# existence and uniqueness solution Cauchy problem $y'=y(y+1)e^{-y}$

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 $R$?

• 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 $t_\beta=e^{-\beta}$. – Did Jun 22 '16 at 18:54

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 $t$).
• for $\alpha<-1$ I want calculate $lim_{t \rightarrow t_{\alpha}}y(t)$. It's $-\infty$? – Giulia B. Jun 23 '16 at 9:52