Linear integral equations

CHAPI'ER3

Method of Successive
Approximations

3.1. Iterative Scheme
Ordinary first-order differential equations can be solved by the well-known Pi-
card method of successive approximations. An iterative scheme based on the
same principle is also available for linear integral equations of the second kind:

g(s) = f(s) +).. j K(s, t)g(t)dt . (3.1.1)

In this chapter, we present this method. We assume the functions f (s) and
K(s, t) to be .C2-functions as defined in Chapter 1.
As a zero-order approximation to the desired function g(s), the solution
go(s),
go(s) = f(s) , (3.1.2)
is taken. This is substituted into the right side of Equation (3.1.1) to give the
first-order approximation

g1(s) = f(s) +).. j K(s, t)go(t)dt. (3.1.3)

This function, when substituted into Equation (3.1.1), yields the second approx-
imation. This process is then repeated; the (n + 1)th approximation is obtained
by substituting the nth approximation in the right side of (3.1.1). There results
the recurrence relation

gn+l(s)=f(s)+J.. J K(s,t)gn(t)dt. (3.1.4)

If gn (s) tends uniformly to a limit as n ---+ oo, then this limit is the required
solution. To study such a limit, let us examine the iterative procedure (3.1.4) in
detail. The first- and second-order approximations are

g1(s) = f(s)+J.. j K(s,t)f(t)dt (3.1.5)

J
and
g2(s) =f(s)+J.. K(s,t)f(t)dt

j [j
(3.1.6)
+.A? K(s, t) K(t, x)f(x)dx Jdt.

R.P. Kanwal, Linear Integral Equations: Theory & Technique, Modern Birkhäuser Classics, 25
DOI 10.1007/978-1-4614-6012-1_3, © Springer Science+Business Media New York 2013

26 3. Method of Successive Approximations

This formula can be simplified by setting

K2(s, t) = J K(s, x)K(x, t)dx (3.1.7)

and by changing the order of integration. The result is

g2(s) = f(s) +). J K(s, t)f(t)dt + ),,Z J K2(s, t)f(t)dt. (3.1.8)

Similarly,

g3(s) =f(s) +). j K(s, t)f(t)dt

+ ). 2 f K2(S, t)f(t)dt + ). 3 f K3(S, t)f(t)dt'
(3.1.9)

f
where
K3(s, t) = K(s, x)K2 (x, t)dx. (3.1.10)

By continuing this process, and denoting

Km(s,t) = J K(s,x)Km-t(X,t)dx, (3.1.11)

we get the nth approximate solution of integral equation (3.1.1) as

gn(s) = f(s) + ~).m J Km(S, t)f(t)dt. (3.1.12)

We call the expression Km(s, t) the mth iterate, where Kt(S, t) = K(s, t).
Passing to the limit as n --* oo, we obtain the so-called Neumann series

g(s) = lim gn(s) = f(s)
n-:,.oo
~).m
+ L....,
m=l
J Km(s, t)f(t)dt. (3.1.13)

It remains to determine the conditions under which this convergence is
achieved. For this purpose, we attend to the partial sum (3.1.12) and apply the
Schwartz inequality (1.6.3) to the general term of this sum. This gives

If Km(s,t)f(t)d{::;; ( / 1Km(s,t)i 2 dt) J if(t)i 2 dt. (3.1.14)

Let D be the norm of f,
(3.1.15)

3.1. Iterative Scheme 27

and let c;. denote the upper bound of the integral
ff Km(s, t)i 2 dt.

Then, the inequality (3.1.14) becomes

If Km(S, t)f(t)dtl
2
~ c;, D 2 • (3.1.16)

The next step is to connect the estimate c;.
with the estimate c;. This is achieved
by applying the Schwarz inequality to the relation (3.1.11):

which, when integrated with respect to t, yields

(3.1.17)

ff
where
B2 = IK(x, t)i 2dx dt. (3.1.18)
The inequality (3.1.17) sets up the recurrence relation
cz ~ BZm-zcz (3.1.19)
m '-" 1 •

From Equations (3.1.16) and (3.1.19), we have the inequality

If Km(s, t)f(t)dtl
2
~ CfD 2 B2m-Z. (3.1.20)

