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## de.tex \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} ## de.sty \mathtoolsset{showonlyrefs,showmanualtags} %%% Sections, subsections, and header/footers styles. % Stolen from memman.pdf. \newcommand{\ruledsec}[1]{ \Large\bfseries\scshape\centering\MakeLowercase{#1} } \setsecheadstyle{\ruledsec} \setsubsecheadstyle{\normalfont\large\itshape\raggedright} % Number subsections. \setsecnumdepth{subsection} % No chapter numbers. \renewcommand{\thesection}{\arabic{section}} % Slap a page number in the right place. \makepagestyle{sexy} \makeoddfoot{sexy}{}{}{\thepage} %%% Stop self from printing drafts. \usepackage{ifdraft} \ifdraft{ \makeoddhead{sexy}{DRAFT}{DRAFT}{DRAFT} }{ \makeoddhead{sexy}{}{}{Hao Lian, programmer-poet.} } \pagestyle{sexy} %%% Math stylings. \DeclareMathOperator{\poly}{Poly} \newcommand{\lap}[1]{\mathcal{L}\!\{#1\}} \newcommand{\ilap}[1]{\mathcal{L}^{-1}\!\{#1\}} \renewcommand{\vec}{\mathbf} % Non-asterisk align thanks to `showonlyrefs` from mathools. \newenvironment{al}{ \setkeys{EmphEqEnv}{align} \EmphEqMainEnv }{\endEmphEqMainEnv} \theoremstyle{plain}% default \newtheorem{thm}{Theorem}[section] \newtheorem{lem}[thm]{Lemma} \theoremstyle{definition} \newtheorem{defn}{Definition}[section]

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