Page 288 - Handbook Of Integral Equations
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1 . Solution with λ ≠ λ 1,2 :
                      ◦
                                              y(x)= f(x)+ λ(A 1 e βx  + A 2 ),
                     where the constants A 1 and A 2 are given by

                                    f 1 + λ f 1 ∆ β – (b – a)f 2  –f 2 + λ(f 2 ∆ β – f 1 ∆ 2β )
                               A 1 =  
                 ,  A 2 =  
                 ,
                                    λ (b – a)∆ 2β – ∆ 2  +1     λ (b – a)∆ 2β – ∆ 2  +1
                                                                 2
                                     2
                                                   β                          β
                                      b               b                 1

                                                                                βa
                                f 1 =  f(x) dx,  f 2 =  f(x)e βx  dx,  ∆ β =  (e βb  – e ).
                                     a               a                  β
                      ◦
                     2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
                                    y(x)= f(x)+ Cy 1 (x),  y 1 (x)= e βx  +  1 – λ 1 ∆ β  ,
                                                                      λ 1 (b – a)
                     where C is an arbitrary constant and y 1 (x) is an eigenfunction of the equation corresponding
                     to the characteristic value λ 1 .
                      ◦
                     3 . The solution with λ = λ 2 ≠ λ 1 and f 1 = f 2 = 0 is given by the formulas of item 2 in
                                                                                           ◦
                     which one must replace λ 1 and y 1 (x)by λ 2 and y 2 (x), respectively.
                     4 . The equation has no multiple characteristic values.
                      ◦
                                b
                                          βt
               3.    y(x) – λ  (Ae βx  + Be )y(t) dt = f(x).
                             a
                     The characteristic values of the equation:

                                                  2  2
                               (A + B)∆ β ±  (A – B) ∆ +4AB(b – a)∆ 2β
                                                     β                         1
                                                                                       βa
                         λ 1,2 =            
                        ,    ∆ β =  (e βb  – e ).
                                               2
                                        2AB ∆ – (b – a)∆ 2β                    β
                                               β
                      ◦
                     1 . Solution with λ ≠ λ 1,2 :
                                              y(x)= f(x)+ λ(A 1 e βx  + A 2 ),
                     where the constants A 1 and A 2 are given by

                                               Af 1 – ABλ f 1 ∆ β – (b – a)f 2
                                     A 1 =     
  2                          ,
                                              2
                                          ABλ ∆ – (b – a)∆ 2β – (A + B)λ∆ β +1
                                                 β
                                                Bf 2 – ABλ(f 2 ∆ β – f 1 ∆ 2β )
                                     A 2 =     
                             ,
                                              2
                                                 2
                                          ABλ ∆ – (b – a)∆ 2β – (A + B)λ∆ β +1
                                                 β
                                                 b               b
                                          f 1 =  f(x) dx,  f 2 =  f(x)e βx  dx.
                                               a               a
                      ◦
                     2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
                                    y(x)= f(x)+ Cy 1 (x),  y 1 (x)= e βx  +  1 – Aλ 1 ∆ β  ,
                                                                      A(b – a)λ 1
                     where C is an arbitrary constant and y 1 (x) is an eigenfunction of the equation corresponding
                     to the characteristic value λ 1 .
                     3 . The solution with λ = λ 2 ≠ λ 1 and f 1 = f 2 = 0 is given by the formulas of item 2 in
                      ◦
                                                                                           ◦
                     which one must replace λ 1 and y 1 (x)by λ 2 and y 2 (x), respectively.
                                                                                         2
                      ◦
                     4 . Solution with λ=λ 1,2 =λ ∗ and f 1 =f 2 =0, where the characteristic value λ ∗ =
                                                                                     (A + B)∆ β
                     is double:
                                    y(x)= f(x)+ Cy ∗ (x),  y ∗ (x)= e βx  –  (A – B)∆ β  ,
                                                                      2A(b – a)
                     where C is an arbitrary constant and y ∗ (x) is an eigenfunction of the equation corresponding
                     to λ ∗ .
                 © 1998 by CRC Press LLC








               © 1998 by CRC Press LLC
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