Therefore, the general term of the partial sum (3.1.12) has a magnitude less
than the quantity DC 1 1Aim Bm- 1, and it follows that the infinite series (3.1.13)
converges faster than the geometric series with common ratio IAlB. Hence, if

IAIB < 1, (3.1.21)
the uniform convergence of this series is assured.
We now prove that, for given A, Equation (3.1.1) has a unique solution.
Suppose the contrary, and let g 1 (s) and g 2 (s) be two solutions of Equation

f
(3.1.1):
g1(s) = f(s) +A K(s, t)g1(t)dt,

gz(s) = f(s) +A f K(s, t)gz(t)dt.


By subtracting these equations and setting g 1 (s)- g2(s) = </J(s), there results
the homogeneous integral equation

</J(s) =A I K(s, t)</J(t)dt.

Apply the Schwarz inequality to this equation and get

I</J(s)l 2 ~ IAI 2 I IK(s, t)l 2 dt I I</J(t)l 2dt,

or
(3.1.22)

In view of the inequality (3.1.21) and the nature of the function <P (s) = g 1 (s) -
g2 (s), we readily conclude that <P (s) =
0, that is, g 1 (s) g2 (s). =
What is the estimate of the error for neglecting terms after the nth term in
the Neumann series (3.1.13)? This is found by writing this series as

g(s) = f(s) + 1; IAm Km(S, t) f(t)dt + Rn(s). (3.1.23)

Then, it follows from the preceding analysis that

(3.1.24)
Finally, we can evaluate the resolvent kernel as defined in the previous
chapter, in terms of the iterated kernels Km (s, t). Indeed, by changing the order
of integration and summation in the Neumann series (3.1.13), we obtain

g(s) = f(s) + '- Jlt '-"-' K.(s, I)] f(t)dt.
Comparing this with (2.3. 7),

g(s) = f(s) +A I f(s, t; A)j(t)dt, (3.1.25)

we have
L Am- Km(S, t).
00

f(s, t; A)= 1 (3.1.26)
m=1
From the previous discussion, we can infer (see Section 3.5) that the series
(3.1.26) is also convergent at least for IAIB < 1. Hence, the resolvent kernel is
an analytic function of A, regular at least inside the circle IAI < B- 1 •
From the uniqueness of the solution of (3.1.1), we can prove that the resol-
vent kernel f(s, t; A) is unique. In fact, let Equation (3.1.1) have, with A = Ao,

3.2. Examples 29

two resolvent kernels r 1 (s, t; .A) and r 2 (s, t; .Ao). In view of the uniqueness of
the solution of (3.1.1), an arbitrary function f(s) satisfies the identity

f(s) + Ao I rt(S, t; Ao)f(t)dt =f(s) + Ao I rz(S, t; .A.o)f(t)dt. (3.1.27)
Setting ll(s, t; .Ao) = rt (s, t; .A.o)-rz(s, t; .Ao), we have, fromEquation(3.1.27),
I ll(s, t; .A.o)f(t)dt = 0,
for an arbitrary function f(t). Let us take f(t) = ll*(s, t; .A), with fixed s. This
implies that

I lll(s, t; .Ao)l 2 dt =0,
=
which means that II (s, t; A.o) 0, proving the uniqueness of the resolvent kernel.
The preceding analysis can be summed up in the following basic theorem.

Theorem. To each .C2 -kernel K(s, t), there corresponds a unique resolvent
kernel r(s, t; .A) which is an analytic function of .A, regular at least inside the
circle I.AI < n- 1, and represented by the power series (3.1.26). Furthermore,
if f(s) is also an .C2 -function, then the unique .Cz-solution of the Fredholm
equation (3.1.1) valid in the circle 1>..1 < n- 1, is given by the formula (3.1.25).

The method of successive approximations has many drawbacks. In addition
to being a cumbersome process, the Neumann series, in general, cannot be
summed in closed form. Furthermore, a solution of the integral Equation (3.1.1)
may exist even if I.AIB > 1, as evidenced in the previous chapter. In fact, we
saw that the resolvent kernel is a quotient of two polynomials of nth degree in
.A, and therefore the only possible singular points of r(s, t; .A) are the roots of
the denominator D(.A) = 0. But, for I.AIB > 1, the Neumann series does not
converge and as such does not provide the desired solution. We have more to
say about these ideas in the next chapter.

