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step4.tex
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step4.tex
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\documentclass{article}
\usepackage{ctex}
\usepackage{upgreek}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{xfrac}
\usepackage{color}
\usepackage{subfigure}
\usepackage{geometry}
\geometry{a4paper,centering,scale=0.9}
\usepackage[colorlinks,linkcolor=blue]{hyperref}
\begin{document}\zihao{-4}
\noindent 方程：
\begin{equation}\label{eq:1}
v_x\frac{\partial n_m(x,v_x)}{\partial x}=-\frac{n_m(x,v_x)-n_m^0(v_x)}{\uptau'_{th}}-\frac{n_m(x,v_x)-\bar{n}_m(x)}{\uptau_m}-\beta v_x\frac{\mathrm{d}\bar{n}_m(x)}{\mathrm{d}x}
\end{equation}
整理即：
\begin{equation}\label{eq:2}
\frac{\partial n_m}{\partial x}+\frac{1}{v_x\uptau_c}n_m=\frac{1}{v_x\uptau'_{th}}n_m^0+\frac{1}{v_x\uptau_m}\bar{n}_m-\beta\frac{\mathrm{d}\bar{n}_m}{\mathrm{d}x}
\end{equation}
其中，$\dfrac{1}{\uptau_c}=\dfrac{1}{\uptau'_{th}}+\dfrac{1}{\uptau_m}$.
\noindent 且
\[\upepsilon=\upepsilon_m\sin^2(\frac{aq}{2})+\upepsilon_H\]
\[v_x=\frac{\partial\upepsilon}{\hbar\partial q}=v_m\sin(aq)\]
可得：
\begin{equation}\label{eq:3}
\begin{aligned}
n_m(x,v_x>0)=&\textbf{N}_{injection}+\textbf{N}_{thermal-relaxation}+\textbf{N}_{magnon_relaxtion}+\textbf{N}_{pressure}+\textbf{N}_{chemical}\\
=&C_0^>\exp(-\frac{1}{v_x\uptau_c}x)\\
&+pn^0_m[1-\exp(-\frac{1}{v_x\uptau_c}x)]\\
&+\int_0^x\frac{\mathrm{d}\beta}{\mathrm{d}x'}\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\bar{n}_m\mathrm{d}x'\\
&+\int_0^x\beta[\frac{1}{v_x\uptau_c}+\delta(x')-\delta(x-x')]\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\bar{n}_m\mathrm{d}x'\\
&+\int_0^x\frac{1}{v_x\uptau_m}\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
其中，$p=\dfrac{\uptau_c}{\uptau'_{th}}.$
\begin{equation}\label{eq:4}
\begin{aligned}
n_m(x,v_x<0)=&C_0^<\exp[\frac{1}{v_x\uptau_c}(d-x)]\\
&+pn^0_m\{1-\exp[\frac{1}{v_x\uptau_c}(d-x)]\}\\
&-\int_x^d\frac{\mathrm{d}\beta}{\mathrm{d}x'}\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\bar{n}_m\mathrm{d}x'\\
&-\int_x^d\beta[\frac{1}{v_x\uptau_c}+\delta(x-x')-\delta(d-x')]\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\bar{n}_m\mathrm{d}x'\\
&-\int_x^d\frac{1}{v_x\uptau_m}\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
从而：
\begin{equation}\label{eq:5}
\begin{aligned}
\int_{-\infty}^{\infty}n_m\mathrm{d}\vec{\textbf{v}}=&\int_0^\infty \{C_0^>\exp(-\frac{1}{v_x\uptau_c}x)+C_0^<\exp[-\frac{1}{v_x\uptau_c}(d-x)]\}\mathrm{d}\vec{\textbf{v}}\\
&+\int_0^\infty pn^0_m\{2-\exp(-\frac{1}{v_x\uptau_c}x)-\exp[-\frac{1}{v_x\uptau_c}(d-x)]\}\mathrm{d}\vec{\textbf{v}}\\
&+\int_0^d[\int_0^\infty\exp(-\frac{1}{v_x\uptau_{c}}|x-x'|)\mathrm{d}\vec{\textbf{v}}]\frac{\mathrm{d}\beta}{\mathrm{d}x'}\mathrm{sgn}(x-x')\bar{n}_m\mathrm{d}x'\\
