, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. main.tex 2011/08/13( )
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1 , 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,.
2 ζ t + V (ζ + βy) = 0 (4.2.1), V = 0 (4.2.2). (4.2.1), (3.3.66) R 1 Φ / Z, Γ., F 1 ( 3.2 ). 7,., ( )., (4.2.1) 500 hpa., 500 hpa (4.2.1) 1949,., U = U(y).. ζ = ζ(y) + ζ, u = U(y) + u v = v, ζ = dū/dy. (4.2.1), ζ t + U ζ ( ) x + ζ v y + β = 0 (4.2.3). ψ, u = ψ/ y, v = ψ/ x., (4.2.2)., (4.2.3) t 2 ψ + U x 2 ψ + ( ) β d2 U ψ dy 2 x = 0 (4.2.4)., ψ = Ψ(y) e ik(x ct) ( ). (4.2.4), ( ) ( ) d 2 Ψ d 2 (U c) dy U 2 k2 Ψ dy β Ψ = 0 (4.2.5) 2
3 83., y = 0, (y = ±d)., v(±d) = ψ x = ik Ψ(±d) e ik(x ct) = 0 y=±d., Ψ(d) = Ψ( d) = 0 (4.2.6),. (4.2.6) (4.2.5), c. U = U(y), U., (4.2.5) ( d 2 Ψ dy k 2 β ) Ψ = 0 (4.2.7) 2 U c. (4.2.6),. [ ] (2n 1)πy Ψ n = A cos n = 1, 2,... (4.2.8) 2d, [(2n 1)π/2d] 2 = k 2 + β/(u c)., c = U β k 2 + [(2n 1)π/2d] 2 (4.2.9)., 2.5. (4.2.8) Ψ n,. n = 1, d, c = U β/k 2 (4.2.10)., q/ y (2.6.53), (2.6.52). 2, (2.6.52) (2.6.53) (4.2.10)., (2.6.52) L *1., U k q *1 (2.6.52) (2.6.53) c = U β + U/L2 R k 2 (1 + L 2 /L 2 R ) (4.2.11)., L R = (gh) 1/2 /f, L = k 1. U 10 m, L R L 10 6 m, (4.2.10).
4 84 ((2.6.52) ). q = (f + ζ)/h 0 = f/h 0 (H 0 ), q.,, (4.2.10). (4.2.10) Rossby(1939),, (Hough, 1898). U > 0 λ = λ s = 2π(U/β) 1/2,. L < λ s, L > λ s. 500 hpa (4.2.10), (4.2.10). 4.3,. ψ = Ψ(y)e ik(x ct),. c = c r + ic i, Ψ., (4.2.5) Ψ (Ψ ), ) (U c) ( Ψ d2 Ψ dy 2 k2 Ψ Ψ ( d 2 U dy 2 β., Ψ d2 Ψ dy = d ( Ψ dψ ) dψ 2 dy dy dy ) Ψ Ψ = 0 (4.3.12)., 2. ΨΨ = Ψ 2 and dψ dψ dy dy = dψ dy, (4.3.12) U c ±d [ d ( d Ψ dψ ) k 2 Ψ 2 dψ 2] d dy dy dy dy (4.3.13) d (d 2 U/dy 2 β) Ψ 2 = dy U c. Ψ(±d) = 0, Ψ(±d)., Ψ Ψ, d dψ dy
5 85 1. (U c) = (U c r )+ic i, [ d k 2 Ψ 2 + dψ 2] d (d 2 U/dy 2 β)(u c r ) Ψ 2 dy = dy d dy d U c 2 (4.3.14) d (d 2 U/dy 2 β) Ψ 2 ic i dy U c 2 d. 1,, i. c i d d (d 2 U/dy 2 β) Ψ 2 U c 2 dy = 0 (4.3.15) c i 0, c i., d 2 U/dy 2 β d < y < d., y ( y k ) ( ) d 2 U dy β = 0 d < y 2 k < d (4.3.16) y k. Lord Rayleigh(1880), H.L.Kuo(1951),. (4.3.16). d dy ( du dy + f) = 0 or dζ a dy = 0 at y = y k (4.3.17),,.,. (4.2.4) ψ, x, y d [ ψ ψ Uψ t x 2 ψ ( ) ( )] ψ 2 β d2 U dy = 0 (4.3.18) x 2 dy 2 d., ( ) = L 1 L ( ) dx., 0., ψ(x + L, y, t) = ψ(x, y, t) (4.3.19) ψ(x, d, t) = ψ(x, d, t) = 0 (4.3.20)
6 86. (4.3.18), (4.3.19), *2. d d u 2 + v 2 dt d 2 d dy = u v d du dy dy (4.3.21), u = ψ/ y, v = ψ/ x., ψ. ψ(x, y, t) = Φ(y, t) cos[kx θ(y, t)] (4.3.22), Φ, θ. u v. u v = ψ ψ y x = Ψ 2 θ y k sin2 (kx θ) + k Ψ Ψ sin(kx θ) cos(kx θ) y (4.3.23), L 2π/k, u v = k θ Ψ2 (4.3.24) 2 y., u v θ/ y. (4.3.24), (4.3.21) d d u 2 + v 2 dt d 2 dy = k 2 d d Ψ 2 θ du y dy dy (4.3.25), ( θ/ y)(du/dy) < 0 ( ), ( θ/ y)(du/dy) > 0 ( ). 4.4, 2. U = U 0 sech 2 (y/y 0 ), (4.4.26) *2 : (4.3.21).
