Page 514 - Handbook Of Integral Equations
P. 514

we can rewrite Eq. (7) in the form
                                               ˆ
                                                            ˆ
                                               K(s) ˆy(1 – s)= f(s).                        (8)
               Replacing 1 – s by s in (8) and solving the resulting relation for ˆy(s), we obtain the transform
                                                       ˆ
                                                      f(1 – s)
                                                 ˆ y(s)=                                    (9)
                                                       ˆ
                                                      K(1 – s)
               of the desired solution.
                   Applying the Mellin inversion formula, we obtain the solution of the integral equation (6) in the
               form
                                                     c+i∞ ˆ
                                                1        f(1 – s)  –s
                                         y(x)=                  x ds.
                                                         ˆ
                                               2πi  c–i∞ K(1 – s)
                ◦
               2 . Now we consider the more complicated equation

                                          ∞

                                            K ϕ(x)ψ(t) g(t)y(t) dt = f(x).                 (10)
                                         0
               Assume that the conditions ϕ(0) = 0, ϕ(∞)= ∞, ϕ >0, ψ(0) = 0, ψ(∞)= ∞, and ψ > 0 are


                                                          x                            x
               satisfied.
                   The transform
                                                                  g(t)
                                    z = ϕ(x),   τ = ψ(t),   y(t)=     w(τ)
                                                                 ψ (t)

                                                                   t
               takes (10) to the following equation of the form (6):

                                              ∞
                                                K(zτ)w(τ) dτ = F(z),
                                             0
               where the function F(z)isdefined parametrically by F = f(x), z = ϕ(x). In many cases, on
               eliminating x from these relations, we obtain the dependence F = F(z) in an explicit form.
                •
                 References for Section 10.3: V. A. Ditkin and A. P. Prudnikov (1965), M. L. Krasnov, A. I. Kiselev, and G. I. Makarenko (1971).
               10.4. The Riemann Problem for the Real Axis

               The Riemann boundary value problem is one of the main tools for constructing solutions of integral
               equations provided that various integral transforms can be applied to a given equation and the
               corresponding convolution-type theorems can be applied. This problem is investigated by an
               example of the Fourier integral transform.



                 10.4-1. Relationships Between the Fourier Integral and the Cauchy Type Integral
               Let Y(τ) be a function integrable on a closed or nonclosed contour L on the complex plane of the
               variable z = u + iv (τ is the complex coordinate of the contour points). Consider the integral of the
               Cauchy type (see Section 12.2):

                                                 1     Y(τ)
                                                           dτ.
                                                2πi  L  τ – z
               This integral defines a function that is analytic on the complex plane with a cut along the contour L.
               If L is a closed curve, then the integral is a function that is analytic on each of the connected parts
               of the plane bounded by L. If the contour L is the real axis, then we have


                                                          +

                                      1    ∞  Y(τ)      Y (z)if Im z >0,
                                                  dτ =    –                                 (1)
                                     2πi     τ – z      Y (z)  if Im z <0.
                                          –∞
                 © 1998 by CRC Press LLC








               © 1998 by CRC Press LLC
                                                                                                             Page 496
   509   510   511   512   513   514   515   516   517   518   519