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.

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 $$\left\{\begin{aligned}&X’=e^{-2t}XY\\&Y’=-X^2\end{aligned}\right.. $$