Is there a vector identity that can rewrite a product of type
$$ \mathbf{a}\cdot(\mathbf{a}\cdot\nabla\mathbf{a}), $$
where $\mathbf{a}$ is a vector, into a divergence, for example of the form
$$ \mathbf{a}\cdot(\mathbf{a}\cdot\nabla\mathbf{a})= \nabla\cdot(|\mathbf{a}|^2 \mathbf{a}) + … ? $$
I have had no luck so far in trying to employ any of the well known vector identities, but it feels like something like this should be possible
Best Answer
I am interpretting your product as $$\mathbf{a}\cdot\left[\left(\mathbf{a}\cdot\nabla\right)\mathbf{a}\right]=\mathbf{a}\cdot\left(\sum_i a_i\partial_i \mathbf{a}\right)=\sum_{ij}a_i a_j\partial_i a_j.$$ The divergence you use as an example can be written as $$\nabla\cdot\left(|\mathbf{a}|^2\mathbf{a}\right)=\sum_{ij}\partial_i\left(a_j^2 a_i\right)=2\sum_{ij}a_ia_j\partial_ia_j+\sum_{ij}a_j^2\partial_ia_i.$$ Identifying $$\sum_{ij}a_j^2\partial_ia_i=|\mathbf{a}|^2\left(\nabla\cdot\mathbf{a}\right),$$ your desired identity could be $$\mathbf{a}\cdot\left[\left(\mathbf{a}\cdot\nabla\right)\mathbf{a}\right]=\frac{1}{2}\left[\nabla\cdot\left(|\mathbf{a}|^2\mathbf{a}\right) - |\mathbf{a}|^2\left(\nabla\cdot\mathbf{a}\right)\right].$$
Here I have used the shorthand $\partial_i=\partial/\partial i$.