Page 130 - Handbook Of Integral Equations
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2.1-2. Kernels Quadratic in the Arguments x and t

                              x

                                 2
               7.    y(x)+ A    x y(t) dt = f(x).
                              a
                     This is a special case of equation 2.1.50 with λ = 2 and µ =0.
                        Solution:
                                                     x

                                                                 3
                                                        2
                                                                     3

                                      y(x)= f(x) – A   x exp    1  A(t – x ) f(t) dt.
                                                             3
                                                    a
                                x
               8.    y(x)+ A    xty(t) dt = f(x).
                              a
                     This is a special case of equation 2.1.50 with λ = 1 and µ =1.
                        Solution:
                                                      x
                                                              1  3   3
                                      y(x)= f(x) – A   xt exp  A(t – x ) f(t) dt.
                                                             3
                                                    a
                              x

                                 2
               9.    y(x)+ A    t y(t) dt = f(x).
                              a
                     This is a special case of equation 2.1.50 with λ = 0 and µ =2.
                        Solution:
                                                     x

                                                                    3
                                                                 3
                                                        2

                                      y(x)= f(x) – A   t exp   1  A(t – x ) f(t) dt.
                                                             3
                                                     a
                                x
                                     2
               10.   y(x)+ λ   (x – t) y(t) dt = f(x).
                             a
                     This is a special case of equation 2.1.34 with n =2.
                        Solution:
                                                         x

                                            y(x)= f(x) –   R(x – t)f(t) dt,
                                                        a
                                                    √      √      √      
            1/3
                                 2
                                          2
                           R(x)= ke –2kx  – ke kx  cos  3 kx –  3 sin  3 kx ,  k =  1  λ  .
                                 3        3                                       4
                              x

                                  2
                                     2
               11.   y(x)+ A    (x – t )y(t) dt = f(x).
                              a
                                                                2
                     This is a special case of equation 2.9.5 with g(x)= Ax .
                        Solution:                   x
                                                1



                                    y(x)= f(x)+       u (x)u (t) – u (x)u (t) f(t) dt,

                                                            2
                                                                 2
                                                                      1
                                                       1
                                               W   a
                     where the primes denote differentiation with respect to the argument specified in the parenthe-
                     ses; u 1 (x), u 2 (x) is a fundamental system of solutions of the second-order linear homogeneous
                     ordinary differential equation u     +2Axu = 0; and the functions u 1 (x) and u 2 (x) are ex-
                                              xx
                     pressed in terms of Bessel functions or modified Bessel functions, depending on the sign of
                     the parameter A:
                        For A >0,
                                           √      
	                  √
                           W =3/π, u 1 (x)=  xJ 1/3  8  Ax 3/2  , u 2 (x)=  xY 1/3  8 Ax 3/2  .
                                                     9                          9
                        For A <0,
                                          √     
	  8    3/2         √      
	  8    3/2
                                3
                          W =– , u 1 (x)=   xI 1/3   |A| x  , u 2 (x)=  xK 1/3   |A| x  .
                                2                   9                           9
                 © 1998 by CRC Press LLC
               © 1998 by CRC Press LLC
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