WEBSTER’S HORN EQUATION REVISITED

Sjoerd W. Rienstra

Department of Mathematics and Computing Science,Eindhoven University of Technology,

The Netherlands,s.w.rienstra@tue.nl

Abstract

The problem of low-frequency sound propagation in slowly varying ducts is systematically analysed as aperturbation problem of slow variation. Webster’s Horn equation and variants in bent ducts, in ducts withnon-uniform soundspeed, and in ducts with irrotational mean flow, with and without lining, are derived,and the entrance/exit plane boundary layer is given. It is shown why a varying lined duct in general doesnot have an (acoustic) solution.

1 Introduction

Sound of long wavelength, propagating in ducts of varying diameter like horns, is suitably described by anapproximate equation, known as Webster’s horn equation or just Webster’s equation. This is an ordinarydifferential equation in the axial coordinate, and therefore forms a significant simplification of the problem[1, 2, 3].

The usual derivation is based on the assumption of a crosswise uniform acoustic pressure field, suchthat by averaging over a duct cross section the spatial dimensions of the problem are reduced from three toone.

Although it shows a remarkable evidence of ingenuity and physical insight, this derivation is mathemat-ically not satisfying. It is not clear (i) what exactlyis the small parameter underlying the approximation,(ii) why the pressure may be assumed to be uniform, (iii) what the error is of the approximation, (iv) whatthe conditions are on the duct geometry and on the frequency of the field, (v) how to generalize to similarproblems, (vi) how to generate higher order corrections, and (vii) what happens near the source or ductentrance or exit plane.

An asymptotically systematic derivation of the 3D classic problem was given by Lesser & Crighton [4],extending the derivation of Lesser & Lewis in [5, 6]. They also showed for a number of 2D configurationshow abrupt changes of the geometry (open end, slit in the wall) can be incorporated as boundary layerregions in a setting of matched asymptotic expansion. Their approach, based on introducing differentlongitudinal and lateral scales, is a special case of the method of slow variation put forward by Van Dyke[7]. Although only an asymptotically sound derivation is able to indicate the range of validity and the orderof the error of the approximation, we found, however,in the literature no variants of this problem (e.g. withmean flow [8, 9, 10, 11, 12]) that strictly follow this approach.

Particularly interesting would be an investigation of the related problems of lined ducts without andwith flow, as this would form a natural long wave length closure of the multiple scales theory of soundpropagation in slowly varying ducts [13, 14, 15, 16].

Another problem of practical interest that is directly related to a systematic set up is the entranceproblem for a 3D duct of arbitrary cross section. The structure of the boundary layer was indicated byLesser & Crighton [4] but they gave only explicit examples for 2D geometries.

All in all, while the problem of long wave sound propagation in slowly varying ducts, in various gen-eralisations, is practically important, it still has a lot of open ends.

We will consider various cases in detail. First, we show how a systematic approach, known as themethod of slow variation coupled with ideas of matched asymptotic expansions, leads to the classic Web-ster’s equation for hard-walled ducts with the entrance boundary layer. The small parameterε is equal

1 08:40, 02-08-2004

to the Helmholtz number, the ratio between a typical wave length and the duct diameter, while a typicallength scale of duct variation is of the same order of magnitude as the wave length. Using similar resultsfor the related problem of heat conduction [17], this entrance problem will be solved explicitly. It leads viamatching conditions to conclusions on the way theO(1) duct field error (O(ε) or O(ε2)) depends on thesource.

Then we will show that the problem is not essentially different in other coordinate systems (like spher-ical coordinates) although special coordinates may be helpful to obtain a more efficient approximation.Curved ducts, with a curvature radius of no more than the typical length scale of diameter variation, areshown to produce still the same equation.

The same type of analysis can be applied to ducts with lined walls. It is found that forZ = O(1)only the trivial solution exists, while forZ = O(ε) there are only non-trivial solutions possible for certain,geometry dependent, values of the wall impedance. As these impedance values vary along the duct, thereare in general no solutions possible for the full duct. A subtle functional analytic result is used due toProfessor Jan de Graaf (TU Eindhoven), which is not available in the literature. Therefore, Prof. de Graafwas kind enough to attach his derivation as an appendix to this paper.

We continue with more general analyses of the problem in a stagnant medium with slowly varyingsound speed, and of sound in an irrotational isentropic mean flow, leading to generalised forms of Webster’sHorn equation.

We finish with the same problem with mean flow but now extended with lined walls. Using a recentresult obtained for the related problem for high frequency sound propagation in lined flow ducts [16] weare able to show forZ = O(1) that also here only a special hydrodynamic (non-acoustic) wave is possible.

2 The physical models

2.1 The equations

In the acoustic realm of a perfect gas that we will consider, we have for pressurep, velocity v, densityρ,entropys, and soundspeedc

ddt ρ = −ρ∇· v, ρ d

dt v = −∇ p, ddt s = 0,

s = CV log p − CP log ρ, c2 = γ p

ρ, γ = CP

CV.

(1)

whereγ , CP andCV are gas constants. The flow is assumed to originate from a thermodynamically uni-form state. When we have a stationary mean flow with unsteady time-harmonic perturbations of frequencyω, given, in the usual complex notation, by

v = V + Re(v eiωt ), p = P + Re(p eiωt ), ρ = D + Re(ρ eiωt ), s = S + Re(s eiωt ), (2)

(ω > 0) and linearize for small amplitude, we obtain for the mean flow

∇·(DV ) = 0, D(V ·∇)V = −∇ P,

(V ·∇)S = 0, S = CV log P − CP log D, C2 = γ P

D

(3)

and the perturbations

iωρ + ∇·(Vρ + vD) = 0 (4a)

D(iω + V ·∇)

v + D(v·∇)

V + ρ(V ·∇)V = −∇ p (4b)

(iω + V ·∇)s + v·∇S = 0 (4c)

2 08:40, 02-08-2004

while

s = CV

Pp − CP

Dρ = CV

P

(p − C2ρ

). (4d)

Without mean flow, such thatV = ∇ P = 0, the equations may be reduced to (section 7)

∇·(C2∇ p) + ω2 p = 0. (5)

If, in addition, the ambient medium is uniform, with a constant soundspeedC and densityD, the acousticfield becomes isentropic and irrotational and we may introduce a potentialv = ∇φ. Furthermore, equation(5) reduces to the Helmholtz equation. After introducing the free field wave numberk = ω/C we have(sections 3, 4, 5, 6)

∇2φ + k2φ = 0. (6)

If the original flow field v is irrotational and isentropic everywhere (homentropic), we can introduce apotential for the velocity, wherev = ∇φ, and expressp as a function ofρ only, such that we can integratethe momentum equation (Bernoulli’s law, with constantE), to obtain for the mean flow

12V 2 + C2

γ − 1= E, ∇·(DV ) = 0,

P

Dγ= constant (7)

and for the acoustic perturbations(iω + V ·∇)

ρ + ρ∇·V + ∇·(D∇φ) = 0, D(iω + V ·∇)

φ + p = 0, p = C2ρ. (8)

These last equations are further simplified (eliminatep andρ and use the fact that∇·(DV ) = 0) to therather general convected wave equation (section 8)

D−1∇·(D∇φ) − (iω + V ·∇)[

C−2(iω + V ·∇)φ]

= 0. (9)

2.2 Nondimensionalisation

Without further change of notation, we will assumethroughout this paper that the problem is made dimen-sionless: lengths on a typical duct radius, time on typical sound speed / typical duct radius,etc.

