A problem in Apostol's Calculus [Tom M. Apostol. Calculus, Volume 1, Second Edition (Wiley, 1967). Section 2.19 Exercises, Problem 21, page 125.]

Suppose the existence of a function $f$ with the properties that $f$ is odd, $f'$ is even, $f(5)=7$, and $f'(x)=f(x+5)$. This led me to modify this a bit and study it.

Suppose $a$ is real, $f$ is infinitely differentiable everywhere on $\mathbb{R}, f$ is an odd function, $f(a)=1$, and $f'(x)=f(x+a)$. It is easy to prove from the definition of the derivative that $f'$ is even. If $a=0$, there is no solution, since if there were, $f(0)=-f(0)$, so $f(0)=0$, but $f(0)=1$ by assumption.

In general, note that: $$f'(-x)=f(a-x)=f'(x)=f(x+a)$$ $$=-f(x-a)=-f(-x-a)$$ We also have: $$f(x+2a)=-f(x) \space \text{and }f(2a)=0$$ $$\text{and }f(x+4a)=-f(x+2a)=f(x)$$ so $f$ is periodic of period $4a$. $f'$ is the same as $f$ shifted $|a|$ units to the left or right, so $f'(x+2a)=-f'(x)$ and $f'$ is periodic of period $4a$. $$f'(0)=f(a)=1, f'(a)=f(2a)=0,$$ $$\text{ and }f'(2a)=-f'(0)=-1$$ Since $f'(x) = f(x+a) = f(x-3a)$, we can assume WLOG that $a > 0$ (or that $a < 0$). If $a=\pi/2+2\pi n, f(x)=\sin x$ is an obvious solution.

Now for my questions, none of which I can answer:

  1. If $a=\pi/2+2\pi n$, is there any solution besides $\sin x$?

  2. If $a \ne \pi/2+2\pi n$, is there a solution? Is it unique? Is there a simple formula for it?

The questions $f'(x) = f(x-1)$ and $f'(x) = f(x+1)$ may be pertinent, but so far they haven't helped me much. I have tried to apply Rolle's Theorem to prove uniqueness without success.


1 Answer 1


If $f'(x) = f(x+a)$ and $f(x+4a) = f(x)$, then $f''''(x) = f(x+4a) = f(x)$. The general solution of the differential equation $f''''(x) = f(x)$ is $$f(x) = c_1 e^x + c_2 e^{-x} + c_3 \cos(x) + c_4 \sin(x) $$ For $f$ to be periodic, we must have $c_1 = c_2 = 0$. For $f$ to be odd, $c_3 = 0$. So the only solutions can be $c_4 \sin(x)$, which is periodic with period $2\pi$. The only possibilities for $a$ are then $\pi/2 + 2 \pi n$, and if you want $f(a) = 1$ then $c_4 = 1$ and $f(x) = \sin(x)$.

  • $\begingroup$ Thanks. Stupid of me not to realize this. It would be a little easier to consider $f''(x)=f'(x+a)=f(x+2a)=-f(x)$ with solutions $c_1 \cos x + c_2 \sin x$. This also means that Apostol's problem concerns a function that does not exist! $\endgroup$ Commented Mar 3, 2023 at 14:45

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