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) Euler..................................... 12 (3) θ, -............................. 13 6 15 7 16 (1) (Landau) O( ), o( )................... 16 (2)................................. 17 (3)......................... 17 (4) 1 (46)......................... 19 (5)........................... 20 (6) (Gauss)...................... 24 (7) (35)...................................... 25 8 25 ( [1]) 7 ( ) 1
1 4 1 ( ) ( ) 1. 2. ( ) ( ) () 1 4 2
2 f x f (x) (1) f (x) = lim h 0 f(x + h) f(x) h h f (x) (2) f(x + h) f(x) h f (x) f Taylor f x (3) f(x + h) = f(x) + f (x)h + f (x) 2 h h (4) f(x h) = f(x) f (x)h + f (x) 2 (3) (5) f (x) = f(x + h) f(x) h h 2 + f (x) 3! h 2 f (x) 3! + O(h) (h 0) h 3 + f (4) (x) h 4 + 4! h 3 + f (4) (x) h 4 +. 4! O(h) O(h) Landau ( (1) (p. 16) ) (4) (6) f (x) = f(x) f(x h) h + O(h) (h 0) 1 f (x) O(h) (3) (4) (7) f (x) = f(x + h) f(x h) 2h + O(h 2 ) (h 0) 1 1 f (x) 1 O(h 2 ) (3) (4) : (8) f (x) = f(x + h) 2f(x) + f(x h) h 2 + O(h 2 ) (h 0). 1 2 Landau C 3
t O 1 x 1: [0, 1] [0, ) f x C 2 f(x + h) 2f(x) + f(x h) lim h 0 h 2 = f (x) 3 6 () (9) u t (x, t) = 2 u (x, t) x2 (t > 0, 0 < x < 1) (10) u(0, t) = u(1, t) = 0 (t > 0) (11) u(x, 0) = f(x) (0 x 1) u = u(x, t) [0, 1] [0, + ) 1 (x i, t n ) u (12) u n i := u(x i, t n ) Ui n x [0, 1] N x i (i = 0,, N) (stepsize) h = 1/N (13) x i = ih (i = 0, 1,, N). t τ (> 0) 1 (14) t n = nτ (n = 0, 1, 2, ) (x i, t n ) u u t 2 2 x2 (15) (16) u t (x i, t n ) = un+1 i u n i + O(τ) (τ +0), τ 2 u x (x i, t 2 n ) = un i+1 2u n i + u n i 1 + O(h 2 ) (h +0) h 2 4
u (9) h τ (17) U n i (18) u n+1 i U n+1 i u n i τ Ui n τ un i+1 2u n i + u n i 1 h 2 = U n i+1 2U n i + U n i 1 h 2 2 n 0 i U n i 1, U n i+1 i = 0, N (18) 1 i N 1 λ = τ/h 2 (19) U n+1 i U n i = λ(u n i+1 2U n i + U n i 1). (20) U n+1 i = (1 2λ)Ui n + λ ( ) Ui 1 n + Ui+1 n (10) (21) U n 0 = U n N = 0 (n = 1, 2, ), (11) (22) U 0 i = f(x i ) (0 i N) (1 i N 1; n = 0, 1, ) (9), (10), (11) {Ui n ; 0 i N, n 0} (23) (24) (25) U n+1 i = (1 2λ)Ui n + λ ( Ui 1 n + Ui+1 n U0 n = UN n = 0 (n = 1, 2, ) U 0 i = f(x i ) (0 i N) ) (1 i N 1; n = 0, 1, ) {U n i ; 0 i N, n 0} 1. n 0 {U n i ; 0 i N} (25) 2. n = 0, 1, {U n+1 i ; 0 i N} (a) 1 i N 1 U n+1 i (23) (b) i = 0, N (24) U n+1 0 = U n+1 N = 0. Euler U n+1 i ( ) 2 F n = F n 1 + F n 2 Fibonacci 5
