Prove that $\begin{cases}x'(t)=-x^{3}(t)+\sin(t)x^{2}(t)+1 & t\geq 0,\\x(0)=x_0\in \Bbb{R}&\end{cases}$ has a global solution on $\Bbb{R}^+$

Question

We consider the following ODE

\begin{align}(a)\begin{cases}x'(t)=-x^{3}(t)+\sin(t)x^{2}(t)+1 & t\geq 0,\\x(0)=x_0\in \Bbb{R}&\end{cases}\end{align} I want to prove $(a)$ that has a global solution on $\Bbb{R}^+.$

MY TRIAL

Let $f(t,x(t))=-x^{3}(t)+\sin(t)x^{2}(t)+1.$ I'm thinking of proving that $f$ is continuous and Lipschitz.

Let $x,y\in\Bbb{R},$ THEN \begin{align}\Vert f(t,x(t))-f(t,y(t)) \Vert&=\Vert -x^{3}(t)+\sin(t)x^{2}(t)+1 -\left[-y^{3}(t)+\sin(t)y^{2}(t)+1 \right]\Vert \\&=\Vert -x^{3}(t)+\sin(t)x^{2}(t) +y^{3}(t)-\sin(t)y^{2}(t) \Vert\\&=\Vert y^{3}(t) -x^{3}(t)+\sin(t)\left(x^{2}(t) -y^{2}(t) \right)\Vert\\&\leq\Vert y^{3}(t) -x^{3}(t)\Vert+\Vert x^{2}(t) -y^{2}(t) \Vert\end{align} But I'm finding it hard to bound this inequality to get a constant.

Questions:

I'm I going the right way?

If yes, could you help me with the inequality?

If no, could you please, provide an alternative solution?

Answer 1

For $x > 2$, $\dot{x} = x^2 (\sin(t)-x) + 1 < -3$, and similarly for $x < -1$, $\dot{x} > 2$. Thus for any initial $x_0$, the solution will eventually be trapped in the interval $-1 \le x \le 2$. This implies that it will exist for all future time.