2.3 The geometry

The domain of interest consists of a ductV of arbitrary cross section, slowly varying in axial direction (seefigure 1).

θ

A(εx)

nn⊥

�

v·n = 0

x-axis

r = R(εx, θ)

Figure 1. Sketch of geometry

For definiteness, it is given by the functionS in cylindrical coordinates as follows

S(X, r, θ) = r − R(X, θ) ≤ 0 (10)

whereX = εx � 0 is a so-called slow variable whileε is small. A cross sectionA(X) at axial positionXhas surface areaA(X). Whenever relevant∗, we assume lengths made dimensionless such that

A(0) = 1.∗ in particular in section 4

3 08:40, 02-08-2004

At the duct surfaceS = 0 the gradient∇S is a vector normal to the surface (i.e. ∇S ∝ n), while thetransverse gradient∇⊥S

∇⊥ = er∂

∂r+ eθ

1

r

∂

∂θ, with ∇⊥S = er − eθ

1

rRθ , (11)

(where an index denotes a partial derivative)is directed in the plane of a cross sectionA(X), and normalto the duct circumference∂A. So if n⊥ is the component of the surface normal vectorn in the plane of across section, we have∇⊥S ∝ n⊥.

2.4 Frequency

The frequencies considered are low, such that the corresponding typical wave number is of the same orderof magnitude as the length scale of the duct variations,i.e. dimensionlessO(ε−1). In order to quantify this,we will rescalek = εκ andω = ε.

3 The classical problem

3.1 Equations and boundary conditions.

The duct is semi-infinite and hard-walled. The solution is determined by a source at entrance planex = 0,and radiation conditions forx → ∞. Other conditions, like a reflecting impedance plane at some exitplanex = L (e.g. modelling a radiating open end [5], or a slit in the wall [4]), are also possible but they donot essentially alter the present analysis.

InsideV we have for acoustic potentialφ (eq. 6)

∇2φ + ε2κ2φ = 0 if x ∈ V, with ∇φ·n = 0 at x ∈ ∂V. (12)

At the entrance interfacex = 0 we have a suitable boundary condition, say,

φ(0, r, θ) = F(r, θ). (13)

The boundary condition of hard walls atr = R(X, θ) may be given by

∇⊥φ·∇⊥S = φr − RθR2φθ = εRXφx . (14)

Except for the immediate neighbourhood of the entrance plane, the typical axial variations of the acousticfield scale on the slow variableX , so we rewrite the equations and boundary conditions

ε2φX X + ∇2⊥φ + ε2κ2φ = 0, with ∇φ·∇S = −ε2φX RX + ∇⊥φ·∇⊥S = 0 at r = R. (15)

This rewriting in a slow variable is known as the method of slow variation [7]. Note that this equation hasa small parameter multiplied with the highest derivative inX-direction, suggesting a singular perturbationproblem [4, 18, 19, 20] with boundary layers inX .

3.2 Asymptotic analysis: outer solution.

The following outer solution analysis will largely follow Lesser & Crighton [4], but we will give it insome detail for two reasons. First, we will have to define the solution for the inner solution at the entranceboundary layer to be discussed later. Second, it explicates the method of integration along a cross sectionthat will be used in the various other configurations later.

Based on the observation thatε2 is the only small parameter that occurs, we might be tempted toexpand the solution in a Poincaré asymptotic power series inε2. However, we will see that this is notexactly true. Depending on the behaviour of the solution near the entrance, the correction term should in

4 08:40, 02-08-2004

general beO(ε) for matching. Nevertheless, the leading and first order equations will be equivalent. Withthe assumed Poincaré expansion ofφ, expressed inX ,

φ(X, r, θ; ε) = φ0(X, r, θ)+ εφ1(X, r, θ)+ ε2φ2(X, r, θ)+ . . . (16)

we obtain to leading order

∇2⊥φ0 = 0, with ∇⊥φ0·n⊥ = 0 at r = R, (17)

with a solutionφ0 = 0. As the solution of a Neumann problem is unique up to a constant,φ0 = φ0(X), afunction to be determined. To first order we have

∇2⊥φ1 = 0, with ∇⊥φ1·n⊥ = 0 at r = R, (18)

also with a constant solution, and soφ1 = φ1(X), a function to be determined. To second order we nowhave

∇2⊥φ2 + φ0,X X + κ2φ0 = 0, with ∇⊥φ2·n⊥ = φ0,XRRX√

R2 + R2θ

at r = R. (19)

The assumption (16) that there exists a Poincaré expansion forφ, expressed in this slow variableX , is nottrivial (Poincaré expansions are critically dependent on the variables chosen!). It requires certain solvabilityconditions for,e.g. φ2, yielding an equation forφ0. To obtain this, we integrate along a cross sectionA(X)and apply Gauss’ theorem∫∫

A∇2⊥φ2 dσ =

∫∂A

∇⊥φ2·n⊥ d� =∫∂Aφ0,X

RRX√R2 + R2

θ

d� = . . .

Then we parametrize∂A with θ , such that d� =√

R2 + R2θ dθ , and we continue

=∫ 2π

0φ0,X RRX dθ = φ0,X

∫ 2π

0RRX dθ = φ0,X AX . (20)

On the other hand, we also have∫∫A

[φ0,X X + κ2φ0

]dσ = A

(φ0,X X + κ2φ0

)(21)

Altogether we have forφ0 the equation

A−1(Aφ0,X)

X + κ2φ0 = 0, (22)

which is indeed Webster’s Horn equation [1, 2] in properly scaled coordinates.Evidently, the first order solution follows the same pattern and satisfies also

A−1(Aφ1,X)

X + κ2φ1 = 0, (23)

For completeness we note from [21, 22, 23, 24, 3]that Webster’s equation can be recast into a moretransparent form by the transformation

A(X) = d(X)2, φ = d−1ψ, (24)

leading to

ψ ′′ +(κ2 − d ′′

d

)ψ = 0. (25)

Depending on the sign ofκ2 − d ′′/d, the solutions behave like propagating or exponentially decayingwaves. Elementary solutions are readily found for geometries withd ′′/d = m2, a constant, yieldingSalmon’s family of exponential and conical horns [21, 22].

