Chapter 2 Linear Systems

Transcription

1 Chapter 2 Linear Systems This chapter deals with linear systems of ordinary differential equations (ODEs, both homogeneous and nonhomogeneous equations Linear systems are etremely useful for analyzing nonlinear systems The main emphasis is given for finding solutions of linear systems with constant coefficients so that the solution methods could be etended to higher dimensional systems easily The well-known methods such as eigenvalue eigenvector method and the fundamental matri method have been described in detail The properties of fundamental matri, the fundamental theorem, and important properties of eponential matri function are given in this chapter It is important to note that the set of all solutions of a linear system forms a vector space The eigenvectors constitute the solution space of the linear system The general solution procedure for linear systems using fundamental matri, the concept of generalized eigenvector, solutions of multiple eigenvalues, both real and comple, are discussed 2 Linear Systems Consider a linear system of ordinary differential equations as follows: 9 d ¼ _ ¼ a þ a 2 2 þþa n n þ b dt d 2 ¼ _ 2 ¼ a 2 þ a 22 2 þþa 2n n þ b 2 >= dt d n ¼ _ n ¼ a n þ a n2 2 þþa nn n þ b n >; dt ð2:þ where a ij ; b j ði; j ¼ ; 2; ; nþ are all given constants The system (2 canbe written in matri notation as Springer India 25 GC Layek, An Introduction to Dynamical Systems and Chaos, DOI 7/ _2 37

2 38 2 Linear Systems _ ¼ A þ b ð2:2þ where ðtþ ¼ð ðtþ; 2 ðtþ; ; n ðtþþ t ; b ¼ ðb ; b 2 ; ; b n Þ t are the column vectors and A =[a ij ] n n is the square matri of order n, known as the coefficient matri of the system The system (22 is said to be homogeneous if b ¼, that is, if all b i s are identically zero On the other hand, if b 6¼, that is, if at least one b i is nonzero, then the system is called nonhomogeneous We consider first linear homogeneous system as _ ¼ A ð2:3þ A differentiable function ðtþ is said to be a solution of (23 if it satisfies the equation _ ¼ A Let ðtþ and 2ðtÞ be two solutions of (23 Then any linear combination ðtþ ¼c ðtþþc 2 2ðtÞ of ðtþ and 2ðtÞ is also a solution of (23 This can be shown very easily as below _ ¼ c _ þ c 2 _ 2 and so A ¼ Aðc þ c 2 2Þ¼c A þ c 2 A 2 ¼ c _ þ c 2 _ 2 ¼ _ : The solution ¼ c þ c 2 2 is known as general solution of the system (23 Thus the general solution of a system is the linear combination of the set of all solutions of that system (superposition principle Since the system is linear, we may consider a nontrivial solution of (23 as ðtþ ¼a ekt ð2:4þ where a is a column vector with components a ¼ ða ; a 2 ; ; a n Þ t and λ is a number Substituting (24 into (23 we obtain ka e kt ¼ Aa e kt or; ða kiþa ¼ ð2:5þ where I is the identity matri of order n Equation (25 gives a nontrivial solution if and only if det(a kiþ ¼ ð2:6þ

3 2 Linear Systems 39 On epansion, Eq (26 gives a polynomial equation of degree n in λ, known as the characteristic equation of matri A The roots of the characteristic equation (26 are called the characteristic roots or eigenvalues or latent roots of A The vector a, which is a nontrivial solution of (25, is known as an eigenvector of A corresponding to the eigenvalue λ If a is an eigenvector of a matri A corresponding to an eigenvalue λ, then ðtþ ¼e kt a is a solution of the system _ ¼ A The set of linearly independent eigenvectors constitutes a solution space of the linear homogeneous ordinary differential equations which is a vector space All properties of vector space hold good for the solution space We now discuss the general solution of a linear system below 22 Eigenvalue Eigenvector Method As we know, the solution of a linear system constitutes a linear space and the solution is formed by the eigenvectors of the matri There may have four possibilities according to the eigenvalues and corresponding eigenvectors of matri A We proceed now case-wise as follows Case I: Eigenvalues of A are real and distinct If the coefficient matri A has real distinct eigenvalues, then it has linearly independent (LI eigenvectors Let a; a2; ; a n be the eigenvectors corresponding to the eigenvalues k ; k 2 ; k n of matri A Then each jðtþ ¼a je kjt, j =,2,, n is a solution of _ ¼ A The general solution is a linear combination of the solutions j ðtþ and is given by ðtþ ¼Xn c j jðtþ j¼ where c ; c 2 ; ; c n are arbitrary constants In R 2, the solution can be written as ðtþ ¼X2 c j a je kjt ¼ c a e kt þ c 2 a 2e k2t : j¼ Case II: Eigenvalues of A are real but repeated In this case matri A may have either n linearly independent eigenvectors or only one or many (<n linearly independent eigenvectors corresponding to the repeated

4 4 2 Linear Systems eigenvalues The generalized eigenvectors have been used for linearly independent eigenvectors We discuss this case in the following two sub-cases Sub-case : Matri A has linearly independent eigenvectors Let a; a2; ; an be n linearly independent eigenvectors corresponding to the repeated real eigenvalue λ of matri A In this case the general solution of the linear system is given by ðtþ ¼Xn c i a ie kt : i¼ Sub-case 2 Matri A has only one or many (<n linearly independent eigenvectors First, we give the definition of generalized eigenvector of A Let λ be an eigenvalue of the n n matri A of multiplicity m n Then for k =,2,, m, any nonzero solution of the equation ða kiþ k v ¼ is called a generalized eigenvector of A For simplicity consider a two dimensional system Let the eigenvalues be repeated but only one eigenvector, say a be linearly independent Let a 2 be a generalized eigenvector of the 2 2 matri A Then a 2 can be obtained from the relation ða kiþa 2 ¼ a Aa 2 ¼ ka 2 þ a So the general solution of the system is given by ðtþ ¼c a e kt þ c 2 ðta e kt þ a 2 e kt Þ: Similarly, for an n n matri A, the general solution may be written as ðtþ ¼P n i¼ c i iðtþ, where ðtþ ¼a e kt ; 2 ðtþ ¼ta e kt þ a 2e kt ; 3 ðtþ ¼ t2 2! a e kt þ ta 2e kt þ a 3e kt ; n ðtþ ¼ tn ðn Þ! a e kt þ þ t2 2! a n 2 e kt þ ta n e kt þ a ne kt : Case III: Matri A has non-repeated comple eigenvalues Suppose the real n n matri A has m-pairs of comple eigenvalues a j ib j ; j ¼ ; 2; ; m Let a j ib ; j ¼ ; 2; ; m denote the corresponding j eigenvectors Then the solution of the system _ ðtþ ¼A ðtþ for these comple eigenvalues is given by

