Solutions and Properties of the Pridmore-Brown Equation

AIAA-2019-259425th AIAA/CEAS Aeroacoustics Conference, 20 - 24 May 2019, Delft, The Netherlands

Solutions and Properties ofthe Pridmore-Brown Equation

Sjoerd W. Rienstra∗

A study is made of acoustic duct modes in 2D and axi-symmetric3D lined ducts withisentropic inviscid transversely non-uniform mean flow andsound speed. These modesare described by a one-dimensional eigenvalue problem consisting of a Pridmore-Brownequation complemented by hard-wall or impedance-wall boundary conditions.

A numerical solution, based on a Galerkin projection and an efficient method for theresulting non-linear eigenvalue problem, is compared withanalytical approximationsfor low and high frequencies. A collection of results is presented and discussed.

Modal wave numbers are traced in the complex plane for varying impedance, show-ing the usual regular modes and surface waves. A study of a vanishing boundary layer(the Ingard limit) showed that, in contrast to the smoothly converging acoustic modesand downstream running acoustic surface wave, the convergence of the other surfacewaves is difficult to reproduce numerically.

Effects of (transverse) turning points and exponential decay are discussed. Espe-cially the occurrence of modes insensitive to the wall impedance is pointed out.

Cut-on wave numbers of hard-wall modes are presented as a function of frequency.A strongly non-uniform mean flow gives rise to strong differences between modal be-haviour for low and for high frequencies.

I. Introduction

Duct modes are time harmonic solutions of the linearised compressible Euler equations in aduct, which are self-similar in axial direction. This is possible when duct and medium

are constant in axial direction. They are interesting in aeroacoustics not only because they forma set of building blocks to construct by linear combination more general solutions, but moreimportantly they are simple enough to reveal behaviour of sound propagation in ducts.

For example: the mechanism of cut-off explains that sufficiently far away from any harmonicsource or scatterer its sound field consists of only a finite number of modes [1, 2]; the rotor-alonenoise of a fan, consisting of certain specified frequencies and circumferential mode numbers, onlyexists when the blade tips rotate (approx.) supersonically[1]; the Tyler-Sofrin selection rule forrotor-stator interaction tells what rotor blades, stator vanes combinations radiate sound [3]; bothmechanisms may be spoiled by distortion modes due to non-uniformities of mean flow [4, 5], forexample due to upwash; the imaginary parts of the prevailingmodal wave numbers constitute asimple measure for the quality of acoustic lining; the indirect determination of a liner’s impedanceby resolving a measured sound field into its modal spectrum isbased on modes [6, 7].

∗Associate Professor, Department of Mathematics & ComputerScience, Eindhoven University of Technology,The Netherlands, Senior Member AIAA.

Copyright c© 2019 by S.W. Rienstra. Published by the American Institute of Aeronautics and Astronautics, Inc.with permission.

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American Institute of Aeronautics and Astronautics Paper AIAA-2019-2594

In the simplest configuration of a lined duct with uniform medium without mean flow, modescome in two types: acoustic modes and acoustic surface waves[8]. With uniform mean flow, twoother types of surface waves appear: upstream and downstream running hydrodynamic surfacewaves. Under certain conditions the upstream running surface wave is really downstream runningand in fact an instability. Brambley showed that the problemin the limit of a vanishing boundarylayer (slip flow) is in time domain ill-posed [9] and therefore only well interpretable if the problemis recast into some regularised form, for example by taking the boundary layer thickness smallbut non-zero [10].

Almost all our working knowledge and comprehension of modesis based on the basic config-uration of a duct with uniform mean flow. Any interaction withshear flow (refraction, couplingof acoustic energy with mean flow energy) is not covered. It istherefore useful to refine ourknowledge of modes by including non-uniform flows.

We will consider here modes in a medium with isentropic mean flow and sound speed varyingin transverse direction. They constitute a class of simple but non-trivial sound waves in shearflow and and a variable sound speed, albeit without entropy variations. For typical 2D and axi-symmetric 3D configurations they are governed by what has become known as the 2D and 3DPridmore-Brown equations. In the following we will presenta collection of results obtained bynumerical and analytical methods.

II. Problem Description

Consider in a duct of constant cross section, with a medium and boundary conditions inde-pendent of the axial position, time-harmonic acoustic perturbations of plane parallel isentropicinviscid mean flow. This mean flow has a uniformly constant mean pressurep0 = ρ0c2

0/γ , other-wise it varies in transverse direction with mean velocityv0 = u0(y, z)ex , mean densityρ0(y, z)and mean sound speedc0(y, z). The (linear) boundary conditions are of impedance type, inde-pendent of the axial coordinatex . The ratio of heat capacities is denoted1 by γ = Cp/Cv.

The sound fieldp may be described by an infinite sum of self-similar solutions(see sectionC), called modes, that retain their shape when travelling down the duct. With given frequencyωand the usual complex notation they are of the form

p(x, t) = p(y, z) f (x − vt) = p(y, z)eiω(t− xv) = p(y, z)eiωt−iκx, (1)

and consist of an exponential function multiplied by a shapefunction p, being an eigenfunctionwith eigenvalueκ of a Laplace-type operator valid on a duct cross section, andsatisfying anequation also known as a preform of the Pridmore-Brown equation [2]

�2∇·( 1

�2∇ p

)

+(

�2 − κ2)

p = 0, with � = ω − κu0

c0, (2a)

ρ0c0�u + ∇ p ·∇u0

c0�− κ p = 0, (2b)

ρ0c0�v − i∂ p

∂y= 0, (2c)

ρ0c0�w − i∂ p

∂z= 0. (2d)

1Following tradition, we use hereγ to denote the ratio of heat capacities, but this variable will only appear in theintroduction of the problem. Later we will useγ to denote a certain square root. Both notations are not related.

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See equation (90). Eigenvalueκ corresponds to axial wave numberκ, and is probably the mostimport hallmark of a mode. In general, the set{ pn, κn} is infinite and countable. Due to their self-similarity modes are relatively easy to study and help the understanding of sound propagationin, and interaction with, mean flow. From linearity they can be used as basis (building blocks)to construct more general solutions of the equations. Without uniform or no mean flow the setof modes is rich enough to represent all solutions [11]. This is almost, but not exactly, the casewith non-uniform mean flow, where the factorω − κu0 in (2a) may be zero at so-called “criticallayers” for a continuum of values ofκ [12], and some solutions cannot be described by discretemodes. Except when the possible modes are cut-off (exponentially decaying inx direction) andvanishingly small, this critical layers contribution is small and can be ignored in most practicalsituations. Altogether, this makes these modes interesting and useful to study.

The partial differential equation (2a) that remains for shape functionp(y, z) is still difficult toanalyse. The two most important simplifications are in 2D a duct for 06 y 6 a and a mean flowonly depending ony, and in 3D a cylindrical hollow duct 06 r 6 a and an axi-symmetric meanflow only depending onr .

In 2D withp(x, t) = p(y)eiωt−iκx, (3)

andu0 = u0(y), c0 = c0(y), we have

�2( 1

�2p′

)′+

(

�2 − κ2) p = 0, (4a)

ρ0c0�u + p′u′0

c0�− κ p = 0, (4b)

ρ0c0�v − i p′ = 0, (4c)

(where the prime denote a derivative toy). For the boundary conditions we adopt the Ingard-Myers conditions [13, 14] for a wall with impedanceZ and a vanishingly thin boundary layer

iωZ(v·n) = ic0� p, (5)

with normal unit vectorn pointing into the wall. In terms ofp and p′ this becomes then[

iωZ0 p′ = −ρ0c20�

2 p]

y=0,

[

iωZ1 p′ = ρ0c20�

2 p]

y=a. (6)

(Note thatρ0c20 = γ p0 constant.) For a mean flow vanishing at the wall, this condition reduces

to the commonZ(v·n) = p. In time domain, the Ingard-Myers condition has some fundamentalissues [9] and a regularised version should be used. In frequency domain the problems are mild.Only in case of a thin but non-zero boundary layer the accuracy of surface modes [8] may be less,since they are more sensitive to details of the boundary layer than other modes [15, 16].

In an axi-symmetric 3D configuration with

p(x, t) = pm(r)eiωt−iκx−imθ, (7)

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wherem ∈ Z andu0 = u0(r), c0 = c0(r), we have

�2

r

( r

�2p′

m

)′+

(

�2 − κ2 − m2

r2

)

pm = 0, (8a)

ρ0c0�um + p′mu′

0

c0�− κ pm = 0, (8b)

ρ0c0�vm − i p′m = 0, (8c)

ρ0c0�wm − m

rpm = 0, (8d)

(where the prime denote a derivative tor). The boundary condition at the wall is then[

iωZ p′m = ρ0c2

0�2 pm

]

r=a, (9)

while the solution is regular atr = 0. With an annular duct we would have had a wall conditionat the inner wall, but we did not include that generalisationhere.

Equations (4a) and (8a) are respectively known as the 2D and 3D Pridmore-Brown equation[2, 17].

For uniform mean flow,� is constant and above equations reduce to the standard harmonicequation and the (scaled) Bessel equation, providing analytically exact descriptions of duct modes[1, 2, 8].

In non-uniform flow where� is a function ofy, respectivelyr , the Pridmore-Brown equationsare in general not solvable in terms of standard functions [11, 12, 17, 18, 19, 20] and has to besolved numerically. Even in the 2D case of linear shear flow (and constant sound speed) wherethere is an exact solution known [21] in terms of Weber’s Parabolic Cylinder functions [22], thenumerical evaluation [23, 24] appears to be so problematic [25] due to subtraction of exponentiallylarge terms of the constituting parts, that it was only feasible for a rather limited parameter range.Especially the upstream running modes deteriorate quickly. Numerical solutions are therefore ourmain source of information [11, 19], but they are not equally applicable (shooting is only possibleif the mode is not vanishingly small near the wall) or flexible(if auxiliary libraries are requirednot available for MATLABTM). Therefore new methods that are both fast, simple, flexibleanduniversally applicable are always of interest to study.

