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$sin(x)$ and $cos(x)$ both satisfy the Delay Differential Equation

$f'(x) = f(x+a)$ with $a = \pi/2$

I have consulted other questions on this forum and understood there are more solutions to the Delay Differential Equation $f'(x) = f(x+a)$. But are there any others that oscillate, and is the period always $4a$?

For this particular solution "$2\pi$" and thus "$4a$" emerges as the period $w$. Obviously many solutions just keep going up or down, related to the real exponential function $e^{cx}$, but some solutions seem to "oscillate".

A function like $\sin(5x)$ then has a period which is 5 times smaller but still satisfies $w = 4a$.

I am trying to understand what this means in terms of "the meaning of $\pi$". Is "$\pi$" special in the sense that it's the only period where you can construct a function that oscillates in a certain manner?

I also have no idea how this relates to the second order derivative of $f$, which for $\sin$ obviously has a relationship with itself, in the form of a simple second order differential equation.

I spent time with all answers regarding delay differentials and the Wikipedia page for them, but couldn't really get any further in understanding whether or not this insight reveals something about pi - something I realise that will most likely be tautological to its geometric meaning.

John B
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buddhabrot
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  • For any ( non constant) polynomials $f$ and $g$ it is not true also for $ \frac{f}{g}$. For $e^x$ and $a=0$. – amir bahadory Dec 15 '19 at 15:57
  • Start by considering all $\lambda\in\mathbb C$ such that $f(x)=e^{\lambda x}$ is a solution. Actually, all solutions are generated by these ones (you really need to have a look at a good source because this is somewhat technical). Hint: take $\lambda=i$ and $\lambda=-i$. – John B Dec 16 '19 at 00:41

1 Answers1

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Taking for convenience

$$ f'(t) = f(t+a) $$

and transforming Laplace we have

$$ sF(s)-e^{a s}F(s) = f(0) $$

or

$$ F(s) = \frac{f(0)}{s-e^{as}} $$

now if $s_k = x_k + j y_k$ are such that $s_k - e^{a s_k} = 0$ we have

$$ f(t) = f(0)\sum_{k=0}^{\infty}\left(\alpha_k\sin (y_k t)+\beta_k \cos (y_k t)\right)e^{x_k t} $$

where $\alpha_k,\ \beta_k$ are coefficients of the residue expansion for $\frac{1}{s-e^{as}}$.

The zeroes for $s-e^{a s}=0$ can be located by solving

$$ \cases{ x-e^{a x}\cos(a y) = 0\\ y-e^{a x}\sin(a y) = 0 } $$

The zeroes are contained also in the trace of

$$ x^2+y^2 = e^{2a x} $$

for $a = \frac{\pi}{2}$ the left-most zero is located at

$$ 0^2+y^2 = 1\Rightarrow \alpha_0\cos t+\beta_0 \sin t $$

It is the first expansion pole, and the only pure periodic term. The following terms are exponentially weighted and then are not purely periodic.

Follow a plot showing in blue the locus for $x-e^{a x}\cos(a y) = 0$ in red the locus for $y-e^{a x}\sin(a y) = 0$ and in black the locus $x^2+y^2 = e^{2a x}$

enter image description here

This graphic was made for $a = \frac{\pi}{2}$

NOTE

The parametric values $a = (4k+1)\frac{\pi}{2}$ for $k = 0, 1, 2,\cdots,$ are the values such that there exists a pole which gives pure oscillations.

Cesareo
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  • Thanks for this, I need to unwrap this. I still need to play with it to understand whether or not $\pi$ has a special emergence here, or if there are other values of $a$ that cause periods – buddhabrot Dec 16 '19 at 10:03
  • As long as $x = 0$ in $x^2+y^2=e^{2x}$ we have necessarily $y = \pm 1$ then the unique values for a pure oscillating component are at $a=(4k+1)\frac{\pi}{2}$. – Cesareo Dec 16 '19 at 10:56
  • Thanks, that part of the reduction wasn't clear to me but it is clear now. – buddhabrot Dec 16 '19 at 11:21
  • Due to my limited grasp of maths I don't yet see the link between this and the second order differential equation y'' = -y From a geometrical perspective it seems to be tightly linked to the construction of a curve that "looks ahead" (or back) for its derivative – buddhabrot Dec 16 '19 at 11:25