M257-316Notes_Lecture30

# M257-316Notes_Lecture30 - Chapter 26 Lecture 30 Wedges with...

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Chapter 26 Lecture 30 Wedges with cut-outs, circles, holes and annuli Example 26.1 A circular wedge with a cut-out: u rr + 1 r u r + 1 r 2 u θθ =0 (26.1) u θ ( r, 0) = 0 u θ ( r, α )=0 u ( b, θ u ( a, θ )= f ( θ ) (26.2) Let u ( r, θ R ( r )Θ( θ ). r 2 ( R 0 + 1 r R ) R ( r ) = Θ 0 ( θ ) Θ( θ ) = λ 2 ½ r 2 R 0 + rR 0 λ 2 R Θ 0 + λ 2 Θ=0 (26.3) Θ equation i Θ 0 + λ 2 Θ 0 (0)=0=Θ 0 ( α ) ¾ Θ= A cos λθ + B sin λθ Θ 0 (0) = B or λ , (26.4) Θ 0 = sin λθ + cos λθ Θ 0 ( α sin λα = α n , 1 ,... (26.5) 151

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Lecture 30 Wedges with cut-outs, circles, holes and annuli R equation i n = 0 :( rR 0 0 ) 0 =0 0 0 = B 0 R 0 = A 0 + B 0 ln r . Note u 0 ( b, θ )= R 0 ( b 0 ( θ )=0 R 0 ( b A 0 + B 0 ln b ,A 0 = B 0 ln b. (26.6) Therefore R 0 = B 0 ln( r/b ). Choose B 0 =1. n 1 : r 2 R 0 n + 0 n λ 2 R n R ( r A n r λ n + B n r λ n R n ( b A n b λ n + B n b λ n B n = A n b 2 λ n (26.7) R n ( r A n [ r λ n b 2 λ n r λ n ] Choose A n (26.8) u n ( r, θ h r ( α ) b 2 ( α ) r ( α ) i cos µ nπθ α (26.9) u 0 ( r, θ )=l n ³ r b ´ · 1 (26.10) Therefore u ( r, θ c 0 ln ³ r b ´ + ± n =1 c n h r ( α ) b ( 2 α ) r ( α ) i cos µ nπθ α (26.11) u ( a, θ f ( θ )=2 ( c 0 ln ( a b )) 2 + ± n =1 c n h a ( α ) b ( 2 α ) r ( α ) i cos µ nπθ α (26.12) = a 0 2 + ± n =1 a n cos µ nπθ α . (26.13) Therefore 2 c 0 ln( a/b 2 α α ² 0 f ( θ ) dθ. (26.14) c n = 2 α h a ( α ) b ( 2 α ) a α i α ² 0 f ( θ ) cos µ nπθ α (26.15) c 0 = 1 α ln( a/b ) α ² 0 f ( θ ) dθ. (26.16) Note: In the special case f ( θ
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## This note was uploaded on 05/04/2011 for the course MATH 25 taught by Professor Lo during the Spring '11 term at BC.

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M257-316Notes_Lecture30 - Chapter 26 Lecture 30 Wedges with...

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