AN ABSTRACT OF THE THESIS OF
JEROME JESS JACOBY
presented on
Mathematics
in
Title:
Master of Science
for the
/;(
IÌC
NUMERICAL INTEGRATION OF LINEAR INTEGRAL
EQUATIONS WITH WEAKLY DISCONTINUOUS KERNELS
Abstract approved:
(Joel Davis)
This thesis outlines a technique for accurately
approximating a linear Volterra integral operator
cs
(Kx)(s)
k(s,t)x(t)dt.
=
An approximation,
Kn,
to
J0
K,
the Volterra operator, is made by replacing the in-
tegral in
K
depend on
s.
by a quadrature rule on [0,1] whose weights
The special case of a modified Simpson's
rule is worked out in detail.
for the uniform convergence of
Error bounds are derived
Knx
to
Kx
when
x
is
a continuous function.
The technique is generalized to linear Fredholm in-
tegral operators that have kernels with finite jump dis-
continuities along a continuous curve.
ditions on the function
x
Under optimum con-
and the kernel the convergence
for Simpson's rule is improved from first to fourth order.
Numerical Integration of Linear
Integral Equations with Weakly
Discontinuous Kernels
by
Jerome Jess Jacoby
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Master of Science
June 1968
APPROVED:
Assistant Professor of Mathematics
In Charge of Major
Chairman of Department of Math matics
Dean of Graduate School
Date thesis is presented
Nky 2,
1968
Typed by Mrs. Charlene Williams for Jerome Jess Jacoby
ACKNOWLEDGEMENT
I
wish to express my thanks to Prof. Joel Davis for
his interest and encouragement during the preparation of
this thesis.
The initial idea was his, as were several
others that led to fruitfull results.
I
also express thanks to my wife, Sally, who has
earned her Ph. T. with honors.
This work was supported in part by the A. E. C.
under contract no. ÁT(45 -1) -1947.
TABLE OF CONTENTS
I.
Introduction
I.1
I.2
II.
Review of Numerical Integration of
Integral Equations
Theorems to be Used in the Thesis
The Special Integration Rule and the Volterra
Case
Generating a Smooth Kernel
The Quadrature Rule for Indefinite
Integrals
II.3 The Sum of the Absolute Values of the
Weights
II.4 The Remainder for the Special Integration
Rule
II.5 Applying the Formula
Interpolation Using the Special Rule
II.6
II.7 Convergence of the Approximation for
Smooth Solutions
II.8 Convergence of the Approximation for
Continuous Solutions
Convergence of the Solution
II.9
II.10 Examples
II.1
II.2
III.
IV.
1
1
3
9
9
14
16
19
24
30
32
35
37
39
Applications to Fredholm Equations
III.1 Fredholm Equations with Jump Discontinuities
III.2 An Example
Generalizations
III.3
45
Summary and Conclusions
50
Bibliography
52
Appendix A
53
Appendix B
55
45
47'
49
NUMERICAL INTEGRATION
OF LINEAR INTEGRAL EQUATIONS
WITH WEAKLY DISCONTINUOUS KERNELS
INTRODUCTION
I.
Review of Numerical Integration of Integral Equations
I.1
In this thesis we consider integral equations of the
Consider first a Fredholm equation of
second kind only.
the form
C1
x(s)
K(s,t)x(t)dt = f(s),
-
0
<
s
<
1,
(1-1)
0
where
and
K
are given real valued functions, the
f
is the
x
is the unit square, and
unknown
domain of
K
function.
One technique to find an approximation,
to
xn,
is to first approximate the integral with a finite
x
sum:
n
f(s)
-J¿1 wnjK(s,tnj )xn(tnj
= xn(s)
j
)
(1 -2)
,
=1
n
where
n3
j
the
wnj
and where
forms a partition of [0,1],
{tnj}
=1
are weights that depend on
tion formula used.
n
Note that for each
and the integras,
equation
(1 -2)
n
n
is an equation in the
unknowns
{xn(tnj)}
.
j
let
s
take on the
n
We
=1
values of the partition, genera-
2
.
Finally, to approximate
at the remaining points in
x(s)
(1 -2),
at the
xn(s)
ving the system of equations we obtain
partition points in [0,1]
By sol-
unknowns.
n
equations in
n
ting a system of
[0,1]
we use equation
,
which thus serves as an interpolation formula.
Consider next a Volterra integral equation of the
form
s
K(s,t)x(t)dt = f(s),
x(s) -
s
<
0
<
(1-3)
1,
S 0
and
f
are as in the Fredholm case, but now
the domain of
K
is the triangle
where
x
One technique to approximate
with
s
<
<
t
<
s
,
0
<
is to first replace
s
<
1.
K
Fredholm kernel which is identically zero for
a
t
x
0
<
1.
The resulting Fredholm equation would then be
solved as above.
The difficulty with treating Volterra equations as
though they are Fredholm equations is that the approximation corresponding to equation
good;
(1 -2)
is usually not very
that is, to guarantee a reasonably accurate answer
the number of partition points, and hence the size of the
algebraic system, usually must be very large.
The source
of the difficulty is the discontinuity of the new kernel
along the line
s
= t.
Disposal of this difficulty is
the motivation behind this thesis.
3
I.2
Theorems to be Used in the Thesis
In this section we develop the theorems that will be
used later, so as not to interrupt the continuity of the
The first three theorems parallel the
later discussion.
work of Milne (1949).
For any nonnegative integer, n, define
Definition.
a function
G
as follows:
n
t
tn,
Gn(t)
Note that for
G0(t)
=
n
>
0,
t
0,
>
0.
<
is continuous,
Gn
1,
is discontinuous at
and for
n = 0,
Two important pro-
t = O.
()
perties of
Theorem
vative of
are as follows:
Gn
G
For
1.
as defined above,
Gn(t)
is
n
(nGn_1(t), n
n(t)
>
1,
=
0,
n = 0,
except for the case when
n = 1
the anti -derivative of
Proof:
the deri-
Gn
is
with
n+
t = O.
n+1.
Differentiating formally, we have
d
d
ntn-1
=
n)
dt(tn)
,
t
n (t) =
át(0)
=
0
,
t
<
O.
>
0,
Further,
4
This formula obviously holds except when
and
n = 1
for then the difference quotient
t = 0,
h[Gn(0+h)
Gn(0)]
-
has two different limits according as
h -
0
or
h
0 +.
Hence, with this single exception
dn(t)
=
nGn-1(t)'
The second part of the theorem follows directly from
the proof of the first.
Theorem
If the
on
Let
2.
(n +l)th
be a function defined on
derivative of
then with
[a,b],
f
G
n
[a,b].
exists and is integrable
f
defined as above, and for
in
y
[a,b]
f(y)
= f(a) +
/
(yká)
kf (k) (a)
+
k=1
Proof:
n
bGn
!
Ja
(y-u) f
(n+l) (a) du.
By the fundamental theorem of integral cal-
culus,
f (y)
for
y
= f(a)
in
+
[a,b].
yf' (u) du = f(a)
a
+
y
Integration by parts
(y-u) Of' (u) du
f
a
n
times yields
5
f (y)
+
= f(a)
(yká) )
11
kf
)
(a)
( a )
(
k=1
J
J
From the definition of
V(17-ton
f
(n+1)
(yñu) n (n+l)
+
Gn(y -u)
a
(u)du.
we can write
b Gn(yu-)
(u)du =
n.
y
a
f
(n+l)
(u)du,
a
since the integrand is zero for
u
>
The desired re-
y.
sult follows immediately.
When approximating a definite integral by a finite
sum, one can define a remainder operator which acts on
functions:
n
pb
R(f)
=
af t) dt J
a
(
1wn7
f( tn
(1949), or Hildebrand
See, for example, Milne
(1956).
If
the integral is indefinite we do the analogous thing:
Let
Definition.
be in [a,b]
be defined on
f
If the integral from
.