On the other hand, a disadvantage of a numerical solution is the lack of insight, that a suf-ficiently transparent analytical solution provides, possibly (in case of approximations) at the ex-pense of some accuracy.

Well-known analytical solutions (their origin dating backas far as the early days of quan-tum mechanics) are WKB-approximations for high frequencyω [18]. They provide insight into(transverse2) turning point behaviour and give for high enoughω excellent approximations ofmodal wave numberκ. For general mean flow profiles, this requires numerical evaluation of in-tegrals that provide the dispersion relation, but for simple mean flows (linear, parabolic profiles)analytical evaluation is possible.

In the other direction for low frequency (smallω), approximations are also possible, both forhard and for soft walls. These solutions seem to be not available in the literature, at least not on asystematic basis.

2In this paper only turning point behaviour in transverse direction is considered. Other turning point behaviour mayexist in axial direction, but only if duct diameter or mean flow vary gradually in axial direction. In that case,κ turnsfrom real to complex. Strictly speaking, we have then only approximate modes.

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In the present paper we will explore a numerical solution based on Galerkin projection on aChebyshev basis, with a remarkably efficient non-linear eigenvalue search routine. We will usethis method to analyse the behaviour of the modal wave numbers as a function ofω, and as afunction of the impedance for the behaviour and occurrence of surface waves. Furthermore, wewill compare the results with high and low frequency approximations.

II.A. Non-Dimensional Form

For later use, it is convenient to have the problems in non-dimensional form. Assuming a charac-teristic sound speedc∞ we write

y = ay , r = ar , c0 = c∞c0, u0 = c∞u0,

ω = c∞aω, � = 1

a�, κ = 1

aκ , Z = γ p0

c∞Z .

(10)

Since the problem is linear, it is not necessary to rescalep nor pm . For notational convenience wewill drop tildes and hat. Then we obtain in 2D for 06 y 6 1

�2( 1

�2p′

)′+

(

�2 − κ2)

p = 0, (11a)[

iωZ0p′ = −�2 p]

y=0,

[

iωZ1p′ = �2 p]

y=1. (11b)

x-axisy = 0

y = 1

u0(y)

Z0

Z1

2D configuration with duct of height 1, wall impedancesZ0 and Z1, mean flowu0(y) and soundspeedc0(y).

In 3D for 06 r 6 1 we have

�2

r

( r

�2p′

m

)′+

(

�2 − κ2 − m2

r2

)

pm = 0, (12a)

pm regular inr = 0,[

iωZp′m = �2 pm

]

r=1. (12b)

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x-axisr = 0

r = 1

u0(r)

Z

3D axi-symmetric configuration with circular duct of radius1, wall impedanceZ , mean flowu0(r) and sound speedc0(r).

III. Numerical Solution

III.A. Galerkin Projection

We start with the 2D problem (11) with impedancesZ0,1 6= 0. We can rewrite this equation withboundary conditions in the equivalent form that forevery test functionw

∫ 1

0

[

( p′

�2

)′+

(

1 − κ2

�2

)

p

]

w dy = 1

�2p′w

∣

∣

∣

∣

1

0

+∫ 1

0− 1

�2p′w′ +

(

1 − κ2

�2

)

pw dy =

pw

iωZ1

∣

∣

∣

∣

1

+ pw

iωZ0

∣

∣

∣

∣

0

+∫ 1

0− 1

�2p′w′ +

(

1 − κ2

�2

)

pw dy = 0. (13)

If we assume thatp can be written as a sum over a function basis{φn},

p =∞

∑

n=0

anφn , (14)

and we use the same basis for the test functions,i.e. w = φl , l = 0,1, . . ., then (13) becomesequivalent with

∞∑

n=0

an

[

φnφl

iωZ1

∣

∣

∣

∣

1

+ φnφl

iωZ0

∣

∣

∣

∣

0

+∫ 1

0

{

− 1

�2φ′

nφ′l +

(

1 − κ2

�2

)

φnφl

}

dy

]

= 0. (15)

This can also be written in matrix form

M(κ)a = 0, a = (a0, a1, . . . )T , (16)

for the[0,∞)× [0,∞) matrixM with elements

Mn,l(κ) = φnφl

iωZ1

∣

∣

∣

∣

1

+ φnφl

iωZ0

∣

∣

∣

∣

0

+∫ 1

0

{

− 1

�2φ′

nφ′l +

(

1 − κ2

�2

)

φnφl

}

dy . (17)

The non-linear eigenvalue problem is then approximated by cutting off M to an N × N matrixand finding those discrete values ofκ, and its corresponding solution spacea, for which M has

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a non-empty null space. Only the first few, absolutely smallest, κ will be of interest as well asnumerically accurate.

We choose Chebyshev polynomials on[0,1]

φn(y) = Tn(2y − 1) = cos(

n arccos(2y − 1))

. (18)

Another option would be to take a Fourier basis (sines and cosines). For the 2D problem thisappears to work out equally well.

We have a serious problem if�(y0) = 0 for somey0 ∈ [0,1], but this can be tackled (as-suming analyticity of the integrand) by slightly deformingthe integration contour around thesingularity in a way that respects causality of the solution. However, this occurs very rarely; neverin our examples. See [12].

The approach for the 3D problem (12) is similar, although the singularity atr = 0 due tothe polar coordinates require special consideration. We can rewrite the equation with boundarycondition (Z 6= 0) in the equivalent form that forevery test functionw

∫ 1

0

[

( r

�2p′

m

)′+ r

�2

(

�2 − κ2 − m2

r2

)

pm

]

w dr =

r

�2p′

mw

∣

∣

∣

∣

1

0

+∫ 1

0− r

�2p′

mw′ + r

�2

(

�2 − κ2 − m2

r2

)

pmw dr =

pmw

iωZ

∣

∣

∣

∣

1

+∫ 1

0

{

− 1

�2p′

mw′ +

(

1 − κ2

�2− m2

r2�2

)

pmw

}

r dr = 0. (19)

An important difference with the 2D problem (13) is the singularity inr = 0 if m 6= 0. Following[26], this has been dealt with by formally extending the solution symmetrically ifm is even, andanti-symmetrically ifm is odd. As a result we know in advance thatpm can be written as a sumof even basis functions ifm is even, and odd basis functions ifm is odd. For the same reason weneed only even, resp. odd test functions ifm is even, resp. odd. This is seen as follows.

If u0(r) andc0(r) are smooth inr = 0 (physically the only configuration that makes sense)and symmetric inr , it is easily checked by a formal Taylor expansion ofpm(r) that a regularsolution behaves like

pm(r) = O(rm) for r → 0. (20)

So for a smooth solution it is necessary to extendpm symmetrically,i.e. pm(r) = pm(−r), if m iseven, and anti-symmetrically,i.e. pm(r) = −pm(−r), if m is odd. Now consider the singularityat r = 0. If m = 0 there is no singularity, so this case is not a problem. Ifm is odd andw(r) = rw′(0)+ . . . anti-symmetric, the integrand behaves like

(

−mrm−1w′(0)− m2

r2rmrw′(0)

)

r + · · · = −m(m + 1)rmw′(0)+ . . . . (21)

So there is no singularity and the integrand vanishes atr = 0. If m is even> 2 andw(r) =w(0)+ r2w′′(0)+ . . . symmetric, the integrand behaves like

(

−mrm−12rw′′(0)− m2

r2rmw(0)

)

r + · · · = −m2rm−1w(0)+ . . . . (22)

Sincem − 1 > 1 there is also here no singularity and the integrand again vanishes.So there appears to be no singularity at the origin withpm(r) itself. That does not immediately

guarantee smooth behaviour whenpm is written as a sum over basis functions, which individually

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do not have the smoothO(rm) behaviour. Nevertheless, partly because the Gauss-Legendre nodes(see below) do not include the origin, there were never problems and the convergence was smoothand fast.

Again we choose a basis of Chebyshev polynomials, but now on[−1,1]. So

φn(r) =

T2n(r) = cos(2n arccosr) if m is even,

T2n+1(r) = cos((2n + 1)arccosr) if m is odd.(23)

A Fourier basis is also possible, but comparison with exact solutions in uniform flow suggests thatresults with Chebyshev bases are slightly smoother.

While taking the above into consideration, we assume thatpm can be written as a sum overthis function basis{φn},

pm =∞

∑

n=0

anφn. (24)

Equation (19) becomes equivalent (withw = φl , andl = 0,1, . . .) with

∞∑

n=0

an

[

φnφl

iωZ

∣

∣

∣

∣

1

+∫ 1

0

{

− 1

�2φ′

nφ′l +

(

1 − κ2

�2− m2

r2�2

)

φnφl

}

r dr

]

= 0. (25)

This can also be written in matrix form

M(κ)a = 0, a = (a0, a1, . . . )T , (26)

for the[0,∞)× [0,∞) matrixM with elements

Mn,l(κ) = φnφl

iωZ

∣

∣

∣

∣

1

+∫ 1

0

{

− 1

�2φ′

nφ′l +

(

1 − κ2

�2− m2

r2�2

)

φnφl

}

r dr . (27)

Again, for the non-linear eigenvalue problem we cut offM to an N × N matrix and find thosediscrete values ofκ for whichM has a non-empty null space and corresponding solution spacea.

III.B. Special Case ofZ = 0

The problem with boundary conditions corresponding to a pressure release wall (Z = 0) cannotbe formulated in weak form like above. So any Galerkin procedure has to follow another strategy.The usual route is to utilise the fact that we may expect uniform convergence, so if we choose abasis that vanish at the interval ends, also their sum will vanish there.

If we consider the weak formulation withnatural boundary conditions (13, 19) (hard walls),then it is common for this type of problems (standard theory [27, 28, 29] cannot be applied di-rectly) that the Galerkin projection converges in a SobolevspaceH 1, which means both the func-tion and its derivative. This implies for one dimensional problems uniform convergence. So bychoosing a basis that vanishes at one or both ends of the interval (spanning a Sobolev spaceH 1

0 , asubspace ofH 1) we have guaranteed a solution that also vanishes at the boundary or boundaries.

