Файл: Ufimtsev P. Fundamentals of the physical theory of diffraction (Wiley 2007)(348s) PEo .pdf

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ВНИМАНИЕ! Если данный файл нарушает Ваши авторские права, то обязательно сообщите нам.

52 Chapter 2

Wedge Diffraction: Exact Solution and Asymptotics

 

 

and obtain

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

 

π

 

 

∞

 

 

 

 

 

 

e−krs

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

v(kr, ψ )

 

n

ei(kr+π/4)

f (s, ψ )

 

 

 

ds,

(2.90)

 

 

 

 

 

 

 

 

 

= − nπ √

 

−∞

(s2 − s12)(s2 − s22)

 

2

 

 

 

 

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f (s, ψ )

=

 

 

(cos ζ + cos ψ )[cos ζ + cos(2α − ψ )]

.

 

(2.91)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n −

 

ψ

n

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

cos

 

+

 

 

cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This function is finite and continuous in the vicinity of the saddle point s = 0 and therefore it can be expanded into the Taylor series. The integration of this series in Equation (2.90) leads to the uniform asymptotic expansion for v(kr, ψ ) valid for any angles ϕ and ϕ0. One should note that only the even terms with factors s2m(m = 1, 2, 3, . . .) give nonzero contributions to the integral in Equation (2.90). We retain and calculate here the two first terms of this asymptotic expansion. The second term is retained because for α → 2π it partially contains the quantity of the same order as the first term. The related asymptotic expression for the function v(kr, ψ ) is determined by

 

 

 

 

 

 

 

v(kr, ψ ) = v1(kr, ψ ) + v2(kr, ψ )

 

 

 

 

 

 

 

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

∞

 

 

e−krs

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ds

 

 

 

 

 

 

 

 

v

(kr, ψ )

n

ei(kr+π/4)f (0, ψ )

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= − nπ

√

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

2

 

 

 

 

 

 

−∞ (s2 − s12)(s2 − s22)

 

 

 

 

 

 

 

sin

π

 

 

 

d2f (0,

ψ ) ∞

 

 

 

2

s2ds

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

v

(kr, ψ )

 

 

 

 

 

ei(kr+π/4)

e−krs

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= − nπ 2√

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

2

 

 

 

ds2

 

−∞ (s2 − s12)(s2 − s22)

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

∞

e−krs2

 

 

 

 

 

 

 

 

 

 

2

 

 

∞ e−krs2 ds

 

 

 

∞ e−krs2 ds

,

−∞

 

 

ds =

 

 

 

 

− 0

 

 

 

(s2 − s12)(s2 − s22)

(s12 − s22)

0

(s2 − s12)

 

 

(s2 − s22)

∞

e−krs2 s2

 

 

 

 

 

 

 

 

 

 

2

 

 

∞ e−krs2 s2ds

 

 

∞ e−krs2 s2ds

 

−∞

 

ds =

 

 

 

 

− 0

 

 

 

(s2 − s12)(s2 − s22)

(s12 − s22)

0

(s2 − s12)

 

 

 

(s2 − s22)

(2.92)

(2.93)

(2.94)

(2.95)

,

 

 

 

 

 

 

 

 

 

 

 

(2.96)

∞ e−krs2 s2

∞ 2

∞

 

e−krs2

 

 

 

 

 

ds =

e−krs ds + s1,22

 

 

 

 

ds.

(2.97)

0 s

2

2

0

s

2

2

 

− s1,2

0

 

− s1,2

 

TEAM LinG

2.6 Uniform Asymptotics: Extension of the Pauli Technique 53

According to Equation (2.73), these integrals are reduced to the Fresnel integrals. As a result we obtain the following asymptotic expression

 

 

 

 

 

 

 

 

1

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

e−i π4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

v(kr, ψ ) =

 

 

 

 

n

 

n

 

 

 

 

 

 

·

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

·

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

√

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos π

 

 

 

cos ψ

 

 

sin α sin(α − ψ )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P(α,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

∞ cos ψ2

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ ) cos

3

 

 

ψ

e−

ikr cos

ψ

 

 

 

 

iq2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

√

 

 

 

 

 

cos ψ

 

e

 

 

dq

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2kr

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

α −

ψ

 

 

 

 

 

ikr cos(2α

 

 

 

ψ )

 

∞ cos(α− ψ2 )

 

 

 

 

iq2

dq

 

 

 

 

 

 

− Q(α, ψ ) cos

 

 

 

e−

 

 

 

 

 

 

 

 

 

 

 

−

 

 

 

 

 

 

 

√

 

 

 

cos(α− ψ2 ) e

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2kr

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(2.98)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 + sin2

