I have problem with the alignment of equations and sometimes their line-break.
Here's my code and a sample pdf file is attached.
I'll appreciate any help.
Thanks
Code: Select all
\documentclass[10pt,a4paper,fleqn]{book}
\usepackage[LAE]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[english,farsi]{babel}
\usepackage[fleqn]{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\marginparwidth=10pt
\begin{document}
\subsection*{بخش چهارم: توابع هنکل}
1- درستی فرمولهای رونسکین زیر را تحقیق کنید.\\
الف)
$$ J_{\nu}(x)H_{\nu}^{(1)'}(x)-J_{\nu}^{'}(x)H_{\nu}^{(1)}(x)=\frac{2i}{\pi x} $$\\
ب)
$$ J_{\nu}(x)H_{\nu}^{(2)'}(x)-J_{\nu}^{'}(x)H_{\nu}^{(2)}(x)=-\frac{2i}{\pi x} $$\\
ج)
$$ N_{\nu}(x)H_{\nu}^{(1)'}(x)-N_{\nu}^{'}(x)H_{\nu}^{(1)}(x)=-\frac{2}{\pi x} $$\\
د)
$$ N_{\nu}(x)H_{\nu}^{(2)'}(x)-N_{\nu}^{'}(x)H_{\nu}^{(2)}(x)=-\frac{2}{\pi x} $$\\
ه)
$$H_{\nu}^{(1)}(x)H_{\nu}^{(2)'}(x)-H_{\nu}^{(1)'}(x)H_{\nu}^{(2)}(x)=-\frac{4i}{\pi x}$$
و)
$$H_{\nu}^{(2)}(x)H_{\nu+1}^{(1)}(x)-H_{\nu}^{(1)}(x)H_{\nu+1}^{(2)}(x)=-\frac{4i}{\pi x}$$
ز)
$$J_{\nu-1}(x)H_{\nu}^{(1)}(x)-J_{\nu}(x)H_{\nu-1}^{(1)}(x)=-\frac{2i}{\pi x}$$
حل: الف)\\
$$ H_{\nu}^{(1)}(x)=J_{\nu}(x)+iN_{\nu}(x) \ , \ H_{\nu}^{(2)}(x)=J_{\nu}(x)-iN_{\nu}(x)$$
$J_{\nu}(x)[J^{'}_{\nu}(x)+iN^{'}_{\nu}(x)]-J^{'}_{\nu}(x)[J_{\nu}(x)+iN_{\nu}(x)]=\\J_{\nu}(x)J^{'}_{\nu}(x)+iJ_{\nu}(x)N^{'}_{\nu}(x)-J^{'}_{\nu}(x)J_{\nu}(x)-iJ^{'}_{\nu}(x)N_{\nu}(x)=i[J_{\nu}(x)N^{'}_{\nu}(x)-J^{'}_{\nu}(x)N_{\nu}(x)]$\\
$ \left\{\begin{array}{cc} N_{\nu}(x)=\frac{\cos{\nu \pi}J_{\nu}(x)-J_{-\nu}(x)}{\sin{\nu \pi}}\ , \ N^{'}_{\nu}(x)=\frac{\cos{\nu \pi}J^{'}_{\nu}(x)-J^{'}_{-\nu}(x)}{\sin{\nu \pi}}
\\ N_{\nu-1}(x)-N_{\nu+1}(x)=2N^{'}_{\nu}(x) \end{array}\right.$\\\\\\
