# solving coupled discrete systems

Suppose you have a discrete system, whose evolution is governed by the following equations:

$\mathbf{x}[k+1] = f_1(\mathbf{F}[k], \mathbf{x}[k])$
$\mathbf{F}[k+1] = f_2(\mathbf{F}[k], \mathbf{x}[k+1], \mathbf{x}[k])$

where $\mathbf{x}$ is a vector and $\mathbf{F}$ is a matrix. What is the best way to show they both converge to some fixed point? Shall I look at the difference between two consecutive steps, $k+1$ and $k$, or is there a more effective way to solve it? For simplicity, suppose $f_1$ and $f_2$ are linear (and/or affine) functions.

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For a fixed point you need to have $$\mathbf{x}[k+1]=\mathbf{x}[k]=\mathbf{x}^*$$ and $$\mathbf{F}[k+1]=\mathbf{F}[k]=\mathbf{F}^*.$$ However, for linear systems of difference equations there is typically no fixed point apart from $\mathbf{x}^*=\mathbf{F}^*=0$ (as sums of solutions are also solutions you would have a whole line of fixed points given one). For nonlinear (or affine systems) you can find the fixed points (the fixed point) and linearize the system around these points to see whether the points are attractive or repulsive.
The rate of convergence is related to the largest eigenvalue. That is, suppose your system is $X[k+1] = A X[k] + B$, and all eigenvalues $\lambda$ of $A$ have $|\lambda| < r < 1$, then any solution $X[k]$ converges to $Y = (I - A)^{-1} B$, with $\|X[k] - Y\| < c r^k$ for some constant $c$ (depending on $X[0]$). – Robert Israel Jun 6 '11 at 22:46
the two equations above are coupled, meaning that $\mathbf{x}[k+1]$, for example, depends on $\mathbf{F}[k]$. $\mathbf{F}[k+1]$ depends on both $\mathbf{x}[k]$ and $\mathbf{x}[k+1]$. – ACAC Jun 6 '11 at 23:19