I was toying with equations of the type f(x+α)=f′(x) where f is a real function....

kuhse4461a

kuhse4461a

Answered

2022-01-15

I was toying with equations of the type f(x+α)=f(x) where f is a real function. For example if α=π2 then the solutions include the function fλ,μ(x)=λcos(x+μ). Are there more solutions?
On the other hand, if I want to solve the equation for any α, I can assume a solution of the form f(x)=eλx, and find λ 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 ϕ:f(x)f(x+1) adds another one? Is there some litterature about this kind of equations?

Answer & Explanation

Melissa Moore

Melissa Moore

Expert

2022-01-16Added 32 answers

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 α 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,gR is any smooth function satisfying g(n)(α)=g(n+1)(0) for all n0, then there exists a unique extension of g to a smooth solution f defined on the entire real line. For x0, this solution is defined by
f(x)=g(n)(xnα) where n=[xα]
while for x0 it involves iterated integrals of g.
Esther Phillips

Esther Phillips

Expert

2022-01-17Added 34 answers

Integers without large
alenahelenash

alenahelenash

Expert

2022-01-24Added 366 answers

Note that f(x)=eλx is solution of f(x)=f(x+α) if λ=eλα, and this equation is satisfied by infinitely many complex λ if α0, namely λ=wn/α where wn are the branches of Lambert W(α)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?