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I have the following equation:

$(aq^n+b+s)C_n(s)=aq^nC_{n+1}(s)+bC_{n-1}(s)$ , for $n\geq1$.

$C_0(s)=1$ , and for all $s\geq 0$ we have $0\leq C_n(s)\leq1$.

$a>0$, $b>0$, $0\leq q\leq1$.

In fact, $C_n(s)$ is a Laplace Transform of a non-negative RV, for all n.

Any thoughts on how to solve this equation?

Thanks!

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Try let $C_n(s)=\int_{-\infty}^\infty q^{nt}K(t,s)~dt$ or let $C_n(s)=\sum_{-\infty}^\infty q^{nt}K(t,s)$ . –  doraemonpaul Sep 6 '12 at 11:54

1 Answer 1

Case $1$: $q=0$

Then $(b+s)C_n(s)=bC_{n-1}(s)$

$C_n(s)=\dfrac{bC_{n-1}(s)}{b+s}$

$C_n(s)=\dfrac{b^n\Theta(n)}{(b+s)^n}$ , where $\Theta(n)$ is an arbitrary periodic functions with unit period

$C_0(s)=1$ :

$\Theta(0)=1$

$\therefore C_n(s)=\dfrac{b^n\Theta(n)}{(b+s)^n}$ , where $\Theta(n)$ is an arbitrary periodic functions with unit period and $\Theta(0)=1$

Case $2$: $q=1$

Then $(a+b+s)C_n(s)=aC_{n+1}(s)+bC_{n-1}(s)$

$aC_{n+1}(s)-(a+b+s)C_n(s)+bC_{n-1}(s)=0$

The auxiliary equation is

$a\lambda^2-(a+b+s)\lambda+bC_{n-1}(s)=0$

$\lambda=\dfrac{a+b+s\pm\sqrt{(a+b+s)^2-4ab}}{2a}$

$\therefore C_n(s)=\begin{cases}\Theta_1(n)\left(\dfrac{a+b+s+\sqrt{(a+b+s)^2-4ab}}{2a}\right)^n+\Theta_2(n)\left(\dfrac{a+b+s-\sqrt{(a+b+s)^2-4ab}}{2a}\right)^n&\text{when}~(a+b+s)^2-4ab\neq0\\\Theta_1(n)n\left(\dfrac{a+b+s}{2a}\right)^n+\Theta_2(n)\left(\dfrac{a+b+s}{2a}\right)^n&\text{when}~(a+b+s)^2-4ab=0\end{cases}$ , where $\Theta_1(n)$ and $\Theta_2(n)$ are arbitrary periodic functions with unit period

$C_0(s)=1$ :

$\begin{cases}\Theta_1(0)+\Theta_2(0)=1&\text{when}~(a+b+s)^2-4ab\neq0\\\Theta_2(0)=1&\text{when}~(a+b+s)^2-4ab=0\end{cases}$

$\therefore C_n(s)=\begin{cases}\Theta_1(n)\left(\dfrac{a+b+s+\sqrt{(a+b+s)^2-4ab}}{2a}\right)^n+\Theta_2(n)\left(\dfrac{a+b+s-\sqrt{(a+b+s)^2-4ab}}{2a}\right)^n&\text{when}~(a+b+s)^2-4ab\neq0\\\Theta_1(n)n\left(\dfrac{a+b+s}{2a}\right)^n+\Theta_2(n)\left(\dfrac{a+b+s}{2a}\right)^n&\text{when}~(a+b+s)^2-4ab=0\end{cases}$ , where $\Theta_1(n)$ and $\Theta_2(n)$ are arbitrary periodic functions with unit period and $\begin{cases}\Theta_1(0)+\Theta_2(0)=1&\text{when}~(a+b+s)^2-4ab\neq0\\\Theta_2(0)=1&\text{when}~(a+b+s)^2-4ab=0\end{cases}$

Case $3$: $q\neq0$ and $q\neq1$

Then $(aq^n+b+s)C_n(s)=aq^nC_{n+1}(s)+bC_{n-1}(s)$

Let $C_n(s)=\int_{-\infty}^\infty q^{nt}K(t,s)~dt$ ,

Then $(aq^n+b+s)\int_{-\infty}^\infty q^{nt}K(t,s)~dt=aq^n\int_{-\infty}^\infty q^{(n+1)t}K(t,s)~dt+b\int_{-\infty}^\infty q^{(n-1)t}K(t,s)~dt$

$aq^n\int_{-\infty}^\infty q^{nt}K(t,s)~dt+(b+s)\int_{-\infty}^\infty q^{nt}K(t,s)~dt-aq^n\int_{-\infty}^\infty q^{(n+1)t}K(t,s)~dt-b\int_{-\infty}^\infty q^{(n-1)t}K(t,s)~dt=0$

$\int_{-\infty}^\infty aq^{n(t+1)}K(t,s)~dt+\int_{-\infty}^\infty(b+s)q^{nt}K(t,s)~dt-\int_{-\infty}^\infty aq^{n(t+1)}q^tK(t,s)~dt-\int_{-\infty}^\infty bq^{nt}q^{-t}K(t,s)~dt=0$

$\int_{-\infty}^\infty aq^{nt}K(t-1,s)~dt+\int_{-\infty}^\infty(b+s)q^{nt}K(t,s)~dt-\int_{-\infty}^\infty aq^{nt}q^{t-1}K(t-1,s)~dt-\int_{-\infty}^\infty bq^{nt}q^{-t}K(t,s)~dt=0$

$\int_{-\infty}^\infty((b+s-bq^{-t})K(t,s)+a(1-q^{t-1})K(t-1,s))q^{nt}~dt=0$

$\therefore(b+s-bq^{-t})K(t,s)+a(1-q^{t-1})K(t-1,s)=0$

$K(t,s)=-\dfrac{a(1-q^{t-1})K(t-1,s)}{b+s-bq^{-t}}$

$K(t,s)=\theta(t)(-1)^ta^t\prod_{k=0}^\infty\dfrac{1-q^kq^t}{b+s-bq^{-k-1}q^{-t}}$ , where $\theta(t)$ is an arbitrary periodic functions with unit period

$\therefore C_n(s)=\Theta(n)\int_{-\infty}^\infty(-1)^ta^tq^{nt}\prod_{k=0}^\infty\dfrac{1-q^kq^t}{b+s-bq^{-k-1}q^{-t}}dt$ , where $\Theta(n)$ is an arbitrary periodic functions with unit period

But this is only one of the group of the linear independent solution. I have no idea to find another group of the linear independent solution, since second order linear recurrence relations unlike second order linear differential equations which have reduction of order.

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There was a suggested edit by an anonymous user with the following request: Can you please give a reference for the solution, to explain the selected solution (this integral with this K function...? Or an intuition? why does the integral starts from $-\infty$ and not from 0? $C_n(s) \geq 0$ for all $n$, so maybe the integral should start from 0? thanks! –  t.b. Sep 8 '12 at 20:15
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@t.b.: The idea that assuming a suitable kernel is just for the intuition. It seems that assuming a suitable kernel is the only way to solving this exponential function coefficient recurrence relation. Because some of the terms should be unavoidable to perform linear shifting of the kernel, $(-\infty,\infty)$ is the only way that allow freely linear shifting. –  doraemonpaul Sep 9 '12 at 2:07
    
Thanks! (I was only the messenger; the comment was likely from the OP) –  t.b. Sep 9 '12 at 2:33
    
Does anyone know a good book where this kind of equations are handeled? –  Stan Sep 9 '12 at 7:04

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