What type equation Navier-Stokes is: Elliptic, parabolic, or hyperbolic? Should it give always the same answer no matter what is the initial condition? How these statements could be proved?


1 Answer 1


The question assumes that there is a strict delineation between equations, but there isn't. On paper, of course, the Navier-Stokes equations have a parabolic character because there is a non-zero diffusion term. But, in reality, we say that equations are "hyperbolic" when we mean that they are advection dominated, and "parabolic" when they are diffusion dominated, and the Navier-Stokes equations can be either depending on whether your Reynolds number is large or small.

I have tried to make this point in lecture #26 at http://www.math.tamu.edu/~bangerth/videos.html in more detail.

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    $\begingroup$ And in the limits: for negligible viscosity you get the Euler equations which are purely hyperbolic, for negligible advection (large viscosity) you get Stokes flow which is elliptic. $\endgroup$ Commented Jun 5, 2014 at 14:29
  • $\begingroup$ Wolfgang probably mentions this in his lecture, but I'd add that a standard general approach to numerically solving NS is treating the advection operator as hyperbolic, the diffusion as elliptic, and the time stepping as parabolic. $\endgroup$
    – Aurelius
    Commented Jun 5, 2014 at 19:54
  • $\begingroup$ @DougLipinski: Yes, basically. Though it's not that simple: in the large viscosity case, you get the time dependent Stokes equations which are, at best, parabolic. But, if you applied the strict conditions based on the operator involved, it's not actually that simple any more because you have a system of equations and it's not so obvious whether the eigenvalues of the system operator actually satisfy the conditions for parabolic-ness. In the limit of vanishing viscosity, it's the same: it doesn't quite fit the categories because there is no pressure time derivative. $\endgroup$ Commented Jun 6, 2014 at 3:49
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    $\begingroup$ @WolfgangBangerth The compressible Euler equations are purely hyperbolic (they are diagonalizable and the eigenvalues are all real). Incompressibility is a further assumption and yes, it complicates things a bit. Conversely, It's my understanding that the Stokes equations are purely elliptic and there is no time dependence aside from the possibility of time-dependent boundary conditions. There is no memory in the system and information propagates instantly through the entire domain, the hallmark of an elliptic PDE. $\endgroup$ Commented Jun 12, 2014 at 1:06

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