The question is to show that: $$ (\nabla \cdot \rho \mathbf {V V})\cdot \mathbf V = (\nabla \cdot \frac {\rho \mathbf V V^2 }{2} ) $$
I have been stuck a while and would appreciate any help (solutions or hints). Thank you very much for your time.
Grad double dot product vector simplification
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2what are those terms? – 2017-02-07
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0What conditions do you have on the terms included? I suspect, for instance, that you know something about $\rho\mathbf{V}$. – 2017-02-07
1 Answers
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In order to prove this, you have to assume that $\bf V$ is divergenceless (for example because it is the velocity field of a stationary flow).
It is then second relation here. In particular, it states that $$\nabla \cdot (\psi {\bf A}) = \psi \nabla \cdot {\bf A}+ {\bf A}\cdot \nabla \psi.$$
Applying this to $$ {\bf A} = {\bf V}, \qquad \psi = \frac{\rho V^2}2,$$ we obtain $$\nabla \cdot \left(\frac{\rho V^2}2 {\bf V} \right) =\frac{\rho V^2}2 \nabla \cdot {\bf V}+ {\bf V}\cdot \nabla \left(\frac{\rho V^2}2 \right) = {\bf V}\cdot \nabla \left(\frac{\rho V^2}2 \right)$$ provided that $\nabla \cdot {\bf V}=0$.