For this problem it is not convenient to choose a Chebyshev basis since Chebyshev polyno-mials don’t vanish at the interval ends, but a Fourier basis of the right type works well.

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III.C. Non-Linear Eigenvalue Problem

Since the eigenvalue problem (16, 26) is non-linear, its numerical solution is not straightforward.For example, to find numerically the zeros of det(M(κ)) is a very unstable process for matriceslarger than (say) 20× 20 and therefore not an option. A number of approaches, working on thematrix itself, have been reviewed in [30]. Algorithm 3, “Method of successive linear problems”,is both simple to implement and efficient in operation (undernormal conditions quadraticallyconvergent), and appeared indeed suitable for our purposes. The method is qua structure similarto Newton’s method, and runs as follows.

Let a starting valueκ = κ0 be given. The corresponding vectora0 is not important. Findinga good starting value is a problem on its own, but tracing withsmall steps inu0 and Z from aknown configuration (uniform flow, hard walls) works well.

Suppose we have thej -th iterated eigenvalue approximationκ j with amplitude vectora j .Linearise forκ nearκ j

M(κ) = M(κ j)+ (κ − κ j )M′(κ j)+ . . . . (28)

Assuming a convergent process, we construct thej + 1-st iterated eigenvalueκ j+1 with vectora j+1 by solving the approximate equation

M(κ j+1)a j+1 ≃{

M(κ j)+ (κ j+1 − κ j )M′(κ j )

}

a j+1 = 0. (29)

This can be solved by standard methods as follows. Solve the generalised eigenvalue problem

M(κ j )x = λM′(κ j )x , (30)

with solutionsλ1, λ2, . . . andx1, x2, . . . . Choose from allλ’s the absolutely smallest,λmin, andidentify κ j+1 = κ j −λmin anda j+1 = x, wherex corresponds to thisλmin. Iterate until‖M(κ j )a j‖is small enough. This simple and elegant iteration appears to converge quickly.

The derivativesM′ are easily obtained analytically. For the 2D case (17) it is

d

dκMn,l(κ) = −2

∫ 1

0

1

c0�3

{

u0φ′nφ

′l + κωφnφl

}

dy . (31)

For the 3D case (27) it is

d

dκMn,l(κ) = −2

∫ 1

0

r

c0�3

{

u0φ′nφ

′l +

(

κω + m2

r2u0

)

φnφl

}

dr . (32)

An exact evaluation of these integrals is theoretically possible for a linear velocity profile andpolynomial basis functions, but the numerical evaluation of higher order polynomials is veryunstable. Much better is it to determine the integral numerically by the exceptionally efficientGauss-Legendre integration rule. This method amounts to anapproximation of the form

∫ b

af (x)dx ≃

J∑

j=0

w j f (x j ) (33)

with nodesx j and weightsw j chosen such that all polynomials of degree 2J + 1 or less areevaluated exactly. The method appears to be very accurate already for a moderate number of nodes

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(typically between 20 and 60, but at least 2N ), except when the resolution of steep variations ofthe mean flow asks for a finer grid.

The number of terms of the series over basisφn are typicallyN = 20 or 40. Lower values(N = 5 or N = 10) are sometimes sufficient, but for high cut-off modes, certain surface waves,higher frequencies, and sharp changes in the mean flow like thin boundary layers, higher valuesof N (100− 200) may become necessary. To assess the accuracy and confirmthe correctnesscomparisons with exact solutions with constantu0 andc0 were carried out. Other useful checkswere the various exact integrals that can be obtained of the form as given in SectionD.

IV. Asymptotic Solution for ω Small

In 2D, and 3D withm = 0, and hard walls, there is forω = 0 a real eigenvalueκ = 0 andinfinitely many imaginary. We consider here the neighbourhood of κ = 0 in the limit of smallω.By scaling and balancing terms in the differential equations, it is quickly found that eigenvalueκ = O(ω). Therefore, we rescaleκ = ωµ,� = ω(1 − µu0)/c0 = ωW .

IV.A. 2D and Hard Walls

We have for 2D( 1

W 2p′

)′+ ω2

(

1 − µ2

W 2

)

p = 0, (34)

with p′(0) = p′(1) = 0, andp andµ to be found. We expand in powers ofω2

p = p0 + ω2p1 + ω4p2 + . . . , µ = µ0 + ω2µ1 + . . . , (35)

After collecting equal orders of magnitude we find to leadingorders

( c20

(1 − µ0u0)2p′

0

)′= 0,

( c20

(1 − µ0u0)2

p′1

)′= −2

( µ1u0c20

(1 − µ0u0)3

p′0

)′−

(

1 − c20µ

20

(1 − µ0u0)2

)

p0,

and boundary conditionsp′0 = p′

1 = 0 at y = 0 andy = 1. It immediately follows thatp′0 = 0

and sop0 = 1 (or any other constant, but this is unimportant for an eigensolution). As a result wefind for the next order

p′1 = (1 − µ0u0)

2

c20

[

−y + µ20

∫ y

0

c20

(1 − µ0u0)2

dy′]

. (36)

The boundary condition aty = 1 yields an algebraic equation forµ0, valid for anyu0 andc0,

µ20

∫ 1

0

c20

(1 − µ0u0)2dy = 1. (37)

This is a general result, but findingµ0 requires numerical evaluation. For linear flow profile andconstant sound speed

u0 = τ + σ y , c0 = 1, (38)

the analysis can be performed analytically exactly. We have∫

1

(1 − µ0u0)2dy = 1

µ0σ (1 − µ0u0), (39)

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provided 1− µ0u0 6= 0 for 06 y 6 1, which is indeed confirmed by the final result

µ±0 = ±1

√

1 + 14σ

2 ± (τ + 12σ )

. (40)

The corresponding wave numbers and eigenfunctions with normalisationp(0) = 1 are

κ± = ±ω√

1 + 14σ

2 ± (τ + 12σ )

+ O(ω3), (41a)

p± = 1 − ω2µ±0 σ y2

[

(1 − µ±0 τ)

(

12 − 1

3 y)

− µ±0 σ

(

13 y − 1

4 y2)]

+ O(ω4). (41b)

This result is exact for a uniform profile whereσ = 0.

IV.B. 2D and Impedance Walls

For impedance conditions the asymptotics depend on the order of magnitude ofZ0,1 in terms ofsmall parameterω. There are many possible limits and we will not consider themall but just afew that seem the more interesting ones. With againκ = ωµ we see from the boundary condition

iωZp′ = ±ω2W 2 p, (42)

that any impedanceZ ≫ O(ω) yields a perturbation of hard wall conditions. In order not to mixup the expansion (35) in powers ofω2 suggested by equation (34), we choose3 Z0,1 = O(ω−1)

and rescaleZ0 = ω−1ζ0, Z1 = ω−1ζ1, (43)

with ζ0,1 = O(1) and consider the boundary conditions

iζ0p′ = −ω2W 2 p at y = 0, iζ1p′ = ω2W 2 p at y = 1. (44)

Expansion ofp results to leading orderp0 = 1. Higher orders can in principle be found for anyprofiles, but for the linear profile (38) we find analytically exactly

κ± = ±ω√

1

1 − i(ζ−10 + ζ−1

1 )+ 1

4σ2 ± (τ + 1

2σ )

+ O(ω3), (45a)

with eigensolutionsp = p0 + ω2p1 + . . . (normalised byp(0) = 1)

p±1 (y) = iζ−1

0 y(

13(yµ

±0 σ )

2 − yµ±0 σ (1 − µ±

0 τ)+ (1 − µ±0 τ)

2)

− y2

(

14(yµ

±0 σ )

2 − 13 yµ±

0 σ(

2(1 − µ±0 τ)− µ±

02

1 − µ±0 τ

)

+ 12

(

(1 − µ±0 τ)

2 − µ±0

2)

)

. (45b)

This approximation remains valid if one or both sides are hard. In that case simplyζ−10 = 0 or

ζ−11 = 0. If Z0 or Z1 are zero, we need a separate analysis. For example, ifZ0 = Z1 = 0, we find

to leading order

κ± = ω

τ + (12 ± i 1

2

√3)σ

+ O(ω3), (46a)

with (arbitrarily normalised) eigensolution

p±0 = y(1 − y)

(

y − 12 ∓ i 1

2

√3)

. (46b)

3Other scalings can be analysed on a case-by-case basis.

11 of 38


IV.C. 3D and Hard Walls

For the 3D,m = 0, problem we can follow a similar strategy to find

( r

W 2p′

)′+ ω2r

(

1 − µ2

W 2

)

p = 0, (47)

leading to

( rc20

(1 − µ0u0)2p′

0

)′= 0,

( rc20

(1 − µ0u0)2p′

1

)′= −2

( rµ1u0c20

(1 − µ0u0)3p′

0

)′− r

(

1 − c20µ

20

(1 − µ0u0)2

)

p0.

