Уравнение непрерывности поля скоростей

Last modified by Alexey Popov on 2019/02/27 15:27

$p=\frac{{1}}{3}\rho_0 \mathbf v^2$

$\mathcal L(x,t) = KE - PE = \frac{1}{2}\rho_0 \mathbf v^2 - \frac{1}{2}\frac{p^2}{ \rho_0 \, c_s^2} =$

$=\frac{1}{2}\rho_0 \mathbf v^2 - \frac{1}{18}\frac{\rho_0^2\mathbf v^4}{ \rho_0 \, c_s^2}$

$\frac{\partial \mathcal L}{\partial \mathbf v} = \rho_0 \mathbf v - \frac{2}{9}\frac{\rho_0\mathbf v^3}{ c_s^2}$

$\frac{d}{dt} \frac{\partial^2 \mathcal L}{\partial \mathbf v^2} = - \frac{2}{3}\frac{\rho_0}{ c_s^2}\frac{\partial \mathbf v^2}{\partial t} - \frac{2}{3}\frac{\rho_0 }{ c_s^2}\mathbf v \cdot \nabla \mathbf v^2$

$\nabla \cdot \frac{\partial \mathcal L}{\partial \mathbf v} - \frac{d}{dt} \frac{\partial^2 \mathcal L}{\partial \mathbf v^2} = \rho_0 \nabla \cdot \mathbf v + \frac{2}{3}\frac{\rho_0}{ c_s^2}\frac{\partial \mathbf v^2}{\partial t} + \frac{4}{9}\frac{\rho_0 }{ c_s^2} \nabla\cdot (\mathbf v\mathbf v^2) - \frac{2}{3}\frac{\rho_0 }{ c_s^2}(\nabla \cdot \mathbf v) \mathbf v^2 = 0$

$\mathbf a = \frac{\partial v}{\partial t} + \frac{1}{2}\nabla \mathbf v^2 + \mathbf v \times (\nabla \times \mathbf v)$

$\frac{\partial}{\partial t} \frac{\partial^2 \mathcal L}{\partial \mathbf v^2} = - \frac{2}{3}\frac{\rho_0}{ c_s^2}\frac{\partial \mathbf v^2}{\partial t}$

$\nabla \cdot \frac{\partial \mathcal L}{\partial \mathbf v} - \frac{\partial}{\partial t} \frac{\partial^2 \mathcal L}{\partial \mathbf v^2} = \rho_0 \nabla \cdot \mathbf v + \frac{1}{2}\frac{4}{3}\frac{\rho_0}{ c_s^2}\frac{\partial \mathbf v^2}{\partial t} = 0$

$\nabla \mathcal L - \frac{d}{dt}\frac{\partial \mathcal L}{\partial \mathbf v} + \frac{d^2}{dt^2}\frac{\partial \mathcal L}{\partial \mathbf a} = 0$

$\frac{1}{2}\rho_0 \nabla \mathbf v^2 - \frac{1}{18}\frac{\rho_0}{c_s^2} \nabla \mathbf v^4 - \rho_0\frac{d}{dt} \mathbf v + \frac{2}{9}\frac{\rho_0}{c_s^2}\frac{d}{dt}\mathbf v^3= 0$

$\rho_0 \nabla \cdot \mathbf v - \frac{2}{9}\frac{\rho_0}{c_s^2}\nabla \cdot \mathbf v^3 + \frac{2}{3}\frac{\rho_0}{c_s^2}\frac{d}{dt}\mathbf v^2= 0$

$\nabla \cdot \mathbf v = -\frac{2}{3}\frac{1}{c_s^2}\frac{\partial \mathbf v^2}{\partial t} + \frac{2}{9}\frac{\rho_0}{c_s^2}\nabla \cdot \mathbf v^3- \frac{2}{3}\frac{1}{c_s^2}\mathbf v \cdot \nabla\mathbf v^2$

$\nabla \cdot \mathbf v = -\frac{1}{2}\frac{1}{c_s^2}\frac{\partial \mathbf v^2}{\partial t} + \frac{1}{2}\frac{1}{c_s^2}\mathbf v^2 \nabla\cdot \mathbf v$

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Created by Alexey Popov on 2019/02/12 13:11