Second order evolution ODE I'm studying an article about the decay of solutions to some evolution equations of second order and i'm trying to understand the simple ODE case :
$y'' +  y' + f(y) = 0 $ 
with $f(y)=y^2$ or $y^3$ or more generally $f(y)=c|y|^{a-1}y$, $c \in \mathbb{R}, a>1$ 
I would love to understand just the case $y'' +  y' + y^2 = 0$. 
Any help please ? I thought about using Cauchy-Lipschitz for the existence and uniqueness of the solution. 
 A: Ok, i've been working on it and here's what i found, i'm not quite sure about it so i would love to know if this works.
Let $0<\beta <1$ and consider $X_{\beta} = \lbrace f \in \mathcal{C}^{0}(\mathbb{R}^+), \sup_{t \geq 0}|f(t)| \leq \beta \rbrace$. $(X_{\beta}, || . ||_{\infty})$ is a complete metric space, for the distance induced from the norm. 
Let $z \in X_\beta$ such that $|z(0)|=|z_0|<\frac{\beta}{2}$, and define $T : X_{\beta} \longrightarrow \mathcal{C}^0_b(\mathbb{R}^+)$ such that for all $t>0$ 
$Tz(t):= u_0 + \int_0^{t} e^r \int_{r}^{+\infty} e^{-2s} z(s)^2 ds dr$


*

*$T$ is well defined and $T(X_\beta) \subset X_{\beta}$:


$ |Tz(t)| \leq |z_0| + \int_0^{t} e^r \int_{r}^{+\infty} e^{-2s} |z(s)|^2 ds dr \leq \frac{\beta}{2} + \frac{\beta^2}{2} \int_{0}^{t} e^{-r} dr \leq \beta $
hence $T(X_\beta) \subset X_{\beta}$


*

*$T$ is $\beta$-Lipchitz continuous in $X_{\beta}$: for $z_1,z_2 \in  X_{\beta}$


$ |Tz_1(t) - Tz_2(t)| \leq \int_0^t e^r \int_{r}^{+\infty} e^{-2s} |z_{1}(s)^{2} - z_{2}(s)^2| ds dr \leq \beta||z_{1}-z_{2}||_{\infty} \int_0^t e^{-r} dr $
$\qquad \qquad \leq \beta||z_{1}-z_{2}||_{\infty} $
Hence, $ ||Tz_1 - Tz_2||_{\infty} \leq \beta||z_{1}-z_{2}||_{\infty} $
We apply Banach fixed point theorem, $\exists! z \in X_\beta$ such that  $Tz=z$, by posing for all $t>0$ $u(t):=z(t)e^{-t}$ we obtain an $u \in C^2(\mathbb{R}^+)$ such that 
$\sup_{t>0} \lbrace e^t|u(t)| \rbrace \leq \beta $, $|u(0)|=|z_0|< \frac{\beta}{2}$ and 
$e^t u(t) - e^t u(0) = \int_0^t e^r \int_{r}^{+\infty} u(s)^2 ds \Leftrightarrow u'(t) + u(t) = \int_{t}^{+\infty} u(s)^2 ds$
Now we show that $u$ is solution to the second order ODE. In a sense, if we differentiate the relation above in $t$ we get 
$u''(t) + u'(t) = \lim_{A \rightarrow +\infty}u(A)^2 - u(t)^2$, since $|u(t)|\leq \beta e^{-t} \longrightarrow 0$ when $t\rightarrow+\infty$ we obtain finally that $u$ satisfies $u''(t) + u'(t) + u(t)^2 = 0 $
In the other sense, if we have a solution $u$ of the second order equation such that $u(0)=u_0$ and $\sup_{t>0}\lbrace e^t|u(t)|\rbrace<\beta$
$u''(t) + u'(t) = \frac{d}{dt}(u'+u) =  - u(t)^2$, by integrating from $t$ to $A$ for all $A>t$ we get
$u'(t) + u(t) = u'(A) + u(A) + \int_{t}^{A} u(s)^2 ds$, by letting $A\rightarrow+\infty$ :
$u'(t) + u(t) = \lim_{A \rightarrow +\infty}u'(A) - \int_{t}^{+\infty}u(s)^2 ds$ 
We can easily show that $\lim_{A \rightarrow +\infty}u'(A)$ when we have that $\lim_{A \rightarrow +\infty}u(A) = 0$. I thought about what Ian said in the comment. By multiplying by $u'(t)$ in both sides in the second order equation, we have 
$u'(t) u''(t) + u'(t) u(t)^2 = \frac{d}{dt}(\frac{3u'(t)^2 + 2u(t)^3}{6}) = -2(u'(t)^2 + \frac{1}{3}u(t)^3) + \frac{2}{3}u(t)^3$
By posing $x(t) = \frac{3u'(t)^2 + 2u(t)^3}{6}$, we obtain the first order differential equation $x' = -2x + \frac{2}{3}u(t)^3$ 
which has for a solution the function 
$x(t) = (x_0-\frac{u_{0}^4}{6} + \frac{1}{6} u(t)^{4})e^{-t} \underset{t\to+\infty}{\longrightarrow} 0$ hence $u'(t) \underset{t\to+\infty}{\longrightarrow} 0$ and we have the wanted result.
We showed that any solution of $u'' + u' + u^2 = 0$ satisfying the initial condition properties satisfies $u'(t) + u(t) = \int_{t}^{+\infty} u(s)^2 ds$
Since there's just one, we have a unique solution of $u'' + u' + u^2 = 0$ that decreases exponentially fast. 
