Vector Calculus and the Navier Stokes Equation I am reading through a paper that derives the thin-film equation from the Navier Stokes equation; however, am having some trouble understanding the vector calculus notation used. For example, the material derivative is defined as:
$$ \frac{D\mathbf{v}}{dt} = \frac{\partial\mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}$$
where in $2D$ we have $\mathbf{v}=(u,w)$ and $\nabla=(\partial_x,\partial_z)$. Hence, the gradient of $\mathbf{v}$ is the transpose of the Jacobian given by:
$$ \nabla\mathbf{v} = \begin{bmatrix} u_x & w_x \\ u_z & w_z \end{bmatrix} $$
Now my question is what to make of:
$$ \mathbf{v}\cdot\nabla\mathbf{v}=\begin{bmatrix} u \\ w \end{bmatrix} \cdot \begin{bmatrix} u_x & w_x \\ u_z & w_z \end{bmatrix} $$
As far as I am aware the dot product can be defined for two vectors and can be extended to two square matrices but I don't quite know what to make of this when you have a vector and a square matrix.
Specifically, for this case I already know from past experience that:
$$ \frac{D\mathbf{v}}{dt} = \frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v} = (u_t+uu_x+wu_z,w_t+uw_x+ww_z) $$
but upon coming back to this I am a little confused as to how to show this given the notation above.
 A: It's better to write $(\mathbf{v}\cdot\nabla)\mathbf{v}.$ That makes it clearer:
$$
(\mathbf{v}\cdot\nabla)\mathbf{v}
= (u \partial_x + w \partial_y) \begin{bmatrix}u\\w\end{bmatrix}
= \begin{bmatrix}(u \partial_x + w \partial_y)u\\(u \partial_x + w \partial_y)w\end{bmatrix}.
$$
A: The correct way to parse this, in my opinion, is $(\mathbf{v}\cdot\nabla)\mathbf{v}$, which sidesteps the matrix issue. However, $\nabla \mathbf{v}$ gets used as a tensor a lot in fluid dynamics, so we should probably still address that. In that case, you should see the dot product as symbolizing tensor contraction, which generalizes the dot product to tensors of higher rank. In particular, it also includes standard matrix multiplication as the special case for rank 2 tensors.
Whenever I get confused by exactly how expressions with $\nabla$ and dot products shake out, it can be useful to rely on Einstein notation, which keeps track of the dot products without having to worry about noncommutativity of tensor contractions.
