The fundamental theorem of autonomous ODE states that if $V:\Bbb R^n\to\Bbb R^n$ is a smooth map, then the initial value problem $$ \begin{aligned} \dot{y}^i(t) &= V^i(y^1(t),\ldots,y^n(t)),&i=1,\ldots,n \\ y^i(t_0) &= c^i, &i=1,\ldots,n \end{aligned}\tag{1} $$ for $t_0\in\Bbb R$ and $c=(c^1,\ldots,c^n)\in\Bbb R^n$ has the following existence property:

Existence: For any $t_0\in\Bbb R$ and $x_0\in\Bbb R^n$, there exist an open interval $J$ containing $t_0$ and an open subset $U$ containing $x_0$ such that for each $c\in U$, there is a smooth map $y:J\to\Bbb R^n$ that solves $(1)$.

Now here is my question:

Question: Suppose we already know that a solution exists with initial value $y(t_0)=x_0$ on an interval $J_0$ containing $t_0$. Does the interval $J$ above can be assumed to contain $J_0$?

A priori, there is noting telling us that in the statement of the theorem. My question can be rephrased as follows.

Reformulation of the Question: Let $y:J\to\Bbb R^n$ be a smooth solution to $(1)$ with initial value $y(t_0)=x_0$. Is there an open set $U$ containing $x_0$ such that for all $c\in U$ there is a smooth solution $z:J\to\Bbb R^n$ to $(1)$ with initial value $z(t_0)=c$?

Edit: And what about the case where $J$ is a compact interval?

  • $\begingroup$ You should keep reading until you come up with the extension theorem and the dependence on initial conditions theorem. $\endgroup$ – Pragabhava Jan 7 '16 at 13:11
  • $\begingroup$ @Pragabhava This is the only theorem I have. Where did you read about this extension theorem and dependence on initial conditions theorem? Is the answer to my question yes? $\endgroup$ – user303366 Jan 7 '16 at 13:19
  • $\begingroup$ In any reasonable book on odes you'll find all the existence, continuation and parameter dependence theorems. The example given by Giuseppe is how a typical section on continuation of solutions begins. $\endgroup$ – Pragabhava Jan 7 '16 at 15:45
  • $\begingroup$ @Pragabhava Could you give me an example of a reasonable ODE book? Many thanks. $\endgroup$ – user303366 Jan 7 '16 at 15:52
  • $\begingroup$ For a first course, two classics are Boyce and DiPrima, and Braun. For a more advanced course, Brauer or Coddington and Levinson, which is the bible (or was in my time). $\endgroup$ – Pragabhava Jan 7 '16 at 15:57

The answer to the reformulation is negative. Consider the problem \begin{equation} \begin{cases} y'=y^2 \\ y(0)=c \end{cases} \end{equation} Its solution is $y(t)=\frac{1}{c^{-1}-t}$ and it is defined on $J_c=(-\infty, c^{-1})$. So, for example, the solution with initial datum $c=1$ is defined on $(-\infty, 1)$ while the solution with initial datum $1+\epsilon$ is defined on a strictly smaller interval, no matter how small $\epsilon$ is.

  • $\begingroup$ What about the case where $J$ is a closed interval? $\endgroup$ – user303366 Jan 7 '16 at 13:35
  • $\begingroup$ In that case the answer is positive. You just want to exclude "blow-up points" lying at the boundary of your interval of existence. $\endgroup$ – Giuseppe Negro Jan 7 '16 at 13:41
  • $\begingroup$ Why is it true? Is it easy to prove? $\endgroup$ – user303366 Jan 7 '16 at 13:43
  • $\begingroup$ Well, I don't know how to write down a rigorous proof right now, but the idea is the following: if a solution is defined on $[a,b]$, then there exists an open interval $(A, B)\supset [a,b]$ such that such solution is defined on $(A,B)$. If you perturb the initial datum slightly, this results in a slight perturbation of $A$ and $B$, and you can make it so small that $[a,b]$ is still included in this perturbed interval. $\endgroup$ – Giuseppe Negro Jan 7 '16 at 21:56

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