Linear diferential equations

HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS ———o0o——— PHAN THI THUY CHINH LINEAR DIFFERENTIAL EQUATIONS BACHELOR THESIS Hanoi – 2019 HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS ———o0o——— PHAN THI THUY CHINH LINEAR DIFFERENTIAL EQUATIONS BACHELOR THESIS Major: Analysis SUPERVISOR: Dr Tran Van Bang Hanoi – 2019 Bachelor thesis PHAN THI THUY CHINH Thesis Acknowledgement I would like to express my gratitude to the teachers of the Department of Mathematics, Hanoi Pedagogical University 2, the teachers in the analytic group as well as the teachers involved The lecturers have imparted valuable knowledge and facilitated for me to complete the course and the thesis In particular, I would like to express my deep respect and gratitude to Dr Tran Van Bang, who has direct guidance, help me complete this thesis Due to time, capacity and conditions are limited, so the thesis can not avoid errors Then, I look forward to receiving valuable comments from teachers and friends Ha Noi, May 5, 2019 Student Phan Thi Thuy Chinh i Bachelor thesis PHAN THI THUY CHINH Thesis Assurance I assure that the data and the results of this thesis are true and not identical to other topics I also assure that all the help for this thesis has been acknowledge and that the results presented in the thesis has been identified clearly Ha Noi, May 5, 2019 Student Phan Thi Thuy Chinh ii Contents Notation iv Preface 1 Preliminaries 1.1 Matrix theory 1.2 Differential equations 1.3 Solving differential equations 1.4 Existence and uniqueness theorems 10 1.5 Phase space and flows 11 1.6 Limit sets and trajectories 13 LINEAR DIFFERENTIAL EQUATIONS 15 2.1 Autonomous linear differential equations 17 2.2 Normal forms 19 2.3 Invariant manifolds 30 2.4 Geometry of phase space 33 2.5 Floquet Theory 35 Conclusion 40 References 41 iii Bachelor thesis PHAN THI THUY CHINH Notation To study the contents of the thesis, we need to understand the mathematical symbols We will give some common symbols used in the thesis, helping readers read and understand easier x˙ Derivates with respect to time t γ(x) The trajectory through x γ + (x) The positive semi-trajectory through x γ − (x) The negative semi-trajectory through x Λ(x) The w−limit set of x A(x) The α-limit set of x T Period A The square n × n matrix A−1 The invertible matrix of A diag (a1 , , an ) The diagonal matrix ρ ± iω A pair of complex conjugate eigenvalues Im (z) Imaginary part of z Re (z) Real part of z E u (0) The unstable manifold E c (0) The centre manifold E s (0) The stable manifold ϕ (T ) The Floquet multiplier σ A Floquet exponent Φ (t) The fundamental matrix iv Bachelor thesis PHAN THI THUY CHINH Preface Differential equation is an important research content in Mathematics In addition, it is widely applied in engineering, physics, economics and some other industries Differential equations are used throughout the sciences to model dynamic processes In our university program, we have been studying how to solve some classes of specific linear differential equations, a few qualitative properties such as existence, uniqueness of the solution Another way is to use the properties of matrix analysis that can help us get a good presentation of the results of linear differential equations That is the reason why we would like to study linear differential equations Chapter Preliminaries 1.1 Matrix theory We consider A is the n × n matrix, we have some basic definitions, operations and its properties Definition 1.1.1 In linear algebra, a square matrix A is called invertible, if existing a square matrix B such that: AB = BA = In where In = diag (1, 1, , 1) is denoted the n × n identity matrix In this case, the matrix B is uniquely determined by A and is called the inverse of A, denoted by A−1 Proposition 1.1.2 i) A is invertible ⇔ det A = 0; ii) If A