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\documentclass[twocolumn, article, oneside]{memoir}
\usepackage[margin={0.7in}]{geometry}
\usepackage[charter]{mathdesign}
\usepackage[overload]{empheq}
\usepackage{microtype, mathtools, amsthm}
\usepackage{de}
\begin{document}
\noindent
This review of an introduction to differential equations course
assumes in various, tiny ways that you've been exposed to a formal
class. Otherwise, you probably should read a textbook first.
\section{First-order linear}
When using the separable method, losing zeroes occurs. For example,
the na\"{i}ve solution to $y' = x^2(y - 13)$ misses $y(x) = 13$.
Given $a_1(x)y'(x) + a_0(x)y(x) = b(x)$, there are two trivial cases:
$a_0(x) = 0$ and $a_0(x) = a_1'(x)$, the latter due the product rule.
We can reduce all such equations to this case. Let $\mu(x)$ such that
$\mu' = \mu P$, implying $\mu = \exp(\int \! P \, dx)$.
\begin{al}
\mu Q &= \mu(y' + Py) = \frac{d(\mu y)}{dx}
\\ \mu y &= \int \! \mu Q \, dx + C.
\end{al}
\begin{defn}[Exact equation]
An \emph{exact equation} on rectangle $R$ is of the form $M(x,y)\,dx +
N(x,y) \, dy = 0$ such that there exists $F$ such that $F_x = M$ and
$F_y = N$.
\end{defn}
\begin{thm}
An equation is exact iff $M_y = N_x$. The solution follows from the
properties of a total derivative:
\begin{al}
M \, dx + N \, dy &= 0
& F_x \, dx + F_y \, dy &= 0
& F(x,y) &= C.
\end{al}
\end{thm}
\section{Second-order linear}
A guess of $\exp(rt)$ for $ay'' + by' + c = 0$, the equation resolves
to $\exp(rt)(ar^2 + br + c) = 0$. With two distinct $r$ values as
solutions, then by superposition $y(t) = c_1\exp(r_1t) +
c_2\exp(r_2t)$.
\begin{thm}[Existence and uniqueness]
Given the universe of reals, there exists a unique solution to
second-order linear equations with initial conditions $y(t_0) = Y_0$
and $y'(t_0) = Y_1$ \emph{valid for all $t \in \mathbb R$}.
\end{thm}
\begin{thm}[Representation]
For $y_1, y_2$ that are lindep on $I$ with $t_0 \in
I$, there exists $c_1, c_2$ such that $c_1y_1 + c_2y_2$ satisfies the
IVP on $I$ with the initial $t_0$ condition.
\end{thm}
The proof follows from knowing $W[y_1, y_2] = 0$ iff the two solutions
are linindep and some non-obvious algebra.
\subsection{Working from the characteristic polynomial roots}
For two identical roots, use a $t\exp(rt)$ term. For
characteristic equations with complex conjugates $\alpha \pm \beta i$,
there exists two: $\exp(\alpha t)(\cos \beta t, \sin \beta t)$. Note
that complex conjugates only occur if $a,b,c \in \mathbb R$.
Otherwise, two non-conjugate complex roots may arise.
\subsection{Non-homogeneous}
\begin{thm}[Existence and uniqueness.]
For $a,b,c,t_0,Y_0,Y_1 \in \mathbb R$ with $y_P$ on $I$ with $t_0 \in
I$ with $y_1,y_2$ lindep, there exists a unique solution to the
non-homogeneous IVP.
\end{thm}
\begin{proof}
We use the EUT for the homogeneous case and superposition principle to
construct the the solution; therefore, it exists. To prove the
solution ($y_g = y_p + c_1y_1 + c_2y_2$) is unique, assume there
exists another solution $\psi$ such that $\psi \neq y_g$. Let $y = y_g
- \psi$. Trivially, $y$ satisfies the IVP where $ay'' + by' + cy = y_g
- \psi = 0$ with initial values $y(0) = 0$ and $y'(0) = 0$. Note that
this is a homogeneous equation. However, $f(t) = 0$ is already a
unique solution to this IVP. By the homogeneous EUT, it must be the
same solution as $y$, implying $y = y_g - \psi = 0$ or $y_g = \psi$, a
contradiction.
