Masujima M. Applied Mathematical Methods in Theoretical Physics
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3.3 Problems for Chapter 3 115
3.17. (due to H. C.). The convolution of f
1
(x), f
2
(x),..., f
n
(x)isdefinedas
C(x) ≡
∞
0
dx
n
···
∞
0
dx
1
n
i=1
f
i
(x
i
)δ
x −
n
i=1
x
i
.
(a) Verify that for n = 2, this is the convolution defined in the text, and that
˜
C (s) =
n
i=1
˜
f
i
(s).
(b) If f
1
(x) = f (x)andf
2
(x) = f
3
(x) =···=f
n
(x) = 1, show that C(x)isthe
nth integral of f (x). Show also that
˜
C (s) =
˜
f (s)s
1−n
,
and hence
C(x) =
x
0
(x − y)
n−1
(n − 1)!
f (y)dy.
(c) With the result in (b), can you define the ‘‘one-third integral’’ of f (x)?
117
4
Integral Equations of the Fredholm Type
4.1
Iterative Solution to the Fredholm Integral Equation of the Second Kind
We consider the inhomogeneous Fredholm Integral Equation of the second kind,
φ(x) = f (x) + λ
h
0
dx
K(x, x
)φ(x
), 0 ≤ x ≤ h, (4.1.1)
and assume that f (x)andK(x, x
)arebothsquare-integrable,
f
2
< ∞, (4.1.2)
K
2
≡
h
0
dx
h
0
dx
K(x, x
)
2
< ∞. (4.1.3)
Suppose that we look for an iterative solution in λ,
φ(x) = φ
0
(x) + λφ
1
(x) +λ
2
φ
2
(x) +···+λ
n
φ
n
(x) +···. (4.1.4)
We substitute Eq. (4.1.4) into Eq. (4.1.1).
We obtain
φ
0
(x) = f (x),
φ
1
(x) =
h
0
dy
1
K(x, y
1
)φ
0
(y
1
) =
h
0
dy
1
K(x, y
1
)f (y
1
),
φ
2
(x) =
h
0
dy
2
K(x, y
2
)φ
1
(y
2
) =
h
0
dy
2
h
0
dy
1
K(x, y
2
)K(y
2
, y
1
)f (y
1
).
In general, we have
φ
n
(x) =
h
0
dy
n
K(x, y
n
)φ
n−1
(y
n
) =
h
0
dy
n
h
0
dy
n−1
···
h
0
dy
1
×K(x, y
n
)K(y
n
, y
n−1
) ···K(y
2
, y
1
)f (y
1
). (4.1.5)
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
118 4 Integral Equations of the Fredholm Type
Bounds: First, in order to establish the bound on
φ
n
(x)
,define
A(x) ≡
h
0
dy
K(x, y)
2
. (4.1.6)
ThenthesquareofthenormofthekernelK(x, y)isgivenby
K
2
=
h
0
dxA(x). (4.1.7)
Applying the Schwarz inequality, each term in the iteration series (4.1.4) is bounded
as follows:
φ
1
(x) =
h
0
dy
1
K(x, y
1
)f (y
1
) ⇒
φ
1
(x)
2
≤ A(x)
f
2
,
φ
2
(x) =
h
0
dy
2
K(x, y
2
)φ
1
(y
2
) ⇒
φ
2
(x)
2
≤ A(x)
φ
1
2
≤ A(x)
f
2
K
2
,
φ
n
(x)
2
≤ A(x)
f
2
K
2(n−1)
.
Thus the bound on
φ
n
(x)
is established as
φ
n
(x)
≤
A(x)
f
K
n−1
, n = 1, 2, 3, .... (4.1.8)
Now examine the whole series (4.1.4):
φ(x) −f (x) = λφ
1
(x) + λ
2
φ
2
(x) + λ
3
φ
3
(x) +···.
Taking the absolute value of both sides, applying the triangular inequality on the
right-hand side, and using the bound on
φ
n
(x)
established as in Eq. (4.1.8), we
have
φ(x) −f (x)
≤
|
λ
|
φ
1
(x)
+
|
λ
|
2
φ
2
(x)
+···=
∞
n=1
|
λ
|
n
φ
n
(x)
≤
∞
n=1
|
λ
|
n
A(x) ·
f
·
K
n−1
=
|
λ
|
A(x) ·
f
·
∞
n=0
|
λ
|
n
·
K
n
. (4.1.9)
Now, the series on the right-hand side of inequality (4.1.9),
∞
n=0
|
λ
|
n
·
K
n
,con-
verges as long as
|
λ
|
< 1
K
, (4.1.10)
4.1 Iterative Solution to the Fredholm Integral Equation of the Second Kind 119
converging to
1(1 −
|
λ
|
·
K
).
