ON SOLVING A CLASS OF SINGULAR
FREDHOLM INTEGRAL EQUATIONS
by
Stephen L. Skinner
B.A., University of Kansas, 1970
A thesis submitted to the
Faculty of the Graduate School of the
University of Colorado in partial fulfillment
of the requirements for the degree of
Master of Science
Department of Mathematics
This thesis for the Master of Science degree by
Stephen Lee Skinner
has been approved for the
Department of
Mathematics
by
Weldon A. Lodwick
Date
iii
Skinner, Stephen Lee (M.S., Applied Mathematics)
On Solving a Class of Singular Fredholm Integral Equations
Thesis directed by Assistant Professor Weldon A. Lodwick
A known procedure for obtaining closed form solutions to a
class of singular Fredholm integral equations with quotient
kernels is shown to be applicable to a more general class of such
equations. Conditions under which the procedure is valid are
given.
The procedure relies on application of the Fourier
transform to the integral equation, yielding a functional relation
between the transforms of the known functions and transform of the
solution in Fourier space. A mapping in the complex plane is then
utilized to arrive at the wellknown RiemannHilbert problem for a
sectionally analytic function satisfying a given jump condition.
The RiemannHilbert problem is. solved using standard techniques.
Inverse mappings and inverse Fourier transforms are then applied
to its solution, thereby yielding a solution to the original
integral equation.
IV
ACKNOWLEDGEMENTS
I would like to express my thanks to Professor Weldon A.
Lodwick for serving as my thesis advisor, and to Professors
William L. Briggs and Stephen McCormick for serving on my
committee. In addition, I would like to thank Marvin E. Goldstein
of NASA Lewis Research Center and Professor Henry W. Wyld of the
University of Illinois for providing information regarding
applications of the integral equations discussed herein.
CONTENTS
CHAPTER
I. INTRODUCTION . ........................................ 1
1.1 Purpose............................................ 1
1 .2 Assumptions....................................... 2
1.3 Previous Results and Applications................... 3
1.4 Solution Procedure ................................. 3
II. PRELIMINARIES............................................ 7
2.1 Introduction ....................................... 7
2.2 Definitions....................................... 7
2.3 Theorems............................................ 9
2.4 Plemelj Formulas .................................. 11
2.5 The RiemannHilbert Problem.........................13
2.6 Solution of the RiemannHilbert Problem
on an Open Contour ................................15
2.7 Uniqueness of the Solution of the Riemann
Hilbert Problem. .................................. 24
III. SOLUTION OF THE INTEGRAL EQUATION........................26
3.1 Introduction........................................26
3.2 Change of Variables.................................26
33 Application of the Fourier Transform .............. 28
3.4 Complex Mapping.....................................32
3.5 Solution of the RiemannHilbert Problem.............40
VI
CONTENTS (continued)
CHAPTER
3.6 Inversion...........................................'44
3.7 Uniqueness......................................... 49
IV. CONCLUSIONS. /.....................................'. . 50
4.1 Conditions of Validity...............................50
4.2 Topics for Further Study.............................52
REFERENCES
53
vii
FIGURES
Figure
2.4 1 Limiting values of W on approaching an open
contour..................................................12
3.3_1 Strip of analyticity in Fourier space....................31
3.4 1 Image under $ of opposite points on boundary of
strip................................................ 35
3.4 2 Image under $ of points approaching boundary of
strip. '..............................................37.
CHAPTER I
INTRODUCTION
1.1 Purpose
The purpose of this study is to obtain a closed form
solution to the following class of integral equations:
f(x) = h(x) + xk(^]ymf(y)dy (0 < x < )
o
(1.1.1)
where
f(x) is an unknown function of the real (1.1.2)
variable x
\ is a given nonzero constant
m is a given nonnegative integer
(1.1.3)
(1.1.4)
h(x) is a given function of the real
variable x, and h e L2 on (o,).
is a given function of the real
quotient and k e L2 on (o,).
(1.1.5)
(1.1.6)
2
In solving Eq. (1.1.1) we wish to obtain an expression for the
unknown function f(x) in terms of the known quantities X, m, h(x)
and kf).
V
Since the upper limit of integration is fixed at infinity,
Eq. (1.1.1) represents a singular Fredholm integral equation. The
coefficient of f(x) on the left hand side of (1.1.1) is unity, so
the integral equation is of the second kind. Furthermore, the
given function h(x) will not in general be identically zero, so
the equation is nonhomogeneous. The function k(^J is known as the
kernel.
1.2 Assumptions
The only a priori assumptions that are made regarding
Eq. (1.1.1) are those stated in (1.1.2)(1.1.6). We make the
assumption h e L2 and k e L2 on (o,) in order to justify the
initial application of the Fourier transform to the integral
equation (after first invoking a change of variable).
In the course of carrying out the solution it will be
necessary to impose additional conditions on certain functions
that are related to f, h and k. These conditions are sufficient
to guarantee the validity of the solution method, and are
summarized in Section 4.1. In general, these conditions arise out
of the need to evaluate Fourier transforms and their inverses, as
well as Cauchytype integrals as part of the solution procedure.
The reader who is interested in applying the solution method
described herein for specific forms of h and k may find it useful
to review Section 4.1 before proceeding.
3
1.3 Previous Results and Applications
Upon setting m = 0 in Eq. (1.1.1) we obtain the following
integral equation:
f(x) = h(x) + (y)dy
o
(0 < x'< ) .
(1.3.1)
A procedure for solving the above equation in closedform has been
given by Goldstein and Leitman [H]. They noted that their
procedure for solving (1.3.1) could be used to solve the more
general case (1.1.1) where m is allowed to be any real constant.
The application of their method for nonnegative integral values of
m is particularly straightforward, as shown herein.
Integral equations of the type given above are encountered
in both fluid mechanics and high energy physics. Illustrations of
how equations of this type arise can be found in Goldstein [3], as
well as in Horn and Zachariasen [5], page 212.