and B are invertible then AB is also invertible and (AB)−1 = B −1 A−1 ; iii) At −1 t = A−1 In linear algebra, we had learnt how to find the invertible matrix of A (n × n matrix) We can list three method as follow: Method 1.1.3 (Using determinant method) Bachelor thesis PHAN THI THUY CHINH To start, we recall the complement of a element Let A is the n×n matrix If we omit the ith row and the j th column of A, we will get a submatrix (n − 1)×(n − 1) of A, denoted by Mij Then, A = (−1)i+j det Mij × Mij is said to be the complement algebra of the element set on the ith row and the j th column of A Hence, we get the formula to find out the invertible matrix of A: • If det A = 0, then A is not invertible • If det A = 0, then A is invertible and A−1 = PA det A Example 1.1.4 Let the matrix A     0 1   Find the invertible matrix of A We get, det A = 1 = = Thus, A is invertible Applying determinant method above, we get: A11 = (−1)1+1 1 A12 = (−1)1+2 3 = 1, = 1, Bachelor thesis PHAN THI THUY CHINH A13 = (−1)1+3 = −1, Similarly, A22 = 2, A23 = 0, A31 = 1, A32 = −1, A32 = Thus, the invertible matrix of A is:   −4    −1   −1 Example 1.1.5 Let the matrix −2 Therefore, we apply the formula −1 a b c d −1 = a b = c d ad − bc d −b −c a We get the invertible matrix is: −2 −1 = 10 Method 1.1.6 (Using Gauss method) 4 −3 Bachelor thesis PHAN THI THUY CHINH for all x ∈ R2 Now either (A − λI) x = for all x or there exists e2 such that (A − λI) e2 = and e2 = In the first case, A = diag (λ, λ) for any choice of basis vectors e1 and e2 , since Ax = λx for all x ∈ R2 In the second case, we define e1 = (A − λI) e2 (2.19) Then (A − λI)2 e2 = = (A − λI) e1 and so Ae1 = λe1 , whilst Ae2 = e1 + λe2 from the definition of e1 , (2.19) Hence A [e1 , e2 ] = [λe1 , λe2 + e1 ] = [e1 , e2 ] λ λ (2.20) In other words, if λ is a double eigenvalue of A, then there are two cases of matrix A by the change of coordinate: λ 0 λ or λ λ (2.21) In two cases above, it is not difficult to find the solutions of the differential equation x˙ = Ax in this choice of coordinate system and back to find the solutions in the original coordinate system We will consider two specific examples in R2 and R2 In R2 : Solving the differential equations x˙ = λx, y˙ = λy, with the initial conditions (x (0) , y (0)) = (x0 , y0 ), the the solution is x (t) = x0 eλt , y (t) = y0 eλt 27 Bachelor thesis PHAN THI THUY CHINH Similarly, with differential equations dotx = λx + y, y˙ = λy, with the same initial condition, the the solution is x (t) = eλt (x0 + y0 t) , y (t) = y0 eλt λ Show from first principles that if Λ = e tΛ = λ etλ tetλ etλ , then Example 2.2.5 Consider the matrix A= −1 The characteristic polynomial is s2 − 6s + = (s − 3)2 = so λ = is a double eigenvalue It is easy to see that e2 = (1, −1)T does not satisfy the equation (A − 2I) e2 = and we get e1 = (A − 2I) e2 = (−2, −2)T Hence, we set P = −2 −2 −1 and P −1 = −1 28 1 −2 , Bachelor thesis PHAN THI THUY CHINH so P −1 AP = Λ = tΛ and e e3t te3t = e3t Now, we find etA : tA e tΛ = Pe P −1 = e3t (1 − t) te3t −te3t e3t (1 + t) , which allow us to write down the solution to the differential equation x˙ = Ax with the initial condition x (0) = (x0 , y0 )T as x0 e3t (1 − t) + y0 te3t −x0 te3t + y0 e3t (1 + t) In R3 the case of three repeated real eigenvalues is similar, but there are three cases Firstly, we get the characteristic polynomial is (s − λ)3 = and so (A − λI)3 x for all x ∈ Rn Hence, we consider three cases: i) There exists e3 = such that (A − λI)2 e3 = 0; ii) (A − λI)2 x = for all x but there exists e2 = such that (A − λI) e2 = 0; and iii) (A − λI) x = for all x There are three cases given respectively to     λ λ      λ  ,  λ  , and     0 λ 0 λ the normal forms:   λ 0   0 λ 0   0 λ The last of three situations is as obvious as the diagonal case in R2 In case (i) define e1 = (A − λI)2 e3 and e2 = (A − λI) e3 