Chapter 4

Solution to Linear Integral

Equations

This section studies the solutions to linear Fredholm integral equations with special kernels.

Definition 1. A kernel K(s, t) is called degenerate or separable if it can be expressed as the

sum of a finite number of terms, each of which is the product of function of s only and a
function of t only, i.e.
n
X
K(s, t) = ai (s)bi (t) = a1 (s)b1 (t) + a2 (s)b2 (t) + . . . + an (s)bn (t).
i=1

For example, the kernels K(s, t) = st+s2 t2 and K(s, t) = sin(s+t) = sin s cos t+cos s sin t
are degenerate.

Definition 2. A complex-values kernel K(s, t) is called symmetric kernel or Hermitian kernel

if
K(s, t) = K ∗ (t, s),
where K ∗ denotes the complex conjugate of K. For real kernel, we have

K(s, t) = K(t, s).

For instance, K(s, t) = cos(s+t) = cos(t+s) and K(s, t) = es+t are examples of symmetric
kernels.

Definition 3. For a homogeneous Fredholm integral equation

Zb
f (s) = λ K(s, t)f (t)dt,
a

λ is the eigenvalue or characteristic value and f (t) is the eigenfunction or the characteristic
function.

1
Definition 4. A solution of an integral equation is a function f (s) which when substituted
in the given integral equation reduces it to an identity, i.e. f (s) satisfies the given equation.

Example 1. Show that f (s) = 1 − s is a solution of the Volterra integral equation

Zs
s= es−t f (t)dt.
0

Solution. Since f (s) = 1 − s, we have f (t) = 1 − t. Putting it in the given integral equation,
we have
Zs
s= es−t (1 − t)dt
0
Zs
s−t s
=− e (1 − t) − 0
es−t dt
0
s−s s s
+ es−t
 
=− e (1 − s) − e 0
= −(1 − s − es ) + es−s − es
= −1 + s + es + 1 − es
=s
Rs
Which shows that f (s) = 1 − s is a solution to s = es−t f (t)dt.
0

4.1 Integral equations with degenerate kernel

Let us consider the non-homogeneous Fredholm integral equations with degenerate kernel
Zb
g(s) + λ K(s, t)f (t)dt = f (s) (1)
a

Since the kernel K(s, t) is degenerate, so let

n
X
K(s, t) = ai (s)bi (t).
i=1

Putting this value in (1), we obtain

Zb X
n
g(s) + λ ai (s)bi (t)f (t)dt = f (s)
a i=1

n
X Zb
g(s) + λ ai (s) bi (t)f (t)dt = f (s) (2)
i=1 a

2
Let ci be the unknown constants defined by

Zb
ci = bi (t)f (t)dt (3)
a

Making these substitutions in (3), we have

n
X
g(s) + λ ai (s)ci = f (s) (4)
i=1

From (4), we obtain

n
X
g(t) + λ ak (t)ck = f (t) (5)
k=1

Putting these values of f (t) in (3), we have

Zb " n
X
#
ci = bi (t) g(t) + λ ak (t)ck dt
a k=1

Zb n
X Zb
=⇒ ci = bi (t)g(t)dt + λ ck bi (t)ak (t)dt (6)
a k=1 a

Let
Zb
gi = bi (t)g(t)dt, (7)
a

and
Zb
aik = bi (t)ak (t)dt, i = 1, 2, . . . , n (8)
a

Using these values from (7) and (8) in (6), we obtain

n
X
ci = gi + λ ck aik (9)
k=1

Note that the Kronecker delta δik is defined as

(
0, if i 6= k;
δik =
1, if i = k.

Using this definition, ci can be written as

n
X
δik ck = ci (10)
k=1

3
Putting this value of ci in (9), we obtain
n
X n
X
δik ck = gi + λ ck aik
k=1 k=1
Xn n
X
δik ck − λ ck aik = gi
k=1 k=1
Xn
(δik − λaik )ck = gi , i = 1, 2, . . . , n (11)
k=1

The system (11) consists of n nonhomogeneous linear equations in n unknowns c1 , c2 , . . . , cn .

