A genus-3 Riemann-Hilbert problem and diﬀraction of a …yantipov/CMFTAntipov.pdf00 (0000), No. 0...

Computational Methods and Function TheoryVolume 00 (0000), No. 0, 1–? CMFT-MS XXYYYZZ

A genus-3 Riemann-Hilbert problem and diffraction of a

wave by two orthogonal resistive half-planes

Yuri A. Antipov

Abstract. Diffraction of a plane electromagnetic wave (E-polarization) bytwo orthogonal electrically resistive half-planes is analyzed. The physical prob-lem reduces to a Riemann-Hilbert problem in the real axis for four pairs of an-alytic functions Φ+

j (η) (η ∈ C+) and Φ−

j (η) = Φ+

j (−η) (η ∈ C−), j = 1, 2, 3, 4,

and C+ and C− are the upper and lower half-planes. It is shown that theproblem is equivalent to two scalar Riemann-Hilbert problems on a plane anda Riemann-Hilbert problem on a genus-3 hyperelliptic surface subject to acertain symmetry condition. A closed-form solution is derived in terms ofsingular integrals and the genus-3 Riemann Theta function.

Keywords. Riemann-Hilbert problem, Riemann surfaces, matrix factoriza-tion, electromagnetic diffraction.

2000 MSC. Primary 30Exx; Secondary 35P05, 45E05, 78A45.

1. Introduction

Many model problems of elasticity, fluid mechanics and diffraction theory requirethe solution of a scalar Riemann-Hilbert problem on the Riemann surface ofan algebraic function. The Riemann-Hilbert problem on a hyperelliptic surfaceassociated with a static mixed boundary-value problem for an elastic isotropicplane with a system of cuts along a line is analyzed and solved by quadratures in[21]. This problem generalizes the formulation [11] by removing any restrictionon the location of the points where the boundary conditions are changed. Thisapproach is applied in [18], [19], [17] for some other contact problems of fracturemechanics. Method of conformal mappings and the Riemann-Hilbert problem ona Riemann surface is developed in [8], [9] for nonlinear free-boundary problemsof supercavitating flow in multiply connected domains. An approach for theRiemann-Hilbert problem on a hyperelliptic surface based on the use of theBaker-Akhieser function is employed in the theory of finite gap integration [10].

Version January 3, 2011.The work was partly funded by National Science Foundation through grant DMS0707724.

ISSN 1617-9447/$ 2.50 c© 20XX Heldermann Verlag

2 Y.A. Antipov CMFT

Another application of the theory of the Riemann-Hilbert problem on a Rie-mann surface arises in the Wiener-Hopf factorization of matrices of the followingstructure:

(1) G(t) = f0(t)I + f1(t)A + . . . + fn−1(t)An−1,

fj(t) (j = 0, 1, . . . , n − 1) are Holder functions, I is the n × n unit matrix, andA is an n × n polynomial matrix with the characteristic polynomial ϕ(z, w) =det(wI − A(z)) irreducible over the field of rational functions. The idea [15] ofreducing the vector Riemann-Hilbert problem with the matrix coefficient (1) to ascalar Riemann-Hilbert problem on a Riemann surface of the algebraic functionϕ(z, w) = 0 is developed and implemented for the hyperelliptic case in [16], [2]to solve the governing systems of singular integral equations arising in contactand fracture mechanics.

Diffraction problems in acoustics and electromagnetics for a half-plane and awedge can be treated by means of the Sommerfeld integral and require the solu-tion of a system of difference equations with periodic coefficients and, ultimately,a Riemann-Hilbert problem on two segments in a Riemann surface [7], [4], [5],[6]. Another approach for solving difference equations with periodic coefficientsin diffraction theory using the Riemann bilinear relations is proposed in [14]. Ithas recently been shown [1] that the use of the method [12] and the Laplaceintegral substantially simplifies the solution procedure. Instead of a system ofMaliuzhinets-type difference equations this technique leads to a system of sym-metric Wiener-Hopf equations. This method was applied [1] to the diffractionproblem for two orthogonal half-planes, an electrically resistive screen and aperfectly magnetically conductive one, when the structure was illuminated by aline electric current of uniform excitation. The problem was reduced to a inho-mogeneous Riemann-Hilbert problem on an elliptic surface subject to a certainsymmetry condition. The condition was verified for the homogeneous case. Inthis paper we generalize the result [1] by considering the case when both screensare electrically resistive half-planes with the surface resistivity being neither zero,nor infinity. The structure is illuminated by a plane wave Ei

z = e−ikx cos ϕ0−iky sinϕ0

(ϕ0 is the angle of incidence). We show that the physical problem is equivalentto a certain homogeneous Riemann-Hilbert problem for four pairs of functionssubject to a symmetry condition and solve it exactly by reducing the problem toa scalar Riemann-Hilbert problem on a genus-3 hyperelliptic surface.

2. Formulation

Let W be a planar structure consisting of two orthogonal electrically resistivehalf-planes, W1 = (x, y) : 0 < x < +∞, y = ±0 and W2 = (x, y) : x =±0, 0 < y < +∞, of resistivity R1 and R2, respectively. On these half-planes,the electric and magnetic fields, E = (Ex, Ey, Ez)

⊤ and H = (Hx, Hy, Hz)⊤,

00 (0000), No. 0 A genus-3 Riemann-Hilbert problem 3

satisfy the transition conditions

(2) n ×E = Rjn× [n× H]+−, [n ×E]+− = 0, (x, y) ∈ Wj , j = 1, 2,

where n is the unit normal external to the positive side of the half-planes, and[f ]+− denotes the jump of a function f . Let µ be the magnetic permeability, εthe electric permittivity, ω the angular frequency, k = ω

√εµ the wave number,

and Z =√

µ/ε the intrinsic impedance of the medium. It is assumed that µ and

ν are complex numbers and the branches√

εµ and√

µ/ε are chosen such thatk = k1 + ik2, k2 > 0, kZ = ωµ, and k/Z = ωε. The structure is illuminated bya plane wave (the factor e−iωt is suppressed)

(3) Eiz = e−ikx cos ϕ0−iky sin ϕ0,

where the angle ϕ0 defines the direction of incidence.

