Masujima M. Applied Mathematical Methods in Theoretical Physics
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7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 247
Recalling the definition of the beta function B(n, m),
B(n, m) =
1
0
dρρ
n−1
(1 −ρ)
m−1
= (n)(m)/(n + m),
we have
ˆ
K(k) = 2
ik +
1
2
−ik +
1
2
/
ik +
1
2
−ik +
1
2
= 2
ik +
1
2
−ik +
1
2
.
Recalling the property of the gamma function,
(z)(1 −z) = π/ sin π z, (7.2.39)
we thus obtain the Fourier transform of K(x)as
ˆ
K(k) = 2π/ cosh π k. (7.2.40)
Defining ψ(x)by
ψ(x) = λ
+∞
0
1
cosh
1
2
(x − y)
φ(y)dy, x < 0, (7.2.41)
we obtain the following equation as usual:
(1 −λ
ˆ
K(k))
ˆ
φ
−
(k) =−
ˆ
ψ
+
(k), (7.2.42)
where
1 −λ
ˆ
K(k) = 1 −
2πλ
cosh π k
= Y
−
(k)/Y
+
(k), (7.2.43)
and the regions of the analyticity of
ˆ
φ
−
(k)and
ˆ
ψ
+
(k)aresuchthat
(i)
ˆ
φ
−
(k) is analytic in the lower half plane ( Im k ≤−1/2),
(ii)
ˆ
ψ
+
(k)isanalyticintheupperhalfplane(Imk > −1/2).
(7.2.44)
Rewriting Eq. (7.2.42) in terms of Y
±
(k)’s, we have
Y
−
(k)
ˆ
φ
−
(k) =−Y
+
(k)
ˆ
ψ
+
(k) ≡ G(k), (7.2.45)
where G(k)isentireink.
248 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation
Factorization of 1 − λ
ˆ
K(k):
Y
−
(k)/Y
+
(k) = 1 −
2πλ
cosh π k
=
cosh π k − 2πλ
cosh π k
. (7.2.46)
Case 1. 0 < 2πλ ≤ 1.
Setting
cos πα ≡ 2πλ,0≤ α<1/2, (7.2.47)
we have
Y
−
(k)/Y
+
(k) = [cos(iπ k) − cos πα]/ sin π
ik +
1
2
= 2sin[
π
2
(α + ik)] sin
π
2
(α − ik)
/ sin π
ik +
1
2
. (7.2.48)
Replacing all sine functions in Eq. (7.2.48) with the appropriate product of the
gamma functions through the use of the formula
sin π z = π/(z)(1 −z), (7.2.49)
we obtain
Y
−
(k)/Y
+
(k)
= 2π
1
2
+ ik
1
2
− ik
/
α + ik
2
α − ik
2
1 −
α + ik
2
1 −
α − ik
2
. (7.2.50)
We note that
(z) has simple pole at z = 0, −1, −2, ...,
(1 −z) has simple pole at z = 1, 2, 3, ....
(7.2.51)
(1)
1
2
+ ik
has simple poles at k = i
1
2
, i
3
2
, i
5
2
, i
7
2
, ..., all of which are assigned
to Y
−
(k).
(2)
1
2
− ik
has simple poles at k =−i
1
2
, −i
3
2
, −i
5
2
, −i
7
2
, ...,allofwhichare
assigned to Y
+
(k).
(3)
α+ik
2
has simple poles at k = iα, i(2 +α), i(4 +α), ...,allofwhichare
assigned to Y
−
(k).
(4)
α−ik
2
has simple poles at k =−iα, −i(2 +α), −i(4 +α), ....Since
0 <α<1/2, the first pole at k =−iα is assigned to Y
−
(k), while the
remaining poles are assigned to Y
+
(k). Using the property of the gamma
function, (z) = (z + 1)/z,werewrite
7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 249
α − ik
2
=
2
α − ik
1 +
α − ik
2
,
(5) where (α − ik)/2 is assigned to Y
−
(k)while(1 +
α−ik
2
) is assigned to Y
+
(k).
(6)
1 −
α+ik
2
has simple poles at k =−i(2 − α), −i(4 − α), −i(6 − α), ...,all
of which are assigned to Y
+
(k).
(7)
1 −
α−ik
2
has simple poles at k = i(2 −α), i(4 −α), i(6 −α), ...,allof
which are assigned to Y
−
(k).
