Tham khảo tài liệu 'mixed boundary value problems episode 3', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Historical Background 47 Using separation of variables or transform methods the solution to Equation is tt u 0 e kzl B k e-klz-LI 1 J0 kr . k This solution also satisfies the boundary conditions given by Equation and Equation . Substituting Equation into Equation and Equation we obtain the dual integral equations for A k and B k and dk J A k B k e-kL j0 kr dk _ rỊk e kL B k j0 kr dk k r-tt A k J0 kr dk 0 0 r-tt B k J0 kr dk 0 0 V1 V2 0 r 1 0 r 1 r x r x . To solve Equation and Equation we introduce . 2k . A k f t cos kt dt n J0 and _ 2k f1. B k g t cos kt dt. n J0 To show that the A k given by Equation satisfies Equation we evaluate A k J0 kr dk f t kcos kt dt J0 kr dk J0 n J0 _J0 J 1 2 1 - f t sin kt Jo kr dk n 0 0 0 _ oo sin kt J0 kr dk dt 0 2 r I f 1 sin k J0 kr dk n J 0 I-1 n J 0 0 r-tt sin kt J0 kr dk dt 0 tt 0 1 1 1 0 tt 0 1 0 tt 0 2008 by Taylor Francis Group LLC 48 Mixed Boundary Value Problems because r 1 and we used Equation . A similar demonstration holds for B k . We now turn to the solution of Equation . Substituting Equation and Equation into Equation we find that cos kt Jo kr dk dt rc e kL cos kt Jo kr dk _ nVi dt ---. 2 Using Equation we can evaluate the first integral and obtain h r y g r y e kL cos kr Jo kr dk dr 1 where r h fo d 2-2 19 If we now multiply Equation by 2r dr n f2 r2 integrate from 0 to t and then taking the derivative with respect to t we have f t Jo g T 2 d r r r _ndt Jo Jo kL cos kT Jo kr dk r dr Vt2 2 dT d r ft r dr V dt jo V V1ị ỉ2 r21 dt o V1 e since the solution to Equation is f t 2 d r t rh r ---77 dr n dt Jo ỵ t2 r2 . Defining K t T e tL co kT ắ k rJo kr Vt 2 dr dk rc o we substitute Equation into Equation and K t T simplifies to K t T e kL cos kT cos kt dk n J o 1 1 L2 t T 2 L2 t T 2 .
