Very similar to that of theorem 310 steps a and c are

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Unformatted text preview: hen if must satisfy. □ we also require leading to the modified inequality that Now prove the important Grönwall inequality. As a preliminary remark, note that if then is increasing less rapidly that . Integrate both sides to obtain , provided . Lemma 3.13 (Grönwall). Suppose obeys the inequality are continuous, , and [2] for all . Then for all . Proof. Notice that is differentiable. an integrating factor to give , or . Multiply by 13 Chapter 3 . Integrate both sides between and : , from which the result follows. ■ Example. Suppose two solutions of with initial conditions compared on a common time interval of existence. They satisfy If is Lipschitz with constant the inequalities and and are being , then the distance between the trajectories satisfies . This has the form of [2] with and . Then obtain .■ Theorem 3.14 (Lipschitz Dependence on Initial Conditions). Let a such that is Lipschitz with constant and that interval of existence for solutions . Then in with Lipschitz constant . Proof. Based on the example above and a similar example with Recall that , and suppose there is is the common is uniformly Lipschitz is differentiable at . This can be rewritten . if the following limit exists .□ 14 Chapter 3 is differentiable at if there is such that . This means where . Th Ty [3] ’ u wh such that Let d Note that [3] is equivalent to: . be the solution of the initial value problem [4] The linearization of about satisfies . Example. If has period , obtain subject of Floquet theory in section 2.8. ■ The Fundamental Matrix Solution , where has period . This was the satisfies . [5] Theorem 3.15 (Smooth Dependence on Initial Conditions). Suppose is on an open set . Then there exists an such that the solution of [1] is a function of for . Proof. Do not give. See in text. Remark. The proof is an excellent illustration of the use of G ö w ’ inequality. It shows that is by demonstrating that is its derivative: . If we assume that , this can be demonstrated directly. . 15 Chapter 3 Thus satisfies [5]. ■ Let be the solution of the initial value problem , where [6] is a vector of parameters. [6] is equivalent to . In the integrand, depends on [7] in two ways. Theorem 3.16 (Continuous Dependence on Parameters). Suppose is Lipschitz in and uniformly continuous in . Then [6] has a unique solution for that is a uniformly continuous function of on some interval . Proof. By theorem 3.10, and for some . Consider will have a common interval of existence . From [7] . Write Since is a uniformly continuous in , for any there is a . Then assume to obtain such that implies , wh ch g v , by G w’ qu y, . This expression can be made as small as we wish by making sufficiently small. A similar argument holds for .□ 16 Chapter 3 3.5 Maximal Interval of Existence Consider the initial value problem . [1] If is locally Lipschitz, then theorem 3.10 gua...
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This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .

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