Clearly, Ae1 = λe1 (as (A − λI)3 x = 0, for all x) and Ae3 = λe3 + e2 (from the definition of e2 ) Furthermore, the definition of e1 can be 29 Bachelor thesis PHAN THI THUY CHINH rewritten as e1 = (A − λI) e2 and Ae2 = λe2 + e1 Putting these three relationships together and forming the matrix [e1 , e2 , e3 ] as before giving   λ    A [e1 , e2 , e3 ] = [λe1 , e1 + λe2 , e2 + λe3 ] = [e1 , e2 , e3 ]  0 λ 1 0 λ and so the matrix P = [e1 , e2 , e3 ] gives the transformation for the matrix A to have the required form The second case is the most difficult to establish We have that (A − λI)2 x = and there exists e2 such that (A − λI) e2 = As in the two dimensional case define e1 = (A − λI) e2 , so Ae1 = λe1 and Ae2 = λe2 + e1 Now, we claim that there is another vector, e3 , which is independent of e1 and e2 and which also satisfies with Ae3 = λe3 The matrix P = [e1 , e2 , e3 ] the produces the desired form 2.3 Invariant manifolds The applications of the previous section show that given the system x˙ = Ax, we can get a simple change of coordinates to bring the equation into the normal form y˙ = Λy In the simplest case, where A has distinct eigenvalues, the matrix Λ is Λ = diag (λ1 , , λk , B1 , , Bm ) (2.22) Where (λi ) are the real eigenvalues and Bj are the matrices ρj −ωj ωj ρj associated with the complex conjugate pairs of eigenvalues ρj + iωj In component form, with y = (y1 , , yk , ω1 , , ω2m ) the equation y˙ = Λy 30 Bachelor thesis PHAN THI THUY CHINH is y˙ i = λi yi , ≤ i ≤ k (2.23) and ω˙ 2j−1 ω˙ 2j = ρj −ωj ωj ρj ω2j−1 ω2j ,1 ≤ j ≤ m (2.24) and since they are uncoupled each of these k +m equations can be solved separately It follows immediately that the real eigenspaces of Λ are invariant, since if y0 = (0, , Y0 , , 0) then y (t) = 0, , Y0 eλi t , , , and similarly the two-dimensional eigenspaces corresponding to the complex conjugate pair of eigenvalues ρj ± iωj are also invariant Returning to the original x˙ = Ax this implies that corresponding eigenspaces of A are invariant Since all the eigenvalues of A are distinct, these eigenspaces are either one- dimensional or two-dimensional depending on whether the corresponding eigenvalue is real or not This proves the following theorem Theorem 2.3.1 If the eigenvalues of the ntimesn real matrix A are distinct then Rn decomposes into a direct sum of one-dimensional spaces and two-dimensional spaces Each of these eigenspaces is invariant under the flow defined by x˙ = Ax Example 2.3.2 Consider the matrix   0 0   0 −3 0      0 −3 0    0 0 −2 −1   0 −2 A has blocks: a one-dimensional eigenspace corresponding to the eigen31 Bachelor thesis PHAN THI THUY CHINH value 2, a two-dimensional eigenspace corresponding to the multiple eigenvalue −3 and a a two-dimensional eigenspace corresponding to the complex conjugate pair of eigenvalues −2 ± i To solve the differential equation x˙ = Ax with x (0) = (y1 , y2 , y3 , y4 , y5 ) we can work independently in these three eigenspaces Firstly, x˙1 = 2x1 , x1 (0) = y1 and so x1(t) = x0 e2t In the degenerate eigenspace x˙2 x3 = −3 x2 −3 x3 , x2 (0) x3(0) = y2 y3 and so x2 (t) x3 (t) = y2 e−3t + y3 te−3t y3 e−3t Lastly x˙4 x˙5 = −2 −1 x4 −2 x5 , x4 (0) x5 (0) = y4 y5 , with solution x4 (t) x5 (t) = e−2t (y4 cos t − y5 sin t) e−2t (y4 sin t + y5 cos t) Because each generalized eigenspace A is invariant under the flow x˙ = Ax, so the spaces spanned by several generalized eigenspaces is also invariant We can divide into three dynamically important sets of eigenspaces Let (ui ), ≤ k ≤ ns , denote the generalized eigenvectors associated with eigenvalues of A which have strictly negative real parts 32 Bachelor thesis PHAN THI THUY CHINH Then every generalized eigenvector is in one of these sets and so nu + nc + ns = n The origin x = is a stationary point of the flow, we define: • The unstable manifold of x = 0, E u (0) is