Chapter 8 - Ordinary Differential Equations

Reference "Mathematical Methods in the Physical Sciences" 3e by Mary L. Boas.

Differential equations are equations that contain both a function and it's derivative. $$m\frac{d^{2}r}{d{t}^2}=-kr$$ Partial differential equations contain partial derivatives: $$x\frac{\partial y}{\partial x}+z\frac{\partial y}{\partial z}=y$$ If there are no partial derivatives, it's an ordinary differential equation. Some properties:

• Order is the highest derivative in the equation (i.e. $y^{\prime \prime }(x)\equiv \frac{d^{2}y}{d{x}^2}$ is second-order).
• Linear equations will take the form $$a_{0}y+a_1{y}^{\prime }+a_2{y}^{\prime \prime }+\cdots =b$$ where $a_n$ and $b$ are constants or functions of the independent variable ($x$), but not of the dependent variable ($y$). $y^{\prime 2}$ is nonlinear.
• Solutions usually mean to solve for the dependent variable: i.e. $y(x)$.

A linear differential equation of order $n$ will have a solution (the general solution) containing $n$ independent constants $a_n$, from which all solutions of the diffeq. can be found by changing the values of the independent constants $a_n$.

The values of these constants can be found using boundary conditions or initial conditions (at $t=0$).

Separable Equations

Separable differential equations are those for which we can separate the variables, putting the dependent variable on one side, independent on the other: $$\begin{array}{rl}\frac{dy}{dx}& =g(y)f(x)\\ \frac{dy}{g(y)}& =f(x)dx\\ \int \frac{1}{g(y)}dy& =\int f(x)dx\end{array}$$ The general solution will often have some form $$y=Cf(x)$$ which represents a family of solutions (by varying $C$). A particular solution is therefore some single value of $C$.

Nonlinear equations

Separable equations may also be nonlinear, such as $$y^{\prime }=\sqrt{1-y^2}$$ Solving it as a separable equation yields $$y=\sin (x+a)$$ which allows for only non-negative solutions for $y$, and is not valid for $y=\pm 1$ ($\int \frac{1}{\sqrt{1-y^2}}dy$ for $y=\pm 1$ is undefined).

First-Order Equations & Int. Factor

Linear first-order ODEs take the form $$y^{\prime }+Py=Q$$ and are most easily solved via an integrating factor.

1. Let $I=\int Pdx$
2. Write the equation as $$y{e}^I=\int Q{e}^{I}dx+C$$
3. Integrate and solve for $y$.

Nonlinear First-Order Equations

There are other methods to solve specific (but not all) first order equations.

Bernoulli equation: for equations of the form $$y^{\prime }+Py=Q{y}^n$$

1. Let: $$z=y^{1-n}{z}^{\prime }=(1-n)y^{-n}{y}^{\prime }$$
2. Multiply both sides by $(1-n)y^{-n}$: $$(1-n)y^{-n}{y}^{\prime }+(1-n)P{y}^{1-n}=(1-n)Q$$
3. Substitute in $z$ and $z^{\prime }$: $$z^{\prime }+(1-n)Pz=(1-n)Q$$
4. Solve for $z$ with integrating factor.
5. Substitute (again) $z=y^{1-n}$ and find $y$.

Exact equation: exact differentials (or "perfect" differentials) of some function $F(x,y)$ (even an unknown one) have the property $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$ Integrals over them are path-independent.

1. Let some function $F(x,y)$ exist, such that $$P=\frac{\partial F}{\partial x}Q=\frac{\partial F}{\partial y},Pdx+Qdy=dF$$
2. By the definition of exact differentials, $Pdx+Qdy=dF=0$, and $F(x,y)=\text{const}.$

TODO: Review chapter 6, section 8 for more information on finding $F$.

Homogenous equation: a homogenous function of $x$ and $y$ of degree $n$ is one that can be written $$x^{n}f(y/x)$$ i.e. $x^3-x{y}^2=x^3[1-(y/x{)}^2]$ is homogenous of degree 3.

The equation $$P(x,y)+Q(x,y)dy=0$$ where $P$ and $Q$ are same-degree homogenous functions, is also homogenous. Dividing them cancels the $x^n$ terms as in the definition.

1. Write $$y^{\prime }=\frac{dy}{dx}=-\frac{P(x,y)}{Q(x,y)}=f\left(\frac{y}{x}\right)$$
2. Make the change of variable $v=y/x$, such that $$y=xv$$
3. Solve the resulting separable equation in $x$ and $v$.
4. Reverse-substitute $v$ to solve for $y$.

Constant-Coefficient SOLDEs (Homogenous)

Second-order linear differential equations (or SOLDEs) with constant coefficients tend to look like $$a_2{y}^{\prime \prime }+a_1{y}^{\prime }+a_{0}y=f(x)$$ and are homogenous if $f(x)=0$. They are solved using the differential operator $D^n=\frac{d^n}{d{x}^n}$, such that $$a_2{y}^{\prime \prime }+a_1{y}^{\prime }+a_{0}y\equiv y(a_2{D}^2+a_{1}D+a_0)=0$$ Solve for the roots $r=D$ as a second-order polynomial. $$r=\frac{-b\pm \sqrt{b^2-4(a)(a)}}{2a}$$ Solutions vary based on multiplicity and complexity.