&+\int_0^d\int_0^\infty\beta[\frac{1}{v_x\uptau_c}+\delta(x')+\delta(d-x')-2\delta(x-x')]\exp(-\frac{1}{v_x\uptau_{c}}|x-x'|)\mathrm{d}\vec{\textbf{v}}\bar{n}_m\mathrm{d}x'\\
&+\int_0^d\int_0^\infty\frac{1}{v_x\uptau_m}\exp(-\frac{1}{v_x\uptau_{c}}|x-x'|)\mathrm{d}\vec{\textbf{v}}\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
因为$n_m=n_m^0+\delta n_m$，故：
\begin{equation}\label{eq:6}
\begin{aligned}
n_m(x,v_x>0)=&C_0^>\exp(-\frac{1}{v_x\uptau_c}x)\\
&+pn^0_m[1-\exp(-\frac{1}{v_x\uptau_c}x)]\\
&+(1-p)\bar{n}^0_m[1-\exp(-\frac{1}{v_x\uptau_c}x)]\\
&+\int_0^x\frac{\mathrm{d}\beta}{\mathrm{d}x'}\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\delta\bar{n}_m\mathrm{d}x'\\
&+\int_0^x\beta[\frac{1}{v_x\uptau_c}+\delta(x')-\delta(x-x')]\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\delta\bar{n}_m\mathrm{d}x'\\
&+\int_0^x\frac{1}{v_x\uptau_m}\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\delta\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
\begin{equation}\label{eq:7}
\begin{aligned}
n_m(x,v_x<0)=&C_0^<\exp[\frac{1}{v_x\uptau_c}(d-x)]\\
&+pn^0_m\{1-\exp[\frac{1}{v_x\uptau_c}(d-x)]\}\\
&+(1-p)\bar{n}^0_m\{1-\exp[\frac{1}{v_x\uptau_c}(d-x)]\}\\
&-\int_x^d\frac{\mathrm{d}\beta}{\mathrm{d}x'}\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\delta\bar{n}_m\mathrm{d}x'\\
&-\int_x^d\beta[\frac{1}{v_x\uptau_c}+\delta(x-x')-\delta(d-x')]\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\delta\bar{n}_m\mathrm{d}x'\\
&-\int_x^d\frac{1}{v_x\uptau_m}\exp[-\frac{1}{v_x\uptau_{c}}(x-x')]\delta\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
从而：
\begin{equation}\label{eq:8}
\begin{aligned}
\int_{-\infty}^{\infty}n_m\mathrm{d}\vec{\textbf{v}}=&\int_0^\infty \{(C_0^>+\frac{\mathrm{d}\beta}{\mathrm{d}x'}\delta\bar{n}_m(0^-))\exp(-\frac{1}{v_x\uptau_c}x)+C_0^<\exp[-\frac{1}{v_x\uptau_c}(d-x)]\}\mathrm{d}\vec{\textbf{v}}\\
&+\int_0^\infty pn^0_m\{2-\exp(-\frac{1}{v_x\uptau_c}x)-\exp[-\frac{1}{v_x\uptau_c}(d-x)]\}\mathrm{d}\vec{\textbf{v}}\\
&+\int_0^\infty (1-p)\bar{n}^0_m\{2-\exp(-\frac{1}{v_x\uptau_c}x)-\exp[-\frac{1}{v_x\uptau_c}(d-x)]\}\mathrm{d}\vec{\textbf{v}}\\
&+\int_0^d\int_0^\infty\beta[\frac{1}{v_x\uptau_c}+\delta(x')+\delta(d-x')-2\delta(x-x')]\exp(-\frac{1}{v_x\uptau_{c}}|x-x'|)\mathrm{d}\vec{\textbf{v}}\delta\bar{n}_m\mathrm{d}x'\\
&+\int_0^d\int_0^\infty\frac{1}{v_x\uptau_m}\exp(-\frac{1}{v_x\uptau_{c}}|x-x'|)\mathrm{d}\vec{\textbf{v}}\delta\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
边界条件为:
\begin{equation}\label{eq:9}
\begin{aligned}
&\delta n_m(0,v_x>0)=g(v_x)+R(v_x)\cdot \delta n_m(0,v'_x<0) \\
&n_m(0,v_x<0)=\sum T(v_x)\cdot n_m(0,v_x<0)\\
&n_m(d,v_x<0)=T'(v_x)\cdot g'(v_x)+R'(v_x)\cdot n_m(d,v'_x>0)
\end{aligned}
\end{equation}
不妨令$v_x=-v'_x$，将边界条件带入式\eqref{eq:6}\eqref{eq:7}可得：
\begin{equation}\label{eq:10}
\begin{aligned}
C_0^>=&g\\
&+RC_0^<\exp(-\frac{1}{v_x\uptau_c}d)\\
&+R\cdot pn^0_m[1-\exp(-\frac{1}{v_x\uptau_c}d)]\\
