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4.1-4. Kernels Containing Higher-Order Polynomials in x and t

                                b
                                 n
                                     n
               31.   y(x) – λ  (x + t )y(t) dt = f(x),  n =1, 2, ...
                             a
                     The characteristic values of the equation:
                                           1                         1    n+1  n+1
                                λ 1,2 =    √       ,   where  ∆ n =     (b   – a  ).
                                      ∆ n ±  ∆ 0 ∆ 2n               n +1
                     1 . Solution with λ ≠ λ 1,2 :
                      ◦
                                                               n
                                              y(x)= f(x)+ λ(A 1 x + A 2 ),
                     where the constants A 1 and A 2 are given by
                                   f 1 – λ(f 1 ∆ n – f 2 ∆ 0 )    f 2 – λ(f 2 ∆ n – f 1 ∆ 2n )
                           A 1 =                        ,  A 2 =                       ,
                                                                    2
                                    2
                                 2
                                                                 2
                                λ (∆ – ∆ 0 ∆ 2n ) – 2λ∆ n +1   λ (∆ – ∆ 0 ∆ 2n ) – 2λ∆ n +1
                                    n
                                                                    n
                                       b               b                  1

                                                         n
                                 f 1 =  f(x) dx,  f 2 =  x f(x) dx,  ∆ n =   (b n+1  – a n+1 ).
                                      a               a                  n +1
                      ◦
                     2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:

                                                                  n
                                    y(x)= f(x)+ Cy 1 (x),  y 1 (x)= x +  ∆ 2n /∆ 0 ,
                     where C is an arbitrary constant and y 1 (x) is an eigenfunction of the equation corresponding
                     to the characteristic value λ 1 .
                     3 . Solution with λ = λ 2 ≠ λ 1 and f 1 = f 2 =0:
                      ◦

                                                                  n
                                    y(x)= f(x)+ Cy 2 (x),  y 2 (x)= x –  ∆ 2n /∆ 0 ,
                     where C is an arbitrary constant and y 2 (x) is an eigenfunction of the equation corresponding
                     to the characteristic value λ 2 .
                      ◦
                     4 . The equation has no multiple characteristic values.
                                b
                                     n
                                 n
               32.   y(x) – λ  (x – t )y(t) dt = f(x),  n =1, 2, ...
                             a
                     The characteristic values of the equation:
                                                                                	 –1/2
                                         1    n+1   n+1 2   1    2n+1  2n+1
                              λ 1,2 = ±     2  (b  – a  ) –     (b   – a   )(b – a)  .
                                      (n +1)              2n +1
                      ◦
                     1 . Solution with λ ≠ λ 1,2 :
                                                               n
                                              y(x)= f(x)+ λ(A 1 x + A 2 ),
                     where the constants A 1 and A 2 are given by
                                      f 1 + λ(f 1 ∆ n – f 2 ∆ 0 )  –f 2 + λ(f 2 ∆ n – f 1 ∆ 2n )
                                 A 1 =                 ,  A 2 =                   ,
                                       2
                                                  2
                                                                            2
                                                                 2
                                      λ (∆ 0 ∆ 2n – ∆ )+1       λ (∆ 0 ∆ 2n – ∆ )+1
                                                  n                         n
                                     b               b
                                                      n                1    n+1  n+1
                              f 1 =  f(x) dx,  f 2 =  x f(x) dx,  ∆ n =   (b   – a  ).
                                                                      n +1
                                   a               a
                     2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
                      ◦
                                                                  n
                                     y(x)= f(x)+ Cy 1 (x),  y 1 (x)= x +  1 – λ 1 ∆ n  ,
                                                                       λ 1 ∆ 0
                     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.
                      ◦
                 © 1998 by CRC Press LLC


               © 1998 by CRC Press LLC
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