Page 253 - Handbook Of Integral Equations

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

A = –F (b), B = 1 (a + b)F (b) – F(a) – F(b) .
x 2 x

◦
4 .For g (a) = 0, the right-hand side of the integral equation must satisfy the conditions
x

f (a)=0, g(b) – g(a) f (b)= f(a)+ f(b) g (b).

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

A = –F (a), B = 1 (a + b)F (a) – F(a) – F(b) + g(b) – g(a) F (b) – F (a) .

x 2 x x x
2g (b)
x
a

4. g(x) – g(λt) y(t) dt = f(x), λ >0.

0

Assume that 0 ≤ x ≤ a,0 ≤ t ≤ a and 0 < g (x)< ∞.
x
1 . Let us remove the modulus in the integrand:
◦
x/λ a

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

g (x) y(t) dt – g (x) y(t) dt = f (x). (2)
x
x
x
0 x/λ
Let us divide both sides of (2) by g (x) and differentiate the resulting equation to obtain

x
1

y(x/λ)= λ f (x)/g (x) . Substituting x by λx yields the solution
2 x x x

λ d f (z)

z
y(x)= , z = λx. (3)
2 dz g (z)

z
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

g(λt) – g(0) y(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):

f (0)g (λa)+ f (λa)g (0) = 0,

x x x x
f (λa) (5)

x
g(λa) – g(0) = f(0) + f(λa).
g (λa)

x
Conditions (5) make it possible to find the admissible general form of the right-hand side
of the integral equation:
f(x)= F(x)+ Ax + B, (6)
where F(x) is an arbitrary bounded twice differentiable function (with bounded first deriva-
tive), and the coefficients A and B are given by

g (0)F (λa)+ g (λa)F (0)

x
x
x
x
A = – ,
g (0) + g (λa)

x x

1
B = – Aaλ – 1 F(0) + F(λa) – g(λa) – g(0) A + F (0) .

2 2 2g (0) x

x
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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