Consider the Cauchy momentum equation:

$$\rho \frac{d^2 \mathbf{u}}{d t^2} = \nabla \cdot \boldsymbol{\sigma} + \rho \mathbf{f}$$

where $\rho(\mathbf{x},t)$ is the density, $\mathbf{u}(\mathbf{x},t)$ is the displacement vector, $\boldsymbol{\sigma}(\mathbf{x},t)$ is the Cauchy stress tensor, and $\mathbf{f}(\mathbf{x},t)$ represents the body force per unit mass.

Let us focus on the left-hand side of the equation. The material derivative of $\mathbf{u}$ is:

$$
\frac{d \mathbf{u}}{d t} = \frac{\partial \mathbf{u}}{\partial t} + \frac{d\mathbf{x}}{dt} \cdot \nabla \mathbf{u},
$$

if you proceed to take the second derivative of this expression you get something nasty looking. However, if $d\mathbf{x}/dt=0$, you get great simplification:

$$
\frac{d^2\mathbf{u}}{dt^2}=\frac{\partial^2\mathbf{u}}{\partial t^2}
$$

What I am confused with is that for the wave equation this is pretty much what is assumed, but for Navier-Stokes equation we need to keep the $d\mathbf{x}/dt$ term. 

**I do not see why this difference arrises.**

The problem I have at hand is to derive that for an elastic string which is held fixed at endpoints, the governing equation of motion is $\partial^2 \mathbf{u}/\partial t^2=c^2\partial^2 \mathbf{u}/\partial x^2$. For the right hand side it is relatively easy: assume $\mathbf{f}=0$ and use Hooke's law to calculate $\mathbf{\sigma}$. Again, the lack of terms containg $d\mathbf{x}/dt$ on the left hand side is surpriding to me.