Scattering by a Perforated Sandwich Panel: Method of Riemann Surfaces

Summary

The model problem of scattering of a sound wave by an infinite plane structure formed by a semi-infinite acoustically hard screen and a semi-infinite sandwich panel perforated from one side and covered by a membrane from the other is exactly solved. The model is governed by two Helmholtz equations for the velocity potentials in the upper and lower half-planes coupled by the Leppington effective boundary condition and the equation of vibration of a membrane in a fluid. Two methods of solution are proposed and discussed. Both methods reduce the problem to an order-2 vector Riemann–Hilbert problem. The matrix coefficients have different entries, have the Chebotarev–Khrapkov structure and share the same order-4 characteristic polynomial. Exact Wiener–Hopf matrix factorization requires solving a scalar Riemann–Hilbert on an elliptic surface and the associated genus-1 Jacobi inversion problem solved in terms of the associated Riemann θ-function. Numerical results for the absolute value of the total velocity potentials are reported and discussed.

1 Introduction

The effect of perforation on the transmission of sound waves through single-leaf and double-leaf panels was analyzed in (1,2). When an elastic double-leaf honeycomb panel is perforated from one or both sides the transmission of sound is significantly reduced (3). Leppington (3) applied the method of matched asymptotic expansions to analyze this effect for an infinite honeycomb cellular structure in the cases of acoustically hard or acoustically transparent cell walls. One of the main results of this work is the derivation of the effective boundary conditions. In particular, in the case of cells with acoustically hard walls, the effective boundary condition on the panel surface $S={x∈I,y=0}$ has the form

where ψ₀ and ψ₁ are the velocity potentials in the lower and upper half-planes, ψ_jy is the normal derivative of ψ_j, k is the wave number and τ is a parameter that accounts for perforations. The condition (1.1) is to be complemented by the classical boundary condition

where D_x is the differential operator with respect to x of order 2 or 4 depending on whether the lower unperforated skin of the elastic structure is a membrane or an elastic plate and α is a parameter. The model problem of scattering of a plane sound wave by an infinite plane structure formed by a semi-infinite acoustically hard screen and semi-infinite honeycomb elastic panel with acoustically hard walls perforated from one side was reduced (4) to a Riemann–Hilbert problem for two pairs of functions. This work does not factorize the matrix coefficient. Instead, it applies the asymptotic method of small τ with the leading-order term determined by the decoupled problem.

The vector Riemann–Hilbert (4) was analyzed (5) by the method of factorization on a Riemann surface (6,7). It was shown that the matrix coefficient of the vector Riemann–Hilbert problem has the Chebotarev–Khrapkov structure (8,9) with the characteristic polynomial f(s) of degree 8. The Khrapkov methodology is applicable when $degf(s)≤2$⁠. For a particular case of the Chebotarev–Khrapkov matrix and when $degf(s)=4$⁠, a method of elliptic functions eliminating the essential singularity at infinity was proposed in (10). A numerical approach of Padé approximants for factorization of the Chebotarev–Khrapkov matrix was developed in (11). If $degf(s)=8$⁠, then the exact representation of the Wiener–Hopf matrix factors includes exponents of functions having an order-3 pole at infinity and therefore has an unacceptable essential singularity. This singularity was eliminated (5) by reducing the matrix factorization problem to a scalar Riemann–Hilbert problem on a hyperelliptic surface (12) and solving the associated genus-3 Jacobi inversion problem (13,14). The solution (5) was designed for the case of real wave numbers. In this case, all eight branch points lie on the same circle and are symmetric with respect to the origin. That is why the brunch cuts and the A- and B-cross-sections (5, 14) of the Riemann surface are two-sided arcs of a circle. The full solution of the model problem requires to determine two unknown constants by satisfying the additional conditions, to analyze the behavior of the Wiener–Hopf matrix factors at infinity, and based on the solution obtained develop an efficient numerical procedure for the velocity potentials. These aspects were not a scope of the investigation (5).

In this work, we analyze the model problem of scattering of a plane sound wave by a structure similar to the one considered in (4,5). The only one difference is that the lower skin of the structure is a membrane, not a thin plate. Mathematically, this means that instead of the fourth-order differential operator D_x in (1.2) we have a second-order operator. We have succeeded in deriving the full solution, finding unknown constants, analyzing the behavior of the matrix factors at infinity and obtaining numerical results. In Section 2, we formulate the model problem and write down the governing boundary value problem for two Helmholtz equations coupled by the boundary conditions. We apply the Laplace transform and convert the problem into an order-2 vector Riemann–Hilbert problem in Section 3. We show that the problem of matrix factorization is equivalent to a scalar Riemann–Hilbert problem on an elliptic surface. To eliminate the essential singularity of the factors at infinity, we solve a genus-1 Jacobi inversion problem in terms of the Riemann θ-function. At the end of Section 3, we compute the partial indices of factorization and show that they are stable. In Section 4, we propose another method of reduction of the model problem to a vector Riemann–Hilbert problem in order to understand if the method has an advantage over the method applied in Section 3. The Riemann–Hilbert problem is different from the one derived by the first method. Both of the problems require solving a scalar Riemann–Hilbert problem stated on the same Riemann surface but having distinct coefficients. Our derivations show that the analysis of the Wiener–Hopf matrix-factors at infinity is more complicated in the case of the second vector Riemann–Hilbert problem. This is due to the logarithmic growth at infinity of the densities of the singular integrals in the representations formulas for the solution. Section 5 presents numerical results for the velocity potentials obtained on the basis of the solution derived in Section 3.

