Existence of a real valued function satisfying $f'(x)=f(x+1)$ where $f(x)\ne0$

Question

I am wondering about the existence of a real valued function that satisfies $f'(x)=f(x+1)$ other than the trivial solution $f(x)=0$. I thought a likely solution would be a repeating (perhaps sinusoidal) function, where $f(0)=0, f'(0)=1,f'(1)=0,f'(2)=-1,f'(3)=0$, etc.

I tried to find such a function on Desmos, however I failed after realizing that for a Sin function the derivative at $0$ is always greater than the value of the maximum since between the zero and the maximum the function is strictly decreasing.

In my exploration, I was able to find a solution to the similar differential equation, $f'(x)=f(x+1)*{\pi\over2}$, with the solution $f(x)=\sin{({\pi\over2}x)}$. This was easier to find due to the coefficient of $\pi\over2$, which removes the obstacle of the previous inequality. Perhaps this provides hope for a solution to the other one.

Overall, the question is, is there any functional solution for the equation $f'(x)=f(x+1)$ where $f(x)\ne0$?

Answer 1

Let $f$ be a $C^{\infty} $ function with compact support in $(0,1)$ and define $f(0)$ and $f(1)$ to be $0$. For $n \leq x \leq n+1$ define $f(x)= f^{(n)}(x-n)$.

Answer 2

This is an example of delay differential equations. Make an ansatz that $f(x) = e^{ax}\cos(bx)$ is a solution. Plugging this to the equation,

$$ e^{ax}(a\cos(bx) - b\sin(bx)) = e^{ax+a}(\cos(bx)\cos(b)-\sin(bx)\sin(b)). $$

So it suffices to find a pair of numbers $(a, b)$ for which

$$ a = e^a \cos(b) \quad \text{and} \quad b = e^a \sin (b). $$

Or, if we write $\alpha = a + ib$, then $\alpha = e^{\alpha}$. It can be proved that this equation has many solutions. (This is an instance of the method using characteristic equation.) One such solution is numerically given by

$$ (a, b) = (0.576412723031435283\cdots, 0.374699020737117494\cdots).$$

although it is likely impossible to provide a closed form in terms of elementary functions.

Answer 3

This is an example of a linear delay differential equation. Given a finite set of solutions, then any linear combination is also a solution. Now define $\, f_k(x) := \exp(\Re(t_k)x) \cos(\Im(t_k)x)\,$ where $\, t_k := -W_k(-1) \,$ and where $\, W_k(z) \,$ is the $k$-th branch of the Lambert W function. It is easy to check that $\, f_k(x), \,$ for any integer $\,k,\,$ is a solution of $\, f'(x) = f(x+1).\,$