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Consider the following initial value problem, $$ \dot x = tx^3 \\ x(0) = x_{0} $$

We have the following theorem,

enter image description here. Since the hypotheses of the theorem are satisfied, we must have a solution on $[-a, a]$. To calculate $M$, it is easy to see that the maximum value attained by $t$ is $a$ and the one attained by $x$ in the ball of radius $b$ with center $x_0$ is $(x_0 + b)$. Hence, $$M = a(x_0 +b)^3$$ Solving for $a$ we get that $$ a = \pm \sqrt { \frac {b}{(x_0 + b)^3} } $$ Here is where I am stuck. I suppose that there exists a $b$ that maximizes $a$. Then, by maximizing $a$ we have thus found the maximal interval of existence for the solution. However, I cannot find a way to maximize the above expression for a fixed $x_0$. Is there something that I am missing or should another approach be taken to find the maximal interval of existence?

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  • $\begingroup$ This IVP is separable and can be solved explicitly without too much trouble, that is, you can simply compute the solution. $\endgroup$
    – Travis Willse
    Nov 21, 2014 at 1:51
  • $\begingroup$ Note also that maximal intervals of existence in general are not of the form $[t_0 - a, t_0 + a]$. $\endgroup$
    – Travis Willse
    Nov 21, 2014 at 1:53
  • $\begingroup$ The theorem statement seems slightly circular ($a$ in particular). $\endgroup$
    – copper.hat
    Nov 21, 2014 at 2:58

1 Answer 1

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The solution to the initial problem is $$x(t) = \frac{x_0}{\sqrt{1-x_0^2t^2 }}$$

The max interval is determined by: $$1-x_0^2T^2> 0\implies 0<T< |x_0|^{-1}$$

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