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9.4-4. Solution of Operator Equations of Polynomial Form

               The method described in Subsection 9.4-3 can be generalized to the case of operator equations of
               polynomial form. Suppose that the solution of the linear nonhomogeneous equation (21) is given
               by formula (22) and that the corresponding homogeneous equation has only the trivial solution.
                   Let us construct the solution of the more complicated equation with polynomial left-hand side
               with respect to the operator L:
                                          n
                                                k               k      k–1
                                    y(x) –  A k L [y]= f(x),  L ≡ L L     ,                (30)
                                         k=1

               where A k are some numbers and f(x) is an arbitrary function.
                   We denote by λ 1 , ... , λ n the roots of the characteristic equation

                                                   n
                                               n
                                              λ –     A k λ n–k  = 0.                      (31)
                                                   k=1
               The left-hand side of Eq. (30) can be expressed in the form of a product of operators:

                                             n            n
                                                   k
                                       y(x) –  A k L [y] ≡  1 – λ k L [y].                 (32)
                                            k=1          k=1
               The solution of the auxiliary equation (26), in which we use the substitution w → y n–1 and λ 2 → λ n ,
               is given by the formula y n–1 (x)= Y (f, λ n ). Reasoning similar to that in Subsection 9.4-3 shows
               that the solution of Eq. (30) is reduced to the solution of the simpler equation

                                            n–1


                                               1 – λ k L [y]= y n–1 (x),                   (33)
                                            k=1
               whose degree is less by one than that of the original equation with respect to the operator L. We can
               show in a similar way that Eq. (33) can be reduced to the solution of the simpler equation

                                n–2


                                   1 – λ k L [y]= y n–2 (x),  y n–2 (x)= Y (y n–1 , λ n–1 ).
                                k=1
               Successively reducing the order of the equation, we eventually arrive at an equation of the form (28)
               whose right-hand side contains the function y 1 (x)= Y (y 2 , λ 2 ). The solution of this equation is given
               by the formula y(x)= Y (y 1 , λ 1 ).
                   The solution of the original equation (30) is defined recursively by the following formulas:

                       y k–1 (x)= Y (y k , λ k );  k = n, ... , 1,  where  y n (x) ≡ f(x),  y 0 (x) ≡ y(x).

               Note that here the decreasing sequence k = n, ... , 1 is used.


                 9.4-5. A Generalization
               Suppose that the left-hand side of a linear (integral) equation

                                               y(x) – Q [y]= f(x)                          (34)




                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
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