7 87 U = U 0 tanh 2 (y/y 0 ) (4.4.27). (4.4.26), Bickley (4.4.26) (U 0 > 0),. (U 0 < 0), (4.4.27) d 2 U/dy 2, β = 0, (4.3.16)., ( ) ( ) d 2 U d 2 U > β and < β (4.4.28) dy 2 dy 2 max. U 3 (d 2 U/dy 2 ) max (d 2 U/dy 2 ) min, Bickley, min 2 < b < 2/3 (4.4.29) 4.4.1: U = U 0 sech 2 y/y : U = U 0 tanh y/y 0
8 88., b = βy 2 0/U 0 (4.4.30)., b < 4/(3 3) (4.4.31). (4.4.26) (4.4.27) (4.2.5), b ky 0. Kuo(1973), (4.2.5), (4.2.6) c (4.4.26) (4.4.27) , Kuo(1973), (d ) Bickley (4.4.26) c r /U 0 δ = ky 0 c i /U 0. b > 0, b < (a). (b > 0), U 0 50%., (b < 0), U ((4.2.11) ), b (k 0), c r (b), 2 < b < 2/3. β,., (b )., b , (4.4.27) (Kuo(1973) ). (4.4.31), (a), c r = 0 b = 0,. c r U 0 90%, b c r (b), b = 0, b = 4/(3 3)., b. Bickley.,.., Williams et al(1971) U 0 > 0, b = 0, d = 5y , Bickley (ky 0 = 0.9)
9 : Kuo(1973) U = U 0 sech 2 (y/y 0 ). (a)ky 0 c r /U 0 b ; (b)b ky 0 δ. (H.L.Kuo, Dynamics of Quasi-geostrophic Flows. Academic Press, 1973.)
10 : Kuo(1973) U = U 0 tanh (y/y 0 ). (a)b ky 0 c r / U 0 ; (b)b ky 0 δ. (H.L.Kuo, Dynamics of Quasi-geostrophic Flows. Academic Press, 1973.)
11 : Bickley. (a) P ; (b) θ ( ). (Haltiner,G.J., Williams,R.T., 1979 : NUMER- ICAL PREDICTION AND DYNAMIC METEOROLOGY SECOND EDITION. John Wiley & Sons, 477pp). ψ = P (y) cos(k(x c r t) θ(y)), P (y), θ. y = 0,., (4.5)., , U 0 > 0, b = 0 (ky 0 = 0.45)., y = ±y 0, y = 0., ( ).