Therefore, in that case (i.e., for Eq. (4.1.10)), the assumed series (4.1.4) is a
convergent series, giving us a solution φ(x) to the integral equation (4.1.1), which
is analytic inside the disk (4.1.10). Symbolically the integral equation (4.1.1) can be
written as if it is an algebraic equation,
φ = f + λKφ ⇒ (1 − λK)φ = f
⇒ φ =
f
1 −λK
= (1 +λK + λ
2
K
2
+···)f ,
which converges only for
|
λK
|
< 1.
Uniqueness: Inside the disk, Eq. (4.1.10), we can establish the uniqueness of
the solution by showing that the corresponding homogeneous problem has no
nontrivial solutions. Consider the homogeneous problem,
φ
H
(x) = λ
h
0
K(x, y)φ
H
(y). (4.1.11)
Applying the Schwarz inequality to Eq. (4.1.11),
φ
H
(x)
2
≤
|
λ
|
2
A(x)
φ
H
2
.
Integrating both sides with respect to x from 0 to h,
φ
H
2
≤
|
λ
|
2
·
φ
H
2
h
0
A(x)dx
=
|
λ
|
2
·
K
2
·
φ
H
2
,
i.e.,
φ
H
2
·(1 −
|
λ
|
2
·
K
2
) ≤ 0. (4.1.12)
Since
|
λ
|
·
K
< 1, inequality (4.1.12) can only be satisfied if and only if
φ
H
= 0
or
φ
H
≡ 0. (4.1.13)
Thus the homogeneous problem has only the trivial solution. Hence the inhomo-
geneous problem has a unique solution.
120 4 Integral Equations of the Fredholm Type
4.2
Resolvent Kernel
Returning to the series solution (4.1.4), we find that upon making the following
definitions of the iterated kernels:
K
1
(x, y) = K(x, y), (4.2.1)
K
2
(x, y) =
h
0
dy
2
K(x, y
2
)K(y
2
, y), (4.2.2)
K
3
(x, y) =
h
0
dy
3
h
0
dy
2
K(x, y
3
)K(y
3
, y
2
)K(y
2
, y), (4.2.3)
K
n
(x, y) =
h
0
dy
n
h
0
dy
n−1
···
h
0
dy
2
K(x, y
n
)K(y
n
, y
n−1
) ···K(y
2
, y), (4.2.4)
we may write each term in the series (4.1.4) as
φ
1
(x) =
h
0
dyK
1
(x, y)f (y), (4.2.5)
φ
2
(x) =
h
0
dyK
2
(x, y)f (y), (4.2.6)
φ
3
(x) =
h
0
dyK
3
(x, y)f (y), (4.2.7)
φ
n
(x) =
h
0
dyK
n
(x, y)f (y). (4.2.8)
As such, we have
φ(x) = f (x) +
∞
n=1
λ
n
h
0
dyK
n
(x, y)f (y). (4.2.9)
Now, define the resolvent kernel H(x, y;λ)tobe
−H(x, y;λ) ≡ K
1
(x, y) +λK
2
(x, y) +λ
2
K
3
(x, y) +···=
∞
n=1
λ
n−1
K
n
(x, y).
