Page 173 - Handbook Of Integral Equations
P. 173

Eliminating the integral I n from (4) and (5), we obtain

                                                    n–1
                                                                              2
                                             2
                                                                2
                                                            2
                                y (x)+(σ n + λ )y(x)+    A k (λ – λ )I k = f (x)+ λ f(x).   (6)


                                 xx          n              n   k      xx     n
                                                    k=1
                     Differentiating (6) with respect to x twice followed by eliminating I n–1 from the resulting
                     expression with the aid of (6) yields a similar equation whose left-hand side is a fourth-
                                                                                       n–2
                     order differential operator (acting on y) with constant coefficients plus the sum     B k I k .
                                                                                       k=1
                     Successively eliminating the terms I n–2 , I n–3 , ... using double differentiation and formula (3),
                     we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2n with
                     constant coefficients.
                        The initial conditions for y(x) can be obtained by setting x = a in the integral equation
                     and all its derivative equations.
                     2 . Let us find the roots z k of the algebraic equation
                      ◦
                                                   n
                                                     λ k A k

                                                           + 1 = 0.                         (7)
                                                     z + λ 2
                                                  k=1    k
                     By reducing it to a common denominator, we arrive at the problem of determining the roots
                     of an nth-degree characteristic polynomial.
                        Assume that all z k are real, different, and nonzero. Let us divide the roots into two groups
                                   z 1 >0,  z 2 >0,  ... ,  z s > 0  (positive roots);
                                   z s+1 <0,  z s+2 <0,  ... ,  z n < 0  (negative roots).
                        Then the solution of the integral equation can be written in the form

                                    s                   n
                                  x


                      y(x)=f(x)+       B k sinh µ k (x–t) +  C k sin µ k (x–t)  f(t) dt,  µ k =  |z k |. (8)
                                 a
                                    k=1                k=s+1
                     The coefficients B k and C k are determined from the following system of linear algebraic
                     equations:
                            s            n

                                B k µ k      C k µ k
                                      +             – 1=0,    µ k =  |z k |  m =1, 2, ... , n.  (9)
                                             2
                                2
                              λ + µ 2       λ – µ 2
                                             m
                                m
                           k=0      k   k=s+1     k
                        In the case of a nonzero root z s = 0, we can introduce the new constant D = B s µ s and
                     proceed to the limit µ s → 0. As a result, the term D(x – t) appears in solution (8) instead of
                                                              –2
                     B s sinh µ s (x – t) and the corresponding terms Dλ  appear in system (9).
                                                              m
                              x
                                sin(λx)
               20.   y(x) – A          y(t) dt = f(x).
                             a  sin(λt)
                     Solution:
                                                        x      sin(λx)
                                        y(x)= f(x)+ A    e A(x–t)    f(t) dt.
                                                       a       sin(λt)
                              x
                                sin(λt)
               21.   y(x) – A          y(t) dt = f(x).
                             a sin(λx)
                     Solution:
                                                        x      sin(λt)
                                        y(x)= f(x)+ A    e A(x–t)    f(t) dt.
                                                       a       sin(λx)
                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
                                                                                                             Page 152
   168   169   170   171   172   173   174   175   176   177   178