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step2.tex
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step2.tex
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\documentclass{article}
\usepackage{ctex}
\usepackage{upgreek}
\usepackage{xfrac}
\usepackage{amsmath}
\usepackage{geometry}
\geometry{a4paper,centering,scale=0.8}
\usepackage[colorlinks,linkcolor=blue]{hyperref}
\begin{document}\zihao{-4}
\noindent 将上一次的方程增加一个散射项,可得:
\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)-\overline{N}_m(x)}{\uptau_m}
\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}\overline{N}_m
\end{equation}
其中,$\dfrac{1}{\uptau_c}=\dfrac{1}{\uptau'_{th}}+\dfrac{1}{\uptau_m}$.
\\[0.5cm]
令$P(x)=\dfrac{1}{v_x\uptau_c},Q(x)=\dfrac{1}{v_x\uptau'_{th}}N_m^0+\dfrac{1}{v_x\uptau_m}\overline{N}_m$,则:
\begin{equation}\label{eq:3}
\begin{aligned}
N_m(x,v_x>0)&=C_0^>\exp(-\int_0^xP(x)\mathrm{d}x)+\exp(-\int_0^xP(x)\mathrm{d}x)\int_0^xQ(x)\exp(\int_0^xP(x)\mathrm{d}x)\mathrm{d}x\\
&=\frac{\uptau_c}{\uptau'_{th}}N_m^0+(C_0^>-\frac{\uptau_c}{\uptau'_{th}}N_m^0)\exp(-\frac{1}{v_x\uptau_c}x)+\int_0^x\frac{1}{v_x\uptau_m}\overline{N}_m(x')\exp[-\frac{1}{v_x\uptau_c}(x-x')]\mathrm{d}x'
\end{aligned}
\end{equation}
\begin{equation}\label{eq:4}
\begin{aligned}
N_m(x,v_x<0)&=C_0^<\exp(-\int_d^xP(x)\mathrm{d}x)+\exp(-\int_d^xP(x)\mathrm{d}x)\int_d^xQ(x)\exp(\int_d^xP(x)\mathrm{d}x)\mathrm{d}x\\
&=\frac{\uptau_c}{\uptau'_{th}}N_m^0+(C_0^<-\frac{\uptau_c}{\uptau'_{th}}N_m^0)\exp[\frac{1}{v_x\uptau_c}(d-x)]-\int_x^d\frac{1}{v_x\uptau_m}\overline{N}_m(x')\exp[-\frac{1}{v_x\uptau_c}(x-x')]\mathrm{d}x'
\end{aligned}
\end{equation}
于是可得:
\begin{equation}\label{eq:5}
\begin{aligned}
\overline{N}_m(x)&=\frac{\displaystyle\int_{-\infty}^{\infty}N_m(x,v_x)\mathrm{d}v_x}{\displaystyle\int_{-\infty}^{\infty}\mathrm{d}v_x}\\
\int_{-\infty}^{\infty}\mathrm{d}v_x\cdot\overline{N}_m(x)&=\int_{-\infty}^0N_m(x,v_x<0)\mathrm{d}v_x+\int_0^{\infty}N_m(x,v_x>0)\mathrm{d}v_x\\
&=\int_0^{\infty}[N_m(x,-v_x<0)+N_m(x,v_x>0)]\mathrm{d}v_x\\
&=\int_{-\infty}^{\infty}\mathrm{d}v_x\cdot s_1+s_2+s_3
\end{aligned}
\end{equation}
其中:
\begin{equation}\label{eq:6}
\begin{aligned}
\int_{-\infty}^{\infty}\mathrm{d}v_x\cdot s_1&=\int_0^{\infty}\{2\frac{\uptau_c}{\uptau'_{th}}N_m^0+(C_0^>-\frac{\uptau_c}{\uptau'_{th}}N_m^0)\exp(-\frac{1}{v_x\uptau_c}x)+(C_0^<-\frac{\uptau_c}{\uptau'_{th}}N_m^0)\exp[-\frac{1}{v_x\uptau_c}(d-x)]\}\mathrm{d}v_x\\
&=\frac{\uptau_c}{\uptau'_{th}}\overline{N}_m^{\phantom{0}0}+\int_0^{\infty}\{(C_0^>-\frac{\uptau_c}{\uptau'_{th}}N_m^0)\exp(-\frac{1}{v_x\uptau_c}x)+(C_0^<-\frac{\uptau_c}{\uptau'_{th}}N_m^0)\exp[-\frac{1}{v_x\uptau_c}(d-x)]\}\mathrm{d}v_x
\end{aligned}
\end{equation}
\phantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}(注:$\displaystyle\int_0^{\infty}2N_m^0\mathrm{d}v_x=\overline{N}_m^{\phantom{0}0}$).
\[
\begin{aligned}
s_2&=\int_0^{\infty}\int_0^x\frac{1}{v_x\uptau_m}\overline{N}_m(x')\exp[-\frac{1}{v_x\uptau_c}(x-x')]\mathrm{d}x'\mathrm{d}v_x \\
s_3&=\int_0^{\infty}\int_x^d\frac{1}{v_x\uptau_m}\overline{N}_m(x')\exp[\frac{1}{v_x\uptau_c}(x-x')]\mathrm{d}x'\mathrm{d}v_x
\end{aligned}
\]
合并$s_2$和$s_3$:
\begin{equation}\label{eq:7}
s_2+s_3=\int_0^d\begin{bmatrix}\displaystyle\int_0^{\infty}\frac{1}{v_x\uptau_m}\exp(-\frac{1}{v_x\uptau_c}|x-x'|)\mathrm{d}v_x\end{bmatrix}\mathrm{d}x'\cdot\overline{N}_m(x')
\end{equation}
将式\eqref{eq:6}和式\eqref{eq:7}带回式\eqref{eq:5},可得矩阵方程:
\begin{equation}\label{eq:8}
\begin{pmatrix}
\overline{N}_m(0)\\ \vdots\\ \overline{N}_m(x)\\ \vdots \\ \overline{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}
\overline{N}_m(0)\\ \vdots\\ \overline{N}_m(x')\\ \vdots \\ \overline{N}_m(\infty)
\end{pmatrix}
+
\begin{pmatrix}
s_1(0)\\ \vdots\\s_1(x)\\ \vdots\\s_1(\infty)
\end{pmatrix}
\end{equation}
其中:$f_{ak}(x,x')=\dfrac{\displaystyle\int_0^{\infty}\frac{1}{v_x\uptau_m}\exp(-\frac{1}{v_x\uptau_c}|x-x'|)\mathrm{d}v_x\mathrm{d}x'}{\displaystyle\int_{-\infty}^{\infty}\mathrm{d}v_x}$\\[0.7cm]
利用计算机求其数值解绘图可得:
\begin{figure}[ht]
\centering
\includegraphics[width=10cm]{Nmb.png}
\caption{有两个散射项所得$\overline{N}_m(x)$}
\label{fg:1}
\end{figure}
\end{document}