Continuous/differentiable functions

[fk378]

Don't really know how to think about these...

(1) Give an example of a 1-dimensional ODE of form x' = f(x), x(0)=x* where f: R->R is continuous but there exists more than one differentiable solution. Prove your assertion.

(2) Is it true that a 1-dimensional ODE of the same form as (1) where f(x) is differentiable in x for all x has a differential solution x(t) defined for all t>0? Why?

[HallsofIvy]

General existance and uniqueness theorem for first order equations:
If f(x,y) is continuous in both variables and "Lipschitz" in y ("Lipschitz" lies between "continuous" and "differentiable". If f(x,y) is differentiable with respect to y, that is sufficient but not necessary) in some neighborhood of (x_0, y_0), then there is a unique solution to y'= f(x,y) in some sub-neighborhood satisfying y(x_0)= x_0.

In particular, in order that there NOT be a unique solution (more than one) f(x,y) must not be differentiable with respect to y. In the form x'= f(x), you must have a function that is not differentiable with respect to x. Try f(x)= x^{1/2}. It's derivative is f'(x)= (1/2)x^{-1/2} which does not exist at x= 0.

By simple integration, dx/dt= x^{1/2} becomes x^{1/2}dx= dt so that (2/3)x^{3/2}= t+ C and, solving for x, x= ((3/2)(t+ C))^{2/3}[/tex]. In particular, taking C= 0, [itex]x= ((3/2)t)^{2/3} satisfies that equation as well as x(0)=0.

But, obviously, x(t)= 0 for all x also satisfies x'= x^{1/2} as well as x(0)= 0. Once you have two different solutions for that, we can extend it. We can find C such that x(t)= ((3/2)(t+ C))^{2/3} satisfies x(t_0)= 0 for any given t_0. But then, since x'= x^{1/2}= 0 whenever x= 0, we can "smoothly" attach x= 0 to the left of t_0 and then, at some t_1< t_0, attach another ((3/2)(t+ C))^{2/3} to the left of t_1. In other words, there exist an infinite number of solutions to x'= x^{1/2} that satisfy x(0)= 0.