How to write a function which depends on another function which in turn depends on the first function... Here’s something I can’t figure out, and I need some insight—let me know what you think.
Two boxers are in a ring. 
Boxer A can deliver 20 hits each round.
Boxer B can deliver 15 hits each round.
Boxer A can take 80 hits before being knocked out.
Boxer B can take 100 hits before being knocked out.
Here’s the tricky part:
The closer a boxer gets to being knocked out, the fewer hits he can deliver.
The percentage of hits per round he can deliver is equal to the percentage of hits he can take before being knocked out.
For example, if boxer A has been hit 40% of the amount it would take to knock him out, his ability to deliver hits would also be reduced by 40%.   He would have taken 32 out of 80 hits and would now only be able to deliver 12 hits per round instead of 20.
Both boxers are wearing each other out at different rates.  Is there a set of functions to show how this fight or any like it would pan out?
 A: Naively, $B$ can last $5$ rounds with A dealing full damage, while $A$ can last $\frac {16}3$ rounds with B at full strength.  The decreasing strength is in $A$'s favor, so he will win the fight.  What matters is the ratio of damage dealt/damage you can take.
To solve it explicitly, define $a$ as the health of  boxer $A$, ranging from $1$ at the start of the fight to $0$ when he is knocked out.  Similarly, $b$ is the health of boxer $B$.  Then we can write differential equations for the health. Each hit that $A$ takes decreases his health by $\frac 1{80}$ while each hit that $B$ takes decreases his health by $\frac 1{100}$.  So $$a'=-\frac {15b}{80},a(0)=1\\b'=-\frac {20a}{100},b(0)=1$$  Already it should be obvious that $B$ starts off losing health faster than $A$, will continue to do so and will lose the bout.  What matters is the ratio of the damage you can do to the damage you can take.  We can solve it explicitly if we want.  $$a''=-\frac 3{16}b'=\frac 3{80}a\\b''=-\frac 1{5}a'=\frac 3{80}b\\k=\sqrt{\frac 3{80}}\\a=a_1e^{kt}+a_2e^{-kt},b=b_1e^{kt}+b_2e^{-kt}$$  Alpha gives $$a=\frac 18 \exp(-kt)((4-\sqrt{15})\exp(2kt)+4+\sqrt{15})\\b=\frac 1{30}\exp(-kt)((15-4\sqrt{15})\exp(2kt)+15+4\sqrt{15})$$ and $B$ lasts about $10.6555$ rounds at which time $A$ is at about $\frac 14$ strength.
