1
$\begingroup$

I have this Cauchy problem ($\alpha \in \mathbb{R}$)

\begin{cases} y'(t) = y(y+1)e^{-y} \\ y(0) = \alpha \end{cases}

$f(t,y)=y(y+1)e^{-y} \in C^1(\mathbb{R}^2)$ so for Cauchy-Lipschitz theorem exists and is unique the solution of locally $\forall \alpha$.

I know $y=0$ and $y=-1$ are stationary solutions.

How can I prove the other solutions are defined on all $\mathbb{R}$? Can I use existence and global theorem?

If I consider $\alpha>0$, $y_\alpha>0$ and $\partial_yf=e^{-y}(-y^2+y+1)$.

|$\partial_yf|=e^{-y}|-y^2+y+1|\rightarrow 0$ for $y\rightarrow+\infty$.

So $\partial_yf$ is limited and I have global existence and uniqueness on all $\mathbb{R}$ for $\alpha>0$?

Also for $-1<\alpha<0$ $\partial_yf$ is limited but for $\alpha<-1$ the solution exists for all $R$?

$\endgroup$
  • $\begingroup$ If $\alpha<-1$, the maximal solution $y$ is defined on the interval $(t_\alpha,+\infty)$ for some finite negative $t_\alpha$. The situation is similar to the case $z'=e^z$ with $z(0)=\beta$ for some $\beta>0$, which can be solved explicitely as $e^{-\beta}-e^{-z(t)}=t$, hence the maximal solution is defined on $(-\infty,t_\beta)$ with $t_\beta=e^{-\beta}$. $\endgroup$ – Did Jun 22 '16 at 18:54
0
$\begingroup$

For $t < 0$, there is no problem. The reason is that $y$ in increasing for $y(0) > 0$ and decreasing for $y(0) < -1$. Hence we only consider the case $t > 0$.

For $y(0) > 0$, $y(t) > 0$. Since $f(t,y)$ is Lipschitz in $y$ for $y\geq 0$, you have $$ y'(t) \leq Cy(t) $$ for some $C$. This implies that $y(t)\leq y(0)e^{Ct}$.

For $y(0) < 1$, $y(t) < -1$. Put $u = -y-1$. Then $$ u'(t) = u(u+1) e^{u+1} \geq u^2e^u \geq u^4. $$ By integration, $$ \frac{1}{u(0)^3} - \frac{1}{u(t)^3} \geq 3t. $$ Which means the solution can't be global (otherwise $u(t) < 0$ for large $t$).

$\endgroup$
  • $\begingroup$ for $\alpha<-1$ I want calculate $lim_{t \rightarrow t_{\alpha}}y(t)$. It's $-\infty$? $\endgroup$ – Giulia B. Jun 23 '16 at 9:52
  • $\begingroup$ @GiuliaB. A solution either blows up at finite time or is global. $\endgroup$ – Qiyu Wen Jun 23 '16 at 14:58

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.