N u(1/2, 0.1) (%) 10 0.36654433 1.653709 20 0.37118820 0.407728 40 0.37232923 0.101583 80 0.37261327 0.025374 160 0.37268420 0.006342 320 0.37270193 0.001585 640 0.37270636 0.000396 1: (x, t) = (1/2, 0.1) (λ = 1/2). 4 (1) ( ) (9)-(11) u f f(x) = f 0 (x) sin πx u (26) u(x, t) = e π2t sin πx 6 (27) u(x, t) = b k e k2 π 2t sin kπx, b k = 2 1 k=1 0 f(x) sin kπx dx (k = 1, 2, ) b k b 1 = 1, b k = 0 (k = 2, 3, ) (26) u (9), (10), (11) 1 (x, t) = (1/2, 0.1) u(1/2, 0.1) = exp( 0.1π 2 ) sin(π/2) = 0.372707838 N 2 ( (2) ) N N 2 (2) t t + u(, t) 6
10 1 0.1 Error 0.01 0.001 0.0001 10 100 1000 N 2: (x, t) = (1/2, 0.1) (λ = 1/2). (Dirichlet) (9)-(11) (28) u(x, t) = b k e k2 π 2t sin kπx, b k = 2 k=1 t u 0 : 1 (29) lim t u(x, t) = 0 (0 x 1). ( (29) (3) ) f x (30) f 1 (x) = 1 x 0 f(x) sin kπx dx (k = 1, 2, ) ( 0 x 1 ) 2 ( ) 1 2 x 1 ( 3) (9)-(11) N = 50, λ = 1/2 4 7 n {U n i ; 0 i N} t = nτ = 0.01, 0.02, 0.03, 0.04 x u(x, t) 8 t = 0 t = 0.3 0.01 0 (Neumann) (10) (31) u x (0, t) = u x (1, t) = 0 (t > 0) 6 (32) u(x, t) = 1 2 a 0 + a k e k2 π 2t cos kπx, a k = 2 1 k=1 0 7 f(x) cos kπx dx (k = 0, 1, )
theta=0, N=50, lambda=0.5 theta=0, N=50, lambda=0.5 t = 0.01 theta=0, N=50, lambda=0.5 t = 0.02 3: f = f 1 4: t = 0.01 5: t = 0.02 theta=0, N=50, lambda=0.5 t = 0.03 theta=0, N=50, lambda=0.5 t = 0.04 theta=0, N=50, lambda=0.5 Tmax = 0.3, interval = 0.01 6: t = 0.03 7: t = 0.04 8:, t = 0 0.3, t = 0.01 8
heat equation, Neumann B.C. theta=0, N=50, lambda=0.5 heat equation, Neumann B.C. theta=0, N=50, lambda=0.5 t = 0.01 heat equation, Neumann B.C. theta=0, N=50, lambda=0.5 t = 0.02 9: t = 0 10: t = 0.01 11: t = 0.02 heat equation, Neumann B.C. theta=0, N=50, lambda=0.5 t = 0.03 heat equation, Neumann B.C. theta=0, N=50, lambda=0.5 t = 0.04 heat equation, Neumann B.C. theta=0, N=50, lambda=0.5 Tmax = 0.2, interval = 0.01 12: t = 0.04 13: t = 0.05 14: t = 0 0.2 t u : (33) lim t u(x, t) = a 0 2 (0 x 1). 9 13 f 1 N = 50, λ = 1/2 u(, t) f = f 1 (34) a 0 2 = 1 0 f 1 (x) dx = 3 x = 1 4 2 1/4 = 0.25 ( ) (3) t 9