For hard walls withp′0(1) = 0 it follows that p0 = 1. Then

p′1 = −1

2r(1 − µ0u0)

2

c20

+ µ20(1 − µ0u0)

2

rc20

∫ r

0

r ′c20

(1 − µ0u0)2dr ′ . (48)

p′1(1) = 0 yields (ignoring the singular solution 1− µ0u0(1) = 0) the algebraic equation forµ0

µ20

∫ 1

0

r ′c20

(1 − µ0u0)2dr ′ = 1

2 . (49)

This is again a general result, valid for any (physical) profilesu0 andc0, but findingµ0 requiresnumerical evaluation. For parabolic profiles

u0 = τ − σr2, c0 = 1, (50)

we can derive analytically exact results by using∫

r

(1 − µ0u0)2

dr = − 1

2µ0σ (1− µ0u0), (51)

providedµ0u0 6= 1 for 06 r 6 1. This is indeed satisfied by the resulting modal wave numbers

κ± = ±ω√

1 + 14σ

2 ± (τ − 12σ )

+ O(ω3). (52a)

(This result is exact for a uniform profile whereσ = 0.) The derivativep′1 is

p′1 = 1

2στ(

1 − µ±0 τ

)

− µ±0

(

1 − µ±0 τ

)

(τ 2 − στ − 1)r(1 − r2)

(

1 − µ±0 u0(r)

)

,

and so, with normalisationp(0) = 1, the corresponding eigensolutions are

p± = 1 + 14ω

2στ(

1 − µ±0 τ

)

− µ±0

(

1 − µ±0 τ

)

(τ 2 − στ − 1)· · ·

· · · r2

[

(

1 − µ±0 τ

)(

1 − 12r2) − µ±

0 σ(

1 − r2 + 13r4)

]

+ O(ω4). (52b)

12 of 38


IV.D. 3D and Impedance Walls

The effect of the regularity condition at the origin is that the leading order solutionp0 can only bea constant ifω is small andκ = O(ω). So for a manageable and interesting problem we rescale

Z = ω−1ζ , (53)

with ζ = O(1), leading for the parabolic mean flow profile (50) to the boundary condition

iζ p′1 = (1 − µ0(τ − σ ))2 at r = 1. (54)

By using (48, 51) we arrive at the algebraic equation forµ0

12 iζ(1 − µ0(τ − σ ))

(

µ20

1 − µ0τ− 1 + µ0(τ − σ )

)

= (1 − µ0(τ − σ ))2,

with eventually the solutions

κ± = ±ω√

1

1 − 2iζ−1+ 1

4σ2 ± (τ − 1

2σ )

+ O(ω3), (55)

and a hydrodynamic modeκ ≃ ωµ0 = ω/(τ − σ ) (to leading order no pressure, only velocitycomponents). The corresponding eigensolutions can be constructed by integration ofp′

1.If Z = 0 we need a separate analysis. It transpires thatκ remainsO(1) for ω → 0 (in par-

ticular its imaginary part), and the equation forp does not simplify unlessu0 andc0 are constant.As a result there is no simple analytical approximation of the present type.

V. WKB Solution for ω Large

The used WKB solution [31, 32] is comparable to what has been studied in [18], so we willgive only a concise derivation.

V.A. 2D and Hard Walls

Consider

�2( 1

�2p′

)′+

(

�2 − κ2)

p = 0, p′(0) = p′(1) = 0, (56)

for largeω. We postulate cut-on modes for realκ not near resonance,i.e. κ = O(ω) and� =O(ω). We make the WKB-Ansatz (with phase functionγ = O(ω) to be determined)

p(y) = A(y)e−i∫ yγ (z) dz , (57)

such that(

−γ 2 +�2 − κ2)

A − i�2

A

(γ A2

�2

)′+�2

( A′

�2

)′= 0. (58)

We find to leading orderO(1)(−γ 2 +�2 − κ2)A = 0, (59)

or (with a suitably chosen sign)γ =

√

�2 − κ2. (60)

13 of 38


With thisγ , we have then to next orderO(ω−1)

(γ A2

�2

)′= 0, (61)

or (with Q0 a constant)

A = Q0�

√γ

= Q0�

(

�2 − κ2)

14

. (62)

We consider first the case ofγ is real everywhere (that is:γ 2 > 0). Then we have along the wholeinterval the solution

p = �(

�2 − κ2)

14

(

Q e−i∫ y

0 γ dz +R ei∫ y

0 γ dz)

, γ =√

�2 − κ2. (63)

To leading order this is aty = 0

p′(0) ≃ −i�√γ (Q − R) = 0,

such thatR = Q. We can write

p ≃ �√γ

cos(

∫ y

0γ dz

)

, p′ ≃ −�√γ sin

(

∫ y

0γ dz

)

, (64)

and we retain aty = 1

p′(1) ∼ √γ sin

(

∫ 1

0γ dy

)

= 0. (65)

In other words, the (approximate) condition to find eigenvalueκ is∫ 1

0γ dy = nπ , n ∈ N. (66)

Note that to respect the asymptotics ofω → ∞ andγ = O(ω), this equation is only valid forsufficiently largen, such thatω ∼ nπ . Usually, however, WKB is very forgiving.

In general this equation is to be solved numerically, even ifthe mean flow profile is linear,i.e.u0(y) = τ + σ y andc0(y) = 1, and the integral can be found analytically exactly.

Interesting special cases are those with turning points,i.e. there whereγ 2 changes sign alongthe interval[0,1]. There whereγ is imaginary, the wave becomes exponentially small and (apartfrom a small region beyond the turning point) is practicallynot present. Consider for conveniencethe situation of a single turning point on the interval(0,1). Then we have, taking into account asmall region beyond the turning point4, for a turning pointy1 on the left the condition

∫ 1

y1

γ dy = (n + 14)π , n ∈ N, (67)

and a turning pointy2 on the right the condition∫ y2

0γ dy = (n + 1

4)π , n ∈ N. (68)

More complicated configurations work in a similar way. For example, for two turning points0< y1 < y2 < 1 andγ real along[y1, y2] we have

∫ y2

y1

γ dy = (n + 12)π , n ∈ N. (69)

4This is a result of matching between outer solutions at either side of the turning point and an inner solution in theneighbourhood of the turning point [32].

14 of 38


V.B. 3D and Hard Walls

Consider�2

r

( r

�2p′

m

)′+

(

�2 − κ2 − m2

r2

)

pm = 0, p′(1) = 0, (70)

for largeω, r = O(1) andm = O(1). Largem, in particularm = O(ω), is also possible but withsome adaptations. We postulate cut-on modes for realκ not near resonance,i.e. κ = O(ω) and� = O(ω). We make the WKB-Ansatz (with phase functionγ = O(ω) to be determined)

pm(r) = A(r)e−i∫ rγ (z) dz, (71)

such that(

−γ 2 +�2 − κ2)

A − i�2

r A

(rγ A2

�2

)′+ �2

r

(r A′

�2

)′− m2

r2A = 0. (72)

For O(ω2) this leads toγ =

√

�2 − κ2. (73)

(with a suitably chosen sign) and to orderO(ω)

A = Q�

√rγ, (74)

with Q a constant. Altogether we have the approximation

pm(r) ≃ �√

rγ

(

Q e−i∫ r

0 γ (z) dz +R ei∫ r

0 γ (z) dz)

, (75)

with Q, R andκ to be determined.We start with the situation thatγ 2 = �2 − κ2 is positive along the interval[0,1], or has a

single zero at radial turning pointr2(κ) < 1, which is such thatγ 2 is positive along[0, r2] andnegative along the remaining[r2,1].

Near the origin the solution breaks down, and we have to consider a local analysis. Scale

r = ω−1z, γ (0) = ωα, z = O(1), pm(r) = P(z). (76)

and substitute in the equation, we get to leading order

ω2Pzz + ω21

zPz + ω2

(

α2 − m2

z2

)

P + · · · = 0. (77)

This has the regular solution (a multiple of)

P(z) = Jm(αz), (78)

which has to match with the WKB approximation. By takingu′0(r) 6 O(r), z large andr small,

we find

Jm(αz) ≃ 1√

12παz

cos(αz − 14π − 1

2mπ) ∼ �(0)√rωα

(

Q e−iωαr +R eiωαr)

. (79)

So withQ ∼ e

14π i+ 1

2mπ i , R ∼ e− 14π i− 1

2mπ i , (80)

15 of 38


the WKB approximation forr > O(ω−1) becomes

pm(r) ≃ �(r)√

rγcos

(

∫ r

0γ (z)dz − 1

4π − 12mπ

)

. (81)

An eigenvalueκ for a hard wall mode is then (to leading order) given byp′m(1) = 0 or

∫ 1

0γ (z)dz = (n + 1

4 + 12m)π , (82)

for integern and provided there is no radial turning pointr2 along[0,1]. With a turning pointr2

we have to take into account the fact that the solution is exponentially small along(r2,1] whereγ is imaginary. As a result, only the first interval is considered, Furthermore, matching with theexponentially decaying solution in[r2,1] yields a slight change of the effective interval to theeffect that we have an extra14π in the condition

∫ r2

0γ (z)dz = (n + 1

2 + 12m)π , (83)

for integern and r2 = r2(κ). It is important to realise that this condition is (asymptotically)independent of the behaviour nearr = 1. So for any other boundary conditions at the wall thesolution, including the value ofκ, is the same.

If γ 2 has a zeror1(κ) along [0,1] such, thatγ 2 is negative along[0, r1] and positive along[r1,1], the singularity atr = 0 plays no role because the solution is exponentially small anyway.We have to consider the second interval[r1,1], including the same slight change of the effectiveinterval due to matching with the exponentially decaying solution in [0, r1] as before. We obtainthe approximation

pm(r) ≃ �√

rγsin

(

∫ r

r1

γ (z)dz + 14π

)

, (84)

which leads, due to the boundary conditionp′(1) = 0, to the eigenvalue condition

∫ 1

r1

γ (z)dz = (n + 14)π , (85)

for integern andr1 = r1(κ). For other configurations similar arguments apply. Note that theapproximation deteriorates ifω is not large,m is too large, andr1 or r2 are too close to 0, 1 oreach other. All these bordering situations can be treated ona case-by-case basis, but on the wholethat is not necessary to understand the major trends and global behaviour that appear from thenumerical solution.

