Maximal unique solution to an IVP.

Question

In class we learned the existence and uniqueness theorems for differential equations. The weaker Picard-Lindelof states that for any IVP, $$ \begin{cases} x'(t) = f(t, x(t))\\ x(t_0) = x_0 \end{cases} $$ where $f$ is continuous in the first argument and locally Lipschitz in the second, there is a unique solution in some neighborhood around $t_0$ (in a open interval $I_0$ containing $t_0$). This result was extended to: there is a maximal unique solution to all IVP with the above form (Basically means there is biggest possible interval on which the solution is unique). More precisely, it means if $x:(a, b) \to \mathbb{R}^n$ is the maximal solution and $y:(a', b') \to \mathbb{R}^n$ is any other solution to the same IVP, then $(a', b') \subset (a, b)$ and $x = y$ on $(a', b')$.

My question is: can we find the maximal interval where there is a unique solution for any given IVP (also for any function $f$)?

Answer 1

It depends on what you mean by "find". You can approximate it arbitrarily closely, using numerical methods. But in general you won't be able to find a closed-form expression for it.

EDIT: To be more precise: suppose that for each $(a,b,R)$ with $a < b$ you know (or can compute) a modulus of continuity and a Lipschitz constant for $f(t,y)$ on $\{(t,y): a \le t \le b, |y| \le R\}$. There is (computable) $\delta > 0$ such that if $a \le s-\delta < s + \delta \le b$ and $0 < \epsilon < 1$ and $|Y| < R - 2 \epsilon$, then we can calculate (e.g. by the methods used in the proof of Picard-Lindelof) $\tilde{y}(t)$ for $s-\delta \le t \le s+\delta$ such that $|y(t) - \tilde{y}(t)| < 2 \epsilon$ for every solution $y(t)$ with $|y(s) - Y| < \epsilon$. If your initial value problem does have a solution defined on $[a,b]$, there is some $R$ that bounds $|y(t)|$ on $[a,b]$, and by appropriate choice of $\epsilon$ we can get an approximate solution accurate enough to show that $|y(t)| < R + 1$ on $[a,b]$, and in particular that the solution exists on this interval (i.e. a maximal solution that ceases to exist at some point must cross the region $R < |y| < R + 1$ before it ceases to exist). So you can approximate the maximal interval from below. Approximating it from above seems to be more difficult, and I don't know if it can be done in general.