二阶微分方程解法总结 Summary of Second Order Equations

Second Order Equations

Definitions

Second-Order Differential Equation

$$y^{\prime \prime }=f(t,y,y^{\prime })$$

Solution

$$y^{\prime \prime }(t)=f(t,y(t),y^{\prime }(t))$$

Linear Equations

$$y^{\prime \prime }+p(t)y^{\prime }+q(t)y=g(t)$$

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:
$$y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0$$

Existence and Uniqueness

Suppose the functions p(t), q(t), and g(t) are continuous on the interval $(\alpha ,\beta )$. Let $t_0$ be any point in $(\alpha ,\beta )$. Then for any real numbers $y_0$ and $y_1$ there is one and only one function $y(t)$ defined on $(\alpha ,\beta )$, which is a solution to
$$y^{\prime \prime }+p(t)y^{\prime }+q(t)y=g(t)for\alpha <t<\beta$$
and $y(t)$ satisfies the initial conditions
$$\begin{gathered} y(t_0)=y_0 \\ {y}^{\prime }(t_0)=y_1 \end{gathered}$$

Structure of General Solutions

Suppose that $y_1$ and $y_2$ are both solutions to the homogeneous, linear equation
$$y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0.$$
Then the function
$$y=C_1{y}_1+C_2{y}_2$$
is also a solution.

Linear Combination

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

Linear Independent

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

General Solution

Suppose that $y_1$ and $y_2$ are linearly independent solutions to the homogeneous, linear equation
$$y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0.$$
Then the general solution is
$$y=C_1{y}_1+C_2{y}_2,$$
where $C_1$ and $C_2$ are arbitrary constants.

$y_1$ and $y_2$ form a fundamental set of solutions.

Wronskian

The Wronskian of two functions $u$ and $v$ 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 $u$ and $v$ are sulutions to the linear, homogeneous equation
$$y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0$$
in the interval $(\alpha ,\beta )$. Then the Wronskian of $u$ and $v$ is either identically equal to zero on $(\alpha ,\beta )$ or it is never equal to zero there.

Use Wronskian to Check Linear Dependency

Suppose the functions $u$ and $v$ are solutions to the linear, homogeneous equation
$$y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0$$
in the interval $(\alpha ,\beta )$. Then $u$ and $v$ are linearly dependent if and only if their Wronskian is identically zero in $(\alpha ,\beta )$.

If $W(t_0)\ne 0$ for some $t_0$ in the interval $(\alpha ,\beta )$, then u and v are linearly independent in $(\alpha ,\beta )$. On the other hand, if u and v are linearly independent in $(\alpha ,\beta )$, then $W(t)$ never vanishes in $(\alpha ,\beta )$

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
$$\begin{gathered} x^{\prime }=f(t,x,y) \\ {y}^{\prime }=g(t,x,y), \end{gathered}$$
where $f$ and $g$ are functions of the independent variable $t$ and the two unknowns $x$ and $y$.

二阶方程$y^{\prime \prime }=F(t,y,y^{\prime })$可以写成以下一阶系统：
$$\begin{gathered} y^{\prime }=v \\ {v}^{\prime }=F(t,y,v) \end{gathered}$$
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 $y^{\prime \prime }+p{y}^{\prime }+qy=0$ is
$${\lambda }^2+p\lambda +q=0.$$
The polynomial ${\lambda }^2+p\lambda +q$ is called the characteristic polynomial for the equation. The root(s) of the characteristic equation is called characteristic root.

There are three cases:

• $p^2-4q>0$, the characteristic equation has two distinct, real roots ${\lambda }_1$ and ${\lambda }_2$. A fundamental set of solutions is
  $$y_1(t)=e^{{\lambda }_{1}t}and{y}_2(t)=e^{{\lambda }_{2}t}$$

• $p^2-4q=0$, one repeated real root $\lambda$. Fundamental set of solutions:
  $$y_1(t)=e^{\lambda t}and{y}_2(t)=t{e}^{\lambda t}$$

• $p^2-4q<0$, two complex conjugate roots $a\pm ib$. Fundamental set of solutions:
  $$y_1(t)=e^{at}\cos (bt)and{y}_2(t)=e^{at}\sin (bt)$$

Inhomogeneous Equations

$$y^{\prime \prime }+p{y}^{\prime }+qy=f$$

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

General solution to inhomogeneous equation

Suppose that $y_p$ is a particular solution to the inhomogeneous equation, and that $y_1$ and $y_2$ form a fundamental set of solutions to the associated homogeneous equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$.