t \ x 0 0.1 0.2 0.3 0.4 0.5 0.00 0.000000 0.100000 0.200000 0.300000 0.400000 0.500000 0.05 0.221849 0.227225 0.241301 0.258699 0.272775 0.278151 0.10 0.246121 0.246862 0.248801 0.251199 0.253138 0.253879 0.15 0.249466 0.249568 0.249835 0.250165 0.250432 0.250534 0.20 0.249926 0.249940 0.249977 0.250023 0.250060 0.250074 0.25 0.249990 0.249992 0.249997 0.250003 0.250008 0.250010 0.30 0.249999 0.249999 0.250000 0.250000 0.250001 0.250001 0.35 0.250000 0.250000 0.250000 0.250000 0.250000 0.250000 0.40 0.250000 0.250000 0.250000 0.250000 0.250000 0.250000 2: (, θ = 0, N = 40, λ = 0.5) 6 f a 0 2 + (a k cos kπx + b k sin kπx), k=1 f ( ) f a k, b k a k, b k k : (35) f C l (l ) = a k, b k = o(k l ) (k + ). f f (= ) (36) k l ( a k + b k ) < = f C l. k=1 f 15 f 2 f k + a k, b k 3 f = f 2 (9)-(11) (27) k exp ( k 2 π 2 t) t > 0 k f = f 2 u () f = f 2 15 20 t = 0.001 () 3 a k, b k = o(1) (k ) k ( a k + b k ) = k=1 10
k sin kπx k exp ( k 2 π 2 t) t (27) t k = 2, 3, k = 1 b 1 exp( π 2 t) sin πx t (37) u(x, t) b 1 e π2t sin πx. 20 theta=0, N=100, lambda=0.25 theta=0, N=100, lambda=0.25 t = 0.001 theta=0, N=100, lambda=0.25 t = 0.002 15: t = 0 theta=0, N=100, lambda=0.25 t = 0.003 16: t = 0.001 theta=0, N=100, lambda=0.25 t = 0.004 17: t = 0.002 theta=0, N=100, lambda=0.25 Tmax = 0.2, interval = 0.01 18: t = 0.003 19: t = 0.004 20: t = 0 0.2, t = 0.01 5 (1) λ = τ/h 2 1/2 λ = 1/4 λ = 0.51 f = f 1 (9)-(11) 21 24 11
4 7 (N ) 3 λ 0 < λ 1/2 0 < λ 1/2 (38) Ui n max f(x) (0 i N; n = 0, 1, ) x [0,1] ( (5) 3 ) 0 < λ 1/2 T (39) lim max u(x i, t n ) Ui n = 0 N 0 i N 0 n T /τ ( (5) 2 ) 0 i N, 0 n T/τ (x i, t n ) [0, 1] [0, T ] theta=0, N=50, lambda=0.51 t = 0.009996 theta=0, N=50, lambda=0.51 t = 0.019992 theta=0, N=50, lambda=0.51 t = 0.029988 21: t = 0.009996 22: t = 0.019992 theta=0, N=50, lambda=0.51 t = 0.039984 23: t = 0.029988 24: t = 0.039984 (2) Euler 3 0 < λ 1/2 (= ) 12
