Page 298 - Handbook Of Integral Equations
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b
                                   3
               28.   y(x)+ A    sinh (λ|x – t|)y(t) dt = f(x).
                              a
                                       3
                     Using the formula sinh β =  1  sinh 3β –  3  sinh β, we arrive at an equation of the form 4.3.29
                                            4        4
                     with n =2:
                                        b
                                        
  1              3
                                y(x)+     A sinh(3λ|x – t|) – A sinh(λ|x – t|) y(t) dt = f(x).
                                         4                4
                                      a
                                n
                             b
               29.   y(x)+        A k sinh(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)=  sinh(λ k |x–t|)y(t) dt =  sinh[λ k (x–t)]y(t) dt+  sinh[λ k (t–x)]y(t) dt. (1)
                             a                    a                     x
                     Differentiating (1) with respect to x twice yields
                                     x                        b

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

                           I =2λ k y(x)+ λ 2  sinh[λ k (x – t)]y(t) dt + λ 2  sinh[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 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.)
                 © 1998 by CRC Press LLC


               © 1998 by CRC Press LLC
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