Examples include: (a) the Lorenz system for climate and weather modeling (b) the Hodgkin-Huxley system for neuron modeling

Transcription

1 1 Introduction Many natural processes can be viewed as dynamical systems, where the system is represented by a set of state variables and its evolution governed by a set of differential equations. Examples include: (a) the Lorenz system for climate and weather modeling (b) the Hodgkin-Huxley system for neuron modeling The purpose of this course is to develop necessary mathematical tools to facilitate the study of dynamical systems.

2 2 1 Linear Systems The goal of this chapter is to study linear systems of ordinary differential equations: ẋ = Ax, x(0) = x 0, (1) where x R n, A is an n n matrix and ẋ = dx ( dt = dx1 dt,..., dx ) T n. dt It will be shown that the unique solution of Eq. (1) is given by x(t) = e At x 0, where e At is an n n matrix defined by its Taylor series. A good portion of this chapter is concerned with the computation of e At in terms of the eigenvalues and eigenvectors of A.

3 3 1.1 Uncoupled Linear Systems Let s start the solution of the linear system (1) with the simplest case, where the system contains only one equation (i.e. n = 1): ẋ = ax, x(0) = c. The method of separation of variables immediately gives x(t) = c e at.

4 4 2 2 Uncoupled Systems: an Example To move one step forward, let s consider the following 2 2 uncoupled system: ẋ 1 = x 1, ẋ 2 = 2x 2, x 1 (0) = c 1, x 2 (0) = c 2. Its solution is easily found to be: x 1 (t) = c 1 e t, x 2 (t) = c 2 e 2t, or in matrix form: x(t) = [ e t 0 0 e 2t ] c =: e At c.

5 5 The General Case Clearly, the same procedure can be applied to solve uncoupled systems of any size. For example, the solution of the following 3 3 system: ẋ 1 = x 1, ẋ 2 = x 2, ẋ 3 = x 3, x 1 (0) = c 1, x 2 (0) = c 2, x 3 (0) = c 3. is given by x(t) = e t e t 0 c =: eat c. 0 0 e t

6 6 Phase Plane Analysis Before wrapping up this section, let s introduce some notations that will be useful in the study of the linear system ẋ = Ax. Let s first consider the 2 2 system: [ 1 0 ẋ = Ax, A = 0 2 ]. Recall it has the solution: x(t) = [ e t 0 0 e 2t ] c. Clearly, the above formula describes the dynamics of each of the system components x 1 and x 2.

7 7 Phase Portrait In many cases, however, we are interested in the dynamics of the entire system x = (x 1, x 2 ) T besides that of the individual components x 1 or x 2. To gain a better understanding of the entire system, we eliminate the variable t from the solution representation so that a single formula involving only x 1 and x 2 results: x 2 = c2 1c 2 x 2 1 For any fixed c 1, c 2, the above equation defines a curve in the x 1 x 2 -plane the so-called phase plane. The set of all solution curves for all possible values of c 1 and c 2 constitutes a phase portrait of the linear system ẋ = Ax..

8 8 The Phase Portrait of the 2 2 System The phase portrait of the 2 2 system is shown below. Figure 1: The phase portrait of the 2 2 system.

9 9 Vector Field The direction of the motion of the solution x = (x 1, x 2 ) T along the solution curves can be read off from the explicit formula x 1 (t) = c 1 e t, x 2 (t) = c 2 e 2t. On the other hand, it can also be directly determined from the right-hand side [ ] 1 0 f(x) = Ax = 0 2 x of the system, which defines a vector field on the phase plane. The vector field must be tangent to the solution curves at every point x on the phase plane.

10 10 The Vector Field of the 2 2 System The vector field of the 2 2 system is shown below. Figure 2: The vector field of the 2 2 system.

11 11 The Phase Portrait of the 3 3 System Similar analysis can be carried out for more general linear systems. For example, the following figure shows the phase portrait of the 3 3 system ẋ = Ax where A = diag[1, 1, 1]: Figure 3: The phase portrait of the 3 3 system.

