Mixing
Solution of advanced functional differential equation
$begingroup$
Statement
Consider an advanced functional differential equation
$$
Lf(x) = f(2x+pi)+f(2x-pi),quad
Lequivfrac{d^2}{dx^2}+1. tag{1}
$$
Let's construct a solution of Eq. $(1)$ with finite support
$Iequivoperatorname{supp}f=bigl[-pi,,pibigr].$
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
$$
hat{lambda}(t)=prodlimits_{k=0}^{infty}frac{cosBigl(frac{pi}{2}cdot tcdot 2^{-k}bigr)}{1-(tcdot 2^{-k})^2}. tag{2}
$$
Expression $(2)$ transforms into a simpler one as well
$$
hat{lambda}(t) = operatorname{sinc}(picdot t)cdotprodlimits_{k=0}^{infty}frac{1}{1-(tcdot 2^{-k})^2}. tag{3}
$$
Solution of $(1)$ is defined by inverse Fourier transform
$$
lambda(x)=frac{1}{2pi}intlimits_{mathbb{R}}e^{itx}cdothat{lambda}(t),dt. tag{4}
$$
Computation
Consider an approximation of $lambda(x)$ by lacunary Fourier series
$$
lambda(x) = frac{a_0}{2} + sumlimits_{k=0}^{infty}a_{2^k}cos(2^kx).
tag{5}
$$
Note, that
$$
a_n = dfrac{1}{pi}intlimits_{I}lambda(x)cos(nx),dx = dfrac{1}{pi}hat{lambda}(n).
tag{6}
$$
Function $hat{lambda}(t)$ does not vanish only at points $n=2^k,,k=0,1,ldots,$ and
$$
a_{2^k} = dfrac{1}{pi}lim_{trightarrow 2^k}hat{lambda}(t).
tag{7}
$$
Values of first several coefficients are as follows
$$
mathbf{a}=(a_0,,a_1,,a_2,,a_4,,a_8)=biggl(frac1pi,,2.3cdot10^{-1},,7.7cdot10^{-2},,-5.1cdot10^{-3},, 8.1cdot10^{-5},,-3.2cdot10^{-7}biggr)
$$
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
Questions
- Does a rectangular function function $chi_{I}(x)$, which is also a characteristic function of the interval $I$ satisfy the Eq. $(1)?$
- How to construct a fast convergence algorithm to compute values of $f(x)$, like a proposed one?
- 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
integration fourier-analysis recurrence-relations functional-equations fractals
asked Jan 24 at 11:58
Oleg KravchenkoOleg Kravchenko
465
$endgroup$
|
show 2 more comments
$begingroup$
Statement
Consider an advanced functional differential equation
$$
Lf(x) = f(2x+pi)+f(2x-pi),quad
Lequivfrac{d^2}{dx^2}+1. tag{1}
$$
Let's construct a solution of Eq. $(1)$ with finite support
$Iequivoperatorname{supp}f=bigl[-pi,,pibigr].$
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
$$
hat{lambda}(t)=prodlimits_{k=0}^{infty}frac{cosBigl(frac{pi}{2}cdot tcdot 2^{-k}bigr)}{1-(tcdot 2^{-k})^2}. tag{2}
$$
Expression $(2)$ transforms into a simpler one as well
$$
hat{lambda}(t) = operatorname{sinc}(picdot t)cdotprodlimits_{k=0}^{infty}frac{1}{1-(tcdot 2^{-k})^2}. tag{3}
$$
Solution of $(1)$ is defined by inverse Fourier transform
$$
lambda(x)=frac{1}{2pi}intlimits_{mathbb{R}}e^{itx}cdothat{lambda}(t),dt. tag{4}
$$
Computation
Consider an approximation of $lambda(x)$ by lacunary Fourier series
$$
lambda(x) = frac{a_0}{2} + sumlimits_{k=0}^{infty}a_{2^k}cos(2^kx).
tag{5}
$$
Note, that
$$
a_n = dfrac{1}{pi}intlimits_{I}lambda(x)cos(nx),dx = dfrac{1}{pi}hat{lambda}(n).
tag{6}
$$
Function $hat{lambda}(t)$ does not vanish only at points $n=2^k,,k=0,1,ldots,$ and
$$
a_{2^k} = dfrac{1}{pi}lim_{trightarrow 2^k}hat{lambda}(t).
tag{7}
$$
Values of first several coefficients are as follows
$$
mathbf{a}=(a_0,,a_1,,a_2,,a_4,,a_8)=biggl(frac1pi,,2.3cdot10^{-1},,7.7cdot10^{-2},,-5.1cdot10^{-3},, 8.1cdot10^{-5},,-3.2cdot10^{-7}biggr)
$$
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
Questions
- Does a rectangular function function $chi_{I}(x)$, which is also a characteristic function of the interval $I$ satisfy the Eq. $(1)?$
- How to construct a fast convergence algorithm to compute values of $f(x)$, like a proposed one?
