Taylor Series Method for Second-Order Polynomial Differential Equations

Transcription

1 Taylor Series Method for Second-Order Polynomial Differential Equations Viktor N. Latypov Sergey V. Sokolov Saint-Petersburg State University SCP-2015

2 Introduction New numerical algorithm to solve second-order polynomial ODEs Based on series expansion

3 Introduction New numerical algorithm to solve second-order polynomial ODEs Based on series expansion Uses only basic linear algebra

4 Introduction New numerical algorithm to solve second-order polynomial ODEs Based on series expansion Uses only basic linear algebra Easy to implement

5 Lagrange Equations Mechanics/Lagrange equations D(q) q + C(q, q) q + K(q) = 0

6 Lagrange Equations Mechanics/Lagrange equations D(q) q + C(q, q) q + K(q) = 0 General second-order system A(q) q + b(q, q) = 0

7 Lagrange Equations Mechanics/Lagrange equations General second-order system D(q) q + C(q, q) q + K(q) = 0 A(q) q + b(q, q) = 0 Goal: Derive series expansion for the solution q k (t) = + p=0 q p k (t t 0) p

8 First-Order systems For the first-order Cauchy problem i N n j 1 ẋ k = a k +... a k;i1,i 2,...,i j x i1... x ij. j=1 i 1=1 i j =1 x k (t 0 ) = x k,0, k = 1,..., n.

9 First-Order systems For the first-order Cauchy problem i N n j 1 ẋ k = a k +... a k;i1,i 2,...,i j x i1... x ij. j=1 i 1=1 i j =1 x k (t 0 ) = x k,0, k = 1,..., n. we can directly substitute x k (t) = + q=0 x q k (t t 0) q

10 First-Order systems For the first-order Cauchy problem i N n j 1 ẋ k = a k +... a k;i1,i 2,...,i j x i1... x ij. j=1 i 1=1 i j =1 x k (t 0 ) = x k,0, k = 1,..., n. we can directly substitute to get (for N = 2) x q+1 k = 1 q + 1 x k (t) = + q=0 x q k (t t 0) q δ q a k + n i=1 a k;ix q i + + n i i=1 j=1 a ] q k;ij l=0 x i lx q l j. (1)

11 First-Order system: an example For the non-polynomial equation ẋ 1 = sin(x 1 ) + e x1

12 First-Order system: an example For the non-polynomial equation ẋ 1 = sin(x 1 ) + e x1 one may find a set of auxiliary variables x 2 = sin(x 1 ) x 3 = cos(x 1 ) x 4 = e x1

13 First-Order system: an example For the non-polynomial equation ẋ 1 = sin(x 1 ) + e x1 one may find a set of auxiliary variables x 2 = sin(x 1 ) x 3 = cos(x 1 ) x 4 = e x1 such that the new system becomes polynomial: ẋ 1 = x 2 + x 4 ẋ 2 = x 3 (x 2 + x 4 ) ẋ 3 = x 2 (x 2 + x 4 ) ẋ 4 = x 4 (x 2 + x 4 )

14 Second-Order system: an example For the double pendulum we have a non-polynomial equation with ( ) (m1 + m D = 2 ) L 2 1 m 2 L 1 L 2 c 1 2 m 2 L 1 L 2 c 1 2 m 2 L 2, 2

15 Second-Order system: an example For the double pendulum we have a non-polynomial equation with ( ) (m1 + m D = 2 ) L 2 1 m 2 L 1 L 2 c 1 2 m 2 L 1 L 2 c 1 2 m 2 L 2, 2 ( ) m2 L C = 1 L 2 s 1 2 θ2 0, m 2 L 1 L 2 s 1 2 θ1 0

16 Second-Order system: an example For the double pendulum we have a non-polynomial equation with ( ) (m1 + m D = 2 ) L 2 1 m 2 L 1 L 2 c 1 2 m 2 L 1 L 2 c 1 2 m 2 L 2, 2 ( ) m2 L C = 1 L 2 s 1 2 θ2 0, m 2 L 1 L 2 s 1 2 θ1 0 ( ) L1 (m K = g 1 + m 2 ) s 1. m 2 L 2 s 2

