AN ABSTRACT OF THE THESIS OF
Ralph Leland James
for the
(Name)
Date thesis is presented
,.
<<
Mathematics
(Major)
in
M.S.
(Degree)
,fue
.
Title THE SOLUTION OF SINGULAR VOLTERRA INTEGRAL
EQUATIONS B
S
E APPROXIMATIONS
E
Abstract approved
t
The integral equation
tR(t)
=
SK(t-u)R(u)du, where the
0
kernel function K(t)
solution
R(t)
satisfies certain conditions, has a unique
which satisfies appropriate auxiliary conditions.
Successive approximations to
R(t)
are derived by means of a
They converge uni-
trapezoidal numerical integration scheme.
formly and in the mean to
properties of R(t)
R(t)
for
0 <
t
<
oo
.
A
number of
and the approximations are derived.
THE SOLUTION OF SINGULAR VOLTERRA INTEGRAL
EQUATIONS BY SUCCESSIVE APPROXIMATIONS
by
RALPH LELAND JAMES
A THESIS
submitted to
OREGON STATE UNIVERSITY
in
partial fulfillment of
the requirements for the
degree of
MASTER OF SCIENCE
June 1965
APPROVAL:
Professor
of Mathematics
In Charge of Major
Ch rman of Mathematics Department
Dean of Graduate School
Date thesis is presented
Typed by Carol Baker
r
/
('..
TABLE OF CONTENTS
Page
Chapter
I
INTRODUCTION
1
UNIQUENESS AND OTHER PROPERTIES OF
SOLUTIONS
4
III
THE METHOD OF SUCCESSIVE APPROXIMATIONS
9
IV
PROPERTIES OF THE APPROXIMATIONS
18
V
CONVERGENCE OF THE
Rh(t)
23
VI
CONVERGENCE OF THE
Ph(t)
26
II
BIBLIOGRAPHY
33
THE SOLUTION OF SINGULAR VOLTERRA INTEGRAL EQUATIONS
BY SUCCESSIVE APPROXIMATIONS
CHAPTER
I
INTRODUCTION
In
this paper we shall study the integral equation
t
tR(t)
SK(t-u)R(u)du,
=
0 <
t <
cc
.
0
The kernel function
K(t),
t
0 <
< oo,
satisfies the following
conditions:
(1.2)
K(t) >
0
(1. 3)
K(t)
is continuous for
(1. 4)
lim K(t)
t
for
+ 00
=
-0
0 <
t <
00
0 <
t <
co
,
1
J SK(u
(1. 5)
SK(u)du
<
+co,
0
K(u)
(1. 6)
du <
Sco
1
u
(1. 7)
+00
,
is monotone non- increasing for
K(t)
These conditions imply
(1. 8)
K(t)
t
1
-i0, t
1
0
K(u)du --0
as t
.
O
< t <
co
.
2
There are several ways in which
(1.
1)
can be classified. For
example it is a Volterra integral equation. In Fredholm's classifica-
tion,
(1.
1)
is an integral equation of the third kind.
Moreover, (1.
1)
is homogeneous and the kernel has a weak singularity at zero.
In general it is usually difficult to construct non -trivial solu-
tions to homogeneous integral equations.
Moreover, a solution is not
unique since any constant multiple of it is also a solution.
We
shall show that (1.
has a real solution
1)
unique a. e. to within a constant factor, such that
integrable (Lebesgue) for some x
R(t)
> 0;
R(t),
0
e- xtR(t)
<t<
oo,
is
moreover, such a solution
will be shown to be bounded, continuous, non -negative and
integrable on
(0,
oc)
.
The existence of
R(t)
will be established
by constructing successive approximations and proving that they
converge uniformly to
The uniqueness will be established
R(t).
by Laplace transform analysis.
Since a unique solution is desired it is essential that
satisfy some normalization condition.
It is convenient to
R(t)
require
that
max R(t)
0<t<00
(1. 9)
A
=
1
different normalization condition may be desired for some
3
purposes.
Thus consider the same integral equation, but with the
unknown function
P(t):
tP(t)
(1. 10)
('t
=
J K(t-u)P(u)du,
0
<t
<oo
,
0
where
oo
(1. 11)
Si
P(t)dt
=
1
.
0
The functions
P(t) and
are related by
R(t)
P(t) -
(1. 12)
R(t)
('
J R(t)dt
0
R(t)
(1. 13)
P(t)
max P(t)
=
0 <t <oo
A
K(t)
=
special case of
µt -X
tained in
considered.
,
(1)
µ > 0,
0 <
(1.
X
was studied in (1).
1)
<
1
.
In that paper
All of the relevant results ob-
are established here in the more general case
Further generalizations can be obtained by slightly
weakening some of the conditions (1. 2)
-
(1.7) on
the analysis then becomes more complicated.
K(t)
.
