# Lipschitz Condition, Uniqueness and Existence of ODE

1. Feb 7, 2013

### BrainHurts

1. The problem statement, all variables and given/known data

Find a solution of the IVP

$\frac{dy}{dt}$ = t(1-y2)$\frac{1}{2}$ and y(0)=0 (*)

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

2. Relevant equations

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

3. The attempt at a solution

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

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

f(t,y) = t(1-y2)$\frac{1}{2}$
So $\frac{∂f}{∂y}$ = ty(1-y2)$\frac{-1}{2}$

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)$\inB$|$\frac{∂f}{∂y}$|, this is the Lipschitz condition and f satisfies this condition.

2. Feb 7, 2013

### 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 $\partial f/\partial y$ is not.

3. Feb 7, 2013

### 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?

Last edited: Feb 7, 2013
4. Feb 7, 2013

### HallsofIvy

Yes, the Lipschitz condition fails so the existance 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.)