Tham khảo tài liệu 'mixed boundary value problems episode 16', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 440 Mixed Boundary Value Problems or u r e z ia i n Jo Jo arctan Ệ- . R z _ R3 f p d ddpdp where R2 r2 p2 2rpcos ớ d z2 p2 x ịự x r 2 z2 y x r 2 z2 the operator L k ơ r 6 is given by 1 f2n L k ơ r 6 X k e d ơ r d dd 2 oo r 1 2n V k1 n 1 eine e-inứơ r d dd 2n .Io n -o 1- J0 -1 1 - k2 . 2 d k e 63221 v n -o and z aa - p2 2 a 2 pV12 a 12 p J Ể2 a á2 d2 a Ja2 p2y a2 2 a ựa2 p2 y 2 a r2 a 2 a Equation is recommended in those cases when the integrals can be evaluated exactly while Equation is more convenient when the integrals must be computed numerically. To illustrate Fabrikant s results consider the case when f r 6 w0 a constant. In this case 00 L p f p 6 p1 n 1 énỡ n -o 2n e-intìwo dd wo o because all of the terms in the summation vanish except n 0. Therefore d i L p f p 6 - dp w0 dP Jo ỵ ri2 p2 and l 1 L n L p f p 6 L r n2 dn do Vn2 p2 y p2 n rn2 n -o L n ein 2n e-intìwo dd wo. o 2008 by Taylor Francis Group LLC Green s Function 441 From Equation . 2wo r Mi n u r e z - n Jo Ợr- - 2 n - 2w Ri n l a - arcsin n L r J 0 2wo . tfia - arcsin n r Fabrikant also considered the case when the mixed boundary condition reads i uz r e 0 ơ r e u r e 0 0 0 r a a r X 0 e 2n 0 e 2n In this case he showed that the solution is u r e z 4C S x p dp Vg2 x - p x dx ựx2 r2 2 u r e z 4C ỉ J f pdp Jo I Jo yjrp pr p L e2 n ơp e . tM I v 2 n r2 l or . . . . . u r e z 2C i i arctan 4 ơ 7 dê p dp n Jo Jo R R where g x xyj1 z2 r2 x2 and R and 4 have been defined earlier. The constant coefficient C equals 1 2n in classical potential problems and different values in other applications. Equation and Equation are best when the integrals can be evaluated exactly while Equation should be used otherwise. To illustrate Equation through Equation consider the case when ơ r e ơ0 a constant. Then L Fr ơ p e Ễ ein4 í e-intìơo dê ơo vx 7 n
