Fundamental Matrices and the Matrix Exponential: Solving Linear Systems via Diagonalization

Fundamental Matrices Suppose that x(1) (t), . . . , x(n) (t) form a fundamental set of solutions for the equation x0 = P(t)x (1) on some interval α < t < β. Then the matrix Ψ(t) = (x(1) (t), . . . , x(n) (t)) =  (1) x1 (t)  .  .  . (1) xn (t) ··· ···  (n) x1 (t) ..   . , (n) xn (t) is said to be a fundamental matrix for the system (1). Note: a fundamental matrix is nonsingular since its columns are linearly independent vectors. (2) Find a fundamental matrix for the system  1 x0 = 4  1 x. −2 (3) The solution of an IVP in terms of a fundamental matrix The general solution of (1) is x = = c1 x(1) (t) + · · · + cn x(n) (t) Ψ(t)c, (4) where c = (c1 , · · · , cn )T . For an IVP of Eq. (1) and the initial condition x(t0 ) = x0 , (5) where t0 is a given point in α < t < β. x0 is a given initial vector. We need to determine the vector c using (5) as follows Ψ(t0 )c = x0 =⇒ c = Ψ−1 (t0 )x0 . (6) x = Ψ(t)Ψ−1 (t0 )x0 (7) Then is the solution of the IVP (1) and (5). Note: Ψ0 = P(t)Ψ(t). The fundamental matrix Φ(t) Recall that Theorem 7.4.4 Let e(1)       1 0 0 0 0 1         .   = 0 , e(2) = 0 , · · · , e(3) =  ..  .   .  ..  0  ..  0 0 1 further, let x(1) , . . . , x(n) be the solutions of the system (1) that satisfy the initial conditions x(1) (t0 ) = e(1) , . . . , x(n) (t0 ) = e(n) , (1) respectively, where t0 is any point in α < t < β. Then x , . . . , x set of solutions of the system (1). (n) The fundamental matrix Φ(t): Φ(t) = (x(1) (t), . . . , x(n) (t)). Note: we use Ψ to denote an arbitrary fundamental matrix. (8) form a fundamental The fundamental matrix Φ(t) cont’d Note:  1 0  Φ(t0 ) =  .  .. 0 0 1 .. . 0 ··· ··· ···  0 0  ..  = I. . (9) 1 Remark: The solution of the IVP (1) and (5) is x = Φ(t)Φ−1 (t)x0 = Φ(t)x0 . (10) Note: If the fundamental matrix Φ(t) has been determined, then the solution for each set of initial conditions can be found simply by matrix multiplication, as indicated by Eq. (10). The fundamental matrix Φ(t) represents a transformation of the initial condition x0 into the solution x(t) at an arbitrary time t. Φ(t) = Ψ(t)Ψ−1 (t0 ). For the system (3) x0 =  1 4  1 x −2 in Example 1, find the fundamental matrix Φ(t) such that Φ(0) = I. (11) The Matrix exp(At) Recall that the solution of the scalar IVP x 0 = ax, x(0) = x0 (12) Where a is a constant, is Now consider the corresponding IVP for an n × n system, x0 = Ax, x(0) = x0 , where A is a constant matrix. Its solution is The comparison of these two solutions suggests that the matrix Φ(t) might have an exponential character! (13) The Matrix exp(At) cont’d The scalar exponential function exp(at) can be represented by the power series exp(at) = 1 + ∞ X an t n , n! n=1 (14) which converges for all t. Now, replace the scalar a by the n × n constant matrix A and consider the corresponding series I+ ∞ X A2 e 2 An t n An t n = I + At + + ··· + + ··· . n! 2! n! n=1 (15) Thus, the series (15) defines as its sum a new matrix (denoted by exp(At)); that is exp(At) = I + d [exp(At)] = dt exp(At)|t=0 = ∞ X An t n n! n=1 (16) Φ(t) = exp(At) Recall that the fundamental matrix Φ(t) satisfies the same IVP as exp(At), namely Φ0 = AΦ, Φ(0) = I. (17) Then, by the uniqueness part of Theorem 7.2.1 (extended to matrix differential equations), we conclude that Φ(t) = exp(At). Therefore, the solution of the IVP (13) is in the form of x = exp(At)x0 . Note: The matrix function exp(At) does have the properties we associate with the exponential function. (See Problem 15.) (18) Diagonalizable Matrices coupled system: some or all of the equations involve more than one– typically–of the unknown variables. (must be solved simultaneously.) uncoupled system: each equation involves only a single variable. (can be solved independently of all the others.) This suggests that transform a system of equations into an equivalent uncoupled system in which each equation contains only one unknown variable. transform the coefficient matrix A into a diagonal matrix. Diagonalizable Matrices Cont’d Suppose that the n × n matrix A has a full set of n linearly independent eigenvectors. (This can be the case if the eigenvalues of A are all different, or if A is Hermitian. ) ξ (1) , . . . , ξ (n) : eigenvectors λ1 , . . . , λn : the corresponding eigenvalues. Now to form the matrix T whose columns are the eigenvectors,   (1) (n) ξ1 · · · ξ1  . ..   T= .   .. (n) ξn (1) · · · ξn Note: the columns of T are linearly independent vectors, det T 6= 0 =⇒ T−1 exists. AT = (19) Diagonalizable Matrices Cont’d AT = TD (20) ⇓ T−1 AT = D. (21) Similarity Transformation: the process shown in (21) which can be used to transform A into a diagonal matrix if the eigenvalues and eigenvectors of A are known. Note: (21) means that A is similar to the diagonal matrix D. (A is diagonalizable.) The similarity transformation leaves the eigenvalues of A unchanged and transform its eigenvectors into the coordinate vectors e(1) , . . . , e(n) . Remark: If A has fewer than n linearly independent eigenvectors, then there is no matrix T such that T−1 AT = D. (A is not similar to a diagonal matrix and is not diagonalizable.) Consider the matrix  A= 1 4  1 , −2 find the similarity transformation matrix T and show that A can be diagonalized. (22) Solving x0 = Ax by diagonalizing the coefficient matrix A For the system x0 = Ax, (23) where A is a constant matrix. In Sec. 7.5 and 7.6, Now, to solve the system based on diagonalizing the coefficient matrix A. If A has a full set of n linearly independent eigenvector, ξ (1) , · · · , ξ (n) eigenvectors of A r1 , · · · , rn the corresponding eigenvalues
T = ξ (1) , · · · , ξ (n) the transformation matrix Define a new dependent variable y by x = Ty =⇒ (24) Solving x0 = Ax by diagonalizing the coefficient matrix A cont’d A fundamental matrix for the system y0 = Dy is the diagonal matrix Q(t) = exp(Dt) = (25) A fundamental matrix for the system x0 = Ax is then found from Q by the transformation x = Ty Ψ = TQ = (26) Solve the system of DE (in Example 1) x0 = by diagonalizing the coefficient matrix.  1 4  1 x −2 (27)