Convergence of the solution and their derivative of the equation $y''+y'+y^{3}=0$. I have the ODE $y''+y'+y^{3}=0$ and I must prove that the solution $y(t)$ and $y'(t)$ converges to zero when $t\to \infty$.
I try to write the associated system of two equations, this is one form
\begin{equation*}
y'=z;\qquad z'=-z-y^3
\end{equation*}
and this is the other
\begin{equation*}
y'=z-y;\qquad z'=-y^3.
\end{equation*}
I try to use Lyapounov method with the function $V(y,z)=z^2+\frac{1}{2} y^4$ and obtain that $\nabla V\cdot (y',z') <0$. But I don't know how to conclude this proof.
On the other hand, using the associated matrix for the linear system, the real part of the eigenvalues is not negative, in fact is zero and I don't know how to continue.
I accept any suggestion, hint or book to read.
 A: We have $y''+y'+y^3=0$. We can choose the state variables as $y,z$ and the dynamic equation of system becomes $z=y'$ and $z'=y''=-y'-y^3=-z-y^3$. Thus
$$\left\{
\begin{array}{l}
\frac{dy}{dt}=z\\
\frac{dz}{dt}=-z-y^3
\end{array}\right.$$
We must prove that for some energy function $V(z,y) >0$ the rate of change of energy in system is always negative $\frac{\partial V}{\partial t} <0$. A good choice for energy function is usually $V(z,y)=ay^{2n}+bz^{2m}$ where $a,b\in \mathbb{R}_+$ and $ m,n \in \mathbb{Z}_+$ Therefore
$$\frac{\partial V}{\partial t} = \frac{\partial V}{\partial z}\frac{d z}{d t} + \frac{\partial V}{\partial y}\frac{d y}{d t} = 2mbz^{2m-1}(-z-y^3) + 2nay^{2n-1}(z)<0 \Rightarrow -mbz^{2m} +zy^3(-mbz^{2m-2}+nay^{2n-4})$$
The first term $-mbz^{2m}$ is always negative. The second term however can be positive and negative due to $zy^3$. Therefore we must choose $m,n,a,b$ to make it zero. To do this we can choose $m=1$ and $n=2$ to get real numbers and then $-b+2a=0$ which gives $b=2a$. One possible solution is $a=\frac{1}{2}$ and $b=1$. Therefore the Energy function of the system is $V(z,y)=\frac{1}{2}y^4+z^2$.
