Mathematic Methods HW Solutions 63

Mathematic Methods HW Solutions 63 - Chapter 13 63 2 2 2 2...

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Unformatted text preview: Chapter 13 63 2 2 2 2 7.21 ψn (x) = e−α (x +y +z )/2 Hnx (αx)Hny (αy )Hnz (αz ), α = mω/¯ , h 1 1 1 3 En = (nx + 2 + ny + 2 + nz + 2 )¯ ω = (n + 2 )¯ ω. h h (n + 2)(n + 1) , n = 0 to ∞. Degree of degeneracy of En is C (n + 2, n) = 2 M e4 +1 2r 7.22 Ψ(r, θ, φ) = R(r)Ylm (θ, φ), R(r) = rl e−r/(na) L2l−l−1 na , En = − 2 2 n 2¯ n h 8.3 8.4 The second terms in (8.20) and (8.21) are replaced by al r l −qR/a −q Pl (cos θ) = R2l+1 l r2 − 2(rR2 /a) cos θ + (R2 /a)2 Image charge -qR/a at(0, 0, R2 /a) LetK = line charge per unit length. Then V = −K ln(r2 + a2 − 2ra cos θ) + K ln a2 − K ln R2 + K ln r2 + R2 a 2 −2 8.5 K at (a, 0), −K at (R2 /a, 0) 9.2 u= 9.4 200 u(x, t) = π R2 r cos θ a 9.7 200 π ∞ 0 k −2 (1 − cos 2k )e−ky cos kx dk ∞ 0 u(x, t) = 100 erf 10.1 T = ∞ 2 π 1 ∞ 1 − cos k −k2 α2 t e sin kx dk k x x−1 √ − 50 erf √ − 50 erf 2α t 2α t x+1 √ 2α t 1 sinh nπ (2 − y ) sin nπx n sinh 2nπ 2 1 10.2 T = sinh nπy sin nπx π 1 n sinh 2nπ 1 4 1 10.3 T = (2 − y ) + 2 sinh nπ (2 − y ) cos nπx 4 π n2 sinh 2nπ o dd n nπy 40 nπx 1 10.4 T = 20 + 3nπ sinh 5 sin 5 π n sinh 5 o dd n 1 40 nπ (5 − x) nπy sinh sin π 3 3 n sinh 5nπ 3 o dd n 80 1 −(nπα/l)2 t nπx 10.5 u = 20 − e sin π n l + o dd n ∞ 80 (−1)n −[(2n+1)πα/(2l)]2 t 2n + 1 e cos πx π n=0 2n + 1 2l 4 1 1 sinh nπy cos nπx 10.7 u = y + 2 2 sinh 2nπ 4 π n 10.6 u = 20 − o dd n 10.8 u = 20 − x − 40 π ∞ 2 even n 2 1 −(nπα/10)2 t nπx e sin n 10 8l nπvt nπx 1 cos sin 3 3 π n l l o dd n mπr mπz 1 1600 sin nθ sin In 10.10 u = 2 π nmI n (3mπ/20) 20 20 10.9 y = o dd n o dd m ...
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