The Maxwell equations imply

(4) Ex = −Z

ik

∂Hz

∂y, Ey =

Z

ik

∂Hz

∂x, Ez = −Z

ik

(

∂Hy

∂x− ∂Hx

∂y

)

,

Hx =1

ikZ

∂Ez

∂y, Hy = − 1

ikZ

∂Ez

∂x, Hz =

1

ikZ

(

∂Ey

∂x− ∂Ex

∂y

)

,

and reduce to the Helmholtz equation for the component Ez

(5)

(

∂2

∂x2+

∂2

∂y2+ k2

)

Ez = 0, (x, y) ∈ R2 \ W.

For E-polarization, the transition conditions (2) read

(6) Ez|y=0± = −R1[Hx|y=0+ − Hx|y=0− ], 0 < x < +∞,

Ez|x=0± = R2[Hy|x=0+ − Hy|x=0−], 0 < y < +∞.

To solve the problem, we convert it to a vector Riemann-Hilbert problem. First,we introduce the following Laplace transforms:

(7)

(

E+

H+

)

(x, η) =

∞∫

0

eiηy

(

E

H

)

(x, y)dy,

(

E−

H−

)

(x, η) =

0∫

−∞

eiηy

(

E

H

)

(x, y)dy,

(

E+

H+

)

(ζ, y) =

∞∫

0

eiζx

(

E

H

)

(x, y)dx,

and obtain from equations (4) and (5) [1]

(8) kZHy+(0+, η) + iξEz+(0+, η) = kZHx+(iξ, 0+) − ηEz+(iξ, 0+),

kZHy+(0+,−η) + iξEz+(0+,−η) = kZHx+(iξ, 0+) + ηEz+(iξ, 0+).

4 Y.A. Antipov CMFT

kZHy−(0+, η) + iξEz−(0+, η) = −kZHx+(iξ, 0−) + ηEz+(iξ, 0−),

kZHy−(0+,−η) + iξEz−(0+,−η) = −kZHx+(iξ, 0−) − ηEz+(iξ, 0−),

−ikZ[Hy+(0−, η) + Hy−(0−, η)] = ξEz+(0−, η) + ξEz−(0−, η),

−ikZ[Hy+(0−,−η) + Hy−(0−,−η)] = ξEz+(0−,−η) + ξEz−(0−,−η).

Here, the function ξ(η) = (η2 − k2)1/2 is the branch of the two-valued functionξ2 = η2 − k2 satisfying the condition ξ(0) = −ik. It is defined in the η-plane cutalong a line joining the branch points η = k and η = −k and passing throughthe infinite point. Applying the Laplace transformations (7) to the the boundaryconditions (6) yields

(9) Ez+(iξ, 0±) = −R1[Hx+(iξ, 0+) − Hx+(iξ, 0−)],

Ez+(0±, η) = R2[Hy+(0+, η) − Hy+(0−, η)].

We next use these conditions and equations (8) to eliminate the functions Ez+(iξ, 0±)

and Hx+(iξ, 0±). This leads to the following result.

Theorem 1. Let Φ+(η) and Φ−(η) be vector-functions whose components are

(10) Φ+(η) =

−ikZHy+(0+, η)

−ikZHy+(0−, η)

−ikZHy−(0±,−η)

Ez−(0±,−η)

.

and

(11) Φ−(η) = Φ+(−η), η ∈ C−.

Then the problem (4), (5), (6) is equivalent to the following vector Riemann-Hilbert problem in the real axis L:

Find two vectors, Φ+(η) and Φ−(η), analytic everywhere in the upper and lowerhalf-planes, C+ and C−, apart from the simple poles (the geometric optics poles)η0 and −η0, respectively, where η0 = k sin ϕ0 if −3π

2< ϕ0 < −π or 0 < ϕ0 < π

2,

and η0 = −k sin ϕ0 if −π < ϕ0 < 0. Furthermore, if 0 < ϕ0 < π2, then

(12) resη=k sinϕ0

Φ+1 (η) = −k cos ϕ0.

If −3π2

< ϕ0 < −π, then

(13) resη=k sinϕ0

Φ+2 (η) = −k cos ϕ0.

If −π < ϕ0 < 0, then

(14) resη=k sinϕ0

Φ−3 (η) = −k cos ϕ0.

As η → ∞, the functions Φ±j (η) (j = 1, 2, 4) vanish, Φ±

j (η) ∼ Njηα (−1 ≤ α <

0), while the functions Φ±3 (η) may grow, Φ±

3 (η) ∼ N3ηα+1 (Nj (j = 1, 2, 3, 4) are


constants). The vectors Φ±(η) are continuous in the half-planes C± up to theboundary L, and their boundary values satisfy the linear relation

(15) Φ+(η) = G(η)Φ−(η), η ∈ L.

Here,(16)

G(η) =1

η − γ

− (γ+iνξ)2

γ+2iνξνξ(iγ−νξ)

γ+2iνξ−ηγ+iνξ(2η−γ)

γ+2iνξ−γξ(η+iνξ)

γ+2iνξ

−νξ(iγ−νξ)γ+2iνξ

− ν2ξ2

γ+2iνξγ(γ−η)−iνξ(2η−γ)

γ+2iνξγξ(η−γ−iνξ)

γ+2iνξ

−η+iνξ2

−η−γ−iνξ2

−γ2

−γξ2

iγνξ−η(γ+2iνξ)2γξ

−γ(γ−η)−iνξ(2η−γ)2γξ

− γ2ξ

−γ2

,

and

(17) γ =kZ

2R1

, ν =R2

2R1

.

Note that the matrix is not singular everywhere in the real axis (it is assumedthat Im γ > 0), and

(18) det G(η) =η + γ

η − γ.

The symmetry and boundary conditions, (11) and (15), imply that the matrixG(η) has to satisfy the condition

(19) G(η)G(−η) = I4, η ∈ L,

where I4 = diag1, 1, 1, 1. It has been verified that this condition is identicallysatisfied.

The conditions (12) for 0 < ϕ0 < π2

follow from the formulas

(20) Ez = Eiz + E

z , Hy = − 1

ikZ

∂Ez

∂x,

(7), (10) and (11). Here, Eiz is the incident wave (3), and E

z is the sum of thereflected, surface, transmitted and diffracted waves. On assuming that Im η >k2 sin ϕ0 we compute the Laplace transform

(21) Φ+1 (η) = Φ+

1 (η) − k cos ϕ0

η − k sin ϕ0

,

where Φ+1 (η) is analytic in the upper half-plane C

+. By analytic continuationwe obtain that the function Φ+

1 (η) is analytic everywhere in C+ except at thesimple pole η = k sin ϕ0, and its residue is equal to −k cos ϕ0.

