Linear stability of parallel flows
Construct a perturbation eigenproblem, interpret its growth convention, and identify what an inviscid stability criterion actually establishes.
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The question for today
What you’ll understand
- Linearize about a specified parallel base flow.
- Translate complex phase speed into an amplitude growth rate.
- Distinguish a necessary instability condition from a sufficient one.
A smooth velocity profile can satisfy the fluid equations perfectly and still be difficult to maintain experimentally. Stability asks a second question: what happens to a small disturbance placed on that solution? We will keep the base state, disturbance class, boundary conditions, and growth convention explicit. Those choices determine the mathematical question; changing one can change the answer even when the same fluid occupies the apparatus.
Separate the reference flow from its disturbance
Consider a steady incompressible flow between parallel boundaries or in an unbounded shear layer. Write the total velocity as this base state plus a disturbance multiplied by a small bookkeeping parameter. Subtract the base-flow equation from Navier–Stokes. Terms linear in the disturbance remain; products of disturbances are discarded. The pressure disturbance is retained because it enforces incompressibility and redistributes momentum between components, even when it does no net work in a closed energy budget.
The result is an approximation to the early evolution of sufficiently small disturbances, not an alternative fluid law. A disturbance eventually amplified to the scale of the base flow invalidates its own linearization. Likewise, an externally maintained base profile requires its actual forcing and boundary conditions to be stated. A calculation about an ideal parallel profile can guide interpretation of a slowly varying experiment, but it does not automatically describe a rapidly developing jet or a separated flow.
Read a mode without reversing the sign
For a two-dimensional inviscid disturbance, introduce a streamfunction with components and . A normal mode has the form , with real positive streamwise wavenumber and complex phase speed . Its real physical part propagates at while its envelope multiplies by . With this convention, positive imaginary phase speed means temporal growth. Write that envelope explicitly before comparing a formula using a different time convention.
Eliminating pressure and neglecting viscosity gives Rayleigh’s equation. Impermeable walls impose zero streamfunction for a nonzero streamwise wavenumber; disturbances in an unbounded domain must satisfy the appropriate decay condition. An admissible eigenvalue therefore belongs to the differential equation together with its boundary conditions. Arbitrarily choosing a complex number for c is useful for practicing the growth convention, but it does not establish that any specified profile actually possesses that mode.
Use criteria within their assumptions
For the classical smooth inviscid problem, an exponentially growing mode requires the base profile to have an inflection point in the domain. One way to understand the result is to multiply Rayleigh’s equation by an appropriate complex conjugate, integrate, and examine the imaginary part. The resulting integral cannot vanish for a growing mode if the curvature retains one strict sign. This is a powerful exclusion test under its hypotheses, rather than a universal rule about every shear-driven transition.
An inflection point does not guarantee instability, and absence of an inflection point does not settle the viscous problem. Plane Poiseuille flow illustrates why retaining viscosity can qualitatively change an eigenproblem. Even negative growth rates for every eigenmode do not prove that every initial disturbance decays monotonically: nonorthogonal modes can cooperate temporarily. Finally, a finite-amplitude perturbation can reach behavior excluded by a linear approximation. A good stability report says which of these questions it has answered and which remain open.
Work it through
Translate an admissible mode into a time scale
Suppose an independently solved temporal eigenproblem gives and . A mode initially has amplitude 1 mm. Find its amplitude after 5 s and its energy amplification factor.
The exponent uses . Its sign is positive with the chosen convention, so the disturbance grows rather than decays.
Over 5 s the amplitude factor is , giving approximately 2.72 mm. The real phase speed translates the pattern without changing this envelope.
For this single fixed-shape mode, perturbation kinetic energy is quadratic in amplitude. Its factor is , provided the disturbance remains small enough for linear theory.
The amplitude e-folding time is 5 s, while energy e-folds in 2.5 s. Neither number alone specifies when nonlinear transition will occur.
The living notebook
Explore the model
Interactive lab · optimization
A scalar amplification factor
Before pressing Play, predict whether a step size η strictly between zero and two will shrink the displayed iterates of xₙ=(1−η)ⁿx₀. Then compare η above two and change the initial x₀. Read amplitude and objective separately: squaring an amplitude changes its growth factor.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThis is gradient descent on x²/2, an illustrative discrete amplification model. It does not solve Rayleigh’s equation, select a physical eigenmode, or reproduce non-normal fluid growth.
Pause & reason
Check your understanding
Your turn
Try it yourself
With the same normal-mode convention, take and . How long does a single mode take to reduce its amplitude to one half, and what fraction of its initial energy remains then?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
First find the signed envelope exponent. Halving an amplitude does not halve a quadratic energy.
Compare with the solution
The envelope is . Setting this equal to gives . A fixed-shape mode’s energy is proportional to amplitude squared, so one quarter remains. This result describes that mode only; it does not prove monotonic decay for every superposition in a non-normal system.
Keep the idea
Take it with you
- A stability claim specifies its base state, perturbations, boundaries, and approximation.
- For the stated temporal convention, amplitude growth rate is α times the imaginary phase speed.
- Necessary inviscid criteria and modal growth rates answer narrower questions than nonlinear transition.
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