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 ( x ) X ( x ) = Y ( y ) Y ( y ) = λ 2 (24.4) Y ( y ) + λ 2 Y ( y ) = 0 Y (0) = 0 = Y ( b ) Y = A cos λy + B sin λy Y = sin λy + cos λy (24.5) 141

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Lecture 28 - Neumann Problem - only ﬂux BC and Circular domains Y (0) = λB = 0 λ = 0 or B = 0 . (24.6) Y ( b ) = sin λb = 0 λ n = ( nπ/b ) n = 0 , 1 , . . . Y n = cos nπy b , Y 0 = 1 (24.7) X n λ 2 X n = 0 (24.8) X n (0) = 0 (24.9) n = 0: X 0 = 0, X 0 = c 0 x + D 0 X 0 = c 0 X 0 (0) = c 0 = 0. Choose D 0 = 1: X 0 = 1 n 1 X n = c n cosh( λ n x ) + D n sinh( λ n x ) X n = c n λ sinh( λ n x ) + D n λ cosh( λ n x ) X n (0) = λ n D n = 0 (24.10) Choose c n = 1: 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 = 1 satisfy homog. BC.(24.11) Therefore u ( x, y ) = A 0 + n =1 A n cosh nπx b cos nπy b . (24.12) Now f ( y ) = u x ( a, y ). u x ( x, y ) = n =1 A n b sinh nπx b cos nπy b (24.13) u x ( a, y ) = n =1 A n b sinh nπa b cos nπy b = f ( y ) . . . (24.14) This is like a Fourier Cosine Series for f ( y ) but without the constant term a 0 .
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