5 08:40, 02-08-2004

3.3 Boundary conditions inX .

The above equation forφ0 andφ1 is of second order and therefore two boundary conditions are requiredto determine the solution. ForX → ∞ we have the condition of radiation. AtX = 0 (figure 2),φ0 andφ1 cannot satisfy the(r, θ)-dependent boundary condition (13). Indeed, as anticipated before, nearx = 0

Figure 2. The entrance

there is a boundary layer ofX = O(ε), i.e. x = O(1), which determines the (outer) solutionsφ0 andφ1via conditions of matching. This will be considered in the next section.

4 Entrance boundary layer

Near the entrance, forX = O(ε), i.e. x = O(1), we have of course equation (12)

∇2φ + ε2κ2φ = 0 if x ∈ V, with ∇⊥φ·n = 0 at x ∈ ∂V. (12)

Up toO(ε2), this Helmholtz equation is equivalent to the Laplace equation. Therefore, the boundary layeranalysis is essentially similar to the one for the heat equation, discussed in Chandra [17]. Expand

φ(X, r, θ; ε) = �0(x, r, θ)+ ε�1(x, r, θ)+ O(ε2) (26)

so we have insideV to leading and first order

O(1) : ∇2�0 = 0, (27a)

O(ε) : ∇2�1 = 0. (27b)

At x = 0 we have from (13) the initial conditions

�0(0, r, θ) = F(r, θ), �1(0, r, θ) = 0. (28)

For x → ∞ conditions of matching with the outer solutionφ0 + εφ1 apply. For the boundary condition atr = R we have to expandR(εx, θ). Note that for any functionf

f (R(εx); ε) = f (R + εx RX + O(ε2); ε) = f0(R)+ ε(

f1(R)+ x f0,r (R)RX) + O(ε2) (29)

whereR without any argument denotes the value atX = 0. Furthermore, we have

Rθ (X, θ)

R2(X, θ)= Rθ

R2+ εx

( RX

R2

)θ

+ O(ε2) (30)

So at the boundary

∇⊥φ·∇⊥S = φr − RθR2φθ = �0,r − Rθ

R2�0,θ + ε

[�1,r − Rθ

R2�1,θ

+ x�0,rr RX − xRθR2 RX�0,rθ − x

( RX

R2

)θ�0,θ

]= εRX�0,x (31)

6 08:40, 02-08-2004

which means atr = R(0, θ) for the leading and first order

∇⊥�0·∇⊥S0 = �0,r − RθR2�0,θ = 0, (32a)

∇⊥�1·∇⊥S0 = �1,r − RθR2�1,θ = RX�0,x − x�0,rr RX + x

RθR2

RX�0,rθ + x( RX

R2

)θ�0,θ , (32b)

whereS0 = S(0, r, θ).It is important for the subsequent matching to note that the solutions of (27) with (32) are only defined

up to a linear termK x . For�0, however, this would result in terms ofO(ε−1) if x = O(ε−1) which donot match with an outer solutionφ0 = O(1). Therefore, we will not include this extra term. For�1, on theother hand, we will have to retain the possibility, and in the end a linear termK1x will be added, whereK1must be determined by the matching.

From the identity atr = Rd

dθ�0,θ = �0,rθ Rθ +�0,θθ , (33)

and with the defining equation applied atr = R while using relation (32a)

−�0,rr = 1

R�0,r + 1

R2�0,θθ +�0,x x = RθR3�0,θ + 1

R2�0,θθ +�0,x x (34)

it follows that equation (32b) is equivalent to

∇⊥�1·∇⊥S0 = Q0(x, θ)def== RX�0,x

∣∣r=R + x

R

{RRX�0,x x

∣∣r=R + d

dθ

( RX

R�0,θ

∣∣r=R

)}(35)

4.1 Leading order.

The right-running solution�0 (only non-increasing exponentials are allowed for matching) may be ex-pressed by the eigenfunction expansion

�0(x) =∞∑

n=0

Fnψn(r, θ) e−λn x (36)

where∇2⊥ψn + λ2

nψn = 0, ∇⊥ψn ·∇⊥S0 = 0 (37)

with λ0 = 0,ψ0 is a constant (normalised to 1), the other eigenvaluesλn are real positive, and the eigen-functionsψn are real, orthogonal and assumed normalized. In general these eigenfunctions are to bedetermined numerically. However, if the duct is cylindrical (i.e. R is independent ofθ ), we have

ψn(r, θ) := ψνµ(r, θ) =

Jν(

j ′νµr/R

)√π

2

(1 − ν2

j ′νµ

2

)R Jν( j ′

νµ)

{cosνθ

sinνθ

}for ν = 0,

J0(

j ′0µr/R

)√π R J0( j ′

0µ)for ν = 0,

(38)

where the indexn is more practically changed into the double index(νµ). Jν is theν-th order ordinaryBessel function of the 1st kind [25], andj ′

νµ is theµ-th (real-valued, positive) zero ofJ ′ν . The correspond-

ing eigenvalue is thusλn := j ′νµ/R.

The amplitudes are determined from the entrance interfacex = 0 as follows

Fn =∫∫

A(0)F(r, θ)ψn(r, θ) dσ. (39)

7 08:40, 02-08-2004

Note that, asψn are orthonormal, the axial flux is to leading order proportional to the imaginary part of∫ 2π

0

∫ R

0�0�

∗0,xr drdθ = −

∞∑n=1

λn |Fn |2 e−2λn x . (40)

As this expression is real, its imaginary part is zero and, thus, the axial flux vanishes to leading order.Indeed, the outer solution is a slowly varying function ofX and therefore the flux, proportional to the axialderivative, isO(ε).

For x → ∞, the exponential terms in�0(x) vanish and we have

�0(x) � F0. (41)

4.2 1st Order.

With the found expression for�0, the right hand side of equation (35),Q0, may be written as

Q0(x, θ) =∞∑

n=1

Fn e−λn x[−RXλnψn

∣∣r=R + x RXλ

2nψn

∣∣r=R + x

R

d

dθ

( RX

Rψn,θ

∣∣r=R

)]

= R−1∞∑

n=1

Fn

[−λn RRX

(x e−λn x)

xψn∣∣r=R + x e−λn x d

dθ

( RX

Rψn,θ

∣∣r=R

)](42)

To solve the problem for�1, we introduce a Green’s functionG(x; ξ) with x = (x, r, θ) andξ = (ξ, ρ, η)

satisfying

∇2⊥G + ∂2

∂x2 G = −δ(x − ξ ), ∂∂n G = 0 at r = R(0, θ), G(x; ξ) = 0 at x = 0,

G(x; ξ) → a constant forx → ∞, x ∂∂x G(x; ξ) → 0 for x → ∞.

}(43)

We determine the Green’s function by applying the Fourier Sine Transform† with respect tox (x → α) to(43), to obtain

∇2⊥G − α2G = −√

2

πsin(αξ)δ(x⊥ − ξ⊥). (44)

wherex⊥ denotes the transverse component ofx, i.e. x⊥ = (r, θ) (similarly for ξ⊥). We assume that theGreen’s function can be expanded by the same basis function as has been used for�0

G(α, r, θ; ξ ) =∞∑

m=0

am(α, ξ )ψm(r, θ).

