Masujima M. Applied Mathematical Methods in Theoretical Physics
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216 6 Singular Integral Equations of the Cauchy Type
Thus f
n
(0, k) is analytic in the upper half plane, Im k > 0. In order to apply the
dispersion relation to f (0, k), we must define f
n
(0, k) for the real negative k.From
Eq. (6.5.21), we have
f
n
(0, −k) = f
∗
n
(0, k)withk real and positive. (6.5.22)
We note that
f
1
(0, 0) =−
1
4π
U(
r
1
)d
3
r
1
= f
1
(0, k)
is a real constant so that we can invoke the dispersion relation,
Re F(x) =
2
π
P
∞
0
x
Im F(x
)
x
2
− x
2
dx
,ImF(x) =−
2
π
P
∞
0
x Re F(x
)
x
2
− x
2
dx
,
(6.5.6)
to the present case with minor modification,
F(k) = f (0, k ) − f
1
(0, k)withk real and positive. (6.5.23)
Thus we have
Re[f (0, k) − f
1
(0, 0)] =
2
π
P
∞
0
k
Im f (0, k
)
k
2
− k
2
dk
. (6.5.24)
We tacitly assumed the square-integrability of [f (0, k) − f
1
(0, 0)] in the present
discussion. If this assumption fails, we may replace
[f (0, k) − f
1
(0, 0)] with [f (0, k) − f
1
(0, 0)]k
2
wherethelatterisassumedtobeboundedatk = 0 and square-integrable. We
obtain the dispersion relation,
Re[f (0, k) − f
1
(0, 0)] =
2k
2
π
P
∞
0
Im f (0, k
)
k
(k
2
− k
2
)
dk
. (6.5.25)
We shall now replace the incident plane wave in Eq. (6.5.19) with the incident
spherical wave originating from the point on the negative z-axis,
r =
r
0
,
ψ
0
(
r) =
r
0
r −
r
0
exp[ik
r −
r
0
− ikr
0
],
and consider the scattered wave. The exponent in Eq. (6.5.21) gets replaced with
−i
k
s
r
1
+ik
r
1
−
r
2
+···+ik
r
n−1
−
r
n
+ ik
r
n
−
r
0
−ikr
0
.
6.5 Dispersion Relations 217
Inthecaseoftheforwardscattering,wehave
k
s
=
k
0
and
k
s
points in the positive
z-direction. Since we have kr
0
=−
k
0
r
0
, the exponent written out above becomes
i
k
0
(
r
0
−
r
1
) + ik{
r
1
−
r
2
+···+
r
n−1
−
r
n
+
r
n
−
r
0
},
so that we have the exponent as
ik × (positive number).
Sincewehave
r
0
on the z-axis, we have
exp[ik
r −
r
0
]
r −
r
0
= ik
∞
l=0
(2l + 1)P
l
(cos θ )j
l
(kr)h
(1)
l
(kr
0
), r
0
> r. (6.5.26)
The total wave must assume the form
ik
∞
l=0
(2l + 1)P
l
(cos θ )h
(1)
l
(kr
0
)r
0
exp[−ikr
0
](j
l
(kr) + i exp[iδ
l
]sinδ
l
h
(1)
l
(kr)),
and the corresponding forward scattering amplitude f
(0, k) becomes
f
(0, k) =
∞
l=0
(2l + 1)h
(1)
l
(kr
0
)r
0
exp[−ikr
0
] · i exp[iδ
l
]sinδ
l
· i
−l
=
1
2ik
∞
l=0
(2l + 1)(exp[iδ
l
] − 1)q
l
(kr
0
). (6.5.27)
Here
q
l
(kr
0
) = kr
0
· h
(1)
l
(kr
0
) · i
−l+1
exp[−ikr
0
] =
l
m=0
i
m
(l + m)!
m!(l − m)!
1
(2kr
0
)
m
. (6.5.28)
We note that f
(0, k) is a function of r
0
.Sincer
0
is arbitrary, we can obtain many
relations from the dispersion relation for f
(0, k) by equating the same power of r
0
.
