Trapezoidal Rule θ = 1/ x n x n 1 t = 1 [f(t n 1, x n 1 ) + f(t n, x n )] (6) 1. dx dt = f(t, x), x(t 0) = x 0 (7) t [t 0, t 1 ] f t [t 0, t 1 ], x x
|
|
|
- あつみね ひのと
- 5 years ago
- Views:
Transcription
1 University of Hyogo d x(t) =f(t, x(t)), dt (1) x(t 0 ) =x 0 () t n = t 0 + n t x x n n x n x 0 x i i = 0,..., n 1 x n x(t) θ 1 θ x n x n 1 t = θf(t n 1, x n 1 ) + (1 θ)f(t n, x n ) () Euler Euler θ = 1 Euler Euler Euler x n x n 1 t = f(t n 1, x n 1 ) (4) Euler Euler θ = 1 Euler x n x n 1 t = f(t n, x n ) (5) x n x n 1
2 Trapezoidal Rule θ = 1/ x n x n 1 t = 1 [f(t n 1, x n 1 ) + f(t n, x n )] (6) 1. dx dt = f(t, x), x(t 0) = x 0 (7) t [t 0, t 1 ] f t [t 0, t 1 ], x x 0 b M t 1 t 0 < b/m t, x, x Lipshitz f(t, x) f(t, x ) L x x (8) N t = t 1 t 0 Euler (x(t)) 1 C x(t n ) x n C t (9) Proof. Euler x n+1 = x n + tf(t n, x n ). (10) (7) t n t n+1 tn+1 x(t n+1 ) = x(t n ) + f(t, x)dt = x(t n ) + tf(t n, x(t n )) + O( t ), (11) t n (f(t) F (t) F (t + t) F (t) = tf(t) + O( t )) x(t n+1 ) x n+1 = x(t n ) x n + t[f(t n, x(t n )) f(t n, x n )] + O( t ). (1) Lipshitz x(t n+1 ) x n+1 x(t n ) x n + t f(t n, x(t n ) f(t n, x n ) + A t. (1) f(t n, x(t n ) f(t n, x n ) L x(t n ) x n (14) e n x(t n ) x n e n+1 e n + tl e n + A t = (1 + tl) e n + A t (15) e 0 = 0 n 1 e n A t (1 + tl) i = A t L [(1 + tl)n 1] (16)
3 (1 + tl) N = exp(l(t 1 t 0 )) (17) x(t n ) x n A t L [exp(l(t 1 t 0 )) 1] (18) 1. (local truncation error) t n 1 x n 1 = x(t n 1 ) t n t n x(t n ) x n e = x(t n ) x n (19) Scheme L(x n 1,, x n k )x n = 0 (0) e = Lx(t n ) (1) Scheme e = O( t m ) or e A t m () Scheme m m ( ) m 1 Euler x(t n+1 ) Taylor x(t n+1 ) = x(t n + t) = x(t n ) + t dx dt + t d x dt + O( t ) () dx/dt = f e Euler = O( t ). (4) Euler 1 Euler 1 Fractional Step Method 1 t n 1 x n 1 f(t, x(t)) x n
4 .1 Runge-Kutta R- (stage) Runge-Kutta R k (r) = f(t n 1 + ta r, x n 1 + t b rs k (s) ) (r = 1,,, R) (5) s=1 R x n = x n 1 + t c r k (r). (6) r=1 a r, b rs, c r R a r = b rs, r = 1,,, R (7) ( X = s=1 ) ( t, F (X) = x(t) ) 1 (8) f(t, x(t)) (5),(6) (consistency condition) (5),(6) R c i = 1. (9) i=1 1. b rs (s r). b rs (s > r) k (1), k (),. R R R 10 m m R R Euler a 1 = b 11 = 0, c 1 = 1 Euler a 1 = b 11 = 1, c 1 = 1 1 Runge-Kutta 1. a = b 1 = 1/, c = 1 k (1) = f(t n 1, x n 1 ) (0) k () = f(t n 1 + t, x n 1 + t k(1) ) (1) x n = x n 1 + tk () () 4
5 . a = b 1 = 1, c 1 = c = 1/ 1 k () = f(t n 1 + t, x n 1 + t f(t n 1, x n 1 )) = f + t = f + t x(t n ) Taylor k (1) = f(t n 1, x n 1 ) () k () = f(t n 1 + t, x n 1 + tk (1) ) (4) x n = x n 1 + t (k(1) + k () ) (5) f t + t f f x + 1 ( t f 4 t + t 4 f f x t + t 4 f f x ( f t + f f ) ( + t f x 8 t + f f x t + f f x ) + O( t ) ) + O( t ) (6) x(t n ) = x(t n 1 ) + t dx dt + t d x dt + t d x 6 dt + O( t4 ) (7) dx = f dt d x dt = df dt = f t + f f x d x dx = d f dt = d ( f dt t + f f ) x = ( f t t + f f ) ( f + x t + f f x = f f + (f + ) f t t x + f f t x + f ) f x + f ( f x t + f f ) x ( f x (8) (9) (40) (41) ) + f f x (4) e rk = x(t n+1 ) x n+1 = O( t ) (4) k (1) + k () = f + t 1 O( t ) ( f t + f f ) ( + t ) f x 4 t + f f x t + f f x + O( t ) (44) 4 4 Runge-Kutta Runge-Kutta 5
6 (a 1 = 0, a = b 1 = 1/, a = b = 1/, a 4 = b 4 = 1, c 1 = c = c = c 4 = 1/) 4 k (1) = f(t n 1, x n 1 ) (45) k () = f(t n 1 + t, x n 1 + t k(1) ) (46) k () = f(t n 1 + t, x n 1 + t k() ) (47) k (4) = f(t n 1 + t, x n 1 + tk () ) (48) x n = x n 1 + t 6 [k(1) + k () + k () + k (4) ] (49) e rk4 = x(t n ) x n = t5 ( ) (50) 880 ( )4 4 Runge-Kutta Gill (Runge-Kutta-GIll ) do i = 1, 4 u = f(x) x = x + t (P i u + Q i v) v = R i u + S i v end do ( = ) u, v Runge-Kutta ((45)-(48) ) Runge-Kutta P 1 = b 1 P = b, P 1 + Q R 1 = b 1 + Q R 1 = b 1 (5) P = b 4, P + Q R = b + Q R = b 4, (5) (51) P 1 + Q R 1 + Q S R 1 = b 1 + Q S R 1 = b 41 (54) P 4 = c 4, P + Q 4 R = b 4 + Q 4 R = c, (55) P + Q R + Q 4 S R = b 4 + Q 4 S R = c, (56) P 1 + Q R 1 + Q S R 1 + Q 4 S S R 1 = b 41 + Q 4 S S R 1 = c 1 (57) Runge-Kutta (51)-(57) (b 41 b 1 )(c b 4 ) = (b 4 b )(c 1 b 41 ). (58) 6
7 a, b, c a 1 = 0, a = 1, b 1 = 1 a = 1, b 1 = c 1 6c, b = 1 6c a 4 = 1, b 41 = 0, b 4 = 1 c, b 4 = c (59) c 1 = 1 6, c = c, c, c 4 = 1 6 c 4 Runge-Kutta a 4 = 1 c = + 6 a 1 = 0, a = 1, b 1 = 1 a = 1, b 1 = 1+, b = a 4 = 1, b 41 = 0, b 4 =, b 4 = + c 1 = 1 6, c = 6, c = + 6, c 4 = 1 6 (60) a, b, c 8 (Q, Q, Q 4, R 1, R, R, S, S ) (51)-(57) 5 R 1 = 1 Q = b 1 b 1 Q = b 4, Q 4 = c 4 P 1 = 1 P = P = + P 4 = 1 6 Q 1 = 0 Q = Q = + Q 4 = 1 R 1 = 1 R = R = + R 4 = 0 S 1 = 0 S = 4 S = +4 S 4 = 0 (61) i = 1 v Q 1 = S 1 = 0 v R 4 = S 4 = 0 Q 1, S 1 Linear Multistep Method ( x f(t, x(t)) x n k β 0 = 0 (8) (6) α i x n i = t β i f(t n i, x n i ) (α 0 0) (6) α i t n i = t β i (6) i i α i x n i = t β i f(t n i, x n i ) (64) i i 7
