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Differential Equations Lecture Work Solutions 81

# Differential Equations Lecture Work Solutions 81 - 14 ut =...

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Unformatted text preview: 14. ut = k 1 1 (rur )r + 2 uθ θ r r ur (a, θ, t) = 0 inside a disk u (r, θ, 0) = f (r, θ) Θ T R = kT Θ 1 (r R ) + r T = kT 1 (rR r ) + R 1 r2 RΘ 1Θ = −λ r2 Θ 1 (rR r T + λkT = 0 ) R + 1Θ = − λ multiply through by r 2 r2 Θ r (rR ) Θ + λ r2 = − =µ R Θ Θ + µΘ = 0; r (rR ) + λ r 2 R − µ R = 0 Θ (0) = Θ (2 π ) | R(0) | < ∞ Θ (0) = Θ (2 π ) ⇓ R (a) = 0 ↓ µn = n2 Θn = µ0 = 0 √ R = Jn ( λ r ) n = 1, 2, · · · √ Jn ( λ a) = 0 sin n θ cos n θ gives λnm Θ0 = 1 Tnm + λnm kTnm = 0 → Tnm = e−λnm k t ∞ a0 + 2 u(r, θ, t) = m=1 ∞ n=1 Tn m Rn m Θn ∞ f (r, θ) = m=1 a0 + 2 λn m r e−k λn m t (an cos n θ + bn sin n θ) Jn ∞ n=1 (an cos n θ + bn sin n θ) Jn Fourier-Bessel expansion of f. See (4.5 later) 81 λn m r ...
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