Consider the equation, $$ x'' + ax^2x' - x + x^3 = 0,\ x ∈ \Bbb R $$ where $0 ≤ a$ is a constant.
It asked me to find all equilibria and determine their stability/instability.
When $a = 0$, I found the conserved quantity $$E(x,y) = 2y^2 - 2x^2 + x^4$$ with $y$ being equal to $y = x'$. However when $0 ≤ a$, I don't know how to find equilibria and their stability.
Also, it asked me to prove there is no periodic orbits or homoclinic orbits and instead prove there exists a heteroclinic orbits. How to prove them?
Thanks in advance for help.
With $v=x'+\frac a3 x^3$ you get the first order system $$ x' = v-\frac a3 x^3\\ v' = x-x^3 $$ This gives you the stationary points at $x_k=-1,0,1$ and $v_k=\frac13x_k^3$. For the stability the first thing to try is to compute the eigenvalues of the Jacobian, if that is not sufficient other methods need to be employed. $$ J=\pmatrix{-ax_k^2&1\\1-3x_k^2&0} $$ with characteristic equation $$0=λ(λ+ax_k^2)-(1-3x_k^2)=(λ+\tfrac12ax_k^2)^2-(1-3x_k^2+\tfrac14a^2x_k^4)$$ which allows to compute the eigenvalues and examine their real parts.
As observed, $E(x,y)=2y^2−2x^2+x^4$ is a conserved quantity of the equation without the friction term. Computing the time derivative of $E(x,x')$ for the equation with friction results in $$ \frac{d}{dt}E(x,x')=4x'(x''-x+x^3)=-4ax^2x'^2 $$ This means that any solution curve constantly intersects the level curves of $E$ in direction of its minimal values. As $E(x,y)=2y^2+(x^2-1)^2-1$, these minima can be found at $x'=0$, $x=\pm 1$.
A consequence is that there can be no homoclinic or periodic orbits, as no solution curve can return to the starting level curve of $E$. Second all solution curves outgoing from $(0,0)$ remain inside the region $E<0$, which is contained in $[-\sqrt2,\sqrt2]\times [-\sqrt{\frac12},\sqrt{\frac12}]$, so can only go to one of the stationary points at the minima of $E$ Which means they are heteroclinic orbits..