17 Second-Order system: an example For the double pendulum we have a non-polynomial equation with ( ) (m1 + m D = 2 ) L 2 1 m 2 L 1 L 2 c 1 2 m 2 L 1 L 2 c 1 2 m 2 L 2, 2 ( ) m2 L C = 1 L 2 s 1 2 θ2 0, m 2 L 1 L 2 s 1 2 θ1 0 ( ) L1 (m K = g 1 + m 2 ) s 1. m 2 L 2 s 2 Here q 1 = θ 1 and q 2 = θ 2, s i = sin θ i, c i = cos θ i, s 1 2 = sin (θ 1 θ 2 ).

18 Second-Order system: polynomial form Introducing q 3 = s 1, q 4 = c 1, q 5 = s 2 and q 6 = c 2 we reduce original equations to the polynomial one.

19 Second-Order system: polynomial form Introducing q 3 = s 1, q 4 = c 1, q 5 = s 2 and q 6 = c 2 we reduce original equations to the polynomial one. First derivative with respect to t: q 3 = q 4 q 1 q 4 = q 3 q 1 q 5 = q 6 q 2 q 6 = q 5 q 2 (2)

20 Second-Order system: polynomial form Introducing q 3 = s 1, q 4 = c 1, q 5 = s 2 and q 6 = c 2 we reduce original equations to the polynomial one. First derivative with respect to t: Second derivative: q 3 = q 4 q 1 + q 4 q 1 q 4 = q 3 q 1 q 3 q 1 q 5 = q 6 q 2 + q 6 q 2 q 6 = q 5 q 2 q 3 q 2 q 3 = q 4 q 1 q 4 = q 3 q 1 q 5 = q 6 q 2 q 6 = q 5 q 2 (2)

21 Second-Order system: polynomial form Introducing q 3 = s 1, q 4 = c 1, q 5 = s 2 and q 6 = c 2 we reduce original equations to the polynomial one. First derivative with respect to t: Second derivative: q 3 = q 4 q 1 + q 4 q 1 q 4 = q 3 q 1 q 3 q 1 q 5 = q 6 q 2 + q 6 q 2 q 6 = q 5 q 2 q 3 q 2 q 3 = q 4 q 1 q 4 = q 3 q 1 q 5 = q 6 q 2 q 6 = q 5 q 2 (2)

22 Second-Order system: polynomial form Introducing q 3 = s 1, q 4 = c 1, q 5 = s 2 and q 6 = c 2 we reduce original equations to the polynomial one. First derivative with respect to t: q 3 = q 4 q 1 q 4 = q 3 q 1 q 5 = q 6 q 2 q 6 = q 5 q 2 (2) Second derivative: q 3 = q 4 q 1 + q 4 q 1 q 4 = q 3 q 1 q 3 q 1 q 5 = q 6 q 2 + q 6 q 2 q 6 = q 5 q 2 q 3 q 2 q 4 q 1 + q 3 = q 4 q 1 q 3 q 1 + q 4 = q 3 q 1 q 6 q 2 + q 5 = q 6 q 2 q 3 q 2 + q 6 = q 5 q 2

23 Second-Order system: polynomial form D = C = d 11 d d 21 d q q q q c 11 c c 21 c q q q q ,.

24 Polynomial system: notations Equation A(q) q + b(q, q) = 0 where A = (a ij ) n i,j=1 and b = (b 1,..., b n ) T.

25 Polynomial system: notations Equation A(q) q + b(q, q) = 0 where A = (a ij ) n i,j=1 and b = (b 1,..., b n ) T. Multi-indices q p = q p qpn n, p = p p n, r = r r n, a p1p 2...,p n,r 1,r 2...,r n;ij = a pr;ij, b p1p 2...,p n,r 1,r 2...,r n;i = b pr;i. (3)