However,
4
CHAPTER II
UNIQUENESS AND OTHER PROPERTIES OF SOLUTIONS
We
proceed to prove a uniqueness theorem and to establish
other properties of a solution R(t)
Suppose that R(t)
of
(1. 1).
is a solution of
e- xtR(t) is integrable for
x
>
x0
.
Assume that
(1. 1).
Then the Laplace transform of
R(t),
co
R(z)
=\ 0 e-ZtR(t)dt,
x > x0, (z
=
x
+
iy),
is defined at least in the indicated half -plane.
R(z)=
e_ztK(t)dt, x
> 0
By (L
5)
and (1. 6),
.
0
Transform
-f.' (z)
where
=
(1. 1) and use the convolution
A(z)(z),
x1= max(x0, 0).
('z
(2. 1) R(z)
=
ft()e
x > xi
theorem to obtain
,
Integration yields
-zt
0o
R(w)dw
=
()e
e
-J
0
t
-e
- K(t)dt
t
with any path of integration in the half -plane for the w- integral.
The standard results from Laplace transform analysis that we
5
have used can be found in (4).
Theorem 2.1.
property that
then
R(t)
is integrable for
e- xtR(t)
factor, so that R(t)
Theorem
R(t)
1)
sufficiently large,
x
determines R(z) to within a constant
1)
is determined a. e. to within a constant factor.
Let
2. 2.
be a non - negative solution of
R(t)
is integrable for
e- xtR(t)
is integrable on
(0,
1-e
t
J
P.(z)
(2. 2)
=
R(0)e
Proof. It follows from
exists at least for x
> O.
limR(x) =
x-.-0
I.( )e
which is finite.
Since
0
K(t)dt
Since
72)
-fit
t
R(t) > 0,
that
=
lm 5°°xtR(t)dt
x
-0
0
RR.(z)
Lebesgue's monotone
K(t) > 0,
yields
)
K(t)dt
another application of the same
theorem yields
lim(x)
Then
-z t
(1. 5), (1. 6) and (2. 1)
convergence theorem (cf. (7,p.
roo
JO
sufficiently large.
x
(1.1)
and
oo)
oo
x-0
with the
is unique a. e. to within a constant factor.
Proof. Equation (2.
such that
is a solution of (1.
R(t)
If
=
5°tdt
0
=
t(0)
.
6
is integrable.
Thus, R(t)
substitute into
tO
Solve for
in terms of
and
1\1(0)
obtain (2. 2).
(2. 1) to
For convenience below, let
(2. 3)
M(t)
sup IR(u)
=
t > 0,
,
I
0 <u<t
for any solution of (1.
Theorem
2. 3.
which is bounded on each finite interval.
1)
is a solution of (1.1) which is
R(t)
If
bounded on each finite interval, then
R(t)
is continuous on (0, co)
and
(2.4)
¡It-s
M(t0)
IR(t) - R(s)
<
I
I
[2J K(u)du+
LL
Proof. Let
0 < s <
ItR(t) - sR(s)
I
=
t <
t
.
Then by (1.
J K(t-u)R(u)du
-
('L
I
1)
s
¡s
J K(s -u)R(u)du
1
K(t-u) - K(s -u) ] R(u)du
[
JO
t -s
<
M(t0)
-
=
M(t0)
I
L
K(u)du
K(u)du - \ K(u)du
+
o
o
t-s
( K(u)du + J
0
t-s
<
and (1. 7),
(°S
K(t-u)R(u)du +
\
0<s, t<t0
J,
0
0
=
t-s
J0
¡
I
I
2M(t0) J K(u)du
0
t-s
t-s
K(u)du
0
-
(t
l
J K(u)du]
s
I
I
.
7
It follows that
ItR(t)
-
tR(s) I< ItR(t)-sR(s)I
2M(t0)
<
('It-ss
+
I
sR(s)-tR(s)
I
I
J K(u)du
It-s
M(t0)
+
s, t
0 <
,
I
<
t0
.
0
Hence,
(2. 4)
holds and
for
By (1. 6),
R(t) is continuous on (0, co)
sufficiently large we have
a
re) K(u)
(2. 5)
.
du
2
<
.
a
Fix
a
such that
satisfied and define
(2. 5) is
a
(2. 6)
A
2
=
.5Th
K(u)du
.
0
Theorem
2. 4.
R(t) is a solution of (1.
If
bounded on each finite interval and
R(t) is bounded on [ 0, oo)
(ii)
R(t)
<
M(A)
(iii)
R(t)
-.
0
as t --
(2.
3)
Proof.
By (1.
1)
R(t)
and
R(t)
I
is continuous on
t > A
,
,
,
.
and Theorem 2. 3,
M(t)
t
(0,
co
which is
then
R(t) 0 0,
(i)
1)
oc)
('t
J K(u)du
,
t
>
0;
0
.
By (2.
5)
and (2. 6),
8
('t
(2. 8)
J K(u)du
t
By (2.
3)
M(t)
M(A)
=
,
I
R(t)
0
K(u)
+
for t
and (1.
> A
5.
8)
R(t) is bounded on every finite
t>
A.
a
t
> A.