2 Formulation

Consider the structure formed by attaching a semi-infinite sandwich panel {$0≤x<∞,0≤y≤d,−∞<z<∞}$ to a semi-infinite acoustically rigid screen {$−∞<x≤0,0≤y≤d*,−∞<z<∞}$ (Fig. 1). The upper side of the panel $M1={{0<x<∞,y=d,−∞<z<∞}$ is periodically perforated, while the lower side $M0={0<x<∞,y=0,−∞<z<∞}$ is a smooth membrane. The unperforated and perforated skins are linked by cells whose sides are assumed to be acoustically hard. The rigid screen $S={−∞<x≤0,0≤y≤d*,−∞<z<∞}$ and the lower side of the sandwich panel are clamped, and without loss of generality, the displacement is equal to zero at the junction line x = 0, $y=0,−∞<z<∞$⁠.

Compressible fluid of wave speed c occupies the regions outside the sandwich panel and the screen. The system is excited by a plane wave of incident velocity potential

where $θ0∈(−π/2,π/2), k=ωc$ is the acoustic wave number, $k=k1+ik2, 0<k2≪k1$ and ω is the radian frequency. The geometry of the structure and the incident wave allows for reduction of the three-dimensional problem to a two-dimensional diffraction model (Fig. 2) (3,4). The velocity potentials $ψ0,1e−iωt$ in the lower and upper half-planes $H0$ and $H1$ satisfy the Helmholtz equation

The two potentials satisfy the standard acoustically hard boundary conditions

The surface deflection $η0e−iωt$ of the membrane M₀ and the pressure fluctuation $pje−iωt$ are given in terms of the potentials by the relations

where ρ_f is the mean density of the fluid. The deflection of the membrane responds to the total surface pressure according to the linearized equation (15)

Here, m₀ is the mass per unit area of the membrane and T is the tension per unit length. Equivalently, this boundary condition can be written in the form

where

The second boundary condition of the sandwich panel is the Leppington (3) effective boundary condition

where

where a is the aperture radius and V is the cell volume, $V=d1d2d$⁠. In the derivation (3) of the boundary condition (2.8), it is assumed that the wavelength is large compared with the spacing parameters d₁, d₂ and d (Fig. 1), so that $|k|d1≪1, |k|d2≪1$ and $|k|d≪1$⁠. Since $|k|d$ is small, the boundary condition can be applied at y = 0 to leading order. The parameter τ is dimensionless, while the parameters μ and α have dimensions $L−1$ and $L−3$⁠, respectively, with L measured in units of length.

It is convenient to introduce a reflected wave of potential $ϕref=eik(x sin θ0−y cos θ0)$ and scattering potentials $ϕ0$ and $ϕ1$⁠. Then the total velocity potentials are expressed through the incident, reflected and scattering potentials by

The scattering potentials satisfy the Helmholtz equation

and the following boundary conditions:

It is also required that the scattering potentials $ϕ0$ and $ϕ1$ satisfy an outgoing wave radiation condition as $x2+y2→∞$⁠.

3 Vector Riemann–Hilbert problem

3.1 Derivation based on the Laplace transform of the velocity potentials

To reduce the boundary value problem defined by (2.11) and (2.12) to a vector Riemann–Hilbert problem (known also as a matrix Wiener–Hopf problem), we introduce the Laplace transforms of the velocity potentials

Their sums $Φj(s;y)=Φj+(s;y)+Φj−(s;y)$ satisfy the equations

Fix a single branch of the function $γ2(s)=(s2−k2)$ in the s-plane cut along the line joining the branch points k and – k and passing through the infinite point. Denote by $θ*=argk∈(0,π2)$ and select

where $θ±=arg(s±k)$⁠. Then $γ(0)=−ik$ and $Reγ(s)>0$ on the line $L=(−∞+iκ0,∞+iκ0)$ with κ₀ being a real number such that

On disregarding the solution exponentially growing at infinity we write the general solution of equation (3.2)

Applying the Laplace transforms (3.1) to the boundary conditions (2.12) gives four equations. They are

Here, we used the fact that, owing to the continuity of the displacements at the point x = 0,

and denoted $N=d2dxdyϕ0(0+;0)$ (N is a free constant at this stage). On fixing y = 0 in the general solutions defined by (3.5) and their derivatives and also employing the first two equations in the system (3.7) we find

Analysis at infinity of the fourth equation in (3.6) and equation (3.8) shows that

As for the asymptotics at infinity of the other functions in the system (3.6), they ensue from the Laplace representations (3.1) and the asymptotics (3.9). We have

Now, the relations (3.8) enable us to eliminate the derivatives $ddyΦj+(s;0)$ from the third and fourth equations of the system (3.6) and obtain

To rewrite this system in the matrix–vector form, we denote $Φj±(s)=Φj±(s,0)$ and introduce the vectors

and the matrix

where b(s) and c(s) are Hölder functions on the contour L, l(s) is a polynomial and m is a constant given by

In these notations, the system (3.11) can be reformulated as the following vector Riemann–Hilbert problem of the theory of analytic functions.

