How long until equilibrium between two recursive functions?

Question

Problem

I'm trying to figure out how long it takes Particle A and Particle B to reach an equilibrium in their temperatures. I've simplified this down to a pair of recursive functions. Let's say that Particle A starts at $1100°K$, Particle B starts at $100°K$, and they transfer $10\%$ of their temperature at each "step". Everything about them is identical other than temperature, so we also know that their equilibrium temperature would be at $600°K$. So I have these functions:

$$ \begin{array}{lll} f_a(0) & = & 1100°K \\ f_b(0) & = & 100°K \\ f_a(t) & = & f_a(t-1) + 0.1\bigl(f_b(t-1) - f_a(t-1)\bigr) \\ f_b(t) & = & f_b(t-1) + 0.1\bigl(f_a(t-1) - f_b(t-1)\bigr) \end{array} $$

After 1 "step" the temperature for Particle A would be:

$$ \begin{array}{llll} f_a(1) & = & f_a(0) + 0.1\bigl(f_b(0) - f_a(0)\bigr) \\ & = & 1100°K + 0.1\bigl(100°K - 1100°K\bigr) \\ & = & 1000°K \end{array} $$

And for Particle B:

$$ \begin{array}{llll} f_b(1) & = & f_b(0) + 0.1\bigl(f_a(0) - f_b(0)\bigr) \\ & = & 100°K + 0.1\bigl(1100°K - 100°K\bigr) \\ & = & 200°K \end{array} $$

Question 1

How many steps does it take until Particle A and Particle B are within $x°K$ of each other, for some $x$? In other words, what is the minimum value of $t$ such that: $|f_a(t) - f_b(t)| \le x°K$?

(Use $x = 1$ if you need a specific value)

Question 2

Can $f_a(t)$ be defined without $f_b(t)$ using the fact that we know it will approach $600°K$? If so, will this approach hold if they transfer heat at different rates, perhaps $0.1$ and $0.2$ instead of both being $0.1$?

Answer 1

After some more work I was able to simplify even further and work out a solution.

The fact that the overall temperature is constant ($C_h$) helps, it gives:

$$ f_a(t) + f_b(t) = C_h $$

Which can be used to rewrite both equations without referencing the other (transfer ratio replaced with $C_r$):

$$ \begin{array}{lll} f(t) & = & f(t-1) + C_r\bigl(C_h - f(t-1) - f(t-1)\bigr) \\ & = & C_rC_h + f(t-1)(1-2C_r) \end{array} $$

Simplifying the constants gives a really simply recurrence relation:

$$ f(n) = a + b \cdot f(n-1) $$

Plugging in a few values for n gives the pattern:

$$ f(n) = a \cdot \sum_{k=1}^nb^{k-1} + b^n\cdot f(0) $$

Which has the nice closed form:

$$ f(n) = a \cdot \frac{1-b^n}{1-b} + b^n\cdot f(0) $$

Answer 2

From

$$ \left\{ \begin{array}{c} f_a(k)=f_a(k-1)+\mu \left(f_b(k-1)-f_a(k-1)\right) \\ f_b(k)=f_b(k-1)-\mu \left(f_b(k-1)-f_a(k-1)\right) \\ \end{array} \right. $$

Using the $\mathcal{Z}$ transform we have

$$ \left\{ \begin{array}{c} f_a(z)=f_a(-1)+\frac{f_a(k)}{z}+\mu \left(f_b(-1)+\frac{f_b(k)}{z}-f_a(-1)-\frac{f_a(k)}{z}\right) \\ f_b(z)=f_b(-1)+\frac{f_b(k)}{z}-\mu \left(f_b(-1)+\frac{f_b(k)}{z}-f_a(-1)-\frac{f_a(k)}{z}\right) \\ \end{array} \right. $$

and solving for $f_a(z), f_b(z)$

$$ \left\{ \begin{array}{c} f_a(z) = \frac{z \left((-\mu (z-2)+z-1) f_a(-1)+\mu z f_b(-1)\right)}{(z-1) (2 \mu +z-1)} \\ f_b(z) = \frac{z \left(\mu z f_a(-1)+(-\mu (z-2)+z-1) f_b(-1)\right)}{(z-1) (2 \mu +z-1)} \\ \end{array} \right. $$

and anti transforming

$$ \left\{ \begin{array}{c} f_a(t) = \frac{1}{2} \left(\left(f_a(-1)-f_b(-1)\right) (1-2 \mu )^{t+1}+f_a(-1)+f_b(-1)\right) \\ f_b(t) = \frac{1}{2} \left(\left(f_b(-1)-f_a(-1)\right) (1-2 \mu )^{t+1}+f_a(-1)+f_b(-1)\right) \\ \end{array} \right. $$