Then we obtain Y
±
(k) as follows:
Y
−
(k) =−2π
1
2
+ ik
α + ik
2
−
α − ik
2
, (7.2.52a)
Y
+
(k) =
1 +
α − ik
2
1 −
α + ik
2
1
2
− ik
. (7.2.52b)
Now G(k) is determined by the asymptotic behavior of Y
±
(k)ask →∞.Making
use of the Duplication formula and the Stirling formula,
(2z) = 2
2z−1
(z)
z +
1
2
/
√
π, (7.2.53)
lim
|
z
|
→∞
(z + β)/(z) ∼ z
β
, (7.2.54)
we find the asymptotic behavior of Y
−
(k)tobegivenby
Y
−
(k) ∼−i
π
2
2
ik
· k. (7.2.55)
Defining
Z
±
(k) ≡ 2
−ik
· Y
±
(k), (7.2.56)
we find
Z
±
(k) ∼−i
π
2
k, (7.2.57)
since
Z
−
(k)/Z
+
(k) = Y
−
(k)/Y
+
(k) = 1 −λ
ˆ
K(k) → 1ask →∞.
Then Eq. (7.2.45) becomes
Z
−
(k)
ˆ
φ
−
(k) =−Z
+
(k)
ˆ
ψ
+
(k) = 2
−ik
G(k) ≡ g(k), (7.2.58)
250 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation
where g(k)isnowentire.Since
ˆ
φ
−
(k),
ˆ
ψ
+
(k) → 0ask →∞,
and Eq. (7.2.57) for Z
±
(k), g(k) cannot grow as fast as k. By Liouville’s theorem, we
then have
g(k) = C
,constant.
Thus we obtain
ˆ
φ
−
(k) = C
/Z
−
(k) = C
2
ik
α + ik
2
−
α − ik
2
/
1
2
+ik
. (7.2.59)
We now invert
ˆ
φ
−
(k)toobtainφ(x),
φ(x) = C
+∞
−∞
dk
2π
e
ikx
2
ik
α + ik
2
−
α − ik
2
/
1
2
+ik
, x ≥ 0,
(7.2.60)
φ(x) = 0, x < 0.
For x > 0, we close the contour in the upper half plane, picking up the pole
contributions from
α+ik
2
and
−
α−ik
2
. Poles of
α+ik
2
are located at
k = i(2n + α), n = 0, 1, 2, .... Poles of
−
α−ik
2
are located at k = i(2n − α), n =
0, 1, 2, ....Since
Res.(z)
|
z=−n
= (−1)
n
1
n!
, (7.2.61)
we have
φ(x) = C
∞
n=0
e
−(2n+α)x
2
−(2n+α)
(−1)
n
n!
(−n − α)
1
2
− 2n − α
+ (α →−α)
.
(7.2.62)
Since
(z) =
π
sin π z
1
(1 −z)
,
we have
(−n − α) =
(−1)
n+1
π
sin απ
1
(n + 1 +α)
,
and with the use of the duplication formula,
1
2
− 2n − α
=
π
cos απ
·
√
2π2
−(2n+α)
·
1
(
1
4
+
α
2
+ n)
3
4
+
α
2
+ n
,
7.2 Homogeneous Wiener–Hopf Integral Equation of the Second Kind 251
we have
(−n − α)
1
2
−2n − α
=
(−1)
n+1
√
2π
· 2
(2n+α)
·
cos απ
sin απ
·
1
4
+
α
2
+n
3
4
+
α
2
+ n
(1 +α + n)
.
Our solution φ(x)isgivenby
φ(x)
= C
cos απ
sin απ
·
∞
n=0
e
−αx
(e
−2x
)
n
n!
(
1
4
+
α
2
+ n)
3
4
+
α
2
+ n
× (1 + α +n)
− (α →−α)
.
(7.2.63)
Recall that the hypergeometric function F(a, b, c;z)isgivenby
F(a, b, c;z) =
∞
n=0
(α + n)
(a)
·
(b + n)
(b)
·
(c)
(c + n)
·
z
n
n!
. (7.2.64)
Setting
a =
1
4
+
α
2
, b =
3
4
+
α
2
, c = 1 +α,
we have
φ(x) = C
cos απ
sin απ
e
−αx
(
1
4
+
α
2
)(
3
4
+
α
2
)
(1 +α)
× F
1
4
+
α
2
,
3
4
+
α
2
,1+ α;e
−2x
−(α →−α)
. (7.2.65)
Recall that the Legendre function of the second kind, Q
α−
1
2
(z)isgivenby
Q
α−
1
2
(z) =
√
π
2
α+
1
2
·
(
1
2
+ α)
(1 +α)
· z
−(
1
2
+α)
· F
1
4
+
α
2
,
3
4
+ α,1+α;z
−2
=
1
2
·
(
1
4
+
α
2
)(
3
4
+
α
2
)
(1 +α)
·z
−(
1
2
+α)
·F
1
4
+
α
2
,
3
4
+
α
2
,1+ α;z
−2
,
(7.2.66)
so our solution given above can be simplified as
φ(x) = C
cos απ
sin απ
e
x
2
Q
α−
1
2
(e
x
) −Q
−α−
1
2
(e
x
)
. (7.2.67)
252 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation
Recall that the Legendre function of the first kind, P
β
(z), is given by
P
β
(z) =
1
π
sin βπ
cos βπ
Q
β
(z) −Q
−β−1
(z)
, (7.2.68)
so
P
α−
1
2
(z) =−
1
π
cos απ
sin απ
Q
α−
1
2
(z) − Q
−α−
1
2
(z)
. (7.2.69)
So, the final expression for φ(x )isgivenby
φ(x) = C ·
exp
x
2
·P
α−
1
2
e
x
, x ≥ 0, 0 ≤ α<1/2, 2πλ = cos απ.
