Hello,


I'm looking for help with the following ODE:


f'(t) = x f(1 - at)


for 0 < a < 1, x in any interval (though 0 < x < 1 would be best), and f(0) = 1. There should be a solution for $0 \leq t \leq 1$...


My rather weak repertoire of techniques from undergrad & an introductory textbook on time-shifted ODE's hasn't gotten me very far at all. 


I don't know if the origins will be helpful, but just in case any combinatorics people are drifting through, this comes from calculating the volume of polytopes related to alternating permutations. Stanley has the answer when a = 1 in his survey of the subject, but none of the 3-4 easy ways of getting the answer in that case seem to generalize very well.


Thanks!



time-shifted ODEs/volume of polytopes
Consider the mapping $g:t\to 1- at$. It has a fixed point $t\_0=1/(1+a)\;$. Denote $g\_n$ the $n$-th iteration of $g$. Then $g\_n(t)=(-a)^n t+1-a+\ldots+(-a)^{n-1}\;$, $g'\_n(t)=(-a)^n$, $n=0,1,\ldots\;$. From the equation we have 
$$
f^{(n)}(t)=x^n g\_1'(t)g\_2'(t)\ldots g\_{n-1}'(t)f(g\_n(t)),
$$
so $f^{(n)}(t\_0)=(-a)^{n(n-1)/2}x^n f(t\_0)$. The expansion of $f$ in the Taylor series at $t=t\_0$ is
$$
f(t)=f(t\_0)\sum\_{n=0}^\infty\frac{(-a)^{n(n-1)/2}x^n(t-t\_0)^n}{n!}.
$$
The value of $f(t\_0)$ can be obtained form the initial condition. For $a=1$ the solution has an explicit form:
$$
f(t)=\frac{\cos
 \left(\left(t-\frac{1}{2}\right) x\right)+\sin \left(\left(t-\frac{1}{2}\right) x\right)}{\cos \left(\frac{x}{2}\right)-\sin
 \left(\frac{x}{2}\right)}.
$$


