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.

  • $\begingroup$ @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. $\endgroup$
    – tpg2114
    Commented Feb 6, 2017 at 16:34

2 Answers 2


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.


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:


$$ \frac{D \rho}{Dt} + \rho \nabla \cdot \vec{u} = 0$$


$$ \frac{D\vec{u}}{Dt} = -\frac{1}{\rho} \nabla p - \nabla f_g + \nu \left( \nabla^2 \vec{u} + \frac{1}{3} \nabla \left(\nabla \cdot \vec{u}\right)\right) $$

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.

  • $\begingroup$ +1: Thank you a lot, so it is just the Navier-Stokes equation as we know. $\endgroup$
    – MrYouMath
    Commented Feb 6, 2017 at 17:12
  • $\begingroup$ @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. $\endgroup$
    – tpg2114
    Commented Feb 6, 2017 at 17:13
  • 7
    $\begingroup$ 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. $\endgroup$
    – Deep
    Commented Feb 7, 2017 at 4:03
  • $\begingroup$ 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! $\endgroup$
    – tpg2114
    Commented Jun 3, 2021 at 23:46

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