While I was studying on analysis of the NSE and MHD problem, I encountered the following identity.
\begin{eqnarray} \nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big)=Q({\bf u},{\bf u})-\dfrac{1}{2} {\bf u} \cdot \nabla (\nabla \cdot {\bf u})+\dfrac{1}{2}|\nabla \cdot {\bf u}|^2, \hspace{2cm}(1) \end{eqnarray} where $$Q({\bf u},{\bf u}):=\nabla {\bf u}: (\nabla {\bf u})^T=\sum_{i,j} \dfrac{\partial u_i}{\partial x_j}\dfrac{\partial u_j}{\partial x_i}, \quad d=2,3.$$ I need to generalize the previous identity as
\begin{eqnarray} \nabla \cdot \Big(({\bf u}\cdot\nabla) {\bf v}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf v})\Big)&=&Q({\bf u},{\bf v})+C_1{\bf u} \cdot \nabla (\nabla \cdot {\bf v})\nonumber\\&&+C_2{\bf v} \cdot \nabla (\nabla \cdot {\bf u})+C_3(\nabla \cdot {\bf u}) (\nabla \cdot {\bf v}),\hspace{2.5cm} (2) \end{eqnarray} where $C_1, C_2, C_3$ are unknown coefficients. I need to find them. I believe that, if I can prove the equation (1), I can get the equation (2) in a similar way. Hence, first I tried to prove the equation (1).
My attempt at a solution: \begin{eqnarray} \nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big)=\nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}\Big)+ \nabla \cdot \Big(\dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big). \hspace{2.5cm} (3) \end{eqnarray} By using $\nabla \cdot \Big(\psi {\bf u}\Big)=\psi \nabla \cdot {\bf u} +(\nabla \psi)\cdot {\bf u}$, the first term on the right hand side of (3) can be organized as \begin{eqnarray} \nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}\Big)= ({\bf u}\cdot\nabla ) \nabla \cdot {\bf u} +\Big(\nabla ({\bf u}\cdot\nabla )\Big)\cdot {\bf u} \qquad \qquad \quad \end{eqnarray} Utilizing $\nabla ({\bf u}\cdot\nabla )=\nabla (\nabla \cdot {\bf u} )$ and $\nabla (\nabla \cdot {\bf u} )=\nabla \times (\nabla \times {\bf u} )+\Delta {\bf u}$, we get \begin{eqnarray} \nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}\Big)= ({\bf u}\cdot\nabla ) \nabla \cdot {\bf u} +\nabla \times (\nabla \times {\bf u} )\cdot{\bf u} +\Delta {\bf u} \cdot {\bf u}\hspace{3.5cm} (4) \end{eqnarray}
In a similar manner, the second term on the right-hand side of (3) can be written as \begin{eqnarray} \nabla \cdot \Big(\dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big)&=&\dfrac{1}{2} (\nabla \cdot {\bf u}) \nabla \cdot {\bf u} +\dfrac{1}{2}(\nabla (\nabla \cdot {\bf u}))\cdot {\bf u} \nonumber\\ &=&\dfrac{1}{2} |\nabla \cdot {\bf u})|^2 +\dfrac{1}{2}\nabla \times (\nabla \times {\bf u} )\cdot {\bf u} +\dfrac{1}{2}\Delta {\bf u}\cdot {\bf u} \hspace{3cm}(5) \end{eqnarray} Inserting (4) and (5) into (3) produces
\begin{eqnarray} \nabla \cdot \Big(({\bf u}\cdot\nabla ){\bf u}+ \dfrac{1}{2}(\nabla \cdot {\bf u}){\bf u})\Big)&=& ({\bf u}\cdot\nabla ) \nabla \cdot {\bf u} +\nabla \times (\nabla \times {\bf u} )\cdot{\bf u} +\Delta {\bf u} \cdot {\bf u}\nonumber\\ &&+\dfrac{1}{2} |\nabla \cdot {\bf u})|^2 +\dfrac{1}{2}\nabla \times (\nabla \times {\bf u} )\cdot {\bf u} +\dfrac{1}{2}\Delta {\bf u}\cdot {\bf u} \end{eqnarray}
I'm stuck here. How can I continue?
@Sal says the identity may need a sign change for the first term on each side, so I'll prove that modified version. (I'll delete this answer if it becomes clear the real issue isn't those signs, but they would fix it.) Summing over repeated indices (all but the leftmost denoting a derivative), the left-hand side is$$(-u_ju_{ij}+\tfrac12u_{jj}u_i)_i=-u_{ji}u_{ij}-u_ju_{iji}+\tfrac12u_{jji}u_i+\tfrac12u_{jj}u_{ii},$$while the right-hand side is$$-u_{ij}u_{ji}-\tfrac12u_iu_{jji}+\tfrac12u_{ii}u_{jj},$$so left minus right is$$-u_ju_{iji}+u_{jji}u_i.$$But $u_{jji}u_i=u_{iij}u_j=u_{iji}u_j$, first by relabelling, then commuting derivatives.