Therefore

∇2G = −∞∑

m=0

amλ2mψm(r, θ).

Substituting this into (44) yields

∞∑m=0

amψm(λ2m + α2) =

√2

πsin(αξ)δ(x⊥ − ξ⊥). (45)

Next, we multiply (45) withψn and integrate over the cross sectionA(0) to obtain∫∫A(0)

∞∑m=0

amψnψm(λ2m + α2) dσ =

√2

π

∫∫A(0)

ψn(r, θ) sin(αξ)δ(x⊥ − ξ⊥) dσ. (46)

†where f (α) =√

2π

∫ ∞0 sin(αx) f (x) dx, f (x) =

√2π

∫ ∞0 sin(αx) f (α) dα.

8 08:40, 02-08-2004

Orthonormality of the basis functions yields

am =√

2

π

sin(αξ)

λ2m + α2ψm(ρ, η). (47)

Therefore,

G(α, r, θ; ξ, ρ, η) =√

2

π

∞∑m=0

sin(αξ)

λ2m + α2

ψm(ρ, η)ψm(r, θ). (48)

The inverse Fourier Sine Transform yields

G(x; ξ) = 2

π

∞∑m=0

ψm(ρ, η)ψm (r, θ)∫ ∞

0

sin(αx) sin(αξ)

λ2m + α2 dα, (49)

where [25] forλ0 = 0 ∫ ∞

0

sin(αx) sin(αξ)

α2dα = 1

2π min(x, ξ), (50)

and forλm > 0, ∫ ∞

0

sin(αx) sin(αξ)

λ2m + α2 dα = 1

2π e−λm max(x,ξ ) 1

λmsinh(λm min(x, ξ)). (51)

Therefore, them = 0-term can be taken apart and the Green’s function becomes

G(x; ξ) = x +∞∑

m=1

ψm(ρ, η)ψm(r, θ) e−λmξsinh(λm x)

λmif 0 ≤ x ≤ ξ, (52a)

= ξ +∞∑

m=1

ψm(ρ, η)ψm(r, θ) e−λm x sinh(λmξ)

λmif 0 ≤ ξ ≤ x . (52b)

Note that asx → ∞, G tends toξ and ∂G∂x tends to zero exponentially.

Using this Green’s function, we obtain for�1 the following relation, to be integrated over domainV,

�1δ(x − ξ ) = G∇2�1 −�1∇2G. (53)

However, since�1 ∼ K1ξ for largeξ (see the remark below equations 32), this yields a divergent integralas the domain here is a semi-infinite duct. Therefore, we consider a regionV ′ with a finite length 0≤ x ≤x0, wherex0 is small compared toε−1, but large enough for all exponential terms to practically vanish.Integrate (53) along domainV ′ and by using Green’s second identity we get

�1(ξ ) =∫∫∫V ′

(G∇2�1 −�1∇2G

)dx =

∫∫x=0

(−G∂�1

∂x+�1

∂G

∂x

)dσ

+∫∫

r=R(0,η)

(G∇⊥�1 −�1∇⊥G

)·n⊥ dσ +∫∫

x=x0

(G∂�1

∂x−�1

∂G

∂x

)dσ

=∫∫

r=R(0,η)

GQ0(x, θ)

|∇⊥S| d�dξ + K1ξ. (54)

Since|∇⊥S| = 1R

√R2 + R2

θ and d� =√

R2 + R2θ dθ , we obtain

�1(ξ ) =∫ 2π

0

∫ ∞

0Q0(x, θ)G(x; ξ)|r=R R dxdθ + K1ξ. (55)

9 08:40, 02-08-2004

As we haveQ0 in the form of a series expansion, we can write

�1(ξ) = K1ξ +∞∑

n=1

Fn

∫ 2π

0

[−RRXλnψn

∣∣r=R

∫ ∞

0e−λn x G(x; ξ)

∣∣r=R dx

+{

RRXλ2nψn

∣∣r=R + d

dθ

( RX

Rψn,θ

∣∣r=R

)}∫ ∞

0x e−λn x G(x; ξ )

∣∣r=R dx

]dθ (56)

As the series forQ0 converges uniformly forx > 0, we may exchange summation and integration. On theother hand, the fact that all basis functions have vanishing normal derivatives at the wall,i.e. ∇⊥ψn ·n⊥ =0, whereas∇⊥�1·n⊥ = 0 suggests that this series does not converge uniformly near the wall.

The expression for�1 is further specified by removing thex-integration∫ ∞

0e−λn x G(x; ξ)

∣∣∣r=R

dx = 1 − e−λnξ

λ2n

−∞∑

m=1

ψm(R, θ)ψm(ρ, η)e−λnξ − e−λmξ

λ2n − λ2

m, (57)

∫ ∞

0x e−λn x G(x; ξ )

∣∣∣r=R

dx = 2 − (2 + λnξ) e−λnξ

λ3n

−∞∑

m=1

ψm(R, θ)ψm(ρ, η)2λn(e−λnξ − e−λmξ )+ ξ(λ2

n − λ2m) e−λnξ

(λ2n − λ2

m)2

. (58)

If m = n, the limitλm → λn should be taken. Now we are able to recognize the nature of the non-uniformconvergence better. The dominating term is (we ignore for the moment theθ -integration)

�1(ξ ) ∼∞∑

m=1

ψm(R, θ)ψm(ρ, η)

λ2m

.

For a circular duct this may be compared, nearρ = R, to the proto-type series

∼∞∑

m=1

cos(2πmρ/R)

m2 .

The normal derivative yields the well-known saw-tooth function that vanishes (pointwise) atρ = R, butconverges to a finite non-zero value for anyρ = R.

For x → ∞, the exponential terms in�1(x) vanish and we have (we exchange the variablesx andξ )

�1(x) � K1x +∞∑

n=1

Fn

∫ 2π

0

[RRXλ

−1n ψn

∣∣ρ=R + 2

λn

d

dη

( RX

Rψn,η

∣∣ρ=R

)]dη

By using the periodicity ofψn in its circumferential argumentη, we have finally

�1(x) � K1x +∞∑

n=1

Fn

λn

∫ 2π

0RRXψn

∣∣ρ=Rdη for x → ∞. (59)

4.3 Matching.

Both the initial conditions forφ0 andφ1 and the constantK1 are determined from matching with the outersolution. From equations (41) and (59) we have

φ0(0)+ Xφ0,X (0)+ εφ1(0) ∼ F0 + εK1x + ε

∞∑n=1

Fn

λn

∫ 2π

0RRXψn

∣∣ρ=Rdη (60)

10 08:40, 02-08-2004

and so we findφ0(0) = F0

K1 = φ0,X (0)

φ1(0) =∞∑

n=1

Fn

λn

∫ 2π

0RRXψn

∣∣ρ=Rdη

(61)

This determines the outer solutionφ0 + εφ1 (together with the radiation condition). It wouldn’t be toodifficult to guess thatφ0 depends on the average source excitationF0, but the initial value forφ1 is reallysubtle. The constant term in (59) is therefore probably the most important result of this tour de force todetermine�1.

