M257-316Notes_Lecture28

M257-316Notes_Lecture28 - Chapter 24 Lecture 28 - Neumann...

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Chapter 24 Lecture 28 - Neumann Problem - only flux BC and Circular domains Example 24.1 Eg. 5 Neumann Problem: u xx + u yy =0 , 0 <x<a 0 <y<b (24.1) u x (0 ,y )=0 u x ( a, y )= f ( y ) (24.2) u y ( x, 0 )=0= u y ( x, b ) . (24.3) Let u ( x, y X ( x ) Y ( y ). X 0 ( x ) X ( x ) = Y 0 ( y ) Y ( y ) = λ 2 (24.4) Y 0 ( y )+ λ 2 Y ( y Y 0 (0) = 0 = Y 0 ( b ) ¾ Y = A cos λy + B sin λy Y 0 = sin λy + cos λy (24.5) 141
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Lecture 28 - Neumann Problem - only flux BC and Circular domains Y 0 (0) = λB =0 λ =0or B . (24.6) Y 0 ( b )= sin λb λ n =( nπ/b ) n , 1 ,... Y n = cos ³ nπy b ´ ,Y 0 =1 (24.7) X 0 n λ 2 X n (24.8) X 0 n ( 0 )=0 (24.9) n =0: X 0 0 =0, X 0 = c 0 x + D 0 X 0 0 = c 0 X 0 0 (0) = c 0 =0. Choose D 0 =1: X 0 n 1 X n = c n cosh( λ n x )+ D n sinh( λ n x ) X 0 n = c n λ sinh( λ n x D n λ cosh( λ n x ) X 0 n (0) = λ n D n (24.10) Choose c n X n = cosh( λ n x ). Thus u n ( x, y X n Y n = cosh( λ n x ) cos( λ n y ) u 0 ( x, y X 0 Y 0 ¾ satisfy homog. BC.(24.11) Therefore u ( x, y A 0 + X n =1 A n cosh ³ nπx b ´ cos ³ nπy b ´ . (24.12) Now f ( y u x ( a, y ). u x ( x, y X n =1 A n ³ b ´ sinh ³ nπx b ´ cos ³ nπy b ´ (24.13) u x ( a, y X n =1 n A n ³ b ´ sinh ³ nπa b ´o cos ³ nπy b ´ = f ( y ) ...
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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_Lecture28 - Chapter 24 Lecture 28 - Neumann...

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