3.2. Examples

Example 1. Solve the integral equation

g(s) = f(s) +A 1 1
es-t g(t)dt . (3.2.1)


Following the method of the previous section, we have

K1(s, t) =es-t,

K2(s, t) = 11 es-xex-tdx =es-t.

Proceeding in this way, we find that all the iterated kernels coincide with K (s, t).
Using Equation (3.1.26), we obtain the resolvent kernel as

r(s, t; A.)= K(s, t)(1 +).. + A. 2 +···)=es-t /(1- A.). (3.2.2)

Although the series (1 +A.+ A. 2 + · · ·)converges only for lA. I < 1, the resolvent
kernel is, in fact, an analytic function of A., regular in the whole plane except at
the point A. = 1, which is a simple pole of the kernel r. The solution g(s) then
follows from (3.1.25):

g(s) = f(s)- [A./(A.- 1)] 1 1
es-t f(t)dt. (3.2.3)

Example 2. Solve the Fredholm integral equation

g(s) = 1 +A. 1 1
(1 - 3st)g(t)dt (3.2.4)

and evaluate the resolvent kernel.
Starting with g0 (s) = 1, we have

g1(s) = 1 +A. 1
1
(1 - 3st)dt = 1 +A. ( 1 - ~s) ,

g 2 (s) = 1 +A. 1
1
(1- 3st) [ 1 +A. ( 1- ~t) Jdt = 1 +A. ( 1- ~s) + ~A. 2 ,

or
(3.2.5)

The geometric series in Equation (3.2.4) is convergent provided lA. I < 2. Then,

g(s) = [4 + 2A.(2- 3s)]/(4- A. 2 ) , (3.2.6)

3.2. Examples 31

and precisely the same remarks apply to the region of the validity of this solution
as given in Example 1.
To evaluate the resolvent kernel, we find the iterated kernels.

Kt(s,t) = 1-3st,

K2(s, t) = 1 1 3
(1 - 3sx)(1 - 3xt)dx = 1 - -(s + t) + 3st ,
2
1
0

K3(s, t) =
1
(1- 3sx) [ 1- ~(x + t) + 3xt] dx
1 1
= -(1- 3st) = -K1 (s, t),
4 4

Similarly,

and

Hence,

f(s, t; A.)= Kt + lK2 + l 2K3 + ...

= (1+~l2 + 1~l4 + .. .)Kt+l(l+~l2 + 1~l4 + .. .)K2
= [o+A.)-~l(s+t)-3(1-l)st]/ (t-~l 2 ), Ill <2.
(3.2.7)


g(s) = 1 + l 1" [sin(s + t)]g(t)dt . (3.2.8)


Let us first evaluate the iterated kernels in this example:
K1(s,t) =K(s,t) =sin(s+t),

Kz(s, t) = 1rr [sin(s +;)]sin(x + t)dx
= 1 -Jr( sms sm
2
[ 0 0 1
t + coss cost = -Jl' cos(s - t) ,
]

2
K 3 (s, t) = ~Jl' t [sin(s + x)] cos(x- t)dx
2 Jo

= 1 -Jr 1<sms
2 0
1
0
cosx + smx sms)
0 0

x (cosx cost +sin x sin t)dx

= (~Jr Y [sins cost+ coss sin t] = (~Jl' Ysin(s + t) .

