Given $$\ddot{x}+\lambda\dot{x}=x-x^3,$$ which I rewrote to \begin{align} \dot{x}&=y \\ \dot{y} &= -\lambda y + x - x^3, \end{align} For the exercise we take $\lambda\geq 0$. I have determined that the equilibrium points are given by $(x,y)$ equal to $(0,0)$, $(-1, 0)$ and $(1,0)$. I now need to determine what their stability type is. From phase planes I saw that $(0,0)$ was a saddle point, thus unstable, and $(-1,0)$ and $(1,0)$ are spiral sinks, thus stable, but how do I prove this? I tried proving they were Lyaponov stable, because I think that determines stability or instability, but I don't know how to do that with a system of differential equations:
What I tried for the point $e=(-1,0)$: Let $a$ be a point close to $e$, so $||a-e||<\delta$, with $\delta>0$. I then want to find a solution, which I'll call $z(t)$, corresponding to this point $a$ and show that for all $t$ there exists an $\epsilon>0$ such that $||z(t)-e||<\epsilon$. I don't know how to find such a solution or how to proceed.
Let's utilize the Jacobian matrix $A$ at each equilibrium point and check its eigenvalues: $$ A= \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1-3x^2 & -\gamma \end{bmatrix} $$ For the origin, we get $$ A= \begin{bmatrix} 0 & 1 \\ 1-3x^2 & -\gamma \end{bmatrix}_{x=y=0} = \begin{bmatrix} 0 & 1 \\ 1 & -\gamma \end{bmatrix} $$ The eigenvalues of $A$ are $$ \lambda_{1,2} = \frac{-\gamma\pm \sqrt{\gamma^2+4}}{2} $$ They are real and opposite signs, therefore, the origin is saddle (i.e unstable).
For the equilibrium point (-1,0), we have
$$ A= \begin{bmatrix} 0 & 1 \\ 1-3x^2 & -\gamma \end{bmatrix}_{x=-1,y=0} = \begin{bmatrix} 0 & 1 \\ -2 & -\gamma \end{bmatrix} $$
The eigenvalues of $A$ are $$ \lambda_{1,2} = \frac{-\gamma\pm \sqrt{\gamma^2-8}}{2} $$
If $\gamma < \sqrt{8}$, the eigenvalues are complex with negative real part, hence, the equilibrium point is stable focus. If $\gamma \geq \sqrt{8}$, the eigenvalues are real and negative, therefore, the equilibrium point is stable node. Apply same procedure for the equilibrium point (1,0), you will end up with the same result.