20 1. Differential Equations and Their Solutions If a constant map σ : I Rn, σ(t) = p for all t I, is a solution of the equation dx dt = V (x, t), then V (p, t) = σ (t) = 0 for all t, and conversely this implies σ(t) p is a solution. In particular, in the autonomous case, the maximal solution σp is a constant map if and only if V (p) = 0. Such points p are of course called zeros of the time- independent vector field V , but because of their great importance they have also been given many more aliases, including critical point, singularity, stationary point, rest point, equilibrium point, and fixed point. A related but more interesting type of solution of dx dt = V (x, t) is a so-called closed orbit, also referred to as a periodic solution. To define these, we start with an arbitrary solution σ defined on the whole real line. A real number T is called a period of σ if σ(t) = σ(t + T ) for all t R, and we will denote by Per(σ) the set of all periods of σ. Of course 0 is always a period of σ, and one possibility is that it is the only period, in which case σ is called a nonperiodic orbit. At the other extreme, σ is a constant solution if and only if every real number is a period of σ. What other possibilities are there for Per(σ)? To answer that, let us look at some obvious properties of the set of periods. First, Per(σ) is clearly a closed subset of R—this follows from the continuity of σ. Secondly, if T1 and T2 are both periods of σ, then σ(t + (T1 T2)) = σ((t T2) + T1) = σ(t T2) = σ(t T2 + T2) = σ(t), so we see that the difference of any two periods is another period. Thus Per(σ) is a closed subgroup of the group of real numbers under addition. But the structure of such groups is well known. 1.3.1. Proposition. If Γ is a closed subgroup of R, then either Γ = R, or Γ = {0}, or else there is a smallest positive element γ in Γ and Γ consists of all integer multiples of γ. Exercise 1–16. Prove this proposition. (Hint: If Γ is nontrivial, then the set of positive elements of Γ is nonempty and hence has a greatest lower bound γ which is in Γ since Γ is closed. If γ = 0, show that Γ is dense in R and hence it is all of R. If γ = 0, it is the smallest positive element of Γ. In this case, if n Γ, then dividing n
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