Find two vectors $Φ+(s)$and $Φ−(s)$analytic in the upper and lower half-planes $C+$and $C−$, respectively, Hölder-continuous up to the contour L, such that their limit values on the boundary satisfy the vector equation

The solution vanishes at infinity, $Φ±(s)=O(s−1), s→∞, s∈C±$, and

Also, due to the condition(3.7),

As the wave number k tends to the critical Helmholtz resonance value $kres$ (3,16)

the parameter τ and the function b(s) become infinite, and there in no way to pass to the limit $τ→∞$ in the vector problem (3.15). On letting $τ→∞$ in the system (3.6) we find $Φ1+(s;0)=0$⁠. This brings us to the following two scalar Riemann–Hilbert problems:

The non-trivial solution of the first problem with the lowest growth of the function $ddyΦ1+(s;0)$ at infinity has the form

where C₀ is an arbitrary constant. The Tauberian theorem for one-sided Fourier transforms yields $ϕ1(x,0)=O(x−1/2), x→0−$⁠. This means that the pressure distribution at the junction point $x=y=0$ has an unacceptable square root singularity. Therefore, for the critical value $k=kres$ the solution of the model problem does not exist. Finally notice that for the complex wave numbers k, the condition (3.18) is never satisfied. However, when $Rek≫Imk>0$ and $|k|→kres$⁠, we have $|τ|→∞$⁠.

3.2 Matrix factorization

The matrix coefficient G(s) of the vector Riemann–Hilbert problem is a Chebotarev–Khrapkov matrix (8,9). Its characteristic polynomial is a degree-4 polynomial

In this case, the problem of factorization reduces to a scalar Riemann–Hilbert problem on a two-sheeted genus-1 Riemann surface $R$ of the algebraic function $w2=f(s)$⁠. Fix a single branch of this function by the condition $f1/2(s)∼s2, s→∞$⁠, in the plane $C^$ cut along the segments $Γ1=[s1,s2]$ and $Γ2=[−s2,−s1]$⁠. Here, $±s1$ and $±s2$ are the four zeros of the function f(s), $s12=μ2+im, s22=μ2−im$⁠, s₁ and s₂ are the zeros lying in the second and first quadrant, respectively (Fig. 3).

Denote the two sheets of the surface $R$ glued along the cuts $Γ1$ and $Γ2$ by $C1$ and $C2$⁠. Let $w=f1/2(s), (s,w)∈C1$ and $w=−f1/2(s), (s,w)∈C2$⁠.

A meromorphic solution of the factorization problem on the contour $L=(−∞+iκ0,∞+iκ0)$

the matrix X(s) and its inverse, has the form (5–7)

where κ₀ is a real number described in (3.4) and

The function F(s, w) solves the following scalar Riemann–Hilbert problem on the contour $L=L1∪L2,$ where $L1⊂C1$ and $L2⊂C2$ are two copies of the contour L.

Find a function F(s, w) piece-wise meromorphic on the surface $R$, Hölder-continuous up to the contour $L$, bounded at the two infinite points of the surface $R$, such that the limit values of the function F(s, w) on the contour $L$satisfy the relation

where $ξ=w(t)$, $λ(t,ξ)=λj(t)$on $Cj$, j = 1, 2, and $λ1(t)=b(t)+c(t)f1/2(t)$and $λ2(t)=b(t)−c(t)f1/2(t)$are the two eigenvalues of the matrix G(t).

It is directly verified that the increments of the arguments of the eigenvalues $λ1(t)$ and $λ2(t)$ when t traverses the contour L from $−∞+iκ0$ to $+∞+iκ0$ are equal to zero. Therefore, the general solution of the problem (3.25) has the form

where

dW is the Weierstrass analog of the Cauchy kernel on an elliptic surface,

and Γ is a contour on the surface $R$ whose starting point q₀ is fixed arbitrarily say, $q0=(ζ0,f1/2(ζ0))∈C1$⁠, while the terminal point $q1=(ζ1,w(ζ1))∈R$ cannot be fixed a priori and has to be determined. In formula (3.27), there are two more undetermined quantities, m_a and m_b. They are integers and have also to be determined. The contours of integration a and b are canonical cross-sections of the surface $R$ (Fig. 2). The cross-section $a=a+∪a−$ is a two-sided loop, $a+∈C1$ and $a−∈C2$⁠. The closed contour b consists of the segment $[s2,−s1]∈C1$ and the segment $[−s1,s2]∈C2$ (the dashed line in Fig. 2). The loop a intersects the loop b at the branch point s₂ from left to right. Note that the contour Γ has to be chosen such that it intersects neither the contour a nor the contour b.

Because of the logarithmic singularities of the second integral in (3.27) at the endpoints of the contour Γ the solution F(s, w) has a simple pole at the point $q0∈C1$ and a simple zero at the point $q1∈R$⁠. The kernel dW has an order-2 pole at the infinite points of the surface. That is why the solution F(s, w) has unacceptable essential singularities at the infinite points. To remove them, we first rewrite $χ(s,w)$ in the form

where

The function $χ(s,w)$ is bounded at infinity if and only if $wχ2(s)=O(1), s→∞$⁠, or, equivalently,

If the function $χ2(s)$ satisfies this condition, then because of the identity

it admits an alternative representation

This formula is valid for all $s≠0$ and can be used for numerical calculations when $|s|>1$ instead of formula (3.30) that is more convenient when $|s|≤1$⁠.

3.3 Jacobi inversion problem

We have shown that the Wiener–Hopf factors do not have the exponential growth at infinity if and only if the terminal point of the contour Γ and the two integers m_a and m_b are selected such that the condition (3.31) is fulfilled. Our next step is to demonstrate that the condition (3.31) is equivalent to a genus-1 Jacobi inversion problem (13,14). To show this, consider the abelian (elliptic) integral

Then the third and fourth integrals in equation (3.31),

are the A- and B-periods of the integral $ω(q)$⁠. Thus, the condition (3.31) constitute the following Jacobi inversion problem.

