Equilibria of 2x2 linear system

Question

Consider the $2\times2$ linear system given by $$A = \begin{pmatrix} 0 & 3\\-1 & 0\end{pmatrix}.$$ We wish to look at the behavior of the plane under iterations of $A$. The eigenvalues of this matrix are purely imaginary. In general, solutions of such a map are given by $X_n = c_1 \lambda_1^n v_1 + c_2 \lambda_2^n v_2$. How do we reach the solution of such a system?

Answer 1

Your system is $$ \begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} = \begin{pmatrix} 0 & 3 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x_n \\ y_n \end{pmatrix}, $$ or $$ x_{n+1} = 3y_n \qquad \text{ and } \qquad y_{n+1} = -x_n . $$ These two difference equations can be combined $$ x_{n+2} = 3y_{n+1} = -3 x_n. $$ To solve $x_{n+2} + 3x_n = 0$ we make the guess $x \sim r^n$ where $r$ is some number we will find. Plugging this in gives the characteristic equation: $$ r^{n+2} + 3 r^n = 0$$ implying that $r^2 = -3$ or $r = \pm \sqrt{-3}$. Plugging this back in for our guess for $x$ gives $$ x_n = c_1 ( \sqrt{-3} )^n + c_2 ( - \sqrt{-3} )^n,$$ which can be written in a variety of ways, each helpful in different cases. Then we can easily find $y_n$ from this too since $$ y_n = - x_{n-1} = -c_1 (\sqrt{-3})^{n-1} - c_2 ( - \sqrt{-3})^{n-1}. $$ $c_1$ and $c_2$ are constants which depend on the initial state. The long-time behaviour is found by letting $n \to \infty$. You can see here that since $\sqrt{3}>1$ you will have solutions unbounded, as you mentioned (in your original post) you found numerically.