- 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
integration fourier-analysis recurrence-relations functional-equations fractals
asked Jan 24 at 11:58
Oleg KravchenkoOleg Kravchenko
465
$endgroup$
$begingroup$
Small, pedantic FYI: the TeX commandoperatornamecan 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 thanmathrm).
$endgroup$
– Xander Henderson
Jan 24 at 16:43
$begingroup$
@XanderHenderson Thank you for the notice.
$endgroup$
– Oleg Kravchenko
Jan 24 at 17:53
$begingroup$
@reuns I'm not getting why there is no multiplier 2 in terms $e^{pm ipiomega}?$
$endgroup$
– Oleg Kravchenko
Jan 25 at 13:27
1
$begingroup$
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)
$endgroup$
– reuns
Jan 25 at 13:31
1
$begingroup$
$F$ 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 $1/2$
$endgroup$
– reuns
Jan 25 at 13:38
|
show 2 more comments
$begingroup$
Statement
Consider an advanced functional differential equation
$$
Lf(x) = f(2x+pi)+f(2x-pi),quad
Lequivfrac{d^2}{dx^2}+1. tag{1}
$$
Let's construct a solution of Eq. $(1)$ with finite support
$Iequivoperatorname{supp}f=bigl[-pi,,pibigr].$
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
$$
hat{lambda}(t)=prodlimits_{k=0}^{infty}frac{cosBigl(frac{pi}{2}cdot tcdot 2^{-k}bigr)}{1-(tcdot 2^{-k})^2}. tag{2}
$$
Expression $(2)$ transforms into a simpler one as well
$$
hat{lambda}(t) = operatorname{sinc}(picdot t)cdotprodlimits_{k=0}^{infty}frac{1}{1-(tcdot 2^{-k})^2}. tag{3}
$$
Solution of $(1)$ is defined by inverse Fourier transform
$$
lambda(x)=frac{1}{2pi}intlimits_{mathbb{R}}e^{itx}cdothat{lambda}(t),dt. tag{4}
$$
Computation
Consider an approximation of $lambda(x)$ by lacunary Fourier series
$$
lambda(x) = frac{a_0}{2} + sumlimits_{k=0}^{infty}a_{2^k}cos(2^kx).
tag{5}
$$
Note, that
$$
a_n = dfrac{1}{pi}intlimits_{I}lambda(x)cos(nx),dx = dfrac{1}{pi}hat{lambda}(n).
tag{6}
$$
Function $hat{lambda}(t)$ does not vanish only at points $n=2^k,,k=0,1,ldots,$ and
$$
a_{2^k} = dfrac{1}{pi}lim_{trightarrow 2^k}hat{lambda}(t).
tag{7}
$$
Values of first several coefficients are as follows
$$
mathbf{a}=(a_0,,a_1,,a_2,,a_4,,a_8)=biggl(frac1pi,,2.3cdot10^{-1},,7.7cdot10^{-2},,-5.1cdot10^{-3},, 8.1cdot10^{-5},,-3.2cdot10^{-7}biggr)
$$
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
Questions
- Does a rectangular function function $chi_{I}(x)$, which is also a characteristic function of the interval $I$ satisfy the Eq. $(1)?$
- How to construct a fast convergence algorithm to compute values of $f(x)$, like a proposed one?
- 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
integration fourier-analysis recurrence-relations functional-equations fractals
asked Jan 24 at 11:58
Oleg KravchenkoOleg Kravchenko
465
$endgroup$
Statement
Consider an advanced functional differential equation
$$
Lf(x) = f(2x+pi)+f(2x-pi),quad
Lequivfrac{d^2}{dx^2}+1. tag{1}
$$
Let's construct a solution of Eq. $(1)$ with finite support
$Iequivoperatorname{supp}f=bigl[-pi,,pibigr].$
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
$$
hat{lambda}(t)=prodlimits_{k=0}^{infty}frac{cosBigl(frac{pi}{2}cdot tcdot 2^{-k}bigr)}{1-(tcdot 2^{-k})^2}. tag{2}
$$
Expression $(2)$ transforms into a simpler one as well
$$
hat{lambda}(t) = operatorname{sinc}(picdot t)cdotprodlimits_{k=0}^{infty}frac{1}{1-(tcdot 2^{-k})^2}. tag{3}
$$
Solution of $(1)$ is defined by inverse Fourier transform
$$
lambda(x)=frac{1}{2pi}intlimits_{mathbb{R}}e^{itx}cdothat{lambda}(t),dt. tag{4}
$$
Computation
Consider an approximation of $lambda(x)$ by lacunary Fourier series
$$
lambda(x) = frac{a_0}{2} + sumlimits_{k=0}^{infty}a_{2^k}cos(2^kx).
tag{5}
$$
Note, that
$$
a_n = dfrac{1}{pi}intlimits_{I}lambda(x)cos(nx),dx = dfrac{1}{pi}hat{lambda}(n).