Proceeding in this manner, we obtain

K4(s, t) = (~Jr) 3 cos(s- t) ,

Ks(s, t) = (~Jr) 4
sin(s + t) ,

K6(s, t) = (~Jr) 5 cos(s- t), etc.
Substituting these values in the formula (3.1.13) and integrating, there results
the solution

g(s) = 1+ 2A(coss) [ 1+ G")' + G" + .. .J 1.2 )'A4

+J.2 ,.(sins) [1+ (H'+ (H'<'+ ···] , (3.2.9)
or

(3.2.10)
Since

3.2. Examples 33

the interval of convergence of the series (3.2.9) lies between -J2/n and J2;n.
Example 4. Prove that the mth iterated kernel Km(s, t) satisfies the following
relation:
Km(S, t) = J K,(s, x)Km-r(X, t)dx , (3.2.11)

where r is any positive integer less than m.
By successive applications of Equation (3.1.11 ), we have

Km(s, t) = J···J K(s, xi)K(Xt. x2) · · · K(Xm-t. t)dxm-1 · · · dx1 .
(3.2.12)
Thus, Km (s, t) is an (m - 1)-fold integral. Similarly, K, (s, x) and Km_,(x, t)
are (r - 1)- and (m - r - 1)-fold integrals. This means that

J K,(s, x)Km_,(x, t)dx

is an (m - 1)-fold integral, and the result follows.
One can take the go(s) approximation different from f(s), as we demon-
strate by the following example.
Example 5. Solve the inhomogeneous Fredholm integral equation of the second
kind,

g(s) = 2s + >..1 1 (s + t)g(t)dt, (3.2.13)

by the method of successive approximations to the third order.
For this equation, we take g 0 (s) = 1. Then,

gl (s) = 2s +A. t (s + t)dt = 2s + A.(s +!) ,
lo 2

gz(s) = 2s + A.1 1 (s + t){2t + A.[t + (1/2)]}dt
= 2s + A.[s + (2/3)] +A. 2 [s + (7 /12)],
g3(s) = 2s + A.1 1 (s + t){2t + A.[t + (2/3)] + A. 2 [t + (7/12)]}dt
= 2s + A.[s + (2/3)] +). 2 [(7 /6)s + (2/3)] + .A 3 [(13/12)s + (5/8)].
(3.2.14)
From Example 2 of Section 2.2, we find the exact solution to be

g(s) = [12(2- .A)s + 8.A]/(12- 12).- .A 2), (3.2.15)
and the comparison of Equations (3.2.14) and (3.2.15) is left to the reader.


3.3. Volterra Integral Equation

The same iterative scheme is applicable to the Volterra integral equation of the
second kind. In fact, the formulas corresponding to Equations (3.1.13) and
(3.1.25) are, respectively,

ges) = Jes)+ ~Am is Kmes,t)Jet)dt, (3.3.1)

ges) = Jes) +A. is res, t; A.) Jet)dt , (3.3.2)

where the iterated kernel Kmes, t) satisfies the recurrence formula

Kmes,t) = js Kes,x)Km-lex,t)dx (3.3.3)

with Ktes, t) = Kes, t), as before. The resolvent kernel res, t; A.) is given by
the same formula as (3.1.26), and it is an entire function of A. for any given es, t)
(see Exercise 8).
We illustrate this concept by the following examples.

3.4. Examples

Example 1. Find the Neumann series for the solution of the integral equation

ges) = e1 + s) +A. loses - t)get)dt . (3.4.1)

From the formula (3.3.3), we have
K 1 (s, t) = (s- t),

K 2 (s, t) =f 1
s
(s- x)(x- t)dx =
es- t)3
3! ,

=f
s (s - x)(x - t) 3 (s - t) 5
K 3 (s, t) 1 3! dx = 5! ,

and so on. Thus,

g(s) = 1 + s +A. (
s2
- +-
5 + A. 2 ( 5 +5 + ...
3) -
4 -
5) (3.4.2)
2! 3! 4! 5!
For A.= 1, g(s) = e'.

3.4. Examples 35


g(s) = f(s) +.l.. los t!-'g(t)dt (3.4.3)

and evaluate the resolvent kernel.
For this case,

(s- t)m-1 -t
Km(S, t} = (m -1)!
tf .

