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Ch07 7
The document discusses fundamental matrices and their properties. A fundamental matrix Ψ(t) is a matrix whose columns are fundamental solutions to the system x' = P(t)x. Ψ(t) satisfies the differential equation Ψ' = P(t)Ψ and is nonsingular. The general solution to the system can be written as x = Ψ(t)c, where c is a constant vector. For an initial value problem, the solution is x = Ψ(t)Ψ-1(t0)x0. The fundamental matrix Φ(t) corresponding to a set of fundamental solutions satisfying initial conditions is also discussed. Matrix exponential functions are introduced as the fundamental matrix
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Ch07 7
- 1. Ch 7.7: FundamentalMatrices Suppose that x(1) (t),…, x(n) (t) form a fundamental set of solutions for x' = P(t)x on α < t < β. The matrix whose columns are x(1) (t),…, x(n) (t), is a fundamental matrix for the system x' = P(t)x. This matrix is nonsingular since its columns are linearly independent, and hence detΨ ≠ 0. Note also that since x(1) (t),…, x(n) (t) are solutions of x' = P(t)x, Ψ satisfies the matrix differential equation Ψ' = P(t)Ψ. , )()( )()( )( )()1( )( 1 )1( 1 = txtx txtx t n nn n Ψ
- 2. Example 1: Consider thehomogeneous equation x' = Ax below. In Chapter 7.5, we found the following fundamental solutions for this system: Thus a fundamental matrix for this system is xx =′ 14 11 tt etet − − = = 2 1 )(, 2 1 )( )2(3)1( xx − = − − tt tt ee ee t 22 )( 3 3 Ψ
- 3. Fundamental Matrices andGeneral Solution The general solution of x' = P(t)x can be expressed x = Ψ(t)c, where c is a constant vector with components c1,…, cn: )()1( 1 )( n nctc xxx ++= == n n nn n c c txtx txtx t 1 )()1( )( 1 )1( 1 )()( )()( )( cΨx
- 4. Fundamental Matrix &Initial Value Problem Consider an initial value problem x' = P(t)x, x(t0) = x0 where α < t0 < β and x0 is a given initial vector. Now the solution has the form x = Ψ(t)c, hence we choose c so as to satisfy x(t0) = x0 . Recalling Ψ(t0) is nonsingular, it follows that Thus our solution x = Ψ(t)c can be expressed as 0 0 10 0 )()( xΨcxcΨ tt − =⇒= 0 0 1 )()( xΨΨx tt − =
- 5. Recall: Theorem 7.4.4 Let Letx(1) ,…, x(n) be solutions of x' = P(t)x on I: α < t < β that satisfy the initial conditions Then x(1) ,…, x(n) are fundamental solutions of x' = P(t)x. = = = 1 0 0 0 ,, 0 0 1 0 , 0 0 0 1 )()2()1( n eee βα <<== 0 )( 0 )()1( 0 )1( ,)(,,)( ttt nn exex
- 6. Fundamental Matrix &Theorem 7.4.4 Suppose x(1) (t),…, x(n) (t) form the fundamental solutions given by Thm 7.4.4. Denote the corresponding fundamental matrix by Φ(t). Then columns of Φ(t) are x(1) (t),…, x(n) (t), and hence Thus Φ-1 (t0) = I, and the hence general solution to the corresponding initial value problem is It follows that for any fundamental matrix Ψ(t), I= =Φ 100 010 001 )( 0 t 00 0 1 )()()( xΦxΦΦx ttt == − )()()()()()( 0 100 0 1 tttttt −− =⇒== ΨΨΦxΦxΨΨx
- 7. The Fundamental MatrixΦ and Varying Initial Conditions Thus when using the fundamental matrix Φ(t), the general solution to an IVP is This representation is useful if same system is to be solved for many different initial conditions, such as a physical system that can be started from many different initial states. Also, once Φ(t) has been determined, the solution to each set of initial conditions can be found by matrix multiplication, as indicated by the equation above. Thus Φ(t) represents a linear transformation of the initial conditions x0 into the solution x(t) at time t. 00 0 1 )()()( xΦxΦΦx ttt == −
- 8. Example 2: FindΦ(t) for 2 x 2 System (1 of 5) Find Φ(t) such that Φ(0) = I for the system below. Solution: First, we must obtain x(1) (t) and x(2) (t) such that We know from previous results that the general solution is Every solution can be expressed in terms of the general solution, and we use this fact to find x(1) (t) and x(2) (t). xx =′ 14 11 tt ecec − − + = 2 1 2 1 2 3 1x = = 1 0 )0(, 0 1 )0( )2()1( xx