called the invariant space spanned by the eigenvector (ui ) • The centre manifold, E c (0) is called the invariant manifold spanned by the eigenvectors cj • The stable manifold, E s (0) is called the invariant manifold spanned by the eigenvectors sk The above types of manifold is defined in linear algebra, but in there we will use some properties of flows to represent the unstable manifold and the stable manifold Theorem 2.3.3 Suppose x˙ = Ax where A is an constant n × n If nc = then E u (0) = x ∈ Rn |etA x → as t → −∞ E s (0) = x ∈ Rn |etA x → as t → ∞ 2.4 Geometry of phase space Corresponding to generalized eigenspaces, each invariant manifold is simply surface in Rn And the motion on each of these surfaces can be obtained by one of a set of easy differential equations By superposing the solutions within each invariant subspace, we can find out the general solution or flow Consider the equation in the example: x˙ = ρx − ωy, y˙ = ωx + ρy, 33 (2.25) Bachelor thesis PHAN THI THUY CHINH with the initial condition (x (0) , y((0)) = (x0 , y0 ) (cf (2.15) and (2.24)) In the polar coordinates this equation becomes r˙ = ρr, θ˙ = ω (2.26) and so the motion is uniform rotation about the origin together with exponential growth (if ρ > 0) or decay (if ρ < 0) of radial component Trajectories of this two-dimensional system are therefore logarithmic spirals if ρ = since dr ρ = r dθ ω and so ρθ ω r (θ) = r (0) exp or θ+ϕ= ω log r, ρ (2.27) where ρ = 0, ϕ is a constant If ρ = 0, then trajectories are concentric circles Example 2.4.1 Consider the flow x˙ = Ax for x ∈ R3 where A has a pair of complex conjugate eigenvalues with negative real part and a positive real eigenvalue In normal form this equation becomes y˙ = Λy, where  ρ −ω   Λ= ω ρ  0 , 0 λ where ρ < 0, λ > and ω = Then, the component form is      y˙1 ρ −ω y1      y˙2  = ω ρ  y2       y3 y˙3 0 λ 34 Bachelor thesis PHAN THI THUY CHINH and we immediately see that the unstable manifold of the origin is the y3 -axis and the stable manifold of the origin is the two-dimensional plane defined by y3 = On this plane, the motion is given by logarithmic spirals into the origin and on the y3 -axis the motion is away from the origin Thus, putting these two motions together we obtain the phase space diagram depicted in figure 2.1 The general motion of the original system x˙ = Ax can be obtained from this by applying a linear transformation to the whole picture, giving a phase portrait like the on sketched in figure 2.1 2.5 Floquet Theory In this section, we want to develop a treatment of non-autonomous linear differential equations with periodic coefficients, i.e x˙ = A (t) x, A (t) = A (t + T ) (2.28) The book of Iooss and Joseph (1980), which used to find the treatment in this book Let us begin with the simplest case, when x ∈ R Then x˙ = a (t) x, a (t) = a (t + T ) (2.29) and we can integrate immediately to find t x (t) = x0 exp a (s) ds (2.30) t Hence, ϕ (t) = exp a (s) ds 35 is a fundamental matrix for the Bachelor thesis PHAN THI THUY CHINH Figure 2.1: Phase portraits for Example 2.4.1: (a) in normal coordinates; (b) in general coordinates problem with x (t) = exp ϕ (t) x0 Now, since t+T T a (s) ds = t+T a (s) ds + a (s) ds T and using the periodicity of a (s) , a (s) = a (s + T ) t+T t a (s) ds = T a (s) ds We get: T ϕ (t + T ) = exp t+T a (s) ds exp a (s) ds = ϕ (T ) ϕ (t) T (2.31) and in particular ϕ (nT ) = ϕ (T )n (2.32) The number ϕ (T ) = eσT is called the Floquet multiplier, whilst σ is called a Floquet exponent Note that σ is only determined up to a con- 36 Bachelor thesis PHAN THI THUY CHINH stant: σ= 2kπi log ϕ (T ) + , T T (2.33) for any integer k To determine the stability of the origin define v (t) = ϕ (t) e−σt Then, v (t + T ) = ϕ (t + T ) x0 = v (t) eσt x0 , since ϕ (t + T ) e−σT = ϕ (t) Hence, v (t) = v (t + T ),i.e v (t) is periodic and bounded