I need to study the behavior of this discrete dynamical system:
$$\left\{
\begin{array}{c}
x_{n+1}= y_n \\
y_{n+1}= \frac{1}{a}x_n-\frac{1}{a}y_n \\
\end{array}
\right.$$
in function of the parameter $a>0$.
I studied before that the stability of the origin is given by the eigenvalues: $\lambda_1 , \lambda_2$
If $\lambda_1 , \lambda_2 < 1$, the origin is stable, hence, attractor.
If $\lambda_1 , \lambda_2 > 1$, the origin is unstable.
If $\lambda_1 > 1 > \lambda_2$ or $\lambda_2 > 1 > \lambda_1$, the origin is a saddle point.
First I found the matrix $A$ associated to this system
$A := \begin{bmatrix} 0 & 1 \\ \frac{1}{a} & \frac{-1}{a} \end{bmatrix}$
But I have problems when I'm studying the eigenvalues for stability. After analyzing the three cases, I determinated that the origin is a saddle point if $a \in (0,2)$ or the origin is stable if $a \in (2,+\infty)$. The problem is that if I graph this here, I always get a saddle point (at least in all the values of $a>0$ that i tested.
Any hints?
Best Answer
This system can be represented as
$$ U_n = M U_{n-1} $$
with $U_n = (n_n, y_n)^{\dagger}$ and $M = \left(\begin{array}{cc}0 & 1\\ \frac 1a & -\frac 1a\end{array}\right)$
but $M = T^{-1}\Lambda T$ then
$$ U_n = T^{-1}\Lambda T U_{n-1}\Rightarrow TU_n = \Lambda T U_{n-1} $$
calling now $V_n = TU_n$ we have
$$ V_{n+1} = \Lambda V_n $$
Here $\Lambda = \left(\begin{array}{cc}-\frac 12\left(\frac{1+\sqrt{1+4a}}{2a}\right) & 0\\ 0 & \frac 12\left(\frac{-1+\sqrt{1+4a}}{2a}\right)\end{array}\right)$ the diagonal eigenvalues matrix. Those eigenvalues have opposite signs characterizing a saddle point.
$$ V_n = \Lambda^n V_0 $$
Attached a plot showing in blue the two eigenvalues as well as the stability region when both eigenvalues are between the black limits $-1 < \lambda_1(a),\lambda_2(a) < 1$