Then the general solution to the inhomogeneous equation is given by
$$y=y_p+C_1{y}_1+C_2{y}_2,$$
where $C_1$ and $C_2$ are arbitrary constants.

The general solution can also be written as
$$y=y_p+y_h$$
where $y_h=C_1{y}_1+C_2{y}_2$

The Method of Undetermined Coefficients

$$y^{\prime \prime }+p{y}^{\prime }+qy=f$$

where $p$ and $q$ are constants.

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

特殊情况

$f(t)$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$a{e}^{rt}$，带入方程后造成两边不相等，那就尝试$at{e}^{rt}$，如果还不行，再试$a{t}^2{e}^{rt}$。

The Method of Variation of Parameters

$$y^{\prime \prime }+p{y}^{\prime }+qy=f$$

1. Find a fundamental set of solutions $y_1$, $y_2$ to the associated homogeneous equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$

2. Form $y_p=v_1{y}_1+v_2{y}_2$, where $v_1$ and $v_2$ are functions to be determined.

3. Find $v_1$ and $v_2$ by solving the equations and integrating:
   $$\begin{gathered} v_1^{\prime }{y}_1+v_2^{\prime }{y}_2=0 \\ {v}_1^{\prime }{y}_1^{\prime }+v_2^{\prime }{y}_2^{\prime }=f(t) \end{gathered}$$

4. Substitute $v_1$ and $v_2$ into $y_p=v_1{y}_1+v_2{y}_2$

Harmonic Motion

The equation for the motion of a vibrating spring is
$$m{y}^{\prime \prime }+\mu y^{\prime }+ky=F(t),$$
where $m$ is mass, $\mu$ is damping constant, and $k$ is spring constant, F(t) is the external force.

Rewrite the equation as
$$\frac{d^{2}y}{d{t}^2}+\frac{\mu }{m}\frac{dy}{dt}+\frac{k}{m}y=\frac{1}{m}F(t),$$
and make $c=\frac{\mu }{2m}$, ${\omega }_0=\sqrt{\frac{k}{m}}$, $f(t)=\frac{F(t)}{m}$, and $x=y$, we get the equation
$$x^{\prime \prime }+2c{x}^{\prime }+{\omega }_0^{2}x=f(t).$$
We refer to this equation as the equation for harmonic motion. $c$ is called damping constant, and $f$ is the forcing term.

Unforced Harmonic Motion

$$x^{\prime \prime }+2c{x}^{\prime }+{\omega }_0^{2}x=0$$

where $c\ge 0$ and ${\omega }_0>0$ are constants.

Simple harmonic motion

When there is no damping, that is, $c=0$
$$x^{\prime \prime }+{\omega }_0^{2}x=0.$$
The general solution is
$$x(t)=a\cos ({\omega }_{0}t)+b\sin ({\omega }_{0}t),$$
where a and b are constants.

${\omega }_0$ is called the natural frequency.

Periodicity
$$T=\frac{2\pi }{{\omega }_0}$$

The general solution to simple harmonic motion can be written as
$$x(t)=A\cos ({\omega }_{0}t-\varphi )$$
where A is the amplitude, $\varphi$ is the phase
$$\begin{gathered} A=\sqrt{a^2+b^2} \\ \tan \varphi =\frac{b}{a} \end{gathered}$$
To solve for $\varphi$, we take $arctan(\frac{b}{a})$, but it takes value between $-\pi /2$ and $\pi /2$, while $\varphi$ can be from $-\pi$ to $\pi$. Therefore, we take
$$\varphi =\{\begin{array}{cc}\arctan (b/a),& ifa>0;\\ \arctan (b/a)+\pi ,& ifa<0andb>0;\\ \arctan (b/a)-\pi ,& ifa<0andb<0.\end{array}$$

Damped Harmonic Motion

Now $c>0$,
$$x^{\prime \prime }+2c{x}^{\prime }+{\omega }_0^{2}x=0.$$
The characteristic equation
$${\lambda }^2+2c\lambda +{\omega }_0^2=0$$
has roots
$${\lambda }_1=-c-\sqrt{c^2-{\omega }_0^2}and{\lambda }_2=-c+\sqrt{c^2-{\omega }_0^2}.$$
There are three cases for the sign of $c^2-{\omega }_0^2$, and thus for the general solution:

