Let us consider the following Cauchy problem:
\begin{cases} y''(x)=-y^3(x) \\ y(0)=0 \\ y'(0) = 1 \end{cases}
I want to show that there exists a global solution for all $x \in \mathbb{R}$, but I can't prove it.
The function $f$ is defined as $f(x,y_1,y_2)=[y_2,-y_1^3]^T$. I would like to show that $f$ is sublinear, i.e. $$||f(t,y)|| \leq h+k||y||$$ where $y = [y_1,y_2]^T$ but I can't show it.
I have $$||f|| = y_2^2 + y_1^6$$ but I have no clue on how to bound the last term. Any hint or answer is highly appreciated
$$y''=-y^3$$ $$2y''y'=-2y^3y'$$ $$y'^2=-\frac12 y^4+c_1$$ Conditions $y(0)=0$ and $y'(0)=1$ imply $c_1=\frac12$ $$y'^2=\frac12-\frac12 y^4$$ $$y'=\sqrt{\frac12-\frac12 y^4}\qquad \text{with sign according to }y'(0)=1$$ This is a Jacobi elliptic integral.
$$y=\text{sn}\left( \frac{x}{\sqrt{2}}+c_2\:\Big|\:-1\right)$$ sn is the Jacobi sn elliptic function https://mathworld.wolfram.com/JacobiEllipticFunctions.html
The condition $y(0)=0$ implies $c_2=0$. The solution is : $$y=\text{sn}\left( \frac{x}{\sqrt{2}}\:\Big|\:-1\right)$$