% vim: tw=50 % 11/11/2022 10AM \subsubsection*{Eigenfunction expansion of $G(x, \xi)$} Suppose $\mathcal{L}$ is in Sturm Liouville form (2.7) with eigenfunctions $y_n(x)$ and eigenvalues $\lambda_n$, then seek \[ G(x, \xi) = \sum_{n = 1}^\infty A_n y_n(x) \] satisfying $\mathcal{L} G = \delta(x - \xi)$. \begin{align*} \mathcal{L} G &= \sum_n A_n \mathcal{L} y_n(x) \\ &= \sum_n A_n \lambda_n \omega(x) y_n(x) \\ &= \delta(x - \xi) \\ &= \omega(x) \sum_n \frac{y_n(\xi) y_n(x)}{\mathcal{N}_n} \end{align*} with $\mathcal{N}_n = \int \omega y_n^2 \dd x$. So $A_n(\xi) = \frac{y_n(\xi)}{\lambda_n \mathcal{N}_n}$ by orthogonality (2.13). Thus \[ \boxed{G(x, \xi) = \sum_{n = 1}^\infty \frac{y_n(\xi) y_n(x)}{\lambda_n \mathcal{N}_n} = \sum_{n = 1}^\infty \frac{Y_n(\xi) Y_n(x)}{\lambda_n}} \tag{7.23} \] which obtained without $\delta(x - \xi)$ in (2.31): refer to section 2.6 Sturm Liouville theory. \subsection{Constructing $G(t, \tau)$: Initial Value Problem} Solve $\mathcal{L} y = f(t)$ for $t \ge a$ with $y(a) = y'(a) = 0$ using $G(t, c)$ satisfying $\mathcal{L} G = \delta(t - \tau)$ with same boundary conditions. \begin{itemize} \item For $t < \tau$, $G_1 = A y_1(t) + B y_2(t)$ with $W(a) \neq a$, $y_1, y_2$ independent. \[ A y_1(a) + B y_2(a) = 0\] and \[ A y_1'(a) + B y_2'(a) = 0 \] implies \[ y_1 y_2' - y_2 y_1' = 0 \] unless $A = B = 0$. So $G_1(t, \tau) \equiv 0$, $a \le t < \tau$, i.e. no change until impulse at $t = \tau$. \item For $t > \tau$, by $G$ continuity (7.12), $G_2(\tau, \tau) = 0$ so choose $G_2 = D y_+(t)$, $y_+(t) = A y_1(t) + B y_2(t)$ such that $y_+(\tau) = 0$. \end{itemize} But by discontinuity in $G'$ (7.13): \[ [G']_{\tau_-}^{\tau_+} = G_2'(\tau, \tau) - G_1'(\tau, \tau) = D y_+'(\tau) = \frac{1}{\alpha(\tau)} \] i.e. $A y_1'(\tau) + B y_2'(\tau) = \frac{1}{\alpha(\tau)}$ so \[ D(\tau) = \frac{1}{\alpha(\tau) y_+'(\tau)} \] or solve for $A, B$. Hence, we have \[ \boxed{G(t, \tau) = \begin{cases} 0 & t < c \\ \frac{y_+(t)}{\alpha(\tau) y_+'(\tau)} & t > \tau \end{cases}} \tag{7.25} \] \begin{center} \includegraphics[width=0.6\linewidth] {images/137c992a61ac11ed.png} \end{center} The initial value problem is \begin{align*} y(t) &= \int_a^t G(t, \tau) f(\tau) \dd \tau \\ &= \int_a^t \frac{y_+(t) f(\tau)}{\alpha(\tau) y_+'(\tau)} \dd \tau \end{align*} Causality is ``built in'', as only forces acting prior to $t$ affect the solution at $t$. \begin{example*} Solve \[ y'' - y = f(t) \] with $y(0) = y'(0) = 0$. \begin{enumerate} \item[1\&2] Homogeneous solutions and initial conditions \begin{itemize} \item $t < \tau$, $G_1 = 0$, \item $t > \tau$, $G_2 = A e^t + B e^{-t}$ \end{itemize} \item[3] Continuity implies $G_2(\tau, \tau) = 0 \implies G_2 = D \sinh(t - \tau)$ \item $[G'] = \frac{1}{\alpha} = 1 \implies G_2'(\tau, \tau) = D \cosh(0) = D = 1$. \end{enumerate} Hence, solution (7.26) is \[ y(t) = \int_0^t f(\tau) \sinh(t - \tau) \dd t \] \end{example*} \newpage \section{Fourier Transforms} \subsection{Introduction} \begin{flashcard}[FourierTransform] \begin{definition*} The \emph{Fourier transform} (FT) of a function $f(x)$ is \begin{align*} \tilde{f}(k) &= \mathcal{F}(f)(k) \\ &= \cloze{\int_{-\infty}^\infty f(x) e^{-ikx} \dd x} \tag{8.1} \end{align*} \fcscrap{and the \emph{inverse Fourier transform} is \begin{align*} f(x) &= \mathcal{F}^{-1}(\tilde{f})(x) \\ &= \frac{1}{2\pi} \int_{-\infty}^\infty \tilde{f}(k) e^{ikx} \dd k \tag{8.2} \end{align*} Beware there are several conventions.