8 t n t n 1 = t t n α i = t (β i + iα i ) (65) i i t n, t α i = 0 (66) iα i = (66),(67) (consistency condition) k β i (67) γ i x n i = ϕ n, n = n 0, n 0 + 1, (68) γ n γ 0 0, γ k 0 x n0, x n0+1, {x n } γ i x n i = 0, n = n 0, n 0 + 1, (69) {ˆx n } (68) {ψ n } x n = ˆx n + ψ n ϕ n k γ i 0 ψ n = ϕ/ γ i (70) (69) {x n,t }, t = 1,, K a 1 x n,1 + a x n, + + a K x n,k = 0, n = n 0, n 0 + 1, (71) a i (69) k {ˆx n,t }, t = 1,, k fundamental system (69) k t=1 d tx n,t (69) x n,t = rt n γ i rt i = 0. (7) r t P (r) = k γ ir i = 0 P (r) = 0 k r t {r n t } (69) fundamental system (68) x n = d t rt n + ψ n (7) t=1 P (r) = 0 r j, j = 1,, p µ j ( p j=1 µ j = k) x n = [d 1,1 + d 1, n + d 1, n(n 1) + + d 1,µ1 n(n 1) (n µ 1 1)]r n 1 + [d,1 + d, n + d, n(n 1) + + d,µ n(n 1) (n µ 1)]r n + + [d p,1 + d p, n + d p, n(n 1) + + d p,µ n(n 1) (n µ p 1)]r n p + ψ n (74) 8
9 trivial dx/dt = 0, x(0) = 0 α i x n i = 0 (75) ρ ζ ζ α i ζ n i = 0 (76) x n = t[d 1,1 + d 1, n + d 1, n(n 1) + + d 1,µ1 n(n 1) (n µ 1 1)]ζ1 n + t[d,1 + d, n + d, n(n 1) + + d,µ n(n 1) (n µ 1)]ζ n + + t[d p,1 + d p, n + d p, n(n 1) + + d p,µ n(n 1) (n µ p 1)]ζp n (77) t 0 x(t) = 0 lim t 0,n t=x tnq ζ n i = X lim n nq 1 ζ n i = 0 (78) ζ i < 1 Definition 1. ρ(ζ) 1 (zero stable) n e = 1 [ α i x(t n i ) t β i f(x(t n i ))] (79) α 0 = 1 α 0 [ α i x(t n i ) t β i x (t n i )] (80) m x m m p m p m (t) = (t t n ) m p 0,, p m m p j (j = 1,, m) e = 0 m (80) p j (t t n ) j = 0 α 0 e = α i = 0 (81) 9
10 j 0 α 0 e = = [α i (t n i t n ) j tβ i j(t n i t n ) j 1 ] (8) ( i t) j 1 [ i tα i j tβ i ] (8) = ( t) j [i j α i + ji j 1 β i ] = 0 (84) j=0 α i = 0 (85) i j α i = j i j 1 β i (j 1) (86) m α i, β i (i = 0,, k)k + 1 k + 1 m + 1 k m = k k > m = k + 1 [Dahlquist(1956)] m t n Taylor x(t) = x (t) = m j=0 p j (t)x (j) (t n ) j! m p j (t)x(j) (t n ) j=1 j! + p m+1(t)x (m+1) (ξ(t)) (m + 1)! + p m+1(t)x (m+1) (η(t)) (m + 1)! (87) (88) x (j) (t) x t j ξ(t),η(t) t n,t L e = Lx Lp 0 = Lp 1 = = Lp m = 0. [ ] p m+1 (t n i )x (m+1) (ξ i (t n i )) p Lx = α i tβ m+1(t n i )x (m+1) (η i (t n i )) i (89) (m + 1)! (m + 1)! i=1 i=1 t n i ξ i (t n i ), η i (t n i ) t n 1 p m+1 (t n i )x (m+1) (ξ i (t n i )) α i (m + 1)! α ( i t) m+1 x (m+1) (ξ i (t n i )) i (m + 1)! (90) i=1 tm+1 (m + 1)! α i i m+1 x (m+1) (ξ i (t n i)) (91) i=1 10
11 p β m+1(t n i )x (m+1) (η i (t n i )) i (m + 1)! β ( i t) m x (m+1) (η i (t n i )) i m! (9) i=1 i=1 tm m! β i i m x (m+1) (η i (t n i)) (9) i=1 Lx Lx = sup t [t n k,t n] x (m+1) (t) M t m+1 (94) M = 1 α i i m β i i m (m + 1)! m! (95) i=1 i=1 Lx = O( t m+1 ) (96) k = 1 k = 1 k+1 = m+1 m = k = m m = 1 j = 1 α 0 + α 1 =0, (97) α 1 = (β 0 + β 1 ) (98) α 0 = 1 α 1 = 1, β 1 = 1 β 0 x n x n 1 = t (β 0 f n + (1 β 0 )f n 1 ). (99) () m = j = α 1 = β 1 (100) β 1 = 1/, β 0 = 1/ 1.1. Adams α 0 = 1 α i (i = 1,, k) 1 (α l ) ( α l = α 0 = 1) l = 1 Adams 1 Adams 1 β 0 11
12 Adams-Bashforth ( ( ) x n = x n 1 + t f(t n 1, x n 1 ) 1 ) f(t n, x n ), (101) ( ( ) x n = x n 1 + t 1 f(t n 1, x n 1 ) 4 f(t n, x n ) + 5 ) 1 f(t n, x n ). (10) k Adams-Bashforth α i =,..., k β β 0 = 0 k (k 1) 1 = k Adams-Moulton ( 5 ( ) x n = x n 1 + t 1 f(t n, x n ) + f(t n 1, x n 1 ) 1 ) 1 f(t n, x n ) (10) ( ( )4 x n = x n 1 + t 8 f(t n, x n ) f(t n 1, x n 1 ) (104) 5 4 f(t n, x n ) f(t n, x n ) Adams-Moulton β 0 k + 1 Nystöm: l = generalized Milen-Simpson: l = Cotes: k = l ) (105).1. BDF (Backward Differentiation Formulae) β i = 0 (i 0) Gear (BDF) (Stiffly-Stable Method) [Gear] β 0 = 1 α (85)(86) α i x n i = β 0 f(t n, x n ) (106) x n x n x n = f(t n, x n ) (107) 11 6 x n x n 1 + x n 1 x n = f(t n, x n ). (108) k + 1 k k 6 1 BDF f(t, x) f [7]( BDF[explicit BDF] ) α i (i = 0,, k) β 0 (85)(86) β i (i = 1,, k ) k < i k i α i = 0 k = k k = 1 Adams-Bashforth 1
13 1: Coefficients of Gear s BDF Method Coefficient k = 1 k = k = k = 4 k = 5 k = α α α 0 5 α α α α Coefficient k = 1 k = 1 : Coefficients of explicit BDF Method k = k = k = k = k = k = 4 k = k = k = 4 k = α α α 0 α α β β β β β