So
R(t) be a solution of (1.
Let
< 1,
imply the theorem.
(2. 7)
,
du
S(2.
cannot attain a maximum at
I
Theorem 2.
R(0 +)
¡a
t J 0 K(u)du
<
interval, and
such that
1)
exists. Then
R(0 +)
= O.
Assume
Proof.
exists
5 > 0
and
e
By (1.
1)
and (1.
7)
I
such that
for
> 5 > 0
0
<t <e
.
,
R(t)
In view of (1.4),
I
> 0
R(t)
Then, by Theorem 2. 3, there
IR(0 +) >0.
SK(t)
I
t
O
t<
<
'
e
.
R(t) on every
this contradicts the boundedness of
finite interval.
We have shown that if
e
xtR(t)
R(t)
is a solution of (1.
is integrable for some x, then
R(t)
bounded, then
is integrable on
R(t)
(0,
is continuous on
00).
(0,
is non -nega-
R(t)
Moreover,
00)
such that
is unique a. e. to
R(t)
within a multiplicative factor and if, in addition,
tive, then
1)
if
R(t)
is
and tends to zero as t
tends to infinity. Finally, if R(t) is continuous on [ 0,
00)
,
then R(0)
= O.
9
CHAPTER III
THE M THOD OF SUCCESSIVE APPROXIMATIONS
For h
let
> 0,
Rh(t),
< t <
O
oo,
denote a non -trivial, con-
tinuous, non- negative, piecewise- linear function with possible changes
in slope only at the points
t > 0,
nh, n
=
=
1, 2,
.
is determined in terms of the values
linear interpolation
(3.1)
t
Rh(t)
Rh(nh),
n > 0,
(n+ 1)h-t Rh(nh) +
_
<
t
<
(n+ 1)h,
satisfies equation
Rh(t)
t hnh Rh( (n+ 1)h)
n>
(1. 1)
0
nhRh(nh)
at the points
=
J K(nh
- u)Rl_
(u)du,
n >
0
.
0
Ultimately, we shall let
By (3.
1)
and (3.
h
2)
nhR(nh)
_
)
k=0
k=0rJ
n-1
+
"h
k
kh
K(nh-u)
(k+ 1)h
kh
.
K(nh.-u)Rh(u)du
K(nh-u)
0
0
(k+ 1)h
J
(k+ 1)h
)
k
-
,
n-1
n-1.
by
(k+
1h
,
)h-u Rh(kh)du -
Z
u-kh Rh
h
(
,
.
nSh
(3. 2)
Rh(t),
:
nh
Assume that
Then
.
(k+ 1)h) du
.
=
nh
:
10
Replace
in the second sum,
k- 1
by
k
collect terms and simplify
to obtain
n-1
(3. 3) (n-nh)Rh(nh)
=
12 bnRh(0)
-
+ 12
h
h
L
cn-kRh(kh),
n > 1,
k=0
where
(n+1)h
(3. 4)
b
n
=
b (h)
n
=
(h)
=
K(u)
1
[
(n+l)h-u] du,
-
n >
nh
h
(3.
5)
c
n
= c
n
1
,
(n+1)h
K(u) [
u-
(n-1)h
(n-1)h] du
+ SI
K(u)
[
(n+l)h-u] du, n
>1,
nh
and
(3.
6)
=
h
We
from
(1. 2)
1
(h
nh
2
(h-u)K(u)du
1
=
(1-t)K(ht)dt
0
0
shall need some properties of these quantities. It follows
,
(1. 7) and (3. 6)
that
nh
is a non-negative, monotone
decreasing function of h and
K(h)
<
Hence, by (1. 4), (1.
nh
5)
<
(h
J gK(u)du.
h
0
and (1. 8),
nh--4- +00
as
h-
0
as
h--
00 ,
(3. 7)
-h
and
.
0
,
11
hnh-#
(3. 8)
as
0
h
--
0
.
It follows from (1.2), (1.4), (3.4) and (3.
(3. 9)
>
C1
and from (3.
that
5)
0,
101-2.
5)
ç
(3. 10)
(n+ 1)h
2h2
<
n
1)h
K(u)
du,
u
-
n > 2
.
By (3. 3),
(3.11)
(1-nh)Rh(h)
=
2
(0)(c1 - b1)
First assume that nh is not an integer.
termines
(3. 1)
Rh(nh), n
Rh(t)
then by (3.11) and (3. 9),
>
1
by
Since we desire
assume that nh
Rh(t)
Since
O.
and, hence
= 0
(3. 7)
j
nh
If
(3. 2).
of
Rh(h)
sis, it follows that Rh(0)
small that nh
inductively in terms of
> 1,
yields a solution
Then (3.
Rh(t)
*
=
by hypothe-
> 0
-0
Rh(0) < 0,
if
h
is so
.
is an integer.
H
and
1
0
and intend to let
We
h
{h: nh
--
0,
we must
shall, henceforth, restrict h
to the bounded and countable set
(3. 12)
de-
Rh(0), and
>
Rh(t)
3)
=
1,2,
.