Find a point $q1=(ζ1,w1)∈R$and two integers m_a and m_b such that

where

By dividing equation (3.36) by $A$ we arrive at the canonical form of the Jacobi problem

for the canonical abelian integral $ω^(q)=ω(q)A$⁠. It has the unit A-period, while its B-period $B^=BA$ has a positive imaginary part, $ImB^>0$⁠. Here, $e1=d0A+k1$ and k₁ is the Riemann constant of the surface $R˜$ cut along the loops a and b. It is computed in (17), $k1=−12+12B^$⁠.

The unknown point $q1=(ζ1,w1)$ is the single zero of the genus-1 Riemann θ-function

To find ζ₁, consider the integral

where v is an arbitrary fixed point not lying on the cuts $Γ1=[s1,s2]$ and $Γ2=[−s1,−s2]$⁠, and $∂R^$ is the boundary of the surface $R$ cut along the cross-section $a$ only. The procedure we are going to apply is a modification of the method (5) that instead of the poles at the two points $v1=(v,f1/2(v))∈C1$ and $v2=(v,−f1/2(v))∈C2$ in (3.40) uses an integrand with two poles at the infinite points of the surface $R$⁠. Without loss we assume that $F(vn)≠0$⁠, n = 1, 2, and compute the integral M by applying the theory of residues. We have

where the derivative of the Riemann θ-function is a rapidly convergent series

The integral (3.40) can also be represented as a contour integral,

where $a+$ and $a−$ have opposite directions and pointwise coincide with the loop $a$ from the side of the sheets $C1$ and $C2$⁠, respectively, while the positive direction of the loops $a+$ and $a$ are chosen such that the exterior of the cut $[s1,s2]$ on the first sheet is on the left. Using the relation between the boundary values of the Riemann θ-function on the loop $a$

where

we derive another formula for the integral M

Combining formulas (3.41) and (3.46) and introducing the quantity

we find the parameter ζ₁

We have verified numerically that the position of the point ζ₁ is independent of the choice of v indeed. Evaluate now the abelian integral at the point ζ₁ lying on the first and second sheets,

and denote

Taking the imaginary and real parts of the complex equation (3.44), we determine the constants $mb±$ and $ma±$

If it turns out that both of the integers $ma+$ and $mb+$ are integers, then $ma=ma+$⁠, $mb=mb+$ and the point $q1∈C1$⁠. Otherwise, $ma=ma−$⁠, $mb=mb−$ are integers, and $q1∈C2$⁠.

3.4 Solution of the vector Riemann–Hilbert problem

On having factorized the matrix $G(t)=X+(t)[X−(t)]−1=[X−(t)]−1X+(t), t∈L$⁠, and eliminated the essential singularity of the factors $X±(s)$ at infinity, we proceed with the solution of the vector Riemann–Hilbert problem (3.15) by rewriting the boundary relation as

Here, $Ψ±(n)(t)$ are the limit values of the Cauchy integrals

defined by the Sokhotski–Plemelj formulas

By the continuity principle and the generalized Liouville’s theorem of the theory of analytic functions,

where R(s) is a rational vector-function to be determined. Since the matrices $X±(s)$ are bounded at infinity, while the vectors $Φ±(s), Ψ±(1)(s)$ and $Ψ±(2)(s)$ behave as $s−1const$ for large s, we have $R(s)=O(s−1), s→∞$⁠.

Shown next that the rational vector-function R(s) has a pole at the point $s=ζ0$⁠. Indeed, due to the logarithmic singularity of the function $χ(s,w)$ at the starting point $(ζ0,f1/2(ζ0))∈C1$ of the contour Γ, the function F(s, w) has a simple pole at this point and is bounded at the point $(ζ0,−f1/2(ζ0))∈C2$⁠,

Employing the first formula in (3.23) for the matrix X(s) we obtain

where

This enables us to find the vector R(s). We have

where C is a free constant. By substituting this expression into equation (3.55) we get the solution of the vector Riemann–Hilbert problem (3.15)

Now, the function F(s, w) has a simple zero at the point $q1∈R$ caused by the logarithmic singularity of the function $χ(s,w)$ at the terminal point of the contour Γ. If $q1=(ζ1,w1)∈C1$⁠, then $w1=f1/2(ζ1)$ and

Otherwise, is $q1=(ζ1,w1)∈C2$⁠, then $w1=−f1/2(ζ1)$ and

Here, D₁ and D₂ are non-zero constants. From the second formula in (3.23) for the inverse matrix $[X(s)]−1$⁠, it becomes evident that the matrix $[X(s)]−1$ has a pole at the point $s=ζ1$⁠. Since $rankB(s,w)=1$⁠, the vector-function F(s, w) has a removable singularity at the point $s=ζ1$ if and only if

where

and $Ψ0(n)(s)$ and $Ψ1(n)(s)$ (n = 1, 2) are the two components of the vectors $Ψ(n)(s)$ given by (3.53). Equation (3.63) constitutes the first equation for the unknown constants C and N.

The solution derived has to be restricted to the class of functions satisfying the conditions (3.16). To verify these conditions, we examine the asymptotics of the vectors $Φ+(t)$ and $Φ−(t)$ as $|t|→∞, t∈L$⁠. We start with the analysis of the functions $Δ(t)$ and $ϵ(t)$⁠. From formulas (3.14) and (3.30), we have

By applying the Sokhotski–Plemelj formulas to the integral representations of the functions $χ1(s)$ and $χ2(s)$ given by (3.30) and (3.33) we deduce

We focus our attention on the principal terms of the asymptotic expansions at infinity of the two functions in (3.66) and observe that

where

and h₁ and h₂ are some constants. Their values do not affect the asymptotics we aim to derive.