Finally, we note that the requirements Φ±j (η) ∼ Njη

α as η → ∞ (j = 1, 2, andthe parameter α ∈ [−1, 0) is to be determined), guarantees that the functionsHy(0

±, y) are integrable as y → 0+. From the relations (8) we conclude that as

η → ∞, Φ±3 (η) ∼ N3η

α+1 (the Laplace transform Hy−(0±, η) is understood inthe generalized sense), and Φ±

4 (η) ∼ N4ηα.

6 Y.A. Antipov CMFT

3. Reduction of the Riemann-Hilbert problem dimension

We aim to reduce the Riemann-Hilbert problem for four pairs of functions toproblems whose coefficients are either scalars or matrices of lower dimension. Byfollowing the scheme [1] we split the matrix G(η) as follows:

(22) G(η) =R(η)

1 + 2iνξ/γ

[

I4 +B(η)

γξ(γ2 − 16ν2ξ2)

]

,

where(23)

R(η) =1

γ(η − γ)

−γ2 + ν2ξ2 −ν2ξ2 −ηγ −iγνξ2

ν2ξ2 −ν2ξ2 γ(γ − η) −iγνξ2

12(−ηγ + 2ν2ξ2) 1

2(−ηγ + γ2 − 2ν2ξ2) −γ2

2−iγνξ2

− iν2(4η − γ) i

2(4ην − 3γν) −iγν −γ2

2

,

and B(η) is a 4 × 4 polynomial matrix. Its elements are

(24) b11 = 2iνξ2(γ2 − 10ν2ξ2), b12 = −2iνξ2(γ2 − 2ν2ξ2),

b13 = −2iνξ2(γ2 − 8ν2ξ2), b14 = 2ξ2[6γν2ξ2 + η(γ2 − 8ν2ξ2)],

b21 = −2iνξ2(γ2 + 2ν2ξ2), b22 = 6iνξ2(γ2 − 2ν2ξ2),

b23 = 2iνξ2(γ2 + 8ν2ξ2), b24 = −2ξ2[γ(γ2 + 2ν2ξ2) + η(γ2 + 8ν2ξ2)],

b31 = −iνξ2(γ2 − 4ν2ξ2), b32 = iνξ2(γ2 + 12ν2ξ2),

b33 = 2iνξ2(γ2 − 8ν2ξ2), b34 = −ξ2[γ3 + 4ν2ξ2(4η − γ)],

b41 = η(γ2 − 4ν2ξ2) − 8ν2ξ2

(

γ +2

γν2ξ2

)

,

b42 = γ2(γ − η) + 4ν2ξ2

(

γ − 3η +4

γν2ξ2

)

,

b43 = γ3 − 4ν2ξ2(4η + γ), b44 = −2iνξ2(8η2 − 5γ2 + 8ν2ξ2).

The possibility of the reduction of the dimension of the vector Riemann-Hilbertproblem is determined by the algebraic structure of the characteristic polynomialof the matrix B. This polynomial, irreducible over the field of rational functionsof η, can be factorized as

(25) det[B(η) − µI4] = φ1(µ)φ2(µ),

where

(26) φ1(µ) = [µ − 2iνξ2(γ2 − 16ν2ξ2)]2,

φ2(µ) = µ2 + 16i(η2 − γ2)νξ2µ + ξ2(γ2 − 16ν2ξ2)[3γ4 − 4η2(γ2 − 4ν2ξ2)].

Because of the structure of the characteristic polynomial there exists a rationalmatrix of transformation, T (η), such that

(27)1

1 + 2iνξ/γ

[

I4 +B(η)

γξ(γ2 − 16ν2ξ2)

]

= T (η)Γ(η)T−1(η),


where Γ(η) has a block-diagonal structure. The first two columns of the matrix Tare eigenvectors of the matrix B(η) corresponding to the multiplicity-2 eigenvalueµ = 2iνξ2(γ2 − 16ν2ξ2), while the other two columns comprise a basis of thekernel of the operator φ2(B). The matrix T (η) is not unique. By comparingthe complexity of the matrix Γ and the symmetry condition (11) for a newRiemann-Hilbert problem for order-2 vectors to be derived, we have found thatit is preferable to work with

(28) T (η) =

1 − iν(η − γ) 1 0

1 − iην

0 1−1 0 1

212

0 1 i(ηγ+γ2+2ν2ξ2)2γνξ2

i(ηγ−2ν2ξ2)2γνξ2

.

This choice uniquely identifies the matrix Γ(η)

(29) Γ(η) =1

2ν

2ν 0 0 00 2ν 0 00 0 b + cl0 cl10 0 cl2 b − cl0

,

where b(η) and c(η) are Holder functions in the real axis, and l0(η), l1(η) andl2(η) are polynomials defined by

(30) b(η) = 2νγ3 − 8iν(η2 − γ2)ξ − 16γν2ξ2

(γ + 2iνξ)(γ2 − 16ν2ξ2),

c(η) = − 2i

γξ(γ + 2iνξ)(γ2 − 16ν2ξ2),

l0 = γ3(ν2ξ2 − η2) − η(γ4 − 16ν4ξ4),

l1(η) = ν2ξ2(3γ3 − 4ηγ2 − 16ν2ηξ2) − η2(γ3 − 8γν2ξ2),

l2(η) = γ3(γ2 + 5ν2ξ2) + η2(γ3 + 8γν2ξ2) + 2η(γ4 + 6γ2ν2ξ2 + 8ν4ξ4).

To complete the reduction of the order-4 problem (15) to scalar and order-2Riemann-Hilbert problems, introduce two new vector-functions

(31) Ω+(η) = U0(η)Φ+(η), η ∈ C+,

Ω−(η) = U1(η)Φ−(η), η ∈ C−,

where

(32) U0(η) = γd(η)T−1(η)R−1(η), U1(η) = γd(η)T−1(η),

d(η) = 3γ2 − 4η2 + 16ν2ξ2.

Since the original vector-functions Φ±(η) have to satisfy the symmetry condition(11), we observe that

(33) Ω+(η) = S(η)Ω−(−η), η ∈ C+,

8 Y.A. Antipov CMFT

where

(34) S(η) = T−1(η)R−1(η)T (−η) =

1 iην

0 00 1 0 00 0 s33(η) s34(η)0 0 s43(η) s44(η)

,

s33(η) = −s44(η) = −2(η − γ)(γ2 − 4ν2ξ2)

γ(γ2 − 16ν2ξ2),

s34(η) = − 8(η − γ)ν2ξ2

γ(γ2 − 16ν2ξ2), s43(η) =

4(η − γ)(γ2 + 2ν2ξ2)

γ(γ2 − 16ν2ξ2).