Now, x (t) = ϕ (t) x0 = v (t) eσt x0 , (2.34) so if Reσ < solution tend to zero, whilst if Reσ > solutions are unbounded as t → ∞ Example 2.5.1 Consider the differential equation x˙ = (δ + cos t) x Then a (t) = δ + cos t with T = 2π Furthermore, 2π 2π a (s) ds = (δ + cos s) ds = 2πδ and so the Floquet exponent is σ = δ and the origin is stable if δ < and unstable if δ > To generalize the idea of Floquet exponents to Rn We consider a similar procedure, but now the Floquet multipliers be the eigenvalues of the matrix Φ (T ) In general, consider the equation x˙ = A (t) x, x ∈ Rn , A (t) = A (t + T ) (2.35) Let Φ (t) be the fundamental matrix which satisfies Φ (0) = I So, we get Φ˙ (t) = A (t) Φ (t) Hence Φ˙ (t + T ) = A (t + T ) Φ (t + T ) = A (t) Φ (t + T ) 37 (2.36) Bachelor thesis PHAN THI THUY CHINH and so Φ (t + T ) and since the discussion of fundamental matrices on the first part of the chapter Φ (t + T ) = Φ (t) C, (2.37) where C is a constant matrix Putting t = 0, we see that C = Φ (T ) i.e., Φ (t + T ) = Φ (t) Φ (T ) (2.38) Especially, by introduction on m Φ (mT ) = Φ (T )m (2.39) Let λi be the distinct eigenvalues of Φ (T ) and ei is the corresponding eigenvectors The (λi ) are said to be the Floquet multipliers and if λi = eσi T the σi is a Floquet exponent As in the one-dimensional case, σi is 2πi Now, let x0 = ei and remember only defined up to multiples of T k that x (t) = Φ (t) x0 Hence x (t + T ) = Φ (t + T ) x0 = Φ (t) Φ (T ) ei k eσk T ei = Φ (t) k (2.40) Let xk (t) = Φ (t) ak ek , ≤ k ≤ n, so xk (t + T ) = Φ (t) ak eσk T ek Similar with the one-dimensional case, define vk (t) = e−σk t xk (t) Then, vk (t + T ) = e−σk (t+T ) xk (t + T ) = e−σk (t) xk (t) = vk (t) Thus, each of the components vk (t) is periodic with period T Furthermore, x (t) = eσk t vk (t) and so we conclude, similar with xk (t) = k k one-dimensional case, that if all the Floquet exponents have strictly negative real parts, then the origin is asymptotically stable, whilst if any of 38 Bachelor thesis PHAN THI THUY CHINH the Floquet exponents has positive real parts then solutions diverge to infinity 39 Bachelor thesis PHAN THI THUY CHINH CONCLUSION Bachelor thesis research about Linear differential equations The results of the thesis include: The thesis has recalled some definitions and basic properties of matrix For differential equations, we have given solving methods and existence and uniqueness theorem Moreover, we have referred to some basic concepts of phase space and flows Lastly, limit set and trajectory have been also represented in the preliminaries We have showed solving the linear differential equations x˙ = Ax by transforming the matrices into normal form We have developed a treatment of non-autonomous linear differential equations with periodic coefficients through Floquet theory We have represented invariant manifolds and types of manifold spaned by the eigenvectors Then, we have given geometry of phase space by solving differential equations 40 Bibliography [A] References in Vietnamese [1] Nguyễn Thế Hoàn - Phạm Phu, Cơ sở phương trình vi phân lí thuyết ổn định, NXB Giáo dục Việt Nam, 2010 [2] Cung Thế Anh, Cơ sở lí thuyết phương trình vi phân, NXB Đai học sư phạm, 2015 [B] References in English [3] Paul Glendinning, Stability and Instability and Chaos, USA, 1994 41 ... 11 1.6 Limit sets and trajectories 13 LINEAR DIFFERENTIAL EQUATIONS 15 2.1 Autonomous linear differential equations 17 2.2 Normal forms ... help us get a good presentation of the results of linear differential equations That is the reason why we would like to study linear differential equations Chapter Preliminaries 1.1 Matrix theory... tn ) → y as n → ∞ Hence, Λ(x) = B(0, x ) = A(x) 14 Chapter LINEAR DIFFERENTIAL EQUATIONS It is really easy to solve linear differential equations We can write down solutions in terms of the exponential