Next, by virtue of the relation (3.29) we can derive the asymptotics of the solution of the Riemann–Hilbert problem on the surface $R$ as follows:

where $h±=h1±h2$⁠. Substituting these expressions into formula (3.23) for the inverse matrix $[X(s)]−1$ gives

The above result, together with formulas (3.54) and (3.60), enables us to derive formulas that describe the behavior of the solution $Φ±(t)$ of the vector Riemann–Hilbert problem on the contour L when $|t|→∞$

where C₀ is a constant. From here we immediately get $Φ0+(t)+Φ0−(t)=O(t−3)$ and $Φ1+(t)+Φ1−(t)=O(t−2)$⁠, $|t|→∞, t∈L$⁠, and the conditions (3.16) are fulfilled as required.

Finally, we satisfy the condition (3.17) that guarantees the continuity of the displacement at the junction point x = 0. The asymptotics of the sum $Φ0+(t)+Φ0−(t)=O(t−3), t→∞, t∈L$⁠, we just verified, is sufficient for the convergence of the integral in (3.17). To transform the condition (3.17) into an equation with respect to the constants C and N, we recast the formula (3.23) and express $[X±(t)]−1$ through functions on the complex plane

It is convenient to introduce the functions

In terms of these functions, the functions $Φ0±(t)$ can be written as follows:

where $Ψ0±(n)(t)$ and $Ψ1±(n)(t)$ are the two components of the vectors $Ψ±(n)(t)$⁠, n = 1, 2, given by (3.54). On substituting these expressions into the relation (3.17) we obtain the second equation for the constants C and N

where

The system of two equations (3.63) and (3.76) determines the constants C and N

where $Δ0=Π0Ω1−Π1Ω0.$ The determination of these constants completes the solution of the vector Riemann–Hilbert problem (3.15)–(3.17).

3.5 Partial indices of factorization

In this section, we wish to determine the partial indices of factorization, κ₁ and κ₂, defined as the orders of the columns of the canonical matrix of factorization (18–20). To compute them, we apply the method (18) used in the genus-3 case in (5). If it turns out that $|κ2−κ1|>1$⁠, then the partial indices are unstable (21). This means that in any neighborhood of the matrix G(t), there exists a matrix $Gϵ(t)$ having different partial indices $κ˜1$ and $κ˜2$ (22). In this case, the associated Wiener–Hopf matrix factors $Xϵ+(t)$ and $Xϵ−(t)$ cannot be close to matrices $X+(t)$ and $X−(t)$⁠, and an approximate solution may not converge to the exact one.

The canonical matrix is a matrix X(s) that solves the factorization problem $X+(t)=G(t)X−(t), t∈L$⁠, and satisfies the conditions

• at any finite point $s∈C$⁠, X(s) is in normal form,

• $detX(s)$ does not have zeros at any finite point in the complex plane, and

• the matrix X(s) is in normal form at infinity.

We recall that a matrix Y(s) is in normal form at a point (finite or infinite) if the order of the determinant of the matrix at this point is equal to the sum of the orders of the matrix columns.

Assume $Yj(s)=Yj*(s)(s−s0)αj, s→s0, j=1,…,n$⁠, where α_j is real, $Yj*(s)$ is bounded at s = s₀ and $Yj*(s0)≠0$⁠. Then α_j is called the order of the function $Yj(s)$ at s = s₀ and $α=min{α1,…,αn}$ is called the order of the vector $Y(s)=(Y1(s),…,Yn(s))$ at the point s = s₀.

Let $Yj(s)=Yj*(s)s−αj, s→∞, j=1,…,n$⁠, where α_j is real, $Yj*(s)$ is bounded at infinity and $Yj*(∞)≠0$⁠. Then the numbers α_j and $α=min{α1,…,αn}$ are called the order at infinity of the function $Yj(s)$ and vector Y(s), respectively.

Since the properties of the inverse matrix $[X(s)]−1$ have been studied in Section 3.4, we shall convert the matrix $[X(s)]−1$⁠, not the matrix X(s) itself, into the canonical form. The matrix $[X(s)]−1$ has three singular points, ζ₀, ζ₁ and $∞$⁠. At the point $s=ζ0$⁠, it admits the representation

where $l0=l(ζ0), w0=f1/2(ζ0)$ and F₀ and $Fnj$ are non-zero constants. It is clear that $det{[X(s)]−1}∼0$ as $s→ζ0$⁠, and the order of the determinant at the point ζ₀ is equal to 1, while both columns have zero-orders. To transform the matrix into normal form, we multiply it from the right by the matrix

The new matrix $[X(s)]−1T0(s)$ is in normal form at the point $s=ζ0$⁠,

Proceed now with converting the new matrix into normal form at the point $ζ=ζ1$⁠. The original matrix $[X(s)]−1$ behaves at the point $(ζ1,w1)$ as

where $w1=(−1)n−1f1/2(ζ1)$⁠, n = 1 if the point $(ζ1,w1)$ lies on the first sheet $C1$ and n = 2 otherwise, $l1=l(ζ1)$⁠, and F₁ and $F^nj$ are non-zero constants. By multiplying the new matrix $[X(s)]−1T0(s)$ from the right by the matrix

we obtain the matrix $[X(s)]−1T^(s)$⁠, where

It is directly verified that the matrix $[X(s)]−1T^(s)$ is in normal form at both points, $s=ζ0$ and ζ₁. At the point $(s,w)=(ζ1,w1)$⁠, we have

