TITLE: Why is $(\mathbf{v} \cdot \nabla)\mathbf{v} = (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2)$?
QUESTION [5 upvotes]: The Convective Derivative or Material Derivative is usually written as $\frac{D}{Dt}=\frac{\partial}{\partial t} + \mathbf{v} \cdot \nabla$.  According to MathWorld, this equation, multiplied with ${\bf{v}}$ equals:
$$
    \frac{D \mathbf{v}}{Dt} = \frac{\partial \mathbf{v}}{\partial t} + (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2)
$$
Clearly, it must hold that;
$$
    (\mathbf{v} \cdot \nabla)\mathbf{v} = (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2)
$$
However, I do not spot why this is true. What is the (trivial) identity that I am missing?

REPLY [4 votes]: Here is another approach using the BAC-CAB identity and the definition of tensor product. I remind you of the BAC-CAB identity
$$\begin{align}
A \times (B \times C) &= B(A \cdot C)-C(A \cdot B) \\
&= (B \otimes A) \cdot C - A \cdot (B \otimes C) 
\end{align}$$
So we can write
$$\begin{align}
(\nabla \times \mathbf{v}) \times \mathbf{v} &= -\mathbf{v} \times (\nabla \times \mathbf{v}) \\
&= - [(\nabla \otimes {\bf{v}}) \cdot {\bf{v}} - {\bf{v}} \cdot (\nabla \otimes {\bf{v}})] \\
&\equiv - [(\nabla {\bf{v}}) \cdot {\bf{v}} - {\bf{v}} \cdot (\nabla  {\bf{v}})] \\
&=-\nabla {\bf{v}} \cdot {\bf{v}} + {\bf{v}} \cdot \nabla {\bf{v}} \\
&=-\frac{1}{2}\nabla ({\bf{v}} \cdot {\bf{v}}) + {\bf{v}} \cdot \nabla {\bf{v}}
\end{align}$$
or equivalently we get
$${\bf{v}} \cdot \nabla {\bf{v}}=(\nabla \times \mathbf{v}) \times \mathbf{v}+\frac{1}{2}\nabla ({\bf{v}} \cdot {\bf{v}})$$