$J_{\nu}(x)[\frac{\cos{\nu \pi}J^{'}_{\nu}(x)-J^{'}_{-\nu}(x)}{\sin{\nu \pi}}]-J^{'}_{\nu}(x)[\frac{\cos{\nu \pi}J_{\nu}(x)-J_{-\nu}(x)}{\sin{\nu \pi}}]=\\\frac{\cos{\nu\pi}}{\sin{\nu\pi}}J_{\nu}(x)J^{'}_{\nu}(x)-\frac{1}{\sin{\nu\pi}}J_{\nu}(x)J^{'}_{-\nu}(x)-\frac{\cos{\nu\pi}}{\sin{\nu\pi}}J^{'}_{\nu}(x)J_{\nu}(x)+\frac{1}{\sin{\nu\pi}}J^{'}_{\nu}(x)J_{-\nu}(x)=\\\frac{1}{\sin{\nu\pi}}(J_{\nu}(x)J_{-\nu}(x)-J_{\nu}(x)J^{'}_{-\nu}(x))=-\frac{1}{\sin{\nu\pi}}(J_{\nu}(x)J^{'}_{\nu}(x)-J^{'}_{\nu}(x)J_{-\nu}(x))=-\frac{1}{\sin{\nu\pi}}(-2\frac{\sin{\nu\pi}}{\pi x})=\frac{2}{\pi x}$\\\\
ب)\\
$J_{\nu}(x)[J^{'}_{\nu}(x)-iN^{'}_{\nu}(x)]-J^{'}_{\nu}(x)[J_{\nu}(x)-iN_{\nu}(x)]=\\
J_{\nu}(x)J^{'}_{\nu}(x)-iJ_{\nu}(x)N^{'}_{\nu}(x)-J^{'}_{\nu}(x)J_{\nu}(x)+iJ^{'}_{\nu}(x)N_{\nu}(x)=\\
-i(J_{\nu}(x)N^{'}_{\nu}(x)-J^{'}_{\nu}(x)N_{\nu}(x))=-i(-\frac{1}{\sin{\nu\pi}}\frac{-2\sin{\nu\pi}}{\pi x})=-\frac{2i}{\pi x}$
\\\\
ج)\\
$
N_{\nu}(x)[J^{'}_{\nu}(x)+iN^{'}_{\nu}(x)]-N^{'}_{\nu}(x)[J_{\nu}(x)+iN_{\nu}(x)]=\\
N_{\nu}(x)J^{'}_{\nu}(x)+iN_{\nu}(x)N^{'}_{\nu}(x)-N^{'}_{\nu}(x)J_{\nu}(x)-iN^{'}_{\nu}(x)N_{\nu}(x)=\\
N_{\nu}(x)J^{'}_{\nu}(x)-N^{'}_{\nu}(x)J_{\nu}(x)=\frac{\cos{\nu \pi}J_{\nu}(x)-J_{-\nu}(x)}{\sin{\nu \pi}}J^{'}_{\nu}(x)-\frac{\cos{\nu \pi}J^{'}_{\nu}(x)-J^{'}_{-\nu}(x)}{\sin{\nu \pi}}J_{\nu}(x)=-\frac{2}{\pi x}
$\\ \\
د)\\
$N_{\nu}(x)[J^{'}_{\nu}(x)-iN^{'}_{\nu}(x)]-N^{'}_{\nu}(x)[J_{\nu}(x)-iN_{\nu}(x)]=\\
N_{\nu}(x)J^{'}_{\nu}(x)-iN_{\nu}(x)N^{'}_{\nu}(x)-N^{'}_{\nu}(x)J_{\nu}(x)+iN^{'}_{\nu}(x)N_{\nu}(x)=\\
N_{\nu}(x)J^{'}_{\nu}(x)-N^{'}_{\nu}(x)J_{\nu}(x)=-\frac{2}{\pi x}
$\\ \\
ه)\\
$[J_{\nu}(x)+iN_{\nu}(x)][J^{'}_{\nu}(x)-iN^{'}_{\nu}(x)]-[J^{'}_{\nu}(x)+iN^{'}_{\nu}(x)][J_{\nu}(x)-iN_{\nu}(x)]=\\
J_{\nu}(x)J^{'}_{\nu}(x)-iJ_{\nu}(x)N^{'}_{\nu}(x)+iN_{\nu}(x)J^{'}_{\nu}(x)-J^{'}_{\nu}(x)J_{\nu}(x)+\\iJ^{'}_{\nu}(x)N_{\nu}(x)-iN^{'}_{\nu}(x)J_{\nu}(x)-N^{'}_{\nu}(x)N_{\nu}(x)+N_{\nu}(x)N^{'}_{\nu}(x)=\\