26 Polynomial system: notations Equation A(q) q + b(q, q) = 0 where A = (a ij ) n i,j=1 and b = (b 1,..., b n ) T. Multi-indices q p = q p qpn n, p = p p n, r = r r n, a p1p 2...,p n,r 1,r 2...,r n;ij = a pr;ij, b p1p 2...,p n,r 1,r 2...,r n;i = b pr;i. (3) Components a ij, b i polynomials of degree α ij and β i respectively. a ij (q, q) = b i (q, q) = α ij s=0 p + r =s β i s=0 p + r =s a pr;ij q p q r, b pr;i q p q r

27 Series expansion: shortcuts Substituting q i (t) = + m=0 q i m(t t 0 ) m into A(q) q + b(q, q) = 0 requires

28 Series expansion: shortcuts Substituting q i (t) = + m=0 q i m(t t 0 ) m into A(q) q + b(q, q) = 0 requires 1. Series products h = f g = + m=0 k=0 m (f k g m k ) (t t 0 ) m = h m = (f m g m ) = + m=0 m f k g m k. k=0 h m (t t 0 ) m.

29 Series expansion: shortcuts 2. Series powers Arbitrary power of q i (t) is q p i (t) = + q p m=0 i m = m k=0 q p i m (t t 0) m, ( q p 1 i k q i m k ).

30 Series expansion: shortcuts 3. Derivatives Derivatives of q i (t): + q i r (t) = (m + 1)qi m+1 r (t t 0) m, q i (t) = m=0 + m=0 (m + 1)(m + 2)q i m+2 (t t 0 ) m.

31 Series expansion: equations for coefficients Components of A(q) q + b(q, q) = 0 are n (a ij m ((m + 1)(m + 2)q j m+2 )) + b i m = 0. j=1

32 Series expansion: equations for coefficients Components of A(q) q + b(q, q) = 0 are n (a ij m ((m + 1)(m + 2)q j m+2 )) + b i m = 0. j=1 a ij m = b i m = α ij s=0 p + r =s β i s=0 p + r =s a pr;ij q p m [ (m + 1) q m+1 ] r. (4) b pr;i q p m [ (m + 1) q m+1 ] r. (5)

33 Series expansion: equations for coefficients Components of A(q) q + b(q, q) = 0 are n (a ij m ((m + 1)(m + 2)q j m+2 )) + b i m = 0. j=1 a ij m = b i m = α ij s=0 p + r =s β i s=0 p + r =s a pr;ij q p m [ (m + 1) q m+1 ] r. (4) b pr;i q p m [ (m + 1) q m+1 ] r. (5) q p m = q p1 1 m q p2 2 m... q pn n m (6)

34 Series expansion: equations for coefficients Components of A(q) q + b(q, q) = 0 are n (a ij m ((m + 1)(m + 2)q j m+2 )) + b i m = 0. j=1 a ij m = b i m = q p α ij s=0 p + r =s β i s=0 p + r =s i m = m a pr;ij q p m [ (m + 1) q m+1 ] r. (4) b pr;i q p m [ (m + 1) q m+1 ] r. (5) q p m = q p1 1 m q p2 2 m... q pn n m (6) k=0 q p 1 i k q i m k = q i m... q i m. }{{} p

35 Series expansion: equations for coefficients For q m+2 we get (m + 1)(m + 2)A 0 q m+2 + c m = 0,

36 Series expansion: equations for coefficients For q m+2 we get (m + 1)(m + 2)A 0 q m+2 + c m = 0, where A 0 = { a ij 0 } n i,j=1, c i;0 = b i 0, and for m > 0

37 Series expansion: equations for coefficients For q m+2 we get (m + 1)(m + 2)A 0 q m+2 + c m = 0, where A 0 = { a ij 0 } n i,j=1, c i;0 = b i 0, and for m > 0 c i;m = m 1 k=0 j=1 n (a ij m k ((k + 1)(k + 2)q j k+2 )) + b i m. (7)

38 Summary Given initial conditions q 0 and q 1 for each m q m+2 = A 1 0 c m (m + 1) (m + 2), A 0 = { a ij 0 } n i,j.