How to solve following linear differential-difference equation?
$$\frac{da_{n}(t)}{dt}=i k na_{n}(t)+G\left\{n(n-1)a_{n-2}(t)-a_{n+2}(t) \right\},~n=0,1,2,\ldots~~~~~(1)$$
where, k and G is a constant. And $i$ is the imaginary unit. The initial conditions are $$a_{n}(0)=\delta_{mn}=\begin{cases} C_{0}~~(n=m) \\ 0~~(n\neq m) \end{cases}$$ where, $C_{0}$ is a constant. $a_{0}(t)$ and $a_{1}(t)$ is unknown function. I want to find $a_{n}(t)$.
If $k=0$, then equation $(1)$ reduce to following equation: $$ \frac{da_{n}(t)}{dt}=G\left\{n(n-1)a_{n-2}(t)-a_{n+2}(t) \right\},~n=0,1,2,\ldots.~~~~~(\mathrm{A}) $$
The general solution of equation $(\mathrm{A})$ is $$ a_{n}(t)=C_{0}\frac{1}{\sqrt{\mathstrut 2\pi}}\left(\frac{n}{2} \right)!~\sqrt[]{\mathstrut 2^{n}}\left\{1+(-1)^{n+1} \right\}(\cosh{2Gt})^{-\frac{3}{2}}(\tanh{2Gt})^{-\frac{1}{2}(n-1)}.~~~~~(\mathrm{B}) $$ Equation $(\mathrm{B})$ satisfy following initial conditions: $$ a_{n}(0)=\begin{cases} C_{0}~~(n=1) \\ 0~~~~~(n\neq 1) \end{cases} ~~~~~~~~~(\mathrm{C}) $$
I have tried Laplace transform to equaton $(1)$.
Multiplying both sides of equation $(1)$ by $\mathrm{e}^{-st}$ and then Integrating for the interval $0$ to $\infty$ to obtain $$ \int_{0}^{\infty}dt~\mathrm{e}^{-st} \frac{da_{n}(t)}{dt}=i k n\int_{0}^{\infty}dt~\mathrm{e}^{-st}a_{n}(t)+G\left\{n(n-1)\int_{0}^{\infty}dt~\mathrm{e}^{-st}a_{n-2}(t)-\int_{0}^{\infty}dt~\mathrm{e}^{-st}a_{n+2}(t) \right\}.~~~~~(2) $$
We define the $U_{n}(s)$ $$ U_{n}(s):=\int_{0}^{\infty}dt~\mathrm{e}^{-st}a_{n}(t).~~~~~(3) $$ Then equation $(2)$ to be $$ sU_{n}(s)-a_{n}(0)=iknU_{n}(s)+ G\left\{n(n-1)U_{n-2}(s)-U_{n+2}(s) \right\}.~~~~~(4) $$ where, we use following integration by parts $$ \int_{0}^{\infty}dt~\mathrm{e}^{-st}\frac{d a_{n}(t)}{dt}=sU_{n}(s)-a_{n}(0). $$
Let's solve equation $(4)$ by using Z-transform.
First, we define unilateral Z-transform $W(s,z)$ as follows:
$$ \mathcal{Z}[U_{n}(s)]= W(s,z):=\sum_{n=0}^{\infty} U_{n}(s)z^{-n}.~~~~~(5) $$
Noting following relations
Differentiation & Time delay $$ \mathcal{Z}[n(n-1)U_{n-2}(s)]= 2z^{-2}W(s,z)-2z^{-1}\frac{\partial W(s,z)}{\partial z}+\frac{\partial^{2}W(s,z)}{\partial z^{2}}, ~~~~~~~~(6) $$ Time advance $$ \mathcal{Z}[U_{n+2}(s)]=z^{2}W(s,z)-z^{2}U_{0}(s)-zU_{1}(s), ~~~~~~~~~(7) $$ Using propaty of $a_{n}(0)=\delta_{mn}$ $$ \mathcal{Z}[a_{n}(0)]=C_{0}z^{-m}, ~~~~~~~~~~(8) $$
we can transform equation $(4)$ as follows: $$ Gz^{2}\frac{\partial^{2}W(s,z)}{\partial z^{2}}+(-2G-ikz^{2})z\frac{\partial W(s,z)}{\partial z} +\left\{z^{2}(-Gz^{2}-s)+2G \right\}W(s,z)+Gz^{3}\left\{zU_{0}(s)+U_{1}(s) \right\} +C_{0}z^{-m+2}=0. ~~~~~~(9) $$ Equation $(9)$ is some kind of Bessel equation.
I'm trying to solve equation $(9)$.
Hint
If: $$ \mathcal{Z}[U_{n}(s)]= W(s,z):=\sum_{n=0}^{\infty} U_{n}(s)z^{-n} $$
Then:
$$ \mathcal{Z}[U_{n-2}(s)]= \sum_{n=0}^{\infty} U_{n-2}(s)z^{-n}=z^{-2}\sum_{n=0}^{\infty} U_{n-2}(s)z^{-(n-2)}=z^{-2}\sum_{n=0}^{\infty} U_{n-2}(s)z^{-(n-2)} $$
Now because $a_{n-1}=0$ and $a_{n-2}0$ (please check this assumption) $$ \mathcal{Z}[U_{n-2}(s)]= \sum_{n=0}^{\infty} U_{n-2}(s)z^{-n}=z^{-2}\sum_{n=0}^{\infty} U_{n-2}(s)z^{-(n-2)}=z^{-2}\mathcal{Z}[U_{n}(s)] = z^{-2}W(s,z) = \Gamma(s,z) $$ And this is the time-shift property. I'll call the result $\Gamma$ because it will soon be convenient. Then: $$ \gamma(s,n) =U_{n-2}(s) $$
Now look at $$ \mathcal{Z}[n(n-1)U_{n-2}(s)]=\mathcal{Z}[n(n-1)\gamma(s,n)] =\mathcal{Z}[n^2\gamma(s,n)]- \mathcal{Z}[n\gamma(s,n)] $$
From the differentiation property: $$ \mathcal{Z}[n\gamma(s,n)] = -\frac{d}{dz} \left[ \Gamma(s,z)\right] =-\frac{\partial}{\partial z}\left[ \Gamma(s,z)\right] =-\frac{\partial}{\partial z}\left[ z^{-2}W(s,z)\right] =\\ 2z^{-3}W(s,z)-z^{-2}\frac{\partial W(s,z)}{\partial z} $$
Now do it again to get $\mathcal{Z}[n^2\gamma(s,n)] $. Then rebuild your equation 6.