5 22 Eigenvalue-Eigenvector Method 4 ðtþ ¼Xm c j u j þ d j v j j¼ where u j ¼ epða j tþfa j cosðb jtþ b j sinðb jtþg, v j ¼ epða j tþfa j sinðb jtþþb j cosðb j tþg and c j ; d j ðj ¼ ; 2; ; mþ are arbitrary constants We discuss each of the above cases through specific eamples below Eample 2 Find the general solution of the following linear homogeneous system using eigenvalue-eigenvector method: _ ¼ 5 þ 4y _y ¼ þ 2y: Solution In matri notation, the system can be written as _ ¼ A, where ¼ and A ¼ 5 4 The eigenvalues of A satisfy the equation y 2 det(a kiþ ¼ 5 k 4 2 k ¼ ð5 kþð2 kþ 4 ¼ k 2 7k þ 6 ¼ : The roots of the characteristic equation k 2 7k þ 6 ¼ are λ =, 6 So the eigenvalues of A are real and distinct We shall now find the eigenvectors corresponding to these eigenvalues Let e ¼ e be the eigenvector corresponding to the eigenvalue k ¼ Then e 2 ða IÞe ¼ 5 4 e ¼ 2 e 2 4e þ 4e 2 ¼ e þ e 2 4e þ 4e 2 ¼ ; e þ e 2 ¼ :

6 42 2 Linear Systems We can choose e ¼ ; e 2 ¼ So, the eigenvector corresponding to the eigenvalue k ¼ ise ¼ : Again, let e ¼ e be the eigenvector corresponding to the eigenvalue k 2 ¼ 6 Then e 2 ða 6IÞe ¼ e 2 6 e ¼ 2 e þ 4e 2 e ¼ 4e 2 e þ 4e 2 ¼ e 4e 2 ¼ : We can choose e ¼ 4; e 2 ¼ So, the eigenvector corresponding to theeigenvalue k 2 ¼ 6 is e ¼ 4 The eigenvectors e, e are linearly independent Hence the general solution of the system is given as ðþ¼c t e e t þ c 2 e e 6 t ¼ c e t 4 þ c 2 e 6 t t ðþ¼c or, e t þ 4c 2 e 6 t yt ðþ¼ c e t þ c 2 e 6 t, where c ; c 2 are arbitrary constants Eample 22 Find the general solution of the linear system d dt y y ¼ 3 3 Solution The characteristic equation of matri A is det(a kiþ ¼ 3 k or; 3 k ¼ or; ð3 kþ 2 ¼ or; k ¼ 3; 3: So, the eigenvalues of A are 3, 3, which are real and repeated Clearly, e ¼ and e 2 ¼ are two linearly independent eigenvectors corresponding to the repeated eigenvalue λ = 3 Thus, the general solution of the system is

7 22 Eigenvalue-Eigenvector Method 43 t ðþ yt ðþ ðþ¼c t e e k t þ c 2 e 2e k t ¼ c e 3t þ c 2 e 3t ¼ c e 3t c 2e 3t t ðþ¼c e 3t ; yt ðþ¼c 2 e 3t ; where c ; c 2 are arbitrary constants: : Eample 23 Find the general solution of the system _ ¼ 3 4y _y ¼ y using eigenvalue-eigenvector method Solution The characteristic equation of matri A is det(a kiþ ¼ 3 k 4 k ¼ k 2 2k þ ¼ k ¼ ; So matri A has repeated real eigenvalues λ =, Let e ¼ e be the eigenvector corresponding to the eigenvalue λ = Then e 2 ða IÞe ¼ 3 4 e ¼ e 2 2e 4e 2 ¼ e 2e 2 2e 4e 2 ¼ ; e 2e 2 ¼ We can choose e ¼ 2; e 2 ¼ Therefore, e ¼ 2 Let g ¼ g be the generalized eigenvector corresponding to the eigenvalue λ = Then g 2

8 44 2 Linear Systems ða IÞg ¼ e 3 4 g ¼ 2 g 2 2g 4g 2 ¼ 2 g 2g 2 2g 4g 2 ¼ 2; g 2g 2 ¼ We can choose g 2 ¼ ; g ¼ 3 Therefore g ¼ 3 Therefore the general solution of the system is e e t þ c 2 e te t þ g e t t ðþ 2 or; ¼ c e t 2 þ c 2 te t 3 þ c 2 e t yt ðþ t ðþ¼f2c or, þ ð2t þ 3Þc 2 ge t yt ðþ¼fc þ ðt þ Þc 2 ge t, where c and c 2 are arbitrary constants Eample 24 Find the general solution of the linear system _ ¼ y _y ¼ 25 þ 2y Solution Given system can be written as _ ; where A ¼ 25 2 and ¼ y : The characteristic equation of matri A is det(a kiþ ¼ k 25 2 k ¼ k 2 2k þ 45 ¼ k ¼ 6 3i: Therefore, matri A has a pair of comple conjugate eigenvalues 6 ± 3i Let e ¼ e be the eigenvector corresponding to the eigenvalue λ =6+3i Then e 2

9 22 Eigenvalue-Eigenvector Method 45 ða ð6 þ 3iÞIÞe ¼ 6 3i e ¼ i e 2 ð4 3iÞe e 2 ¼ 25e ð4 þ 3iÞe 2 ð4 3iÞe e 2 ¼ ; 25e ð4 þ 3iÞe 2 ¼ : A nontrivial solution of this system is e ¼ ; e 2 ¼ 4 3i: Therefore e ¼ ¼ þ i ¼ a 4 3i 4 3 þ ia 2, where a ¼ 4 and a 2 ¼ 3 Similarly, the eigenvector corresponding to the eigenvalue λ = 6 3i is e ¼ ¼ a 4 þ 3i ia 2 Therefore, u ¼ e at a cos bt a 2 sin bt ¼ e 6t 4 cos 3t sin 3t 3 and v ¼ e at a sin bt þ a 2 cos bt ¼ e 6t 4 sin 3t þ cos 3t : 3 Therefore, the general solution is ðþ¼c t u þ d v ¼ e 6t c cos 3t c sin 3t 4 3 þ e 6t d sin 3t þ d cos 3t 4 3 ¼ e 6t c cos 3t þ d sin 3t ð4c 3d Þcos 3t þ ð3c þ 4d Þsin 3t t ðþ¼e 6t ðc cos 3t þ d sin 3tÞ; yt ðþ¼e 6t ½ð4c 3d Þcos 3t þ ð3c þ 4d Þsin 3tŠ where c and d are arbitrary constants