(7.2.70)
Similar analysis can be carried out for the cases 2πλ > 1andλ<0. We only list
the final answers for all cases.
Summary of Example 7.4:
Case 1:0< 2πλ ≤ 1, 2πλ = cos απ,0≤ α<1/2.
φ(x) = C
1
· (exp
x
2
) ·P
α−
1
2
(e
x
), x ≥ 0.
Case 2:2πλ > 1, 2πλ = cosh απ, α>0.
φ(x) = C
2
· (exp
x
2
) · P
iα−
1
2
(e
x
), x ≥ 0.
Case 3:2πλ ≤ 0.
φ(x) = 0, x ≥ 0.
7.3
General Decomposition Problem
In the original Wiener–Hopf problem we examined in Section 7.1,
φ
−
(k) = ψ
+
(k) + F(k), (7.3.1)
we need to make the decomposition
F(k) = F
+
(k) + F
−
(k). (7.3.2)
7.3 General Decomposition Problem 253
k
k
2
k
1
t
+
t
−
Fig. 7.12 Sum splitting of F(k). φ
−
(k)isanalyticinthe
lower half plane, Im k <τ
−
. ψ
+
(k) is analytic in the upper
half plane, Im k >τ
+
. F(k) is analytic inside the strip, τ
+
<
Im k <τ
−
.
In the problems we just examined in Section 7.2, i.e., the homogeneous
Wiener–Hopf integral equation of the second kind, we need to make the
decomposition,
1 −λ
ˆ
K(k) = Y
−
(k)/Y
+
(k). (7.3.3)
Here we discuss how this can be done in general, as opposed to by inspection.
Consider the first problem (sum splitting) first (Figure 7.12):
φ
−
(k) = ψ
+
(k) +F(k).
Assume that
φ
−
(k), ψ
+
(k), F(k) → 0ask →∞. (7.3.4)
Examine the decomposition of F(k). Since F(k) is analytic inside the strip,
τ
+
< Im k = k
2
<τ
−
, (7.3.5)
bytheCauchyintegralformula,wehave
F(k) =
1
2πi
C
F(ζ )
ζ − k
dζ (7.3.6)
where the complex integration contour C consists of the following path as in
Figure 7.13:
C = C
1
+ C
2
+ C
↑
+C
↓
. (7.3.7)
254 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation
k
k
2
k
1
k
2
= t
−
k
2
= t
+
C
2
C
1
C
↑
C
↓
Fig. 7.13 Sum splitting contour C of Eq. (7.3.6) for F(k)insidethestrip,τ
+
< Im k <τ
−
.
The contributions from C
↑
and C
↓
vanish as these contours tend to infinity, since
F(ζ )
is bounded (actually→0), and
1/(ζ − k)
→ 0asζ →∞.
Thus we have
F(k) =
1
2πi
C
1
F(ζ )
ζ − k
dζ +
1
2πi
C
2
F(ζ )
ζ − k
dζ , (7.3.8)
where the contribution from C
1
is a + function, analytic for Im k = k
2
>τ
+
,while
the contribution from C
2
is a − function, analytic for Im k = k
2
<τ
−
, i.e.,
F
+
(k) =
1
2πi
+∞+iτ
+
−∞+iτ
+
F(ζ )
ζ − k
dζ , (7.3.9a)
F
−
(k) =
1
2πi
+∞+iτ
−
−∞+iτ
−
F(ζ )
ζ − k
dζ. (7.3.9b)
Consider now the factorization of 1 −λ
ˆ
K(k) into a ratio ofthe − functiontothe +
function. The function 1 − λ
ˆ
K(k) is analytic inside the strip,
−a < Im k = k
2
< b, (7.3.10)
and the inversion contour is somewhere inside the strip,
−a < Im k = k
2
< −a + ε. (7.3.11)
The analytic function 1 −λ
ˆ
K(k) may have some zeros inside the strip, −a < Im k =
k
2
< b. Choose a rectangular contour, as indicated in Figure 7.14, below all zeros
of 1 − λ
ˆ
K(k) inside the strip, −a < Im k = k
2
< b.
