Interpretation of $\nabla$ operator in an expression

Question

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, \begin{equation} 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) \end{equation}

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.

Answer 1

Your method should be preceded by some caution, replacing a vector by the del operator is dangerous in the sense that the del operator does not commute the way a vector does, in inner products for instance. I think this is what causes your problem.

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.