Page 254 - Handbook Of Integral Equations

k
3 .If g(x) is representable in the form g(x)= O(x) with 0 < k < 1 in the vicinity of the
◦
point x = 0 (in particular, the derivative g is unbounded as x → 0), then the solution of

x
the integral equation is given by formula (3) as well. In this case, the right-hand side of the
integral equation must satisfy the conditions
f(0) + f(λa)=0, f (λa)=0. (7)

x
As before, the right-hand side of the integral equation is given by (6), with

A = –F (λa), B = 1 aλF (λa) – F(0) – F(λa) .
x 2 x
a

5. g(x) – t y(t) dt = f(x).

0
Assume that 0 ≤ x ≤ a,0 ≤ t ≤ a; g(0) = 0, and 0 < g (x)< ∞.

x
◦
1 . Let us remove the modulus in the integrand:
g(x) a

g(x) – t y(t) dt + t – g(x) y(t) dt = f(x). (1)
0 g(x)
Differentiating (1) with respect to x yields
g(x) a

g (x) y(t) dt – g (x) y(t) dt = f (x). (2)

x x x
0 g(x)

Let us divide both sides of (2) by g (x) and differentiate the resulting equation to obtain
x

2g (x)y g(x) = f (x)/g (x) . Hence, we find the solution:

x
x
x
x
1 d f (z) –1

z
y(x)= , z = g (x), (3)
2g (z) dz g (z)

z
z
where g –1 is the inverse of g.
2 . Let us demonstrate that the right-hand side f(x) of the integral equation must satisfy
◦
certain relations. By setting x = 0 in (1) and (2), we obtain two corollaries
a a

ty(t) dt = f(0), g (0) y(t) dt = –f (0). (4)
x
x
0 0
Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f(x):
–1

f (0)g (x a )+ f (x a )g (0) = 0, x a = g (a);

x x x x

f (x a ) (5)
x
g(x a ) = f(0) + f(x a ).
g (x a )

x
Conditions (5) make it possible to find the admissible general form of the right-hand side
of the integral equation in question:
f(x)= F(x)+ Ax + B, (6)
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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