material derivative of displacement I am slightly confused about what the material derivative of displacement is. 
$$\frac{D}{Dt}=\frac{\partial}{\partial t}+ v\frac{ \partial}{\partial x}$$
which means that for the displacement we should have,
$$\frac{Du}{Dt}=\frac{\partial u}{\partial t}+ v\frac{ \partial u}{\partial x}$$
but we also have that,
$$v=\frac{Du}{Dt}=\frac{\partial u}{\partial t}$$
Does it mean $\frac{\partial u}{\partial x}=0$?
I noticed someone edited and changed my notations. Please do not do that without explaining why. I need an answer to the question, editing just creates more questions in my head plus I don't see anything wrong with my notations so I have got them back to their original form.
Thanks. 
 A: The answer to your question depends upon whether you are taking an Eulerian or Lagrangian point of view. A discussion of this can be found here
http://en.wikipedia.org/wiki/Continuum_mechanics
In particular, section 7 which discusses Kinematics.
A: I agree one hundred percent with @Hooman.  This equation has not been explained well in the books. I myself have the same question and here is how I understand the problem.
There are three different velocities involved in the problem. One is the velocity of propagation (transport) $v_p$ (I call this the velocity of the reference frame)  another is the velocity within the particle in the system $V$ and finally the velocity of displacement ($v_u$).
This is better understood with an example 1D (as you indicate). It is a good idea to use 1D since it explains the problem without considering multiple dimensions and complicating more the problem.
Assume an object (a fly for example) traveling in the space ( in the $x$ axis
in this example) and one particle in the fly wings would have a trajectory  $x=x(t)$. The displacement $u$ is a function
of two variables $u(x, t)$. The displacement measures the change on position with respect to the (probably moving) reference frame. That is, we would like to see, for example, how the wings of the fly are moving but we do not care about the speed of the fly, just the wings motion as if the origin of our reference frame is located in the heart of the fly.
That is, if the reference frame  (the fly) is moving
according to the equation to $X(t)=2 t$, then the displacement is 
\begin{equation}
  u(x,t) = x(t)-X(t)=x(t)-2 t.
\end{equation}
Then we use the chain rule with partial derivatives and find
\begin{equation}
     \frac{D u}{Dt} = 
   \frac{\partial u}{\partial x} \frac{\partial x}{\partial t}
  + \frac{\partial u}{\partial t}.
\end{equation}
That is, 
\begin{equation}
   v_u = v_p \frac{\partial u}{\partial x} + V
\end{equation}
Let us study the three velocities involved here.
$V=\displaystyle{\frac{\partial u}{\partial t}}=-2$. This is
actually $-\displaystyle{\frac{\partial X}{\partial t}}$ since $x$ is seen as a whole in $u$ and the dependence on $t$ is on the term $X(t)$. We are subtracting
the speed of the reference frame and now we look at the fly's wing from the center of his(her) hearth.
$v_p=\displaystyle{ \frac{\partial x}{\partial t}}$: (for example if $x(t)=t+
\sin \omega t$, then $x'(t)=1+ \omega \, \cos(\omega t)$) This is the speed of the wing but observed from a fixed origin somewhere in a frozen reference frame (sitting on my comfy chair observing the fly). 
$v_u=Du/Dt$ this is the displacement velocity. We substract the reference (transport
velocity) and observe the wing from sitting  in the center of the fly's hearth.
if $x(t)=t+ \sin \omega t$ then $Du/Dt=-1+\omega \cos \omega t$.
In the context of Euler or Lagrange the reference frame $X$ is the  $material$
coordinate system (Lagrangian) and the coordinates $x$ indicate the $spatial$ (Eulerian) coordinates.
As a final remark, the factor $\partial u/\partial x$ (spatial gradient) accounts for stretching. In this example there is no stretching. Just translation and swinging. However if, instead of the example above, $X(t)=2t-0.1 
\sin(x)$ there would be some stretching of the wings of the fly.
