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Differential Equations and Dynamical Systems Classnotes for Math 645 University of Massachusetts V3

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Differential Equations and Dynamical Systems Classnotes for Math 645 University of Massachusetts V3 Differential Equations and Dynamical Systems Classnotes for Math 645 University of Massachusetts v3: Fall 2008 Luc Rey-Bellet Department of Mathematics and Statistics University of Massachusetts Amherst, MA 01003 Contents 1 Existence and Uniqueness 4 1.1 Introduction . 4 1.2 Banach fixed point theorem . 7 1.3 Existence and uniqueness for the Cauchy problem . 12 1.4 Peano Theorem . 16 1.5 Continuation of solutions . 19 1.6 Global existence . 21 1.7 Wellposedness and dynamical systems . 26 1.8 Exercises . 29 2 Linear Differential Equations 37 2.1 General theory . 37 2.2 The exponential of a linear map A .................... 42 2.3 Linear systems with constant coefficients . 48 2.4 Stability of linear systems . 56 2.5 Floquet theory . 61 2.6 Linearization . 65 2.7 Exercises . 70 3 Stability analysis 77 3.1 Stability of critical points of nonlinear systems . 77 3.2 Stable and unstable manifold theorem . 84 3.3 Center manifolds . 90 3.4 Stability by Liapunov functions . 93 3.5 Gradient and Hamiltonian systems . 98 3.5.1 Gradient systems . 98 3.5.2 Hamiltonian systems . 100 4 Poincar´e-Bendixson Theorem 106 4.1 Limit sets and attractors . 106 4.2 Poincar´emaps and stability of periodic solutions . 108 2 CONTENTS 3 4.3 Bendixson criterion . 114 4.4 Poincar´e-BendixsonTheorem . 114 4.5 Examples . 118 Chapter 1 Existence and Uniqueness 1.1 Introduction An ordinary differential equation (ODE) is given by a relation of the form F (t; x; x0; x00; ··· ; x(m)) = 0 ; (1.1) where t 2 R, x; x0; ··· ; x(m) 2 Rn and the function F is defined on some open set of R × Rn × · · · × Rn. A function x : I ! Rn, where I is an interval in R, is a solution of (1.1) if x(t) is of class Cm (i.e., m-times continuously differentiable) and if F (t; x(t); x0(t); x00(t); ··· ; x(m)(t)) = 0 for all t 2 I: (1.2) We say that the ODE is of order m if the maximal order of the derivative occurring in (1.1) is m. Example 1.1.1 Clairaut equation (1734) Let us consider the first order equation x − tx0 + f(x0) = 0 ; (1.3) where f is some given function. It is given, in implicit form, by a nonlinear equation in x0. It is easy to verify that the lines x(t) = Ct − f(C) are solutions of (1.3) for any 2 C. Consider for example f(z) = z + z, then one sees easily that given a point (t0; x0) there exists either 0 or 2 such solutions passing by the point (t0; x0) (see Figure 1.1). As we see from this example, it is in general very difficult to obtain results on the uniqueness or existence of solutions for general equations of the form (1.1). We will therefore restrict ourselves to situations where (1.1) can be solved as a function of x(m), x(m) = g(t; x; x0; ··· ; x(m−1)) : (1.4) CHAPTER 1. EXISTENCE AND UNIQUENESS 5 Figure 1.1: Some solutions for Clairault equation for f(z) = z2 + z. Such an equation is called an explicit ODE of order m. One can always reduce an ODE of order m to a first order ODE for a vector in a space of larger dimension. For example we can introduce the new variables 0 00 (m−1) x1 = x ; x2 = x ; x3 = x ··· ; xm = x ; (1.5) and rewrite (1.4) as the system 0 x1 = x2 ; 0 x2 = x3 ; . (1.6) 0 xm−1 = xm ; 0 xm = g(t; x1; x2; ··· ; xm) : nm This is an equation of order 1 for the supervector x = (x1; ··· ; xm) 2 R (each xi is in Rn) and it has the form x0 = f(t; x). Therefore, in general, it is sufficient to consider the case of first order equations (m=1). If f does not depend explicitly on t, i.e., f(t; x) = f(x), the ODE x0 = f(x) is called autonomous. The function f : U ! Rn, where U is an open set of Rn, defines a vector field. A solution of x0 = f(x) is then a parametrized curve x(t) which is tangent to the vector field f(x) at any point, see figures 1.2 and (1.3). Note a non-autonomous ODE x0 = f(t; x) with x 2 Rn can be written as an autonomous ODE in Rn+1 by setting x ! x0 ! f(t; x) ! y = y0 = = ≡ F (y) : (1.7) t t0 1 CHAPTER 1. EXISTENCE AND UNIQUENESS 6 Example 1.1.2 