By Gerald Teschl
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Extra resources for Ordinary Differential Equations and Dynamical Systems (draft)
The rest follows from induction. 5. Show that f ∈ C 1 (Rm , Rn ) is locally Lipschitz continuous. 3. Some extensions 37 (x0 ) where ∂f∂x denotes the Jacobi matrix at x0 and . denotes the matrix norm (cf. 8)). 2. 6. Are the following functions Lipschitz continuous near 0? If yes, find a Lipschitz constant for some interval containing 0. (i) f (x) = (ii) f (x) = 1 . 1−x2 |x|1/2 . (iii) f (x) = x2 sin( x1 ). 7. Apply the Picard iteration to the first-order linear equation x˙ = x, x(0) = 1. 8. Apply the Picard iteration to the first-order equation x˙ = 2t − 2 max(0, x), x(0) = 0.
0 T ˜ Then Gronwall’s inequality implies |θ(t, x)| ≤ R(x) exp( 0 A(s) ds). Since limy→x |R(y, x)| = 0 uniformly in x in some neighborhood of 0, we have 0 ˜ limx→0 R(x) = 0 and hence limx→0 θ(t, x) = 0. Moreover, ∂φ ∂x (t, x) is C as the solution of the first variational equation. 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.
Discuss the equation x˙ = x2 − t. 6. Qualitative analysis of first-order periodic equations Some of the most interesting examples are periodic ones, where f (t + 1, x) = f (t, x) (without loss we have assumed the period to be one). 69) where h ≥ 0 is some positive constant. In fact, we could replace 1 − sin(2πt) by any nonnegative periodic function g(t) and the analysis below will still hold. 2 are depicted below. It looks like all solutions starting above some value x1 converge to a periodic solution starting at some other value x2 > x1 , while solutions starting below x1 diverge to −∞.