123
Irnat. J. Math. & Math. Sci.
(1982) 123-131
Vol 5 No.
SOLUTIONS OF SINGULAR INTEGRAL EQUATIONS
RINA LING
Department of Mathematics, California State University
Los Angeles, California
(Received August 7, 1980)
ABSTRACT.
Qualitative behavior of solutions of possibly singluar integral
equations is studied.
It includes properties such as positivity, boundedness
and monotonicity of the solutions of the infinite interval.
KEY WORDS AND PHRASES.
Singula integral equations, Positivity of solutions, and
Boundedness and monotonicity of solutions.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
I.
INTRODUCTION.
Both qualitative and quantitative analyses of solutions to integral
equations of Volterra type have been done in the past; see, for example, [i-ii].
The purpose of this paper is to extend some of the results in
[5]
on properties
of solutions to integral equations of the form
t
f(t)
i
I
K(t
T)f(T)dx
(i.i)
0
to cases where the kernel K(t) or its derivatives may be infinite at the origin.
It includes properties such as positivity, boundedness and monotonicity of the
solution on the infinite interval.
2.
DECREASING KERNELS
In this section, properties of the solutions to (i.i) with the possibility
R. LING
124
of the kernel being monotonically decreasing are studied.
If (I) K(O)
THEOREM 2.1.
constant satisfying
K’ (t)
f(n)(t)
0, n
> 0 for t >
PROOF.
I.i.i in
a < 0
K’ for all
<
2
4K’ < a
(2)
and
is a
(K’ (0) needs not be finite), then
t > 0
0,1,2.
The equation under consideration is f
[5], f also
K’
where
* f.
k
i
By theorem
satisfies
f(t)
e
-yt
L
(2.1)
* f,
where y is any constant and
L(t)
(a- )e
-yt
e-"t
K’ *
(2.2)
Differentiation of (2.2) leads to
If a < y, then L(0) < 0.
(y2
L’(t)
ay)e-’t
ye -Yt
+ K’
If y < 0, then
* K’.
(2.3)
t
L’ (t) <_
(y2
ay)e-
(2
.yt
K’
+ K’
e -Yz dr
a +’)e -xt
.
a
The polynomial in (2.3) is minimal for y
(2.4)
Taking y
,
a
(2.4) becomes
2
e’(t) <_
a
yt
(---+ K’)e-
and so L(t) < 0 for all t > 0.
Differentiation of (2.1) leads to
f’(t) =-ye -Yt
L(t)
L * f’
and
f"(t)
y
2e-Yt
L’(t) + aL(t)
From the Neumann Series [12], we conclude that
n
L * f".
f(n)(t)
> 0 for all
t >
O,
0,1,2.
The following theorem on positive decreasing kernels has been obtained in
[5].
SOLUTIONS OF SINGULAR INTEGRAL EQUATIONS
125
THEOREM A. If (i) K(t) > 0 on (0,) (K(0) needs not be finite) and (2)
K’(t) < 0 on (0,=), then f(t) > 0 for t > 0.
Under different conditions, monotonicity of the solution can be obtained,
using Theorem 2.1.
THEORENZ2 If (i) K(0)
be finite) and (3)
’
<
a > 0, (2)
aK(T)
K’(t) < 0 for 0
K’
where
a
<_
t
<_
T
(K’(0) needs not
is a constant satisfying
for 0 < t < T, then (I) f(t) > 0 and (2) f’(t) has at most one zero on
PROOF.
K’ (t) <_ K’
[0,T].
Consider the fundamental equation corresponding to (2.1), namely,
h(t)
If a < y, then L(0) < 0.
-.
L * h.
i
(2.5)
If y is positive, then from (2.3),
L’ (t)
<
(y2
ay) +
(72
t
K’ (T)dT
y
0
ay) + -K-T- yK(t) + ay
< y
-
2
K(T)y +
(2.6)
Let the bound on L’(t), in (2.6), be denoted by L’.
be chosen so that a < y
and
4(y
L2(0).
4
<
2
K(T)y +
To apply Theorem 2.1, y must
The last inequality leads to
T)
)2
< (a-
or
3y
2
2K(T)]y +
+ 2[a
(4-r
Let the polynomial in (2.7) be denoted by p(y).
y
[2K(T)
a]
_+ 2
/a 2
a
2
(2,7)
< 0.
The roots of p(y)
aK(T) +
K2(T)
0 are
3-T
(2.8)
3
Let the large and small roots in (2.8) be denoted by
Clearly,
y_
< a, and in requiring that a <
y+,
y+
we obtain
and
y_
respectively.
R. LING
126
3a < 2K(T)
[2a- K(T)]
a
-
Ja
2 < 2
a
aK(T) +
2
0,1,2.
aK(T) +
2
aK(T) + K (T)
K2(T)
-,
3
-
3K
< 0.
a, L(0) < 0 and p(y) < 0.