1.U Solution Procedure
The extension of the procedure developed by Goldstein and
Leitman which we will use to solve Eq. (1.1.1) is summarized
briefly in the steps' below. These steps will be carried out in
more detail in Chapter III.
4
Step (i) Change of Variables
The variables x and y in Eq. (1.1.1) are replaced by
x = lnx
( < X < oo)
(1.4.1)
y = lny
(oo < y < oo).
(1.4.2)
In terms of these new variables, Eq. (1.1.1) becomes
F(x)e ^m+1^x = h(x) + xjK(xy)F(y)dy
(1.4.3)
where
FU) e(m*1)x f(ex)
(1.4.4)
H(x) = h(e )
(1.4.5)
K(xy) = k(eX y)
(1.4.6)
Step (i'i) Fourier Transform
The Fourier transform with complex argument a is applied
to Eq. (1.4.3) giving formally
F(a + (m+1)i] = H(a) + (2i:) ^ XK(a)F(a)
(1.4.7)
5
A A A
where F, H and K denote the Fourier transforms of F, H and
K. To avoid possible complications when computing inverse
transforms, we will impose the following condition:
AAA
F(a), H(a), K(a) must be analytic in the (1.4.8)
strip in the complex
Fourier plane given by
0 Â£ Im(ct) Â£ (m+1), where
Im denotes imaginary part.
Step (iii) Complex Mapping
The mapping
z = $(<*) = exp[(^y)a] (1.4.9)
is then applied to the strip 0 Â£ Im(ct) < (m+1), and the
equivalent form of the relation (1.4.7) under $ is
derived. In doing so, the task of determining the unknown
transform F is found to be equivalent to determining a
function W(z) = F(^lnz) which is analytic everywhere in
the complex plane, except on the positive real axis
(o,0. W(z) is thus a sectionally analytic function.
In addition, W(z) must satisfy a 'jump condition across
the positive real axis (o,), a direct consequence of the
6
relation (1.4.7) in Fourier space. This jump condition
will be shown to be of the form
W+(s) p(s)W~(s) = q(s) (0 < s < oo) (1.4.10)
where the functions p(s) and q(s) are known and W+,W
denote limiting values of the function W on approaching
the positive real axis from opposite directions. The task
of determining a sectionally analytic function subject to
a given jump condition is the wellknown RiemannHilbert
problem (Carrier, Krook and Pearson [1], Gakhov [2]).
Step (iv) Solve the RiemannHilbert Problem
' Standard techniques are then used to solve the above
RiemannHilbert problem, thus obtaining the unknown
function W. In carrying out this step, extensive use is
made of the Plemelj formulas (Section 2.4).
Step (v) Inversion
The final step consists of using inverse mappings and
inverse Fourier transforms to determine an expression for
f(x) in terms of the known quantities.
CHAPTER II
PRELIMINARIES
2.1 Introduction
In this chapter a few wellknown definitions and theorems
from analysis which will be used later are summarized. In
addition, a brief discussion of the Plemelj formulas and the
RiemannHilbert problem is given.
2.2 Definitions
A. Holder Condition (Gakhov [2])
A function f(z) defined on a smooth contour C in the
complex plane is said to satisfy a Holder condition on C. if for
all pairs zt,z2 on C there exist constants 'A' and 'a' such that
f(z2) f(z j)  < A z2z!  (2.2.1)
where
A > 0 and 0 < a Â£ 1 .
Eq. (2.2.1) is also commonly referred to as a Lipschitz condition
of order a'.
8
B. Complex Fourier Transform (Morse and Feshbach [6])
A
Let a be a complex variable. The Fourier transform f(ot)
of a function f(x) is given by
f(a) = (2ir) 1/2 jelaXf(x)dx
(2.2.2)
C. Order of a Function (Gakhov [2])
By the order of an analytic function f(z) at a point z0 in
the finite complex plane we mean the exponent of the lowest power
of (zz0) in the Taylor series expansion of f(z).
To determine the order of f(z) in the vicinity of the
point at infinity, we expand f(z) in powers of The exponent of
Z
the lowest power of ^ in the expansion is termed the order of f at
infinity. If the order is negative, f has a pole at infinity.
To illustrate the above, consider the order of
f(z) = z + z2 at the point z0 = 0. Since f is already in
the correct Taylor series form, we see directly that the order
1
at z0= 0 is one. Expanding f(z) in powers of gives
z
1 ~ 1 i 2
f(z) = () + () so its order at infinity is minus two.
D. Cauchy Integral (Gakhov [2])
Let L be a smooth contour in the complex plane, and let
f(t) be a complexvalued function prescribed on L. If z is not a
point on L, then the following is known as a Cauchy integral:
9
1 ff(t)
2iri Jtz
dt
L
(2.2.3)
2.3 Theorems
A. Fourier Inversion Formula (Morse and Feshbach [6])
/v
Let f(a) be the Fourier transform of a function f(x).
Then
f(x) = (2rr)_ 1/2 je'iaXf(a)dcx
(2.3.1)
B. Convolution Theorem (Morse and Feshbach [6])
Let f(x),g(x) be Lz integrable functions on the interval
A /S
(,). In addition, let f(a),g(a) be the Fourier transforms of
f(x),g(x). Define
00
h(x) = (2ir) 1/2 Jf (s)g(xs)ds
(2.3.2)
Then
h(a) = f(a)g(a)
(2.3.3)
10
C. Function Having a Given Fourier Transform (Noble [7])
Let a = a + it be a complex variable, and let F(a) be a
function 'of a analytic in a strip in the complex plane given by
a < t < b. In addition, suppose F(a)  * 0 as  a  * uniformly
in the strip a+e Â£ t Â£ be, where e>0. For a given t such that
a < t < b, and any real x, define f(x) as follows:
iT+co
f (x) = (2tt)"1/2 J F(a)eiaXda . (2.3.4)
i t~
Then
F(a) = (2tt) 1/2 jeiaXf(x)dx
(2.3.5)
In other words, F(a) is the Fourier transform of f(x):
f(a) = F(a) . (2.3.6)
D. Generalized LiouvilleTs Theorem (Gakhov [2])
Let the function f(z) of the complex variable z be
analytic in the finite complex plane and let f(z) have a pole of
order n at infinity. Then f(z) is a polynomial of order n.
