Mathematic Methods HW Solutions 61

Mathematic Methods HW Solutions 61 - 16 n odd the six...

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Unformatted text preview: Chapter 13 61 16 , n odd, the six solutions on (0, π ) are nπ (4 − n2 ) 1. T = bn e−ny sin nx bn sinh n(H − y ) sin nx 2. T = sinh nH 2 3. u = bn e(−nα) t sin nx h ¯ 2 n2 ¯ 4. Ψ = bn sin nx e−iEn t/h , En = 2m 5. y = bn sin nx cos nvt bn 6. y = sin nx sin nvt nv 12(−1)n+1 4.14 Same as 4.12 with bn = , all n, on (0, 1) n3 π 3 4.13 With bn = 5.1 5.2 (a) u ∼ 9.76◦ = (a) 2 J1 (km r ) e−km z sin θ, km = zeros of J1 km J2 (km ) m=1 ∞ 2a J1 (km r/a) e−km z/a sin θ, km = zeros of J1 km J2 (km ) m=1 u(r = 1, z = 1, θ = π/2) ∼ 0.211 = ∞ 200 J0 (km r ) sinh km (10 − z ), km = zeros of J0 (a) u = km J1 (km ) sinh(10km ) m=1 (b) 5.3 (b) u = 5.4 (b) u ∼ 9.76◦ = ∞ ∞ 200 km (H − z ) J0 (km r/a) sinh , k J (k ) sinh(km H/a) a m=1 m 1 m km = zeros of J0 ∞ 2 200 u= J0 (km r/a)e−(km α/a) t , km = zeros of J0 km J1 (km ) m=1 ∞ 5.5 2 200a J1 (km r/a)e−(km α/a) t sin θ, km = zeros of J1 km J2 (km ) m=1 5.6 amn = 5.7 5.8 2π f (r, θ)Jn (kmn r/a) cos nθ r dr dθ 2 πa2 Jn+1 (kmn ) 0 0 a 2π 2 f (r, θ)Jn (kmn r/a) sin nθ r dr dθ bmn = 2 πa2 Jn+1 (kmn ) 0 0 400 nπz 1 nπr u= sin + I0 π nI0 (3nπ/20) 20 20 u = 40 + o dd n ∞ 2 2 120 J0 (km r )e−km α t , where J0 (km ) = 0 k J (k ) m=1 m 1 m 1600 π2 o dd m 6400 5.10 u = 3 π 5.9 a 2 u= n sin(nπx/10) sin(mπy/10) sinh[π (n2 + m2 )1/2 (10 − z )/10] mn sinh[π (n2 + m2 )1/2 ] o dd n nπx mπy pπz −(απ/l)2 (n2 +m2 +p2 )t 1 sin sin sin e nmp l l l o dd n o dd m o dd p −n 5.11 R = r , r , n = 0; R = ln r, const., n = 0 R = r l , r − l− 1 200 r n sin nθ 5.12 u = 50 + π a n o dd n 1 r 4n 400 sin 4nθ 5.13 u = π n 10 o ddn ...
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This note was uploaded on 02/29/2012 for the course MHF 2312 taught by Professor Dr.chet during the Fall '11 term at UNF.

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