Picking up the 1r
m
0
term, we obtain
f
m
(k) =
1
2ik
2m+1
∞
l=m
(2l + 1)
(l + m)!
(l − m)!
(exp[iδ
l
] − 1). (6.5.29)
Using the identity
l
m=l
(−1)
m−l
(l
+ m)!
(m − l)!(m + l + 1)!(l
− m)!
=
1(2l + 1), for l
= l,
0, for l
> l,
218 6 Singular Integral Equations of the Cauchy Type
we invert Eq. (6.5.29) for exp[iδ
l
] − 1, obtaining
exp[iδ
l
] − 1
2i
=
∞
m=l
(−1)
m−l
k
2m+1
(m − l)!(m + l + 1)!
f
m
(k).
The dispersion relation for the phase shift δ
l
becomes
Im exp[2iδ
l
] =
∞
m=l
(−1)
m−l
k
2m+1
(m − l)!(m + l + 1)!
×
2
π
P
∞
l
=m
[(2l
+ 1)(l
+ m)!(l
− m)!] Re(1 − exp[iδ
l
(k
)])
k
2
−k
2
dk
+ C
m
,
(6.5.30)
where
C
m
=
1
2
d
m
dk
m
Im exp[2iδ
m
(k)]
k=0
.
We refer the reader to the following book for the mathematical details of the dis-
persion relations in classical electrodynamics, quantum mechanics and relativistic
quantum field theory.
Goldberger, M.L. and Watson, K.M.: Collision Theory, John Wiley & Sons, New York,
(1964). Chapter 10 and Appendix G.2.
6.6
Problems for Chapter 6
6.1. Solve
1
πi
P
+1
−1
1
y − x
φ(y)dy = 0, −1 ≤ x ≤ 1.
6.2. (due to H. C.). Solve
φ(x) =
tan x
π
P
1
0
1
y − x
φ(y)dy + f (x), 0 ≤ x ≤ 1.
6.3. (due to H. C.). Solve
φ(x) =
tan(x
3
)
π
P
1
0
1
y − x
φ(y)dy + f (x), 0 ≤ x ≤ 1.
6.6 Problems for Chapter 6 219
6.4. (due to H. C.). Solve
x
0
1
√
x − y
φ(y)dy + A
1
x
1
√
y − x
φ(y)dy = 1,
where
0 ≤ x ≤ 1.
6.5. (due to H. C.). Solve
λP
+∞
0
1
y − x
φ(y)dy = φ(x), 0 ≤ x < ∞.
Find the eigenvalues and the corresponding eigenfunctions.
6.6. (due to H. C.). Solve
P
+1
−1
1 + α(y − x)
y − x
φ(y)dy = 0, −1 ≤ x ≤ 1.
6.7. (due to H. C.). Obtain the eigenvalues and the eigenfunctions of
φ(x) =
λ
π
P
+1
−1
1
y − x
φ(y)dy, −1 ≤ x ≤ 1.
6.8. (due to H. C.). Solve
√
x ψ (x) =
1
π
P
+∞
0
1
y − x
ψ(y)dy,0≤ x < ∞.
6.9. (due to H. C.). Solve
1
π
P
+∞
−∞
1
y − x
φ(y)dy = f (x), −∞ < x < ∞.
6.10. (due to H. C.). Solve
1
π
P
+∞
0
1
y − x
φ(y)dy = f (x), 0 ≤ x < ∞.
6.11. (due to H. C.). Solve
1
0
ln
x − y
φ(y)dy = 1, 0 ≤ x ≤ 1.
220 6 Singular Integral Equations of the Cauchy Type
6.12. Solve
(x)
B(x)
+
1
π
P
+a
−a
(y)
x − y
dy = f (x)
with (x), B(x), and f (x)evenand
(−a) = (a) = 0.
Hint: The unknown (x) is the circulation of the thin wing in the Prandtl’s
thin wing theory.
Kondo, J. : Integral Equations, Kodansha Ltd., Tokyo, (1991), p. 412.
6.13. (due to H. C.). Prove that
P
∞
0
dy
√
y(y − x)
= 0forx > 0 .