14 .1.4 (Predictor-Corrector) 1 1) (P) )f(x n ) (E) ) (C) 4) x n f(x n ) (E) ᾱ i x n i = t α i x n i = t β i f(t n i, x n i ) (109) i=1 β i f(t n i, x n i ) (110) 4 PEC α 0 x (0) n + ᾱ i x i = t β i f(t i, x (0) i ) (111) i=1 α i x i = t i=1 β i f(t i, x (0) i ) (11) 1 PEC(E) P(EC) (E).1.5 Lagrange Polynomial Interpolation x f N +1 (x j, f(x j )), j = 0,, N f(x) N P N P N (x) = l 0 0(x) =1, N f(x j )l N j (x) (11) j=0 l N j (x) = (x x 0)(x x 1 ) (x x j 1 )(x x j+1 ) (x x N ) (x j x 0 )(x j x 1 ) (x j x j 1 )(x j x j+1 ) (x j x N ) (114) (N 1). (115) (t n i, f(t n i, x n i )) i = 1,..., N P e N (t n, f(t n, x n )) e explicit (t n 1, f(t n 1, x n 1 )) P1 e f x n x n 1 + tn t n 1 P e 1 (t)dt = x n 1 + tn t n 1 f(t n 1, x n 1 )dt = x n 1 + (t n t n 1 )f(t n 1, x n 1 ) (116) Euler P e, P e Adams-Bashforth P e t t n t t n 1 (t) =f(t n 1, x n 1 ) + f(t n, x n ) (117) t n 1 t n t n t n 1 P e (t t n )(t t n ) (t) =f(t n 1, x n 1 ) (t n 1 t n )(t n 1 t n ) + f(t (t t n 1 )(t t n ) n, x n ) (t n t n 1 )(t n t n ) (t t n 1 )(t t n ) + f(t n, x n ) (t n t n 1 )(t n t n ). (118) 14
15 tn x n x n 1 P e (t)dt t n 1 tn ( ) t t n t t n 1 = f(t n 1, x n 1 ) + f(t n, x n ) dt t n 1 t n 1 t n t n t n 1 [ ( ) f(tn 1, x n 1 ) t = t n 1 t n t n t f(t ( )] n, x n ) t tn t n 1 t n t n 1t t n 1 = f(t ( ) n 1, x n 1 ) t0 (t n + t n 1 ) t n t 0 f(t ( ) n, x n 1 ) t0 (t n + t n 1 ) t n 1 t 0 t 1 t 1 =f(t n 1, x n 1 ) t 0 t 1 ((t n t n 1 ) + (t n 1 t n )) f(t n, x n ) t 0 t 1 ((t n t n 1 )) =f(t n 1, x n 1 ) t 0 ( t 0 + t 1 ) t 1 f(t n, x n ) t 0 t 1. (119) t i = t n i t n i 1 i t = t i ( x n x n 1 t f(t n 1, x n 1 ) 1 ) f(t n, x n ) (10) 15
16 tn x n x n 1 P e (t)dt t n 1 tn ( (t t n )(t t n ) = f(t n 1, x n 1 ) t n 1 (t n 1 t n )(t n 1 t n ) + f(t (t t n 1 )(t t n ) n, x n ) (t n t n 1 )(t n t n ) ) (t t n 1 )(t t n ) +f(t n, x n ) dt (t n t n 1 )(t n t n ) [ ( )] f(t n 1, x n 1 ) t tn = (t n 1 t n )(t n 1 t n ) (t n + t n ) t + t n t n t t n 1 [ ( )] f(t n, x n ) t tn + (t n t n 1 )(t n t n ) (t n 1 + t n ) t + t n 1t n t t n 1 [ ( )] f(t n, x n ) t tn + (t n t n 1 )(t n t n ) (t n 1 + t n ) t + t n 1t n t t n 1 ( = f(t ( ) n 1, x n 1 ) t0 t n + t n t n 1 + tn 1) (t n + t n ) t 0 (t n + t n 1 ) + t n t n t 0 t 1 ( t 1 + t ) ( + f(t ( ) n, x n ) t0 t n + t n t n 1 + tn 1) (t n 1 + t n ) t 0 (t n + t n 1 ) + t n 1 t n t 0 t 1 t ( + f(t ( ) n, x n ) t0 t n + t n t n 1 + tn 1) (t n 1 + t n ) t 0 (t n + t n 1 ) + t n 1 t n t 0 ( t 1 + t ) t = f(t n 1, x n 1 ) t 0 ( ( t 6 t 1 ( t 1 + t ) n + t n t n 1 + t ) n 1) (tn + t n ) (t n + t n 1 ) + 6t n t n f(t n, x n ) t 0 6 t 1 t + f(t n, x n ) t 0 6 t ( t 1 + t ) ( ( t n + t n t n 1 + t ) ) n 1 (tn 1 + t n ) (t n + t n 1 ) + 6t n 1 t n ( ( t n + t n t n 1 + t ) ) n 1 (tn 1 + t n ) (t n + t n 1 ) + 6t n 1 t n. (11) t i = t n t n i = i 1 j=0 t j (t n + t n t n 1 + t n 1) (t n i + t n j ) (t n + t n 1 ) + 6t n i t n j ( = t n + t n (t n t 0 ) + (t n t 0 ) ) ( t n t i t j) (tn t 0 ) + 6 (t n t i) ( t n t ) j = t 0 t 0 ( t i + t j) + 6 t i t j (1) 16
17 t n x n x n 1 t ( 0 t 0 t 0 ( t 0 + t 1 + t ) + 6 ( t 0 + t 1 ) ( t 0 + t 1 + t ) ) f(t n 1, x n 1 ) 6 t 1 ( t 1 + t ) t ( 0 t 0 t 0 ( t 0 + t 1 + t ) + 6 t 0 ( t 0 + t 1 + t ) ) f(t n, x n ) 6 t 1 ( t 1 + t ) + t ( 0 t 0 t 0 ( t 0 + t 1 ) + 6 t 0 ( t 0 + t 1 ) ) f(t n, x n ) 6 t ( t 1 + t ) = t 0 + t 0 ( t 1 + t ) + 6 t 0 t 1 ( t 1 + t ) f(t n 1, x n 1 ) 6 t 1 ( t 1 + t ) t 0 + t 0 ( t 1 + t ) 6 t 1 t f(t n, x n ) + t 0 + t 0 t 1 6 t ( t 1 + t ) f(t n, x n ) (1) x n x n 1 1 f(t n 1.x n 1 ) 4 f(t n, x n ) f(t n, x n ) (14) Adams-Bashforth (t n, f(t n, x n )) N + 1 P i N P i 0 =f(t n, x n ), (15) P1 i =f(t n, x n ) t t n 1 t t n + f(t n 1 ), (16) t n t n 1 t n 1 t n P i =f(t n, x n ) (t t n 1)(t t n ) (t n t n 1 )(t n t n ) + f(t (t t n )(t t n ) n 1, x n 1 ) (t n 1 t n )(t n 1 t n ) (t t n )(t t n 1 ) + f(t n, x n ) (17) (t n t n )(t n t n 1 ) x n x n 1 t 0 f(t n, x n ), (18) x n x n 1 t 0 (f(t n, x n ) + f(t n 1, x n 1 )), (19) x n x n t 0 t 1 6( t 0 + t 1 ) f(t n, x n ) + t 0 + t 0 t 1 f(t n 1, x n 1 ) 6 t 1 Euler Adams-Moulton t 0 6 ( t 0 + t 1 ) t 1 f(t n, x n ). f(t n, x n ) x BDF (t n i, x n i ) (i = 0,..., N) x Q N (t) (10) t t n 1 t t n Q 1 (t) = x n + x n 1 (11) t n t n 1 t n 1 t n dx dt dq 1 = dt x n t n t n 1 + x n 1 t n 1 t n = x n x n 1 t 0 = f(t n, x n ). (1) 17
