By Ivan Erdelyi

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Then ∆(t) > 0 for t ∈ (t0 , t1 ), which contradicts ˙ 1 ) > 0. ∆(t Similar results hold for t < t0 . 28). Now we are able to answer our remaining questions. Since we were already successful by considering the curves given by f (t, x) = 0, let us look at the isoclines f (t, x) = const. 66) for t > 2 2/3. Thus, as soon as a solution x(t) enters the region between y+ (t) and x− (t) it must stay there and hence converge to the line x = −t since y+ (t) does. 24 1. Introduction But will every solution in region II eventually end up between y+ (t) and x− (t)?

This settles the case k = 1 since all partial derivatives (including the one with respect to t) are continuous. For the general case k ≥ 1 we use induction: Suppose the claim holds for k and let f ∈ C k+1 . Then φ(t, x) ∈ C 1 and the partial derivative ∂φ k ∂x (t, x) solves the first variational equation. But A(t, x) ∈ C and hence ∂φ k k+1 . 3, shows φ(t, x) ∈ C In fact, we can also handle the dependence on parameters. 61) with corresponding solution φ(t, x0 , λ). 11. Suppose f ∈ C k (U ×Λ, Rn ), x0 ∈ C k (Λ, V ), k ≥ 1.

28 1. 5 1 1 1 1 2 1 2 1 2 1 2 1 1 2 1 1 2 1 It seems to indicate that P (x) is decreasing as a function of h. To prove this we proceed as before. 10) to obtain ψ(t, x) = ˙ x) = (1 − 2φ(t, x))ψ(t, x) + (1 − sin(2πt)). 84) ∂h where we have added h as a subscript to emphasize the dependence on the parameter h. 85) 1 − (1 − x)e and there are two fixed points x1 = 0 and x1 = 1. As h increases these points will approach each other and collide at some critical value hc . Above this value there are no periodic orbits and all orbits converge to −∞ since P (x) < x for all x ∈ R (show this).