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               2 . A representation of a nonvanishing function in the form of the ratio of boundary values of
               analytic functions (factorization).
                   Here the second operation can be reduced to the first by taking the logarithm. Some complications
               related to the case of a nonzero index are due to the multivaluedness of the logarithm only. The first
               operation for arbitrary functions is equivalent to the calculation of a Cauchy type integral. In this
               connection, the solution to the problem by formulas (17) and (23)–(25) is explicitly expressed (in
               the closed form) via Cauchy type integrals.


                 12.3-7. The Riemann Problem With Rational Coefficients
               Consider the Riemann boundary value problem with a contour that consists of finitely many simple
               curves and with coefficient D(t) a rational function that has neither zeros nor poles on the contour.
               Note that an arbitrary continuous function (and all the functions satisfying the H¨ older condition) can
               be approximated with arbitrary accuracy by rational functions, and the solution of problems with
               rational coefficients can serve as a basis for the approximate solution in the general case. Assume
               that the Riemann problem has the form
                                                   p(t)  –
                                              +
                                            Φ (t)=     Φ (t)+ H(t),                        (30)
                                                   q(t)
               and the polynomials p(z) and q(z) can be factorized as follows:

                                       p(z)= p + (z)p – (z),  q(z)= q + (z)q – (z),        (31)
                                                                 +
               where p + (z) and q + (z) are polynomials whose roots belong to Ω and p – (z) and q – (z) are polynomials
                           –
               with roots in Ω . It readily follows from property 4 of the index (Subsection 12.3-3) that ν = m + –n + ,
                                                       ◦
               where m + and n + are the numbers of zeros of the polynomials p + (z) and q + (z).
                   Since the coefficient of the problem is a function that can be analytically continued to the
                       ±
               domain Ω , it follows that in this case it is reasonable to avoid using the general formulas and obtain
                                                                                          ν
               a solution directly by analytic continuation; here the role of the standard function of the type t that
               is used in the reduction of the index to zero can be played by the product  ν    (t–a j ), where a 1 , ... , a ν
                                                                        j=1
                                            +
               are arbitrary points of the domain Ω . On representing the boundary condition in the form
                                        q – (t)  +  p + (t)  –  q – (t)
                                            Φ (t) –    Φ (t)=     H(t),
                                        p – (t)    q + (t)    p – (t)
               where the canonical function is determined by the expressions
                                                p – (z)    –     q + (z)
                                          +
                                        X (z)=      ,    X (z)=      ,                     (32)
                                                q – (z)          p + (z)
               we obtain the solution by the same reasoning as in Subsection 12.3-6 in the following form:

                                   p – (z)               –     q + (z)
                              +
                            Φ (z)=      [Ψ(z)+ P ν–1 (z)],  Φ (z)=  [Ψ(z)+ P ν–1 (z)],     (33)
                                    q – (z)                    p + (z)
               where
                                            1     q – (τ) H(τ) dτ  –
                                    Ψ(z)=                    ,  Φ (∞)=0.
                                           2πi  L  p – (τ)  τ – z
                   If the index is negative, then we must set P ν–1 (z) ≡ 0 and add the solvability conditions

                                      q – (τ)   k–1

                                           H(τ)τ   dτ =0,    k =1, 2, ... , –ν,            (34)
                                      p – (τ)
                                     L

                 © 1998 by CRC Press LLC








               © 1998 by CRC Press LLC
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