Differential Equations Lecture Work Solutions 119

Differential Equations Lecture Work Solutions 119 - 1 d. Θ...

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Unformatted text preview: 1 d. Θ + µΘ = 0 Z − λZ = 0 Z (0) = Z (H ) = 0 ⇓ r (r R ) + (λ r 2 − µ) R = 0 | R (0) | < ∞ ⇓ ⇓ n− 1 2 π H Zn = sin n − 1 2 π H λn = − as before ∞ ∞ u (r, θ, z ) = m=0 n=1 ∞ ur (a, θ, z ) = ∞ m=0 n=1 2 z (anm cos m θ + bnm sin m θ) sin n − Rnm = Im n− 1 2 π z Im H n− 1π r 2H 1 2 (anm cos m θ + bnm sin m θ) sin n − 1 2 π z· H n− Im π H r 1π 2H n− 1π a 2H Since ur (a, θ, z ) = γ (z ) is independent of θ, we must have no terms with θ in the above expansion, that is bnm = 0 for all n, m and anm = 0 for all n, m ≥ 1. Thus a1 0 = 0 γ (z ) = a1 0 sin a1 0 = H 0 π 2H I0 π π π z I0 a 2H 2H 2H π z dz 2H H 2π 0 sin 2H z γ (z ) sin π a 2H dz And the solution is u(r, θ, z ) = a1 0 sin π π z I0 r 2H 2H 119 ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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