tailieunhanh - HANDBOOK OFINTEGRAL EQUATIONS phần 9

Tham khảo tài liệu 'handbook ofintegral equations phần 9', ngoại ngữ, ngữ pháp tiếng anh phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | where ao 1 . an are the coefficients of the polynomial Un z and the a-k are the coefficients of the expansion of the function Ỷ z which are given by the obvious formula a-k -h T -3j p H rr dr 2ni J L 1 X r The solvability conditions acquire the form an an-1 an-p v 2 0. If a solution must satisfy the additional condition to 0 then for V-p 0 in formulas 53 we must take the polynomial Pv-p-1 z and for V -p 0 p - V conditions must be satisfied. . The Riemann Problem for a Multiply Connected Domain Let L L0 L1 Lm be a collection of m 1 disjoint contours and let the interior of the contour L0 contain the other contours. By Q we denote the m 1 -connected domain interior for L0 and exterior for L1 . Lm. By Q- we denote the complement of Q L in the entire complex plane. To be definite we assume that the origin lies in Q . The positive direction of the contour L is that for which the domain Q remains to the left . the contour L0 must be traversed counterclockwise and the contours L1 . Lm clockwise. We first note that the jump problem ln - t H t is solved by the same formula z H t dr T - z as in the case of a simply connected domain. This follows from the Sokhotski-Plemelj formulas which have the same form for a multiply connected domain as for a simply connected domain. The Riemann problem homogeneous and nonhomogeneous can be posed in the same way as for a simply connected domain. We write Vk 2n arg D t Lk all contours are passed in the positive direction . By the index of the problem we mean the number m V Vk. 54 k 0 If Vk k 1 . m are zero for the inner contours then the solution of the problem has just the same form as for a simply connected domain. To reduce the general case to the simplest one we introduce the function m J - zk Vk k 1 where the zk are some points inside the contours Lk k 1 . m . Taking into account the fact that arg t - zk Lj 0 for k j and arg t - Zj Lj -2n we obtain m 1m argTT t - Zk v 2n 44 k 1 - Lj 2n arg - zj j L Vj j 1 . m. 1998 by .