TITLE: The IVP $\begin{cases}\dot{x}=x^3+e^{-t^2}\\x(0)=1\end{cases}$ possesses a solution in $I=(-1/9,1/9)$
QUESTION [6 upvotes]: I'm stuck with this problem:

Prove that the IVP
$$\begin{cases}\dot{x}=x^3+e^{-t^2}\\x(0)=1\end{cases}$$
has an unique solution defined on $I=(-1/9,1/9)$. Which is the largest interval of definition of the solution? We can extend that solution to $t=1$?

Okay, so it's obvious that $f(t,x)=x^3+e^{-t^2}$ is locally Lipschitz with respect to $x$, and it's continuous too. So by Picard we can find a unique solution defined on $\mathbb{R}$.
But then the exercise wouldn't have any sense, so I think that I'm doing something wrong.
Thanks for your time.

REPLY [4 votes]: Fact 1. The flux function $f(t,x)=\mathrm{e}^{-t^2}+x^3$ of our ODE is $C^1$ in $x$, and hence locally Lipschitz, and therefore our IVP enjoys uniqueness. 
Fact 2. Since the flux function is smooth in the whole $\mathbb R^2$, then our IVP possesses a solution $\varphi$ defined on a maximal open interval $(a,b)$. If $a$ is finite then $\lim_{t\to a^+}\varphi(t)\in\{-\infty,\infty\}$, and similarly, if $b$ if finite then $\lim_{t\to b^-}\varphi(t)\in\{-\infty,\infty\}$.
Fact 3. The unique and maximally defined solution of 
$$
x'=g(t,x)=x^3,\,\, x(0)=1,
$$
is $\psi(t)=(1-2t)^{-1/2}$, defined in the maximal interval 
$\big(-\infty,\frac{1}{2}\big)$.
Fact 4. Clearly $\,g(t,x)<f(t,x),\,$ and since $\varphi(0)=\psi(0),\,$ then 
$$
\varphi(t)>\psi(t), \quad \text{for all t>0},
$$ 
thus $\varphi$ blows up for some $t\in \big(0,\frac{1}{2}\big]$, and hence $\varphi$ CAN NOT be extended until $t=1$.
Fact 5. 
Consider now the Picard iterative sequence which converges to $\varphi$:
$$
\varphi_0(t)=1, \quad \varphi_{n+1}(t)=1+\int_0^t \big(\varphi_{n}(s)^3+\mathrm{e}^{-s^2}\big)\,ds.
$$
We shall show that, 
$$
|\varphi_n(t)|\le 2, \quad \text{for all $n\in\mathbb N$ and $|t|\le 1/9$.}
$$
For $n=0$ is given. Assume that this holds for $n=k$. Then for $n=k+1$ and $|t|\le 1/9$, we have that
$$
|\varphi_{k+1}(t)|=\left|\,1+\int_0^t \big(\varphi_{k}(s)^3+\mathrm{e}^{-s^2}\big)\,ds\,\right|\le 1+
\int_0^t\big| \,\varphi_{k}(s)^3+\mathrm{e}^{-s^2}\big|\,ds\le 1+\frac{1}{9}\big(2^3+1\big)=2,
$$
and hence
$$
\Big[-\frac{1}{9},\frac{1}{9}\Big]\subset (a,b),
$$
and hence $\varphi$ is defined for $|t|\le 1/9.$
Fact 6. Let $h(t,x)=x^3+1\ge f(t,x)$. Then the solution $\vartheta$ of 
$$
x'=h(t,x), \quad x(0)=1,
$$
is definable in the whole $(-\infty,0]$, and since $\varphi(t)\ge \vartheta(t)$, for $t\le 0$, and $\varphi$ is increasing, then clearly $\varphi$ is also definable in the whole $(-\infty,0]$.