The resolvent kernel is
es-t L...m=1 .... - 1 (s-t)'"-l =
{
"'00 e<'-+1)(s-t)
~""'( t ~ s'
1 s, t,• II.}
' _
-
(m-1)! '
(3.4.4)
0, t > s.
Hence, the solution is

g(s) = /(s) + .l.. los e<A+1)(s-t) f(t)dt. (3.4.5)

Example 3. Solve the Volterra equation

g(s) = 1 +los stg(t)dt . (3.4.6)

For this example, K 1 (s, t) = K(s, t) = st,

Kz(s, t) =is sx 2 t dx = (s 4t- st 4)J3,

K 3 (s, t) =is [(sx)(x 4t - xt 4)j3]dx = (s 1 t- 2s 4t 4 + st 1 )j18,

K4(s, t) =is [(sx)(x 1 t - 2x 4t 4 +xt 1 )j18]dx

= (s 10 t - 3s 1 t 4 + 3s4t 1 - st 10 )j162,
and so on. Thus,
53 59 59 5 12
g(s) = 1 + 3 + 2 · 5 + 2 · 5 · 8 + 2 · 5 · 8 ·11 + · · · · (3.4.7)


3.5. Some Results about the Resolvent Kernel

The series for the resolvent kernel f(s, t; A.),

L A.m-l Km(S, t) ,
00

f(s, t; A.)= (3.5.1)
m=l

can be proved to be absolutely and uniformly convergent for all values of s and
t in the circle lA. I < 1/B. In addition to the assumptions of Section 3.1, we need
the additional inequality

E = const. (3.5.2)

Recall that this is one of the conditions for the kernel K to be an £ 2 -kernel.
Applying the Schwarz inequality to the recurrence formula

Km(s, t) =I Km-1(s,x)K(x, t)dx (3.5.3)

(I I
yields

!Km(S, 1)1 2 ~ !Km-1 (s, x)! 2dx) IK(x, t)! 2 dx ,

which, with the help of Equation (3.1.19), becomes

!Km(s, t)! ~ C1E Bm- 1 . (3.5.4)

Thus, the series (3.5.1) is dominated by the geometric series with the general
term C 1E(A.m- 1Bm- 1), and that completes the proof.
Next, we prove that the resolvent kernel satisfies the integral equation

f(s, t; A.)= K(s, t) +A. I f(s, x; A.)K(x, t) dx . (3.5.5)

This follows by replacing Km(s, t) in the series (3.5.1) by the integral relation
(3.5.3). Then,

f(s, t; A.)= K1(s, t) + f>m- I 1 Km-1(s, x)K(x, t)dx

I
m=2

= K(s, t) +A. f>..m- 1 Km(s,x)K(x, t)dx
m=l

= K(s, t) +A J[.~A"- 1 K.(s,x)] K(x, t)dx,

3.5. Some Results about the Resolvent Kernel 37

and the integral Equation (3.5.5) follows immediately. The change of order of
integration and summation is legitimate in view of the uniform convergence of
the series involved.
By proceeding in a similar manner we can prove that

f'(s,t;A) =K(s,t)+A I K(s,x)f'(x,t;A)dx. (3.5.6)

The two Equations (3.5.5) and (3.5.6) are called the Fredholm identities. From
the preceding analysis it follows that they are valid, at least, within the circle
IAlB < 1. Actually, they hold through the whole domain of the existence of the
resolvent kernel in the complex A-plane.
Another interesting result for the resolvent kernel is that it satisfies the
integro-differential equation

of'(s,t;A)/oA= I f'(s,X;A)f'(x,t;A)dx. (3.5.7)

In fact

I f'(s, x; A)f'(x, t; A)dx = I f : Am- 1 Km(s, x)
m=l
f:
n=l
An-l K,(x, t)dx.

On account of the absolute and uniform convergence of the series (3.5.1), we
can multiply the series under the integral sign and integrate it term by term.
Therefore,

I 00 00

f'(s,x;A)f'(x,t;A)dx = LL~>m+n-ZKm+n(s,t).
m=l n=l
(3.5.8)

Now, set m +n = p and change the order of summation; there results the relation
00 00 00 p-1

