I was trying to prove the solution of the following system is globally unbounded and globally uniformly ultimately bounded: \begin{equation} \dot{x} = x^2 - x^3 +d(t) \end{equation}
And then compute the uniform bound supposing $|d(t)| \le D \le \infty$.
I tried using the following Lyapunov function: \begin{equation} V(x) = \frac{1}{2}x^2 \end{equation}
Which is definite positive and radially unbounded, and we have (see Lemma at the bottom of this question): \begin{align} \alpha_1(||x||) = \alpha_2(||x||) = \frac{1}{2}|x|^2 \end{align}
Then, taking the derivative with respect to time and introducing $\theta$ at the second step ($0 < \theta < 1$): \begin{align} \dot{V} &= x\left(x^2 - x^3 + d(t)\right) \\ &= x^3 - x^4 +x \cdot d(t) +\theta x^3 -\theta x^3 \\ &\le -x^4 + (1 + \theta)x^3 \end{align}
Therefore, we have: \begin{equation} W_3(x) = x^4 - (1+\theta)x^3 \end{equation}
And the inequality is true whenever: \begin{equation} |x| \ge \left(\frac{|d(t)|}{\theta}\right)^{\frac{1}{2}} = \alpha_4(||x||) \end{equation}
So $\alpha_1$, $\alpha_2$, $W_3$, and $\alpha_4$ are known, but $W_3$ is not positive definite so something must be wrong and I can't proceed with computing the uniform bound $\delta$.
Any guidance would be of great help, thank you :)
APPENDIX
Lemma:
You can cut through the theorem by observing that for $x=1+D$ at some time $t$ you get $$ \dot x=x^2-x^2(1+D)+d(t)\le D(1-x^2)<0 $$ and similarly at $(t,-1-D)$ one has $\dot x>0$. This also extends for the intervals $|x|\ge 1+D$.
This means that the interval $[-1-D,1+D]$ is trapping in forward direction, and additionally all initial values outside this interval have solutions that fall inside this interval in finite time. This implies bounded solutions existing for all times after the initial time $t_0$.
If the task was to identify the components of the construction in the theorem, this answer only helps in giving an impression what the solution and constants should look like. Usually, a more careful construction will lead to tighter bounds.