(4.2.10)
Then solution (4.2.9) can be expressed compactly as
φ(x) = f (x) −λ
h
0
dyH(x, y;λ)f (y), (4.2.11)
with the resolvent H(x, y;λ) defined by Eq. (4.2.10). We have in effect shown that
H(x, y;λ) exists and is analytic for
|
λ
|
< 1
K
. (4.2.12)
4.2 Resolvent Kernel 121
Example 4.1. Solve the Fredholm Integral Equation of the second kind,
φ(x) = f (x) + λ
1
0
e
x−y
φ(y)dy. (4.2.13)
Solution. We have, for the iterated kernels,
K
1
(x, y) = K(x, y) = e
x−y
,
K
2
(x, y) =
1
0
dξ K(x, ξ)K(ξ , y) = e
x−y
,
K
n
(x, y) = e
x−y
for all n. (4.2.14)
Then we have as the resolvent kernel of this problem
−H(x, y;λ) =
∞
n=1
λ
n−1
e
x−y
= e
x−y
(1 +λ + λ
2
+λ
3
+···). (4.2.15)
For
|
λ
|
< 1, (4.2.16)
we have
H (x, y;λ) =−e
x−y
(1 −λ). (4.2.17)
Thus the solution to this problem is given by
φ(x) = f (x) +
λ
1 − λ
1
0
dye
x−y
f (y). (4.2.18)
We remark that in this case not only do we know the radius of convergence for
the series solution (or for the resolvent kernel) as in Eq. (4.2.16), we also know the
nature of the singularity (a simple pole at λ = 1). In fact, our solution (4.2.18) is
valid for all values of λ,eventhosewhichhave
|
λ
|
> 1, with the exception of λ = 1.
Properties of the resolvent: We now derive some properties of the resolvent
H(x, y;λ). Consider the original integral operator written as
˜
K =
h
0
dyK(x, y), (4.2.19)
and the operator corresponding to the resolvent as
˜
H =
h
0
dyH(x, y;λ). (4.2.20)
122 4 Integral Equations of the Fredholm Type
We note that the integrals in Eqs. (4.2.19) and (4.2.20) are with respect to the second
argument. The operators,
˜
K
2
,
˜
H
2
,
˜
K
˜
H,or
˜
H
˜
K are defined in the usual way:
˜
K
2
=
h
0
dy
2
h
0
dy
1
K(x, y
2
)K(y
2
, y
1
). (4.2.21)
We wish to show that the two operators
˜
K and
˜
H commute, i.e.,
˜
K
˜
H =
˜
H
˜
K. (4.2.22)
The original integral equation can be written as
φ = f + λ
˜
Kφ (4.2.23)
while the solution by the resolvent takes the form
φ = f − λ
˜
Hf . (4.2.24)
With the identity operator
˜
I,
˜
I =
h
0
dyδ(x − y), (4.2.25)
Eqs. (4.2.23) and (4.2.24) can be written as
f = (
˜
I − λ
˜
K)φ, φ = (
˜
I − λ
˜
H)f . (4.2.26,27)
Then, combining Eqs. (4.2.26) and (4.2.27), we obtain
f = (
˜
I − λ
˜
K)(
˜
I − λ
˜
H )f and φ = (
˜
I − λ
˜
H )(
˜
I − λ
˜
K)φ.
In other words, we have
(
˜
I − λ
˜
K)(
˜
I − λ
˜
H) =
˜
I and (
˜
I − λ
˜
H )(
˜
I − λ
˜
K) =
˜
I.
Thus we obtain
˜
K +
˜
H = λ
˜
K
˜
H and
˜
K +
˜
H = λ
˜
H
˜
K.
Hencewehaveestablishedtheidentity
˜
K
˜
H =
˜
H
˜
K, (4.2.28a)
i.e.,
˜
K and
˜
H commute. This can be written explicitly as
h
0
dy
2
h
0
dy
1
K(x, y
2
)H(y
2
, y
1
) =
h
0
dy
2
h
0
dy
1
H (x, y
2
)K(y
2
, y
1
).
4.3 Pincherle–Goursat Kernel 123
Let both of these operators act on the function δ(y
1
− y). Then we find
h
0
dξ K(x, ξ)H(ξ , y) =
h
0
dξ H (x, ξ )K(ξ, y). (4.2.28b)
Similarly, the operator equation
˜
K +
˜
H = λ
˜
K
˜
H (4.2.29a)
may be written as
H (x, y;λ) =−K(x, y) +λ
h
0
K(x, ξ)H(ξ , y;λ)dξ. (4.2.29b)
We will find this to be a useful relation later.
4.3
Pincherle–Goursat Kernel
Let us now examine the problem of determining a more explicit formula for the
resolvent which points out more clearly the nature of the singularities of H(x, y;λ)
in the complex λ plane. We examine two types of kernels in sequence. First, we
look at the case of a kernel which is given by a finite sum of separable terms (the
so-called Pincherle–Goursat kernel). Secondly, we examine the case of a general
kernel which we decompose into a sum of a Pincherle–Goursat kernel and a
remainder which can be made as small as possible.