[a,b]
to
a
s
and let
is
s
approx-
imated by a finite sum with weights that are functions of
then we define a remainder operator acting on
s,
s
by
n
rs
R(f,$)
-
`f (t)dt
a
w
-
j=1
(s) f
(tn
)
.
f
and
6
Note that if we construct the weights so that a poly-
if
such a polynomial, then
is
f
or less is integrated exactly, and
m
nomial of degree
R(f,$) =
0
.
Further,
since the integral and summation are linear operators,
R
will be a linear operator with respect to its first
argument.
Theorem
R(tk,$) =
for
0
M =
and let
Let
3.
R
be a remainder operator such that
0,
1,
k =
sup
f (m+l)
I
and
..., m,
If
.
.
I
f
s
(m+1)
in
(t)
[a,b],
is Riemann
a <t <b
integrable, then
R
(
f
,
s
=
)
mR
b
rI
Gm(t-u)f(m+l) (u)du,
J a
s
,
and further,
b
IR(f-s)
where
and
R
s
I
<
m!
R [ Gm
T
operates on
(
t-u
s]
du,
a
as a function of
Gm(t -u),
t,
only.
Proof:
By the previous theorem, if
s
is
in
[a,b],
then we have
f(t)
= f(a)
+
L
(tká)f (k)
k=1
Letting
R
operate on
f(t)
(a)
+
m!
we have
5:Gm(t_u)f(m±U
a
(u)du.
7
b
mR
R(f,$) =
Gm(t-u)f
(m+1)
(u)du,
s
,
a
since
R
linear in its first argument, and
is
m
polynomial of degree
nition of
=
of a
From the defi-
we have
R
R(f,$)
or less is zero.
R
m
Os rb
J Gm(t-u)f(m+l) (u)dudt
l
1
a
a
n
(s)
-
b
(m+1)
(u)du
aGm(tnj°u) f
.
7lwnj
From the theorem on iterated Riemann integrals (Fulks,
the order of integration may be interchanged, lead-
1961),
ing to
n.
R(f,$)
=
Sa(SaGm(t_l1t
a
a
m
b
=
S
m,
a
R[Gm(t-u), s]
wnj(s)Gm(tnj-u)
-
f(m+l) (u)du
j=1
f(m+l) (u)du,
and the final result follows directly.
Theorem
f
[Weierstrass Approximation Theorem]
4.
Let
be a real valued continuous function on a compact interThen, given any
val [a,b].
nomial
If (t)
p
-
E
(which may depend on
p (t)
I
<
s
for every
t
0,
>
s
in
),
there is a polysuch that
[a,b].
8
Proof:
See,
e.g.,
Goldberg (1964), page 261.
9
The Special Integration Rule and the Volterra
II.
II.1
Case
Generating a Smooth Kernel
Consider the Volterra integral equation of the sec-
ond kind:
s
x(s)
K(s,t)x(t)dt = f(s),
-
0
<
s
1,
<
0
with the kernel,
at least
r +
and the known function,
K,
f,
having
continuous derivatives on their respec-
1
tive (closed) domains.
A Volterra equation may be thought
of as a Fredholm equation with a kernel
K(s,t)
N(s,t) =
0, s
<
0
,
t
<
<
t
<
s,
1.
As mentioned in I.1, the discontinuity of
line
s
= t
along the
N
is the principal source of difficulty when
approximating the integral by a sum.
In order to eliminate the difficulty, we subtract the
discontinuity as follows:
's
f(s) = x(s)
-
{K(s,t) - K(s,$) + K(s,$)
J
}
x(t)dt
0
1
= x(s)
(s,t)x(t)dt -
N
S
0
0
K(s,$)
cs
J0
x(t)dt,
(2-1)
10
where
K(s,t) t)
NO (s
0,
t
<
s
is continuous across the line
ate a kernel,
N1(s,t)
{K(s,t)
-
0
t
-
s,
<
1
<
=
s
-<
t
We can also gener-
which has a continuous partial
,
derivative with respect to
f(s) = x(s)
K(s,$),
=
across the line
t
K(s,$) +
-
s =
t
:
x(t)dt
(s-t)K(1) (s,$)}
0
K(s,$) Ssx(t)dt + K(1) (s,$)
-
s(s-t)x(t)dt
0
0
= x(s)
1
-SN1(s,t)x(t)dt
-
K(Q) (s,$)
-
a
QK(s,t)
at
that
K(0) (s,$)
QK(R,)
(s,$)Js(s-t)x(t)dt,
Q=0
0
where
(-1n)
= K(s,$)
0
with the convention
,
t=s
and where
,
K(s,t) - K(s,$)
+
(s-t)K(1) (s,$)
,
t<
0 <
s,
N1 (s,t) =
0,
s
In general then,
<
t
1
<
,
subtracting
terms of Taylor's
a kernel with
series generates Nr(s,t),
partials with respect to
r
t
across the line
continuous
r
s
= t
:
11
1
f(s)
= x(s)
-
Nr(s,t)x(t)dt
S
0
-
(
Q=0
(
kQ
Q)
(s,$)
(2 -2)
Ss(s-t)Rx(t)dt,
0
where
(-1)IS-t) QK(Q)(s,$)
K(s,t)
0
,
<
t
s
= t.
<
s,
2=0
Nr (s,t) =
0, s
<
Clearly Nr(s,t)
If
x(t)
then we
also has
r
t
<
1.
is smooth across the line
continuous derivatives on [0,1]
can accurately approximate the first integral in
by a finite sum of the form
(2 -2)
(1 -2).
It is easy to
show that with the differentiability assumptions on
and
that
f
,
x
is forced to
have
r
K
continuous deriva-
tives.
We deal with the indefinite integrals in
section II.2.
(2 -2)
Presently, we consider the choice of
in
r --
the number of terms of Taylor's series subtracted from
K(s,t)
Integration rules under consideration here have
.
(p +1)th
error bounds which depend on a derivative,
of the integrand, assuming the
uous.
than
derivative is contin-
pth
Therefore, it makes sense to choose
p.
r
For example, with Simpson's rule
(Hildebrand, 1956).
Of course if
say,
K
and
no greater
p =
f
3
don't have
12
sufficiently many continuous derivatives, then we must
choose a smaller
r.
We elaborate on the above by applying a low "order"
rule repeatedly
n
For fixed
times.
continuity of the kernel
occurs in at most one of the
K
subintervals of application.
fixed
then
Let
If we apply Simpson's
s.
the jump dis-
s
cP(t)
rule
= K(s,t)x(t)
for
times on [0,1],
n
is discontinuous on at most one subinterval.
4
Consequently, the error will be bounded by
4
5
ñM0 + (n-1) (T)
where
M0
°C
=
M4
01I(t)
ñ
[MO +
and
1
(n-1)
M4
c[
(ñ)
T
0
M4]
I
,
qlv (t)
.
1
We
therefore have first order convergence for piecewise continuous functions
(Note:
which have only one discontinuity.
4
kth order convergence if the error
we would have
bound was divided by
same way, if
0
2k
when
n
was doubled.)
In the
is continuous, then we would have second
The extension is obvious,
order convergence.
for if
(I)"
is continuous then we will have fourth order convergence.
For Simpson's rule (which has fourth order convergence if the integrand is sufficiently differentiable),
a good choice for
this case,
r
in the general case is
N2(s,t)x(t)
r = 2.
In
will have a continuous second
derivative with respect to
t,
a piecewise continuous,
hence bounded, third derivative with respect to
t,
and
13
we will have fourth
order convergence.
Choosing
r =
as predicted in the earlier discussion, will not give
higher order convergence.
3,
14
The Quadrature Rule for Indefinite Integrals
II.2
We now develop
formula to approximate the integral
a
(s-t)x(t)dt.
0
We start with a quadrature rule to approximate the inte-
gral with
Nr
in
This rule will be called the
(2 -2).
"regular" rule as contrasted with the "special" rule to
Using the same abscissas as in the regular
be developed.
rule, we obtain a special rule with weights that are
functions of
s
and
It is necessary to use the
Z.
same abscissas in both rules so that the same
n
{xn(tn3)}
j
s
Thus, when
arise from both approximations.
