Masujima M. Applied Mathematical Methods in Theoretical Physics

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7.3 General Decomposition Problem 257

number of z

= number of z

, number of p

= number of p

Thus the index of 1 −λ

K(k) onthereallineisequaltozero(ν ≡ 0) for

K(k) even.

Suppose we now lift the path above the real line (Im k = 0) into the upper half

plane. As the path C (Im k = A) passes by a zero of 1 −λ

K(k)inImk > 0, index ν

of 1 −λ

K(k) with respect to the path C (Im k = A) decreases by 1. This is because

the point k = z

is the zero in the upper half plane with respect to the path C

(Im k = A

−

) while it is the zero in the lower half plane with respect to the path C

(Im k = A

), and hence

ν = ν(z

) − ν(z

) =−

−

=−1. (7.3.19z)

Likewise, for a pole of 1 − λ

K(k)inImk > 0, we ﬁnd

ν = ν(p

) − ν(p

) =+

=+1. (7.3.19p)

Consider ﬁrst the case when index ν is equal to zero,

ν = 0. (7.3.20)

We choose a branch so that ln[1 −λ

K(ζ )] vanishes on C

↑

and C

↓

.Wehave

ln Y

−

(k) − ln Y

(k) =

2πi





1 −λ

K(ζ )



ζ − k

dζ −

2πi



−C



1 −λ

K(ζ )



ζ − k

dζ.

(7.3.21)

In the ﬁrst integral on the right-hand side of Eq. (7.3.21), we let k be anywhere

above C

where C

is arbitrarily close to Im k = k

=−a from above. Then

ln Y

(k) =−

2πi





1 −λ

K(ζ )



ζ − k

dζ ,Imk = k

> −a, (7.3.22)

is analytic in the upper half plane, and hence is identiﬁed to be a + function. It

also vanishes as



→∞in the upper half plane (Im k > −a).

In the second integral on the right-hand side of Eq. (7.3.21), we let k be anywhere

below −C

where −C

is arbitrarily close to Im k = k

=−a from above. Then

ln Y

−

(k) =−

2πi



−C



1 −λ

K(ζ )



ζ − k

dζ ,Imk = k

≤−a, (7.3.23)

is analytic in the lower half plane, and hence is identiﬁed to be a − function. It also

vanishes as



→∞inthelowerhalfplane(Imk ≤−a).

258 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation

Thus

(k) = exp[−

2πi





1 −λ

K(ζ )



ζ − k

dζ ], (7.3.24)

−

(k) = exp





−

2πi



−C



1 −λ

K(ζ )



ζ − k

dζ





. (7.3.25)

We also note that

(k) → 1as



→∞ in



Im k > −a,

Im k ≤−a.

(7.3.26)

ThentheentirefunctionG(k) in the following equation:

−

(k)

−

(k) =−Y

(k)

(k) = G(k),

must vanish identically by Liouville’s theorem . Hence

−

(k) = 0orφ(x) = 0whenν = 0. (7.3.27)

Consider next the case when index ν is positive,

ν>0. (7.3.28)

Instead of dealing with C

↑

and C

↓

of the integral (7.3.13), we construct the object

whose index is equal to zero,

i=1



k − z

(i)

k − p

(i)





1 −λ

K(k)



= Z

−

(k)/Z

(k). (7.3.29)

Here z

(i) is a point in the lower half plane (Im k ≤ A) which contributes −ν/2in

its totality (i = 1,..., ν) to the index and p

(i) is a point in the upper half plane

(Im k > A) which contributes −ν/2 in its totality (i = 1,..., ν)totheindex.Then

expression (7.3.29) has the index equal to zero with respect to Im k = A,

−

(from z

(i)



s) −

(from p

(i)



s) +ν(from 1 −λ

K(k)) = 0. (7.3.30)

By factoring of Eq. (7.3.29), using Eqs. (7.3.24) and (7.3.25), we obtain

−

(k) = exp







−

2πi



−C



(1 −λ

K(ζ ))

i=1



ζ −z

(i)

ζ −p

(i)





ζ − k

dζ







, (7.3.31)

7.3 General Decomposition Problem 259

(k) = exp







−

2πi





(1 −λ

K(ζ ))

i=1



ζ −z

(i)

ζ −p

(i)





ζ − k

dζ







, (7.3.32)

with the following properties:

(1) Z

(k) → 1as



→∞,

(2) Z

−

(k)(Z

(k)) is analytic in the lower (upper) half plane,

(3) Z

−

(k)(Z

(k)) has no zero in the lower (upper) half plane.

