Page 599 - Handbook Of Integral Equations

This coincidence on 2n straight lines permits us to expect that K(x, t) is close to Q(x, t) and the
solution of the equation with kernel K(x, t) is close to the solution of the equation with kernel Q(x, t).
It should be noted that if Q(x, t) is degenerate, i.e., has the form

n

Q(x, t)= g k (x)h k (t), (11)
k=1
then the determinant in the numerator is identically zero, and hence in this case we have

K(x, t) ≡ Q(x, t). (12)

For the kernel K(x, t), the resolvent can be evaluated on the basis of the following relations:
r(x, t; λ)=0, ϕ i (x)= g i (x)= Q(x, t i ), ψ j (t)= h j (t)= –Q(x j , t),
b (13)

b ij = – Q(x, t j )Q(x i , x) dx = –Q 2 (x i , t j ), i, j =1, ... , n,
a
where Q 2 (x, t) is the second iterated kernel for Q(x, t):

b

Q 2 (x, y)= Q(x, s)Q(s, t) ds,
a
and hence

0 Q(x, t 1 ) ··· Q(x, t n )

1 Q(x 1 , t) Q(x 1 , t 1 ) – λQ 2 (x 1 , t 1 ) ··· Q(x 1 , t n ) – λQ 2 (x 1 , t n )
R(x, t; λ)= . . . . , (14)
. . . .
. . . .
D – λD 2

Q(x n , t) Q(x n , t 1 ) – λQ 2 (x n , t 1 ) ··· Q(x n , t n ) – λQ 2 (x n , t n )
where
Q 2 (x 1 , t 1 ) ··· Q 2 (x 1 , t n )
. . .
D 2 = . . . . .
. .
Q 2 (x n , t 1 ) ··· Q 2 (x n , t n )

By using the resolvent R(x, t; λ), we can obtain an approximate solution of the equation with
˜
kernel Q(x, t). In particular, approximate characteristic values λ of this kernel can be found by
equating the determinant in the denominator of (14) with zero.
Example. Consider the equation
1

y(x) – λ Q(x, t)y(t) dt =0, 0 ≤ x ≤ 1, (15)
0
x(t – 1) for x ≤ t,
"
Q(x, t)=
t(x – 1) for x ≥ t.
Let us find its characteristic values. To this end, we apply formula (14), where for the second iterated kernel we have
%
1 x(1 – t)(2t – x – t ) for x ≤ t,
1 2 2
Q 2 (x, t)= Q(x, s)Q(s, t) ds = 6 1
2
2
0 t(1 – x)(2x – x – t ) for x ≥ t.
6
We choose equidistant points x i and t j and take n = 5. This implies
x 1 = t 1 = 1 , x 2 = t 2 = 2 , x 3 = t 3 = 3 , x 4 = t 4 = 4 , x 5 = t 5 = 5 .
6 6 6 6 6
Let us equate the determinant in the denominator of (14) with zero. After some algebraic manipulations, we obtain the
following equation:
2
5
4
3
130µ – 441µ + 488µ – 206µ +30µ – 1=0 ( ˜ λ = 216µ),
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 582