The system (11) will have a unique solution if the matrix of coefficients δik − λaik is nonsin-
gular, i.e.,
D(λ) = |δik − λaik | =
6 0.
where
1 − λa11 −λa12 · · · −λa1n
−λa21 1 − λa22 · · · −λa2n
D(λ) = .. .. .. ..
. . . .
−λan1 −λan2 · · · 1 − λann
is a polynomial in λ if degree n.
For D(λ) = 0, the n roots λ1 , λ2 , . . . , λn are the eigenvalues of the given nonhomogeneous
Fredholm integral equation or of the homogeneous Fredholm integral equation
Zb
λ K(s, t)f (t)dt = f (s),
a

then, the given nonhomogeneous integral equation has a solution if and only if
Zb
φai (s)f (s)ds = 0,
a

for all φai (s) which are linearly independent (nontrivial) solutions of
Zb
λ K(s, t)f (t)dt = f (s).
a

When λ 6= λi (i = 1, 2, . . . , n), i.e. λ is not an eigenvalues of the nonhomogeneous integral

equation, them system (11) will have a unique solution. On substituting those values of λ
which are not the eigenvalues in (5), we obtain the unique solution of the given nonhomoge-
neous integral equation.

Next, let us employ this method to solve some nonhomogeneous Fredholm integral equa-
tions.

4
Example 2. Solve the following nonhomogeneous Fredholm integral equation

Z1
s+λ (st + s2 t2 )f (t)dt = f (s)
0

Solution. The given integral equation is

Z1
s+λ (st + s2 t2 )f (t)dt = f (s) (12)
0

Here the kernel K(s, t) = st + s2 t2 is degenerate. Eq. (12) can be written as

Z1 Z1
2
s + sλ tf (t)dt + λs t2 f (t)dt = f (s) (13)
0 0

Let
Z1 Z1
c1 = tf (t)dt, c2 = t2 f (t)dt (14)
0 0

Making these substitutions in (13), we obtain

s + λsc1 + λs2 c2 = f (s) (15)

If (15) is the solution of (12), then it must satisfy (14), i.e.,

t + λtc1 + λt2 c2 = f (t), (16)

must satisfy (14), so putting the value of f (t) from (16) in c1 of (14), we obtain

Z1
c1 = t(t + λtc1 + λt2 c2 )dt
0
Z1
= (t2 + λt2 c1 + λt3 c2 )dt
0
1
t3 t3 t4
=
+ λ c1 + λ c2
3 3 4 0
1 λc1 λc2
= + +
3 3 4
=⇒ 12c1 = 4 + 4λc1 + 3λc2 = 4
=⇒ (12 − 4λ)c1 − 3λc2 = 4 (17)

5
Similarly, putting the value of f (t) from (16) in c2 of (14), we obtain

Z1
c2 = t2 (t + λtc1 + λt2 c2 )dt
0
Z1
= (t3 + λt3 c1 + λt4 c2 )dt
0
1
t4 t4 t5
= + λ c1 + λ c2
4 4 5 0
1 λc1 λc2
= + +
4 4 5
=⇒ 20c2 = 5 + 5λc1 + 4λc2
=⇒ − 5λc1 + (20 − 4λ)c2 = 5 (18)

Eqs. (17) & (18) is a system of two linear equations with unknowns c1 and c2 . Next, we solve
this system whose augmented matrix à is
" # " #
−3λ 4
12 − 4λ −3λ 4 1 12−4λ 12−4λ 1
A
e= ∼ , R1
−5λ 20 − 4λ 5 −5λ 20 − 4λ 5 12 − 4λ
" # " #
−3λ 4 −3λ 4
1 12−4λ 12−4λ 1 12−4λ 12−4λ
∼ 15λ2 20λ , R2 + 5λR1 , = 2
0 20 − 4λ − 12−4λ 5 + 12−4λ 0 240−128λ+λ
12−4λ
60
12−4λ
" #
−3λ 4
1 12−4λ 12−4λ 12 − 4λ
∼ 60 , R2
0 1 240−128λ+λ2
240 − 128λ + λ2
" # " #
4 180λ 80−λ
1 0 12−4λ + (12−4λ)(240−128λ+λ 2) 3λ 1 0 240−128λ+λ 2
∼ 60
, R1 + R2 , = 60
0 1 240−128λ+λ2
12 − 4λ 0 1 240−128λ+λ 2

which is in the reduced echelon form, providing the solution

80 − λ 60
c1 = 2
, c2 =
240 − 128λ + λ 240 − 128λ + λ2
Putting this value of c1 and c2 in (15), we obtain

80 − λ 60
f (s) = s + λs + λs2
240 − 128λ + λ2 240 − 128λ + λ2
s(240 − 128λ + λ2 ) + λs(80 − λ) + 60λs2
=
240 − 128λ + λ2
(240 − 48λ)s + 60λs2
=
240 − 128λ + λ2
This is the required solution of the given nonhomogeneous Fredholm integral equation.

Let us solve another nonhomogeneous Fredholm integral equation.

