I recently encountered a particular delay PDE in my work, the solution of which corresponds to the Laplace transform of some probability distribution. I'm having trouble to solve this equation. The solution $f(q,s)$ defined on $q\geq 0, s>1$ should verify $f(0,s) = 1 \forall s$ and $f(q, s\to \infty) = \delta\_{q,0}$ (the Kronecker delta). It should also the verify the following PDE
$$\partial\_q f(q,s-1) = -\sqrt{s(2+s)}f(q,s) $$
It looks a bit like a "deformed" Hurwitz zeta function relation (which is $\partial\_q \zeta(q,s) = -s\zeta(q,s+1)$). I looked at some simple ansätze but couldn't come up with anything useful. I also am not sure if the boundary conditions I specified above are enough to obtain a unique solution...


I would be glad if you have any suggestions.
Thanks a lot !



Solving a particular delay PDE $\partial_q f(q,s-1) = -\sqrt{s(2+s)}f(q,s)$
Inspired by your comparison with the Hurwitz zeta function, which has the integral representation
$$ \zeta (s,a)={\frac {1}{\Gamma (s)}}\int \_{0}^{\infty }{\frac {x^{s-1}e^{-ax}}{1-e^{-x}}}\mathrm{d}x \:, $$
I tried to come up with an integral representation for your function $f$. The factor $s$ that comes out when taking the $q$ derivative of $\zeta(s,q)$ arises from the $\Gamma(s)$ in the denominator. To get your factor $\sqrt{s(2+s)}$, we need to change this expression into something of the form
$$ f(q,s) = \frac{1}{\sqrt{\Gamma(s+1)\Gamma(s+3)}} \int\_0^\infty x^{s-1} e^{-q x} G(x) \mathrm{d} x \:, $$
with an unknown function $G$. This expression satisfies your PDE.
The function $G$ can be determined from your condition $f(0,s) = 1$, which actually gives the Mellin transform of $G$,
$$ \int\_0^\infty x^{s-1} G(x) \mathrm{d}x = \sqrt{\Gamma(s+1)\Gamma(s+3)}
= \Gamma(s) s\sqrt{(s+2)(s+1)} \:. $$
In this form, you can obtain $G$ as a power series via Ramanujan's master theorem, which gives
$$ G(x) = -\sum\_{k=3}^\infty \frac{(-x)^k}{(k-1)!}\sqrt{(k-1)(k-2)} \:. $$


