# 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.

• @AccidentalFourierTransform Lagrangian form is different from the Lagrangian of a system. Lagrangian Navier-Stokes is written following a fluid particle as it moves, as opposed to Eulerian form which tracks the variables at fixed locations in space as the flow moves through them. Commented Feb 6, 2017 at 16:34

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.

• +1: Thank you a lot, so it is just the Navier-Stokes equation as we know. Commented Feb 6, 2017 at 17:12
• @MrYouMath Maybe the ones you know -- but I only ever work with the Eulerian formulation, so I never see the Lagrangian form outside of exam questions that make sure I know the difference. Commented Feb 6, 2017 at 17:13
• This is Eulerian description of flow. But the Lagrangian? No offence, but it is far from trivial to switch between the two descriptions, and I seriously doubt whether you can begin with the equations given above and go over to Lagrangian description. See Lagrangian Fluid Dynamics by A. Bennett.
– Deep
Commented Feb 7, 2017 at 4:03
• There seems to be some confusion out there about a Lagrangian description of fluid dynamics, which is what I am talking about here (following a fluid particle's trajectory), and the Lagrangian of a fluid particle in the mechanical sense. What is written here in this answer, and in the question, is about going between a fixed-in-space view (Eulerian) and a following-fluid-particle view (Lagrangian) -- and again, this is not the same thing as the "Lagrangian" of some dynamical system. It's not my fault people re-use words! Commented Jun 3, 2021 at 23:46

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.

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

2. 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 $$\partial^x_i\equiv F^{-1}_{\alpha i}\partial^X_{\alpha}$$

3. 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.