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What functions $f$ satisfy the equation $$f'(x)=\frac{f(\sqrt{1-x^2}) - f(x)}x$$

Could you kindly show how you arrived at the function?

Bumblebee
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David Okogbenin
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1 Answers1

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$$f'(x)=\frac{f(\sqrt{1-x^2}) - f(x)}x$$ $f(x)=g(x^2)$

$f'(x)=2xg'(x^2)$ $$2xg'(x^2)=\frac{g(1-x^2) - g(x^2)}{x}$$ $$2x^2g'(x^2)=g(1-x^2) - g(x^2)$$ $x^2=s$ $$2sg'(s)=g(1-s) - g(s)$$ $Dg(s)=g'(s)$

$Ag(s)=g(1-s)$ $$(2sD-A+1)g=0$$

Operator Analysis

$$\left(2sD-\exp\left(\pi isD\right)\exp(D)+1\right)g=0$$ Baker-Campbell-Hausdorff Formula

This solves $\ln(\exp(X)\exp(Y))$ which is usually $X+Y$ except when $XY-YX\neq 0$

$\pi isD*D-D*\pi isD=\pi isD^2-\pi i(1+sD)D=\pi isD^2-\pi iD-\pi isD^2=-\pi iD$

$\exp(\pi isD)\exp(D)=\exp(\pi i(s-\frac{1}{2})D)$

$$\left(2sD-\exp\left(\pi i\left(s-\frac{1}{2}\right)D\right)+1\right)g=0$$

There are infinitely many solutions to this but the easiest is $g=0$.

empCarnivore
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  • I’ll have to read up the links you gave as I’m a high school student and I got lost from “Operator Analysis”. What I do know though is that g=k satisfies the equation where k is any real number. Are u saying there are still infinite solutions other than g=k – David Okogbenin Mar 21 '20 at 22:38
  • Yep. Although why is a bit complicated. It partially has to do with how f(x+1)=f(x) has infinitely many solutions. – empCarnivore Mar 21 '20 at 23:20
  • f(x) for f(x+1)=f(x) is a periodic function of a periodicity( don’t know if that’s the word) of 1 right? So something like f(x) = sin(2pix) or f(x)=k. How does this relate to my question? If it’s still too complex then it’s okay. I’ve found out what I wanted to. But I’d still like to get it though. – David Okogbenin Mar 21 '20 at 23:31