M257-316Notes_Lecture27

# M257-316Notes_Lecture27 - Chapter 23 Lecture 27 - Problem...

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Chapter 23 Lecture 27 - Problem (1B) 23.1 Problem (1B): Δ u = u xx + u yy =0 (23.1) 0= u ( x, 0) = u ( x, b )= u (0 ,y ); u ( b, y g 2 ( y ) (23.2) Let u ( x, y X ( x ) Y ( y ) (23.3) X 0 ( x ) X ( x ) = Y 0 ( y ) Y ( y ) = ± λ 2 . (23.4) Since we have homogeneous BC at y = 0 and y = b we want the function Y ( y ) to behave like sines and cosines. So we choose const = + λ 2 X 0 λ 2 X Y 0 + λ 2 Y X = c 1 cosh λx + c 2 sinh λx Y = A cos( λx )+ B sin( λx ) (23.5) u ( x, 0) = X ( x ) Y (0) = 0 Y (0) = 0 Y (0) = A (23.6) u ( x, b X ( x ) Y ( b )=0 Y ( b Y = B sin( λb n = b n =1 , 2 ,... (23.7) Y n = sin ³ nπy b ´ 135

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Lecture 27 - Problem (1B) u (0 ,y )= X (0) Y ( y )=0 X (0) = c 1 =0. Therefore X n ( x c 2 sinh ³ nπx b ´ . Therefore u n ( x, y ) = sin ³ nπy b ´ sinh ³ nπx b ´ satisfy the homogeneous BC. Therefore u ( x, y X n =1 c n sin h ³ nπx b ´ sin ³ nπy b ´ . Now to satisfy the inhomogeneous BC g 2 ( y u ( a, y X n =1 c n sin h ³ nπa b ´ | {z } b n sin ³ nπy b ´ (23.8) where c n sinh ³ nπa b ´ = 2 b b Z 0 g 2 ( y ) sin ³ nπy b ´ dy. (23.9) 23.2 Summarizing: u ( x, y X n =1 c n sinh ³ nπx b ´ sin ³ nπy b ´ ; c n = 2 b sin h ( nπa b ) b Z 0 g 2 ( y ) sin ³ nπy b ´ dy. (23.10) Example 23.1
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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_Lecture27 - Chapter 23 Lecture 27 - Problem...

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