12 : (4.4.27) (Haltiner,G.J., Williams,R.T., 1979 : NUMERICAL PREDICTION AND DYNAMIC METE- OROLOGY SECOND EDITION. John Wiley & Sons, 477pp) 4.5,.,., β = 0, (4.2.6) (4.2.5).,. (4.2.3) (β ), U = U 0 + Sy., (4.2.4) 2 ψ t.. + U 2 ψ x = 0 (4.5.32) 2 ψ = F (x Ut) (4.5.33), F, F (x) t = 0.,. v = v 0 sin kx
13 93,, t 2 ψ = ζ = kv 0 cos kx at t = 0 2 ψ = kv 0 cos k(x Ut) (4.5.34).,., ψ = A cos k(x Ut),. 2 ψ = Ak 2 cos k(x Ut) AS 2 t 2 k 2 cos k(x Ut) (4.5.34) A.,., ψ = v 0 cos k(x Ut) k(1 + S 2 t 2 ) u = U ψ y = U v 0kSt sin k(x Ut) k(1 + S 2 t 2 ) v = ψ x = v 0 sin k(x Ut) 1 + S 2 t 2 (4.5.35). ψ v t 2, u t. U = U 0 + Sy,.,.,., y = y, c(y ) = U(y )..,.,,., (4.2.8),. Case(1960), (4.2.5) (4.2.6),.,,.,,,.
14 94 4.6,,.,., (Case, 1960).,,. -,., u v.,,,,.,,.,.,, ( ).
15 95 4.7,., Charney(1947) Eady(1949).,., (3.3.66). ( ) t + k ψ q = 0 (4.7.36), q q = 2 ψ + e Z ( f 2 0 e Z Z Γ ) ψ + βy (4.7.37) Z, (potential vorticity)., ψ = f0 1 Φ, Γ (3.3.47) *3., (3.3.55) ( ) ψ t + k ψ Z + f 0 1 ΓŻ = 0 (4.7.38). Z, T/ z Γ., Z., *4,. ψ,. ψ (x, y, Z, t) = ψ(y, z) + ψ(x, y, Z, t) (4.7.39), ψ / ψ., U = ψ/ y. (4.7.36) (4.7.39), ( t + U ) q + ψ q x x y = 0 (4.7.40) *3 (3.3.47). Z, Γ(Z) = ( ) ( ) ( Φ Z Z + κ Φ T = R Z + κ T = H2 g g + 1 T ) T c p H Z. *4 σ(p),. σ(p) = Γ/p 2
16 96., q = 2 ψ + e Z ( f 2 0 e Z ψ Z Γ Z q y = β 2 U y ( f 2 0 e Z 2 ez Z Γ ), (4.7.41) ) U Z (4.7.42). (4.7.42) ( 2 ) ( 3 )., ( t + U ) ψ x Z ψ U x Z + f 0 1 ΓŻ = 0 (4.7.43)., Ż.,. ψ = Ψ(y, Z)e ik(x ct), (4.7.40) [ 2 Ψ (U c) y + ( ) ] f 2 0 e Z Ψ 2 ez k 2 Ψ Z Γ Z., + q y Ψ = 0 (4.7.44) (U c) Ψ Z U Z Ψ ik 1 f 1 0 ΓW = 0 (4.7.45)., Ż = W (y, Z)e ik(x ct). Charney and Stern(1962) Pedlosky(1964a), (4.3.16)., q/ y U/ Z. 3, (4.3.16), (Holton, 1975 ). 4.8 Eady, Eady(1949).. β = 0 (e Z const)
17 97 (Γ = const) y ( / y = 0) rigid-lid(w = 0 at Z = 1), Z U = SZ (4.8.46)., S.,., e Z, Γ, (4.7.44) (U c) ( f 2 0 Γ ) d 2 Ψ dz 2 k2 Ψ = 0 (4.8.47). U c 0,. ( ) ( ) Z Z Ψ = A sinh + B cosh ε 1/2 ε 1/2 (4.8.48), ε = f 2 0 /k 2 Γ ((3.3.48) ). Z = 0, 1 W = 0, (4.7.45), (U c) dψ dz SΨ = 0 at Z = 0, 1 (4.8.49). (4.8.48) Z, (4.8.49) cε 1/2 A SB = 0, (S c)ε 1/2 [A cosh(ε 1/2 ) + B sinh(ε 1/2 )] S[A sinh(ε 1/2 ) + B cosh(ε 1/2 )] = 0 (4.8.50). A, B,.,. c 2 Sc + S 2 (ε 1/2 coth ε 1/2 ε) = 0 (4.8.51) c, c = S 2 ± S [ 1/4 (ε 1/2 coth ε 1/2 ε) ] 1/2 = S {[ ( )] [ ε 1/2 ε 1/2 ε 1/2 2 ± Sε1/2 tanh ( )]} ε 1/2 1/2 (4.8.52) coth 2