An interesting question is then whenφ1 is present at all in the outer solution. (Or put in another way:what is the error if we only considerφ0). For example,φ1 is zero when the source consists of a simplepiston with justF(r, θ) = F0, or when the duct entrance starts smoothly withRX = 0, or whenRRXψn

for all n > 0 are periodic along the circumference.Although this last condition is not very likely to be possible, for a cylindrical duct at least the non-

symmetric modes vanish. In this case the eigenfunctions are given by equation (38). The integrals in (59)vanish for allν = 0. As a result we have

φ1(0) = 2√π RRX

∞∑µ=2

F0µ

j ′0µ. (62)

In other words, the first, constant mode determinesφ0, while only the non-constant, symmetric modesdetermineφ1. For example, a piston tilting along a diagonal likeF ∼ r sinθ would produce a fieldvanishing toO(ε2), while a “piston” that is symmetrically folded likeF ∼ r2 would produce bothO(1)andO(ε) terms.

4.4 Other coordinate systems

It was shown by Agullo et al. [26] that if the shape of the hard-walled duct is described in an orthogonalcoordinate system(u, v,w) by the surfaceS(v,w) = 0, while the Helmholtz equation allows separablesolutions of the formφ(u, v,w) = F(u)G(v,w), then there exist unidimensional (i.e. self-similar) wavesin u of the typeφ(u, v,w) = F(u). In this way it is possible to produce exact solutions of certain hornshapes, like the straight and exponential cone and others.

Although these solutions are interesting on their own, they have little to do with the present lowkasymptotic problem, where the duct wall is never outside the lateral near field of the wave. Without this,there is no built-in mechanism that enforces the self-similarity, so any defect of symmetry in source orsurface will produce deviations in the wave field that propagate without attenuation in other directions.Also the generalisations that will be discussed below are not possible at all or only in very limited form.

On the other hand, if the duct shape considered is close to one that allows such an exact solution, it maybe advantageous, in terms of practical accuracy of the final result, to reformulate the problem in the otherset of coordinates. The essence of the asymptotic problem remains the same.

We will illustrate this for spherical coordinates(r, θ, ϕ), where we temporarily redefinedx = r cosϕ,y = r sinϕ cosθ, z = r sinϕ sinθ . (Note that we will use these coordinatesonly in this section.) A circularcone around the positivex-axis is given byϕ = constant, and a general cone of constant cross section byϕ = f (θ).

In order to maintain the slender shape, necessary for the asymptotics, the duct will be long inr , com-pensated by a small opening angle inϕ. We therefore introduce the scaled variables

τ = 2 sin12ϕ

ε, R = εr (63)

and write the general duct geometry as

S(R, τ, θ) = τ − T (R, θ) = 0 (64)

11 08:40, 02-08-2004

whereT is by assumption independent ofε. By this choice the surface areaA(R) of any spherical crosssectionR = constant is now exactly (i.e. independent ofε) equal to

A(R) =∫ 2π

0

∫ ϕ(R,θ)

0r2 sinϕ dϕdθ =

∫ 2π

0

∫ T

0r2ε2τ dτdθ

= 12 R2

∫ 2π

0T 2(R, θ) dθ. (65)

Other choices of describing the duct shape are not essentially different, other thanT , and thereforeA,becoming dependent onε. This gives complications in the form of extra asymptotic terms in the higherorders, which are irrelevant now.

The Helmholtz equation is given by

ε2

R2

∂

∂R

(R2 ∂φ

∂R

)+ 1

R2τ

∂

∂τ

(τ(1 − 1

4ε2τ2)∂φ

∂τ

)+ 1

R2τ2(1 − 14ε

2τ2)

∂2φ

∂θ2 + ε2κφ = 0, (66)

while the hard-wall boundary condition becomes

∇φ·∇ S = 1 − 14ε

2T 2

R2

∂φ

∂τ− ε2∂T

∂R

∂φ

∂R− 1

R2T 2(1 − 14ε

2T 2)

∂T

∂θ

∂φ

∂θ= 0. (67)

We expand, like before,

φ(R, τ, θ; ε) = φ0(R, τ, θ) + ε2φ2(R, τ, θ) + . . .

(skipping for now theO(ε)-term) to obtain to leading order

φ0,ττ + 1

τφ0,τ + 1

τ2φ0,θθ = 0, with φ0,τ − TθT 2φ0,θ = 0 at τ = T . (68)

If τ andθ are read as polar coordinates this problem is qua form the same as (17), so we have the solutionφ0 = φ0(R) to be determined at the next order. We have

φ2,ττ + 1

τφ2,τ + 1

τ2φ2,θθ + (

R2φ0,R)

R + R2κ2φ0 = 0, with φ2,τ − TθT 2φ2,θ = R2TRφ0,R at τ = T .

This can be written as

∇2φ2 + (R2φ0,R

)R + R2κ2φ0 = 0, with ∇φ2· n = R2φ0,R

T TR√T 2 + T 2

θ

, (69)

where∇ andn denote gradient and normal in(τ, θ)-plane. As a result we have virtually the same equationas (19), and after integration along a spherical surfaceA(R) in (τ, θ) and using (65) we obtain

− A

R2

(R2φ0,R

)R

− Aκ2φ0 = 12 R2φ0,R

d

dR

∫ 2π

0T 2(R, θ) dθ = R2φ0,R

( A

R2

)R

orA−1(Aφ0,R

)R + κ2φ0 = 0. (70)

We see that changing from the axial coordinateX to R and from the transverse cross sectionA to thespherical cross sectionA leaves the final equation forφ0 unchanged. Indeed, to the order considered,XandR, andA and A are the same.

12 08:40, 02-08-2004

5 Curved ducts

The present results remain valid for the slightly more general problem of curved ducts (like certain musicalinstruments) if the curvature of the duct axis (and its derivative) isO(ε). Together with the assumed slowvariation in the axial coordinate, the associated orthogonal coordinate system (based on the tangent and –possibly– the normal and binormal of the curve that describes the duct axis) leave the Laplacian unchangedup toO(ε3).

