# Interpretation of $\nabla$ operator in an expression

Let a tensor (3x3) be of the form $$U = \mathbf{u}\mathbf{v}$$ ($$\mathbf{u}$$ and $$\mathbf{v}$$) being two fluid velocity vectors (of dimension 1x3).

In my analysis, for such a tensor $$U$$, following expression arises, $$$$E = \frac{1}{2}\nabla^2(\mathbf{u}\cdot\mathbf{v}) + \frac{1}{2}\left(\nabla\cdot(\mathbf{u}\cdot\nabla\mathbf{v}) + \nabla\cdot(\mathbf{v}\cdot\nabla\mathbf{u})\right)$$$$

Above expression remains the same when we replace $$\mathbf{v}$$ with $$\mathbf{u}$$, although the original tensor $$U$$ changes to its transpose with such replacement.

Now, I would like to write down the expression $$E$$ above for the case where $$\mathbf{u}$$ is replaced with gradient operator $$\nabla$$. In this case $$U = \nabla\mathbf{v}$$. I proceeded with blindly replacing $$\mathbf{u}$$ with $$\nabla$$ to arrive at, $$\begin{eqnarray} E &=& \frac{1}{2}\nabla^2(\nabla\cdot\mathbf{v}) + \frac{1}{2}\left(\nabla\cdot(\nabla\cdot\nabla\mathbf{v}) + \nabla\cdot(\mathbf{v}\cdot\nabla\nabla)\right) \\ &=& \frac{1}{2}\nabla^2(\nabla\cdot\mathbf{v}) + \frac{1}{2}\nabla^2(\nabla\cdot\mathbf{v}) + \nabla\cdot(\mathbf{v}\cdot\nabla\nabla) \\ &=& \nabla^2(\nabla\cdot\mathbf{v})+\nabla\cdot(\mathbf{v}\cdot\nabla\nabla) \end{eqnarray}$$

I am not sure how to interpret the last term in the above expression $$\nabla\cdot(\mathbf{v}\cdot\nabla\nabla)$$. This term does not have an argument on the right hand side of tensor $$\nabla\nabla$$. Any help in the interpretation is highly appreciated.

• The part $\nabla\cdot(\mathbf{v}\cdot\nabla\mathbf{u})$ in the original equation doesn't make any sense. If $\mathbf u$ and $\mathbf v$ are both vectors, then what is $\nabla \mathbf u$ and what is $\mathbf{v}\cdot\nabla\mathbf{u}$? – polfosol Mar 30 '19 at 12:00
• @polfosol The gradient of a vector is defined in this Wikipedia page. en.wikipedia.org/wiki/Gradient#Gradient_of_a_vector . I think that $\bf{v} .\nabla u$ is the vector ${\bf {u}}^T{\bf{J}}^T$, where $\bf J$ is the Jacobian matrix for $\bf u$. – Angela Pretorius Apr 1 '19 at 19:37

You mention tensors so I assume you are familiar with tensor notation. This is the notation which is almost necessary to use in these situations. Assuming you are dealing with affine coordinates we write (for $$\mathbb{u}=u_i$$) that $$\nabla \mathbb{u}=\partial_ju_i$$, while $$\nabla\bullet\mathbb{u}=\partial_iu_i=\sum_i\partial_iu_i$$. Then we see that $$\nabla\nabla=\partial_j\partial_i$$.
Now, I think the key observation in your example here is to notice the expression $$\mathbb{v}\bullet\nabla\mathbb{u}=\mathbb{v}^T\nabla\mathbb{u}=(\nabla\mathbb{u})^Tv=(\nabla\mathbb{u})^T\bullet v$$. This gives us the interpretation \begin{align*} \mathbb{v}\bullet\nabla\nabla=(\nabla\nabla)^T\bullet v = \partial_i\partial_j v_j. \end{align*} From here you can try to figure out the rest.