Consider the discrete dynamical system given by $x_{n+1} = A x_n$, where $A = \begin {pmatrix} a & -b\\c &d\end {pmatrix}$ and $x_n = \begin {pmatrix} u_n\\v_n\end {pmatrix}$. Are there nonconstant solutions for certain $a,b,c,d > 0$ such that $\lim_{k \to \infty} (u_k,v_k)_k$ is finite?
Since the recurrence relation is linear and homogeneous, the solution is given by $x_{n} = A^n x_0$. Now I wonder if here the Perron-Frobenius theorem could be applied. Unfortunately, I never learned this theory, so I am grateful for any hints. The idea would be to assume $A$ has a largest eigenvalue which is real (by putting appropriate conditions on the coefficients of $A$). But how exactly could I use this theorem to answer the above question? The eigenvalues of a generic $A$ are given by $\lambda_{1,2} = \frac{tr(A)}{2} \pm \frac{1}{2}\sqrt{tr(A)^2-4 \det(A)}$.
Perron-Frobenius only works when all the entries of the matrix are positive. It guarantees a greatest modulus, real, eigenvalue, but we can work out what's going on without it.
In particular, assume that $A$ has two distinct eigenvalues, $\lambda_1$ and $\lambda_2$, and eigenvectors, $v_1$ and $v_2$. Then, if $x_0 = \alpha v_1 + \beta v_2$, then $x_n = A^n x_0 = \alpha \lambda_1^n v_1 + \beta \lambda_2^n v_2$. The solutions which are finite and nonzero correspond to eigenvalues whose modulus is $1$. However, being $1$ leads to constant solutions, and being $-1$ is almost equally boring, and in fact inaccessible here (the sum of the eigenvalues is $\operatorname{tr} (A) > 0$, and their product is $\det(A)>0$). So, we need at least one eigenvalue of the form $e^{i\theta}$, and because we've got a real valued matrix, the second will be $e^{-i\theta}$.
So, assume we have these eigenvalues, where $0< \theta <\pi/2$, and we have $\operatorname{tr} (A) = 2\cos(\theta)$ and $\det(A) = 1$. One matrix that satisfies this is $$\begin{pmatrix} \cos(\theta) & -sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}$$ and there are several others. For this matrix, we have one solution for each radius, and the matrix acts by rotating the point around the origin on a circular trajectory. The other maps with the same eigenvalues correspond to very nearly the same thing, except that the orbits are elliptical.
I invite you to determine what happens when the eigenvalues are identical, and there aren't eigenvectors that span $\mathbb{R}^2$ by yourself. Its not that exciting.