Đang chuẩn bị liên kết để tải về tài liệu:
Mixed Boundary Value Problems Episode 9

Tham khảo tài liệu 'mixed boundary value problems episode 9', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 228 Mixed Boundary Value Problems Substituting for A k from Equation 4.3.118 into Equation 4.3.119 we find after interchanging the order of integration that Ớ r f . J f rf r d g C Q t de rf dr 4.3.120 dt _Ja J Jt Vr2 -12 where mg p Í J k cos kt Yi k sin kt Q e t Vo J2 k Y2 k X Y1 k cos ke - Ji k sin k 4.3.121 k or A ư ỉ . . dk f . . dk Q e t - ựo sin k e - t k - Ịữ sin k e t k 3 jr 1 d L 4.3.122 Ự0 p b k J k J Our final task is to evaluate the contour integral H 2 z H 1 z ei Ị t z ị z where the contour r consists of the real axis from the origin to R an arc in the first quadrant z R 0 6 n 2 and imaginary axis from iR to the origin. As R TO we find that H 2 k H 1 k i i t k edk k p Ii k Jo Ki k Therefore Equation 4.3.120 becomes -k t Ị dk k . 4.3.123 4.3.124 t t Ii k k t p dk Ki k k rf r Vĩ2-2 dr 1 or a t Ỉ a kJ r e-k t V dk de 2 r rf r 9 Ja ưo Ki k e dk e nJt Vr -t2dr 4.3.125 2008 by Taylor Francis Group LLC Transform Methods 229 Figure 4.3.4 The solution u r z to the mixed boundary value problem governed by Equation 4.3.108 through Equation 4.3.111 with a 2. for a t . Figure 4.3.4 illustrates the solution when a 2. Example 4.3.6 Let us solve44 2 1 du. TT G 1 0 0 r 0 z 4.3.126 dr2 r dr z ơz dz subject to the boundary conditions lim 2 r z x lim u r z 0 r 0 r 0 z x 4.3.127 lim u r z 0 0 r x 4.3.128 and z u r 0 1 0 znUz r z z 0 0 1 r 1 r x 4.3.129 where K 0. 44 Taken from Brutsaert W. 1967 Evaporation from a very small water surface at ground level Three-dimensional turbulent diffusion without convection. J. Geophys. Res. 72 5631 5639. 1967 American Geophysical Union. Reproduced modified by permission of American Geophysical Union. 2008 by Taylor Francis Group LLC 230 Mixed Boundary Value Problems Using transform methods or separation of variables the general solution to Equation 4.3.126 Equation 4.3.127 and Equation 4.3.128 is u r z z 1-n 2 A k J-2LUz 1-n 2v Jo L1 - n Jo kr dk 4.3.130 where V 1 n p n 2 . Substituting Equation 4.3.130 into Equation 4.3.129 we have that