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Chapter 8 Stability Theory Chapter 8 Stability theory We discuss properties of solutions of a first order two dimensional system, and stability theory for a special class of linear systems. We denote the independent variable by ‘t’ in place of ‘x’, and x,y denote dependent variables. Let I ⊆ R be an interval, and Ω ⊆ R2 be a domain. Let us consider the system dx = F (t, x, y), dt (8.1) dy = G(t, x, y), dt where the functions are defined on I × Ω, and are locally Lipschitz w.r.t. variable (x, y) ∈ Ω. Definition 8.1 (Autonomous system) A system of ODE having the form (8.1) is called an autonomous system if the functions F (t, x, y) and G(t, x, y) are constant w.r.t. variable t. That is, dx = F (x, y), dt (8.2) dy = G(x, y), dt Definition 8.2 A point (x0, y0) ∈ Ω is said to be a critical point of the autonomous system (8.2) if F (x0, y0) = G(x0, y0) = 0. (8.3) A critical point is also called an equilibrium point, a rest point. Definition 8.3 Let (x(t), y(t)) be a solution of a two-dimensional (planar) autonomous system (8.2). The trace of (x(t), y(t)) as t varies is a curve in the plane. This curve is called trajectory. Remark 8.4 (On solutions of autonomous systems) (i) Two different solutions may represent the same trajectory. For, (1) If (x1(t), y1(t)) defined on an interval J is a solution of the autonomous system (8.2), then the pair of functions (x2(t), y2(t)) defined by (x2(t), y2(t)) := (x1(t − s), y1(t − s)), for t ∈ s + J (8.4) is a solution on interval s + J, for every arbitrary but fixed s ∈ R. (2) However, traces of both the solutions on the respective intervals is the same. If the independent variable t is interpreted as time, then we note that the two different solutions visit every point on the trajectory with a time lag s. Also see Example 8.5 67 68 §8.0 (ii) The trajectories do not cross. For, (1) Suppose two trajectories γ1 and γ2 cross at a point (x0, y0) ∈ Ω. Let (x1(t), y1(t)) defined on interval I1 be a solution whose trace is γ1, and (x2(t), y2(t)) defined on interval I2 be a solution whose trace is γ2. By the assumption of crossing, there exist t1 ∈ I1, and t2 ∈ I2 such that (x1(t1), y1(t1)) = (x0, y0) = (x2(t2), y2(t2)). (8.5) Let us define a pair of functions (x3(t), y3(t)) on interval t1 − t2 + I2 by (x3(t), y3(t)) := (x2(t − t1 + t2), y2(t − t1 + t2)) (8.6) It can be easily checked, via Chain rule, that (x3(t), y3(t)) is a solution of (8.2) on the interval t1 − t2 + I2. Note that (x3(t1), y3(t1)) = (x2(t1 − t1 + t2), y2(t1 − t1 + t2)) = (x2(t2), y2(t2)) = (x0, y0) (8.7) Thus, (x3(t), y3(t)) and (x1(t), y1(t)) are solutions of the same initial value prob- lem, contradicting the uniqueness of solutions to IVP. Therefore two trajectories do not cross each other. (iii) The trajectories fill the domain Ω, since through every point a trajectory passes. This is a consequence of existence theorem. (iv) Through every point in the phase space Ω, exactly one trajecory passes. This is a consequence of uniqueness of solutions to IVPs. (v) From the last two remarks, it follows that the trajectories partition the phase space Ω. In fact, defining a relation on Ω by saying that two points (x1, y1), (x2, y2) ∈ Ω are related if (x1, y1) and (x2, y2) lie on the same trajectory, it is easy to verify that this relation is an equivalence relation and thereby giving rise to a partition of Ω in terms of equivalnece classes. Each equivalence class is a trajectory. (vi) Note that trajectories consisting of single point correspond to critical points. (vii) types of trajectories: For autonomous systems with two dimensional phase space, three types of trajectories are possible. A trajectory consisting of single point (corresponding to equilib- rium solutions), and if trajectory has more than one point then it could be a closed curve (corresponding to periodic solutions), or a curve without self-intersection. (viii) For linear autonomous systems, a special class of systems (8.2) for which F and G are linear in x, y, note that saturated solutions are global, i.e., saturated solutions are defined on the entire real line R. Hence for linear autonomous systems, we do not