3 u t (40) U n i U n 1 i τ = U n i+1 2U n i + U n i 1 h 2 (1 i N 1, n 1) n (41) U n+1 i Ui n τ = U n+1 i+1 2U n+1 i + U n+1 i 1 h 2 (1 i N 1, n 0) n n + 1 (42) (1 + 2λ)U n+1 i λ ( U n+1 i 1 + U ) n+1 i+1 = U n i (1 i N 1, n 0) Euler (42) 3 (23) n + 1 (23) U n+1 i (i = 1, 2,, N 1) (42) i (1 i N 1) 1 (23) ( ) (42) ( ) λ = 0.51 25 28 (42) λ (> 0) 29 32 λ = 5 (3) θ, - (41) (18) (41) 1 θ : θ (43) U n+1 i Ui n τ = (1 θ) U i+1 n 2Ui n + Ui 1 n + θ U n+1 i+1 h 2 2U n+1 i + U n+1 i 1 h 2 θ 0 θ 1 θ = 0 (18) θ = 1 (41) n + 1 (44) (1 + 2θλ)U n+1 i θλ ( U n+1 i 1 + U ) n+1 i+1 = {1 2(1 θ)λ} U n i + (1 θ)λ ( Ui 1 n + Ui+1) n (1 i N 1, n 0) ({Ui n ; 0 i N} ) N +1 {U n+1 i ; 0 i N} N 1 (45) U n+1 0 = U n+1 N = 0 13
theta=1, N=50, lambda=0.51 t = 0.009996 theta=1, N=50, lambda=0.51 t = 0.019992 theta=1, N=50, lambda=0.51 t = 0.029988 25: t = 0.009996 26: t = 0.019992 theta=1, N=50, lambda=0.51 t = 0.039984 27: t = 0.029988 28: t = 0.039984 theta=1, N=50, lambda=5 t = 0.01 theta=1, N=50, lambda=5 t = 0.02 theta=1, N=50, lambda=5 t = 0.03 29: t = 0.01 30: t = 0.02 theta=1, N=50, lambda=5 t = 0.04 31: t = 0.03 32: t = 0.04 14
2 U0 n+1, U n+1 N N 1 {U n+1 i ; 1 i N 1} N 1 1 (46) 1 + 2θλ θλ 0 U 1 n+1 θλ 1 + 2θλ θλ U......... 2. n+1 θλ 1 + 2θλ θλ U......... i. n+1 θλ 1 + 2θλ θλ U n+1 N 2 0 θλ 1 + 2θλ U n+1 N 1 {1 2(1 θ)λ}u1 n + (1 θ)λ(u0 n + U2 n ) {1 2(1 θ)λ}u2 n + (1 θ)λ(u1 n + U3 n ) = {1 2(1 θ)λ}ui n + (1 θ)λ(ui 1 n + Ui+1) n {1 2(1 θ)λ}un 2 n + (1 θ)λ(un 3 n + U N 1 n ) {1 2(1 θ)λ}un 1 n + (1 θ)λ(un 2 n + U N n ) θ θ = 1/2 Crank-Nicolson (- ) ( Crank-Nicolson [2]). 6 6 (47) u t (x, t) = 2 u (x, t) + ku(x, t)(1 u(x, t)) x2 (t > 0, 0 < x < 1) u x (t, 0) = 1, u(t, 1) = 0 (t > 0) u(0, x) = f(x) (0 x 1) (48) U n+1 i Ui n τ = (1 θ) U n i+1 2U n i + U n i 1 + θ U n+1 i+1 h 2 2U n+1 i + U n+1 i 1 h 2 + ku n i (1 U n i ) 15
7 (1) (Landau) O( ), o( ) O(h) (h 0) o(n k ) (n ) O( ) C α x (49) f(x) C g(x) 4 (50) f(x) = O(g(x)) (x α) 5 α f g o( ) (51) lim x α f(x) g(x) = 0 (52) f(x) = o(g(x)) (x α) f a A (53) A = lim h 0 f(a + h) f(a) h (54) f(a + h) f(a) Ah = o(h) (h 0) f C k (55) f(a + h) = f(a) + f (a)h + f (a) h 2 + + f (k 1) (a) 2! (k 1)! hk 1 + f (k) (a + θh) k! θ (0 < θ < 1) (56) f(a + h) = f(a) + f (a)h + f (a) h 2 + + f (k 1) (a) 2! (k 1)! hk 1 + O(h k ) (h 0) 4 D f C > 0, ε > 0, x B(α; ε) D f(x) C g(x). 5 : f(x) O(g(x)) O(g(x)) 1 h k 16
