9 P T A H > T 2 4 D 0 C LinearSystemswithConstant - PDF document

R D E T − < T 2 4 D 0 9 P T A H − > T 2 4 D 0 C LinearSystemswithConstant Coefficients W e are now ready to turn to the task of finding the general solution to linear, homogeneous equations and systems. Remember the strategy we developed in the previous chapter. For a system of dimension n , we need to find a fundamental set of solutions, which is a set of n linearly independent solutions. The general solution is a linear combination of these solutions. We will also find the general solution for higher-order equations. However, this will be an easy job, since we need only use the results for the associated first-order system. 9.1 Overview of the Technique We are looking for a solution to the initial value problem y ′ = A y y ( 0 ) = y 0 , with (1.1) where A is a matrix with constant entries. For motivation, let’s look at a system of dimension 1, which we already understand. The initial value problem has the form y ′ = ay y ( 0 ) = y 0 . with Using the techniques developedin Chapter 2, we find that the solution to this problem is the exponential function y ( t ) = e at y 0 . (1.2) Is it possible that we can write down the solution to the general initial value problem in equation (1.1) by analogy to the solution for dimension 1 in (1.2)? This would mean that the solution is given by y ( t ) = e t A y 0 . (1.3) 444

445 9.1 Overview of the Technique At the moment, the term e t A , the exponential of a matrix, makes no sense. Our task, therefore, is to make sense of the exponential of a matrix, to learn as much about it as we can, to show that (1.3) actually gives us a solution to the initial value problem in (1.1), and to learn how to compute it. The exponential of a matrix and solutions to differential equations The correct way to define the exponential of a matrix is not at all obvious. We will do so using a power series, in analogy to the power series for the exponential function, ∞ e a = 1 + a + 1 2 ! a 2 + 1 1 3 ! a 3 + · · · = k ! a k . � (1.4) k = 0 DEFINITION 1.5 The exponential of the matrix A is defined to be ∞ e A = I + A + 1 2 ! A 2 + 1 1 3 ! A 3 + · · · = � k ! A k . (1.6) k = 0 In this formula, A 2 = AA , A 3 = AAA , etc., all of the products being matrix products of the n × n matrix A with itself. By convention, A 0 = I . Consequently, all of the terms make good sense. They are all n × n matrices, so e A is also an n × n matrix, provided that the series converges. Convergence of this infinite series with matrix terms means that we consider the partial sum matrices N 1 k ! A k . � S N = k = 0 The components of S N are very complicated expressions involving the entries of A . Convergence of the infinite series means that each component of the partial sum matrices converges. We will not worry about convergence. It is a fact, although we will not prove it, that the series in (1.6) converges for every matrix A . Furthermore, the convergenceis rapid enoughthat all operations done on this series in what follows are justified. r 1 � � 0 EXAMPLE 1.7 � Show that the exponential of the diagonal matrix A = is the diagonal r 2 0 e r 1 � � 0 matrix e A = . e r 2 0 Since A is a diagonal matrix, the powers of A are easy to compute: � r 1 � � r 1 � � r 2 � 0 0 0 A 2 = A · A = 1 · = , r 2 r 2 r 2 0 0 0 2 r 2 r 1 r 3 � � � � � � 0 0 0 A 3 = A 2 · A = 1 1 · = , r 2 r 2 r 3 0 0 0 2 2

446 Chapter 9 Linear Systems with Constant Coefficients and so forth. Therefore, e A = I + A + A 2 2 ! + A 3 3 ! + · · · r 1 r 2 r 3 � � � � � � � � + 1 + 1 1 0 0 0 0 1 1 = + + · · · r 2 r 2 r 3 0 1 0 0 0 2 ! 3 ! 2 2 � 1 + r 1 + r 2 1 / 2 ! + r 3 � 1 / 3 ! + · · · 0 = 1 + r 2 + r 2 2 / 2 ! + r 3 0 2 / 3 ! + · · · e r 1 � � 0 � = . e r 2 0 Obviously, there is nothing special about the dimension 2 in this example. In general, the exponential of a diagonal matrix is the diagonal matrix containing the exponentials of the diagonal entries. In particular, we have e rI = e r I . (1.8) When r = 0, we have the important special case e 0 I = e 0 I = I . (1.9) Soltuion to the intial value problem We will be looking at e t A = I + t A + t 2 2 ! A 2 + t 3 3 ! A 3 + · · · . (1.10) for a fixed n × n matrix A and a real number t . Consider e t A as a function of t with values that are n × n matrices. If v ∈ R n , we will also be looking at the function e t A v = v + t A v + t 2 2 ! A 2 v + t 3 3 ! A 3 v + · · · , (1.11) which has values in R n . Remember that it is our goal to compute e t A v for every v in a basis of R n . Let’s prove immediately that the exponential can be used to solve the initial value problem. Suppose A is an n × n matrix. PROPOSITION 1.12 1. Then d dt e t A = Ae t A . 2. If v ∈ R n , the function x ( t ) = e t A v is the solution to the initial-value problem x ′ = A x x ( 0 ) = v . with