To study the other properties of the vectors Ω±(η), we explicitly express theelements of the vectors Φ±(η) through the functions Ω±

j (η). We have

(35) Φ+1 (η) =

1

γd(η)

[

Ω+1 (η) +

iγ

νΩ+

2 (η) +(4ν2ξ2 − γ2)Ω+

3 (η) − 4ν2ξ2Ω+4 (η)

2γ(η − γ)

]

,

Φ+2 (η) =

1

γd(η)

[

Ω+1 (η) +

2(2ν2ξ2 + γ2)Ω+3 (η) − (4ν2ξ2 − γ2)Ω+

4 (η)

2γ(η − γ)

]

,

Φ+3 (η) =

1

γd(η)

[

−Ω+1 (η) +

iη

νΩ+

2 (η) +(8ν2ξ2 + γ2)Ω+

3 (η) − (8ν2ξ2 − γ2)Ω+4 (η)

4γ(η − γ)

]

,

Φ+4 (η) =

1

γd(η)

Ω+2 (η) − i

4γνξ2(η − γ)

×[

(γ(γ2 + 2ν2ξ2) + η(8ν2ξ2 + γ2))Ω+3 (η) + (6γν2ξ2 − η(8ν2ξ2 − γ2))Ω+

4 (η)]

,

and

(36) Φ−1 (η) =

1

γd(η)

[

Ω−1 (η) − i(η − γ)

νΩ−

2 (η) + Ω−3 (η)

]

,

Φ−2 (η) =

1

γd(η)

[

Ω−1 (η) − iη

νΩ−

2 (η) + Ω−4 (η)

]

,

Φ−3 (η) =

1

γd(η)

[

−Ω−1 (η) +

1

2Ω−

3 (η) +1

2Ω−

4 (η)

]

,

Φ−4 (η) =

1

γd(η)

Ω−2 (η) +

i

2γνξ2[(ηγ + γ2 + 2ν2ξ2)Ω−

3 (η) − (2ν2ξ2 − ηγ)Ω−4 (η)]

.

We are now ready to establish a one-to-one correspondence between the Riemann-Hilbert problem (15) and the problems for the components of the vectors Ω+(η)and Ω−(η).


Theorem 2. Let Ω±(η) be the general form of dimension-4 vector-functionswhich

(i) are analytic everywhere in the half-planes C± apart from simple poles at the

points ±η0,

(ii) satisfy the symmetry condition (33),

(iii) satisfy the boundary condition

(37) Ω+(η) = Γ(η)Ω−(η), η ∈ L,

(iv) grow at infinity as

(38) Ω±j (η) = K±

j η3+α + O(η2+α), j = 1, 3, 4, η → ∞, η ∈ C±,

Ω±2 (η) = K±

2 η1+α + O(ηα), η → ∞, η ∈ C±,

where −1 ≤ α < 0, K±j are nonzero constants, and

(39) K+3 = K+

4 ,

(40) limη→∞

Ω+1 (η)

η2+

2ν2

ηγ[Ω+

3 (η) − Ω+4 (η)]

= 0,

(v) satisfy the following seven conditions:

(41) Ω+3 (γ) = 0, Ω+

4 (γ) = 0,

(42) (k + γ)Ω+3 (k) + kΩ+

4 (k) = 0,

(43) Ω−1 (η−) − i(η− − γ)

νΩ−

2 (η−) + Ω−3 (η−) = 0,

Ω−1 (η−) − iη−

νΩ−

2 (η−) + Ω−4 (η−) = 0,

Ω−1 (η−) − 1

2Ω−

3 (η−) − 1

2Ω−

4 (η−) = 0,

(44) 3 resη=γ0

Ω+3 (η) + res

η=γ0

Ω+4 (η) = 0,

where

(45) η− = −1

2

√

3γ2 − 16k2ν2

1 − 4ν2, γ0 =

√

k2 +γ2

16ν2,

and the branches of the square roots are chosen such that η− ∈ C− and γ0 ∈ C

+.

Then the vector-functions Φ+(η) and Φ−(η) whose components are defined by(35) and (36) are the general solution of the problem (12) - (17) provided theconditions (12) - (14) are satisfied.

10 Y.A. Antipov CMFT

Proof. It follows from (33), (37) and (31) that the functions (35) satisfy theboundary and symmetry conditions (15) and (11). Analysis of the expressions(35) shows that if the functions Ω±

j (η) grow at infinity as in (38) and satisfy the

conditions (39), then the functions Φ±j (η) (j = 1, 2, 4) vanish, while Φ±

3 (η) ∼K±ηα+1 (K± are constants) when η → ∞, η ∈ C±.

According to (35), the functions Φ±j (η) (j = 1, 2, 3, 4) have simple poles at the

points ∓η− ∈ C±. It is directly verified that the system

(46) resη=η−

Φ−j (η) = 0, j = 1, 2, 3, 4,

with respect to Ω−s (η−) (s = 1, 2, 3, 4) has rank 3, and because of the symmetry

condition (11), the conditions (43) are necessary and sufficient to these pointsbeing removable singularities of the vector-functions Φ±(η).

The point η = γ ∈ C+ is a simple pole of the functions Φ+j (η) (j = 1, 2, 3, 4), and

the points η = ±k ∈ C± are simple poles of the functions Φ±4 (η). The conditions

(41), (42) remove these singularities.

