Masujima M. Applied Mathematical Methods in Theoretical Physics
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146 4 Integral Equations of the Fredholm Type
(b) Let
φ
n
(x) = exp[−x
2
2]H
n
(x).
Assume that φ
n
= λ
n
Kφ
n
. Show that φ
n+1
= λ
n+1
Kφ
n+1
with
λ
n
= tλ
n+1
. This means that the original integral equation has
eigenvalues λ
n
= t
−n
√
π, with the corresponding eigenfunctions
φ
n
(x).
4.10. (due to H. C.). Find the eigenvalues and eigenfunctions of the integral
equation,
φ(x) = λ
∞
0
exp[−xy]φ(y)dy.
Hint: Consider the Mellin transform,
(p) =
∞
0
x
ip−
1
2
φ(x)dx with φ(x) =
1
2π
∞
−∞
x
−ip−
1
2
(p)dp.
4.11. (due to H. C.). Solve
ψ(x) = e
bx
+ λ
x
0
ψ(y)dy + 2λ
1
x
ψ(y)dy.
4.12. Solve
φ(x) = λ
+1
−1
K(x, y)φ(y)dy −
1
2
+1
−1
φ(y)dy with φ(±1) = finite,
where
K(x, y) =
1
2
ln
1 +x
<
1 −x
>
,
x
<
=
1
2
(x + y) −
1
2
x − y
and x
>
=
1
2
(x + y) +
1
2
x − y
.
4.13. Solve
φ(x) = λ
+∞
−∞
K(x, y)φ(y)dy with φ(±∞) = finite,
where
K(x, y) =
α
π
exp
α
2
(x
2
+y
2
)
x
<
−∞
exp[−ατ
2
]dτ ·
+∞
x
>
exp[−ατ
2
]dτ
,
x
<
=
1
2
(x + y) −
1
2
x − y
and x
>
=
1
2
(x + y) +
1
2
x − y
.
4.8 Problems for Chapter 4 147
4.14. Solve
φ(x) = λ
∞
0
K(x, y)φ(y)dy,
with
φ(x)
< ∞ for 0 ≤ x < ∞,
where
K(x, y) =
exp[−β
x − y
]
2βxy
.
4.15. (due to D. M.) Show that the nontrivial solutions of the homogeneous
integral equation,
φ(x) = λ
π
−π
1
4π
(x − y)
2
−
1
2
x − y
φ(y)dy,
are cos(mx) and sin(mx), where λ = m
2
and m is any integer.
Hint for Problems 4.12 through 4.15: The kernels change their forms
continuously as x passes through y. Differentiate the given integral
equations with respect to x and reduce them to the ordinary differential
equations.
4.16. (due to D. M.) Solve the inhomogeneous integral equation,
φ(x) = f (x) +λ
∞
0
cos(2xy)φ(y)dy, x ≥ 0,
where λ
2
= 4π.
Hint: Multiply both sides of the integral equation by cos(2xξ )andintegrate
over x.Usetheidentity
cos(2xy)cos(2xξ ) =
1
2
cos[2x(y + ξ )] + cos[2x(y − ξ )]
,
and observe that
∞
0
cos(αx)dx =
1
2
∞
0
(exp[iαx] +exp[−iαx])dx
=
1
2
∞
−∞
exp[iαx]dx
=
1
2
· 2πδ(α) = πδ(α).
148 4 Integral Equations of the Fredholm Type
4.17. (due to D. M.) In the theoretical search for supergain antennas, maximizing
the directivity in the far field of axially invariant currents j(φ) that flow along
the surface of infinitely long, circular cylinders of radius a leads to the
following Fredholm integral equation for the current density j(φ):
j(φ) = exp[ika sin φ] −α
2π
0
dφ
2π
J
0
2ka sin
φ − φ
2
j(φ
),
0 ≤ φ<2π,
where φ is the polar angle of the circular cross section, k is a positive wave
number, α is a parameter (Lagrange multiplier) that expresses a constraint
on the current magnitude, α ≥ 0, and J
0
(x) is the 0th-order Bessel function of
the first kind.
