How to solve this recurrence relation of x and g?

Question

$$x_{t+1}=x_t-2g_t\\ g_t=0.9g_{t-1}+0.2x_t\\ g_{-1}=0\\$$ Suppose $x_0$ will be given. How to solve this recurrence relation? Write it as $s_t=\begin{bmatrix} x_t\\x_{t-1}\end{bmatrix}, s_{t+1}=Ms_t$, where M is a 2x2 matrix?

Answer 1

Here are some observations:

\begin{eqnarray*} x_{t+1}&=&x_t-2g_t = x_t-2[0.2x_t-0.9]g_{t-1} = (0.6)x_t+(-1.8)g_{t-1}\\ g_t &=& (0.2)x_t+(0.9)g_{t-1} \end{eqnarray*} Hence, $$ \begin{bmatrix} x_{t+1}\\ g_t \end{bmatrix} = \begin{bmatrix} 0.6 & -1.8\\ 0.2 & 0.9 \end{bmatrix} \begin{bmatrix} x_t\\ g_{t-1} \end{bmatrix} = \begin{bmatrix} 0.6 & -1.8\\ 0.2 & 0.9 \end{bmatrix}^2 \begin{bmatrix} x_{t-1}\\ g_{t-2} \end{bmatrix} =\cdots = \begin{bmatrix} 0.6 & -1.8\\ 0.2 & 0.9 \end{bmatrix}^t \begin{bmatrix} x_1\\ g_0 \end{bmatrix} = \begin{bmatrix} 0.6 & -1.8\\ 0.2 & 0.9 \end{bmatrix}^{t+1} \begin{bmatrix} x_0\\ g_{-1} \end{bmatrix} = \begin{bmatrix} 0.6 & -1.8\\ 0.2 & 0.9 \end{bmatrix}^{t+1} \begin{bmatrix} x_0\\ 0 \end{bmatrix} $$

and good old Wolfram Alpha gives $$ \begin{bmatrix} x_{500}\\ g_{499} \end{bmatrix} = \begin{bmatrix} 0.6 & -1.8\\ 0.2 & 0.9 \end{bmatrix}^{500} \begin{bmatrix} x_0\\ 0 \end{bmatrix} = x_0 \begin{bmatrix} -3.7\times 10^{-12}\\ 4.1\times 10^{-13} \end{bmatrix} $$ The $t=5000$ values of $x$ and $g$ are of the order of $10^{-115}$ and the $t=10000$ values are like $10^{-229}$ so it looks like both $x$ and $g$ converge to zero. There is a theorem about $A^n\to 0$ for loose conditions on a matrix $A$ but I can't find it easily. One can see easily that $\det(A^n)=(\det A)^n = (0.9)^n$ so the determinant of our coefficient matrix converges to zero.

Edit: OK, here is the result: If the spectral radius of a matrix $A$ is less than 1 then $A^n\to 0$ as $n\to\infty$. The eigenvalues of our matrix are $\frac{1}{20}(15\pm i\sqrt{135})$ which have norm $0.9<1$ so our matrix converges to zero.

Answer 2

The recurrence can be written as

$$ \cases{ x_{k+1}= x_k-2g_k\\ g_{k+1}= \frac{1}{2} x_k-\frac{1}{5} g_k} $$

or also as

$$ \phi_{k+1}=M\phi_k $$

with $M = \left( \begin{array}{cc} 1 & -2 \\ \frac{1}{5} & \frac{1}{2} \\ \end{array} \right)$ and $\phi_k = \left( \begin{array}{c} x_k \\ g_k \\ \end{array}\right)$

Now

$M = T \Lambda T^{-1}$ with

$$\Lambda = \left( \begin{array}{cc} \frac{1}{20} \left(15+3 i \sqrt{15}\right) & 0 \\ 0 & \frac{1}{20} \left(15-3 i \sqrt{15}\right) \\ \end{array} \right) =\left( \begin{array}{cc} \rho e^{i\theta} & 0 \\ 0 & \rho e^{-i\theta} \\ \end{array} \right)\\ T = \left( \begin{array}{cc} \frac{1}{4} \left(5+3 i \sqrt{15}\right) & \frac{1}{4} \left(5-3 i \sqrt{15}\right) \\ 1 & 1 \\ \end{array} \right) $$

then, from $\phi_{k+1}=T\Lambda T^{-1}\phi_k$ we have $T^{-1} \phi_{k+1}= \Lambda T^{-1}\phi_k$ and now calling $\eta_k = T^{-1}\phi_k$ we follow with

$$ \eta_{k+1}=\Lambda \eta_k \Rightarrow \eta_k = \Lambda^k\eta_0 $$

and

$$ \eta_k = \left( \begin{array}{cc} \rho^k e^{i k\theta} & 0 \\ 0 & \rho^k e^{-ik\theta} \\ \end{array} \right)\eta_0 $$

Here $\rho = \frac{3}{\sqrt{10}} < 1$ so the $\eta_k$ sequence converges to zero with $k$.