18 (t t n 1 )(t t n ) Q (t) = x n (t n t n 1 )(t n t n ) + x (t t n )(t t n ) n 1 (t n 1 t n )(t n 1 t n ) + x (t t n )(t t n 1 ) n (t n t n )(t n t n 1 ) dx dt dq dt t (t n 1 + t n ) =x n (t n t n 1 )(t n t n ) + x t (t n + t n ) n 1 (t n 1 t n )(t n 1 t n ) + x t (t n + t n 1 ) n (t n t n )(t n t n 1 ). (14) t = t n dx t n t n 1 t n dt x n t=tn (t n t n 1 )(t n t n ) + x t n t n t n n 1 (t n 1 t n )(t n 1 t n ) + x t n t n t n 1 n (t n t n )(t n t n 1 ) t 0 = t 1 (1) t 0 + t 1 =x n t 0 ( t 0 + t 1 ) x t 0 + t 1 t 0 n 1 + x n = f(t n, x n ). (15) t 0 t 1 ( t 0 + t 1 ) t 1 4 Q x n x n x n = f(t n, x n ). (16) (t t n 1 )(t t n )(t t n ) Q (t) =x n (t n t n 1 )(t n t n )(t n t n ) + x (t t n )(t t n )(t t n ) n 1 (t n 1 t n )(t n 1 t n )(t n 1 t n ) (t t n )(t t n 1 )(t t n ) + x n (t n t n )(t n t n 1 )(t n t n ) + x (t t n )(t t n 1 )(t t n ) n (t n t n )(t n t n 1 )(t n t n ). (17) x Q t = t n dx x n ( dt t t=tn t 0 ( t 0 + t 1 )( t 0 + t 1 + t ) n t n (t n 1 + t n + t n ) + (t n 1 t n + t n t n + t n t n 1 ) ) x n 1 ( t t 0 t 1 ( t 1 + t ) n t n (t n + t n + t n ) + (t n t n + t n t n + t n t n ) ) x n ( + t ( t 0 + t 1 ) t 1 t n t n (t n + t n 1 + t n ) + (t n t n 1 + t n 1 t n + t n t n ) ) x n ( t ( t 0 + t 1 + t )( t 1 + t ) t n t n (t n + t n 1 + t n ) + (t n t n 1 + t n 1 t n + t n t n ) ) x n = t 0 ( t 0 + t 1 )( t 0 + t 1 + t ) ( t 0( t 0 + t 1 ) + ( t 0 + t 1 )( t 0 + t 1 + t ) + t 0 ( t 0 + t 1 + t )) x n 1 t 0 t 1 ( t 1 + t ) ( t 0 + t 1 )( t 0 + t 1 + t ) x n + t 0 ( t 0 + t 1 + t ) ( t 0 + t 1 ) t 1 t x n t 0 ( t 0 + t 1 ) ( t 0 + t 1 + t )( t 1 + t ) t =x n t 0 + (4 t 1 + t ) t 0 + t 1 ( t 1 + t ) t 0 ( t 0 + t 1 )( t 0 + t 1 + t ) x n 1 ( t 0 + t 1 )( t 0 + t 1 + t ) t 0 t 1 ( t 1 + t ) + x n t 0 ( t 0 + t 1 + t ) ( t 0 + t 1 ) t 1 t x n t 0 ( t 0 + t 1 ) ( t 0 + t 1 + t )( t 1 + t ) t =f(t n, x n ). (18) 18
19 t 0 = t 1 = t 11 6 x n x n 1 + x n 1 x n = f(t n, x n ) (19) dq dt t 0 + t 1 x n t 0 ( t 0 + t 1 ) x t 0 + t 1 t 0 n 1 + x n t 0 t 1 ( t 0 + t 1 ) t 1 P e (t n ) (140) tn = f(t n 1, x n 1 ) t n t n t n 1 t n + f(t n, x n ) t n t n 1 t n t n 1 dq dt x n t 0 + (4 t 1 + t ) t 0 + t 1 ( t 1 + t ) t 0 ( t 0 + t 1 ) ( t 0 + t 1 + t ) = f(t n 1, x n 1 ) t 0 + t 1 t 1 f(t n, x n ) t 0 t 1, (141) P e (t n ) (14) t=tn x n 1 ( t 0 + t 1 ) ( t 0 + t 1 + t ) t 0 t 1 ( t 1 + t ) t 0 ( t 0 + t 1 + t ) t 0 ( t 0 + t 1 ) + x n x n ( t 0 + t 1 ) t 1 t ( t 0 + t 1 + t ) ( t 1 + t ) t (t n t n )(t n t n ) = f(t n 1, x n 1 ) (t n 1 t n )(t n 1 t n ) + f(t (t n t n 1 )(t n t n ) n, x n ) (t n t n 1 )(t n t n ) (t n t n 1 )(t n t n ) + f(t n, x n ) (t n t n 1 )(t n t n ) = f(t n 1, x n 1 ) ( t 0 + t 1 )( t 0 + t 1 + t ) t 1 ( t 1 + t ) t 0 = t 1 = t f(t n, x n ) t 0( t 0 + t 1 + t ) t 1 t + f(t n, x n ) t 0( t 0 + t 1 ) ( t 1 + t ) t (14) 11 6 x n x n 1 + x n 1 x n = f(t n 1, x n 1 ) f(t n, x n ) + f(t n, x n ). (144) dq dt t 0 + t 1 x n t 0 ( t 0 + t 1 ) x t 0 + t 1 t 0 n 1 + x n t 0 t 1 ( t 0 + t 1 ) t 1 = f(t n 1, x n 1 ) ( t 0 + t 1 )( t 0 + t 1 + t ) t 1 ( t 1 + t ) P e (t n ) (145) t=tn f(t n, x n ) t 0( t 0 + t 1 + t ) t 1 t + f(t n, x n ) t 0( t 0 + t 1 ) ( t 1 + t ) t (146) 19
20 (85), (86) dq dt c 1 P e (t n ) + c P e (t n ) (147) t=tn (c 1 + c = 1) t 0 + t 1 x n t 0 ( t 0 + t 1 ) x t 0 + t 1 t 0 n 1 + x n t 0 t 1 ( t 0 + t 1 ) t 1 = f(t n 1, x n 1 ) ( t 0 + t 1 )(c t 0 + t 1 + t ) t 1 ( t 1 + t ) c = / t 0 = t 1 = t f(t n, x n ) t 0(c ( t 0 + t 1 ) + t ) t 1 t + f(t n, x n ) c t 0 ( t 0 + t 1 ) ( t 1 + t ) t. (148) x n x n x n = 8 f(t n 1, x n 1 ) 7 f(t n, x n ) f(t n, x n ) (149) c = /. Taylor Hamilton ( ) Symplectic(leap-flog) 4 bound bound ( (standard test problem) ) dx dt = λx, λ C. (150) x Rλ 0 bound (Rλ < 0 ) Euler x n = x 0 (1 + λ t) n λ t 1 [bounded] ( λ t < 1 [convergent]) λ t bound λ t (stability region) ( strict stability region) Definition (A-stability (Dahlquist)). C {λ : Rλ < 0} A 0
21 A Runge-Kutta A A A A R R Runge-Kutta A A C(α) = {λ C : arg(λ) α} A(α) [Widlund] (Stiff Stability)[Gear] Definition (Widlund). α (0, π/) W α = {λ t : π arg(λ t) < α} A(α) α A(0) A(0) k k + 1 α [0, π/] k = m =, k = m = 4 k m A(α) Definition 4 (Gear). R 1, R [stiffly stable] R 1 = {λ t : Rλ t < a} (151) R = {λ t : a Rλ t b, Iλ t c} (15) a, b, c λ t R Runge-Kutta x 0 = 1 Runge-Kutta x n = ξ(λ t) n K = (k (1), k (),, k (R) ) T, B = (b rs ), C = (c r ), E = (1, 1, ) K = λ(x n 1 E + tbk) (15) x n = x n 1 + tc T K (154) K x n = x n 1 + tc T (I λ tb) 1 λx n 1 E. (155) ξ(λ t) = 1 + λ tc T (I λ tb) 1 E. (156) ξ(λ t) Runge-Kutta (R 4) R > Runge-Kutta Lawson[9] 1
22 Im(λ t) R1 c R -a b Re(λ t) -c 4. (150) (α i λ tβ i )x n i = 0. (157) ρ(ζ) = σ(ζ) = α i ζ i (158) β i ζ i (159) π(ζ, λ t) = ρ(ζ) λ tσ(ζ) = 0 (160) ζ i (157) x n = t[d 1,1 + d 1, n + d 1, n(n 1) + + d 1,µ1 n(n 1) (n µ 1 1)]ζ1 n + t[d,1 + d, n + d, n(n 1) + + d,µ n(n 1) (n µ 1)]ζ n + + t[d p,1 + d p, n + d p, n(n 1) + + d p,µ n(n 1) (n µ p 1)]ζp n (161) ζ i < 4. 4.