α ± x 2 + x + 1 x = O(x) (x ). (2) 4 (1) x, y y = ax + b (a, b ) 1 a y = bx a (a, b, b > 0) y = bx a (57) log y = log bx a = a log x + log b. Y = log y, B = log b, X = log x Y = ax + B. X, Y 1 x, y X, Y a N E E = C/N 2 (C ) log N, log E 2 (3) u (58) lim t u(x, t) = 0, u (59) lim t u(x, t) = a 0 2 1 0 f(x) dx (58) (0 ) (60) u(x, t) = b n e n2 π 2t sin nπx lim t e n2 π 2t = 0 (61) lim t n=1 b n e n2 π 2t sin nπx = n=1 17 lim b ne n2 π 2t sin nπx t n=1
n=1 lim t 0 1 t 0 > 0 (62) u(x, t) Ce π2 t (t t 0, 0 x 1) C f(t) C, τ (63) f(t) Ce t/τ ( t) Ce t/τ f(t) t u(x, t) 0 f M := max x [0,1] f(x) b n (64) b n = 2 (65) b n 2 1 0 1 f(x) sin nπx dx 2 0 f(x) sin nπx dx 1 0 M 1 dx = 2M 1 0 dx = 2M. b n n 2M (66) u(x, t) = (67) u(x, t) = b n e n2 π 2t sin nπx n=1 b n e n2 π 2t sin nπx n=1 ( t t 0 2M n=1 e (n2 1)π 2 t (68) u(x, t) (69) ( ) 2M n=1 n=1 e π2t. ( ) 2M e n2 π 2t 1 = 2M ) e (n2 1)π 2 t 0 e π2t. n=1 e (n2 1)π 2 t 0 18 n=1 e n2 π 2 t
(70) e (n2 1)π 2 t 0 e (n 1)π2 t 0 = r n 1, r = e π2 t 0 < 1 r n 1, 0 < r < 1 n=1 (4) 1 (46) 5 (3) 1 (46) {U n+1 i } (, ) N 1 N 1 1 n A = (a ij ) (71) a ii > a ij (i = 1, 2,, n) 1 j n,j i A 5.3 1 1 + 2θλ 0 θλ, θλ (72) 1 + 2θλ θλ θλ = 1 + 2θλ 2(θλ) = 1 > 0 1 1 A = (a ij ) Ax = 0 x ( 0) 6 (73) x = x k Ax = 0 k x 1. x n (74) n a kj x j = 0. j=1 6 n A A f : R n x Ax R n (i) A (A 1 ), (ii) f, (iii) f, (iv) f, (v) ker f = {0}, (vi) rank f = n, (vii) det A 0. n dim ker f = rank f ( ) 19
k (75) a kk x k = j k a kj x j. (76) a kk x k = a kj x j ( ) a kj x j a kj x k. j k j k j k x k (> 0) (77) a kk a jk. j k A (5) (9)-(11) 2 ( ) T (9)-(11) 0 x 1, 0 t T u θ λ 2 (i) 0 θ < 1, 0 < λ (ii) θ = 1, λ > 0. 1 2(1 θ). θ {U n i } 0 i N,n 0 : (78) max Ui n u n i = O(τ + h 2 ) (N ). 0 i N 0 n J u n i = u(x i, t n ) J T/τ N (79) max Ui n u n i 0. 0 i N 0 n J (79) 7 7 20
3 ( ) {U n i ; 0 i N, 0 n J} (80) (1 + 2θλ)U n+1 i θλ(u n+1 i 1 + U n+1 i+1 ) = [1 2(1 θ)λ]ui n + (1 θ)λ(ui 1 n + Ui+1) n (1 i N 1, 0 n J 1) θ, λ 2 (i) 0 θ < 1, 0 < λ (ii) θ = 1, λ > 0. 1 2(1 θ). : { = max (81) max Ui n 0 i N 0 n J max Uk 0, max U0, l max UN l 0 k N 0 l J 0 l J }. 