10 46 2 Linear Systems Eample 25 Find the solution of the system _ ¼ 5y; _y ¼ 3y satisfying the initial condition ( =, y( = Describe the behavior of the solution as t Solution The characteristic equation of matri A is det(a kiþ ¼ k 5 3 k ¼ k 2 þ 2k þ 2 ¼ k ¼ i: So, matri A has a pair of comple conjugate eigenvalues ( ± i Let e ¼ e be the eigenvector corresponding to the eigenvalue e 2 λ = +i Then ða ð þ iþiþe ¼ 2 i 5 e ¼ 2 i e 2 ð2 iþe 5e 2 ¼ e ð2 þ iþe 2 ð2 iþe 5e 2 ¼ ; e ð2 þ iþe 2 ¼ : A nontrivial solution of this system is a 2 ¼ e Therefore e ¼ 2 þ i ¼ 2 þ i e ¼ 2 þ i; e 2 ¼ : ¼ a þ ia 2, where a ¼ 2 Similarly, the eigenvector corresponding to the eigenvalue λ = i is ¼ 2 i ¼ a ia 2 and

11 22 Eigenvalue-Eigenvector Method 47 u ¼ e at a cos bt a2 sin bt v ¼ e at a sin bt þ a2 cos bt Therefore, the solution of the system is ðþ¼ t ðþu þ yðþv ¼ e t 2 ¼ e t 2 cos t ¼ e t 2 ¼ e t 2 sin t ðcos t þ sin tþþ þ e t 2 cos t sin t sin t þ cos t ðcos t sin tþ and sin t þ cos t When t, e t! So, in this case ðþ! t, that is, the solution of the system is stable in the usual sense Eample 26 Find the solution of the system _ ¼ A, where 2 3 A 2 A: 3 Solution The characteristic equation of A is det(a kiþ ¼ k k ¼ 3 k ðk þ Þðk Þðk þ 3Þ ¼ k ¼ ; ; 3 Therefore the eigenvalues of matri A are λ =,, 3 We shall now find the eigenvector corresponding to each of the eigenvalues e Let e e 2 A be the eigenvector corresponding to the eigenvalue λ = e 3 Then :

12 48 2 Linear Systems ða þ Þe ¼ þ 2 3 e B CB C B 2 þ A@ e 2 A A 3 e 3 2e 2 þ 3e 3 ¼ ; e 2 þ e 3 ¼ ; 3e 2 þ e 3 ¼ e 2 ¼ e 3 ¼ and e is arbitrary: We choose e ¼ Therefore, the eigenvector corresponding to the eigenvalue λ = ise A Similarly, the eigenvectors corresponding to λ =andλ = 3 are, respectively, g ¼ is A and a ¼ 3 A Therefore the general solution ðtþ ¼c e e t þ c 2 g e t þ c 3 a e 3t =2 =2 B C ¼ Ae t B C þ c Ae t B C þ c 3 where c,c 2 and c 3 are arbitrary constants Eample 27 Solve the system _ ¼ A, where 3 3 A 3 5 3A: Solution The characteristic equation of matri A is det(a kiþ ¼ k k 3 ¼ k k ¼ 4; 2; 2: Ae 3t

13 22 Eigenvalue-Eigenvector Method 49 So ( 2 is a repeated eigenvalue of A The eigenvector for the eigenvalue k ¼ 4 is given A The eigenvector corresponding to the repeated eigenvalue 2 k 2 ¼ k 3 ¼ 2isðe e 2 e 3 Þ T such that A e 2 A A e 3 which is equivalent to 3e 3e 2 þ 3e 3 ¼ ; 3e 3e 2 þ 3e 3 ¼ ; 6e 6e 2 þ 6e 3 ¼ ; that is, e e 2 þ e 3 ¼ We can choose e ¼, e 2 ¼ and e 3 ¼, and so we can take one eigenvector A Again, we can choose e ¼, e 2 ¼ and e 3 ¼ Then we obtain another A Clearly, these two eigenvectors are linearly independent Thus, we have two linearly independent eigenvectors corresponding to the repeated eigenvalue 2 Hence, the general solution of the system is given by ðtþ Ae 4t þ c Ae 2t þ c 2 where c, c 2 and c 3 are arbitrary constants Eample 28 Solve the system _ ¼ A where A ¼ Ae 2t

14 5 2 Linear Systems Solution Here matri A has two pair of comple conjugate eigenvalues k ¼ i and k 2 ¼ i The corresponding pair of eigenvectors is w ¼ a ib ¼ w 2 ¼ a 2 ib 2 ¼ i C A ¼ i B A i B A and C A ¼ C A i B A Therefore, the general solution of the system is epressed as ðtþ ¼X2 c j u j þ d j v j j¼ >< ¼ c e t >= >< B C cos t B A sin t þ c 2 e t >= B C cos t B A sin t >: >; >: >; 8 9 >< þ d e t B A sin t þ >= B A cos t >: >; 8 9 >< þ d 2 e t B A sin t þ >= B A cos t >: >; e t ðd cos t c sin tþ e t ðc cos t þ d sin tþ ¼ B e t fðd 2 c 2 Þ cos t ðd 2 þ c 2 Þ sin tg A e t ðc 2 cos t þ d 2 sin tþ where c j ; d j ðj ¼ ; 2Þ are arbitrary constants

