第1章 微分方程式と近似解法

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1 April 12, / 52

2 1.1 ( ) 2 / 52

3 : 3 / 52

4 1.3 Poisson Poisson Poisson 1 d {2, 3} 4 / 52

5 u,b b(t,x) u(t,x) x=0 1.1: 1 a x=l (0, t T ) (0, l) 1 a b : (0, t T ) (0, l) R, u : (0, t T ) (0, l) R b u 5 / 52

6 ( ) u (t, x) (t, x) (0, t T ) (0, l) w (t, x) = c V (x) u (t, x) (1.3.1) c V : (0, l) R Fourier ( 1.2) 6 / 52

7 (Fourier (1 )) u (t, x) x ( ) q (t, x) = λ (x) u (t, x) x (1.3.2) λ : (0, l) R u u(t,x) q(t,x) x 1.2: Fourier 7 / 52

8 1 (t, x) (0, t T ) (0, l) adxdt (w (t + dt, x) w (t, x)) adx = (b (t, x) dx q (t, x + dx) + q (t, x)) adt ( 1.3) dx 0, dt 0 w t q (t, x) = b (t, x) (t, x) x (1.3.1) (1.3.2) u c V t ( λ u ) = b x x 8 / 52

9 1 b(t,x)adxdt a q(t,x)adt q(t,x+dx)adt dx 1.3: 2 1 u 1 u D : (0, t T ) R x = 0 u (t, 0) = u D (t) u 1 Dirichlet 9 / 52

10 1 2 p N : (0, t T ) R ( 1.3.2) x = l λ u x (t, l) = p N (t) u 2 Neumann 3 u 0 : (0, l) R t = 0 u (0, x) = u 0 (x) u 10 / 52

11 1 u 2 (1.4 ) b (t, x) = b (x) u (t, x) = u (x) d ( λ du ) = b dx dx (1.3.3) (1.3.3) d (1.3.5) (1.4 ) 11 / 52

12 d d d {2, 3} Ω R d Γ D Ω Ω Γ N = Ω \ Γ D b : (0, t T ) Ω R, u : (0, t T ) Ω R Fourier u,b u(t,x) x 2 u D Ω b(t,x) νp(t,x) x 1 Γ D 1.4: 2 12 / 52

13 d (Fourier (d )) u : (0, t T ) Ω R ( ) q : (0, t T ) Ω R d 1 λ 11 λ 1d q = q. =..... q d λ d1 λ dd x 1. x d u = Λ u Λ = (λ ij ) ij : Ω R d d ( 2.4.5) 13 / 52

14 d λ : Ω R Λ = λi (I ) q = λ u (1.3.4) 14 / 52

15 d (t, x) (0, t T ) Ω dx 1 dx d dt e i x i (w (t + dt, x) w (t, x)) dx 1 dx 2 dx d = b (t, x) (q i (t, x + e i dx i ) q i (t, x)) dt i {1,,d} ( 1.5) dx 1,, dx d 0, dt 0 w t = b i {1,,d} q i x i 15 / 52

16 d q(t,x+e 3 dx 3 )dx 1 dx 2 dt q(t,x)dx 2 dx 3 dt q(t,x)dx 1 dx 3 dt b(t,x)dx 1 dx 2 dx 3 dt q(t,x+e 2 dx 2 )dx 1 dx 3 dt dx 3 q(t,x+e 1 dx 1 )dx 2 dx 3 dt x 3 dx 1 x 2 x 1 dx 2 q(t,x)dx 1 dx 2 dt 1.5: 3 16 / 52

17 d (1.3.1), (1.3.4) c V u t ( x 1 x d ) = c V u t (Λ u) = b λ 11 λ 1d λ d1 λ dd x 1.. x 3 u d c V u t (λ u) = b λ c V u t λ u = b 17 / 52

18 d = Laplace 2 18 / 52

19 d d u 1 u D : (0, t T ) Γ D R u = u D on (0, t T ) Γ D ( ) 2 p N : (0, t T ) Γ N R ν (Λ u) = p N on (0, t T ) Γ N ( ) λ ν u = p N on (0, t T ) Γ N ν ( ) ( ( )/ x) ν 3 u 0 : Ω R t 0 (0, t T ) u (t 0, x) = u 0 (x) in x Ω ( ) 19 / 52

20 d b (t, x) = b (x) u (t, x) = u (x) (Λ u) = b (1.3.5) Ω u Γ D Ω d {2, 3}, Γ D Ω Γ N = Ω \ Γ D c V : Ω R Λ : Ω R d d 20 / 52

