5

Statement

Consider an advanced functional differential equation $$ Lf(x) = f(2x+\pi)+f(2x-\pi),\quad L\equiv\frac{d^2}{dx^2}+1. \tag{1} $$ Let's construct a solution of Eq. $(1)$ with finite support $I\equiv\operatorname{supp}f=\bigl[-\pi,\,\pi\bigr].$

Notation: Denote solution as $\lambda(x),$ i.e. $\lambda(x)\equiv f(x).$

Solution

Applying Fourier transform to $(1)$ after some algebra an expression of a spectrum can be obtained $$ \hat{\lambda}(t)=\prod\limits_{k=0}^{\infty}\frac{\cos\Bigl(\frac{\pi}{2}\cdot t\cdot 2^{-k}\bigr)}{1-(t\cdot 2^{-k})^2}. \tag{2} $$ Expression $(2)$ transforms into a simpler one as well $$ \hat{\lambda}(t) = \operatorname{sinc}(\pi\cdot t)\cdot\prod\limits_{k=0}^{\infty}\frac{1}{1-(t\cdot 2^{-k})^2}. \tag{3} $$ Solution of $(1)$ is defined by inverse Fourier transform $$ \lambda(x)=\frac{1}{2\pi}\int\limits_{\mathbb{R}}e^{itx}\cdot\hat{\lambda}(t)\,dt. \tag{4} $$

Computation

Consider an approximation of $\lambda(x)$ by lacunary Fourier series $$ \lambda(x) = \frac{a_0}{2} + \sum\limits_{k=0}^{\infty}a_{2^k}\cos(2^kx). \tag{5} $$ Note, that $$ a_n = \dfrac{1}{\pi}\int\limits_{I}\lambda(x)\cos(nx)\,dx = \dfrac{1}{\pi}\hat{\lambda}(n). \tag{6} $$ Function $\hat{\lambda}(t)$ does not vanish only at points $n=2^k,\,k=0,1,\ldots,$ and $$ a_{2^k} = \dfrac{1}{\pi}\lim_{t\rightarrow 2^k}\hat{\lambda}(t). \tag{7} $$ Values of first several coefficients are as follows $$ \mathbf{a}=(a_0,\,a_1,\,a_2,\,a_4,\,a_8)=\biggl(\frac1\pi,\,2.3\cdot10^{-1},\,7.7\cdot10^{-2},\,-5.1\cdot10^{-3},\, 8.1\cdot10^{-5},\,-3.2\cdot10^{-7}\biggr) $$

Plot of $\lambda(x)$ graph by $(5)$ with $5$ terms approximation $\mathbf{a}$ is a blue line, first derivative $\lambda'(x)$ (orange), and second derivative $\lambda''(x)$ (green) are shown in the figure la2 plot

Questions

  1. Does a rectangular function function $\chi_{I}(x)$, which is also a characteristic function of the interval $I$ satisfy the Eq. $(1)?$
  2. How to construct a fast convergence algorithm to compute values of $f(x)$, like a proposed one?
  3. Derivation an exact expression of $a_{2^k}$ in (7)?

Discussion

The problem above is related to the problem of Recursive Integration over Piecewise Polynomials: Closed form? and the form of Eq. (1) close to the Fabius equation.

Reference

Kolodyazhny, V.M., Rvachov, V.A. Cybern Syst Anal (2007) 43: 893 (page 898). DOI: https://doi.org/10.1007/s10559-007-0114-y

Oleg Kravchenko
  • 183
  • 2
  • 9
  • Small, pedantic FYI: the TeX command \operatorname can be used to properly format operators like $\operatorname{supp}$ and $\operatorname{sinc}$. In particular, it will put the right amount of space after the operator, which improves readability (and works much better than \mathrm). – Xander Henderson Jan 24 '19 at 16:43
  • @XanderHenderson Thank you for the notice. – Oleg Kravchenko Jan 24 '19 at 17:53
  • @reuns I'm not getting why there is no multiplier 2 in terms $e^{\pm i\pi\omega}?$ – Oleg Kravchenko Jan 25 '19 at 13:27
  • 1
    Ah I missed that sorry. Then it is not a convolution equation, with the Fourier transform you get $F(\omega) = F(\omega/2) \frac{\cos(\pi \omega/2)}{1-\omega^2}$ which has solutions in the sense of tempered distributions, bounded on $[-1/2,1/2]$ and away from simple poles at $\pm 2^k \omega$ (those needing a principal value in the inverse Fourier transform) – reuns Jan 25 '19 at 13:31
  • @reuns Completely agree, that is how I obtained (2) indeed. It seems to be a standard way. – Oleg Kravchenko Jan 25 '19 at 13:34
  • 1
    $F$ is fully determined by its value on $[-2/3,-1/3] \cup [1/3,2/3]$ and the extension to a distribution on $\mathbb{R}$ by $F(\omega) = F(\omega/2) \frac{\cos(\pi \omega/2)}{1-\omega^2}$ is well-defined whenever $\frac{F(x)-F(1/2)}{x-1/2}$ is $L^1$ around $1/2$ – reuns Jan 25 '19 at 13:38
  • @reuns Added question 3, any ideas? Seems it's possible to achieve an exact representation of Fourier coefficient. – Oleg Kravchenko Jan 25 '19 at 14:49

0 Answers0