Can this difference operator be factorised? If a difference operator is defined as $$LY_i=\left(-\epsilon\dfrac{D^+ -D^-}{h_1}+aD^-\right)Y_i,\quad 1\leq i\leq N$$ Suppose $Y_N$ and $Y_0$ are given and that the difference operators are defined as follows $D^+V_i=(V_{i+1}-V_i)/h_1$, $D^-V_i=(V_i-V_{i-1})/h_1$. How is it possible to write the difference operator as $$LY_i=(Y_N-Y_0)\left(-\epsilon\dfrac{D^+ -D^-}{h_1}+aD^-\right)\psi_i?$$ 
I am thinking that telescoping trick is used, but am failing to see how.
If my question is not clear could someone clarify the second $L_\epsilon^NY_i$ on page 57 of the excerpt which is attached below.




$-\frac{\epsilon}{h^2} (Y_{i+1}-2Y_i + Y_{i-1}) + \alpha(Y_i-Y_{i-1}) = 0 \;\;\; (1)$
$-\frac{\epsilon}{h^2} Y_{N+1} + \left ( \frac{2\epsilon}{h^2} + \alpha \right )Y_N - \left ( \frac{\epsilon}{h^2} + \alpha \right ) Y_{N-1} = 0$
$\dots$
$Y_{N} = \frac{\epsilon Y_{N+1}}{2\epsilon + \alpha h^2}+\frac{\epsilon + \alpha h^2}{2\epsilon + \alpha h^2} Y_{N} \equiv aY_{N+1} + (1-a)Y_{N-1} = aY_{N+1}(1+(1-a) + (1-a)^2 + \dots+ (1-a)^N)+(1-a)^{N+1}Y_0 = a\frac{1-(1-a)^{N+1}}{a}Y_{N+1} +(1-a)^{N+1}Y_0 = (1-(1-a)^{N+1})Y_{N+1}+(1-a)^{N+1}Y_0$
$Y_{N-1} = (1-(1-a)^{N})Y_{N+1}+(1-a)^{N}Y_0$
Hence, (1) can be rewritten as a function of $Y_0$ and $Y_{N+1}$. Collecting the terms by $Y_0$ and $Y_{N+1}$ and defining your $\phi$ appropriately, you should get the desired.
Not worth the bounty but should point you in the right direction.
