Has $f'(x)=f(x+1/2)$ non-oscillating solutions?

Question

Question: Is there a non-oscillating solution of $f'(x)=f(x+1/2)$ ?

With oscillating I mean "having infinitely many zeros". This question builds on this other question which discusses the differential equations $$(*)\quad f'(x)=f(x+\lambda)$$ for general $\lambda\in\Bbb R$. It was found that if $z=a+bi$ is a solution of $z=\exp(\lambda z)$, then

$$f(x)=e^{ax}\cos(bx)$$

is a solution of $(*)$. For $\lambda\le1/e$ the value $z$ can be chosen real, so the corresponding solution is not oscillating. For $\lambda>1/e$ however, no real valued solution $z$ exists, so this examplary solution contains $\cos(bx)$, therefore oscillates.

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For the interested reader

Why I am asking for $\lambda=1/2$? It is just an example. I am actually interested in a general argument for $\lambda\in(1/e,1)$. For $\lambda\ge 1$ I already know that there must be infinitely many zeros. The argument is pretty straight forward and uses the mean value theorem.

Proof.

Let $\lambda> 0$. If there are only finitely many zeros, then $f$ finally becomes positive/negative. With every solution $f$ there is another solution $-f$, hence we can say $f$ becomes positive in the end. So we have $f(x)>0$ for all $x\ge x_0$. Note that this implies $f^{(n)}(x_0)=f(x_0+n\lambda)>0$. So the function is not only increasing, but increasing with an increasing rate and so on.

The mean value theorem states

$$f'(\xi)=\frac{f(x_0+\lambda)-f(x_0)}\lambda$$

for some $\xi\in[x_0,x_0+\lambda]$. Because $f''(x_0)>0$ we have

$$f(x_0+\lambda)=f'(x_0)\le f'(\xi)=\frac{f(x_0+\lambda)-f(x_0)}\lambda$$

which rearranges to $(1-\lambda)f(x_0+\lambda)\ge f(x_0)$. Assuming $\lambda > 1\Rightarrow1-\lambda<0$ gives the immediate contradiction

$$f(x_0+\lambda)\le \frac{f(x_0)}{1-\lambda}<0.$$

The case $\lambda =1$ gives a similar contradiction. $\square$

This proof yields absolutely no hint on any magical barrier for non-oscillating solutions at $\lambda=1/e$. For $\lambda < 1$ we find $f(x_0+\lambda)\ge f(x_0)/(1-\lambda)$ which is no problem for at least exponentially growing functions.

Answer 1

There is no non-oscillating solution for $\lambda>1/e$. My solution is a straightforward generalisation of the considerations in the problem showing that there is no non-oscillating solution for $\lambda\geq1$.

As in that proof, we assume that there exists a solution $f$ having only finitely many zeros and can conclude that $f$ and all its derivatives are positive for $x\geq x_0$ if $x_0$ is sufficiently large. Then we can use Taylor's formula with remainder and obtain for any positive $n$ $$\tag{1}\label{ineq}f(x_0+n\lambda)-f(x_0)=\sum_{k=1}^n\frac{f^{(k)}(x_0)}{k!}(n\lambda)^k+\frac{f^{(n+1)}(\xi)}{(n+1)!}(n\lambda)^{n+1} \geq \frac{f^{(n)}(x_0)}{n!}(n\lambda)^n=f(x_0+n\lambda)\frac{(n\lambda)^n}{n!}.$$ Now, by Stirling's formula $n!\sim (n/e)^n\sqrt{2\pi n}$ as $n\to\infty.$ Hence $$\frac{(n\lambda)^n}{n!}\sim \frac1{\sqrt{2\pi n}} (e\lambda)^n.$$ Therefore, if $\lambda>1/e$ then there exists a positive $n$ such that $\frac{(n\lambda)^n}{n!}>1$. This leads to a contradiction in (\ref{ineq}).

So the "magical barrier" $1/e$ comes from Stirling's formula!

Actually, this solution also gives an upper bound on the lengths of intervals on which solutions of $f'(x)=f(x+\lambda)$ with $\lambda>1/e$ can have constant sign. In the case of $\lambda=1/2$, we calculate that $(n\lambda)^n/n!>1$ for $n=6$ already. Hence there can be no solution with constant sign on any interval of length $6.5=13*0.5$, because otherwise the above proof would lead to a contradiction on that interval. (Observe that the proof needs that $f^{(n+1)}(\xi)>0$ on the interval $]x_0,x_0+n\lambda[.$)

My colleague found a different solution. Consider the real number $a\geq1$ which is the infimum over all quotients $y(\lambda)/y(0)$, where $y:[0,\infty[\to]0,\infty[$ is a positive solution of the equation. Then for any such solution $f$, we have $f((n+1)\lambda)\geq a\, f(n\lambda)$, because $\tilde y(x)=f(x+n\lambda)$ also is a solution. By induction, we obtain $f^{(n)}(0)=f(n\lambda)\geq a^n f(0)$ for all $n$. By Taylor's formula, we obtain similarly to (\ref{ineq}) that $$f(\lambda)\geq\sum_{n=0}^\infty \frac{a^n}{n!}\lambda^n\,f(0)=\exp(a\lambda)f(0).$$ This proves that for all positive solutions $f$ on $[0,\infty[$, we have $f(\lambda)/f(0)\geq \exp(a\lambda)$. This inequality then also holds for the infimum $a$ of all the quotients $f(\lambda)/f(0)$, where $f$ is a positive solution on $[0,\infty[$. Hence we have proved that $a\geq\exp(a\lambda)$ and therefore $\lambda\leq \log(a)/a\leq \max\{\frac1a\log(a)\mid a\geq1\}=1/e.$

Observe that this second proof needs an infinite interval on which $f$ is positive. Hence it does not provide an upper bound on the lengths of intervals on which solutions of $f'(x)=f(x+\lambda)$ with $\lambda>1/e$ can have constant sign.