We write Eq. (7.3.29) as

1 −λ

K(k) =

−

(k)

−

(k)

i=1

(k − p

(i))

i=1

(k − z

(i))

. (7.3.33)

By the prescription stated in Eq. (7.2.17), we obtain

−

(k) = Z

−

(k) ·

i=1

(k − p

(i)), (7.3.34)

(k) = Z

(k) ·

i=1

(k − z

(i)). (7.3.35)

We observe that

(k) → k



→∞. (7.3.36)

Thus the entire function G(k) in the following equation:

−

(k)

−

(k) =−Y

(k)

(k) = G(k),

cannot grow as fast as k

as k →∞. By Liouville’s theorem , we have

G(k) =

ν−1



j=0

,0≤ j ≤ ν − 1, (7.3.37)

where C

’s are arbitrary ν constants. Then we obtain

−

(k) = G(k)/Y

−

(k) =

ν−1



j=0

−

(k), Im k ≤ A. (7.3.38)

Inverting this expression along Im k = A,weobtain

φ(x) =

2πi



+∞+iA

−∞+iA

dke

ikx

−

(k) =

ν−1



j=0

(j)

(x), when ν>0, (7.3.39)

260 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation

where

(j)

(x) =

2πi



+∞+iA

−∞+iA

dke

ikx

−

(k), j = 0, ..., ν − 1. (7.3.40)

We have ν independent homogeneous solutions, φ

(j)

(x), j = 0,..., ν −1, which are

related to each other by differentiation,



−i



(j)

(x) = φ

(j+1)

(x), 0 ≤ j ≤ ν −2.

Thus it is sufﬁcient to compute φ

(0)

(x),

(j)

(x) =



−i



(0)

(x), j = 0, 1, ..., ν − 1. (7.3.41)

Differentiation under the integral, Eq. (7.3.40), is justiﬁed by

j+1

−

(k)

→

j+1

→ 0ask →∞, j ≤ ν − 2.

Consider thirdly the case when index ν is negative,

ν<0. (7.3.42)

As before, we construct the object whose index is equal to zero:

i=1

(k − z

(i))

(k − p

(i))



1 −λ

K(k)



= Z

−

(k)/Z

(k), (7.3.43)

which has indeed index zero as shown below.

(fromz

(i)



s) +

(fromp

(i)



s) + ν(from1 − λ

K(k)) = 0. (7.3.44)

We apply the factorization to the left-hand side of Eq. (7.3.43). Then we write

1 −λ

K(k) =

−

(k)

−

(k)

i=1

(k − p

(i))

i=1

(k − z

(i))

. (7.3.45)

By the prescription stated in Eq. (7.2.17), we obtain

−

(k) = Z

−

(k)/

i=1

(k − z

(i)), (7.3.46)

(k) = Z

(k)/

i=1

(k − p

(i)). (7.3.47)

7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 261

Then we have

(k) → 1andY

(k) → 1/k

,ask →∞. (7.3.48)

Thus the entire function G(k) in the following equation:

−

(k)

−

(k) =−Y

(k)

(k) = G(k),

must vanish identically by Liouville’s theorem. Hence we obtain

φ(x) = 0, when ν<0. (7.3.49)

7.4

Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind

Let us consider the inhomogeneous Wiener–Hopf integral equation of the second kind,

φ(x) = f (x) + λ



+∞

K(x − y)φ(y)dy, x ≥ 0, (7.4.1)

where we assume as in Section 7.3 that the asymptotic behavior of the kernel K(x)

is given by

K(x) ∼



O(e

)asx →−∞,

O(e

−bx

)asx →+∞,

a, b > 0, (7.4.2)

and the asymptotic behavior of the inhomogeneous term f (x)isgivenby

f (x) → O(e

)asx →+∞. (7.4.3)

We deﬁne ψ(x)forx < 0asbefore

ψ(x) = λ



+∞

K(x − y)φ(y)dy, x < 0. (7.4.4)