1. If the roots are real and distinct ($r=(r_1,r_2)$): $$y=c_1{e}^{r_{1}t}+c_2{e}^{r_{2}t}$$
2. If the roots are real and have multiplicity of 2 (i.e. $r_1=r_2$): $$y=c_1{e}^{rt}+c_{2}t{e}^{rt}$$
3. If the roots are imaginary ($r=\alpha \pm i\beta$): $$\begin{array}{rl}y& =e^{\alpha t}(c_1{e}^{i\beta t}+c_2{e}^{-i\beta t})\\ & =e^{\alpha t}(c_1\sin (\beta t)+c_2\cos (\beta t))\\ & =c{e}^{\alpha t}\sin (\beta t+\gamma )\end{array}$$ (I tend to use the second solution (5) - it's easiest and simplest).

Oscillators

Hooke's law + Newton's 2nd say that, for some simple harmonic oscillator, $$m\frac{d^{2}y}{d{t}^2}=-ky-l{y}^{\prime }$$ where $-k$ is the spring constant and $l$ is the "retarding force".

Let $\omega =\frac{k}{m}$ and $2b=\frac{l}{m}$. This turns into $$\frac{d^{2}y}{d{t}^2}+2b\frac{dy}{dt}+{\omega }^{2}y=0$$ The roots to the differential operator are $-b\pm \sqrt{b^2-{\omega }^2}$, with three scenarios for answers: $$\{\begin{array}{ll}\text{overdamped}& b^2>{\omega }^2\\ \text{critically damped}& b^2={\omega }^2\\ \text{underdamped}& b^2<{\omega }^2\end{array}$$

Overdamped: $\sqrt{b^2-{\omega }^2}$ is real and has two distinct solutions, so $$y=A{e}^{-(b+\sqrt{b^2-{\omega }^2})t}+B{e}^{-(b-\sqrt{b^2-{\omega }^2})t}$$

Critically damped: $b=\omega$, so real roots with one distinct solution. $$y=(A+Bt)e^{-bt}$$

Underdamped: $b^2<{\omega }^2$, so imaginary roots $$y=e^{-bt}(A\sin (\beta t)+B\cos (\beta t))$$

Method of Undetermined Coefficients

Non-homogenous constant-coefficient second-order linear differential equations $$a_2{y}^{\prime \prime }+a_1{y}^{\prime }+a_{0}y=f(x)$$ The general solution to this equation is the sum of the homogenous and particular solutions: $$y=y_h+y_p$$ The particular solution can be found primarily through the method of undetermined coefficients.

1. Let $y_p$ take the form of one of the solutions below, based on what $g(x)$ looks like.
2. Calculate $y_p^{\prime }$ and $y_p^{\prime \prime }$.
3. Substitute in the guesses into the diffeq, then solve for the constants.

The textbook provides some additional solution options for equations of the form $(D-a)(D-b)y=e^{cx}{P}_n(x)$, where $P_n(x)$ is some polynomial of degree $n$. $$y_p=\{\begin{array}{ll}{e}^{cx}{Q}_n(x)& \text{if c=a∣b}\\ x{e}^{cx}{Q}_n(x)& \text{if c=a∣b and a=b}\\ x^2{e}^{cx}{Q}_n(x)& \text{if c=a=b}\end{array}$$

Note: if there are several terms on the right-hand-size, we can use superposition with each term to find $y_p=y_{p1}+y_{p2}+y_{p3}$.

Other Second-Order Equations

Some other special second-order equations with methods to solve them are shown below.

(a). Missing dependent variable If $y$ is missing from the equation, substitute $y^{\prime }=p$ and $y^{\prime \prime }=p^{\prime }$. Solve for $p(x)$, then reverse-substitute $p=y^{\prime }$ and solve the first order equation for $y$.

(b). Missing independent variable If $x$ is missing, let $$y^{\prime }=p,y^{\prime \prime }=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p\frac{dp}{dy}$$ (c). $y^{\prime \prime }+f(y)=0$. Multiply by $y^{\prime }$, then integrate to get $$\frac{1}{2}{y}^{\prime 2}=\int f(y)dy=C$$ (d). Cauchy/Euler equations if equation has the form $$a_2{x}^2\frac{d^{2}y}{d{x}^2}+a_{1}x\frac{dy}{dx}+a_{0}y=f(x)$$ can be reduced to SOLDE C.C. by substituting $x=e^z$ to get $$a_2\frac{d^{2}y}{d{z}^2}+(a_1-a_2)\frac{dy}{dz}+a_{0}y=f(e^z)$$ (e). Reduction of order to find a second solution of $$y^{\prime \prime }+f(x)y^{\prime }+g(x)y=0$$ given one solution $u(x)$, substitute $$y=u(x)v(x)$$ into the differential equation and solve for $v(x)$.

