做个公式记录
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经常有些公式简单到不能包含在任何文章里，但是每次一步一步推导的时候还不能升略，专门写个文档还很麻烦，在这开个贴记录一下把，随时随地可以访问。
\begin{equation}
\frac{1}{\tau_0 \xi^{2/3}}\frac{\partial vn}{\partial v}
\end{equation}\begin{equation}
\int_{-\infty}^{+\infty}\int_{0}^{+\infty}\frac{v^j \xi^k}{ \tau_0 \xi^{2/3}}\frac{\partial vn}{\partial v}\mathrm{d}v\mathrm{d}\xi= \frac{1}{\tau_0}\int_{-\infty}^{+\infty}\int_{0}^{+\infty}v^j \xi^{k-2/3} \frac{\partial vn}{\partial v}\mathrm{d}v\mathrm{d}\xi
\end{equation}\begin{equation}
=\frac{1}{\tau_0}\int_{-\infty}^{+\infty}\int_{0}^{+\infty}v^j \xi^{k-2/3} \mathrm{d}(vn)\mathrm{d}\xi
\end{equation}\begin{equation}
=\frac{1}{\tau_0}\int_{-\infty}^{+\infty}\xi^{k-2/3} \left(\int_{0}^{+\infty}v^j \mathrm{d}(vn)\right) \mathrm{d}\xi
\end{equation}\begin{equation}
=\frac{1}{\tau_0}\int_{-\infty}^{+\infty}\xi^{k-2/3} \left(v^{j+1}n_{-\infty}^{+\infty} - \int_{0}^{+\infty}j v^j n \mathrm{d}(v)\right) \mathrm{d}\xi
\end{equation}\begin{equation}
=-\frac{1}{\tau_0}\int_{-\infty}^{+\infty}\int_{0}^{+\infty}j v^j \xi^{k-2/3} n \mathrm{d}(v) \mathrm{d}\xi
\end{equation}\begin{equation}
=-\frac{j}{\tau_0}m_{j,k-2/3}
\end{equation} -
\begin{equation}
\int_0^\infty\int_{-\infty}^\infty u_d^j\xi^k \frac{\p n}{\p t} \rd\xi\rd u_\rd=\frac{\p m_{j,k}}{\p t}
\end{equation}\begin{equation}
\int_0^\infty\int_{-\infty}^\infty u_d^j\xi^k \frac{\p nu_\rd}{\p x} \rd \xi \rd u_\rd=\frac{\p m_{j+1,k}}{\p x}
\end{equation}\begin{equation}
\int_0^\infty\int_{-\infty}^\infty u_d^j\xi^k \frac{\p}{\p u_\rd}\left(\frac{1}{\rho_\rd}\frac{\p p_c}{\p x}n\right) \rd\xi\rd u_\rd = \frac{1}{\rho_\rd}\frac{\p p_c}{\p x}\int_0^\infty\int_{-\infty}^\infty u_d^j\xi^k \frac{\p n}{\p u_\rd}\rd\xi\rd u_\rd = \frac{1}{\rho_\rd}\frac{\p p_c}{\p x}\int_0^\infty\int_{-\infty}^\infty u_d^j\xi^k \rd n\rd\xi =\frac{1}{\rho_\rd}\frac{\p p_c}{\p x}\int_0^\infty \xi^k\left( \int_{-\infty}^\infty u_d^j \rd n \right)\rd\xi
\end{equation}
\begin{equation}
=\frac{1}{\rho_\rd}\frac{\p p_c}{\p x}\int_0^\infty \xi^k\left( u_d^j n_{-\infty}^\infty-\int_{-\infty}^\infty n\rd u^j \right)\rd\xi
=-\frac{1}{\rho_\rd}\frac{\p p_c}{\p x}\int_0^\infty \xi^k\left( j\int_{-\infty}^\infty nu^{j-1}\rd u \right)\rd\xi =-j\frac{1}{\rho_\rd}\frac{\p p_c}{\p x}\int_0^\infty \int_{-\infty}^\infty nu^{j-1}\xi^k\rd u \rd\xi =-j\frac{1}{\rho_\rd}\frac{\p p_c}{\p x}m_{j-1,k}
\end{equation} -
\begin{equation}
\frac{\mathrm{d}u_d}{\mathrm{d}t}=g+\frac{1}{\tau}(u_c-u_d)
\end{equation}\begin{equation}
\frac{\mathrm{d} u_d}{u_d-\tau g-u_c}=-\frac{1}{\tau}\rd t
\end{equation}\begin{equation}
\mathrm{ln}\left|u_d-\tau g-u_c\right|=-\frac{t}{\tau}+C
\end{equation}\begin{equation}
u_d-\tau g-u_c=e^{-t/\tau+C},t=0,u_d=u_d^0
\end{equation}\begin{equation}
u_d^0-\tau g-u_c=e^{C}
\end{equation}\begin{equation}
C=\mathrm{ln}\left|u_d^0-\tau g-u_c\right|
\end{equation}\begin{equation}
u_d=e^{-t/\tau}\left(u_d^0-\tau g-u_c\right)+\tau g+u_c \\ =e^{-t/\tau}u_d^0+(1-e^{-t/\tau})(\tau g+u_c )
\end{equation}
\begin{equation}
S=\int _0^t u_d\rd t=\left(u_d^0-\tau g-u_c\right)\int _0^t e^{-t/\tau}\rd t+(\tau g+u_c)t
\end{equation} -
\begin{equation}
