# Is there a proof of this identity: $(\mathbf{\nabla} \cdot \textbf{u})P = (\mathbf{\nabla} P) \cdot \textbf{u}$

I want to verify if this vector identity is true and if so is there a simple proof of it? $\textbf{u}$ is a vector and $P$ is a scalar field, namely the pressure.

$$(\mathbf{\nabla} \cdot \textbf{u})P = (\mathbf{\nabla} P) \cdot \textbf{u}$$

• It does not seem this identity holds for any vector field $\textbf{u}$ and scalar $P$. – Reza Jan 10 '18 at 22:52

In Cartesian coordinates and using index notation one finds $$(\nabla \cdot u) P = (\partial_i u_i) P$$ but on the other hand $$(\nabla P)\cdot u = (\partial_i P) u_i$$ In general these are not equal.
Notice that $$(\nabla \cdot u) P = (\partial_i u_i) P = \partial_i (u_i P) - (\partial_i P) u_i = \nabla\cdot(P u) - u\cdot\nabla P$$ Even if $\nabla\cdot(P u) = 0$, there would be a sign difference with your statement.