Functional differential equation (FDE): $f'(x)=f(x+e)$

I would like to solve the functional differential equation $$f'(x)=f(x-e)$$ I've solved functional equations before, and I've solved differential equations, but I've never solved a functional differential equation before, so please be patient with me.

I've found a class of solutions through guesswork: $$f(x)=Ce^{x/e}$$ If this were just a differential equation, I'd be done, because the highest derivative of $f$ in the DE is $f'$, and I have one arbitrary constant in my class of solutions.

Are there solutions not in this form? When I'm dealing with an FDE, how many arbitrary constants I need to have an exhaustive solution?

• you seem to have en.wikipedia.org/wiki/… and they have reference Schmitt, Klaus, ed. (1972). Delay and Functional Differential Equations and Their Applications. However, so far i do not see an obvious text for beginners – Will Jagy Apr 5 '18 at 23:15
• @WillJagy From personal knowledge, can you tell me if I have the correct number of constants to have an exhaustive solution? – Frpzzd Apr 5 '18 at 23:22
• No, I don't think I have ever studied these. I also am not sure about some important facts: if we have an ordinary differential equation of first order, solution curves, graphed in the plane, are not allowed to cross each other. If two curves touch they are actually the same. For delay differential equations, I really do not know if that applies. – Will Jagy Apr 5 '18 at 23:24
• Check out the paper by Falbo: Some Elementary Methods for Solving Functional Differential Equations. On page 2, section 1.1.1 he describes the method of characteristics to solve your problem. Here is the link: mathfile.net/hicstat_FDE.pdf – JEM Apr 6 '18 at 3:47

The initial value problem has the form $$\tag{*} \begin{cases} f'(x) = f(x - e) & \text{ for } x \in [0, \infty) \\ f(x) = u_0(x) & \text{ for } x \in [-e, 0] \end{cases}$$ where $u_0 \colon [-e, 0] \to \mathbb{R}$ is a continuous (say) function (initial condition).
How to find a solution to $(*)$? Notice that the derivative of the solution for $x \in [0, e]$ equals just the value of $u_0$ at $x - e$, therefore $$f(x) = u_0(0) + \int\limits_{0}^{x} f(u_0(\xi-e)) \, d\xi, \quad x \in [0, e].$$ We repeat this procedure for $x \in [e, 2e]$, taking the restriction of the solution to $[0, e]$ instead of $u_0$. And so on (Myshkis' method of steps).
But if you are interested in the equation $$f'(x) = f(x + e)$$ (as in the title), the situation changes dramatically.