Дивергенция вектора скорости

Last modified by Alexey Popov on 2019/02/27 13:28

$\mathbf u = 2\frac{\partial \mathbf r}{\partial t}\,+\,2\mathbf {u} + \nabla \left(\mathbf {r} \cdot \mathbf {u} \right)- \mathbf {r} \left(\nabla \cdot \mathbf {u} \right) \,+\,\nabla \times \left(\mathbf {r} \times \mathbf {u} \right) - \mathbf {r} \times \boldsymbol \omega$

$\nabla \cdot \mathbf u = 2\nabla \cdot \mathbf {u} + \nabla \cdot (\nabla \left(\mathbf {r} \cdot \mathbf {u} \right))- \nabla \cdot (\mathbf {r} \left(\nabla \cdot \mathbf {u} \right)) - \nabla \cdot (\mathbf {r} \times \boldsymbol \omega)$

$\nabla \cdot \left(\left(\nabla \cdot \mathbf {u} \right) \mathbf {r} \right)\ =\ \left(\nabla \cdot \mathbf {u} \right) \,\nabla \cdot \mathbf {r} \,+\,\mathbf {r} \cdot (\nabla \left(\nabla \cdot \mathbf {u} \right)) =\ 3\left(\nabla \cdot \mathbf {u} \right) \,+\,\mathbf {r} \cdot (\nabla \left(\nabla \cdot \mathbf {u} \right))$

$2 \nabla \cdot \mathbf u = \nabla \cdot (\nabla \left(\mathbf {r} \cdot \mathbf {u} \right)) -\,\mathbf {r} \cdot (\nabla \left(\nabla \cdot \mathbf {u} \right)) - \nabla \cdot (\mathbf {r} \times \boldsymbol \omega)$

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Created by Alexey Popov on 2019/02/27 13:22