kuhse4461a

Answered

2022-01-15

I was toying with equations of the type $f(x+\alpha )={f}^{\prime}\left(x\right)$ where f is a real function. For example if $\alpha =\frac{\pi}{2}$ then the solutions include the function ${f}_{\lambda ,\mu}\left(x\right)=\lambda \mathrm{cos}(x+\mu )$ . 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$\varphi :f\left(x\right)\to f(x+1)$ adds another one? Is there some litterature about this kind of equations?

On the other hand, if I want to solve the equation for any

I was wondering: is the set of the solutions of dimension 2 because the derivative operator creates one dimension and the operator

Answer & Explanation

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 $\alpha$ 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 ${g}^{\left(n\right)}\left(\alpha \right)={g}^{(n+1)}\left(0\right)$ 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

$f\left(x\right)={g}^{\left(n\right)}(x-n\alpha )$ where $n=\left[\frac{x}{\alpha}\right]$

while for$x\le 0$ it involves iterated integrals of g.

The equation you have given has an infinite dimensional space of solutions. In particular, if

while for

Esther Phillips

Expert

2022-01-17Added 34 answers

Integers without large

alenahelenash

Expert

2022-01-24Added 366 answers

Note that $f(x)={e}^{\lambda x}$ is solution of ${f}^{\prime}(x)=f(x+\alpha )\text{}\text{if}\text{}\lambda ={e}^{\lambda \alpha}$ , and this equation is satisfied by infinitely many complex $\lambda \text{}\text{if}\text{}\alpha \ne 0$ , namely $\lambda =-{w}_{n}/\alpha $ where ${w}_{n}$ are the branches of Lambert $W(-\alpha )$

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