Questions: Existence and Uniqueness Theorems (Picard-Lindelöf Theorem)

Question 1

Consider the IVP dy/dx = y^(2/3), y(0) = 0. What does the Picard-Lindelöf theorem say about this problem?

• A. It guarantees a unique solution because f(x,y) = y^(2/3) is continuous everywhere
• B. It guarantees at least one solution but not uniqueness, because ∂f/∂y is not continuous at y = 0
• C. It guarantees no solution exists because the conditions fail
• D. It is irrelevant because the trivial solution y ≡ 0 is obvious by inspection

Answer: (B)

The theorem requires both f continuous AND ∂f/∂y continuous near the initial point. Here f = y^(2/3) is continuous, but ∂f/∂y = (2/3)y^(−1/3) blows up at y = 0 — the Lipschitz condition fails. The theorem gives no uniqueness guarantee. In fact, this IVP has multiple solutions: y ≡ 0 and y = (x/3)³. Option A is the tempting mistake: continuity of f alone is not enough.

Question 2

An IVP satisfies both conditions of Picard-Lindelöf near the initial point. What does the theorem actually guarantee?

• A. The solution is valid on the entire real line because the conditions hold everywhere near the initial point
• B. A unique solution exists in some neighborhood of the initial point — the theorem says nothing about global existence
• C. The solution is unique globally because it is the only one found by standard methods
• D. The theorem guarantees the solution is defined for all initial conditions, not just this one

Answer: (B)

The Picard-Lindelöf theorem is a local result: it guarantees a unique solution in some neighborhood of x₀, not for all x. Global existence requires separate analysis. For example, dy/dx = y², y(0) = 1 satisfies the theorem's conditions near (0,1), yet the solution y = 1/(1−x) blows up at x = 1. The conditions guarantee local existence and uniqueness; how far the solution extends is a separate question.

Question 3

If the Picard-Lindelöf conditions fail for an IVP, the IVP has no solution.

• T. True
• F. False

Answer: False

Failing the theorem's conditions means the theorem provides no guarantee — it says nothing about what actually happens. The IVP may still have a solution (or even multiple solutions). The classic example dy/dx = y^(2/3), y(0) = 0 fails the Lipschitz condition but has multiple solutions. The theorem is a sufficient condition for existence and uniqueness, not a necessary one.

Question 4

The Picard-Lindelöf theorem guarantees that when its conditions hold, the solution to an IVP is unique in some neighborhood of the initial point.

• T. True
• F. False

Answer: True

This is precisely what the theorem states. Continuity of f and continuity of ∂f/∂y (the Lipschitz condition) near (x₀, y₀) together guarantee both existence and uniqueness of a solution near x₀. The Picard iteration argument constructs the solution as a converging sequence, and the Lipschitz condition is what prevents divergence — and therefore what prevents multiple solutions from branching off the same initial point.

Question 5

Why does the Lipschitz condition on ∂f/∂y prevent non-uniqueness of solutions, while continuity of f alone is not sufficient?

Model answer: Continuity of f ensures the right-hand side doesn't jump discontinuously, which helps existence, but it doesn't control how rapidly f can change in the y-direction. If f varies too steeply in y, two solution curves starting at the same point can diverge — the equation pushes them apart. The Lipschitz condition (bounded ∂f/∂y) limits this y-sensitivity: nearby solutions can't diverge faster than exponentially, and the Picard iteration converges to a unique limit. Without it, multiple solution curves can pass through the same initial point, as in the y^(2/3) example.

The intuition is that the Lipschitz condition bounds the spread rate of nearby solutions. If ∂f/∂y is unbounded, solutions can branch at the initial point. Continuity of f is necessary to construct a candidate solution, but uniqueness requires the additional control on how f varies with y.