7.3 General Decomposition Problem 255
k
k
2
k
1
C
1
C
2
C
↑
C
↓
k
2
=
A
k
2
=
b
k
2
=−
a
Fig. 7.14 Rectangularcontourforthefactorizationof
1 −λ
ˆ
K(k). This contour is chosen below all the zeros of
1 −λ
ˆ
K(k)insidethestrip,−a < Im k < b.
Note if 1 −λ
ˆ
K(k)hasazeroonk
2
=−a, it is all right, since it just remains in the
lower half plane. The inversion contour k
2
= A will be chosen inside this rectangle.
Now, 1 −λ
ˆ
K(k) is analytic inside the rectangle
C
1
+ C
↑
+ C
2
+C
↓
and has no zeros inside this rectangle. Also since
ˆ
K(k) → 0ask →∞,
we know
1 −λ
ˆ
K(k) → 1ask →∞.
In order to express 1 −λ
ˆ
K(k)astheratioY
−
(k)/Y
+
(k), we shall take the logarithm
of (7.3.3) to find
ln
1 −λ
ˆ
K(k)
= ln
Y
−
(k)/Y
+
(k)
= ln Y
−
(k) − ln Y
+
(k). (7.3.12)
Now, ln[1 − λ
ˆ
K(k)] is itself analytic in the rectangle (because it has no branch points
since 1 −λ
ˆ
K(k) has no zeros there), so we can apply the Cauchy integral formula
ln
1 −λ
ˆ
K(k)
=
1
2πi
C
ln
1 −λ
ˆ
K(ζ )
ζ − k
dζ ,
with k inside the rectangle and C consisting of C
1
+ C
2
+ C
↑
+ C
↓
.Thuswewrite
256 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation
ln
1 − λ
ˆ
K(k)
=
1
2πi
C
1
ln
1 −λ
ˆ
K(ζ )
ζ − k
dζ −
1
2πi
−C
2
ln
1 −λ
ˆ
K(ζ )
ζ − k
dζ
+
1
2πi
C↑+C↓
ln
1 −λ
ˆ
K(ζ )
ζ − k
dζ. (7.3.13)
In Eq. (7.3.13), it is tempting to drop the contributions from C
↑
and C
↓
altogether.
It is, however, not always possible to do so. Because of the multivaluedness of the
logarithm, we may have, in the limit
|
ζ
|
→∞,
ln
1 − λ
ˆ
K(ζ )
→ ln e
2πin
= 2πin (n = 0, ±1, ±2, ...) (7.3.14)
and we have no guarantee that the contributions from C
↑
and C
↓
cancel each other.
In other words, 1 −λ
ˆ
K(ζ ) may develop a phase angle as ζ ranges from −∞ + iA
to +∞+ iA.
Let us define index ν of 1 −λ
ˆ
K(ζ )by
ν ≡
1
2πi
ln
1 −λ
ˆ
K(ζ )
ζ =+∞+iA
ζ =−∞+iA
. (7.3.15)
Graphically what we do is the following: plot z = [1 − λ
ˆ
K(ζ )] as ζ ranges from
−∞ + iA to +∞ + iA in the complex z plane, and count the number of counter-
clockwise revolutions z makesabouttheorigin.Theindexν is equal to the number
of these revolutions.
We now examine the properties of index ν;inparticular,arelationship between
index ν and the zeros and the poles of 1 −λ
ˆ
K(k) in the complex k plane. Suppose
1 − λ
ˆ
K(k) has a zero in the upper half plane, say, 1 −λ
ˆ
K(k) = k − z
u
,Imz
u
> −a.
Then the contribution from this z
u
to the index ν is ν(z
u
) =
1
2πi
0 − (−iπ)
=
1
2
.
Similar analysis yields the following results:
zero in the upper half plane ⇒ ν(z
u
) =+
1
2
,
pole in the upper half plane ⇒ ν(p
u
) =−
1
2
,
zero in the lower half plane ⇒ ν(z
l
) =−
1
2
,
pole in the lower half plane ⇒ ν(p
l
) =+
1
2
.
(7.3.16)
In many cases, the translation kernel K(x − y)isoftheform
K(x − y) = K(
x − y
). (7.3.17)
Then
ˆ
K(k)isevenink,
ˆ
K(k) =
ˆ
K(−k). (7.3.18)
Then 1 −λ
ˆ
K(k) (which is even) has an equal number of zeros (poles) in the upper
half plane and in the lower half plane,