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How does one find all the differentiable functions $f\colon \mathbb{R} \to \mathbb{R}$ which satisfy the equation $$ f'(x)=af(x+b),\quad \text{for}\quad a,b \in \mathbb{R}? $$ I see that functions of the form $\alpha e^{\beta x}$ and $\alpha \sin(\beta x + \gamma)$ will be solutions, but I'm not sure how to exhaustively find all of the solutions.

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  • $\begingroup$ Isn't this equivalent to the initial value problem $$\begin{cases} y'=ay\\ y(0)=f(b)\end{cases}$$? $\endgroup$
    – leo
    Dec 31, 2013 at 19:55
  • $\begingroup$ @leo No. For example, that would not have $\sin(x)$ as a solution. $\endgroup$
    – Malper
    Dec 31, 2013 at 21:08

2 Answers 2

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This equation is a Delay Differential Equation for $b<0$, and it can be solved with special type "initial data" - meaning that, instead of looking for a solution satisfying an initial condition of the form $f(t_0)=x_0$, i.e., prescribing the value of the solution at one point, we need to prescribe the solution in a whole interval, in this case an interval of length $|b|$.

For example: If $f$ is known in the interval $[-|b|,0]$, then using the equation we can obtain its values in the interval $[0,|b|]$, next in $[|b|,2|b|]$, and recursively in $[0,\infty)$.

Simple example: Assume that $b<0$, and $\beta=-b$. Let $f\rvert_{[0,\beta]}=1$. Then \begin{align} f\rvert_{[\beta,2\beta]} &=a(x-\beta)+\beta, \\ f\rvert_{[2\beta,3\beta]} &=\frac{a^2}{2!}(x-2\beta)^2+a\beta (x-2\beta)+a\beta+\beta, \\ f\rvert_{[3\beta,4\beta]} &=\frac{a^2}{3!}(x-3\beta)^3+\frac{a^2\beta}{2} (x-3\beta)^2+(a\beta+\beta)(x-3\beta)+\frac{a^2b^2}{2}+a\beta^2++a\beta+\beta, \end{align} and in general, once $f$ is known in $[(k-1)\beta,k\beta]$, then $$ f(x)=a\int_{k\beta}^x f(s-\beta)\,ds \,\,\,\,\,\text{for all $\,\,\,x\in[k\beta,(k+1)\beta]$.} $$

If $b>0$, we can do the same thing but backwards.

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Try working with the Fourier transform. The Fourier transform of a translated function like $f(x+b)$ is the function's transform multiplied by a unit-magnitude phase factor, something like $\exp(-i b k)$ while the derivative $f'$ is described in the Fourier world as multiplication by k. I imagine a bit of fooling around with that will turn up useful solutions.

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    $\begingroup$ try it! what happens? $\endgroup$
    – GEdgar
    Dec 31, 2013 at 20:02

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