}
.
12
determines
Now (3. 3)
in terms of
imply that
for all h
Rh(0). As before,
Rh(0)
e
Rh(nh) for
=
0
for
(3.13)
nh
but not for n
,
(3.9), (3.11) and
(3. 7),
sufficiently small.
h
Rh(nh)
Let
>
nh,
Rh(t)
> 0
Rh(0)
= 0
0,
=
0
<n <nh
.
assume that
(3. 14)
Rh(nhh)
Then (3.
< n <
Then by (3. 3),
H,
We now
1
> 0
.
reduces to
3)
n-1
(n-nh)Rh(nh)
(3. 15)
=
12
h
Li
cn-kRh(kh)
n > nh
,
,
k- nh
which determines
Then (3.
1)
Rh(nh),
:n >
nh, inductively in terms of
yields a solution Rh(t)
of (3. 2).
Since (3.
Rh(nhh).
and
1)
(3.15) are linear relations,
(3.16)
where
Rh(t)
R1(t)
=
Rh(nhh)Rh(t)
is the particular solution with
,
Rh(nhh)
=
1
.
Equations (3.1) and (3.13) yield
(3. 17)
where
Rh(t)
(nh -1)h
-0
as
h -0
=
0,
0 <
by (3. 8).
t < (nh-1)h
,
We have by (3. 5),
(3.9)
13
and (1.
2)
that
(3. 18)
> 0;
c
c
1
n > 1;
n-> 0,
so that by (3. 1), (3. 14) and (3.15),
Rh(t)
(3. 19)
Hence, the hypothesis that
Theorem
some point
t
=
Proof.
3.
1
kh
Thus far,
(3. 1)
<
is bounded and attains its maximum at
A, where
2. 4,
Rh(t)
parts
(i)
and (ii)
will
It is analogous
.
Rh(t)
is bounded
assume that
max Rh(t)
0<t<oo
Rh(t)
(2. 6).
and (3, 2).
(3. 1)
and (3. 2). Since by Theorem 3. 1,
20)
Then
is defined by
A
is the general non- trivial, non -negative solu-
and attains its maximum, we
(3.
.
is satisfied.
> 0
The proof is based on
to that of Theorem
tion of
Rh(t)
Rh(t)
.
t > (nh-1)h
> 0,
=
h
1,
e
H
.
is determined completely and is given explicitly by
R1 (t)
(3.
21)
Rh(t) -
Theorem
3.
2.
As
Proof. The proof
t --oo
is
,
max Rh1 (u)
0 <u <oo
Rh(t)
based on
-
(3. 1),
0
uniformly in
(3. 2)
h.
and Theorem 3.1.
.
14
It is analogous to that of Theorem 2. 4, part (iii)
Theorem
(0,
)
For each h
3. 3.
is integrable on
Rh(t)
H,
e
.
,
Proof.
The proof is analogous to that of Theorem 2. 2, with
generating functions used in place of Laplace transforms.
is piecewise -linear and
the function Rh(t)
Rh(0)
=
Since
0,
co
¡¡oo
1Rh(t)dt
(3.22)
Rh(nh)
h
=
n=
if the sum is finite.
0
Let
00
(3. 23)
Ah(w)
w > 0,
Rh(nh)wn,,
)
=
n=
when the series converges.
fined at least for
Ah(1)
0 <
Since
0 <
Note that
w < 1.
Rh(nh) < 1,
Ah(w)
is de-
is integrable if
Rh(t)
exists, in which case
co
(3. 24)
JO SRh(t)dt =
o
Let
c0
=
0
and use
hAh(1)
.
(3. 5) to define
co
(3.25)
Bh(w)
=
!
c
0<w
w
<
1
.
,
n=0
By (3.10) and (1.6),
Bh(w)
exists.
Since
c0
= 0
and Rh(0)
=
0,
15
yields
(3. 3)
0o
n
0o
) (n-nh)Rh(nh)wn
=
12
h
n=0
By (3. 23)
(3. 25) and the
wAh(w)
-
) cn-kRb (kh)wn
0 <
,
w <
1
.
n=0 k=0
convolution theorem for sums (cf. (6,p.1.79))
1
nhAh(w)
(w) Bh(w) ,
=
0
<w <1
0
<w
.
h
Integration yields, with the use of (3.25),
00
Ah(w)
(3. 26)
=
Cw
nh
exp
h
where
C
is a constant of
c
{
ri
wn},
< 1,
n=1
integration
c
By (3.10) and (1.6), the
series
n
converges. Hence,
n
00
n=1
by Abel's theorem (cf. (6, p. 177) ),
applied to
c
n
n
w
n
,
n=
CO
Ah(1-)
=
C
exp
2h 2, ñcn },
{
1
n=1
which is finite. Since
Rh(nh)
> 0,
an elementary Tauberian theorem
(cf. (6, p. 189)) applied to (3. 23) yields
exists, so that Rh(t)
We
also obtained
Ah(1)
=
Ah(1 -).
