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Consider the following equation (with $f \in C^{\infty}(\mathbb{R})$): $$f'(x)=f(x+\pi/2)$$

This equation is satisfied by $f(x) = A\cos(x) +B\sin(x)$, for any $A,B \in \mathbb{R}$.

Question: What are all the (other) solutions of this equation (if any)?

Harry Peter
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Sebastien Palcoux
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2 Answers2

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The "Ansatz" $f(x):=e^{\lambda x}$ leads to the equation $$\lambda=e^{\lambda\pi/2}$$ with the obvious solutions $\lambda=\pm i$. But there are (probably an infinity) other complex solutions, one of them being $$a\pm ib:=1.0214 \pm 4.86821 \, i$$ (found numerically). The functions $$f(x):=e^{ax}\bigl(A\cos(bx)+B\sin(bx)\bigr)$$ are then new solutions of your delay-differential equation.

Christian Blatter
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    Yes! Your first equation is a particular case of the one solved here. It follows that $\lambda = -\frac{2}{\pi}W_k(-\frac{\pi}{2})$, with $W_k$ a branch of the Lambert W-function. – Sebastien Palcoux Sep 18 '17 at 15:10
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    I was considering this way of solving it too using Fourier transformation, but I haven't found a proof that the solution need to be Fourier transformable. This means that I couldn't rule out the possibility of yet even more solutions to the equation. – skyking Sep 19 '17 at 05:08
  • Using an ansatz indeed allows you to find some solutions, but is not necessarily sufficient to exhaust them, is it ? –  Sep 19 '17 at 08:06
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    @YvesDaoust: These are just the solutions that come immediately to mind. I'm sure the theory alluded to in my answer has a lot to say about this problem. E.g., one could choose $f$ arbitrarily in $[0,\pi/2]$, just making sure that the ends meet in a $C^\infty$ way, and then extend $f$ to all of ${\mathbb R}$ using the DDE. – Christian Blatter Sep 19 '17 at 08:25
  • @ChristianBlatter: this is the interesting part of the question. –  Sep 19 '17 at 08:28
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A heuristic comment. The ansatz above works because the delay operator $f(\cdot) \mapsto f(\cdot+a)$ can be formally written as $e^{aD}$, where $D = \frac{d}{dx}$. This is simply the operator-side manifestation of the Taylor's theorem.

So your equation can be formally written as $(D - e^{\frac{\pi}{2}D}) f = 0$. Considering that the eigenfunctions of $D$ are exponential functions, this operator diagonalizes over the space of exponentials:

$$(\lambda - e^{\frac{\pi}{2}\lambda}) e^{\lambda x} = 0.$$

So it suffices to hunt for values $\lambda$ for which $\lambda - e^{\frac{\pi}{2}\lambda} = 0$ holds. Of course, all this discussion is not so rigorous and the possibility of other form of solutions is still open.

(Though I humbly suspect that the span of the solutions of the form above is in some sense 'dense in $C^{\infty}$' so that any solution is a possibly infinite linear combination of the solutions of the form above.)

Sangchul Lee
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    A similar argument: taking the Laplace transform leads to an expression that has for poles the solutions of the same equation. –  Sep 19 '17 at 08:26
  • For more details on "operator-side manifestation of the Taylor's theorem" I found https://math.stackexchange.com/q/278365/84284 – Sebastien Palcoux Sep 19 '17 at 12:03
  • @YvesDaoust: the assumption $C^{\infty}$ is not strong enough for defining the delay operator with derivative: https://mathoverflow.net/a/275319/34538 – Sebastien Palcoux Sep 21 '17 at 17:14
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    @SebastienPalcoux, I am not claiming that this is the solution, precisely for the reason you mentioned: operator-theoretic approach does not work well in general. Hence it is heuristics. On the other hand, the situation is not that bad considering that $iD$ is a self-adjoint operator on $C_c^{\infty}(\mathbb{R})$ and hence the functional calculus allows $e^{aD}$ to be defined on a broader range of functions, say $L^2(\mathbb{R})$. Of course, for application to our problem this is still not enough. (Any solution of the form above does not lie in $L^2(\mathbb{R})$.) – Sangchul Lee Sep 21 '17 at 17:29