 

n − cos

n cos

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ

 

 

 

2

 

 

 

 

 

4

 

 

 

P(α, ψ ) = 2 − cos α −

 

 

 

1 + n2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ

 

 

 

π

 

 

ψ

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos π

 

ψ 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(2.99)

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4 1 + sin2

 

ψ

 

− cos

π

 

 

ψ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

n

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q(α, ψ ) = 2 − cos

2

 

 

 

+ n2

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

−

 

 

 

 

 

 

n

 

2

 

 

 

 

 

 

(2.100)

 

 

 

 

1

 

 

 

cos π

 

 

 

 

 

 

 

cos ψ

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

One can show that away from the boundaries of reflected plane waves, where

√

 

 

ψ

 

 

 

 

 

 

 

 

√

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos 2

1 and

 

 

 

cos(α

 

 

 

2 )

 

 

 

 

1, the asymptotics (2.98) transforms into

 

kr

 

kr

−

 

 

 

 

 

the ray-type

asymptotics

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

sin

π

 

 

 

 

ei(kr+π/4)

 

 

 

1

 

sin

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ )

 

 

 

n

n

 

 

 

 

 

 

 

n

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

v(kr,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

√

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

ψ

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

− cos

 

 

2π kr

 

 

 

− cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

n

 

 

ei(kr−π/4)

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

1 + sin2

 

n

− cos n cos

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 +

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ψ

 

 

 

 

 

 

π

 

 

 

 

ψ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

×

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4√

 

(2kr)3/2

.

(2.101)

 

 

 

 

 

n2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

 

π

 

 

 

cos

ψ

 

 

 

 

 

 

 

 

 

 

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

nn

TEAM LinG

54 Chapter 2 Wedge Diffraction: Exact Solution and Asymptotics

We emphasize that only the first term here can be interpreted in terms of diffracted rays, and the second term, as well as all high-order asymptotic terms not presented in Equation (2.101), have a pronounced wave nature.

At the boundaries of the reflected waves, the function (2.98) is discontinuous,

v(kr, π ± 0) = v(kr, 2α − π 0)

= ±

1

 

ikr

1

 

π

ikr cos 2α e−iπ/4

∞ sin α

iq2

 

 

e

 

−

 

cot

 

sin αe

 

√

 

 

√

 

sin α e

dq, (2.102)

2

 

n

n

 

 

 

π

 

 

 

 

 

 

 

2kr

 

and compensates the discontinuities in the geometrical optics part of the total field.

√

 

|sin α|

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Under the condition

2kr

1, it follows from Equation (2.102) that

 

 

 

 

1

 

ikr

1

 

π ei(kr+π/4)

 

v(kr, π ± 0) = v(kr, 2α − π 0) = ±

 

e

 

−

 

cot

 

 

√

 

 

.

(2.103)

2

 

2n

n

 

 

2π kr

 

It is easy to check that the

asymptotics (2.98)

provides the

correct result

v(kr, ψ ) = 0 in the limiting case α = π , when the wedge transforms into the infinite plane and the diffracted field vanishes. Taking into account that P(2π , ψ ) = Q(2π , ψ ) = 0 and applying L’Hospital’s rule, one can show that, in the other limiting case when α → 2π , the asymptotics (2.98) converts into the function (2.82) related to the exact solution of the half-plane diffraction problem.

Now let us estimate the accuracy of asymptotics (2.98). This expression represents the sum of the two first terms in the asymptotic series resulting from the term-by-term integration of the Taylor series for the integrand in Equation (2.90). Therefore, the error of Equation (2.98) is the magnitude of the order of the neglected third asymptotic term. It is determined by the integral

∞ e−krs2 s4

 

O

(kr)−3/2

]

with s1,2

=

0

 

 

 

 

 

 

ds

[

 

 

 

 

 

 

(2.104)

 

 

2

2

 

5/2

 

 

 

 

−∞ s

 

 

(kr)−

 

with √kr

 

s1,2

 

1,

 

− s1,2

= O

 

]

 

 

 

 

 

 

 

 

[

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where the values s1,2 = 0 relate to the geometrical optics boundaries of the incident and reflected waves. Note also that the asymptotic expressions for the total field based on Equation (2.98) satisfy the boundary conditions√ (2.2) and (2.3) not rigorously, but only asymptotically under the condition kr|s1,2| 1.

In Equation (2.104) we have used the symbol O[(kr)m] to show the behavior of the integral under the condition kr 1. This is an ordinary definition accepted in the asymptotic theory: The expression f (x) = O(xm) means that lim[ f (x)/xm] = const with x → ∞. This common asymptotic terminology is used throughout the book.