23 4 Stability Diagram of Runge-Kutta Method Stage[ R]& Order[ m ] R=1, m=1 R=, m= R=, m= R=4, m=4 R=6, m=5[ Lawson] Im( λ t) Re( λ t) 1: Stability diagram of Runge-Kutta method.
24 1.5 1 Stability Diagram of Adams-Bashforth Method Number of Steps k=1 k= k= k=4 k=5 k=6 0.5 Im( λ t) Re( λ t) : Stability diagram of Adams-Bashfortht method. 4
25 4 Stability Diagram of Adams-Moulton Method Number of Steps k=1 k= k= k=4 k=5 k=6 1 Im( λ t) Re( λ t) : Stability diagram of Adams-Moulton method. 5
26 Stability Diagram of Gear s BDF Method Number of Steps k=1 k= k= k=4 k=5 k=6 5 Im( λ t) Re( λ t) 4: Stability diagram of BDF method. 6
27 1.5 1 Stability Diagram of explicit BDF Method Number of Steps k=1 k= k= k=4 k=[ α k=[ =0] =0, α 4 =0] 0.5 Im( λ t) Re( λ t) 5: Stability diagram of explicit BDF method. 7
28 4. stiffness Definition 5. λ t, t = 1,,, m dx dt = Ax + Φ(t) (16) Rλ t < 0, t = 1,,, m max t=1,,,m Rλ t min t=1,,,m Rλ t (stiff) stiffness ratio [1] J.C. Butcher, The Numerical Analysis of Ordinary Differential Equations (John Wiley & Sons, Chichester/New York, 1987). [] J.D. Lambert, Computational Methods in Ordinary Differential Equations (John Wiley & Sons, London, 197). [] C.W. Gear, Numerical Initial Value Problems in Ordinary Differential Equations (Prentice-Hall, Englewood Cliffs, N.J., 197). [4] R.D. Richtmyer, K.W. Morton, Difference Methods for Initial Value Problems (Interscience, New York, 1957). [5] P. Hartman, Ordinary Differential Equations (John Wiley & Sons, New York, 1964). [6] C.A.J. Fletcher, Computational Techniques for Fluid Dynamics (Springer, Berlin, 1991). [7] G.E. Karniadakis et al, J. Comput. Phys 97, 414 (1991); M.A. Hulsen, MEAH-18 (1996). [8] S. Gill, Proc. Cambridge Philos. Soc. 47, 95 (1951). [9] J.D. Lawson, SIAM J. Numer. Anal. 9??(197). 8
p12.dvi
301 12 (2) : 1 (1) dx dt = f(x,t) ( (t 0,t 1,...,t N ) ) h k = t k+1 t k. h k k h. x(t k ) x k. : 2 (2) :1. step. 1 : explicit( ) : ξ k+1 = ξ k +h k Ψ(t k,ξ k,h k ) implicit( ) : ξ k+1 = ξ k +h k Ψ(t k,t
d dt A B C = A B C d dt x = Ax, A 0 B 0 C 0 = mm 0 mm 0 mm AP = PΛ P AP = Λ P A = ΛP P d dt x = P Ax d dt (P x) = Λ(P x) d dt P x =
3 MATLAB Runge-Kutta Butcher 3. Taylor Taylor y(x 0 + h) = y(x 0 ) + h y (x 0 ) + h! y (x 0 ) + Taylor 3. Euler, Runge-Kutta Adams Implicit Euler, Implicit Runge-Kutta Gear y n+ y n (n+ ) y n+ y n+ y n+
December 28, 2018
e-mail : kigami@i.kyoto-u.ac.jp December 28, 28 Contents 2............................. 3.2......................... 7.3..................... 9.4................ 4.5............. 2.6.... 22 2 36 2..........................
Z: Q: R: C: sin 6 5 ζ a, b
Z: Q: R: C: 3 3 7 4 sin 6 5 ζ 9 6 6............................... 6............................... 6.3......................... 4 7 6 8 8 9 3 33 a, b a bc c b a a b 5 3 5 3 5 5 3 a a a a p > p p p, 3,
: 1g99p038-8
16 17 : 1g99p038-8 1 3 1.1....................................... 4 1................................... 5 1.3.................................. 5 6.1..................................... 7....................................
86 6 r (6) y y d y = y 3 (64) y r y r y r ϕ(x, y, y,, y r ) n dy = f(x, y) (6) 6 Lipschitz 6 dy = y x c R y(x) y(x) = c exp(x) x x = x y(x ) = y (init
8 6 ( ) ( ) 6 ( ϕ x, y, dy ), d y,, dr y r = (x R, y R n ) (6) n r y(x) (explicit) d r ( y r = ϕ x, y, dy ), d y,, dr y r y y y r (6) dy = f (x, y) (63) = y dy/ d r y/ r 86 6 r (6) y y d y = y 3 (64) y
1 4 1 ( ) ( ) ( ) ( ) () 1 4 2
7 1995, 2017 7 21 1 2 2 3 3 4 4 6 (1).................................... 6 (2)..................................... 6 (3) t................. 9 5 11 (1)......................................... 11 (2)
Numerical Analysis II, Exam End Term Spring 2017
H. Ammari W. Wu S. Yu Spring Term 2017 Numerical Analysis II ETH Zürich D-MATH End Term Spring 2017 Problem 1 Consider dx = f(t, x), t [0, T ] dt x(0) = x 0 R [28 Marks] with f C subject to the Lipschitz
error_g1.eps
Runge-Kutta Method Runge-Kutta Method x n+ = x n +hf(t n x n ) dx dt dx x(t+h) x(t) (t) = lim dt h h t = t n h > dx dt (t x(t n +h) x(t n ) n) = lim x(t n+) x(t n ) h h h x = f(tx) x(t n+ ) x(t n ) f(t
1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2
2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6
ver.1 / c /(13)
1 -- 11 1 c 2010 1/(13) 1 -- 11 -- 1 1--1 1--1--1 2009 3 t R x R n 1 ẋ = f(t, x) f = ( f 1,, f n ) f x(t) = ϕ(x 0, t) x(0) = x 0 n f f t 1--1--2 2009 3 q = (q 1,..., q m ), p = (p 1,..., p m ) x = (q,
untitled
3 3. (stochastic differential equations) { dx(t) =f(t, X)dt + G(t, X)dW (t), t [,T], (3.) X( )=X X(t) : [,T] R d (d ) f(t, X) : [,T] R d R d (drift term) G(t, X) : [,T] R d R d m (diffusion term) W (t)
201711grade1ouyou.pdf
2017 11 26 1 2 52 3 12 13 22 23 32 33 42 3 5 3 4 90 5 6 A 1 2 Web Web 3 4 1 2... 5 6 7 7 44 8 9 1 2 3 1 p p >2 2 A 1 2 0.6 0.4 0.52... (a) 0.6 0.4...... B 1 2 0.8-0.2 0.52..... (b) 0.6 0.52.... 1 A B 2
I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
17 1 25 http://grape.astron.s.u-tokyo.ac.jp/pub/people/makino/kougi/stellar_dynamics/index.html http://grape.astron.s.u-tokyo.ac.jp/pub/people/makino/talks/index-j.html d 2 x i dt 2 = j i Gm j x j x i
x i [, b], (i 0, 1, 2,, n),, [, b], [, b] [x 0, x 1 ] [x 1, x 2 ] [x n 1, x n ] ( 2 ). x 0 x 1 x 2 x 3 x n 1 x n b 2: [, b].,, (1) x 0, x 1, x 2,, x n
