ASTR/ATOC-5410: Fluid Instabilities, Waves, and Turbulence September 16, 2016, Axel Brandenburg

Handout 7: Inflection point instability

We consider the stability of a parallel shear flow U = ŷU(x) and write the velocity as (u_{x}, U + u_{y}, 0). Recall that the nonlinearity associated with the background flow, U · ∇U, vanishes. To appreciate the essence of the instability, we restrict ourselves to the inviscid case in 2-D.

1 Linearized equations

We consider the incompressible Euler equations in the form

\frac{∂u}{∂t} = −u · ∇u − ∇p,  (1)
∇ · u = 0.  (2)

where p corresponds to the reduced pressure, P/ρ_{00}. Linearize around U and write out in component form

\frac{∂u_{x}}{∂t} + U \frac{∂u_{x}}{∂y} = −\frac{∂p}{∂x}  (3)
\frac{∂u_{y}}{∂t} + U \frac{∂u_{y}}{∂y} + u_{x} \frac{∂U}{∂x} = −\frac{∂p}{∂y}  (4)
\frac{∂u_{x}}{∂x} + \frac{∂u_{y}}{∂y} = 0  (5)

Assume

u_{x}(x, y, t) = û_{x}(x) e^{iky+σt}
u_{y}(x, y, t) = û_{y}(x) e^{iky+σt}
p(x, y, t) = p̂(x) e^{iky+σt}  (6)

so

(σ + ikU)û_{x} = −p̂′,  (7)
(σ + ikU)û_{y} + û_{x}U′ = −ikp̂,  (8)
û′_{x} + ikû_{y} = 0.  (9)

or in matrix form

σ ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} ) ( û_{x} ; û_{y} ; p̂ ) = − ( \begin{matrix}ikU & 0 & ∂_{x} \\ U′ & ikU & ik \\ ∂_{x} & ik & 0 \end{matrix} ) ( û_{x} ; û_{y} ; p̂ )  (10)

Applying impenetrable boundary conditions on x = ±L_{x}/2 implies û_{x} = 0 on both boundaries. (Since there is no viscosity, there is also no viscous stress, and so stress-free boundary conditions cannot be applied.)

2 Numerical treatment

By discretizing in the x direction with n meshpoints, we have a 3n×3n eigenmatix, where the differential operator, discretized to second order, is written as

∂_{x} = ( \begin{matrix}0 & (2δ_{x})^{−1} & ... & ... & ... \\ −(2δ_{x})^{−1} & 0 & (2δ_{x})^{−1} & ... & ... \\ ... & −(2δ_{x})^{−1} & 0 & (2δ_{x})^{−1} & ... \\ ... & ... & ... & ... & ... \end{matrix} )  (11)

Figure 1: Numerical solutions of the discretized eigenvalue problem using 60 meshpoints.

where δx is the mesh width. It is often convenient to incorporate the boundary condition into the matrix itself, so the boundary points are therefore not part of the matrix. The above matrix is unchanged if the relevant variable has a vanishing value on the boundary. In the case of a vanishing derivative, e.g. for p̂(x) we use a one-sided finite difference formula, so we have

0 = +p̂(x_{0}) −4p̂(x_{1}) +3p̂(x_{2})
(2δ_{x}) p̂′(x_{1}) = −p̂(x_{0})  +p̂(x_{2})  (12)

which allows us to eliminate the boundary point, p̂(x_{0}), so

(2δ_{x}) p̂′(x_{1}) = −4p̂(x_{1}) + 4p̂(x_{2})  (13)

or

p̂′(x_{1}) = −2δ_{x}^{−1}p̂(x_{1}) + 2δ_{x}^{−1}p̂(x_{2})  (14)

One obtains 3n eigenvalues from the 3n × 3n eigenmatix problems. It has to be solved for each value of k. The result is plotted in Figure 1.

3 Analytical treatment

It is convenient to define a stream function so that Equation (9) is automatically obeyed, so

u = ∇ × (ψẑ)  (15)

so that

u_{x} = ∂_{x}ψ, u_{y} = −∂_{y}ψ,  (16)

or, for û_{x} and û_{y},

û_{x} = ikψ̂, û_{y} = −ψ̂′.  (17)

Inserting this into Equations (7) and (8), we have

(σ + ikU)ikψ̂ = −p̂′,  (18)
−(σ + ikU)ψ̂′ + ikψ̂U′ = −ikp̂,  (19)

Furthermore, it is convenient to work with a wave speed c and write σ = −ikc, so that

ik(U − c)ikψ̂ = −p̂′,  (20)
−ik(U − c)ψ̂′ + ikψ̂U′ = −ikp̂,  (21)

In the last equation, the ik term cancels, so

−k^{2}(U − c)ψ̂ = −p̂′,  (22)
−(U − c)ψ̂′ + ψ̂U′ = −p̂,  (23)

Eliminating now p̂ yields

−U′ψ̂′ − (U − c)ψ̂′′ + ψ̂′U′ + ψ̂U′′ = −k^{2}(U − c)ψ̂.  (24)

The U′ψ̂′ term cancels, so

−(U − c)ψ̂′′ + ψ̂U′′ = −k^{2}(U − c)ψ̂.  (25)

which can also be written as

(U − c) (∂_{x}^{2} − k^{2}) ψ̂ − U′′ψ̂ = 0,  (26)

which is known as Rayleigh’s instability equation.

4 Rayleigh’s inflection point theorem

Writing the above equation as

ψ′′ − k^{2}ψ − \frac{U′′}{U − c}ψ = 0  (27)

where we have dropped the hat. Multiplying by the ψ^{∗} and integrating over the x interval yields

∫ (|ψ′| + k^{2}|ψ|^{2}) dz + ∫ \frac{U′′}{U − c}|ψ|^{2} = 0.  (28)

The imaginary part of this equation is

Im(c) ∫ \frac{U′′}{|U − c|^{2}} ψ|^{2} = 0.  (29)

which shows that U′′ must change sign at least once in the interval.

$Header: /var/cvs/brandenb/tex/teach/ASTR_5410/7_Inflection_Pt_Inst/notes.tex,v 1.3 2016/09/16 16:42:05 brandenb Exp $