,
unknowns,
n
=1
is allowed to take on the values of the partition,
in I.1, the system of equations will be
nonsingular, can be solved.
n
x
as
and, if
n
The theory for the special
Simpson's rule is developed below.
For fixed
k,
and for
0
<
-
s
<
-
2h,
where
h =
l
2n
we require
rs
2
(s-t)
J
0
where
R
quire that
x(t)dt =
w. (s,k)x(ih)
i=0 1
is defined by equation
(2 -3).
w0(2h,0) = w2(2h,O) = 3,
+ R(x,$)
,
(2 -3)
We naturally re-
and that
15
wl(2h,0) =
4h
that is, the special rule reduces to the
;
regular rule for
weights let
= 2h
s
and
1,
To find the three
= O.
and then let
R(x,$) = 0,
sively the functions
Q
t,
x(t)
be succes-
We arrive at a
and t2.
system of three equations whose solution is
w0
E
w0(s,Q) = CsQ+1[2s2-3sh(Q+3)+2h2(Q+2)(Q+3)],
wl
=
wl(s,Q) =
w2
=
w
where
2
(s
C =
tion of the
-
4CsQ+2[s
-
h(Q+3)].
= CsQ+2 [2s - h(Q+3)
[2h2(Q +1)(Q +2)(Q +3)] -1.
wi,
R(x,$)
]
,
(2-4a)
(2 -4b)
(2 -4c)
Note that by construc-
for a polynomial of degree two
or less will be identically zero.
16
The Sum of the Absolute Values of the Weights
II.3
V =
Let
+
Iw11
1w21
+
Iw31
This sum will be used
.
It is also important in that it provides a mea-
in 1I.8.
sure of the amplification of the errors in the ordinates
(Hildebrand, 1956).
From equations
for
s
in
(2 -4)
note that
and for all
[0,2h]
w0
R
>
0.
<
s
<
>
and
0
For
wl
0
>
we have
w2
three cases:
w2(s,Q)
>0,
R=
0,
<0,
Q=
0,
0
<
s
<
2h
(case 2),
>
1,
0
<
s
<
2h
(case 3).
<0,
2,
22
2h
(case 1),
le
Case
1:
Since
w2
>
V = w0 + wl + w2 =
0,
s
<
2h
by construction of the weights.
Case
2:
V =
Here
w2
s
-
(
<
0,
V = w0 + wl
so
-
w2
and
4s2 + 6sh + 12h2).
12h2
Taking the derivative of
V
with respect to
we have
s
2
V'
=
1 +
h
- s2,
h
which is a parabola that is concave downward.
V'(0) =
1
and
V'
zh)
(
increasing throughout
=
T'
[0, 2h]
we
conclude that
so that
Since
V
must be
17
V
Case
Again
3:
C =
-
w2
V = C[- 4sk +3+ 2h
where
V(2h= áh
<
(k +3)
so
sk +2+ 2h2 (k +2) (k +3) sk
V'
Since the product
C(k +3)s9'
where
2h2(k+l)(k+2).
is always non- negative, the
is determined by the sign of
V'
+1]
Taking the derivatives
= C(k +3)ska(s),
= - 4s2 + 2h(k+2)s +
a(s)
a(s).
And since
is a parabola that is concave downward with a(0)
a(s)
and
0,
2h.
[2h2(k +l)(k +2)(k +3)] -1.
as in case 2, we have
sign of
<
<
a(2h)
out [0,2h].
>
0,
we conclude that
V
<
V(2h) = g(k) (2h)
k+1
where
g(k)
for
k
>
1.
Result
_
k2 + 72, + 4
(k+1) (k+2) (k+3)
We have thus established
I.
0
is increasing through-
Thus,
V
>
The sum of the absolute values of the
weights of the special Simpson's rule is given by
18
V
<
g(9) (2h)
,Z+1
,
where
,
g(R)
k
-
0
_
2,2 + 7!Z + 4
(Q+l) (lZ+2) (Q+3)
Note that
for
2
>
0
Q
'
1.
>
is decreasing for
g(R)
the maximum of
g(R)
2
is one at
>
so that
0
R,
=
O.
Therefore, the sum of the absolute values of the weights
is bounded for all
Z
and for all
s
in
[0,2h].
19
II.4
The Remainder for the Special Integration Rule
From equation
(2 -3)
arbitrary function
x
we have the remainder of an
:
2
cs
R(x,$)
(s-t)
E
x(t)dt
50
where
<
0
s
<
2h
i=0
From theorem
.
(2-5)
wo (s,Q)x(ih),
-
1
2
of I.3,
+
2
x(t)
may be
(t-u)x"'
(u)du.
written
x(t)
=
x(0)
2h
2
tx'
+
2x" (0)
(0) +
G2
0
Applying the remainder function to this equation, we-have
2h
R(x,$)
= R
`
JO
2.
and applying theorem
G2(t-u)x"' (u)du,
s
we can interchange
3
,
and the
R
integral to get
M3
IR(x,$)I
<
2h IR[G2(t-u), s]Idu,
2
(2 -6)
0
M
where
=
sup
Os<llx
To determine
placed by
G2(t -u).
t1I
'I
R(G2,$)
With
apply
W = Ii
(2 -5)
wi
with
(s
R,)
we have
(s-t) kG2 (t-u)dt - W.
R(G2,$) =
'00
x(t)
re-
G2 (ih -u),
20
Integrating by parts
integrating
R(G
2
Gk
directly yields
+2
k1
_ -
,$)
0
u
<
2k(k-1)...(k-i) xk-iG.
2k1
Gn( -u)
Result II.
If
- G
(-u)
[G
k+3)T
2h,
<
i+3
(i+2)!
i =0
-
Since
(using theorem 1), and then
times
k
E
k+3
(s-u)
]
(-u)
-
W
.
so we have
0,
has a continuous second derivative
x
and a bounded third derivative, then
2h
M3
IR(x,$)I
--
<
2
where
M3 =
2
wi(s,k)G2(ih-u)Idu,
ILGk+3(s-u) -
sup
(2-7)
i=1
0
I
x "(
)
I
and
L = 2[(k +l) (k +2) (k +3)] -1.
0<s <2h
Analytic integration of
(2 -7)
for arbitrary
k
is
difficult due to the absolute value inside the integral.
The case for
eral
k,
k
=
0
will be worked out below.
For a gen-
note that the integrand is continuous and its
first derivative is piecewise continuous.
Therefore, we
are assured that the integral exists and can be accurately
approximated by the repeated midpoint rule, say.
A com-
puter program to estimate the remainder at selected points
in
[0,2h] was written and is presented in Appendix A.
results for
in figure
1
k
= 0, 1,
2,
and
on the next page.
3
The
are presented graphically
21
in units of
IR1
M hQ+4
3
0.15
Q
max
0
0.04166... = 1/24
1
0.04444... = 2/45
2
0.08888... = 4/45
3
0.15238...
0.1
0.05
/
Q
= 1
R
= 0
2h
h
Figure
1
Sketch graph of the remainder for
Q
= 0, 1,
2,
3.
22
For
=
2,
we have from
0
(2 -7)
M3 ç2h
IR(x,$)I
<
2
13G3(s-u) - wl(s,0)G2(h-u)
J
0
This splits into two cases:
For
-w2(s,0) (2h-u)2,
-wl(s,0)(h -u)2
R(G2,$) =
-
u
-
w2(s,0)G2(2
0
<
s
<
)Idu.
we have
h
[h,2h]
e
w2(s,0)(2h -u)2,
u
e
[s,h]
ue[0,s].
3(s-u)3 -wl(s,0) (h-u)2-w2(s,0) (2h-u)2,
And for
h
<
s
<
we have
2h
-w2 (s,0) (2h-u) 2,
4(s_u)3
R(G2,$) =
3(s-u)
-
u
[s,2h]
e
w2 (s,0) (2h-u) 2,
3- wl
(s,0) (h-u)
2-- w2
u
[h,s]
E
(s,0) (2h-u)2, u e[O,h].