![x i [, b], (i 0, 1, 2,, n),, [, b], [, b] [x 0, x 1 ] [x 1, x 2 ] [x n 1, x n ] ( 2 ). x 0 x 1 x 2 x 3 x n 1 x n b 2: [, b].,, (1) x 0, x 1, x 2,, x n](/thumbs/92/108343288.jpg "x i [, b], (i 0, 1, 2,, n),, [, b], [, b] [x 0, x 1 ] [x 1, x 2 ] [x n 1, x n ] ( 2 ). x 0 x 1 x 2 x 3 x n 1 x n b 2: [, b].,, (1) x 0, x 1, x 2,, x n")
1, R f : R R,.,, b R < b, f(x) [, b] f(x)dx,, [, b] f(x) x ( ) ( 1 ). y y f(x) f(x)dx b x 1: f(x)dx, [, b] f(x) x ( ).,,,,,., f(x)dx,,,, f(x)dx. 1.1 Riemnn,, [, b] f(x) x., x 0 < x 1 < x 2 < < x n 1
Feynman Encounter with Mathematics 52, [1] N. Kumano-go, Feynman path integrals as analysis on path space by time slicing approximation. Bull
![Feynman Encounter with Mathematics 52, [1] N. Kumano-go, Feynman path integrals as analysis on path space by time slicing approximation. Bull](/thumbs/94/118365979.jpg "Feynman Encounter with Mathematics 52, [1] N. Kumano-go, Feynman path integrals as analysis on path space by time slicing approximation. Bull")
Feynman Encounter with Mathematics 52, 200 9 [] N. Kumano-go, Feynman path integrals as analysis on path space by time slicing approximation. Bull. Sci. Math. vol. 28 (2004) 97 25. [2] D. Fujiwara and
untitled
2 : n =1, 2,, 10000 0.5125 0.51 0.5075 0.505 0.5025 0.5 0.4975 0.495 0 2000 4000 6000 8000 10000 2 weak law of large numbers 1. X 1,X 2,,X n 2. µ = E(X i ),i=1, 2,,n 3. σi 2 = V (X i ) σ 2,i=1, 2,,n ɛ>0
Design of highly accurate formulas for numerical integration in weighted Hardy spaces with the aid of potential theory 1 Ken ichiro Tanaka 1 Ω R m F I = F (t) dt (1.1) Ω m m 1 m = 1 1 Newton-Cotes Gauss
() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.
![() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.](/thumbs/89/98384840.jpg "() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.")
() 6 f(x) [, b] 6. Riemnn [, b] f(x) S f(x) [, b] (Riemnn) = x 0 < x < x < < x n = b. I = [, b] = {x,, x n } mx(x i x i ) =. i [x i, x i ] ξ i n (f) = f(ξ i )(x i x i ) i=. (ξ i ) (f) 0( ), ξ i, S, ε >
n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................
II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2
![II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2](/thumbs/101/149159646.jpg "II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2")
II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%
1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n
09 8 9 3 Chebyshev 5................................. 5........................................ 5.3............................. 6.4....................................... 8.4...................................
Note.tex 2008/09/19( )
1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................
構造と連続体の力学基礎
II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
2011 8 26 3 I 5 1 7 1.1 Markov................................ 7 2 Gau 13 2.1.................................. 13 2.2............................... 18 2.3............................ 23 3 Gau (Le vy
4 2016 3 8 2.,. 2. Arakawa Jacobin., 2 Adams-Bashforth. Re = 80, 90, 100.. h l, h/l, Kármán, h/l 0.28,, h/l.., (2010), 46.2., t = 100 t = 2000 46.2 < Re 46.5. 1 1 4 2 6 2.1............................
I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google
I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59
meiji_resume_1.PDF
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
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
1 R n (x (k) = (x (k) 1,, x(k) n )) k 1 lim k,l x(k) x (l) = 0 (x (k) ) 1.1. (i) R n U U, r > 0, r () U (ii) R n F F F (iii) R n S S S = { R n ; r > 0
III 2018 11 7 1 2 2 3 3 6 4 8 5 10 ϵ-δ http://www.mth.ngoy-u.c.jp/ ymgmi/teching/set2018.pdf http://www.mth.ngoy-u.c.jp/ ymgmi/teching/rel2018.pdf n x = (x 1,, x n ) n R n x 0 = (0,, 0) x = (x 1 ) 2 +
v_-3_+2_1.eps
I 9-9 (3) 9 9, x, x (t)+a(t)x (t)+b(t)x(t) = f(t) (9), a(t), b(t), f(t),,, f(t),, a(t), b(t),,, x (t)+ax (t)+bx(t) = (9),, x (t)+ax (t)+bx(t) = f(t) (93), b(t),, b(t) 9 x (t), x (t), x (t)+a(t)x (t)+b(t)x(t)
2 1 x 2 x 2 = RT 3πηaN A t (1.2) R/N A N A N A = N A m n(z) = n exp ( ) m gz k B T (1.3) z n z = m = m ρgv k B = erg K 1 R =
1 1 1.1 1827 *1 195 *2 x 2 t x 2 = 2Dt D RT D = RT N A 1 6πaη (1.1) D N A a η 198 *3 ( a =.212µ) *1 Robert Brown (1773-1858. *2 Albert Einstein (1879-1955 *3 Jean Baptiste Perrin (187-1942 2 1 x 2 x 2
No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2
No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j
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. 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.
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)
,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
takei.dvi
0 Newton Leibniz ( ) α1 ( ) αn (1) a α1,...,α n (x) u(x) = f(x) x 1 x n α 1 + +α n m 1957 Hans Lewy Lewy 1970 1 1.1 Example 1.1. (2) d 2 u dx 2 Q(x)u = f(x), u(0) = a, 1 du (0) = b. dx Q(x), f(x) x = 0
2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................
f(x) = f(x ) + α(x)(x x ) α(x) x = x. x = f (y), x = f (y ) y = f f (y) = f f (y ) + α(f (y))(f (y) f (y )) f (y) = f (y ) + α(f (y)) (y y ) ( (2) ) f
22 A 3,4 No.3 () (2) (3) (4), (5) (6) (7) (8) () n x = (x,, x n ), = (,, n ), x = ( (x i i ) 2 ) /2 f(x) R n f(x) = f() + i α i (x ) i + o( x ) α,, α n g(x) = o( x )) lim x g(x) x = y = f() + i α i(x )
chap1.dvi
1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f
30
3 ............................................2 2...........................................2....................................2.2...................................2.3..............................
