# Hint needed for lipschitz problem.

1. Oct 7, 2011

### JakobReed

a sufficient condition for uniqueness is the Lipschitz condition:

On a domain D of the plane, the function f (x, y) is said to satisfy the Lipschitz condition for a constant k > 0 if:

|f(x,y1)−f(x,y2)|≤k|y1−y2|

for all points (x,y1) and (x,y2) in D.

Give an example of an IVP with two solutions on a domain (say, a rectangle) and show that the function f(x,y) appearing in the differential equation fails to be Lipschitz for any k > 0.

i really have no idea where to begin. Can i get a hint or a suggestion from someone?

2. Oct 7, 2011

### HallsofIvy

Staff Emeritus
"Lipschitz" is intermediate between "continuous" and "differentiable" in strength (any Lipschitz function is continuous but not vice versa; any differentiable function is Lipschitz but not vice versa). Start by looking at dy/dx= f(x,y) where f is continuous but not differentiable with respect to y. Try fractional powers of y.