Page 597 - Handbook Of Integral Equations

Hence, instead of Eq. (11) we obtain

4 4
2 2

y 2 (x)=1 + 1/2 1 – x t + 1 2 x t y 2 (t) dt. (12)
0
Therefore,
4
2
y 2 (x)=1 + A 1 + A 2 x + A 3 x , (13)
where
1/2 1/2 1 1/2

2
4
A 1 = y 2 (x) dx, A 2 = – x y 2 (x) dx, A 3 = x y 2 (x) dx. (14)
0 0 2 0
From (13) and (14) we obtain a system of three equations with three unknowns; to the fourth decimal place, the solution
is
A 1 = 0.9930, A 2 = –0.0833, A 3 = 0.0007.
Hence,
4
2
y(x) ≈ y 2 (x) = 1.9930 – 0.0833 x + 0.0007 x , 0 ≤ x ≤ 1 . (15)
2
An error estimate for the approximate solution (15) can be performed by formula (10).
•
References for Section 11.13: L. V. Kantorovich and V. I. Krylov (1958), S. G. Mikhlin (1960), B. P. Demidovich,
I. A. Maron, and E. Z. Shuvalova (1963), M. L. Krasnov, A. I. Kiselev, and G. I. Makarenko (1971).
11.14. The Bateman Method
11.14-1. The General Scheme of the Method
In some cases it is useful, instead of replacing a given kernel by a degenerate kernel, to represent
the given kernel approximately as the sum of a kernel whose resolvent is known and a degenerate
kernel. For the latter, the resolvent can be written out in a closed form.
Consider the Fredholm integral equation of the second kind
b

y(x) – λ k(x, t)y(t) dt = f(x) (1)
a
with kernel k(x, t) whose resolvent r(x, t; λ) is known; thus, the solution of (1) can be represented
in the form
b
y(x)= f(x)+ λ r(x, t; λ)f(t) dt. (2)
a
Then, for the integral equation with kernel

k(x, t) g 1 (x) ··· g n (x) a 11 a 12 ··· a 1n

1 h 1 (t) a 11 ··· a 1n a 21 a 22 ··· a 2n
K(x, t)= . . . . , ∆(a ij )= . . . . , (3)
∆(a ij ) . . . . . . . . . . . . . .
.
.
h n (t) a n1 ··· a nn a n1 a n2 ··· a nn

where g k (x) and h k (t)(k =1, ... , n) are arbitrary functions and a ij (i, j =1, ... , n) are arbitrary
numbers, the resolvent has the form

r(x, t; λ) ϕ 1 (x) ··· ϕ n (x)

1 ψ 1 (t) a 11 + λb 11 ··· a 1n + λb 1n
R(x, t; λ)= . . . . , (4)
∆(a ij + λb ij ) . . . . . . . .

ψ n (t) a n1 + λb n1 ··· a nn + λb nn

where
b b

ϕ k (x)= g k (x)+ λ r(x, t; λ)g k (t) dt, ψ k (x)= h k (x)+ λ r(x, t; λ)h k (t) dt,
a a (5)
b

b ij = g j (x)h i (x) dx, k, i, j =1, ... , n.
a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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