The elements of the matrix U1(η) are polynomials of η, and the vector Ω−(η)does not have poles in any finite part of the half-plane C

−. As for U0(η), it is arational matrix, and

(47) Ω+1 (η) = 2[ηγ(γ − η) + ν2ξ2(γ + 2η)]Φ+

1 (η) + [γ(γ2 − 2η2)

+2ν2ξ2(3γ − 2η)]Φ+2 (η) + 2γ(γη − γ2 − 4ν2ξ2)Φ+

3 (η) − 2iγ(γ − 2η)νξ2Φ+4 (η),

Ω+2 (η) = −iν(−2ηγ + 3γ2 + 8ν2ξ2)Φ+

1 (η) + iν(2ηγ + γ2 + 8ν2ξ2)Φ+2 (η)

−2iγν(γ − 2η)Φ+3 (η) + 8γν2ξ2Φ+

4 (η),

Ω+3 (η) =

4(η − γ)

γ2 − 16ν2ξ2[ν2ξ2(6ηγ − 3γ2 − 8η2 + 16ν2ξ2) + η2γ2]Φ+

1 (η)

−[ν2ξ2(5γ2 + 16ν2ξ2 − 8η2 − 2ηγ) + ηγ2(η + γ)]Φ+2 (η)

−γ[8ν2ξ2(γ − η) + ηγ2]Φ+3 (η) + iγνξ2(2ηγ + 3γ2 − 16ν2ξ2)Φ+

4 (η),

Ω+4 (η) =

4(η − γ)

γ2 − 16ν2ξ2[ν2ξ2(5γ2 + 16ν2ξ2 − 8η2 − 2ηγ) − ηγ2(η + γ)]Φ+

1 (η)

+[ν2ξ2(10ηγ + 3γ2 + 8η2 − 16ν2ξ2) + γ2(η + γ)2] Φ+2 (η)

+γ(η + γ)(γ2 + 8ν2ξ2)Φ+3 (η) − iγνξ2(6ηγ + 5γ2 + 16ν2ξ2)Φ+

4 (η).Thus, the functions Ω+

3 (η) and Ω+4 (η) have a simple pole at the point η = γ0.

This point is a removable singularity of the functions Φ+j (η) if

(48) Ωj = res

η=η0

Ω+j (η), j = 3, 4,

solve the system of the four equations

(49) resη=η0

Φ+j (η) = 0, j = 1, 2, 3, 4,

It is a matter of simple algebra to show that its rank is 1, and it reduces to theequation 3Ω

3 + Ω4 = 0 that is the condition (44).


4. Vectors Ω+(η) and Ω−(η)

It follows from (37) that the first two components of the vectors Ω±(η), thefunctions Ω1(η) and Ω2(η), are rational function which have simple poles at thepoints η0 and −η0. Because of the asymptotic relations (38) these functions havethe form (α = −1)

(50) Ω1(η) = C1 + C2η + C3η2 +

D1

η − η0+

D2

η + η0,

Ω2(η) = C ′1 +

D′1

η − η0

+D′

2

η + η0

.

where C1, C2, C3, D1 and D2 are constants to be determined a posteriori. Theother constants are fixed by the symmetry condition (33) as

(51) C ′1 = −2iνC2, D′

1 = −D′2 = −iν

η0(D1 + D2).

4.1. Riemann-Hilbert problem on a genus-3 Riemann surface. In what

follows we shall define the vectors Ω±(η) = (Ω±

3 (η), Ω±4 (η))⊤ from the Riemann-

Hilbert problem:

(52) Ω+(η) =

1

2νΓ(η)Ω

−(η), η ∈ L,

subject to the symmetry condition

(53) Ω+(η) = S(η)Ω

−(−η), s ∈ C

+,

where

(54) Γ(η) =

(

b + cl0 cl1cl2 b − cl0

)

, S(η) =

(

s33(η) s34(η)s43(η) s44(η)

)

.

Notice that

(55) det Γ(η) =η2 − γ2

η2 − γ20

.

Since the matrix coefficient Γ(η) has the structure (54), and f(η) = l20 + l1l2 is adegree-8 polynomial:

(56) f(η) = ν20(η

2 − k2)

3∏

j=1

(η2 − η2j ),

the problem of matrix factorization

(57) Γ(η) = X+(η)[X−(η)]−1, η ∈ L,


can be solved in terms of the solution of a Riemann-Hilbert problem on a genus-3Riemann surface. Here,

(58) ν0 = 8ν2γ√

1 − 4ν2, η1 = i

√

−k2

2+

√

k4

4+

γ4

16ν2,

η2 =

√

k2

2+

√

k4

4+

γ4

16ν2, η3 =

1

2

√

16k2ν2 − 3γ2

4ν2 − 1,

and the branches of the square roots are chosen such that ηj ∈ C+, j = 1, 2, 3.

Consider the algebraic function w2 = f(η). To fix its branch,√

f(η), we cut theextended η-plane along the segments l±1 = [±η1,±k] ∈ C± and l±2 = [±η2,±η3] ∈C± and enforce the condition f 1/2(η) ∼ −ν0η

4, η → ∞.

Let R be the hyperelliptic surface generated by the algebraic function w2 = f(η)and formed by gluing two copies, C1 and C2, of the extended η-plane C ∪ ∞cut along the segments l±1 and l±2 according to the rule

(59) w =

√

f(η), η ∈ C1,

−√

f(η), η ∈ C2.

On following the results in [15], [3], allows us to reduce the problem of matrixfactorization to a scalar Riemann-Hilbert problem on the surface R.

Theorem 3. Let λ(η, w) be a function defined on the surface R as

(60) λ(η, w) =

λ1(η), η ∈ C1,λ2(η), η ∈ C2,

where λj(η) = b(η) + (−1)j−1c(η)√

f(η) (j = 1, 2) are the eigenvalues of the

matrix Γ(η), and let χ(η, w) be a solution to the Riemann-Hilbert problem

(61) χ+(η, w) = λ(η, w)χ−(η, w), (η, w) ∈ L,

where L = (L ⊂ C1) ∪ (L ⊂ C2). Then the factors

(62) X±(η) = eχ±(η,w)Y (η, w) + eχ±(η,−w)Y (η,−w), η ∈ C±,

solve the factorization problem (57). Here,

(63) Y (η, w) =1

2

[

I2 +1

wQ(η)

]

, Q(η) =

(

l0(η) l1(η)l2(η) −l0(η)

)

.

The solution to the factorization problem is not unique. We aim to construct avector-function that satisfies not only the boundary condition (52) but also thesymmetry condition (53). Since the winding numbers of the eigenvalues vanish:

(64)1

2π[arg λj(t)]|L = 0, j = 1, 2,


we choose the solution to the problem (61) in the form

(65) χ(η, w) =1

2πi

∫

L

log λ(t, u)dW + β(η, w)− β(−η, w),

where the branches of the logarithmic functions log λj(η) = log |λj(η)| + iαj arefixed by the conditions −π ≤ αj ≤ π, j = 1, 2,

(66) β(η, w) =3∑

j=1

q1j∫

q0j

dW + mj

∮

aj

dW + nj

∮

bj

dW

,

mj and nj (j = 1, 2, 3) are integers to be fixed, q0j = (σ0j , w0j) ∈ C1, q1j =(σ1j , w1j) ∈ R, wmj = w(σmj), m = 0, 1. The points q0j are arbitrary fixedpoints in the first sheet, while the points q1j may lie in either sheet of the surfaceand are to be determined,

(67) dW (η) =w + u

2u

dt

t − η

is the Weierstrass kernel, and u = w(t), (t, u) ∈ L.