447 9.1 Overview of the Technique Proof Let’s prove part (2) first, since it easily follows from part (1). First, by part (1), dt x ( t ) = d d = d e t A v e t A � v = Ae t A v = A x ( t ). � � � dt dt To finish, we evaluate (1.11) at t = 0 to show that x ( 0 ) = v . To prove part (1), we differentiate the series (1.10) term by term, dt e t A = d d I + t A + t 2 2 ! A 2 + t 3 � � 3 ! A 3 + · · · dt = A + t 1 ! A 2 + t 2 2 ! A 3 + · · · I + t A + t 2 � � 2 ! A 2 + · · · = A = Ae t A . The second part of Proposition 1.12 must seem like a panacea for readers who have been laboriously computing solutions to initial-value problems. It would be one, indeed, if the exponential of a matrix were easy to compute. However, the exponential of a matrix is usually difficult to compute. Truncation There is one other situation where we can easily compute the exponential e t A v . If most of the terms in the series in (1.11) are equal to the zero matrix, the infinite series becomes a finite sum. For example, if A 2 v = 0 and if p > 2, then A p v = A p − 2 A 2 v = A p − 2 0 = 0 . Therefore, using (1.11) we have e t A v = v + t A v + t 2 2 ! A 2 v + t 3 3 ! A 3 v + · · · = v + t A v . When this happens we will say that the series for e t A v truncates . Since we will refer to it often, we will state the result as a proposition. PROPOSITION 1.13 Suppose A is an n × n matrix and v is an n -vector. 1. If A v = 0 , then e t A v = v for all t . 2. If A 2 v = 0 , then e t A v = v + t A v for all t . 3. More generally, If A k v = 0 , then t k − 1 e t A v = v + t A v + · · · + ( k − 1 ) ! A k − 1 v for all t . According to Proposition 1.13, we can compute e t A v whenever the vector v is in the nullspace of A or in the nullspace of a power of A . Let’s use the result to compute some examples.

448 Chapter 9 Linear Systems with Constant Coefficients � � − 3 − 6 Compute e t A v where A = and v = ( − 2 , 1 ) T . EXAMPLE 1.14 � 1 2 We compute that � � � � � � − 3 − 6 − 2 0 A v = = . 1 2 1 0 Therefore by part (1) of Proposition 1.13, e t A v = v . � EXAMPLE 1.15 � Consider ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − 4 − 2 1 − 1 0 A = v = w = ⎠ , ⎠ , ⎠ . 4 2 − 2 2 and 0 ⎝ ⎝ ⎝ 8 4 0 0 1 Compute e t A v and e t A w . We compute that ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − 4 − 2 1 − 1 0 A v = ⎠ = ⎠ . 4 2 − 2 2 0 ⎝ ⎠ ⎝ ⎝ 8 4 0 0 0 Hence by part (1) of Proposition 1.13, e t A v = v . On the other hand, ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − 4 − 2 1 0 1 A w = ⎠ = ⎠ = − v . 4 2 − 2 0 − 2 ⎝ ⎠ ⎝ ⎝ 8 4 0 1 0 so A 2 w = 0 . Therefore by part (2) of Proposition 1.13, e t A w = w + t A w = w − t v ⎛ t ⎞ − 2 t ⎠ . � = ⎝ 1 Proposition 1.13, together with the examples illustrating its use, provides a very modest beginning to the computations of e t A v . However, this modest beginning will bear fruit when we learn some properties of the exponential. The law of exponents The key property of the ordinary exponential function is the laaw of exponents e a + b = e a e b . It’s time for us to explore the extent to which this property remains true for the exponential of a matrix.