damping, that is, $c=0$c=0
$x'' + \omega_0^2 x = 0.$x′′+ω02​x=0.
The general solution is
$x(t) = a\cos(\omega_0 t)+b\sin(\omega_0 t),$x(t)=acos(ω0​t)+bsin(ω0​t),
where a and b are constants.

$\omega_0$ω0​ is called the natural frequency.

Periodicity
$T = \frac{2\pi}{\omega_0}$T=ω0​2π​

The general solution to simple harmonic motion can be written as
$x(t) = A\cos(\omega_0 t - \phi)$x(t)=Acos(ω0​t−ϕ)
where A is the amplitude, $\phi$ϕ is the phase
$A = \sqrt{a^2 + b^2} \\ \tan\phi = \frac{b}{a}$A=a2+b2​tanϕ=ab​
To solve for $\phi$ϕ, we take $arctan(\frac{b}{a})$arctan(ab​), but it takes value between $-\pi/2$−π/2 and $\pi/2$π/2, while $\phi$ϕ can be from $-\pi$−π to $\pi$π. Therefore, we take
$\phi = \left\{ \begin{matrix} \arctan(b/a), & if\ a>0; \\ \arctan(b/a)+\pi, & if\ a<0\ and\ b>0; \\ \arctan(b/a)-\pi, & if\ a<0\ and\ b<0. \end{matrix} \right.$ϕ=⎩⎨⎧​arctan(b/a),arctan(b/a)+π,arctan(b/a)−π,​if a>0;if a<0 and b>0;if a<0 and b<0.​

Damped Harmonic Motion

Now $c>0$c>0,
$x'' + 2cx' + \omega_0^2 x = 0.$x′′+2cx′+ω02​x=0.
The characteristic equation
$\lambda^2 + 2c\lambda + \omega_0^2 = 0$λ2+2cλ+ω02​=0
has roots
$\lambda_1 = -c-\sqrt{c^2-\omega_0^2}\quad and\quad \lambda_2 = -c+\sqrt{c^2-\omega_0^2}.$λ1​=−c−c2−ω02​​andλ2​=−c+c2−ω02​​.
There are three cases for the sign of $c^2-\omega_0^2$c2−ω02​, and thus for the general solution:

1. Underdamped ($c^2-\omega_0^2 < 0$c2−ω02​<0, that is, $c < \omega_0$c<ω0​)
   $x(t) = e^{-ct}[C_1\cos(\omega t)+C_2\sin(\omega t)],$x(t)=e−ct[C1​cos(ωt)+C2​sin(ωt)],
   where $\omega = \sqrt{\omega_0^2 - c^2}$ω=ω02​−c2​

2. Overdamped ($c^2-\omega_0^2 > 0$c2−ω02​>0, that is, $c > \omega_0$c>ω0​)
   $x(t) = C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t}$x(t)=C1​eλ1​t+C2​eλ2​t
   where $\lambda_1 < \lambda_2 < 0$λ1​<λ2​<0

3. Critically damped ($c^2-\omega_0^2 = 0$c2−ω02​=0, that is, $c = \omega_0$c=ω0​, $\lambda = -c$λ=−c)
   $x(t) = C_1 e^{-ct} + C_2 te^{-ct}$x(t)=C1​e−ct+C2​te−ct

Forced Harmonic Motion

$x''+2cx'+\omega_0^2 x = A\cos(\omega t)$x′′+2cx′+ω02​x=Acos(ωt)

where $A$A is the amplitude of the driving force, and $\omega$ω is the driving frequency

Forced Undamped Harmonic Motion

$x''+\omega_0^2 x = A\cos(\omega t)$x′′+ω02​x=Acos(ωt)

The associated homogeneous equation is
$x''+\omega_0^2x = 0$x′′+ω02​x=0
with general solution
$x_h = C_1\cos(\omega_0 t) + C_2\sin(\omega_0 t)$xh​=C1​cos(ω0​t)+C2​sin(ω0​t)
There are two cases for the driving frequency $\omega$ω:

1. $\omega \ne \omega_0$ω​=ω0​, that is, the driving frequency is not equal to the natural freq

   The equation becomes $x''+\omega_0^2 x = A\cos(\omega_0 t)$x′′+ω02​x=Acos(ω0​t)

   Using undetermined coefficient method, $x_p=a\cos(\omega t)+b\sin(\omega t)$xp​=acos(ωt)+bsin(ωt)

   We get
   $a = \frac{A}{\omega_0^2 - \omega^2} \quad and \quad b=0.$a=ω02​−ω2A​andb=0.
   The particular solution is
   $x_p(t) = \frac{A}{\omega_0^2 - \omega}\cos(\omega t)$xp​(t)=ω02​−ωA​cos(ωt)

2. $\omega = \omega_0$ω=ω0​

   In this case, $x_p=a\ cos(\omega_0 t)+b\ sin(\omega_0 t)$xp​=a cos(ω0​t)+b sin(ω0​t) is the solution of homogeneous equation $x_h = C_1cos(\omega_0 t) + C_2sin(\omega_0 t)$xh​=C1​cos(ω0​t)+C2​sin(ω0​t). Therefore, we look for a particular solution
   $x_p = t(a\cos(\omega_0 t) + b\sin(\omega_0 t)).$xp​=t(acos(ω0​t)+bsin(ω0​t)).
   We get
   $b = \frac{A}{2\omega_0} \quad and \quad a = 0.$b=2ω0​A​anda=0.
   The particular solution is
   $x_p = \frac{A}{2\omega_0}t\sin (\omega_0 t).$xp​=2ω0​A​tsin(ω0​t).
   As $t$t increases, $x_p$xp​ grows larger.

   This case is called resonance.

Forced Damped Harmonic Motion

$x''+2cx'+\omega_0^2 x = A\cos(\omega t)$x′′+2cx′+ω02​x=Acos(ωt)

The associated homogeneous equation:
$x'' + 2cx' + \omega_0^2 x = 0$x′′+2cx′+ω02​x=0
The characteristic roots are
$\lambda = -c \pm \sqrt{c^2 - \omega_0^2}$λ=−c±c2−ω02​​
和unforced的时候一样，有三种情况。但是homogeneous equation的通解$x_h$xh​是一样的，不一样的是特解$x_p$xp​.

对于Underdamped情况，即$c<\omega_0$c<ω0​，
$x(t) = e^{-ct}[C_1\cos(\eta t)+C_2\sin(\eta t)],$x(t)=e−ct[C1​cos(ηt)+C2​sin(ηt)],
where
$\eta = \sqrt{\omega_0^2 - c^2}.$η=ω02​−c2​.
就这里与之前unforced不一样，之前用的是$\omega$ω表示$\eta$η，但是因为在forced这里已经表示了driving frequency，所以只能换成另外的字母表示，即$\eta$η。

特解还是用待定系数法，通过一系列地计算我们可以得到
$x_p = G(\omega)A\cos(\omega t-\phi).$xp​=G(ω)Acos(ωt−ϕ).
where
$G(\omega) = \frac{1}{R}$G(ω)=R1​

$R\cos\phi = \omega_0^2 - \omega^2 \quad and \quad R\sin\phi=2c\omega \\ R = \sqrt{(\omega_0^2-\omega)^2 + 4c^2\omega^2}$Rcosϕ=ω02​−ω2andRsinϕ=2cωR=(ω02​−ω)2+4c2ω2​

$\cos\phi=\frac{\omega_0^2 - \omega^2}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4c^2\omega^2}} \quad \sin\phi = \frac{2c\omega}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4c^2\omega^2}}$cosϕ=(ω02​−ω2)2+4c2ω2​ω02​−ω2​sinϕ=(ω02​−ω2)2+4c2ω2​2cω​

$cot\phi = \frac{\omega_0^2 - \omega^2}{2c\omega}$cotϕ=2cωω02​−ω2​