The contours aj and bj form a system of canonical cross-sections of the hyperel-liptic surface R. The cross-sections aj are closed contours formed by the branchcuts l−1 , l+1 and l+2 . The positive direction is chosen in the standard way suchthat the sheet C1 is on the left. The canonical cross-sections bj are chosen to be

(68) b1 = [−η2,−η1]C2∪ [−η1,−η2]C1

,

b2 = [−η2,−η1]C2∪[−η1,−k]+

C2∪[−k, η1]C2

∪[η1,−k]C1∪[−k,−η1]

−C1∪[−η1,−η2]C1

,

b3 = [−η2,−η1]C2∪ [−η1,−k]+

C2∪ [−k, η1]C2

∪ [η1, k]+C2

∪ [k, η2]C2

∪[η2, k]C1∪ [k, η1]

−C1

∪ [η1,−k]C1∪ [−k,−η1]

−C1

∪ [−η1,−η2]C1.

The loop bj crosses the contour aj from right to left and does not cross the otherloops am and bm (m 6= j).

To analyze the behavior of the function χ(η, w) at infinity, it is convenient towrite it in the form

(69) χ(η, w) = χ1(η) + wχ2(η),

where χ1(η) and χ2(η) are odd functions given by

(70) χ1(η) =η

πi

∞∫

0

ε+(t)dt

t2 − η2+ η

3∑

j=1

σ1j∫

σ0j

dt

t2 − η2,

(71) χ2(η) =η

πi

∞∫

0

ε−(t)dt√

f(t)(t2 − η2)+ η

3∑

j=1

q1j∫

q0j

dt

u(t)(t2 − η2)


a

a

a

b

k

k0

−η

η

η

η

1

1

2

3

−

1

1

2

3

b3b

2

−η2

−η3

Figure 1. Canonical cross-sections.

+mj

∮

aj

dt

u(t)(t2 − η2)+ nj

∮

bj

dt

u(t)(t2 − η2)

,

ε±(t) =1

2[log λ1(t) ± log λ2(t)].

Since dW (η) = O(η4), (η, w) → (∞,∞)j ∈ Cj , j = 1, 2, and χ2(η) = O(η−1),η → ∞, in general, the function χ(η, w) has an order-3 pole at the two infinitepoints of the surface, and the factors X±(η) have an inadmissible essential sin-gularity at the infinite point in the η-plane. For the function expχ(η, w) beingbounded as (η, w) → (∞,∞)j, j = 1, 2, it is necessary and sufficient that

(72)1

πi

∞∫

0

ts−1ε−(t)dt√

f(t)+

3∑

j=1

q1j∫

q0j

ts−1dt

u(t)+ mjAsj + njBsj

= 0, s = 1, 3,


where

(73) Asj =

∮

aj

ts−1dt

u(t), Bsj =

∮

bj

ts−1dt

u(t)

are the A- and B-periods of the abelian integrals

(74) ωs = ωs(η, w) =

(η,w)∫

(−η3,0)

ts−1dt

u(t), s = 1, 2, 3.

It will be convenient to add a third equation to the system (72), equation (72)for s = 2 (it does not affect the properties of the function χ2(η)). This gives

(75)

3∑

j=1

[ωs(q1j) + mjAsj + njBsj ] = ρs, s = 1, 2, 3,

where

(76) ρs =

3∑

j=1

ωs(q0j) −1

πi

∞∫

0

ts−1ε−(t)dt√

f(t).

To reduce the nonlinear system (75) to the classical genus-3 Jacobi inversionproblem, we normalize the basis of abelian integrals. Let A be the matrix whoseelements are the A-periods Asj (s, j = 1, 2, 3) and A = A

sj be its inverse,

A−1. Denote the normalized (canonical) basis as ω = ωs (s = 1, 2, 3) with theA- and B-periods

(77) Asj =

∮

aj

dωs = δsj,

Bsj =

∮

bj

dωs =3∑

n=1

AsnBnj .

Here δsj is the Kronecker symbol. Thus, to eliminate the essential singularity ofthe solution, one needs to solve the following Jacobi problem:

Find three points q1j ∈ R and six integers mj and nj (j = 1, 2, 3) such that(78)

3∑

j=1

ωs(q1j) = es−ks−3∑

j=1

njBsj−mi ≡ es−ks (modulo the periods), s = 1, 2, 3,

where es = ρs + ks,

(79) ρs =

3∑

j=1

Asjρj , s = 1, 2, 3,


ks (s = 1, 2, 3) are the Riemann constants

(80) ks = −1 +s

2− 1

2

3∑

j=1

Bsj .

The unknown points q1j (j = 1, 2, 3) coincide with the three zeros of the RiemannTheta function [13], [20]

(81) Θ(q) = θ(ω1(q) − e1, ω2(q) − e2, ω3(q) − e3) =

∞∑

t1,t2,t3=−∞

θt(q),

where

(82) θt(q) = exp

πi

3∑

r=1

3∑

s=1

Brstrts + 2πi

3∑

s=1

ts[ωs(q) − es]

, t = (t1, t2, t3).

Since the matrix Im B is positive definite, the series (81) converges exponentially.The affixes of the points q11, q12, and q13 can be expressed through the roots ofthe cubic equation [7]

(83) (ε31 − 3ε1ε2 + 2ε3)σ

3 + 3(ε2 − ε21)σ

2 + 6ε1σ − 6 = 0,

where the coefficients of the equation are

(84) εs = 23∑

j=1

3∑

r=1

Ajr

∫

γj

τ r−s−1dτ

f 1/2(τ)−

2∑

m=1

resq=0m∈Cm

η−sΘ′(q)

Θ(q),

γ1 = [−k,−η1]+, γ2 = [η1, k]+, γ3 = [η2, η3]

+, 0m = (0, (−1)m−1f 1/2(0)) ∈ Cm,and without loss of generality, at the two zero points of the Riemann surface theTheta function Θ(q) does not vanish. By computing the three roots of the cubic

equations we identify three pairs of points, (σ1s,√

f(σ1s)) and (σ1s,−√

f(σ1j))

(s = 1, 2, 3). However, one and only one point, q1s = (σ1s, (−1)m−1√

f(σ1s)) ∈Cm, is a zero of the Theta function Θ(q). To complete the solution of the Jacobiinversion problem, one needs to identify the integers ms and ns. The integers ns

solve the system of linear algebraic equations

(85)3∑

j=1

njIm(Bsj) = Im vs, s = 1, 2, 3,

and the integers ms are defined explicitly through the integers ns by

(86) ms = Re vs −3∑

j=1

njRe(Bsj).