VI. Examples and Applications

The first results are centered around the 2D problem with linear mean flow profileu0 =0.5 + 0.3y, a relatively high frequencyω = 20 and no sound speed variation,c0 = 1. In table 1a comparison is made of the 2× 9 cut-on modal wave numbers, found numerically (withN = 40Chebyshev basis functions andJ = 80 Gauss-Legendre nodes) and by WKB. The agreementis very promising, although depends on secondary effects like the distance of the turning pointsy1, y2 from the walls. Observe that the first 2 right-running modes have turning pointsy2 anddecay neary = 1, and the first 7 left-running modes have turning points aty1 and decay near

16 of 38


y = 0. This is confirmed byfigure 1, where the modal shapes are given. The first right-runningmode practically vanishes aty = 1, and the first 5 or 6 left-running modes practically vanish aty = 0. This may cause these modes to be insensitive to an impedance placed at the side wherethe mode vanishes. This happens for example in the two cases given in figure 2, where oneside is hard and the other soft, and vice versa. The modes are traced from their hard-wall valuesfor varying impedances, with fixed real parts (0.4) and imaginary parts varying from∞ to −∞.We see that the first right-running mode is (practically) insensitive to the impedance aty = 1,and the first 6 left-running modes are insensitive to the impedance aty = 0. Associated to therespective impedance, 4 surface waves develop. Since they live in different mean flow velocities,they develop differently.

A study, similar to the one above but now in 3D, is given intables2, 3 andfigures 3, 4, and5. A parabolic mean flow profileu0 = 0.8− 0.5r2 and constant sound speedc0 = 1 is consideredfor modes of relatively high frequencyω = 20 andm = 0 andm = 1. The agreement betweennumerical and WKB modal wave numbers (tables2, 3) is very promising. The effect of the localapproximation nearr = 0 (required form = 1) is well captured. The first 2 right-running modeshave turning points atr1 and decay towardsr = 0, and the first 7 or 8 left-running modes haveturning points atr2 and decay towards the wallr = 1. This is confirmed by the modal shapeplots in figures3 and4. The first 6 or 7 (left-running) upstream running modes practically vanishat r = 1, with the result that these modes are almost insensitive tothe wall impedance. This iswell illustrated in the modal trace plots of figure5 for Re(Z) = 0.1. The insensitive modal wavenumbers vary along very small circles. In the plot these appear like stationary points.

Although modes remain self-similar in axial direction and by themselves do not refract, thisinsensitivity may be interpreted as a manifestation in the modal spectrum of rays that refract toregions with an effectively lower sound speed, and away fromregions with a higher sound speed[1, 31]. With sheared mean flow vanishing at the wall and a constant sound speed, the wall regionhas an effective sound speed ofc0 + u0 ≃ c0 < c0 + u0,max in case of downstream propagatingsound, while the center region has an effective sound speed of c0 − u0 ≃ c0 − u0,max< c0 in casethe sound propagates in upstream direction.

Since the present analysis is equally well valid for no flow and a varying sound speed, we haveincluded examples with a convex and concave sound speed profile, showing the same effects ofturning points and modes vanishing near the wall, but now independent of propagation direction.Consider intables 4 and 5 numerically and by WKB produced modal wave numbers for 3D,ω = 20, m = 0 problems with a convex sound speed profilec0 = 1 − 2

5r2 and with a concaveprofile c0 = 3

5 + 25r2. Both profiles are very comparable, since they have the same average of

25 and the same difference between the maximum and minimum values, and have very similar 8cut-on right-running modes (left-running modes are the same with minus sign). Only the concaveprofile has 5 modes with turning points atr2 and decay towardsr = 1. As a result (figure 6) thefirst 4 concave modes are practically insensitive to the wallimpedance.

In figure 7, 2D modes ofω = 5, constantc0 = 1, linearu0 = 0.45 + 0.25y, are traced forvarying Im(Z) and fixed Re(Z) = 0.45. In the left figure only the wall aty = 1 has an impedanceZ . In the right figure both walls have the same impedanceZ . With only one soft side, there are4 surface waves, but with both sides soft 2× 4. Since they live in different Mach numbers (0.45and 0.7) their behaviour is different. The surface waves aty = 1 are present in both cases andare very similar, because they don’t see the other side. The regular acoustic modes, on the otherhand, live in the whole duct and differ in the two cases.

In figure 8, 3D modes ofω = 5, m = 0, c0 = 1 and parabolic profilesu0 = τ − σr2,τ = 3

4 andσ = 34 (no slip flow; left) andσ = 1

4 (slip flow; right), are traced with Im(Z) and

17 of 38


fixed Re(Z) = 0.01. The relatively small value of Re(Z) is not a problem numerically, evenwith the moderate number of basis functions (N = 40) and Gauss-Legendre nodes (J = 80)that has been used. The well-known structure of regular modes, and acoustic and hydrodynamicsurface waves for uniform mean flow [8] is also present for parabolic profiles with slip flow, butwithout slip the hydrodynamic surface waves are not present, as there is indeed no mean flow atthe wall. However, the presence of the mean flow in the rest of the duct requires a left-runningfirst modal wave number ofκ−1 ≃ −14.795 which makes the square root

√�2 − κ2 imaginary

nearr = 1 and therefore the mode shape exponentially small. Indeed this mode is insensitive tothe impedance and the corresponding acoustic surface wave develops fromκ−2, rather thanκ−1.

Since the boundary condition for vanishing boundary layer (11b) and (12b) (the so-calledIngard limit [13]) is known to be sensitive, at least for the surface waves [15, 16], it is of interestto check the behaviour of the numerical solution for decreasing boundary layer thickness. A seriesof cases is presented infigure 9, where 3D modes ofω = 5, m = 0, are traced with Im(Z) alongRe(Z) = 0.1 in an almost uniform mean flow ofu0 ≃ 0.5 (given by equation107), a boundarylayer of typical thicknessδ = 0.1, = 0.05, 0.01 and= 0. The regular acoustic modes and thesurface wave in the 4th quadrant are easily reproduced. The other three surface waves convergepoorly. For the surface waves in the first and third quadrant we cannot expect simple convergence,because the real parts ofκ tends to infinity, and this implies a non-uniform limit withδ → 0. Forexample, neary = 0 a first quadrant surface wave behaves like

p ∼ e±i√�2−κ2y ≃ e−

√1−u2

0κy .

The Ingard limit assumes a boundary layer thickness much less than the typical width of the mode,which meansδκ ≪ 1, so the limit is not valid forκ & δ−1. Probably a much smaller value ofδ is necessary, but unfortunately a solution appears difficult to realise with the present numericalmethod. As mentioned above, this is only true for the surfacewaves in the first, second and thirdquadrant. All other modes are produced without a serious problem for these thin boundary layers.

Another investigation of the effect of a boundary layer is presented infigure 10, where thetwo cut-on hard wall modes of rotor-alone noise of a 22 blade rotor with periodicitym = 22and blade passing frequencyω varying aroundω = 22 is studied in almost uniform mean flow(u0 ≃ 0.6, eq.107) as a function of boundary layerδ = 0 · · · 0.1 in steps of 0.01. It is seen that atω = 22 andδ = 0 the modes are well cut-on, but whenδ is increased, the modes become cut-offfor δ > 0.055. So the presence of rotor-alone noise greatly depends ondetails of the boundarylayer (and the mean flow Mach number, of course).

In figures 11 and12, 3D cut-on hard-wall modes are plotted as a function of frequency and(for clarity) relatively high centerline Mach number (∼ 0.9). We see that with a parabolic profile(right) there is a quantitative difference between low and high frequencies, which is not presentwith a uniform profile (left). This is clearly caused by the fact that with parabolic profile and highω the first few (left and right-running) modes only live in the wall region with low Mach number.This is not possible with lowω. In figure11 right we see the lowω approximation (dotted line)given in equation (52a). Forω < 2 the agreement is very reasonable.

Finally, in figure 13 some (relatively) low frequency 3D modes with impedance walls andparabolic mean flow profiles are traced inZ for various fixed Re(Z), while the lowω approxi-mation, given by equation (55), are included as dotted lines. Consistent with the assumedscalingZ = O(ω−1), the approximation is especially very good near the hard wall limits. Otherwise, aslong as the mode does not become a surface wave the agreement is very reasonable.

18 of 38


VII. Conclusions

Duct acoustics is an important part of aeroacoustics, as it relates to the generation, propaga-tion, radiation and attenuation of turbine and rotor-stator interaction noise of aero-engines. Manyfeatures are described, or at least understood and clarified, by the concept of modes, a wave formthat emerges naturally if the duct can be modelled as straight with a plane parallel mean flow andconstant wall properties.

In the present paper we have studied duct modes for non-uniform mean flow profiles, de-scribed by Pridmore-Brown equations. The 2D Pridmore-Brown equation has been studied in aduct with lined walls both at the top and at the bottom; the 3D Pridmore-Brwon equation in ahollow cylindrical duct with a lined wall.

Numerical solutions have been constructed based on Galerkin projection with a Chebyshevfunction basis (except for the situation withZ = 0), leading to a non-linear eigenvalue problemthat has been solved by a very effective Newton-type routine.

Analytical approximations have been constructed for low and high frequenciesω.Except forZ = 0, the wave number of the first modes areO(ω) for smallω, and after scaling

a formal asymptotic solution could be given. For simple enough mean flow profiles this solutioncan be evaluated entirely analytically, with very favourable agreement with the numerical solution.The found expressions for the wave numbers in 2D with linear profile and in 3D with parabolicprofile show interesting similarities.

For high frequencies the well-known WKB method can be invoked successfully. A most in-teresting feature that is clearly displayed by the WKB solution is the possibility of a transverseturning point that separates the acoustic part of the solution from an exponentially small part nearthe wall. In this way the solution may in some cases be insensitive to the impedance condition.This happens with parabolic flow profiles for upstream running modes, and for both left and rightrunning mods in a medium with concave type sound speed profiles. This is by and large in agree-ment with the ray-acoustic rule that sound refracts in the direction of the lowest effective soundspeed. If this occurs, it is clear that eduction of the wall impedance from acoustic measurementsis impossible, at least if the insensitive modes dominate the sound field.