Pincherle–Goursat kernel: Suppose that we are given the kernel which is a finite
sum of separable terms,
K(x, y) =
N
n=1
g
n
(x)h
n
(y), (4.3.1)
i.e., we are given the following integral equation:
φ(x) = f (x) + λ
h
0
N
n=1
g
n
(x)h
n
(y)φ(y)dy. (4.3.2)
Define β
n
to be
β
n
≡
h
0
h
n
(y)φ(y)dy. (4.3.3)
Then the integral equation (4.3.2) takes the form
φ(x) = f (x) + λ
N
k=1
g
k
(x)β
k
. (4.3.4)
124 4 Integral Equations of the Fredholm Type
Substituting Eq. (4.3.4) into expression (4.3.3) for β
n
,wehave
β
n
=
h
0
dyh
n
(y)f (y) +
h
0
dyh
n
(y) ·λ
N
k=1
g
k
(y)β
k
. (4.3.5)
Hence, we let
A
nk
=
h
0
dyh
n
(y)g
k
(y), α
n
=
h
0
dyh
n
(y)f (y). (4.3.6,7)
Equation (4.3.5) takes the form
β
n
= α
n
+ λ
N
k=1
A
nk
β
k
,
or
N
k=1
(δ
nk
−λA
nk
)β
k
= α
n
, (4.3.8a)
which is equivalent to the N × N matrix equation
(I − λA)
β =α, (4.3.8b)
where the α’s are known and the β
s are unknown. If the determinant of the matrix
(I − λA) is denoted by
˜
D(λ),
˜
D(λ) = det(I − λA),
the inverse of (I − λA) can be written as
(I − λA)
−1
= (1
˜
D(λ)) ·D,
with D amatrixwhoseijth element is the cofactor of the jith element of I − λA.
(We recall that the cofactor of a
ij
is given by (−1)
i+j
det M
ij
,withdetM
ij
the
minor determinant obtained by deleting the row and column to which a
ij
belongs.)
Therefore,
β
n
= (1
˜
D(λ)) ·
N
k=1
D
nk
α
k
. (4.3.9)
From Eqs. (4.3.4) and (4.3.9), we obtain the solution φ(x)as
φ(x) = f (x) +(λ
˜
D(λ))
N
n=1
N
k=1
g
n
(x)D
nk
α
k
. (4.3.10)
4.3 Pincherle–Goursat Kernel 125
Writing out α
k
explicitly, we have
φ(x) = f (x) + (λ
˜
D(λ))
N
n=1
N
k=1
g
n
(x)D
nk
h
0
dyh
k
(y)f (y). (4.3.11)
Comparing Eq. (4.3.11) with the definition of the resolvent H(x , y;λ),
φ(x) = f (y) −λ
h
0
dyH(x, y;λ)f (y), (4.3.12)
we obtain the resolvent for the case of the Pincherle–Goursat kernel as
−H(x, y;λ) = (1
˜
D(λ))
N
n=1
N
k=1
g
n
(x)D
nk
h
k
(y). (4.3.13)
Note that this is a ratio of two polynomials in λ.
We remark that the cofactors of the matrix (I − λA) are polynomials in λ
and hence have no singularities in λ. Thus the numerator of H(x, y;λ)has
no singularities. Then the only singularities of H(x, y;λ) occur at the zeros of
the denominator,
˜
D(λ) = det(I − λA), which is a polynomial of degree N in λ.
Therefore, H(x, y;λ) in this case has at most N singularities which are poles in the
complex λ plane.
General kernel: By approximating a general kernel as a sum of a Pincherle–Goursat
kernel, we can now prove that in any finite region of the complex λ plane, there
can be at most finitely many singularities. Consider the integral equation
φ(x) = f (x) + λ
h
0
K(x, y)φ(y)dy, (4.3.14)
with a general square-integrable kernel K(x, y). Suppose we are interested in
examining the singularities of H(x, y;λ)intheregion
|
λ
|
< 1ε in the complex
λ plane (with ε quite small). We can always find an approximation to the kernel
K(x, y)intheform(withN quite large)
K(x, y) =
N
n=1
g
n
(x)h
n
(y) + R(x, y), (4.3.15)
with
R
<ε. (4.3.16)
The integral equation (4.3.14) then becomes
φ(x) = f (x) + λ
h
0
N
n=1
g
n
(x)h
n
(y)φ(y)dy + λ
h
0
R(x , y)φ(y)dy.