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