6.14. (due to H. C.). Prove that
P
1
0
dy
y(1 − y)(y − x)
= 0, 0 < x < 1.
6.15. (due to H. C.). Prove that
1
π
P
1
0
y(1 − y)
y − x
dy =
1
2
− x,0< x < 1.
6.16. (due to H. C.). Prove that
1
π
P
1
0
ln y
y(1 − y)(y − x)
dy =
π
x (1 − x)
−
∞
1
1
y(y − 1)(y − x)
dy
0 ≤ x ≤ 1.
6.17. Solve the integral equation of the Cauchy type,
F(x) =
1
π
F(x
)h
∗
(x
)
x
− x − iε
dx
,
where h(x)isgivenby
h(x ) = exp[iδ(x)] sin δ(x),
and δ(x)isagivenrealfunction.
6.6 Problems for Chapter 6 221
Hint: Consider f (z)definedby
f (z) =
1
π
F(x
)h
∗
(x
)
x
− z
dx
,
and compute the discontinuity across the real x-axis. Also use the identity
1 − 2ih
∗
(x) = exp[−2iδ(x)].
The function h(x) originates from the phase shift analysis of the potential
scattering problem in quantum mechanics.
223
7
Wiener–Hopf Method and Wiener–Hopf Integral Equation
7.1
The Wiener–Hopf Method for Partial Differential Equations
In Sections 6.3, 6.4, and 6.5, we reduced the singular integral equations of
Cauchy type and their variants to the inhomogeneous Hilbert problem through the
introduction of the function (z) appropriately defined.
Suppose now we are given one linear equation involving two unknown functions,
φ
−
(k)andψ
+
(k), in the complex k plane,
φ
−
(k) = ψ
+
(k) +F(k), (7.1.1)
where φ
−
(k) is analytic in the lower half plane (Im k <τ
−
)andψ
+
(k)isanalytic
in the upper half plane (Im k ≥ τ
+
). Can we solve Eq. (7.1.1) for φ
−
(k)andψ
+
(k)?
As long as φ
−
(k)andψ
+
(k) have a common region of analyticity as in Figure 7.1,
namely
τ
+
≤ τ
−
, (7.1.2)
we can solve for φ
−
(k)andψ
+
(k). In the most stringent case, the common region
of analyticity can be an arc below which (excluding the arc) φ
−
(k) is analytic and
above which (including the arc) ψ
+
(k)isanalytic.
We proceed to split F(k) into a sum of two functions, one analytic in the upper
half plane and the other analytic in the lower half plane,
F(k) = F
+
(k) +F
−
(k). (7.1.3)
This sum splitting can be carried out either by inspection or by the general method
utilizing the Cauchy integral formula to be discussed in Section 7.3. Once the sum
splitting is accomplished, we write Eq. (7.1.1) in the following form:
φ
−
(k) −F
−
(k) = ψ
+
(k) + F
+
(k) ≡ G(k). (7.1.4)
We immediately note that G(k)isentire in k. If the asymptotic behaviors of F
±
(k)as
k
→∞are such that
F
±
(k) → 0as
k
→∞, (7.1.5)
Applied Mathematical Methods in Theoretical Physics, Second Edition. Michio Masujima
Copyright
2009 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim
ISBN: 978-3-527-40936-5
224 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation
k
t
+
t
−
k
1
k
2
Fig. 7.1 The common region of analyticity for φ
−
(k)and
ψ
+
(k)inthecomplexk plane. ψ
+
(k)andF
+
(k) are ana-
lytic in the upper half plane, Im k ≥ τ
+
. φ
−
(k)andF
−
(k)are
analytic in the lower half plane, Im k <τ
−
. The common re-
gion of analyticity for φ
−
(k)andψ
+
(k)isinsidethestrip,
τ
+
≤ Im k <τ
−
in the complex k plane.
and on some physical ground,
φ
−
(k), ψ
+
(k) → 0as
k
→∞, (7.1.6)
then the entire function G(k) must vanish by Liouville’s theorem. Thus we obtain
φ
−
(k) = F
−
(k), (7.1.7a)
ψ
+
(k) =−F
+
(k). (7.1.7b)
We call this method the Wiener–Hopf method.