18 98., ε 1/2 < 1.20 (4.8.53) 2,. c,. (3.3.48) ε = L 2 /L 2 R ( L R = Γ 1/2 /f 0 ), (4.8.53). λ 2π = L > L R 2.40, kc i = ± S {[ ( )] [ ( ) ]} ε 1/2 ε 1/2 ε 1/2 tanh coth ε 1/2 L R (4.8.54) (4.8.55)., S., L (4.9), S 0. L, L = L R (4.8.56) 1.61., L L R ε 1. ε 1,., c r = S 2 (4.8.57),. Z = 1/2, Z = 1/2 *5.,. *5, Eady.,. (4.8.50), (4.8.48) B, Ψ(Z) Ψ = A [{ sinh ( Z ε ) c r S ε cosh ( Z ε )} c i i S ε cosh ( Z ε., Ψ Ψ kθ (, Ψ(Z) = Ψ(Z) e kθ ). ( ) Z Ψ(Z) 2 = A {sinh 2 c ( )} 2 ( ) r Z ε S ε cosh + A 2 c2 i Z ε S 2 ε cosh2 ε tan kθ = c ( ) [ ( ) i Z Z S ε cosh sinh c ( )] 1 r Z ε ε S ε cosh ε. A. )]
19 99, (4.8.49). Ψ = 0,. Eady,., k 1. Pedlosky(1964b),. Charney Charney(1947) Eady, Eady. Charney,.,., (4.7.42), q y = β + f 2 0 S/Γ (4.8.59), Γ. (S > 0),., Eady., (4.7.44). (SZ c) [ f 2 0 Γ ( d 2 Ψ dz dψ 2 dz ) ] k 2 Ψ + (β + f 0 2 S )Ψ = 0 (4.8.60) Γ q/ y, (4.8.47).,., r = Γ(β + Sf 2 0 /Γ) f 2 0 S(1 + 4Γf 2 0 k 2 ) 1/2 (4.8.61)., c. Gambo(1950), Kuo(1952), Green(1960), Burger(1962), Kuo(1973) ,, ψ, ψ(x, Z, t) = Ψ(Z) e kc it exp [ik(x c r t + θ(z))] (4.8.58). (4.8.53), Ψ θ, Z = 1/2, Z = 1/2.
20 : Green(1960) Charney., Z = 1. (Green, J.S.A, 1960 : A Problem in Baroclinic Stability. Royal Meteorological Socitey) Green(1960), Z = 1. 1/k, S., kc i. ( ), r = 1 Charney, Gambo, Kuo.. Burger, r, r. 2, r = 2, 3.,., 0.1,. S, Eady.
21 Charney, , 1,3, 2. 2, Charney.,. (3.3.76) (3.3.77) : 2
22 102 ( ) t + k ψ ( 2 ψ ω + βy) f 0 p = 0 (4.9.62) ( ) ψ t + k ψ p + σ ω = 0 (4.9.63) f 0 2, (4.9.62) (4.9.63), 2.,. (4.9.62) (4.9.63),. ψ = U(p)y + ψ(x, p, t) ω = ω (x, p, t) (4.9.64), U(p) y. (4.9.64) (4.9.62) (4.9.63), ( t + U ) 2 ψ x x + β ψ 2 x f ω 0 p = 0 (4.9.65) ( t + U ) ψ x p du ψ dp x σ ω = 0 (4.9.66) f 0. (4.9.64), ψ ω. p = p s p = p T ω = 0. p = p T,,., (4.9.65) 1, 3, ( ) t + U 2 ψ 1 1 x x + β ψ 1 2 x f ω 0 p = 0 (4.9.67) ( ) t + U 2 ψ 3 3 x x + β ψ 3 2 x + f ω 0 p = 0 (4.9.68) (4.9.69)., (4.9.66) 2., 2 ψ U 1,3 ψ U, [ t + ( ) ] U1 + U 3 (ψ 1 ψ 3 ) 2 x ( U1 U 3 2 ) ( ψ1 x + ψ 3 x ) ( pσ2 f 0 ) ω 2 = 0 (4.9.70)
23 103. (4.9.67), (4.9.68), (4.9.70), 3 ψ 1, ψ 3, ω 2.,. (4.9.67) ψ 1, (4.9.68) ψ 3 2, x, [ ] d ( ψ1 / x) 2 + ( ψ 3 / x) 2 dt 2 = f 0 p ω 2(ψ 1 ψ 3 ) (4.9.71).., (4.9.70) λ(ψ 1 ψ 3 ), x. [ ] d λ(ψ1 ψ 3 ) 2 dt 4 = λ ( 4 (U ψ1 1 U 3 ) (ψ 1 ψ 3 ) x + ψ ) 3 x + f 0 p ω 2(ψ 1 ψ 3 ) (4.9.72), λ = 2f 2 0 /( p 2 σ 2 ) (4.9.73). ψ 1 ψ 3, *6. (4.9.72), *6, Φ p = RT p., Φ. Φ = Φ(p) + Φ (x, p, t), Φ p = RT p., Φ = f 0 ψ. 1 3, T 2 = f 0 ψ 1 ψ 3 R ln(p 3 /p 1 )., 2 T p3 T p 2 = 1 (T/p) dp p3 (4.9.74) dlnp p 1, ψ 1 ψ 3, 1 3 T 2.
