Continuous/differentiable functions

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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?
 

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HallsofIvy
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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 [itex](x_0, y_0)[/itex], then there is a unique solution to y'= f(x,y) in some sub-neighborhood satisfying [itex]y(x_0)= x_0[/itex].

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 [itex]f(x)= x^{1/2}[/itex]. It's derivative is [itex]f'(x)= (1/2)x^{-1/2}[/itex] which does not exist at x= 0.

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

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

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