The full problem reads: Suppose that the motion of a mass $m$ is described by the nonlinear differential equation $m \frac{d^2x}{dt^2} = -\alpha x^3 - \sigma \left( \frac{dx}{dt} \right)^3$ where $\alpha >0$ and $\sigma \geq 0$.
(a) Show that if $\sigma = 0$ then every nonzero solution $x(t)$ of the system is a periodic solution.
(b) Show that if $\sigma > 0$ then as $t \to \infty$, $x(t) \to 0$.
I have attempted part (a) by setting $\dot x = y$ and thus $\dot y = \ddot x = \frac{-\alpha}{m} x^3$. But then the problem I have run into is that there are no equilibrium points where $x(t) > 0$, the only one that exists for this situation is $(x,y) = (0,0)$.
So my question is: how do I go about showing that every nonzero solution here is periodic when I can't use something like Poincare-Bendixson. Any tips on part (b) are also welcome. Thanks in advace!
Considering the dynamic system
$$ \left\{ \begin{array}{rcl} \dot x & = & y\\ \dot y & = & -\alpha x^3 \end{array} \right. \Rightarrow \left\{ \begin{array}{rcl} \alpha x^3\dot x & = & \alpha x^3 y\\ y\dot y & = & -\alpha x^3 y \end{array} \right. \Rightarrow \left(\frac 14\alpha x^4+\frac 12 y^2\right)' = 0 $$
hence the orbits are described by the curves
$$ \frac 14\alpha x^4+\frac 12 y^2 = C_0 $$
characterizing a center. Attached a stream plot for $\alpha = 1, \sigma = 0\;$. The same procedure can be used to show that when $\sigma > 0\;\;$ those orbits collapse to zero with time.
NOTE
$$ \left(\frac 14\alpha x^4+\frac 12 y^2\right)' = -\sigma y^4 < 0,\;\;\forall y \ne 0 $$
Considering now $\sigma=\alpha=1\;$ we have the stream plot
MATHEMATICA scripts attached
alpha = 1; sigma = 0; StreamPlot[{y, -alpha x^3 - sigma y^3}, {x, -1, 1}, {y, -1, 1}] alpha = 1; sigma = 1; StreamPlot[{y, -alpha x^3 - sigma y^3}, {x, -1, 1}, {y, -1, 1}]