24 104,. (4.9.71), (4.9.72) 2,., 2 (ψ 1 ψ 3 > 0) (ω 2 < 0), (ψ 1 ψ 3 < 0) (ω 2 > 0).,. U 1 U 3,. (4.9.72) 1, (U 1 U 3 )., 2 (( ψ 1 / x + ψ 3 / x)/2 > 0) (ψ 1 ψ 3 > 0), (( ψ 1 / x + ψ 3 / x)/2 < 0) (ψ 1 ψ 3 < 0),., U 1 U 3.,. ψ 1 = Ψ 1 e ik(x ct) ψ 3 = Ψ 3 e ik(x ct) (4.9.75) ω 2 = ( pf0 1 )W e ik(x ct) (4.9.67), (4.9.68), (4.9.70), k[β k 2 (U 1 c)]ψ 1 + iw = 0 k[β k 2 (U 3 c)]ψ 3 iw = 0 k[u 3 c]ψ 1 k[u 1 c]ψ 3 + i2w/λ = 0 (4.9.76a) (4.9.76b) (4.9.76c). Ψ 1, Ψ 3, W,. c 2. 2k 2 (k 2 + λ)c 2 + [2β(2k 2 + λ) 2k 2 (k 2 + λ)(u 1 + U 3 )]c + 2β 2 β(2k 2 + λ)(u 1 + U 3 ) + 2k 4 U 1 U 3 + k 2 λ(u U 2 3 ) = 0 2 c, c = (U 1 + U 3 )/2 β(2k2 + λ) 2k 2 (k 2 + λ) ± [λ2 β 2 k 4 (λ 2 k 4 )(U 1 U 3 ) 2 ] 1/2 2k 2 (k 2 + λ) (4.9.77). (4.9.77) c,., k 4 λ 2 (U 1 U 3 ) 2
25 105. c,. (k 1 ) (U 1 U 3 ),.. (U 1 U 3 ) 2 λ 2 β 2 = (4.9.78) k 4 (λ 2 k 4 ) λ β,, n = 0.,. k 1 = λ 1/2 = pσ 1/2 2 /(f 0 / 2) (4.9.79) 4.9.9: λ = m 2, β = m 2 s 1 2 n = k c i. n U 1 U 3 k 1., k 1, U 1 U 3, n 10 6 m, ms 1, 10 5 s 1.
26 106, k 2 λ, (4.9.78). U 1 U 3 = β/k 2 (4.9.80), (4.9.78). (4.9.78) k 1, k 1 = 2 1/4 λ 1/2, U 1 U 3 (U 1 U 3 ) min = 2β/λ = β p2 σ 2 f 2 0 (4.9.81). 2 (two level model),, 2 2 (two layer model)(phillips, 1951). (4.9.79),.,, (4.9.80),., β.,,., β = 0 k 1 > σ 1/2 2 /(f 0 2) U1 U 3 (4.9.82)., Eady k 1 > Γ/(f 0 2.4) S = du/dz 0 (4.9.83).,. β 0, 2 ( 4.9.9), Charney ( 4.8.7).,,.,.,., , (4.9.78)., 2., km.