15 23 Fundamental Matri 5 23 Fundamental Matri A set f ðtþ; 2ðtÞ; ; nðtþg of solutions of a linear homogeneous system _ ¼ A is said to be a fundamental set of solutions of that system if it satisfies the following two conditions: (i The set f ðtþ; 2ðtÞ; ; nðtþg is linearly independent, that is, for c ; c 2 ; ; c n 2 R; c þ c 2 2 þþc n n ¼ c ¼ c 2 ¼¼c n ¼ : (ii For any solution ðtþ of the system _ ¼ A, there eist c ; c 2 ; ; c n 2 R such that ðtþ ¼c ðtþþc 2 2ðtÞþ þc n nðtþ; 8t 2 R The solution, epressed as a linear combination of a fundamental set of solutions of a system, is called a general solution of the system Let f ðtþ; 2ðtÞ; ; nðtþg be a fundamental set of solutions of the system _ ¼ A for t 2 I ¼ ½ a; bš; a; b 2 R Then the matri UðtÞ ¼ ðtþ; 2ðtÞ; ; nðtþ is called a fundamental matri of the system _ ¼ A, 2 R n Since the set f ðtþ; 2ðtÞ; ; nðtþg is linearly independent, the fundamental matri UðtÞ is nonsingular Now the general solution of the system is ðtþ ¼c ðtþþc 2 2ðtÞþ þc n nðtþ c c 2 ¼ ðtþ; 2ðtÞ; ; nðtþ B A ¼ UðtÞ c c n where c ¼ðc ; c 2 ; c n Þ t is a constant column vector If the initial condition is ðþ ¼, then UðÞ c ¼ c ¼ U ðþ ½Since UðtÞ is nonsingular for all tš:

16 52 2 Linear Systems Thus the solution of the initial value problem _ ¼ A with the initial conditions ðþ ¼ can be epressed in terms of the fundamental matri Φ(t as ðtþ ¼UðtÞU ðþ ð2:7þ Note that two different homogeneous systems cannot have the same fundamental matri Again, if UðtÞ is a fundamental matri of _ ¼ A, then for any constant C, CUðtÞ is also a fundamental matri of the system Eample 29 Find the fundamental matri of the system _ ¼ A, where 2 A ¼ Hence find its solution 3 2 Solution The characteristic equation of matri A is ja kij ¼ k k ¼ ð kþð2 kþ 6 ¼ k 2 3k 4 ¼ k ¼ ; 4: So, the eigenvalues of matri A are, 4, which are real and distinct Let e ¼ e be the eigenvector corresponding to the eigenvalue k ¼ Then e 2 ða þ I Þe ¼ þ 2 e ¼ 3 2þ e 2 2e 2e 2 ¼ ; 3e þ 3e 2 ¼ : A nontrivial solution of this system is e ¼ ; e 2 ¼ e ¼ : Again, let g ¼ g k 2 ¼ 4 Then g 2 be the eigenvector corresponding to the eigenvalue

17 23 Fundamental Matri 53 ða 4IÞg ¼ 4 2 g g 2 ¼ 3g þ 2g 2 ¼ 2 Choose g ¼ 2; g 2 ¼ 3 Therefore, g ¼ 3 Therefore the eigenvectors corresponding to the eigenvalues λ =, 4 are 2 respectively and, which are linearly independent So two fundamental solutions of the system are 3 ðþ¼ t e t 2 ; 2ðÞ¼ t e 4t 3 and a fundamental matri of the system is UðtÞ ¼ ðtþ 2ðtÞ ¼ e t 2e 4t e t 3e 4t Now UðÞ ¼ 2 and so U ðþ ¼ Therefore the general solution of the system is given by : ðtþ ¼UðtÞU ðþ ¼ e t 2e 4t e t 3e 4t ¼ 3e t þ 2e 4t 2e t 2e 4t 5 3e t 3e 4t 2e t þ 3e 4t : 23 General Solution of Linear Systems Consider a simple linear equation _ ¼ a ð2:8þ with initial condition ðþ ¼, where a and are certain constants The solution of this initial value problem (IVP is given as ðtþ ¼ e at Then we may epect that the solution of the initial value problem for n n system

18 54 2 Linear Systems _ ¼ A with ðþ ¼ ð2:9þ can be epressed in term of eponential matri function as ðþ¼eat t ð2:þ where A is an n n matri Comparing (2 with the solution obtained by the fundamental matri, we have the relation e At ¼ UðtÞU ðþ ð2:þ Thus we see that if UðtÞ is a fundamental matri of the system _ ¼ A, then UðÞ is invertible and e At ¼ UðtÞU ðþ Note that if UðÞ ¼I, then U ðþ ¼I and so, e At ¼ UðtÞI ¼ UðtÞ 2e Eample 2 Does UðtÞ ¼ t e 3t 4e t 2e 3t a fundamental matri for a system _ ¼ A? Solution We know that if UðtÞ is a fundamental matri, then UðÞ is invertible 2e Here UðtÞ ¼ t e 3t 2 4e t 2e 3t So, UðÞ ¼ 4 2 Since det(uðþþ ¼ 4 4 ¼, UðÞ is not invertible and hence the given matri is not a fundamental matri for the system _ ¼ A Eample 2 Find e At for the system _ ¼ A, where A ¼ 4 Solution The characteristic equation of A is ja kij ¼ k 4 k ¼ ðk Þ 4 ¼ k ¼ 3; : So, the eigenvalue of A are λ =3, The eigenvector corresponding to the eigenvalues λ = 3, are, respectively, and, which are linearly 2 2 independent So, two fundamental solutions of the system are e 3t ; 2ðÞ¼ t e t Therefore a fundamental matri of the ðþ¼ t 2 system is 2

19 23 Fundamental Matri 55 UðtÞ ¼ ðtþ 2ðtÞ ¼ e3t e t 2e 3t 2e t Now, UðÞ ¼ and U ðþ ¼ Therefore, : ¼ e At ¼ UðtÞU ðþ ¼ e3t e t 2 2e 3t 2e t 4 2 4! ¼! 2 ð e3t þ e t Þ 4 ð e3t e t Þ ðe 3t e t Þ 2 ð e3t þ e t : Þ 232 Fundamental Matri Method The fundamental matri can be used to obtain the general solution of a linear system The fundamental theorem gives the eistence and uniqueness of solution of a linear system _ ¼ A, 2 R n subject to the initial conditions 2 R n We now present the fundamental theorem Theorem 2 (Fundamental theorem Let A be an n n matri Then for given any initial condition 2 R n, the initial value problem _ ¼ A with ðþ ¼ has the unique solution ðtþ ¼e At Proof The initial value problem is We have _ ¼ A ; ðþ ¼ ð2:2þ Differentiating (23 wrto t, d e At d ¼ dt dt e At ¼ I þ At þ A2 t 2 2! þ A3 t 3 3! þ I þ At þ A2 t 2 þ A3 t 3 þ 2! 3! ¼ d ðþþ I d ðatþþ d A 2 t 2 þ d dt dt dt 2! dt A 3 t 3 3! þ ð2:3þ