&+R\cdot (1-p)\bar{n}^0_m[1-\exp(-\frac{1}{v_x\uptau_c}d)]\\
&-R\int_0^d\frac{\mathrm{d}\beta}{\mathrm{d}x'}\exp(-\frac{1}{v_x\uptau_{c}}x')\delta\bar{n}_m\mathrm{d}x'\\
&+R\int_0^d\beta[\frac{1}{v_x\uptau_c}-\delta(x')+\delta(d-x')]\exp(-\frac{1}{v_x\uptau_{c}}x')\delta\bar{n}_m\mathrm{d}x'\\
&+R\int_0^d\frac{1}{v_x\uptau_m}\exp(-\frac{1}{v_x\uptau_{c}}x')\delta\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
\begin{equation}\label{eq:11}
\begin{aligned}
C_0^<=&T'\cdot g'\\
&+R'C_0^>\exp(-\frac{1}{v_x\uptau_c}d)\\
&+R'\cdot pn^0_m[1-\exp(-\frac{1}{v_x\uptau_c}d)]\\
&+R'\cdot (1-p)\bar{n}^0_m[1-\exp(-\frac{1}{v_x\uptau_c}d)]\\
&+R'\int_0^d\frac{\mathrm{d}\beta}{\mathrm{d}x'}\exp[-\frac{1}{v_x\uptau_{c}}(d-x')]\delta\bar{n}_m\mathrm{d}x'\\
&+R'\int_0^d\beta[\frac{1}{v_x\uptau_c}+\delta(x')-\delta(d-x')]\exp[-\frac{1}{v_x\uptau_{c}}(d-x')]\delta\bar{n}_m\mathrm{d}x'\\
&+R'\int_0^d\frac{1}{v_x\uptau_m}\exp[-\frac{1}{v_x\uptau_{c}}(d-x')]\delta\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
先令$C_0^<=0$，假设$T(v_x),R(v_x),T'(v_x),R'(v_x)$都为常数,且$T(v_x)=1$，然后将$C_0^>$代回式\eqref{eq:8}
\begin{equation}\label{eq:12}
\begin{aligned}
\int_{-\infty}^{\infty}\delta n_m\mathrm{d}\vec{\textbf{v}}=&\int_0^\infty [g(v_x)+R\frac{\mathrm{d}\beta}{\mathrm{d}x'}\delta\bar{n}_m(0^-)]\exp(-\frac{1}{v_x\uptau_c}x)\mathrm{d}\vec{\textbf{v}}\\
&+\int_0^\infty pn^0_m\{R\exp(-\frac{1}{v_x\uptau_c}x)-\exp(-\frac{1}{v_x\uptau_c}x)\}\mathrm{d}\vec{\textbf{v}}\\
&+\int_0^\infty (1-p)\bar{n}^0_m\{R\exp(-\frac{1}{v_x\uptau_c}x)-\exp(-\frac{1}{v_x\uptau_c}x)\}\mathrm{d}\vec{\textbf{v}}\\
&+R\int_0^d\int_0^\infty\beta[\frac{1}{v_x\uptau_c}-\delta(x')]\exp[-\frac{1}{v_x\uptau_{c}}(x+x')]\mathrm{d}\vec{\textbf{v}}\delta\bar{n}_m\mathrm{d}x'\\
&+R\int_0^d\int_0^\infty\frac{1}{v_x\uptau_m}\exp[-\frac{1}{v_x\uptau_{c}}(x+x')]\mathrm{d}\vec{\textbf{v}}\delta\bar{n}_m\mathrm{d}x'\\
&+\int_0^d\int_0^\infty\beta[\frac{1}{v_x\uptau_c}+\delta(x')-2\delta(x-x')]\exp(-\frac{1}{v_x\uptau_{c}}|x-x'|)\mathrm{d}\vec{\textbf{v}}\delta\bar{n}_m\mathrm{d}x'\\
&+\int_0^d\int_0^\infty\frac{1}{v_x\uptau_m}\exp(-\frac{1}{v_x\uptau_{c}}|x-x'|)\mathrm{d}\vec{\textbf{v}}\delta\bar{n}_m\mathrm{d}x'\\
\end{aligned}
\end{equation}
得矩阵方程：
\begin{equation}\label{eq:13}
\begin{pmatrix}
\delta\bar{n}_m(0)\\ \vdots\\ \delta\bar{n}_m(x)\\ \vdots \\ \delta\bar{n}_m(\infty)
\end{pmatrix}
=
\begin{pmatrix}
f_{ak}(0,0)&\dots&f_{ak}(0,x')&\dots&f_{ak}(0,\infty)\\
\vdots&\ddots&\vdots&\ddots&\vdots\\
f_{ak}(x,0)&\dots&f_{ak}(x,x')&\dots&f_{ak}(x,\infty)\\
\vdots&\ddots&\vdots&\ddots&\vdots\\
f_{ak}(\infty,0)&\dots&f_{ak}(\infty,x')&\dots&f_{ak}(\infty,\infty)\\
\end{pmatrix}
\begin{pmatrix}
\delta\bar{n}_m(0)\\ \vdots\\ \delta\bar{n}_m(x')\\ \vdots \\ \delta\bar{n}_m(\infty)
\end{pmatrix}
+
\begin{pmatrix}
S(0)\\ \vdots\\S(x)\\ \vdots\\S(\infty)
\end{pmatrix}
\end{equation}
由此可利用程序算出$\delta\bar n_m(x)$的数值解.
\end{document}