Thus, Ah(1)
is integrable and the theorem is proved.
16
CO
A (1)
Cexp{
=
h
c
n
2
nL=J
}
n=
This result and (3. 26) yield
00
(3. 27)
Ah(w)
Ah(l)w h exp
=
{
h
\)
2
c
n
(wn-1)},
0<w<1
.
n=1
Define
Rh(t)
(3. 28)
Ph(t) -
,
00
t
> 0
.
J Rh(u)du
,J
0
Then
nh
(3.29)
nhPh(nh)
n > 0,
SK(nh-u)Ph(u)du,
=
0
and
00
,00
Ph(t)dt
(3. 30)
h )
=
0
Each function
Ph(nh)
=
1,
h
e
H
.
n=0
Ph(t),
h
E
H,
is a non -trivial, continuous,
piecewise -linear, non -negative, approximation to the solution
of (1.10) and (1.11).
(3. 31)
Rh(t)
is determined in terms of
Ph(t)
Rh(t) - max Ph(u)
0 <u<oo
Ph(t)
P(t)
by
17
In this
and
chapter we have constructed approximations to
R(t)
P(t), and have derived some of their properties. Other proper-
ties are derived in the next chapter.
18
CHAPTER IV
PROPERTIES OF THE APPROXIMATIONS
We now
establish some important properties of the functions
Rh(t). A property of the constants
c
n
,
defined by (3. 5), will be
needed in the proof of the first theorem.
By (3. 5), (1.
cn+
1
7)
and a change of variable we have,
nh
K(u+h) [ u- (n- 1)h] du
-
(n+ 1)h
en
n > 1, whence
(4. 1)
-
{c
n
}
is a monotone non- increasing sequence for each fixed
Theorem
Rh(t), h
0 <
<
nh
(n-1)h
for
K(u+h) [ (n+ 1)h-ú] du
+
t
<
E
t0
H,
4. 1.
For h sufficiently small, the functions
are monotone non- decreasing in some interval
such that
K(t0 +h)
<
1
.
Proof. There exists a unique integer
Nh >
such that
.{-Rh(kh) < Rh( (k+ 1)h)
nh-1
,
(4. 2)
Rh(Nhh) ? Rh((Nh+ 1)h)
.
<
k < Nh
nh,
hE H,
h.
19
Then
Rh(t) is non -decreasing for
By (3. 15),
t < Nhh
0 <
(Nh+ 1 -nh)Rh((Nh+ 1)h) - (Nh-nh)Rh(Nhh)
Nh-1
Nh
cN
2
k=
We
h
0
+
replace k by k+ 1
1-k Rh(kh)
in the
-
2
h
) cN _kRh(kh)
k=
.
h
0
first sum and use
Rh(0)
=
0
to
obtain
(Nh+l-nh)[Rh((Nh+1)h)-Rh(Nhh)]
+
h(Nhh)
Nh-1
cN _k [ Rh((k+
k= 0 h
Hence, by (4.1), (4.2) and Rh(0)
1)h) - Rh(kh) ]
=
.
0,
Nh-1
h(Nhh)
h
2
cN
h
Therefore, cN
JO
k=0
<
h
cN > K((Nh+ 1)h)
h
(
[Rh((k+1)h)-Rh(kh)]
h2
By (3.
.
<
h
cN Rh(Nhh)
.
h
(Nh+ 1)h
S[u(Nh1)h] du
Therefore, K((Nh +1)h)
Z
and (1.7),
Nhh
Nh -1) h
(1. 7).
5)
=
+ J
[
(Nh+ 1)h-u]du
= h2. K((Nh+ 1)h)
Nhh
1
and the theorem follows from (1. 4) and
.
20
Lemma 4. 1.
Let
ba
a <
0 <
s K(u)du
and
1
ab
<
y K(u)du
0
.
0
Proof. The lemma is obvious if either
For
0 <
a <
t
Define
f(x)
and
1
=
t
0
-
f' (x)
K(u)du,
=
x
x
4. 2.
K(x) -
> 0
Proof. For n
that
0.
0.
.
Then by (1.
7)
,
The functions
I
<
>
m
> 0
and the lemma follows.
< 0
0
Rh(t), h
t1 < t <
6
t
1
2)
=
assertion is
t > s >
12 J K(u)du
x
IRh(t) - Rh( S)
and (1.
or b
0
O
equicontinuous on each interval
(4. 3)
K(u)du,
=
0
J
Theorem
(s
s
<
a
an equivalent
b < 1,
0 <
K(u)du
Then
b < 1.
0 <
co
t -s
E
H, are uniformly
with t1
> 0
.
Moreover,
lI
1
Su)du+
It -s
I
}
t,s>t1,heH.
0
it follows from (3. 2), (3. 20), (1.