2i(N_{\nu}(x)J^{'}_{\nu}(x)-N^{'}_{\nu}(x)J_{\nu}(x))=2i(-\frac{2}{\pi x})=-\frac{4i}{\pi x}
$\\ \\
و)\\
$
[J_{\nu}(x)-iN_{\nu}(x)][J_{\nu+1}(x)+iN_{\nu+1}(x)]-[J_{\nu}(x)+iN_{\nu}(x)][J_{\nu+1}(x)-iN_{\nu+1}(x)]=\\
J_{\nu}(x)J_{\nu+1}(x)+iJ_{\nu}(x)N_{\nu+1}(x)-iN_{\nu}(x)J_{\nu+1}(x)+N_{\nu}(x)N_{\nu+1}(x)-\\
J_{\nu}(x)J_{\nu+1}(x)+iJ_{\nu}(x)N_{\nu+1}(x)-iN_{\nu}(x)J_{\nu+1}(x)-N_{\nu}(x)N_{\nu+1}(x)=\\
2i[J_{\nu}(x)N_{\nu+1}(x)-N_{\nu}(x)J_{\nu+1}(x)]=\\2i[J_{\nu}(x)\frac{\cos{\nu \pi}J_{\nu+1}(x)+J_{-\nu-1}(x)}{\sin{\nu \pi}}-\frac{\cos{\nu \pi}J_{\nu}(x)-J_{-\nu}(x)}{\sin{\nu \pi}}J_{\nu+1}(x)]=\\\frac{2i}{\sin{\nu\pi}}[J_{\nu}(x)J_{-\nu-1}(x)+J_{-\nu}(x)J_{\nu+1}(x)]=\frac{2i}{\sin{\nu\pi}}[J_{\nu}(x)J^{'}_{-\nu}(x)-\frac{\nu}{x}J_{\nu}(x)J_{-\nu}(x)+\\
\frac{\nu}{x}J_{-\nu}(x)J_{\nu}(x)-J_{-\nu}(x)J^{'}_{\nu}(x)]=\frac{2i}{\sin{\nu\pi}}[J_{\nu}(x)J^{'}_{-\nu}(x)-J_{-\nu}(x)J^{'}_{\nu}(x)]=\frac{2i}{\sin{\nu\pi}}(-2\frac{\sin{\nu\pi}}{\pi x})=-\frac{4i}{\pi x}
$
ز)\\
$
J_{\nu-1}(x)[J_{\nu}(x)+iN_{\nu}(x)]-J_{\nu}(x)[J_{\nu-1}(x)+iN_{\nu-1}(x)]=\\
J_{\nu-1}(x)J_{\nu}(x)+iJ_{\nu-1}(x)N_{\nu}(x)-J_{\nu}(x)J_{\nu-1}(x)-iJ_{\nu}(x)N_{\nu-1}(x)=i[J_{\nu-1}(x)N_{\nu}(x)-J_{\nu}(x)N_{\nu-1}(x)]=\\i[J_{\nu-1}(x)\frac{\cos{\nu \pi}J_{\nu}(x)-J_{-\nu}(x)}{\sin{\nu \pi}}-J_{\nu}(x)\frac{\cos{\nu \pi}J_{\nu-1}(x)+J_{-(\nu-1)}(x)}{\sin{\nu \pi}}]=i[\frac{\cos{\nu\pi}}{\sin{\nu\pi}}J_{\nu-1}(x)J_{\nu}(x)-\frac{1}{\sin{\nu\pi}}J_{\nu-1}(x)J_{-\nu}(x)\\-\frac{\cos{\nu\pi}}{\sin{\nu\pi}}J_{\nu}(x)J_{\nu-1}(x)-\frac{1}{\sin{\nu\pi}}J_{\nu}(x)J_{-(\nu-1)}(x)]=-\frac{i}{\sin{\nu\pi}}[J_{\nu-1}(x)J_{-\nu}(x)+J_{\nu}(x)J_{-\nu+1)}(x)]=\\