} \end{definition*} \end{flashcard} \begin{hiddenflashcard}[InvFourierTransform] \begin{definition*} The \emph{inverse Fourier transform} is \begin{align*} f(x) &= \mathcal{F}^{-1}(\tilde{f})(x) \\ &= \cloze{\frac{1}{2\pi} \int_{-\infty}^\infty \tilde{f}(k) e^{ikx} \dd k} \end{align*} \end{definition*} \end{hiddenflashcard} \noindent The \emph{Fourier inversion theorem} states that \[ \mathcal{F}^{-1}(\mathcal{F}(f))(x) = f(x) \tag{8.3} \] with a sufficient condition that $f$ and $\tilde{f}$ are absolutely integrable. That is, \[ \int_{-\infty}^\infty |f(x)| \dd x = M < \infty \] so $f \to 0$ as $x \to \pm \infty$. \subsubsection*{Gaussian example} Find the Fourier transform of \[ f(x) = \frac{1}{\sigma \sqrt{\pi}} e^{-x^2/\sigma^2} \tag{8.4} \] \begin{align*} \tilde{f}(k) &= \frac{1}{\sigma \sqrt{\pi}} \int_{-\infty}^\infty e^{-x^2/\sigma^2} e^{-ikx} \dd x \\ &= \frac{1}{\sigma \sqrt{\pi}} \int_{-\infty}^\infty e^{-x^2/\sigma^2} \cos kx \dd x \end{align*} Consider $\dfrac{\tilde{f}}{k}$: \begin{align*} \tilde{f}'(k) &= -\frac{1}{\sigma\sqrt{\pi}}\int_{-\infty}^\infty x e^{-x^2/\sigma^2} \sin kx \dd x \\ &= \frac{1}{\sigma \sqrt{\pi}} \left[ \frac{\sigma^2}{2} e^{-x^2/\sigma^2} \sin kx \right]_{-\infty}^\infty - \frac{1}{\sigma \sqrt{\pi}} \int_{-\infty}^\infty \left( \frac{k\sigma^2}{2} \right) e^{-x^2/\sigma^2} \cos kx \dd x \\ &= -\frac{k\sigma^2}{2} \tilde{f}(k) \end{align*} Integrate $\frac{\tilde{f}'}{\tilde{f}} = -\frac{k\sigma^2}{2}$ to find \[ \tilde{f}(k) = C e^{-k^2 \sigma^2/4} \] But put $k = 0$ into (8.4), $\tilde{f}(0) = 1 \implies C = 1$ \[ \boxed{\tilde{f}(k) = e^{-k^2 \sigma^2/4}} \tag{8.5} \] Exercise: Show that $\mathcal{F}^{-1}(e^{-k^2 \sigma^2/4}) = f(x)$. \subsubsection*{Exponential exercise:} Show that $f(x) = e^{-a|x|}$, $a > 0$ has Fourier Transform \[ \tilde{f}(k) = \frac{2a}{a^2 + k^2} \tag{8.6} \] in two ways: \begin{enumerate}[(i)] \item Integrate $2\int_0^\infty e^{-ax} \cos kx \dd x$ by parts twice. \item Integrate $\int_0^\infty e^{-(a - ik)x} \dd x + \int_{-\infty}^0 e^{(a + ik)x} \dd x$ directly. \end{enumerate} Note if \[ f(x) = \begin{cases} e^{-ax} & x > 0 \\ 0 & x \le 0 \end{cases} \] ($a > 0$) then \[ \boxed{\tilde{f}(k) = \frac{1}{ik + a}} \tag{8.6a} \] \begin{hiddenflashcard}[inverse-FT-of-1-over-ik-plus-a] If $\tilde{f}(x) = \frac{1}{ik + a}$ then \[ f(x) = \cloze{ \begin{cases} e^{-ax} & x > 0 \\ 0 & x \le 0 \end{cases}} \] \end{hiddenflashcard} \subsection{Fourier Transform relation to Fourier series} We can write Fourier series (1.13) as \[ f(x) = \sum_{n = -\infty}^\infty c_n e^{ik_n x} \tag{$*$} \] where $k_n = \frac{n\pi}{L}$, so write $k_n = n\Delta k$ with $\Delta k = \frac{\pi}{L}$, then \begin{align*} c_n &= \frac{1}{2L} \int_{-L}^L f(x) e^{-ik_n x} \dd x \\ &= \frac{\Delta k}{2\pi} \int_{-L}^L f(x) e^{-ik_n x} \dd x \end{align*}