# Fundamental Existence and Uniqueness Theorem

This question is for anybody who has a copy of the ODE book named "Differential Equations and Dynamical Systems (3rd ed.)" by Lawrence Perko (feel free to comment if you don't have it either). In section 2.4, Perko states that, "The fundamental existence-uniqueness theorem of section 2.2 established that if $f \in C^1(E)$ (where $E$ is an open set containing $x_0$) then the initial value problem $x^{\prime} = f(x), x(0) = x_0$ has a unique solution defined on some interval $(-a,a)$ (where as the fundamental existence-uniqueness theorem claims that the initial value problem $x^{\prime}=f(x), x(0)=x_0$ has a unique solution $x(t)$ on the interval $[-a,a]$). I'm quite sure that this is not a typo. My question is why is there a unique solution to the ivp on $(-a,a)$? I'm following the proof of the fundamental existence-uniqueness theorem in section 2.2 and can not conclude why the unique solution to the ivp is on $(-a,a)$.

• If you have a solution on $[-a,a]$, you can restrict it to $(-a,a)$. Conversely, if you have a solution on an open interval, you can restrict it to a slightly smaller closed interval. Since it is all about existence (and uniqueness) on some interval, both are equivalent. – Daniel Fischer Oct 11 '13 at 23:34
• @Daniel: Thanks for the comment and commenting on the converse (I think that's the main issue of my question). – Mr Curious Oct 11 '13 at 23:37
• However, if you have a solution $x$ on $[-a,a]$, then $x(a) \in E$, and you can invoke the existence theorem on the initial value problem $y(a) = x(a)$ to get a solution on some $[a-\varepsilon,a+\varepsilon]$. Uniqueness then says $y \equiv x$ on $[a-\varepsilon,a]$, so you can extend the solution. Similar at $-a$. That yields a maximal solution defined on an open interval (if $E$ is open, and $f$ is $C^1$, at least). – Daniel Fischer Oct 11 '13 at 23:46