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7.4 Transformation of Triple Integrals

181

When all poles of the integrand of Ju are inside the region closed by the contour C = D + F+ + F−, the Cauchy theorem states that

4

Ju(C) = 2π i

Resm.

(7.46)

 

m=1

 

Here, the quantities Res1 and Res2 are the residues at the poles η1 = ϕ0 and η2 = −ϕ0:

 

Res1

= Res2 =

α

·

ε(ϕ0) sin ϕ0

 

 

 

 

 

 

 

 

 

,

with

 

(7.47)

 

iπ

p2

+

k2 cos ϕ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1,

if

 

0 ≤ x ≤ π

 

 

 

 

 

 

 

 

ε(x) = 0,

if

 

π < x,

 

 

 

 

 

 

 

(7.48)

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

[w−1(σ

− ϕ0) − w−1(σ + ϕ0)],

 

 

 

Res3

=

 

 

(7.49)

 

k2

 

 

 

 

 

 

1

[w−1(−σ − ϕ0) − w−1(−σ + ϕ0)]

 

 

 

Res4

=

 

 

(7.50)

 

k2

 

are the residues at the poles η3 and η4. In addition,

 

 

 

 

 

 

3

+

 

 

4 = k2

 

 

 

 

 

2α

 

 

−

 

 

2α

 

Res

 

 

Res

 

 

 

i

 

cot

 

π(σ

− ϕ0)

 

 

cot

π(σ + ϕ0)

.

(7.51)

 

 

 

 

 

 

 

 

 

 

 

 

 

One can verify that the residues Res3 and Res4 tend to zero when Im(σ ) → ±∞. Notice also that for certain large values of the parameter | p|, the poles η3 and η4

can reach the contours F+ and F−. When they are exactly on these lines, the integral Ju(C) is determined as

Ju(C) = 2π i Res1 + Res2 +

1

(Res3

+ Res4) .

(7.52)

2

Our goal is to calculate the integral Ju(D) over the contour D = C − (F+ + F−) when the contours F+ and F− are shifted to infinity (Im(η) = ±∞). According to Equations (7.46) and (7.52), the integral Ju(D) is determined by

4

Ju(D) = 2π i

Resm − Ju(F+ + F−), if |Im(σ )| < A,

(7.53)

m=1

 

 

 

 

and

 

 

 

 

 

Ju(D) = 2π i Res1 + Res2 +

1

(Res3

+ Res4) − Ju(F+ + F−),

 

 

 

2

 

if

|Im(σ )| = A,

 

 

 

(7.54)

TEAM LinG

Ju(F+) =

 

lim

 

2π i

·

1

 

 

+

Res

)

 

 

 

 

 

(Res

 

 

 

 

σ →+i∞

 

 

2

 

3

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

δ+iA

 

−π +δ+iA

 

−3π/2+iA

 

 

 

 

 

 

 

 

lim

lim

π/2+iA + −δ+iA

 

+ −π −δ+iA

 

 

 

 

 

 

+ A→∞

δ→0

 

 

 

 

 

 

 

×

 

w−1(η + ϕ0) − w−1(η − ϕ0) sin η dη

(7.61)

 

 

 

 

 

 

 

 

 

p2 + k2 cos η

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ju(F−) = σ

 

limi

 

 

2π i ·

1

(Res3 + Res4)

 

 

 

182 Chapter 7

 

 

 

 

→−

∞

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−δ−iA

 

 

π −δ−iA

3π/2−iA

 

 

 

 

 

 

lim

 

lim

 

 

 

 

+ +δ−iA

+ π +δ−iA

 

 

 

 

+ A→∞

δ→0 −π/2−iA

 

 

 

 

×

w−1

(η + ϕ0) − w−1(η − ϕ0) sin η dη .

(7.62)

 

 

 

 

 

 

p2

+ k2 cos η

 

 

 

 

 

 

Elementary Acoustic and Electromagnetic Edge Waves

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ju(F+ + F−) = Ju(F+) + Ju(F−)

 

(7.55)

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

J

(F

+

)

=

−3π/2+iA w−1(η + ϕ0) − w−1(η − ϕ0) sin η dη,

(7.56)

u

 

 

π/2+iA

 

 

 

 

p2 + k2 cos η

 

 

 

J (F

−

)

=

3π/2−iA

w−1(η + ϕ0) − w−1(η − ϕ0) sin η dη.

(7.57)

u

 

 

−π/2−iA

 

 

p2 + k2 cos η

 

 

 

 

When the poles η3 and η4 are inside the region closed by the contour D + F+ +

F− and A → ∞, the integrals Ju(F±) equal zero due to the relationships

 

 

 

 

 

w−1(η ±

 

 

 

1,

 

 

with Im(η)

−→ +∞.

(7.58)

 

 

 

 

ϕ0) −→ 0,

 

 

with Im(η)

w−1(η + ϕ0) − w−1(η − ϕ0) −→ 0,

 

 

 

 

 

 

−→ −∞

 

 

with Im(η) −→ ±∞,

(7.59)

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

tan η −→ i, i,

 

with Im(η)

−→ +∞.

