Анализ новых уравнений движения жидкости

Last modified by Alexey Popov on 2020/11/11 14:58

$\frac{D\mathbf v}{Dt}=-\frac{1}{\rho}\nabla p + \mathbf g$

$\frac{D p}{D t} = - \rho v_s^2(\nabla \cdot \mathbf v)$

$\frac{D\mathbf v}{Dt}=-\frac{1}{\rho}\nabla p + \nu \Delta \mathbf v + \mathbf g$

$\frac{D p}{D t}= - \rho v_s^2(\nabla \cdot \mathbf v) + \rho\nu(\nabla \cdot \mathbf a)$

$\frac{D\mathbf a}{Dt}=-\frac{1}{\rho}\nabla\frac{D p}{Dt} + \nu \Delta \mathbf a$

$\frac{D\mathbf a}{Dt}=v_s^2\nabla(\nabla \cdot \mathbf v) - \nu\nabla(\nabla \cdot \mathbf a) + \nu \Delta \mathbf a$

$\frac{D\mathbf a}{Dt}=v_s^2\nabla \times (\nabla \times \mathbf v)+v_s^2\Delta\mathbf v - \nu\nabla \times (\nabla \times \mathbf a)$

$\frac{D\boldsymbol \omega}{Dt} = \nu \nabla \times \Delta \mathbf v$

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Created by Alexey Popov on 2020/11/11 14:57