Thus in both cases we can split the integral into three
parts and deal with each part separately.
wl(s,0)
=
and noting the signs of
w2(s,0)
it can be shown
w2
Q
and
By writing out
(the algebra is tedious)
wl
and
that for
0
M3h4
I
for all
lated for
in
s
9,
=
[0,2h],
R(x)
<
I
24
which is what the program calcu-
O.
The fact that the integrand of
on
[0,2h]
'
(2 -7)
shows that there is a number
e2
is continuous
depending
23
on
k
and
suplxt(s)I
for
IR(x,$)I
s
c
<
e
(0,2h)
k
hQ+4
such that
;
that is, the bound on the remainder can be made independent
of
s.
This fact will be used in II.7 to prove a uniform
convergence theorem.
.
24
Applying the Formula
II.5
Let us reconsider the original problem:
cs
f(s)
= x(s)
-
J
K(s,t)x(t)dt,
0
(2-8)
1.
<
s
<
0
Recall that to approximate the function
we choose an
x
integration rule and a corresponding partition of [0,1],
convert
(2 -8)
to a Fredholm equation with a kernel
which is zero for
t >
and approximate the Fredholm
s,
integral with a finite sum.
an
n
x
This approximation leads to
system of equations whose solution determines
n
an approximation to
at the partition points.
x
Suppose we choose Simpson's rule repeated
so that the partition is
in
(2
x(0)
-8)
N(s,t)
that if
j
=
2n
}i =0'
=
0,
1,
then the integral is zero, so
0
2,
times
First note
h = 1/2n.
Consequently, we assume
= f(0).
for some
s
{ih
n
...,
n -1.
2jh
<
s
<
(2j +2)h,
The Fredholm problem can
then be written
2jh
f(s)
= x(s)
K(s,t)x(t)dt -
0
If
K
and
x
1
N(s,t)x(t)dt.(2-9)
S
2jh
have continuous third and bounded
fourth derivatives with respect to
t
on [0,2jh], then
we can approximate the first integral in
regular Simpson's rule
j
times:
(2 -9)
using the
25
2jh
hj-1
K(s,t)x(t)dt
1
3
0
{K(s,t2i)x2i + 4K(s,t2i+1)x2i+1
i=0
+ K(s,t2i+2)x2i+2},
xk = x(tk)
where
and
We use the regular rule
tk = kh.
because it has a higher order remainder than the special
rule and, hence, is likely to be more accurate.
Next, consider the second integral in
N(s,t)
=
for
0
c1
t
>
s,
N(s,t)x(t)dt
('
(2j+2)h
J
=
N(s,t)x(t)dt
.
2jh
takes on the partition points there will be two
s
cases:
N(s,t)
Since
we have
2jh
When
(2 -9).
s
=
=
(2j+l)h
on
K(s,t)
In the second case
(2j +2)h.
and
so the integral can
[2jh,(2j +2)h],
be approximated by the regular Simpson's rule applied once.
In the first case we use the results of II.1 as
follows:
('
(2j+2)h
(2j+2)h
K(s,t)x(t)dt =
J
2jh r
2jh
-
(-1)
!
i=0
where
s
=
Nr(s,t)x(t)dt
S
(2j+l)h.
!
k
K(Q)
2jh
To approximate this integral use the
regular rule on the integral with
rule on the integral with
for
t
>
s,
(s-t)Qx(t)dt,
s
(s,$)
and letting
(s
-t)
K2 +1
Nr
.
=
and the special
Recalling that
K()(s2j +l's2j +1)'
Nr
we
E
0
26
have
(2j+1)h
K(s,t)x(t)dt
5
-3Nr(s,t2j)x2j
°
2jh
2
L
Q=0
(
Q!
1
wi (h' Q) s2j+1
QK2j4 i=0
j.
Summarizing the above work, we have three cases:
(' S
s
= 0:
K(s,t)x(t)dt =
J
(2 -10a)
0,
0
s
=
(s
(2j+l)h:
K(s,t)x(t)dt
h
-
3
J 0
j-1
{K(s,t2i)x2i
i=0
+ 4K(s,t2i+1)x2i+1 + K(s,t2i+2)x21+2}
r
+
3Nr
(s't2
j
)
x2 j
(-1)
i=0
Q
2Q
+1
Q.
Q
2
X
L
wi
(h, Q)
(2
x2j+i'
-10b)
i=0
s
s
=
(2j+2)h:
SK(s,t)x(t)dt
3
0
i=0
{K(s,t2i)x2i
+ 4K(s,t2i+1)x2i+1 + K(s,t2i+2)x2i+2}.
Substituting approximations
generates a system of
2n +
1
(2 -10)
(2-10c)
into equation
equations in the
2n +
unknowns that will be of the form indicated in Figure
The *'s indicate generally nonzero entries.
(2 -8)
1
2.
The elements
27
inside the dashed lines are given in Figure
3.
The ele-
ments are shown with a general subscript, and as indicated,
the two -by -two system is repeated
r--I
*
0
0
0
0
0
0
0
xl
fl
0
0
0
0
x2
f2
*¡
0
0
x3
0
0
x4
f4
*
*
x5
f5
*
*
x6
f6
0
**
*
*
*
*
*
*
r*
*
*
*
*
*
*
*
*
*!
*!
*
*
*
*
*
;
Hence, the
f0
0
*
i
times.
n
0
0
;
f3
=
AX = F
Figure
A Schematic for the Array A for n =
2.
3.
system is reducible; that is, from the zeroth equation
x0 = f0.
Substitution of
into the first through
x0 = f0
(2n)th equation reduces the system to
equations one and two
xl
and
Cramer's rule; substitution for
three through
2n
From
can be found using
x2
xl
and
reduces the system to
Repeating this process on the pairs
(x2n- l,x2n)
(2n)x(2n).
x2
in equations
(2n- 2)x(2n -2).
(x3,x4),
(x5,x6)...,
will solve the system.
It is important to note that since a general formula
for the elements of the system can be written, it is unnec-
essary to store the whole array.
In fact, only the two -by-
two subsystems are stored as the elements are calculated.
A computer program using Simpson's rule is presented in
28
column 2j +2
column 2j +1
row
0)
0
0
2j)
0
0
2j+1)
1-
Q=0
(-1)
Q!
2j+3)
-TT K2j+3,2j+1
2j+4)
-
Key:
3
K2j+4,2j+1
3
_2h
4h
K2n,2j+1
3
= 1;
A1,0 = -3x1,0
Ku
= K(su,sv),
3.
K2j+3,
2j+2 + T2j+3
K2j+4,2j+2
K
K2n,2j+2
is a special case:
v
T
Figure
2h
_2h
4h
3
- 3
0
(h,Q)
K2j+2,2j+1
3
1
4h
2n)
(-1)Q K(Q) w
2j+2 2
Q!
h
-3 K2j+2,2j+1
A0,0
Q=0
4h
2j+2)
Column
_
K(Q) w (h,Q)
2j+1 1
u
=
rCC
L
Q,=O
=
Av,0
v,0
v = 2,3,...,n.
su = uh
(-1)
k
+ T1;
i
2
K(k) hQ+1
3
u
-
w0 (h, Q)
General Formulas for the Elements of the
Array A.
29
Appendix B.
When solving an actual problem, one needs the
following information:
and
1)
The known functions
2)
The partial derivatives of
to
f
and
K,
K
with respect
t.
These formulas are used as indicated-by-the-comment state-
ments in the program.
termines
h
The only- inputs are
and hence the error bound) and
ber of terms of Taylor's series to be used).
Some examples are given in II.10.
(which de-
n
r
(the num-
30
Interpolation Using the Special Rule
II.6
Further consideration of the approximations in the
previous section and the development-of-the special rule
in II.1 and II.2 indicates that an interpolation formula
may be easily obtained.
Suppose we have approximated the solution,
points of some partition of
an arbitrary
is desired.
x(s)
x(0)
= 0,
j
equation
If
s
Further, suppose
[0,1].
0
= 0,
<
s
<
2,
1,
...,
(n -1).
