How to prove: Uniqueness of solution to first order autonomous ODE

[Jösus]

Hello!

I would like to prove the following statement: Assume f\in C^{1}(\mathbb{R}). Then the initial value problem \dot{x} = f(x),\quad x(0) = x_{0} has a unique solution, on any interval on which a solution may be defined.

I haven't been able to come up with a proof myself, but would really like to see a direct proof, not using too serious tools from a sophisticated theory of ODE's. I would very much appreciate it if someone could help me out.

Thanks in advance!

 

[tiny-tim]

Hello Jösus!

… the initial value problem \dot{x} = f(x),\quad x(0) = x_{0} has a unique solution, on any interval on which a solution may be defined.

So you can assume that there is a solution g, and you have to prove that there can't be two solutions, g and h.

So suppose dg/dx = dh/dx = f.

Then … ? ​

 

[Jösus]

Correct me if I'm wrong, but shouldn't the equation read dg/dt = f(g(t)), \quad dh/dt = f(h(t)), and thus there would be no apparent reason for these derivatives to be equal?

I have thought about it some more, and found that if f(x_{0}) \neq 0 then there is an interval containing x_{0} on which a unique solution exists (the equation is separable, so a simple integration trick works). When f reaches a zero in finite time, so that the inteval on which this solution is defined is cut off there may be extensions to the solutions beyond the problematic points. If, say, f(a) = 0 then setting x(t) = a for t larger than (or smaller than if a is on the right of our starting point) will, I believe, do the trick. It is solutions of this type that aren't always unique, but with the requirement f \in C^{1}(\mathbb{R}) it should work. Any new ideas?