6
Example 3. Solve the following nonhomogeneous Fredholm integral equation

Zπ/2
1+λ sin(s + t)f (t)dt = f (s)
0

Solution. The given integral equation is

Zπ/2
1 + λ (sin s cos t + cos s sin t)f (t)dt = f (s)
0
Zπ/2 Zπ/2
1 + λ sin s cos tf (t)dt + λ cos s sin tf (t)dt = f (s) (19)
0 0

Let
Zπ/2 Zπ/2
c1 = cos tf (t)dt, c2 = sin tf (t)dt (20)
0 0

Making these substitutions in (19), we have

1 + λ sin sc1 + λ cos sc2 = f (s) (21)

If (21) is the solution of (19), then it must satisfy (20), i.e.,

1 + λ sin tc1 + λ cos tc2 = f (t) (22)

must satisfy (20), so putting the value of f (t) from (22) in c1 of (20), we have

Zπ/2
c1 = cos t(1 + λ sin tc1 + λ cos tc2 )dt
0
Zπ/2
= (cos t + λ cos t sin tc1 + λ cos2 tc2 )dt
0
Zπ/2 
1 + cos 2t
= cos t + λ cos t sin tc1 + λ( )c2 dt
2
0
π/2
sin2 t λc2
 
sin 2t
= sin t + λc1 + t+
2 2 2 0
λc1 λπc2
=1+ +
2 4
=⇒ 4c1 = 4 + 2λc1 + λπc2
=⇒ (4 − 2λ)c1 − λπc2 = 4 (23)

7
Similarly, putting the value of f (t) from (22) in c2 of (20), we obtain

Zπ/2
c2 = sin t(1 + λ sin tc1 + λ cos tc2 )dt
0
Zπ/2
= (sin t + λ sin2 tc1 + λ cos t sin tc2 )dt
0
π/2
1 − cos 2t λc2
= sin t + λc1 + sin2 t
2 2 0
λπc1 λc2
=1+ +
4 2
=⇒ 4c2 = 4 + λπc1 + 2λc2
=⇒ − λπc1 + (4 − 2λ)c2 = 4 (24)

Eqs. (23) & (25) is a system of two linear equations with unknowns c1 and c2 . Next, we solve
this system whose augmented matrix à is
" # " #
−λπ 4
4 − 2λ −λπ 4 1 4−2λ 4−2λ 1
A
e= ∼ , R1
−λπ 4 − 2λ 4 −λπ 4 − 2λ 4 4 − 2λ
" # " #
−λπ 4 −λπ 4
1 4−2λ 4−2λ
1 4−2λ 4−2λ
∼ λ2 π 2 , R2 + λπR1 , = 2 −λ2 π 2 ,
0 4 − 2λ − 4−2λ 4λπ
4 + 4−2λ 0 (4−2λ)
4−2λ
8−4λ+2λπ
2−λ
" #
−λπ 4
1 4−2λ 4−2λ 4 − 2λ
∼ 16−8λ+4λπ , R
2 − λ2 π 2 2
0 1 (4−2λ)2 −λ2 π 2 (4 − 2λ)
λπ(16−8λ+4λπ)
" #
4
1 0 4−2λ + (4−2λ)[(4−2λ) 2 −λ2 π 2 ] λπ
∼ 16−8λ+4λπ , R1 + R2 .
0 1 (4−2λ)2 −λ2 π 2
4 − 2λ

After some routine simplifications, we obtain the reduced echelon form as follows:
" #
16−8λ+4λπ
1 0 (4−2λ) 2 −λ2 π 2
Ae∼
16−8λ+4λπ
0 1 (4−2λ) 2 −λ2 π 2

This gives us the following solution

16 − 8λ + 4λπ
c1 = c2 =
(4 − 2λ)2 − λ2 π 2
Putting these values of c1 and c2 in (21), we obtain
λ(16 − 8λ + 4λπ) sin s λ(16 − 8λ + 4λπ) cos s
f (s) = 1 + +
(4 − 2λ)2 − λ2 π 2 (4 − 2λ)2 − λ2 π 2
λ(16 − 8λ + 4λπ) sin s + λ(16 − 8λ + 4λπ) cos s
=1+
(4 − 2λ)2 − λ2 π 2
This is the required solution of the given nonhomogeneous Fredholm integral equation.

8
4.1.1 Eigenvalues and eigenfunctions
Given a homogenous Fredholm integral equation with degenerate kernel, this subsection com-
putes its eigenvalues and eigenfunctions.
Let us illustrate this method with the help of some examples.