(a) Determine the eigenvalues of the homogeneous equation.
(b) Solve the given inhomogeneous equation in terms of Fourier series,
j(φ) =
∞
n=−∞
f
n
exp[inφ].
Hint: Use the formulas,
exp[ika sin φ] =
∞
n=−∞
J
n
(ka)exp[inφ],
and
J
0
(2ka sin
φ −φ
2
) =
∞
m=−∞
J
m
(ka)
2
exp[im(φ − φ
)],
where J
n
(x)isthenth-order Bessel function of t he first kind. Substitution of
Fourier series for j(φ),
j(φ) =
∞
n=−∞
f
n
exp[inφ],
yields the decoupled equation for f
n
,
f
n
=
J
n
(ka)
1 +αJ
n
(ka)
2
, n =−∞, ..., ∞.
4.18. (due to D. M.) Problem 4.17 corresponds to the circular loop in two
dimensions. For the circular disk in two dimensions, we have the following
4.8 Problems for Chapter 4 149
Fredholm integral equation for the current density j(φ):
j(r, φ) = exp[ikr sin φ] −
2α
a
2
2π
0
dφ
2π
a
0
r
dr
×J
0
(k
r
2
+ r
2
−2rr
cos(φ − φ
))j(r
, φ
),
with
0 ≤ r ≤ a,0≤ φ<2π.
Solve the given inhomogeneous equation in terms of Fourier series,
j(r, φ) =
n=∞
n=−∞
f
n
(r)exp[inφ].
Hint: By using the addition formula
J
0
(k
r
2
+ r
2
− 2rr
cos(φ −φ
)) =
∞
m=−∞
J
m
(kr)J
m
(kr
)exp[im(φ − φ
)],
it is found that f
n
(r) satisfy the following integral equation:
f
n
(r) =
1 −
2α
a
2
a
0
r
dr
f
n
(r
)J
n
(kr
)
J
n
(kr), n =−∞, ..., ∞.
Substitution of
f
n
(r) = λ
n
J
n
(kr)
yields
λ
n
=
1 +
2α
a
2
a
0
r
dr
J
n
(kr
)
2
−1
=
1 +α[J
n
(ka)
2
− J
n+1
(ka)J
n−1
(ka)]
−1
.
4.19. (due to D. M.) Problem 4.17 corresponds to the circular loop in two
dimensions. For the circular loop in three dimensions, we have the
following Fredholm integral equation for the current density j(φ):
j(φ) = exp[ika sin φ] − α
2π
0
dφ
2π
K(φ − φ
)j(φ
), 0 ≤ φ<2π,
150 4 Integral Equations of the Fredholm Type
with
K(φ) =
sin w
w
+
cos w
w
2
−
sin w
w
3
=
1
4
1
−1
(1 +ξ
2
)exp[iwξ ]dξ ,
and
w = w(φ) = 2ka sin
φ
2
.
Solve the given inhomogeneous equation in terms of Fourier series,
j(φ) =
n=∞
n=−∞
f
n
exp[inφ].
Hint: Following the step employed in Problem 4.17, substitute the Fourier
series into the integral equation. The decoupled equation for f
n
,
f
n
=
J
n
(ka)
1 +αU
n
(ka)
, n =−∞, ..., ∞,
results, where
U
n
(ka) =
π
−π
dφ
2π
K(φ)exp[−inφ]
=
1
8π
1
−1
dξ (1 +ξ
2
)
π
−π
dφ exp[iw(φ)ξ]cos(nφ)
=
1
2
1
0
dξ (1 +ξ
2
)J
2n
(2kaξ ).
The integral for U
n
(ka) can be further simplified by the use of Lommel’s
function S
µ,ν
(x)andWeber’s function E
ν
(x).