mention the interval on which a given solution is defined. The above remark is illustrated by the following example. Example 8.5 dx dy = y, = −4x. (8.8) dt dt Note that (x1(t), y1(t)) := (cos 2t, −2 sin 2t) is a solution of (8.8). The trajectory passing through 2 2 y2 the point (1, 0) ∈ R is the ellipse x + 4 = 1 travelled counterclockwise. Consider the solution π π (x2(t), y2(t)) := (cos(t − π/2), −2 sin 2(t − π/2)). This solution satisfies (x2( 2 ), y2( 2 )) = (1, 0), and hence has the same trajectory as (x1(t), y1(t)). Draw figure MA 417: Ordinary Differential Equations Sivaji Ganesh Sista Chapter 8 : Stability theory 69 8.1 Solving linear planar systems with constant coefficients Consider the system of ODE µ ¶ µ ¶ µ ¶ µ ¶ x0 a b x x = =: A . (8.9) y0 c d y y 8.1.1 Fundamental matrix Definition 8.6 (Fundamental matrix) A matrix valued function Φ whose columns are solu- tions of the system of ODE (8.9) is called a solution matrix. A solution matrix Φ is called a fundamental matrix if the columns of Φ form a fundamental pair of solutions for the system (8.9). A fundamental matrix Φ is called the standard fundamental matrix if Φ(0) is the identity matrix. Remark 8.7 Since the columns of a solution matrixΦ are solutions of (8.9), the matrix valued function Φ satisfies the system of ODE Φ0 = AΦ. (8.10) Exercise 8.8 A solution matrix is a fundamental matrix if and only if its determinant is not zero. Exercise 8.9 Prove that if Ψ is a fundamental matrix then ΨC is also a fundamental matrix for every constant invertible matrix C. Prove that all fundamental matrices occur this way. Computation of fundamental matrix By definition of a fundamental matrix, finding a fundamental pair of solutions to system of ODE (8.9) is equivalent to finding a fundamental matrix. In view of Exercise 8.9, fundamental matrix is not unique but the standard fundamental matrix is unique. µ ¶ a (1) Observe that eλt is a non-trivial solution of (8.9) if and only if b µ ¶ µ ¶ µ ¶ µ ¶ a 0 a a 6= ,A = λ . (8.11) b 0 b b µ ¶ a That is, λ is an eigenvalue and is an eigenvector corresponding to λ. b (2) Questionµ ¶ Is it possible to find a fundamental pair, both of which are of the form a eλt ? b µ ¶ µ ¶ a c Answer Supposing that φ (t) = eλ1t and φ (t) = eλ2t are two solutions of 1 b 2 d (8.9), φ1, φ2 form a fundamental pair if and only if ¯ ¯ ¯ a c ¯ ¯ ¯ 6= 0, (8.12) ¯ b d ¯ since the above determinant is the Wronskian of φ1, φ2 at t = 0. That is, the matrix A should have two linearly independent eigenvectors. Note that this is equivalent to saying that A is diagonalisable. Sivaji Ganesh Sista MA 417: Ordinary Differential Equations 70 8.1. Solving linear planar systems with constant coefficients (3) Question What if the matrix A does not have two linearly independent eigenvectors? This can happen when A has only one eigenvalue of multiplicity two. Inspired by a similar situation in the contextµ of¶ constant coefficientµ second¶ order linear ODE, we are a a tempted to try φ (t) = eλ1t and φ (t) = teλ1t as a fundamental pair. But 1 b 2 b note that φ1, φ2 does not form a fundamental pair since Wronskian at t = 0 will be zero, also note that φ2 is not even a solution of the linear system (8.9). Nevertheless, we can find a solution having the form of φ1. Therefore, we try a variant of above suggestion to find another solution that together φ1 constitutes a fundamental pair. Let µ ¶ µ ¶ a c φ (t) = teλ1t + eλ1t (8.13) 2 b d Then φ2(t) solves the system (8.9) if and only if µ ¶ µ ¶ c a (A − λ I) = . (8.14) 1 d b µ ¶ µ ¶ a c One can easily verify that , are linearly independent. Thus, φ , φ defined b d 1 2 by µ ¶ µ ¶ µ ¶ a a c φ (t) = eλ1t , φ (t) = teλ1t + eλ1t (8.15) 1 b 2 b d µ ¶ µ ¶ a c is a fundamental pair, where , are related by the equation (8.14). b d (4) In case the matrix A does not have real eigenvalues, then eigenvalues are complex conjugates of each other. In this case, (λ, v) is an eigen pair if and only if (λ, v) is also an eigen pair for A. (8.16) µ ¶ α + iβ Denoting λ = r + iq (note q 6= 0), v = , define γ + iδ µ ¶ µ ¶ α cos qt − β sin qt α sin qt + β cos qt φ (t) = ert , φ (t) = ert .
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