Surface waves are equally present with non-uniform flow, except that the combined limitof the boundary layer tending to zero and the hydrodynamic surface wave number tending toinfinity is non-uniform and difficult to capture numerically. Perhaps for this particular problem –hydrodynamic surface waves in a very thin boundary layer – another numerical method, with thepossibility to refine the accuracy locally in the boundary layer region, is preferable. For a meanflow profile that vanishes at the wall (a no-slip profile), without steep shear, the hydrodynamicsurface waves are not present and we have only the acoustic surface waves. The presence of themean flow elsewhere, however, may cause the first upstream running modes to be insensitive tothe impedance, and the upstream acoustic surface wave to develop from a higher order mode.

VIII. Acknowledgements

We want to acknowledge the interest, advice and help of colleagues Han Slot (to suggestthe Galerkin projection), Michiel Hochstenbach (to suggest the non-linear eigenvalue solver),Mark Peletier (who made useful suggestions for theZ = 0-problem), Pieter Sijtsma (to suggestthe rotor-alone modes) and students Rik Rutjens (to suggestthe Gauss-Legendre method) andMatthijs van Raaij.

19 of 38


numerical WKB

n κn κn y2

1 +12.7404 +12.7203 0.2410

2 +11.6482 +11.6186 0.7379

(a) [0, y2]

numerical WKB

n κn κn

3 +10.9908 +11.0352

4 +9.7991 +9.8030

5 +7.9130 +7.9130

6 +5.2757 +5.2755

7 +1.6580 +1.6579

8 -3.4857 -3.4858

9 -12.2566 -12.2572

-9 -31.6342 -31.6756

-8 -39.1405 -39.4023

(b) [0,1]

numerical WKB

n κn κn y1

-1 -91.7189 -91.0914 0.9348

-2 -75.6769 -75.6259 0.7851

-3 -65.7266 -65.7160 0.6522

-4 -58.1112 -58.1116 0.5194

-5 -51.9320 -51.9366 0.3830

-6 -46.7717 -46.7752 0.2414

-7 -42.5897 -42.3839 0.0937

(c) [y1,1]

Table 1: Comparison of 18 cut-on hard-wall modes: WKB approximation and numerical solution(40 Chebyshev polynomials, 80 Gauss-Legendre points); 2D duct of height unity;ω = 20; linearmean flow profileu0 = 0.5 + 0.3y andc0 = 1; turning points aty1 = ω+κ−κτ

κσ, y2 = ω−κ−κτ

κσ

numerical WKB

n κmn κmn r1

1 +13.9525 +13.8400 0.8425

2 +11.7647 +11.6915 0.4227

(a) [r1,1]

numerical WKB

n κmn κmn

3 +10.8217 +10.8247

4 +9.5047 +9.4845

5 +7.4557 +7.4323

6 +4.5837 +4.5588

7 +0.5924 +0.5636

8 -5.3314 -5.3720

(b) [0,1]

numerical WKB

n κmn κmn r2

-1 -94.9605 -95.0098 0.1449

-2 -85.0424 -85.0975 0.2647

-3 -75.2396 -75.3017 0.3622

-4 -65.5951 -65.6660 0.4573

-5 -56.1736 -56.2556 0.5577

-6 -47.0736 -47.1695 0.6693

-7 -38.4487 -38.5576 0.7984

-8 -31.3117 -30.6360 0.9517

(c) [0, r2]

Table 2: Comparison of 16 cut-on hard-wall modes: WKB approximation and numerical solution(40 even Chebyshev polynomials, 80 Gauss-Legendre points); 3D circular duct of radius unity;ω = 20, m = 0; parabolic mean flow profileu0 = 0.8 − 0.5r2 andc0 = 1; turning points atr1 = ( κτ+κ−ω

κσ)

12 , r2 = ( κτ−κ−ω

κσ)

12

20 of 38


numerical WKB

n κmn κmn r1

1 +13.9204 +13.8400 0.8425

2 +11.6581 +11.6915 0.4227

(a) [r1,1]

numerical WKB

n κmn κmn

3 +10.3427 +10.2436

4 +8.6215 +8.5506

5 +6.1758 +6.1112

6 +2.7973 +2.7321

7 -1.9877 -2.0617

8 -9.8657 -9.9806

(b) [0,1]

numerical WKB

n κmn κmn r2

-1 -89.9857 -90.0412 0.2103

-2 -80.1206 -80.1827 0.3144

-3 -70.3901 -70.4604 0.4095

-4 -60.8468 -60.9274 0.5065

-5 -51.5706 -51.6641 0.6117

-6 -42.6829 -42.7924 0.7313

-7 -34.4697 -34.4942 0.8716

-8 -28.4241 -27.0191 1.0394

(c) [0, r2]. n = −8 is questionable sincer2 ≃ 1

Table 3: The same as table2, but withm = 1.

numerical WKB

n κmn κmn r1

1 +30.3029 +30.0617 0.9147

2 +24.8656 +24.8079 0.6961

3 +21.9627 +21.9006 0.4658

4 +20.2593 +20.1724 0.1462

(a) [r1, 1]

numerical WKB

n κmn κmn

5 +18.8495 +18.8587

6 +16.5214 +16.5059

7 +12.8806 +12.8550

8 +6.0694 +6.0137

(b) [0,1]Table 4: Comparison of 8 right-running5) cut-on hard-wall modes: WKB approximation andnumerical solution; 3D circular duct of radius unity;ω = 20, m = 0; no mean flowu0 = 0,convex sound speed profilec0 = 1 − 2

5r2; turning points atr1 =(

52 − 5ω

2κ

)12 .

numerical WKB

n κmn κmn r2

1 +32.1479 +32.1812 0.2317

2 +29.8583 +29.8931 0.4155

3 +25.3544 +25.3922 0.6849

4 +23.1696 +23.1862 0.8102

5 +21.2544 +21.0157 0.9376

(a) [0, r2]

numerical WKB

n κmn κmn

6 +19.2399 +19.3132

7 +15.8955 +15.8878

8 +10.3510 +10.3196

(b) [0,1]

Table 5: The same as table4, but with concave profilec0 = 35 + 2

5r2; turning points atr2 =(

5ω2κ − 3

2

)12 .

5The left-running modes are the same with opposite signs

21 of 38


0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p1

1 = 12.740386

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p2

2 = 11.648158

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p3

3 = 10.990838

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p4

4 = 9.799080

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p5

5 = 7.913044

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p6

6 = 5.275705

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p7

7 = 1.657967

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p8

8 = -3.485691

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p9

9 = -12.256554

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-9

-9 = -31.634220

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-8

-8 = -39.140469

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-7

-7 = -42.589690

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-6

-6 = -46.771710

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-5

-5 = -51.931970

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-4

-4 = -58.111172

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-3

-3 = -65.726552

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-2

-2 = -75.676900

0 0.2 0.4 0.6 0.8 1

y

-1

-0.5

0

0.5

1

p-1

-1 = -91.718885

Figure 1: 18 Eigenfunction profiles for 2D, hard wall,ω = 20, linear mean flowu0 = 0.5+ 0.3y.Right running (positive index) and left running (negative index). First 2 right-running, and first 7left-running modes have turning point in interval, resulting in the first right-running mode havingno contact withy = 1, and the first 5 or 6 left-running mode having no contact withy = 0.Compare with table1.

22 of 38


-150 -100 -50 0

-60

-40

-20

0

20

40

60

= 20.0, = 0.50, = 0.30, Re( Z ) = 0.401

-150 -100 -50 0

-60

-40

-20

0

20

40

60

= 20.0, = 0.50, = 0.30, Re(Z ) = 0.400

Figure 2: 2D, complex wave numbersκn for ω = 20, linear profileu0 = 0.5 + 0.3y. One sidehard, other side impedanceZ . Top: impedance aty = 1. Bottom: impedance aty = 0. Tracingfor varying Im(Z) and Re(Z) = 0.4 fixed. 4 surface waves are associated to respective impedancewall. The 1st right-running mode is insensitive to impedance aty = 1 (top), the first 6 left-runningmodes are insensitive to impedance aty = 0 (bottom). Compare with table1 and figure1.

23 of 38


0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p1

1 = 13.952536

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p2

2 = 11.764687

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p3

3 = 10.821737

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p4

4 = 9.504711

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p5

5 = 7.455668

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p6

6 = 4.583742

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p7

7 = 0.592429

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p8

8 = -5.331368

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-8

-8 = -31.310013

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-7

-7 = -38.440692

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-6

-6 = -47.058096

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-5

-5 = -56.158474

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-4

-4 = -65.587349

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-3

-3 = -75.237679

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-2

-2 = -85.042235

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-1

-1 = -94.960449

Figure 3: Eigenfunction profiles for 3D, hard wall,ω = 20, m = 0, parabolicu0 = 0.8 − 0.5r2.Right running (positive index) and left running (negative index). First 7 left-running modes haveno contact withr = 1. Compare with table2.

24 of 38


0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p1

1 = 13.920388

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p2

2 = 11.658116

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p3

3 = 10.342656

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p4

4 = 8.621492

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p5

5 = 6.175829

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p6

6 = 2.797292

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p7

7 = -1.987749

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p8

8 = -9.865744

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-8

-8 = -28.424060

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-7

-7 = -34.469665

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-6

-6 = -42.682856

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-5

-5 = -51.570555

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-4

-4 = -60.846820

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-3

-3 = -70.390069

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-2

-2 = -80.120597

0 0.2 0.4 0.6 0.8 1

r

-1

-0.5

0

0.5

1

p-1

-1 = -89.985729

Figure 4: The same as figure3, but withm = 1. First 6 left-running modes have no contact withr = 1. Compare with table3.

25 of 38


-100 -50 0 50

-80

-60

-40

-20

0

20

40

60

80

= 20.0, m = 0, = 0.80, = 0.50, Re(Z) = 0.10

-100 -50 0 50

-80

-60

-40

-20

0

20

40

60

80

= 20.0, m = 1, = 0.80, = 0.50, Re(Z) = 0.10

Figure 5: 3D, complex wave numbersκmn for ω = 20, m = 0 (top), m = 1 (bottom), andparabolic profileu0 = 0.8 − 0.5r2. Tracing for varying Im(Z) and fixed Re(Z) = 0.1. The first6 left-running modes are practically independent ofZ . Compare with tables2, 3 and figures3, 4.

