Why is the Kelvin-Helmholtz Instability so regular across space? If any perturbations anywhere in the boundary can lead to instability, why doesn't the pattern appear randomly across the boundary?

Does the appearance of the pattern at one location influence the dynamics of neighboring patterns?

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    $\begingroup$ The answer to the third question is yes, kind of. The resulting vortices from the instability have specific scale sizes dependent upon the medium and flow properties. For instance, under some circumstances one can get little vortices on the outer edge of a big vortex (e.g., relates to fractals in turbulence relating to scale invariance). $\endgroup$ Jul 15, 2017 at 19:47

1 Answer 1


The boundary conditions, under which K-H instability and the resultant patterns come into existence, are spatially homogeneous along the interface between the two fluids. That is to say, there is no special point along the interface of the two fluids. Therefore it is only to be expected that the resulting pattern will be (statistically) homogeneous in that direction, meaning you could shift the origin of your coordinate system anywhere along the interface and the pattern would still look the same (statistically speaking).

P.S. The word "interface" is used here in the sense of being the common surface between two fluid layers in relative motion to each other.

  • $\begingroup$ If I read correctly, you're saying the boundary is spatial homogeneous. But that doesn't say anything about the homogeneity of the perturbations, right? So, if the perturbations leading to the instability aren't spatially homogeneous, what sort of stabilizing mechanism produces spatially homogeneous patterns? $\endgroup$ Jul 16, 2017 at 11:34
  • $\begingroup$ @KooZhengqun What type of inhomogeneous perturbation do you have in mind? In the theory of K-H instability, a sinusoidal perturbation with constant wavelength is applied. Any other perturbation may be decomposed into a sum of sinusoids. $\endgroup$
    – Deep
    Jul 17, 2017 at 8:07
  • $\begingroup$ To my knowledge, the sum of sinusoids may still be spatially inhomogeneous. The inverse Fourier transform may recover a spatially inhomogeneous function, e.g. many spikes of non-uniform spacing. $\endgroup$ Jul 17, 2017 at 12:51
  • $\begingroup$ @KooZhengqun That is certainly possible, and in real world I think the disturbance would certainly not be spatially homogeneous (that's a very artificial boundary condition). But then the K-H patterns that form are not perfectly symmetric either. Perhaps whatever inhomogeneities that were present in the disturbance is to a large extent forgotten once the interface becomes unstable. However this is an interesting question and only experiments or simulations can answer this question conclusively, I suppose. $\endgroup$
    – Deep
    Jul 18, 2017 at 5:13
  • $\begingroup$ Was hoping for some equation governing the dynamics, but from your answer, I don't think that's known yet? Anyways, thank you for your time. $\endgroup$ Jul 18, 2017 at 8:40

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