I was looking for some advice in regard to incorporating a system of difference equations with a delay (sort of?). I'm not a mathematician and just started to play a little bit around with systems of difference equations so if something does not make any sense for you, let me know it's normally done ;)
So, I'm trying to modell the following system (that damn arrows :D):
I have three nodes $(A,B,C)$ with a start value of $A=5$,$B=2$,$C=4$. Every node is connected through an edge which represents the distance
$$A\rightarrow B=2$$ $$B\rightarrow C=2$$ $$C\rightarrow A=1$$
For some discrete time steps the "information", "flows" around, that is $B$ becomes updated with the value of $A$, $C$ with the value of $B$ and so forth. The equations are (assuming no time delay/ignoring distances!):
$$ A_{n+1}=-A_n+C_n $$ $$ B_{n+1}=-B_n+A_n $$ $$ C_{n+1}=-C_n+B_n$$
What I want do consider now in the equations, is that the information from $A$ arrives at $C$ only after two timesteps (see picture), whereas the information from $C$ to $A$ takes only one timestep. So there should be some sort of delay. And I'm not sure how to incorporate this in the equations and hope you can give me some advices on how to proceed.
Thanks!
It might help if you write out the values of each node at each time period to see the pattern.
The subscript will be used to denote the time period.
Also, I assumed you meant that the information leaves each node at every time period (i.e. $ n = 1,2,..$ )
We have
$$ A_0 = 5 \ B_0 = 2 \ C_0 = 4 $$
At time $ 1 $, the only information that has moved is from $ C \rightarrow A $
$$ A_1 = 4 \ B_1 = 2 \ C_1 = 4 $$
At time $ 2 $, the information from $ A_0 $ and $ B_0 $ has arrived at $ B $ and $ C $ respectively, whilst the information from $ C_1 $ has arrived at $ A $
$$ A_2 = 4 \ B_2 = 5 \ C_2 = 2 $$
Iterating this pattern, we see that
$$ A_3 = 2 \ B_3 = 5 \ C_3 = 2 $$ $$ A_4 = 2 \ B_4 = 4 \ C_4 = 5 $$ $$ A_5 = 5 \ B_5 = 4 \ C_5 = 5 $$ $$ A_6 = 5 \ B_6 = 2 \ C_6 = 4 $$ $$ A_7 = 4 \ B_7 = 2 \ C_7 = 4 $$ $$ A_8 = 2 \ B_8 = 5 \ C_8 = 2 $$ $$ ... $$
Here, you might notice that for $n \ge 0$
$$ A_{n+1} = C_n $$
$$ B_{m} = \begin{cases} B_{2n} & m = 2n+1 \\ A_n & m = 2n+2 \\ \end{cases} $$
$$ C_{m} = \begin{cases} C_{2n} & m = 2n+1 \\ B_n & m = 2n+2 \\ \end{cases} $$
Let me know if you find any mistakes, it's late at night where I am.