Does Non-Lipschitz f(x,t) Ensure No Global Unique Solution?

[fortune]

Hi,

For a first order Diff Equa. x'=f(x,t) and the IC: x(0)=x_0.
with t from [0 to infinity)
If f(x,t) doesn't satisfy the Lipschitz condition, can I say for sure that there doesn't exist a global unique solution?
I think the answer is "no" but I am not sure. Can you all confirm?

Also, can I use the Lipschitz condition to check the existence of a local solution around the IC? I see somebody often check the continuity of f(x) and df(x)/dx around the IC. Is this equavilent to the Lipschitz?

Thanks

 

[HallsofIvy]

No, even if the Lipschitz condition is not satisfied there may still exist a unique solution. You just can't be certain that the solution is unique.

You need Lipschitz to guarantee uniqueness. The fact that the function f(x,y) is continuous is sufficient to give existence.

Showing that \frac{\partial f}{\partial x} is "overkill". You can use the mean value theorem to show that if a function is differentiable on an interval, then it is Lipschitz there so differentiable is sufficient. But there exist Lipschitz functions that are not differentiable so it is not necessary.