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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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