real dynamical systems

Question

I am a bit confused on how to solve this system or generally these kind of systems $$x(t+1)= x(t) -2y(t)$$ $$y(t+1) = 2x(t) +y(t)$$ My confusion derives from that fact that I am working with $(t+1)$ we did these kind of tasks with derivatives and differential equations earlier.

Answer 1

The system you have considered is in the class of Discrete-Time, Linear Time Invariant systems. That is discrete time systems of the form $ z[k+1] = A z[k]. $ where $A$ is a real matrix. Given an initial condition $z[0]\in\mathbb{R},$ such systems have closed form solutions of the form $$ z[k] = A^{k} z[0] $$ Under some appropriate conditions on $A$ you can easily find simpler expressions. For example, defining the state vector $z = (x,y)^\top,$ your system is solved by $$ z[k] = \begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix}^k z[0] $$ Your matrix $A$ is diagonalizable over $\mathbb{C}$ which connects with the observation made by @Lutz. In particular if we define a new state $$ w = \begin{pmatrix} i\sqrt{2} & -\sqrt{2}\\ -i\sqrt{2} & -\sqrt{2}\end{pmatrix} z $$ we find the dynamics reduce to $$ w[k + 1] = \begin{pmatrix} (1 + 2i) & 0 \\ 0 & (1 - 2i) \end{pmatrix} w[k],$$ which admits solutions of the form $$\begin{aligned} w[k] &= \begin{pmatrix} (1 + 2i)^k & 0 \\ 0 & (1 - 2i)^k \end{pmatrix} w[0]\\ &= 5^{\frac{k}{2}}\begin{pmatrix} e^{i\theta k} & 0 \\ 0 & e^{-i\theta k} \end{pmatrix} w[0] \end{aligned}$$ where $\theta = \arctan(2).$ If $A$ is not diagonalizable, you can still find a change of coordinates that puts the matrix in Jordan Normal Form. This technique generalizes for higher dimensional discrete time systems.