Page 219 - Handbook Of Integral Equations
P. 219

Substitute y(x) of (3) into (4). Integration by parts yields f (1) = f(1)+f(0) and f (1)–f (0) =



                                                                                    x
                                                                  x
                                                                                          x
                     2f(1)+2f(0). Hence, we obtain the desired constraints for f(x):
                                        f (1) = f(0) + f(1),  f (0) + f (1) = 0.            (5)



                                         x                   x      x
                        Conditions (5) make it possible to find the admissible general form of the right-hand side
                     of the integral equation:
                                                f(x)= F(x)+ Ax + B,


                                   A =– 1    F (1) + F (0) ,  B =  1    F (1) – F(1) – F(0) ,



                                        2  x      x          2  x
                     where F(x) is an arbitrary bounded twice differentiable function with bounded first derivative.
                         b
               2.       |x – t| y(t) dt = f(x),  0 ≤ a < b < ∞.
                      a
                     This is a special case of equation 3.8.3 with g(x)= x.
                        Solution:
                                                         1

                                                   y(x)= f (x).
                                                         2 xx
                        The right-hand side f(x) of the integral equation must satisfy certain relations. The
                     general form of f(x) is as follows:
                                                f(x)= F(x)+ Ax + B,





                               A =–  1    F (a)+ F (b) ,  B =  1    aF (a)+ bF (b)– F(a)– F(b) ,

                                    2  x      x          2   x      x
                     where F(x) is an arbitrary bounded twice differentiable function (with bounded first deriva-
                     tive).
                         a
               3.       |λx – t| y(t) dt = f(x),  λ >0.
                      0
                     Here 0 ≤ x ≤ a and 0 ≤ t ≤ a.
                      ◦
                     1 . Let us remove the modulus in the integrand:
                                        λx                a

                                          (λx – t)y(t) dt +  (t – λx)y(t) dt = f(x).        (1)
                                       0                 λx
                     Differentiating (1) with respect to x, we find that
                                              λx           a


                                          λ     y(t) dt – λ  y(t) dt = f (x).               (2)
                                                                     x
                                             0            λx
                                           2
                     Differentiating (2) yields 2λ y(λx)= f (x). Hence, we obtain the solution

                                                    xx
                                                        1
                                                               x

                                                 y(x)=    f xx   .                          (3)
                                                       2λ 2    λ
                      ◦
                     2 . Let us demonstrate that the right-hand side f(x) of the integral equation must satisfy
                     certain relations. By setting x = 0 in (1) and (2), we obtain two corollaries
                                         a                    a


                                          ty(t) dt = f(0),  λ  y(t) dt =–f (0),             (4)
                                                                         x
                                        0                    0
                     Substitute y(x) from (3) into (4). Integrating by parts yields the desired constraints for f(x):



                                   (a/λ)f (a/λ)= f(0) + f(a/λ),  f (0) + f (a/λ)=0.         (5)
                                                                       x
                                        x
                                                                 x
                        Conditions (5) make it possible to establish the admissible general form of the right-hand
                     side of the integral equation:
                                           f(x)= F(z)+ Az + B,    z = λx;





                                  A =–  1    F (a)+ F (0) ,  B =  1    aF (a)– F(a)– F(0) ,
                                       2   z     z          2   z
                     where F(x) is an arbitrary bounded twice differentiable function (with bounded first deriva-
                     tive).
                 © 1998 by CRC Press LLC
               © 1998 by CRC Press LLC
                                                                                                             Page 198
   214   215   216   217   218   219   220   221   222   223   224