Stability of equilibrium point origin for the equation $x'' + x + x^3(4 + \sin x) = 0$

Question

The question asks to discuss the stability of equilibrium point origin for the equation $x'' + x + x^3(4 + \sin x) = 0$

Attempt: Given equation $x'' + x + x^3(4 + \sin x) = 0$

now taking $x' = y$ and $y' = -x -x^3(4 + \sin x) $

Let $v = (x,y)$

$\dot v = f(v) = f(x,y) = (y , -y -x^3(4 + \sin x))$

Now (0,0) is a critical point of the system $\dot v = f(v)$ as $f(0,0) = (0,0)$

$\ $

What would be the required $C^1$ function '"V" (which is positive in a deleted neighbourhood of the point?) that is to be taken here so that I can use Lyapunov to get a set of conditions then to solve the problem?

Answer 1

Your equation is of the form $$x''+P'(x)=0$$ of a conservative mechanical system in one dimension with potential function $P(x)$. All solutions will have constant energy, that is, follow level curves $$\frac12x'^2+P(x)=E.$$ This, along with the linearization $x''+x=0$ close to the zero solution, completely determines the stability of that solution.