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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 ) ∂u ∂x ( L, t ) = ∂u ∂x ( L, t ) Periodic BC IC: u ( x, 0) = f ( x ) Assume u ( x, t ) = X ( x ) T ( t ). As before: X ( 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 ∂r 2 + 1 r ∂u ∂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 + λ 2 X = 0 X ( L ) = X ( L ) X ( L ) = X ( 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 ( x ) = sin λx + cos( λx ) (8.24) X ( L ) = + sin( λL ) + cos( λL ) = sin( λL ) + cos( λL ) = X ( L ) therefore 2 sin( λL ) = 0 Therefore λ n L = ( ) n = 0 , 1 , . . . .
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M257-316Notes_Lecture12 - 8.3 HEAT EQ ON A CIRCULAR RING...

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