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Part CCSIR NET June 2025peano-gives-existence-from-continuity-lipschitz-is-only-needed-for-uniqueness

Peano gives existence from continuity lipschitz is only needed for uniqueness

Let D = {(x,y)∈R2:−1≤x≤1,−1≤y≤1(x, y) \in \mathbb{R}^{2} : -1 \le x \le 1, -1 \le y \le 1}, and f:D→Rf : D \to \mathbb{R} be the function defined by f(x,y)=1+√(yf(x, y) = 1 + \surd(y₊), where y₊ = max{y, 0}. Consider the initial value problem (IVP) dy/dx = f(x, y), y(0) = 0. Then which of the following statements are true?

  1. A.f is a Lipschitz continuous function on D
  2. B.f is NOT a Lipschitz continuous function on D✓ Correct
  3. C.IVP has at least one solution✓ Correct
  4. D.IVP has NO solution

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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