A simple example is the inside of a perturbed torus, described by a fixed torus radiusε−1 and slowlyvarying tube radiusR. With local (polar-type) coordinatesξ, r, ϕ, we define

x = ε−1(1 + εr cosθ) cos(εξ), y = ε−1(1 + εr cosθ) sin(εξ), z = r sinθ, (71)

where 0≤ r ≤ R(εξ, θ), 0 ≤ θ < 2π , 0 ≤ εξ < 2π . If we write X = εξ , we get (cf. equation 6)

∇2φ + ε2κ2φ

= ∇2⊥φ + ε2(1 + εr cosθ)−2 ∂2

∂X2φ + ε(1 + εr cosθ

)−1[cosθ ∂∂r φ − 1

r∂∂θφ] + ε2κ2φ = 0. (72)

Boundary conditions atS = r − R(X, θ) = 0 are

∇⊥φ·∇⊥S − ε2RXφX

(1 + εr cosθ)2= 0. (73)

If we expandφ = φ0 + εφ1 + ε2φ2 + . . . , we get to leading order

∇2⊥φ0 = 0, with ∇⊥φ0·n⊥ = 0, (74)

soφ0 = φ0(X). Then ∂∂r φ0 = ∂

∂θφ0 = 0 and we have also

∇2⊥φ1 = 0, with ∇⊥φ1·n⊥ = 0, (75)

leading toφ1 = φ1(X). So again∂∂r φ1 = ∂

∂θφ1 = 0 and we obtain again

∇2⊥φ2 + φ0,X X + κ2φ0 = 0, with ∇⊥φ2·∇⊥S = φ0,X RX ,

yielding thus, after a similar argument as before, Webster’s Horn equation.

6 Impedance walls

If the duct walls is equipped with an impedance-type acoustic lining, we will in general (at least ifRe(Z) > 0) expect solutions, which will decay exponentially in axial direction. Therefore, in the com-pressed variableX , only trivial (i.e. zero) solutions will exist. We will see that this is by and large the case,not only for dissipative walls with Re(Z) > 0, but for anyZ < ∞. Only for a purely imaginary impedancein a straight duct there are exceptions.

The impedance-wall boundary condition atr = R is given by

∇φ·n = − iεκ

Zφ = ζφ (76)

with specific impedanceZ . As before, we assume the Poincaré expansionφ = φ0 + εφ1 + ε2φ2 + . . . .First we note that it is easily verified, that ifZ = 0 only the trivial solutionsφ0 = φ1 = 0 occur. Then weconsider two possibilities:Z = O(1) andZ = O(ε).

13 08:40, 02-08-2004

6.1 Z = O(1)

As ζ = O(ε), we write ζ = εζ1. In this case we have only trivial solutions. Expand equations andboundary conditions as before, to get to leading order

∇2⊥φ0 = 0, with ∇⊥φ0·n⊥ = 0 (77)

with solutionφ0 = φ0(X), a function to be determined. To first order we have

∇2⊥φ1 = 0, with ∇⊥φ1·n⊥ = ζ1φ0. (78)

Since ∫∫A

∇2⊥φ1 dσ = ζ1φ0

∫∂A

d� = 0 (79)

we must haveφ0 = 0, and soφ1 = φ1(X). Nothing changes when we continue, and so all terms of theexpansion vanish. Note that this is true for anyZ .

6.2 Z = O(ε)

Now we haveζ = O(1), which changes the boundary condition expansion. To leading order we have

∇2⊥φ0 = 0 in A, with ∇φ0·n⊥ = ζφ0 at ∂A. (80)

We would be tempted to assume that this problem has a solution or solutions for any givenZ , but his isnot true. Non-trivial solutions exist only for certainζ . From Green’s 2nd identity, applied toφ0 and itscomplex conjugate, it can be deduced that any possibleζ is real. Furthermore, from Green’s 1st identityapplied toφ0 it follows that any possibleζ is positive, andZ is thus negative imaginary.

But even withζ real positive, there are only certain discrete values that allow a solution. This is bestseen as follows. The problem described in (80) is an eigenvalue problem for the Dirichlet-to-Neumannoperator�: f �→ g, that maps a given Dirichlet boundary valuef to the normal derivativeg of f ’sharmonic extension intoA ([17]). In other words,�( f ) = ∂

∂nψ∣∣∂A

whereψ is the solution of

∇2ψ = 0 in A, with ψ = f at ∂A. (81)

As we are looking for�(φ0) = ζφ0, equation (80) corresponds to the eigenvalue problem of�. Forthe present discussion it is most relevant to know that this spectrum of eigenvalues of� is discrete. Asthis result appears to be not available in the literature, it is considered concisely, but in great depth, in theAppendix.

An example that illustrates this behaviour explicitly is the circular ductr = R(X), where

φ0 = f (X)( r

R(X)

)m{

cosmθ

sinmθ

}, with ζ = m

R(X)(82)

andm is a non-negative integer. As the shape of the cross sectionA(X) changes withX , the discretenessof the spectrum of� implies that the values ofζ that allow a solution also change withX , and in generalthere are no (non-zero) solutions possible along a varying duct for a fixed, givenζ .

This is of course not true for a duct of constant cross section,r = R(θ), although now the asymptoticsfor smallε loses its meaning because there is no axial length scale for the acoustic wave to be comparedwith. The problem simplifies further for the circular ductr = R where (without approximation)

φ(x, r, θ) = Jm(αr) e−imθ−iγ x , α2 + γ 2 = k2 (83)

and the boundary condition requires

αR J ′m (αR)

Jm(αR)= m − αR Jm+1(αR)

Jm(αR)= ζ R. (84)

14 08:40, 02-08-2004

This equation has infinitely many solutions, but the wave is guaranteed unattenuated (γ real) if α is imagi-nary, sayα = iτ . Such solutions exist for realζ ≥ m/R, because

ζ R = m − iτ R Jm+1(iτ R)

Jm(iτ R)= m + τ RIm+1(τ R)

Im(τ R)≥ m (85)

(see [27]). Note that for smallk, γ , α solutions we recover (82):

ζ = m

R− α2R

2m + 2+ O(α4). (86)

In other words, only solutions of this type exist near special values ofζ .

7 Variable mean soundspeed and density

If soundspeedC = C(X, r, θ) and mean densityD = D(X, r, θ) are not uniformly constant, but vary inr ,θ and slowly inx , we have the reduced wave equation (5), rewritten in slowly varying coordinates,

ε2 ∂∂X

(C2 pX

) + ∇⊥·(C2∇⊥ p) + ε22 p = 0, (87)

where the dimensionless frequencyω = ε is small. The hard-wall boundary condition is the same asequation (14). When we expandp = p0 + εp1 + ε2 p2 . . . , we get to leading order

∇⊥·(C2∇⊥ p0) = 0, with ∇⊥ p0·n⊥ = 0, (88)

which has a constant as the solution, sop0 = p0(X), a function to be determined. We can derive the sameequation forp1, to get the same resultφ1 = φ1(X). For the second order we have

∇⊥·(C2∇⊥ p2) + ∂

∂X

(C2 p0,X

) +2 p0 = 0, with ∇⊥ p2·n⊥ = p0,XRRX√

R2 + R2θ

. (89)

We go on to find a solvability condition forp2 by integrating this equation along a cross sectionA. Utilizingthe following identity for any differentiable functionf

d

dX

∫∫A

f (X) dσ = d

dX

∫ 2π

0

∫ R

0f (X, r, θ)r drθ =

∫ 2π

0

∫ R

0fX r drdθ +

∫ 2π

0f (X, R, θ)RRX dθ, (90)

we have∫∫A∇⊥·(C2∇⊥ p2) dσ = p0,X

∫ 2π

0C2RRX dθ = p0,X

[d

dX

∫∫A

C2 dσ −∫∫

A

∂∂X C2 dσ

]. (91a)

