Behaviour of the solutions of a system of differential equations at infinity. Consider the following system of differential equations:
$$
\begin{cases}
u'=-u+uv\\
v'=-2v-u^2
\end{cases}
$$
I'm able to prove that solutions must tend to $0$ if $t\to 0$ by the use of Lyapunov function $L(x,y)=x^2+y^2$ but I'm unable to prove that the function:
$$
I(t)=\frac{x^2(t)+2y^2(t)}{x^2(t)+y^2(t)}
$$
must admit limit for $t\to+\infty$
Any hint for a solution?
Thanks in advance.
 A: If you compute the time derivative of $I$, you obtain
$$I’(t)=-\frac{2x^2y^2}{(x^2+y^2)^2}+\frac{-2x^2y^3-4x^4y}{(x^2+y^2)^2}.$$
The idea is that the first part is negative, while the second part goes to zero exponentially.
More precisely, integrating the above equality in the time variable from $0$ to $t$, you have
$$I(t)=I(0)+f(t)+g(t),$$
where
$$f(t)=\int_0^t -\frac{2x^2 (s) y^2 (s)}{(x^2 (s) +y^2 (s))^2} ds$$
is a non-increasing function, and
$$g(t)= \int_0^t -\frac{-2x^2 (s) y^3 (s)-4x^4 (s) y (s)}{(x^2 (s) +y^2 (s))^2} ds$$
is such that $g(t)$ converges to some finite value for $x\to+\infty$ thanks to the exponential decay of the integrand (see below).
This means that both functions admit limits, hence $I$ admits a limit. In principle, the limit of $f$ might be infinite, but then you can use the fact that $1\leq I(t)\leq 2$ to conclude that the limit is finite.

Concerning the exponential decay of the integrand of $g$, this essentially follows from the calculations you did for the Lyapunov functions. In particular, you have
$$\frac{1}{2}\frac{d}{dt}(x^2(t)+y^2(t))=-x^2(t)-2y^2(t)\leq -x^2(t)-y^2(t).$$
By Gronwall’s lemma, you have
$$(x^2(t)+y^2(t))\leq (x^2(0)+y^2(0))e^{-2t}.$$
Now you can use the algebraic inequalities
$$\left|\frac{-2x^2y^3}{(x^2+y^2)^2}\right|\leq 2\sqrt{x^2+y^2},$$
$$\left|\frac{-4x^4y}{(x^2+y^2)^2}\right|\leq 4\sqrt{x^2+y^2}$$
to conclude that the integrand of $g$ goes to zero at lest as $e^{-t}$.

There is another, more standard proof which uses a bootstrap argument to show that $x(t)\sim e^{-t}$ if $x(0)\neq 0$ and $|y(t)|\leq C(1+t)e^{-2t}$. If you prove this, it’s easy to show that if $x(0)=0$ you have $I\equiv 2$, while if $x(0)\neq 0$ you have $I(t)\to 1$.
The first step for the proof would be to make a change of variables $X=e^{t}x$, $Y=e^{2t}y$, so that the system becomes
$$\left\{\begin{aligned}&X’=e^{-2t}XY\\&Y’=-X^2\end{aligned}\right.. $$
A: The Lyapunov function allows one to conclude that the origin is asymptotically stable. Therefore, eventually one can approximate the dynamics with the dynamics linearized at the origin, which gives
\begin{cases}
u'=-u \\
v'=-2v
\end{cases}
from which one can see that $u(t)$ and $v(t)$ can eventually be approximated with $e^{-t} C_1$ and $e^{-2t} C_2$ respectively. One could substitute these expressions in the expression for $I(t)$ and find it's limit as $t\to\infty$ or use that $e^{-2t}$ goes faster to zero than $e^{-t}$ directly.

It can also be noted that the sign analysis approach of the time derivative of $I(t)$ might give inconclusive results. For example consider the following system
\begin{cases}
x' = -2\,x + x\,z - y\,z \\
y' = -y + x\,z - z^2 \\
z' = -3\,z - x^2 + y\,z
\end{cases}
for which it can also be shown to be asymptotically stable using the Lyapunov function $V(x,y,z) = x^2 + y^2 + z^2$.
The limiting behavior as $t\to\infty$ of
$$
I(x,y,z) = \frac{x^2 + 2\,y^2 + 3\,z^2}{x^2 + y^2 + z^2}
$$
will probably be inconclusive when looking at the time derivative of $I(x,y,z)$. But using the linearization one can say that it will go to $2$ for most initial conditions.
