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j0215 - 1 Part a Jackson Problem 2.15 B J Mattson In...

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Unformatted text preview: 1 Part a Jackson Problem 2.15 B. J. Mattson In general, the Green function must satisfy V’2GD : —47r6(f—a;’ For Dirichlet boundary conditions, the Green function has boundary conditions: G(b0undaries) = 0. So I need to solve: V'QGD = —4776(:c — $’)(5(y — y’) GD(.'E, Z Z GD(.TI Z I GD(yl Z Z GD(yl : Z (1) Assume GD 2 gm($, x’)gy(y, y’). Then the equation can be written: 829:5 829;, , , 9;, 856,2 + 9;; aylz = —47T6(ac — ac )6(y — 3/) Look at 35 7E 35’ and 3/ 7E y’. Then, iaZQw + £8291; : 0 9w (gm/2 9y all/2 And, since there is no x’ or y’ depencence on the RHS, 82g 829 at: = Solving for gm, 935(93, x') = sin(mm:) sin(mrx’) Now let (leaving the = 47? with the 6(y — y’)), 6(30 — :L") : Z c” sin(mr:c) sin(mmc') 11:1 And 0,, can be found using orthoganality considerations, 1 / 6(20 — 33') sin(n7rac')d:v' cnsinmmc) - 0 sin(mm:) cnsmmfix) - [\D|>—\N>|>—\ So c“ = 2. And 00 gm 2 2 Z sin(mrx) sin(mr:c') (12) n21 So, 0(56, 96', y, 3/) = 2 2 97M, 2/) 8111mm) siIIWWI) (13) n21 Where gn satisfies 829" — 2 2 — —4 6< — '> (14) 831,2 fingn— 7T y y 2 Part b To satisfy the boundary conditions (gn(y, 0) = gn(y, 1) = 0), let _ A sinh(7my’) sinh[7m(1 — W] 3/ < y g“ _ sinh[7m(1 — y')]SiIlh(7T71y) Z/ < 3/, This satisfies both the boundary conditions and one of the jump conditions. However, I still need A, which I can find by using the discontinuity in slope. y+ 69” = —47r (15) ay’ y_ Or, A[7m cosh(7my) sinh[7m(1 — y)] — 7m sinh(7my) cosh[7m(1 — y)] = 47r (16) After a bit of algebra, 4 A : ; [cosh(7my) [sinh(7rn) cosh(—7my) — cosh(7m) sinh(—7Tny)] — (17) sinh(7my) [cosh(7m) cosh(—7my) — sinh(7m) sinh(—7my)]]_1 (18) And a bit more magic (hyperbolic identities, etc.) gives: 4 A : — 19 n sinh(7m) ( ) SO, 9” = sinh(7rny<) sinh[7rn(1 — y>)] (20) Finally, plugging this into the earlier expression for 0(55, y, 90’, y’), 1 G(:c, y, xla 3/) = 8 Z nsinh(7rn) 71:1 sin(7mx) sin(7m:c’) sinh(7my<) sinh[7m(1 — y>)] (21) ...
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