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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

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"...not explicitly solvable..." - it can actually be solved in terms of Bessel functions. –  J. M. Aug 6 '12 at 10:57
    
Let W|A help you... –  draks ... Aug 6 '12 at 11:41
    
As is, tags and text contradict each other: The text claims it is not homework, the tags claim it is. –  celtschk Aug 6 '12 at 13:29
1  
@celtschk: if I were a robot, I would now self-destruct because of the contradiction in the input... –  Fabian Aug 6 '12 at 14:47
    
It should be better to find methods to solving this question without requiring to directly solving the ODE. –  doraemonpaul Aug 7 '12 at 7:56

1 Answer 1

up vote 2 down vote accepted

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 $\varepsilon>0$.

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