Tham khảo tài liệu 'mixed boundary value problems episode 4', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Historical Background 77 or .TO - Ị A k Jỵx dk 21 2 wx3 3ựn after using integral tables. Differentiating both sides of Equation with respect to x VkA k Ji kx dk Jo 2 vn 0 x a. We now turn to Equation . Multiplying this equation by 2x n 1 2 Ạ r2 x2 and integrating the resulting equation over r from x a to TO we find that k 1 aXo k A k di ựxJi kr ọ 0 dr r2 x2 dk 0 or o i Vk 1 aXok A k J kx dk 0 a x TO. Jo 1 Let us now replace x with r in Equation and Equation and express J1 z ỵ 2 nz sin z . This yields i A k sin kr dk 2wr 0 r a o and o I 1 aXok A k sin kr dk 0 a r TO. o We can rewrite Equation as A k sin kr dk Xo akA k sin kr dk oo a r TO. From the theory of Fourier integrals a r sin kr dr 0 -oo 0 i akA k sin kr dk n Ja Uo sin kr dr a 2xa2 oo 0 f00 r f00 z aỆA Ệ sin r dỆ n Jo Uo 2X za n Jo o oo sin kr dr sin kr dr. o 2008 by Taylor Francis Group LLC 78 Mixed Boundary Value Problems The second term in Equation equals aAokA k . Therefore 1 aXok A k 2 A J2 ak tt f n Jo o sin kr dr. Interchanging the order of integration we finally have that 1 aXok A k 2wa2 J2 ak Aojfa aỊA Ị K k Ị dỊ where Ky 4 2 i sin ry s n r dr 1 I -fll - al 1 . nJ0 n y- Ị y Ị J One of the intriguing aspects of Equation is that the unknown is the Hankel transform A k . In general the integral equations that we will see will involve an unknown which is related to A k via an integral definition. See Equation Equation and Equation . In Goodrich s paper he solved the integral equation as a variational problem and found an approximate solution by the optimization of suitable solutions. However we shall shortly outline an alternative method for any value of Ao. In 1978 Shail27 reexamined Goodrich s paper for two reasons. First the solution for the n 0 case Equation creates a divergent integral when it is substituted back into dual equations Equation .