21 d ( ) b : (0, t T ) Ω R, p N : (0, t T ) Γ N R, u D : (0, t T ) Γ D R, u 0 : Ω R c V u t (Λ u) = b in (0, t T) Ω, ν (Λ u) = p N on (0, t T ) Γ N, u = u D on (0, t T ) Γ D, u = u 0 in Ω at t = 0 u : (0, t T ) Ω R 21 / 52

22 d ( ) b : Ω R, p N : Γ N R, u D : Γ D R (Λ u) = b in Ω, ν (Λ u) = p N on Γ N, u = u D on Γ D u : Ω R Λ = I Poisson 22 / 52

23 ( 2 ) 23 / 52

24 ( 2 ) / x i (i {1,, d}) ξ i ( ) f (ξ 1, ξ 2,, ξ d ) = 0 1 (ξ 1,..., ξ d ) = (0,..., 0) 2 (ξ 1,..., ξ d ) (0,..., 0) 3 f (ξ 1, ξ 2,, ξ d ) = 0 ξ 1 f 1 (ξ 2,, ξ d ) = 0 f 1 (ξ 2,, ξ d ) = 0 (ξ 2,..., ξ d ) = (0,..., 0) Laplace ( ) 2 u = x x 2 u = 0 d 24 / 52

25 2 f (ξ 1,, ξ d ) = ξ ξ 2 d = 0 (x 1,..., x d ) = (0,..., 0) Laplace Poisson u = b Helmholz u + ω 2 u = 0 b ω 1 (Dirichlet ) 2 (Neumann ) 3 (Robin ) ( ) ( ) ( ) ( ) Stokes ( ) 25 / 52

26 2 ( ) ü c 2 u = 2 u 2 t 2 c2 x x x 2 u = 0 3 c f (ξ 1,, ξ d ) = ξ 2 1 c 2 ( ξ ξ 2 d) = 0 (x 1,..., x d ) (0,..., 0) 26 / 52

27 2 ( 2 u a u = u t a x x x 2 3 ) u = 0 a f (ξ 1,, ξ d ) = ξ 1 a ( ξ ξ 2 d) = 0 27 / 52

28 ( ) 1 u : R d R n α, β R, x, y R d u (αx + βy) = αu (x) + βu (y) u (linear function), (linear operator) 2 U, V D : U V α, β R, x, y U D (αx + βy) = αdx + βdy D (linear operator) ( ) D (x) Dx 28 / 52

29 2 u : R R du u (x + ϵ) u (x) (x) = lim dx ϵ 0 ϵ α, β R, u, v d αu (x + ϵ) + βv (x + ϵ) αu (x) βv (x) (αu + βv) = lim dx ϵ 0 ϵ αu (x + ϵ) αu (x) βv (x + ϵ) βv (x) = lim + lim ϵ 0 ϵ ϵ 0 ϵ = α du dx + β dv dx 29 / 52

30 2 : µ mg 1.1: 30 / 52

31 2 1.1 l d2 θ dt 2 + g sin θ = 0 sin θ θ = 0 sin θ = θ θ3 3! + θ5 5! + θ π sin θ θ 31 / 52

32 2 Van der Pol d 2 u dt 2 µ ( 1 u 2) du dt = 0 u / 52

33 2 1.2: Van der Pol 33 / 52

34 2 f(u) f(u) f(u) u u u (a) (b) (c) 1.3: 34 / 52

35 2 l θ l(1 cosθ) (a) mg (b) 1.4: 35 / 52

36 2 f(u ) f(u ) u u (a) Coulomb (b) 1.5: 36 / 52

37 Poisson 1.6 Poisson Poisson (1 Poisson ) b : (0, 1) R, u D R, p N R d2 u = b in (0, 1), dx2 u(0) = u D, du dx (1) = p N u : (0, 1) R 37 / 52

38 Poisson Γ D Ω Γ N = Ω \ Γ D (d Poisson ) Ω R d, d {2, 3}, Γ D Ω, b : Ω R, p N : Γ N R, u D : Γ D R u = b in Ω, u ν = p N on Γ N, u = u D on Γ D u : Ω R 38 / 52

