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Part CCSIR NET June 2023bounded-rhs-global-existence

Bounded rhs global existence

Let f∈C1(R)f \in C^{1}(\mathbb{R}) be bounded. Consider the initial-value problem (P): x′(t) = f(x(t)), t > 0, x(0) = 0. Which of the following statements are true?

  1. A.(P) has solution(s) defined for all t > 0.✓ Correct
  2. B.(P) has a unique solution.✓ Correct
  3. C.(P) has infinitely many solutions.
  4. D.The solution(s) of (P) is/are Lipschitz.✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests existence vs uniqueness.

See pricing

50 are analysed free — try those first.

The trap it tests

Existence vs uniqueness

A theorem giving one was read as giving both.

Drill statements like this

Related counterexample: Every IVP has a unique solution

More on this topic

The chapter behind this: Existence, uniqueness and how far the solution extends — free to read

From Ordinary Differential Equations › Existence–uniqueness, Picard, Lipschitz

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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