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Mixed Boundary Value Problems Episode 15

Tham khảo tài liệu 'mixed boundary value problems episode 15', 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 409 use the differential equation and first two boundary conditions to show that U k y A k e yVk2 . Step 2 Taking the Fourier transform of the boundary condition along y 0 show that A k U k 11 ki and U k Vk2 1 A k 1 where r 0 U k u x 0 eikx dx J tt and U k i uy x 0 eikx dx. 0 Here we have assumed that lu x 0 is bounded by e ex 0 e c 1 as x so that U is analytic in the upper half-plane 3 k e while U is analytic in the lower half-plane 3 k 0. Step 3 Show that 1 can be rewritten U k T1 ki 2 1 ki w ki u_ k V T ki . 1 ki Note that the right side of this equation is analytic in the lower half-plane A k 0 while the left side is analytic in the upper half-plane 3 k e. Step 4 Use Liouville s theorem and deduce that U k V2 1 1 ki T1 ki 1 ki Step 5 Show that ự2 e yVk2 U k y k -7ĨỈTĨ 1 i e yy k2 1 k i Tk i y k2 1 Step 6 Finish the problem by retracing Step 6 through Step 9 of the previous problem and show that you recover the same solution. Gramberg and van de Ven24 found an alternative representation rc u x y e x ú 2 1 0 1. 7 sin ny dp 1 x 0 24 Gramberg H. J. J. and A. A. F. van de Ven 2005 Temperature distribution in a Newtonian fluid injected between two semi-infinite plates. Eur. J. Mech. Ser. B. 24 767-787. 2008 by Taylor Francis Group LLC 410 Mixed Boundary Value Problems and 1 2 rtt x r 1 u x y . cos ny dn x 0 n Jo n2 ly vV 1 1 by evaluating the inverse Fourier transform via contour integration. 3. Use the Wiener-Hopf technique to solve the mixed boundary value problem 2 Ww u TO x x 0 y ox2 ay2 with the boundary conditions limy -x. u x y 0 J u x 0 1 x 0 uy x 0 0 0 x. Step 1 Assuming that lu x 0 is bounded by e x as x TO where 0 e c 1 let us define the following Fourier transforms U k y u x y eikx dx U k y u x y eikx dx J tt J 0 and U k y u x y eikx dx J tt so that U k y U k y U k y . Here U k y is analytic in the halfspace A k e while U k y is analytic in the half-space A k 0. Then show that the partial differential equation becomes dU- m2U 0 0