Lipschitz Condition, Uniqueness and Existence of ODE

[BrainHurts]

Homework Statement

Find a solution of the IVP

[itex]\frac{dy}{dt}[/itex] = t(1-y²)^{[itex]\frac{1}{2}[/itex]} and y(0)=0 (*)

other than y(t) = 1. Does this violate the uniqueness part of the Existence/Uniqueness Theorem. Explain.

Homework Equations

Initial Value Problem [itex]\frac{dy}{dt}[/itex]=f(t,y) y(t₀)=y₀ has a solution if f is continuous on B = [t₀,t₀ + a] x [y₀-b,y₀+b]

The Attempt at a Solution

So when I solved * by separation of variables, my solution

y(t)= sin(t²+[itex]\frac{πk}{2}[/itex]) where k = ±1,±2,...

f(t,y) = t(1-y²)^{[itex]\frac{1}{2}[/itex]}
So [itex]\frac{∂f}{∂y}[/itex] = ty(1-y²)[itex]\frac{-1}{2}[/itex]

I want to say that that as long as |y|≤ 1, f(t,y) is continuous and because the set B is closed, there exists a Maximum value M = max_{(t,y)[itex]\inB[/itex]}|[itex]\frac{∂f}{∂y}[/itex]|, this is the Lipschitz condition and f satisfies this condition.

 

[HallsofIvy]

No, that's not true. The partial derivative is what you say (you mean that last "-1/2" to be an exponent) but that goes to infinity at y= 1 so there is no such maximum value.

f is continuous, and so has a maximum on a closed and bounded set, but [itex]\partial f/\partial y[/itex] is not.

 

[BrainHurts]

yes sorry that should be raised to the -1/2. I just saw that. So because ∂f/∂y goes to infinity at y=±1, does this mean that the Lipschitz condition failed and that means and that the existence and uniqueness theorem is violated?

 

[HallsofIvy]

Yes, the Lipschitz condition fails so the existence and uniqueness theorem does NOT APPLY. (I would not say it is "violated". That would imply a situation where the hypotheses are true and the conclusion is false- a counterexample.)