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[Analysis] LDE
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\section{Differential Equations}
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\input{parts/01_diffeq.tex}
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\newpage
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\section{Differential Calculus in $\R^n$}
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\input{parts/02_diff.tex}
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\end{document}
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@@ -26,12 +26,8 @@ DE s.t. $f: I^d \to \R$ is in multiple variables.
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\subsection{Linear Differential Equations}
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\definition \textbf{Linear Differential Equation} (LDE)\\
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$$
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y^{(k)} + a_{k-1}y^{(k-1)} + \ldots + a_1y' + a_0y = b
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$$
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\subtext{
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$I \subset \R$ is open$,\quad k \geq 1,\quad \forall i < k: a_i: I \to \C$
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}
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$$y^{(k)} + a_{k-1}y^{(k-1)} + \ldots + a_1y' + a_0y = b$$
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\subtext{$I \subset \R$ is open$,\quad k \geq 1,\quad \forall i < k: a_i: I \to \C$}
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\definition Homogeneity of LDEs\\
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\begin{tabular}{ll}
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@@ -40,12 +36,8 @@ $$
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\end{tabular}
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\remark $D(y) := y^{(k)} + \ldots + a_0y$ is a linear operation:
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$$
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D(z_1f_1 + z_2f_2) = z_1D(f_1) + z_2D(f_2)
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$$
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\subtext{
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$\forall z_1,z_2 \in \C,\quad f_1,f_2\ k$-times differentiable:
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}
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$$D(z_1f_1 + z_2f_2) = z_1D(f_1) + z_2D(f_2)$$
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\subtext{$\forall z_1,z_2 \in \C,\quad f_1,f_2\ k$-times differentiable}
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\definition \textbf{Homogeneous Solution Space}\\
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$\S(F) := \{ f: I \to \C \sep f \text{ solves } F, f \text{ is } k \text{-times diff.} \}$
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@@ -94,24 +86,122 @@ If $f_1$ solves $F$ for $b_1$, and $f_2$ for $b_2$: $f_1 + f_2$ solves $b_1 + b_
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Follows from: $D(f_1) + D(f_2) = b_1 + b_2$.
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\newpage
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\subsection{Finding Solutions: First Order}
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\subsection{Linear Solutions: First Order}
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\subtext{ $I \subset \R, \quad a,b: I \to \R$ }
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$$ y' + ay = b $$
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Approach:
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\begin{enumerate}
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\item Hom. Solution: $y' + ay = 0$ using $f_1 = ke^{-A(x)}$\\
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\subtext{Note that $\S$ has $\dim(\S) = 1$, so $f_1 \neq 0$ is a Basis for $\S$}
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\item Part. Solution: $f_0 \in \S_b$ using Variance of Parameters
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\end{enumerate}
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Solutions: $ f_0 + zf_1 \quad \text{ for } z \in \C $
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\begin{subbox}{Explicit Solution for 1st Order LDEs}
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\textbf{Form:}
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$$ y' + ay = b $$
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\textbf{Approach:}
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\begin{enumerate}
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\item Hom. Solution $f_1$ for: $y' + ay = 0$\\
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\subtext{Note that $\S$ has $\dim(\S) = 1$, so $f_1 \neq 0$ is a Basis for $\S$}
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\item Part. Solution $f_0$ for $y' + ay = b$
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\end{enumerate}
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\textbf{Solutions:} $ f_0 + zf_1 \quad \text{ for } z \in \C $
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\begin{subbox}{Explicit Homogeneous Solution}
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\smalltext{$A(x)$ is a primitive of $a$, $f(x_0) = y_0$}
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\begin{align*}
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f(x) &= z \cdot \exp(-A(x)) \\
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f(x) &= y_0 \cdot \exp(A(x_0) - a(x))
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f_1(x) &= z \cdot \exp(-A(x)) \\
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f_1(x) &= y_0 \cdot \exp(A(x_0) - a(x))
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\end{align*}
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\end{subbox}
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Variation of Constants: Treating $z$ as $z(x)$ yields:
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\begin{subbox}{Explicit Inhomogeneous Solution}
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\smalltext{$A(x)$ is a primitive of $a$}
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$$
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f_0(x) = \underbrace{\left(\int b(x)\cdot\exp(A(x)) \right)}_{z(x)} \cdot \exp\left(-A(x)\right)
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$$
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\end{subbox}
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\method \textbf{Educated Guess}\\
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Usually, $y$ has a similar form to $b$:
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\begin{tabular}{ll}
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\hline
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$b(x)$ & \text{Guess} \\
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\hline
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$a \cdot e^{\alpha x}$ & $b \cdot e^{\alpha x}$ \\
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$a \cdot \sin(\beta x)$ & $c\sin(\beta x) + d\cos(\beta x)$\\
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$b \cdot \cos(\beta x)$ & $c\sin(\beta x) + d\cos(\beta x)$\\
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$ae^{\alpha x} \cdot \sin(\beta x)$ & $e^{\alpha x}\left(c\sin(\beta x) + d\cos(\beta x)\right)$\\
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$be^{\alpha x} \cdot \cos(\beta x)$ & $e^{\alpha x}\left(c\sin(\beta x) + d\cos(\beta x)\right)$\\
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$P_n(x) \cdot e^{\alpha x}$ & $R_n(x) \cdot e^{\alpha x}$\\
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$P_n(x) \cdot e^{\alpha x}\sin(\beta x)$ & $e^{\alpha x}\left( R_n(x) \sin(\beta x) + S_n(x) \cos(\beta x) \right)$\\
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$P_n(x) \cdot e^{\alpha x}\cos(\beta x)$ & $e^{\alpha x}\left( R_n(x) \sin(\beta x) + S_n(x) \cos(\beta x) \right)$\\
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\hline
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\end{tabular}
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\remark If $\alpha, \beta$ are roots of $P(X)$ with multiplicity $j$, multiply guess with a $P_j(x)$.