26 of 38


-50 0 50-30

-20

-10

0

10

20

30 = 20.0, m = 0, convex, Re(Z) = 2.00

-50 0 50-30

-20

-10

0

10

20

30 = 20.0, m = 0, concav, Re(Z) = 2.00

-50 0 50-30

-20

-10

0

10

20

30 = 20.0, m = 0, convex, Re(Z) = 1.00

-50 0 50-30

-20

-10

0

10

20

30 = 20.0, m = 0, concav, Re(Z) = 1.00

-50 0 50-30

-20

-10

0

10

20

30 = 20.0, m = 0, convex, Re(Z) = 0.50

-50 0 50-30

-20

-10

0

10

20

30 = 20.0, m = 0, concav, Re(Z) = 0.50

Figure 6: 3D, complex wave numbersκmn for ω = 20,m = 0,u0 = 0. Left: convexc0 = 1 − 25r2,

right: concavec0 = 35 + 2

5r2 sound speed profiles. Tracing for varying Im(Z) and fixed Re(Z).For the concave profile, the first 4 modes are practically insensitive to the impedance (comparewith table5).

27 of 38


-40 -30 -20 -10 0 10 20 30-40

-30

-20

-10

0

10

20

30

40 = 5.0, = 0.45, = 0.25, Re(Z) = 0.40

-40 -30 -20 -10 0 10 20

-40

-30

-20

-10

0

10

20

30

40

= 5.0, = 0.45, = 0.25, Re(Z) = 0.40

Figure 7: 2D, complex wave numbersκn for ω = 5 and linear profileu0 = 0.45+ 0.25y. Left:hard wall aty = 0 and impedanceZ at y = 1. Right: impedanceZ at bothy = 0 andy = 1.Tracing for varying Im(Z) and fixed Re(Z) = 0.4. Left: 4 surface waves associated to the rightwall. Right: 2× 4 = 8 surface waves, but living in different Mach numbers: 0.45 and 0.7.

-40 -30 -20 -10 0 10 20 30-40

-30

-20

-10

0

10

20

30

40 = 5.0, m = 0, = 0.75, = 0.75, Re(Z) = 0.01

-40 -30 -20 -10 0 10 20 30-40

-30

-20

-10

0

10

20

30

40 = 5.0, m = 0, = 0.75, = 0.25, Re(Z) = 0.01

Figure 8: 3D, complex wave numbersκmn for ω = 5,m = 0, and parabolic profilesu0 = 34(1−r2)

(left) andu0 = 34 − 1

4r2 (right). Tracing for varying Im(Z) and fixed small Re(Z) = 0.01. SmallRe(Z) gives no problems with the surface waves. The no-slip flow (left) shows no hydrodynamicsurface waves, while the first upstream modeκ−1 = −14.795 is insensitive to the impedance.

28 of 38


-40 -30 -20 -10 0 10 20

-30

-20

-10

0

10

20

30

= 5.0, m = 0, = 0.50, = 0.50, = 0.1, Re(Z) = 0.10

-40 -30 -20 -10 0 10 20

-30

-20

-10

0

10

20

30

= 5.0, m = 0, = 0.50, = 0.50, = 0.05, Re(Z) = 0.10

-40 -30 -20 -10 0 10 20

-30

-20

-10

0

10

20

30

= 5.0, m = 0, = 0.50, = 0.50, = 0.01, Re(Z) = 0.10

-40 -30 -20 -10 0 10 20

-30

-20

-10

0

10

20

30

= 5.0, m = 0, = 0.50, = 0.00, Re(Z) = 0.10

Figure 9: 3D, complex wave numbersκmn for ω = 5, m = 0 and almost uniform profile (SectionE, eq.107) with τ = σ = 0.5 and boundary layers ofδ = 0.1, = 0.05,= 0.01,= 0.0 (δ = 0 usesthe Ingard limit). Tracing for varying Im(Z) and fixed Re(Z) = 0.1. Surface waves convergepoorly, except in 4th quadrant.

29 of 38


19 20 21 22 23 24 25-50

-40

-30

-20

-10

0

10

mn

mn( ), m = 22, = 0.60, = 0.60, = 0.0-0.1

Figure 10: 3D, cut-on wave numbersκm,1 (red) andκm,−1 (blue) vs frequencyω and boundarylayerδ. 196 ω 6 25,m = 22. Almost uniform profileu0 (sectionE, eq.107) with τ = σ = 0.6and boundary layer varying fromδ = 0.0 (left) to δ = 0.1 (right) in steps of 0.01.Rotor-alone modeω = m = 22 is only cut-on forδ < 0.055.

30 of 38


0 5 10 15 20-200

-150

-100

-50

0

mn

mn( ), m = 0, = 0.90, = 0.00, fla

0 5 10 15 20-200

-150

-100

-50

0

mn

mn( ), m = 0, = 0.90, = 0.90, par

Figure 11: 3D, cut-on wave numbersκmn vs frequencyω. 0 6 ω 6 20.Comparison of uniform profileu0 = 0.9 (left), and parabolic profileu0 = 0.9(1 − r2) (right);red: right-running, blue: left-running. Dashed lines:ω ≪ 1 approximation (eq.52a), which isexact for the uniform profile.

0 5 10 15 20-200

-150

-100

-50

0

mn

mn( ), m = 10, = 0.90, = 0.00, fla

0 5 10 15 20-60

-50

-40

-30

-20

-10

0

10

20

mn

mn( ), m = 10, = 0.90, = 0.90, par

Figure 12: 3D, cut-on wave numbersκmn vs frequencyω. 0 6 ω 6 20,m = 10.Comparison of uniform profileu0 = 0.9 (left) and parabolic profileu0 = 0.9(1 − r2) (right);red: right-running, blue: left-running.

31 of 38


-4 -3 -2 -1 0 1 2-5

0

5 = 1.0, m = 0, = 0.50, = 0.50, Re(Z) = 0.50

-4 -3 -2 -1 0 1 2-5

0

5 = 1.0, m = 0, = 0.50, = 0.50, Re(Z) = 1.00

-6 -4 -2 0 2 4

-4

-2

0

2

4

6

8

10

12

= 0.2, m = 0, = 0.50, = 0.50, Re(Z) = 0.10

-4 -3 -2 -1 0 1 2

-4

-2

0

2

4

6 = 0.5, m = 0, = 0.50, = 0.50, Re(Z) = 0.25

Figure 13: 3D, complex wave numbersκmn of smallω, m = 0, and parabolicu0 = 12(1 − r2).

Tracing for varying Im(Z) and fixed Re(Z). Dashed lines:ω ≪ 1 approximation (eq.55).Consistent with assumed scalingZ = O(ω−1), the approximation is best near hard wall limit.

32 of 38


A. Wronskian

Two independent solutionsf and g of (11a), respectively (12a), are related by their Wron-skians. By integrating the differencesg(equation for f )− f (equation for g) in the usual way, wefind respectively

f g′ − g f ′ = C�2, f g′ − g f ′ = C�2

r. (86)

whereC are constants depending on the chosen normalisation off andg.

B. Preparatory Manipulations to the Pridmore-Brown Equati ons

The isentropic Euler equations, with velocityv, pressurep, densityρ, soundspeedc, entropys, small mass sourceq ′ and small forcef ′, are

dρ

dt+ ρ∇·v = q ′, ρ

dv

dt+ ∇ p = f ′,

ds

dt= CV

p

(dp

dt− c2 dρ

dt

)

= 0, c2 = γp

ρ. (87)

Consider a mean flowv0, p0, ρ0, c0 with perturbationsv′, p′, ρ ′ due toq ′ and f ′. If the mean flowis plane parallel withp0 = γ −1ρ0c2

0 a constant,v0 = u0ex , while u0, ρ0, c0 only depend ony, z,then the mean flow satisfies the stationary Euler equations. For the perturbations we have afterlinearisation

D0ρ′ + v

′·∇ρ0 + ρ0(∇·v′) = q ′, (88a)

ρ0D0v′ + ρ0(v

′·∇u0)ex + ∇ p′ = f ′, (88b)

D0p′ − c20D0ρ

′ − c20(v

′·∇ρ0) = 0, (88c)

where the convective derivative isD0 = ∂∂t + u0

∂∂x . Following [2, 33], we take the convective

derivative of the divergence of (88b), which becomes, with (88a) and they, z-components of (88b)

−c20D3

0ρ′ − 2c2

0∂

∂x

(

∇ p′ ·∇u0)

+ D0(c20∇2p′) = c2

0D0∇· f ′ − c20D2

0q ′ − 2c20∂

∂x

(

f ′·∇u0)

.

Further elaboration yields the wave equation

D0∇·(c20∇ p′)− D3

0 p′−2c20∂

∂x

(

∇ p′ ·∇u0)

= −c20D2

0q ′ + D0∇·(c20 f ′)−2c2

0∂

∂x

(

f ′·∇u0)

. (89)

Without the forcing terms we can assume modes of the formp′(x, t) = eiωt−iκx p(y, z). With theforcing terms we apply Fourier transformation

p′(x, y, z, t) = 1

2π

∫ ∞

−∞p(y, z)eiωt−iκx dκ ,

such that∂∂x → −iκ andD0 → i(ω − κu0) = ic0�, while for clarity we denote by∇ the nabla-

operator restricted toy andz. We get eventually the preform of the Pridmore-Brown equations

�2∇·( 1

�2∇ p

)

+ (�2 − κ2) p = −i(ω − κu0)q′ − iκ f · ex +�2∇·

( 1

�2f)

. (90)

33 of 38


C. Green’s Function

A natural scenario that shows how modes may constitute a solution of the acoustic problem, isthe construction of a solution of an inhomogeneous problem via Fourier transformation inx . Justas an example (we will not elaborate on this here further) we will describe the procedure formallyfor the 2D problem, but 3D is similar. Since it is very much like [12, 34], we will follow the samelines of argument.