= f(s)
2jh
<
s
(2j +2)h
<
x(s):
(2j+2)h
K(s,t)x(t)dt +
+
for
As before, we arrive at
2jh
x(s)
that
and the number
1,
on page 24 but now we solve for
(2 -9)
at the
then we have immediately
Consequently, assume
= f(0).
some
given,
is
s
x,
0
S 2jh
N(s,t)x(t)dt.
(2-11)
The first of these integrals is approximated as before
using the regular Simpson's rule
integral we generate
tegral with
Nr
r
on the interval
(s -t)2'
a
j
times.
In the second
smooth kernel as in II.5.
The in-
is approximated with Simpson's rule once
[2jh,2jh +2h],
and the integral with
is approximated with the special rule.
approximation is as follows:
The final
31
JCl{K(s't2i)x2i
3
JK(s,t)x(t)dt
0
+ 4K(s,t2i+1)
o{K(s,t2i)x2i
i=0
+ K(s't2i+2)x2i+2}
+ 3Nr(s't2j)x2j +
3
2
+
G
(
-tl K(Q) (s,$)
i=0
i =0
wi(s- 2jh,Q)x2j +i
(2 -12)
2jh
Notice that for
<
s
<
(2j
+2)h
the approximation
involves the previous calculated values of
x(2jh +2h).
x2j +2 =
Hence,
the formula for
x
up through
x(s)
an extrapolation but an interpolation:
x(s)
-
f
(
s
)
+
[approximation (2-12)].
is not
32
II.7
Convergence of the Approximation for Smooth Functions
We now investigate convergence of the approximations
of the last two sections to their respective integrals as
n +
Our hope, to be established in I1.9, is that if
cc.
these approximations converge to their integrals then the
corresponding approximate solutions,
(in
n
some sense) to the correct solution,
Let us fix
for if
(0,1]
then
= 0,
s
x,
and choose an arbitrary
x
half -open interval
will converge
xn,
n
as
s
_.
-
in the
[There is no loss of generality,
.
x(0) = f(0)
by
In order
(2- 10a).]
to investigate convergence, we write
rs
E
E.
50
K(s,t)x(t)dt
Assume
= min {3,
[Approximation (2-12)].
is a non -negative integer and let
r
Assume
r +1 }.
with respect to
is well -known
-
and
t
has
K
x
(Hildebrand,
p + 1
has
p + 1
(2-13)
p
partial derivatives
derivatives.
It
1956) that for each application
of Simpson's rule, the error bound is
hp
+2MP +1
where
in this case
ap+l
Mp+l
If
p = 3,
=
p+1 OstPl
atp+1
for example,
three terms of
(2 -12)
to the integral with
of
Nr(s,t)x(t)
.
then
K(s,t)x(t)
C4 = 1/90.
In the first
we applied Simpson's rule
K(s,t)x(t)
j
times
and once to the integral
33
If
M
*
= C
p
p
a
N (s,
sup
r
0 <t<1 atp
t) x (t)
then the magnitude of the error from these three terms will
be bounded by
JMp+lhp+2 + Mp*hp+1
p+1
<
n
By the results of II.4, the error bound for approxi-
s
J
2jh
Apkhk +p +1,
to
will be less than or equal
(s- t)x(t)dt
mation of
Apk
where
=
Qp"k
0
sup (x(p)(t).
<t<1
for example are the numbers from equation
gested in figure
1.
_
=
Q Ap,kNkhk
[C +
2
r
P-1
¿
k=0
where
r
<
N
By the above work we have established
3.)
Result III.
If
0
<
s
<
1,
vatives are bounded on
are bounded for
t
and its first
x(t)
and if
0
<
p +
and its first
K(s,t)
derivatives with respect to
and
Q!Ap ,kNk hk] np +l
(Note that this requires
sup IK(k)(s,$)I.
<t<1
0<t
=
and sug-
(2 -7)
Therefore,
hp+1
np+l +
IEI
Q3,k
t
<
1,
where
r
partial
1
0
p + 1
<
t
<
s
deri-
is the high-
34
est order derivative subtracted from the kernel and
= min {3,
r +l },
then there is a constant
°,
p
depending
on the above bounds, such that the error in approximating
VoK(s,t)x(t)dt,
0
<
s
<
1,
o
with
n
tion on
applications of Simpson's rule and its modifica[0,1]
is bounded by
iEl
np+l
That is, the approxiamtion converges uniformly to the
integral as
n ±
00.
Note that by Result III there is no gain in taking
r
>
2
for the Simpson's rule case.
prediction in II.1 that
r = 2
This confirms the
is a good choice.
35
II.8
Convergence of the Approximation for Continuous
Functions
In the previous section strong assumptions were made
Uniform convergence of the approximation
on the solution.
operator to the integral operator can also be established
if the solution,
is only known to be continuous.
x,
The starting point is the Weierstrass approximation
theorem (see I.2 Theorem 4):
polynomial
in
s
such that
q(s)
lq(s)
e
there is a
0,
>
- x(s)1
<
e
for every
Knowing that such a polynomial exists, we
[0,1].
look to equation
in equation
given
(2 -12).
(2 -13)
-
Recalling the definition of
we subtract and add
q
wherever
E
x
Using the associative and distributive laws and
occurs.
the linearity of the integral and summand, we have,
for
example,
JsK(s,t)x(t)dt = sK(s,t)g(t)dt + JsK(s,t) [x(t) -q(t)]dt.
0
0
0
Next, we use the triangle inequality and the property of
integrals
tion
If
(t)dti
<
!J
(t)ldt
for any integrable func-
to write
4,
s
IEI
K(s,t)q(t)dt
<
-
[Approximation of K(s,t)q(t)]
0
+
+
[The remaining terms of
[(2 -12)
(2 -12)
with q(t)]
with lx(t) -q(t)1 wherever x occurs].
(2 -14)
36
Note that the first two terms of
K(s,t)q(t).
the error in approximating the integral of
the error of this approximation is less
So, by Result III,
than or equal to
for some
n-13-1,
%
q
q.
q
Finally, we use the assumption that
that
lx(t)-q(t)1
constitute
(2 -14)
<
depending on
1K(s,t)1
<
M0,
and that the sum of the weights
c,
(2h)2
of the special rule is less than or equal to
+1
to
assert that the sum of the absolute values of all the
weights will be bounded by
2
1 +
(2h)
2
+1
<
2, say,
2 =0
for
n
sufficiently large, and using this to conclude that
1E1
uniformly.
c
<-7Jc+i'jnp-l-0-/7c, as
q
Further, 97 is independent of
is arbitrary
1E1
-
0
as
n +
co.
00
c,
and since
37
Convergence to the Solution
II.9
In II.7 and II.8 we showed that the Simpson's rule
approximations converged to the integral as
lowing the lead of Anselone
K
and
K
n
n
-}
Fol-
co.
(1965), define the operators
as follows:
's
(Kx) (s)
E
\
JO
(Knx)(s) =
K(s,t)x(t)dt
[the approximation
(2 -12)].
Then our results have shown that
Knx
->
Kx uniformly as n +
00.
The integral equation and the approximate equation
have their corresponding operator equations:
Solving for
x
x - Kx = f,
(2-15a)
xn - Knxn = f.
(2 -15b)
in equation
(2 -15a)
and subtracting
(2 -15b)
from the result we have
x - xn - Knx + Knxn = Kx - Knx,
after subtracting
to
Knx
from both sides.
This equivalent
38
(I-Kn)(x-xn) = Kx
where
Knx,
-
If the operator
is the identity operator.
I
has an inverse, that is, if the determinant of the
(I -Kn)
matrix
A
represented in Figure
operate on the left by
(I -Kn)
x - xn =
then we can
is non -zero,
2
to get
-1
(I-Kn)-1 (Kx-Knx),
and taking norms we have that
(l
Now, if
H
x -xn
(
I
II
<
x-xn
(I -Kn)
II
Kx -Knx
uniformly as
<
I
-1
n
B
I
n +
<
II
II
-1II
-1
II
II
B
for some
0
as
-9-
II
(I-
I
Kx-Knx
and for all
B
n }
since
co,
.
