In a little calculation I'm doing for fun, I've come across this equation involving a function of two arguments which is nested on the right side:

$$f(t_1 + t_2, K) = f\bigl(t_2, f(t_1, K)\bigr)$$

I'm looking for functions $f$ which satisfy this equation for all $t_1,t_2 \ge 0$, $K > 0$, given the condition $f(0, K) = K$ and that $f$ is (EDIT:) piecewise continuously differentiable in $t$.

If I assume that the solution takes the form $f(t, K) = Kg(t)$ then I can reduce this to Cauchy's functional equation and from there demonstrate that the only solution is $f(t, K) = K e^{at}$ for any constant $a$. But I would like to relax that assumption if possible; I'm curious about other possible forms, where $f(t, K)$ may have a different dependence on $K$. Is there a method I could use to find other such solutions if they exist, or to prove that they don't?

  • $\begingroup$ $f(t,K)=at+K$. $ $ $\endgroup$ – Did Sep 2 '11 at 23:05
  • $\begingroup$ @Didier: ha, silly me I completely missed that one. Although I will have to think about how I've defined this problem because the entire reason I'm doing this calculation is supposed to be showing that a linear function does not work... $\endgroup$ – David Z Sep 2 '11 at 23:08
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    $\begingroup$ Which leads us to the often repeated admonestation on this site, that you would probably optimize the time people spend on your question if you showed the actual problem you have in mind. $\endgroup$ – Did Sep 2 '11 at 23:12
  • $\begingroup$ I guess so. Basically I'm trying to show that radioactive decay has to be exponential, but by translating directly from the argument I gave in the first paragraph in this physics.SE answer, rather than using the easy proof shown in e.g. Anonymous Coward's answer. (I did say this was just for fun :-P) Evidently there are some other conditions I've neglected to mention. $\endgroup$ – David Z Sep 2 '11 at 23:48

Algebraically, your equation is equivalent to saying that $f$ defines a continuous group homomorphism from $(\mathbb R,+)$ to the group of invertible (and continuous?) functions $\mathbb R_+\to\mathbb R_+$ under function composition.

The arithmetic structure of $\mathbb R_+$ does not matter -- only the topology (and smoothness) does. So you can take your solution and conjugate it with any smooth bijection $g: \mathbb R_+\to\mathbb R_+$: $$f_g(t,K) = g(g^{-1}(K)e^t)$$

(Hmm, missed the $t\ge 0$, so it only has to be a monoid homomorphism. This allows even more solution, such as Didier's $at+K$).


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