微分方程式の解を見る

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としみさんきち
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1 ( ) norikazu[at]ms.u-tokyo.ac.jp NS ( ) / 5

2 Newton 2 3 Navier-Stokes NS ( ) / 5

3 2 x 2 + bx + c = x = b ± b 2 4c 2 3 x 3 + px + q = x = 3 q q p q q p3 27 e x 2x =? f (x) = e x 2x x = < x < 2 NS ( ) / 5

4 Newton y y = f(x) y f(x k ) = f (x k )(x x k ) O a x k+2 x k+ x k x NS ( ) / 5

5 Newton x x, x 2,... y = f (x) x = x k y f (x k ) = f (x k )(x x k ) x x k+ x k+ = x k f (x k) f (x k ) f (x) = e x 2x x x x x x x x NS ( ) / 5

6 f (x) = x k+ = φ(x k ) (Newton φ(x) = x f (x)/f (x)) a x + a 2 x 2 + a 3 x 3 = b, a 2 x + a 22 x 2 + a 23 x 3 = b 2, a 3 x + a 32 x 2 + a 33 x 3 = b 3 x 2 + e2x2 + x 3 = b, x + x 2 + x3 4 = b 2, x 3 + sin x 2 + x 3 = b 3 NS ( ) / 5

7 Newton 2 3 Navier-Stokes NS ( ) / 5

8 Newton 2 3 Navier-Stokes NS ( ) / 5

9 L u t (x, t) = u α 2 (x, t) ( < x < L, t > ) x 2 (Neumann ) u x u (, t) = (L, t) = (t > ) x u = u(x, t): x t α > : u(x, ) = a(x) (+ ) (, linear) v(x, t), w(x, t): C v(x, t) + C 2 w(x, t): NS ( ) / 5

10 Fourier (/2) u(x, t) = φ(x)η(t) u t (x, t) = u α 2 (x, t) x 2 η (t)φ(x) = αφ (x)η(t) η (t) αη(t) = φ (x) = = λ φ(x) ( ) η (t) = αλη(t) ( ) φ (x) = λφ(x), φ () = φ (L) = NS ( ) / 5

11 Fourier (2/5) ( ) φ (x) + λφ(x) = s 2 + λ = s = ± λ. λ < φ(x) = C e λx + C 2e λx 2 λ = φ(x) = C + C 2x 3 λ > φ(x) = C cos λx + C 2 sin λx φ () = φ (L) = φ ( nπ ) 2 λ = λ n = (n =,,...) L ( ) φ(x) = φ n (x) = cos λn x ( nπ ) = cos L x (n =,,...) ( ) NS ( ) / 5

12 Fourier (3/5) λ n ( ) η(t) = e αλnt = exp( αλ n t). ( nπx ) u n (x, t) = ( ) e αλnt cos L ( nπx ) u(x, t) = c n u n (x, t) = c n e αλnt cos L n= n= c, c, c 2,... NS ( ) / 5

13 Fourier (4/5) t = a(x) = L cos ( nπx ) c n cos L n= ( mπx ) ( nπx ) cos L L cos(mπx/l) L L a(x) cos ( mπx ) L dx = (m n) dx = L/2 (m = n ) L (m = n = ). L c n cos n= = c m L 2 ( nπx ) cos L (m ) ( mπx ) L dx NS ( ) / 5

14 Fourier (5/5) u(x, t) = c + ( c n e α λ nπx ) nt cos. L n= c = L, c n = 2 L L a(x) cos ( nπx ) L dx (n =, 2,...). u(x, t) = L a(y) L [ + 2 n= exp ( α n2 π 2 ) L 2 t cos ( nπx ) cos L ( nπy ) ] dy. L NS ( ) / 5

15 2 Gray-Scott ( ) u t = α u u xx + u 2 v (β + γ)u v t = α v v xx u 2 v + β( v) u x (, t) = u x (L, t) = v x (L, t) = v x (L, t) = u = u(x, t), v = v(x, t) 2 ( ): 26 α u, α v > : (u, v ) u t = u/ t u xx = 2 u/ x 2 β, γ > ( ): Fourier ( ) ( )!! NS ( ) / 5

16 ( ) k > dg dt g(t + k) g(t) (t). k 2 h > (Taylor ) d 2 f f (x + h) 2f (x) + f (x h) (x) dx 2 h 2. g(t + k) = g(t) + g (t)k + 2 g (s)k 2, s = t + θk, < θ <, f (x + h) = f (x) + f (x)h + 2 f (x)h 2 + 3! f (3) (x)h 3 + 4! f (4) (y)h 4, f (x h) = f (x) f (x)h + 2 f (x)h 2 3! f (3) (x)h 3 + 4! f (4) (z)h 4 NS ( ) / 5

17 h = L/N (N ) ( x i = i ) h (i =,,, N + ) 2 k > t n = nk (n =,, ) t L x NS ( ) / 5

18 h = L/N (N ) ( x i = i ) h (i =,,, N + ) 2 k > t n = nk (n =,, ) t h x x x 2 x 3 x 4 x N L x N+ x NS ( ) / 5

