Tham khảo tài liệu 'mixed boundary value problems episode 6', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 138 Mixed Boundary Value Problems Example Let us solve38 d 2 du 1 d r du Pr r dĩ sin 6 dê _sin 6 ỡỡ subject to the boundary conditions lim u r 6 w lim u r 6 0 r 0 r and u a- 6 u a 6 1 u a- 6 u a 6 ur a- 6 ur a 6 u a- ê u a ê 1 m 0 r w 0 6 n 0 6 n 0 6 a a 6 n a n a 6 n where m 0 or 1 and 0 a n 2. The solution must be finite at the 6 0 n. Separation of variables yields the solution u r ê y A2n m n 0 r 2n m a P2n m cos e 0 r a and u r ê A2n m P2n m cos 6 n 0 r a r w. We have written the solution in this form so that we can take advantage of symmetry and limit 6 between 0 and n 2 rather than 0 6 n. Equation and Equation satisfy not only Equation but also Equation . Substituting Equation and Equation into Equation yields the dual Fourier-Legendre series 00 A2n mP2n m cos 6 1 0 6 a n 0 and 2n m 2 A2n m P2n m cos ớ 0 a 6 n 2. n 0 38 Minkov I. M. 1963 Electrostatic field of a sectional spherical capacitor. Sov. Tech. Phys. 7 1041-1043. 2008 by Taylor Francis Group LLC Separation of Variables 139 At this point we introduce ra . . _ A2n m g t cos 2n m 2 t dt Jo _ si 2n m 2 7 _ i sin 2n m 2 t d g 2n m 2 Jo 9 2n m 2 Now 5 2n m 2 A2n mP2n m cos 6 n 0 g a 2 p2n m cos 6 sin 2n m 2 a n o g t I 57 P2n m cos ớ sin 2n m 2 t I dt 0. o n o Equation follows from Problem 4 at the end of Section as well as the facts that 0 t a 6 n 2. Therefore our choice for A2n m satisfies Equation identically. Turning to Equation fa .J g t i 57 P2npm cos ớ cos 2n m 2 t dt 1. o n o Again using the results from Problem 4 at the end of Section we have g t dt 2 1 m g dr Jo ự2 cos t cos 6 Jo ạ 2 cos t cos 6 where 0 6 a. Applying Equation and Equation g t 4dI z wn mi d4 a K t T g T dT n dt y J0 yf 2 cos 6 cos t J 2n Jo where t K t T 2d i t x in6 d6 dt Jo y cos r cos 0 ựcos ớ cos t sec t T 2 sec t T 2 . 2008 by .