Here,

(87) vs = es − ks −3∑

j=1

ωs(qj).

4.2. Symmetry condition. We have found the general form of the vectors

Ω±

0 (η) satisfying the boundary condition (52). It is

(88) Ω+

0 (η) =1

2νX+(η)r(η), η ∈ C

+,

Ω−

0 (η) = X−(η)r(η), η ∈ C−,

where r(η) is a rational vector-function to be determined. In general, this so-lution does not satisfy the symmetry condition (53). To meet this condition,we reconstruct the vector r(η). First, we rewrite the factors (62) in terms offunctions defined in the complex η-plane, not in the surface R, as(89)

X±(η) = eχ±

1(η)

[

cosh(f 1/2(η)χ±2 (η))I2 +

1

f 1/2(η)sinh(f 1/2(η)χ±

2 (η))Q(η)

]

,

where η ∈ C±, and I2 = diag1, 1. The functions χ±

1 and χ±2 possess the

symmetry property

(90) χ−j (−η) = −χ+

j (η), j = 1, 2,

and the integral (70) can be computed explicitly

(91) χ±1 (η) =

1

2log

3∏

j=1

(η + σ0j)(η − σ1j)

(η − σ0j)(η + σ1j)± 1

2log

η ± γ

η ± γ0, η ∈ C

±.

We are now ready to recover the rational vector-function r(η). Because of theproperty (90) we have(92)

X−(−η) = e−χ+

1(η)

[

cosh(f 1/2(η)χ+2 (η))I2 −

1

f 1/2(η)sinh(f 1/2(η)χ±

2 (η))Q(−η)

]

,

where η ∈ C+. Rewrite the symmetry condition (53) as

(93) r(η) = S(η)r(−η),

where

(94) S(η) = − η − γ

4νγ(η − γ0)(η + γ)

3∏

j=1

(η − σ0j)(η + σ1j)

(η + σ0j)(η − σ1j)

×(

−(γ2 − 4ν2ξ2) −4ν2ξ2

2(γ2 + 2ν2ξ2) γ2 − 4ν2ξ2

)

.


Recall that the functions Ω+j (η) (j = 3, 4) have simple zeros at the point η =

γ ∈ C+ and simple poles at the point η = γ0 ∈ C+. Therefore, the points η = γand η = γ0 are a simple zero and a simple pole, respectively, of the rationalvector r(η). Analysis of the function (91) and the singular integral (71) showsthat the matrix X(η) has simple zeros at the points η = σ1j and η = −σ0j . Thismeans that the components of the vector r(η), the functions r1(η) and r2(η),have simple poles at these points. At the geometric optics poles of the vector-functions Φ±(η), the vector r(η) cannot have poles, otherwise the symmetry

condition (93) fails. As η → ∞, the vectors Ω±(η) grow as in (38) (α = −1).

Having established all the properties of the rational vector r(η) we can write itsmost general form

(95) r(η) =(η − γ)p(η)

(η − γ0)∏3

j=1(η + σ0j)(η − σ1j)

where p(η) = (p1(η), p2(η))⊤, and p1(η) and p2(η) are degree-8 polynomials

(96) p1(η) =8∑

j=0

ajηj, p2(η) =

8∑

j=0

bjηj

with the coefficients to be determined from the symmetry condition (93) thatreduces to

(97) p(η) = S(η)p(−η),

where S(η) is a 2× 2 matrix. It satisfies the identity S(η) = [S(−η)]−1 and isgiven by

(98) S(η) =1

4νγ(η + γ0)

(

γ2 − 4ν2ξ2 4ν2ξ2

−2(γ2 + 2ν2ξ2) −(γ2 − 4ν2ξ2)

)

.

It is convenient to rewrite the vector equation (97) as

(99) p1(η) = 0, p2(η) = 0,

where

(100) p1(η) =

8∑

j=0

ηj[4νγ(η + γ0) − (−1)j(γ2 − 4ν2ξ2]aj − 4(−1)jν2ξ2bj

p2(η) =8∑

j=0

ηj[4νγ(η + γ0) + (−1)j(γ2 − 4ν2ξ2]bj + 2(−1)j(γ2 + 2ν2ξ2)aj

By equating the coefficients of the degree-10 polynomial p1(η) to zero we obtain11 equations. Only 10 of them are linearly independent. They express thecoefficients bj through aj (j = 0, 1, . . . , 8) as

(101) b8 = a8,

b7 = a7 − ν1a8,


b6 = a6 + ν1a7 + ν−a8,

b5 = a5 − ν1a6 − ν+a7 − ν1k2a8,

b4 = a4 + ν1a5 + ν−a6 + ν1k2a7 + ν−k2a8,

b3 = a3 − ν1a4 − ν+a5 − ν1k2a6 − ν+k2a7 − ν1k

4a8,

b2 = a2 + ν1a3 + ν−a4 + ν1k2a5 + ν−k2a6 + ν1k

4a7 + ν−k4a8,

b1 = a1 − ν1a2 − ν+a3 − ν1k2a4 − ν+k2a5 − ν1k

4a6 − ν+k4a7 − ν1k6a8,

b0 = a0 + ν1a1 + ν−a2 + ν1k2a3 + ν−k2a4 + ν1k

4a5 + ν−k4a6 + ν1k6a7 + ν−k6a8,

and a0 through aj (j = 1, 2, . . . , 8) as

(102) a0 = −ν+

ν1a1 − k2a2 −

ν+

ν1k2a3 − k4a4 −

ν+

ν1k4a5 − k6a6 −

ν+

ν1k6a7 − k8a8.

Here,

(103) ν1 =γ

ν, ν± =

γ(4γ0ν ± γ)

4ν2.