\end{proof}
\subsection{Miscellaneous}
Let
\begin{al}
v_1 &= \int \! \frac{-gy_2}{aW[y_1,y_2]} \, dt
& v_2 &= \int \! \frac{gy_1}{aW[y_1,y_2]} \, dt
\end{al}
where $ay'' + by' + c = g$. Then $y_p = v_1y_1 + v_2y_2$. This is the
\emph{method of variations}.
Linear equations always obey the superposition principle,
distinguishing them from non-linear equations. Non-linear equations
also do not guarantee uniqueness of solutions.
\section{Higher-order linear}
We can easily generalize the homogeneous linear equation to higher
powers using the characteristic equation
We can also generalize the method of undetermined coefficients by
finding a linear differential operator $A$ (an annihilator) such that
$A[f](x) = 0$. So suppose we have a linear equation in operator form
$L[y](x) = f(x)$. Then $AL[y](x) = A[f](x) = 0$. We have reduced the
non-homogeneous equation to homogeneous form. Boo-ya.
Constructing an annhilator is like constructing a guess using the
method of undetermined coefficients: Memorize a brief table. If
$x^k\exp(rx)$, then $(D - r)^m$ where $m$ is any integer such that $m
> k$. $\sin \beta x$ or $\cos \beta x$, then $(D^2 + \beta^2)$.
$x^k\exp(rx)\sin \beta x$, then $[(D-r)^2 + \beta^2]^m$. For example,
\begin{al}
0 &= (D-1)^3(D^2+1)(D+1)[y]
\\ y &= (C_1e^x + C_2xe^x + C_3xe^x) + (C_4 \sin x + C_5 \cos x) + C_6e^{-x}.
\end{al}
\section{Matrices}
The following statements are equivalent: (1) $\vec A$ is singular (has
no inverse); (2) $|\vec A| = 0$; (3) $\vec{Ax}$ has non-trivial
solutions, i.e. where $\vec x \neq \vec 0$; and (4) rows of $\vec A$
are lindep.
If $\vec A$ is singular, $\vec{Ax} = \vec 0$ has infinitely many
solutions, all a scalar multiple of some $\vec{x}_0 \neq \vec 0$.
Furthermore, $\vec{Ax} = \vec b$ either has no solutions or infinitely
many of the form $\vec x = \vec x_p + \vec x_h$. Another property,
which is significant in the light that matrix multiplication is not commutative:
\begin{al}
\frac{d}{dt} \vec{AB} &= \vec A \frac{d\vec B}{dt} + \frac{d\vec A}{dt}
\vec B.
\end{al}
\begin{thm}[Existence and uniqueness]
If $\vec A, \vec f$ are continuous on an open interval $I$, $t_0 \in
I$, and $\vec x_0$, then there exists a unique $\vec x(t)$ on $I$ to
the IVP $\vec x'(t) = \vec A \vec x + \vec f$.
\end{thm}
\begin{thm}
Let $\vec x_i$ be linindep solutions to $\vec A \vec x = \vec x'$.
If they are linindep, then the Wronskian is never zero on
$I$.
\end{thm}
\begin{proof}
Suppose for a contradiction that $W(t_0) = 0$ for some $t_0$. $\vec
X(t_0) \vec c = 0$ for some $\vec c \neq \vec 0$ because columns
become linearly dependent when the matrix has a nonzero determinant
per the above theorem. Another solution to $\vec A \vec x = \vec 0$ is
$\vec z(t) = 0$. These solutions are identical on $I$ by the EUT,
implying $\vec X(t) \vec c = 0$ for \emph{all $t$}. This implies the
$\vec x_i$ are lindep, contradicting the initial assumption.
