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n
                             b
               16.   y(x)+        A k exp(λ k |x – t|) y(t) dt = f(x),  –∞ < a < b < ∞.
                            a
                               k=1
                     1 . Let us remove the modulus in the kth summand of the integrand:
                      ◦

                              b                    x                     b
                      I k (x)=  exp(λ k |x – t|)y(t) dt =  exp[λ k (x – t)]y(t) dt +  exp[λ k (t – x)]y(t) dt. (1)
                             a                    a                     x
                     Differentiating (1) with respect to x twice yields

                                     x                       b

                            I = λ k   exp[λ k (x – t)]y(t) dt – λ k  exp[λ k (t – x)]y(t) dt,
                             k
                                   a                        x
                                              x                       b                     (2)
                           I =2λ k y(x)+ λ 2  exp[λ k (x – t)]y(t) dt + λ 2  exp[λ k (t – x)]y(t) dt,

                            k             k                       k
                                            a                       x
                     where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2),
                     we find the relation between I and I k :

                                             k
                                                        2
                                          I =2λ k y(x)+ λ I k ,  I k = I k (x).             (3)

                                           k            k
                     2 . With the aid of (1), the integral equation can be rewritten in the form
                      ◦
                                                      n

                                                y(x)+   A k I k = f(x).                     (4)
                                                      k=1
                     Differentiating (4) with respect to x twice and taking into account (3), we find that
                                                n                           n
                                                      2
                                y (x)+ σ n y(x)+  A k λ I k = f (x),  σ n =2  A k λ k .     (5)


                                 xx                   k     xx
                                               k=1                         k=1
                     Eliminating the integral I n from (4) and (5) yields
                                                    n–1
                                                            2   2             2
                                             2

                                y (x)+(σ n – λ )y(x)+  A k (λ – λ )I k = f (x) – λ f(x).    (6)

                                 xx          n              k   n      xx     n
                                                    k=1
                     Differentiating (6) with respect to x twice and eliminating I n–1 from the resulting equation
                     with the aid of (6), we obtain a similar equation whose left-hand side is a second-order linear
                                                                                 n–2

                     differential operator (acting on y) with constant coefficients plus the sum  B k I k .If we
                                                                                 k=1
                     successively eliminate I n–2 , I n–3 , ... , with the aid of double differentiation, then we finally
                     arrive at a linear nonhomogeneous ordinary differential equation of order 2n with constant
                     coefficients.
                     3 . The boundary conditions for y(x) can be found by setting x = a in the integral equation
                      ◦
                     and all its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b
                     in the integral equation and all its derivatives obtained by means of double differentiation.)
                 4.2-2. Kernels Containing Power-Law and Exponential Functions

                                b
                                     γt
               17.   y(x) – λ  (x – t)e y(t) dt = f(x).
                             a
                                                                             γt
                     This is a special case of equation 4.9.8 with A =0, B = 1, and h(t)= e .



                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
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