Second Order Equations

Definitions

Second-Order Differential Equation

y′′=f(t,y,y′)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)) 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) 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=0y''+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 (α,β)(\alpha, \beta)(α,β). Let t0t_0t0​ be any point in (α,β)(\alpha, \beta)(α,β). Then for any real numbers y0y_0y0​ and y1y_1y1​ there is one and only one function y(t)y(t)y(t) defined on (α,β)(\alpha, \beta)(α,β), which is a solution to
y′′+p(t)y′+q(t)y=g(t)forα<t<βy''+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)y(t) satisfies the initial conditions
y(t0)=y0y′(t0)=y1y(t_0)=y_0 \\ y'(t_0)=y_1 y(t0​)=y0​y′(t0​)=y1​

Structure of General Solutions

Suppose that y1y_1y1​ and y2y_2y2​ are both solutions to the homogeneous, linear equation
y′′+p(t)y′+q(t)y=0.y''+p(t)y'+q(t)y=0. y′′+p(t)y′+q(t)y=0.
Then the function
y=C1y1+C2y2y=C_1y_1+C_2y_2 y=C1​y1​+C2​y2​
is also a solution.

Linear Combination

A linear combination of the two functions uuu and vvv is any function of the form
w=Au+Bv,w=Au+Bv, w=Au+Bv,
where AAA and BBB are constants.

Linear Independent

Two functions u and v are said to be linearly independent on the interval (α,β)(\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 (α,β)(\alpha, \beta)(α,β) they are said to be linearly dependent there.

General Solution

Suppose that y1y_1y1​ and y2y_2y2​ 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. y′′+p(t)y′+q(t)y=0.
Then the general solution is
y=C1y1+C2y2,y=C_1y_1+C_2y_2, y=C1​y1​+C2​y2​,
where C1C_1C1​ and C2C_2C2​ are arbitrary constants.

y1y_1y1​ and y2y_2y2​ form a fundamental set of solutions.

Wronskian

The Wronskian of two functions uuu and vvv 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 uuu and vvv are sulutions to the linear, homogeneous equation
y′′+p(t)y′+q(t)y=0y''+p(t)y'+q(t)y=0 y′′+p(t)y′+q(t)y=0
in the interval (α,β)(\alpha, \beta)(α,β). Then the Wronskian of uuu and vvv is either identically equal to zero on (α,β)(\alpha, \beta)(α,β) or it is never equal to zero there.

Use Wronskian to Check Linear Dependency

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

If W(t0)≠0W(t_0)\neq 0W(t0​)​=0 for some t0t_0t0​ in the interval (α,β)(\alpha, \beta)(α,β), then u and v are linearly independent in (α,β)(\alpha, \beta)(α,β). On the other hand, if u and v are linearly independent in (α,β)(\alpha, \beta)(α,β), then W(t)W(t)W(t) never vanishes in (α,β)(\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), x′=f(t,x,y)y′=g(t,x,y),
where fff and ggg are functions of the independent variable ttt and the two unknowns xxx and yyy.

二阶方程y′′=F(t,y,y′)y''=F(t,y,y')y′′=F(t,y,y′)可以写成以下一阶系统：
y′=vv′=F(t,y,v)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=0y''+py'+qy=0y′′+py′+qy=0 is
λ2+pλ+q=0.\lambda^2+p\lambda + q=0. λ2+pλ+q=0.
The polynomial λ2+pλ+q\lambda^2+p\lambda + qλ2+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:

• p2−4q>0p^2-4q \gt 0p2−4q>0, the characteristic equation has two distinct, real roots λ1\lambda_1λ1​ and λ2\lambda_2λ2​. A fundamental set of solutions is
  y1(t)=eλ1tandy2(t)=eλ2ty_1(t)=e^{\lambda_1 t} \quad and \quad y_2(t)=e^{\lambda_2 t} y1​(t)=eλ1​tandy2​(t)=eλ2​t

• p2−4q=0p^2-4q = 0p2−4q=0, one repeated real root λ\lambdaλ. Fundamental set of solutions:
  y1(t)=eλtandy2(t)=teλty_1(t)=e^{\lambda t} \quad and \quad y_2(t)=te^{\lambda t} y1​(t)=eλtandy2​(t)=teλt

• p2−4q<0p^2-4q \lt 0p2−4q<0, two complex conjugate roots a±iba \pm iba±ib. Fundamental set of solutions:
  y1(t)=eatcos⁡(bt)andy2(t)=eatsin⁡(bt)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=fy''+py'+qy = f y′′+py′+qy=f

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

General solution to inhomogeneous equation

Suppose that ypy_pyp​ is a particular solution to the inhomogeneous equation, and that y1y_1y1​ and y2y_2y2​ form a fundamental set of solutions to the associated homogeneous equation y′′+py′+qy=0y''+py'+qy=0y′′+py′+qy=0.