\end{proof}
Two implications: $W(t)$ is always zero or never zero on $I$. And the
set of solutions are linindep iff $W(t)$ is never zero, whence the
representation theorem:
\begin{thm}[Representation]
Let $x_i$ be linindep solutions to the homogeneous system $\vec x' =
\vec A \vec x(t)$ on $\vec I$. Then every solution is in the form
$\vec x(t) = \vec X(t) \vec c$. Denote $\vec X$ as the
\emph{fundamental matrix} for the \emph{fundamental solution set}.
\end{thm}
\subsection{Non-homogeneous}
\begin{thm}[Superposition principle]
Let $L[\vec x] \coloneq \vec x' - \vec A \vec x$. If $\vec x_1, \vec x_2$
are solutions to $L[\vec x] = \vec g_1$ and $L[\vec x] = \vec g_2$,
then $c_1\vec x_1 + c_2\vec x_2$ are solutions to $L[\vec x] = c_1\vec
g_1 + c_2\vec g_2$.
\end{thm}
\begin{thm}[Representation]
Let $\vec x_p$ solve $\vec x' = \vec A\vec x + \vec f$ on $I$ and
$\vec X$ be the fundamental matrix for the homogeneous system $\vec x
= \vec A \vec x$. Then every solution on $I$ is of the form $\vec x =
\vec x_p + \vec X \vec c$ by the superposition principle. Denote this
as the \emph{general solution}.
\end{thm}
\subsection{Eigenapalooza}
For the system $\vec x' = \vec{Ax}$, guess $\vec x = \exp(rt)\vec u$,
implying $r\exp(rt)\vec u = \exp(rt) \vec{Au}$ or $(\vec A - r\vec
I)\vec u = \vec 0$. \emph{Eigenvalues} are numbers $r$ such that that
equation has a nontrivial ($\vec u \neq \vec 0$) solution; the
corresponding $\vec u$ are \emph{eigenvectors}. Nontrivial solutions,
from a previous theorem, occurs iff $|\vec A - r \vec I| = 0$. The
determinant is the \emph{characteristic polynomial}, and this is the
\emph{characteristic equation}.
\begin{thm}
$\exp(r_it) \vec u_i$ is a fundamental set for linearly
dependent $\{ \vec u_i \}$.
\end{thm}
\begin{proof}
The Wronskian over the set of those solutions is $\exp(t\sum r_i)
|\vec U|$ where $\vec U = [\vec u_1, \ldots, \vec u_n]$. The
determinant is never zero because the eigenvectors are linearly
independent; therefore, the Wronskian is never zero for all $t$.
\end{proof}
\begin{thm}
If $r_1, r_2$ are distinct eigenvalues, then $\vec u_1, \vec u_2$ are
linearly independent.
\end{thm}
\begin{proof}
Suppose $\vec u_1 = c \vec u_2$. Then $(r_1 - r_2)\vec u_1 = \vec 0$,
implying for a contradiction that $r_1 = r_2$ because we know $\vec
u_1 \neq \vec 0$.
\end{proof}
As a corollary, $n$ distinct eigenvalues and eigenvectors imply a
fundamental solution set for $\vec x' = \vec{Ax}$.
The conjugates for complex eigenvalues and eigenvectors create two
linearly independent real vector solutions:
\begin{al}
\exp(\alpha t)(\vec a, \vec b) \cdot ((\cos \beta t, -\sin \beta t),
(\sin \beta t, \cos \beta t)).
\end{al}
\subsection{Non-homogeneous}
Method of undetermined coefficients works similarly. However,
variation of parameters yields the equation
\begin{al}
\vec x &= \vec X \vec c + \vec X \int \! \vec X^{-1}\vec f \, dt.
\end{al}
\subsection{Matrix exponential}
\begin{defn}[Matrix exponential]
\begin{align}
\exp(\vec A t) &= \vec I + \vec A t + \vec A^2 \frac{t^2}{2} + \cdots.
