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13.4-6. The Carleman–Vekua Approach to the Regularization

               Let us transfer the regular part of a singular equation to the right-hand side and rewrite the equation
               as follows:
                                         b(t)   ϕ(τ)
                                 a(t)ϕ(t)+          dτ = f(t) –  K(t, τ)ϕ(τ) dτ,           (34)
                                          πi  L  τ – t         L
               or, in the operator form,
                                              ◦
                                            K [ϕ(t)] = f(t) – K r [ϕ(t)].                  (35)
                   We regard the last equation as a characteristic one and solve it by temporarily assuming that the
               right-hand side is a known function. In this case (see Subsection 13.2-1)


                                    b(t)Z(t)  f(τ) dτ
                     ϕ(t)= a(t)f(t) –                  + b(t)Z(t)P ν–1 (t)
                                      πi    L  Z(τ) τ – t

                                                 b(t)Z(t)      dτ 1

                          – a(t)   K(t, τ)ϕ(τ) dτ –                      K(τ 1 , τ)ϕ(τ) dτ ,  (36)
                                 L                 πi    L  Z(τ 1 )(τ 1 – t)  L
               where for ν ≤ 0 we must set P ν–1 (t) ≡ 0. Let us reverse the order of integration in the iterated integral
               and rewrite the expression in the last parentheses as follows:
                                              b(t)Z(t)    K(τ 1 , τ)

                                   a(t)K(t, τ) –                   dτ 1 ϕ(τ) dτ.
                                 L               πi    L  Z(τ 1 )(τ 1 – t)
               Since Z(t) satisfies the H¨ older condition (and hence is bounded) and does not vanish and since
               K(τ 1 , τ) satisfies the estimate |K(τ 1 , τ)| < A|τ 1 – τ| –λ  (with 0 ≤ λ < 1) near the point τ 1 = τ, we can
               see that the entire integral

                                                   K(τ 1 , τ)
                                                            dτ 1
                                                  Z(τ 1 )(τ 1 – t)
                                                L
               satisfies an estimate similar to that for K(τ 1 , τ). Hence, the kernel
                                                     b(t)Z(t)     K(τ 1 , τ)
                                 N(t, τ)= a(t)K(t, τ) –                   dτ 1             (37)
                                                       πi    L  Z(τ 1 )(τ 1 – t)
               is a Fredholm kernel. On transferring the terms with ϕ(t) to the right-hand side, we obtain

                                          ϕ(t)+   N(t, τ)ϕ(τ) dτ = f 1 (t),                (38)
                                                L
               where N(t, τ) is the Fredholm kernel defined by formula (37) and f 1 (t) has the form
                                             b(t)Z(t)     f(τ) dτ
                               f 1 (t)= a(t)f(t) –               + b(t)Z(t)P ν–1 (t).      (39)
                                                πi    L  Z(τ) τ – t
                   If the index of Eq. (34) ν is negative, then the function must satisfy not only the Fredholm
               equation (38) but also the relations
                                 K(t, τ)  k–1            f(t)  k–1

                                       t  dt ϕ(τ) dτ =      t   dt,   k =1, 2, ... , –ν.   (40)
                                  Z(t)                  Z(t)
                           L   L                       L
                   Thus, if ν ≥ 0, then the solution of a complete singular integral equation (34) is reduced to the
               solution of the Fredholm integral equation (38). If ν < 0, then Eq. (34) can be reduced to Eq. (38)
               (where we must set P ν–1 (t) ≡ 0) together with conditions (40), which can be rewritten in the form

                                      ρ k (τ)ϕ(τ) dτ = f k ,  k =1, 2, ... , –ν,
                                    L
                                                                                           (41)
                                            K(t, τ)  k–1          f(t)  k–1
                                   ρ k (τ)=       t   dt,  f k =     t   dt,
                                          L  Z(t)               L  Z(t)
               where the ρ k (τ) are known functions and the f k are known constants.



                 © 1998 by CRC Press LLC







               © 1998 by CRC Press LLC

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