research36 min studyLesson 185 in Physics

Orr–Sommerfeld equation and transient growth

Retain viscosity in the parallel-flow eigenproblem and show concretely why stable eigenvalues do not exclude temporary disturbance-energy growth.

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What you’ll understand

  • State the viscous mode equation with its no-slip wall conditions.
  • Explain the role of nonorthogonal disturbance directions.
  • Compute transient amplification in a stable two-component model.

Viscosity usually removes kinetic energy, but its presence in a shear-flow stability problem is more subtle than adding a negative number to every inviscid growth rate. It changes the differential order and the wall conditions. Even after solving the resulting modal problem, eigenvalues describe only part of disturbance evolution. A finite-time initial-value calculation is needed to determine whether a combination of decaying directions can first become much larger.

A fourth-order equation needs four wall conditions

Nondimensionalize a steady parallel channel flow using a specified speed and length, so Re=UL/ν\mathrm{Re}=UL/\nu. For a two-dimensional streamfunction mode ϕ(y)eiα(x−ct)\phi(y)e^{i\alpha(x-ct)}, eliminate the pressure from the linearized viscous equations. With D=d/dyD=d/dy, the result is the Orr–Sommerfeld equation. The viscous term contains four derivatives of the disturbance profile. It therefore cannot generally be appended as a small algebraic damping correction to an inviscid eigenvalue while retaining the old boundary data.

At a stationary rigid wall, no penetration requires ϕ=0\phi=0 for nonzero α\alpha, while no slip requires Dϕ=0D\phi=0. Two walls provide four homogeneous conditions. In an unbounded problem, decay conditions replace these wall conditions. Changing the Reynolds-number length convention also changes quoted numerical critical values, so a number without its normalization is ambiguous. For three-dimensional disturbances, the wall-normal vorticity dynamics must also be included through the coupled Orr–Sommerfeld–Squire formulation.

(D2−α2)2ϕ=iαRe[(U−c)(D2−α2)ϕ−U′′ϕ],ϕ=Dϕ=0 at rigid walls(D^2-\alpha^2)^2\phi=i\alpha\mathrm{Re}\big[(U-c)(D^2-\alpha^2)\phi-U''\phi\big],\qquad \phi=D\phi=0\ \text{at rigid walls}
The displayed nondimensional equation uses the same e^{iα(x−ct)} temporal convention as the previous stability lesson and describes two-dimensional incompressible disturbances of a parallel flow.↔ Scroll wide equations horizontally.

Stable eigenvalues need not mean monotonic energy decay

Represent a discretized initial-value problem as q˙=Lq\dot{\mathbf q}=L\mathbf q. If L is normal in the physically chosen energy inner product, orthogonal eigenvectors make the eigenvalue picture particularly direct. If it is non-normal, its eigenvectors can be strongly nonorthogonal. Contributions initially canceling in one component can decay at different rates and stop canceling later. Their sum can grow temporarily even while each eigenmode eventually decays. The energy norm and the allowed initial conditions are part of the question.

A transparent example has matrix entries L11=L22=−1L_{11}=L_{22}=-1, L12=KL_{12}=K, and L21=0L_{21}=0. Both eigenvalues are negative, but initial data (0,1)(0,1) produce a first component Kte−tKt e^{-t}. For sufficiently large K, the Euclidean norm grows far above its initial value before decaying. The repeated-eigenvalue example is deliberately simple; the same general finite-time issue occurs for nonorthogonal eigenvectors with distinct eigenvalues. It is the propagator’s singular values in an energy norm, rather than eigenvalues alone, that quantify optimal gain.

L=(−1K0−1),q(0)=(01),q(t)=e−t(Kt1),G(t)=∥q(t)∥2∥q(0)∥2=e−2t(K2t2+1)L=\begin{pmatrix}-1&K\\0&-1\end{pmatrix},\quad \mathbf q(0)=\binom01,\quad \mathbf q(t)=e^{-t}\binom{Kt}{1},\quad G(t)=\frac{\|\mathbf q(t)\|^2}{\|\mathbf q(0)\|^2}=e^{-2t}(K^2t^2+1)
This dimensionless two-component linear model demonstrates transient growth with stable eigenvalues. It is not a discretization of a specific channel flow and does not supply its optimal fluid disturbance.↔ Scroll wide equations horizontally.