(7.60)

 

 

 

 

 

with Im(η)

 

 

 

 

 

 

 

 

 

 

−

 

 

 

 

 

 

−→ −∞

 

In the case when the poles η3 and η4 are exactly on the contours F+ and F−,

and

Here, the terms with the double limits represent the principal values of the integrals Ju(F±). In view of Equations (7.59) and (7.60) and the relationship Res3,4 → 0 (when Im(η) → ±i∞), the integrals Ju(F±) again equal zero.

TEAM LinG

7.5 General Asymptotics for Elementary Edge Waves 183

Thus, for any positions of η3 and η4, it turns out that Ju(F+ + F−) = 0 with A → ∞. Because of that, and due to Equations (7.53) and (7.54), we obtain the following rigorous result

4

Ju(D) = 2π i

Resm,

(7.63)

 

m=1

 

where the residues are defined above by Equations (7.47) and (7.49), (7.50).

By application of the same procedure to the integral Jv (D), one can show that it also can be expressed in the form of Equation (7.63), but with other expressions for the residues. We omit all routine intermediate manipulations and bring the final expressions for the field integrals (7.32) and (7.33):

Iu =

1 ∞

i

 

−∞

 

 

 

 

+

2π

 

 

k2

 

 

 

(1)

(qh1) +

4αε(ϕ0) sin ϕ0

 

 

eipx1 H0

 

 

 

 

p2 + k2 cos ϕ0

,

 

 

 

2α

−

 

2α

 

cot

π(σ

+ ϕ0)

 

 

cot

π(σ − ϕ0)

dp

(7.64)

 

 

 

 

 

and

1

∞

(1)

(qh1) +−

 

4αε(ϕ0)

 

 

Iv =

 

 

−∞ eipx1 qH1

 

 

 

 

i

p2 + k2 cos ϕ0

,

 

 

+ k2 sin σ

 

 

2α

+

 

2α

 

 

 

 

2π

cot

π(σ + ϕ0)

 

cot

π(σ − ϕ0)

dp.

(7.65)

 

 

 

 

 

 

 

Integrals Iu,v determine the fields (7.25), (7.26) generated by elementary strip 1. In the next section we develop the asymptotic estimations for these fields.

7.5 GENERAL ASYMPTOTICS FOR ELEMENTARY EDGE WAVES

A main contribution to the integrals Iu,v is given by the vicinity of the stationary point pst . To determine this point we use the asymptotic expressions for Hankel functions

H0(1)

(qh1)

2

eiqh1

−iπ/4,

H1(1)(qh1)

2

eiqh1

−i3π/4

(7.66)

π qh1

π qh1

with qh1

1.

 

 

 

 

 

 

 

 

The integrands of Iu,v have the fast oscillating factor exp[i ( p)] = exp[i( px1 + qh1)]. The stationary point pst is found from the condition d ( p)/dp = 0, which

TEAM LinG

184 Chapter 7 Elementary Acoustic and Electromagnetic Edge Waves

leads to the equation

(

x1 k2 − p2st − h1pst = 0. (7.67)

By substituting here x1 = R cos β1 and h1 = R sin β1, one obtains pst = k cos β1. Now one could apply the standard procedure of the stationary phase technique to derive the asymptotic expressions for Iu,v , but we use the following idea, which leads to the same result, but faster.

Let us consider the integral

 

 

 

 

 

I0 =

∞ eipx1 H0(1)(qh1)F( p)dp,

(7.68)

 

 

 

 

 

 

 

−∞

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where F( p) is a slowly varying function. It is clear that

 

 

 

 

 

 

I0 ≈ F( pst )

 

 

∞ eipx1 H0(1)(qh1)dp,

(7.69)

 

 

 

 

 

 

 

 

 

 

 

−∞

 

 

 

 

 

 

 

 

 

and, according to Equation (7.21),

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I0 ≈ −2iF( pst )

eikR

,

 

 

 

(7.70)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

R

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where R =

 

x12 + h12.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In a

similar way one can evaluate the integral

 

 

 

 

 

 

(

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I1 =

∞ eipx1 qH1(1)(qh1)F( p)dp

 

 

 

 

 

 

−∞

∞

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= −

d

e

ipx

 

 

(1)

(qh1)F( p)dp

 

 

 

 

 

 

 

 

 

 

1 H0

 

 

 

 

 

dh1

−∞

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2iF( pst )

d

 

 

 

eikR

≈ −2F( pst )k

h1 eikR

 

 

 

 

 

dh1

 

R

 

R

 

R

 

 

 

 

 

= −2k sin β1F( pst )

eikR

 

 

 

 

 

 

 

 

 

 

 

 

,

 

 

 

 

 

(7.71)

 

 

 

 

R

 

 

 

 

 

 

with kR

 

1.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

There is another advantage of this approach. It is valid under the condition kR 1

and is free from the restriction qst h1 = kR sin β1

1 in the standard stationary

technique.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TEAM LinG

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