Полная система уравнений несжимаемой вязкой жидкости

Last modified by Alexey Popov on 2019/09/16 08:02

\frac  {D{\vec  {v}}}{D t}+\frac  {1}{\rho }\nabla p + \frac{1}{\rho}\nabla \times\rho \vec{m}=\nu \square {\vec  {v}}+{\vec  {f}}

-\frac{1}{c^2}\frac  {1}{\rho }\frac  {D{\rho \vec  {m}}}{D t}-\frac{1}{c^2}\frac  {1}{\rho }\nabla (\rho(\vec{v}\cdot\vec{m})) + \nabla \times \vec{v}=\frac{1}{c^2}\frac{1}{\rho}\nu \square \rho {\vec  {m}}+{\vec  {\omega}}

\nabla \cdot \vec{v} + \frac{1}{c^2}\frac{D(\vec{v}\cdot\vec{v})}{D t} = 0

\nabla \cdot \rho \vec{m} + \frac{1}{c^2}\frac{D\rho(\vec{v}\cdot \vec{m})}{D t} = 0

\vec{m} = \vec{r} \times \vec{a}

Интегральная форма

\oint{\rho\vec{m}}\cdot \vec{dl}+\int \nabla p \cdot \vec{ds}-\int{\rho\vec{g}}\cdot \vec{ds}+\int{\frac{d \rho \vec{v}}{d t }}\cdot \vec{ds}=0

\nabla \cdot \rho\vec{v}+\frac{1}{c^2}\frac{dp}{dt}+\frac{\partial \rho}{\partial t}=0

\oint{\rho\vec{v}}\cdot \vec{dl}-\frac{1}{c^2} \int \nabla \rho(\vec{v}\cdot\vec{m}) \cdot \vec{ds}-\int{\rho\vec{\omega}}\cdot \vec{ds}-\frac{1}{c^2}\int{\frac{d \rho \vec{m}}{d t }}\cdot \vec{ds}=0

\nabla \cdot \rho \vec{m}+\frac{1}{c^2}\frac{d(\rho (\vec{v}\cdot\vec{m}))}{dt}=0

Tags:
Created by Alexey Popov on 2019/09/13 11:16