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
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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- Spring '14
- Boundary value problem, lipschitz condition, IVPs, di erential equations