# Continuous/differentiable functions

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
Homework Helper
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.