11
E. Theorem on Analytic Continuation (Gakhov [2])
Suppose that the two domains Dj and D2 in the complex
plane have a common smooth boundary L. In the domains Dj and D2
two analytic functions ft(z) and f2(z) are given. Assume that
when a point z in the complex plane tends to the curve L both
functions tend to limiting values which are continuous on the
curve L and equal to each other. Under these conditions, the
functions fx(z) and f2(z) constitute the analytic continuation of
one another.
2.4 Plemelj Formulas
Let C be a smooth contour (closed or open) in the complex
plane. Let f(a) be a function of the complex variable a defined
on the contour C, and let f satisfy the Holder condition (2.2.1)
on C.
Let t denote an arbitrary point on the contour C, with the
stipulation that t not be an endpoint of C in the case where C is
open. Now let a point z (not on C) approach t along an arbitrary
path from the left or right. Here, left and right correspond to
the regions to one's immediate left or right on traversing C in a
positive direction (see Figure 2.41). If C is closed, the
positive traversal will always be taken to be counterclockwise.
In that case, approaching t from the left or right corresponds to
approaching t from inside or outside C, respectively.
Under the above conditions, the Cauchy integral
W(z)
1 rf(ct)
2ir i Jaz
da
C
(2.4.1)
Positive
Direction
w*(t)
Fi gur e 2.41
Limiting values of W on approaching
an open contour
13
tends to the following limits as z + t from the inside/left ( + )
and the outside/right () of C:
W+(t) = lf(t) + ^r X ^^da (2.4.2)
2 2iri J a~t
C
W(t) = f(t) + 5T (fT^a (2.4.3)
2 2iri J at
C
In the above (and in the sequel) ^ denotes the Cauchy principal
value. Also, subtracting (2.4.3) from (2.4.2) gives the useful
result
W+(t) W(t) = f(t) . (2.4.4)
Eqs. (2.4.2), (2.4.3) are known as the Plemelj formulas,
or the Sokhotski formulas. They are derived in Gakhov [2].
2.5 The RiemannHilbert Problem
The RiemannHilbert problem consists of finding a function
W(z) analytic in the complex plane except along a contour L (open
or closed), across which W satisfies a prescribed'jump condition.
It should be noted that some authors refer to the above as
Riemann's problem. We shall adopt the nomenclature of Carrier,
Krook and Pearson [1] in which Riemann's problem' refers
exclusively to the case where the contour L is closed, and
'RiemannHilbert problem' refers to the more general case where L
may be open or closed. In the sequel we will be concerned with
the case where L is open.
14
The jump condition for the RiemannHilbert problem takes
the following form:
W+(t) p(t)W(t) = q(t) t e L
(2.5.1)
where
L is an open contour
W+(t),W(t) denote the limiting values of
W(z) as z approaches the point
t on the contour from the left
(+) or right (),.respectively.
p(t),q(t) are given complexvalued (2.5.2)
functions defined on L,
both satisfying a Holder
condition on L, and
p(t) 4 0 on L
(2.5.3)
As can be seen from Eq. (2.5.1), the jump condition constitutes a
linear relationship between W+(t) and W"(t). The function p(t) is
known as the coefficient of the RiemannHilbert problem while q(t)
is referred to as the free term (Gakhov [2]). The problem is
termed homogeneous if q(t) h 0 on L, and nonhomogeneous otherwise.
In order for the RiemannHilbert problem to be completely
specified, the behavior of W(z) at infinity and near the ends of
15
the open contour L must be prescribed (Carrier, Krook and Pearson
[1], page 419). Also, we will require that the solution of the
problem be such that W+(t) and W(t) are integrable on L.
2.6 Solution of the RiemannHilbert Problem on an Open
Contour
Methods for solving the RiemannHilbert problem on open
contours are developed fully in Chapter VI of Gakhov [2]. A brief
and informative discussion is also given on pages 418421 of
Carrier, Krook and Pearson [1]. The main aspects of the solution
are summarized here. The solution proceeds by first solving the
homogeneous case obtained by setting q(t) = 0 in (2.5.1), and then
solving the nonhomogeneous case.
In the following discussion we will assume that all
contours are smooth and do not intersect themselves. The initial
point 'a' and final point 'b' of the open contour are defined in
such as way that one goes from 'a' to 'b' when traversing the
contour in the positive direction. Before proceeding, we
introduce the necessary concept of the index of the problem.
A. Index of the Problem
The solution of the RiemannHilbert problem will involve
an integer n known as the index of the problem. To compute the
value of n we first express in polar form the value of the
coefficient p(t) in (2.5.1) evaluated at the initial point 'a' of
L:
/ \ i 0
p(a) = r e
(2.6.1)
16
The range of 0 in (2.6.1) is taken as
2tt < 0 Â£ 0 if the solution W(z)
is required to be bounded
at z = a, or (2.6.2)
0 < 6 < 2ir if the solution W(z) is
to be unbounded at z = a. (2.6.3)
Next, we express the value of p(t) at the final point b1
of L as
p(b) = r"el(0+A) . (2.6.4)
In (2.6.4), k denotes the change in argument of p(t) on L.
Now, if [x] denotes the largest integer not exceeding x,
we define
n = if the solution is to be (2.6.5)
L 2tt J
bounded at z = b, or
n = [^r^l + 1 if the solution is to be (2.6.6)
unbounded at z = b.
The integer n given by (2.6.5) or (2.6.6) is termed the index of
the problem (Gakhov [2], pages 421422).
17
B. Solution of the Homogeneous Problem
Setting q(t) = 0 in (2.5.1) and denoting the unknown
function by V(z) gives
V+(t) = p(t)V"(t) t e L (2.6.7)
where V(z) is required to be a nontrivial solution which is
analytic everywhere in the complex plane, except on the open
contour L where the jump condition (2.6.7) must be satisfied.