- 9. Example 2: UseGeneral Solution (2 of 5) Thus, to find x(1) (t), express it terms of the general solution and then find the coefficients c1 and c2. To do so, use the initial conditions to obtain or equivalently, tt ecect − − + = 2 1 2 1 )( 2 3 1 )1( x = − + = 0 1 2 1 2 1 )0( 21 )1( ccx = − 0 1 22 11 2 1 c c
- 10. Example 2: Solvefor x(1) (t) (3 of 5) To find x(1) (t), we therefore solve by row reducing the augmented matrix: Thus 2/1 2/1 2/110 2/101 2/110 111 240 111 022 111 2 1 = = → → → −− → − c c − + = − + = − − − tt tt tt ee ee eet 3 3 3)1( 2 1 2 1 2 1 2 1 2 1 2 1 )(x = − 0 1 22 11 2 1 c c
- 11. Example 2: Solvefor x(2) (t) (4 of 5) To find x(2) (t), we similarly solve by row reducing the augmented matrix: Thus 4/1 4/1 4/110 4/101 4/110 011 140 011 122 011 2 1 −= = → − → − → − → − c c + − = − − = − − − tt tt tt ee ee eet 2 1 2 1 4 1 4 1 2 1 4 1 2 1 4 1 )( 3 3 3)2( x = − 1 0 22 11 2 1 c c
- 12. Example 2: ObtainΦ(t) (5 of 5) The columns of Φ(t) are given by x(1) (t) and x(2) (t), and thus from the previous slide we have Note Φ(t) is more complicated than Ψ(t) found in Ex 1. However, now that we have Φ(t), it is much easier to determine the solution to any set of initial conditions. +− −+ = −− −− tttt tttt eeee eeee t 2 1 2 1 4 1 4 1 2 1 2 1 )( 33 33 Φ − = − − tt tt ee ee t 22 )( 3 3 Ψ
- 13. Matrix Exponential Functions Considerthe following two cases: The solution to x' = ax, x(0) = x0, is x = x0eat , where e0 = 1. The solution to x' = Ax, x(0) = x0 , is x = Φ(t)x0 , where Φ(0) = I. Comparing the form and solution for both of these cases, we might expect Φ(t) to have an exponential character. Indeed, it can be shown that Φ(t) = eAt , where is a well defined matrix function that has all the usual properties of an exponential function. See text for details. Thus the solution to x' = Ax, x(0) = x0 , is x = eAt x0 . ∑∑ ∞ = ∞ = +== 10 !! n nn n nn t n t n t e A I AA
- 14. Example 3: MatrixExponential Function Consider the diagonal matrix A below. Then In general, Thus = == ∑∑ ∞ = ∞ = t t n n n n nn t e e t n n n t e 2 00 0 0 !/20 0!/1 ! AA = 20 01 A , 20 01 20 01 20 01 , 20 01 20 01 20 01 32 3 2 2 = = = = AA = n n 20 01 A
- 15. Coupled Systems ofEquations Recall that our constant coefficient homogeneous system written as x' = Ax with is a system of coupled equations that must be solved simultaneously to find all the unknown variables. ,2211 12121111 nnnnnn nn xaxaxax xaxaxax +++=′ +++=′ ,, )( )( )( 1 1111 = = nnn n n aa aa tx tx t Ax
- 16. Uncoupled Systems &Diagonal Matrices In contrast, if each equation had only one variable, solved for independently of other equations, then task would be easier. In this case our system would have the form or x' = Dx, where D is a diagonal matrix: ,00 00 00 21 21112 21111 nnnn n n xdxxx xxdxx xxxdx +++=′ +++=′ +++=′ = = nn n d d d tx tx t 00 00 00 , )( )( )( 22 11 1 Dx
- 17. Uncoupling: Transform MatrixT In order to explore transforming our given system x' = Ax of coupled equations into an uncoupled system x' = Dx, where D is a diagonal matrix, we will use the eigenvectors of A. Suppose A is n x n with n linearly independent eigenvectors ξ(1) ,…, ξ(n) , and corresponding eigenvalues λ1,…, λn. Define n x n matrices T and D using the eigenvalues & eigenvectors of A: Note that T is nonsingular, and hence T-1 exists. = = n n nn n λ λ λ ξξ ξξ 00 00 00 , 2 1 )()1( )( 1 )1( 1 DT
- 18. Uncoupling: T-1 AT =D Recall here the definitions of A, T and D: Then the columns of AT are Aξ(1) ,…, Aξ(n) , and hence It follows that T-1 AT = D. = = = n n nn n nnn n aa aa λ λ λ ξξ ξξ 00 00 00 ,, 2 1 )()1( )( 1 )1( 1 1 111 DTA TDAT = = )()1( 1 )( 1 )1( 11 n nnn n n ξλξλ ξλξλ