12 Diagonalization In the last section, we have seen how to solve uncoupled linear systems of the form ẋ = Ax = diag[λ 1,...,λ n ]x. The purpose of this and the following sections is to develop solution techniques for general, coupled linear system where the matrix A is not necessarily diagonal. The key is to reduce A to its diagonal form, or in more general situations, to its Jordan form.

13 13 Matrices with Real Distinct Eigenvalues Let s start with the simple case where A has real, distinct eigenvalues. The following theorem provides the basis for the solution of the linear system ẋ = Ax. Theorem 1 If the eigenvalues λ 1, λ 2,...,λ n of an n n matrix A are real and distinct, then any set of corresponding eigenvectors {v 1,v 2,...,v n } forms a basis for R n, the matrix P = [v 1,v 2,...,v n ] is invertible and P 1 AP = diag[λ 1, λ 2,...,λ n ]. The proof of the theorem can be found in any standard linear algebra text, for example Lowenthal [Lo].

14 14 Matrices with Real Distinct Eigenvalues (Cont d) Using the above theorem, we may solve the linear system ẋ = Ax by introducing the change of variable: y = P 1 x. It reduces the original system to an uncoupled linear system: ẏ = diag[λ 1,...,λ n ]y, and the solution of the original system can then be easily found: x(t) = PE(t)P 1 x(0), where E(t) is the diagonal matrix E(t) = diag [ e λ 1t,...,e λ nt ].

15 15 Example As an example, consider the linear system ẋ 1 = x 1 3x 2, ẋ 2 = 2x 2, x 1 (0) = c 1, x 2 (0) = c 2. Using the procedure described above, the solution is found to be x 1 (t) = c 1 e t + c 2 (e t e 2t ), x 2 (t) = c 2 e 2t.

16 16 Example (Cont d) The phase portrait of the above system is shown below. Figure 4: The phase portrait of the example.

17 17 Stable and Unstable Subspaces Note that the subspaces spanned by the eigenvectors v 1 and v 2 of the matrix A determine the stable and unstable subspaces of the linear system ẋ = Ax, according to the following definition. Definition 2 Suppose that the n n matrix A has k negative eigenvalues λ 1,...,λ k and n k positive eigenvalues λ k+1,...,λ n and that these eigenvalues are distinct. Let {v 1,...,v n } be a corresponding set of eigenvectors. Then the stable and unstable subspaces of the linear system, E s and E u, are the linear subspaces spanned by {v 1,...,v k } and {v k+1,...,v n } respectively; i.e., E s = span{v 1,...,v k }, E u = span{v k+1,...,v n }.

18 Exponentials of Operators In the last section, we have seen how to solve the linear system ẋ = Ax when A has real distinct eigenvalues, or more generally when A is diagonalizable. The purpose of this and the following sections is to study the general case where A is not necessarily diagonalizable. The key is to define the matrix exponential e At and verify the identity d dt eat = Ae At.

19 19 Matrices as Linear Operators We shall define e At through the Taylor series e At = k=0 1 k! Ak t k, but first we need to make sure that the series converges in appropriate norms. To introduce a norm (i.e. a measure ) for an n n matrix A, we view it as a linear operator T that maps an element in R n (i.e. an n-vector) to another element in R n : T : R n R n, T(x) = Ax. It can be shown that the converse is also true, i.e. any linear operator that maps R m to R n can be identified with an n m matrix. So matrices are indeed synonyms for linear operators.

20 20 Operator Norm For a linear operator T : R n R n, we define the operator norm: T = sup x =0 T(x), x where x denotes the Euclidean norm of x R n : x = x x2 n. It can be readily verified that the operator norm has the following equivalent definitions: T = sup T(x) or T = sup T(x). x 1 x =1 Remark. The induced norm of the matrix representation A of the operator T is called the 2-norm of A.

21 21 Example To illustrate the concept of operator norm, let s compute A for the 2 2 matrix [ ] a b A =. b a For any x R 2, it is easy to see that Ax 2 = (a 2 + b 2 ) x 2. Hence the induced norm (or 2-norm) of A is given by A = a 2 + b 2. In general, it can be shown that the 2-norm of an m n matrix A is the largest singular value of A, i.e., A = λ max (A H A).