The polynomial p2(η) identically vanishes provided the coefficients a0 and bj (j =0, 1, . . . , 8) are determined by (101) and (102). Thus, the rational vector-functionr(η) has eight free parameters a1, a2, . . . , a8. In total, there are 13 parameters,Cj (j = 1, 2, 3), Dj (j = 1, 2) and aj (j = 1, 2, . . . , 8), to be determined.

Analysis of the function (91) and the singular integral (71) shows that the matrixX(η) has simple poles at the points η = −σ1j and η = σ0j (j = 1, 2, 3)

(104) X(η) ∼ χ∗(σj)

η − σjX∗(σj), η → σj , j = 1, 2, . . . , 6,

where σj = −σ1j and σj+3 = σ0j (j = 1, 2, 3), χ∗(η) is a function bounded at thepoints σj and having a certain nonzero limit at these points, and X∗(η) is a 2×2rank-1 matrix

(105) X∗(η) =

1 − l0(η)√f(η)

− l1(η)√f(η)

− l2(η)√f(η)

1 + l0(η)√f(η)

.

These six poles are removable singularities of the vectors Ω±(η) if and only if thefollowing six conditions are satisfied:

(106)

[

√

f(σj) − l0(σj)

]

p1(σj) − l1(σj)p2(σj) = 0, j = 1, 2, . . . , 6.

Since r(γ) = 0, the functions Ω+3 (η) and Ω+

4 (η) satisfy the condition (41) au-tomatically. The conditions (42) - (44) bring five additional equations for theunknown constants. One extra equation follows from the physical conditions (12)- (14) in Theorem 1. Finally, we verify the asymptotics of the functions Φ±

j (η)at infinity and derive the last, the 13th, equation. Since

(107) cosh(f 1/2(η)χ2(η)) ∼ 1,


sinh(f 1/2(η)χ2(η)) ∼ s0

η, η → ∞, s0 = const,

the matrix X+(η) admits the following asymptotic expansion:

(108) X+(η) ∼(

1 − 16ν4s0

ν0

16ν4s0

ν0

−16ν4s0

ν01 + 16ν4s0

ν0

)

− 8γν2s0

ν0η

(

0 11 0

)

, η → ∞,

and therefore

(109) Ω+(η) ∼ a8η

2

2ν

(

11

)

+η

2ν

(

d0

d0 − γνa8

)

, η → ∞,

where d0 is a constant. Thus, Ω+3 (η) − Ω+

4 (η) ∼ (2ν2)−1γa8η, η → ∞. Asη → ∞, the functions Φ±

j (η) (j = 1, 2, 4) vanish, Φ±j (η) = O(η−1), and the

functions Φ±3 (η) are bounded if and only if the condition (40) is satisfied. It

ultimately gives

(110) C3 = −a8.

This equation excludes the constant C3 from the representations (50) of thefunctions Ω1(η) and Ω2(η). The definition of the 12 constants Cj (j = 1, 2), Dj

(j = 1, 2) and aj (j = 1, 2, . . . , 8) from the 12 linear equations (12) - (14), (42) -(44) and (106) completes the solution of the Riemann-Hilbert problem (15).

5. Solution of the discontinuous boundary-value problem

We have established that the solution to the vector Riemann-Hilbert problem(15), the functions Φ±

j (η) (j = 1, 2, 3, 4), have the following asymptotics at in-finity:

(111) Φ±j (η) = O(η−1), Φ±

3 (η) = O(1), η → ∞, η ∈ C±.

Since

(112)

∞∫

−∞

Hy(0−, y)eiηydy = Hy+(0−, η) + Hy−(0−, η),

Hy+(0−, η) = i(kZ)−1Φ+2 (η) and Hy−(0−, η) = i(kZ)−1Φ−

3 (η), the asymptotics(111) implies that the function Hy(0

−, y) can be represented as

(113) Hy(0−, y) = H

y (0−, y) + Cδ(y),

where C = const, δ(y) is the generalized δ-function, and the function Hy (0−, y)

is bounded as y → 0. Also,

(114)

∞∫

0

Hy (0−, y)eiηydy = H

y+(0−, η) = O(η−1), η → ∞, η ∈ C+,


0∫

−∞

Hy(0

−, y)eiηydy = Hy−(0−, η) = O(η−1), η → ∞, η ∈ C

−.

This leads to the relations

(115) Hy+(0−, η) = Hy+(0−, η) =

i

kZΦ+

2 (η) = O(η−1), η → ∞, η ∈ C+,

Hy−(0−, η) = C + Hy−(0−, η) =

i

kZΦ−

3 (η) = C + O(η−1), η → ∞, η ∈ C−.

Finally, we write down the solution to the discontinuous boundary value problemfor the Helmholtz equation (5), (6). We have

(116) Ez(x, y) =1

2π

∞∫

−∞

Ez(x, η)e−iηydη,

where

(117) Ez(x, η) = A(η)e−ξx − 1

2ξ

∞∫

0

F (x1, η)e−ξ|x−x1|dx1,

A(η) = Ez+(0+, η) − iη

2ξEz+(iξ, 0+) +

ikZ

2ξHx+(iξ, 0+),

F (x, η) = −iηEz(x, 0+) + ikZHx(x, 0+).

By using the relations (8) and (10) we express the functions A(η), Ez(x, 0+)and Hx(x, 0+) through the solution of the Riemann-Hilbert problem (15). Thefunction A(η) is given by

(118) A(η) = −γ − iνξ

2γξΦ+

1 (η) − iν

2γΦ+

2 (η),

while the other two functions are the inverse Laplace transforms

(119) Ez(x, 0+) =1

2πi

+i∞∫

−i∞

Ez(iξ, 0+)e−ξxdξ,

Hx(x, 0+) =1

2πi

+i∞∫

−i∞

Hx(iξ, 0+)e−ξxdξ,

and

(120) Ez(iξ, 0+) = − i

2ηγ(γ + iνξ)[Φ+

1 (η) + Φ−1 (η)] − iνξ[Φ+

2 (η) + Φ−2 (η)],

Hx(iξ, 0+) =

i

2γkZ(γ + iνξ)[Φ+

1 (η) − Φ−1 (η)] − iνξ[Φ+

2 (η) − Φ−2 (η)].


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Yuri A. Antipov E-mail: [email protected]: Department of Mathematics, Louisiana State University, LA 70803 Baton Rouge,

United States

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