time-shifted ODEs/volume of polytopes

Question

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!

Answer 1

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: $$ f(t)=\frac{\cos \left(\left(t-\frac{1}{2}\right) x\right)+\sin \left(\left(t-\frac{1}{2}\right) x\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}. $$

Answer 2

I will assume that $x$ is a constant. Your equation is equivalent to the integral equation $$ f(t)=\frac{x}{a}\,\int_{1-at}^1f(s)\,ds,\quad 0\le t\le 1. $$ It is clear that $f\equiv0$ is a solution. The question is whether there are other solutions. If $f\colon[0,1]\to\mathbb{R}$ is continuous, define $$ Kf(t)=\frac{x}{a}\,\int_{1-at}^1f(s)\,ds. $$ It is clear that $K$ is a linear operator on the Banach space $X$ of all continuous real functions defined on $[0,1]$ with the supremum norm $\|f\|=\sup_{0\le t\le1}|f(t)|$. The original equation is equivalent to $Kf=f$. It is also easy to show that fir $f,g\in X$, then $$ \|Kf-Kg\|\le x\,\|f-g\|. $$ Thus, if $|x|<1$, the Banach fixed point theorem implies that $f\equiv0$ is the unique solution.

Edit

The initial condition is $f(0)=1$, not $f(0)=0$ as I wrongly assumed above. The original problem is the equivalent to $Kf=f$ with $$ Kf(t)=1+\frac{x}{a}\,\int_{1-at}^1f(s)\,ds,\quad 0\le t\le 1. $$ Again, we see that there is a unique solution. How to find it? One possibility is through successive approximations. Start with $f_0\equiv1$ and let $$ f_{n+1}(t)=1+\frac{x}{a}\,\int_{1-at}^1f_n(s)\,ds,\quad n\ge0. $$ Then $f_n$ converges uniformly in $[0,1]$ to the unique solution.