= Solution
Let $h_i=\delta F/\delta p_i$ be the <polar molecular field>, with the positive <functional derivative> convention used in the paper. During a pure advective displacement $u_j$, the <polar order parameter> changes by
$$
\delta p_i=-u_j\partial_jp_i-\Omega^u_{ij}p_j+\xi D^u_{ij}p_j,
\quad \Omega^u_{ij}=\frac{\partial_i u_j-\partial_j u_i}{2},
\quad D^u_{ij}=\frac{\partial_i u_j+\partial_j u_i}{2}.
$$
The <free energy> change is $\delta F=\int h_i\delta p_i\,d^dr$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and <incompressibility> $\partial_j u_j=0$. <Integration by parts> in the advective term gives
$$
-\int h_i u_j\partial_jp_i\,d^dr
=\int u_jp_i\partial_jh_i\,d^dr
=\int\Sigma^{(1)}_{ij}\partial_i u_j\,d^dr,
\qquad \partial_i\Sigma^{(1)}_{ij}=-p_k\partial_jh_k.
$$
Relabelling the dummy indices in the rotational and <flow alignment of a polar order parameter> terms gives
$$
-h_i\Omega^u_{ij}p_j
=\frac{p_i h_j-p_jh_i}{2}\partial_i u_j,
\qquad
\xi h_iD^u_{ij}p_j
=\frac{\xi(p_i h_j+p_jh_i)}2\partial_i u_j.
$$
Therefore $\delta F=\int\Sigma^p_{ij}\partial_i u_j\,d^dr$, with the <reversible stress of a polar liquid crystal>
$$
\boxed{\Sigma^p_{ij}=\Sigma^{(1)}_{ij}
+\frac{p_i h_j-p_jh_i}{2}
+\frac{\xi(p_i h_j+p_jh_i)}2,
\quad \partial_i\Sigma^{(1)}_{ij}=-p_k\partial_jh_k}.
$$
For a local <free-energy density> $f(\mathbf p,\nabla\mathbf p)$, an explicit choice is
$$
\Sigma^{(1)}_{ij}=(f-p_kh_k)\delta_{ij}
-\frac{\partial f}{\partial(\partial_i p_k)}\partial_jp_k.
$$
Its <divergence> is exactly $-p_k\partial_jh_k$, by the <Euler-Lagrange equation> for $h_k$. A <pressure> term can be reassigned in an <incompressible flow>; the chosen representative realizes the printed divergence without that ambiguity. The force density $\partial_i\Sigma^p_{ij}$ has mechanical power $-\int\Sigma^p_{ij}\partial_i v_j$, the negative of the advective <free energy> rate, which checks the stress sign.
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