Taking the logarithm of both sides of (2.6.7) gives
In V+(t) in V"(tj = In p(t) t e L (2.6.8)
where any branch of In p(t) is acceptable (Gakhov [2],' page 92).
If we assume for the moment that In V(z) vanishes at infinity then
a solution of (2.6.8) is
In V(z) = ^f1 P(^~ dx . (2.6.9)
2irlJ TZ
L
We first note that the function defined in (2.6.9) is indeed
analytic for z not on L (Wyld [8], page 479). To verify that
(2.6.9) satisfies (2.6.8) we simply evaluate the limiting values
of the integral (2.6.9) from left and right using the Plemelj
formulas (2.4.2) and (2.4.3). In doing so we obtain
In V+(t) = 1 In p(t) + ^r { PpdT
2 2iu J tt
L
(2.6.10)
18
In V~(t) = in p(t) + 2^nr f (2.6.11)
L
Subtracting (2.6.11) from (2.6.10) gives
In V+(t) In V"(t) = In p(t) (2.6.12)
which validates (2.6.9).
If we define
r(z) = In V(z) (2.6.13)
then the nonvanishing function
V(z).er(z) (26'11,)
is analytic except on L (being a composition of analytic
functions) and is the required solution of the homogeneous problem
(2.6.7), except possibly at the endpoints of L, assuming that
In V(z) vanishes at infinity.
We now recall that the Plemelj formulas used to obtain
(2.6.14) are not in general valid at the endpoints of an open
contour (Section 2.4). Thus, (2.6.14) may not satisfy the jump
condition at the initial and final points of L. The following
slight modification of (2.6.14) is used to obtain a solution
having the proper behavior at the endpoints of L (Gakhov [2], page
429):
19
V(z) = (zbfn er(z)Pn(z) (n i 0)
(2.6.15)
where
n is the index of the problem, computed by
(2.6.5) or (2.6.6)
b is the final point of the open contour L
Pn(z) is an arbitrary polynomial of degree n
with coefficients that may be complex.
It is important to note that Eq. (2.6.15) is valid only
for n > 0. A nontrivial solution of the homogeneous problem does
not exist if n < 0 (although the nonhomogeneous problem may still
be solvable in such cases). It is also appropriate to note that
(2.6.15) is valid only for an open contour having only one initial
point a and one final point b. This will be suitable for our
purposes since we will be specifically interested in the case
where L is the positive real axis (o,). For more general types
of open contours, Eq. (2.6.15) must be modified slightly (Gakhov
[2], page 429).
It is customary to rewrite (2.6.15) as follows:
V(z) = X(z)Pn(z)
(n > 0)
(2.6.16)
20
where
X(z) = (zb)ner(z) . (2.6.17)
The function X(z) is known as the canonical function of the
homogeneous problem. X(z) has order zero in the finite complex
plane and order n at infinity.
The special case in which V(z) is required to vanish at
infinity is worthy of note. Under these conditions (2.6.16)
becomes
V(z) = X(z)Pn_1(z) (nil), V() = 0 (2.6.18)
where
P^_^(z) is an arbitrary polynomial of (2.6.19)
degree n1.
C. Solution of the Nonhomogeneous Problem
With the general homogeneous solution (2.6.16) in hand, we
now return to the original nonhomogeneous problem with jump
condition (2.5.1)
W+(t) p(t)W"(t) = q(t)
t e L
(2.5.1)
21
To solve the above, Eqs. (2.6.7) and (2.6.16) are used to
replace p(t) by the ratio of the limiting values of the canonical
function X(z) on L to get
W+(t)  W(t) = q(t)
X(t)
Rewriting the above yields
W+(t) W~(t) + q(t)
X+(t) X~(t) X+(t)
(2.6.20)
(2.6.21 )
It can be shown that the function ^^ satisfies a Holder condi
X+(t)
tion on L (Gakhov [2], page 38).
We now introduce the function t(z) (Carrier, Kpook and
Pearson [1], page 419),
'V(z)
1 r q(x) dx
2it1x+(x) t_Z
(2.6.22)
A direct application of the Plemelj formulas to the above gives
X+(t)
where as usual Â¥'f(t), 4,_(t) denote the limiting values of i'(z) on
approaching t e L from the left and right.
Substituting (2.6.23) into (2.6.21) gives the jump
condition in the revised form
22
W+(t)
X+(t)
 Â¥+(t) = " Â¥(t) .
X"(t)
(2.6.24)
By the above equation we see that the function
(2.6.25)
has the same boundary values on each of the two sides of L. By
the theorem on analytic continuation (2.3E) it can be argued that
the function (2.6.25) is analytic in the whole complex plane,
except possibly at infinity (Gakhov [2], page 90). If the
function (2.6.25) is prescribed to have order n at infinity, where
n is the index of the problem, it must be equal to an arbitrary
polynomial of degree n by the generalized version of Liouville's
theorem (2.3_D). If n < 0 this polynomial is taken to be
identically zero. Prescribing such behavior at infinity, we
obtain
(2.6.26)
where
n
is the index (2.6.5) or (2.6.6)
Pn(z) is an arbitrary polynomial of degree n
with possibly complex coefficients
(n i 0)
23
Pn(z) =0 (n < 0)
X(z)
is the canonical function (2.6.17)
Â¥(z)
is given by (2.6.22).
Rewriting (2.6.26) gives
W(z) = X(z) ['J'(z) + Pn(z) ]
(2.6.27)
which is the desired solution of the nonhomogeneous Riemann
Hilbert problem (2.5.1). It is observed that this solution
contains the term X(z)Pn(z), which was the solution of the
homogeneous case.
For the important special case in which the solution W(z)
is required to vanish at infinity, (2.6.27) takes the form
additional condition must be satisfied in order for (2.6.27) to
constitute a solution to the nonhomogeneous problem (Gakhov [2],
page 430):
W(z) = X(z) [l'(z) + Pn1(z)]
(2.6.28)
W(>) = 0
(2.6.29)
Lastly, it should be noted that if n < 0 the following
n)
(2.6.30)
24
where
n < 0 .
