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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 (x a )+ g (x a )F (0) –1
x
x
x
x
A = – , x a = g (a),
g (0) + g (x a )
x
x
1
B = – Ax a – 1 F(0) + F(x a ) – g(x a ) A + F (0) .
2 2 2g (0) x
x
k
◦
3 .If g(x) is representable in the vicinity of the point x = 0 in the form g(x)= O(x) with
0< k < 1 (i.e., the derivative g is unbounded as x → 0), then the solution of the integral
x
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(x a )=0, f (x a )=0. (7)
x
As before, the right-hand side of the integral equation is given by (6), with
A = –F (x a ), B = 1 x a F (x a ) – F(0) – F(x a ) .
x 2 x
a
6. x – g(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:
–1
g (x) a
x – g(t) y(t) dt + g(t) – x y(t) dt = f(x), (1)
0 g –1 (x)
where g –1 is the inverse of g. Differentiating (1) with respect to x yields
–1
g (x) a
y(t) dt – y(t) dt = f (x). (2)
x
0 g –1 (x)
–1
Differentiating the resulting equation yields 2y g (x) = g (x)f (x). Hence, we obtain the
xx
x
solution
1
y(x)= g (z)f (z), z = g(x). (3)
zz
2 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(t)y(t) dt = f(0), y(t) dt = –f (0). (4)
x
0 0
Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f(x):
x a f (x a )= f(0) + f(x a ), f (0) + f (x a )=0, x a = g(a). (5)
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,
A = – 1 F (0) + F (x a ) , B = 1 x a F (0) – F(x a ) – F(0) , x a = g(a),
2 x x 2 x
where F(x) is an arbitrary bounded twice differentiable function (with bounded first deriva-
tive).
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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