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Tham khảo tài liệu 'mixed boundary value problems episode 13', 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 349 where Figure 5.0.2 Primarily known for his work on topology and ergodic theory Eberhard Frederich Ferdinand Hopf 1902-1983 received his formal education in Germany. It was during an extended visit to the United States that he worked with Norbert Wiener on what we now know as the Wiener-Hopf technique. Returning to Germany in 1936 he would eventually become an American citizen 1949 and a professor at Indiana University. Photo courtesy of the MIT Museum. g x i p ift x 2k i x eiK x 2 k x 0 x 0 5.0.3 and A k A k Ỗ 0. This integral equation was constructed by Grzesik and Lee7 to illustrate how various transform methods can be applied to electromagnetic scattering problems. We intend to solve Equation 5.0.2 via Fourier transforms. An important aspect of the Wiener-Hopf method is the splitting of the Fourier transform into two parts F k F k F k where F k i f x e ikx dx 5.0.4 0 7 Grzesik J. A. and S. C. Lee 1995 The dielectric half space as a test bed for transform methods. Radio Sci. 30 853-862. @1995 American Geophysical Union. Repro-duced modified by permission of the American Geophysical Union. 2008 by Taylor Francis Group LLC 350 Mixed Boundary Value Problems and Ỗ -Ỗ Re k Figure 5.0.3 The location of the half-planes F k and F k as well as K and K in the complex k-plane used in solving Equation 5.0.2 by the Wiener-Hopf technique. f 0 F k f x e ikx dx. J tt 5.0.5 Because the integral in Equation 5.0.4 converges only if 3 k 0 when x 0 we have added the subscript to denote its analyticity in the halfspace below 3 k 0 in the k-plane. Similarly the integral in Equation 5.0.5 converges only where 3 k 0 and the plus sign denotes its analyticity in the half-space above 3 k 0 in the k-plane. We will refine these definitions shortly. Direct computation of G k gives G k . 2 k K 1 1 2 k K k K 2 5.0.6 Note that this transform is analytic in the strip 3 k Ỗ. Next taking the Fourier transform of Equation 5.0.2 we find that F k F k G k K K F k k2 K .