As was mentioned in the previous section, the Pauli asymptotics (2.75) is not invariant with respect to the choice of the coordinate system. A similar situation happens with asymptotic expressions (2.98):

v[kr, α − ϕ − (α − ϕ0)] =v(kr, ϕ − ϕ0)

(2.105)

and

 

v[kr, α − ϕ + (α − ϕ0)] = v(kr, ϕ + ϕ0).

(2.106)

 

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2.7 Comments on Alternative Asymptotics 55

In addition, the asymptotics (2.98) does not satisfy the reciprocity principle:

v(kr, ϕ − ϕ0) =v(kr, ϕ0 − ϕ).

(2.107)

However, these shortcomings are admissible for asymptotic expressions, which satisfy the rigorous laws only approximately. Deviations from these laws are asymptotically small and they are beyond the accuracy of these asymptotics. For example, asymptotics (2.102) and (2.103) for the field at the geometrical optics boundaries are indeed invariant with respect to the choice of polar coordinates and according to estimations (2.104) their error is a small value of the order of (kr)−3/2. In addition, the ray-type asymptotics (2.101) is invariant in the same sense and it rigorously satisfies the reciprocity principle.

The found asymptotics (2.98) is convenient for field analysis. In the particular case when α = 2π and the wedge transforms into the half-plane, it provides the exact solution. This asymptotics is simple and applicable for all observation angles 0 ≤ ϕ ≤ α. However, because of its asymptotic origination, it satisfies the boundary conditions, reciprocity, and invariance principles not rigorously, but only approximately.

The asymptotics (2.98) represents the combination of the two first terms in the total asymptotic series for the function (2.90). Calculation of higher-order terms in this asymptotic series is straightforward, but we did not derive them because of their small practical value, as mentioned in the beginning of Section 2.5.

2.7COMMENTS ON ALTERNATIVE ASYMPTOTICS

Oberhettinger and Tuzhilin (Bowmam et al., 1987) derived alternative asymptotic solutions for the wedge diffraction problem. These authors used another technique to treat the singularities in the exact solution. They included into the rigorous integrals two special identical terms with opposite signs. One of these terms completely cancels the singularity of the original integrand. Therefore, a regular technique can be applied to derive the asymptotic expansion for this part of the field. The second additional term after integration generates the Fresnel integral that describes the basic features of the field in the vicinity of the geometrical optics boundaries. These asymptotics have a structure similar to that of Equation (2.80). The only difference is that in Equation (2.80) the Fresnel integral and the term 1/(ψ − π ) relate to diffraction at a black half-plane, and in the Oberhettinger and Tuzhilin expressions the similar quantities relate to diffraction at a perfectly reflecting half-plane.

The Oberhettinger asymptotics (Bowman et al., 1987) is the alternative to the Pauli expansion. Just like the Pauli expansion, it also fails in the direction of the plane wave reflected from the face ϕ = α and does not satisfy rigorously the invariance principle. The asymptotic solution by Tuzhilin (Bowman et al., 1987) is the alternative to the asymptotics (2.98). With an appropriate choice of different numbers n in Equations (6.21) and (6.25) of (Bowman et al., 1987), the Tuzhilin asymptotics can properly treat different singularities but do not satisfy rigorously the boundary conditions on the face ϕ = 0.

TEAM LinG

56 Chapter 2 Wedge Diffraction: Exact Solution and Asymptotics

Notice also that the Pauli technique was extended by Clemmow (1950) to the case when several poles of the integrand can approach the saddle point. However, he did not apply his theory to the wedge diffraction problem. Instead he utilized it for the investigation of scattering at a black half-plane that does not represent a boundary value problem.

This section studies the edge waves scattered from a wedge with semi-infinite planar faces. A review on edge waves arising in two-dimensional structures with curved concave and convex faces is given in the paper by Molinet (2005).

PROBLEMS

2.1The incident wave uinc = u0 exp[−ikr cos(ϕ − ϕ0)] generates the field inside a 2-D soft corner (a wedge with the angle α < π between its faces). Derive an exact expression for the total field inside the corner with the angle α = π/2 . Does this field contain the edge wave or not. Explain why?

2.2Solve the problem similar to Problem 2.1, but for the hard corner with α = π/2.

2.3The incident wave Ezinc = E0z exp[−ikr cos(ϕ − ϕ0)] generates the field inside a 2-D perfectly conducting corner (a wedge with the angle α < π between its faces). Derive an exact expression for the total field inside the corner with the angle α = π/2. Does this field contain the edge wave or not. Explain why?

