Navier-Stokes equations of motion in the Lagrangian description In general, the Navier-Stokes equations of motion are derived in the Eulerian description. I tried to find the Navier-Stokes in the Lagrangian description but was not very successful.
I would be glad if someone could state the Navier-Stokes equation in the Lagrangian description or give me at least a reference where I can find it.
 A: Let $$x=\chi(X,t)$$ be the deformation map that maps points in the reference configuration (denoted by $X$) to points in the deformed configuration at time $t$ (denoted by $x$). The inverse map is given by
$$X=\chi^{-1}(x,t);$$
here we are assuming that the map $\chi$ is bijective. We have the corresponding deformation gradients
$$F_{i\alpha}=\dfrac{\partial\chi_i}{\partial X_{\alpha}}$$
and the inverse
$$F^{-1}_{\alpha i}=\dfrac{\partial\chi^{-1}_{\alpha}}{\partial x_i}.$$
The above is a simple application of the inverse function theorem.
Let us consider the incompressible Navier-Stokes in Eulerian form
$$\partial_tu_i+u_j\partial^x_{j}u_i=-\dfrac{1}{\rho}\partial^x_ip+\nu\partial^{x}_{k}\partial^{x}_ku_i$$
where $u,p,\nu,\rho$ are variables which are described with respect to $x,t$. Moreover, $\partial_t:=\dfrac{\partial}{\partial t}$ and $\partial^x_i:=\dfrac{\partial}{\partial x_i}$; notice that $x$ is the eulerian descriptor.
To map back to Lagrangian, we need to express everything in $X$ and $t$. Given a function $f(x,t)$, we can define its lagrangian counterpart as
$$f(\chi(X,t),t)=\tilde{f}(X,t).$$
We will drop the tilde and assume that all variables are expressed in their lagrangian format.

*

*The first term becomes a simple
$$Du/Dt.$$


*The 2nd term becomes
$$\partial^x_ip=\partial^X_{\alpha}p~\partial^x_i\chi_{\alpha}^{-1}=F^{-1}_{\alpha i}\partial^X_{\alpha}p.$$
As a side note, observe the equivalence of operators from the above relation
$$\partial^x_i\equiv F^{-1}_{\alpha i}\partial^X_{\alpha}$$


*The third term consists of mapping the laplacian back to the reference configuration. We just apply the above calculation twice, to get
$$\partial^x_j\partial^x_j u_i=F^{-1}_{\alpha j}\partial^X_{\alpha}\Big(F^{-1}_{\beta j}\partial^X_{\beta}\Big)u_i$$
Putting the three together we get the following expression
$$\dfrac{Du_i}{Dt}+\dfrac{1}{\rho}F^{-1}_{\alpha i}\dfrac{\partial p}{\partial X_{\alpha}}=\nu F^{-1}_{\alpha j}\dfrac{\partial}{\partial X_{\alpha}}\Big(F^{-1}_{\beta j}\dfrac{\partial u_i}{\partial X_{\beta}}\Big).$$
I suspect that there is a way of writing this without using indicial notations, but i didn't have time to figure it out.
Also, I would recommend the book by Andrew Bennet "Lagrangian Fluid Dynamics". Check equations 5.1 and 5.6 and the discussion in between. I have however used notation from CS Jog's "Continuum Mechanics--Foundations and Applications of Mechanics". I had spoken to CS Jog, and he had shown me the expression for Navier Stokes in his book, but I couldn't find it ergo I cite Bennet's book.
A: It's pretty straight forward to compute, but even easier to locate using search engines -- but here is the mass and momentum equations, you can figure out the energy on your own. The key is using the material, or substantial, derivative:
Mass:
$$ \frac{D \rho}{Dt} + \rho \nabla \cdot \vec{u} = 0$$
Momentum:
$$ \frac{D\vec{u}}{Dt} = -\frac{1}{\rho} \nabla p - \nabla f_g + \nu \left( \nabla^2 \vec{u} + \frac{1}{3} \nabla \left(\nabla \cdot \vec{u}\right)\right) $$
It is easy to go back and forth between the Eulerian and Lagrangian forms, using the definition of the material derivative. That is left as an exercise. 
