Over the past few weeks, I have been teaching myself fluid mechanics using these notes. The picture shows an extract from page 45, which deals with a plane Couette flow between two parallel, horizontal plates. The diagram in the centre has confused me - what exactly is this trying to show? It isn't the horizontal velocity profile, as we have just used incompressibility to prove that $u$ is discontinuous (the profile shown is clearly continuous). Also, how can we tell that the mode is 'marginally stable' as claimed? Any help is much appreciated.

enter image description here


1 Answer 1


It is the vertical velocity profile $w = \hat w$ (type in "plot -1/3*sinh(z+1), sinh(z-1), z=-1..1" on WolframAlpha and rotate the resulting graph by +90, for a positive outcome change inverse the sign of the leading constants).

Two-dimensional stability analysis of viscid and inviscid flow assumes flow that is perturbed according to

$$ u(x,z,t) = U(z) + \hat u(x,z,t)$$ $$ w(x,z,t) = \hat w(x,z,t)$$ $$ p(x,z,t) = P + \hat p(x,z,t)$$

For an incompressible fluid the continuity equation can be fulfilled by a stream function

$$\psi(x,z,t) = \Psi(z) e^{i k (x - ct) }$$

where $k$ and $c$ can be complex numbers. For judging the stability we have to consider the imaginary part of these complex numbers.

  • The flow is stable in space ($k$) and time ($c$) if the corresponding imaginary part leads to damping $Im < 0$.
  • It is unstable if the imaginary part is instigating $Im > 0$.
  • It behaves neutrally if $Im = 0$: This behaviour is called marginally stable.

The corresponding velocities are given by the general solution

$$ \hat u = \frac{\partial \psi}{\partial z} = \frac{\partial \Psi}{\partial z} e^{i k (x - ct) } $$

$$ \hat w = - \frac{\partial \psi}{\partial x} = - i k \Psi e^{i k (x - ct) } $$

Popping this into the linearised momentum equation for an inviscid fluid yields the Rayleigh's equation

$$ (U - c) (\Psi'' - k^2 \Psi) - U'' \Psi = 0 $$

In your example the base velocity of the unperturbed flow in x-direction is given by a Couette flow $U(z) = z$. Due to ($\frac{\partial^2 U(z)}{\partial z^2} = 0$ the Rayleigh's equation in your case simplifies to

$$ (U - c) (\Psi'' - k^2 \Psi) = 0 $$

If there was no perturbation in your system due to the walls on both side there would be no vertical motion at all. While inviscid flow does not have to fulfilled the no-slip condition near walls (in your case the flow is a developed profile and due to absence to viscosity it will stay that way but inviscid flow itself would not result in such a flow profile), the perturbation has nonetheless to respect the top and bottom boundary condition $w(z=-1)=0$ and $w(z=1)=0$ - it can't penetrate the wall. Due to continuity for an incompressible fluid

$$ \frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0 $$

we can express a correlation between the two velocities

$$ \frac{\partial u}{\partial x} = - \frac{\partial w}{\partial z} = - i k \underbrace{\frac{\partial \Psi}{\partial z} e^{i k (x - ct) }}_{\hat u} = - i k \hat u.$$

In your case the perturbation is given by

$$ \hat w_+ = A_+ sinh(k(z-1)) \, \, \, \text{for} \, \, \, 1 \geq z > z_c $$ $$ \hat w_- = A_- sinh(k(z+1)) \, \, \, \text{for} \, \, \, 1 \leq z < z_c $$

which are both of type $\hat w = A sinh(y(z)) = \frac{A}{2} (e^{y(z)} - e ^{-y(z)})$ where $y = k(z \pm 1)$.

  • $\begingroup$ Thank you very much! This is most helpful $\endgroup$
    – wrb98
    Nov 27, 2019 at 15:44
  • $\begingroup$ @2b-t In the en.wikipedia.org/wiki/…. Is z in that link also vertical velocity ? Wiki says it is perpendicular to the flow. $\endgroup$
    – gansub
    Oct 11, 2020 at 12:46
  • $\begingroup$ @gansub Hey there, it has been a while since I have answered this question. Do you want to know if the coordinate system used in the paragraph "Derivation" in your Wikipedia link is the same as in the derivation above? $x$, the main flow direction, is assumed to be horizontal in both cases and $z$ is perpendicular to it, pointing in vertical direction. So it is no contradiction: $z$ is perpendicular to the main flow as well as vertical. Please let me know if that make sense and answers your question. $\endgroup$
    – 2b-t
    Oct 11, 2020 at 13:30
  • $\begingroup$ @2b-t Yes it does answer my question ! So your equation in the answer is basically the Rayleigh Kuo stability criterion ? $\endgroup$
    – gansub
    Oct 11, 2020 at 13:38
  • 1
    $\begingroup$ @gansub No, in the derivation you basically have a main flow profile termed $U(z)$ that is constant along any cross-section $(x = const, z)$ and only varies with the height and perturbations $\hat u(x,z,t)$ and $\hat w(x,z,t)$ that are superimposed to the main flow. This results in the velocity profiles $u(x,z,t)$ and $w(x,z,t)$. $\endgroup$
    – 2b-t
    Oct 11, 2020 at 14:19

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