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While working on my answer to another question, I noticed that the function $$ f(x) = \int_0^\infty t^{x-t} dt $$ satisfies the delay differential equation $$ f'(x) = x f(x-1) - f(x) $$ for $x>0$. To prove this, observe that $f'(x) = \int_0^\infty (\log t) t^{x-t} dt$. The equation rearranges to give $0 = xf(x-1) - f(x) - f'(x) = \int_0^\infty t^{x-t}\left(\frac{x}{t} - 1 - \log t)\right)dt$. This integrates to 0 because the integrand is $\frac{d}{dt} t^{x-t}$.

Is there a way that you could construct the function $f$ from this equation? What about other analytic solutions?

In the case of ordinary differential equations, I know there are some tricks, and in general $f'(x) = g(x,f(x))$ has a 1-parameter solution space. For DDEs it's a lot more complicated. Typically you're given an initial condition and you solve by piecewise extensions. If you're interested in analytic solutions and you don't care about initial conditions, that's not much help though. My question is: can you construct an analytic solution to a delay differential equation? In this example, given the equation $f'(x) = x f(x-1) - f(x)$, is there a way to arrive at an analytic solution that's not guess-and-check? Furthermore, it seems intuitive that if one analytic solution exists, there should be an infinite dimensional family of solutions (similar to how the family of solutions to the difference equation $f(x) = x f(x-1)$ is essentially parametrized by 1-periodic functions). For example, solutions to $f'(x) = f(x-1)$ have a Fourier series-like form $$f(x) = \sum_{\{c \space:\space c e^c = 1\}} a_c e^{c x}$$

For $f'(x) = xf(x-1) - f(x)$, I have discovered one particular solution. How would you going about constructing other analytic solutions? Is there anything analogous to $e^{c x}$ for $f'(x) = f(x-1)$?

Update (5/2/2020): As I suspected, there is an infinite dimensional space of solutions, which can be naturally parametrized by 1-periodic functions. Given any 1-periodic function $g$, then $$ f_g(x) = \int_0^\infty t^{x-t} g(x-t) dt $$ solves the same DDE. (This can be proven similarly to above). It's clear that $f_g$ is going to be analytic if $g$ is. I suppose showing only if wouldn't be difficult either. So that answers one part of the question. I'm still curious how one would arrive at this solution from the equation directly. It's easy to show this function solves the DDE, but how would one construct $f$ given $f'(x) = xf(x-1) - f(x)$.

Dark Malthorp
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    Are you sure that it satisfies this equation, $t^{x-t}$ does not? In any case, there is no "the function" for DDE. You can specify $f$ on $[0,1]$ arbitrarily and then find a solution for $x\geq1$ by integrating the equation, see Falbo. – Conifold Jan 11 '20 at 00:13
  • Yes it satisfies the equation. I am aware that there is no "the function" for DDEs, let me update the question to clarify. – Dark Malthorp Jan 11 '20 at 00:17
  • I do not see how, maybe you can explain that too. – Conifold Jan 11 '20 at 00:19
  • As for how you can construct other solutions, notice that the equation is linear, so $Cf'$ satisfies it for any constant $C$. – Simply Beautiful Art Jan 11 '20 at 02:30
  • Yes, obviously. I suspect that (analogous to $f'(x) = f(x-1)$) there should be some linearly independent solutions, but I don't know how to prove that or prove false. – Dark Malthorp Jan 11 '20 at 07:27
  • $f'(x) +f(x) = f(x-1)$ and $f'(x)+f(x)=xf(x-1)$ seem to have very different analytic structure. For the former the Laplace transform of solutions is of the form $\frac{N(p)}{p+1-e^{-p}}$, and the residue expansion is where the "Fourier series" comes from. But for the latter it is $N(p)\exp(p(1-e^p))$ and the factor has no poles, only essential singularity at $\infty$. You want some suitable expansion for $\exp(p(1-e^p))$ to invert, maybe some sort of transseries. This seems rather advanced, so you may want to ask on Math Overflow. – Conifold Jan 12 '20 at 00:22
  • I'm a little confused why you would even be able to take a Laplace transform here, since this $f(t)\rightarrow\infty$ faster than exponentially. – Dark Malthorp Jan 12 '20 at 15:06
  • One can always take it in a distributional sense, and images provide clues about originals regardless of convergence. But if you want convergence you can treat $f(x)$ as images rather than originals, i.e. apply the inverse transform. The difference remains. De Bruijn has a paper that may be useful to you. He discusses finding analytic solutions to equations of the form $F'(x)=K(x)F(x-1)$, although only exponential $K(x)$ is considered explicitly. Yours is convertible into this form with $K(x)=ex$ by a substitution $F(x)=e^{x}f(x)$. – Conifold Jan 14 '20 at 01:41

1 Answers1

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Consider instead:

$$f'(t)=tf(t-1)H(t-1)-f(t)$$

where $H$ is the Heaviside step function.

Take the Laplace transform:

$$sF(s)-f(0)=-e^{-s}F'(s)+e^{-s}F(s)-F(s)$$

Solving this gives

$$F(s)=f(0)g(s)\exp(s(1-e^s))\\g(s)=\int\exp(se^s)~\mathrm ds\color{gray}{+C}$$

Take the inverse Laplace transform:

$$f(t)=\frac{f(0)}{2\pi i}\oint g(s)\exp(s(t+1-e^s))~\mathrm ds$$

Substitute $u=e^s$:

$$f(t)=\frac{f(0)}{2\pi i}\oint g(\ln(u))u^{t-u}~\mathrm du$$

which is remarkably close to your solution. It is probably the case that the integral is vanishing, aside from the constant of integration from $g$, leaving us with

$$f(t)=Cf(0)\oint u^{t-u}~\mathrm du$$

but I've yet to prove such.

Simply Beautiful Art
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  • What is odd is that doing the Laplace transform when the equation has constant coefficients also seemingly leads to a single solution. But one can explicitly find infinitely many of them (exponentials with exponents given by branches of the Lambert function). One can even adjust the coefficients so that they are all integrable on $[0,\infty)$. I wonder why. – Conifold Jan 11 '20 at 04:09
  • I'm skeptical that $\oint u^{t-u} du$ converges. The integrand behaves very poorly in the complex plane. – Dark Malthorp Jan 11 '20 at 16:28