Systems of Differential Equations Matrix Methods

Transcription

1 Characteristic Equation Cayley-Hamilton Systems of Differential Equations Matrix Methods Cayley-Hamilton Theorem An Example The Cayley-Hamilton-Ziebur Method for u = A u A Working Rule for Solving u = A u Solving 2 2 u = A u Finding d 1 and d 2 A Matrix Method for Finding d 1 and d 2 Other Representations of the Solution u Another General Solution of u = A u Change of Basis Equation

2 Characteristic Equation Definition 1 (Characteristic Equation) Given a square matrix A, the characteristic equation of A is the polynomial equation det(a ri) = 0. The determinant det(a ri) is formed by subtracting r from the diagonal of A. The polynomial p(r) = det(a ri) is called the characteristic polynomial. If A is 2 2, then p(r) is a quadratic. If A is 3 3, then p(r) is a cubic. The determinant is expanded by the cofactor rule, in order to preserve factorizations.

3 Characteristic Equation Examples Create det(a ri) by subtracting r from the diagonal of A. Evaluate by the cofactor rule. ( ) 2 3 A =, p(r) = r r = (2 r)(4 r) A = , p(r) = 2 r r r = (2 r)(5 r)(7 r)

4 Cayley-Hamilton Theorem 1 (Cayley-Hamilton) A square matrix A satisfies its own characteristic equation. If p(r) = ( r) n + a n 1 ( r) n 1 + a 0, then the result is the equation ( A) n + a n 1 ( A) n a 1 ( A) + a 0 I = 0, where I is the n n identity matrix and 0 is the n n zero matrix.

5 Cayley-Hamilton Example Assume Then p(r) = A = 2 r r r and the Cayley-Hamilton Theorem says that (2I A)(5I A)(7I A) = = (2 r)(5 r)(7 r)

6 Cayley-Hamilton-Ziebur Theorem Theorem 2 (Cayley-Hamilton-Ziebur Structure Theorem for u = A u) A component function u k (t) of the vector solution u(t) for u (t) = A u(t) is a solution of the nth order linear homogeneous constant-coefficient differential equation whose characteristic equation is det(a ri) = 0. The theorem implies that the vector solution u(t) of u = A u is a vector linear combination of atoms constructed from the roots of the characteristic equation det(a ri) = 0.

7 Proof of the Cayley-Hamilton-Ziebur Theorem Consider the case n = 2, because the proof details are similar in higher dimensions. r 2 + a 1 r + a 0 = 0 A 2 + a 1 A + a 0 I = 0 Expanded characteristic equation Cayley-Hamilton matrix equation A 2 u + a 1 A u + a 0 u = 0 Right-multiply by u = u(t) u = A u = A 2 u Differentiate u = A u u + a 1 u + a 0 u = 0 Replace A 2 u u, A u u Then the components x(t), y(t) of u(t) satisfy the two differential equations x (t) + a 1 x (t) + a 0 x(t) = 0, y (t) + a 1 y (t) + a 0 y(t) = 0. This system implies that the components of u(t) are solutions of the second order DE with characteristic equation det(a ri) = 0.

8 Cayley-Hamilton-Ziebur Method The Cayley-Hamilton-Ziebur Method for u = A u Let atom 1,..., atom n denote the atoms constructed from the nth order characteristic equation det(a ri) = 0 by Euler s Theorem. The solution of u = A u is given for some constant vectors d 1,..., d n by the equation u(t) = (atom 1 ) d (atom n ) d n

9 Cayley-Hamilton-Ziebur Method Conclusions Solving u = A u is reduced to finding the constant vectors d 1,..., d n. The vectors d j are not arbitrary. They are uniquely determined by A and u(0)! A general method to find them is to differentiate the equation u(t) = (atom 1 ) d (atom n ) d n n 1 times, then set t = 0 and replace u (k) (0) by A k u(0) [because u = A u, u = A u = AA u, etc]. The resulting n equations in vector unknowns d1,..., d n can be solved by elimination. If all atoms constructed are base atoms constructed from real roots, then each d j is a constant multiple of a real eigenvector of A. Atom e rt corresponds to the eigenpair equation Av = rv.

10 A 2 2 Illustration ( ) ( ) Let s solve u = u, u(0) = ( ) 1 2 The characteristic polynomial of the non-triangular matrix A = is r r = (1 r)2 4 = (r + 1)(r 3). Euler s theorem implies solution atoms are e t, e 3t. Then u is a vector linear combination of the solution atoms, u = e t d 1 + e 3t d 2.