22 22 Properties of the Operator Norm The operator norm has all of the usual properties of a norm, namely for any linear operators S, T : R n R n, (a) T 0 and T = 0 iff T = 0 (positive definiteness) (b) at = a T for any a R (positive homogeneity) (c) S + T S + T (triangle inequality or subadditivity) It can be shown that the space L(R n ) of linear operators T : R n R n equipped with the norm is a complete normed space, or in other words, a Banach space. The convergence of a sequence of operators T k L(R n ) can then be defined in terms of the norm.

23 23 Convergence in Operator Norm Definition 3 A sequence of linear operators T k L(R n ) is said to converge to a linear operator T L(R n ) as k, i.e., lim T k = T, k if for any ǫ > 0, there exists an N such that T T k < ǫ for all k N. Now we can show that the infinite Taylor series e Tt = k=0 1 k! T k t k, converges in the operator norm.

24 24 The Operator Exponential e Tt Theorem 4 Given T L(R n ) and t 0 > 0, the series e Tt := k=0 1 k! T k t k is absolutely and uniformly convergent for all t t 0. Moreover, e Tt e Tt. To prove this theorem, we need the following lemma. Lemma 5 For S, T L(R n ) and x R n, (a) T(x) T x (b) TS T S (c) T k T k for k = 0, 1, 2,...

25 25 The Matrix Exponential e At By identifying an n n matrix A with a linear operator T L(R n ) via the relation T(x) = Ax, we may define the matrix exponential e At as follows. Definition 6 Let A be an n n matrix. Then for t R, e At is the n n matrix defined by the Taylor series e At = k=0 1 k! Ak t k. As will be shown later, the matrix exponential e At can be computed in terms of the eigenvalues and eigenvectors of A.

26 26 Properties of the Matrix Exponential We next establish some basic properties of the operator exponential e T in order to facilitate the computation of the corresponding matrix exponential e A. Proposition 7 If P and T are linear operators on R n and S = PTP 1, then e S = Pe T P 1. Corollary 8 If P 1 AP = diag[λ j ], then e At = P diag[e λ jt ]P 1. Proposition 9 If S and T are linear operators on R n which commute, i.e., ST = TS, then e S+T = e S e T = e T e S. Corollary 10 If T is a linear operator on R n, the inverse of the linear operator e T is given by (e T ) 1 = e T.

27 27 Properties of the Matrix Exponential (Cont d) Corollary 11 (Complex Conjugate Eigenvalues) If [ ] a b A =, b a then e A = e a [ cosb sinb sinb cosb ]. Corollary 12 (Nontrivial Jordan Block) If [ ] a b A =, 0 a then e A = e a [ 1 b 0 1 ].

28 28 Matrix Exponential for 2 2 Matrices It will be shown in Section 1.8 that, for any 2 2 matrix A, there is an invertible 2 2 matrix P (whose columns consist of generalized eigenvectors of A) such that the matrix B = P 1 AP has one of the following forms: [ ] [ ] [ ] λ 0 λ 1 a b B =, B =, or B =. 0 µ 0 λ b a It then follows that e At = Pe Bt P 1 where [ ] [ e λt e Bt 0 1 t =, e Bt = e λt 0 e µt 0 1 [ ] cosbt sinbt or e Bt = e at. sinbt cosbt ],

29 The Fundamental Theorem for Linear Systems In the last section, we have seen how to define the matrix exponential e At through the Taylor series e At = k=0 1 k! Ak t k. The purpose of this section is to verify the identity d dt eat = Ae At and solve the linear system ẋ = Ax in terms of the matrix exponential e At : x(t) = e At x(0).

30 30 The Fundamental Theorem We first compute the derivative of e At in the following lemma. Lemma 13 Let A be a square matrix, then d dt eat = Ae At. Now we are ready to formulate the main theorem. Theorem 14 Let A be an n n matrix. Then for a given x 0 R n, the initial value problem has a unique solution given by ẋ = Ax, x(0) = x 0, x(t) = e At x 0.