In the following examples, we apply this method to the mixed boundary value
problem of the partial differential equation.
Example 7.1. Find the solution to the Laplace equation in the half-plane.
∂
2
∂x
2
+
∂
2
∂y
2
φ(x, y) = 0, y ≥ 0, −∞ < x < ∞, (7.1.8)
7.1 The Wiener–Hopf Method for Partial Differential Equations 225
∂f
∂
y
(
x
,0) =
ce
bx
f(
x
,0) =
e
x
x
y
Fig. 7.2 Boundary conditions of φ(x, y)fortheLaplace
equation (7.1.8) in the half plane, y ≥ 0and−∞ < x < ∞.
subject to the boundary condition on y = 0, as displayed in Figure 7.2,
φ(x,0)= e
−x
, x > 0, (7.1.9a)
φ
y
(x,0) = ce
bx
, b > 0, x < 0. (7.1.9b)
We further assume
φ(x, y) → 0asx
2
+y
2
→∞. (7.1.9c)
Solution. Since −∞ < x < ∞, we may take the Fourier transform with respect to x,
i.e.,
ˆ
φ(k, y) ≡
+∞
−∞
dxe
−ikx
φ(x, y), φ(x , y) =
1
2π
+∞
−∞
dke
ikx
ˆ
φ(k, y). (7.1.10)
So, if we can obtain
ˆ
φ(k, y), we will be done. Taking the Fourier transform of the
partial differential equation (7.1.8) with respect to x,wehave
−k
2
+
∂
2
∂y
2
ˆ
φ(k, y) = 0,
i.e.,
∂
2
∂y2
ˆ
φ(k, y) = k
2
ˆ
φ(k, y), (7.1.11)
from which we obtain
ˆ
φ(k, y) = C
1
(k)e
ky
+ C
2
(k)e
−ky
.
226 7 Wiener–Hopf Method and Wiener–Hopf Integral Equation
The boundary condition at infinity,
ˆ
φ(k, y) → 0asy →+∞, implies
C
1
(k) = 0fork > 0,
C
2
(k) = 0fork < 0.
We can write more generally
ˆ
φ(k, y) = A(k)e
−
|
k
|
y
. (7.1.12)
To obtain A(k), we need to apply the boundary conditions at y = 0.
Take the Fourier transform of φ(x,0)tofind
ˆ
φ(k,0)≡
+∞
−∞
dxe
−ikx
φ(x,0)
=
0
−∞
dxe
−ikx
φ(x,0)+
+∞
0
dxe
−ikx
e
−x
. (7.1.13)
Here the first term is unknown for x < 0, while the second term is equal to
1/(1 +ik). Recall that
e
−ikx
=
e
−i(k
1
+ik
2
)x
= e
k
2
x
with k = k
1
+ ik
2
,
which vanishes in the upper half plane (k
2
> 0) for x < 0. So, define
φ
+
(k) ≡
0
−∞
dxe
−ikx
φ(x, 0), (7.1.14)
which is a + function. (Assuming that φ(x,0)∼ O(e
bx
)asx →−∞, φ
+
(k)is
analytic for k
2
> −b.) So, we have from Eqs. (7.1.12) and (7.1.13)
ˆ
φ(k,0)= φ
+
(k) +
1
1 +ik
or A(k) = φ
+
(k) +
1
1 +ik
. (7.1.15)
Also, take the Fourier transform of ∂φ(x,0)/∂y to find
∂
ˆ
φ
∂y
(k,0) ≡
+∞
−∞
dxe
−ikx
∂φ(x,0)
∂y
=
0
−∞
dxe
−ikx
ce
bx
+
+∞
0
dxe
−ikx
∂φ(x,0)
∂y
, (7.1.16)
where the first term is equal to c/(b − ik), and the second term is unknown for
x > 0. So, define
ψ
−
(k) ≡
+∞
0
dxe
−ikx
∂φ(x,0)
∂y
, (7.1.17)