Maximum interval for every initial condition

Question

Consider the IVP given by:

$\frac{dx}{dt}=-\frac{x^3}{1-t},\; x(0)=x_{0}$

WITHOUT solving the equation, show that:

a) For every $x_{0}$, the maximal interval is [0,1)

b) $\underset{t\uparrow 1} {\lim}\;x(t)$ exists and calculate it's value

My attempt: I know that for item a I need to show that f is continuous on the interval $[0,1)$, but do I need to show that it's locally lipschitz for every $x_{0}$?

Answer 1

The answer to both of these questions is contained in the proof in Appendix D.$6$ of a more general result. The limit in part $b)$ is zero.

I can offer a partial proof from scratch, not a complete answer, but too long for a comment. I think it is labored, though, so I would like to see a better, complete one.

If $r>0$ and $0<a<1$ then, $|f(t,x)|\le \frac{(r+x_0)^3}{1-a}$ on $\overline B(x_0,r)$ and for all $t\in [0,a)$, and $$ |f(t,x)-f(t,x')|=\left(\frac{x^2+xx'+x'^2}{1-t}\right)\cdot |x-x'|\le \frac{3(r+x_0)^2)}{1-a}\cdot |x-x'|. $$ Now apply the Picard–Lindelöf theorem: there is a unique solution at least for times in the interval $\left[0,\min\left\{a,\frac{r(1-a)}{(r+x_0)^3}\right\}\right)$. To conclude, we would like to show that there is an $r$ for which $$ a\le\frac{r(1-a)}{(r+x_0)^3} $$ but unfortunately, the lone maximum of the functions $r\mapsto \frac{r(1-a)}{(r+x_0)^3}$ depends heavily on $x_0.$

For part $b).$ note that if $x_0=0$ then the trivial solution works. Otherwise, take the case $x_0>0$. The other case $x_0<0$ should be similar. Then, $x$ is decreasing at $t=0.$ If $x(\tau)<0$ for some $0<\tau<1,$ then $x$ is increasing at $\tau$, which implies that

1. there is a $\tau_1$ such that $0<\tau_1<\tau$ and $x(\tau_1)<x(\tau)$

and

2. $x$ has a local minimum $\tau_2$ on $(0,\tau_1)$ that satisfies $x(\tau_2)<x(\tau_1)<x(\tau)<0.$

But this would imply that $x'(\tau_2)>0$ which is a contradiction.

We conclude that $x$ is nonegative on $[0,1)$, and this implies also that $x$ is nonincreasing there. Therefore, if there is a $\tau\in (0,1)$ such that $x(\tau)=0$ then $x(t)=0$ for all $\tau<t<1$ and we may conclude that $\underset{t\to 1^-}\lim x(t)=0.$

In any case, the limit exists because $x$ is monotone and bounded. and $\underset{t\to 1^-}\lim x(t)=l\ge 0$ and it remains to show that $l=0.$