7)
21
nh
InhRh(nh)-mhRh(mh)
=
I
I
¡* u
K(nh-u)Rh(u)du- K(mh-u)Rh(u)du
1
0
nh
K(nh-u)du
mh
<
=
By
2
+
I
0
mh
[ K(mh-u) -K(nh-u) ] du
0
nh - mh
h
K(u)du - J K (u)du
J0
< 2
mh
¡nh - mh
K(u)du
J0
.
symmetry
(4. 4)
I
nhRh(nh) -mhRh(mh)
I
<
¡lnh-mhl
2
K(u)du
1
n, m >
,
0
.
0
By (3. 20) and (4. 4
I
,
nhRh(nh) -nhRh(mh)
I
<
I
nhRh(nh) -mhRh(mh) I+ ImhRh(mh) nhRh(1rn.N
r lnh-mhl
(4. 5) IRh(nh)-Rh(mh)I<
K(u)du
+
_
Inh-mh
0
Let
nh
<
s, t
-Rh(s)
IRh(t)-Rh(s)
I
=
1
Iths
- (n+ 1)h {
(4. 6)
IRh(t)-Rh(s)
Then by (3.
1)h.
< (n+
I
it-s
h
(
I
(n+ 1)h) -Rh(nh)
('h
K(u)du
J0
It-s
I<
(n+l)h
+
It-s
,
m
> 0, n > 0
Lemma 4.
.
1,
I
Il
I
2J K(u)du+
SK(u)du
0
I
J
(4. 5) and
1)
2J
I
It-s
l
t < s , t<(n+1)h,
1-
,
22
and (4.
3)
holds in this case.
Now assume that
n
such that
s,t
we have for
i
< s <
> t
mh
+
3)
tl
2
Define
n.
nh < t < (n+ 1)h.
<
m
and
By (4. 5) , (4. b)
,
,
I
{2
for some
nh < t
Rh(t) -Rh(s) I< Rh(t) -Rh(nh)
<
and (4.
(m- 1)h
s <
ft-nh
K(u)du
1+
I
Rh(nh) -Rh(mh)
1+
I
Rh(mh) -Rh(s)
nh-mh
+
(t-nh)
+ 2
0
J K(u)du
+
(nh-mh)
0
('rnh-s
JK(u)du
11ff
+ n111.-s
0
<
t-s
.1 16SK(u)du+ t-s }
0
tl
,
is true in general.
Theorem 4. 2 allows us to use in the next chapter, the powerful
Arzelá- Asco1i Theorem. Each nonvoid, bounded, equicontinuous
family of real functions defined on a closed and bounded interval con-
tains a uniformly convergent sequence.
For a proof, see
(5, p. 59).
I
23
CHAPTER
V
CONVERGENCE OF THE
We
h --
O
shall consider the convergence of the functions
through
H.
We
shall regard {Rh: h
ordered by letting h decrease through
Theorem 5.1.
of {Rh: h
t
0,
>
Rh(t)
e
H}
E
Rh(t)
as
as a sequence
H}
H.
Every infinite subsequence
{Rh: h E
H'cH}
has a further subsequence which converges for all
uniformly on any interval
Proof. Since
0 <
0
< t1 < <:t < t2 <
Rh(t) < 1,
Theorem 4.
oo
2
.
implies that the
Arzelá- Ascoli theorem is applicable, and by that theorem there
exists successive (nested) subsequences of {Rh: h H'
E
respectively converge uniformly on the intervals
m
2, 3,
=
,
..
.
[
}
which,
m , m]
,
Then the usual diagonal procedure yields a single
subsequence which converges uniformly on each of the intervals
1
[
nz
,
m]
.
Since
for all he H, this subsequence con-
Rh(0) =0
verges pointwise for t
> 0
.
Theorem 5.2. Suppose that Rh,(t) --R(t)
some H'CH uniformly on each interval
is the unique solution of (1.
(5.1)
:
1)
,
(1.
<
9)
R(t)
t1
0 <
and
1
.
<
as h'--'-0 through
t < t2
<
oo
.
Then R(t)
24
Proof. In
hEH
,
let n-'
00,
theorem of Lebesgue (cf.
A
.
(3. 2)
h-' 0
and
with
t
implies that
(3, p. 22 -23))
¡h
J K(nh-u)Rh(u)du--
nh-t
t
K(t-u)R(u)du
J
.
0
0
Thus, R(t) satisfies (1.
1)
imply (1.9) and (5.
By
1).
.
Theorem
3.
Theorem 2.
1,
(3. 20) and
Rh(t) >
0
R(t) is unique.
1,
Auxiliary Lemma. If a sequence of monotone functions con-
verges pointwise to a continuous function on a closed and bounded
interval, then the convergence is uniform.
The proof can be found in
Theorem
of (1.
1)
5. 3.
such that
(i)
(2, p. 90)
.
There exists a unique solution
R(t)
,
0 <
:
R(t) is bounded and
max R(t)
=
1,
0 <t <oc
(ii) R(t)
is non -negative,
(iii) R(t) is continuous on
(iv) R(t)
Moreover,
Rh(t)
R(t) ,
0, oc)
is integrable on (0,00)
- R(t)
0 .<
t
< oc,
,
.
uniformly as h--"
Proof. By Theorems
function
[
5.