\end{align}
\end{defn}
The inverse---$\exp(\vec A t)^{-1}$---is $\exp(-\vec At)$. The derivative
is $\vec A\exp(\vec A t)$ by virtue of differentiating that
pseudo-Taylor series pseudo-polynomial. Therefore, the exponential is
a solution to $\vec{X' = AX}$. Because $\exp(\vec A t)$ is invertible,
the columns are linindep solutions to the system. The general solution
is then $\vec x(t) = \exp(\vec At)\vec c$, and $\exp(\vec At)$ is the
fundamental matrix for the system.
\begin{defn}[Nilpotent]
A matrix $\vec B$ is \emph{nilpotent} iff there exists $k > 0$ such
that $\vec B^k = \vec 0$. In such a case, the exponential has a finite
number of terms.
\end{defn}
If $\vec A - r\vec I$ is nilpotent, then there is a finite expansion:
\begin{al}
\label{eqn:at-decomp}
e^{\vec At} &= e^{rt}e^{(\vec A - r\vec I)t} = e^{rt}\left(\cdots +
\frac{(\vec A - r\vec I)^{n-1}t^{n-1}}{(n-1)!}\right).
\end{al}
\subsection{Generalized eigenvectors}
With repeated eigenvalues, we need a strategy to find additional
eigenvectors.
\emph{Lemma.} If $\vec X, \vec Y$ are fundamental matrices for the
system $\vec{x' = Ax}$, then there exists a constant matrix such that
$\vec{X(t) = Y(t)C}$.
What is the relationship between $\exp(\vec At)$ and a given
fundamental matrix $\vec X$? By the lemma, $\exp(\vec At) = \vec
X(t)\vec C$ for some $\vec C$. Plugging $t = 0$, we find $\vec I =
\vec X(0) \vec C$. Therefore, $\exp(\vec At) = \vec X(t)\vec
X^{-1}(0)$.
So, to obtain the fundamental solution matrix $\vec X$, assume the
requirement that columns must be of the form $\exp(\vec At)\vec u$,
which can be decomposed by \eqref{eqn:at-decomp}. Therefore, we need
$n$ vectors $\vec u$ whose calculations are feasible. To do so, find
$p(r) = |\vec A - r\vec I|$ and therefore the eigenvalues. For each
$r_i$ with multiplicity $m_i$, find $m_i$ linindep generalized
eigenvectors. (A nonzero vector $u$ that satisfies $(\vec A - r\vec
I)^k\vec u = \vec 0$ for some positive $k$ is a generalized
eigenvector.)
Compute $\vec x = \exp(rt )\exp(t(\vec A - r\vec I))$, using the Taylor
expansion for the last exponential. It becomes one of the $n$ linindep
solutions to the system. The expansion will be finite because $(\vec A
- r\vec I)^i\vec u_i = \vec 0$ for some $i \in [1, m_i]$. The hardest
part is finding such a $\vec u_i$ each time.
\section{Laplace transform}
\begin{defn}[Laplace transform]
If $f$ is defined on $[0, \infty)$, then
\begin{al}
\lap{f}(s) = \int_0^\infty \! e^{-st}f(t) \, dt.
\end{al}
\end{defn}
The power lies within its ability (usually) to replace differential
equations with algebriac equations by recursively applying $\lap{x'} =
s\lap{x} - x(0)$.
\begin{defn}
A function is \emph{piecewise continuous} on the interval $[a,b]$ iff
it is continous on that interval except at a finite number of points
at which a jump discontinuity occurs.
\end{defn}
The jump at $f(x) = 1/x$ for $x = 0$ would, for example, count as an
``infinite'' jump and not a jump discontinuity.
\begin{thm}
Roughly, $\lap{f}$ exists for $s > \alpha$ if $f$ does not grow faster
than an exponential function of some positive order $\alpha$ and $f$
is piecewise continuous on $[0, \infty)$.
