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Consider the ODE $$y''+y'+y^3=0$$ I need to prove that $$\lim_{x\rightarrow \infty} y(x) = 0$$

and $$\lim_{x\rightarrow \infty} y'(x) = 0.$$

Well, introducing the change of variables such as $x_1=y,x_2=y'$ I get the system of equations nonlinear in $x_1, x_2$. My question is, if I linearize this system around $(0,0)$ and analyze the behaviour of the linearized system there, would I be correct to infer that the behaviour is the same for a nonlinear (original) system? Say, for a solution to the linearized system the limits above hold true. Would they hold true for the original system as well then?

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    @imranfat okay, haven't tried actually solving it, not sure now :) – 2017-02-14
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    It can't be solved analytically as far as I know. One way to show it is to define the "energy" $E(x) = \frac{1}{2}y'^2 + \frac{1}{4}y^4$. By multiplying the ODE by $y'$ and integrating we get $E(x) = E(0) - \int_0^x y'^2{\rm d}x$ which decays in time (but is bounded below by $0$). The limit $\lim_{x\to\infty} E(x)$ therefore exists from which the result follows. – 2017-02-14
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    @Winther thank you, I will look into that – 2017-02-14

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Here is one possible way to derive the result: the related ODE $y'' + y^3 = 0$ can be though of as describing the path of a particle that moves in a potential $V(y) = \frac{1}{4}y^4$. This ODE is the equation of motion of the Lagrangian $L = \frac{1}{2}y'^2 - \frac{1}{4}y^4$ for which the corresponding Hamiltonian (the energy of the system) is $H = \frac{1}{2}y'^2 + \frac{1}{4}y^4$. The energy is conserved under the evolution: $\frac{dH}{dx} = 0$. The addition of $y'$ in the ODE acts as a friction term which removes energy from the particle so we would expect the particle to eventually end up at the bottom of the potential with zero energy. This is the physical reasoning behind the result. To prove it we can multiply the ODE $y'' + y' + y^3 = 0$ by $y'$ and integrate to get an equation for the evolution of $H$:

$$y''y' + y'^2 + y^3y' = 0 \implies \frac{dH}{dx} = -y'^2 \implies H(x) = H(0) - \int_0^xy'(t)^2{\rm d}t$$

This shows that $H$ is decreasing with $x$ and since it's bounded below by $0$ it must converge as $x\to \infty$ from which $\lim_{x\to\infty}y'(x) = 0$ follows. There is only one possible fixpoint $(y,y') = (0,0)$ of the corresponding dynamical system $(y,y')' = (y',-y'-y^3)$ for it to converge to so we must have $\lim_{x\to\infty}y(x) = 0$.

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Thank you, for comments and ideas,I will post a sketch of the solution here.

One of the ways to solve this is to represent this a system of two first-order ODE's and then use Lyapunov function in the for that @Winther has proposed. Then the stability of a rest point (0,0) will follow by Lyapunov theorem directly.