Therefore at y
0 < t < T, n
2
+ 2
a
By Theorem 2.1, h (n) (t) > 0 for
By a convolution theorem [13], the solutions f and h are related by
e -Yt
f(t)
+ e -Yt * h’,
from which it follows that f(t) > 0 for 0 < t < T.
(2.9)
Differentiation of (2.9)
leads to
f’(t)
-ye -Yt
+ h’(t)
ye
-Tt
* h’,
(2.0)
and from (2.9) and (2.10), we obtain
yf(t)
+ f’(t)
h’(t),
so
yf’(t) + f"(t)
h"(t).
(2.11)
Since f’(O) =-a < 0 and h"(t) > O, it follows from (2.11) that f’ has at most
one zero.
INCREASING KERNELS
3.
Equations of the form (I.I) with monotonically increasing kernels are now
A result from [5] will be stated first.
studied.
If (i) K(t) > 0, K’(t) > 0 and K"(t) < 0 for 0 < t <
THEOREM B.
4b
<_
a
2
K(0) and b
where a
K’(0), then
If(t)
< 1 for 0 < t <
.
and (2)
Boundedness of the solutions for the next class of increasing kernels can be
obtained, using Theorem B.
THEOREM 3.1.
0 < t <
b
,
(2) a
K’(0) and c
2
If (i) K(t) > 0, K’(t) > 0, K"(t) > 0 and K"’(t) < 0 for
< 4b, (3) 3b < a
K"(0), then
2
If(t)
and (4) 2 a
3
9ab +27c < 0, where a
< 2 for0< t <
.
k(0),
SOLUTIONS OF SINGULAR INTEGRAL EQUATIONS
PROOF.
I
The equation is f(t)
K * f.
127
As before, let h be the solution
(2.5) corresponding to the equivalent equation (2.1).
to the fundamental equation
Differentiation of (2.2) leads to
(y2
L’(t)
ay
+ b)e -Yt + K" *
e -Yt
(3.1)
and
(_y3
L"(t)
2
Since a < 4b,
L’ (t)
+
> 0 for any
ay2
+ c)e -Yt + K"’ * e -Yt
by
(3.2)
The requirement that 4L’ (0) <
y.
L2(0)
leads to
4(y
2
+ b)
< (a
+ (4b
a
ay
y)
2
or
3y
2
2ay
Let the polynomial in (3.3) be p(y).
16(a
y
2
,
a
2)
< 0.
,
[3.3)
The discriminant of p(y)
3b) > 0 and the vertex of p(y) is at y
so
p()
0 is
< 0.
Taking
L(t) > 0 for all t, and
-y
so from (3.2), L"(t)
_<
3
0.
+
ay
2
by
+
i
c
By Theorem B,
(2a
lh(t)
3
9ab
< i
+ 27c)
for0<_
t <
=.
By the convolution theorem in [13], f(t) and h(t) are related by
f(t)
e -Yt
h(t)
so
* h,
t
If(t) <_ i +
y
I
e-Yr
d
0
1- (e
2
The case of 4b < a is examined
THEOREM 3.2.
-yt
l)
in the following theorem.
If (I) K(t) > 0, K’(t) > 0, K"(t) > 0 and K"’(t) < 0 for
R. LING
128
0it
a
!
L’ (t)
-4b
If(t) <_
K"(0), then
c
(3) 2a
and
2
3
9ab
+
27c < 0, where
2 for 0 < t < =.
2
As in Theorem 3.1, taking y
leads to 4L’(0) <_ L (0), L(t) > 0
2
0. But since 4b <_ a mow, condition (2) in this theorem guarantees
a
PROOF.
that
a
a- Va
<
K’(0),
K(0), b
and L"(t)
/ 2
i
(2)
<=
> 0.
The proof can be complete as before.
The next class of increasing kernels can be studied accordingly.
THEOREM 3.3.
K(4) (t)
< 0 for 0
4
(5) -3a +
d
16a2b
8
2 <
b
3b, (3)
t <
=, (2)
a
64ac
+ 256d
< 0, where a
<_
If(t)
K"’(0), then
PROOF.
If (I) K(t) > 0, K’(t) > 0, K"(t) > 0, K"’(t) > 0 and
<_
a
2
(4) a 3
K(0), b
+
4ab
8c
!
0 and
K"(0) and
K’(0), c
< 4 for 0 < t < =.
Differentiation of (3.2) leads to
L"’(t)
(T
4
aT
+ bT 2
3
L2(0)
The requirement that 3L’(0) <
3(T
cT
2
aT
+ d)e -Tt +
K(4)
*
e-Tt
(3.4)
in Theorem 3.1 leads to
+ b)
< (a
+ (3b
a
T)
2
or
2T
2
aT
Let the polynomial in (3.5) be P(T).
2
33a
8b) > 0, we have
p(a)
<_
2
(3.5)
< 0.