20 56 2 Linear Systems The term by term differentiation is valid because the series of e At is convergent for all t under the operator or; d dt e At ¼ u þ A þ A 2 t þ A3 t 2 2! ¼ A Iþ At þ A2 t 2 ¼ Ae At : 2! þ A4 t 3 3! þ A3 t 3 3! þ þ Therefore, d e At ¼ Ae At dt ð2:4þ This shows that the matri ¼ e At is a solution of the matri differential equation _ ¼ A The matri e At is known as the fundamental matri of the system (22 Now using (24 d e At dt ¼ d e At ¼ Ae At dt _ ¼ d dt ð Þ¼A ; where ¼ e At Also, ðþ ¼ e At t¼ ¼ ½e At Š t¼ ¼ I ¼ Thus ðtþ ¼e At is a solution of (22 We prove the uniqueness of solution as follows Let ðtþ be a solution of (22 andy ðtþ ¼e At ðtþ be its another solution Then _y ðtþ ¼ Ae At ðtþþe At _ ðtþ ¼ Ae At ðtþþae At ðtþ ¼: This implies yðtþ is constant At t =, for t 2 R, it shows that yðtþ ¼ Therefore any solution of the IVP (22 is given as ðtþ ¼e At yðtþ ¼e At This completes the proof 233 Matri Eponential Function From the fundamental theorem, the general solution of a linear system can be obtained using the eponential matri function The eponential matri function has

21 23 Fundamental Matri 57 some interesting properties in which the general solution can be obtained easily For an n n matri A, the matri eponential function e A of A is defined as e A ¼ X n¼ A n n! ¼ I þ A þ A2 þ ð2:5þ 2! Note that the infinite series (25 converges for all n n matri A IfA =[a], a matri, then e A ¼ ½e a Š(see the book by L Perko [] We now discuss some of the important properties of matri eponential function e A Property If A = φ, the null matri, then e At ¼ I Proof By definition þ A3 t 3 3! e At ¼ I þ At þ A2 t 2 2! ¼ I þ ut þ u2 t 2 2! ¼ I: So, e At ¼ I for A = φ Property 2 Let A = I, the identity matri Then þ u3 t 3 3! e At ¼ et e t ¼ Ie t : þ þ Proof We know that e At ¼ I þ At þ A2 t 2 2! þ A3 t 3 3! þ Therefore e It ¼ I þ It þ I2 t 2 þ I3 t 3 þ 2! 3! ¼ I þ It þ It2 2! þ It3 3! þ ¼ I þ t þ t2 2! þ t3 3! þ ¼ Ie t ¼ e t ¼ et e t : Note If A = αi, α being a scalar, then e At ¼ e ait ¼ Ie at ¼ eat e at :

22 58 2 Linear Systems Property 3 Suppose D ¼ k, a diagonal matri Then k 2 Proof By definition e Dt ¼ ek t e k 2t e Dt ¼ I þ Dt þ D2 t 2 þ D3 t 3 þ 2! 3! ¼ þ k t þ k 2 t 2 k 2 k 2 2! þ ¼ þ k " # t þ k2 t 2 k 2 k 2 2! þ ¼ þ k t þ k 2 t2 2! þ 4 5 þ k 2 t þ k2 2 t2 2! þ " # ¼ ek t : e k 2t Property 4 Let P AP ¼ D, D being a diagonal matri Then e At ¼ Pe Dt P ¼ P ek t e k P ; where D ¼ k : 2t k 2 Proof We have X n e At A k t k ¼ lim n! k! k¼ X n ðpdp Þ k t k ¼ lim ½* D ¼ P AP; so A ¼ PDP Š n! k! k¼ 2 X n PD k P t k ðpdp Þ k 3 ¼ ðpdp ÞðPDP ÞðPDP Þ 6 ¼ lim n! ¼ PDðP k! PÞDP ð PÞðP PÞDP k¼ ¼ PD k P! X n D k t k ¼ P lim P n! k! k¼ ¼ Pe Dt P " # ¼ P ek t P e k 2t

23 23 Fundamental Matri 59 Property 5 Let N be a nilpotent matri of order k Then e Nt is a series containing finite terms only Proof A matri N is said to be a nilpotent matri of order or inde k if k is the least positive integer such that N k ¼ u but N k 6¼ u, φ being the null matri Since N is a nilpotent matri of order k, N k 6¼ u but N k ¼ u Therefore e Nt ¼ I þ Nt þ N2 t 2 2! ¼ I þ Nt þ N2 t 2 2! þ N3 t 3 3! þ N3 t 3 3! þþ Nk t k ðk Þ! þ Nk t k k! þþ Nk t k ðk Þ! þ which is a series of finite terms only Property 6 If A ¼ a b, then e b a At ¼ e ait ½I cosðbtþþj sinðbtþš, where I ¼ and J ¼ Proof We have A ¼ a b b a J ¼ : Therefore ¼ a þ b ¼ ai þ bj; where I ¼ ; e At aitþbjt ¼ e " # ¼ e ait e bjt ¼ e ait I þ bjt þ ðbjtþ2 þ ðbjtþ3 þ 2! 3! ¼ e ait I b2 t 2 þ b4 t 4 þ þ J bt b3 t 3 þ 2! 4! 3! 2 * J 2 ¼ 3 ¼ e ait ¼ ¼ I ½I cosðbtþþj sinðbtþš J 3 ¼ J 2 J ¼ ð IÞJ ¼ J J 4 ¼ J 3 J ¼ ð JÞJ ¼ J 2 ¼ I 5 etc:

24 6 2 Linear Systems Property 7 e A þ B ¼ e A e B, provided AB = BA Proof Suppose AB = BA Then by Binomial theorem, Therefore ða þ BÞ n ¼ Xn k¼ n! ðn kþ!k! An k B k ¼ n! X j þ k¼n A j B k j!k! : e A þ B ¼ X n¼ ¼ X j¼ ¼ e A e B : ða þ B n! A j X B k j! k! k¼ Þ n ¼ X X n¼ j þ k¼n A j B k j!k! It is true that e A þ B ¼ e A e B if AB = BA But in general e A þ B 6¼ e A e B d Property 8 For any n n matri A, dt ðe At Þ ¼ Ae At Proof By definition d dt e At ¼ I þ At þ A2 t 2 e At ¼ d dt þ A3 t 3 þ 2! 3! I þ At þ A2 t 2 þ A3 t 3 þ 2! 3! ¼ d ðþþ I d ðatþþ d dt dt dt A 2 t 2 2! þ d dt A 3 t 3 3! þ The term by term differentiation is valid because the series of e At is convergent for all t under the operator or; d dt e At ¼ u þ A þ A 2 t þ A3 t 2 2! ¼ A Iþ At þ A2 t 2 ¼ Ae At : Therefore, d ð Þ¼ Ae At dt We now establish the important result below e At 2! þ A4 t 3 3! þ A3 t 3 3! þ þ