11 How to Find d 1 and d 2 We solve for vectors d 1, d 2 in the equation Advice: Define d 0 = u = e t d1 + e 3t d2. ( ) 1 2. Differentiate the above relation. Replace u via u = A u, then set t = 0 and replace u(0) by d 0 in the two formulas to obtain the relations d0 = e 0 d1 + e 0 d2 A d 0 = e 0 d1 + 3e 0 d2 We solve for ( d 1, d) 2 by elimination. Adding the equations gives d 0 + A d 0 = 4 d 2 and then 1 d 0 = implies 2 ( ) d1 = 3 3/2 d A d 4 0 =, 3/2 ( ) d2 = 1 1/2 d A d 4 0 =. 1/2

12 Summary of the 2 2 Illustration The solution of the dynamical system ( 1 2 u = 2 1 ) u, u(0) = ( 1 2 ) is a vector linear combination of solution atoms e t, e 3t given by the equation ( ) ( ) 3/2 1/2 u = e t + e 3/2 3t. 1/2

13 A Matrix Method for Finding d 1 and d 2 The Cayley-Hamilton-Ziebur Method produces a unique solution for d 1, d 2 because the coefficient matrix ( e 0 e 0 e 0 3e 0 ) is exactly the Wronskian W of the basis of atoms e t, e 3t evaluated at t = 0. This same fact applies no matter the number of coefficients d 1, d 2,... to be determined. The answer for d 1 and d 2 can be written in matrix form in terms of the transpose W T of the Wronskian matrix as aug( d 1, d 2 ) = aug( d 0, A d 0 )(W T ) 1.

14 Solving a 2 2 Initial Value Problem by the Matrix Method Then d 0 = u = A u, u(0) = ( 1 2 ) aug( d 1, d 2 ) =, A d 0 = ( ( ) 1, A = 2 ( ) ( 1 2 ) ( ( ) = ( 3 0 ) ) T ) 1 = ( and ). ( 3/2 1/2 3/2 1/2 ). The solution of the initial value problem is ( ) ( 3/2 1/2 u(t) = e t + e 3/2 3t 1/2 ) = ( 3 2 e t e3t 3 2 e t e3t ).

15 Other Representations of the Solution u Let y 1 (t),..., y n (t) be a solution basis for the nth order linear homogeneous constantcoefficient differential equation whose characteristic equation is det(a ri) = 0. Consider the solution basis atom 1, atom 2,..., atom n. Each atom is a linear combination of y 1,..., y n. Replacing the atoms in the formula u(t) = (atom 1 ) d (atom n ) d n by these linear combinations implies there are constant vectors d 1,..., d n such that u(t) = y 1 (t) d y n (t) d n

16 Another General Solution of u = A u Theorem 3 (General Solution) The unique solution of u = Au, u(0) = d 0 is u(t) = φ 1 (t)u 0 + φ 2 (t)au φ n (t)a n 1 u 0 where φ 1,..., φ n are linear combinations of atoms constructed from roots of the characteristic equation det(a ri) = 0, such that Wronskian(φ 1 (t),..., φ n (t)) t=0 = I.

17 Proof of the theorem Proof: Details will be given for n = 3. The details for arbitrary matrix dimension n is a routine modification of this proof. The Wronskian condition implies φ 1, φ 2, φ 3 are independent. Then each atom constructed from the characteristic equation is a linear combination of φ 1, φ 2, φ 3. It follows that the unique solution u can be written for some vectors d 1, d 2, d 3 as u(t) = φ 1 (t) d 1 + φ 2 (t) d 2 + φ 3 (t) d 3. Differentiate this equation twice and then set t = 0 in all 3 equations. The relations u = A u and u = A u = AA u imply the 3 equations d 0 = φ 1 (0) d 1 + φ 2 (0) d 2 + φ 3 (0) d 3 A d 0 = φ 1 (0) d 1 + φ 2 (0) d 2 + φ 3 (0) d 3 A 2 d 0 = φ 1 (0) d 1 + φ 2 (0) d 2 + φ 3 (0) d 3 Because the Wronskian is the identity matrix I, then these equations reduce to d0 = 1 d d d 3 A d 0 = 0 d d d 3 A 2 d0 = 0 d d d 3 which implies d 1 = d 0, d 2 = A d 0, d 3 = A 2 d0. The claimed formula for u(t) is established and the proof is complete.

18 Change of Basis Equation Illustrated here is the change of basis formula for n = 3. The formula for general n is similar. Let φ 1 (t), φ 2 (t), φ 3 (t) denote the linear combinations of atoms obtained from the vector formula ( φ1 (t), φ 2 (t), φ 3 (t) ) = ( atom 1 (t), atom 2 (t), atom 3 (t) ) C 1 where C = Wronskian(atom 1, atom 2, atom 3 )(0). The solutions φ 1 (t), φ 2 (t), φ 3 (t) are called the principal solutions of the linear homogeneous constant-coefficient differential equation constructed from the characteristic equation det(a ri) = 0. They satisfy the initial conditions Wronskian(φ 1, φ 2, φ 3 )(0) = I.