Корректный вывод уравнения Эйлера для сжимаемой невязкой жидкости

Last modified by Alexey Popov on 2019/06/27 19:29

$L = \frac{1}{2}\rho (\mathbf v \cdot \mathbf v) - \rho c^2$

$\nabla L - \frac{d}{dt}\frac{\partial L}{\partial \mathbf v} = 0$

$\frac{D\rho\mathbf v(t,\mathbf r(t))}{Dt} = \frac{\partial \rho\mathbf v}{\partial t} + (\mathbf v \cdot \nabla)\rho \mathbf v = \rho\frac{\partial \mathbf v}{\partial t}+ \frac{1}{2}\rho \nabla (\mathbf v \cdot \mathbf v) -\rho(\mathbf v \times \boldsymbol\omega) =\rho \mathbf a+ \frac{\partial \rho}{\partial t}\mathbf v$

$\rho\frac{\partial \mathbf v}{\partial t} + \frac{1}{2}\rho\nabla ( \mathbf v \cdot \mathbf v)= \rho \mathbf a+ \frac{\partial \rho\mathbf v}{\partial t} - \frac{1}{2}(\mathbf v \cdot \mathbf v)\nabla \rho + \nabla \rho c^2$

$\frac{1}{c^2 }\frac{\partial \rho\mathbf v}{\partial t} = \frac{1}{c^2 }\rho(\mathbf v \times \boldsymbol \omega)+ \frac{1}{2}\frac{(\mathbf v \cdot \mathbf v)}{c^2 }\nabla \rho - \nabla \rho$

$\frac{\partial \mathbf v}{\partial t} + \frac{1}{2}\nabla ( \mathbf v \cdot \mathbf v)= \mathbf a+ \frac{1}{\rho }\frac{\partial \rho\mathbf v}{\partial t} - \frac{1}{\rho }\frac{\mathbf v \cdot \mathbf v}{2}\nabla \rho + \frac{1}{\rho }\nabla \rho c^2$

$\frac{\partial \mathbf v}{\partial t} + \frac{1}{2}\nabla ( \mathbf v \cdot \mathbf v)- (\mathbf v \times \boldsymbol \omega) = \mathbf a+ \frac{1}{\rho }\frac{\partial \rho}{\partial t}\mathbf v- \frac{1}{2}\frac{1}{\rho }\nabla (\rho(\mathbf v \cdot \mathbf v)) + \frac{1}{\rho }\nabla \rho c^2$

$\frac{D\rho}{Dt} = \frac{\partial \rho}{\partial t} + \mathbf v \cdot \nabla\rho = \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf v )$

$L = \rho (\boldsymbol \omega\cdot \mathbf v) - \rho f c$

$\nabla L - \frac{d}{dt}\frac{\partial L}{\partial \mathbf v} = 0$

$\nabla L = \nabla (\rho (\boldsymbol \omega\cdot \mathbf v)) - \nabla \rho f c= \rho \nabla (\boldsymbol \omega\cdot \mathbf v) + (\boldsymbol \omega\cdot \mathbf v)\nabla (\rho ) - \nabla \rho f c$

$\frac{d}{dt}\frac{\partial L}{\partial \mathbf v} = \frac{D\rho\boldsymbol \omega}{Dt} = \rho\frac{D\boldsymbol \omega}{Dt} + \boldsymbol \omega\frac{D\rho}{Dt} = \rho\boldsymbol \epsilon + \boldsymbol \omega( \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf v ) )$

$\rho \nabla (\boldsymbol \omega\cdot \mathbf v) + 2(\boldsymbol \omega\cdot \mathbf v)\nabla \rho - 2\nabla \rho f c - \rho \frac{\partial \boldsymbol \omega}{\partial t} \, -\rho (\nabla \times \mathbf a) + \rho (\mathbf {v} \times \left(\nabla \times \boldsymbol \omega \right)) = 0$

$\nabla (\boldsymbol \omega\cdot \mathbf v) - \frac{\partial \boldsymbol \omega}{\partial t} -\nabla \times \mathbf a + \mathbf {v} \times (\nabla \times \boldsymbol \omega ) = f c \frac{2}{\rho} \nabla \rho - \frac{2}{ \rho} \nabla \rho (\boldsymbol \omega\cdot \mathbf v)$