Furthermore, we have∫∫A

∂∂X

(C2 p0,X

)dσ = p0,X

∫∫A

∂∂X C2 dσ + p0,X X

∫∫A

C2 dσ, and∫∫

A2 p0 dσ = 2 p0A. (91b)

Then, after introducing the cross-sectional averaged squared sound speed

C2 = 1

A

∫∫A

C2 dσ, (92)

a generalisation of Webster’s Horn equation is obtained

A−1(AC2 p0,X)

X +2 p0 = 0. (93)

This may be further simplified by the transformation

A(X)C2(X) = d(X)2, p0 = d−1ψ (94)

into

ψ ′′ +(2

C2− d ′′

d

)ψ = 0. (95)

15 08:40, 02-08-2004

8 Irrotational and isentropic mean flow

To analyse asymptotically low frequency acoustic perturbations in a slowly varying duct with an irrotationalisentropic mean flow, as described by equations (7) and (9), we need to approximate both mean flow andacoustic field to the same order of accuracy.

We start here with the mean flow. In the dimensionless variables used, we haveC2 = Dγ−1, soequations (7) simplify to

12V 2 + Dγ−1

γ − 1= E, ∇·(DV ) = 0. (96)

The mass flux at any cross sectionA is given by∫∫A

DU dσ = F . (97)

Due to the non-dimensionalisation,U , D, A, F andE areO(1). Introduce the slow variableX = εx , andassumeV andD to depend essentially onX , rather thanx . We write the velocity as

V = U ex + V⊥ (98)

to distinguish between axial and cross-wise components. If fluxF and thermodynamical constantE aregiven and independent ofε, we can expandU = U0 + O(ε2) and D = D0 + O(ε2). As the flow is apotential flow, we can derive, in the same way as in Rienstra [14, 16], thatD0 = D0(X), U0 = U0(X) andV⊥ = εV⊥0 + O(ε3), satisfying the equations (to be solved numerically)

D0U0A = F ,F 2

2D20 A2

+ Dγ−10

γ − 1= E . (99)

8.1 Mean flow and hard walls.

Next we consider the acoustic field. Using the above results for the mean flow, equation (9) becomes toleading orders

∇2⊥φ + ε2D−10

(D0φX

)X = ε2

(i+ U0

∂∂X + V⊥0·∇⊥

)[C−2

0

(i+ U0

∂∂X + V⊥0·∇⊥

)φ]

with hard wall boundary condition∇φ·n = 0 at r = R.

We expandφ = φ0 + εφ1 + ε2φ2 + . . . . To leading order we have

∇2⊥φ0 = 0, ∇⊥φ0·n⊥ = 0 (100)

yielding the constant solution,i.e. φ0 = φ0(X).To first order we have the same equation. To second order we have

∇2⊥φ2 + D−10

(D0φ0,X

)X =

(i+ U0

∂∂X + V⊥0·∇⊥

)[C−2

0

(i+ U0

∂∂X + V⊥0·∇⊥

)φ0

]with boundary conditions given by equation (19). After integration across a cross sectionA(X), we obtain,similar to before, Webster’s Horn equation generalised for irrotational isentropic mean flow

(D0A)−1(D0Aφ0,X )X = (i+ U0

∂∂X

)[C−2

0

(i+ U0

∂∂X

)φ0

]. (101)

This result seems to be equivalent to equations given by [8, 9, 10, 11, 12] and (apart from a factor12) [3,

p. 422].

16 08:40, 02-08-2004

8.2 Mean flow and impedance walls.

The problem with mean flow and an impedance wall is more intricate. Instead of the duct wall boundarycondition given in equation (76), we have Myers’ condition [28], rewritten (see [29, 30]) as follows

iωD(v·n

) = iωDp

Z+ M

( DV p

Z

), (102)

whereZ = Z(X, θ) may be function of position, and operatorM is defined by

M(F) = ∇·F − n·(n·∇ F). (103)

SinceM(DV p

Z ) = O(ε), we writeM(DV p

Z ) = εM(DV p

Z ). After expandingφ = φ0 + εφ1 + . . . , andp = εp0 + . . . with

p0 = −D0(i+ U0∂

∂X+ V⊥0·∇⊥)φ0, (104)

we get

iD0(∇⊥φ0·n⊥)+ iεD0

(∇⊥φ1·n⊥) = εiD0 p0

Z+ εM

( D0V 0 p0

Z

)+ O(ε2), (105)

whereV 0 = U0ex + εV⊥0.

8.2.1 Z = O(1).

As before, we get leading order

∇2⊥φ0 = 0, with ∇⊥φ0·n⊥ = 0,

soφ0 = φ0(X) and thereforep0 = p0(X). To first order we have the same equation∇2⊥φ1 = 0 for φ1, butthe boundary condition is now

iD0(∇⊥φ1·n⊥) = iD0 p0

Z+ M

( D0V 0 p0

Z

). (106)

In order to continue, we need from [16] the following property of the operatorM.

For any sufficiently smooth vectorfield with f ·n = 0 at r = R, we have∫∂A

[∇· f − n·(n·∇ f

)]∥∥∥ ∂ r∂x

×∂ r∂�

∥∥∥d� = d

dx

∫∂A

(f ×n

)·d� .

where (x, �) �→ r(x, �) is a parameterisation of the surface.

Since ∥∥∥ ∂ r∂x

×∂ r∂�

∥∥∥ =√

1 + ε2R2R2

X

R2 + R2θ

= 1 + O(ε2),

we have as a result ∫∂A

M( D0V 0 p0

Z

)d� = d

dX

∫∂A

D0U0 p0

Zd�+ O(ε).

We apply this to the equation forφ1, in order to obtain an equation forφ0. From∫∫A

∇2⊥φ1 dσ =∫∂A

∇⊥φ1·n⊥ d� = 0,

together with equation (106) and noting that most functions depend onX only, it follows that

iD0 p0L + d

dX

(U0D0 p0L

) = 0, where L(X) =∫∂A

1

Zd�,

17 08:40, 02-08-2004

(L may be interpreted as the “total admittance” atX) with solution

p0 = constant1

U0D0Lexp

(−i

∫ X

U0(ξ)dξ

).

φ0 follows from equation (104) but is more difficult to obtain in explicit form. Note that this pressure fieldis not an acoustic wave, but it is of hydrodynamic nature. It does not propagate with the sound speed, butwith the mean flow velocity.

8.2.2 Z = O(ε).

WhenZ = εZ0 we get forφ0 the apparently difficult boundary condition

iD0(∇⊥φ0·n⊥) = iD0 p0

Z0+ M

( D0V 0 p0

Z0

).

which is, analogous to the no-flow case, likely to be an eigenvalue problem with discrete eigenvaluesZ0

(apart from trivial solutionsp0 = 0, φ0 = φ0(X) ∝ exp(−i∫ X U0(ξ)

−1dξ , i.e. hydrodynamicallyconvected pressureless perturbations). If this conjecture is true, the possible eigenvalues vary with thegeometry, and no other than the trivial solution is possible in a varying duct.

