Using the reversibility arguments alone, show that the system
$\dot{x}=y$
$\dot{y}=x-x^{2}$
has a homoclinic orbit in the half-plane $x\leq 0$
This is a non-linear system.
A linear approximation requires the Jacobian:
$J=\begin{bmatrix}
0 &1 \\
1-2x &0
\end{bmatrix}$
The fixed points are $(0,0) \text{and} (1,0)$.
It is easy to check that $(0,0)$ is a saddle point.
Mathematica gives the Eigenvalues:
$\left\{\left\{\lambda \to -\sqrt{1-2 x}\right\},\left\{\lambda \to \sqrt{1-2
x}\right\}\right\}$
There doesn't seem to be any reliable way, at least to the best of my knowledge, to determine the Eigenvalues and Eigenvectors for these sort of crazy non-linear examples.
It doesn't help that in the text "Nonlinear dynamics and Chaos" by Strogaz
the explanation is as
"Consider the unstable manifold of the saddle point at the origin, The manifold leaves the origin along the vector $(1,1),\cdot\cdot\cdot$
Can someone assists me with determining the Eigenvectors? It is giving me a lot of problems.
Best Answer
As stated, the eigenvalues are $\lambda_1=-\sqrt{1-2x},\ \lambda_2=\sqrt{1-2x}$.
To find the eigenvalues, you need to solve $(A-\lambda I)v=0$.
$$\left[ \begin{matrix} \sqrt{1-2x} & 1\\ 1-2x & 0\\ \end{matrix} \right] v_1=0$$
(a) Letting $v_1=\left[ \begin{matrix} v_1^{(1)}\\ v_1^{(2)}\\ \end{matrix} \right]$, we see that $$\sqrt{1-2x}v_1^{(1)}+v_1^{(2)}=0,$$ giving us $$v_1^{(2)}=-\sqrt{1-2x}v_1^{(1)},$$ and we see that the first eigenvector $v_1$ is $$v_1=\left[ \begin{matrix} 1\\ -\sqrt{1-2x}\\ \end{matrix} \right]$$ up to scalar multiplication.
(b) Similarly, $v_2$ satisfies $$\left[ \begin{matrix} -\sqrt{1-2x} & 1\\ 1-2x & 0\\ \end{matrix} \right]v_2=0,$$ yielding $$v_2^{(2)}=\sqrt{1-2x}v_2^{(1)},$$ showing that the second eigenvector is $$v_2=\left[ \begin{matrix} 1\\ \sqrt{1-2x}\\ \end{matrix} \right].$$
Essentially you find them the same way you would a linear system.