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Unformatted text preview: The Exponential Series 1 Section 1 We consider the initial value problem X = AX X (0) = [1 , 1] t (1) where A = 2 1 4 2 Then (as you can check) det( A λI ) = λ 2 so the only eigenvalue is λ = 0. The equation AX o = 0 X o is equivalent with the system x o + 2 y o = 0 4 x o 2 y o = 0 The corresponding eigenspace is spanned by [ 2 , 1] t and the straight line solution is Y 1 ( t ) = e t 2 1 To solve our initial value problem (1), we attempt to find a constant C such that 1 1 = CY 1 (0) = 2 C C No such C exists! The problem is that it takes two linearly independent vectors to span R 2 . We cannot hope to solve for the general initial value using only Y 1 (0). We will solve our initial value problem in another, quite cleaver, technique. To explain this, consider, for a moment, the following initial value problem. y = 2 y, y (0) = y o 1 1 SECTION 1 2 This is a separable equation and the solution is y ( t ) = y o e 2 t Now, recall that e x is given by the power series e x = 1 + x + x 2 2! + x 3 3! + ··· + x n n ! + ... Hence, y ( t ) = y o e 2 t = y o + 2 ty o + 2 2 t 2 2! y o + 2 3 t 3 3! y o + ··· + 2 n t n n ! y o + ... = y o + t 2 y o + t 2 2! 2 2 y o + t 3 3! 2 3 y o + ··· + t n n ! 2 n y o + ... Is it conceivable that the solution to our initial value problem (1) can be expressed in exactly the same manner? i.e. X ( t ) = e At X o = X o + tAX o + t 2 2! A 2 X o + t 3 3! A 3 X + ··· + t n n ! A n X o + ... (2) (We interpret quantities such as A 3 X o as A ( A ( AX o )).) To check this, we compute AX o = 2 1 4 2 1 1 = 3 6 A 2 X o = A ( AX o ) = 2 1 4 2 3 6 = Thus, A n X o = for all n ≥ 2, so our reasoning suggests that X ( t ) = X o + tAX o = 1 1 + t 3 6 = 1 + 3 t 1 6 t As a check, we compute that AX ( t ) = 2 1 4 2 1 + 3 t 1 6 t = 3 6 = X ( t ) 1 SECTION 1 3 It is also clear that X (0) = [1 , 1] t , so we indeed have solved our initial value problem. In general, it turns out that the series represented by formula (2) con verges for any n × n matrix A and n × 1 column vector X o . Granted this, and granted that we may differentiate this series term by term, we can show that this series always represents a solution to the initial value problem X = AX , X (0) = X o . Specifically, from formula (2), we compute that X (0) = X o and X ( t ) = AX o + 2 t 2! A 2 X o + 3 t 2 3! A 3 X + ··· + n t n 1 n ! A n X o + ... = A ( X o + tAX o + t 2 2! A 2 X o + t 3 3! A 3 X + ··· + t n 1 ( n 1)! A n 1 X o + ... = AX ( t ) Usually, this infinite series of vectors is difficult to explicitly sum. In our example, we were aided by the fact that the seemingly infinite series (2) actually turned out to be a finite series because A ( AX o ) = . Was this due to our choice of initial data? Suppose instead we had set X o = [ a,b ] t . Then AX o = 2 1 4 2 a b = a 2 4 + b 1 2 (3) Hence A ( AX o ) = aA 2 4 + bA 1 2 = a + b = Thus, for the given system, the same technique would work for any initial data....
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This note was uploaded on 04/25/2011 for the course MA 366 taught by Professor Cho during the Spring '08 term at Purdue.
 Spring '08
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