-\frac{i}{\sin{\nu\pi}}[J_{-\nu}(x)(J^{'}_{\nu}(x)+\frac{\nu}{x}J_{\nu}(x))+J_{\nu}(x)(-\frac{\nu}{x}J_{-\nu}(x)-J^{'}_{-\nu}(x))]=\frac{i}{\sin{\nu\pi}}[J_{\nu}(x)J^{'}_{-\nu}(x)-J_{-\nu}(x)J^{'}_{\nu}(x)]=\\\frac{i}{\sin{\nu\pi}}\frac{-2\sin{\nu \pi}}{\pi x}=-\frac{2i}{\pi x}
$
\\ \\
2- نشان دهید که صورتهای انتگرالی زیر، در معادله ی بسل صدق می کنند.\\
$$\frac{1}{i\pi}\int_{0_{C_1}}^{\infty e^{i\pi}} e^{\frac{x}{2}(t-\frac{1}{t})} \frac{dt}{t^{\nu+1}}=H_{\nu}^{(1)}(x)$$
$$\frac{1}{i\pi}\int_{\infty e^{-i\pi}_{C_2}}^{0} e^{\frac{x}{2}(t-\frac{1}{t})} \frac{dt}{t^{\nu+1}}=H_{\nu}^{(2)}(x)$$
حل:\\
$ \int_{C_1\ ,\ C_2} \frac{dt}{t^{\nu+1}}[x^2 \frac{d^2}{dx^2}e^{\frac{x}{2}(t-\frac{1}{t})}+x\frac{d}{dx}e^{\frac{x}{2}(t-\frac{1}{t})}+(x^2-\nu^2)e^{\frac{x}{2}(t-\frac{1}{t})}]=\\\int_{C_1\ ,\ C_2} \frac{dt}{t^{\nu+1}}[x^2 (\frac{t-\frac{1}{t}}{2})^2+x\frac{t-\frac{1}{t}}{2}+(x^2-\nu^2)]e^{\frac{x}{2}(t-\frac{1}{t})}=\\\int_{C_1\ ,\ C_2} dt \frac{d}{dt}[\frac{e^{\frac{x}{2}(t-\frac{1}{t})}}{t^{\nu}}(\frac{x}{2}[t+\frac{1}{t}]+\nu)]=\frac{ e^{\frac{x}{2}(t-\frac{1}{t})}}{t^{\nu}}(\frac{x}{2}[t+\frac{1}{t}]+\nu)|_{C_1\ ,\ C_2}=0
$\\ \\
3- با استفاده از انتگرالها و پربندهای مسئله ی 2 نشان دهید:\\
$$\frac{1}{2i}[H_{\nu}^{(1)}(x)-H_{\nu}^{(2)}(x)]=N_{\nu}(x)$$
حل:\\
$$H_{\nu}^{(1)}(x)=\frac{1}{i\pi}\int_{0_{C_1}}^{\infty e^{i\pi}} e^{\frac{x}{2}(t-\frac{1}{t})} \frac{dt}{t^{\nu+1}}$$
$$t=\frac{e^{i\pi}}{s} \Rightarrow dt=-\frac{e^{i \pi}}{s^2}ds \ \ , \ \ \left\{\begin{array}{cc} t=0 \rightarrow s=\infty e^{i\pi}\\ t=\infty e^{i\pi} \rightarrow s=0 \end{array}\right.$$\\
$ H_{\nu}^{(1)}(x)=\frac{1}{i \pi}\int_{-C_1} e^{\frac{x}{2}(\frac{e^{i\pi}}{s}-se^{-i\pi})}s^{\nu+1}e^{-i\pi (\nu+1)}(-\frac{e^{i\pi}}{s^2}ds)=\frac{e^{-i\pi\nu}}{i\pi}\int_{C_1}e^{\frac{x}{2}(s-\frac{1}{s})}\frac{ds}{s^{-\nu+1}}=e^{-i\pi\nu} H_{-\nu}^{(1)}(x)
$