2.7 Uniqueness of the Solution of the RiemannHilbert Problem
In this section we briefly discuss the question of
uniqueness of the solution of the general RiemannHilbert problem.
For a more complete discussion of this topic,, see pages 93~99 of
Gakhov [2].
A. Homogeneous RiemannHilbert Problem
We consider first the uniqueness of the solution of the
homogeneous problem discussed in Section 2.6B. If the index
n = 0, the solution of the homogeneous problem is unique up to an
arbitrary multiplicative constant. Once the behavior of the
homogeneous solution at infinity is specified, the value of this
constant becomes determined and the solution then becomes unique.
If the index n is strictly positive (n > 0) in the
homogeneous problem then there exist (n + 1) linearly independent
solutions. The general homogeneous solution contains (n + 1 )
arbitrary constants which enter as the coefficients of the
polynomial Pn(z) (see Eq. 2.6.15).
Lastly, if the index n is strictly negative (n < 0) then
the homogeneous RiemannHilbert problem has' no solution.
25
B. NonHomogeneous RiemannHilbert Problem
Next, we examine uniqueness of the solution of the
nonhomogeneous problem discussed in Section 2.6C. If n > 0 then
the solution of the nonhomogeneous problem is not unique since
this solution contains the polynomial Pn(z) with (n + 1) arbitrary
coefficients (Eq. 2.6.27). For the special case n = 0, Pn(z) =
P0(z) is a constant and this constant is determined once the
behavior of the solution at infinity is specified. Thus, for n = .
0 the nonhomogeneous solution is unique provided that behavior at
infinity is given.
If the index n is strictly negative (n < 0) then the
polynomial pn(z) is by definition indentically zero and the
arbitrariness in the solution is removed (provided, of course,
that a solution exists). If n = 1 the nonhomogeneous problem has
a unique solution. If n < 1 then a necessary and sufficient
condition for solubility of the nonhomogeneous problem is that the
conditions (2.6.30) be satisfied. If this is the case, then the
nonhomogeneous solution is unique.
CHAPTER III
SOLUTION OF THE INTEGRAL EQUATION
31 Introduction
We now derive a solution of the integral equation (1.1.1),
namely
f(x) = h(x) + xk(^)ymf(y)dy (0 < x < <) . (1.1.1)
o
As stated in Section 1.1 the function f is to be determined, h and
k are known functions such that h e L2 and k e L2 on (o,), X is a
nonzero constant, and m is a nonnegative integer.
We will transform the above integral equation into a
nonhomogeneous RiemannHilbert problem which can be solved using
the results of the previous chapter. The solution proceeds
according to the steps outlined in Section 1.4.
3.2 Change of Variables
As in (1.4.1) and (1.4.2) we make the substitution
x = lnx (1.4.1)
y = lny .
(1.4.2)
27
Then
dy = eydy
Also,
oo < y < oo
and
< x < <*>
Substituting (3.2.1)(32.5) into (1.1.1) gives
(32.1)
(32.2)
(3.2.3)
(32.4)
(3.2.5)
(32.6)
(32.7)
f(ex) = h(ex) + AJk(eX y)e(m+1 )yf(ey)dy . (32.8)
00
28
We now introduce the functions
FU> e(m*,)xf(ex>
KCxy) = k(ex y)
H(x) = h(e ) .
Substituting (3.2.9)(32.11) into (3.2.8) gives
F(x)e (m+1)x = h(x) + AK(xy)F(y)dy
where
OT < X <
33 Application of the Fourier Transform
At this point the Fourier transform with complex
ct is applied to (3.2.12). The transforms of H(x) and K(x
since we have assumed
h e L2 on (o,)
(3.2.9)
(3.2.10)
(32.11)
(3.2.12)
argument
y) exist
(3.3.1)
k e L2
on
(o,)
(3.3.2)
29
Applying the Fourier transform formally to the left side
of (3.2.12) gives
(2ir)
1/2
riax
e e
(m+i )x.
F(x)dx =
(2,rr/s 
00
A
F[a + (m+1 )i] (3.3.3)
A
where F(a) denotes the Fourier transform of F(x).
Taking the transform of the right side of (3.2.12) and
using the convolution theorem (2.3~B) gives formally
H( a) + (2Tr)l/2AK(a)F(a) (3 3 ^)
where, as before, carets denote Fourier transforms of the
respective functions.
Equating (3.33) and (3.3 4) gives
F[a + (m+1)i] = H(a) + (2u)l/2AK(ct)F(a) . (335)
The above relationship allows one to compute the value of F
outside the strip 0 Â£ Im(a) Â£ (m+1) if its value within this strip
is known.
30
At this point we would like to solve (3.3.5) for F(a) in
^ A
terms of the known transforms H(a) and K(a). Inspection of
(33.5) shows however that this cannot be done, except in the very
special case m = 1. Since we are assuming m to be a nonnegative
integer, a different approach will, be required to obtain F(a) in
terms of known quantities.
We can make some important observations before proceeding.
First of all, the integrals representing Fourier transforms of
most functions do not converge in the entire complex Fourier
plane. Instead, convergence normally occurs only in a strip
parallel to the real axis extending from  to + in the real
direction, and of finite extent in the imaginary direction. Also,
when dealing with expressions relating the Fourier transforms, of
several functions, it is prudent to require that their respective
strips of analyticity overlap (Morse and Feshbach [6], page 964).
In view of the above comment and the fact that (335) is
A A A
an expression relating the transforms F, H and K, we will require
that their strips of analyticity overlap. Also, in view of the
nature of the relationship (3.3.5), we will require that this
strip of analyticity be given by 0 < Im(ct) < (m+1). We thus
arrive at the following condition:
A A A
F(a), H(a), K(a) must be analytic in the strip (3*3.6)
0 C Im(ct) < (m+1) in the
complex Fourier plane.