Since the discriminant of
P(T)
0 is
In order for the condition (4) in Theorem 3.1
0.
to be satiisfied, we must have
2(a -T)
2(a
3
3
9(a -y)(T
3a2T + 3aT 2
T
3
2
+ b) + 27(-T 3 +
aT
9a(T
2
aT
aT
2
+ c)
by
+ b) + 18(-T 3 +
aT
2
< 0,
by)
+ 27c
< 0
or
2
2
3
-20T + 15aT + 3(a
Let the polynomial in (3.6) be q(T).
3
6b)T + 2a
Then
9ab
+ 27c
< 0.
(3.6)
123
SOLUTIONS OF SINGULAR INTEGRAL EQUATIONS
--
15 a 3
+ 3(a 2
a
+
27 a 3
27
Note that a 3
4ab
27
(a 3
ab
3
2a
9ab
+ 27c
+ 27c
+ 8c)
4ab
+ 8c <_
6b)+
0 implies that a 2
<_
4b
0.
The remaining inequality
required in Theorem (3.1) leads to
_y)2
(a
<
4(y2
ay
+ (4b
a
+ b)
or
3y
2
2ay
Let the polynomial in (3.7) be r(y).
16(a
2
, ,
2)
> 0.
(3.7)
The dlscrlmlnant of r(y)
0 is
3b) < 0, so r(y) > 0 for any y.
we have from (2.2) that L(t) > 0.
Taking y
leads to L’(t) > 0.
-y
If
y
3
ay2
+
Since a
2
! 45, (3.1)
then
by
+
a
3
6- +
c
a
3
3a3
64
ab
+
4
16
c
+ 64c)
16ab
>0
by condition (5) in this theorem, and
y
4
ay
3
+ by 2
cy
+
d
a
4
a
4
25- 6- +
4
3a
256
+ a2b
16
4
i
_-m(-3a
ZDb-
+
a2b
16
ac
4 + d
ac
+d
4
16a2b
64ac + 256d)
It follows from (3.2) and (3.4) that L"(t) > 0 and L"’ (t) < 0.
]h(t)[
<
2 for 0 < t <
(R).
By theorem (3.1),
R. LING
130
As before, f(t) and h(t) are related by
f(t)
ye -Yt
h(t)
and so
*
h
t
f(t)
<_
2
+ 2y
I e-YT
dr
0
2- 2(e -Yt
i)
< 4.
The case of 3b <
is examined in the following theorem.
THEOREM 3.4.
K
If (i) K(t) > 0, K’(t) > 0, K"(t) > O, K"’(t) > 0 and
2
a- 2 /a
a
3b
3
< 0 for 0 < t <
<
(2)
(3) a
4ab + 8c < 0 and
3
,
(4)(t)
(4) -3a
d
4
+
16a2b
K"’(0), then
PROOF.
64ac + 256d < O, where a
If(t)
< 4 for 0 < t <
K(0), b
K’(0)
=.
As in the proof of Theorem 3.3, the choice of
conditions (3) and (4) in Theorem 3.1.
K" (0) and
c
y
a
satisfies
The zeroes of the polynomial r(y) in
(3.7) are at
Y-so
r()
a-2
a
-3b
3
> 0 and the remaining inequality (2) of Theorem 3.1 is true.
The rest of the proof is the same as that of Theorem 3.3.
REFERENCES
i.
FRIEDMAN, A. On Integral Equations of Volterra Type, J. Analyse Math. ii
(1963), 381-413.
2.
HANSON, F.B., KLIMAS, A., RAMANATHAN, G.V., and SANDRI, G.
3.
HANSON, F.B., KLIMAS, A., RAMANATHAN, G.V.,
4.
LEVIN, J.J.
Analysis of a
Model for Transport of Charged Particles in a Random Magnetic Field,
J. Math. Anal. Appl. 44 (1973), 786-798.
and SANDRI, G. Uniformly Valid
Asymptotic Solution to a Volterra Equation on an infinite Interval,
J. Math. Phys. 14 (1973), 1592-1600.
On a Nonlinear Volterra Equation, J. Math. Anal. App. 39 (1972)
458-476.
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LING, R.
Integral Equations of Volterra Type, J. Math. Anal. Appl. 64 (197
381-397.
SOLUTIONS OF SINGULAR INTEGRAL EQUATIONS
131
6.
LING, R. Uniformly Valid Solutions to Volterra Integral Equations, J. Math
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LING, R.
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LING, R.
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LONDEN, S.O.
i0.
MILLER, R.K.
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On the Solutions of a Nonlinear Volterra Equation, J. Math.
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Anal.
On the Linearizatlon of Volterra Integral Equations, J. Math.
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WEIS, DENNIS G.
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BELLMAN, R. and COOKE, K.L.
New York, 1963.
198-208,
Asymptotic Behavior of Integral Equations Using
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