Study the behavior of a 2x2 dynamical system with parameters

Question

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?

Answer 1

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 $-1 < \lambda_1(a),\lambda_2(a) < 1$

Answer 2

Where does the $1$ in your conditions come from? It is supposed to be a $0$, as far as I know. The eigenvalues show "where your system tends to" after a slight disturbence of the equilibrium. So positive ( = growing) or negative ( = decaying) is a sensible property to discuss.

To get an idea what is supposed to happen, you can in a 2x2 system always use that $$-\dfrac{1}{a} = \mathrm{det}(A)=\lambda_1\cdot\lambda_2,$$ $$-\dfrac{1}{a} = \mathrm{trace}(A)=\lambda_1 + \lambda_2.$$

From the first equation, you see that $\lambda_1$ and $\lambda_2$ have to have different signs. Consquently, $0$ is always a saddle point.

Answer 3

If you eliminate $x$, you get a second order recurrence on $y$, with the characteristic polynomial

$$a\lambda^2+\lambda-1.$$

The roots are

$$\lambda=\frac{-1\pm\sqrt{1+4a}}{2a}=-\frac2{1\pm\sqrt{1+4a}}.$$

They are smaller than one in module, except for the "minus" one before $a=2$.