2 (82) 2 u n+1 i u n i τ = u t (x i, t n ) + O(τ), (83) (84) u n i 1 2u n i + u n i+1 h 2 = u xx (x i, t n ) + O(h 2 ). u n+1 i u n i un i 1 2u n i + u n i+1 τ h 2 = u t (x i, t j ) u xx (x i, t j ) + O(τ + h 2 ) = O(τ + h 2 ). u u t = u xx (85) u n+1 i u n i τ un+1 i 1 2un+1 i + u n+1 i+1 = O(τ + h 2 ). h 2 {u n i } : (86) (1 + 2θλ)u n+1 i θλ(u n+1 i 1 + un+1 i+1 ) = [1 2(1 θ)λ]u n i + (1 θ)λ(u n i 1 + u n i+1) + O(τ 2 + τh 2 ). ( 1 i N 1, 0 n J 1 (i, n) ) Ui n ( ) (87) (1 + 2θλ)U n+1 i θλ(u n+1 i 1 + U n+1 i+1 ) = [1 2(1 θ)λ]u n i + (1 θ)λ(u n i 1 + U n i+1) 21
e n i := U n i u n i e n i (88) (1 + 2θλ)e n+1 i θλ(e n+1 i 1 + en+1 i+1 ) = [1 2(1 θ)λ]e n i + (1 θ)λ(e n i 1 + e n i+1) + O(τ 2 + τh 2 ) θ, λ 1 + 2θλ, θλ, 1 2(1 θ)λ, (1 θ)λ 0 (89) (1 + 2θλ) e n+1 i θλ( e n+1 i 1 + en+1 i+1 ) [1 2(1 θ)λ] e n i + (1 θ)λ( e n i 1 + e n i+1 ) + C(τ 2 + τh 2 ). C E n := max e n i 0 i N (1 + 2θλ) e n+1 i 2θλE n+1 (1 + 2θλ) e n+1 i θλ( e n+1 i 1 + en+1 i+1 ) (90) [1 2(1 θ)λ] e n i + (1 θ)λ( e n i 1 + e n i+1 ) + C(τ 2 + τh 2 ) [1 2(1 θ)λ]e n + 2(1 θ)λe n + C(τ 2 + τh 2 ) = E n + C(τ 2 + τh 2 ). e n+1 0 = e n+1 N = 0 i 1 i N 1 (91) (1 + 2θλ)E n+1 2θλE n+1 E n + C(τ 2 + τh 2 ). (92) E n+1 E n + C(τ 2 + τh 2 ). (93) E n E n 1 + C(τ 2 + τh 2 ) E n 2 + 2C(τ 2 + τh 2 ) E 0 + nc(τ 2 + τh 2 ) = nc(τ 2 + τh 2 ). i e 0 i = U 0 i u 0 i = f(x i ) f(x i ) = 0 E 0 = 0 nτ Jτ T (94) E n CT (τ + h 2 ). (95) max Ui n u n i CT (τ + h 2 ). 0 i N 0 n J 22
3 (81) U n i 0 n J 1 n (96) max U n+1 i max 0 i N { } max Uk n, U0 n+1, U n+1 N. 0 k N (97) max U n+1 i max Uk n 1 i N 1 0 k N i = i 0 (80) i = i 0 : max U n+1 i = U n+1 i 1 i N 1 0 (98) (1 + 2θλ)U n+1 i 0 θλ(u n+1 i 0 1 + U n+1 i 0 +1 ) = (1 2(1 θ)λ)u n i 0 + (1 θ)λ(u n i 0 1 + U n i 0 +1). θ, λ 1 + 2θλ, θλ, 1 2(1 θ)λ, (1 θ)λ 0 (99) (98) {1 2(1 θλ)} max Uk n + (1 θ)λ( max Uk n + max Uk n ) 0 k N 0 k N 0 k N = {[1 2(1 θλ)] + 2(1 θ)λ} max Uk n 0 k N = max Uk n. 0 k N (100) (98) (1 + 2θλ)U n+1 i 0 θλ{u n+1 i 0 + U n+1 i 0 } = U n+1 i 0. (99), (100) U n+1 i 0 max Uk n. (97) : 0 k N (101) max U n+1 i max Uk n. 1 i N 1 0 k N (96) (96) n = 0 { } (102) max Ui 1 max max Uk 0, U0 1, UN 1. 0 i N 0 k N (96) n = 1 { max (103) max Ui 2 0 i N (102), (103) { max (104) max Ui 2 0 i N max Uk 0, U0 1, U0 2, UN, 1 UN 2 0 k N max Uk 1, U0 2, UN 2 0 k N n = 1, 2,, J { } (105) max Ui n max max Uk 0, max U0, l max UN l 0 i N 0 k N 1 l n 1 l n (81) 23 } }. { } = max max Uk 0, max U0, l max UN l 0 k N 1 l 2 1 l 2