I
Knx
--
n,
then
Kx
Thus we have
co.
IIx-xnll<B
;AT
n
for some
B
as defined in Result III.
Note that since
is a Volterra operator
K
(I -K)
exists and it can be shown (Anselone, 1965) that
a
II(I
-111
for all
This implies that
II(I
II
x
continuous on
Kn) -1II< B
n
as
[0,1]
for some
B.
II(I
-1
-Kn)
n +
00.
1II
39
Examples
II.10
section we present four examples that illus-
In this
trate the results predicted in section II.9.
presented in TABLES
through
1
The data are
The numbers are the max-
4.
imum of the errors listed on the computer print out, and it
the norm of the error,
is suggested that they "close to"
using the max norm.
The numbers in parentheses are the
factors- -the
convergence
error at one value of
n
is di-
vided by the convergence factor to obtain the error for
twice the value of
values of
and
0
r,
with
n.
*
indication that
The columns are for the various
indicating regular Simpson's rule
K(s,$)
only was subtracted.
Examples one and two are initial value problems:
1)
x' (s)
= x(s), x(0) = 1 => x(s)
2)
x' (s)
-
= es.
(3s2-4s+1)x(s) = 0, x(0) =
3
1
2
es -2s +s®
=>'x(s) =
The results speak largely for themselves --using the best
value for
predicted.
r
we obtain a convergence rate of
16
just as
Note also that the error with just two appli-
cations of the modified rule is many times better than the
error using 32 applications of the regular rule.
This sug-
gests a significant savings in computer time to obtain a
desired degree of accuracy by using the modification.
40
Worst Error at Grid Points
0
n
*
2
0.763
0.000492
(13.7)
(2.0)
4
0.374
8
0.188
0.0000360
(14.7)
(2.0)
0.00000244
(15.3)
(2.0)
0.000000159
0.0941
16
(15.9)
(2.0)
0.0000000101
0.0471
32
Worst error of grid points and 0.01, 0.02, ...,
0.99, 1.0
0
n
*
2
0.764
0.000497
(13.8)
(2.0)
4
0.375
8
0.188
0.0000361
(14.7)
(2.0)
0.00000244
(15.3)
(2.0)
16
0.0941
32
0.0471
0.000000159
(15.9)
(2.0)
0.0000000101
s
x' (s)
= x(s)
,
x(0)
= 0,
x(t)dt = 1.0
or x(s) 0
TABLE
1.
Example
1:
A listing of the worst error
at the grid points and at the points
j=
1,
2,
...,
100.
j(0.01), for
Worst Error at Grid Points
n
2
*
1
0
0.0442
0.0332
(<
4
0.0457
8
0.0336
32
0.00999
0.00148
(18.9)
0.00992
0.0000809
(16.0)
0.00258
0.00000506
(16.0)
0.00000472
(16.2)
(4.0)
0.000651
(1.9)
(19.5)
0.0000756
(3.8)
(1.8)
0.0190
0.00153
(3.3)
1)
(1.4)
16
2
0.000000312
(16.1)
0.000000292
(16.1)
(4.0)
0.000163
0.0000000194
(15.9)
0.0000000183
Worst of Grid and interpolation points
n
2
0
*
0.0735
1
0.0332
(1.3)
4
0.0557
8
0.0336
16
0.0190
32
0.00999
0.00403
0.00148
(13.7)
(3.3)
0.00992
(1.7
0.000394
(1.8)
0.00000472
(4.0)
(4.0)
Example
2:
0.000000292
(15.9)
(8.1)
0.000000658
x'(s)
(16.1)
(8.3)
0.00000536
0.000163
2.
(16.0)
(8.8)
0.0000447
0.000651
(1.9)
(19.5)
0.0000756
(3.8)
0.00258
TABLE
2
-
(3s2- 4s +1)x(s) = 0,
0.0000000183
x(0)
= 1 or
s
x(s)
(3t2-4t+1)x(t)dt = 1.
0
x(s)
= exp(s3-2s2+s).
Worst Error at Grid Points
n
*
2
0.211
0.113
8
0.0572
(2.0)
32
*
2
0.211
0.113
8
0.0572
16
0.0287
32
0.0144
TABLE
3.
(2.8)
0.000274
(13.6)
0.0000194
(15.1)
(9.5)
0.00000128
(15.7)
(8.9)
(3.7)
0.000000894
0.0000000816
(15.2)
(8.4)
(3.9)
0.0000743
2
0.00000795
0.000288
(2.0)
0.99, 1.00.
(5.3)
(3.4)
0.00107
(2.0)
...,
0.0000765
0.00369
Example
0.01, 0.02,
0.000404
0.0103
(2.0)
0.00000000536
1
0
4
(15.2)
(16.1)
0.00000000722
0.0000743
(1.9)
0.0000000816
0.000000116
(3.9)
Worst error of grid points and
n
(15.6)
(15.8)
(3.7)
0.0144
0.00000127
0.00000183
0.000288
(2.0)
(15.0)
(15.5)
(3.4)
0.0287
(13.7)
0.0000191
0.0000284
0.00107
(2.0)
0.000262
(14.2)
(2.8)
0.00369
4
2
0.000404
0.0103
(1.9)
16
1
0
0.00000000536
0.000000106
3:
s
x(s)
sin(s)cos(t)x(t)dt = sin(s) + (1-sin(s) )exp(sin(s)
S
0
)
.
x(s)
= exp [sin(s)
]
.
V
43
Example
cs
x(s)
-
3
is the problem
sin(s)cos(t)x(t)dt = sin(s) + (1-sin(s))e
sin(s)
,
50
and the solution is
The results are
= exp[sin(s)].
x(s)
very similiar to those of examples
1
and
An unexpected
2.
result is fourth order convergence on the grid points for
r = 1.
Notice that this occurred in example
2
originally expected the convergence factors to be
16
for
r =
*,
0,
1,
2,
4,
8,
No adequate expla-
respectively.
2
We
also.
nation has been presented as to why this phenomenon occurs.
Example
4
is the problem
s+ cos Tr(Trs2)
s
x(s)
sin
'
0
Trs 1ff5)
(Tr-1)
with solution
0
<
s
-<
Tr
s<
1
1
(l+cos (Trs2)
sin(s)-i-sin(Trs2)
ts,
0<-
Tr
sin(Trst)x(t)dt =
-
(Trs2)
2s2
s
-
<
1
s
<
7s
2
1
Tr
<
-
x(s) =
1-s
-1
Tr
1
Tr
-
- 1,
which was chosen to test the results of II.8.
Note that
the orderly convergence patterns are disrupted and we are
no longer quite so sure convergence is occurring.
But re-
call that no convergence rate was predicted for this case,
simply that convergence would occur.
44
Worst Error at Grid Points
n
2
1
0
*
0.0234
0.000720
0.00321
(4.5)
(1.4)
0.0166
8
0.00860
(<
(1.9)
0.00406
32
0.00203
64
0.00102
1)
1)
(z.
1)
0.000714
0.000771
(9.5)
(10.0)
(2.1)
16
(z
0.000670
0.000683
4
-
0.0000744
0.0000761
(3.3)
(3.3)
(2.0)
0.0000227
0.0000230
(7.4)
(5.2)
(2.0)
0.00000307
0.00000441
Worst error of Grid Points
0.01, 0.02,
and
n
2
...,
0.000870
0.00321
0.0234
0.000676
0.000683
0.0166
(<
0.000777
0.00860
0.00000307
0.00000441
Example
's
0
<
x(s) -
4:
sin(frst)x(t)dt
=
f(s)
5 0
<
s,
where
(7.4)
(5.2)
0.00102
4.
0.0000227
0.0000230
0.00203
TABLE
(3.8)
(3.3)
(2.0)
64
0.0000744
0.0000761
0.00406
1)
(9.6)
(10.0)
(2.0)
32
(<
1)
0.000716
(2.1)
16
(1.3)
(4.8)
(1.9)
8
1.00.
