How can I prove that $u$ is defined for all $x \in \mathbb{R}$? Problem: Suppose $f:\mathbb{R} \to \mathbb{R}$ is defined as follow: $f(t)=\frac{1}{1+t^4}-t^3$.
How can I prove that for all $y_0,y_1 \in \mathbb{R}$ the problem:
$$\begin{cases}
y''(x)=f(y(x))\\
y(0)=y_0 \\
y'(0)=y_1
\end{cases}$$
has a solution $y$ defined for all $x \in \mathbb{R}$?
Attempt: I tried to reduce to the following:
$$\begin{cases}
v'=f(y) \\
y'=v
\end{cases}$$
but I cannot go further because I would like to estimate $|f(y_1)-f(y_2)|$ but the term $t^3$ gives me problems.
 A: We have
$$y^{\prime \prime}(x)y^\prime(x) = f(y(x)) y^\prime(x)$$ and therefore
$$\frac{1}{2}\left(y^\prime(x)\right)^2-\frac{1}{2}y_1^2 = \int_{y_0}^{y(x)}f(t) \ dt=\int_{y_0}^{y(x)}\left(\frac{1}{1+t^4}-t^3\right) \ dt. \text{(E)}$$
Suppose that the solution exists only on the right in a bounded interval $[0, a)$ with $a \gt 0$. The limit $\lim\limits_{x \to a^-} y(x)$ if it exists can't be finite as otherwise, $\lim\limits_{x \to a^-} y^\prime(x)$ would also exists according to the equation (E) and we could extend the solution $y$ beyond $a$. If $\lim\limits_{x \to a^-} y(x) \neq \pm \infty$, then it exists an increasing sequence $\{a_n\}$ converging to $a$ such that $\lim\limits_{x \to a^-} y^\prime(x) =\infty$. A contradiction with the equation (E) if $\{y(a_n)\}$ can also be chosen to be bounded. Therefore we have $\lim\limits_{x \to a^-} y(x) \in \{\infty, - \infty\}$.
We then get a contradiction with equation (E) making $x \to a^-$. Namely that the RHS moves towards $-\infty$ while the LHS is bounded below by $-\frac{1}{2}y_1^2 $.
Therefore the solution exists in $[0, \infty)$. We can have similar arguments to prove that the solution exists on the left in the interval $(-\infty,0]$ to get the desired conclusion.