Connect amplification to transition cautiously

Shear flows can convert cross-stream motion into streamwise velocity variations by displacing fluid across the base gradient. This lift-up mechanism is one route to strong finite-time amplification. In other settings, disturbance tilting and pressure-mediated effects matter. A finite-time optimal perturbation depends on the target time and norm; the direction that maximizes one observable may not maximize another. Numerical calculations also require convergence tests, since a poorly resolved operator can manufacture sensitive eigenvalues or distorted optimal gains.

Transient growth is neither a proof that a particular experiment becomes turbulent nor a way to ignore nonlinear effects. It can bring a sufficiently large initial disturbance into a regime where nonlinear feedback matters, creating a possible route to transition. Whether that route sustains motion depends on the full dynamics and disturbance environment. Conversely, a linearly unstable mode can grow without immediately creating fully developed turbulence. Report modal instability, transient amplification, and observed nonlinear transition as distinct pieces of evidence.

Amplification despite two negative eigenvalues

Use the dimensionless two-component model with K=4K=4 and initial vector (0,1)(0,1). Evaluate its amplitude and energy at t=1, then determine whether this temporary increase contradicts its eigenvalues.

  1. The exact state is q(1)=e−1(4,1)\mathbf q(1)=e^{-1}(4,1). Its squared norm is (16+1)e−2=17/e2≈2.30(16+1)e^{-2}=17/e^2\approx2.30, compared with initial squared norm one.

  2. The amplitude factor is 17/e≈1.52\sqrt{17}/e\approx1.52. Energy, represented by the squared Euclidean norm in this example, increases by approximately 130 percent despite both eigenvalues being −1.

  3. As time tends to infinity, both e−te^{-t} and te−tt e^{-t} vanish. The growth is temporary. The eigenvalues correctly describe the eventual decay but do not guarantee monotonic decay of this initial vector.

The result

Stable modal eigenvalues and finite-time amplification are compatible. The two-component calculation exposes the missing information rather than predicting a fluid transition Reynolds number.

Explore the model

Interactive lab

Interactive lab · vectors

A measured component depends on direction

Predict how the x projection of a vector changes as its angle passes through 90 degrees while its magnitude stays fixed. Use the angle and magnitude controls, then compare the component readouts with total length. This illustrates why a chosen component and a total disturbance norm answer different questions.

A measured component depends on directionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forThe lab draws one vector and its projections. It contains no non-normal matrix, evolving eigenvector superposition, or Orr–Sommerfeld solver; use the worked two-component calculation to establish transient growth.

Check your understanding

Every eigenvalue of a finite-dimensional linear disturbance operator has negative real part. Which behavior can still occur?

Try it yourself

For the same initial vector and t=1, compare G(1)G(1) when K=1 and K=3. Does the result change the eigenvalues? What does it imply about relying only on the spectrum?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the exact gain expression and remember that the off-diagonal coupling does not alter this triangular matrix’s diagonal eigenvalues.

Compare with the solution

For K=1, G(1)=2/e2≈0.271G(1)=2/e^2\approx0.271; for K=3, G(1)=10/e2≈1.35G(1)=10/e^2\approx1.35. Both matrices still have eigenvalues −1 and −1. The larger off-diagonal coupling produces gain for this initial condition at this time without changing the spectrum. This establishes a finite-time difference, not a nonlinear instability or a statement that t=1 is the globally optimal observation time.

Take it with you

  • Viscosity raises the order of the mode equation and introduces no-slip boundary information.
  • Non-normal evolution can temporarily amplify an energy norm despite asymptotically stable eigenvalues.
  • Transient gain, modal instability, and sustained nonlinear turbulence require different analyses.

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