A similar remedy is to be used to make the matrix $[X(s)]−1T^(s)$ in normal form at infinity. Analysis of the matrix $[X(s)]−1T^(s)$ at infinity shows

where h₀ is given by (3.68) and h₁₂ is a non-zero constant. The orders of the columns of the matrix $[X(s)]−1T^(s)$ at infinity are equal to 0 and –1, and their sum does not equal the order 0 of the determinant of this matrix at infinity. We multiply it from the right by the matrix

and arrive at the matrix $X˜(s)=[X(s)]−1T(s)$ that is in normal form at the infinite point if the parameter ν is chosen to be $ν=(ν0−ν1)−1$⁠. Then

where d₂₂ is a constant. The transformation matrix $T(s)=T^(s)U(s)$ is

The orders of the columns of the matrix $X˜(s)$ and its determinant at infinity are equal to 0. The determinant of the matrix $X˜(s)$ does not have zeros in any finite complex plane. Therefore, the matrix $X˜(s)$ is the canonical matrix of factorization and the partial indices $κ1=κ2=0$⁠.

Notice that the original vector Riemann–Hilbert problem can be rewritten as

Since

and

we see that

Therefore, $X˜(s)$ is the canonical matrix of factorization of the coefficient $[G(t)]−1$ of the Riemann–Hilbert problem (3.89). We may conclude now that the vector Riemann–Hilbert problem (3.15) has zero partial indices and according to the stability criterion (20,21) they are stable.

4 Vector Riemann–Hilbert problem associated with the direct extension of the boundary conditions

In the preceding section, we solved the vector Riemann–Hilbert problem derived by employing the Laplace transforms of the velocity potentials. In this section, we wish to apply a different approach for its derivation that employs the Fourier transform and another way of extension of the boundary conditions on the whole real axis. We aim to understand whether one method has the advantage of the other. For simplicity, we confine ourselves to the case $0<θ0<π/2$ and take L as the real axis. We start with writing the general integral representation of the scattering potentials

where $γ=γ(s)$ is the branch fixed by (3.3). The first derivative $∂∂yϕ0(x,0)$ is continuous at the point x = 0 and equals 0, while the mixed derivative $∂2∂x∂yϕ0(x,0)$ is bounded at the point x = 0 and discontinuous,

where N is a non-zero constant. Extend now the boundary conditions (2.12) onto the whole real axis $L={−∞<x<∞,y=0}$ except for the point x = 0,

where

To derive the associated vector Riemann–Hilbert problem, we introduce the Laplace transforms (one-sided Fourier integrals)

apply the Fourier integral transform to the boundary conditions (4.3) and observe that

Then the boundary condition (4.3) can be rewritten in terms of the functions $Φj±(s)$ and $Aj(s)$ as

We now eliminate the functions $A0(s)$ and $A1(s)$ to have the vector Riemann–Hilbert problem

where

To transform the matrix coefficient of the problem to the form (3.13), we replace the functions $Φ^0−(s)$ and $Φ^1−(s)$ by two new functions,

In terms of the vector-functions $Φ±(s)=(Φ0±(s),Φ1±(s))⊤$⁠, the Riemann–Hilbert problem has the form

with the matrix coefficient G(s) defined by

It is seen that the functions $b0(s), c0(s)$ and $l0(s)$ differ from the corresponding functions b(s), c(s) and l(s) appeared in Section 3, while the characteristic polynomial $f(s)=(s2−μ2)2+m2$ is the same. This means that both vector Riemann–Hilbert problems reduce to the scalar Riemann–Hilbert problem (3.25) on the same genus-1 Riemann surface $R$⁠. The solution of the problem (3.25), the matrix factorization and Jacobi inversion problems are given by the same formulas as in Section 3 if the functions b(s), c(s) and l(s) are replaced by $b0(s), c0(s)$ and $l0(s)$⁠, respectively.

Substantial differences between the two problems begin when we start analyzing the behavior of the functions $Δ(s)=λ1(s)λ2(s)$ and $ϵ(s)=λ1(s)/λ2(s)$ at infinity. These asymptotics are required to determine the asymptotics of the the factorization matrix that is crucial for the application of the Liouville’s theorem. It is not hard to show that

This results in logarithmic singularities of the functions $log Δ(t)$ and $log ϵ(t)$ at infinity which make the analysis of the behavior of the functions $χ1(s)$ and $χ2(s)$ at infinity harder. Consider first the limit values $χ1±(t)$ on the contour L of the function $χ1(s)$ given by (3.66)

where

Except for the last integral $I3(t)$ the asymptotics at infinity of the terms in (4.14) and (4.15) can be written immediately. For the integrals $I1(t)$ and $I2(t)$⁠, we have

As for the integral $I3(t)$⁠, it can be expressed through the limit

On making the substitutions $t0=y−1/2$ and $|t|=x−1/2$⁠, we represent the integral $Iν(t)$ as a Mellin convolution integral

where

By the Mellin convolution theorem, we recast the integral to write

and compute it by the theory of residues. In the case $|t|>1$⁠, we have

If we substitute this expression into (4.18) and compute the limit, we obtain

When we combine the asymptotics we derived with those of the other terms in (4.14) and (4.15) we get the following representation of the functions $χ1±(t)$ on the real axis for large $|t|$⁠:

where

Analyze now the behavior of the function $χ1(s)$ as $s→∞$ and $s∈C±∖L$⁠. We have

where $Ij(s)$ are defined by (4.15). As before, we focus our attention on the integral $I3(s)$⁠. On making the substitution $s=±is0,−12π<args0<12π, s∈C±$⁠, we represent the integral $I3(s)$ in the form

Substitute next t by $y−1/2$ and $s02$ by $x−1$ and write the integral $Jν(s0)$ as a Mellin convolution integral

We apply the convolution theorem and convert this integral into the following one:

After the theory of residues is employed, we have a series representation of the integral for $|s0|>1$

On coming back to s (⁠$s0=∓is, s∈C±$⁠) and computing the limit in (4.26), we obtain

The asymptotics of the other terms in (4.25) is derived in a simple manner, and we have

where r₁ is given by (4.24). It becomes evident that when $s→t∈L$ in (4.31), then for both cases t > 0 and t < 0 treated separately, the asymptotics deduced coincide with formula (4.23).