$$H_{\nu}^{(2)}(x)=\frac{1}{i\pi}\int_{\infty e^{-i\pi}_{C_2}}^{0} e^{\frac{x}{2}(t-\frac{1}{t})} \frac{dt}{t^{\nu+1}}$$
$$t=\frac{e^{-i\pi}}{s} \Rightarrow dt=-\frac{e^{-i \pi}}{s^2}ds \ \ , \ \ \left\{\begin{array}{cc} t=0 \rightarrow s=\infty e^{-i\pi}\\ t=\infty e^{-i\pi} \rightarrow s=0 \end{array}\right.$$\\
$ H_{\nu}^{(2)}(x)=\frac{1}{i \pi}\int_{-C_2} e^{\frac{x}{2}(\frac{e^{-i\pi}}{s}-se^{i\pi})}s^{\nu+1}e^{i\pi (\nu+1)}(-\frac{e^{-i\pi}}{s^2}ds)=\frac{e^{i\pi\nu}}{i\pi}\int_{C_2}e^{\frac{x}{2}(s-\frac{1}{s})}\frac{ds}{s^{-\nu+1}}=e^{i\pi\nu} H_{-\nu}^{(2)}(x)
$
$$ J_{\nu}(x)=\frac{1}{2}[H_{\nu}^{(1)}(x)+H_{\nu}^{(2)}(x)] \ , \ J_{-\nu}(x)=\frac{1}{2}[e^{i\nu\pi}H_{\nu}^{(1)}(x)+e^{-i\nu\pi}H_{\nu}^{(2)}(x)]$$
$$ N_{\nu}(x)=\frac{\cos{\nu \pi}J_{\nu}(x)-J_{-\nu}(x)}{\sin{\nu \pi}} \Rightarrow N_{\nu}(x)=\frac{1}{2i}[H_{\nu}^{(1)}(x)-H_{\nu}^{(2)}(x)]$$
4- نشان دهید که با تبدیل انتگرالهای مسئله ی 2 می توان به عبارتهای زیر رسید.\\
$$H_{\nu}^{(1)}(x)=\frac{1}{\pi i}\int_{C_3}e^{x \sinh{\gamma}-\nu\gamma} d\gamma$$
$$H_{\nu}^{(2)}(x)=\frac{1}{\pi i}\int_{C_4}e^{x \sinh{\gamma}-\nu\gamma} d\gamma$$
حل:\\
$$\frac{1}{i\pi}\int_{0_{C_1}}^{\infty e^{i\pi}} e^{\frac{x}{2}(t-\frac{1}{t})} \frac{dt}{t^{\nu+1}}=H_{\nu}^{(1)}(x)$$
$$
t=e^{\gamma} \Rightarrow dt=d^{\gamma} d\gamma\ , \ \frac{x}{2}(t-\frac{1}{t})=\frac{x}{2}(e^{\gamma}-e^{-\gamma})=x \sinh{\gamma}$$
$$ t=0 \rightarrow \gamma=-\infty $$
$$ t=\infty e^{i\pi} \rightarrow \gamma=\infty+i\pi $$
\begin{equation}
\begin{split}
H_{\nu}^{(1)}(x)=\frac{1}{i \pi} \int_{0_{C_1}}^{\infty e^{i\pi}} e^{\frac{x}{2}(t-\frac{1}{t})}\frac{dt}{t^{\nu+1}}=\frac{1}{i\pi} \int_{C_3} e^{x\sinh{\gamma}}\frac{e^{\gamma}d\gamma}{e^{(\nu+1)\gamma}}=\frac{1}{i\pi}\int_{C_3} e^{x\sinh{\gamma}-\nu \gamma} d\gamma
\end{split}
\end{equation}
\end{document}