12 are unique as expressed by the following theorem

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Unformatted text preview: ) be continuous and satisfy a Lipschitz condition on f(t y) j 0 t T ;1 < y < 1g: Then the IVP (1.1.2) has a unique continuously di erentiable solution on 0 T ] for all y0 2 (;1 1). Proof. The proof appears in most books on ODE theory, e.g., 2]. Remark 1. A Lipschitz conditions guarantees unique solutions and is not needed for esistence. For example, the IVP y = 3y2=3 y(0) = 0 0 has the two solution y(t) = 0 and y(t) = t3 . This f (t y) = 3y2=3 does not satisfy a Lipschitz condition. 9 Remark 2. Theorem 1.2.1 applies when f satisfys a Lipschitz condition on a compact, rather than an unbounded, domain D as long as the solution y(t) remains in D 2]. In addition to existence and uniqueness, we will want to know something about the stability of solutions of the IVP. In particular, we will usually be interested in the sensitivity of the solution to small changes in the data. Perturbations arise naturally in numerical computation due to discretization and round o errors. A formal study of sensitivity would lead us to the following notion of a well-posed...
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This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

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