M257-316Notes_Lecture12

M257-316Notes_Lecture12 - 8.3. HEAT EQ ON A CIRCULAR RING -...

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8.3. HEAT EQ ON A CIRCULAR RING - FULL FOURIER SERIES Lecture 12 8.3 Heat Eq on a Circular Ring - Full Fourier Se- ries Physical Interpretation: Consider a thin circular wire in which there is no radial temperature dependence. ∂u ∂r =0. u t = α 2 u xx (8.22) BC: u ( L, t )= u ( L, t ) ∂x ( L, t ( L, t ) ) Periodic BC IC: u ( x, 0) = f ( x ) Assume u ( x, t X ( x ) T ( t ). As before: X 0 ( x ) X ( x ) = ˙ T ( t ) α 2 T ( t ) = λ 2 . IVP: ˙ T ( t ) α 2 T ( t ) = λ 2 T ( t c e λ 2 t . r = 2 L 2 π = L π = Constant. The Laplacian becomes Δ u = 2 u 2 + 1 r + 1 r 2 2 u ∂θ 2 = 2 u ( ) 2 (8.23) 57
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Separation of Variables if we let x = we obtain (1.1). BVP: X 0 + λ 2 X =0 X ( L )= X ( L ) X 0 ( L X 0 ( L ) Eigenvalue Problem look for λ such that nontrivial x can be found. X ( x A cos( λx )+ B sin( λx ) X ( L A cos( λL ) B sin( λL A cos( λL B sin( λL X ( L ) therefore 2 B sin( λL )=0 X 0 ( x sin λx + cos( λx ) (8.24) X 0 ( L )=+ sin( λL cos( λL sin( λL cos( λL X 0 ( L ) therefore 2 sin( λL Therefore λ n L =( ) n , 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_Lecture12 - 8.3. HEAT EQ ON A CIRCULAR RING -...

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