Consider the initial value problem
$$
\begin{align*}
\dot x&=x(y^2-x^2),\\\\
\dot y&=-y^2, \\\\
x(0)&=(x_0,y_0)
\end{align*}
$$
where $y_0>x_0>0$. Say $\phi(t)$ is the solution of this IVP (with maximal domain). Since $\dot y=-y^2$ yields a maximal solution $y(t)$ that exists on $(a,\infty)$ with $a>\infty$, one can argue that $\phi(t)$ has also lower bound $a$. I want to argue that
$$
\lim_{t\to b^{+}}x(t)=0.
$$
We know that $x(t)$ descreases monotonically as $t$ decreases. Therefore the limit $\lim_{t\to a^+}x(t)$ exists and has to be at least $0$:
$$
L=\lim_{t\to a^+}x(t)\geq 0.
$$
I want to arrive at a contradiction if $\lim_{t\to a^+}x(t)>0$.
My attempt:
Let's consider the system
$$
\begin{align*}
\dot x&=-x(y^2-x^2),\\\\
\dot y&=y^2, \\\\
x(0)&=(x_0,y_0)
\end{align*}
$$
whose solutions are the same as for the original solution, except the direction of the time has been reversed. We can compute the solution to $\dot y=y^2$ explicitly: $y(t)=\frac{y_0}{1-y_0t}$ for $t<y_0^{-1}$. Note that if the solution $(x(t),y(t))$ doesn't not exist on $[0,y_0^{-1})$, then $(x(t),y(t))$ would be contained in a compact set for positive time, and since it converges (by monotonicity), it would have to converge to a fixed point, which is not possible. Hence $(x(t),y(t))$ exists on $[0,y_0^{-1})$ and
$$
L=\lim_{t\to (v_0^{-1})^-}x(t)\geq 0
$$
since the solution can't cross the $y$-axis (which already contains solutions).
Assume for a contradiction that $L>0$. Then
$$
\lim_{t\to (v_0^{-1})^-}\dot x(t)=-\infty,
$$
since $\lim_{t\to (v_0^{-1})^-} y(t)=\infty$. Now I somehow need to arrive at a contradiction, knowing that $x(t)$ is bounded, monotonic and continuously differentiable on $(\delta,v_0^{-1})$ (for some $\delta$ close enough to $v_0^{-1}$).
Best Answer
Not all solutions behave as you describe, and this can be seen by solving the system explicitly.
Allow me to change $t$ to $-t$, so that the system becomes $$ \dot x = x(x^2-y^2) ,\qquad \dot y = y^2 ,\qquad (x(0),y(0))=(x_0,y_0) \in \mathbf{R}_+^2 . $$ Since we're looking at solutions in the positive quadrant, we get an equivalent system by setting $(u,v)=(x^2,y)$: $$ \dot u = 2u(u-v^2) ,\qquad \dot v = v^2 ,\qquad (u(0),v(0))=(u_0,v_0) \in \mathbf{R}_+^2 . $$ From $\dot v = v^2$ with $v(0)=v_0>0$ we get $$ v(t) = \frac{v_0}{1-v_0 t} ,\qquad \text{for} \quad t < \frac{1}{v_0} . $$ We can insert this into the equation for $u$: $$ \dot u + 2v(t)^2 u = 2u^2 ,\qquad u(0) = u_0 > 0 . $$ This is a Bernoulli ODE, which is linearized by the change of dependent variable $w(t) = 1/u(t)$ (where we don't have to worry about division by zero, since $u$ is positive): $$ \dot u + 2v(t)^2 u = 2u^2 \iff \frac{\dot u}{u^2} + \frac{2v(t)^2}{u} = 2 \iff \dot w - 2v(t)^2 w = -2 . $$ Since $v^2=\dot v$ we can write this as $$ \dot w + (-2 \dot v(t)) w = -2 ,\qquad w(0) = 1/u_0 > 0 , $$ and multiply by the integrating factor $\exp(-2 v(t))$, which gives $$ \frac{d}{dt} \Bigl( w(t) e^{-2v(t)} \Bigr) = -2 e^{-2v(t)} ,\qquad w(0) = 1/u_0 > 0 . $$ Integrating, we obtain $$ e^{-2v(t)} w(t) - e^{-2v(0)} w(0) = -2 \int_0^t e^{-2v(s)} ds , $$ or in other words $$ w(t) = e^{2v(t)} \biggl( \frac{e^{-2 v_0}}{u_0} - 2 \int_0^t e^{-2v(s)} ds \biggr) . $$ Going back to $u=1/w$, the solution of the system for $(u,v)$ is therefore $$ u(t) = \frac{u_0}{\displaystyle e^{2v(t)} \biggl( e^{-2v_0} - 2u_0 \int_0^t e^{-2v(s)} ds \biggr)} ,\qquad v(t) = \frac{v_0}{1-v_0 t} . $$ The interval of existence for this solution depends on the relation between $u_0 > 0$ and $v_0 > 0$. If $$ u_0 \le \frac{e^{-2v_0}}{\displaystyle 2 \int_0^{1/v_0} e^{-2v(s)} ds} = \frac{e^{-2v_0}}{\displaystyle 2 \biggl( \frac{1}{v_0 e^{2v_0}} + 2 \operatorname{Ei}(-2v_0) \biggr)} = \frac{v_0}{\displaystyle 2 \bigl( 1 + 2 v_0 e^{2v_0} \operatorname{Ei}(-2v_0) \bigr)} $$ (where the integral is evaluated in terms of the special function $\operatorname{Ei}(z) = -\int_{-z}^\infty \frac{e^{-t}}{t} dt$ by substituting $r=2v_0/(1-v_0s)$ and then integrating by parts), then the solution lives for as long as $v(t)$ does, i.e., for $t < 1/v_0$, since the bracket in the expression for $u(t)$ remains positive for those $t$. And if the strict inequality $u_0 < (\cdots)$ holds, then indeed $u(t) \to 0$ as $t \to (1/v_0)^-$, since the bracket tends to a positive value, while $e^{2v(t)} \to +\infty$.
But if the reverse inequality holds, $u_0 > (\cdots)$, then the denominator in the expression for $u(t)$ becomes zero at some earlier time, say $t_0$ with $0 < t_0 < 1/v_0$, so that the lifespan of the solution is just $t<t_0$, with $u(t) \to +\infty$ as $t \to t_0^-$.
So there is a separatrix $$ u = \frac{v}{\displaystyle 2 \bigl( 1 + 2 v e^{2v} \operatorname{Ei}(-2v) \bigr)} $$ in the phase portrait, and the fate of the solution is determined by on which side of this curve that the starting point $(u_0,v_0)$ lies.
Here's the phase portrait in the $(u,v)$-plane, with the $u$-nullcline $u=v^2$ in purple, and the separatrix in red:
Curves starting below the separatrix go off to the right ($u \to +\infty$, $v$ tends to a constant). And curves starting above it approach the $v$-axis ($u \to 0$, $v \to +\infty$).
(And everything is of course easily translated back to the original variables $(x,y)=(\sqrt{u},v)$, where the nullcline becomes the straight line $y=x$, and the separatrix becomes some curve below that line.)