Riccati type equation Recently I've encountered the following equation in a book (don't remember the source, I just took some notes down on my scrap paper) 
$$\dot x=t-x^2,\quad x(1)=1.$$ Then I've recognized that this is a Riccati type equation, but not explicitly solvable. The exercise proposed there was however to make a qualitative study of the equation.
In particular (and I am not able to solve any of the points, my bad):


*

*show that the solution to the Cauchy Problem is defined on $[1,+\infty)$,

*show that $\lim_{t\to+\infty}x(t)=+\infty,$

*show that $\lim_{t\to+\infty}x(t)-t^{1/2}=0.$
Can anybody help me please? It is not homework. Thank you in advance
Guido
 A: First point. 


*

*Set $f(t)=-\sqrt t$. Then $f(t)$ is strictly decreasing and convex on $[1,+\infty)$. Moreover, it goes to $-\infty$ in an infinite time.

*$\forall t=\bar t$ such that $x(t)=-\sqrt t$, $x(t)$ has a local minimum. Indeed $x'(\bar t)=0$ and $x''(\bar t)=1-2x(\bar t)(t-x^2(\bar t))=1>0$.

*The solution $t\mapsto x(t)$ cannot cross the graph of $f(t)$. For, if this were the case, setting $g(t)=x(t)-f(t)$, we would have $g(1)=2>0$, and therefore by continuity, around a small neighborhood of $t=\bar t$ we would get $g(t)<0\;\forall t\in (\bar t, \bar t+\varepsilon)$. This would in turn imply $x(t)<-\sqrt t<-\sqrt{\bar t}=x(\bar t)\; \forall t\in (\bar t,\bar t+\varepsilon)$, contradicting the hypothesis of $t=\bar t$ being a local minimum for $x(t)$. In particular, also $t\mapsto x(t)$ goes to $-\infty$ in an infinite time. 

*Suppose now that the solution $t\mapsto x(t)$ is bounded in time. Then it cannot be bounded also in space. Supposing $t\mapsto x(t)$ to be defined on $[1,b),\; b<+\infty$, we would have $$\lim_{t\to b^-}|x(t)|=+\infty.$$ If it were that $\lim_{t\to b^-}x(t)=+\infty$, there would exist $a<b$ such that $x(t)>\sqrt{b+1}\;\forall t\in(a,b)$.
But then $x'(t)<b-(b+1)=-1\;\forall t\in(a,b)$, and therefore, integrating over $(a,b)$, we would obtain $$\lim_{x\to b^-}x(t)<x(a)<+\infty,$$ which is absurd. 
It then follows $$\lim_{t\to b^-}x(t)=-\infty.$$

*We have now the contradiction: if the solution were bounded in time, then it must go to $-\infty$ in a finite time, which is not possible by what we have said; the solution $t\mapsto x(t)$ is therefore defined on $[1,+\infty)$. 
Second point.


*

*Suppose the solution to be bounded as $t\to+\infty$. In particular $x^2$ is bounded say by $A>0$ as $t\to\infty.$ Then $$x'(t)=t-x^2(t)>t-A>1,\;\forall t\in[A+1,+\infty).$$ Integrating this last relation on $[A+1,+\infty)$, one gets $$\lim_{t\to+\infty}x(t)>(\lim_{t\to+\infty} t)-(A+1)+x(A+1)=+\infty,$$ which is absurd.

*Set now $h(t)=\sqrt t$, and suppose that the solution crosses the graph of $h(t)$ at some point $t=\bar t$. Then $\exists\, p:=\min Q$ where $$Q:=\{t\in[1,+\infty)\,\colon\, x(t)\text{ crosses the graph of } h(t)\}.$$

*$p>1$. Indeed $x'(1)=0$ and $x''(1)=1$. Then $x'$ is strictly greater than $0$ in a right neighborhood of $1$, i.e. $\sqrt t\geq x(t)\,\forall t\in(1,1+\varepsilon)$. Notice that, in taking the square roots, we have used the fact that $x(1)=1$ hence, possibly shrinking the neighborhood, we may assume $x(t)>0$ in $(1,1+\varepsilon)$.

*Again $p$ would be of local minimum for $x(t)$; however, there would exist a neighborhood of the form $(p-\varepsilon, p)$ in which, by continuity, $$x(t)<\sqrt t<\sqrt p=x(p),\,\forall t\in(p-\varepsilon, p)$$ which is absurd.

*It follows that $$\lim_{t\to+\infty} x(t)=+\infty,$$ combining the fact that $x$ is unbounded and $x'(t)\geq 0$ on $[1,+\infty)$.
Third point


*

*Observe that $\lim_{t\to+\infty}h'(t)=0$. Suppose that there existed $\varepsilon>0$ such that $$\lim_{t\to+\infty}\left(t^{1/2}-x(t)\right)>\varepsilon.$$ Then $$\lim_{t\to+\infty} x'(t)>\varepsilon\lim_{t\to+\infty}\left(t^{1/2}+x(t)\right)=+\infty.$$

*We may choose $T\in [1,+\infty)$ so big that $x'(t)>1$ and $h'(t)<1/2$ for every $t\in(T,+\infty)$. Now, considering $$q(t):=h(t)-x(t),$$ by our assumptions it would follow that $q'(t)<1/2$ for every $t\in(T,+\infty)$, then, by integration over $(T,+\infty)$ we would have $$\lim_{t\to+\infty} q(t)<-\frac 12\left(\lim_{t\to+\infty}t-T\right)+q(T)=-\infty.$$ Combining this with the fact that $q(t)>0$ in a right neighborhood of $1$, then it follows that there is a point in which the graph of $x(t)$ crosses the graph of $t\mapsto \sqrt t$, which is impossible as shown in the second point.

*Therefore, for any $\varepsilon>0$, we have that $$\varepsilon>\lim_{t\to+\infty} t^{1/2}-x(t)\geq 0.$$ This is sufficient to complete the proof due to the arbitrarity of $\varepsilon>0$.
A: I'd just like to tell you my approach to the first point. Let $u(t)=1/t$ and $v(t)=t^2$. Those are the solutions of
$$ \begin{cases} u'(t)= -u(t)^2 \\ u(1)=1
\end{cases} $$
and
$$ \begin{cases} v'(t)= t \\ v(1)=1
\end{cases}.$$
Hence, we have for $t\geq 1$
$$ 1/t=u(t) \leq f(t) \leq v(t)= t^2,$$
which tells you that $f$ exists for all positive time.
