Total Time Derivative Identity $\frac{d \nabla \phi}{dt} = \nabla \frac{d\phi}{dt} - \nabla \mathbf{u} \nabla{\phi}$ I'm reading a paper where $\phi: (\Omega\subset \mathbb{R}^3) \times [0,T] \to \mathbb{R}$ is assumed to be continuously differentiable with respect to all variables and $\mathbf{u} = (\dot{x},\dot{y},\dot{z})$ is the velocity of a fluid. The authors write the total time derivative of $ \frac{d \nabla \phi}{dt}$ as
$$ \frac{d \nabla \phi}{dt} = \nabla \frac{d\phi}{dt} - \nabla \mathbf{u} \nabla{\phi}.$$
To the best of my knowledge, the left-hand side of the equation above should read as $\frac{d\nabla \phi}{dt} = \frac{\partial \nabla \phi}{\partial t} + \mathbf{u}\cdot \nabla \nabla \phi$. I have no idea where the authors of this paper are getting this expression from. Any ideas?
 A: It is heuristically easier to start from
$$ \nabla \frac{d\phi}{dt} \overset?=\frac{d\nabla\phi}{dt}+\nabla u \nabla \phi$$
because calculus rules do not introduce minus signs (except for inverse function theorem). Assuming $\phi$ is a vector $\phi=\phi^i$,
$$\text{LHS}_{ij}= \partial_j (\partial_t \phi^i + u^k\partial_k\phi^i)=\partial_t\partial_j\phi^i + (\partial_j u^k) \partial_k \phi^i + u^k \partial_k \partial_j\phi^i$$
the first and last terms are $d(\nabla \phi)/dt$ and the remaining middle term is $\nabla \phi\nabla u$ (note: order is different...)
If $\phi$ is a scalar then you can ignore the $i$s and we get
for the middle term $ (\partial_j u^k) \partial_k \phi = \nabla\phi^T\nabla u$.
NB I'm using the convention that $\nabla \phi$ is a column vector if $\phi$ is a scalar, but if $\phi=\phi^i$ is a vector then the Jacobian matrix $(\nabla \phi)_{ij}=\partial_j \phi^i$ i.e. the rows of $\nabla \phi$ are $(\nabla\phi^i)^T$.  Since i wrote out the coordinate expressions directly it should be easy to adapt to other conventions.