31 31 Example Solve the initial value problem [ ] 2 1 ẋ = x, x(0) = 1 2 [ 1 0 ], and sketch the solution curves in the phase plane R 2. By applying the fundamental theorem and the results from the last section, the solution is easily found to be [ ] cost x(t) = e 2t. sint

32 32 Example (Cont d) The phase portrait of the above system is shown below. Figure 5: The phase portrait of the example.

33 Linear Systems in R 2 Now we have learned how to solve the linear system ẋ = Ax in terms of the matrix exponential e At : x(t) = e At x(0), and the only task that remains is to compute e At for arbitrary square matrices A. Before moving on, we shall take a short discursion and discuss in this section the various phase portraits that are possible for linear systems in R 2, i.e. systems with 2 2 matrices A.

34 34 Reduction to Jordan Form Based on the results summarized at the end of Section 1.3, for any 2 2 matrix A, there is an invertible 2 2 matrix P such that the matrix B = P 1 AP has one of the following forms: [ ] [ ] [ ] λ 0 λ 1 a b B =, B =, or B =. 0 µ 0 λ b a It suffices to describe the phase portraits for the linear system ẏ = By since the phase portrait for ẋ = Ax can be obtained from that for ẋ = Bx under the linear transformation x = Py.

35 35 The Solution It follows from the fundamental theorem in Section 1.4 and the form of the matrix e Bt computed in Section 1.3 that the solution of ẏ = By with y(0) = y 0 is given by [ ] [ ] e λt 0 1 t y(t) = y 0, y(t) = e λt y 0, 0 e µt 0 1 [ cosbt sinbt or y(t) = e at sinbt cosbt We now list the various phase portraits that result from these solutions. ] y 0.

36 36 Case I: Saddle [ ] λ 0 Consider the case B = with λ < 0 < µ. The system 0 µ ẏ = By has a saddle at the origin in this case. Figure 6: A saddle at the origin.

37 37 Case II: Stable/Unstable Node [ ] λ 0 Consider the case B = with λ µ < 0 or B = 0 µ [ ] λ 1 with λ < 0. The system ẏ = By has a stable node 0 λ at the origin in these cases. If λ µ > 0 or λ > 0, the node is called an unstable node. Figure 7: A stable node at the origin.

38 38 Case III: Stable/Unstable Focus [ ] a b Consider the case B = with a < 0. The system b a ẏ = By has a stable focus at the origin in this case. If a > 0, the focus is called an unstable focus. Figure 8: A stable focus at the origin.

39 39 Case IV: Center [ 0 b Consider the case B = b 0 center at the origin in this case. ]. The system ẏ = By has a Figure 9: A center at the origin. Remark. If one (or both) of the eigenvalues of B is zero, i.e., if detb = 0, the origin is called a degenerate equilibrium point of the system ẏ = By.

40 40 Example As an example, consider the linear system [ ] 0 4 ẋ = Ax = x, x(0) = c. 1 0 It can be reduced to the Jordan form [ 0 2 ẏ = By = 2 0 ] y, and its solution can be found to be [ cos2t x(t) = 2 sin2t sin2t cos2t 1 2 ] c.

41 41 Example (Cont d) The phase portrait of the above system consists of concentric ellipses as shown in the following figure. Figure 10: The phase portrait of the example.

42 42 Classification of Linear Systems in R 2 Now we may classify the linear systems ẋ = Ax in R 2. Definition 15 The linear system ẋ = Ax is said to have a saddle, a node, a focus or a center at the origin if the matrix A is similar to one of the matrices B in Cases I, II, III or IV respectively. Remark. We say the matrix A is similar to the matrix B if there is a nonsingular matrix P such that P 1 AP = B. Furthermore, the direction of rotation of trajectories in the phase portraits for the systems ẋ = Ax and ẏ = By will be the same if detp > 0 (i.e., if P is orientation preserving) and it will be opposite if detp < 0 (i.e., if P is orientation reversing).