1. Underdamped ($c^2-{\omega }_0^2<0$, that is, $c<{\omega }_0$)
   $$x(t)=e^{-ct}[C_1\cos (\omega t)+C_2\sin (\omega t)],$$
   where $\omega =\sqrt{{\omega }_0^2-c^2}$

2. Overdamped ($c^2-{\omega }_0^2>0$, that is, $c>{\omega }_0$)
   $$x(t)=C_1{e}^{{\lambda }_{1}t}+C_2{e}^{{\lambda }_{2}t}$$
   where ${\lambda }_1<{\lambda }_2<0$

3. Critically damped ($c^2-{\omega }_0^2=0$, that is, $c={\omega }_0$, $\lambda =-c$)
   $$x(t)=C_1{e}^{-ct}+C_{2}t{e}^{-ct}$$

Forced Harmonic Motion

$$x^{\prime \prime }+2c{x}^{\prime }+{\omega }_0^{2}x=A\cos (\omega t)$$

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

Forced Undamped Harmonic Motion

$$x^{\prime \prime }+{\omega }_0^{2}x=A\cos (\omega t)$$

The associated homogeneous equation is
$$x^{\prime \prime }+{\omega }_0^{2}x=0$$
with general solution
$$x_h=C_1\cos ({\omega }_{0}t)+C_2\sin ({\omega }_{0}t)$$
There are two cases for the driving frequency $\omega$:

1. $\omega \ne {\omega }_0$, that is, the driving frequency is not equal to the natural freq

   The equation becomes $x^{\prime \prime }+{\omega }_0^{2}x=A\cos ({\omega }_{0}t)$

   Using undetermined coefficient method, $x_p=a\cos (\omega t)+b\sin (\omega t)$

   We get
   $$a=\frac{A}{{\omega }_0^2-{\omega }^2}andb=0.$$
   The particular solution is
   $$x_p(t)=\frac{A}{{\omega }_0^2-\omega }\cos (\omega t)$$

2. $\omega ={\omega }_0$

   In this case, $x_p=acos({\omega }_{0}t)+bsin({\omega }_{0}t)$ is the solution of homogeneous equation $x_h=C_{1}cos({\omega }_{0}t)+C_{2}sin({\omega }_{0}t)$. Therefore, we look for a particular solution
   $$x_p=t(a\cos ({\omega }_{0}t)+b\sin ({\omega }_{0}t)).$$
   We get
   $$b=\frac{A}{2{\omega }_0}anda=0.$$
   The particular solution is
   $$x_p=\frac{A}{2{\omega }_0}t\sin ({\omega }_{0}t).$$
   As $t$ increases, $x_p$ grows larger.

   This case is called resonance.

Forced Damped Harmonic Motion

$$x^{\prime \prime }+2c{x}^{\prime }+{\omega }_0^{2}x=A\cos (\omega t)$$

The associated homogeneous equation:
$$x^{\prime \prime }+2c{x}^{\prime }+{\omega }_0^{2}x=0$$
The characteristic roots are
$$\lambda =-c\pm \sqrt{c^2-{\omega }_0^2}$$
和unforced的时候一样，有三种情况。但是homogeneous equation的通解$x_h$是一样的，不一样的是特解$x_p$.

对于Underdamped情况，即$c<{\omega }_0$，
$$x(t)=e^{-ct}[C_1\cos (\eta t)+C_2\sin (\eta t)],$$
where
$$\eta =\sqrt{{\omega }_0^2-c^2}.$$
就这里与之前unforced不一样，之前用的是$\omega$表示$\eta$，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$\eta$。

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

$$\begin{gathered} R\cos \varphi ={\omega }_0^2-{\omega }^{2}andR\sin \varphi =2c\omega \\ R=\sqrt{({\omega }_0^2-\omega {)}^2+4c^2{\omega }^2} \end{gathered}$$

$$\cos \varphi =\frac{{\omega }_0^2-{\omega }^2}{\sqrt{({\omega }_0^2-{\omega }^2{)}^2+4c^2{\omega }^2}}\sin \varphi =\frac{2c\omega }{\sqrt{({\omega }_0^2-{\omega }^2{)}^2+4c^2{\omega }^2}}$$

$$cot\varphi =\frac{{\omega }_0^2-{\omega }^2}{2c\omega }$$