Then the general solution to the inhomogeneous equation is given by
y=yp+C1y1+C2y2,y = y_p + C_1 y_1 + C_2 y_2, y=yp​+C1​y1​+C2​y2​,
where C1C_1C1​ and C2C_2C2​ are arbitrary constants.

The general solution can also be written as
y=yp+yhy=y_p + y_h y=yp​+yh​
where yh=C1y1+C2y2y_h = C_1 y_1 + C_2 y_2yh​=C1​y1​+C2​y2​

The Method of Undetermined Coefficients

y′′+py′+qy=fy''+py'+qy=f y′′+py′+qy=f

where ppp and qqq are constants.

因为通解的形式是y=yp+yhy=y_p + y_hy=yp​+yh​，而我们之前已经知道了yhy_hyh​的解法，所以现在只要找到一个特解ypy_pyp​就行。怎么找，看f(t)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)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aertae^{rt}aert，带入方程后造成两边不相等，那就尝试atertate^{rt}atert，如果还不行，再试at2ertat^2e^{rt}at2ert。

The Method of Variation of Parameters

y′′+py′+qy=fy''+py'+qy = f y′′+py′+qy=f

1. Find a fundamental set of solutions y1y_1y1​, y2y_2y2​ to the associated homogeneous equation y′′+py′+qy=0y''+py'+qy=0y′′+py′+qy=0

2. Form yp=v1y1+v2y2y_p=v_1 y_1 + v_2 y_2yp​=v1​y1​+v2​y2​, where v1v_1v1​ and v2v_2v2​ are functions to be determined.

3. Find v1v_1v1​ and v2v_2v2​ by solving the equations and integrating:
   v1′y1+v2′y2=0v1′y1′+v2′y2′=f(t)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 v1v_1v1​ and v2v_2v2​ into yp=v1y1+v2y2y_p=v_1 y_1 + v_2 y_2yp​=v1​y1​+v2​y2​

Harmonic Motion

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

Rewrite the equation as
d2ydt2+μmdydt+kmy=1mF(t),\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=μ2mc = \frac{\mu}{2m}c=2mμ​, ω0=km\omega_0 = \sqrt{\frac{k}{m}}ω0​=mk​​, f(t)=F(t)mf(t)=\frac{F(t)}{m}f(t)=mF(t)​, and x=yx=yx=y, we get the equation
x′′+2cx′+ω02x=f(t).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. ccc is called damping constant, and fff is the forcing term.

Unforced Harmonic Motion

x′′+2cx′+ω02x=0x'' + 2cx' + \omega_0^2 x = 0 x′′+2cx′+ω02​x=0

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

Simple harmonic motion

When there is no damping, that is, c=0c=0c=0
x′′+ω02x=0.x'' + \omega_0^2 x = 0. x′′+ω02​x=0.
The general solution is
x(t)=acos⁡(ω0t)+bsin⁡(ω0t),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.

ω0\omega_0ω0​ is called the natural frequency.

Periodicity
T=2πω0T = \frac{2\pi}{\omega_0} T=ω0​2π​

The general solution to simple harmonic motion can be written as
x(t)=Acos⁡(ω0t−ϕ)x(t) = A\cos(\omega_0 t - \phi) x(t)=Acos(ω0​t−ϕ)
where A is the amplitude, ϕ\phiϕ is the phase
A=a2+b2tan⁡ϕ=baA = \sqrt{a^2 + b^2} \\ \tan\phi = \frac{b}{a} A=a2+b2​tanϕ=ab​
To solve for ϕ\phiϕ, we take arctan(ba)arctan(\frac{b}{a})arctan(ab​), but it takes value between −π/2-\pi/2−π/2 and π/2\pi/2π/2, while ϕ\phiϕ can be from −π-\pi−π to π\piπ. Therefore, we take
ϕ={arctan⁡(b/a),ifa>0;arctan⁡(b/a)+π,ifa<0andb>0;arctan⁡(b/a)−π,ifa<0andb<0.\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>0c>0c>0,
x′′+2cx′+ω02x=0.x'' + 2cx' + \omega_0^2 x = 0. x′′+2cx′+ω02​x=0.
The characteristic equation
λ2+2cλ+ω02=0\lambda^2 + 2c\lambda + \omega_0^2 = 0 λ2+2cλ+ω02​=0
has roots
λ1=−c−c2−ω02andλ2=−c+c2−ω02.\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 c2−ω02c^2-\omega_0^2c2−ω02​, and thus for the general solution:

