Matrix Difference Equation Solution

Question

I have a system of recurrence relations in the following form:

$$ \begin{pmatrix} f(n+1)\\ g(n+1)\\ \end{pmatrix} = \textbf{A} \begin{pmatrix} f(n)\\ g(n)\\ \end{pmatrix} +\vec{b} $$

which hold for all $n \in \{ 0,1,2,...,N\}.$ I also have the conditions: $f(0) = g(N+1) = 0.$ I've been trying to find a way to solve this but I'm really not sure how to proceed. Any help would be appreciated.

Answer 1

$$\begin{pmatrix} f(1)\\ g(1)\\ \end{pmatrix} = \textbf{A} \begin{pmatrix} 0\\ g(0)\\ \end{pmatrix} +\vec{b} $$ $$\begin{pmatrix} f(2)\\ g(2)\\ \end{pmatrix} = \textbf{A} \begin{pmatrix} f(1)\\ g(1)\\ \end{pmatrix} +\vec{b} = \textbf{A}(\textbf{A} \begin{pmatrix} 0\\ g(0)\\ \end{pmatrix} +\vec{b} )+\vec{b} = \textbf{A}^2\begin{pmatrix} 0\\ g(0)\\ \end{pmatrix} +\textbf{A}\vec{b}+\vec{b} $$ $$\vdots$$ $$\vdots$$

$$\begin{pmatrix} f(n)\\ g(n)\\ \end{pmatrix} = \textbf{A}^n \begin{pmatrix} 0\\ g(0)\\ \end{pmatrix} +\sum_{k=0}^{n-1} A^{k}\vec{b} \qquad \forall n = 0 \ldots N$$

however $$\begin{pmatrix} f(N+1)\\ g(N+1)\\ \end{pmatrix} = \begin{pmatrix} f(N+1)\\ 0\\ \end{pmatrix} = \textbf{A}^{N+1} \begin{pmatrix} 0\\ g(0)\\ \end{pmatrix} +\sum_{k=0}^{N} A^{k}\vec{b} $$ You could use the last equation to get $g(0)$ as a function of $f(N+1)$.

Answer 2

Hints:

You can get rid of the constant vector $b$ by writing

$$v_{n+1}=A v_n+b\iff u_{n+1}+t=A(u_n+t)+b.$$

Then if $t$ is a solution of $$At+b-t=0$$ the relation reduces to

$$u_{n+1}=Au_n,$$ or by induction

$$u_{N}=A^{N}u_0.$$

Now by diagonalization of $A=P^{-1}\Lambda P$, so that $A^N=P^{-1}\Lambda^NP$, you have

$$Pu_n=\Lambda^NPu_0.$$

Denoting $u',u''$ the components of $u$,

$$au'_N+bu''_N=\lambda_0^n(au'_0+bu''_0), \\cu'_N+du''_N=\lambda_1^n(cu'_0+du''_0).$$

As you already know $u'_0$ and $u''_N$, this is a system of two equations in two unknowns.