\frac{{\p \left( {{\alpha_k }{\rho_k}{\bfU_k }} \right)}}{{\p t}} + \nabla \cdot \left( {{\alpha_k}{\rho_k } {{\bfU_k} {\bfU_k}} } \right) - \nabla \cdot \left( {{\alpha_k}{ \rho_k}{\tau_k}} \right)
= - {\alpha_k} \nabla p_k + {\alpha_k}{\rho_k} \bfg + \sum {\bfM_{ij}},
\end{equation}\begin{equation}
\frac{{\p \left( {{\alpha_k }{\rho_k}{ }} \right)}}{{\p t}} + \nabla \cdot \left( {{\alpha_k}{\rho_k } { {\bfU_k}} } \right) =0
\end{equation}\begin{equation}
\nabla \cdot \left( {{\alpha_k}{\rho_k } { {\bfU_k}} } \right) =0
\end{equation}\begin{equation}
\sum {{\alpha_k}{\rho_k } { {\bfU_k}} }\cdot\bfS_f =0
\end{equation}\begin{equation}
\tau_k=-\nu_\mathrm{k,eff}\left(\nabla \bfU_k+\nabla^\rT \bfU_k\right)+\frac{2}{3}\nu_\mathrm{k,eff}\nabla \cdot \left(\bfU_k \cdot\bfI\right),
\label{taud}
\end{equation}\begin{equation}
\bfM_{\mathrm{drag}}=\frac{3}{4}\alpha_k\rho_\rc C_\rD\frac{1}{d_k} \left|\bfU_\rc-\bfU_k\right| \left(\bfU_\rc-\bfU_k\right),
\end{equation}\begin{equation}
Re=\frac{d_k|\bfU_k-\bfU_\rc|}{\nu_\rc}
\end{equation}\begin{equation}
\bfM_\lift=\alpha_\rd C_\rL\rho_\rc\bfU_\rr\times\left(\nabla\times\bfU_\rc\right),
\end{equation}\begin{equation}
\bfM_\wall=C_\wall\rho_\rc\alpha_k|\bfU_\rc-\bfU_k|^2\cdot\bfn
\end{equation}\begin{equation}
\bfM_\turb=C_\rT\rho_\rc k_\rc\nabla\alpha_\rd,
\end{equation}\begin{equation}\label{m1}
\frac{{\p \left( {{\alpha_k }{\rho_k }{\bfU_k}} \right)}}{{\p t}} + \nabla \cdot \left( {{\alpha_k}{\rho_k} {{\bfU_k} {\bfU_k}} } \right) - \nabla \cdot \left( {{\alpha_\rd}{ \rho_\rd}{\tau_\rd}} \right)
= -\Kd_k\bfU_k+\bfM_{\lift,k}+\bfM_{\turb,k}+\bfM_{\wall,k},
\end{equation}\begin{equation}\label{Kd}
\Kd=\frac{3}{4}\alpha_k\rho_\rc C_{\rD,k}\frac{1}{d_k} \left|\bfU_\rc-\bfU_k\right|.
\end{equation}\begin{equation}
{A_{k,\mathrm{P}}}\mathbf{U}_{k,\mathrm{P}}{\rm{ + }}\sum {A_{k,\mathrm{N}}\mathbf{U}_{k,\mathrm{N}}} = S_{k,\mathrm{P}},
\label{apanmomrd}
\end{equation}\begin{equation}
\mathbf{HbyA}_{k,\mathrm{P}} = \frac{1}{{{A_{k,\mathrm{P}}}}}\left( { - \sum {{A_{k,\mathrm{N}}}\mathbf{U}_{k,\mathrm{N}}} + S_{k,\mathrm{P}}} \right),
\label{hbyad}
\end{equation}\begin{equation}
\bfU_{k,\rP} = \bfHbyA_{k,\rP}+\frac{\alpha_{k,\rP}}{A_{k,\rP}}\left(\nabla p_{\mathrm{rgh},\rP}-\alpha_{\rc,\rP}\left(\rho_\rc-\rho_k\right)\bfg-\bfg\cdot\bfh_\rP\nabla\rho_\rP\right)+\frac{\Kd_k}{A_{k,\rP}}\bfU_{\rc,\rP},
\label{hbyad2}
\end{equation}\begin{equation}\label{incompressiblep}
\sum\alpha_{k,f}\phi_{k}+\alpha_{\rc,f}\phi_{\rc}=\nabla\cdot\left(\left(\sum\alpha_{k,\rP}\frac{\alpha_{k,\rP}}{A_{k,\rP}}+\alpha_{\rc,\rP}\frac{\alpha_{\rc,\rP}}{A_{\rc,\rP}}
\right)\nabla p_{\mathrm{rgh},\rP}\right),
\end{equation}\begin{equation}
\phi_{k}=\left(\bfHbyA_{k,f}+\frac{\alpha_{k,f}}{A_{k,f}}\left(-\alpha_{\rc,f}\left(\rho_\rc-\rho_\rd\right)\bfg-\bfg\cdot\bfh_f\nabla\rho_f\right)+\frac{\Kd_f}{A_{k,f}}\bfU_{\rc,f}\right)\cdot\bfS_f
\end{equation}\begin{equation}
\phi_{\rc}=\left(\bfHbyA_{\rc,f}+\frac{\alpha_{\rc,f}}{A_{\rc,f}}\left(-\alpha_{\rd,f}\left(\rho_\rd-\rho_\rc\right)\bfg-\bfg\cdot\bfh_f\nabla\rho_f\right)+\frac{\Kd_f}{A_{\rc,f}}\bfU_{\rd,f}\right)\cdot\bfS_f
\end{equation} -
李 李东岳 被引用 于这个主题