Tham khảo tài liệu 'mixed boundary value problems episode 10', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 258 Mixed Boundary Value Problems lim u r z 0 0 r tt z and auz r 0 - 3u r 0 f r u r 0 0 0 r 1 1 r tt. Both a and 3 are nonzero. In line with previous examples the solution that satisfies Equation through Equation is u r z A k e-kz J0 kr dk. Jo Substituting u r z into the mixed boundary condition i ak 3 A k J0 kr dk f r 0 r 1 Jo and TO i A k J0 kr dk 0 1 r tt. 0 At this point we introduce an integral definition for A k A k i h t sin kt dt h 0 0. 0 The demonstration that this definition of A k satisfies Equation is left as part of Problem 6. Turning to Equation we substitute A k into Equation and find that poo k sin kt J0 kr dk dt 0 3 sin kt J0 kr dk dt f r 0 r 1. At this point we would normally manipulate Equation into a Fredholm integral equation. This is left as an exercise in Problem 6. Here we introduce an alternative method developed by Gladwell et The derivation begins by showing that kA k J0 kt dk dt y x2 poo A k sin kx dk. 0 62 Taken from Gladwell G. M. L. J. R. Barber and Z. Olesiak 1983 Thermal problems with radiation boundary conditions. Quart. J. Mech. Appl. Math. 36 387 401 by permission of Oxford University Press see also Lemczyk T. F. and M. M. Yovanovich 1988 Thermal constriction resistance with convective boundary conditions 1. Half-space contacts. Int. J. Heat Mass Transfer 31 1861 1872. Í1 a 0 t 2008 by Taylor Francis Group LLC Transform Methods 259 This follows from interchanging the order of integration and applying Equation . Next we view the quantity within the square brackets on the left side of Equation as the unknown in an integral equation of the Abel type. From Equation and Equation we have that kA k Jo kt dk A k sin kT dk ndt Jo Jo J J2 - T12 2 d f r Jf T sin kT Ji ndt Jo A k ưo VT dJd7 d dt t i A k J1 kt dk o t J h T y sin kT J1 kt dk d ự 4hj dT. dt Jo yft2 T2 _ dJ .