Example 4. Find the eigenvalues and eigenfunctions of the integral equation

Zπ/4
λ sin2 xφ(t)dt = φ(x)
0

Solution. Note that the kernel K(x, t) = sin2 x is separable. The given equation can be
written as
Zπ/4
2
λ sin x φ(t)dt = φ(x) (25)
0
Let
Zπ/4
c= φ(t)dt. (26)
0
Making this substitution in (25), we have

λc sin2 x = φ(x) (27)

If (27) is the solution to the given equation, then it must satisfy (26), i.e.

λc sin2 t = φ(t) (28)

must satisfy (26), so putting the value of φ(t) from (28) in (26), we have

Zπ/4 Zπ/4
λc
c= λc sin2 tdt = (1 − cos 2t)dt
2
0 0
2 π/4  
λc sin t λc π 1
= t− = −
2 2 0 2 2 2
=⇒ 8c = λc(π − 2) =⇒ [λ(π − 2) − 8] c = 0
=⇒ λ(π − 2) − 8 = 0, ∵ c 6= 0
8
=⇒ λ =
π−2
This is the required eigenvalue. Putting this values of λ in (27), we have
8
φ(x) = c sin2 x. (29)
π−2
8c
Since c 6= 0 is constant, we consider A = π−2 . Putting this (29), we obtain φ(x) = A sin2 x
which is the required eigenfunction.

9
 Eigenvalues of an integral equation do not always exist. The next example illustrates this
fact.

Example 5. Show that the homogeneous Fredholm integral equation

Zπ
λ sin s sin 2tf (t)dt = f (s)
0

has no eigenvalues.

Solution. Here the kernel K(s, t) = sin s sin 2t is separable. The given equation can be
written as
Zπ
λ sin s sin 2tf (t)dt = f (s) (30)
0

Let
Zπ
c= sin 2tf (t)dt. (31)
0

Making this substitution in (30), we have

λc sin s = f (s) (32)

If (32) is the solution to the given equation, then it must satisfy (31), i.e.

λc sin t = f (t) (33)

must satisfy (31), so putting the value of f (t) from (33) in (31), we have
Zπ Zπ
λc
c= sin 2tλ sin tcdt = sin 2t sin tdt
2
0 0
Zπ Zπ
λc λc
=− −2 sin 2t sin tdt = − [cos(2t + t) − cos(2t − 2)] dt
2 2
0 0
Zπ
λc
=− (cos 3t − cos t) dt
2
0
π
λc sin 3t
=⇒ − − sin t =0
2 3 0

This shows that the given integral equation has no eigenvalues.

10
4.1.2 Resolvent kernel
n
P
A nonhomogeneous Fredholm integral equation with degenerate kernel K(s, t) = ai (s)bi (t)
i=1
given as
Zb
g(s) + λ K(s, t)f (t)dt = f (s)
a
has a solution of the form
Zb
g(s) + λ Γ(s, t; λ)f (t)dt = f (s),
a

where λ is the eigenvalue and Γ(s, t; λ) is called the resolvent kernel of the given integral
equation. Here, we explain this method of finding the solution by resolvent kernel with an
example.

Example 6. Find the eigenvalues and the resolvent kernel of the integral equation
Z1
g(s) + λ (s + t)f (t)dt = f (s).
0

Solution. The given integral equation is

Z1
g(s) + λ (s + t)f (t)dt = f (s), (34)
0

where the kernel K(s, t) = s + t is degenerate. Comparing this with the integral equation
Z1 2
X
! Z1
g(s) + λ ai (s)bi (t) f (t)dt = g(s) + λ (a1 (s)b1 (t) + a2 (s)b2 (t)) f (t)dt = f (s)
0 i=1 0

we have a1 (s) = s, b1 (t) = 1, a2 (s) = 1, b2 (t) = t. We know that

Z1
aik = bi (t)ak (s)dt (35)
0

For 1 ≤ i, k ≤ 2, we calculate aik from (35) as follows:

Z1 Z1
1 1 1
a11 = b1 (t)a1 (s)dt = tdt = , and a12 = 1, a21 = , a22 =
2 3 2
0 0

We also know that

Z1
fi = bi (t)f (t)dt, i = 1, 2.
0

11
By putting b1 (t) = 1, b2 (t) = t, we obtain

Z1 Z1
f1 = f (t)dt, f2 = tf (t)dt.
0 0

Moreover, we have
2
X 2
X
ci = fi + λ ck aik , ci − λ ck aik = fi (36)
k=1 k=1

Putting i = 1, 2 in (36), we obtain

c1 − λ(c1 a11 + c2 a12 ) = f1

 
1 1
=⇒ c1 − λ c1 + c2 = f1 ∵ a11 = , a12 = 1
2 2
 
λ
=⇒ 1 − c1 − λc2 = f1 (37)
2
and, similarly,

c2 − λ(c1 a21 + c2 a22 ) = f2

 
1 1 1 1
=⇒ c2 − λ c1 + c2 = f2 ∵ a21 = , a22 =
3 2 3 2
(38)
 
λ λ
=⇒ − c1 + 1 − c2 = f2 (39)
3 2

For eigenvalues, we must have D(λ) = 0 from (37) & (38), i.e.

