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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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 Spring '08
 smith

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