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3.2. Equations Whose Kernels Contain Exponential
                      Functions

                 3.2-1. Kernels Containing Exponential Functions
                       b

               1.       e λ|x–t| y(t) dt = f(x),  –∞ < a < b < ∞.
                      a
                     1 . Let us remove the modulus in the integrand:
                      ◦
                                          x              b

                                           e λ(x–t) y(t) dt +  e λ(t–x) y(t) dt = f(x).     (1)
                                         a              x
                     Differentiating (1) with respect to x twice yields
                                             x                 b

                                 2λy(x)+ λ 2  e λ(x–t) y(t) dt + λ 2  e λ(t–x) y(t) dt = f (x).  (2)

                                                                             xx
                                            a                 x
                        By eliminating the integral terms from (1) and (2), we obtain the solution
                                                    1          2
                                              y(x)=    f (x) – λ f(x) .                     (3)

                                                        xx
                                                    2λ
                     2 . The right-hand side f(x) of the integral equation must satisfy certain relations. By setting
                      ◦
                     x = a and x = b in (1), we obtain two corollaries
                                     b                       b

                                                 λa
                                       λt
                                      e y(t) dt = e f(a),     e –λt y(t) dt = e –λb f(b).   (4)
                                    a                       a
                     On substituting the solution y(x) of (3) into (4) and then integrating by parts, we see that
                                                            λa
                                                                      λb
                                         λb
                                        e f (b) – e f (a)= λe f(a)+ λe f(b),
                                                 λa
                                           x        x
                                      e –λb   x  –λa   x    –λa f(a)+ λe –λb f(b).
                                                   f (a)= λe
                                          f (b) – e
                     Hence, we obtain the desired constraints for f(x):

                                         f (a)+ λf(a)=0,    f (b) – λf(b) = 0.              (5)

                                                             x
                                          x
                     The general form of the right-hand side satisfying conditions (5) is given by
                                                f(x)= F(x)+ Ax + B,
                             1                                       1

                     A =           F (a)+ F (b)+ λF(a) – λF(b) ,  B = –  F (a)+ λF(a)+ Aaλ + A ,


                                           x
                                     x
                                                                         x
                         bλ – aλ – 2                                 λ
                     where F(x) is an arbitrary bounded, twice differentiable function.
                       b

                           λ|x–t|    µ|x–t|
               2.        Ae     + Be      y(t) dt = f(x),  –∞ < a < b < ∞.
                      a
                     Let us remove the modulus in the integrand and differentiate the resulting equation with
                     respect to x twice to obtain
                                                 b

                                                     2 λ|x–t|   2 µ|x–t|

                                2(Aλ + Bµ)y(x)+    Aλ e    + Bµ e     y(t) dt = f (x).      (1)
                                                                              xx
                                                a
                                                µ|x–t|
                     Eliminating the integral term with e  from (1) with the aid of the original integral equation,
                     we find that
                                                          b
                                                                               2
                                                  2
                                                      2

                               2(Aλ + Bµ)y(x)+ A(λ – µ )  e λ|x–t| y(t) dt = f (x) – µ f(x).  (2)
                                                                        xx
                                                        a
                     For Aλ+Bµ = 0, this is an equation of the form 3.2.1, and for Aλ+Bµ ≠ 0, this is an equation
                     of the form 4.2.15.
                        The right-hand side f(x) must satisfy certain relations, which can be obtained by setting
                     x = a and x = b in the original equation (a similar procedure is used in 3.2.1).
                 © 1998 by CRC Press LLC
               © 1998 by CRC Press LLC
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