19 h = L/N (N ) ( x i = i ) h (i =,,, N + ) 2 k > t n = nk (n =,, ) t h t 5 t 4 t 3 U n i u(x i, t n ) t 2 t k x x x 2 x 3 x 4 x N L x N+ x NS ( ) / 5

20 u t (x i, t n ) = α 2 u x 2 (x i, t n ) Un+ i Ui n k = α Un i+ 2Un i + U n i h 2 u x (, t n) = u x (L, t n) Un Un h = Un N+ Un N h = t h t 5 t 4 t 3 t 2 t k x x L x x 2 x 3 x 4 x N x N+ NS ( ) / 5

21 ( ) λ = αk/h 2 U n+ = ( λ)u n + λu2 n, U n+ i = ( 2λ)Ui n + λ(ui+ n + Ui ) n (i = 2,, N ), U n+ N = ( λ)un n + λun n t h t 5 t 4 t 3 t 2 t k x x x 2 x 3 x 4 x N L x N+ x NS ( ) / 5

22 ( ) λ = αk/h 2 U n+ = ( λ)u n + λu2 n, U n+ i = ( 2λ)Ui n + λ(ui+ n + Ui ) n (i = 2,, N ), U n+ N = ( λ)un n + λun n t h t 5 t 4 t 3 t 2 t k x x x 2 x 3 x 4 x N L x N+ x NS ( ) / 5

23 ( ) λ = αk/h 2 U n+ = ( λ)u n + λu2 n, U n+ i = ( 2λ)Ui n + λ(ui+ n + Ui ) n (i = 2,, N ), U n+ N = ( λ)un n + λun n t h t 5 t 4 t 3 t 2 t k x x x 2 x 3 x 4 x N L x N+ x NS ( ) / 5

24 ( ) λ = αk/h 2 U n+ = ( λ)u n + λu2 n, U n+ i = ( 2λ)Ui n + λ(ui+ n + Ui ) n (i = 2,, N ), U n+ N = ( λ)un n + λun n t h t 5 t 4 t 3 t 2 t k x x x 2 x 3 x 4 x N L x N+ x NS ( ) / 5

25 GS U n+ i V n+ i Ui n k k V n i = α u U n i+ 2Un i + U n i h 2 + (U n i ) 2 V n i (β + γ)u n i Vi+ n = α 2V i n + Vi n v h 2 (Ui n ) 2 Vi n + β( Vi n ) U n Un h = Un N+ Un N h = V n V n h = V n N+ V n N h = t h α u = 5, α v = 2 5 h = L 28, k = 6 h2 α v t = T 2 L =.5 t 5 t 4 t 3 t 2 t k x x L x x 2 x 3 x 4 x N x N+ NS ( ) / 5

26 GS (/3) time x (a) (β, γ) = (.54,.4) time x (b) (β, γ) = (.54,.392) NS ( ) / 5

27 GS (2/3) time x (c) (β, γ) = (.54,.38) x time (d) (β, γ) = (.54,.56) NS ( ) / 5

28 GS (3/3) x time (e) (β, γ) = (.92,.448) x time (f) (β, γ) = (.96,.38) NS ( ) / 5

29 GS (= ) n = 2, 3 ( ) ( ) ( ) NO ( ) NS ( ) / 5

30 Newton 2 3 Navier-Stokes NS ( ) / 5

31 Navier-Stokes Navier Stokes equations: L. M. H. Navier (827), G. G. Stokes (845) u i t + 3 u i u j = ν u i p + f i, (i =, 2, 3) x j ρ x i j= 3 j= u j x j =. u(x, x 2, x 3, t) = (u, u 2, u 3 ): p(x, x 2, x 3, t): f (x, x 2, x 3, t) = (f, f 2, f 3 ): ρ: ( ); ν: ( ) NS ( ) / 5

32 Navier-Stokes Navier Stokes equations: L. M. H. Navier (827), G. G. Stokes (845) u i t + 3 u i u j = ν u i p + f i, (i =, 2, 3) x j ρ x i j= 3 j= u j x j =. NS ( ) / 5

33 Navier-Stokes tool NS ( Sobolev...) Leray, Hopf, Ladyzhenskaya, Kato,... ( ) NS R 3 f = R 3 u(x, t) 2 dx <?? MathSciNet (AMS ) Title: Navier Stokes 643 (24 ) 965 (26 ) 9467 (27 ) 63 (28 2 ) NS ( ) / 5

34 Navier-Stokes ( t ) ( ) NS ( ) / 5

35 Newton 2 3 Navier-Stokes NS ( ) / 5

36 ( ) etc. NS ( ) / 5

37 ( ) etc. NS ( ) / 5

38 ( ) etc. (?) NS ( ) / 5

39 ( ) etc. NS ( ) / 5

40 (functional analysis ) NS ( ) / 5

41 Newton 2 3 Navier-Stokes NS ( ) / 5

42 : Poisson xy Γ F = F (x, y) u = u(x, y) 2 u x u [ y 2 = F (x, y) u = F ] u = u(x, y) F (x, y) Siméon Denis Poisson (78 84) NS ( ) / 5

