kuhse4461a

2022-01-15

I was toying with equations of the type $f\left(x+\alpha \right)={f}^{\prime }\left(x\right)$ where f is a real function. For example if $\alpha =\frac{\pi }{2}$ then the solutions include the function ${f}_{\lambda ,\mu }\left(x\right)=\lambda \mathrm{cos}\left(x+\mu \right)$. Are there more solutions?
On the other hand, if I want to solve the equation for any $\alpha$, I can assume a solution of the form $f\left(x\right)={e}^{\lambda x}$, and find $\lambda$ as a complex number that enables me to solve the equation...
I was wondering: is the set of the solutions of dimension 2 because the derivative operator creates one dimension and the operator $\varphi :f\left(x\right)\to f\left(x+1\right)$ adds another one? Is there some litterature about this kind of equations?

Melissa Moore

Expert

This sort of thing is known as a delay differential equation, and there are lots and lots of books on this subject. They're called ''delay'' because the constant $\alpha$ is usually chosen to be negative, so that f′(x) depends on the values of f at earlier times.
The equation you have given has an infinite dimensional space of solutions. In particular, if $g|0,g\mid \to R$ is any smooth function satisfying ${g}^{\left(n\right)}\left(\alpha \right)={g}^{\left(n+1\right)}\left(0\right)$ for all $n\ge 0$, then there exists a unique extension of g to a smooth solution f defined on the entire real line. For $x\ge 0$, this solution is defined by
$f\left(x\right)={g}^{\left(n\right)}\left(x-n\alpha \right)$ where $n=\left[\frac{x}{\alpha }\right]$
while for $x\le 0$ it involves iterated integrals of g.

Esther Phillips

Expert

Integers without large

alenahelenash

Expert

Note that $f\left(x\right)={e}^{\lambda x}$ is solution of , and this equation is satisfied by infinitely many complex , namely $\lambda =-{w}_{n}/\alpha$ where ${w}_{n}$ are the branches of Lambert $W\left(-\alpha \right)$