1 − λ2 λ 2 λ2
 
−λ
= 0, =⇒ 1 − − = 0 =⇒ λ2 + 12λ − 12 = 0
− λ3 1 − λ2 2 3
√
12 ± 144 + 48 √
=⇒ λ = = −6 ± 4 3
2
√
Thus, the required eigenvalues are λ = −6 ± 4 3. Next, we solve (37) & (38) for c1 and c2
using Crammer’s rule as follows:
Here
1 − λ2 λ 2 λ2 λ2 + 12λ − 12
 
−λ
|A| = λ λ = 1 − − =−
−3 1− 2 2 3 12
Now,

12f1 1 − λ2 + 12f2 λ


1 f1 −λ 12 f1 −λ
c1 = =− 2 =−
|A| f2 1 − λ2 λ + 12λ − 12 f2 1 − λ2 λ2 + 12λ − 12

−12f1 + λ(6f1 − 12f2 )

=⇒ c1 =
λ2 + 12λ − 12

12
Similarly,

12f2 1 − λ2 − 12 f13λ


1 1 − λ2 f1 12 1 − λ2 f1
c2 = =− 2 =−
|A| − λ3 f2 λ + 12λ − 12 − λ3 f2 λ2 + 12λ − 12

−12f2 − λ(4f1 − 6f2 )

=⇒ c2 =
λ2 + 12λ − 12
Finally, using the relation
2
X
g(s) + λ ai (s)ci = f (s), =⇒ g(s) + λ [a1 (s)c1 + a2 (s)c2 ]
i=1

=⇒ g(s) + λ [sc1 + c2 ] = f (s), ∵ a1 (s) = s, a2 (s) = 1

Putting the values of c1 and c2 , we obtain
 
[−12f1 + λ(6f1 − 12f2 )]s −12f2 − λ(4f1 − 6f2 )
g(s) + λ + = f (s)
λ2 + 12λ − 12 λ2 + 12λ − 12
Plugging back the values of f1 and f2 , we obtain
 
R1
 1
R1 R1
 1
R1

R R
 −12 f (t)dt + λ 6 f (t)dt − 12 tf (t)dt s −12 tf (t)dt − λ 4 f (t)dt − 6 tf (t)dt 
0 0 0 0 0 0
g(s)+λ  +  = f (s)
 
 λ2 + 12λ − 12 λ2 + 12λ − 12 

Routine simplification gives us

Z1
6(λ − 2)(s + t) − 12λst − 4λ
g(s) + λ f (t)dt = f (s).
λ2 + 12λ − 12
0

This is the solution of the given integral equation (34). The resolvent kernel is
6(λ − 2)(s + t) − 12λst − 4λ
Γ(s, t; λ) =
λ2 + 12λ − 12

4.1.3 Fredholm alternative theorem

Here is the integral equations analog of the Fredholm alternative theorem.

Theorem 4.1.1. Either the integral equation

Z
g(s) + λ K(s, t)f (t)dt = f (s) (40)

with fixed λ possesses one and only one solution f (s) for arbitrary functions g(s) and K(s, t),
in particular, the solution f = 0 for g = 0; or the homogeneous equation
Z
λ K(s, t)f (t)dt = f (s)

13
possesses a finite number of r linearly independent solution fi0 , i = 1, 2, . . . , r. In the first
case, the transposed nonhomogeneous equation
Z
g(s) + λ K(t, s)ψ(t)dt = ψ(s)

also possesses a unique solution. In the second case, the transposed homogeneous equation
Z
λ K(t, s)ψ(t)dt = ψ(s)

also has r linearly independent solution ψi0 , i = 1, 2, . . . , r; the nonhomogeneous integral

equation Z
g(s) + λ Γ(s, t; λ)g(t)dt = f (s)

has a solution if and only if the given function g(s) satisfies the r conditions
Z
(g, ψi ) = g(s)ψi0 ds = 0, i = 1, 2, . . . , r
0