9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x
2009 9 6 16 7 1 7.1 1 1 1 9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x(cos y y sin y) y dy 1 sin
A
A05-132 2010 2 11 1 1 3 1.1.......................................... 3 1.2..................................... 3 1.3..................................... 3 2 4 2.1............................... 4 2.2
量子力学 問題
3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,
Grushin 2MA16039T
Grushin 2MA1639T 3 2 2 R d Borel α i k (x, bi (x, 1 i d, 1 k N d N α R d b α = α(x := (αk(x i 1 i d, 1 k N b = b(x := (b i (x 1 i d X = (X t t x R d dx t = α(x t db t + b(x t dt ( 3 u t = Au + V u, u(,
第5章 偏微分方程式の境界値問題
October 5, 2018 1 / 113 4 ( ) 2 / 113 Poisson 5.1 Poisson ( A.7.1) Poisson Poisson 1 (A.6 ) Γ p p N u D Γ D b 5.1.1: = Γ D Γ N 3 / 113 Poisson 5.1.1 d {2, 3} Lipschitz (A.5 ) Γ D Γ N = \ Γ D Γ p Γ N Γ
1: 3.3 1/8000 1/ m m/s v = 2kT/m = 2RT/M k R 8.31 J/(K mole) M 18 g 1 5 a v t πa 2 vt kg (
1905 1 1.1 0.05 mm 1 µm 2 1 1 2004 21 2004 7 21 2005 web 2 [1, 2] 1 1: 3.3 1/8000 1/30 3 10 10 m 3 500 m/s 4 1 10 19 5 6 7 1.2 3 4 v = 2kT/m = 2RT/M k R 8.31 J/(K mole) M 18 g 1 5 a v t πa 2 vt 6 6 10
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
f f x, y, u, v, r, θ r > = x + iy, f = u + iv C γ D f f D f f, Rm,. = x + iy = re iθ = r cos θ + i sin θ = x iy = re iθ = r cos θ i sin θ x = + = Re, y = = Im i r = = = x + y θ = arg = arctan y x e i =
9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =
5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =
9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P
9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)
x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s
... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z
keisoku01.dvi
2.,, Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 4 Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 5 Mon, 2006, 401, SAGA, JAPAN Dept.
Hanbury-Brown Twiss (ver. 2.0) van Cittert - Zernike mutual coherence
Hanbury-Brown Twiss (ver. 2.) 25 4 4 1 2 2 2 2.1 van Cittert - Zernike..................................... 2 2.2 mutual coherence................................. 4 3 Hanbury-Brown Twiss ( ) 5 3.1............................................
xia2.dvi
Journal of Differential Equations 96 (992), 70-84 Melnikov method and transversal homoclinic points in the restricted three-body problem Zhihong Xia Department of Mathematics, Harvard University Cambridge,
,,,17,,, ( ),, E Q [S T F t ] < S t, t [, T ],,,,,,,,
![,,,17,,, ( ),, E Q [S T F t ] < S t, t [, T ],,,,,,,,](/thumbs/91/105754403.jpg ",,,17,,, ( ),, E Q [S T F t ] < S t, t [, T ],,,,,,,,")
14 5 1 ,,,17,,,194 1 4 ( ),, E Q [S T F t ] < S t, t [, T ],,,,,,,, 1 4 1.1........................................ 4 5.1........................................ 5.........................................
数学Ⅱ演習（足助・09夏）
II I 9/4/4 9/4/2 z C z z z z, z 2 z, w C zw z w 3 z, w C z + w z + w 4 t R t C t t t t t z z z 2 z C re z z + z z z, im z 2 2 3 z C e z + z + 2 z2 + 3! z3 + z!, I 4 x R e x cos x + sin x 2 z, w C e z+w
20 6 4 1 4 1.1 1.................................... 4 1.1.1.................................... 4 1.1.2 1................................ 5 1.2................................... 7 1.2.1....................................
2 (March 13, 2010) N Λ a = i,j=1 x i ( d (a) i,j x j ), Λ h = N i,j=1 x i ( d (h) i,j x j ) B a B h B a = N i,j=1 ν i d (a) i,j, B h = x j N i,j=1 ν i
1. A. M. Turing [18] 60 Turing A. Gierer H. Meinhardt [1] : (GM) ) a t = D a a xx µa + ρ (c a2 h + ρ 0 (0 < x < l, t > 0) h t = D h h xx νh + c ρ a 2 (0 < x < l, t > 0) a x = h x = 0 (x = 0, l) a = a(x,
, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. main.tex 2011/08/13( )
81 4 2 4.1, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. 82 4.2. 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,.,
4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.
A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c
( ) ) ) ) 5) 1 J = σe 2 6) ) 9) 1955 Statistical-Mechanical Theory of Irreversible Processes )
( 3 7 4 ) 2 2 ) 8 2 954 2) 955 3) 5) J = σe 2 6) 955 7) 9) 955 Statistical-Mechanical Theory of Irreversible Processes 957 ) 3 4 2 A B H (t) = Ae iωt B(t) = B(ω)e iωt B(ω) = [ Φ R (ω) Φ R () ] iω Φ R (t)
I
I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
三石貴志.indd
流通科学大学論集 - 経済 情報 政策編 - 第 21 巻第 1 号,23-33(2012) SIRMs SIRMs Fuzzy fuzzyapproximate approximatereasoning reasoningusing using Lukasiewicz Łukasiewicz logical Logical operations Operations Takashi Mitsuishi
Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)
![Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)](/thumbs/94/118373946.jpg "Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)")
3 3 22 Z[i] Z[i] π 4, (x) π 4,3 (x) x (x ) 2 log x π m,a (x) x ϕ(m) log x. ( ). π(x) x (a, m) = π m,a (x) x modm a π m,a (x) ϕ(m) π(x) ϕ(m) x log x ϕ(m) m f(x) g(x) (x α) lim f(x)/g(x) = x α mod m (a,
1 (Contents) (1) Beginning of the Universe, Dark Energy and Dark Matter Noboru NAKANISHI 2 2. Problem of Heat Exchanger (1) Kenji
8 4 2018 6 2018 6 7 1 (Contents) 1. 2 2. (1) 22 3. 31 1. Beginning of the Universe, Dark Energy and Dark Matter Noboru NAKANISHI 2 2. Problem of Heat Exchanger (1) Kenji SETO 22 3. Editorial Comments Tadashi
数学の基礎訓練I
I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
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
I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345
u = u(t, x 1,..., x d ) : R R d C λ i = 1 := x 2 1 x 2 d d Euclid Laplace Schrödinger N := {1, 2, 3,... } Z := {..., 3, 2, 1,, 1, 2, 3
2 2 1 5 5 Schrödinger i u t + u = λ u 2 u. u = u(t, x 1,..., x d ) : R R d C λ i = 1 := 2 + + 2 x 2 1 x 2 d d Euclid Laplace Schrödinger 3 1 1.1 N := {1, 2, 3,... } Z := {..., 3, 2, 1,, 1, 2, 3,... } Q
( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c) yoshioka/education-09.html pdf 1
2009 1 ( ) ( 40 )+( 60 ) 1 1. 2. Schrödinger 3. (a) (b) (c) http://goofy.phys.nara-wu.ac.jp/ yoshioka/education-09.html pdf 1 1. ( photon) ν λ = c ν (c = 3.0 108 /m : ) ɛ = hν (1) p = hν/c = h/λ (2) h
II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )
II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11
( ) a C n ( R n ) R a R C n. a C n (or R n ) a 0 2. α C( R ) a C n αa = α a 3. a, b C n a + b a + b ( ) p 8..2 (p ) a = [a a n ] T C n p n a
![( ) a C n ( R n ) R a R C n. a C n (or R n ) a 0 2. α C( R ) a C n αa = α a 3. a, b C n a + b a + b ( ) p 8..2 (p ) a = [a a n ] T C n p n a](/thumbs/91/106692724.jpg "( ) a C n ( R n ) R a R C n. a C n (or R n ) a 0 2. α C( R ) a C n αa = α a 3. a, b C n a + b a + b ( ) p 8..2 (p ) a = [a a n ] T C n p n a")
9 8 m n mn N.J.Nigham, Accuracy and Stability of Numerical Algorithms 2nd ed., (SIAM) x x = x2 + y 2 = x + y = max( x, y ) x y x () (norm) (condition number) 8. R C a, b C a b 0 a, b a = a 0 0 0 n C n
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z
Introduction to Numerical Analysis of Differential Equations Naoya Enomoto (Kyoto.univ.Dept.Science(math))
Introduction to Numerical Analysis of Differential Equations Naoya Enomoto (Kyoto.univ.Dept.Science(math)) 2001 1 e-mail:s00x0427@ip.media.kyoto-u.ac.jp 1 1 Van der Pol 1 1 2 2 Bergers 2 KdV 2 1 5 1.1........................................