We take the Fourier transform of φ(x)andψ(x)forx ≥ 0andx < 0 and add the

results together,

−

(k) +

(k) =

−

(k) + λ

K(k)

−

(k), (7.4.5)

where

K(k) is analytic inside the strip,

−a < Im k = k

< b. (7.4.6)

262 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation

Difﬁculty may arise when the inhomogeneous term f (x)growstoofastasx →∞

so there may not exist a common region of analyticity for Eq. (7.4.5) to hold. The

Fourier transform

−

(k)isdeﬁnedby

−

(k) =



+∞

dxe

−ikx

f (x), (7.4.7)

where



−ikx

f (x)



∼ e

+c)x

as x →∞ with k = k

+ ik

That is,

−

(k) is analytic in the lower half plane,

Im k = k

< −c. (7.4.8)

We require that a and c satisfy

a > c. (7.4.9)

In other words, f (x) grows at most as fast as

f (x) ∼ e

(a−ε)x

, ε>0, as x →∞.

We try to solve Eq. (7.4.5) in the narrower strip,

−a < Im k = k

< min(−c, b). (7.4.10)

Writing Eq. (7.4.5) as

(1 −λ

K(k))

−

(k) =

−

(k) −

(k), (7.4.11)

we are content to obtain one particular solution of Eq. (7.4.1). We factor 1 −λ

K(k)

as before,

1 −λ

K(k) = Y

−

(k)/Y

(k). (7.4.12)

Thus we have from Eqs. (7.4.11) and (7.4.12)

−

(k)

−

(k) = Y

(k)

−

(k) − Y

(k)

(k), (7.4.13)

where Y

−

(k)

−

(k) is analytic in the lower half plane and Y

(k)

(k)isanalyticin

the upper half plane. We split Y

(k)

−

(k) into a sum of two functions, one analytic

in the upper half plane and the other analytic in the lower half plane,

7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 263

=−

↑

↓

Fig. 7.15 Region of analyticity and the integration contour

C for

F(k)insidethestrip,−a < Im k < min(−c, b).

(k)

−

(k) = (Y

(k)

−

(k))

+ (Y

(k)

−

(k))

−

Inordertodothis,wemustconstructY

(k)suchthat

(k)

−

(k) → 0ask →∞,

(k) → constant as k →∞. (7.4.14)

Suppose

F (k) is analytic inside the strip,

−a < Im k = k

< min(−c, b). (7.4.15)

By choosing the contour C inside the strip as in Figure 7.15, we apply the Cauchy

integral formula.

F(k) =

2πi



F(ζ )

ζ − k

dζ =

2πi



F(ζ )

ζ − k

dζ −

2πi



−C

F(ζ )

ζ − k

dζ. (7.4.16)

By the same argument as in the previous section, we identify

−

(k) =−

2πi



−C

F(ζ )

ζ − k

dζ , (7.4.17)

(k) =

2πi



F(ζ )

ζ − k

dζ. (7.4.18)

264 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation

Thus, under the assumption that Y

(k) satisfy the above-stipulated condition

(7.4.14), we obtain

(k)

−

(k))

−

=−

2πi



−C

(ζ )

−

(ζ )/(ζ − k)]dζ , (7.4.19)

(k)

−

(k))

2πi



(ζ )

−

(ζ )/(ζ − k)]dζ. (7.4.20)

Then we write

−

(k)

−

(k) −(Y

(k)

−

(k))

−

= (Y

(k)

−

(k))

− Y

(k)

(k) ≡ G(k), (7.4.21)

where G(k)isentireink. If we are looking for the most general homogeneous

solutions, we set

−

(k) ≡ 0 and determine the most general form of the entire

function G(k). Now we are just looking for one particular solution to the inhomogeneous

equation,sothatwesetG(k) = 0. Then we have

−

(k) = (Y

(k)

−

(k))

−

(k), (7.4.22)

(k) = (Y

(k)

−

(k))

(k). (7.4.23)

Choices of Y

(k) for an inhomogeneous solution are different from those for a

homogeneous solution. In view of Eqs. (7.4.22) and (7.4.23), we require that

(1) 1/Y

−

(k) is analytic in the lower half plane,

Im k < c



, −a < c



< b, (7.4.24)

and 1/Y

(k) is analytic in the upper half plane,

Im k > c



, −a < c



< b. (7.4.25)

