Page 238 - Handbook Of Integral Equations
P. 238

3.3-4. Kernels Containing Hyperbolic Cotangent


                       b


               21.        coth(λx) – coth(λt) y(t) dt = f(x).

                      a
                     This is a special case of equation 3.8.3 with g(x) = coth(λx).
                       b

                            k
               22.        coth x – coth t y(t) dt = f(x),  0 < k <1.
                                    k
                      0
                                                                 k
                     This is a special case of equation 3.8.3 with g(x) = coth x.
               3.4. Equations Whose Kernels Contain Logarithmic
                      Functions

                 3.4-1. Kernels Containing Logarithmic Functions


                       b


               1.         ln(x/t) y(t) dt = f(x).

                      a
                     This is a special case of equation 3.8.3 with g(x)=ln x.
                        Solution:
                                                       1 d

                                                y(x)=       xf (x) .
                                                              x
                                                       2 dx
                                                                                           ◦
                     The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2 of
                     equation 3.8.3).
                         b
               2.       ln |x – t| y(t) dt = f(x).
                      a
                     Carleman’s equation.
                      ◦
                     1 . Solution with b – a ≠ 4:
                                              √                                  b
                                            b


                                 1             (t – a)(b – t) f (t) dt  1           f(t) dt
                                                          t
                     y(x)=  √                                   +                √            .
                          π 2  (x – a)(b – x)  a     t – x        π ln  1 4 (b – a)  a  (t – a)(b – t)
                      ◦
                     2 .If b – a = 4, then for the equation to be solvable, the condition
                                               b
                                               f(t)(t – a) –1/2 (b – t) –1/2  dt =0
                                             a
                     must be satisfied. In this case, the solution has the form
                                                           √
                                                         b

                                              1             (t – a)(b – t) f (t) dt

                                                                       t
                                 y(x)=   √                                   + C ,
                                       π 2  (x – a)(b – x)  a    t – x
                     where C is an arbitrary constant.
                     •
                       Reference: F. D. Gakhov (1977).


                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
                                                                                                             Page 217
   233   234   235   236   237   238   239   240   241   242   243