1. Underdamped (c2−ω02<0c^2-\omega_0^2 < 0c2−ω02​<0, that is, c<ω0c < \omega_0c<ω0​)
   x(t)=e−ct[C1cos⁡(ωt)+C2sin⁡(ωt)],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 ω=ω02−c2\omega = \sqrt{\omega_0^2 - c^2}ω=ω02​−c2​

2. Overdamped (c2−ω02>0c^2-\omega_0^2 > 0c2−ω02​>0, that is, c>ω0c > \omega_0c>ω0​)
   x(t)=C1eλ1t+C2eλ2tx(t) = C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t} x(t)=C1​eλ1​t+C2​eλ2​t
   where λ1<λ2<0\lambda_1 < \lambda_2 < 0λ1​<λ2​<0

3. Critically damped (c2−ω02=0c^2-\omega_0^2 = 0c2−ω02​=0, that is, c=ω0c = \omega_0c=ω0​, λ=−c\lambda = -cλ=−c)
   x(t)=C1e−ct+C2te−ctx(t) = C_1 e^{-ct} + C_2 te^{-ct} x(t)=C1​e−ct+C2​te−ct

Forced Harmonic Motion

x′′+2cx′+ω02x=Acos⁡(ωt)x''+2cx'+\omega_0^2 x = A\cos(\omega t) x′′+2cx′+ω02​x=Acos(ωt)

where AAA is the amplitude of the driving force, and ω\omegaω is the driving frequency

Forced Undamped Harmonic Motion

x′′+ω02x=Acos⁡(ωt)x''+\omega_0^2 x = A\cos(\omega t) x′′+ω02​x=Acos(ωt)

The associated homogeneous equation is
x′′+ω02x=0x''+\omega_0^2x = 0 x′′+ω02​x=0
with general solution
xh=C1cos⁡(ω0t)+C2sin⁡(ω0t)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 ω\omegaω:

1. ω≠ω0\omega \ne \omega_0ω​=ω0​, that is, the driving frequency is not equal to the natural freq

   The equation becomes x′′+ω02x=Acos⁡(ω0t)x''+\omega_0^2 x = A\cos(\omega_0 t)x′′+ω02​x=Acos(ω0​t)

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

   We get
   a=Aω02−ω2andb=0.a = \frac{A}{\omega_0^2 - \omega^2} \quad and \quad b=0. a=ω02​−ω2A​andb=0.
   The particular solution is
   xp(t)=Aω02−ωcos⁡(ωt)x_p(t) = \frac{A}{\omega_0^2 - \omega}\cos(\omega t) xp​(t)=ω02​−ωA​cos(ωt)

2. ω=ω0\omega = \omega_0ω=ω0​

   In this case, xp=acos(ω0t)+bsin(ω0t)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 xh=C1cos(ω0t)+C2sin(ω0t)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
   xp=t(acos⁡(ω0t)+bsin⁡(ω0t)).x_p = t(a\cos(\omega_0 t) + b\sin(\omega_0 t)). xp​=t(acos(ω0​t)+bsin(ω0​t)).
   We get
   b=A2ω0anda=0.b = \frac{A}{2\omega_0} \quad and \quad a = 0. b=2ω0​A​anda=0.
   The particular solution is
   xp=A2ω0tsin⁡(ω0t).x_p = \frac{A}{2\omega_0}t\sin (\omega_0 t). xp​=2ω0​A​tsin(ω0​t).
   As ttt increases, xpx_pxp​ grows larger.

   This case is called resonance.

Forced Damped Harmonic Motion

x′′+2cx′+ω02x=Acos⁡(ωt)x''+2cx'+\omega_0^2 x = A\cos(\omega t) x′′+2cx′+ω02​x=Acos(ωt)

The associated homogeneous equation:
x′′+2cx′+ω02x=0x'' + 2cx' + \omega_0^2 x = 0 x′′+2cx′+ω02​x=0
The characteristic roots are
λ=−c±c2−ω02\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xhx_hxh​是一样的，不一样的是特解xpx_pxp​.

对于Underdamped情况，即c<ω0c<\omega_0c<ω0​，
x(t)=e−ct[C1cos⁡(ηt)+C2sin⁡(ηt)],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
η=ω02−c2.\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ω\omegaω表示η\etaη，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η\etaη。

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

Rcos⁡ϕ=ω02−ω2andRsin⁡ϕ=2cωR=(ω02−ω)2+4c2ω2R\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⁡ϕ=ω02−ω2(ω02−ω2)2+4c2ω2sin⁡ϕ=2cω(ω02−ω2)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​

cotϕ=ω02−ω22cωcot\phi = \frac{\omega_0^2 - \omega^2}{2c\omega} cotϕ=2cωω02​−ω2​

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