In my research, I encountered the following recurrence relation:
\begin{align} g(t) &= (\beta-1) \; g(t-1) + \beta \; f(t)\\ f(t) &=\min\{f(t-1)+g(t-1), \, c \cdot \lambda^t \} \end{align}
where $1<\beta<2$ is a constant parameter, $\lambda=\sqrt{1-(2/\beta -1)^2}$, and $c >1$ is another constant parameter. Also, $f(0) = g(0) = 1$.
I am familiar with linear recurrences, but not with sets of recurrences like this, so any help to direct me toward the answer is appreciated. If necessary, solving a relaxed version of the recurrence set where $f(t) =f(t-1)+g(t-1)$ would be helpful too.
Well, I can at least get you to generating functions for the two sequences, in the relaxed case:
Let $F(z)$ and $G(z)$ be the generating functions for $f$ and $g$, respectively: $$ F(z):=\sum_{t=0}^{\infty}f(t)z^t\qquad G(z):=\sum_{t=0}^{\infty}g(t)z^t $$ Then $$\begin{align*} F(z)&=\sum_{t=0}^{\infty}f(t)z^t\\ &=1+\sum_{t=1}^{\infty}f(t-1)z^{t}+\sum_{t=1}^{\infty}g(t-1)z^t\\ &=1+z\cdot F(z)+z\cdot G(z).\tag{1} \end{align*}$$ and $$\begin{align*} G(z)&=\sum_{t=0}^{\infty}g(t)z^t\\ &=1+(\beta-1)\sum_{t=1}^{\infty}g(t-1)z^t+\beta\sum_{t=1}^{\infty}f(t)z^t\\ &=1+(\beta-1)z\cdot G(z)+\beta(F(z)-1).\tag{2} \end{align*}$$ From $(1)$, we see that $$ F(z)=\frac{1+zG(z)}{1-z}. $$ Plugging this in to $(2)$ and solving yields $$ G(z)=\frac{1+(\beta-1)z}{(\beta-1)z^2-2\beta z+1}, $$ from which we can also deduce that $$ F(z)=\frac{1-2(\beta-1)z}{(\beta-1)z^2-2\beta z+1}. $$ From here, your best bet is to factor the denominators and do a partial fraction decomposition; this will let you rewrite in terms of geometric series for explicit solutions (much like when using generating functions to find an explicit description of the Fibonacci sequence).