9 Conclusions

Generalisations of Webster’s classic horn equation for non-uniform media, lined walls and mean flow havebeen derived systematically, as an asymptotic perturbation problem for low Helmholtz number and slowlyvarying duct diameter. The conditions on frequency, acoustic medium and duct geometry are explicitlyindicated in terms of small parameterε, the ratio between a typical length of duct variation and the ductdiameter. The error and higher order corrections are also explicitly stated.

The presence of lining in a varying duct is shown to allow in general only trivial or merely hydrody-namic solutions. A curved duct is shown to produce the same equation if the radius of curvature is notsmaller than the typical wave length or duct length scale.

The approximation is non-uniform near source or entrance. The prevailing boundary layer solution foran arbitrary duct cross section is given, together with theO(1) andO(ε) matching conditions to the outer(“Webster”) region. From these expressions conditions are derived for which theO(ε)-outer field is absent.

10 Acknowledgement

This work was carried out in the context of the “TurboNoiseCFD” project of the European Union’s 5thFramework “Growth” Programme.

We gratefully acknowledge the contribution on the discreteness of the spectrum of� (see below) byJ. de Graaf and the cooperation with T.D. Chandra on theanalysis of the entrance boundary layer. We thankN.C. Ovenden for helpful discussions and critical reading of the text.

Appendix. On the spectrum of the Dirichlet-to-Neumann operator� on smooth bounded domains inR2.

by Jan de Graaf, TUE.

We will show that the Dirichlet-to-Neumann operator�, introduced in section 6, equation (80), has adiscrete spectrum of finite multiplicity. The basic idea is to relate the problem for the general, simplyconnected, open domain ⊂ R

2 (which has apparently no explicit solution), via conformal mapping, tothe corresponding problem for the unit diskD, which does have a simple, explicit, solution.

Note that the related result for an annular shaped domain is entirely analogous.

18 08:40, 02-08-2004

Step 1. Consider the open unit-diskD ⊂ R2. Its boundary∂D, the unit circle, is parametrized by the

angleθ , with 0 ≤ θ < 2π . The set of functions

en : θ �→ en(θ) = 1√2π

einθ , n ∈ Z, (A.1)

establishes an orthonormal basis inL2(∂D; dθ)‡. We introduce for reala the linear operatorNa inL2(∂D,dθ), defined via the way it acts on the basis{en}

Na : en �→ Naen, with Naen(θ) = (|n| + a)en(θ), (A.2)

followed by linear extension and closure.Let u : ∂D → C be a sufficiently smooth function. LetuH , the harmonic extension ofu, denote the

(unique) solution of the Dirichlet problem

∇2uH(x) = 0 for x ∈ D, while uH(x) = u for x ∈ ∂D. (A.3)

The normal derivative at the boundary∂D produces a function

∂∂n uH : ∂D → C. (A.4)

Altogether this defines the linear mappingu �→ ∂∂n uH , wich is called theDirichlet-to-Neumann operator

in L2(∂D; dθ). By noting that enH (x) = (x ± i y)|n| = r |n| e±i|n|θ , and hence∂∂n enH = |n|en at ∂D, it is

easily verified that this operator is just equal toN0.

Step 2. Consider the bounded open domain ⊂ R2 with piecewise smooth boundary∂. Letv : ∂ →

C be a sufficiently smooth function. As in the previous section (just replaceD by), we introduce

� : v �→ �v = ∂

∂nvH , (A.5)

the ‘Dirichlet-to-Neumann’ operator inL2(∂; dθ). So� = N0 if = D. We want to show that�is non-negative, self-adjoint withpure point spectrum of finite multiplicity. In the previous paragraph weshowed this to be true inL2(∂D; dθ).

The self-adjointness and non-negativity follows, formally, from Green’s first and second identity (seesection 6). In order to achieve some spectral results, we invoke the Riemann mapping theorem and considera conformal mappingF : D → . The supposed smoothness of∂ implies that the parametrizationθ �→ F(eiθ ) for ∂ is such, that both|F ′(eiθ )| and its reciprocal are bounded.

Standard results from conformal mapping theory and harmonic functions onR2 lead to

�v(F(eiθ )

) = ( ∂∂nvH

)(F(eiθ )

) = ∣∣F ′(eiθ )∣∣−1 ∂

∂n(v◦ F)H(eiθ ). (A.6)

This means that, instead of the original problem, we could study he eigenvalue problem

MN0u = λu (A.7)

in L2(∂D,dθ), with M the multiplication operator defined by

(Mw)(θ) = M(θ)w(θ) = ∣∣F ′(eiθ )∣∣−1w(θ). (A.8)

(Although the inverseM−1 involves no more than just division by the functionM(θ), we retain for claritythe operator symbolism.)

‡L2(U;w(x)dx) denotes the space of square integrable functions, defined onU , with inner product∫

U f (x)g(x)w(x)dx.

19 08:40, 02-08-2004

Step 3. In order to turn the operatorMN0 into a self-adjoint one, we consider the eigenvalue problem inL2(∂D; M−1(θ)dθ), which is topologically equivalent withL2(∂D; dθ). Note that{

θ �→ u(θ)} �→ {

θ �→ M− 12 (θ)u(θ)

}(A.9)

furnishes a unitary transformation fromL2(∂D; M−1(θ)dθ) to L2(∂D; dθ), because∫ 2π

0u(θ)v(θ)M−1(θ) dθ =

∫ 2π

0

(M− 1

2 (θ)u(θ))(

M− 12 (θ)v(θ)

)dθ. (A.10)

At the same time this implies that the eigenvalue problem (A.7) is unitary equivalent to the eigenvalueproblem

M12 N0M

12ϕ = λϕ, (A.11)

with ϕ = M− 12 u.

Step 4. If we can show that(I + M

12 N0M

12)−1 (whereI is the identity) is a compact self-adjoint

operator, we are ready. In that case it has a discrete spectrum with finite multiplicity [31] and the same

holds,a fortiori, for M12 N0M

12 .

Takea positive and sufficiently small, such thatθ �→ M−1(θ) − a is still positive and uniformlybounded away from zero. By noting thatNa = N0 + aI, we can rewrite

I + M12 N0M

12 = M

12 N

12

a{N

− 12

a (M−1 − aI)N− 1

2a + I

}N

12

a M12 . (A.12)

The operator between brackets,{ }, is bounded, positive and self-adjoint and has an inverse with the sameproperties. We thus find

(I + M

12 N0M

12)−1 = M− 1

2 N− 1

2a

{N

− 12

a (M−1 − aI)N− 1

2a + I

}−1N

− 12

a M− 12 , (A.13)

which is a composition of operators. Since the factorN− 1

2a is compact, also(I + M

12 N0M

12 )−1 is a

compact operator. �

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