In this case, the solution of (40) is determined only up to an additive linear combination
r
ci fi0 , for some ci ∈ R.
P
i=1

The following example elaborates the existence of solutions of the Fredholm integral equa-
tions provided by the Fredholm alternative theorem i.e. Theorem 4.1.1.
Example 7. Show that the integral equation
Z2π
1
g(s) + sin(s + t)f (t)dt = f (s)
π
0

possesses no solution for g(s) = s, but that it possesses infinitely many solutions when
g(s) = 1.
Solution. The given integral equation is
Z2π
1
g(s) + (sin s cos t + sin t cos s)f (t)dt = f (s) (41)
π
0

First, we find the eigenvalues of the homogeneous integral equation

Z2π
λ (sin s cos t + sin t cos s)f (t)dt = f (s) (42)
0

Here the kernel K(s, t) = sin s cos t + sin t cos s is symmetric and degenerate. The equation
(42) can be written as
Z2π Z2π
λ sin s cos tf (t)dt + λ cos s sin tf (t)dt = f (s) (43)
0 0

14
Let
Z2π
c1 = cos tf (t)dt (44)
0
and
Z2π
c2 = sin tf (t)dt (45)
0
Making these substitutions in (43), we have

λ sin sc1 + λ cos sc2 = f (s) (46)

If (46) is the solution of (42), then it must satisfy (44) & (45), i.e.,

λ sin tc1 + λ cos tc2 = f (t) (47)

must satisfy (44) & (45), so putting the value of f (t) from (47) in (44), we have
Z2π
c1 = cos t(λ sin tc1 + λ cos tc2 )dt
0
Z2π Z2π
= λc1 cos t sin tdt + λc2 cos2 tdt
0 0
2π Z2π
sin2 t λc2
= λc1 + (1 + cos 2t)dt
2 0 2
0
2π
λc2 sin 2t
=0+ t+ = λπc2
2 2 0
=⇒ c1 − λπc2 = 0 (48)

Similarly, putting the value of f (t) from (47) in (45), we obtain

Z2π
c2 = sin t(λ sin tc1 + λ cos tc2 )dt
0
Z2π Z2π
2
= λc1 sin tdt + λc2 s ∈ t cos tdt
0 0
Z2π 2π
λc1 sin2 t
= (1 − cos 2t)dt + λc2
2 2 0
0
2π
λc2 sin 2t
= t− + 0 = λπc1
2 2 0
=⇒ − λπc1 + c2 = 0 (49)

15
For the nontrivial solution of (48) & (49), we have

1 −λπ 1
= 0, =⇒ 1 − λ2 π 2 = 0 =⇒ λ2 π 2 = 1 =⇒ λ = ±
−λπ 1 π

Thus, the eigenvalues are λ = ± π1 . For λ = π1 , we have c2 = c1 . Putting these values in (46),
we have
1 1 1
sin sc1 + cos sc2 = f (s), =⇒ f (s) = c1 (sin s + cos s) (50)
π π π
The nonhomogeneous integral equation (41) will have solution if and only if

Z2π
f (s)g(s)ds = 0
0

Now for g(s) = s, we have

Z2π Z2π
1
f (s)g(s)ds = c1 (sin s + cos s)sds
π
0 0
Z2π
1
= c1 (sin s + cos s)sds
π
0
Z2π
 
1
= c1 |(− cos s + sin s)s|2π
0 − (− cos s + sin s)ds
π
0
1 h i 1
= c1 −2π − |− sin s − cos s|2π
0 = c1 (−2π) = −2c1 = 6 0, ∵ c1 6= 0
π π
This shows that the integral equation (41) has no solution for g(s) = s.
Now for f (s) = 1, we obtain

Z2π Z2π
1
f (s)g(s)ds = c1 (sin s + cos s)ds
π
0 0
Z2π
1
= c1 (sin s + cos s)ds
π
0
1
= c1 |(cos s + sin s)|2π
0 =0
π
This shows that (41) has infinitely many solutions for g(s) = 1.

