Page 252 - Handbook Of Integral Equations

3.8-2. Equations Containing Modulus

b
3. |g(x) – g(t)| y(t) dt = f(x).
a
◦
◦
Let a ≤ x ≤ b and a ≤ t ≤ b; it is assumed in items 1 and 2 that 0 < g (x)< ∞.

x
1 . Let us remove the modulus in the integrand:
◦
x b

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

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

x
x
x
a x
Divide both sides of (2) by g (x) and differentiate the resulting equation to obtain the solution

x
1 d f (x)

x
y(x)= . (3)
2 dx g (x)

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

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

f (b)
x
g(b) – g(a) = f(a)+ f(b),
g (b)

x (5)

f (a)
x
g(a) – g(b) = f(a)+ f(b).
g (a)

x

Let us point out a useful property of these constraints: f (b)g (a)+ f (a)g (b)=0.

x
x
x
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 (a)F (b)+ g (b)F (a)

x
x
x
x
A = – ,
g (a)+ g (b)

x
x
1

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

2 2 x
2g (a)

x
k
◦
3 .If g(x) is representable in the form g(x)= O(x – a) with 0 < k < 1 in the vicinity of

the point x = a (in particular, the derivative g is unbounded as x → a), 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(a)+ f(b)=0, f (b) = 0. (7)
x
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Page 231