I'm analyzing the stability of such system difference equation with the constraint that $y_n \geq 0$ $\forall n \geq 0$ :
$y_n = B y_{n-1} + D y_{n-2} \enspace (1)$
Using variable transform, the above system can be transformed to:
$z_n = L z_{n-1}$
where
$L = \left(\begin{matrix}B & D\\ I&0\end{matrix}\right)$,
and $z_{n} = \left(\begin{matrix}y_n \\ y_{n-1} \end{matrix}\right)$
I can analyze the stability of system (1) depending on the eigenvalues and the eigenvectors of $L$.
BUT, eigenvalues do not consider the strong constraint $y_n \geq 0$ $\forall n \geq 0$. How this constraint could be integrated in the stability analysis?
I'm new to this, too. This approach works for me. I hope it is of some help.
Solve the system:
This is a relatively simple equation; I can use it to figure out how to relate the system to the matrix.
Say we have some initial values, $\;y_{-2}\! = k_2/D, \;\; y_{-1} \!\!= k_1,\;$ so that $$y_0 = Bk_1 + k_2$$ Then$$y_1 = (B^2+D)k_1 + Bk_2\\ y_2 = (B^3+2BD)k_1 + (B^2 + D)k_2\\ y_3 = (B^4+3B^2D + D^2)k_1 + (B^3 + 2BD)k_2\\ ...$$ Setting $y_{-2}\!\! = k_2$ gets there, too. Now I want to show that $$ y_n = P_n k_1 + P_{n-1} k_2$$ That is, the coefficient $P_n$ of $k_1$ in $y_n$, it also the coefficient of $k_2$ in $y_{n+1}$. The system is determined by a single polynomial sequence.
$k_1$ and $k_2$ are arbitrary initial values (!scalar parameters!) which must satisfy the linear system $\;P_n k_1 + P_{n-1}k_2 \ge 0\;\; \forall n \ge 0.$
But we can represent a polynomial pA as, say, $\;p_A(\lambda) = \det(\lambda I-A)$. And I can also, if I wish, write $k_1$ in the form $a k_2 D + b$.
Now consider the given
Conditions:
B party! If $|B|>1$, there is some $n$ past which the sum $y_n$, (and therefore its sign), is dominated by the term $B^{n+1}$, irrespective of the other initial values. This term alternates even/odd powers, so we have the pair of conditions,
$\;\;\;B \ge 0,$
$\;\;\;B>1 \implies k_1 >0.$
In a similar way, we can shake more conditions out.
Depending on what's left, we probably want at least $\;y_n \ge 0 \implies y_{n+1} \ge 0$. While I'm stuck, I work the slow way (regular algebra) and then figure out how to abstract. Takes longer at first, but then I know what I'm up to.
Lastly, I find strong conditions make it easier to characterize a system. Break the conditions, hem in the solution. IMO, breaking "Positive for all n" is a good plan of attack for figuring out how to satisfy it.
Hope that helps.
I read this as a question about process. Let me know if I should finish the problem.