Laplace Transforms

Laplace transforms are way to solve differential equations - for some function $f(t)$,

$$L(f)={\int }_0^{\infty }f(t)e^{-pt}dt$$ For example, if $f(t)=1$, $L(f)={\int }_0^{\infty }1\ast e^{-pt}dt=\frac{1}{p}$. This can be converted back into the original function by referencing a Laplace transform table.

Paul's Math Notes (of course) has a wonderful table here.

To use it,

1. Apply the Laplace transform to all terms in the differential equation (both left and right).
2. Plug in initial conditions.
3. Group $p$ (or $s$) terms and try to isolate terms on each side (all $p$ terms on one side, $Y=L(y)$ terms on the other).
4. Use table to find inverse Laplace transform of $L(y)$.
   • If there isn't an exact equivalent on the table, try using partial fraction decomposition or other techniques to separate into individual terms, and sum them.

$$\begin{array}{rl}y:& L(y)\\ y^{\prime }:& L(y^{\prime })=pL(y)-y_0\\ y^{\prime \prime }:& L(y^{\prime \prime })=p^{2}L(y)-y_0-y_0^{\prime }\end{array}$$

For higher orders of differential equations, plug $y^{(n)}$ into $L(f)$ and use integration by parts.

Dirac Delta Functions

The Dirac delta function lets us model functions which change rapidly over a very short ($dt\approx 0$) period of time, such as unit impulses (a hammer strike, or the impact of an air particle).

$d(mv)/dt$ at $t=t_1$ is mathematically infinite - not very useful at all. Instead, we can represent this increase by a delta function: $${\int }_a^{b}f(t)\delta (t-t_0)dt=\{\begin{array}{ll}f(t_0),& a<t<b\\ 0,& \text{otherwise}\end{array}$$ This is a generalized function, with the following properties:

• Derivatives of the $\delta$ function turn into derivatives of $f(t)$: $${\int }_{-\infty }^{\infty }f(t){\delta }^{(n)}(x-a)dx=(-1{)}^n{f}^{(n)}(a)$$
• $\delta$ is an operator. Let $u^{\prime }=\delta$. $$\begin{array}{rl}{u}^{\prime }(x-a)& =\delta (x-a)\\ u(x-a)& =\{\begin{array}{l}1,x>a\\ 0,x<a\end{array}\end{array}$$

Some other properties: $$\begin{array}{rl}\delta (-x)& =\delta (x)\\ {\delta }^{\prime }(-x)& =-{\delta }^{\prime }(x)\\ \delta (ax)& =\frac{1}{|a|}\delta (x)\\ \delta [(x-a)(x-b)]& =\frac{1}{|a-b|}[\delta (x-a)+\delta (x-b)]\\ \delta (f(x))& =\sum _i\frac{1}{|f^{\prime }(x_i)|}\delta (x-x_i)\text{if f(xi)=0 and f′(xi)=0}\end{array}$$ $\delta$ can also work in two and three dimensions, and vectors:

$$\begin{array}{rl}& \int \int f(x,y)\delta (x-x_0)\delta (y-y_0)dxdy=f(x_0,y_0)\\ & \int \int \int f(r,\theta ,\varphi )\delta (\vec{r}-\overrightarrow{r_0})d\tau =f(\overrightarrow{r_0})=f(r_0,{\theta }_0,{\varphi }_0)\end{array}$$ Finally, in three dimensions we have two useful operator equations:

$$\begin{array}{r}\vec{\nabla }\cdot \frac{{\vec{e}}_r}{r^2}=4\pi \delta (\vec{r})\\ {\nabla }^2\frac{1}{r}=-4\pi \delta (\vec{r})\end{array}$$

Green's Function

I'm not confident in my knowledge of these - Boas 8.12 has more info on them, but the presented examples seem surprisingly complicated.

As far as I can tell, just as for $A\vec{x}=\vec{b}$ there is some $A^{-1}$ (for invertible matrices) such that $\vec{x}=A^{-1}\vec{b}$, $G(t,t^{\prime })$ is a similar transform for differential equations such that $y(t)=G(t,t^{\prime })f(t)$ - see this Stack Exchange post.

The Wikipedia page on Green's function says that, if $L$ is some differential operator, then

• $G$ is the solution of the equation $LG=\delta$.
• The solution of the IVP $Ly=f$ is the convolution $(G\ast f)$.

From Boas: for some differential equation $y^{\prime \prime }+{\omega }^{2}y=f(t)$, we can represent $f(t)$ as a sum of unit impulses: $$f(t)={\int }_0^{\infty }f(t^{\prime })\delta (t^{\prime }-t)d{t}^{\prime }$$ For instance, if $f(t)$ models the air resistance on some oscillator, a unit impulse might be a single air particle bouncing off the oscillating object.

The differential equation in terms of unit impulses can be written with $G$ being a differential operator (like $D$ or $L$): $$\frac{d^2}{d{t}^2}G(t,t^{\prime })+{\omega }^{2}G(t,t^{\prime })=\delta (t^{\prime }-t)$$ and the solution is $$y(t)={\int }_0^{\infty }G(t,t^{\prime })f(t^{\prime })d{t}^{\prime }$$

The example on the Wikipedia page for Green's function provides (in my belief) a much better explanation than Boas does.