27 , (4.9.71) (4.9.72). (4.9.77) 2, c +, c. 2.6,, ( ). 2, [ ] ψ 1 = Re ψ 3 = Re ω 2 = pf 1 0 Re a + Ψ + 1 e ik(x c+ t) + a Ψ 1 e ik(x c t) [ a + Ψ + 3 e ik(x c+ t) + a Ψ 3 e ik(x c t) [ a + W + e ik(x c+ t) + a W e ik(x c t) ] ] ( a) ( b) ( c)., Re. ψ = Ψ e iθ, ( ) e ix = cos x + i sin x ( ). ω 2, ψ 1 ψ 3, ( c). c + c, (4.9.76). (4.9.76a) (4.9.76b), Ψ ± 1 = c± (U 3 β/k 2 ) c ± (U 1 β/k 2 ) Ψ± 3 ( )., (4.9.76b). W ± = ik 3 [c ± (U 3 β/k 2 )]Ψ ± 3 ( ) ψ 1 ψ 3, a + a ( a) ( b) , c ±, c + > c., ( ), Ψ ± 1 Ψ± , ( ), 90.
28 108, (4.9.71) (4.9.72), *7. ( ) 2,., (4.9.77) β/k 2., (4.9.77) c ± = U 1 + U 3 2 ± 1 2 [ ] k 2 1/2 λ k 2 + λ (U 1 U 3 ) 2 ( )., ( ) Ψ + 1 /Ψ + 3 > 1, Ψ 1 /Ψ 3 < 1 ( ).,,. W = W e iθ (θ ), Ψ 3 > 0, ( ) ( ) θ = π/2., (ω < 0), (ω > 0)., 4.8 Eady.,. U 1 U 3, (4.9.77) c + = (U 1 + U 3 )/2 β/(k 2 + λ) c = (U 1 + U 3 )/2 β/k 2 ( ). c, ( ) ( ) W = 0 Ψ 1 = Ψ 3, ((4.2.10) )., c +, ( ) ( ) W 0 Ψ + 1 = Ψ + 3,., (U 1 U 3 ) 2, ,. c ± = c r ± in/k *7 Ψ + 1 Ψ+ 3 ( Ψ 1 Ψ 3 ) 180, Ψ+ 1 Ψ 3 ( Ψ 1 Ψ+ 3 )., Ψ+ 1, Ψ+ 3 W ( Ψ 1, Ψ 3 W + ) 90. ( ) 2,.
29 109, c r = (U 1 + U 3 )/2 β(2k 2 + λ)/[2k 2 (k 2 + λ)] n = [k 4 (λ 2 k 2 )(U 1 U 3 ) 2 λ 2 β 2 ] 1/2 /[2k(k 2 + λ)] ( ), n., ψ 1,3 = Re [ a + Ψ + 1,3e ik(x crt) e nt + a Ψ 1,3e ik(x crt) e nt] ( ). a + = 0,,.,. c, ( ) ( ),. ( ), ( ) ( ). Ψ ± 1 = {c r [(U 1 + U 3 )/2 β/k 2 ]} 2 (U 1 U 3 ) 2 /4 + n 2 /k 2 in(u 1 U 3 )/k [c r (U 1 β/k 2 )] 2 + n 2 /k 2 Ψ ± 3 ( ) W ± = k 3 {±n/k i[c r (U 3 β/k 2 )]}Ψ ± 3 ( ) ( ),. 2, *6. T 2 = f 0 (ψ 1 ψ 3 )/[R ln(p 3 /p 1 )] ( ),. f 0 = 10 4 s 1, β = m 1 s 1, 2π/k = 4000 km, λ = m 2, p s = 1000hPa, p = 400hPa, U 1 U 3 = 20 ms 1, a = 0, a + Ψ + 3 = 10 ms 1 /k, 3. ( ), ( ) ψ 1 ψ 3. ψ 1 = (14.2 ms 1 )k 1 cos[k(x c r t) + 64 ] e nt ψ 3 = (10 ms 1 )k 1 cos[k(x c r t)] e nt ( ),, ( c), ( ). ω 2 = hpa s 1 cos[k(x c r t) ] e nt ( )
30 110, ψ 1, ψ 3 ( ), T 2 = (4.24 K) cos[k(x c r t) ] e nt ( )., y k. ( ), ( ), ( ), : ( ), ( ), ( ).
