Homework Template - Methods of Applied Mathematics Problem Set 1
Autor:
Qi Lei
Last Updated:
hace 10 años
License:
Creative Commons CC BY 4.0
Resumen:
homework template
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\title{
\normalfont \normalsize
\textsc{university of texas at austin} \\ [25pt] % Your university, school and/or department name(s)
\horrule{0.5pt} \\[0.4cm] % Thin top horizontal rule
\huge Methods of Applied Mathematics Problem Set 1 \\ % The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
}
\author{Qi Lei} % Your name
\date{\normalsize\today} % Today's date or a custom date
\begin{document}
\maketitle % Print the title
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% PROBLEM 1
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\section{Exercise 6.1}
Compute he Fourier transform of $e^{-|x|}$ for $x\in \mathbb{R}$.\\
Solution: \begin{eqnarray*}
\hat{f}(\xi)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-|x|}e^{-ix\xi}dx\\
&=&\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-x-ix\xi}dx+\int_{-\infty}^0e^{x-ix\xi}dx\\
&=&\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}(e^{-x-ix\xi}-e^{-x+ix\xi})dx\\
&=&\frac{1}{\sqrt{2\pi}}[\frac{1}{-(1+i\xi)}(-1)-\frac{1}{-1+i\xi}(-1)]\\
&=&\frac{1}{\sqrt{2\pi}}[\frac{1-i\xi}{1+\xi^2}+\frac{-(1+i\xi)}{1+\xi^2}]\\
&=&\frac{1}{\sqrt{2\pi}}\frac{-2i\xi}{1+\xi^2}\\
&=&-\sqrt{\frac{2}{\pi}}\frac{i\xi}{1+\xi^2}
\end{eqnarray*}
\\
\section{Exercise 6.2}
Compute the Fourier transform of $e^{-a|x|^2},~a>0$, directly, where $x\in \mathbb{R}$.\\
Solution:
\begin{eqnarray*}
\hat{f}(\xi)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a|x|^2}e^{-ix\xi}dx\\
&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a(x+\frac{i\xi}{2a})^2+\frac{-\xi^2}{4a}}dx~~~~~~~~x'\doteq x+\frac{i\xi}{2a}\\
&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{\xi^2}{4a}}\int_{-\infty}^{\infty}e^{-ax^2}dx\\
&=&\frac{e^{-\frac{\xi^2}{4a}}}{2a}
\end{eqnarray*}
\\
\section{Exercise 6.4}
$f\in L_1(\mathbb{R}^d),$ and $f(x)=g(|x|)$ for some $g$, show that $\tilde{f}(\xi)=h(|\xi|)$ for some $h$.
Solution:
\begin{eqnarray}
\nonumber
\hat{f}(\xi)&=&\frac{1}{(2\pi)^{d/2}}\int_{\mathbb{R}^d}f(x)e^{-ix\xi}dx~~~~~~~~~~~~~(\text{polar coordinate transformation})\\
\nonumber
&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}\int_{\partial S^d}f(r\alpha)e^{-ir\alpha\cdot\xi} r^{d-1}d\alpha dr\\
\label{1.1}
&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}g(r)r^{d-1}\int_{\partial S^d}e^{-ir\alpha\cdot\xi}d\alpha dr
\end{eqnarray}
For any $\xi$, $\exists$ rotation $Q$, such that $\xi=|\xi|Qe_1$, so that $\alpha\cdot \xi=\alpha^T\xi=\alpha^T(Qe_1)|\xi|=|\xi|(Q^T\alpha)^Te_1$\\
\begin{eqnarray*}
\therefore~~(\ref{1.1})&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}g(r)r^{d-1}\int_{\partial S^d}e^{-ir(Q^T\alpha)\cdot e_1|\xi|}d\alpha dr\\
&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}g(r)r^{d-1}\int_{\partial S^d}e^{-ir(Q^T\alpha)\cdot e_1|\xi|}d(Q^T\alpha) dr~~~~\beta\doteq Q^T\alpha\\
&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}g(r)r^{d-1}\int_{\partial S^d}e^{-ir\beta_1|\xi|}d\beta dr\\
&=&h(|\xi|)
\end{eqnarray*}
only depends on $|\xi|$\\
\section{Exercise 6.10}
Let the field be complex and define $T: L^2(\mathbb{R}^d)\rightarrow L^2(\mathbb{R}^d)$ by
\begin{equation*}
Tf(x)=\int e^{-|x-y|^2/2}f(y)dy
\end{equation*}
Show that $T$ is positive, injective operator, but not surjective.\\
Solution:
\begin{eqnarray*}
\text{Positive}&\Longleftrightarrow& \langle Tf(x),f(x)\rangle\geq 0, \forall f\in L^2(\mathbb{R}^d)\\
&\Longleftrightarrow& \int \int e^{-|x-y|^2/2}f(y)dy f(x)dx\geq 0,\forall f\in L^2(\mathbb{R}^d)\\
\end{eqnarray*}
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\end{document}