25 23 Fundamental Matri 6 Result Multiplying both sides of d e At ¼ Ae At by UðÞ in right, we have dt d e At UðÞ ¼Ae At UðÞ dt d e At UðÞ ¼ Ae At UðÞ dt d UðtÞU ðþuðþ ¼ AUðtÞU ðþuðþ [since e At ¼ UðtÞU ðþš dt d ðuðtþþ ¼ _UðtÞ ¼AUðtÞ: dt This shows that the fundamental matri UðtÞ must satisfy the system _ ¼ A This is true for all t So, it is true for t = Putting t =in _ UðtÞ ¼AUðtÞ, we get _UðÞ ¼AUðÞ A ¼ _UðÞU ðþ: This gives that the coefficient matri A can be epressed in terms of the fundamental matri UðtÞ Eample 22 Does UðtÞ ¼ et e 2t 2e t 3e 2t a fundamental matri for the system _ ¼ A? If so, then find the matri A Solution We know that if UðtÞ is a fundamental matri, then UðÞ is invertible Here UðtÞ ¼ et e 2t 2e t 3e 2t So,UðÞ ¼ 2 3 Since detðuðþþ ¼ 3 2 ¼ 6¼, UðÞ is invertible Hence the given matri is a fundamental matri for the system _ ¼ A We shall now find the coefficient matri A We have UðÞ ¼ SoU 3 ðþ ¼ Also _UðtÞ ¼ et 2e 2t 2e t 6e 2t, and _UðÞ ¼ Therefore the matri A is A ¼ UðÞU _ ðþ ¼ ¼ 7 3 : 8 8 Eample 23 Find e At for the matri A ¼ 3 3 Hence find the solution of the system _ ¼ A

26 62 2 Linear Systems Solution We see that the eigenvectors corresponding to the eigenvalues λ = 2,4of A are respectively e ¼ and g ¼, which are linearly independent Therefore, two fundamental solutions of the system are ðtþ ¼ e 2t and 2 ðtþ ¼ e 4t So a fundamental matri of the system is UðtÞ ¼ ðtþ 2ðtÞ ¼ We find UðÞ ¼ and U ðþ ¼ 2 e 2t e 2t e 4 t e 4 t : Therefore e At ¼ UðtÞU ðþ ¼ e 2t e 4t 2 e 2t e 4t ¼ e 2t þ e 4 t 2 e 4t e 2t e 4t e 2t e 2t þ e 4 t : By fundamental theorem, the solution of the system _ ¼ A is ðtþ ¼eAt ¼ e 2t þ e 4 t e 4t e 2t c 2 e 4t e 2t e 2t þ e 4 t where ¼ c is an arbitrary constant column vector c 2 c 2 24 Solution Procedure of Linear Systems The general solution of a linear homogeneous system can be easily deduced from the fundamental theorem According to this theorem the solution of _ ¼ A with ðþ ¼ is given as ðtþ ¼e At and this solution is unique For a simple change of coordinates ¼ Py where P is an invertible matri, the equation _ ¼ A is transformed as _ ¼ A P _y ¼ APy _y ¼ P AP y _y ¼ Cy ; where C ¼ P AP:

27 24 Solution Procedure of Linear Systems 63 The initial conditions ðþ ¼ become y ðþ ¼P ðþ ¼P ¼ y So, the new system is _y ¼ Cy It has the solution with y ðþ ¼y, where C ¼ P AP y ðtþ ¼eCt y : Hence the solution of the original system is ðtþ ¼Py ðtþ ¼ PeCt y ¼ Pe Ct P : We see that e At ¼ Pe Ct P The matri P is chosen in such a way that matri C takes a simple form We now discuss three cases (i Matri A has distinct real eigenvalues Let P ¼ a; a2; ; an so that, P eists The matri C is obtained as C ¼ P AP which is a diagonal matri Hence the eponential function of C becomes e Ct ¼ diagðe k t ; e k 2t ; ; e k nt Þ: Therefore we can write the solution of _ ¼ A e At ¼ Pe Ct P So with ðþ ¼ as ðtþ ¼ ðtþ ¼Pdiagðek t ; e k 2t ; ; e k nt ÞP where ¼ðc ; c 2 ; ; c n Þ t is an arbitrary constant (ii Matri A has real repeated eigenvalues In this case the following theorems are relevant (proofs are available in the book Hirsch and Smale [2] for finding general solution of a linear system when matri A has repeated eigenvalues Theorem 22 Let the n n matri A have real eigenvalues λ, λ 2,, λ n repeated according to their multiplicity Then there eists a basis of generalized eigenvectors fa ; a 2; ; a ng such that the matri P ¼ða ; a 2; ; a nþ is invertible and A = S + N, where P SP ¼ diagðk ; k ; ; k n Þ and N(=A S is nilpotent of order k n, and S and N commute Using the theorem the linear system subject to the initial conditions ðþ ¼ has the solution

28 64 2 Linear Systems ðtþ ¼Pdiagðek jt ÞP I þ Nt þ þ Nk t k ðk Þ! : (iii Matri A has comple eigenvalues Theorem 23 Let A be a 2n 2n matri with comple eigenvalues a j ± ib j, j =,2,, n Then there eists generalized comple eigenvectors ða j ib jþ; j ¼ ; 2; n such that the matri P ¼ðb ; a ; b 2; a 2; ; b n; a nþ is invertible and A = S + N, where P SP ¼ diag a j b j, and N(=A S is a nilpotent matri of order b j a j k 2n, and S and N commute Using the theorem the linear system of equations subject to the initial conditions ðþ ¼ has the solution ðtþ ¼Pdiagðea jt Þ cosðb jtþ sinðb j tþ P sinðb j tþ cosðb j tþ I þ Nt þ þ Nk t k k! : For a 2 2 matri A with comple eigenvalues ða ibþ the solution is given by ðtþ cos bt sin bt ¼Peat P sin bt cos bt : Eample 24 Solve the initial value problem _ ¼ þ y; _y ¼ 4 2y 2 with initial condition ðþ ¼ 3 Solution The characteristic equation of matri A is ja kij ¼ k 4 2 k ¼ ðk Þðk þ 2Þ 4 ¼ k 2 þ k 6 ¼ k ¼ 2; 3 So the eigenvalues of matri A are 2, 3, which are real and distinct