1
through
0
H.
and 5.2 there exists a unique
which satisfies (1.
1)
,
(i)
and (ii) .
t <00,
25
By
Theorem 4.
so that
t
=
0,
Then Theorems 2.3 and 2.5 imply (iii).
exists.
R(0 +)
is monotone in some neighborhood of
R(t)
1,
Theorem 2.2 implies (iv).
Fix t
point,
{Rh(t)
as
h-+
arbitrarily and let r(t)
co)
through
0
of the numerical sequence
H,
has a subsequence which converges at
Theorems
5.
converges to
Rh(t)--. R(t)
By
1
and 5.2,
as
= 0
[0,
on
R
h
1,
t
to the value
r(t)
{Rh: hEH}
.
By
there is a further subsequence which
Therefore,
co).
through
0
Theorem 4.
R(t), we have that
t
be any accumulation
Then by (3.20), the sequence of functions
hEH }.
:
0,
e [
r(t)
=
R(t)
and, hence,
H.
the Auxiliary Lemma and the continuity of
Rh(t)-. R(t)
uniformly in some neighborhood of
and the pointwise convergence and uniform equicontinuity of
the functions
Rh(t)
imply uniform convergence on every finite
interval. Then by Theorem 3.2, the convergence in uniform on
[0,
00).
.
In this chapter we have shown that the approximations con-
verge uniformly to a function
R(t)
with the prescribed properties.
26
CHAPTER VI
Ph(t)
CONVERGENCE OF THE
We
as
shall investigate the convergence of the functions
h --0 through H.
ordered by letting
shall regard {Ph: h
We
decrease through
h
h(w, t)
We define a function
(6.1) tiih(w, t)
=
wt for t
Lemma 6. 1. As
uniformly for h
E
H
Proof. Since
ph(0)
J
(w, 0)
=
0,
=
w
nh, n
1-
=
as a sequence
H}
H.
<w
0
,
E
0,1,
. . .
< 1,
,
t > 0, by
and linear between.
oo
,
o0
SPh(t)h(w,t)dt_Ph(t)dt=
.
Ph(t)Jh(w, t)
is piecewise- linear and
we have by (3. 28) and (3. 23),
00
oo
Ph(t)Lph(w, t)dt
=
h 2/- Ph(nh)wnh
=
hEhAh(wh)
,
n=
where
Eh
1
=
is a constant.
SPh(t)dt
=
Let
hEhAh(1),
Therefore, by (3.27),
Ph(t)
w
=
1
whence
above to obtain
hEh
1
Ah(1)
h
1
27
h)
(6.2)
SPh(t)h(w,t)dt
A
=
We want to show
w--1-
.
- whnh exp
h(
0
00
)
cn
(1-wnh)
}
n=
that this converges to
In view of (3. 8),
{-
l
uniformly in
1
h
as
it suffices to show that
c
Çn
n (1-wnh) -4-0
(6. 3)
uniformly in
h
as
w
-1-
.
n=2
By (3.10),
-
0 <
nh
cn
1
h2
<
n
2
(
J
1)h
(n+ 1)h
K(u)
(('n+
K(u)
2J
nh
2u
n>2.
(n -1)h
(n -1)h
Hence,
00
nh)
ri (1-w
(6.4)
h n=
4
J
2
JK(u)'
0
lim l -w2u
u
u 0
exists by (1.
<
n=
<
Since
5)
(n+ 1)h
¡
co
c
(1
2
Ku(u)
(1-w2u)du
(n-1)h
-w2u)du
.
u
- -2 log w, which is finite, the last integral
and (1. 6), and does not depend on
the integrand decreases everywhere to zero.
Lebesgue's monotone convergence theorem,
h.
As
Therefore, by
1
-,
28
J(u)
-w2u)du--0
(1
as
w -.4-1-
u
0
This result and (6.4) implies (6,
3)
which yields the desired result.
Lemma 6.2. There exists a constant
0
for all
<t <oo,
Proof.
h
E
and Theorem 3.
(3. 20)
Rh(t)
=
hoe H such
<00,
0 <
A
Rh(u)du
< Mo
oo,
he H.
0
> Z
0
Ph(t)
t<
Rh(t),
there
that
A
1
-< M,
1,
Theorem 2.4 and the uniform convergence of the
exists
Ph(t)
rRh(u)du
Rh(u)du
0
So
1
<
00
Sb
such that
M
H.
By (3. 28),
Ph(t)
By
.
for
J0
R(u)du
h<
> 0,
h < h0, he H, where
h0, h e H
Mo
2
=
R(u)du
Since each
Ph(t), h
E
H,
is bounded and the set
{he H
:
h >h0}
finite, the lemma follows. (Another proof uses Fotou's lemma).
Lemma 6.3.
Let
w,
0 <
w < 1,
be fixed.