Predator-Prey equation Let us consider the equation x0 = x(α − βy) ; y0 = y(γx − δ) ; (1.8) where α; β; γ; δ are given positive constants. Here x(t) is the population of the preys and y(t) is the population of the predators. If the population of predators y is below the threshold α/β then x is increasing while if y is above α/β then x is decreasing. The opposite holds for the population y. In order to study the solutions, let us divide the first equation, by the second one and consider x as a function of y. We obtain dx x (α − βy) (γx − δ) (α − βy) = or dx = dy : (1.9) dy y (γx − δ) x y Integrating gives γx − δ log x = α log y − βy + Const: (1.10) One can verify that the level curves (1.10) are closed bounded curves and each solution (x(t); y(t)) stays on a level curve of (1.10) for any t 2 R. This suggests that the solutions are periodic (see Figure 1.2). Figure 1.2: The vector field for the predator-prey equation with α = 1, β = 2, γ = 3, δ = 2 and the solutions passing through the point (1; 1) and (0:5; 0:5). Example 1.1.3 van der Pol equation. The van der Pol equation x00 = (1 − x2)x0 − x : (1.11) CHAPTER 1. EXISTENCE AND UNIQUENESS 7 It can written as a first order system by setting y = x0 x0 = y ; y0 = (1 − x2)y − x : (1.12) It is a perturbation of the harmonic oscillator ( = 0) x00 + x = 0 whose solutions are the periodic solution x(t) = A cos(t − φ) and y(t) = x0(t) = −A sin(t − φ) (circles). When > 0 one observes that one periodic solution survives which is the deformation of a circle of radius 2 and all other solution are attracted to this periodic solution (limit cycle), see Figure 1.3. Figure 1.3: The vector field for the van der Pol equation with = 0:1 as well as two solutions passing through the points (:1;:2) and (2; 3). We will discuss these examples in more details later. For now we observe that, in both cases, the solutions curves never intersect. This means that there are never two solutions passing by the same point. Our first goal will be to find sufficient conditions for the problem 0 x = f(t; x) ; x(t0) = x0 ; (1.13) to have a unique solution. We say that t0 and x0 are the initial values and the problem (1.13) is called a Cauchy Problem or an initial value problem (IVP). 1.2 Banach fixed point theorem We will need some (simple) tools of functional analysis. Let E be a vector space with addition + and multiplication by scalar λ in R or C.A norm on E is a map k · k : E ! R which satisfies the following three properties • N1 kxk ≥ 0 and kxk = 0 if and only if x = 0 ; CHAPTER 1. EXISTENCE AND UNIQUENESS 8 • N2 kλxk = jλjkxk ; • N3 kx + yk ≤ kxk + kyk (triangle inequality) : A vector space E equipped with a norm k · k is called a normed vector space. In a normed vector space E we can define the convergence of sequence fxng. We say that the sequence fxng converges to x 2 E, if for any > 0, there exists N ≥ 1 such that, for all n ≥ N, we have kxn − xk ≤ . We say that fxng is a Cauchy sequence if for any > 0, there exists N ≥ 1 such that, for all n; m ≥ N, we have kxn − xmk ≤ . Definition 1.2.1 A normed vector space E is said to be complete if every Cauchy sequence in E converges to an element of E. A complete normed vector space E is called a Banach space. Let k · k and k · k? denote two norms on the vector space E. We say that the norms k · k and k · k? are equivalent if there exist positive constants c and C such that ckxk ≤ kxk? ≤ Ckxk for all x 2 E: It is easy to check that the equivalence of norm defines an equivalence relation. Further- more if a Cauchy sequence for a norm k · k is also a Cauchy sequence for an equivalent norm k · k?. n n Example 1.2.2 The vector space E = R or C with the euclidean norm kxk2 = P 2 1=2 P ( i xi ) is a Banach space. Other examples of norms are kxk1 = i jxij or x1 = n n supi jxij. In any case R or C equipped with any norm is a Banach space, since all norm are equivalent in a finite-dimensional space (see exercises). The previous example shows that for finite dimensional vector spaces the choice of a norm does not matter much. For infinite-dimensional vector spaces the situation is very different as the following example demonstrate.
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