29 24 Solution Procedure of Linear Systems 65 Let e ¼ e be the eigenvector corresponding to the eigenvalue k ¼ 2 Then e 2 ða 2IÞe ¼ 2 e ¼ e 2 e þ e 2 ¼ ; 4e 4e 2 ¼ A nontrivial solution of this system is e ¼ ; e 2 ¼ e ¼ : Again let g ¼ g k 2 ¼ 3 Then g 2 be the eigenvector corresponding to the eigenvalue ða þ 3IÞg ¼ þ 3 g 4 2 þ 3 g 2 4g þ g 2 ¼ ; 4g þ g 2 ¼ ¼ A nontrivial solution of this system is g ¼ ; g 2 ¼ 4 Let P ¼ e ; g g ¼ : 4 ¼ 4 Then P ¼ 5 4 C ¼ P AP ¼ ¼ 8 2 ¼ e Ct ¼ e2t e 3t ¼ 5 4 ¼ 2 3

30 66 2 Linear Systems Therefore by the fundamental theorem, the solution of the system is ðtþ ¼eAt ¼ Pe Ct P ¼ e 2t 4 4 e 3t 5 ¼ 4e 2t þ e 3t e 2t e 3t 5 4e 2t 4e 3t e 2t þ 4e 3t ðtþ ¼ yðtþ 4 5 e2t þ 5 e 3t 5 e2t 5 e 3t 4 5 e2t 4 5 e 3t 5 e2t þ 4 5 e 3t ðtþ ¼e 2t þ e 3t ; yðtþ ¼e 2t 4e 3t :! 2 ¼ e2t þ e 3t 3 e 2t 4e 3t Eample 25 Solve the system _ ¼ 3 2 ; _ 2 ¼ 2 2 : Also sketch the phase portrait Solution The characteristic equation of matri A is ja kij ¼ k 3 2 k ¼ ðk þþðk 2Þ ¼ k ¼ ; 2 The eigenvalues of matri A are, 2, which are real and distinct Let e ¼ e be the eigenvector corresponding to the eigenvalue k ¼ Then e 2 Choose e = so that e ¼ ða þ IÞe ¼ þ 3 e ¼ 2þ e 2 3e 2 ¼ ; 3e 2 ¼ e 2 ¼ and e is arbitrary:

31 24 Solution Procedure of Linear Systems 67 Again, let g ¼ g be the eigenvector corresponding to the eigenvalue k 2 ¼ 2 Then g 2 ða 2IÞg ¼ 2 3 g 2 2 g þ g 2 ¼ Choose g ¼ ; g 2 ¼ Then g ¼ Let P ¼ e ; g ¼ Therefore g 2 ¼ Then P ¼ C ¼ P AP ¼ 3 2 ¼ ¼ 2 2 and so e Ct ¼ e t e 2t Therefore by fundamental theorem, the solution of the system is ðtþ ¼eAt ¼ Pe Ct P ¼ e t e 2t ¼ e t e t e 2t c e 2t c 2 ðtþ ¼ e t e t e 2t c ðc þ c 2 Þe t c 2 e 2t ¼ 2 ðtþ e 2t c 2 c 2 e 2t ðtþ ¼c e t þ c 2 e t e 2t ; 2 ðtþ ¼c 2 e 2t where c ; c 2 are arbitrary constants The phase diagram is presented in Fig 2 Eample 26 Solve the following system using the fundamental theorem _ ¼ 5 þ 4y _y ¼ þy

32 68 2 Linear Systems Fig 2 A typical phase portrait of the system Solution The characteristic equation of matri A is ja kij ¼ 5 k 4 k ¼ ðk Þðk 5Þþ4 ¼ k 2 6k þ 9 ¼ k ¼ 3; 3: This shows that matri A has an eigenvalue λ = 3 of multiplicity 2 Then S ¼ 3 and N ¼ A S ¼ 2 4 Clearly, matri N is a nilpotent matri 3 2 of order 2 So, the general solution of the system is given by ðtþ ¼eAt ¼ e ðs þ NÞt ¼ e St e Nt ¼ e3t e 3t ½I þ NtŠ ¼ e3t þ 2t 4t e 3t t 2t : Eample 27 Find the general solution of the system of linear equations _ ¼ 4 2y _y ¼ 5 þ 2y

33 24 Solution Procedure of Linear Systems 69 Solution The characteristic equation of matri A is ja kij ¼ 4 k k ¼ ðk 4Þðk 2Þþ ¼ k 2 6k þ 8 ¼ k ¼ 6 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 3 3i: 2 So matri A has a pair of comple conjugate eigenvalues 3 ± 3i Let e ¼ e be the eigenvector corresponding to the eigenvalue k ¼ 3 þ 3i Then e 2 ða ð3 þ 3iÞIÞe ¼ 4 ð3 þ 3iÞ 2 e ¼ 5 2 ð3 þ 3iÞ e 2 3i 2 e ¼ 5 3i ð 3iÞe 2e 2 ¼ ; 5e þ ð þ 3iÞe 2 ¼ A nontrivial solution of this system is e ¼ 2; e 2 ¼ 3i e ¼ e 2 2 : 3i Similarly, the eigenvector corresponding to the eigenvalue k 2 ¼ 3 3i is g ¼ 2 þ 3i 2 Let P ¼ Then P 3 ¼ Let C ¼ P AP Then C ¼ P AP ¼ ¼ : 3 3 So, e Ct ¼ e 3t cos 3t sin 3t : sin 3t cos 3t

34 7 2 Linear Systems Therefore, the solution of the system is ðtþ ¼eAt ¼ Pe Ct P ¼ 2 cos 3t sin 3t 2 6 e3t 3 sin 3t cos 3t 3 : Eample 28 Solve the initial value problem _ ¼ A, with ðþ ¼, where A ¼ Also sketch the solution curve in the phase plane 2 2 ; ¼ y R 2 Solution The characteristic equation of matri A is ja kij ¼ 2 k 2 k ¼ ðk þ 2Þ 2 þ ¼ k 2 þ 4k þ 5 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k ¼ ¼ 2i: 2 So matri A has a pair of comple conjugate eigenvalues 2 ± i Let e ¼ e be the eigenvector corresponding to the eigenvalue k ¼ 2þi Then e 2 ða ð 2 þ iþiþe ¼ 2 ð 2 þ iþ e ¼ 2 ð 2 þ iþ e 2 i e ¼ i e 2 ie e 2 ¼ ; e ie 2 ¼ A nontrivial solution of this system is e ¼ ; e 2 ¼ i e ¼ Similarly, the eigenvector corresponding to the eigenvalue k i 2 ¼ 2 i is g ¼ Let P ¼ Then P i ¼ and C ¼ P AP ¼ 2 2 ¼ : 2 2