2.4Solve the problem similar to Problem 2.3, but with the incident wave

Hzinc = H0z exp[−ikr cos(ϕ − ϕ0)].

2.5Use the exact solution (2.40) for a soft wedge (α > π ) and derive asymptotic approxi-

mations for the scattering sources js = ∂u/∂n induced on both faces of the wedge, close to the edge (kr 1).

(a) Apply the asymptotics of the Bessel functions:

J0(x) ≈ 1 − %

x

&

2

Jv (x) ≈

1

%

x

&

v

 

,

 

 

 

2

(v + 1)

2

 

(b) Derive the asymptotic expression for the source function js(kr, α). Realize the different behavior of this function for the angle α > π and α < π .

2.6Analyze the problem similar to Problem 2.5, but for a hard wedge.

2.7Explore the electromagnetic version of Problem 2.5 for the incident wave Ezinc = E0z exp[−ikr cos(ϕ − ϕ0)] exciting a perfectly conducting wedge. Analyze the surface currents close to the edge (kr 1).

2.8 Explore the problem similar to Problem 2.7, but for the incident wave Hzinc = H0z exp[−ikr cos(ϕ − ϕ0)] exciting a perfectly conducting wedge. Analyze the surface currents.

2.9Use the Sommerfeld ray asymptotics (2.61) and (2.63). Calculate the surface sources

js and jh far from the edge (kr 1). They are the waves running from the edge. At which wedge (soft or hard) do these waves decrease faster? Explain the origination of the phase factor exp(iπ/4).

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Problems 57

2.10Use the Sommerfeld ray asymptotics (2.61) and (2.63) for electromagnetic waves with E- and H-polarization, respectively. Calculate the surface currents far from the edge (kr 1). They are the waves running from the edge. Which waves (with component jz or jr ) decrease faster? Explain the origination of the phase factor exp(iπ/4).

2.11Find the first two terms of the asymptotc expansion (with p → ∞) for the integrals

∞

e

ix2

∞

e

ix2 dx

 

xdx,

 

 

 

with n ≥ 1.

p

p

 

xn

2.12An acoustic wave hits a wedge barrier (Fig. 2.7). Behind which barrier (soft or hard) is the intensity of the diffracted acoustic wave higher? Explore this problem utilizing the Sommerfeld ray asymptotics (2.61) and (2.63). Compute and plot the ratio

f (ϕ, ϕ0, α)/g(ϕ, ϕ0, α). Explore these two examples: (a) α = 315◦, ϕ0 = 45◦, 225◦ ≤ ϕ ≤ α,

(b) α = 355◦, ϕ0 = 45◦, 225◦ ≤ ϕ ≤ α.

Prove analytically that this ratio equals unity at the shadow boundary ϕ = 180◦ + ϕ0.

2.13An electromagnetic wave hits a perfectly conducting wedge (Fig. 2.7). Compare the intensity of edge diffracted waves with Ez- and Hz-polarization. Which of them is more intensive in the shadow region? Use the Sommerfeld ray asymptotics (2.61) and (2.63).

Compute and plot the ratio f (ϕ, ϕ0, α)/g(ϕ, ϕ0, α). Explore these two examples: (a) α = 300◦, ϕ0 = 45◦, 225◦ ≤ ϕ ≤ α,

(b) α = 360◦, ϕ0 = 45◦, 225◦ ≤ ϕ ≤ α.

Prove analytically that this ratio equals unity at the shadow boundary ϕ = 180◦ + ϕ0.

2.14Suppose that the incident wave Ezinc = E0z exp(ikx) undergoes grazing diffraction at a perfectly conducting wedge. This is a special case of diffraction illustrated in Figure 2.8 when ϕ0 → π . The problem is to calculate the surface currents on the face ϕ = 0. (The acoustic version of this problem is the grazing diffraction of a plane wave at a soft wedge.) In this case, the geometrical optics part of the field above the face ϕ = 0 equals zero. According to the exact solution (2.40), the total field in the vicinity of this face equals

Ez = E0z[v(kr, ϕ − π ) − v(kr, ϕ + π )]

with function v(kr, ψ ) defined by the integral (2.51). To calculate the surface current, complete the following steps:

(a) Write the integral expression for the current

 

jz = −Hr = Y0

i dEz

,

with ϕ = 0 and Y0 = 1/Z0,

 

kr

 

dϕ

(b)

In the integrand, replace the differential operator d/dϕ, by d/dζ , where ζ is the

 

integration variable,

 

 

(c)

Integrate by parts,

 

 

(d) Apply the stationary phase technique and show that

eikr+iπ/4

 

jz ≈ 2E0zY0 √2π kr

with kr 1.

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