43 NS ( ) / 5

44 [] P, P 2,..., P N N = 76 NS ( ) / 5

45 [] P, P 2,..., P N N = [2] u, u 2,..., u N 2 NS ( ) / 5

46 [] P, P 2,..., P N N = [2] u, u 2,..., u N 2 u32 u4 P32 P4 u3 P3 [3] P 4, P 3, P 32 NS ( ) / 5

47 [] P, P 2,..., P N N = [2] u, u 2,..., u N 2 u32 [4] u, u 2,..., u N u4 P4 u3 P32 a a N u F =. P3 a N a NN u N F N [3] P 4, P 3, P 32 (???) NS ( ) / 5

48 (finite element method, FEM) Poisson 3 NS ( ) / 5

49 F (x, x 2 ) = NS ( ) / 5

50 NS ( ) / 5

51 Poisson u = F (x, y) Ω, u = (x, y) Γ. J(v) = 2 Ω J(u) = min J(u) v V v 2 dxdy Fv dxdy, V = {v C(Ω) v: C, v Γ = } Ω Poisson u V, (u x ϕ x + u y ϕ y ) dxdy = Ω Ω F ϕ dxdy ( ϕ V ) Poisson NS ( ) / 5

52 V ϕ, ϕ 2,..., ϕ n n n u(x, y) = u k ϕ k (x, y), ϕ(x, y) = c k ϕ k (x, y) k= k= (u x ϕ x + u y ϕ y ) dxdy = F ϕ dxdy ( ϕ = ϕ,..., ϕ n ) Ω Ω u = (u i ) i n ( ) f = F ϕ i dxdy, A = Ω i n Au = f ( (ϕ i,x ϕ j,x + ϕ i,y ϕ j,y ) dxdy Ω ). i,j n NS ( ) / 5

53 (finite element method, FEM) Ω = element U ϕ i (x, y) NS ( ) / 5

54 (finite element method, FEM) Ω = element U ϕ i (x, y) NS ( ) / 5

55 (finite element method, FEM) Ω = element U ϕ i (x, y) NS ( ) / 5

56 u = (2 x) 4 + ( y) 2, (x, y) Ω u =, (x, y) Ω. "elliptic.plt" "elliptic.plt" n = 673 n = 677 NS ( ) / 5

57 ( ) (94 ) Turner, Clough, Martin and Topp (956) Clough (96) finite-element, (96 97 ) Galerkin (Ritz )? R. Courant NS ( ) / 5

58 R. Courant Richard Courant ( ) D. Hilbert (936 ) Courant Institute of Mathematical Sciences Methoden der Mathematischen Physik, & What is Mathematics? ( ) (Wikipedia) C., ( ) 978 (From Wikipedia) NS ( ) / 5

59 R. Courant: Variational methods for the solution of problems of equilibrium and vibrations, Bull. Amer. Math. Soc. 49 (943) 23. As Henri Poincaré once remarked, solution of a mathematical problem is a phrase of indefinite meaning. Pure mathematicians sometimes are satisfied with showing that the non-existence of a solution implies a logical contradiction, while engineers might consider a numerical result as the only reasonable goal. Such one sided views seem to reflect human limitations rather than objective values. In itself mathematics is an indivisible organism uniting theoretical contemplation and active application. This address will deal with a topic in which such a synthesis of theoretical and applied mathematics has become particularly convincing. Usually the solution of a difficult problem in analysis proceeds according to a general scheme: The given problem P with the solution is replaced by a related problem P n so simple that its solution S n can be found with comparative ease. Then by improving the approximation P n to P we may expect, or we may assume, or we may prove, that S n tends to the desired solution S of P. The essential point in an individual case is to choose the sequence P n in a suitable manner. NS ( ) / 5

60 ( ) [Chap. II, ] Suppose we seek the minimum d of an integral expression or any other variational expression I (ϕ) (for example, our quadratic functionals of the preceding section). We then start with a minimizing sequence ϕ, ϕ 2, ϕ 3,, ϕ n,, () that is, a sequence of functions, admissible in our variational problem, for which lim I (ϕn) = d (2) n being the lower bound of the functional I (ϕ). The existence of the lower bound d is obvious or may be easily proved in all relevant problems and the existence of the minimizing sequence () is then a logical consequence. However, the problem in applications is one, not of the existence, but of the practical construction of such a minimizing sequence. Ritz s method is nothing but a recipe for such a construction. A minimizing sequence immediately furnishes an approximation to d (sometimes this is all we wish to know, for example, if we are interested in the natural frequencies of a vibrating system). NS ( ) / 5

61 NS ( ) / 5

62 Newton 2 3 Navier-Stokes NS ( ) / 5

63 ( ) etc. NS ( ) / 5

64 ( ) [ ] ,8 8 e , ,5 NS ( ) / 5

65 R. Courant: Variational methods for the solution of problems of equilibrium and vibrations, Bull. Amer. Math. Soc. 49 (943) 23 AMS : 29 8 : 5 7A 2 9 norikazu[at]ms.u-tokyo.ac.jp NS ( ) / 5

第5章偏微分方程式の境界値問題

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