35 24 Solution Procedure of Linear Systems 7 So, e Ct ¼ e 2t cos t sin t : sin t cos t Hence the solution of the system is ðtþ ¼eAt ¼ Pe Ct P ¼ e 2t cos t sin t sin t cos t : ¼ e 2t sin t cos t cos t sin t ¼ e 2t cos t sin t sin t cos t ¼ e 2t cos t sin t ðtþ ¼e 2t cos t; yðtþ ¼e 2t sin t: Phase Portrait The phase portrait of the solution curve is shown in Fig 22 Eample 29 Solve the system _ ¼ A with ðþ ¼, where A ¼ B 2 5 A : 2 Fig 22 Phase portrait of the solution curve

36 72 2 Linear Systems Solution Clearly, matri A has the eigenvalue λ = 2 with multiplicity 4 Therefore, S ¼ B C 2 A and N ¼ A S ¼ B 5 A : 2 It is easy to check that the matri N is nilpotent of order 4 Therefore, the solution of the system is ðtþ ¼eSt I þ Nt þ N2 t 2 þ N3 t 3 2! 3! : 25 Nonhomogeneous Linear Systems The most general form of a nonhomogeneous linear system is given as _ ðtþ ¼AðtÞ ðtþþb ðtþ ð2:6þ where A(tisann n matri, usually depends on time and b ðtþ is a time dependent column vector Here we consider matri A(t to be time independent, that is, A (t A Then (26 becomes _ ðtþ ¼A ðtþþb ðtþ ð2:7þ The corresponding homogeneous system is given as _ ðtþ ¼A ðtþ ð2:8þ We have described solution techniques for homogeneous system (28 We now find the solution of the nonhomogeneous system (27, subject to initial conditions ðþ ¼ As discussed earlier if Φ(t be the fundamental matri of (28 with ðþ ¼, then the solution of (28 is given by We assume that ðtþ ¼UðtÞU ðþ ðtþ ¼UðtÞU ðþ þ UðtÞU ðþu ðtþ ð2:9þ

37 25 Nonhomogeneous Linear Systems 73 be the solution of the nonhomogeneous linear system (27 Then the initial conditions are obtained as u ðþ ¼ Differentiating (29 with respect to t, we get _ ðtþ ¼ _UðtÞU ðþ þ _UðtÞU ðþu ðtþþuðtþu ðþ _u ðtþ ð2:2þ Substituting (22 and (29 into (27, _UðtÞU ðþ þ _ UðtÞU ðþu ðtþþuðtþu ðþ _u ðtþ ¼ AUðtÞU ðþ þ AUðtÞU ðþu ðtþþb ðtþ ð2:2þ Since Φ(t is a fundamental matri solution of (28, _UðtÞ ¼AUðtÞ: Using this in (22, we get UðtÞU ðþ _u ðtþ ¼b ðtþ _u ðtþ ¼UðÞ _U ðtþb ðtþ Integrating wrto t and using u ðþ ¼, we get Z t u ðtþ ¼ UðÞU ðtþb ðtþdt: Hence the general solution of the nonhomogeneous system (27 subject to ðþ ¼ is given by Z t ðtþ ¼UðtÞU ðþ þ UðtÞ U ðaþb ðaþda ð2:22þ Eample 22 Find the solution of the nonhomogeneous system _ ¼ þ y þ t; _y ¼ y þ with the initial conditions ( =, y( = Solution In matri notation, the system takes the form _ ðtþ ¼A ðtþþb ðtþ, where A ¼ and b ðtþ ¼ t :

38 74 2 Linear Systems The initial conditions become ðþ ¼, where ¼ Matri A has eigenvalues k ¼, k 2 ¼ with corresponding eigenvectors and 2 Therefore UðtÞ ¼ et e t 2e t : This gives U ðtþ ¼ 2 2e t e t e t Therefore the required solution is Z t ðtþ ¼UðtÞU ðþ þ UðtÞ ; UðÞ ¼ and U ðþ ¼ 2 : 2 2 U ðaþb ðaþda 8 9 ¼ < Z 2 UðtÞ 2 t 2e a e a a = þ e a da : ; ¼ 2 UðtÞ 2 3 ð2tþ3þe t þ e t ¼ e t e t 5 ð2tþ3þe t 2 2e t e t ¼ 5e t 2t 4 þ e t 2 2 2e 2t : Eample 22 Prove that the flow evolution operator / t ð Þ¼e At following properties: (i / ð Þ¼, (ii / t / t ð Þ¼, (iii / t / s ð Þ¼/ t þ s ð Þ satisfies the for all s, t 2 R and 2 R n Is/ t / s ¼ / s / t? Solution We have (i / ð Þ¼e A ¼ : (ii / t / t ð Þ¼/ t ðy Þ¼e At y ¼ e At e At ¼, where y ¼ e At (iii / t / s ð Þ¼/ t ðy Þ¼e At y ¼ e At e As ¼ e Aðt þ sþ ¼ / t þ s ð Þ:

39 25 Nonhomogeneous Linear Systems 75 Now, / t / s ð Þ¼/ t ðy Þ¼e At y ¼ e At e As ¼ e As e At ¼ / s ð z Þ¼/ s / t ð Þ for all 2 R n, where z ¼ e As Hence / t / s ¼ / s / t This indicates that the given flow evolution operator is commutative 26 Eercises

40 76 2 Linear Systems

41 26 Eercises 77

42 78 2 Linear Systems

43 26 Eercises 79

44 8 2 Linear Systems

45 26 Eercises 8

46 82 2 Linear Systems References Perko, L: Differential Equations and Dynamical Systems, 3rd edn Springer, New York (2 2 Hirsch, MW, Smale, S: Differential Equations, Dynamical Systems and Linear Algebra Academic Press, London (974 3 Robinson, RC: An Introduction to Dynamical Systems: Continuous and Discrete American Mathematical Society (22 4 Jordan, DW, Smith, P: Non-linear Ordinary Differential Equations Oford University Press, Oford (27 5 Friedberg, SH, Insel, AJ, Spence, LE: Linear Algebra Prentice Hall (23 6 Hoffman, K, Kunze, R: Linear Algebra Prentice Hall (97

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