Consider the following equation: $$ x_{n+2}-2ax_{n+1}+x_n=0$$ a) Define a new auxiliary variable $y_n = x_{n+1}$ and rewrite the previous equation as a discrete, two-equation dynamical system. b) What can be said about the asymptotic behaviour of $x_n$ when $n \to \infty$ depending on the parameter $a \in \mathbb{R}$?
When doing this, I'm left with the system:
$$ \left\{ \begin{array}{c} x_{n+1}= y_n \\ y_{n+1}=-x_n+2ay_n \\ \end{array} \right. $$
Which can be written as $$\pmatrix{x_{n+1}\\y_{n+1}}=\pmatrix{0&1\\-1&2a}\pmatrix{x_n\\y_n}$$ $$\begin{matrix}\end{matrix}$$
But here's my question: the characteristic polynomial p($\lambda$)=$\lambda^2-2a\lambda+1$, and therefore $\lambda_1+\lambda_2=2a$ and $\lambda_1\lambda_2 =1$. (I don't know if I'm skipping any restrictions on the eigenvalues, apart from the fact that they've got to have the same sign). The eigenvalues are:
$$\lambda_{1,2}=\frac{2a\pm\sqrt{4a^2-4}}{2}={a\pm\sqrt{a^2-1}}$$.
So, if i let $a \in ]-1,1[$, then the eigenvalues take the form $\lambda_{1,2}={a\pm i\sqrt{1-a^2} = \alpha \pm i\beta}$. Since the eigenvalues are complex numbers, I can take the matrix to one of Jordan canonical forms:
$$A=\pmatrix{0&1\\-1&2a}=r\pmatrix{\alpha&\beta\\-\beta&\alpha}$$
The problem is that I get that $r = a^2 + (\sqrt{1-a^2})^2 = a^2+1-a^2=1$ for all $a \in ]-1,1[$, but when I plot the dynamical system when $a \neq 0$, I get that $(0,0)$ is a repulsor fixed point for the system: as $n \to \infty$, when I raise $A^n$ I get a $r^n$ factor that diverges, which means $|r|>1$. Any ideas of why this happens? I've been struggling with this all day and I have absolutely no idea where to begin.
Let $x_n=t^n$, then you get $$t^2-2at+1=0 \Rightarrow _{1,2}=a\pm\sqrt{a^2-1}.$$ So the solution of your recurrence relation (equation) is $$x_n=C_1 t_1^n + C_2 t_2^n \Rightarrow x_n= C_1 (a+\sqrt{a^2-1})^n+ C_2 (a-\sqrt{a^2-1})^n.$$ Suppose $a^2<1$, then let $\sqrt{a^2-1}=ib =i \sqrt{1-a^2}.$ So $$x_n=C_1 (a+ib)^n + C_2 (a-ib)^n.$$ As $|a+ib|=1$,we write $a\pm ib= \exp(i\theta)$, then $$x_n=C_1 e^{in\theta} + C_2 e^{-in\theta}=(C_1+C_2) \cos n\theta + i(C_1-C_2) \sin n\theta. $$ Finally $$x_n= D_1 \cos n\theta+ C_2 \sin n\theta$$, where $$\theta =\tan^{-1} \left( \frac{\sqrt{1-a^2}}{a} \right)$$