Reference for Problems 4.17, 4.18, and 4.19:
We cite the following article for the Fredholm integral equations of the
second kind in the theoretical search for ‘‘supergain antennas.’’
Margetis, D., Fikioris, G. , Myers, J.M., and Wu, T.T. : Phys. Rev. E58.,
2531, (1998).
We cite the following article for the Fredholm integral equations of the
second kind for the two dimensional, highly directive currents on large
circular loops.
4.8 Problems for Chapter 4 151
Margetis,D.andFikioris,G.:J.Math.Phys.41, 6130, (2000).
We can derive the above-stated Fredholm integral equations of the second
kind for the localized, monochromatic, and highly directive classical current
distributions in two and three dimensions by maximizing the directivity D
in the far field while constraining C = NT,whereN is the integral of the
square of the magnitude of the current density and T is proportional to the
total radiated power. This derivation is the application of the calculus of
variations. We derive the homogeneous Fredholm integral equations of the
second kind and the inhomogeneous Fredholm integral equations of the
second kind in their general forms in Section 9.6 of Chapter 9.
4.20. Consider the S-wave scattering off a spherically symmetric potential U(r).
The governing Schr
¨
odinger equation is given by
d
2
dr
2
u(r) + k
2
u(r) = U(r)u(r),
with
u(0) = 0andu(r) ∼
sin(kr + δ)
sin δ
as r →∞.
(a) Convert this differential equation into a Fredholm integral equation of
the second kind,
u(r) = exp[−ikr] − exp[ikr] +
∞
0
g(r, r
)U(r
)u(r
),
with Green’s function g(r, r
)givenby
g(r, r
) =−
1
2ik
exp
ik(r + r
)
− exp
ik
r − r
.
(b) Setting
K(r, r
) = g(r, r
)U(r
),
rewrite the integral equation above as
u(r) = exp[−ikr] − exp[ikr] +
∞
0
K(r, r
)u(r
).
Apply Fredholm theory for a bounded kernel to this integral equation
to obtain a formal solution,
u(r) = exp[−ikr] −exp[ikr] +
1
D(k)
∞
0
D(k;r, r
)(exp[−ikr
]
−exp[ikr
])dr
,
152 4 Integral Equations of the Fredholm Type
where
D(k) = 1 +
∞
n=1
(−1)
n
n!
∞
0
dr
1
·
∞
0
dr
n
K(r
1
, r
1
) K(r
1
, r
2
) · K(r
1
, r
n
)
·
K(r
n
, r
1
) K(r
n
, r
2
) · K(r
n
, r
n
)
,
and
D(k;r, r
) = K(r, r
) +
∞
n=1
(−1)
n
n!
∞
0
dr
1
·
∞
0
dr
n
×
K(r, r
) K(r, r
1
) · K(r, r
n
)
K(r
1
, r
) K(r
1
, r
1
) K(r
1
, r
n
)
·
K(r
n
, r
) K(r
n
, r
1
) · K(r
n
, r
n
)
.
(c) Obtain the condition on the potential U(r) for this formal solution to
converge.
Reference for Problem 4.20:
We cite the following book for the application of Fredholm theory for a
bounded kernel to potential scattering problem.
Nishijima, K.: Relativistic Quantum Mechanics, Baifuu-kan, Tokyo, (1973),
Chapter 4, Section 4.7 (In Japanese).
153
5
Hilbert–Schmidt Theory of Symmetric Kernel
5.1
Real and Symmetric Matrix
We now would like to examine the case of a symmetric kernel (self-adjoint integral
operator)whichisalsosquare-integrable. Recalling from our earlier discussions in
Chapter 1 that self-adjoint operators can be diagonalized, our principal aim is to
accomplish the same goal for the case of symmetric kernels.
