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3.1-5. Kernels Containing Arbitrary Powers

                         a
                              k
                          k
               24.      |x – t | y(t) dt = f(x),  0 < k <1,  0 < a < ∞.
                      0
                     1 . Let us remove the modulus in the integrand:
                      ◦
                                         x                a
                                           k  k             k   k
                                         (x – t )y(t) dt +  (t – x )y(t) dt = f(x).         (1)
                                       0                x
                     Differentiating (1) with respect to x yields

                                               x               a

                                       kx k–1   y(t) dt – kx k–1  y(t) dt = f (x).          (2)
                                                                       x
                                              0              x
                     Let us divide both sides of (2) by kx k–1  and differentiate the resulting equation. As a result,
                     we obtain the solution
                                                     1 d    1–k
                                               y(x)=       x  f (x) .                       (3)
                                                               x
                                                     2k dx
                     2 . Let us demonstrate that the right-hand side f(x) of the integral equation must satisfy
                      ◦
                                                                                 a

                                                                                   k
                     certain relations. By setting x=0 and x=a, in (1), we obtain two corollaries  t y(t) dt=f(0)
                                                                                0
                          a

                                k
                            k
                     and   (a – t )y(t) dt = f(a), which can be rewritten in the form
                         0
                                      a                      a

                                        k                 k
                                       t y(t) dt = f(0),  a   y(t) dt = f(0) + f(a).        (4)
                                     0                      0
                     Substitute y(x) of (3) into (4). Integrating by parts yields the relations af (a)= kf(a)+kf(0)

                                                                              x

                     and af (a)=2kf(a)+2kf(0). Hence, the desired constraints for f(x) have the form
                          x

                                             f(0) + f(a)=0,   f (a)=0.                      (5)
                                                               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 = –F (a),  B =  1 2    aF (a) – F(a) – F(0) ,


                                                         x
                                                                       x
                     where F(x) is an arbitrary bounded twice differentiable function with bounded first derivative.
                                                                                    1–k
                     The first derivative may be unbounded at x = 0, in which case the conditions x  F     =0
                                                                                       x x=0
                     must hold.
                       a

                          k
                               k
               25.      |x – βt | y(t) dt = f(x),  0 < k <1,  β >0.
                      0
                                                                        k
                                                               k
                     This is a special case of equation 3.8.4 with g(x)= x and β = λ .
                         a
                          k m
               26.      |x t  – t k+m | y(t) dt = f(x),  0 < k <1,  0 < a < ∞.
                      0
                                        m
                     The substitution w(t)= t y(t) leads to an equation of the form 3.1.24:
                                                 a

                                                   k   k
                                                  |x – t |w(t) dt = f(x).
                                                0
                 © 1998 by CRC Press LLC

               © 1998 by CRC Press LLC
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