simple resurgent.tex

\documentclass{article}

\usepackage{url}
\usepackage[hmargin=1.5in]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{xcolor}
% convenience aliases
\newcommand{\maps}{\colon}

% symbology
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\laplace}{\mathcal{L}}
\include{command2}

\title{Resurgence of modified Bessel functions of second kind}
\author{Veronica Fantini}

\begin{document}
\maketitle

\section{Simple resurgent functions}

We are in the same assumption of Theorem 3.1. Our aim is to prove that $\hat{\varphi}_{\alpha}(\zeta)$ is a simple resurgent function: we expect it is a consequence of the half derivatives formula. In the proof of Theorem 3.1 we have seen

\begin{align}
&\hat{\varphi}_{\alpha}(\zeta)=\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\hat{I}_{\alpha} \\
&\hat{I}_{\alpha}(\zeta)=\sum_{n\geq 0}a_n^{\alpha}(\zeta-\zeta_{\alpha})^{n-1/2}\,\qquad a_n\defeq 2^{n+1/2}b_{2n}^{\alpha}
\end{align}

\begin{align*}
\hat{\varphi}_{\alpha}(\zeta)&=\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{-1/2}a_0^{\alpha}+(\zeta-\zeta_\alpha)^{1/2}\sum_{n\geq 0}a_{n+1}^{\alpha}(\zeta-\zeta_{\alpha})^{n}\right]\\
&=a_0^{\alpha}\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{-1/2}\right]+\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{1/2}\right]\sum_{n\geq 0}a_{n+1}^{\alpha}(\zeta-\zeta_{\alpha})^{n}\\
&\qquad+(\zeta-\zeta_\alpha)^{1/2}\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[\sum_{n\geq 0}a_{n+1}^{\alpha}(\zeta-\zeta_{\alpha})^{n}\right]\\
&=a_0^{\alpha}\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{-1/2}\right]+\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{1/2}\right]\sum_{n\geq 0}a_{n+1}^{\alpha}(\zeta-\zeta_{\alpha})^{n}\\
&\qquad +(\zeta-\zeta_\alpha)^{1/2}\sum_{n\geq 0}a_{n+1}^{\alpha}\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_{\alpha})^{n}\right]\\
&=a_0^{\alpha}\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{-1/2}\right]+\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{1/2}\right]\sum_{n\geq 0}a_{n+1}^{\alpha}(\zeta-\zeta_{\alpha})^{n}\\
&\qquad +(\zeta-\zeta_\alpha)^{1/2}\sum_{n\geq 0}a_{n+1}^{\alpha}\left[\sqrt{\pi}\tfrac{\Gamma(n+1)}{\Gamma(n+\tfrac{1}{2})}(\zeta-\zeta_{\alpha})^{n-1/2}\right]\\
&=a_0^{\alpha}\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{-1/2}\right]+\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{1/2}\right]\sum_{n\geq 0}a_{n+1}^{\alpha}(\zeta-\zeta_{\alpha})^{n}\\
&\qquad +\sqrt{\pi}\sum_{n\geq 0}a_{n+1}^{\alpha}\tfrac{\Gamma(n+1)}{\Gamma(n+\tfrac{1}{2})}(\zeta-\zeta_{\alpha})^{n}
\end{align*}

Notice that $\sum_{n\geq 0}a_{n+1}^{\alpha}\tfrac{\Gamma(n+1)}{\Gamma(n+\tfrac{1}{2})}(\zeta-\zeta_{\alpha})^{n}$ has a finite radius of convergence, hence the second series is a germ of holomorphic function. We are left with the half--derivative of $(\zeta-\zeta_\alpha)^{\pm 1/2}$:

\begin{align*}
\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{1/2}\right]&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\partial_\zeta\left[\int_{\zeta_\alpha}^{\zeta}(\zeta-\zeta')^{-1/2}(\zeta'-\zeta_\alpha)^{1/2}d\zeta'\right]\\
&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\partial_{\zeta}\left[\left[-2(\zeta-\zeta')^{1/2}(\zeta'-\zeta_\alpha)^{1/2}\right]_{\zeta_\alpha}^{\zeta}+\int_{\zeta_\alpha}^{\zeta}2(\zeta-\zeta')^{1/2}\frac{1}{2}(\zeta'-\zeta_{\alpha})^{-1/2}d\zeta'\right]\\
&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\partial_{\zeta}\left[\int_{\zeta_\alpha}^{\zeta}(\zeta-\zeta')^{1/2}(\zeta'-\zeta_{\alpha})^{-1/2}d\zeta'\right]\\
&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\int_{\zeta_\alpha}^{\zeta}\frac{1}{2}(\zeta-\zeta')^{-1/2}(\zeta'-\zeta_\alpha)^{-1/2}d\zeta'\\
&=\frac{1}{2\Gamma\left(\tfrac{1}{2}\right)}\int_{\zeta_\alpha}^{\zeta}(\zeta\zeta'-\zeta\zeta_\alpha-\zeta'^2+\zeta'\zeta_\alpha)^{-1/2}d\zeta'\\
&=\frac{1}{2\Gamma\left(\tfrac{1}{2}\right)}\int_{\zeta_\alpha}^{\zeta}\left(\frac{(\zeta-\zeta_\alpha)^2}{4}-\left(\zeta'-\frac{\zeta+\zeta_\alpha}{2}\right)^2\right)^{-1/2}d\zeta'\\
&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\int_{-\tfrac{\zeta-\zeta_{\alpha}}{2}}^{\tfrac{\zeta-\zeta_{\alpha}}{2}}\frac{1}{\sqrt{\tfrac{(\zeta-\zeta_{\alpha})^2}{4}-y^2}}dy\\
&=\frac{\sqrt{\pi}}{2}
\end{align*}

\begin{align*}
\partial_{\zeta \text{ from } \zeta_\alpha}^{1/2}\left[(\zeta-\zeta_\alpha)^{-1/2}\right]&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\partial_\zeta\left[\int_{\zeta_\alpha}^{\zeta}(\zeta-\zeta')^{-1/2}(\zeta'-\zeta_\alpha)^{-1/2}d\zeta'\right]\\
&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\partial_{\zeta}\left[\int_{\zeta_\alpha}^{\zeta}(\zeta\zeta'-\zeta\zeta_\alpha-\zeta'^2+\zeta'\zeta_\alpha)^{-1/2}d\zeta'\right]\\
&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\partial_{\zeta}\left[\int_{\zeta_\alpha}^{\zeta}\left(\frac{(\zeta-\zeta_\alpha)^2}{4}-\left(\zeta'-\frac{\zeta+\zeta_\alpha}{2}\right)^2\right)^{-1/2}d\zeta'\right]\\
&=\frac{1}{\Gamma\left(\tfrac{1}{2}\right)}\partial_{\zeta}\left[\int_{-\tfrac{\zeta-\zeta_{\alpha}}{2}}^{\tfrac{\zeta-\zeta_{\alpha}}{2}}\frac{1}{\sqrt{\tfrac{(\zeta-\zeta_{\alpha})^2}{4}-y^2}}dy\right]\\
&=0
\end{align*}

Therefore, collecting all the contributions we get

\begin{align*}
\hat{\varphi}_{\alpha}(\zeta)&=-\frac{4a_0^{\alpha}}{\Gamma\left(\tfrac{1}{2}\right)}\frac{1}{(\zeta-\zeta_\alpha)\sqrt{(\zeta-\zeta_{\alpha})^2-4}}-\frac{i}{\Gamma\left(\tfrac{1}{2}\right)}\log(2i+\sqrt{(\zeta-\zeta_\alpha)^2-4})\sum_{n\geq 0}a_{n+1}^{\alpha}(\zeta-\zeta_{\alpha})^{n}\\
&\qquad +\frac{i}{\Gamma\left(\tfrac{1}{2}\right)}\log(\zeta-\zeta_{\alpha})\sum_{n\geq 0}a_{n+1}^{\alpha}(\zeta-\zeta_{\alpha})^{n}+\sqrt{\pi}\sum_{n\geq 0}a_{n+1}^{\alpha}\tfrac{\Gamma(n+1)}{\Gamma(n+\tfrac{1}{2})}(\zeta-\zeta_{\alpha})^{n}
\end{align*}


\bibliographystyle{utphys}
\bibliography{airy-resurgence}
\end{document}