Consider the dynamical system : $$x_1' = x_2, \space \space x_2' = -2x_1 + x_2(2-5x_1^2 - 3x_2^2)$$ where $x_1,x_2 \in \mathbb R$. Using the functional $V(x_1,x_2) = \frac{1}{2} x_1^2 + \frac{1}{2} x_2^2$, prove that for any given inital value, the initial value problem has a bounded solution.
Attempt :
Take the derivative over the solution curves for the given functional, which is :
$$\dot{V}(x_1,x_2) = -x_1x_2 + x_2^2(2-5x_1^2-3x_2^2)$$
I will try "increase" this expression via inequalities (which is a standard way of handling such problems), so that I can provide a compact form with $V(x_1,x_2)$ in expressions which we know it's positive :
$$\dot{V}(x_1,x_2)\leq-x_1x_2 + x_2^2(2-3x_1^2-3x_2^2)\leq\frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 + x_2^2(2-x_1^2 - x_2^2)$$
Now, I have reduced the coefficients inside the parenthesis, since $x_1^2,x_2^2 \geq0$ which means that a smaller coefficient leads to a bigger quality. The inequality derived for the expression at front is a simple application of $(x_1+x_2)^2 \geq 0$.
However, I cannot seem how I should continue from now on, since that $x_2^2$ in front of the parenthesis is bugging me and I do not know how to convert it while keeping the inequality in a form that would produce an expression of $V$.
Any tips ?
The functional $V$ is ill-adapted to this system. On the other hand, a one-line computation shows that $$W(x_1,x_2)=2x_1^2+x_2^2$$ is such that $\dot W(x_1,x_2)$ has the sign of $$Z(x_1,x_2)=2-5x_1^2-3x_2^2$$ Thus $\dot W\leqslant0$ on the set $\{W\geqslant3\}\subset\{Z\leqslant0\}$ and $\dot W\geqslant0$ on the set $\left\{W\leqslant\frac23\right\}\subset\{Z\geqslant0\}$.
This shows that every solution starting from $(x_1,x_2)\ne(0,0)$ ends in the annulus $$\left\{\tfrac23\leqslant W\leqslant3\right\}$$ which implies the desired boundedness result. The behaviour we just described is confirmed by the rough figure below.