1
0
*
(1.4)
4
0.99,
s
<
-
is a continuous function.
x(s) =
1-s
Tr-
r,
1
<
s
<
1,
45
Applications to Fredholm Equations
III.
Fredholm Equations with Jump Discontinuities
III.1
An easy extension of II.1 applies to Fredholm integral
equations of the type
1
x(s)
N(s,t)x(t)dt = f(s)
0
where
K(s,t)
,
0
<
t
<
s,
R(s,t)
,
s
<
t
<
1,
N(s,t) _
and where both
K
and
are continuous on their closed
R
That is, along the line
triangles of definition.
there is a jump discontinuity in
K(s,$)
-
R(s,$)
N
s
= t
with magnitude
.
If in addition we assume that the
vatives with respect to
t
of
K
and
rth
R
partial deriare at least
piecewise continuous on their closed domains, then we can
define a jump function
R
J(Q) (s)
[K(s,t) - R(s,t)lt=s,
E
a
and proceed as in II.1:
tQ
46
f(s)
= x(s)
1
-
Nr(s,t)x(t)dt
J
0
Q-(k)
-
(s)
(3 -1)
(s-t)%c(t)dt,
Q=0
0
where
QiQ(s t)kJ(R,) (s)
K(s,t) - C(
Nr(s,t) =
R(s,t)
,
s
<
t
is a continuous kernel and has
across the line
<
,
t
<
s
k=0
s =
t.
<
1
r
continuous derivatives
Since both integrals in
(3 -1)
can
be well approximated by finite sums, it is reasonable to
expect that the corresponding approximations,
solution,
x,
will also be good.
If
r
is
xn,
to the
large enough
we would also expect fourth order convergence using repeated Simpson's rule and its modification.
47
An Example
III.2
The one -dimensional Green's function
t(1 -s),
0
<
t
<
s
s(1 -t),
s
<
t
<
1
G(s,t) =
arises in two -point boundary value problems.
G(s,t)
is
As seen,
continuous, but its first derivative has a jump
discontinuity along the line
x(s) -
1
10
S
s =
G(s,t)x(t)dt
The problem
t.
=
(1
0
- 12)sin(Trs)
r
was solved by first using the regular Simpson's rule ap-
proximation of the integral and then using the special rule.
Since
J(Q)(s)
to be 1.
=
0
for
Q =
0,
and
Q
>
r
2,
The data are presented in table
5.
was chosen
Note that as
predicted in II.7 there is fourth order convergence using
the special rule, which is a significant improvement over
the first order convergence of the regular Simpson's rule.
48
The worst error at the Grid Points
n
Regular
2
0.869
Special with r =
0.735
(24.2)
(1.5)
4
0.0303
0.582
(16.0)
(2.3)
8
0.00189
0.252
(16.0)
(3.4)
16
0.000119
0.0770
TABLE
5.
1
The worst error at the
partition points of the problem
x(s) - 1' G(s,t)x(t)dt
0
=
x(s) = sin(Trs).
(1
-
)sin(ffs),
n
49
Generalizations
III.3
A generalization of the problem in III.1 is one in
which the discontinuity in
g(s),
0
<
s
<
is along a continuous curve,
N
The procedure is nearly the same, except
1.
that
J(k) (s)
ak
-
[K(s,t)
R(s,t)
-
]
at
=g(s)
and
-1(
K(s,t)
Nr
r
(s, t)
Q)k[g(s)-t]kJ(k) (s),
0<
t < g(s),
k=0
E
R(s,t), g(s)
<
t
1.
<
The equation corresponding to
(2 -2)
is
1
f(s) = x(s)
C
k =0
(
Nr(s,t)x(t)dt
-
S
Q)
R!
0
J(k)
\g(s)
(s)
[g(s)- t]kx(t)dt.
0
Note that now the upper limit of integration is
of
s,
a
function
so that the special rule developed in II.2 will no
longer work (unless g(s)
E
of course).
s
It should be
relatively straight forward, however, to develop weights
that are functions of both
technique in II.2.
2,
and
g(s), using the same
50
Summary and Conclusions
IV.
We started with the Volterra integral equation
(1 -3)
and generated a Fredholm integral equation with a continuous kernel.
value of the discontinuity at
s
= t,
K(s,$),
namely
leaving a Fredholm kernel which is zero for
the
K(s,t)
This was done by subtracting from
t S 1.
5
s
The Fredholm kernel was made smoother across the line
s
= t
by subtracting terms of Taylor's series; the result-
ing kernel was
K(s,t)
-
_(-1) R (s-t)
L
Nr(s,t) =
0, s
Assuming that
K
<
and
t
<
x
K(R)
(s,$)
,
0
<
t
<
s
!'
k=0
1.
have
continuous partials with
r
1
respect to
t,
Nr(s,t)x(t)dt
we showed that
S
was well
0
approximated by existing integration rules.
In order to preserve equality,
(-1)
Q
K(Q)
R!
were added.
s,$)
rs
JJ
terms of the form
(s-t
£
x (t) dt
0
In II.2 we developed a generalization of Simp-
son's rule which approximated the indefinite integral.
In
II.5 we combined this new rule with the regular Simpson's
rule,
and showed in II.7 that the resulting approximation
has fourth order convergence assuming enough differentia-
bility of the integrand.
We also showed in II.8 that with-
out differentiability the approximation still converges
51
uniformly to the integral as
n }
00.
The work of II.5
developed an algorithm for solving linear Volterra integral
equations of the second kind.
In the last chapter we extended the results to Fred -
holm integral equations with jump discontinuities in the
kernel or one of its partial derivatives with respect to
t
along the line
s
= t.
52
BIBLIOGRAPHY
1.
Anselone, P. M. Convergence and error bounds for
approximate solutions of integral and operator
equations. In: Error in digital computation, ed.
by L. B. Rall. Vol. 2. New York, John Wiley, 1965.
p.
231 -252.
2.
Fulks, Watson. Advanced calculus; an introduction to
analysis. New York, John Wiley, 1961. 521 p.
3.
Goldberg, Richard R. Methods of real analysis. Waltham, Mass., Blaisdell, 1964. 359 p.
4.
Hildebrand, F. B. Introduction to numerical analysis.
New York, McGraw -Hill, 1956. 511 p.
5.
Milne, William E. Numerical calculus. Princeton,
Princeton University, 1949. 393 p.
6.
Tricorni,
F. G. Integral equations. New York, Inter
science, 1957. 238 p.
-
APPENDICES
53
Appendix A
begin
comment a program to estimate
IR[x(s)]
I
<(M3/2)hQ +4
s
(
-11)] Idu
JO
has been parameterized out of the
where the
h
integral.
The integral is approximated by the
repeated midpoint rule;
integer n,tn,i,j,l;
real s,t,rem,prod;
real procedure Gn(a,b);
value a,b;
integer
real b;
a;
Gn:= if b >0 then b +a else 0.0;
real procedure WB(b,c);
value b,c;
real b; integer c;
WB:= b +(c +2)
x
((2- b) /(c +1) +
(2xb)- 2) /(c +2)- b /(c +3));
real procedure WC(b,c);
value b,c;
integer
real b;
WC: =
-0.5 x
b+
c;
(c +2) x
(
(l-b)/(c+l) +
(2xb-1)/(c+2)-b/(c+3)
comment the program body appears below;
inreal
(60,n);
inreal
(60,k);
comment a statement to print
n
and
R,
and set up a
table for the output would occur here;
tn: =2xn;
prod =
:
(R,
+1)
x
(i +2)
x
(k+3);
for j: =0 step 1 until to do
begin
real temb, temc;
s: =j /n;
temb: =WB (s,
R,)
;
)
;
54
temc:= WC(s,k);
rem: =0.0;
for i: =1 step
1
until to do
begin
t:= (2xi- 1) /tn;
rem: =rem + abs(2xGn(Z +3,s -t) /prod
-temc
x
Gn(2,2 -t)
-
temb
x
Gn(2,1 -t));
end;
rem:= rem /n;
comment a statement to output
rem, at
end;
end of program;
s
s
would go here;
and the remainder,
55
Appendix B
begin
comment a program to solve a Volterra integral equation
of the second kind using a modified Simpson's rule.
