lipschitz condition solved examples


  • How do you solve Lipschitz condition?

    Let f(t, y) = ty2. Then since f(t, y2) ? f(t, y1) = ty2 + y1y2 ? y1 is not bounded by any constant times y2 ? y1, f is not Lipschitz with respect to y on the domain R × R. However f is Lipschitz on any rectangle R = [a, b] × [c, d] since we have ty1 + y2 ? 2 max{a, b} · max{c,d} on R.
  • What is an example of a function that fails to satisfy a Lipschitz condition at a point of continuity?

    f(x)=?x at x=0.
  • How do you prove a function is Lipschitz?

    A function is called locally Lipschitz continuous if for every x in X there exists a neighborhood U of x such that f restricted to U is Lipschitz continuous. Equivalently, if X is a locally compact metric space, then f is locally Lipschitz if and only if it is Lipschitz continuous on every compact subset of X.
  • Lipschitz condition. Definition: function f (t,y) satisfies a Lipschitz condition in the. variable y on a set D ? R2 if a constant L > 0 exists with. f (t,y1) ? f (t,y2) ? Ly1 ? y2, whenever (t,y1),(t,y2) are in D.
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