L LAm+n- 2 Km+n(s, t) = L LV- 2 Kp(S, t)
m=ln=1 p=2n=l

= f:<P- 1)V-2 Kp(s, t) = of'(s, t; A)
p=Z oA
(3.5.9)
Combining Equations (3.5.8) and (3.5.9), we have the result (3.5.7).
By a similar process we can obtain the integral equation

f'(s, t; A)- f'(s, t; J.-L) =(A- J.-L) I f'(s, x; J.-L)f'(x, t; A)dx , (3.5.10)

which is called the resolvent equation. Then Equation (3.5.7) can be derived
from (3.5.10) by dividing it by (A - J.-L) and letting J.-L ~ A. Equation (3.5.10)


can also be derived by the help of the Fredholm identities (3.5.5) and (3.5.6) (see
Exercise 9).
A Volterra integral equation of the second kind satisfies the identities

f(s,t;'A)=K(s,t)+>.. { f/s f(s, x; 'A)K(x, t)dx, (3.5.11)
J, K(s,x)f(x,t;'A)dx,
and are proved in a similar fashion [4].
In Section 1.6 we defined the orthogonal kernels L(s, t) and M(s, t). If we
let r t and r m denote the resolvent kernels for L and M, so that we have
00 00

ft(S, t; A)= L'An-lLn(S, t) = L'AnLn+t(S, t), (3.5.12)
n=l n=O

and
00 00

fm(s,t;>..) = L'An-lMn(s,t) = L'AnMn+l(s,t), (3.5.13)
n=l n=O
where Ln and Mn denote the iterated kernels. For the case that L and Mare
orthogonal it follows that

J L(s, x)f(x, t; 'A)dx = 0, (3.5.14)

and
I M(s, x)ft(X, t; >..)dx = 0. (3.5.15)

Furthermore, by applying the Fredholm identity (3.5.6) to the resolvent
kernels r l and r m, it follows that

ft(S, t; A)+ f m(s, t; 'A)= L(s, t) + M(s, t)

+'A I[L(s,x)ft(x,t;>..)+M(s,x)fm(X,t;>..)dx, (3.5.16)

which, in view of Equations (3.5.14) and (3.5.15), can be written as

ft(s, t; A)+ f m(s, t; A)= L(s, t) + M(s, t)

+)..I [L (s, x) + M(s, x)][ft(S, t; A)+ r m(X, t; >..)]dx . (3.5.17)

This shows that the resolvent kernel of the sum L(s, t) + M(s, t) of two orthog-
onal kernels is the sum of their resolvent kernels, that is,

ft+m(s, t; A)= ft(S, t; A)+ f m(s, t; A) . (3.5.18)

3.5. Some Results about the Resolvent Kernel 39

Exercises

1. Solve the following Fredholm integral equations by the method of successive
approximations:

1
1
g(s) = es - -e + - +-
1 11 g(t)dt .
2 2 2 0

1 111C/2
g(s) =(sins)- -s +- stg(t)dt.
4 4 0

2. Consider the integral equation

g(s) = 1 +A 1 1
stg(t)dt .

(a) Make use of the relation IAI < B- 1 to show that the iterative procedure
is valid for IAI < 3.
(b) Show that the iterative procedure leads formally to the solution

g(s) = 1 +s[(A/2) + (A 2 /6) + (A 3 j18) + ... ].

(c) Use the method of the previous chapter to obtain the exact solution

g(s) = 1 + [3As/2(3- A)],

3. Solve the integral equation

g(s) = 1 +A 1 1
(s + t)g(t)dt ,

by the method of successive approximations and show that the estimate afforded
by the relation IAI < B- 1 is conservative in this case.

4. Find the resolvent kernel associated with the following kernels: (i) Is - t I,
in the interval (0, 1); (ii) exp(-ls- tl), in the interval (0, 1); (iii) cos(s + t),
in the interval (0, 2rr).

5. Solve the following Volterra integral equations by the method of this chapter:

g(s) = 1 + 1s (s - t)g(t)dt , (i)

g(s) = 29 + 6s + 1s (6s - 6t + S)g(t)dt . (ii)


6. Find an approximate solution of the integral equation

g(s) = (sinh s) + 1s er-s g(t) dt ,

by the method of iteration.

7. Obtain the radius of convergence of the Neumann series when the functions
f(s) and the kernel K(s, t) are continuous in the interval (a, b).

8. Prove that the resolvent kernel for a Volterra integral equation of the second
kind is an entire function of .1.. for any given (s, t).

9. Derive the resolvent equation (3.5.10) by appealing to the Fredholm identities
(3.5.5) and (3.5.6).

Linear integral equations

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