Solving a particular delay PDE $\partial_q f(q,s-1) = -\sqrt{s(2+s)}f(q,s)$

Question

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 !

Answer 1

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 $$ G(x) = -\sum_{k=3}^\infty \frac{(-x)^k}{(k-1)!}\sqrt{(k-1)(k-2)} \:. $$