Consider the inhomogeneous preform of the Pridmore-Brown equation (90), but restricted tovariabley. Rather than an arbitrary source term, we take a point sourceδ(x, y − y0), y0 ∈ (0,1)such that we have for dimensionless solutionp (the Green’s function) with Fourier representation

p(x, y) = 1

2π

∫ ∞

−∞p(y; κ)e−iκx dκ , (91)

the following equation and boundary conditions

( 1

�2p′

)′+

(

1 − κ2

�2

)

p = δ(y − y0), (92a)[

iωZ0 p′ = −�2 p]

y=0,

[

iωZ1 p′ = �2 p]

y=1. (92b)

Suppose we have two independent solutionsψ andχ of the homogeneous Pridmore-Brown equa-tion with generalκ. χ is arbitrary andψ satisfies the required boundary condition aty = 0 (anda less important normalisation)

iωZ0ψ′(0)+�2(0)ψ(0) = 0, ψ(0) = 1. (93)

The Wronskian ofψ andχ isψχ ′ − χψ ′ = Cw�

2, (94)

whereCw is a constant; see (86). By suitable normalisation ofψ andχ we may assume thatCw

is the same constant for allκ. For example, ifχ(0) = 0 andCw = 1, we have

χ = ψ

∫ y

0

�2

ψ2dy′ . (95)

With this relation we may verify the general solution

p(y) = A(κ)ψ + H(y − y0)(

ψ(y0)χ(y)− χ(y0)ψ(y))

, (96)

where H(·) denotes the Heaviside function andA is to be determined by applying the boundarycondition aty = 1. We find6

A(κ) = χ(y0)− H (κ)

E(κ)ψ(y0),

E(κ) = iωZ1ψ′(1)−�2(1)ψ(1),

H (κ) = iωZ1χ′(1)−�2(1)χ(1).

(97)

It is important to note thatE(κ) = 0 is effectively the eigenvalue equation for Pridmore-Brownmodes. IfE(κn) = 0 thenψ(y; κn) is a mode. By taking the pieces together we have

p(y; κ) = ψ(y<)(

χ(y>)− H (κ)

E(κ)ψ(y>)

)

, (98)

6Note: we have Heaviside function H and auxiliary functionH .

34 of 38


wherey< = min(y, y0) andy> = max(y, y0). Neglecting any possible critical layer contribution(branch cuts in the complexκ plane [12]) and assuming that bothψ andχ are defined such thatthey are analytic inκ, the physical solution found by inverse Fourier transformation (91) in κ,consists of sums over the residues in the polesκ±

n given byE(κ) = 0. If x > 0 we can close thecontour via the lower complex half plane and obtain the sum over the polesκ+

n with Im(κ+n ) < 0.

If x < 0 we can close via the upper half plane to capture the polesκ−n . We find

p(x, y) = ±i∞

∑

n=0

H (κ±n )

ddκ E(κ±

n )ψ(y0; κ±

n )ψ(y; κ±n )e−iκ±

n x , (99)

whereψ(y; κn) satisfies now both boundary conditions and is therefore a Pridmore-Brown mode.ddκ E(κn) can be constructed by first considering

dE

dκ= iωZ1ψ

′κ(1)−�2(1)ψκ(1)+ 2

u0(1)

c0(1)�(1)ψ(1), (100)

and then noting thatψκ = ddκψ is a solution of the inhomogeneous Pridmore-Brown equation

�2( 1

�2ψ ′κ

)′+

(

�2 − κ2)ψκ = 2(

κ + u0

c0�

)

ψ − 2ωu′0

(ω − κu0)2ψ ′, (101a)

iωZ0ψ′κ(0)+�2(0)ψκ(0) = 2

u0(0)

c0(0)�(0)ψ(0), ψκ(0) = 0. (101b)

D. Exact Integrals

Among other things, exact results for the Pridmore-Brown modes are very useful to check andverify the numerical routines. Therefore we give here a few.

Suppose we have two different solutionsp1, p2 of the 2D Pridmore-Brown equation, with allparameters the same, except the eigenvaluesκ1, κ2 and impedancesZ01 andZ02 at y = 0, andZ11

andZ12 at y = 1, then we can derive in the same way as in [11] that in 2D

−∫ 1

0Fp2

[

( 1

�21

p′1

)′+

(

1 − κ21

�21

)

p1

]

− Gp1

[

( 1

�22

p′2

)′+

(

1 − κ22

�22

)

p2

]

dy =∫ 1

0

(

Fκ2

1

�21

− Gκ2

2

�22

− F + G

)

p1 p2 +(

F ′

�21

p′1 p2 − G ′

�22

p1 p′2

)

+(

F

�21

− G

�22

)

p′1 p′

2 dy

−[

F

�21

p′1 p2 − G

�22

p1 p′2

]1

0

= 0,

such that

∫ 1

0

(

F

�21

− G

�22

)

p′1 p′

2 +(

F ′

�21

p′1 p2 − G ′

�22

p1 p′2

)

+(

Fκ2

1

�21

− Gκ2

2

�22

− F + G

)

p1 p2 dy =

1

iω

[

( F

Z11− G

Z12

)

p1 p2

]

y=1

+ 1

iω

[

( F

Z01− G

Z02

)

p1 p2

]

y=0

.

35 of 38


In a similar vein in 3D

−∫ 1

0Fp2

[

( r

�21

p′1

)′+

(

1 − κ21

�21

− m2

r2

)

p1

]

− Gp1

[

( 1

�22

p′2

)′+

(

1 − κ22

�22

− m2

r2

)

p2

]

dr =∫ 1

0

( F

�21

− G

�22

)

rp′1 p′

2 +( F ′r

�21

p′1 p2 − G ′r

�22

p1 p′2

)

+( κ2

1

�21

+ m2

r2�21

− 1)

Fp1 p2

−( κ2

2

�22

+ m2

r2�22

− 1)

Gp1 p2 dr −[

F

�21

p′1 p2 − G

�22

p1 p′2

]

r=1

= 0,

such that

∫ 1

0

( F

�21

− G

�22

)

rp′1 p′

2 +( F ′r

�21

p′1 p2 − G ′r

�22

p1 p′2

)

+( κ2

1

�21

+ m2

r2�21

− 1)

Fp1 p2

−( κ2

2

�22

+ m2

r2�22

− 1)

Gp1 p2 dr =[

( F

iωZ1− G

iωZ2

)

p1 p2

]

r=1

.

By choosingF andG we can generate exact integrals. For exampleF = G = 1 yields

∫ 1

0

(

κ21

�21

− κ22

�22

)

p1 p2 +(

1

�21

− 1

�22

)

p′1 p′

2 dy =

1

iω

[

( 1

Z11− 1

Z12

)

p1 p2

]

y=1

+ 1

iω

[

( 1

Z01− 1

Z02

)

p1 p2

]

y=0

, (102)

and

∫ 1

0

(

κ21

�21

+ m2

r2�21

− κ22

�22

− m2

r2�22

)

p1 p2 +(

1

�21

− 1

�22

)

rp′1 p′

2 dr =

1

iω

[

( 1

Z1− 1

Z2

)

p1 p2

]

r=1

. (103)

Another integral, not of the above form and essentially the equivalent of the result derived in [11],is given by

∫ 1

0

1

�2

[

(

(

1 + κ1κ2

�1�2

)u0

c0+ κ1

�1+ κ2

�2

)

p1 p2 − ωu′0

c20�

21�2

p′1 p2 + u0

c0�1�2p′

1 p′2

]

dy =

1

iω(κ1 − κ2)

[

[

( �1

Z11− �2

Z12

) p1 p2

�2

]

y=1

+[

( �1

Z01− �2

Z02

) p1 p2

�2

]

y=0

]

. (104)

In a similar way we can find for the 3D equation, with impedances Z1 andZ2 the integral

∫ 1

0

r

�2

[

(

(

1+ κ1κ2

�1�2− m2

r2�1�2

)u0

c0+ κ1

�1+ κ2

�2

)

p1 p2− ωu′0

c20�

21�2

p′1 p2+ u0

c0�1�2p′

1 p′2

]

dr =

1

iω(κ1 − κ2)

[

(�1

Z1− �2

Z2

) p1 p2

�2

]

r=1

. (105)

36 of 38


E. A Family of Flow Profiles with Boundary Layer

In order to obtain a flow profile in a cylindrical duct such thatthe boundary layer has exponen-tial decay, the behaviour in the origin is smooth, and the displacement and momentum thicknessesare explicitly available, we could try [16]

U (r) = tanh(1 − r

δ

)

+ a + br + cr2.

With the conditionsU (0) = 1, U (1) = 0 andU ′(0) = 0, we obtain7 then

U (r) = tanh(1 − r

δ

)

+(

1 − tanh(δ−1))

(

1 + r + 1 + tanh(δ−1)

δr

)

(1 − r) (106)

with the properties

U ′′(0) = −8δ−2 e−2δ−1(1 + δ + 1

2δ2)+ · · · = O(δ−2 e−2δ−1

),

U ′(1) = −δ−1 − 4δ−1 e−2δ−1 + · · · ≃ −δ−1,

U (1 − δ) = tanh(1)− 4e−2δ−1 + · · · ≃ 0.7616,

U (1 − 3δ) = tanh(3)− 12e−2δ−1 + · · · ≃ 0.995,

δdisp =∫ 1

0

(

1 − U (r))

r dr = ln(2)δ − 124π

2δ2 + · · · ≃ 0.69315δ,

δmom =∫ 1

0U (r)

(

1 − U (r))

r dr = (1 − ln(2))δ − (ln(2)− 124π

2)δ2 + · · · ≃ 0.30685δ.

With u0(0) = τ andu0(1) = τ − σ , more general profiles are created by

u0(r) = τ − σ + σU (r). (107)

References

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Solutions and Properties of the Pridmore-Brown Equation

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