How do I expand: $\langle u , \nabla\rangle u$? I'm studying the Landau (VI) and when he "introduces" the material derivative (he is building up the continuity equation), something like this appears: $(u,\operatorname{grad})u$ (sometimes referred as: $\langle u , \nabla\rangle u$ ) which is, if I understood correctly, the derivative of the vector field $u$ in the direction of the vector $u$, divided by "$dt$". (I still cannot understand how a generic $dr$ can be equal to $(dx,dy,dz)$ so that, right after, he can divide by $dt$ and $\frac{dr}{dt} = u$.)
How can i expand:  $\langle u , \nabla\rangle u = (u, \operatorname{grad})$ ? Does it give a tensor?
What sort of operation is that? I can recognize $\operatorname{div} u = \langle\nabla,u\rangle$, but that's not the case of interest, and $\operatorname{grad} p = \nabla p $ necessarily takes a scalar field $p$.
So what is " $\langle u , \nabla\rangle u$" asking me to compute, and where can I learn the theory behind it?
 A: The object $\langle \vec{u}, \nabla\rangle$ is an operator defined by $\langle \vec{u}, \nabla\rangle = u_x \frac{\partial}{\partial x} + u_y \frac{\partial}{\partial y} + u_z \frac{\partial}{\partial z}$. It acts on whatever is on the right. In your case it acts on the vector field $\vec{u}(x,y,z,t)$. The action is on each component of $\vec{u}$ separately, so $\langle \vec{u}, \nabla\rangle \vec{u}$ is a vector field:
$$\langle \vec{u}, \nabla\rangle \vec{u} = \left( {\begin{array}{*{20}{c}}
  \langle \vec{u}, \nabla\rangle u_x\\ 
   \langle \vec{u}, \nabla\rangle u_y \\ 
  \langle \vec{u}, \nabla\rangle u_z
\end{array}} \right)
=\left( {\begin{array}{*{20}{c}}
  u_x \frac{\partial u_x}{\partial x} + u_y \frac{\partial u_x}{\partial y} + u_z \frac{\partial u_x}{\partial z}\\ 
   u_x \frac{\partial u_y}{\partial x} + u_y \frac{\partial u_y}{\partial y} + u_z \frac{\partial u_y}{\partial z} \\ 
  u_x \frac{\partial u_z}{\partial x} + u_y \frac{\partial u_z}{\partial y} + u_z \frac{\partial u_z}{\partial z}
\end{array}}\right).$$
I think the easiest way of understanding where this operator comes from is by imagining following a small parcel of material as it moves along with the flow. For instance, how does the temperature of the parcel change over the short time $dt$?
The parcel is located at position $\vec{r}_1=\left(\matrix{x\\y\\z}\right)$ at time $t_1$ and after a time $dt$ has traveled to $\vec{r}_2$. Since $\vec{u}$ is the parcel's velocity, at time $t_2=t_1+dt$ the parcel is at $\vec{r}_2=\vec{r}_1+\vec{u}dt$.
The temperature field $T(\vec{r},t)$ is a scalar function of $x,y,z$ and $t$. The difference in temperature of the parcel between $t_1$ and $t_2$ is (to lowest order in $dt$),
$$\begin{align}
T(\vec{r}_2, t_2) - T(\vec{r}_1, t_1) &= T(\vec{r}_1+\vec{u}dt, t_1+dt) - T(\vec{r}_1, t_1)\\
&=T(x+u_xdt,y+u_ydt,z+u_zdt,t_1+dt)-T(x,y,z,t_1)\\
&=T(x,y,z,t) + \frac{\partial T}{\partial x} u_x dt + \frac{\partial T}{\partial y} u_y dt + \frac{\partial T}{\partial z} u_z dt + \frac{\partial T}{\partial t} dt - T(x,y,z,t)\\
&= \left( \frac{\partial T}{\partial x} u_x + \frac{\partial T}{\partial y} u_y + \frac{\partial T}{\partial z} u_z + \frac{\partial T}{\partial t} \right) dt\\
&= \left[ \left(\langle \vec{u}, \nabla \rangle + \frac{\partial}{\partial t} \right)T \right] dt.
\end{align}
$$
The rate of change of the temperature of the parcel is the above divided by $dt$. If you wanted to get the rate of change of the $x$-component of the parcel's velocity you would replace $T$ by $u_x$, and similarly for the $y$ and $z$ components.
