# Is turbulence more likely to form with the Euler equation as opposed to Navier-Stokes?

The Euler equation models perfectly inviscid fluids. Under this assumption, with $$\nu = 0$$, the Reynolds number should be infinite. I would guess that this implies the Euler equation is always turbulent, but this is not the case as in practice it is used to model regular (low viscous) fluids. Why does this occur?

And second, does this imply that the Euler equation is always more turbulent than the Navier-Stokes, which accounts for viscosity?

• This is a nice question (+1). As it well known there is one million prize for the right answer to this question. Commented Jan 29, 2023 at 14:52

What do you mean by "more turbulent"?

Anyway, to qualitatively answer your question, you have to keep in mind that:

• viscosity, beside damping smallest scales of turbulence, could trigger instability that can evolve in turbulent flows or regions in the flow. As an example, think as pipe flows, stable in the inviscid limits, unstable for viscous flows (you can have a look at historically noteworthy criteria of stability criteria for oarallel flows, as the Rayleigh, Fjortroft crieteria, and Orr-Sommerfeld equations); or you can think at boundary transition to turbulent regime in real-life flows on solid boundaries at high Reynolds number; in a very extreme case, you can compare the results of a Euler and Navier-Stokes numerical simulation of an airfoil at large angle of incidence: with Euler inviscid equations you have no separation, while with Navier-Stokes taking into account viscosity influence you have large separation regions, turbulent at large Reynolds number;
• Euler equations and Navier-Stokes equations are mathematical models to describe the experience and they have their limits of validity within they give us good results, consistent with experiments and real-life;
• using Euler equations and other inviscid models at high Reynolds number within they're limits of validity, it's like you're lumping all the viscosity and rotational effects in infinitesimally thin regiona of the domain, like vortex sheets or vortex lines. So you can't really appreciate the small-scale turbulence in these thin regions, while it's likely that you can appreciate larger scales coming from the instabilities and interactions of these rotational regions of the fluid
• By more turbulent I did indeed mean qualitatively, as in usually you picture turbulent activity to increase with the Reynold's number. I had thought that since in the inviscid case the Reynold's number is infinite then we will have a very turbulent flow. But it seems the viscosity itself is what invokes some turbulence, and without it you achieve laminar flow in some cases? Commented Aug 30, 2022 at 21:06
• @CBBAM "usually you picture turbulent activity to increase with the Reynold's number" That really depends on the way you are looking at it. many flows are pretty much Reynolds number independent after above certain Reynolds number limit. And their turbulent kinetic energies or similar measures do not increase with increasing the Reynolds number any more. Commented Aug 31, 2022 at 12:07

Turbulence is the time-dependent chaotic behaviour seen in many fluid flows. It is generally believed that it is due to the inertia of the fluid as a whole: the culmination of time-dependent and convective acceleration;

So every continuum equation which includes material derivative of flow velocity term :

$${\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}}= {\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u}$$

includes time-dependent flow velocity change and convective acceleration term (flow velocity gradient) and is able to describe how flow parcel speed changes along travel trajectory with passage of time. Thus by definition, continuum equations with flow velocity material derivatives can describe flow turbulence, including but not limited to Navier–Stokes equations and Euler equations. So your intuition based on high Reynolds number in Euler equations was right.

• So the Euler equation is almost always turbulent then, given that the Reynold's number is infinite? Commented Aug 30, 2022 at 21:11
• The equations are not turbulent themselves. You can write equations for a turbulent flow (like the RANS, Reynolds Averaged Navier-Stokes equations) but it means that you decided not to solve the full details (especially small scales) of the flow, and model it. Navier-Stokes eqns aims at solving all the details of the flow. On the other sides, despite the "infinite Reynolds number"-limit of Euler equations, in many cases of interest (e.g. airflow around an airplane) it can be proved mathematically that Euler model doesn't leads to turbulent flows, or large turbulent regions in the domain. Commented Aug 30, 2022 at 22:24
• @basics I see, so despite having infinite Reynold's number, because the Euler equation does not resolve the full fluid picture we might not observe turbulence in a Eulerian inviscid flow? Commented Aug 30, 2022 at 22:31
• @basics, Can you post a sketch of such proof ? Because it's strange why wikipedia says that all continuum equations having $\frac {\mathrm {D} \mathbf {u} }{\mathrm {D} t}$ can lead to turbulence description. I may agree that Euler equations may not describe full turbulence model (or as you say "large turbulent regions"). However if Euler equations can describe even small turbulence regions a bit,- it's still "turbulence enabled" modeling. Commented Aug 31, 2022 at 6:15
• Now, I'm going to work. Please, open a new question. I'll try to give a more detailed answer in my spare time. To cut a long story short, who wrote the wiki page is right: it's hard to give very specific details about generic problems in fluid dynamics, but the non linear term $D\mathbf{u}/Dt = \partial \mathbf{u}/\partial \ t + (\mathbf{u}\cdot \nabla)\mathbf{u}$ contains the mechanism allowing for "turbulence". Meanwhile, you can look for vortex stretching and tilting and the equations of the kinetic energy, of the flow, the mean flow and the fluctuations, e.g. in Pope, Turbulent Flows Commented Aug 31, 2022 at 6:33

This is a very good question. What you're talking about in a certain sense is known as the 'dissipative anomaly': in the limit of infinite Reynolds number (or viscosity to zero) the dissipation of energy is finite, but of course the Euler equations are inviscid and its solutions don't dissipate energy, or at least not the solutions in the usual sense. Then, if when $$\nu \to 0$$ the solutions to Navier-Stokes equations dissipate energy, Euler equations can't describe fully developed turbulence?

This is related to what is known as 'Onsager's conjecture': there are solutions to the Euler equations that dissipate energy but they must have a Hölder exponent $$\leq1/3$$. What can these 'weak' Euler solutions tell us about fully developed turbulence?. For more information about this you can check the papers by Eyink "Review of the Onsager ideal turbulence theory" and "Onsager and the theory of hydrodinamic turbulence". I would also recommend "Do dissipative weak Euler solutions dream of turbulence?" by Takeshi Matsumoto.

• So the Euler equations do not dissipate energy, but the Navier-Stokes equations do even in the limit as $\nu \rightarrow 0$? If so, how does turbulence end in Euler equations where the Hölder exponent $> 1/3$ (in the sense that in NS the turbulence ends when the energy is dissipated in the Kolmogorov range)? Commented Oct 4, 2022 at 20:42