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Mixed Boundary Value Problems Episode 14

Tham khảo tài liệu 'mixed boundary value problems episode 14', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | The Wiener-Hopf Technique 379 Next we note that a k cosh M_ a a k cosh M_ a 2k cosh M_ k cosh a k cosh a k cosh 2k cosh @ M- k cosh ---------------------------. 5.1.155 a k cosh Therefore Equation 5.1.154 becomes 2 a k cosh M_ a Q_ a i exp kb sinh a k cosh M_ a 2k cosh M_ k cosh 1 a k cosh Ị cosh2 P a a k cosh M a r i2k cosh M- k cosh . _ i exp kb sinh ---2LLL-L . 5.1.156 a k cosh The fundamental reason for the factorization and the subsequent algebraic manipulation is the fact that the left side of Equation 5.1.156 is analytic in To T while the right side is analytic in T T0. Hence both sides are analytic on the strip tI T0. Then by analytic continuation it follows that Equation 5.1.156 is defined in the entire a-plane and both sides equal an entire function p a . To determine p a we examine the asymptotic value of Equation 5.1.156 as a x as well as using the edge conditions Equation 5.1.129 and Equation 5.1.130. Applying Liouville s theorem p a is a constant. Because in the limit of a TO p a 0 then p a 0. Therefore from Equation 5.1.156 p. - 2ikM k cosh exp kb sinh P a ----------- ----------M 5.1.157 Knowing P a we find from Equation 5.1.140 through Equation 5.1.144 that 4 EM a 5.1.158 7 a tanh sin b B EM a 7 a tanh sin b 5.1.159 C EM a eYa 7 a tanh sin 7 b 5.1.160 and EM a e-Ya 7 a tanh sin 7 a b 5.1.161 2008 by Taylor Francis Group LLC 380 Mixed Boundary Value Problems ikM k cosh exp -kb sinh E -----------------------------------. 5.1.162 cosh q With these values of A B C and D we have found T a y . Therefore . x y follows from the inversion of T a y . For example for x x 0 y b sxy E r MM .a . __ 2n J-ix -ei sinh yb _Y a tanh Y a tanh e- ax da. 5.1.163 For x 0 we evaluate Equation 5.1.163 by closing the integration along the real axis with an infinite semicircle in the upper half of the a-plane by Jordan s lemma and using the residue theorem. The integrand of Equation 5.1.163 has simple poles at yb nn where n 1 2 . and the zeros of Y a tanh . .