1.4. Continuity with Respect to Initial Conditions 23 Exercise 1–18. Prove Gronwall’s Inequality. Hint: Since u ≤ U, it is enough to show that U(t) ≤ CeKt, or equivalently that e−KtU(t) ≤ C, and since U(0) = C, it will suﬃce to show that e−KtU(t) is nonincreasing, i.e., that (e−KtU(t)) ≤ 0. Since (e−KtU(t)) = e−Kt(U (t) − KU) and U = Ku, this just says that Ke−Kt(u − U) ≤ 0. 1.4.2. Theorem on Continuity w.r.t. Initial Conditions. Let V be a C1 vector field on Rn and let σp(t) denote the maxi- mal solution curve of dx dt = V (x) with initial condition p. Then as q tends to p, σq(t) approaches σp(t), and the convergence is uniform for t in any bounded interval I on which σp is defined. Proof. We have seen that σp(t) = p+ t 0 V (σp(s),s) ds, and it follows that σp(t) − σq(t)≤ p − q + t 0 V (σp(s),s) − V (σq(s),s) ds. On the other hand, it is proved in Appendix A that on any bounded set (and in particular on a bounded neighborhood of σp(I) × I) V satisfies a Lipschitz condition V (x, t) − V (y, t)≤ K x − y , so it follows that σp(t) − σq(t)≤ p − q + K t t0 σp(s) − σq(s) ds. It now follows from Gronwall’s Inequality that σp(t) − σq(t)≤ p − q eKt. 1.4.3. Remark. For the differential equation dx dt = kx, the max- imal solution is σp(t) = ektp, so σp(t) − σq(t) = ekt p − q . Thus if k is positive, then any two solutions diverge from each other expo- nentially fast, while if k is negative, all solutions approach the origin (and hence each other) exponentially fast. But continuity with respect to initial conditions is not the whole story. 1.4.4. Theorem on Smoothness w.r.t. Initial Conditions. Let V be a Cr vector field on Rn, r ≥ 1, and let σp(t) denote the maximal solution curve of dx dt = V (x) with initial condition p. Then the map (p, t) → σp(t) is Cr. The proof of this theorem is one of the most diﬃcult in elementary ODE theory, and we have deferred it to Appendix F.

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