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3.4-2. Kernels Containing Power-Law and Logarithmic Functions

                         a

               9.         k ln(1 + λx) – t y(t) dt = f(x).

                      0
                     This is a special case of equation 3.8.5 with g(x)= k ln(1 + λx).
                       a


               10.        x – k ln(1 + λt) y(t) dt = f(x).

                      0
                     This is a special case of equation 3.8.6 with g(x)= k ln(1 + λt).
                       ∞  1    x + t

               11.         ln        y(t) dt = f(x).
                         t      x – t
                      0
                     Solution:
                                                                     2
                                                x d     ∞  df(t)       x
                                          y(x)=               ln 1 –     dt.

                                                 2
                                                π dx  0    dt       t 2
                     •
                       Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).
                       ∞  ln x – ln t

               12.                y(t) dt = f(x).
                      0    x – t
                     The left-hand side of this equation is the iterated Stieltjes transform.
                        Under some assumptions, the solution of the integral equation can be represented in the
                     form
                                        1          4n  n 2n  2n 2n  n          d
                                                 e
                                 y(x)=      lim      D x D x D f(x),      D =    .
                                       4π 2  n→∞ n                             dx
                     To calculate the solution approximately, one should restrict oneself to a specific value of n in
                     this formula instead of taking the limit.
                     •
                       Reference: I. I. Hirschman and D. V. Widder (1955).
                         b
                            β
                                β
               13.      ln |x – t | y(t) dt = f(x),  β >0.
                      a
                     The transformation
                                                       β
                                                β
                                           z = x ,  τ = t ,  w(τ)= t 1–β y(t)
                     leads to Carleman’s equation 3.4.2:
                                      B

                                                                             β
                                                                     β
                                        ln |z – τ|w(τ) dτ = F(z),  A = a ,  B = b ,
                                     A
                                      1/β
                     where F(z)= βf z   .
                       1

                            β
                                µ
               14.      ln |x – t | y(t) dt = f(x),  β >0, µ >0.
                      0
                     The transformation
                                                       µ
                                                β
                                           z = x ,  τ = t ,  w(τ)= t 1–µ y(t)
                     leads to an equation of the form 3.4.2:
                                        1
                                                                           1/β
                                        ln |z – τ|w(τ) dτ = F(z),  F(z)= µf z  .
                                      0

                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
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