state for what value(s) of t0 the existence and uniqueness theorem fails to apply


NOTES ON THE EXISTENCE AND UNIQUENESS THEOREM. FOR FIRST ORDER DIFFERENTIAL EQUATIONS. I. Statement of the theorem. We consider the initial value problem.
  • Why does the existence uniqueness theorem not apply to this IVP?

    The uniqueness theorem does not apply because the function f (y) = y 23 has an infinite slope at y = 0 and therefore is not Lipschitz continuous, violating the hypothesis of the theorem.
  • What is the existence and uniqueness theorem for initial value problem?

    Hence the existence and uniqueness theorem ensures that in some open interval centred at 0, the solution of the given ODE exists. Thus, the solution of given ODE is y = 1/ (1 – x), which exists for all x ? ( – ?, 1).
  • How do you know if existence and uniqueness theorem applies?

    Existence and Uniqueness Theorem (EUT)
    If f, ? f ? y , and ? f ? y ? are continuous in a closed box B in three-dimensional space (t-y- y ? space) and the point ( t 0 , y 0 , y ? 0 ) lies inside B, then the IVP has a unique solution y ( t ) on some t-interval I containing t 0 .
  • an existence and uniqueness theorem, asserting that by the prescription of some freely chosen functions, one can single out one specific solution of the PDE. by continuously changing the free choices, one continuously changes the corresponding solution.
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