(6) (Gauss) 1 (Cramer) ( Jordan ) 8 2x 1 + 3x 2 x 3 = 5 (106) 4x 1 + 4x 2 3x 3 = 3 2x 1 + 3x 2 x 3 = 1. 1. 0 2. 2 3. 3 2 3 1 5 1 1 5 3 1 1 5 2 2 2 2 2 2 4 4 3 3 4 4 3 3 0 2 1 7 2 3 1 1 2 3 1 1 0 6 2 6 1 3 1 5 2 2 2 1 7 0 1 2 2 0 6 2 6 1 0 0 1 0 1 0 2 0 0 1 3 1 0 5 11 4 4 1 7 0 1 2 2 0 0 5 15 2 2, x 1 x 2 x 3 = 1 0 5 11 4 4 1 7 0 1 2 2 0 0 1 3 1 2 3. 0 2 3 1 5 2 3 1 5 2 3 1 5 4 4 3 3 0 2 1 7 0 2 1 7. 2 3 1 1 0 6 2 6 0 0 5 15 2x 1 + 3x 2 x 3 = 5, 2x 2 x 3 = 7, 5x 3 = 15 x 3 = 15 5 = 3, x 2 = 7 + x 3 2 = 2, x 1 = 5 3x 2 + x 3 2 = 5 3 2 + 3 2 = 1 8 () ( ) 24
0 (forward elimination) (backward substitution) 3 n + 1 () n ( n ) 5 5 n 5, 6 θλ = 1 (7) (35) ( ) 2π f a k (f) = 1 π π π f(x) cos kx dx, b k (f) = 1 π f C l { a k (f) = ± 1 a k (f (l) ) (l ) k l b k (f (l) ) (l ), π π b k (f) = ± 1 k l { f (l) Riemann-Lebesgue lim a k(f (l) ) = lim b k (f (l) ) = 0 k k f(x) sin kx dx b k (f (l) ) a k (f (l) ) (l ) (l ) lim k kl a k (f) = lim k l b k (f) = 0. k 8 1. Neumann I [3] 25
2., von Neumann [4], I [3] 3. LU Gauss [4], 1 1 [5] 4. [6] 5., [7] 6., [8] 7. (?) Richtmyer-Morton [9], [10], Smith [11], [12] [1], (2000), 7 http://nalab.mind.meiji.ac.jp/~mk/labo/text/heat-fdm-0.pdf [2] Crank, J. and Nicolson, P.: A Practical Method for Numerical Evaluation of Solutions of Partial Differential Equations of the Heat-Conduction Type, Proc. Camb. Phil. Soc., Vol. 43, pp. 50 67 (1947), http://www.springerlink.com/content/ g43649440307j807/ reprint. [3] I, http://nalab.mind. meiji.ac.jp/~mk/labo/text/heat-fdm-1.pdf (1998 ). [4], http://nalab.mind.meiji.ac.jp/~mk/labo/text/ heat-fdm-0-add.pdf (1997 ). [5] 1 I, http://nalab.mind.meiji.ac.jp/~mk/labo/text/ linear-eq-1.pdf (2002 ). [6], http://nalab.mind.meiji.ac.jp/~mk/labo/ text/wave.pdf (1999? ). [7],, (1991). [8], http://infsup.jp/ (2006). [9] Richtmyer, R. D. and Morton, K. W.: Difference Methods for Initial-Value Problems, Krieger Pub Co, 2nd edition (1994). 26
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