I have to resolve the following discrete dynamical system:
$$x_{n+1}=0.5x_n+0.2y_n+0.5z_n$$
$$y_{n+1}=0.1x_n+0.8y_n+0.1z_n$$
$$z_{n+1}=0.4x_n+0.6z_n$$
with the starting conditions:
$$x_0=12$$
$$y_0=7$$
$$z_0=1$$
solve the system for n=5 and determine the solution when $n \rightarrow \infty$
I already searched in the internet but didn't find anything useful and we never really saw it in class, so I honestly have no clue of how to solve this problem.
I'll appreciate any help.
$\endgroup$ 04 Answers
$\begingroup$These are recurrence equations. The solution is to think that you recurrently substitute each solution to solve n steps:
STEP 1:
$x_1=0.5(12)+0.2(7)+0.5(1)$
$y_1=0.1(12)+0.8(7)+0.1(1)$
$z_1=0.4(12)+0.6(1)$
so
$x_1=6+1.4+0.5=7.9$
$y_1=1.2+5.6+0.1=6.9$
$z_1=4.8+0.6=5.4$
These values are STEP 1.
To get STEP 2, you increment $n$, you use same equations and just make $x_0=x_1$, $y_0=y_1$, $z_0=z_1$ you will get the new $x_2, y_2$ and $z_2$. This is a recurrence.
You can use a simple spreadsheet to solve for n=5 and also for a large n to see what happens when $n$ approaches infinity.
$\endgroup$ $\begingroup$General procedure: your system can be written in matrix form: $$v_{n+1} = M v_n$$ with $$v_n = \pmatrix{x_n\cr y_n\cr z_n},\qquad M=\cdots$$ So $$v_n = M^n v_0.$$ Diagonalizing/finding the Jordan form of $M$ you can find easily the required limit.
$\endgroup$ $\begingroup$For the solution when $n \to \infty$, if there is a stable one you can find it by setting $x_{n+1}=x_n$ and so on. That gives you three equations in three unknowns which you should be able to solve. If the sum of the coefficients of the $x_{n+1}$ equation were $1$, the total of the three variables would be a constant and you would know there was a stable solution. In this case the solution is constantly growing.
$\endgroup$ $\begingroup$Discrete dynamic system
$$ x_{n+1} = \mathbf{A} \, x_{n} $$ with $$ \mathbf{A} = \frac{1}{10} \left[ \begin{array}{ccc} 5 & 2 & 5 \\ 1 & 8 & 1 \\ 4 & 0 & 6 \\ \end{array} \right], \qquad x_{0} = \left[ \begin{array}{ccc} 12 \\ 7 \\ 1 \\ \end{array} \right] $$
Reduce matrix to Jordan form
This matrix can be diagonalized: $$ \mathbf{A} = \mathbf{P} \, \Lambda \, \mathbf{P}^{-1} \\ $$ with $$ \begin{align} % P \mathbf{P} & = \left[ \begin{array}{lll} 1.16782 & \phantom{-}0.340305 & -1.25813 \\ 0.811527 & -2.09823 & \phantom{-}0.0367049 \\ 1 & \phantom{-}1 & \phantom{-}1 \\ \end{array} \right] \\[5pt] % L \Lambda & = \left[ \begin{array}{ccc} 1.06713 & 0 & 0 \\ 0 & 0.736122 & 0 \\ 0 & 0 & 0.0967491 \\ \end{array} \right] \\[5pt] % Pinv \mathbf{P}^{-1} & = \left[ \begin{array}{lll} \phantom{-}0.332661 & \phantom{-}0.249064 & 0.409388 \\ \phantom{-}0.120731 & -0.378006 & 0.16577 \\ -0.453392 & \phantom{-}0.128942 & 0.424842 \\ \end{array} \right] % \end{align} $$
Solution
$$ x_{5} = \mathbf{A} \left( \mathbf{A} \left( \mathbf{A} \left( \mathbf{A} \left( \mathbf{A} x_{0} \right) \right) \right) \right) = \mathbf{A}^{5} x_{0} $$ $$ x_{5} = \mathbf{A}^{5} x_{0} = \mathbf{P} \, \Lambda^{5}\, \mathbf{P}^{-1} x_{0} = \left[ \begin{array}{lll} 2.17157 & 0.83869 & 1.44319 \\ 0.746115 & 1.79575 & 0.9095 \\ 1.02384 & 0.653539 & 2.26414 \\ \end{array} \right] % \left[ \begin{array}{ccc} 12 \\ 7 \\ 1 \\ \end{array} \right] % = % \left[ \begin{array}{l} 33.3728 \\ 22.4332 \\ 19.125 \\ \end{array} \right] % $$
Large $n$ behaviour
To understand the asymptotic behavior, compute $ \lim_{n\to\infty} \mathbf{A}^{n}, $ and examine $\lim_{n\to\infty} \Lambda^{n}$. Eigenvalues less than unity will disappear, and the largest eigenvalues will grow exponentially. The three eigenvalues are plotted below.