The derivation of the asymptotics of the function $f1/2(s)χ2(s)$ as $s→∞$ is analogous to the deduction of the asymptotics of the function $χ1(s)$⁠. We have

where

As before for the function $χ1(s)$⁠, the two asymptotics for large t when $t∈L$ derived by means of the Sokhotski–Plemelj formulas and directly from the asymptotics (4.32) as $s→t±0$ coincide.

Our next step is to write down the asymptotics of the solution

of the scalar Riemann–Hilbert problem on the surface $R$⁠. We have

where

On substituting the asymptotics (4.35) into the expressions (3.23) and (3.24), where l(s) needs to be replaced by $l0(s)$⁠, we deduce

where

Now we replace G(s) by $[X−(s)]−1X+(s), s∈L$⁠, in (4.11) and represent $X−(s)g(s)$ as

where

and

Return next to the boundary condition of the Riemann–Hilbert problem. We have

Using the asymptotics (4.37) of the matrices $X±(s)$ at infinity, we conclude from (4.42) that

The asymptotics of $Φ+(s)=O(s−2), s→∞, s∈C+$ results in $ϕ+(x)→0, x→0+$⁠, and the continuity condition for the displacement at x = 0, $y=0−$ is fulfilled.

As in Section 3, the vector $X(s)Φ(s)$ has a simple pole at the point ζ₀ and admits the representation (3.57). By applying the continuity principle and the generalized Liouville theorem we find the solution of the vector Riemann–Hilbert problem

where C is an arbitrary constant. Now, the solution we derived has an unacceptable simple pole at the point $(ζ1,w1)∈R$⁠. It becomes a removable point if the constant C is fixed by the condition

Finally, we verify the asymptotics of the solution (4.44) at infinity. On substituting the representation of the matrix $[X−(s)]−1$ from (4.37) the two formulas in (4.44) we determine

where

It is clear that the function $Φ0−(s)$ vanishes at infinity, $Φ0−(s)=O(s−1), s→∞, s∈C−$⁠, if and only if the constant $N=∂2∂x∂yϕ0(0+,0)$ is fixed as

This completes the solution of the vector Riemann–Hilbert problem defined by (4.11) and (4.12).

A comparative analysis of the asymptotic formulas (3.67) and (3.70) and their counterparts (4.31), (4.32) and (4.37) shows that the derivations by the first method are simpler. The disadvantage of the second method is explained by the differences between formulas (3.65) and (4.13) which give rise to the logarithmic growth at infinity of the functions $log Δ(t)$ and $log ϵ(t)$ in the second method and their boundedness in the first method. These differences make numerical computations of the solution obtained by the second method harder.

5 Numerical results

For our computations, we shall use the solution obtained in Section 3 and consider the case $0<θ0<π/2$⁠. In this case, we may choose $κ0=0$ and the contour L is the real axis. By inverting the integrals (3.5) and employing formula (3.8), we express the two potential $ϕ0(x,y)$ and $ϕ1(x,y)$ through the solution of the Riemann–Hilbert problem defined by (3.15)–(3.17)

On substituting formulas (3.60) and (3.72) into these expressions we transform them as

where $(r,θ)$ are the polar coordinates of a point (x, y), $x=r cos θ, y=r sin θ$⁠,

and the functions $Λj±(t)$ are given by (3.73).

The improper integrals over the real axis L are computed by mapping the contour L onto the unit, positively oriented circle $C={|u|=1}$ and applying the standard Simpson rule,

where $t=−2i(u+i)(u−i)−1$ and $u=eiθ$⁠. The principal value of the Cauchy integrals over the contour L is computed by employing the same mapping to the unit circle $C$ and the following formula:

where

and n is the number of knots of the integration formula. For computing integrals (3.35), (3.47) and (3.49), we apply the Gauss quadrature formula with Chebyshev’s weights and abscissas.

In our numerical tests, we focus our attention on the absolute values of the full potentials $ψ0=ψinc+ϕref+ϕ0$ in the lower half-plane $H0$ and $ψ1=ϕ1$ in the upper half-plane $H1$⁠. Denote by $P(r,θ)=|ψ1(x,y)|, (x,y)∈H1$ (⁠$0<θ<π$⁠) and $P(r,θ)=|ψ0(x,y)|, (x,y)∈H0$ (⁠$π<θ<2π$⁠). For all tests, we choose water’s density $ρf=997$ kg/m³. Except for Figs 9 and 10, we choose $argk= tan −10.1, |k|=1$ m^{– 1}, the cell measurements $d=d1=d2=0.01$ m and the aperture radius a = 0.001 m. In this case, the parameter τ is complex and its magnitude is small, $τ=5.0356×10−2+i5.1531×10−3$⁠. It is worth to mention that for all problem parameters tested the two integers m_a and m_b are equal to zero when the point q₁ falls on the first sheet of the surface. They are non-zero fractions if $q1∈C2$⁠.

