Estimating convergence with Liapunov function

we have the system: $x' = 2y(z-1) - x^3, y' = -x(z-1) -y^3, z' = - z^3$ and consider its equilibrium at $\vec{o}$. We want to prove that for every non-trivial solution there exists two positive constants such that for $t$ big enough, we have:

$$\frac{c_1}{\sqrt{t}} < \big|\vec{y}(t)\big| < \frac{c_2}{\sqrt{t}}$$

I found the left inequality using the particular Liapunov function $V = x^2+2y^2+z^2$ and by noticing that $-\dot{V} < 2V^2$, which when integrated and using $V < 2|\vec{y}|^2$ yields $\frac{c_1}{\sqrt{t}} < |\vec{y}(t)|$, but I am having some problems finding the other inequality. I tried using the fact that the solution to the 3rd equation is $z \propto \frac{1}{\sqrt{t}}$ for large $t$ but that still does not exclude weaker convergence for $x$ and $y$.

Edit: The steps to the left inequality are $\frac{d}{dt}\left(\frac{1}{2V}\right) <1 \Rightarrow \frac{1}{V}-\frac{1}{V_0} < 2t \Rightarrow 2|\vec y|^2 > V > \frac{V_0}{2V_0t+1}$ and then for a big enough $t$ the conclusion follows.

• There are typos in your question: you probably mean that $\dot V\gt-2V^2$ and that $V\gt2y^2$, then I do not see how to deduce directly the bound on $y(t)$. Likewise, $z(t)$ is not what you say but $z(t)=z_0(1+2tz_0^2)^{-1/2}$. Please correct these.
– Did
Dec 8, 2014 at 10:20
• @Did You're right, I'm sorry. I have corrected them. Dec 8, 2014 at 11:25
• Re the "Edit", the inequality $2y^2\gt V$ should be reversed hence the lower bound on $V$ does not allow to conclude directly (as mentioned in my previous comment).
– Did
Dec 8, 2014 at 12:07
• @Did $V = x^2 + 2y^2 + z^2 = 2|\vec{y}|^2 - x^2 - z^2 < 2|\vec{y}|$. I just realised that my notations are maybe a bit confusing $\vec{y}$ is the solution to the system, whereas $y$ in $V$ just represents the second coordinate. Dec 8, 2014 at 13:05
• "I just realised that my notations are maybe a bit confusing" Indeed!
– Did
Dec 8, 2014 at 13:31

Let $V(t)=v(x(t),y(t),z(t))$ with $v(x,y,z)=x^2+2y^2+z^2$ and $W(t)=w(x(t),y(t),z(t))$ with $w(x,y,z)=x^4+2y^4+z^4$ then $$V'(t)=-2W(t),$$ for every $t$, and$^{*}$ $$v^2(x,y,z)\leqslant4w(x,y,z)\leqslant4v^2(x,y,z),$$ for every $(x,y,z)$, hence, for every $t$, $$-4V^2(t)\leqslant2V'(t)\leqslant-V^2(t).$$ Integrating this on $[0,t]$ yields $$\frac{V_0}{1+2V_0t}\leqslant V(t)\leqslant\frac{2V_0}{2+V_0t}.$$ Now, $x^2+y^2+z^2\leqslant v(x,y,z)\leqslant2(x^2+y^2+z^2)$ for every $(x,y,z)$ hence $$\frac{V_0}{2(1+2V_0t)}\leqslant x^2(t)+y^2(t)+z^2(t)\leqslant\frac{2V_0}{2+V_0t},$$ in particular, when $t\to\infty$, $$x^2(t)+y^2(t)+z^2(t)=\Theta\left(\frac1t\right).$$ $^*$ To check the inequality $4w-v^2\geqslant0$, one considers the matrix $\begin{pmatrix}3&-2&-1\\-2&4&-2\\-1&-2&3\end{pmatrix}$ of the quadratic form $(x^2,y^2,z^2)\mapsto4w(x,y,z)-v^2(x,y,z)$ and one checks that its eigenvalues (namely, $6$, $4$ and $0$) are nonnegative.