Define the operator $$F(\boldsymbol{\xi})=\nabla(\boldsymbol{\xi}\cdot\nabla p)$$ where $p$ is a given smooth function.
Show that $F$ is self-adjoint, i.e., there exists a vector field $P$ such that
$$\int_V \boldsymbol{\eta}\cdot F(\boldsymbol{\xi})-\int_V \boldsymbol{\xi}\cdot F(\boldsymbol{\eta})=\int_{\partial V} P(\boldsymbol{\xi},\boldsymbol{\eta})\cdot\mathrm{d}S.$$
holds for all bounded area $V$.
My attempts
The operator being self-adjoint is equivalent to RHS being determined by boundary condition.
Thus, I tried to integrate by parts: $$\int_V v\frac{\partial u}{\partial x_i}=-\int_V u\frac{\partial v}{\partial x_i}+\text{Boundary term}.$$
From now on, I use the Einstein summation convention:
The explicit formula for $\nabla(f\cdot g)$ is rather complex, so I decide to use brute force:
\begin{align} &\boldsymbol{\eta}\cdot F(\boldsymbol{\xi})-\boldsymbol{\xi}\cdot F(\boldsymbol{\eta})\\\\ =&\eta_i\frac{\partial }{\partial x_i}\left(\xi_j\frac{\partial p}{\partial x_j}\right)-\xi_j\frac{\partial }{\partial x_j}\left(\eta_i\frac{\partial p}{\partial x_i}\right)\\\\ =&\eta_i\frac{\partial\xi_j}{\partial x_i}\frac{\partial p}{\partial x_j}-\xi_j\frac{\partial p}{\partial x_i}\frac{\partial\eta_i}{\partial\xi_j}. \end{align} Thus, \begin{align*} &\int_V \boldsymbol{\eta}\cdot F(\boldsymbol{\xi})-\int_V \boldsymbol{\xi}\cdot F(\boldsymbol{\eta})\\\\ =&\int _V\eta_i\frac{\partial\xi_j}{\partial x_i}\frac{\partial p}{\partial x_j}-\int_V\xi_j\frac{\partial p}{\partial x_i}\frac{\partial\eta_i}{\partial\xi_j}\\\\ =&-\int_V\xi_j\left(\frac{\partial\eta_i}{\partial x_i}\frac{\partial p}{\partial x_j}+\frac{\partial^2 p}{\partial x_i\partial x_j}\eta_i\right)+\int_V\eta_i\left(\frac{\partial^2 p}{\partial x_i\partial x_j}\xi_j+\frac{\partial p}{\partial x_i}\frac{\partial\xi_j}{\partial x_j}\right)+\text{Boundary term}\\\\ =&\int_V\eta_i\frac{\partial p}{\partial x_i}\frac{\partial \xi_j}{\partial x_j}-\xi_j\frac{\partial\eta_i}{\partial x_i}\frac{\partial p}{\partial x_j}+\text{Boundary term}. \end{align*}
If we integrate by parts again, the second-order derivative term $\partial^2\xi_j/\partial x_i\partial x_j$ will occur. I cannot proceed anymore.
Other facts
This problem comes from our physics course concerning fluid mechanics. One of my friends majoring in mathematics told me to try integration by parts, which turned out to be less fruitful than we had imagined. As I mentioned above, the expression of $\nabla(f\cdot g)$ uses cross product and has four terms (in fact, one term is zero but not helping at all), so I cannot directly apply divergence theorem. The example dicussed in class is the Laplacian $\nabla^2(\boldsymbol{\xi})=(\nabla^2\xi_1,\nabla^2\xi_2,\nabla^2\xi_3)$, which is easily worked out by $v\Delta u=\nabla\cdot(v\nabla u)-\nabla u\cdot\nabla v$ and the divergence theroem. According to my limited knowledge, this exercise is far more difficult than the example.
Thanks for reading such a long question written in unauthentic English. Any idea is welcomed! (I can do the calculation once the idea is outlined and you can post some instruction instead of a detailed proof)