25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3
Part () () Γ Part ,
Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35
(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fou
(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fourier) (Fourier Bessel).. V ρ(x, y, z) V = 4πGρ G :.
http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg
??
( ) 2014 2014 1/119 = (ISS) ISS ISS ISS iss-clf iss-clf ISS = (ISS) FB 2014 2/119 = (ISS) ISS ISS ISS iss-clf iss-clf ISS R + : 0 K: γ: R + R + K γ γ(0) = 0 K : γ: R + R + K γ K γ(r) (r ) FB K K K K R
ohpmain.dvi
fujisawa@ism.ac.jp 1 Contents 1. 2. 3. 4. γ- 2 1. 3 10 5.6, 5.7, 5.4, 5.5, 5.8, 5.5, 5.3, 5.6, 5.4, 5.2. 5.5 5.6 +5.7 +5.4 +5.5 +5.8 +5.5 +5.3 +5.6 +5.4 +5.2 =5.5. 10 outlier 5 5.6, 5.7, 5.4, 5.5, 5.8,
grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )
2 9 2 5 2.2.3 grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = g () g () (3) grad φ(p ) p grad φ φ (P, φ(p )) y (, y) = (ξ(t), η(t)) ( ) ξ (t) (t) := η (t) grad f(ξ(t), η(t)) (t) g(t) := f(ξ(t), η(t))
SFGÇÃÉXÉyÉNÉgÉãå`.pdf
SFG 1 SFG SFG I SFG (ω) χ SFG (ω). SFG χ χ SFG (ω) = χ NR e iϕ +. ω ω + iγ SFG φ = ±π/, χ φ = ±π 3 χ SFG χ SFG = χ NR + χ (ω ω ) + Γ + χ NR χ (ω ω ) (ω ω ) + Γ cosϕ χ NR χ Γ (ω ω ) + Γ sinϕ. 3 (θ) 180
2 1 κ c(t) = (x(t), y(t)) ( ) det(c (t), c x (t)) = det (t) x (t) y (t) y = x (t)y (t) x (t)y (t), (t) c (t) = (x (t)) 2 + (y (t)) 2. c (t) =
1 1 1.1 I R 1.1.1 c : I R 2 (i) c C (ii) t I c (t) (0, 0) c (t) c(i) c c(t) 1.1.2 (1) (2) (3) (1) r > 0 c : R R 2 : t (r cos t, r sin t) (2) C f : I R c : I R 2 : t (t, f(t)) (3) y = x c : R R 2 : t (t,
v er.1/ c /(21)
12 -- 1 1 2009 1 17 1-1 1-2 1-3 1-4 2 2 2 1-5 1 1-6 1 1-7 1-1 1-2 1-3 1-4 1-5 1-6 1-7 c 2011 1/(21) 12 -- 1 -- 1 1--1 1--1--1 1 2009 1 n n α { n } α α { n } lim n = α, n α n n ε n > N n α < ε N {1, 1,
chap9.dvi
9 AR (i) (ii) MA (iii) (iv) (v) 9.1 2 1 AR 1 9.1.1 S S y j = (α i + β i j) D ij + η j, η j = ρ S η j S + ε j (j =1,,T) (1) i=1 {ε j } i.i.d(,σ 2 ) η j (j ) D ij j i S 1 S =1 D ij =1 S>1 S =4 (1) y j =
数値計算：常微分方程式
( ) 1 / 82 1 2 3 4 5 6 ( ) 2 / 82 ( ) 3 / 82 C θ l y m O x mg λ ( ) 4 / 82 θ t C J = ml 2 C mgl sin θ θ C J θ = mgl sin θ = θ ( ) 5 / 82 ω = θ J ω = mgl sin θ ω J = ml 2 θ = ω, ω = g l sin θ = θ ω ( )
¼§À�ÍýÏÀ – Ê×ÎòÅŻҼ§À�¤È¥¹¥Ô¥ó¤æ¤é¤® - No.7, No.8, No.9
No.7, No.8, No.9 email: takahash@sci.u-hyogo.ac.jp Spring semester, 2012 Introduction (Critical Behavior) SCR ( b > 0) Arrott 2 Total Amplitude Conservation (TAC) Global Consistency (GC) TAC 2 / 25 Experimental
untitled
18 1 2,000,000 2,000,000 2007 2 2 2008 3 31 (1) 6 JCOSSAR 2007pp.57-642007.6. LCC (1) (2) 2 10mm 1020 14 12 10 8 6 4 40,50,60 2 0 1998 27.5 1995 1960 40 1) 2) 3) LCC LCC LCC 1 1) Vol.42No.5pp.29-322004.5.
x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ
W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)
3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)
1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2
filename=mathformula58.tex ax + bx + c =, x = b ± b 4ac, (.) a x + x = b a, x x = c a, (.) ax + b x + c =, x = b ± b ac. a (.3). sin(a ± B) = sin A cos B ± cos A sin B, (.) cos(a ± B) = cos A cos B sin
KENZOU
KENZOU 2008 8 9 5 1 2 3 4 2 5 6 2 6.1......................................... 2 6.2......................................... 2 6.3......................................... 4 7 5 8 6 8.1.................................................
II Brown Brown
II 16 12 5 1 Brown 3 1.1..................................... 3 1.2 Brown............................... 5 1.3................................... 8 1.4 Markov.................................... 1 1.5
1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =
1 1.1 ( ). z = + bi,, b R 0, b 0 2 + b 2 0 z = + bi = ( ) 2 + b 2 2 + b + b 2 2 + b i 2 r = 2 + b 2 θ cos θ = 2 + b 2, sin θ = b 2 + b 2 2π z = r(cos θ + i sin θ) 1.2 (, ). 1. < 2. > 3. ±,, 1.3 ( ). A
2011de.dvi
211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37
prime number theorem
For Tutor MeBio ζ Eite by kamei MeBio 7.8.3 : Bernoulli Bernoulli 4 Bernoulli....................................................................................... 4 Bernoulli............................................................................
磁性物理学 - 遷移金属化合物磁性のスピンゆらぎ理論
email: takahash@sci.u-hyogo.ac.jp May 14, 2009 Outline 1. 2. 3. 4. 5. 6. 2 / 262 Today s Lecture: Mode-mode Coupling Theory 100 / 262 Part I Effects of Non-linear Mode-Mode Coupling Effects of Non-linear
Z: Q: R: C:
0 Z: Q: R: C: 3 4 4 4................................ 4 4.................................. 7 5 3 5...................... 3 5......................... 40 5.3 snz) z)........................... 4 6 46 x
nosenote3.dvi
i 1 1 2 5 3 Verlet 9 4 18 5 23 6 26 1 1 1 MD t N r 1 (t), r 2 (t),, r N (t) ṙ 1 (t), ṙ 2 (t),, ṙ N (t) MD a 1, a 2, a 3 r i (i =1,,n) 1 2 T =0K r i + m 1 a 1 + m 2 a 2 + m 3 a 3 (m 1,m 2,m 3 =0, ±1, ±2,,