Page 592 - Handbook Of Integral Equations
P. 592

◦
               1 . Let the equation
                                           b

                                    y(x) –  K(x, t)y(t) dt = f(x),  a ≤ x ≤ b,              (1)
                                          a
               be given. Along with (1), we consider two auxiliary equations depending on a parameter ξ (a≤ξ ≤b):

                                                   ξ
                                        w(x, ξ) –  K(x, t)w(t, ξ) dt =1,
                                                 a
                                                    ξ                                       (2)
                                        w (x, ξ) –  K(t, x)w (t, ξ) dt =1,
                                          ∗
                                                           ∗
                                                  a
               where a ≤ x ≤ ξ. Assume that for any ξ the auxiliary equations (2) have unique continuous solutions
                                                                       ∗
               w(x, ξ) and w (x, ξ), respectively, which satisfy the condition w(ξ, ξ)w (ξ, ξ) ≠ 0(a ≤ ξ ≤ b). In this
                          ∗
               case, for any continuous function f(x), the unique continuous solution of Eq. (1) can be obtained
               by the formula
                                          b                        1   d     ξ
                      y(x)= F(b)w(x, b) –  w(x, ξ)F (ξ) dξ,  F(ξ)=          w (t, ξ)f(t) dt,  (3)
                                                                              ∗

                                                 ξ
                                        x                        m(ξ) dξ  a
               where
                                                           ∗
                                             m(ξ)= w(ξ, ξ)w (ξ, ξ).
                   Formula (3) permits one to construct a solution of Eq. (1) with an arbitrary right-hand side f(x)
               by means of solutions to the two simpler auxiliary equations (2) (depending on the parameter ξ)
               with a constant right-hand side equal to 1.

               2 . Consider now an equation with the kernel depending on the difference of the arguments:
                ◦
                                           b

                                   y(x)+    K(x – t)y(t) dt = f(x),  a ≤ x ≤ b.             (4)
                                          a
               It is assumed that K(x) is an even function integrable on [a – b, b – a]. Along with (4) we consider
               the following auxiliary equation depending on a parameter ξ (a ≤ ξ ≤ b):


                                             ξ
                                  w(x, ξ)+   K(x – t)w(t, ξ) dt =1,  a ≤ x ≤ ξ.             (5)
                                           a
                   Assume that for an arbitrary ξ the auxiliary equation (5) has a unique continuous solution w(x, ξ).
               In this case, for any continuous function f(x), a solution of Eq. (4) can be obtained from formula (3)
               by setting w (x, t)= w(x, t) in this formula.
                         ∗
                   Now let us indicate another useful formula for equations whose kernel depends on the difference
               of the arguments:
                                          a

                                  y(x)+    K(x – t)y(t) dt = f(x),  –a ≤ x ≤ a.             (6)
                                         –a
               It is assumed that K(x) is an even function that is integrable on the segment [–2a,2a]. Along with
               (6) we consider an auxiliary equation depending on a parameter ξ (0 < ξ ≤ a):

                                            ξ
                                  w(x, ξ)+   K(x – t)w(t, ξ) dt =1,  –ξ ≤ x ≤ ξ.            (7)
                                           –ξ



                 © 1998 by CRC Press LLC








               © 1998 by CRC Press LLC
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