1.1. First-Order ODE: Existence and Uniqueness 9 in O there is a solution xp : (t0 − , t0 + ) → Rn of the differential equation dx dt = V (x, t) satisfying the initial condition xp(t0) = p. The proofs of existence and uniqueness have been greatly simpli- fied over time, but understanding the details still requires nontrivial effort. Here we will sketch some of the most important ideas and con- structs that go into the complete proof, but in order not to interrupt the flow of our exposition, we will defer the details to Appendix B. But even if you choose not to study these proofs now, we urge you to do so at some later time. We think you will find that these proofs are so elegant, and the ideas and constructions that enter into them are of such interest in their own right, that studying them is well worth the time and effort it requires. We begin with a simple but very important reformulation of the ODE initial value problem x (s) = V (x(s),s) and x(t0) = x0. Namely, if we integrate both sides of the first of these equations from t0 to t, we find that x(t) = x0 + t t0 V (x(s),s) ds, and we refer to this equation as the integral form of the initial value problem. Note that by substituting t = t0 in the integral form and by differentiating it, we get back the two original equations, so the integral form and the ODE form are equivalent. This suggests that we make the following definition. 1.1.3. Definition. Associated to each time-dependent vector field V on Rn and x0 ∈ Rn, we define a mapping F V,x0 that transforms a continuous function x : I → Rn (where I is any interval containing t0) to another such function F V,x0 (x) : I → Rn defined by F V,x0 (x)(t) = x0 + t t0 V (x(s),s) ds. Exercise 1–2. Show that any y of the form F V,x0 (x) satisfies the initial condition y(t0) = x0, and moreover y is continuously differen- tiable with derivative y (t) = V (x(t),t). Recall that if f is any mapping, then a point in the domain of f such that f(p) = p is called a fixed point of f. Thus we can rephrase the integral form of the initial value problem as follows:

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