Then
is
29
( co
lim J Ph(t)LP h(w, t)dt
T--oo J T
Proof.
By Lemma 6.
2
uniformly for h
0
=
e
H.
we have
co
P(t)h(w, t)dt
\
By the definition of
tlih(w, t)
,
so that
wt ^h >
i
t
,
h
H
e
.
T
T
s < nh <
(w, t)dt,
< lvi
we have
(W
t)
ws
wnh
>
for all t
> O.
>
h(w
t)
for
Therefore
t
h(w, t) < wt-h
Then, for all
<
for all h
w2
<
.
Let
T >
he H,
t
¡oo
w2
T
¡+00
JJ(wt)dt<3
T
dt--0 as
T
-
which implies the desired result.
Lemma 6. 4.
As
co
Ph(t)dt
T
uniformly for
Proof.
2
he H
0
,
We have
T -e
oo,
and
J0
Ph(t)dt
..
1
co
,
max {h:hEH
.
30
0 <
(
oo
Ph(t)dt
T
+
T
Ph(t)tph(w, t)dt
T
¡oo
<
¡co
J Ph(t) (1-h(w, t))dt
=
00
J Ph(t) (14
SPh(t)h(w, t)dt,
(w, t) )dt +
0
Then the lemma follows from Lemmas 6.
Lemma 6. 5.
max Ph(t) >X
0 <t<oo
<w<1,
such that
X
sufficiently large we have
T
T
SPht)dt
for all
> -z
Then
hE H.
0
0
max Ph(t) >
2T
<t <oo
Theorem
6.
for all h
1
,
0o
As
h
1
2
0
t
o0
.
-- 0,
.
0
00
(o0
J Rh(t)dt
R(t)dt
-
0
(T
<
J (Rh(t)-R(t))dt
0
0
oo
+
(3. 31) and Lemma 6. 5,
00
Rh(t)dt
J
T
By
max Ph(t) and
< T
¡00
0
Proof.
H
E
J SRh(tdt_ j R(t)dt
E
.
Proof. By Lemma 6. 4, for
J
h H
and 6. 3, and (3. 30)
1
There exists a constant
for all hE H
> 0
0
+
J
R(t)dt
T
.
31
co
max Ph(u) J
0 <u <oo
('=
SRh(t)dt
(t)dt
Ph(t)dt
5
<
T
Ph(t)dt
h
.
Therefore,
co
0o
rT
J R(t)dt
0
<
I
0
J0
oo
(Rh(t)-R(t))dt
+
)
Ph(t)dt
X
I
J R(t)dt.
+
The theorem follows from Lemma 6. 4 and the uniform convergence
of
Rh(t)
to
R(t)
Theorem 6.
There exists a unique bounded, continuous, non-
2.
negative solution
P(t) of (1.
uniformly for
t
Proof.
0 <
<
oo
as
10)
h
and (1.11), and
0
through
functions
h
0
Ph(t)
through
converge uniformly to
P(t).
By
Theorem 6.
Rh(t)
on
0 <
R(t), the
to
t
1,
<
co
as
H.
Theorem 6.
mean to
P(t)
follow from
P(t)
(1.12) and the existence and uniqueness of R(t).
uniform convergence of
P(t)
H.
The existence and uniqueness of
(1. 12), (3.28) and the
Ph(t)
3.
The functions {Ph(t)
:
hE H}
converge in the
32
Proof.
We have
oo
J
Ph(t) -P(t) dt
<
I
I
(T
J
Ph(t) -P(t) dt
I
+
J Ph(t)dt
T
0
0
¡`o0
o0
+
J P(t)dt
T
The theorem follows from Lemma 6.4 and Theorem 6. 2.
Corollary.
As
h- 0,
b
J SPh(tdt -i SP(t)dt.
a
a < b <
0
oo
.
a
Proof.
b
b
a
as
h
`
I
<
SIPh(t)-P(t) Idt
a
a
<
J
-- o
ShtitIdt
0
by Theorem 6.3.
0
We
b
SPh(t)dt - J P(t)dt
0 <
have shown that the functions
and in the mean to a solution
P(t) of
Ph(t)
converge uniformly
(1.10)
with the prescribed
properties.
We have by
Ph(t)
=
and
S2h -S2
SZhRh(t),
.
(1.12), (3, 28) and Theorem 6.1 that
P(t)
=
In view of
S2R(t),
where
S2
and
211
are constants,
this, the two sets of approximations are
very closely related.
In
set of approximations
Ph or Rh, will also be true for the other set
with trivial modifications,
fact, any property derived above for either
.
33
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2.
Boas, Ralph P. Jr. A primer of real functions. Rahway, New
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3.
Bochner, S. and K. Chandrasekharen. Fourier transforms.
Princeton, New Jersey, Princeton University Press, 1949.
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219 p.
New York,
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Churchill, Ruel V. Operational mathematics.
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Courant, R. and D Hilbert. Methods of mathematical physics.
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Royden, H. L.
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