$L = \rho (\mathbf r \cdot \mathbf v) - \rho r c$

$\nabla L - \frac{d}{dt}\frac{\partial L}{\partial \mathbf v} = 0$

$\nabla L = \nabla (\rho (\mathbf r \cdot \mathbf v)) - \nabla \rho r c= \rho \nabla (\mathbf r \cdot \mathbf v) + (\mathbf r \cdot \mathbf v)\nabla \rho - \nabla \rho r c$

$\frac{d}{dt}\frac{\partial L}{\partial \mathbf v} = \frac{D\rho\mathbf r }{Dt} = \rho\frac{D\mathbf r }{Dt} + \mathbf r \frac{D\rho}{Dt} = \rho\mathbf v + \mathbf r ( \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf v ) )$

$\rho \nabla (\mathbf r \cdot \mathbf v) + (\mathbf r \cdot \mathbf v)\nabla \rho - \nabla \rho r c - \rho\mathbf v = 0$

$\mathbf v - \nabla (\mathbf r \cdot \mathbf v) = -\frac{1}{ \rho} r c \nabla \rho + \frac{1}{ \rho} (\mathbf r \cdot \mathbf v)\nabla \rho$

Уравнение движения с учётом ускорения

$L = \rho (\mathbf a \cdot \mathbf v) - \rho a c$

$\nabla L - \frac{d}{dt}\frac{\partial L}{\partial \mathbf v} = 0$

$\nabla L = \nabla (\rho (\mathbf a \cdot \mathbf v)) - \nabla \rho a c= \rho \nabla (\mathbf a \cdot \mathbf v) + (\mathbf a \cdot \mathbf v)\nabla \rho - \nabla \rho a c$

$\frac{d}{dt}\frac{\partial L}{\partial \mathbf v} = \frac{D\rho\mathbf a }{Dt} = \rho\frac{D\mathbf a }{Dt} + \mathbf a \frac{D\rho}{Dt} = \rho\mathbf j + \mathbf a ( \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf v ) )$

$\mathbf j = \frac{\partial \mathbf a}{\partial t} + \frac{1}{2} \left ( \nabla \left(\mathbf {a} \cdot \mathbf {v} \right)\,+\,\mathbf {v} \left(\nabla \cdot \mathbf {a} \right) \,+\,\nabla \times \left(\mathbf {a} \times \mathbf {v} \right) - \mathbf {a} \times \boldsymbol \omega-\mathbf {v} \times \left(\nabla \times \mathbf {a} \right) \right)$

$\frac{1}{2} \nabla (\mathbf a \cdot \mathbf v) - \frac{\partial \mathbf a}{\partial t}-\,\frac{1}{2}(\mathbf {v} \left(\nabla \cdot \mathbf {a} \right)) \,-\,\frac{1}{2} (\nabla \times \left(\mathbf {a} \times \mathbf {v} \right)) + \frac{1}{2} (\mathbf {a} \times \boldsymbol \omega)+\frac{1}{2} (\mathbf {v} \times \left(\nabla \times \mathbf {a} \right))= - \frac{1}{\rho} (\mathbf a \cdot \mathbf v)\nabla \rho + \frac{1}{\rho} \nabla \rho a c$

$L = \frac{1}{2}\rho (\boldsymbol \omega \cdot \boldsymbol \omega) - \rho f^2$

$\frac{\partial L}{\partial \boldsymbol \phi} - \frac{d}{dt}\frac{\partial L}{\partial \boldsymbol \omega} = 0$

$\frac{D\rho\boldsymbol \omega}{Dt} = \rho\frac{D\boldsymbol \omega}{Dt} + \boldsymbol \omega\frac{D\rho}{Dt} = \rho\boldsymbol \epsilon + \boldsymbol \omega( \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf v ) )$

$\boldsymbol \varepsilon =\frac{1}{2} \left ( \frac{\partial \boldsymbol \omega}{\partial t} + \nabla \left(\boldsymbol \omega \cdot \mathbf {v} \right)\, + \nabla \times \mathbf a -\mathbf {v} \times \left(\nabla \times \boldsymbol \omega \right) \right)$

$\frac{D\rho(\mathbf v \cdot \mathbf v)}{Dt} = \rho\frac{D(\mathbf v \cdot \mathbf v)}{Dt} + (\mathbf v \cdot \mathbf v)\frac{D\rho}{Dt} = \rho\frac{\partial(\mathbf v \cdot \mathbf v)}{\partial t} + \rho(\mathbf v \cdot \nabla (\mathbf v \cdot\mathbf v )) + (\mathbf v \cdot \mathbf v)\frac{D\rho}{Dt}$

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Created by Alexey Popov on 2019/06/27 13:02