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\subsection{Linear Solutions: Constant Coefficients}
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\textbf{Form:}
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$$
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y^{(k)} + a_{k-1}y^{(k-1)} + \ldots + a_1y' + a_0y = b
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$$
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\subtext{Where $a_0, \ldots, a_{k-1} \in \C$ are constants, $b(x)$ is continuous.}
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\subsubsection{Homogeneous Equations}
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The idea is to find a Basis of $\S$:
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\definition \textbf{Characteristic Polynomial} $P(X) = \prod_{i=1}^{k} (X-\alpha_i)$
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\remark The unique roots $\alpha_1,\ldots,\alpha_l$ form a Basis:
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$$
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\text{span}(\S) = \{ x^je^{\alpha_i x} \sep i \leq l,\quad 0 \leq j \leq v_i \}
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$$
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\subtext{$v_1,\ldots,v_k$ are the Multiplicities of $\alpha_1,\ldots,\alpha_k$}
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\remark If $\alpha_j = \beta + \gamma i \in \C$ is a root, $\bar{\alpha_j} = \beta - \gamma i$ is too.\\
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To get a real-valued solution, apply:
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$$
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e^{\alpha_j x} = e^{\beta x}\left( \cos(\gamma x) + i \sin(\gamma x) \right)
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$$
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\begin{subbox}{Explicit Homogeneous Solution}
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\smalltext{Using $\alpha_1,\ldots,\alpha_k$ from $P(X)$ s.t. $\alpha_i \neq \alpha_j$, $z_i \in \C$ arbitrary}
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$$
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f(x) = \prod_{i=1}^{k} z_i \cdot e^{\alpha_i x} \quad\text{with}\quad f^{(j)(x)} = \prod_{i=1}^{k} z_i \cdot \alpha_i^j e^{\alpha_i x}
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$$
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\smalltext{Multiple roots: same scheme, using the basis vectors of $\S$}
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\end{subbox}
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\subtext{Solutions exist $\forall Z = (z_1,\ldots,z_k)$ since that system's $\det(M_Z) \neq 0$.}
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\newpage
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\subsubsection{Inhomogeneous Equations}
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\method \textbf{Undetermined Coefficients}: An educated guess.
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\begin{enumerate}
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\item $b(x) = cx^d \cdot e^{\alpha x} \implies f_p(x) = Q(x)e^{\alpha x}$\\
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\subtext{$\deg(Q) \leq d + v_\alpha$, where $v_\alpha$ is $\alpha$'s multiplicity in $P(X)$}
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\item $\begin{rcases*}
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b(x) = cx^d \cdot \cos(\alpha x) \\
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b(x) = cx^d \cdot \sin(\alpha x)
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\end{rcases*} f_p = Q_1(x)\cos(\alpha x) + Q_2(x)\sin(\alpha x)$
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\subtext{$\deg(Q_{1,2}) \leq d + v_\alpha$, where $v_\alpha$ is $\alpha$'s multiplicity in $P(X)$}
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\end{enumerate}
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\remark \textbf{Applying Linearity}\\
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If $b(x) = \sum_{i=1}^{n} b_i(x)$, A solution for $b(x)$ is $f(x) = \sum_{i=1}^{n} f_i(x)$\\
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\subtext{Sometimes called \textit{Superposition Principle} in this context}
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\subsection{Other Methods}
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\method \textbf{Change of Variable}\\
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If $f(x)$ is replaced by $h(y) = f(g(y))$, then $h$ is a sol. too.\\
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\subtext{Changes like $h(t) = f(e^t)$ may lead to useful properties.}
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\begin{subbox}{Separation of Variables}
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Form:
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$$
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y' = a(y)\cdot b(x)
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$$
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Solve using:
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$$
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\int \frac{1}{a(y)}\ \text{d}y = \int b(x) \dx + c
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$$
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\end{subbox}
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\subtext{Usually $\int 1/a(y)\ \text{d}y$ can be solved directly for $\ln|a(y)|+c$.}
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@@ -52,6 +52,7 @@
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\def \notation{\colorbox{lightgray}{Notation} }
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\def \remark{\colorbox{lightgray}{Remark} }
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\def \theorem{\colorbox{lightgray}{Th.} }
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\def \method{\colorbox{lightgray}{Method} }
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% For intuiton and less important notes
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\def \subtext#1{
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