16
4.2 Bounded kernels, iterated kernels, resolvent & the method
of successive approximations
Let K : [a, b] × [a, b] → C be a continuous function. The function K := K(s, t) is said to be
bounded if
Zb
sup |K(s, t)|dt < M, 0 < M ∈ C. (51)
s∈[a,b]
a
The following theorem delivers the condition for the existence solution(s) to the nonhomoge-
neous Fredholm integral equation,
Zb
g(s) + K(s, t)f (t)dt = f (s). (52)
a

Theorem 4.2.1. If the kernel K(s, t) of (52) is bounded with M = 1 i.e. K satisfies the
condition (51) with M = 1. Then the linear integral equation (52) is uniquely solvable for
any g and its solution can be written in the form

f (s) = g(s) + (K 1 g)(s) + (K 2 g)(s) + . . . (53)

where
Zb
n
(K g)(s) = K(s, t)(K n−1 g)(t)dt, n≥2
a
and
Zb
1
(K g)(s) = (Kg)(s) = K(s, t)g(t)dt.
a

Proof: Since inequality (51) with M = 1 implies that kKk < 1, the operator I −K is invertible
and
(I − K)−1 = I + K + K 2 + . . . (54)
Since C[a, b] is a Banach space, so the series (54) converges. In this case, the integral
equation (52) has unique solution

f = (I − K)−1 g
= (I + K + K 2 + . . .)g
= g + Kg + K 2 g + . . .

which completes the proof. 

When the kernel is bounded, the method of successive approximation is employed in

finding the unique solution of nonhomogeneous linear Volterra/Fredholm integral equation
with parameter λ i.e., Z
f (s) = g(s) + λ K(s, t)f (t)dt. (55)

17
Here we describe the method of successive approximation by using the iterated kernels pro-
posed herewith.
Let f0 (s) = g(s) be the zeroth-order approximation to the desired solution f (s). Using
this zeroth-order approximation in (55), we have the first-order approximation of f (s) as
follows: Z
f1 (s) = g(s) + λ K(s, t)f0 (t)dt (56)

Similarly, the second-order approximation of f (s) is

Z
f2 (s) = g(s) + λ K(s, t)f1 (t)dt (57)

Continuing this way, we have the (n + 1)th approximation

Z
fn+1 (s) = g(s) + λ K(s, t)fn (t)dt (58)

If fn (s) tends uniformly to a limit as n → ∞, then this limit is the required solution.
Using f0 (s) = g(s) and (56), we have
Z
f1 (s) = g(s) + λ K(s, t)g(t)dt (59)

Equation (57) can be rewritten as

Z
f2 (s) = g(s) + λ K(s, x)f1 (x)dx (60)

Using (59) & (60), we have

Z  Z 
f2 (s) = g(s) + λ K(s, x) g(x) + λ K(x, t)g(t)dt dx (61)

Z Z Z
2
f2 (s) = g(s) + λ K(s, x)g(x)dx + λ
K(s, x) K(x, t)g(t)dtdx
Z Z Z 
2
=⇒ = g(s) + λ K(s, x)g(x)dx + λ K(s, x)K(x, t)dx g(t)dt
Z Z Z 
=⇒ f2 (s) = g(s) + λ K(s, t)g(t)dt + λ2 K(s, x)K(x, t)dx g(t)dt (62)

R
Let K2 (s, t) = K(s, x)K(x, t)dx. Using this value in (62), we obtain
Z Z
2
f2 (s) = g(s) + λ K(s, t)g(t)dt + λ K2 (s, t)g(t)dt

Similarly,
Z Z Z
2 3
f3 (s) = g(s) + λ K(s, t)g(t)dt + λ K2 (s, t)g(t)dt + λ K3 (s, t)g(t)dt,

18
 R
where K3 (s, t) = K(s, x)K2 (x, t)dx. Continuing this way, and denoting
Z
Km (s, t) = K(s, x)Km−1 (x, t)dx

we have the (n + 1)th approximate solution of the integral equation (55) as

n
X Z
fn (s) = g(s) + λm Km (s, t)g(t)dt (63)
m=1

where Km (s, t) is known as the mth iterated kernel and K1 (s, t) = K(s, t).
Taking limit n → ∞ of (63), we have the Neumann series.
X∞ Z
m
lim fn (s) = g(s) + λ Km (s, t)g(t)dt
n→∞
m=1
X∞ Z
m
f (s) = g(s) + λ Km (s, t)g(t)dt,
m=1

where f (s) = lim fn (s) is the required solution (55). The function
n→∞
∞
X
Γ(s, t; λ) = λm Km (s, t)
m=1

is called the resolvent kernel.

Next, we apply this method of successive approximation in order to find a solution of a
Volterra integral equation
Example 8. Solve the Volterra integral equation
Zs
1+ stf (t)dt = f (s)
0

Solution. The given integral equation is

Zs
1+ stf (t)dt = f (s)
0

Let us first compute the iterated kernels. Here

K1 (s, t) = K(s, t) = st.

