# Study the behavior of a 2x2 dynamical system with parameters

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?

• Is $a > 0$ only? Aug 20, 2019 at 5:35
• @user7440 yes, edited. Sorry about that. Aug 20, 2019 at 5:54
• Note: This question and this question Aug 20, 2019 at 17:38

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

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.

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 $$a=2$$.