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?