Zs Zs
K2 (s, t) = K(s, x)K1 (x, t)dx = K(s, x)K(x, t)dx
t t
Zs Zs s
x3
= (sx)(xt)dx = stx2 dx = st
3 t
t t
s3 t3 s4 t − st4
 
=⇒ K2 (s, t) = st − = .
3 3 3

19
 Zs Zs
x4 t − xt4
 
K3 (s, t) = K(s, x)K2 (x, t)dx = (sx) dx
3
t t
Zs  5 s
x t x2 t4 x6 t x3 t4

= s − dx = s −
3 3 18 9 t
t
s
x6 t − 2x3 t4 s6 t − 2s3 t4
 6
s t − 2s3 t4
  
=s =s −
18 t 18 18
s7 t − 2s4 t4 + st7
=⇒ K3 (s, t) =
18

Zs Zs
x7 t − 2x4 t4 + xt7
 
K4 (s, t) = K(s, x)K3 (x, t)dx = (sx) dx
18
t t
Zs s
1 s x9 t x6 t4 x3 t2
s x8 t − 2x5 t4 + x2 t7 dx =


= −2 +
18 18 9 6 3 t
t
s
x9 t − 3x6 t4 + 3x3 t7
 9
s t − 3s6 t4 + 3s3 t7
  10
t − 3t10 + 3t10

=s =s −
162 t 162 162
s10 t − 3s7 t4 + 3s4 t7 − st10
=⇒ K4 (s, t) =
162
Similarly, we continue to find the iterated kernels. Next, using
X∞ Z s
f (s) = g(s) + λm Km (s, t)g(t)dt
m=1 0

We have
∞ Z
X s
f (s) = 1 + Km (s, t)dt, ∵ λ = 1, g(s) = 1, g(t) = 1
m=1 0

Z s Z s Z s Z s
f (s) = 1 + K1 (s, t)dt + K2 (s, t)dt + K3 (s, t)dt + K4 (s, t)dt + . . .
0 0 0 0
Z s Z s 4 Z s 7 Z s 10
s t − st4 s t − 2s4 t4 + st7 s t − 3s7 t4 + 3s4 t7 − st10
=1+ stdt + dt + dt + dt + . . .
0 0 3 0 18 0 162
s s s s
st2 1 s4 t2 st5 1 s7 t2 2s4 t5 st8 1 s10 t2 3s7 t5 3s4 t8 st11
=1+ + − + − + + − + − + ...
2 0 3 2 5 0 18 2 5 8 0 162 2 5 8 11 0
s3 s6 s9 s12
=1+ + + + + ...
2 2.5 2.5.8 2.5.8.11
∞
X s3k
=⇒ f (s) = 1 +
2.5.8.11. . . . .(3k − 2)(3k − 1)
k=1

This is the required solution of the given integral equation.

20
 Next, we apply the method of successive approximation to find the resolvent kernel and
solve a nonhomogeneous Fredholm integral equation.

Example 9. Solve the following integral equation

Z1
g(s) + λ es−t f (t)dt = f (s)
0

by the method of successive approximation and evaluate the resolvent kernel.

Solution. The given integral equation is

Z1
g(s) + λ es−t f (t)dt = f (s)
0

Next, we find the iterated kernels.

K1 (s, t) = K(s, t) = es−t

Zs Zs
K2 (s, t) = K(s, x)K1 (x, t)dx = K(s, x)K(x, t)dx
t t
Zs Zs
= es−x ex−t dx = es−t dx = es−t [x]st = (s − t)es−t
t t

Zs Zs
K3 (s, t) = K(s, x)K2 (x, t)dx = es−x (x − t)ex−t dx
t t
Zs s
(x − t)2 (s − t)2 s−t
= (x − t)es−t dx = es−t = e
2 t 2!
t

Continuing this way, we have

(s − t)m−1 s−t
Km (s, t) = e
(m − 1)!

The resolvent kernel is

 ∞
 es−t P λm−1 (s−t)m−1 es−t , t ≤ s;
(m−1)!
Γ(s, t; λ) = m=1
0, t > s.


21
Since
∞
λm−1 (s − t)m−1 s−t λ2 (s − t)2
X  
s−t s−t
e e =e 1 + λ(s − t) + + ...
(m − 1)! 2!
m=1
λ2 (s − t)2
= es−t eλ(s−t) , ∵ 1 + λ(s − t) + + . . . = eλ(s−t)
2!
= e(1−λ)(s−t)

Therefore, the resolvent kernel

(
e(1−λ)(s−t) , t ≤ s;
Γ(s, t; λ) =
0, t > s.

The solution of the given integral equation

Z1
f (s) = g(s) + λ e(1−λ)(s−t) g(t)dt.
0

22