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2.2. Equations Whose Kernels Contain Exponential
                      Functions

                 2.2-1. Kernels Containing Exponential Functions

                              x

               1.    y(x)+ A    e λ(x–t) y(t) dt = f(x).
                              a
                     Solution:
                                                         x

                                          y(x)= f(x) – A   e (λ–A)(x–t) f(t) dt.
                                                         a
                              x

               2.    y(x)+ A    e λx+βt y(t) dt = f(x).
                              a
                     For β = –λ, see equation 2.2.1. This is a special case of equation 2.9.2 with g(x)= –Ae λx
                              βt
                     and h(t)= e .
                        Solution:
                                    x

                       y(x)= f(x) –   R(x, t)f(t) dt,  R(x, t)= Ae λx+βt  exp  A   e (λ+β)t  – e (λ+β)x     .
                                   a                                   λ + β
                                x

               3.    y(x)+ A      e λ(x–t)  – 1 y(t) dt = f(x).
                              a
                     1 . Solution with D ≡ λ(λ – 4A)>0:
                      ◦
                                      2Aλ     x                           1       √
                                                                                  1
                          y(x)= f(x) – √     R(x – t)f(t) dt,  R(x)=exp  2 λx sinh  2  Dx .
                                        D  a
                     2 . Solution with D ≡ λ(λ – 4A)<0:
                      ◦
                                            x
                                      2Aλ                                 1

                                                                                 1
                          y(x)= f(x) – √      R(x – t)f(t) dt,  R(x)=exp  λx sin    |D| x .
                                       |D|  a                            2       2
                     3 . Solution with λ =4A:
                      ◦
                                                      x



                                     y(x)= f(x) – 4A 2  (x – t)exp 2A(x – t) f(t) dt.
                                                     a
                             x

                                  λ(x–t)
               4.    y(x)+     Ae      + B y(t) dt = f(x).
                            a
                     This is a special case of equation 2.2.10 with A 1 = A, A 2 = B, λ 1 = λ, and λ 2 =0.
                     1 . The structure of the solution depends on the sign of the discriminant
                      ◦
                                                             2
                                               D ≡ (A – B – λ) +4AB                         (1)
                     of the square equation
                                               2
                                              µ +(A + B – λ)µ – Bλ = 0.                     (2)
                      ◦
                     2 .If D > 0, then equation (2) has the real different roots
                                                    √                        √
                                                               1
                                       1
                                  µ 1 = (λ – A – B)+  1  D,  µ 2 = (λ – A – B) –  1  D.
                                       2           2           2            2
                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
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