This strip of analyticity is illustrated in Figure 33~1.
31
Im a
32
3.4 Complex Mapping
We now define the following mapping from the complex
Fourier domain into the complex zplane:
z = *(a) = exp[(~)a]
(3.^.1)
where in general both a and z are complex. Now let
z = s + it (s,t real) . (3.4.2)
Taking the logarithm of (3.4.1) gives
0 Â£ Arg z < 2ir
(34.3)
(3.4.4)
We now express the transforms F(a), H(a), K(a) in terms of
the new complex variable z by defining the following functions:
W(z) = F(^lnz) = F( a)
(3.4.5)
q(z) = H(^lnz) = H(o)
(3.4.6)
p (z) = ( 2tt) AK(^ilnz)
l/ 2 ~
(2ir) AK(a) .
(3.4.7)
33
It can now be observed that in the zplane the functions
W(z), p(z), and q(z) will in general be discontinuous across the
positive real axis, by virtue of the discontinuity in lnz there.
Thus, these functions are not analytic on the positive real axis.
However, the functions W(z), p(z) and q(z) are analytic at
any point z0 in the zplane not on the positive real axis (o,).
To see this, consider W(z0) (p(z0) and q(z0) are similar).
W(z0) = F[(~)lnzD] = F(ct0) (3.4.8)
where
o
(^)[ln z0  + i Arg z0]
0 C Arg z0 < 2ir
(3.4.9)
From (3.4.9)
0 < Im(a0) < m+1
so by (33.6) F(a0) is analytic, so W(z0) is analytic by (3.4.8).
To summarize, we have found that the new functions W(z),
p(z) and q(z) introduced in (3.4.5)(34.7) above are analytic
everywhere in the zplane, except along the positive real axis
(o,).
We now investigate the behavior of the relation (3.3.5)
under the mapping (3.4.1). To do this we first note that any
34
point on the boundary of the strip 0 Â£ Im(a) Â£ m+1 in Fourier
space, is mapped onto the positive real axis by $. Furthermore,
any two points ct^c^ on opposite boundaries of the strip such that
Re a1 = Re a2 are mapped to the same point on the real axis (see
Figure 3.41).
To verify the above, let
(x real)
(34.10)
a2 = x + i(m+1)
(3.4.11)
Then by (3.4.1)
$(0 = exp[(Jf)x]
(3.4.1)
which is real and positive. Also
$(a2) = exp[(^y)(x + i(m+1 )) ]
= $(dj)
So $(aj) = $(a2) as conjectured.
35
Fourier space zplane
F i gur e 3.41
Image under $ of opposite points
on boundary of strip.
36
We now show that approaching the points alta2 on the strip
boundaries from above and below, respectively, corresponds to
approaching their common image point (on the positive real axis)
under $ from above and below the real axis (see Figure 3.42).
To see this, let a1 and a2 denote points directly opposite
each other on the boundaries of the strip 0 Â£ Im(a) < m+1 as
before, that is
Now consider points a'x and a2 inside the strip, lying just above
and below ax and ct2. We can then write
a
x
(x real)
(3.10)
a2 = x + i(m+1)
(3.4.11)
af = x + ie
0 < E <<
(34.12)
ct2 = x + i(m+1e)
(3.4.13)
Then
4>(<0 = exp[(^)(x + ie)]
(34.14)
37
Fourier space zplane
Figure 3.42
Image under $ of points approaching
boundary of strip.
38
In the above equation, we note that is small and positive,
so $(cti) lies above the positive real axis. As e tends to zero in
(3.4.14) the cosine and sine terms tend to one and zero, respec
tively, so $(a0 tends to $(0^) = exp[(^y)x] from above the
real axis.
Similarly
$(ct2) = exp[(^y)(x + i (m+1e) ]
(3.4.15)
In this case the imaginary part of $>(a2) is negative, so $(a2)
approaches $(a2) from below as e tends to zero.
Bearing the above in mind, consider the relation (3.3*5)
where a = x is real. Then
f[x + (m+1)i] = H(x) + (2tt) l/2AK(x)F(x) (34.16)
which is equivalent to
lim {F[x+(m+1e)i ] H(x+ie) (2ir) AK(x+ie)F(x+ie)} = 0
e+0
(34.17)
39
If we denote $(x) = s (where s is real) and rewrite (3.4.17) in
terms of the functions W(z), p(z), q(z) defined above we get by
analogy with (3.4.14) and (3.4.15)
lim {w(s(1e") ie") q[s(1e~) + ie"] 
e"+0
z">0
p[s('le) + ie']w[s(1e~) + is']} = 0 (34.18)
where
e
2ire
m+1
(34.19)
E .
(3.4.20)
In (3.4.19), (3.4.20) we have expanded the functions sin(^y] and
cos(^y) and retained only the lowest order term in e.
m+1
Letting z" tend to zero in (3.4.18) gives
lim fw(sie'') qts+ie') p(s+ie')W(s+ie'*)} = 0 . (3.4.21)
E"+0
If we define
W+(s) = lim W(sie^)
e'*0
(3.4.22)
W~(s) = lim W(s+ie')
Â£'+0
(3.4.23)
40
and note from their definitions that p(z) and q(z) are continuous
on approaching z = s from above, then (3.4.21) becomes
W+(s) p(s)W(s) = q(s) (0 < s < oo) . (3,4.24)
We immediately recognize (3.4.24) as a jump condition on the
positive real axis (o,) for the unknown function W(z) defined by
(3.4.5). Furthermore, we showed above that W(z) must be analytic
everywhere in the zplane except on the positive real axis (o,).
Thus, solving the integral equation (1.1.1) is seen to be equi
valent to solving a RiemannHilbert problem for W(z) in the z
plane, subject to the jump condition (3.4.24).
3.5 Solution of the RiemannHilbert Problem
We can immediately apply the results of Chapter II to find
the unknown function W(z), provided that the functions p(s) and
q(s) in (3.4.24) satisfy the Holder condition (2.2A) on (o,) and
that p(s) is nonvanishing. We will assume that this is the case.
