I am trying to work through a paper Instability in Parallel Flows Revisited by Friedlander and Howard, and there are a couple steps in the beginning that I do not understand. I apologize in advance for being long-winded, but I want to be candid.

Consider a steady two-dimensional parallel-shear incompressible flow $$\nabla\cdot u=0,$$ $$u\cdot\nabla u=-\nabla p+\frac{1}{Re}\Delta u+\frac{1}{Re}f$$ for $u=(U(y),0)$, pressure $p$ and (steady) force density $f=(h(y),0)$. If we consider a (incompressible) perturbation of this base flow $\tilde{u}=u+w$ (and $\tilde{p}=p+q$), then when plugging $\tilde{u},\tilde{p}$ into the incompressible NSE and dropping the quadratic term, we have the linearized equations for the perturbed flow $w=(w_1,w_2)$ as $$\nabla\cdot w=0,$$ $$\partial_t w+U\partial_x w+(U'(y)w_2,0)=\nabla q+\frac{1}{Re}\Delta w.$$ Next, taking the curl ($\nabla^\perp\cdot(v_1,v_2)=\partial_y v_1-\partial_x v_2$) of this equation, along with sufficient smoothness so derivatives commute, we have $$\partial_t(\nabla^\perp\cdot w)+U(y)\partial_x (\nabla^\perp\cdot w)+\partial_y(U'(y)w_2)=\frac{1}{Re}\Delta (\nabla^\perp\cdot w).$$ At this point, by incompressibility we can switch to using the stream function for $w$, calling it $\phi$, writing $\nabla^\perp\cdot w=\Delta \phi$, and renaming in the last equation $w_2=-\partial_x\phi$, we should achieve the equation for the perturbed stream function as $$\partial_t\Delta\phi+U(y)\partial_x\Delta\phi-\left[U'(y)\partial^2_{xy}\phi+U''(y)\partial_x\phi\right]=\frac{1}{Re}\Delta^2\phi.$$

My first question involves the bracketed terms: in the paper, they present the linearized stream function equation as $$\partial_t\Delta\phi+U(y)\partial_x\Delta\phi-U''(y)\partial_x\phi=\frac{1}{Re}\Delta^2\phi,$$ where the $-U'(y)\partial^2_{xy}\phi$ term seems to be omitted. Have I missed something completely obvious, or have they made an assumption about $\phi$ (common enough to not be mentioned in the paper)?

I have worked through the derivation a couple times now, and always seem to be getting the "extra term" from the product rule...

For my second question, more of a soft question, the authors proceed to provide motivation for deriving the Orr-Sommerfeld equation by suggesting $\phi$ be of the form $$\phi=\Psi(y)e^{i\alpha(x-ct)}$$ Is there an intuitive reason for the motivation behind this ansatz? This assumption seems common in linear stability analyses, but I haven't come across any reasoning behind it. Obviously it works, but is there a physical interpretation of this "travelling phase" perturbation?


1 Answer 1


First compare your derivation to the following. Consider the Navier-Stokes equation with the plane symmetric ansatz, which reduces to $v = (v_1,v_2)$ solving $$ \partial_t v + (v\cdot \nabla)v + \nabla p = \epsilon \triangle v $$ (We ignore the external force for now, since we are only concerned with the linearized equation.) Taking the two-dimensional curl of $v$ we have the vorticity equation $$\tag{V} \partial_t \omega + (v \cdot\nabla)\omega = \epsilon \triangle \omega $$ using, in its derivation, the incompressibility assumption. The stream function satisfies $$\tag{1} \triangle \phi = \omega $$ and $$\tag{2} v = w + (\partial_2 \phi, - \partial_1 \phi) $$ where $w$ is both divergence free and curl free. This $w$ corresponds to the freedom of replacing $\phi\mapsto \phi + \psi$ where $\psi$ is a harmonic function so doesn't change (1).

Plugging into (V) we arrive at the equation for stream function $$ \tag{S} \partial_t \triangle\phi + (w\cdot \nabla) \triangle\phi + (\nabla\phi)\wedge(\nabla\triangle \phi) = \epsilon \triangle^2 \phi $$

Now write $\phi = \phi_0 + \tilde{\phi}$, where $\phi_0$ satisfies that $\triangle \phi_0 = U'$ (the stream function corresponding to the background solution) and that $w$ is defined so that $(U,0) = w + (\partial_2 \phi_0, -\partial_1\phi_0)$, we see that the linearized equation is exactly as you quoted:

$$\tag{SL} \partial_t \triangle \tilde{\phi} + U \partial_x \triangle \tilde{\phi} - U'' \partial_x \tilde{\phi} = \epsilon \triangle^2\tilde{\phi} $$

So what's the difference? The key difference is in the phrase "linearized vorticity equation" on page 2 of the article you mentioned. Converting from velocity to vorticity you are losing a bit of information! The vorticity only captures the curl part of the velocity, and the curl-free part is left un-prescribed! So while you considered "perturbations of velocity" from which you derived the "perturbations of vorticity", the article starts from considering "perturbations of vorticity"; and this additional degree of freedom allows you to essentially assume that the quantity "$\partial_y w_2$" in your computation vanishes.

  • $\begingroup$ Thank you very much for your thoughtful response. What you've said makes sense. I awarded the bounty, though I am still curious what the motivation behind choosing $\phi=\Psi(y)e^{ia(x-ct)}$. Is there a nice intuitive connection between this ansatz and why it works? $\endgroup$ Feb 3, 2016 at 0:38
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    $\begingroup$ @charlestoncrabb: "why it works" is basically answered by "you plug it in and see what happens". The motivation behind the choice is that parallel shear flow can be imagined to be a flow between two plates, where the fluid is flowing in the $x$ direction with velocity depending on the $y$ direction only. If you add a perturbation, you sort of intuitively expect the perturbation to be carried down stream in a traveling wave. So a first order approximation is precisely of the form $\Psi(y) \exp (i \alpha (x - ct))$. (A traveling wave solution more generally should depend on $f(x-ct)$ for $\endgroup$ Feb 3, 2016 at 14:13
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    $\begingroup$ some arbitrary function $f$ [e.g. the case of traveling solitons]. But if you expect wave like behaviour than it is natural to postulate that $f$ takes the form of $\exp i \alpha$ and see what happens.) $\endgroup$ Feb 3, 2016 at 14:15

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