31 111. ψ 1 ψ 3, T, R.., ( ) 1, *8., W C.,,.,., 2.,.,, ,,. (4.9.71) (4.9.72), ( ), ( ), ( ).,,,.,.,.,. Oort(1964), , K, P., K ( ), P.,. { P P }, {P K }, (4.9.72), (4.9.71)., {K K} (4.3.21). { P K ω T, , Q P, P K.,.,,,. K K *8 β = 0, ( ) 1.
32 : Oort(1964)., J m 2., Wm 2. (Oort, A.H., 1964: On Estimates of the Atmospheric Energy Cycle., American Meteorological Society )
33 113, 4.6. P K.,.,.,,.,,.,. (4.9.62), (, ) ( ω/ p > 0),.,.,. Eady Charney,. Gall(1976), Simmons and Hoskins(1977),., 2000 km.,, 4000 km.,. Staley and Gall(1977),. Gall, etal(1979),.,.
34 ,.,,. (2.4.11), ρ 1/2,.,,.,.,., Eliassen and Palm(1960), Charney and Drazin(1961). c, (4.7.43), (U c) 2 ψ x Z ψ U x Z + f 0 1 ΓŻ = 0 ( )., ψ x, ψż = f 0Γ 1 (U c) ψ ψ x Z ( ). ψ Φ,., ψ/ x ψ/ Z. ( ),,.,. U(Z),., Charney and Drazin(1961), (4.7.40). U y, ψ = Ψ(Z) cos αy e ik(x ct) (4.7.40), [ (U c) e Z ( ) ] f 2 0 e Z Ψ (k 2 + α 2 )Ψ + q Z Γ Z y Ψ = 0 ( )
35 115.,., q y = β ( ) f 2 0 e Z U ez Z Γ Z Q = f 0 (e Z /Γ) 1/2 Ψ ( ), ( )., ( )., d 2 Q dz 2 + n2 Q = 0. ( ) n 2 = Γ f 2 0 (k 2 + α 2 ) Γ 1/2 Z/2 d2 e dz 2 (e Z/2 Γ 1/2 Γ q/ y ) + f0 2 (U c) ( ). ( ), n(z).,. n 2 > 0 : Q(Z) Z,. n 2 < 0 : Q(Z) Z,., U Γ,., ( ), n 2 = Γ [ k 2 + α 2 + f ] 0 2 f0 2 4Γ β ( ) (U c), n 2., ( ) e inz e inz. n,, e Z/2 (( ) ). n, e n Z e n Z.,, (e n Z ). Charney and Drazin(1961),, ( ). ( ) c = 0, (n 2 > 0), 0 < U < β k 2 + α 2 + f 2 0 /4Γ ( )
36 116.,,,.,, ( ) 38 m/s.. ( ), U Z,.,,., Matsuno(1970),,. ( ) ( k),. Charney and Pedlosky(1963),. Holton(1975),.,., 7 10.
37 ,.,.,.,., Matsuno(1966a).,, (f = βy).,., φ = gh, Φ = gh. u t + φ βyv = 0, x v t + φ + βyu = 0, ( ) y ( φ u + Φ x + v ) = 0. y
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Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT
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4 1 1.1 ( ) 5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 da n i n da n i n + 3 A ni n n=1 3 n=1
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
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β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E
chap03.dvi
99 3 (Coriolis) cm m (free surface wave) 3.1 Φ 2.5 (2.25) Φ 100 3 r =(x, y, z) x y z F (x, y, z, t) =0 ( DF ) Dt = t + Φ F =0 onf =0. (3.1) n = F/ F (3.1) F n Φ = Φ n = 1 F F t Vn on F = 0 (3.2) Φ (3.1)
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Note.tex 2008/09/19( )
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II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j
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) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4
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1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev
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145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2
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微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
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1 13 Fall Semester N. Yamada Version:13.9.3 Chapter. Preliminalies (1 3) Chapter 1. (4 16) Chapter. (17 9) Chapter 3. (3 49) Chapter 4. (5 63) Chapter 5. (64 7) Chapter 6. (71 8) 11, ISBN 978-4-535-618-4.
IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................
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量子力学 問題
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5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
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27 3 27 ) Ising Landau Synergetics Fokker-Planck F-P Landau F-P Gizburg-Landau G-L G-L Bénard/ Hopfield H.Haken Synergetics 2nd (1978) (1) Ising m T T C 1: m h Hamiltonian H = J ij S i S j h i S
(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x
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