As seen by (3.4.24), the open contour L for the Riemann
Hilbert problem which we are about to solve is the positive real
axis (o,). In order to proceed we must specify the positive
direction of traversal on L. To be consistent with (3.4.22) and
(34.23) the positive direction is taken as being toward the
origin. This gives a = as the initial point and b = 0 as the
final point of L.
Now, proceeding as in Section 2.6 we solve the homogeneous
problem obtained by setting q(s) = 0 in (34.24)
41
V+(s) p(s)V (s) = 0
(3.5.1)
The solution of the above equation can be obtained directly by
setting b = 0 in Eq. (2.6.15):
V(z) = z ner(z)Pn(z)
(35.2)
where
n is the index of p(s), computed using (2.6.5)
or (2.6.6)
r(z) = ^f1i}.o.P(s) ds (by (2.6.9),(2.6.13))
2ttij sz
o
(3.5.3)
Pn(z) is an arbitrary polynomial of degree n.
The leading minus sign in (3.5.3) is required since the
contour is being traversed in a negative direction by the
convention established above. Also, we assume that the integral
(3.5.3) converges.
If V(z) is required to vanish at infinity, (3*5.2) is
replaced by the following (using (2.6.18)):
V(z) = z ner^ZVn_^ (z)
(3.5.4)
42
Rewriting (3.5.2) and (3.5.4) in terms of the canonical function
(2.6.17) gives
At this juncture it is worth recalling that the index n
appearing above is computed by expressing p(s) in polar form at
the endpoints of L, that is at s = 0 and s = . To insure that
this can be done, we will require that p(s) tend to a finite limit
as these endpoints are approached along L.
The nonhomogeneous solution is now obtained from (2.6.27),
V(z) = X(z)Pn(z)
(3.5.5)
V(z) = X(z)Pn_1(z) (V() = 0)
(3.5.6)
where
(3.5.7)
W(z) = X(z)[t(z) + Pn(z)]
(3.5.8)
where
X(z) is given by (35.7)
Pn(z) is an arbitrary polynomial of degree n
43
__1r q(s) ds
X+(s) 3z
(using (2.6.22)) .
(3.5.9)
We assume that the above integral converges.
As before, if W(z) is required to vanish at infinity then
(3.5.8) is replaced by
W(z) = X(z) ['{'(z) + Pn1(z)]
Recalling (2.6.30), if p(s) is such that the index n is
negative, then the following additional condition must be
satisfied in order for a nonhomogeneous solution to exist:
0
f _aiÂ£lgj1C[S = 0 (j = 1,2........n) (3.5.10)
^ X+(s)
if
n < 0 .
For later use, we rewrite (35.8), substituting for {' from
(3.5.9) to get
W(z)
ds
sz
Pn(z)]
+
(3.5.11)
44
where
X+(s) is the limiting value of the canonical
function X(z) on approaching the point
s on L from the left (which, by our
convention, will be from below the
positive real axis).
The expression for W(z) in (3.5.11) can in principle be
evaluated once the functions k(~) and h(x> in the original
integral equation are given. These functions determine p(s) and
q(s) in (3.5.11). The index n appearing in (3.5.11) is computable
from p(s)' and the specified behavior of W(z) at infinity using
(2.6.5) or (2.6.6). If n is negative, condition (3.5.10) must
also be verified.
36 Inversion
Inverse Fourier transforms and inverse mappings can now be
used in conjunction with the expression (3.5.8) for W(z) to obtain
the integral equation solution f(x).
Since W(z) = F(a) by (3.4.5), we must first obtain the
inverse Fourier transform of W. Recalling Eq. (2.3.1) we see that
this will involve integrating W along the entire real axis, and we
must thus determine whether to integrate W+(s) or W(s) on (o,).
To do this, we recall that W(z) = F(a) and for the case at hand 0
< x < (m+1) in the inversion formula (2.3.4). As ix in (2.3.4)
approaches the real axis from above, so does its image point $(ix)
45
in the zplane (Figure 3*42). The correct function to integrate
on (o,) is therefore W(s).
We now need an expression for W_(s) containing only known
functions. Applying the Plemelj formula (2.4.3) to (3*5.11) gives
W~ (s) = X~(s)[^)
2X+(s)
00
o
q(t)
X+(t)
dt
ts
Pn(s)]
(3*6.1)
Noting that X+(s) = p(s)X (s), the above becomes
W"(s)
X (s) Â£ q(t) dt +
2*iri J /1. \ v / 4 \ t s
o p(t)x (t)
1 q(s)
2 p(s)
+ X(s)Pn(s)
(3*6.2)
Similarly, applying the Plemelj formula (2.4.3) to (3.5*7) gives
X (s)
n
= s
exp[
lnp(s)
2
.1 r lnp(t)
2iri J ts
o
dt]
(3*6.3)
Simplifying the above yields
X (s) = s
[p(s)]
o
lnp(t)
ts
dt]
(3*6.4)
46
Insertion of X(s) from (3.6.4) into (3.6.2) gives an expression
for W~(s) in terms of the known functions p(s) and q(s), the
computable index n, and the arbitrary polynomial Pn(s).
By the Fourier inversion theorem
F(x) = (2tt) 1/2 jF(t)e ltxdt
00
(3.6.5)
Letting
t
m+1
2ir
Ins
(3.6.6)
gives
dt
m+1
2tts
ds
(3.6.7)
and 0 < s <
(36.8)
Substituting the above change of variable into (3.6.5) gives
F(x) = (m+1)(2ir) 3/2 F(^j;lns)s 1 exp[i(^^jxlns]ds
(3.6.9)
47
In view of the comments made at the beginning of this section and
Eq. (3.4.5), we replace F(^jjplns) with W(s) in the above to get
00
F(x) = (m+1 ) (2tt) 3/2 Jw(s)s 1 exp[i lns]ds
0
(3.6.10)
A somewhat more.compact expression for the above is obtained by
defining the new constant
c0
m+1
2ir
(3.6.11)
and reverting to the original integration variable t given in
(3.6.6):
CO
(3.6.12)
By (3.2.9), (3.6.11), and (3.6.12)
2irc0x
f(eX) = e (2ir)
1/2
. itx
)e
dt
(3.6.13)
CO
48
By (1.4.1) and the above
2irc0lnx
f(x) = e
(2ir)
1/2
w"(e
c0t
1 itlnx,,
Je dt
(3.6.14)
Simplifying the above gives
2ttc0 . c0t ...