- 19. Similarity Transformations Thus, ifthe eigenvalues and eigenvectors of A are known, then A can be transformed into a diagonal matrix D, with T-1 AT = D This process is known as a similarity transformation, and A is said to be similar to D. Alternatively, we could say that A is diagonalizable. = = = n n nn n nnn n aa aa λ λ λ ξξ ξξ 00 00 00 ,, 2 1 )()1( )( 1 )1( 1 1 111 DTA
- 20. Similarity Transformations: HermitianCase Recall: Our similarity transformation of A has the form T-1 AT = D where D is diagonal and columns of T are eigenvectors of A. If A is Hermitian, then A has n linearly independent orthogonal eigenvectors ξ(1) ,…, ξ(n) , normalized so that (ξ(i) , ξ(i) ) =1 for i = 1,…, n, and (ξ(i) , ξ(k) ) = 0 for i ≠ k. With this selection of eigenvectors, it can be shown that T-1 = T* . In this case we can write our similarity transform as T* AT = D
- 21. Nondiagonalizable A Finally, ifA is n x n with fewer than n linearly independent eigenvectors, then there is no matrix T such that T-1 AT = D. In this case, A is not similar to a diagonal matrix and A is not diagonlizable. = = = n n nn n nnn n aa aa λ λ λ ξξ ξξ 00 00 00 ,, 2 1 )()1( )( 1 )1( 1 1 111 DTA
- 22. Example 4: Find TransformationMatrix T (1 of 2) For the matrix A below, find the similarity transformation matrix T and show that A can be diagonalized. We already know that the eigenvalues are λ1 = 3, λ2 = -1 with corresponding eigenvectors Thus = 14 11 A − = = 2 1 )(, 2 1 )( )2()1( tt ξξ − = − = 10 03 , 22 11 DT
- 23. Example 4: SimilarityTransformation (2 of 2) To find T-1 , augment the identity to T and row reduce: Then Thus A is similar to D, and hence A is diagonalizable. − =→ − → − → −− → − − 4/12/1 4/12/1 4/12/110 4/12/101 4/12/110 0111 1240 0111 1022 0111 1 T D ATT = − = − − = − − =− 10 03 26 13 4/12/1 4/12/1 22 11 14 11 4/12/1 4/12/11
- 24. Fundamental Matrices forSimilar Systems (1 of 3) Recall our original system of differential equations x' = Ax. If A is n x n with n linearly independent eigenvectors, then A is diagonalizable. The eigenvectors form the columns of the nonsingular transform matrix T, and the eigenvalues are the corresponding nonzero entries in the diagonal matrix D. Suppose x satisfies x' = Ax, let y be the n x 1 vector such that x = Ty. That is, let y be defined by y = T-1 x. Since x' = Ax and T is a constant matrix, we have Ty' = ATy, and hence y' = T-1 ATy = Dy. Therefore y satisfies y' = Dy, the system similar to x' = Ax. Both of these systems have fundamental matrices, which we examine next.
- 25. Fundamental Matrix forDiagonal System (2 of 3) A fundamental matrix for y' = Dy is given by Q(t) = eDt . Recalling the definition of eDt , we have ( ) ( ) = = == ∑ ∑ ∑∑ ∞ = ∞ =∞ = ∞ = t t n n n n n n n n n n n nn n e e n t n t n t n t t λ λ λ λ λ λ 00 00 00 ! 00 00 00 ! ! 00 00 00 ! )( 1 0 0 1 0 1 0 D Q
- 26. Fundamental Matrix forOriginal System (3 of 3) To obtain a fundamental matrix Ψ(t) for x' = Ax, recall that the columns of Ψ(t) consist of fundamental solutions x satisfying x' = Ax. We also know x = Ty, and hence it follows that The columns of Ψ(t) given the expected fundamental solutions of x' = Ax. = == tn n t n tnt t t n nn n n n n ee ee e e λλ λλ λ λ ξξ ξξ ξξ ξξ )()1( )( 1 )1( 1 )()1( )( 1 )1( 1 1 11 00 00 00 TQΨ
- 27. Example 5: Fundamental Matricesfor Similar Systems We now use the analysis and results of the last few slides. Applying the transformation x = Ty to x' = Ax below, this system becomes y' = T-1 ATy = Dy: A fundamental matrix for y' = Dy is given by Q(t) = eDt : Thus a fundamental matrix Ψ(t) for x' = Ax is yyxx − =′⇒ =′ 10 03 14 11 = −t t e e t 0 0 )( 3 Q − = − == − − − tt tt t t ee ee e e t 220 0 22 11 )( 3 33 TQΨ