tag{6}
$$
Function $hat{lambda}(t)$ does not vanish only at points $n=2^k,,k=0,1,ldots,$ and
$$
a_{2^k} = dfrac{1}{pi}lim_{trightarrow 2^k}hat{lambda}(t).
tag{7}
$$
Values of first several coefficients are as follows
$$
mathbf{a}=(a_0,,a_1,,a_2,,a_4,,a_8)=biggl(frac1pi,,2.3cdot10^{-1},,7.7cdot10^{-2},,-5.1cdot10^{-3},, 8.1cdot10^{-5},,-3.2cdot10^{-7}biggr)
$$
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
Questions
- Does a rectangular function function $chi_{I}(x)$, which is also a characteristic function of the interval $I$ satisfy the Eq. $(1)?$
- How to construct a fast convergence algorithm to compute values of $f(x)$, like a proposed one?
- 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
integration fourier-analysis recurrence-relations functional-equations fractals
integration fourier-analysis recurrence-relations functional-equations fractals
asked Jan 24 at 11:58
Oleg KravchenkoOleg Kravchenko
465
asked Jan 24 at 11:58
Oleg KravchenkoOleg Kravchenko
465
edited Jan 25 at 14:47
Oleg Kravchenko
asked Jan 24 at 11:58
Oleg KravchenkoOleg Kravchenko
465
asked Jan 24 at 11:58
Oleg KravchenkoOleg Kravchenko
465
asked Jan 24 at 11:58
Oleg KravchenkoOleg Kravchenko
465
465
$begingroup$
Small, pedantic FYI: the TeX commandoperatornamecan 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 thanmathrm).
$endgroup$
– Xander Henderson
Jan 24 at 16:43
$begingroup$
@XanderHenderson Thank you for the notice.
$endgroup$
– Oleg Kravchenko
Jan 24 at 17:53
$begingroup$
@reuns I'm not getting why there is no multiplier 2 in terms $e^{pm ipiomega}?$
$endgroup$
– Oleg Kravchenko
Jan 25 at 13:27
1
$begingroup$
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)
$endgroup$
– reuns
Jan 25 at 13:31
1
$begingroup$
$F$ 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 $1/2$
$endgroup$
– reuns
Jan 25 at 13:38
|
show 2 more comments
$begingroup$
Small, pedantic FYI: the TeX commandoperatornamecan 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 thanmathrm).
$endgroup$
– Xander Henderson
Jan 24 at 16:43
$begingroup$
@XanderHenderson Thank you for the notice.
$endgroup$
– Oleg Kravchenko
Jan 24 at 17:53
$begingroup$
@reuns I'm not getting why there is no multiplier 2 in terms $e^{pm ipiomega}?$
$endgroup$
– Oleg Kravchenko
Jan 25 at 13:27
1
$begingroup$
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)
$endgroup$
– reuns
Jan 25 at 13:31
1
$begingroup$
$F$ 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 $1/2$
$endgroup$
– reuns
Jan 25 at 13:38
$begingroup$
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).$endgroup$
– Xander Henderson
Jan 24 at 16:43
$begingroup$
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).$endgroup$
– Xander Henderson
Jan 24 at 16:43
$begingroup$
@XanderHenderson Thank you for the notice.
$endgroup$
– Oleg Kravchenko
Jan 24 at 17:53
$begingroup$
@XanderHenderson Thank you for the notice.
$endgroup$
– Oleg Kravchenko
Jan 24 at 17:53
$begingroup$
@reuns I'm not getting why there is no multiplier 2 in terms $e^{pm ipiomega}?$
$endgroup$
– Oleg Kravchenko
Jan 25 at 13:27
$begingroup$
@reuns I'm not getting why there is no multiplier 2 in terms $e^{pm ipiomega}?$
$endgroup$
– Oleg Kravchenko
Jan 25 at 13:27
1
1
$begingroup$
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)
$endgroup$
– reuns
Jan 25 at 13:31
$begingroup$
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)
$endgroup$
– reuns
Jan 25 at 13:31
1
1
$begingroup$
$F$ 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 $1/2$
$endgroup$
– reuns
Jan 25 at 13:38
$begingroup$
$F$ 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 $1/2$
$endgroup$
– reuns
Jan 25 at 13:38
|
show 2 more comments
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$begingroup$
Small, pedantic FYI: the TeX command
operatornamecan 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 thanmathrm).$endgroup$
– Xander Henderson
Jan 24 at 16:43
$begingroup$
@XanderHenderson Thank you for the notice.
$endgroup$
– Oleg Kravchenko
Jan 24 at 17:53
$begingroup$
@reuns I'm not getting why there is no multiplier 2 in terms $e^{pm ipiomega}?$
$endgroup$
– Oleg Kravchenko
Jan 25 at 13:27
1
$begingroup$
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)
$endgroup$
– reuns
Jan 25 at 13:31
1
$begingroup$
$F$ 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 $1/2$
$endgroup$
– reuns
Jan 25 at 13:38