MAT1193 – 6d Solving Linear DTDS So far, we have only one discrete-­‐time dynamical system where we can write down a formula for the solution given the updating function. If the updating procedure is to just multiply by a constant, st +1 = r * st , then the solution function is just exponential growth or decay, s(t) = s0 * r t = s0 * e ln(r )*t . In this section we’re going to use a typical mathematicians trick to find the solution €
function for a more complicate updating function st +1 = a* st + b . In this system, the €
previous is state is multiplied by a constant and then a constant number is added. This constant will contribute to the growth or decay of the system, and as the updating gets repeated over many time steps, the addition and the multiplication get mixed up. How are we ever going to sort this out? €
The strategy that we’ll use is to convert this problem to an easier problem, solve that easy problem, and then convert back to find the solution to the harder problem. It turns out that finding the easy problem in the case of st +1 = a* st + b , starts with finding the equilibrium of this system. To find the equilibrium, we set b
s* = a* s* + b →s* − a* s* = b →s* (1− a) = b →s* =
€ (1− a)
Notice that we can only do the last step if a≠1 so that we don’t divide by 0. So a=1 is a special case that we’ll discuss below (if I have time). €
Now lets look at the value Δst, which is the distance from the equilibrium to the state *
st. What is the updating function for Δst? Using Δst +1 = st +1 − s and st
get €
Δst +1 = st +1 − s* = ( a* st + b) − s* = a*(Δst + s* ) + b − s* = a* Δst + a* s* + b − s* = a* Δst + (a −1)* s* + b
€
€
b
*
Now we plug in s =
(1− a)
⎛ b ⎞
Δst +1 = a* Δst + (a −1)* ⎜
⎟ + b = a* Δst − b + b = a* Δst ⎝1− a ⎠
€
€
€
= Δst + s* we So the multiplication and addition in terms of the state variable st, becomes just a multiplication if we focus on the distance from the equilibrium Δst. But we know the solution function for Δst: Δs(t) = Δs0 * a t = Δs0 *e ln(a)*t Now we can use this and the relationships st
*
= Δst + s* , Δs0 = s0 − s* , and s =
b
(1− a)
to write down the solution function for the original system ⎛
b ⎞ t b ⎛
b ⎞
b
s(t) = ⎜ s0 −
= ⎜ s0 −€ ⎟ *e ln(a )*t€+
⎟ * a +
⎝ 1− a ⎠
1− a ⎝ 1− a ⎠
1− a
€
This looks like a very complicated function! But if you think about the problem in the right way, it’s actually a translation of the simple equation €
€
Δs(t) = Δs0 * a t = Δs0 *e ln(a)*t Lets do an example. Suppose your old beater car has an oil leak and it loses 10% of its oil every week. To keep your old car rolling you put in .25 quarts of oil every week. Suppose that after an oil change, the engine is at its full capacity of 4 quarts of oil. If you keep this up over the long term, how much oil will there in the engine? How much oil will you have in the engine after 12 weeks? First, lets derive the updating function. Let ot be the amount of oil in the engine after t weeks. If you knew the amount of oil this week (=ot) then the amount of oil next week should be 90% of that amount (10% leaked out) plus the .25 quarts that you put in: ot +1 = 0.9*ot + 0.25 €
In the long term, the amount of oil in the engine should reach a point where the oil that leaks out should balance the oil that you put in. If the system is stable, the amount of oil in the long term will be equal to that equilibrium value. To find the *
equilibrium for this system we let o
= ot +1 = ot and plug that in to get o* = 0.9*o* + 0.25 →0.1*o* = .25 →o* = 2.5 So in the long term, adding .25 quarts per week will keep 2.5 quarts of oil in your engine. €
€
To find the amount of oil in the engine after 12 weeks, we could just use o0=4 quarts and crank through the updating function 12 times. But we know how to solve this problem. Let Δo be the amount of oil in your engine above or below 2.5 quarts. Since 4 quarts is 1.5 quarts more than 2.5, Δo0=1.5. From the argument above, the updating function for Δot is Δot +1 = 0.9* Δot and that means the solution function is Δo(t) = Δo0 *0.9 t =1.5*e ln(0.9)*t Plugging in t=12 we find Δo12=0.4326 quarts. In other words, after 12 weeks the engine has 0.4326 €
quarts more than 2.5 quarts or 2.9326 quarts. €
One final topic to consider is the case where the slope of the updating function is negative. For example, suppose we have an updating function st+1 = -­‐.5*st+15 and the initial condition s0=12. To solve this DTDS we proceed as before and first find the equilibrium: s* = -­‐.5*s*+15 -­‐>1.5*s* = 15 -­‐> s*=10. So Δs0 = s0-­‐s* = 12-­‐10 = 2. The solution in terms of Δst is Δst = 2*(-­‐.9)t. Now we’d like to write it in the form of e to some power, but we can’t take the natural log of a negative number. So what we do is to write -­‐.9 as (-­‐1)*(.9) so -­‐.9t= ((-­‐1)*(.9) )t = (-­‐1)t *(.9)t . The (-­‐1)t just makes the sign of the solution alternate in sign: (-­‐1)t = +1 for t even and (-­‐1)t = -­‐1 for t odd. If we bring that out front we can write Δst = (-­‐1)t *2*eln(.9)*t. So Δs will alternate back an forth between being positive and negative and will have a magnitude |Δst| = 2*eln(.9)*t. To find st we just use st = s*+Δst to find st = 10+(-­‐1)t *2*eln(.9)*t.