integer n,r,tn,i,Q,fac,j,P;
real s,h,th,hsq,D,sn,hpw,tl,t,xs;
real array X,F[O:UL];
comment UL is the upper limit of the array, i.e., UL > 2xn;
comment the three weight functions are below. s =a, Q =b,
h =c and hsq = e;
real procedure WA(a,b,c,e);
value a,b,c,e;
real a,c,e;
integer b;
WA: =a +(b +l)
x
(b +3))
x
(a
(2.0xa- 3.0xcx(b +3))
x
/(2.Oxex(b +l)
x
(b +2)
x
+ 2.0xex(b +2)
(b +3));
real procedure WB(a,b,c,e);
value a,b,c,e;
real a,c,e;
integer b;
WB: =2.0
x
a +(b +2)
x
{ex(b +l)
(cx(b +3)- a) /(ex(b
(b +2)
x
x
(b +3));
real procedure WC(a,b,c,e);
value a,b,c,e;
real a,c,e;
integer b;
WC: =a+ (b +2)
x
(2.0xa -cx
(b +3)) /
(2.0xex (b +l)
x
(b +2) x (b +3))
real procedure KER(a,b);
value a,b;
real a,b;
comment evaluates the kernel K(s,t) at
KER: =[the kernel];
s =a
and t =b;
;
56
real procedure PK(a,b);
value a,b;
integer b;
real a;
comment evaluate the bth partial of K(s,t) with respect to
t at s =t =a;
begin
switch Q:= PO,Pl,P2;
real RL;
20 to Q[b +l];
PO:
RL:=[the zeroth partial];
to fin;
coo
Pl:
RL: =[the first partial];
go to fin;
P2:
fin:
RL: =[the second partial];
PK: =RL;
end;
real procedure NR(a,b,c);
value a,b,c;
real a,b;
integer
c;
comment evaluates the kernel Nr at
s =a,
t =b,
begin integer LL;
real temp, nn,ff,dd,pp;
if b >a then
begin temp: =0.0; go to BBB end;
temp: = KER(a,b);
nn: = -1.0;
dd: =a -b;
for LL: =0 step
1
until
c do
begin
ff:= if LL =0 then 1.0 else ffxLL;
nn: = -nn;
and r =c;
57
pp:= if LL =0 then 1.0 else dd
temp := temp - nn
x
pp
X
pp;
x
PK(a,LL) /ff;
end;
BBB: NR: =temp;
end of procedure Nr;
real procedure FUN(a);
value a;
real a;
comment evaluates the known function at
s =a;
FUN: =[the known function];
comment the program begins here.
the number of
is
n
times the rule is applied and
r
of derivatives to be subtracted;
eof (finished)
inreal
(60,n);
inreal
(60,r);
tn: =2
;
n;
X
for is =0 step 1 until to do
begin
s:
=i /tn;
F[i]
:
=FUN
(s)
;
end;
X[0]
:
=F
[0]
;
h:= 1.0 /tn;
the =2.0
X
h;
hsq:= 1.0 /(tnxtn);
D: = -h
X
KER(h,0.0) /3.0;
sn: = -1.0;
hpw: =1.0;
for Q: =0 step 1 until r do
begin
fac:= if
2
=0 then
sn: = -sn;
hpw: =h
X
hpw;
1
else fac
X
2;
is
the number
58
D: =D + sn
PK(h,k)
x
x
(hpw/3.0 - WA(h,R,,h,hsg) )/fac;
end;
f[1]:=f[1]
for
i:
- D
=2 step
X[0];
x
until to do
1
begin
s:
=i /tn;
F[i] :=F[i]
+ h
x
KER(s,0.0)
X[0]/3.0;
x
end;
for j: =0 step
1
until n -1 do
begin
real temp,D1,D2,D3,D4,DET;
integer p,q;
p: =2
x
j
+ 1;
=P + 1;
s: =p /tn;
q:
D1:= if r
<
4.0
x
h
else fac
x
k;
then 1.0
0
-
x
KER(s,$) /3.0
else 1.0;
sn: = -1.0;
for
2,:
=0 step 1 until r do
begin
fac:= if
then
k =0
1
sn: = -sn;
D1: =D1 - sn
x
PK(s,k)
x
WB(h,k,h,hsq) /fac;
end;
D2 =0.0;
:
sn: = -1.0;
for
k:
=0 step 1 until r do
begin
fac:= if
then
k =0
1
else fac
x
k;
sn:=-sn;
D2:=D2
end;
s:=q/tn;
t:=p/tn;
-
sn
x
PK(s,k)
x
WC(h,i,h,hsq) /fac;
59
D3: = -4.0
h
X
D4: =1.0 - h
DET: =D1
KER(s,$) /3.0;
x
D4
X
if abs(DET)
KER(s,t) /3.0;
X
D2
<
D3;
X
then
10 -7
begin
comment a statement here would print "The problem
generated a singular matrix. ";
go to finished;
end;
- F [q] XD2)
/DET;
X [p]
:
= (F [p] XD4
X [q]
:
= (F [q] XDl - F [p] XD3) /DET;
if
j
s:=
22 to otpt;
= n -1 then
(q +l)
/tn;
temp: =4.0
t:
=q /tn;
D:
=2.0
X
h
X
h
KER(s,t) /3.0;
X
KER(s,t) /3.0;
X
sn: = -1.0;
hpw: =1.0;
for k: =0 step 1 until r do
begin
fact= if
then
k =0
else fac
1
X
k;
sn: = -sn;
hpw: =h
X
D: =D
sn
-
hpw;
PK(s,k)
X
(hpw/3.0- WA(h,Q,,h,hsq)) /fac;
x
end;
-
F[q+l]:=F[q+1] + temp
X[p]
x
+ D
X
X[q];
tl: =p /tn;
for
step
i:=441.1-2
1
until to do
begin
s:
=i /tn;
F[i]: =F[i]
X
end;
end;
+
4.0'x h
KER(s,t)
X
X
KER(s,tl)
X[q] /3.0;
X
X[p] /3.0 + 2.0
x
h
60
comment a statement to print the problem, the
values of n and r, and column headings
otpt:
for i
:
would go here;
=0 step 1 until to do
begin
s
:
=i /tn;
comment a statement to output s and
X[i] = x(s) would go here;
end;
Values of
comment the interpolation begins here.
s
are
input until there are no more, whereupon the
eof (finished) command halts the program;
begin real xs,tl,t2,t3; integer Il;
inter: comment a statement goes here to input s;
if s =0.0 then begin j: =0;
for is =0 step
if
2
begin
until n -1 do
l
h<
x
i
x
to AAA end;
coo
s
j: =i;
A
coo
(2xi +2)
h>
x
s
then
to AAA end;
xs:= FUN(s);
AAA:
for
step
is =0
until
1
j
-1 do
begin
I1 =2
i;
x
tl: =ll
h;
x
t2:= (I1 +1)
x
h;
t3:= (I1 +2)
x
h;
xs: =xs +
(
KER( s, tl)xX[I1] +4.0xKER(s,t2)XX[I1 +1]
+KER(s,t3)xX[I1 +2])
x
h /3.0;
end;
11;=2
x
j;
tl:=Il
x
h;
t2:=(I1+1)
x
h;
xs:=xs + (NR(s,tl,r)xX[I1]+4.0xNR(s,t2,r)xX[11+1])
sn:=-1.0;
tl:=s
-
2.0
for i:=0
-
begin
x
h
x
step
1
until r do
j;
x
h/3.0;
61
fac:= if
then
i =0
1
else fac
x
i;
sn:=-sn;
xs:=xs + sn
x
PK(s,i)
x
(WA(tl,i,h,hsq)xX[Il]
+WB(tl,i,h,hsq)xX[I1+1]+WC(tl,i,h,hsq)
xX[I1+2] )/fac;
end;
comment a statement here would output
end;
finished: end of program;
s
and
xs = x(s);