Thus corollary 37 if is on an open set then it is

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Unformatted text preview: l convex, by lemma 3.6 is Lipschitz on it. □ . Since each 3.3 Existence and Uniqueness Theorem Let be open and . Consider the initial value problem is compact and 7 Chapter 3 . [1] Integrate to obtain . [2] The initial value problem [1] is equivalent to the integral equation [2]. We will solve Eq. [2] on for some , which will give . There is a relationship between the smoothness of we are given , we have from [2] that Now assume [2] gives and . Assume only . , Then if and . From the reasoning above obtain . But then , which means . This can be generalized to obtain Lemma 3.9. Suppose . for and is a solution of [2]. Then Proof. By induction in Meiss. Let be an element of a suitable function space . Define an operator . [3] A solution of [2] is a fixed point of : . The contraction mapping theorem implies that if contraction then exists and is unique. is a complete metric space and is a Theorem (3.10) (Existence and Uniqueness) Suppose that for there is a such that is Lipschitz with constant . Then the initial value problem [1] has a unique solution, for , provided that where Partial Proof. Let . Divide the proof into three parts. 8 Chapter 3 (A). Show complete. Use the sup norm. Consider a Cauchy sequence of functions in . Since convergence is uniform, the limit is a continuous function is closed, . Therefore and is complete. Since (B) Show that . From [3], closed, attains a maximum on . is continuous. Since is continuous and . Let . From [3] is , where it is understood in expressions like these that if reversed. If then . (C) Show is a contraction. Consider . Then since the limits of integration should be is Lipschitz . Taking the max of both sides over obtain contraction provided that . . Then Since the hypotheses of the contraction mapping theorem are satisfied, point that solves [2] and hence [1]. □ has a unique fixed Remark. The only deficiency in this proof is the additional condition Remark. If and is locally Lipschitz on , then for every that is Lipschitz on and theorem 3.10 applies. is a . there is a such Remark. Geometrically, this theorem implies that trajectories never cross. sketch trajectories crossing at . This can never happen. Remark. A non-autonomous system may be converted into an autonomous one. Is equivalent to , , , . where , and . Then theorem 3.10 gives existence and uniqueness as long as is Lipschitz in as well as . However, the requirement that is Lipschitz in may be relaxed. 9 Chapter 3 Theorem 3.11 (Nonautonomous Existence and Uniqueness). Suppose uniformly Lipschitz function of with constant , and a continuous function on . Then the initial value problem is a , has a unique solution for and Where the maximum is taken over , where and . The contraction mapping theorem helps prove the existence and uniqueness of a solution to [1] but does not show us how to find a solution. Picard iteration constructs succ...
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