43 43 A Simple Criterion For deta 0, the following simple criterion can be used to determine the type of the linear system ẋ = Ax without reducing A to its Jordan form. Theorem 16 Let δ = deta and τ = tracea and consider the linear system ẋ = Ax. (a) If δ < 0 then the system has a saddle at the origin. (b) If δ > 0 and τ 2 4δ 0 then the system has a node at the origin; it is stable if τ < 0 and unstable if τ > 0. (c) If δ > 0, τ 2 4δ < 0, and τ 0 then the system has a focus at the origin; it is stable if τ < 0 and unstable if τ > 0. (d) If δ > 0 and τ = 0 then the system has a center at the origin.

44 44 The Bifurcation Diagram The above results can be summarized in a bifurcation diagram as shown in the following figure. Note that a stable node or focus of the linear system ẋ = Ax is called a sink and an unstable node or focus is called a source. Figure 11: A bifurcation diagram for the linear system ẋ = Ax.

45 Complex Eigenvalues Having studied the classification of the linear system ẋ = Ax in R 2, we return in this section to the computation of the matrix exponential e At for arbitrary square matrices A. We have learned how to compute e At for matrices A that (1) have real eigenvalues and (2) have a complete set of eigenvectors, in which case A is diagonalizable (in R). We still need to handle the case where A (1) has complex eigenvalues and/or (2) has an incomplete set of eigenvectors, in which case A is not diagonalizable (in R). In this section, we shall first handle the case where A has complex eigenvalues. This has been done for 2 2 systems but we need to generalize the results to arbitrary dimensions.

46 46 Matrices with Distinct Complex Eigenvalues Let s start with the special case where A has distinct, complex (only) eigenvalues. The following theorem provides the basis for the solution of the linear system ẋ = Ax (recall that complex eigenvalues occur in complex conjugate pairs). Theorem 17 If the 2n 2n real matrix A has 2n distinct complex eigenvalues λ j = a j +ib j and λ j = a j ib j and corresponding complex eigenvectors w j = u j + iv j, and w j = u j + iv j, j = 1,...,n, then {u 1,v 1,...,u n,v n } is a basis for R 2n, the matrix P = [v 1 u 1 v 2 u 2 v n u n ] is invertible and P 1 AP = diag [ aj b j b j a j ], a real 2n 2n matrix with 2 2 blocks along the diagonal.

47 47 Matrices with Distinct Complex Eigenvalues (Cont d) Using the above theorem, we may solve the linear system ẋ = Ax by introducing the change of variable y = P 1 x, as we have done before. Corollary 18 Under the hypotheses of the above theorem, the solution of the initial value problem ẋ = Ax, x(0) = x 0 is given by x(t) = P diag e a jt [ cosbj t sinb j t sinb j t cos b j t ] P 1 x 0.

48 48 Example Solve the initial value problem ẋ = x, x(0) = x By applying the above theorem, the solution is found to be e t cost e t sint 0 0 e t sint e t cost 0 0 x(t) = x 0 0 e 2t (cos t + sint) 2e 2t sint e 2t sint e 2t (cos t sint)

49 49 Matrices with Distinct Mixed Eigenvalues In case A has both real and complex eigenvalues and they are distinct, we have the following result. Theorem 19 If A has distinct real eigenvalues λ j and corresponding eigenvectors v j, j = 1,...,k and distinct complex eigenvalues λ j = a j + ib j and λ j = a j ib j and corresponding eigenvectors w j = u j +iv j and w j = u j iv j, j = k +1,...,n, then the matrix P = [v 1 v k v k+1 u k+1 v n u n ] is invertible and P 1 AP = diag[λ 1,...,λ k, B k+1,...,b n ], B j = [ aj b j b j a j ].

50 50 Example Solve the initial value problem ẋ = x, x(0) = x By applying the above theorem, the solution is found to be e 3t 0 0 x(t) = 0 e 2t (cos t + sint) 2e 2t sint x 0. 0 e 2t sint e 2t (cost sint)

51 51 Example (Cont d) The phase portrait of the above system is shown below. Figure 12: The phase portrait of the example.