Page 223 - Handbook Of Integral Equations
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3.1-4. Kernels Containing Square Roots


                       a

                         √    √
               18.         x –  t y(t) dt = f(x),  0 < a < ∞.

                      0
                                                              √
                     This is a special case of equation 3.8.3 with g(x)=  x.
                        Solution:
                                                       d  √

                                                y(x)=      xf (x) .
                                                              x
                                                      dx
                        The right-hand side f(x) of the equation must satisfy certain conditions. The general
                     form of the right-hand side is


                            f(x)= F(x)+ Ax + B,    A = –F (a),  B =  1    aF (a) – F(a) – F(0) ,

                                                         x         2   x
                     where F(x) is an arbitrary bounded twice differentiable function (with bounded first deriva-
                     tive).
                       a

                         √     √
               19.         x – β t y(t) dt = f(x),  β >0.

                      0
                                                              √         √
                     This is a special case of equation 3.8.4 with g(x)=  x and β =  λ.
                       a

                         √
               20.         x – t y(t) dt = f(x).

                      0
                                                              √
                                                                          ◦
                     This is a special case of equation 3.8.5 with g(x)=  x (see item 3 of 3.8.5).
                       a

                            √
               21.        x –  t y(t) dt = f(x).

                      0
                                                             √
                     This is a special case of equation 3.8.6 with g(t)=  t (see item 3 of 3.8.6).
                                                                         ◦
                       a
                          y(t)
               22.       √      dt = f(x),   0 < a ≤ ∞.
                      0   |x – t|
                                                             1
                     This is a special case of equation 3.1.29 with k = .
                                                             2
                        Solution:
                                    A  d      a  dt      t  f(s) ds  	           1
                            y(x)= –                                  ,  A = √          .
                                                                                 2
                                   x 1/4  dx  x  (t – x) 1/4  0  s 1/4 (t – s) 1/4  8π Γ (3/4)
                       ∞   y(t)

               23.       √       dt = f(x).
                      –∞   |x – t|
                                                             1
                     This is a special case of equation 3.1.34 with λ = .
                                                             2
                        Solution:
                                                    1     ∞  f(x) – f(t)
                                             y(x)=                  dt.
                                                   4π      |x – t| 3/2
                                                       –∞


                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
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