TITLE: Domain of a solution of differential equation
QUESTION [5 upvotes]: Given the following system of differential equations
\begin{equation}
\begin{cases}
\frac{dx}{dt}=-x+xy
\\\frac{dy}{dt}=-2y-x^2
\end{cases}
\end{equation}
I want to prove that the maximal solutions of the system are defined on all $\mathbb{R}$ and that
$$\lim_{t\rightarrow +\infty } {(x(t),y(t))}=0$$
As shown here I tried proceeding same way but I don't have any information about $t=0$.
$$x′x+y′y=-x^2-2y^2\leq 0$$
$$x′x+y′y= x'+y'\cdot x+y =\frac{1}{2}\frac{d}{dt}|| x+y ||^2$$
$$\frac{1}{2}\frac{d}{dt}|| x+y ||^2\leq 0$$
Hence I have information about the monotony, but I don't know how to show that, for example, the solution is limited on a subset of $\mathbb{R}$ (I could use this hypothesis to prolong the solution on all $\mathbb{R}$).

REPLY [2 votes]: Note that you already have
$$x′x+y′y=-x^2-2y^2$$
which implies
$$ \frac{1}{2}(x^2+y^2)'=-x^2-2y^2\le-x^2-y^2=-(x^2+y^2). $$
So
$$ (x^2+y^2)'+2(x^2+y^2)\le0$$
or
$$ e^{2t}(x^2+y^2)'+2e^{2t}(x^2+y^2)\le0$$
or
$$ [e^{2x}(x^2+y^2)]'\le0. $$
Integrating from $0$ to $t$ gives
$$ e^{2t}(x^2(t)+y^2(t))\le x^2(0)+y^2(0) $$
or
$$ x^2(t)+y^2(t)\le e^{-2t}[x^2(0)+y^2(0)]. $$
So
$$ \lim_{t\to\infty}(x^2(t)+y^2(t))=0. $$