\label{rs2conven}
\hat{G}_{RS2}[a, b; E]&=& \hat{G}_0^D[a, b; E]
+ \frac{\sqrt{a b}}{c v}
\frac{K_{\gamma}(\sqrt{2E} a) K_{\gamma}(\sqrt{2E} b)}
     {K_{\gamma}^2(\sqrt{2E} c)}   \\  \nonumber
& &
\hspace{3.0cm} \times
\left[ \left( \frac{\gamma - \frac{1}{2}}{2\xi c v} - 1 \right)
       + \frac{\sqrt{2E}}{2 \xi v}
        \frac{K_{\gamma - 1}(\sqrt{2E} c)}
             {K_{\gamma}(\sqrt{2E} c)}         \right]^{-1}.\end{aligned}$$ Since $\hat{G}_0^D$ satisfies the usual Dirichlet BC, [*i.e.*]{} $\hat{G}_0^D[a, c; E] = \hat{G}_0^D[c, b; E] = 0$, the fixed-energy amplitude on the brane is simply reduced to $$\label{rs2onbrane}
\hat{G}_{RS2}[c, c; E] =
\left[ \left( \frac{\gamma - \frac{1}{2}}{2 \xi c} - v \right)
       + \frac{\sqrt{2E}}{2 \xi}
        \frac{K_{\gamma - 1}(\sqrt{2E} c)}
             {K_{\gamma}(\sqrt{2E} c)}         \right]^{-1}.$$ If $(\gamma - 1/2) / (2 \xi c) - v = 0$, $\hat{G}_{RS2}[c, c; E]$ becomes $$\label{rs2simple}
\hat{G}_{RS2}[c, c; E] = \Delta_0 + \Delta_{KK}$$ where $$\begin{aligned}
\label{d0dkk}
\Delta_0&=&\frac{2 \xi (\gamma - 1)}{c E}    \\   \nonumber
\Delta_{KK}&=&\frac{2 \xi}{\sqrt{2E}}
              \frac{K_{\gamma - 2}(\sqrt{2E} c)}
                   {K_{\gamma - 1}(\sqrt{2E} c)}.\end{aligned}$$ Of course, $\Delta_0$ and $\Delta_{KK}$ represent the zero mode and the higher Kaluza-Klein excitations respectively. This means that the condition for the localized graviton to be massless is $$\label{rs2condi}
\frac{\gamma - \frac{1}{2}}{2 \xi c} - v = 0.$$ At the RS limit $\gamma = 2$, $c = R$ and $v = 3 / 2 R$ this condition really holds 