# time-shifted ODEs/volume of polytopes

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!

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$x$ is a constant? – Gerry Myerson Jul 31 '11 at 5:42
Sorry, yes x and a are both constants. – user146 Jul 31 '11 at 15:01

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

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This is quite nice, thank you! Accepting the answer, and I'd be interested in hearing how you recognized the (slightly complicated) explicit form below from the sum above. – user146 Aug 1 '11 at 18:22
Actually, a little confused by the result. If $t_{0} = 0$ (as we might as well assume it is), the sum above is completely symmetric in x and t, but the equation below certainly isn't. Strange. – user146 Aug 1 '11 at 18:23
(Finally, please let me know if you'd like a mention if/when the note using this is put up. The series seems to be enough for what I need) – user146 Aug 1 '11 at 18:27
@unknown (google) Please use it as you like. The value of $t_0=1/(1+a)$ is fixed, it cannot be given any value. For arbitrary $y\in [0,1]$ the formula becomes $$f(t)=\sum_{n=0}^\infty\frac{(-a)^{n(n-1)/2}x^n(t-y)^n f(g_n(y))}{n!}.$$ Only for the fixed point $t_0$ it simplifies as above. – Andrew Aug 1 '11 at 18:59
As for the case $a=1$ the series splits in two: $$\sum_{n=0}^\infty\frac{(-1)^{n(n-1)/2}x^n(t-1/2)^n}{n!}= \sum_{n=0}^\infty\frac{(-1)^{n}x^{2n}(t-1/2)^{2n}}{(2n)!}+ \sum_{n=0}^\infty\frac{(-1)^{n}x^{2n+1}(t-1/2)^{2n+1}}{(2n+1)!}=$$ $$\cos \left(\left(t-\frac{1}{2}\right) x\right)+\sin \left(\left(t-\frac{1}{2}\right) x\right).$$ Taking into account the initial condition gets $$f(t_0)=\frac1{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}.$$ – Andrew Aug 1 '11 at 18:59

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 thar $f\equiv0$ is a solution. The question is wether tere 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 $X$ be 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 trhogh succesive 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.

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Thanks for the comment. I think that you have shown that there is at most one solution for a given initial condition. However, I'm interested in the initial condition f(0) = 1, not f(0) = 0. For f(0) = 1, there is certainly a unique nonzero solution when a = 1, and your proof seems to not depend on a < 1 (which is really the only thing that initially made me suspicious, I have little intuition). – user146 Jul 31 '11 at 16:42
I misread the initial condition. I have edited my answer. – Julián Aguirre Jul 31 '11 at 19:50