two special cases.

The first case we will consider is $\xi_1 = \xi_2 = 1/2$. In this case Eq.(\[toy2schul1\]) and Eq.(\[toy2schul2\]) yield $$\begin{aligned}
\label{toy2xi1/2}
\hat{G}_{\delta, \xi_1=\xi_2=1/2}^{Box}[a, b; E]&=&
\frac{e^{-\sqrt{2E}|a - b|}}{\sqrt{2E}} + 
\left( \frac{\sqrt{2E}}{v_1} - 1 \right)^{-1}
\frac{e^{-\sqrt{2E}(a + b)}}{\sqrt{2E}}
                                         \\    \nonumber
& & -
\left[ \left( \frac{\sqrt{2E}}{v_1} - 1 \right)
       \left( \frac{\sqrt{2E}}{v_2} + 1 \right)
       + e^{-2 \sqrt{2E} L}                     \right]^{-1}
                                                       \\  \nonumber
& & \times
\Bigg[ \left( \frac{\sqrt{2E}}{v_1} - 1 \right)
       \frac{e^{\sqrt{2E}(a + b)}}{\sqrt{2E}} +
       \frac{e^{\sqrt{2E}(a - b)}}{\sqrt{2E}} +
       \frac{e^{-\sqrt{2E}(a - b)}}{\sqrt{2E}} +
                                                       \\   \nonumber
& & \hspace{6.0cm}
       \left( \frac{\sqrt{2E}}{v_1} - 1 \right)^{-1}
       \frac{e^{-\sqrt{2E}(a + b)}}{\sqrt{2E}}
                                                \Bigg].\end{aligned}$$

The second case we will consider is $\xi_1 = \xi_2 = 0$. In this case $\hat{G}_0^{Box}$ is $\hat{G}_0^{DD}$ in Eq.(\[toy2special\]). As expected $\hat{G}_0^{DD}$ satisfies the usual Dirichlet-Dirichlet BCs; $$\label{toy2ddbcs}
\hat{G}_0^{DD}[0, b; E] = \hat{G}_0^{DD}[a, 0; E]
= \hat{G}_0^{DD}[L, b; E] = \hat{G}_0^{DD}[a, L; E] = 0.$$ If, therefore, $v_1$ and $v_2$ are finite, we arrive at a conclusion $\hat{G}_{\delta}^{Box}[a, b; E] = \hat{G}_0^{Box}[a, b; E]$.

If however, $v_1$ and $v_2$ are infinite and unphysical bare quantities, one can arrive at different conclusion via the coupling constant renormalization as we have seen in section 2 and 3. To adopt the coupling constant renormalization 