3
$\begingroup$

I am interested in the differential equation

$$ (\mathcal{S})\left\{ \begin{array}{l} y'(t) = \displaystyle \frac{y^{2}(t)}{1+t^{2}+y^{2}(t)} \\[2mm] y(0) = \displaystyle \frac{3}{4} \end{array} \right. $$

Let $y$ be a maximal solution of $(\mathcal{S})$. I already proved that $y$ is defined on $\mathbb{R}$. However, I do not see how to prove that $\displaystyle \lim \limits_{t \to +\infty} y(t)$ is lower than $14$ and $\displaystyle \lim \limits_{t \to -\infty} y(t)$ is greater than $\frac{1}{3}$. A hint would be appreciated !

$\endgroup$
  • $\begingroup$ Where do the constants $14$ and $1/3$ come from? The text of the exercise? $\endgroup$ – TZakrevskiy Mar 13 '15 at 22:48
  • $\begingroup$ @TZakrevskiy : Yes, from the text of the exercise! $\endgroup$ – Odile Mar 13 '15 at 23:14
  • $\begingroup$ $\frac13$ is odd. $\endgroup$ – Did Mar 14 '15 at 15:45
3
$\begingroup$

From the equation $0<y'\le1$, so that $y$ is increasing and $$ \frac34\le y(t)\le\frac34+t,\quad t\ge0. $$ Substituting into the equation we get $$ y'\le\frac{y^2}{1+t^2+(3/4)^2}=\frac{y^2}{t^2+(5/4)^2}. $$ Integrate to obtain $$ -\frac{1}{y}+\frac43\le\frac45\arctan\frac{4\,t}{5},\quad t\ge0, $$ and from here $$ y(t)\le\frac{1}{4/3-(4/5)\arctan t}<\frac{1}{4/3-(4/5)(\pi/2)}=13.038\dots $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.