. 0,o .1/2 f , 0 itlnx..
f(x) = x (2tt) W (e Je dt
(0 < x < )
(3.6.15)
which is the solution to the integral equation (1.1.1). The ex
plicit dependence on the parameter m is shown by replacing c0 in
the above with its value from (3.6.11):
m+1.
2ir 1 itlnx
f(x) = x'(m+1)(2lr)_l/2 Jwte^ )e
dt
(0 < x < )
(3.6.16)
49
3.7 Uniqueness
The comments made in Section 2.7 regarding uniqueness of
the solution of the RiemannHilbert problem can be used to
determine the uniqueness of the solution of the integral equation
(1.1.1) under various conditions. Since the primary purpose of
this study is to develop a method for solving the integral
equation rather than to verify the uniqueness of its solution, we
only discuss the latter for some special cases.
First of all, consider the case where the index of the
RiemannHilbert problem associated with the integral equation
(1.1.1) is strictly negative. If, in addition, h(x) = 0 in the
integral equation then the associated RiemannHilbert problem is
homogeneous and unsolvable. We thus conclude that under these
conditions the integral equation is not solvable by the method
described herein. If, on the other hand, n < 0 and h(x) is
nonzero then the solution of the associated RiemannHilbert
problem is unique (if it exists), and thus so is the solution of
the integral equation.
Next, consider the case where n = 0 in the associated
RiemannHilbert problem and h(x) = 0 in the integral equation.
This yields a homogeneous RiemannHilbert problem whose solution
is unique only up to a multiplicative constant. Hence, the
function W" appearing in Eq. (36.16) is unique only up to a
multiplicative constant, and likewise for the solution f(x) of the
integral equation. Thus, under these conditions, the integral
equation solution is not unique.
CHAPTER IV
CONCLUSIONS
As shown in the preceding chapter, the method of Goldstein
and Leitman [4] can be used to solve the integral equation (1.1.1)
for nonnegative integral values of m, subject to suitable
restrictions. These restrictions are summarized in the next
section, followed by some suggestions for further study.
4.1 Conditions of Validity
The solution method described in the preceding chapter is
valid if the following conditions are satisfied:
\ t 0
(1.1.3)
m
is a nonnegative integer
(1.1.4)
h e L2
on (o,)
(1.1.5)
k e L2
on (o,)
(1.1.6)
p(s), q(s) satisfy a Holder condition
on (o, )
(2.5.2)
51
p(s) t 0 on (o,)
(2.5.3)
p(s) tends to a finite limit as
s tends to zero or infinity (4,1.1)
AAA
F(a),H(a),K(a) are analytic in the strip
0 Â£ Imct C m+1 (336)
fln_Â£(s>
J sz
(3.5.3)
r q(s) ds ,
] X+(s) s_z
(3.5.9)
The conditions h e Lz and k e Lz are sufficient to
guarantee the existence of the Fourier transforms of the functions
H and K defined in Eqs. (1.4.5) and (1.4.5). The condition
(2.5.2) justifies the use of the Plemelj formulas in solving the
homogeneous and nonhomogeneous RiemannHilbert problems (Eqs.
(2.6.10) , (2.6.11), (2.6.23)). The condition p(s) M 0 on (o,)
allows us to take the logarithm in Eq. (2.6.8). The analyticity
conditions in (33.6) above insure that the inversion integral
(3.6.10) is wellbehaved. Lastly, convergence of the integrals
(3.5.3) and (35.9) is necessary for the solution of the Riemann
Hilbert problem on the open contour (0,) to be welldefined.
52
IJ.2 Topics for Further Study
Further analysis could be performed with the intent of
relaxing some of the conditions stated above. For example, it
would be worthwhile to extend the preceding solution method to the
case where m is any real number.
As another generalization, one might consider cases where
p(s) = 0 at isolated points on (o,<=) or where p(s) fails to.
satisfy a Holder condition everywhere on (o,>). Such extensions
are discussed on pages 107113 of Gakhov [2].
Also, one would want to consider the effect of lifting
^ /s.
the restriction that H(a) and K(a) be analytic in the strip
0 i. Ima C m+1 Removing this condition would broaden the class of
problems for which the method is applicable.
Finally, precise conditions under which the integrals
(3.5.3) and (3.5.9) converge should be defined. Convergence
criteria for Cauchy integrals similar to (35.3) and (3.5.9) are
given on pages 333^ of Gakhov [2].
53
REFERENCES
1. Carrier, G., M. Krook and C. Pearson, Functions of a
Complex Variable, McGrawHill, New York, 1966.
2. Gakhov, F., Boundary Value Problems, Pergamon, Oxford,
1966.
3. Goldstein, M., Sound Generation and Upstream Influence Due
to Instability Waves Interacting with NonUniform Mean
Flows, J. Fluid Mech., 149 (1984), pp. 161177.
4. Goldstein, M. and M. Leitman, A Procedure for Obtaining
Explicit Solutions to a Class of Fredholm Integral
Equations, J. Integral Eqs., 8 (1985), pp. 8793 
5. Horn, D. and F. Zachariasen, Hadron Physics at Very High
Energies, W.A. Benjamin, Reading, Massachusetts, 1973
6. Morse, P. and H. Feshbach, Methods of Theoretical Physics,
Part I, McGrawHill, New York, 1953.
7. Noble, B., Methods Based on the WienerHopf Technique,
Pergamon, New York, 1958.
8. Wyld, H. Mathematical Methods for Physics, Benjamin
Cummings, Menlo Park, California, 1976.
