做个公式记录


  • 网格教授 OpenFOAM教授 管理员

    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。


    Third term in Eq. 8.122:

    \begin{equation}
    \frac{1}{\tau_0 \xi^{2/3}}\frac{\partial vn}{\partial v}
    \end{equation}
    Transform it to be moments:

    \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}
    \frac{\mathrm{d}u}{\mathrm{d}t}=\frac{u_c-u_d}{A}
    \end{equation}

    \begin{equation}
    \frac{\mathrm{d}(u_d-u_c)}{u_d-u_c}=-\frac{\mathrm{d}t}{A}
    \end{equation}

    \begin{equation}
    \mathrm{ln}(u_d-u_c)=-\frac{1}{A}+C
    \end{equation}

    \begin{equation}
    u_d-u_c=e^{-t/A+C}
    \end{equation}

    $t=0$, $u_d=u_d^*$

    \begin{equation}
    u_d^*-u_c=e^{C}
    \end{equation}

    \begin{equation}
    C=\mathrm{ln}(u_d^*-u_c)
    \end{equation}

    for next time step:

    \begin{equation}
    u_d=\mathrm{exp}(-\frac{\Delta t}{A}+\mathrm{ln}(u_d^* -u_c))=e^{-\Delta t/A}(u_d^*-u_c)
    \end{equation}


  • 网格教授 OpenFOAM教授 管理员

    \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}


  • 网格教授 OpenFOAM教授 管理员

    $$\rho=\begin{matrix} a & b \\ c & d \end{matrix}$$

    $$\rho=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$$