Solve $\frac{d}{dx}f(x)=f(x-1)$

Question

I am trying to find a function such that

$\dfrac{d}{dx}f(x)=f(x-1)$

Is there such function other than $0$ ?

Answer 1

Certainly. Let $b$ be the unique real number such that $b=e^{-b}.$ Then for any real $a,$ the function $$f(x)=ae^{bx}$$ satisfies the desired property. In fact, for real-valued functions on the reals, only functions of this form will satisfy the desired property. (As achillehui points out in the comments, there are other alternatives if we consider complex-valued functions.)

In particular, $b=W(1)$, where $W$ is the Lambert W function. $W(1)$ is also sometimes known as the Omega constant.

Answer 2

$f (x)=ce^{ax}$ will give solutions as well. Both differentiation and horizontal shift of the function will act to scale the initial function and you can solve for $a $

Answer 3

If $f : [0;1]$ is any smooth function with $f^{(k+1)}(1) = f^{(k)}(0)$ for all $k$, then we can extend it to a function $f : \Bbb R \to \Bbb R$ by induction with $f(x) = f(1) + \int_0^{x-1} f(t)dt$ if $x \ge 1$ and $f(x) = f'(x+1)$ if $x \le 0$. This new function is a solution to $f'(x) = f(x-1)$.

Conversely, if $f$ is a solution, its restriction to $[0;1]$ is smooth and satifies $f^{(k+1)}(1) = f^{(k)}(0)$ for all $k$, and $f$ is the extension of its restriction. So you actually have a bijection between solutions and smooth functions on $[0;1]$ satisfying $f^{(k+1)}(1) = f^{(k)}(0)$ for all $k$.

Answer 4

Here is an intuition(!): Let us write $D = \frac{d}{dx}$. If $f$ is analytic,

$$ f(x+h) = \sum_{n=0}^{\infty} \frac{h^{n}}{n!} D^{n} f(x) = e^{hD} f(x) $$

at least in a formal sense (and a genuine sense in view of functional-analysis context). So if you want to solve the equation

$$ Df(x) = f(x-1) = e^{-D}f(x), $$

you are tempted to solve the equation $D = e^{-D}$, or equivalently, $De^{D} = 1$. Since an eigenfunction of $D$ corresponding to an eigenvalue $\lambda$ is $e^{\lambda x}$, passing to this eigenfunction, it amounts to solve the equation $\lambda e^{\lambda} = 1$.