(2)

−

(k) → 0,

(k) → 0ask →∞. (7.4.26)

According to this requirement, Y

−

(k) for an inhomogeneous solution can have a

pole in the lower half plane. This is all right because then 1/Y

−

(k)hasazeroin

the lower half plane. Once requirements (1) and (2) are satisﬁed,

−

(k)and

(k)

given by Eqs. (7.4.22) and (7.4.23) are analytic in the respective half plane. Then we

construct the following expression from Eqs. (7.4.22) and (7.4.23):

[1 −λ

K(k)]

−

(k) =

−

(k) −

(k). (7.4.27)

7.4 Inhomogeneous Wiener–Hopf Integral Equation of the Second Kind 265

Inverting for x ≥ 0, we obtain

φ(x) −λ



+∞

K(x − y)φ(y)dy = f (x), x ≥ 0.

Thus

−

(k)and

(k) derived in Eqs. (7.4.22) and (7.4.23) under the requirements

(1) and (2) do provide a particular solution to Eq. (7.4.1).

Case 1:Indexν = 0.

When index ν of 1 − λ

K(k) with respect to the line Im k = A is equal to zero,no

nontrivial homogeneous solution exists. From Eqs. (7.3.22) and (7.3.23), we have

1 −λ

K(k) = Y

−

(k)/Y

(k), (7.4.28)

where

−

(k) = exp





−

2πi



−C

dζ



1 −λ

K(ζ )



ζ − k





,Imk ≤ A, (7.4.29)

(k) = exp





−

2πi



dζ



1 −λ

K(ζ )



ζ − k





,Imk > A, (7.4.30)

and

(k) → 1as



→∞. (7.4.31)

Since Y

(k)(Y

−

(k)) has no zeros in the upper half plane (the lower half plane),

1/Y

(k)(1/Y

−

(k)) is analytic in the upper half plane (the lower half plane). Then

we have

(k)

−

(k) → 0ask →∞,

so that Y

(k)

−

(k) can be split into the + part and the − part as in Eqs. (7.4.19) and

(7.4.20).

Using Y

(k)givenfortheν ≡ 0 case, Eqs. (7.4.29) and (7.4.30), we construct a

resolvent kernel H(x , y). Since 1/Y

−

(k) is analytic in the lower plane and approaches

to 1 as k →∞,wedeﬁney

−

(x)by

−

(k)

− 1 ≡



+∞

dxe

−ikx

−

(x), (7.4.32)

266 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation

where the left-hand side is analytic in the lower half plane and vanishes as k →∞.

Inverting Eq. (7.4.32) for y

−

(x), we have

−

(x) =



+∞+iA

−∞+iA

2π

ikx



−

(k)

− 1



for x ≥ 0, (7.4.33)

−

(x) = 0forx < 0. (7.4.34)

Similarly, we deﬁne y

(x)by

(k) −1 ≡



−∞

dxe

−ikx

(x), (7.4.35)

where the left-hand side is analytic in the upper half plane and vanishes as k →∞.

Inverting Eq. (7.4.35) for y

(x), we have

(x) =



+∞+iA

−∞+iA

2π

ikx

(k) − 1] for x < 0, (7.4.36)

(x) = 0forx ≥ 0. (7.4.37)

We deﬁne

(k)by

−

(k)

≡ 1 +



+∞

dxe

−ikx

−

(x) ≡ 1 +

−

(k), (7.4.38)

(k) ≡ 1 +



−∞

dxe

−ikx

(x) ≡ 1 +

(k). (7.4.39)

Then

−

(k) given by Eq. (7.4.23) becomes

−

(k) =

−

(k)

−

(k))

−

= (1 +

−

(k))(

−

(k) +

(k)

−

(k))

−

(k) +

−

(k)

−

(k) + (

(k)

−

(k))

−

(k)(

(k)

−

(k))

−

. (7.4.40)

Inverting Eq. (7.4.40) for x > 0,

φ(x) = f (x) +



+∞

−

(x − y)f (y)dy +



+∞

(x − y)f (y)dy



+∞

−

(x − z)dz



+∞

(z − y)f (y)dy

= f (x) +



+∞

H (x, y)f (y)dy, x ≥ 0, (7.4.41)