39 m (0, 1) {x 0, x 1, x 2,, x m }, x 0 = 0, x m = 1, h = 1/m u h : (0, 1) R u h (x i ) = u i u 1 u i{1 u i u m u 0=u D u i+1 u m{1 h x 0=0 x 1 x i{1 x i x i+1 x m{1 x m=1 1.1: u h (x) x 39 / 52

40 Taylor u h (x i+1 ) = u h (x i ) + h du h dx (x i) + h2 2 + h3 6 d 3 u h dx 3 (x i) + O ( h 4) u h (x i 1 ) = u h (x i ) h du h dx (x i) + h2 2 h3 6 d 3 u h dx 3 (x i) + O ( h 4) 2 d 2 u h dx 2 (x i) d 2 u h dx 2 (x i) u h (x i+1 ) + u h (x i 1 ) = 2u h (x i ) + h 2 d2 u h dx 2 (x i) + O ( h 4) 40 / 52

41 h 2 O ( h 2) d2 u h dx 2 (x i) = u h (x i+1 ) 2u h (x i ) + u h (x i 1 ) h 2 41 / 52

42 Taylor u h (x m 1 ) = u h (x m ) h du h dx (x m) + O ( h 2) h O (h) du h dx (x m) = u h (x m ) u h (x m 1 ) h 42 / 52

43 u h (x i ) = u i, b (x i ) = b i u 0, u 1, u 2,, u m m u 0 = u D, u i+1 2u i + u i 1 h 2 = b i i {1, 2,, m 1}, u m u m 1 = p N h 1 43 / 52

44 1.6.2 m n Ω D {x 00, x 01, x 02,, x mn } h = 1/m = 1/n u h : D R u h (x ij ) = u ij D x mn x ij x i j+1 x i{1 j x ij x i+1 j x 00 h D x i j{1 1.2: 2 44 / 52

45 Taylor u h (x i+1 j ) = u h (x ij ) + h u h (x ij ) + h2 2 u h x 1 2 x 2 (x ij ) 1 + h3 6 3 u h x 3 1 (x ij ) + O ( h 4) u h (x i 1 j ) = u h (x ij ) h u h (x ij ) + h2 2 u h x 1 2 x 2 (x ij ) 1 h3 6 3 u h x 3 1 (x ij ) + O ( h 4) u h (x i j+1 ) = u h (x ij ) + h u h (x ij ) + h2 2 u h x 2 2 x 2 (x ij ) 2 + h3 6 3 u h x 3 2 (x ij ) + O ( h 4) u h (x i j 1 ) = u h (x ij ) h u h (x ij ) + h2 2 u h x 2 2 x 2 (x ij ) 2 45 / 52

46 h3 6 3 u h x 3 2 (x ij ) + O ( h 4) 4 h 2 O ( h 2) 2 u h x 2 1 (x ij ) + 2 u h x 2 2 (x ij ) = u h (x i+1 j ) + u h (x i j+1 ) + u h (x i 1 j ) + u h (x i j 1 ) 4u h (x ij ) h 2 Γ D x ij u h (x ij ) = u D (x ij ) Γ D x 1 x 2 (x ij1, x i+1 j 1 ), (x ij1, x i j+1 1 ) Taylor u h (x i+1 j ) = u h (x ij ) + h u h x 1 (x ij ) + O ( h 2) 46 / 52

47 u h (x i j+1 ) = u h (x ij ) + h u h x 2 (x ij ) + O ( h 2) h O (h) u h (x ij ) = u h (x i+1 j ) u h (x ij ) x 1 h u h (x ij ) = u h (x i j+1 ) u h (x ij ) x 2 h Ω Ω x ij u h (x ij ) = u ij, b (x ij ) = b ij, p N (x ij ) = p Nij, u D (x ij ) = u D ij x ij ν ij {u ij } ij 1 u ij = u D ij, u i+1 j + u i j+1 + u i 1 j + u i j 1 4u ij h 2 = b ij, 47 / 52

48 u i+1 j u ij h ν ij1 + u i j+1 u ij ν ij2 = p Nij h 1 48 / 52

49 x ij ν ij (weak form) 49 / 52

50 1.8 1 ( ) / 2 / / d2 u dx 2 + u = b in (0, 1), u(0) = u D, du dx (1) = p N 1 (0, 1) {x 0 = 0, x 1, x 2,, x m = 1} h = 1/m u h : (0, 1) R u h (x i ) = u i, b (x i ) = b i 50 / 52

51 ( ) Fourier Van der Pol Poisson 51 / 52

52 [1]. :., [2],.., [3].,., [4]. :., / 52

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

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