tailieunhanh - Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 5

Tham khảo tài liệu 'advanced mathematical methods for scientists and engineers episode 6 part 5', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Thus the solution of the integral equation is x - X si 60 Ợ X - 24 2 X 60 2. For x 1 the integral equation reduces to ộ x x. For x 1 the integral equation becomes ự x x X sin xy y dy. Jo We could solve this problem by writing down the Neumann series. Instead we will use an eigenfunction expansion. Let X and ộn be the eigenvalues and orthonormal eigenfunctions of 0 x X o 1 sin xy ệ y dy. We expand ộ x and x in terms of the eigenfunctions. x 2 anộn x n 1 x 2 bnộn x n 1 bn x 0n x We determine the coefficients an by substituting the series expansions into the Fredholm equation and equating 2134 coefficients of the eigenfunctions. ộ x x A sin xy y dy Jo an n x y2 bn n x A sin xy an n y dy r 1 r 1 V 0 n 1 V an n x V bn n x A V a 1 n x 1 1 1 An an n If A is not an eigenvalue then we can solve for the an to obtain the unique solution. _ bn _ Anbn _ b Abn an 1 - A An An - A bn An - A .A Abn _ x x y2 n x for x 1. An - A n 1 If A Am and x Jm 0 then there is the one parameter family of solutions z . . x Abn . r x x cộm x X n x for x 1. -1 An A n 1 n m If A Am and x ộm 0 then there is no solution. Solution 1. Kx L1L2x Ax .