\end{thm}
The proof follows from expanding out the exponential order test and
then performing an integral comparison test. For $s > 0$: $1$ to
$1/s$, $t^n$ to $n!/s^{n+1}$, $\sin bt$ to $b/(s^2 + b^2)$, and $\cos
bt$ to $s^2/(s^2 + b^2)$. For $s > a$, $e^{at}$ to $1/(s-a)$ and
$e^{at}t^n$ to $n!/(s-a)^{n+1}$. Finally, $f^{(n)}$ to $s^nF(s) -
s^{n-1}f(0) - \cdots - f^{(n-1)}(0)$. Also,
\begin{al}
\lap{e^{at}f(t)}(s) &= F(s-a)
\\ \lap{t^nf(t)}(s) &= (-1)^nF^{(n)}(s).
\end{al}
\subsection{Inverse Laplace}
$\mathcal{L}^{-1}$, like $\mathcal{L}$, is a linear operator, implying
$\ilap{f_1 + f_2} = \ilap{f_1} + \ilap{f_2}$ and $\ilap{cf} =
c\ilap{f}$. To take the inverse, you basically consult the table. For
rational polynomial fractions, partial function decomposition is
needed with a small twist:
\begin{al}
\frac{2s^2 + 10s}{(s^2 - 2s + 5)} &\Rightarrow \frac{A(s-1) +
2B}{(s-1)^2 + 2^2}.
\end{al}
Completing the square is necessary to remove any $cs^n$ terms. Then,
those factors have to go into the numerator so the inverse Laplace
rules for $\sin$ and $\cos$ work correctly.
For $t^1$ coefficients to linear equations, the transform turns these
non-constant equations into constant coefficient first-order linear
equations.
\begin{al}
y'' + 2ty' - 4y &= 1 \text{ with } y(0) = y'(0) = 0
\\ Y' + \left(\frac{3}{s} - \frac{s}{2}\right)Y &= -\frac{1}{2s^2}
\Rightarrow \frac{d(\mu Y)}{ds} = -\frac{s}{2} e^{-s^2/4}
\\ Y(s) &= \frac{1}{s^2} + C\frac{e^{s^2/4}}{s^3}. \label{eqn:t1-soln}
\end{al}
\begin{thm}
If $f(t)$ is piecewise continuous on $[0, \infty)$ and of exponential
order, then $\lim_{s \to \infty} \lap{f}(s) = 0$.
\end{thm}
How to determine $C$? Using the above theorem, we take the limit of solution at
\eqref{eqn:t1-soln} and set it equal to zero, finding easily that $C =
0$. Therefore, with $C$, we can now find $y$. From $Y(s) = 1/s^3$, it
follows that $y(t) = t^2/2$. Magic.
\subsection{Discontinuous and periodic functions}
\begin{defn}[Heaviside step function]
\begin{al}
u(t|t < 0) &= 0
\\u(t|0 < t) &= 1.
\end{al}
\end{defn}
We can then express any piecewise function in terms of the step
function for easier transformation. For example,
\begin{al}
f(t | t < 2) &= 3
\\f(t | 2 < t < 5) &= 1
\\f(t | 5 < t < 8) &= t
\\f(t | 8 < t) &= t^2/10
\\f(t) &= 3 - (1 - 3)u(t-2)+ (t-1)u(t-5)
\\&\quad + (t^2/10 - t)u(t-8).
\end{al}
Properties:
\begin{al}
\lap{u(t-a)}(s) &= e^{-as}/s
\\ \ilap{e^{-as}F(s)}(t) &= f(t-a)u(t-a)
\\ \text{N.B. } \lap{e^{at}f(t)}(s) &= F(s-a).
\end{al}
The solution to a constant-coefficient linear second-order
differential equation with a stepped non-homogeneous expression can
be, magically, a continous function. However, its second derivative is
instead discontinuous.
\end{document}
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