Page 681 - Handbook Of Integral Equations
P. 681

If y 0 (x) ≡ 0, then
                                x  dt
                       y 1 (x)=      = arctan x,
                                1+ t 2
                              0
                                      2
                               x  1 + arctan t

                                                        3
                       y 2 (x)=          dt = arctan x +  1 3  arctan x,
                              0   1+ t 2
                               x  1 + arctan t +  arctan t
                                         1    3
                                                                    2
                                                                3
                                                                                    7
                                                                              1
                                                                          5
                       y 3 (x)=          3       dt = arctan x +  1 3  arctan x +  3⋅5  arctan x +  7⋅9  arctan x.
                              0       1+ t 2
                   On continuing this process, we can observe that y k (x) → tan(arctan x)= x as k →∞, i.e., y(x)= x. The substitution
               of this result into the original equation shows the validity of the result.
                   Example 3. For the nonlinear equation
                                                      x
                                                        2
                                               y(x)=  [ty (t) – 1] dt
                                                    0
               we must obtain the first three approximations. If we set y 0 (x) = 0, then
                                        x
                                y 1 (x)=  (–1) dt = –x,
                                      0
                                        x
                                         3
                                                      4
                                y 2 (x)=  (t – 1) dt = –x +  1 4  x ,
                                      0

                                                                 4
                                            8
                                                                            10
                                           1
                                                                          1
                                                                      7

                                         t
                                                t + t
                                y 3 (x)=  x 	    16 t –  1 5  2    – 1 dt = –x +  1 4  x –  14 1  x +  160  x .
                                               2
                                      0
                ◦
               2 . The successive approximation method can be applied to solve other forms of nonlinear equations,
               for instance, equations of the form
                                                      x
                                          y(x)= F x,    K(x, t)y(t) dt
                                                     a
               solved for y(x) in which the integral has x as the upper integration limit. This makes it possible to
               obtain a numerical solution by applying small steps with respect to x and by linearization at each
               step, which usually provides the uniqueness of the result of the iterations for an arbitrary initial
               approximation.
               3 . The initial approximation substantially influences the number of iterations necessary to obtain
                ◦
               the result with prescribed accuracy, and therefore when choosing this approximation, some additional
               arguments are usually applied. Namely, for the equation
                                                x


                                       Ay(x) –   Q(x – t)Φ y(t) dt = f(x),
                                               0
               where A is a constant, a good initial approximation y 0 (x) can sometimes be found from the solution
               of the following (in general, transcendental) equation for ˜y 0 (p):
                                                                ˜
                                                  ˜


                                          A ˜y 0 (p) – Q(p)Φ ˜y 0 (p) = f(p),
                                   ˜
                          ˜
               where ˜y 0 (p), Q(p), and f(p) are the transforms of the corresponding functions obtained by means of
               the Laplace transform. If ˜y 0 (p)isdefined, then the initial approximation can be found by applying
                                                 –1
               the Laplace inversion formula: y 0 (x)= L { ˜y 0 (p)}.
                 © 1998 by CRC Press LLC






               © 1998 by CRC Press LLC
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