The curves drawn in Figs 4–7 show the variation of the function $P(r,θ)$ with change of θ when r is kept constant. In Fig. 4, we use $θ0=π/4, |α|=10$ m^{– 3}, $ρf/m0=100$ m^{– 1} and r = 5 m. For Fig. 5, we choose the same parameters as for Fig. 4 except for $θ0=π16$⁠. In Fig. 6, we increase the ratio $ρf/m0$ from 100 to 500 that results in a fivefold decrease of the panel surface density. The other parameters coincide with those employed for computations portrayed in Fig. 4. It is possible to infer from this figure that as the membrane surface density increases, the absolute value of the potential ψ₁ decreases. In Fig. 7, we decrease r and select it to be 3 m and keep the other parameters of Fig. 4 unchanged. Figure 8 shows how the function P(r) varies with change of r when the polar angle θ equals $π/16, π/8, π/4, π/3$ and $5π/12$⁠, while the other parameters are selected in the same way as in Fig. 4. In Fig. 9, we increase the value of $|k|$ from 1 to 5, change its argument, $argk= tan −10.02$⁠, and because of formula (2.7), increase $|α|$ from 10 to 250. The other parameters of Fig. 4 are the same.

As $k2→0+$ and $k1→kres=2a/V$⁠, the parameter $τ→∞$⁠. In Fig. 10, we change the cell measurements, the aperture radius and $argk$⁠, $d=d1=d2=0.215$ m, a = 0.005 m and $argk= tan −10.02$⁠, and keep the other parameters the same as in Fig. 4. For the parameters chosen we have $τ=0.81394+i5.26690, |k|=1$ (⁠$k1=0.99980, k2=0.019996$⁠) and $kres=1.11803>|k|$⁠. For these values of the parameters τ and $kres$⁠, the accuracy of the solution is approximately the same as that for smaller values of the parameter $|τ|$⁠. For example, the constants m_a and m_b in the Jacobi problem have the values $ma=−1.2×10−6$ and $mb=−3.6×10−6$ when the point q₁ is on the first sheet and $ma=−0.29$ and $mb=−0.23$ when the point $q1∈C2$⁠. The integers m_a and m_b are accepted to be 0, and $q1∈C1$⁠. However, for $d=d1=d2≥0.2156$ m, a = 0.005 m, the real part of the parameter τ and the difference $|k|−kres$ change their sign. As $d=d1=d2=0.2156$ m, $τ=−0.29247+i5.36927, kres=0.99891$⁠, while k is still the same $k=0.99980+i0.019996$⁠. The errors of approximation of the integrals involved in the Riemann θ-function accumulate, and the recovering of the correct numerical values of ζ₁, n_a and n_b is not easy.

6 Conclusions

A closed-form solution has been given for the model problem of the scattering of a plane sound wave by an infinite thin structure formed by a semi-infinite acoustically hard screen attached to a sandwich panel with acoustically hard walls. The upper side of the sandwich panel is perforated, while the lower side is an unperforated membrane. We have applied two methods of extension of the boundary conditions to the whole real axis and deduce two order-2 vector Riemann–Hilbert problems. The matrix coefficients of both problems have the Chebotarev–Khrapkov structure with the same order-4 characteristic polynomial but with distinct entries. Wiener–Hopf matrix factors for both problems have been derived by quadratures by solving a scalar Riemann–Hilbert problem on the same elliptic surface. The coefficient of the scalar problem is equal to the first eigenvalue of the matrix on the upper sheet of the surface and the second eigenvalue on the lower sheet. We have eliminated the essential singularity caused by simple poles of the Cauchy analog at the two infinite points of the surface by solving a genus-1 Jacobi inversion problems in terms of the Riemann θ-function.

We have found that the analysis of the Wiener–Hopf matrix factors at infinity is simpler for the first method that sets the Riemann–Hilbert problem for the one-sided Fourier transforms of the velocity potentials on the upper and lower sides of the infinite structure. The second method extends the four boundary conditions to the whole real axis by means of unknown functions and employs the one-sided Fourier transforms of these functions. The advantage of the first method over the second one is explained by the logarithmic growth at infinity of the densities of the singular integrals involved in the solution obtained by the second method. Both methods lead to the solution having two arbitrary constants. The constants have been fixed by additional conditions of the problem. For the first method, in addition to the meromorphic Wiener–Hopf factors, we constructed the canonical matrix of factorization and computed the partial indices of factorization. It turns out that they both are equal to zero and therefore stable.

Numerical tests have been implemented for the solution derived by the first method. The integrals involved are rapidly convergent for all values of the parameters tested except for the case when $|τ|→∞$⁠, when the method is not numerically efficient. We have computed the absolute values of the full velocity potentials, the function $P(r,θ)=|ϕ1|$⁠, $0<θ<π$⁠, and $P(r,θ)=|ϕinc+ϕref+ϕ0|, π<θ<2π$⁠. We have found that the presence of the sandwich panel perforated from the upper side reduces the transmission of sound, and when the membrane surface density m₀ is growing the function $P(r,θ)$ (⁠$0<θ<π)$ decreases. We have also discovered that when the absolute value $|k|$ of the complex wave number approaches the resonance value $kres$⁠, then the parameter $|τ|$ and the function $P(r,θ)$ are growing to infinity. In the limiting case $|τ|=∞$⁠, the diffraction model problem reduces to two separate scalar Riemann–Hilbert problems, and one of them does not have a physical solution.

References