For this purpose, let us first examine the corresponding problem for an n × nreal
and symmetric matrix A. Suppose A has eigenvalues λ
k
and normalized eigenvectors
v
k
, i.e.,
Av
k
= λ
k
v
k
, k = 1, 2, ..., n;v
T
k
v
m
= δ
km
. (5.1.1)
We may thus write
A
[
v
1
, v
2
, ·, v
n
]
=
[
λ
1
v
1
, λ
2
v
2
, ·, λ
n
v
n
]
=
[
v
1
, ·, v
n
]
λ
1
0 · 0
0 λ
2
·
·· ·
····0
0 · 0 λ
n
.
(5.1.2)
Define the matrix S by
S =
[
v
1
, v
2
, ..., v
n
]
(5.1.3a)
and consider S
T
,
S
T
=
v
T
1
·
·
v
T
n
. (5.1.3b)
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
154 5 Hilbert–Schmidt Theory of Symmetric Kernel
Then we have
S
T
S =
v
T
1
·
·
v
T
n
[
v
1
, ..., v
n
]
=
10 0
01
· 0
001
= I, (5.1.4a)
sincewehaveEq.(5.1.1).Hencewehave
S
T
= S
−1
. (5.1.5)
Define
D =
λ
1
00
0 ·
· 0
00λ
n
. (5.1.6)
From Eq. (5.1.2), we have
AS = SD. (5.1.7a)
Hencewehave
A = SDS
−1
= SDS
T
. (5.1.7b)
The above relation can also be written as
A =
n
k=1
λ
k
v
T
k
v
k
. (5.1.8)
This represents the diagonalization of a symmetric matrix. Equation (5.1.7b) is
really convenient for calculation of functions of A. For example, we compute
A
2
=
SDS
−1
SDS
−1
= SD
2
S
−1
= S
λ
2
1
00
0 ·
· 0
00λ
2
n
S
−1
.
e
A
= I + A +
1
2
A
2
+
1
3!
A
3
+···
= S
e
λ
1
00
0 ·
· 0
00e
λ
n
S
−1
.
f (A) = S
f (λ
1
)0 0
0 ·
· 0
00f (λ
n
)
S
−1
. (5.1.9)
5.2 Real and Symmetric Kernel 155
Finally we have
det A = det SDS
−1
= det S det D det S
−1
= det D =
n
k=1
λ
k
, (5.1.10a)
tr(A) = tr
SDS
−1
= tr
DS
−1
S
= tr
(
D
)
=
n
k=1
λ
k
. (5.1.10b)
5.2
Real and Symmetric Kernel
Symmetric kernels have the property that when transposed they remain the same
as the original kernel. We denote the transposed kernel K
T
by
K
T
(x, y) = K(y, x), (5.2.1)
and note that when the kernel K is symmetric, we have
K
T
(x, y) = K(y, x) = K(x, y). (5.2.2)
The eigenvalues of K(x, y) and eigenvalues of K
T
(x, y) are the same. This is because
an eigenvalue λ
n
is a zero of D(λ). From the definition of D(λ), we find that, since
a determinant remains the same as we exchange its rows and columns,
D(λ)forK(x, y) = D(λ)forK
T
(x, y). (5.2.3)
Thus the spectrum of K(x, y) coincides with that of K
T
(x, y).
We will need to make use of the orthogonality property held by the eigenfunctions
belonging to each eigenvalue. To show this property, we start with the eigenvalue
equations,
φ
n
(x) = λ
n
h
0
K(x, y)φ
n
(y)dy, (5.2.4)
ψ
n
(x) = λ
n
h
0
K
T
(x, y)ψ
n
(y)dy, (5.2.5)
and from the definition of K
T
(x, y),
ψ
n
(x) = λ
n
h
0
K(y, x)ψ
n
(y)dy. (5.2.6)
Multiplying by ψ
m
(x) in Eq. (5.2.4) and integrating over x,weget
h
0
ψ
m
(x)φ
n
(x)dx = λ
n
h
0
dxψ
m
(x)
h
0
K(x, y)φ
n
(y)dy
=
λ
n
λ
m
h
0
ψ
m
(y)φ
n
(y)dy.